{"text": ["... 1523 A Atomic Masses.................................................... 1531 B Selected Radioactive Isotopes............................................ 1537 C Useful Information.................................................. 1541 D Glossary of Key Symbols and Notation....................................... 1545 Index........................................................... 1659 This content is available for free at http://cnx.org/content/col11844/1.13 Preface 1 PREFACE The OpenStax College Physics: AP\u00ae Edition program has been developed with several goals in mind: accessibility, customization, and student engagement\u2014all while encouraging science students toward high levels of academic scholarship. Instructors and students alike will find that this program offers a strong foundation in physics in an accessible format. Welcome! About OpenStax College OpenStax College is a nonprofit organization committed to improving student access to quality learning materials. Our free textbooks are developed and peer-reviewed by educators to ensure they are readable, accurate, and meet the scope and sequence requirements of today\u2019s high school courses. Unlike traditional textbooks, OpenStax College resources live online and are owned by the community of educators using them. Through our partnerships with companies and foundations committed to reducing costs for students, OpenStax College is working to improve access to education for all. OpenStax College is an initiative of Rice University and is made possible through the generous support of several philanthropic foundations. OpenStax College resources provide quality academic instruction. Three key features set our materials apart from others: they can be customized by instructors for each class, they are a \u201cliving\u201d resource that grows online through contributions from science educators, and they are available for free or at minimal cost. Customization OpenStax College learning resources are designed to be customized for each course. Our textbooks provide a solid foundation on which instructors can build, and our resources are conceived and written with flexibility in mind. Instructors can simply select the sections most relevant to their curricula and create a textbook that speaks directly to the needs of their classes and students. Teachers are encouraged to expand on existing examples by adding unique context via geographically localized applications and topical connections. This customization feature will help bring physics to life for students and will ensure that your textbook truly reflects the goals of your course. Curation To broaden access and encourage community curation, OpenStax College Physics: AP\u00ae Edition is \u201copen source\u201d licensed under a Creative Commons Attribution (CC-BY) license. The scientific community is invited to submit examples, emerging research, and other feedback to enhance and strengthen the material and keep it current and relevant for today\u2019s students. Submit your suggestions to info@openstaxcollege.org, and find information on edition status, alternate versions, errata, and news on the StaxDash at http://openstaxcollege.org (http://openstaxcollege.org). Cost Our textbooks are available for free online and in low-cost print and e-book editions. About OpenStax College Physics: AP\u00ae Edition In 2012, OpenStax College published College Physics as part of a series that offers free and open college textbooks for higher education. College Physics was quickly adopted for science courses all around the country, and as word about this valuable resource spread, advanced placement teachers around the country started utilizing the book in AP\u00ae courses too. Physics: AP\u00ae Edition is the result of an effort to better serve these teachers and students. Based on College Physics\u2014a program based on the teaching and research experience of numerous physicists\u2014Physics: AP\u00ae Edition focuses on and emphasizes the new AP\u00ae curriculum's concepts and practices. Alignment to the AP\u00ae curriculum The new AP\u00ae Physics curriculum framework outlines the two full-year physics courses AP\u00ae Physics 1: Algebra-Based and AP\u00ae Physics 2: Algebra-Based. These two courses replaced the one-year AP\u00ae Physics B course, which over the years had become a fast", "-paced survey of physics facts and formulas that did not provide in-depth conceptual understanding of major physics ideas and the connections between them. The new AP\u00ae Physics 1 and 2 courses focus on the big ideas typically included in the first and second semesters of an algebrabased, introductory college-level physics course, providing students with the essential knowledge and skills required to support future advanced course work in physics. The AP\u00ae Physics 1 curriculum includes mechanics, mechanical waves, sound, and electrostatics. The AP\u00ae Physics 2 curriculum focuses on thermodynamics, fluid statics, dynamics, electromagnetism, geometric and physical optics, quantum physics, atomic physics, and nuclear physics. Seven unifying themes of physics called the Big Ideas each include three to seven Enduring Understandings (EU), which are themselves composed of Essential Knowledge (EK) that provides details and context for students as they explore physics. AP\u00ae Science Practices emphasize inquiry-based learning and development of critical thinking and reasoning skills. Inquiry usually uses a series of steps to gain new knowledge, beginning with an observation and following with a hypothesis to explain the observation; then experiments are conducted to test the hypothesis, gather results, and draw conclusions from data. The 2 Preface AP\u00ae framework has identified seven major science practices, which can be described by short phrases: using representations and models to communicate information and solve problems; using mathematics appropriately; engaging in questioning; planning and implementing data collection strategies; analyzing and evaluating data; justifying scientific explanations; and connecting concepts. The framework\u2019s Learning Objectives merge content (EU and EK) with one or more of the seven science practices that students should develop as they prepare for the AP\u00ae Physics exam. Each chapter of OpenStax College Physics: AP\u00ae Edition begins with a Connection for AP\u00ae Courses introduction that explains how the content in the chapter sections align to the Big Ideas, Enduring Understandings, and Essential Knowledge in the AP\u00ae framework. Physics: AP\u00ae Edition contains a wealth of information and the Connection for AP\u00ae Courses sections will help you distill the required AP\u00ae content from material that, although interesting, exceeds the scope of an introductory-level course. Each section opens with the program\u2019s learning objectives as well as the AP\u00ae learning objectives and science practices addressed. We have also developed Real World Connections features and Applying the Science Practices features that highlight concepts, examples, and practices in the framework. Pedagogical Foundation and Features OpenStax College Physics: AP\u00ae Edition is organized such that topics are introduced conceptually with a steady progression to precise definitions and analytical applications. The analytical aspect (problem solving) is tied back to the conceptual before moving on to another topic. Each introductory chapter, for example, opens with an engaging photograph relevant to the subject of the chapter and interesting applications that are easy for most students to visualize. Our features include: \u2022 Connections for AP\u00ae Courses introduce each chapter and explain how its content addresses the AP\u00ae curriculum. \u2022 Worked examples promote both analytical and conceptual skills. They are introduced using an application of interest followed by a strategy that emphasizes the concepts involved, a mathematical solution, and a discussion. \u2022 Problem-solving strategies are presented independently and subsequently appear at crucial points in the text where students can benefit most from them. \u2022 Misconception Alerts address common misconceptions that students may bring to class. \u2022 Take Home Investigations provide the opportunity for students to apply or explore what they have learned with a hands- on activity. \u2022 Real World Connections highlight important concepts and examples in the AP\u00ae framework. \u2022 Applying the Science Practices includes activities and challenging questions that engage students while they apply the AP\u00ae science practices. \u2022 Things Great and Small explain macroscopic phenomena (such as air pressure) with submicroscopic phenomena (such as atoms bouncing off walls). \u2022 Simulations direct students to further explore the physics concepts they have learned about in the module through the interactive PHeT physics simulations developed by the University of Colorado. Assessment Physics: AP\u00ae Edition offers a wealth of assessment options that include: \u2022 End-of-Module Problems include conceptual questions that challenge students\u2019 ability to explain what they have learned conceptually, independent of the mathematical details, and problems and exercises that challenge students to apply both concepts and skills to solve mathematical physics problems. Integrated Concept Problems challenge students to apply concepts and skills to solve a problem. \u2022 \u2022 Unreasonable Results encourage students to analyze the answer with respect to how likely or realistic it really is. \u2022 Construct Your Own Problem requires students to construct the details of a problem, justify their starting assumptions, show specific steps in the problem\u2019s solution, and finally discuss the meaning of the result. \u2022 Test Prep for AP\u00ae Courses consists of end-of-module problems that include assessment items with the format and rigor found in the AP\u00ae exam to help prepare students. About Our Team Physics: AP\u00ae Edition would not be possible if not for the tremendous contributions of the authors and community reviewing team. Contributors to Open", "Stax College Physics: AP\u00ae Edition This content is available for free at http://cnx.org/content/col11844/1.13 Preface Senior Contributors 3 Irna Lyublinskaya CUNY College of Staten Island, Staten Island, NY Gregg Wolfe Avonworth High School, Pittsburgh, PA Douglas Ingram TCU Department of Physics and Astronomy, Fort Worth, TX Liza Pujji Manukau Institute of Technology (MIT), New Zealand Sudhi Oberoi Visiting Research Student, QuIC Lab, Raman Research Institute, India Nathan Czuba Sabio Academy, Chicago, IL Julie Kretchman Science Writer, BS, University of Toronto, Canada John Stoke Science Writer, MS, University of Chicago, IL David Anderson Science Writer, PhD, College of William and Mary, Williamsburg, VA Erika Gasper Science Writer, MA, University of California, Santa Cruz, CA Advanced Placement Teacher Reviewers Faculty Reviewers Michelle Burgess Avon Lake High School, Avon Lake, OH Alexander Lavy Xavier High School, New York, NY Brian Hastings Spring Grove Area School District, York, PA John Boehringer Prosper High School, Prosper, TX Victor Brazil Petaluma High School, Petaluma, CA Jerome Mass Glastonbury Public Schools, Glastonbury, CT Bryan Callow Lindenwold High School, Lindenwold, NJ Anand Batra Howard University, Washington, DC John Aiken Georgia Institute of Technology, Atlanta, GA Robert Arts University of Pikeville, Pikeville, KY Ulrich Zurcher Cleveland State University, Cleveland, OH Michael Ottinger Missouri Western State University, Kansas City, MO James Smith Caldwell University, Caldwell, NJ Additional Resources Preparing for the AP\u00ae Physics 1 Exam Rice Online\u2019s dynamic new course, available on edX, is fully integrated with Physics for AP\u00ae Courses for free. Developed by nationally recognized Rice Professor Dr. Jason Hafner and AP\u00ae Physics teachers Gigi Nevils-Noe and Matt Wilson the course combines innovative learning technologies with engaging, professionally-produced Concept Trailers\u2122, inquiry based labs, practice problems, lectures, demonstrations, assessments, and other compelling resources to promote engagement and longterm retention of AP\u00ae Physics 1 concepts and application. Learn more at online.rice.edu. Other learning resources (powerpoint slides, testbanks, online homework etc) are updated frequently and can be viewed by going to https://openstaxcollege.org. To the AP\u00ae Physics Student The fundamental goal of physics is to discover and understand the \u201claws\u201d that govern observed phenomena in the world around us. Why study physics? If you plan to become a physicist, the answer is obvious\u2014introductory physics provides the foundation for your career; or if you want to become an engineer, physics provides the basis for the engineering principles used to solve applied and practical problems. For example, after the discovery of the photoelectric effect by physicists, engineers developed photocells that are used in solar panels to convert sunlight to electricity. What if you are an aspiring medical doctor? Although the applications of the laws of physics may not be obvious, their understanding is tremendously valuable. Physics is involved in medical diagnostics, such as x-rays, magnetic resonance imaging (MRI), and ultrasonic blood flow measurements. Medical therapy sometimes directly involves physics; cancer radiotherapy uses ionizing radiation. What if you are planning a nonscience career? Learning physics provides you with a well-rounded education and the ability to make important decisions, such as evaluating the pros and cons of energy production sources or voting on decisions about nuclear waste disposal. 4 Preface This AP\u00ae Physics 1 course begins with kinematics, the study of motion without considering its causes. Motion is everywhere: from the vibration of atoms to the planetary revolutions around the Sun. Understanding motion is key to understanding other concepts in physics. You will then study dynamics, which considers the forces that affect the motion of moving objects and systems. Newton\u2019s laws of motion are the foundation of dynamics. These laws provide an example of the breadth and simplicity of the principles under which nature functions. One of the most remarkable simplifications in physics is that only four distinct forces account for all known phenomena. Your journey will continue as you learn about energy. Energy plays an essential role both in everyday events and in scientific phenomena. You can likely name many forms of energy, from that provided by our foods, to the energy we use to run our cars, to the sunlight that warms us on the beach. The next stop is learning about oscillatory motion and waves. All oscillations involve force and energy: you push a child in a swing to get the motion started and you put energy into a guitar string when you pluck it. Some oscillations create waves. For example, a guitar creates sound waves. You will conclude this first physics course with the study of static electricity and electric currents. Many of the characteristics of static electricity can be explored by rubbing things together. Rubbing creates the spark you get", " from walking across a wool carpet, for example. Similarly, lightning results from air movements under certain weather conditions. In AP\u00ae Physics 2 course you will continue your journey by studying fluid dynamics, which explains why rising smoke curls and twists and how the body regulates blood flow. The next stop is thermodynamics, the study of heat transfer\u2014energy in transit\u2014that can be used to do work. Basic physical laws govern how heat transfers and its efficiency. Then you will learn more about electric phenomena as you delve into electromagnetism. An electric current produces a magnetic field; similarly, a magnetic field produces a current. This phenomenon, known as magnetic induction, is essential to our technological society. The generators in cars and nuclear plants use magnetism to generate a current. Other devices that use magnetism to induce currents include pickup coils in electric guitars, transformers of every size, certain microphones, airport security gates, and damping mechanisms on sensitive chemical balances. From electromagnetism you will continue your journey to optics, the study of light. You already know that visible light is the type of electromagnetic waves to which our eyes respond. Through vision, light can evoke deep emotions, such as when we view a magnificent sunset or glimpse a rainbow breaking through the clouds. Optics is concerned with the generation and propagation of light. The quantum mechanics, atomic physics, and nuclear physics are at the end of your journey. These areas of physics have been developed at the end of the 19th and early 20th centuries and deal with submicroscopic objects. Because these objects are smaller than we can observe directly with our senses and generally must be observed with the aid of instruments, parts of these physics areas may seem foreign and bizarre to you at first. However, we have experimentally confirmed most of the ideas in these areas of physics. AP\u00ae Physics is a challenging course. After all, you are taking physics at the introductory college level. You will discover that some concepts are more difficult to understand than others; most students, for example, struggle to understand rotational motion and angular momentum or particle-wave duality. The AP\u00ae curriculum promotes depth of understanding over breadth of content, and to make your exploration of topics more manageable, concepts are organized around seven major themes called the Big Ideas that apply to all levels of physical systems and interactions between them (see web diagram below). Each Big Idea identifies Enduring Understandings (EU), Essential Knowledge (EK), and illustrative examples that support key concepts and content. Simple descriptions define the focus of each Big Idea. \u2022 Big Idea 1: Objects and systems have properties. \u2022 Big Idea 2: Fields explain interactions. \u2022 Big Idea 3: The interactions are described by forces. \u2022 Big Idea 4: Interactions result in changes. \u2022 Big Idea 5: Changes are constrained by conservation laws. \u2022 Big Idea 6: Waves can transfer energy and momentum. \u2022 Big Idea 7: The mathematics of probability can to describe the behavior of complex and quantum mechanical systems. Doing college work is not easy, but completion of AP\u00ae classes is a reliable predictor of college success and prepares you for subsequent courses. The more you engage in the subject, the easier your journey through the curriculum will be. Bring your enthusiasm to class every day along with your notebook, pencil, and calculator. Prepare for class the day before, and review concepts daily. Form a peer study group and ask your teacher for extra help if necessary. The AP\u00ae lab program focuses on more open-ended, student-directed, and inquiry-based lab investigations designed to make you think, ask questions, and analyze data like scientists. You will develop critical thinking and reasoning skills and apply different means of communicating information. By the time you sit for the AP\u00ae exam in May, you will be fluent in the language of physics; because you have been doing real science, you will be ready to show what you have learned. Along the way, you will find the study of the world around us to be one of the most relevant and enjoyable experiences of your high school career. Irina Lyublinskaya, PhD Professor of Science Education To the AP\u00ae Physics Teacher The AP\u00ae curriculum was designed to allow instructors flexibility in their approach to teaching the physics courses. OpenStax College Physics: AP\u00ae Edition helps you orient students as they delve deeper into the world of physics. Each chapter includes a Connection for AP\u00ae Courses introduction that describes the AP\u00ae Physics Big Ideas, Enduring Understandings, and Essential Knowledge addressed in that chapter. Each section starts with specific AP\u00ae learning objectives and includes essential concepts, illustrative examples, and science practices, along with suggestions for applying the learning objectives through take home experiments, virtual lab investigations, and activities and questions for preparation and review. At the end of each section, students will find the Test Prep for AP\u00ae courses with multiple-choice and open-response questions addressing AP\u00ae learning objectives to help them prepare for the AP\u00ae exam. This content is available for free at http://cnx.org/content/col11844/1.", "13 Preface 5 OpenStax College Physics: AP\u00ae Edition has been written to engage students in their exploration of physics and help them relate what they learn in the classroom to their lives outside of it. Physics underlies much of what is happening today in other sciences and in technology. Thus, the book content includes interesting facts and ideas that go beyond the scope of the AP\u00ae course. The AP\u00ae Connection in each chapter directs students to the material they should focus on for the AP\u00ae exam, and what content\u2014although interesting\u2014is not part of the AP\u00ae curriculum. Physics is a beautiful and fascinating science. It is in your hands to engage and inspire your students to dive into an amazing world of physics, so they can enjoy it beyond just preparation for the AP\u00ae exam. Irina Lyublinskaya, PhD Professor of Science Education The concept map showing major links between Big Ideas and Enduring Understandings is provided below for visual reference. 6 Preface This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 1 | Introduction: The Nature of Science and Physics 7 INTRODUCTION: THE NATURE OF 1 SCIENCE AND PHYSICS Figure 1.1 Galaxies are as immense as atoms are small. Yet the same laws of physics describe both, and all the rest of nature\u2014an indication of the underlying unity in the universe. The laws of physics are surprisingly few in number, implying an underlying simplicity to nature's apparent complexity. (credit: NASA, JPL-Caltech, P. Barmby, Harvard-Smithsonian Center for Astrophysics) Chapter Outline 1.1. Physics: An Introduction 1.2. Physical Quantities and Units 1.3. Accuracy, Precision, and Significant Figures 1.4. Approximation Connection for AP\u00ae Courses What is your first reaction when you hear the word \u201cphysics\u201d? Did you imagine working through difficult equations or memorizing formulas that seem to have no real use in life outside the physics classroom? Many people come to the subject of physics with a bit of fear. But as you begin your exploration of this broad-ranging subject, you may soon come to realize that physics plays a much larger role in your life than you first thought, no matter your life goals or career choice. For example, take a look at the image above. This image is of the Andromeda Galaxy, which contains billions of individual stars, huge clouds of gas, and dust. Two smaller galaxies are also visible as bright blue spots in the background. At a staggering 2.5 million light years from Earth, this galaxy is the nearest one to our own galaxy (which is called the Milky Way). The stars and planets that make up Andromeda might seem to be the furthest thing from most people's regular, everyday lives. But Andromeda is a great starting point to think about the forces that hold together the universe. The forces that cause Andromeda to act as it does are the same forces we contend with here on Earth, whether we are planning to send a rocket into space or simply raise the walls for a new home. The same gravity that causes the stars of Andromeda to rotate and revolve also causes water to flow over hydroelectric dams here on Earth. Tonight, take a moment to look up at the stars. The forces out there are the same as the ones here on Earth. Through a study of physics, you may gain a greater understanding of the interconnectedness of everything we can see and know in this universe. Think now about all of the technological devices that you use on a regular basis. Computers, smart phones, GPS systems, MP3 players, and satellite radio might come to mind. Next, think about the most exciting modern technologies that you have heard about in the news, such as trains that levitate above tracks, \u201cinvisibility cloaks\u201d that bend light around them, and microscopic robots that fight cancer cells in our bodies. All of these groundbreaking advancements, commonplace or unbelievable, rely on the principles of physics. Aside from playing a significant role in technology, professionals such as engineers, pilots, physicians, physical therapists, electricians, and computer programmers apply physics concepts in their daily work. For example, a pilot must understand how wind forces affect a flight path and a physical therapist must understand how the muscles in the body experience forces as they move and bend. As you will learn in this text, physics principles are propelling new, exciting technologies, and these principles are applied in a wide range of careers. In this text, you will begin to explore the history of the formal study of physics, beginning with natural philosophy and the ancient Greeks, and leading up through a review of Sir Isaac Newton and the laws of physics that bear his name. You will also be introduced to the standards scientists use when they study physical quantities and the interrelated system of measurements most of the scientific community uses to communicate in a single mathematical language. Finally, you will study the limits of our ability to be accurate and precise, and", " the reasons scientists go to painstaking lengths to be as clear as possible regarding their own limitations. 8 Chapter 1 | Introduction: The Nature of Science and Physics Chapter 1 introduces many fundamental skills and understandings needed for success with the AP\u00ae Learning Objectives. While this chapter does not directly address any Big Ideas, its content will allow for a more meaningful understanding when these Big Ideas are addressed in future chapters. For instance, the discussion of models, theories, and laws will assist you in understanding the concept of fields as addressed in Big Idea 2, and the section titled \u2018The Evolution of Natural Philosophy into Modern Physics' will help prepare you for the statistical topics addressed in Big Idea 7. This chapter will also prepare you to understand the Science Practices. In explicitly addressing the role of models in representing and communicating scientific phenomena, Section 1.1 supports Science Practice 1. Additionally, anecdotes about historical investigations and the inset on the scientific method will help you to engage in the scientific questioning referenced in Science Practice 3. The appropriate use of mathematics, as called for in Science Practice 2, is a major focus throughout sections 1.2, 1.3, and 1.4. 1.1 Physics: An Introduction Figure 1.2 The flight formations of migratory birds such as Canada geese are governed by the laws of physics. (credit: David Merrett) Learning Objectives By the end of this section, you will be able to: \u2022 Explain the difference between a principle and a law. \u2022 Explain the difference between a model and a theory. The physical universe is enormously complex in its detail. Every day, each of us observes a great variety of objects and phenomena. Over the centuries, the curiosity of the human race has led us collectively to explore and catalog a tremendous wealth of information. From the flight of birds to the colors of flowers, from lightning to gravity, from quarks to clusters of galaxies, from the flow of time to the mystery of the creation of the universe, we have asked questions and assembled huge arrays of facts. In the face of all these details, we have discovered that a surprisingly small and unified set of physical laws can explain what we observe. As humans, we make generalizations and seek order. We have found that nature is remarkably cooperative\u2014it exhibits the underlying order and simplicity we so value. It is the underlying order of nature that makes science in general, and physics in particular, so enjoyable to study. For example, what do a bag of chips and a car battery have in common? Both contain energy that can be converted to other forms. The law of conservation of energy (which says that energy can change form but is never lost) ties together such topics as food calories, batteries, heat, light, and watch springs. Understanding this law makes it easier to learn about the various forms energy takes and how they relate to one another. Apparently unrelated topics are connected through broadly applicable physical laws, permitting an understanding beyond just the memorization of lists of facts. The unifying aspect of physical laws and the basic simplicity of nature form the underlying themes of this text. In learning to apply these laws, you will, of course, study the most important topics in physics. More importantly, you will gain analytical abilities that will enable you to apply these laws far beyond the scope of what can be included in a single book. These analytical skills will help you to excel academically, and they will also help you to think critically in any professional career you choose to pursue. This module discusses the realm of physics (to define what physics is), some applications of physics (to illustrate its relevance to other disciplines), and more precisely what constitutes a physical law (to illuminate the importance of experimentation to theory). Science and the Realm of Physics Science consists of the theories and laws that are the general truths of nature as well as the body of knowledge they encompass. Scientists are continually trying to expand this body of knowledge and to perfect the expression of the laws that describe it. Physics is concerned with describing the interactions of energy, matter, space, and time, and it is especially interested in what fundamental mechanisms underlie every phenomenon. The concern for describing the basic phenomena in nature essentially defines the realm of physics. Physics aims to describe the function of everything around us, from the movement of tiny charged particles to the motion of people, cars, and spaceships. In fact, almost everything around you can be described quite accurately by the laws of physics. Consider a smart phone (Figure 1.3). Physics describes how electricity interacts with the various circuits inside the device. This This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 1 | Introduction: The Nature of Science and Physics 9 knowledge helps engineers select the appropriate materials and circuit layout when building the smart phone. Next, consider a GPS system. Physics describes the relationship between the speed of an object, the distance over which it travels, and the time it takes to travel that distance. When you use a GPS device in a vehicle, it utilizes these physics equations to determine", " the travel time from one location to another. Figure 1.3 The Apple \u201ciPhone\u201d is a common smart phone with a GPS function. Physics describes the way that electricity flows through the circuits of this device. Engineers use their knowledge of physics to construct an iPhone with features that consumers will enjoy. One specific feature of an iPhone is the GPS function. GPS uses physics equations to determine the driving time between two locations on a map. (credit: @gletham GIS, Social, Mobile Tech Images) Applications of Physics You need not be a scientist to use physics. On the contrary, knowledge of physics is useful in everyday situations as well as in nonscientific professions. It can help you understand how microwave ovens work, why metals should not be put into them, and why they might affect pacemakers. (See Figure 1.4 and Figure 1.5.) Physics allows you to understand the hazards of radiation and rationally evaluate these hazards more easily. Physics also explains the reason why a black car radiator helps remove heat in a car engine, and it explains why a white roof helps keep the inside of a house cool. Similarly, the operation of a car's ignition system as well as the transmission of electrical signals through our body's nervous system are much easier to understand when you think about them in terms of basic physics. Physics is the foundation of many important disciplines and contributes directly to others. Chemistry, for example\u2014since it deals with the interactions of atoms and molecules\u2014is rooted in atomic and molecular physics. Most branches of engineering are applied physics. In architecture, physics is at the heart of structural stability, and is involved in the acoustics, heating, lighting, and cooling of buildings. Parts of geology rely heavily on physics, such as radioactive dating of rocks, earthquake analysis, and heat transfer in the Earth. Some disciplines, such as biophysics and geophysics, are hybrids of physics and other disciplines. Physics has many applications in the biological sciences. On the microscopic level, it helps describe the properties of cell walls and cell membranes (Figure 1.6 and Figure 1.7). On the macroscopic level, it can explain the heat, work, and power associated with the human body. Physics is involved in medical diagnostics, such as x-rays, magnetic resonance imaging (MRI), and ultrasonic blood flow measurements. Medical therapy sometimes directly involves physics; for example, cancer radiotherapy uses ionizing radiation. Physics can also explain sensory phenomena, such as how musical instruments make sound, how the eye detects color, and how lasers can transmit information. It is not necessary to formally study all applications of physics. What is most useful is knowledge of the basic laws of physics and a skill in the analytical methods for applying them. The study of physics also can improve your problem-solving skills. Furthermore, physics has retained the most basic aspects of science, so it is used by all of the sciences, and the study of physics makes other sciences easier to understand. Figure 1.4 The laws of physics help us understand how common appliances work. For example, the laws of physics can help explain how microwave ovens heat up food, and they also help us understand why it is dangerous to place metal objects in a microwave oven. (credit: MoneyBlogNewz) 10 Chapter 1 | Introduction: The Nature of Science and Physics Figure 1.5 These two applications of physics have more in common than meets the eye. Microwave ovens use electromagnetic waves to heat food. Magnetic resonance imaging (MRI) also uses electromagnetic waves to yield an image of the brain, from which the exact location of tumors can be determined. (credit: Rashmi Chawla, Daniel Smith, and Paul E. Marik) Figure 1.6 Physics, chemistry, and biology help describe the properties of cell walls in plant cells, such as the onion cells seen here. (credit: Umberto Salvagnin) Figure 1.7 An artist's rendition of the the structure of a cell membrane. Membranes form the boundaries of animal cells and are complex in structure and function. Many of the most fundamental properties of life, such as the firing of nerve cells, are related to membranes. The disciplines of biology, chemistry, and physics all help us understand the membranes of animal cells. (credit: Mariana Ruiz) Models, Theories, and Laws; The Role of Experimentation The laws of nature are concise descriptions of the universe around us; they are human statements of the underlying laws or rules that all natural processes follow. Such laws are intrinsic to the universe; humans did not create them and so cannot change them. We can only discover and understand them. Their discovery is a very human endeavor, with all the elements of mystery, imagination, struggle, triumph, and disappointment inherent in any creative effort. (See Figure 1.8 and Figure 1.9.) The cornerstone of discovering natural laws is observation; science must describe the universe as it is, not as we may imagine it to be. This content is", " available for free at http://cnx.org/content/col11844/1.13 Chapter 1 | Introduction: The Nature of Science and Physics 11 Figure 1.8 Isaac Newton (1642\u20131727) was very reluctant to publish his revolutionary work and had to be convinced to do so. In his later years, he stepped down from his academic post and became exchequer of the Royal Mint. He took this post seriously, inventing reeding (or creating ridges) on the edge of coins to prevent unscrupulous people from trimming the silver off of them before using them as currency. (credit: Arthur Shuster and Arthur E. Shipley: Britain's Heritage of Science. London, 1917.) Figure 1.9 Marie Curie (1867\u20131934) sacrificed monetary assets to help finance her early research and damaged her physical well-being with radiation exposure. She is the only person to win Nobel prizes in both physics and chemistry. One of her daughters also won a Nobel Prize. (credit: Wikimedia Commons) We all are curious to some extent. We look around, make generalizations, and try to understand what we see\u2014for example, we look up and wonder whether one type of cloud signals an oncoming storm. As we become serious about exploring nature, we become more organized and formal in collecting and analyzing data. We attempt greater precision, perform controlled experiments (if we can), and write down ideas about how the data may be organized and unified. We then formulate models, theories, and laws based on the data we have collected and analyzed to generalize and communicate the results of these experiments. A model is a representation of something that is often too difficult (or impossible) to display directly. While a model is justified with experimental proof, it is only accurate under limited situations. An example is the planetary model of the atom in which electrons are pictured as orbiting the nucleus, analogous to the way planets orbit the Sun. (See Figure 1.10.) We cannot observe electron orbits directly, but the mental image helps explain the observations we can make, such as the emission of light from hot gases (atomic spectra). Physicists use models for a variety of purposes. For example, models can help physicists analyze a scenario and perform a calculation, or they can be used to represent a situation in the form of a computer simulation. A theory is an explanation for patterns in nature that is supported by scientific evidence and verified multiple times by various groups of researchers. Some theories include models to help visualize phenomena, whereas others do not. Newton's theory of gravity, for example, does not require a model or mental image, because we can observe the objects directly with our own senses. The kinetic theory of gases, on the other hand, is a model in which a gas is viewed as being composed of atoms and molecules. Atoms and molecules are too small to be observed directly with our senses\u2014thus, we picture them mentally to understand what our instruments tell us about the behavior of gases. A law uses concise language to describe a generalized pattern in nature that is supported by scientific evidence and repeated experiments. Often, a law can be expressed in the form of a single mathematical equation. Laws and theories are similar in that they are both scientific statements that result from a tested hypothesis and are supported by scientific evidence. However, the designation law is reserved for a concise and very general statement that describes phenomena in nature, such as the law that 12 Chapter 1 | Introduction: The Nature of Science and Physics energy is conserved during any process, or Newton's second law of motion, which relates force, mass, and acceleration by the simple equation F = a. A theory, in contrast, is a less concise statement of observed phenomena. For example, the Theory of Evolution and the Theory of Relativity cannot be expressed concisely enough to be considered a law. The biggest difference between a law and a theory is that a theory is much more complex and dynamic. A law describes a single action, whereas a theory explains an entire group of related phenomena. And, whereas a law is a postulate that forms the foundation of the scientific method, a theory is the end result of that process. Less broadly applicable statements are usually called principles (such as Pascal's principle, which is applicable only in fluids), but the distinction between laws and principles often is not carefully made. Figure 1.10 What is a model? This planetary model of the atom shows electrons orbiting the nucleus. It is a drawing that we use to form a mental image of the atom that we cannot see directly with our eyes because it is too small. Models, Theories, and Laws Models, theories, and laws are used to help scientists analyze the data they have already collected. However, often after a model, theory, or law has been developed, it points scientists toward new discoveries they would not otherwise have made. The models, theories, and laws we devise sometimes imply the existence of objects or phenomena as yet unobserved. These predictions are remarkable triumphs and tributes to the power of science", ". It is the underlying order in the universe that enables scientists to make such spectacular predictions. However, if experiment does not verify our predictions, then the theory or law is wrong, no matter how elegant or convenient it is. Laws can never be known with absolute certainty because it is impossible to perform every imaginable experiment in order to confirm a law in every possible scenario. Physicists operate under the assumption that all scientific laws and theories are valid until a counterexample is observed. If a good-quality, verifiable experiment contradicts a well-established law, then the law must be modified or overthrown completely. The study of science in general and physics in particular is an adventure much like the exploration of uncharted ocean. Discoveries are made; models, theories, and laws are formulated; and the beauty of the physical universe is made more sublime for the insights gained. The Scientific Method As scientists inquire and gather information about the world, they follow a process called the scientific method. This process typically begins with an observation and question that the scientist will research. Next, the scientist typically performs some research about the topic and then devises a hypothesis. Then, the scientist will test the hypothesis by performing an experiment. Finally, the scientist analyzes the results of the experiment and draws a conclusion. Note that the scientific method can be applied to many situations that are not limited to science, and this method can be modified to suit the situation. Consider an example. Let us say that you try to turn on your car, but it will not start. You undoubtedly wonder: Why will the car not start? You can follow a scientific method to answer this question. First off, you may perform some research to determine a variety of reasons why the car will not start. Next, you will state a hypothesis. For example, you may believe that the car is not starting because it has no engine oil. To test this, you open the hood of the car and examine the oil level. You observe that the oil is at an acceptable level, and you thus conclude that the oil level is not contributing to your car issue. To troubleshoot the issue further, you may devise a new hypothesis to test and then repeat the process again. The Evolution of Natural Philosophy into Modern Physics Physics was not always a separate and distinct discipline. It remains connected to other sciences to this day. The word physics comes from Greek, meaning nature. The study of nature came to be called \u201cnatural philosophy.\u201d From ancient times through the Renaissance, natural philosophy encompassed many fields, including astronomy, biology, chemistry, physics, mathematics, and medicine. Over the last few centuries, the growth of knowledge has resulted in ever-increasing specialization and branching of natural philosophy into separate fields, with physics retaining the most basic facets. (See Figure 1.11, Figure 1.12, and Figure 1.13.) Physics as it developed from the Renaissance to the end of the 19th century is called classical physics. It was transformed into modern physics by revolutionary discoveries made starting at the beginning of the 20th century. This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 1 | Introduction: The Nature of Science and Physics 13 Figure 1.11 Over the centuries, natural philosophy has evolved into more specialized disciplines, as illustrated by the contributions of some of the greatest minds in history. The Greek philosopher Aristotle (384\u2013322 B.C.) wrote on a broad range of topics including physics, animals, the soul, politics, and poetry. (credit: Jastrow (2006)/Ludovisi Collection) Figure 1.12 Galileo Galilei (1564\u20131642) laid the foundation of modern experimentation and made contributions in mathematics, physics, and astronomy. (credit: Domenico Tintoretto) Figure 1.13 Niels Bohr (1885\u20131962) made fundamental contributions to the development of quantum mechanics, one part of modern physics. (credit: United States Library of Congress Prints and Photographs Division) Classical physics is not an exact description of the universe, but it is an excellent approximation under the following conditions: Matter must be moving at speeds less than about 1% of the speed of light, the objects dealt with must be large enough to be seen with a microscope, and only weak gravitational fields, such as the field generated by the Earth, can be involved. Because humans live under such circumstances, classical physics seems intuitively reasonable, while many aspects of modern physics seem bizarre. This is why models are so useful in modern physics\u2014they let us conceptualize phenomena we do not ordinarily experience. We can relate to models in human terms and visualize what happens when objects move at high speeds or imagine what objects too small to observe with our senses might be like. For example, we can understand an atom's properties because we can picture it in our minds, although we have never seen an atom with our eyes. New tools, of course, allow us to better picture phenomena we cannot see. In fact,", " new instrumentation has allowed us in recent years to actually \u201cpicture\u201d the atom. 14 Chapter 1 | Introduction: The Nature of Science and Physics Limits on the Laws of Classical Physics For the laws of classical physics to apply, the following criteria must be met: Matter must be moving at speeds less than about 1% of the speed of light, the objects dealt with must be large enough to be seen with a microscope, and only weak gravitational fields (such as the field generated by the Earth) can be involved. Figure 1.14 Using a scanning tunneling microscope (STM), scientists can see the individual atoms that compose this sheet of gold. (credit: Erwinrossen) Some of the most spectacular advances in science have been made in modern physics. Many of the laws of classical physics have been modified or rejected, and revolutionary changes in technology, society, and our view of the universe have resulted. Like science fiction, modern physics is filled with fascinating objects beyond our normal experiences, but it has the advantage over science fiction of being very real. Why, then, is the majority of this text devoted to topics of classical physics? There are two main reasons: Classical physics gives an extremely accurate description of the universe under a wide range of everyday circumstances, and knowledge of classical physics is necessary to understand modern physics. Modern physics itself consists of the two revolutionary theories, relativity and quantum mechanics. These theories deal with the very fast and the very small, respectively. Relativity must be used whenever an object is traveling at greater than about 1% of the speed of light or experiences a strong gravitational field such as that near the Sun. Quantum mechanics must be used for objects smaller than can be seen with a microscope. The combination of these two theories is relativistic quantum mechanics, and it describes the behavior of small objects traveling at high speeds or experiencing a strong gravitational field. Relativistic quantum mechanics is the best universally applicable theory we have. Because of its mathematical complexity, it is used only when necessary, and the other theories are used whenever they will produce sufficiently accurate results. We will find, however, that we can do a great deal of modern physics with the algebra and trigonometry used in this text. Check Your Understanding A friend tells you he has learned about a new law of nature. What can you know about the information even before your friend describes the law? How would the information be different if your friend told you he had learned about a scientific theory rather than a law? Solution Without knowing the details of the law, you can still infer that the information your friend has learned conforms to the requirements of all laws of nature: it will be a concise description of the universe around us; a statement of the underlying rules that all natural processes follow. If the information had been a theory, you would be able to infer that the information will be a large-scale, broadly applicable generalization. PhET Explorations: Equation Grapher Learn about graphing polynomials. The shape of the curve changes as the constants are adjusted. View the curves for the individual terms (e.g. = ) to see how they add to generate the polynomial curve. Figure 1.15 Equation Grapher (http://cnx.org/content/m54764/1.2/equation-grapher_en.jar) This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 1 | Introduction: The Nature of Science and Physics 15 1.2 Physical Quantities and Units Figure 1.16 The distance from Earth to the Moon may seem immense, but it is just a tiny fraction of the distances from Earth to other celestial bodies. (credit: NASA) By the end of this section, you will be able to: Learning Objectives \u2022 Perform unit conversions both in the SI and English units. \u2022 Explain the most common prefixes in the SI units and be able to write them in scientific notation. The range of objects and phenomena studied in physics is immense. From the incredibly short lifetime of a nucleus to the age of Earth, from the tiny sizes of sub-nuclear particles to the vast distance to the edges of the known universe, from the force exerted by a jumping flea to the force between Earth and the Sun, there are enough factors of 10 to challenge the imagination of even the most experienced scientist. Giving numerical values for physical quantities and equations for physical principles allows us to understand nature much more deeply than does qualitative description alone. To comprehend these vast ranges, we must also have accepted units in which to express them. And we shall find that (even in the potentially mundane discussion of meters, kilograms, and seconds) a profound simplicity of nature appears\u2014all physical quantities can be expressed as combinations of only four fundamental physical quantities: length, mass, time, and electric current. We define a physical quantity either by specifying how it is measured or by stating how it is calculated from other measurements. For example, we define distance and time", " by specifying methods for measuring them, whereas we define average speed by stating that it is calculated as distance traveled divided by time of travel. Measurements of physical quantities are expressed in terms of units, which are standardized values. For example, the length of a race, which is a physical quantity, can be expressed in units of meters (for sprinters) or kilometers (for distance runners). Without standardized units, it would be extremely difficult for scientists to express and compare measured values in a meaningful way. (See Figure 1.17.) Figure 1.17 Distances given in unknown units are maddeningly useless. There are two major systems of units used in the world: SI units (also known as the metric system) and English units (also known as the customary or imperial system). English units were historically used in nations once ruled by the British Empire and are still widely used in the United States. Virtually every other country in the world now uses SI units as the standard; the metric system is also the standard system agreed upon by scientists and mathematicians. The acronym \u201cSI\u201d is derived from the French Syst\u00e8me International. 16 Chapter 1 | Introduction: The Nature of Science and Physics SI Units: Fundamental and Derived Units Table 1.1 gives the fundamental SI units that are used throughout this textbook. This text uses non-SI units in a few applications where they are in very common use, such as the measurement of blood pressure in millimeters of mercury (mm Hg). Whenever non-SI units are discussed, they will be tied to SI units through conversions. Table 1.1 Fundamental SI Units Length Mass Time Electric Charge meter (m) kilogram (kg) second (s) coulomb (c) It is an intriguing fact that some physical quantities are more fundamental than others and that the most fundamental physical quantities can be defined only in terms of the procedure used to measure them. The units in which they are measured are thus called fundamental units. In this textbook, the fundamental physical quantities are taken to be length, mass, time, and electric charge. (Note that electric current will not be introduced until much later in this text.) All other physical quantities, such as force and electric current, can be expressed as algebraic combinations of length, mass, time, and current (for example, speed is length divided by time); these units are called derived units. Units of Time, Length, and Mass: The Second, Meter, and Kilogram The Second The SI unit for time, the second(abbreviated s), has a long history. For many years it was defined as 1/86,400 of a mean solar day. More recently, a new standard was adopted to gain greater accuracy and to define the second in terms of a non-varying, or constant, physical phenomenon (because the solar day is getting longer due to very gradual slowing of the Earth's rotation). Cesium atoms can be made to vibrate in a very steady way, and these vibrations can be readily observed and counted. In 1967 the second was redefined as the time required for 9,192,631,770 of these vibrations. (See Figure 1.18.) Accuracy in the fundamental units is essential, because all measurements are ultimately expressed in terms of fundamental units and can be no more accurate than are the fundamental units themselves. Figure 1.18 An atomic clock such as this one uses the vibrations of cesium atoms to keep time to a precision of better than a microsecond per year. The fundamental unit of time, the second, is based on such clocks. This image is looking down from the top of an atomic fountain nearly 30 feet tall! (credit: Steve Jurvetson/Flickr) The Meter The SI unit for length is the meter (abbreviated m); its definition has also changed over time to become more accurate and precise. The meter was first defined in 1791 as 1/10,000,000 of the distance from the equator to the North Pole. This measurement was improved in 1889 by redefining the meter to be the distance between two engraved lines on a platinum-iridium bar now kept near Paris. By 1960, it had become possible to define the meter even more accurately in terms of the wavelength of light, so it was again redefined as 1,650,763.73 wavelengths of orange light emitted by krypton atoms. In 1983, the meter was given its present definition (partly for greater accuracy) as the distance light travels in a vacuum in 1/299,792,458 of a second. (See Figure 1.19.) This change defines the speed of light to be exactly 299,792,458 meters per second. The length of the meter will change if the speed of light is someday measured with greater accuracy. The Kilogram The SI unit for mass is the kilogram (abbreviated kg); it is defined to be the mass of a platinum-iridium cylinder kept with the old meter standard at the International Bureau of Weights", " and Measures near Paris. Exact replicas of the standard kilogram are also kept at the United States' National Institute of Standards and Technology, or NIST, located in Gaithersburg, Maryland outside of Washington D.C., and at other locations around the world. The determination of all other masses can be ultimately traced to a comparison with the standard mass. This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 1 | Introduction: The Nature of Science and Physics 17 Figure 1.19 The meter is defined to be the distance light travels in 1/299,792,458 of a second in a vacuum. Distance traveled is speed multiplied by time. Electric current and its accompanying unit, the ampere, will be introduced in Introduction to Electric Current, Resistance, and Ohm's Law when electricity and magnetism are covered. The initial modules in this textbook are concerned with mechanics, fluids, heat, and waves. In these subjects all pertinent physical quantities can be expressed in terms of the fundamental units of length, mass, and time. Metric Prefixes SI units are part of the metric system. The metric system is convenient for scientific and engineering calculations because the units are categorized by factors of 10. Table 1.2 gives metric prefixes and symbols used to denote various factors of 10. Metric systems have the advantage that conversions of units involve only powers of 10. There are 100 centimeters in a meter, 1000 meters in a kilometer, and so on. In nonmetric systems, such as the system of U.S. customary units, the relationships are not as simple\u2014there are 12 inches in a foot, 5280 feet in a mile, and so on. Another advantage of the metric system is that the same unit can be used over extremely large ranges of values simply by using an appropriate metric prefix. For example, distances in meters are suitable in construction, while distances in kilometers are appropriate for air travel, and the tiny measure of nanometers are convenient in optical design. With the metric system there is no need to invent new units for particular applications. The term order of magnitude refers to the scale of a value expressed in the metric system. Each power of 10 in the metric system represents a different order of magnitude. For example, 101 102 103 magnitude. All quantities that can be expressed as a product of a specific power of 10 are said to be of the same order of magnitude. For example, the number 800 can be written as 8\u00d7102, and the number 450 can be written as 4.5\u00d7102. Thus, the numbers 800 and 450 are of the same order of magnitude: 102. Order of magnitude can be thought of as a ballpark estimate for the scale of a value. The diameter of an atom is on the order of 10\u22129 m, while the diameter of the Sun is on the order of 109 m., and so forth are all different orders of The Quest for Microscopic Standards for Basic Units The fundamental units described in this chapter are those that produce the greatest accuracy and precision in measurement. There is a sense among physicists that, because there is an underlying microscopic substructure to matter, it would be most satisfying to base our standards of measurement on microscopic objects and fundamental physical phenomena such as the speed of light. A microscopic standard has been accomplished for the standard of time, which is based on the oscillations of the cesium atom. The standard for length was once based on the wavelength of light (a small-scale length) emitted by a certain type of atom, but it has been supplanted by the more precise measurement of the speed of light. If it becomes possible to measure the mass of atoms or a particular arrangement of atoms such as a silicon sphere to greater precision than the kilogram standard, it may become possible to base mass measurements on the small scale. There are also possibilities that electrical phenomena on the small scale may someday allow us to base a unit of charge on the charge of electrons and protons, but at present current and charge are related to large-scale currents and forces between wires. 18 Chapter 1 | Introduction: The Nature of Science and Physics Table 1.2 Metric Prefixes for Powers of 10 and their Symbols Value[1] Symbol Prefix Example (some are approximate) exa peta tera giga mega kilo hecto deka \u2014 deci centi milli micro nano pico femto atto E P T G M k h da \u2014 d c m \u00b5 n p f a 1018 1015 1012 109 106 103 102 101 100 (=1) 10\u22121 10\u22122 10\u22123 10\u22126 10\u22129 10\u221212 10\u221215 10\u221218 exameter Em 1018 m distance light travels in a century petasecond Ps 1015 s 30 million years terawatt TW 1012 W powerful laser output gigahertz GHz 109 Hz a microwave frequency megacurie MCi 106 Ci high radioactivity kilometer km 103 m about 6/", "10 mile hectoliter hL 102 L 26 gallons dekagram dag 101 g teaspoon of butter deciliter dL 10\u22121 L less than half a soda centimeter cm 10\u22122 m fingertip thickness millimeter mm 10\u22123 m flea at its shoulders micrometer \u00b5m 10\u22126 m detail in microscope nanogram ng 10\u22129 g small speck of dust picofarad pF 10\u221212 F small capacitor in radio femtometer fm 10\u221215 m size of a proton attosecond as 10\u221218 s time light crosses an atom Known Ranges of Length, Mass, and Time The vastness of the universe and the breadth over which physics applies are illustrated by the wide range of examples of known lengths, masses, and times in Table 1.3. Examination of this table will give you some feeling for the range of possible topics and numerical values. (See Figure 1.20 and Figure 1.21.) Figure 1.20 Tiny phytoplankton swims among crystals of ice in the Antarctic Sea. They range from a few micrometers to as much as 2 millimeters in length. (credit: Prof. Gordon T. Taylor, Stony Brook University; NOAA Corps Collections) 1. See Appendix A for a discussion of powers of 10. This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 1 | Introduction: The Nature of Science and Physics 19 Figure 1.21 Galaxies collide 2.4 billion light years away from Earth. The tremendous range of observable phenomena in nature challenges the imagination. (credit: NASA/CXC/UVic./A. Mahdavi et al. Optical/lensing: CFHT/UVic./H. Hoekstra et al.) Unit Conversion and Dimensional Analysis It is often necessary to convert from one type of unit to another. For example, if you are reading a European cookbook, some quantities may be expressed in units of liters and you need to convert them to cups. Or, perhaps you are reading walking directions from one location to another and you are interested in how many miles you will be walking. In this case, you will need to convert units of feet to miles. Let us consider a simple example of how to convert units. Let us say that we want to convert 80 meters (m) to kilometers (km). The first thing to do is to list the units that you have and the units that you want to convert to. In this case, we have units in meters and we want to convert to kilometers. Next, we need to determine a conversion factor relating meters to kilometers. A conversion factor is a ratio expressing how many of one unit are equal to another unit. For example, there are 12 inches in 1 foot, 100 centimeters in 1 meter, 60 seconds in 1 minute, and so on. In this case, we know that there are 1,000 meters in 1 kilometer. Now we can set up our unit conversion. We will write the units that we have and then multiply them by the conversion factor so that the units cancel out, as shown: 80m\u00d7 1 km 1000m = 0.080 km. (1.1) Note that the unwanted m unit cancels, leaving only the desired km unit. You can use this method to convert between any types of unit. Click Appendix C for a more complete list of conversion factors. 20 Chapter 1 | Introduction: The Nature of Science and Physics Table 1.3 Approximate Values of Length, Mass, and Time Lengths in meters Masses in kilograms (more precise values in parentheses) Times in seconds (more precise values in parentheses) 10\u221218 Present experimental limit to smallest observable detail 10\u221230 Mass of an electron 9.11\u00d710\u221231 kg 10\u221223 Time for light to cross a proton 10\u221215 Diameter of a proton 10\u221227 Mass of a hydrogen atom 1.67\u00d710\u221227 kg 10\u221222 Mean life of an extremely unstable nucleus 10\u221214 Diameter of a uranium nucleus 10\u221215 Mass of a bacterium 10\u221210 Diameter of a hydrogen atom 10\u22125 Mass of a mosquito 10\u221215 Time for one oscillation of visible light 10\u221213 Time for one vibration of an atom in a solid 10\u22128 Thickness of membranes in cells of living organisms 10\u22122 Mass of a hummingbird 10\u22128 Time for one oscillation of an FM radio wave Mass of a liter of water (about a quart) 10\u22123 Duration of a nerve impulse 10\u22126 Wavelength of visible light 10\u22123 Size of a grain of sand Height of a 4-year-old child Length of a football field 1 102 103 108 Mass of a person Mass of a car Mass of a large ship Greatest ocean depth 1012 Mass of a large iceberg Diameter of the Earth 1015 Mass of the nucleus of a comet 1011 Recorded history 1 105 107 109 Time for one heartbeat One day", " 8.64\u00d7104 s One year (y) 3.16\u00d7107 s About half the life expectancy of a human 1017 1018 Age of the Earth Age of the universe 1 102 104 107 1011 Distance from the Earth to the Sun 1023 Mass of the Moon 7.35\u00d71022 kg 1016 Distance traveled by light in 1 year (a light year) 1025 Mass of the Earth 5.97\u00d71024 kg 1021 1022 1026 Diameter of the Milky Way galaxy 1030 Mass of the Sun 1.99\u00d71030 kg Distance from the Earth to the nearest large galaxy (Andromeda) Distance from the Earth to the edges of the known universe 1042 Mass of the Milky Way galaxy (current upper limit) 1053 Mass of the known universe (current upper limit) Example 1.1 Unit Conversions: A Short Drive Home Suppose that you drive the 10.0 km from your university to home in 20.0 min. Calculate your average speed (a) in kilometers per hour (km/h) and (b) in meters per second (m/s). (Note: Average speed is distance traveled divided by time of travel.) Strategy First we calculate the average speed using the given units. Then we can get the average speed into the desired units by picking the correct conversion factor and multiplying by it. The correct conversion factor is the one that cancels the unwanted unit and leaves the desired unit in its place. Solution for (a) (1) Calculate average speed. Average speed is distance traveled divided by time of travel. (Take this definition as a given for now\u2014average speed and other motion concepts will be covered in a later module.) In equation form, average speed =distance time. (1.2) (2) Substitute the given values for distance and time. This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 1 | Introduction: The Nature of Science and Physics average speed = 10.0 km 20.0 min = 0.500 km min. (3) Convert km/min to km/h: multiply by the conversion factor that will cancel minutes and leave hours. That conversion factor is 60 min/hr. Thus, average speed =0.500 km min \u00d7 60 min 1 h = 30.0 km h. Discussion for (a) To check your answer, consider the following: 21 (1.3) (1.4) (1) Be sure that you have properly cancelled the units in the unit conversion. If you have written the unit conversion factor upside down, the units will not cancel properly in the equation. If you accidentally get the ratio upside down, then the units will not cancel; rather, they will give you the wrong units as follows: = 1 60 \u00d7 1 hr 60 min km min, km \u22c5 hr min2 (1.5) which are obviously not the desired units of km/h. (2) Check that the units of the final answer are the desired units. The problem asked us to solve for average speed in units of km/h and we have indeed obtained these units. (3) Check the significant figures. Because each of the values given in the problem has three significant figures, the answer should also have three significant figures. The answer 30.0 km/hr does indeed have three significant figures, so this is appropriate. Note that the significant figures in the conversion factor are not relevant because an hour is defined to be 60 minutes, so the precision of the conversion factor is perfect. (4) Next, check whether the answer is reasonable. Let us consider some information from the problem\u2014if you travel 10 km in a third of an hour (20 min), you would travel three times that far in an hour. The answer does seem reasonable. Solution for (b) There are several ways to convert the average speed into meters per second. (1) Start with the answer to (a) and convert km/h to m/s. Two conversion factors are needed\u2014one to convert hours to seconds, and another to convert kilometers to meters. (2) Multiplying by these yields Average speed = 30.0km h 3,600 s Average speed = 8.33m s. \u00d7 1 h \u00d7 1,000 m 1 km, (1.6) (1.7) Discussion for (b) If we had started with 0.500 km/min, we would have needed different conversion factors, but the answer would have been the same: 8.33 m/s. You may have noted that the answers in the worked example just covered were given to three digits. Why? When do you need to be concerned about the number of digits in something you calculate? Why not write down all the digits your calculator produces? The module Accuracy, Precision, and Significant Figures will help you answer these questions. Nonstandard Units While there are numerous types of units that we are all familiar with, there are others that are much more", " obscure. For example, a firkin is a unit of volume that was once used to measure beer. One firkin equals about 34 liters. To learn more about nonstandard units, use a dictionary or encyclopedia to research different \u201cweights and measures.\u201d Take note of any unusual units, such as a barleycorn, that are not listed in the text. Think about how the unit is defined and state its relationship to SI units. Check Your Understanding Some hummingbirds beat their wings more than 50 times per second. A scientist is measuring the time it takes for a hummingbird to beat its wings once. Which fundamental unit should the scientist use to describe the measurement? Which factor of 10 is the scientist likely to use to describe the motion precisely? Identify the metric prefix that corresponds to this factor of 10. Solution 22 Chapter 1 | Introduction: The Nature of Science and Physics The scientist will measure the time between each movement using the fundamental unit of seconds. Because the wings beat so fast, the scientist will probably need to measure in milliseconds, or 10\u22123 20 milliseconds per beat.) seconds. (50 beats per second corresponds to Check Your Understanding One cubic centimeter is equal to one milliliter. What does this tell you about the different units in the SI metric system? Solution The fundamental unit of length (meter) is probably used to create the derived unit of volume (liter). The measure of a milliliter is dependent on the measure of a centimeter. 1.3 Accuracy, Precision, and Significant Figures Figure 1.22 A double-pan mechanical balance is used to compare different masses. Usually an object with unknown mass is placed in one pan and objects of known mass are placed in the other pan. When the bar that connects the two pans is horizontal, then the masses in both pans are equal. The \u201cknown masses\u201d are typically metal cylinders of standard mass such as 1 gram, 10 grams, and 100 grams. (credit: Serge Melki) Figure 1.23 Many mechanical balances, such as double-pan balances, have been replaced by digital scales, which can typically measure the mass of an object more precisely. Whereas a mechanical balance may only read the mass of an object to the nearest tenth of a gram, many digital scales can measure the mass of an object up to the nearest thousandth of a gram. (credit: Karel Jakubec) By the end of this section, you will be able to: Learning Objectives \u2022 Determine the appropriate number of significant figures in both addition and subtraction, as well as multiplication and division calculations. \u2022 Calculate the percent uncertainty of a measurement. Accuracy and Precision of a Measurement Science is based on observation and experiment\u2014that is, on measurements. Accuracy is how close a measurement is to the correct value for that measurement. For example, let us say that you are measuring the length of standard computer paper. The This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 1 | Introduction: The Nature of Science and Physics 23 packaging in which you purchased the paper states that it is 11.0 inches long. You measure the length of the paper three times and obtain the following measurements: 11.1 in., 11.2 in., and 10.9 in. These measurements are quite accurate because they are very close to the correct value of 11.0 inches. In contrast, if you had obtained a measurement of 12 inches, your measurement would not be very accurate. The precision of a measurement system is refers to how close the agreement is between repeated measurements (which are repeated under the same conditions). Consider the example of the paper measurements. The precision of the measurements refers to the spread of the measured values. One way to analyze the precision of the measurements would be to determine the range, or difference, between the lowest and the highest measured values. In that case, the lowest value was 10.9 in. and the highest value was 11.2 in. Thus, the measured values deviated from each other by at most 0.3 in. These measurements were relatively precise because they did not vary too much in value. However, if the measured values had been 10.9, 11.1, and 11.9, then the measurements would not be very precise because there would be significant variation from one measurement to another. The measurements in the paper example are both accurate and precise, but in some cases, measurements are accurate but not precise, or they are precise but not accurate. Let us consider an example of a GPS system that is attempting to locate the position of a restaurant in a city. Think of the restaurant location as existing at the center of a bull's-eye target, and think of each GPS attempt to locate the restaurant as a black dot. In Figure 1.24, you can see that the GPS measurements are spread out far apart from each other, but they are all relatively close to the actual location of the restaurant at the center of the target. This indicates a low precision,", " high accuracy measuring system. However, in Figure 1.25, the GPS measurements are concentrated quite closely to one another, but they are far away from the target location. This indicates a high precision, low accuracy measuring system. Figure 1.24 A GPS system attempts to locate a restaurant at the center of the bull's-eye. The black dots represent each attempt to pinpoint the location of the restaurant. The dots are spread out quite far apart from one another, indicating low precision, but they are each rather close to the actual location of the restaurant, indicating high accuracy. (credit: Dark Evil) Figure 1.25 In this figure, the dots are concentrated rather closely to one another, indicating high precision, but they are rather far away from the actual location of the restaurant, indicating low accuracy. (credit: Dark Evil) Accuracy, Precision, and Uncertainty The degree of accuracy and precision of a measuring system are related to the uncertainty in the measurements. Uncertainty is a quantitative measure of how much your measured values deviate from a standard or expected value. If your measurements are not very accurate or precise, then the uncertainty of your values will be very high. In more general terms, uncertainty can be thought of as a disclaimer for your measured values. For example, if someone asked you to provide the mileage on your car, you might say that it is 45,000 miles, plus or minus 500 miles. The plus or minus amount is the uncertainty in your value. That is, you are indicating that the actual mileage of your car might be as low as 44,500 miles or as high as 45,500 miles, or anywhere in between. All measurements contain some amount of uncertainty. In our example of measuring the length of the paper, we might say that the length of the paper is 11 in., plus or minus 0.2 in. The uncertainty in a measurement,, is often denoted as (\u201cdelta \u201d), so the measurement result would be recorded as \u00b1. In our paper example, the length of the paper could be expressed as 11 in. \u00b1 0.2. The factors contributing to uncertainty in a measurement include: 24 Chapter 1 | Introduction: The Nature of Science and Physics 1. Limitations of the measuring device, 2. The skill of the person making the measurement, 3. Irregularities in the object being measured, 4. Any other factors that affect the outcome (highly dependent on the situation). In our example, such factors contributing to the uncertainty could be the following: the smallest division on the ruler is 0.1 in., the person using the ruler has bad eyesight, or one side of the paper is slightly longer than the other. At any rate, the uncertainty in a measurement must be based on a careful consideration of all the factors that might contribute and their possible effects. Making Connections: Real-World Connections\u2014Fevers or Chills? Uncertainty is a critical piece of information, both in physics and in many other real-world applications. Imagine you are caring for a sick child. You suspect the child has a fever, so you check his or her temperature with a thermometer. What if the uncertainty of the thermometer were 3.0\u00baC? If the child's temperature reading was 37.0\u00baC (which is normal body temperature), the \u201ctrue\u201d temperature could be anywhere from a hypothermic 34.0\u00baC to a dangerously high 40.0\u00baC. A thermometer with an uncertainty of 3.0\u00baC would be useless. Percent Uncertainty One method of expressing uncertainty is as a percent of the measured value. If a measurement is expressed with uncertainty,, the percent uncertainty (%unc) is defined to be % unc = \u00d7100%. (1.8) Example 1.2 Calculating Percent Uncertainty: A Bag of Apples A grocery store sells 5 lb bags of apples. You purchase four bags over the course of a month and weigh the apples each time. You obtain the following measurements: \u2022 Week 1 weight: 4.8 lb \u2022 Week 2 weight: 5.3 lb \u2022 Week 3 weight: 4.9 lb \u2022 Week 4 weight: 5.4 lb You determine that the weight of the 5 lb bag has an uncertainty of \u00b10.4 lb. What is the percent uncertainty of the bag's weight? Strategy First, observe that the expected value of the bag's weight,, is 5 lb. The uncertainty in this value,, is 0.4 lb. We can use the following equation to determine the percent uncertainty of the weight: Solution Plug the known values into the equation: Discussion % unc = \u00d7100%. % unc =0.4 lb 5 lb \u00d7100% = 8%. (1.9) (1.10) We can conclude that the weight of the apple bag is 5 lb \u00b1 8%. Consider how this percent uncertainty would change if the bag of apples were half as heavy, but the uncertainty in the weight remained the same. Hint for future calculations:", " when calculating percent uncertainty, always remember that you must multiply the fraction by 100%. If you do not do this, you will have a decimal quantity, not a percent value. Uncertainties in Calculations There is an uncertainty in anything calculated from measured quantities. For example, the area of a floor calculated from measurements of its length and width has an uncertainty because the length and width have uncertainties. How big is the uncertainty in something you calculate by multiplication or division? If the measurements going into the calculation have small uncertainties (a few percent or less), then the method of adding percents can be used for multiplication or division. This method says that the percent uncertainty in a quantity calculated by multiplication or division is the sum of the percent This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 1 | Introduction: The Nature of Science and Physics 25 uncertainties in the items used to make the calculation. For example, if a floor has a length of 4.00 m and a width of 3.00 m, with uncertainties of 2% and 1%, respectively, then the area of the floor is 12.0 m2 and has an uncertainty of 3%. (Expressed as an area this is 0.36 m2, which we round to 0.4 m2 since the area of the floor is given to a tenth of a square meter.) Check Your Understanding A high school track coach has just purchased a new stopwatch. The stopwatch manual states that the stopwatch has an uncertainty of \u00b10.05 s. Runners on the track coach's team regularly clock 100 m sprints of 11.49 s to 15.01 s. At the school's last track meet, the first-place sprinter came in at 12.04 s and the second-place sprinter came in at 12.07 s. Will the coach's new stopwatch be helpful in timing the sprint team? Why or why not? Solution No, the uncertainty in the stopwatch is too great to effectively differentiate between the sprint times. Precision of Measuring Tools and Significant Figures An important factor in the accuracy and precision of measurements involves the precision of the measuring tool. In general, a precise measuring tool is one that can measure values in very small increments. For example, a standard ruler can measure length to the nearest millimeter, while a caliper can measure length to the nearest 0.01 millimeter. The caliper is a more precise measuring tool because it can measure extremely small differences in length. The more precise the measuring tool, the more precise and accurate the measurements can be. When we express measured values, we can only list as many digits as we initially measured with our measuring tool. For example, if you use a standard ruler to measure the length of a stick, you may measure it to be 36.7 cm. You could not express this value as 36.71 cm because your measuring tool was not precise enough to measure a hundredth of a centimeter. It should be noted that the last digit in a measured value has been estimated in some way by the person performing the measurement. For example, the person measuring the length of a stick with a ruler notices that the stick length seems to be somewhere in between 36.6 cm and 36.7 cm, and he or she must estimate the value of the last digit. Using the method of significant figures, the rule is that the last digit written down in a measurement is the first digit with some uncertainty. In order to determine the number of significant digits in a value, start with the first measured value at the left and count the number of digits through the last digit written on the right. For example, the measured value 36.7 cm has three digits, or significant figures. Significant figures indicate the precision of a measuring tool that was used to measure a value. Zeros Special consideration is given to zeros when counting significant figures. The zeros in 0.053 are not significant, because they are only placekeepers that locate the decimal point. There are two significant figures in 0.053. The zeros in 10.053 are not placekeepers but are significant\u2014this number has five significant figures. The zeros in 1300 may or may not be significant depending on the style of writing numbers. They could mean the number is known to the last digit, or they could be placekeepers. So 1300 could have two, three, or four significant figures. (To avoid this ambiguity, write 1300 in scientific notation.) Zeros are significant except when they serve only as placekeepers. Check Your Understanding Determine the number of significant figures in the following measurements: a. 0.0009 b. 15,450.0 c. 6\u00d7103 d. 87.990 e. 30.42 Solution (a) 1; the zeros in this number are placekeepers that indicate the decimal point (b) 6; here, the zeros indicate that a measurement was made to the 0.1 decimal point, so the zeros", " are significant (c) 1; the value 103 signifies the decimal place, not the number of measured values (d) 5; the final zero indicates that a measurement was made to the 0.001 decimal point, so it is significant (e) 4; any zeros located in between significant figures in a number are also significant 26 Chapter 1 | Introduction: The Nature of Science and Physics Significant Figures in Calculations When combining measurements with different degrees of accuracy and precision, the number of significant digits in the final answer can be no greater than the number of significant digits in the least precise measured value. There are two different rules, one for multiplication and division and the other for addition and subtraction, as discussed below. 1. For multiplication and division: The result should have the same number of significant figures as the quantity having the least significant figures entering into the calculation. For example, the area of a circle can be calculated from its radius using = 2. Let us see how many significant figures the area has if the radius has only two\u2014say, = 1.2 m. Then, = 2 = (3.1415927...)\u00d7(1.2 m)2 = 4.5238934 m2 (1.11) is what you would get using a calculator that has an eight-digit output. But because the radius has only two significant figures, it limits the calculated quantity to two significant figures or =4.5 m2, (1.12) even though is good to at least eight digits. 2. For addition and subtraction: The answer can contain no more decimal places than the least precise measurement. Suppose that you buy 7.56 kg of potatoes in a grocery store as measured with a scale with precision 0.01 kg. Then you drop off 6.052 kg of potatoes at your laboratory as measured by a scale with precision 0.001 kg. Finally, you go home and add 13.7 kg of potatoes as measured by a bathroom scale with precision 0.1 kg. How many kilograms of potatoes do you now have, and how many significant figures are appropriate in the answer? The mass is found by simple addition and subtraction: 7.56 kg - 6.052 kg kg +13.7 15.208 kg = 15.2 kg. (1.13) Next, we identify the least precise measurement: 13.7 kg. This measurement is expressed to the 0.1 decimal place, so our final answer must also be expressed to the 0.1 decimal place. Thus, the answer is rounded to the tenths place, giving us 15.2 kg. Significant Figures in this Text In this text, most numbers are assumed to have three significant figures. Furthermore, consistent numbers of significant figures are used in all worked examples. You will note that an answer given to three digits is based on input good to at least three digits, for example. If the input has fewer significant figures, the answer will also have fewer significant figures. Care is also taken that the number of significant figures is reasonable for the situation posed. In some topics, particularly in optics, more accurate numbers are needed and more than three significant figures will be used. Finally, if a number is exact, such as the two in the formula for the circumference of a circle, = 2\u03c0, it does not affect the number of significant figures in a calculation. Check Your Understanding Perform the following calculations and express your answer using the correct number of significant digits. (a) A woman has two bags weighing 13.5 pounds and one bag with a weight of 10.2 pounds. What is the total weight of the bags? (b) The force on an object is equal to its mass multiplied by its acceleration. If a wagon with mass 55 kg accelerates at a rate of 0.0255 m/s2, what is the force on the wagon? (The unit of force is called the newton, and it is expressed with the symbol N.) Solution (a) 37.2 pounds; Because the number of bags is an exact value, it is not considered in the significant figures. (b) 1.4 N; Because the value 55 kg has only two significant figures, the final value must also contain two significant figures. PhET Explorations: Estimation Explore size estimation in one, two, and three dimensions! Multiple levels of difficulty allow for progressive skill improvement. Figure 1.26 Estimation (http://cnx.org/content/m54766/1.7/estimation_en.jar) This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 1 | Introduction: The Nature of Science and Physics 27 1.4 Approximation Learning Objectives By the end of this section, you will be able to: \u2022 Make reasonable approximations based on given data. On many occasions, physicists, other scientists, and engineers need to make approximations or \u201cguesstimates\u201d for a particular quantity. What is", " the distance to a certain destination? What is the approximate density of a given item? About how large a current will there be in a circuit? Many approximate numbers are based on formulae in which the input quantities are known only to a limited accuracy. As you develop problem-solving skills (that can be applied to a variety of fields through a study of physics), you will also develop skills at approximating. You will develop these skills through thinking more quantitatively, and by being willing to take risks. As with any endeavor, experience helps, as well as familiarity with units. These approximations allow us to rule out certain scenarios or unrealistic numbers. Approximations also allow us to challenge others and guide us in our approaches to our scientific world. Let us do two examples to illustrate this concept. Example 1.3 Approximate the Height of a Building Can you approximate the height of one of the buildings on your campus, or in your neighborhood? Let us make an approximation based upon the height of a person. In this example, we will calculate the height of a 39-story building. Strategy Think about the average height of an adult male. We can approximate the height of the building by scaling up from the height of a person. Solution Based on information in the example, we know there are 39 stories in the building. If we use the fact that the height of one story is approximately equal to about the length of two adult humans (each human is about 2 m tall), then we can estimate the total height of the building to be 2 m 1 person \u00d7 2 person 1 story \u00d739 stories = 156 m. (1.14) Discussion You can use known quantities to determine an approximate measurement of unknown quantities. If your hand measures 10 cm across, how many hand lengths equal the width of your desk? What other measurements can you approximate besides length? Example 1.4 Approximating Vast Numbers: a Trillion Dollars Figure 1.27 A bank stack contains one-hundred $100 bills, and is worth $10,000. How many bank stacks make up a trillion dollars? (credit: Andrew Magill) The U.S. federal deficit in the 2008 fiscal year was a little greater than $10 trillion. Most of us do not have any concept of how much even one trillion actually is. Suppose that you were given a trillion dollars in $100 bills. If you made 100-bill stacks and used them to evenly cover a football field (between the end zones), make an approximation of how high the money pile 28 Chapter 1 | Introduction: The Nature of Science and Physics would become. (We will use feet/inches rather than meters here because football fields are measured in yards.) One of your friends says 3 in., while another says 10 ft. What do you think? Strategy When you imagine the situation, you probably envision thousands of small stacks of 100 wrapped $100 bills, such as you might see in movies or at a bank. Since this is an easy-to-approximate quantity, let us start there. We can find the volume of a stack of 100 bills, find out how many stacks make up one trillion dollars, and then set this volume equal to the area of the football field multiplied by the unknown height. Solution (1) Calculate the volume of a stack of 100 bills. The dimensions of a single bill are approximately 3 in. by 6 in. A stack of 100 of these is about 0.5 in. thick. So the total volume of a stack of 100 bills is: volume of stack = length\u00d7width\u00d7height, volume of stack = 6 in.\u00d73 in.\u00d70.5 in. volume of stack = 9 in.3. (1.15) (2) Calculate the number of stacks. Note that a trillion dollars is equal to $1\u00d71012, and a stack of one-hundred $100 bills is equal to $10000 or $1\u00d7104. The number of stacks you will have is: $1\u00d71012(a trillion dollars)/ $1\u00d7104 per stack = 1\u00d7108 stacks. (3) Calculate the area of a football field in square inches. The area of a football field is 100 yd\u00d750 yd, which gives 5,000 yd2. Because we are working in inches, we need to convert square yards to square inches: Area = 5,000 yd2\u00d7 3 ft 1 yd \u00d7 3 ft 1 yd \u00d7 12 in. 1 ft \u00d7 12 in. 1 ft = 6,480000 in.2 Area \u2248 6\u00d7106 in.2. (1.16) (1.17) This conversion gives us 6\u00d7106 in.2 for the area of the field. (Note that we are using only one significant figure in these calculations.) (4) Calculate the total volume of the bills. The volume of all the $100 -bill stacks is 9 in.3 / stack\u00d7108 stacks = 9\u00d7108 in.3", ". (5) Calculate the height. To determine the height of the bills, use the equation: = area of field\u00d7height of money: volume of bills Height of money = volume of bills area of field Height of money = 9\u00d7108in.3 6\u00d7106in.2 = 1.33\u00d7102 in., Height of money \u2248 1\u00d7102 in. = 100 in. The height of the money will be about 100 in. high. Converting this value to feet gives 100 in.\u00d7 1 ft 12 in. = 8.33 ft \u2248 8 ft. Discussion (1.18) (1.19) The final approximate value is much higher than the early estimate of 3 in., but the other early estimate of 10 ft (120 in.) was roughly correct. How did the approximation measure up to your first guess? What can this exercise tell you in terms of rough \u201cguesstimates\u201d versus carefully calculated approximations? Check Your Understanding Using mental math and your understanding of fundamental units, approximate the area of a regulation basketball court. Describe the process you used to arrive at your final approximation. Solution An average male is about two meters tall. It would take approximately 15 men laid out end to end to cover the length, and about 7 to cover the width. That gives an approximate area of 420 m2. This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 1 | Introduction: The Nature of Science and Physics 29 Glossary accuracy: the degree to which a measured value agrees with correct value for that measurement approximation: an estimated value based on prior experience and reasoning classical physics: physics that was developed from the Renaissance to the end of the 19th century conversion factor: a ratio expressing how many of one unit are equal to another unit derived units: units that can be calculated using algebraic combinations of the fundamental units English units: system of measurement used in the United States; includes units of measurement such as feet, gallons, and pounds fundamental units: units that can only be expressed relative to the procedure used to measure them kilogram: the SI unit for mass, abbreviated (kg) law: a description, using concise language or a mathematical formula, a generalized pattern in nature that is supported by scientific evidence and repeated experiments meter: the SI unit for length, abbreviated (m) method of adding percents: the percent uncertainty in a quantity calculated by multiplication or division is the sum of the percent uncertainties in the items used to make the calculation metric system: a system in which values can be calculated in factors of 10 model: representation of something that is often too difficult (or impossible) to display directly modern physics: the study of relativity, quantum mechanics, or both order of magnitude: refers to the size of a quantity as it relates to a power of 10 percent uncertainty: the ratio of the uncertainty of a measurement to the measured value, expressed as a percentage physical quantity : a characteristic or property of an object that can be measured or calculated from other measurements physics: the science concerned with describing the interactions of energy, matter, space, and time; it is especially interested in what fundamental mechanisms underlie every phenomenon precision: the degree to which repeated measurements agree with each other quantum mechanics: the study of objects smaller than can be seen with a microscope relativity: the study of objects moving at speeds greater than about 1% of the speed of light, or of objects being affected by a strong gravitational field scientific method: a method that typically begins with an observation and question that the scientist will research; next, the scientist typically performs some research about the topic and then devises a hypothesis; then, the scientist will test the hypothesis by performing an experiment; finally, the scientist analyzes the results of the experiment and draws a conclusion second: the SI unit for time, abbreviated (s) SI units : the international system of units that scientists in most countries have agreed to use; includes units such as meters, liters, and grams significant figures: express the precision of a measuring tool used to measure a value theory: an explanation for patterns in nature that is supported by scientific evidence and verified multiple times by various groups of researchers uncertainty: a quantitative measure of how much your measured values deviate from a standard or expected value units : a standard used for expressing and comparing measurements Section Summary 1.1 Physics: An Introduction \u2022 Science seeks to discover and describe the underlying order and simplicity in nature. 30 Chapter 1 | Introduction: The Nature of Science and Physics \u2022 Physics is the most basic of the sciences, concerning itself with energy, matter, space and time, and their interactions. \u2022 Scientific laws and theories express the general truths of nature and the body of knowledge they encompass. These laws of nature are rules that all natural processes appear to follow. 1.2 Physical Quantities and Units \u2022 Physical quantities are a characteristic or property of an object that can be measured or calculated from other measurements. \u2022 Units are standards for expressing and comparing the measurement of physical quantities. All units", " can be expressed as combinations of four fundamental units. \u2022 The four fundamental units we will use in this text are the meter (for length), the kilogram (for mass), the second (for time), and the ampere (for electric current). These units are part of the metric system, which uses powers of 10 to relate quantities over the vast ranges encountered in nature. \u2022 The four fundamental units are abbreviated as follows: meter, m; kilogram, kg; second, s; and ampere, A. The metric system also uses a standard set of prefixes to denote each order of magnitude greater than or lesser than the fundamental unit itself. \u2022 Unit conversions involve changing a value expressed in one type of unit to another type of unit. This is done by using conversion factors, which are ratios relating equal quantities of different units. 1.3 Accuracy, Precision, and Significant Figures \u2022 Accuracy of a measured value refers to how close a measurement is to the correct value. The uncertainty in a measurement is an estimate of the amount by which the measurement result may differ from this value. \u2022 Precision of measured values refers to how close the agreement is between repeated measurements. \u2022 The precision of a measuring tool is related to the size of its measurement increments. The smaller the measurement increment, the more precise the tool. \u2022 Significant figures express the precision of a measuring tool. \u2022 When multiplying or dividing measured values, the final answer can contain only as many significant figures as the least precise value. \u2022 When adding or subtracting measured values, the final answer cannot contain more decimal places than the least precise value. 1.4 Approximation Scientists often approximate the values of quantities to perform calculations and analyze systems. Conceptual Questions 1.1 Physics: An Introduction 1. Models are particularly useful in relativity and quantum mechanics, where conditions are outside those normally encountered by humans. What is a model? 2. How does a model differ from a theory? 3. If two different theories describe experimental observations equally well, can one be said to be more valid than the other (assuming both use accepted rules of logic)? 4. What determines the validity of a theory? 5. Certain criteria must be satisfied if a measurement or observation is to be believed. Will the criteria necessarily be as strict for an expected result as for an unexpected result? 6. Can the validity of a model be limited, or must it be universally valid? How does this compare to the required validity of a theory or a law? 7. Classical physics is a good approximation to modern physics under certain circumstances. What are they? 8. When is it necessary to use relativistic quantum mechanics? 9. Can classical physics be used to accurately describe a satellite moving at a speed of 7500 m/s? Explain why or why not. 1.2 Physical Quantities and Units 10. Identify some advantages of metric units. 1.3 Accuracy, Precision, and Significant Figures 11. What is the relationship between the accuracy and uncertainty of a measurement? 12. Prescriptions for vision correction are given in units called diopters (D). Determine the meaning of that unit. Obtain information (perhaps by calling an optometrist or performing an internet search) on the minimum uncertainty with which corrections in diopters are determined and the accuracy with which corrective lenses can be produced. Discuss the sources of uncertainties in both the prescription and accuracy in the manufacture of lenses. This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 1 | Introduction: The Nature of Science and Physics 31 Problems & Exercises 1.2 Physical Quantities and Units 1. The speed limit on some interstate highways is roughly 100 km/h. (a) What is this in meters per second? (b) How many miles per hour is this? 2. A car is traveling at a speed of 33 m/s. (a) What is its speed in kilometers per hour? (b) Is it exceeding the 90 km/h speed limit? 3. Show that 1.0 m/s = 3.6 km/h. Hint: Show the explicit steps involved in converting 1.0 m/s = 3.6 km/h. 4. American football is played on a 100-yd-long field, excluding the end zones. How long is the field in meters? (Assume that 1 meter equals 3.281 feet.) 5. Soccer fields vary in size. A large soccer field is 115 m long and 85 m wide. What are its dimensions in feet and inches? (Assume that 1 meter equals 3.281 feet.) 6. What is the height in meters of a person who is 6 ft 1.0 in. tall? (Assume that 1 meter equals 39.37 in.) 7. Mount Everest, at 29,028 feet, is the tallest mountain on the Earth. What is its height in kilometers? (Assume that 1 kilometer equals 3,281 feet.) 8. The speed of sound is", " measured to be 342 m/s on a certain day. What is this in km/h? 9. Tectonic plates are large segments of the Earth's crust that move slowly. Suppose that one such plate has an average speed of 4.0 cm/year. (a) What distance does it move in 1 s at this speed? (b) What is its speed in kilometers per million years? 10. (a) Refer to Table 1.3 to determine the average distance between the Earth and the Sun. Then calculate the average speed of the Earth in its orbit in kilometers per second. (b) What is this in meters per second? 1.3 Accuracy, Precision, and Significant Figures Express your answers to problems in this section to the correct number of significant figures and proper units. 11. Suppose that your bathroom scale reads your mass as 65 kg with a 3% uncertainty. What is the uncertainty in your mass (in kilograms)? 12. A good-quality measuring tape can be off by 0.50 cm over a distance of 20 m. What is its percent uncertainty? 13. (a) A car speedometer has a 5.0% uncertainty. What is the range of possible speeds when it reads 90 km/h? (b) Convert this range to miles per hour. (1 km = 0.6214 mi) 14. An infant's pulse rate is measured to be 130 \u00b1 5 beats/ min. What is the percent uncertainty in this measurement? 15. (a) Suppose that a person has an average heart rate of 72.0 beats/min. How many beats does he or she have in 2.0 y? (b) In 2.00 y? (c) In 2.000 y? (106.7)(98.2) / (46.210)(1.01) (b) (18.7)2 (c) 1.60\u00d710\u221219 (3712). 18. (a) How many significant figures are in the numbers 99 and 100? (b) If the uncertainty in each number is 1, what is the percent uncertainty in each? (c) Which is a more meaningful way to express the accuracy of these two numbers, significant figures or percent uncertainties? 19. (a) If your speedometer has an uncertainty of 2.0 km/h at a speed of 90 km/h, what is the percent uncertainty? (b) If it has the same percent uncertainty when it reads 60 km/h, what is the range of speeds you could be going? 20. (a) A person's blood pressure is measured to be 120 \u00b1 2 mm Hg. What is its percent uncertainty? (b) Assuming the same percent uncertainty, what is the uncertainty in a blood pressure measurement of 80 mm Hg? 21. A person measures his or her heart rate by counting the number of beats in 30 s. If 40 \u00b1 1 beats are counted in 30.0 \u00b1 0.5 s, what is the heart rate and its uncertainty in beats per minute? 22. What is the area of a circle 3.102 cm in diameter? 23. If a marathon runner averages 9.5 mi/h, how long does it take him or her to run a 26.22 mi marathon? 24. A marathon runner completes a 42.188 km course in 2 h, 30 min, and 12 s. There is an uncertainty of 25 m in the distance traveled and an uncertainty of 1 s in the elapsed time. (a) Calculate the percent uncertainty in the distance. (b) Calculate the uncertainty in the elapsed time. (c) What is the average speed in meters per second? (d) What is the uncertainty in the average speed? 25. The sides of a small rectangular box are measured to be 1.80 \u00b1 0.01 cm, 2.05 \u00b1 0.02 cm, and 3.1 \u00b1 0.1 cm long. Calculate its volume and uncertainty in cubic centimeters. 26. When non-metric units were used in the United Kingdom, a unit of mass called the pound-mass (lbm) was employed, where 1 lbm = 0.4539 kg. (a) If there is an uncertainty of 0.0001 kg in the pound-mass unit, what is its percent uncertainty? (b) Based on that percent uncertainty, what mass in pound-mass has an uncertainty of 1 kg when converted to kilograms? 27. The length and width of a rectangular room are measured to be 3.955 \u00b1 0.005 m and 3.050 \u00b1 0.005 m. Calculate the area of the room and its uncertainty in square meters. 28. A car engine moves a piston with a circular cross section of 7.500 \u00b1 0.002 cm diameter a distance of 3.250 \u00b1 0.001 cm to compress the gas in the cylinder. (a) By what amount is the gas decreased in volume in cubic centimeters? (b) Find the uncertainty in this", " volume. 16. A can contains 375 mL of soda. How much is left after 308 mL is removed? 1.4 Approximation 17. State how many significant figures are proper in the results of the following calculations: (a) 29. How many heartbeats are there in a lifetime? 32 Chapter 1 | Introduction: The Nature of Science and Physics 30. A generation is about one-third of a lifetime. Approximately how many generations have passed since the year 0 AD? 31. How many times longer than the mean life of an extremely unstable atomic nucleus is the lifetime of a human? (Hint: The lifetime of an unstable atomic nucleus is on the order of 10\u221222 s.) 32. Calculate the approximate number of atoms in a bacterium. Assume that the average mass of an atom in the bacterium is ten times the mass of a hydrogen atom. (Hint: The mass of a hydrogen atom is on the order of 10\u221227 kg and the mass of a bacterium is on the order of 10\u221215 kg. ) Figure 1.28 This color-enhanced photo shows Salmonella typhimurium (red) attacking human cells. These bacteria are commonly known for causing foodborne illness. Can you estimate the number of atoms in each bacterium? (credit: Rocky Mountain Laboratories, NIAID, NIH) 33. Approximately how many atoms thick is a cell membrane, assuming all atoms there average about twice the size of a hydrogen atom? 34. (a) What fraction of Earth's diameter is the greatest ocean depth? (b) The greatest mountain height? 35. (a) Calculate the number of cells in a hummingbird assuming the mass of an average cell is ten times the mass of a bacterium. (b) Making the same assumption, how many cells are there in a human? 36. Assuming one nerve impulse must end before another can begin, what is the maximum firing rate of a nerve in impulses per second? This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 2 | Kinematics 33 2 KINEMATICS Figure 2.1 The motion of an American kestrel through the air can be described by the bird's displacement, speed, velocity, and acceleration. When it flies in a straight line without any change in direction, its motion is said to be one dimensional. (credit: Vince Maidens, Wikimedia Commons) Chapter Outline 2.1. Displacement 2.2. Vectors, Scalars, and Coordinate Systems 2.3. Time, Velocity, and Speed 2.4. Acceleration 2.5. Motion Equations for Constant Acceleration in One Dimension 2.6. Problem-Solving Basics for One Dimensional Kinematics 2.7. Falling Objects 2.8. Graphical Analysis of One Dimensional Motion Connection for AP\u00ae Courses Objects are in motion everywhere we look. Everything from a tennis game to a space-probe flyby of the planet Neptune involves motion. When you are resting, your heart moves blood through your veins. Even in inanimate objects, there is a continuous motion in the vibrations of atoms and molecules. Questions about motion are interesting in and of themselves: How long will it take for a space probe to get to Mars? Where will a football land if it is thrown at a certain angle? Understanding motion will not only provide answers to these questions, but will be key to understanding more advanced concepts in physics. For example, the discussion of force in Chapter 4 will not fully make sense until you understand acceleration. This relationship between force and acceleration is also critical to understanding Big Idea 3. Additionally, this unit will explore the topic of reference frames, a critical component to quantifying how things move. If you have ever waved to a departing friend at a train station, you are likely familiar with this idea. While you see your friend move away from you at a considerable rate, those sitting with her will likely see her as not moving. The effect that the chosen reference frame has on your observations is substantial, and an understanding of this is needed to grasp both Enduring Understanding 3.A and Essential Knowledge 3.A.1. Our formal study of physics begins with kinematics, which is defined as the study of motion without considering its causes. In one- and two-dimensional kinematics we will study only the motion of a football, for example, without worrying about what forces cause or change its motion. In this chapter, we examine the simplest type of motion\u2014namely, motion along a straight line, or one-dimensional motion. Later, in two-dimensional kinematics, we apply concepts developed here to study motion along curved paths (two- and three-dimensional motion), for example, that of a car rounding a curve. The content in this chapter supports: Big Idea 3 The interactions of an object with other objects can be described by forces. 34 Chapter 2 | Kinematics End", "uring Understanding 3.A All forces share certain common characteristics when considered by observers in inertial reference frames. Essential Knowledge 3.A.1 An observer in a particular reference frame can describe the motion of an object using such quantities as position, displacement, distance, velocity, speed, and acceleration. 2.1 Displacement Figure 2.2 These cyclists in Vietnam can be described by their position relative to buildings and a canal. Their motion can be described by their change in position, or displacement, in the frame of reference. (credit: Suzan Black, Fotopedia) Learning Objectives By the end of this section, you will be able to: \u2022 Define position, displacement, distance, and distance traveled in a particular frame of reference. \u2022 Explain the relationship between position and displacement. \u2022 Distinguish between displacement and distance traveled. \u2022 Calculate displacement and distance given initial position, final position, and the path between the two. The information presented in this section supports the following AP\u00ae learning objectives and science practices: \u2022 3.A.1.1 The student is able to express the motion of an object using narrative, mathematical, and graphical representations. (S.P. 1.5, 2.1, 2.2) \u2022 3.A.1.3 The student is able to analyze experimental data describing the motion of an object and is able to express the results of the analysis using narrative, mathematical, and graphical representations. (S.P. 5.1) Position In order to describe the motion of an object, you must first be able to describe its position\u2014where it is at any particular time. More precisely, you need to specify its position relative to a convenient reference frame. Earth is often used as a reference frame, and we often describe the position of an object as it relates to stationary objects in that reference frame. For example, a rocket launch would be described in terms of the position of the rocket with respect to the Earth as a whole, while a professor's position could be described in terms of where she is in relation to the nearby white board. (See Figure 2.3.) In other cases, we use reference frames that are not stationary but are in motion relative to the Earth. To describe the position of a person in an airplane, for example, we use the airplane, not the Earth, as the reference frame. (See Figure 2.4.) Displacement If an object moves relative to a reference frame (for example, if a professor moves to the right relative to a white board or a passenger moves toward the rear of an airplane), then the object's position changes. This change in position is known as displacement. The word \u201cdisplacement\u201d implies that an object has moved, or has been displaced. Displacement Displacement is the change in position of an object: where \u0394 is displacement, f is the final position, and 0 is the initial position. \u0394 = f \u2212 0, (2.1) This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 2 | Kinematics 35 In this text the upper case Greek letter \u0394 (delta) always means \u201cchange in\u201d whatever quantity follows it; thus, \u0394 means change in position. Always solve for displacement by subtracting initial position 0 from final position f. Note that the SI unit for displacement is the meter (m) (see Physical Quantities and Units), but sometimes kilometers, miles, feet, and other units of length are used. Keep in mind that when units other than the meter are used in a problem, you may need to convert them into meters to complete the calculation. Figure 2.3 A professor paces left and right while lecturing. Her position relative to the blackboard is given by. The +2.0 m displacement of the professor relative to the blackboard is represented by an arrow pointing to the right. Figure 2.4 A passenger moves from his seat to the back of the plane. His location relative to the airplane is given by. The \u22124 m displacement of the passenger relative to the plane is represented by an arrow toward the rear of the plane. Notice that the arrow representing his displacement is twice as long as the arrow representing the displacement of the professor (he moves twice as far) in Figure 2.3. Note that displacement has a direction as well as a magnitude. The professor's displacement is 2.0 m to the right, and the airline passenger's displacement is 4.0 m toward the rear. In one-dimensional motion, direction can be specified with a plus or minus sign. When you begin a problem, you should select which direction is positive (usually that will be to the right or up, but you are free to select positive as being any direction). The professor's initial position is 0 = 1.5 m and her final position is f = 3.5 m. Thus her displacement is \u0394 = f \u22120 = 3.5 m \u2212 1.", "5 m = + 2.0 m. (2.2) In this coordinate system, motion to the right is positive, whereas motion to the left is negative. Similarly, the airplane passenger's initial position is 0 = 6.0 m and his final position is f = 2.0 m, so his displacement is 36 Chapter 2 | Kinematics His displacement is negative because his motion is toward the rear of the plane, or in the negative direction in our coordinate system. \u0394 = f \u22120 = 2.0 m \u2212 6.0 m = \u22124.0 m. (2.3) Distance Although displacement is described in terms of direction, distance is not. Distance is defined to be the magnitude or size of displacement between two positions. Note that the distance between two positions is not the same as the distance traveled between them. Distance traveled is the total length of the path traveled between two positions. Distance has no direction and, thus, no sign. For example, the distance the professor walks is 2.0 m. The distance the airplane passenger walks is 4.0 m. Misconception Alert: Distance Traveled vs. Magnitude of Displacement It is important to note that the distance traveled, however, can be greater than the magnitude of the displacement (by magnitude, we mean just the size of the displacement without regard to its direction; that is, just a number with a unit). For example, the professor could pace back and forth many times, perhaps walking a distance of 150 m during a lecture, yet still end up only 2.0 m to the right of her starting point. In this case her displacement would be +2.0 m, the magnitude of her displacement would be 2.0 m, but the distance she traveled would be 150 m. In kinematics we nearly always deal with displacement and magnitude of displacement, and almost never with distance traveled. One way to think about this is to assume you marked the start of the motion and the end of the motion. The displacement is simply the difference in the position of the two marks and is independent of the path taken in traveling between the two marks. The distance traveled, however, is the total length of the path taken between the two marks. Check Your Understanding A cyclist rides 3 km west and then turns around and rides 2 km east. (a) What is her displacement? (b) What distance does she ride? (c) What is the magnitude of her displacement? Solution Figure 2.5 (a) The rider's displacement is \u0394 = f \u2212 0 = \u22121 km. (The displacement is negative because we take east to be positive and west to be negative.) (b) The distance traveled is 3 km + 2 km = 5 km. (c) The magnitude of the displacement is 1 km. This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 2 | Kinematics 37 2.2 Vectors, Scalars, and Coordinate Systems Figure 2.6 The motion of this Eclipse Concept jet can be described in terms of the distance it has traveled (a scalar quantity) or its displacement in a specific direction (a vector quantity). In order to specify the direction of motion, its displacement must be described based on a coordinate system. In this case, it may be convenient to choose motion toward the left as positive motion (it is the forward direction for the plane), although in many cases, the -coordinate runs from left to right, with motion to the right as positive and motion to the left as negative. (credit: Armchair Aviator, Flickr) Learning Objectives By the end of this section, you will be able to: \u2022 Define and distinguish between scalar and vector quantities. \u2022 Assign a coordinate system for a scenario involving one-dimensional motion. The information presented in this section supports the following AP\u00ae learning objectives and science practices: \u2022 3.A.1.2 The student is able to design an experimental investigation of the motion of an object. What is the difference between distance and displacement? Whereas displacement is defined by both direction and magnitude, distance is defined only by magnitude. Displacement is an example of a vector quantity. Distance is an example of a scalar quantity. A vector is any quantity with both magnitude and direction. Other examples of vectors include a velocity of 90 km/h east and a force of 500 newtons straight down. The direction of a vector in one-dimensional motion is given simply by a plus ( + ) or minus ( \u2212 ) sign. Vectors are represented graphically by arrows. An arrow used to represent a vector has a length proportional to the vector's magnitude (e.g., the larger the magnitude, the longer the length of the vector) and points in the same direction as the vector. Some physical quantities, like distance, either have no direction or none is specified. A scalar is any quantity that has a magnitude, but no direction. For example, a", " 20\u00baC temperature, the 250 kilocalories (250 Calories) of energy in a candy bar, a 90 km/h speed limit, a person's 1.8 m height, and a distance of 2.0 m are all scalars\u2014quantities with no specified direction. Note, however, that a scalar can be negative, such as a \u221220\u00baC temperature. In this case, the minus sign indicates a point on a scale rather than a direction. Scalars are never represented by arrows. Coordinate Systems for One-Dimensional Motion In order to describe the direction of a vector quantity, you must designate a coordinate system within the reference frame. For one-dimensional motion, this is a simple coordinate system consisting of a one-dimensional coordinate line. In general, when describing horizontal motion, motion to the right is usually considered positive, and motion to the left is considered negative. With vertical motion, motion up is usually positive and motion down is negative. In some cases, however, as with the jet in Figure 2.6, it can be more convenient to switch the positive and negative directions. For example, if you are analyzing the motion of falling objects, it can be useful to define downwards as the positive direction. If people in a race are running to the left, it is useful to define left as the positive direction. It does not matter as long as the system is clear and consistent. Once you assign a positive direction and start solving a problem, you cannot change it. 38 Chapter 2 | Kinematics Figure 2.7 It is usually convenient to consider motion upward or to the right as positive ( + ) and motion downward or to the left as negative ( \u2212 ). Check Your Understanding A person's speed can stay the same as he or she rounds a corner and changes direction. Given this information, is speed a scalar or a vector quantity? Explain. Solution Speed is a scalar quantity. It does not change at all with direction changes; therefore, it has magnitude only. If it were a vector quantity, it would change as direction changes (even if its magnitude remained constant). Switching Reference Frames A fundamental tenet of physics is that information about an event can be gathered from a variety of reference frames. For example, imagine that you are a passenger walking toward the front of a bus. As you walk, your motion is observed by a fellow bus passenger and by an observer standing on the sidewalk. Both the bus passenger and sidewalk observer will be able to collect information about you. They can determine how far you moved and how much time it took you to do so. However, while you moved at a consistent pace, both observers will get different results. To the passenger sitting on the bus, you moved forward at what one would consider a normal pace, something similar to how quickly you would walk outside on a sunny day. To the sidewalk observer though, you will have moved much quicker. Because the bus is also moving forward, the distance you move forward against the sidewalk each second increases, and the sidewalk observer must conclude that you are moving at a greater pace. To show that you understand this concept, you will need to create an event and think of a way to view this event from two different frames of reference. In order to ensure that the event is being observed simultaneously from both frames, you will need an assistant to help out. An example of a possible event is to have a friend ride on a skateboard while tossing a ball. How will your friend observe the ball toss, and how will those observations be different from your own? Your task is to describe your event and the observations of your event from both frames of reference. Answer the following questions below to demonstrate your understanding. For assistance, you can review the information given in the \u2018Position' paragraph at the start of Section 2.1. 1. What is your event? What object are both you and your assistant observing? 2. What do you see as the event takes place? 3. What does your assistant see as the event takes place? 4. How do your reference frames cause you and your assistant to have two different sets of observations? This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 2 | Kinematics 39 2.3 Time, Velocity, and Speed Figure 2.8 The motion of these racing snails can be described by their speeds and their velocities. (credit: tobitasflickr, Flickr) By the end of this section, you will be able to: Learning Objectives \u2022 Explain the relationships between instantaneous velocity, average velocity, instantaneous speed, average speed, displacement, and time. \u2022 Calculate velocity and speed given initial position, initial time, final position, and final time. \u2022 Derive a graph of velocity vs. time given a graph of position vs. time. \u2022 Interpret a graph of velocity vs. time. The information presented in this section supports the following AP\u00ae learning objectives and science practices: \u2022 3.A.1", ".1 The student is able to express the motion of an object using narrative, mathematical, and graphical representations. (S.P. 1.5, 2.1, 2.2) \u2022 3.A.1.3 The student is able to analyze experimental data describing the motion of an object and is able to express the results of the analysis using narrative, mathematical, and graphical representations. (S.P. 5.1) There is more to motion than distance and displacement. Questions such as, \u201cHow long does a foot race take?\u201d and \u201cWhat was the runner's speed?\u201d cannot be answered without an understanding of other concepts. In this section we add definitions of time, velocity, and speed to expand our description of motion. Time As discussed in Physical Quantities and Units, the most fundamental physical quantities are defined by how they are measured. This is the case with time. Every measurement of time involves measuring a change in some physical quantity. It may be a number on a digital clock, a heartbeat, or the position of the Sun in the sky. In physics, the definition of time is simple\u2014 time is change, or the interval over which change occurs. It is impossible to know that time has passed unless something changes. The amount of time or change is calibrated by comparison with a standard. The SI unit for time is the second, abbreviated s. We might, for example, observe that a certain pendulum makes one full swing every 0.75 s. We could then use the pendulum to measure time by counting its swings or, of course, by connecting the pendulum to a clock mechanism that registers time on a dial. This allows us to not only measure the amount of time, but also to determine a sequence of events. How does time relate to motion? We are usually interested in elapsed time for a particular motion, such as how long it takes an airplane passenger to get from his seat to the back of the plane. To find elapsed time, we note the time at the beginning and end of the motion and subtract the two. For example, a lecture may start at 11:00 A.M. and end at 11:50 A.M., so that the elapsed time would be 50 min. Elapsed time \u0394t is the difference between the ending time and beginning time, \u0394 = f \u2212 0, (2.4) where \u0394 is the change in time or elapsed time, f is the time at the end of the motion, and 0 is the time at the beginning of the motion. (As usual, the delta symbol, \u0394, means the change in the quantity that follows it.) Life is simpler if the beginning time 0 is taken to be zero, as when we use a stopwatch. If we were using a stopwatch, it would simply read zero at the start of the lecture and 50 min at the end. If 0 = 0, then \u0394 = f \u2261. In this text, for simplicity's sake, \u2022 motion starts at time equal to zero (0 = 0) \u2022 the symbol is used for elapsed time unless otherwise specified (\u0394 = f \u2261 ) 40 Velocity Chapter 2 | Kinematics Your notion of velocity is probably the same as its scientific definition. You know that if you have a large displacement in a small amount of time you have a large velocity, and that velocity has units of distance divided by time, such as miles per hour or kilometers per hour. Average Velocity Average velocity is displacement (change in position) divided by the time of travel, f \u2212 0 f \u2212 0 - is the average (indicated by the bar over the ) velocity, \u0394 is the change in position (or displacement), and f where and 0 are the final and beginning positions at times f and 0, respectively. If the starting time 0 is taken to be zero, then the average velocity is simply - = \u0394 \u0394 = (2.5) - = \u0394. (2.6) Notice that this definition indicates that velocity is a vector because displacement is a vector. It has both magnitude and direction. The SI unit for velocity is meters per second or m/s, but many other units, such as km/h, mi/h (also written as mph), and cm/s, are in common use. Suppose, for example, an airplane passenger took 5 seconds to move \u22124 m (the minus sign indicates that displacement is toward the back of the plane). His average velocity would be - = \u0394 = \u22124 m 5 s = \u2212 0.8 m/s. (2.7) The minus sign indicates the average velocity is also toward the rear of the plane. The average velocity of an object does not tell us anything about what happens to it between the starting point and ending point, however. For example, we cannot tell from average velocity whether the airplane passenger stops momentarily or backs up before he goes to the back of the plane. To get more details, we must consider smaller segments of the trip over smaller time intervals. Figure", " 2.9 A more detailed record of an airplane passenger heading toward the back of the plane, showing smaller segments of his trip. The smaller the time intervals considered in a motion, the more detailed the information. When we carry this process to its logical conclusion, we are left with an infinitesimally small interval. Over such an interval, the average velocity becomes the instantaneous velocity or the velocity at a specific instant. A car's speedometer, for example, shows the magnitude (but not the direction) of the instantaneous velocity of the car. (Police give tickets based on instantaneous velocity, but when calculating how long it will take to get from one place to another on a road trip, you need to use average velocity.) Instantaneous velocity is the average velocity at a specific instant in time (or over an infinitesimally small time interval). Mathematically, finding instantaneous velocity,, at a precise instant can involve taking a limit, a calculus operation beyond the scope of this text. However, under many circumstances, we can find precise values for instantaneous velocity without calculus. Speed In everyday language, most people use the terms \u201cspeed\u201d and \u201cvelocity\u201d interchangeably. In physics, however, they do not have the same meaning and they are distinct concepts. One major difference is that speed has no direction. Thus speed is a scalar. Just as we need to distinguish between instantaneous velocity and average velocity, we also need to distinguish between instantaneous speed and average speed. This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 2 | Kinematics 41 Instantaneous speed is the magnitude of instantaneous velocity. For example, suppose the airplane passenger at one instant had an instantaneous velocity of \u22123.0 m/s (the minus meaning toward the rear of the plane). At that same time his instantaneous speed was 3.0 m/s. Or suppose that at one time during a shopping trip your instantaneous velocity is 40 km/h due north. Your instantaneous speed at that instant would be 40 km/h\u2014the same magnitude but without a direction. Average speed, however, is very different from average velocity. Average speed is the distance traveled divided by elapsed time. We have noted that distance traveled can be greater than displacement. So average speed can be greater than average velocity, which is displacement divided by time. For example, if you drive to a store and return home in half an hour, and your car's odometer shows the total distance traveled was 6 km, then your average speed was 12 km/h. Your average velocity, however, was zero, because your displacement for the round trip is zero. (Displacement is change in position and, thus, is zero for a round trip.) Thus average speed is not simply the magnitude of average velocity. Figure 2.10 During a 30-minute round trip to the store, the total distance traveled is 6 km. The average speed is 12 km/h. The displacement for the round trip is zero, since there was no net change in position. Thus the average velocity is zero. Another way of visualizing the motion of an object is to use a graph. A plot of position or of velocity as a function of time can be very useful. For example, for this trip to the store, the position, velocity, and speed-vs.-time graphs are displayed in Figure 2.11. (Note that these graphs depict a very simplified model of the trip. We are assuming that speed is constant during the trip, which is unrealistic given that we'll probably stop at the store. But for simplicity's sake, we will model it with no stops or changes in speed. We are also assuming that the route between the store and the house is a perfectly straight line.) 42 Chapter 2 | Kinematics Figure 2.11 Position vs. time, velocity vs. time, and speed vs. time on a trip. Note that the velocity for the return trip is negative. Making Connections: Take-Home Investigation\u2014Getting a Sense of Speed If you have spent much time driving, you probably have a good sense of speeds between about 10 and 70 miles per hour. But what are these in meters per second? What do we mean when we say that something is moving at 10 m/s? To get a better sense of what these values really mean, do some observations and calculations on your own: \u2022 calculate typical car speeds in meters per second \u2022 estimate jogging and walking speed by timing yourself; convert the measurements into both m/s and mi/h \u2022 determine the speed of an ant, snail, or falling leaf This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 2 | Kinematics 43 Check Your Understanding A commuter train travels from Baltimore to Washington, DC, and back in 1 hour and 45 minutes. The distance between the two stations is approximately 40 miles. What is (a) the average velocity of the train, and (b", ") the average speed of the train in m/s? Solution (a) The average velocity of the train is zero because f = 0 ; the train ends up at the same place it starts. (b) The average speed of the train is calculated below. Note that the train travels 40 miles one way and 40 miles back, for a total distance of 80 miles. = 80 miles 105 minutes distance time \u00d7 5280 feet 1 mile \u00d7 1 meter 3.28 feet \u00d7 1 minute 60 seconds = 20 m/s (2.8) (2.9) 80 miles 105 minutes 2.4 Acceleration Figure 2.12 A plane decelerates, or slows down, as it comes in for landing in St. Maarten. Its acceleration is opposite in direction to its velocity. (credit: Steve Conry, Flickr) Learning Objectives By the end of this section, you will be able to: \u2022 Define and distinguish between instantaneous acceleration and average acceleration. \u2022 Calculate acceleration given initial time, initial velocity, final time, and final velocity. The information presented in this section supports the following AP\u00ae learning objectives and science practices: \u2022 3.A.1.1 The student is able to express the motion of an object using narrative, mathematical, and graphical representations. (S.P. 1.5, 2.1, 2.2) \u2022 3.A.1.3 The student is able to analyze experimental data describing the motion of an object and is able to express the results of the analysis using narrative, mathematical, and graphical representations. (S.P. 5.1) In everyday conversation, to accelerate means to speed up. The accelerator in a car can in fact cause it to speed up. The greater the acceleration, the greater the change in velocity over a given time. The formal definition of acceleration is consistent with these notions, but more inclusive. Average Acceleration Average Acceleration is the rate at which velocity changes, where is average acceleration, is velocity, and is time. (The bar over the means average acceleration.) - = 2.10) Because acceleration is velocity in m/s divided by time in s, the SI units for acceleration are m/s2, meters per second squared or meters per second per second, which literally means by how many meters per second the velocity changes every second. 44 Chapter 2 | Kinematics Recall that velocity is a vector\u2014it has both magnitude and direction. This means that a change in velocity can be a change in magnitude (or speed), but it can also be a change in direction. For example, if a car turns a corner at constant speed, it is accelerating because its direction is changing. The quicker you turn, the greater the acceleration. So there is an acceleration when velocity changes either in magnitude (an increase or decrease in speed) or in direction, or both. Acceleration as a Vector Acceleration is a vector in the same direction as the change in velocity, \u0394. Since velocity is a vector, it can change either in magnitude or in direction. Acceleration is therefore a change in either speed or direction, or both. Keep in mind that although acceleration is in the direction of the change in velocity, it is not always in the direction of motion. When an object's acceleration is in the same direction of its motion, the object will speed up. However, when an object's acceleration is opposite to the direction of its motion, the object will slow down. Speeding up and slowing down should not be confused with a positive and negative acceleration. The next two examples should help to make this distinction clear. Figure 2.13 A subway train in Sao Paulo, Brazil, decelerates as it comes into a station. It is accelerating in a direction opposite to its direction of motion. (credit: Yusuke Kawasaki, Flickr) Making Connections: Car Motion Figure 2.14 Above are arrows representing the motion of five cars (A\u2013E). In all five cases, the positive direction should be considered to the right of the page. Consider the acceleration and velocity of each car in terms of its direction of travel. Figure 2.15 Car A is speeding up. Because the positive direction is considered to the right of the paper, Car A is moving with a positive velocity. Because it is speeding up while moving with a positive velocity, its acceleration is also considered positive. This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 2 | Kinematics 45 Figure 2.16 Car B is slowing down. Because the positive direction is considered to the right of the paper, Car B is also moving with a positive velocity. However, because it is slowing down while moving with a positive velocity, its acceleration is considered negative. (This can be viewed in a mathematical manner as well. If the car was originally moving with a velocity of +25 m/s, it is finishing with a speed less than that, like +5 m/s. Because the change in velocity is negative, the", " acceleration will be as well.) Figure 2.17 Car C has a constant speed. Because the positive direction is considered to the right of the paper, Car C is moving with a positive velocity. Because all arrows are of the same length, this car is not changing its speed. As a result, its change in velocity is zero, and its acceleration must be zero as well. Figure 2.18 Car D is speeding up in the opposite direction of Cars A, B, C. Because the car is moving opposite to the positive direction, Car D is moving with a negative velocity. Because it is speeding up while moving in a negative direction, its acceleration is negative as well. Figure 2.19 Car E is slowing down in the same direction as Car D and opposite of Cars A, B, C. Because it is moving opposite to the positive direction, Car E is moving with a negative velocity as well. However, because it is slowing down while moving in a negative direction, its acceleration is actually positive. As in example B, this may be more easily understood in a mathematical sense. The car is originally moving with a large negative velocity (\u221225 m/s) but slows to a final velocity that is less negative (\u22125 m/s). This change in velocity, from \u221225 m/s to \u22125 m/s, is actually a positive change ( \u2212 = \u2212 5 m/s \u2212 \u2212 25 m/s of 20 m/s. Because the change in velocity is positive, the acceleration must also be positive. Making Connection - Illustrative Example The three graphs below are labeled A, B, and C. Each one represents the position of a moving object plotted against time. 46 Chapter 2 | Kinematics Figure 2.20 Three position and time graphs: A, B, and C. As we did in the previous example, let's consider the acceleration and velocity of each object in terms of its direction of travel. Figure 2.21 Graph A of Position (y axis) vs. Time (x axis). Object A is continually increasing its position in the positive direction. As a result, its velocity is considered positive. This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 2 | Kinematics 47 Figure 2.22 Breakdown of Graph A into two separate sections. During the first portion of time (shaded grey) the position of the object does not change much, resulting in a small positive velocity. During a later portion of time (shaded green) the position of the object changes more, resulting in a larger positive velocity. Because this positive velocity is increasing over time, the acceleration of the object is considered positive. Figure 2.23 Graph B of Position (y axis) vs. Time (x axis). As in case A, Object B is continually increasing its position in the positive direction. As a result, its velocity is considered positive. Figure 2.24 Breakdown of Graph B into two separate sections. During the first portion of time (shaded grey) the position of the object changes a large amount, resulting in a large positive velocity. During a later portion of time (shaded green) the position of the object does not change as much, resulting in a smaller positive velocity. Because this positive velocity is decreasing over time, the acceleration of the object is considered negative. 48 Chapter 2 | Kinematics Figure 2.25 Graph C of Position (y axis) vs. Time (x axis). Object C is continually decreasing its position in the positive direction. As a result, its velocity is considered negative. Figure 2.26 Breakdown of Graph C into two separate sections. During the first portion of time (shaded grey) the position of the object does not change a large amount, resulting in a small negative velocity. During a later portion of time (shaded green) the position of the object changes a much larger amount, resulting in a larger negative velocity. Because the velocity of the object is becoming more negative during the time period, the change in velocity is negative. As a result, the object experiences a negative acceleration. Example 2.1 Calculating Acceleration: A Racehorse Leaves the Gate A racehorse coming out of the gate accelerates from rest to a velocity of 15.0 m/s due west in 1.80 s. What is its average acceleration? Figure 2.27 (credit: Jon Sullivan, PD Photo.org) Strategy This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 2 | Kinematics 49 First we draw a sketch and assign a coordinate system to the problem. This is a simple problem, but it always helps to visualize it. Notice that we assign east as positive and west as negative. Thus, in this case, we have negative velocity. Figure 2.28 We can solve this problem by identifying \u0394 and \u0394 from the given information and then calculating the average f \u2212 0 f \u2212 0 acceleration directly from the equation - = \u0394", " \u0394 =. Solution 1. Identify the knowns. 0 = 0, f = \u221215.0 m/s (the minus sign indicates direction toward the west), \u0394 = 1.80 s. 2. Find the change in velocity. Since the horse is going from zero to \u2212 15.0 m/s, its change in velocity equals its final velocity: \u0394 = f = \u221215.0 m/s. -. 3. Plug in the known values ( \u0394 and \u0394 ) and solve for the unknown - = \u0394 \u0394 = \u221215.0 m/s 1.80 s = \u22128.33 m/s2. (2.11) Discussion The minus sign for acceleration indicates that acceleration is toward the west. An acceleration of 8.33 m/s2 due west means that the horse increases its velocity by 8.33 m/s due west each second, that is, 8.33 meters per second per second, which we write as 8.33 m/s2. This is truly an average acceleration, because the ride is not smooth. We shall see later that an acceleration of this magnitude would require the rider to hang on with a force nearly equal to his weight. Instantaneous Acceleration Instantaneous acceleration, or the acceleration at a specific instant in time, is obtained by the same process as discussed for instantaneous velocity in Time, Velocity, and Speed\u2014that is, by considering an infinitesimally small interval of time. How do we find instantaneous acceleration using only algebra? The answer is that we choose an average acceleration that is representative of the motion. Figure 2.29 shows graphs of instantaneous acceleration versus time for two very different motions. In Figure 2.29(a), the acceleration varies slightly and the average over the entire interval is nearly the same as the instantaneous acceleration at any time. In this case, we should treat this motion as if it had a constant acceleration equal to the average (in this case about 1.8 m/s2 ). In Figure 2.29(b), the acceleration varies drastically over time. In such situations it is best to consider smaller time intervals and choose an average acceleration for each. For example, we could consider motion over the time intervals from 0 to 1.0 s and from 1.0 to 3.0 s as separate motions with accelerations of +3.0 m/s2 and \u20132.0 m/s2, respectively. 50 Chapter 2 | Kinematics Figure 2.29 Graphs of instantaneous acceleration versus time for two different one-dimensional motions. (a) Here acceleration varies only slightly and is always in the same direction, since it is positive. The average over the interval is nearly the same as the acceleration at any given time. (b) Here the acceleration varies greatly, perhaps representing a package on a post office conveyor belt that is accelerated forward and backward as it bumps along. It is necessary to consider small time intervals (such as from 0 to 1.0 s) with constant or nearly constant acceleration in such a situation. The next several examples consider the motion of the subway train shown in Figure 2.30. In (a) the shuttle moves to the right, and in (b) it moves to the left. The examples are designed to further illustrate aspects of motion and to illustrate some of the reasoning that goes into solving problems. Figure 2.30 One-dimensional motion of a subway train considered in Example 2.2, Example 2.3, Example 2.4, Example 2.5, Example 2.6, and Example 2.7. Here we have chosen the -axis so that + means to the right and \u2212 means to the left for displacements, velocities, and accelerations. (a) The subway train moves to the right from 0 to f. Its displacement \u0394 is +2.0 km. (b) The train moves to the left from \u20320 to \u2032f. Its displacement \u0394\u2032 is \u22121.5 km. (Note that the prime symbol (\u2032) is used simply to distinguish between displacement in the two different situations. The distances of travel and the size of the cars are on different scales to fit everything into the diagram.) Example 2.2 Calculating Displacement: A Subway Train What are the magnitude and sign of displacements for the motions of the subway train shown in parts (a) and (b) of Figure 2.30? Strategy This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 2 | Kinematics 51 A drawing with a coordinate system is already provided, so we don't need to make a sketch, but we should analyze it to make sure we understand what it is showing. Pay particular attention to the coordinate system. To find displacement, we use the equation \u0394 = f \u2212 0. This is straightforward since the initial and final positions are given. Solution 1. Identify the knowns. In the figure we see that f = 6.70 km", " and 0 = 4.70 km for part (a), and \u2032f = 3.75 km and \u20320 = 5.25 km for part (b). 2. Solve for displacement in part (a). \u0394 = f \u2212 0 = 6.70 km \u2212 4.70 km= +2.00 km 3. Solve for displacement in part (b). \u0394\u2032 = \u2032f \u2212 \u20320 = 3.75 km \u2212 5.25 km = \u2212 1.50 km Discussion (2.12) (2.13) The direction of the motion in (a) is to the right and therefore its displacement has a positive sign, whereas motion in (b) is to the left and thus has a minus sign. Example 2.3 Comparing Distance Traveled with Displacement: A Subway Train What are the distances traveled for the motions shown in parts (a) and (b) of the subway train in Figure 2.30? Strategy To answer this question, think about the definitions of distance and distance traveled, and how they are related to displacement. Distance between two positions is defined to be the magnitude of displacement, which was found in Example 2.2. Distance traveled is the total length of the path traveled between the two positions. (See Displacement.) In the case of the subway train shown in Figure 2.30, the distance traveled is the same as the distance between the initial and final positions of the train. Solution 1. The displacement for part (a) was +2.00 km. Therefore, the distance between the initial and final positions was 2.00 km, and the distance traveled was 2.00 km. 2. The displacement for part (b) was \u22121.5 km. Therefore, the distance between the initial and final positions was 1.50 km, and the distance traveled was 1.50 km. Discussion Distance is a scalar. It has magnitude but no sign to indicate direction. Example 2.4 Calculating Acceleration: A Subway Train Speeding Up Suppose the train in Figure 2.30(a) accelerates from rest to 30.0 km/h in the first 20.0 s of its motion. What is its average acceleration during that time interval? Strategy It is worth it at this point to make a simple sketch: Figure 2.31 This problem involves three steps. First we must determine the change in velocity, then we must determine the change in time, and finally we use these values to calculate the acceleration. Solution 52 Chapter 2 | Kinematics 1. Identify the knowns. 0 = 0 (the trains starts at rest), f = 30.0 km/h, and \u0394 = 20.0 s. 2. Calculate \u0394. Since the train starts from rest, its change in velocity is \u0394= +30.0 km/h, where the plus sign means velocity to the right. -. 3. Plug in known values and solve for the unknown, - = \u0394 \u0394 = +30.0 km/h 20.0 s 4. Since the units are mixed (we have both hours and seconds for time), we need to convert everything into SI units of meters and seconds. (See Physical Quantities and Units for more guidance.) - = +30 km/h 20.0 s 103 m 1 km 1 h 3600 s = 0.417 m/s2 (2.14) (2.15) Discussion The plus sign means that acceleration is to the right. This is reasonable because the train starts from rest and ends up with a velocity to the right (also positive). So acceleration is in the same direction as the change in velocity, as is always the case. Example 2.5 Calculate Acceleration: A Subway Train Slowing Down Now suppose that at the end of its trip, the train in Figure 2.30(a) slows to a stop from a speed of 30.0 km/h in 8.00 s. What is its average acceleration while stopping? Strategy Figure 2.32 In this case, the train is decelerating and its acceleration is negative because it is toward the left. As in the previous example, we must find the change in velocity and the change in time and then solve for acceleration. Solution 1. Identify the knowns. 0 = 30.0 km/h, f = 0 km/h (the train is stopped, so its velocity is 0), and \u0394 = 8.00 s. 2. Solve for the change in velocity \u2212 30.0 km/h = \u221230.0 km/h -. 3. Plug in the knowns, \u0394 and \u0394, and solve for - = \u0394 \u0394 = \u221230.0 km/h 8.00 s 4. Convert the units to meters and seconds. - = \u0394 \u0394 = \u221230.0 km/h 8.00 s 103 m 1 km 1 h 3600 s = \u22121.04 m/s2. (2.16) (2", ".17) (2.18) Discussion The minus sign indicates that acceleration is to the left. This sign is reasonable because the train initially has a positive velocity in this problem, and a negative acceleration would oppose the motion. Again, acceleration is in the same direction as the change in velocity, which is negative here. This acceleration can be called a deceleration because it has a direction opposite to the velocity. The graphs of position, velocity, and acceleration vs. time for the trains in Example 2.4 and Example 2.5 are displayed in Figure 2.33. (We have taken the velocity to remain constant from 20 to 40 s, after which the train decelerates.) This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 2 | Kinematics 53 Figure 2.33 (a) Position of the train over time. Notice that the train's position changes slowly at the beginning of the journey, then more and more quickly as it picks up speed. Its position then changes more slowly as it slows down at the end of the journey. In the middle of the journey, while the velocity remains constant, the position changes at a constant rate. (b) Velocity of the train over time. The train's velocity increases as it accelerates at the beginning of the journey. It remains the same in the middle of the journey (where there is no acceleration). It decreases as the train decelerates at the end of the journey. (c) The acceleration of the train over time. The train has positive acceleration as it speeds up at the beginning of the journey. It has no acceleration as it travels at constant velocity in the middle of the journey. Its acceleration is negative as it slows down at the end of the journey. Example 2.6 Calculating Average Velocity: The Subway Train What is the average velocity of the train in part b of Example 2.2, and shown again below, if it takes 5.00 min to make its trip? 54 Chapter 2 | Kinematics Figure 2.34 Strategy Average velocity is displacement divided by time. It will be negative here, since the train moves to the left and has a negative displacement. Solution 1. Identify the knowns. \u2032f = 3.75 km, \u20320 = 5.25 km, \u0394 = 5.00 min. 2. Determine displacement, \u0394\u2032. We found \u0394\u2032 to be \u2212 1.5 km in Example 2.2. 3. Solve for average velocity. 4. Convert units. - = \u0394\u2032 \u0394 = \u22121.50 km 5.00 min - = \u0394\u2032 \u0394 = \u22121.50 km 5.00 min 60 min 1 h = \u221218.0 km/h Discussion The negative velocity indicates motion to the left. Example 2.7 Calculating Deceleration: The Subway Train Finally, suppose the train in Figure 2.34 slows to a stop from a velocity of 20.0 km/h in 10.0 s. What is its average acceleration? Strategy Once again, let's draw a sketch: Figure 2.35 As before, we must find the change in velocity and the change in time to calculate average acceleration. Solution 1. Identify the knowns. 0 = \u221220 km/h, f = 0 km/h, \u0394 = 10.0 s. 2. Calculate \u0394. The change in velocity here is actually positive, since \u0394 = f \u2212 0 = 0 \u2212 (\u221220 km/h)=+20 km/h. -. 3. Solve for 4. Convert units. - = \u0394 \u0394 = +20.0 km/h 10.0 s This content is available for free at http://cnx.org/content/col11844/1.13 (2.19) (2.20) (2.21) (2.22) Chapter 2 | Kinematics Discussion - = +20.0 km/h 10.0 s 103 m 1 km 1 h 3600 s = +0.556 m/s2 55 (2.23) The plus sign means that acceleration is to the right. This is reasonable because the train initially has a negative velocity (to the left) in this problem and a positive acceleration opposes the motion (and so it is to the right). Again, acceleration is in the same direction as the change in velocity, which is positive here. As in Example 2.5, this acceleration can be called a deceleration since it is in the direction opposite to the velocity. Sign and Direction Perhaps the most important thing to note about these examples is the signs of the answers. In our chosen coordinate system, plus means the quantity is to the right and minus means it is to the left. This is easy to imagine for displacement and velocity. But it is a little less obvious for acceleration. Most people interpret negative acceleration as the slowing of an object. This was not the case in Example 2.7, where a positive acceleration slowed a", " negative velocity. The crucial distinction was that the acceleration was in the opposite direction from the velocity. In fact, a negative acceleration will increase a negative velocity. For example, the train moving to the left in Figure 2.34 is sped up by an acceleration to the left. In that case, both and are negative. The plus and minus signs give the directions of the accelerations. If acceleration has the same sign as the velocity, the object is speeding up. If acceleration has the opposite sign as the velocity, the object is slowing down. Check Your Understanding An airplane lands on a runway traveling east. Describe its acceleration. Solution If we take east to be positive, then the airplane has negative acceleration, as it is accelerating toward the west. It is also decelerating: its acceleration is opposite in direction to its velocity. PhET Explorations: Moving Man Simulation Learn about position, velocity, and acceleration graphs. Move the little man back and forth with the mouse and plot his motion. Set the position, velocity, or acceleration and let the simulation move the man for you. Figure 2.36 Moving Man (http://cnx.org/content/m54772/1.3/moving-man_en.jar) 2.5 Motion Equations for Constant Acceleration in One Dimension Figure 2.37 Kinematic equations can help us describe and predict the motion of moving objects such as these kayaks racing in Newbury, England. (credit: Barry Skeates, Flickr) By the end of this section, you will be able to: Learning Objectives 56 Chapter 2 | Kinematics \u2022 Calculate displacement of an object that is not accelerating, given initial position and velocity. \u2022 Calculate final velocity of an accelerating object, given initial velocity, acceleration, and time. \u2022 Calculate displacement and final position of an accelerating object, given initial position, initial velocity, time, and acceleration. The information presented in this section supports the following AP\u00ae learning objectives and science practices: \u2022 3.A.1.1 The student is able to express the motion of an object using narrative, mathematical, or graphical representations. (S.P. 1.5, 2.1, 2.2) \u2022 3.A.1.3 The student is able to analyze experimental data describing the motion of an object and is able to express the results of the analysis using narrative, mathematical, and graphical representations. (S.P. 5.1) We might know that the greater the acceleration of, say, a car moving away from a stop sign, the greater the displacement in a given time. But we have not developed a specific equation that relates acceleration and displacement. In this section, we develop some convenient equations for kinematic relationships, starting from the definitions of displacement, velocity, and acceleration already covered. Notation: t, x, v, a First, let us make some simplifications in notation. Taking the initial time to be zero, as if time is measured with a stopwatch, is a great simplification. Since elapsed time is \u0394 = f \u2212 0, taking 0 = 0 means that \u0394 = f, the final time on the stopwatch. When initial time is taken to be zero, we use the subscript 0 to denote initial values of position and velocity. That is, 0 is the initial position and 0 is the initial velocity. We put no subscripts on the final values. That is, is the final time, is the final position, and is the final velocity. This gives a simpler expression for elapsed time\u2014now, \u0394 =. It also simplifies the expression for displacement, which is now \u0394 = \u2212 0. Also, it simplifies the expression for change in velocity, which is now \u0394 = \u2212 0. To summarize, using the simplified notation, with the initial time taken to be zero2.24) where the subscript 0 denotes an initial value and the absence of a subscript denotes a final value in whatever motion is under consideration. We now make the important assumption that acceleration is constant. This assumption allows us to avoid using calculus to find instantaneous acceleration. Since acceleration is constant, the average and instantaneous accelerations are equal. That is, - = = constant, (2.25) so we use the symbol for acceleration at all times. Assuming acceleration to be constant does not seriously limit the situations we can study nor degrade the accuracy of our treatment. For one thing, acceleration is constant in a great number of situations. Furthermore, in many other situations we can accurately describe motion by assuming a constant acceleration equal to the average acceleration for that motion. Finally, in motions where acceleration changes drastically, such as a car accelerating to top speed and then braking to a stop, the motion can be considered in separate parts, each of which has its own constant acceleration. Solving for Displacement ( \u0394 ) and Final Position ( ) from Average Velocity when Acceleration ( ) is Constant To get our first two new equations, we start with the definition of average velocity: - = \u0394 \u0394.", " Substituting the simplified notation for \u0394 and \u0394 yields - = \u2212 0. Solving for yields where the average velocity is = 0 +, - = 0 + 2 (constant ). This content is available for free at http://cnx.org/content/col11844/1.13 (2.26) (2.27) (2.28) (2.29) Chapter 2 | Kinematics The equation - = 0 + 2 reflects the fact that, when acceleration is constant, is just the simple average of the initial and 57 final velocities. For example, if you steadily increase your velocity (that is, with constant acceleration) from 30 to 60 km/h, then your average velocity during this steady increase is 45 km/h. Using the equation - = 0 + 2 to check this, we see that (2.30) - = 0 + 2 = 30 km/h + 60 km/h 2 = 45 km/h, which seems logical. Example 2.8 Calculating Displacement: How Far does the Jogger Run? A jogger runs down a straight stretch of road with an average velocity of 4.00 m/s for 2.00 min. What is his final position, taking his initial position to be zero? Strategy Draw a sketch. Figure 2.38 The final position is given by the equation To find, we identify the values of 0,, and from the statement of the problem and substitute them into the equation. Solution = 0 +. (2.31) 1. Identify the knowns. - = 4.00 m/s, \u0394 = 2.00 min, and 0 = 0 m. 2. Enter the known values into the equation. = 0 + = 0 + (4.00 m/s)(120 s) = 480 m (2.32) Discussion Velocity and final displacement are both positive, which means they are in the same direction. gives insight into the relationship between displacement, average velocity, and time. It shows, for The equation = 0 + example, that displacement is a linear function of average velocity. (By linear function, we mean that displacement depends on rather than on raised to some other power, such as. When graphed, linear functions look like straight lines with a constant slope.) On a car trip, for example, we will get twice as far in a given time if we average 90 km/h than if we average 45 km/h. 58 Chapter 2 | Kinematics Figure 2.39 There is a linear relationship between displacement and average velocity. For a given time, an object moving twice as fast as another object will move twice as far as the other object. Solving for Final Velocity We can derive another useful equation by manipulating the definition of acceleration. Substituting the simplified notation for \u0394 and \u0394 gives us = \u0394 \u0394 Solving for yields = \u2212 0 (constant ). = 0 + (constant ). (2.33) (2.34) (2.35) Example 2.9 Calculating Final Velocity: An Airplane Slowing Down after Landing An airplane lands with an initial velocity of 70.0 m/s and then decelerates at 1.50 m/s2 for 40.0 s. What is its final velocity? Strategy Draw a sketch. We draw the acceleration vector in the direction opposite the velocity vector because the plane is decelerating. Figure 2.40 Solution 1. Identify the knowns. 0 = 70.0 m/s, = \u22121.50 m/s2, = 40.0 s. 2. Identify the unknown. In this case, it is final velocity, f. This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 2 | Kinematics 3. Determine which equation to use. We can calculate the final velocity using the equation = 0 +. 4. Plug in the known values and solve. = 0 + = 70.0 m/s + \u22121.50 m/s2 (40.0 s) = 10.0 m/s 59 (2.36) Discussion The final velocity is much less than the initial velocity, as desired when slowing down, but still positive. With jet engines, reverse thrust could be maintained long enough to stop the plane and start moving it backward. That would be indicated by a negative final velocity, which is not the case here. Figure 2.41 The airplane lands with an initial velocity of 70.0 m/s and slows to a final velocity of 10.0 m/s before heading for the terminal. Note that the acceleration is negative because its direction is opposite to its velocity, which is positive. In addition to being useful in problem solving, the equation = 0 + gives us insight into the relationships among velocity, acceleration, and time. From it we can see, for example, that \u2022 \u2022 \u2022 final velocity depends on how large the acceleration is and how long", " it lasts if the acceleration is zero, then the final velocity equals the initial velocity ( = 0), as expected (i.e., velocity is constant) if is negative, then the final velocity is less than the initial velocity (All of these observations fit our intuition, and it is always useful to examine basic equations in light of our intuition and experiences to check that they do indeed describe nature accurately.) Making Connections: Real-World Connection Figure 2.42 The Space Shuttle Endeavor blasts off from the Kennedy Space Center in February 2010. (credit: Matthew Simantov, Flickr) An intercontinental ballistic missile (ICBM) has a larger average acceleration than the Space Shuttle and achieves a greater velocity in the first minute or two of flight (actual ICBM burn times are classified\u2014short-burn-time missiles are more difficult for an enemy to destroy). But the Space Shuttle obtains a greater final velocity, so that it can orbit the earth rather than come directly back down as an ICBM does. The Space Shuttle does this by accelerating for a longer time. Solving for Final Position When Velocity is Not Constant ( \u2260 0 ) We can combine the equations above to find a third equation that allows us to calculate the final position of an object experiencing constant acceleration. We start with Adding 0 to each side of this equation and dividing by 2 gives =. (2.37) (2.38) 60 Chapter 2 | Kinematics Since 0 + 2 - for constant acceleration, then = Now we substitute this expression for - = 0 + 1 2. - into the equation for displacementconstant )., yielding (2.39) (2.40) Example 2.10 Calculating Displacement of an Accelerating Object: Dragsters Dragsters can achieve average accelerations of 26.0 m/s2. Suppose such a dragster accelerates from rest at this rate for 5.56 s. How far does it travel in this time? Figure 2.43 U.S. Army Top Fuel pilot Tony \u201cThe Sarge\u201d Schumacher begins a race with a controlled burnout. (credit: Lt. Col. William Thurmond. Photo Courtesy of U.S. Army.) Strategy Draw a sketch. Figure 2.44 We are asked to find displacement, which is if we take 0 to be zero. (Think about it like the starting line of a race. It can be anywhere, but we call it 0 and measure all other positions relative to it.) We can use the equation = 0 + 0 + 1 2 2 once we identify 0,, and from the statement of the problem. Solution 1. Identify the knowns. Starting from rest means that 0 = 0, is given as 26.0 m/s2 and is given as 5.56 s. 2. Plug the known values into the equation to solve for the unknown : 2 2. = 0 + 0 + 1 Since the initial position and velocity are both zero, this simplifies to Substituting the identified values of and gives = 1 2 2. (2.41) (2.42) This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 2 | Kinematics yielding Discussion = 1 2 26.0 m/s2 (5.56 s)2 = 402 m. 61 (2.43) (2.44) If we convert 402 m to miles, we find that the distance covered is very close to one quarter of a mile, the standard distance for drag racing. So the answer is reasonable. This is an impressive displacement in only 5.56 s, but top-notch dragsters can do a quarter mile in even less time than this. What else can we learn by examining the equation = 0 + 0 + 1 2 2? We see that: \u2022 displacement depends on the square of the elapsed time when acceleration is not zero. In Example 2.10, the dragster \u2022 covers only one fourth of the total distance in the first half of the elapsed time if acceleration is zero, then the initial velocity equals average velocity ( 0 = = 0 + 0 - ) and = 0 + 0 + 1 2 2 becomes Solving for Final Velocity when Velocity Is Not Constant ( \u2260 0 ) A fourth useful equation can be obtained from another algebraic manipulation of previous equations. If we solve = 0 + for, we get Substituting this and - = 0 + 2 into = 0 +, we get = \u2212 0. 2 = 0 2 + 2( \u2212 0) (constant). Example 2.11 Calculating Final Velocity: Dragsters Calculate the final velocity of the dragster in Example 2.10 without using information about time. Strategy Draw a sketch. Figure 2.45 The equation 2 = 0 displacement, and no time information is required. 2 + 2( \u2212 0) is ideally suited to this task because it relates velocities, acceleration, and Solution 1. Identify the known values. We know", " that 0 = 0, since the dragster starts from rest. Then we note that \u2212 0 = 402 m (this was the answer in Example 2.10). Finally, the average acceleration was given to be = 26.0 m/s2. 2. Plug the knowns into the equation 2 = 0 2 + 2( \u2212 0) and solve for. 2 = 0 + 2 26.0 m/s2 (402 m). (2.45) (2.46) (2.47) 62 Thus To get, we take the square root: Discussion 2 = 2.09\u00d7104 m2 /s2. = 2.09\u00d7104 m2 /s2 = 145 m/s. Chapter 2 | Kinematics (2.48) (2.49) 145 m/s is about 522 km/h or about 324 mi/h, but even this breakneck speed is short of the record for the quarter mile. Also, note that a square root has two values; we took the positive value to indicate a velocity in the same direction as the acceleration. An examination of the equation 2 = 0 physical quantities: 2 + 2( \u2212 0) can produce further insights into the general relationships among \u2022 The final velocity depends on how large the acceleration is and the distance over which it acts \u2022 For a fixed deceleration, a car that is going twice as fast doesn't simply stop in twice the distance\u2014it takes much further to stop. (This is why we have reduced speed zones near schools.) Putting Equations Together In the following examples, we further explore one-dimensional motion, but in situations requiring slightly more algebraic manipulation. The examples also give insight into problem-solving techniques. The box below provides easy reference to the equations needed. Summary of Kinematic Equations (constant ) = ( \u2212 0) (2.50) (2.51) (2.52) (2.53) (2.54) Example 2.12 Calculating Displacement: How Far Does a Car Go When Coming to a Halt? On dry concrete, a car can decelerate at a rate of 7.00 m/s2, whereas on wet concrete it can decelerate at only 5.00 m/s2. Find the distances necessary to stop a car moving at 30.0 m/s (about 110 km/h) (a) on dry concrete and (b) on wet concrete. (c) Repeat both calculations, finding the displacement from the point where the driver sees a traffic light turn red, taking into account his reaction time of 0.500 s to get his foot on the brake. Strategy Draw a sketch. Figure 2.46 This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 2 | Kinematics 63 In order to determine which equations are best to use, we need to list all of the known values and identify exactly what we need to solve for. We shall do this explicitly in the next several examples, using tables to set them off. Solution for (a) 1. Identify the knowns and what we want to solve for. We know that 0 = 30.0 m/s ; = 0 ; = \u22127.00 m/s2 ( is negative because it is in a direction opposite to velocity). We take 0 to be 0. We are looking for displacement \u0394, or \u2212 0. 2. Identify the equation that will help up solve the problem. The best equation to use is 2 = 0 2 + 2( \u2212 0). (2.55) This equation is best because it includes only one unknown,. We know the values of all the other variables in this equation. (There are other equations that would allow us to solve for, but they require us to know the stopping time,, which we do not know. We could use them but it would entail additional calculations.) 3. Rearrange the equation to solve for. 4. Enter known values. Thus = 02 \u2212 (30.0 m/s)2 \u22127.00 m/s2 2 = 64.3 m on dry concrete. (2.56) (2.57) (2.58) Solution for (b) This part can be solved in exactly the same manner as Part A. The only difference is that the deceleration is \u2013 5.00 m/s2. The result is wet = 90.0 m on wet concrete. (2.59) Solution for (c) Once the driver reacts, the stopping distance is the same as it is in Parts A and B for dry and wet concrete. So to answer this question, we need to calculate how far the car travels during the reaction time, and then add that to the stopping time. It is reasonable to assume that the velocity remains constant during the driver's reaction time. 1. Identify the knowns and what we want to solve for. We know", " that We take 0 \u2212 reaction to be 0. We are looking for reaction. - = 30.0 m/s ; reaction = 0.500 s ; reaction = 0. 2. Identify the best equation to use. = 0 + works well because the only unknown value is, which is what we want to solve for. 3. Plug in the knowns to solve the equation. = 0 + (30.0 m/s)(0.500 s) = 15.0 m. (2.60) This means the car travels 15.0 m while the driver reacts, making the total displacements in the two cases of dry and wet concrete 15.0 m greater than if he reacted instantly. 4. Add the displacement during the reaction time to the displacement when braking. braking + reaction = total (2.61) a. 64.3 m + 15.0 m = 79.3 m when dry b. 90.0 m + 15.0 m = 105 m when wet 64 Chapter 2 | Kinematics Figure 2.47 The distance necessary to stop a car varies greatly, depending on road conditions and driver reaction time. Shown here are the braking distances for dry and wet pavement, as calculated in this example, for a car initially traveling at 30.0 m/s. Also shown are the total distances traveled from the point where the driver first sees a light turn red, assuming a 0.500 s reaction time. Discussion The displacements found in this example seem reasonable for stopping a fast-moving car. It should take longer to stop a car on wet rather than dry pavement. It is interesting that reaction time adds significantly to the displacements. But more important is the general approach to solving problems. We identify the knowns and the quantities to be determined and then find an appropriate equation. There is often more than one way to solve a problem. The various parts of this example can in fact be solved by other methods, but the solutions presented above are the shortest. Example 2.13 Calculating Time: A Car Merges into Traffic Suppose a car merges into freeway traffic on a 200-m-long ramp. If its initial velocity is 10.0 m/s and it accelerates at 2.00 m/s2, how long does it take to travel the 200 m up the ramp? (Such information might be useful to a traffic engineer.) Strategy Draw a sketch. Figure 2.48 We are asked to solve for the time. As before, we identify the known quantities in order to choose a convenient physical relationship (that is, an equation with one unknown, ). Solution 1. Identify the knowns and what we want to solve for. We know that 0 = 10 m/s ; = 2.00 m/s2 ; and = 200 m. 2. We need to solve for. Choose the best equation. = 0 + 0 + 1 equation is the variable for which we need to solve. 2 2 works best because the only unknown in the 3. We will need to rearrange the equation to solve for. In this case, it will be easier to plug in the knowns first. 200 m = 0 m + (10.0 m/s) + 1 2 This content is available for free at http://cnx.org/content/col11844/1.13 2.00 m/s2 2 (2.62) Chapter 2 | Kinematics 65 4. Simplify the equation. The units of meters (m) cancel because they are in each term. We can get the units of seconds (s) to cancel by taking = s, where is the magnitude of time and s is the unit. Doing so leaves 5. Use the quadratic formula to solve for. (a) Rearrange the equation to get 0 on one side of the equation. 200 = 10 + 2. This is a quadratic equation of the form 2 + 10 \u2212 200 = 0 2 + + = 0, where the constants are = 1.00, = 10.0, and = \u2212200. (b) Its solutions are given by the quadratic formula: This yields two solutions for, which are = \u2212 \u00b1 2 \u2212 4 2. = 10.0 and\u221220.0. In this case, then, the time is = in seconds, or = 10.0 s and \u2212 20.0 s. (2.63) (2.64) (2.65) (2.66) (2.67) (2.68) A negative value for time is unreasonable, since it would mean that the event happened 20 s before the motion began. We can discard that solution. Thus, = 10.0 s. (2.69) Discussion Whenever an equation contains an unknown squared, there will be two solutions. In some problems both solutions are meaningful, but in others, such as the above, only one solution is reasonable. The 10.0 s answer seems reasonable for a typical freeway on-r", "amp. With the basics of kinematics established, we can go on to many other interesting examples and applications. In the process of developing kinematics, we have also glimpsed a general approach to problem solving that produces both correct answers and insights into physical relationships. Problem-Solving Basics discusses problem-solving basics and outlines an approach that will help you succeed in this invaluable task. Making Connections: Take-Home Experiment\u2014Breaking News We have been using SI units of meters per second squared to describe some examples of acceleration or deceleration of cars, runners, and trains. To achieve a better feel for these numbers, one can measure the braking deceleration of a car - = \u0394 / \u0394. While traveling in a car, slowly apply the doing a slow (and safe) stop. Recall that, for average acceleration, brakes as you come up to a stop sign. Have a passenger note the initial speed in miles per hour and the time taken (in seconds) to stop. From this, calculate the deceleration in miles per hour per second. Convert this to meters per second squared and compare with other decelerations mentioned in this chapter. Calculate the distance traveled in braking. Check Your Understanding A manned rocket accelerates at a rate of 20 m/s2 during launch. How long does it take the rocket reach a velocity of 400 m/s? Solution To answer this, choose an equation that allows you to solve for time, given only, 0, and. Rearrange to solve for. = 0 + = \u2212 = 400 m/s \u2212 0 m/s 20 m/s2 = 20 s (2.70) (2.71) 66 Chapter 2 | Kinematics 2.6 Problem-Solving Basics for One Dimensional Kinematics Figure 2.49 Problem-solving skills are essential to your success in Physics. (credit: scui3asteveo, Flickr) By the end of this section, you will be able to: Learning Objectives \u2022 Apply problem-solving steps and strategies to solve problems of one-dimensional kinematics. \u2022 Apply strategies to determine whether or not the result of a problem is reasonable, and if not, determine the cause. Problem-solving skills are obviously essential to success in a quantitative course in physics. More importantly, the ability to apply broad physical principles, usually represented by equations, to specific situations is a very powerful form of knowledge. It is much more powerful than memorizing a list of facts. Analytical skills and problem-solving abilities can be applied to new situations, whereas a list of facts cannot be made long enough to contain every possible circumstance. Such analytical skills are useful both for solving problems in this text and for applying physics in everyday and professional life. Problem-Solving Steps While there is no simple step-by-step method that works for every problem, the following general procedures facilitate problem solving and make it more meaningful. A certain amount of creativity and insight is required as well. Step 1 Examine the situation to determine which physical principles are involved. It often helps to draw a simple sketch at the outset. You will also need to decide which direction is positive and note that on your sketch. Once you have identified the physical principles, it is much easier to find and apply the equations representing those principles. Although finding the correct equation is essential, keep in mind that equations represent physical principles, laws of nature, and relationships among physical quantities. Without a conceptual understanding of a problem, a numerical solution is meaningless. Step 2 Make a list of what is given or can be inferred from the problem as stated (identify the knowns). Many problems are stated very succinctly and require some inspection to determine what is known. A sketch can also be very useful at this point. Formally identifying the knowns is of particular importance in applying physics to real-world situations. Remember, \u201cstopped\u201d means velocity is zero, and we often can take initial time and position as zero. Step 3 Identify exactly what needs to be determined in the problem (identify the unknowns). In complex problems, especially, it is not always obvious what needs to be found or in what sequence. Making a list can help. Step 4 Find an equation or set of equations that can help you solve the problem. Your list of knowns and unknowns can help here. It is easiest if you can find equations that contain only one unknown\u2014that is, all of the other variables are known, so you can easily solve for the unknown. If the equation contains more than one unknown, then an additional equation is needed to solve the problem. In some problems, several unknowns must be determined to get at the one needed most. In such problems it is especially important to keep physical principles in mind to avoid going astray in a sea of equations. You may have to use two (or more) different equations to get the final answer. Step 5 Substitute the knowns along with their units into the appropriate equation, and obtain", " numerical solutions complete with units. This step produces the numerical answer; it also provides a check on units that can help you find errors. If the units of the answer are incorrect, then an error has been made. However, be warned that correct units do not guarantee that the numerical part of the answer is also correct. This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 2 | Kinematics Step 6 67 Check the answer to see if it is reasonable: Does it make sense? This final step is extremely important\u2014the goal of physics is to accurately describe nature. To see if the answer is reasonable, check both its magnitude and its sign, in addition to its units. Your judgment will improve as you solve more and more physics problems, and it will become possible for you to make finer and finer judgments regarding whether nature is adequately described by the answer to a problem. This step brings the problem back to its conceptual meaning. If you can judge whether the answer is reasonable, you have a deeper understanding of physics than just being able to mechanically solve a problem. When solving problems, we often perform these steps in different order, and we also tend to do several steps simultaneously. There is no rigid procedure that will work every time. Creativity and insight grow with experience, and the basics of problem solving become almost automatic. One way to get practice is to work out the text's examples for yourself as you read. Another is to work as many end-of-section problems as possible, starting with the easiest to build confidence and progressing to the more difficult. Once you become involved in physics, you will see it all around you, and you can begin to apply it to situations you encounter outside the classroom, just as is done in many of the applications in this text. Unreasonable Results Physics must describe nature accurately. Some problems have results that are unreasonable because one premise is unreasonable or because certain premises are inconsistent with one another. The physical principle applied correctly then produces an unreasonable result. For example, if a person starting a foot race accelerates at 0.40 m/s2 for 100 s, his final speed will be 40 m/s (about 150 km/h)\u2014clearly unreasonable because the time of 100 s is an unreasonable premise. The physics is correct in a sense, but there is more to describing nature than just manipulating equations correctly. Checking the result of a problem to see if it is reasonable does more than help uncover errors in problem solving\u2014it also builds intuition in judging whether nature is being accurately described. Use the following strategies to determine whether an answer is reasonable and, if it is not, to determine what is the cause. Step 1 Solve the problem using strategies as outlined and in the format followed in the worked examples in the text. In the example given in the preceding paragraph, you would identify the givens as the acceleration and time and use the equation below to find the unknown final velocity. That is, = 0 + = 0 + 0.40 m/s2 (100 s) = 40 m/s. (2.72) Step 2 Check to see if the answer is reasonable. Is it too large or too small, or does it have the wrong sign, improper units, \u2026? In this case, you may need to convert meters per second into a more familiar unit, such as miles per hour. 40 m s 3.28 ft m 1 mi 5280 ft 60 s min 60 min 1 h = 89 mph (2.73) This velocity is about four times greater than a person can run\u2014so it is too large. Step 3 If the answer is unreasonable, look for what specifically could cause the identified difficulty. In the example of the runner, there are only two assumptions that are suspect. The acceleration could be too great or the time too long. First look at the acceleration and think about what the number means. If someone accelerates at 0.40 m/s2, their velocity is increasing by 0.4 m/s each second. Does this seem reasonable? If so, the time must be too long. It is not possible for someone to accelerate at a constant rate of 0.40 m/s2 for 100 s (almost two minutes). 2.7 Falling Objects Learning Objectives By the end of this section, you will be able to: \u2022 Describe the effects of gravity on objects in motion. \u2022 Describe the motion of objects that are in free fall. \u2022 Calculate the position and velocity of objects in free fall. The information presented in this section supports the following AP\u00ae learning objectives and science practices: \u2022 3.A.1.1 The student is able to express the motion of an object using narrative, mathematical, or graphical representations. (S.P. 1.5, 2.1, 2.2) \u2022 3.A.1.2 The student is able to design an experimental investigation of the motion of an object. (S.P. 4.2) \u2022 3.", "A.1.3 The student is able to analyze experimental data describing the motion of an object and is able to express the results of the analysis using narrative, mathematical, and graphical representations. (S.P. 5.1) 68 Chapter 2 | Kinematics Falling objects form an interesting class of motion problems. For example, we can estimate the depth of a vertical mine shaft by dropping a rock into it and listening for the rock to hit the bottom. By applying the kinematics developed so far to falling objects, we can examine some interesting situations and learn much about gravity in the process. Gravity The most remarkable and unexpected fact about falling objects is that, if air resistance and friction are negligible, then in a given location all objects fall toward the center of Earth with the same constant acceleration, independent of their mass. This experimentally determined fact is unexpected, because we are so accustomed to the effects of air resistance and friction that we expect light objects to fall slower than heavy ones. Figure 2.50 A hammer and a feather will fall with the same constant acceleration if air resistance is considered negligible. This is a general characteristic of gravity not unique to Earth, as astronaut David R. Scott demonstrated on the Moon in 1971, where the acceleration due to gravity is only 1.67 m/s2. In the real world, air resistance can cause a lighter object to fall slower than a heavier object of the same size. A tennis ball will reach the ground after a hard baseball dropped at the same time. (It might be difficult to observe the difference if the height is not large.) Air resistance opposes the motion of an object through the air, while friction between objects\u2014such as between clothes and a laundry chute or between a stone and a pool into which it is dropped\u2014also opposes motion between them. For the ideal situations of these first few chapters, an object falling without air resistance or friction is defined to be in free-fall. The force of gravity causes objects to fall toward the center of Earth. The acceleration of free-falling objects is therefore called the acceleration due to gravity. The acceleration due to gravity is constant, which means we can apply the kinematics equations to any falling object where air resistance and friction are negligible. This opens a broad class of interesting situations to us. The acceleration due to gravity is so important that its magnitude is given its own symbol,. It is constant at any given location on Earth and has the average value = 9.80 m/s2. (2.74) Although varies from 9.78 m/s2 to 9.83 m/s2, depending on latitude, altitude, underlying geological formations, and local topography, the average value of 9.80 m/s2 will be used in this text unless otherwise specified. The direction of the acceleration due to gravity is downward (towards the center of Earth). In fact, its direction defines what we call vertical. Note that whether the acceleration in the kinematic equations has the value + or \u2212 depends on how we define our coordinate system. If we define the upward direction as positive, then = \u2212 = \u22129.80 m/s2, and if we define the downward direction as positive, then = = 9.80 m/s2. One-Dimensional Motion Involving Gravity The best way to see the basic features of motion involving gravity is to start with the simplest situations and then progress toward more complex ones. So we start by considering straight up and down motion with no air resistance or friction. These assumptions mean that the velocity (if there is any) is vertical. If the object is dropped, we know the initial velocity is zero. Once the object has left contact with whatever held or threw it, the object is in free-fall. Under these circumstances, the motion is onedimensional and has constant acceleration of magnitude. We will also represent vertical displacement with the symbol and use for horizontal displacement. Kinematic Equations for Objects in Free-Fall where Acceleration = -( \u2212 0) (2.75) (2.76) (2.77) This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 2 | Kinematics 69 Example 2.14 Calculating Position and Velocity of a Falling Object: A Rock Thrown Upward A person standing on the edge of a high cliff throws a rock straight up with an initial velocity of 13.0 m/s. The rock misses the edge of the cliff as it falls back to Earth. Calculate the position and velocity of the rock 1.00 s, 2.00 s, and 3.00 s after it is thrown, neglecting the effects of air resistance. Strategy Draw a sketch. Figure 2.51 We are asked to determine the position at various times. It is reasonable to take the initial position 0 to be zero. This problem involves one-dimensional motion in the vertical direction. We use plus and minus signs to indicate direction, with up being positive", " and down negative. Since up is positive, and the rock is thrown upward, the initial velocity must be positive too. The acceleration due to gravity is downward, so is negative. It is crucial that the initial velocity and the acceleration due to gravity have opposite signs. Opposite signs indicate that the acceleration due to gravity opposes the initial motion and will slow and eventually reverse it. Since we are asked for values of position and velocity at three times, we will refer to these as 1 and 1 ; and 2 ; and 3 and 3. Solution for Position 1 1. Identify the knowns. We know that 0 = 0 ; 0 = 13.0 m/s ; = \u2212 = \u22129.80 m/s2 ; and = 1.00 s. 2. Identify the best equation to use. We will use = 0 + 0 + 1 here), which is the value we want to find. 3. Plug in the known values and solve for 1. 2 2 because it includes only one unknown, (or 1, = 0 + (13.0 m/s)(1.00 s) + 1 2 \u22129.80 m/s2 (1.00 s)2 = 8.10 m (2.78) Discussion The rock is 8.10 m above its starting point at = 1.00 s, since 1 > 0. It could be moving up or down; the only way to tell is to calculate 1 and find out if it is positive or negative. Solution for Velocity 1 1. Identify the knowns. We know that 0 = 0 ; 0 = 13.0 m/s ; = \u2212 = \u22129.80 m/s2 ; and = 1.00 s. We also know from the solution above that 1 = 8.10 m. 2. Identify the best equation to use. The most straightforward is = 0 \u2212 (from = 0 +, where = gravitational acceleration = \u2212 ). 3. Plug in the knowns and solve. 1 = 0 \u2212 = 13.0 m/s \u2212 9.80 m/s2 (1.00 s) = 3.20 m/s (2.79) Discussion The positive value for 1 means that the rock is still heading upward at = 1.00 s. However, it has slowed from its original 13.0 m/s, as expected. Solution for Remaining Times The procedures for calculating the position and velocity at = 2.00 s and 3.00 s are the same as those above. The results are summarized in Table 2.1 and illustrated in Figure 2.52. 70 Chapter 2 | Kinematics Table 2.1 Results Time, t Position, y Velocity, v Acceleration, a 1.00 s 8.10 m 3.20 m/s 2.00 s 6.40 m \u22126.60 m/s 3.00 s \u22125.10 m \u221216.4 m/s \u22129.80 m/s2 \u22129.80 m/s2 \u22129.80 m/s2 Graphing the data helps us understand it more clearly. Figure 2.52 Vertical position, vertical velocity, and vertical acceleration vs. time for a rock thrown vertically up at the edge of a cliff. Notice that velocity changes linearly with time and that acceleration is constant. Misconception Alert! Notice that the position vs. time graph shows vertical position only. It is easy to get the impression that the graph shows some horizontal motion\u2014the shape of the graph looks like the path of a projectile. But this is not the case; the horizontal axis is time, not space. The actual path of the rock in space is straight up, and straight down. Discussion The interpretation of these results is important. At 1.00 s the rock is above its starting point and heading upward, since 1 and 1 are both positive. At 2.00 s, the rock is still above its starting point, but the negative velocity means it is moving downward. At 3.00 s, both 3 and 3 are negative, meaning the rock is below its starting point and continuing to move This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 2 | Kinematics 71 downward. Notice that when the rock is at its highest point (at 1.5 s), its velocity is zero, but its acceleration is still \u22129.80 m/s2. Its acceleration is \u22129.80 m/s2 for the whole trip\u2014while it is moving up and while it is moving down. Note that the values for are the positions (or displacements) of the rock, not the total distances traveled. Finally, note that freefall applies to upward motion as well as downward. Both have the same acceleration\u2014the acceleration due to gravity, which remains constant the entire time. Astronauts training in the famous Vomit Comet, for example, experience free-fall while arcing up as well as down, as we will", " discuss in more detail later. Making Connections: Take-Home Experiment\u2014Reaction Time A simple experiment can be done to determine your reaction time. Have a friend hold a ruler between your thumb and index finger, separated by about 1 cm. Note the mark on the ruler that is right between your fingers. Have your friend drop the ruler unexpectedly, and try to catch it between your two fingers. Note the new reading on the ruler. Assuming acceleration is that due to gravity, calculate your reaction time. How far would you travel in a car (moving at 30 m/s) if the time it took your foot to go from the gas pedal to the brake was twice this reaction time? Example 2.15 Calculating Velocity of a Falling Object: A Rock Thrown Down What happens if the person on the cliff throws the rock straight down, instead of straight up? To explore this question, calculate the velocity of the rock when it is 5.10 m below the starting point, and has been thrown downward with an initial speed of 13.0 m/s. Strategy Draw a sketch. Figure 2.53 Since up is positive, the final position of the rock will be negative because it finishes below the starting point at 0 = 0. Similarly, the initial velocity is downward and therefore negative, as is the acceleration due to gravity. We expect the final velocity to be negative since the rock will continue to move downward. Solution 1. Identify the knowns. 0 = 0 ; 1 = \u2212 5.10 m ; 0 = \u221213.0 m/s ; = \u2212 = \u22129.80 m/s2. 2. Choose the kinematic equation that makes it easiest to solve the problem. The equation 2 = 0 well because the only unknown in it is. (We will plug 1 in for.) 2 + 2( \u2212 0) works 3. Enter the known values 2 = (\u221213.0 m/s)2 + 2 \u22129.80 m/s2 (\u22125.10 m \u2212 0 m) = 268.96 m2 /s2, where we have retained extra significant figures because this is an intermediate result. Taking the square root, and noting that a square root can be positive or negative, gives = \u00b116.4 m/s. The negative root is chosen to indicate that the rock is still heading down. Thus, = \u221216.4 m/s. Discussion (2.80) (2.81) (2.82) Note that this is exactly the same velocity the rock had at this position when it was thrown straight upward with the same initial speed. (See Example 2.14 and Figure 2.54(a).) This is not a coincidental result. Because we only consider the acceleration due to gravity in this problem, the speed of a falling object depends only on its initial speed and its vertical position relative to the starting point. For example, if the velocity of the rock is calculated at a height of 8.10 m above the starting point (using the method from Example 2.14) when the initial velocity is 13.0 m/s straight up, a result of \u00b13.20 m/s 72 Chapter 2 | Kinematics is obtained. Here both signs are meaningful; the positive value occurs when the rock is at 8.10 m and heading up, and the negative value occurs when the rock is at 8.10 m and heading back down. It has the same speed but the opposite direction. Figure 2.54 (a) A person throws a rock straight up, as explored in Example 2.14. The arrows are velocity vectors at 0, 1.00, 2.00, and 3.00 s. (b) A person throws a rock straight down from a cliff with the same initial speed as before, as in Example 2.15. Note that at the same distance below the point of release, the rock has the same velocity in both cases. Another way to look at it is this: In Example 2.14, the rock is thrown up with an initial velocity of 13.0 m/s. It rises and then falls back down. When its position is = 0 on its way back down, its velocity is \u221213.0 m/s. That is, it has the same speed on its way down as on its way up. We would then expect its velocity at a position of = \u22125.10 m to be the same whether we have thrown it upwards at +13.0 m/s or thrown it downwards at \u221213.0 m/s. The velocity of the rock on its way down from = 0 is the same whether we have thrown it up or down to start with, as long as the speed with which it was initially thrown is the same. Example 2.16 Find g from Data on a Falling Object The acceleration due to gravity on Earth differs slightly from place to place, depending on topography (e.g., whether you are on a hill or in a valley) and subsurface", " geology (whether there is dense rock like iron ore as opposed to light rock like salt beneath you.) The precise acceleration due to gravity can be calculated from data taken in an introductory physics This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 2 | Kinematics 73 laboratory course. An object, usually a metal ball for which air resistance is negligible, is dropped and the time it takes to fall a known distance is measured. See, for example, Figure 2.55. Very precise results can be produced with this method if sufficient care is taken in measuring the distance fallen and the elapsed time. Figure 2.55 Positions and velocities of a metal ball released from rest when air resistance is negligible. Velocity is seen to increase linearly with time while displacement increases with time squared. Acceleration is a constant and is equal to gravitational acceleration. Suppose the ball falls 1.0000 m in 0.45173 s. Assuming the ball is not affected by air resistance, what is the precise acceleration due to gravity at this location? Strategy Draw a sketch. 74 Chapter 2 | Kinematics Figure 2.56 We need to solve for acceleration. Note that in this case, displacement is downward and therefore negative, as is acceleration. Solution 1. Identify the knowns. 0 = 0 ; = \u20131.0000 m ; = 0.45173 ; 0 = 0. 2. Choose the equation that allows you to solve for using the known values. Substitute 0 for 0 and rearrange the equation to solve for. Substituting 0 for 0 yields Solving for gives 4. Substitute known values yields = 0 + 1 2 2. = 2( \u2212 0) 2. = 2( \u2212 1.0000 m \u2013 0) (0.45173 s)2 = \u22129.8010 m/s2 so, because = \u2212 with the directions we have chosen, = 9.8010 m/s2. Discussion (2.83) (2.84) (2.85) (2.86) (2.87) The negative value for indicates that the gravitational acceleration is downward, as expected. We expect the value to be somewhere around the average value of 9.80 m/s2, so 9.8010 m/s2 makes sense. Since the data going into the calculation are relatively precise, this value for is more precise than the average value of 9.80 m/s2 ; it represents the local value for the acceleration due to gravity. Applying the Science Practices: Finding Acceleration Due to Gravity While it is well established that the acceleration due to gravity is quite nearly 9.8 m/s2 at all locations on Earth, you can verify this for yourself with some basic materials. Your task is to find the acceleration due to gravity at your location. Achieving an acceleration of precisely 9.8 m/s2 will be difficult. However, with good preparation and attention to detail, you should be able to get close. Before you begin working, consider the following questions. What measurements will you need to take in order to find the acceleration due to gravity? What relationships and equations found in this chapter may be useful in calculating the acceleration? What variables will you need to hold constant? What materials will you use to record your measurements? Upon completing these four questions, record your procedure. Once recorded, you may carry out the experiment. If you find that your experiment cannot be carried out, you may revise your procedure. Once you have found your experimental acceleration, compare it to the assumed value of 9.8 m/s2. If error exists, what were the likely sources of this error? How could you change your procedure in order to improve the accuracy of your findings? This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 2 | Kinematics 75 Check Your Understanding A chunk of ice breaks off a glacier and falls 30.0 meters before it hits the water. Assuming it falls freely (there is no air resistance), how long does it take to hit the water? Solution We know that initial position 0 = 0, final position = \u221230.0 m, and = \u2212 = \u22129.80 m/s2. We can then use the equation = 0 + 0 + 1 (2.88) 2 2 to solve for. Inserting = \u2212, we obtain = ( \u2212 30.0 m) \u22129.80 m/s2 = \u00b1 6.12 s2 = 2.47 s \u2248 2.5 s where we take the positive value as the physically relevant answer. Thus, it takes about 2.5 seconds for the piece of ice to hit the water. PhET Explorations: Equation Grapher Learn about graphing polynomials. The shape of the curve changes as the constants are adjusted. View the curves for the individual terms (e.g. = ) to see how they", " add to generate the polynomial curve. Figure 2.57 Equation Grapher (http://cnx.org/content/m54775/1.5/equation-grapher_en.jar) 2.8 Graphical Analysis of One Dimensional Motion By the end of this section, you will be able to: Learning Objectives \u2022 Describe a straight-line graph in terms of its slope and y-intercept. \u2022 Determine average velocity or instantaneous velocity from a graph of position vs. time. \u2022 Determine average or instantaneous acceleration from a graph of velocity vs. time. \u2022 Derive a graph of velocity vs. time from a graph of position vs. time. \u2022 Derive a graph of acceleration vs. time from a graph of velocity vs. time. A graph, like a picture, is worth a thousand words. Graphs not only contain numerical information; they also reveal relationships between physical quantities. This section uses graphs of displacement, velocity, and acceleration versus time to illustrate onedimensional kinematics. Slopes and General Relationships First note that graphs in this text have perpendicular axes, one horizontal and the other vertical. When two physical quantities are plotted against one another in such a graph, the horizontal axis is usually considered to be an independent variable and the vertical axis a dependent variable. If we call the horizontal axis the -axis and the vertical axis the -axis, as in Figure 2.58, a straight-line graph has the general form = +. (2.89) Here is the slope, defined to be the rise divided by the run (as seen in the figure) of the straight line. The letter is used for the y-intercept, which is the point at which the line crosses the vertical axis. 76 Chapter 2 | Kinematics Figure 2.58 A straight-line graph. The equation for a straight line is = +. Graph of Displacement vs. Time (a = 0, so v is constant) Time is usually an independent variable that other quantities, such as displacement, depend upon. A graph of displacement versus time would, thus, have on the vertical axis and on the horizontal axis. Figure 2.59 is just such a straight-line graph. It shows a graph of displacement versus time for a jet-powered car on a very flat dry lake bed in Nevada. Figure 2.59 Graph of displacement versus time for a jet-powered car on the Bonneville Salt Flats. Using the relationship between dependent and independent variables, we see that the slope in the graph above is average velocity - and the intercept is displacement at time zero\u2014that is, 0. Substituting these symbols into = + gives + 0 = (2.90) or Thus a graph of displacement versus time gives a general relationship among displacement, velocity, and time, as well as giving detailed numerical information about a specific situation. = 0 +. (2.91) The Slope of x vs. t The slope of the graph of displacement vs. time is velocity. Notice that this equation is the same as that derived algebraically from other motion equations in Motion Equations for Constant Acceleration in One Dimension. slope = \u0394 \u0394 = (2.92) From the figure we can see that the car has a displacement of 400 m at time 0.650 m at = 1.0 s, and so on. Its displacement at times other than those listed in the table can be read from the graph; furthermore, information about its velocity and acceleration can also be obtained from the graph. This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 2 | Kinematics 77 Example 2.17 Determining Average Velocity from a Graph of Displacement versus Time: Jet Car Find the average velocity of the car whose position is graphed in Figure 2.59. Strategy The slope of a graph of vs. is average velocity, since slope equals rise over run. In this case, rise = change in displacement and run = change in time, so that slope = \u0394 \u0394 -. = (2.93) Since the slope is constant here, any two points on the graph can be used to find the slope. (Generally speaking, it is most accurate to use two widely separated points on the straight line. This is because any error in reading data from the graph is proportionally smaller if the interval is larger.) Solution 1. Choose two points on the line. In this case, we choose the points labeled on the graph: (6.4 s, 2000 m) and (0.50 s, 525 m). (Note, however, that you could choose any two points.) 2. Substitute the and values of the chosen points into the equation. Remember in calculating change (\u0394) we always use final value minus initial value. yielding Discussion - = \u0394 \u0394 = 2000 m \u2212 525 m 6.4 s \u2212 0.50 s, - = 250 m/s.", " (2.94) (2.95) This is an impressively large land speed (900 km/h, or about 560 mi/h): much greater than the typical highway speed limit of 60 mi/h (27 m/s or 96 km/h), but considerably shy of the record of 343 m/s (1234 km/h or 766 mi/h) set in 1997. Graphs of Motion when is constant but \u2260 0 The graphs in Figure 2.60 below represent the motion of the jet-powered car as it accelerates toward its top speed, but only during the time when its acceleration is constant. Time starts at zero for this motion (as if measured with a stopwatch), and the displacement and velocity are initially 200 m and 15 m/s, respectively. 78 Chapter 2 | Kinematics Figure 2.60 Graphs of motion of a jet-powered car during the time span when its acceleration is constant. (a) The slope of an vs. graph is velocity. This is shown at two points, and the instantaneous velocities obtained are plotted in the next graph. Instantaneous velocity at any point is the slope of the tangent at that point. (b) The slope of the vs. graph is constant for this part of the motion, indicating constant acceleration. (c) Acceleration has the constant value of 5.0 m/s2 over the time interval plotted. This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 2 | Kinematics 79 Figure 2.61 A U.S. Air Force jet car speeds down a track. (credit: Matt Trostle, Flickr) The graph of displacement versus time in Figure 2.60(a) is a curve rather than a straight line. The slope of the curve becomes steeper as time progresses, showing that the velocity is increasing over time. The slope at any point on a displacement-versustime graph is the instantaneous velocity at that point. It is found by drawing a straight line tangent to the curve at the point of interest and taking the slope of this straight line. Tangent lines are shown for two points in Figure 2.60(a). If this is done at every point on the curve and the values are plotted against time, then the graph of velocity versus time shown in Figure 2.60(b) is obtained. Furthermore, the slope of the graph of velocity versus time is acceleration, which is shown in Figure 2.60(c). Example 2.18 Determining Instantaneous Velocity from the Slope at a Point: Jet Car Calculate the velocity of the jet car at a time of 25 s by finding the slope of the vs. graph in the graph below. Figure 2.62 The slope of an vs. graph is velocity. This is shown at two points. Instantaneous velocity at any point is the slope of the tangent at that point. Strategy The slope of a curve at a point is equal to the slope of a straight line tangent to the curve at that point. This principle is illustrated in Figure 2.62, where Q is the point at = 25 s. Solution 1. Find the tangent line to the curve at = 25 s. 2. Determine the endpoints of the tangent. These correspond to a position of 1300 m at time 19 s and a position of 3120 m at time 32 s. 3. Plug these endpoints into the equation to solve for the slope,. slope = Q = \u0394Q \u0394 Q = (3120 m \u2212 1300 m) (32 s \u2212 19 s) Q = 1820 m 13 s = 140 m/s. (2.96) (2.97) Thus, Discussion 80 Chapter 2 | Kinematics This is the value given in this figure's table for at = 25 s. The value of 140 m/s for Q is plotted in Figure 2.62. The entire graph of vs. can be obtained in this fashion. Carrying this one step further, we note that the slope of a velocity versus time graph is acceleration. Slope is rise divided by run; on a vs. graph, rise = change in velocity \u0394 and run = change in time \u0394. The Slope of v vs. t The slope of a graph of velocity vs. time is acceleration. slope = \u0394 \u0394 = (2.98) Since the velocity versus time graph in Figure 2.60(b) is a straight line, its slope is the same everywhere, implying that acceleration is constant. Acceleration versus time is graphed in Figure 2.60(c). Additional general information can be obtained from Figure 2.62 and the expression for a straight line, = +. In this case, the vertical axis is, the intercept is 0, the slope is, and the horizontal axis is. Substituting these symbols yields = 0 +. (2.99) A general relationship for velocity, acceleration, and time has", " again been obtained from a graph. Notice that this equation was also derived algebraically from other motion equations in Motion Equations for Constant Acceleration in One Dimension. It is not accidental that the same equations are obtained by graphical analysis as by algebraic techniques. In fact, an important way to discover physical relationships is to measure various physical quantities and then make graphs of one quantity against another to see if they are correlated in any way. Correlations imply physical relationships and might be shown by smooth graphs such as those above. From such graphs, mathematical relationships can sometimes be postulated. Further experiments are then performed to determine the validity of the hypothesized relationships. Graphs of Motion Where Acceleration is Not Constant Now consider the motion of the jet car as it goes from 165 m/s to its top velocity of 250 m/s, graphed in Figure 2.63. Time again starts at zero, and the initial displacement and velocity are 2900 m and 165 m/s, respectively. (These were the final displacement and velocity of the car in the motion graphed in Figure 2.60.) Acceleration gradually decreases from 5.0 m/s2 to zero when the car hits 250 m/s. The slope of the vs. graph increases until = 55 s, after which time the slope is constant. Similarly, velocity increases until 55 s and then becomes constant, since acceleration decreases to zero at 55 s and remains zero afterward. This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 2 | Kinematics 81 Figure 2.63 Graphs of motion of a jet-powered car as it reaches its top velocity. This motion begins where the motion in Figure 2.60 ends. (a) The slope of this graph is velocity; it is plotted in the next graph. (b) The velocity gradually approaches its top value. The slope of this graph is acceleration; it is plotted in the final graph. (c) Acceleration gradually declines to zero when velocity becomes constant. Example 2.19 Calculating Acceleration from a Graph of Velocity versus Time Calculate the acceleration of the jet car at a time of 25 s by finding the slope of the vs. graph in Figure 2.63(b). Strategy The slope of the curve at = 25 s is equal to the slope of the line tangent at that point, as illustrated in Figure 2.63(b). Solution Determine endpoints of the tangent line from the figure, and then plug them into the equation to solve for slope,. slope = \u0394 \u0394 = (260 m/s \u2212 210 m/s) (51 s \u2212 1.0 s) (2.100) 82 Discussion = 50 m/s 50 s = 1.0 m/s2. Chapter 2 | Kinematics (2.101) Note that this value for is consistent with the value plotted in Figure 2.63(c) at = 25 s. A graph of displacement versus time can be used to generate a graph of velocity versus time, and a graph of velocity versus time can be used to generate a graph of acceleration versus time. We do this by finding the slope of the graphs at every point. If the graph is linear (i.e., a line with a constant slope), it is easy to find the slope at any point and you have the slope for every point. Graphical analysis of motion can be used to describe both specific and general characteristics of kinematics. Graphs can also be used for other topics in physics. An important aspect of exploring physical relationships is to graph them and look for underlying relationships. Check Your Understanding A graph of velocity vs. time of a ship coming into a harbor is shown below. (a) Describe the motion of the ship based on the graph. (b)What would a graph of the ship's acceleration look like? Figure 2.64 Solution (a) The ship moves at constant velocity and then begins to decelerate at a constant rate. At some point, its deceleration rate decreases. It maintains this lower deceleration rate until it stops moving. (b) A graph of acceleration vs. time would show zero acceleration in the first leg, large and constant negative acceleration in the second leg, and constant negative acceleration. Figure 2.65 Glossary acceleration: the rate of change in velocity; the change in velocity over time acceleration due to gravity: acceleration of an object as a result of gravity average acceleration: the change in velocity divided by the time over which it changes average speed: distance traveled divided by time during which motion occurs average velocity: displacement divided by time over which displacement occurs deceleration: acceleration in the direction opposite to velocity; acceleration that results in a decrease in velocity dependent variable: the variable that is being measured; usually plotted along the -axis displacement: the change in position of an object distance: the magnitude of displacement between two positions distance traveled: the total length of the path traveled between two positions This content is available for free at http://cn", "x.org/content/col11844/1.13 Chapter 2 | Kinematics 83 elapsed time: the difference between the ending time and beginning time free-fall: the state of movement that results from gravitational force only independent variable: the variable that the dependent variable is measured with respect to; usually plotted along the -axis instantaneous acceleration: acceleration at a specific point in time instantaneous speed: magnitude of the instantaneous velocity instantaneous velocity: velocity at a specific instant, or the average velocity over an infinitesimal time interval kinematics: the study of motion without considering its causes model: simplified description that contains only those elements necessary to describe the physics of a physical situation position: the location of an object at a particular time scalar: a quantity that is described by magnitude, but not direction slope: the difference in -value (the rise) divided by the difference in -value (the run) of two points on a straight line time: change, or the interval over which change occurs vector: a quantity that is described by both magnitude and direction y-intercept: the - value when = 0, or when the graph crosses the -axis Section Summary 2.1 Displacement \u2022 Kinematics is the study of motion without considering its causes. In this chapter, it is limited to motion along a straight line, called one-dimensional motion. \u2022 Displacement is the change in position of an object. In symbols, displacement \u0394 is defined to be \u2022 \u0394 = f \u2212 0, where 0 is the initial position and f is the final position. In this text, the Greek letter \u0394 (delta) always means \u201cchange in\u201d whatever quantity follows it. The SI unit for displacement is the meter (m). Displacement has a direction as well as a magnitude. \u2022 When you start a problem, assign which direction will be positive. \u2022 Distance is the magnitude of displacement between two positions. \u2022 Distance traveled is the total length of the path traveled between two positions. 2.2 Vectors, Scalars, and Coordinate Systems \u2022 A vector is any quantity that has magnitude and direction. \u2022 A scalar is any quantity that has magnitude but no direction. \u2022 Displacement and velocity are vectors, whereas distance and speed are scalars. \u2022 In one-dimensional motion, direction is specified by a plus or minus sign to signify left or right, up or down, and the like. 2.3 Time, Velocity, and Speed \u2022 Time is measured in terms of change, and its SI unit is the second (s). Elapsed time for an event is \u0394 = f \u2212 0, where f is the final time and 0 is the initial time. The initial time is often taken to be zero, as if measured with a stopwatch; the elapsed time is then just. - is defined as displacement divided by the travel time. In symbols, average velocity is \u2022 Average velocity - = \u0394 \u0394 = f \u2212 0 f \u2212 0. \u2022 The SI unit for velocity is m/s. \u2022 Velocity is a vector and thus has a direction. \u2022 Instantaneous velocity is the velocity at a specific instant or the average velocity for an infinitesimal interval. Instantaneous speed is the magnitude of the instantaneous velocity. Instantaneous speed is a scalar quantity, as it has no direction specified. \u2022 \u2022 84 Chapter 2 | Kinematics \u2022 Average speed is the total distance traveled divided by the elapsed time. (Average speed is not the magnitude of the average velocity.) Speed is a scalar quantity; it has no direction associated with it. 2.4 Acceleration \u2022 Acceleration is the rate at which velocity changes. In symbols, average acceleration - is - = \u0394 \u0394 = f \u2212 0 f \u2212 0. \u2022 The SI unit for acceleration is m/s2. \u2022 Acceleration is a vector, and thus has a both a magnitude and direction. \u2022 Acceleration can be caused by either a change in the magnitude or the direction of the velocity. \u2022 Instantaneous acceleration is the acceleration at a specific instant in time. \u2022 Deceleration is an acceleration with a direction opposite to that of the velocity. 2.5 Motion Equations for Constant Acceleration in One Dimension \u2022 To simplify calculations we take acceleration to be constant, so that \u2022 We also take initial time to be zero. \u2022 Initial position and velocity are given a subscript 0; final values have no subscript. Thus, - = at all times. \u2022 The following kinematic equations for motion with constant are useful( \u2212 0) \u2022 In vertical motion, is substituted for. 2.6 Problem-Solving Basics for One Dimensional Kinematics \u2022 The six basic problem solving steps for physics are: Step 1. Examine the situation to determine which physical principles are involved. Step 2. Make a list of what is given or can be inferred from the problem as stated (identify the knowns). Step 3. Identify exactly what needs to be determined in the problem (identify the unknowns). Step 4. Find an equation or set", " of equations that can help you solve the problem. Step 5. Substitute the knowns along with their units into the appropriate equation, and obtain numerical solutions complete with units. Step 6. Check the answer to see if it is reasonable: Does it make sense? 2.7 Falling Objects \u2022 An object in free-fall experiences constant acceleration if air resistance is negligible. \u2022 On Earth, all free-falling objects have an acceleration due to gravity, which averages \u2022 Whether the acceleration a should be taken as + or \u2212 is determined by your choice of coordinate system. If you = 9.80 m/s2. choose the upward direction as positive, = \u2212 = \u22129.80 m/s2 is negative. In the opposite case, = +g = 9.80 m/s2 is positive. Since acceleration is constant, the kinematic equations above can be applied with the appropriate + or \u2212 substituted for. \u2022 For objects in free-fall, up is normally taken as positive for displacement, velocity, and acceleration. 2.8 Graphical Analysis of One Dimensional Motion \u2022 Graphs of motion can be used to analyze motion. This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 2 | Kinematics 85 \u2022 Graphical solutions yield identical solutions to mathematical methods for deriving motion equations. \u2022 The slope of a graph of displacement vs. time is velocity. \u2022 The slope of a graph of velocity vs. time graph is acceleration. \u2022 Average velocity, instantaneous velocity, and acceleration can all be obtained by analyzing graphs. Conceptual Questions 2.1 Displacement 1. Give an example in which there are clear distinctions among distance traveled, displacement, and magnitude of displacement. Specifically identify each quantity in your example. 2. Under what circumstances does distance traveled equal magnitude of displacement? What is the only case in which magnitude of displacement and displacement are exactly the same? 3. Bacteria move back and forth by using their flagella (structures that look like little tails). Speeds of up to 50 \u03bcm/s have been observed. The total distance traveled by a bacterium is large for its size, while its 50\u00d710\u22126 m/s displacement is small. Why is this? 2.2 Vectors, Scalars, and Coordinate Systems 4. A student writes, \u201cA bird that is diving for prey has a speed of \u2212 /.\u201d What is wrong with the student's statement? What has the student actually described? Explain. 5. What is the speed of the bird in Exercise 2.4? 6. Acceleration is the change in velocity over time. Given this information, is acceleration a vector or a scalar quantity? Explain. 7. A weather forecast states that the temperature is predicted to be \u22125\u00baC the following day. Is this temperature a vector or a scalar quantity? Explain. 2.3 Time, Velocity, and Speed 8. Give an example (but not one from the text) of a device used to measure time and identify what change in that device indicates a change in time. 9. There is a distinction between average speed and the magnitude of average velocity. Give an example that illustrates the difference between these two quantities. 10. Does a car's odometer measure position or displacement? Does its speedometer measure speed or velocity? 11. If you divide the total distance traveled on a car trip (as determined by the odometer) by the time for the trip, are you calculating the average speed or the magnitude of the average velocity? Under what circumstances are these two quantities the same? 12. How are instantaneous velocity and instantaneous speed related to one another? How do they differ? 2.4 Acceleration 13. Is it possible for speed to be constant while acceleration is not zero? Give an example of such a situation. 14. Is it possible for velocity to be constant while acceleration is not zero? Explain. 15. Give an example in which velocity is zero yet acceleration is not. 16. If a subway train is moving to the left (has a negative velocity) and then comes to a stop, what is the direction of its acceleration? Is the acceleration positive or negative? 17. Plus and minus signs are used in one-dimensional motion to indicate direction. What is the sign of an acceleration that reduces the magnitude of a negative velocity? Of a positive velocity? 2.6 Problem-Solving Basics for One Dimensional Kinematics 18. What information do you need in order to choose which equation or equations to use to solve a problem? Explain. 19. What is the last thing you should do when solving a problem? Explain. 2.7 Falling Objects 20. What is the acceleration of a rock thrown straight upward on the way up? At the top of its flight? On the way down? 21. An object that is thrown straight up falls back to Earth. This is one-dimensional motion. (a) When is its velocity zero? (b) Does its", " velocity change direction? (c) Does the acceleration due to gravity have the same sign on the way up as on the way down? 22. Suppose you throw a rock nearly straight up at a coconut in a palm tree, and the rock misses on the way up but hits the coconut on the way down. Neglecting air resistance, how does the speed of the rock when it hits the coconut on the way down compare with what it would have been if it had hit the coconut on the way up? Is it more likely to dislodge the coconut on the way up or down? Explain. 86 Chapter 2 | Kinematics 23. If an object is thrown straight up and air resistance is negligible, then its speed when it returns to the starting point is the same as when it was released. If air resistance were not negligible, how would its speed upon return compare with its initial speed? How would the maximum height to which it rises be affected? 24. The severity of a fall depends on your speed when you strike the ground. All factors but the acceleration due to gravity being the same, how many times higher could a safe fall on the Moon be than on Earth (gravitational acceleration on the Moon is about 1/6 that of the Earth)? 25. How many times higher could an astronaut jump on the Moon than on Earth if his takeoff speed is the same in both locations (gravitational acceleration on the Moon is about 1/6 of on Earth)? 2.8 Graphical Analysis of One Dimensional Motion 26. (a) Explain how you can use the graph of position versus time in Figure 2.66 to describe the change in velocity over time. Identify (b) the time ( a, b, c, d, or e ) at which the instantaneous velocity is greatest, (c) the time at which it is zero, and (d) the time at which it is negative. Figure 2.66 27. (a) Sketch a graph of velocity versus time corresponding to the graph of displacement versus time given in Figure 2.67. (b) Identify the time or times ( a, b, c, etc.) at which the instantaneous velocity is greatest. (c) At which times is it zero? (d) At which times is it negative? Figure 2.67 28. (a) Explain how you can determine the acceleration over time from a velocity versus time graph such as the one in Figure 2.68. (b) Based on the graph, how does acceleration change over time? This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 2 | Kinematics 87 Figure 2.68 29. (a) Sketch a graph of acceleration versus time corresponding to the graph of velocity versus time given in Figure 2.69. (b) Identify the time or times ( a, b, c, etc.) at which the acceleration is greatest. (c) At which times is it zero? (d) At which times is it negative? Figure 2.69 30. Consider the velocity vs. time graph of a person in an elevator shown in Figure 2.70. Suppose the elevator is initially at rest. It then accelerates for 3 seconds, maintains that velocity for 15 seconds, then decelerates for 5 seconds until it stops. The acceleration for the entire trip is not constant so we cannot use the equations of motion from Motion Equations for Constant Acceleration in One Dimension for the complete trip. (We could, however, use them in the three individual sections where acceleration is a constant.) Sketch graphs of (a) position vs. time and (b) acceleration vs. time for this trip. Figure 2.70 31. A cylinder is given a push and then rolls up an inclined plane. If the origin is the starting point, sketch the position, velocity, and acceleration of the cylinder vs. time as it goes up and then down the plane. 88 Chapter 2 | Kinematics Problems & Exercises 2.1 Displacement Figure 2.71 1. Find the following for path A in Figure 2.71: (a) The distance traveled. (b) The magnitude of the displacement from start to finish. (c) The displacement from start to finish. 2. Find the following for path B in Figure 2.71: (a) The distance traveled. (b) The magnitude of the displacement from start to finish. (c) The displacement from start to finish. 3. Find the following for path C in Figure 2.71: (a) The distance traveled. (b) The magnitude of the displacement from start to finish. (c) The displacement from start to finish. 4. Find the following for path D in Figure 2.71: (a) The distance traveled. (b) The magnitude of the displacement from start to finish. (c) The displacement from start to finish. 2.3 Time, Velocity, and Speed 5. (a) Calculate Earth", "'s average speed relative to the Sun. (b) What is its average velocity over a period of one year? 6. A helicopter blade spins at exactly 100 revolutions per minute. Its tip is 5.00 m from the center of rotation. (a) Calculate the average speed of the blade tip in the helicopter's frame of reference. (b) What is its average velocity over one revolution? 7. The North American and European continents are moving apart at a rate of about 3 cm/y. At this rate how long will it take them to drift 500 km farther apart than they are at present? 8. Land west of the San Andreas fault in southern California is moving at an average velocity of about 6 cm/y northwest relative to land east of the fault. Los Angeles is west of the fault and may thus someday be at the same latitude as San Francisco, which is east of the fault. How far in the future will this occur if the displacement to be made is 590 km northwest, assuming the motion remains constant? 9. On May 26, 1934, a streamlined, stainless steel diesel train called the Zephyr set the world's nonstop long-distance speed record for trains. Its run from Denver to Chicago took 13 hours, 4 minutes, 58 seconds, and was witnessed by more than a million people along the route. The total distance traveled was 1633.8 km. What was its average speed in km/h and m/s? 10. Tidal friction is slowing the rotation of the Earth. As a result, the orbit of the Moon is increasing in radius at a rate of approximately 4 cm/year. Assuming this to be a constant rate, how many years will pass before the radius of the Moon's orbit increases by 3.84\u00d7106 m (1%)? This content is available for free at http://cnx.org/content/col11844/1.13 11. A student drove to the university from her home and noted that the odometer reading of her car increased by 12.0 km. The trip took 18.0 min. (a) What was her average speed? (b) If the straight-line distance from her home to the university is 10.3 km in a direction 25.0\u00ba south of east, what was her average velocity? (c) If she returned home by the same path 7 h 30 min after she left, what were her average speed and velocity for the entire trip? 12. The speed of propagation of the action potential (an electrical signal) in a nerve cell depends (inversely) on the diameter of the axon (nerve fiber). If the nerve cell connecting the spinal cord to your feet is 1.1 m long, and the nerve impulse speed is 18 m/s, how long does it take for the nerve signal to travel this distance? 13. Conversations with astronauts on the lunar surface were characterized by a kind of echo in which the earthbound person's voice was so loud in the astronaut's space helmet that it was picked up by the astronaut's microphone and transmitted back to Earth. It is reasonable to assume that the echo time equals the time necessary for the radio wave to travel from the Earth to the Moon and back (that is, neglecting any time delays in the electronic equipment). Calculate the distance from Earth to the Moon given that the echo time was 2.56 s and that radio waves travel at the speed of light (3.00\u00d7108 m/s). 14. A football quarterback runs 15.0 m straight down the playing field in 2.50 s. He is then hit and pushed 3.00 m straight backward in 1.75 s. He breaks the tackle and runs straight forward another 21.0 m in 5.20 s. Calculate his average velocity (a) for each of the three intervals and (b) for the entire motion. 15. The planetary model of the atom pictures electrons orbiting the atomic nucleus much as planets orbit the Sun. In this model you can view hydrogen, the simplest atom, as having a single electron in a circular orbit 1.06\u00d710\u221210 m in diameter. (a) If the average speed of the electron in this orbit is known to be 2.20\u00d7106 m/s, calculate the number of revolutions per second it makes about the nucleus. (b) What is the electron's average velocity? 2.4 Acceleration 16. A cheetah can accelerate from rest to a speed of 30.0 m/s in 7.00 s. What is its acceleration? 17. Professional Application Dr. John Paul Stapp was U.S. Air Force officer who studied the effects of extreme deceleration on the human body. On December 10, 1954, Stapp rode a rocket sled, accelerating from rest to a top speed of 282 m/s (1015 km/h) in 5.00 s, and was brought jarringly back to rest in only 1.40 s! Calculate his (a)", " acceleration and (b) deceleration. Express each in multiples of (9.80 m/s2) by taking its ratio to the acceleration of gravity. 18. A commuter backs her car out of her garage with an acceleration of 1.40 m/s2. (a) How long does it take her to reach a speed of 2.00 m/s? (b) If she then brakes to a stop in 0.800 s, what is her deceleration? 19. Assume that an intercontinental ballistic missile goes from rest to a suborbital speed of 6.50 km/s in 60.0 s (the actual speed and time are classified). What is its average acceleration in m/s2 and in multiples of (9.80 m/s2)? Chapter 2 | Kinematics 89 2.5 Motion Equations for Constant Acceleration in One Dimension 20. An Olympic-class sprinter starts a race with an acceleration of 4.50 m/s2. (a) What is her speed 2.40 s later? (b) Sketch a graph of her position vs. time for this period. 21. A well-thrown ball is caught in a well-padded mitt. If the deceleration of the ball is 2.10\u00d7104 m/s2, and 1.85 ms (1 ms = 10\u22123 s) elapses from the time the ball first touches the mitt until it stops, what was the initial velocity of the ball? 22. A bullet in a gun is accelerated from the firing chamber to the end of the barrel at an average rate of 6.20\u00d7105 m/s2 for 8.10\u00d710\u22124 s. What is its muzzle velocity (that is, its final velocity)? 23. (a) A light-rail commuter train accelerates at a rate of 1.35 m/s2. How long does it take to reach its top speed of 80.0 km/h, starting from rest? (b) The same train ordinarily decelerates at a rate of 1.65 m/s2. How long does it take to come to a stop from its top speed? (c) In emergencies the train can decelerate more rapidly, coming to rest from 80.0 km/h in 8.30 s. What is its emergency deceleration in m/s2? 24. While entering a freeway, a car accelerates from rest at a rate of 2.40 m/s2 for 12.0 s. (a) Draw a sketch of the situation. (b) List the knowns in this problem. (c) How far does the car travel in those 12.0 s? To solve this part, first identify the unknown, and then discuss how you chose the appropriate equation to solve for it. After choosing the equation, show your steps in solving for the unknown, check your units, and discuss whether the answer is reasonable. (d) What is the car's final velocity? Solve for this unknown in the same manner as in part (c), showing all steps explicitly. 25. At the end of a race, a runner decelerates from a velocity of 9.00 m/s at a rate of 2.00 m/s2. (a) How far does she travel in the next 5.00 s? (b) What is her final velocity? (c) Evaluate the result. Does it make sense? 26. Professional Application: Blood is accelerated from rest to 30.0 cm/s in a distance of 1.80 cm by the left ventricle of the heart. (a) Make a sketch of the situation. (b) List the knowns in this problem. (c) How long does the acceleration take? To solve this part, first identify the unknown, and then discuss how you chose the appropriate equation to solve for it. After choosing the equation, show your steps in solving for the unknown, checking your units. (d) Is the answer reasonable when compared with the time for a heartbeat? 27. In a slap shot, a hockey player accelerates the puck from a velocity of 8.00 m/s to 40.0 m/s in the same direction. If this shot takes 3.33\u00d710\u22122 s, calculate the distance over which the puck accelerates. 28. A powerful motorcycle can accelerate from rest to 26.8 m/ s (100 km/h) in only 3.90 s. (a) What is its average acceleration? (b) How far does it travel in that time? 29. Freight trains can produce only relatively small accelerations and decelerations. (a) What is the final velocity of a freight train that accelerates at a rate of 0.0500 m/s2 for 8.00 min, starting with an initial velocity of 4.00 m/s? (b) If the train can", " slow down at a rate of 0.550 m/s2, how long will it take to come to a stop from this velocity? (c) How far will it travel in each case? 30. A fireworks shell is accelerated from rest to a velocity of 65.0 m/s over a distance of 0.250 m. (a) How long did the acceleration last? (b) Calculate the acceleration. 31. A swan on a lake gets airborne by flapping its wings and running on top of the water. (a) If the swan must reach a velocity of 6.00 m/s to take off and it accelerates from rest at an average rate of 0.350 m/s2, how far will it travel before becoming airborne? (b) How long does this take? 32. Professional Application: A woodpecker's brain is specially protected from large decelerations by tendon-like attachments inside the skull. While pecking on a tree, the woodpecker's head comes to a stop from an initial velocity of 0.600 m/s in a distance of only 2.00 mm. (a) Find the acceleration in m/s2 and in multiples of. (b) Calculate the stopping time. (c) The tendons cradling the brain stretch, making its stopping distance 4.50 mm (greater than the head and, hence, less deceleration of the brain). What is the brain's deceleration, expressed in multiples of? = 9.80 m/s2 33. An unwary football player collides with a padded goalpost while running at a velocity of 7.50 m/s and comes to a full stop after compressing the padding and his body 0.350 m. (a) What is his deceleration? (b) How long does the collision last? 34. In World War II, there were several reported cases of airmen who jumped from their flaming airplanes with no parachute to escape certain death. Some fell about 20,000 feet (6000 m), and some of them survived, with few lifethreatening injuries. For these lucky pilots, the tree branches and snow drifts on the ground allowed their deceleration to be relatively small. If we assume that a pilot's speed upon impact was 123 mph (54 m/s), then what was his deceleration? Assume that the trees and snow stopped him over a distance of 3.0 m. 35. Consider a grey squirrel falling out of a tree to the ground. (a) If we ignore air resistance in this case (only for the sake of this problem), determine a squirrel's velocity just before hitting the ground, assuming it fell from a height of 3.0 m. (b) If the squirrel stops in a distance of 2.0 cm through bending its limbs, compare its deceleration with that of the airman in the previous problem. 36. An express train passes through a station. It enters with an initial velocity of 22.0 m/s and decelerates at a rate of 0.150 m/s2 as it goes through. The station is 210 m long. (a) How long is the nose of the train in the station? (b) How fast is it going when the nose leaves the station? (c) If the train is 130 m long, when does the end of the train leave the station? (d) What is the velocity of the end of the train as it leaves? 37. Dragsters can actually reach a top speed of 145 m/s in only 4.45 s\u2014considerably less time than given in Example 2.10 and Example 2.11. (a) Calculate the average acceleration for such a dragster. (b) Find the final velocity of this dragster starting from rest and accelerating at the rate found in (a) for 402 m (a quarter mile) without using any 90 Chapter 2 | Kinematics information on time. (c) Why is the final velocity greater than that used to find the average acceleration? Hint: Consider whether the assumption of constant acceleration is valid for a dragster. If not, discuss whether the acceleration would be greater at the beginning or end of the run and what effect that would have on the final velocity. known and identify its value. Then identify the unknown, and discuss how you chose the appropriate equation to solve for it. After choosing the equation, show your steps in solving for the unknown, checking units, and discuss whether the answer is reasonable. (c) How long is the dolphin in the air? Neglect any effects due to his size or orientation. 38. A bicycle racer sprints at the end of a race to clinch a victory. The racer has an initial velocity of 11.5 m/s and accelerates at the rate of 0.500 m/s2 for 7.00 s. (a) What is his final velocity? (b) The racer", " continues at this velocity to the finish line. If he was 300 m from the finish line when he started to accelerate, how much time did he save? (c) One other racer was 5.00 m ahead when the winner started to accelerate, but he was unable to accelerate, and traveled at 11.8 m/s until the finish line. How far ahead of him (in meters and in seconds) did the winner finish? 39. In 1967, New Zealander Burt Munro set the world record for an Indian motorcycle, on the Bonneville Salt Flats in Utah, with a maximum speed of 183.58 mi/h. The one-way course was 5.00 mi long. Acceleration rates are often described by the time it takes to reach 60.0 mi/h from rest. If this time was 4.00 s, and Burt accelerated at this rate until he reached his maximum speed, how long did it take Burt to complete the course? 40. (a) A world record was set for the men's 100-m dash in the 2008 Olympic Games in Beijing by Usain Bolt of Jamaica. Bolt \u201ccoasted\u201d across the finish line with a time of 9.69 s. If we assume that Bolt accelerated for 3.00 s to reach his maximum speed, and maintained that speed for the rest of the race, calculate his maximum speed and his acceleration. (b) During the same Olympics, Bolt also set the world record in the 200-m dash with a time of 19.30 s. Using the same assumptions as for the 100-m dash, what was his maximum speed for this race? 2.7 Falling Objects Assume air resistance is negligible unless otherwise stated. 41. Calculate the displacement and velocity at times of (a) 0.500, (b) 1.00, (c) 1.50, and (d) 2.00 s for a ball thrown straight up with an initial velocity of 15.0 m/s. Take the point of release to be 0 = 0. 42. Calculate the displacement and velocity at times of (a) 0.500, (b) 1.00, (c) 1.50, (d) 2.00, and (e) 2.50 s for a rock thrown straight down with an initial velocity of 14.0 m/s from the Verrazano Narrows Bridge in New York City. The roadway of this bridge is 70.0 m above the water. 43. A basketball referee tosses the ball straight up for the starting tip-off. At what velocity must a basketball player leave the ground to rise 1.25 m above the floor in an attempt to get the ball? 44. A rescue helicopter is hovering over a person whose boat has sunk. One of the rescuers throws a life preserver straight down to the victim with an initial velocity of 1.40 m/s and observes that it takes 1.8 s to reach the water. (a) List the knowns in this problem. (b) How high above the water was the preserver released? Note that the downdraft of the helicopter reduces the effects of air resistance on the falling life preserver, so that an acceleration equal to that of gravity is reasonable. 45. A dolphin in an aquatic show jumps straight up out of the water at a velocity of 13.0 m/s. (a) List the knowns in this problem. (b) How high does his body rise above the water? To solve this part, first note that the final velocity is now a This content is available for free at http://cnx.org/content/col11844/1.13 46. A swimmer bounces straight up from a diving board and falls feet first into a pool. She starts with a velocity of 4.00 m/ s, and her takeoff point is 1.80 m above the pool. (a) How long are her feet in the air? (b) What is her highest point above the board? (c) What is her velocity when her feet hit the water? 47. (a) Calculate the height of a cliff if it takes 2.35 s for a rock to hit the ground when it is thrown straight up from the cliff with an initial velocity of 8.00 m/s. (b) How long would it take to reach the ground if it is thrown straight down with the same speed? 48. A very strong, but inept, shot putter puts the shot straight up vertically with an initial velocity of 11.0 m/s. How long does he have to get out of the way if the shot was released at a height of 2.20 m, and he is 1.80 m tall? 49. You throw a ball straight up with an initial velocity of 15.0 m/s. It passes a tree branch on the way up at a height of 7.00 m. How much additional time will pass", " before the ball passes the tree branch on the way back down? 50. A kangaroo can jump over an object 2.50 m high. (a) Calculate its vertical speed when it leaves the ground. (b) How long is it in the air? 51. Standing at the base of one of the cliffs of Mt. Arapiles in Victoria, Australia, a hiker hears a rock break loose from a height of 105 m. He can't see the rock right away but then does, 1.50 s later. (a) How far above the hiker is the rock when he can see it? (b) How much time does he have to move before the rock hits his head? 52. An object is dropped from a height of 75.0 m above ground level. (a) Determine the distance traveled during the first second. (b) Determine the final velocity at which the object hits the ground. (c) Determine the distance traveled during the last second of motion before hitting the ground. 53. There is a 250-m-high cliff at Half Dome in Yosemite National Park in California. Suppose a boulder breaks loose from the top of this cliff. (a) How fast will it be going when it strikes the ground? (b) Assuming a reaction time of 0.300 s, how long will a tourist at the bottom have to get out of the way after hearing the sound of the rock breaking loose (neglecting the height of the tourist, which would become negligible anyway if hit)? The speed of sound is 335 m/s on this day. 54. A ball is thrown straight up. It passes a 2.00-m-high window 7.50 m off the ground on its path up and takes 1.30 s to go past the window. What was the ball's initial velocity? 55. Suppose you drop a rock into a dark well and, using precision equipment, you measure the time for the sound of a splash to return. (a) Neglecting the time required for sound to travel up the well, calculate the distance to the water if the sound returns in 2.0000 s. (b) Now calculate the distance taking into account the time for sound to travel up the well. The speed of sound is 332.00 m/s in this well. 56. A steel ball is dropped onto a hard floor from a height of 1.50 m and rebounds to a height of 1.45 m. (a) Calculate its velocity just before it strikes the floor. (b) Calculate its velocity just after it leaves the floor on its way back up. (c) Calculate its acceleration during contact with the floor if that contact lasts 0.0800 ms (8.00\u00d710\u22125 s). (d) How much did the ball Chapter 2 | Kinematics 91 compress during its collision with the floor, assuming the floor is absolutely rigid? 57. A coin is dropped from a hot-air balloon that is 300 m above the ground and rising at 10.0 m/s upward. For the coin, find (a) the maximum height reached, (b) its position and velocity 4.00 s after being released, and (c) the time before it hits the ground. 58. A soft tennis ball is dropped onto a hard floor from a height of 1.50 m and rebounds to a height of 1.10 m. (a) Calculate its velocity just before it strikes the floor. (b) Calculate its velocity just after it leaves the floor on its way back up. (c) Calculate its acceleration during contact with the floor if that contact lasts 3.50 ms (3.50\u00d710\u22123 s). (d) How much did the ball compress during its collision with the floor, assuming the floor is absolutely rigid? 2.8 Graphical Analysis of One Dimensional Motion Note: There is always uncertainty in numbers taken from graphs. If your answers differ from expected values, examine them to see if they are within data extraction uncertainties estimated by you. 59. (a) By taking the slope of the curve in Figure 2.72, verify that the velocity of the jet car is 115 m/s at = 20 s. (b) By taking the slope of the curve at any point in Figure 2.73, verify that the jet car's acceleration is 5.0 m/s2. Figure 2.74 61. Using approximate values, calculate the slope of the curve in Figure 2.74 to verify that the velocity at = 30.0 s is 0.238 m/s. Assume all values are known to 3 significant figures. 62. By taking the slope of the curve in Figure 2.75, verify that the acceleration is 3.2 m/s2 at = 10 s. Figure 2.72 Figure 2.75 63. Construct the displacement graph for the subway shuttle train as shown in Figure 2.30(a). Your graph should", " show the position of the train, in kilometers, from t = 0 to 20 s. You will need to use the information on acceleration and velocity given in the examples for this figure. 64. (a) Take the slope of the curve in Figure 2.76 to find the jogger's velocity at = 2.5 s. (b) Repeat at 7.5 s. These values must be consistent with the graph in Figure 2.77. Figure 2.73 60. Using approximate values, calculate the slope of the curve in Figure 2.74 to verify that the velocity at = 10.0 s is 0.208 m/s. Assume all values are known to 3 significant figures. Figure 2.76 92 Chapter 2 | Kinematics Figure 2.77 Figure 2.80 Figure 2.78 65. A graph of () is shown for a world-class track sprinter in a 100-m race. (See Figure 2.79). (a) What is his average velocity for the first 4 s? (b) What is his instantaneous velocity at = 5 s? (c) What is his average acceleration between 0 and 4 s? (d) What is his time for the race? Figure 2.79 66. Figure 2.80 shows the displacement graph for a particle for 5 s. Draw the corresponding velocity and acceleration graphs. Test Prep for AP\u00ae Courses 2.1 Displacement 1. Which of the following statements comparing position, distance, and displacement is correct? This content is available for free at http://cnx.org/content/col11844/1.13 a. An object may record a distance of zero while recording a non-zero displacement. b. An object may record a non-zero distance while recording a displacement of zero. c. An object may record a non-zero distance while maintaining a position of zero. Chapter 2 | Kinematics 93 d. An object may record a non-zero displacement while maintaining a position of zero. velocity v as a function of time t is shown in the graph. The five labeled points divide the graph into four sections. 2.2 Vectors, Scalars, and Coordinate Systems 2. A student is trying to determine the acceleration of a feather as she drops it to the ground. If the student is looking to achieve a positive velocity and positive acceleration, what is the most sensible way to set up her coordinate system? a. Her hand should be a coordinate of zero and the upward direction should be considered positive. b. Her hand should be a coordinate of zero and the downward direction should be considered positive. c. The floor should be a coordinate of zero and the upward direction should be considered positive. d. The floor should be a coordinate of zero and the downward direction should be considered positive. 2.3 Time, Velocity, and Speed 3. A group of students has two carts, A and B, with wheels that turn with negligible friction. The two carts travel along a straight horizontal track and eventually collide. Before the collision, cart A travels to the right and cart B is initially at rest. After the collision, the carts stick together. a. Describe an experimental procedure to determine the velocities of the carts before and after the collision, including all the additional equipment you would need. You may include a labeled diagram of your setup to help in your description. Indicate what measurements you would take and how you would take them. Include enough detail so that another student could carry out your procedure. b. There will be sources of error in the measurements taken in the experiment both before and after the collision. Which velocity will be more greatly affected by this error: the velocity prior to the collision or the velocity after the collision? Or will both sets of data be affected equally? Justify your answer. 2.4 Acceleration 4. Figure 2.81 Graph showing Velocity vs. Time of a cart. A cart is constrained to move along a straight line. A varying net force along the direction of motion is exerted on the cart. The cart's Which of the following correctly ranks the magnitude of the average acceleration of the cart during the four sections of the graph? a. aCD > aAB > aBC > aDE b. aBC > aAB > aCD > aDE c. aAB > aBC > aDE > aCD d. aCD > aAB > aDE > aBC 5. Push a book across a table and observe it slow to a stop. Draw graphs showing the book's position vs. time and velocity vs. time if the direction of its motion is considered positive. Draw graphs showing the book's position vs. time and velocity vs. time if the direction of its motion is considered negative. 2.5 Motion Equations for Constant Acceleration in One Dimension 6. A group of students is attempting to determine the average acceleration of a marble released from the top of a long ramp. Below is a set of data representing the marble's position with respect to time. Position (", "cm) Time (s) 0.0 0.3 1.25 2.8 5.0 7.75 11.3 0.0 0.5 1.0 1.5 2.0 2.5 3.0 Use the data table above to construct a graph determining the acceleration of the marble. Select a set of data points from the table and plot those points on the graph. Fill in the blank column in the table for any quantities you graph other than the given data. Label the axes and indicate the scale for each. Draw a best-fit line or curve through your data points. Using the best-fit line, determine the value of the marble's acceleration. 2.7 Falling Objects 7. Observing a spacecraft land on a distant asteroid, scientists notice that the craft is falling at a rate of 5 m/s. When it is 100 m closer to the surface of the asteroid, the craft reports a velocity of 8 m/s. According to their data, what is the approximate gravitational acceleration on this asteroid? a. 0 m/s2 b. 0.03 m/s2 c. 0.20 m/s2 d. 0.65 m/s2 e. 33 m/s2 94 Chapter 2 | Kinematics This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 3 | Two-Dimensional Kinematics 95 3 TWO-DIMENSIONAL KINEMATICS Figure 3.1 Everyday motion that we experience is, thankfully, rarely as tortuous as a rollercoaster ride like this\u2014the Dragon Khan in Spain's Universal Port Aventura Amusement Park. However, most motion is in curved, rather than straight-line, paths. Motion along a curved path is two- or threedimensional motion, and can be described in a similar fashion to one-dimensional motion. (credit: Boris23/Wikimedia Commons) Chapter Outline 3.1. Kinematics in Two Dimensions: An Introduction 3.2. Vector Addition and Subtraction: Graphical Methods 3.3. Vector Addition and Subtraction: Analytical Methods 3.4. Projectile Motion 3.5. Addition of Velocities Connection for AP\u00ae Courses Most instances of motion in everyday life involve changes in displacement and velocity that occur in more than one direction. For example, when you take a long road trip, you drive on different roads in different directions for different amounts of time at different speeds. How can these motions all be combined to determine information about the trip such as the total displacement and average velocity? If you kick a ball from ground level at some angle above the horizontal, how can you describe its motion? To what maximum height does the object rise above the ground? How long is the object in the air? How much horizontal distance is covered before the ball lands? To answer questions such as these, we need to describe motion in two dimensions. Examining two-dimensional motion requires an understanding of both the scalar and the vector quantities associated with the motion. You will learn how to combine vectors to incorporate both the magnitude and direction of vectors into your analysis. You will learn strategies for simplifying the calculations involved by choosing the appropriate reference frame and by treating each dimension of the motion separately as a one-dimensional problem, but you will also see that the motion itself occurs in the same way regardless of your chosen reference frame (Essential Knowledge 3.A.1). 96 Chapter 3 | Two-Dimensional Kinematics This chapter lays a necessary foundation for examining interactions of objects described by forces (Big Idea 3). Changes in direction result from acceleration, which necessitates force on an object. In this chapter, you will concentrate on describing motion that involves changes in direction. In later chapters, you will apply this understanding as you learn about how forces cause these motions (Enduring Understanding 3.A). The concepts in this chapter support: Big Idea 3 The interactions of an object with other objects can be described by forces. Enduring Understanding 3.A All forces share certain common characteristics when considered by observers in inertial reference frames. Essential Knowledge 3.A.1 An observer in a particular reference frame can describe the motion of an object using such quantities as position, displacement, distance, velocity, speed, and acceleration. 3.1 Kinematics in Two Dimensions: An Introduction Learning Objectives By the end of this section, you will be able to: \u2022 Observe that motion in two dimensions consists of horizontal and vertical components. \u2022 Understand the independence of horizontal and vertical vectors in two-dimensional motion. The information presented in this section supports the following AP\u00ae learning objectives and science practices: \u2022 3.A.1.1 The student is able to express the motion of an object using narrative, mathematical, and graphical representations. (S.P. 1.5, 2.1, 2.2) \u2022 3.A.1.2 The student is able to", " design an experimental investigation of the motion of an object. (S.P. 4.2) \u2022 3.A.1.3 The student is able to analyze experimental data describing the motion of an object and is able to express the results of the analysis using narrative, mathematical, and graphical representations. (S.P. 5.1) Figure 3.2 Walkers and drivers in a city like New York are rarely able to travel in straight lines to reach their destinations. Instead, they must follow roads and sidewalks, making two-dimensional, zigzagged paths. (credit: Margaret W. Carruthers) Two-Dimensional Motion: Walking in a City Suppose you want to walk from one point to another in a city with uniform square blocks, as pictured in Figure 3.3. Figure 3.3 A pedestrian walks a two-dimensional path between two points in a city. In this scene, all blocks are square and are the same size. This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 3 | Two-Dimensional Kinematics 97 The straight-line path that a helicopter might fly is blocked to you as a pedestrian, and so you are forced to take a twodimensional path, such as the one shown. You walk 14 blocks in all, 9 east followed by 5 north. What is the straight-line distance? An old adage states that the shortest distance between two points is a straight line. The two legs of the trip and the straight-line path form a right triangle, and so the Pythagorean theorem, 2 + 2 = 2, can be used to find the straight-line distance. Figure 3.4 The Pythagorean theorem relates the length of the legs of a right triangle, labeled and, with the hypotenuse, labeled. The. This can be rewritten, solving for : = 2 + 2 relationship is given by: 2 + 2 = 2. The hypotenuse of the triangle is the straight-line path, and so in this case its length in units of city blocks is (9 blocks)2+ (5 blocks)2= 10.3 blocks, considerably shorter than the 14 blocks you walked. (Note that we are using three significant figures in the answer. Although it appears that \u201c9\u201d and \u201c5\u201d have only one significant digit, they are discrete numbers. In this case \u201c9 blocks\u201d is the same as \u201c9.0 or 9.00 blocks.\u201d We have decided to use three significant figures in the answer in order to show the result more precisely.) Figure 3.5 The straight-line path followed by a helicopter between the two points is shorter than the 14 blocks walked by the pedestrian. All blocks are square and the same size. The fact that the straight-line distance (10.3 blocks) in Figure 3.5 is less than the total distance walked (14 blocks) is one example of a general characteristic of vectors. (Recall that vectors are quantities that have both magnitude and direction.) As for one-dimensional kinematics, we use arrows to represent vectors. The length of the arrow is proportional to the vector's magnitude. The arrow's length is indicated by hash marks in Figure 3.3 and Figure 3.5. The arrow points in the same direction as the vector. For two-dimensional motion, the path of an object can be represented with three vectors: one vector shows the straight-line path between the initial and final points of the motion, one vector shows the horizontal component of the motion, and one vector shows the vertical component of the motion. The horizontal and vertical components of the motion add together to give the straight-line path. For example, observe the three vectors in Figure 3.5. The first represents a 9-block displacement east. The second represents a 5-block displacement north. These vectors are added to give the third vector, with a 10.3-block total displacement. The third vector is the straight-line path between the two points. Note that in this example, the vectors that we are adding are perpendicular to each other and thus form a right triangle. This means that we can use the Pythagorean theorem to calculate the magnitude of the total displacement. (Note that we cannot use the Pythagorean theorem to add vectors that are not perpendicular. We will develop techniques for adding vectors having any direction, not just those perpendicular to one another, in Vector Addition and Subtraction: Graphical Methods and Vector Addition and Subtraction: Analytical Methods.) The Independence of Perpendicular Motions The person taking the path shown in Figure 3.5 walks east and then north (two perpendicular directions). How far he or she walks east is only affected by his or her motion eastward. Similarly, how far he or she walks north is only affected by his or her motion northward. Independence of Motion The horizontal and vertical components of two-dimensional motion are", " independent of each other. Any motion in the horizontal direction does not affect motion in the vertical direction, and vice versa. 98 Chapter 3 | Two-Dimensional Kinematics This is true in a simple scenario like that of walking in one direction first, followed by another. It is also true of more complicated motion involving movement in two directions at once. For example, let's compare the motions of two baseballs. One baseball is dropped from rest. At the same instant, another is thrown horizontally from the same height and follows a curved path. A stroboscope has captured the positions of the balls at fixed time intervals as they fall. Figure 3.6 This shows the motions of two identical balls\u2014one falls from rest, the other has an initial horizontal velocity. Each subsequent position is an equal time interval. Arrows represent horizontal and vertical velocities at each position. The ball on the right has an initial horizontal velocity, while the ball on the left has no horizontal velocity. Despite the difference in horizontal velocities, the vertical velocities and positions are identical for both balls. This shows that the vertical and horizontal motions are independent. Applying the Science Practices: Independence of Horizontal and Vertical Motion or Maximum Height and Flight Time Choose one of the following experiments to design: Design an experiment to confirm what is shown in Figure 3.6, that the vertical motion of the two balls is independent of the horizontal motion. As you think about your experiment, consider the following questions: \u2022 How will you measure the horizontal and vertical positions of each ball over time? What equipment will this require? \u2022 How will you measure the time interval between each of your position measurements? What equipment will this require? If you were to create separate graphs of the horizontal velocity for each ball versus time, what do you predict it would look like? Explain. If you were to compare graphs of the vertical velocity for each ball versus time, what do you predict it would look like? Explain. If there is a significant amount of air resistance, how will that affect each of your graphs? \u2022 \u2022 \u2022 Design a two-dimensional ballistic motion experiment that demonstrates the relationship between the maximum height reached by an object and the object's time of flight. As you think about your experiment, consider the following questions: \u2022 How will you measure the maximum height reached by your object? \u2022 How can you take advantage of the symmetry of an object in ballistic motion launched from ground level, reaching maximum height, and returning to ground level? \u2022 Will it make a difference if your object has no horizontal component to its velocity? Explain. \u2022 Will you need to measure the time at multiple different positions? Why or why not? \u2022 Predict what a graph of travel time versus maximum height will look like. Will it be linear? Parabolic? Horizontal? Explain the shape of your predicted graph qualitatively or quantitatively. If there is a significant amount of air resistance, how will that affect your measurements and your results? \u2022 It is remarkable that for each flash of the strobe, the vertical positions of the two balls are the same. This similarity implies that the vertical motion is independent of whether or not the ball is moving horizontally. (Assuming no air resistance, the vertical motion of a falling object is influenced by gravity only, and not by any horizontal forces.) Careful examination of the ball thrown horizontally shows that it travels the same horizontal distance between flashes. This is due to the fact that there are no additional forces on the ball in the horizontal direction after it is thrown. This result means that the horizontal velocity is constant, and affected neither by vertical motion nor by gravity (which is vertical). Note that this case is true only for ideal conditions. In the real world, air resistance will affect the speed of the balls in both directions. The two-dimensional curved path of the horizontally thrown ball is composed of two independent one-dimensional motions (horizontal and vertical). The key to analyzing such motion, called projectile motion, is to resolve (break) it into motions along perpendicular directions. Resolving two-dimensional motion into perpendicular components is possible because the components are independent. We shall see how to resolve vectors in Vector Addition and Subtraction: Graphical Methods and Vector Addition and Subtraction: Analytical Methods. We will find such techniques to be useful in many areas of physics. This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 3 | Two-Dimensional Kinematics 99 PhET Explorations: Ladybug Motion 2D Learn about position, velocity and acceleration vectors. Move the ladybug by setting the position, velocity or acceleration, and see how the vectors change. Choose linear, circular or elliptical motion, and record and playback the motion to analyze the behavior. Figure 3.7 Ladybug Motion 2D (http://cnx.org/content/m54779/1.2/ladybug-motion-2d_en.jar) 3.2 Vector Addition and Subtraction", ": Graphical Methods By the end of this section, you will be able to: Learning Objectives \u2022 Understand the rules of vector addition, subtraction, and multiplication. \u2022 Apply graphical methods of vector addition and subtraction to determine the displacement of moving objects. The information presented in this section supports the following AP\u00ae learning objectives and science practices: \u2022 3.A.1.1 The student is able to express the motion of an object using narrative, mathematical, and graphical representations. (S.P. 1.5, 2.1, 2.2) \u2022 3.A.1.3 The student is able to analyze experimental data describing the motion of an object and is able to express the results of the analysis using narrative, mathematical, and graphical representations. (S.P. 5.1) Figure 3.8 Displacement can be determined graphically using a scale map, such as this one of the Hawaiian Islands. A journey from Hawai'i to Moloka'i has a number of legs, or journey segments. These segments can be added graphically with a ruler to determine the total two-dimensional displacement of the journey. (credit: US Geological Survey) Vectors in Two Dimensions A vector is a quantity that has magnitude and direction. Displacement, velocity, acceleration, and force, for example, are all vectors. In one-dimensional, or straight-line, motion, the direction of a vector can be given simply by a plus or minus sign. In two dimensions (2-d), however, we specify the direction of a vector relative to some reference frame (i.e., coordinate system), using an arrow having length proportional to the vector's magnitude and pointing in the direction of the vector. Figure 3.9 shows such a graphical representation of a vector, using as an example the total displacement for the person walking in a city considered in Kinematics in Two Dimensions: An Introduction. We shall use the notation that a boldface symbol, such as D, stands for a vector. Its magnitude is represented by the symbol in italics,, and its direction by. 100 Chapter 3 | Two-Dimensional Kinematics Vectors in this Text In this text, we will represent a vector with a boldface variable. For example, we will represent the quantity force with the vector F, which has both magnitude and direction. The magnitude of the vector will be represented by a variable in italics, such as, and the direction of the variable will be given by an angle. Figure 3.9 A person walks 9 blocks east and 5 blocks north. The displacement is 10.3 blocks at an angle 29.1\u00b0 north of east. Figure 3.10 To describe the resultant vector for the person walking in a city considered in Figure 3.9 graphically, draw an arrow to represent the total displacement vector D. Using a protractor, draw a line at an angle relative to the east-west axis. The length of the arrow is proportional to the vector's magnitude and is measured along the line with a ruler. In this example, the magnitude of the vector is 10.3 units, and the direction is 29.1\u00b0 north of east. Vector Addition: Head-to-Tail Method The head-to-tail method is a graphical way to add vectors, described in Figure 3.11 below and in the steps following. The tail of the vector is the starting point of the vector, and the head (or tip) of a vector is the final, pointed end of the arrow. This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 3 | Two-Dimensional Kinematics 101 Figure 3.11 Head-to-Tail Method: The head-to-tail method of graphically adding vectors is illustrated for the two displacements of the person walking in a city considered in Figure 3.9. (a) Draw a vector representing the displacement to the east. (b) Draw a vector representing the displacement to the north. The tail of this vector should originate from the head of the first, east-pointing vector. (c) Draw a line from the tail of the east-pointing vector to the head of the north-pointing vector to form the sum or resultant vector D. The length of the arrow D is proportional to the vector's magnitude and is measured to be 10.3 units. Its direction, described as the angle with respect to the east (or horizontal axis) is measured with a protractor to be 29.1\u00b0. Step 1. Draw an arrow to represent the first vector (9 blocks to the east) using a ruler and protractor. Figure 3.12 Step 2. Now draw an arrow to represent the second vector (5 blocks to the north). Place the tail of the second vector at the head of the first vector. Figure 3.13 Step 3. If there are more than two vectors, continue this process for each vector to be added.", " Note that in our example, we have only two vectors, so we have finished placing arrows tip to tail. Step 4. Draw an arrow from the tail of the first vector to the head of the last vector. This is the resultant, or the sum, of the other vectors. 102 Chapter 3 | Two-Dimensional Kinematics Figure 3.14 Step 5. To get the magnitude of the resultant, measure its length with a ruler. (Note that in most calculations, we will use the Pythagorean theorem to determine this length.) Step 6. To get the direction of the resultant, measure the angle it makes with the reference frame using a protractor. (Note that in most calculations, we will use trigonometric relationships to determine this angle.) The graphical addition of vectors is limited in accuracy only by the precision with which the drawings can be made and the precision of the measuring tools. It is valid for any number of vectors. Example 3.1 Adding Vectors Graphically Using the Head-to-Tail Method: A Woman Takes a Walk Use the graphical technique for adding vectors to find the total displacement of a person who walks the following three paths (displacements) on a flat field. First, she walks 25.0 m in a direction 49.0\u00b0 north of east. Then, she walks 23.0 m heading 15.0\u00b0 north of east. Finally, she turns and walks 32.0 m in a direction 68.0\u00b0 south of east. Strategy Represent each displacement vector graphically with an arrow, labeling the first A, the second B, and the third C, making the lengths proportional to the distance and the directions as specified relative to an east-west line. The head-to-tail method outlined above will give a way to determine the magnitude and direction of the resultant displacement, denoted R. Solution (1) Draw the three displacement vectors. Figure 3.15 (2) Place the vectors head to tail retaining both their initial magnitude and direction. This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 3 | Two-Dimensional Kinematics 103 Figure 3.16 (3) Draw the resultant vector, R. Figure 3.17 (4) Use a ruler to measure the magnitude of R, and a protractor to measure the direction of R. While the direction of the vector can be specified in many ways, the easiest way is to measure the angle between the vector and the nearest horizontal or vertical axis. Since the resultant vector is south of the eastward pointing axis, we flip the protractor upside down and measure the angle between the eastward axis and the vector. Figure 3.18 In this case, the total displacement R is seen to have a magnitude of 50.0 m and to lie in a direction 7.0\u00b0 south of east. By using its magnitude and direction, this vector can be expressed as = 50.0 m and = 7.0\u00b0 south of east. Discussion The head-to-tail graphical method of vector addition works for any number of vectors. It is also important to note that the resultant is independent of the order in which the vectors are added. Therefore, we could add the vectors in any order as illustrated in Figure 3.19 and we will still get the same solution. 104 Chapter 3 | Two-Dimensional Kinematics Figure 3.19 Here, we see that when the same vectors are added in a different order, the result is the same. This characteristic is true in every case and is an important characteristic of vectors. Vector addition is commutative. Vectors can be added in any order. A + B = B + A. (3.1) (This is true for the addition of ordinary numbers as well\u2014you get the same result whether you add 2 + 3 or 3 + 2, for example). Vector Subtraction Vector subtraction is a straightforward extension of vector addition. To define subtraction (say we want to subtract B from A, written A \u2013 B, we must first define what we mean by subtraction. The negative of a vector B is defined to be \u2013B ; that is, graphically the negative of any vector has the same magnitude but the opposite direction, as shown in Figure 3.20. In other words, B has the same length as \u2013B, but points in the opposite direction. Essentially, we just flip the vector so it points in the opposite direction. Figure 3.20 The negative of a vector is just another vector of the same magnitude but pointing in the opposite direction. So B is the negative of \u2013B ; it has the same length but opposite direction. The subtraction of vector B from vector A is then simply defined to be the addition of \u2013B to A. Note that vector subtraction is the addition of a negative vector. The order of subtraction does not affect the results. A \u2013 B = A + (\u2013B). (3.2) This is analogous to the", " subtraction of scalars (where, for example, 5 \u2013 2 = 5 + (\u20132) ). Again, the result is independent of the order in which the subtraction is made. When vectors are subtracted graphically, the techniques outlined above are used, as the following example illustrates. Example 3.2 Subtracting Vectors Graphically: A Woman Sailing a Boat A woman sailing a boat at night is following directions to a dock. The instructions read to first sail 27.5 m in a direction 66.0\u00b0 north of east from her current location, and then travel 30.0 m in a direction 112\u00b0 north of east (or 22.0\u00b0 west of This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 3 | Two-Dimensional Kinematics 105 north). If the woman makes a mistake and travels in the opposite direction for the second leg of the trip, where will she end up? Compare this location with the location of the dock. Figure 3.21 Strategy We can represent the first leg of the trip with a vector A, and the second leg of the trip with a vector B. The dock is located at a location A + B. If the woman mistakenly travels in the opposite direction for the second leg of the journey, she will travel a distance (30.0 m) in the direction 180\u00b0 \u2013 112\u00b0 = 68\u00b0 south of east. We represent this as \u2013B, as shown below. The vector \u2013B has the same magnitude as B but is in the opposite direction. Thus, she will end up at a location A + (\u2013B), or A \u2013 B. Figure 3.22 We will perform vector addition to compare the location of the dock, A + B, with the location at which the woman mistakenly arrives, A + (\u2013B). Solution (1) To determine the location at which the woman arrives by accident, draw vectors A and \u2013B. (2) Place the vectors head to tail. (3) Draw the resultant vector R. (4) Use a ruler and protractor to measure the magnitude and direction of R. Figure 3.23 In this case, = 23.0 m and = 7.5\u00b0 south of east. 106 Chapter 3 | Two-Dimensional Kinematics (5) To determine the location of the dock, we repeat this method to add vectors A and B. We obtain the resultant vector R': Figure 3.24 In this case = 52.9 m and = 90.1\u00b0 north of east. We can see that the woman will end up a significant distance from the dock if she travels in the opposite direction for the second leg of the trip. Discussion Because subtraction of a vector is the same as addition of a vector with the opposite direction, the graphical method of subtracting vectors works the same as for addition. Multiplication of Vectors and Scalars If we decided to walk three times as far on the first leg of the trip considered in the preceding example, then we would walk 3 \u00d7 27.5 m, or 82.5 m, in a direction 66.0\u00b0 north of east. This is an example of multiplying a vector by a positive scalar. Notice that the magnitude changes, but the direction stays the same. If the scalar is negative, then multiplying a vector by it changes the vector's magnitude and gives the new vector the opposite direction. For example, if you multiply by \u20132, the magnitude doubles but the direction changes. We can summarize these rules in the following way: When vector A is multiplied by a scalar, \u2022 \u2022 \u2022 the magnitude of the vector becomes the absolute value of, if is positive, the direction of the vector does not change, if is negative, the direction is reversed. In our case, = 3 and = 27.5 m. Vectors are multiplied by scalars in many situations. Note that division is the inverse of multiplication. For example, dividing by 2 is the same as multiplying by the value (1/2). The rules for multiplication of vectors by scalars are the same for division; simply treat the divisor as a scalar between 0 and 1. Resolving a Vector into Components In the examples above, we have been adding vectors to determine the resultant vector. In many cases, however, we will need to do the opposite. We will need to take a single vector and find what other vectors added together produce it. In most cases, this involves determining the perpendicular components of a single vector, for example the x- and y-components, or the north-south and east-west components. For example, we may know that the total displacement of a person walking in a city is 10.3 blocks in a direction 29.0\u00b0 north of east and want to find out how many blocks east and north had to be walked. This method is called finding the components (or parts) of the displacement in the east", " and north directions, and it is the inverse of the process followed to find the total displacement. It is one example of finding the components of a vector. There are many applications in physics where this is a useful thing to do. We will see this soon in Projectile Motion, and much more when we cover forces in Dynamics: Newton's Laws of Motion. Most of these involve finding components along perpendicular axes (such as north and east), so that right triangles are involved. The analytical techniques presented in Vector Addition and Subtraction: Analytical Methods are ideal for finding vector components. PhET Explorations: Maze Game Learn about position, velocity, and acceleration in the \"Arena of Pain\". Use the green arrow to move the ball. Add more walls to the arena to make the game more difficult. Try to make a goal as fast as you can. This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 3 | Two-Dimensional Kinematics 107 Figure 3.25 Maze Game (http://cnx.org/content/m54781/1.2/maze-game_en.jar) 3.3 Vector Addition and Subtraction: Analytical Methods Learning Objectives By the end of this section, you will be able to: \u2022 Understand the rules of vector addition and subtraction using analytical methods. \u2022 Apply analytical methods to determine vertical and horizontal component vectors. \u2022 Apply analytical methods to determine the magnitude and direction of a resultant vector. The information presented in this section supports the following AP\u00ae learning objectives and science practices: \u2022 3.A.1.1 The student is able to express the motion of an object using narrative, mathematical, and graphical representations. (S.P. 1.5, 2.1, 2.2) Analytical methods of vector addition and subtraction employ geometry and simple trigonometry rather than the ruler and protractor of graphical methods. Part of the graphical technique is retained, because vectors are still represented by arrows for easy visualization. However, analytical methods are more concise, accurate, and precise than graphical methods, which are limited by the accuracy with which a drawing can be made. Analytical methods are limited only by the accuracy and precision with which physical quantities are known. Resolving a Vector into Perpendicular Components Analytical techniques and right triangles go hand-in-hand in physics because (among other things) motions along perpendicular directions are independent. We very often need to separate a vector into perpendicular components. For example, given a vector like A in Figure 3.26, we may wish to find which two perpendicular vectors, A and A, add to produce it. Figure 3.26 The vector A, with its tail at the origin of an x, y-coordinate system, is shown together with its x- and y-components, A and A. These vectors form a right triangle. The analytical relationships among these vectors are summarized below. A and A are defined to be the components of A along the x- and y-axes. The three vectors A, A, and A form a right triangle: A + Ay = A. (3.3) Note that this relationship between vector components and the resultant vector holds only for vector quantities (which include both magnitude and direction). The relationship does not apply for the magnitudes alone. For example, if A = 3 m east, A = 4 m north, and A = 5 m north-east, then it is true that the vectors A + Ay = A. However, it is not true that the sum of the magnitudes of the vectors is also equal. That is3.4) Thus, 108 Chapter 3 | Two-Dimensional Kinematics If the vector A is known, then its magnitude (its length) and its angle (its direction) are known. To find and, its xand y-components, we use the following relationships for a right triangle. + \u2260 (3.5) and = cos = sin. (3.6) (3.7) Figure 3.27 The magnitudes of the vector components A and A can be related to the resultant vector A and the angle with trigonometric identities. Here we see that = cos and = sin. Suppose, for example, that A is the vector representing the total displacement of the person walking in a city considered in Kinematics in Two Dimensions: An Introduction and Vector Addition and Subtraction: Graphical Methods. Figure 3.28 We can use the relationships = cos and = sin to determine the magnitude of the horizontal and vertical component vectors in this example. Then = 10.3 blocks and = 29.1\u00ba, so that = cos = = sin = 10.3 blocks cos 29.1\u00ba sin 29.1\u00ba 10.3 blocks = 9.0 blocks = 5.0 blocks. (3.8) (3.9) Calculating a Resultant Vector If the perpendicular components A", " and A of a vector A are known, then A can also be found analytically. To find the magnitude and direction of a vector from its perpendicular components A and A, we use the following relationships: = 2 + 2 = tan\u22121( / ). (3.10) (3.11) This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 3 | Two-Dimensional Kinematics 109 Figure 3.29 The magnitude and direction of the resultant vector can be determined once the horizontal and vertical components and have been determined. Note that the equation = hypotenuse. For example, if and are 9 and 5 blocks, respectively, then = 92 +52=10.3 blocks, again consistent with the example of the person walking in a city. Finally, the direction is = tan\u20131(5/9)=29.1\u00ba, as before. 2 is just the Pythagorean theorem relating the legs of a right triangle to the length of the 2 + Determining Vectors and Vector Components with Analytical Methods Equations = cos and = sin are used to find the perpendicular components of a vector\u2014that is, to go 2 and = tan\u20131( / ) are used to find a vector from its from and to and. Equations = perpendicular components\u2014that is, to go from and to and. Both processes are crucial to analytical methods of vector addition and subtraction. 2 + Adding Vectors Using Analytical Methods To see how to add vectors using perpendicular components, consider Figure 3.30, in which the vectors A and B are added to produce the resultant R. Figure 3.30 Vectors A and B are two legs of a walk, and R is the resultant or total displacement. You can use analytical methods to determine the magnitude and direction of R. If A and B represent two legs of a walk (two displacements), then R is the total displacement. The person taking the walk ends up at the tip of R. There are many ways to arrive at the same point. In particular, the person could have walked first in the x-direction and then in the y-direction. Those paths are the x- and y-components of the resultant, R and R. If we know R 2 and = tan\u20131( / ). When you use the analytical and R, we can find and using the equations = method of vector addition, you can determine the components or the magnitude and direction of a vector. 2 + Step 1. Identify the x- and y-axes that will be used in the problem. Then, find the components of each vector to be added along the chosen perpendicular axes. Use the equations = cos and = sin to find the components. In Figure 3.31, 110 Chapter 3 | Two-Dimensional Kinematics these components are,,, and. The angles that vectors A and B make with the x-axis are A and B, respectively. Figure 3.31 To add vectors A and B, first determine the horizontal and vertical components of each vector. These are the dotted vectors A, A, B and B shown in the image. Step 2. Find the components of the resultant along each axis by adding the components of the individual vectors along that axis. That is, as shown in Figure 3.32, and = + = +. (3.12) (3.13) Figure 3.32 The magnitude of the vectors A and B add to give the magnitude of the resultant vector in the horizontal direction. Similarly, the magnitudes of the vectors A and B add to give the magnitude of the resultant vector in the vertical direction. Components along the same axis, say the x-axis, are vectors along the same line and, thus, can be added to one another like ordinary numbers. The same is true for components along the y-axis. (For example, a 9-block eastward walk could be taken in two legs, the first 3 blocks east and the second 6 blocks east, for a total of 9, because they are along the same direction.) So resolving vectors into components along common axes makes it easier to add them. Now that the components of R are known, its magnitude and direction can be found. Step 3. To get the magnitude of the resultant, use the Pythagorean theorem: Step 4. To get the direction of the resultant: = 2. 2 + = tan\u22121( / ). The following example illustrates this technique for adding vectors using perpendicular components. (3.14) (3.15) This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 3 | Two-Dimensional Kinematics 111 Example 3.3 Adding Vectors Using Analytical Methods Add the vector A to the vector B shown in Figure 3.33, using perpendicular components along the x- and y-axes. The xand y-", "axes are along the east\u2013west and north\u2013south directions, respectively. Vector A represents the first leg of a walk in which a person walks 53.0 m in a direction 20.0\u00ba north of east. Vector B represents the second leg, a displacement of 34.0 m in a direction 63.0\u00ba north of east. Figure 3.33 Vector A has magnitude 53.0 m and direction 20.0 \u00ba north of the x-axis. Vector B has magnitude 34.0 m and direction 63.0\u00ba north of the x-axis. You can use analytical methods to determine the magnitude and direction of R. Strategy The components of A and B along the x- and y-axes represent walking due east and due north to get to the same ending point. Once found, they are combined to produce the resultant. Solution Following the method outlined above, we first find the components of A and B along the x- and y-axes. Note that = 53.0 m, A = 20.0\u00ba, = 34.0 m, and B = 63.0\u00ba. We find the x-components by using = cos, which gives and = cos A = (53.0 m)(cos 20.0\u00ba) = (53.0 m)(0.940) = 49.8 m = cos B = (34.0 m)(cos 63.0\u00ba) = (34.0 m)(0.454) = 15.4 m. Similarly, the y-components are found using = sin A : and = sin A = (53.0 m)(sin 20.0\u00ba) = (53.0 m)(0.342) = 18.1 m = sin B = (34.0 m)(sin 63.0 \u00ba ) = (34.0 m)(0.891) = 30.3 m. The x- and y-components of the resultant are thus and = + = 49.8 m + 15.4 m = 65.2 m = + = 18.1 m+30.3 m = 48.4 m. Now we can find the magnitude of the resultant by using the Pythagorean theorem: = 2 + 2 = (65.2)2 + (48.4)2 m so that = 81.2 m. (3.16) (3.17) (3.18) (3.19) (3.20) (3.21) (3.22) (3.23) 112 Chapter 3 | Two-Dimensional Kinematics Finally, we find the direction of the resultant: Thus, = tan\u22121( / )=+tan\u22121(48.4 / 65.2). = tan\u22121(0.742) = 36.6 \u00ba. (3.24) (3.25) Figure 3.34 Using analytical methods, we see that the magnitude of R is 81.2 m and its direction is 36.6\u00ba north of east. Discussion This example illustrates the addition of vectors using perpendicular components. Vector subtraction using perpendicular components is very similar\u2014it is just the addition of a negative vector. Subtraction of vectors is accomplished by the addition of a negative vector. That is, A \u2212 B \u2261 A + (\u2013B). Thus, the method for the subtraction of vectors using perpendicular components is identical to that for addition. The components of \u2013B are the negatives of the components of B. The x- and y-components of the resultant A \u2212 B = R are thus and = + \u2013 = + \u2013 and the rest of the method outlined above is identical to that for addition. (See Figure 3.35.) (3.26) (3.27) Analyzing vectors using perpendicular components is very useful in many areas of physics, because perpendicular quantities are often independent of one another. The next module, Projectile Motion, is one of many in which using perpendicular components helps make the picture clear and simplifies the physics. Figure 3.35 The subtraction of the two vectors shown in Figure 3.30. The components of \u2013B are the negatives of the components of B. The method of subtraction is the same as that for addition. This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 3 | Two-Dimensional Kinematics 113 PhET Explorations: Vector Addition Learn how to add vectors. Drag vectors onto a graph, change their length and angle, and sum them together. The magnitude, angle, and components of each vector can be displayed in several formats. Figure 3.36 Vector Addition (http://cnx.org/content/m54783/1.2/vector-addition_en.jar) 3.4 Projectile Motion By the end of this section, you will be able to: Learning Objectives \u2022 Identify and explain the properties", " of a projectile, such as acceleration due to gravity, range, maximum height, and trajectory. \u2022 Determine the location and velocity of a projectile at different points in its trajectory. \u2022 Apply the principle of independence of motion to solve projectile motion problems. The information presented in this section supports the following AP\u00ae learning objectives: \u2022 3.A.1.1 The student is able to express the motion of an object using narrative, mathematical, and graphical representations. (S.P. 1.5, 2.1, 2.2) \u2022 3.A.1.3 The student is able to analyze experimental data describing the motion of an object and is able to express the results of the analysis using narrative, mathematical, and graphical representations. (S.P. 5.1) Projectile motion is the motion of an object thrown or projected into the air, subject to only the acceleration of gravity. The object is called a projectile, and its path is called its trajectory. The motion of falling objects, as covered in Problem-Solving Basics for One-Dimensional Kinematics, is a simple one-dimensional type of projectile motion in which there is no horizontal movement. In this section, we consider two-dimensional projectile motion, such as that of a football or other object for which air resistance is negligible. The most important fact to remember here is that motions along perpendicular axes are independent and thus can be analyzed separately. This fact was discussed in Kinematics in Two Dimensions: An Introduction, where vertical and horizontal motions were seen to be independent. The key to analyzing two-dimensional projectile motion is to break it into two motions, one along the horizontal axis and the other along the vertical. (This choice of axes is the most sensible, because acceleration due to gravity is vertical\u2014thus, there will be no acceleration along the horizontal axis when air resistance is negligible.) As is customary, we call the horizontal axis the x-axis and the vertical axis the y-axis. Figure 3.37 illustrates the notation for displacement, where s is defined to be the total displacement and x and y are its components along the horizontal and vertical axes, respectively. The magnitudes of these vectors are s, x, and y. (Note that in the last section we used the notation A to represent a vector with components A and A. If we continued this format, we would call displacement s with components s and s. However, to simplify the notation, we will simply represent the component vectors as x and y.) Of course, to describe motion we must deal with velocity and acceleration, as well as with displacement. We must find their components along the x- and y-axes, too. We will assume all forces except gravity (such as air resistance and friction, for example) are negligible. The components of acceleration are then very simple: = \u2013 = \u2013 9.80 m/s2. (Note that this definition assumes that the upwards direction is defined as the positive direction. If you arrange the coordinate system instead such that the downwards direction is positive, then acceleration due to gravity takes a positive value.) Because gravity is vertical, = 0. Both accelerations are constant, so the kinematic equations can be used. Review of Kinematic Equations (constant ) = ( \u2212 0). (3.28) (3.29) (3.30) (3.31) (3.32) 114 Chapter 3 | Two-Dimensional Kinematics Figure 3.37 The total displacement s of a soccer ball at a point along its path. The vector s has components x and y along the horizontal and vertical axes. Its magnitude is, and it makes an angle with the horizontal. Given these assumptions, the following steps are then used to analyze projectile motion: Step 1. Resolve or break the motion into horizontal and vertical components along the x- and y-axes. These axes are perpendicular, so = cos and = sin are used. The magnitude of the components of displacement s along these axes are and The magnitudes of the components of the velocity v are = cos and = sin \u03b8, where is the magnitude of the velocity and is its direction, as shown in Figure 3.38. Initial values are denoted with a subscript 0, as usual. Step 2. Treat the motion as two independent one-dimensional motions, one horizontal and the other vertical. The kinematic equations for horizontal and vertical motion take the following forms: Horizontal Motion( = 0) = 0 + = 0 = = velocity is a constant. Vertical Motion(assuming positive is up = \u2212 = \u22129.80m/s2) (( \u2212 0). (3.33) (3.34) (3.35) (3.36) (3.37) (3.38) (3.39) (3.40) Step 3. Solve for the unknowns in the two separate motions\u2014one horizontal and one vertical. Note that the only common variable between the motions is", " time. The problem solving procedures here are the same as for one-dimensional kinematics and are illustrated in the solved examples below. Step 4. Recombine the two motions to find the total displacement s and velocity v. Because the x - and y -motions are perpendicular, we determine these vectors by using the techniques outlined in the Vector Addition and Subtraction: Analytical Methods and employing = displacement s and is the direction of the velocity v : 2 and = tan\u22121( / ) in the following form, where is the direction of the 2 + Total displacement and velocity = 2 + 2 = tan\u22121( / ) 2 2 + = = tan\u22121( / ). (3.41) (3.42) (3.43) (3.44) This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 3 | Two-Dimensional Kinematics 115 Figure 3.38 (a) We analyze two-dimensional projectile motion by breaking it into two independent one-dimensional motions along the vertical and horizontal axes. (b) The horizontal motion is simple, because = 0 and is thus constant. (c) The velocity in the vertical direction begins to decrease as the object rises; at its highest point, the vertical velocity is zero. As the object falls towards the Earth again, the vertical velocity increases again in magnitude but points in the opposite direction to the initial vertical velocity. (d) The x - and y -motions are recombined to give the total velocity at any given point on the trajectory. Example 3.4 A Fireworks Projectile Explodes High and Away During a fireworks display, a shell is shot into the air with an initial speed of 70.0 m/s at an angle of 75.0\u00b0 above the horizontal, as illustrated in Figure 3.39. The fuse is timed to ignite the shell just as it reaches its highest point above the ground. (a) Calculate the height at which the shell explodes. (b) How much time passed between the launch of the shell and the explosion? (c) What is the horizontal displacement of the shell when it explodes? Strategy Because air resistance is negligible for the unexploded shell, the analysis method outlined above can be used. The motion can be broken into horizontal and vertical motions in which = 0 and = \u2013. We can then define 0 and 0 to be zero and solve for the desired quantities. Solution for (a) 116 Chapter 3 | Two-Dimensional Kinematics By \u201cheight\u201d we mean the altitude or vertical position above the starting point. The highest point in any trajectory, called the apex, is reached when = 0. Since we know the initial and final velocities as well as the initial position, we use the following equation to find : 2 = 0 2 \u2212 2( \u2212 0). (3.45) Figure 3.39 The trajectory of a fireworks shell. The fuse is set to explode the shell at the highest point in its trajectory, which is found to be at a height of 233 m and 125 m away horizontally. Because 0 and are both zero, the equation simplifies to Solving for gives 3.46) (3.47) Now we must find 0, the component of the initial velocity in the y-direction. It is given by 0 = 0 sin, where 0 is the initial velocity of 70.0 m/s, and 0 = 75.0\u00b0 is the initial angle. Thus, 0 = 0 sin 0 = (70.0 m/s)(sin 75\u00b0) = 67.6 m/s. and is so that Discussion for (a) = (67.6 m/s)2 2(9.80 m/s2), = 233m. (3.48) (3.49) (3.50) Note that because up is positive, the initial velocity is positive, as is the maximum height, but the acceleration due to gravity is negative. Note also that the maximum height depends only on the vertical component of the initial velocity, so that any projectile with a 67.6 m/s initial vertical component of velocity will reach a maximum height of 233 m (neglecting air resistance). The numbers in this example are reasonable for large fireworks displays, the shells of which do reach such heights before exploding. In practice, air resistance is not completely negligible, and so the initial velocity would have to be somewhat larger than that given to reach the same height. Solution for (b) As in many physics problems, there is more than one way to solve for the time to the highest point. In this case, the easiest method is to use = 0 + 1 (0 + ). Because 0 is zero, this equation reduces to simply 2 This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 3 | Two-Dimensional Kinematics Note that the final vertical velocity,, at", " the highest point is zero. Thus, = 1 2 (0 + ). = 2 (0y + ) = 2(233 m) (67.6 m/s) = 6.90 s. 117 (3.51) (3.52) Discussion for (b) This time is also reasonable for large fireworks. When you are able to see the launch of fireworks, you will notice several seconds pass before the shell explodes. (Another way of finding the time is by using = 0 + 0 \u2212 1 the quadratic equation for.) 2 2, and solving Solution for (c) Because air resistance is negligible, = 0 and the horizontal velocity is constant, as discussed above. The horizontal displacement is horizontal velocity multiplied by time as given by = 0 +, where 0 is equal to zero: =, where is the x-component of the velocity, which is given by = 0 cos 0. Now, = 0 cos 0 = (70.0 m/s)(cos 75.0\u00b0) = 18.1 m/s. The time for both motions is the same, and so is = (18.1 m/s)(6.90 s) = 125 m. Discussion for (c) (3.53) (3.54) (3.55) The horizontal motion is a constant velocity in the absence of air resistance. The horizontal displacement found here could be useful in keeping the fireworks fragments from falling on spectators. Once the shell explodes, air resistance has a major effect, and many fragments will land directly below. In solving part (a) of the preceding example, the expression we found for is valid for any projectile motion where air resistance is negligible. Call the maximum height = ; then, = 2 0 2. (3.56) This equation defines the maximum height of a projectile and depends only on the vertical component of the initial velocity. Defining a Coordinate System It is important to set up a coordinate system when analyzing projectile motion. One part of defining the coordinate system is to define an origin for the and positions. Often, it is convenient to choose the initial position of the object as the origin such that 0 = 0 and 0 = 0. It is also important to define the positive and negative directions in the and directions. Typically, we define the positive vertical direction as upwards, and the positive horizontal direction is usually the direction of the object's motion. When this is the case, the vertical acceleration,, takes a negative value (since it is directed downwards towards the Earth). However, it is occasionally useful to define the coordinates differently. For example, if you are analyzing the motion of a ball thrown downwards from the top of a cliff, it may make sense to define the positive direction downwards since the motion of the ball is solely in the downwards direction. If this is the case, takes a positive value. Example 3.5 Calculating Projectile Motion: Hot Rock Projectile Kilauea in Hawaii is the world's most continuously active volcano. Very active volcanoes characteristically eject red-hot rocks and lava rather than smoke and ash. Suppose a large rock is ejected from the volcano with a speed of 25.0 m/s and at an angle 35.0\u00b0 above the horizontal, as shown in Figure 3.40. The rock strikes the side of the volcano at an altitude 20.0 m lower than its starting point. (a) Calculate the time it takes the rock to follow this path. (b) What are the magnitude and direction of the rock's velocity at impact? 118 Chapter 3 | Two-Dimensional Kinematics Figure 3.40 The trajectory of a rock ejected from the Kilauea volcano. Strategy Again, resolving this two-dimensional motion into two independent one-dimensional motions will allow us to solve for the desired quantities. The time a projectile is in the air is governed by its vertical motion alone. We will solve for first. While the rock is rising and falling vertically, the horizontal motion continues at a constant velocity. This example asks for the final velocity. Thus, the vertical and horizontal results will be recombined to obtain and at the final time determined in the first part of the example. Solution for (a) While the rock is in the air, it rises and then falls to a final position 20.0 m lower than its starting altitude. We can find the time for this by using = 0 + 0 \u2212 1 2 2. (3.57) If we take the initial position 0 to be zero, then the final position is = \u221220.0 m. Now the initial vertical velocity is the vertical component of the initial velocity, found from 0 = 0 sin 0 = ( 25.0 m/s )( sin 35.0\u00b0 ) = 14.3 m/s. Substituting known values yields Rearranging terms gives a quadratic equation in : \u221220.0 m = (14.3 m/s) \u2212 4.90 m/s2 2. 4.90 m/s2 2", " \u2212 (14.3 m/s) \u2212 (20.0 m) = 0. (3.58) (3.59) This expression is a quadratic equation of the form 2 + + = 0, where the constants are = 4.90, = \u2013 14.3, and = \u2013 20.0. Its solutions are given by the quadratic formula3.60) This equation yields two solutions: = 3.96 and = \u2013 1.03. (It is left as an exercise for the reader to verify these solutions.) The time is = 3.96 s or \u2013 1.03 s. The negative value of time implies an event before the start of motion, and so we discard it. Thus, = 3.96 s. (3.61) Discussion for (a) The time for projectile motion is completely determined by the vertical motion. So any projectile that has an initial vertical velocity of 14.3 m/s and lands 20.0 m below its starting altitude will spend 3.96 s in the air. Solution for (b) From the information now in hand, we can find the final horizontal and vertical velocities and and combine them to find the total velocity and the angle 0 it makes with the horizontal. Of course, is constant so we can solve for it at any horizontal location. In this case, we chose the starting point since we know both the initial velocity and initial angle. Therefore: = 0 cos 0 = (25.0 m/s)(cos 35\u00b0) = 20.5 m/s. The final vertical velocity is given by the following equation: = 0 \u2212 (3.62) (3.63) This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 3 | Two-Dimensional Kinematics 119 where 0y was found in part (a) to be 14.3 m/s. Thus, so that = 14.3 m/s \u2212 (9.80 m/s2)(3.96 s) = \u221224.5 m/s. To find the magnitude of the final velocity we combine its perpendicular components, using the following equation: = 2 + 2 = (20.5 m/s)2 + ( \u2212 24.5 m/s)2, which gives The direction is found from the equation: = 31.9 m/s. = tan\u22121( / ) so that Thus, Discussion for (b) = tan\u22121( \u2212 24.5 / 20.5) = tan\u22121( \u2212 1.19). = \u221250.1 \u00b0. (3.64) (3.65) (3.66) (3.67) (3.68) (3.69) (3.70) The negative angle means that the velocity is 50.1\u00b0 below the horizontal. This result is consistent with the fact that the final vertical velocity is negative and hence downward\u2014as you would expect because the final altitude is 20.0 m lower than the initial altitude. (See Figure 3.40.) One of the most important things illustrated by projectile motion is that vertical and horizontal motions are independent of each other. Galileo was the first person to fully comprehend this characteristic. He used it to predict the range of a projectile. On level ground, we define range to be the horizontal distance traveled by a projectile. Galileo and many others were interested in the range of projectiles primarily for military purposes\u2014such as aiming cannons. However, investigating the range of projectiles can shed light on other interesting phenomena, such as the orbits of satellites around the Earth. Let us consider projectile range further. Figure 3.41 Trajectories of projectiles on level ground. (a) The greater the initial speed 0, the greater the range for a given initial angle. (b) The effect of initial angle 0 on the range of a projectile with a given initial speed. Note that the range is the same for 15\u00b0 and 75\u00b0, although the maximum heights of those paths are different. 120 Chapter 3 | Two-Dimensional Kinematics How does the initial velocity of a projectile affect its range? Obviously, the greater the initial speed 0, the greater the range, as shown in Figure 3.41(a). The initial angle 0 also has a dramatic effect on the range, as illustrated in Figure 3.41(b). For a fixed initial speed, such as might be produced by a cannon, the maximum range is obtained with 0 = 45\u00b0. This is true only for conditions neglecting air resistance. If air resistance is considered, the maximum angle is approximately 38\u00b0. Interestingly, for every initial angle except 45\u00b0, there are two angles that give the same range\u2014the sum of those angles is 90\u00b0. The range also depends on the value of the acceleration of gravity. The lunar astronaut Alan Shepherd was able to drive a golf ball a great distance on the Moon because gravity is weaker there. The range of a projectile on", " level ground for which air resistance is negligible is given by = 2 sin 2 0 0, (3.71) where 0 is the initial speed and 0 is the initial angle relative to the horizontal. The proof of this equation is left as an end-ofchapter problem (hints are given), but it does fit the major features of projectile range as described. When we speak of the range of a projectile on level ground, we assume that is very small compared with the circumference of the Earth. If, however, the range is large, the Earth curves away below the projectile and acceleration of gravity changes direction along the path. The range is larger than predicted by the range equation given above because the projectile has farther to fall than it would on level ground. (See Figure 3.42.) If the initial speed is great enough, the projectile goes into orbit. This possibility was recognized centuries before it could be accomplished. When an object is in orbit, the Earth curves away from underneath the object at the same rate as it falls. The object thus falls continuously but never hits the surface. These and other aspects of orbital motion, such as the rotation of the Earth, will be covered analytically and in greater depth later in this text. Once again we see that thinking about one topic, such as the range of a projectile, can lead us to others, such as the Earth orbits. In Addition of Velocities, we will examine the addition of velocities, which is another important aspect of two-dimensional kinematics and will also yield insights beyond the immediate topic. Figure 3.42 Projectile to satellite. In each case shown here, a projectile is launched from a very high tower to avoid air resistance. With increasing initial speed, the range increases and becomes longer than it would be on level ground because the Earth curves away underneath its path. With a large enough initial speed, orbit is achieved. PhET Explorations: Projectile Motion Blast a Buick out of a cannon! Learn about projectile motion by firing various objects. Set the angle, initial speed, and mass. Add air resistance. Make a game out of this simulation by trying to hit a target. Figure 3.43 Projectile Motion (http://cnx.org/content/m54787/1.2/projectile-motion_en.jar) This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 3 | Two-Dimensional Kinematics 121 3.5 Addition of Velocities Learning Objectives By the end of this section, you will be able to: \u2022 Apply principles of vector addition to determine relative velocity. \u2022 Explain the significance of the observer in the measurement of velocity. The information presented in this section supports the following AP\u00ae learning objectives and science practices: \u2022 3.A.1.1 The student is able to express the motion of an object using narrative, mathematical, and graphical representations. (S.P. 1.5, 2.1, 2.2) \u2022 3.A.1.3 The student is able to analyze experimental data describing the motion of an object and is able to express the results of the analysis using narrative, mathematical, and graphical representations. (S.P. 5.1) Relative Velocity If a person rows a boat across a rapidly flowing river and tries to head directly for the other shore, the boat instead moves diagonally relative to the shore, as in Figure 3.44. The boat does not move in the direction in which it is pointed. The reason, of course, is that the river carries the boat downstream. Similarly, if a small airplane flies overhead in a strong crosswind, you can sometimes see that the plane is not moving in the direction in which it is pointed, as illustrated in Figure 3.45. The plane is moving straight ahead relative to the air, but the movement of the air mass relative to the ground carries it sideways. Figure 3.44 A boat trying to head straight across a river will actually move diagonally relative to the shore as shown. Its total velocity (solid arrow) relative to the shore is the sum of its velocity relative to the river plus the velocity of the river relative to the shore. 122 Chapter 3 | Two-Dimensional Kinematics Figure 3.45 An airplane heading straight north is instead carried to the west and slowed down by wind. The plane does not move relative to the ground in the direction it points; rather, it moves in the direction of its total velocity (solid arrow). In each of these situations, an object has a velocity relative to a medium (such as a river) and that medium has a velocity relative to an observer on solid ground. The velocity of the object relative to the observer is the sum of these velocity vectors, as indicated in Figure 3.44 and Figure 3.45. These situations are only two of many in which it is useful to add velocities. In this module, we first re-examine how", " to add velocities and then consider certain aspects of what relative velocity means. How do we add velocities? Velocity is a vector (it has both magnitude and direction); the rules of vector addition discussed in Vector Addition and Subtraction: Graphical Methods and Vector Addition and Subtraction: Analytical Methods apply to the addition of velocities, just as they do for any other vectors. In one-dimensional motion, the addition of velocities is simple\u2014they add like ordinary numbers. For example, if a field hockey player is moving at 5 m/s straight toward the goal and drives the ball in the same direction with a velocity of 30 m/s relative to her body, then the velocity of the ball is 35 m/s relative to the stationary, profusely sweating goalkeeper standing in front of the goal. In two-dimensional motion, either graphical or analytical techniques can be used to add velocities. We will concentrate on analytical techniques. The following equations give the relationships between the magnitude and direction of velocity ( and ) and its components ( and ) along the x- and y-axes of an appropriately chosen coordinate system: = cos = sin 2 2 + = = tan\u22121( / ). (3.72) (3.73) (3.74) (3.75) Figure 3.46 The velocity,, of an object traveling at an angle to the horizontal axis is the sum of component vectors v and v. These equations are valid for any vectors and are adapted specifically for velocity. The first two equations are used to find the components of a velocity when its magnitude and direction are known. The last two are used to find the magnitude and direction of velocity when its components are known. This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 3 | Two-Dimensional Kinematics 123 Take-Home Experiment: Relative Velocity of a Boat Fill a bathtub half-full of water. Take a toy boat or some other object that floats in water. Unplug the drain so water starts to drain. Try pushing the boat from one side of the tub to the other and perpendicular to the flow of water. Which way do you need to push the boat so that it ends up immediately opposite? Compare the directions of the flow of water, heading of the boat, and actual velocity of the boat. Example 3.6 Adding Velocities: A Boat on a River Figure 3.47 A boat attempts to travel straight across a river at a speed 0.75 m/s. The current in the river, however, flows at a speed of 1.20 m/s to the right. What is the total displacement of the boat relative to the shore? Refer to Figure 3.47, which shows a boat trying to go straight across the river. Let us calculate the magnitude and direction of the boat's velocity relative to an observer on the shore, vtot. The velocity of the boat, vboat, is 0.75 m/s in the direction relative to the river and the velocity of the river, vriver, is 1.20 m/s to the right. Strategy We start by choosing a coordinate system with its -axis parallel to the velocity of the river, as shown in Figure 3.47. Because the boat is directed straight toward the other shore, its velocity relative to the water is parallel to the -axis and perpendicular to the velocity of the river. Thus, we can add the two velocities by using the equations tot = = tan\u22121( / ) directly. 2 + 2 and Solution The magnitude of the total velocity is where and Thus, yielding tot = 2, 2 + = river = 1.20 m/s = boat = 0.750 m/s. tot = (1.20 m/s)2 + (0.750 m/s)2 tot = 1.42 m/s. The direction of the total velocity is given by: (3.76) (3.77) (3.78) (3.79) (3.80) 124 Chapter 3 | Two-Dimensional Kinematics This equation gives Discussion = tan\u22121( / ) = tan\u22121(0.750 / 1.20). = 32.0\u00ba. (3.81) (3.82) Both the magnitude and the direction of the total velocity are consistent with Figure 3.47. Note that because the velocity of the river is large compared with the velocity of the boat, it is swept rapidly downstream. This result is evidenced by the small angle (only 32.0\u00ba ) the total velocity has relative to the riverbank. Example 3.7 Calculating Velocity: Wind Velocity Causes an Airplane to Drift Calculate the wind velocity for the situation shown in Figure 3.48. The plane is known to be moving at 45.0 m/s due north relative to the air mass, while its velocity relative to the ground (", "its total velocity) is 38.0 m/s in a direction 20.0\u00ba west of north. Figure 3.48 An airplane is known to be heading north at 45.0 m/s, though its velocity relative to the ground is 38.0 m/s at an angle west of north. What is the speed and direction of the wind? Strategy In this problem, somewhat different from the previous example, we know the total velocity vtot and that it is the sum of two other velocities, vw (the wind) and vp (the plane relative to the air mass). The quantity vp is known, and we are asked to find vw. None of the velocities are perpendicular, but it is possible to find their components along a common set of perpendicular axes. If we can find the components of vw, then we can combine them to solve for its magnitude and direction. As shown in Figure 3.48, we choose a coordinate system with its x-axis due east and its y-axis due north (parallel to vp ). (You may wish to look back at the discussion of the addition of vectors using perpendicular components in Vector Addition and Subtraction: Analytical Methods.) Solution Because vtot is the vector sum of the vw and vp, its x- and y-components are the sums of the x- and y-components of the wind and plane velocities. Note that the plane only has vertical component of velocity so p = 0 and p = p. That is, tot = w (3.83) and This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 3 | Two-Dimensional Kinematics 125 tot = w + p. We can use the first of these two equations to find w : Because tot = 38.0 m / s and cos 110\u00ba = \u2013 0.342 we have w = tot = totcos 110\u00ba. w = (38.0 m/s)(\u20130.342)=\u201313.0 m/s. The minus sign indicates motion west which is consistent with the diagram. Now, to find w we note that tot = w + p Here tot = totsin 110\u00ba ; thus, w = (38.0 m/s)(0.940) \u2212 45.0 m/s = \u22129.29 m/s. (3.84) (3.85) (3.86) (3.87) (3.88) This minus sign indicates motion south which is consistent with the diagram. Now that the perpendicular components of the wind velocity w and w are known, we can find the magnitude and direction of vw. First, the magnitude is w = w 2 2 + w = ( \u2212 13.0 m/s)2 + ( \u2212 9.29 m/s)2 w = 16.0 m/s. = tan\u22121(w / w) = tan\u22121( \u2212 9.29 / \u221213.0) = 35.6\u00ba. so that The direction is: giving Discussion (3.89) (3.90) (3.91) (3.92) The wind's speed and direction are consistent with the significant effect the wind has on the total velocity of the plane, as seen in Figure 3.48. Because the plane is fighting a strong combination of crosswind and head-wind, it ends up with a total velocity significantly less than its velocity relative to the air mass as well as heading in a different direction. Note that in both of the last two examples, we were able to make the mathematics easier by choosing a coordinate system with one axis parallel to one of the velocities. We will repeatedly find that choosing an appropriate coordinate system makes problem solving easier. For example, in projectile motion we always use a coordinate system with one axis parallel to gravity. Relative Velocities and Classical Relativity When adding velocities, we have been careful to specify that the velocity is relative to some reference frame. These velocities are called relative velocities. For example, the velocity of an airplane relative to an air mass is different from its velocity relative to the ground. Both are quite different from the velocity of an airplane relative to its passengers (which should be close to zero). Relative velocities are one aspect of relativity, which is defined to be the study of how different observers moving relative to each other measure the same phenomenon. Nearly everyone has heard of relativity and immediately associates it with Albert Einstein (1879\u20131955), the greatest physicist of the 20th century. Einstein revolutionized our view of nature with his modern theory of relativity, which we shall study in later chapters. The relative velocities in this section are actually aspects of classical relativity, first discussed correctly by Galileo and Isaac Newton. Classical relativity is limited to situations where speeds are less than about 1% of the speed of light\u2014that is", ", less than 3,000 km/s. Most things we encounter in daily life move slower than this speed. Let us consider an example of what two different observers see in a situation analyzed long ago by Galileo. Suppose a sailor at the top of a mast on a moving ship drops his binoculars. Where will it hit the deck? Will it hit at the base of the mast, or will it hit behind the mast because the ship is moving forward? The answer is that if air resistance is negligible, the binoculars will hit at the base of the mast at a point directly below its point of release. Now let us consider what two different observers see when the binoculars drop. One observer is on the ship and the other on shore. The binoculars have no horizontal velocity relative to the observer on the ship, and so he sees them fall straight down the mast. (See Figure 3.49.) To the observer on shore, the 126 Chapter 3 | Two-Dimensional Kinematics binoculars and the ship have the same horizontal velocity, so both move the same distance forward while the binoculars are falling. This observer sees the curved path shown in Figure 3.49. Although the paths look different to the different observers, each sees the same result\u2014the binoculars hit at the base of the mast and not behind it. To get the correct description, it is crucial to correctly specify the velocities relative to the observer. Figure 3.49 Classical relativity. The same motion as viewed by two different observers. An observer on the moving ship sees the binoculars dropped from the top of its mast fall straight down. An observer on shore sees the binoculars take the curved path, moving forward with the ship. Both observers see the binoculars strike the deck at the base of the mast. The initial horizontal velocity is different relative to the two observers. (The ship is shown moving rather fast to emphasize the effect.) Example 3.8 Calculating Relative Velocity: An Airline Passenger Drops a Coin An airline passenger drops a coin while the plane is moving at 260 m/s. What is the velocity of the coin when it strikes the floor 1.50 m below its point of release: (a) Measured relative to the plane? (b) Measured relative to the Earth? Figure 3.50 The motion of a coin dropped inside an airplane as viewed by two different observers. (a) An observer in the plane sees the coin fall straight down. (b) An observer on the ground sees the coin move almost horizontally. Strategy Both problems can be solved with the techniques for falling objects and projectiles. In part (a), the initial velocity of the coin is zero relative to the plane, so the motion is that of a falling object (one-dimensional). In part (b), the initial velocity is 260 m/ This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 3 | Two-Dimensional Kinematics 127 s horizontal relative to the Earth and gravity is vertical, so this motion is a projectile motion. In both parts, it is best to use a coordinate system with vertical and horizontal axes. Solution for (a) Using the given information, we note that the initial velocity and position are zero, and the final position is 1.50 m. The final velocity can be found using the equation: 2 = 0 2 \u2212 2( \u2212 0). Substituting known values into the equation, we get 2 = 02 \u2212 2(9.80 m/s2)( \u2212 1.50 m \u2212 0 m) = 29.4 m2 /s2 yielding = \u22125.42 m/s. (3.93) (3.94) (3.95) We know that the square root of 29.4 has two roots: 5.42 and -5.42. We choose the negative root because we know that the velocity is directed downwards, and we have defined the positive direction to be upwards. There is no initial horizontal velocity relative to the plane and no horizontal acceleration, and so the motion is straight down relative to the plane. Solution for (b) Because the initial vertical velocity is zero relative to the ground and vertical motion is independent of horizontal motion, the final vertical velocity for the coin relative to the ground is = \u2212 5.42 m/s, the same as found in part (a). In contrast to part (a), there now is a horizontal component of the velocity. However, since there is no horizontal acceleration, the initial and final horizontal velocities are the same and = 260 m/s. The x- and y-components of velocity can be combined to find the magnitude of the final velocity: Thus, yielding The direction is given by: so that Discussion = 2 + 2. = (260 m/s)2 + ( \u2212 5.42 m/s)2 = 260.06 m/s. = tan\u22121(", " / ) = tan\u22121( \u2212 5.42 / 260) = tan\u22121( \u2212 0.0208) = \u22121.19\u00ba. (3.96) (3.97) (3.98) (3.99) (3.100) In part (a), the final velocity relative to the plane is the same as it would be if the coin were dropped from rest on the Earth and fell 1.50 m. This result fits our experience; objects in a plane fall the same way when the plane is flying horizontally as when it is at rest on the ground. This result is also true in moving cars. In part (b), an observer on the ground sees a much different motion for the coin. The plane is moving so fast horizontally to begin with that its final velocity is barely greater than the initial velocity. Once again, we see that in two dimensions, vectors do not add like ordinary numbers\u2014the final velocity v in part (b) is not (260 \u2013 5.42) m/s ; rather, it is 260.06 m/s. The velocity's magnitude had to be calculated to five digits to see any difference from that of the airplane. The motions as seen by different observers (one in the plane and one on the ground) in this example are analogous to those discussed for the binoculars dropped from the mast of a moving ship, except that the velocity of the plane is much larger, so that the two observers see very different paths. (See Figure 3.50.) In addition, both observers see the coin fall 1.50 m vertically, but the one on the ground also sees it move forward 144 m (this calculation is left for the reader). Thus, one observer sees a vertical path, the other a nearly horizontal path. Making Connections: Relativity and Einstein Because Einstein was able to clearly define how measurements are made (some involve light) and because the speed of light is the same for all observers, the outcomes are spectacularly unexpected. Time varies with observer, energy is stored as increased mass, and more surprises await. 128 Chapter 3 | Two-Dimensional Kinematics PhET Explorations: Motion in 2D Try the new \"Ladybug Motion 2D\" simulation for the latest updated version. Learn about position, velocity, and acceleration vectors. Move the ball with the mouse or let the simulation move the ball in four types of motion (2 types of linear, simple harmonic, circle). Figure 3.51 Motion in 2D (http://cnx.org/content/m54798/1.2/motion-2d_en.jar) Glossary air resistance: a frictional force that slows the motion of objects as they travel through the air; when solving basic physics problems, air resistance is assumed to be zero analytical method: the method of determining the magnitude and direction of a resultant vector using the Pythagorean theorem and trigonometric identities classical relativity: the study of relative velocities in situations where speeds are less than about 1% of the speed of light\u2014that is, less than 3000 km/s commutative: refers to the interchangeability of order in a function; vector addition is commutative because the order in which vectors are added together does not affect the final sum component (of a 2-d vector): a piece of a vector that points in either the vertical or the horizontal direction; every 2-d vector can be expressed as a sum of two vertical and horizontal vector components direction (of a vector): the orientation of a vector in space head (of a vector): the end point of a vector; the location of the tip of the vector's arrowhead; also referred to as the \u201ctip\u201d head-to-tail method: a method of adding vectors in which the tail of each vector is placed at the head of the previous vector kinematics: the study of motion without regard to mass or force magnitude (of a vector): the length or size of a vector; magnitude is a scalar quantity motion: displacement of an object as a function of time projectile: an object that travels through the air and experiences only acceleration due to gravity projectile motion: the motion of an object that is subject only to the acceleration of gravity range: the maximum horizontal distance that a projectile travels relative velocity: the velocity of an object as observed from a particular reference frame relativity: the study of how different observers moving relative to each other measure the same phenomenon resultant: the sum of two or more vectors resultant vector: the vector sum of two or more vectors scalar: a quantity with magnitude but no direction tail: the start point of a vector; opposite to the head or tip of the arrow trajectory: the path of a projectile through the air vector: a quantity that has both magnitude and direction; an arrow used to represent quantities with both magnitude and direction vector addition: the rules that apply to adding vectors together velocity: speed in a given direction This content is available for free at http://cnx.org/", "content/col11844/1.13 Chapter 3 | Two-Dimensional Kinematics 129 Section Summary 3.1 Kinematics in Two Dimensions: An Introduction \u2022 The shortest path between any two points is a straight line. In two dimensions, this path can be represented by a vector with horizontal and vertical components. \u2022 The horizontal and vertical components of a vector are independent of one another. Motion in the horizontal direction does not affect motion in the vertical direction, and vice versa. 3.2 Vector Addition and Subtraction: Graphical Methods \u2022 The graphical method of adding vectors A and B involves drawing vectors on a graph and adding them using the head-to-tail method. The resultant vector R is defined such that A + B = R. The magnitude and direction of R are then determined with a ruler and protractor, respectively. \u2022 The graphical method of subtracting vector B from A involves adding the opposite of vector B, which is defined as \u2212B. In this case, A \u2013 B = A + (\u2013B) = R. Then, the head-to-tail method of addition is followed in the usual way to obtain the resultant vector R. \u2022 Addition of vectors is commutative such that A + B = B + A. \u2022 The head-to-tail method of adding vectors involves drawing the first vector on a graph and then placing the tail of each subsequent vector at the head of the previous vector. The resultant vector is then drawn from the tail of the first vector to the head of the final vector. If a vector A is multiplied by a scalar quantity, the magnitude of the product is given by. If is positive, the direction of the product points in the same direction as A ; if is negative, the direction of the product points in the opposite direction as A. \u2022 3.3 Vector Addition and Subtraction: Analytical Methods \u2022 The analytical method of vector addition and subtraction involves using the Pythagorean theorem and trigonometric identities to determine the magnitude and direction of a resultant vector. \u2022 The steps to add vectors A and B using the analytical method are as follows: Step 1: Determine the coordinate system for the vectors. Then, determine the horizontal and vertical components of each vector using the equations and = cos = cos = sin = sin. Step 2: Add the horizontal and vertical components of each vector to determine the components and of the resultant vector, R : and = + = +. Step 3: Use the Pythagorean theorem to determine the magnitude,, of the resultant vector R : Step 4: Use a trigonometric identity to determine the direction,, of R : = tan\u22121( / ). = 2. 2 + 3.4 Projectile Motion \u2022 Projectile motion is the motion of an object through the air that is subject only to the acceleration of gravity. \u2022 To solve projectile motion problems, perform the following steps: 1. Determine a coordinate system. Then, resolve the position and/or velocity of the object in the horizontal and vertical components. The components of position s are given by the quantities and, and the components of the velocity v are given by = cos and = sin, where is the magnitude of the velocity and is its direction. 2. Analyze the motion of the projectile in the horizontal direction using the following equations: 130 Chapter 3 | Two-Dimensional Kinematics Horizontal motion( = 0) = 0 + 3. Analyze the motion of the projectile in the vertical direction using the following equations: = 0 = vx = velocity is a constant. Vertical motion(Assuming positive direction is up; = \u2212 = \u22129.80 m/s20 + ) 2 = 0 4. Recombine the horizontal and vertical components of location and/or velocity using the following equations( \u2212 0). = 2 + 2 = tan\u22121( / ) = 2 2 + v = tan\u22121( / ). \u2022 The maximum height of a projectile launched with initial vertical velocity 0 is given by = 2 0 2. \u2022 The maximum horizontal distance traveled by a projectile is called the range. The range of a projectile on level ground launched at an angle 0 above the horizontal with initial speed 0 is given by = 2 sin 2 0 0. 3.5 Addition of Velocities \u2022 Velocities in two dimensions are added using the same analytical vector techniques, which are rewritten as = cos = sin 2 2 + = = tan\u22121( / ). \u2022 Relative velocity is the velocity of an object as observed from a particular reference frame, and it varies dramatically with reference frame. \u2022 Relativity is the study of how different observers measure the same phenomenon, particularly when the observers move relative to one another. Classical relativity is limited to situations where speed is less than about 1% of the speed of light (3000 km/s). Conceptual Questions 3.2 Vector Addition and Subtraction: Graphical Methods 1. Which of the following is a vector: a person's height, the altitude on Mt. Everest,", " the age of the Earth, the boiling point of water, the cost of this book, the Earth's population, the acceleration of gravity? 2. Give a specific example of a vector, stating its magnitude, units, and direction. 3. What do vectors and scalars have in common? How do they differ? 4. Two campers in a national park hike from their cabin to the same spot on a lake, each taking a different path, as illustrated below. The total distance traveled along Path 1 is 7.5 km, and that along Path 2 is 8.2 km. What is the final displacement of each camper? This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 3 | Two-Dimensional Kinematics 131 Figure 3.52 5. If an airplane pilot is told to fly 123 km in a straight line to get from San Francisco to Sacramento, explain why he could end up anywhere on the circle shown in Figure 3.53. What other information would he need to get to Sacramento? Figure 3.53 6. Suppose you take two steps A and B (that is, two nonzero displacements). Under what circumstances can you end up at your starting point? More generally, under what circumstances can two nonzero vectors add to give zero? Is the maximum distance you can end up from the starting point A + B the sum of the lengths of the two steps? 7. Explain why it is not possible to add a scalar to a vector. 8. If you take two steps of different sizes, can you end up at your starting point? More generally, can two vectors with different magnitudes ever add to zero? Can three or more? 3.3 Vector Addition and Subtraction: Analytical Methods 9. Suppose you add two vectors A and B. What relative direction between them produces the resultant with the greatest magnitude? What is the maximum magnitude? What relative direction between them produces the resultant with the smallest magnitude? What is the minimum magnitude? 10. Give an example of a nonzero vector that has a component of zero. 11. Explain why a vector cannot have a component greater than its own magnitude. 12. If the vectors A and B are perpendicular, what is the component of A along the direction of B? What is the component of B along the direction of A? 3.4 Projectile Motion 13. Answer the following questions for projectile motion on level ground assuming negligible air resistance (the initial angle being neither 0\u00b0 nor 90\u00b0 ): (a) Is the velocity ever zero? (b) When is the velocity a minimum? A maximum? (c) Can the velocity ever be the same as the initial velocity at a time other than at = 0? (d) Can the speed ever be the same as the initial speed at a time other than at = 0? 132 Chapter 3 | Two-Dimensional Kinematics 14. Answer the following questions for projectile motion on level ground assuming negligible air resistance (the initial angle being neither 0\u00b0 nor 90\u00b0 ): (a) Is the acceleration ever zero? (b) Is the acceleration ever in the same direction as a component of velocity? (c) Is the acceleration ever opposite in direction to a component of velocity? 15. For a fixed initial speed, the range of a projectile is determined by the angle at which it is fired. For all but the maximum, there are two angles that give the same range. Considering factors that might affect the ability of an archer to hit a target, such as wind, explain why the smaller angle (closer to the horizontal) is preferable. When would it be necessary for the archer to use the larger angle? Why does the punter in a football game use the higher trajectory? 16. During a lecture demonstration, a professor places two coins on the edge of a table. She then flicks one of the coins horizontally off the table, simultaneously nudging the other over the edge. Describe the subsequent motion of the two coins, in particular discussing whether they hit the floor at the same time. 3.5 Addition of Velocities 17. What frame or frames of reference do you instinctively use when driving a car? When flying in a commercial jet airplane? 18. A basketball player dribbling down the court usually keeps his eyes fixed on the players around him. He is moving fast. Why doesn't he need to keep his eyes on the ball? 19. If someone is riding in the back of a pickup truck and throws a softball straight backward, is it possible for the ball to fall straight down as viewed by a person standing at the side of the road? Under what condition would this occur? How would the motion of the ball appear to the person who threw it? 20. The hat of a jogger running at constant velocity falls off the back of his head. Draw a sketch showing the path of the hat in the jogger's frame of reference. Draw its path as viewed by a stationary observer.", " 21. A clod of dirt falls from the bed of a moving truck. It strikes the ground directly below the end of the truck. What is the direction of its velocity relative to the truck just before it hits? Is this the same as the direction of its velocity relative to ground just before it hits? Explain your answers. This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 3 | Two-Dimensional Kinematics 133 Problems & Exercises 3.2 Vector Addition and Subtraction: Graphical Methods Use graphical methods to solve these problems. You may assume data taken from graphs is accurate to three digits. 1. Find the following for path A in Figure 3.54: (a) the total distance traveled, and (b) the magnitude and direction of the displacement from start to finish. Figure 3.54 The various lines represent paths taken by different people walking in a city. All blocks are 120 m on a side. 2. Find the following for path B in Figure 3.54: (a) the total distance traveled, and (b) the magnitude and direction of the displacement from start to finish. 3. Find the north and east components of the displacement for the hikers shown in Figure 3.52. 4. Suppose you walk 18.0 m straight west and then 25.0 m straight north. How far are you from your starting point, and what is the compass direction of a line connecting your starting point to your final position? (If you represent the two legs of the walk as vector displacements A and B, as in Figure 3.55, then this problem asks you to find their sum R = A + B.) Figure 3.56 6. Repeat the problem above, but reverse the order of the two legs of the walk; show that you get the same final result. That is, you first walk leg B, which is 20.0 m in a direction exactly 40\u00b0 south of west, and then leg A, which is 12.0 m in a direction exactly 20\u00b0 west of north. (This problem shows that A + B = B + A.) 7. (a) Repeat the problem two problems prior, but for the second leg you walk 20.0 m in a direction 40.0\u00b0 north of east (which is equivalent to subtracting B from A \u2014that is, to finding R\u2032 = A \u2212 B ). (b) Repeat the problem two problems prior, but now you first walk 20.0 m in a direction 40.0\u00b0 south of west and then 12.0 m in a direction 20.0\u00b0 east of south (which is equivalent to subtracting A from B \u2014that is, to finding R\u2032\u2032 = B - A = - R\u2032 ). Show that this is the case. 8. Show that the order of addition of three vectors does not affect their sum. Show this property by choosing any three vectors A, B, and C, all having different lengths and directions. Find the sum A + B + C then find their sum when added in a different order and show the result is the same. (There are five other orders in which A, B, and C can be added; choose only one.) 9. Show that the sum of the vectors discussed in Example 3.2 gives the result shown in Figure 3.24. 10. Find the magnitudes of velocities A and B in Figure 3.57 Figure 3.55 The two displacements A and B add to give a total displacement R having magnitude and direction. 5. Suppose you first walk 12.0 m in a direction 20\u00b0 west of north and then 20.0 m in a direction 40.0\u00b0 south of west. How far are you from your starting point, and what is the compass direction of a line connecting your starting point to your final position? (If you represent the two legs of the walk as vector displacements A and B, as in Figure 3.56, then this problem finds their sum R = A + B.) Figure 3.57 The two velocities vA and vB add to give a total vtot. 11. Find the components of tot along the x- and y-axes in Figure 3.57. 134 Chapter 3 | Two-Dimensional Kinematics 12. Find the components of tot along a set of perpendicular axes rotated 30\u00b0 counterclockwise relative to those in Figure 3.57. 3.3 Vector Addition and Subtraction: Analytical Methods 13. Find the following for path C in Figure 3.58: (a) the total distance traveled and (b) the magnitude and direction of the displacement from start to finish. In this part of the problem, explicitly show how you follow the steps of the analytical method of vector addition. Figure 3.58 The various lines represent paths taken by different people walking in a city. All blocks are 120 m on a side. 14. Find the following for path D", " in Figure 3.58: (a) the total distance traveled and (b) the magnitude and direction of the displacement from start to finish. In this part of the problem, explicitly show how you follow the steps of the analytical method of vector addition. 15. Find the north and east components of the displacement from San Francisco to Sacramento shown in Figure 3.59. Figure 3.59 16. Solve the following problem using analytical techniques: Suppose you walk 18.0 m straight west and then 25.0 m straight north. How far are you from your starting point, and what is the compass direction of a line connecting your starting point to your final position? (If you represent the two legs of the walk as vector displacements A and B, as in Figure 3.60, then this problem asks you to find their sum R = A + B.) This content is available for free at http://cnx.org/content/col11844/1.13 Figure 3.60 The two displacements A and B add to give a total displacement R having magnitude and direction. Note that you can also solve this graphically. Discuss why the analytical technique for solving this problem is potentially more accurate than the graphical technique. 17. Repeat Exercise 3.16 using analytical techniques, but reverse the order of the two legs of the walk and show that you get the same final result. (This problem shows that adding them in reverse order gives the same result\u2014that is, B + A = A + B.) Discuss how taking another path to reach the same point might help to overcome an obstacle blocking you other path. 18. You drive 7.50 km in a straight line in a direction 15\u00ba east of north. (a) Find the distances you would have to drive straight east and then straight north to arrive at the same point. (This determination is equivalent to find the components of the displacement along the east and north directions.) (b) Show that you still arrive at the same point if the east and north legs are reversed in order. 19. Do Exercise 3.16 again using analytical techniques and change the second leg of the walk to 25.0 m straight south. (This is equivalent to subtracting B from A \u2014that is, finding R\u2032 = A \u2013 B ) (b) Repeat again, but now you first walk 25.0 m north and then 18.0 m east. (This is equivalent to subtract A from B \u2014that is, to find A = B + C. Is that consistent with your result?) 20. A new landowner has a triangular piece of flat land she wishes to fence. Starting at the west corner, she measures the first side to be 80.0 m long and the next to be 105 m. These sides are represented as displacement vectors A from B in Figure 3.61. She then correctly calculates the length and orientation of the third side C. What is her result? Chapter 3 | Two-Dimensional Kinematics 135 Figure 3.61 21. You fly 32.0 km in a straight line in still air in the direction 35.0\u00ba south of west. (a) Find the distances you would have to fly straight south and then straight west to arrive at the same point. (This determination is equivalent to finding the components of the displacement along the south and west directions.) (b) Find the distances you would have to fly first in a direction 45.0\u00ba south of west and then in a direction 45.0\u00ba west of north. These are the components of the displacement along a different set of axes\u2014one rotated 45\u00ba. 22. A farmer wants to fence off his four-sided plot of flat land. He measures the first three sides, shown as A B and C in Figure 3.62, and then correctly calculates the length and orientation of the fourth side D. What is his result? Figure 3.62 23. In an attempt to escape his island, Gilligan builds a raft and sets to sea. The wind shifts a great deal during the day, and he is blown along the following straight lines: 2.50 km 45.0\u00ba north of west; then 4.70 km 60.0\u00ba south of east; then 1.30 km 25.0\u00ba south of west; then 5.10 km straight east; then 1.70 km 5.00\u00ba east of north; then 7.20 km 55.0\u00ba south of west; and finally 2.80 km 10.0\u00ba north of east. What is his final position relative to the island? 24. Suppose a pilot flies 40.0 km in a direction 60\u00ba north of east and then flies 30.0 km in a direction 15\u00ba north of east as shown in Figure 3.63. Find her total distance from the starting point and the direction of the straight-line path to the final position. Discuss qualitatively how this flight would be altered by a wind from the north and how the effect of the wind would depend on both wind speed and the speed of the", " plane relative to the air mass. Figure 3.63 3.4 Projectile Motion 25. A projectile is launched at ground level with an initial speed of 50.0 m/s at an angle of 30.0\u00b0 above the horizontal. It strikes a target above the ground 3.00 seconds later. What are the and distances from where the projectile was launched to where it lands? 26. A ball is kicked with an initial velocity of 16 m/s in the horizontal direction and 12 m/s in the vertical direction. (a) At what speed does the ball hit the ground? (b) For how long does the ball remain in the air? (c)What maximum height is attained by the ball? 27. A ball is thrown horizontally from the top of a 60.0-m building and lands 100.0 m from the base of the building. Ignore air resistance. (a) How long is the ball in the air? (b) What must have been the initial horizontal component of the velocity? (c) What is the vertical component of the velocity just before the ball hits the ground? (d) What is the velocity (including both the horizontal and vertical components) of the ball just before it hits the ground? 28. (a) A daredevil is attempting to jump his motorcycle over a line of buses parked end to end by driving up a 32\u00b0 ramp at a speed of 40.0 m/s (144 km/h). How many buses can he clear if the top of the takeoff ramp is at the same height as the bus tops and the buses are 20.0 m long? (b) Discuss what your answer implies about the margin of error in this act\u2014that is, consider how much greater the range is than the horizontal distance he must travel to miss the end of the last bus. (Neglect air resistance.) 29. An archer shoots an arrow at a 75.0 m distant target; the bull's-eye of the target is at same height as the release height of the arrow. (a) At what angle must the arrow be released to hit the bull's-eye if its initial speed is 35.0 m/s? In this part of the problem, explicitly show how you follow the steps involved in solving projectile motion problems. (b) There is a large tree halfway between the archer and the target with an overhanging horizontal branch 3.50 m above the release height of the arrow. Will the arrow go over or under the branch? 30. A rugby player passes the ball 7.00 m across the field, where it is caught at the same height as it left his hand. (a) At what angle was the ball thrown if its initial speed was 12.0 m/ s, assuming that the smaller of the two possible angles was used? (b) What other angle gives the same range, and why would it not be used? (c) How long did this pass take? 31. Verify the ranges for the projectiles in Figure 3.41(a) for = 45\u00b0 and the given initial velocities. 32. Verify the ranges shown for the projectiles in Figure 3.41(b) for an initial velocity of 50 m/s at the given initial angles. 33. The cannon on a battleship can fire a shell a maximum distance of 32.0 km. (a) Calculate the initial velocity of the shell. (b) What maximum height does it reach? (At its highest, the shell is above 60% of the atmosphere\u2014but air resistance is not really negligible as assumed to make this problem easier.) (c) The ocean is not flat, because the Earth is curved. Assume that the radius of the Earth is 6.37\u00d7103 km. How many meters lower will its surface be 32.0 km from the ship along a horizontal line parallel to the surface at the ship? Does your answer imply that error introduced by the assumption of a flat Earth in projectile motion is significant here? 136 Chapter 3 | Two-Dimensional Kinematics 34. An arrow is shot from a height of 1.5 m toward a cliff of height. It is shot with a velocity of 30 m/s at an angle of 60\u00b0 above the horizontal. It lands on the top edge of the cliff 4.0 s later. (a) What is the height of the cliff? (b) What is the maximum height reached by the arrow along its trajectory? (c) What is the arrow's impact speed just before hitting the cliff? 35. In the standing broad jump, one squats and then pushes off with the legs to see how far one can jump. Suppose the extension of the legs from the crouch position is 0.600 m and the acceleration achieved from this position is 1.25 times the acceleration due to gravity,. How far can they jump? State your assumptions. (Increased range can be achieved by swinging the arms in the direction of the jump.) 36. The world long jump record", " is 8.95 m (Mike Powell, USA, 1991). Treated as a projectile, what is the maximum range obtainable by a person if he has a take-off speed of 9.5 m/s? State your assumptions. 37. Serving at a speed of 170 km/h, a tennis player hits the ball at a height of 2.5 m and an angle below the horizontal. The service line is 11.9 m from the net, which is 0.91 m high. What is the angle such that the ball just crosses the net? Will the ball land in the service box, whose out line is 6.40 m from the net? 38. A football quarterback is moving straight backward at a speed of 2.00 m/s when he throws a pass to a player 18.0 m straight downfield. (a) If the ball is thrown at an angle of 25\u00b0 relative to the ground and is caught at the same height as it is released, what is its initial speed relative to the ground? (b) How long does it take to get to the receiver? (c) What is its maximum height above its point of release? 39. Gun sights are adjusted to aim high to compensate for the effect of gravity, effectively making the gun accurate only for a specific range. (a) If a gun is sighted to hit targets that are at the same height as the gun and 100.0 m away, how low will the bullet hit if aimed directly at a target 150.0 m away? The muzzle velocity of the bullet is 275 m/s. (b) Discuss qualitatively how a larger muzzle velocity would affect this problem and what would be the effect of air resistance. 40. An eagle is flying horizontally at a speed of 3.00 m/s when the fish in her talons wiggles loose and falls into the lake 5.00 m below. Calculate the velocity of the fish relative to the water when it hits the water. 41. An owl is carrying a mouse to the chicks in its nest. Its position at that time is 4.00 m west and 12.0 m above the center of the 30.0 cm diameter nest. The owl is flying east at 3.50 m/s at an angle 30.0\u00b0 below the horizontal when it accidentally drops the mouse. Is the owl lucky enough to have the mouse hit the nest? To answer this question, calculate the horizontal position of the mouse when it has fallen 12.0 m. 42. Suppose a soccer player kicks the ball from a distance 30 m toward the goal. Find the initial speed of the ball if it just passes over the goal, 2.4 m above the ground, given the initial direction to be 40\u00b0 above the horizontal. 43. Can a goalkeeper at her/ his goal kick a soccer ball into the opponent's goal without the ball touching the ground? The distance will be about 95 m. A goalkeeper can give the ball a speed of 30 m/s. 44. The free throw line in basketball is 4.57 m (15 ft) from the basket, which is 3.05 m (10 ft) above the floor. A player standing on the free throw line throws the ball with an initial This content is available for free at http://cnx.org/content/col11844/1.13 speed of 7.15 m/s, releasing it at a height of 2.44 m (8 ft) above the floor. At what angle above the horizontal must the ball be thrown to exactly hit the basket? Note that most players will use a large initial angle rather than a flat shot because it allows for a larger margin of error. Explicitly show how you follow the steps involved in solving projectile motion problems. 45. In 2007, Michael Carter (U.S.) set a world record in the shot put with a throw of 24.77 m. What was the initial speed of the shot if he released it at a height of 2.10 m and threw it at an angle of 38.0\u00b0 above the horizontal? (Although the maximum distance for a projectile on level ground is achieved at 45\u00b0 when air resistance is neglected, the actual angle to achieve maximum range is smaller; thus, 38\u00b0 will give a longer range than 45\u00b0 in the shot put.) 46. A basketball player is running at 5.00 m/s directly toward the basket when he jumps into the air to dunk the ball. He maintains his horizontal velocity. (a) What vertical velocity does he need to rise 0.750 m above the floor? (b) How far from the basket (measured in the horizontal direction) must he start his jump to reach his maximum height at the same time as he reaches the basket? 47. A football player punts the ball at a 45.0\u00b0 angle. Without an effect from the wind, the ball would travel 60.0 m horizontally. (a) What is the initial speed of the ball? (b) When the", " ball is near its maximum height it experiences a brief gust of wind that reduces its horizontal velocity by 1.50 m/s. What distance does the ball travel horizontally? 48. Prove that the trajectory of a projectile is parabolic, having the form = + 2. To obtain this expression, solve the equation = 0 for and substitute it into the expression for = 0 \u2013 (1 / 2) 2 (These equations describe the and positions of a projectile that starts at the origin.) You should obtain an equation of the form = + 2 where and are constants. 49. Derive = 2 sin 2\u03b80 0 for the range of a projectile on level ground by finding the time at which becomes zero and substituting this value of into the expression for \u2212 0, noting that = \u2212 0 50. Unreasonable Results (a) Find the maximum range of a super cannon that has a muzzle velocity of 4.0 km/s. (b) What is unreasonable about the range you found? (c) Is the premise unreasonable or is the available equation inapplicable? Explain your answer. (d) If such a muzzle velocity could be obtained, discuss the effects of air resistance, thinning air with altitude, and the curvature of the Earth on the range of the super cannon. 51. Construct Your Own Problem Consider a ball tossed over a fence. Construct a problem in which you calculate the ball's needed initial velocity to just clear the fence. Among the things to determine are; the height of the fence, the distance to the fence from the point of release of the ball, and the height at which the ball is released. You should also consider whether it is possible to choose the initial speed for the ball and just calculate the angle at which it is thrown. Also examine the possibility of multiple solutions given the distances and heights you have chosen. Chapter 3 | Two-Dimensional Kinematics 137 3.5 Addition of Velocities 52. Bryan Allen pedaled a human-powered aircraft across the English Channel from the cliffs of Dover to Cap Gris-Nez on June 12, 1979. (a) He flew for 169 min at an average velocity of 3.53 m/s in a direction 45\u00ba south of east. What was his total displacement? (b) Allen encountered a headwind averaging 2.00 m/s almost precisely in the opposite direction of his motion relative to the Earth. What was his average velocity relative to the air? (c) What was his total displacement relative to the air mass? 53. A seagull flies at a velocity of 9.00 m/s straight into the wind. (a) If it takes the bird 20.0 min to travel 6.00 km relative to the Earth, what is the velocity of the wind? (b) If the bird turns around and flies with the wind, how long will he take to return 6.00 km? (c) Discuss how the wind affects the total round-trip time compared to what it would be with no wind. 54. Near the end of a marathon race, the first two runners are separated by a distance of 45.0 m. The front runner has a velocity of 3.50 m/s, and the second a velocity of 4.20 m/s. (a) What is the velocity of the second runner relative to the first? (b) If the front runner is 250 m from the finish line, who will win the race, assuming they run at constant velocity? (c) What distance ahead will the winner be when she crosses the finish line? 55. Verify that the coin dropped by the airline passenger in the Example 3.8 travels 144 m horizontally while falling 1.50 m in the frame of reference of the Earth. 56. A football quarterback is moving straight backward at a speed of 2.00 m/s when he throws a pass to a player 18.0 m straight downfield. The ball is thrown at an angle of 25.0\u00ba relative to the ground and is caught at the same height as it is released. What is the initial velocity of the ball relative to the quarterback? 57. A ship sets sail from Rotterdam, The Netherlands, heading due north at 7.00 m/s relative to the water. The local ocean current is 1.50 m/s in a direction 40.0\u00ba north of east. What is the velocity of the ship relative to the Earth? 58. (a) A jet airplane flying from Darwin, Australia, has an air speed of 260 m/s in a direction 5.0\u00ba south of west. It is in the jet stream, which is blowing at 35.0 m/s in a direction 15\u00ba south of east. What is the velocity of the airplane relative to the Earth? (b) Discuss whether your answers are consistent with your expectations for the effect of the wind on the plane's path. 59. (a) In what direction would the ship in Exercise 3.57 have to travel in", " order to have a velocity straight north relative to the Earth, assuming its speed relative to the water remains 7.00 m/s? (b) What would its speed be relative to the Earth? 60. (a) Another airplane is flying in a jet stream that is blowing at 45.0 m/s in a direction 20\u00ba south of east (as in Exercise 3.58). Its direction of motion relative to the Earth is 45.0\u00ba south of west, while its direction of travel relative to the air is 5.00\u00ba south of west. What is the airplane's speed relative to the air mass? (b) What is the airplane's speed relative to the Earth? 61. A sandal is dropped from the top of a 15.0-m-high mast on a ship moving at 1.75 m/s due south. Calculate the velocity of the sandal when it hits the deck of the ship: (a) relative to the ship and (b) relative to a stationary observer on shore. (c) Discuss how the answers give a consistent result for the position at which the sandal hits the deck. 62. The velocity of the wind relative to the water is crucial to sailboats. Suppose a sailboat is in an ocean current that has a velocity of 2.20 m/s in a direction 30.0\u00ba east of north relative to the Earth. It encounters a wind that has a velocity of 4.50 m/s in a direction of 50.0\u00ba south of west relative to the Earth. What is the velocity of the wind relative to the water? 63. The great astronomer Edwin Hubble discovered that all distant galaxies are receding from our Milky Way Galaxy with velocities proportional to their distances. It appears to an observer on the Earth that we are at the center of an expanding universe. Figure 3.64 illustrates this for five galaxies lying along a straight line, with the Milky Way Galaxy at the center. Using the data from the figure, calculate the velocities: (a) relative to galaxy 2 and (b) relative to galaxy 5. The results mean that observers on all galaxies will see themselves at the center of the expanding universe, and they would likely be aware of relative velocities, concluding that it is not possible to locate the center of expansion with the given information. Figure 3.64 Five galaxies on a straight line, showing their distances and velocities relative to the Milky Way (MW) Galaxy. The distances are in millions of light years (Mly), where a light year is the distance light travels in one year. The velocities are nearly proportional to the distances. The sizes of the galaxies are greatly exaggerated; an average galaxy is about 0.1 Mly across. 64. (a) Use the distance and velocity data in Figure 3.64 to find the rate of expansion as a function of distance. (b) If you extrapolate back in time, how long ago would all of the galaxies have been at approximately the same position? The two parts of this problem give you some idea of how the Hubble constant for universal expansion and the time back to the Big Bang are determined, respectively. 65. An athlete crosses a 25-m-wide river by swimming perpendicular to the water current at a speed of 0.5 m/s relative to the water. He reaches the opposite side at a distance 40 m downstream from his starting point. How fast is the water in the river flowing with respect to the ground? What is the speed of the swimmer with respect to a friend at rest on the ground? 66. A ship sailing in the Gulf Stream is heading 25.0\u00ba west of north at a speed of 4.00 m/s relative to the water. Its velocity relative to the Earth is 4.80 m/s 5.00\u00ba west of north. What is the velocity of the Gulf Stream? (The velocity obtained is typical for the Gulf Stream a few hundred kilometers off the east coast of the United States.) 67. An ice hockey player is moving at 8.00 m/s when he hits the puck toward the goal. The speed of the puck relative to the player is 29.0 m/s. The line between the center of the goal and the player makes a 90.0\u00ba angle relative to his path as shown in Figure 3.65. What angle must the puck's velocity make relative to the player (in his frame of reference) to hit the center of the goal? 138 Chapter 3 | Two-Dimensional Kinematics Figure 3.65 An ice hockey player moving across the rink must shoot backward to give the puck a velocity toward the goal. 68. Unreasonable Results Suppose you wish to shoot supplies straight up to astronauts in an orbit 36,000 km above the surface of the Earth. (a) At what velocity must the supplies be launched? (b) What is unreasonable about this velocity? (c) Is there a problem with the relative velocity between the supplies and the astronauts", " when the supplies reach their maximum height? (d) Is the premise unreasonable or is the available equation inapplicable? Explain your answer. 69. Unreasonable Results A commercial airplane has an air speed of 280 m/s due east and flies with a strong tailwind. It travels 3000 km in a direction 5\u00ba south of east in 1.50 h. (a) What was the velocity of the plane relative to the ground? (b) Calculate the magnitude and direction of the tailwind's velocity. (c) What is unreasonable about both of these velocities? (d) Which premise is unreasonable? 70. Construct Your Own Problem Consider an airplane headed for a runway in a cross wind. Construct a problem in which you calculate the angle the airplane must fly relative to the air mass in order to have a velocity parallel to the runway. Among the things to consider are the direction of the runway, the wind speed and direction (its velocity) and the speed of the plane relative to the air mass. Also calculate the speed of the airplane relative to the ground. Discuss any last minute maneuvers the pilot might have to perform in order for the plane to land with its wheels pointing straight down the runway. Test Prep for AP\u00ae Courses 3.1 Kinematics in Two Dimensions: An Introduction 1. A ball is thrown at an angle of 45 degrees above the horizontal. Which of the following best describes the acceleration of the ball from the instant after it leaves the thrower's hand until the time it hits the ground? a. Always in the same direction as the motion, initially b. positive and gradually dropping to zero by the time it hits the ground Initially positive in the upward direction, then zero at maximum height, then negative from there until it hits the ground c. Always in the opposite direction as the motion, initially positive and gradually dropping to zero by the time it hits the ground d. Always in the downward direction with the same constant value 2. In an experiment, a student launches a ball with an initial horizontal velocity at an elevation 2 meters above ground. The ball follows a parabolic trajectory until it hits the ground. Which of the following accurately describes the graph of the ball's vertical acceleration versus time (taking the downward direction to be negative)? a. A negative value that does not change with time b. A gradually increasing negative value (straight line) This content is available for free at http://cnx.org/content/col11844/1.13 c. An increasing rate of negative values over time (parabolic curve) d. Zero at all times since the initial motion is horizontal 3. A student wishes to design an experiment to show that the acceleration of an object is independent of the object's velocity. To do this, ball A is launched horizontally with some initial speed at an elevation 1.5 meters above the ground, ball B is dropped from rest 1.5 meters above the ground, and ball C is launched vertically with some initial speed at an elevation 1.5 meters above the ground. What information would the student need to collect about each ball in order to test the hypothesis? 3.2 Vector Addition and Subtraction: Graphical Methods 4. A ball is launched vertically upward. The vertical position of the ball is recorded at various points in time in the table shown. Chapter 3 | Two-Dimensional Kinematics 139 Table 3.1 Height (m) Time (sec) 0.490 0.882 1.176 1.372 1.470 1.470 1.372 0.1 0.2 0.3 0.4 0.5 0.6 0.7 Which of the following correctly describes the graph of the ball's vertical velocity versus time? a. Always positive, steadily decreasing b. Always positive, constant c. Initially positive, steadily decreasing, becoming negative at the end Initially zero, steadily getting more and more negative d. 5. Table 3.2 Height (m) Time (sec) 0.490 0.882 1.176 1.372 1.470 1.470 1.372 0.1 0.2 0.3 0.4 0.5 0.6 0.7 A ball is launched at an angle of 60 degrees above the horizontal, and the vertical position of the ball is recorded at various points in time in the table shown, assuming the ball was at a height of 0 at time t = 0. a. Draw a graph of the ball's vertical velocity versus time. b. Describe the graph of the ball's horizontal velocity. c. Draw a graph of the ball's vertical acceleration versus time. 3.4 Projectile Motion 6. In an experiment, a student launches a ball with an initial horizontal velocity of 5.00 meters/sec at an elevation 2.00 meters above ground. Draw and clearly label with appropriate values and units a graph of the ball's horizontal velocity vs. time and the ball's vertical velocity vs. time. The graph should cover the motion from the instant after the ball is launched until the", " instant before it hits the ground. Assume the downward direction is negative for this problem. 140 Chapter 3 | Two-Dimensional Kinematics This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 4 | Dynamics: Force and Newton's Laws of Motion 141 4 DYNAMICS: FORCE AND NEWTON'S LAWS OF MOTION Figure 4.1 Newton\u2019s laws of motion describe the motion of the dolphin\u2019s path. (credit: Jin Jang) Chapter Outline 4.1. Development of Force Concept 4.2. Newton's First Law of Motion: Inertia 4.3. Newton's Second Law of Motion: Concept of a System 4.4. Newton's Third Law of Motion: Symmetry in Forces 4.5. Normal, Tension, and Other Examples of Force 4.6. Problem-Solving Strategies 4.7. Further Applications of Newton's Laws of Motion 4.8. Extended Topic: The Four Basic Forces\u2014An Introduction Connection for AP\u00ae Courses Motion draws our attention. Motion itself can be beautiful, causing us to marvel at the forces needed to achieve spectacular motion, such as that of a jumping dolphin, a leaping pole vaulter, a bird in flight, or an orbiting satellite. The study of motion is kinematics, but kinematics only describes the way objects move\u2014their velocity and their acceleration. Dynamics considers the forces that affect the motion of moving objects and systems. Newton\u2019s laws of motion are the foundation of dynamics. These laws provide an example of the breadth and simplicity of principles under which nature functions. They are also universal laws in that they apply to situations on Earth as well as in space. Isaac Newton\u2019s (1642\u20131727) laws of motion were just one part of the monumental work that has made him legendary. The development of Newton\u2019s laws marks the transition from the Renaissance into the modern era. This transition was characterized by a revolutionary change in the way people thought about the physical universe. For many centuries natural philosophers had debated the nature of the universe based largely on certain rules of logic, with great weight given to the thoughts of earlier 142 Chapter 4 | Dynamics: Force and Newton's Laws of Motion classical philosophers such as Aristotle (384\u2013322 BC). Among the many great thinkers who contributed to this change were Newton and Galileo Galilei (1564\u20131647). Figure 4.2 Isaac Newton\u2019s monumental work, Philosophiae Naturalis Principia Mathematica, was published in 1687. It proposed scientific laws that are still used today to describe the motion of objects. (credit: Service commun de la documentation de l'Universit\u00e9 de Strasbourg) Galileo was instrumental in establishing observation as the absolute determinant of truth, rather than \u201clogical\u201d argument. Galileo\u2019s use of the telescope was his most notable achievement in demonstrating the importance of observation. He discovered moons orbiting Jupiter and made other observations that were inconsistent with certain ancient ideas and religious dogma. For this reason, and because of the manner in which he dealt with those in authority, Galileo was tried by the Inquisition and punished. He spent the final years of his life under a form of house arrest. Because others before Galileo had also made discoveries by observing the nature of the universe and because repeated observations verified those of Galileo, his work could not be suppressed or denied. After his death, his work was verified by others, and his ideas were eventually accepted by the church and scientific communities. Galileo also contributed to the formulation of what is now called Newton\u2019s first law of motion. Newton made use of the work of his predecessors, which enabled him to develop laws of motion, discover the law of gravity, invent calculus, and make great contributions to the theories of light and color. It is amazing that many of these developments were made by Newton working alone, without the benefit of the usual interactions that take place among scientists today. Newton\u2019s laws are introduced along with Big Idea 3, that interactions can be described by forces. These laws provide a theoretical basis for studying motion depending on interactions between the objects. In particular, Newton's laws are applicable to all forces in inertial frames of references (Enduring Understanding 3.A). We will find that all forces are vectors; that is, forces always have both a magnitude and a direction (Essential Knowledge 3.A.2). Furthermore, we will learn that all forces are a result of interactions between two or more objects (Essential Knowledge 3.A.3). These interactions between any two objects are described by Newton's third law, stating that the forces exerted on these objects are equal in magnitude and opposite in direction to each other (Essential Knowledge 3.A.4). We will discover that there is an empirical cause-effect relationship between the net force exerted on an object of mass m and its acceleration, with this relationship described by Newton's second law (Enduring Understanding", " 3.B). This supports Big Idea 1, that inertial mass is a property of an object or a system. The mass of an object or a system is one of the factors affecting changes in motion when an object or a system interacts with other objects or systems (Essential Knowledge 1.C.1). Another is the net force on an object, which is the vector sum of all the forces exerted on the object (Essential Knowledge 3.B.1). To analyze this, we use free-body diagrams to visualize the forces exerted on a given object in order to find the net force and analyze the object's motion (Essential Knowledge 3.B.2). Thinking of these objects as systems is a concept introduced in this chapter, where a system is a collection of elements that could be considered as a single object without any internal structure (Essential Knowledge 5.A.1). This will support Big Idea 5, that changes that occur to the system due to interactions are governed by conservation laws. These conservation laws will be the focus of later chapters in this book. They explain whether quantities are conserved in the given system or change due to transfer to or from another system due to interactions between the systems (Enduring Understanding 5.A). Furthermore, when a situation involves more than one object, it is important to define the system and analyze the motion of a whole system, not its elements, based on analysis of external forces on the system. This supports Big Idea 4, that interactions between systems cause changes in those systems. All kinematics variables in this case describe the motion of the center of mass of the system (Essential Knowledge 4.A.1, Essential Knowledge 4.A.2). The internal forces between the elements of the system do not affect the velocity of the center of mass (Essential Knowledge 4.A.3). The velocity of the center of mass will change only if there is a net external force exerted on the system (Enduring Understanding 4.A). We will learn that some of these interactions can be explained by the existence of fields extending through space, supporting Big Idea 2. For example, any object that has mass creates a gravitational field in space (Enduring Understanding 2.B). Any material object (one that has mass) placed in the gravitational field will experience gravitational force (Essential Knowledge 2.B.1). This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 4 | Dynamics: Force and Newton's Laws of Motion 143 Forces may be categorized as contact or long-distance (Enduring Understanding 3.C). In this chapter we will work with both. An example of a long-distance force is gravitation (Essential Knowledge 3.C.1). Contact forces, such as tension, friction, normal force, and the force of a spring, result from interatomic electric forces at the microscopic level (Essential Knowledge 3.C.4). It was not until the advent of modern physics early in the twentieth century that it was discovered that Newton\u2019s laws of motion produce a good approximation to motion only when the objects are moving at speeds much, much less than the speed of light and when those objects are larger than the size of most molecules (about 10\u20139 m in diameter). These constraints define the realm of classical mechanics, as discussed in Introduction to the Nature of Science and Physics. At the beginning of the twentieth century, Albert Einstein (1879\u20131955) developed the theory of relativity and, along with many other scientists, quantum theory. Quantum theory does not have the constraints present in classical physics. All of the situations we consider in this chapter, and all those preceding the introduction of relativity in Special Relativity, are in the realm of classical physics. The development of special relativity and empirical observations at atomic scales led to the idea that there are four basic forces that account for all known phenomena. These forces are called fundamental (Enduring Understanding 3.G). The properties of gravitational (Essential Knowledge 3.G.1) and electromagnetic (Essential Knowledge 3.G.2) forces are explained in more detail. Big Idea 1 Objects and systems have properties such as mass and charge. Systems may have internal structure. Essential Knowledge 1.C.1 Inertial mass is the property of an object or a system that determines how its motion changes when it interacts with other objects or systems. Big Idea 2 Fields existing in space can be used to explain interactions. Enduring Understanding 2.A A field associates a value of some physical quantity with every point in space. Field models are useful for describing interactions that occur at a distance (long-range forces) as well as a variety of other physical phenomena. Essential Knowledge 2.A.1 A vector field gives, as a function of position (and perhaps time), the value of a physical quantity that is described by a vector. Essential Knowledge 2.A.2 A scalar field gives the value of a physical quantity. Enduring Understanding 2.B", " A gravitational field is caused by an object with mass. Essential Knowledge 2.B.1 A gravitational field g at the location of an object with mass m causes a gravitational force of magnitude mg to be exerted on the object in the direction of the field. Big Idea 3 The interactions of an object with other objects can be described by forces. Enduring Understanding 3.A All forces share certain common characteristics when considered by observers in inertial reference frames. Essential Knowledge 3.A.2 Forces are described by vectors. Essential Knowledge 3.A.3 A force exerted on an object is always due to the interaction of that object with another object. Essential Knowledge 3.A.4 If one object exerts a force on a second object, the second object always exerts a force of equal magnitude on the first object in the opposite direction. Enduring Understanding 3.B Classically, the acceleration of an object interacting with other objects can be predicted by using = \u2211 /. Essential Knowledge 3.B.1 If an object of interest interacts with several other objects, the net force is the vector sum of the individual forces. Essential Knowledge 3.B.2 Free-body diagrams are useful tools for visualizing the forces being exerted on a single object and writing the equations that represent a physical situation. Enduring Understanding 3.C At the macroscopic level, forces can be categorized as either long-range (action-at-a-distance) forces or contact forces. Essential Knowledge 3.C.1 Gravitational force describes the interaction of one object that has mass with another object that has mass. Essential Knowledge 3.C.4 Contact forces result from the interaction of one object touching another object, and they arise from interatomic electric forces. These forces include tension, friction, normal, spring (Physics 1), and buoyant (Physics 2). Enduring Understanding 3.G Certain types of forces are considered fundamental. Essential Knowledge 3.G.1 Gravitational forces are exerted at all scales and dominate at the largest distance and mass scales. Essential Knowledge 3.G.2 Electromagnetic forces are exerted at all scales and can dominate at the human scale. Big Idea 4 Interactions between systems can result in changes in those systems. Enduring Understanding 4.A The acceleration of the center of mass of a system is related to the net force exerted on the system, where = \u2211 /. Essential Knowledge 4.A.1 The linear motion of a system can be described by the displacement, velocity, and acceleration of its center of mass. Essential Knowledge 4.A.2 The acceleration is equal to the rate of change of velocity with time, and velocity is equal to the rate of change of position with time. 144 Chapter 4 | Dynamics: Force and Newton's Laws of Motion Essential Knowledge 4.A.3 Forces that systems exert on each other are due to interactions between objects in the systems. If the interacting objects are parts of the same system, there will be no change in the center-of-mass velocity of that system. Big Idea 5 Changes that occur as a result of interactions are constrained by conservation laws. Enduring Understanding 5.A Certain quantities are conserved, in the sense that the changes of those quantities in a given system are always equal to the transfer of that quantity to or from the system by all possible interactions with other systems. Essential Knowledge 5.A.1 A system is an object or a collection of objects. The objects are treated as having no internal structure. 4.1 Development of Force Concept By the end of this section, you will be able to: \u2022 Understand the definition of force. Learning Objectives The information presented in this section supports the following AP\u00ae learning objectives and science practices: \u2022 3.A.2.1 The student is able to represent forces in diagrams or mathematically using appropriately labeled vectors with magnitude, direction, and units during the analysis of a situation. (S.P. 1.1) \u2022 3.A.3.2 The student is able to challenge a claim that an object can exert a force on itself. (S.P. 6.1) \u2022 3.A.3.3 The student is able to describe a force as an interaction between two objects and identify both objects for any force. (S.P. 1.4) \u2022 3.B.2.1 The student is able to create and use free-body diagrams to analyze physical situations to solve problems with motion qualitatively and quantitatively. (S.P. 1.1, 1.4, 2.2) Dynamics is the study of the forces that cause objects and systems to move. To understand this, we need a working definition of force. Our intuitive definition of force\u2014that is, a push or a pull\u2014is a good place to start. We know that a push or pull has both magnitude and direction (therefore, it is a vector quantity) and can vary considerably in each regard. For example, a cannon exerts a strong force on a cannonball that is", " launched into the air. In contrast, Earth exerts only a tiny downward pull on a flea. Our everyday experiences also give us a good idea of how multiple forces add. If two people push in different directions on a third person, as illustrated in Figure 4.3, we might expect the total force to be in the direction shown. Since force is a vector, it adds just like other vectors, as illustrated in Figure 4.3(a) for two ice skaters. Forces, like other vectors, are represented by arrows and can be added using the familiar head-to-tail method or by trigonometric methods. These ideas were developed in Two-Dimensional Kinematics. By definition, force is always the result of an interaction of two or more objects. No object possesses force on its own. For example, a cannon does not possess force, but it can exert force on a cannonball. Earth does not possess force on its own, but exerts force on a football or on any other massive object. The skaters in Figure 4.3 exert force on one another as they interact. No object can exert force on itself. When you clap your hands, one hand exerts force on the other. When a train accelerates, it exerts force on the track and vice versa. A bowling ball is accelerated by the hand throwing it; once the hand is no longer in contact with the bowling ball, it is no longer accelerating the bowling ball or exerting force on it. The ball continues moving forward due to inertia. Figure 4.3 Part (a) shows an overhead view of two ice skaters pushing on a third. Forces are vectors and add like other vectors, so the total force on the third skater is in the direction shown. In part (b), we see a free-body diagram representing the forces acting on the third skater. Figure 4.3(b) is our first example of a free-body diagram, which is a technique used to illustrate all the external forces acting on a body. The body is represented by a single isolated point (or free body), and only those forces acting on the body from the outside (external forces) are shown. (These forces are the only ones shown, because only external forces acting on the body affect its motion. We can ignore any internal forces within the body.) Free-body diagrams are very useful in analyzing forces acting on a system and are employed extensively in the study and application of Newton\u2019s laws of motion. This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 4 | Dynamics: Force and Newton's Laws of Motion 145 A more quantitative definition of force can be based on some standard force, just as distance is measured in units relative to a standard distance. One possibility is to stretch a spring a certain fixed distance, as illustrated in Figure 4.4, and use the force it exerts to pull itself back to its relaxed shape\u2014called a restoring force\u2014as a standard. The magnitude of all other forces can be stated as multiples of this standard unit of force. Many other possibilities exist for standard forces. (One that we will encounter in Magnetism is the magnetic force between two wires carrying electric current.) Some alternative definitions of force will be given later in this chapter. Figure 4.4 The force exerted by a stretched spring can be used as a standard unit of force. (a) This spring has a length when undistorted. (b) When stretched a distance \u0394, the spring exerts a restoring force, Frestore, which is reproducible. (c) A spring scale is one device that uses a spring to measure force. The force Frestore is exerted on whatever is attached to the hook. Here Frestore has a magnitude of 6 units in the force standard being employed. Take-Home Experiment: Force Standards To investigate force standards and cause and effect, get two identical rubber bands. Hang one rubber band vertically on a hook. Find a small household item that could be attached to the rubber band using a paper clip, and use this item as a weight to investigate the stretch of the rubber band. Measure the amount of stretch produced in the rubber band with one, two, and four of these (identical) items suspended from the rubber band. What is the relationship between the number of items and the amount of stretch? How large a stretch would you expect for the same number of items suspended from two rubber bands? What happens to the amount of stretch of the rubber band (with the weights attached) if the weights are also pushed to the side with a pencil? 4.2 Newton's First Law of Motion: Inertia Learning Objectives By the end of this section, you will be able to: \u2022 Define mass and inertia. \u2022 Understand Newton's first law of motion. Experience suggests that an object at rest will remain at rest if left alone, and that an object in motion tends to slow down and stop unless some", " effort is made to keep it moving. What Newton\u2019s first law of motion states, however, is the following: Newton\u2019s First Law of Motion There exists an inertial frame of reference such that a body at rest remains at rest, or, if in motion, remains in motion at a constant velocity unless acted on by a net external force. Note the repeated use of the verb \u201cremains.\u201d We can think of this law as preserving the status quo of motion. The first law of motion postulates the existence of at least one frame of reference which we call an inertial reference frame, relative to which the motion of an object not subject to forces is a straight line at a constant speed. An inertial reference frame is any reference frame that is not itself accelerating. A car traveling at constant velocity is an inertial reference frame. A car slowing down for a stoplight, or speeding up after the light turns green, will be accelerating and is not an inertial reference frame. Finally, when the car goes around a turn, which is due to an acceleration changing the direction of the velocity vector, it is not an inertial reference frame. Note that Newton\u2019s laws of motion are only valid for inertial reference frames. Rather than contradicting our experience, Newton\u2019s first law of motion states that there must be a cause (which is a net external force) for there to be any change in velocity (either a change in magnitude or direction) in an inertial reference frame. We will define net external force in the next section. An object sliding across a table or floor slows down due to the net force of friction acting on the object. If friction disappeared, would the object still slow down? The idea of cause and effect is crucial in accurately describing what happens in various situations. For example, consider what happens to an object sliding along a rough horizontal surface. The object quickly grinds to a halt. If we spray the surface with talcum powder to make the surface smoother, the object slides farther. If we make the surface even smoother by rubbing 146 Chapter 4 | Dynamics: Force and Newton's Laws of Motion lubricating oil on it, the object slides farther yet. Extrapolating to a frictionless surface, we can imagine the object sliding in a straight line indefinitely. Friction is thus the cause of the slowing (consistent with Newton\u2019s first law). The object would not slow down at all if friction were completely eliminated. Consider an air hockey table. When the air is turned off, the puck slides only a short distance before friction slows it to a stop. However, when the air is turned on, it creates a nearly frictionless surface, and the puck glides long distances without slowing down. Additionally, if we know enough about the friction, we can accurately predict how quickly the object will slow down. Friction is an external force. Newton\u2019s first law is completely general and can be applied to anything from an object sliding on a table to a satellite in orbit to blood pumped from the heart. Experiments have thoroughly verified that any change in velocity (speed or direction) must be caused by an external force. The idea of generally applicable or universal laws is important not only here\u2014it is a basic feature of all laws of physics. Identifying these laws is like recognizing patterns in nature from which further patterns can be discovered. The genius of Galileo, who first developed the idea for the first law, and Newton, who clarified it, was to ask the fundamental question, \u201cWhat is the cause?\u201d Thinking in terms of cause and effect is a worldview fundamentally different from the typical ancient Greek approach when questions such as \u201cWhy does a tiger have stripes?\u201d would have been answered in Aristotelian fashion, \u201cThat is the nature of the beast.\u201d True perhaps, but not a useful insight. Mass The property of a body to remain at rest or to remain in motion with constant velocity is called inertia. Newton\u2019s first law is often called the law of inertia. As we know from experience, some objects have more inertia than others. It is obviously more difficult to change the motion of a large boulder than that of a basketball, for example. The inertia of an object is measured by its mass. An object with a small mass will exhibit less inertia and be more affected by other objects. An object with a large mass will exhibit greater inertia and be less affected by other objects. This inertial mass of an object is a measure of how difficult it is to alter the uniform motion of the object by an external force. Roughly speaking, mass is a measure of the amount of \u201cstuff\u201d (or matter) in something. The quantity or amount of matter in an object is determined by the numbers of atoms and molecules of various types it contains. Unlike weight, mass does not vary with location. The mass of an object is the same on Earth, in orbit, or on the surface of the Moon. In", " practice, it is very difficult to count and identify all of the atoms and molecules in an object, so masses are not often determined in this manner. Operationally, the masses of objects are determined by comparison with the standard kilogram. Check Your Understanding Which has more mass: a kilogram of cotton balls or a kilogram of gold? Solution They are equal. A kilogram of one substance is equal in mass to a kilogram of another substance. The quantities that might differ between them are volume and density. 4.3 Newton's Second Law of Motion: Concept of a System Learning Objectives By the end of this section, you will be able to: \u2022 Define net force, external force, and system. \u2022 Understand Newton\u2019s second law of motion. \u2022 Apply Newton\u2019s second law to determine the weight of an object. Newton\u2019s second law of motion is closely related to Newton\u2019s first law of motion. It mathematically states the cause and effect relationship between force and changes in motion. Newton\u2019s second law of motion is more quantitative and is used extensively to calculate what happens in situations involving a force. Before we can write down Newton\u2019s second law as a simple equation giving the exact relationship of force, mass, and acceleration, we need to sharpen some ideas that have already been mentioned. First, what do we mean by a change in motion? The answer is that a change in motion is equivalent to a change in velocity. A change in velocity means, by definition, that there is an acceleration. Newton\u2019s first law says that a net external force causes a change in motion; thus, we see that a net external force causes acceleration. Another question immediately arises. What do we mean by an external force? An intuitive notion of external is correct\u2014an external force acts from outside the system of interest. For example, in Figure 4.5(a) the system of interest is the wagon plus the child in it. The two forces exerted by the other children are external forces. An internal force acts between elements of the system. Again looking at Figure 4.5(a), the force the child in the wagon exerts to hang onto the wagon is an internal force between elements of the system of interest. Only external forces affect the motion of a system, according to Newton\u2019s first law. (The internal forces actually cancel, as we shall see in the next section.) You must define the boundaries of the system before you can determine which forces are external. Sometimes the system is obvious, whereas other times identifying the boundaries of a system is more subtle. The concept of a system is fundamental to many areas of physics, as is the correct application of Newton\u2019s laws. This concept will be revisited many times on our journey through physics. This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 4 | Dynamics: Force and Newton's Laws of Motion 147 When we describe the acceleration of a system, we are modeling the system as a single point which contains all of the mass of that system. The point we choose for this is the point about which the system\u2019s mass is evenly distributed. For example, in a rigid object, this center of mass is the point where the object will stay balanced even if only supported at this point. For a sphere or disk made of homogenous material, this point is of course at the center. Similarly, for a rod made of homogenous material, the center of mass will be at the midpoint. For the rider in the wagon in Figure 4.5, the center of mass is probably between the rider\u2019s hips. Due to internal forces, the rider\u2019s hand or hair may accelerate slightly differently, but it is the acceleration of the system\u2019s center of mass that interests us. This is true whether the system is a vehicle carrying passengers, a bowl of grapes, or a planet. When we draw a free-body diagram of a system, we represent the system\u2019s center of mass with a single point and use vectors to indicate the forces exerted on that center of mass. (See Figure 4.5.) Figure 4.5 Different forces exerted on the same mass produce different accelerations. (a) Two children push a wagon with a child in it. Arrows representing all external forces are shown. The system of interest is the wagon and its rider. The weight w of the system and the support of the ground N are also shown for completeness and are assumed to cancel. The vector f represents the friction acting on the wagon, and it acts to the left, opposing the motion of the wagon. (b) All of the external forces acting on the system add together to produce a net force, Fnet. The free-body diagram shows all of the forces acting on the system of interest. The dot represents the center of mass of the system. Each force vector extends from this dot. Because there are two forces", " acting to the right, we draw the vectors collinearly. (c) A larger net external force produces a larger acceleration ( a\u2032 > a ) when an adult pushes the child. Now, it seems reasonable that acceleration should be directly proportional to and in the same direction as the net (total) external force acting on a system. This assumption has been verified experimentally and is illustrated in Figure 4.5. In part (a), a smaller force causes a smaller acceleration than the larger force illustrated in part (c). For completeness, the vertical forces are also shown; they are assumed to cancel since there is no acceleration in the vertical direction. The vertical forces are the weight w and the support of the ground N, and the horizontal force f represents the force of friction. These will be discussed in more detail in later sections. For now, we will define friction as a force that opposes the motion past each other of objects that are touching. Figure 4.5(b) shows how vectors representing the external forces add together to produce a net force, Fnet. To obtain an equation for Newton\u2019s second law, we first write the relationship of acceleration and net external force as the proportionality a \u221d Fnet (4.1) where the symbol \u221d means \u201cproportional to,\u201d and Fnet is the net external force. (The net external force is the vector sum of all external forces and can be determined graphically, using the head-to-tail method, or analytically, using components. The techniques are the same as for the addition of other vectors, and are covered in Two-Dimensional Kinematics.) This proportionality states what we have said in words\u2014acceleration is directly proportional to the net external force. Once the system of interest is chosen, it is important to identify the external forces and ignore the internal ones. It is a tremendous simplification not to have to consider the numerous internal forces acting between objects within the system, such as muscular forces within the child\u2019s body, let alone the myriad of forces between atoms in the objects, but by doing so, we can easily solve some very complex problems with only minimal error due to our simplification 148 Chapter 4 | Dynamics: Force and Newton's Laws of Motion Now, it also seems reasonable that acceleration should be inversely proportional to the mass of the system. In other words, the larger the mass (the inertia), the smaller the acceleration produced by a given force. And indeed, as illustrated in Figure 4.6, the same net external force applied to a car produces a much smaller acceleration than when applied to a basketball. The proportionality is written as a \u221d 1 (4.2) where is the mass of the system. Experiments have shown that acceleration is exactly inversely proportional to mass, just as it is exactly linearly proportional to the net external force. Figure 4.6 The same force exerted on systems of different masses produces different accelerations. (a) A basketball player pushes on a basketball to make a pass. (The effect of gravity on the ball is ignored.) (b) The same player exerts an identical force on a stalled SUV and produces a far smaller acceleration (even if friction is negligible). (c) The free-body diagrams are identical, permitting direct comparison of the two situations. A series of patterns for the free-body diagram will emerge as you do more problems. Both of these proportionalities have been experimentally verified repeatedly and consistently, for a broad range of systems and scales. Thus, it has been experimentally found that the acceleration of an object depends only on the net external force and the mass of the object. Combining the two proportionalities just given yields Newton's second law of motion. Applying the Science Practices: Testing the Relationship Between Mass, Acceleration, and Force Plan three simple experiments using objects you have at home to test relationships between mass, acceleration, and force. (a) Design an experiment to test the relationship between mass and acceleration. What will be the independent variable in your experiment? What will be the dependent variable? What controls will you put in place to ensure force is constant? (b) Design a similar experiment to test the relationship between mass and force. What will be the independent variable in your experiment? What will be the dependent variable? What controls will you put in place to ensure acceleration is constant? (c) Design a similar experiment to test the relationship between force and acceleration. What will be the independent variable in your experiment? What will be the dependent variable? Will you have any trouble ensuring that the mass is constant? What did you learn? Newton\u2019s Second Law of Motion The acceleration of a system is directly proportional to and in the same direction as the net external force acting on the system, and inversely proportional to its mass. In equation form, Newton\u2019s second law of motion is This is often written in the more familiar form a = Fnet. Fnet = a. When only the magnitude of", " force and acceleration are considered, this equation is simply net =. (4.3) (4.4) (4.5) Although these last two equations are really the same, the first gives more insight into what Newton\u2019s second law means. The law is a cause and effect relationship among three quantities that is not simply based on their definitions. The validity of the second law is completely based on experimental verification. This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 4 | Dynamics: Force and Newton's Laws of Motion 149 Applying the Science Practices: Systems and Free-Body Diagrams First, consider a person on a sled sliding downhill. What is the system in this situation? Try to draw a free-body diagram describing this system, labeling all the forces and their directions. Which of the forces are internal? Which are external? Next, consider a person on a sled being pushed along level ground by a friend. What is the system in this situation? Try to draw a free-body diagram describing this system, labelling all the forces and their directions. Which of the forces are internal? Which are external? Units of Force Fnet = a is used to define the units of force in terms of the three basic units for mass, length, and time. The SI unit of force is called the newton (abbreviated N) and is the force needed to accelerate a 1-kg system at the rate of 1m/s2. That is, since Fnet = a, 1 N = 1 kg \u22c5 m/s2. (4.6) While almost the entire world uses the newton for the unit of force, in the United States the most familiar unit of force is the pound (lb), where 1 N = 0.225 lb. Weight and the Gravitational Force When an object is dropped, it accelerates toward the center of Earth. Newton\u2019s second law states that a net force on an object is responsible for its acceleration. If air resistance is negligible, the net force on a falling object is the gravitational force, commonly called its weight w. Weight can be denoted as a vector w because it has a direction; down is, by definition, the direction of gravity, and hence weight is a downward force. The magnitude of weight is denoted as. Galileo was instrumental in showing that, in the absence of air resistance, all objects fall with the same acceleration. Using Galileo\u2019s result and Newton\u2019s second law, we can derive an equation for weight. Consider an object with mass falling downward toward Earth. It experiences only the downward force of gravity, which has magnitude. Newton\u2019s second law states that the magnitude of the net external force on an object is net =. Since the object experiences only the downward force of gravity, net =. We know that the acceleration of an object due to gravity is, or =. Substituting these into Newton\u2019s second law gives Weight This is the equation for weight\u2014the gravitational force on a mass : =. Since = 9.80 m/s2 on Earth, the weight of a 1.0 kg object on Earth is 9.8 N, as we see: = = (1.0 kg)(9.80 m/s2 ) = 9.8 N. (4.7) (4.8) Recall that can take a positive or negative value, depending on the positive direction in the coordinate system. Be sure to take this into consideration when solving problems with weight. When the net external force on an object is its weight, we say that it is in free-fall. That is, the only force acting on the object is the force of gravity. In the real world, when objects fall downward toward Earth, they are never truly in free-fall because there is always some upward force from the air acting on the object. The acceleration due to gravity varies slightly over the surface of Earth, so that the weight of an object depends on location and is not an intrinsic property of the object. Weight varies dramatically if one leaves Earth\u2019s surface. On the Moon, for example, the acceleration due to gravity is only 1.67 m/s2. A 1.0-kg mass thus has a weight of 9.8 N on Earth and only about 1.7 N on the Moon. The broadest definition of weight in this sense is that the weight of an object is the gravitational force on it from the nearest large body, such as Earth, the Moon, the Sun, and so on. This is the most common and useful definition of weight in physics. It differs dramatically, however, from the definition of weight used by NASA and the popular media in relation to space travel and exploration. When they speak of \u201cweightlessness\u201d and \u201cmicrogravity,\u201d they are really referring to the phenomenon we call \u201cfreefall\u201d in physics. We shall use the", " above definition of weight, and we will make careful distinctions between free-fall and actual weightlessness. It is important to be aware that weight and mass are very different physical quantities, although they are closely related. Mass is the quantity of matter (how much \u201cstuff\u201d) and does not vary in classical physics, whereas weight is the gravitational force and 150 Chapter 4 | Dynamics: Force and Newton's Laws of Motion does vary depending on gravity. It is tempting to equate the two, since most of our examples take place on Earth, where the weight of an object only varies a little with the location of the object. Furthermore, the terms mass and weight are used interchangeably in everyday language; for example, our medical records often show our \u201cweight\u201d in kilograms, but never in the correct units of newtons. Common Misconceptions: Mass vs. Weight Mass and weight are often used interchangeably in everyday language. However, in science, these terms are distinctly different from one another. Mass is a measure of how much matter is in an object. The typical measure of mass is the kilogram (or the \u201cslug\u201d in English units). Weight, on the other hand, is a measure of the force of gravity acting on an object. Weight is equal to the mass of an object ( ) multiplied by the acceleration due to gravity ( ). Like any other force, weight is measured in terms of newtons (or pounds in English units). Assuming the mass of an object is kept intact, it will remain the same, regardless of its location. However, because weight depends on the acceleration due to gravity, the weight of an object can change when the object enters into a region with stronger or weaker gravity. For example, the acceleration due to gravity on the Moon is 1.67 m/s2 (which is much less than the acceleration due to gravity on Earth, 9.80 m/s2 ). If you measured your weight on Earth and then measured your weight on the Moon, you would find that you \u201cweigh\u201d much less, even though you do not look any skinnier. This is because the force of gravity is weaker on the Moon. In fact, when people say that they are \u201closing weight,\u201d they really mean that they are losing \u201cmass\u201d (which in turn causes them to weigh less). Take-Home Experiment: Mass and Weight What do bathroom scales measure? When you stand on a bathroom scale, what happens to the scale? It depresses slightly. The scale contains springs that compress in proportion to your weight\u2014similar to rubber bands expanding when pulled. The springs provide a measure of your weight (for an object which is not accelerating). This is a force in newtons (or pounds). In most countries, the measurement is divided by 9.80 to give a reading in mass units of kilograms. The scale measures weight but is calibrated to provide information about mass. While standing on a bathroom scale, push down on a table next to you. What happens to the reading? Why? Would your scale measure the same \u201cmass\u201d on Earth as on the Moon? Example 4.1 What Acceleration Can a Person Produce when Pushing a Lawn Mower? Suppose that the net external force (push minus friction) exerted on a lawn mower is 51 N (about 11 lb) parallel to the ground. The mass of the mower is 24 kg. What is its acceleration? Figure 4.7 The net force on a lawn mower is 51 N to the right. At what rate does the lawn mower accelerate to the right? Strategy Since Fnet and are given, the acceleration can be calculated directly from Newton\u2019s second law as stated in Fnet = a. Solution The magnitude of the acceleration is = net. Entering known values gives = 51 N 24 kg (4.9) This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 4 | Dynamics: Force and Newton's Laws of Motion Substituting the units kg \u22c5 m/s2 for N yields = 51 kg \u22c5 m/s2 24 kg = 2.1 m/s2. 151 (4.10) Discussion The direction of the acceleration is the same direction as that of the net force, which is parallel to the ground. There is no information given in this example about the individual external forces acting on the system, but we can say something about their relative magnitudes. For example, the force exerted by the person pushing the mower must be greater than the friction opposing the motion (since we know the mower moves forward), and the vertical forces must cancel if there is to be no acceleration in the vertical direction (the mower is moving only horizontally). The acceleration found is small enough to be reasonable for a person pushing a mower. Such an effort would not last too long because the person\u2019s top speed would soon be reached. Example 4", ".2 What Rocket Thrust Accelerates This Sled? Prior to manned space flights, rocket sleds were used to test aircraft, missile equipment, and physiological effects on human subjects at high speeds. They consisted of a platform that was mounted on one or two rails and propelled by several rockets. Calculate the magnitude of force exerted by each rocket, called its thrust T, for the four-rocket propulsion system shown in Figure 4.8. The sled\u2019s initial acceleration is 49 m/s2, opposing the motion is known to be 650 N. the mass of the system is 2100 kg, and the force of friction Figure 4.8 A sled experiences a rocket thrust that accelerates it to the right. Each rocket creates an identical thrust T. As in other situations where there is only horizontal acceleration, the vertical forces cancel. The ground exerts an upward force N on the system that is equal in magnitude and opposite in direction to its weight, w. The system here is the sled, its rockets, and rider, so none of the forces between these objects are considered. The arrow representing friction ( f ) is drawn larger than scale. Strategy Although there are forces acting vertically and horizontally, we assume the vertical forces cancel since there is no vertical acceleration. This leaves us with only horizontal forces and a simpler one-dimensional problem. Directions are indicated with plus or minus signs, with right taken as the positive direction. See the free-body diagram in the figure. Solution Since acceleration, mass, and the force of friction are given, we start with Newton\u2019s second law and look for ways to find the thrust of the engines. Since we have defined the direction of the force and acceleration as acting \u201cto the right,\u201d we need to consider only the magnitudes of these quantities in the calculations. Hence we begin with net =, (4.11) where net is the net force along the horizontal direction. We can see from Figure 4.8 that the engine thrusts add, while friction opposes the thrust. In equation form, the net external force is net = 4 \u2212. (4.12) 152 Chapter 4 | Dynamics: Force and Newton's Laws of Motion Substituting this into Newton\u2019s second law gives net = = 4 \u2212. Using a little algebra, we solve for the total thrust 4T: 4 = +. Substituting known values yields 4 = + = (2100 kg)(49 m/s2 ) + 650 N. So the total thrust is and the individual thrusts are Discussion 4 = 1.0\u00d7105 N, = 1.0\u00d7105 N 4 = 2.6\u00d7104 N. (4.13) (4.14) (4.15) (4.16) (4.17) The numbers are quite large, so the result might surprise you. Experiments such as this were performed in the early 1960s to test the limits of human endurance and the setup designed to protect human subjects in jet fighter emergency ejections. Speeds of 1000 km/h were obtained, with accelerations of 45's. (Recall that, the acceleration due to gravity, is 9.80 m/s2. When we say that an acceleration is 45's, it is 459.80 m/s2, which is approximately 440 m/s2.) While living subjects are not used any more, land speeds of 10,000 km/h have been obtained with rocket sleds. In this example, as in the preceding one, the system of interest is obvious. We will see in later examples that choosing the system of interest is crucial\u2014and the choice is not always obvious. Newton\u2019s second law of motion is more than a definition; it is a relationship among acceleration, force, and mass. It can help us make predictions. Each of those physical quantities can be defined independently, so the second law tells us something basic and universal about nature. The next section introduces the third and final law of motion. Applying the Science Practices: Sums of Forces Recall that forces are vector quantities, and therefore the net force acting on a system should be the vector sum of the forces. (a) Design an experiment to test this hypothesis. What sort of a system would be appropriate and convenient to have multiple forces applied to it? What features of the system should be held constant? What could be varied? Can forces be arranged in multiple directions so that, while the hypothesis is still tested, the resulting calculations are not too inconvenient? (b) Another group of students has done such an experiment, using a motion capture system looking down at an air hockey table to measure the motion of the 0.10-kg puck. The table was aligned with the cardinal directions, and a compressed air hose was placed in the center of each side, capable of varying levels of force output and fixed so that it was aimed at the center of the table. Table 4.1 Forces Measured acceleration (magnitudes) 3 N north, 4 N west 5", " N south, 12 N east 48 \u00b1 4 m/s2 132 \u00b1 6 m/s2 6 N north, 12 N east, 4 N west 99 \u00b1 3 m/s2 Given the data in the table, is the hypothesis confirmed? What were the directions of the accelerations? 4.4 Newton's Third Law of Motion: Symmetry in Forces Learning Objectives By the end of this section, you will be able to: \u2022 Understand Newton's third law of motion. \u2022 Apply Newton's third law to define systems and solve problems of motion. This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 4 | Dynamics: Force and Newton's Laws of Motion 153 The information presented in this section supports the following AP\u00ae learning objectives and science practices: \u2022 3.A.2.1 The student is able to represent forces in diagrams or mathematically using appropriately labeled vectors with magnitude, direction, and units during the analysis of a situation. (S.P. 1.1) \u2022 3.A.3.1 The student is able to analyze a scenario and make claims (develop arguments, justify assertions) about the forces exerted on an object by other objects for different types of forces or components of forces. (S.P. 6.4, 7.2) \u2022 3.A.3.3 The student is able to describe a force as an interaction between two objects and identify both objects for any force. (S.P. 1.4) \u2022 3.A.4.1 The student is able to construct explanations of physical situations involving the interaction of bodies using Newton's third law and the representation of action-reaction pairs of forces. (S.P. 1.4, 6.2) \u2022 3.A.4.2 The student is able to use Newton's third law to make claims and predictions about the action-reaction pairs of forces when two objects interact. (S.P. 6.4, 7.2) \u2022 3.A.4.3 The student is able to analyze situations involving interactions among several objects by using free-body diagrams that include the application of Newton's third law to identify forces. (S.P. 1.4) \u2022 3.B.2.1 The student is able to create and use free-body diagrams to analyze physical situations to solve problems with motion qualitatively and quantitatively. (S.P. 1.1, 1.4, 2.2) \u2022 4.A.2.1 The student is able to make predictions about the motion of a system based on the fact that acceleration is equal to the change in velocity per unit time, and velocity is equal to the change in position per unit time. (S.P. 6.4) \u2022 4.A.2.2 The student is able to evaluate using given data whether all the forces on a system or whether all the parts of a system have been identified. (S.P. 5.3) \u2022 4.A.3.1 The student is able to apply Newton's second law to systems to calculate the change in the center-of-mass velocity when an external force is exerted on the system. (S.P. 2.2) There is a passage in the musical Man of la Mancha that relates to Newton\u2019s third law of motion. Sancho, in describing a fight with his wife to Don Quixote, says, \u201cOf course I hit her back, Your Grace, but she\u2019s a lot harder than me and you know what they say, \u2018Whether the stone hits the pitcher or the pitcher hits the stone, it\u2019s going to be bad for the pitcher.\u2019\u201d This is exactly what happens whenever one body exerts a force on another\u2014the first also experiences a force (equal in magnitude and opposite in direction). Numerous common experiences, such as stubbing a toe or throwing a ball, confirm this. It is precisely stated in Newton\u2019s third law of motion. Newton\u2019s Third Law of Motion Whenever one body exerts a force on a second body, the first body experiences a force that is equal in magnitude and opposite in direction to the force that it exerts. This law represents a certain symmetry in nature: Forces always occur in pairs, and one body cannot exert a force on another without experiencing a force itself. We sometimes refer to this law loosely as \u201caction-reaction,\u201d where the force exerted is the action and the force experienced as a consequence is the reaction. Newton\u2019s third law has practical uses in analyzing the origin of forces and understanding which forces are external to a system. We can readily see Newton\u2019s third law at work by taking a look at how people move about. Consider a swimmer pushing off from the side of a pool, as illustrated in Figure 4.9. She pushes against the pool wall with her feet and accelerates", " in the direction opposite to that of her push. The wall has exerted an equal and opposite force back on the swimmer. You might think that two equal and opposite forces would cancel, but they do not because they act on different systems. In this case, there are two systems that we could investigate: the swimmer or the wall. If we select the swimmer to be the system of interest, as in the figure, then Fwall on feet is an external force on this system and affects its motion. The swimmer moves in the direction of Fwall on feet. In contrast, the force Ffeet on wall acts on the wall and not on our system of interest. Thus Ffeet on wall does not directly affect the motion of the system and does not cancel Fwall on feet. Note that the swimmer pushes in the direction opposite to that in which she wishes to move. The reaction to her push is thus in the desired direction. 154 Chapter 4 | Dynamics: Force and Newton's Laws of Motion Figure 4.9 When the swimmer exerts a force Ffeet on wall on the wall, she accelerates in the direction opposite to that of her push. This means the net external force on her is in the direction opposite to Ffeet on wall. This opposition occurs because, in accordance with Newton\u2019s third law of motion, the wall exerts a force Fwall on feet on her, equal in magnitude but in the direction opposite to the one she exerts on it. The line around the swimmer indicates the system of interest. Note that Ffeet on wall does not act on this system (the swimmer) and, thus, does not cancel Fwall on feet. Thus the free-body diagram shows only Fwall on feet, w, the gravitational force, and BF, the buoyant force of the water supporting the swimmer\u2019s weight. The vertical forces w and BF cancel since there is no vertical motion. Similarly, when a person stands on Earth, the Earth exerts a force on the person, pulling the person toward the Earth. As stated by Newton\u2019s third law of motion, the person also exerts a force that is equal in magnitude, but opposite in direction, pulling the Earth up toward the person. Since the mass of the Earth is so great, however, and =, the acceleration of the Earth toward the person is not noticeable. Other examples of Newton\u2019s third law are easy to find. As a professor paces in front of a whiteboard, she exerts a force backward on the floor. The floor exerts a reaction force forward on the professor that causes her to accelerate forward. Similarly, a car accelerates because the ground pushes forward on the drive wheels in reaction to the drive wheels pushing backward on the ground. You can see evidence of the wheels pushing backward when tires spin on a gravel road and throw rocks backward. In another example, rockets move forward by expelling gas backward at high velocity. This means the rocket exerts a large backward force on the gas in the rocket combustion chamber, and the gas therefore exerts a large reaction force forward on the rocket. This reaction force is called thrust. It is a common misconception that rockets propel themselves by pushing on the ground or on the air behind them. They actually work better in a vacuum, where they can more readily expel the exhaust gases. Helicopters similarly create lift by pushing air down, thereby experiencing an upward reaction force. Birds and airplanes also fly by exerting force on air in a direction opposite to that of whatever force they need. For example, the wings of a bird force air downward and backward in order to get lift and move forward. An octopus propels itself in the water by ejecting water through a funnel from its body, similar to a jet ski. In a situation similar to Sancho\u2019s, professional cage fighters experience reaction forces when they punch, sometimes breaking their hand by hitting an opponent\u2019s body. Example 4.3 Getting Up To Speed: Choosing the Correct System A physics professor pushes a cart of demonstration equipment to a lecture hall, as seen in Figure 4.10. Her mass is 65.0 kg, the cart\u2019s is 12.0 kg, and the equipment\u2019s is 7.0 kg. Calculate the acceleration produced when the professor exerts a backward force of 150 N on the floor. All forces opposing the motion, such as friction on the cart\u2019s wheels and air resistance, total 24.0 N. This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 4 | Dynamics: Force and Newton's Laws of Motion 155 Figure 4.10 A professor pushes a cart of demonstration equipment. The lengths of the arrows are proportional to the magnitudes of the forces (except for f, since it is too small to draw to scale). Different questions are asked in each example; thus, the system of interest must be defined differently for each. System 1", " is appropriate for Example 4.4, since it asks for the acceleration of the entire group of objects. Only Ffloor and f are external forces acting on System 1 along the line of motion. All other forces either cancel or act on the outside world. System 2 is chosen for this example so that Fprof will be an external force and enter into Newton\u2019s second law. Note that the free-body diagrams, which allow us to apply Newton\u2019s second law, vary with the system chosen. Strategy Since they accelerate as a unit, we define the system to be the professor, cart, and equipment. This is System 1 in Figure 4.10. The professor pushes backward with a force Ffoot of 150 N. According to Newton\u2019s third law, the floor exerts a forward reaction force Ffloor of 150 N on System 1. Because all motion is horizontal, we can assume there is no net force in the vertical direction. The problem is therefore one-dimensional along the horizontal direction. As noted, f opposes the motion and is thus in the opposite direction of Ffloor. Note that we do not include the forces Fprof or Fcart because these are internal forces, and we do not include Ffoot because it acts on the floor, not on the system. There are no other significant forces acting on System 1. If the net external force can be found from all this information, we can use Newton\u2019s second law to find the acceleration as requested. See the free-body diagram in the figure. Solution Newton\u2019s second law is given by = net. The net external force on System 1 is deduced from Figure 4.10 and the discussion above to be net = floor \u2212 = 150 N \u2212 24.0 N = 126 N. The mass of System 1 is = (65.0 + 12.0 + 7.0) kg = 84 kg. These values of net and produce an acceleration of net = = 126 N 84 kg = 1.5 m/s2. Discussion (4.18) (4.19) (4.20) (4.21) 156 Chapter 4 | Dynamics: Force and Newton's Laws of Motion None of the forces between components of System 1, such as between the professor\u2019s hands and the cart, contribute to the net external force because they are internal to System 1. Another way to look at this is to note that forces between components of a system cancel because they are equal in magnitude and opposite in direction. For example, the force exerted by the professor on the cart results in an equal and opposite force back on her. In this case both forces act on the same system and, therefore, cancel. Thus internal forces (between components of a system) cancel. Choosing System 1 was crucial to solving this problem. Example 4.4 Force on the Cart\u2014Choosing a New System Calculate the force the professor exerts on the cart in Figure 4.10 using data from the previous example if needed. Strategy If we now define the system of interest to be the cart plus equipment (System 2 in Figure 4.10), then the net external force on System 2 is the force the professor exerts on the cart minus friction. The force she exerts on the cart, Fprof, is an external force acting on System 2. Fprof was internal to System 1, but it is external to System 2 and will enter Newton\u2019s second law for System 2. Solution Newton\u2019s second law can be used to find Fprof. Starting with = net and noting that the magnitude of the net external force on System 2 is net = prof \u2212, we solve for prof, the desired quantity: prof = net +. (4.22) (4.23) (4.24) The value of is given, so we must calculate net net. That can be done since both the acceleration and mass of System 2 are known. Using Newton\u2019s second law we see that net =, (4.25) where the mass of System 2 is 19.0 kg ( = 12.0 kg + 7.0 kg) and its acceleration was found to be = 1.5 m/s2 in the previous example. Thus, Now we can find the desired force: net =, net = (19.0 kg)(1.5 m/s2 ) = 29 N. prof = net +, prof = 29 N+24.0 N = 53 N. (4.26) (4.27) (4.28) (4.29) Discussion It is interesting that this force is significantly less than the 150-N force the professor exerted backward on the floor. Not all of that 150-N force is transmitted to the cart; some of it accelerates the professor. The choice of a system is an important analytical step both in solving problems and in thoroughly understanding the physics of the situation (which is not necessarily the same thing). PhET Explorations: Gravity Force Lab Visualize the", " gravitational force that two objects exert on each other. Change properties of the objects in order to see how it changes the gravity force. This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 4 | Dynamics: Force and Newton's Laws of Motion 157 Figure 4.11 Gravity Force Lab (http://cnx.org/content/m54849/1.2/gravity-force-lab_en.jar) 4.5 Normal, Tension, and Other Examples of Force Learning Objectives By the end of this section, you will be able to: \u2022 Define normal and tension forces. \u2022 Apply Newton's laws of motion to solve problems involving a variety of forces. \u2022 Use trigonometric identities to resolve weight into components. The information presented in this section supports the following AP\u00ae learning objectives and science practices: \u2022 2.B.1.1 The student is able to apply = to calculate the gravitational force on an object with mass m in a gravitational field of strength g in the context of the effects of a net force on objects and systems. (S.P. 2.2, 7.2) \u2022 3.A.2.1 The student is able to represent forces in diagrams or mathematically using appropriately labeled vectors with magnitude, direction, and units during the analysis of a situation. (S.P. 1.1) \u2022 3.A.3.1 The student is able to analyze a scenario and make claims (develop arguments, justify assertions) about the forces exerted on an object by other objects for different types of forces or components of forces. (S.P. 6.4, 7.2) \u2022 3.A.3.3 The student is able to describe a force as an interaction between two objects and identify both objects for any force. (S.P. 1.4) \u2022 3.A.4.1 The student is able to construct explanations of physical situations involving the interaction of bodies using Newton's third law and the representation of action-reaction pairs of forces. (S.P. 1.4, 6.2) \u2022 3.A.4.2 The student is able to use Newton's third law to make claims and predictions about the action-reaction pairs of forces when two objects interact. (S.P. 6.4, 7.2) \u2022 3.A.4.3 The student is able to analyze situations involving interactions among several objects by using free-body diagrams that include the application of Newton's third law to identify forces. (S.P. 1.4) \u2022 3.B.1.3 The student is able to re-express a free-body diagram representation into a mathematical representation and solve the mathematical representation for the acceleration of the object. (S.P. 1.5, 2.2) \u2022 3.B.2.1 The student is able to create and use free-body diagrams to analyze physical situations to solve problems with motion qualitatively and quantitatively. (S.P. 1.1, 1.4, 2.2) Forces are given many names, such as push, pull, thrust, lift, weight, friction, and tension. Traditionally, forces have been grouped into several categories and given names relating to their source, how they are transmitted, or their effects. The most important of these categories are discussed in this section, together with some interesting applications. Further examples of forces are discussed later in this text. Normal Force Weight (also called force of gravity) is a pervasive force that acts at all times and must be counteracted to keep an object from falling. You definitely notice that you must support the weight of a heavy object by pushing up on it when you hold it stationary, as illustrated in Figure 4.12(a). But how do inanimate objects like a table support the weight of a mass placed on them, such as shown in Figure 4.12(b)? When the bag of dog food is placed on the table, the table actually sags slightly under the load. This would be noticeable if the load were placed on a card table, but even rigid objects deform when a force is applied to them. Unless the object is deformed beyond its limit, it will exert a restoring force much like a deformed spring (or trampoline or diving board). The greater the deformation, the greater the restoring force. So when the load is placed on the table, the table sags until the restoring force becomes as large as the weight of the load. At this point the net external force on the load is zero. That is the situation when the load is stationary on the table. The table sags quickly, and the sag is slight so we do not notice it. But it is similar to the sagging of a trampoline when you climb onto it. 158 Chapter 4 | Dynamics: Force and Newton's Laws of Motion Figure 4.12 (a) The person holding the bag of dog", " food must supply an upward force Fhand equal in magnitude and opposite in direction to the weight of the food w. (b) The card table sags when the dog food is placed on it, much like a stiff trampoline. Elastic restoring forces in the table grow as it sags until they supply a force N equal in magnitude and opposite in direction to the weight of the load. We must conclude that whatever supports a load, be it animate or not, must supply an upward force equal to the weight of the load, as we assumed in a few of the previous examples. If the force supporting a load is perpendicular to the surface of contact between the load and its support, this force is defined to be a normal force and here is given the symbol N. (This is not the unit for force N.) The word normal means perpendicular to a surface. The normal force can be less than the object\u2019s weight if the object is on an incline, as you will see in the next example. Common Misconception: Normal Force (N) vs. Newton (N) In this section we have introduced the quantity normal force, which is represented by the variable N. This should not be confused with the symbol for the newton, which is also represented by the letter N. These symbols are particularly important to distinguish because the units of a normal force ( N ) happen to be newtons (N). For example, the normal force N that the floor exerts on a chair might be N = 100 N. One important difference is that normal force is a vector, while the newton is simply a unit. Be careful not to confuse these letters in your calculations! You will encounter more similarities among variables and units as you proceed in physics. Another example of this is the quantity work ( ) and the unit watts (W). Example 4.5 Weight on an Incline, a Two-Dimensional Problem Consider the skier on a slope shown in Figure 4.13. Her mass including equipment is 60.0 kg. (a) What is her acceleration if friction is negligible? (b) What is her acceleration if friction is known to be 45.0 N? This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 4 | Dynamics: Force and Newton's Laws of Motion 159 Figure 4.13 Since motion and friction are parallel to the slope, it is most convenient to project all forces onto a coordinate system where one axis is parallel to the slope and the other is perpendicular (axes shown to left of skier). N is perpendicular to the slope and f is parallel to the slope, but w has components along both axes, namely w\u22a5 and w \u2225. N is equal in magnitude to w\u22a5, so that there is no motion perpendicular to the slope, but is less than \u2225, so that there is a downslope acceleration (along the parallel axis). Strategy This is a two-dimensional problem, since the forces on the skier (the system of interest) are not parallel. The approach we have used in two-dimensional kinematics also works very well here. Choose a convenient coordinate system and project the vectors onto its axes, creating two connected one-dimensional problems to solve. The most convenient coordinate system for motion on an incline is one that has one coordinate parallel to the slope and one perpendicular to the slope. (Remember that motions along mutually perpendicular axes are independent.) We use the symbols \u22a5 and \u2225 to represent perpendicular and parallel, respectively. This choice of axes simplifies this type of problem, because there is no motion perpendicular to the slope and because friction is always parallel to the surface between two objects. The only external forces acting on the system are the skier\u2019s weight, friction, and the support of the slope, respectively labeled w, f, and N in Figure 4.13. N is always perpendicular to the slope, and f is parallel to it. But w is not in the direction of either axis, and so the first step we take is to project it into components along the chosen axes, defining \u2225 to be the component of weight parallel to the slope and \u22a5 the component of weight perpendicular to the slope. Once this is done, we can consider the two separate problems of forces parallel to the slope and forces perpendicular to the slope. Solution The magnitude of the component of the weight parallel to the slope is \u2225 = sin (25\u00ba) = sin (25\u00ba), and the magnitude of the component of the weight perpendicular to the slope is \u22a5 = cos (25\u00ba) = cos (25\u00ba). (a) Neglecting friction. Since the acceleration is parallel to the slope, we need only consider forces parallel to the slope. (Forces perpendicular to the slope add to zero, since there is no acceleration in that direction.) The forces parallel to the slope are the amount of the skier\u2019s weight parallel to the", " slope \u2225 and friction. Using Newton\u2019s second law, with subscripts to denote quantities parallel to the slope, where net \u2225 = \u2225 = sin (25\u00ba), assuming no friction for this part, so that \u2225 = net \u2225 \u2225 = net \u2225 = sin (25\u00ba) (9.80 m/s2)(0.4226) = 4.14 m/s2 = sin (25\u00ba) (4.30) (4.31) (4.32) is the acceleration. (b) Including friction. We now have a given value for friction, and we know its direction is parallel to the slope and it opposes motion between surfaces in contact. So the net external force is now net \u2225 = \u2225 \u2212, (4.33) and substituting this into Newton\u2019s second law, \u2225 = net \u2225, gives 160 Chapter 4 | Dynamics: Force and Newton's Laws of Motion \u2225 = net \u2223 \u2223 = \u2225 \u2212 = sin (25\u00ba) \u2212. We substitute known values to obtain \u2225 = (60.0 kg)(9.80 m/s2)(0.4226) \u2212 45.0 N 60.0 kg, which yields \u2225 = 3.39 m/s2, (4.34) (4.35) (4.36) which is the acceleration parallel to the incline when there is 45.0 N of opposing friction. Discussion Since friction always opposes motion between surfaces, the acceleration is smaller when there is friction than when there is none. In fact, it is a general result that if friction on an incline is negligible, then the acceleration down the incline is = sin, regardless of mass. This is related to the previously discussed fact that all objects fall with the same acceleration in the absence of air resistance. Similarly, all objects, regardless of mass, slide down a frictionless incline with the same acceleration (if the angle is the same). Resolving Weight into Components Figure 4.14 An object rests on an incline that makes an angle \u03b8 with the horizontal. When an object rests on an incline that makes an angle with the horizontal, the force of gravity acting on the object is divided into two components: a force acting perpendicular to the plane, w\u22a5, and a force acting parallel to the plane, w \u2225. The perpendicular force of weight, w\u22a5, is typically equal in magnitude and opposite in direction to the normal force, N. The force acting parallel to the plane, w \u2225, causes the object to accelerate down the incline. The force of friction, f, opposes the motion of the object, so it acts upward along the plane. It is important to be careful when resolving the weight of the object into components. If the angle of the incline is at an angle to the horizontal, then the magnitudes of the weight components are and \u2225 = sin () = sin () \u22a5 = cos () = cos (). (4.37) (4.38) Instead of memorizing these equations, it is helpful to be able to determine them from reason. To do this, draw the right triangle formed by the three weight vectors. Notice that the angle of the incline is the same as the angle formed between w and w\u22a5. Knowing this property, you can use trigonometry to determine the magnitude of the weight components: cos () = \u22a5 \u22a5 = cos () = cos () \u2225 sin () = \u2225 = sin () = sin () (4.39) (4.40) This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 4 | Dynamics: Force and Newton's Laws of Motion 161 Take-Home Experiment: Force Parallel To investigate how a force parallel to an inclined plane changes, find a rubber band, some objects to hang from the end of the rubber band, and a board you can position at different angles. How much does the rubber band stretch when you hang the object from the end of the board? Now place the board at an angle so that the object slides off when placed on the board. How much does the rubber band extend if it is lined up parallel to the board and used to hold the object stationary on the board? Try two more angles. What does this show? Tension A tension is a force along the length of a medium, especially a force carried by a flexible medium, such as a rope or cable. The word \u201ctension\u201d comes from a Latin word meaning \u201cto stretch.\u201d Not coincidentally, the flexible cords that carry muscle forces to other parts of the body are called tendons. Any flexible connector, such as a string, rope, chain, wire, or cable, can exert pulls only parallel to its length; thus, a force carried by a flexible connector is a tension with direction parallel to the connector", ". It is important to understand that tension is a pull in a connector. In contrast, consider the phrase: \u201cYou can\u2019t push a rope.\u201d The tension force pulls outward along the two ends of a rope. Consider a person holding a mass on a rope as shown in Figure 4.15. Figure 4.15 When a perfectly flexible connector (one requiring no force to bend it) such as this rope transmits a force T, that force must be parallel to the length of the rope, as shown. The pull such a flexible connector exerts is a tension. Note that the rope pulls with equal force but in opposite directions on the hand and the supported mass (neglecting the weight of the rope). This is an example of Newton\u2019s third law. The rope is the medium that carries the equal and opposite forces between the two objects. The tension anywhere in the rope between the hand and the mass is equal. Once you have determined the tension in one location, you have determined the tension at all locations along the rope. Tension in the rope must equal the weight of the supported mass, as we can prove using Newton\u2019s second law. If the 5.00-kg mass in the figure is stationary, then its acceleration is zero, and thus Fnet = 0. The only external forces acting on the mass are its weight w and the tension T supplied by the rope. Thus, net = \u2212 = 0, (4.41) where and are the magnitudes of the tension and weight and their signs indicate direction, with up being positive here. Thus, just as you would expect, the tension equals the weight of the supported mass: For a 5.00-kg mass, then (neglecting the mass of the rope) we see that = = (5.00 kg)(9.80 m/s2 ) = 49.0 N. = =. (4.42) (4.43) If we cut the rope and insert a spring, the spring would extend a length corresponding to a force of 49.0 N, providing a direct observation and measure of the tension force in the rope. Flexible connectors are often used to transmit forces around corners, such as in a hospital traction system, a finger joint, or a bicycle brake cable. If there is no friction, the tension is transmitted undiminished. Only its direction changes, and it is always parallel to the flexible connector. This is illustrated in Figure 4.16 (a) and (b). 162 Chapter 4 | Dynamics: Force and Newton's Laws of Motion Figure 4.16 (a) Tendons in the finger carry force T from the muscles to other parts of the finger, usually changing the force\u2019s direction, but not its magnitude (the tendons are relatively friction free). (b) The brake cable on a bicycle carries the tension T from the handlebars to the brake mechanism. Again, the direction but not the magnitude of T is changed. Example 4.6 What Is the Tension in a Tightrope? Calculate the tension in the wire supporting the 70.0-kg tightrope walker shown in Figure 4.17. Figure 4.17 The weight of a tightrope walker causes a wire to sag by 5.0 degrees. The system of interest here is the point in the wire at which the tightrope walker is standing. Strategy As you can see in the figure, the wire is not perfectly horizontal (it cannot be!), but is bent under the person\u2019s weight. Thus, the tension on either side of the person has an upward component that can support his weight. As usual, forces are vectors represented pictorially by arrows having the same directions as the forces and lengths proportional to their magnitudes. The system is the tightrope walker, and the only external forces acting on him are his weight w and the two tensions TL (left tension) and TR (right tension), as illustrated. It is reasonable to neglect the weight of the wire itself. The net external force is zero since the system is stationary. A little trigonometry can now be used to find the tensions. One conclusion is possible at the outset\u2014we can see from part (b) of the figure that the magnitudes of the tensions L and R must be equal. This is because there is no horizontal acceleration in the rope, and the only forces acting to the left and right are L and. Thus, the magnitude of those forces must be equal so that they cancel each other out. Whenever we have two-dimensional vector problems in which no two vectors are parallel, the easiest method of solution is to pick a convenient coordinate system and project the vectors onto its axes. In this case the best coordinate system has one axis horizontal and the other vertical. We call the horizontal the -axis and the vertical the -axis. This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 4 |", " Dynamics: Force and Newton's Laws of Motion 163 Solution First, we need to resolve the tension vectors into their horizontal and vertical components. It helps to draw a new free-body diagram showing all of the horizontal and vertical components of each force acting on the system. Figure 4.18 When the vectors are projected onto vertical and horizontal axes, their components along those axes must add to zero, since the tightrope walker is stationary. The small angle results in being much greater than. Consider the horizontal components of the forces (denoted with a subscript ): net = L \u2212 R. The net external horizontal force net = 0, since the person is stationary. Thus, net = 0 = L \u2212 R L = R. Now, observe Figure 4.18. You can use trigonometry to determine the magnitude of L and R. Notice that: cos (5.0\u00ba) = L L L = L cos (5.0\u00ba) cos (5.0\u00ba) = R R R = R cos (5.0\u00ba). L cos (5.0\u00ba) = R cos (5.0\u00ba). L = R =, Equating L and R : Thus, (4.44) (4.45) (4.46) (4.47) (4.48) as predicted. Now, considering the vertical components (denoted by a subscript ), we can solve for. Again, since the person is stationary, Newton\u2019s second law implies that net = 0. Thus, as illustrated in the free-body diagram in Figure 4.18, net = L + R \u2212 = 0. Observing Figure 4.18, we can use trigonometry to determine the relationship between L, R, and. As we determined from the analysis in the horizontal direction, L = R = : sin (5.0\u00ba) = L L L = L sin (5.0\u00ba) = sin (5.0\u00ba) sin (5.0\u00ba) = R R R = R sin (5.0\u00ba) = sin (5.0\u00ba). (4.49) (4.50) 164 Chapter 4 | Dynamics: Force and Newton's Laws of Motion Now, we can substitute the values for L and R, into the net force equation in the vertical direction: = L + R \u2212 = 0 = sin (5.0\u00ba) + sin (5.0\u00ba) \u2212 = 0 net net 2 sin (5.0\u00ba) \u2212 = 0 = 2 sin (5.0\u00ba) and so that and the tension is Discussion = 2 sin (5.0\u00ba) = 2 sin (5.0\u00ba), = (70.0 kg)(9.80 m/s2) 2(0.0872), = 3900 N. (4.51) (4.52) (4.53) (4.54) Note that the vertical tension in the wire acts as a normal force that supports the weight of the tightrope walker. The tension is almost six times the 686-N weight of the tightrope walker. Since the wire is nearly horizontal, the vertical component of its tension is only a small fraction of the tension in the wire. The large horizontal components are in opposite directions and cancel, and so most of the tension in the wire is not used to support the weight of the tightrope walker. If we wish to create a very large tension, all we have to do is exert a force perpendicular to a flexible connector, as illustrated in Figure 4.19. As we saw in the last example, the weight of the tightrope walker acted as a force perpendicular to the rope. We saw that the tension in the roped related to the weight of the tightrope walker in the following way: = 2 sin (). (4.55) We can extend this expression to describe the tension created when a perpendicular force ( F\u22a5 ) is exerted at the middle of a flexible connector: = \u22a5 2 sin (). (4.56) Note that is the angle between the horizontal and the bent connector. In this case, becomes very large as approaches zero. Even the relatively small weight of any flexible connector will cause it to sag, since an infinite tension would result if it were horizontal (i.e., = 0 and sin = 0 ). (See Figure 4.19.) Figure 4.19 We can create a very large tension in the chain by pushing on it perpendicular to its length, as shown. Suppose we wish to pull a car out of the mud when no tow truck is available. Each time the car moves forward, the chain is tightened to keep it as nearly straight as possible. The tension in the chain is given by = \u22a5 2 sin () ; since is small, is very large. This situation is analogous to the tightrope walker shown in Figure 4.17, except that the tensions shown here are those transmitted to", " the car and the tree rather than those acting at the point where F\u22a5 is applied. This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 4 | Dynamics: Force and Newton's Laws of Motion 165 Figure 4.20 Unless an infinite tension is exerted, any flexible connector\u2014such as the chain at the bottom of the picture\u2014will sag under its own weight, giving a characteristic curve when the weight is evenly distributed along the length. Suspension bridges\u2014such as the Golden Gate Bridge shown in this image\u2014are essentially very heavy flexible connectors. The weight of the bridge is evenly distributed along the length of flexible connectors, usually cables, which take on the characteristic shape. (credit: Leaflet, Wikimedia Commons) Extended Topic: Real Forces and Inertial Frames There is another distinction among forces in addition to the types already mentioned. Some forces are real, whereas others are not. Real forces are those that have some physical origin, such as the gravitational pull. Contrastingly, fictitious forces are those that arise simply because an observer is in an accelerating frame of reference, such as one that rotates (like a merry-go-round) or undergoes linear acceleration (like a car slowing down). For example, if a satellite is heading due north above Earth\u2019s northern hemisphere, then to an observer on Earth it will appear to experience a force to the west that has no physical origin. Of course, what is happening here is that Earth is rotating toward the east and moves east under the satellite. In Earth\u2019s frame this looks like a westward force on the satellite, or it can be interpreted as a violation of Newton\u2019s first law (the law of inertia). An inertial frame of reference is one in which all forces are real and, equivalently, one in which Newton\u2019s laws have the simple forms given in this chapter. Earth\u2019s rotation is slow enough that Earth is nearly an inertial frame. You ordinarily must perform precise experiments to observe fictitious forces and the slight departures from Newton\u2019s laws, such as the effect just described. On the large scale, such as for the rotation of weather systems and ocean currents, the effects can be easily observed. The crucial factor in determining whether a frame of reference is inertial is whether it accelerates or rotates relative to a known inertial frame. Unless stated otherwise, all phenomena discussed in this text are considered in inertial frames. All the forces discussed in this section are real forces, but there are a number of other real forces, such as lift and thrust, that are not discussed in this section. They are more specialized, and it is not necessary to discuss every type of force. It is natural, however, to ask where the basic simplicity we seek to find in physics is in the long list of forces. Are some more basic than others? Are some different manifestations of the same underlying force? The answer to both questions is yes, as will be seen in the next (extended) section and in the treatment of modern physics later in the text. PhET Explorations: Forces in 1 Dimension Explore the forces at work when you try to push a filing cabinet. Create an applied force and see the resulting friction force and total force acting on the cabinet. Charts show the forces, position, velocity, and acceleration vs. time. View a free-body diagram of all the forces (including gravitational and normal forces). Figure 4.21 Forces in 1 Dimension (http://cnx.org/content/m54857/1.4/forces-1d_en.jar) 4.6 Problem-Solving Strategies By the end of this section, you will be able to: Learning Objectives 166 Chapter 4 | Dynamics: Force and Newton's Laws of Motion \u2022 Apply a problem-solving procedure to solve problems using Newton's laws of motion The information presented in this section supports the following AP\u00ae learning objectives and science practices: \u2022 3.A.2.1 The student is able to represent forces in diagrams or mathematically using appropriately labeled vectors with magnitude, direction, and units during the analysis of a situation. (S.P. 1.1) \u2022 3.A.3.3 The student is able to describe a force as an interaction between two objects and identify both objects for any force. (S.P. 1.4) \u2022 3.B.1.1 The student is able to predict the motion of an object subject to forces exerted by several objects using an application of Newton's second law in a variety of physical situations with acceleration in one dimension. (S.P. 6.4, 7.2) \u2022 3.B.1.3 The student is able to re-express a free-body diagram representation into a mathematical representation and solve the mathematical representation for the acceleration of the object. (S.P. 1.5, 2.2) \u2022 3.B.2.1 The", " student is able to create and use free-body diagrams to analyze physical situations to solve problems with motion qualitatively and quantitatively. (S.P. 1.1, 1.4, 2.2) Success in problem solving is obviously necessary to understand and apply physical principles, not to mention the more immediate need of passing exams. The basics of problem solving, presented earlier in this text, are followed here, but specific strategies useful in applying Newton\u2019s laws of motion are emphasized. These techniques also reinforce concepts that are useful in many other areas of physics. Many problem-solving strategies are stated outright in the worked examples, and so the following techniques should reinforce skills you have already begun to develop. Problem-Solving Strategy for Newton\u2019s Laws of Motion Step 1. As usual, it is first necessary to identify the physical principles involved. Once it is determined that Newton\u2019s laws of motion are involved (if the problem involves forces), it is particularly important to draw a careful sketch of the situation. Such a sketch is shown in Figure 4.22(a). Then, as in Figure 4.22(b), use arrows to represent all forces, label them carefully, and make their lengths and directions correspond to the forces they represent (whenever sufficient information exists). Figure 4.22 (a) A sketch of Tarzan hanging from a vine. (b) Arrows are used to represent all forces. T is the tension in the vine above Tarzan, FT is the force he exerts on the vine, and w is his weight. All other forces, such as the nudge of a breeze, are assumed negligible. (c) Suppose we are given the ape man\u2019s mass and asked to find the tension in the vine. We then define the system of interest as shown and draw a free-body diagram. FT is no longer shown, because it is not a force acting on the system of interest; rather, FT acts on the outside world. (d) Showing only the arrows, the head-to-tail method of addition is used. It is apparent that T = - w, if Tarzan is stationary. Step 2. Identify what needs to be determined and what is known or can be inferred from the problem as stated. That is, make a list of knowns and unknowns. Then carefully determine the system of interest. This decision is a crucial step, since Newton\u2019s second law involves only external forces. Once the system of interest has been identified, it becomes possible to determine which forces are external and which are internal, a necessary step to employ Newton\u2019s second law. (See Figure 4.22(c).) Newton\u2019s third law may be used to identify whether forces are exerted between components of a system (internal) or between the system and something outside (external). As illustrated earlier in this chapter, the system of interest depends on what This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 4 | Dynamics: Force and Newton's Laws of Motion 167 question we need to answer. This choice becomes easier with practice, eventually developing into an almost unconscious process. Skill in clearly defining systems will be beneficial in later chapters as well. A diagram showing the system of interest and all of the external forces is called a free-body diagram. Only forces are shown on free-body diagrams, not acceleration or velocity. We have drawn several of these in worked examples. Figure 4.22(c) shows a free-body diagram for the system of interest. Note that no internal forces are shown in a free-body diagram. Step 3. Once a free-body diagram is drawn, Newton\u2019s second law can be applied to solve the problem. This is done in Figure 4.22(d) for a particular situation. In general, once external forces are clearly identified in free-body diagrams, it should be a straightforward task to put them into equation form and solve for the unknown, as done in all previous examples. If the problem is one-dimensional\u2014that is, if all forces are parallel\u2014then they add like scalars. If the problem is two-dimensional, then it must be broken down into a pair of one-dimensional problems. This is done by projecting the force vectors onto a set of axes chosen for convenience. As seen in previous examples, the choice of axes can simplify the problem. For example, when an incline is involved, a set of axes with one axis parallel to the incline and one perpendicular to it is most convenient. It is almost always convenient to make one axis parallel to the direction of motion, if this is known. Applying Newton\u2019s Second Law Before you write net force equations, it is critical to determine whether the system is accelerating in a particular direction. If the acceleration is zero in a particular direction, then the net force is zero in that direction. Similarly, if the acceleration is nonzero in a particular direction, then", " the net force is described by the equation: net =. For example, if the system is accelerating in the horizontal direction, but it is not accelerating in the vertical direction, then you will have the following conclusions: net =, net = 0. (4.57) (4.58) You will need this information in order to determine unknown forces acting in a system. Step 4. As always, check the solution to see whether it is reasonable. In some cases, this is obvious. For example, it is reasonable to find that friction causes an object to slide down an incline more slowly than when no friction exists. In practice, intuition develops gradually through problem solving, and with experience it becomes progressively easier to judge whether an answer is reasonable. Another way to check your solution is to check the units. If you are solving for force and end up with units of m/s, then you have made a mistake. 4.7 Further Applications of Newton's Laws of Motion Learning Objectives By the end of this section, you will be able to: \u2022 Apply problem-solving techniques to solve for quantities in more complex systems of forces. \u2022 Integrate concepts from kinematics to solve problems using Newton's laws of motion. The information presented in this section supports the following AP\u00ae learning objectives and science practices: \u2022 3.A.2.1 The student is able to represent forces in diagrams or mathematically using appropriately labeled vectors with magnitude, direction, and units during the analysis of a situation. (S.P. 1.1) \u2022 3.A.3.1 The student is able to analyze a scenario and make claims (develop arguments, justify assertions) about the forces exerted on an object by other objects for different types of forces or components of forces. (S.P. 6.4, 7.2) \u2022 3.A.3.3 The student is able to describe a force as an interaction between two objects and identify both objects for any force. (S.P. 1.4) \u2022 3.B.1.1 The student is able to predict the motion of an object subject to forces exerted by several objects using an application of Newton's second law in a variety of physical situations with acceleration in one dimension. (S.P. 6.4, 7.2) \u2022 3.B.1.3 The student is able to re-express a free-body diagram representation into a mathematical representation and solve the mathematical representation for the acceleration of the object. (S.P. 1.5, 2.2) \u2022 3.B.2.1 The student is able to create and use free-body diagrams to analyze physical situations to solve problems with motion qualitatively and quantitatively. (S.P. 1.1, 1.4, 2.2) There are many interesting applications of Newton\u2019s laws of motion, a few more of which are presented in this section. These serve also to illustrate some further subtleties of physics and to help build problem-solving skills. 168 Chapter 4 | Dynamics: Force and Newton's Laws of Motion Example 4.7 Drag Force on a Barge Suppose two tugboats push on a barge at different angles, as shown in Figure 4.23. The first tugboat exerts a force of 2.7\u00d7105 N in the x-direction, and the second tugboat exerts a force of 3.6\u00d7105 N in the y-direction. Figure 4.23 (a) A view from above of two tugboats pushing on a barge. (b) The free-body diagram for the ship contains only forces acting in the plane of the water. It omits the two vertical forces\u2014the weight of the barge and the buoyant force of the water supporting it cancel and are not shown. Since the applied forces are perpendicular, the x- and y-axes are in the same direction as F and F. The problem quickly becomes a one-dimensional problem along the direction of Fapp, since friction is in the direction opposite to Fapp. If the mass of the barge is 5.0\u00d7106 kg and its acceleration is observed to be 7.5\u00d710\u22122 m/s2 in the direction shown, what is the drag force of the water on the barge resisting the motion? (Note: drag force is a frictional force exerted by fluids, such as air or water. The drag force opposes the motion of the object.) Strategy The directions and magnitudes of acceleration and the applied forces are given in Figure 4.23(a). We will define the total force of the tugboats on the barge as Fapp so that: Fapp =F + F (4.59) Since the barge is flat bottomed, the drag of the water FD will be in the direction opposite to Fapp, as shown in the freebody diagram in Figure 4.23(b). The system of interest here is the barge, since the forces on", " it are given as well as its acceleration. Our strategy is to find the magnitude and direction of the net applied force Fapp, and then apply Newton\u2019s second law to solve for the drag force FD. Solution Since F and F are perpendicular, the magnitude and direction of Fapp are easily found. First, the resultant magnitude is given by the Pythagorean theorem: 2 2 + F app = F app = (2.7\u00d7105 N)2 + (3.6\u00d7105 N)2 = 4.5\u00d7105 N. The angle is given by = tan\u22121 3.6\u00d7105 N = tan\u22121 2.7\u00d7105 N = 53\u00ba (4.60) (4.61) which we know, because of Newton\u2019s first law, is the same direction as the acceleration. FD is in the opposite direction of Fapp, since it acts to slow down the acceleration. Therefore, the net external force is in the same direction as Fapp, but its magnitude is slightly less than Fapp. The problem is now one-dimensional. From Figure 4.23(b), we can see that net = app \u2212 D. (4.62) But Newton\u2019s second law states that This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 4 | Dynamics: Force and Newton's Laws of Motion Thus, net =. app \u2212 D =. This can be solved for the magnitude of the drag force of the water D in terms of known quantities: D = app \u2212. 169 (4.63) (4.64) (4.65) Substituting known values gives FD = (4.5\u00d7105 N) \u2212 (5.0\u00d7106 kg)(7.5\u00d710\u20132 m/s2 ) = 7.5\u00d7104 N. The direction of FD has already been determined to be in the direction opposite to Fapp, or at an angle of 53\u00ba south of west. (4.66) Discussion The numbers used in this example are reasonable for a moderately large barge. It is certainly difficult to obtain larger accelerations with tugboats, and small speeds are desirable to avoid running the barge into the docks. Drag is relatively small for a well-designed hull at low speeds, consistent with the answer to this example, where D is less than 1/600th of the weight of the ship. In the earlier example of a tightrope walker we noted that the tensions in wires supporting a mass were equal only because the angles on either side were equal. Consider the following example, where the angles are not equal; slightly more trigonometry is involved. Example 4.8 Different Tensions at Different Angles Consider the traffic light (mass 15.0 kg) suspended from two wires as shown in Figure 4.24. Find the tension in each wire, neglecting the masses of the wires. 170 Chapter 4 | Dynamics: Force and Newton's Laws of Motion Figure 4.24 A traffic light is suspended from two wires. (b) Some of the forces involved. (c) Only forces acting on the system are shown here. The free-body diagram for the traffic light is also shown. (d) The forces projected onto vertical (y) and horizontal (x) axes. The horizontal components of the tensions must cancel, and the sum of the vertical components of the tensions must equal the weight of the traffic light. (e) The free-body diagram shows the vertical and horizontal forces acting on the traffic light. Strategy The system of interest is the traffic light, and its free-body diagram is shown in Figure 4.24(c). The three forces involved are not parallel, and so they must be projected onto a coordinate system. The most convenient coordinate system has one axis vertical and one horizontal, and the vector projections on it are shown in part (d) of the figure. There are two unknowns in this problem ( 1 and 2 ), so two equations are needed to find them. These two equations come from applying Newton\u2019s second law along the vertical and horizontal axes, noting that the net external force is zero along each axis because acceleration is zero. Solution First consider the horizontal or x-axis: Thus, as you might expect, net = 2 \u2212 1 = 0. (4.67) This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 4 | Dynamics: Force and Newton's Laws of Motion This gives us the following relationship between 1 and 2 : 1 = 2. Thus, 1 cos (30\u00ba) = 2 cos (45\u00ba). 2 = (1.225)1. 171 (4.68) (4.69) (4.70) Note that 1 and 2 are not equal in this case, because the angles on either side are not equal. It is reasonable that 2 ends up being greater than 1, because it is exerted more", " vertically than 1. Now consider the force components along the vertical or y-axis: This implies net = 1 + 2 \u2212 = 0. 1 + 2 =. Substituting the expressions for the vertical components gives 1 sin (30\u00ba) + 2 sin (45\u00ba) =. There are two unknowns in this equation, but substituting the expression for 2 in terms of 1 reduces this to one equation with one unknown: which yields 1(0.500) + (1.2251)(0.707) = =, (1.366)1 = (15.0 kg)(9.80 m/s2). Solving this last equation gives the magnitude of 1 to be 1 = 108 N. (4.71) (4.72) (4.73) (4.74) (4.75) (4.76) Finally, the magnitude of 2 is determined using the relationship between them, 2 = 1.225 1, found above. Thus we obtain 2 = 132 N. (4.77) Discussion Both tensions would be larger if both wires were more horizontal, and they will be equal if and only if the angles on either side are the same (as they were in the earlier example of a tightrope walker). The bathroom scale is an excellent example of a normal force acting on a body. It provides a quantitative reading of how much it must push upward to support the weight of an object. But can you predict what you would see on the dial of a bathroom scale if you stood on it during an elevator ride? Will you see a value greater than your weight when the elevator starts up? What about when the elevator moves upward at a constant speed: will the scale still read more than your weight at rest? Consider the following example. Example 4.9 What Does the Bathroom Scale Read in an Elevator? Figure 4.25 shows a 75.0-kg man (weight of about 165 lb) standing on a bathroom scale in an elevator. Calculate the scale reading: (a) if the elevator accelerates upward at a rate of 1.20 m/s2, and (b) if the elevator moves upward at a constant speed of 1 m/s. 172 Chapter 4 | Dynamics: Force and Newton's Laws of Motion Figure 4.25 (a) The various forces acting when a person stands on a bathroom scale in an elevator. The arrows are approximately correct for when the elevator is accelerating upward\u2014broken arrows represent forces too large to be drawn to scale. T is the tension in the supporting cable, w is the weight of the person, ws is the weight of the scale, we is the weight of the elevator, Fs is the force of the scale on the person, Fp is the force of the person on the scale, Ft is the force of the scale on the floor of the elevator, and N is the force of the floor upward on the scale. (b) The free-body diagram shows only the external forces acting on the designated system of interest\u2014the person. Strategy If the scale is accurate, its reading will equal p, the magnitude of the force the person exerts downward on it. Figure 4.25(a) shows the numerous forces acting on the elevator, scale, and person. It makes this one-dimensional problem look much more formidable than if the person is chosen to be the system of interest and a free-body diagram is drawn as in Figure 4.25(b). Analysis of the free-body diagram using Newton\u2019s laws can produce answers to both parts (a) and (b) of this example, as well as some other questions that might arise. The only forces acting on the person are his weight w and the upward force of the scale Fs. According to Newton\u2019s third law Fp and Fs are equal in magnitude and opposite in direction, so that we need to find s in order to find what the scale reads. We can do this, as usual, by applying Newton\u2019s second law, net =. From the free-body diagram we see that net = s \u2212, so that s \u2212 =. Solving for s gives an equation with only one unknown: or, because =, simply s = +, s = +. (4.78) (4.79) (4.80) (4.81) No assumptions were made about the acceleration, and so this solution should be valid for a variety of accelerations in addition to the ones in this exercise. Solution for (a) This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 4 | Dynamics: Force and Newton's Laws of Motion 173 In this part of the problem, = 1.20 m/s2, so that s = (75.0 kg)(1.20 m/s2 ) + (75.0 kg)(9.80 m/s2), yielding Discussion for (a)", " s = 825 N. (4.82) (4.83) This is about 185 lb. What would the scale have read if he were stationary? Since his acceleration would be zero, the force of the scale would be equal to his weight: net = = 0 = s \u2212 s = = = (75.0 kg)(9.80 m/s2) = 735 N. s s (4.84) So, the scale reading in the elevator is greater than his 735-N (165 lb) weight. This means that the scale is pushing up on the person with a force greater than his weight, as it must in order to accelerate him upward. Clearly, the greater the acceleration of the elevator, the greater the scale reading, consistent with what you feel in rapidly accelerating versus slowly accelerating elevators. Solution for (b) Now, what happens when the elevator reaches a constant upward velocity? Will the scale still read more than his weight? For any constant velocity\u2014up, down, or stationary\u2014acceleration is zero because = \u0394 \u0394, and \u0394 = 0. Thus, Now which gives Discussion for (b75.0 kg)(9.80 m/s2), s = 735 N. (4.85) (4.86) (4.87) The scale reading is 735 N, which equals the person\u2019s weight. This will be the case whenever the elevator has a constant velocity\u2014moving up, moving down, or stationary. The solution to the previous example also applies to an elevator accelerating downward, as mentioned. When an elevator accelerates downward, is negative, and the scale reading is less than the weight of the person, until a constant downward velocity is reached, at which time the scale reading again becomes equal to the person\u2019s weight. If the elevator is in free-fall and accelerating downward at, then the scale reading will be zero and the person will appear to be weightless. Integrating Concepts: Newton\u2019s Laws of Motion and Kinematics Physics is most interesting and most powerful when applied to general situations that involve more than a narrow set of physical principles. Newton\u2019s laws of motion can also be integrated with other concepts that have been discussed previously in this text to solve problems of motion. For example, forces produce accelerations, a topic of kinematics, and hence the relevance of earlier chapters. When approaching problems that involve various types of forces, acceleration, velocity, and/or position, use the following steps to approach the problem: Problem-Solving Strategy Step 1. Identify which physical principles are involved. Listing the givens and the quantities to be calculated will allow you to identify the principles involved. Step 2. Solve the problem using strategies outlined in the text. If these are available for the specific topic, you should refer to them. You should also refer to the sections of the text that deal with a particular topic. The following worked example illustrates how these strategies are applied to an integrated concept problem. 174 Chapter 4 | Dynamics: Force and Newton's Laws of Motion Example 4.10 What Force Must a Soccer Player Exert to Reach Top Speed? A soccer player starts from rest and accelerates forward, reaching a velocity of 8.00 m/s in 2.50 s. (a) What was his average acceleration? (b) What average force did he exert backward on the ground to achieve this acceleration? The player\u2019s mass is 70.0 kg, and air resistance is negligible. Strategy 1. To solve an integrated concept problem, we must first identify the physical principles involved and identify the chapters in which they are found. Part (a) of this example considers acceleration along a straight line. This is a topic of kinematics. Part (b) deals with force, a topic of dynamics found in this chapter. 2. The following solutions to each part of the example illustrate how the specific problem-solving strategies are applied. These involve identifying knowns and unknowns, checking to see if the answer is reasonable, and so forth. Solution for (a) We are given the initial and final velocities (zero and 8.00 m/s forward); thus, the change in velocity is \u0394 = 8.00 m/s. We are given the elapsed time, and so \u0394 = 2.50 s. The unknown is acceleration, which can be found from its definition: = \u0394 \u0394. (4.88) Substituting the known values yields = 8.00 m/s 2.50 s = 3.20 m/s2. (4.89) Discussion for (a) This is an attainable acceleration for an athlete in good condition. Solution for (b) Here we are asked to find the average force the player exerts backward to achieve this forward acceleration. Neglecting air resistance, this would be equal in magnitude to the net external force on the player, since this force causes his acceleration. Since we now know the", " player\u2019s acceleration and are given his mass, we can use Newton\u2019s second law to find the force exerted. That is, Substituting the known values of and gives net =. net = (70.0 kg)(3.20 m/s2) = 224 N. Discussion for (b) This is about 50 pounds, a reasonable average force. (4.90) (4.91) This worked example illustrates how to apply problem-solving strategies to situations that include topics from different chapters. The first step is to identify the physical principles involved in the problem. The second step is to solve for the unknown using familiar problem-solving strategies. These strategies are found throughout the text, and many worked examples show how to use them for single topics. You will find these techniques for integrated concept problems useful in applications of physics outside of a physics course, such as in your profession, in other science disciplines, and in everyday life. The following problems will build your skills in the broad application of physical principles. 4.8 Extended Topic: The Four Basic Forces\u2014An Introduction By the end of this section, you will be able to: \u2022 Understand the four basic forces that underlie the processes in nature. Learning Objectives The information presented in this section supports the following AP\u00ae learning objectives and science practices: \u2022 3.C.4.1 The student is able to make claims about various contact forces between objects based on the microscopic cause of those forces. (S.P. 6.1) \u2022 3.C.4.2 The student is able to explain contact forces (tension, friction, normal, buoyant, spring) as arising from interatomic electric forces and that they therefore have certain directions. (S.P. 6.2) \u2022 3.G.1.1 The student is able to articulate situations when the gravitational force is the dominant force and when the electromagnetic, weak, and strong forces can be ignored. (S.P. 7.1) This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 4 | Dynamics: Force and Newton's Laws of Motion 175 One of the most remarkable simplifications in physics is that only four distinct forces account for all known phenomena. In fact, nearly all of the forces we experience directly are due to only one basic force, called the electromagnetic force. (The gravitational force is the only force we experience directly that is not electromagnetic.) This is a tremendous simplification of the myriad of apparently different forces we can list, only a few of which were discussed in the previous section. As we will see, the basic forces are all thought to act through the exchange of microscopic carrier particles, and the characteristics of the basic forces are determined by the types of particles exchanged. Action at a distance, such as the gravitational force of Earth on the Moon, is explained by the existence of a force field rather than by \u201cphysical contact.\u201d The four basic forces are the gravitational force, the electromagnetic force, the weak nuclear force, and the strong nuclear force. Their properties are summarized in Table 4.2. Since the weak and strong nuclear forces act over an extremely short range, the size of a nucleus or less, we do not experience them directly, although they are crucial to the very structure of matter. These forces determine which nuclei are stable and which decay, and they are the basis of the release of energy in certain nuclear reactions. Nuclear forces determine not only the stability of nuclei, but also the relative abundance of elements in nature. The properties of the nucleus of an atom determine the number of electrons it has and, thus, indirectly determine the chemistry of the atom. More will be said of all of these topics in later chapters. Concept Connections: The Four Basic Forces The four basic forces will be encountered in more detail as you progress through the text. The gravitational force is defined in Uniform Circular Motion and Gravitation, electric force in Electric Charge and Electric Field, magnetic force in Magnetism, and nuclear forces in Radioactivity and Nuclear Physics. On a macroscopic scale, electromagnetism and gravity are the basis for all forces. The nuclear forces are vital to the substructure of matter, but they are not directly experienced on the macroscopic scale. Table 4.2 Properties of the Four Basic Forces[1] Force Approximate Relative Strengths Range Attraction/Repulsion Carrier Particle Gravitational 10\u221238 Electromagnetic 10 \u2013 2 Weak nuclear 10 \u2013 13 Strong nuclear 1 \u221e \u221e attractive only Graviton attractive and repulsive Photon < 10\u201318 m attractive and repulsive W+, W \u2013, Z0 < 10\u201315 m attractive and repulsive gluons The gravitational force is surprisingly weak\u2014it is only because gravity is always attractive that we notice it at all. Our weight is the gravitational force due to the entire Earth acting on us. On the very large scale, as in astronomical systems, the gravitational force is the", " dominant force determining the motions of moons, planets, stars, and galaxies. The gravitational force also affects the nature of space and time. As we shall see later in the study of general relativity, space is curved in the vicinity of very massive bodies, such as the Sun, and time actually slows down near massive bodies. Take a good look at the ranges for the four fundamental forces listed in Table 4.2. The range of the strong nuclear force, 10\u221215 m, is approximately the size of the nucleus of an atom; the weak nuclear force has an even shorter range. At scales on the order of 10\u221210 m, approximately the size of an atom, both nuclear forces are completely dominated by the electromagnetic force. Notice that this scale is still utterly tiny compared to our everyday experience. At scales that we do experience daily, electromagnetism tends to be negligible, due to its attractive and repulsive properties canceling each other out. That leaves gravity, which is usually the strongest of the forces at scales above ~10\u22124 m, and hence includes our everyday activities, such as throwing, climbing stairs, and walking. Electromagnetic forces can be either attractive or repulsive. They are long-range forces, which act over extremely large distances, and they nearly cancel for macroscopic objects. (Remember that it is the net external force that is important.) If they did not cancel, electromagnetic forces would completely overwhelm the gravitational force. The electromagnetic force is a combination of electrical forces (such as those that cause static electricity) and magnetic forces (such as those that affect a compass needle). These two forces were thought to be quite distinct until early in the 19th century, when scientists began to discover that they are different manifestations of the same force. This discovery is a classical case of the unification of forces. Similarly, friction, tension, and all of the other classes of forces we experience directly (except gravity, of course) are due to electromagnetic interactions of atoms and molecules. It is still convenient to consider these forces separately in specific applications, however, because of the ways they manifest themselves. 1. The graviton is a proposed particle, though it has not yet been observed by scientists. See the discussion of gravitational waves later in this section. The particles W+ are called vector bosons; these were predicted by theory and first observed in 1983. There are eight types of gluons proposed by scientists, and their existence is indicated by meson exchange in the nuclei of atoms., W\u2212, and Z0 176 Chapter 4 | Dynamics: Force and Newton's Laws of Motion Concept Connections: Unifying Forces Attempts to unify the four basic forces are discussed in relation to elementary particles later in this text. By \u201cunify\u201d we mean finding connections between the forces that show that they are different manifestations of a single force. Even if such unification is achieved, the forces will retain their separate characteristics on the macroscopic scale and may be identical only under extreme conditions such as those existing in the early universe. Physicists are now exploring whether the four basic forces are in some way related. Attempts to unify all forces into one come under the rubric of Grand Unified Theories (GUTs), with which there has been some success in recent years. It is now known that under conditions of extremely high density and temperature, such as existed in the early universe, the electromagnetic and weak nuclear forces are indistinguishable. They can now be considered to be different manifestations of one force, called the electroweak force. So the list of four has been reduced in a sense to only three. Further progress in unifying all forces is proving difficult\u2014especially the inclusion of the gravitational force, which has the special characteristics of affecting the space and time in which the other forces exist. While the unification of forces will not affect how we discuss forces in this text, it is fascinating that such underlying simplicity exists in the face of the overt complexity of the universe. There is no reason that nature must be simple\u2014it simply is. Action at a Distance: Concept of a Field All forces act at a distance. This is obvious for the gravitational force. Earth and the Moon, for example, interact without coming into contact. It is also true for all other forces. Friction, for example, is an electromagnetic force between atoms that may not actually touch. What is it that carries forces between objects? One way to answer this question is to imagine that a force field surrounds whatever object creates the force. A second object (often called a test object) placed in this field will experience a force that is a function of location and other variables. The field itself is the \u201cthing\u201d that carries the force from one object to another. The field is defined so as to be a characteristic of the object creating it; the field does not depend on the test object placed in it. Earth\u2019s gravitational field, for example, is a function of the mass of Earth and the distance from its center, independent of the presence of other masses", ". The concept of a field is useful because equations can be written for force fields surrounding objects (for gravity, this yields = at Earth\u2019s surface), and motions can be calculated from these equations. (See Figure 4.26.) Figure 4.26 The electric force field between a positively charged particle and a negatively charged particle. When a positive test charge is placed in the field, the charge will experience a force in the direction of the force field lines. Concept Connections: Force Fields The concept of a force field is also used in connection with electric charge and is presented in Electric Charge and Electric Field. It is also a useful idea for all the basic forces, as will be seen in Particle Physics. Fields help us to visualize forces and how they are transmitted, as well as to describe them with precision and to link forces with subatomic carrier particles. Making Connections: Vector and Scalar Fields These fields may be either scalar or vector fields. Gravity and electromagnetism are examples of vector fields. A test object placed in such a field will have both the magnitude and direction of the resulting force on the test object completely defined by the object\u2019s location in the field. We will later cover examples of scalar fields, which have a magnitude but no direction. The field concept has been applied very successfully; we can calculate motions and describe nature to high precision using field equations. As useful as the field concept is, however, it leaves unanswered the question of what carries the force. It has been proposed in recent decades, starting in 1935 with Hideki Yukawa\u2019s (1907\u20131981) work on the strong nuclear force, that all forces are transmitted by the exchange of elementary particles. We can visualize particle exchange as analogous to macroscopic phenomena such as two people passing a basketball back and forth, thereby exerting a repulsive force without touching one another. (See Figure 4.27.) This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 4 | Dynamics: Force and Newton's Laws of Motion 177 Figure 4.27 The exchange of masses resulting in repulsive forces. (a) The person throwing the basketball exerts a force Fp1 on it toward the other person and feels a reaction force FB away from the second person. (b) The person catching the basketball exerts a force Fp2 on it to stop the ball and feels a reaction force F\u2032B away from the first person. (c) The analogous exchange of a meson between a proton and a neutron carries the strong nuclear forces Fexch and F\u2032exch between them. An attractive force can also be exerted by the exchange of a mass\u2014if person 2 pulled the basketball away from the first person as he tried to retain it, then the force between them would be attractive. This idea of particle exchange deepens rather than contradicts field concepts. It is more satisfying philosophically to think of something physical actually moving between objects acting at a distance. Table 4.2 lists the exchange or carrier particles, both observed and proposed, that carry the four forces. But the real fruit of the particle-exchange proposal is that searches for Yukawa\u2019s proposed particle found it and a number of others that were completely unexpected, stimulating yet more research. All of this research eventually led to the proposal of quarks as the underlying substructure of matter, which is a basic tenet of GUTs. If successful, these theories will explain not only forces, but also the structure of matter itself. Yet physics is an experimental science, so the test of these theories must lie in the domain of the real world. As of this writing, scientists at the CERN laboratory in Switzerland are starting to test these theories using the world\u2019s largest particle accelerator: the Large Hadron Collider. This accelerator (27 km in circumference) allows two high-energy proton beams, traveling in opposite directions, to collide. An energy of 14 trillion electron volts will be available. It is anticipated that some new particles, possibly force carrier particles, will be found. (See Figure 4.28.) One of the force carriers of high interest that researchers hope to detect is the Higgs boson. The observation of its properties might tell us why different particles have different masses. Figure 4.28 The world\u2019s largest particle accelerator spans the border between Switzerland and France. Two beams, traveling in opposite directions close to the speed of light, collide in a tube similar to the central tube shown here. External magnets determine the beam\u2019s path. Special detectors will analyze particles created in these collisions. Questions as broad as what is the origin of mass and what was matter like the first few seconds of our universe will be explored. This accelerator began preliminary operation in 2008. (credit: Frank Hommes) 178 Chapter 4 | Dynamics: Force and Newton's Laws of Motion Tiny particles also have wave-like behavior, something we will explore more in a later chapter. To better understand", " force-carrier particles from another perspective, let us consider gravity. The search for gravitational waves has been going on for a number of years. Almost 100 years ago, Einstein predicted the existence of these waves as part of his general theory of relativity. Gravitational waves are created during the collision of massive stars, in black holes, or in supernova explosions\u2014like shock waves. These gravitational waves will travel through space from such sites much like a pebble dropped into a pond sends out ripples\u2014except these waves move at the speed of light. A detector apparatus has been built in the U.S., consisting of two large installations nearly 3000 km apart\u2014one in Washington state and one in Louisiana! The facility is called the Laser Interferometer Gravitational-Wave Observatory (LIGO). Each installation is designed to use optical lasers to examine any slight shift in the relative positions of two masses due to the effect of gravity waves. The two sites allow simultaneous measurements of these small effects to be separated from other natural phenomena, such as earthquakes. Initial operation of the detectors began in 2002, and work is proceeding on increasing their sensitivity. Similar installations have been built in Italy (VIRGO), Germany (GEO600), and Japan (TAMA300) to provide a worldwide network of gravitational wave detectors. International collaboration in this area is moving into space with the joint EU/US project LISA (Laser Interferometer Space Antenna). Earthquakes and other Earthly noises will be no problem for these monitoring spacecraft. LISA will complement LIGO by looking at much more massive black holes through the observation of gravitational-wave sources emitting much larger wavelengths. Three satellites will be placed in space above Earth in an equilateral triangle (with 5,000,000-km sides) (Figure 4.29). The system will measure the relative positions of each satellite to detect passing gravitational waves. Accuracy to within 10% of the size of an atom will be needed to detect any waves. The launch of this project might be as early as 2018. \u201cI\u2019m sure LIGO will tell us something about the universe that we didn\u2019t know before. The history of science tells us that any time you go where you haven\u2019t been before, you usually find something that really shakes the scientific paradigms of the day. Whether gravitational wave astrophysics will do that, only time will tell.\u201d \u2014David Reitze, LIGO Input Optics Manager, University of Florida Figure 4.29 Space-based future experiments for the measurement of gravitational waves. Shown here is a drawing of LISA\u2019s orbit. Each satellite of LISA will consist of a laser source and a mass. The lasers will transmit a signal to measure the distance between each satellite\u2019s test mass. The relative motion of these masses will provide information about passing gravitational waves. (credit: NASA) The ideas presented in this section are but a glimpse into topics of modern physics that will be covered in much greater depth in later chapters. Glossary acceleration: the rate at which an object\u2019s velocity changes over a period of time carrier particle: a fundamental particle of nature that is surrounded by a characteristic force field; photons are carrier particles of the electromagnetic force dynamics: the study of how forces affect the motion of objects and systems external force: a force acting on an object or system that originates outside of the object or system force: a push or pull on an object with a specific magnitude and direction; can be represented by vectors; can be expressed as a multiple of a standard force force field: a region in which a test particle will experience a force free-body diagram: a sketch showing all of the external forces acting on an object or system; the system is represented by a dot, and the forces are represented by vectors extending outward from the dot free-fall: a situation in which the only force acting on an object is the force due to gravity friction: a force past each other of objects that are touching; examples include rough surfaces and air resistance This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 4 | Dynamics: Force and Newton's Laws of Motion 179 inertia: the tendency of an object to remain at rest or remain in motion inertial frame of reference: a coordinate system that is not accelerating; all forces acting in an inertial frame of reference are real forces, as opposed to fictitious forces that are observed due to an accelerating frame of reference law of inertia: see Newton\u2019s first law of motion mass: the quantity of matter in a substance; measured in kilograms net external force: the vector sum of all external forces acting on an object or system; causes a mass to accelerate Newton\u2019s first law of motion: in an inertial frame of reference, a body at rest remains at rest, or, if in motion, remains in motion at a constant velocity unless acted on by a net external force; also known as the law of inertia", " Newton\u2019s second law of motion: the net external force Fnet on an object with mass is proportional to and in the same direction as the acceleration of the object, a, and inversely proportional to the mass; defined mathematically as a = Fnet Newton\u2019s third law of motion: whenever one body exerts a force on a second body, the first body experiences a force that is equal in magnitude and opposite in direction to the force that the first body exerts normal force: the force that a surface applies to an object to support the weight of the object; acts perpendicular to the surface on which the object rests system: defined by the boundaries of an object or collection of objects being observed; all forces originating from outside of the system are considered external forces tension: the pulling force that acts along a medium, especially a stretched flexible connector, such as a rope or cable; when a rope supports the weight of an object, the force on the object due to the rope is called a tension force thrust: a reaction force that pushes a body forward in response to a backward force; rockets, airplanes, and cars are pushed forward by a thrust reaction force weight: the force w due to gravity acting on an object of mass ; defined mathematically as: w g, where g is the magnitude and direction of the acceleration due to gravity Section Summary 4.1 Development of Force Concept \u2022 Dynamics is the study of how forces affect the motion of objects. \u2022 Force is a push or pull that can be defined in terms of various standards, and it is a vector having both magnitude and direction. \u2022 External forces are any outside forces that act on a body. A free-body diagram is a drawing of all external forces acting on a body. 4.2 Newton's First Law of Motion: Inertia \u2022 Newton\u2019s first law of motion states that in an inertial frame of reference a body at rest remains at rest, or, if in motion, remains in motion at a constant velocity unless acted on by a net external force. This is also known as the law of inertia. Inertia is the tendency of an object to remain at rest or remain in motion. Inertia is related to an object\u2019s mass. \u2022 \u2022 Mass is the quantity of matter in a substance. 4.3 Newton's Second Law of Motion: Concept of a System \u2022 Acceleration, a, is defined as a change in velocity, meaning a change in its magnitude or direction, or both. \u2022 An external force is one acting on a system from outside the system, as opposed to internal forces, which act between components within the system. \u2022 Newton\u2019s second law of motion states that the acceleration of a system is directly proportional to and in the same direction as the net external force acting on the system, and inversely proportional to its mass. \u2022 In equation form, Newton\u2019s second law of motion is a = Fnet. \u2022 This is often written in the more familiar form: Fnet = a. \u2022 The weight w of an object is defined as the force of gravity acting on an object of mass. The object experiences an acceleration due to gravity g : \u2022 If the only force acting on an object is due to gravity, the object is in free fall. w = g. 180 Chapter 4 | Dynamics: Force and Newton's Laws of Motion \u2022 Friction is a force that opposes the motion past each other of objects that are touching. 4.4 Newton's Third Law of Motion: Symmetry in Forces \u2022 Newton\u2019s third law of motion represents a basic symmetry in nature. It states: Whenever one body exerts a force on a second body, the first body experiences a force that is equal in magnitude and opposite in direction to the force that the first body exerts. \u2022 A thrust is a reaction force that pushes a body forward in response to a backward force. Rockets, airplanes, and cars are pushed forward by a thrust reaction force. 4.5 Normal, Tension, and Other Examples of Force \u2022 When objects rest on a surface, the surface applies a force to the object that supports the weight of the object. This supporting force acts perpendicular to and away from the surface. It is called a normal force, N. \u2022 When objects rest on a non-accelerating horizontal surface, the magnitude of the normal force is equal to the weight of the object: \u2022 When objects rest on an inclined plane that makes an angle with the horizontal surface, the weight of the object can be resolved into components that act perpendicular ( w\u22a5 ) and parallel ( w \u2225 ) to the surface of the plane. These components can be calculated using: =. \u2022 The pulling force that acts along a stretched flexible connector, such as a rope or cable, is called tension, T. When a rope \u2225 = sin () = sin () \u22a5 = cos () = cos (). supports the weight of an object that is at rest, the tension in the rope is equal to the weight of the object: \u2022", " =. In any inertial frame of reference (one that is not accelerated or rotated), Newton\u2019s laws have the simple forms given in this chapter and all forces are real forces having a physical origin. 4.6 Problem-Solving Strategies \u2022 To solve problems involving Newton\u2019s laws of motion, follow the procedure described: 1. Draw a sketch of the problem. 2. Identify known and unknown quantities, and identify the system of interest. Draw a free-body diagram, which is a sketch showing all of the forces acting on an object. The object is represented by a dot, and the forces are represented by vectors extending in different directions from the dot. If vectors act in directions that are not horizontal or vertical, resolve the vectors into horizontal and vertical components and draw them on the free-body diagram. 3. Write Newton\u2019s second law in the horizontal and vertical directions and add the forces acting on the object. If the object does not accelerate in a particular direction (for example, the -direction) then net = 0. If the object does accelerate in that direction, net =. 4. Check your answer. Is the answer reasonable? Are the units correct? 4.7 Further Applications of Newton's Laws of Motion \u2022 Newton\u2019s laws of motion can be applied in numerous situations to solve problems of motion. \u2022 Some problems will contain multiple force vectors acting in different directions on an object. Be sure to draw diagrams, resolve all force vectors into horizontal and vertical components, and draw a free-body diagram. Always analyze the direction in which an object accelerates so that you can determine whether net = or net = 0. \u2022 The normal force on an object is not always equal in magnitude to the weight of the object. If an object is accelerating, the normal force will be less than or greater than the weight of the object. Also, if the object is on an inclined plane, the normal force will always be less than the full weight of the object. \u2022 Some problems will contain various physical quantities, such as forces, acceleration, velocity, or position. You can apply concepts from kinematics and dynamics in order to solve these problems of motion. 4.8 Extended Topic: The Four Basic Forces\u2014An Introduction \u2022 The various types of forces that are categorized for use in many applications are all manifestations of the four basic forces in nature. \u2022 The properties of these forces are summarized in Table 4.2. \u2022 Everything we experience directly without sensitive instruments is due to either electromagnetic forces or gravitational forces. The nuclear forces are responsible for the submicroscopic structure of matter, but they are not directly sensed because of their short ranges. Attempts are being made to show all four forces are different manifestations of a single unified force. \u2022 A force field surrounds an object creating a force and is the carrier of that force. This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 4 | Dynamics: Force and Newton's Laws of Motion 181 Conceptual Questions 4.1 Development of Force Concept 1. Propose a force standard different from the example of a stretched spring discussed in the text. Your standard must be capable of producing the same force repeatedly. 2. What properties do forces have that allow us to classify them as vectors? 4.2 Newton's First Law of Motion: Inertia 3. How are inertia and mass related? 4. What is the relationship between weight and mass? Which is an intrinsic, unchanging property of a body? 4.3 Newton's Second Law of Motion: Concept of a System 5. Which statement is correct? (a) Net force causes motion. (b) Net force causes change in motion. Explain your answer and give an example. 6. Why can we neglect forces such as those holding a body together when we apply Newton\u2019s second law of motion? 7. Explain how the choice of the \u201csystem of interest\u201d affects which forces must be considered when applying Newton\u2019s second law of motion. 8. Describe a situation in which the net external force on a system is not zero, yet its speed remains constant. 9. A system can have a nonzero velocity while the net external force on it is zero. Describe such a situation. 10. A rock is thrown straight up. What is the net external force acting on the rock when it is at the top of its trajectory? 11. (a) Give an example of different net external forces acting on the same system to produce different accelerations. (b) Give an example of the same net external force acting on systems of different masses, producing different accelerations. (c) What law accurately describes both effects? State it in words and as an equation. 12. If the acceleration of a system is zero, are no external forces acting on it? What about internal forces? Explain your answers. 13. If a constant, nonzero force is applied to an object, what can you say about the velocity and acceleration of the", " object? 14. The gravitational force on the basketball in Figure 4.6 is ignored. When gravity is taken into account, what is the direction of the net external force on the basketball\u2014above horizontal, below horizontal, or still horizontal? 4.4 Newton's Third Law of Motion: Symmetry in Forces 15. When you take off in a jet aircraft, there is a sensation of being pushed back into the seat. Explain why you move backward in the seat\u2014is there really a force backward on you? (The same reasoning explains whiplash injuries, in which the head is apparently thrown backward.) 16. A device used since the 1940s to measure the kick or recoil of the body due to heart beats is the \u201cballistocardiograph.\u201d What physics principle(s) are involved here to measure the force of cardiac contraction? How might we construct such a device? 17. Describe a situation in which one system exerts a force on another and, as a consequence, experiences a force that is equal in magnitude and opposite in direction. Which of Newton\u2019s laws of motion apply? 18. Why does an ordinary rifle recoil (kick backward) when fired? The barrel of a recoilless rifle is open at both ends. Describe how Newton\u2019s third law applies when one is fired. Can you safely stand close behind one when it is fired? 19. An American football lineman reasons that it is senseless to try to out-push the opposing player, since no matter how hard he pushes he will experience an equal and opposite force from the other player. Use Newton\u2019s laws and draw a free-body diagram of an appropriate system to explain how he can still out-push the opposition if he is strong enough. 20. Newton\u2019s third law of motion tells us that forces always occur in pairs of equal and opposite magnitude. Explain how the choice of the \u201csystem of interest\u201d affects whether one such pair of forces cancels. 4.5 Normal, Tension, and Other Examples of Force 21. If a leg is suspended by a traction setup as shown in Figure 4.30, what is the tension in the rope? 182 Chapter 4 | Dynamics: Force and Newton's Laws of Motion Figure 4.30 A leg is suspended by a traction system in which wires are used to transmit forces. Frictionless pulleys change the direction of the force T without changing its magnitude. 22. In a traction setup for a broken bone, with pulleys and rope available, how might we be able to increase the force along the tibia using the same weight? (See Figure 4.30.) (Note that the tibia is the shin bone shown in this image.) 4.7 Further Applications of Newton's Laws of Motion 23. To simulate the apparent weightlessness of space orbit, astronauts are trained in the hold of a cargo aircraft that is accelerating downward at. Why will they appear to be weightless, as measured by standing on a bathroom scale, in this accelerated frame of reference? Is there any difference between their apparent weightlessness in orbit and in the aircraft? 24. A cartoon shows the toupee coming off the head of an elevator passenger when the elevator rapidly stops during an upward ride. Can this really happen without the person being tied to the floor of the elevator? Explain your answer. 4.8 Extended Topic: The Four Basic Forces\u2014An Introduction 25. Explain, in terms of the properties of the four basic forces, why people notice the gravitational force acting on their bodies if it is such a comparatively weak force. 26. What is the dominant force between astronomical objects? Why are the other three basic forces less significant over these very large distances? 27. Give a detailed example of how the exchange of a particle can result in an attractive force. (For example, consider one child pulling a toy out of the hands of another.) This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 4 | Dynamics: Force and Newton's Laws of Motion 183 Problems & Exercises 4.3 Newton's Second Law of Motion: Concept of a System You may assume data taken from illustrations is accurate to three digits. 1. A 63.0-kg sprinter starts a race with an acceleration of 4.20 m/s2. What is the net external force on him? Figure 4.32 2. If the sprinter from the previous problem accelerates at that rate for 20 m, and then maintains that velocity for the remainder of the 100-m dash, what will be his time for the race? 8. What is the deceleration of the rocket sled if it comes to rest in 1.1 s from a speed of 1000 km/h? (Such deceleration caused one test subject to black out and have temporary blindness.) 3. A cleaner pushes a 4.50-kg laundry cart in such a way that the net external force on it is 60.0 N. Calculate the magnitude", " of its acceleration. 4. Since astronauts in orbit are apparently weightless, a clever method of measuring their masses is needed to monitor their mass gains or losses to adjust diets. One way to do this is to exert a known force on an astronaut and measure the acceleration produced. Suppose a net external force of 50.0 N is exerted and the astronaut\u2019s acceleration is measured to be 0.893 m/s2. (a) Calculate her mass. (b) By exerting a force on the astronaut, the vehicle in which they orbit experiences an equal and opposite force. Discuss how this would affect the measurement of the astronaut\u2019s acceleration. Propose a method in which recoil of the vehicle is avoided. 5. In Figure 4.7, the net external force on the 24-kg mower is stated to be 51 N. If the force of friction opposing the motion is 24 N, what force (in newtons) is the person exerting on the mower? Suppose the mower is moving at 1.5 m/s when the force is removed. How far will the mower go before stopping? 6. The same rocket sled drawn in Figure 4.31 is decelerated at a rate of 196 m/s2. What force is necessary to produce this deceleration? Assume that the rockets are off. The mass of the system is 2100 kg. 9. Suppose two children push horizontally, but in exactly opposite directions, on a third child in a wagon. The first child exerts a force of 75.0 N, the second a force of 90.0 N, friction is 12.0 N, and the mass of the third child plus wagon is 23.0 kg. (a) What is the system of interest if the acceleration of the child in the wagon is to be calculated? (b) Draw a free-body diagram, including all forces acting on the system. (c) Calculate the acceleration. (d) What would the acceleration be if friction were 15.0 N? 10. A powerful motorcycle can produce an acceleration of 3.50 m/s2 while traveling at 90.0 km/h. At that speed the forces resisting motion, including friction and air resistance, total 400 N. (Air resistance is analogous to air friction. It always opposes the motion of an object.) What is the magnitude of the force the motorcycle exerts backward on the ground to produce its acceleration if the mass of the motorcycle with rider is 245 kg? 11. The rocket sled shown in Figure 4.33 accelerates at a rate of 49.0 m/s2. Its passenger has a mass of 75.0 kg. (a) Calculate the horizontal component of the force the seat exerts against his body. Compare this with his weight by using a ratio. (b) Calculate the direction and magnitude of the total force the seat exerts against his body. Figure 4.31 7. (a) If the rocket sled shown in Figure 4.32 starts with only one rocket burning, what is the magnitude of its acceleration? Assume that the mass of the system is 2100 kg, the thrust T is 2.4\u00d7104 N, and the force of friction opposing the motion is known to be 650 N. (b) Why is the acceleration not onefourth of what it is with all rockets burning? Figure 4.33 12. Repeat the previous problem for the situation in which the rocket sled decelerates at a rate of 201 m/s2. In this problem, the forces are exerted by the seat and restraining belts. 13. The weight of an astronaut plus his space suit on the Moon is only 250 N. How much do they weigh on Earth? What is the mass on the Moon? On Earth? 14. Suppose the mass of a fully loaded module in which astronauts take off from the Moon is 10,000 kg. The thrust of its engines is 30,000 N. (a) Calculate its the magnitude of acceleration in a vertical takeoff from the Moon. (b) Could it lift off from Earth? If not, why not? If it could, calculate the magnitude of its acceleration. 184 Chapter 4 | Dynamics: Force and Newton's Laws of Motion 4.4 Newton's Third Law of Motion: Symmetry in Forces 15. What net external force is exerted on a 1100-kg artillery shell fired from a battleship if the shell is accelerated at 2.40104 m/s2? What is the magnitude of the force exerted on the ship by the artillery shell? 16. A brave but inadequate rugby player is being pushed backward by an opposing player who is exerting a force of 800 N on him. The mass of the losing player plus equipment is 90.0 kg, and he is accelerating at 1.20 m/s2 backward. (a) What is the force of friction between the losing player\u2019s feet and the grass? (b) What force does the winning player exert on the ground to", " move forward if his mass plus equipment is 110 kg? (c) Draw a sketch of the situation showing the system of interest used to solve each part. For this situation, draw a free-body diagram and write the net force equation. 4.5 Normal, Tension, and Other Examples of Force 17. Two teams of nine members each engage in a tug of war. Each of the first team\u2019s members has an average mass of 68 kg and exerts an average force of 1350 N horizontally. Each of the second team\u2019s members has an average mass of 73 kg and exerts an average force of 1365 N horizontally. (a) What is magnitude of the acceleration of the two teams? (b) What is the tension in the section of rope between the teams? 18. What force does a trampoline have to apply to a 45.0-kg gymnast to accelerate her straight up at 7.50 m/s2? Note that the answer is independent of the velocity of the gymnast\u2014she can be moving either up or down, or be stationary. 19. (a) Calculate the tension in a vertical strand of spider web if a spider of mass 8.00\u00d710\u22125 kg hangs motionless on it. (b) Calculate the tension in a horizontal strand of spider web if the same spider sits motionless in the middle of it much like the tightrope walker in Figure 4.17. The strand sags at an angle of 12\u00ba below the horizontal. Compare this with the tension in the vertical strand (find their ratio). 20. Suppose a 60.0-kg gymnast climbs a rope. (a) What is the tension in the rope if he climbs at a constant speed? (b) What is the tension in the rope if he accelerates upward at a rate of 1.50 m/s2? 21. Show that, as stated in the text, a force F\u22a5 exerted on a flexible medium at its center and perpendicular to its length (such as on the tightrope wire in Figure 4.17) gives rise to a \u22a5 2 sin () tension of magnitude =. 22. Consider the baby being weighed in Figure 4.34. (a) What is the mass of the child and basket if a scale reading of 55 N is observed? (b) What is the tension 1 in the cord attaching the baby to the scale? (c) What is the tension 2 in the cord attaching the scale to the ceiling, if the scale has a mass of 0.500 kg? (d) Draw a sketch of the situation indicating the system of interest used to solve each part. The masses of the cords are negligible. This content is available for free at http://cnx.org/content/col11844/1.13 Figure 4.34 A baby is weighed using a spring scale. 4.6 Problem-Solving Strategies 23. A 5.00\u00d7105-kg rocket is accelerating straight up. Its engines produce 1.250\u00d7107 N of thrust, and air resistance is 4.50\u00d7106 N. What is the rocket\u2019s acceleration? Explicitly show how you follow the steps in the Problem-Solving Strategy for Newton\u2019s laws of motion. 24. The wheels of a midsize car exert a force of 2100 N backward on the road to accelerate the car in the forward direction. If the force of friction including air resistance is 250 N and the acceleration of the car is 1.80 m/s2, what is the mass of the car plus its occupants? Explicitly show how you follow the steps in the Problem-Solving Strategy for Newton\u2019s laws of motion. For this situation, draw a free-body diagram and write the net force equation. 25. Calculate the force a 70.0-kg high jumper must exert on the ground to produce an upward acceleration 4.00 times the acceleration due to gravity. Explicitly show how you follow the steps in the Problem-Solving Strategy for Newton\u2019s laws of motion. 26. When landing after a spectacular somersault, a 40.0-kg gymnast decelerates by pushing straight down on the mat. Calculate the force she must exert if her deceleration is 7.00 times the acceleration due to gravity. Explicitly show how you follow the steps in the Problem-Solving Strategy for Newton\u2019s laws of motion. Chapter 4 | Dynamics: Force and Newton's Laws of Motion 185 27. A freight train consists of two 8.00\u00d7104 -kg engines and 45 cars with average masses of 5.50\u00d7104 kg. (a) What force must each engine exert backward on the track to accelerate the train at a rate of 5.00\u00d710\u20132 m/s2 if the force of friction is 7.50\u00d7105 N, assuming the engines exert identical forces? This is not a large frictional force for such a massive system", ". Rolling friction for trains is small, and consequently trains are very energy-efficient transportation systems. (b) What is the force in the coupling between the 37th and 38th cars (this is the force each exerts on the other), assuming all cars have the same mass and that friction is evenly distributed among all of the cars and engines? 28. Commercial airplanes are sometimes pushed out of the passenger loading area by a tractor. (a) An 1800-kg tractor exerts a force of 1.75\u00d7104 N backward on the pavement, and the system experiences forces resisting motion that total 2400 N. If the acceleration is 0.150 m/s2, what is the mass of the airplane? (b) Calculate the force exerted by the tractor on the airplane, assuming 2200 N of the friction is experienced by the airplane. (c) Draw two sketches showing the systems of interest used to solve each part, including the free-body diagrams for each. 29. A 1100-kg car pulls a boat on a trailer. (a) What total force resists the motion of the car, boat, and trailer, if the car exerts a 1900-N force on the road and produces an acceleration of 0.550 m/s2? The mass of the boat plus trailer is 700 kg. (b) What is the force in the hitch between the car and the trailer if 80% of the resisting forces are experienced by the boat and trailer? 30. (a) Find the magnitudes of the forces F1 and F2 that add to give the total force Ftot shown in Figure 4.35. This may be done either graphically or by using trigonometry. (b) Show graphically that the same total force is obtained independent of the order of addition of F1 and F2. (c) Find the direction and magnitude of some other pair of vectors that add to give Ftot. Draw these to scale on the same drawing used in part (b) or a similar picture. Figure 4.35 31. Two children pull a third child on a snow saucer sled exerting forces F1 and F2 as shown from above in Figure 4.36. Find the acceleration of the 49.00-kg sled and child system. Note that the direction of the frictional force is unspecified; it will be in the opposite direction of the sum of F1 and F2. Figure 4.36 An overhead view of the horizontal forces acting on a child\u2019s snow saucer sled. 32. Suppose your car was mired deeply in the mud and you wanted to use the method illustrated in Figure 4.37 to pull it out. (a) What force would you have to exert perpendicular to the center of the rope to produce a force of 12,000 N on the car if the angle is 2.00\u00b0? In this part, explicitly show how you follow the steps in the Problem-Solving Strategy for Newton\u2019s laws of motion. (b) Real ropes stretch under such forces. What force would be exerted on the car if the angle increases to 7.00\u00b0 and you still apply the force found in part (a) to its center? Figure 4.37 33. What force is exerted on the tooth in Figure 4.38 if the tension in the wire is 25.0 N? Note that the force applied to the tooth is smaller than the tension in the wire, but this is necessitated by practical considerations of how force can be applied in the mouth. Explicitly show how you follow steps in the Problem-Solving Strategy for Newton\u2019s laws of motion. Figure 4.38 Braces are used to apply forces to teeth to realign them. Shown in this figure are the tensions applied by the wire to the protruding tooth. The total force applied to the tooth by the wire, Fapp, points straight toward the back of the mouth. 34. Figure 4.39 shows Superhero and Trusty Sidekick hanging motionless from a rope. Superhero\u2019s mass is 90.0 kg, while Trusty Sidekick\u2019s is 55.0 kg, and the mass of the rope is negligible. (a) Draw a free-body diagram of the situation showing all forces acting on Superhero, Trusty Sidekick, and the rope. (b) Find the tension in the rope above Superhero. (c) Find the tension in the rope between 186 Chapter 4 | Dynamics: Force and Newton's Laws of Motion Superhero and Trusty Sidekick. Indicate on your free-body diagram the system of interest used to solve each part. is unreasonable about the result? (c) Which premise is unreasonable, and why is it unreasonable? 39. Unreasonable Results (a) What is the initial acceleration of a rocket that has a mass of 1.50\u00d7106 kg at takeoff, the engines of which produce a thrust of 2.00\u00d7106 N? Do not neglect gravity. (b) What is unreasonable about the", " result? (This result has been unintentionally achieved by several real rockets.) (c) Which premise is unreasonable, or which premises are inconsistent? (You may find it useful to compare this problem to the rocket problem earlier in this section.) 4.7 Further Applications of Newton's Laws of Motion 40. A flea jumps by exerting a force of 1.20\u00d710\u22125 N straight down on the ground. A breeze blowing on the flea parallel to the ground exerts a force of 0.500\u00d710\u22126 N on the flea. Find the direction and magnitude of the acceleration of the flea if its mass is 6.00\u00d710\u22127 kg. Do not neglect the gravitational force. 41. Two muscles in the back of the leg pull upward on the Achilles tendon, as shown in Figure 4.40. (These muscles are called the medial and lateral heads of the gastrocnemius muscle.) Find the magnitude and direction of the total force on the Achilles tendon. What type of movement could be caused by this force? Figure 4.40 Achilles tendon 42. A 76.0-kg person is being pulled away from a burning building as shown in Figure 4.41. Calculate the tension in the two ropes if the person is momentarily motionless. Include a free-body diagram in your solution. Figure 4.39 Superhero and Trusty Sidekick hang motionless on a rope as they try to figure out what to do next. Will the tension be the same everywhere in the rope? 35. A nurse pushes a cart by exerting a force on the handle at a downward angle 35.0\u00ba below the horizontal. The loaded cart has a mass of 28.0 kg, and the force of friction is 60.0 N. (a) Draw a free-body diagram for the system of interest. (b) What force must the nurse exert to move at a constant velocity? 36. Construct Your Own Problem Consider the tension in an elevator cable during the time the elevator starts from rest and accelerates its load upward to some cruising velocity. Taking the elevator and its load to be the system of interest, draw a free-body diagram. Then calculate the tension in the cable. Among the things to consider are the mass of the elevator and its load, the final velocity, and the time taken to reach that velocity. 37. Construct Your Own Problem Consider two people pushing a toboggan with four children on it up a snowcovered slope. Construct a problem in which you calculate the acceleration of the toboggan and its load. Include a freebody diagram of the appropriate system of interest as the basis for your analysis. Show vector forces and their components and explain the choice of coordinates. Among the things to be considered are the forces exerted by those pushing, the angle of the slope, and the masses of the toboggan and children. 38. Unreasonable Results (a) Repeat Exercise 4.29, but assume an acceleration of 1.20 m/s2 is produced. (b) What This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 4 | Dynamics: Force and Newton's Laws of Motion 187 average force on the shell in the mortar? Express your answer in newtons and as a ratio to the weight of the shell. 48. Integrated Concepts Repeat Exercise 4.47 for a shell fired at an angle 10.0\u00ba from the vertical. 49. Integrated Concepts An elevator filled with passengers has a mass of 1700 kg. (a) The elevator accelerates upward from rest at a rate of 1.20 m/s2 for 1.50 s. Calculate the tension in the cable supporting the elevator. (b) The elevator continues upward at constant velocity for 8.50 s. What is the tension in the cable during this time? (c) The elevator decelerates at a rate of 0.600 m/s2 for 3.00 s. What is the tension in the cable during deceleration? (d) How high has the elevator moved above its original starting point, and what is its final velocity? 50. Unreasonable Results (a) What is the final velocity of a car originally traveling at 50.0 km/h that decelerates at a rate of 0.400 m/s2 for 50.0 s? (b) What is unreasonable about the result? (c) Which premise is unreasonable, or which premises are inconsistent? 51. Unreasonable Results A 75.0-kg man stands on a bathroom scale in an elevator that accelerates from rest to 30.0 m/s in 2.00 s. (a) Calculate the scale reading in newtons and compare it with his weight. (The scale exerts an upward force on him equal to its reading.) (b) What is unreasonable about the result? (c) Which premise is unreasonable, or which premises are inconsistent? 4.8 Extended Topic: The Four Basic Forces\u2014An Introduction", " 52. (a) What is the strength of the weak nuclear force relative to the strong nuclear force? (b) What is the strength of the weak nuclear force relative to the electromagnetic force? Since the weak nuclear force acts at only very short distances, such as inside nuclei, where the strong and electromagnetic forces also act, it might seem surprising that we have any knowledge of it at all. We have such knowledge because the weak nuclear force is responsible for beta decay, a type of nuclear decay not explained by other forces. 53. (a) What is the ratio of the strength of the gravitational force to that of the strong nuclear force? (b) What is the ratio of the strength of the gravitational force to that of the weak nuclear force? (c) What is the ratio of the strength of the gravitational force to that of the electromagnetic force? What do your answers imply about the influence of the gravitational force on atomic nuclei? 54. What is the ratio of the strength of the strong nuclear force to that of the electromagnetic force? Based on this ratio, you might expect that the strong force dominates the nucleus, which is true for small nuclei. Large nuclei, however, have sizes greater than the range of the strong nuclear force. At these sizes, the electromagnetic force begins to affect nuclear stability. These facts will be used to explain nuclear fusion and fission later in this text. Figure 4.41 The force T2 needed to hold steady the person being rescued from the fire is less than her weight and less than the force T1 in the other rope, since the more vertical rope supports a greater part of her weight (a vertical force). 43. Integrated Concepts A 35.0-kg dolphin decelerates from 12.0 to 7.50 m/s in 2.30 s to join another dolphin in play. What average force was exerted to slow him if he was moving horizontally? (The gravitational force is balanced by the buoyant force of the water.) 44. Integrated Concepts When starting a foot race, a 70.0-kg sprinter exerts an average force of 650 N backward on the ground for 0.800 s. (a) What is his final speed? (b) How far does he travel? 45. Integrated Concepts A large rocket has a mass of 2.00\u00d7106 kg at takeoff, and its engines produce a thrust of 3.50\u00d7107 N. (a) Find its initial acceleration if it takes off vertically. (b) How long does it take to reach a velocity of 120 km/h straight up, assuming constant mass and thrust? (c) In reality, the mass of a rocket decreases significantly as its fuel is consumed. Describe qualitatively how this affects the acceleration and time for this motion. 46. Integrated Concepts A basketball player jumps straight up for a ball. To do this, he lowers his body 0.300 m and then accelerates through this distance by forcefully straightening his legs. This player leaves the floor with a vertical velocity sufficient to carry him 0.900 m above the floor. (a) Calculate his velocity when he leaves the floor. (b) Calculate his acceleration while he is straightening his legs. He goes from zero to the velocity found in part (a) in a distance of 0.300 m. (c) Calculate the force he exerts on the floor to do this, given that his mass is 110 kg. 47. Integrated Concepts A 2.50-kg fireworks shell is fired straight up from a mortar and reaches a height of 110 m. (a) Neglecting air resistance (a poor assumption, but we will make it for this example), calculate the shell\u2019s velocity when it leaves the mortar. (b) The mortar itself is a tube 0.450 m long. Calculate the average acceleration of the shell in the tube as it goes from zero to the velocity found in (a). (c) What is the 188 Chapter 4 | Dynamics: Force and Newton's Laws of Motion Test Prep for AP\u00ae Courses 4.1 Development of Force Concept 1. Figure 4.42 The figure above represents a racetrack with semicircular sections connected by straight sections. Each section has length d, and markers along the track are spaced d/4 apart. Two people drive cars counterclockwise around the track, as shown. Car X goes around the curves at constant speed vc, increases speed at constant acceleration for half of each straight section to reach a maximum speed of 2vc, then brakes at constant acceleration for the other half of each straight section to return to speed vc. Car Y also goes around the curves at constant speed vc, increases its speed at constant acceleration for one-fourth of each straight section to reach the same maximum speed 2vc, stays at that speed for half of each straight section, then brakes at constant acceleration for the remaining fourth of each straight section to return to speed vc. (a) On the figures below, draw an arrow showing the direction of", " the net force on each of the cars at the positions noted by the dots. If the net force is zero at any position, label the dot with 0. Figure 4.43 The position of the six dots on the Car Y track on the right are as follows: The first dot on the left center of the track is at the same position as it is on the Car X track. The second dot is just slight to the right of the Car X dot (less than a dash) past three perpendicular hash marks moving to the right. The third dot is about one and two-thirds perpendicular hash marks to the right of the center top perpendicular has mark. The fourth dot is in the same position as the Car X figure (one perpendicular hash mark above the center right perpendicular hash mark). The fifth dot is about one and two-third perpendicular hash marks to the right of the center bottom perpendicular hash mark. The sixth dot is in the same position as the Car Y dot (one and two third perpendicular hash marks to the left of the center bottom hash mark). (b) i. Indicate which car, if either, completes one trip around the track in less time, and justify your answer qualitatively without using equations. This content is available for free at http://cnx.org/content/col11844/1.13 ii. Justify your answer about which car, if either, completes one trip around the track in less time quantitatively with appropriate equations. 2. Which of the following is an example of a body exerting a force on itself? a. a person standing up from a seated position b. a car accelerating while driving c. both of the above d. none of the above 3. A hawk accelerates as it glides in the air. Does the force causing the acceleration come from the hawk itself? Explain. 4. What causes the force that moves a boat forward when someone rows it? a. The force is caused by the rower\u2019s arms. b. The force is caused by an interaction between the oars and gravity. c. The force is caused by an interaction between the oars and the water the boat is traveling in. d. The force is caused by friction. 4.4 Newton's Third Law of Motion: Symmetry in Forces 5. What object or objects commonly exert forces on the following objects in motion? (a) a soccer ball being kicked, (b) a dolphin jumping, (c) a parachutist drifting to Earth. 6. A ball with a mass of 0.25 kg hits a gym ceiling with a force of 78.0 N. What happens next? a. The ball accelerates downward with a force of 80.5 N. b. The ball accelerates downward with a force of 78.0 N. c. The ball accelerates downward with a force of 2.45 N. d. It depends on the height of the ceiling. 7. Which of the following is true? a. Earth exerts a force due to gravity on your body, and your body exerts a smaller force on the Earth, because your mass is smaller than the mass of the Earth. b. The Moon orbits the Earth because the Earth exerts a force on the Moon and the Moon exerts a force equal in magnitude and direction on the Earth. c. A rocket taking off exerts a force on the Earth equal to the force the Earth exerts on the rocket. d. An airplane cruising at a constant speed is not affected by gravity. 8. Stationary skater A pushes stationary skater B, who then accelerates at 5.0 m/s2. Skater A does not move. Since forces act in action-reaction pairs, explain why Skater A did not move? 9. The current in a river exerts a force of 9.0 N on a balloon floating in the river. A wind exerts a force of 13.0 N on the balloon in the opposite direction. Draw a free-body diagram to show the forces acting on the balloon. Use your free-body diagram to predict the effect on the balloon. 10. A force is applied to accelerate an object on a smooth icy surface. When the force stops, which of the following will be true? (Assume zero friction.) a. The object\u2019s acceleration becomes zero. b. The object\u2019s speed becomes zero. c. The object\u2019s acceleration continues to increase at a constant rate. d. The object accelerates, but in the opposite direction. 11. A parachutist\u2019s fall to Earth is determined by two opposing forces. A gravitational force of 539 N acts on the parachutist. After 2 s, she opens her parachute and experiences an air resistance of 615 N. At what speed is the parachutist falling after 10 s? Chapter 4 | Dynamics: Force and Newton's Laws of Motion 189 12. A flight attendant pushes a cart down the aisle of a plane in flight. In", " determining the acceleration of the cart relative to the plane, which factor do you not need to consider? a. The friction of the cart\u2019s wheels. b. The force with which the flight attendant\u2019s feet push on the floor. c. The velocity of the plane. d. The mass of the items in the cart. 13. A landscaper is easing a wheelbarrow full of soil down a hill. Define the system you would analyze and list all the forces that you would need to include to calculate the acceleration of the wheelbarrow. 14. Two water-skiers, with masses of 48 kg and 61 kg, are preparing to be towed behind the same boat. When the boat accelerates, the rope the skiers hold onto accelerates with it and exerts a net force of 290 N on the skiers. At what rate will the skiers accelerate? a. 10.8 m/s2 b. 2.7 m/s2 c. 6.0 m/s2 and 4.8 m/s2 d. 5.3 m/s2 15. A figure skater has a mass of 40 kg and her partner's mass is 50 kg. She pushes against the ice with a force of 120 N, causing her and her partner to move forward. Calculate the pair\u2019s acceleration. Assume that all forces opposing the motion, such as friction and air resistance, total 5.0 N. 4.5 Normal, Tension, and Other Examples of Force 16. An archer shoots an arrow straight up with a force of 24.5 N. The arrow has a mass of 0.4 kg. What is the force of gravity on the arrow? a. 9.8 m/s2 b. 9.8 N c. 61.25 N d. 3.9 N 17. A cable raises a mass of 120.0 kg with an acceleration of 1.3 m/s2. What force of tension is in the cable? 18. A child pulls a wagon along a grassy field. Define the system, the pairs of forces at work, and the results. 19. Two teams are engaging in a tug\u2013of-war. The rope suddenly snaps. Which statement is true about the forces involved? a. The forces exerted by the two teams are no longer equal; the teams will accelerate in opposite directions as a result. b. The forces exerted by the players are no longer balanced by the force of tension in the rope; the teams will accelerate in opposite directions as a result. c. The force of gravity balances the forces exerted by the players; the teams will fall as a result d. The force of tension in the rope is transferred to the players; the teams will accelerate in opposite directions as a result. 20. The following free-body diagram represents a toboggan on a hill. What acceleration would you expect, and why? Figure 4.44 a. Acceleration down the hill; the force due to being pushed, together with the downhill component of gravity, overcomes the opposing force of friction. b. Acceleration down the hill; friction is less than the opposing component of force due to gravity. c. No movement; friction is greater than the force due to being pushed. It depends on how strong the force due to friction is. p d. 21. Draw a free-body diagram to represent the forces acting on a kite on a string that is floating stationary in the air. Label the forces in your diagram. 22. A car is sliding down a hill with a slope of 20\u00b0. The mass of the car is 965 kg. When a cable is used to pull the car up the slope, a force of 4215 N is applied. What is the car\u2019s acceleration, ignoring friction? 4.6 Problem-Solving Strategies 23. A toboggan with two riders has a total mass of 85.0 kg. A third person is pushing the toboggan with a force of 42.5 N at the top of a hill with an angle of 15\u00b0. The force of friction on the toboggan is 31.0 N. Which statement describes an accurate free-body diagram to represent the situation? a. An arrow of magnitude 10.5 N points down the slope of the hill. b. An arrow of magnitude 833 N points straight down. c. An arrow of magnitude 833 N points perpendicular to the slope of the hill. d. An arrow of magnitude 73.5 N points down the slope of the hill. 24. A mass of 2.0 kg is suspended from the ceiling of an elevator by a rope. What is the tension in the rope when the elevator (i) accelerates upward at 1.5 m/s2? (ii) accelerates downward at 1.5 m/s2? a. b. Because the mass is hanging from the elevator itself, the tension in the rope will not change in either case. (", "i) 22.6 N; (ii) 19.6 N (i) 16.6 N; (ii) 19.6 N (i) 22.6 N; (ii) 16.6 N c. d. 25. Which statement is true about drawing free-body diagrams? 190 Chapter 4 | Dynamics: Force and Newton's Laws of Motion a. Drawing a free-body diagram should be the last step in solving a problem about forces. b. Drawing a free-body diagram helps you compare forces 30. Explain which of the four fundamental forces is responsible for a ball bouncing off the ground after it hits, and why this force has this effect. quantitatively. c. The forces in a free-body diagram should always balance. d. Drawing a free-body diagram can help you determine the net force. 4.7 Further Applications of Newton's Laws of Motion 26. A basketball player jumps as he shoots the ball. Describe the forces that are acting on the ball and on the basketball player. What are the results? 27. Two people push on a boulder to try to move it. The mass of the boulder is 825 kg. One person pushes north with a force of 64 N. The other pushes west with a force of 38 N. Predict the magnitude of the acceleration of the boulder. Assume that friction is negligible. 28. 31. Which of the basic forces best explains tension in a rope being pulled between two people? Is the acting force causing attraction or repulsion in this instance? a. gravity; attraction b. electromagnetic; attraction c. weak and strong nuclear; attraction d. weak and strong nuclear; repulsion 32. Explain how interatomic electric forces produce the normal force, and why it has the direction it does. 33. The gravitational force is the weakest of the four basic forces. In which case can the electromagnetic, strong, and weak forces be ignored because the gravitational force is so strongly dominant? a. a person jumping on a trampoline b. a rocket blasting off from Earth c. a log rolling down a hill d. all of the above 34. Describe a situation in which gravitational force is the dominant force. Why can the other three basic forces be ignored in the situation you described? Figure 4.45 The figure shows the forces exerted on a block that is sliding on a horizontal surface: the gravitational force of 40 N, the 40 N normal force exerted by the surface, and a frictional force exerted to the left. The coefficient of friction between the block and the surface is 0.20. The acceleration of the block is most nearly a. 1.0 m/s2 to the right b. 1.0 m/s2 to the left c. 2.0 m/s2 to the right d. 2.0 m/s2 to the left 4.8 Extended Topic: The Four Basic Forces\u2014An Introduction 29. Which phenomenon correctly describes the direction and magnitude of normal forces? a. electromagnetic attraction b. electromagnetic repulsion c. gravitational attraction d. gravitational repulsion This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 5 | Further Applications of Newton's Laws: Friction, Drag, and Elasticity 191 FURTHER APPLICATIONS OF NEWTON'S 5 LAWS: FRICTION, DRAG, AND ELASTICITY Figure 5.1 Total hip replacement surgery has become a common procedure. The head (or ball) of the patient's femur fits into a cup that has a hard plastic-like inner lining. (credit: National Institutes of Health, via Wikimedia Commons) Chapter Outline 5.1. Friction 5.2. Drag Forces 5.3. Elasticity: Stress and Strain Connection for AP\u00ae Courses Have you ever wondered why it is difficult to walk on a smooth surface like ice? The interaction between you and the surface is a result of forces that affect your motion. In the previous chapter, you learned Newton's laws of motion and examined how net force affects the motion, position and shape of an object. Now we will look at some interesting and common forces that will provide further applications of Newton's laws of motion. The information presented in this chapter supports learning objectives covered under Big Idea 3 of the AP Physics Curriculum Framework, which refer to the nature of forces and their roles in interactions among objects. The chapter discusses examples of specific contact forces, such as friction, air or liquid drag, and elasticity that may affect the motion or shape of an object. It also discusses the nature of forces on both macroscopic and microscopic levels (Enduring Understanding 3.C and Essential Knowledge 3.C.4). In addition, Newton's laws are applied to describe the motion of an object (Enduring Understanding 3.B) and to examine relationships between contact forces and other forces exerted on an object (Enduring Understanding 3.A, 3.A.3 and 192 Chapter 5 | Further Applications of Newton's Laws: Fr", "iction, Drag, and Elasticity Essential Knowledge 3.A.4). The examples in this chapter give you practice in using vector properties of forces (Essential Knowledge 3.A.2) and free-body diagrams (Essential Knowledge 3.B.2) to determine net force (Essential Knowledge 3.B.1). Big Idea 3 The interactions of an object with other objects can be described by forces. Enduring Understanding 3.A All forces share certain common characteristics when considered by observers in inertial reference frames. Essential Knowledge 3.A.2 Forces are described by vectors. Essential Knowledge 3.A.3 A force exerted on an object is always due to the interaction of that object with another object. Essential Knowledge 3.A.4 If one object exerts a force on a second object, the second object always exerts a force of equal magnitude on the \ufb01rst object in the opposite direction. Enduring Understanding 3.B Classically, the acceleration of an object interacting with other objects can be predicted by using \u2192 = \u2192 \u2211. Essential Knowledge 3.B.1 If an object of interest interacts with several other objects, the net force is the vector sum of the individual forces. Essential Knowledge 3.B.2 Free-body diagrams are useful tools for visualizing forces being exerted on a single object and writing the equations that represent a physical situation. Enduring Understanding 3.C At the macroscopic level, forces can be categorized as either long-range (action-at-a-distance) forces or contact forces. Essential Knowledge 3.C.4 Contact forces result from the interaction of one object touching another object, and they arise from interatomic electric forces. These forces include tension, friction, normal, spring (Physics 1), and buoyant (Physics 2). 5.1 Friction Learning Objectives By the end of this section, you will be able to: \u2022 Discuss the general characteristics of friction. \u2022 Describe the various types of friction. \u2022 Calculate the magnitudes of static and kinetic frictional forces. The information presented in this section supports the following AP\u00ae learning objectives and science practices: \u2022 3.C.4.1 The student is able to make claims about various contact forces between objects based on the microscopic cause of those forces. (S.P. 6.1) \u2022 3.C.4.2 The student is able to explain contact forces (tension, friction, normal, buoyant, spring) as arising from interatomic electric forces and that they therefore have certain directions. (S.P. 6.2) Friction is a force that is around us all the time that opposes relative motion between systems in contact but also allows us to move (which you have discovered if you have ever tried to walk on ice). While a common force, the behavior of friction is actually very complicated and is still not completely understood. We have to rely heavily on observations for whatever understandings we can gain. However, we can still deal with its more elementary general characteristics and understand the circumstances in which it behaves. Friction Friction is a force that opposes relative motion between systems in contact. One of the simpler characteristics of friction is that it is parallel to the contact surface between systems and always in a direction that opposes motion or attempted motion of the systems relative to each other. If two systems are in contact and moving relative to one another, then the friction between them is called kinetic friction. For example, friction slows a hockey puck sliding on ice. But when objects are stationary, static friction can act between them; the static friction is usually greater than the kinetic friction between the objects. Kinetic Friction If two systems are in contact and moving relative to one another, then the friction between them is called kinetic friction. Imagine, for example, trying to slide a heavy crate across a concrete floor\u2014you may push harder and harder on the crate and not move it at all. This means that the static friction responds to what you do\u2014it increases to be equal to and in the opposite direction of your push. But if you finally push hard enough, the crate seems to slip suddenly and starts to move. Once in motion it is easier to keep it in motion than it was to get it started, indicating that the kinetic friction force is less than the static friction This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 5 | Further Applications of Newton's Laws: Friction, Drag, and Elasticity 193 force. If you add mass to the crate, say by placing a box on top of it, you need to push even harder to get it started and also to keep it moving. Furthermore, if you oiled the concrete you would find it to be easier to get the crate started and keep it going (as you might expect). Figure 5.2 is a crude pictorial representation of how friction occurs at the interface between two objects. Close-up inspection of these surfaces shows them to be rough. So", " when you push to get an object moving (in this case, a crate), you must raise the object until it can skip along with just the tips of the surface hitting, break off the points, or do both. A considerable force can be resisted by friction with no apparent motion. The harder the surfaces are pushed together (such as if another box is placed on the crate), the more force is needed to move them. Part of the friction is due to adhesive forces between the surface molecules of the two objects, which explain the dependence of friction on the nature of the substances. Adhesion varies with substances in contact and is a complicated aspect of surface physics. Once an object is moving, there are fewer points of contact (fewer molecules adhering), so less force is required to keep the object moving. At small but nonzero speeds, friction is nearly independent of speed. Figure 5.2 Frictional forces, such as, always oppose motion or attempted motion between objects in contact. Friction arises in part because of the roughness of the surfaces in contact, as seen in the expanded view. In order for the object to move, it must rise to where the peaks can skip along the bottom surface. Thus a force is required just to set the object in motion. Some of the peaks will be broken off, also requiring a force to maintain motion. Much of the friction is actually due to attractive forces between molecules making up the two objects, so that even perfectly smooth surfaces are not friction-free. Such adhesive forces also depend on the substances the surfaces are made of, explaining, for example, why rubber-soled shoes slip less than those with leather soles. The magnitude of the frictional force has two forms: one for static situations (static friction), the other for when there is motion (kinetic friction). When there is no motion between the objects, the magnitude of static friction fs is s \u2264 s, (5.1) where s is the coefficient of static friction and is the magnitude of the normal force (the force perpendicular to the surface). Magnitude of Static Friction Magnitude of static friction s is where s is the coefficient of static friction and is the magnitude of the normal force. s \u2264 s, (5.2) The symbol \u2264 means less than or equal to, implying that static friction can have a minimum and a maximum value of s. Static friction is a responsive force that increases to be equal and opposite to whatever force is exerted, up to its maximum limit. Once the applied force exceeds s(max), the object will move. Thus Once an object is moving, the magnitude of kinetic friction fk is given by k = k, s(max) = s. (5.3) (5.4) where k is the coefficient of kinetic friction. A system in which k = k is described as a system in which friction behaves simply. Magnitude of Kinetic Friction The magnitude of kinetic friction k is given by 194 Chapter 5 | Further Applications of Newton's Laws: Friction, Drag, and Elasticity where k is the coefficient of kinetic friction. k = k, (5.5) As seen in Table 5.1, the coefficients of kinetic friction are less than their static counterparts. That values of in Table 5.1 are stated to only one or, at most, two digits is an indication of the approximate description of friction given by the above two equations. Table 5.1 Coefficients of Static and Kinetic Friction System Static friction \u03bcs Kinetic friction \u03bck Rubber on dry concrete Rubber on wet concrete Wood on wood Waxed wood on wet snow Metal on wood Steel on steel (dry) Steel on steel (oiled) Teflon on steel 1.0 0.7 0.5 0.14 0.5 0.6 0.05 0.04 Bone lubricated by synovial fluid 0.016 Shoes on wood Shoes on ice Ice on ice Steel on ice 0.9 0.1 0.1 0.4 0.7 0.5 0.3 0.1 0.3 0.3 0.03 0.04 0.015 0.7 0.05 0.03 0.02 The equations given earlier include the dependence of friction on materials and the normal force. The direction of friction is always opposite that of motion, parallel to the surface between objects, and perpendicular to the normal force. For example, if the crate you try to push (with a force parallel to the floor) has a mass of 100 kg, then the normal force would be equal to its weight, = = (100 kg)(9.80 m/s2) = 980 N, perpendicular to the floor. If the coefficient of static friction is 0.45, you would have to exert a force parallel to the floor greater than s(max) = s = (0.45)(980 N) = 440 N to move the crate. Once there is motion, friction is less and the coefficient of kinetic friction might be 0.", "30, so that a force of only 290 N ( k = k = (0.30)(980 N) = 290 N ) would keep it moving at a constant speed. If the floor is lubricated, both coefficients are considerably less than they would be without lubrication. Coefficient of friction is a unit less quantity with a magnitude usually between 0 and 1.0. The coefficient of the friction depends on the two surfaces that are in contact. Take-Home Experiment Find a small plastic object (such as a food container) and slide it on a kitchen table by giving it a gentle tap. Now spray water on the table, simulating a light shower of rain. What happens now when you give the object the same-sized tap? Now add a few drops of (vegetable or olive) oil on the surface of the water and give the same tap. What happens now? This latter situation is particularly important for drivers to note, especially after a light rain shower. Why? Many people have experienced the slipperiness of walking on ice. However, many parts of the body, especially the joints, have much smaller coefficients of friction\u2014often three or four times less than ice. A joint is formed by the ends of two bones, which are connected by thick tissues. The knee joint is formed by the lower leg bone (the tibia) and the thighbone (the femur). The hip is a ball (at the end of the femur) and socket (part of the pelvis) joint. The ends of the bones in the joint are covered by cartilage, which provides a smooth, almost glassy surface. The joints also produce a fluid (synovial fluid) that reduces friction and wear. A damaged or arthritic joint can be replaced by an artificial joint (Figure 5.3). These replacements can be made of metals (stainless steel or titanium) or plastic (polyethylene), also with very small coefficients of friction. This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 5 | Further Applications of Newton's Laws: Friction, Drag, and Elasticity 195 Figure 5.3 Artificial knee replacement is a procedure that has been performed for more than 20 years. In this figure, we see the post-op x rays of the right knee joint replacement. (credit: Mike Baird, Flickr) Other natural lubricants include saliva produced in our mouths to aid in the swallowing process, and the slippery mucus found between organs in the body, allowing them to move freely past each other during heartbeats, during breathing, and when a person moves. Artificial lubricants are also common in hospitals and doctor's clinics. For example, when ultrasonic imaging is carried out, the gel that couples the transducer to the skin also serves to to lubricate the surface between the transducer and the skin\u2014thereby reducing the coefficient of friction between the two surfaces. This allows the transducer to mover freely over the skin. Example 5.1 Skiing Exercise A skier with a mass of 62 kg is sliding down a snowy slope. Find the coefficient of kinetic friction for the skier if friction is known to be 45.0 N. Strategy The magnitude of kinetic friction was given in to be 45.0 N. Kinetic friction is related to the normal force N as k = k ; thus, the coefficient of kinetic friction can be found if we can find the normal force of the skier on a slope. The normal force is always perpendicular to the surface, and since there is no motion perpendicular to the surface, the normal force should equal the component of the skier's weight perpendicular to the slope. (See the skier and free-body diagram in Figure 5.4.) 196 Chapter 5 | Further Applications of Newton's Laws: Friction, Drag, and Elasticity Figure 5.4 The motion of the skier and friction are parallel to the slope and so it is most convenient to project all forces onto a coordinate system where one axis is parallel to the slope and the other is perpendicular (axes shown to left of skier). N (the normal force) is perpendicular to the slope, and f (the friction) is parallel to the slope, but w (the skier's weight) has components along both axes, namely w\u22a5 and W//. N is equal in magnitude to w\u22a5, so there is no motion perpendicular to the slope. However, f is less than W// in magnitude, so there is acceleration down the slope (along the x-axis). That is, = \u22a5 = cos 25\u00ba = cos 25\u00ba. Substituting this into our expression for kinetic friction, we get k = k cos 25\u00ba, which can now be solved for the coefficient of kinetic friction k. Solution Solving for k gives k = k = k = k cos 25\u00ba cos 25\u00ba. Substituting known values on the right-hand side of the equation", ", k = 45.0 N (62 kg)(9.80 m/s2)(0.906) = 0.082. Discussion (5.6) (5.7) (5.8) (5.9) This result is a little smaller than the coefficient listed in Table 5.1 for waxed wood on snow, but it is still reasonable since values of the coefficients of friction can vary greatly. In situations like this, where an object of mass slides down a slope that makes an angle with the horizontal, friction is given by k = k cos. All objects will slide down a slope with constant acceleration under these circumstances. Proof of this is left for this chapter's Problems and Exercises. Take-Home Experiment An object will slide down an inclined plane at a constant velocity if the net force on the object is zero. We can use this fact to measure the coefficient of kinetic friction between two objects. As shown in Example 5.1, the kinetic friction on a slope k = k cos. The component of the weight down the slope is equal to sin (see the free-body diagram in Figure 5.4). These forces act in opposite directions, so when they have equal magnitude, the acceleration is zero. Writing these out: Solving for k, we find that k = k cos = sin. k = sin cos = tan. (5.10) (5.11) (5.12) This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 5 | Further Applications of Newton's Laws: Friction, Drag, and Elasticity 197 Put a coin on a book and tilt it until the coin slides at a constant velocity down the book. You might need to tap the book lightly to get the coin to move. Measure the angle of tilt relative to the horizontal and find k. Note that the coin will not start to slide at all until an angle greater than is attained, since the coefficient of static friction is larger than the coefficient of kinetic friction. Discuss how this may affect the value for k and its uncertainty. We have discussed that when an object rests on a horizontal surface, there is a normal force supporting it equal in magnitude to its weight. Furthermore, simple friction is always proportional to the normal force. Making Connections: Submicroscopic Explanations of Friction The simpler aspects of friction dealt with so far are its macroscopic (large-scale) characteristics. Great strides have been made in the atomic-scale explanation of friction during the past several decades. Researchers are finding that the atomic nature of friction seems to have several fundamental characteristics. These characteristics not only explain some of the simpler aspects of friction\u2014they also hold the potential for the development of nearly friction-free environments that could save hundreds of billions of dollars in energy which is currently being converted (unnecessarily) to heat. Figure 5.5 illustrates one macroscopic characteristic of friction that is explained by microscopic (small-scale) research. We have noted that friction is proportional to the normal force, but not to the area in contact, a somewhat counterintuitive notion. When two rough surfaces are in contact, the actual contact area is a tiny fraction of the total area since only high spots touch. When a greater normal force is exerted, the actual contact area increases, and it is found that the friction is proportional to this area. Figure 5.5 Two rough surfaces in contact have a much smaller area of actual contact than their total area. When there is a greater normal force as a result of a greater applied force, the area of actual contact increases as does friction. But the atomic-scale view promises to explain far more than the simpler features of friction. The mechanism for how heat is generated is now being determined. In other words, why do surfaces get warmer when rubbed? Essentially, atoms are linked with one another to form lattices. When surfaces rub, the surface atoms adhere and cause atomic lattices to vibrate\u2014essentially creating sound waves that penetrate the material. The sound waves diminish with distance and their energy is converted into heat. Chemical reactions that are related to frictional wear can also occur between atoms and molecules on the surfaces. Figure 5.6 shows how the tip of a probe drawn across another material is deformed by atomic-scale friction. The force needed to drag the tip can be measured and is found to be related to shear stress, which will be discussed later in this chapter. The variation in shear stress is remarkable (more than a factor of 1012 ) and difficult to predict theoretically, but shear stress is yielding a fundamental understanding of a large-scale phenomenon known since ancient times\u2014friction. 198 Chapter 5 | Further Applications of Newton's Laws: Friction, Drag, and Elasticity Figure 5.6 The tip of a probe is deformed sideways by frictional force as the probe is dragged across a surface. Measurements of how the force varies for different materials are yielding fundamental insights into the atomic", " nature of friction. PhET Explorations: Forces and Motion Explore the forces at work when you try to push a filing cabinet. Create an applied force and see the resulting friction force and total force acting on the cabinet. Charts show the forces, position, velocity, and acceleration vs. time. Draw a free-body diagram of all the forces (including gravitational and normal forces). Figure 5.7 Forces and Motion (http://cnx.org/content/m54899/1.2/forces-and-motion_en.jar) 5.2 Drag Forces Learning Objectives By the end of this section, you will be able to: \u2022 Define drag force and model it mathematically. \u2022 Discuss the applications of drag force. \u2022 Define terminal velocity. \u2022 Perform calculations to find terminal velocity. Another interesting force in everyday life is the force of drag on an object when it is moving in a fluid (either a gas or a liquid). You feel the drag force when you move your hand through water. You might also feel it if you move your hand during a strong wind. The faster you move your hand, the harder it is to move. You feel a smaller drag force when you tilt your hand so only the side goes through the air\u2014you have decreased the area of your hand that faces the direction of motion. Like friction, the drag force always opposes the motion of an object. Unlike simple friction, the drag force is proportional to some function of the velocity of the object in that fluid. This functionality is complicated and depends upon the shape of the object, its size, its velocity, and the fluid it is in. For most large objects such as bicyclists, cars, and baseballs not moving too slowly, the magnitude of the drag force D is found to be proportional to the square of the speed of the object. We can write this relationship mathematically as. When taking into account other factors, this relationship becomes D = 1 2 C\u03c1 2, (5.13) where is the drag coefficient, is the area of the object facing the fluid, and is the density of the fluid. (Recall that density is mass per unit volume.) This equation can also be written in a more generalized fashion as D = 2, where is a constant equivalent to 0.5. We have set the exponent for these equations as 2 because, when an object is moving at high velocity This content is available for free at http://cnx.org/content/col11844/1.13 212 Chapter 5 | Further Applications of Newton's Laws: Friction, Drag, and Elasticity Although measurable, this is not a significant decrease in volume considering that the force per unit area is about 500 atmospheres (1 million pounds per square foot). Liquids and solids are extraordinarily difficult to compress. Conversely, very large forces are created by liquids and solids when they try to expand but are constrained from doing so\u2014which is equivalent to compressing them to less than their normal volume. This often occurs when a contained material warms up, since most materials expand when their temperature increases. If the materials are tightly constrained, they deform or break their container. Another very common example occurs when water freezes. Water, unlike most materials, expands when it freezes, and it can easily fracture a boulder, rupture a biological cell, or crack an engine block that gets in its way. Other types of deformations, such as torsion or twisting, behave analogously to the tension, shear, and bulk deformations considered here. Glossary deformation: change in shape due to the application of force drag force: D, found to be proportional to the square of the speed of the object; mathematically D \u221d 2 D = 1 where is the drag coefficient, is the area of the object facing the fluid, and is the density of the fluid 2 2, friction: a force that opposes relative motion or attempts at motion between systems in contact Hooke's law: proportional relationship between the force on a material and the deformation \u0394 it causes, = \u0394 kinetic friction: a force that opposes the motion of two systems that are in contact and moving relative to one another magnitude of kinetic friction: k = k, where k is the coefficient of kinetic friction magnitude of static friction: s \u2264 s, where s is the coefficient of static friction and is the magnitude of the normal force shear deformation: deformation perpendicular to the original length of an object static friction: a force that opposes the motion of two systems that are in contact and are not moving relative to one another Stokes' law: s = 6, where is the radius of the object, is the viscosity of the fluid, and is the object's velocity strain: ratio of change in length to original length stress: ratio of force to area tensile strength: measure of deformation for a given tension or compression Section Summary 5.1 Friction \u2022 Friction is a contact force between systems that opposes the motion or attempted motion between them. Simple friction is proportional", " to the normal force pushing the systems together. (A normal force is always perpendicular to the contact surface between systems.) Friction depends on both of the materials involved. The magnitude of static friction s between systems stationary relative to one another is given by s \u2264 s, where s is the coefficient of static friction, which depends on both of the materials. \u2022 The kinetic friction force k between systems moving relative to one another is given by where k is the coefficient of kinetic friction, which also depends on both materials. k = k, 5.2 Drag Forces This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 5 | Further Applications of Newton's Laws: Friction, Drag, and Elasticity 213 \u2022 Drag forces acting on an object moving in a fluid oppose the motion. For larger objects (such as a baseball) moving at a velocity in air, the drag force is given by D = 1 2 2, where is the drag coefficient (typical values are given in Table 5.2), is the area of the object facing the fluid, and is the fluid density. \u2022 For small objects (such as a bacterium) moving in a denser medium (such as water), the drag force is given by Stokes' law, s = 6, where is the radius of the object, is the fluid viscosity, and is the object's velocity. 5.3 Elasticity: Stress and Strain \u2022 Hooke's law is given by = \u0394, where \u0394 is the amount of deformation (the change in length), is the applied force, and is a proportionality constant that depends on the shape and composition of the object and the direction of the force. The relationship between the deformation and the applied force can also be written as \u0394 = 1 0, where is Young's modulus, which depends on the substance, is the cross-sectional area, and 0 is the original length. \u2022 The ratio of force to area,, is defined as stress, measured in N/m2. \u2022 The ratio of the change in length to length, \u0394 0, is defined as strain (a unitless quantity). In other words, \u2022 The expression for shear deformation is stress = \u00d7strain. \u0394 = 1 where is the shear modulus and is the force applied perpendicular to 0 and parallel to the cross-sectional area. 0, \u2022 The relationship of the change in volume to other physical quantities is given by \u0394 = 1 where is the bulk modulus, 0 is the original volume, and surfaces. 0, is the force per unit area applied uniformly inward on all Conceptual Questions 5.1 Friction 1. Define normal force. What is its relationship to friction when friction behaves simply? 2. The glue on a piece of tape can exert forces. Can these forces be a type of simple friction? Explain, considering especially that tape can stick to vertical walls and even to ceilings. 3. When you learn to drive, you discover that you need to let up slightly on the brake pedal as you come to a stop or the car will stop with a jerk. Explain this in terms of the relationship between static and kinetic friction. 4. When you push a piece of chalk across a chalkboard, it sometimes screeches because it rapidly alternates between slipping and sticking to the board. Describe this process in more detail, in particular explaining how it is related to the fact that kinetic friction is less than static friction. (The same slip-grab process occurs when tires screech on pavement.) 5.2 Drag Forces 5. Athletes such as swimmers and bicyclists wear body suits in competition. Formulate a list of pros and cons of such suits. 6. Two expressions were used for the drag force experienced by a moving object in a liquid. One depended upon the speed, while the other was proportional to the square of the speed. In which types of motion would each of these expressions be more applicable than the other one? 7. As cars travel, oil and gasoline leaks onto the road surface. If a light rain falls, what does this do to the control of the car? Does a heavy rain make any difference? 8. Why can a squirrel jump from a tree branch to the ground and run away undamaged, while a human could break a bone in such a fall? 214 Chapter 5 | Further Applications of Newton's Laws: Friction, Drag, and Elasticity 5.3 Elasticity: Stress and Strain 9. The elastic properties of the arteries are essential for blood flow. Explain the importance of this in terms of the characteristics of the flow of blood (pulsating or continuous). 10. What are you feeling when you feel your pulse? Measure your pulse rate for 10 s and for 1 min. Is there a factor of 6 difference? 11. Examine different types of shoes, including sports shoes and thongs. In terms of physics, why are the bottom surfaces designed as they are? What differences will dry and wet conditions make", " for these surfaces? 12. Would you expect your height to be different depending upon the time of day? Why or why not? 13. Why can a squirrel jump from a tree branch to the ground and run away undamaged, while a human could break a bone in such a fall? 14. Explain why pregnant women often suffer from back strain late in their pregnancy. 15. An old carpenter's trick to keep nails from bending when they are pounded into hard materials is to grip the center of the nail firmly with pliers. Why does this help? 16. When a glass bottle full of vinegar warms up, both the vinegar and the glass expand, but vinegar expands significantly more with temperature than glass. The bottle will break if it was filled to its tightly capped lid. Explain why, and also explain how a pocket of air above the vinegar would prevent the break. (This is the function of the air above liquids in glass containers.) This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 5 | Further Applications of Newton's Laws: Friction, Drag, and Elasticity 215 Problems & Exercises 5.1 Friction 1. A physics major is cooking breakfast when he notices that the frictional force between his steel spatula and his Teflon frying pan is only 0.200 N. Knowing the coefficient of kinetic friction between the two materials, he quickly calculates the normal force. What is it? 2. (a) When rebuilding her car's engine, a physics major must exert 300 N of force to insert a dry steel piston into a steel cylinder. What is the magnitude of the normal force between the piston and cylinder? (b) What is the magnitude of the force would she have to exert if the steel parts were oiled? 3. (a) What is the maximum frictional force in the knee joint of a person who supports 66.0 kg of her mass on that knee? (b) During strenuous exercise it is possible to exert forces to the joints that are easily ten times greater than the weight being supported. What is the maximum force of friction under such conditions? The frictional forces in joints are relatively small in all circumstances except when the joints deteriorate, such as from injury or arthritis. Increased frictional forces can cause further damage and pain. 4. Suppose you have a 120-kg wooden crate resting on a wood floor. (a) What maximum force can you exert horizontally on the crate without moving it? (b) If you continue to exert this force once the crate starts to slip, what will the magnitude of its acceleration then be? 5. (a) If half of the weight of a small 1.00\u00d7103 kg utility truck is supported by its two drive wheels, what is the magnitude of the maximum acceleration it can achieve on dry concrete? (b) Will a metal cabinet lying on the wooden bed of the truck slip if it accelerates at this rate? (c) Solve both problems assuming the truck has four-wheel drive. 6. A team of eight dogs pulls a sled with waxed wood runners on wet snow (mush!). The dogs have average masses of 19.0 kg, and the loaded sled with its rider has a mass of 210 kg. (a) Calculate the magnitude of the acceleration starting from rest if each dog exerts an average force of 185 N backward on the snow. (b) What is the magnitude of the acceleration once the sled starts to move? (c) For both situations, calculate the magnitude of the force in the coupling between the dogs and the sled. 7. Consider the 65.0-kg ice skater being pushed by two others shown in Figure 5.21. (a) Find the direction and magnitude of Ftot, the total force exerted on her by the others, given that the magnitudes 1 and 2 are 26.4 N and 18.6 N, respectively. (b) What is her initial acceleration if she is initially stationary and wearing steel-bladed skates that point in the direction of Ftot? (c) What is her acceleration assuming she is already moving in the direction of Ftot? (Remember that friction always acts in the direction opposite that of motion or attempted motion between surfaces in contact.) Figure 5.21 8. Show that the acceleration of any object down a frictionless incline that makes an angle with the horizontal is = sin. (Note that this acceleration is independent of mass.) 9. Show that the acceleration of any object down an incline where friction behaves simply (that is, where k = k ) is = ( sin \u2212 kcos ). Note that the acceleration is independent of mass and reduces to the expression found in the previous problem when friction becomes negligibly small ( k = 0). 10. Calculate the deceleration of a snow boarder going up a 5.0\u00ba, slope assuming the coefficient of friction for waxed wood on wet snow. The result of Exercise 5.9 may be useful", ", but be careful to consider the fact that the snow boarder is going uphill. Explicitly show how you follow the steps in Problem-Solving Strategies. 11. (a) Calculate the acceleration of a skier heading down a 10.0\u00ba slope, assuming the coefficient of friction for waxed wood on wet snow. (b) Find the angle of the slope down which this skier could coast at a constant velocity. You can neglect air resistance in both parts, and you will find the result of Exercise 5.9 to be useful. Explicitly show how you follow the steps in the Problem-Solving Strategies. 12. If an object is to rest on an incline without slipping, then friction must equal the component of the weight of the object parallel to the incline. This requires greater and greater friction for steeper slopes. Show that the maximum angle of an incline above the horizontal for which an object will not slide down is = tan\u20131 \u03bcs. You may use the result of the previous problem. Assume that = 0 and that static friction has reached its maximum value. 13. Calculate the maximum deceleration of a car that is heading down a 6\u00ba slope (one that makes an angle of 6\u00ba with the horizontal) under the following road conditions. You may assume that the weight of the car is evenly distributed on all four tires and that the coefficient of static friction is involved\u2014that is, the tires are not allowed to slip during the deceleration. (Ignore rolling.) Calculate for a car: (a) On dry concrete. (b) On wet concrete. (c) On ice, assuming that s = 0.100, the same as for shoes on ice. 14. Calculate the maximum acceleration of a car that is heading up a 4\u00ba slope (one that makes an angle of 4\u00ba with the horizontal) under the following road conditions. Assume that only half the weight of the car is supported by the two drive wheels and that the coefficient of static friction is involved\u2014that is, the tires are not allowed to slip during the acceleration. (Ignore rolling.) (a) On dry concrete. (b) On wet 216 Chapter 5 | Further Applications of Newton's Laws: Friction, Drag, and Elasticity concrete. (c) On ice, assuming that \u03bcs = 0.100, the same as for shoes on ice. 15. Repeat Exercise 5.14 for a car with four-wheel drive. 16. A freight train consists of two 8.00\u00d7105-kg engines and 45 cars with average masses of 5.50\u00d7105 kg. (a) What force must each engine exert backward on the track to accelerate the train at a rate of 5.00\u00d710\u22122 m / s2 if the force of friction is 7.50\u00d7105 N, assuming the engines exert identical forces? This is not a large frictional force for such a massive system. Rolling friction for trains is small, and consequently trains are very energy-efficient transportation systems. (b) What is the magnitude of the force in the coupling between the 37th and 38th cars (this is the force each exerts on the other), assuming all cars have the same mass and that friction is evenly distributed among all of the cars and engines? 17. Consider the 52.0-kg mountain climber in Figure 5.22. (a) Find the tension in the rope and the force that the mountain climber must exert with her feet on the vertical rock face to remain stationary. Assume that the force is exerted parallel to her legs. Also, assume negligible force exerted by her arms. (b) What is the minimum coefficient of friction between her shoes and the cliff? Figure 5.22 Part of the climber's weight is supported by her rope and part by friction between her feet and the rock face. 18. A contestant in a winter sporting event pushes a 45.0-kg block of ice across a frozen lake as shown in Figure 5.23(a). (a) Calculate the minimum force he must exert to get the block moving. (b) What is the magnitude of its acceleration once it starts to move, if that force is maintained? 19. Repeat Exercise 5.18 with the contestant pulling the block of ice with a rope over his shoulder at the same angle above the horizontal as shown in Figure 5.23(b). This content is available for free at http://cnx.org/content/col11844/1.13 Figure 5.23 Which method of sliding a block of ice requires less force\u2014(a) pushing or (b) pulling at the same angle above the horizontal? 5.2 Drag Forces 20. The terminal velocity of a person falling in air depends upon the weight and the area of the person facing the fluid. Find the terminal velocity (in meters per second and kilometers per hour) of an 80.0-kg skydiver falling in a pike (headfirst) position with", " a surface area of 0.140 m2. 21. A 60-kg and a 90-kg skydiver jump from an airplane at an altitude of 6000 m, both falling in the pike position. Make some assumption on their frontal areas and calculate their terminal velocities. How long will it take for each skydiver to reach the ground (assuming the time to reach terminal velocity is small)? Assume all values are accurate to three significant digits. 22. A 560-g squirrel with a surface area of 930 cm2 falls from a 5.0-m tree to the ground. Estimate its terminal velocity. (Use a drag coefficient for a horizontal skydiver.) What will be the velocity of a 56-kg person hitting the ground, assuming no drag contribution in such a short distance? 23. To maintain a constant speed, the force provided by a car's engine must equal the drag force plus the force of friction of the road (the rolling resistance). (a) What are the magnitudes of drag forces at 70 km/h and 100 km/h for a Toyota Camry? (Drag area is 0.70 m2 ) (b) What is the magnitude of drag force at 70 km/h and 100 km/h for a Hummer H2? (Drag area is 2.44 m2 ) Assume all values are accurate to three significant digits. 24. By what factor does the drag force on a car increase as it goes from 65 to 110 km/h? 25. Calculate the speed a spherical rain drop would achieve falling from 5.00 km (a) in the absence of air drag (b) with air drag. Take the size across of the drop to be 4 mm, the density to be 1.00\u00d7103 kg/m3, and the surface area to be 2. Chapter 5 | Further Applications of Newton's Laws: Friction, Drag, and Elasticity 217 26. Using Stokes' law, verify that the units for viscosity are kilograms per meter per second. 27. Find the terminal velocity of a spherical bacterium (diameter 2.00 \u03bcm ) falling in water. You will first need to note that the drag force is equal to the weight at terminal velocity. Take the density of the bacterium to be 1.10\u00d7103 kg/m3. 28. Stokes' law describes sedimentation of particles in liquids and can be used to measure viscosity. Particles in liquids achieve terminal velocity quickly. One can measure the time it takes for a particle to fall a certain distance and then use Stokes' law to calculate the viscosity of the liquid. Suppose a steel ball bearing (density 7.8\u00d7103 kg/m3 3.0 mm ) is dropped in a container of motor oil. It takes 12 s to fall a distance of 0.60 m. Calculate the viscosity of the oil., diameter 5.3 Elasticity: Stress and Strain 29. During a circus act, one performer swings upside down hanging from a trapeze holding another, also upside-down, performer by the legs. If the upward force on the lower performer is three times her weight, how much do the bones (the femurs) in her upper legs stretch? You may assume each is equivalent to a uniform rod 35.0 cm long and 1.80 cm in radius. Her mass is 60.0 kg. 30. During a wrestling match, a 150 kg wrestler briefly stands on one hand during a maneuver designed to perplex his already moribund adversary. By how much does the upper arm bone shorten in length? The bone can be represented by a uniform rod 38.0 cm in length and 2.10 cm in radius. 31. (a) The \u201clead\u201d in pencils is a graphite composition with a Young's modulus of about 1\u00d7109 N / m2. Calculate the change in length of the lead in an automatic pencil if you tap it straight into the pencil with a force of 4.0 N. The lead is 0.50 mm in diameter and 60 mm long. (b) Is the answer reasonable? That is, does it seem to be consistent with what you have observed when using pencils? 32. TV broadcast antennas are the tallest artificial structures on Earth. In 1987, a 72.0-kg physicist placed himself and 400 kg of equipment at the top of one 610-m high antenna to perform gravity experiments. By how much was the antenna compressed, if we consider it to be equivalent to a steel cylinder 0.150 m in radius? 33. (a) By how much does a 65.0-kg mountain climber stretch her 0.800-cm diameter nylon rope when she hangs 35.0 m below a rock outcropping? (b) Does the answer seem to be consistent with what you have observed for nylon ropes? Would it make sense if the rope were actually a bungee cord? 34. A", " 20.0-m tall hollow aluminum flagpole is equivalent in strength to a solid cylinder 4.00 cm in diameter. A strong wind bends the pole much as a horizontal force of 900 N exerted at the top would. How far to the side does the top of the pole flex? 35. As an oil well is drilled, each new section of drill pipe supports its own weight and that of the pipe and drill bit beneath it. Calculate the stretch in a new 6.00 m length of steel pipe that supports 3.00 km of pipe having a mass of 20.0 kg/m and a 100-kg drill bit. The pipe is equivalent in strength to a solid cylinder 5.00 cm in diameter. 36. Calculate the force a piano tuner applies to stretch a steel piano wire 8.00 mm, if the wire is originally 0.850 mm in diameter and 1.35 m long. 37. A vertebra is subjected to a shearing force of 500 N. Find the shear deformation, taking the vertebra to be a cylinder 3.00 cm high and 4.00 cm in diameter. 38. A disk between vertebrae in the spine is subjected to a shearing force of 600 N. Find its shear deformation, taking it to have the shear modulus of 1\u00d7109 N / m2. The disk is equivalent to a solid cylinder 0.700 cm high and 4.00 cm in diameter. 39. When using a pencil eraser, you exert a vertical force of 6.00 N at a distance of 2.00 cm from the hardwood-eraser joint. The pencil is 6.00 mm in diameter and is held at an angle of 20.0\u00ba to the horizontal. (a) By how much does the wood flex perpendicular to its length? (b) How much is it compressed lengthwise? 40. To consider the effect of wires hung on poles, we take data from Example 4.8, in which tensions in wires supporting a traffic light were calculated. The left wire made an angle 30.0\u00ba below the horizontal with the top of its pole and carried a tension of 108 N. The 12.0 m tall hollow aluminum pole is equivalent in strength to a 4.50 cm diameter solid cylinder. (a) How far is it bent to the side? (b) By how much is it compressed? 41. A farmer making grape juice fills a glass bottle to the brim and caps it tightly. The juice expands more than the glass when it warms up, in such a way that the volume increases by 0.2% (that is, \u0394 / 0 = 2\u00d710\u22123 available. Calculate the magnitude of the normal force exerted by the juice per square centimeter if its bulk modulus is 1.8\u00d7109 N/m2, assuming the bottle does not break. In view of your answer, do you think the bottle will survive? ) relative to the space 42. (a) When water freezes, its volume increases by 9.05% (that is, \u0394 / 0 = 9.05\u00d710\u22122 ). What force per unit area is water capable of exerting on a container when it freezes? (It is acceptable to use the bulk modulus of water in this problem.) (b) Is it surprising that such forces can fracture engine blocks, boulders, and the like? 43. This problem returns to the tightrope walker studied in Example 4.6, who created a tension of 3.94\u00d7103 N in a wire making an angle 5.0\u00ba below the horizontal with each supporting pole. Calculate how much this tension stretches the steel wire if it was originally 15 m long and 0.50 cm in diameter. 44. The pole in Figure 5.24 is at a 90.0\u00ba bend in a power line and is therefore subjected to more shear force than poles in straight parts of the line. The tension in each line is 4.00\u00d7104 N, at the angles shown. The pole is 15.0 m tall, has an 18.0 cm diameter, and can be considered to have half the strength of hardwood. (a) Calculate the compression of the pole. (b) Find how much it bends and in what direction. (c) Find the tension in a guy wire used to keep the pole straight if it is attached to the top of the pole at an angle of 30.0\u00ba with the vertical. (Clearly, the guy wire must be in the opposite direction of the bend.) 218 Chapter 5 | Further Applications of Newton's Laws: Friction, Drag, and Elasticity Figure 5.24 This telephone pole is at a 90\u00ba bend in a power line. A guy wire is attached to the top of the pole at an angle of 30\u00ba with the vertical. Test Prep for AP\u00ae Courses 5.1 Friction 1. When a force of 20 N is applied to a stationary box weighing 40 N, the box", " does not move. This means the coefficient of static friction is equal to 0.5. is greater than 0.5. is less than 0.5. a. b. c. d. cannot be determined. 2. A 2-kg block slides down a ramp which is at an incline of 25\u00ba. If the frictional force is 4.86 N, what is the coefficient of friction? At what incline will the box slide at a constant velocity? Assume g = 10 m/s2. 3. A block is given a short push and then slides with constant friction across a horizontal floor. Which statement best explains the direction of the force that friction applies on the moving block? a. Friction will be in the same direction as the block's motion because molecular interactions between the block and the floor will deform the block in the direction of motion. b. Friction will be in the same direction as the block's motion because thermal energy generated at the interface between the block and the floor adds kinetic energy to the block. c. Friction will be in the opposite direction of the block's motion because molecular interactions between the block and the floor will deform the block in the opposite direction of motion. d. Friction will be in the opposite direction of the block's motion because thermal energy generated at the interface between the block and the floor converts some of the block's kinetic energy to potential energy. 4. A student pushes a cardboard box across a carpeted floor and afterwards notices that the bottom of the box feels warm. Explain how interactions between molecules in the cardboard and molecules in the carpet produced this heat. This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 6 | Gravitation and Uniform Circular Motion 219 6 GRAVITATION AND UNIFORM CIRCULAR MOTION Figure 6.1 This Australian Grand Prix Formula 1 race car moves in a circular path as it makes the turn. Its wheels also spin rapidly\u2014the latter completing many revolutions, the former only part of one (a circular arc). The same physical principles are involved in each. (credit: Richard Munckton) Chapter Outline 6.1. Rotation Angle and Angular Velocity 6.2. Centripetal Acceleration 6.3. Centripetal Force 6.4. Fictitious Forces and Non-inertial Frames: The Coriolis Force 6.5. Newton's Universal Law of Gravitation 6.6. Satellites and Kepler's Laws: An Argument for Simplicity Connection for AP\u00ae Courses Many motions, such as the arc of a bird's flight or Earth's path around the Sun, are curved. Recall that Newton's first law tells us that motion is along a straight line at constant speed unless there is a net external force. We will therefore study not only motion along curves, but also the forces that cause it, including gravitational forces. This chapter supports Big Idea 3 that interactions between objects are described by forces, and thus change in motion is a result of a net force exerted on an object. In this chapter, this idea is applied to uniform circular motion. In some ways, this chapter is a continuation of Dynamics: Newton's Laws of Motion as we study more applications of Newton's laws of motion. This chapter deals with the simplest form of curved motion, uniform circular motion, which is motion in a circular path at constant speed. As an object moves on a circular path, the magnitude of its velocity remains constant, but the direction of the velocity is changing. This means there is an acceleration that we will refer to as a \u201ccentripetal\u201d acceleration caused by a net external force, also called the \u201ccentripetal\u201d force (Enduring Understanding 3.B). The centripetal force is the net force totaling all 220 Chapter 6 | Gravitation and Uniform Circular Motion external forces acting on the object (Essential Knowledge 3.B.1). In order to determine the net force, a free-body diagram may be useful (Essential Knowledge 3.B.2). Studying this topic illustrates most of the concepts associated with rotational motion and leads to many new topics we group under the name rotation. This motion can be described using kinematics variables (Essential Knowledge 3.A.1), but in addition to linear variables, we will introduce angular variables. We use various ways to describe motion, namely, verbally, algebraically and graphically (Learning Objective 3.A.1.1). Pure rotational motion occurs when points in an object move in circular paths centered on one point. Pure translational motion is motion with no rotation. Some motion combines both types, such as a rotating hockey puck moving over ice. Some combinations of both types of motion are conveniently described with fictitious forces which appear as a result of using a non-inertial frame of reference (Enduring Understanding 3.A). Furthermore, the properties of uniform circular motion can be applied", " to the motion of massive objects in a gravitational field. Thus, this chapter supports Big Idea 1 that gravitational mass is an important property of an object or a system. We have experimental evidence that gravitational and inertial masses are equal (Enduring Understanding 1.C), and that gravitational mass is a measure of the strength of the gravitational interaction (Essential Knowledge 1.C.2). Therefore, this chapter will support Big Idea 2 that fields existing in space can be used to explain interactions, because any massive object creates a gravitational field in space (Enduring Understanding 2.B). Mathematically, we use Newton's universal law of gravitation to provide a model for the gravitational interaction between two massive objects (Essential Knowledge 2.B.2). We will discover that this model describes the interaction of one object with mass with another object with mass (Essential Knowledge 3.C.1), and also that gravitational force is a long-range force (Enduring Understanding 3.C). The concepts in this chapter support: Big Idea 1 Objects and systems have properties such as mass and charge. Systems may have internal structure. Enduring Understanding 1.C Objects and systems have properties of inertial mass and gravitational mass that are experimentally verified to be the same and that satisfy conservation principles. Essential Knowledge 1.C.2 Gravitational mass is the property of an object or a system that determines the strength of the gravitational interaction with other objects, systems, or gravitational fields. Essential Knowledge 1.C.3 Objects and systems have properties of inertial mass and gravitational mass that are experimentally verified to be the same and that satisfy conservation principles. Big Idea 2 Fields existing in space can be used to explain interactions. Enduring Understanding 2.B A gravitational field is caused by an object with mass. Essential Knowledge 2.B.2. The gravitational field caused by a spherically symmetric object with mass is radial and, outside the object, varies as the inverse square of the radial distance from the center of that object. Big Idea 3 The interactions of an object with other objects can be described by forces. Enduring Understanding 3.A All forces share certain common characteristics when considered by observers in inertial reference frames. Essential Knowledge 3.A.1. An observer in a particular reference frame can describe the motion of an object using such quantities as position, displacement, distance, velocity, speed, and acceleration. Essential Knowledge 3.A.3. A force exerted on an object is always due to the interaction of that object with another object. Enduring Understanding 3.B Classically, the acceleration of an object interacting with other objects can be predicted by using = \u2211 /. Essential Knowledge 3.B.1 If an object of interest interacts with several other objects, the net force is the vector sum of the individual forces. Essential Knowledge 3.B.2 Free-body diagrams are useful tools for visualizing forces being exerted on a single object and writing the equations that represent a physical situation. Enduring Understanding 3.C At the macroscopic level, forces can be categorized as either long-range (action-at-a-distance) forces or contact forces. Essential Knowledge 3.C.1. Gravitational force describes the interaction of one object that has mass with another object that has mass. 6.1 Rotation Angle and Angular Velocity Learning Objectives By the end of this section, you will be able to: \u2022 Define arc length, rotation angle, radius of curvature, and angular velocity. \u2022 Calculate the angular velocity of a car wheel spin. In Kinematics, we studied motion along a straight line and introduced such concepts as displacement, velocity, and acceleration. Two-Dimensional Kinematics dealt with motion in two dimensions. Projectile motion is a special case of two-dimensional This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 6 | Gravitation and Uniform Circular Motion 221 kinematics in which the object is projected into the air, while being subject to the gravitational force, and lands a distance away. In this chapter, we consider situations where the object does not land but moves in a curve. We begin the study of uniform circular motion by defining two angular quantities needed to describe rotational motion. Rotation Angle When objects rotate about some axis\u2014for example, when the CD (compact disc) in Figure 6.2 rotates about its center\u2014each point in the object follows a circular arc. Consider a line from the center of the CD to its edge. Each pit used to record sound along this line moves through the same angle in the same amount of time. The rotation angle is the amount of rotation and is analogous to linear distance. We define the rotation angle \u0394 to be the ratio of the arc length to the radius of curvature: \u0394 = \u0394. (6.1) Figure 6.2 All points on a CD travel in circular arcs. The pits along a line from the center to the edge all move through the same angle \u0394", " in a time \u0394. Figure 6.3 The radius of a circle is rotated through an angle \u0394. The arc length \u0394s is described on the circumference. The arc length \u0394 is the distance traveled along a circular path as shown in Figure 6.3 Note that is the radius of curvature of the circular path. We know that for one complete revolution, the arc length is the circumference of a circle of radius. The circumference of a circle is 2\u03c0. Thus for one complete revolution the rotation angle is \u0394 = 2\u03c0 = 2\u03c0. This result is the basis for defining the units used to measure rotation angles, \u0394 to be radians (rad), defined so that A comparison of some useful angles expressed in both degrees and radians is shown in Table 6.1. 2\u03c0 rad = 1 revolution. (6.2) (6.3) 222 Chapter 6 | Gravitation and Uniform Circular Motion Table 6.1 Comparison of Angular Units Degree Measures Radian Measure 30\u00ba 60\u00ba 90\u00ba 120\u00ba 135\u00ba 180\u00ba 6 3 2 2\u03c0 3 3\u03c0 4 Figure 6.4 Points 1 and 2 rotate through the same angle ( \u0394 ), but point 2 moves through a greater arc length (\u0394) because it is at a greater distance from the center of rotation (). If \u0394 = 2 rad, then the CD has made one complete revolution, and every point on the CD is back at its original position. Because there are 360\u00ba in a circle or one revolution, the relationship between radians and degrees is thus so that Angular Velocity 2 rad = 360\u00ba 1 rad = 360\u00ba 2\u03c0 \u2248 57.3\u00ba. How fast is an object rotating? We define angular velocity as the rate of change of an angle. In symbols, this is = \u0394 \u0394, (6.4) (6.5) (6.6) where an angular rotation \u0394 takes place in a time \u0394. The greater the rotation angle in a given amount of time, the greater the angular velocity. The units for angular velocity are radians per second (rad/s). Angular velocity is analogous to linear velocity. To get the precise relationship between angular and linear velocity, we again consider a pit on the rotating CD. This pit moves an arc length \u0394 in a time \u0394, and so it has a linear velocity = \u0394 \u0394. From \u0394 = \u0394 we see that \u0394 = \u0394. Substituting this into the expression for gives = \u0394 \u0394 =. (6.7) (6.8) This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 6 | Gravitation and Uniform Circular Motion We write this relationship in two different ways and gain two different insights: = or =. 223 (6.9) states that the linear velocity is proportional to the distance from the center of The first relationship in = or = rotation, thus, it is largest for a point on the rim (largest ), as you might expect. We can also call this linear speed of a point on the rim the tangential speed. The second relationship in = or = moving car. Note that the speed of a point on the rim of the tire is the same as the speed of the car. See Figure 6.5. So the faster the car moves, the faster the tire spins\u2014large means a large, because =. Similarly, a larger-radius tire rotating at the same angular velocity ( ) will produce a greater linear speed ( ) for the car. can be illustrated by considering the tire of a Figure 6.5 A car moving at a velocity to the right has a tire rotating with an angular velocity.The speed of the tread of the tire relative to the axle is, the same as if the car were jacked up. Thus the car moves forward at linear velocity =, where is the tire radius. A larger angular velocity for the tire means a greater velocity for the car. Example 6.1 How Fast Does a Car Tire Spin? Calculate the angular velocity of a 0.300 m radius car tire when the car travels at 15.0 m/s (about 54 km/h ). See Figure 6.5. Strategy Because the linear speed of the tire rim is the same as the speed of the car, we have = 15.0 m/s. The radius of the tire is given to be = 0.300 m. Knowing and, we can use the second relationship in =, = to calculate the angular velocity. Solution To calculate the angular velocity, we will use the following relationship: Substituting the knowns, Discussion =. = 15.0 m/s 0.300 m = 50.0 rad/s. (6.10) (6.11) When we cancel units in the above calculation, we get 50.0/s. But the angular velocity must have units of rad/s. Because radians are actually unitless (radians are defined as a ratio of distance), we can simply insert them into the answer for", " the angular velocity. Also note that if an earth mover with much larger tires, say 1.20 m in radius, were moving at the same speed of 15.0 m/s, its tires would rotate more slowly. They would have an angular velocity = (15.0 m/s) / (1.20 m) = 12.5 rad/s. (6.12) 224 Chapter 6 | Gravitation and Uniform Circular Motion Both and have directions (hence they are angular and linear velocities, respectively). Angular velocity has only two directions with respect to the axis of rotation\u2014it is either clockwise or counterclockwise. Linear velocity is tangent to the path, as illustrated in Figure 6.6. Take-Home Experiment Tie an object to the end of a string and swing it around in a horizontal circle above your head (swing at your wrist). Maintain uniform speed as the object swings and measure the angular velocity of the motion. What is the approximate speed of the object? Identify a point close to your hand and take appropriate measurements to calculate the linear speed at this point. Identify other circular motions and measure their angular velocities. Figure 6.6 As an object moves in a circle, here a fly on the edge of an old-fashioned vinyl record, its instantaneous velocity is always tangent to the circle. The direction of the angular velocity is clockwise in this case. PhET Explorations: Ladybug Revolution Figure 6.7 Ladybug Revolution (http://cnx.org/content/m54992/1.2/rotation_en.jar) Join the ladybug in an exploration of rotational motion. Rotate the merry-go-round to change its angle, or choose a constant angular velocity or angular acceleration. Explore how circular motion relates to the bug's x,y position, velocity, and acceleration using vectors or graphs. 6.2 Centripetal Acceleration Learning Objectives By the end of this section, you will be able to: \u2022 Establish the expression for centripetal acceleration. \u2022 Explain the centrifuge. We know from kinematics that acceleration is a change in velocity, either in its magnitude or in its direction, or both. In uniform circular motion, the direction of the velocity changes constantly, so there is always an associated acceleration, even though the magnitude of the velocity might be constant. You experience this acceleration yourself when you turn a corner in your car. (If you hold the wheel steady during a turn and move at constant speed, you are in uniform circular motion.) What you notice is a sideways acceleration because you and the car are changing direction. The sharper the curve and the greater your speed, the more noticeable this acceleration will become. In this section we examine the direction and magnitude of that acceleration. Figure 6.8 shows an object moving in a circular path at constant speed. The direction of the instantaneous velocity is shown at two points along the path. Acceleration is in the direction of the change in velocity, which points directly toward the center of rotation (the center of the circular path). This pointing is shown with the vector diagram in the figure. We call the acceleration of This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 6 | Gravitation and Uniform Circular Motion 225 an object moving in uniform circular motion (resulting from a net external force) the centripetal acceleration( c ); centripetal means \u201ctoward the center\u201d or \u201ccenter seeking.\u201d Figure 6.8 The directions of the velocity of an object at two different points are shown, and the change in velocity \u0394v is seen to point directly toward the center of curvature. (See small inset.) Because ac = \u0394v / \u0394, the acceleration is also toward the center; a is called centripetal acceleration. (Because \u0394 is very small, the arc length \u0394 is equal to the chord length \u0394 for small time differences.) The direction of centripetal acceleration is toward the center of curvature, but what is its magnitude? Note that the triangle formed by the velocity vectors and the one formed by the radii and \u0394 are similar. Both the triangles ABC and PQR are isosceles triangles (two equal sides). The two equal sides of the velocity vector triangle are the speeds 1 = 2 =. Using the properties of two similar triangles, we obtain Acceleration is \u0394 / \u0394, and so we first solve this expression for \u0394 : \u0394 = \u0394. Then we divide this by \u0394, yielding \u0394 = \u0394. \u0394 \u0394 = \u00d7 \u0394 \u0394. (6.13) (6.14) (6.15) Finally, noting that \u0394 / \u0394 = c and that \u0394 / \u0394 =, the linear or tangential speed, we see that the magnitude of the centripetal acceleration is c = 2, which is the acceleration of an object in a circle of radius at a speed", ". So, centripetal acceleration is greater at high speeds and in sharp curves (smaller radius), as you have noticed when driving a car. But it is a bit surprising that c is proportional to speed squared, implying, for example, that it is four times as hard to take a curve at 100 km/h than at 50 km/h. A sharp corner has a small radius, so that c is greater for tighter turns, as you have probably noticed. (6.16) It is also useful to express c in terms of angular velocity. Substituting = into the above expression, we find c = ()2 / = 2. We can express the magnitude of centripetal acceleration using either of two equations: c = 2 ; c = 2. (6.17) Recall that the direction of c is toward the center. You may use whichever expression is more convenient, as illustrated in examples below. 226 Chapter 6 | Gravitation and Uniform Circular Motion A centrifuge (see Figure 6.9b) is a rotating device used to separate specimens of different densities. High centripetal acceleration significantly decreases the time it takes for separation to occur, and makes separation possible with small samples. Centrifuges are used in a variety of applications in science and medicine, including the separation of single cell suspensions such as bacteria, viruses, and blood cells from a liquid medium and the separation of macromolecules, such as DNA and protein, from a solution. Centrifuges are often rated in terms of their centripetal acceleration relative to acceleration due to gravity () ; maximum centripetal acceleration of several hundred thousand is possible in a vacuum. Human centrifuges, extremely large centrifuges, have been used to test the tolerance of astronauts to the effects of accelerations larger than that of Earth's gravity. Example 6.2 How Does the Centripetal Acceleration of a Car Around a Curve Compare with That Due to Gravity? What is the magnitude of the centripetal acceleration of a car following a curve of radius 500 m at a speed of 25.0 m/s (about 90 km/h)? Compare the acceleration with that due to gravity for this fairly gentle curve taken at highway speed. See Figure 6.9(a). Strategy Because and are given, the first expression in c = 2 ; c = 2 is the most convenient to use. Solution Entering the given values of = 25.0 m/s and = 500 m into the first expression for c gives c = 2 = (25.0 m/s)2 500 m = 1.25 m/s2. (6.18) Discussion To compare this with the acceleration due to gravity ( = 9.80 m/s2), we take the ratio of c / = seat belt. 9.80 m/s2 1.25 m/s2 / = 0.128. Thus, c = 0.128 g and is noticeable especially if you were not wearing a This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 6 | Gravitation and Uniform Circular Motion 227 Figure 6.9 (a) The car following a circular path at constant speed is accelerated perpendicular to its velocity, as shown. The magnitude of this centripetal acceleration is found in Example 6.2. (b) A particle of mass in a centrifuge is rotating at constant angular velocity. It must be accelerated perpendicular to its velocity or it would continue in a straight line. The magnitude of the necessary acceleration is found in Example 6.3. Example 6.3 How Big Is the Centripetal Acceleration in an Ultracentrifuge? Calculate the centripetal acceleration of a point 7.50 cm from the axis of an ultracentrifuge spinning at 7.5 \u00d7 104 rev/min. Determine the ratio of this acceleration to that due to gravity. See Figure 6.9(b). Strategy The term rev/min stands for revolutions per minute. By converting this to radians per second, we obtain the angular velocity. Because is given, we can use the second expression in the equation c = 2 c = 2 to calculate the centripetal acceleration. Solution To convert 7.50\u00d7104 rev / min to radians per second, we use the facts that one revolution is 2\u03c0rad and one minute is 60.0 s. Thus, = 7.50\u00d7104 rev min \u00d7 2\u03c0 rad 1 rev \u00d7 1 min 60.0 s = 7854 rad/s. Now the centripetal acceleration is given by the second expression in c = 2 ; c = 2 as Converting 7.50 cm to meters and substituting known values gives c = (0.0750 m)(7854 rad/s)2 = 4.63\u00d7106 m/s2. c = 2. (6.19) (6.20) (", "6.21) 228 Chapter 6 | Gravitation and Uniform Circular Motion Note that the unitless radians are discarded in order to get the correct units for centripetal acceleration. Taking the ratio of c to yields = 4.63\u00d7106 c 9.80 = 4.72\u00d7105. (6.22) Discussion This last result means that the centripetal acceleration is 472,000 times as strong as. It is no wonder that such high centrifuges are called ultracentrifuges. The extremely large accelerations involved greatly decrease the time needed to cause the sedimentation of blood cells or other materials. Of course, a net external force is needed to cause any acceleration, just as Newton proposed in his second law of motion. So a net external force is needed to cause a centripetal acceleration. In Centripetal Force, we will consider the forces involved in circular motion. PhET Explorations: Ladybug Motion 2D Learn about position, velocity and acceleration vectors. Move the ladybug by setting the position, velocity or acceleration, and see how the vectors change. Choose linear, circular or elliptical motion, and record and playback the motion to analyze the behavior. Figure 6.10 Ladybug Motion 2D (http://cnx.org/content/m54995/1.2/ladybug-motion-2d_en.jar) 6.3 Centripetal Force Learning Objectives By the end of this section, you will be able to: \u2022 Calculate coefficient of friction on a car tire. \u2022 Calculate ideal speed and angle of a car on a turn. Any force or combination of forces can cause a centripetal or radial acceleration. Just a few examples are the tension in the rope on a tether ball, the force of Earth's gravity on the Moon, friction between roller skates and a rink floor, a banked roadway's force on a car, and forces on the tube of a spinning centrifuge. Any net force causing uniform circular motion is called a centripetal force. The direction of a centripetal force is toward the center of curvature, the same as the direction of centripetal acceleration. According to Newton's second law of motion, net force is mass times acceleration: net F =. For uniform circular motion, the acceleration is the centripetal acceleration\u2014 =. Thus, the magnitude of centripetal force Fc is Fc = c. (6.23) By using the expressions for centripetal acceleration from = 2 force Fc in terms of mass, velocity, angular velocity, and radius of curvature: = 2, we get two expressions for the centripetal = 2. You may use whichever expression for centripetal force is more convenient. Centripetal force c is always perpendicular to the path and pointing to the center of curvature, because a is perpendicular to the velocity and pointing to the center of curvature. = 2 (6.24) Note that if you solve the first expression for, you get This implies that for a given mass and velocity, a large centripetal force causes a small radius of curvature\u2014that is, a tight curve. = 2. (6.25) This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 6 | Gravitation and Uniform Circular Motion 229 Figure 6.11 The frictional force supplies the centripetal force and is numerically equal to it. Centripetal force is perpendicular to velocity and causes uniform circular motion. The larger the Fc, the smaller the radius of curvature and the sharper the curve. The second curve has the same, but a larger Fc produces a smaller \u2032. Example 6.4 What Coefficient of Friction Do Car Tires Need on a Flat Curve? (a) Calculate the centripetal force exerted on a 900 kg car that negotiates a 500 m radius curve at 25.0 m/s. (b) Assuming an unbanked curve, find the minimum static coefficient of friction, between the tires and the road, static friction being the reason that keeps the car from slipping (see Figure 6.12). Strategy and Solution for (a) We know that c = 2. Thus, Strategy for (b) c = 2 = (900 kg)(25.0 m/s)2 (500 m) = 1125 N. (6.26) Figure 6.12 shows the forces acting on the car on an unbanked (level ground) curve. Friction is to the left, keeping the car from slipping, and because it is the only horizontal force acting on the car, the friction is the centripetal force in this case. We know that the maximum static friction (at which the tires roll but do not slip) is s, where s is the static coefficient of friction and N", " is the normal force. The normal force equals the car's weight on level ground, so that =. Thus the centripetal force in this situation is c = = s = s. (6.27) Now we have a relationship between centripetal force and the coefficient of friction. Using the first expression for c from the equation c = 2 c = 2 2 = s. (6.28) (6.29) We solve this for s, noting that mass cancels, and obtain 230 Chapter 6 | Gravitation and Uniform Circular Motion Solution for (b) Substituting the knowns, s = 2. s = (25.0 m/s)2 (500 m)(9.80 m/s2) = 0.13. (6.30) (6.31) (Because coefficients of friction are approximate, the answer is given to only two digits.) Discussion We could also solve part (a) using the first expression in c = 2 c = 2 because and are given. The coefficient of friction found in part (b) is much smaller than is typically found between tires and roads. The car will still negotiate the curve if the coefficient is greater than 0.13, because static friction is a responsive force, being able to assume a value less than but no more than s. A higher coefficient would also allow the car to negotiate the curve at a higher speed, but if the coefficient of friction is less, the safe speed would be less than 25 m/s. Note that mass cancels, implying that in this example, it does not matter how heavily loaded the car is to negotiate the turn. Mass cancels because friction is assumed proportional to the normal force, which in turn is proportional to mass. If the surface of the road were banked, the normal force would be less as will be discussed below. Figure 6.12 This car on level ground is moving away and turning to the left. The centripetal force causing the car to turn in a circular path is due to friction between the tires and the road. A minimum coefficient of friction is needed, or the car will move in a larger-radius curve and leave the roadway. Let us now consider banked curves, where the slope of the road helps you negotiate the curve. See Figure 6.13. The greater the angle, the faster you can take the curve. Race tracks for bikes as well as cars, for example, often have steeply banked curves. In an \u201cideally banked curve,\u201d the angle is such that you can negotiate the curve at a certain speed without the aid of friction between the tires and the road. We will derive an expression for for an ideally banked curve and consider an example related to it. For ideal banking, the net external force equals the horizontal centripetal force in the absence of friction. The components of the normal force N in the horizontal and vertical directions must equal the centripetal force and the weight of the car, respectively. In cases in which forces are not parallel, it is most convenient to consider components along perpendicular axes\u2014in this case, the vertical and horizontal directions. Figure 6.13 shows a free body diagram for a car on a frictionless banked curve. If the angle is ideal for the speed and radius, then the net external force will equal the necessary centripetal force. The only two external forces acting on the car are its weight w and the normal force of the road N. (A frictionless surface can only exert a force perpendicular to the surface\u2014that is, a normal force.) These two forces must add to give a net external force that is horizontal toward the center of curvature and has This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 6 | Gravitation and Uniform Circular Motion 231 magnitude mv2 /r. Because this is the crucial force and it is horizontal, we use a coordinate system with vertical and horizontal axes. Only the normal force has a horizontal component, and so this must equal the centripetal force\u2014that is, sin = 2. (6.32) Because the car does not leave the surface of the road, the net vertical force must be zero, meaning that the vertical components of the two external forces must be equal in magnitude and opposite in direction. From the figure, we see that the vertical component of the normal force is cos, and the only other vertical force is the car's weight. These must be equal in magnitude; thus, Now we can combine the last two equations to eliminate and get an expression for, as desired. Solving the second equation for = / (cos ), and substituting this into the first yields cos =. = 2 sin cos tan() = 2 2 tan = Taking the inverse tangent gives = tan\u22121 2 (ideally banked curve, no friction). (6.33) (6.34) (6.35) (", "6.36) This expression can be understood by considering how depends on and. A large will be obtained for a large and a small. That is, roads must be steeply banked for high speeds and sharp curves. Friction helps, because it allows you to take the curve at greater or lower speed than if the curve is frictionless. Note that does not depend on the mass of the vehicle. Figure 6.13 The car on this banked curve is moving away and turning to the left. Example 6.5 What Is the Ideal Speed to Take a Steeply Banked Tight Curve? Curves on some test tracks and race courses, such as the Daytona International Speedway in Florida, are very steeply banked. This banking, with the aid of tire friction and very stable car configurations, allows the curves to be taken at very high speed. To illustrate, calculate the speed at which a 100 m radius curve banked at 65.0\u00b0 should be driven if the road is frictionless. Strategy We first note that all terms in the expression for the ideal angle of a banked curve except for speed are known; thus, we need only rearrange it so that speed appears on the left-hand side and then substitute known quantities. Solution Starting with tan = 2 (6.37) 232 we get Noting that tan 65.0\u00ba = 2.14, we obtain = ( tan )1 / 2. = (100 m)(9.80 m/s2)(2.14) Chapter 6 | Gravitation and Uniform Circular Motion 1 / 2 (6.38) (6.39) Discussion = 45.8 m/s. This is just about 165 km/h, consistent with a very steeply banked and rather sharp curve. Tire friction enables a vehicle to take the curve at significantly higher speeds. Calculations similar to those in the preceding examples can be performed for a host of interesting situations in which centripetal force is involved\u2014a number of these are presented in this chapter's Problems and Exercises. Take-Home Experiment Ask a friend or relative to swing a golf club or a tennis racquet. Take appropriate measurements to estimate the centripetal acceleration of the end of the club or racquet. You may choose to do this in slow motion. PhET Explorations: Gravity and Orbits Move the sun, earth, moon and space station to see how it affects their gravitational forces and orbital paths. Visualize the sizes and distances between different heavenly bodies, and turn off gravity to see what would happen without it! Figure 6.14 Gravity and Orbits (http://cnx.org/content/m55002/1.2/gravity-and-orbits_en.jar) 6.4 Fictitious Forces and Non-inertial Frames: The Coriolis Force Learning Objectives By the end of this section, you will be able to: \u2022 Discuss the inertial frame of reference. \u2022 Discuss the non-inertial frame of reference. \u2022 Describe the effects of the Coriolis force. What do taking off in a jet airplane, turning a corner in a car, riding a merry-go-round, and the circular motion of a tropical cyclone have in common? Each exhibits fictitious forces\u2014unreal forces that arise from motion and may seem real, because the observer\u2019s frame of reference is accelerating or rotating. When taking off in a jet, most people would agree it feels as if you are being pushed back into the seat as the airplane accelerates down the runway. Yet a physicist would say that you tend to remain stationary while the seat pushes forward on you, and there is no real force backward on you. An even more common experience occurs when you make a tight curve in your car\u2014say, to the right. You feel as if you are thrown (that is, forced) toward the left relative to the car. Again, a physicist would say that you are going in a straight line but the car moves to the right, and there is no real force on you to the left. Recall Newton\u2019s first law. This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 6 | Gravitation and Uniform Circular Motion 233 Figure 6.15 (a) The car driver feels herself forced to the left relative to the car when she makes a right turn. This is a fictitious force arising from the use of the car as a frame of reference. (b) In the Earth\u2019s frame of reference, the driver moves in a straight line, obeying Newton\u2019s first law, and the car moves to the right. There is no real force to the left on the driver relative to Earth. There is a real force to the right on the car to make it turn. We can reconcile these points of view by examining the frames of reference used. Let us concentrate on people in a car. Passengers instinctively use the", " car as a frame of reference, while a physicist uses Earth. The physicist chooses Earth because it is very nearly an inertial frame of reference\u2014one in which all forces are real (that is, in which all forces have an identifiable physical origin). In such a frame of reference, Newton\u2019s laws of motion take the form given in Dynamics: Newton's Laws of Motion The car is a non-inertial frame of reference because it is accelerated to the side. The force to the left sensed by car passengers is a fictitious force having no physical origin. There is nothing real pushing them left\u2014the car, as well as the driver, is actually accelerating to the right. Let us now take a mental ride on a merry-go-round\u2014specifically, a rapidly rotating playground merry-go-round. You take the merry-go-round to be your frame of reference because you rotate together. In that non-inertial frame, you feel a fictitious force, named centrifugal force (not to be confused with centripetal force), trying to throw you off. You must hang on tightly to counteract the centrifugal force. In Earth\u2019s frame of reference, there is no force trying to throw you off. Rather you must hang on to make yourself go in a circle because otherwise you would go in a straight line, right off the merry-go-round. Figure 6.16 (a) A rider on a merry-go-round feels as if he is being thrown off. This fictitious force is called the centrifugal force\u2014it explains the rider\u2019s motion in the rotating frame of reference. (b) In an inertial frame of reference and according to Newton\u2019s laws, it is his inertia that carries him off and not a real force (the unshaded rider has net = 0 and heads in a straight line). A real force, centripetal, is needed to cause a circular path. This inertial effect, carrying you away from the center of rotation if there is no centripetal force to cause circular motion, is put to good use in centrifuges (see Figure 6.17). A centrifuge spins a sample very rapidly, as mentioned earlier in this chapter. Viewed from the rotating frame of reference, the fictitious centrifugal force throws particles outward, hastening their sedimentation. The greater the angular velocity, the greater the centrifugal force. But what really happens is that the inertia of the particles carries them along a line tangent to the circle while the test tube is forced in a circular path by a centripetal force. 234 Chapter 6 | Gravitation and Uniform Circular Motion Figure 6.17 Centrifuges use inertia to perform their task. Particles in the fluid sediment come out because their inertia carries them away from the center of rotation. The large angular velocity of the centrifuge quickens the sedimentation. Ultimately, the particles will come into contact with the test tube walls, which will then supply the centripetal force needed to make them move in a circle of constant radius. Let us now consider what happens if something moves in a frame of reference that rotates. For example, what if you slide a ball directly away from the center of the merry-go-round, as shown in Figure 6.18? The ball follows a straight path relative to Earth (assuming negligible friction) and a path curved to the right on the merry-go-round\u2019s surface. A person standing next to the merry-go-round sees the ball moving straight and the merry-go-round rotating underneath it. In the merry-go-round\u2019s frame of reference, we explain the apparent curve to the right by using a fictitious force, called the Coriolis force, that causes the ball to curve to the right. The fictitious Coriolis force can be used by anyone in that frame of reference to explain why objects follow curved paths and allows us to apply Newton\u2019s Laws in non-inertial frames of reference. Figure 6.18 Looking down on the counterclockwise rotation of a merry-go-round, we see that a ball slid straight toward the edge follows a path curved to the right. The person slides the ball toward point B, starting at point A. Both points rotate to the shaded positions (A\u2019 and B\u2019) shown in the time that the ball follows the curved path in the rotating frame and a straight path in Earth\u2019s frame. Up until now, we have considered Earth to be an inertial frame of reference with little or no worry about effects due to its rotation. Yet such effects do exist\u2014in the rotation of weather systems, for example. Most consequences of Earth\u2019s rotation can be qualitatively understood by analogy with the merry-go-round. Viewed from above the North Pole, Earth rotates counterclockwise, as does the merry-go-round in Figure 6.18. As on the merry", "-go-round, any motion in Earth\u2019s northern hemisphere experiences a This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 6 | Gravitation and Uniform Circular Motion 235 Coriolis force to the right. Just the opposite occurs in the southern hemisphere; there, the force is to the left. Because Earth\u2019s angular velocity is small, the Coriolis force is usually negligible, but for large-scale motions, such as wind patterns, it has substantial effects. The Coriolis force causes hurricanes in the northern hemisphere to rotate in the counterclockwise direction, while the tropical cyclones (what hurricanes are called below the equator) in the southern hemisphere rotate in the clockwise direction. The terms hurricane, typhoon, and tropical storm are regionally-specific names for tropical cyclones, storm systems characterized by low pressure centers, strong winds, and heavy rains. Figure 6.19 helps show how these rotations take place. Air flows toward any region of low pressure, and tropical cyclones contain particularly low pressures. Thus winds flow toward the center of a tropical cyclone or a low-pressure weather system at the surface. In the northern hemisphere, these inward winds are deflected to the right, as shown in the figure, producing a counterclockwise circulation at the surface for low-pressure zones of any type. Low pressure at the surface is associated with rising air, which also produces cooling and cloud formation, making low-pressure patterns quite visible from space. Conversely, wind circulation around high-pressure zones is clockwise in the northern hemisphere but is less visible because high pressure is associated with sinking air, producing clear skies. The rotation of tropical cyclones and the path of a ball on a merry-go-round can just as well be explained by inertia and the rotation of the system underneath. When non-inertial frames are used, fictitious forces, such as the Coriolis force, must be invented to explain the curved path. There is no identifiable physical source for these fictitious forces. In an inertial frame, inertia explains the path, and no force is found to be without an identifiable source. Either view allows us to describe nature, but a view in an inertial frame is the simplest and truest, in the sense that all forces have real origins and explanations. Figure 6.19 (a) The counterclockwise rotation of this northern hemisphere hurricane is a major consequence of the Coriolis force. (credit: NASA) (b) Without the Coriolis force, air would flow straight into a low-pressure zone, such as that found in tropical cyclones. (c) The Coriolis force deflects the winds to the right, producing a counterclockwise rotation. (d) Wind flowing away from a high-pressure zone is also deflected to the right, producing a clockwise rotation. (e) The opposite direction of rotation is produced by the Coriolis force in the southern hemisphere, leading to tropical cyclones. (credit: NASA) 6.5 Newton's Universal Law of Gravitation Learning Objectives By the end of this section, you will be able to: \u2022 Explain Earth's gravitational force. \u2022 Describe the gravitational effect of the Moon on Earth. \u2022 Discuss weightlessness in space. \u2022 Understand the Cavendish experiment. The information presented in this section supports the following AP\u00ae learning objectives and science practices: \u2022 2.B.2.1 The student is able to apply = 2 to calculate the gravitational field due to an object with mass M, where the field is a vector directed toward the center of the object of mass M. (S.P. 2.2) \u2022 2.B.2.2 The student is able to approximate a numerical value of the gravitational field (g) near the surface of an object from its radius and mass relative to those of the Earth or other reference objects. (S.P. 2.2) \u2022 3.A.3.4. The student is able to make claims about the force on an object due to the presence of other objects with the same property: mass, electric charge. (S.P. 6.1, 6.4) 236 Chapter 6 | Gravitation and Uniform Circular Motion What do aching feet, a falling apple, and the orbit of the Moon have in common? Each is caused by the gravitational force. Our feet are strained by supporting our weight\u2014the force of Earth's gravity on us. An apple falls from a tree because of the same force acting a few meters above Earth's surface. And the Moon orbits Earth because gravity is able to supply the necessary centripetal force at a distance of hundreds of millions of meters. In fact, the same force causes planets to orbit the Sun, stars to orbit the center of the galaxy, and galaxies to cluster together. Gravity is another example of underlying simplicity in nature. It is the weakest of the four basic forces found in nature, and in", " some ways the least understood. It is a force that acts at a distance, without physical contact, and is expressed by a formula that is valid everywhere in the universe, for masses and distances that vary from the tiny to the immense. Sir Isaac Newton was the first scientist to precisely define the gravitational force, and to show that it could explain both falling bodies and astronomical motions. See Figure 6.20. But Newton was not the first to suspect that the same force caused both our weight and the motion of planets. His forerunner Galileo Galilei had contended that falling bodies and planetary motions had the same cause. Some of Newton's contemporaries, such as Robert Hooke, Christopher Wren, and Edmund Halley, had also made some progress toward understanding gravitation. But Newton was the first to propose an exact mathematical form and to use that form to show that the motion of heavenly bodies should be conic sections\u2014circles, ellipses, parabolas, and hyperbolas. This theoretical prediction was a major triumph\u2014it had been known for some time that moons, planets, and comets follow such paths, but no one had been able to propose a mechanism that caused them to follow these paths and not others. This was one of the earliest examples of a theory derived from empirical evidence doing more than merely describing those empirical results; it made claims about the fundamental workings of the universe. Figure 6.20 According to early accounts, Newton was inspired to make the connection between falling bodies and astronomical motions when he saw an apple fall from a tree and realized that if the gravitational force could extend above the ground to a tree, it might also reach the Sun. The inspiration of Newton's apple is a part of worldwide folklore and may even be based in fact. Great importance is attached to it because Newton's universal law of gravitation and his laws of motion answered very old questions about nature and gave tremendous support to the notion of underlying simplicity and unity in nature. Scientists still expect underlying simplicity to emerge from their ongoing inquiries into nature. The gravitational force is relatively simple. It is always attractive, and it depends only on the masses involved and the distance between them. Stated in modern language, Newton's universal law of gravitation states that every particle in the universe attracts every other particle with a force along a line joining them. The force is directly proportional to the product of their masses and inversely proportional to the square of the distance between them. This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 6 | Gravitation and Uniform Circular Motion 237 Figure 6.21 Gravitational attraction is along a line joining the centers of mass of these two bodies. The magnitude of the force is the same on each, consistent with Newton's third law. Misconception Alert The magnitude of the force on each object (one has larger mass than the other) is the same, consistent with Newton's third law. The bodies we are dealing with tend to be large. To simplify the situation we assume that the body acts as if its entire mass is concentrated at one specific point called the center of mass (CM), which will be further explored in Linear Momentum and Collisions. For two bodies having masses and with a distance between their centers of mass, the equation for Newton's universal law of gravitation is = 2, (6.40) where is the magnitude of the gravitational force and is a proportionality factor called the gravitational constant. is a universal gravitational constant\u2014that is, it is thought to be the same everywhere in the universe. It has been measured experimentally to be = 6.673\u00d710\u221211N \u22c5 m2 kg2 (6.41) in SI units. Note that the units of are such that a force in newtons is obtained from = 2, when considering masses in kilograms and distance in meters. For example, two 1.000 kg masses separated by 1.000 m will experience a gravitational attraction of 6.673\u00d710\u221211 N. This is an extraordinarily small force. The small magnitude of the gravitational force is consistent with everyday experience. We are unaware that even large objects like mountains exert gravitational forces on us. In fact, our body weight is the force of attraction of the entire Earth on us with a mass of 6\u00d71024 kg. The experiment to measure G was first performed by Cavendish, and is explained in more detail later. The fundamental concept it is based on is having a known mass on a spring with a known force (or spring) constant. Then, a second known mass is placed at multiple known distances from the first, and the amount of stretch in the spring resulting from the gravitational attraction of the two masses is measured. Recall that the acceleration due to gravity is about 9.80 m/s2 on Earth. We can now determine why this is so. The weight of an object mg is the gravitational force between it and Earth. Substituting mg for in Newton's universal law of gravitation gives = 2 where is the mass", " of the object, is the mass of Earth, and is the distance to the center of Earth (the distance between the centers of mass of the object and Earth). See Figure 6.22. The mass of the object cancels, leaving an equation for :, (6.42) 238 Chapter 6 | Gravitation and Uniform Circular Motion = 2. Substituting known values for Earth's mass and radius (to three significant figures), 5.98\u00d71024 kg (6.38\u00d7106 m)2 6.67\u00d710\u221211N \u22c5 m2 kg2 \u00d7 = and we obtain a value for the acceleration of a falling body: = 9.80 m/s2., (6.43) (6.44) (6.45) Figure 6.22 The distance between the centers of mass of Earth and an object on its surface is very nearly the same as the radius of Earth, because Earth is so much larger than the object. This is the expected value and is independent of the body's mass. Newton's law of gravitation takes Galileo's observation that all masses fall with the same acceleration a step further, explaining the observation in terms of a force that causes objects to fall\u2014in fact, in terms of a universally existing force of attraction between masses. Gravitational Mass and Inertial Mass Notice that, in Equation 6.40, the mass of the objects under consideration is directly proportional to the gravitational force. More mass means greater forces, and vice versa. However, we have already seen the concept of mass before in a different context. In Chapter 4, you read that mass is a measure of inertia. However, we normally measure the mass of an object by measuring the force of gravity (F) on it. How do we know that inertial mass is identical to gravitational mass? Assume that we compare the mass of two objects. The objects have inertial masses m1 and m2. If the objects balance each other on a pan balance, we can conclude that they have the same gravitational mass, that is, that they experience the same force due to gravity, F. Using Newton's second law of motion, F = ma, we can write m1 a1 = m2 a2. If we can show that the two objects experience the same acceleration due to gravity, we can conclude that m1 = m2, that is, that the objects' inertial masses are equal. In fact, Galileo and others conducted experiments to show that, when factors such as wind resistance are kept constant, all objects, regardless of their mass, experience the same acceleration due to gravity. Galileo is famously said to have dropped two balls of different masses off the leaning tower of Pisa to demonstrate this. The balls accelerated at the same rate. Since acceleration due to gravity is constant for all objects on Earth, regardless of their mass or composition, i.e., a1 = a2, then m1 = m2. Thus, we can conclude that inertial mass is identical to gravitational mass. This allows us to calculate the acceleration of free fall due to gravity, such as in the orbits of planets. Take-Home Experiment Take a marble, a ball, and a spoon and drop them from the same height. Do they hit the floor at the same time? If you drop a piece of paper as well, does it behave like the other objects? Explain your observations. Making Connections: Gravitation, Other Forces, and General Relativity Attempts are still being made to understand the gravitational force. As we shall see in Particle Physics, modern physics is exploring the connections of gravity to other forces, space, and time. General relativity alters our view of gravitation, leading us to think of gravitation as bending space and time. This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 6 | Gravitation and Uniform Circular Motion 239 Applying the Science Practices: All Objects Have Gravitational Fields We can use the formula developed above, = 2, to calculate the gravitational fields of other objects. For example, the Moon has a radius of 1.7 \u00d7 106 m and a mass of 7.3 \u00d7 1022 kg. The gravitational field on the surface of the Moon can be expressed as = 2 = 6.67\u00d710\u221211 N\u00b7m2 kg2 \u00d7 = 1.685 m/s2 7.3\u00d71022 kg 2 1.7\u00d7106 m This is about 1/6 of the gravity on Earth, which seems reasonable, since the Moon has a much smaller mass than Earth does. A person has a mass of 50 kg. The gravitational field 1.0 m from the person's center of mass can be expressed as = 2 = 6.67\u00d710\u221211 N\u00b7m2 kg2 \u00d7 50 kg (1 m)2 = 3.34\u00d710\u22129 m/s2 This is less than one millionth of the gravitational field at the surface of Earth.", " In the following example, we make a comparison similar to one made by Newton himself. He noted that if the gravitational force caused the Moon to orbit Earth, then the acceleration due to gravity should equal the centripetal acceleration of the Moon in its orbit. Newton found that the two accelerations agreed \u201cpretty nearly.\u201d Example 6.6 Earth's Gravitational Force Is the Centripetal Force Making the Moon Move in a Curved Path (a) Find the acceleration due to Earth's gravity at the distance of the Moon. (b) Calculate the centripetal acceleration needed to keep the Moon in its orbit (assuming a circular orbit about a fixed Earth), and compare it with the value of the acceleration due to Earth's gravity that you have just found. Strategy for (a) This calculation is the same as the one finding the acceleration due to gravity at Earth's surface, except that is the distance from the center of Earth to the center of the Moon. The radius of the Moon's nearly circular orbit is 3.84\u00d7108 m. Solution for (a) Substituting known values into the expression for found above, remembering that is the mass of Earth not the Moon, yields = = 2 6.67\u00d710\u221211N \u22c5 m2 kg2 = 2.70\u00d710\u22123 m/s.2 Strategy for (b) Centripetal acceleration can be calculated using either form of We choose to use the second form: = 2 = 2. \u00d7 5.98\u00d71024 kg (3.84\u00d7108 m)2 (6.46) (6.47) 240 Chapter 6 | Gravitation and Uniform Circular Motion where is the angular velocity of the Moon about Earth. Solution for (b) = 2, Given that the period (the time it takes to make one complete rotation) of the Moon's orbit is 27.3 days, (d) and using we see that The centripetal acceleration is 1 d\u00d724hr d \u00d760min hr \u00d760 s min = 86,400 s = \u0394 \u0394 = 2\u03c0 rad (27.3 d)(86,400 s/d) = 2.66\u00d710\u22126rad s. = 2 = (3.84\u00d7108 m)(2.66\u00d710\u22126 rad/s)2 = 2.72\u00d710\u22123 m/s.2 (6.48) (6.49) (6.50) (6.51) The direction of the acceleration is toward the center of the Earth. Discussion The centripetal acceleration of the Moon found in (b) differs by less than 1% from the acceleration due to Earth's gravity found in (a). This agreement is approximate because the Moon's orbit is slightly elliptical, and Earth is not stationary (rather the Earth-Moon system rotates about its center of mass, which is located some 1700 km below Earth's surface). The clear implication is that Earth's gravitational force causes the Moon to orbit Earth. Why does Earth not remain stationary as the Moon orbits it? This is because, as expected from Newton's third law, if Earth exerts a force on the Moon, then the Moon should exert an equal and opposite force on Earth (see Figure 6.23). We do not sense the Moon's effect on Earth's motion, because the Moon's gravity moves our bodies right along with Earth but there are other signs on Earth that clearly show the effect of the Moon's gravitational force as discussed in Satellites and Kepler's Laws: An Argument for Simplicity. Figure 6.23 (a) Earth and the Moon rotate approximately once a month around their common center of mass. (b) Their center of mass orbits the Sun in an elliptical orbit, but Earth's path around the Sun has \u201cwiggles\u201d in it. Similar wiggles in the paths of stars have been observed and are considered direct evidence of planets orbiting those stars. This is important because the planets' reflected light is often too dim to be observed. Tides Ocean tides are one very observable result of the Moon's gravity acting on Earth. Figure 6.24 is a simplified drawing of the Moon's position relative to the tides. Because water easily flows on Earth's surface, a high tide is created on the side of Earth nearest to the Moon, where the Moon's gravitational pull is strongest. Why is there also a high tide on the opposite side of Earth? The answer is that Earth is pulled toward the Moon more than the water on the far side, because Earth is closer to the Moon. So the water on the side of Earth closest to the Moon is pulled away from Earth, and Earth is pulled away from water on the far side. As Earth rotates, the tidal bulge (an effect of the tidal forces between an orbiting natural satellite and the primary planet that it orbits) keeps its orientation with the Moon. Thus there are two tides per day (the actual tidal period is about 12 hours and 25.", "2 minutes), because the Moon moves in its orbit each day as well). This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 6 | Gravitation and Uniform Circular Motion 241 Figure 6.24 The Moon causes ocean tides by attracting the water on the near side more than Earth, and by attracting Earth more than the water on the far side. The distances and sizes are not to scale. For this simplified representation of the Earth-Moon system, there are two high and two low tides per day at any location, because Earth rotates under the tidal bulge. The Sun also affects tides, although it has about half the effect of the Moon. However, the largest tides, called spring tides, occur when Earth, the Moon, and the Sun are aligned. The smallest tides, called neap tides, occur when the Sun is at a 90\u00ba angle to the Earth-Moon alignment. Figure 6.25 (a, b) Spring tides: The highest tides occur when Earth, the Moon, and the Sun are aligned. (c) Neap tide: The lowest tides occur when the Sun lies at 90\u00ba to the Earth-Moon alignment. Note that this figure is not drawn to scale. Tides are not unique to Earth but occur in many astronomical systems. The most extreme tides occur where the gravitational force is the strongest and varies most rapidly, such as near black holes (see Figure 6.26). A few likely candidates for black holes have been observed in our galaxy. These have masses greater than the Sun but have diameters only a few kilometers across. The tidal forces near them are so great that they can actually tear matter from a companion star. 242 Chapter 6 | Gravitation and Uniform Circular Motion Figure 6.26 A black hole is an object with such strong gravity that not even light can escape it. This black hole was created by the supernova of one star in a two-star system. The tidal forces created by the black hole are so great that it tears matter from the companion star. This matter is compressed and heated as it is sucked into the black hole, creating light and X-rays observable from Earth. \u201dWeightlessness\u201d and Microgravity In contrast to the tremendous gravitational force near black holes is the apparent gravitational field experienced by astronauts orbiting Earth. What is the effect of \u201cweightlessness\u201d upon an astronaut who is in orbit for months? Or what about the effect of weightlessness upon plant growth? Weightlessness doesn't mean that an astronaut is not being acted upon by the gravitational force. There is no \u201czero gravity\u201d in an astronaut's orbit. The term just means that the astronaut is in free-fall, accelerating with the acceleration due to gravity. If an elevator cable breaks, the passengers inside will be in free fall and will experience weightlessness. You can experience short periods of weightlessness in some rides in amusement parks. Figure 6.27 Astronauts experiencing weightlessness on board the International Space Station. (credit: NASA) Microgravity refers to an environment in which the apparent net acceleration of a body is small compared with that produced by Earth at its surface. Many interesting biology and physics topics have been studied over the past three decades in the presence of microgravity. Of immediate concern is the effect on astronauts of extended times in outer space, such as at the International Space Station. Researchers have observed that muscles will atrophy (waste away) in this environment. There is also a corresponding loss of bone mass. Study continues on cardiovascular adaptation to space flight. On Earth, blood pressure is usually higher in the feet than in the head, because the higher column of blood exerts a downward force on it, due to gravity. When standing, 70% of your blood is below the level of the heart, while in a horizontal position, just the opposite occurs. What difference does the absence of this pressure differential have upon the heart? Some findings in human physiology in space can be clinically important to the management of diseases back on Earth. On a somewhat negative note, spaceflight is known to affect the human immune system, possibly making the crew members more vulnerable to infectious diseases. Experiments flown in space also have shown that some bacteria grow faster in microgravity than they do on Earth. However, on a positive note, studies indicate that microbial antibiotic production can increase by a factor of two in space-grown cultures. One hopes to be able to understand these mechanisms so that similar successes can be achieved on the ground. In another area of physics space research, inorganic crystals and protein crystals have been grown in outer space that have much higher quality than any grown on Earth, so crystallography studies on their structure can yield much better results. This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 6 | Gravitation and Uniform Circular Motion 243 Plants have evolved with the stimulus of gravity and with gravity sensors. Roots grow downward and shoots grow upward. Plants might be able to", " provide a life support system for long duration space missions by regenerating the atmosphere, purifying water, and producing food. Some studies have indicated that plant growth and development are not affected by gravity, but there is still uncertainty about structural changes in plants grown in a microgravity environment. The Cavendish Experiment: Then and Now As previously noted, the universal gravitational constant is determined experimentally. This definition was first done accurately by Henry Cavendish (1731\u20131810), an English scientist, in 1798, more than 100 years after Newton published his universal law of gravitation. The measurement of is very basic and important because it determines the strength of one of the four forces in nature. Cavendish's experiment was very difficult because he measured the tiny gravitational attraction between two ordinary-sized masses (tens of kilograms at most), using apparatus like that in Figure 6.28. Remarkably, his value for differs by less than 1% from the best modern value. One important consequence of knowing was that an accurate value for Earth's mass could finally be obtained. This was done by measuring the acceleration due to gravity as accurately as possible and then calculating the mass of Earth from the relationship Newton's universal law of gravitation gives = 2 where is the mass of the object, is the mass of Earth, and is the distance to the center of Earth (the distance between the centers of mass of the object and Earth). See Figure 6.21. The mass of the object cancels, leaving an equation for :, (6.52) Rearranging to solve for yields = 2. = 2. (6.53) (6.54) So can be calculated because all quantities on the right, including the radius of Earth, are known from direct measurements. We shall see in Satellites and Kepler's Laws: An Argument for Simplicity that knowing also allows for the determination of astronomical masses. Interestingly, of all the fundamental constants in physics, is by far the least well determined. The Cavendish experiment is also used to explore other aspects of gravity. One of the most interesting questions is whether the gravitational force depends on substance as well as mass\u2014for example, whether one kilogram of lead exerts the same gravitational pull as one kilogram of water. A Hungarian scientist named Roland von E\u00f6tv\u00f6s pioneered this inquiry early in the 20th century. He found, with an accuracy of five parts per billion, that the gravitational force does not depend on the substance. Such experiments continue today, and have improved upon E\u00f6tv\u00f6s' measurements. Cavendish-type experiments such as those of Eric Adelberger and others at the University of Washington, have also put severe limits on the possibility of a fifth force and have verified a major prediction of general relativity\u2014that gravitational energy contributes to rest mass. Ongoing measurements there use a torsion balance and a parallel plate (not spheres, as Cavendish used) to examine how Newton's law of gravitation works over sub-millimeter distances. On this small-scale, do gravitational effects depart from the inverse square law? So far, no deviation has been observed. Figure 6.28 Cavendish used an apparatus like this to measure the gravitational attraction between the two suspended spheres ( ) and the two on the stand ( ) by observing the amount of torsion (twisting) created in the fiber. Distance between the masses can be varied to check the dependence of the force on distance. Modern experiments of this type continue to explore gravity. Chapter 6 | Gravitation and Uniform Circular Motion 249 Figure 6.31(b) represents the modern or Copernican model. In this model, a small set of rules and a single underlying force explain not only all motions in the solar system, but all other situations involving gravity. The breadth and simplicity of the laws of physics are compelling. As our knowledge of nature has grown, the basic simplicity of its laws has become ever more evident. Figure 6.31 (a) The Ptolemaic model of the universe has Earth at the center with the Moon, the planets, the Sun, and the stars revolving about it in complex superpositions of circular paths. This geocentric model, which can be made progressively more accurate by adding more circles, is purely descriptive, containing no hints as to what are the causes of these motions. (b) The Copernican model has the Sun at the center of the solar system. It is fully explained by a small number of laws of physics, including Newton's universal law of gravitation. Glossary angular velocity:, the rate of change of the angle with which an object moves on a circular path arc length: \u0394, the distance traveled by an object along a circular path banked curve: the curve in a road that is sloping in a manner that helps a vehicle negotiate the curve center of mass: the point where the entire mass of an object can be thought to be concentrated centrifugal force: a fictitious force that tends to throw an object off when the object is rotating in a non-", "inertial frame of reference centripetal acceleration: the acceleration of an object moving in a circle, directed toward the center centripetal force: any net force causing uniform circular motion Coriolis force: reference the fictitious force causing the apparent deflection of moving objects when viewed in a rotating frame of fictitious force: a force having no physical origin gravitational constant, G: a proportionality factor used in the equation for Newton's universal law of gravitation; it is a universal constant\u2014that is, it is thought to be the same everywhere in the universe ideal angle: the angle at which a car can turn safely on a steep curve, which is in proportion to the ideal speed ideal banking: the sloping of a curve in a road, where the angle of the slope allows the vehicle to negotiate the curve at a certain speed without the aid of friction between the tires and the road; the net external force on the vehicle equals the horizontal centripetal force in the absence of friction ideal speed: the maximum safe speed at which a vehicle can turn on a curve without the aid of friction between the tire and the road microgravity: an environment in which the apparent net acceleration of a body is small compared with that produced by Earth at its surface Newton's universal law of gravitation: every particle in the universe attracts every other particle with a force along a line joining them; the force is directly proportional to the product of their masses and inversely proportional to the square of the distance between them non-inertial frame of reference: an accelerated frame of reference pit: a tiny indentation on the spiral track moulded into the top of the polycarbonate layer of CD 250 Chapter 6 | Gravitation and Uniform Circular Motion radians: a unit of angle measurement radius of curvature: radius of a circular path the ratio of the arc length to the radius of curvature on a circular path: rotation angle: \u0394 = \u0394 ultracentrifuge: a centrifuge optimized for spinning a rotor at very high speeds uniform circular motion: the motion of an object in a circular path at constant speed Section Summary 6.1 Rotation Angle and Angular Velocity \u2022 Uniform circular motion is motion in a circle at constant speed. The rotation angle \u0394 is defined as the ratio of the arc length to the radius of curvature: \u0394 = \u0394, where arc length \u0394 is distance traveled along a circular path and is the radius of curvature of the circular path. The quantity \u0394 is measured in units of radians (rad), for which \u2022 The conversion between radians and degrees is 1 rad = 57.3\u00ba. \u2022 Angular velocity is the rate of change of an angle, 2\u03c0 rad = 360\u00ba= 1 revolution. = \u0394 \u0394, where a rotation \u0394 takes place in a time \u0394. The units of angular velocity are radians per second (rad/s). Linear velocity and angular velocity are related by = or =. 6.2 Centripetal Acceleration \u2022 Centripetal acceleration c is the acceleration experienced while in uniform circular motion. It always points toward the center of rotation. It is perpendicular to the linear velocity and has the magnitude \u2022 The unit of centripetal acceleration is m / s2. 6.3 Centripetal Force c = 2 c = 2. \u2022 Centripetal force Fc is any force causing uniform circular motion. It is a \u201ccenter-seeking\u201d force that always points toward the center of rotation. It is perpendicular to linear velocity and has magnitude which can also be expressed as c = c, c = 2 or c = 2 6.4 Fictitious Forces and Non-inertial Frames: The Coriolis Force \u2022 Rotating and accelerated frames of reference are non-inertial. \u2022 Fictitious forces, such as the Coriolis force, are needed to explain motion in such frames. 6.5 Newton's Universal Law of Gravitation \u2022 Newton's universal law of gravitation: Every particle in the universe attracts every other particle with a force along a line joining them. The force is directly proportional to the product of their masses and inversely proportional to the square of the distance between them. In equation form, this is = 2, This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 6 | Gravitation and Uniform Circular Motion 251 where F is the magnitude of the gravitational force. is the gravitational constant, given by = 6.673\u00d710\u201311 N \u22c5 m2/kg2. \u2022 Newton's law of gravitation applies universally. 6.6 Satellites and Kepler's Laws: An Argument for Simplicity \u2022 Kepler's laws are stated for a small mass orbiting a larger mass in near-isolation. Kepler's laws of planetary motion are then as follows: Kepler's first law The orbit of each planet about the Sun is an ellipse with the Sun at one focus. Kepler's second law Each planet moves so that an imaginary line drawn from", " the Sun to the planet sweeps out equal areas in equal times. Kepler's third law The ratio of the squares of the periods of any two planets about the Sun is equal to the ratio of the cubes of their average distances from the Sun: where is the period (time for one orbit) and is the average radius of the orbit. \u2022 The period and radius of a satellite's orbit about a larger body are related by 2 1 2 2 = 3 1 3 2, or Conceptual Questions 6.1 Rotation Angle and Angular Velocity 2 = 4\u03c02 3 3 2 = 4\u03c02. 1. There is an analogy between rotational and linear physical quantities. What rotational quantities are analogous to distance and velocity? 6.2 Centripetal Acceleration 2. Can centripetal acceleration change the speed of circular motion? Explain. 6.3 Centripetal Force 3. If you wish to reduce the stress (which is related to centripetal force) on high-speed tires, would you use large- or smalldiameter tires? Explain. 4. Define centripetal force. Can any type of force (for example, tension, gravitational force, friction, and so on) be a centripetal force? Can any combination of forces be a centripetal force? 5. If centripetal force is directed toward the center, why do you feel that you are \u2018thrown' away from the center as a car goes around a curve? Explain. 6. Race car drivers routinely cut corners as shown in Figure 6.32. Explain how this allows the curve to be taken at the greatest speed. 252 Chapter 6 | Gravitation and Uniform Circular Motion Figure 6.32 Two paths around a race track curve are shown. Race car drivers will take the inside path (called cutting the corner) whenever possible because it allows them to take the curve at the highest speed. 7. A number of amusement parks have rides that make vertical loops like the one shown in Figure 6.33. For safety, the cars are attached to the rails in such a way that they cannot fall off. If the car goes over the top at just the right speed, gravity alone will supply the centripetal force. What other force acts and what is its direction if: (a) The car goes over the top at faster than this speed? (b)The car goes over the top at slower than this speed? Figure 6.33 Amusement rides with a vertical loop are an example of a form of curved motion. 8. What is the direction of the force exerted by the car on the passenger as the car goes over the top of the amusement ride pictured in Figure 6.33 under the following circumstances: (a) The car goes over the top at such a speed that the gravitational force is the only force acting? (b) The car goes over the top faster than this speed? (c) The car goes over the top slower than this speed? 9. As a skater forms a circle, what force is responsible for making her turn? Use a free body diagram in your answer. 10. Suppose a child is riding on a merry-go-round at a distance about halfway between its center and edge. She has a lunch box resting on wax paper, so that there is very little friction between it and the merry-go-round. Which path shown in Figure 6.34 will the lunch box take when she lets go? The lunch box leaves a trail in the dust on the merry-go-round. Is that trail straight, curved to the left, or curved to the right? Explain your answer. This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 6 | Gravitation and Uniform Circular Motion 253 Figure 6.34 A child riding on a merry-go-round releases her lunch box at point P. This is a view from above the clockwise rotation. Assuming it slides with negligible friction, will it follow path A, B, or C, as viewed from Earth's frame of reference? What will be the shape of the path it leaves in the dust on the merry-go-round? 11. Do you feel yourself thrown to either side when you negotiate a curve that is ideally banked for your car's speed? What is the direction of the force exerted on you by the car seat? 12. Suppose a mass is moving in a circular path on a frictionless table as shown in figure. In the Earth's frame of reference, there is no centrifugal force pulling the mass away from the centre of rotation, yet there is a very real force stretching the string attaching the mass to the nail. Using concepts related to centripetal force and Newton's third law, explain what force stretches the string, identifying its physical origin. Figure 6.35 A mass attached to a nail on a frictionless table moves in a circular path. The force stretching the string is real and not", " fictional. What is the physical origin of the force on the string? 6.4 Fictitious Forces and Non-inertial Frames: The Coriolis Force 13. When a toilet is flushed or a sink is drained, the water (and other material) begins to rotate about the drain on the way down. Assuming no initial rotation and a flow initially directly straight toward the drain, explain what causes the rotation and which direction it has in the northern hemisphere. (Note that this is a small effect and in most toilets the rotation is caused by directional water jets.) Would the direction of rotation reverse if water were forced up the drain? 14. Is there a real force that throws water from clothes during the spin cycle of a washing machine? Explain how the water is removed. 15. In one amusement park ride, riders enter a large vertical barrel and stand against the wall on its horizontal floor. The barrel is spun up and the floor drops away. Riders feel as if they are pinned to the wall by a force something like the gravitational force. This is a fictitious force sensed and used by the riders to explain events in the rotating frame of reference of the barrel. Explain in an inertial frame of reference (Earth is nearly one) what pins the riders to the wall, and identify all of the real forces acting on them. 254 Chapter 6 | Gravitation and Uniform Circular Motion 16. Action at a distance, such as is the case for gravity, was once thought to be illogical and therefore untrue. What is the ultimate determinant of the truth in physics, and why was this action ultimately accepted? 17. Two friends are having a conversation. Anna says a satellite in orbit is in freefall because the satellite keeps falling toward Earth. Tom says a satellite in orbit is not in freefall because the acceleration due to gravity is not 9.80 m/s2. Who do you agree with and why? 18. A non-rotating frame of reference placed at the center of the Sun is very nearly an inertial one. Why is it not exactly an inertial frame? 6.5 Newton's Universal Law of Gravitation 19. Action at a distance, such as is the case for gravity, was once thought to be illogical and therefore untrue. What is the ultimate determinant of the truth in physics, and why was this action ultimately accepted? 20. Two friends are having a conversation. Anna says a satellite in orbit is in freefall because the satellite keeps falling toward Earth. Tom says a satellite in orbit is not in freefall because the acceleration due to gravity is not 9.80 m/s2. Who do you agree with and why? 21. Draw a free body diagram for a satellite in an elliptical orbit showing why its speed increases as it approaches its parent body and decreases as it moves away. 22. Newton's laws of motion and gravity were among the first to convincingly demonstrate the underlying simplicity and unity in nature. Many other examples have since been discovered, and we now expect to find such underlying order in complex situations. Is there proof that such order will always be found in new explorations? 6.6 Satellites and Kepler's Laws: An Argument for Simplicity 23. In what frame(s) of reference are Kepler's laws valid? Are Kepler's laws purely descriptive, or do they contain causal information? This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 6 | Gravitation and Uniform Circular Motion 255 Problems & Exercises 6.1 Rotation Angle and Angular Velocity 1. Semi-trailer trucks have an odometer on one hub of a trailer wheel. The hub is weighted so that it does not rotate, but it contains gears to count the number of wheel revolutions\u2014it then calculates the distance traveled. If the wheel has a 1.15 m diameter and goes through 200,000 rotations, how many kilometers should the odometer read? 2. Microwave ovens rotate at a rate of about 6 rev/min. What is this in revolutions per second? What is the angular velocity in radians per second? 3. An automobile with 0.260 m radius tires travels 80,000 km before wearing them out. How many revolutions do the tires make, neglecting any backing up and any change in radius due to wear? 4. (a) What is the period of rotation of Earth in seconds? (b) What is the angular velocity of Earth? (c) Given that Earth has a radius of 6.4\u00d7106 m at its equator, what is the linear velocity at Earth's surface? 5. A baseball pitcher brings his arm forward during a pitch, rotating the forearm about the elbow. If the velocity of the ball in the pitcher's hand is 35.0 m/s and the ball is 0.300 m from the elbow joint, what is the angular velocity of the forearm? 6. In lacrosse, a ball is thrown from a net on the end of a stick by", " rotating the stick and forearm about the elbow. If the angular velocity of the ball about the elbow joint is 30.0 rad/s and the ball is 1.30 m from the elbow joint, what is the velocity of the ball? 7. A truck with 0.420-m-radius tires travels at 32.0 m/s. What is the angular velocity of the rotating tires in radians per second? What is this in rev/min? 8. Integrated Concepts When kicking a football, the kicker rotates his leg about the hip joint. (a) If the velocity of the tip of the kicker's shoe is 35.0 m/s and the hip joint is 1.05 m from the tip of the shoe, what is the shoe tip's angular velocity? (b) The shoe is in contact with the initially stationary 0.500 kg football for 20.0 ms. What average force is exerted on the football to give it a velocity of 20.0 m/s? (c) Find the maximum range of the football, neglecting air resistance. 9. Construct Your Own Problem Consider an amusement park ride in which participants are rotated about a vertical axis in a cylinder with vertical walls. Once the angular velocity reaches its full value, the floor drops away and friction between the walls and the riders prevents them from sliding down. Construct a problem in which you calculate the necessary angular velocity that assures the riders will not slide down the wall. Include a free body diagram of a single rider. Among the variables to consider are the radius of the cylinder and the coefficients of friction between the riders' clothing and the wall. 6.2 Centripetal Acceleration 10. A fairground ride spins its occupants inside a flying saucer-shaped container. If the horizontal circular path the riders follow has an 8.00 m radius, at how many revolutions per minute will the riders be subjected to a centripetal acceleration whose magnitude is 1.50 times that due to gravity? 11. A runner taking part in the 200 m dash must run around the end of a track that has a circular arc with a radius of curvature of 30 m. If he completes the 200 m dash in 23.2 s and runs at constant speed throughout the race, what is the magnitude of his centripetal acceleration as he runs the curved portion of the track? 12. Taking the age of Earth to be about 4\u00d7109 years and assuming its orbital radius of 1.5 \u00d71011 has not changed and is circular, calculate the approximate total distance Earth has traveled since its birth (in a frame of reference stationary with respect to the Sun). 13. The propeller of a World War II fighter plane is 2.30 m in diameter. (a) What is its angular velocity in radians per second if it spins at 1200 rev/min? (b) What is the linear speed of its tip at this angular velocity if the plane is stationary on the tarmac? (c) What is the centripetal acceleration of the propeller tip under these conditions? Calculate it in meters per second squared and convert to multiples of. 14. An ordinary workshop grindstone has a radius of 7.50 cm and rotates at 6500 rev/min. (a) Calculate the magnitude of the centripetal acceleration at its edge in meters per second squared and convert it to multiples of. (b) What is the linear speed of a point on its edge? 15. Helicopter blades withstand tremendous stresses. In addition to supporting the weight of a helicopter, they are spun at rapid rates and experience large centripetal accelerations, especially at the tip. (a) Calculate the magnitude of the centripetal acceleration at the tip of a 4.00 m long helicopter blade that rotates at 300 rev/min. (b) Compare the linear speed of the tip with the speed of sound (taken to be 340 m/s). 16. Olympic ice skaters are able to spin at about 5 rev/s. (a) What is their angular velocity in radians per second? (b) What is the centripetal acceleration of the skater's nose if it is 0.120 m from the axis of rotation? (c) An exceptional skater named Dick Button was able to spin much faster in the 1950s than anyone since\u2014at about 9 rev/ s. What was the centripetal acceleration of the tip of his nose, assuming it is at 0.120 m radius? (d) Comment on the magnitudes of the accelerations found. It is reputed that Button ruptured small blood vessels during his spins. 17. What percentage of the acceleration at Earth's surface is the acceleration due to gravity at the position of a satellite located 300 km above Earth? 18. Verify that the linear speed of an ultracentrifuge is about 0.50 km/s, and Earth in its orbit is about 30 km/s by calculating: (", "a) The linear speed of a point on an ultracentrifuge 0.100 m from its center, rotating at 50,000 rev/min. (b) The linear speed of Earth in its orbit about the Sun (use data from the text on the radius of Earth's orbit and approximate it as being circular). 256 Chapter 6 | Gravitation and Uniform Circular Motion 19. A rotating space station is said to create \u201cartificial gravity\u201d\u2014a loosely-defined term used for an acceleration that would be crudely similar to gravity. The outer wall of the rotating space station would become a floor for the astronauts, and centripetal acceleration supplied by the floor would allow astronauts to exercise and maintain muscle and bone strength more naturally than in non-rotating space environments. If the space station is 200 m in diameter, what angular velocity would produce an \u201cartificial gravity\u201d of 9.80 m/s2 at the rim? 20. At takeoff, a commercial jet has a 60.0 m/s speed. Its tires have a diameter of 0.850 m. (a) At how many rev/min are the tires rotating? (b) What is the centripetal acceleration at the edge of the tire? (c) With what force must a determined 1.00\u00d710\u221215 kg bacterium cling to the rim? (d) Take the ratio of this force to the bacterium's weight. 21. Integrated Concepts Riders in an amusement park ride shaped like a Viking ship hung from a large pivot are rotated back and forth like a rigid pendulum. Sometime near the middle of the ride, the ship is momentarily motionless at the top of its circular arc. The ship then swings down under the influence of gravity. (a) Assuming negligible friction, find the speed of the riders at the bottom of its arc, given the system's center of mass travels in an arc having a radius of 14.0 m and the riders are near the center of mass. (b) What is the centripetal acceleration at the bottom of the arc? (c) Draw a free body diagram of the forces acting on a rider at the bottom of the arc. (d) Find the force exerted by the ride on a 60.0 kg rider and compare it to her weight. (e) Discuss whether the answer seems reasonable. 22. Unreasonable Results A mother pushes her child on a swing so that his speed is 9.00 m/s at the lowest point of his path. The swing is suspended 2.00 m above the child's center of mass. (a) What is the magnitude of the centripetal acceleration of the child at the low point? (b) What is the magnitude of the force the child exerts on the seat if his mass is 18.0 kg? (c) What is unreasonable about these results? (d) Which premises are unreasonable or inconsistent? 6.3 Centripetal Force 23. (a) A 22.0 kg child is riding a playground merry-go-round that is rotating at 40.0 rev/min. What centripetal force must she exert to stay on if she is 1.25 m from its center? (b) What centripetal force does she need to stay on an amusement park merry-go-round that rotates at 3.00 rev/min if she is 8.00 m from its center? (c) Compare each force with her weight. 24. Calculate the centripetal force on the end of a 100 m (radius) wind turbine blade that is rotating at 0.5 rev/s. Assume the mass is 4 kg. This content is available for free at http://cnx.org/content/col11844/1.13 25. What is the ideal banking angle for a gentle turn of 1.20 km radius on a highway with a 105 km/h speed limit (about 65 mi/h), assuming everyone travels at the limit? 26. What is the ideal speed to take a 100 m radius curve banked at a 20.0\u00b0 angle? 27. (a) What is the radius of a bobsled turn banked at 75.0\u00b0 and taken at 30.0 m/s, assuming it is ideally banked? (b) Calculate the centripetal acceleration. (c) Does this acceleration seem large to you? 28. Part of riding a bicycle involves leaning at the correct angle when making a turn, as seen in Figure 6.36. To be stable, the force exerted by the ground must be on a line going through the center of gravity. The force on the bicycle wheel can be resolved into two perpendicular components\u2014friction parallel to the road (this must supply the centripetal force), and the vertical normal force (which must equal the system's weight). (a) Show that (as defined in the figure) is related to the speed and radius", " of curvature of the turn in the same way as for an ideally banked roadway\u2014that is, = tan\u20131 2/ (b) Calculate for a 12.0 m/s turn of radius 30.0 m (as in a race). Figure 6.36 A bicyclist negotiating a turn on level ground must lean at the correct angle\u2014the ability to do this becomes instinctive. The force of the ground on the wheel needs to be on a line through the center of gravity. The net external force on the system is the centripetal force. The vertical component of the force on the wheel cancels the weight of the system while its horizontal component must supply the centripetal force. This process produces a relationship among the angle, the speed, and the radius of curvature of the turn similar to that for the ideal banking of roadways. 29. A large centrifuge, like the one shown in Figure 6.37(a), is used to expose aspiring astronauts to accelerations similar to those experienced in rocket launches and atmospheric reentries. (a) At what angular velocity is the centripetal acceleration 10 if the rider is 15.0 m from the center of rotation? Chapter 6 | Gravitation and Uniform Circular Motion 257 (b) The rider's cage hangs on a pivot at the end of the arm, allowing it to swing outward during rotation as shown in Figure 6.37(b). At what angle below the horizontal will the cage hang when the centripetal acceleration is 10? (Hint: The arm supplies centripetal force and supports the weight of the cage. Draw a free body diagram of the forces to see what the angle should be.) Figure 6.38 Teardrop-shaped loops are used in the latest roller coasters so that the radius of curvature gradually decreases to a minimum at the top. This means that the centripetal acceleration builds from zero to a maximum at the top and gradually decreases again. A circular loop would cause a jolting change in acceleration at entry, a disadvantage discovered long ago in railroad curve design. With a small radius of curvature at the top, the centripetal acceleration can more easily be kept greater than so that the passengers do not lose contact with their seats nor do they need seat belts to keep them in place. 32. Unreasonable Results (a) Calculate the minimum coefficient of friction needed for a car to negotiate an unbanked 50.0 m radius curve at 30.0 m/ s. (b) What is unreasonable about the result? (c) Which premises are unreasonable or inconsistent? 6.5 Newton's Universal Law of Gravitation 33. (a) Calculate Earth's mass given the acceleration due to gravity at the North Pole is 9.830 m/s2 and the radius of the Earth is 6371 km from pole to pole. (b) Compare this with the accepted value of 5.979\u00d71024 kg. 34. (a) Calculate the magnitude of the acceleration due to gravity on the surface of Earth due to the Moon. (b) Calculate the magnitude of the acceleration due to gravity at Earth due to the Sun. (c) Take the ratio of the Moon's acceleration to the Sun's and comment on why the tides are predominantly due to the Moon in spite of this number. 35. (a) What is the acceleration due to gravity on the surface of the Moon? (b) On the surface of Mars? The mass of Mars is 6.418\u00d71023 kg and its radius is 3.38\u00d7106 m. 36. (a) Calculate the acceleration due to gravity on the surface of the Sun. (b) By what factor would your weight increase if you could stand on the Sun? (Never mind that you cannot.) Figure 6.37 (a) NASA centrifuge used to subject trainees to accelerations similar to those experienced in rocket launches and reentries. (credit: NASA) (b) Rider in cage showing how the cage pivots outward during rotation. This allows the total force exerted on the rider by the cage to be along its axis at all times. 30. Integrated Concepts If a car takes a banked curve at less than the ideal speed, friction is needed to keep it from sliding toward the inside of the curve (a real problem on icy mountain roads). (a) Calculate the ideal speed to take a 100 m radius curve banked at 15.0\u00ba. (b) What is the minimum coefficient of friction needed for a frightened driver to take the same curve at 20.0 km/h? 31. Modern roller coasters have vertical loops like the one shown in Figure 6.38. The radius of curvature is smaller at the top than on the sides so that the downward centripetal acceleration at the top will be greater than the acceleration due to gravity, keeping the passengers pressed firmly into their seats. What is the speed of the roller coaster at the top of", " the loop if the radius of curvature there is 15.0 m and the downward acceleration of the car is 1.50 g? 258 Chapter 6 | Gravitation and Uniform Circular Motion 37. The Moon and Earth rotate about their common center of mass, which is located about 4700 km from the center of Earth. (This is 1690 km below the surface.) (a) Calculate the magnitude of the acceleration due to the Moon's gravity at that point. (b) Calculate the magnitude of the centripetal acceleration of the center of Earth as it rotates about that point once each lunar month (about 27.3 d) and compare it with the acceleration found in part (a). Comment on whether or not they are equal and why they should or should not be. 38. Solve part (b) of Example 6.6 using = 2 /. 39. Astrology, that unlikely and vague pseudoscience, makes much of the position of the planets at the moment of one's birth. The only known force a planet exerts on Earth is gravitational. (a) Calculate the magnitude of the gravitational force exerted on a 4.20 kg baby by a 100 kg father 0.200 m away at birth (he is assisting, so he is close to the child). (b) Calculate the magnitude of the force on the baby due to Jupiter if it is at its closest distance to Earth, some 6.29\u00d71011 m away. How does the force of Jupiter on the baby compare to the force of the father on the baby? Other objects in the room and the hospital building also exert similar gravitational forces. (Of course, there could be an unknown force acting, but scientists first need to be convinced that there is even an effect, much less that an unknown force causes it.) 40. The existence of the dwarf planet Pluto was proposed based on irregularities in Neptune's orbit. Pluto was subsequently discovered near its predicted position. But it now appears that the discovery was fortuitous, because Pluto is small and the irregularities in Neptune's orbit were not well known. To illustrate that Pluto has a minor effect on the orbit of Neptune compared with the closest planet to Neptune: (a) Calculate the acceleration due to gravity at Neptune due to Pluto when they are 4.50\u00d71012 m apart, as they are at present. The mass of Pluto is 1.4\u00d71022 kg. (b) Calculate the acceleration due to gravity at Neptune due to Uranus, presently about 2.50\u00d71012 m apart, and compare it with that due to Pluto. The mass of Uranus is 8.62\u00d71025 kg. 41. (a) The Sun orbits the Milky Way galaxy once each 2.60 x 108 y, with a roughly circular orbit averaging 3.00 x 104 light years in radius. (A light year is the distance traveled by light in 1 y.) Calculate the centripetal acceleration of the Sun in its galactic orbit. Does your result support the contention that a nearly inertial frame of reference can be located at the Sun? (b) Calculate the average speed of the Sun in its galactic orbit. Does the answer surprise you? 42. Unreasonable Result A mountain 10.0 km from a person exerts a gravitational force on him equal to 2.00% of his weight. (a) Calculate the mass of the mountain. (b) Compare the mountain's mass with that of Earth. (c) What is unreasonable about these results? This content is available for free at http://cnx.org/content/col11844/1.13 (d) Which premises are unreasonable or inconsistent? (Note that accurate gravitational measurements can easily detect the effect of nearby mountains and variations in local geology.) 6.6 Satellites and Kepler's Laws: An Argument for Simplicity 43. A geosynchronous Earth satellite is one that has an orbital period of precisely 1 day. Such orbits are useful for communication and weather observation because the satellite remains above the same point on Earth (provided it orbits in the equatorial plane in the same direction as Earth's rotation). Calculate the radius of such an orbit based on the data for the moon in Table 6.2. 44. Calculate the mass of the Sun based on data for Earth's orbit and compare the value obtained with the Sun's actual mass. 45. Find the mass of Jupiter based on data for the orbit of one of its moons, and compare your result with its actual mass. 46. Find the ratio of the mass of Jupiter to that of Earth based on data in Table 6.2. 47. Astronomical observations of our Milky Way galaxy indicate that it has a mass of about 8.0\u00d71011 solar masses. A star orbiting on the galaxy's periphery is about 6.0\u00d7104 light years from its center. (a) What should the orbital period of that star be? (b) If its period is 6.0\u00d7107 instead, what", " is the mass of the galaxy? Such calculations are used to imply the existence of \u201cdark matter\u201d in the universe and have indicated, for example, the existence of very massive black holes at the centers of some galaxies. 48. Integrated Concepts Space debris left from old satellites and their launchers is becoming a hazard to other satellites. (a) Calculate the speed of a satellite in an orbit 900 km above Earth's surface. (b) Suppose a loose rivet is in an orbit of the same radius that intersects the satellite's orbit at an angle of 90\u00ba relative to Earth. What is the velocity of the rivet relative to the satellite just before striking it? (c) Given the rivet is 3.00 mm in size, how long will its collision with the satellite last? (d) If its mass is 0.500 g, what is the average force it exerts on the satellite? (e) How much energy in joules is generated by the collision? (The satellite's velocity does not change appreciably, because its mass is much greater than the rivet's.) 49. Unreasonable Results (a) Based on Kepler's laws and information on the orbital characteristics of the Moon, calculate the orbital radius for an Earth satellite having a period of 1.00 h. (b) What is unreasonable about this result? (c) What is unreasonable or inconsistent about the premise of a 1.00 h orbit? 50. Construct Your Own Problem On February 14, 2000, the NEAR spacecraft was successfully inserted into orbit around Eros, becoming the first artificial satellite of an asteroid. Construct a problem in which you determine the orbital speed for a satellite near Eros. You will need to find the mass of the asteroid and consider such things as a safe distance for the orbit. Although Eros is not spherical, calculate the acceleration due to gravity on its surface at a point an average distance from its center of mass. Your instructor may also wish to have you calculate the escape velocity from this point on Eros. Chapter 6 | Gravitation and Uniform Circular Motion 259 Test Prep for AP\u00ae Courses 6.5 Newton's Universal Law of Gravitation 1. Jupiter has a mass approximately 300 times greater than Earth's and a radius about 11 times greater. How will the gravitational acceleration at the surface of Jupiter compare to that at the surface of the Earth? a. Greater b. Less c. About the same d. Not enough information 2. Given Newton's universal law of gravitation (Equation 6.40), under what circumstances is the force due to gravity maximized? 3. In the formula = 2, what does G represent? a. The acceleration due to gravity b. A gravitational constant that is the same everywhere in the universe c. A gravitational constant that is inversely proportional to the radius d. The factor by which you multiply the inertial mass to obtain the gravitational mass 4. Saturn's moon Titan has a radius of 2.58 \u00d7 106 m and a measured gravitational field of 1.35 m/s2. What is its mass? 5. A recently discovered planet has a mass twice as great as Earth's and a radius twice as large as Earth's. What will be the approximate size of its gravitational field? a. 19 m/s2 b. 4.9 m/s2 c. 2.5 m/s2 d. 9.8 m/s2 6. 4. Earth is 1.5 \u00d7 1011 m from the Sun. Mercury is 5.7 \u00d7 1010 m from the Sun. How does the gravitational field of the Sun on Mercury (gSM) compare to the gravitational field of the Sun on Earth (gSE)? 260 Chapter 6 | Gravitation and Uniform Circular Motion This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 7 | Work, Energy, and Energy Resources 261 7 WORK, ENERGY, AND ENERGY RESOURCES Figure 7.1 How many forms of energy can you identify in this photograph of a wind farm in Iowa? (credit: J\u00fcrgen from Sandesneben, Germany, Wikimedia Commons) Chapter Outline 7.1. Work: The Scientific Definition 7.2. Kinetic Energy and the Work-Energy Theorem 7.3. Gravitational Potential Energy 7.4. Conservative Forces and Potential Energy 7.5. Nonconservative Forces 7.6. Conservation of Energy 7.7. Power 7.8. Work, Energy, and Power in Humans 7.9. World Energy Use Connection for AP\u00ae Courses Energy plays an essential role both in everyday events and in scientific phenomena. You can no doubt name many forms of energy, from that provided by our foods to the energy we use to run our cars and the sunlight that warms us on the beach. You can also cite examples of what people call \u201cenergy\u201d that may not be scientific, such as someone having an energetic personality. Not only", " does energy have many interesting forms, it is involved in almost all phenomena, and is one of the most important concepts of physics. There is no simple and accurate scientific definition for energy. Energy is characterized by its many forms and the fact that it is conserved. We can loosely define energy as the ability to do work, admitting that in some circumstances not all energy is available to do work. Because of the association of energy with work, we begin the chapter with a discussion of work. Work is intimately related to energy and how energy moves from one system to another or changes form. The work-energy theorem supports Big Idea 3, that interactions between objects are described by forces. In particular, exerting a force on an object may do work on it, changing it's energy (Enduring Understanding 3.E). The work-energy theorem, introduced in this chapter, establishes the relationship between work done on an object by an external force and changes in the object\u2019s kinetic energy (Essential Knowledge 3.E.1). Similarly, systems can do work on each other, supporting Big Idea 4, that interactions between systems can result in changes in those systems\u2014in this case, changes in the total energy of the system (Enduring Understanding 4.C). The total energy of the system is the sum of its kinetic energy, potential energy, and microscopic internal energy (Essential Knowledge 4.C.1). In this chapter students learn how to calculate kinetic, gravitational, and elastic potential energy in order to determine the total mechanical energy of a system. The transfer of mechanical energy into or out of a system is equal to the work done on the system by an external force with a nonzero component parallel to the displacement (Essential Knowledge 4.C.2). An important aspect of energy is that the total amount of energy in the universe is constant. Energy can change forms, but it cannot appear from nothing or disappear without a trace. Energy is thus one of a handful of physical quantities that we say is 262 Chapter 7 | Work, Energy, and Energy Resources \u201cconserved.\u201d Conservation of energy (as physicists call the principle that energy can neither be created nor destroyed) is based on experiment. Even as scientists discovered new forms of energy, conservation of energy has always been found to apply. Perhaps the most dramatic example of this was supplied by Einstein when he suggested that mass is equivalent to energy (his famous equation E = mc2). This is one of the most important applications of Big Idea 5, that changes that occur as a result of interactions are constrained by conservation laws. Specifically, there are many situations where conservation of energy (Enduring Understanding 5.B) is both a useful concept and starting point for calculations related to the system. Note, however, that conservation doesn\u2019t necessarily mean that energy in a system doesn\u2019t change. Energy may be transferred into or out of the system, and the change must be equal to the amount transferred (Enduring Understanding 5.A). This may occur if there is an external force or a transfer between external objects and the system (Essential Knowledge 5.A.3). Energy is one of the fundamental quantities that are conserved for all systems (Essential Knowledge 5.A.2). The chapter introduces concepts of kinetic energy and potential energy. Kinetic energy is introduced as an energy of motion that can be changed by the amount of work done by an external force. Potential energy can only exist when objects interact with each other via conservative forces according to classical physics (Essential Knowledge 5.B.3). Because of this, a single object can only have kinetic energy and no potential energy (Essential Knowledge 5.B.1). The chapter also introduces the idea that the energy transfer is equal to the work done on the system by external forces and the rate of energy transfer is defined as power (Essential Knowledge 5.B.5). From a societal viewpoint, energy is one of the major building blocks of modern civilization. Energy resources are key limiting factors to economic growth. The world use of energy resources, especially oil, continues to grow, with ominous consequences economically, socially, politically, and environmentally. We will briefly examine the world\u2019s energy use patterns at the end of this chapter. The concepts in this chapter support: Big Idea 3 The interactions of an object with other objects can be described by forces. Enduring Understanding 3.E A force exerted on an object can change the kinetic energy of the object. Essential Knowledge 3.E.1 The change in the kinetic energy of an object depends on the force exerted on the object and on the displacement of the object during the interval that the force is exerted. Big Idea 4 Interactions between systems can result in changes in those systems. Enduring Understanding 4.C Interactions with other objects or systems can change the total energy of a system. Essential Knowledge 4.C.1 The energy of a system includes its kinetic energy, potential energy, and microscopic internal energy. Examples should include gravitational potential energy, elastic potential energy, and", " kinetic energy. Essential Knowledge 4.C.2 Mechanical energy (the sum of kinetic and potential energy) is transferred into or out of a system when an external force is exerted on a system such that a component of the force is parallel to its displacement. The process through which the energy is transferred is called work. Big Idea 5 Changes that occur as a result of interactions are constrained by conservation laws. Enduring Understanding 5.A Certain quantities are conserved, in the sense that the changes of those quantities in a given system are always equal to the transfer of that quantity to or from the system by all possible interactions with other systems. Essential Knowledge 5.A.2 For all systems under all circumstances, energy, charge, linear momentum, and angular momentum are conserved. Essential Knowledge 5.A.3 An interaction can be either a force exerted by objects outside the system or the transfer of some quantity with objects outside the system. Enduring Understanding 5.B The energy of a system is conserved. Essential Knowledge 5.B.1 Classically, an object can only have kinetic energy since potential energy requires an interaction between two or more objects. Essential Knowledge 5.B.3 A system with internal structure can have potential energy. Potential energy exists within a system if the objects within that system interact with conservative forces. Essential Knowledge 5.B.5 Energy can be transferred by an external force exerted on an object or system that moves the object or system through a distance; this energy transfer is called work. Energy transfer in mechanical or electrical systems may occur at different rates. Power is de\ufb01ned as the rate of energy transfer into, out of, or within a system. 7.1 Work: The Scientific Definition By the end of this section, you will be able to: Learning Objectives \u2022 Explain how an object must be displaced for a force on it to do work. \u2022 Explain how relative directions of force and displacement of an object determine whether the work done on the object is positive, negative, or zero. The information presented in this section supports the following AP\u00ae learning objectives and science practices: \u2022 5.B.5.1 The student is able to design an experiment and analyze data to examine how a force exerted on an object or system does work on the object or system as it moves through a distance. (S.P. 4.2, 5.1) This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 7 | Work, Energy, and Energy Resources 263 \u2022 5.B.5.2 The student is able to design an experiment and analyze graphical data in which interpretations of the area under a force-distance curve are needed to determine the work done on or by the object or system. (S.P. 4.5, 5.1) \u2022 5.B.5.3 The student is able to predict and calculate from graphical data the energy transfer to or work done on an object or system from information about a force exerted on the object or system through a distance. (S.P. 1.5, 2.2, 6.4) What It Means to Do Work The scientific definition of work differs in some ways from its everyday meaning. Certain things we think of as hard work, such as writing an exam or carrying a heavy load on level ground, are not work as defined by a scientist. The scientific definition of work reveals its relationship to energy\u2014whenever work is done, energy is transferred. For work, in the scientific sense, to be done on an object, a force must be exerted on that object and there must be motion or displacement of that object in the direction of the force. Formally, the work done on a system by a constant force is defined to be the product of the component of the force in the direction of motion and the distance through which the force acts. For a constant force, this is expressed in equation form as = \u2223 F \u2223 (cos ) \u2223 d \u2223, (7.1) where is work, d is the displacement of the system, and is the angle between the force vector F and the displacement vector d, as in Figure 7.2. We can also write this as To find the work done on a system that undergoes motion that is not one-way or that is in two or three dimensions, we divide the motion into one-way one-dimensional segments and add up the work done over each segment. = cos. (7.2) What is Work? The work done on a system by a constant force is the product of the component of the force in the direction of motion times the distance through which the force acts. For one-way motion in one dimension, this is expressed in equation form as = cos, (7.3) where is work, is the magnitude of the force on the system, is the magnitude of the displacement of the system, and is the angle between the force vector F and the displacement vector d. 264", " Chapter 7 | Work, Energy, and Energy Resources Figure 7.2 Examples of work. (a) The work done by the force F on this lawn mower is cos. Note that cos is the component of the force in the direction of motion. (b) A person holding a briefcase does no work on it, because there is no motion. No energy is transferred to or from the briefcase. (c) The person moving the briefcase horizontally at a constant speed does no work on it, and transfers no energy to it. (d) Work is done on the briefcase by carrying it up stairs at constant speed, because there is necessarily a component of force F in the direction of the motion. Energy is transferred to the briefcase and could in turn be used to do work. (e) When the briefcase is lowered, energy is transferred out of the briefcase and into an electric generator. Here the work done on the briefcase by the generator is negative, removing energy from the briefcase, because F and d are in opposite directions. To examine what the definition of work means, let us consider the other situations shown in Figure 7.2. The person holding the briefcase in Figure 7.2(b) does no work, for example. Here = 0, so = 0. Why is it you get tired just holding a load? The answer is that your muscles are doing work against one another, but they are doing no work on the system of interest (the \u201cbriefcase-Earth system\u201d\u2014see Gravitational Potential Energy for more details). There must be displacement for work to be done, and there must be a component of the force in the direction of the motion. For example, the person carrying the briefcase This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 7 | Work, Energy, and Energy Resources 265 on level ground in Figure 7.2(c) does no work on it, because the force is perpendicular to the motion. That is, cos 90\u00ba = 0, and so = 0. In contrast, when a force exerted on the system has a component in the direction of motion, such as in Figure 7.2(d), work is done\u2014energy is transferred to the briefcase. Finally, in Figure 7.2(e), energy is transferred from the briefcase to a generator. There are two good ways to interpret this energy transfer. One interpretation is that the briefcase\u2019s weight does work on the generator, giving it energy. The other interpretation is that the generator does negative work on the briefcase, thus removing energy from it. The drawing shows the latter, with the force from the generator upward on the briefcase, and the displacement downward. This makes = 180\u00ba, and cos 180\u00ba = \u20131 ; therefore, is negative. Real World Connections: When Work Happens Note that work as we define it is not the same as effort. You can push against a concrete wall all you want, but you won\u2019t move it. While the pushing represents effort on your part, the fact that you have not changed the wall\u2019s state in any way indicates that you haven\u2019t done work. If you did somehow push the wall over, this would indicate a change in the wall\u2019s state, and therefore you would have done work. This can also be shown with Figure 7.2(a): as you push a lawnmower against friction, both you and friction are changing the lawnmower\u2019s state. However, only the component of the force parallel to the movement is changing the lawnmower\u2019s state. The component perpendicular to the motion is trying to push the lawnmower straight into Earth; the lawnmower does not move into Earth, and therefore the lawnmower\u2019s state is not changing in the direction of Earth. Similarly, in Figure 7.2(c), both your hand and gravity are exerting force on the briefcase. However, they are both acting perpendicular to the direction of motion, hence they are not changing the condition of the briefcase and do no work. However, if the briefcase were dropped, then its displacement would be parallel to the force of gravity, which would do work on it, changing its state (it would fall to the ground). Calculating Work Work and energy have the same units. From the definition of work, we see that those units are force times distance. Thus, in SI units, work and energy are measured in newton-meters. A newton-meter is given the special name joule (J), and 1 J = 1 N \u22c5 m = 1 kg \u22c5 m2/s2. One joule is not a large amount of energy; it would lift a small 100-gram apple a distance of about 1 meter. Example 7.1 Calculating the Work You Do to Push a Lawn", " Mower Across a Large Lawn How much work is done on the lawn mower by the person in Figure 7.2(a) if he exerts a constant force of 75.0 N at an angle 35\u00ba below the horizontal and pushes the mower 25.0 m on level ground? Convert the amount of work from joules to kilocalories and compare it with this person\u2019s average daily intake of 10000 kJ (about 2400 kcal ) of food energy. One calorie (1 cal) of heat is the amount required to warm 1 g of water by 1\u00baC, and is equivalent to 4.184 J, while one food calorie (1 kcal) is equivalent to 4184 J. Strategy We can solve this problem by substituting the given values into the definition of work done on a system, stated in the equation = cos. The force, angle, and displacement are given, so that only the work is unknown. Solution The equation for the work is Substituting the known values gives = cos. = (75.0 N)(25.0 m) cos (35.0\u00ba) = 1536 J = 1.54\u00d7103 J. (7.4) (7.5) Converting the work in joules to kilocalories yields = (1536 J)(1 kcal / 4184 J) = 0.367 kcal. The ratio of the work done to the daily consumption is 2400 kcal = 1.53\u00d710\u22124. (7.6) Discussion This ratio is a tiny fraction of what the person consumes, but it is typical. Very little of the energy released in the consumption of food is used to do work. Even when we \u201cwork\u201d all day long, less than 10% of our food energy intake is used to do work and more than 90% is converted to thermal energy or stored as chemical energy in fat. 266 Chapter 7 | Work, Energy, and Energy Resources Applying the Science Practices: Boxes on Floors Plan and design an experiment to determine how much work you do on a box when you are pushing it over different floor surfaces. Make sure your experiment can help you answer the following questions: What happens on different surfaces? What happens if you take different routes across the same surface? Do you get different results with two people pushing on perpendicular surfaces of the box? What if you vary the mass in the box? Remember to think about both your effort in any given instant (a proxy for force exerted) and the total work you do. Also, when planning your experiments, remember that in any given set of trials you should only change one variable. You should find that you have to exert more effort on surfaces that will create more friction with the box, though you might be surprised by which surfaces the box slides across easily. Longer routes result in your doing more work, even though the box ends up in the same place. Two people pushing on perpendicular sides do less work for their total effort, due to the forces and displacement not being parallel. A more massive box will take more effort to move. Applying the Science Practices: Force-Displacement Diagrams Suppose you are given two carts and a track to run them on, a motion detector, a force sensor, and a computer that can record the data from the two sensors. Plan and design an experiment to measure the work done on one of the carts, and compare your results to the work-energy theorem. Note that the motion detector can measure both displacement and velocity versus time, while the force sensor measures force over time, and the carts have known masses. Recall that the work-energy theorem states that the work done on a system (force over displacement) should equal the change in kinetic energy. In your experimental design, describe and compare two possible ways to calculate the work done. Sample Response: One possible technique is to set up the motion detector at one end of the track, and have the computer record both displacement and velocity over time. Then attach the force sensor to one of the carts, and use this cart, through the force sensor, to push the second cart toward the motion detector. Calculate the difference between the final and initial kinetic energies (the kinetic energies after and before the push), and compare this to the area of a graph of force versus displacement for the duration of the push. They should be the same. 7.2 Kinetic Energy and the Work-Energy Theorem Learning Objectives By the end of this section, you will be able to: \u2022 Explain work as a transfer of energy and net work as the work done by the net force. \u2022 Explain and apply the work-energy theorem. The information presented in this section supports the following AP\u00ae learning objectives and science practices: \u2022 3.E.1.1 The student is able to make predictions about the changes in kinetic energy of an object based on considerations of the direction of the net force on the object as the object moves. (S.P. 6.4, 7.2) \u2022 3", ".E.1.2 The student is able to use net force and velocity vectors to determine qualitatively whether kinetic energy of an object would increase, decrease, or remain unchanged. (S.P. 1.4) \u2022 3.E.1.3 The student is able to use force and velocity vectors to determine qualitatively or quantitatively the net force exerted on an object and qualitatively whether kinetic energy of that object would increase, decrease, or remain unchanged. (S.P. 1.4, 2.2) \u2022 3.E.1.4 The student is able to apply mathematical routines to determine the change in kinetic energy of an object given the forces on the object and the displacement of the object. (S.P. 2.2) \u2022 4.C.1.1 The student is able to calculate the total energy of a system and justify the mathematical routines used in the calculation of component types of energy within the system whose sum is the total energy. (S.P. 1.4, 2.1, 2.2) \u2022 4.C.2.1 The student is able to make predictions about the changes in the mechanical energy of a system when a component of an external force acts parallel or antiparallel to the direction of the displacement of the center of mass. (S.P. 6.4) \u2022 4.C.2.2 The student is able to apply the concepts of conservation of energy and the work-energy theorem to determine qualitatively and/or quantitatively that work done on a two-object system in linear motion will change the kinetic energy of the center of mass of the system, the potential energy of the systems, and/or the internal energy of the system. (S.P. 1.4, 2.2, 7.2) \u2022 5.B.5.3 The student is able to predict and calculate from graphical data the energy transfer to or work done on an object or system from information about a force exerted on the object or system through a distance. (S.P. 1.5, 2.2, 6.4) Work Transfers Energy What happens to the work done on a system? Energy is transferred into the system, but in what form? Does it remain in the system or move on? The answers depend on the situation. For example, if the lawn mower in Figure 7.2(a) is pushed just hard enough to keep it going at a constant speed, then energy put into the mower by the person is removed continuously by friction, and eventually leaves the system in the form of heat transfer. In contrast, work done on the briefcase by the person carrying it up stairs in Figure 7.2(d) is stored in the briefcase-Earth system and can be recovered at any time, as shown in Figure 7.2(e). In fact, the building of the pyramids in ancient Egypt is an example of storing energy in a system by doing work on the system. This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 7 | Work, Energy, and Energy Resources 267 Some of the energy imparted to the stone blocks in lifting them during construction of the pyramids remains in the stone-Earth system and has the potential to do work. In this section we begin the study of various types of work and forms of energy. We will find that some types of work leave the energy of a system constant, for example, whereas others change the system in some way, such as making it move. We will also develop definitions of important forms of energy, such as the energy of motion. Net Work and the Work-Energy Theorem We know from the study of Newton\u2019s laws in Dynamics: Force and Newton's Laws of Motion that net force causes acceleration. We will see in this section that work done by the net force gives a system energy of motion, and in the process we will also find an expression for the energy of motion. Let us start by considering the total, or net, work done on a system. Net work is defined to be the sum of work done by all external forces\u2014that is, net work is the work done by the net external force Fnet. In equation form, this is net = net cos where is the angle between the force vector and the displacement vector. Figure 7.3(a) shows a graph of force versus displacement for the component of the force in the direction of the displacement\u2014that is, an cos vs. graph. In this case, cos is constant. You can see that the area under the graph is cos, or the work done. Figure 7.3(b) shows a more general process where the force varies. The area under the curve is divided into strips, each having an average force ( cos ) (ave) for each strip, and the (ave) total work done is the sum of the. Thus the total work done is the", " total area under the curve, a useful property to which we shall refer later.. The work done is ( cos ) Figure 7.3 (a) A graph of cos vs., when cos is constant. The area under the curve represents the work done by the force. (b) A graph of cos vs. in which the force varies. The work done for each interval is the area of each strip; thus, the total area under the curve equals the total work done. Real World Connections: Work and Direction Consider driving in a car. While moving, you have forward velocity and therefore kinetic energy. When you hit the brakes, they exert a force opposite to your direction of motion (acting through the wheels). The brakes do work on your car and reduce the kinetic energy. Similarly, when you accelerate, the engine (acting through the wheels) exerts a force in the direction of motion. The engine does work on your car, and increases the kinetic energy. Finally, if you go around a corner at a constant speed, you have the same kinetic energy both before and after the corner. The force exerted by the engine was perpendicular to the direction of motion, and therefore did no work and did not change the kinetic energy. 268 Chapter 7 | Work, Energy, and Energy Resources Net work will be simpler to examine if we consider a one-dimensional situation where a force is used to accelerate an object in a direction parallel to its initial velocity. Such a situation occurs for the package on the roller belt conveyor system shown in Figure 7.4. Figure 7.4 A package on a roller belt is pushed horizontally through a distance d. The force of gravity and the normal force acting on the package are perpendicular to the displacement and do no work. Moreover, they are also equal in magnitude and opposite in direction so they cancel in calculating the net force. The net force arises solely from the horizontal applied force Fapp and the horizontal friction force f. Thus, as expected, the net force is parallel to the displacement, so that = 0\u00ba and cos = 1, and the net work is given by net = net. (7.7) The effect of the net force Fnet is to accelerate the package from 0 to. The kinetic energy of the package increases, indicating that the net work done on the system is positive. (See Example 7.2.) By using Newton\u2019s second law, and doing some algebra, we can reach an interesting conclusion. Substituting net = from Newton\u2019s second law gives net = (7.8) To get a relationship between net work and the speed given to a system by the net force acting on it, we take = \u2212 0 and use the equation studied in Motion Equations for Constant Acceleration in One Dimension for the change in speed over a 2 + 2 (note that appears in the expression for distance if the acceleration has the constant value ; namely, 2 = 0 the net work). Solving for acceleration gives = 2 2 \u2212 0 2 obtain. When is substituted into the preceding expression for net, we The cancels, and we rearrange this to obtain net = 2 2 \u2212 0 2. = 1 22 \u2212 1 2. 20 (7.9) (7.10) This expression is called the work-energy theorem, and it actually applies in general (even for forces that vary in direction and magnitude), although we have derived it for the special case of a constant force parallel to the displacement. The theorem implies that the net work on a system equals the change in the quantity 1 22. This quantity is our first example of a form of energy. The Work-Energy Theorem The net work on a system equals the change in the quantity 1 22. 22 \u2212 1 2 20 net = 1 (7.11) The quantity 1 22 in the work-energy theorem is defined to be the translational kinetic energy (KE) of a mass moving at a speed. (Translational kinetic energy is distinct from rotational kinetic energy, which is considered later.) In equation form, the translational kinetic energy, KE = 1 22 (7.12) This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 7 | Work, Energy, and Energy Resources 269 is the energy associated with translational motion. Kinetic energy is a form of energy associated with the motion of a particle, single body, or system of objects moving together. We are aware that it takes energy to get an object, like a car or the package in Figure 7.4, up to speed, but it may be a bit surprising that kinetic energy is proportional to speed squared. This proportionality means, for example, that a car traveling at 100 km/h has four times the kinetic energy it has at 50 km/h, helping to explain why high-speed collisions are so devastating. We will now consider a series of examples to illustrate various aspects of work and energy. Applying the Science Practices: Cars on a Hill Assemble", " a ramp suitable for rolling some toy cars up or down. Then plan a series of experiments to determine how the direction of a force relative to the velocity of an object alters the kinetic energy of the object. Note that gravity will be pointing down in all cases. What happens if you start the car at the top? How about at the bottom, with an initial velocity that is increasing? If your ramp is wide enough, what happens if you send the toy car straight across? Does varying the surface of the ramp change your results? Sample Response: When the toy car is going down the ramp, with a component of gravity in the same direction, the kinetic energy increases. Sending the car up the ramp decreases the kinetic energy, as gravity is opposing the motion. Sending the car sideways should result in little to no change. If you have a surface that generates more friction than a smooth surface (carpet), note that the friction always opposed the motion, and hence decreases the kinetic energy. Example 7.2 Calculating the Kinetic Energy of a Package Suppose a 30.0-kg package on the roller belt conveyor system in Figure 7.4 is moving at 0.500 m/s. What is its kinetic energy? Strategy Because the mass and speed are given, the kinetic energy can be calculated from its definition as given in the equation KE = 1 22. Solution The kinetic energy is given by Entering known values gives which yields Discussion KE = 1 22. KE = 0.5(30.0 kg)(0.500 m/s)2 KE = 3.75 kg \u22c5 m2/s2 = 3.75 J. (7.13) (7.14) (7.15) Note that the unit of kinetic energy is the joule, the same as the unit of work, as mentioned when work was first defined. It is also interesting that, although this is a fairly massive package, its kinetic energy is not large at this relatively low speed. This fact is consistent with the observation that people can move packages like this without exhausting themselves. Real World Connections: Center of Mass Suppose we have two experimental carts, of equal mass, latched together on a track with a compressed spring between them. When the latch is released, the spring does 10 J of work on the carts (we\u2019ll see how in a couple of sections). The carts move relative to the spring, which is the center of mass of the system. However, the center of mass stays fixed. How can we consider the kinetic energy of this system? By the work-energy theorem, the work done by the spring on the carts must turn into kinetic energy. So this system has 10 J of kinetic energy. The total kinetic energy of the system is the kinetic energy of the center of mass of the system relative to the fixed origin plus the kinetic energy of each cart relative to the center of mass. We know that the center of mass relative to the fixed origin does not move, and therefore all of the kinetic energy must be distributed among the carts relative to the center of mass. Since the carts have equal mass, they each receive an equal amount of kinetic energy, so each cart has 5.0 J of kinetic energy. In our example, the forces between the spring and each cart are internal to the system. According to Newton\u2019s third law, these internal forces will cancel since they are equal and opposite in direction. However, this does not imply that these internal forces will not do work. Thus, the change in kinetic energy of the system is caused by work done by the force of the spring, and results in the motion of the two carts relative to the center of mass. 270 Chapter 7 | Work, Energy, and Energy Resources Example 7.3 Determining the Work to Accelerate a Package Suppose that you push on the 30.0-kg package in Figure 7.4 with a constant force of 120 N through a distance of 0.800 m, and that the opposing friction force averages 5.00 N. (a) Calculate the net work done on the package. (b) Solve the same problem as in part (a), this time by finding the work done by each force that contributes to the net force. Strategy and Concept for (a) This is a motion in one dimension problem, because the downward force (from the weight of the package) and the normal force have equal magnitude and opposite direction, so that they cancel in calculating the net force, while the applied force, friction, and the displacement are all horizontal. (See Figure 7.4.) As expected, the net work is the net force times distance. Solution for (a) The net force is the push force minus friction, or net = 120 N \u2013 5.00 N = 115 N. Thus the net work is net = net = (115 N)(0.800 m) = 92.0 N \u22c5 m = 92.0 J. (7.16) Discussion for (a) This value is", " the net work done on the package. The person actually does more work than this, because friction opposes the motion. Friction does negative work and removes some of the energy the person expends and converts it to thermal energy. The net work equals the sum of the work done by each individual force. Strategy and Concept for (b) The forces acting on the package are gravity, the normal force, the force of friction, and the applied force. The normal force and force of gravity are each perpendicular to the displacement, and therefore do no work. Solution for (b) The applied force does work. app = app cos(0\u00ba) = app = (120 N)(0.800 m) = 96.0 J The friction force and displacement are in opposite directions, so that = 180\u00ba, and the work done by friction is fr = fr cos(180\u00ba) = \u2212fr = \u2212(5.00 N)(0.800 m) = \u22124.00 J. So the amounts of work done by gravity, by the normal force, by the applied force, and by friction are, respectively, gr = 0, N = 0, app = 96.0 J, fr = \u2212 4.00 J. The total work done as the sum of the work done by each force is then seen to be total = gr + N + app + fr = 92.0 J. Discussion for (b) (7.17) (7.18) (7.19) (7.20) The calculated total work total as the sum of the work by each force agrees, as expected, with the work net done by the net force. The work done by a collection of forces acting on an object can be calculated by either approach. Example 7.4 Determining Speed from Work and Energy Find the speed of the package in Figure 7.4 at the end of the push, using work and energy concepts. Strategy Here the work-energy theorem can be used, because we have just calculated the net work, net, and the initial kinetic energy, 1 2. These calculations allow us to find the final kinetic energy, 1 22, and thus the final speed. 20 This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 7 | Work, Energy, and Energy Resources 271 Solution The work-energy theorem in equation form is net = 1 22 \u2212 1 20 2. Solving for 1 22 gives Thus, 1 22 = net + 1 20 2. 1 22 = 92.0 J+3.75 J = 95.75 J. Solving for the final speed as requested and entering known values gives = 2(95.75 J) = 191.5 kg \u22c5 m2/s2 30.0 kg = 2.53 m/s. (7.21) (7.22) (7.23) (7.24) Discussion Using work and energy, we not only arrive at an answer, we see that the final kinetic energy is the sum of the initial kinetic energy and the net work done on the package. This means that the work indeed adds to the energy of the package. Example 7.5 Work and Energy Can Reveal Distance, Too How far does the package in Figure 7.4 coast after the push, assuming friction remains constant? Use work and energy considerations. Strategy We know that once the person stops pushing, friction will bring the package to rest. In terms of energy, friction does negative work until it has removed all of the package\u2019s kinetic energy. The work done by friction is the force of friction times the distance traveled times the cosine of the angle between the friction force and displacement; hence, this gives us a way of finding the distance traveled after the person stops pushing. Solution The normal force and force of gravity cancel in calculating the net force. The horizontal friction force is then the net force, and it acts opposite to the displacement, so = 180\u00ba. To reduce the kinetic energy of the package to zero, the work fr by friction must be minus the kinetic energy that the package started with plus what the package accumulated due to the pushing. Thus fr = \u221295.75 J. Furthermore, fr = \u2032 cos = \u2013 \u2032, where \u2032 is the distance it takes to stop. Thus, and so Discussion \u2032 = \u2212 fr = \u2212 \u221295.75 J 5.00 N, \u2032 = 19.2 m. (7.25) (7.26) This is a reasonable distance for a package to coast on a relatively friction-free conveyor system. Note that the work done by friction is negative (the force is in the opposite direction of motion), so it removes the kinetic energy. Some of the examples in this section can be solved without considering energy, but at the expense of missing out on gaining insights about what work and energy are doing in this situation. On the whole, solutions involving energy are generally shorter and easier than those using kinematics", " and dynamics alone. 7.3 Gravitational Potential Energy By the end of this section, you will be able to: \u2022 Explain gravitational potential energy in terms of work done against gravity. Learning Objectives 272 Chapter 7 | Work, Energy, and Energy Resources \u2022 Show that the gravitational potential energy of an object of mass m at height h on Earth is given by PEg = mgh. \u2022 Show how knowledge of potential energy as a function of position can be used to simplify calculations and explain physical phenomena. The information presented in this section supports the following AP\u00ae learning objectives and science practices: \u2022 4.C.1.1 The student is able to calculate the total energy of a system and justify the mathematical routines used in the calculation of component types of energy within the system whose sum is the total energy. (S.P. 1.4, 2.1, 2.2) \u2022 5.B.1.1 The student is able to set up a representation or model showing that a single object can only have kinetic energy and use information about that object to calculate its kinetic energy. (S.P. 1.4, 2.2) \u2022 5.B.1.2 The student is able to translate between a representation of a single object, which can only have kinetic energy, and a system that includes the object, which may have both kinetic and potential energies. (S.P. 1.5) Work Done Against Gravity Climbing stairs and lifting objects is work in both the scientific and everyday sense\u2014it is work done against the gravitational force. When there is work, there is a transformation of energy. The work done against the gravitational force goes into an important form of stored energy that we will explore in this section. Let us calculate the work done in lifting an object of mass through a height, such as in Figure 7.5. If the object is lifted straight up at constant speed, then the force needed to lift it is equal to its weight. The work done on the mass is then. We define this to be the gravitational potential energy (PEg) put into (or gained by) the object-Earth system. This energy is associated with the state of separation between two objects that attract each other by the gravitational force. For convenience, we refer to this as the PEg gained by the object, recognizing that this is energy stored in the gravitational field of Earth. Why do we use the word \u201csystem\u201d? Potential energy is a property of a system rather than of a single object\u2014due to its physical position. An object\u2019s gravitational potential is due to its position relative to the surroundings within the Earth-object system. The force applied to the object is an external force, from outside the system. When it does positive work it increases the gravitational potential energy of the system. Because gravitational potential energy depends on relative position, we need a reference level at which to set the potential energy equal to 0. We usually choose this point to be Earth\u2019s surface, but this point is arbitrary; what is important is the difference in gravitational potential energy, because this difference is what relates to the work done. The difference in gravitational potential energy of an object (in the Earth-object system) between two rungs of a ladder will be the same for the first two rungs as for the last two rungs. Converting Between Potential Energy and Kinetic Energy Gravitational potential energy may be converted to other forms of energy, such as kinetic energy. If we release the mass, gravitational force will do an amount of work equal to on it, thereby increasing its kinetic energy by that same amount (by the work-energy theorem). We will find it more useful to consider just the conversion of PEg to KE without explicitly considering the intermediate step of work. (See Example 7.7.) This shortcut makes it is easier to solve problems using energy (if possible) rather than explicitly using forces. This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 7 | Work, Energy, and Energy Resources 273 Figure 7.5 (a) The work done to lift the weight is stored in the mass-Earth system as gravitational potential energy. (b) As the weight moves downward, this gravitational potential energy is transferred to the cuckoo clock. More precisely, we define the change in gravitational potential energy \u0394PEg to be \u0394PEg =, (7.27) where, for simplicity, we denote the change in height by rather than the usual \u0394. Note that is positive when the final height is greater than the initial height, and vice versa. For example, if a 0.500-kg mass hung from a cuckoo clock is raised 1.00 m, then its change in gravitational potential energy is = 0.500 kg 9.80 m/s2 (1.00 m) (7.28) = 4.90 kg \u22c5 m2/s2 = 4.90 J. Note that the units of", " gravitational potential energy turn out to be joules, the same as for work and other forms of energy. As the clock runs, the mass is lowered. We can think of the mass as gradually giving up its 4.90 J of gravitational potential energy, without directly considering the force of gravity that does the work. Using Potential Energy to Simplify Calculations The equation \u0394PEg = applies for any path that has a change in height of, not just when the mass is lifted straight up. (See Figure 7.6.) It is much easier to calculate (a simple multiplication) than it is to calculate the work done along a complicated path. The idea of gravitational potential energy has the double advantage that it is very broadly applicable and it makes calculations easier. From now on, we will consider that any change in vertical position of a mass is accompanied by a change in gravitational potential energy, and we will avoid the equivalent but more difficult task of calculating work done by or against the gravitational force. 274 Chapter 7 | Work, Energy, and Energy Resources Figure 7.6 The change in gravitational potential energy (\u0394PEg) between points A and B is independent of the path. \u0394PEg = for any path between the two points. Gravity is one of a small class of forces where the work done by or against the force depends only on the starting and ending points, not on the path between them. Example 7.6 The Force to Stop Falling A 60.0-kg person jumps onto the floor from a height of 3.00 m. If he lands stiffly (with his knee joints compressing by 0.500 cm), calculate the force on the knee joints. Strategy This person\u2019s energy is brought to zero in this situation by the work done on him by the floor as he stops. The initial PEg is transformed into KE as he falls. The work done by the floor reduces this kinetic energy to zero. Solution The work done on the person by the floor as he stops is given by = cos = \u2212, (7.29) with a minus sign because the displacement while stopping and the force from floor are in opposite directions (cos = cos 180\u00ba = \u2212 1). The floor removes energy from the system, so it does negative work. The kinetic energy the person has upon reaching the floor is the amount of potential energy lost by falling through height : KE = \u2212\u0394PEg = \u2212, (7.30) The distance that the person\u2019s knees bend is much smaller than the height of the fall, so the additional change in gravitational potential energy during the knee bend is ignored. The work done by the floor on the person stops the person and brings the person\u2019s kinetic energy to zero: Combining this equation with the expression for gives \u2212 =. = \u2212KE =. Recalling that is negative because the person fell down, the force on the knee joints is given by (7.31) (7.32) This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 7 | Work, Energy, and Energy Resources = \u2212 = \u2212 Discussion 275 (7.33) = 3.53\u00d7105 N. 60.0 kg 9.80 m/s2 5.00\u00d710\u22123 m (\u22123.00 m) Such a large force (500 times more than the person\u2019s weight) over the short impact time is enough to break bones. A much better way to cushion the shock is by bending the legs or rolling on the ground, increasing the time over which the force acts. A bending motion of 0.5 m this way yields a force 100 times smaller than in the example. A kangaroo's hopping shows this method in action. The kangaroo is the only large animal to use hopping for locomotion, but the shock in hopping is cushioned by the bending of its hind legs in each jump.(See Figure 7.7.) Figure 7.7 The work done by the ground upon the kangaroo reduces its kinetic energy to zero as it lands. However, by applying the force of the ground on the hind legs over a longer distance, the impact on the bones is reduced. (credit: Chris Samuel, Flickr) Example 7.7 Finding the Speed of a Roller Coaster from its Height (a) What is the final speed of the roller coaster shown in Figure 7.8 if it starts from rest at the top of the 20.0 m hill and work done by frictional forces is negligible? (b) What is its final speed (again assuming negligible friction) if its initial speed is 5.00 m/s? Figure 7.8 The speed of a roller coaster increases as gravity pulls it downhill and is greatest at its lowest point. Viewed in terms of energy, the roller-coaster-Earth system\u2019s gravitational potential energy is converted to kinetic energy. If work done by friction is negligible, all \u0394PEg is converted to KE. Strategy The roller", " coaster loses potential energy as it goes downhill. We neglect friction, so that the remaining force exerted by the track is the normal force, which is perpendicular to the direction of motion and does no work. The net work on the roller 276 Chapter 7 | Work, Energy, and Energy Resources coaster is then done by gravity alone. The loss of gravitational potential energy from moving downward through a distance equals the gain in kinetic energy. This can be written in equation form as \u2212\u0394PEg = \u0394KE. Using the equations for PEg and KE, we can solve for the final speed, which is the desired quantity. Solution for (a) Here the initial kinetic energy is zero, so that \u0394KE = 1 22. The equation for change in potential energy states that \u0394PEg =. Since is negative in this case, we will rewrite this as \u0394PEg = \u2212 \u2223 \u2223 clearly. Thus, to show the minus sign becomes \u2212\u0394PEg = \u0394KE \u2223 \u2223 = 1 22. Solving for, we find that mass cancels and that = 2 \u2223 \u2223. Substituting known values, = 2 9.80 m/s2 (20.0 m) = 19.8 m/s. Solution for (b) Again \u2212 \u0394PEg = \u0394KE. In this case there is initial kinetic energy, so \u0394KE = 1 22 \u2212 1 20 \u2223 \u2223 = 1 22 \u2212 1 20 2. Rearranging gives 22 = \u2223 \u2223 + 1 1 20 2. (7.34) (7.35) (7.36) (7.37) (7.38) (7.39) 2. Thus, This means that the final kinetic energy is the sum of the initial kinetic energy and the gravitational potential energy. Mass again cancels, and = 2 \u2223 \u2223 + 0 2. (7.40) This equation is very similar to the kinematics equation = 0 valid only for constant acceleration, whereas our equation above is valid for any path regardless of whether the object moves with a constant acceleration. Now, substituting known values gives 2 + 2, but it is more general\u2014the kinematics equation is = 2(9.80 m/s2)(20.0 m) + (5.00 m/s)2 = 20.4 m/s. (7.41) Discussion and Implications First, note that mass cancels. This is quite consistent with observations made in Falling Objects that all objects fall at the same rate if friction is negligible. Second, only the speed of the roller coaster is considered; there is no information about its direction at any point. This reveals another general truth. When friction is negligible, the speed of a falling body depends only on its initial speed and height, and not on its mass or the path taken. For example, the roller coaster will have the same final speed whether it falls 20.0 m straight down or takes a more complicated path like the one in the figure. Third, and perhaps unexpectedly, the final speed in part (b) is greater than in part (a), but by far less than 5.00 m/s. Finally, note that speed can be found at any height along the way by simply using the appropriate value of at the point of interest. We have seen that work done by or against the gravitational force depends only on the starting and ending points, and not on the path between, allowing us to define the simplifying concept of gravitational potential energy. We can do the same thing for a few other forces, and we will see that this leads to a formal definition of the law of conservation of energy. This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 7 | Work, Energy, and Energy Resources 277 Making Connections: Take-Home Investigation\u2014Converting Potential to Kinetic Energy One can study the conversion of gravitational potential energy into kinetic energy in this experiment. On a smooth, level surface, use a ruler of the kind that has a groove running along its length and a book to make an incline (see Figure 7.9). Place a marble at the 10-cm position on the ruler and let it roll down the ruler. When it hits the level surface, measure the time it takes to roll one meter. Now place the marble at the 20-cm and the 30-cm positions and again measure the times it takes to roll 1 m on the level surface. Find the velocity of the marble on the level surface for all three positions. Plot velocity squared versus the distance traveled by the marble. What is the shape of each plot? If the shape is a straight line, the plot shows that the marble\u2019s kinetic energy at the bottom is proportional to its potential energy at the release point. Figure 7.9 A marble rolls down a ruler, and its speed on the level surface is measured. 7.4 Conservative Forces and Potential Energy Learning Objectives By the", " end of this section, you will be able to: \u2022 Define conservative force, potential energy, and mechanical energy. \u2022 Explain the potential energy of a spring in terms of its compression when Hooke\u2019s law applies. \u2022 Use the work-energy theorem to show how having only conservative forces leads to conservation of mechanical energy. The information presented in this section supports the following AP\u00ae learning objectives and science practices: \u2022 4.C.1.1 The student is able to calculate the total energy of a system and justify the mathematical routines used in the calculation of component types of energy within the system whose sum is the total energy. (S.P. 1.4, 2.1, 2.2) \u2022 4.C.2.1 The student is able to make predictions about the changes in the mechanical energy of a system when a component of an external force acts parallel or antiparallel to the direction of the displacement of the center of mass. (S.P. 6.4) \u2022 5.B.1.1 The student is able to set up a representation or model showing that a single object can only have kinetic energy and use information about that object to calculate its kinetic energy. (S.P. 1.4, 2.2) \u2022 5.B.1.2 The student is able to translate between a representation of a single object, which can only have kinetic energy, and a system that includes the object, which may have both kinetic and potential energies. (S.P. 1.5) \u2022 5.B.3.1 The student is able to describe and make qualitative and/or quantitative predictions about everyday examples of systems with internal potential energy. (S.P. 2.2, 6.4, 7.2) \u2022 5.B.3.2 The student is able to make quantitative calculations of the internal potential energy of a system from a description or diagram of that system. (S.P. 1.4, 2.2) \u2022 5.B.3.3 The student is able to apply mathematical reasoning to create a description of the internal potential energy of a system from a description or diagram of the objects and interactions in that system. (S.P. 1.4, 2.2) Potential Energy and Conservative Forces Work is done by a force, and some forces, such as weight, have special characteristics. A conservative force is one, like the gravitational force, for which work done by or against it depends only on the starting and ending points of a motion and not on the path taken. We can define a potential energy (PE) for any conservative force, just as we did for the gravitational force. For example, when you wind up a toy, an egg timer, or an old-fashioned watch, you do work against its spring and store energy in it. (We treat these springs as ideal, in that we assume there is no friction and no production of thermal energy.) This stored energy is recoverable as work, and it is useful to think of it as potential energy contained in the spring. Indeed, the reason that the spring has this characteristic is that its force is conservative. That is, a conservative force results in stored or potential energy. Gravitational potential energy is one example, as is the energy stored in a spring. We will also see how conservative forces are related to the conservation of energy. Potential Energy and Conservative Forces Potential energy is the energy a system has due to position, shape, or configuration. It is stored energy that is completely recoverable. 278 Chapter 7 | Work, Energy, and Energy Resources A conservative force is one for which work done by or against it depends only on the starting and ending points of a motion and not on the path taken. We can define a potential energy (PE) for any conservative force. The work done against a conservative force to reach a final configuration depends on the configuration, not the path followed, and is the potential energy added. Real World Connections: Energy of a Bowling Ball How much energy does a bowling ball have? (Just think about it for a minute.) If you are thinking that you need more information, you\u2019re right. If we can measure the ball\u2019s velocity, then determining its kinetic energy is simple. Note that this does require defining a reference frame in which to measure the velocity. Determining the ball\u2019s potential energy also requires more information. You need to know its height above the ground, which requires a reference frame of the ground. Without the ground\u2014in other words, Earth\u2014the ball does not classically have potential energy. Potential energy comes from the interaction between the ball and the ground. Another way of thinking about this is to compare the ball\u2019s potential energy on Earth and on the Moon. A bowling ball a certain height above Earth is going to have more potential energy than the same bowling ball the same height above the surface of the Moon, because Earth has greater mass than the Moon and therefore exerts more gravity on the ball", ". Thus, potential energy requires a system of at least two objects, or an object with an internal structure of at least two parts. Potential Energy of a Spring First, let us obtain an expression for the potential energy stored in a spring ( PEs ). We calculate the work done to stretch or compress a spring that obeys Hooke\u2019s law. (Hooke\u2019s law was examined in Elasticity: Stress and Strain, and states that the magnitude of force on the spring and the resulting deformation \u0394 are proportional, = \u0394.) (See Figure 7.10.) For our spring, we will replace \u0394 (the amount of deformation produced by a force ) by the distance that the spring is stretched or compressed along its length. So the force needed to stretch the spring has magnitude, where is the spring\u2019s force constant. The force increases linearly from 0 at the start to in the fully stretched position. The average force is / 2. Thus the work done in stretching or compressing the spring is s = = 22. Alternatively, we noted in Kinetic Energy = 1 2 and the Work-Energy Theorem that the area under a graph of vs. is the work done by the force. In Figure 7.10(c) we see that this area is also 1 22. We therefore define the potential energy of a spring, PEs, to be 22, where is the spring\u2019s force constant and is the displacement from its undeformed position. The potential energy represents the work done on the spring and the energy stored in it as a result of stretching or compressing it a distance. The potential energy of the spring PEs does not depend on the path taken; it depends only on the stretch or squeeze in the final configuration. PEs = 1 (7.42) Figure 7.10 (a) An undeformed spring has no PEs stored in it. (b) The force needed to stretch (or compress) the spring a distance has a magnitude =, and the work done to stretch (or compress) it is 1 energy (PEs) in the spring, and it can be fully recovered. (c) A graph of vs. has a slope of, and the area under the graph is 1 the work done or potential energy stored is 1. Because the force is conservative, this work is stored as potential 22 22.. Thus 22 This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 7 | Work, Energy, and Energy Resources 279 The equation PEs = 1 22 has general validity beyond the special case for which it was derived. Potential energy can be stored in any elastic medium by deforming it. Indeed, the general definition of potential energy is energy due to position, shape, or configuration. For shape or position deformations, stored energy is PEs = 1 system and is its deformation. Another example is seen in Figure 7.11 for a guitar string. 22, where is the force constant of the particular Figure 7.11 Work is done to deform the guitar string, giving it potential energy. When released, the potential energy is converted to kinetic energy and back to potential as the string oscillates back and forth. A very small fraction is dissipated as sound energy, slowly removing energy from the string. Conservation of Mechanical Energy Let us now consider what form the work-energy theorem takes when only conservative forces are involved. This will lead us to the conservation of energy principle. The work-energy theorem states that the net work done by all forces acting on a system equals its change in kinetic energy. In equation form, this is net = 1 22 \u2212 1 20 If only conservative forces act, then where c is the total work done by all conservative forces. Thus, c = \u0394KE. net = c, 2 = \u0394KE. (7.43) (7.44) (7.45) Now, if the conservative force, such as the gravitational force or a spring force, does work, the system loses potential energy. That is, c = \u2212\u0394PE. Therefore, or \u2212\u0394PE = \u0394KE \u0394KE + \u0394PE = 0. (7.46) (7.47) This equation means that the total kinetic and potential energy is constant for any process involving only conservative forces. That is, KE + PE = constant or KEi + PEi = KEf + PEf (conservative forces only), (7.48) where i and f denote initial and final values. This equation is a form of the work-energy theorem for conservative forces; it is known as the conservation of mechanical energy principle. Remember that this applies to the extent that all the forces are conservative, so that friction is negligible. The total kinetic plus potential energy of a system is defined to be its mechanical 280 Chapter 7 | Work, Energy, and Energy Resources energy, (KE + PE). In a system that experiences only conservative forces, there is a potential energy associated with each force", ", and the energy only changes form between KE and the various types of PE, with the total energy remaining constant. The internal energy of a system is the sum of the kinetic energies of all of its elements, plus the potential energy due to all of the interactions due to conservative forces between all of the elements. Real World Connections Consider a wind-up toy, such as a car. It uses a spring system to store energy. The amount of energy stored depends only on how many times it is wound, not how quickly or slowly the winding happens. Similarly, a dart gun using compressed air stores energy in its internal structure. In this case, the energy stored inside depends only on how many times it is pumped, not how quickly or slowly the pumping is done. The total energy put into the system, whether through winding or pumping, is equal to the total energy conserved in the system (minus any energy loss in the system due to interactions between its parts, such as air leaks in the dart gun). Since the internal energy of the system is conserved, you can calculate the amount of stored energy by measuring the kinetic energy of the system (the moving car or dart) when the potential energy is released. Example 7.8 Using Conservation of Mechanical Energy to Calculate the Speed of a Toy Car A 0.100-kg toy car is propelled by a compressed spring, as shown in Figure 7.12. The car follows a track that rises 0.180 m above the starting point. The spring is compressed 4.00 cm and has a force constant of 250.0 N/m. Assuming work done by friction to be negligible, find (a) how fast the car is going before it starts up the slope and (b) how fast it is going at the top of the slope. Figure 7.12 A toy car is pushed by a compressed spring and coasts up a slope. Assuming negligible friction, the potential energy in the spring is first completely converted to kinetic energy, and then to a combination of kinetic and gravitational potential energy as the car rises. The details of the path are unimportant because all forces are conservative\u2014the car would have the same final speed if it took the alternate path shown. Strategy The spring force and the gravitational force are conservative forces, so conservation of mechanical energy can be used. Thus, or KEi +PEi = KEf + PEf 1 2i 2 + i + 1 2i 2 = 1 2f 2 + f + 1 2f 2, (7.49) (7.50) where is the height (vertical position) and is the compression of the spring. This general statement looks complex but becomes much simpler when we start considering specific situations. First, we must identify the initial and final conditions in a problem; then, we enter them into the last equation to solve for an unknown. Solution for (a) This part of the problem is limited to conditions just before the car is released and just after it leaves the spring. Take the initial height to be zero, so that both i and f are zero. Furthermore, the initial speed i is zero and the final compression of the spring f is zero, and so several terms in the conservation of mechanical energy equation are zero and it simplifies to 1 2i 2 = 1 2f 2. (7.51) In other words, the initial potential energy in the spring is converted completely to kinetic energy in the absence of friction. Solving for the final speed and entering known values yields This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 7 | Work, Energy, and Energy Resources f = i = 250.0 N/m 0.100 kg = 2.00 m/s. (0.0400 m) 281 (7.52) Solution for (b) One method of finding the speed at the top of the slope is to consider conditions just before the car is released and just after it reaches the top of the slope, completely ignoring everything in between. Doing the same type of analysis to find which terms are zero, the conservation of mechanical energy becomes 2 = 1 1 2i 2f 2 + f. (7.53) This form of the equation means that the spring\u2019s initial potential energy is converted partly to gravitational potential energy and partly to kinetic energy. The final speed at the top of the slope will be less than at the bottom. Solving for f and substituting known values gives f = 2 i \u2212 2f 250.0 N/m 0.100 kg (0.0400 m)2 \u2212 2(9.80 m/s2)(0.180 m) (7.54) = Discussion = 0.687 m/s. Another way to solve this problem is to realize that the car\u2019s kinetic energy before it goes up the slope is converted partly to potential energy\u2014that is, to take the final conditions in part (a) to be the initial conditions in part (b). Applying the", " Science Practices: Potential Energy in a Spring Suppose you are running an experiment in which two 250 g carts connected by a spring (with spring constant 120 N/m) are run into a solid block, and the compression of the spring is measured. In one run of this experiment, the spring was measured to compress from its rest length of 5.0 cm to a minimum length of 2.0 cm. What was the potential energy stored in this system? Answer Note that the change in length of the spring is 3.0 cm. Hence we can apply Equation 7.42 to find that the potential energy is PE = (1/2)(120 N/m)(0.030 m)2 = 0.0541 J. Note that, for conservative forces, we do not directly calculate the work they do; rather, we consider their effects through their corresponding potential energies, just as we did in Example 7.8. Note also that we do not consider details of the path taken\u2014only the starting and ending points are important (as long as the path is not impossible). This assumption is usually a tremendous simplification, because the path may be complicated and forces may vary along the way. PhET Explorations: Energy Skate Park Learn about conservation of energy with a skater dude! Build tracks, ramps and jumps for the skater and view the kinetic energy, potential energy and friction as he moves. You can also take the skater to different planets or even space! Figure 7.13 Energy Skate Park (http://cnx.org/content/m55076/1.4/energy-skate-park_en.jar) 282 Chapter 7 | Work, Energy, and Energy Resources 7.5 Nonconservative Forces By the end of this section, you will be able to: Learning Objectives \u2022 Define nonconservative forces and explain how they affect mechanical energy. \u2022 Show how the principle of conservation of energy can be applied by treating the conservative forces in terms of their potential energies and any nonconservative forces in terms of the work they do. The information presented in this section supports the following AP\u00ae learning objectives and science practices: \u2022 4.C.1.2 The student is able to predict changes in the total energy of a system due to changes in position and speed of objects or frictional interactions within the system. (S.P. 6.4) \u2022 4.C.2.1 The student is able to make predictions about the changes in the mechanical energy of a system when a component of an external force acts parallel or antiparallel to the direction of the displacement of the center of mass. (S.P. 6.4) Nonconservative Forces and Friction Forces are either conservative or nonconservative. Conservative forces were discussed in Conservative Forces and Potential Energy. A nonconservative force is one for which work depends on the path taken. Friction is a good example of a nonconservative force. As illustrated in Figure 7.14, work done against friction depends on the length of the path between the starting and ending points. Because of this dependence on path, there is no potential energy associated with nonconservative forces. An important characteristic is that the work done by a nonconservative force adds or removes mechanical energy from a system. Friction, for example, creates thermal energy that dissipates, removing energy from the system. Furthermore, even if the thermal energy is retained or captured, it cannot be fully converted back to work, so it is lost or not recoverable in that sense as well. Figure 7.14 The amount of the happy face erased depends on the path taken by the eraser between points A and B, as does the work done against friction. Less work is done and less of the face is erased for the path in (a) than for the path in (b). The force here is friction, and most of the work goes into thermal energy that subsequently leaves the system (the happy face plus the eraser). The energy expended cannot be fully recovered. How Nonconservative Forces Affect Mechanical Energy Mechanical energy may not be conserved when nonconservative forces act. For example, when a car is brought to a stop by friction on level ground, it loses kinetic energy, which is dissipated as thermal energy, reducing its mechanical energy. Figure 7.15 compares the effects of conservative and nonconservative forces. We often choose to understand simpler systems such as that described in Figure 7.15(a) first before studying more complicated systems as in Figure 7.15(b). This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 7 | Work, Energy, and Energy Resources 283 Figure 7.15 Comparison of the effects of conservative and nonconservative forces on the mechanical energy of a system. (a) A system with only conservative forces. When a rock is dropped onto a spring, its mechanical energy remains constant (neglecting air resistance) because the force in the spring is conservative. The spring can propel the rock back to its original height", ", where it once again has only potential energy due to gravity. (b) A system with nonconservative forces. When the same rock is dropped onto the ground, it is stopped by nonconservative forces that dissipate its mechanical energy as thermal energy, sound, and surface distortion. The rock has lost mechanical energy. How the Work-Energy Theorem Applies Now let us consider what form the work-energy theorem takes when both conservative and nonconservative forces act. We will see that the work done by nonconservative forces equals the change in the mechanical energy of a system. As noted in Kinetic Energy and the Work-Energy Theorem, the work-energy theorem states that the net work on a system equals the change in its kinetic energy, or net = \u0394KE. The net work is the sum of the work by nonconservative forces plus the work by conservative forces. That is, so that net = nc + c, nc + c = \u0394KE, (7.55) (7.56) where nc is the total work done by all nonconservative forces and c is the total work done by all conservative forces. Figure 7.16 A person pushes a crate up a ramp, doing work on the crate. Friction and gravitational force (not shown) also do work on the crate; both forces oppose the person\u2019s push. As the crate is pushed up the ramp, it gains mechanical energy, implying that the work done by the person is greater than the work done by friction. Consider Figure 7.16, in which a person pushes a crate up a ramp and is opposed by friction. As in the previous section, we note that work done by a conservative force comes from a loss of gravitational potential energy, so that c = \u2212\u0394PE. Substituting this equation into the previous one and solving for nc gives nc = \u0394KE + \u0394PE. (7.57) This equation means that the total mechanical energy (KE + PE) changes by exactly the amount of work done by nonconservative forces. In Figure 7.16, this is the work done by the person minus the work done by friction. So even if energy is not conserved for the system of interest (such as the crate), we know that an equal amount of work was done to cause the change in total mechanical energy. We rearrange nc = \u0394KE + \u0394PE to obtain KEi +PEi + nc = KEf + PEf. (7.58) 284 Chapter 7 | Work, Energy, and Energy Resources This means that the amount of work done by nonconservative forces adds to the mechanical energy of a system. If nc is positive, then mechanical energy is increased, such as when the person pushes the crate up the ramp in Figure 7.16. If nc is negative, then mechanical energy is decreased, such as when the rock hits the ground in Figure 7.15(b). If nc is zero, then mechanical energy is conserved, and nonconservative forces are balanced. For example, when you push a lawn mower at constant speed on level ground, your work done is removed by the work of friction, and the mower has a constant energy. Applying Energy Conservation with Nonconservative Forces When no change in potential energy occurs, applying KEi +PEi + nc = KEf + PEf amounts to applying the work-energy theorem by setting the change in kinetic energy to be equal to the net work done on the system, which in the most general case includes both conservative and nonconservative forces. But when seeking instead to find a change in total mechanical energy in situations that involve changes in both potential and kinetic energy, the previous equation KEi + PEi + nc = KEf + PEf says that you can start by finding the change in mechanical energy that would have resulted from just the conservative forces, including the potential energy changes, and add to it the work done, with the proper sign, by any nonconservative forces involved. Example 7.9 Calculating Distance Traveled: How Far a Baseball Player Slides Consider the situation shown in Figure 7.17, where a baseball player slides to a stop on level ground. Using energy considerations, calculate the distance the 65.0-kg baseball player slides, given that his initial speed is 6.00 m/s and the force of friction against him is a constant 450 N. Figure 7.17 The baseball player slides to a stop in a distance. In the process, friction removes the player\u2019s kinetic energy by doing an amount of work equal to the initial kinetic energy. Strategy Friction stops the player by converting his kinetic energy into other forms, including thermal energy. In terms of the workenergy theorem, the work done by friction, which is negative, is added to the initial kinetic energy to reduce it to zero. The work done by friction is negative, because f is in the opposite direction of the motion (that is, = 180\u00ba, and so cos = \u22121 ). Thus n", "c = \u2212. The equation simplifies to or 1 2i 2 \u2212 = 0 = 1 2i 2. This equation can now be solved for the distance. Solution Solving the previous equation for and substituting known values yields (7.59) (7.60) This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 7 | Work, Energy, and Energy Resources = = 2 i 2 (65.0 kg)(6.00 m/s)2 (2)(450 N) = 2.60 m. 285 (7.61) Discussion The most important point of this example is that the amount of nonconservative work equals the change in mechanical energy. For example, you must work harder to stop a truck, with its large mechanical energy, than to stop a mosquito. Example 7.10 Calculating Distance Traveled: Sliding Up an Incline Suppose that the player from Example 7.9 is running up a hill having a 5.00\u00ba incline upward with a surface similar to that in the baseball stadium. The player slides with the same initial speed. Determine how far he slides. Figure 7.18 The same baseball player slides to a stop on a 5.00\u00ba slope. Strategy In this case, the work done by the nonconservative friction force on the player reduces the mechanical energy he has from his kinetic energy at zero height, to the final mechanical energy he has by moving through distance to reach height along the hill, with = sin 5.00\u00ba. This is expressed by the equation KE + PEi + nc = KEf + PEf. (7.62) Solution The work done by friction is again nc = \u2212 ; initially the potential energy is PEi = \u22c5 0 = 0 and the kinetic energy is KEi = 1 2 ; the final energy contributions are KEf = 0 for the kinetic energy and PEf = = sin for the potential energy. 2i Substituting these values gives Solve this for to obtain 1 2i 2 + 0 + \u2212 = 0 + sin = 2 1 2 i + sin (7.63) (7.64) = (0.5)(65.0 kg)(6.00 m/s)2 450 N+(65.0 kg)(9.80 m/s2) sin (5.00\u00ba) = 2.31 m. Discussion As might have been expected, the player slides a shorter distance by sliding uphill. Note that the problem could also have been solved in terms of the forces directly and the work energy theorem, instead of using the potential energy. This method would have required combining the normal force and force of gravity vectors, which no longer cancel each other because they point in different directions, and friction, to find the net force. You could then use the net force and the net work to find the distance that reduces the kinetic energy to zero. By applying conservation of energy and using the potential energy 286 Chapter 7 | Work, Energy, and Energy Resources instead, we need only consider the gravitational potential energy, without combining and resolving force vectors. This simplifies the solution considerably. Making Connections: Take-Home Investigation\u2014Determining Friction from the Stopping Distance This experiment involves the conversion of gravitational potential energy into thermal energy. Use the ruler, book, and marble from Making Connections: Take-Home Investigation\u2014Converting Potential to Kinetic Energy. In addition, you will need a foam cup with a small hole in the side, as shown in Figure 7.19. From the 10-cm position on the ruler, let the marble roll into the cup positioned at the bottom of the ruler. Measure the distance the cup moves before stopping. What forces caused it to stop? What happened to the kinetic energy of the marble at the bottom of the ruler? Next, place the marble at the 20-cm and the 30-cm positions and again measure the distance the cup moves after the marble enters it. Plot the distance the cup moves versus the initial marble position on the ruler. Is this relationship linear? With some simple assumptions, you can use these data to find the coefficient of kinetic friction k of the cup on the table. The force of friction on the cup is k, where the normal force is just the weight of the cup plus the marble. The normal force and force of gravity do no work because they are perpendicular to the displacement of the cup, which moves horizontally. The work done by friction is. You will need the mass of the marble as well to calculate its initial kinetic energy. It is interesting to do the above experiment also with a steel marble (or ball bearing). Releasing it from the same positions on the ruler as you did with the glass marble, is the velocity of this steel marble the same as the velocity of the marble at the bottom of the ruler? Is the distance the cup moves proportional to the mass of the steel and glass marbles? Figure 7.19 Rolling a marble down a ruler into a", " foam cup. PhET Explorations: The Ramp Explore forces, energy and work as you push household objects up and down a ramp. Lower and raise the ramp to see how the angle of inclination affects the parallel forces acting on the file cabinet. Graphs show forces, energy and work. Figure 7.20 The Ramp (http://cnx.org/content/m55047/1.4/the-ramp_en.jar) 7.6 Conservation of Energy By the end of this section, you will be able to: Learning Objectives \u2022 Explain the law of the conservation of energy. \u2022 Describe some of the many forms of energy. \u2022 Define efficiency of an energy conversion process as the fraction left as useful energy or work, rather than being transformed, for example, into thermal energy. The information presented in this section supports the following AP\u00ae learning objectives and science practices: \u2022 4.C.1.2 The student is able to predict changes in the total energy of a system due to changes in position and speed of objects or frictional interactions within the system. (S.P. 6.4) This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 7 | Work, Energy, and Energy Resources 287 \u2022 4.C.2.1 The student is able to make predictions about the changes in the mechanical energy of a system when a component of an external force acts parallel or antiparallel to the direction of the displacement of the center of mass. (S.P. 6.4) \u2022 4.C.2.2 The student is able to apply the concepts of conservation of energy and the work-energy theorem to determine qualitatively and/or quantitatively that work done on a two-object system in linear motion will change the kinetic energy of the center of mass of the system, the potential energy of the systems, and/or the internal energy of the system. (S.P. 1.4, 2.2, 7.2) \u2022 5.A.2.1 The student is able to define open and closed systems for everyday situations and apply conservation concepts for energy, charge, and linear momentum to those situations. (S.P. 6.4, 7.2) \u2022 5.B.5.4 The student is able to make claims about the interaction between a system and its environment in which the environment exerts a force on the system, thus doing work on the system and changing the energy of the system (kinetic energy plus potential energy). (S.P. 6.4, 7.2) \u2022 5.B.5.5 The student is able to predict and calculate the energy transfer to (i.e., the work done on) an object or system from information about a force exerted on the object or system through a distance. (S.P. 2.2, 6.4) Law of Conservation of Energy Energy, as we have noted, is conserved, making it one of the most important physical quantities in nature. The law of conservation of energy can be stated as follows: Total energy is constant in any process. It may change in form or be transferred from one system to another, but the total remains the same. We have explored some forms of energy and some ways it can be transferred from one system to another. This exploration led to the definition of two major types of energy\u2014mechanical energy (KE + PE) and energy transferred via work done by nonconservative forces (nc). But energy takes many other forms, manifesting itself in many different ways, and we need to be able to deal with all of these before we can write an equation for the above general statement of the conservation of energy. Other Forms of Energy than Mechanical Energy At this point, we deal with all other forms of energy by lumping them into a single group called other energy ( OE ). Then we can state the conservation of energy in equation form as KEi + PEi + nc + OEi = KEf + PEf + OEf. (7.65) All types of energy and work can be included in this very general statement of conservation of energy. Kinetic energy is KE, work done by a conservative force is represented by PE, work done by nonconservative forces is nc, and all other energies are included as OE. This equation applies to all previous examples; in those situations OE was constant, and so it subtracted out and was not directly considered. Making Connections: Usefulness of the Energy Conservation Principle The fact that energy is conserved and has many forms makes it very important. You will find that energy is discussed in many contexts, because it is involved in all processes. It will also become apparent that many situations are best understood in terms of energy and that problems are often most easily conceptualized and solved by considering energy. When does OE play a role? One example occurs when a person eats. Food is", " oxidized with the release of carbon dioxide, water, and energy. Some of this chemical energy is converted to kinetic energy when the person moves, to potential energy when the person changes altitude, and to thermal energy (another form of OE ). Some of the Many Forms of Energy What are some other forms of energy? You can probably name a number of forms of energy not yet discussed. Many of these will be covered in later chapters, but let us detail a few here. Electrical energy is a common form that is converted to many other forms and does work in a wide range of practical situations. Fuels, such as gasoline and food, carry chemical energy that can be transferred to a system through oxidation. Chemical fuel can also produce electrical energy, such as in batteries. Batteries can in turn produce light, which is a very pure form of energy. Most energy sources on Earth are in fact stored energy from the energy we receive from the Sun. We sometimes refer to this as radiant energy, or electromagnetic radiation, which includes visible light, infrared, and ultraviolet radiation. Nuclear energy comes from processes that convert measurable amounts of mass into energy. Nuclear energy is transformed into the energy of sunlight, into electrical energy in power plants, and into the energy of the heat transfer and blast in weapons. Atoms and molecules inside all objects are in random motion. This internal mechanical energy from the random motions is called thermal energy, because it is related to the temperature of the object. These and all other forms of energy can be converted into one another and can do work. 288 Chapter 7 | Work, Energy, and Energy Resources Real World Connections: Open or Closed System? Consider whether the following systems are open or closed: a car, a spring-operated dart gun, and the system shown in Figure 7.15(a). A car is not a closed system. You add energy in the form of more gas in the tank (or charging the batteries), and energy is lost due to air resistance and friction. A spring-operated dart gun is not a closed system. You have to initially compress the spring. Once that has been done, however, the dart gun and dart can be treated as a closed system. All of the energy remains in the system consisting of these two objects. Figure 7.15(a) is an example of a closed system, once it has been started. All of the energy in the system remains there; none is brought in from outside or leaves. Table 7.1 gives the amount of energy stored, used, or released from various objects and in various phenomena. The range of energies and the variety of types and situations is impressive. Problem-Solving Strategies for Energy You will find the following problem-solving strategies useful whenever you deal with energy. The strategies help in organizing and reinforcing energy concepts. In fact, they are used in the examples presented in this chapter. The familiar general problem-solving strategies presented earlier\u2014involving identifying physical principles, knowns, and unknowns, checking units, and so on\u2014continue to be relevant here. Step 1. Determine the system of interest and identify what information is given and what quantity is to be calculated. A sketch will help. Step 2. Examine all the forces involved and determine whether you know or are given the potential energy from the work done by the forces. Then use step 3 or step 4. Step 3. If you know the potential energies for the forces that enter into the problem, then forces are all conservative, and you can apply conservation of mechanical energy simply in terms of potential and kinetic energy. The equation expressing conservation of energy is KEi + PEi = KEf + PEf. (7.66) Step 4. If you know the potential energy for only some of the forces, possibly because some of them are nonconservative and do not have a potential energy, or if there are other energies that are not easily treated in terms of force and work, then the conservation of energy law in its most general form must be used. KEi + PEi + nc + OEi = KEf + PEf + OEf. (7.67) In most problems, one or more of the terms is zero, simplifying its solution. Do not calculate c, the work done by conservative forces; it is already incorporated in the PE terms. Step 5. You have already identified the types of work and energy involved (in step 2). Before solving for the unknown, eliminate terms wherever possible to simplify the algebra. For example, choose = 0 at either the initial or final point, so that PEg is zero there. Then solve for the unknown in the customary manner. Step 6. Check the answer to see if it is reasonable. Once you have solved a problem, reexamine the forms of work and energy to see if you have set up the conservation of energy equation correctly. For example, work done against friction should be negative, potential energy at the bottom of a hill should be less than that at the top, and so on", ". Also check to see that the numerical value obtained is reasonable. For example, the final speed of a skateboarder who coasts down a 3-mhigh ramp could reasonably be 20 km/h, but not 80 km/h. Transformation of Energy The transformation of energy from one form into others is happening all the time. The chemical energy in food is converted into thermal energy through metabolism; light energy is converted into chemical energy through photosynthesis. In a larger example, the chemical energy contained in coal is converted into thermal energy as it burns to turn water into steam in a boiler. This thermal energy in the steam in turn is converted to mechanical energy as it spins a turbine, which is connected to a generator to produce electrical energy. (In all of these examples, not all of the initial energy is converted into the forms mentioned. This important point is discussed later in this section.) Another example of energy conversion occurs in a solar cell. Sunlight impinging on a solar cell (see Figure 7.21) produces electricity, which in turn can be used to run an electric motor. Energy is converted from the primary source of solar energy into electrical energy and then into mechanical energy. This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 7 | Work, Energy, and Energy Resources 289 Figure 7.21 Solar energy is converted into electrical energy by solar cells, which is used to run a motor in this solar-power aircraft. (credit: NASA) 290 Chapter 7 | Work, Energy, and Energy Resources Table 7.1 Energy of Various Objects and Phenomena Object/phenomenon Energy in joules Big Bang Energy released in a supernova Fusion of all the hydrogen in Earth\u2019s oceans Annual world energy use Large fusion bomb (9 megaton) 1 kg hydrogen (fusion to helium) 1 kg uranium (nuclear fission) Hiroshima-size fission bomb (10 kiloton) 90,000-ton aircraft carrier at 30 knots 1 barrel crude oil 1 ton TNT 1 gallon of gasoline 1068 1044 1034 41020 3.81016 6.41014 8.01013 4.21013 1.11010 5.9109 4.2109 1.2108 Daily home electricity use (developed countries) 7107 Daily adult food intake (recommended) 1000-kg car at 90 km/h 1 g fat (9.3 kcal) ATP hydrolysis reaction 1 g carbohydrate (4.1 kcal) 1 g protein (4.1 kcal) Tennis ball at 100 km/h Mosquito 10\u20132 g at 0.5 m/s Single electron in a TV tube beam Energy to break one DNA strand 1.2107 3.1105 3.9104 3.2104 1.7104 1.7104 22 1.310\u22126 4.010\u221215 10\u221219 Efficiency Even though energy is conserved in an energy conversion process, the output of useful energy or work will be less than the energy input. The efficiency of an energy conversion process is defined as Efficiency( ) = useful energy or work output total energy input = out in. (7.68) Table 7.2 lists some efficiencies of mechanical devices and human activities. In a coal-fired power plant, for example, about 40% of the chemical energy in the coal becomes useful electrical energy. The other 60% transforms into other (perhaps less useful) energy forms, such as thermal energy, which is then released to the environment through combustion gases and cooling towers. This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 7 | Work, Energy, and Energy Resources 291 Table 7.2 Efficiency of the Human Body and Mechanical Devices Activity/device Efficiency (%)[1] Cycling and climbing Swimming, surface Swimming, submerged Shoveling Weightlifting Steam engine Gasoline engine Diesel engine Nuclear power plant Coal power plant Electric motor Compact fluorescent light Gas heater (residential) Solar cell 20 2 4 3 9 17 30 35 35 42 98 20 90 10 PhET Explorations: Masses and Springs A realistic mass and spring laboratory. Hang masses from springs and adjust the spring stiffness and damping. You can even slow time. Transport the lab to different planets. A chart shows the kinetic, potential, and thermal energies for each spring. Figure 7.22 Masses and Springs (http://cnx.org/content/m55049/1.3/mass-spring-lab_en.jar) 7.7 Power What is Power? Learning Objectives By the end of this section, you will be able to: \u2022 Calculate power by calculating changes in energy over time. \u2022 Examine power consumption and calculations of the cost of energy consumed. Power\u2014the word conjures up many images: a professional football player muscling aside his opponent, a dragster roaring away from the starting line, a volcano blowing its lava into the atmosphere, or a rocket blasting off,", " as in Figure 7.23. 1. Representative values 292 Chapter 7 | Work, Energy, and Energy Resources Figure 7.23 This powerful rocket on the Space Shuttle Endeavor did work and consumed energy at a very high rate. (credit: NASA) These images of power have in common the rapid performance of work, consistent with the scientific definition of power ( ) as the rate at which work is done. Power Power is the rate at which work is done. The SI unit for power is the watt ( W ), where 1 watt equals 1 joule/second (1 W = 1 J/s). = (7.69) Because work is energy transfer, power is also the rate at which energy is expended. A 60-W light bulb, for example, expends 60 J of energy per second. Great power means a large amount of work or energy developed in a short time. For example, when a powerful car accelerates rapidly, it does a large amount of work and consumes a large amount of fuel in a short time. Calculating Power from Energy Example 7.11 Calculating the Power to Climb Stairs What is the power output for a 60.0-kg woman who runs up a 3.00 m high flight of stairs in 3.50 s, starting from rest but having a final speed of 2.00 m/s? (See Figure 7.24.) Figure 7.24 When this woman runs upstairs starting from rest, she converts the chemical energy originally from food into kinetic energy and gravitational potential energy. Her power output depends on how fast she does this. Strategy and Concept This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 7 | Work, Energy, and Energy Resources 293 The work going into mechanical energy is = KE + PE. At the bottom of the stairs, we take both KE and PEg as initially zero; thus, = KEf + PEg = 1 given, we can calculate and then divide it by time to get power. 2 +, where is the vertical height of the stairs. Because all terms are 2f Solution Substituting the expression for into the definition of power given in the previous equation, = / yields = = 1 2f 2 +. Entering known values yields = 0.5 60.0 kg (2.00 m/s)2 + 60.0 kg 9.80 m/s2 (3.00 m) 3.50 s (7.70) (7.71) = 120 J + 1764 J 3.50 s = 538 W. Discussion The woman does 1764 J of work to move up the stairs compared with only 120 J to increase her kinetic energy; thus, most of her power output is required for climbing rather than accelerating. It is impressive that this woman\u2019s useful power output is slightly less than 1 horsepower (1 hp = 746 W)! People can generate more than a horsepower with their leg muscles for short periods of time by rapidly converting available blood sugar and oxygen into work output. (A horse can put out 1 hp for hours on end.) Once oxygen is depleted, power output decreases and the person begins to breathe rapidly to obtain oxygen to metabolize more food\u2014this is known as the aerobic stage of exercise. If the woman climbed the stairs slowly, then her power output would be much less, although the amount of work done would be the same. Making Connections: Take-Home Investigation\u2014Measure Your Power Rating Determine your own power rating by measuring the time it takes you to climb a flight of stairs. We will ignore the gain in kinetic energy, as the above example showed that it was a small portion of the energy gain. Don\u2019t expect that your output will be more than about 0.5 hp. Examples of Power Examples of power are limited only by the imagination, because there are as many types as there are forms of work and energy. (See Table 7.3 for some examples.) Sunlight reaching Earth\u2019s surface carries a maximum power of about 1.3 kilowatts per square meter (kW/m2). A tiny fraction of this is retained by Earth over the long term. Our consumption rate of fossil fuels is far greater than the rate at which they are stored, so it is inevitable that they will be depleted. Power implies that energy is transferred, perhaps changing form. It is never possible to change one form completely into another without losing some of it as thermal energy. For example, a 60-W incandescent bulb converts only 5 W of electrical power to light, with 55 W dissipating into thermal energy. Furthermore, the typical electric power plant converts only 35 to 40% of its fuel into electricity. The remainder becomes a huge amount of thermal energy that must be dispersed as heat transfer, as rapidly as it is created. A coal-fired power plant may produce 1000 megawatts; 1 megawatt (MW) is 106 W of electric power. But", " the power plant consumes chemical energy at a rate of about 2500 MW, creating heat transfer to the surroundings at a rate of 1500 MW. (See Figure 7.25.) 294 Chapter 7 | Work, Energy, and Energy Resources Figure 7.25 Tremendous amounts of electric power are generated by coal-fired power plants such as this one in China, but an even larger amount of power goes into heat transfer to the surroundings. The large cooling towers here are needed to transfer heat as rapidly as it is produced. The transfer of heat is not unique to coal plants but is an unavoidable consequence of generating electric power from any fuel\u2014nuclear, coal, oil, natural gas, or the like. (credit: Kleinolive, Wikimedia Commons) Table 7.3 Power Output or Consumption Object or Phenomenon Power in Watts Supernova (at peak) Milky Way galaxy Crab Nebula pulsar The Sun Volcanic eruption (maximum) Lightning bolt Nuclear power plant (total electric and heat transfer) Aircraft carrier (total useful and heat transfer) Dragster (total useful and heat transfer) Car (total useful and heat transfer) Football player (total useful and heat transfer) Clothes dryer Person at rest (all heat transfer) 51037 1037 1028 41026 41015 21012 3109 108 2106 8104 5103 4103 100 Typical incandescent light bulb (total useful and heat transfer) 60 Heart, person at rest (total useful and heat transfer) Electric clock Pocket calculator 8 3 10\u22123 Power and Energy Consumption We usually have to pay for the energy we use. It is interesting and easy to estimate the cost of energy for an electrical appliance if its power consumption rate and time used are known. The higher the power consumption rate and the longer the appliance is used, the greater the cost of that appliance. The power consumption rate is = / = /, where is the energy supplied by the electricity company. So the energy consumed over a time is This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 7 | Work, Energy, and Energy Resources = 295 (7.72) Electricity bills state the energy used in units of kilowatt-hours (kW \u22c5 h) which is the product of power in kilowatts and time in hours. This unit is convenient because electrical power consumption at the kilowatt level for hours at a time is typical. Example 7.12 Calculating Energy Costs What is the cost of running a 0.200-kW computer 6.00 h per day for 30.0 d if the cost of electricity is $0.120 per kW \u22c5 h? Strategy Cost is based on energy consumed; thus, we must find from = and then calculate the cost. Because electrical energy is expressed in kW \u22c5 h, at the start of a problem such as this it is convenient to convert the units into kW and hours. Solution The energy consumed in kW \u22c5 h is = = (0.200 kW)(6.00 h/d)(30.0 d) = 36.0 kW \u22c5 h, and the cost is simply given by cost = (36.0 kW \u22c5 h)($0.120 per kW \u22c5 h) = $4.32 per month. Discussion (7.73) (7.74) The cost of using the computer in this example is neither exorbitant nor negligible. It is clear that the cost is a combination of power and time. When both are high, such as for an air conditioner in the summer, the cost is high. The motivation to save energy has become more compelling with its ever-increasing price. Armed with the knowledge that energy consumed is the product of power and time, you can estimate costs for yourself and make the necessary value judgments about where to save energy. Either power or time must be reduced. It is most cost-effective to limit the use of highpower devices that normally operate for long periods of time, such as water heaters and air conditioners. This would not include relatively high power devices like toasters, because they are on only a few minutes per day. It would also not include electric clocks, in spite of their 24-hour-per-day usage, because they are very low power devices. It is sometimes possible to use devices that have greater efficiencies\u2014that is, devices that consume less power to accomplish the same task. One example is the compact fluorescent light bulb, which produces over four times more light per watt of power consumed than its incandescent cousin. Modern civilization depends on energy, but current levels of energy consumption and production are not sustainable. The likelihood of a link between global warming and fossil fuel use (with its concomitant production of carbon dioxide), has made reduction in energy use as well as a shift to non-fossil fuels of the utmost importance. Even though energy in an isolated system is a conserved quantity, the final result of most energy transformations", " is waste heat transfer to the environment, which is no longer useful for doing work. As we will discuss in more detail in Thermodynamics, the potential for energy to produce useful work has been \u201cdegraded\u201d in the energy transformation. 7.8 Work, Energy, and Power in Humans By the end of this section, you will be able to: Learning Objectives \u2022 Explain the human body\u2019s consumption of energy when at rest versus when engaged in activities that do useful work. \u2022 Calculate the conversion of chemical energy in food into useful work. Energy Conversion in Humans Our own bodies, like all living organisms, are energy conversion machines. Conservation of energy implies that the chemical energy stored in food is converted into work, thermal energy, and/or stored as chemical energy in fatty tissue. (See Figure 7.26.) The fraction going into each form depends both on how much we eat and on our level of physical activity. If we eat more than is needed to do work and stay warm, the remainder goes into body fat. Chapter 7 | Work, Energy, and Energy Resources 301 bookkeeping device, and no exceptions have ever been found. It states that the total amount of energy in an isolated system will always remain constant. Related to this principle, but remarkably different from it, is the important philosophy of energy conservation. This concept has to do with seeking to decrease the amount of energy used by an individual or group through (1) reduced activities (e.g., turning down thermostats, driving fewer kilometers) and/or (2) increasing conversion efficiencies in the performance of a particular task\u2014such as developing and using more efficient room heaters, cars that have greater miles-pergallon ratings, energy-efficient compact fluorescent lights, etc. Since energy in an isolated system is not destroyed or created or generated, one might wonder why we need to be concerned about our energy resources, since energy is a conserved quantity. The problem is that the final result of most energy transformations is waste heat transfer to the environment and conversion to energy forms no longer useful for doing work. To state it in another way, the potential for energy to produce useful work has been \u201cdegraded\u201d in the energy transformation. (This will be discussed in more detail in Thermodynamics.) Glossary basal metabolic rate: the total energy conversion rate of a person at rest chemical energy: the energy in a substance stored in the bonds between atoms and molecules that can be released in a chemical reaction conservation of mechanical energy: the rule that the sum of the kinetic energies and potential energies remains constant if only conservative forces act on and within a system conservative force: followed a force that does the same work for any given initial and final configuration, regardless of the path efficiency: a measure of the effectiveness of the input of energy to do work; useful energy or work divided by the total input of energy electrical energy: the energy carried by a flow of charge energy: the ability to do work fossil fuels: oil, natural gas, and coal friction: the force between surfaces that opposes one sliding on the other; friction changes mechanical energy into thermal energy gravitational potential energy: the energy an object has due to its position in a gravitational field horsepower: an older non-SI unit of power, with 1 hp = 746 W joule: SI unit of work and energy, equal to one newton-meter kilowatt-hour: (kW \u22c5 h) unit used primarily for electrical energy provided by electric utility companies kinetic energy: the energy an object has by reason of its motion, equal to 1 22 for the translational (i.e., non-rotational) motion of an object of mass moving at speed law of conservation of energy: the general law that total energy is constant in any process; energy may change in form or be transferred from one system to another, but the total remains the same mechanical energy: the sum of kinetic energy and potential energy metabolic rate: the rate at which the body uses food energy to sustain life and to do different activities net work: work done by the net force, or vector sum of all the forces, acting on an object nonconservative force: a force whose work depends on the path followed between the given initial and final configurations nuclear energy: energy released by changes within atomic nuclei, such as the fusion of two light nuclei or the fission of a heavy nucleus potential energy: energy due to position, shape, or configuration potential energy of a spring: the stored energy of a spring as a function of its displacement; when Hooke\u2019s law applies, it is given by the expression 1 22 where is the distance the spring is compressed or extended and is the spring constant power: the rate at which work is done 302 Chapter 7 | Work, Energy, and Energy Resources radiant energy: the energy carried by electromagnetic waves renewable forms of energy: those sources that cannot be used up, such as water, wind, solar, and biomass thermal energy: the energy within an object due to the random", " motion of its atoms and molecules that accounts for the object's temperature useful work: work done on an external system watt: (W) SI unit of power, with 1 W = 1 J/s work: the transfer of energy by a force that causes an object to be displaced; the product of the component of the force in the direction of the displacement and the magnitude of the displacement work-energy theorem: the result, based on Newton\u2019s laws, that the net work done on an object is equal to its change in kinetic energy Section Summary 7.1 Work: The Scientific Definition \u2022 Work is the transfer of energy by a force acting on an object as it is displaced. \u2022 The work that a force F does on an object is the product of the magnitude of the force, times the magnitude of the displacement, times the cosine of the angle between them. In symbols, \u2022 The SI unit for work and energy is the joule (J), where 1 J = 1 N \u22c5 m = 1 kg \u22c5 m2/s2. \u2022 The work done by a force is zero if the displacement is either zero or perpendicular to the force. \u2022 The work done is positive if the force and displacement have the same direction, and negative if they have opposite direction. = cos. 7.2 Kinetic Energy and the Work-Energy Theorem \u2022 The net work net is the work done by the net force acting on an object. \u2022 Work done on an object transfers energy to the object. \u2022 The translational kinetic energy of an object of mass moving at speed is KE = 1 22. \u2022 The work-energy theorem states that the net work net on a system changes its kinetic energy, net = 1 22 \u2212 1 20 2. 7.3 Gravitational Potential Energy \u2022 Work done against gravity in lifting an object becomes potential energy of the object-Earth system. \u2022 The change in gravitational potential energy, \u0394PEg, is \u0394PEg =, with being the increase in height and g the acceleration due to gravity. \u2022 The gravitational potential energy of an object near Earth\u2019s surface is due to its position in the mass-Earth system. Only differences in gravitational potential energy, \u0394PEg, have physical significance. \u2022 As an object descends without friction, its gravitational potential energy changes into kinetic energy corresponding to increasing speed, so that \u0394KE= \u2212\u0394PEg. 7.4 Conservative Forces and Potential Energy \u2022 A conservative force is one for which work depends only on the starting and ending points of a motion, not on the path taken. \u2022 We can define potential energy (PE) for any conservative force, just as we defined PEg for the gravitational force. \u2022 The potential energy of a spring is PEs = 1 22, where is the spring\u2019s force constant and is the displacement from its undeformed position. \u2022 Mechanical energy is defined to be KE + PE for a conservative force. \u2022 When only conservative forces act on and within a system, the total mechanical energy is constant. In equation form, This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 7 | Work, Energy, and Energy Resources 303 KE + PE = constant or KEi + PEi = KEf + PEf where i and f denote initial and final values. This is known as the conservation of mechanical energy. 7.5 Nonconservative Forces \u2022 A nonconservative force is one for which work depends on the path. \u2022 Friction is an example of a nonconservative force that changes mechanical energy into thermal energy. \u2022 Work nc done by a nonconservative force changes the mechanical energy of a system. In equation form, nc = \u0394KE + \u0394PE or, equivalently, KEi + PEi + nc = KEf + PEf. \u2022 When both conservative and nonconservative forces act, energy conservation can be applied and used to calculate motion in terms of the known potential energies of the conservative forces and the work done by nonconservative forces, instead of finding the net work from the net force, or having to directly apply Newton\u2019s laws. 7.6 Conservation of Energy \u2022 The law of conservation of energy states that the total energy is constant in any process. Energy may change in form or be transferred from one system to another, but the total remains the same. \u2022 When all forms of energy are considered, conservation of energy is written in equation form as KEi + PEi + nc + OEi = KEf + PEf + OEf, where OE is all other forms of energy besides mechanical energy. \u2022 Commonly encountered forms of energy include electric energy, chemical energy, radiant energy, nuclear energy, and thermal energy. \u2022 Energy is often utilized to do work, but it is not possible to convert all the energy of a system to work. \u2022 The efficiency of a machine or human is defined to be = out in, where out is useful work output and in is the energy consumed. 7.", "7 Power \u2022 Power is the rate at which work is done, or in equation form, for the average power for work done over a time, = /. \u2022 The SI unit for power is the watt (W), where 1 W = 1 J/s. \u2022 The power of many devices such as electric motors is also often expressed in horsepower (hp), where 1 hp = 746 W. 7.8 Work, Energy, and Power in Humans \u2022 The human body converts energy stored in food into work, thermal energy, and/or chemical energy that is stored in fatty tissue. \u2022 The rate at which the body uses food energy to sustain life and to do different activities is called the metabolic rate, and the corresponding rate when at rest is called the basal metabolic rate (BMR) \u2022 The energy included in the basal metabolic rate is divided among various systems in the body, with the largest fraction going to the liver and spleen, and the brain coming next. \u2022 About 75% of food calories are used to sustain basic body functions included in the basal metabolic rate. \u2022 The energy consumption of people during various activities can be determined by measuring their oxygen use, because the digestive process is basically one of oxidizing food. 7.9 World Energy Use \u2022 The relative use of different fuels to provide energy has changed over the years, but fuel use is currently dominated by oil, although natural gas and solar contributions are increasing. \u2022 Although non-renewable sources dominate, some countries meet a sizeable percentage of their electricity needs from renewable resources. \u2022 The United States obtains only about 10% of its energy from renewable sources, mostly hydroelectric power. \u2022 Economic well-being is dependent upon energy use, and in most countries higher standards of living, as measured by GDP (Gross Domestic Product) per capita, are matched by higher levels of energy consumption per capita. \u2022 Even though, in accordance with the law of conservation of energy, energy can never be created or destroyed, energy that can be used to do work is always partly converted to less useful forms, such as waste heat to the environment, in all of our uses of energy for practical purposes. Conceptual Questions 7.1 Work: The Scientific Definition 1. Give an example of something we think of as work in everyday circumstances that is not work in the scientific sense. Is energy transferred or changed in form in your example? If so, explain how this is accomplished without doing work. 304 Chapter 7 | Work, Energy, and Energy Resources 2. Give an example of a situation in which there is a force and a displacement, but the force does no work. Explain why it does no work. 3. Describe a situation in which a force is exerted for a long time but does no work. Explain. 7.2 Kinetic Energy and the Work-Energy Theorem 4. The person in Figure 7.33 does work on the lawn mower. Under what conditions would the mower gain energy? Under what conditions would it lose energy? Figure 7.33 5. Work done on a system puts energy into it. Work done by a system removes energy from it. Give an example for each statement. 6. When solving for speed in Example 7.4, we kept only the positive root. Why? 7.3 Gravitational Potential Energy 7. In Example 7.7, we calculated the final speed of a roller coaster that descended 20 m in height and had an initial speed of 5 m/s downhill. Suppose the roller coaster had had an initial speed of 5 m/s uphill instead, and it coasted uphill, stopped, and then rolled back down to a final point 20 m below the start. We would find in that case that it had the same final speed. Explain in terms of conservation of energy. 8. Does the work you do on a book when you lift it onto a shelf depend on the path taken? On the time taken? On the height of the shelf? On the mass of the book? 7.4 Conservative Forces and Potential Energy 9. What is a conservative force? 10. The force exerted by a diving board is conservative, provided the internal friction is negligible. Assuming friction is negligible, describe changes in the potential energy of a diving board as a swimmer dives from it, starting just before the swimmer steps on the board until just after his feet leave it. 11. Define mechanical energy. What is the relationship of mechanical energy to nonconservative forces? What happens to mechanical energy if only conservative forces act? 12. What is the relationship of potential energy to conservative force? 7.6 Conservation of Energy 13. Consider the following scenario. A car for which friction is not negligible accelerates from rest down a hill, running out of gasoline after a short distance. The driver lets the car coast farther down the hill, then up and over a small crest. He then coasts down that hill into a gas station, where he brakes to a stop and fills the tank with gasoline. Identify the forms of energy the car has, and how they are changed and transferred", " in this series of events. (See Figure 7.34.) This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 7 | Work, Energy, and Energy Resources 305 Figure 7.34 A car experiencing non-negligible friction coasts down a hill, over a small crest, then downhill again, and comes to a stop at a gas station. 14. Describe the energy transfers and transformations for a javelin, starting from the point at which an athlete picks up the javelin and ending when the javelin is stuck into the ground after being thrown. 15. Do devices with efficiencies of less than one violate the law of conservation of energy? Explain. 16. List four different forms or types of energy. Give one example of a conversion from each of these forms to another form. 17. List the energy conversions that occur when riding a bicycle. 7.7 Power 18. Most electrical appliances are rated in watts. Does this rating depend on how long the appliance is on? (When off, it is a zerowatt device.) Explain in terms of the definition of power. 19. Explain, in terms of the definition of power, why energy consumption is sometimes listed in kilowatt-hours rather than joules. What is the relationship between these two energy units? 20. A spark of static electricity, such as that you might receive from a doorknob on a cold dry day, may carry a few hundred watts of power. Explain why you are not injured by such a spark. 7.8 Work, Energy, and Power in Humans 21. Explain why it is easier to climb a mountain on a zigzag path rather than one straight up the side. Is your increase in gravitational potential energy the same in both cases? Is your energy consumption the same in both? 22. Do you do work on the outside world when you rub your hands together to warm them? What is the efficiency of this activity? 23. Shivering is an involuntary response to lowered body temperature. What is the efficiency of the body when shivering, and is this a desirable value? 24. Discuss the relative effectiveness of dieting and exercise in losing weight, noting that most athletic activities consume food energy at a rate of 400 to 500 W, while a single cup of yogurt can contain 1360 kJ (325 kcal). Specifically, is it likely that exercise alone will be sufficient to lose weight? You may wish to consider that regular exercise may increase the metabolic rate, whereas protracted dieting may reduce it. 7.9 World Energy Use 25. What is the difference between energy conservation and the law of conservation of energy? Give some examples of each. 26. If the efficiency of a coal-fired electrical generating plant is 35%, then what do we mean when we say that energy is a conserved quantity? 306 Chapter 7 | Work, Energy, and Energy Resources Problems & Exercises 7.1 Work: The Scientific Definition 1. How much work does a supermarket checkout attendant do on a can of soup he pushes 0.600 m horizontally with a force of 5.00 N? Express your answer in joules and kilocalories. 2. A 75.0-kg person climbs stairs, gaining 2.50 meters in height. Find the work done to accomplish this task. 3. (a) Calculate the work done on a 1500-kg elevator car by its cable to lift it 40.0 m at constant speed, assuming friction averages 100 N. (b) What is the work done on the lift by the gravitational force in this process? (c) What is the total work done on the lift? 4. Suppose a car travels 108 km at a speed of 30.0 m/s, and uses 2.0 gal of gasoline. Only 30% of the gasoline goes into useful work by the force that keeps the car moving at constant speed despite friction. (See Table 7.1 for the energy content of gasoline.) (a) What is the magnitude of the force exerted to keep the car moving at constant speed? (b) If the required force is directly proportional to speed, how many gallons will be used to drive 108 km at a speed of 28.0 m/s? 5. Calculate the work done by an 85.0-kg man who pushes a crate 4.00 m up along a ramp that makes an angle of 20.0\u00ba with the horizontal. (See Figure 7.35.) He exerts a force of 500 N on the crate parallel to the ramp and moves at a constant speed. Be certain to include the work he does on the crate and on his body to get up the ramp. Figure 7.35 A man pushes a crate up a ramp. shopper exerts, using energy considerations. (e) What is the total work done on the cart? 8. Suppose the ski patrol lowers a rescue sled and victim, having a total mass of 90.0 kg, down a 60.0", "\u00ba slope at constant speed, as shown in Figure 7.37. The coefficient of friction between the sled and the snow is 0.100. (a) How much work is done by friction as the sled moves 30.0 m along the hill? (b) How much work is done by the rope on the sled in this distance? (c) What is the work done by the gravitational force on the sled? (d) What is the total work done? Figure 7.37 A rescue sled and victim are lowered down a steep slope. 7.2 Kinetic Energy and the Work-Energy Theorem 9. Compare the kinetic energy of a 20,000-kg truck moving at 110 km/h with that of an 80.0-kg astronaut in orbit moving at 27,500 km/h. 10. (a) How fast must a 3000-kg elephant move to have the same kinetic energy as a 65.0-kg sprinter running at 10.0 m/ s? (b) Discuss how the larger energies needed for the movement of larger animals would relate to metabolic rates. 6. How much work is done by the boy pulling his sister 30.0 m in a wagon as shown in Figure 7.36? Assume no friction acts on the wagon. 11. Confirm the value given for the kinetic energy of an aircraft carrier in Table 7.1. You will need to look up the definition of a nautical mile (1 knot = 1 nautical mile/h). 12. (a) Calculate the force needed to bring a 950-kg car to rest from a speed of 90.0 km/h in a distance of 120 m (a fairly typical distance for a non-panic stop). (b) Suppose instead the car hits a concrete abutment at full speed and is brought to a stop in 2.00 m. Calculate the force exerted on the car and compare it with the force found in part (a). 13. A car\u2019s bumper is designed to withstand a 4.0-km/h (1.1-m/s) collision with an immovable object without damage to the body of the car. The bumper cushions the shock by absorbing the force over a distance. Calculate the magnitude of the average force on a bumper that collapses 0.200 m while bringing a 900-kg car to rest from an initial speed of 1.1 m/s. 14. Boxing gloves are padded to lessen the force of a blow. (a) Calculate the force exerted by a boxing glove on an opponent\u2019s face, if the glove and face compress 7.50 cm during a blow in which the 7.00-kg arm and glove are brought to rest from an initial speed of 10.0 m/s. (b) Calculate the force exerted by an identical blow in the gory old days when no gloves were used and the knuckles and face would compress only 2.00 cm. (c) Discuss the magnitude of the Figure 7.36 The boy does work on the system of the wagon and the child when he pulls them as shown. 7. A shopper pushes a grocery cart 20.0 m at constant speed on level ground, against a 35.0 N frictional force. He pushes in a direction 25.0\u00ba below the horizontal. (a) What is the work done on the cart by friction? (b) What is the work done on the cart by the gravitational force? (c) What is the work done on the cart by the shopper? (d) Find the force the This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 7 | Work, Energy, and Energy Resources 307 force with glove on. Does it seem high enough to cause damage even though it is lower than the force with no glove? 15. Using energy considerations, calculate the average force a 60.0-kg sprinter exerts backward on the track to accelerate from 2.00 to 8.00 m/s in a distance of 25.0 m, if he encounters a headwind that exerts an average force of 30.0 N against him. 7.3 Gravitational Potential Energy 16. A hydroelectric power facility (see Figure 7.38) converts the gravitational potential energy of water behind a dam to electric energy. (a) What is the gravitational potential energy relative to the generators of a lake of volume 50.0 km3 mass = 5.00\u00d71013 kg), given that the lake has an average height of 40.0 m above the generators? (b) Compare this with the energy stored in a 9-megaton fusion bomb. ( Figure 7.38 Hydroelectric facility (credit: Denis Belevich, Wikimedia Commons) 17. (a) How much gravitational potential energy (relative to the ground on which it is built) is stored in the Great Pyramid of Cheops, given that its mass is about 7", " \u00d7 109 kg and its center of mass is 36.5 m above the surrounding ground? (b) How does this energy compare with the daily food intake of a person? 18. Suppose a 350-g kookaburra (a large kingfisher bird) picks up a 75-g snake and raises it 2.5 m from the ground to a branch. (a) How much work did the bird do on the snake? (b) How much work did it do to raise its own center of mass to the branch? 19. In Example 7.7, we found that the speed of a roller coaster that had descended 20.0 m was only slightly greater when it had an initial speed of 5.00 m/s than when it started from rest. This implies that \u0394PE >> KEi. Confirm this statement by taking the ratio of \u0394PE to KEi. (Note that mass cancels.) 20. A 100-g toy car is propelled by a compressed spring that starts it moving. The car follows the curved track in Figure 7.39. Show that the final speed of the toy car is 0.687 m/s if its initial speed is 2.00 m/s and it coasts up the frictionless slope, gaining 0.180 m in altitude. Figure 7.39 A toy car moves up a sloped track. (credit: Leszek Leszczynski, Flickr) 21. In a downhill ski race, surprisingly, little advantage is gained by getting a running start. (This is because the initial kinetic energy is small compared with the gain in gravitational potential energy on even small hills.) To demonstrate this, find the final speed and the time taken for a skier who skies 70.0 m along a 30\u00ba slope neglecting friction: (a) Starting from rest. (b) Starting with an initial speed of 2.50 m/s. (c) Does the answer surprise you? Discuss why it is still advantageous to get a running start in very competitive events. 7.4 Conservative Forces and Potential Energy 22. A 5.00\u00d7105-kg subway train is brought to a stop from a speed of 0.500 m/s in 0.400 m by a large spring bumper at the end of its track. What is the force constant of the spring? 23. A pogo stick has a spring with a force constant of 2.50104 N/m, which can be compressed 12.0 cm. To what maximum height can a child jump on the stick using only the energy in the spring, if the child and stick have a total mass of 40.0 kg? Explicitly show how you follow the steps in the Problem-Solving Strategies for Energy. 7.5 Nonconservative Forces 24. A 60.0-kg skier with an initial speed of 12.0 m/s coasts up a 2.50-m-high rise as shown in Figure 7.40. Find her final speed at the top, given that the coefficient of friction between her skis and the snow is 0.0800. (Hint: Find the distance traveled up the incline assuming a straight-line path as shown in the figure.) Figure 7.40 The skier\u2019s initial kinetic energy is partially used in coasting to the top of a rise. 25. (a) How high a hill can a car coast up (engine disengaged) if work done by friction is negligible and its initial speed is 110 km/h? (b) If, in actuality, a 750-kg car with an initial speed of 110 km/h is observed to coast up a hill to a height 22.0 m above its starting point, how much thermal energy was generated by friction? (c) What is the average force of friction if the hill has a slope 2.5\u00ba above the horizontal? 308 Chapter 7 | Work, Energy, and Energy Resources 7.6 Conservation of Energy 26. Using values from Table 7.1, how many DNA molecules could be broken by the energy carried by a single electron in the beam of an old-fashioned TV tube? (These electrons were not dangerous in themselves, but they did create dangerous x rays. Later model tube TVs had shielding that absorbed x rays before they escaped and exposed viewers.) 27. Using energy considerations and assuming negligible air resistance, show that a rock thrown from a bridge 20.0 m above water with an initial speed of 15.0 m/s strikes the water with a speed of 24.8 m/s independent of the direction thrown. 28. If the energy in fusion bombs were used to supply the energy needs of the world, how many of the 9-megaton variety would be needed for a year\u2019s supply of energy (using data from Table 7.1)? This is not as far-fetched as it may sound\u2014there are thousands of nuclear bombs, and their energy can be trapped in underground explosions and converted to electricity, as natural geothermal energy is. 29.", " (a) Use of hydrogen fusion to supply energy is a dream that may be realized in the next century. Fusion would be a relatively clean and almost limitless supply of energy, as can be seen from Table 7.1. To illustrate this, calculate how many years the present energy needs of the world could be supplied by one millionth of the oceans\u2019 hydrogen fusion energy. (b) How does this time compare with historically significant events, such as the duration of stable economic systems? 7.7 Power 30. The Crab Nebula (see Figure 7.41) pulsar is the remnant of a supernova that occurred in A.D. 1054. Using data from Table 7.3, calculate the approximate factor by which the power output of this astronomical object has declined since its explosion. Figure 7.41 Crab Nebula (credit: ESO, via Wikimedia Commons) 31. Suppose a star 1000 times brighter than our Sun (that is, emitting 1000 times the power) suddenly goes supernova. Using data from Table 7.3: (a) By what factor does its power output increase? (b) How many times brighter than our entire Milky Way galaxy is the supernova? (c) Based on your answers, discuss whether it should be possible to observe supernovas in distant galaxies. Note that there are on the This content is available for free at http://cnx.org/content/col11844/1.13 order of 1011 observable galaxies, the average brightness of which is somewhat less than our own galaxy. 32. A person in good physical condition can put out 100 W of useful power for several hours at a stretch, perhaps by pedaling a mechanism that drives an electric generator. Neglecting any problems of generator efficiency and practical considerations such as resting time: (a) How many people would it take to run a 4.00-kW electric clothes dryer? (b) How many people would it take to replace a large electric power plant that generates 800 MW? 33. What is the cost of operating a 3.00-W electric clock for a year if the cost of electricity is $0.0900 per kW \u22c5 h? 34. A large household air conditioner may consume 15.0 kW of power. What is the cost of operating this air conditioner 3.00 h per day for 30.0 d if the cost of electricity is $0.110 per kW \u22c5 h? 35. (a) What is the average power consumption in watts of an appliance that uses 5.00 kW \u22c5 h of energy per day? (b) How many joules of energy does this appliance consume in a year? 36. (a) What is the average useful power output of a person who does 6.00106 J of useful work in 8.00 h? (b) Working at this rate, how long will it take this person to lift 2000 kg of bricks 1.50 m to a platform? (Work done to lift his body can be omitted because it is not considered useful output here.) 37. A 500-kg dragster accelerates from rest to a final speed of 110 m/s in 400 m (about a quarter of a mile) and encounters an average frictional force of 1200 N. What is its average power output in watts and horsepower if this takes 7.30 s? 38. (a) How long will it take an 850-kg car with a useful power output of 40.0 hp (1 hp = 746 W) to reach a speed of 15.0 m/ s, neglecting friction? (b) How long will this acceleration take if the car also climbs a 3.00-m-high hill in the process? 39. (a) Find the useful power output of an elevator motor that lifts a 2500-kg load a height of 35.0 m in 12.0 s, if it also increases the speed from rest to 4.00 m/s. Note that the total mass of the counterbalanced system is 10,000 kg\u2014so that only 2500 kg is raised in height, but the full 10,000 kg is accelerated. (b) What does it cost, if electricity is $0.0900 per kW \u22c5 h? 40. (a) What is the available energy content, in joules, of a battery that operates a 2.00-W electric clock for 18 months? (b) How long can a battery that can supply 8.00104 J run a pocket calculator that consumes energy at the rate of 1.0010\u22123 W? 41. (a) How long would it take a 1.50105 engines that produce 100 MW of power to reach a speed of 250 m/s and an altitude of 12.0 km if air resistance were negligible? (b) If it actually takes 900 s, what is the power? (c) Given this power, what is the average force of air resistance if the airplane takes 1200 s", "? (Hint: You must find the distance the plane travels in 1200 s assuming constant acceleration.) -kg airplane with 42. Calculate the power output needed for a 950-kg car to climb a 2.00\u00ba slope at a constant 30.0 m/s while encountering wind resistance and friction totaling 600 N. Explicitly show how you follow the steps in the ProblemSolving Strategies for Energy. 43. (a) Calculate the power per square meter reaching Earth\u2019s upper atmosphere from the Sun. (Take the power output of Chapter 7 | Work, Energy, and Energy Resources 309 the Sun to be 4.00\u00d71026 W.) (b) Part of this is absorbed and reflected by the atmosphere, so that a maximum of 1.30 kW/m2 reaches Earth\u2019s surface. Calculate the area in km2 of solar energy collectors needed to replace an electric power plant that generates 750 MW if the collectors convert an average of 2.00% of the maximum power into electricity. (This small conversion efficiency is due to the devices themselves, and the fact that the sun is directly overhead only briefly.) With the same assumptions, what area would be needed to meet the United States\u2019 energy needs (1.05\u00d71020 J)? Australia\u2019s energy needs (5.4\u00d71018 J)? China\u2019s energy needs (6.3\u00d71019 J)? (These energy consumption values are from 2006.) 7.8 Work, Energy, and Power in Humans 44. (a) How long can you rapidly climb stairs (116/min) on the 93.0 kcal of energy in a 10.0-g pat of butter? (b) How many flights is this if each flight has 16 stairs? 45. (a) What is the power output in watts and horsepower of a 70.0-kg sprinter who accelerates from rest to 10.0 m/s in 3.00 s? (b) Considering the amount of power generated, do you think a well-trained athlete could do this repetitively for long periods of time? 46. Calculate the power output in watts and horsepower of a shot-putter who takes 1.20 s to accelerate the 7.27-kg shot from rest to 14.0 m/s, while raising it 0.800 m. (Do not include the power produced to accelerate his body.) 3.00 h, and studies for 6.00 h. (Studying consumes energy at the same rate as sitting in class.) 50. What is the efficiency of a subject on a treadmill who puts out work at the rate of 100 W while consuming oxygen at the rate of 2.00 L/min? (Hint: See Table 7.5.) 51. Shoveling snow can be extremely taxing because the arms have such a low efficiency in this activity. Suppose a person shoveling a footpath metabolizes food at the rate of 800 W. (a) What is her useful power output? (b) How long will it take her to lift 3000 kg of snow 1.20 m? (This could be the amount of heavy snow on 20 m of footpath.) (c) How much waste heat transfer in kilojoules will she generate in the process? 52. Very large forces are produced in joints when a person jumps from some height to the ground. (a) Calculate the magnitude of the force produced if an 80.0-kg person jumps from a 0.600\u2013m-high ledge and lands stiffly, compressing joint material 1.50 cm as a result. (Be certain to include the weight of the person.) (b) In practice the knees bend almost involuntarily to help extend the distance over which you stop. Calculate the magnitude of the force produced if the stopping distance is 0.300 m. (c) Compare both forces with the weight of the person. 53. Jogging on hard surfaces with insufficiently padded shoes produces large forces in the feet and legs. (a) Calculate the magnitude of the force needed to stop the downward motion of a jogger\u2019s leg, if his leg has a mass of 13.0 kg, a speed of 6.00 m/s, and stops in a distance of 1.50 cm. (Be certain to include the weight of the 75.0-kg jogger\u2019s body.) (b) Compare this force with the weight of the jogger. 54. (a) Calculate the energy in kJ used by a 55.0-kg woman who does 50 deep knee bends in which her center of mass is lowered and raised 0.400 m. (She does work in both directions.) You may assume her efficiency is 20%. (b) What is the average power consumption rate in watts if she does this in 3.00 min? 55. Kanellos Kanellopoulos flew 119 km from Crete to Santorini, Greece, on April 23, 1988, in", " the Daedalus 88, an aircraft powered by a bicycle-type drive mechanism (see Figure 7.43). His useful power output for the 234-min trip was about 350 W. Using the efficiency for cycling from Table 7.2, calculate the food energy in kilojoules he metabolized during the flight. Figure 7.42 Shot putter at the Dornoch Highland Gathering in 2007. (credit: John Haslam, Flickr) 47. (a) What is the efficiency of an out-of-condition professor who does 2.10105 J of useful work while metabolizing 500 kcal of food energy? (b) How many food calories would a well-conditioned athlete metabolize in doing the same work with an efficiency of 20%? 48. Energy that is not utilized for work or heat transfer is converted to the chemical energy of body fat containing about 39 kJ/g. How many grams of fat will you gain if you eat 10,000 kJ (about 2500 kcal) one day and do nothing but sit relaxed for 16.0 h and sleep for the other 8.00 h? Use data from Table 7.5 for the energy consumption rates of these activities. Figure 7.43 The Daedalus 88 in flight. (credit: NASA photo by Beasley) 49. Using data from Table 7.5, calculate the daily energy needs of a person who sleeps for 7.00 h, walks for 2.00 h, attends classes for 4.00 h, cycles for 2.00 h, sits relaxed for 56. The swimmer shown in Figure 7.44 exerts an average horizontal backward force of 80.0 N with his arm during each 1.80 m long stroke. (a) What is his work output in each 310 Chapter 7 | Work, Energy, and Energy Resources stroke? (b) Calculate the power output of his arms if he does 120 strokes per minute. long time? Discuss why exercise is necessary but may not be sufficient to cause a person to lose weight. Figure 7.44 57. Mountain climbers carry bottled oxygen when at very high altitudes. (a) Assuming that a mountain climber uses oxygen at twice the rate for climbing 116 stairs per minute (because of low air temperature and winds), calculate how many liters of oxygen a climber would need for 10.0 h of climbing. (These are liters at sea level.) Note that only 40% of the inhaled oxygen is utilized; the rest is exhaled. (b) How much useful work does the climber do if he and his equipment have a mass of 90.0 kg and he gains 1000 m of altitude? (c) What is his efficiency for the 10.0-h climb? 58. The awe-inspiring Great Pyramid of Cheops was built more than 4500 years ago. Its square base, originally 230 m on a side, covered 13.1 acres, and it was 146 m high, with a mass of about 7\u00d7109 kg. (The pyramid\u2019s dimensions are slightly different today due to quarrying and some sagging.) Historians estimate that 20,000 workers spent 20 years to construct it, working 12-hour days, 330 days per year. (a) Calculate the gravitational potential energy stored in the pyramid, given its center of mass is at one-fourth its height. (b) Only a fraction of the workers lifted blocks; most were involved in support services such as building ramps (see Figure 7.45), bringing food and water, and hauling blocks to the site. Calculate the efficiency of the workers who did the lifting, assuming there were 1000 of them and they consumed food energy at the rate of 300 kcal/h. What does your answer imply about how much of their work went into block-lifting, versus how much work went into friction and lifting and lowering their own bodies? (c) Calculate the mass of food that had to be supplied each day, assuming that the average worker required 3600 kcal per day and that their diet was 5% protein, 60% carbohydrate, and 35% fat. (These proportions neglect the mass of bulk and nondigestible materials consumed.) 7.9 World Energy Use 60. Integrated Concepts (a) Calculate the force the woman in Figure 7.46 exerts to do a push-up at constant speed, taking all data to be known to three digits. (b) How much work does she do if her center of mass rises 0.240 m? (c) What is her useful power output if she does 25 push-ups in 1 min? (Should work done lowering her body be included? See the discussion of useful work in Work, Energy, and Power in Humans. Figure 7.46 Forces involved in doing push-ups. The woman\u2019s weight acts as a force exerted downward on her center of gravity (CG). 61. Integrated Concepts A 75.0-kg cross-country skier is climbing a 3.0\u00ba slope at", " a constant speed of 2.00 m/s and encounters air resistance of 25.0 N. Find his power output for work done against the gravitational force and air resistance. (b) What average force does he exert backward on the snow to accomplish this? (c) If he continues to exert this force and to experience the same air resistance when he reaches a level area, how long will it take him to reach a velocity of 10.0 m/s? 62. Integrated Concepts The 70.0-kg swimmer in Figure 7.44 starts a race with an initial velocity of 1.25 m/s and exerts an average force of 80.0 N backward with his arms during each 1.80 m long stroke. (a) What is his initial acceleration if water resistance is 45.0 N? (b) What is the subsequent average resistance force from the water during the 5.00 s it takes him to reach his top velocity of 2.50 m/s? (c) Discuss whether water resistance seems to increase linearly with velocity. 63. Integrated Concepts A toy gun uses a spring with a force constant of 300 N/m to propel a 10.0-g steel ball. If the spring is compressed 7.00 cm and friction is negligible: (a) How much force is needed to compress the spring? (b) To what maximum height can the ball be shot? (c) At what angles above the horizontal may a child aim to hit a target 3.00 m away at the same height as the gun? (d) What is the gun\u2019s maximum range on level ground? 64. Integrated Concepts (a) What force must be supplied by an elevator cable to produce an acceleration of 0.800 m/s2 against a 200-N frictional force, if the mass of the loaded elevator is 1500 kg? (b) How much work is done by the cable in lifting the elevator 20.0 m? (c) What is the final speed of the elevator if it starts from rest? (d) How much work went into thermal energy? Figure 7.45 Ancient pyramids were probably constructed using ramps as simple machines. (credit: Franck Monnier, Wikimedia Commons) 59. (a) How long can you play tennis on the 800 kJ (about 200 kcal) of energy in a candy bar? (b) Does this seem like a 65. Unreasonable Results A car advertisement claims that its 900-kg car accelerated from rest to 30.0 m/s and drove 100 km, gaining 3.00 km in altitude, on 1.0 gal of gasoline. The average force of friction This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 7 | Work, Energy, and Energy Resources 311 including air resistance was 700 N. Assume all values are known to three significant figures. (a) Calculate the car\u2019s efficiency. (b) What is unreasonable about the result? (c) Which premise is unreasonable, or which premises are inconsistent? 66. Unreasonable Results Body fat is metabolized, supplying 9.30 kcal/g, when dietary intake is less than needed to fuel metabolism. The manufacturers of an exercise bicycle claim that you can lose 0.500 kg of fat per day by vigorously exercising for 2.00 h per day on their machine. (a) How many kcal are supplied by the metabolization of 0.500 kg of fat? (b) Calculate the kcal/min that you would have to utilize to metabolize fat at the rate of 0.500 kg in 2.00 h. (c) What is unreasonable about the results? (d) Which premise is unreasonable, or which premises are inconsistent? 67. Construct Your Own Problem Consider a person climbing and descending stairs. Construct a problem in which you calculate the long-term rate at which stairs can be climbed considering the mass of the person, his ability to generate power with his legs, and the height of a single stair step. Also consider why the same person can descend stairs at a faster rate for a nearly unlimited time in spite of the fact that very similar forces are exerted going down as going up. (This points to a fundamentally different process for descending versus climbing stairs.) 68. Construct Your Own Problem Consider humans generating electricity by pedaling a device similar to a stationary bicycle. Construct a problem in which you determine the number of people it would take to replace a large electrical generation facility. Among the things to consider are the power output that is reasonable using the legs, rest time, and the need for electricity 24 hours per day. Discuss the practical implications of your results. 69. Integrated Concepts A 105-kg basketball player crouches down 0.400 m while waiting to jump. After exerting a force on the floor through this 0.400 m, his feet leave the floor and his center of gravity rises 0.950 m above its normal standing erect position. (a) Using energy considerations", ", calculate his velocity when he leaves the floor. (b) What average force did he exert on the floor? (Do not neglect the force to support his weight as well as that to accelerate him.) (c) What was his power output during the acceleration phase? Test Prep for AP\u00ae Courses c. 450 N d. 600 N 7.1 Work: The Scientific Definition 1. Given Table 7.7 about how much force does the rocket engine exert on the 3.0-kg payload? Table 7.7 Distance traveled with rocket engine firing (m) Payload final velocity (m/s) 500 490 1020 505 a. 150 N b. 300 N 310 300 450 312 2. You have a cart track, a cart, several masses, and a position-sensing pulley. Design an experiment to examine how the force exerted on the cart does work as it moves through a distance. 3. Look at Figure 7.10(c). You compress a spring by x, and then release it. Next you compress the spring by 2x. How much more work did you do the second time than the first? a. Half as much b. The same c. Twice as much d. Four times as much 4. You have a cart track, two carts, several masses, a position-sensing pulley, and a piece of carpet (a rough surface) that will fit over the track. Design an experiment to examine how the force exerted on the cart does work as the cart moves through a distance. 312 Chapter 7 | Work, Energy, and Energy Resources 5. A crane is lifting construction materials from the ground to an elevation of 60 m. Over the first 10 m, the motor linearly increases the force it exerts from 0 to 10 kN. It exerts that constant force for the next 40 m, and then winds down to 0 N again over the last 10 m, as shown in the figure. What is the total work done on the construction materials? Figure 7.47 a. 500 kJ b. 600 kJ c. 300 kJ d. 18 MJ 7.2 Kinetic Energy and the Work-Energy Theorem 6. A toy car is going around a loop-the-loop. Gravity ____ the kinetic energy on the upward side of the loop, ____ the kinetic energy at the top, and ____ the kinetic energy on the downward side of the loop. a. increases, decreases, has no effect on b. decreases, has no effect on, increases c. increases, has no effect on, decreases d. decreases, increases, has no effect on 7. A roller coaster is set up with a track in the form of a perfect cosine. Describe and graph what happens to the kinetic energy of a cart as it goes through the first full period of the track. 8. If wind is blowing horizontally toward a car with an angle of 30 degrees from the direction of travel, the kinetic energy will ____. If the wind is blowing at a car at 135 degrees from the direction of travel, the kinetic energy will ____. a. increase, increase increase, decrease b. c. decrease, increase d. decrease, decrease 9. In what direction relative to the direction of travel can a force act on a car (traveling on level ground), and not change the kinetic energy? Can you give examples of such forces? 10. A 2000-kg airplane is coming in for a landing, with a velocity 5 degrees below the horizontal and a drag force of 40 kN acting directly rearward. Kinetic energy will ____ due to the net force of ____. a. increase, 20 kN b. decrease, 40 kN c. increase, 45 kN d. decrease, 45 kN 11. You are participating in the Iditarod, and your sled dogs are pulling you across a frozen lake with a force of 1200 N while a 300 N wind is blowing at you at 135 degrees from your direction of travel. What is the net force, and will your kinetic energy increase or decrease? 12. A model drag car is being accelerated along its track from rest by a motor with a force of 75 N, but there is a drag force of 30 N due to the track. What is the kinetic energy after 2 m of travel? a. 90 J This content is available for free at http://cnx.org/content/col11844/1.13 b. 150 J c. 210 J d. 60 J 13. You are launching a 2-kg potato out of a potato cannon. The cannon is 1.5 m long and is aimed 30 degrees above the horizontal. It exerts a 50 N force on the potato. What is the kinetic energy of the potato as it leaves the muzzle of the potato cannon? 14. When the force acting on an object is parallel to the direction of the motion of the center of mass, the mechanical energy ____. When the force acting on an object is antiparallel to", " the direction of the center of mass, the mechanical energy ____. increases, increases a. b. increases, decreases c. decreases, increases d. decreases, decreases 15. Describe a system in which the main forces acting are parallel or antiparallel to the center of mass, and justify your answer. 16. A child is pulling two red wagons, with the second one tied to the first by a (non-stretching) rope. Each wagon has a mass of 10 kg. If the child exerts a force of 30 N for 5.0 m, how much has the kinetic energy of the two-wagon system changed? a. 300 J b. 150 J c. 75 J d. 60 J 17. A child has two red wagons, with the rear one tied to the front by a (non-stretching) rope. If the child pushes on the rear wagon, what happens to the kinetic energy of each of the wagons, and the two-wagon system? 18. Draw a graph of the force parallel to displacement exerted on a stunt motorcycle going through a loop-the-loop versus the distance traveled around the loop. Explain the net change in energy. 7.3 Gravitational Potential Energy 19. A 1.0 kg baseball is flying at 10 m/s. How much kinetic energy does it have? Potential energy? a. 10 J, 20 J b. 50 J, 20 J c. unknown, 50 J d. 50 J, unknown 20. A 2.0-kg potato has been launched out of a potato cannon at 9.0 m/s. What is the kinetic energy? If you then learn that it is 4.0 m above the ground, what is the total mechanical energy relative to the ground? a. 78 J, 3 J b. 160 J, 81 J c. 81 J, 160 J d. 81 J, 3 J 21. You have a 120-g yo-yo that you are swinging at 0.9 m/s. How much energy does it have? How high can it get above the lowest point of the swing without your doing any additional work, on Earth? How high could it get on the Moon, where gravity is 1/6 Earth\u2019s? 7.4 Conservative Forces and Potential Energy 22. Two 4.0 kg masses are connected to each other by a spring with a force constant of 25 N/m and a rest length of 1.0 m. If the spring has been compressed to 0.80 m in length and Chapter 7 | Work, Energy, and Energy Resources 313 the masses are traveling toward each other at 0.50 m/s (each), what is the total energy in the system? a. 1.0 J b. 1.5 J c. 9.0 J d. 8.0 J 23. A spring with a force constant of 5000 N/m and a rest length of 3.0 m is used in a catapult. When compressed to 1.0 m, it is used to launch a 50 kg rock. However, there is an error in the release mechanism, so the rock gets launched almost straight up. How high does it go, and how fast is it going when it hits the ground? 24. What information do you need to calculate the kinetic energy and potential energy of a spring? Potential energy due to gravity? How many objects do you need information about for each of these cases? 25. You are loading a toy dart gun, which has two settings, the more powerful with the spring compressed twice as far as the lower setting. If it takes 5.0 J of work to compress the dart gun to the lower setting, how much work does it take for the higher setting? a. 20 J b. 10 J c. 2.5 J d. 40 J 26. Describe a system you use daily with internal potential energy. 27. Old-fashioned pendulum clocks are powered by masses that need to be wound back to the top of the clock about once a week to counteract energy lost due to friction and to the chimes. One particular clock has three masses: 4.0 kg, 4.0 kg, and 6.0 kg. They can drop 1.3 meters. How much energy does the clock use in a week? a. 51 J b. 76 J c. 127 J d. 178 J 28. A water tower stores not only water, but (at least part of) the energy to move the water. How much? Make reasonable estimates for how much water is in the tower, and other quantities you need. 29. Old-fashioned pocket watches needed to be wound daily so they wouldn\u2019t run down and lose time, due to the friction in the internal components. This required a large number of turns of the winding key, but not much force per turn, and it was possible to overwind and break the watch. How was the energy stored? a. A small mass raised a long distance b. A large mass", " raised a short distance c. A weak spring deformed a long way d. A strong spring deformed a short way 30. Some of the very first clocks invented in China were powered by water. Describe how you think this was done. 7.5 Nonconservative Forces 31. You are in a room in a basement with a smooth concrete floor (friction force equals 40 N) and a nice rug (friction force equals 55 N) that is 3 m by 4 m. However, you have to push a very heavy box from one corner of the rug to the opposite corner of the rug. Will you do more work against friction going around the floor or across the rug, and how much extra? a. Across the rug is 275 J extra b. Around the floor is 5 J extra c. Across the rug is 5 J extra d. Around the floor is 280 J extra 32. In the Appalachians, along the interstate, there are ramps of loose gravel for semis that have had their brakes fail to drive into to stop. Design an experiment to measure how effective this would be. 7.6 Conservation of Energy 33. You do 30 J of work to load a toy dart gun. However, the dart is 10 cm long and feels a frictional force of 10 N while going through the dart gun\u2019s barrel. What is the kinetic energy of the fired dart? a. 30 J b. 29 J c. 28 J d. 27 J 34. When an object is lifted by a crane, it begins and ends its motion at rest. The same is true of an object pushed across a rough surface. Explain why this happens. What are the differences between these systems? 35. A child has two red wagons, with the rear one tied to the front by a stretchy rope (a spring). If the child pulls on the front wagon, the ____ increases. a. kinetic energy of the wagons b. potential energy stored in the spring c. both A and B d. not enough information 36. A child has two red wagons, with the rear one tied to the front by a stretchy rope (a spring). If the child pulls on the front wagon, the energy stored in the system increases. How do the relative amounts of potential and kinetic energy in this system change over time? 37. Which of the following are closed systems? a. Earth b. a car c. a frictionless pendulum d. a mass on a spring in a vacuum 38. Describe a real-world example of a closed system. 39. A 5.0-kg rock falls off of a 10 m cliff. If air resistance exerts a force of 10 N, what is the kinetic energy when the rock hits the ground? a. 400 J b. 12.6 m/s c. 100 J d. 500 J 40. Hydroelectricity is generated by storing water behind a dam, and then letting some of it run through generators in the dam to turn them. If the system is the water, what is the environment that is doing work on it? If a dam has water 100 m deep behind it, how much energy was generated if 10,000 kg of water exited the dam at 2.0 m/s? 41. Before railroads were invented, goods often traveled along canals, with mules pulling barges from the bank. If a mule is exerting a 1200 N force for 10 km, and the rope connecting the mule to the barge is at a 20 degree angle from the direction of travel, how much work did the mule do on the barge? a. 12 MJ b. 11 MJ c. 4.1 MJ d. 6 MJ 42. Describe an instance today in which you did work, by the scientific definition. Then calculate how much work you did in that instance, showing your work. 314 Chapter 7 | Work, Energy, and Energy Resources This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 8 | Linear Momentum and Collisions 315 8 LINEAR MOMENTUM AND COLLISIONS Figure 8.1 Each rugby player has great momentum, which will affect the outcome of their collisions with each other and the ground. (credit: ozzzie, Flickr) Chapter Outline 8.1. Linear Momentum and Force 8.2. Impulse 8.3. Conservation of Momentum 8.4. Elastic Collisions in One Dimension 8.5. Inelastic Collisions in One Dimension 8.6. Collisions of Point Masses in Two Dimensions 8.7. Introduction to Rocket Propulsion Connection for AP\u00ae courses In this chapter, you will learn about the concept of momentum and the relationship between momentum and force (both vector quantities) applied over a time interval. Have you ever considered why a glass dropped on a tile floor will often break, but a glass dropped on carpet will often remain intact? Both involve changes in momentum, but the actual collision with", " the floor is different in each case, just as an automobile collision without the benefit of an airbag can have a significantly different outcome than one with an airbag. You will learn that the interaction of objects (like a glass and the floor or two automobiles) results in forces, which in turn result in changes in the momentum of each object. At the same time, you will see how the law of momentum conservation can be applied to a system to help determine the outcome of a collision. The content in this chapter supports: Big Idea 3 The interactions of an object with other objects can be described by forces. Enduring Understanding 3.D A force exerted on an object can change the momentum of the object. Essential Knowledge 3.D.2 The change in momentum of an object occurs over a time interval. Big Idea 4: Interactions between systems can result in changes in those systems. Enduring Understanding 4.B Interactions with other objects or systems can change the total linear momentum of a system. Essential Knowledge 4.B.1 The change in linear momentum for a constant-mass system is the product of the mass of the system and the change in velocity of the center of mass. Big Idea 5 Changes that occur as a result of interactions are constrained by conservation laws. Enduring Understanding 5.A Certain quantities are conserved, in the sense that the changes of those quantities in a given system are always equal to the transfer of that quantity to or from the system by all possible interactions with other systems. 316 Chapter 8 | Linear Momentum and Collisions Essential Knowledge 5.A.2 For all systems under all circumstances, energy, charge, linear momentum, and angular momentum are conserved. Essential Knowledge 5.D.1 In a collision between objects, linear momentum is conserved. In an elastic collision, kinetic energy is the same before and after. Essential Knowledge 5.D.2 In a collision between objects, linear momentum is conserved. In an inelastic collision, kinetic energy is not the same before and after the collision. 8.1 Linear Momentum and Force Learning Objectives By the end of this section, you will be able to: \u2022 Define linear momentum. \u2022 Explain the relationship between linear momentum and force. \u2022 State Newton\u2019s second law of motion in terms of linear momentum. \u2022 Calculate linear momentum given mass and velocity. The information presented in this section supports the following AP\u00ae learning objectives and science practices: \u2022 3.D.1.1 The student is able to justify the selection of data needed to determine the relationship between the direction of the force acting on an object and the change in momentum caused by that force. (S.P. 4.1) Linear Momentum The scientific definition of linear momentum is consistent with most people\u2019s intuitive understanding of momentum: a large, fastmoving object has greater momentum than a smaller, slower object. Linear momentum is defined as the product of a system\u2019s mass multiplied by its velocity. In symbols, linear momentum is expressed as p = v. (8.1) Momentum is directly proportional to the object\u2019s mass and also its velocity. Thus the greater an object\u2019s mass or the greater its velocity, the greater its momentum. Momentum p is a vector having the same direction as the velocity v. The SI unit for momentum is kg \u00b7 m/s. Linear Momentum Linear momentum is defined as the product of a system\u2019s mass multiplied by its velocity: p = v. (8.2) Example 8.1 Calculating Momentum: A Football Player and a Football (a) Calculate the momentum of a 110-kg football player running at 8.00 m/s. (b) Compare the player\u2019s momentum with the momentum of a hard-thrown 0.410-kg football that has a speed of 25.0 m/s. Strategy No information is given regarding direction, and so we can calculate only the magnitude of the momentum,. (As usual, a symbol that is in italics is a magnitude, whereas one that is italicized, boldfaced, and has an arrow is a vector.) In both parts of this example, the magnitude of momentum can be calculated directly from the definition of momentum given in the equation, which becomes = (8.3) when only magnitudes are considered. Solution for (a) To determine the momentum of the player, substitute the known values for the player\u2019s mass and speed into the equation. player = 110 kg (8.00 m/s) = 880 kg \u00b7 m/s Solution for (b) To determine the momentum of the ball, substitute the known values for the ball\u2019s mass and speed into the equation. ball = 0.410 kg (25.0 m/s) = 10.3 kg \u00b7 m/s (8.4) (8.5) The ratio of the player\u2019s momentum to that of the ball is This content is available for free at http://", "cnx.org/content/col11844/1.13 Chapter 8 | Linear Momentum and Collisions player ball = 880 10.3 = 85.9. 317 (8.6) Discussion Although the ball has greater velocity, the player has a much greater mass. Thus the momentum of the player is much greater than the momentum of the football, as you might guess. As a result, the player\u2019s motion is only slightly affected if he catches the ball. We shall quantify what happens in such collisions in terms of momentum in later sections. Momentum and Newton\u2019s Second Law The importance of momentum, unlike the importance of energy, was recognized early in the development of classical physics. Momentum was deemed so important that it was called the \u201cquantity of motion.\u201d Newton actually stated his second law of motion in terms of momentum: The net external force equals the change in momentum of a system divided by the time over which it changes. Using symbols, this law is Fnet = \u0394p \u0394 where Fnet is the net external force, \u0394p is the change in momentum, and \u0394 is the change in time. Newton\u2019s Second Law of Motion in Terms of Momentum The net external force equals the change in momentum of a system divided by the time over which it changes. Fnet = \u0394p \u0394 (8.7) (8.8) Making Connections: Force and Momentum Force and momentum are intimately related. Force acting over time can change momentum, and Newton\u2019s second law of motion, can be stated in its most broadly applicable form in terms of momentum. Momentum continues to be a key concept in the study of atomic and subatomic particles in quantum mechanics. This statement of Newton\u2019s second law of motion includes the more familiar Fnet =a as a special case. We can derive this form as follows. First, note that the change in momentum \u0394p is given by If the mass of the system is constant, then \u0394p = \u0394 v. \u0394(v) = \u0394v. So that for constant mass, Newton\u2019s second law of motion becomes Because \u0394v \u0394 = a, we get the familiar equation when the mass of the system is constant. Fnet = \u0394p \u0394 = \u0394v \u0394. Fnet =a (8.9) (8.10) (8.11) (8.12) Newton\u2019s second law of motion stated in terms of momentum is more generally applicable because it can be applied to systems where the mass is changing, such as rockets, as well as to systems of constant mass. We will consider systems with varying mass in some detail; however, the relationship between momentum and force remains useful when mass is constant, such as in the following example. Example 8.2 Calculating Force: Venus Williams\u2019 Racquet During the 2007 French Open, Venus Williams hit the fastest recorded serve in a premier women\u2019s match, reaching a speed of 58 m/s (209 km/h). What is the average force exerted on the 0.057-kg tennis ball by Venus Williams\u2019 racquet, assuming that the ball\u2019s speed just after impact is 58 m/s, that the initial horizontal component of the velocity before impact is negligible, and that the ball remained in contact with the racquet for 5.0 ms (milliseconds)? Strategy 318 Chapter 8 | Linear Momentum and Collisions This problem involves only one dimension because the ball starts from having no horizontal velocity component before impact. Newton\u2019s second law stated in terms of momentum is then written as As noted above, when mass is constant, the change in momentum is given by \u0394 = \u0394 = (f \u2212 i). Fnet = \u0394p \u0394. In this example, the velocity just after impact and the change in time are given; thus, once \u0394 is calculated, net = can be used to find the force. Solution To determine the change in momentum, substitute the values for the initial and final velocities into the equation above. \u0394 = (f \u2013 i) 0.057 kg = = 3.306 kg \u00b7 m/s \u2248 3.3 kg \u00b7 m/s (58 m/s \u2013 0 m/s) Now the magnitude of the net external force can determined by using net = \u0394 \u0394 : net = \u0394 \u0394 = 3.306 kg \u22c5 m/s 5.0\u00d710\u22123 s = 661 N \u2248 660 N, (8.13) (8.14) \u0394 \u0394 (8.15) (8.16) where we have retained only two significant figures in the final step. Discussion This quantity was the average force exerted by Venus Williams\u2019 racquet on the tennis ball during its brief impact (note that the ball also experienced the 0.56-N force of gravity, but that force was not due to the racquet). This problem could also be solved by first finding the acceleration and then using net =,", " but one additional step would be required compared with the strategy used in this example. Making Connections: Illustrative Example Figure 8.2 A puck has an elastic, glancing collision with the edge of an air hockey table. In Figure 8.2, a puck is shown colliding with the edge of an air hockey table at a glancing angle. During the collision, the edge of the table exerts a force F on the puck, and the velocity of the puck changes as a result of the collision. The change in momentum is found by the equation: This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 8 | Linear Momentum and Collisions \u0394p = \u0394v = v' - v = (v' + ( - v)) 319 (8.17) As shown, the direction of the change in velocity is the same as the direction of the change in momentum, which in turn is in the same direction as the force exerted by the edge of the table. Note that there is only a horizontal change in velocity. There is no difference in the vertical components of the initial and final velocity vectors; therefore, there is no vertical component to the change in velocity vector or the change in momentum vector. This is consistent with the fact that the force exerted by the edge of the table is purely in the horizontal direction. 8.2 Impulse Learning Objectives By the end of this section, you will be able to: \u2022 Define impulse. \u2022 Describe effects of impulses in everyday life. \u2022 Determine the average effective force using graphical representation. \u2022 Calculate average force and impulse given mass, velocity, and time. The information presented in this section supports the following AP\u00ae learning objectives and science practices: \u2022 3.D.2.1 The student is able to justify the selection of routines for the calculation of the relationships between changes in momentum of an object, average force, impulse, and time of interaction. (S.P. 2.1) \u2022 3.D.2.2 The student is able to predict the change in momentum of an object from the average force exerted on the object and the interval of time during which the force is exerted. (S.P. 6.4) \u2022 3.D.2.3 The student is able to analyze data to characterize the change in momentum of an object from the average force exerted on the object and the interval of time during which the force is exerted. (S.P. 5.1) \u2022 3.D.2.4 The student is able to design a plan for collecting data to investigate the relationship between changes in momentum and the average force exerted on an object over time. (S.P. 4.1) \u2022 4.B.2.1 The student is able to apply mathematical routines to calculate the change in momentum of a system by analyzing the average force exerted over a certain time on the system. (S.P. 2.2) \u2022 4.B.2.2 The student is able to perform analysis on data presented as a force-time graph and predict the change in momentum of a system. (S.P. 5.1) The effect of a force on an object depends on how long it acts, as well as how great the force is. In Example 8.1, a very large force acting for a short time had a great effect on the momentum of the tennis ball. A small force could cause the same change in momentum, but it would have to act for a much longer time. For example, if the ball were thrown upward, the gravitational force (which is much smaller than the tennis racquet\u2019s force) would eventually reverse the momentum of the ball. Quantitatively, the effect we are talking about is the change in momentum \u0394p. By rearranging the equation Fnet = \u0394p \u0394 to be we can see how the change in momentum equals the average net external force multiplied by the time this force acts. The quantity Fnet \u0394 is given the name impulse. Impulse is the same as the change in momentum. \u0394p = Fnet\u0394, (8.18) Impulse: Change in Momentum Change in momentum equals the average net external force multiplied by the time this force acts. \u0394p = Fnet\u0394 (8.19) The quantity Fnet \u0394 is given the name impulse. There are many ways in which an understanding of impulse can save lives, or at least limbs. The dashboard padding in a car, and certainly the airbags, allow the net force on the occupants in the car to act over a much longer time when there is a sudden stop. The momentum change is the same for an occupant, whether an air bag is deployed or not, but the force (to bring the occupant to a stop) will be much less if it acts over a larger time. Cars today have many plastic components. One advantage of plastics is their lighter weight, which results in better gas mileage. Another advantage is that", " a car will crumple in a collision, especially in the event of a head-on collision. A longer collision time means the force on the car will be less. Deaths during car races decreased dramatically when the rigid frames of racing cars were replaced with parts that could crumple or collapse in the event of an accident. Bones in a body will fracture if the force on them is too large. If you jump onto the floor from a table, the force on your legs can be immense if you land stiff-legged on a hard surface. Rolling on the ground after jumping from the table, or landing with a parachute, extends the time over which the force (on you from the ground) acts. 320 Chapter 8 | Linear Momentum and Collisions Making Connections: Illustrations of Force Exerted Figure 8.3 This is a graph showing the force exerted by a fixed barrier on a block versus time. A 1.2-kg block slides across a horizontal, frictionless surface with a constant speed of 3.0 m/s before striking a fixed barrier and coming to a stop. In Figure 8.3, the force exerted by the barrier is assumed to be a constant 15 N during the 0.24-s collision. The impulse can be calculated using the area under the curve. \u0394 = \u0394 = (15 N)(0.24 s) = 3.6 kg\u2022m/s Note that the initial momentum of the block is: = = (1.2 kg)( \u2212 3.0 m/s) = \u2212 3.6 kg\u2022m/s (8.20) (8.21) We are assuming that the initial velocity is \u22123.0 m/s. We have established that the force exerted by the barrier is in the positive direction, so the initial velocity of the block must be in the negative direction. Since the final momentum of the block is zero, the impulse is equal to the change in momentum of the block. Suppose that, instead of striking a fixed barrier, the block is instead stopped by a spring.Consider the force exerted by the spring over the time interval from the beginning of the collision until the block comes to rest. Figure 8.4 This is a graph showing the force exerted by a spring on a block versus time. In this case, the impulse can be calculated again using the area under the curve (the area of a triangle): = 1 2 (base)(height) = 1 2 (0.24 s)(30 N) = 3.6 kg\u2022m/s (8.22) Again, this is equal to the difference between the initial and final momentum of the block, so the impulse is equal to the change in momentum. This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 8 | Linear Momentum and Collisions 321 Example 8.3 Calculating Magnitudes of Impulses: Two Billiard Balls Striking a Rigid Wall Two identical billiard balls strike a rigid wall with the same speed, and are reflected without any change of speed. The first ball strikes perpendicular to the wall. The second ball strikes the wall at an angle of 30\u00ba from the perpendicular, and bounces off at an angle of 30\u00ba from perpendicular to the wall. (a) Determine the direction of the force on the wall due to each ball. (b) Calculate the ratio of the magnitudes of impulses on the two balls by the wall. Strategy for (a) In order to determine the force on the wall, consider the force on the ball due to the wall using Newton\u2019s second law and then apply Newton\u2019s third law to determine the direction. Assume the -axis to be normal to the wall and to be positive in the initial direction of motion. Choose the -axis to be along the wall in the plane of the second ball\u2019s motion. The momentum direction and the velocity direction are the same. Solution for (a) The first ball bounces directly into the wall and exerts a force on it in the direction. Therefore the wall exerts a force on the ball in the direction. The second ball continues with the same momentum component in the direction, but reverses its -component of momentum, as seen by sketching a diagram of the angles involved and keeping in mind the proportionality between velocity and momentum. These changes mean the change in momentum for both balls is in the direction, so the force of the wall on each ball is along the direction. Strategy for (b) Calculate the change in momentum for each ball, which is equal to the impulse imparted to the ball. Solution for (b) Let be the speed of each ball before and after collision with the wall, and the mass of each ball. Choose the -axis and -axis as previously described, and consider the change in momentum of the first ball which strikes perpendicular to the wall. xi = ; yi = 0 xf = \u2212; yf = 0 Impulse is", " the change in momentum vector. Therefore the -component of impulse is equal to \u22122 and the component of impulse is equal to zero. Now consider the change in momentum of the second ball. xi = cos 30\u00ba; yi = sin 30\u00ba xf = \u2013 cos 30\u00ba; yf = \u2212 sin 30\u00ba It should be noted here that while x changes sign after the collision, y does not. Therefore the -component of impulse is equal to \u22122 cos 30\u00ba and the -component of impulse is equal to zero. The ratio of the magnitudes of the impulse imparted to the balls is 2 2 cos 30\u00ba = 2 3 = 1.155. (8.23) (8.24) (8.25) (8.26) (8.27) Discussion The direction of impulse and force is the same as in the case of (a); it is normal to the wall and along the negative -direction. Making use of Newton\u2019s third law, the force on the wall due to each ball is normal to the wall along the positive -direction. Our definition of impulse includes an assumption that the force is constant over the time interval \u0394. Forces are usually not constant. Forces vary considerably even during the brief time intervals considered. It is, however, possible to find an average effective force eff that produces the same result as the corresponding time-varying force. Figure 8.5 shows a graph of what an actual force looks like as a function of time for a ball bouncing off the floor. The area under the curve has units of momentum and is equal to the impulse or change in momentum between times 1 and 2. That area is equal to the area inside the 322 Chapter 8 | Linear Momentum and Collisions rectangle bounded by eff, 1, and 2. Thus the impulses and their effects are the same for both the actual and effective forces. Figure 8.5 A graph of force versus time with time along the -axis and force along the -axis for an actual force and an equivalent effective force. The areas under the two curves are equal. Making Connections: Baseball In most real-life collisions, the forces acting on an object are not constant. For example, when a bat strikes a baseball, the force is very small at the beginning of the collision since only a small portion of the ball is initially in contact with the bat. As the collision continues, the ball deforms so that a greater fraction of the ball is in contact with the bat, resulting in a greater force. As the ball begins to leave the bat, the force drops to zero, much like the force curve in Figure 8.5. Although the changing force is difficult to precisely calculate at each instant, the average force can be estimated very well in most cases. Suppose that a 150-g baseball experiences an average force of 480 N in a direction opposite the initial 32 m/s speed of the baseball over a time interval of 0.017 s. What is the final velocity of the baseball after the collision? \u0394 = \u0394 = (480)(0.017) = 8.16 kg\u2022m/s \u2212 = 8.16 kg \u2022 m/s (0.150 kg) \u2212 (0.150 kg)( \u2212 32 m/s) = 8.16 kg \u2022 m/s = 22 m/s (8.28) (8.29) (8.30) (8.31) Note in the above example that the initial velocity of the baseball prior to the collision is negative, consistent with the assumption we initially made that the force exerted by the bat is positive and in the direction opposite the initial velocity of the baseball. In this case, even though the force acting on the baseball varies with time, the average force is a good approximation of the effective force acting on the ball for the purposes of calculating the impulse and the change in momentum. Making Connections: Take-Home Investigation\u2014Hand Movement and Impulse Try catching a ball while \u201cgiving\u201d with the ball, pulling your hands toward your body. Then, try catching a ball while keeping your hands still. Hit water in a tub with your full palm. After the water has settled, hit the water again by diving your hand with your fingers first into the water. (Your full palm represents a swimmer doing a belly flop and your diving hand represents a swimmer doing a dive.) Explain what happens in each case and why. Which orientations would you advise people to avoid and why? Making Connections: Constant Force and Constant Acceleration The assumption of a constant force in the definition of impulse is analogous to the assumption of a constant acceleration in kinematics. In both cases, nature is adequately described without the use of calculus. Applying the Science Practices: Verifying the Relationship between Force and Change in Linear Momentum Design an experiment in order to experimentally verify the relationship between the impulse of a force and change in linear momentum. For simplicity, it would be best to ensure that frictional forces are very small or zero in", " your experiment so that the effect of friction can be neglected. As you design your experiment, consider the following: \u2022 Would it be easier to analyze a one-dimensional collision or a two-dimensional collision? \u2022 How will you measure the force? This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 8 | Linear Momentum and Collisions 323 \u2022 Should you have two objects in motion or one object bouncing off a rigid surface? \u2022 How will you measure the duration of the collision? \u2022 How will you measure the initial and final velocities of the object(s)? \u2022 Would it be easier to analyze an elastic or inelastic collision? \u2022 Should you verify the relationship mathematically or graphically? 8.3 Conservation of Momentum Learning Objectives By the end of this section, you will be able to: \u2022 Describe the law of conservation of linear momentum. \u2022 Derive an expression for the conservation of momentum. \u2022 Explain conservation of momentum with examples. \u2022 Explain the law of conservation of momentum as it relates to atomic and subatomic particles. The information presented in this section supports the following AP\u00ae learning objectives and science practices: \u2022 5.A.2.1 The student is able to define open and closed systems for everyday situations and apply conservation concepts for energy, charge, and linear momentum to those situations. (S.P. 6.4, 7.2) \u2022 5.D.1.4 The student is able to design an experimental test of an application of the principle of the conservation of linear momentum, predict an outcome of the experiment using the principle, analyze data generated by that experiment whose uncertainties are expressed numerically, and evaluate the match between the prediction and the outcome. (S.P. 4.2, 5.1, 5.3, 6.4) \u2022 5.D.2.1 The student is able to qualitatively predict, in terms of linear momentum and kinetic energy, how the outcome of a collision between two objects changes depending on whether the collision is elastic or inelastic. (S.P. 6.4, 7.2) \u2022 5.D.2.2 The student is able to plan data collection strategies to test the law of conservation of momentum in a two- object collision that is elastic or inelastic and analyze the resulting data graphically. (S.P.4.1, 4.2, 5.1) \u2022 5.D.3.1 The student is able to predict the velocity of the center of mass of a system when there is no interaction outside of the system but there is an interaction within the system (i.e., the student simply recognizes that interactions within a system do not affect the center of mass motion of the system and is able to determine that there is no external force). (S.P. 6.4) Momentum is an important quantity because it is conserved. Yet it was not conserved in the examples in Impulse and Linear Momentum and Force, where large changes in momentum were produced by forces acting on the system of interest. Under what circumstances is momentum conserved? The answer to this question entails considering a sufficiently large system. It is always possible to find a larger system in which total momentum is constant, even if momentum changes for components of the system. If a football player runs into the goalpost in the end zone, there will be a force on him that causes him to bounce backward. However, the Earth also recoils \u2014conserving momentum\u2014because of the force applied to it through the goalpost. Because Earth is many orders of magnitude more massive than the player, its recoil is immeasurably small and can be neglected in any practical sense, but it is real nevertheless. Consider what happens if the masses of two colliding objects are more similar than the masses of a football player and Earth\u2014for example, one car bumping into another, as shown in Figure 8.6. Both cars are coasting in the same direction when the lead car (labeled 2) is bumped by the trailing car (labeled 1). The only unbalanced force on each car is the force of the collision. (Assume that the effects due to friction are negligible.) Car 1 slows down as a result of the collision, losing some momentum, while car 2 speeds up and gains some momentum. We shall now show that the total momentum of the two-car system remains constant. 324 Chapter 8 | Linear Momentum and Collisions Figure 8.6 A car of mass 1 moving with a velocity of 1 bumps into another car of mass 2 and velocity 2 that it is following. As a result, the first car slows down to a velocity of v\u20321 and the second speeds up to a velocity of v\u20322. The momentum of each car is changed, but the total momentum tot of the two cars is the same before and after the collision (if you assume friction is negligible). Using the definition of impulse, the change", " in momentum of car 1 is given by \u03941 = 1\u0394, (8.32) where 1 is the force on car 1 due to car 2, and \u0394 is the time the force acts (the duration of the collision). Intuitively, it seems obvious that the collision time is the same for both cars, but it is only true for objects traveling at ordinary speeds. This assumption must be modified for objects travelling near the speed of light, without affecting the result that momentum is conserved. Similarly, the change in momentum of car 2 is \u03942 = 2\u0394 (8.33) where 2 is the force on car 2 due to car 1, and we assume the duration of the collision \u0394 is the same for both cars. We know from Newton\u2019s third law that 2 = \u2013 1, and so Thus, the changes in momentum are equal and opposite, and \u03941 + \u03942 = 0. \u03942 = \u22121\u0394 = \u2212\u03941. Because the changes in momentum add to zero, the total momentum of the two-car system is constant. That is, 1 + 2 = constant, 1 + 2 = \u20321 + \u20322, where \u20321 and \u20322 are the momenta of cars 1 and 2 after the collision. (We often use primes to denote the final state.) This result\u2014that momentum is conserved\u2014has validity far beyond the preceding one-dimensional case. It can be similarly shown that total momentum is conserved for any isolated system, with any number of objects in it. In equation form, the conservation of momentum principle for an isolated system is written or ptot = constant, ptot = p\u2032tot, (8.34) (8.35) (8.36) (8.37) (8.38) (8.39) This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 8 | Linear Momentum and Collisions 325 where ptot is the total momentum (the sum of the momenta of the individual objects in the system) and p\u2032tot is the total momentum some time later. (The total momentum can be shown to be the momentum of the center of mass of the system.) An isolated system is defined to be one for which the net external force is zero Fnet = 0. Conservation of Momentum Principle ptot = constant ptot = p\u2032tot (isolated system) (8.40) Isolated System An isolated system is defined to be one for which the net external force is zero Fnet = 0. Making Connections: Cart Collisions Consider two air carts with equal mass (m) on a linear track. The first cart moves with a speed v towards the second cart, which is initially at rest. We will take the initial direction of motion of the first cart as the positive direction. The momentum of the system will be conserved in the collision. If the collision is elastic, then the first cart will stop after the collision. Conservation of momentum therefore tells us that the second cart will have a final velocity v after the collision in the same direction as the initial velocity of the first cart. The kinetic energy of the system will be conserved since the masses are equal and the final velocity of cart 2 is equal to the initial velocity of cart 1. What would a graph of total momentum vs. time look like in this case? What would a graph of total kinetic energy vs. time look like in this case? Consider the center of mass of this system as the frame of reference. As cart 1 approaches cart 2, the center of mass remains exactly halfway between the two carts. The center of mass moves toward the stationary cart 2 at a speed 2. After the collision, the center of mass continues moving in the same direction, away from (now stationary) cart 1 at a speed 2. How would a graph of center-of-mass velocity vs. time compare to a graph of momentum vs. time? Suppose instead that the two carts move with equal speeds v in opposite directions towards the center of mass. Again, they have an elastic collision, so after the collision, they exchange velocities (each cart moving in the opposite direction of its initial motion with the same speed). As the two carts approach, the center of mass is exactly between the two carts, at the point where they will collide. In this case, how would a graph of center-of-mass velocity vs. time compare to a graph of the momentum of the system vs. time? Let us return to the example where the first cart is moving with a speed v toward the second cart, initially at rest. Suppose the second cart has some putty on one end so that, when the collision occurs, the two carts stick together in an inelastic collision. In this case, conservation of momentum tells us that the final velocity of the two-cart system will be half the initial velocity of the first cart, in the", " same direction as the first cart\u2019s initial motion. Kinetic energy will not be conserved in this case, however. Compared to the moving cart before the collision, the overall moving mass after the collision is doubled, but the velocity is halved. The initial kinetic energy of the system is: = 1 22(1st cart)+0(2nd cart)=1 22 The final kinetic energy of the two carts (2m) moving together (at speed v/2) is: = 1 2 (2) 2 2 =1 42 (8.41) (8.42) What would a graph of total momentum vs. time look like in this case? What would a graph of total kinetic energy vs. time look like in this case? Consider the center of mass of this system. As cart 1 approaches cart 2, the center of mass remains exactly halfway between the two carts. The center of mass moves toward the stationary cart 2 at a speed 2. After the collision, the two carts move together at a speed 2 momentum vs. time?. How would a graph of center-of-mass velocity vs. time compare to a graph of Suppose instead that the two carts move with equal speeds v in opposite directions towards the center of mass. They have putty on the end of each cart so that they stick together after the collision. As the two carts approach, the center of mass is exactly between the two carts, at the point where they will collide. In this case, how would a graph of center-of-mass velocity vs. time compare to a graph of the momentum of the system vs. time? 326 Chapter 8 | Linear Momentum and Collisions Perhaps an easier way to see that momentum is conserved for an isolated system is to consider Newton\u2019s second law in terms of momentum, Fnet = \u0394ptot \u0394. For an isolated system, Fnet = 0 ; thus, \u0394ptot = 0, and ptot is constant. We have noted that the three length dimensions in nature\u2014,, and \u2014are independent, and it is interesting to note that momentum can be conserved in different ways along each dimension. For example, during projectile motion and where air resistance is negligible, momentum is conserved in the horizontal direction because horizontal forces are zero and momentum is unchanged. But along the vertical direction, the net vertical force is not zero and the momentum of the projectile is not conserved. (See Figure 8.7.) However, if the momentum of the projectile-Earth system is considered in the vertical direction, we find that the total momentum is conserved. Figure 8.7 The horizontal component of a projectile\u2019s momentum is conserved if air resistance is negligible, even in this case where a space probe separates. The forces causing the separation are internal to the system, so that the net external horizontal force \u2013 net is still zero. The vertical component of the momentum is not conserved, because the net vertical force \u2013 net is not zero. In the vertical direction, the space probe-Earth system needs to be considered and we find that the total momentum is conserved. The center of mass of the space probe takes the same path it would if the separation did not occur. The conservation of momentum principle can be applied to systems as different as a comet striking Earth and a gas containing huge numbers of atoms and molecules. Conservation of momentum is violated only when the net external force is not zero. But another larger system can always be considered in which momentum is conserved by simply including the source of the external force. For example, in the collision of two cars considered above, the two-car system conserves momentum while each one-car system does not. Making Connections: Take-Home Investigation\u2014Drop of Tennis Ball and a Basketball Hold a tennis ball side by side and in contact with a basketball. Drop the balls together. (Be careful!) What happens? Explain your observations. Now hold the tennis ball above and in contact with the basketball. What happened? Explain your observations. What do you think will happen if the basketball ball is held above and in contact with the tennis ball? Making Connections: Take-Home Investigation\u2014Two Tennis Balls in a Ballistic Trajectory Tie two tennis balls together with a string about a foot long. Hold one ball and let the other hang down and throw it in a ballistic trajectory. Explain your observations. Now mark the center of the string with bright ink or attach a brightly colored sticker to it and throw again. What happened? Explain your observations. Some aquatic animals such as jellyfish move around based on the principles of conservation of momentum. A jellyfish fills its umbrella section with water and then pushes the water out resulting in motion in the opposite direction to that of the jet of water. Squids propel themselves in a similar manner but, in contrast with jellyfish, are able to control the direction in which they move by aiming their nozzle forward or backward. Typical squids can move at speeds of 8 to 12 km/h. The ballistocardi", "ograph (BCG) was a diagnostic tool used in the second half of the 20th century to study the strength of the heart. About once a second, your heart beats, forcing blood into the aorta. A force in the opposite direction is exerted on the rest of your body (recall Newton\u2019s third law). A ballistocardiograph is a device that can measure this reaction force. This measurement is done by using a sensor (resting on the person) or by using a moving table suspended from the ceiling. This technique can gather information on the strength of the heart beat and the volume of blood passing from the heart. However, the electrocardiogram (ECG or EKG) and the echocardiogram (cardiac ECHO or ECHO; a technique that uses ultrasound to see an image of the heart) are more widely used in the practice of cardiology. This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 8 | Linear Momentum and Collisions 327 Applying Science Practices: Verifying the Conservation of Linear Momentum Design an experiment to verify the conservation of linear momentum in a one-dimensional collision, both elastic and inelastic. For simplicity, try to ensure that friction is minimized so that it has a negligible effect on your experiment. As you consider your experiment, consider the following questions: \u2022 Predict how the final momentum of the system will compare to the initial momentum of the system that you will measure. Justify your prediction. \u2022 How will you measure the momentum of each object? \u2022 Should you have two objects in motion or one object bouncing off a rigid surface? \u2022 Should you verify the relationship mathematically or graphically? \u2022 How will you estimate the uncertainty of your measurements? How will you express this uncertainty in your data? When you have completed each experiment, compare the outcome to your prediction about the initial and final momentum of the system and evaluate your results. Making Connections: Conservation of Momentum and Collision Conservation of momentum is quite useful in describing collisions. Momentum is crucial to our understanding of atomic and subatomic particles because much of what we know about these particles comes from collision experiments. Subatomic Collisions and Momentum The conservation of momentum principle not only applies to the macroscopic objects, it is also essential to our explorations of atomic and subatomic particles. Giant machines hurl subatomic particles at one another, and researchers evaluate the results by assuming conservation of momentum (among other things). On the small scale, we find that particles and their properties are invisible to the naked eye but can be measured with our instruments, and models of these subatomic particles can be constructed to describe the results. Momentum is found to be a property of all subatomic particles including massless particles such as photons that compose light. Momentum being a property of particles hints that momentum may have an identity beyond the description of an object\u2019s mass multiplied by the object\u2019s velocity. Indeed, momentum relates to wave properties and plays a fundamental role in what measurements are taken and how we take these measurements. Furthermore, we find that the conservation of momentum principle is valid when considering systems of particles. We use this principle to analyze the masses and other properties of previously undetected particles, such as the nucleus of an atom and the existence of quarks that make up particles of nuclei. Figure 8.8 below illustrates how a particle scattering backward from another implies that its target is massive and dense. Experiments seeking evidence that quarks make up protons (one type of particle that makes up nuclei) scattered high-energy electrons off of protons (nuclei of hydrogen atoms). Electrons occasionally scattered straight backward in a manner that implied a very small and very dense particle makes up the proton\u2014this observation is considered nearly direct evidence of quarks. The analysis was based partly on the same conservation of momentum principle that works so well on the large scale. Figure 8.8 A subatomic particle scatters straight backward from a target particle. In experiments seeking evidence for quarks, electrons were observed to occasionally scatter straight backward from a proton. 328 Chapter 8 | Linear Momentum and Collisions 8.4 Elastic Collisions in One Dimension Learning Objectives By the end of this section, you will be able to: \u2022 Describe an elastic collision of two objects in one dimension. \u2022 Define internal kinetic energy. \u2022 Derive an expression for conservation of internal kinetic energy in a one-dimensional collision. \u2022 Determine the final velocities in an elastic collision given masses and initial velocities. The information presented in this section supports the following AP\u00ae learning objectives and science practices: \u2022 4.B.1.1 The student is able to calculate the change in linear momentum of a two-object system with constant mass in linear motion from a representation of the system (data, graphs, etc.). (S.P. 1.4, 2.2) \u2022 4.B.1.", "2 The student is able to analyze data to find the change in linear momentum for a constant-mass system using the product of the mass and the change in velocity of the center of mass. (S.P. 5.1) \u2022 5.A.2.1 The student is able to define open and closed systems for everyday situations and apply conservation concepts for energy, charge, and linear momentum to those situations. (S.P. 6.4, 7.2) \u2022 5.D.1.1 The student is able to make qualitative predictions about natural phenomena based on conservation of linear momentum and restoration of kinetic energy in elastic collisions. (S.P. 6.4, 7.2) \u2022 5.D.1.2 The student is able to apply the principles of conservation of momentum and restoration of kinetic energy to reconcile a situation that appears to be isolated and elastic, but in which data indicate that linear momentum and kinetic energy are not the same after the interaction, by refining a scientific question to identify interactions that have not been considered. Students will be expected to solve qualitatively and/or quantitatively for one-dimensional situations and only qualitatively in two-dimensional situations. (S.P. 2.2, 3.2, 5.1, 5.3) \u2022 5.D.1.5 The student is able to classify a given collision situation as elastic or inelastic, justify the selection of conservation of linear momentum and restoration of kinetic energy as the appropriate principles for analyzing an elastic collision, solve for missing variables, and calculate their values. (S.P. 2.1, 2.2) \u2022 5.D.1.6 The student is able to make predictions of the dynamical properties of a system undergoing a collision by application of the principle of linear momentum conservation and the principle of the conservation of energy in situations in which an elastic collision may also be assumed. (S.P. 6.4) \u2022 5.D.1.7 The student is able to classify a given collision situation as elastic or inelastic, justify the selection of conservation of linear momentum and restoration of kinetic energy as the appropriate principles for analyzing an elastic collision, solve for missing variables, and calculate their values. (S.P. 2.1, 2.2) \u2022 5.D.2.1 The student is able to qualitatively predict, in terms of linear momentum and kinetic energy, how the outcome of a collision between two objects changes depending on whether the collision is elastic or inelastic. (S.P. 6.4, 7.2) \u2022 5.D.2.2 The student is able to plan data collection strategies to test the law of conservation of momentum in a two- object collision that is elastic or inelastic and analyze the resulting data graphically. (S.P.4.1, 4.2, 5.1) \u2022 5.D.3.2 The student is able to make predictions about the velocity of the center of mass for interactions within a defined one-dimensional system. (S.P. 6.4) Let us consider various types of two-object collisions. These collisions are the easiest to analyze, and they illustrate many of the physical principles involved in collisions. The conservation of momentum principle is very useful here, and it can be used whenever the net external force on a system is zero. We start with the elastic collision of two objects moving along the same line\u2014a one-dimensional problem. An elastic collision is one that also conserves internal kinetic energy. Internal kinetic energy is the sum of the kinetic energies of the objects in the system. Figure 8.9 illustrates an elastic collision in which internal kinetic energy and momentum are conserved. Truly elastic collisions can only be achieved with subatomic particles, such as electrons striking nuclei. Macroscopic collisions can be very nearly, but not quite, elastic\u2014some kinetic energy is always converted into other forms of energy such as heat transfer due to friction and sound. One macroscopic collision that is nearly elastic is that of two steel blocks on ice. Another nearly elastic collision is that between two carts with spring bumpers on an air track. Icy surfaces and air tracks are nearly frictionless, more readily allowing nearly elastic collisions on them. Elastic Collision An elastic collision is one that conserves internal kinetic energy. Internal Kinetic Energy Internal kinetic energy is the sum of the kinetic energies of the objects in the system. This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 8 | Linear Momentum and Collisions 329 Figure 8.9 An elastic one-dimensional two-object collision. Momentum and internal kinetic energy are conserved. Now, to solve problems involving one-dimensional elastic collisions between two objects we can use the equations for conservation of momentum and conservation of internal kinetic energy. First, the equation for conservation of momentum for two objects in a one-dimensional collision is or 1 + 2 = \u20321", "+ \u20322 net = 0 1 1 + 22 = 1\u20321 + 2\u20322 net = 0, (8.43) (8.44) where the primes (') indicate values after the collision. By definition, an elastic collision conserves internal kinetic energy, and so the sum of kinetic energies before the collision equals the sum after the collision. Thus, 2 = 1 2 (two-object elastic collision) 2 + 1 2 + 1 (8.45) 22 2 21 \u20321 22 \u20322 1 21 1 expresses the equation for conservation of internal kinetic energy in a one-dimensional collision. Making Connections: Collisions Suppose data are collected on a collision between two masses sliding across a frictionless surface. Mass A (1.0 kg) moves with a velocity of +12 m/s, and mass B (2.0 kg) moves with a velocity of \u221212 m/s. The two masses collide and stick together after the collision. The table below shows the measured velocities of each mass at times before and after the collision: Table 8.1 Time (s) Velocity A (m/s) Velocity B (m/s) 0 1.0 s 2.0 s 3.0 s +12 +12 \u22124.0 \u22124.0 \u221212 \u221212 \u22124.0 \u22124.0 The total mass of the system is 3.0 kg. The velocity of the center of mass of this system can be determined from the conservation of momentum. Consider the system before the collision: ( + ) = + (3.0) = (1)(12) + (2)( \u2212 12) = \u2212 4.0 m/s (8.46) (8.47) (8.48) 330 Chapter 8 | Linear Momentum and Collisions After the collision, the center-of-mass velocity is the same: ( + ) = ( + ) (3.0) = (3)( \u2212 4.0) = \u2212 4.0 m/s The total momentum of the system before the collision is: + = (1)(12) + (2)( \u2212 12) = \u2212 12 kg \u2022 m/s The total momentum of the system after the collision is: ( + ) = (3)( \u2212 4) = \u2212 12 kg \u2022 m/s (8.49) (8.50) (8.51) (8.52) (8.53) Thus, the change in momentum of the system is zero when measured this way. We get a similar result when we calculate the momentum using the center-of-mass velocity. Since the center-of-mass velocity is the same both before and after the collision, we calculate the same momentum for the system using this method both before and after the collision. Example 8.4 Calculating Velocities Following an Elastic Collision Calculate the velocities of two objects following an elastic collision, given that 1 = 0.500 kg, 2 = 3.50 kg, 1 = 4.00 m/s, and 2 = 0. (8.54) Strategy and Concept First, visualize what the initial conditions mean\u2014a small object strikes a larger object that is initially at rest. This situation is slightly simpler than the situation shown in Figure 8.9 where both objects are initially moving. We are asked to find two unknowns (the final velocities \u20321 and \u20322 ). To find two unknowns, we must use two independent equations. Because this collision is elastic, we can use the above two equations. Both can be simplified by the fact that object 2 is initially at rest, and thus 2 = 0. Once we simplify these equations, we combine them algebraically to solve for the unknowns. Solution For this problem, note that 2 = 0 and use conservation of momentum. Thus, or 1 = \u20321 + \u20322 1 1 = 1\u20321 + 2\u20322. Using conservation of internal kinetic energy and that 2 = 0, Solving the first equation (momentum equation) for \u20322, we obtain 1 21 1 2 = 1 21 \u20321 2 + 1 22 \u20322 2. \u20322 = 1 2 1 \u2212 \u20321. (8.55) (8.56) (8.57) (8.58) Substituting this expression into the second equation (internal kinetic energy equation) eliminates the variable \u20322, leaving only \u20321 as an unknown (the algebra is left as an exercise for the reader). There are two solutions to any quadratic equation; in this example, they are and \u20321 = 4.00 m/s \u20321 = \u22123.00 m/s. (8.59) (8.60) As noted when quadratic equations were encountered in earlier chapters, both solutions may or may not be meaningful. In this case, the first solution is the same as the initial condition. The first", " solution thus represents the situation before the collision and is discarded. The second solution (\u20321 = \u22123.00 m/s) is negative, meaning that the first object bounces backward. When this negative value of \u20321 is used to find the velocity of the second object after the collision, we get This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 8 | Linear Momentum and Collisions \u20322 = 1 2 1 \u2212 \u20321 = 0.500 kg 3.50 kg 4.00 \u2212 (\u22123.00) m/s or Discussion \u20322 = 1.00 m/s. 331 (8.61) (8.62) The result of this example is intuitively reasonable. A small object strikes a larger one at rest and bounces backward. The larger one is knocked forward, but with a low speed. (This is like a compact car bouncing backward off a full-size SUV that is initially at rest.) As a check, try calculating the internal kinetic energy before and after the collision. You will see that the internal kinetic energy is unchanged at 4.00 J. Also check the total momentum before and after the collision; you will find it, too, is unchanged. The equations for conservation of momentum and internal kinetic energy as written above can be used to describe any onedimensional elastic collision of two objects. These equations can be extended to more objects if needed. Making Connections: Take-Home Investigation\u2014Ice Cubes and Elastic Collision Find a few ice cubes which are about the same size and a smooth kitchen tabletop or a table with a glass top. Place the ice cubes on the surface several centimeters away from each other. Flick one ice cube toward a stationary ice cube and observe the path and velocities of the ice cubes after the collision. Try to avoid edge-on collisions and collisions with rotating ice cubes. Have you created approximately elastic collisions? Explain the speeds and directions of the ice cubes using momentum. PhET Explorations: Collision Lab Investigate collisions on an air hockey table. Set up your own experiments: vary the number of discs, masses and initial conditions. Is momentum conserved? Is kinetic energy conserved? Vary the elasticity and see what happens. Figure 8.10 Collision Lab (http://cnx.org/content/m55171/1.3/collision-lab_en.jar) 8.5 Inelastic Collisions in One Dimension Learning Objectives By the end of this section, you will be able to: \u2022 Define inelastic collision. \u2022 Explain perfectly inelastic collisions. \u2022 Apply an understanding of collisions to sports. \u2022 Determine recoil velocity and loss in kinetic energy given mass and initial velocity. The information presented in this section supports the following AP\u00ae learning objectives and science practices: \u2022 4.B.1.1 The student is able to calculate the change in linear momentum of a two-object system with constant mass in linear motion from a representation of the system (data, graphs, etc.). (S.P. 1.4, 2.2) \u2022 5.A.2.1 The student is able to define open and closed systems for everyday situations and apply conservation concepts for energy, charge, and linear momentum to those situations. (S.P. 6.4, 7.2) \u2022 5.D.1.3 The student is able to apply mathematical routines appropriately to problems involving elastic collisions in one dimension and justify the selection of those mathematical routines based on conservation of momentum and restoration of kinetic energy. (S.P. 2.1, 2.2) \u2022 5.D.1.5 The student is able to classify a given collision situation as elastic or inelastic, justify the selection of conservation of linear momentum and restoration of kinetic energy as the appropriate principles for analyzing an elastic collision, solve for missing variables, and calculate their values. (S.P. 2.1, 2.2) \u2022 5.D.2.1 The student is able to qualitatively predict, in terms of linear momentum and kinetic energy, how the outcome of a collision between two objects changes depending on whether the collision is elastic or inelastic. (S.P. 6.4, 7.2) \u2022 5.D.2.2 The student is able to plan data collection strategies to test the law of conservation of momentum in a two- object collision that is elastic or inelastic and analyze the resulting data graphically. (S.P.4.1, 4.2, 5.1) \u2022 5.D.2.3 The student is able to apply the conservation of linear momentum to a closed system of objects involved in an inelastic collision to predict the change in kinetic energy. (S.P. 6.4, 7.2) 332 Chapter 8 | Linear Momentum and Collisions \u2022 5.D.2.4 The student is able to", " analyze data that verify conservation of momentum in collisions with and without an external friction force. (S.P. 4.1, 4.2, 4.4, 5.1, 5.3) \u2022 5.D.2.5 The student is able to classify a given collision situation as elastic or inelastic, justify the selection of conservation of linear momentum as the appropriate solution method for an inelastic collision, recognize that there is a common final velocity for the colliding objects in the totally inelastic case, solve for missing variables, and calculate their values. (S.P. 2.1 2.2) \u2022 5.D.2.6 The student is able to apply the conservation of linear momentum to an isolated system of objects involved in an inelastic collision to predict the change in kinetic energy. (S.P. 6.4, 7.2) We have seen that in an elastic collision, internal kinetic energy is conserved. An inelastic collision is one in which the internal kinetic energy changes (it is not conserved). This lack of conservation means that the forces between colliding objects may remove or add internal kinetic energy. Work done by internal forces may change the forms of energy within a system. For inelastic collisions, such as when colliding objects stick together, this internal work may transform some internal kinetic energy into heat transfer. Or it may convert stored energy into internal kinetic energy, such as when exploding bolts separate a satellite from its launch vehicle. Inelastic Collision An inelastic collision is one in which the internal kinetic energy changes (it is not conserved). Figure 8.11 shows an example of an inelastic collision. Two objects that have equal masses head toward one another at equal speeds and then stick together. Their total internal kinetic energy is initially 1 22 + 1 22 = 2. The two objects come to rest after sticking together, conserving momentum. But the internal kinetic energy is zero after the collision. A collision in which the objects stick together is sometimes called a perfectly inelastic collision because it reduces internal kinetic energy more than does any other type of inelastic collision. In fact, such a collision reduces internal kinetic energy to the minimum it can have while still conserving momentum. Perfectly Inelastic Collision A collision in which the objects stick together is sometimes called \u201cperfectly inelastic.\u201d Figure 8.11 An inelastic one-dimensional two-object collision. Momentum is conserved, but internal kinetic energy is not conserved. (a) Two objects of equal mass initially head directly toward one another at the same speed. (b) The objects stick together (a perfectly inelastic collision), and so their final velocity is zero. The internal kinetic energy of the system changes in any inelastic collision and is reduced to zero in this example. Example 8.5 Calculating Velocity and Change in Kinetic Energy: Inelastic Collision of a Puck and a Goalie (a) Find the recoil velocity of a 70.0-kg ice hockey goalie, originally at rest, who catches a 0.150-kg hockey puck slapped at him at a velocity of 35.0 m/s. (b) How much kinetic energy is lost during the collision? Assume friction between the ice and the puck-goalie system is negligible. (See Figure 8.12 ) This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 8 | Linear Momentum and Collisions 333 Figure 8.12 An ice hockey goalie catches a hockey puck and recoils backward. The initial kinetic energy of the puck is almost entirely converted to thermal energy and sound in this inelastic collision. Strategy Momentum is conserved because the net external force on the puck-goalie system is zero. We can thus use conservation of momentum to find the final velocity of the puck and goalie system. Note that the initial velocity of the goalie is zero and that the final velocity of the puck and goalie are the same. Once the final velocity is found, the kinetic energies can be calculated before and after the collision and compared as requested. Solution for (a) Momentum is conserved because the net external force on the puck-goalie system is zero. Conservation of momentum is or 1 + 2 = \u20321 + \u20322 1 1 + 22 = 1\u20321 + 2\u20322. (8.63) (8.64) Because the goalie is initially at rest, we know 2 = 0. Because the goalie catches the puck, the final velocities are equal, or \u20321 = \u20322 = \u2032. Thus, the conservation of momentum equation simplifies to Solving for \u2032 yields 1 1 = (1 + 2)\u2032. \u2032 = 1 1 + 2 1. Entering known values in this equation, we get \u2032 = 0.150 kg 70.0 kg + 0.150 kg (", "35.0 m/s) = 7.48\u00d710\u22122 m/s. Discussion for (a) (8.65) (8.66) (8.67) This recoil velocity is small and in the same direction as the puck\u2019s original velocity, as we might expect. Solution for (b) Before the collision, the internal kinetic energy KEint of the system is that of the hockey puck, because the goalie is initially at rest. Therefore, KEint is initially KEint = 1 22 = 1 2 = 91.9 J. 0.150 kg (35.0 m/s)2 After the collision, the internal kinetic energy is KE\u2032int = 1 2 ( + )2 = 1 2 = 0.196 J. The change in internal kinetic energy is thus 70.15 kg 7.48\u00d710\u22122 m/s (8.68) 2 (8.69) 334 Chapter 8 | Linear Momentum and Collisions KE\u2032int \u2212 KEint = 0.196 J \u2212 91.9 J = \u2212 91.7 J (8.70) where the minus sign indicates that the energy was lost. Discussion for (b) Nearly all of the initial internal kinetic energy is lost in this perfectly inelastic collision. KEint is mostly converted to thermal energy and sound. During some collisions, the objects do not stick together and less of the internal kinetic energy is removed\u2014such as happens in most automobile accidents. Alternatively, stored energy may be converted into internal kinetic energy during a collision. Figure 8.13 shows a one-dimensional example in which two carts on an air track collide, releasing potential energy from a compressed spring. Example 8.6 deals with data from such a collision. Figure 8.13 An air track is nearly frictionless, so that momentum is conserved. Motion is one-dimensional. In this collision, examined in Example 8.6, the potential energy of a compressed spring is released during the collision and is converted to internal kinetic energy. Collisions are particularly important in sports and the sporting and leisure industry utilizes elastic and inelastic collisions. Let us look briefly at tennis. Recall that in a collision, it is momentum and not force that is important. So, a heavier tennis racquet will have the advantage over a lighter one. This conclusion also holds true for other sports\u2014a lightweight bat (such as a softball bat) cannot hit a hardball very far. The location of the impact of the tennis ball on the racquet is also important, as is the part of the stroke during which the impact occurs. A smooth motion results in the maximizing of the velocity of the ball after impact and reduces sports injuries such as tennis elbow. A tennis player tries to hit the ball on the \u201csweet spot\u201d on the racquet, where the vibration and impact are minimized and the ball is able to be given more velocity. Sports science and technologies also use physics concepts such as momentum and rotational motion and vibrations. Take-Home Experiment\u2014Bouncing of Tennis Ball 1. Find a racquet (a tennis, badminton, or other racquet will do). Place the racquet on the floor and stand on the handle. Drop a tennis ball on the strings from a measured height. Measure how high the ball bounces. Now ask a friend to hold the racquet firmly by the handle and drop a tennis ball from the same measured height above the racquet. Measure how high the ball bounces and observe what happens to your friend\u2019s hand during the collision. Explain your observations and measurements. 2. The coefficient of restitution () is a measure of the elasticity of a collision between a ball and an object, and is defined as the ratio of the speeds after and before the collision. A perfectly elastic collision has a of 1. For a ball bouncing off the floor (or a racquet on the floor), can be shown to be = ( / )1 / 2 which the ball bounces and is the height from which the ball is dropped. Determine for the cases in Part 1 and where is the height to This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 8 | Linear Momentum and Collisions 335 for the case of a tennis ball bouncing off a concrete or wooden floor ( = 0.85 for new tennis balls used on a tennis court). Example 8.6 Calculating Final Velocity and Energy Release: Two Carts Collide In the collision pictured in Figure 8.13, two carts collide inelastically. Cart 1 (denoted 1 carries a spring which is initially compressed. During the collision, the spring releases its potential energy and converts it to internal kinetic energy. The mass of cart 1 and the spring is 0.350 kg, and the cart and the spring together have an initial velocity of 2.00 m/s. Cart 2 (denoted 2 in Figure 8.13) has a mass of 0.500 kg and an initial velocity of \u22120.500", " m/s. After the collision, cart 1 is observed to recoil with a velocity of \u22124.00 m/s. (a) What is the final velocity of cart 2? (b) How much energy was released by the spring (assuming all of it was converted into internal kinetic energy)? Strategy We can use conservation of momentum to find the final velocity of cart 2, because net = 0 (the track is frictionless and the force of the spring is internal). Once this velocity is determined, we can compare the internal kinetic energy before and after the collision to see how much energy was released by the spring. Solution for (a) As before, the equation for conservation of momentum in a two-object system is The only unknown in this equation is \u20322. Solving for \u20322 and substituting known values into the previous equation yields 1 1 + 22 = 1\u20321 + 2\u20322. (8.71) \u20322 = = 1 1 + 22 \u2212 1 \u20321 2 (2.00 m/s) + 0.350 kg 0.500 kg (\u22120.500 m/s) 0.500 kg \u2212 0.350 kg (\u22124.00 m/s) 0.500 kg = 3.70 m/s. Solution for (b) The internal kinetic energy before the collision is 2 + 1 KEint = 1 = 1 2 21 1 0.350 kg 2 22 2 (2.00 m/s)2 + 1 2 0.500 kg ( \u2013 0.500 m/s)2 After the collision, the internal kinetic energy is = 0.763 J. 2 + 1 KE\u2032int = 1 = 1 2 = 6.22 J. 21 \u20321 0.350 kg 2 22 \u20322 (-4.00 m/s)2 + 1 2 0.500 kg (3.70 m/s)2 The change in internal kinetic energy is thus KE\u2032int \u2212 KEint = 6.22 J \u2212 0.763 J = 5.46 J. (8.72) (8.73) (8.74) (8.75) Discussion The final velocity of cart 2 is large and positive, meaning that it is moving to the right after the collision. The internal kinetic energy in this collision increases by 5.46 J. That energy was released by the spring. 8.6 Collisions of Point Masses in Two Dimensions By the end of this section, you will be able to: \u2022 Discuss two-dimensional collisions as an extension of one-dimensional analysis. Learning Objectives 336 Chapter 8 | Linear Momentum and Collisions \u2022 Define point masses. \u2022 Derive an expression for conservation of momentum along the x-axis and y-axis. \u2022 Describe elastic collisions of two objects with equal mass. \u2022 Determine the magnitude and direction of the final velocity given initial velocity and scattering angle. The information presented in this section supports the following AP\u00ae learning objectives and science practices: \u2022 5.D.1.2 The student is able to apply the principles of conservation of momentum and restoration of kinetic energy to reconcile a situation that appears to be isolated and elastic, but in which data indicate that linear momentum and kinetic energy are not the same after the interaction, by refining a scientific question to identify interactions that have not been considered. Students will be expected to solve qualitatively and/or quantitatively for one-dimensional situations and only qualitatively in two-dimensional situations. \u2022 5.D.3.3 The student is able to make predictions about the velocity of the center of mass for interactions within a defined two-dimensional system. In the previous two sections, we considered only one-dimensional collisions; during such collisions, the incoming and outgoing velocities are all along the same line. But what about collisions, such as those between billiard balls, in which objects scatter to the side? These are two-dimensional collisions, and we shall see that their study is an extension of the one-dimensional analysis already presented. The approach taken (similar to the approach in discussing two-dimensional kinematics and dynamics) is to choose a convenient coordinate system and resolve the motion into components along perpendicular axes. Resolving the motion yields a pair of one-dimensional problems to be solved simultaneously. One complication arising in two-dimensional collisions is that the objects might rotate before or after their collision. For example, if two ice skaters hook arms as they pass by one another, they will spin in circles. We will not consider such rotation until later, and so for now we arrange things so that no rotation is possible. To avoid rotation, we consider only the scattering of point masses\u2014that is, structureless particles that cannot rotate or spin. We start by assuming that Fnet = 0, so that momentum p is conserved. The simplest collision is one in which one of the particles is initially at rest. (See Figure 8.14.) The best choice for a coordinate system is one with an axis parallel to the velocity of the incoming", " particle, as shown in Figure 8.14. Because momentum is conserved, the components of momentum along the - and -axes ( and ) will also be conserved, but with the chosen coordinate system, is initially zero and is the momentum of the incoming particle. Both facts simplify the analysis. (Even with the simplifying assumptions of point masses, one particle initially at rest, and a convenient coordinate system, we still gain new insights into nature from the analysis of twodimensional collisions.) Figure 8.14 A two-dimensional collision with the coordinate system chosen so that 2 is initially at rest and 1 is parallel to the -axis. This coordinate system is sometimes called the laboratory coordinate system, because many scattering experiments have a target that is stationary in the laboratory, while particles are scattered from it to determine the particles that make-up the target and how they are bound together. The particles may not be observed directly, but their initial and final velocities are. Along the -axis, the equation for conservation of momentum is Where the subscripts denote the particles and axes and the primes denote the situation after the collision. In terms of masses and velocities, this equation is 1 + 2 = \u20321 + \u20322. (8.76) 1 1 + 22 = 11 + 22. But because particle 2 is initially at rest, this equation becomes 1 1 = 1\u20321 + 2\u20322. (8.77) (8.78) This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 8 | Linear Momentum and Collisions 337 The components of the velocities along the -axis have the form cos. Because particle 1 initially moves along the -axis, we find 1 = 1. Conservation of momentum along the -axis gives the following equation: 1 1 = 1\u20321 cos 1 + 2\u20322 cos 2, where 1 and 2 are as shown in Figure 8.14. Conservation of Momentum along the -axis 1 1 = 1\u20321 cos 1 + 2\u20322 cos 2 Along the -axis, the equation for conservation of momentum is or 1 + 2 = \u20321 + \u20322 1 1 + 22 = 1\u20321 + 2\u20322. (8.79) (8.80) (8.81) (8.82) But 1 is zero, because particle 1 initially moves along the -axis. Because particle 2 is initially at rest, 2 is also zero. The equation for conservation of momentum along the -axis becomes 0 = 1\u20321 + 2\u20322. The components of the velocities along the -axis have the form sin. Thus, conservation of momentum along the -axis gives the following equation: 0 = 1\u20321 sin 1 + 2\u20322 sin 2. Conservation of Momentum along the -axis 0 = 1\u20321 sin 1 + 2\u20322 sin 2 (8.83) (8.84) (8.85) The equations of conservation of momentum along the -axis and -axis are very useful in analyzing two-dimensional collisions of particles, where one is originally stationary (a common laboratory situation). But two equations can only be used to find two unknowns, and so other data may be necessary when collision experiments are used to explore nature at the subatomic level. Making Connections: Real World Connections We have seen, in one-dimensional collisions when momentum is conserved, that the center-of-mass velocity of the system remains unchanged as a result of the collision. If you calculate the momentum and center-of-mass velocity before the collision, you will get the same answer as if you calculate both quantities after the collision. This logic also works for twodimensional collisions. For example, consider two cars of equal mass. Car A is driving east (+x-direction) with a speed of 40 m/s. Car B is driving north (+y-direction) with a speed of 80 m/s. What is the velocity of the center-of-mass of this system before and after an inelastic collision, in which the cars move together as one mass after the collision? Since both cars have equal mass, the center-of-mass velocity components are just the average of the components of the individual velocities before the collision. The x-component of the center of mass velocity is 20 m/s, and the y-component is 40 m/s. Using momentum conservation for the collision in both the x-component and y-component yields similar answers: (40) + (0) = (2)final() final() = 20 m/s (0) + (80) = (2)final() final() = 40 m/s (8.86) (8.87) (8.88) (8.89) 338 Chapter 8 | Linear Momentum and Collisions Since the two masses move together after the collision, the velocity of this combined object is equal to the center-of", "-mass velocity. Thus, the center-of-mass velocity before and after the collision is identical, even in two-dimensional collisions, when momentum is conserved. Example 8.7 Determining the Final Velocity of an Unseen Object from the Scattering of Another Object Suppose the following experiment is performed. A 0.250-kg object (1) is slid on a frictionless surface into a dark room, where it strikes an initially stationary object with mass of 0.400 kg (2). The 0.250-kg object emerges from the room at an angle of 45.0\u00ba with its incoming direction. The speed of the 0.250-kg object is originally 2.00 m/s and is 1.50 m/s after the collision. Calculate the magnitude and direction of the velocity (\u20322 and 2) of the 0.400-kg object after the collision. Strategy Momentum is conserved because the surface is frictionless. The coordinate system shown in Figure 8.15 is one in which 2 is originally at rest and the initial velocity is parallel to the -axis, so that conservation of momentum along the - and -axes is applicable. Everything is known in these equations except \u20322 and 2, which are precisely the quantities we wish to find. We can find two unknowns because we have two independent equations: the equations describing the conservation of momentum in the - and -directions. Solution Solving 1 1 = 1\u20321 cos 1 + 2\u20322 cos 2 for 2\u2032 cos 2 and 0 = 1\u20321 sin 1 + 2\u20322 sin 2 for \u20322 sin 2 and taking the ratio yields an equation (in which \u03b82 is the only unknown quantity. Applying the identity tan = sin cos, we obtain: tan 2 = \u20321 sin 1 \u20321 cos 1 \u2212 1. Entering known values into the previous equation gives tan 2 = (1.50 m/s)(0.7071) (1.50 m/s)(0.7071) \u2212 2.00 m/s = \u22121.129. Thus, 2 = tan\u22121(\u22121.129) = 311.5\u00ba \u2248 312\u00ba. (8.90) (8.91) (8.92) Angles are defined as positive in the counter clockwise direction, so this angle indicates that 2 is scattered to the right in Figure 8.15, as expected (this angle is in the fourth quadrant). Either equation for the - or -axis can now be used to solve for \u20322, but the latter equation is easiest because it has fewer terms. Entering known values into this equation gives \u20322 = \u2212 1 2 \u20321 sin 1 sin 2 \u20322 = \u2212 0.250 kg 0.400 kg (1.50 m/s) 0.7071 \u22120.7485. \u20322 = 0.886 m/s. Thus, Discussion (8.93) (8.94) (8.95) It is instructive to calculate the internal kinetic energy of this two-object system before and after the collision. (This calculation is left as an end-of-chapter problem.) If you do this calculation, you will find that the internal kinetic energy is less after the collision, and so the collision is inelastic. This type of result makes a physicist want to explore the system further. This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 8 | Linear Momentum and Collisions 339 Figure 8.15 A collision taking place in a dark room is explored in Example 8.7. The incoming object 1 is scattered by an initially stationary object. Only the stationary object\u2019s mass 2 is known. By measuring the angle and speed at which 1 emerges from the room, it is possible to calculate the magnitude and direction of the initially stationary object\u2019s velocity after the collision. Elastic Collisions of Two Objects with Equal Mass Some interesting situations arise when the two colliding objects have equal mass and the collision is elastic. This situation is nearly the case with colliding billiard balls, and precisely the case with some subatomic particle collisions. We can thus get a mental image of a collision of subatomic particles by thinking about billiards (or pool). (Refer to Figure 8.14 for masses and angles.) First, an elastic collision conserves internal kinetic energy. Again, let us assume object 2 (2) is initially at rest. Then, the internal kinetic energy before and after the collision of two objects that have equal masses is 1 21 2 = 1 2\u20321 2 + 1 2\u20322 2. (8.96) Because the masses are equal, 1 = 2 =. Algebraic manipulation (left to the reader) of conservation of momentum in the - and -directions can show that 2\u20321 (Remember that 2 is negative here.) The two preceding equations can both", " be true only if 1 \u2212 2 2\u20322 2 + \u20321\u20322 cos 1 21 2 = 1 2 + 1. \u20321 \u20322 cos 1 \u2212 2 = 0. (8.97) (8.98) \u2022 There are three ways that this term can be zero. They are \u20321 = 0 : head-on collision; incoming ball stops \u20322 = 0 : no collision; incoming ball continues unaffected cos(1 \u2212 2) = 0 : angle of separation (1 \u2212 2) is 90\u00ba after the collision \u2022 \u2022 All three of these ways are familiar occurrences in billiards and pool, although most of us try to avoid the second. If you play enough pool, you will notice that the angle between the balls is very close to 90\u00ba after the collision, although it will vary from this value if a great deal of spin is placed on the ball. (Large spin carries in extra energy and a quantity called angular momentum, which must also be conserved.) The assumption that the scattering of billiard balls is elastic is reasonable based on the correctness of the three results it produces. This assumption also implies that, to a good approximation, momentum is conserved for the two-ball system in billiards and pool. The problems below explore these and other characteristics of two-dimensional collisions. 340 Chapter 8 | Linear Momentum and Collisions Connections to Nuclear and Particle Physics Two-dimensional collision experiments have revealed much of what we know about subatomic particles, as we shall see in Medical Applications of Nuclear Physics and Particle Physics. Ernest Rutherford, for example, discovered the nature of the atomic nucleus from such experiments. 8.7 Introduction to Rocket Propulsion Learning Objectives By the end of this section, you will be able to: \u2022 State Newton\u2019s third law of motion. \u2022 Explain the principle involved in propulsion of rockets and jet engines. \u2022 Derive an expression for the acceleration of the rocket. \u2022 Discuss the factors that affect the rocket\u2019s acceleration. \u2022 Describe the function of a space shuttle. Rockets range in size from fireworks so small that ordinary people use them to immense Saturn Vs that once propelled massive payloads toward the Moon. The propulsion of all rockets, jet engines, deflating balloons, and even squids and octopuses is explained by the same physical principle\u2014Newton\u2019s third law of motion. Matter is forcefully ejected from a system, producing an equal and opposite reaction on what remains. Another common example is the recoil of a gun. The gun exerts a force on a bullet to accelerate it and consequently experiences an equal and opposite force, causing the gun\u2019s recoil or kick. Making Connections: Take-Home Experiment\u2014Propulsion of a Balloon Hold a balloon and fill it with air. Then, let the balloon go. In which direction does the air come out of the balloon and in which direction does the balloon get propelled? If you fill the balloon with water and then let the balloon go, does the balloon\u2019s direction change? Explain your answer. Figure 8.16 shows a rocket accelerating straight up. In part (a), the rocket has a mass and a velocity relative to Earth, and hence a momentum. In part (b), a time \u0394 has elapsed in which the rocket has ejected a mass \u0394 of hot gas at a velocity e relative to the rocket. The remainder of the mass ( \u2212 \u0394) now has a greater velocity ( + \u0394). The momentum of the entire system (rocket plus expelled gas) has actually decreased because the force of gravity has acted for a time \u0394, producing a negative impulse \u0394 = \u2212\u0394. (Remember that impulse is the net external force on a system multiplied by the time it acts, and it equals the change in momentum of the system.) So, the center of mass of the system is in free fall but, by rapidly expelling mass, part of the system can accelerate upward. It is a commonly held misconception that the rocket exhaust pushes on the ground. If we consider thrust; that is, the force exerted on the rocket by the exhaust gases, then a rocket\u2019s thrust is greater in outer space than in the atmosphere or on the launch pad. In fact, gases are easier to expel into a vacuum. By calculating the change in momentum for the entire system over \u0394, and equating this change to the impulse, the following expression can be shown to be a good approximation for the acceleration of the rocket. \u201cThe rocket\u201d is that part of the system remaining after the gas is ejected, and is the acceleration due to gravity. = e \u0394 \u0394 \u2212 Acceleration of a Rocket Acceleration of a rocket is = e \u0394 \u0394 \u2212 (8.99) (8.100) where is the acceleration of the rocket, e is the escape velocity, is the mass of the rocket, \u0394 is the mass of the ejected gas, and \u0394 is the time in which the gas is ejected. This content is available for free at http://cn", "x.org/content/col11844/1.13 Chapter 8 | Linear Momentum and Collisions 341 Figure 8.16 (a) This rocket has a mass and an upward velocity. The net external force on the system is \u2212, if air resistance is neglected. (b) A time \u0394 later the system has two main parts, the ejected gas and the remainder of the rocket. The reaction force on the rocket is what overcomes the gravitational force and accelerates it upward. A rocket\u2019s acceleration depends on three major factors, consistent with the equation for acceleration of a rocket. First, the greater the exhaust velocity of the gases relative to the rocket, e, the greater the acceleration is. The practical limit for e is about 2.5\u00d7103 m/s for conventional (non-nuclear) hot-gas propulsion systems. The second factor is the rate at which mass is ejected from the rocket. This is the factor \u0394 / \u0394 in the equation. The quantity (\u0394 / \u0394)e, with units of newtons, is called \"thrust.\u201d The faster the rocket burns its fuel, the greater its thrust, and the greater its acceleration. The third factor is the mass of the rocket. The smaller the mass is (all other factors being the same), the greater the acceleration. The rocket mass decreases dramatically during flight because most of the rocket is fuel to begin with, so that acceleration increases continuously, reaching a maximum just before the fuel is exhausted. Factors Affecting a Rocket\u2019s Acceleration \u2022 The greater the exhaust velocity e of the gases relative to the rocket, the greater the acceleration. \u2022 The faster the rocket burns its fuel, the greater its acceleration. \u2022 The smaller the rocket\u2019s mass (all other factors being the same), the greater the acceleration. Example 8.8 Calculating Acceleration: Initial Acceleration of a Moon Launch A Saturn V\u2019s mass at liftoff was 2.80\u00d7106 kg, its fuel-burn rate was 1.40\u00d7104 kg/s, and the exhaust velocity was 2.40\u00d7103 m/s. Calculate its initial acceleration. Strategy This problem is a straightforward application of the expression for acceleration because is the unknown and all of the terms on the right side of the equation are given. Solution Substituting the given values into the equation for acceleration yields 342 Chapter 8 | Linear Momentum and Collisions = e \u2212 \u0394 \u0394 = 2.40\u00d7103 m/s 2.80\u00d7106 kg 1.40\u00d7104 kg/s \u2212 9.80 m/s2 (8.101) Discussion = 2.20 m/s2. This value is fairly small, even for an initial acceleration. The acceleration does increase steadily as the rocket burns fuel, because decreases while e and \u0394 \u0394 show that the thrust of the engines was 3.36\u00d7107 N. remain constant. Knowing this acceleration and the mass of the rocket, you can To achieve the high speeds needed to hop continents, obtain orbit, or escape Earth\u2019s gravity altogether, the mass of the rocket other than fuel must be as small as possible. It can be shown that, in the absence of air resistance and neglecting gravity, the final velocity of a one-stage rocket initially at rest is = e ln 0 r (8.102) where ln 0 / r is the natural logarithm of the ratio of the initial mass of the rocket (0) to what is left (r) after all of the fuel is exhausted. (Note that is actually the change in velocity, so the equation can be used for any segment of the flight. If we start from rest, the change in velocity equals the final velocity.) For example, let us calculate the mass ratio needed to escape Earth\u2019s gravity starting from rest, given that the escape velocity from Earth is about 11.2\u00d7103 m/s, and assuming an exhaust velocity e = 2.5\u00d7103 m/s. Solving for 0 / r gives Thus, the mass of the rocket is ln 0 r = e = 11.2\u00d7103 m/s 2.5\u00d7103 m/s = 4.48 0 r = 4.48 = 88. r = 0 88. (8.103) (8.104) (8.105) This result means that only 1 / 88 of the mass is left when the fuel is burnt, and 87 / 88 of the initial mass was fuel. Expressed as percentages, 98.9% of the rocket is fuel, while payload, engines, fuel tanks, and other components make up only 1.10%. Taking air resistance and gravitational force into account, the mass r remaining can only be about 0 / 180. It is difficult to build a rocket in which the fuel has a mass 180 times everything else. The solution is multistage rockets. Each stage only needs to achieve part of the final velocity and is discarded after it burns its fuel. The result is", " that each successive stage can have smaller engines and more payload relative to its fuel. Once out of the atmosphere, the ratio of payload to fuel becomes more favorable, too. The space shuttle was an attempt at an economical vehicle with some reusable parts, such as the solid fuel boosters and the craft itself. (See Figure 8.17) The shuttle\u2019s need to be operated by humans, however, made it at least as costly for launching satellites as expendable, unmanned rockets. Ideally, the shuttle would only have been used when human activities were required for the success of a mission, such as the repair of the Hubble space telescope. Rockets with satellites can also be launched from airplanes. Using airplanes has the double advantage that the initial velocity is significantly above zero and a rocket can avoid most of the atmosphere\u2019s resistance. This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 8 | Linear Momentum and Collisions 343 Figure 8.17 The space shuttle had a number of reusable parts. Solid fuel boosters on either side were recovered and refueled after each flight, and the entire orbiter returned to Earth for use in subsequent flights. The large liquid fuel tank was expended. The space shuttle was a complex assemblage of technologies, employing both solid and liquid fuel and pioneering ceramic tiles as reentry heat shields. As a result, it permitted multiple launches as opposed to single-use rockets. (credit: NASA) PhET Explorations: Lunar Lander Can you avoid the boulder field and land safely, just before your fuel runs out, as Neil Armstrong did in 1969? Our version of this classic video game accurately simulates the real motion of the lunar lander with the correct mass, thrust, fuel consumption rate, and lunar gravity. The real lunar lander is very hard to control. Figure 8.18 Lunar Lander (http://cnx.org/content/m55174/1.2/lunar-lander_en.jar) Glossary change in momentum: the difference between the final and initial momentum; the mass times the change in velocity conservation of momentum principle: when the net external force is zero, the total momentum of the system is conserved or constant elastic collision: a collision that also conserves internal kinetic energy impulse: the average net external force times the time it acts; equal to the change in momentum inelastic collision: a collision in which internal kinetic energy is not conserved internal kinetic energy: the sum of the kinetic energies of the objects in a system isolated system: a system in which the net external force is zero linear momentum: the product of mass and velocity perfectly inelastic collision: a collision in which the colliding objects stick together point masses: structureless particles with no rotation or spin quark: fundamental constituent of matter and an elementary particle 344 Chapter 8 | Linear Momentum and Collisions second law of motion: physical law that states that the net external force equals the change in momentum of a system divided by the time over which it changes Section Summary 8.1 Linear Momentum and Force \u2022 Linear momentum (momentum for brevity) is defined as the product of a system\u2019s mass multiplied by its velocity. \u2022 In symbols, linear momentum p is defined to be where is the mass of the system and v is its velocity. \u2022 The SI unit for momentum is kg \u00b7 m/s. p = v, \u2022 Newton\u2019s second law of motion in terms of momentum states that the net external force equals the change in momentum of a system divided by the time over which it changes. In symbols, Newton\u2019s second law of motion is defined to be \u2022 Fnet is the net external force, \u0394p is the change in momentum, and \u0394 is the change time. Fnet = \u0394p \u0394, 8.2 Impulse \u2022 Impulse, or change in momentum, equals the average net external force multiplied by the time this force acts: \u2022 Forces are usually not constant over a period of time. \u0394p = Fnet\u0394. 8.3 Conservation of Momentum \u2022 The conservation of momentum principle is written or ptot = constant ptot = p\u2032tot (isolated system), ptot is the initial total momentum and p\u2032tot is the total momentum some time later. \u2022 An isolated system is defined to be one for which the net external force is zero Fnet = 0. \u2022 During projectile motion and where air resistance is negligible, momentum is conserved in the horizontal direction because horizontal forces are zero. \u2022 Conservation of momentum applies only when the net external force is zero. \u2022 The conservation of momentum principle is valid when considering systems of particles. 8.4 Elastic Collisions in One Dimension \u2022 An elastic collision is one that conserves internal kinetic energy. \u2022 Conservation of kinetic energy and momentum together allow the final velocities to be calculated in terms of initial velocities and masses in one dimensional two-body collisions. 8.5 Inelastic Collisions in One Dimension", " \u2022 An inelastic collision is one in which the internal kinetic energy changes (it is not conserved). \u2022 A collision in which the objects stick together is sometimes called perfectly inelastic because it reduces internal kinetic energy more than does any other type of inelastic collision. \u2022 Sports science and technologies also use physics concepts such as momentum and rotational motion and vibrations. 8.6 Collisions of Point Masses in Two Dimensions \u2022 The approach to two-dimensional collisions is to choose a convenient coordinate system and break the motion into components along perpendicular axes. Choose a coordinate system with the -axis parallel to the velocity of the incoming particle. \u2022 Two-dimensional collisions of point masses where mass 2 is initially at rest conserve momentum along the initial direction of mass 1 (the -axis), stated by 1 1 = 1\u20321 cos 1 + 2\u20322 cos 2 and along the direction perpendicular to the initial direction (the -axis) stated by 0 = 1\u20321 +2\u20322. \u2022 The internal kinetic before and after the collision of two objects that have equal masses is 2 = 1 2 + 1 2 + \u20321\u20322 cos 1 \u2212 2. 2\u20321 2\u20322 1 21 This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 8 | Linear Momentum and Collisions 345 \u2022 Point masses are structureless particles that cannot spin. 8.7 Introduction to Rocket Propulsion \u2022 Newton\u2019s third law of motion states that to every action, there is an equal and opposite reaction. \u2022 Acceleration of a rocket is = e \u2212. \u0394 \u0394 \u2022 A rocket\u2019s acceleration depends on three main factors. They are 1. The greater the exhaust velocity of the gases, the greater the acceleration. 2. The faster the rocket burns its fuel, the greater its acceleration. 3. The smaller the rocket's mass, the greater the acceleration. Conceptual Questions 8.1 Linear Momentum and Force 1. An object that has a small mass and an object that has a large mass have the same momentum. Which object has the largest kinetic energy? 2. An object that has a small mass and an object that has a large mass have the same kinetic energy. Which mass has the largest momentum? 3. Professional Application Football coaches advise players to block, hit, and tackle with their feet on the ground rather than by leaping through the air. Using the concepts of momentum, work, and energy, explain how a football player can be more effective with his feet on the ground. 4. How can a small force impart the same momentum to an object as a large force? 8.2 Impulse 5. Professional Application Explain in terms of impulse how padding reduces forces in a collision. State this in terms of a real example, such as the advantages of a carpeted vs. tile floor for a day care center. 6. While jumping on a trampoline, sometimes you land on your back and other times on your feet. In which case can you reach a greater height and why? 7. Professional Application Tennis racquets have \u201csweet spots.\u201d If the ball hits a sweet spot then the player's arm is not jarred as much as it would be otherwise. Explain why this is the case. 8.3 Conservation of Momentum 8. Professional Application If you dive into water, you reach greater depths than if you do a belly flop. Explain this difference in depth using the concept of conservation of energy. Explain this difference in depth using what you have learned in this chapter. 9. Under what circumstances is momentum conserved? 10. Can momentum be conserved for a system if there are external forces acting on the system? If so, under what conditions? If not, why not? 11. Momentum for a system can be conserved in one direction while not being conserved in another. What is the angle between the directions? Give an example. 12. Professional Application Explain in terms of momentum and Newton\u2019s laws how a car\u2019s air resistance is due in part to the fact that it pushes air in its direction of motion. 13. Can objects in a system have momentum while the momentum of the system is zero? Explain your answer. 14. Must the total energy of a system be conserved whenever its momentum is conserved? Explain why or why not. 8.4 Elastic Collisions in One Dimension 15. What is an elastic collision? 8.5 Inelastic Collisions in One Dimension 16. What is an inelastic collision? What is a perfectly inelastic collision? 17. Mixed-pair ice skaters performing in a show are standing motionless at arms length just before starting a routine. They reach out, clasp hands, and pull themselves together by only using their arms. Assuming there is no friction between the blades of their skates and the ice, what is their velocity after their bodies meet? 346 Chapter 8 | Linear Momentum and Collisions 18. A small pickup truck that has", " a camper shell slowly coasts toward a red light with negligible friction. Two dogs in the back of the truck are moving and making various inelastic collisions with each other and the walls. What is the effect of the dogs on the motion of the center of mass of the system (truck plus entire load)? What is their effect on the motion of the truck? 8.6 Collisions of Point Masses in Two Dimensions 19. Figure 8.19 shows a cube at rest and a small object heading toward it. (a) Describe the directions (angle 1 ) at which the small object can emerge after colliding elastically with the cube. How does 1 depend on, the so-called impact parameter? Ignore any effects that might be due to rotation after the collision, and assume that the cube is much more massive than the small object. (b) Answer the same questions if the small object instead collides with a massive sphere. Figure 8.19 A small object approaches a collision with a much more massive cube, after which its velocity has the direction 1. The angles at which the small object can be scattered are determined by the shape of the object it strikes and the impact parameter. 8.7 Introduction to Rocket Propulsion 20. Professional Application Suppose a fireworks shell explodes, breaking into three large pieces for which air resistance is negligible. How is the motion of the center of mass affected by the explosion? How would it be affected if the pieces experienced significantly more air resistance than the intact shell? 21. Professional Application During a visit to the International Space Station, an astronaut was positioned motionless in the center of the station, out of reach of any solid object on which he could exert a force. Suggest a method by which he could move himself away from this position, and explain the physics involved. 22. Professional Application It is possible for the velocity of a rocket to be greater than the exhaust velocity of the gases it ejects. When that is the case, the gas velocity and gas momentum are in the same direction as that of the rocket. How is the rocket still able to obtain thrust by ejecting the gases? This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 8 | Linear Momentum and Collisions 347 Problems & Exercises 8.1 Linear Momentum and Force 1. (a) Calculate the momentum of a 2000-kg elephant charging a hunter at a speed of 7.50 m/s. (b) Compare the elephant\u2019s momentum with the momentum of a 0.0400-kg tranquilizer dart fired at a speed of 600 m/s. (c) What is the momentum of the 90.0-kg hunter running at 7.40 m/s after missing the elephant? 2. (a) What is the mass of a large ship that has a momentum of 1.60\u00d7109 kg \u00b7 m/s, when the ship is moving at a speed of 48.0 km/h? (b) Compare the ship\u2019s momentum to the momentum of a 1100-kg artillery shell fired at a speed of 1200 m/s. 3. (a) At what speed would a 2.00\u00d7104-kg airplane have to fly to have a momentum of 1.60\u00d7109 kg \u00b7 m/s (the same as the ship\u2019s momentum in the problem above)? (b) What is the plane\u2019s momentum when it is taking off at a speed of 60.0 m/s? (c) If the ship is an aircraft carrier that launches these airplanes with a catapult, discuss the implications of your answer to (b) as it relates to recoil effects of the catapult on the ship. 4. (a) What is the momentum of a garbage truck that is 1.20\u00d7104 kg and is moving at 10.0 m/s? (b) At what speed would an 8.00-kg trash can have the same momentum as the truck? 5. A runaway train car that has a mass of 15,000 kg travels at a speed of 5.4 m/s down a track. Compute the time required for a force of 1500 N to bring the car to rest. 6. The mass of Earth is 5.972\u00d71024 kg and its orbital radius is an average of 1.496\u00d71011 m. Calculate its linear momentum. 8.2 Impulse 7. A bullet is accelerated down the barrel of a gun by hot gases produced in the combustion of gun powder. What is the average force exerted on a 0.0300-kg bullet to accelerate it to a speed of 600 m/s in a time of 2.00 ms (milliseconds)? 8. Professional Application A car moving at 10 m/s crashes into a tree and stops in 0.26 s. Calculate the force the seat belt exerts on a passenger in the car to bring him to a halt. The mass of the passenger is 70 kg. 9", ". A person slaps her leg with her hand, bringing her hand to rest in 2.50 milliseconds from an initial speed of 4.00 m/s. (a) What is the average force exerted on the leg, taking the effective mass of the hand and forearm to be 1.50 kg? (b) Would the force be any different if the woman clapped her hands together at the same speed and brought them to rest in the same time? Explain why or why not. 10. Professional Application A professional boxer hits his opponent with a 1000-N horizontal blow that lasts for 0.150 s. (a) Calculate the impulse imparted by this blow. (b) What is the opponent\u2019s final velocity, if his mass is 105 kg and he is motionless in midair when struck near his center of mass? (c) Calculate the recoil velocity of the opponent\u2019s 10.0-kg head if hit in this manner, assuming the head does not initially transfer significant momentum to the boxer\u2019s body. (d) Discuss the implications of your answers for parts (b) and (c). 11. Professional Application Suppose a child drives a bumper car head on into the side rail, which exerts a force of 4000 N on the car for 0.200 s. (a) What impulse is imparted by this force? (b) Find the final velocity of the bumper car if its initial velocity was 2.80 m/s and the car plus driver have a mass of 200 kg. You may neglect friction between the car and floor. 12. Professional Application One hazard of space travel is debris left by previous missions. There are several thousand objects orbiting Earth that are large enough to be detected by radar, but there are far greater numbers of very small objects, such as flakes of paint. Calculate the force exerted by a 0.100-mg chip of paint that strikes a spacecraft window at a relative speed of 4.00\u00d7103 m/s, given the collision lasts 6.00\u00d710 \u2013 8 s. 13. Professional Application A 75.0-kg person is riding in a car moving at 20.0 m/s when the car runs into a bridge abutment. (a) Calculate the average force on the person if he is stopped by a padded dashboard that compresses an average of 1.00 cm. (b) Calculate the average force on the person if he is stopped by an air bag that compresses an average of 15.0 cm. 14. Professional Application Military rifles have a mechanism for reducing the recoil forces of the gun on the person firing it. An internal part recoils over a relatively large distance and is stopped by damping mechanisms in the gun. The larger distance reduces the average force needed to stop the internal part. (a) Calculate the recoil velocity of a 1.00-kg plunger that directly interacts with a 0.0200-kg bullet fired at 600 m/s from the gun. (b) If this part is stopped over a distance of 20.0 cm, what average force is exerted upon it by the gun? (c) Compare this to the force exerted on the gun if the bullet is accelerated to its velocity in 10.0 ms (milliseconds). 15. A cruise ship with a mass of 1.00\u00d7107 kg strikes a pier at a speed of 0.750 m/s. It comes to rest 6.00 m later, damaging the ship, the pier, and the tugboat captain\u2019s finances. Calculate the average force exerted on the pier using the concept of impulse. (Hint: First calculate the time it took to bring the ship to rest.) 16. Calculate the final speed of a 110-kg rugby player who is initially running at 8.00 m/s but collides head-on with a padded goalpost and experiences a backward force of 1.76\u00d7104 N for 5.50\u00d710\u20132 s. 17. Water from a fire hose is directed horizontally against a wall at a rate of 50.0 kg/s and a speed of 42.0 m/s. Calculate the magnitude of the force exerted on the wall, assuming the water\u2019s horizontal momentum is reduced to zero. 18. A 0.450-kg hammer is moving horizontally at 7.00 m/s when it strikes a nail and comes to rest after driving the nail 1.00 cm into a board. (a) Calculate the duration of the impact. (b) What was the average force exerted on the nail? 19. Starting with the definitions of momentum and kinetic energy, derive an equation for the kinetic energy of a particle expressed as a function of its momentum. 348 Chapter 8 | Linear Momentum and Collisions 20. A ball with an initial velocity of 10 m/s moves at an angle 60\u00ba above the -direction. The ball hits a vertical wall and bounces off so that it is moving 60\u00ba above the \u2212", " -direction with the same speed. What is the impulse delivered by the wall? 21. When serving a tennis ball, a player hits the ball when its velocity is zero (at the highest point of a vertical toss). The racquet exerts a force of 540 N on the ball for 5.00 ms, giving it a final velocity of 45.0 m/s. Using these data, find the mass of the ball. 22. A punter drops a ball from rest vertically 1 meter down onto his foot. The ball leaves the foot with a speed of 18 m/s at an angle 55\u00ba above the horizontal. What is the impulse delivered by the foot (magnitude and direction)? 8.3 Conservation of Momentum 23. Professional Application Train cars are coupled together by being bumped into one another. Suppose two loaded train cars are moving toward one another, the first having a mass of 150,000 kg and a velocity of 0.300 m/s, and the second having a mass of 110,000 kg and a velocity of \u22120.120 m/s. (The minus indicates direction of motion.) What is their final velocity? 24. Suppose a clay model of a koala bear has a mass of 0.200 kg and slides on ice at a speed of 0.750 m/s. It runs into another clay model, which is initially motionless and has a mass of 0.350 kg. Both being soft clay, they naturally stick together. What is their final velocity? 25. Professional Application Consider the following question: A car moving at 10 m/s crashes into a tree and stops in 0.26 s. Calculate the force the seatbelt exerts on a passenger in the car to bring him to a halt. The mass of the passenger is 70 kg. Would the answer to this question be different if the car with the 70-kg passenger had collided with a car that has a mass equal to and is traveling in the opposite direction and at the same speed? Explain your answer. 26. What is the velocity of a 900-kg car initially moving at 30.0 m/s, just after it hits a 150-kg deer initially running at 12.0 m/s in the same direction? Assume the deer remains on the car. 27. A 1.80-kg falcon catches a 0.650-kg dove from behind in midair. What is their velocity after impact if the falcon\u2019s velocity is initially 28.0 m/s and the dove\u2019s velocity is 7.00 m/s in the same direction? which it came. What would their final velocities be in this case? 8.5 Inelastic Collisions in One Dimension 31. A 0.240-kg billiard ball that is moving at 3.00 m/s strikes the bumper of a pool table and bounces straight back at 2.40 m/s (80% of its original speed). The collision lasts 0.0150 s. (a) Calculate the average force exerted on the ball by the bumper. (b) How much kinetic energy in joules is lost during the collision? (c) What percent of the original energy is left? 32. During an ice show, a 60.0-kg skater leaps into the air and is caught by an initially stationary 75.0-kg skater. (a) What is their final velocity assuming negligible friction and that the 60.0-kg skater\u2019s original horizontal velocity is 4.00 m/s? (b) How much kinetic energy is lost? 33. Professional Application Using mass and speed data from Example 8.1 and assuming that the football player catches the ball with his feet off the ground with both of them moving horizontally, calculate: (a) the final velocity if the ball and player are going in the same direction and (b) the loss of kinetic energy in this case. (c) Repeat parts (a) and (b) for the situation in which the ball and the player are going in opposite directions. Might the loss of kinetic energy be related to how much it hurts to catch the pass? 34. A battleship that is 6.00\u00d7107 kg and is originally at rest fires a 1100-kg artillery shell horizontally with a velocity of 575 m/s. (a) If the shell is fired straight aft (toward the rear of the ship), there will be negligible friction opposing the ship\u2019s recoil. Calculate its recoil velocity. (b) Calculate the increase in internal kinetic energy (that is, for the ship and the shell). This energy is less than the energy released by the gun powder\u2014significant heat transfer occurs. 35. Professional Application Two manned satellites approaching one another, at a relative speed of 0.250 m/s, intending to dock. The first has a mass of 4.00\u00d7103 kg, and the second a mass of 7.50\u00d7103 kg. (a) Calcul", "ate the final velocity (after docking) by using the frame of reference in which the first satellite was originally at rest. (b) What is the loss of kinetic energy in this inelastic collision? (c) Repeat both parts by using the frame of reference in which the second satellite was originally at rest. Explain why the change in velocity is different in the two frames, whereas the change in kinetic energy is the same in both. 8.4 Elastic Collisions in One Dimension 36. Professional Application 28. Two identical objects (such as billiard balls) have a onedimensional collision in which one is initially motionless. After the collision, the moving object is stationary and the other moves with the same speed as the other originally had. Show that both momentum and kinetic energy are conserved. 29. Professional Application Two manned satellites approach one another at a relative speed of 0.250 m/s, intending to dock. The first has a mass of 4.00\u00d7103 kg, and the second a mass of 7.50\u00d7103 kg. If the two satellites collide elastically rather than dock, what is their final relative velocity? A 30,000-kg freight car is coasting at 0.850 m/s with negligible friction under a hopper that dumps 110,000 kg of scrap metal into it. (a) What is the final velocity of the loaded freight car? (b) How much kinetic energy is lost? 37. Professional Application Space probes may be separated from their launchers by exploding bolts. (They bolt away from one another.) Suppose a 4800-kg satellite uses this method to separate from the 1500-kg remains of its launcher, and that 5000 J of kinetic energy is supplied to the two parts. What are their subsequent velocities using the frame of reference in which they were at rest before separation? 30. A 70.0-kg ice hockey goalie, originally at rest, catches a 0.150-kg hockey puck slapped at him at a velocity of 35.0 m/ s. Suppose the goalie and the ice puck have an elastic collision and the puck is reflected back in the direction from 38. A 0.0250-kg bullet is accelerated from rest to a speed of 550 m/s in a 3.00-kg rifle. The pain of the rifle\u2019s kick is much worse if you hold the gun loosely a few centimeters from your shoulder rather than holding it tightly against your shoulder. This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 8 | Linear Momentum and Collisions 349 (a) Calculate the recoil velocity of the rifle if it is held loosely away from the shoulder. (b) How much kinetic energy does the rifle gain? (c) What is the recoil velocity if the rifle is held tightly against the shoulder, making the effective mass 28.0 kg? (d) How much kinetic energy is transferred to the rifleshoulder combination? The pain is related to the amount of kinetic energy, which is significantly less in this latter situation. (e) Calculate the momentum of a 110-kg football player running at 8.00 m/s. Compare the player\u2019s momentum with the momentum of a hard-thrown 0.410-kg football that has a speed of 25.0 m/s. Discuss its relationship to this problem. 39. Professional Application One of the waste products of a nuclear reactor is plutonium-239. This nucleus is radioactive and 239 Pu decays by splitting into a helium-4 nucleus and a uranium-235 nucleus, the latter of which is also 4 He + 235 U radioactive and will itself decay some time later. The energy emitted in the plutonium decay is 8.40\u00d710 \u2013 13 J and is entirely converted to kinetic energy of the helium and uranium nuclei. The mass of the helium nucleus is 6.68\u00d710 \u2013 27 kg, while that of the uranium is 3.92\u00d710 \u2013 25 kg (note that the ratio of the masses is 4 to 235). (a) Calculate the velocities of the two nuclei, assuming the plutonium nucleus is originally at rest. (b) How much kinetic energy does each nucleus carry away? Note that the data given here are accurate to three digits only. 40. Professional Application The Moon\u2019s craters are remnants of meteorite collisions. Suppose a fairly large asteroid that has a mass of 5.00\u00d71012 kg (about a kilometer across) strikes the Moon at a speed of 15.0 km/s. (a) At what speed does the Moon recoil after the perfectly inelastic collision (the mass of the Moon is 7.36\u00d71022 kg )? (b) How much kinetic energy is lost in the collision? Such an event may have been observed by medieval English monks who reported observing a red glow and subsequent haze about the Moon. (c) In October 2009, NASA crashed a rocket into the Moon, and analyzed the plume produced", " by the impact. (Significant amounts of water were detected.) Answer part (a) and (b) for this real-life experiment. The mass of the rocket was 2000 kg and its speed upon impact was 9000 km/h. How does the plume produced alter these results? 41. Professional Application Two football players collide head-on in midair while trying to catch a thrown football. The first player is 95.0 kg and has an initial velocity of 6.00 m/s, while the second player is 115 kg and has an initial velocity of \u20133.50 m/s. What is their velocity just after impact if they cling together? 42. What is the speed of a garbage truck that is 1.20\u00d7104 kg and is initially moving at 25.0 m/s just after it hits and adheres to a trash can that is 80.0 kg and is initially at rest? 43. During a circus act, an elderly performer thrills the crowd by catching a cannon ball shot at him. The cannon ball has a mass of 10.0 kg and the horizontal component of its velocity is 8.00 m/s when the 65.0-kg performer catches it. If the performer is on nearly frictionless roller skates, what is his recoil velocity? 44. (a) During an ice skating performance, an initially motionless 80.0-kg clown throws a fake barbell away. The clown\u2019s ice skates allow her to recoil frictionlessly. If the clown recoils with a velocity of 0.500 m/s and the barbell is thrown with a velocity of 10.0 m/s, what is the mass of the barbell? (b) How much kinetic energy is gained by this maneuver? (c) Where does the kinetic energy come from? 8.6 Collisions of Point Masses in Two Dimensions 45. Two identical pucks collide on an air hockey table. One puck was originally at rest. (a) If the incoming puck has a speed of 6.00 m/s and scatters to an angle of 30.0\u00ba,what is the velocity (magnitude and direction) of the second puck? (You may use the result that 1 \u2212 2 = 90\u00ba for elastic collisions of objects that have identical masses.) (b) Confirm that the collision is elastic. 46. Confirm that the results of the example Example 8.7 do conserve momentum in both the - and -directions. 47. A 3000-kg cannon is mounted so that it can recoil only in the horizontal direction. (a) Calculate its recoil velocity when it fires a 15.0-kg shell at 480 m/s at an angle of 20.0\u00ba above the horizontal. (b) What is the kinetic energy of the cannon? This energy is dissipated as heat transfer in shock absorbers that stop its recoil. (c) What happens to the vertical component of momentum that is imparted to the cannon when it is fired? 48. Professional Application A 5.50-kg bowling ball moving at 9.00 m/s collides with a 0.850-kg bowling pin, which is scattered at an angle of 85.0\u00ba to the initial direction of the bowling ball and with a speed of 15.0 m/s. (a) Calculate the final velocity (magnitude and direction) of the bowling ball. (b) Is the collision elastic? (c) Linear kinetic energy is greater after the collision. Discuss how spin on the ball might be converted to linear kinetic energy in the collision. 49. Professional Application Ernest Rutherford (the first New Zealander to be awarded the Nobel Prize in Chemistry) demonstrated that nuclei were very small and dense by scattering helium-4 nuclei from 4 He 197 Au gold-197 nuclei. The energy of the incoming helium nucleus was 8.00\u00d710\u221213 J, and the masses of the helium and gold nuclei were 6.68\u00d710\u221227 kg and 3.29\u00d710\u221225 kg, respectively (note that their mass ratio is 4 to 197). (a) If a helium nucleus scatters to an angle of 120\u00ba during an elastic collision with a gold nucleus, calculate the helium nucleus\u2019s final speed and the final velocity (magnitude and direction) of the gold nucleus. (b) What is the final kinetic energy of the helium nucleus? 50. Professional Application Two cars collide at an icy intersection and stick together afterward. The first car has a mass of 1200 kg and is approaching at 8.00 m/s due south. The second car has a mass of 850 kg and is approaching at 17.0 m/s due west. (a) Calculate the final velocity (magnitude and direction) of the cars. (b) How much kinetic energy is lost in the collision? (This energy goes into deformation of the cars.) Note that 350 Chapter 8 | Linear Momentum and Collisions gravity.", " The mass of the rocket just as it runs out of fuel is 75,000-kg, and its exhaust velocity is 2.40\u00d7103 m/s. Assume that the acceleration of gravity is the same as on Earth\u2019s surface. (b) Why might it be necessary 9.80 m/s2 to limit the acceleration of a rocket? 59. Given the following data for a fire extinguisher-toy wagon rocket experiment, calculate the average exhaust velocity of the gases expelled from the extinguisher. Starting from rest, the final velocity is 10.0 m/s. The total mass is initially 75.0 kg and is 70.0 kg after the extinguisher is fired. 60. How much of a single-stage rocket that is 100,000 kg can be anything but fuel if the rocket is to have a final speed of 8.00 km/s, given that it expels gases at an exhaust velocity of 2.20\u00d7103 m/s? 61. Professional Application (a) A 5.00-kg squid initially at rest ejects 0.250-kg of fluid with a velocity of 10.0 m/s. What is the recoil velocity of the squid if the ejection is done in 0.100 s and there is a 5.00-N frictional force opposing the squid\u2019s movement. (b) How much energy is lost to work done against friction? 62. Unreasonable Results Squids have been reported to jump from the ocean and travel 30.0 m (measured horizontally) before re-entering the water. (a) Calculate the initial speed of the squid if it leaves the water at an angle of 20.0\u00ba, assuming negligible lift from the air and negligible air resistance. (b) The squid propels itself by squirting water. What fraction of its mass would it have to eject in order to achieve the speed found in the previous part? The water is ejected at 12.0 m/s ; gravitational force and friction are neglected. (c) What is unreasonable about the results? (d) Which premise is unreasonable, or which premises are inconsistent? 63. Construct Your Own Problem Consider an astronaut in deep space cut free from her space ship and needing to get back to it. The astronaut has a few packages that she can throw away to move herself toward the ship. Construct a problem in which you calculate the time it takes her to get back by throwing all the packages at one time compared to throwing them one at a time. Among the things to be considered are the masses involved, the force she can exert on the packages through some distance, and the distance to the ship. 64. Construct Your Own Problem Consider an artillery projectile striking armor plating. Construct a problem in which you find the force exerted by the projectile on the plate. Among the things to be considered are the mass and speed of the projectile and the distance over which its speed is reduced. Your instructor may also wish for you to consider the relative merits of depleted uranium versus lead projectiles based on the greater density of uranium. because both cars have an initial velocity, you cannot use the equations for conservation of momentum along the -axis and -axis; instead, you must look for other simplifying aspects. 51. Starting with equations 1 1 = 1\u20321 cos 1 + 2\u20322 cos 2 and 0 = 1\u20321 sin 1 + 2\u20322 sin 2 for conservation of momentum in the - and -directions and assuming that one object is originally stationary, prove that for an elastic collision of two objects of equal masses, 2+ 1 1 21 as discussed in the text. 2+\u20321\u20322 cos 1 \u2212 2 2\u20322 2\u20321 2 = 1 52. Integrated Concepts A 90.0-kg ice hockey player hits a 0.150-kg puck, giving the puck a velocity of 45.0 m/s. If both are initially at rest and if the ice is frictionless, how far does the player recoil in the time it takes the puck to reach the goal 15.0 m away? 8.7 Introduction to Rocket Propulsion 53. Professional Application Antiballistic missiles (ABMs) are designed to have very large accelerations so that they may intercept fast-moving incoming missiles in the short time available. What is the takeoff acceleration of a 10,000-kg ABM that expels 196 kg of gas per second at an exhaust velocity of 2.50\u00d7103 m/s? 54. Professional Application What is the acceleration of a 5000-kg rocket taking off from the Moon, where the acceleration due to gravity is only 1.6 m/s2, if the rocket expels 8.00 kg of gas per second at an exhaust velocity of 2.20\u00d7103 m/s? 55. Professional Application Calculate the increase in velocity of a 4000-kg space probe that expels 3500 kg of its mass at an exhaust velocity of 2.00\u00d7103 m/s. You may assume the gravitational force is", " negligible at the probe\u2019s location. 56. Professional Application Ion-propulsion rockets have been proposed for use in space. They employ atomic ionization techniques and nuclear energy sources to produce extremely high exhaust velocities, perhaps as great as 8.00\u00d7106 m/s. These techniques allow a much more favorable payload-to-fuel ratio. To illustrate this fact: (a) Calculate the increase in velocity of a 20,000-kg space probe that expels only 40.0-kg of its mass at the given exhaust velocity. (b) These engines are usually designed to produce a very small thrust for a very long time\u2014the type of engine that might be useful on a trip to the outer planets, for example. Calculate the acceleration of such an engine if it expels 4.50\u00d710\u22126 kg/s at the given velocity, assuming the acceleration due to gravity is negligible. 57. Derive the equation for the vertical acceleration of a rocket. 58. Professional Application (a) Calculate the maximum rate at which a rocket can expel gases if its acceleration cannot exceed seven times that of This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 8 | Linear Momentum and Collisions 351 Test Prep for AP\u00ae Courses 8.1 Linear Momentum and Force 1. A boy standing on a frictionless ice rink is initially at rest. He throws a snowball in the +x-direction, and it travels on a ballistic trajectory, hitting the ground some distance away. Which of the following is true about the boy while he is in the act of throwing the snowball? a. He feels an upward force to compensate for the downward trajectory of the snowball. b. He feels a backward force exerted by the snowball he is throwing. c. He feels no net force. d. He feels a forward force, the same force that propels the snowball. 2. A 150-g baseball is initially moving 80 mi/h in the \u2013xdirection. After colliding with a baseball bat for 20 ms, the baseball moves 80 mi/h in the +x-direction. What is the magnitude and direction of the average force exerted by the bat on the baseball? 8.2 Impulse 3. A 1.0-kg ball of putty is released from rest and falls vertically 1.5 m until it strikes a hard floor, where it comes to rest in a 0.045-s time interval. What is the magnitude and direction of the average force exerted on the ball by the floor during the collision? a. 33 N, up b. 120 N, up c. 120 N, down d. 240 N, down 4. A 75-g ball is dropped from rest from a height of 2.2 m. It bounces off the floor and rebounds to a maximum height of 1.7 m. If the ball is in contact with the floor for 0.024 s, what is the magnitude and direction of the average force exerted on the ball by the floor during the collision? 5. A 2.4-kg ceramic bowl falls to the floor. During the 0.018-s impact, the bowl experiences an average force of 750 N from the floor. The bowl is at rest after the impact. From what initial height did the bowl fall? a. 1.6 m b. 2.8 m c. 3.2 m d. 5.6 m 6. Whether or not an object (such as a plate, glass, or bone) breaks upon impact depends on the average force exerted on that object by the surface. When a 1.2-kg glass figure hits the floor, it will break if it experiences an average force of 330 N. When it hits a tile floor, the glass comes to a stop in 0.015 s. From what minimum height must the glass fall to experience sufficient force to break? How would your answer change if the figure were falling to a padded or carpeted surface? Explain. 7. A 2.5-kg block slides across a frictionless table toward a horizontal spring.As the block bounces off the spring, a probe measures the velocity of the block (initially negative, moving away from the probe) over time as follows: Table 8.2 Velocity (m/s) Time (s) \u221212.0 \u221210.0 \u22126.0 0 6.0 10.0 12.0 0 0.10 0.20 0.30 0.40 0.50 0.60 What is the average force exerted on the block by the spring over the entire 0.60-s time interval of the collision? a. 50 N b. 60 N c. 100 N d. 120 N 8. During an automobile crash test, the average force exerted by a solid wall on a 2500-kg car that hits the wall is measured to be 740,000 N over a 0.22-s time interval. What was the initial speed of the car", " prior to the collision, assuming the car is at rest at the end of the time interval? 9. A test car is driving toward a solid crash-test barrier with a speed of 45 mi/h. Two seconds prior to impact, the car begins to brake, but it is still moving when it hits the wall. After the collision with the wall, the car crumples somewhat and comes to a complete stop. In order to estimate the average force exerted by the wall on the car, what information would you need to collect? a. The (negative) acceleration of the car before it hits the wall and the distance the car travels while braking. b. The (negative) acceleration of the car before it hits the wall and the velocity of the car just before impact. c. The velocity of the car just before impact and the duration of the collision with the wall. d. The duration of the collision with the wall and the distance the car travels while braking. 10. Design an experiment to verify the relationship between the average force exerted on an object and the change in momentum of that object. As part of your explanation, list the equipment you would use and describe your experimental setup. What would you measure and how? How exactly would you verify the relationship? Explain. 11. A 22-g puck hits the wall of an air hockey table perpendicular to the wall with an initial speed of 14 m/s.The puck is in contact with the wall for 0.0055 s, and it rebounds from the wall with a speed of 14 m/s in the opposite direction.What is the magnitude of the average force exerted by the wall on the puck? a. 0.308 N b. 0.616 N c. 56 N d. 112 N 12. A 22-g puck hits the wall of an air hockey table perpendicular to the wall with an initial speed of 7 m/s. The puck is in contact with the wall for 0.011 s, and the wall exerts an average force of 28 N on the puck during that time. Calculate the magnitude and direction of the change in momentum of the puck. 13. 352 Chapter 8 | Linear Momentum and Collisions Which of the following will be true about the total momentum of the two cars? It will be greater before the collision. It will be equal before and after the collision. It will be greater after the collision. a. b. c. d. The answer depends on whether the collision is elastic or inelastic. 18. A group of students has two carts, A and B, with wheels that turn with negligible friction. The carts can travel along a straight horizontal track. Cart A has known mass mA. The students are asked to use a one-dimensional collision between the carts to determine the mass of cart B. Before the collision, cart A travels to the right and cart B is initially at rest. After the collision, the carts stick together. a. Describe an experimental procedure to determine the velocities of the carts before and after a collision, including all the additional equipment you would need. You may include a labeled diagram of your setup to help in your description. Indicate what measurements you would take and how you would take them. Include enough detail so that another student could carry out your procedure. b. There will be sources of error in the measurements taken in the experiment, both before and after the collision. For your experimental procedure, will the uncertainty in the calculated value of the mass of cart B be affected more by the error in the measurements taken before the collision or by those taken after the collision, or will it be equally affected by both sets of measurements? Justify your answer. A group of students took measurements for one collision. A graph of the students\u2019 data is shown below. Figure 8.22 The image shows a graph with position in meters on the vertical axis and time in seconds on the horizontal axis. c. Given mA = 0.50 kg, use the graph to calculate the mass of cart B. Explicitly indicate the principles used in your calculations. d. The students are now asked to Consider the kinetic energy changes in an inelastic collision, specifically whether the initial values of one of the physical quantities affect the fraction of mechanical energy dissipated in the collision. How could you modify the experiment to investigate this question? Be sure to explicitly describe the calculations you would make, specifying all equations you would use (but do not actually do any algebra or arithmetic). 19. Cart A is moving with an initial velocity +v (in the positive direction) toward cart B, initially at rest. Both carts have equal mass and are on a frictionless surface. Which of the following Figure 8.20 This is a graph showing the force exerted by a rigid wall versus time. The graph in Figure 8.20 represents the force exerted on a particle during a collision. What is the magnitude of the change in momentum of the particle as a result of the collision? a. 1.2 kg \u2022 m/s b.", " 2.4 kg \u2022 m/s c. 3.6 kg \u2022 m/s d. 4.8 kg \u2022 m/s 14. Figure 8.21 This is a graph showing the force exerted by a rigid wall versus time. The graph in Figure 8.21 represents the force exerted on a particle during a collision. What is the magnitude of the change in momentum of the particle as a result of the collision? 8.3 Conservation of Momentum 15. Which of the following is an example of an open system? a. Two air cars colliding on a track elastically. b. Two air cars colliding on a track and sticking together. c. A bullet being fired into a hanging wooden block and becoming embedded in the block, with the system then acting as a ballistic pendulum. d. A bullet being fired into a hillside and becoming buried in the earth. 16. A 40-kg girl runs across a mat with a speed of 5.0 m/s and jumps onto a 120-kg hanging platform initially at rest, causing the girl and platform to swing back and forth like a pendulum together after her jump. What is the combined velocity of the girl and platform after the jump? What is the combined momentum of the girl and platform both before and after the collision? A 50-kg boy runs across a mat with a speed of 6.0 m/s and collides with a soft barrier on the wall, rebounding off the wall and falling to the ground. The boy is at rest after the collision. What is the momentum of the boy before and after the collision? Is momentum conserved in this collision? Explain. Which of these is an example of an open system and which is an example of a closed system? Explain your answer. 17. A student sets up an experiment to measure the momentum of a system of two air cars, A and B, of equal mass, moving on a linear, frictionless track. Before the collision, car A has a certain speed, and car B is at rest. This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 8 | Linear Momentum and Collisions 353 before, \u2212 2 statements correctly characterizes the velocity of the center of mass of the system before and after the collision? a. + 2 b. + 2 c. + 2 before, 0 after after after before, + 2 d. 0 before, 0 after 20. Cart A is moving with a velocity of +10 m/s toward cart B, which is moving with a velocity of +4 m/s. Both carts have equal mass and are moving on a frictionless surface. The two carts have an inelastic collision and stick together after the collision. Calculate the velocity of the center of mass of the system before and after the collision. If there were friction present in this problem, how would this external force affect the center-of-mass velocity both before and after the collision? 8.4 Elastic Collisions in One Dimension 21. Two cars (A and B) of mass 1.5 kg collide. Car A is initially moving at 12 m/s, and car B is initially moving in the same direction with a speed of 6 m/s. The two cars are moving along a straight line before and after the collision. What will be the change in momentum of this system after the collision? a. \u221227 kg \u2022 m/s b. zero c. +27 kg \u2022 m/s d. It depends on whether the collision is elastic or inelastic. 22. Two cars (A and B) of mass 1.5 kg collide. Car A is initially moving at 24 m/s, and car B is initially moving in the opposite direction with a speed of 12 m/s. The two cars are moving along a straight line before and after the collision. (a) If the two cars have an elastic collision, calculate the change in momentum of the two-car system. (b) If the two cars have a completely inelastic collision, calculate the change in momentum of the two-car system. 23. Puck A (200 g) slides across a frictionless surface to collide with puck B (800 g), initially at rest. The velocity of each puck is measured during the experiment as follows: Table 8.3 Time Velocity A Velocity B 0 1.0 s 2.0 s 3.0 s +8.0 m/s +8.0 m/s 0 0 \u22122.0 m/s +2.5 m/s \u22122.0 m/s +2.5 m/s What is the change in momentum of the center of mass of the system as a result of the collision? a. +1.6 kg\u2022m/s b. +0.8 kg\u2022m/s c. 0 d. \u22121.6 kg\u2022m/s 24. For the table above, calculate the center-of-mass", "g the collision. c. Energy was lost due to friction between the ball and the floor. d. Energy was lost due to the work done by gravity during the motion. 28. A tennis ball strikes a wall with an initial speed of 15 m/s. The ball bounces off the wall but rebounds with slightly less speed (14 m/s) after the collision. Explain (a) what else changed its momentum in response to the ball\u2019s change in momentum so that overall momentum is conserved, and (b) how some of the ball\u2019s kinetic energy was lost. 29. Two objects, A and B, have equal mass. Prior to the collision, mass A is moving 10 m/s in the +x-direction, and mass B is moving 4 m/s in the +x-direction. Which of the following results represents an inelastic collision between A and B? a. After the collision, mass A is at rest, and mass B moves 14 m/s in the +x-direction. b. After the collision, mass A moves 4 m/s in the \u2013x- direction, and mass B moves 18 m/s in the +x-direction. c. After the collision, the two masses stick together and move 7 m/s in the +x-direction. d. After the collision, mass A moves 4 m/s in the +x- direction, and mass B moves 10 m/s in the +x-direction. 30. Mass A is three times more massive than mass B. Mass A is initially moving 12 m/s in the +x-direction. Mass B is initially moving 12 m/s in the \u2013x-direction. Assuming that the collision is elastic, calculate the final velocity of both masses after the collision. Show that your results are consistent with conservation of momentum and conservation of kinetic energy. 31. Two objects (A and B) of equal mass collide elastically. Mass A is initially moving 5.0 m/s in the +x-direction prior to the collision. Mass B is initially moving 3.0 m/s in the \u2013xdirection prior to the collision. After the collision, mass A will be moving with a velocity of 3.0 m/s in the \u2013x-direction. What will be the velocity of mass B after the collision? a. 3.0 m/s in the +x-direction b. 5.0 m/s in the +x-direction c. 3.0 m/s in the \u2013x-direction 354 Chapter 8 | Linear Momentum and Collisions d. 5.0 m/s in the \u2013x-direction 32. Two objects (A and B) of equal mass collide elastically. Mass A is initially moving 4.0 m/s in the +x-direction prior to the collision. Mass B is initially moving 8.0 m/s in the \u2013xdirection prior to the collision. After the collision, mass A will be moving with a velocity of 8.0 m/s in the \u2013x-direction. (a) Use the principle of conservation of momentum to predict the velocity of mass B after the collision. (b) Use the fact that kinetic energy is conserved in elastic collisions to predict the velocity of mass B after the collision. 33. Two objects of equal mass collide. Object A is initially moving in the +x-direction with a speed of 12 m/s, and object B is initially at rest. After the collision, object A is at rest, and object B is moving away with some unknown velocity. There are no external forces acting on the system of two masses. What statement can we make about this collision? a. Both momentum and kinetic energy are conserved. b. Momentum is conserved, but kinetic energy is not conserved. a. There will be no change in the center-of-mass velocity. b. The center-of-mass velocity will decrease by 2 m/s. c. The center-of-mass velocity will decrease by 6 m/s. d. The center-of-mass velocity will decrease by 8 m/s. 40. Mass A (1.0 kg) slides across a frictionless surface with a velocity of 4 m/s in the positive direction. Mass B (1.0 kg) slides across the same surface in the opposite direction with a velocity of \u22128 m/s. The two objects collide and stick together after the collision. Predict how the center-of-mass velocity will change as a result of the collision, and explain your prediction. Calculate the center-of-mass velocity of the system both before and after the collision and explain why it remains the same or why it has changed. 8.5 Inelastic Collisions in One Dimension 41. Mass A (2.0 kg) has an initial velocity of 4 m/s in the +xdirection. Mass B (2.0 kg) has an initial velocity", " of 5 m/s in the \u2013x-direction. If the two masses have an elastic collision, what will be the final velocities of the masses after the collision? c. Neither momentum nor kinetic energy is conserved. d. More information is needed in order to determine which a. Both will move 0.5 m/s in the \u2013x-direction. b. Mass A will stop; mass B will move 9 m/s in the +x- is conserved. 34. Two objects of equal mass collide. Object A is initially moving with a velocity of 15 m/s in the +x-direction, and object B is initially at rest. After the collision, object A is at rest. There are no external forces acting on the system of two masses. (a) Use momentum conservation to deduce the velocity of object B after the collision. (b) Is this collision elastic? Justify your answer. 35. Which of the following statements is true about an inelastic collision? a. Momentum is conserved, and kinetic energy is conserved. b. Momentum is conserved, and kinetic energy is not conserved. c. Momentum is not conserved, and kinetic energy is conserved. d. Momentum is not conserved, and kinetic energy is not conserved. 36. Explain how the momentum and kinetic energy of a system of two colliding objects changes as a result of (a) an elastic collision and (b) an inelastic collision. 37. Figure 8.9 shows the positions of two colliding objects measured before, during, and after a collision. Mass A is 1.0 kg. Mass B is 3.0 kg. Which of the following statements is true? a. This is an elastic collision, with a total momentum of 0 kg \u2022 m/s. b. This is an elastic collision, with a total momentum of 1.67 kg \u2022 m/s. c. This is an inelastic collision, with a total momentum of 0 kg \u2022 m/s. d. This is an inelastic collision, with a total momentum of 1.67 kg \u2022 m/s. 38. For the above graph, determine the initial and final momentum for both objects, assuming mass A is 1.0 kg and mass B is 3.0 kg. Also, determine the initial and final kinetic energies for both objects. Based on your results, explain whether momentum is conserved in this collision, and state whether the collision is elastic or inelastic. 39. Mass A (1.0 kg) slides across a frictionless surface with a velocity of 8 m/s in the positive direction. Mass B (3.0 kg) is initially at rest. The two objects collide and stick together. What will be the change in the center-of-mass velocity of the system as a result of the collision? This content is available for free at http://cnx.org/content/col11844/1.13 direction. c. Mass B will stop; mass A will move 9 m/s in the \u2013x- direction. d. Mass A will move 5 m/s in the \u2013x-direction; mass B will move 4 m/s in the +x-direction. 42. Mass A has an initial velocity of 22 m/s in the +x-direction. Mass B is three times more massive than mass A and has an initial velocity of 22 m/s in the \u2013x-direction. If the two masses have an elastic collision, what will be the final velocities of the masses after the collision? 43. Mass A (2.0 kg) is moving with an initial velocity of 15 m/s in the +x-direction, and it collides with mass B (5.0 kg), initially at rest. After the collision, the two objects stick together and move as one. What is the change in kinetic energy of the system as a result of the collision? a. no change b. decrease by 225 J c. decrease by 161 J d. decrease by 64 J 44. Mass A (2.0 kg) is moving with an initial velocity of 15 m/s in the +x-direction, and it collides with mass B (4.0 kg), initially moving 7.0 m/s in the +x-direction. After the collision, the two objects stick together and move as one. What is the change in kinetic energy of the system as a result of the collision? 45. Mass A slides across a rough table with an initial velocity of 12 m/s in the +x-direction. By the time mass A collides with mass B (a stationary object with equal mass), mass A has slowed to 10 m/s. After the collision, the two objects stick together and move as one. Immediately after the collision, the velocity of the system is measured to be 5 m/s in the +xdirection, and the system eventually slides", " to a stop. Which of the following statements is true about this motion? a. Momentum is conserved during the collision, but it is not conserved during the motion before and after the collision. b. Momentum is not conserved at any time during this analysis. c. Momentum is conserved at all times during this analysis. d. Momentum is not conserved during the collision, but it is conserved during the motion before and after the collision. 46. Mass A is initially moving with a velocity of 12 m/s in the +x-direction. Mass B is twice as massive as mass A and is Chapter 8 | Linear Momentum and Collisions 355 initially at rest. After the two objects collide, the two masses move together as one with a velocity of 4 m/s in the +xdirection. Is momentum conserved in this collision? 47. Mass A is initially moving with a velocity of 24 m/s in the +x-direction. Mass B is twice as massive as mass A and is initially at rest. The two objects experience a totally inelastic collision. What is the final speed of both objects after the collision? a. A is not moving; B is moving 24 m/s in the +x-direction. b. Neither A nor B is moving. c. A is moving 24 m/s in the \u2013x-direction. B is not moving. d. Both A and B are moving together 8 m/s in the +x- direction. 48. Mass A is initially moving with some unknown velocity in the +x-direction. Mass B is twice as massive as mass A and initially at rest. The two objects collide, and after the collision, they move together with a speed of 6 m/s in the +x-direction. (a) Is this collision elastic or inelastic? Explain. (b) Determine the initial velocity of mass A. 49. Mass A is initially moving with a velocity of 2 m/s in the +x-direction. Mass B is initially moving with a velocity of 6 m/s in the \u2013x-direction. The two objects have equal masses. After they collide, mass A moves with a speed of 4 m/s in the \u2013xdirection. What is the final velocity of mass B after the collision? a. 6 m/s in the +x-direction b. 4 m/s in the +x-direction c. zero d. 4 m/s in the \u2013x-direction 50. Mass A is initially moving with a velocity of 15 m/s in the +x-direction. Mass B is twice as massive and is initially moving with a velocity of 10 m/s in the \u2013x-direction. The two objects collide, and after the collision, mass A moves with a speed of 15 m/s in the \u2013x-direction. (a) What is the final velocity of mass B after the collision? (b) Calculate the change in kinetic energy as a result of the collision, assuming mass A is 5.0 kg. 8.6 Collisions of Point Masses in Two Dimensions 51. Two cars of equal mass approach an intersection. Car A is moving east at a speed of 45 m/s. Car B is moving south at a speed of 35 m/s. They collide inelastically and stick together after the collision, moving as one object. Which of the following statements is true about the center-of-mass velocity of this system? a. The center-of-mass velocity will decrease after the collision as a result of lost energy (but not drop to zero). b. The center-of-mass velocity will remain the same after the collision since momentum is conserved. c. The center-of-mass velocity will drop to zero since the two objects stick together. d. The magnitude of the center-of-mass velocity will remain the same, but the direction of the velocity will change. 52. Car A has a mass of 2000 kg and approaches an intersection with a velocity of 38 m/s directed to the east. Car B has a mass of 3500 kg and approaches the intersection with a velocity of 53 m/s directed 63\u00b0 north of east. The two cars collide and stick together after the collision. Will the center-of-mass velocity change as a result of the collision? Explain why or why not. Calculate the center-of-mass velocity before and after the collision. 356 Chapter 8 | Linear Momentum and Collisions This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 9 | Statics and Torque 357 9 STATICS AND TORQUE Figure 9.1 On a short time scale, rocks like these in Australia's Kings Canyon are static, or motionless relative to the Earth. (credit: freeaussiestock.com) Chapter Outline 9.1. The First Condition for Equilibrium 9.2", ". The Second Condition for Equilibrium 9.3. Stability 9.4. Applications of Statics, Including Problem-Solving Strategies 9.5. Simple Machines 9.6. Forces and Torques in Muscles and Joints Connection for AP\u00ae Courses What might desks, bridges, buildings, trees, and mountains have in common? What do these objects have in common with a car moving at a constant velocity? While it may be apparent that the objects in the first group are all motionless relative to Earth, they also share something with the moving car and all objects moving at a constant velocity. All of these objects, stationary and moving, share an acceleration of zero. How can this be? Consider Newton's second law, F = ma. When acceleration is zero, as is the case for both stationary objects and objects moving at a constant velocity, the net external force must also be zero (Big Idea 3). Forces are acting on both stationary objects and on objects moving at a constant velocity, but the forces are balanced. That is, they are in equilibrium. In equilibrium, the net force is zero. The first two sections of this chapter will focus on the two conditions necessary for equilibrium. They will not only help you to distinguish between stationary bridges and cars moving at constant velocity, but will introduce a second equilibrium condition, this time involving rotation. As you explore the second equilibrium condition, you will learn about torque, in support of both Enduring Understanding 3.F and Essential Knowledge 3.F.1. Much like a force, torque provides the capability for acceleration; however, with careful attention, torques may also be balanced and equilibrium can be reached. The remainder of this chapter will discuss a variety of interesting equilibrium applications. From the art of balancing, to simple machines, to the muscles in your body, the world around you relies upon the principles of equilibrium to remain stable. This chapter will help you to see just how closely related these events truly are. 358 Chapter 9 | Statics and Torque The content in this chapter supports: Big Idea 3 The interactions of an object with other objects can be described by forces. Enduring Understanding 3.F A force exerted on an object can cause a torque on that object. Essential Knowledge 3.F.1 Only the force component perpendicular to the line connecting the axis of rotation and the point of application of the force results in a torque about that axis. 9.1 The First Condition for Equilibrium Learning Objectives By the end of this section, you will be able to: \u2022 State the first condition of equilibrium. \u2022 Explain static equilibrium. \u2022 Explain dynamic equilibrium. The first condition necessary to achieve equilibrium is the one already mentioned: the net external force on the system must be zero. Expressed as an equation, this is simply net F = 0 Note that if net is zero, then the net external force in any direction is zero. For example, the net external forces along the typical x- and y-axes are zero. This is written as net = 0 and = 0 Figure 9.2 and Figure 9.3 illustrate situations where net = 0 for both static equilibrium (motionless), and dynamic equilibrium (constant velocity). (9.1) (9.2) Figure 9.2 This motionless person is in static equilibrium. The forces acting on him add up to zero. Both forces are vertical in this case. Figure 9.3 This car is in dynamic equilibrium because it is moving at constant velocity. There are horizontal and vertical forces, but the net external force in any direction is zero. The applied force app between the tires and the road is balanced by air friction, and the weight of the car is supported by the normal forces, here shown to be equal for all four tires. However, it is not sufficient for the net external force of a system to be zero for a system to be in equilibrium. Consider the two situations illustrated in Figure 9.4 and Figure 9.5 where forces are applied to an ice hockey stick lying flat on ice. The net external force is zero in both situations shown in the figure; but in one case, equilibrium is achieved, whereas in the other, it is not. In Figure 9.4, the ice hockey stick remains motionless. But in Figure 9.5, with the same forces applied in different places, This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 9 | Statics and Torque 359 the stick experiences accelerated rotation. Therefore, we know that the point at which a force is applied is another factor in determining whether or not equilibrium is achieved. This will be explored further in the next section. Figure 9.4 An ice hockey stick lying flat on ice with two equal and opposite horizontal forces applied to it. Friction is negligible, and the gravitational force is balanced by the support of the ice (a normal force). Thus, net = 0. Equilibrium is achieved, which is static equilibrium in this case. Figure 9.5 The same forces are applied", " at other points and the stick rotates\u2014in fact, it experiences an accelerated rotation. Here net = 0 but the system is not at equilibrium. Hence, the net = 0 is a necessary\u2014but not sufficient\u2014condition for achieving equilibrium. PhET Explorations: Torque Investigate how torque causes an object to rotate. Discover the relationships between angular acceleration, moment of inertia, angular momentum and torque. Figure 9.6 Torque (http://cnx.org/content/m55176/1.2/torque_en.jar) 9.2 The Second Condition for Equilibrium Learning Objectives By the end of this section, you will be able to: \u2022 State the second condition that is necessary to achieve equilibrium. \u2022 Explain torque and the factors on which it depends. \u2022 Describe the role of torque in rotational mechanics. The information presented in this section supports the following AP\u00ae learning objectives and science practices: \u2022 3.F.1.1 The student is able to use representations of the relationship between force and torque. (S.P. 1.4) \u2022 3.F.1.2 The student is able to compare the torques on an object caused by various forces. (S.P. 1.4) \u2022 3.F.1.3 The student is able to estimate the torque on an object caused by various forces in comparison to other situations. (S.P. 2.3) 360 Torque Chapter 9 | Statics and Torque The second condition necessary to achieve equilibrium involves avoiding accelerated rotation (maintaining a constant angular velocity. A rotating body or system can be in equilibrium if its rate of rotation is constant and remains unchanged by the forces acting on it. To understand what factors affect rotation, let us think about what happens when you open an ordinary door by rotating it on its hinges. Several familiar factors determine how effective you are in opening the door. See Figure 9.7. First of all, the larger the force, the more effective it is in opening the door\u2014obviously, the harder you push, the more rapidly the door opens. Also, the point at which you push is crucial. If you apply your force too close to the hinges, the door will open slowly, if at all. Most people have been embarrassed by making this mistake and bumping up against a door when it did not open as quickly as expected. Finally, the direction in which you push is also important. The most effective direction is perpendicular to the door\u2014we push in this direction almost instinctively. Figure 9.7 Torque is the turning or twisting effectiveness of a force, illustrated here for door rotation on its hinges (as viewed from overhead). Torque has both magnitude and direction. (a) Counterclockwise torque is produced by this force, which means that the door will rotate in a counterclockwise due to F. Note that \u22a5 is the perpendicular distance of the pivot from the line of action of the force. (b) A smaller counterclockwise torque is produced by a smaller force F\u2032 acting at the same distance from the hinges (the pivot point). (c) The same force as in (a) produces a smaller counterclockwise torque when applied at a smaller distance from the hinges. (d) The same force as in (a), but acting in the opposite direction, produces a clockwise torque. (e) A smaller counterclockwise torque is produced by the same magnitude force acting at the same point but in a different direction. Here, is less than 90\u00ba. (f) Torque is zero here since the force just pulls on the hinges, producing no rotation. In this case, = 0\u00ba. The magnitude, direction, and point of application of the force are incorporated into the definition of the physical quantity called torque. Torque is the rotational equivalent of a force. It is a measure of the effectiveness of a force in changing or accelerating a rotation (changing the angular velocity over a period of time). In equation form, the magnitude of torque is defined to be = sin (9.3) where (the Greek letter tau) is the symbol for torque, is the distance from the pivot point to the point where the force is applied, is the magnitude of the force, and is the angle between the force and the vector directed from the point of application to the pivot point, as seen in Figure 9.7 and Figure 9.8. An alternative expression for torque is given in terms of the perpendicular lever arm \u22a5 as shown in Figure 9.7 and Figure 9.8, which is defined as so that \u22a5 = sin = \u22a5. (9.4) (9.5) This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 9 | Statics and Torque 361 Figure 9.8 A force applied to an object can produce a torque, which depends on the location of the pivot point. (a) The", " three factors,, and for pivot point A on a body are shown here\u2014 is the distance from the chosen pivot point to the point where the force is applied, and is the angle between F and the vector directed from the point of application to the pivot point. If the object can rotate around point A, it will rotate counterclockwise. This means that torque is counterclockwise relative to pivot A. (b) In this case, point B is the pivot point. The torque from the applied force will cause a clockwise rotation around point B, and so it is a clockwise torque relative to B. The perpendicular lever arm \u22a5 is the shortest distance from the pivot point to the line along which F acts; it is shown as a dashed line in Figure 9.7 and Figure 9.8. Note that the line segment that defines the distance \u22a5 is perpendicular to F, as its name implies. It is sometimes easier to find or visualize \u22a5 than to find both and. In such cases, it may be more convenient to use \u03c4 = r \u22a5 F rather than = sin for torque, but both are equally valid. The SI unit of torque is newtons times meters, usually written as N \u00b7 m. For example, if you push perpendicular to the door with a force of 40 N at a distance of 0.800 m from the hinges, you exert a torque of 32 N\u00b7m(0.800 m\u00d740 N\u00d7sin 90\u00ba) relative to the hinges. If you reduce the force to 20 N, the torque is reduced to 16 N\u00b7m, and so on. The torque is always calculated with reference to some chosen pivot point. For the same applied force, a different choice for the location of the pivot will give you a different value for the torque, since both and depend on the location of the pivot. Any point in any object can be chosen to calculate the torque about that point. The object may not actually pivot about the chosen \u201cpivot point.\u201d Note that for rotation in a plane, torque has two possible directions. Torque is either clockwise or counterclockwise relative to the chosen pivot point, as illustrated for points B and A, respectively, in Figure 9.8. If the object can rotate about point A, it will rotate counterclockwise, which means that the torque for the force is shown as counterclockwise relative to A. But if the object can rotate about point B, it will rotate clockwise, which means the torque for the force shown is clockwise relative to B. Also, the magnitude of the torque is greater when the lever arm is longer. Making Connections: Pivoting Block A solid block of length d is pinned to a wall on its right end. Three forces act on the block as shown below: FA, FB, and FC. While all three forces are of equal magnitude, and all three are equal distances away from the pivot point, all three forces will create a different torque upon the object. FA is vectored perpendicular to its distance from the pivot point; as a result, the magnitude of its torque can be found by the equation \u03c4=FA*d. Vector FB is parallel to the line connecting the point of application of force and the pivot point. As a result, it does not provide an ability to rotate the object and, subsequently, its torque is zero. FC, however, is directed at an angle \u03f4to the line connecting the point of application of force and the pivot point. In this instance, only the component perpendicular to this line is exerting a torque. This component, labeled F\u22a5, can be found using the equation F\u22a5=FCsin\u03b8. The component of the force parallel to this line, labeled F\u2225, does not provide an ability to rotate the object and, as a result, does not provide a torque. Therefore, the resulting torque created by FC is \u03c4=F\u22a5*d. 362 Chapter 9 | Statics and Torque Figure 9.9 Forces on a block pinned to a wall. A solid block of length d is pinned to a wall on its right end. Three forces act on the block: FA, FB, and FC. Now, the second condition necessary to achieve equilibrium is that the net external torque on a system must be zero. An external torque is one that is created by an external force. You can choose the point around which the torque is calculated. The point can be the physical pivot point of a system or any other point in space\u2014but it must be the same point for all torques. If the second condition (net external torque on a system is zero) is satisfied for one choice of pivot point, it will also hold true for any other choice of pivot point in or out of the system of interest. (This is true only in an inertial frame of reference.) The second condition necessary to achieve equilibrium is stated in equation form as net = 0 (9.6) where", " net means total. Torques, which are in opposite directions are assigned opposite signs. A common convention is to call counterclockwise (ccw) torques positive and clockwise (cw) torques negative. When two children balance a seesaw as shown in Figure 9.10, they satisfy the two conditions for equilibrium. Most people have perfect intuition about seesaws, knowing that the lighter child must sit farther from the pivot and that a heavier child can keep a lighter one off the ground indefinitely. Figure 9.10 Two children balancing a seesaw satisfy both conditions for equilibrium. The lighter child sits farther from the pivot to create a torque equal in magnitude to that of the heavier child. Example 9.1 She Saw Torques On A Seesaw The two children shown in Figure 9.10 are balanced on a seesaw of negligible mass. (This assumption is made to keep the example simple\u2014more involved examples will follow.) The first child has a mass of 26.0 kg and sits 1.60 m from the pivot.(a) If the second child has a mass of 32.0 kg, how far is she from the pivot? (b) What is p, the supporting force exerted by the pivot? Strategy Both conditions for equilibrium must be satisfied. In part (a), we are asked for a distance; thus, the second condition (regarding torques) must be used, since the first (regarding only forces) has no distances in it. To apply the second condition for equilibrium, we first identify the system of interest to be the seesaw plus the two children. We take the supporting pivot to be the point about which the torques are calculated. We then identify all external forces acting on the system. This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 9 | Statics and Torque 363 Solution (a) The three external forces acting on the system are the weights of the two children and the supporting force of the pivot. Let us examine the torque produced by each. Torque is defined to be = sin. (9.7) Here = 90\u00ba, so that sin = 1 for all three forces. That means \u22a5 = for all three. The torques exerted by the three forces are first, second, and third, 1 = 11 2 = \u2013 22 p = pp = 0 \u22c5 p = 0. (9.8) (9.9) (9.10) Note that a minus sign has been inserted into the second equation because this torque is clockwise and is therefore negative by convention. Since p acts directly on the pivot point, the distance p is zero. A force acting on the pivot cannot cause a rotation, just as pushing directly on the hinges of a door will not cause it to rotate. Now, the second condition for equilibrium is that the sum of the torques on both children is zero. Therefore or 2 = \u2013 1, 2 2 = 11. Weight is mass times the acceleration due to gravity. Entering for, we get Solve this for the unknown. The quantities on the right side of the equation are known; thus, 2 is 2 = (1.60 m) 26.0 kg 32.0 kg = 1.30 m. As expected, the heavier child must sit closer to the pivot (1.30 m versus 1.60 m) to balance the seesaw. Solution (b) This part asks for a force p. The easiest way to find it is to use the first condition for equilibrium, which is The forces are all vertical, so that we are dealing with a one-dimensional problem along the vertical axis; hence, the condition can be written as net = 0 net F = 0. (9.11) (9.12) (9.13) (9.14) (9.15) (9.16) (9.17) where we again call the vertical axis the y-axis. Choosing upward to be the positive direction, and using plus and minus signs to indicate the directions of the forces, we see that p \u2013 1 \u2013 2 = 0. This equation yields what might have been guessed at the beginning: p = 1 + 2. So, the pivot supplies a supporting force equal to the total weight of the system: p = 1 + 2. (9.18) (9.19) (9.20) 364 Chapter 9 | Statics and Torque Entering known values gives p = 26.0 kg 9.80 m/s2 + 32.0 kg 9.80 m/s2 (9.21) Discussion = 568 N. The two results make intuitive sense. The heavier child sits closer to the pivot. The pivot supports the weight of the two children. Part (b) can also be solved using the second condition for equilibrium, since both distances are known, but only if the pivot point is chosen to be somewhere other than the location of the seesaw's actual", " pivot! Several aspects of the preceding example have broad implications. First, the choice of the pivot as the point around which torques are calculated simplified the problem. Since p is exerted on the pivot point, its lever arm is zero. Hence, the torque exerted by the supporting force p is zero relative to that pivot point. The second condition for equilibrium holds for any choice of pivot point, and so we choose the pivot point to simplify the solution of the problem. Second, the acceleration due to gravity canceled in this problem, and we were left with a ratio of masses. This will not always be the case. Always enter the correct forces\u2014do not jump ahead to enter some ratio of masses. Third, the weight of each child is distributed over an area of the seesaw, yet we treated the weights as if each force were exerted at a single point. This is not an approximation\u2014the distances 1 and 2 are the distances to points directly below the center of gravity of each child. As we shall see in the next section, the mass and weight of a system can act as if they are located at a single point. Finally, note that the concept of torque has an importance beyond static equilibrium. Torque plays the same role in rotational motion that force plays in linear motion. We will examine this in the next chapter. Take-Home Experiment Take a piece of modeling clay and put it on a table, then mash a cylinder down into it so that a ruler can balance on the round side of the cylinder while everything remains still. Put a penny 8 cm away from the pivot. Where would you need to put two pennies to balance? Three pennies? 9.3 Stability Learning Objectives By the end of this section, you will be able to: \u2022 State the types of equilibrium. \u2022 Describe stable and unstable equilibriums. \u2022 Describe neutral equilibrium. The information presented in this section supports the following AP\u00ae learning objectives and science practices: \u2022 3.F.1.1 The student is able to use representations of the relationship between force and torque. (S.P. 1.4) \u2022 3.F.1.2 The student is able to compare the torques on an object caused by various forces. (S.P. 1.4) \u2022 3.F.1.3 The student is able to estimate the torque on an object caused by various forces in comparison to other situations. (S.P. 2.3) \u2022 3.F.1.4 The student is able to design an experiment and analyze data testing a question about torques in a balanced rigid system. (S.P. 4.1, 4.2, 5.1) \u2022 3.F.1.5 The student is able to calculate torques on a two-dimensional system in static equilibrium, by examining a representation or model (such as a diagram or physical construction). (S.P. 1.4, 2.2) It is one thing to have a system in equilibrium; it is quite another for it to be stable. The toy doll perched on the man's hand in Figure 9.11, for example, is not in stable equilibrium. There are three types of equilibrium: stable, unstable, and neutral. Figures throughout this module illustrate various examples. Figure 9.11 presents a balanced system, such as the toy doll on the man's hand, which has its center of gravity (cg) directly over the pivot, so that the torque of the total weight is zero. This is equivalent to having the torques of the individual parts balanced about the pivot point, in this case the hand. The cgs of the arms, legs, head, and torso are labeled with smaller type. This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 9 | Statics and Torque 365 Figure 9.11 A man balances a toy doll on one hand. A system is said to be in stable equilibrium if, when displaced from equilibrium, it experiences a net force or torque in a direction opposite to the direction of the displacement. For example, a marble at the bottom of a bowl will experience a restoring force when displaced from its equilibrium position. This force moves it back toward the equilibrium position. Most systems are in stable equilibrium, especially for small displacements. For another example of stable equilibrium, see the pencil in Figure 9.12. Figure 9.12 This pencil is in the condition of equilibrium. The net force on the pencil is zero and the total torque about any pivot is zero. A system is in unstable equilibrium if, when displaced, it experiences a net force or torque in the same direction as the displacement from equilibrium. A system in unstable equilibrium accelerates away from its equilibrium position if displaced even slightly. An obvious example is a ball resting on top of a hill. Once displaced, it accelerates away from the crest. See the next several figures for examples of unstable equilibrium. Figure 9.13 If the pencil is displaced slightly to", " the side (counterclockwise), it is no longer in equilibrium. Its weight produces a clockwise torque that returns the pencil to its equilibrium position. 366 Chapter 9 | Statics and Torque Figure 9.14 If the pencil is displaced too far, the torque caused by its weight changes direction to counterclockwise and causes the displacement to increase. Figure 9.15 This figure shows unstable equilibrium, although both conditions for equilibrium are satisfied. Figure 9.16 If the pencil is displaced even slightly, a torque is created by its weight that is in the same direction as the displacement, causing the displacement to increase. A system is in neutral equilibrium if its equilibrium is independent of displacements from its original position. A marble on a flat horizontal surface is an example. Combinations of these situations are possible. For example, a marble on a saddle is stable for displacements toward the front or back of the saddle and unstable for displacements to the side. Figure 9.17 shows another example of neutral equilibrium. This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 9 | Statics and Torque 367 Figure 9.17 (a) Here we see neutral equilibrium. The cg of a sphere on a flat surface lies directly above the point of support, independent of the position on the surface. The sphere is therefore in equilibrium in any location, and if displaced, it will remain put. (b) Because it has a circular cross section, the pencil is in neutral equilibrium for displacements perpendicular to its length. When we consider how far a system in stable equilibrium can be displaced before it becomes unstable, we find that some systems in stable equilibrium are more stable than others. The pencil in Figure 9.12 and the person in Figure 9.18(a) are in stable equilibrium, but become unstable for relatively small displacements to the side. The critical point is reached when the cg is no longer above the base of support. Additionally, since the cg of a person's body is above the pivots in the hips, displacements must be quickly controlled. This control is a central nervous system function that is developed when we learn to hold our bodies erect as infants. For increased stability while standing, the feet should be spread apart, giving a larger base of support. Stability is also increased by lowering one's center of gravity by bending the knees, as when a football player prepares to receive a ball or braces themselves for a tackle. A cane, a crutch, or a walker increases the stability of the user, even more as the base of support widens. Usually, the cg of a female is lower (closer to the ground) than a male. Young children have their center of gravity between their shoulders, which increases the challenge of learning to walk. Figure 9.18 (a) The center of gravity of an adult is above the hip joints (one of the main pivots in the body) and lies between two narrowly-separated feet. Like a pencil standing on its eraser, this person is in stable equilibrium in relation to sideways displacements, but relatively small displacements take his cg outside the base of support and make him unstable. Humans are less stable relative to forward and backward displacements because the feet are not very long. Muscles are used extensively to balance the body in the front-to-back direction. (b) While bending in the manner shown, stability is increased by lowering the center of gravity. Stability is also increased if the base is expanded by placing the feet farther apart. Animals such as chickens have easier systems to control. Figure 9.19 shows that the cg of a chicken lies below its hip joints and between its widely separated and broad feet. Even relatively large displacements of the chicken's cg are stable and result in 368 Chapter 9 | Statics and Torque restoring forces and torques that return the cg to its equilibrium position with little effort on the chicken's part. Not all birds are like chickens, of course. Some birds, such as the flamingo, have balance systems that are almost as sophisticated as that of humans. Figure 9.19 shows that the cg of a chicken is below the hip joints and lies above a broad base of support formed by widelyseparated and large feet. Hence, the chicken is in very stable equilibrium, since a relatively large displacement is needed to render it unstable. The body of the chicken is supported from above by the hips and acts as a pendulum between the hips. Therefore, the chicken is stable for front-to-back displacements as well as for side-to-side displacements. Figure 9.19 The center of gravity of a chicken is below the hip joints. The chicken is in stable equilibrium. The body of the chicken is supported from above by the hips and acts as a pendulum between them. Engineers and architects strive to achieve extremely stable equilibriums for buildings and other systems that must withstand wind, earthquakes, and other forces that displace them", " from equilibrium. Although the examples in this section emphasize gravitational forces, the basic conditions for equilibrium are the same for all types of forces. The net external force must be zero, and the net torque must also be zero. Take-Home Experiment Stand straight with your heels, back, and head against a wall. Bend forward from your waist, keeping your heels and bottom against the wall, to touch your toes. Can you do this without toppling over? Explain why and what you need to do to be able to touch your toes without losing your balance. Is it easier for a woman to do this? 9.4 Applications of Statics, Including Problem-Solving Strategies Learning Objectives By the end of this section, you will be able to: \u2022 Discuss the applications of statics in real life. \u2022 State and discuss various problem-solving strategies in statics. The information presented in this section supports the following AP\u00ae learning objectives and science practices: \u2022 3.F.1.1 The student is able to use representations of the relationship between force and torque. (S.P. 1.4) \u2022 3.F.1.2 The student is able to compare the torques on an object caused by various forces. (S.P. 1.4) \u2022 3.F.1.3 The student is able to estimate the torque on an object caused by various forces in comparison to other situations. (S.P. 2.3) \u2022 3.F.1.4 The student is able to design an experiment and analyze data testing a question about torques in a balanced rigid system. (S.P. 4.1, 4.2, 5.1) \u2022 3.F.1.5 The student is able to calculate torques on a two-dimensional system in static equilibrium, by examining a representation or model (such as a diagram or physical construction). (S.P. 1.4, 2.2) Statics can be applied to a variety of situations, ranging from raising a drawbridge to bad posture and back strain. We begin with a discussion of problem-solving strategies specifically used for statics. Since statics is a special case of Newton's laws, both the general problem-solving strategies and the special strategies for Newton's laws, discussed in Problem-Solving Strategies, still apply. Problem-Solving Strategy: Static Equilibrium Situations 1. The first step is to determine whether or not the system is in static equilibrium. This condition is always the case when the acceleration of the system is zero and accelerated rotation does not occur. 2. It is particularly important to draw a free body diagram for the system of interest. Carefully label all forces, and note their relative magnitudes, directions, and points of application whenever these are known. 3. Solve the problem by applying either or both of the conditions for equilibrium (represented by the equations net = 0 and net = 0, depending on the list of known and unknown factors. If the second condition is involved, This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 9 | Statics and Torque 369 choose the pivot point to simplify the solution. Any pivot point can be chosen, but the most useful ones cause torques by unknown forces to be zero. (Torque is zero if the force is applied at the pivot (then = 0 ), or along a line through the pivot point (then = 0 )). Always choose a convenient coordinate system for projecting forces. 4. Check the solution to see if it is reasonable by examining the magnitude, direction, and units of the answer. The importance of this last step never diminishes, although in unfamiliar applications, it is usually more difficult to judge reasonableness. These judgments become progressively easier with experience. Now let us apply this problem-solving strategy for the pole vaulter shown in the three figures below. The pole is uniform and has a mass of 5.00 kg. In Figure 9.20, the pole's cg lies halfway between the vaulter's hands. It seems reasonable that the force exerted by each hand is equal to half the weight of the pole, or 24.5 N. This obviously satisfies the first condition for equilibrium (net = 0). The second condition (net = 0) is also satisfied, as we can see by choosing the cg to be the pivot point. The weight exerts no torque about a pivot point located at the cg, since it is applied at that point and its lever arm is zero. The equal forces exerted by the hands are equidistant from the chosen pivot, and so they exert equal and opposite torques. Similar arguments hold for other systems where supporting forces are exerted symmetrically about the cg. For example, the four legs of a uniform table each support one-fourth of its weight. In Figure 9.20, a pole vaulter holding a pole with its cg halfway between his hands is", " shown. Each hand exerts a force equal to half the weight of the pole, = = / 2. (b) The pole vaulter moves the pole to his left, and the forces that the hands exert are no longer equal. See Figure 9.20. If the pole is held with its cg to the left of the person, then he must push down with his right hand and up with his left. The forces he exerts are larger here because they are in opposite directions and the cg is at a long distance from either hand. Similar observations can be made using a meter stick held at different locations along its length. Figure 9.20 A pole vaulter holds a pole horizontally with both hands. Figure 9.21 A pole vaulter is holding a pole horizontally with both hands. The center of gravity is near his right hand. 370 Chapter 9 | Statics and Torque Figure 9.22 A pole vaulter is holding a pole horizontally with both hands. The center of gravity is to the left side of the vaulter. If the pole vaulter holds the pole as shown in Figure 9.20, the situation is not as simple. The total force he exerts is still equal to the weight of the pole, but it is not evenly divided between his hands. (If =, then the torques about the cg would not be equal since the lever arms are different.) Logically, the right hand should support more weight, since it is closer to the cg. In fact, if the right hand is moved directly under the cg, it will support all the weight. This situation is exactly analogous to two people carrying a load; the one closer to the cg carries more of its weight. Finding the forces and is straightforward, as the next example shows. If the pole vaulter holds the pole from near the end of the pole (Figure 9.22), the direction of the force applied by the right hand of the vaulter reverses its direction. Example 9.2 What Force Is Needed to Support a Weight Held Near Its CG? For the situation shown in Figure 9.20, calculate: (a), the force exerted by the right hand, and (b), the force exerted by the left hand. The hands are 0.900 m apart, and the cg of the pole is 0.600 m from the left hand. Strategy Figure 9.20 includes a free body diagram for the pole, the system of interest. There is not enough information to use the first condition for equilibrium (net = 0 ), since two of the three forces are unknown and the hand forces cannot be assumed to be equal in this case. There is enough information to use the second condition for equilibrium (net = 0) if the pivot point is chosen to be at either hand, thereby making the torque from that hand zero. We choose to locate the pivot at the left hand in this part of the problem, to eliminate the torque from the left hand. Solution for (a) There are now only two nonzero torques, those from the gravitational force ( w ) and from the push or pull of the right hand ( ). Stating the second condition in terms of clockwise and counterclockwise torques, or the algebraic sum of the torques is zero. Here this is net cw = \u2013net ccw. = \u2013\u03c4w (9.22) (9.23) since the weight of the pole creates a counterclockwise torque and the right hand counters with a clockwise toque. Using the definition of torque, = sin, noting that = 90\u00ba, and substituting known values, we obtain (0.900 m) = (0.600 m)(). Thus, Solution for (b) = (0.667) = 32.7 N. 5.00 kg 9.80 m/s2 (9.24) (9.25) The first condition for equilibrium is based on the free body diagram in the figure. This implies that by Newton's second law: + \u2013 = 0 (9.26) This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 9 | Statics and Torque From this we can conclude: Solving for, we obtain + = = = \u2212 = \u2212 32.7 N 371 (9.27) (9.28) = 5.00 kg 9.80 m/s2 \u2212 32.7 N Discussion FL is seen to be exactly half of, as we might have guessed, since is applied twice as far from the cg as. = 16.3 N If the pole vaulter holds the pole as he might at the start of a run, shown in Figure 9.22, the forces change again. Both are considerably greater, and one force reverses direction. Take-Home Experiment This is an experiment to perform while standing in a bus or a train. Stand facing sideways. How do you", " move your body to readjust the distribution of your mass as the bus accelerates and decelerates? Now stand facing forward. How do you move your body to readjust the distribution of your mass as the bus accelerates and decelerates? Why is it easier and safer to stand facing sideways rather than forward? Note: For your safety (and those around you), make sure you are holding onto something while you carry out this activity! PhET Explorations: Balancing Act Play with objects on a teeter totter to learn about balance. Test what you've learned by trying the Balance Challenge game. Figure 9.23 Balancing Act (http://phet.colorado.edu/en/simulation/balancing-act) 9.5 Simple Machines By the end of this section, you will be able to: \u2022 Describe different simple machines. \u2022 Calculate the mechanical advantage. Learning Objectives The information presented in this section supports the following AP\u00ae learning objectives and science practices: \u2022 3.F.1.1 The student is able to use representations of the relationship between force and torque. (S.P. 1.4) \u2022 3.F.1.2 The student is able to compare the torques on an object caused by various forces. (S.P. 1.4) \u2022 3.F.1.3 The student is able to estimate the torque on an object caused by various forces in comparison to other situations. (S.P. 2.3) \u2022 3.F.1.5 The student is able to calculate torques on a two-dimensional system in static equilibrium, by examining a representation or model (such as a diagram or physical construction). (S.P. 1.4, 2.2) Simple machines are devices that can be used to multiply or augment a force that we apply \u2013 often at the expense of a distance through which we apply the force. The word for \u201cmachine\u201d comes from the Greek word meaning \u201cto help make things easier.\u201d Levers, gears, pulleys, wedges, and screws are some examples of machines. Energy is still conserved for these devices because a machine cannot do more work than the energy put into it. However, machines can reduce the input force that is needed to perform the job. The ratio of output to input force magnitudes for any simple machine is called its mechanical advantage (MA). MA = o i (9.29) 380 Chapter 9 | Statics and Torque Figure 9.31 This figure shows that large forces are exerted by the back muscles and experienced in the vertebrae when a person lifts with their back, since these muscles have small effective perpendicular lever arms. The data shown here are analyzed in the preceding example, Example 9.5. What are the benefits of having most skeletal muscles attached so close to joints? One advantage is speed because small muscle contractions can produce large movements of limbs in a short period of time. Other advantages are flexibility and agility, made possible by the large numbers of joints and the ranges over which they function. For example, it is difficult to imagine a system with biceps muscles attached at the wrist that would be capable of the broad range of movement we vertebrates possess. There are some interesting complexities in real systems of muscles, bones, and joints. For instance, the pivot point in many joints changes location as the joint is flexed, so that the perpendicular lever arms and the mechanical advantage of the system change, too. Thus the force the biceps muscle must exert to hold up a book varies as the forearm is flexed. Similar mechanisms operate in the legs, which explain, for example, why there is less leg strain when a bicycle seat is set at the proper height. The methods employed in this section give a reasonable description of real systems provided enough is known about the dimensions of the system. There are many other interesting examples of force and torque in the body\u2014a few of these are the subject of endof-chapter problems. Glossary center of gravity: the point where the total weight of the body is assumed to be concentrated dynamic equilibrium: velocity are zero a state of equilibrium in which the net external force and torque on a system moving with constant mechanical advantage: the ratio of output to input forces for any simple machine neutral equilibrium: a state of equilibrium that is independent of a system's displacements from its original position perpendicular lever arm: the shortest distance from the pivot point to the line along which F lies SI units of torque: newton times meters, usually written as N\u00b7m stable equilibrium: displacement a system, when displaced, experiences a net force or torque in a direction opposite to the direction of the static equilibrium: a state of equilibrium in which the net external force and torque acting on a system is zero static equilibrium: equilibrium in which the acceleration of the system is zero and accelerated rotation does not occur torque: turning or twisting effectiveness of a force unstable equilibrium: from equilibrium a system, when displaced, experiences a", " net force or torque in the same direction as the displacement Section Summary 9.1 The First Condition for Equilibrium \u2022 Statics is the study of forces in equilibrium. This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 9 | Statics and Torque 381 \u2022 Two conditions must be met to achieve equilibrium, which is defined to be motion without linear or rotational acceleration. \u2022 The first condition necessary to achieve equilibrium is that the net external force on the system must be zero, so that net F = 0. 9.2 The Second Condition for Equilibrium \u2022 The second condition assures those torques are also balanced. Torque is the rotational equivalent of a force in producing a rotation and is defined to be = sin where is torque, is the distance from the pivot point to the point where the force is applied, is the magnitude of the force, and is the angle between F and the vector directed from the point where the force acts to the pivot point. The perpendicular lever arm \u22a5 is defined to be so that \u22a5 = sin = \u22a5. \u2022 The perpendicular lever arm \u22a5 is the shortest distance from the pivot point to the line along which acts. The SI unit for torque is newton-meter (N\u00b7m). The second condition necessary to achieve equilibrium is that the net external torque on a system must be zero: By convention, counterclockwise torques are positive, and clockwise torques are negative. net = 0 9.3 Stability \u2022 A system is said to be in stable equilibrium if, when displaced from equilibrium, it experiences a net force or torque in a direction opposite the direction of the displacement. \u2022 A system is in unstable equilibrium if, when displaced from equilibrium, it experiences a net force or torque in the same direction as the displacement from equilibrium. \u2022 A system is in neutral equilibrium if its equilibrium is independent of displacements from its original position. 9.4 Applications of Statics, Including Problem-Solving Strategies \u2022 Statics can be applied to a variety of situations, ranging from raising a drawbridge to bad posture and back strain. We have discussed the problem-solving strategies specifically useful for statics. Statics is a special case of Newton's laws, both the general problem-solving strategies and the special strategies for Newton's laws, discussed in Problem-Solving Strategies, still apply. 9.5 Simple Machines \u2022 Simple machines are devices that can be used to multiply or augment a force that we apply \u2013 often at the expense of a distance through which we have to apply the force. \u2022 The ratio of output to input forces for any simple machine is called its mechanical advantage \u2022 A few simple machines are the lever, nail puller, wheelbarrow, crank, etc. 9.6 Forces and Torques in Muscles and Joints \u2022 Statics plays an important part in understanding everyday strains in our muscles and bones. \u2022 Many lever systems in the body have a mechanical advantage of significantly less than one, as many of our muscles are attached close to joints. \u2022 Someone with good posture stands or sits in such as way that their center of gravity lies directly above the pivot point in their hips, thereby avoiding back strain and damage to disks. Conceptual Questions 9.1 The First Condition for Equilibrium 1. What can you say about the velocity of a moving body that is in dynamic equilibrium? Draw a sketch of such a body using clearly labeled arrows to represent all external forces on the body. 2. Under what conditions can a rotating body be in equilibrium? Give an example. 9.2 The Second Condition for Equilibrium 3. What three factors affect the torque created by a force relative to a specific pivot point? 4. A wrecking ball is being used to knock down a building. One tall unsupported concrete wall remains standing. If the wrecking ball hits the wall near the top, is the wall more likely to fall over by rotating at its base or by falling straight down? Explain your answer. How is it most likely to fall if it is struck with the same force at its base? Note that this depends on how firmly the wall is attached at its base. 382 Chapter 9 | Statics and Torque 5. Mechanics sometimes put a length of pipe over the handle of a wrench when trying to remove a very tight bolt. How does this help? (It is also hazardous since it can break the bolt.) 9.3 Stability 6. A round pencil lying on its side as in Figure 9.14 is in neutral equilibrium relative to displacements perpendicular to its length. What is its stability relative to displacements parallel to its length? 7. Explain the need for tall towers on a suspension bridge to ensure stable equilibrium. 9.4 Applications of Statics, Including Problem-Solving Strategies 8. When visiting some countries, you may see a person balancing a load on the head. Explain why the center of mass of the load needs to be directly above the person's neck vertebrae. 9", ".5 Simple Machines 9. Scissors are like a double-lever system. Which of the simple machines in Figure 9.24 and Figure 9.25 is most analogous to scissors? 10. Suppose you pull a nail at a constant rate using a nail puller as shown in Figure 9.24. Is the nail puller in equilibrium? What if you pull the nail with some acceleration \u2013 is the nail puller in equilibrium then? In which case is the force applied to the nail puller larger and why? 11. Why are the forces exerted on the outside world by the limbs of our bodies usually much smaller than the forces exerted by muscles inside the body? 12. Explain why the forces in our joints are several times larger than the forces we exert on the outside world with our limbs. Can these forces be even greater than muscle forces (see previous Question)? 9.6 Forces and Torques in Muscles and Joints 13. Why are the forces exerted on the outside world by the limbs of our bodies usually much smaller than the forces exerted by muscles inside the body? 14. Explain why the forces in our joints are several times larger than the forces we exert on the outside world with our limbs. Can these forces be even greater than muscle forces? 15. Certain types of dinosaurs were bipedal (walked on two legs). What is a good reason that these creatures invariably had long tails if they had long necks? 16. Swimmers and athletes during competition need to go through certain postures at the beginning of the race. Consider the balance of the person and why start-offs are so important for races. 17. If the maximum force the biceps muscle can exert is 1000 N, can we pick up an object that weighs 1000 N? Explain your answer. 18. Suppose the biceps muscle was attached through tendons to the upper arm close to the elbow and the forearm near the wrist. What would be the advantages and disadvantages of this type of construction for the motion of the arm? 19. Explain one of the reasons why pregnant women often suffer from back strain late in their pregnancy. This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 9 | Statics and Torque 383 Problems & Exercises 9.2 The Second Condition for Equilibrium 1. (a) When opening a door, you push on it perpendicularly with a force of 55.0 N at a distance of 0.850m from the hinges. What torque are you exerting relative to the hinges? (b) Does it matter if you push at the same height as the hinges? 2. When tightening a bolt, you push perpendicularly on a wrench with a force of 165 N at a distance of 0.140 m from the center of the bolt. (a) How much torque are you exerting in newton \u00d7 meters (relative to the center of the bolt)? (b) Convert this torque to footpounds. 3. Two children push on opposite sides of a door during play. Both push horizontally and perpendicular to the door. One child pushes with a force of 17.5 N at a distance of 0.600 m from the hinges, and the second child pushes at a distance of 0.450 m. What force must the second child exert to keep the door from moving? Assume friction is negligible. 4. Use the second condition for equilibrium (net \u03c4 = 0) to calculate p in Example 9.1, employing any data given or solved for in part (a) of the example. 5. Repeat the seesaw problem in Example 9.1 with the center of mass of the seesaw 0.160 m to the left of the pivot (on the side of the lighter child) and assuming a mass of 12.0 kg for the seesaw. The other data given in the example remain unchanged. Explicitly show how you follow the steps in the Problem-Solving Strategy for static equilibrium. 9.3 Stability 6. Suppose a horse leans against a wall as in Figure 9.32. Calculate the force exerted on the wall assuming that force is horizontal while using the data in the schematic representation of the situation. Note that the force exerted on the wall is equal in magnitude and opposite in direction to the force exerted on the horse, keeping it in equilibrium. The total mass of the horse and rider is 500 kg. Take the data to be accurate to three digits. 8. (a) Calculate the magnitude and direction of the force on each foot of the horse in Figure 9.32 (two are on the ground), assuming the center of mass of the horse is midway between the feet. The total mass of the horse and rider is 500kg. (b) What is the minimum coefficient of friction between the hooves and ground? Note that the force exerted by the wall is horizontal. 9. A person carries a plank of wood 2 m long with one hand pushing down on it at one end with a force F1 and the other hand holding it", " up at 50 cm from the end of the plank with force F2. If the plank has a mass of 20 kg and its center of gravity is at the middle of the plank, what are the magnitudes of the forces F1 and F2? 10. A 17.0-m-high and 11.0-m-long wall under construction and its bracing are shown in Figure 9.33. The wall is in stable equilibrium without the bracing but can pivot at its base. Calculate the force exerted by each of the 10 braces if a strong wind exerts a horizontal force of 650 N on each square meter of the wall. Assume that the net force from the wind acts at a height halfway up the wall and that all braces exert equal forces parallel to their lengths. Neglect the thickness of the wall. Figure 9.33 11. (a) What force must be exerted by the wind to support a 2.50-kg chicken in the position shown in Figure 9.34? (b) What is the ratio of this force to the chicken's weight? (c) Does this support the contention that the chicken has a relatively stable construction? Figure 9.32 7. Two children of mass 20 kg and 30 kg sit balanced on a seesaw with the pivot point located at the center of the seesaw. If the children are separated by a distance of 3 m, at what distance from the pivot point is the small child sitting in order to maintain the balance? Figure 9.34 12. Suppose the weight of the drawbridge in Figure 9.35 is supported entirely by its hinges and the opposite shore, so that its cables are slack. (a) What fraction of the weight is supported by the opposite shore if the point of support is directly beneath the cable attachments? (b) What is the direction and magnitude of the force the hinges exert on the bridge under these circumstances? The mass of the bridge is 2500 kg. 384 Chapter 9 | Statics and Torque Figure 9.35 A small drawbridge, showing the forces on the hinges ( F ), its weight ( w ), and the tension in its wires ( T ). 13. Suppose a 900-kg car is on the bridge in Figure 9.35 with its center of mass halfway between the hinges and the cable attachments. (The bridge is supported by the cables and hinges only.) (a) Find the force in the cables. (b) Find the direction and magnitude of the force exerted by the hinges on the bridge. 14. A sandwich board advertising sign is constructed as shown in Figure 9.36. The sign's mass is 8.00 kg. (a) Calculate the tension in the chain assuming no friction between the legs and the sidewalk. (b) What force is exerted by each side on the hinge? Figure 9.36 A sandwich board advertising sign demonstrates tension. 15. (a) What minimum coefficient of friction is needed between the legs and the ground to keep the sign in Figure 9.36 in the position shown if the chain breaks? (b) What force is exerted by each side on the hinge? 16. A gymnast is attempting to perform splits. From the information given in Figure 9.37, calculate the magnitude and direction of the force exerted on each foot by the floor. This content is available for free at http://cnx.org/content/col11844/1.13 Figure 9.37 A gymnast performs full split. The center of gravity and the various distances from it are shown. 9.4 Applications of Statics, Including ProblemSolving Strategies 17. To get up on the roof, a person (mass 70.0 kg) places a 6.00-m aluminum ladder (mass 10.0 kg) against the house on a concrete pad with the base of the ladder 2.00 m from the house. The ladder rests against a plastic rain gutter, which we can assume to be frictionless. The center of mass of the ladder is 2 m from the bottom. The person is standing 3 m from the bottom. What are the magnitudes of the forces on the ladder at the top and bottom? 18. In Figure 9.22, the cg of the pole held by the pole vaulter is 2.00 m from the left hand, and the hands are 0.700 m apart. Calculate the force exerted by (a) his right hand and (b) his left hand. (c) If each hand supports half the weight of the pole in Figure 9.20, show that the second condition for equilibrium (net = 0) is satisfied for a pivot other than the one located at the center of gravity of the pole. Explicitly show how you follow the steps in the Problem-Solving Strategy for static equilibrium described above. 9.5 Simple Machines 19. What is the mechanical advantage of a nail puller\u2014similar to the one shown in Figure 9.24 \u2014where you exert a force 45 cm from the pivot and the nail is 1.8 cm", " on the other side? What minimum force must you exert to apply a force of 1250 N to the nail? 20. Suppose you needed to raise a 250-kg mower a distance of 6.0 cm above the ground to change a tire. If you had a 2.0-m long lever, where would you place the fulcrum if your force was limited to 300 N? 21. a) What is the mechanical advantage of a wheelbarrow, such as the one in Figure 9.25, if the center of gravity of the wheelbarrow and its load has a perpendicular lever arm of 5.50 cm, while the hands have a perpendicular lever arm of 1.02 m? (b) What upward force should you exert to support the wheelbarrow and its load if their combined mass is 55.0 kg? (c) What force does the wheel exert on the ground? 22. A typical car has an axle with 1.10 cm radius driving a tire with a radius of 27.5 cm. What is its mechanical advantage assuming the very simplified model in Figure 9.26(b)? 23. What force does the nail puller in Exercise 9.19 exert on the supporting surface? The nail puller has a mass of 2.10 kg. Chapter 9 | Statics and Torque 385 equivalent lever system. Calculate the force exerted by the upper leg muscle to lift the mass at a constant speed. Explicitly show how you follow the steps in the ProblemSolving Strategy for static equilibrium in Applications of Statistics, Including Problem-Solving Strategies. 24. If you used an ideal pulley of the type shown in Figure 9.27(a) to support a car engine of mass 115 kg, (a) What would be the tension in the rope? (b) What force must the ceiling supply, assuming you pull straight down on the rope? Neglect the pulley system's mass. 25. Repeat Exercise 9.24 for the pulley shown in Figure 9.27(c), assuming you pull straight up on the rope. The pulley system's mass is 7.00 kg. 9.6 Forces and Torques in Muscles and Joints 26. Verify that the force in the elbow joint in Example 9.4 is 407 N, as stated in the text. 27. Two muscles in the back of the leg pull on the Achilles tendon as shown in Figure 9.38. What total force do they exert? Figure 9.40 A mass is connected by pulleys and wires to the ankle in this exercise device. 30. A person working at a drafting board may hold her head as shown in Figure 9.41, requiring muscle action to support the head. The three major acting forces are shown. Calculate the direction and magnitude of the force supplied by the upper vertebrae FV to hold the head stationary, assuming that this force acts along a line through the center of mass as do the weight and muscle force. Figure 9.38 The Achilles tendon of the posterior leg serves to attach plantaris, gastrocnemius, and soleus muscles to calcaneus bone. 28. The upper leg muscle (quadriceps) exerts a force of 1250 N, which is carried by a tendon over the kneecap (the patella) at the angles shown in Figure 9.39. Find the direction and magnitude of the force exerted by the kneecap on the upper leg bone (the femur). Figure 9.39 The knee joint works like a hinge to bend and straighten the lower leg. It permits a person to sit, stand, and pivot. 29. A device for exercising the upper leg muscle is shown in Figure 9.40, together with a schematic representation of an Figure 9.41 31. We analyzed the biceps muscle example with the angle between forearm and upper arm set at 90\u00ba. Using the same numbers as in Example 9.4, find the force exerted by the biceps muscle when the angle is 120\u00ba and the forearm is in a downward position. 32. Even when the head is held erect, as in Figure 9.42, its center of mass is not directly over the principal point of support (the atlanto-occipital joint). The muscles at the back of the neck should therefore exert a force to keep the head 386 Chapter 9 | Statics and Torque erect. That is why your head falls forward when you fall asleep in the class. (a) Calculate the force exerted by these muscles using the information in the figure. (b) What is the force exerted by the pivot on the head? Figure 9.44 A child being lifted by a father's lower leg. 35. Unlike most of the other muscles in our bodies, the masseter muscle in the jaw, as illustrated in Figure 9.45, is attached relatively far from the joint, enabling large forces to be exerted by the back teeth. (a) Using the information in the figure, calculate the force exerted by the", " lower teeth on the bullet. (b) Calculate the force on the joint. Figure 9.42 The center of mass of the head lies in front of its major point of support, requiring muscle action to hold the head erect. A simplified lever system is shown. 33. A 75-kg man stands on his toes by exerting an upward force through the Achilles tendon, as in Figure 9.43. (a) What is the force in the Achilles tendon if he stands on one foot? (b) Calculate the force at the pivot of the simplified lever system shown\u2014that force is representative of forces in the ankle joint. Figure 9.45 A person clenching a bullet between his teeth. 36. Integrated Concepts Suppose we replace the 4.0-kg book in Exercise 9.31 of the biceps muscle with an elastic exercise rope that obeys Hooke's Law. Assume its force constant = 600 N/m. (a) How much is the rope stretched (past equilibrium) to provide the same force B as in this example? Assume the rope is held in the hand at the same location as the book. (b) What force is on the biceps muscle if the exercise rope is pulled straight up so that the forearm makes an angle of 25\u00ba with the horizontal? Assume the biceps muscle is still perpendicular to the forearm. 37. (a) What force should the woman in Figure 9.46 exert on the floor with each hand to do a push-up? Assume that she moves up at a constant speed. (b) The triceps muscle at the back of her upper arm has an effective lever arm of 1.75 cm, and she exerts force on the floor at a horizontal distance of 20.0 cm from the elbow joint. Calculate the magnitude of the force in each triceps muscle, and compare it to her weight. (c) Figure 9.43 The muscles in the back of the leg pull the Achilles tendon when one stands on one's toes. A simplified lever system is shown. 34. A father lifts his child as shown in Figure 9.44. What force should the upper leg muscle exert to lift the child at a constant speed? This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 9 | Statics and Torque 387 How much work does she do if her center of mass rises 0.240 m? (d) What is her useful power output if she does 25 pushups in one minute? Figure 9.46 A woman doing pushups. 38. You have just planted a sturdy 2-m-tall palm tree in your front lawn for your mother's birthday. Your brother kicks a 500 g ball, which hits the top of the tree at a speed of 5 m/s and stays in contact with it for 10 ms. The ball falls to the ground near the base of the tree and the recoil of the tree is minimal. (a) What is the force on the tree? (b) The length of the sturdy section of the root is only 20 cm. Furthermore, the soil around the roots is loose and we can assume that an effective force is applied at the tip of the 20 cm length. What is the effective force exerted by the end of the tip of the root to keep the tree from toppling? Assume the tree will be uprooted rather than bend. (c) What could you have done to ensure that the tree does not uproot easily? 39. Unreasonable Results Suppose two children are using a uniform seesaw that is 3.00 m long and has its center of mass over the pivot. The first child has a mass of 30.0 kg and sits 1.40 m from the pivot. (a) Calculate where the second 18.0 kg child must sit to balance the seesaw. (b) What is unreasonable about the result? (c) Which premise is unreasonable, or which premises are inconsistent? 40. Construct Your Own Problem Consider a method for measuring the mass of a person's arm in anatomical studies. The subject lies on her back, extends her relaxed arm to the side and two scales are placed below the arm. One is placed under the elbow and the other under the back of her hand. Construct a problem in which you calculate the mass of the arm and find its center of mass based on the scale readings and the distances of the scales from the shoulder joint. You must include a free body diagram of the arm to direct the analysis. Consider changing the position of the scale under the hand to provide more information, if needed. You may wish to consult references to obtain reasonable mass values. Test Prep for AP\u00ae Courses 9.2 The Second Condition for Equilibrium 1. Which of the following is not an example of an object undergoing a torque? a. A car is rounding a bend at a constant speed. b. A merry-go-round increases from rest to a constant rotational speed. c. A pend", "ulum swings back and forth. d. A bowling ball rolls down a bowling alley. 2. Five forces of equal magnitude, labeled A\u2013E, are applied to the object shown below. If the object is anchored at point P, which force will provide the greatest torque? Figure 9.47 Five forces acting on an object. a. Force A 388 b. Force B c. Force C d. Force D e. Force E 9.3 Stability 3. Using the concept of torque, explain why a traffic cone placed on its base is in stable equilibrium, while a traffic cone placed on its tip is in unstable equilibrium. 9.4 Applications of Statics, Including ProblemSolving Strategies 4. A child sits on the end of a playground see-saw. Which of the following values is the most appropriate estimate of the torque created by the child? a. 6 N\u2022m b. 60 N\u2022m c. 600 N\u2022m d. 6000 N\u2022m 5. A group of students is stacking a set of identical books, each one overhanging the one below it by 1 inch. They would like to estimate how many books they could place on top of each other before the stack tipped. What information below would they need to know to make this calculation? Figure 9.48 3 overlapping stacked books. I. The mass of each book II. The width of each book III. The depth of each book a. b. c. d. e. I only I and II only I and III only II only I, II, and III 6. A 10 N board of uniform density is 5 meters long. It is supported on the left by a string bearing a 3 N upward force. In order to prevent the string from breaking, a person must place an upward force of 7 N at a position along the bottom surface of the board. At what distance from its left edge would they need to place this force in order for the board to be in static equilibrium? Chapter 9 | Statics and Torque a. If a 1000 kg car comes to rest at a point 5 meters from the left pier, how much force will the bridge provide to the left and right piers? b. How will FL and FR change as the car drives to the right side of the bridge? 8. An object of unknown mass is provided to a student. Without using a scale, design an experimental procedure detailing how the magnitude of this mass could be experimentally found. Your explanation must include the concept of torque and all steps should be provided in an orderly sequence. You may include a labeled diagram of your setup to help in your description. Include enough detail so that another student could carry out your procedure. 9.5 Simple Machines 9. As a young student, you likely learned that simple machines are capable of increasing the ability to lift and move objects. Now, as an educated AP Physics student, you are aware that this capability is governed by the relationship between force and torque. In the space below, explain why torque is integral to the increase in force created by a simple machine. You may use an example or diagram to assist in your explanation. Be sure to cite the mechanical advantage in your explanation as well. 10. Figure 9.24(a) shows a wheelbarrow being lifted by an applied force Fi. If the wheelbarrow is filled with twenty bricks massing 3 kg each, estimate the value of the applied force Fi. Provide an explanation behind the total weight w and any reasoning toward your final answer. Additionally, provide a range of values over which you feel the force could exist. 9.6 Forces and Torques in Muscles and Joints 11. When you use your hand to raise a 20 lb dumbbell in a curling motion, the force on your bicep muscle is not equal to 20 lb. a. Compare the size of the force placed on your bicep muscle to the force of the 20 lb dumbbell lifted by your hand. Using the concept of torque, which force is greater and explain why the two forces are not identical. b. Does the force placed on your bicep muscle change as you curl the weight closer toward your body? (In other words, is the force on your muscle different when your forearm is 90\u00b0 to your upper arm than when it is 45\u00b0 to your upper arm?) Explain your answer using torque. m m a. b. 3 7 5 2 25 7 30 7 e. 5 m d. c. m m 7. A bridge is supported by two piers located 20 meters apart. Both the left and right piers provide an upward force on the bridge, labeled FL and FR respectively. This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 10 | Rotational Motion and Angular Momentum 389 10 ROTATIONAL MOTION AND ANGULAR MOMENTUM Figure 10.1 The mention of a tornado conjures up images of raw destructive power. Tornadoes blow houses away as if they were made of paper and", " have been known to pierce tree trunks with pieces of straw. They descend from clouds in funnel-like shapes that spin violently, particularly at the bottom where they are most narrow, producing winds as high as 500 km/h. (credit: Daphne Zaras, U.S. National Oceanic and Atmospheric Administration) Chapter Outline 10.1. Angular Acceleration 10.2. Kinematics of Rotational Motion 10.3. Dynamics of Rotational Motion: Rotational Inertia 10.4. Rotational Kinetic Energy: Work and Energy Revisited 10.5. Angular Momentum and Its Conservation 10.6. Collisions of Extended Bodies in Two Dimensions 10.7. Gyroscopic Effects: Vector Aspects of Angular Momentum Connection for AP\u00ae Courses Why do tornados spin? And why do tornados spin so rapidly? The answer is that the air masses that produce tornados are themselves rotating, and when the radii of the air masses decrease, their rate of rotation increases. An ice skater increases her spin in an exactly analogous manner, as seen in Figure 10.2. The skater starts her rotation with outstretched limbs and increases her rate of spin by pulling them in toward her body. The same physics describes the exhilarating spin of a skater and the wrenching force of a tornado. We will find that this is another example of the importance of conservation laws and their role in determining how changes happen in a system, supporting Big Idea 5. The idea that a change of a conserved quantity is always equal to the transfer of that quantity between interacting systems (Enduring Understanding 5.A) is presented for both energy and angular momentum (Enduring Understanding 5.E). The conservation of angular momentum in relation to the external net torque (Essential Knowledge 5.E.1) parallels that of linear momentum conservation in relation to the external net force. The concept of rotational inertia is introduced, a concept that takes into account not only the mass of an object or a system, but also the distribution of mass within the object or system. Therefore, changes in the rotational inertia of a system could lead to changes in the motion (Essential Knowledge 5.E.2) of the system. We shall see that all important aspects of rotational motion either have already been defined for linear motion or have exact analogues in linear motion. Clearly, therefore, force, energy, and power are associated with rotational motion. This supports Big Idea 3, that interactions are described by forces. The ability of forces to cause torques (Enduring Understanding 3.F) is extended to the interactions between objects that result in nonzero net torque. This nonzero net torque in turn causes changes in the rotational motion of an object (Essential Knowledge 3.F.2) and results in changes of the angular momentum of an object (Essential Knowledge 3.F.3). 390 Chapter 10 | Rotational Motion and Angular Momentum Similarly, Big Idea 4, that interactions between systems cause changes in those systems, is supported by the empirical observation that when torques are exerted on rigid bodies these torques cause changes in the angular momentum of the system (Enduring Understanding 4.D). Again, there is a clear analogy between linear and rotational motion in this interaction. Both the angular kinematics variables (angular displacement, angular velocity, and angular acceleration) and the dynamics variables (torque and angular momentum) are vectors with direction depending on whether the rotation is clockwise or counterclockwise with respect to an axis of rotation (Essential Knowledge 4.D.1). The angular momentum of the system can change due to interactions (Essential Knowledge 4.D.2). This change is defined as the product of the average torque and the time interval during which torque is exerted (Essential Knowledge 4.D.3), analogous to the impulse-momentum theorem for linear motion. The concepts in this chapter support: Big Idea 3. The interactions of an object with other objects can be described by forces. Enduring Understanding 3.F. A force exerted on an object can cause a torque on that object. Extended Knowledge 3.F.2. The presence of a net torque along any axis will cause a rigid system to change its rotational motion or an object to change its rotational motion about that axis. Extended Knowledge 3.F.3. A torque exerted on an object can change the angular momentum of an object. Big Idea 4. Interactions between systems can result in changes in those systems. Enduring Understanding 4.D. A net torque exerted on a system by other objects or systems will change the angular momentum of the system. Extended Knowledge 4.D.1. Torque, angular velocity, angular acceleration, and angular momentum are vectors and can be characterized as positive or negative depending upon whether they give rise to or correspond to counterclockwise or clockwise rotation with respect to an axis. Extended Knowledge 4.D.2. The angular momentum of a system may", " change due to interactions with other objects or systems. Extended Knowledge 4.D.3. The change in angular momentum is given by the product of the average torque and the time interval during which the torque is exerted. Big Idea 5. Changes that occur as a result of interactions are constrained by conservation laws. Enduring Understanding 5.A. Certain quantities are conserved, in the sense that the changes of those quantities in a given system are always equal to the transfer of that quantity to or from the system by all possible interactions with other systems. Extended Knowledge 5.A.2. For all systems under all circumstances, energy, charge, linear momentum, and angular momentum are conserved. Enduring Understanding 5.E. The angular momentum of a system is conserved. Extended Knowledge 5.E.1. If the net external torque exerted on the system is zero, the angular momentum of the system does not change. Extended Knowledge 5.E.2. The angular momentum of a system is determined by the locations and velocities of the objects that make up the system. The rotational inertia of an object or system depends upon the distribution of mass within the object or system. Changes in the radius of a system or in the distribution of mass within the system result in changes in the system's rotational inertia, and hence in its angular velocity and linear speed for a given angular momentum. Examples should include elliptical orbits in an Earth-satellite system. Mathematical expressions for the moments of inertia will be provided where needed. Students will not be expected to know the parallel axis theorem. Figure 10.2 This figure skater increases her rate of spin by pulling her arms and her extended leg closer to her axis of rotation. (credit: Luu, Wikimedia Commons) This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 10 | Rotational Motion and Angular Momentum 391 10.1 Angular Acceleration Learning Objectives By the end of this section, you will be able to: \u2022 Describe uniform circular motion. \u2022 Explain nonuniform circular motion. \u2022 Calculate angular acceleration of an object. \u2022 Observe the link between linear and angular acceleration. Uniform Circular Motion and Gravitation discussed only uniform circular motion, which is motion in a circle at constant speed and, hence, constant angular velocity. Recall that angular velocity was defined as the time rate of change of angle : = \u0394 \u0394, (10.1) where is the angle of rotation as seen in Figure 10.3. The relationship between angular velocity and linear velocity was also defined in Rotation Angle and Angular Velocity as = (10.2) or =, where is the radius of curvature, also seen in Figure 10.3. According to the sign convention, the counter clockwise direction is considered as positive direction and clockwise direction as negative (10.3) Figure 10.3 This figure shows uniform circular motion and some of its defined quantities. Angular velocity is not constant when a skater pulls in her arms, when a child starts up a merry-go-round from rest, or when a computer's hard disk slows to a halt when switched off. In all these cases, there is an angular acceleration, in which changes. The faster the change occurs, the greater the angular acceleration. Angular acceleration is defined as the rate of change of angular velocity. In equation form, angular acceleration is expressed as follows: where \u0394 is the change in angular velocity and \u0394 is the change in time. The units of angular acceleration are (rad/s)/s, or rad/s2. If increases, then is positive. If decreases, then is negative. = \u0394 \u0394, (10.4) Example 10.1 Calculating the Angular Acceleration and Deceleration of a Bike Wheel Suppose a teenager puts her bicycle on its back and starts the rear wheel spinning from rest to a final angular velocity of 250 rpm in 5.00 s. (a) Calculate the angular acceleration in rad/s2. (b) If she now slams on the brakes, causing an angular acceleration of \u2013 87.3 rad/s2, how long does it take the wheel to stop? Strategy for (a) The angular acceleration can be found directly from its definition in = \u0394 \u0394 are given. We see that \u0394 is 250 rpm and \u0394 is 5.00 s. Solution for (a) Entering known information into the definition of angular acceleration, we get because the final angular velocity and time 392 Chapter 10 | Rotational Motion and Angular Momentum = \u0394 \u0394 250 rpm 5.00 s =. (10.5) Because \u0394 is in revolutions per minute (rpm) and we want the standard units of rad/s2 for angular acceleration, we need to convert \u0394 from rpm to rad/s: \u0394 = 250 rev min = 26.2rad s. \u22c5 2\u03c0 rad rev \u22c5 1 min 60 sec Entering this quantity into the expression for, we get = \u0394 \u0394 = 26", ".2 rad/s 5.00 s = 5.24 rad/s2. (10.6) (10.7) Strategy for (b) In this part, we know the angular acceleration and the initial angular velocity. We can find the stoppage time by using the definition of angular acceleration and solving for \u0394, yielding \u0394 = \u0394. (10.8) Solution for (b) Here the angular velocity decreases from 26.2 rad/s (250 rpm) to zero, so that \u0394 is \u2013 26.2 rad/s, and is given to be \u2013 87.3 rad/s2. Thus, \u0394 = \u2013 26.2 rad/s \u2013 87.3 rad/s2 = 0.300 s. (10.9) Discussion Note that the angular acceleration as the girl spins the wheel is small and positive; it takes 5 s to produce an appreciable angular velocity. When she hits the brake, the angular acceleration is large and negative. The angular velocity quickly goes to zero. In both cases, the relationships are analogous to what happens with linear motion. For example, there is a large deceleration when you crash into a brick wall\u2014the velocity change is large in a short time interval. If the bicycle in the preceding example had been on its wheels instead of upside-down, it would first have accelerated along the ground and then come to a stop. This connection between circular motion and linear motion needs to be explored. For example, it would be useful to know how linear and angular acceleration are related. In circular motion, linear acceleration is tangent to the circle at the point of interest, as seen in Figure 10.4. Thus, linear acceleration is called tangential acceleration t. Figure 10.4 In circular motion, linear acceleration, occurs as the magnitude of the velocity changes: is tangent to the motion. In the context of circular motion, linear acceleration is also called tangential acceleration t. This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 10 | Rotational Motion and Angular Momentum 393 Linear or tangential acceleration refers to changes in the magnitude of velocity but not its direction. We know from Uniform Circular Motion and Gravitation that in circular motion centripetal acceleration, c, refers to changes in the direction of the velocity but not its magnitude. An object undergoing circular motion experiences centripetal acceleration, as seen in Figure 10.5. Thus, t and c are perpendicular and independent of one another. Tangential acceleration t is directly related to the angular acceleration and is linked to an increase or decrease in the velocity, but not its direction. Figure 10.5 Centripetal acceleration c occurs as the direction of velocity changes; it is perpendicular to the circular motion. Centripetal and tangential acceleration are thus perpendicular to each other. Now we can find the exact relationship between linear acceleration t and angular acceleration. Because linear acceleration is proportional to a change in the magnitude of the velocity, it is defined (as it was in One-Dimensional Kinematics) to be t = \u0394 \u0394 (10.10). For circular motion, note that =, so that t = \u0394() \u0394. The radius is constant for circular motion, and so \u0394() = (\u0394). Thus, By definition, = \u0394 \u0394. Thus, or 10.11) (10.12) (10.13) (10.14) These equations mean that linear acceleration and angular acceleration are directly proportional. The greater the angular acceleration is, the larger the linear (tangential) acceleration is, and vice versa. For example, the greater the angular acceleration of a car's drive wheels, the greater the acceleration of the car. The radius also matters. For example, the smaller a wheel, the smaller its linear acceleration for a given angular acceleration. Example 10.2 Calculating the Angular Acceleration of a Motorcycle Wheel A powerful motorcycle can accelerate from 0 to 30.0 m/s (about 108 km/h) in 4.20 s. What is the angular acceleration of its 0.320-m-radius wheels? (See Figure 10.6.) 394 Chapter 10 | Rotational Motion and Angular Momentum Figure 10.6 The linear acceleration of a motorcycle is accompanied by an angular acceleration of its wheels. Strategy We are given information about the linear velocities of the motorcycle. Thus, we can find its linear acceleration t. Then, the expression = t can be used to find the angular acceleration. Solution The linear acceleration is t = \u0394 \u0394 = 30.0 m/s 4.20 s = 7.14 m/s2. We also know the radius of the wheels. Entering the values for t and into = t, we get = t = 7.14 m/s2 0.320 m = 22.3 rad/s2. (10.15) (10.16) Discussion Units of radians are dimension", "less and appear in any relationship between angular and linear quantities. So far, we have defined three rotational quantities\u2014,, and. These quantities are analogous to the translational quantities,, and. Table 10.1 displays rotational quantities, the analogous translational quantities, and the relationships between them. Table 10.1 Rotational and Translational Quantities Rotational Translational Relationship = = = Making Connections: Take-Home Experiment Sit down with your feet on the ground on a chair that rotates. Lift one of your legs such that it is unbent (straightened out). Using the other leg, begin to rotate yourself by pushing on the ground. Stop using your leg to push the ground but allow the chair to rotate. From the origin where you began, sketch the angle, angular velocity, and angular acceleration of your leg as a function of time in the form of three separate graphs. Estimate the magnitudes of these quantities. Check Your Understanding Angular acceleration is a vector, having both magnitude and direction. How do we denote its magnitude and direction? Illustrate with an example. This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 10 | Rotational Motion and Angular Momentum 395 Solution The magnitude of angular acceleration is and its most common units are rad/s2. The direction of angular acceleration along a fixed axis is denoted by a + or a \u2013 sign, just as the direction of linear acceleration in one dimension is denoted by a + or a \u2013 sign. For example, consider a gymnast doing a forward flip. Her angular momentum would be parallel to the mat and to her left. The magnitude of her angular acceleration would be proportional to her angular velocity (spin rate) and her moment of inertia about her spin axis. PhET Explorations: Ladybug Revolution Join the ladybug in an exploration of rotational motion. Rotate the merry-go-round to change its angle, or choose a constant angular velocity or angular acceleration. Explore how circular motion relates to the bug's x,y position, velocity, and acceleration using vectors or graphs. Figure 10.7 Ladybug Revolution (http://cnx.org/content/m55183/1.2/rotation_en.jar) 10.2 Kinematics of Rotational Motion Learning Objectives By the end of this section, you will be able to: \u2022 Observe the kinematics of rotational motion. \u2022 Derive rotational kinematic equations. \u2022 Evaluate problem solving strategies for rotational kinematics. Just by using our intuition, we can begin to see how rotational quantities like,, and are related to one another. For example, if a motorcycle wheel has a large angular acceleration for a fairly long time, it ends up spinning rapidly and rotates through many revolutions. In more technical terms, if the wheel's angular acceleration is large for a long period of time, then the final angular velocity and angle of rotation are large. The wheel's rotational motion is exactly analogous to the fact that the motorcycle's large translational acceleration produces a large final velocity, and the distance traveled will also be large. Kinematics is the description of motion. The kinematics of rotational motion describes the relationships among rotation angle, angular velocity, angular acceleration, and time. Let us start by finding an equation relating,, and. To determine this equation, we recall a familiar kinematic equation for translational, or straight-line, motion: = 0 + (constant ) (10.17) Note that in rotational motion = t, and we shall use the symbol for tangential or linear acceleration from now on. As in linear kinematics, we assume is constant, which means that angular acceleration is also a constant, because =. Now, let us substitute = and = into the linear equation above: The radius cancels in the equation, yielding = 0 + (constant ) = 0 +. (10.18) (10.19) where 0 is the initial angular velocity. This last equation is a kinematic relationship among,, and \u2014that is, it describes their relationship without reference to forces or masses that may affect rotation. It is also precisely analogous in form to its translational counterpart. Making Connections Kinematics for rotational motion is completely analogous to translational kinematics, first presented in One-Dimensional Kinematics. Kinematics is concerned with the description of motion without regard to force or mass. We will find that translational kinematic quantities, such as displacement, velocity, and acceleration have direct analogs in rotational motion. Starting with the four kinematic equations we developed in One-Dimensional Kinematics, we can derive the following four rotational kinematic equations (presented together with their translational counterparts): 396 Chapter 10 | Rotational Motion and Angular Momentum Table 10.2 Rotational Kinematic Equations Rotational Translational", " \u00af = = 0 + = = 0 + (constant, ) = constant, ) 2 + 2 (constant, ) In these equations, the subscript 0 denotes initial values ( 0, 0, and 0 are initial values), and the average angular velocity - and average velocity - are defined as follows: \u00af = 0 + 2 and \u00af = 0 + 2. (10.20) The equations given above in Table 10.2 can be used to solve any rotational or translational kinematics problem in which and are constant. Problem-Solving Strategy for Rotational Kinematics 1. Examine the situation to determine that rotational kinematics (rotational motion) is involved. Rotation must be involved, but without the need to consider forces or masses that affect the motion. 2. Identify exactly what needs to be determined in the problem (identify the unknowns). A sketch of the situation is useful. 3. Make a list of what is given or can be inferred from the problem as stated (identify the knowns). 4. Solve the appropriate equation or equations for the quantity to be determined (the unknown). It can be useful to think in terms of a translational analog because by now you are familiar with such motion. 5. Substitute the known values along with their units into the appropriate equation, and obtain numerical solutions complete with units. Be sure to use units of radians for angles. 6. Check your answer to see if it is reasonable: Does your answer make sense? Example 10.3 Calculating the Acceleration of a Fishing Reel A deep-sea fisherman hooks a big fish that swims away from the boat pulling the fishing line from his fishing reel. The whole system is initially at rest and the fishing line unwinds from the reel at a radius of 4.50 cm from its axis of rotation. The reel is given an angular acceleration of 110 rad/s2 for 2.00 s as seen in Figure 10.8. (a) What is the final angular velocity of the reel? (b) At what speed is fishing line leaving the reel after 2.00 s elapses? (c) How many revolutions does the reel make? (d) How many meters of fishing line come off the reel in this time? Strategy In each part of this example, the strategy is the same as it was for solving problems in linear kinematics. In particular, known values are identified and a relationship is then sought that can be used to solve for the unknown. Solution for (a) Here and are given and needs to be determined. The most straightforward equation to use is = 0 + because the unknown is already on one side and all other terms are known. That equation states that We are also given that 0 = 0 (it starts from rest), so that = 0 +. = 0 + 110 rad/s2 (2.00s) = 220 rad/s. Solution for (b) Now that is known, the speed can most easily be found using the relationship =, This content is available for free at http://cnx.org/content/col11844/1.13 (10.21) (10.22) (10.23) Chapter 10 | Rotational Motion and Angular Momentum 397 where the radius of the reel is given to be 4.50 cm; thus, = (0.0450 m)(220 rad/s) = 9.90 m/s. (10.24) Note again that radians must always be used in any calculation relating linear and angular quantities. Also, because radians are dimensionless, we have m\u00d7rad = m. Solution for (c) Here, we are asked to find the number of revolutions. Because 1 rev = 2\u03c0 rad, we can find the number of revolutions by finding in radians. We are given and, and we know 0 is zero, so that can be obtained using = 0.500) 110 rad/s2 (2.00 s)2 = 220 rad. Converting radians to revolutions gives = (220 rad) 1 rev 2\u03c0 rad = 35.0 rev. Solution for (d) The number of meters of fishing line is, which can be obtained through its relationship with : = = (0.0450 m)(220 rad) = 9.90 m. Discussion (10.25) (10.26) (10.27) This example illustrates that relationships among rotational quantities are highly analogous to those among linear quantities. We also see in this example how linear and rotational quantities are connected. The answers to the questions are realistic. After unwinding for two seconds, the reel is found to spin at 220 rad/s, which is 2100 rpm. (No wonder reels sometimes make high-pitched sounds.) The amount of fishing line played out is 9.90 m, about right for when the big fish bites. Figure 10.8 Fishing line coming off a rotating reel moves", " linearly. Example 10.3 and Example 10.4 consider relationships between rotational and linear quantities associated with a fishing reel. Example 10.4 Calculating the Duration When the Fishing Reel Slows Down and Stops Now let us consider what happens if the fisherman applies a brake to the spinning reel, achieving an angular acceleration of \u2013 300 rad/s2. How long does it take the reel to come to a stop? Strategy We are asked to find the time for the reel to come to a stop. The initial and final conditions are different from those in the previous problem, which involved the same fishing reel. Now we see that the initial angular velocity is 0 = 220 rad/s and the final angular velocity is zero. The angular acceleration is given to be = \u2212300 rad/s2. Examining the available equations, we see all quantities but t are known in = 0 +, making it easiest to use this equation. Solution The equation states 398 Chapter 10 | Rotational Motion and Angular Momentum = 0 +. We solve the equation algebraically for t, and then substitute the known values as usual, yielding = \u2212 0 = 0 \u2212 220 rad/s \u2212300 rad/s2 = 0.733 s. (10.28) (10.29) Discussion Note that care must be taken with the signs that indicate the directions of various quantities. Also, note that the time to stop the reel is fairly small because the acceleration is rather large. Fishing lines sometimes snap because of the accelerations involved, and fishermen often let the fish swim for a while before applying brakes on the reel. A tired fish will be slower, requiring a smaller acceleration. Example 10.5 Calculating the Slow Acceleration of Trains and Their Wheels Large freight trains accelerate very slowly. Suppose one such train accelerates from rest, giving its 0.350-m-radius wheels an angular acceleration of 0.250 rad/s2. After the wheels have made 200 revolutions (assume no slippage): (a) How far has the train moved down the track? (b) What are the final angular velocity of the wheels and the linear velocity of the train? Strategy In part (a), we are asked to find, and in (b) we are asked to find and. We are given the number of revolutions, the radius of the wheels, and the angular acceleration. Solution for (a) The distance is very easily found from the relationship between distance and rotation angle: Solving this equation for yields =. = Before using this equation, we must convert the number of revolutions into radians, because we are dealing with a relationship between linear and rotational quantities: = (200 rev)2\u03c0 rad 1 rev = 1257 rad. Now we can substitute the known values into = to find the distance the train moved down the track: = = (0.350 m)(1257 rad) = 440 m. (10.30) (10.31) (10.32) (10.33) Solution for (b) We cannot use any equation that incorporates to find, because the equation would have at least two unknown values. 2 + 2 will work, because we know the values for all variables except : The equation 2 = 0 Taking the square root of this equation and entering the known values gives (0.250 rad/s2)(1257 rad) 1 / 2 = 25.1 rad/s. We can find the linear velocity of the train,, through its relationship to : = = (0.350 m)(25.1 rad/s) = 8.77 m/s. Discussion The distance traveled is fairly large and the final velocity is fairly slow (just under 32 km/h). (10.34) (10.35) (10.36) There is translational motion even for something spinning in place, as the following example illustrates. Figure 10.9 shows a fly on the edge of a rotating microwave oven plate. The example below calculates the total distance it travels. This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 10 | Rotational Motion and Angular Momentum 399 Figure 10.9 The image shows a microwave plate. The fly makes revolutions while the food is heated (along with the fly). Example 10.6 Calculating the Distance Traveled by a Fly on the Edge of a Microwave Oven Plate A person decides to use a microwave oven to reheat some lunch. In the process, a fly accidentally flies into the microwave and lands on the outer edge of the rotating plate and remains there. If the plate has a radius of 0.15 m and rotates at 6.0 rpm, calculate the total distance traveled by the fly during a 2.0-min cooking period. (Ignore the start-up and slow-down times.) Strategy \u00af First, find the total number of revolutions, and then the linear distance traveled. = because - is given to be 6.0 rpm. can", " be used to find Solution \u00af Entering known values into = gives = = 6.0 rpm (2.0 min) = 12 rev. As always, it is necessary to convert revolutions to radians before calculating a linear quantity like from an angular quantity like : = (12 rev) 2 rad 1 rev = 75.4 rad. Now, using the relationship between and, we can determine the distance traveled: = = (0.15 m)(75.4 rad) = 11 m. Discussion (10.37) (10.38) (10.39) Quite a trip (if it survives)! Note that this distance is the total distance traveled by the fly. Displacement is actually zero for complete revolutions because they bring the fly back to its original position. The distinction between total distance traveled and displacement was first noted in One-Dimensional Kinematics. Check Your Understanding Rotational kinematics has many useful relationships, often expressed in equation form. Are these relationships laws of physics or are they simply descriptive? (Hint: the same question applies to linear kinematics.) Solution Rotational kinematics (just like linear kinematics) is descriptive and does not represent laws of nature. With kinematics, we can describe many things to great precision but kinematics does not consider causes. For example, a large angular acceleration describes a very rapid change in angular velocity without any consideration of its cause. 400 Chapter 10 | Rotational Motion and Angular Momentum 10.3 Dynamics of Rotational Motion: Rotational Inertia Learning Objectives By the end of this section, you will be able to: \u2022 Understand the relationship between force, mass, and acceleration. \u2022 Study the turning effect of force. \u2022 Study the analogy between force and torque, mass and moment of inertia, and linear acceleration and angular acceleration. The information presented in this section supports the following AP\u00ae learning objectives and science practices: \u2022 4.D.1.1 The student is able to describe a representation and use it to analyze a situation in which several forces exerted on a rotating system of rigidly connected objects change the angular velocity and angular momentum of the system. (S.P. 1.2, 1.4) \u2022 4.D.1.2 The student is able to plan data collection strategies designed to establish that torque, angular velocity, angular acceleration, and angular momentum can be predicted accurately when the variables are treated as being clockwise or counterclockwise with respect to a well-defined axis of rotation, and refine the research question based on the examination of data. (S.P. 3.2, 4.1, 5.1, 5.3) \u2022 5.E.2.1 The student is able to describe or calculate the angular momentum and rotational inertia of a system in terms of the locations and velocities of objects that make up the system. Students are expected to do qualitative reasoning with compound objects. Students are expected to do calculations with a fixed set of extended objects and point masses. (S.P. 2.2) If you have ever spun a bike wheel or pushed a merry-go-round, you know that force is needed to change angular velocity as seen in Figure 10.10. In fact, your intuition is reliable in predicting many of the factors that are involved. For example, we know that a door opens slowly if we push too close to its hinges. Furthermore, we know that the more massive the door, the more slowly it opens. The first example implies that the farther the force is applied from the pivot, the greater the angular acceleration; another implication is that angular acceleration is inversely proportional to mass. These relationships should seem very similar to the familiar relationships among force, mass, and acceleration embodied in Newton's second law of motion. There are, in fact, precise rotational analogs to both force and mass. Figure 10.10 Force is required to spin the bike wheel. The greater the force, the greater the angular acceleration produced. The more massive the wheel, the smaller the angular acceleration. If you push on a spoke closer to the axle, the angular acceleration will be smaller. To develop the precise relationship among force, mass, radius, and angular acceleration, consider what happens if we exert a force on a point mass that is at a distance from a pivot point, as shown in Figure 10.11. Because the force is perpendicular to, an acceleration = is obtained in the direction of. We can rearrange this equation such that = and then look for ways to relate this expression to expressions for rotational quantities. We note that =, and we substitute this expression into =, yielding Recall that torque is the turning effectiveness of a force. In this case, because F is perpendicular to, torque is simply =. So, if we multiply both sides of the equation above by, we get torque on the left-hand side. That is, =. or = 2 = 2\u03b1. (10.40) (10.41) (10.42)", " This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 10 | Rotational Motion and Angular Momentum 401 This last equation is the rotational analog of Newton's second law ( = ), where torque is analogous to force, angular acceleration is analogous to translational acceleration, and 2 is analogous to mass (or inertia). The quantity 2 is called the rotational inertia or moment of inertia of a point mass a distance from the center of rotation. Figure 10.11 An object is supported by a horizontal frictionless table and is attached to a pivot point by a cord that supplies centripetal force. A force is applied to the object perpendicular to the radius, causing it to accelerate about the pivot point. The force is kept perpendicular to. Making Connections: Rotational Motion Dynamics Dynamics for rotational motion is completely analogous to linear or translational dynamics. Dynamics is concerned with force and mass and their effects on motion. For rotational motion, we will find direct analogs to force and mass that behave just as we would expect from our earlier experiences. Rotational Inertia and Moment of Inertia Before we can consider the rotation of anything other than a point mass like the one in Figure 10.11, we must extend the idea of rotational inertia to all types of objects. To expand our concept of rotational inertia, we define the moment of inertia of an object to be the sum of 2 for all the point masses of which it is composed. That is, = \u2211 2. Here is analogous to in translational motion. Because of the distance, the moment of inertia for any object depends on the chosen axis. Actually, calculating is beyond the scope of this text except for one simple case\u2014that of a hoop, which has all its mass at the same distance from its axis. A hoop's moment of inertia around its axis is therefore 2, where is its total mass and its radius. (We use and for an entire object to distinguish them from and for point masses.) In all other cases, we must consult Figure 10.12 (note that the table is piece of artwork that has shapes as well as formulae) for formulas for that have been derived from integration over the continuous body. Note that has units of mass multiplied by distance squared ( kg \u22c5 m2 ), as we might expect from its definition. The general relationship among torque, moment of inertia, and angular acceleration is or net \u03c4 = = net \u03c4, (10.43) (10.44) where net is the total torque from all forces relative to a chosen axis. For simplicity, we will only consider torques exerted by forces in the plane of the rotation. Such torques are either positive or negative and add like ordinary numbers. The relationship in =, = net \u03c4 is the rotational analog to Newton's second law and is very generally applicable. This equation is actually valid for any torque, applied to any object, relative to any axis. As we might expect, the larger the torque is, the larger the angular acceleration is. For example, the harder a child pushes on a merry-go-round, the faster it accelerates. Furthermore, the more massive a merry-go-round, the slower it accelerates for the same torque. The basic relationship between moment of inertia and angular acceleration is that the larger the moment of inertia, the smaller is the angular acceleration. But there is an additional twist. The moment of inertia depends not only on the mass of an object, but also on its distribution of mass relative to the axis around which it rotates. For example, it will be much easier to accelerate a merry-go-round full of children if they stand close to its axis than if they all stand at the outer edge. The mass is the same in both cases; but the moment of inertia is much larger when the children are at the edge. Take-Home Experiment Cut out a circle that has about a 10 cm radius from stiff cardboard. Near the edge of the circle, write numbers 1 to 12 like hours on a clock face. Position the circle so that it can rotate freely about a horizontal axis through its center, like a wheel. (You could loosely nail the circle to a wall.) Hold the circle stationary and with the number 12 positioned at the top, attach a 402 Chapter 10 | Rotational Motion and Angular Momentum lump of blue putty (sticky material used for fixing posters to walls) at the number 3. How large does the lump need to be to just rotate the circle? Describe how you can change the moment of inertia of the circle. How does this change affect the amount of blue putty needed at the number 3 to just rotate the circle? Change the circle's moment of inertia and then try rotating the circle by using different amounts of blue putty. Repeat this process several times. In what direction did the circle rotate when you added putty at the number 3 (clockwise or counterclockwise)? In which of these directions was the", " resulting angular velocity? Was the angular velocity constant? What can we say about the direction (clockwise or counterclockwise) of the angular acceleration? How could you change the placement of the putty to create angular velocity in the opposite direction? Problem-Solving Strategy for Rotational Dynamics 1. Examine the situation to determine that torque and mass are involved in the rotation. Draw a careful sketch of the situation. 2. Determine the system of interest. 3. Draw a free body diagram. That is, draw and label all external forces acting on the system of interest. 4. Apply, the rotational equivalent of Newton's second law, to solve the problem. Care must be taken to use the correct moment of inertia and to consider the torque about the point of rotation. 5. As always, check the solution to see if it is reasonable. Making Connections In statics, the net torque is zero, and there is no angular acceleration. In rotational motion, net torque is the cause of angular acceleration, exactly as in Newton's second law of motion for rotation. This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 10 | Rotational Motion and Angular Momentum 403 Figure 10.12 Some rotational inertias. Example 10.7 Calculating the Effect of Mass Distribution on a Merry-Go-Round Consider the father pushing a playground merry-go-round in Figure 10.13. He exerts a force of 250 N at the edge of the 50.0-kg merry-go-round, which has a 1.50 m radius. Calculate the angular acceleration produced (a) when no one is on the merry-go-round and (b) when an 18.0-kg child sits 1.25 m away from the center. Consider the merry-go-round itself to be a uniform disk with negligible retarding friction. Figure 10.13 A father pushes a playground merry-go-round at its edge and perpendicular to its radius to achieve maximum torque. Strategy Angular acceleration is given directly by the expression = net \u03c4 : 404 Chapter 10 | Rotational Motion and Angular Momentum To solve for, we must first calculate the torque (which is the same in both cases) and moment of inertia (which is greater in the second case). To find the torque, we note that the applied force is perpendicular to the radius and friction is negligible, so that =. (10.45) \u03c4 = sin \u03b8 = (1.50 m)(250 N) = 375 N \u22c5 m. Solution for (a) The moment of inertia of a solid disk about this axis is given in Figure 10.12 to be 1 22, where = 50.0 kg and = 1.50 m, so that Now, after we substitute the known values, we find the angular acceleration to be = (0.500)(50.0 kg)(1.50 m)2 = 56.25 kg \u22c5 m2. = = 375 N \u22c5 m 56.25 kg \u22c5 m2 = 6.67rad s2. Solution for (b) (10.46) (10.47) (10.48) (10.49) We expect the angular acceleration for the system to be less in this part, because the moment of inertia is greater when the child is on the merry-go-round. To find the total moment of inertia, we first find the child's moment of inertia c by considering the child to be equivalent to a point mass at a distance of 1.25 m from the axis. Then, c = 2 = (18.0 kg)(1.25 m)2 = 28.13 kg \u22c5 m2. (10.50) The total moment of inertia is the sum of moments of inertia of the merry-go-round and the child (about the same axis). To justify this sum to yourself, examine the definition of : = 28.13 kg \u22c5 m2 + 56.25 kg \u22c5 m2 = 84.38 kg \u22c5 m2. Substituting known values into the equation for gives = \u03c4 = 375 N \u22c5 m 84.38 kg \u22c5 m2 = 4.44rad s2. (10.51) (10.52) Discussion The angular acceleration is less when the child is on the merry-go-round than when the merry-go-round is empty, as expected. The angular accelerations found are quite large, partly due to the fact that friction was considered to be negligible. If, for example, the father kept pushing perpendicularly for 2.00 s, he would give the merry-go-round an angular velocity of 13.3 rad/s when it is empty but only 8.89 rad/s when the child is on it. In terms of revolutions per second, these angular velocities are 2.12 rev/s and 1.41 rev", "/s, respectively. The father would end up running at about 50 km/h in the first case. Summer Olympics, here he comes! Confirmation of these numbers is left as an exercise for the reader. Making Connections: Multiple Forces on One System A large potter's wheel has a diameter of 60.0 cm and a mass of 8.0 kg. It is powered by a 20.0 N motor acting on the outer edge. There is also a brake capable of exerting a 15.0 N force at a radius of 12.0 cm from the axis of rotation, on the underside. What is the angular acceleration when the motor is in use? The torque is found by = sin = (0.300 m)(20.0 N) = 6.00 N\u00b7m. The moment of inertia is calculated as = 1 2 2 = 1 2 8.0 kg (0.300 m)2 = 0.36 kg \u22c5 m2. Thus, the angular acceleration would be = = 6.00 N \u22c5 m 0.36 kg \u22c5 m2 = 17 rad/s2. Note that the friction is always acting in a direction opposite to the rotation that is currently happening in this system. If the potter makes a mistake and has both the brake and motor on simultaneously, the friction force of the brake will exert a torque opposite that of the motor. This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 10 | Rotational Motion and Angular Momentum 405 The torque from the brake is = sin = (0.120 m)(15.0 N) = 1.80 N\u22c5m. Thus, the net torque is 6.00 N\u22c5m \u2212 1.80 \u039d\u22c5m = 4.20 \u039d\u22c5m. And the angular acceleration is = = 4.20 N \u22c5 m 0.36 kg \u22c5 m2 = 12 rad/s2. Check Your Understanding Torque is the analog of force and moment of inertia is the analog of mass. Force and mass are physical quantities that depend on only one factor. For example, mass is related solely to the numbers of atoms of various types in an object. Are torque and moment of inertia similarly simple? Solution No. Torque depends on three factors: force magnitude, force direction, and point of application. Moment of inertia depends on both mass and its distribution relative to the axis of rotation. So, while the analogies are precise, these rotational quantities depend on more factors. 10.4 Rotational Kinetic Energy: Work and Energy Revisited Learning Objectives By the end of this section, you will be able to: \u2022 Derive the equation for rotational work. \u2022 Calculate rotational kinetic energy. \u2022 Demonstrate the law of conservation of energy. The information presented in this section supports the following AP\u00ae learning objectives and science practices: \u2022 3.F.2.1 The student is able to make predictions about the change in the angular velocity about an axis for an object when forces exerted on the object cause a torque about that axis. (S.P. 6.4) \u2022 3.F.2.2 The student is able to plan data collection and analysis strategies designed to test the relationship between a torque exerted on an object and the change in angular velocity of that object about an axis. (S.P. 4.1, 4.2, 5.1) In this module, we will learn about work and energy associated with rotational motion. Figure 10.14 shows a worker using an electric grindstone propelled by a motor. Sparks are flying, and noise and vibration are created as layers of steel are pared from the pole. The stone continues to turn even after the motor is turned off, but it is eventually brought to a stop by friction. Clearly, the motor had to work to get the stone spinning. This work went into heat, light, sound, vibration, and considerable rotational kinetic energy. Figure 10.14 The motor works in spinning the grindstone, giving it rotational kinetic energy. That energy is then converted to heat, light, sound, and vibration. (credit: U.S. Navy photo by Mass Communication Specialist Seaman Zachary David Bell) Work must be done to rotate objects such as grindstones or merry-go-rounds. Work was defined in Uniform Circular Motion and Gravitation for translational motion, and we can build on that knowledge when considering work done in rotational motion. The simplest rotational situation is one in which the net force is exerted perpendicular to the radius of a disk (as shown in Figure 10.15) and remains perpendicular as the disk starts to rotate. The force is parallel to the displacement, and so the net work done is the product of the force times the arc length traveled: net = (net )\u0394. (10.53) To get torque and other rotational", " quantities into the equation, we multiply and divide the right-hand side of the equation by, and gather terms: 406 Chapter 10 | Rotational Motion and Angular Momentum We recognize that net = net \u03c4 and \u0394 / =, so that net = (net \u03c4). net = ( net )\u0394. (10.54) (10.55) This equation is the expression for rotational work. It is very similar to the familiar definition of translational work as force multiplied by distance. Here, torque is analogous to force, and angle is analogous to distance. The equation net = (net \u03c4) is valid in general, even though it was derived for a special case. To get an expression for rotational kinetic energy, we must again perform some algebraic manipulations. The first step is to note that net \u03c4 =, so that net =. (10.56) Figure 10.15 The net force on this disk is kept perpendicular to its radius as the force causes the disk to rotate. The net work done is thus (net )\u0394. The net work goes into rotational kinetic energy. Making Connections qWork and energy in rotational motion are completely analogous to work and energy in translational motion, first presented in Uniform Circular Motion and Gravitation. Now, we solve one of the rotational kinematics equations for. We start with the equation Next, we solve for : 2 = 0 2 + 2. = 2 \u2212 0 2 2. Substituting this into the equation for net and gathering terms yields net = 1 22 \u2212 1 20 (10.57) (10.58) (10.59) 2. This equation is the work-energy theorem for rotational motion only. As you may recall, net work changes the kinetic energy of a system. Through an analogy with translational motion, we define the term 2 to be rotational kinetic energy KErot for 1 2 an object with a moment of inertia and an angular velocity : KErot = 1 22. (10.60) The expression for rotational kinetic energy is exactly analogous to translational kinetic energy, with being analogous to and to. Rotational kinetic energy has important effects. Flywheels, for example, can be used to store large amounts of rotational kinetic energy in a vehicle, as seen in Figure 10.16. This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 10 | Rotational Motion and Angular Momentum 407 Figure 10.16 Experimental vehicles, such as this bus, have been constructed in which rotational kinetic energy is stored in a large flywheel. When the bus goes down a hill, its transmission converts its gravitational potential energy into KErot. It can also convert translational kinetic energy, when the bus stops, into KErot. The flywheel's energy can then be used to accelerate, to go up another hill, or to keep the bus from going against friction. Example 10.8 Calculating the Work and Energy for Spinning a Grindstone Consider a person who spins a large grindstone by placing her hand on its edge and exerting a force through part of a revolution as shown in Figure 10.17. In this example, we verify that the work done by the torque she exerts equals the change in rotational energy. (a) How much work is done if she exerts a force of 200 N through a rotation of 1.00 rad(57.3\u00ba)? The force is kept perpendicular to the grindstone's 0.320-m radius at the point of application, and the effects of friction are negligible. (b) What is the final angular velocity if the grindstone has a mass of 85.0 kg? (c) What is the final rotational kinetic energy? (It should equal the work.) Strategy To find the work, we can use the equation net = (net \u03c4). We have enough information to calculate the torque and are given the rotation angle. In the second part, we can find the final angular velocity using one of the kinematic relationships. In the last part, we can calculate the rotational kinetic energy from its expression in KErot = 1 22. Solution for (a) The net work is expressed in the equation net = (net \u03c4), where net is the applied force multiplied by the radius () because there is no retarding friction, and the force is perpendicular to. The angle is given. Substituting the given values in the equation above yields net = = (0.320 m)(200 N)(1.00 rad) = 64.0 N \u22c5 m. Noting that 1 N \u00b7 m = 1 J, net = 64.0 J. (10.61) (10.62) (10.63) Figure 10.17 A large grindstone is given a spin by a person grasping its outer edge. Solution for (b) To find from the given information requires more than one step. We start with the kinematic relationship", " in the equation 2 = 0 2 + 2. (10.64) 408 Chapter 10 | Rotational Motion and Angular Momentum Note that 0 = 0 because we start from rest. Taking the square root of the resulting equation gives Now we need to find. One possibility is = (2)1 / 2. = net \u03c4, where the torque is The formula for the moment of inertia for a disk is found in Figure 10.12: net \u03c4 = = (0.320 m)(200 N) = 64.0 N \u22c5 m. = 1 22 = 0.5 85.0 kg (0.320 m)2 = 4.352 kg \u22c5 m2. Substituting the values of torque and moment of inertia into the expression for, we obtain = 64.0 N \u22c5 m 4.352 kg \u22c5 m2 = 14.7rad s2. Now, substitute this value and the given value for into the above expression for : = (2)1 / 2 = 2 14.7rad s2 1 / 2 (1.00 rad) = 5.42rad s. Solution for (c) The final rotational kinetic energy is Both and were found above. Thus, KErot = 1 22. KErot = (0.5) 4.352 kg \u22c5 m2 (5.42 rad/s)2 = 64.0 J. (10.65) (10.66) (10.67) (10.68) (10.69) (10.70) (10.71) (10.72) Discussion The final rotational kinetic energy equals the work done by the torque, which confirms that the work done went into rotational kinetic energy. We could, in fact, have used an expression for energy instead of a kinematic relation to solve part (b). We will do this in later examples. Helicopter pilots are quite familiar with rotational kinetic energy. They know, for example, that a point of no return will be reached if they allow their blades to slow below a critical angular velocity during flight. The blades lose lift, and it is impossible to immediately get the blades spinning fast enough to regain it. Rotational kinetic energy must be supplied to the blades to get them to rotate faster, and enough energy cannot be supplied in time to avoid a crash. Because of weight limitations, helicopter engines are too small to supply both the energy needed for lift and to replenish the rotational kinetic energy of the blades once they have slowed down. The rotational kinetic energy is put into them before takeoff and must not be allowed to drop below this crucial level. One possible way to avoid a crash is to use the gravitational potential energy of the helicopter to replenish the rotational kinetic energy of the blades by losing altitude and aligning the blades so that the helicopter is spun up in the descent. Of course, if the helicopter's altitude is too low, then there is insufficient time for the blade to regain lift before reaching the ground. Take-Home Experiment Rotational motion can be observed in wrenches, clocks, wheels or spools on axels, and seesaws. Choose an object or system that exhibits rotational motion and plan an experiment to test how torque affects angular velocity. How will you create and measure different amounts of torque? How will you measure angular velocity? Remember that net =, \u221d 2, and =. Problem-Solving Strategy for Rotational Energy 1. Determine that energy or work is involved in the rotation. 2. Determine the system of interest. A sketch usually helps. 3. Analyze the situation to determine the types of work and energy involved. This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 10 | Rotational Motion and Angular Momentum 409 4. For closed systems, mechanical energy is conserved. That is, KEi + PEi = KEf + PEf. Note that KEi and KEf may each include translational and rotational contributions. 5. For open systems, mechanical energy may not be conserved, and other forms of energy (referred to previously as ), such as heat transfer, may enter or leave the system. Determine what they are, and calculate them as necessary. 6. Eliminate terms wherever possible to simplify the algebra. 7. Check the answer to see if it is reasonable. Example 10.9 Calculating Helicopter Energies A typical small rescue helicopter, similar to the one in Figure 10.18, has four blades, each is 4.00 m long and has a mass of 50.0 kg. The blades can be approximated as thin rods that rotate about one end of an axis perpendicular to their length. The helicopter has a total loaded mass of 1000 kg. (a) Calculate the rotational kinetic energy in the blades when they rotate at 300 rpm. (b) Calculate the translational kinetic energy of the helicopter when it flies at 20", ".0 m/s, and compare it with the rotational energy in the blades. (c) To what height could the helicopter be raised if all of the rotational kinetic energy could be used to lift it? Strategy Rotational and translational kinetic energies can be calculated from their definitions. The last part of the problem relates to the idea that energy can change form, in this case from rotational kinetic energy to gravitational potential energy. Solution for (a) The rotational kinetic energy is 22. We must convert the angular velocity to radians per second and calculate the moment of inertia before we can find KErot. The angular velocity is KErot = 1 (10.73) = 300 rev 1.00 min \u22c5 2\u03c0 rad 1 rev \u22c5 1.00 min 60.0 s = 31.4rad s. (10.74) The moment of inertia of one blade will be that of a thin rod rotated about its end, found in Figure 10.12. The total is four times this moment of inertia, because there are four blades. Thus, = 42 3 = 4\u00d7 (4.00 m)2 50.0 kg 3 = 1067 kg \u22c5 m2. Entering and into the expression for rotational kinetic energy gives KErot = 0.5(1067 kg \u22c5 m2)(31.4 rad/s)2 = 5.26\u00d7105 J (10.75) (10.76) Solution for (b) Translational kinetic energy was defined in Uniform Circular Motion and Gravitation. Entering the given values of mass and velocity, we obtain KEtrans = 1 22 = (0.5) 1000 kg (20.0 m/s)2 = 2.00\u00d7105 J. To compare kinetic energies, we take the ratio of translational kinetic energy to rotational kinetic energy. This ratio is 2.00\u00d7105 J 5.26\u00d7105 J = 0.380. (10.77) (10.78) Solution for (c) At the maximum height, all rotational kinetic energy will have been converted to gravitational energy. To find this height, we equate those two energies: or KErot = PEgrav 1 22 =. (10.79) (10.80) 410 Chapter 10 | Rotational Motion and Angular Momentum We now solve for and substitute known values into the resulting equation 2 1 2 = = 5.26\u00d7105 J 1000 kg 9.80 m/s2 = 53.7 m. (10.81) Discussion The ratio of translational energy to rotational kinetic energy is only 0.380. This ratio tells us that most of the kinetic energy of the helicopter is in its spinning blades\u2014something you probably would not suspect. The 53.7 m height to which the helicopter could be raised with the rotational kinetic energy is also impressive, again emphasizing the amount of rotational kinetic energy in the blades. Figure 10.18 The first image shows how helicopters store large amounts of rotational kinetic energy in their blades. This energy must be put into the blades before takeoff and maintained until the end of the flight. The engines do not have enough power to simultaneously provide lift and put significant rotational energy into the blades. The second image shows a helicopter from the Auckland Westpac Rescue Helicopter Service. Over 50,000 lives have been saved since its operations beginning in 1973. Here, a water rescue operation is shown. (credit: 111 Emergency, Flickr) Making Connections Conservation of energy includes rotational motion, because rotational kinetic energy is another form of KE. Uniform Circular Motion and Gravitation has a detailed treatment of conservation of energy. How Thick Is the Soup? Or Why Don't All Objects Roll Downhill at the Same Rate? One of the quality controls in a tomato soup factory consists of rolling filled cans down a ramp. If they roll too fast, the soup is too thin. Why should cans of identical size and mass roll down an incline at different rates? And why should the thickest soup roll the slowest? The easiest way to answer these questions is to consider energy. Suppose each can starts down the ramp from rest. Each can starting from rest means each starts with the same gravitational potential energy PEgrav, which is converted entirely to KE, provided each rolls without slipping. KE, however, can take the form of KEtrans or KErot, and total KE is the sum of the two. If a can rolls down a ramp, it puts part of its energy into rotation, leaving less for translation. Thus, the can goes slower than it would if it slid down. Furthermore, the thin soup does not rotate, whereas the thick soup does, because it sticks to the can. The thick soup thus puts more of the can's original gravitational potential energy into rotation than the thin soup, and the can rolls more slowly, as seen in Figure 10.19. This content is available for free at http://cnx.org/content/col11844/", "1.13 Chapter 10 | Rotational Motion and Angular Momentum 411 Figure 10.19 Three cans of soup with identical masses race down an incline. The first can has a low friction coating and does not roll but just slides down the incline. It wins because it converts its entire PE into translational KE. The second and third cans both roll down the incline without slipping. The second can contains thin soup and comes in second because part of its initial PE goes into rotating the can (but not the thin soup). The third can contains thick soup. It comes in third because the soup rotates along with the can, taking even more of the initial PE for rotational KE, leaving less for translational KE. Assuming no losses due to friction, there is only one force doing work\u2014gravity. Therefore the total work done is the change in kinetic energy. As the cans start moving, the potential energy is changing into kinetic energy. Conservation of energy gives PEi = KEf. (10.82) More specifically, or PEgrav = KEtrans + KErot (10.83) 22 + 1 So, the initial is divided between translational kinetic energy and rotational kinetic energy; and the greater is, the less energy goes into translation. If the can slides down without friction, then = 0 and all the energy goes into translation; thus, the can goes faster. = 1 22. (10.84) Take-Home Experiment Locate several cans each containing different types of food. First, predict which can will win the race down an inclined plane and explain why. See if your prediction is correct. You could also do this experiment by collecting several empty cylindrical containers of the same size and filling them with different materials such as wet or dry sand. Example 10.10 Calculating the Speed of a Cylinder Rolling Down an Incline Calculate the final speed of a solid cylinder that rolls down a 2.00-m-high incline. The cylinder starts from rest, has a mass of 0.750 kg, and has a radius of 4.00 cm. Strategy We can solve for the final velocity using conservation of energy, but we must first express rotational quantities in terms of translational quantities to end up with as the only unknown. Solution Conservation of energy for this situation is written as described above: 22 + 1 Before we can solve for, we must get an expression for from Figure 10.12. Because and are related (note here that the cylinder is rolling without slipping), we must also substitute the relationship = / into the expression. These substitutions yield = 1 22. (10.85) = 1 22 + 1 2 1 22 2 2. Interestingly, the cylinder's radius and mass cancel, yielding 42 = 3 22 + 1 Solving algebraically, the equation for the final velocity gives = 1 42. (10.86) (10.87) 412 Chapter 10 | Rotational Motion and Angular Momentum = 4 3 1 / 2. Substituting known values into the resulting expression yields = 9.80 m/s2 4 3 (2.00 m) 1 / 2 = 5.11 m/s. (10.88) (10.89) Discussion Because and cancel, the result = 1 / 2 4 3 is valid for any solid cylinder, implying that all solid cylinders will roll down an incline at the same rate independent of their masses and sizes. (Rolling cylinders down inclines is what Galileo actually did to show that objects fall at the same rate independent of mass.) Note that if the cylinder slid without friction down the incline without rolling, then the entire gravitational potential energy would go into translational kinetic energy. Thus, 1 22 = and = (2)1 / 2, which is 22% greater than (4 / 3)1 / 2. That is, the cylinder would go faster at the bottom. Check Your Understanding Analogy of Rotational and Translational Kinetic Energy Is rotational kinetic energy completely analogous to translational kinetic energy? What, if any, are their differences? Give an example of each type of kinetic energy. Solution Yes, rotational and translational kinetic energy are exact analogs. They both are the energy of motion involved with the coordinated (non-random) movement of mass relative to some reference frame. The only difference between rotational and translational kinetic energy is that translational is straight line motion while rotational is not. An example of both kinetic and translational kinetic energy is found in a bike tire while being ridden down a bike path. The rotational motion of the tire means it has rotational kinetic energy while the movement of the bike along the path means the tire also has translational kinetic energy. If you were to lift the front wheel of the bike and spin it while the bike is stationary, then the wheel would have only rotational kinetic energy relative to the Earth. PhET Explorations: My Solar System Build your own system of heavenly bodies and watch the gravitational ballet. With this orbit", " simulator, you can set initial positions, velocities, and masses of 2, 3, or 4 bodies, and then see them orbit each other. Figure 10.20 My Solar System (http://cnx.org/content/m55188/1.4/my-solar-system_en.jar) 10.5 Angular Momentum and Its Conservation Learning Objectives By the end of this section, you will be able to: \u2022 Understand the analogy between angular momentum and linear momentum. \u2022 Observe the relationship between torque and angular momentum. \u2022 Apply the law of conservation of angular momentum. The information presented in this section supports the following AP\u00ae learning objectives and science practices: \u2022 4.D.2.1 The student is able to describe a model of a rotational system and use that model to analyze a situation in which angular momentum changes due to interaction with other objects or systems. (S.P. 1.2, 1.4) \u2022 4.D.2.2 The student is able to plan a data collection and analysis strategy to determine the change in angular momentum of a system and relate it to interactions with other objects and systems. (S.P. 2.2) This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 10 | Rotational Motion and Angular Momentum 413 \u2022 4.D.3.1 The student is able to use appropriate mathematical routines to calculate values for initial or final angular momentum, or change in angular momentum of a system, or average torque or time during which the torque is exerted in analyzing a situation involving torque and angular momentum. (S.P. 2.2) \u2022 4.D.3.2 The student is able to plan a data collection strategy designed to test the relationship between the change in angular momentum of a system and the product of the average torque applied to the system and the time interval during which the torque is exerted. (S.P. 4.1, 4.2) \u2022 5.E.1.1 The student is able to make qualitative predictions about the angular momentum of a system for a situation in which there is no net external torque. (S.P. 6.4, 7.2) \u2022 5.E.1.2 The student is able to make calculations of quantities related to the angular momentum of a system when the net external torque on the system is zero. (S.P. 2.1, 2.2) \u2022 5.E.2.1 The student is able to describe or calculate the angular momentum and rotational inertia of a system in terms of the locations and velocities of objects that make up the system. Students are expected to do qualitative reasoning with compound objects. Students are expected to do calculations with a fixed set of extended objects and point masses. (S.P. 2.2) Why does Earth keep on spinning? What started it spinning to begin with? And how does an ice skater manage to spin faster and faster simply by pulling her arms in? Why does she not have to exert a torque to spin faster? Questions like these have answers based in angular momentum, the rotational analog to linear momentum. By now the pattern is clear\u2014every rotational phenomenon has a direct translational analog. It seems quite reasonable, then, to define angular momentum as =. (10.90) This equation is an analog to the definition of linear momentum as =. Units for linear momentum are kg \u22c5 m/s while units for angular momentum are kg \u22c5 m2/s. As we would expect, an object that has a large moment of inertia, such as Earth, has a very large angular momentum. An object that has a large angular velocity, such as a centrifuge, also has a rather large angular momentum. Making Connections Angular momentum is completely analogous to linear momentum, first presented in Uniform Circular Motion and Gravitation. It has the same implications in terms of carrying rotation forward, and it is conserved when the net external torque is zero. Angular momentum, like linear momentum, is also a property of the atoms and subatomic particles. Example 10.11 Calculating Angular Momentum of the Earth Strategy No information is given in the statement of the problem; so we must look up pertinent data before we can calculate =. First, according to Figure 10.12, the formula for the moment of inertia of a sphere is so that = 22 5 = = 22 5. (10.91) (10.92) Earth's mass is 5.979\u00d71024 kg and its radius is 6.376\u00d7106 m. The Earth's angular velocity is, of course, exactly one revolution per day, but we must covert to radians per second to do the calculation in SI units. Solution Substituting known information into the expression for and converting to radians per second gives = 0.4 5.979\u00d71024 kg = 9.72\u00d71037 kg \u22c5", " m2 \u22c5 rev/d. 6.376\u00d7106 m 2 1 rev d Substituting 2\u03c0 rad for 1 rev and 8.64\u00d7104 s for 1 day gives = 2\u03c0 rad/rev 9.72\u00d71037 kg \u22c5 m2 8.64\u00d7104 s/d = 7.07\u00d71033 kg \u22c5 m2/s. (1 rev/d) (10.93) (10.94) 414 Chapter 10 | Rotational Motion and Angular Momentum Discussion This number is large, demonstrating that Earth, as expected, has a tremendous angular momentum. The answer is approximate, because we have assumed a constant density for Earth in order to estimate its moment of inertia. When you push a merry-go-round, spin a bike wheel, or open a door, you exert a torque. If the torque you exert is greater than opposing torques, then the rotation accelerates, and angular momentum increases. The greater the net torque, the more rapid the increase in. The relationship between torque and angular momentum is net = \u0394 \u0394. (10.95) This expression is exactly analogous to the relationship between force and linear momentum, = \u0394 / \u0394. The equation net = \u0394 \u0394 is very fundamental and broadly applicable. It is, in fact, the rotational form of Newton's second law. Example 10.12 Calculating the Torque Putting Angular Momentum Into a Lazy Susan Figure 10.21 shows a Lazy Susan food tray being rotated by a person in quest of sustenance. Suppose the person exerts a 2.50 N force perpendicular to the lazy Susan's 0.260-m radius for 0.150 s. (a) What is the final angular momentum of the lazy Susan if it starts from rest, assuming friction is negligible? (b) What is the final angular velocity of the lazy Susan, given that its mass is 4.00 kg and assuming its moment of inertia is that of a disk? Figure 10.21 A partygoer exerts a torque on a lazy Susan to make it rotate. The equation net = \u0394 \u0394 gives the relationship between torque and the angular momentum produced. Strategy We can find the angular momentum by solving net = \u0394 \u0394 for \u0394, and using the given information to calculate the torque. The final angular momentum equals the change in angular momentum, because the lazy Susan starts from rest. That is, \u0394 =. To find the final velocity, we must calculate from the definition of in =. Solution for (a) Solving net = \u0394 \u0394 for \u0394 gives Because the force is perpendicular to, we see that net =, so that \u0394 = (net \u03c4)\u0394t. = rF\u0394 = (0.260 m)(2.50 N)(0.150 s) = 9.75\u00d710\u22122 kg \u22c5 m2 / s. Solution for (b) The final angular velocity can be calculated from the definition of angular momentum, =. Solving for and substituting the formula for the moment of inertia of a disk into the resulting equation gives And substituting known values into the preceding equation yields = = 1 22. (10.96) (10.97) (10.98) (10.99) This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 10 | Rotational Motion and Angular Momentum = Discussion 9.75\u00d710\u22122 kg \u22c5 m2/s 4.00 kg (0.260 m) (0.500) = 0.721 rad/s. 415 (10.100) Note that the imparted angular momentum does not depend on any property of the object but only on torque and time. The final angular velocity is equivalent to one revolution in 8.71 s (determination of the time period is left as an exercise for the reader), which is about right for a lazy Susan. Take-Home Experiment Plan an experiment to analyze changes to a system's angular momentum. Choose a system capable of rotational motion such as a lazy Susan or a merry-go-round. Predict how the angular momentum of this system will change when you add an object to the lazy Susan or jump onto the merry-go-round. What variables can you control? What are you measuring? In other words, what are your independent and dependent variables? Are there any independent variables that it would be useful to keep constant (angular velocity, perhaps)? Collect data in order to calculate or estimate the angular momentum of your system when in motion. What do you observe? Collect data in order to calculate the change in angular momentum as a result of the interaction you performed. Using your data, how does the angular momentum vary with the size and location of an object added to the rotating system? Example 10.13 Calculating the Torque in a Kick The person whose leg is shown in Figure 10.22 kicks his leg by exerting a 2000-N force with his upper leg muscle. The", " effective perpendicular lever arm is 2.20 cm. Given the moment of inertia of the lower leg is 1.25 kg \u22c5 m2, (a) find the angular acceleration of the leg. (b) Neglecting the gravitational force, what is the rotational kinetic energy of the leg after it has rotated through 57.3\u00ba (1.00 rad)? Figure 10.22 The muscle in the upper leg gives the lower leg an angular acceleration and imparts rotational kinetic energy to it by exerting a torque about the knee. F is a vector that is perpendicular to. This example examines the situation. Strategy The angular acceleration can be found using the rotational analog to Newton's second law, or = net /. The moment of inertia is given and the torque can be found easily from the given force and perpendicular lever arm. Once the angular acceleration is known, the final angular velocity and rotational kinetic energy can be calculated. Solution to (a) From the rotational analog to Newton's second law, the angular acceleration is = net. (10.101) Because the force and the perpendicular lever arm are given and the leg is vertical so that its weight does not create a torque, the net torque is thus net = \u22a5 = (0.0220 m)(2000 N) = 44.0 N \u22c5 m. (10.102) Substituting this value for the torque and the given value for the moment of inertia into the expression for gives 416 Chapter 10 | Rotational Motion and Angular Momentum = 44.0 N \u22c5 m 1.25 kg \u22c5 m2 = 35.2 rad/s2. Solution to (b) The final angular velocity can be calculated from the kinematic expression or 2 = 0 2 + 2 2 = 2 because the initial angular velocity is zero. The kinetic energy of rotation is (10.103) (10.104) (10.105) 22 so it is most convenient to use the value of 2 just found and the given value for the moment of inertia. The kinetic energy is then KErot = 1 (10.106) KErot = 0.5 1.25 kg \u22c5 m2 70.4 rad2 / s2. (10.107) Discussion = 44.0 J These values are reasonable for a person kicking his leg starting from the position shown. The weight of the leg can be neglected in part (a) because it exerts no torque when the center of gravity of the lower leg is directly beneath the pivot in the knee. In part (b), the force exerted by the upper leg is so large that its torque is much greater than that created by the weight of the lower leg as it rotates. The rotational kinetic energy given to the lower leg is enough that it could give a ball a significant velocity by transferring some of this energy in a kick. Making Connections: Conservation Laws Angular momentum, like energy and linear momentum, is conserved. This universally applicable law is another sign of underlying unity in physical laws. Angular momentum is conserved when net external torque is zero, just as linear momentum is conserved when the net external force is zero. Conservation of Angular Momentum We can now understand why Earth keeps on spinning. As we saw in the previous example, \u0394 = (net )\u0394. This equation means that, to change angular momentum, a torque must act over some period of time. Because Earth has a large angular momentum, a large torque acting over a long time is needed to change its rate of spin. So what external torques are there? Tidal friction exerts torque that is slowing Earth's rotation, but tens of millions of years must pass before the change is very significant. Recent research indicates the length of the day was 18 h some 900 million years ago. Only the tides exert significant retarding torques on Earth, and so it will continue to spin, although ever more slowly, for many billions of years. What we have here is, in fact, another conservation law. If the net torque is zero, then angular momentum is constant or conserved. We can see this rigorously by considering net = \u0394 \u0394 for the situation in which the net torque is zero. In that case, implying that net = 0 \u0394 \u0394 = 0. If the change in angular momentum \u0394 is zero, then the angular momentum is constant; thus, or = constant (net = 0) = \u2032(net = 0). (10.108) (10.109) (10.110) (10.111) These expressions are the law of conservation of angular momentum. Conservation laws are as scarce as they are important. An example of conservation of angular momentum is seen in Figure 10.23, in which an ice skater is executing a spin. The net torque on her is very close to zero, because there is relatively little friction between her skates and the ice and because the friction is exerted very close to the pivot point. (Both and are small, and so", " is negligibly small.) Consequently, she can This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 10 | Rotational Motion and Angular Momentum 417 spin for quite some time. She can do something else, too. She can increase her rate of spin by pulling her arms and legs in. Why does pulling her arms and legs in increase her rate of spin? The answer is that her angular momentum is constant, so that = \u2032. (10.112) Expressing this equation in terms of the moment of inertia, = \u2032\u2032, (10.113) where the primed quantities refer to conditions after she has pulled in her arms and reduced her moment of inertia. Because \u2032 is smaller, the angular velocity \u2032 must increase to keep the angular momentum constant. The change can be dramatic, as the following example shows. Figure 10.23 (a) An ice skater is spinning on the tip of her skate with her arms extended. Her angular momentum is conserved because the net torque on her is negligibly small. In the next image, her rate of spin increases greatly when she pulls in her arms, decreasing her moment of inertia. The work she does to pull in her arms results in an increase in rotational kinetic energy. Example 10.14 Calculating the Angular Momentum of a Spinning Skater Suppose an ice skater, such as the one in Figure 10.23, is spinning at 0.800 rev/ s with her arms extended. She has a moment of inertia of 2.34 kg \u22c5 m2 with her arms extended and of 0.363 kg \u22c5 m2 with her arms close to her body. (These moments of inertia are based on reasonable assumptions about a 60.0-kg skater.) (a) What is her angular velocity in revolutions per second after she pulls in her arms? (b) What is her rotational kinetic energy before and after she does this? Strategy In the first part of the problem, we are looking for the skater's angular velocity \u2032 after she has pulled in her arms. To find this quantity, we use the conservation of angular momentum and note that the moments of inertia and initial angular velocity are given. To find the initial and final kinetic energies, we use the definition of rotational kinetic energy given by KErot = 1 (10.114) 22. Solution for (a) Because torque is negligible (as discussed above), the conservation of angular momentum given in = \u2032\u2032 is applicable. Thus, or = \u2032 = \u2032\u2032 Solving for \u2032 and substituting known values into the resulting equation gives \u2032 = \u2032 = = 5.16 rev/s. 2.34 kg \u22c5 m2 0.363 kg \u22c5 m2 (0.800 rev/s) Solution for (b) Rotational kinetic energy is given by KErot = 1 22. The initial value is found by substituting known values into the equation and converting the angular velocity to rad/s: (10.115) (10.116) (10.117) (10.118) 418 Chapter 10 | Rotational Motion and Angular Momentum KErot = (0.5) = 29.6 J. 2.34 kg \u22c5 m2 (0.800 rev/s)(2\u03c0 rad/rev) 2 (10.119) The final rotational kinetic energy is KErot \u2032 = 1 2\u2032\u20322. (10.120) Substituting known values into this equation gives rot\u2032 = (0.5) = 191 J. 0.363 kg \u22c5 m2 (5.16 rev/s)(2\u03c0 rad/rev) 2 (10.121) Discussion In both parts, there is an impressive increase. First, the final angular velocity is large, although most world-class skaters can achieve spin rates about this great. Second, the final kinetic energy is much greater than the initial kinetic energy. The increase in rotational kinetic energy comes from work done by the skater in pulling in her arms. This work is internal work that depletes some of the skater's food energy. There are several other examples of objects that increase their rate of spin because something reduced their moment of inertia. Tornadoes are one example. Storm systems that create tornadoes are slowly rotating. When the radius of rotation narrows, even in a local region, angular velocity increases, sometimes to the furious level of a tornado. Earth is another example. Our planet was born from a huge cloud of gas and dust, the rotation of which came from turbulence in an even larger cloud. Gravitational forces caused the cloud to contract, and the rotation rate increased as a result. (See Figure 10.24.) Figure 10.24 The Solar System coalesced from a cloud of gas and dust that was originally rotating. The orbital motions and spins of the planets are in the same direction as the original spin and conserve", " the angular momentum of the parent cloud. In case of human motion, one would not expect angular momentum to be conserved when a body interacts with the environment as its foot pushes off the ground. Astronauts floating in space aboard the International Space Station have no angular momentum relative to the inside of the ship if they are motionless. Their bodies will continue to have this zero value no matter how they twist about as long as they do not give themselves a push off the side of the vessel. Check Your Undestanding Is angular momentum completely analogous to linear momentum? What, if any, are their differences? Solution Yes, angular and linear momentums are completely analogous. While they are exact analogs they have different units and are not directly inter-convertible like forms of energy are. 10.6 Collisions of Extended Bodies in Two Dimensions By the end of this section, you will be able to: Learning Objectives This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 10 | Rotational Motion and Angular Momentum 425 Figure 10.31 As seen in figure (a), the forces on a spinning gyroscope are its weight and the supporting force from the stand. These forces create a horizontal torque on the gyroscope, which create a change in angular momentum \u0394L that is also horizontal. In figure (b), \u0394L and L add to produce a new angular momentum with the same magnitude, but different direction, so that the gyroscope precesses in the direction shown instead of falling over. Check Your Understanding Rotational kinetic energy is associated with angular momentum? Does that mean that rotational kinetic energy is a vector? Solution No, energy is always a scalar whether motion is involved or not. No form of energy has a direction in space and you can see that rotational kinetic energy does not depend on the direction of motion just as linear kinetic energy is independent of the direction of motion. Glossary angular acceleration: the rate of change of angular velocity with time angular momentum: the product of moment of inertia and angular velocity change in angular velocity: the difference between final and initial values of angular velocity kinematics of rotational motion: describes the relationships among rotation angle, angular velocity, angular acceleration, and time law of conservation of angular momentum: angular momentum is conserved, i.e., the initial angular momentum is equal to the final angular momentum when no external torque is applied to the system moment of inertia: mass times the square of perpendicular distance from the rotation axis; for a point mass, it is = 2 and, because any object can be built up from a collection of point masses, this relationship is the basis for all other moments of inertia right-hand rule: direction of angular velocity \u03c9 and angular momentum L in which the thumb of your right hand points when you curl your fingers in the direction of the disk's rotation rotational inertia: resistance to change of rotation. The more rotational inertia an object has, the harder it is to rotate rotational kinetic energy: the kinetic energy due to the rotation of an object. This is part of its total kinetic energy tangential acceleration: the acceleration in a direction tangent to the circle at the point of interest in circular motion torque: the turning effectiveness of a force work-energy theorem: if one or more external forces act upon a rigid object, causing its kinetic energy to change from KE1 to KE2, then the work done by the net force is equal to the change in kinetic energy 426 Chapter 10 | Rotational Motion and Angular Momentum Section Summary 10.1 Angular Acceleration \u2022 Uniform circular motion is the motion with a constant angular velocity = \u0394 \u0394. \u2022 In non-uniform circular motion, the velocity changes with time and the rate of change of angular velocity (i.e. angular acceleration) is = \u0394 \u0394. \u2022 Linear or tangential acceleration refers to changes in the magnitude of velocity but not its direction, given as t = \u0394 \u0394 \u2022 For circular motion, note that =, so that. \u2022 The radius r is constant for circular motion, and so \u0394() = \u0394. Thus, t = \u0394() \u0394. \u2022 By definition, \u0394 / \u0394 =. Thus, or t = \u0394 \u0394. t = = t. 10.2 Kinematics of Rotational Motion \u2022 Kinematics is the description of motion. \u2022 The kinematics of rotational motion describes the relationships among rotation angle, angular velocity, angular acceleration, and time. \u2022 Starting with the four kinematic equations we developed in the One-Dimensional Kinematics, we can derive the four \u2022 rotational kinematic equations (presented together with their translational counterparts) seen in Table 10.2. In these equations, the subscript 0 denotes initial values ( 0 and 0 are initial values), and the average angular velocity - and average velocity - are defined as follows: \u00af = 0 + 2 and \u00af = 0 + 2. 10.3 Dynamics of Rotational Motion: Rotational Inertia \u2022", " The farther the force is applied from the pivot, the greater is the angular acceleration; angular acceleration is inversely \u2022 proportional to mass. If we exert a force on a point mass that is at a distance from a pivot point and because the force is perpendicular to, an acceleration is obtained in the direction of. We can rearrange this equation such that, and then look for ways to relate this expression to expressions for rotational quantities. We note that, and we substitute this expression into, yielding \u2022 Torque is the turning effectiveness of a force. In this case, because is perpendicular to, torque is simply =. If we multiply both sides of the equation above by, we get torque on the left-hand side. That is, or = 2 \u2022 The moment of inertia of an object is the sum of 2 for all the point masses of which it is composed. That is, = 2. \u2022 The general relationship among torque, moment of inertia, and angular acceleration is = \u2211 2. or = = net \u03c4 \u22c5 This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 10 | Rotational Motion and Angular Momentum 427 10.4 Rotational Kinetic Energy: Work and Energy Revisited \u2022 The rotational kinetic energy KErot for an object with a moment of inertia and an angular velocity is given by KErot = 1 \u2022 Helicopters store large amounts of rotational kinetic energy in their blades. This energy must be put into the blades before takeoff and maintained until the end of the flight. The engines do not have enough power to simultaneously provide lift and put significant rotational energy into the blades. 22. \u2022 Work and energy in rotational motion are completely analogous to work and energy in translational motion. \u2022 The equation for the work-energy theorem for rotational motion is, net = 1 22 \u2212 1 20 2. 10.5 Angular Momentum and Its Conservation \u2022 Every rotational phenomenon has a direct translational analog, likewise angular momentum can be defined as =. \u2022 This equation is an analog to the definition of linear momentum as =. The relationship between torque and angular momentum is net = \u0394 \u0394. \u2022 Angular momentum, like energy and linear momentum, is conserved. This universally applicable law is another sign of underlying unity in physical laws. Angular momentum is conserved when net external torque is zero, just as linear momentum is conserved when the net external force is zero. 10.6 Collisions of Extended Bodies in Two Dimensions \u2022 Angular momentum is analogous to linear momentum and is given by =. \u2022 Angular momentum is changed by torque, following the relationship net = \u0394 \u0394. \u2022 Angular momentum is conserved if the net torque is zero = constant (net = 0) or = \u2032 (net = 0). This equation is known as the law of conservation of angular momentum, which may be conserved in collisions. 10.7 Gyroscopic Effects: Vector Aspects of Angular Momentum \u2022 Torque is perpendicular to the plane formed by and F and is the direction your right thumb would point if you curled the fingers of your right hand in the direction of F. The direction of the torque is thus the same as that of the angular momentum it produces. \u2022 The gyroscope precesses around a vertical axis, since the torque is always horizontal and perpendicular to L. If the gyroscope is not spinning, it acquires angular momentum in the direction of the torque ( L = \u0394L ), and it rotates about a horizontal axis, falling over just as we would expect. \u2022 Earth itself acts like a gigantic gyroscope. Its angular momentum is along its axis and points at Polaris, the North Star. Conceptual Questions 10.1 Angular Acceleration 1. Analogies exist between rotational and translational physical quantities. Identify the rotational term analogous to each of the following: acceleration, force, mass, work, translational kinetic energy, linear momentum, impulse. 2. Explain why centripetal acceleration changes the direction of velocity in circular motion but not its magnitude. 3. In circular motion, a tangential acceleration can change the magnitude of the velocity but not its direction. Explain your answer. 4. Suppose a piece of food is on the edge of a rotating microwave oven plate. Does it experience nonzero tangential acceleration, centripetal acceleration, or both when: (a) The plate starts to spin? (b) The plate rotates at constant angular velocity? (c) The plate slows to a halt? 10.3 Dynamics of Rotational Motion: Rotational Inertia 5. The moment of inertia of a long rod spun around an axis through one end perpendicular to its length is 2 /3. Why is this moment of inertia greater than it would be if you spun a point mass at the location of the center of mass of the rod (at / 2 )? (That would be 2 /4.) 428 Chapter 10 | Rotational Motion and Angular Momentum 6. Why is the moment", " of inertia of a hoop that has a mass and a radius greater than the moment of inertia of a disk that has the same mass and radius? Why is the moment of inertia of a spherical shell that has a mass and a radius greater than that of a solid sphere that has the same mass and radius? 7. Give an example in which a small force exerts a large torque. Give another example in which a large force exerts a small torque. 8. While reducing the mass of a racing bike, the greatest benefit is realized from reducing the mass of the tires and wheel rims. Why does this allow a racer to achieve greater accelerations than would an identical reduction in the mass of the bicycle's frame? Figure 10.32 The image shows a side view of a racing bicycle. Can you see evidence in the design of the wheels on this racing bicycle that their moment of inertia has been purposely reduced? (credit: Jes\u00fas Rodriguez) 9. A ball slides up a frictionless ramp. It is then rolled without slipping and with the same initial velocity up another frictionless ramp (with the same slope angle). In which case does it reach a greater height, and why? 10.4 Rotational Kinetic Energy: Work and Energy Revisited 10. Describe the energy transformations involved when a yo-yo is thrown downward and then climbs back up its string to be caught in the user's hand. 11. What energy transformations are involved when a dragster engine is revved, its clutch let out rapidly, its tires spun, and it starts to accelerate forward? Describe the source and transformation of energy at each step. 12. The Earth has more rotational kinetic energy now than did the cloud of gas and dust from which it formed. Where did this energy come from? Figure 10.33 An immense cloud of rotating gas and dust contracted under the influence of gravity to form the Earth and in the process rotational kinetic energy increased. (credit: NASA) 10.5 Angular Momentum and Its Conservation 13. When you start the engine of your car with the transmission in neutral, you notice that the car rocks in the opposite sense of the engine's rotation. Explain in terms of conservation of angular momentum. Is the angular momentum of the car conserved for long (for more than a few seconds)? 14. Suppose a child walks from the outer edge of a rotating merry-go round to the inside. Does the angular velocity of the merrygo-round increase, decrease, or remain the same? Explain your answer. This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 10 | Rotational Motion and Angular Momentum 429 Figure 10.34 A child may jump off a merry-go-round in a variety of directions. 15. Suppose a child gets off a rotating merry-go-round. Does the angular velocity of the merry-go-round increase, decrease, or remain the same if: (a) He jumps off radially? (b) He jumps backward to land motionless? (c) He jumps straight up and hangs onto an overhead tree branch? (d) He jumps off forward, tangential to the edge? Explain your answers. (Refer to Figure 10.34). 16. Helicopters have a small propeller on their tail to keep them from rotating in the opposite direction of their main lifting blades. Explain in terms of Newton's third law why the helicopter body rotates in the opposite direction to the blades. 17. Whenever a helicopter has two sets of lifting blades, they rotate in opposite directions (and there will be no tail propeller). Explain why it is best to have the blades rotate in opposite directions. 18. Describe how work is done by a skater pulling in her arms during a spin. In particular, identify the force she exerts on each arm to pull it in and the distance each moves, noting that a component of the force is in the direction moved. Why is angular momentum not increased by this action? 19. When there is a global heating trend on Earth, the atmosphere expands and the length of the day increases very slightly. Explain why the length of a day increases. 20. Nearly all conventional piston engines have flywheels on them to smooth out engine vibrations caused by the thrust of individual piston firings. Why does the flywheel have this effect? 21. Jet turbines spin rapidly. They are designed to fly apart if something makes them seize suddenly, rather than transfer angular momentum to the plane's wing, possibly tearing it off. Explain how flying apart conserves angular momentum without transferring it to the wing. 22. An astronaut tightens a bolt on a satellite in orbit. He rotates in a direction opposite to that of the bolt, and the satellite rotates in the same direction as the bolt. Explain why. If a handhold is available on the satellite, can this counter-rotation be prevented? Explain your answer. 23. Competitive divers pull their limbs in and curl up their bodies when they", " do flips. Just before entering the water, they fully extend their limbs to enter straight down. Explain the effect of both actions on their angular velocities. Also explain the effect on their angular momenta. Figure 10.35 The diver spins rapidly when curled up and slows when she extends her limbs before entering the water. 24. Draw a free body diagram to show how a diver gains angular momentum when leaving the diving board. 25. In terms of angular momentum, what is the advantage of giving a football or a rifle bullet a spin when throwing or releasing it? 430 Chapter 10 | Rotational Motion and Angular Momentum Figure 10.36 The image shows a view down the barrel of a cannon, emphasizing its rifling. Rifling in the barrel of a canon causes the projectile to spin just as is the case for rifles (hence the name for the grooves in the barrel). (credit: Elsie esq., Flickr) 10.6 Collisions of Extended Bodies in Two Dimensions 26. Describe two different collisions\u2014one in which angular momentum is conserved, and the other in which it is not. Which condition determines whether or not angular momentum is conserved in a collision? 27. Suppose an ice hockey puck strikes a hockey stick that lies flat on the ice and is free to move in any direction. Which quantities are likely to be conserved: angular momentum, linear momentum, or kinetic energy (assuming the puck and stick are very resilient)? 28. While driving his motorcycle at highway speed, a physics student notices that pulling back lightly on the right handlebar tips the cycle to the left and produces a left turn. Explain why this happens. 10.7 Gyroscopic Effects: Vector Aspects of Angular Momentum 29. While driving his motorcycle at highway speed, a physics student notices that pulling back lightly on the right handlebar tips the cycle to the left and produces a left turn. Explain why this happens. 30. Gyroscopes used in guidance systems to indicate directions in space must have an angular momentum that does not change in direction. Yet they are often subjected to large forces and accelerations. How can the direction of their angular momentum be constant when they are accelerated? This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 10 | Rotational Motion and Angular Momentum 431 Problems & Exercises 10.1 Angular Acceleration 1. At its peak, a tornado is 60.0 m in diameter and carries 500 km/h winds. What is its angular velocity in revolutions per second? 2. Integrated Concepts An ultracentrifuge accelerates from rest to 100,000 rpm in 2.00 min. (a) What is its angular acceleration in rad/s2? (b) What is the tangential acceleration of a point 9.50 cm from the axis of rotation? (c) What is the radial acceleration in m/s2 and multiples of of this point at full rpm? 3. Integrated Concepts You have a grindstone (a disk) that is 90.0 kg, has a 0.340-m radius, and is turning at 90.0 rpm, and you press a steel axe against it with a radial force of 20.0 N. (a) Assuming the kinetic coefficient of friction between steel and stone is 0.20, calculate the angular acceleration of the grindstone. (b) How many turns will the stone make before coming to rest? 4. Unreasonable Results You are told that a basketball player spins the ball with an angular acceleration of 100 rad/s2. (a) What is the ball's final angular velocity if the ball starts from rest and the acceleration lasts 2.00 s? (b) What is unreasonable about the result? (c) Which premises are unreasonable or inconsistent? 10.2 Kinematics of Rotational Motion 5. With the aid of a string, a gyroscope is accelerated from rest to 32 rad/s in 0.40 s. (a) What is its angular acceleration in rad/s2? (b) How many revolutions does it go through in the process? 6. Suppose a piece of dust finds itself on a CD. If the spin rate of the CD is 500 rpm, and the piece of dust is 4.3 cm from the center, what is the total distance traveled by the dust in 3 minutes? (Ignore accelerations due to getting the CD rotating.) 7. A gyroscope slows from an initial rate of 32.0 rad/s at a rate of 0.700 rad/s2. (a) How long does it take to come to rest? (b) How many revolutions does it make before stopping? 8. During a very quick stop, a car decelerates at 7.00 m/s2. (a) What is the angular acceleration of its 0.280-m-radius tires, assuming they do not slip on the pavement? (b) How many revolutions do the tires make before", " coming to rest, given their initial angular velocity is 95.0 rad/s? (c) How long does the car take to stop completely? (d) What distance does the car travel in this time? (e) What was the car's initial velocity? (f) Do the values obtained seem reasonable, considering that this stop happens very quickly? Figure 10.37 Yo-yos are amusing toys that display significant physics and are engineered to enhance performance based on physical laws. (credit: Beyond Neon, Flickr) 9. Everyday application: Suppose a yo-yo has a center shaft that has a 0.250 cm radius and that its string is being pulled. (a) If the string is stationary and the yo-yo accelerates away from it at a rate of 1.50 m/s2, what is the angular acceleration of the yo-yo? (b) What is the angular velocity after 0.750 s if it starts from rest? (c) The outside radius of the yo-yo is 3.50 cm. What is the tangential acceleration of a point on its edge? 10.3 Dynamics of Rotational Motion: Rotational Inertia 10. This problem considers additional aspects of example Calculating the Effect of Mass Distribution on a MerryGo-Round. (a) How long does it take the father to give the merry-go-round an angular velocity of 1.50 rad/s? (b) How many revolutions must he go through to generate this velocity? (c) If he exerts a slowing force of 300 N at a radius of 1.35 m, how long would it take him to stop them? 11. Calculate the moment of inertia of a skater given the following information. (a) The 60.0-kg skater is approximated as a cylinder that has a 0.110-m radius. (b) The skater with arms extended is approximately a cylinder that is 52.5 kg, has a 0.110-m radius, and has two 0.900-m-long arms which are 3.75 kg each and extend straight out from the cylinder like rods rotated about their ends. 12. The triceps muscle in the back of the upper arm extends the forearm. This muscle in a professional boxer exerts a force of 2.00\u00d7103 N with an effective perpendicular lever arm of 3.00 cm, producing an angular acceleration of the forearm of 120 rad/s2. What is the moment of inertia of the boxer's forearm? 13. A soccer player extends her lower leg in a kicking motion by exerting a force with the muscle above the knee in the front of her leg. She produces an angular acceleration of 30.00 rad/s2 and her lower leg has a moment of inertia of 0.750 kg \u22c5 m2. What is the force exerted by the muscle if its effective perpendicular lever arm is 1.90 cm? 14. Suppose you exert a force of 180 N tangential to a 0.280-m-radius 75.0-kg grindstone (a solid disk). (a)What torque is exerted? (b) What is the angular acceleration assuming negligible opposing friction? (c) What is the angular acceleration if there is an opposing frictional force of 20.0 N exerted 1.50 cm from the axis? 432 Chapter 10 | Rotational Motion and Angular Momentum 15. Consider the 12.0 kg motorcycle wheel shown in Figure 10.38. Assume it to be approximately an annular ring with an inner radius of 0.280 m and an outer radius of 0.330 m. The motorcycle is on its center stand, so that the wheel can spin freely. (a) If the drive chain exerts a force of 2200 N at a radius of 5.00 cm, what is the angular acceleration of the wheel? (b) What is the tangential acceleration of a point on the outer edge of the tire? (c) How long, starting from rest, does it take to reach an angular velocity of 80.0 rad/s? Figure 10.38 A motorcycle wheel has a moment of inertia approximately that of an annular ring. 16. Zorch, an archenemy of Superman, decides to slow Earth's rotation to once per 28.0 h by exerting an opposing force at and parallel to the equator. Superman is not immediately concerned, because he knows Zorch can only exert a force of 4.00\u00d7107 N (a little greater than a Saturn V rocket's thrust). How long must Zorch push with this force to accomplish his goal? (This period gives Superman time to devote to other villains.) Explicitly show how you follow the steps found in Problem-Solving Strategy for Rotational Dynamics. 17. An automobile engine can produce 200 N \u00b7 m of torque. Calculate the angular acceleration produced if 95.0% of this torque is applied to the drive shaft, axle, and rear wheels of a car", ", given the following information. The car is suspended so that the wheels can turn freely. Each wheel acts like a 15.0 kg disk that has a 0.180 m radius. The walls of each tire act like a 2.00-kg annular ring that has inside radius of 0.180 m and outside radius of 0.320 m. The tread of each tire acts like a 10.0-kg hoop of radius 0.330 m. The 14.0-kg axle acts like a rod that has a 2.00-cm radius. The 30.0-kg drive shaft acts like a rod that has a 3.20-cm radius. = 2 / 3 18. Starting with the formula for the moment of inertia of a rod rotated around an axis through one end perpendicular to its length, prove that the moment of inertia of a rod rotated about an axis through its center perpendicular to its length is = 2 / 12. You will find the graphics in Figure 10.12 useful in visualizing these rotations. 19. Unreasonable Results A gymnast doing a forward flip lands on the mat and exerts a 500-N \u00b7 m torque to slow and then reverse her angular velocity. Her initial angular velocity is 10.0 rad/s, and her moment of inertia is 0.050 kg \u22c5 m2. (a) What time is required for her to exactly reverse her spin? (b) What is This content is available for free at http://cnx.org/content/col11844/1.13 unreasonable about the result? (c) Which premises are unreasonable or inconsistent? 20. Unreasonable Results An advertisement claims that an 800-kg car is aided by its 20.0-kg flywheel, which can accelerate the car from rest to a speed of 30.0 m/s. The flywheel is a disk with a 0.150-m radius. (a) Calculate the angular velocity the flywheel must have if 95.0% of its rotational energy is used to get the car up to speed. (b) What is unreasonable about the result? (c) Which premise is unreasonable or which premises are inconsistent? 10.4 Rotational Kinetic Energy: Work and Energy Revisited 21. This problem considers energy and work aspects of Example 10.7\u2014use data from that example as needed. (a) Calculate the rotational kinetic energy in the merry-go-round plus child when they have an angular velocity of 20.0 rpm. (b) Using energy considerations, find the number of revolutions the father will have to push to achieve this angular velocity starting from rest. (c) Again, using energy considerations, calculate the force the father must exert to stop the merry-goround in two revolutions 22. What is the final velocity of a hoop that rolls without slipping down a 5.00-m-high hill, starting from rest? 23. (a) Calculate the rotational kinetic energy of Earth on its axis. (b) What is the rotational kinetic energy of Earth in its orbit around the Sun? 24. Calculate the rotational kinetic energy in the motorcycle wheel (Figure 10.38) if its angular velocity is 120 rad/s. Assume M = 12.0 kg, R1 = 0.280 m, and R2 = 0.330 m. 25. A baseball pitcher throws the ball in a motion where there is rotation of the forearm about the elbow joint as well as other movements. If the linear velocity of the ball relative to the elbow joint is 20.0 m/s at a distance of 0.480 m from the joint and the moment of inertia of the forearm is 0.500 kg \u22c5 m2, what is the rotational kinetic energy of the forearm? 26. While punting a football, a kicker rotates his leg about the hip joint. The moment of inertia of the leg is 3.75 kg \u22c5 m2 and its rotational kinetic energy is 175 J. (a) What is the angular velocity of the leg? (b) What is the velocity of tip of the punter's shoe if it is 1.05 m from the hip joint? (c) Explain how the football can be given a velocity greater than the tip of the shoe (necessary for a decent kick distance). 27. A bus contains a 1500 kg flywheel (a disk that has a 0.600 m radius) and has a total mass of 10,000 kg. (a) Calculate the angular velocity the flywheel must have to contain enough energy to take the bus from rest to a speed of 20.0 m/s, assuming 90.0% of the rotational kinetic energy can be transformed into translational energy. (b) How high a hill can the bus climb with this stored energy and still have a speed of 3.00 m/s at the top of the hill? Explicitly show how you follow the steps in the Problem-Solving Strategy for", " Rotational Energy. 28. A ball with an initial velocity of 8.00 m/s rolls up a hill without slipping. Treating the ball as a spherical shell, calculate the vertical height it reaches. (b) Repeat the calculation for the same ball if it slides up the hill without rolling. Chapter 10 | Rotational Motion and Angular Momentum 433 (b) How does this angular momentum compare with the angular momentum of the Moon on its axis? Remember that the Moon keeps one side toward Earth at all times. (c) Discuss whether the values found in parts (a) and (b) seem consistent with the fact that tidal effects with Earth have caused the Moon to rotate with one side always facing Earth. 38. Suppose you start an antique car by exerting a force of 300 N on its crank for 0.250 s. What angular momentum is given to the engine if the handle of the crank is 0.300 m from the pivot and the force is exerted to create maximum torque the entire time? 39. A playground merry-go-round has a mass of 120 kg and a radius of 1.80 m and it is rotating with an angular velocity of 0.500 rev/s. What is its angular velocity after a 22.0-kg child gets onto it by grabbing its outer edge? The child is initially at rest. 40. Three children are riding on the edge of a merry-go-round that is 100 kg, has a 1.60-m radius, and is spinning at 20.0 rpm. The children have masses of 22.0, 28.0, and 33.0 kg. If the child who has a mass of 28.0 kg moves to the center of the merry-go-round, what is the new angular velocity in rpm? 41. (a) Calculate the angular momentum of an ice skater spinning at 6.00 rev/s given his moment of inertia is 0.400 kg \u22c5 m2. (b) He reduces his rate of spin (his angular velocity) by extending his arms and increasing his moment of inertia. Find the value of his moment of inertia if his angular velocity decreases to 1.25 rev/s. (c) Suppose instead he keeps his arms in and allows friction of the ice to slow him to 3.00 rev/s. What average torque was exerted if this takes 15.0 s? 42. Consider the Earth-Moon system. Construct a problem in which you calculate the total angular momentum of the system including the spins of the Earth and the Moon on their axes and the orbital angular momentum of the Earth-Moon system in its nearly monthly rotation. Calculate what happens to the Moon's orbital radius if the Earth's rotation decreases due to tidal drag. Among the things to be considered are the amount by which the Earth's rotation slows and the fact that the Moon will continue to have one side always facing the Earth. 10.6 Collisions of Extended Bodies in Two Dimensions 43. Repeat Example 10.15 in which the disk strikes and adheres to the stick 0.100 m from the nail. 44. Repeat Example 10.15 in which the disk originally spins clockwise at 1000 rpm and has a radius of 1.50 cm. 45. Twin skaters approach one another as shown in Figure 10.39 and lock hands. (a) Calculate their final angular velocity, given each had an initial speed of 2.50 m/s relative to the ice. Each has a mass of 70.0 kg, and each has a center of mass located 0.800 m from their locked hands. You may approximate their moments of inertia to be that of point masses at this radius. (b) Compare the initial kinetic energy and final kinetic energy. 29. While exercising in a fitness center, a man lies face down on a bench and lifts a weight with one lower leg by contacting the muscles in the back of the upper leg. (a) Find the angular acceleration produced given the mass lifted is 10.0 kg at a distance of 28.0 cm from the knee joint, the moment of inertia of the lower leg is 0.900 kg \u22c5 m2, the muscle force is 1500 N, and its effective perpendicular lever arm is 3.00 cm. (b) How much work is done if the leg rotates through an angle of 20.0\u00ba with a constant force exerted by the muscle? 30. To develop muscle tone, a woman lifts a 2.00-kg weight held in her hand. She uses her biceps muscle to flex the lower arm through an angle of 60.0\u00ba. (a) What is the angular acceleration if the weight is 24.0 cm from the elbow joint, her forearm has a moment of inertia of 0.250 kg \u22c5 m2, and the net force she exerts is 750 N at an effective perpendicular lever arm of 2.00 cm? (b) How much work does she do? 31. Consider two cylinders that start down identical", " inclines from rest except that one is frictionless. Thus one cylinder rolls without slipping, while the other slides frictionlessly without rolling. They both travel a short distance at the bottom and then start up another incline. (a) Show that they both reach the same height on the other incline, and that this height is equal to their original height. (b) Find the ratio of the time the rolling cylinder takes to reach the height on the second incline to the time the sliding cylinder takes to reach the height on the second incline. (c) Explain why the time for the rolling motion is greater than that for the sliding motion. 32. What is the moment of inertia of an object that rolls without slipping down a 2.00-m-high incline starting from rest, and has a final velocity of 6.00 m/s? Express the moment of inertia as a multiple of 2, where is the mass of the object and is its radius. 33. Suppose a 200-kg motorcycle has two wheels like, the one described in Example 10.15 and is heading toward a hill at a speed of 30.0 m/s. (a) How high can it coast up the hill, if you neglect friction? (b) How much energy is lost to friction if the motorcycle only gains an altitude of 35.0 m before coming to rest? 34. In softball, the pitcher throws with the arm fully extended (straight at the elbow). In a fast pitch the ball leaves the hand with a speed of 139 km/h. (a) Find the rotational kinetic energy of the pitcher's arm given its moment of inertia is 0.720 kg \u22c5 m2 and the ball leaves the hand at a distance of 0.600 m from the pivot at the shoulder. (b) What force did the muscles exert to cause the arm to rotate if their effective perpendicular lever arm is 4.00 cm and the ball is 0.156 kg? 35. Construct Your Own Problem Consider the work done by a spinning skater pulling her arms in to increase her rate of spin. Construct a problem in which you calculate the work done with a \u201cforce multiplied by distance\u201d calculation and compare it to the skater's increase in kinetic energy. 10.5 Angular Momentum and Its Conservation 36. (a) Calculate the angular momentum of the Earth in its orbit around the Sun. (b) Compare this angular momentum with the angular momentum of Earth on its axis. 37. (a) What is the angular momentum of the Moon in its orbit around Earth? 434 Chapter 10 | Rotational Motion and Angular Momentum Figure 10.39 Twin skaters approach each other with identical speeds. Then, the skaters lock hands and spin. 46. Suppose a 0.250-kg ball is thrown at 15.0 m/s to a motionless person standing on ice who catches it with an outstretched arm as shown in Figure 10.40. (a) Calculate the final linear velocity of the person, given his mass is 70.0 kg. (b) What is his angular velocity if each arm is 5.00 kg? You may treat the ball as a point mass and treat the person's arms as uniform rods (each has a length of 0.900 m) and the rest of his body as a uniform cylinder of radius 0.180 m. Neglect the effect of the ball on his center of mass so that his center of mass remains in his geometrical center. (c) Compare the initial and final total kinetic energies. Figure 10.40 The figure shows the overhead view of a person standing motionless on ice about to catch a ball. Both arms are outstretched. After catching the ball, the skater recoils and rotates. 47. Repeat Example 10.15 in which the stick is free to have translational motion as well as rotational motion. 10.7 Gyroscopic Effects: Vector Aspects of Angular Momentum 48. Integrated Concepts The axis of Earth makes a 23.5\u00b0 angle with a direction perpendicular to the plane of Earth's orbit. As shown in Figure 10.41, this axis precesses, making one complete rotation in 25,780 y. (a) Calculate the change in angular momentum in half this time. (b) What is the average torque producing this change in angular momentum? (c) If this torque were created by a single force (it is not) acting at the most effective point on the equator, what would its magnitude be? Test Prep for AP\u00ae Courses This content is available for free at http://cnx.org/content/col11844/1.13 Figure 10.41 The Earth's axis slowly precesses, always making an angle of 23.5\u00b0 with the direction perpendicular to the plane of Earth's orbit. The change in angular momentum for the two shown positions is quite large, although the magnitude L is unchanged. 10.3 Dynamics of Rotational Motion:", " Rotational Inertia Chapter 10 | Rotational Motion and Angular Momentum 435 9. Which measure would not be useful to help you determine the change in angular velocity when the torque on a fishing reel is increased? a. b. c. d. the radius of the reel the amount of line that unspools the angular momentum of the fishing line the time it takes the line to unspool 10. What data could you collect to study the change in angular velocity when two people push a merry-go-round instead of one, providing twice as much torque? How would you use the data you collect? 10.5 Angular Momentum and Its Conservation 11. Which rotational system would be best to use as a model to measure how angular momentum changes when forces on the system are changed? a. a fishing reel b. a planet and its moon c. a figure skater spinning d. a person's lower leg 12. You are collecting data to study changes in the angular momentum of a bicycle wheel when a force is applied to it. Which of the following measurements would be least helpful to you? a. b. c. d. the time for which the force is applied the radius at which the force is applied the angular velocity of the wheel when the force is applied the direction of the force 13. Which torque applied to a disk with radius 7.0 cm for 3.5 s will produce an angular momentum of 25 N\u2022m\u2022s? a. 7.1 N\u2022m b. 357.1 N\u2022m c. 3.6 N\u2022m d. 612.5 N\u2022m 14. Which of the following would be the best way to produce measurable amounts of torque on a system to test the relationship between the angular momentum of the system, the average torque applied to the system, and the time for which the torque is applied? a. having different numbers of people push on a merry-go- round b. placing known masses on one end of a seesaw c. touching the outer edge of a bicycle wheel to a treadmill that is moving at different speeds d. hanging known masses from a string that is wound around a spool suspended horizontally on an axle 15. 1. A piece of wood can be carved by spinning it on a motorized lathe and holding a sharp chisel to the edge of the wood as it spins. How does the angular velocity of a piece of wood with a radius of 0.2 m spinning on a lathe change when a chisel is held to the wood's edge with a force of 50 N? a. b. c. d. It increases by 0.1 N\u2022m multiplied by the moment of inertia of the wood. It decreases by 0.1 N\u2022m divided by the moment of inertia of the wood-and-lathe system. It decreases by 0.1 N\u2022m multiplied by the moment of inertia of the wood. It decreases by 0.1 m/s2. 2. A Ferris wheel is loaded with people in the chairs at the following positions: 4 o'clock, 1 o'clock, 9 o'clock, and 6 o'clock. As the wheel begins to turn, what forces are acting on the system? How will each force affect the angular velocity and angular momentum? 3. A lever is placed on a fulcrum. A rock is placed on the left end of the lever and a downward (clockwise) force is applied to the right end of the lever. What measurements would be most effective to help you determine the angular momentum of the system? (Assume the lever itself has negligible mass.) a. b. c. d. the angular velocity and mass of the rock the angular velocity and mass of the rock, and the radius of the lever the velocity of the force, the radius of the lever, and the mass of the rock the mass of the rock, the length of the lever on both sides of the fulcrum, and the force applied on the right side of the lever 4. You can use the following setup to determine angular acceleration and angular momentum: A lever is placed on a fulcrum. A rock is placed on the left end of the lever and a known downward (clockwise) force is applied to the right end of the lever. What calculations would you perform? How would you account for gravity in your calculations? 5. Consider two sizes of disk, both of mass M. One size of disk has radius R; the other has radius 2R. System A consists of two of the larger disks rigidly connected to each other with a common axis of rotation. System B consists of one of the larger disks and a number of the smaller disks rigidly connected with a common axis of rotation. If the moment of inertia for system A equals the moment of inertia for system B, how many of the smaller disks are in system B? a. 1 b. 2 c. 3 d. 4 6. How do you arrange these objects so that the resulting system has the maximum", " possible moment of inertia? What is that moment of inertia? 10.4 Rotational Kinetic Energy: Work and Energy Revisited 7. Gear A, which turns clockwise, meshes with gear B, which turns counterclockwise. When more force is applied through gear A, torque is created. How does the angular velocity of gear B change as a result? a. b. c. d. It increases in magnitude. It decreases in magnitude. It changes direction. It stays the same. 8. Which will cause a greater increase in the angular velocity of a disk: doubling the torque applied or halving the radius at which the torque is applied? Explain. 436 Chapter 10 | Rotational Motion and Angular Momentum Figure 10.42 A curved arrow lies at the side of a gray disk. There is a point at the center of the disk, and around the point there is a dashed circle. There is a point labeled \u201cChild\u201d on the dashed circle. Below the disc is a label saying \u201cTop View\u201d. The diagram above shows a top view of a child of mass M on a circular platform of mass 2M that is rotating counterclockwise. Assume the platform rotates without friction. Which of the following describes an action by the child that will increase the angular speed of the platformchild system and why? a. The child moves toward the center of the platform, increasing the total angular momentum of the system. b. The child moves toward the center of the platform, decreasing the rotational inertia of the system. c. The child moves away from the center of the platform, increasing the total angular momentum of the system. d. The child moves away from the center of the platform, decreasing the rotational inertia of the system. 16. Figure 10.43 A point labeled \u201cMoon\u201d lies on a dashed ellipse. Two other points, labeled \u201cA\u201d and \u201cB\u201d, lie at opposite ends of the ellipse. A point labeled \u201cPlanet\u201d lies inside the ellipse. A moon is in an elliptical orbit about a planet as shown above. At point A the moon has speed uA and is at distance RA from the planet. At point B the moon has speed uB. Has the moon's angular momentum changed? Explain your answer. 17. A hamster sits 0.10 m from the center of a lazy Susan of negligible mass. The wheel has an angular velocity of 1.0 rev/ s. How will the angular velocity of the lazy Susan change if the hamster walks to 0.30 m from the center of rotation? Assume zero friction and no external torque. a. b. c. d. It will speed up to 2.0 rev/s. It will speed up to 9.0 rev/s. It will slow to 0.01 rev/s. It will slow to 0.02 rev/s. 18. Earth has a mass of 6.0 \u00d7 1024 kg, a radius of 6.4 \u00d7 106 m, and an angular velocity of 1.2 \u00d7 10\u20135 rev/s. How would the planet's angular velocity change if a layer of Earth with mass 1.0 \u00d7 1023 kg broke off of the Earth, decreasing Earth's radius by 0.2 \u00d7 106 m? Assume no friction. 19. Consider system A, consisting of two disks of radius R, with both rotating clockwise. Now consider system B, consisting of one disk of radius R rotating counterclockwise and another disk of radius 2R rotating clockwise. All of the disks have the same mass, and all have the same magnitude of angular velocity. Which system has the greatest angular momentum? a. A b. B c. They're equal. d. Not enough information This content is available for free at http://cnx.org/content/col11844/1.13 20. Assume that a baseball bat being swung at 3\u03c0 rad/s by a batting machine is equivalent to a 1.1 m thin rod with a mass of 1.0 kg. How fast would a 0.15 kg baseball that squarely hits the very tip of the bat have to be going for the net angular momentum of the bat-ball system to be zero? 10.6 Collisions of Extended Bodies in Two Dimensions 21. A box with a mass of 2.0 kg rests on one end of a seesaw. The seesaw is 6.0 m long, and we can assume it has negligible mass. Approximately what angular momentum will the box have if someone with a mass of 65 kg sits on the other end of the seesaw quickly, with a velocity of 1.2 m/s? a. 702 kg\u2022m2/s b. 39 kg\u2022m2/s c. 18 kg\u2022m2/s d. 1.2 kg\u2022m2/s 22. A spinner in a board game can be thought of as a thin rod that", " spins about an axis at its center. The spinner in a certain game is 12 cm long and has a mass of 10 g. How will its angular velocity change when it is flicked at one end with a force equivalent to 15 g travelling at 5.0 m/s if all the energy of the collision is transferred to the spinner? (You can use the table in Figure 10.12 to estimate the rotational inertia of the spinner.) 23. A cyclist pedals to exert a torque on the rear wheel of the bicycle. When the cyclist changes to a higher gear, the torque increases. Which of the following would be the most effective strategy to help you determine the change in angular momentum of the bicycle wheel? a. multiplying the ratio between the two torques by the mass of the bicycle and rider b. adding the two torques together, and multiplying by the time for which both torques are applied c. multiplying the difference in the two torques by the time for which the new torque is applied d. multiplying both torques by the mass of the bicycle and rider 24. An electric screwdriver has two speeds, each of which exerts a different torque on a screw. Describe what calculations you could use to help you compare the angular momentum of a screw at each speed. What measurements would you need to make in order to calculate this? 25. Why is it important to consider the shape of an object when determining the object's angular momentum? a. The shape determines the location of the center of mass. The location of the center of mass in turn determines the angular velocity of the object. b. The shape helps you determine the location of the object's outer edge, where rotational velocity will be greatest. c. The shape helps you determine the location of the center of rotation. d. The shape determines the location of the center of mass. The location of the center of mass contributes to the object's rotational inertia, which contributes to its angular momentum. 26. How could you collect and analyze data to test the difference between the torques provided by two speeds on a tabletop fan? 27. Describe a rotational system you could use to demonstrate the effect on the system's angular momentum of applying different amounts of external torque. Chapter 10 | Rotational Motion and Angular Momentum 437 28. How could you use simple equipment such as balls and string to study the changes in angular momentum of a system when it interacts with another system? 10.7 Gyroscopic Effects: Vector Aspects of Angular Momentum 29. A globe (model of the Earth) is a hollow sphere with a radius of 16 cm. By wrapping a cord around the equator of a globe and pulling on it, a person exerts a torque on the globe of 120 N \u2022 m for 1.2 s. What angular momentum does the globe have after 1.2 s? 30. How could you use a fishing reel to test the relationship between the torque applied to a system, the time for which the torque was applied, and the resulting angular momentum of the system? How would you measure angular momentum? Chapter 16 | Oscillatory Motion and Waves 673 16 OSCILLATORY MOTION AND WAVES Figure 16.1 There are at least four types of waves in this picture\u2014only the water waves are evident. There are also sound waves, light waves, and waves on the guitar strings. (credit: John Norton) Chapter Outline 16.1. Hooke\u2019s Law: Stress and Strain Revisited 16.2. Period and Frequency in Oscillations 16.3. Simple Harmonic Motion: A Special Periodic Motion 16.4. The Simple Pendulum 16.5. Energy and the Simple Harmonic Oscillator 16.6. Uniform Circular Motion and Simple Harmonic Motion 16.7. Damped Harmonic Motion 16.8. Forced Oscillations and Resonance 16.9. Waves 16.10. Superposition and Interference 16.11. Energy in Waves: Intensity Connection for AP\u00ae Courses In this chapter, students are introduced to oscillation, the regular variation in the position of a system about a central point accompanied by transfer of energy and momentum, and to waves. A child\u2019s swing, a pendulum, a spring, and a vibrating string are all examples of oscillations. This chapter will address simple harmonic motion and periods of vibration, aspects of oscillation that produce waves, a common phenomenon in everyday life. Waves carry energy from one place to another.\u201d This chapter will show how harmonic oscillations produce waves that transport energy across space and through time. The information and examples presented support Big Ideas 1, 2, and 3 of the AP\u00ae Physics Curriculum Framework. The chapter opens by discussing the forces that govern oscillations and waves. It goes on to discuss important concepts such as simple harmonic motion, uniform harmonic motion, and damped harmonic motion. You will also learn about energy in simple harmonic motion and how it changes from kinetic to", " potential, and how the total sum, which would be the mechanical energy of the oscillator, remains constant or conserved at all times. The chapter also discusses characteristics of waves, such as their frequency, period of oscillation, and the forms in which they can exist, i.e., transverse or longitudinal. The chapter ends by discussing what happens when two or more waves overlap and how the amplitude of the resultant wave changes, leading to the phenomena of superposition and interference. 674 Chapter 16 | Oscillatory Motion and Waves The concepts in this chapter support: Big Idea 3 The interactions of an object with other objects can be described by forces. Enduring Understanding 3.B Classically, the acceleration of an object interacting with other objects can be predicted by using \u2192 \u2192 = \u2211. Essential Knowledge 3.B.3 Restoring forces can result in oscillatory motion. When a linear restoring force is exerted on an object displaced from an equilibrium position, the object will undergo a special type of motion called simple harmonic motion. Examples should include gravitational force exerted by the Earth on a simple pendulum and a mass-spring oscillator. Big Idea 4 Interactions between systems can result in changes in those systems. Enduring Understanding 4.C Interactions with other objects or systems can change the total energy of a system. Essential Knowledge 4.C.1 The energy of a system includes its kinetic energy, potential energy, and microscopic internal energy. Examples should include gravitational potential energy, elastic potential energy, and kinetic energy. Essential Knowledge 4.C.2 Mechanical energy (the sum of kinetic and potential energy) is transferred into or out of a system when an external force is exerted on a system such that a component of the force is parallel to its displacement. The process through which the energy is transferred is called work. Big Idea 5 Changes that occur as a result of interactions are constrained by conservation laws. Enduring Understanding 5.B The energy of a system is conserved. Essential Knowledge 5.B.2 A system with internal structure can have internal energy, and changes in a system\u2019s internal structure can result in changes in internal energy. [Physics 1: includes mass-spring oscillators and simple pendulums. Physics 2: includes charged object in electric \ufb01elds and examining changes in internal energy with changes in con\ufb01guration.] Big Idea 6 Waves can transfer energy and momentum from one location to another without the permanent transfer of mass and serve as a mathematical model for the description of other phenomena. Enduring Understanding 6.A A wave is a traveling disturbance that transfers energy and momentum. Essential Knowledge 6.A.1 Waves can propagate via different oscillation modes such as transverse and longitudinal. Essential Knowledge 6.A.2 For propagation, mechanical waves require a medium, while electromagnetic waves do not require a physical medium. Examples should include light traveling through a vacuum and sound not traveling through a vacuum. Essential Knowledge 6.A.3 The amplitude is the maximum displacement of a wave from its equilibrium value. Essential Knowledge 6.A.4 Classically, the energy carried by a wave depends on and increases with amplitude. Examples should include sound waves. Enduring Understanding 6.B A periodic wave is one that repeats as a function of both time and position and can be described by its amplitude, frequency, wavelength, speed, and energy. Essential Knowledge 6.B.1 The period is the repeat time of the wave. The frequency is the number of repetitions over a period of time. Essential Knowledge 6.B.2 The wavelength is the repeat distance of the wave. Essential Knowledge 6.B.3 A simple wave can be described by an equation involving one sine or cosine function involving the wavelength, amplitude, and frequency of the wave. Essential Knowledge 6.B.4 The wavelength is the ratio of speed over frequency. Enduring Understanding 6.C Only waves exhibit interference and diffraction. Essential Knowledge 6.C.1 When two waves cross, they travel through each other; they do not bounce off each other. Where the waves overlap, the resulting displacement can be determined by adding the displacements of the two waves. This is called superposition. Enduring Understanding 6.D Interference and superposition lead to standing waves and beats. Essential Knowledge 6.D.1 Two or more wave pulses can interact in such a way as to produce amplitude variations in the resultant wave. When two pulses cross, they travel through each other; they do not bounce off each other. Where the pulses overlap, the resulting displacement can be determined by adding the displacements of the two pulses. This is called superposition. Essential Knowledge 6.D.2 Two or more traveling waves can interact in such a way as to produce amplitude variations in the resultant wave. Essential Knowledge 6.D.3 Standing waves are the result of the addition of incident and re\ufb02ected waves that are con\ufb01ned to a region and have nodes and antinodes. Examples should include waves on a \ufffd", "\ufffdxed length of string, and sound waves in both closed and open tubes. Essential Knowledge 6.D.4 The possible wavelengths of a standing wave are determined by the size of the region to which it is con\ufb01ned. Essential Knowledge 6.D.5 Beats arise from the addition of waves of slightly different frequency. This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 16 | Oscillatory Motion and Waves 675 16.1 Hooke\u2019s Law: Stress and Strain Revisited Learning Objectives By the end of this section, you will be able to: \u2022 Explain Newton\u2019s third law of motion with respect to stress and deformation. \u2022 Describe the restoring force and displacement. \u2022 Use Hooke\u2019s law of deformation, and calculate stored energy in a spring. The information presented in this section supports the following AP\u00ae learning objectives and science practices: \u2022 3.B.3.3 The student can analyze data to identify qualitative or quantitative relationships between given values and variables (i.e., force, displacement, acceleration, velocity, period of motion, frequency, spring constant, string length, mass) associated with objects in oscillatory motion to use that data to determine the value of an unknown. (S.P. 2.2, 5.1) \u2022 3.B.3.4 The student is able to construct a qualitative and/or a quantitative explanation of oscillatory behavior given evidence of a restoring force. (S.P. 2.2, 6.2) Figure 16.2 When displaced from its vertical equilibrium position, this plastic ruler oscillates back and forth because of the restoring force opposing displacement. When the ruler is on the left, there is a force to the right, and vice versa. Newton\u2019s first law implies that an object oscillating back and forth is experiencing forces. Without force, the object would move in a straight line at a constant speed rather than oscillate. Consider, for example, plucking a plastic ruler to the left as shown in Figure 16.2. The deformation of the ruler creates a force in the opposite direction, known as a restoring force. Once released, the restoring force causes the ruler to move back toward its stable equilibrium position, where the net force on it is zero. However, by the time the ruler gets there, it gains momentum and continues to move to the right, producing the opposite deformation. It is then forced to the left, back through equilibrium, and the process is repeated until dissipative forces dampen the motion. These forces remove mechanical energy from the system, gradually reducing the motion until the ruler comes to rest. The simplest oscillations occur when the restoring force is directly proportional to displacement. When stress and strain were covered in Newton\u2019s Third Law of Motion, the name was given to this relationship between force and displacement was Hooke\u2019s law: = \u2212 (16.1) Here, is the restoring force, is the displacement from equilibrium or deformation, and is a constant related to the difficulty in deforming the system. The minus sign indicates the restoring force is in the direction opposite to the displacement. Figure 16.3 (a) The plastic ruler has been released, and the restoring force is returning the ruler to its equilibrium position. (b) The net force is zero at the equilibrium position, but the ruler has momentum and continues to move to the right. (c) The restoring force is in the opposite direction. It stops the ruler and moves it back toward equilibrium again. (d) Now the ruler has momentum to the left. (e) In the absence of damping (caused by frictional forces), the ruler reaches its original position. From there, the motion will repeat itself. 676 Chapter 16 | Oscillatory Motion and Waves The force constant is related to the rigidity (or stiffness) of a system\u2014the larger the force constant, the greater the restoring force, and the stiffer the system. The units of are newtons per meter (N/m). For example, is directly related to Young\u2019s modulus when we stretch a string. Figure 16.4 shows a graph of the absolute value of the restoring force versus the displacement for a system that can be described by Hooke\u2019s law\u2014a simple spring in this case. The slope of the graph equals the force constant in newtons per meter. A common physics laboratory exercise is to measure restoring forces created by springs, determine if they follow Hooke\u2019s law, and calculate their force constants if they do. Figure 16.4 (a) A graph of absolute value of the restoring force versus displacement is displayed. The fact that the graph is a straight line means that the system obeys Hooke\u2019s law. The slope of the graph is the force constant. (b) The data in the graph were generated by measuring the displacement of a spring from equilibrium while supporting various", " weights. The restoring force equals the weight supported, if the mass is stationary. Example 16.1 How Stiff Are Car Springs? Figure 16.5 The mass of a car increases due to the introduction of a passenger. This affects the displacement of the car on its suspension system. (credit: exfordy on Flickr) This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 16 | Oscillatory Motion and Waves 677 What is the force constant for the suspension system of a car that settles 1.20 cm when an 80.0-kg person gets in? Strategy Consider the car to be in its equilibrium position = 0 before the person gets in. The car then settles down 1.20 cm, which means it is displaced to a position = \u22121.20\u00d710\u22122 m. At that point, the springs supply a restoring force equal to the person\u2019s weight = = = 784 N. We take this force to be in Hooke\u2019s law. Knowing and 9.80 m/s2 80.0 kg, we can then solve the force constant. Solution 1. Solve Hooke\u2019s law, = \u2212, for : Substitute known values and solve : = \u2212. = \u2212 784 N \u22121.20\u00d710\u22122 m = 6.53\u00d7104 N/m. (16.2) (16.3) Discussion Note that and have opposite signs because they are in opposite directions\u2014the restoring force is up, and the displacement is down. Also, note that the car would oscillate up and down when the person got in if it were not for damping (due to frictional forces) provided by shock absorbers. Bouncing cars are a sure sign of bad shock absorbers. Energy in Hooke\u2019s Law of Deformation In order to produce a deformation, work must be done. That is, a force must be exerted through a distance, whether you pluck a guitar string or compress a car spring. If the only result is deformation, and no work goes into thermal, sound, or kinetic energy, then all the work is initially stored in the deformed object as some form of potential energy. The potential energy stored in a spring is PEel = 1 described by Hooke\u2019s law. Hence, 2 2. Here, we generalize the idea to elastic potential energy for a deformation of any system that can be 2 2, where PEel is the elastic potential energy stored in any deformed system that obeys Hooke\u2019s law and has a displacement from equilibrium and a force constant. PEel = 1 (16.4) It is possible to find the work done in deforming a system in order to find the energy stored. This work is performed by an applied force app. The applied force is exactly opposite to the restoring force (action-reaction), and so app =. Figure 16.6 shows a graph of the applied force versus deformation for a system that can be described by Hooke\u2019s law. Work done on the system is force multiplied by distance, which equals the area under the curve or (1 / 2) 2 (Method A in the figure). Another way to determine the work is to note that the force increases linearly from 0 to, so that the average force is (1 / 2), the distance moved is, and thus = app = [(1 / 2)]() = (1 / 2) 2 (Method B in the figure). 678 Chapter 16 | Oscillatory Motion and Waves Figure 16.6 A graph of applied force versus distance for the deformation of a system that can be described by Hooke\u2019s law is displayed. The work done on the system equals the area under the graph or the area of the triangle, which is half its base multiplied by its height, or = (1 / 2) 2. Example 16.2 Calculating Stored Energy: A Tranquilizer Gun Spring We can use a toy gun\u2019s spring mechanism to ask and answer two simple questions: (a) How much energy is stored in the spring of a tranquilizer gun that has a force constant of 50.0 N/m and is compressed 0.150 m? (b) If you neglect friction and the mass of the spring, at what speed will a 2.00-g projectile be ejected from the gun? Figure 16.7 (a) In this image of the gun, the spring is uncompressed before being cocked. (b) The spring has been compressed a distance, and the projectile is in place. (c) When released, the spring converts elastic potential energy PEel into kinetic energy. Strategy for a (a): The energy stored in the spring can be found directly from elastic potential energy equation, because and are given. Solution for a Entering the given values for and yields PEel = 1 2 2 = 1 2 =", " 0.563 J Strategy for b (50.0 N/m)(0.150 m)2 = 0.563 N \u22c5 m (16.5) Because there is no friction, the potential energy is converted entirely into kinetic energy. The expression for kinetic energy can be solved for the projectile\u2019s speed. Solution for b This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 16 | Oscillatory Motion and Waves 1. Identify known quantities: 2. Solve for : 3. Convert units: 23.7 m / s Discussion KEf = PEel or 1 / 22 = (1 / 2)2 = PEel = 0.563 J 1 / 2 = 2PEel = 2(0.563 J) 0.002 kg 1 / 2 = 23.7 J/kg 1 / 2 679 (16.6) (16.7) (a) and (b): This projectile speed is impressive for a tranquilizer gun (more than 80 km/h). The numbers in this problem seem reasonable. The force needed to compress the spring is small enough for an adult to manage, and the energy imparted to the dart is small enough to limit the damage it might do. Yet, the speed of the dart is great enough for it to travel an acceptable distance. Check your Understanding Envision holding the end of a ruler with one hand and deforming it with the other. When you let go, you can see the oscillations of the ruler. In what way could you modify this simple experiment to increase the rigidity of the system? Solution You could hold the ruler at its midpoint so that the part of the ruler that oscillates is half as long as in the original experiment. Check your Understanding If you apply a deforming force on an object and let it come to equilibrium, what happened to the work you did on the system? Solution It was stored in the object as potential energy. 16.2 Period and Frequency in Oscillations By the end of this section, you will be able to: Learning Objectives \u2022 Relate recurring mechanical vibrations to the frequency and period of harmonic motion, such as the motion of a guitar string. \u2022 Compute the frequency and period of an oscillation. The information presented in this section supports the following AP\u00ae learning objectives and science practices: \u2022 3.B.3.3 The student can analyze data to identify qualitative or quantitative relationships between given values and variables (i.e., force, displacement, acceleration, velocity, period of motion, frequency, spring constant, string length, mass) associated with objects in oscillatory motion to use that data to determine the value of an unknown. (S.P. 2.2, 5.1) Figure 16.8 The strings on this guitar vibrate at regular time intervals. (credit: JAR) 680 Chapter 16 | Oscillatory Motion and Waves When you pluck a guitar string, the resulting sound has a steady tone and lasts a long time. Each successive vibration of the string takes the same time as the previous one. We define periodic motion to be a motion that repeats itself at regular time intervals, such as exhibited by the guitar string or by an object on a spring moving up and down. The time to complete one oscillation remains constant and is called the period. Its units are usually seconds, but may be any convenient unit of time. The word period refers to the time for some event whether repetitive or not; but we shall be primarily interested in periodic motion, which is by definition repetitive. A concept closely related to period is the frequency of an event. For example, if you get a paycheck twice a month, the frequency of payment is two per month and the period between checks is half a month. Frequency is defined to be the number of events per unit time. For periodic motion, frequency is the number of oscillations per unit time. The relationship between frequency and period is The SI unit for frequency is the cycle per second, which is defined to be a hertz (Hz): 1 Hz = 1 cycle sec or 1 Hz = 1 s = 1. (16.8) (16.9) A cycle is one complete oscillation. Note that a vibration can be a single or multiple event, whereas oscillations are usually repetitive for a significant number of cycles. Example 16.3 Determine the Frequency of Two Oscillations: Medical Ultrasound and the Period of Middle C We can use the formulas presented in this module to determine both the frequency based on known oscillations and the oscillation based on a known frequency. Let\u2019s try one example of each. (a) A medical imaging device produces ultrasound by oscillating with a period of 0.400 \u00b5s. What is the frequency of this oscillation? (b) The frequency of middle C on a typical musical instrument is 264 Hz. What is the time for one complete oscillation? Strategy Both questions (a) and (b)", " can be answered using the relationship between period and frequency. In question (a), the period is given and we are asked to find frequency. In question (b), the frequency period. is given and we are asked to find the Solution a 1. Substitute 0.400 \u03bcs for in = 1 : Solve to find Discussion a = 1 = 1 0.400\u00d710\u22126 s. = 2.50\u00d7106 Hz. (16.10) (16.11) The frequency of sound found in (a) is much higher than the highest frequency that humans can hear and, therefore, is called ultrasound. Appropriate oscillations at this frequency generate ultrasound used for noninvasive medical diagnoses, such as observations of a fetus in the womb. Solution b 1. Identify the known values: The time for one complete oscillation is the period : 2. Solve for : = 1. = 1. 3. Substitute the given value for the frequency into the resulting expression: = 1 = 1 264 Hz = 1 264 cycles/s = 3.79\u00d710\u22123 s = 3.79 ms. Discussion (16.12) (16.13) (16.14) The period found in (b) is the time per cycle, but this value is often quoted as simply the time in convenient units (ms or milliseconds in this case). This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 16 | Oscillatory Motion and Waves 681 Check your Understanding Identify an event in your life (such as receiving a paycheck) that occurs regularly. Identify both the period and frequency of this event. Solution I visit my parents for dinner every other Sunday. The frequency of my visits is 26 per calendar year. The period is two weeks. 16.3 Simple Harmonic Motion: A Special Periodic Motion By the end of this section, you will be able to: Learning Objectives \u2022 Describe a simple harmonic oscillator. \u2022 Relate physical characteristics of a vibrating system to aspects of simple harmonic motion and any resulting waves. The information presented in this section supports the following AP\u00ae learning objectives and science practices: \u2022 3.B.3.1 The student is able to predict which properties determine the motion of a simple harmonic oscillator and what the dependence of the motion is on those properties. (S.P. 6.4, 7.2) \u2022 3.B.3.4 The student is able to construct a qualitative and/or a quantitative explanation of oscillatory behavior given evidence of a restoring force. (S.P. 2.2, 6.2) \u2022 6.A.3.1 The student is able to use graphical representation of a periodic mechanical wave to determine the amplitude of the wave. (S.P. 1.4) \u2022 6.B.1.1 The student is able to use a graphical representation of a periodic mechanical wave (position versus time) to determine the period and frequency of the wave and describe how a change in the frequency would modify features of the representation. (S.P. 1.4, 2.2) The oscillations of a system in which the net force can be described by Hooke\u2019s law are of special importance, because they are very common. They are also the simplest oscillatory systems. Simple Harmonic Motion (SHM) is the name given to oscillatory motion for a system where the net force can be described by Hooke\u2019s law, and such a system is called a simple harmonic oscillator. If the net force can be described by Hooke\u2019s law and there is no damping (by friction or other non-conservative forces), then a simple harmonic oscillator will oscillate with equal displacement on either side of the equilibrium position, as shown for an object on a spring in Figure 16.9. The maximum displacement from equilibrium is called the amplitude. The units for amplitude and displacement are the same, but depend on the type of oscillation. For the object on the spring, the units of amplitude and displacement are meters; whereas for sound oscillations, they have units of pressure (and other types of oscillations have yet other units). Because amplitude is the maximum displacement, it is related to the energy in the oscillation. Take-Home Experiment: SHM and the Marble Find a bowl or basin that is shaped like a hemisphere on the inside. Place a marble inside the bowl and tilt the bowl periodically so the marble rolls from the bottom of the bowl to equally high points on the sides of the bowl. Get a feel for the force required to maintain this periodic motion. What is the restoring force and what role does the force you apply play in the simple harmonic motion (SHM) of the marble? 682 Chapter 16 | Oscillatory Motion and Waves Figure 16.9 An object attached to a spring sliding on a frictionless surface is an uncomplicated simple harmonic oscillator. When displaced from", " equilibrium, the object performs simple harmonic motion that has an amplitude and a period. The object\u2019s maximum speed occurs as it passes through equilibrium. The stiffer the spring is, the smaller the period. The greater the mass of the object is, the greater the period. What is so significant about simple harmonic motion? One special thing is that the period and frequency of a simple harmonic oscillator are independent of amplitude. The string of a guitar, for example, will oscillate with the same frequency whether plucked gently or hard. Because the period is constant, a simple harmonic oscillator can be used as a clock. Two important factors do affect the period of a simple harmonic oscillator. The period is related to how stiff the system is. A very stiff object has a large force constant, which causes the system to have a smaller period. For example, you can adjust a diving board\u2019s stiffness\u2014the stiffer it is, the faster it vibrates, and the shorter its period. Period also depends on the mass of the oscillating system. The more massive the system is, the longer the period. For example, a heavy person on a diving board bounces up and down more slowly than a light one. In fact, the mass and the force constant are the only factors that affect the period and frequency of simple harmonic motion. Period of Simple Harmonic Oscillator The period of a simple harmonic oscillator is given by = 2\u03c0 and, because = 1 /, the frequency of a simple harmonic oscillator is Note that neither nor has any dependence on amplitude. = 1 2\u03c0. Example 16.4 Mechanical Waves (16.15) (16.16) What do sound waves, water waves, and seismic waves have in common? They are all governed by Newton\u2019s laws and they can exist only when traveling in a medium, such as air, water, or rocks. Waves that require a medium to travel are collectively known as \u201cmechanical waves.\u201d This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 16 | Oscillatory Motion and Waves 683 Take-Home Experiment: Mass and Ruler Oscillations Find two identical wooden or plastic rulers. Tape one end of each ruler firmly to the edge of a table so that the length of each ruler that protrudes from the table is the same. On the free end of one ruler tape a heavy object such as a few large coins. Pluck the ends of the rulers at the same time and observe which one undergoes more cycles in a time period, and measure the period of oscillation of each of the rulers. Example 16.5 Calculate the Frequency and Period of Oscillations: Bad Shock Absorbers in a Car If the shock absorbers in a car go bad, then the car will oscillate at the least provocation, such as when going over bumps in the road and after stopping (See Figure 16.10). Calculate the frequency and period of these oscillations for such a car if the car\u2019s mass (including its load) is 900 kg and the force constant ( ) of the suspension system is 6.53\u00d7104 N/m. Strategy The frequency of the car\u2019s oscillations will be that of a simple harmonic oscillator as given in the equation = 1 2\u03c0. The mass and the force constant are both given. Solution 1. Enter the known values of k and m: = 1 2\u03c0 = 1 2\u03c0 6.53\u00d7104 N/m 900 kg. 1 2\u03c0 72.6 / s\u20132 = 1.3656 / s\u20131 \u2248 1.36 / s\u20131 = 1.36 Hz. (16.17) (16.18) to calculate the period, but it is simpler to use the relationship = 1 / and substitute = 1 = 1 1.356 Hz = 0.738 s. (16.19) 2. Calculate the frequency: 3. You could use = 2\u03c0 : the value just found for Discussion The values of and both seem about right for a bouncing car. You can observe these oscillations if you push down hard on the end of a car and let go. The Link between Simple Harmonic Motion and Waves If a time-exposure photograph of the bouncing car were taken as it drove by, the headlight would make a wavelike streak, as shown in Figure 16.10. Similarly, Figure 16.11 shows an object bouncing on a spring as it leaves a wavelike \"trace of its position on a moving strip of paper. Both waves are sine functions. All simple harmonic motion is intimately related to sine and cosine waves. Figure 16.10 The bouncing car makes a wavelike motion. If the restoring force in the suspension system can be described only by Hooke\u2019s law, then the wave is a sine function. (The wave is the trace produced by the headlight as the car", " moves to the right.) 684 Chapter 16 | Oscillatory Motion and Waves Figure 16.11 The vertical position of an object bouncing on a spring is recorded on a strip of moving paper, leaving a sine wave. The displacement as a function of time t in any simple harmonic motion\u2014that is, one in which the net restoring force can be described by Hooke\u2019s law, is given by () = cos2, (16.20) where is amplitude. At = 0, the initial position is 0 =, and the displacement oscillates back and forth with a period. (When =, we get = again because cos 2\u03c0 = 1.). Furthermore, from this expression for, the velocity as a function of time is given by: () = \u2212max sin 2\u03c0, (16.21) where max = 2\u03c0 / = /. The object has zero velocity at maximum displacement\u2014for example, = 0 when = 0, and at that time =. The minus sign in the first equation for () gives the correct direction for the velocity. Just after the start of the motion, for instance, the velocity is negative because the system is moving back toward the equilibrium point. Finally, we can get an expression for acceleration using Newton\u2019s second law. [Then we have () (), and (), the quantities needed for kinematics and a description of simple harmonic motion.] According to Newton\u2019s second law, the acceleration is = / = /. So, () is also a cosine function: () = \u2212 cos2\u03c0. (16.22) Hence, () is directly proportional to and in the opposite direction to (). Figure 16.12 shows the simple harmonic motion of an object on a spring and presents graphs of ()(), and () versus time. This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 16 | Oscillatory Motion and Waves 685 Figure 16.12 Graphs of () () and () versus for the motion of an object on a spring. The net force on the object can be described by Hooke\u2019s law, and so the object undergoes simple harmonic motion. Note that the initial position has the vertical displacement at its maximum value ; is initially zero and then negative as the object moves down; and the initial acceleration is negative, back toward the equilibrium position and becomes zero at that point. The most important point here is that these equations are mathematically straightforward and are valid for all simple harmonic motion. They are very useful in visualizing waves associated with simple harmonic motion, including visualizing how waves add with one another. Check Your Understanding Suppose you pluck a banjo string. You hear a single note that starts out loud and slowly quiets over time. Describe what happens to the sound waves in terms of period, frequency and amplitude as the sound decreases in volume. Solution Frequency and period remain essentially unchanged. Only amplitude decreases as volume decreases. Check Your Understanding A babysitter is pushing a child on a swing. At the point where the swing reaches, where would the corresponding point on a wave of this motion be located? Solution 686 Chapter 16 | Oscillatory Motion and Waves is the maximum deformation, which corresponds to the amplitude of the wave. The point on the wave would either be at the very top or the very bottom of the curve. PhET Explorations: Masses and Springs A realistic mass and spring laboratory. Hang masses from springs and adjust the spring stiffness and damping. You can even slow time. Transport the lab to different planets. A chart shows the kinetic, potential, and thermal energy for each spring. Figure 16.13 Masses and Springs (http://cnx.org/content/m55273/1.2/mass-spring-lab_en.jar) 16.4 The Simple Pendulum By the end of this section, you will be able to: \u2022 Determine the period of oscillation of a hanging pendulum. Learning Objectives The information presented in this section supports the following AP\u00ae learning objectives and science practices: \u2022 3.B.3.1 The student is able to predict which properties determine the motion of a simple harmonic oscillator and what the dependence of the motion is on those properties. (S.P. 6.4, 7.2) \u2022 3.B.3.2 The student is able to design a plan and collect data in order to ascertain the characteristics of the motion of a system undergoing oscillatory motion caused by a restoring force. (S.P. 4.2) \u2022 3.B.3.3 The student can analyze data to identify qualitative or quantitative relationships between given values and variables (i.e., force, displacement, acceleration, velocity, period of motion, frequency, spring constant, string length, mass) associated with objects in oscillatory motion to use that data to determine the value of an unknown. (S.P. 2.2, 5.1) \u2022 3.", "B.3.4 The student is able to construct a qualitative and/or a quantitative explanation of oscillatory behavior given evidence of a restoring force. (S.P. 2.2, 6.2) Figure 16.14 A simple pendulum has a small-diameter bob and a string that has a very small mass but is strong enough not to stretch appreciably. The linear displacement from equilibrium is, the length of the arc. Also shown are the forces on the bob, which result in a net force of \u2212 sin toward the equilibrium position\u2014that is, a restoring force. Pendulums are in common usage. Some have crucial uses, such as in clocks; some are for fun, such as a child\u2019s swing; and some are just there, such as the sinker on a fishing line. For small displacements, a pendulum is a simple harmonic oscillator. A simple pendulum is defined to have an object that has a small mass, also known as the pendulum bob, which is suspended from a light wire or string, such as shown in Figure 16.14. Exploring the simple pendulum a bit further, we can discover the conditions under which it performs simple harmonic motion, and we can derive an interesting expression for its period. We begin by defining the displacement to be the arc length. We see from Figure 16.14 that the net force on the bob is tangent to the arc and equals \u2212 sin. (The weight has components cos along the string and sin tangent to the arc.) Tension in the string exactly cancels the component cos parallel to the string. This leaves a net restoring force back toward the equilibrium position at = 0. This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 16 | Oscillatory Motion and Waves 687 Now, if we can show that the restoring force is directly proportional to the displacement, then we have a simple harmonic oscillator. In trying to determine if we have a simple harmonic oscillator, we should note that for small angles (less than about 15\u00ba ), sin \u2248 ( sin and differ by about 1% or less at smaller angles). Thus, for angles less than about 15\u00ba, the restoring force is \u2248 \u2212\u03b8. (16.23) The displacement is directly proportional to. When is expressed in radians, the arc length in a circle is related to its radius ( in this instance) by: so that =, =. For small angles, then, the expression for the restoring force is: This expression is of the form: \u2248 \u2212 = \u2212, (16.24) (16.25) (16.26) (16.27) where the force constant is given by = / and the displacement is given by =. For angles less than about 15\u00ba, the restoring force is directly proportional to the displacement, and the simple pendulum is a simple harmonic oscillator. Using this equation, we can find the period of a pendulum for amplitudes less than about 15\u00ba. For the simple pendulum: Thus, = 2\u03c0 = 2\u03c0 / = 2\u03c0 (16.28) (16.29) for the period of a simple pendulum. This result is interesting because of its simplicity. The only things that affect the period of a simple pendulum are its length and the acceleration due to gravity. The period is completely independent of other factors, such as mass. As with simple harmonic oscillators, the period for a pendulum is nearly independent of amplitude, especially if is less than about 15\u00ba. Even simple pendulum clocks can be finely adjusted and accurate. Note the dependence of on. If the length of a pendulum is precisely known, it can actually be used to measure the acceleration due to gravity. Consider the following example. Example 16.6 Measuring Acceleration due to Gravity: The Period of a Pendulum What is the acceleration due to gravity in a region where a simple pendulum having a length 75.000 cm has a period of 1.7357 s? Strategy We are asked to find given the period and the length of a pendulum. We can solve = 2\u03c0 for, assuming only that the angle of deflection is less than 15\u00ba. Solution 1. Square = 2\u03c0 and solve for : 2. Substitute known values into the new equation: = 4\u03c02 2. 3. Calculate to find : = 4\u03c02 0.75000 m (1.7357 s)2. = 9.8281 m / s2. (16.30) (16.31) (16.32) 688 Discussion Chapter 16 | Oscillatory Motion and Waves This method for determining can be very accurate. This is why length and period are given to five digits in this example. For the precision of the approximation sin \u03b8 \u2248 to be better than the precision of the pendulum length and period, the maximum displacement angle should be kept below about", " 0.5\u00ba. Making Career Connections Knowing can be important in geological exploration; for example, a map of over large geographical regions aids the study of plate tectonics and helps in the search for oil fields and large mineral deposits. Take Home Experiment: Determining Use a simple pendulum to determine the acceleration due to gravity in your own locale. Cut a piece of a string or dental floss so that it is about 1 m long. Attach a small object of high density to the end of the string (for example, a metal nut or a car key). Starting at an angle of less than 10\u00ba, allow the pendulum to swing and measure the pendulum\u2019s period for 10 oscillations using a stopwatch. Calculate. How accurate is this measurement? How might it be improved? Check Your Understanding An engineer builds two simple pendula. Both are suspended from small wires secured to the ceiling of a room. Each pendulum hovers 2 cm above the floor. Pendulum 1 has a bob with a mass of 10 kg. Pendulum 2 has a bob with a mass of 100 kg. Describe how the motion of the pendula will differ if the bobs are both displaced by 12\u00ba. Solution The movement of the pendula will not differ at all because the mass of the bob has no effect on the motion of a simple pendulum. The pendula are only affected by the period (which is related to the pendulum\u2019s length) and by the acceleration due to gravity. PhET Explorations: Pendulum Lab Play with one or two pendulums and discover how the period of a simple pendulum depends on the length of the string, the mass of the pendulum bob, and the amplitude of the swing. It\u2019s easy to measure the period using the photogate timer. You can vary friction and the strength of gravity. Use the pendulum to find the value of on planet X. Notice the anharmonic behavior at large amplitude. Figure 16.15 Pendulum Lab (http://cnx.org/content/m55274/1.2/pendulum-lab_en.jar) 16.5 Energy and the Simple Harmonic Oscillator By the end of this section, you will be able to: Learning Objectives \u2022 Describe the changes in energy that occur while a system undergoes simple harmonic motion. To study the energy of a simple harmonic oscillator, we first consider all the forms of energy it can have We know from Hooke\u2019s Law: Stress and Strain Revisited that the energy stored in the deformation of a simple harmonic oscillator is a form of potential energy given by: PEel = 1 22. (16.33) Because a simple harmonic oscillator has no dissipative forces, the other important form of energy is kinetic energy KE. Conservation of energy for these two forms is: This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 16 | Oscillatory Motion and Waves or KE + PEel = constant 22 + 1 1 22 = constant. 689 (16.34) (16.35) This statement of conservation of energy is valid for all simple harmonic oscillators, including ones where the gravitational force plays a role Namely, for a simple pendulum we replace the velocity with =, the spring constant with = /, and the displacement term with =. Thus 222 + 1 1 2 2 = constant. (16.36) In the case of undamped simple harmonic motion, the energy oscillates back and forth between kinetic and potential, going completely from one to the other as the system oscillates. So for the simple example of an object on a frictionless surface attached to a spring, as shown again in Figure 16.16, the motion starts with all of the energy stored in the spring. As the object starts to move, the elastic potential energy is converted to kinetic energy, becoming entirely kinetic energy at the equilibrium position. It is then converted back into elastic potential energy by the spring, the velocity becomes zero when the kinetic energy is completely converted, and so on. This concept provides extra insight here and in later applications of simple harmonic motion, such as alternating current circuits. Figure 16.16 The transformation of energy in simple harmonic motion is illustrated for an object attached to a spring on a frictionless surface. The conservation of energy principle can be used to derive an expression for velocity. If we start our simple harmonic motion with zero velocity and maximum displacement ( = ), then the total energy is 1 2 2. (16.37) This total energy is constant and is shifted back and forth between kinetic energy and potential energy, at most times being shared by each. The conservation of energy for this system in equation form is thus: 22 + 1 1 22 = 1 2 2. Solving this equation for yields: Manipulating this expression algebraically gives: = \u00b1 2 \u2212 2. (16.38) (16.39", ") 690 and so where = \u00b1 1 \u2212 2 2 = \u00b1max 1 \u2212 2 2, max =. Chapter 16 | Oscillatory Motion and Waves (16.40) (16.41) (16.42) From this expression, we see that the velocity is a maximum ( max ) at = 0, as stated earlier in () = \u2212 max sin 2\u03c0 Notice that the maximum velocity depends on three factors. Maximum velocity is directly proportional to amplitude. As you might guess, the greater the maximum displacement the greater the maximum velocity. Maximum velocity is also greater for stiffer systems, because they exert greater force for the same displacement. This observation is seen in the expression for max; it is proportional to the square root of the force constant. Finally, the maximum velocity is smaller for objects that have larger masses, because the maximum velocity is inversely proportional to the square root of. For a given force, objects that have large masses accelerate more slowly.. A similar calculation for the simple pendulum produces a similar result, namely: max = max. (16.43) Making Connections: Mass Attached to a Spring Consider a mass m attached to a spring, with spring constant k, fixed to a wall. When the mass is displaced from its equilibrium position and released, the mass undergoes simple harmonic motion. The spring exerts a force = \u2212 on the mass. The potential energy of the system is stored in the spring. It will be zero when the spring is in the equilibrium position. All the internal energy exists in the form of kinetic energy, given by = 1 22. As the system oscillates, which means that the spring compresses and expands, there is a change in the structure of the system and a corresponding change in its internal energy. Its kinetic energy is converted to potential energy and vice versa. This occurs at an equal rate, which means that a loss of kinetic energy yields a gain in potential energy, thus preserving the work-energy theorem and the law of conservation of energy. Example 16.7 Determine the Maximum Speed of an Oscillating System: A Bumpy Road Suppose that a car is 900 kg and has a suspension system that has a force constant = 6.53\u00d7104 N/m. The car hits a bump and bounces with an amplitude of 0.100 m. What is its maximum vertical velocity if you assume no damping occurs? Strategy We can use the expression for max given in max = and are given in the problem statement, and the maximum displacement is 0.100 m. to determine the maximum vertical velocity. The variables Solution 1. Identify known. 2. Substitute known values into max = : max = 6.53\u00d7104 N/m 900 kg (0.100 m). (16.44) 3. Calculate to find max= 0.852 m/s. Discussion This answer seems reasonable for a bouncing car. There are other ways to use conservation of energy to find max. We could use it directly, as was done in the example featured in Hooke\u2019s Law: Stress and Strain Revisited. The small vertical displacement of an oscillating simple pendulum, starting from its equilibrium position, is given as This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 16 | Oscillatory Motion and Waves where is the amplitude, is the angular velocity and is the time taken. Substituting = 2\u03c0, we have () = sin, = sin 2\u03c0. Thus, the displacement of pendulum is a function of time as shown above. Also the velocity of the pendulum is given by () = 2 cos 2\u03c0, so the motion of the pendulum is a function of time. Check Your Understanding 691 (16.45) (16.46) (16.47) Why does it hurt more if your hand is snapped with a ruler than with a loose spring, even if the displacement of each system is equal? Solution The ruler is a stiffer system, which carries greater force for the same amount of displacement. The ruler snaps your hand with greater force, which hurts more. Check Your Understanding You are observing a simple harmonic oscillator. Identify one way you could decrease the maximum velocity of the system. Solution You could increase the mass of the object that is oscillating. 16.6 Uniform Circular Motion and Simple Harmonic Motion Learning Objectives By the end of this section, you will be able to: \u2022 Compare simple harmonic motion with uniform circular motion. Figure 16.17 The horses on this merry-go-round exhibit uniform circular motion. (credit: Wonderlane, Flickr) There is an easy way to produce simple harmonic motion by using uniform circular motion. Figure 16.18 shows one way of using this method. A ball is attached to a uniformly rotating vertical turntable, and its shadow is projected on the floor as shown. The shadow undergoes simple harmonic motion. Hooke\u2019s law usually describes uniform circular motions ( constant", ") rather than systems that have large visible displacements. So observing the projection of uniform circular motion, as in Figure 16.18, is often easier than observing a precise large-scale simple harmonic oscillator. If studied in sufficient depth, simple harmonic motion produced in this manner can give considerable insight into many aspects of oscillations and waves and is very useful mathematically. In our brief treatment, we shall indicate some of the major features of this relationship and how they might be useful. 692 Chapter 16 | Oscillatory Motion and Waves Figure 16.18 The shadow of a ball rotating at constant angular velocity on a turntable goes back and forth in precise simple harmonic motion. Figure 16.19 shows the basic relationship between uniform circular motion and simple harmonic motion. The point P travels around the circle at constant angular velocity. The point P is analogous to an object on the merry-go-round. The projection of the position of P onto a fixed axis undergoes simple harmonic motion and is analogous to the shadow of the object. At the time shown in the figure, the projection has position and moves to the left with velocity. The velocity of the point P around the \u00af circle equals \u00af max.The projection of max on the -axis is the velocity of the simple harmonic motion along the -axis. Figure 16.19 A point P moving on a circular path with a constant angular velocity is undergoing uniform circular motion. Its projection on the x-axis \u00af undergoes simple harmonic motion. Also shown is the velocity of this point around the circle, velocities form a similar triangle to the displacement triangle. max, and its projection, which is. Note that these To see that the projection undergoes simple harmonic motion, note that its position is given by = cos, where =, is the constant angular velocity, and is the radius of the circular path. Thus, = cos. The angular velocity is in radians per unit time; in this case 2\u03c0 radians is the time for one revolution. That is, = 2\u03c0 /. Substituting this expression for, we see that the position is given by: () = cos 2\u03c0. (16.48) (16.49) (16.50) This expression is the same one we had for the position of a simple harmonic oscillator in Simple Harmonic Motion: A Special Periodic Motion. If we make a graph of position versus time as in Figure 16.20, we see again the wavelike character (typical of simple harmonic motion) of the projection of uniform circular motion onto the -axis. This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 16 | Oscillatory Motion and Waves 693 Figure 16.20 The position of the projection of uniform circular motion performs simple harmonic motion, as this wavelike graph of versus indicates. Now let us use Figure 16.19 to do some further analysis of uniform circular motion as it relates to simple harmonic motion. The triangle formed by the velocities in the figure and the triangle formed by the displacements ( and 2 \u2212 2 ) are similar right triangles. Taking ratios of similar sides, we see that We can solve this equation for the speed or max = 2 \u2212 2 = 1 \u2212 2 2. = max 1 \u2212 2 2. (16.51) (16.52) This expression for the speed of a simple harmonic oscillator is exactly the same as the equation obtained from conservation of energy considerations in Energy and the Simple Harmonic Oscillator. You can begin to see that it is possible to get all of the characteristics of simple harmonic motion from an analysis of the projection of uniform circular motion. Finally, let us consider the period of the motion of the projection. This period is the time it takes the point P to complete one revolution. That time is the circumference of the circle 2\u03c0 divided by the velocity around the circle, max. Thus, the period is We know from conservation of energy considerations that = 2\u03c0X max. Solving this equation for / max gives max =. max =. Substituting this expression into the equation for yields = 2\u03c0. (16.53) (16.54) (16.55) (16.56) Thus, the period of the motion is the same as for a simple harmonic oscillator. We have determined the period for any simple harmonic oscillator using the relationship between uniform circular motion and simple harmonic motion. Some modules occasionally refer to the connection between uniform circular motion and simple harmonic motion. Moreover, if you carry your study of physics and its applications to greater depths, you will find this relationship useful. It can, for example, help to analyze how waves add when they are superimposed. 700 Chapter 16 | Oscillatory Motion and Waves Check Your Understanding A famous magic trick involves a performer singing a note toward a crystal glass until the glass shatters. Explain why the trick works in terms of resonance and natural frequency. Solution The performer must be singing a note that corresponds to the", " natural frequency of the glass. As the sound wave is directed at the glass, the glass responds by resonating at the same frequency as the sound wave. With enough energy introduced into the system, the glass begins to vibrate and eventually shatters. 16.9 Waves Learning Objectives By the end of this section, you will be able to: \u2022 Describe various characteristics associated with a wave. \u2022 Differentiate between transverse and longitudinal waves. Figure 16.29 Waves in the ocean behave similarly to all other types of waves. (credit: Steve Jurveston, Flickr) What do we mean when we say something is a wave? The most intuitive and easiest wave to imagine is the familiar water wave. More precisely, a wave is a disturbance that propagates, or moves from the place it was created. For water waves, the disturbance is in the surface of the water, perhaps created by a rock thrown into a pond or by a swimmer splashing the surface repeatedly. For sound waves, the disturbance is a change in air pressure, perhaps created by the oscillating cone inside a speaker. For earthquakes, there are several types of disturbances, including disturbance of Earth\u2019s surface and pressure disturbances under the surface. Even radio waves are most easily understood using an analogy with water waves. Visualizing water waves is useful because there is more to it than just a mental image. Water waves exhibit characteristics common to all waves, such as amplitude, period, frequency and energy. All wave characteristics can be described by a small set of underlying principles. A wave is a disturbance that propagates, or moves from the place it was created. The simplest waves repeat themselves for several cycles and are associated with simple harmonic motion. Let us start by considering the simplified water wave in Figure 16.30. The wave is an up and down disturbance of the water surface. It causes a sea gull to move up and down in simple harmonic motion as the wave crests and troughs (peaks and valleys) pass under the bird. The time for one complete up and down motion is the wave\u2019s period. The wave\u2019s frequency is = 1 /, as usual. The wave itself moves to the right in the figure. This movement of the wave is actually the disturbance moving to the right, not the water itself (or the bird would move to the right). We define wave velocity w to be the speed at which the disturbance moves. Wave velocity is sometimes also called the propagation velocity or propagation speed, because the disturbance propagates from one location to another. Misconception Alert Many people think that water waves push water from one direction to another. In fact, the particles of water tend to stay in one location, save for moving up and down due to the energy in the wave. The energy moves forward through the water, but the water stays in one place. If you feel yourself pushed in an ocean, what you feel is the energy of the wave, not a rush of water. This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 16 | Oscillatory Motion and Waves 701 Figure 16.30 An idealized ocean wave passes under a sea gull that bobs up and down in simple harmonic motion. The wave has a wavelength, which is the distance between adjacent identical parts of the wave. The up and down disturbance of the surface propagates parallel to the surface at a speed w. The water wave in the figure also has a length associated with it, called its wavelength, the distance between adjacent identical parts of a wave. ( is the distance parallel to the direction of propagation.) The speed of propagation w is the distance the wave travels in a given time, which is one wavelength in the time of one period. In equation form, that is or w = w =. (16.66) (16.67) This fundamental relationship holds for all types of waves. For water waves, w is the speed of a surface wave; for sound, w is the speed of sound; and for visible light, w is the speed of light, for example. Applying the Science Practices: Different Types of Waves Consider a spring fixed to a wall with a mass connected to its end. This fixed point on the wall exerts a force on the complete spring-and-mass system, and this implies that the momentum of the complete system is not conserved. Now, consider energy. Since the system is fixed to a point on the wall, it does not do any work; hence, the total work done is conserved, which means that the energy is conserved. Consequently, we have an oscillator in which energy is conserved but momentum is not. Now, consider a system of two masses connected to each other by a spring. This type of system also forms an oscillator. Since there is no fixed point, momentum is conserved as the forces acting on the two masses are equal and opposite. Energy for such a system will be conserved, because there are no external forces acting on", " the spring-twomasses system. It is clear from above that, for momentum to be conserved, momentum needs to be carried by waves. This is a typical example of a mechanical oscillator producing mechanical waves that need a medium in which to propagate. Sound waves are also examples of mechanical waves. There are some waves that can travel in the absence of a medium of propagation. Such waves are called \u201celectromagnetic waves.\u201d Light waves are examples of electromagnetic waves. Electromagnetic waves are created by the vibration of electric charge. This vibration creates a wave with both electric and magnetic field components. Take-Home Experiment: Waves in a Bowl Fill a large bowl or basin with water and wait for the water to settle so there are no ripples. Gently drop a cork into the middle of the bowl. Estimate the wavelength and period of oscillation of the water wave that propagates away from the cork. Remove the cork from the bowl and wait for the water to settle again. Gently drop the cork at a height that is different from the first drop. Does the wavelength depend upon how high above the water the cork is dropped? Example 16.9 Calculate the Velocity of Wave Propagation: Gull in the Ocean Calculate the wave velocity of the ocean wave in Figure 16.30 if the distance between wave crests is 10.0 m and the time for a sea gull to bob up and down is 5.00 s. Strategy 702 Chapter 16 | Oscillatory Motion and Waves We are asked to find w. The given information tells us that = 10.0 m and = 5.00 s. Therefore, we can use w = to find the wave velocity. Solution 1. Enter the known values into w = : 2. Solve for w to find w = 2.00 m/s. Discussion w = 10.0 m 5.00 s. (16.68) This slow speed seems reasonable for an ocean wave. Note that the wave moves to the right in the figure at this speed, not the varying speed at which the sea gull moves up and down. Transverse and Longitudinal Waves A simple wave consists of a periodic disturbance that propagates from one place to another. The wave in Figure 16.31 propagates in the horizontal direction while the surface is disturbed in the vertical direction. Such a wave is called a transverse wave or shear wave; in such a wave, the disturbance is perpendicular to the direction of propagation. In contrast, in a longitudinal wave or compressional wave, the disturbance is parallel to the direction of propagation. Figure 16.32 shows an example of a longitudinal wave. The size of the disturbance is its amplitude X and is completely independent of the speed of propagation w. Figure 16.31 In this example of a transverse wave, the wave propagates horizontally, and the disturbance in the cord is in the vertical direction. Figure 16.32 In this example of a longitudinal wave, the wave propagates horizontally, and the disturbance in the cord is also in the horizontal direction. Waves may be transverse, longitudinal, or a combination of the two. (Water waves are actually a combination of transverse and longitudinal. The simplified water wave illustrated in Figure 16.30 shows no longitudinal motion of the bird.) The waves on the strings of musical instruments are transverse\u2014so are electromagnetic waves, such as visible light. Sound waves in air and water are longitudinal. Their disturbances are periodic variations in pressure that are transmitted in fluids. Fluids do not have appreciable shear strength, and thus the sound waves in them must be longitudinal or compressional. Sound in solids can be both longitudinal and transverse. This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 16 | Oscillatory Motion and Waves 703 Figure 16.33 The wave on a guitar string is transverse. The sound wave rattles a sheet of paper in a direction that shows the sound wave is longitudinal. Earthquake waves under Earth\u2019s surface also have both longitudinal and transverse components (called compressional or Pwaves and shear or S-waves, respectively). These components have important individual characteristics\u2014they propagate at different speeds, for example. Earthquakes also have surface waves that are similar to surface waves on water. Applying the Science Practices: Electricity in Your Home The source of electricity is of a sinusoidal nature. If we appropriately probe using an oscilloscope (an instrument used to display and analyze electronic signals), we can precisely determine the frequency and wavelength of the waveform. Inquire about the maximum voltage current that you get in your house and plot a sinusoidal waveform representing the frequency, wavelength, and period for it. Check Your Understanding Why is it important to differentiate between longitudinal and transverse waves? Solution In the different types of waves, energy can propagate in a different direction relative to the motion of the wave. This is important to understand how different types of waves affect the materials", " around them. PhET Explorations: Wave on a String Watch a string vibrate in slow motion. Wiggle the end of the string and make waves, or adjust the frequency and amplitude of an oscillator. Adjust the damping and tension. The end can be fixed, loose, or open. Figure 16.34 Wave on a String (http://cnx.org/content/m55281/1.2/wave-on-a-string_en.jar) 16.10 Superposition and Interference By the end of this section, you will be able to: Learning Objectives \u2022 Determine the resultant waveform when two waves act in superposition relative to each other. \u2022 Explain standing waves. \u2022 Describe the mathematical representation of overtones and beat frequency. 704 Chapter 16 | Oscillatory Motion and Waves Figure 16.35 These waves result from the superposition of several waves from different sources, producing a complex pattern. (credit: waterborough, Wikimedia Commons) Most waves do not look very simple. They look more like the waves in Figure 16.35 than like the simple water wave considered in Waves. (Simple waves may be created by a simple harmonic oscillation, and thus have a sinusoidal shape). Complex waves are more interesting, even beautiful, but they look formidable. Most waves appear complex because they result from several simple waves adding together. Luckily, the rules for adding waves are quite simple. When two or more waves arrive at the same point, they superimpose themselves on one another. More specifically, the disturbances of waves are superimposed when they come together\u2014a phenomenon called superposition. Each disturbance corresponds to a force, and forces add. If the disturbances are along the same line, then the resulting wave is a simple addition of the disturbances of the individual waves\u2014that is, their amplitudes add. Figure 16.36 and Figure 16.37 illustrate superposition in two special cases, both of which produce simple results. Figure 16.36 shows two identical waves that arrive at the same point exactly in phase. The crests of the two waves are precisely aligned, as are the troughs. This superposition produces pure constructive interference. Because the disturbances add, pure constructive interference produces a wave that has twice the amplitude of the individual waves, but has the same wavelength. Figure 16.37 shows two identical waves that arrive exactly out of phase\u2014that is, precisely aligned crest to trough\u2014producing pure destructive interference. Because the disturbances are in the opposite direction for this superposition, the resulting amplitude is zero for pure destructive interference\u2014the waves completely cancel. Figure 16.36 Pure constructive interference of two identical waves produces one with twice the amplitude, but the same wavelength. Figure 16.37 Pure destructive interference of two identical waves produces zero amplitude, or complete cancellation. This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 16 | Oscillatory Motion and Waves 705 While pure constructive and pure destructive interference do occur, they require precisely aligned identical waves. The superposition of most waves produces a combination of constructive and destructive interference and can vary from place to place and time to time. Sound from a stereo, for example, can be loud in one spot and quiet in another. Varying loudness means the sound waves add partially constructively and partially destructively at different locations. A stereo has at least two speakers creating sound waves, and waves can reflect from walls. All these waves superimpose. An example of sounds that vary over time from constructive to destructive is found in the combined whine of airplane jets heard by a stationary passenger. The combined sound can fluctuate up and down in volume as the sound from the two engines varies in time from constructive to destructive. These examples are of waves that are similar. An example of the superposition of two dissimilar waves is shown in Figure 16.38. Here again, the disturbances add and subtract, producing a more complicated looking wave. Figure 16.38 Superposition of non-identical waves exhibits both constructive and destructive interference. Standing Waves Sometimes waves do not seem to move; rather, they just vibrate in place. Unmoving waves can be seen on the surface of a glass of milk in a refrigerator, for example. Vibrations from the refrigerator motor create waves on the milk that oscillate up and down but do not seem to move across the surface. These waves are formed by the superposition of two or more moving waves, such as illustrated in Figure 16.39 for two identical waves moving in opposite directions. The waves move through each other with their disturbances adding as they go by. If the two waves have the same amplitude and wavelength, then they alternate between constructive and destructive interference. The resultant looks like a wave standing in place and, thus, is called a standing wave. Waves on the glass of milk are one example of standing waves. There are other standing waves, such as on guitar strings and in organ pipes. With the glass of milk, the two waves", " that produce standing waves may come from reflections from the side of the glass. A closer look at earthquakes provides evidence for conditions appropriate for resonance, standing waves, and constructive and destructive interference. A building may be vibrated for several seconds with a driving frequency matching that of the natural frequency of vibration of the building\u2014producing a resonance resulting in one building collapsing while neighboring buildings do not. Often buildings of a certain height are devastated while other taller buildings remain intact. The building height matches the condition for setting up a standing wave for that particular height. As the earthquake waves travel along the surface of Earth and reflect off denser rocks, constructive interference occurs at certain points. Often areas closer to the epicenter are not damaged while areas farther away are damaged. Figure 16.39 Standing wave created by the superposition of two identical waves moving in opposite directions. The oscillations are at fixed locations in space and result from alternately constructive and destructive interference. Standing waves are also found on the strings of musical instruments and are due to reflections of waves from the ends of the string. Figure 16.40 and Figure 16.41 show three standing waves that can be created on a string that is fixed at both ends. Nodes are the points where the string does not move; more generally, nodes are where the wave disturbance is zero in a 706 Chapter 16 | Oscillatory Motion and Waves standing wave. The fixed ends of strings must be nodes, too, because the string cannot move there. The word antinode is used to denote the location of maximum amplitude in standing waves. Standing waves on strings have a frequency that is related to the propagation speed w of the disturbance on the string. The wavelength is determined by the distance between the points where the string is fixed in place. The lowest frequency, called the fundamental frequency, is thus for the longest wavelength, which is seen to be 1 = 2. Therefore, the fundamental frequency is 1 = w / 1 = w / 2. In this case, the overtones or harmonics are multiples of the fundamental frequency. As seen in Figure 16.41, the first harmonic can easily be calculated since 2 =. Thus. Similarly, 3 = 3 1, and so on. All of these frequencies can be changed by adjusting the tension in the string. The greater the tension, the greater w is and the higher the frequencies. This observation is familiar to anyone who has ever observed a string instrument being tuned. We will see in later chapters that standing waves are crucial to many resonance phenomena, such as in sounding boxes on string instruments. Figure 16.40 The figure shows a string oscillating at its fundamental frequency. Figure 16.41 First and second harmonic frequencies are shown. Beats Striking two adjacent keys on a piano produces a warbling combination usually considered to be unpleasant. The superposition of two waves of similar but not identical frequencies is the culprit. Another example is often noticeable in jet aircraft, particularly the two-engine variety, while taxiing. The combined sound of the engines goes up and down in loudness. This varying loudness happens because the sound waves have similar but not identical frequencies. The discordant warbling of the piano and the fluctuating loudness of the jet engine noise are both due to alternately constructive and destructive interference as the two waves go in and out of phase. Figure 16.42 illustrates this graphically. This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 16 | Oscillatory Motion and Waves 707 Figure 16.42 Beats are produced by the superposition of two waves of slightly different frequencies but identical amplitudes. The waves alternate in time between constructive interference and destructive interference, giving the resulting wave a time-varying amplitude. The wave resulting from the superposition of two similar-frequency waves has a frequency that is the average of the two. This wave fluctuates in amplitude, or beats, with a frequency called the beat frequency. We can determine the beat frequency by adding two waves together mathematically. Note that a wave can be represented at one point in space as = cos 2\u03c0 = cos, 2\u03c0 where = 1 / is the frequency of the wave. Adding two waves that have different frequencies but identical amplitudes produces a resultant More specifically, = 1 + 2. = cos 2\u03c0 1 + cos. 2\u03c0 2 Using a trigonometric identity, it can be shown that = 2 cos B cos, 2\u03c0 ave where B = \u2223 1 \u2212 2 \u2223 (16.69) (16.70) (16.71) (16.72) (16.73) is the beat frequency, and ave is the average of amplitude and the average frequency of the two superimposed waves, but it also fluctuates in overall amplitude at the beat frequency B. The first cosine term in the expression effectively causes the amplitude to go up and down. The second cosine term is the wave with frequency ave. This result is valid for all types of waves. However, if it", " is a sound wave, providing the two frequencies are similar, then what we hear is an average frequency that gets louder and softer (or warbles) at the beat frequency. 1 and 2. These results mean that the resultant wave has twice the Real World Connections: Tuning Forks The MIT physics demo (http://openstaxcollege.org/l/31tuningforks/) entitled \u201cTuning Forks: Resonance and Beat Frequency\u201d provides a qualitative picture of how wave interference produces beats. Description: Two identical forks and sounding boxes are placed next to each other. Striking one tuning fork will cause the other to resonate at the same frequency. When a weight is attached to one tuning fork, they are no longer identical. Thus, one will not cause the other to resonate. When two different forks are struck at the same time, the interference of their pitches produces beats. Real World Connections: Jump Rop This is a fun activity with which to learn about interference and superposition. Take a jump rope and hold it at the two ends with one of your friends. While each of you is holding the rope, snap your hands to produce a wave from each side. Record your observations and see if they match with the following: a. One wave starts from the right end and travels to the left end of the rope. b. Another wave starts at the left end and travels to the right end of the rope. c. The waves travel at the same speed. d. The shape of the waves depends on the way the person snaps his or her hands. e. There is a region of overlap. f. The shapes of the waves are identical to their original shapes after they overlap. Now, snap the rope up and down and ask your friend to snap his or her end of the rope sideways. The resultant that one sees here is the vector sum of two individual displacements. 708 Chapter 16 | Oscillatory Motion and Waves This activity illustrates superposition and interference. When two or more waves interact with each other at a point, the disturbance at that point is given by the sum of the disturbances each wave will produce in the absence of the other. This is the principle of superposition. Interference is a result of superposition of two or more waves to form a resultant wave of greater or lower amplitude. While beats may sometimes be annoying in audible sounds, we will find that beats have many applications. Observing beats is a very useful way to compare similar frequencies. There are applications of beats as apparently disparate as in ultrasonic imaging and radar speed traps. Check Your Understanding Imagine you are holding one end of a jump rope, and your friend holds the other. If your friend holds her end still, you can move your end up and down, creating a transverse wave. If your friend then begins to move her end up and down, generating a wave in the opposite direction, what resultant wave forms would you expect to see in the jump rope? Solution The rope would alternate between having waves with amplitudes two times the original amplitude and reaching equilibrium with no amplitude at all. The wavelengths will result in both constructive and destructive interference Check Your Understanding Define nodes and antinodes. Solution Nodes are areas of wave interference where there is no motion. Antinodes are areas of wave interference where the motion is at its maximum point. Check Your Understanding You hook up a stereo system. When you test the system, you notice that in one corner of the room, the sounds seem dull. In another area, the sounds seem excessively loud. Describe how the sound moving about the room could result in these effects. Solution With multiple speakers putting out sounds into the room, and these sounds bouncing off walls, there is bound to be some wave interference. In the dull areas, the interference is probably mostly destructive. In the louder areas, the interference is probably mostly constructive. PhET Explorations: Wave Interference Make waves with a dripping faucet, audio speaker, or laser! Add a second source or a pair of slits to create an interference pattern. Figure 16.43 Wave Interference (http://cnx.org/content/m55282/1.2/wave-interference_en.jar) 16.11 Energy in Waves: Intensity Learning Objectives By the end of this section, you will be able to: \u2022 Calculate the intensity and the power of rays and waves. This content is available for free at http://cnx.org/content/col11844/1.13 712 Chapter 16 | Oscillatory Motion and Waves frequency: number of events per unit of time fundamental frequency: the lowest frequency of a periodic waveform intensity: power per unit area longitudinal wave: a wave in which the disturbance is parallel to the direction of propagation natural frequency: the frequency at which a system would oscillate if there were no driving and no damping forces nodes: the points where the string does not move; more generally, nodes are where the wave disturbance is zero", " in a standing wave oscillate: moving back and forth regularly between two points over damping: the condition in which damping of an oscillator causes it to return to equilibrium without oscillating; oscillator moves more slowly toward equilibrium than in the critically damped system overtones: multiples of the fundamental frequency of a sound period: time it takes to complete one oscillation periodic motion: motion that repeats itself at regular time intervals resonance: the phenomenon of driving a system with a frequency equal to the system's natural frequency resonate: a system being driven at its natural frequency restoring force: force acting in opposition to the force caused by a deformation simple harmonic motion: the oscillatory motion in a system where the net force can be described by Hooke\u2019s law simple harmonic oscillator: a device that implements Hooke\u2019s law, such as a mass that is attached to a spring, with the other end of the spring being connected to a rigid support such as a wall simple pendulum: an object with a small mass suspended from a light wire or string superposition: the phenomenon that occurs when two or more waves arrive at the same point transverse wave: a wave in which the disturbance is perpendicular to the direction of propagation under damping: the condition in which damping of an oscillator causes it to return to equilibrium with the amplitude gradually decreasing to zero; system returns to equilibrium faster but overshoots and crosses the equilibrium position one or more times wave: a disturbance that moves from its source and carries energy wave velocity: the speed at which the disturbance moves. Also called the propagation velocity or propagation speed wavelength: the distance between adjacent identical parts of a wave Section Summary 16.1 Hooke\u2019s Law: Stress and Strain Revisited \u2022 An oscillation is a back and forth motion of an object between two points of deformation. \u2022 An oscillation may create a wave, which is a disturbance that propagates from where it was created. \u2022 The simplest type of oscillations and waves are related to systems that can be described by Hooke\u2019s law: = \u2212, where is the restoring force, is the displacement from equilibrium or deformation, and is the force constant of the system. \u2022 Elastic potential energy PEel stored in the deformation of a system that can be described by Hooke\u2019s law is given by PEel = (1 / 2)2. 16.2 Period and Frequency in Oscillations \u2022 Periodic motion is a repetitious oscillation. \u2022 The time for one oscillation is the period. \u2022 The number of oscillations per unit time is the frequency. \u2022 These quantities are related by This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 16 | Oscillatory Motion and Waves 713 = 1. 16.3 Simple Harmonic Motion: A Special Periodic Motion \u2022 Simple harmonic motion is oscillatory motion for a system that can be described only by Hooke\u2019s law. Such a system is also called a simple harmonic oscillator. \u2022 Maximum displacement is the amplitude. The period and frequency of a simple harmonic oscillator are given by = 2\u03c0 and = 1 2\u03c0, where is the mass of the system. \u2022 Displacement in simple harmonic motion as a function of time is given by () = cos 2\u03c0. \u2022 The velocity is given by () = \u2212 max sin2\u03c0t cos 2\u03c0 \u2022 The acceleration is found to be () = \u2212., where max = /. 16.4 The Simple Pendulum \u2022 A mass suspended by a wire of length is a simple pendulum and undergoes simple harmonic motion for amplitudes less than about 15\u00ba. The period of a simple pendulum is where is the length of the string and is the acceleration due to gravity. = 2\u03c0, 16.5 Energy and the Simple Harmonic Oscillator \u2022 Energy in the simple harmonic oscillator is shared between elastic potential energy and kinetic energy, with the total being constant: \u2022 Maximum velocity depends on three factors: it is directly proportional to amplitude, it is greater for stiffer systems, and it is 22 + 1 1 22 = constant. smaller for objects that have larger masses: max =. 16.6 Uniform Circular Motion and Simple Harmonic Motion A projection of uniform circular motion undergoes simple harmonic oscillation. 16.7 Damped Harmonic Motion \u2022 Damped harmonic oscillators have non-conservative forces that dissipate their energy. \u2022 Critical damping returns the system to equilibrium as fast as possible without overshooting. \u2022 An underdamped system will oscillate through the equilibrium position. \u2022 An overdamped system moves more slowly toward equilibrium than one that is critically damped. 16.8 Forced Oscillations and Resonance \u2022 A system\u2019s natural frequency is the frequency at which the system will oscillate if not affected by driving or damping forces. \u2022 A periodic force driving a harmonic oscillator at its natural frequency produces resonance. The system is said to resonate. \u2022", " The less damping a system has, the higher the amplitude of the forced oscillations near resonance. The more damping a system has, the broader response it has to varying driving frequencies. 16.9 Waves \u2022 A wave is a disturbance that moves from the point of creation with a wave velocity w. \u2022 A wave has a wavelength, which is the distance between adjacent identical parts of the wave. \u2022 Wave velocity and wavelength are related to the wave\u2019s frequency and period by w = or w =. \u2022 A transverse wave has a disturbance perpendicular to its direction of propagation, whereas a longitudinal wave has a disturbance parallel to its direction of propagation. 16.10 Superposition and Interference \u2022 Superposition is the combination of two waves at the same location. 714 Chapter 16 | Oscillatory Motion and Waves \u2022 Constructive interference occurs when two identical waves are superimposed in phase. \u2022 Destructive interference occurs when two identical waves are superimposed exactly out of phase. \u2022 A standing wave is one in which two waves superimpose to produce a wave that varies in amplitude but does not propagate. \u2022 Nodes are points of no motion in standing waves. \u2022 An antinode is the location of maximum amplitude of a standing wave. \u2022 Waves on a string are resonant standing waves with a fundamental frequency and can occur at higher multiples of the fundamental, called overtones or harmonics. \u2022 Beats occur when waves of similar frequencies 1 and 2 are superimposed. The resulting amplitude oscillates with a beat frequency given by B = \u2223 1 \u2212 2 \u2223. 16.11 Energy in Waves: Intensity Intensity is defined to be the power per unit area: = and has units of W/m2. Conceptual Questions 16.1 Hooke\u2019s Law: Stress and Strain Revisited 1. Describe a system in which elastic potential energy is stored. 16.3 Simple Harmonic Motion: A Special Periodic Motion 2. What conditions must be met to produce simple harmonic motion? 3. (a) If frequency is not constant for some oscillation, can the oscillation be simple harmonic motion? (b) Can you think of any examples of harmonic motion where the frequency may depend on the amplitude? 4. Give an example of a simple harmonic oscillator, specifically noting how its frequency is independent of amplitude. 5. Explain why you expect an object made of a stiff material to vibrate at a higher frequency than a similar object made of a spongy material. 6. As you pass a freight truck with a trailer on a highway, you notice that its trailer is bouncing up and down slowly. Is it more likely that the trailer is heavily loaded or nearly empty? Explain your answer. 7. Some people modify cars to be much closer to the ground than when manufactured. Should they install stiffer springs? Explain your answer. 16.4 The Simple Pendulum 8. Pendulum clocks are made to run at the correct rate by adjusting the pendulum\u2019s length. Suppose you move from one city to another where the acceleration due to gravity is slightly greater, taking your pendulum clock with you, will you have to lengthen or shorten the pendulum to keep the correct time, other factors remaining constant? Explain your answer. 16.5 Energy and the Simple Harmonic Oscillator 9. Explain in terms of energy how dissipative forces such as friction reduce the amplitude of a harmonic oscillator. Also explain how a driving mechanism can compensate. (A pendulum clock is such a system.) 16.7 Damped Harmonic Motion 10. Give an example of a damped harmonic oscillator. (They are more common than undamped or simple harmonic oscillators.) 11. How would a car bounce after a bump under each of these conditions? \u2022 overdamping \u2022 underdamping \u2022 critical damping 12. Most harmonic oscillators are damped and, if undriven, eventually come to a stop. How is this observation related to the second law of thermodynamics? 16.8 Forced Oscillations and Resonance 13. Why are soldiers in general ordered to \u201croute step\u201d (walk out of step) across a bridge? 16.9 Waves 14. Give one example of a transverse wave and another of a longitudinal wave, being careful to note the relative directions of the disturbance and wave propagation in each. This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 16 | Oscillatory Motion and Waves 715 15. What is the difference between propagation speed and the frequency of a wave? Does one or both affect wavelength? If so, how? 16.10 Superposition and Interference 16. Speakers in stereo systems have two color-coded terminals to indicate how to hook up the wires. If the wires are reversed, the speaker moves in a direction opposite that of a properly connected speaker. Explain why it is important to have both speakers connected the same way. 16.11 Energy in Waves: Intensity", " 17. Two identical waves undergo pure constructive interference. Is the resultant intensity twice that of the individual waves? Explain your answer. 18. Circular water waves decrease in amplitude as they move away from where a rock is dropped. Explain why. 716 Chapter 16 | Oscillatory Motion and Waves Problems & Exercises 16.1 Hooke\u2019s Law: Stress and Strain Revisited 1. Fish are hung on a spring scale to determine their mass (most fishermen feel no obligation to truthfully report the mass). (a) What is the force constant of the spring in such a scale if it the spring stretches 8.00 cm for a 10.0 kg load? (b) What is the mass of a fish that stretches the spring 5.50 cm? (c) How far apart are the half-kilogram marks on the scale? 2. It is weigh-in time for the local under-85-kg rugby team. The bathroom scale used to assess eligibility can be described by Hooke\u2019s law and is depressed 0.75 cm by its maximum load of 120 kg. (a) What is the spring\u2019s effective spring constant? (b) A player stands on the scales and depresses it by 0.48 cm. Is he eligible to play on this under-85 kg team? 3. One type of BB gun uses a spring-driven plunger to blow the BB from its barrel. (a) Calculate the force constant of its plunger\u2019s spring if you must compress it 0.150 m to drive the 0.0500-kg plunger to a top speed of 20.0 m/s. (b) What force must be exerted to compress the spring? 4. (a) The springs of a pickup truck act like a single spring with a force constant of 1.30\u00d7105 N/m. By how much will the truck be depressed by its maximum load of 1000 kg? (b) If the pickup truck has four identical springs, what is the force constant of each? 5. When an 80.0-kg man stands on a pogo stick, the spring is compressed 0.120 m. (a) What is the force constant of the spring? (b) Will the spring be compressed more when he hops down the road? 6. A spring has a length of 0.200 m when a 0.300-kg mass hangs from it, and a length of 0.750 m when a 1.95-kg mass hangs from it. (a) What is the force constant of the spring? (b) What is the unloaded length of the spring? 16.2 Period and Frequency in Oscillations 7. What is the period of 60.0 Hz electrical power? 8. If your heart rate is 150 beats per minute during strenuous exercise, what is the time per beat in units of seconds? 9. Find the frequency of a tuning fork that takes 2.50\u00d710\u22123 s to complete one oscillation. 10. A stroboscope is set to flash every 8.00\u00d710\u22125 s. What is the frequency of the flashes? 11. A tire has a tread pattern with a crevice every 2.00 cm. Each crevice makes a single vibration as the tire moves. What is the frequency of these vibrations if the car moves at 30.0 m/s? 12. Engineering Application Each piston of an engine makes a sharp sound every other revolution of the engine. (a) How fast is a race car going if its eight-cylinder engine emits a sound of frequency 750 Hz, given that the engine makes 2000 revolutions per kilometer? (b) At how many revolutions per minute is the engine rotating? This content is available for free at http://cnx.org/content/col11844/1.13 16.3 Simple Harmonic Motion: A Special Periodic Motion 13. A type of cuckoo clock keeps time by having a mass bouncing on a spring, usually something cute like a cherub in a chair. What force constant is needed to produce a period of 0.500 s for a 0.0150-kg mass? 14. If the spring constant of a simple harmonic oscillator is doubled, by what factor will the mass of the system need to change in order for the frequency of the motion to remain the same? 15. A 0.500-kg mass suspended from a spring oscillates with a period of 1.50 s. How much mass must be added to the object to change the period to 2.00 s? 16. By how much leeway (both percentage and mass) would you have in the selection of the mass of the object in the previous problem if you did not wish the new period to be greater than 2.01 s or less than 1.99 s? 17. Suppose you attach the object with mass to a vertical spring originally at rest, and let it bounce up and down. You release the object from rest", " at the spring\u2019s original rest length. (a) Show that the spring exerts an upward force of 2.00 on the object at its lowest point. (b) If the spring has a force constant of 10.0 N/m and a 0.25-kg-mass object is set in motion as described, find the amplitude of the oscillations. (c) Find the maximum velocity. 18. A diver on a diving board is undergoing simple harmonic motion. Her mass is 55.0 kg and the period of her motion is 0.800 s. The next diver is a male whose period of simple harmonic oscillation is 1.05 s. What is his mass if the mass of the board is negligible? 19. Suppose a diving board with no one on it bounces up and down in a simple harmonic motion with a frequency of 4.00 Hz. The board has an effective mass of 10.0 kg. What is the frequency of the simple harmonic motion of a 75.0-kg diver on the board? 20. Figure 16.46 This child\u2019s toy relies on springs to keep infants entertained. (credit: By Humboldthead, Flickr) The device pictured in Figure 16.46 entertains infants while keeping them from wandering. The child bounces in a harness suspended from a door frame by a spring constant. Chapter 16 | Oscillatory Motion and Waves 717 (a) If the spring stretches 0.250 m while supporting an 8.0-kg child, what is its spring constant? (b) What is the effect on the period of a pendulum if you decrease its length by 5.00%? (b) What is the time for one complete bounce of this child? (c) What is the child\u2019s maximum velocity if the amplitude of her bounce is 0.200 m? 21. A 90.0-kg skydiver hanging from a parachute bounces up and down with a period of 1.50 s. What is the new period of oscillation when a second skydiver, whose mass is 60.0 kg, hangs from the legs of the first, as seen in Figure 16.47. Figure 16.47 The oscillations of one skydiver are about to be affected by a second skydiver. (credit: U.S. Army, www.army.mil) 16.4 The Simple Pendulum As usual, the acceleration due to gravity in these problems is taken to be = 9.80 m / s2, unless otherwise specified. 22. What is the length of a pendulum that has a period of 0.500 s? 23. Some people think a pendulum with a period of 1.00 s can be driven with \u201cmental energy\u201d or psycho kinetically, because its period is the same as an average heartbeat. True or not, what is the length of such a pendulum? 24. What is the period of a 1.00-m-long pendulum? 25. How long does it take a child on a swing to complete one swing if her center of gravity is 4.00 m below the pivot? 26. The pendulum on a cuckoo clock is 5.00 cm long. What is its frequency? 27. Two parakeets sit on a swing with their combined center of mass 10.0 cm below the pivot. At what frequency do they swing? 28. (a) A pendulum that has a period of 3.00000 s and that is located where the acceleration due to gravity is 9.79 m/s2 is moved to a location where it the acceleration due to gravity is 9.82 m/s2. What is its new period? (b) Explain why so many digits are needed in the value for the period, based on the relation between the period and the acceleration due to gravity. 29. A pendulum with a period of 2.00000 s in one location = 9.80 m/s2 is moved to a new location where the period is now 1.99796 s. What is the acceleration due to gravity at its new location? 30. (a) What is the effect on the period of a pendulum if you double its length? 31. Find the ratio of the new/old periods of a pendulum if the pendulum were transported from Earth to the Moon, where the acceleration due to gravity is 1.63 m/s2. 32. At what rate will a pendulum clock run on the Moon, where the acceleration due to gravity is 1.63 m/s2, if it keeps time accurately on Earth? That is, find the time (in hours) it takes the clock\u2019s hour hand to make one revolution on the Moon. 33. Suppose the length of a clock\u2019s pendulum is changed by 1.000%, exactly at noon one day. What time will it read 24.00 hours later, assuming it the pendulum has kept perfect time before", " the change? Note that there are two answers, and perform the calculation to four-digit precision. 34. If a pendulum-driven clock gains 5.00 s/day, what fractional change in pendulum length must be made for it to keep perfect time? 16.5 Energy and the Simple Harmonic Oscillator 35. The length of nylon rope from which a mountain climber is suspended has a force constant of 1.40\u00d7104 N/m. (a) What is the frequency at which he bounces, given his mass plus and the mass of his equipment are 90.0 kg? (b) How much would this rope stretch to break the climber\u2019s fall if he free-falls 2.00 m before the rope runs out of slack? Hint: Use conservation of energy. (c) Repeat both parts of this problem in the situation where twice this length of nylon rope is used. 36. Engineering Application Near the top of the Citigroup Center building in New York City, there is an object with mass of 4.00\u00d7105 kg on springs that have adjustable force constants. Its function is to dampen wind-driven oscillations of the building by oscillating at the same frequency as the building is being driven\u2014the driving force is transferred to the object, which oscillates instead of the entire building. (a) What effective force constant should the springs have to make the object oscillate with a period of 2.00 s? (b) What energy is stored in the springs for a 2.00-m displacement from equilibrium? 16.6 Uniform Circular Motion and Simple Harmonic Motion 37. (a)What is the maximum velocity of an 85.0-kg person bouncing on a bathroom scale having a force constant of 1.50\u00d7106 N/m, if the amplitude of the bounce is 0.200 cm? (b)What is the maximum energy stored in the spring? 38. A novelty clock has a 0.0100-kg mass object bouncing on a spring that has a force constant of 1.25 N/m. What is the maximum velocity of the object if the object bounces 3.00 cm above and below its equilibrium position? (b) How many joules of kinetic energy does the object have at its maximum velocity? 39. At what positions is the speed of a simple harmonic oscillator half its maximum? That is, what values of / give = \u00b1max / 2, where is the amplitude of the motion? 718 Chapter 16 | Oscillatory Motion and Waves 40. A ladybug sits 12.0 cm from the center of a Beatles music album spinning at 33.33 rpm. What is the maximum velocity of its shadow on the wall behind the turntable, if illuminated parallel to the record by the parallel rays of the setting Sun? 51. Scouts at a camp shake the rope bridge they have just crossed and observe the wave crests to be 8.00 m apart. If they shake it the bridge twice per second, what is the propagation speed of the waves? 16.7 Damped Harmonic Motion 41. The amplitude of a lightly damped oscillator decreases by 3.0% during each cycle. What percentage of the mechanical energy of the oscillator is lost in each cycle? 16.8 Forced Oscillations and Resonance 42. How much energy must the shock absorbers of a 1200-kg car dissipate in order to damp a bounce that initially has a velocity of 0.800 m/s at the equilibrium position? Assume the car returns to its original vertical position. 43. If a car has a suspension system with a force constant of 5.00\u00d7104 N/m, how much energy must the car\u2019s shocks remove to dampen an oscillation starting with a maximum displacement of 0.0750 m? 44. (a) How much will a spring that has a force constant of 40.0 N/m be stretched by an object with a mass of 0.500 kg when hung motionless from the spring? (b) Calculate the decrease in gravitational potential energy of the 0.500-kg object when it descends this distance. (c) Part of this gravitational energy goes into the spring. Calculate the energy stored in the spring by this stretch, and compare it with the gravitational potential energy. Explain where the rest of the energy might go. 45. Suppose you have a 0.750-kg object on a horizontal surface connected to a spring that has a force constant of 150 N/m. There is simple friction between the object and surface with a static coefficient of friction s = 0.100. (a) How far can the spring be stretched without moving the mass? (b) If the object is set into oscillation with an amplitude twice the distance found in part (a), and the kinetic coefficient of friction is k = 0.0850, what total distance does it travel before stopping? Assume it starts at the maximum amplitude. 46. Engineering Application:", " A suspension bridge oscillates with an effective force constant of 1.00\u00d7108 N/m. (a) How much energy is needed to make it oscillate with an amplitude of 0.100 m? (b) If soldiers march across the bridge with a cadence equal to the bridge\u2019s natural frequency and impart 1.00\u00d7104 J of energy each second, how long does it take for the bridge\u2019s oscillations to go from 0.100 m to 0.500 m amplitude? 16.9 Waves 47. Storms in the South Pacific can create waves that travel all the way to the California coast, which are 12,000 km away. How long does it take them if they travel at 15.0 m/s? 48. Waves on a swimming pool propagate at 0.750 m/s. You splash the water at one end of the pool and observe the wave go to the opposite end, reflect, and return in 30.0 s. How far away is the other end of the pool? 49. Wind gusts create ripples on the ocean that have a wavelength of 5.00 cm and propagate at 2.00 m/s. What is their frequency? 50. How many times a minute does a boat bob up and down on ocean waves that have a wavelength of 40.0 m and a propagation speed of 5.00 m/s? This content is available for free at http://cnx.org/content/col11844/1.13 52. What is the wavelength of the waves you create in a swimming pool if you splash your hand at a rate of 2.00 Hz and the waves propagate at 0.800 m/s? 53. What is the wavelength of an earthquake that shakes you with a frequency of 10.0 Hz and gets to another city 84.0 km away in 12.0 s? 54. Radio waves transmitted through space at 3.00\u00d7108 m/s by the Voyager spacecraft have a wavelength of 0.120 m. What is their frequency? 55. Your ear is capable of differentiating sounds that arrive at the ear just 1.00 ms apart. What is the minimum distance between two speakers that produce sounds that arrive at noticeably different times on a day when the speed of sound is 340 m/s? 56. (a) Seismographs measure the arrival times of earthquakes with a precision of 0.100 s. To get the distance to the epicenter of the quake, they compare the arrival times of S- and P-waves, which travel at different speeds. Figure 16.48) If S- and P-waves travel at 4.00 and 7.20 km/s, respectively, in the region considered, how precisely can the distance to the source of the earthquake be determined? (b) Seismic waves from underground detonations of nuclear bombs can be used to locate the test site and detect violations of test bans. Discuss whether your answer to (a) implies a serious limit to such detection. (Note also that the uncertainty is greater if there is an uncertainty in the propagation speeds of the S- and P-waves.) Figure 16.48 A seismograph as described in above problem.(credit: Oleg Alexandrov) 16.10 Superposition and Interference 57. A car has two horns, one emitting a frequency of 199 Hz and the other emitting a frequency of 203 Hz. What beat frequency do they produce? 58. The middle-C hammer of a piano hits two strings, producing beats of 1.50 Hz. One of the strings is tuned to 260.00 Hz. What frequencies could the other string have? 59. Two tuning forks having frequencies of 460 and 464 Hz are struck simultaneously. What average frequency will you hear, and what will the beat frequency be? Chapter 16 | Oscillatory Motion and Waves 719 (a) What is the intensity in W/m2 of a laser beam used to burn away cancerous tissue that, when 90.0% absorbed, puts 500 J of energy into a circular spot 2.00 mm in diameter in 4.00 s? (b) Discuss how this intensity compares to the average intensity of sunlight (about 700 W/m2 ) and the implications that would have if the laser beam entered your eye. Note how your answer depends on the time duration of the exposure. 60. Twin jet engines on an airplane are producing an average sound frequency of 4100 Hz with a beat frequency of 0.500 Hz. What are their individual frequencies? 61. A wave traveling on a Slinky\u00ae that is stretched to 4 m takes 2.4 s to travel the length of the Slinky and back again. (a) What is the speed of the wave? (b) Using the same Slinky stretched to the same length, a standing wave is created which consists of three antinodes and four nodes. At what frequency must the Slinky be oscillating? 62. Three adjacent keys on a piano (F, F-sharp, and", " G) are struck simultaneously, producing frequencies of 349, 370, and 392 Hz. What beat frequencies are produced by this discordant combination? 16.11 Energy in Waves: Intensity 63. Medical Application Ultrasound of intensity 1.50\u00d7102 W/m2 is produced by the rectangular head of a medical imaging device measuring 3.00 by 5.00 cm. What is its power output? 64. The low-frequency speaker of a stereo set has a surface area of 0.05 m2 and produces 1W of acoustical power. What is the intensity at the speaker? If the speaker projects sound uniformly in all directions, at what distance from the speaker is the intensity 0.1 W/m2? 65. To increase intensity of a wave by a factor of 50, by what factor should the amplitude be increased? 66. Engineering Application A device called an insolation meter is used to measure the intensity of sunlight has an area of 100 cm2 and registers 6.50 W. What is the intensity in W/m2? 67. Astronomy Application Energy from the Sun arrives at the top of the Earth\u2019s atmosphere with an intensity of 1.30 kW/m2. How long does it take for 1.8\u00d7109 J to arrive on an area of 1.00 m2? 68. Suppose you have a device that extracts energy from ocean breakers in direct proportion to their intensity. If the device produces 10.0 kW of power on a day when the breakers are 1.20 m high, how much will it produce when they are 0.600 m high? 69. Engineering Application (a) A photovoltaic array of (solar cells) is 10.0% efficient in gathering solar energy and converting it to electricity. If the average intensity of sunlight on one day is 700 W/m2, what area should your array have to gather energy at the rate of 100 W? (b) What is the maximum cost of the array if it must pay for itself in two years of operation averaging 10.0 hours per day? Assume that it earns money at the rate of 9.00 \u00a2 per kilowatt-hour. 70. A microphone receiving a pure sound tone feeds an oscilloscope, producing a wave on its screen. If the sound intensity is originally 2.00\u00d710\u20135 W/m2, but is turned up until the amplitude increases by 30.0%, what is the new intensity? 71. Medical Application 720 Chapter 16 | Oscillatory Motion and Waves Test Prep for AP\u00ae Courses 16.1 Hooke\u2019s Law: Stress and Strain Revisited 1. Which of the following represents the distance (how much ground the particle covers) moved by a particle in a simple harmonic motion in one time period? (Here, A represents the amplitude of the oscillation.) a. 0 cm b. A cm c. 2A cm d. 4A cm 2. A spring has a spring constant of 80 N\u00b7m\u22121. What is the force required to (a) compress the spring by 5 cm and (b) expand the spring by 15 cm? 3. In the formula = \u2212, what does the minus sign indicate? a. b. c. It indicates that the restoring force is in the direction of the displacement. It indicates that the restoring force is in the direction opposite the displacement. It indicates that mechanical energy in the system decreases when a system undergoes oscillation. d. None of the above 4. The splashing of a liquid resembles an oscillation. The restoring force in this scenario will be due to which of the following? a. Potential energy b. Kinetic energy c. Gravity d. Mechanical energy 16.2 Period and Frequency in Oscillations 5. A mass attached to a spring oscillates and completes 50 full cycles in 30 s. What is the time period and frequency of this system? 16.3 Simple Harmonic Motion: A Special Periodic Motion 6. Use these figures to answer the following questions. Figure 16.49 a. Which of the two pendulums oscillates with larger amplitude? b. Which of the two pendulums oscillates at a higher frequency? 7. A particle of mass 100 g undergoes a simple harmonic motion. The restoring force is provided by a spring with a spring constant of 40 N\u00b7m\u22121. What is the period of oscillation? a. 10 s b. 0.5 s c. 0.1 s d. 1 8. The graph shows the simple harmonic motion of a mass m attached to a spring with spring constant k. Figure 16.50 What is the displacement at time 8\u03c0? a. 1 m b. 0 m c. Not defined d. \u22121 m 9. A pendulum of mass 200 g undergoes simple harmonic motion when acted upon by a force of 15 N. The pendulum crosses the point of equilibrium at a speed of 5 m\u00b7s\u22121. What is the energy of", " the pendulum at the center of the oscillation? 16.4 The Simple Pendulum 10. A ball is attached to a string of length 4 m to make a pendulum. The pendulum is placed at a location that is away from the Earth\u2019s surface by twice the radius of the Earth. What is the acceleration due to gravity at that height and what is the period of the oscillations? 11. Which of the following gives the correct relation between the acceleration due to gravity and period of a pendulum? a. b. = 2 2 = 4 2 2 c. = 2 = 2 2 d. 12. Tom has two pendulums with him. Pendulum 1 has a ball of mass 0.1 kg attached to it and has a length of 5 m. Pendulum 2 has a ball of mass 0.5 kg attached to a string of length 1 m. How does mass of the ball affect the frequency of the pendulum? Which pendulum will have a higher frequency and why? 16.5 Energy and the Simple Harmonic Oscillator This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 16 | Oscillatory Motion and Waves 721 13. A mass of 1 kg undergoes simple harmonic motion with amplitude of 1 m. If the period of the oscillation is 1 s, calculate the internal energy of the system. 16.6 Uniform Circular Motion and Simple Harmonic Motion 14. In the equation = sin t, what values can the position take? a. \u22121 to +1 b. \u2013A to +A c. 0 d. \u2013t to t 16.7 Damped Harmonic Motion 15. The non-conservative damping force removes energy from a system in which form? a. Mechanical energy b. Electrical energy c. Thermal energy d. None of the above 16. The time rate of change of mechanical energy for a damped oscillator is always: a. 0 b. Negative c. Positive d. Undefined 17. A 0.5-kg object is connected to a spring that undergoes oscillatory motion. There is friction between the object and the surface it is kept on given by coefficient of friction = 0.06. If the object is released 0.2 m from equilibrium, what is the distance that the object travels? Given that the force constant of the spring is 50 N m-1 and the frictional force between the objects is 0.294 N. 16.8 Forced Oscillations and Resonance 18. How is constant amplitude sustained in forced oscillations? 16.9 Waves 19. What is the difference between the waves coming from a tuning fork and electromagnetic waves? 20. Represent longitudinal and transverse waves in a graphical form. 21. Why is the sound produced by a tambourine different from that produced by drums? 22. A transverse wave is traveling left to right. Which of the following is correct about the motion of particles in the wave? a. The particles move up and down when the wave travels in a vacuum. b. The particles move left and right when the wave travels in a medium. c. The particles move up and down when the wave travels in a medium. d. The particles move right and left when the wave travels in a vacuum. 23. Figure 16.51 The graph shows propagation of a mechanical wave. What is the wavelength of this wave? 16.10 Superposition and Interference 24. A guitar string has a number of frequencies at which it vibrates naturally. Which of the following is true in this context? a. The resonant frequencies of the string are integer multiples of fundamental frequencies. b. The resonant frequencies of the string are not integer multiples of fundamental frequencies. c. They have harmonic overtones. d. None of the above 25. Explain the principle of superposition with figures that show the changes in the wave amplitude. 26. In this figure which points represent the points of constructive interference? Figure 16.52 a. A, B, F b. A, B, C, D, E, F c. A, C, D, E d. A, B, D 27. A string is fixed on both sides. It is snapped from both ends at the same time by applying an equal force. What happens to the shape of the waves generated in the string? Also, will you observe an overlap of waves? 28. In the preceding question, what would happen to the amplitude of the waves generated in this way? Also, consider another scenario where the string is snapped up from one end and down from the other end. What will happen in this situation? 29. Two sine waves travel in the same direction in a medium. The amplitude of each wave is A, and the phase difference between the two is 180\u00b0. What is the resultant amplitude? a. 2A b. 3A c. 0 d. 9A 30. Standing wave patterns consist of nodes and antinodes formed", " by repeated interference between two waves of the same frequency traveling in opposite directions. What are nodes and antinodes and how are they produced? 722 Chapter 16 | Oscillatory Motion and Waves This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 17 | Physics of Hearing 723 17 PHYSICS OF HEARING Figure 17.1 This tree fell some time ago. When it fell, atoms in the air were disturbed. Physicists would call this disturbance sound whether someone was around to hear it or not. (credit: B.A. Bowen Photography) Chapter Outline 17.1. Sound 17.2. Speed of Sound, Frequency, and Wavelength 17.3. Sound Intensity and Sound Level 17.4. Doppler Effect and Sonic Booms 17.5. Sound Interference and Resonance: Standing Waves in Air Columns 17.6. Hearing 17.7. Ultrasound Connection for AP\u00ae Courses In this chapter, the concept of waves is specifically applied to the phenomena of sound. As such, Big Idea 6 continues to be supported, as sound waves carry energy and momentum from one location to another without the permanent transfer of mass. This energy is carried through vibrations caused by disturbances in air pressure (Enduring Understanding 6.A). As air pressure increases, amplitudes of vibration and energy transfer do as well. This idea (Enduring Understanding 6.A.4) explains why a very loud sound can break glass. The chapter continues the fundamental analysis of waves addressed in Chapter 16. Sound waves are periodic, and can therefore be expressed as a function of position and time. Furthermore, sound waves are described by amplitude, frequency, wavelength, and speed (Enduring Understanding 6.B). The relationship between speed and frequency is analyzed further in Section 17.4, as the frequency of sound depends upon the relative motion between the source and observer. This concept, known as the Doppler effect, supports Essential Knowledge 6.B.5. Like all other waves, sound waves can overlap. When they do so, their interaction will produce an amplitude variation within the resultant wave. This amplitude can be determined by adding the displacement of the two pulses, through a process called superposition. This process, covered in Section 17.5, reinforces the content in Enduring Understanding 6.D.1. In situations where the interfering waves are confined, such as on a fixed length of string or in a tube, standing waves can result. These waves are the result of interference between the incident and reflecting wave. Standing waves are described using nodes and antinodes, and their wavelengths are determined by the size of the region to which they are confined. This chapter\u2019s 724 Chapter 17 | Physics of Hearing description of both standing waves and the concept of beats strongly support Enduring Understanding 6.D, as well as Essential Knowledge 6.D.1, 6.D.3, and 6.D.4. The concepts in this chapter support: Big Idea 6 Waves can transfer energy and momentum from one location to another without the permanent transfer of mass and serve as a mathematical model for the description of other phenomena. Enduring Understanding 6.B A periodic wave is one that repeats as a function of both time and position and can be described by its amplitude, frequency, wavelength, speed, and energy. Essential Knowledge 6.B.5 The observed frequency of a wave depends on the relative motion of the source and the observer. This is a qualitative measurement only. Enduring Understanding 6.D Interference and superposition lead to standing waves and beats. Essential Knowledge 6.D.1 Two or more wave pulses can interact in such a way as to produce amplitude variations in the resultant wave. When two pulses cross, they travel through each other; they do not bounce off each other. Where the pulses overlap, the resulting displacement can be determined by adding the displacements of the two pulses. This is called superposition. Essential Knowledge 6.D.3 Standing waves are the result of the addition of incident and reflected waves that are con\ufb01ned to a region and have nodes and antinodes. Examples should include waves on a fixed length of string, and sound waves in both closed and open tubes. Essential Knowledge 6.D.4 The possible wavelengths of a standing wave are determined by the size of the region in which it is con\ufb01ned. 17.1 Sound Learning Objectives By the end of this section, you will be able to: \u2022 Define sound and hearing. \u2022 Describe sound as a longitudinal wave. Figure 17.2 This glass has been shattered by a high-intensity sound wave of the same frequency as the resonant frequency of the glass. While the sound is not visible, the effects of the sound prove its existence. (credit: ||read||, Flickr) Sound can be used as a familiar illustration of waves. Because hearing is one of our most important senses, it is interesting to see how the", " physical properties of sound correspond to our perceptions of it. Hearing is the perception of sound, just as vision is the perception of visible light. But sound has important applications beyond hearing. Ultrasound, for example, is not heard but can be employed to form medical images and is also used in treatment. The physical phenomenon of sound is defined to be a disturbance of matter that is transmitted from its source outward. Sound is a wave. On the atomic scale, it is a disturbance of atoms that is far more ordered than their thermal motions. In many instances, sound is a periodic wave, and the atoms undergo simple harmonic motion. In this text, we shall explore such periodic sound waves. A vibrating string produces a sound wave as illustrated in Figure 17.3, Figure 17.4, and Figure 17.5. As the string oscillates back and forth, it transfers energy to the air, mostly as thermal energy created by turbulence. But a small part of the string\u2019s energy goes into compressing and expanding the surrounding air, creating slightly higher and lower local pressures. These compressions (high pressure regions) and rarefactions (low pressure regions) move out as longitudinal pressure waves having the same frequency as the string\u2014they are the disturbance that is a sound wave. (Sound waves in air and most fluids are longitudinal, because fluids have almost no shear strength. In solids, sound waves can be both transverse and longitudinal.) Figure 17.5 shows a graph of gauge pressure versus distance from the vibrating string. This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 17 | Physics of Hearing 725 Figure 17.3 A vibrating string moving to the right compresses the air in front of it and expands the air behind it. Figure 17.4 As the string moves to the left, it creates another compression and rarefaction as the ones on the right move away from the string. Figure 17.5 After many vibrations, there are a series of compressions and rarefactions moving out from the string as a sound wave. The graph shows gauge pressure versus distance from the source. Pressures vary only slightly from atmospheric for ordinary sounds. The amplitude of a sound wave decreases with distance from its source, because the energy of the wave is spread over a larger and larger area. But it is also absorbed by objects, such as the eardrum in Figure 17.6, and converted to thermal energy by the viscosity of air. In addition, during each compression a little heat transfers to the air and during each rarefaction even less heat transfers from the air, so that the heat transfer reduces the organized disturbance into random thermal motions. (These processes can be viewed as a manifestation of the second law of thermodynamics presented in Introduction to the Second Law of Thermodynamics: Heat Engines and Their Efficiency.) Whether the heat transfer from compression to rarefaction is significant depends on how far apart they are\u2014that is, it depends on wavelength. Wavelength, frequency, amplitude, and speed of propagation are important for sound, as they are for all waves. Figure 17.6 Sound wave compressions and rarefactions travel up the ear canal and force the eardrum to vibrate. There is a net force on the eardrum, since the sound wave pressures differ from the atmospheric pressure found behind the eardrum. A complicated mechanism converts the vibrations to nerve impulses, which are perceived by the person. 726 Chapter 17 | Physics of Hearing PhET Explorations: Wave Interference Make waves with a dripping faucet, audio speaker, or laser! Add a second source or a pair of slits to create an interference pattern. Figure 17.7 Wave Interference (http://cnx.org/content/m55288/1.2/wave-interference_en.jar) 17.2 Speed of Sound, Frequency, and Wavelength Learning Objectives By the end of this section, you will be able to: \u2022 Define pitch. \u2022 Describe the relationship between the speed of sound, its frequency, and its wavelength. \u2022 Describe the effects on the speed of sound as it travels through various media. \u2022 Describe the effects of temperature on the speed of sound. The information presented in this section supports the following AP\u00ae learning objectives and science practices: \u2022 6.B.4.1 The student is able to design an experiment to determine the relationship between periodic wave speed, wavelength, and frequency, and relate these concepts to everyday examples. (S.P. 4.2, 5.1, 7.2) Figure 17.8 When a firework explodes, the light energy is perceived before the sound energy. Sound travels more slowly than light does. (credit: Dominic Alves, Flickr) Sound, like all waves, travels at a certain speed and has the properties of frequency and wavelength. You can observe direct evidence of the speed of sound while watching a fireworks display. The flash of", " an explosion is seen well before its sound is heard, implying both that sound travels at a finite speed and that it is much slower than light. You can also directly sense the frequency of a sound. Perception of frequency is called pitch. The wavelength of sound is not directly sensed, but indirect evidence is found in the correlation of the size of musical instruments with their pitch. Small instruments, such as a piccolo, typically make high-pitch sounds, while large instruments, such as a tuba, typically make low-pitch sounds. High pitch means small wavelength, and the size of a musical instrument is directly related to the wavelengths of sound it produces. So a small instrument creates short-wavelength sounds. Similar arguments hold that a large instrument creates long-wavelength sounds. The relationship of the speed of sound, its frequency, and wavelength is the same as for all waves: w = (17.1) where w is the speed of sound, between adjacent identical parts of a wave\u2014for example, between adjacent compressions as illustrated in Figure 17.9. The frequency is the same as that of the source and is the number of waves that pass a point per unit time. is its frequency, and is its wavelength. The wavelength of a sound is the distance This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 17 | Physics of Hearing 727 Figure 17.9 A sound wave emanates from a source vibrating at a frequency, propagates at w, and has a wavelength. Table 17.4 makes it apparent that the speed of sound varies greatly in different media. The speed of sound in a medium is determined by a combination of the medium\u2019s rigidity (or compressibility in gases) and its density. The more rigid (or less compressible) the medium, the faster the speed of sound. This observation is analogous to the fact that the frequency of a simple harmonic motion is directly proportional to the stiffness of the oscillating object. The greater the density of a medium, the slower the speed of sound. This observation is analogous to the fact that the frequency of a simple harmonic motion is inversely proportional to the mass of the oscillating object. The speed of sound in air is low, because air is compressible. Because liquids and solids are relatively rigid and very difficult to compress, the speed of sound in such media is generally greater than in gases. Applying the Science Practices: Bottle Music When liquid is poured into a small-necked container like a soda bottle, it can make for a fun musical experience! Find a small-necked bottle and pour water into it. When you blow across the surface of the bottle, a musical pitch should be created. This pitch, which corresponds to the resonant frequency of the air remaining in the bottle, can be determined using Equation 17.1. Your task is to design an experiment and collect data to confirm this relationship between the frequency created by blowing into the bottle and the depth of air remaining. 1. Use the explanation above to design an experiment that will yield data on depth of air column and frequency of pitch. Use the data table below to record your data. Table 17.1 Depth of air column (\u03bb) Frequency of pitch generated (f) 2. Construct a graph using the information collected above. The graph should include all five data points and should display frequency on the dependent axis. 3. What type of relationship is displayed on your graph? (direct, inverse, quadratic, etc.) 4. Does your graph align with equation 17.1, given earlier in this section? Explain. Note: For an explanation of why a frequency is created when you blow across a small-necked container, explore Section 17.5 later in this chapter. Answer 1. As the depth of the air column increases, the frequency values must decrease. A sample set of data is displayed below. 728 Chapter 17 | Physics of Hearing Table 17.2 Depth of air column (\u03bb) Frequency of pitch generated (f) 24 cm 22 cm 20 cm 18 cm 16 cm 689.6 Hz 752.3 Hz 827.5 Hz 919.4 Hz 1034.4 Hz 2. The graph drawn should have frequency on the vertical axis, contain five data points, and trend downward and to the right. A graph using the sample data from above is displayed below. Figure 17.10 A graph of the depth of air column versus the frequency of pitch generated. 3. Inverse relationship. Table 17.3 Depth of air column (\u03bb) Frequency of pitch generated (f) Product of wavelength and frequency 24 cm 22 cm 20 cm 18 cm 16 cm 689.6 Hz 752.3 Hz 827.5 Hz 919.4 Hz 1034.4 Hz 165.5 165.5 165.5 165.5 165.5 4. The graph does align with the equation v = f \u03bb. As the wavelength decreases, the frequency of the", " pitch generated increases. This relationship is validated by both the sample data table and the sample graph. Additionally, as Table 17.1 demonstrates, the product of \u03bb and f is constant across all five data points. In addition to these explanations, the student may use the formula as given in the problem statement to show that the product f \u00d7 air column height is consistently 165.5. This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 17 | Physics of Hearing 729 Table 17.4 Speed of Sound in Various Media Medium vw(m/s) Gases at 0\u00baC Air Carbon dioxide Oxygen Helium 331 259 316 965 Hydrogen 1290 Liquids at 20\u00baC Ethanol Mercury Water, fresh Sea water Human tissue 1160 1450 1480 1540 1540 Solids (longitudinal or bulk) Vulcanized rubber 54 Polyethylene Marble Glass, Pyrex Lead Aluminum Steel 920 3810 5640 1960 5120 5960 Earthquakes, essentially sound waves in Earth\u2019s crust, are an interesting example of how the speed of sound depends on the rigidity of the medium. Earthquakes have both longitudinal and transverse components, and these travel at different speeds. The bulk modulus of granite is greater than its shear modulus. For that reason, the speed of longitudinal or pressure waves (Pwaves) in earthquakes in granite is significantly higher than the speed of transverse or shear waves (S-waves). Both components of earthquakes travel slower in less rigid material, such as sediments. P-waves have speeds of 4 to 7 km/s, and S-waves correspondingly range in speed from 2 to 5 km/s, both being faster in more rigid material. The P-wave gets progressively farther ahead of the S-wave as they travel through Earth\u2019s crust. The time between the P- and S-waves is routinely used to determine the distance to their source, the epicenter of the earthquake. The speed of sound is affected by temperature in a given medium. For air at sea level, the speed of sound is given by where the temperature (denoted as ) is in units of kelvin. The speed of sound in gases is related to the average speed of particles in the gas, rms, and that w = (331 m/s) 273 K, rms = 3, (17.2) (17.3) where is the Boltzmann constant ( 1.38\u00d710\u221223 J/K ) and is the mass of each (identical) particle in the gas. So, it is reasonable that the speed of sound in air and other gases should depend on the square root of temperature. While not negligible, this is not a strong dependence. At 0\u00baC, the speed of sound is 331 m/s, whereas at 20.0\u00baC it is 343 m/s, less than a 4% increase. Figure 17.11 shows a use of the speed of sound by a bat to sense distances. Echoes are also used in medical imaging. 730 Chapter 17 | Physics of Hearing Figure 17.11 A bat uses sound echoes to find its way about and to catch prey. The time for the echo to return is directly proportional to the distance. One of the more important properties of sound is that its speed is nearly independent of frequency. This independence is certainly true in open air for sounds in the audible range of 20 to 20,000 Hz. If this independence were not true, you would certainly notice it for music played by a marching band in a football stadium, for example. Suppose that high-frequency sounds traveled faster\u2014then the farther you were from the band, the more the sound from the low-pitch instruments would lag that from the high-pitch ones. But the music from all instruments arrives in cadence independent of distance, and so all frequencies must travel at nearly the same speed. Recall that w = (17.4) In a given medium under fixed conditions, w is constant, so that there is a relationship between and ; the higher the frequency, the smaller the wavelength. See Figure 17.12 and consider the following example. Figure 17.12 Because they travel at the same speed in a given medium, low-frequency sounds must have a greater wavelength than high-frequency sounds. Here, the lower-frequency sounds are emitted by the large speaker, called a woofer, while the higher-frequency sounds are emitted by the small speaker, called a tweeter. Example 17.1 Calculating Wavelengths: What Are the Wavelengths of Audible Sounds? Calculate the wavelengths of sounds at the extremes of the audible range, 20 and 20,000 Hz, in 30.0\u00baC air. (Assume that the frequency values are accurate to two significant figures.) Strategy To find wavelength from frequency, we can use w =. Solution 1. Identify knowns. The value for w, is given by w", " = (331 m/s) 273 K. 2. Convert the temperature into kelvin and then enter the temperature into the equation w = (331 m/s) 303 K 273 K 3. Solve the relationship between speed and wavelength for : = w. = 348.7 m/s. 4. Enter the speed and the minimum frequency to give the maximum wavelength: max = 348.7 m/s 20 Hz = 17 m. This content is available for free at http://cnx.org/content/col11844/1.13 (17.5) (17.6) (17.7) (17.8) Chapter 17 | Physics of Hearing 5. Enter the speed and the maximum frequency to give the minimum wavelength: min = 348.7 m/s 20,000 Hz = 0.017 m = 1.7 cm. Discussion Because the product of multiplied by equals a constant, the smaller is, the larger must be, and vice versa. 731 (17.9) The speed of sound can change when sound travels from one medium to another. However, the frequency usually remains the same because it is like a driven oscillation and has the frequency of the original source. If w changes and same, then the wavelength must change. That is, because w =, the higher the speed of a sound, the greater its wavelength for a given frequency. remains the Making Connections: Take-Home Investigation\u2014Voice as a Sound Wave Suspend a sheet of paper so that the top edge of the paper is fixed and the bottom edge is free to move. You could tape the top edge of the paper to the edge of a table. Gently blow near the edge of the bottom of the sheet and note how the sheet moves. Speak softly and then louder such that the sounds hit the edge of the bottom of the paper, and note how the sheet moves. Explain the effects. Check Your Understanding Imagine you observe two fireworks explode. You hear the explosion of one as soon as you see it. However, you see the other firework for several milliseconds before you hear the explosion. Explain why this is so. Solution Sound and light both travel at definite speeds. The speed of sound is slower than the speed of light. The first firework is probably very close by, so the speed difference is not noticeable. The second firework is farther away, so the light arrives at your eyes noticeably sooner than the sound wave arrives at your ears. Check Your Understanding You observe two musical instruments that you cannot identify. One plays high-pitch sounds and the other plays low-pitch sounds. How could you determine which is which without hearing either of them play? Solution Compare their sizes. High-pitch instruments are generally smaller than low-pitch instruments because they generate a smaller wavelength. 17.3 Sound Intensity and Sound Level Learning Objectives By the end of this section, you will be able to: \u2022 Define intensity, sound intensity, and sound pressure level. \u2022 Calculate sound intensity levels in decibels (dB). The information presented in this section supports the following AP\u00ae learning objectives and science practices: \u2022 6.A.4.1 The student is able to explain and/or predict qualitatively how the energy carried by a sound wave relates to the amplitude of the wave, and/or apply this concept to a real-world example. (S.P. 6.4) 732 Chapter 17 | Physics of Hearing Figure 17.13 Noise on crowded roadways like this one in Delhi makes it hard to hear others unless they shout. (credit: Lingaraj G J, Flickr) In a quiet forest, you can sometimes hear a single leaf fall to the ground. After settling into bed, you may hear your blood pulsing through your ears. But when a passing motorist has his stereo turned up, you cannot even hear what the person next to you in your car is saying. We are all very familiar with the loudness of sounds and aware that they are related to how energetically the source is vibrating. In cartoons depicting a screaming person (or an animal making a loud noise), the cartoonist often shows an open mouth with a vibrating uvula, the hanging tissue at the back of the mouth, to suggest a loud sound coming from the throat Figure 17.14. High noise exposure is hazardous to hearing, and it is common for musicians to have hearing losses that are sufficiently severe that they interfere with the musicians\u2019 abilities to perform. The relevant physical quantity is sound intensity, a concept that is valid for all sounds whether or not they are in the audible range. Intensity is defined to be the power per unit area carried by a wave. Power is the rate at which energy is transferred by the wave. In equation form, intensity is = where is the power through an area. The SI unit for is W/m2. The intensity of a sound wave is related to its amplitude squared by the following relationship:, (17.10", ") = 2 \u0394 2w. (17.11) Here \u0394 is the pressure variation or pressure amplitude (half the difference between the maximum and minimum pressure in the sound wave) in units of pascals (Pa) or N/m2. (We are using a lower case for pressure to distinguish it from power, denoted by above.) The energy (as kinetic energy 2 2 proportional to its amplitude squared. In this equation, is the density of the material in which the sound wave travels, in units of kg/m3, and w is the speed of sound in the medium, in units of m/s. The pressure variation is proportional to the amplitude of the oscillation, and so varies as (\u0394)2 (Figure 17.14). This relationship is consistent with the fact that the sound wave is produced by some vibration; the greater its pressure amplitude, the more the air is compressed in the sound it creates. ) of an oscillating element of air due to a traveling sound wave is Figure 17.14 Graphs of the gauge pressures in two sound waves of different intensities. The more intense sound is produced by a source that has larger-amplitude oscillations and has greater pressure maxima and minima. Because pressures are higher in the greater-intensity sound, it can exert larger forces on the objects it encounters. This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 17 | Physics of Hearing 733 Sound intensity levels are quoted in decibels (dB) much more often than sound intensities in watts per meter squared. Decibels are the unit of choice in the scientific literature as well as in the popular media. The reasons for this choice of units are related to how we perceive sounds. How our ears perceive sound can be more accurately described by the logarithm of the intensity rather than directly to the intensity. The sound intensity level in decibels of a sound having an intensity in watts per meter squared is defined to be (dB) = 10 log10 0, (17.12) where 0 = 10\u201312 W/m2 is a reference intensity. In particular, 0 is the lowest or threshold intensity of sound a person with normal hearing can perceive at a frequency of 1000 Hz. Sound intensity level is not the same as intensity. Because is defined in terms of a ratio, it is a unitless quantity telling you the level of the sound relative to a fixed standard ( 10\u201312 W/m2, in this case). The units of decibels (dB) are used to indicate this ratio is multiplied by 10 in its definition. The bel, upon which the decibel is based, is named for Alexander Graham Bell, the inventor of the telephone. Table 17.5 Sound Intensity Levels and Intensities Sound intensity level \u03b2 (dB) Intensity I(W/m2) Example/effect 0 10 20 30 40 50 60 70 80 90 100 110 120 140 160 1\u00d710\u201312 1\u00d710\u201311 1\u00d710\u201310 1\u00d710\u20139 1\u00d710\u20138 1\u00d710\u20137 1\u00d710\u20136 1\u00d710\u20135 1\u00d710\u20134 1\u00d710\u20133 1\u00d710\u20132 1\u00d710\u20131 1 1\u00d7102 1\u00d7104 Threshold of hearing at 1000 Hz Rustle of leaves Whisper at 1 m distance Quiet home Average home Average office, soft music Normal conversation Noisy office, busy traffic Loud radio, classroom lecture Inside a heavy truck; damage from prolonged exposure[1] Noisy factory, siren at 30 m; damage from 8 h per day exposure Damage from 30 min per day exposure Loud rock concert, pneumatic chipper at 2 m; threshold of pain Jet airplane at 30 m; severe pain, damage in seconds Bursting of eardrums The decibel level of a sound having the threshold intensity of 10 \u2013 12 W/m2 is = 0 dB, because log10 1 = 0. That is, the threshold of hearing is 0 decibels. Table 17.5 gives levels in decibels and intensities in watts per meter squared for some familiar sounds. One of the more striking things about the intensities in Table 17.5 is that the intensity in watts per meter squared is quite small for most sounds. The ear is sensitive to as little as a trillionth of a watt per meter squared\u2014even more impressive when you realize that the area of the eardrum is only about 1 cm2, so that only 10 \u2013 16 W falls on it at the threshold of hearing! Air molecules in a sound wave of this intensity vibrate over a distance of less than one molecular diameter, and the gauge pressures involved are less than 10 \u2013 9 atm. 1. Several government agencies and health-related professional associations recommend that 85 dB not be exceeded for 8-hour daily exposures in the absence of hearing protection. 734 Chapter 17 | Physics of Hearing Another impressive feature of the sounds in", " Table 17.5 is their numerical range. Sound intensity varies by a factor of 1012 from threshold to a sound that causes damage in seconds. You are unaware of this tremendous range in sound intensity because how your ears respond can be described approximately as the logarithm of intensity. Thus, sound intensity levels in decibels fit your experience better than intensities in watts per meter squared. The decibel scale is also easier to relate to because most people are more accustomed to dealing with numbers such as 0, 53, or 120 than numbers such as 1.00\u00d710 \u2013 11. One more observation readily verified by examining Table 17.5 or using = 2 is that each factor of 10 in intensity \u0394 2w corresponds to 10 dB. For example, a 90 dB sound compared with a 60 dB sound is 30 dB greater, or three factors of 10 (that is, 103 times) as intense. Another example is that if one sound is 107 as intense as another, it is 70 dB higher. See Table 17.6. Table 17.6 Ratios of Intensities and Corresponding Differences in Sound Intensity Levels \u03b22 \u2013 \u03b21 I2 / I1 2.0 5.0 10.0 3.0 dB 7.0 dB 10.0 dB Example 17.2 Calculating Sound Intensity Levels: Sound Waves Calculate the sound intensity level in decibels for a sound wave traveling in air at 0\u00baC and having a pressure amplitude of 0.656 Pa. Strategy We are given \u0394, so we can calculate using the equation = from its definition in (dB) = 10 log10 / 0. \u0394 Solution (1) Identify knowns: Sound travels at 331 m/s in air at 0\u00baC. Air has a density of 1.29 kg/m3 at atmospheric pressure and 0\u00baC. 2 / 2w 2. Using, we can calculate straight (2) Enter these values and the pressure amplitude into = \u0394 2 / 2w : = 2 \u0394 2w = (0.656 Pa)2 1.29 kg/m3 2 (331 m/s) = 5.04\u00d710\u22124 W/m2. (17.13) (3) Enter the value for and the known value for 0 into (dB) = 10 log10 level in decibels: / 0. Calculate to find the sound intensity Discussion 10 log10 5.04\u00d7108 = 10 8.70 dB = 87 dB. (17.14) This 87 dB sound has an intensity five times as great as an 80 dB sound. So a factor of five in intensity corresponds to a difference of 7 dB in sound intensity level. This value is true for any intensities differing by a factor of five. Example 17.3 Change Intensity Levels of a Sound: What Happens to the Decibel Level? Show that if one sound is twice as intense as another, it has a sound level about 3 dB higher. Strategy You are given that the ratio of two intensities is 2 to 1, and are then asked to find the difference in their sound levels in decibels. You can solve this problem using of the properties of logarithms. This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 17 | Physics of Hearing 735 Solution (1) Identify knowns: The ratio of the two intensities is 2 to 1, or: 2 1 = 2.00. We wish to show that the difference in sound levels is about 3 dB. That is, we want to show: Note that: (2) Use the definition of to get: 2 \u2212 1 = 3 dB. log10 \u2212 log10 = log10. 2 \u2212 1 = 10 log10 2 1 = 10 log10 2.00 = 10 (0.301) dB. 2 \u2212 1 = 3.01 dB. Thus, Discussion (17.15) (17.16) (17.17) (17.18) (17.19) This means that the two sound intensity levels differ by 3.01 dB, or about 3 dB, as advertised. Note that because only the ratio 2 / 1 is given (and not the actual intensities), this result is true for any intensities that differ by a factor of two. For example, a 56.0 dB sound is twice as intense as a 53.0 dB sound, a 97.0 dB sound is half as intense as a 100 dB sound, and so on. It should be noted at this point that there is another decibel scale in use, called the sound pressure level, based on the ratio of the pressure amplitude to a reference pressure. This scale is used particularly in applications where sound travels in water. It is beyond the scope of most introductory texts to treat this scale because it is not commonly used for sounds in air, but it is important to note that very different decibel levels may be encountered when sound pressure levels are quoted", ". For example, ocean noise pollution produced by ships may be as great as 200 dB expressed in the sound pressure level, where the more familiar sound intensity level we use here would be something under 140 dB for the same sound. Take-Home Investigation: Feeling Sound Find a CD player and a CD that has rock music. Place the player on a light table, insert the CD into the player, and start playing the CD. Place your hand gently on the table next to the speakers. Increase the volume and note the level when the table just begins to vibrate as the rock music plays. Increase the reading on the volume control until it doubles. What has happened to the vibrations? Check Your Understanding Describe how amplitude is related to the loudness of a sound. Solution Amplitude is directly proportional to the experience of loudness. As amplitude increases, loudness increases. Check Your Understanding Identify common sounds at the levels of 10 dB, 50 dB, and 100 dB. Solution 10 dB: Running fingers through your hair. 50 dB: Inside a quiet home with no television or radio. 100 dB: Take-off of a jet plane. 736 Chapter 17 | Physics of Hearing 17.4 Doppler Effect and Sonic Booms Learning Objectives By the end of this section, you will be able to: \u2022 Define Doppler effect, Doppler shift, and sonic boom. \u2022 Calculate the frequency of a sound heard by someone observing Doppler shift. \u2022 Describe the sounds produced by objects moving faster than the speed of sound. The information presented in this section supports the following AP\u00ae learning objectives and science practices: \u2022 6.B.5.1 The student is able to create or use a wave front diagram to demonstrate or interpret qualitatively the observed frequency of a wave, dependent upon relative motions of source and observer. (S.P. 1.4) The characteristic sound of a motorcycle buzzing by is an example of the Doppler effect. The high-pitch scream shifts dramatically to a lower-pitch roar as the motorcycle passes by a stationary observer. The closer the motorcycle brushes by, the more abrupt the shift. The faster the motorcycle moves, the greater the shift. We also hear this characteristic shift in frequency for passing race cars, airplanes, and trains. It is so familiar that it is used to imply motion and children often mimic it in play. The Doppler effect is an alteration in the observed frequency of a sound due to motion of either the source or the observer. Although less familiar, this effect is easily noticed for a stationary source and moving observer. For example, if you ride a train past a stationary warning bell, you will hear the bell\u2019s frequency shift from high to low as you pass by. The actual change in frequency due to relative motion of source and observer is called a Doppler shift. The Doppler effect and Doppler shift are named for the Austrian physicist and mathematician Christian Johann Doppler (1803\u20131853), who did experiments with both moving sources and moving observers. Doppler, for example, had musicians play on a moving open train car and also play standing next to the train tracks as a train passed by. Their music was observed both on and off the train, and changes in frequency were measured. What causes the Doppler shift? Figure 17.15, Figure 17.16, and Figure 17.17 compare sound waves emitted by stationary and moving sources in a stationary air mass. Each disturbance spreads out spherically from the point where the sound was emitted. If the source is stationary, then all of the spheres representing the air compressions in the sound wave centered on the same point, and the stationary observers on either side see the same wavelength and frequency as emitted by the source, as in Figure 17.15. If the source is moving, as in Figure 17.16, then the situation is different. Each compression of the air moves out in a sphere from the point where it was emitted, but the point of emission moves. This moving emission point causes the air compressions to be closer together on one side and farther apart on the other. Thus, the wavelength is shorter in the direction the source is moving (on the right in Figure 17.16), and longer in the opposite direction (on the left in Figure 17.16). Finally, if the observers move, as in Figure 17.17, the frequency at which they receive the compressions changes. The observer moving toward the source receives them at a higher frequency, and the person moving away from the source receives them at a lower frequency. Figure 17.15 Sounds emitted by a source spread out in spherical waves. Because the source, observers, and air are stationary, the wavelength and frequency are the same in all directions and to all observers. Figure 17.16 Sounds emitted by a source moving to the right spread out from the points at which they were emitted. The wavelength is reduced and, consequently, the frequency is increased in the direction of motion, so that the", " observer on the right hears a higher-pitch sound. The opposite is true for the observer on the left, where the wavelength is increased and the frequency is reduced. This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 17 | Physics of Hearing 737 Figure 17.17 The same effect is produced when the observers move relative to the source. Motion toward the source increases frequency as the observer on the right passes through more wave crests than she would if stationary. Motion away from the source decreases frequency as the observer on the left passes through fewer wave crests than he would if stationary. We know that wavelength and frequency are related by w =, where w is the fixed speed of sound. The sound moves in a medium and has the same speed w in that medium whether the source is moving or not. Thus multiplied by is a constant. Because the observer on the right in Figure 17.16 receives a shorter wavelength, the frequency she receives must be higher. Similarly, the observer on the left receives a longer wavelength, and hence he hears a lower frequency. The same thing happens in Figure 17.17. A higher frequency is received by the observer moving toward the source, and a lower frequency is received by an observer moving away from the source. In general, then, relative motion of source and observer toward one another increases the received frequency. Relative motion apart decreases frequency. The greater the relative speed is, the greater the effect. The Doppler Effect The Doppler effect occurs not only for sound but for any wave when there is relative motion between the observer and the source. There are Doppler shifts in the frequency of sound, light, and water waves, for example. Doppler shifts can be used to determine velocity, such as when ultrasound is reflected from blood in a medical diagnostic. The recession of galaxies is determined by the shift in the frequencies of light received from them and has implied much about the origins of the universe. Modern physics has been profoundly affected by observations of Doppler shifts. For a stationary observer and a moving source, the frequency fobs received by the observer can be shown to be obs = s w w \u00b1 s, (17.20) where s is the frequency of the source, s is the speed of the source along a line joining the source and observer, and w is the speed of sound. The minus sign is used for motion toward the observer and the plus sign for motion away from the observer, producing the appropriate shifts up and down in frequency. Note that the greater the speed of the source, the greater the effect. Similarly, for a stationary source and moving observer, the frequency received by the observer obs is given by obs = s w \u00b1 obs w, (17.21) where obs is the speed of the observer along a line joining the source and observer. Here the plus sign is for motion toward the source, and the minus is for motion away from the source. Example 17.4 Calculate Doppler Shift: A Train Horn Suppose a train that has a 150-Hz horn is moving at 35.0 m/s in still air on a day when the speed of sound is 340 m/s. (a) What frequencies are observed by a stationary person at the side of the tracks as the train approaches and after it passes? (b) What frequency is observed by the train\u2019s engineer traveling on the train? Strategy To find the observed frequency in (a), obs = s w w \u00b1 s, must be used because the source is moving. The minus sign is used for the approaching train, and the plus sign for the receding train. In (b), there are two Doppler shifts\u2014one for a moving source and the other for a moving observer. Solution for (a) 738 Chapter 17 | Physics of Hearing (1) Enter known values into obs = s w w \u2013 s. obs = s w w \u2212 s = (150 Hz) 340 m/s 340 m/s \u2013 35.0 m/s (2) Calculate the frequency observed by a stationary person as the train approaches. obs = (150 Hz)(1.11) = 167 Hz (3) Use the same equation with the plus sign to find the frequency heard by a stationary person as the train recedes. obs = s w w + s = (150 Hz) 340 m/s 340 m/s + 35.0 m/s (4) Calculate the second frequency. Discussion on (a) obs = (150 Hz)(0.907) = 136 Hz (17.22) (17.23) (17.24) (17.25) The numbers calculated are valid when the train is far enough away that the motion is nearly along the line joining train and observer. In both cases, the shift is significant and easily noticed. Note that the shift is 17.0 Hz for motion toward and 14.0 Hz for motion away. The shifts are not symmetric.", " Solution for (b) (1) Identify knowns: \u2022 It seems reasonable that the engineer would receive the same frequency as emitted by the horn, because the relative velocity between them is zero. \u2022 Relative to the medium (air), the speeds are s = obs = 35.0 m/s. \u2022 The first Doppler shift is for the moving observer; the second is for the moving source. (2) Use the following equation: obs = s w \u00b1 obs w w w \u00b1 s. (17.26) The quantity in the square brackets is the Doppler-shifted frequency due to a moving observer. The factor on the right is the effect of the moving source. (3) Because the train engineer is moving in the direction toward the horn, we must use the plus sign for obs; however, because the horn is also moving in the direction away from the engineer, we also use the plus sign for s. But the train is carrying both the engineer and the horn at the same velocity, so s = obs. As a result, everything but s cancels, yielding obs = s. (17.27) Discussion for (b) We may expect that there is no change in frequency when source and observer move together because it fits your experience. For example, there is no Doppler shift in the frequency of conversations between driver and passenger on a motorcycle. People talking when a wind moves the air between them also observe no Doppler shift in their conversation. The crucial point is that source and observer are not moving relative to each other. Sonic Booms to Bow Wakes What happens to the sound produced by a moving source, such as a jet airplane, that approaches or even exceeds the speed of sound? The answer to this question applies not only to sound but to all other waves as well. Suppose a jet airplane is coming nearly straight at you, emitting a sound of frequency s. The greater the plane\u2019s speed s, the greater the Doppler shift and the greater the value observed for approaches infinity, because the denominator in obs = s obs. Now, as s approaches the speed of sound, w w \u00b1 s approaches zero. At the speed of sound, this result obs means that in front of the source, each successive wave is superimposed on the previous one because the source moves forward at the speed of sound. The observer gets them all at the same instant, and so the frequency is infinite. (Before airplanes exceeded the speed of sound, some people argued it would be impossible because such constructive superposition would produce pressures great enough to destroy the airplane.) If the source exceeds the speed of sound, no sound is received by the observer until the source has passed, so that the sounds from the approaching source are mixed with those from it when receding. This mixing appears messy, but something interesting happens\u2014a sonic boom is created. (See Figure 17.18.) This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 17 | Physics of Hearing 739 Figure 17.18 Sound waves from a source that moves faster than the speed of sound spread spherically from the point where they are emitted, but the source moves ahead of each. Constructive interference along the lines shown (actually a cone in three dimensions) creates a shock wave called a sonic boom. The faster the speed of the source, the smaller the angle. There is constructive interference along the lines shown (a cone in three dimensions) from similar sound waves arriving there simultaneously. This superposition forms a disturbance called a sonic boom, a constructive interference of sound created by an object moving faster than sound. Inside the cone, the interference is mostly destructive, and so the sound intensity there is much less than on the shock wave. An aircraft creates two sonic booms, one from its nose and one from its tail. (See Figure 17.19.) During television coverage of space shuttle landings, two distinct booms could often be heard. These were separated by exactly the time it would take the shuttle to pass by a point. Observers on the ground often do not see the aircraft creating the sonic boom, because it has passed by before the shock wave reaches them, as seen in Figure 17.19. If the aircraft flies close by at low altitude, pressures in the sonic boom can be destructive and break windows as well as rattle nerves. Because of how destructive sonic booms can be, supersonic flights are banned over populated areas of the United States. Figure 17.19 Two sonic booms, created by the nose and tail of an aircraft, are observed on the ground after the plane has passed by. Sonic booms are one example of a broader phenomenon called bow wakes. A bow wake, such as the one in Figure 17.20, is created when the wave source moves faster than the wave propagation speed. Water waves spread out in circles from the point where created, and the bow wake is the familiar V-shaped wake trailing the source.", " A more exotic bow wake is created when a subatomic particle travels through a medium faster than the speed of light travels in that medium. (In a vacuum, the maximum speed of light will be = 3.00\u00d7108 m/s ; in the medium of water, the speed of light is closer to 0.75. If the particle creates light in its passage, that light spreads on a cone with an angle indicative of the speed of the particle, as illustrated in Figure 17.21. Such a bow wake is called Cerenkov radiation and is commonly observed in particle physics. Figure 17.20 Bow wake created by a duck. Constructive interference produces the rather structured wake, while there is relatively little wave action inside the wake, where interference is mostly destructive. (credit: Horia Varlan, Flickr) 740 Chapter 17 | Physics of Hearing Figure 17.21 The blue glow in this research reactor pool is Cerenkov radiation caused by subatomic particles traveling faster than the speed of light in water. (credit: U.S. Nuclear Regulatory Commission) Doppler shifts and sonic booms are interesting sound phenomena that occur in all types of waves. They can be of considerable use. For example, the Doppler shift in ultrasound can be used to measure blood velocity, while police use the Doppler shift in radar (a microwave) to measure car velocities. In meteorology, the Doppler shift is used to track the motion of storm clouds; such \u201cDoppler Radar\u201d can give velocity and direction and rain or snow potential of imposing weather fronts. In astronomy, we can examine the light emitted from distant galaxies and determine their speed relative to ours. As galaxies move away from us, their light is shifted to a lower frequency, and so to a longer wavelength\u2014the so-called red shift. Such information from galaxies far, far away has allowed us to estimate the age of the universe (from the Big Bang) as about 14 billion years. Check Your Understanding Why did scientist Christian Doppler observe musicians both on a moving train and also from a stationary point not on the train? Solution Doppler needed to compare the perception of sound when the observer is stationary and the sound source moves, as well as when the sound source and the observer are both in motion. Check Your Understanding Describe a situation in your life when you might rely on the Doppler shift to help you either while driving a car or walking near traffic. Solution If I am driving and I hear Doppler shift in an ambulance siren, I would be able to tell when it was getting closer and also if it has passed by. This would help me to know whether I needed to pull over and let the ambulance through. 17.5 Sound Interference and Resonance: Standing Waves in Air Columns By the end of this section, you will be able to: Learning Objectives \u2022 Define antinode, node, fundamental, overtones, and harmonics. \u2022 Identify instances of sound interference in everyday situations. \u2022 Describe how sound interference occurring inside open and closed tubes changes the characteristics of the sound, and how this applies to sounds produced by musical instruments. \u2022 Calculate the length of a tube using sound wave measurements. The information presented in this section supports the following AP\u00ae learning objectives and science practices: \u2022 6.D.1.1 The student is able to use representations of individual pulses and construct representations to model the interaction of two wave pulses to analyze the superposition of two pulses. (S.P. 1.1, 1.4) \u2022 6.D.1.2 The student is able to design a suitable experiment and analyze data illustrating the superposition of mechanical waves (only for wave pulses or standing waves). (S.P. 4.2, 5.1) \u2022 6.D.1.3 The student is able to design a plan for collecting data to quantify the amplitude variations when two or more traveling waves or wave pulses interact in a given medium. (S.P. 4.2) This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 17 | Physics of Hearing 741 \u2022 6.D.3.1 The student is able to refine a scientific question related to standing waves and design a detailed plan for the experiment that can be conducted to examine the phenomenon qualitatively or quantitatively. (S.P. 2.1, 2.2, 4.2) \u2022 6.D.3.2 The student is able to predict properties of standing waves that result from the addition of incident and reflected waves that are confined to a region and have nodes and antinodes. (S.P. 6.4) \u2022 6.D.3.3 The student is able to plan data collection strategies, predict the outcome based on the relationship under test, perform data analysis, evaluate evidence compared to the prediction, explain any discrepancy and, if necessary, revise the relationship among variables responsible", " for establishing standing waves on a string or in a column of air. (S.P. 3.2, 4.1, 5.1, 5.2, 5.3) \u2022 6.D.3.4 The student is able to describe representations and models of situations in which standing waves result from the addition of incident and reflected waves confined to a region. (S.P. 1.2) \u2022 6.D.4.2 The student is able to calculate wavelengths and frequencies (if given wave speed) of standing waves based on boundary conditions and length of region within which the wave is confined, and calculate numerical values of wavelengths and frequencies. Examples should include musical instruments. (S.P. 2.2) Figure 17.22 Some types of headphones use the phenomena of constructive and destructive interference to cancel out outside noises. (credit: JVC America, Flickr) Interference is the hallmark of waves, all of which exhibit constructive and destructive interference exactly analogous to that seen for water waves. In fact, one way to prove something \u201cis a wave\u201d is to observe interference effects. So, sound being a wave, we expect it to exhibit interference; we have already mentioned a few such effects, such as the beats from two similar notes played simultaneously. Figure 17.23 shows a clever use of sound interference to cancel noise. Larger-scale applications of active noise reduction by destructive interference are contemplated for entire passenger compartments in commercial aircraft. To obtain destructive interference, a fast electronic analysis is performed, and a second sound is introduced with its maxima and minima exactly reversed from the incoming noise. Sound waves in fluids are pressure waves and consistent with Pascal\u2019s principle; pressures from two different sources add and subtract like simple numbers; that is, positive and negative gauge pressures add to a much smaller pressure, producing a lower-intensity sound. Although completely destructive interference is possible only under the simplest conditions, it is possible to reduce noise levels by 30 dB or more using this technique. Figure 17.23 Headphones designed to cancel noise with destructive interference create a sound wave exactly opposite to the incoming sound. These headphones can be more effective than the simple passive attenuation used in most ear protection. Such headphones were used on the recordsetting, around the world nonstop flight of the Voyager aircraft to protect the pilots\u2019 hearing from engine noise. Where else can we observe sound interference? All sound resonances, such as in musical instruments, are due to constructive and destructive interference. Only the resonant frequencies interfere constructively to form standing waves, while others interfere destructively and are absent. From the toot made by blowing over a bottle, to the characteristic flavor of a violin\u2019s sounding box, to the recognizability of a great singer\u2019s voice, resonance and standing waves play a vital role. 742 Interference Chapter 17 | Physics of Hearing Interference is such a fundamental aspect of waves that observing interference is proof that something is a wave. The wave nature of light was established by experiments showing interference. Similarly, when electrons scattered from crystals exhibited interference, their wave nature was confirmed to be exactly as predicted by symmetry with certain wave characteristics of light. Applying the Science Practices: Standing Wave Figure 17.24 The standing wave pattern of a rubber tube attached to a doorknob. Tie one end of a strip of long rubber tubing to a stable object (doorknob, fence post, etc.) and shake the other end up and down until a standing wave pattern is achieved. Devise a method to determine the frequency and wavelength generated by your arm shaking. Do your results align with the equation? Do you find that the velocity of the wave generated is consistent for each trial? If not, explain why this is the case. Answer This task will likely require two people. The frequency of the wave pattern can be found by timing how long it takes the student shaking the rubber tubing to move his or her hand up and down one full time. (It may be beneficial to time how long it takes the student to do this ten times, and then divide by ten to reduce error.) The wavelength of the standing wave can be measured with a meter stick by measuring the distance between two nodes and multiplying by two. This information should be gathered for standing wave patterns of multiple different wavelengths. As students collect their data, they can use the equation to determine if the wave velocity is consistent. There will likely be some error in the experiment yielding velocities of slightly different value. This error is probably due to an inaccuracy in the wavelength and/or frequency measurements. Suppose we hold a tuning fork near the end of a tube that is closed at the other end, as shown in Figure 17.25, Figure 17.26, Figure 17.27, and Figure 17.28. If the tuning fork has just the right frequency, the air column in the tube resonates loudly, but at most frequencies it vibrates very little. This observation just means that the air column has only certain natural frequencies. The figures show", " how a resonance at the lowest of these natural frequencies is formed. A disturbance travels down the tube at the speed of sound and bounces off the closed end. If the tube is just the right length, the reflected sound arrives back at the tuning fork exactly half a cycle later, and it interferes constructively with the continuing sound produced by the tuning fork. The incoming and reflected sounds form a standing wave in the tube as shown. Figure 17.25 Resonance of air in a tube closed at one end, caused by a tuning fork. A disturbance moves down the tube. This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 17 | Physics of Hearing 743 Figure 17.26 Resonance of air in a tube closed at one end, caused by a tuning fork. The disturbance reflects from the closed end of the tube. Figure 17.27 Resonance of air in a tube closed at one end, caused by a tuning fork. If the length of the tube is just right, the disturbance gets back to the tuning fork half a cycle later and interferes constructively with the continuing sound from the tuning fork. This interference forms a standing wave, and the air column resonates. Figure 17.28 Resonance of air in a tube closed at one end, caused by a tuning fork. A graph of air displacement along the length of the tube shows none at the closed end, where the motion is constrained, and a maximum at the open end. This standing wave has one-fourth of its wavelength in the tube, so that = 4. The standing wave formed in the tube has its maximum air displacement (an antinode) at the open end, where motion is unconstrained, and no displacement (a node) at the closed end, where air movement is halted. The distance from a node to an antinode is one-fourth of a wavelength, and this equals the length of the tube; thus, = 4. This same resonance can be produced by a vibration introduced at or near the closed end of the tube, as shown in Figure 17.29. It is best to consider this a natural vibration of the air column independently of how it is induced. 744 Chapter 17 | Physics of Hearing Figure 17.29 The same standing wave is created in the tube by a vibration introduced near its closed end. Given that maximum air displacements are possible at the open end and none at the closed end, there are other, shorter wavelengths that can resonate in the tube, such as the one shown in Figure 17.30. Here the standing wave has three-fourths of its wavelength in the tube, or = (3 / 4)\u2032, so that \u2032 = 4 / 3. Continuing this process reveals a whole series of shorterwavelength and higher-frequency sounds that resonate in the tube. We use specific terms for the resonances in any system. The lowest resonant frequency is called the fundamental, while all higher resonant frequencies are called overtones. All resonant frequencies are integral multiples of the fundamental, and they are collectively called harmonics. The fundamental is the first harmonic, the first overtone is the second harmonic, and so on. Figure 17.31 shows the fundamental and the first three overtones (the first four harmonics) in a tube closed at one end. Figure 17.30 Another resonance for a tube closed at one end. This has maximum air displacements at the open end, and none at the closed end. The wavelength is shorter, with three-fourths \u2032 equaling the length of the tube, so that \u2032 = 4 / 3. This higher-frequency vibration is the first overtone. Figure 17.31 The fundamental and three lowest overtones for a tube closed at one end. All have maximum air displacements at the open end and none at the closed end. The fundamental and overtones can be present simultaneously in a variety of combinations. For example, middle C on a trumpet has a sound distinctively different from middle C on a clarinet, both instruments being modified versions of a tube closed at one end. The fundamental frequency is the same (and usually the most intense), but the overtones and their mix of intensities are different and subject to shading by the musician. This mix is what gives various musical instruments (and human voices) their distinctive characteristics, whether they have air columns, strings, sounding boxes, or drumheads. In fact, much of our speech is determined by shaping the cavity formed by the throat and mouth and positioning the tongue to adjust the fundamental and combination of overtones. Simple resonant cavities can be made to resonate with the sound of the vowels, for example. (See Figure 17.32.) In boys, at puberty, the larynx grows and the shape of the resonant cavity changes giving rise to the difference in predominant frequencies in speech between men and women. This content is available for free at http://cnx.org/content/col11844/", "1.13 Chapter 17 | Physics of Hearing 745 Figure 17.32 The throat and mouth form an air column closed at one end that resonates in response to vibrations in the voice box. The spectrum of overtones and their intensities vary with mouth shaping and tongue position to form different sounds. The voice box can be replaced with a mechanical vibrator, and understandable speech is still possible. Variations in basic shapes make different voices recognizable. Now let us look for a pattern in the resonant frequencies for a simple tube that is closed at one end. The fundamental has = 4, and frequency is related to wavelength and the speed of sound as given by: Solving for in this equation gives w = = w = w 4, where w is the speed of sound in air. Similarly, the first overtone has \u2032 = 4 / 3 (see Figure 17.31), so that \u2032 = 3w 4 = 3. Because \u2032 = 3, we call the first overtone the third harmonic. Continuing this process, we see a pattern that can be generalized in a single expression. The resonant frequencies of a tube closed at one end are = w 4, = 1,3,5, (17.28) (17.29) (17.30) (17.31) where 1 is the fundamental, speed of sound and, hence, on temperature. This dependence poses a noticeable problem for organs in old unheated cathedrals, and it is also the reason why musicians commonly bring their wind instruments to room temperature before playing them. 3 is the first overtone, and so on. It is interesting that the resonant frequencies depend on the Example 17.5 Find the Length of a Tube with a 128 Hz Fundamental (a) What length should a tube closed at one end have on a day when the air temperature, is 22.0\u00baC, if its fundamental frequency is to be 128 Hz (C below middle C)? (b) What is the frequency of its fourth overtone? Strategy The length can be found from the relationship in = w 4 Solution for (a) (1) Identify knowns: \u2022 \u2022 the fundamental frequency is 128 Hz the air temperature is 22.0\u00baC, but we will first need to find the speed of sound w. (2) Use = w 4 to find the fundamental frequency ( = 1 ). (3) Solve this equation for length17.32) (17.33) 746 Chapter 17 | Physics of Hearing (4) Find the speed of sound using w = (331 m/s) 273 K. w = (331 m/s) 295 K 273 K = 344 m/s (5) Enter the values of the speed of sound and frequency into the expression for. = w 4 1 = 344 m/s 4(128 Hz) = 0.672 m Discussion on (a) (17.34) (17.35) Many wind instruments are modified tubes that have finger holes, valves, and other devices for changing the length of the resonating air column and hence, the frequency of the note played. Horns producing very low frequencies, such as tubas, require tubes so long that they are coiled into loops. Solution for (b) (1) Identify knowns: \u2022 \u2022 \u2022 \u2022 the first overtone has = 3 the second overtone has = 5 the third overtone has = 7 the fourth overtone has = 9 (2) Enter the value for the fourth overtone into = w 4. 9 = 9w 4 = 9 1 = 1.15 kHz (17.36) Discussion on (b) Whether this overtone occurs in a simple tube or a musical instrument depends on how it is stimulated to vibrate and the details of its shape. The trombone, for example, does not produce its fundamental frequency and only makes overtones. Another type of tube is one that is open at both ends. Examples are some organ pipes, flutes, and oboes. The resonances of tubes open at both ends can be analyzed in a very similar fashion to those for tubes closed at one end. The air columns in tubes open at both ends have maximum air displacements at both ends, as illustrated in Figure 17.33. Standing waves form as shown. Figure 17.33 The resonant frequencies of a tube open at both ends are shown, including the fundamental and the first three overtones. In all cases the maximum air displacements occur at both ends of the tube, giving it different natural frequencies than a tube closed at one end. Based on the fact that a tube open at both ends has maximum air displacements at both ends, and using Figure 17.33 as a guide, we can see that the resonant frequencies of a tube open at both ends are: = w 2, = 1, 2, 3..., (17.37) 2 is the first overtone, where 1 is the fundamental, 3 is the second overtone, and so on.", " Note that a tube open at both ends has a fundamental frequency twice what it would have if closed at one end. It also has a different spectrum of overtones than a tube closed at one end. So if you had two tubes with the same fundamental frequency but one was open at both ends and the other was closed at one end, they would sound different when played because they have different overtones. Middle C, for example, would sound richer played on an open tube, because it has even multiples of the fundamental as well as odd. A closed tube has only odd multiples. This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 17 | Physics of Hearing 747 Applying the Science Practices: Closed- and Open-Ended Tubes Strike an open-ended length of plastic pipe while holding it in the air. Now place one end of the pipe on a hard surface, sealing one opening, and strike it again. How does the sound change? Further investigate the sound created by the pipe by striking pipes of different lengths and composition. Answer When the pipe is placed on the ground, the standing wave within the pipe changes from being open on both ends to being closed on one end. As a result, the fundamental frequency will change from = 2 to = 4. This decrease in frequency results in a decrease in observed pitch. Real-World Applications: Resonance in Everyday Systems Resonance occurs in many different systems, including strings, air columns, and atoms. Resonance is the driven or forced oscillation of a system at its natural frequency. At resonance, energy is transferred rapidly to the oscillating system, and the amplitude of its oscillations grows until the system can no longer be described by Hooke\u2019s law. An example of this is the distorted sound intentionally produced in certain types of rock music. Wind instruments use resonance in air columns to amplify tones made by lips or vibrating reeds. Other instruments also use air resonance in clever ways to amplify sound. Figure 17.34 shows a violin and a guitar, both of which have sounding boxes but with different shapes, resulting in different overtone structures. The vibrating string creates a sound that resonates in the sounding box, greatly amplifying the sound and creating overtones that give the instrument its characteristic flavor. The more complex the shape of the sounding box, the greater its ability to resonate over a wide range of frequencies. The marimba, like the one shown in Figure 17.35 uses pots or gourds below the wooden slats to amplify their tones. The resonance of the pot can be adjusted by adding water. Figure 17.34 String instruments such as violins and guitars use resonance in their sounding boxes to amplify and enrich the sound created by their vibrating strings. The bridge and supports couple the string vibrations to the sounding boxes and air within. (credits: guitar, Feliciano Guimares, Fotopedia; violin, Steve Snodgrass, Flickr) 748 Chapter 17 | Physics of Hearing Figure 17.35 Resonance has been used in musical instruments since prehistoric times. This marimba uses gourds as resonance chambers to amplify its sound. (credit: APC Events, Flickr) We have emphasized sound applications in our discussions of resonance and standing waves, but these ideas apply to any system that has wave characteristics. Vibrating strings, for example, are actually resonating and have fundamentals and overtones similar to those for air columns. More subtle are the resonances in atoms due to the wave character of their electrons. Their orbitals can be viewed as standing waves, which have a fundamental (ground state) and overtones (excited states). It is fascinating that wave characteristics apply to such a wide range of physical systems. Check Your Understanding Describe how noise-canceling headphones differ from standard headphones used to block outside sounds. Solution Regular headphones only block sound waves with a physical barrier. Noise-canceling headphones use destructive interference to reduce the loudness of outside sounds. Check Your Understanding How is it possible to use a standing wave's node and antinode to determine the length of a closed-end tube? Solution When the tube resonates at its natural frequency, the wave's node is located at the closed end of the tube, and the antinode is located at the open end. The length of the tube is equal to one-fourth of the wavelength of this wave. Thus, if we know the wavelength of the wave, we can determine the length of the tube. PhET Explorations: Sound This simulation lets you see sound waves. Adjust the frequency or volume and you can see and hear how the wave changes. Move the listener around and hear what she hears. Figure 17.36 Sound (http://cnx.org/content/m55293/1.2/sound_en.jar) Applying the Science Practices: Variables Affecting Superposition In the PhET Interactive Simulation above, select the tab titled \u2018Two", " Source Interference.\u2019 Within this tab, manipulate the variables present (frequency, amplitude, and speaker separation) to investigate the relationship the variables have with the superposition pattern constructed on the screen. Record all observations. This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 17 | Physics of Hearing 761 (3) Calculate to find the frequency returning to the source: 2,500,649 Hz. Solution for (c) (1) Identify knowns: \u2022 The beat frequency is simply the absolute value of the difference between s and obs, as stated in: (2) Substitute known values: B = \u2223 obs \u2212 s \u2223. \u2223 2,500649 Hz \u2212 2,500000 Hz \u2223 (3) Calculate to find the beat frequency: 649 Hz. Discussion (17.47) (17.48) The Doppler shifts are quite small compared with the original frequency of 2.50 MHz. It is far easier to measure the beat frequency than it is to measure the echo frequency with an accuracy great enough to see shifts of a few hundred hertz out of a couple of megahertz. Furthermore, variations in the source frequency do not greatly affect the beat frequency, because both s and obs would increase or decrease. Those changes subtract out in B = \u2223 obs \u2212 s \u2223. Industrial and Other Applications of Ultrasound Industrial, retail, and research applications of ultrasound are common. A few are discussed here. Ultrasonic cleaners have many uses. Jewelry, machined parts, and other objects that have odd shapes and crevices are immersed in a cleaning fluid that is agitated with ultrasound typically about 40 kHz in frequency. The intensity is great enough to cause cavitation, which is responsible for most of the cleansing action. Because cavitation-produced shock pressures are large and well transmitted in a fluid, they reach into small crevices where even a low-surface-tension cleaning fluid might not penetrate. Sonar is a familiar application of ultrasound. Sonar typically employs ultrasonic frequencies in the range from 30.0 to 100 kHz. Bats, dolphins, submarines, and even some birds use ultrasonic sonar. Echoes are analyzed to give distance and size information both for guidance and finding prey. In most sonar applications, the sound reflects quite well because the objects of interest have significantly different density than the medium in which they travel. When the Doppler shift is observed, velocity information can also be obtained. Submarine sonar can be used to obtain such information, and there is evidence that some bats also sense velocity from their echoes. Similarly, there are a range of relatively inexpensive devices that measure distance by timing ultrasonic echoes. Many cameras, for example, use such information to focus automatically. Some doors open when their ultrasonic ranging devices detect a nearby object, and certain home security lights turn on when their ultrasonic rangers observe motion. Ultrasonic \u201cmeasuring tapes\u201d also exist to measure such things as room dimensions. Sinks in public restrooms are sometimes automated with ultrasound devices to turn faucets on and off when people wash their hands. These devices reduce the spread of germs and can conserve water. Ultrasound is used for nondestructive testing in industry and by the military. Because ultrasound reflects well from any large change in density, it can reveal cracks and voids in solids, such as aircraft wings, that are too small to be seen with x-rays. For similar reasons, ultrasound is also good for measuring the thickness of coatings, particularly where there are several layers involved. Basic research in solid state physics employs ultrasound. Its attenuation is related to a number of physical characteristics, making it a useful probe. Among these characteristics are structural changes such as those found in liquid crystals, the transition of a material to a superconducting phase, as well as density and other properties. These examples of the uses of ultrasound are meant to whet the appetites of the curious, as well as to illustrate the underlying physics of ultrasound. There are many more applications, as you can easily discover for yourself. Check Your Understanding Why is it possible to use ultrasound both to observe a fetus in the womb and also to destroy cancerous tumors in the body? Solution Ultrasound can be used medically at different intensities. Lower intensities do not cause damage and are used for medical imaging. Higher intensities can pulverize and destroy targeted substances in the body, such as tumors. Glossary acoustic impedance: property of medium that makes the propagation of sound waves more difficult antinode: point of maximum displacement bow wake: V-shaped disturbance created when the wave source moves faster than the wave propagation speed 762 Chapter 17 | Physics of Hearing Doppler effect: an alteration in the observed frequency of a sound due to motion of either the source or the observer Doppler shift: the actual change in frequency due to relative motion of source and observer Doppler", "-shifted ultrasound: a medical technique to detect motion and determine velocity through the Doppler shift of an echo fundamental: the lowest-frequency resonance harmonics: the term used to refer collectively to the fundamental and its overtones hearing: the perception of sound infrasound: sounds below 20 Hz intensity: the power per unit area carried by a wave intensity reflection coefficient: a measure of the ratio of the intensity of the wave reflected off a boundary between two media relative to the intensity of the incident wave loudness: the perception of sound intensity node: point of zero displacement note: basic unit of music with specific names, combined to generate tunes overtones: all resonant frequencies higher than the fundamental phon: the numerical unit of loudness pitch: the perception of the frequency of a sound sonic boom: a constructive interference of sound created by an object moving faster than sound sound: a disturbance of matter that is transmitted from its source outward sound intensity level: a unitless quantity telling you the level of the sound relative to a fixed standard sound pressure level: the ratio of the pressure amplitude to a reference pressure timbre: number and relative intensity of multiple sound frequencies tone: number and relative intensity of multiple sound frequencies ultrasound: sounds above 20,000 Hz Section Summary 17.1 Sound \u2022 Sound is a disturbance of matter that is transmitted from its source outward. \u2022 Sound is one type of wave. \u2022 Hearing is the perception of sound. 17.2 Speed of Sound, Frequency, and Wavelength The relationship of the speed of sound w, its frequency, and its wavelength is given by which is the same relationship given for all waves. In air, the speed of sound is related to air temperature by w =, w = (331 m/s) 273 K. w is the same for all frequencies and wavelengths. 17.3 Sound Intensity and Sound Level \u2022 Intensity is the same for a sound wave as was defined for all waves; it is This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 17 | Physics of Hearing 763 where is the power crossing area. The SI unit for is watts per meter squared. The intensity of a sound wave is also related to the pressure amplitude \u0394 =, = 2 \u0394 2w, where is the density of the medium in which the sound wave travels and w is the speed of sound in the medium. \u2022 Sound intensity level in units of decibels (dB) is where 0 = 10\u201312 W/m2 is the threshold intensity of hearing. (dB) = 10 log10 0, 17.4 Doppler Effect and Sonic Booms \u2022 The Doppler effect is an alteration in the observed frequency of a sound due to motion of either the source or the observer. \u2022 The actual change in frequency is called the Doppler shift. \u2022 A sonic boom is constructive interference of sound created by an object moving faster than sound. \u2022 A sonic boom is a type of bow wake created when any wave source moves faster than the wave propagation speed. \u2022 For a stationary observer and a moving source, the observed frequency obs is: obs = s w w \u00b1 s, where s is the frequency of the source, s is the speed of the source, and w is the speed of sound. The minus sign is used for motion toward the observer and the plus sign for motion away. \u2022 For a stationary source and moving observer, the observed frequency is: w \u00b1 obs w obs = s, where obs is the speed of the observer. 17.5 Sound Interference and Resonance: Standing Waves in Air Columns \u2022 Sound interference and resonance have the same properties as defined for all waves. \u2022 In air columns, the lowest-frequency resonance is called the fundamental, whereas all higher resonant frequencies are called overtones. Collectively, they are called harmonics. \u2022 The resonant frequencies of a tube closed at one end are: w 4, = 1, 3, 5..., = 1 is the fundamental and is the length of the tube. \u2022 The resonant frequencies of a tube open at both ends are: w 2 =, = 1, 2, 3... 17.6 Hearing \u2022 The range of audible frequencies is 20 to 20,000 Hz. \u2022 Those sounds above 20,000 Hz are ultrasound, whereas those below 20 Hz are infrasound. \u2022 The perception of frequency is pitch. \u2022 The perception of intensity is loudness. \u2022 Loudness has units of phons. 17.7 Ultrasound \u2022 The acoustic impedance is defined as: =, is the density of a medium through which the sound travels and is the speed of sound through that medium. \u2022 The intensity reflection coefficient, a measure of the ratio of the intensity of the wave reflected off a boundary between two media relative to the intensity of the incident wave, is given by 764 Chapter 17 | Physics of Hearing \u2022 The intensity reflection coefficient is a unitless quantity. Conceptual Questions = 2 \u2212 1 1 + 2 2 2. 17.2 Speed of Sound,", " Frequency, and Wavelength 1. How do sound vibrations of atoms differ from thermal motion? 2. When sound passes from one medium to another where its propagation speed is different, does its frequency or wavelength change? Explain your answer briefly. 17.3 Sound Intensity and Sound Level 3. Six members of a synchronized swim team wear earplugs to protect themselves against water pressure at depths, but they can still hear the music and perform the combinations in the water perfectly. One day, they were asked to leave the pool so the dive team could practice a few dives, and they tried to practice on a mat, but seemed to have a lot more difficulty. Why might this be? 4. A community is concerned about a plan to bring train service to their downtown from the town\u2019s outskirts. The current sound intensity level, even though the rail yard is blocks away, is 70 dB downtown. The mayor assures the public that there will be a difference of only 30 dB in sound in the downtown area. Should the townspeople be concerned? Why? 17.4 Doppler Effect and Sonic Booms 5. Is the Doppler shift real or just a sensory illusion? 6. Due to efficiency considerations related to its bow wake, the supersonic transport aircraft must maintain a cruising speed that is a constant ratio to the speed of sound (a constant Mach number). If the aircraft flies from warm air into colder air, should it increase or decrease its speed? Explain your answer. 7. When you hear a sonic boom, you often cannot see the plane that made it. Why is that? 17.5 Sound Interference and Resonance: Standing Waves in Air Columns 8. How does an unamplified guitar produce sounds so much more intense than those of a plucked string held taut by a simple stick? 9. You are given two wind instruments of identical length. One is open at both ends, whereas the other is closed at one end. Which is able to produce the lowest frequency? 10. What is the difference between an overtone and a harmonic? Are all harmonics overtones? Are all overtones harmonics? 17.6 Hearing 11. Why can a hearing test show that your threshold of hearing is 0 dB at 250 Hz, when Figure 17.39 implies that no one can hear such a frequency at less than 20 dB? 17.7 Ultrasound 12. If audible sound follows a rule of thumb similar to that for ultrasound, in terms of its absorption, would you expect the high or low frequencies from your neighbor\u2019s stereo to penetrate into your house? How does this expectation compare with your experience? 13. Elephants and whales are known to use infrasound to communicate over very large distances. What are the advantages of infrasound for long distance communication? 14. It is more difficult to obtain a high-resolution ultrasound image in the abdominal region of someone who is overweight than for someone who has a slight build. Explain why this statement is accurate. 15. Suppose you read that 210-dB ultrasound is being used to pulverize cancerous tumors. You calculate the intensity in watts per centimeter squared and find it is unreasonably high ( 105 W/cm2 ). What is a possible explanation? This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 17 | Physics of Hearing 765 Problems & Exercises 17.2 Speed of Sound, Frequency, and Wavelength 15. What intensity level does the sound in the preceding problem correspond to? 16. What sound intensity level in dB is produced by earphones that create an intensity of 4.00\u00d710\u22122 W/m2? 1. When poked by a spear, an operatic soprano lets out a 1200-Hz shriek. What is its wavelength if the speed of sound is 345 m/s? 17. Show that an intensity of 10\u201312 W/m2 is the same as 10\u201316 W/cm2. 2. What frequency sound has a 0.10-m wavelength when the speed of sound is 340 m/s? 3. Calculate the speed of sound on a day when a 1500 Hz frequency has a wavelength of 0.221 m. 4. (a) What is the speed of sound in a medium where a 100-kHz frequency produces a 5.96-cm wavelength? (b) Which substance in Table 17.4 is this likely to be? 5. Show that the speed of sound in 20.0\u00baC air is 343 m/s, as claimed in the text. 6. Air temperature in the Sahara Desert can reach 56.0\u00baC (about 134\u00baF ). What is the speed of sound in air at that temperature? 7. Dolphins make sounds in air and water. What is the ratio of the wavelength of a sound in air to its wavelength in seawater? Assume air temperature is 20.0\u00baC. 8. A sonar echo returns to", " a submarine 1.20 s after being emitted. What is the distance to the object creating the echo? (Assume that the submarine is in the ocean, not in fresh water.) 9. (a) If a submarine\u2019s sonar can measure echo times with a precision of 0.0100 s, what is the smallest difference in distances it can detect? (Assume that the submarine is in the ocean, not in fresh water.) (b) Discuss the limits this time resolution imposes on the ability of the sonar system to detect the size and shape of the object creating the echo. 10. A physicist at a fireworks display times the lag between seeing an explosion and hearing its sound, and finds it to be 0.400 s. (a) How far away is the explosion if air temperature is 24.0\u00baC and if you neglect the time taken for light to reach the physicist? (b) Calculate the distance to the explosion taking the speed of light into account. Note that this distance is negligibly greater. 11. Suppose a bat uses sound echoes to locate its insect prey, 3.00 m away. (See Figure 17.11.) (a) Calculate the echo times for temperatures of 5.00\u00baC and 35.0\u00baC. (b) What percent uncertainty does this cause for the bat in locating the insect? (c) Discuss the significance of this uncertainty and whether it could cause difficulties for the bat. (In practice, the bat continues to use sound as it closes in, eliminating most of any difficulties imposed by this and other effects, such as motion of the prey.) 17.3 Sound Intensity and Sound Level 12. What is the intensity in watts per meter squared of 85.0-dB sound? 13. The warning tag on a lawn mower states that it produces noise at a level of 91.0 dB. What is this in watts per meter squared? 14. A sound wave traveling in 20\u00baC air has a pressure amplitude of 0.5 Pa. What is the intensity of the wave? 18. (a) What is the decibel level of a sound that is twice as intense as a 90.0-dB sound? (b) What is the decibel level of a sound that is one-fifth as intense as a 90.0-dB sound? 19. (a) What is the intensity of a sound that has a level 7.00 dB lower than a 4.00\u00d710\u20139 W/m2 sound? (b) What is the intensity of a sound that is 3.00 dB higher than a 4.00\u00d710\u20139 W/m2 sound? 20. (a) How much more intense is a sound that has a level 17.0 dB higher than another? (b) If one sound has a level 23.0 dB less than another, what is the ratio of their intensities? 21. People with good hearing can perceive sounds as low in level as \u20138.00 dB at a frequency of 3000 Hz. What is the intensity of this sound in watts per meter squared? 22. If a large housefly 3.0 m away from you makes a noise of 40.0 dB, what is the noise level of 1000 flies at that distance, assuming interference has a negligible effect? 23. Ten cars in a circle at a boom box competition produce a 120-dB sound intensity level at the center of the circle. What is the average sound intensity level produced there by each stereo, assuming interference effects can be neglected? 24. The amplitude of a sound wave is measured in terms of its maximum gauge pressure. By what factor does the amplitude of a sound wave increase if the sound intensity level goes up by 40.0 dB? 25. If a sound intensity level of 0 dB at 1000 Hz corresponds to a maximum gauge pressure (sound amplitude) of 10\u20139 atm, what is the maximum gauge pressure in a 60-dB sound? What is the maximum gauge pressure in a 120-dB sound? 26. An 8-hour exposure to a sound intensity level of 90.0 dB may cause hearing damage. What energy in joules falls on a 0.800-cm-diameter eardrum so exposed? 27. (a) Ear trumpets were never very common, but they did aid people with hearing losses by gathering sound over a large area and concentrating it on the smaller area of the eardrum. What decibel increase does an ear trumpet produce if its sound gathering area is 900 cm2 and the area of the eardrum is 0.500 cm2, but the trumpet only has an efficiency of 5.00% in transmitting the sound to the eardrum? (b) Comment on the usefulness of the decibel increase found in part (a). 28. Sound is more effectively transmitted into a stethoscope by direct contact than through the air, and it is further intensified by being concentrated on the smaller area of the eardrum. It is reasonable to assume that", " sound is transmitted into a stethoscope 100 times as effectively compared with transmission though the air. What, then, is the gain in decibels produced by a stethoscope that has a sound gathering area of 15.0 cm2, and concentrates the sound 766 Chapter 17 | Physics of Hearing onto two eardrums with a total area of 0.900 cm2 with an efficiency of 40.0%? 29. Loudspeakers can produce intense sounds with surprisingly small energy input in spite of their low efficiencies. Calculate the power input needed to produce a 90.0-dB sound intensity level for a 12.0-cm-diameter speaker that has an efficiency of 1.00%. (This value is the sound intensity level right at the speaker.) 17.4 Doppler Effect and Sonic Booms 30. (a) What frequency is received by a person watching an oncoming ambulance moving at 110 km/h and emitting a steady 800-Hz sound from its siren? The speed of sound on this day is 345 m/s. (b) What frequency does she receive after the ambulance has passed? 31. (a) At an air show a jet flies directly toward the stands at a speed of 1200 km/h, emitting a frequency of 3500 Hz, on a day when the speed of sound is 342 m/s. What frequency is received by the observers? (b) What frequency do they receive as the plane flies directly away from them? 32. What frequency is received by a mouse just before being dispatched by a hawk flying at it at 25.0 m/s and emitting a screech of frequency 3500 Hz? Take the speed of sound to be 331 m/s. 33. A spectator at a parade receives an 888-Hz tone from an oncoming trumpeter who is playing an 880-Hz note. At what speed is the musician approaching if the speed of sound is 338 m/s? 34. A commuter train blows its 200-Hz horn as it approaches a crossing. The speed of sound is 335 m/s. (a) An observer waiting at the crossing receives a frequency of 208 Hz. What is the speed of the train? (b) What frequency does the observer receive as the train moves away? 35. Can you perceive the shift in frequency produced when you pull a tuning fork toward you at 10.0 m/s on a day when the speed of sound is 344 m/s? To answer this question, calculate the factor by which the frequency shifts and see if it is greater than 0.300%. 36. Two eagles fly directly toward one another, the first at 15.0 m/s and the second at 20.0 m/s. Both screech, the first one emitting a frequency of 3200 Hz and the second one emitting a frequency of 3800 Hz. What frequencies do they receive if the speed of sound is 330 m/s? 37. What is the minimum speed at which a source must travel toward you for you to be able to hear that its frequency is Doppler shifted? That is, what speed produces a shift of 0.300% on a day when the speed of sound is 331 m/s? 17.5 Sound Interference and Resonance: Standing Waves in Air Columns 38. A \u201cshowy\u201d custom-built car has two brass horns that are supposed to produce the same frequency but actually emit 263.8 and 264.5 Hz. What beat frequency is produced? 39. What beat frequencies will be present: (a) If the musical notes A and C are played together (frequencies of 220 and 264 Hz)? (b) If D and F are played together (frequencies of 297 and 352 Hz)? (c) If all four are played together? 40. What beat frequencies result if a piano hammer hits three strings that emit frequencies of 127.8, 128.1, and 128.3 Hz? 41. A piano tuner hears a beat every 2.00 s when listening to a 264.0-Hz tuning fork and a single piano string. What are the two possible frequencies of the string? This content is available for free at http://cnx.org/content/col11844/1.13 42. (a) What is the fundamental frequency of a 0.672-m-long tube, open at both ends, on a day when the speed of sound is 344 m/s? (b) What is the frequency of its second harmonic? 43. If a wind instrument, such as a tuba, has a fundamental frequency of 32.0 Hz, what are its first three overtones? It is closed at one end. (The overtones of a real tuba are more complex than this example, because it is a tapered tube.) 44. What are the first three overtones of a bassoon that has a fundamental frequency of 90.0 Hz? It is open at both ends. (", "The overtones of a real bassoon are more complex than this example, because its double reed makes it act more like a tube closed at one end.) 45. How long must a flute be in order to have a fundamental frequency of 262 Hz (this frequency corresponds to middle C on the evenly tempered chromatic scale) on a day when air temperature is 20.0\u00baC? It is open at both ends. 46. What length should an oboe have to produce a fundamental frequency of 110 Hz on a day when the speed of sound is 343 m/s? It is open at both ends. 47. What is the length of a tube that has a fundamental frequency of 176 Hz and a first overtone of 352 Hz if the speed of sound is 343 m/s? 48. (a) Find the length of an organ pipe closed at one end that produces a fundamental frequency of 256 Hz when air temperature is 18.0\u00baC. (b) What is its fundamental frequency at 25.0\u00baC? 49. By what fraction will the frequencies produced by a wind instrument change when air temperature goes from 10.0\u00baC to 30.0\u00baC? That is, find the ratio of the frequencies at those temperatures. 50. The ear canal resonates like a tube closed at one end. (See Figure 17.41.) If ear canals range in length from 1.80 to 2.60 cm in an average population, what is the range of fundamental resonant frequencies? Take air temperature to be 37.0\u00baC, which is the same as body temperature. How does this result correlate with the intensity versus frequency graph (Figure 17.39 of the human ear? 51. Calculate the first overtone in an ear canal, which resonates like a 2.40-cm-long tube closed at one end, by taking air temperature to be 37.0\u00baC. Is the ear particularly sensitive to such a frequency? (The resonances of the ear canal are complicated by its nonuniform shape, which we shall ignore.) 52. A crude approximation of voice production is to consider the breathing passages and mouth to be a resonating tube closed at one end. (See Figure 17.32.) (a) What is the fundamental frequency if the tube is 0.240-m long, by taking air temperature to be 37.0\u00baC? (b) What would this frequency become if the person replaced the air with helium? Assume the same temperature dependence for helium as for air. 53. (a) Students in a physics lab are asked to find the length of an air column in a tube closed at one end that has a fundamental frequency of 256 Hz. They hold the tube vertically and fill it with water to the top, then lower the water while a 256-Hz tuning fork is rung and listen for the first resonance. What is the air temperature if the resonance occurs for a length of 0.336 m? (b) At what length will they observe the second resonance (first overtone)? Chapter 17 | Physics of Hearing 767 54. What frequencies will a 1.80-m-long tube produce in the audible range at 20.0\u00baC if: (a) The tube is closed at one end? (b) It is open at both ends? 17.6 Hearing 55. The factor of 10\u221212 in the range of intensities to which the ear can respond, from threshold to that causing damage after brief exposure, is truly remarkable. If you could measure distances over the same range with a single instrument and the smallest distance you could measure was 1 mm, what would the largest be? 56. The frequencies to which the ear responds vary by a factor of 103. Suppose the speedometer on your car measured speeds differing by the same factor of 103, and the greatest speed it reads is 90.0 mi/h. What would be the slowest nonzero speed it could read? 57. What are the closest frequencies to 500 Hz that an average person can clearly distinguish as being different in frequency from 500 Hz? The sounds are not present simultaneously. 58. Can the average person tell that a 2002-Hz sound has a different frequency than a 1999-Hz sound without playing them simultaneously? 59. If your radio is producing an average sound intensity level of 85 dB, what is the next lowest sound intensity level that is clearly less intense? 60. Can you tell that your roommate turned up the sound on the TV if its average sound intensity level goes from 70 to 73 dB? 61. Based on the graph in Figure 17.38, what is the threshold of hearing in decibels for frequencies of 60, 400, 1000, 4000, and 15,000 Hz? Note that many AC electrical appliances produce 60 Hz, music is commonly 400 Hz, a reference frequency is 1000 Hz, your maximum sensitivity is near 4000 Hz, and many older TVs produce a 15,750 Hz whine. 62. What sound intensity levels must sounds of frequencies 60, 3000, and", " 8000 Hz have in order to have the same loudness as a 40-dB sound of frequency 1000 Hz (that is, to have a loudness of 40 phons)? 63. What is the approximate sound intensity level in decibels of a 600-Hz tone if it has a loudness of 20 phons? If it has a loudness of 70 phons? 64. (a) What are the loudnesses in phons of sounds having frequencies of 200, 1000, 5000, and 10,000 Hz, if they are all at the same 60.0-dB sound intensity level? (b) If they are all at 110 dB? (c) If they are all at 20.0 dB? 65. Suppose a person has a 50-dB hearing loss at all frequencies. By how many factors of 10 will low-intensity sounds need to be amplified to seem normal to this person? Note that smaller amplification is appropriate for more intense sounds to avoid further hearing damage. 66. If a woman needs an amplification of 5.0\u00d71012 times the threshold intensity to enable her to hear at all frequencies, what is her overall hearing loss in dB? Note that smaller amplification is appropriate for more intense sounds to avoid further damage to her hearing from levels above 90 dB. 67. (a) What is the intensity in watts per meter squared of a just barely audible 200-Hz sound? (b) What is the intensity in watts per meter squared of a barely audible 4000-Hz sound? 68. (a) Find the intensity in watts per meter squared of a 60.0-Hz sound having a loudness of 60 phons. (b) Find the intensity in watts per meter squared of a 10,000-Hz sound having a loudness of 60 phons. 69. A person has a hearing threshold 10 dB above normal at 100 Hz and 50 dB above normal at 4000 Hz. How much more intense must a 100-Hz tone be than a 4000-Hz tone if they are both barely audible to this person? 70. A child has a hearing loss of 60 dB near 5000 Hz, due to noise exposure, and normal hearing elsewhere. How much more intense is a 5000-Hz tone than a 400-Hz tone if they are both barely audible to the child? 71. What is the ratio of intensities of two sounds of identical frequency if the first is just barely discernible as louder to a person than the second? 17.7 Ultrasound Unless otherwise indicated, for problems in this section, assume that the speed of sound through human tissues is 1540 m/s. 72. What is the sound intensity level in decibels of ultrasound of intensity 105 W/m2, used to pulverize tissue during surgery? 73. Is 155-dB ultrasound in the range of intensities used for deep heating? Calculate the intensity of this ultrasound and compare this intensity with values quoted in the text. 74. Find the sound intensity level in decibels of 2.00\u00d710\u20132 W/m2 ultrasound used in medical diagnostics. 75. The time delay between transmission and the arrival of the reflected wave of a signal using ultrasound traveling through a piece of fat tissue was 0.13 ms. At what depth did this reflection occur? 76. In the clinical use of ultrasound, transducers are always coupled to the skin by a thin layer of gel or oil, replacing the air that would otherwise exist between the transducer and the skin. (a) Using the values of acoustic impedance given in Table 17.8 calculate the intensity reflection coefficient between transducer material and air. (b) Calculate the intensity reflection coefficient between transducer material and gel (assuming for this problem that its acoustic impedance is identical to that of water). (c) Based on the results of your calculations, explain why the gel is used. 77. (a) Calculate the minimum frequency of ultrasound that will allow you to see details as small as 0.250 mm in human tissue. (b) What is the effective depth to which this sound is effective as a diagnostic probe? 78. (a) Find the size of the smallest detail observable in human tissue with 20.0-MHz ultrasound. (b) Is its effective penetration depth great enough to examine the entire eye (about 3.00 cm is needed)? (c) What is the wavelength of such ultrasound in 0\u00baC air? 79. (a) Echo times are measured by diagnostic ultrasound scanners to determine distances to reflecting surfaces in a patient. What is the difference in echo times for tissues that are 3.50 and 3.60 cm beneath the surface? (This difference is the minimum resolving time for the scanner to see details as small as 0.100 cm, or 1.00 mm. Discrimination of smaller time differences is needed to see smaller details.) (b) Discuss whether the period of this ultrasound must be smaller than the minimum time resolution. If so, what is the minimum 768 Chapter 17 | Physics of Hearing frequency of the ultrasound and", " is that out of the normal range for diagnostic ultrasound? 80. (a) How far apart are two layers of tissue that produce echoes having round-trip times (used to measure distances) that differ by 0.750 \u03bcs? (b) What minimum frequency must the ultrasound have to see detail this small? 81. (a) A bat uses ultrasound to find its way among trees. If this bat can detect echoes 1.00 ms apart, what minimum distance between objects can it detect? (b) Could this distance explain the difficulty that bats have finding an open door when they accidentally get into a house? 82. A dolphin is able to tell in the dark that the ultrasound echoes received from two sharks come from two different objects only if the sharks are separated by 3.50 m, one being that much farther away than the other. (a) If the ultrasound has a frequency of 100 kHz, show this ability is not limited by its wavelength. (b) If this ability is due to the dolphin\u2019s ability to detect the arrival times of echoes, what is the minimum time difference the dolphin can perceive? 83. A diagnostic ultrasound echo is reflected from moving blood and returns with a frequency 500 Hz higher than its original 2.00 MHz. What is the velocity of the blood? (Assume that the frequency of 2.00 MHz is accurate to seven significant figures and 500 Hz is accurate to three significant figures.) 84. Ultrasound reflected from an oncoming bloodstream that is moving at 30.0 cm/s is mixed with the original frequency of 2.50 MHz to produce beats. What is the beat frequency? (Assume that the frequency of 2.50 MHz is accurate to seven significant figures.) Test Prep for AP\u00ae Courses 17.2 Speed of Sound, Frequency, and Wavelength 1. A teacher wants to demonstrate that the speed of sound is not a constant value. Considering her regular classroom voice as the control, which of the following will increase the speed of sound leaving her mouth? I. Submerge her mouth underwater and speak at the same volume. Increase the temperature of the room and speak at the same volume. Increase the pitch of her voice and speak at the same volume. I only I and II only I, II and III II and III III only II. III. a. b. c. d. e. 2. All members of an orchestra begin tuning their instruments at the same time. While some woodwind instruments play high frequency notes, other stringed instruments play notes of lower frequency. Yet an audience member will hear all notes simultaneously, in apparent contrast to the equation. Explain how a student could demonstrate the flaw in the above logic, using a slinky, stopwatch, and meter stick. Make sure to explain what relationship is truly demonstrated in the above equation, in addition to what would be necessary to get the speed of the slinky to actually change. You may include diagrams and equations as part of your explanation. 17.3 Sound Intensity and Sound Level This content is available for free at http://cnx.org/content/col11844/1.13 3. In order to waken a sleeping child, the volume on an alarm clock is tripled. Under this new scenario, how much more energy will be striking the child\u2019s ear drums each second? twice as much three times as much a. b. c. approximately 4.8 times as much d. six times as much e. nine times as much 4. A musician strikes the strings of a guitar such that they vibrate with twice the amplitude. a. Explain why this requires an energy input greater than twice the original value. b. Explain why the sound leaving the string will not result in a decibel level that is twice as great. 17.4 Doppler Effect and Sonic Booms 5. A baggage handler stands on the edge of a runway as a landing plane approaches. Compared to the pitch of the plane as heard by the plane\u2019s pilot, which of the following correctly describes the sensation experienced by the handler? a. The frequency of the plane will be lower pitched according to the baggage handler and will become even lower pitched as the plane slows to a stop. b. The frequency of the plane will be lower pitched according to the baggage handler but will increase in pitch as the plane slows to a stop. c. The frequency of the plane will be higher pitched according to the baggage handler but will decrease in pitch as the plane slows to a stop. d. The frequency of the plane will be higher pitched according to the baggage handler and will further increase in pitch as the plane slows to a stop. 6. The following graph represents the perceived frequency of a car as it passes a student. Chapter 17 | Physics of Hearing 769 c. Figure 17.55 d. Figure 17.56 9. A student sends a transverse wave pulse of amplitude A along a rope attached at one end. As the pulse returns to the student, a second pulse of amplitude 3A is sent", " along the opposite side of the rope. What is the resulting amplitude when the two pulses interact? a. 4A b. A c. 2A, on the side of the original wave pulse d. 2A, on the side of the second wave pulse 10. A student would like to demonstrate destructive interference using two sound sources. Explain how the student could set up this demonstration and what restrictions they would need to place upon their sources. Be sure to consider both the layout of space and the sounds created in your explanation. 11. A student is shaking a flexible string attached to a wooden board in a rhythmic manner. Which of the following choices will decrease the wavelength within the rope? I. The student could shake her hand back and forth with greater frequency. II. The student could shake her hand back in forth with a greater amplitude. III. The student could increase the tension within the rope by stepping backwards from the board. I only I and II I and III II and III I, II, and III a. b. c. d. e. 12. A ripple tank has two locations (L1 and L2) that vibrate in tandem as shown below. Both L1 and L2 vibrate in a plane perpendicular to the page, creating a two-dimensional interference pattern. Figure 17.51 Plot of time versus perceived frequency to illustrate the Doppler effect. a. If the true frequency of the car\u2019s horn is 200 Hz, how fast was the car traveling? b. On the graph above, draw a line demonstrating the perceived frequency for a car traveling twice as fast. Label all intercepts, maximums, and minimums on the graph. 17.5 Sound Interference and Resonance: Standing Waves in Air Columns 7. A common misconception is that two wave pulses traveling in opposite directions will reflect off each other. Outline a procedure that you would use to convince someone that the two wave pulses do not reflect off each other, but instead travel through each other. You may use sketches to represent your understanding. Be sure to provide evidence to not only refute the original claim, but to support yours as well. 8. Two wave pulses are traveling toward each other on a string, as shown below. Which of the following representations correctly shows the string as the two pulses overlap? Figure 17.52 a. b. Figure 17.53 Figure 17.54 770 Chapter 17 | Physics of Hearing c. Using information from the graph, determine the speed of sound within the student\u2019s classroom, and explain what characteristic of the graph provides this evidence. d. Determine the temperature of the classroom. 15. A tube is open at one end. If the fundamental frequency f is created by a wavelength \u03bb, then which of the following describes the frequency and wavelength associated with the tube\u2019s fourth overtone? f \u03bb (a) 4f \u03bb/4 (b) 4f \u03bb (c) 9f \u03bb/9 (d) 9f \u03bb 16. A group of students were tasked with collecting information about standing waves. Table 17.10 a series of their data, showing the length of an air column and a resonant frequency present when the column is struck. Table 17.10 Length (m) Resonant Frequency (Hz) 1 2 3 4 85.75 43 29 21.5 a. From their data, determine whether the air column was open or closed on each end. b. Predict the resonant frequency of the column at a length of 2.5 meters. 17. When a student blows across a glass half-full of water, a resonant frequency is created within the air column remaining in the glass. Which of the following can the student do to increase this resonant frequency? I. Add more water to the glass. II. Replace the water with a more dense fluid. Increase the temperature of the room. III. I only a. I and III b. c. II and III d. all of the above 18. A wooden ruler rests on a desk with half of its length protruding off the desk edge. A student holds one end in place and strikes the protruding end with his other hand, creating a musical sound. a. Explain, without using a sound meter, how the student could experimentally determine the speed of sound that travels within the ruler. b. A sound meter is then used to measure the true frequency of the ruler. It is found that the experimental result is lower than the true value. Explain a factor that may have caused this difference. Also explain what affect this result has on the calculated speed of sound. 19. A musician stands outside in a field and plucks a string on an acoustic guitar. Standing waves will most likely occur in which of the following media? Select two answers. a. The guitar string b. The air inside the guitar c. The air surrounding the guitar Figure 17.57 Describe an experimental procedure to determine the speed of the waves created within the water, including all additional equipment that you", " would need. You may use the diagram below to help your description, or you may create one of your own. Include enough detail so that another student could carry out your experiment. 13. A string is vibrating between two posts as shown above. Students are to determine the speed of the wave within this string. They have already measured the amount of time necessary for the wave to oscillate up and down. The students must also take what other measurements to determine the speed of the wave? a. The distance between the two posts. b. The amplitude of the wave c. The tension in the string d. The amplitude of the wave and the tension in the string e. The distance between the two posts, the amplitude of the wave, and the tension in the string 14. The accepted speed of sound in room temperature air is 346 m/s. Knowing that their school is colder than usual, a group of students is asked to determine the speed of sound in their room. They are permitted to use any materials necessary; however, their lab procedure must utilize standing wave patterns. The students collect the information Table 17.9. Table 17.9 Trial Number Wavelength (m) Frequency (Hz) 1 2 3 4 5 3.45 2.32 1.70 1.45 1.08 95 135 190 240 305 a. Describe an experimental procedure the group of students could have used to obtain this data. Include diagrams of the experimental setup and any equipment used in the process. b. Select a set of data points from the table and plot those points on a graph to determine the speed of sound within the classroom. Fill in the blank column in the table for any quantities you graph other than the given data. Label the axes and indicate the scale for each. Draw a best-fit line or curve through your data points. This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 17 | Physics of Hearing 771 d. The ground beneath the musician a. Based on the information above, what is the speed of 20. the wave within the string? b. The guitarist then slides her finger along the neck of the guitar, changing the string length as a result. Calculate the fundamental frequency of the string and wave speed present if the string length is reduced to 2/3 L. Figure 17.58 This figure shows two tubes that are identical except for their slightly different lengths. Both tubes have one open end and one closed end. A speaker connected to a variable frequency generator is placed in front of the tubes, as shown. The speaker is set to produce a note of very low frequency when turned on. The frequency is then slowly increased to produce resonances in the tubes. Students observe that at first only one of the tubes resonates at a time. Later, as the frequency gets very high, there are times when both tubes resonate. In a clear, coherent, paragraph-length answer, explain why there are some high frequencies, but no low frequencies, at which both tubes resonate. You may include diagrams and/or equations as part of your explanation. Figure 17.59 21. A student connects one end of a string with negligible mass to an oscillator. The other end of the string is passed over a pulley and attached to a suspended weight, as shown above. The student finds that a standing wave with one antinode is formed on the string when the frequency of the oscillator is f0. The student then moves the oscillator to shorten the horizontal segment of string to half its original length. At what frequency will a standing wave with one antinode now be formed on the string? f0/2 f0 a. b. c. 2f0 d. There is no frequency at which a standing wave will be formed. 22. A guitar string of length L is bound at both ends. Table 17.11 shows the string\u2019s harmonic frequencies when struck. Table 17.11 Harmonic Number Frequency 1 2 3 4 225/L 450/L 675/L 900/L Chapter 18 | Electric Charge and Electric Field 773 18 ELECTRIC CHARGE AND ELECTRIC FIELD Figure 18.1 Static electricity from this plastic slide causes the child's hair to stand on end. The sliding motion stripped electrons away from the child's body, leaving an excess of positive charges, which repel each other along each strand of hair. (credit: Ken Bosma/Wikimedia Commons) Chapter Outline 18.1. Static Electricity and Charge: Conservation of Charge 18.2. Conductors and Insulators 18.3. Conductors and Electric Fields in Static Equilibrium 18.4. Coulomb\u2019s Law 18.5. Electric Field: Concept of a Field Revisited 18.6. Electric Field Lines: Multiple Charges 18.7. Electric Forces in Biology 18.8. Applications of Electrostatics Connection for AP\u00ae Courses The image of American politician and scientist Benjamin Franklin (1706\u20131790) flying a kite", " in a thunderstorm (shown in Figure 18.2) is familiar to every schoolchild. In this experiment, Franklin demonstrated a connection between lightning and static electricity. Sparks were drawn from a key hung on a kite string during an electrical storm. These sparks were like those produced by static electricity, such as the spark that jumps from your finger to a metal doorknob after you walk across a wool carpet. Much has been written about Franklin. His experiments were only part of the life of a man who was a scientist, inventor, revolutionary, statesman, and writer. Franklin's experiments were not performed in isolation, nor were they the only ones to reveal connections. 774 Chapter 18 | Electric Charge and Electric Field Figure 18.2 Benjamin Franklin, his kite, and electricity. When Benjamin Franklin demonstrated that lightning was related to static electricity, he made a connection that is now part of the evidence that all directly experienced forces (except gravitational force) are manifestations of the electromagnetic force. For example, the Italian scientist Luigi Galvani (1737-1798) performed a series of experiments in which static electricity was used to stimulate contractions of leg muscles of dead frogs, an effect already known in humans subjected to static discharges. But Galvani also found that if he joined one end of two metal wires (say copper and zinc) and touched the other ends of the wires to muscles; he produced the same effect in frogs as static discharge. Alessandro Volta (1745-1827), partly inspired by Galvani's work, experimented with various combinations of metals and developed the battery. During the same era, other scientists made progress in discovering fundamental connections. The periodic table was developed as systematic properties of the elements were discovered. This influenced the development and refinement of the concept of atoms as the basis of matter. Such submicroscopic descriptions of matter also help explain a great deal more. Atomic and molecular interactions, such as the forces of friction, cohesion, and adhesion, are now known to be manifestations of the electromagnetic force. Static electricity is just one aspect of the electromagnetic force, which also includes moving electricity and magnetism. All the macroscopic forces that we experience directly, such as the sensations of touch and the tension in a rope, are due to the electromagnetic force, one of the four fundamental forces in nature. The gravitational force, another fundamental force, is actually sensed through the electromagnetic interaction of molecules, such as between those in our feet and those on the top of a bathroom scale. (The other two fundamental forces, the strong nuclear force and the weak nuclear force, cannot be sensed on the human scale.) This chapter begins the study of electromagnetic phenomena at a fundamental level. The next several chapters will cover static electricity, moving electricity, and magnetism \u2013 collectively known as electromagnetism. In this chapter, we begin with the study of electric phenomena due to charges that are at least temporarily stationary, called electrostatics, or static electricity. The chapter introduces several very important concepts of charge, electric force, and electric field, as well as defining the relationships between these concepts. The charge is defined as a property of a system (Big Idea 1) that can affect its interaction with other charged systems (Enduring Understanding 1.B). The law of conservation of electric charge is also discussed (Essential Knowledge 1.B.1). The two kinds of electric charge are defined as positive and negative, providing an explanation for having positively charged, negatively charged, or neutral objects (containing equal quantities of positive and negative charges) (Essential Knowledge 1.B.2). The discrete nature of the electric charge is introduced in this chapter by defining the elementary charge as the smallest observed unit of charge that can be isolated, which is the electron charge (Essential Knowledge 1.B.3). The concepts of a system (having internal structure) and of an object (having no internal structure) are implicitly introduced to explain charges carried by the electron and proton (Enduring Understanding 1.A, Essential Knowledge 1.A.1). An electric field is caused by the presence of charged objects (Enduring Understanding 2.C) and can be used to explain interactions between electrically charged objects (Big Idea 2). The electric force represents the effect of an electric field on a charge placed in the field. The magnitude and direction of the electric force are defined by the magnitude and direction of the electric field and magnitude and sign of the charge (Essential Knowledge 2.C.1). The magnitude of the electric field is proportional to the net charge of the objects that created that field (Essential Knowledge 2.C.2). For the special case of a spherically symmetric charged object, the electric field outside the object is radial, and its magnitude varies as the inverse square of the radial distance from the center of that object (Essential Knowledge 2.C.3). The chapter provides examples of vector field maps for various charged systems, including point charges, spherically symmetric charge distributions, and uniformly charged parallel plates (", "Essential Knowledge 2.C.1, Essential Knowledge 2.C.2). For multiple point charges, the chapter explains how to This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 18 | Electric Charge and Electric Field 775 find the vector field map by adding the electric field vectors of each individual object, including the special case of two equal charges having opposite signs, known as an electric dipole (Essential Knowledge 2.C.4). The special case of two oppositely charged parallel plates with uniformly distributed electric charge when the electric field is perpendicular to the plates and is constant in both magnitude and direction is described in detail, providing many opportunities for problem solving and applications (Essential Knowledge 2.C.5). The idea that interactions can be described by forces is also reinforced in this chapter (Big Idea 3). Like all other forces that you have learned about so far, electric force is a vector that affects the motion according to Newton's laws (Enduring Understanding 3.A). It is clearly stated in the chapter that electric force appears as a result of interactions between two charged objects (Essential Knowledge 3.A.3, Essential Knowledge 3.C.2). At the macroscopic level the electric force is a long-range force (Enduring Understanding 3.C); however, at the microscopic level many contact forces, such as friction, can be explained by interatomic electric forces (Essential Knowledge 3.C.4). This understanding of friction is helpful when considering properties of conductors and insulators and the transfer of charge by conduction. Interactions between systems can result in changes in those systems (Big Idea 4). In the case of charged systems, such interactions can lead to changes of electric properties (Enduring Understanding 4.E), such as charge distribution (Essential Knowledge 4.E.3). Any changes are governed by conservation laws (Big Idea 5). Depending on whether the system is closed or open, certain quantities of the system remain the same or changes in those quantities are equal to the amount of transfer of this quantity from or to the system (Enduring Understanding 5.A). The electric charge is one of these quantities (Essential Knowledge 5.A.2). Therefore, the electric charge of a system is conserved (Enduring Understanding 5.C) and the exchange of electric charge between objects in a system does not change the total electric charge of the system (Essential Knowledge 5.C.2). Big Idea 1 Objects and systems have properties such as mass and charge. Systems may have internal structure. Enduring Understanding 1.A The internal structure of a system determines many properties of the system. Essential Knowledge 1.A.1 A system is an object or a collection of objects. Objects are treated as having no internal structure. Enduring Understanding 1.B Electric charge is a property of an object or system that affects its interactions with other objects or systems containing charge. Essential Knowledge 1.B.1 Electric charge is conserved. The net charge of a system is equal to the sum of the charges of all the objects in the system. Essential Knowledge 1.B.2 There are only two kinds of electric charge. Neutral objects or systems contain equal quantities of positive and negative charge, with the exception of some fundamental particles that have no electric charge. Essential Knowledge 1.B.3 The smallest observed unit of charge that can be isolated is the electron charge, also known as the elementary charge. Big Idea 2 Fields existing in space can be used to explain interactions. Enduring Understanding 2.C An electric field is caused by an object with electric charge. Essential Knowledge 2.C.1 The magnitude of the electric force F exerted on an object with electric charge q by an electric field ( \u2192. The direction of the force is determined by the direction of the field and the sign of the charge, with \u2192 = \u2192 is positively charged objects accelerating in the direction of the field and negatively charged objects accelerating in the direction opposite the field. This should include a vector field map for positive point charges, negative point charges, spherically symmetric charge distribution, and uniformly charged parallel plates. Essential Knowledge 2.C.2 The magnitude of the electric field vector is proportional to the net electric charge of the object(s) creating that field. This includes positive point charges, negative point charges, spherically symmetric charge distributions, and uniformly charged parallel plates. Essential Knowledge 2.C.3 The electric field outside a spherically symmetric charged object is radial, and its magnitude varies as the inverse square of the radial distance from the center of that object. Electric field lines are not in the curriculum. Students will be expected to rely only on the rough intuitive sense underlying field lines, wherein the field is viewed as analogous to something emanating uniformly from a source. Essential Knowledge 2.C.4 The electric field around dipoles and other systems of electrically charged objects (that can be modeled as point objects) is found by vector addition of the field", " of each individual object. Electric dipoles are treated qualitatively in this course as a teaching analogy to facilitate student understanding of magnetic dipoles. Essential Knowledge 2.C.5 Between two oppositely charged parallel plates with uniformly distributed electric charge, at points far from the edges of the plates, the electric field is perpendicular to the plates and is constant in both magnitude and direction. Big Idea 3 The interactions of an object with other objects can be described by forces. Enduring Understanding 3.A All forces share certain common characteristics when considered by observers in inertial reference frames. Essential Knowledge 3.A.3 A force exerted on an object is always due to the interaction of that object with another object. Enduring Understanding 3.C At the macroscopic level, forces can be categorized as either long-range (action-at-a-distance) forces or contact forces. Essential Knowledge 3.C.2 Electric force results from the interaction of one object that has an electric charge with another object that has an electric charge. 776 Chapter 18 | Electric Charge and Electric Field Essential Knowledge 3.C.4 Contact forces result from the interaction of one object touching another object, and they arise from interatomic electric forces. These forces include tension, friction, normal, spring (Physics 1), and buoyant (Physics 2). Big Idea 4 Interactions between systems can result in changes in those systems. Enduring Understanding 4.E The electric and magnetic properties of a system can change in response to the presence of, or changes in, other objects or systems. Essential Knowledge 4.E.3 The charge distribution in a system can be altered by the effects of electric forces produced by a charged object. Big Idea 5 Changes that occur as a result of interactions are constrained by conservation laws. Enduring Understanding 5.A Certain quantities are conserved, in the sense that the changes of those quantities in a given system are always equal to the transfer of that quantity to or from the system by all possible interactions with other systems. Essential Knowledge 5.A.2 For all systems under all circumstances, energy, charge, linear momentum, and angular momentum are conserved. Enduring Understanding 5.C The electric charge of a system is conserved. Essential Knowledge 5.C.2 The exchange of electric charges among a set of objects in a system conserves electric charge. 18.1 Static Electricity and Charge: Conservation of Charge Learning Objectives By the end of this section, you will be able to: \u2022 Define electric charge, and describe how the two types of charge interact. \u2022 Describe three common situations that generate static electricity. \u2022 State the law of conservation of charge. The information presented in this section supports the following AP\u00ae learning objectives and science practices: \u2022 1.B.1.1 The student is able to make claims about natural phenomena based on conservation of electric charge. (S.P. 6.4) \u2022 1.B.1.2 The student is able to make predictions, using the conservation of electric charge, about the sign and relative quantity of net charge of objects or systems after various charging processes, including conservation of charge in simple circuits. (S.P. 6.4, 7.2) \u2022 1.B.2.1 The student is able to construct an explanation of the two-charge model of electric charge based on evidence produced through scientific practices. (S.P. 6.4) \u2022 1.B.3.1 The student is able to challenge the claim that an electric charge smaller than the elementary charge has been isolated. (S.P. 1.5, 6.1, 7.2) \u2022 5.A.2.1 The student is able to define open and closed systems for everyday situations and apply conservation concepts for energy, charge, and linear momentum to those situations. (S.P. 6.4, 7.2) \u2022 5.C.2.1 The student is able to predict electric charges on objects within a system by application of the principle of charge conservation within a system. (S.P. 6.4) \u2022 5.C.2.2 The student is able to design a plan to collect data on the electrical charging of objects and electric charge induction on neutral objects and qualitatively analyze that data. (S.P. 4.2, 5.1) \u2022 5.C.2.3 The student is able to justify the selection of data relevant to an investigation of the electrical charging of objects and electric charge induction on neutral objects. (S.P. 4.1) Figure 18.3 Borneo amber was mined in Sabah, Malaysia, from shale-sandstone-mudstone veins. When a piece of amber is rubbed with a piece of silk, the amber gains more electrons, giving it a net negative charge. At the same time, the silk, having lost electrons, becomes positively charged. (credit: Sebakoamber, Wikimedia Commons) This content is available for free at http://cnx.org/content", "/col11844/1.13 Chapter 18 | Electric Charge and Electric Field 777 What makes plastic wrap cling? Static electricity. Not only are applications of static electricity common these days, its existence has been known since ancient times. The first record of its effects dates to ancient Greeks who noted more than 500 years B.C. that polishing amber temporarily enabled it to attract bits of straw (see Figure 18.3). The very word electric derives from the Greek word for amber (electron). Many of the characteristics of static electricity can be explored by rubbing things together. Rubbing creates the spark you get from walking across a wool carpet, for example. Static cling generated in a clothes dryer and the attraction of straw to recently polished amber also result from rubbing. Similarly, lightning results from air movements under certain weather conditions. You can also rub a balloon on your hair, and the static electricity created can then make the balloon cling to a wall. We also have to be cautious of static electricity, especially in dry climates. When we pump gasoline, we are warned to discharge ourselves (after sliding across the seat) on a metal surface before grabbing the gas nozzle. Attendants in hospital operating rooms must wear booties with aluminum foil on the bottoms to avoid creating sparks which may ignite the oxygen being used. Some of the most basic characteristics of static electricity include: \u2022 The effects of static electricity are explained by a physical quantity not previously introduced, called electric charge. \u2022 There are only two types of charge, one called positive and the other called negative. \u2022 Like charges repel, whereas unlike charges attract. \u2022 The force between charges decreases with distance. How do we know there are two types of electric charge? When various materials are rubbed together in controlled ways, certain combinations of materials always produce one type of charge on one material and the opposite type on the other. By convention, we call one type of charge \u201cpositive\u201d, and the other type \u201cnegative.\u201d For example, when glass is rubbed with silk, the glass becomes positively charged and the silk negatively charged. Since the glass and silk have opposite charges, they attract one another like clothes that have rubbed together in a dryer. Two glass rods rubbed with silk in this manner will repel one another, since each rod has positive charge on it. Similarly, two silk cloths so rubbed will repel, since both cloths have negative charge. Figure 18.4 shows how these simple materials can be used to explore the nature of the force between charges. Figure 18.4 A glass rod becomes positively charged when rubbed with silk, while the silk becomes negatively charged. (a) The glass rod is attracted to the silk because their charges are opposite. (b) Two similarly charged glass rods repel. (c) Two similarly charged silk cloths repel. More sophisticated questions arise. Where do these charges come from? Can you create or destroy charge? Is there a smallest unit of charge? Exactly how does the force depend on the amount of charge and the distance between charges? Such questions obviously occurred to Benjamin Franklin and other early researchers, and they interest us even today. Charge Carried by Electrons and Protons Franklin wrote in his letters and books that he could see the effects of electric charge but did not understand what caused the phenomenon. Today we have the advantage of knowing that normal matter is made of atoms, and that atoms contain positive and negative charges, usually in equal amounts. Figure 18.5 shows a simple model of an atom with negative electrons orbiting its positive nucleus. The nucleus is positive due to the presence of positively charged protons. Nearly all charge in nature is due to electrons and protons, which are two of the three building blocks of most matter. (The third is the neutron, which is neutral, carrying no charge.) Other charge-carrying particles are observed in cosmic rays and nuclear decay, and are created in particle accelerators. All but the electron and proton survive only a short time and are quite rare by comparison. 778 Chapter 18 | Electric Charge and Electric Field Figure 18.5 This simplified (and not to scale) view of an atom is called the planetary model of the atom. Negative electrons orbit a much heavier positive nucleus, as the planets orbit the much heavier sun. There the similarity ends, because forces in the atom are electromagnetic, whereas those in the planetary system are gravitational. Normal macroscopic amounts of matter contain immense numbers of atoms and molecules and, hence, even greater numbers of individual negative and positive charges. The charges of electrons and protons are identical in magnitude but opposite in sign. Furthermore, all charged objects in nature are integral multiples of this basic quantity of charge, meaning that all charges are made of combinations of a basic unit of charge. Usually, charges are formed by combinations of electrons and protons. The magnitude of this basic charge is The symbol is commonly used for charge and the subscript indicates the charge of a single electron (or proton). The SI unit of charge is the coulomb (C). The number", " of protons needed to make a charge of 1.00 C is \u2223 \u2223 = 1.60\u00d710\u221219 C. 1.00 C\u00d7 1 proton 1.60\u00d710\u221219 C = 6.25\u00d71018 protons. (18.1) (18.2) Similarly, 6.25\u00d71018 atom), there is a smallest bit of charge. There is no directly observed charge smaller than \u2223 \u2223 Small: The Submicroscopic Origin of Charge), and all observed charges are integral multiples of electrons have a combined charge of \u22121.00 coulomb. Just as there is a smallest bit of an element (an (see Things Great and \u2223 \u2223. Things Great and Small: The Submicroscopic Origin of Charge With the exception of exotic, short-lived particles, all charge in nature is carried by electrons and protons. Electrons carry the charge we have named negative. Protons carry an equal-magnitude charge that we call positive. (See Figure 18.6.) Electron and proton charges are considered fundamental building blocks, since all other charges are integral multiples of those carried by electrons and protons. Electrons and protons are also two of the three fundamental building blocks of ordinary matter. The neutron is the third and has zero total charge. Figure 18.6 shows a person touching a Van de Graaff generator and receiving excess positive charge. The expanded view of a hair shows the existence of both types of charges but an excess of positive. The repulsion of these positive like charges causes the strands of hair to repel other strands of hair and to stand up. The further blowup shows an artist's conception of an electron and a proton perhaps found in an atom in a strand of hair. This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 18 | Electric Charge and Electric Field 779 Figure 18.6 When this person touches a Van de Graaff generator, she receives an excess of positive charge, causing her hair to stand on end. The charges in one hair are shown. An artist's conception of an electron and a proton illustrate the particles carrying the negative and positive charges. We cannot really see these particles with visible light because they are so small (the electron seems to be an infinitesimal point), but we know a great deal about their measurable properties, such as the charges they carry. The electron seems to have no substructure; in contrast, when the substructure of protons is explored by scattering extremely energetic electrons from them, it appears that there are point-like particles inside the proton. These sub-particles, named quarks, have never been directly observed, but they are believed to carry fractional charges as seen in Figure 18.7. Charges on electrons and protons and all other directly observable particles are unitary, but these quark substructures carry charges of either \u2212 1 3. There are continuing attempts to observe fractional charge directly and to learn of the properties of quarks, which or + 2 3 are perhaps the ultimate substructure of matter. Figure 18.7 Artist's conception of fractional quark charges inside a proton. A group of three quark charges add up to the single positive charge on the proton1. Separation of Charge in Atoms Charges in atoms and molecules can be separated\u2014for example, by rubbing materials together. Some atoms and molecules have a greater affinity for electrons than others and will become negatively charged by close contact in rubbing, leaving the other material positively charged. (See Figure 18.8.) Positive charge can similarly be induced by rubbing. Methods other than rubbing can also separate charges. Batteries, for example, use combinations of substances that interact in such a way as to separate charges. Chemical interactions may transfer negative charge from one substance to the other, making one battery terminal negative and leaving the first one positive. 780 Chapter 18 | Electric Charge and Electric Field Figure 18.8 When materials are rubbed together, charges can be separated, particularly if one material has a greater affinity for electrons than another. (a) Both the amber and cloth are originally neutral, with equal positive and negative charges. Only a tiny fraction of the charges are involved, and only a few of them are shown here. (b) When rubbed together, some negative charge is transferred to the amber, leaving the cloth with a net positive charge. (c) When separated, the amber and cloth now have net charges, but the absolute value of the net positive and negative charges will be equal. No charge is actually created or destroyed when charges are separated as we have been discussing. Rather, existing charges are moved about. In fact, in all situations the total amount of charge is always constant. This universally obeyed law of nature is called the law of conservation of charge. Law of Conservation of Charge Total charge is constant in any process. Making Connections: Net Charge Hence if a closed system is", " neutral, it will remain neutral. Similarly, if a closed system has a charge, say, \u221210e, it will always have that charge. The only way to change the charge of a system is to transfer charge outside, either by bringing in charge or removing charge. If it is possible to transfer charge outside, the system is no longer closed/isolated and is known as an open system. However, charge is always conserved, for both open and closed systems. Consequently, the charge transferred to/from an open system is equal to the change in the system's charge. For example, each of the two materials (amber and cloth) discussed in Figure 18.8 have no net charge initially. The only way to change their charge is to transfer charge from outside each object. When they are rubbed together, negative charge is transferred to the amber and the final charge of the amber is the sum of the initial charge and the charge transferred to it. On the other hand, the final charge on the cloth is equal to its initial charge minus the charge transferred out. Similarly when glass is rubbed with silk, the net charge on the silk is its initial charge plus the incoming charge and the charge on the glass is the initial charge minus the outgoing charge. Also the charge gained by the silk will be equal to the charge lost by the glass, which means that if the silk gains \u20135e charge, the glass would have lost \u22125e charge. In more exotic situations, such as in particle accelerators, mass, \u0394, can be created from energy in the amount \u0394 = 2. Sometimes, the created mass is charged, such as when an electron is created. Whenever a charged particle is created, another having an opposite charge is always created along with it, so that the total charge created is zero. Usually, the two particles are \u201cmatter-antimatter\u201d counterparts. For example, an antielectron would usually be created at the same time as an electron. The antielectron has a positive charge (it is called a positron), and so the total charge created is zero. (See Figure 18.9.) All particles have antimatter counterparts with opposite signs. When matter and antimatter counterparts are brought together, they completely annihilate one another. By annihilate, we mean that the mass of the two particles is converted to energy E, again obeying the relationship \u0394 = 2 annihilation; thus, total charge is conserved.. Since the two particles have equal and opposite charge, the total charge is zero before and after the Making Connections: Conservation Laws Only a limited number of physical quantities are universally conserved. Charge is one\u2014energy, momentum, and angular momentum are others. Because they are conserved, these physical quantities are used to explain more phenomena and form more connections than other, less basic quantities. We find that conserved quantities give us great insight into the rules followed by nature and hints to the organization of nature. Discoveries of conservation laws have led to further discoveries, such as the weak nuclear force and the quark substructure of protons and other particles. This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 18 | Electric Charge and Electric Field 781 Figure 18.9 (a) When enough energy is present, it can be converted into matter. Here the matter created is an electron\u2013antielectron pair. ( is the electron's mass.) The total charge before and after this event is zero. (b) When matter and antimatter collide, they annihilate each other; the total charge is conserved at zero before and after the annihilation. The law of conservation of charge is absolute\u2014it has never been observed to be violated. Charge, then, is a special physical quantity, joining a very short list of other quantities in nature that are always conserved. Other conserved quantities include energy, momentum, and angular momentum. PhET Explorations: Balloons and Static Electricity Why does a balloon stick to your sweater? Rub a balloon on a sweater, then let go of the balloon and it flies over and sticks to the sweater. View the charges in the sweater, balloons, and the wall. Figure 18.10 Balloons and Static Electricity (http://cnx.org/content/m55300/1.2/balloons_en.jar) Applying the Science Practices: Electrical Charging Design an experiment to demonstrate the electrical charging of objects, by using a glass rod, a balloon, small bits of paper, and different pieces of cloth (like silk, wool, or nylon). Also show that like charges repel each other whereas unlike charges attract each other. 18.2 Conductors and Insulators Learning Objectives By the end of this section, you will be able to: \u2022 Define conductor and insulator, explain the difference, and give examples of each. \u2022 Describe three methods for charging an object. \u2022 Explain what happens to an electric force as you move farther from the", " source. \u2022 Define polarization. The information presented in this section supports the following AP\u00ae learning objectives and science practices: 782 Chapter 18 | Electric Charge and Electric Field \u2022 1.B.2.2 The student is able to make a qualitative prediction about the distribution of positive and negative electric charges within neutral systems as they undergo various processes. (S.P. 6.4, 7.2) \u2022 1.B.2.3 The student is able to challenge claims that polarization of electric charge or separation of charge must result in a net charge on the object. (S.P. 6.1) \u2022 4.E.3.1 The student is able to make predictions about the redistribution of charge during charging by friction, conduction, and induction. (S.P. 6.4) \u2022 4.E.3.2 The student is able to make predictions about the redistribution of charge caused by the electric field due to other systems, resulting in charged or polarized objects. (S.P. 6.4, 7.2) \u2022 4.E.3.3 The student is able to construct a representation of the distribution of fixed and mobile charge in insulators and conductors. (S.P. 1.1, 1.4, 6.4) \u2022 4.E.3.4 The student is able to construct a representation of the distribution of fixed and mobile charge in insulators and conductors that predicts charge distribution in processes involving induction or conduction. (S.P. 1.1, 1.4, 6.4) \u2022 4.E.3.5 The student is able to plan and/or analyze the results of experiments in which electric charge rearrangement occurs by electrostatic induction, or is able to refine a scientific question relating to such an experiment by identifying anomalies in a data set or procedure. (S.P. 3.2, 4.1, 4.2, 5.1, 5.3) Figure 18.11 This power adapter uses metal wires and connectors to conduct electricity from the wall socket to a laptop computer. The conducting wires allow electrons to move freely through the cables, which are shielded by rubber and plastic. These materials act as insulators that don't allow electric charge to escape outward. (credit: Evan-Amos, Wikimedia Commons) Some substances, such as metals and salty water, allow charges to move through them with relative ease. Some of the electrons in metals and similar conductors are not bound to individual atoms or sites in the material. These free electrons can move through the material much as air moves through loose sand. Any substance that has free electrons and allows charge to move relatively freely through it is called a conductor. The moving electrons may collide with fixed atoms and molecules, losing some energy, but they can move in a conductor. Superconductors allow the movement of charge without any loss of energy. Salty water and other similar conducting materials contain free ions that can move through them. An ion is an atom or molecule having a positive or negative (nonzero) total charge. In other words, the total number of electrons is not equal to the total number of protons. Other substances, such as glass, do not allow charges to move through them. These are called insulators. Electrons and ions in insulators are bound in the structure and cannot move easily\u2014as much as 1023 water and dry table salt are insulators, for example, whereas molten salt and salty water are conductors. times more slowly than in conductors. Pure Figure 18.12 An electroscope is a favorite instrument in physics demonstrations and student laboratories. It is typically made with gold foil leaves hung from a (conducting) metal stem and is insulated from the room air in a glass-walled container. (a) A positively charged glass rod is brought near the tip of the electroscope, attracting electrons to the top and leaving a net positive charge on the leaves. Like charges in the light flexible gold leaves repel, separating them. (b) When the rod is touched against the ball, electrons are attracted and transferred, reducing the net charge on the glass rod but leaving the electroscope positively charged. (c) The excess charges are evenly distributed in the stem and leaves of the electroscope once the glass rod is removed. Charging by Contact Figure 18.12 shows an electroscope being charged by touching it with a positively charged glass rod. Because the glass rod is an insulator, it must actually touch the electroscope to transfer charge to or from it. (Note that the extra positive charges reside on the surface of the glass rod as a result of rubbing it with silk before starting the experiment.) Since only electrons move in This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 18 | Electric Charge and Electric Field 783 metals, we see that they are attracted to the top of the electroscope. There, some are transferred to the positive rod by touch, leaving the electro", "scope with a net positive charge. Electrostatic repulsion in the leaves of the charged electroscope separates them. The electrostatic force has a horizontal component that results in the leaves moving apart as well as a vertical component that is balanced by the gravitational force. Similarly, the electroscope can be negatively charged by contact with a negatively charged object. Charging by Induction It is not necessary to transfer excess charge directly to an object in order to charge it. Figure 18.13 shows a method of induction wherein a charge is created in a nearby object, without direct contact. Here we see two neutral metal spheres in contact with one another but insulated from the rest of the world. A positively charged rod is brought near one of them, attracting negative charge to that side, leaving the other sphere positively charged. This is an example of induced polarization of neutral objects. Polarization is the separation of charges in an object that remains neutral. If the spheres are now separated (before the rod is pulled away), each sphere will have a net charge. Note that the object closest to the charged rod receives an opposite charge when charged by induction. Note also that no charge is removed from the charged rod, so that this process can be repeated without depleting the supply of excess charge. Another method of charging by induction is shown in Figure 18.14. The neutral metal sphere is polarized when a charged rod is brought near it. The sphere is then grounded, meaning that a conducting wire is run from the sphere to the ground. Since the earth is large and most ground is a good conductor, it can supply or accept excess charge easily. In this case, electrons are attracted to the sphere through a wire called the ground wire, because it supplies a conducting path to the ground. The ground connection is broken before the charged rod is removed, leaving the sphere with an excess charge opposite to that of the rod. Again, an opposite charge is achieved when charging by induction and the charged rod loses none of its excess charge. Figure 18.13 Charging by induction. (a) Two uncharged or neutral metal spheres are in contact with each other but insulated from the rest of the world. (b) A positively charged glass rod is brought near the sphere on the left, attracting negative charge and leaving the other sphere positively charged. (c) The spheres are separated before the rod is removed, thus separating negative and positive charge. (d) The spheres retain net charges after the inducing rod is removed\u2014without ever having been touched by a charged object. 784 Chapter 18 | Electric Charge and Electric Field Figure 18.14 Charging by induction, using a ground connection. (a) A positively charged rod is brought near a neutral metal sphere, polarizing it. (b) The sphere is grounded, allowing electrons to be attracted from the earth's ample supply. (c) The ground connection is broken. (d) The positive rod is removed, leaving the sphere with an induced negative charge. Figure 18.15 Both positive and negative objects attract a neutral object by polarizing its molecules. (a) A positive object brought near a neutral insulator polarizes its molecules. There is a slight shift in the distribution of the electrons orbiting the molecule, with unlike charges being brought nearer and like charges moved away. Since the electrostatic force decreases with distance, there is a net attraction. (b) A negative object produces the opposite polarization, but again attracts the neutral object. (c) The same effect occurs for a conductor; since the unlike charges are closer, there is a net attraction. Neutral objects can be attracted to any charged object. The pieces of straw attracted to polished amber are neutral, for example. If you run a plastic comb through your hair, the charged comb can pick up neutral pieces of paper. Figure 18.15 shows how the polarization of atoms and molecules in neutral objects results in their attraction to a charged object. When a charged rod is brought near a neutral substance, an insulator in this case, the distribution of charge in atoms and molecules is shifted slightly. Opposite charge is attracted nearer the external charged rod, while like charge is repelled. Since the electrostatic force decreases with distance, the repulsion of like charges is weaker than the attraction of unlike charges, and so there is a net attraction. Thus a positively charged glass rod attracts neutral pieces of paper, as will a negatively charged rubber rod. Some molecules, like water, are polar molecules. Polar molecules have a natural or inherent separation of charge, although they are neutral overall. Polar molecules are particularly affected by other charged objects and show greater polarization effects than molecules with naturally uniform charge distributions. Check Your Understanding Can you explain the attraction of water to the charged rod in the figure below? This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 18 | Electric Charge and Electric Field 785 Figure 18.16 Solution Water molecules are polarized, giving them slightly positive and slightly negative sides. This makes water even more susceptible to a charged rod's attraction. As the", " water flows downward, due to the force of gravity, the charged conductor exerts a net attraction to the opposite charges in the stream of water, pulling it closer. Applying the Science Practices: Electrostatic Induction Plan an experiment to demonstrate electrostatic induction using household items, like balloons, woolen cloth, aluminum drink cans, or foam cups. Explain the process of induction in your experiment by discussing details of (and making diagrams relating to) the movement and alignment of charges. PhET Explorations: John Travoltage Make sparks fly with John Travoltage. Wiggle Johnnie's foot and he picks up charges from the carpet. Bring his hand close to the door knob and get rid of the excess charge. Figure 18.17 John Travoltage (http://cnx.org/content/m55301/1.2/travoltage_en.jar) 18.3 Conductors and Electric Fields in Static Equilibrium Learning Objectives By the end of this section, you will be able to: \u2022 List the three properties of a conductor in electrostatic equilibrium. \u2022 Explain the effect of an electric field on free charges in a conductor. \u2022 Explain why no electric field may exist inside a conductor. \u2022 Describe the electric field surrounding Earth. \u2022 Explain what happens to an electric field applied to an irregular conductor. \u2022 Describe how a lightning rod works. \u2022 Explain how a metal car may protect passengers inside from the dangerous electric fields caused by a downed line touching the car. The information presented in this section supports the following AP learning objectives: \u2022 2.C.3.1 The student is able to explain the inverse square dependence of the electric field surrounding a spherically symmetric electrically charged object. 786 Chapter 18 | Electric Charge and Electric Field \u2022 2.C.5.1 The student is able to create representations of the magnitude and direction of the electric field at various distances (small compared to plate size) from two electrically charged plates of equal magnitude and opposite signs and is able to recognize that the assumption of uniform field is not appropriate near edges of plates. Conductors contain free charges that move easily. When excess charge is placed on a conductor or the conductor is put into a static electric field, charges in the conductor quickly respond to reach a steady state called electrostatic equilibrium. Figure 18.18 shows the effect of an electric field on free charges in a conductor. The free charges move until the field is perpendicular to the conductor's surface. There can be no component of the field parallel to the surface in electrostatic equilibrium, since, if there were, it would produce further movement of charge. A positive free charge is shown, but free charges can be either positive or negative and are, in fact, negative in metals. The motion of a positive charge is equivalent to the motion of a negative charge in the opposite direction. Figure 18.18 When an electric field E is applied to a conductor, free charges inside the conductor move until the field is perpendicular to the surface. (a) The electric field is a vector quantity, with both parallel and perpendicular components. The parallel component ( E\u2225 ) exerts a force ( F\u2225 ) on the free charge, which moves the charge until F\u2225 = 0. (b) The resulting field is perpendicular to the surface. The free charge has been brought to the conductor's surface, leaving electrostatic forces in equilibrium. A conductor placed in an electric field will be polarized. Figure 18.19 shows the result of placing a neutral conductor in an originally uniform electric field. The field becomes stronger near the conductor but entirely disappears inside it. Figure 18.19 This illustration shows a spherical conductor in static equilibrium with an originally uniform electric field. Free charges move within the conductor, polarizing it, until the electric field lines are perpendicular to the surface. The field lines end on excess negative charge on one section of the surface and begin again on excess positive charge on the opposite side. No electric field exists inside the conductor, since free charges in the conductor would continue moving in response to any field until it was neutralized. Misconception Alert: Electric Field inside a Conductor Excess charges placed on a spherical conductor repel and move until they are evenly distributed, as shown in Figure 18.20. Excess charge is forced to the surface until the field inside the conductor is zero. Outside the conductor, the field is exactly the same as if the conductor were replaced by a point charge at its center equal to the excess charge. This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 18 | Electric Charge and Electric Field 787 Figure 18.20 The mutual repulsion of excess positive charges on a spherical conductor distributes them uniformly on its surface. The resulting electric field is perpendicular to the surface and zero inside. Outside the conductor, the field is identical to that of a point charge at the center equal to the excess charge. Properties of a Conductor in Electrostatic Equilibrium 1.", " The electric field is zero inside a conductor. 2. Just outside a conductor, the electric field lines are perpendicular to its surface, ending or beginning on charges on the surface. 3. Any excess charge resides entirely on the surface or surfaces of a conductor. The properties of a conductor are consistent with the situations already discussed and can be used to analyze any conductor in electrostatic equilibrium. This can lead to some interesting new insights, such as described below. How can a very uniform electric field be created? Consider a system of two metal plates with opposite charges on them, as shown in Figure 18.21. The properties of conductors in electrostatic equilibrium indicate that the electric field between the plates will be uniform in strength and direction. Except near the edges, the excess charges distribute themselves uniformly, producing field lines that are uniformly spaced (hence uniform in strength) and perpendicular to the surfaces (hence uniform in direction, since the plates are flat). The edge effects are less important when the plates are close together. Figure 18.21 Two metal plates with equal, but opposite, excess charges. The field between them is uniform in strength and direction except near the edges. One use of such a field is to produce uniform acceleration of charges between the plates, such as in the electron gun of a TV tube. Earth's Electric Field A near uniform electric field of approximately 150 N/C, directed downward, surrounds Earth, with the magnitude increasing slightly as we get closer to the surface. What causes the electric field? At around 100 km above the surface of Earth we have a layer of charged particles, called the ionosphere. The ionosphere is responsible for a range of phenomena including the electric field surrounding Earth. In fair weather the ionosphere is positive and the Earth largely negative, maintaining the electric field (Figure 18.22(a)). In storm conditions clouds form and localized electric fields can be larger and reversed in direction (Figure 18.22(b)). The exact charge distributions depend on the local conditions, and variations of Figure 18.22(b) are possible. If the electric field is sufficiently large, the insulating properties of the surrounding material break down and it becomes conducting. For air this occurs at around 3\u00d7106 form of lightning sparks and corona discharge. N/C. Air ionizes ions and electrons recombine, and we get discharge in the 788 Chapter 18 | Electric Charge and Electric Field Figure 18.22 Earth's electric field. (a) Fair weather field. Earth and the ionosphere (a layer of charged particles) are both conductors. They produce a uniform electric field of about 150 N/C. (credit: D. H. Parks) (b) Storm fields. In the presence of storm clouds, the local electric fields can be larger. At very high fields, the insulating properties of the air break down and lightning can occur. (credit: Jan-Joost Verhoef) Electric Fields on Uneven Surfaces So far we have considered excess charges on a smooth, symmetrical conductor surface. What happens if a conductor has sharp corners or is pointed? Excess charges on a nonuniform conductor become concentrated at the sharpest points. Additionally, excess charge may move on or off the conductor at the sharpest points. To see how and why this happens, consider the charged conductor in Figure 18.23. The electrostatic repulsion of like charges is most effective in moving them apart on the flattest surface, and so they become least concentrated there. This is because the forces between identical pairs of charges at either end of the conductor are identical, but the components of the forces parallel to the surfaces are different. The component parallel to the surface is greatest on the flattest surface and, hence, more effective in moving the charge. The same effect is produced on a conductor by an externally applied electric field, as seen in Figure 18.23 (c). Since the field lines must be perpendicular to the surface, more of them are concentrated on the most curved parts. Figure 18.23 Excess charge on a nonuniform conductor becomes most concentrated at the location of greatest curvature. (a) The forces between identical pairs of charges at either end of the conductor are identical, but the components of the forces parallel to the surface are different. It is F\u2225 that moves the charges apart once they have reached the surface. (b) F\u2225 producing the electric field shown. (c) An uncharged conductor in an originally uniform electric field is polarized, with the most concentrated charge at its most pointed end. is smallest at the more pointed end, the charges are left closer together, Applications of Conductors On a very sharply curved surface, such as shown in Figure 18.24, the charges are so concentrated at the point that the resulting electric field can be great enough to remove them from the surface. This can be useful. Lightning rods work best when they are most pointed. The large charges created in storm clouds induce an opposite charge on a building that can result in a lightning bolt hitting the building. The", " induced charge is bled away continually by a lightning rod, preventing the more dramatic lightning strike. Of course, we sometimes wish to prevent the transfer of charge rather than to facilitate it. In that case, the conductor should be very smooth and have as large a radius of curvature as possible. (See Figure 18.25.) Smooth surfaces are used on high-voltage transmission lines, for example, to avoid leakage of charge into the air. Another device that makes use of some of these principles is a Faraday cage. This is a metal shield that encloses a volume. All electrical charges will reside on the outside surface of this shield, and there will be no electrical field inside. A Faraday cage is used to prohibit stray electrical fields in the environment from interfering with sensitive measurements, such as the electrical signals inside a nerve cell. During electrical storms if you are driving a car, it is best to stay inside the car as its metal body acts as a Faraday cage with zero electrical field inside. If in the vicinity of a lightning strike, its effect is felt on the outside of the car and the inside is unaffected, provided you remain totally inside. This is also true if an active (\u201chot\u201d) electrical wire was broken (in a storm or an accident) and fell on your car. This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 18 | Electric Charge and Electric Field 789 Figure 18.24 A very pointed conductor has a large charge concentration at the point. The electric field is very strong at the point and can exert a force large enough to transfer charge on or off the conductor. Lightning rods are used to prevent the buildup of large excess charges on structures and, thus, are pointed. Figure 18.25 (a) A lightning rod is pointed to facilitate the transfer of charge. (credit: Romaine, Wikimedia Commons) (b) This Van de Graaff generator has a smooth surface with a large radius of curvature to prevent the transfer of charge and allow a large voltage to be generated. The mutual repulsion of like charges is evident in the person's hair while touching the metal sphere. (credit: Jon \u2018ShakataGaNai' Davis/Wikimedia Commons). 18.4 Coulomb\u2019s Law By the end of this section, you will be able to: Learning Objectives \u2022 State Coulomb's law in terms of how the electrostatic force changes with the distance between two objects. \u2022 Calculate the electrostatic force between two point charges, such as electrons or protons. \u2022 Compare the electrostatic force to the gravitational attraction for a proton and an electron; for a human and the Earth. The information presented in this section supports the following AP\u00ae learning objectives and science practices: \u2022 3.A.3.3 The student is able to describe a force as an interaction between two objects and identify both objects for any force. (S.P. 1.4) \u2022 3.A.3.4 The student is able to make claims about the force on an object due to the presence of other objects with the same property: mass, electric charge. (S.P. 6.1, 6.4) \u2022 3.C.2.1 The student is able to use Coulomb's law qualitatively and quantitatively to make predictions about the interaction between two electric point charges (interactions between collections of electric point charges are not covered in Physics 1 and instead are restricted to Physics 2). (S.P. 2.2, 6.4) \u2022 3.C.2.2 The student is able to connect the concepts of gravitational force and electric force to compare similarities and differences between the forces. (S.P. 7.2) 790 Chapter 18 | Electric Charge and Electric Field Figure 18.26 This NASA image of Arp 87 shows the result of a strong gravitational attraction between two galaxies. In contrast, at the subatomic level, the electrostatic attraction between two objects, such as an electron and a proton, is far greater than their mutual attraction due to gravity. (credit: NASA/HST) Through the work of scientists in the late 18th century, the main features of the electrostatic force\u2014the existence of two types of charge, the observation that like charges repel, unlike charges attract, and the decrease of force with distance\u2014were eventually refined, and expressed as a mathematical formula. The mathematical formula for the electrostatic force is called Coulomb's law after the French physicist Charles Coulomb (1736\u20131806), who performed experiments and first proposed a formula to calculate it. Coulomb's Law = |1 2| 2 Coulomb's law calculates the magnitude of the force between two point charges, 1 and 2, separated by a distance. In SI units, the constant is equal to. (18.3) = 8.988\u00d7109N \u22c5 m2 C2 \u2248 8.99\u00d7109N \ufffd", "\ufffd\ufffd m2 C2. (18.4) The electrostatic force is a vector quantity and is expressed in units of newtons. The force is understood to be along the line joining the two charges. (See Figure 18.27.) Although the formula for Coulomb's law is simple, it was no mean task to prove it. The experiments Coulomb did, with the primitive equipment then available, were difficult. Modern experiments have verified Coulomb's law to great precision. For \u221d 1 / 2 example, it has been shown that the force is inversely proportional to distance between two objects squared to an accuracy of 1 part in 1016. No exceptions have ever been found, even at the small distances within the atom. Figure 18.27 The magnitude of the electrostatic force between point charges 1 and 2 separated by a distance is given by Coulomb's law. Note that Newton's third law (every force exerted creates an equal and opposite force) applies as usual\u2014the force on 1 is equal in magnitude and opposite in direction to the force it exerts on 2. (a) Like charges. (b) Unlike charges. Making Connections: Comparing Gravitational and Electrostatic Forces Recall that the gravitational force (Newton's law of gravitation) quantifies force as = 2. The comparison between the two forces\u2014gravitational and electrostatic\u2014shows some similarities and differences. Gravitational force is proportional to the masses of interacting objects, and the electrostatic force is proportional to the magnitudes of the charges of interacting objects. Hence both forces are proportional to a property that represents the strength of interaction for a given field. In addition, both forces are inversely proportional to the square of the distances between them. It may seem that the two forces are related but that is not the case. In fact, there are huge variations in the magnitudes of the two forces as they depend on different parameters and different mechanisms. For electrons (or protons), electrostatic force is dominant and is much greater than the gravitational force. On the other hand, gravitational force is This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 18 | Electric Charge and Electric Field 791 generally dominant for objects with large masses. Another major difference between the two forces is that gravitational force can only be attractive, whereas electrostatic could be attractive or repulsive (depending on the sign of charges; unlike charges attract and like charges repel). Example 18.1 How Strong is the Coulomb Force Relative to the Gravitational Force? Compare the electrostatic force between an electron and proton separated by 0.530\u00d710\u221210 m with the gravitational force between them. This distance is their average separation in a hydrogen atom. Strategy To compare the two forces, we first compute the electrostatic force using Coulomb's law, = |1 2| 2. We then calculate the gravitational force using Newton's universal law of gravitation. Finally, we take a ratio to see how the forces compare in magnitude. Solution Entering the given and known information about the charges and separation of the electron and proton into the expression of Coulomb's law yields = |1 2| 2 = 8.99\u00d7109 N \u22c5 m2 / C2 \u00d7 (1.60\u00d710\u201319 C)(1.60\u00d710\u201319 C) (0.530\u00d710\u201310 m)2 (18.5) (18.6) Thus the Coulomb force is = 8.19\u00d710\u20138 N. The charges are opposite in sign, so this is an attractive force. This is a very large force for an electron\u2014it would cause an acceleration of 8.99\u00d71022 m / s2 (verification is left as an end-of-section problem).The gravitational force is given by Newton's law of gravitation as: (18.7) = 2 where = 6.67\u00d710\u221211 N \u22c5 m2 / kg2. Here and represent the electron and proton masses, which can be found in the appendices. Entering values for the knowns yields, (18.8) = (6.67\u00d710 \u2013 11 N \u22c5 m2 / kg2)\u00d7 (9.11\u00d710\u201331 kg)(1.67\u00d710\u201327 kg) (0.530\u00d710\u201310 m)2 = 3.61\u00d710\u201347 N (18.9) This is also an attractive force, although it is traditionally shown as positive since gravitational force is always attractive. The ratio of the magnitude of the electrostatic force to gravitational force in this case is, thus, = 2.27\u00d71039. (18.10) Discussion This is a remarkably large ratio! Note that this will be the ratio of electrostatic force to gravitational force for an electron and a proton at any distance (taking the ratio before entering numerical values shows that the distance cancels). This ratio gives", " some indication of just how much larger the Coulomb force is than the gravitational force between two of the most common particles in nature. As the example implies, gravitational force is completely negligible on a small scale, where the interactions of individual charged particles are important. On a large scale, such as between the Earth and a person, the reverse is true. Most objects are nearly electrically neutral, and so attractive and repulsive Coulomb forces nearly cancel. Gravitational force on a large scale dominates interactions between large objects because it is always attractive, while Coulomb forces tend to cancel. Chapter 18 | Electric Charge and Electric Field 801 The Van de Graaff Generator Van de Graaff generators (or Van de Graaffs) are not only spectacular devices used to demonstrate high voltage due to static electricity\u2014they are also used for serious research. The first was built by Robert Van de Graaff in 1931 (based on original suggestions by Lord Kelvin) for use in nuclear physics research. Figure 18.38 shows a schematic of a large research version. Van de Graaffs utilize both smooth and pointed surfaces, and conductors and insulators to generate large static charges and, hence, large voltages. A very large excess charge can be deposited on the sphere, because it moves quickly to the outer surface. Practical limits arise because the large electric fields polarize and eventually ionize surrounding materials, creating free charges that neutralize excess charge or allow it to escape. Nevertheless, voltages of 15 million volts are well within practical limits. Figure 18.38 Schematic of Van de Graaff generator. A battery (A) supplies excess positive charge to a pointed conductor, the points of which spray the charge onto a moving insulating belt near the bottom. The pointed conductor (B) on top in the large sphere picks up the charge. (The induced electric field at the points is so large that it removes the charge from the belt.) This can be done because the charge does not remain inside the conducting sphere but moves to its outside surface. An ion source inside the sphere produces positive ions, which are accelerated away from the positive sphere to high velocities. Take-Home Experiment: Electrostatics and Humidity Rub a comb through your hair and use it to lift pieces of paper. It may help to tear the pieces of paper rather than cut them neatly. Repeat the exercise in your bathroom after you have had a long shower and the air in the bathroom is moist. Is it easier to get electrostatic effects in dry or moist air? Why would torn paper be more attractive to the comb than cut paper? Explain your observations. Xerography Most copy machines use an electrostatic process called xerography\u2014a word coined from the Greek words xeros for dry and graphos for writing. The heart of the process is shown in simplified form in Figure 18.39. A selenium-coated aluminum drum is sprayed with positive charge from points on a device called a corotron. Selenium is a substance with an interesting property\u2014it is a photoconductor. That is, selenium is an insulator when in the dark and a conductor when exposed to light. In the first stage of the xerography process, the conducting aluminum drum is grounded so that a negative charge is induced under the thin layer of uniformly positively charged selenium. In the second stage, the surface of the drum is exposed to the image of whatever is to be copied. Where the image is light, the selenium becomes conducting, and the positive charge is neutralized. In dark areas, the positive charge remains, and so the image has been transferred to the drum. The third stage takes a dry black powder, called toner, and sprays it with a negative charge so that it will be attracted to the positive regions of the drum. Next, a blank piece of paper is given a greater positive charge than on the drum so that it will pull Chapter 18 | Electric Charge and Electric Field 805 electric field is given to be upward, the electric force is upward. We thus have a one-dimensional (vertical direction) problem, and we can state Newton's second law as where net = \u2212. Entering this and the known values into the expression for Newton's second law yields = net. = \u2212 = 9.60\u00d710\u221214 N \u2212 3.92\u00d710\u221214 N 4.00\u00d710\u221215 kg = 14.2 m/s2. (18.24) (18.25) Discussion for (c) This is an upward acceleration great enough to carry the drop to places where you might not wish to have gasoline. This worked example illustrates how to apply problem-solving strategies to situations that include topics in different chapters. The first step is to identify the physical principles involved in the problem. The second step is to solve for the unknown using familiar problem-solving strategies. These are found throughout the text, and many worked examples show how to use them for single topics. In this integrated concepts example, you can", " see how to apply them across several topics. You will find these techniques useful in applications of physics outside a physics course, such as in your profession, in other science disciplines, and in everyday life. The following problems will build your skills in the broad application of physical principles. Unreasonable Results The Unreasonable Results exercises for this module have results that are unreasonable because some premise is unreasonable or because certain of the premises are inconsistent with one another. Physical principles applied correctly then produce unreasonable results. The purpose of these problems is to give practice in assessing whether nature is being accurately described, and if it is not to trace the source of difficulty. Problem-Solving Strategy To determine if an answer is reasonable, and to determine the cause if it is not, do the following. 1. Solve the problem using strategies as outlined above. Use the format followed in the worked examples in the text to solve the problem as usual. 2. Check to see if the answer is reasonable. Is it too large or too small, or does it have the wrong sign, improper units, and so on? 3. If the answer is unreasonable, look for what specifically could cause the identified difficulty. Usually, the manner in which the answer is unreasonable is an indication of the difficulty. For example, an extremely large Coulomb force could be due to the assumption of an excessively large separated charge. Glossary conductor: a material that allows electrons to move separately from their atomic orbits conductor: an object with properties that allow charges to move about freely within it Coulomb force: another term for the electrostatic force Coulomb interaction: the interaction between two charged particles generated by the Coulomb forces they exert on one another Coulomb's law: the mathematical equation calculating the electrostatic force vector between two charged particles dipole: a molecule's lack of symmetrical charge distribution, causing one side to be more positive and another to be more negative electric charge: a physical property of an object that causes it to be attracted toward or repelled from another charged object; each charged object generates and is influenced by a force called an electromagnetic force electric field: a three-dimensional map of the electric force extended out into space from a point charge electric field lines: a series of lines drawn from a point charge representing the magnitude and direction of force exerted by that charge 806 Chapter 18 | Electric Charge and Electric Field electromagnetic force: one of the four fundamental forces of nature; the electromagnetic force consists of static electricity, moving electricity and magnetism electron: a particle orbiting the nucleus of an atom and carrying the smallest unit of negative charge electrostatic equilibrium: an electrostatically balanced state in which all free electrical charges have stopped moving about electrostatic force: the amount and direction of attraction or repulsion between two charged bodies electrostatic precipitators: filters that apply charges to particles in the air, then attract those charges to a filter, removing them from the airstream electrostatic repulsion: the phenomenon of two objects with like charges repelling each other electrostatics: the study of electric forces that are static or slow-moving Faraday cage: a metal shield which prevents electric charge from penetrating its surface field: a map of the amount and direction of a force acting on other objects, extending out into space free charge: an electrical charge (either positive or negative) which can move about separately from its base molecule free electron: an electron that is free to move away from its atomic orbit grounded: when a conductor is connected to the Earth, allowing charge to freely flow to and from Earth's unlimited reservoir grounded: connected to the ground with a conductor, so that charge flows freely to and from the Earth to the grounded object induction: the process by which an electrically charged object brought near a neutral object creates a charge in that object ink-jet printer: small ink droplets sprayed with an electric charge are controlled by electrostatic plates to create images on paper insulator: a material that holds electrons securely within their atomic orbits ionosphere: a layer of charged particles located around 100 km above the surface of Earth, which is responsible for a range of phenomena including the electric field surrounding Earth laser printer: uses a laser to create a photoconductive image on a drum, which attracts dry ink particles that are then rolled onto a sheet of paper to print a high-quality copy of the image law of conservation of charge: is created simultaneously states that whenever a charge is created, an equal amount of charge with the opposite sign photoconductor: a substance that is an insulator until it is exposed to light, when it becomes a conductor point charge: A charged particle, designated, generating an electric field polar molecule: a molecule with an asymmetrical distribution of positive and negative charge polarization: slight shifting of positive and negative charges to opposite sides of an atom or molecule polarized: a state in which the positive and negative charges within an object have collected in separate locations proton: a particle in the nucleus of an atom and carrying a positive charge equal in magnitude and opposite in sign to the amount of negative charge carried by an electron screening: the dilution or blocking of", " an electrostatic force on a charged object by the presence of other charges nearby static electricity: a buildup of electric charge on the surface of an object test charge: A particle (designated ) with either a positive or negative charge set down within an electric field generated by a point charge Van de Graaff generator: a machine that produces a large amount of excess charge, used for experiments with high voltage vector: a quantity with both magnitude and direction vector addition: mathematical combination of two or more vectors, including their magnitudes, directions, and positions xerography: a dry copying process based on electrostatics This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 18 | Electric Charge and Electric Field 807 Section Summary 18.1 Static Electricity and Charge: Conservation of Charge \u2022 There are only two types of charge, which we call positive and negative. \u2022 Like charges repel, unlike charges attract, and the force between charges decreases with the square of the distance. \u2022 The vast majority of positive charge in nature is carried by protons, while the vast majority of negative charge is carried by electrons. \u2022 The electric charge of one electron is equal in magnitude and opposite in sign to the charge of one proton. \u2022 An ion is an atom or molecule that has nonzero total charge due to having unequal numbers of electrons and protons. \u2022 The SI unit for charge is the coulomb (C), with protons and electrons having charges of opposite sign but equal magnitude; the magnitude of this basic charge \u2223 \u2223 is \u2022 Whenever charge is created or destroyed, equal amounts of positive and negative are involved. \u2022 Most often, existing charges are separated from neutral objects to obtain some net charge. \u2022 Both positive and negative charges exist in neutral objects and can be separated by rubbing one object with another. For macroscopic objects, negatively charged means an excess of electrons and positively charged means a depletion of electrons. \u2223 \u2223 = 1.60\u00d710\u221219 C. \u2022 The law of conservation of charge ensures that whenever a charge is created, an equal charge of the opposite sign is created at the same time. 18.2 Conductors and Insulators \u2022 Polarization is the separation of positive and negative charges in a neutral object. \u2022 A conductor is a substance that allows charge to flow freely through its atomic structure. \u2022 An insulator holds charge within its atomic structure. \u2022 Objects with like charges repel each other, while those with unlike charges attract each other. \u2022 A conducting object is said to be grounded if it is connected to the Earth through a conductor. Grounding allows transfer of charge to and from the earth's large reservoir. \u2022 Objects can be charged by contact with another charged object and obtain the same sign charge. \u2022 \u2022 Polarized objects have their positive and negative charges concentrated in different areas, giving them a non-symmetrical If an object is temporarily grounded, it can be charged by induction, and obtains the opposite sign charge. charge. \u2022 Polar molecules have an inherent separation of charge. 18.3 Conductors and Electric Fields in Static Equilibrium \u2022 A conductor allows free charges to move about within it. \u2022 The electrical forces around a conductor will cause free charges to move around inside the conductor until static equilibrium is reached. \u2022 Any excess charge will collect along the surface of a conductor. \u2022 Conductors with sharp corners or points will collect more charge at those points. \u2022 A lightning rod is a conductor with sharply pointed ends that collect excess charge on the building caused by an electrical storm and allow it to dissipate back into the air. \u2022 Electrical storms result when the electrical field of Earth's surface in certain locations becomes more strongly charged, due to changes in the insulating effect of the air. \u2022 A Faraday cage acts like a shield around an object, preventing electric charge from penetrating inside. 18.4 Coulomb\u2019s Law \u2022 Frenchman Charles Coulomb was the first to publish the mathematical equation that describes the electrostatic force between two objects. \u2022 Coulomb's law gives the magnitude of the force between point charges. It is = |1 2| 2, where 1 and 2 are two point charges separated by a distance, and \u2248 8.99\u00d7109 N \u00b7 m2/ C2 \u2022 This Coulomb force is extremely basic, since most charges are due to point-like particles. It is responsible for all electrostatic effects and underlies most macroscopic forces. \u2022 The Coulomb force is extraordinarily strong compared with the gravitational force, another basic force\u2014but unlike gravitational force it can cancel, since it can be either attractive or repulsive. \u2022 The electrostatic force between two subatomic particles is far greater than the gravitational force between the same two particles. 18.5 Electric Field: Concept of a Field Revisited \u2022 The electrostatic force field surrounding a charged object extends out into space in all directions. 808 Chapter 18 | Electric Charge and Electric Field \u2022 The electrostatic force exerted by a point charge on a test charge at a distance depends on the", " charge of both charges, as well as the distance between the two. \u2022 The electric field E is defined to be E = F where F is the Coulomb or electrostatic force exerted on a small positive test charge. E has units of N/C. \u2022 The magnitude of the electric field E created by a point charge is E = || 2. where is the distance from. The electric field E is a vector and fields due to multiple charges add like vectors. 18.6 Electric Field Lines: Multiple Charges \u2022 Drawings of electric field lines are useful visual tools. The properties of electric field lines for any charge distribution are that: \u2022 Field lines must begin on positive charges and terminate on negative charges, or at infinity in the hypothetical case of isolated charges. \u2022 The number of field lines leaving a positive charge or entering a negative charge is proportional to the magnitude of the charge. \u2022 The strength of the field is proportional to the closeness of the field lines\u2014more precisely, it is proportional to the number of lines per unit area perpendicular to the lines. \u2022 The direction of the electric field is tangent to the field line at any point in space. \u2022 Field lines can never cross. 18.7 Electric Forces in Biology \u2022 Many molecules in living organisms, such as DNA, carry a charge. \u2022 An uneven distribution of the positive and negative charges within a polar molecule produces a dipole. \u2022 The effect of a Coulomb field generated by a charged object may be reduced or blocked by other nearby charged objects. \u2022 Biological systems contain water, and because water molecules are polar, they have a strong effect on other molecules in living systems. 18.8 Applications of Electrostatics \u2022 Electrostatics is the study of electric fields in static equilibrium. \u2022 In addition to research using equipment such as a Van de Graaff generator, many practical applications of electrostatics exist, including photocopiers, laser printers, ink-jet printers and electrostatic air filters. Conceptual Questions 18.1 Static Electricity and Charge: Conservation of Charge 1. There are very large numbers of charged particles in most objects. Why, then, don't most objects exhibit static electricity? 2. Why do most objects tend to contain nearly equal numbers of positive and negative charges? 18.2 Conductors and Insulators 3. An eccentric inventor attempts to levitate by first placing a large negative charge on himself and then putting a large positive charge on the ceiling of his workshop. Instead, while attempting to place a large negative charge on himself, his clothes fly off. Explain. 4. If you have charged an electroscope by contact with a positively charged object, describe how you could use it to determine the charge of other objects. Specifically, what would the leaves of the electroscope do if other charged objects were brought near its knob? 5. When a glass rod is rubbed with silk, it becomes positive and the silk becomes negative\u2014yet both attract dust. Does the dust have a third type of charge that is attracted to both positive and negative? Explain. 6. Why does a car always attract dust right after it is polished? (Note that car wax and car tires are insulators.) 7. Describe how a positively charged object can be used to give another object a negative charge. What is the name of this process? 8. What is grounding? What effect does it have on a charged conductor? On a charged insulator? 18.3 Conductors and Electric Fields in Static Equilibrium 9. Is the object in a conductor or an insulator? Justify your answer. This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 18 | Electric Charge and Electric Field 809 Figure 18.43 10. If the electric field lines in the figure above were perpendicular to the object, would it necessarily be a conductor? Explain. 11. The discussion of the electric field between two parallel conducting plates, in this module states that edge effects are less important if the plates are close together. What does close mean? That is, is the actual plate separation crucial, or is the ratio of plate separation to plate area crucial? 12. Would the self-created electric field at the end of a pointed conductor, such as a lightning rod, remove positive or negative charge from the conductor? Would the same sign charge be removed from a neutral pointed conductor by the application of a similar externally created electric field? (The answers to both questions have implications for charge transfer utilizing points.) 13. Why is a golfer with a metal club over her shoulder vulnerable to lightning in an open fairway? Would she be any safer under a tree? 14. Can the belt of a Van de Graaff accelerator be a conductor? Explain. 15. Are you relatively safe from lightning inside an automobile? Give two reasons. 16. Discuss pros and cons of a lightning rod being grounded versus simply being attached to a building. 17. Using the symmetry of the arrangement, show that the net Coulomb force on the charge at the center of the square below (Figure 18.", "44) is zero if the charges on the four corners are exactly equal. Figure 18.44 Four point charges,,, and lie on the corners of a square and is located at its center. 18. (a) Using the symmetry of the arrangement, show that the electric field at the center of the square in Figure 18.44 is zero if the charges on the four corners are exactly equal. (b) Show that this is also true for any combination of charges in which = and = 19. (a) What is the direction of the total Coulomb force on in Figure 18.44 if is negative, = and both are negative, and = and both are positive? (b) What is the direction of the electric field at the center of the square in this situation? 20. Considering Figure 18.44, suppose that = and =. First show that is in static equilibrium. (You may neglect the gravitational force.) Then discuss whether the equilibrium is stable or unstable, noting that this may depend on the signs of the charges and the direction of displacement of from the center of the square. 21. If = 0 in Figure 18.44, under what conditions will there be no net Coulomb force on? 22. In regions of low humidity, one develops a special \u201cgrip\u201d when opening car doors, or touching metal door knobs. This involves placing as much of the hand on the device as possible, not just the ends of one's fingers. Discuss the induced charge and explain why this is done. 23. Tollbooth stations on roadways and bridges usually have a piece of wire stuck in the pavement before them that will touch a car as it approaches. Why is this done? 810 Chapter 18 | Electric Charge and Electric Field 24. Suppose a woman carries an excess charge. To maintain her charged status can she be standing on ground wearing just any pair of shoes? How would you discharge her? What are the consequences if she simply walks away? 18.4 Coulomb\u2019s Law 25. Figure 18.45 shows the charge distribution in a water molecule, which is called a polar molecule because it has an inherent separation of charge. Given water's polar character, explain what effect humidity has on removing excess charge from objects. Figure 18.45 Schematic representation of the outer electron cloud of a neutral water molecule. The electrons spend more time near the oxygen than the hydrogens, giving a permanent charge separation as shown. Water is thus a polar molecule. It is more easily affected by electrostatic forces than molecules with uniform charge distributions. 26. Using Figure 18.45, explain, in terms of Coulomb's law, why a polar molecule (such as in Figure 18.45) is attracted by both positive and negative charges. 27. Given the polar character of water molecules, explain how ions in the air form nucleation centers for rain droplets. 18.5 Electric Field: Concept of a Field Revisited 28. Why must the test charge in the definition of the electric field be vanishingly small? 29. Are the direction and magnitude of the Coulomb force unique at a given point in space? What about the electric field? 18.6 Electric Field Lines: Multiple Charges 30. Compare and contrast the Coulomb force field and the electric field. To do this, make a list of five properties for the Coulomb force field analogous to the five properties listed for electric field lines. Compare each item in your list of Coulomb force field properties with those of the electric field\u2014are they the same or different? (For example, electric field lines cannot cross. Is the same true for Coulomb field lines?) 31. Figure 18.46 shows an electric field extending over three regions, labeled I, II, and III. Answer the following questions. (a) Are there any isolated charges? If so, in what region and what are their signs? (b) Where is the field strongest? (c) Where is it weakest? (d) Where is the field the most uniform? Figure 18.46 18.7 Electric Forces in Biology This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 18 | Electric Charge and Electric Field 811 32. A cell membrane is a thin layer enveloping a cell. The thickness of the membrane is much less than the size of the cell. In a static situation the membrane has a charge distribution of \u22122.5\u00d710\u22126 C/m2 on its outer surface. Draw a diagram of the cell and the surrounding cell membrane. Include on this diagram the charge distribution and the corresponding electric field. Is there any electric field inside the cell? Is there any electric field outside the cell? C/m 2 on its inner surface and +2.5\u00d710\u22126 812 Chapter 18 | Electric Charge and Electric Field Problems & Exercises 18.1 Static Electricity and Charge: Conservation of Charge 1. Common static electricity involves charges ranging from nanocoulombs to microcoulombs.", " (a) How many electrons are needed to form a charge of \u20132.00 nC (b) How many electrons must be removed from a neutral object to leave a net charge of 0.500 C? electrons move through a pocket calculator 2. If 1.80\u00d71020 during a full day's operation, how many coulombs of charge moved through it? 3. To start a car engine, the car battery moves 3.75\u00d71021 electrons through the starter motor. How many coulombs of charge were moved? 4. A certain lightning bolt moves 40.0 C of charge. How many fundamental units of charge \u2223 \u2223 is this? 18.2 Conductors and Insulators 5. Suppose a speck of dust in an electrostatic precipitator has 1.0000\u00d71012 protons in it and has a net charge of \u20135.00 nC (a very large charge for a small speck). How many electrons does it have? 6. An amoeba has 1.00\u00d71016 0.300 pC. (a) How many fewer electrons are there than protons? (b) If you paired them up, what fraction of the protons would have no electrons? 7. A 50.0 g ball of copper has a net charge of 2.00 C. What fraction of the copper's electrons has been removed? (Each copper atom has 29 protons, and copper has an atomic mass of 63.5.) protons and a net charge of 8. What net charge would you place on a 100 g piece of sulfur if you put an extra electron on 1 in 1012 of its atoms? (Sulfur has an atomic mass of 32.1.) 9. How many coulombs of positive charge are there in 4.00 kg of plutonium, given its atomic mass is 244 and that each plutonium atom has 94 protons? 18.3 Conductors and Electric Fields in Static Equilibrium 10. Sketch the electric field lines in the vicinity of the conductor in Figure 18.47 given the field was originally uniform and parallel to the object's long axis. Is the resulting field small near the long side of the object? Figure 18.48 12. Sketch the electric field between the two conducting plates shown in Figure 18.49, given the top plate is positive and an equal amount of negative charge is on the bottom plate. Be certain to indicate the distribution of charge on the plates. Figure 18.49 13. Sketch the electric field lines in the vicinity of the charged insulator in Figure 18.50 noting its nonuniform charge distribution. Figure 18.50 A charged insulating rod such as might be used in a classroom demonstration. 14. What is the force on the charge located at = 8.00 cm in Figure 18.51(a) given that = 1.00 \u03bcC? Figure 18.47 11. Sketch the electric field lines in the vicinity of the conductor in Figure 18.48 given the field was originally uniform and parallel to the object's long axis. Is the resulting field small near the long side of the object? Figure 18.51 (a) Point charges located at 3.00, 8.00, and 11.0 cm along the x-axis. (b) Point charges located at 1.00, 5.00, 8.00, and 14.0 cm along the x-axis. 15. (a) Find the total electric field at = 1.00 cm in Figure 18.51(b) given that = 5.00 nC. (b) Find the total electric field at = 11.00 cm in Figure 18.51(b). (c) If the charges are allowed to move and eventually be brought to rest by friction, what will the final charge configuration be? (That is, This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 18 | Electric Charge and Electric Field 813 will there be a single charge, double charge, etc., and what will its value(s) be?) 16. (a) Find the electric field at = 5.00 cm in Figure 18.51(a), given that = 1.00 \u03bcC. (b) At what position between 3.00 and 8.00 cm is the total electric field the same as that for \u20132 alone? (c) Can the electric field be zero anywhere between 0.00 and 8.00 cm? (d) At very large positive or negative values of x, the electric field approaches zero in both (a) and (b). In which does it most rapidly approach zero and why? (e) At what position to the right of 11.0 cm is the total electric field zero, other than at infinity? (Hint: A graphing calculator can yield considerable insight in this problem.) 17. (a) Find the total Coulomb force on a charge of 2.00 nC located at = 4.", "00 cm in Figure 18.51 (b), given that = 1.00 \u03bcC. (b) Find the x-position at which the electric field is zero in Figure 18.51 (b). 18. Using the symmetry of the arrangement, determine the direction of the force on in the figure below, given that = =+7.50 \u03bcC and = = \u22127.50 \u03bcC. (b) Calculate the magnitude of the force on the charge, given that the square is 10.0 cm on a side and = 2.00 \u03bcC. Figure 18.52 19. (a) Using the symmetry of the arrangement, determine the direction of the electric field at the center of the square in Figure 18.52, given that = = \u22121.00 \u03bcC and = =+1.00 \u03bcC. (b) Calculate the magnitude of the electric field at the location of, given that the square is 5.00 cm on a side. 20. Find the electric field at the location of in Figure 18.52 given that = = =+2.00 nC, = \u22121.00 nC, and the square is 20.0 cm on a side. 21. Find the total Coulomb force on the charge in Figure 18.52, given that = 1.00 \u03bcC, = 2.00 \u03bcC, = \u22123.00 \u03bcC, = \u22124.00 \u03bcC, and =+1.00 \u03bcC. The square is 50.0 cm on a side. 22. (a) Find the electric field at the location of in Figure 18.53, given that b = +10.00 C and c = \u20135.00 C. (b) What is the force on, given that a = +1.50 nC? Figure 18.53 Point charges located at the corners of an equilateral triangle 25.0 cm on a side. 23. (a) Find the electric field at the center of the triangular configuration of charges in Figure 18.53, given that =+2.50 nC, = \u22128.00 nC, and =+1.50 nC. (b) Is there any combination of charges, other than = =, that will produce a zero strength electric field at the center of the triangular configuration? 18.4 Coulomb\u2019s Law 24. What is the repulsive force between two pith balls that are 8.00 cm apart and have equal charges of \u2013 30.0 nC? 25. (a) How strong is the attractive force between a glass rod with a 0.700 C charge and a silk cloth with a \u20130.600 C charge, which are 12.0 cm apart, using the approximation that they act like point charges? (b) Discuss how the answer to this problem might be affected if the charges are distributed over some area and do not act like point charges. 26. Two point charges exert a 5.00 N force on each other. What will the force become if the distance between them is increased by a factor of three? 27. Two point charges are brought closer together, increasing the force between them by a factor of 25. By what factor was their separation decreased? 28. How far apart must two point charges of 75.0 nC (typical of static electricity) be to have a force of 1.00 N between them? 29. If two equal charges each of 1 C each are separated in air by a distance of 1 km, what is the magnitude of the force acting between them? You will see that even at a distance as large as 1 km, the repulsive force is substantial because 1 C is a very significant amount of charge. 30. A test charge of +2 C is placed halfway between a charge of +6 C and another of +4 C separated by 10 cm. (a) What is the magnitude of the force on the test charge? (b) What is the direction of this force (away from or toward the +6 C charge)? 31. Bare free charges do not remain stationary when close together. To illustrate this, calculate the acceleration of two isolated protons separated by 2.00 nm (a typical distance between gas atoms). Explicitly show how you follow the steps in the Problem-Solving Strategy for electrostatics. 32. (a) By what factor must you change the distance between two point charges to change the force between them by a factor of 10? (b) Explain how the distance can either increase or decrease by this factor and still cause a factor of 10 change in the force. 814 Chapter 18 | Electric Charge and Electric Field (b) What magnitude and direction force does this field exert on a proton? 18.6 Electric Field Lines: Multiple Charges 47. (a) Sketch the electric field lines near a point charge +. (b) Do the same for a point charge \u20133.00. 48. Sketch the electric field lines a long distance from", " the charge distributions shown in Figure 18.34 (a) and (b) 49. Figure 18.54 shows the electric field lines near two charges 1 and 2. What is the ratio of their magnitudes? (b) Sketch the electric field lines a long distance from the charges shown in the figure. Figure 18.54 The electric field near two charges. 50. Sketch the electric field lines in the vicinity of two opposite charges, where the negative charge is three times greater in magnitude than the positive. (See Figure 18.54 for a similar situation). 18.8 Applications of Electrostatics 51. (a) What is the electric field 5.00 m from the center of the terminal of a Van de Graaff with a 3.00 mC charge, noting that the field is equivalent to that of a point charge at the center of the terminal? (b) At this distance, what force does the field exert on a 2.00 C charge on the Van de Graaff's belt? 52. (a) What is the direction and magnitude of an electric field that supports the weight of a free electron near the surface of Earth? (b) Discuss what the small value for this field implies regarding the relative strength of the gravitational and electrostatic forces. 53. A simple and common technique for accelerating electrons is shown in Figure 18.55, where there is a uniform electric field between two plates. Electrons are released, usually from a hot filament, near the negative plate, and there is a small hole in the positive plate that allows the electrons to continue moving. (a) Calculate the acceleration of the electron if the field strength is 2.50\u00d7104 N/C. (b) Explain why the electron will not be pulled back to the positive plate once it moves through the hole. 33. Suppose you have a total charge tot that you can split in any manner. Once split, the separation distance is fixed. How do you split the charge to achieve the greatest force? 34. (a) Common transparent tape becomes charged when pulled from a dispenser. If one piece is placed above another, the repulsive force can be great enough to support the top piece's weight. Assuming equal point charges (only an approximation), calculate the magnitude of the charge if electrostatic force is great enough to support the weight of a 10.0 mg piece of tape held 1.00 cm above another. (b) Discuss whether the magnitude of this charge is consistent with what is typical of static electricity. 35. (a) Find the ratio of the electrostatic to gravitational force between two electrons. (b) What is this ratio for two protons? (c) Why is the ratio different for electrons and protons? 36. At what distance is the electrostatic force between two protons equal to the weight of one proton? 37. A certain five cent coin contains 5.00 g of nickel. What fraction of the nickel atoms' electrons, removed and placed 1.00 m above it, would support the weight of this coin? The atomic mass of nickel is 58.7, and each nickel atom contains 28 electrons and 28 protons. 38. (a) Two point charges totaling 8.00 C exert a repulsive force of 0.150 N on one another when separated by 0.500 m. What is the charge on each? (b) What is the charge on each if the force is attractive? 39. Point charges of 5.00 C and \u20133.00 C are placed 0.250 m apart. (a) Where can a third charge be placed so that the net force on it is zero? (b) What if both charges are positive? 40. Two point charges 1 and 2 are 3.00 m apart, and their total charge is 20 C. (a) If the force of repulsion between them is 0.075N, what are magnitudes of the two charges? (b) If one charge attracts the other with a force of 0.525N, what are the magnitudes of the two charges? Note that you may need to solve a quadratic equation to reach your answer. 18.5 Electric Field: Concept of a Field Revisited 41. What is the magnitude and direction of an electric field that exerts a 2.00\u00d710-5 N upward force on a \u20131.75 C charge? 42. What is the magnitude and direction of the force exerted on a 3.50 C charge by a 250 N/C electric field that points due east? 43. Calculate the magnitude of the electric field 2.00 m from a point charge of 5.00 mC (such as found on the terminal of a Van de Graaff). 44. (a) What magnitude point charge creates a 10,000 N/C electric field at a distance of 0.250 m? (b) How large is the field at 10.0 m? 45. Calculate the initial (from rest) acceleration of a proton in a 5.00", "\u00d7106 N/C electric field (such as created by a research Van de Graaff). Explicitly show how you follow the steps in the Problem-Solving Strategy for electrostatics. 46. (a) Find the direction and magnitude of an electric field that exerts a 4.80\u00d710\u221217 N westward force on an electron. This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 18 | Electric Charge and Electric Field 815 because air begins to ionize and charges flow, reducing the field. (a) Calculate the distance a free proton must travel in this field to reach 3.00% of the speed of light, starting from rest. (b) Is this practical in air, or must it occur in a vacuum? 60. Integrated Concepts A 5.00 g charged insulating ball hangs on a 30.0 cm long string in a uniform horizontal electric field as shown in Figure 18.56. Given the charge on the ball is 1.00 C, find the strength of the field. Figure 18.55 Parallel conducting plates with opposite charges on them create a relatively uniform electric field used to accelerate electrons to the right. Those that go through the hole can be used to make a TV or computer screen glow or to produce X-rays. 54. Earth has a net charge that produces an electric field of approximately 150 N/C downward at its surface. (a) What is the magnitude and sign of the excess charge, noting the electric field of a conducting sphere is equivalent to a point charge at its center? (b) What acceleration will the field produce on a free electron near Earth's surface? (c) What mass object with a single extra electron will have its weight supported by this field? 55. Point charges of 25.0 C and 45.0 C are placed 0.500 m apart. (a) At what point along the line between them is the electric field zero? (b) What is the electric field halfway between them? 56. What can you say about two charges 1 and 2, if the electric field one-fourth of the way from 1 to 2 is zero? 57. Integrated Concepts Calculate the angular velocity \u03c9 of an electron orbiting a proton in the hydrogen atom, given the radius of the orbit is 0.530\u00d710\u201310 m. You may assume that the proton is stationary and the centripetal force is supplied by Coulomb attraction. 58. Integrated Concepts An electron has an initial velocity of 5.00\u00d7106 m/s in a uniform 2.00\u00d7105 N/C strength electric field. The field accelerates the electron in the direction opposite to its initial velocity. (a) What is the direction of the electric field? (b) How far does the electron travel before coming to rest? (c) How long does it take the electron to come to rest? (d) What is the electron's velocity when it returns to its starting point? 59. Integrated Concepts The practical limit to an electric field in air is about 3.00\u00d7106 N/C. Above this strength, sparking takes place Figure 18.56 A horizontal electric field causes the charged ball to hang at an angle of 8.00\u00ba. 61. Integrated Concepts Figure 18.57 shows an electron passing between two charged metal plates that create an 100 N/C vertical electric field perpendicular to the electron's original horizontal velocity. (These can be used to change the electron's direction, such as in an oscilloscope.) The initial speed of the electron is 3.00\u00d7106 m/s, and the horizontal distance it travels in the uniform field is 4.00 cm. (a) What is its vertical deflection? (b) What is the vertical component of its final velocity? (c) At what angle does it exit? Neglect any edge effects. Figure 18.57 62. Integrated Concepts The classic Millikan oil drop experiment was the first to obtain an accurate measurement of the charge on an electron. In it, oil drops were suspended against the gravitational force by a vertical electric field. (See Figure 18.58.) Given the oil drop to be 1.00 m in radius and have a density of 920 kg/m3 (a) Find the weight of the drop. (b) If the drop has a single excess electron, find the electric field strength needed to balance its weight. : 816 Chapter 18 | Electric Charge and Electric Field 67. Construct Your Own Problem Consider two insulating balls with evenly distributed equal and opposite charges on their surfaces, held with a certain distance between the centers of the balls. Construct a problem in which you calculate the electric field (magnitude and direction) due to the balls at various points along a line running through the centers of the balls and extending to infinity on either side. Choose interesting points and comment on the meaning of the field at those points. For example, at what points might the field be just that due to one", " ball and where does the field become negligibly small? Among the things to be considered are the magnitudes of the charges and the distance between the centers of the balls. Your instructor may wish for you to consider the electric field off axis or for a more complex array of charges, such as those in a water molecule. 68. Construct Your Own Problem Consider identical spherical conducting space ships in deep space where gravitational fields from other bodies are negligible compared to the gravitational attraction between the ships. Construct a problem in which you place identical excess charges on the space ships to exactly counter their gravitational attraction. Calculate the amount of excess charge needed. Examine whether that charge depends on the distance between the centers of the ships, the masses of the ships, or any other factors. Discuss whether this would be an easy, difficult, or even impossible thing to do in practice. Figure 18.58 In the Millikan oil drop experiment, small drops can be suspended in an electric field by the force exerted on a single excess electron. Classically, this experiment was used to determine the electron charge e by measuring the electric field and mass of the drop. 63. Integrated Concepts (a) In Figure 18.59, four equal charges lie on the corners of a square. A fifth charge is on a mass directly above the center of the square, at a height equal to the length of one side of the square. Determine the magnitude of in terms of,, and, if the Coulomb force is to equal the weight of. (b) Is this equilibrium stable or unstable? Discuss. Figure 18.59 Four equal charges on the corners of a horizontal square support the weight of a fifth charge located directly above the center of the square. 64. Unreasonable Results (a) Calculate the electric field strength near a 10.0 cm diameter conducting sphere that has 1.00 C of excess charge on it. (b) What is unreasonable about this result? (c) Which assumptions are responsible? 65. Unreasonable Results (a) Two 0.500 g raindrops in a thunderhead are 1.00 cm apart when they each acquire 1.00 mC charges. Find their acceleration. (b) What is unreasonable about this result? (c) Which premise or assumption is responsible? 66. Unreasonable Results A wrecking yard inventor wants to pick up cars by charging a 0.400 m diameter ball and inducing an equal and opposite charge on the car. If a car has a 1000 kg mass and the ball is to be able to lift it from a distance of 1.00 m: (a) What minimum charge must be used? (b) What is the electric field near the surface of the ball? (c) Why are these results unreasonable? (d) Which premise or assumption is responsible? This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 18 | Electric Charge and Electric Field 817 Test Prep for AP\u00ae Courses Z is attracted to balloon Y. Which of the following can be the charge on Z? Select two answers. 18.1 Static Electricity and Charge: Conservation of Charge 1. When a glass rod is rubbed against silk, which of the following statements is true? a. Electrons are removed from the silk. b. Electrons are removed from the rod. c. Protons are removed from the silk. d. Protons are removed from the rod. 2. In an experiment, three microscopic latex spheres are sprayed into a chamber and become charged with +3e, +5e, and \u22123e, respectively. Later, all three spheres collide simultaneously and then separate. Which of the following are possible values for the final charges on the spheres? Select two answers. X Y Z (a) +4e \u22124e +5e (b) \u22124e +4.5e +5.5e (c) +5e \u22128e (d) +6e +6e +7e \u22127e 3. If objects X and Y attract each other, which of the following will be false? a. X has positive charge and Y has negative charge. b. X has negative charge and Y has positive charge. c. X and Y both have positive charge. d. X is neutral and Y has a charge. 4. Suppose a positively charged object A is brought in contact with an uncharged object B in a closed system. What type of charge will be left on object B? a. negative b. positive c. neutral d. cannot be determined 5. What will be the net charge on an object which attracts neutral pieces of paper but repels a negatively charged balloon? a. negative b. positive c. neutral d. cannot be determined 6. When two neutral objects are rubbed against each other, the first one gains a net charge of 3e. Which of the following statements is true? a. The second object gains 3e and is negatively charged. b. The second object loses 3e and is negatively charged", ". c. The second object gains 3e and is positively charged. d. The second object loses 3e and is positively charged. 7. In an experiment, a student runs a comb through his hair several times and brings it close to small pieces of paper. Which of the following will he observe? a. Pieces of paper repel the comb. b. Pieces of paper are attracted to the comb. c. Some pieces of paper are attracted and some repel the comb. d. There is no attraction or repulsion between the pieces of paper and the comb. 8. In an experiment a negatively charged balloon (balloon X) is repelled by another charged balloon Y. However, an object a. negative b. positive c. neutral d. cannot be determined 9. Suppose an object has a charge of 1 C and gains 6.88\u00d71018 electrons. a. What will be the net charge of the object? b. If the object has gained electrons from a neutral object, what will be the charge on the neutral object? c. Find and explain the relationship between the total charges of the two objects before and after the transfer. d. When a third object is brought in contact with the first object (after it gains the electrons), the resulting charge on the third object is 0.4 C. What was its initial charge? 10. The charges on two identical metal spheres (placed in a closed system) are -2.4\u00d710\u221217 C and -4.8\u00d710\u221217 C. a. How many electrons will be equivalent to the charge on b. each sphere? If the two spheres are brought in contact and then separated, find the charge on each sphere. c. Calculate the number of electrons that would be equivalent to the resulting charge on each sphere. 11. In an experiment the following observations are made by a student for four charged objects W, X, Y, and Z: \u2022 A glass rod rubbed with silk attracts W. \u2022 W attracts Z but repels X. \u2022 X attracts Z but repels Y. \u2022 Y attracts W and Z. Estimate whether the charges on each of the four objects are positive, negative, or neutral. 18.2 Conductors and Insulators 12. Some students experimenting with an uncharged metal sphere want to give the sphere a net charge using a charged aluminum pie plate. Which of the following steps would give the sphere a net charge of the same sign as the pie plate? a. bringing the pie plate close to, but not touching, the metal sphere, then moving the pie plate away. b. bringing the pie plate close to, but not touching, the metal sphere, then momentarily touching a grounding wire to the metal sphere. c. bringing the pie plate close to, but not touching, the metal sphere, then momentarily touching a grounding wire to the pie plate. touching the pie plate to the metal sphere. d. 13. Figure 18.60 Balloon and sphere. When the balloon is brought closer to the sphere, there will be a redistribution of charges. What is this phenomenon called? a. electrostatic repulsion 818 Chapter 18 | Electric Charge and Electric Field b. conduction c. polarization d. none of the above 14. What will be the charge at Y (i.e., the part of the sphere furthest from the balloon)? a. positive b. negative c. zero d. It can be positive or negative depending on the material. 15. What will be the net charge on the sphere? a. positive b. negative c. zero d. It can be positive or negative depending on the material. second experiment the rod is only brought close to the electroscope but not in contact. However, while the rod is close, the electroscope is momentarily grounded and then the rod is removed. In both experiments the needles of the electroscopes deflect, which indicates the presence of charges. a. What is the charging method in each of the two experiments? b. What is the net charge on the electroscope in the first c. experiment? Explain how the electroscope obtains that charge. Is the net charge on the electroscope in the second experiment different from that of the first experiment? Explain why. 16. If Y is grounded while the balloon is still close to X, which of the following will be true? 18.3 Conductors and Electric Fields in Static Equilibrium a. Electrons will flow from the sphere to the ground. b. Electrons will flow from the ground to the sphere. c. Protons will flow from the sphere to the ground. d. Protons will flow from the ground to the sphere. 21. 17. If the balloon is moved away after grounding, what will be the net charge on the sphere? a. positive b. negative c. zero d. It can be positive or negative depending on the material. 18. A positively charged rod is used to charge a sphere by induction. Which of the following is true? a. The sphere must be a conductor. b.", " The sphere must be an insulator. c. The sphere can be a conductor or insulator but must be connected to ground. d. The sphere can be a conductor or insulator but must be already charged. 19. Figure 18.62 A sphere conductor. An electric field due to a positively charged spherical conductor is shown above. Where will the electric field be weakest? a. Point A b. Point B c. Point C d. Same at all points 22. Figure 18.63 Electric field between two parallel metal plates. The electric field created by two parallel metal plates is shown above. Where will the electric field be strongest? a. Point A b. Point B c. Point C d. Same at all points 23. Suppose that the electric field experienced due to a positively charged small spherical conductor at a certain distance is E. What will be the percentage change in electric field experienced at thrice the distance if the charge on the conductor is doubled? 24. Figure 18.61 Rod and metal balls. As shown in the figure above, two metal balls are suspended and a negatively charged rod is brought close to them. a. If the two balls are in contact with each other what will be the charges on each ball? b. Explain how the balls get these charges. c. What will happen to the charge on the second ball (i.e., the ball further away from the rod) if it is momentarily grounded while the rod is still there? If (instead of grounding) the second ball is moved away and then the rod is removed from the first ball, will the two balls have induced charges? If yes, what will be the charges? If no, why not? d. 20. Two experiments are performed using positively charged glass rods and neutral electroscopes. In the first experiment the rod is brought in contact with the electroscope. In the This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 18 | Electric Charge and Electric Field 819 b. Will this ratio change if the two electrons are replaced by protons? If yes, find the new ratio. 18.5 Electric Field: Concept of a Field Revisited 31. Two particles with charges +2q and +q are separated by a distance r. The +2q particle has an electric field E at distance r and exerts a force F on the +q particle. Use this information to answer questions 31\u201332. What is the electric field of the +q particle at the same distance and what force does it exert on the +2q particle? a. E/2, F/2 b. E, F/2 c. E/2, F d. E, F 32. When the +q particle is replaced by a +3q particle, what will be the electric field and force from the +2q particle experienced by the +3q particle? a. E/3, 3F b. E, 3F c. E/3, F d. E, F 33. The direction of the electric field of a negative charge is inward for both positive and negative charges. a. b. outward for both positive and negative charges. c. inward for other positive charges and outward for other negative charges. d. outward for other positive charges and inward for other negative charges. 34. The force responsible for holding an atom together is frictional a. b. electric c. gravitational d. magnetic 35. When a positively charged particle exerts an inward force on another particle P, what will be the charge of P? a. positive b. negative c. neutral d. cannot be determined 36. Find the force exerted due to a particle having a charge of 3.2\u00d710\u221219 C on another identical particle 5 cm away. 37. Suppose that the force exerted on an electron is 5.6\u00d710\u221217 N, directed to the east. a. Find the magnitude of the electric field that exerts the force. b. What will be the direction of the electric field? c. If the electron is replaced by a proton, what will be the magnitude of force exerted? d. What will be the direction of force on the proton? 18.6 Electric Field Lines: Multiple Charges 38. Figure 18.65 An electric dipole (with +2q and \u20132q as the two charges) is shown in the figure above. A third charge, \u2212q is Figure 18.64 Millikan oil drop experiment. The classic Millikan oil drop experiment setup is shown above. In this experiment oil drops are suspended in a vertical electric field against the gravitational force to measure their charge. If the mass of a negatively charged drop suspended in an electric field of 1.18\u00d710\u22124 N/C strength is 3.85\u00d710\u221221 g, find the number of excess electrons in the drop. 18.4 Coulomb\u2019s Law 25. For questions 25\u201327, suppose that the electrostatics force between two", " charges is F. What will be the force if the distance between them is halved? a. 4F b. 2F c. F/4 d. F/2 26. Which of the following is false? a. b. If the charge of one of the particles is doubled and that of the second is unchanged, the force will become 2F. If the charge of one of the particles is doubled and that of the second is halved, the force will remain F. If the charge of both the particles is doubled, the force will become 4F. d. None of the above. c. 27. Which of the following is true about the gravitational force between the particles? a. b. c. d. It will be 3.25\u00d710\u221238 F. It will be 3.25\u00d71038 F. It will be equal to F. It is not possible to determine the gravitational force as the masses of the particles are not given. 28. Two massive, positively charged particles are initially held a fixed distance apart. When they are moved farther apart, the magnitude of their mutual gravitational force changes by a factor of n. Which of the following indicates the factor by which the magnitude of their mutual electrostatic force changes? a. 1/n2 b. 1/n c. n d. n2 29. a. What is the electrostatic force between two charges of 1 C each, separated by a distance of 0.5 m? b. How will this force change if the distance is increased to 1 m? 30. a. Find the ratio of the electrostatic force to the gravitational force between two electrons. 820 Chapter 18 | Electric Charge and Electric Field placed equidistant from the dipole charges. What will be the direction of the net force on the third charge? negative and that the sign of the charge of object S is positive. a. \u2192 b. \u2190 c. \u2193 d. \u2191 39. ii) Briefly describe the characteristics of the field diagram that indicate that the magnitudes of the charges of objects R and T are equal and that the magnitude of the charge of object S is about twice that of objects R and T. For the following parts, an electric field directed to the right is defined to be positive. (b) On the axes below, sketch a graph of the electric field E along the x-axis as a function of position x. Figure 18.68 An Electric field (E) axis and Position (x) axis. (c) Write an expression for the electric field E along the x-axis as a function of position x in the region between objects S and T in terms of q, d, and fundamental constants, as appropriate. (d) Your classmate tells you there is a point between S and T where the electric field is zero. Determine whether this statement is true, and explain your reasoning using two of the representations from parts (a), (b), or (c). Figure 18.66 Four objects, each with charge +q, are held fixed on a square with sides of length d, as shown in the figure. Objects X and Z are at the midpoints of the sides of the square. The electrostatic force exerted by object W on object X is F. Use this information to answer questions 39\u201340. What is the magnitude of force exerted by object W on Z? a. F/7 b. F/5 c. F/3 d. F/2 40. What is the magnitude of the net force exerted on object X by objects W, Y, and Z? a. F/4 b. F/2 c. 9F/4 d. 3F 41. Figure 18.67 Electric field with three charged objects. The figure above represents the electric field in the vicinity of three small charged objects, R, S, and T. The objects have charges \u2212q, +2q, and \u2212q, respectively, and are located on the x-axis at \u2212d, 0, and d. Field vectors of very large magnitude are omitted for clarity. (a) i) Briefly describe the characteristics of the field diagram that indicate that the sign of the charges of objects R and T is This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 20 | Electric Current, Resistance, and Ohm's Law 867 20 ELECTRIC CURRENT, RESISTANCE, AND OHM'S LAW Figure 20.1 Electric energy in massive quantities is transmitted from this hydroelectric facility, the Srisailam power station located along the Krishna River in India (http://en.wikipedia.org/wiki/Srisailam_Dam), by the movement of charge\u2014that is, by electric current. (credit: Chintohere, Wikimedia Commons) Chapter Outline 20.1. Current 20.2. Ohm\u2019s Law: Resistance and Simple Circuits 20.3. Resistance and Resistivity 20.", "4. Electric Power and Energy 20.5. Alternating Current versus Direct Current 20.6. Electric Hazards and the Human Body 20.7. Nerve Conduction\u2013Electrocardiograms Connection for AP\u00ae Courses In our daily lives, we see and experience many examples of electricity which involve electric current, the movement of charge. These include the flicker of numbers on a handheld calculator, nerve impulses carrying signals of vision to the brain, an ultrasound device sending a signal to a computer screen, the brain sending a message for a baby to twitch its toes, an electric train pulling its load over a mountain pass, and a hydroelectric plant sending energy to metropolitan and rural users. Humankind has indeed harnessed electricity, the basis of technology, to improve the quality of life. While the previous two chapters concentrated on static electricity and the fundamental force underlying its behavior, the next few chapters will be 868 Chapter 20 | Electric Current, Resistance, and Ohm's Law devoted to electric and magnetic phenomena involving electric current. In addition to exploring applications of electricity, we shall gain new insights into its nature \u2013 in particular, the fact that all magnetism results from electric current. This chapter supports learning objectives covered under Big Ideas 1, 4, and 5 of the AP Physics Curriculum Framework. Electric charge is a property of a system (Big Idea 1) that affects its interaction with other charged systems (Enduring Understanding 1.B), whereas electric current is fundamentally the movement of charge through a conductor and is based on the fact that electric charge is conserved within a system (Essential Knowledge 1.B.1). The conservation of charge also leads to the concept of an electric circuit as a closed loop of electrical current. In addition, this chapter discusses examples showing that the current in a circuit is resisted by the elements of the circuit and the strength of the resistance depends on the material of the elements. The macroscopic properties of materials, including resistivity, depend on their molecular and atomic structure (Enduring Understanding 1.E). In addition, resistivity depends on the temperature of the material (Essential Knowledge 1.E.2). The chapter also describes how the interaction of systems of objects can result in changes in those systems (Big Idea 4). For example, electric properties of a system of charged objects can change in response to the presence of, or changes in, other charged objects or systems (Enduring Understanding 4.E). A simple circuit with a resistor and an energy source is an example of such a system. The current through the resistor in the circuit is equal to the difference of potentials across the resistor divided by its resistance (Essential Knowledge 4.E.4). The unifying theme of the physics curriculum is that any changes in the systems due to interactions are governed by laws of conservation (Big Idea 5). This chapter applies the idea of energy conservation (Enduring Understanding 5.B) to electric circuits and connects concepts of electric energy and electric power as rates of energy use (Essential Knowledge 5.B.5). While the laws of conservation of energy in electric circuits are fully described by Kirchoff's rules, which are introduced in the next chapter (Essential Knowledge 5.B.9), the specific definition of power (based on Essential Knowledge 5.B.9) is that it is the rate at which energy is transferred from a resistor as the product of the electric potential difference across the resistor and the current through the resistor. Big Idea 1 Objects and systems have properties such as mass and charge. Systems may have internal structure. Enduring Understanding 1.B Electric charge is a property of an object or system that affects its interactions with other objects or systems containing charge. Essential Knowledge 1.B.1 Electric charge is conserved. The net charge of a system is equal to the sum of the charges of all the objects in the system. Enduring Understanding 1.E Materials have many macroscopic properties that result from the arrangement and interactions of the atoms and molecules that make up the material. Essential Knowledge 1.E.2 Matter has a property called resistivity. Big Idea 4 Interactions between systems can result in changes in those systems. Enduring Understanding 4.E The electric and magnetic properties of a system can change in response to the presence of, or changes in, other objects or systems. Essential Knowledge 4.E.4 The resistance of a resistor, and the capacitance of a capacitor, can be understood from the basic properties of electric fields and forces, as well as the properties of materials and their geometry. Big Idea 5: Changes that occur as a result of interactions are constrained by conservation laws. Enduring Understanding 5.B The energy of a system is conserved. Essential Knowledge 5.B.5 Energy can be transferred by an external force exerted on an object or system that moves the object or system through a distance; this energy transfer is called work. Energy transfer in mechanical or electrical systems may occur at different rates. Power is de\ufb01ned as the rate of energy transfer into,", " out of, or within a system. [A piston \ufb01lled with gas getting compressed or expanded is treated in Physics 2 as a part of thermodynamics.] Essential Knowledge 5.B.9 Kirchhoff's loop rule describes conservation of energy in electrical circuits. [The application of Kirchhoff's laws to circuits is introduced in Physics 1 and further developed in Physics 2 in the context of more complex circuits, including those with capacitors.] 20.1 Current Learning Objectives By the end of this section, you will be able to: \u2022 Define electric current, ampere, and drift velocity. \u2022 Describe the direction of charge flow in conventional current. \u2022 Use drift velocity to calculate current and vice versa. The information presented in this section supports the following AP\u00ae learning objectives and science practices: \u2022 1.B.1.1 The student is able to make claims about natural phenomena based on conservation of electric charge. (S.P. 6.4) \u2022 1.B.1.2 The student is able to make predictions, using the conservation of electric charge, about the sign and relative quantity of net charge of objects or systems after various charging processes, including conservation of charge in simple circuits. (S.P. 6.4, 7.2) This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 20 | Electric Current, Resistance, and Ohm's Law 869 Electric Current Electric current is defined to be the rate at which charge flows. A large current, such as that used to start a truck engine, moves a large amount of charge in a small time, whereas a small current, such as that used to operate a hand-held calculator, moves a small amount of charge over a long period of time. In equation form, electric current is defined to be = \u0394 \u0394, (20.1) where \u0394 is the amount of charge passing through a given area in time \u0394. (As in previous chapters, initial time is often taken to be zero, in which case \u0394 =.) (See Figure 20.2.) The SI unit for current is the ampere (A), named for the French physicist Andr\u00e9-Marie Amp\u00e8re (1775\u20131836). Since = \u0394 / \u0394, we see that an ampere is one coulomb per second: 1 A = 1 C/s (20.2) Not only are fuses and circuit breakers rated in amperes (or amps), so are many electrical appliances. Figure 20.2 The rate of flow of charge is current. An ampere is the flow of one coulomb through an area in one second. Example 20.1 Calculating Currents: Current in a Truck Battery and a Handheld Calculator (a) What is the current involved when a truck battery sets in motion 720 C of charge in 4.00 s while starting an engine? (b) How long does it take 1.00 C of charge to flow through a handheld calculator if a 0.300-mA current is flowing? Strategy We can use the definition of current in the equation = \u0394 / \u0394 to find the current in part (a), since charge and time are given. In part (b), we rearrange the definition of current and use the given values of charge and current to find the time required. Solution for (a) Entering the given values for charge and time into the definition of current gives = \u0394 \u0394 = 180 A. = 720 C 4.00 s = 180 C/s (20.3) Discussion for (a) This large value for current illustrates the fact that a large charge is moved in a small amount of time. The currents in these \u201cstarter motors\u201d are fairly large because large frictional forces need to be overcome when setting something in motion. Solution for (b) Solving the relationship = \u0394 / \u0394 for time \u0394, and entering the known values for charge and current gives \u0394 = \u0394 = 1.00 C 0.30010-3 C/s = 3.33103 s. (20.4) Discussion for (b) This time is slightly less than an hour. The small current used by the hand-held calculator takes a much longer time to move a smaller charge than the large current of the truck starter. So why can we operate our calculators only seconds after turning them on? It's because calculators require very little energy. Such small current and energy demands allow handheld calculators to operate from solar cells or to get many hours of use out of small batteries. Remember, calculators do not have moving parts in the same way that a truck engine has with cylinders and pistons, so the technology requires smaller currents. 870 Chapter 20 | Electric Current, Resistance, and Ohm's Law Figure 20.3 shows a simple circuit and the standard schematic representation of a battery, conducting path, and load (a resistor). Schematics are very useful in visualizing the main features of a", " circuit. A single schematic can represent a wide variety of situations. The schematic in Figure 20.3 (b), for example, can represent anything from a truck battery connected to a headlight lighting the street in front of the truck to a small battery connected to a penlight lighting a keyhole in a door. Such schematics are useful because the analysis is the same for a wide variety of situations. We need to understand a few schematics to apply the concepts and analysis to many more situations. Figure 20.3 (a) A simple electric circuit. A closed path for current to flow through is supplied by conducting wires connecting a load to the terminals of a battery. (b) In this schematic, the battery is represented by the two parallel red lines, conducting wires are shown as straight lines, and the zigzag represents the load. The schematic represents a wide variety of similar circuits. Note that the direction of current in Figure 20.3 is from positive to negative. The direction of conventional current is the direction that positive charge would flow. In a single loop circuit (as shown in Figure 20.3), the value for current at all points of the circuit should be the same if there are no losses. This is because current is the flow of charge and charge is conserved, i.e., the charge flowing out from the battery will be the same as the charge flowing into the battery. Depending on the situation, positive charges, negative charges, or both may move. In metal wires, for example, current is carried by electrons\u2014that is, negative charges move. In ionic solutions, such as salt water, both positive and negative charges move. This is also true in nerve cells. A Van de Graaff generator used for nuclear research can produce a current of pure positive charges, such as protons. Figure 20.4 illustrates the movement of charged particles that compose a current. The fact that conventional current is taken to be in the direction that positive charge would flow can be traced back to American politician and scientist Benjamin Franklin in the 1700s. He named the type of charge associated with electrons negative, long before they were known to carry current in so many situations. Franklin, in fact, was totally unaware of the small-scale structure of electricity. It is important to realize that there is an electric field in conductors responsible for producing the current, as illustrated in Figure 20.4. Unlike static electricity, where a conductor in equilibrium cannot have an electric field in it, conductors carrying a current have an electric field and are not in static equilibrium. An electric field is needed to supply energy to move the charges. Making Connections: Take-Home Investigation\u2014Electric Current Illustration Find a straw and little peas that can move freely in the straw. Place the straw flat on a table and fill the straw with peas. When you pop one pea in at one end, a different pea should pop out the other end. This demonstration is an analogy for an electric current. Identify what compares to the electrons and what compares to the supply of energy. What other analogies can you find for an electric current? Note that the flow of peas is based on the peas physically bumping into each other; electrons flow due to mutually repulsive electrostatic forces. This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 20 | Electric Current, Resistance, and Ohm's Law 871 Figure 20.4 Current is the rate at which charge moves through an area, such as the cross-section of a wire. Conventional current is defined to move in the direction of the electric field. (a) Positive charges move in the direction of the electric field and the same direction as conventional current. (b) Negative charges move in the direction opposite to the electric field. Conventional current is in the direction opposite to the movement of negative charge. The flow of electrons is sometimes referred to as electronic flow. Example 20.2 Calculating the Number of Electrons that Move through a Calculator If the 0.300-mA current through the calculator mentioned in the Example 20.1 example is carried by electrons, how many electrons per second pass through it? Strategy The current calculated in the previous example was defined for the flow of positive charge. For electrons, the magnitude is the same, but the sign is opposite, electrons = \u22120.300\u00d710\u22123 C/s.Since each electron (\u2212) has a charge of \u20131.60\u00d710\u221219 C, we can convert the current in coulombs per second to electrons per second. Solution Starting with the definition of current, we have electrons = \u0394electrons \u0394 = \u20130.300\u00d710\u22123 C s. We divide this by the charge per electron, so that s = \u20130.30010 \u2013 3 C \u2013 = 1.881015 \u2013 s. s 1 \u2013 \u20131.6010\u221219 C (20.5) (20.6) Discussion There are so many charged particles", " moving, even in small currents, that individual charges are not noticed, just as individual water molecules are not noticed in water flow. Even more amazing is that they do not always keep moving forward like soldiers in a parade. Rather they are like a crowd of people with movement in different directions but a general trend to move forward. There are lots of collisions with atoms in the metal wire and, of course, with other electrons. Drift Velocity Electrical signals are known to move very rapidly. Telephone conversations carried by currents in wires cover large distances without noticeable delays. Lights come on as soon as a switch is flicked. Most electrical signals carried by currents travel at speeds on the order of 108 m/s, a significant fraction of the speed of light. Interestingly, the individual charges that make up the current move much more slowly on average, typically drifting at speeds on the order of 10\u22124 m/s. How do we reconcile these two speeds, and what does it tell us about standard conductors? The high speed of electrical signals results from the fact that the force between charges acts rapidly at a distance. Thus, when a free charge is forced into a wire, as in Figure 20.5, the incoming charge pushes other charges ahead of it, which in turn push on charges farther down the line. The density of charge in a system cannot easily be increased, and so the signal is passed on 872 Chapter 20 | Electric Current, Resistance, and Ohm's Law rapidly. The resulting electrical shock wave moves through the system at nearly the speed of light. To be precise, this rapidly moving signal or shock wave is a rapidly propagating change in electric field. Figure 20.5 When charged particles are forced into this volume of a conductor, an equal number are quickly forced to leave. The repulsion between like charges makes it difficult to increase the number of charges in a volume. Thus, as one charge enters, another leaves almost immediately, carrying the signal rapidly forward. Good conductors have large numbers of free charges in them. In metals, the free charges are free electrons. Figure 20.6 shows how free electrons move through an ordinary conductor. The distance that an individual electron can move between collisions with atoms or other electrons is quite small. The electron paths thus appear nearly random, like the motion of atoms in a gas. But there is an electric field in the conductor that causes the electrons to drift in the direction shown (opposite to the field, since they are negative). The drift velocity d is the average velocity of the free charges. Drift velocity is quite small, since there are so many free charges. If we have an estimate of the density of free electrons in a conductor, we can calculate the drift velocity for a given current. The larger the density, the lower the velocity required for a given current. Figure 20.6 Free electrons moving in a conductor make many collisions with other electrons and atoms. The path of one electron is shown. The average velocity of the free charges is called the drift velocity, d, and it is in the direction opposite to the electric field for electrons. The collisions normally transfer energy to the conductor, requiring a constant supply of energy to maintain a steady current. Conduction of Electricity and Heat Good electrical conductors are often good heat conductors, too. This is because large numbers of free electrons can carry electrical current and can transport thermal energy. The free-electron collisions transfer energy to the atoms of the conductor. The electric field does work in moving the electrons through a distance, but that work does not increase the kinetic energy (nor speed, therefore) of the electrons. The work is transferred to the conductor's atoms, possibly increasing temperature. Thus a continuous power input is required to maintain current. An exception, of course, is found in superconductors, for reasons we shall explore in a later chapter. Superconductors can have a steady current without a continual supply of energy\u2014a great energy savings. In contrast, the supply of energy can be useful, such as in a lightbulb filament. The supply of energy is necessary to increase the temperature of the tungsten filament, so that the filament glows. Making Connections: Take-Home Investigation\u2014Filament Observations Find a lightbulb with a filament. Look carefully at the filament and describe its structure. To what points is the filament connected? We can obtain an expression for the relationship between current and drift velocity by considering the number of free charges in a segment of wire, as illustrated in Figure 20.7. The number of free charges per unit volume is given the symbol and depends on the material. The shaded segment has a volume, so that the number of free charges in it is. The charge \u0394 in this segment is thus, where is the amount of charge on each carrier. (Recall that for electrons, is This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 20 | Electric Current, Resistance, and Ohm's Law 873 \u22121.60\u00d710\u2212", "19 C.) Current is charge moved per unit time; thus, if all the original charges move out of this segment in time \u0394, the current is = \u0394 \u0394 = \u0394. Note that / \u0394 is the magnitude of the drift velocity, d, since the charges move an average distance in a time \u0394. Rearranging terms gives = d, (20.7) (20.8) where is the current through a wire of cross-sectional area made of a material with a free charge density. The carriers of the current each have charge and move with a drift velocity of magnitude d. Figure 20.7 All the charges in the shaded volume of this wire move out in a time, having a drift velocity of magnitude d = /. See text for further discussion. Note that simple drift velocity is not the entire story. The speed of an electron is much greater than its drift velocity. In addition, not all of the electrons in a conductor can move freely, and those that do might move somewhat faster or slower than the drift velocity. So what do we mean by free electrons? Atoms in a metallic conductor are packed in the form of a lattice structure. Some electrons are far enough away from the atomic nuclei that they do not experience the attraction of the nuclei as much as the inner electrons do. These are the free electrons. They are not bound to a single atom but can instead move freely among the atoms in a \u201csea\u201d of electrons. These free electrons respond by accelerating when an electric field is applied. Of course as they move they collide with the atoms in the lattice and other electrons, generating thermal energy, and the conductor gets warmer. In an insulator, the organization of the atoms and the structure do not allow for such free electrons. Example 20.3 Calculating Drift Velocity in a Common Wire Calculate the drift velocity of electrons in a 12-gauge copper wire (which has a diameter of 2.053 mm) carrying a 20.0-A current, given that there is one free electron per copper atom. (Household wiring often contains 12-gauge copper wire, and the maximum current allowed in such wire is usually 20 A.) The density of copper is 8.80\u00d7103 kg/m3. Strategy We can calculate the drift velocity using the equation = d. The current = 20.0 A is given, and = \u2013 1.60\u00d710 \u2013 19 C is the charge of an electron. We can calculate the area of a cross-section of the wire using the formula = 2, where is one-half the given diameter, 2.053 mm. We are given the density of copper, 8.80\u00d7103 kg/m3, and the periodic table shows that the atomic mass of copper is 63.54 g/mol. We can use these two quantities along with Avogadro's number, 6.02\u00d71023 atoms/mol, cubic meter. the number of free electrons per to determine, Solution First, calculate the density of free electrons in copper. There is one free electron per copper atom. Therefore, is the same as the number of copper atoms per m3. We can now find as follows: = 1 \u2212 atom\u00d7 6.02\u00d71023 atoms mol \u00d7 1 mol 63.54 g \u00d7 1000 g kg \u00d7 8.80\u00d7103 kg 1 m3 (20.9) = 8.342\u00d71028 \u2212 /m3. The cross-sectional area of the wire is 874 Chapter 20 | Electric Current, Resistance, and Ohm's Law = 2 = 2.053\u00d710\u22123 m 2 2 = 3.310\u00d710\u20136 m2. Rearranging = d to isolate drift velocity gives = d = 20.0 A (8.342\u00d71028/m3)(\u20131.60\u00d710\u201319 C)(3.310\u00d710\u20136 m2) = \u20134.53\u00d710\u20134 m/s. (20.10) (20.11) Discussion The minus sign indicates that the negative charges are moving in the direction opposite to conventional current. The small value for drift velocity (on the order of 10\u22124 m/s ) confirms that the signal moves on the order of 1012 times faster (about 108 m/s ) than the charges that carry it. 20.2 Ohm\u2019s Law: Resistance and Simple Circuits Learning Objectives By the end of this section, you will be able to: \u2022 Explain the origin of Ohm's law. \u2022 Calculate voltages, currents, and resistances with Ohm's law. \u2022 Explain the difference between ohmic and non-ohmic materials. \u2022 Describe a simple circuit. The information presented in this section supports the following AP\u00ae learning objectives and science practices: \u2022 4.E.4.1 The student is able to make predictions about the properties of resistors and/or capacitors when placed in a simple circuit based on the geometry of the", " circuit element and supported by scientific theories and mathematical relationships. (S.P. 2.2, 6.4) What drives current? We can think of various devices\u2014such as batteries, generators, wall outlets, and so on\u2014which are necessary to maintain a current. All such devices create a potential difference and are loosely referred to as voltage sources. When a voltage source is connected to a conductor, it applies a potential difference that creates an electric field. The electric field in turn exerts force on charges, causing current. Ohm's Law The current that flows through most substances is directly proportional to the voltage applied to it. The German physicist Georg Simon Ohm (1787\u20131854) was the first to demonstrate experimentally that the current in a metal wire is directly proportional to the voltage applied: \u221d. (20.12) This important relationship is known as Ohm's law. It can be viewed as a cause-and-effect relationship, with voltage the cause and current the effect. This is an empirical law like that for friction\u2014an experimentally observed phenomenon. Such a linear relationship doesn't always occur. Resistance and Simple Circuits If voltage drives current, what impedes it? The electric property that impedes current (crudely similar to friction and air resistance) is called resistance. Collisions of moving charges with atoms and molecules in a substance transfer energy to the substance and limit current. Resistance is defined as inversely proportional to current, or \u221d 1. (20.13) Thus, for example, current is cut in half if resistance doubles. Combining the relationships of current to voltage and current to resistance gives This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 20 | Electric Current, Resistance, and Ohm's Law =. 875 (20.14) This relationship is also called Ohm's law. Ohm's law in this form really defines resistance for certain materials. Ohm's law (like Hooke's law) is not universally valid. The many substances for which Ohm's law holds are called ohmic. These include good conductors like copper and aluminum, and some poor conductors under certain circumstances. Ohmic materials have a resistance that is independent of voltage and current. An object that has simple resistance is called a resistor, even if its resistance is small. The unit for resistance is an ohm and is given the symbol \u03a9 (upper case Greek omega). Rearranging = gives =, and so the units of resistance are 1 ohm = 1 volt per ampere: Figure 20.8 shows the schematic for a simple circuit. A simple circuit has a single voltage source and a single resistor. The wires connecting the voltage source to the resistor can be assumed to have negligible resistance, or their resistance can be included in. 1 \u03a9 = 1. (20.15) Figure 20.8 A simple electric circuit in which a closed path for current to flow is supplied by conductors (usually metal wires) connecting a load to the terminals of a battery, represented by the red parallel lines. The zigzag symbol represents the single resistor and includes any resistance in the connections to the voltage source. Making Connections: Real World Connections Ohm's law ( = ) is a fundamental relationship that could be presented by a linear function with the slope of the line being the resistance. The resistance represents the voltage that needs to be applied to the resistor to create a current of 1 A through the circuit. The graph (in the figure below) shows this representation for two simple circuits with resistors that have different resistances and thus different slopes. Figure 20.9 The figure illustrates the relationship between current and voltage for two different resistors. The slope of the graph represents the resistance value, which is 2\u03a9 and 4\u03a9 for the two lines shown. 876 Chapter 20 | Electric Current, Resistance, and Ohm's Law Making Connections: Real World Connections The materials which follow Ohm's law by having a linear relationship between voltage and current are known as ohmic materials. On the other hand, some materials exhibit a nonlinear voltage-current relationship and hence are known as nonohmic materials. The figure below shows current voltage relationships for the two types of materials. Figure 20.10 The relationship between voltage and current for ohmic and non-ohmic materials are shown. (a) (b) Clearly the resistance of an ohmic material (shown in (a)) remains constant and can be calculated by finding the slope of the graph but that is not true for a non-ohmic material (shown in (b)). Example 20.4 Calculating Resistance: An Automobile Headlight What is the resistance of an automobile headlight through which 2.50 A flows when 12.0 V is applied to it? Strategy We can rearrange Ohm's law as stated by = and use it to find the resistance. Solution Rearranging = and substituting known values gives = = 12.", "0 V 2.50 A = 4.80 \u03a9. (20.16) Discussion This is a relatively small resistance, but it is larger than the cold resistance of the headlight. As we shall see in Resistance and Resistivity, resistance usually increases with temperature, and so the bulb has a lower resistance when it is first switched on and will draw considerably more current during its brief warm-up period. Resistances range over many orders of magnitude. Some ceramic insulators, such as those used to support power lines, have resistances of 1012 \u03a9 or more. A dry person may have a hand-to-foot resistance of 105 \u03a9, whereas the resistance of the human heart is about 103 \u03a9. A meter-long piece of large-diameter copper wire may have a resistance of 10\u22125 \u03a9, and superconductors have no resistance at all (they are non-ohmic). Resistance is related to the shape of an object and the material of which it is composed, as will be seen in Resistance and Resistivity. Additional insight is gained by solving = for, yielding = (20.17) This expression for can be interpreted as the voltage drop across a resistor produced by the current. The phrase drop is often used for this voltage. For instance, the headlight in Example 20.4 has an drop of 12.0 V. If voltage is measured at various points in a circuit, it will be seen to increase at the voltage source and decrease at the resistor. Voltage is similar to fluid pressure. The voltage source is like a pump, creating a pressure difference, causing current\u2014the flow of charge. The resistor is like a pipe that reduces pressure and limits flow because of its resistance. Conservation of energy has important consequences here. The voltage source supplies energy (causing an electric field and a current), and the resistor converts it to another form (such as thermal energy). In a simple circuit (one with a single simple resistor), the voltage supplied by the source equals the voltage drop across the resistor, since PE = \u0394, and the same flows through each. Thus the energy supplied by the voltage source and the energy converted by the resistor are equal. (See Figure 20.11.) This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 20 | Electric Current, Resistance, and Ohm's Law 877 Figure 20.11 The voltage drop across a resistor in a simple circuit equals the voltage output of the battery. Making Connections: Conservation of Energy In a simple electrical circuit, the sole resistor converts energy supplied by the source into another form. Conservation of energy is evidenced here by the fact that all of the energy supplied by the source is converted to another form by the resistor alone. We will find that conservation of energy has other important applications in circuits and is a powerful tool in circuit analysis. PhET Explorations: Ohm's Law See how the equation form of Ohm's law relates to a simple circuit. Adjust the voltage and resistance, and see the current change according to Ohm's law. The sizes of the symbols in the equation change to match the circuit diagram. Figure 20.12 Ohm's Law (http://cnx.org/content/m55356/1.2/ohms-law_en.jar) 20.3 Resistance and Resistivity Learning Objectives By the end of this section, you will be able to: \u2022 Explain the concept of resistivity. \u2022 Use resistivity to calculate the resistance of specified configurations of material. \u2022 Use the thermal coefficient of resistivity to calculate the change of resistance with temperature. The information presented in this section supports the following AP\u00ae learning objectives and science practices: \u2022 1.E.2.1 The student is able to choose and justify the selection of data needed to determine resistivity for a given material. (S.P. 4.1) \u2022 4.E.4.2 The student is able to design a plan for the collection of data to determine the effect of changing the geometry and/or materials on the resistance or capacitance of a circuit element and relate results to the basic properties of resistors and capacitors. (S.P. 4.1, 4.2) \u2022 4.E.4.3 The student is able to analyze data to determine the effect of changing the geometry and/or materials on the resistance or capacitance of a circuit element and relate results to the basic properties of resistors and capacitors. (S.P. 5.1) Material and Shape Dependence of Resistance The resistance of an object depends on its shape and the material of which it is composed. The cylindrical resistor in Figure 20.13 is easy to analyze, and, by so doing, we can gain insight into the resistance of more complicated shapes. As you might expect, the cylinder's electric resistance is directly proportional to its length, similar to the resistance of a pipe to fluid flow. The longer the cylinder, the", " more collisions charges will make with its atoms. The greater the diameter of the cylinder, the more current it can carry (again similar to the flow of fluid through a pipe). In fact, is inversely proportional to the cylinder's crosssectional area. 878 Chapter 20 | Electric Current, Resistance, and Ohm's Law Figure 20.13 A uniform cylinder of length and cross-sectional area. Its resistance to the flow of current is similar to the resistance posed by a pipe to fluid flow. The longer the cylinder, the greater its resistance. The larger its cross-sectional area, the smaller its resistance. For a given shape, the resistance depends on the material of which the object is composed. Different materials offer different resistance to the flow of charge. We define the resistivity of a substance so that the resistance of an object is directly proportional to. Resistivity is an intrinsic property of a material, independent of its shape or size. The resistance of a uniform cylinder of length, of cross-sectional area, and made of a material with resistivity, is =. (20.18) Table 20.1 gives representative values of. The materials listed in the table are separated into categories of conductors, semiconductors, and insulators, based on broad groupings of resistivities. Conductors have the smallest resistivities, and insulators have the largest; semiconductors have intermediate resistivities. Conductors have varying but large free charge densities, whereas most charges in insulators are bound to atoms and are not free to move. Semiconductors are intermediate, having far fewer free charges than conductors, but having properties that make the number of free charges depend strongly on the type and amount of impurities in the semiconductor. These unique properties of semiconductors are put to use in modern electronics, as will be explored in later chapters. This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 20 | Electric Current, Resistance, and Ohm's Law 879 Table 20.1 Resistivities of Various materials at 20\u00baC Material Conductors Silver Copper Gold Aluminum Tungsten Iron Platinum Steel Lead Resistivity \u03c1 ( \u03a9 \u22c5 m ) 1.59\u00d710\u22128 1.72\u00d710\u22128 2.44\u00d710\u22128 2.65\u00d710\u22128 5.6\u00d710\u22128 9.71\u00d710\u22128 10.6\u00d710\u22128 20\u00d710\u22128 22\u00d710\u22128 Manganin (Cu, Mn, Ni alloy) 44\u00d710\u22128 Constantan (Cu, Ni alloy) Mercury Nichrome (Ni, Fe, Cr alloy) 49\u00d710\u22128 96\u00d710\u22128 100\u00d710\u22128 Semiconductors[1] Carbon (pure) Carbon Germanium (pure) Germanium Silicon (pure) Silicon Insulators Amber Glass Lucite Mica Quartz (fused) Rubber (hard) Sulfur Teflon Wood 3.5\u00d7105 (3.5 \u2212 60)\u00d7105 600\u00d710\u22123 (1 \u2212 600)\u00d710\u22123 2300 0.1\u20132300 5\u00d71014 109 \u2212 1014 >1013 1011 \u2212 1015 75\u00d71016 1013 \u2212 1016 1015 >1013 108 \u2212 1014 880 Chapter 20 | Electric Current, Resistance, and Ohm's Law Example 20.5 Calculating Resistor Diameter: A Headlight Filament A car headlight filament is made of tungsten and has a cold resistance of 0.350 \u03a9. If the filament is a cylinder 4.00 cm long (it may be coiled to save space), what is its diameter? Strategy We can rearrange the equation = to find the cross-sectional area of the filament from the given information. Then its diameter can be found by assuming it has a circular cross-section. Solution The cross-sectional area, found by rearranging the expression for the resistance of a cylinder given in =, is Substituting the given values, and taking from Table 20.1, yields =. = (5.610\u20138 \u03a9 \u22c5 m)(4.0010\u20132 m) 0.350 \u03a9 = 6.4010\u20139 m2. The area of a circle is related to its diameter by = 2 4. Solving for the diameter, and substituting the value found for, gives 1 2 = 2 = 2 = 9.010\u20135 m. 6.4010\u20139 m2 3.14 1 2 (20.19) (20.20) (20.21) (20.22) Discussion The diameter is just under a tenth of a millimeter. It is quoted to only two digits, because is known to only two digits. Temperature Variation of Resistance The resistivity of all materials depends on temperature. Some even become superconductors (zero resistivity) at very low temperatures. (See Figure 20.14.) Conversely, the", " resistivity of conductors increases with increasing temperature. Since the atoms vibrate more rapidly and over larger distances at higher temperatures, the electrons moving through a metal make more collisions, effectively making the resistivity higher. Over relatively small temperature changes (about 100\u00baC or less), resistivity varies with temperature change \u0394 as expressed in the following equation = 0(1 + \u0394), (20.23) where 0 is the original resistivity and is the temperature coefficient of resistivity. (See the values of in Table 20.2 below.) For larger temperature changes, may vary or a nonlinear equation may be needed to find. Note that is positive for metals, meaning their resistivity increases with temperature. Some alloys have been developed specifically to have a small temperature dependence. Manganin (which is made of copper, manganese and nickel), for example, has close to zero (to three digits on the scale in Table 20.2), and so its resistivity varies only slightly with temperature. This is useful for making a temperature-independent resistance standard, for example. 1. Values depend strongly on amounts and types of impurities This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 20 | Electric Current, Resistance, and Ohm's Law 881 Figure 20.14 The resistance of a sample of mercury is zero at very low temperatures\u2014it is a superconductor up to about 4.2 K. Above that critical temperature, its resistance makes a sudden jump and then increases nearly linearly with temperature. Table 20.2 Temperature Coefficients of Resistivity Material Conductors Silver Copper Gold Aluminum Tungsten Iron Platinum Lead Coefficient \u03b1 (1/\u00b0C)[2] 3.8\u00d710\u22123 3.9\u00d710\u22123 3.4\u00d710\u22123 3.9\u00d710\u22123 4.5\u00d710\u22123 5.0\u00d710\u22123 3.93\u00d710\u22123 4.3\u00d710\u22123 Manganin (Cu, Mn, Ni alloy) 0.000\u00d710\u22123 Constantan (Cu, Ni alloy) Mercury Nichrome (Ni, Fe, Cr alloy) Semiconductors Carbon (pure) Germanium (pure) Silicon (pure) 0.002\u00d710\u22123 0.89\u00d710\u22123 0.4\u00d710\u22123 \u22120.5\u00d710\u22123 \u221250\u00d710\u22123 \u221270\u00d710\u22123 Note also that is negative for the semiconductors listed in Table 20.2, meaning that their resistivity decreases with increasing temperature. They become better conductors at higher temperature, because increased thermal agitation increases the number of free charges available to carry current. This property of decreasing with temperature is also related to the type and amount of impurities present in the semiconductors. 2. Values at 20\u00b0C. 882 Chapter 20 | Electric Current, Resistance, and Ohm's Law The resistance of an object also depends on temperature, since 0 is directly proportional to. For a cylinder we know = /, and so, if and do not change greatly with temperature, will have the same temperature dependence as. (Examination of the coefficients of linear expansion shows them to be about two orders of magnitude less than typical temperature coefficients of resistivity, and so the effect of temperature on and is about two orders of magnitude less than on.) Thus, = 0(1 + \u0394) (20.24) is the temperature dependence of the resistance of an object, where 0 is the original resistance and is the resistance after a temperature change \u0394. Numerous thermometers are based on the effect of temperature on resistance. (See Figure 20.15.) One of the most common is the thermistor, a semiconductor crystal with a strong temperature dependence, the resistance of which is measured to obtain its temperature. The device is small, so that it quickly comes into thermal equilibrium with the part of a person it touches. Figure 20.15 These familiar thermometers are based on the automated measurement of a thermistor's temperature-dependent resistance. (credit: Biol, Wikimedia Commons) Example 20.6 Calculating Resistance: Hot-Filament Resistance Although caution must be used in applying = 0(1 + \u0394) and = 0(1 + \u0394) for temperature changes greater than 100\u00baC, for tungsten the equations work reasonably well for very large temperature changes. What, then, is the resistance of the tungsten filament in the previous example if its temperature is increased from room temperature ( 20\u00baC ) to a typical operating temperature of 2850\u00baC? Strategy This is a straightforward application of = 0(1 + \u0394), since the original resistance of the filament was given to be 0 = 0.350 \u03a9, and the temperature change is \u0394 = 2830\u00baC. Solution The hot resistance is obtained by entering known values into the above equation: = 0(1 + \u0394) = (0.350 \u03a9)[1 + (4.510\u20133", " / \u00baC)(2830\u00baC)] = 4.8 \u03a9. (20.25) Discussion This value is consistent with the headlight resistance example in Ohm's Law: Resistance and Simple Circuits. PhET Explorations: Resistance in a Wire Learn about the physics of resistance in a wire. Change its resistivity, length, and area to see how they affect the wire's resistance. The sizes of the symbols in the equation change along with the diagram of a wire. Figure 20.16 Resistance in a Wire (http://cnx.org/content/m55357/1.2/resistance-in-a-wire_en.jar) This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 20 | Electric Current, Resistance, and Ohm's Law 883 Applying the Science Practices: Examining Resistance Using the PhET Simulation \u201cResistance in a Wire\u201d, design an experiment to determine how different variables \u2013 resistivity, length, and area \u2013 affect the resistance of a resistor. For each variable, you should record your results in a table and then create a graph to determine the relationship. 20.4 Electric Power and Energy Learning Objectives By the end of this section, you will be able to: \u2022 Calculate the power dissipated by a resistor and the power supplied by a power supply. \u2022 Calculate the cost of electricity under various circumstances. The information presented in this section supports the following AP\u00ae learning objectives and science practices: \u2022 5.B.9.8 The student is able to translate between graphical and symbolic representations of experimental data describing relationships among power, current, and potential difference across a resistor. (S.P. 1.5) Power in Electric Circuits Power is associated by many people with electricity. Knowing that power is the rate of energy use or energy conversion, what is the expression for electric power? Power transmission lines might come to mind. We also think of lightbulbs in terms of their power ratings in watts. Let us compare a 25-W bulb with a 60-W bulb. (See Figure 20.17(a).) Since both operate on the same voltage, the 60-W bulb must draw more current to have a greater power rating. Thus the 60-W bulb's resistance must be lower than that of a 25-W bulb. If we increase voltage, we also increase power. For example, when a 25-W bulb that is designed to operate on 120 V is connected to 240 V, it briefly glows very brightly and then burns out. Precisely how are voltage, current, and resistance related to electric power? Figure 20.17 (a) Which of these lightbulbs, the 25-W bulb (upper left) or the 60-W bulb (upper right), has the higher resistance? Which draws more current? Which uses the most energy? Can you tell from the color that the 25-W filament is cooler? Is the brighter bulb a different color and if so why? (credits: Dickbauch, Wikimedia Commons; Greg Westfall, Flickr) (b) This compact fluorescent light (CFL) puts out the same intensity of light as the 60-W bulb, but at 1/4 to 1/10 the input power. (credit: dbgg1979, Flickr) Electric energy depends on both the voltage involved and the charge moved. This is expressed most simply as PE =, where is the charge moved and is the voltage (or more precisely, the potential difference the charge moves through). Power is the rate at which energy is moved, and so electric power is = =. (20.26) Recognizing that current is = / (note that \u0394 = here), the expression for power becomes = (20.27) 884 Chapter 20 | Electric Current, Resistance, and Ohm's Law Electric power ( ) is simply the product of current times voltage. Power has familiar units of watts. Since the SI unit for potential energy (PE) is the joule, power has units of joules per second, or watts. Thus, 1 A \u22c5 V = 1 W. For example, cars often have one or more auxiliary power outlets with which you can charge a cell phone or other electronic devices. These outlets may be rated at 20 A, so that the circuit can deliver a maximum power = = (20 A)(12 V) = 240 W. In some applications, electric power may be expressed as volt-amperes or even kilovolt-amperes ( 1 kA \u22c5 V = 1 kW ). To see the relationship of power to resistance, we combine Ohm's law with =. Substituting = gives = ( / ) = 2 /. Similarly, substituting = gives = () = 2. Three expressions for electric power are listed together here for convenience: = = 2 = 2. Note that the first equation is always valid, whereas", " the other two can be used only for resistors. In a simple circuit, with one voltage source and a single resistor, the power supplied by the voltage source and that dissipated by the resistor are identical. (In more complicated circuits, can be the power dissipated by a single device and not the total power in the circuit.) (20.28) (20.29) (20.30) Making Connections: Using Graphs to Calculate Resistance As \u221d 2 and \u221d 2, the graph for power versus current or voltage is quadratic. An example is shown in the figure below. Figure 20.18 The figure shows (a) power versus current and (b) power versus voltage relationships for simple resistor circuits. (a) (b) Using equations (20.29) and (20.30), we can calculate the resistance in each case. In graph (a), the power is 50 W when current is 5 A; hence, the resistance can be calculated as = / 2 = 50/ 52 = 2 \u03a9. Similarly, the resistance value can be calculated in graph (b) as = 2/ = 102/ 50 = 2 \u03a9 Different insights can be gained from the three different expressions for electric power. For example, = 2 / implies that the lower the resistance connected to a given voltage source, the greater the power delivered. Furthermore, since voltage is squared in = 2 /, the effect of applying a higher voltage is perhaps greater than expected. Thus, when the voltage is doubled to a 25-W bulb, its power nearly quadruples to about 100 W, burning it out. If the bulb's resistance remained constant, its power would be exactly 100 W, but at the higher temperature its resistance is higher, too. Example 20.7 Calculating Power Dissipation and Current: Hot and Cold Power (a) Consider the examples given in Ohm's Law: Resistance and Simple Circuits and Resistance and Resistivity. Then find the power dissipated by the car headlight in these examples, both when it is hot and when it is cold. (b) What current does it draw when cold? Strategy for (a) For the hot headlight, we know voltage and current, so we can use = to find the power. For the cold headlight, we know the voltage and resistance, so we can use = 2 / to find the power. Solution for (a) Entering the known values of current and voltage for the hot headlight, we obtain This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 20 | Electric Current, Resistance, and Ohm's Law The cold resistance was 0.350 \u03a9, and so the power it uses when first switched on is = = (2.50 A)(12.0 V) = 30.0 W. = 2 = (12.0 V)2 0.350 \u03a9 = 411 W. 885 (20.31) (20.32) Discussion for (a) The 30 W dissipated by the hot headlight is typical. But the 411 W when cold is surprisingly higher. The initial power quickly decreases as the bulb's temperature increases and its resistance increases. Strategy and Solution for (b) The current when the bulb is cold can be found several different ways. We rearrange one of the power equations, = 2, and enter known values, obtaining = = 411 W 0.350 \u03a9 = 34.3 A. (20.33) Discussion for (b) The cold current is remarkably higher than the steady-state value of 2.50 A, but the current will quickly decline to that value as the bulb's temperature increases. Most fuses and circuit breakers (used to limit the current in a circuit) are designed to tolerate very high currents briefly as a device comes on. In some cases, such as with electric motors, the current remains high for several seconds, necessitating special \u201cslow blow\u201d fuses. The Cost of Electricity The more electric appliances you use and the longer they are left on, the higher your electric bill. This familiar fact is based on the relationship between energy and power. You pay for the energy used. Since = /, we see that = (20.34) is the energy used by a device using power for a time interval. For example, the more lightbulbs burning, the greater used; the longer they are on, the greater is. The energy unit on electric bills is the kilowatt-hour ( kW \u22c5 h ), consistent with the relationship =. It is easy to estimate the cost of operating electric appliances if you have some idea of their power consumption rate in watts or kilowatts, the time they are on in hours, and the cost per kilowatt-hour for your electric utility. Kilowatt-hours, like all other specialized energy units such as food calories, can be converted to joules. You can prove to yourself that", " 1 kW \u22c5 h = 3.6106 J. The electrical energy ( ) used can be reduced either by reducing the time of use or by reducing the power consumption of that appliance or fixture. This will not only reduce the cost, but it will also result in a reduced impact on the environment. Improvements to lighting are some of the fastest ways to reduce the electrical energy used in a home or business. About 20% of a home's use of energy goes to lighting, while the number for commercial establishments is closer to 40%. Fluorescent lights are about four times more efficient than incandescent lights\u2014this is true for both the long tubes and the compact fluorescent lights (CFL). (See Figure 20.17(b).) Thus, a 60-W incandescent bulb can be replaced by a 15-W CFL, which has the same brightness and color. CFLs have a bent tube inside a globe or a spiral-shaped tube, all connected to a standard screw-in base that fits standard incandescent light sockets. (Original problems with color, flicker, shape, and high initial investment for CFLs have been addressed in recent years.) The heat transfer from these CFLs is less, and they last up to 10 times longer. The significance of an investment in such bulbs is addressed in the next example. New white LED lights (which are clusters of small LED bulbs) are even more efficient (twice that of CFLs) and last 5 times longer than CFLs. However, their cost is still high. Making Connections: Energy, Power, and Time The relationship = is one that you will find useful in many different contexts. The energy your body uses in exercise is related to the power level and duration of your activity, for example. The amount of heating by a power source is related to the power level and time it is applied. Even the radiation dose of an X-ray image is related to the power and time of exposure. Example 20.8 Calculating the Cost Effectiveness of Compact Fluorescent Lights (CFL) If the cost of electricity in your area is 12 cents per kWh, what is the total cost (capital plus operation) of using a 60-W incandescent bulb for 1000 hours (the lifetime of that bulb) if the bulb cost 25 cents? (b) If we replace this bulb with a compact fluorescent light that provides the same light output, but at one-quarter the wattage, and which costs $1.50 but lasts 10 times longer (10,000 hours), what will that total cost be? Strategy 886 Chapter 20 | Electric Current, Resistance, and Ohm's Law To find the operating cost, we first find the energy used in kilowatt-hours and then multiply by the cost per kilowatt-hour. Solution for (a) The energy used in kilowatt-hours is found by entering the power and time into the expression for energy: = = (60 W)(1000 h) = 60,000 W \u22c5 h. In kilowatt-hours, this is Now the electricity cost is = 60.0 kW \u22c5 h. cost = (60.0 kW \u22c5 h)($0.12/kW \u22c5 h) = $7.20. The total cost will be $7.20 for 1000 hours (about one-half year at 5 hours per day). Solution for (b) (20.35) (20.36) (20.37) Since the CFL uses only 15 W and not 60 W, the electricity cost will be $7.20/4 = $1.80. The CFL will last 10 times longer than the incandescent, so that the investment cost will be 1/10 of the bulb cost for that time period of use, or 0.1($1.50) = $0.15. Therefore, the total cost will be $1.95 for 1000 hours. Discussion Therefore, it is much cheaper to use the CFLs, even though the initial investment is higher. The increased cost of labor that a business must include for replacing the incandescent bulbs more often has not been figured in here. Making Connections: Take-Home Experiment\u2014Electrical Energy Use Inventory 1) Make a list of the power ratings on a range of appliances in your home or room. Explain why something like a toaster has a higher rating than a digital clock. Estimate the energy consumed by these appliances in an average day (by estimating their time of use). Some appliances might only state the operating current. If the household voltage is 120 V, then use =. 2) Check out the total wattage used in the rest rooms of your school's floor or building. (You might need to assume the long fluorescent lights in use are rated at 32 W.) Suppose that the building was closed all weekend and that these lights were left on from 6 p.m. Friday until 8 a.m. Monday. What would this oversight cost? How", " about for an entire year of weekends? 20.5 Alternating Current versus Direct Current Learning Objectives By the end of this section, you will be able to: \u2022 Explain the differences and similarities between AC and DC current. \u2022 Calculate rms voltage, current, and average power. \u2022 Explain why AC current is used for power transmission. Alternating Current Most of the examples dealt with so far, and particularly those utilizing batteries, have constant voltage sources. Once the current is established, it is thus also a constant. Direct current (DC) is the flow of electric charge in only one direction. It is the steady state of a constant-voltage circuit. Most well-known applications, however, use a time-varying voltage source. Alternating current (AC) is the flow of electric charge that periodically reverses direction. If the source varies periodically, particularly sinusoidally, the circuit is known as an alternating current circuit. Examples include the commercial and residential power that serves so many of our needs. Figure 20.19 shows graphs of voltage and current versus time for typical DC and AC power. The AC voltages and frequencies commonly used in homes and businesses vary around the world. This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 20 | Electric Current, Resistance, and Ohm's Law 901 Figure 20.38 This NASA scientist and NEEMO 5 aquanaut's heart rate and other vital signs are being recorded by a portable device while living in an underwater habitat. (credit: NASA, Life Sciences Data Archive at Johnson Space Center, Houston, Texas) PhET Explorations: Neuron Figure 20.39 Neuron (http://cnx.org/content/m55361/1.2/neuron_en.jar) Stimulate a neuron and monitor what happens. Pause, rewind, and move forward in time in order to observe the ions as they move across the neuron membrane. Glossary AC current: current that fluctuates sinusoidally with time, expressed as I = I0 sin 2\u03c0ft, where I is the current at time t, I0 is the peak current, and f is the frequency in hertz AC voltage: voltage that fluctuates sinusoidally with time, expressed as V = V0 sin 2\u03c0ft, where V is the voltage at time t, V0 is the peak voltage, and f is the frequency in hertz alternating current: (AC) the flow of electric charge that periodically reverses direction ampere: (amp) the SI unit for current; 1 A = 1 C/s bioelectricity: electrical effects in and created by biological systems direct current: (DC) the flow of electric charge in only one direction drift velocity: the average velocity at which free charges flow in response to an electric field electric current: the rate at which charge flows, I = \u0394Q/\u0394t electric power: the rate at which electrical energy is supplied by a source or dissipated by a device; it is the product of current times voltage electrocardiogram (ECG): especially in the heart usually abbreviated ECG, a record of voltages created by depolarization and repolarization, microshock sensitive: a condition in which a person's skin resistance is bypassed, possibly by a medical procedure, rendering the person vulnerable to electrical shock at currents about 1/1000 the normally required level nerve conduction: the transport of electrical signals by nerve cells ohm: the unit of resistance, given by 1\u03a9 = 1 V/A Ohm's law: an empirical relation stating that the current I is proportional to the potential difference V, \u221d V; it is often written as I = V/R, where R is the resistance ohmic: a type of a material for which Ohm's law is valid resistance: the electric property that impedes current; for ohmic materials, it is the ratio of voltage to current, R = V/I 902 resistivity: by \u03c1 an intrinsic property of a material, independent of its shape or size, directly proportional to the resistance, denoted Chapter 20 | Electric Current, Resistance, and Ohm's Law rms current: the root mean square of the current, rms = 0 / 2, where I0 is the peak current, in an AC system rms voltage: the root mean square of the voltage, rms = 0 / 2, where V0 is the peak voltage, in an AC system semipermeable: property of a membrane that allows only certain types of ions to cross it shock hazard: when electric current passes through a person short circuit: also known as a \u201cshort,\u201d a low-resistance path between terminals of a voltage source simple circuit: a circuit with a single voltage source and a single resistor temperature coefficient of resistivity: an empirical quantity, denoted by \u03b1, which describes the change in resistance or resistivity of a", " material with temperature thermal hazard: a hazard in which electric current causes undesired thermal effects Section Summary 20.1 Current \u2022 Electric current is the rate at which charge flows, given by = \u0394 \u0394, where \u0394 is the amount of charge passing through an area in time \u0394. \u2022 The direction of conventional current is taken as the direction in which positive charge moves. \u2022 The SI unit for current is the ampere (A), where 1 A = 1 C/s. \u2022 Current is the flow of free charges, such as electrons and ions. \u2022 Drift velocity d is the average speed at which these charges move. \u2022 Current is proportional to drift velocity d, as expressed in the relationship = d. Here, is the current through a wire of cross-sectional area. The wire's material has a free-charge density, and each carrier has charge and a drift velocity d. \u2022 Electrical signals travel at speeds about 1012 times greater than the drift velocity of free electrons. 20.2 Ohm\u2019s Law: Resistance and Simple Circuits \u2022 A simple circuit is one in which there is a single voltage source and a single resistance. \u2022 One statement of Ohm's law gives the relationship between current, voltage, and resistance in a simple circuit to be =. \u2022 Resistance has units of ohms ( \u03a9 ), related to volts and amperes by 1 \u03a9 = 1 V/A. \u2022 There is a voltage or drop across a resistor, caused by the current flowing through it, given by =. 20.3 Resistance and Resistivity \u2022 The resistance of a cylinder of length and cross-sectional area is =, where is the resistivity of the material. \u2022 Values of in Table 20.1 show that materials fall into three groups\u2014conductors, semiconductors, and insulators. \u2022 Temperature affects resistivity; for relatively small temperature changes \u0394, resistivity is = 0(1 + \u0394), where 0 is the original resistivity and \u03b1 is the temperature coefficient of resistivity. \u2022 Table 20.2 gives values for, the temperature coefficient of resistivity. \u2022 The resistance of an object also varies with temperature: = 0(1 + \u0394), where 0 is the original resistance, and is the resistance after the temperature change. 20.4 Electric Power and Energy \u2022 Electric power is the rate (in watts) that energy is supplied by a source or dissipated by a device. This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 20 | Electric Current, Resistance, and Ohm's Law 903 \u2022 Three expressions for electrical power are = = 2, and \u2022 The energy used by a device with a power over a time is =. = 2. 20.5 Alternating Current versus Direct Current \u2022 Direct current (DC) is the flow of electric current in only one direction. It refers to systems where the source voltage is constant. \u2022 The voltage source of an alternating current (AC) system puts out = 0 sin 2, where is the voltage at time, \u2022 0 is the peak voltage, and In a simple circuit, = and AC current is = 0 sin 2, where is the current at time, and 0 = 0 is the peak current. is the frequency in hertz. \u2022 The average AC power is ave = 1 20 0. \u2022 Average (rms) current rms and average (rms) voltage rms are rms = 0 2 and rms = 0 2, where rms stands for root mean square. \u2022 Thus, ave = rmsrms. \u2022 Ohm's law for AC is rms = rms. \u2022 Expressions for the average power of an AC circuit are ave = rmsrms, ave = rms 2, and ave = rms 2, analogous to the expressions for DC circuits. 20.6 Electric Hazards and the Human Body \u2022 The two types of electric hazards are thermal (excessive power) and shock (current through a person). \u2022 Shock severity is determined by current, path, duration, and AC frequency. \u2022 Table 20.3 lists shock hazards as a function of current. \u2022 Figure 20.28 graphs the threshold current for two hazards as a function of frequency. 20.7 Nerve Conduction\u2013Electrocardiograms \u2022 Electric potentials in neurons and other cells are created by ionic concentration differences across semipermeable membranes. \u2022 Stimuli change the permeability and create action potentials that propagate along neurons. \u2022 Myelin sheaths speed this process and reduce the needed energy input. \u2022 This process in the heart can be measured with an electrocardiogram (ECG). Conceptual Questions 20.1 Current 1. Can a wire carry a current and still be neutral\u2014that is, have a total charge of zero? Explain. 2. Car batteries are rated in ampere-hours ( A \u22c5 h ). To what physical quantity do ampere-hours correspond (voltage, charge", ",...), and what relationship do ampere-hours have to energy content? 3. If two different wires having identical cross-sectional areas carry the same current, will the drift velocity be higher or lower in the better conductor? Explain in terms of the equation d =, by considering how the density of charge carriers relates to whether or not a material is a good conductor. 4. Why are two conducting paths from a voltage source to an electrical device needed to operate the device? 5. In cars, one battery terminal is connected to the metal body. How does this allow a single wire to supply current to electrical devices rather than two wires? 6. Why isn't a bird sitting on a high-voltage power line electrocuted? Contrast this with the situation in which a large bird hits two wires simultaneously with its wings. 20.2 Ohm\u2019s Law: Resistance and Simple Circuits 904 Chapter 20 | Electric Current, Resistance, and Ohm's Law 7. The drop across a resistor means that there is a change in potential or voltage across the resistor. Is there any change in current as it passes through a resistor? Explain. 8. How is the drop in a resistor similar to the pressure drop in a fluid flowing through a pipe? 20.3 Resistance and Resistivity 9. In which of the three semiconducting materials listed in Table 20.1 do impurities supply free charges? (Hint: Examine the range of resistivity for each and determine whether the pure semiconductor has the higher or lower conductivity.) 10. Does the resistance of an object depend on the path current takes through it? Consider, for example, a rectangular bar\u2014is its resistance the same along its length as across its width? (See Figure 20.40.) Figure 20.40 Does current taking two different paths through the same object encounter different resistance? 11. If aluminum and copper wires of the same length have the same resistance, which has the larger diameter? Why? 12. Explain why = 0(1 + \u0394) for the temperature variation of the resistance of an object is not as accurate as = 0(1 + \u0394), which gives the temperature variation of resistivity. 20.4 Electric Power and Energy 13. Why do incandescent lightbulbs grow dim late in their lives, particularly just before their filaments break? 14. The power dissipated in a resistor is given by = 2 /, which means power decreases if resistance increases. Yet this power is also given by = 2, which means power increases if resistance increases. Explain why there is no contradiction here. 20.5 Alternating Current versus Direct Current 15. Give an example of a use of AC power other than in the household. Similarly, give an example of a use of DC power other than that supplied by batteries. 16. Why do voltage, current, and power go through zero 120 times per second for 60-Hz AC electricity? 17. You are riding in a train, gazing into the distance through its window. As close objects streak by, you notice that the nearby fluorescent lights make dashed streaks. Explain. 20.6 Electric Hazards and the Human Body 18. Using an ohmmeter, a student measures the resistance between various points on his body. He finds that the resistance between two points on the same finger is about the same as the resistance between two points on opposite hands\u2014both are several hundred thousand ohms. Furthermore, the resistance decreases when more skin is brought into contact with the probes of the ohmmeter. Finally, there is a dramatic drop in resistance (to a few thousand ohms) when the skin is wet. Explain these observations and their implications regarding skin and internal resistance of the human body. 19. What are the two major hazards of electricity? 20. Why isn't a short circuit a shock hazard? 21. What determines the severity of a shock? Can you say that a certain voltage is hazardous without further information? 22. An electrified needle is used to burn off warts, with the circuit being completed by having the patient sit on a large butt plate. Why is this plate large? 23. Some surgery is performed with high-voltage electricity passing from a metal scalpel through the tissue being cut. Considering the nature of electric fields at the surface of conductors, why would you expect most of the current to flow from the sharp edge of the scalpel? Do you think high- or low-frequency AC is used? 24. Some devices often used in bathrooms, such as hairdryers, often have safety messages saying \u201cDo not use when the bathtub or basin is full of water.\u201d Why is this so? 25. We are often advised to not flick electric switches with wet hands, dry your hand first. We are also advised to never throw water on an electric fire. Why is this so? 26. Before working on a power transmission line, linemen will touch the line with the back of the hand as a final check that the voltage is zero. Why", " the back of the hand? This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 20 | Electric Current, Resistance, and Ohm's Law 905 27. Why is the resistance of wet skin so much smaller than dry, and why do blood and other bodily fluids have low resistances? 28. Could a person on intravenous infusion (an IV) be microshock sensitive? 29. In view of the small currents that cause shock hazards and the larger currents that circuit breakers and fuses interrupt, how do they play a role in preventing shock hazards? 20.7 Nerve Conduction\u2013Electrocardiograms 30. Note that in Figure 20.31, both the concentration gradient and the Coulomb force tend to move Na+ prevents this? ions into the cell. What 31. Define depolarization, repolarization, and the action potential. 32. Explain the properties of myelinated nerves in terms of the insulating properties of myelin. 906 Chapter 20 | Electric Current, Resistance, and Ohm's Law Problems & Exercises 20.1 Current 1. What is the current in milliamperes produced by the solar cells of a pocket calculator through which 4.00 C of charge passes in 4.00 h? 2. A total of 600 C of charge passes through a flashlight in 0.500 h. What is the average current? 3. What is the current when a typical static charge of 0.250 C moves from your finger to a metal doorknob in 1.00 s? 4. Find the current when 2.00 nC jumps between your comb and hair over a 0.500 - s time interval. 5. A large lightning bolt had a 20,000-A current and moved 30.0 C of charge. What was its duration? 6. The 200-A current through a spark plug moves 0.300 mC of charge. How long does the spark last? 7. (a) A defibrillator sends a 6.00-A current through the chest of a patient by applying a 10,000-V potential as in the figure below. What is the resistance of the path? (b) The defibrillator paddles make contact with the patient through a conducting gel that greatly reduces the path resistance. Discuss the difficulties that would ensue if a larger voltage were used to produce the same current through the patient, but with the path having perhaps 50 times the resistance. (Hint: The current must be about the same, so a higher voltage would imply greater power. Use this equation for power: = 2.) Figure 20.41 The capacitor in a defibrillation unit drives a current through the heart of a patient. 8. During open-heart surgery, a defibrillator can be used to bring a patient out of cardiac arrest. The resistance of the path is 500 \u03a9 and a 10.0-mA current is needed. What voltage should be applied? 9. (a) A defibrillator passes 12.0 A of current through the torso of a person for 0.0100 s. How much charge moves? (b) How many electrons pass through the wires connected to the patient? (See figure two problems earlier.) 10. A clock battery wears out after moving 10,000 C of charge through the clock at a rate of 0.500 mA. (a) How long did the clock run? (b) How many electrons per second flowed? This content is available for free at http://cnx.org/content/col11844/1.13 11. The batteries of a submerged non-nuclear submarine supply 1000 A at full speed ahead. How long does it take to move Avogadro's number ( 6.021023 rate? ) of electrons at this 12. Electron guns are used in X-ray tubes. The electrons are accelerated through a relatively large voltage and directed onto a metal target, producing X-rays. (a) How many electrons per second strike the target if the current is 0.500 mA? (b) What charge strikes the target in 0.750 s? 13. A large cyclotron directs a beam of He++ target with a beam current of 0.250 mA. (a) How many He++ nuclei per second is this? (b) How long does it take for 1.00 C to strike the target? (c) How long before 1.00 mol of He++ nuclei strike the target? nuclei onto a 14. Repeat the above example on Example 20.3, but for a wire made of silver and given there is one free electron per silver atom. 15. Using the results of the above example on Example 20.3, find the drift velocity in a copper wire of twice the diameter and carrying 20.0 A. 16. A 14-gauge copper wire has a diameter of 1.628 mm.", " What magnitude current flows when the drift velocity is 1.00 mm/s? (See above example on Example 20.3 for useful information.) 17. SPEAR, a storage ring about 72.0 m in diameter at the Stanford Linear Accelerator (closed in 2009), has a 20.0-A circulating beam of electrons that are moving at nearly the speed of light. (See Figure 20.42.) How many electrons are in the beam? Figure 20.42 Electrons circulating in the storage ring called SPEAR constitute a 20.0-A current. Because they travel close to the speed of light, each electron completes many orbits in each second. 20.2 Ohm\u2019s Law: Resistance and Simple Circuits 18. What current flows through the bulb of a 3.00-V flashlight when its hot resistance is 3.60 \u03a9? 19. Calculate the effective resistance of a pocket calculator that has a 1.35-V battery and through which 0.200 mA flows. 20. What is the effective resistance of a car's starter motor when 150 A flows through it as the car battery applies 11.0 V to the motor? 21. How many volts are supplied to operate an indicator light on a DVD player that has a resistance of 140 \u03a9, given that 25.0 mA passes through it? 22. (a) Find the voltage drop in an extension cord having a 0.0600- \u03a9 resistance and through which 5.00 A is flowing. (b) A cheaper cord utilizes thinner wire and has a resistance of 0.300 \u03a9. What is the voltage drop in it when 5.00 A Chapter 20 | Electric Current, Resistance, and Ohm's Law 907 flows? (c) Why is the voltage to whatever appliance is being used reduced by this amount? What is the effect on the appliance? 23. A power transmission line is hung from metal towers with glass insulators having a resistance of 1.00109 \u03a9. What current flows through the insulator if the voltage is 200 kV? (Some high-voltage lines are DC.) low temperatures. Discuss why and whether this is the case here. (Hint: Resistance can't become negative.) 38. Integrated Concepts (a) Redo Exercise 20.25 taking into account the thermal expansion of the tungsten filament. You may assume a thermal expansion coefficient of 1210\u22126 / \u00baC. (b) By what percentage does your answer differ from that in the example? 20.3 Resistance and Resistivity 39. Unreasonable Results 24. What is the resistance of a 20.0-m-long piece of 12-gauge copper wire having a 2.053-mm diameter? 25. The diameter of 0-gauge copper wire is 8.252 mm. Find the resistance of a 1.00-km length of such wire used for power transmission. 26. If the 0.100-mm diameter tungsten filament in a light bulb is to have a resistance of 0.200 \u03a9 at 20.0\u00baC, how long should it be? 27. Find the ratio of the diameter of aluminum to copper wire, if they have the same resistance per unit length (as they might in household wiring). 28. What current flows through a 2.54-cm-diameter rod of pure silicon that is 20.0 cm long, when 1.00 \u00d7 103 V is applied to it? (Such a rod may be used to make nuclearparticle detectors, for example.) 29. (a) To what temperature must you raise a copper wire, originally at 20.0\u00baC, to double its resistance, neglecting any changes in dimensions? (b) Does this happen in household wiring under ordinary circumstances? 30. A resistor made of Nichrome wire is used in an application where its resistance cannot change more than 1.00% from its value at 20.0\u00baC. Over what temperature range can it be used? 31. Of what material is a resistor made if its resistance is 40.0% greater at 100\u00baC than at 20.0\u00baC? 32. An electronic device designed to operate at any temperature in the range from \u201310.0\u00baC to 55.0\u00baC contains pure carbon resistors. By what factor does their resistance increase over this range? 33. (a) Of what material is a wire made, if it is 25.0 m long with a 0.100 mm diameter and has a resistance of 77.7 \u03a9 at 20.0\u00baC? (b) What is its resistance at 150\u00baC? 34. Assuming a constant temperature coefficient of resistivity, what is the maximum percent decrease in the resistance of a constantan wire starting at 20.0\u00baC? 35. A wire is drawn through a die, stretching it to four times its original length. By what factor does its resistance increase? 36. A copper wire has a resistance of 0.500 \u03a9", " at 20.0\u00baC, and an iron wire has a resistance of 0.525 \u03a9 at the same temperature. At what temperature are their resistances equal? 37. (a) Digital medical thermometers determine temperature by measuring the resistance of a semiconductor device called a thermistor (which has = \u2013 0.0600 / \u00baC ) when it is at the same temperature as the patient. What is a patient's temperature if the thermistor's resistance at that temperature is 82.0% of its value at 37.0\u00baC (normal body temperature)? (b) The negative value for may not be maintained for very (a) To what temperature must you raise a resistor made of constantan to double its resistance, assuming a constant temperature coefficient of resistivity? (b) To cut it in half? (c) What is unreasonable about these results? (d) Which assumptions are unreasonable, or which premises are inconsistent? 20.4 Electric Power and Energy 40. What is the power of a 1.00\u00d7102 MV lightning bolt having a current of 2.00 \u00d7 104 A? 41. What power is supplied to the starter motor of a large truck that draws 250 A of current from a 24.0-V battery hookup? 42. A charge of 4.00 C of charge passes through a pocket calculator's solar cells in 4.00 h. What is the power output, given the calculator's voltage output is 3.00 V? (See Figure 20.43.) Figure 20.43 The strip of solar cells just above the keys of this calculator convert light to electricity to supply its energy needs. (credit: Evan-Amos, Wikimedia Commons) 43. How many watts does a flashlight that has 6.00\u00d7102 C pass through it in 0.500 h use if its voltage is 3.00 V? 44. Find the power dissipated in each of these extension cords: (a) an extension cord having a 0.0600 - \u03a9 resistance and through which 5.00 A is flowing; (b) a cheaper cord utilizing thinner wire and with a resistance of 0.300 \u03a9. 45. Verify that the units of a volt-ampere are watts, as implied by the equation =. 46. Show that the units 1 V2 / \u03a9 = 1W, as implied by the equation = 2 /. 47. Show that the units 1 A2 \u22c5 \u03a9 = 1 W, as implied by the equation = 2. 908 Chapter 20 | Electric Current, Resistance, and Ohm's Law minute? (b) How much water must you put into the vaporizer for 8.00 h of overnight operation? (See Figure 20.45.) 48. Verify the energy unit equivalence that 1 kW \u22c5 h = 3.60106 J. 49. Electrons in an X-ray tube are accelerated through 1.00\u00d7102 kV and directed toward a target to produce Xrays. Calculate the power of the electron beam in this tube if it has a current of 15.0 mA. 50. An electric water heater consumes 5.00 kW for 2.00 h per day. What is the cost of running it for one year if electricity costs 12.0 cents/kW \u22c5 h? See Figure 20.44. Figure 20.44 On-demand electric hot water heater. Heat is supplied to water only when needed. (credit: aviddavid, Flickr) 60. Integrated Concepts Figure 20.45 This cold vaporizer passes current directly through water, vaporizing it directly with relatively little temperature increase. (a) What energy is dissipated by a lightning bolt having a 20,000-A current, a voltage of 1.00\u00d7102 MV, and a length of 1.00 ms? (b) What mass of tree sap could be raised from 18.0\u00baC to its boiling point and then evaporated by this energy, assuming sap has the same thermal characteristics as water? 61. Integrated Concepts What current must be produced by a 12.0-V battery-operated bottle warmer in order to heat 75.0 g of glass, 250 g of baby formula, and 3.00\u00d7102 g of aluminum from 20.0\u00baC to 90.0\u00baC in 5.00 min? 62. Integrated Concepts How much time is needed for a surgical cauterizer to raise the temperature of 1.00 g of tissue from 37.0\u00baC to 100\u00baC and then boil away 0.500 g of water, if it puts out 2.00 mA at 15.0 kV? Ignore heat transfer to the surroundings. 63. Integrated Concepts Hydroelectric generators (see Figure 20.46) at Hoover Dam produce a maximum current of 8.00\u00d7103 A at 250 kV. (a) What is the power output? (b) The water that powers the generators enters and leaves the system at low speed (thus its kinetic energy does not change", ") but loses 160 m in altitude. How many cubic meters per second are needed, assuming 85.0% efficiency? 51. With a 1200-W toaster, how much electrical energy is needed to make a slice of toast (cooking time = 1 minute)? At 9.0 cents/kW \u00b7 h, how much does this cost? 52. What would be the maximum cost of a CFL such that the total cost (investment plus operating) would be the same for both CFL and incandescent 60-W bulbs? Assume the cost of the incandescent bulb is 25 cents and that electricity costs 10 cents/kWh. Calculate the cost for 1000 hours, as in the cost effectiveness of CFL example. 53. Some makes of older cars have 6.00-V electrical systems. (a) What is the hot resistance of a 30.0-W headlight in such a car? (b) What current flows through it? 54. Alkaline batteries have the advantage of putting out constant voltage until very nearly the end of their life. How long will an alkaline battery rated at 1.00 A \u22c5 h and 1.58 V keep a 1.00-W flashlight bulb burning? 55. A cauterizer, used to stop bleeding in surgery, puts out 2.00 mA at 15.0 kV. (a) What is its power output? (b) What is the resistance of the path? 56. The average television is said to be on 6 hours per day. Estimate the yearly cost of electricity to operate 100 million TVs, assuming their power consumption averages 150 W and the cost of electricity averages 12.0 cents/kW \u22c5 h. 57. An old lightbulb draws only 50.0 W, rather than its original 60.0 W, due to evaporative thinning of its filament. By what factor is its diameter reduced, assuming uniform thinning along its length? Neglect any effects caused by temperature differences. 58. 00-gauge copper wire has a diameter of 9.266 mm. Calculate the power loss in a kilometer of such wire when it carries 1.00\u00d7102 A. 59. Integrated Concepts Cold vaporizers pass a current through water, evaporating it with only a small increase in temperature. One such home device is rated at 3.50 A and utilizes 120 V AC with 95.0% efficiency. (a) What is the vaporization rate in grams per This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 20 | Electric Current, Resistance, and Ohm's Law 909 (a) An immersion heater utilizing 120 V can raise the temperature of a 1.00\u00d7102 -g aluminum cup containing 350 g of water from 20.0\u00baC to 95.0\u00baC in 2.00 min. Find its resistance, assuming it is constant during the process. (b) A lower resistance would shorten the heating time. Discuss the practical limits to speeding the heating by lowering the resistance. 68. Integrated Concepts (a) What is the cost of heating a hot tub containing 1500 kg of water from 10.0\u00baC to 40.0\u00baC, assuming 75.0% efficiency to account for heat transfer to the surroundings? The cost of electricity is 9 cents/kW \u22c5 h. (b) What current was used by the 220-V AC electric heater, if this took 4.00 h? 69. Unreasonable Results (a) What current is needed to transmit 1.00\u00d7102 MW of power at 480 V? (b) What power is dissipated by the transmission lines if they have a 1.00 - \u03a9 resistance? (c) What is unreasonable about this result? (d) Which assumptions are unreasonable, or which premises are inconsistent? 70. Unreasonable Results (a) What current is needed to transmit 1.00\u00d7102 MW of power at 10.0 kV? (b) Find the resistance of 1.00 km of wire that would cause a 0.0100% power loss. (c) What is the diameter of a 1.00-km-long copper wire having this resistance? (d) What is unreasonable about these results? (e) Which assumptions are unreasonable, or which premises are inconsistent? 71. Construct Your Own Problem Consider an electric immersion heater used to heat a cup of water to make tea. Construct a problem in which you calculate the needed resistance of the heater so that it increases the temperature of the water and cup in a reasonable amount of time. Also calculate the cost of the electrical energy used in your process. Among the things to be considered are the voltage used, the masses and heat capacities involved, heat losses, and the time over which the heating takes place. Your instructor may wish for you to consider a thermal safety switch (perhaps bimetallic) that will halt the process before damaging temperatures are reached in the immersion unit.", " 20.5 Alternating Current versus Direct Current 72. (a) What is the hot resistance of a 25-W light bulb that runs on 120-V AC? (b) If the bulb's operating temperature is 2700\u00baC, what is its resistance at 2600\u00baC? 73. Certain heavy industrial equipment uses AC power that has a peak voltage of 679 V. What is the rms voltage? 74. A certain circuit breaker trips when the rms current is 15.0 A. What is the corresponding peak current? 75. Military aircraft use 400-Hz AC power, because it is possible to design lighter-weight equipment at this higher frequency. What is the time for one complete cycle of this power? 76. A North American tourist takes his 25.0-W, 120-V AC razor to Europe, finds a special adapter, and plugs it into 240 V AC. Assuming constant resistance, what power does the razor consume as it is ruined? Figure 20.46 Hydroelectric generators at the Hoover dam. (credit: Jon Sullivan) 64. Integrated Concepts (a) Assuming 95.0% efficiency for the conversion of electrical power by the motor, what current must the 12.0-V batteries of a 750-kg electric car be able to supply: (a) To accelerate from rest to 25.0 m/s in 1.00 min? (b) To climb a 2.00\u00d7102 -m high hill in 2.00 min at a constant 25.0-m/s speed while exerting 5.00\u00d7102 N of force to overcome air resistance and friction? (c) To travel at a constant 25.0-m/s speed, exerting a 5.00\u00d7102 N force to overcome air resistance and friction? See Figure 20.47. Figure 20.47 This REVAi, an electric car, gets recharged on a street in London. (credit: Frank Hebbert) 65. Integrated Concepts A light-rail commuter train draws 630 A of 650-V DC electricity when accelerating. (a) What is its power consumption rate in kilowatts? (b) How long does it take to reach 20.0 m/s starting from rest if its loaded mass is 5.30104 kg, assuming 95.0% efficiency and constant power? (c) Find its average acceleration. (d) Discuss how the acceleration you found for the light-rail train compares to what might be typical for an automobile. 66. Integrated Concepts (a) An aluminum power transmission line has a resistance of 0.0580 \u03a9 / km. What is its mass per kilometer? (b) What is the mass per kilometer of a copper line having the same resistance? A lower resistance would shorten the heating time. Discuss the practical limits to speeding the heating by lowering the resistance. 67. Integrated Concepts 910 Chapter 20 | Electric Current, Resistance, and Ohm's Law 90. (a) During surgery, a current as small as 20.0 \u03bcA applied directly to the heart may cause ventricular fibrillation. If the resistance of the exposed heart is 300 \u03a9, what is the smallest voltage that poses this danger? (b) Does your answer imply that special electrical safety precautions are needed? 91. (a) What is the resistance of a 220-V AC short circuit that generates a peak power of 96.8 kW? (b) What would the average power be if the voltage was 120 V AC? 92. A heart defibrillator passes 10.0 A through a patient's torso for 5.00 ms in an attempt to restore normal beating. (a) How much charge passed? (b) What voltage was applied if 500 J of energy was dissipated? (c) What was the path's resistance? (d) Find the temperature increase caused in the 8.00 kg of affected tissue. 93. Integrated Concepts A short circuit in a 120-V appliance cord has a 0.500- \u03a9 resistance. Calculate the temperature rise of the 2.00 g of surrounding materials, assuming their specific heat capacity is 0.200 cal/g\u22c5\u00baC and that it takes 0.0500 s for a circuit breaker to interrupt the current. Is this likely to be damaging? 94. Construct Your Own Problem Consider a person working in an environment where electric currents might pass through her body. Construct a problem in which you calculate the resistance of insulation needed to protect the person from harm. Among the things to be considered are the voltage to which the person might be exposed, likely body resistance (dry, wet, \u2026), and acceptable currents (safe but sensed, safe and unfelt, \u2026). 20.7 Nerve Conduction\u2013Electrocardiograms 95. Integrated Concepts Use the ECG in Figure 20.37 to determine the heart rate in beats per minute assuming a constant time between beats. 96. Integrated Concepts (a) Referring to Figure 20.37, find the time systolic", " pressure lags behind the middle of the QRS complex. (b) Discuss the reasons for the time lag. 77. In this problem, you will verify statements made at the end of the power losses for Example 20.10. (a) What current is needed to transmit 100 MW of power at a voltage of 25.0 kV? (b) Find the power loss in a 1.00 - \u03a9 transmission line. (c) What percent loss does this represent? 78. A small office-building air conditioner operates on 408-V AC and consumes 50.0 kW. (a) What is its effective resistance? (b) What is the cost of running the air conditioner during a hot summer month when it is on 8.00 h per day for 30 days and electricity costs 9.00 cents/kW \u22c5 h? 79. What is the peak power consumption of a 120-V AC microwave oven that draws 10.0 A? 80. What is the peak current through a 500-W room heater that operates on 120-V AC power? 81. Two different electrical devices have the same power consumption, but one is meant to be operated on 120-V AC and the other on 240-V AC. (a) What is the ratio of their resistances? (b) What is the ratio of their currents? (c) Assuming its resistance is unaffected, by what factor will the power increase if a 120-V AC device is connected to 240-V AC? 82. Nichrome wire is used in some radiative heaters. (a) Find the resistance needed if the average power output is to be 1.00 kW utilizing 120-V AC. (b) What length of Nichrome wire, having a cross-sectional area of 5.00mm2, is needed if the operating temperature is 500\u00ba C? (c) What power will it draw when first switched on? 83. Find the time after = 0 when the instantaneous voltage of 60-Hz AC first reaches the following values: (a) 0 / 2 (b) 0 (c) 0. 84. (a) At what two times in the first period following = 0 does the instantaneous voltage in 60-Hz AC equal rms? (b) \u2212rms? 20.6 Electric Hazards and the Human Body 85. (a) How much power is dissipated in a short circuit of 240-V AC through a resistance of 0.250 \u03a9? (b) What current flows? 86. What voltage is involved in a 1.44-kW short circuit through a 0.100 - \u03a9 resistance? 87. Find the current through a person and identify the likely effect on her if she touches a 120-V AC source: (a) if she is standing on a rubber mat and offers a total resistance of 300 k \u03a9 ; (b) if she is standing barefoot on wet grass and has a resistance of only 4000 k \u03a9. 88. While taking a bath, a person touches the metal case of a radio. The path through the person to the drainpipe and ground has a resistance of 4000 \u03a9. What is the smallest voltage on the case of the radio that could cause ventricular fibrillation? 89. Foolishly trying to fish a burning piece of bread from a toaster with a metal butter knife, a man comes into contact with 120-V AC. He does not even feel it since, luckily, he is wearing rubber-soled shoes. What is the minimum resistance of the path the current follows through the person? This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 20 | Electric Current, Resistance, and Ohm's Law 911 Test Prep for AP\u00ae Courses 20.1 Current 1. Which of the following can be explained on the basis of conservation of charge in a closed circuit consisting of a battery, resistor, and metal wires? a. The number of electrons leaving the battery will be equal to the number of electrons entering the battery. b. The number of electrons leaving the battery will be less than the number of electrons entering the battery. c. The number of protons leaving the battery will be equal to the number of protons entering the battery. d. The number of protons leaving the battery will be less than the number of protons entering the battery. 2. When a battery is connected to a bulb, there is 2.5 A of current in the circuit. What amount of charge will flow though the circuit in a time of 0.5 s? a. 0.5 C b. 1 C c. 1.25 C d. 1.5 C 3. If 0.625 \u00d7 1020 electrons flow through a circuit each second, what is the current in the circuit? 4. Two students calculate the charge flowing through a circuit. The first student concludes that 300 C of charge flows in 1 minute. The second student concludes that", " 3.125 \u00d7 1019 electrons flow per second. If the current measured in the circuit is 5 A, which of the two students (if any) have performed the calculations correctly? 20.2 Ohm\u2019s Law: Resistance and Simple Circuits 5. If the voltage across a fixed resistance is doubled, what happens to the current? It doubles. It halves. It stays the same. a. b. c. d. The current cannot be determined. Figure 20.48 If the four wires are made from the same material, which of the following is true? Select two answers. a. Resistance of Wire 3 > Resistance of Wire 2 b. Resistance of Wire 1 > Resistance of Wire 2 c. Resistance of Wire 1 < Resistance of Wire 4 d. Resistance of Wire 4 < Resistance of Wire 3 10. Suppose the resistance of a wire is R \u03a9. What will be the resistance of another wire of the same material having the same length but double the diameter? a. R/2 b. 2R c. R/4 d. 4R 11. The resistances of two wires having the same lengths and cross section areas are 3 \u03a9 and 11 \u03a9. If the resistivity of the 3 \u03a9 wire is 2.65 \u00d7 10\u22128 \u03a9\u00b7m, find the resistivity of the 1 \u03a9 wire. 12. The lengths and diameters of three wires are given below. If they all have the same resistance, find the ratio of their resistivities. Table 20.5 Wire Length Diameter Wire 1 2 m 1 cm Wire 2 1 m 0.5 cm Wire 3 1 m 1 cm 13. Suppose the resistance of a wire is 2 \u03a9. If the wire is stretched to three times its length, what will be its resistance? Assume that the volume does not change. 20.4 Electric Power and Energy 6. The table below gives the voltages and currents recorded across a resistor. 14. Table 20.4 Voltage (V) 2.50 5.00 7.50 10.00 12.50 Current (A) 0.69 1.38 2.09 2.76 3.49 a. Plot the graph and comment on the shape. b. Calculate the value of the resistor. 7. What is the resistance of a bulb if the current in it is 1.25 A when a 4 V voltage supply is connected to it? If the voltage supply is increased to 7 V, what will be the current in the bulb? 20.3 Resistance and Resistivity 8. Which of the following affect the resistivity of a wire? length a. b. area of cross section c. material d. all of the above 9. The lengths and diameters of four wires are given as shown. Figure 20.49 The circuit shown contains a resistor R connected to a voltage supply. The graph shows the total energy E dissipated by the resistance as a function of time. Which of the following shows the corresponding graph for double resistance, i.e., if R is replaced by 2R? 912 Chapter 20 | Electric Current, Resistance, and Ohm's Law a. Figure 20.50 b. Figure 20.51 c. Figure 20.52 d. Figure 20.53 15. What will be the ratio of the resistance of a 120 W, 220 V lamp to that of a 100 W, 110 V lamp? This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 21 | Circuits, Bioelectricity, and DC Instruments 913 21 CIRCUITS, BIOELECTRICITY, AND DC INSTRUMENTS Figure 21.1 Electric circuits in a computer allow large amounts of data to be quickly and accurately analyzed.. (credit: Airman 1st Class Mike Meares, United States Air Force) Chapter Outline 21.1. Resistors in Series and Parallel 21.2. Electromotive Force: Terminal Voltage 21.3. Kirchhoff\u2019s Rules 21.4. DC Voltmeters and Ammeters 21.5. Null Measurements 21.6. DC Circuits Containing Resistors and Capacitors Connection for AP\u00ae Courses Electric circuits are commonplace in our everyday lives. Some circuits are simple, such as those in flashlights while others are extremely complex, such as those used in supercomputers. This chapter takes the topic of electric circuits a step beyond simple circuits by addressing both changes that result from interactions between systems (Big Idea 4) and constraints on such changes due to laws of conservation (Big Idea 5). When the circuit is purely resistive, everything in this chapter applies to both DC and AC. However, matters become more complex when capacitance is involved. We do consider what happens when capacitors are connected to DC voltage sources, but the interaction of capacitors (and other nonresistive devices) with AC sources is left for a later chapter. In addition, a number of important DC instruments, such as meters that", " measure voltage and current, are covered in this chapter. Information and examples presented in the chapter examine cause-effect relationships inherent in interactions involving electrical systems. The electrical properties of an electric circuit can change due to other systems (Enduring Understanding 4.E). More specifically, values of currents and potential differences in electric circuits depend on arrangements of individual circuit components (Essential Knowledge 4.E.5). In this chapter several series and parallel combinations of resistors are discussed and their effects on currents and potential differences are analyzed. In electric circuits the total energy (Enduring Understanding 5.B) and the total electric charge (Enduring Understanding 5.C) are conserved. Kirchoff\u2019s rules describe both, energy conservation (Essential Knowledge 5.B.9) and charge conservation (Essential 914 Chapter 21 | Circuits, Bioelectricity, and DC Instruments Knowledge 5.C.3). Energy conservation is discussed in terms of the loop rule which specifies that the potential around any closed circuit path must be zero. Charge conservation is applied as conservation of current by equating the sum of all currents entering a junction to the sum of all currents leaving the junction (also known as the junction rule). Kirchoff\u2019s rules are used to calculate currents and potential differences in circuits that combine resistors in series and parallel, and resistors and capacitors. The concepts in this chapter support: Big Idea 4 Interactions between systems can result in changes in those systems. Enduring Understanding 4.E The electric and magnetic properties of a system can change in response to the presence of, or changes in, other objects or systems. Essential Knowledge 4.E.5 The values of currents and electric potential differences in an electric circuit are determined by the properties and arrangement of the individual circuit elements such as sources of emf, resistors, and capacitors. Big Idea 5 Changes that occur as a result of interactions are constrained by conservation laws. Enduring Understanding 5.B The energy of a system is conserved. Essential Knowledge 5.B.9 Kirchhoff\u2019s loop rule describes conservation of energy in electrical circuits. Enduring Understanding 5.C The electric charge of a system is conserved. Essential Knowledge 5.C.3 Kirchhoff\u2019s junction rule describes the conservation of electric charge in electrical circuits. Since charge is conserved, current must be conserved at each junction in the circuit. Examples should include circuits that combine resistors in series and parallel. 21.1 Resistors in Series and Parallel By the end of this section, you will be able to: Learning Objectives \u2022 Draw a circuit with resistors in parallel and in series. \u2022 Use Ohm\u2019s law to calculate the voltage drop across a resistor when current passes through it. \u2022 Contrast the way total resistance is calculated for resistors in series and in parallel. \u2022 Explain why total resistance of a parallel circuit is less than the smallest resistance of any of the resistors in that circuit. \u2022 Calculate total resistance of a circuit that contains a mixture of resistors connected in series and in parallel. The information presented in this section supports the following AP\u00ae learning objectives and science practices: \u2022 4.E.5.1 The student is able to make and justify a quantitative prediction of the effect of a change in values or arrangements of one or two circuit elements on the currents and potential differences in a circuit containing a small number of sources of emf, resistors, capacitors, and switches in series and/or parallel. (S.P. 2.2, 6.4) \u2022 4.E.5.2 The student is able to make and justify a qualitative prediction of the effect of a change in values or arrangements of one or two circuit elements on currents and potential differences in a circuit containing a small number of sources of emf, resistors, capacitors, and switches in series and/or parallel. (S.P. 6.1, 6.4) \u2022 4.E.5.3 The student is able to plan data collection strategies and perform data analysis to examine the values of currents and potential differences in an electric circuit that is modified by changing or rearranging circuit elements, including sources of emf, resistors, and capacitors. (S.P. 2.2, 4.2, 5.1) \u2022 5.B.9.3 The student is able to apply conservation of energy (Kirchhoff\u2019s loop rule) in calculations involving the total electric potential difference for complete circuit loops with only a single battery and resistors in series and/or in, at most, one parallel branch. (S.P. 2.2, 6.4, 7.2) Most circuits have more than one component, called a resistor that limits the flow of charge in the circuit. A measure of this limit on charge flow is called resistance. The simplest combinations of resistors are the series and parallel connections illustrated in Figure 21.2. The total resistance of a combination", " of resistors depends on both their individual values and how they are connected. Figure 21.2 (a) A series connection of resistors. (b) A parallel connection of resistors. This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 21 | Circuits, Bioelectricity, and DC Instruments 915 Resistors in Series When are resistors in series? Resistors are in series whenever the flow of charge, called the current, must flow through devices sequentially. For example, if current flows through a person holding a screwdriver and into the Earth, then 1 in Figure 21.2(a) could be the resistance of the screwdriver\u2019s shaft, 2 the resistance of its handle, 3 the person\u2019s body resistance, and 4 the resistance of her shoes. Figure 21.3 shows resistors in series connected to a voltage source. It seems reasonable that the total resistance is the sum of the individual resistances, considering that the current has to pass through each resistor in sequence. (This fact would be an advantage to a person wishing to avoid an electrical shock, who could reduce the current by wearing high-resistance rubbersoled shoes. It could be a disadvantage if one of the resistances were a faulty high-resistance cord to an appliance that would reduce the operating current.) Figure 21.3 Three resistors connected in series to a battery (left) and the equivalent single or series resistance (right). To verify that resistances in series do indeed add, let us consider the loss of electrical power, called a voltage drop, in each resistor in Figure 21.3. According to Ohm\u2019s law, the voltage drop,, across a resistor when a current flows through it is calculated using the equation =, where equals the current in amps (A) and is the resistance in ohms ( \u03a9 ). Another way to think of this is that is the voltage necessary to make a current flow through a resistance. So the voltage drop across 1 is 1 = 1, that across 2 is 2 = 2, and that across 3 is 3 = 3. The sum of these voltages equals the voltage output of the source; that is, = 1 + 2 + 3. (21.1) This equation is based on the conservation of energy and conservation of charge. Electrical potential energy can be described by the equation =, where is the electric charge and is the voltage. Thus the energy supplied by the source is, while that dissipated by the resistors is 1 + 2 + 3. (21.2) Connections: Conservation Laws The derivations of the expressions for series and parallel resistance are based on the laws of conservation of energy and conservation of charge, which state that total charge and total energy are constant in any process. These two laws are directly involved in all electrical phenomena and will be invoked repeatedly to explain both specific effects and the general behavior of electricity. These energies must be equal, because there is no other source and no other destination for energy in the circuit. Thus, = 1 + 2 + 3. The charge cancels, yielding = 1 + 2 + 3, as stated. (Note that the same amount of charge passes through the battery and each resistor in a given amount of time, since there is no capacitance to store charge, there is no place for charge to leak, and charge is conserved.) Now substituting the values for the individual voltages gives = 1 + 2 + 3 = (1 + 2 + 3). Note that for the equivalent single series resistance s, we have = s. This implies that the total or equivalent series resistance s of three resistors is s = 1 + 2 + 3. This logic is valid in general for any number of resistors in series; thus, the total resistance s of a series connection is s = 1 + 2 + 3 +..., (21.3) (21.4) (21.5) 916 Chapter 21 | Circuits, Bioelectricity, and DC Instruments as proposed. Since all of the current must pass through each resistor, it experiences the resistance of each, and resistances in series simply add up. Example 21.1 Calculating Resistance, Current, Voltage Drop, and Power Dissipation: Analysis of a Series Circuit Suppose the voltage output of the battery in Figure 21.3 is 12.0 V, and the resistances are 1 = 1.00 \u03a9, 2 = 6.00 \u03a9, and 3 = 13.0 \u03a9. (a) What is the total resistance? (b) Find the current. (c) Calculate the voltage drop in each resistor, and show these add to equal the voltage output of the source. (d) Calculate the power dissipated by each resistor. (e) Find the power output of the source, and show that it equals the total power dissipated by the resistors. Strategy and Solution for (a) The total resistance is simply the sum", " of the individual resistances, as given by this equation.00 \u03a9 + 6.00 \u03a9 + 13.0 \u03a9 = 20.0 \u03a9. (21.6) Strategy and Solution for (b) The current is found using Ohm\u2019s law, =. Entering the value of the applied voltage and the total resistance yields the current for the circuit: = s = 12.0 V 20.0 \u03a9 = 0.600 A. (21.7) Strategy and Solution for (c) The voltage\u2014or drop\u2014in a resistor is given by Ohm\u2019s law. Entering the current and the value of the first resistance yields Similarly, and 1 = 1 = (0.600 A)(1.0 \u03a9 ) = 0.600 V. 2 = 2 = (0.600 A)(6.0 \u03a9 ) = 3.60 V 3 = 3 = (0.600 A)(13.0 \u03a9 ) = 7.80 V. Discussion for (c) The three drops add to 12.0 V, as predicted: 1 + 2 + 3 = (0.600 + 3.60 + 7.80) V = 12.0 V. (21.8) (21.9) (21.10) (21.11) Strategy and Solution for (d) The easiest way to calculate power in watts (W) dissipated by a resistor in a DC circuit is to use Joule\u2019s law, =, where is electric power. In this case, each resistor has the same full current flowing through it. By substituting Ohm\u2019s law = into Joule\u2019s law, we get the power dissipated by the first resistor as Similarly, and 1 = 21 = (0.600 A)2(1.00 \u03a9 ) = 0.360 W. 2 = 22 = (0.600 A)2(6.00 \u03a9 ) = 2.16 W 3 = 23 = (0.600 A)2(13.0 \u03a9 ) = 4.68 W. (21.12) (21.13) (21.14) Discussion for (d) Power can also be calculated using either = or = 2 full voltage of the source). The same values will be obtained. Strategy and Solution for (e), where is the voltage drop across the resistor (not the This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 21 | Circuits, Bioelectricity, and DC Instruments 917 The easiest way to calculate power output of the source is to use =, where is the source voltage. This gives = (0.600 A)(12.0 V) = 7.20 W. (21.15) Discussion for (e) Note, coincidentally, that the total power dissipated by the resistors is also 7.20 W, the same as the power put out by the source. That is, 1 + 2 + 3 = (0.360 + 2.16 + 4.68) W = 7.20 W. (21.16) Power is energy per unit time (watts), and so conservation of energy requires the power output of the source to be equal to the total power dissipated by the resistors. Major Features of Resistors in Series 1. Series resistances add: s = 1 + 2 + 3 +.... 2. The same current flows through each resistor in series. 3. Individual resistors in series do not get the total source voltage, but divide it. Resistors in Parallel Figure 21.4 shows resistors in parallel, wired to a voltage source. Resistors are in parallel when each resistor is connected directly to the voltage source by connecting wires having negligible resistance. Each resistor thus has the full voltage of the source applied to it. Each resistor draws the same current it would if it alone were connected to the voltage source (provided the voltage source is not overloaded). For example, an automobile\u2019s headlights, radio, and so on, are wired in parallel, so that they utilize the full voltage of the source and can operate completely independently. The same is true in your house, or any building. (See Figure 21.4(b).) 918 Chapter 21 | Circuits, Bioelectricity, and DC Instruments Figure 21.4 (a) Three resistors connected in parallel to a battery and the equivalent single or parallel resistance. (b) Electrical power setup in a house. (credit: Dmitry G, Wikimedia Commons) To find an expression for the equivalent parallel resistance p, let us consider the currents that flow and how they are related to resistance. Since each resistor in the circuit has the full voltage, the currents flowing through the individual resistors are 1 = 1 of these currents:. Conservation of charge implies that the total current produced by the source is the sum, and 3 = 3, 2 = 2 Substit", "uting the expressions for the individual currents gives = 1 + 2 + 3. Note that Ohm\u2019s law for the equivalent single resistance gives = p = 1 p. (21.17) (21.18) (21.19) The terms inside the parentheses in the last two equations must be equal. Generalizing to any number of resistors, the total resistance p of a parallel connection is related to the individual resistances by 3 +.... (21.20) This relationship results in a total resistance p that is less than the smallest of the individual resistances. (This is seen in the next example.) When resistors are connected in parallel, more current flows from the source than would flow for any of them individually, and so the total resistance is lower. This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 21 | Circuits, Bioelectricity, and DC Instruments 919 Example 21.2 Calculating Resistance, Current, Power Dissipation, and Power Output: Analysis of a Parallel Circuit Let the voltage output of the battery and resistances in the parallel connection in Figure 21.4 be the same as the previously considered series connection: = 12.0 V, 1 = 1.00 \u03a9, 2 = 6.00 \u03a9, and 3 = 13.0 \u03a9. (a) What is the total resistance? (b) Find the total current. (c) Calculate the currents in each resistor, and show these add to equal the total current output of the source. (d) Calculate the power dissipated by each resistor. (e) Find the power output of the source, and show that it equals the total power dissipated by the resistors. Strategy and Solution for (a) The total resistance for a parallel combination of resistors is found using the equation below. Entering known values gives.00 \u03a9 + 1 6.00 \u03a9 + 1 13.0 \u03a9. Thus, 1 p = 1.00 \u03a9 + 0.1667 \u03a9 + 0.07692 \u03a9 = 1.2436 \u03a9. (Note that in these calculations, each intermediate answer is shown with an extra digit.) We must invert this to find the total resistance p. This yields The total resistance with the correct number of significant digits is p = 0.804 \u03a9. p = 1 1.2436 \u03a9 = 0.8041 \u03a9. Discussion for (a) p is, as predicted, less than the smallest individual resistance. Strategy and Solution for (b) The total current can be found from Ohm\u2019s law, substituting p for the total resistance. This gives = p = 12.0 V 0.8041 \u03a9 = 14.92 A. (21.21) (21.22) (21.23) (21.24) Discussion for (b) Current for each device is much larger than for the same devices connected in series (see the previous example). A circuit with parallel connections has a smaller total resistance than the resistors connected in series. Strategy and Solution for (c) The individual currents are easily calculated from Ohm\u2019s law, since each resistor gets the full voltage. Thus, Similarly, and 1 = 1 = 12.0 V 1.00 \u03a9 = 12.0 A. 2 = 2 = 12.0 V 6.00 \u03a9 = 2.00 A 3 = 3 = 12.0 V 13.0 \u03a9 = 0.92 A. Discussion for (c) The total current is the sum of the individual currents: 1 + 2 + 3 = 14.92 A. This is consistent with conservation of charge. Strategy and Solution for (d) The power dissipated by each resistor can be found using any of the equations relating power to current, voltage, and resistance, since all three are known. Let us use = 2, since each resistor gets full voltage. Thus, (21.25) (21.26) (21.27) (21.28) 920 Chapter 21 | Circuits, Bioelectricity, and DC Instruments 1 = 2 1 = (12.0 V)2 1.00 \u03a9 = 144 W. 2 = 2 2 = (12.0 V)2 6.00 \u03a9 = 24.0 W 3 = 2 3 = (12.0 V)2 13.0 \u03a9 = 11.1 W. Similarly, and Discussion for (d) (21.29) (21.30) (21.31) The power dissipated by each resistor is considerably higher in parallel than when connected in series to the same voltage source. Strategy and Solution for (e) The total power can also be calculated in several ways. Choosing =, and entering the total current, yields = = (14.92 A)(12.0 V) = 179 W. Discussion for (e) Total power dissipated by the resistors is also", " 179 W: 1 + 2 + 3 = 144 W + 24.0 W + 11.1 W = 179 W. This is consistent with the law of conservation of energy. Overall Discussion Note that both the currents and powers in parallel connections are greater than for the same devices in series. (21.32) (21.33) Major Features of Resistors in Parallel 1. Parallel resistance is found from combination. +..., and it is smaller than any individual resistance in the 2. Each resistor in parallel has the same full voltage of the source applied to it. (Power distribution systems most often use parallel connections to supply the myriad devices served with the same voltage and to allow them to operate independently.) 3. Parallel resistors do not each get the total current; they divide it. Combinations of Series and Parallel More complex connections of resistors are sometimes just combinations of series and parallel. These are commonly encountered, especially when wire resistance is considered. In that case, wire resistance is in series with other resistances that are in parallel. Combinations of series and parallel can be reduced to a single equivalent resistance using the technique illustrated in Figure 21.5. Various parts are identified as either series or parallel, reduced to their equivalents, and further reduced until a single resistance is left. The process is more time consuming than difficult. This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 21 | Circuits, Bioelectricity, and DC Instruments 921 Figure 21.5 This combination of seven resistors has both series and parallel parts. Each is identified and reduced to an equivalent resistance, and these are further reduced until a single equivalent resistance is reached. The simplest combination of series and parallel resistance, shown in Figure 21.6, is also the most instructive, since it is found in many applications. For example, 1 could be the resistance of wires from a car battery to its electrical devices, which are in parallel. 2 and 3 could be the starter motor and a passenger compartment light. We have previously assumed that wire resistance is negligible, but, when it is not, it has important effects, as the next example indicates. Example 21.3 Calculating Resistance, IR Drop, Current, and Power Dissipation: Combining Series and Parallel Circuits Figure 21.6 shows the resistors from the previous two examples wired in a different way\u2014a combination of series and parallel. We can consider 1 to be the resistance of wires leading to 2 and 3. (a) Find the total resistance. (b) What is the drop in 1? (c) Find the current 2 through 2. (d) What power is dissipated by 2? Figure 21.6 These three resistors are connected to a voltage source so that 2 and 3 are in parallel with one another and that combination is in series with 1. Strategy and Solution for (a) To find the total resistance, we note that 2 and 3 are in parallel and their combination p is in series with 1. Thus the total (equivalent) resistance of this combination is tot = 1 + p. First, we find p using the equation for resistors in parallel and entering known values.00 \u03a9 + 1 13.0 \u03a9 = 0.2436 \u03a9. (21.34) (21.35) 922 Chapter 21 | Circuits, Bioelectricity, and DC Instruments Inverting gives So the total resistance is Discussion for (a) p = 1 0.2436 \u03a9 = 4.11 \u03a9. tot = 1 + p = 1.00 \u03a9 + 4.11 \u03a9 = 5.11 \u03a9. (21.36) (21.37) The total resistance of this combination is intermediate between the pure series and pure parallel values ( 20.0 \u03a9 and 0.804 \u03a9, respectively) found for the same resistors in the two previous examples. Strategy and Solution for (b) To find the drop in 1, we note that the full current flows through 1. Thus its drop is We must find before we can calculate 1. The total current is found using Ohm\u2019s law for the circuit. That is, 1 =. = tot = 12.0 V 5.11 \u03a9 = 2.35 A. Entering this into the expression above, we get 1 = = (2.35 A)(1.00 \u03a9 ) = 2.35 V. Discussion for (b) (21.38) (21.39) (21.40) The voltage applied to 2 and 3 is less than the total voltage by an amount 1. When wire resistance is large, it can significantly affect the operation of the devices represented by 2 and 3. Strategy and Solution for (c) To find the current through 2, we must first find the voltage applied to it. We call this voltage p, because it is applied to a parallel combination of resistors. The voltage applied to both 2 and 3 is reduced by the amount 1", ", and so it is Now the current 2 through resistance 2 is found using Ohm\u2019s law: p = \u2212 1 = 12.0 V \u2212 2.35 V = 9.65 V. 2 = p 2 = 9.65 V 6.00 \u03a9 = 1.61 A. (21.41) (21.42) Discussion for (c) The current is less than the 2.00 A that flowed through 2 when it was connected in parallel to the battery in the previous parallel circuit example. Strategy and Solution for (d) The power dissipated by 2 is given by 2 = (2 )22 = (1.61 A)2(6.00 \u03a9 ) = 15.5 W. (21.43) Discussion for (d) The power is less than the 24.0 W this resistor dissipated when connected in parallel to the 12.0-V source. Applying the Science Practices: Circuit Construction Kit (DC only) Plan an experiment to analyze the effect on currents and potential differences due to rearrangement of resistors and variations in voltage sources. Your experimental investigation should include data collection for at least five different scenarios of rearranged resistors (i.e., several combinations of series and parallel) and three scenarios of different voltage sources. This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 21 | Circuits, Bioelectricity, and DC Instruments 923 Practical Implications One implication of this last example is that resistance in wires reduces the current and power delivered to a resistor. If wire resistance is relatively large, as in a worn (or a very long) extension cord, then this loss can be significant. If a large current is drawn, the drop in the wires can also be significant. For example, when you are rummaging in the refrigerator and the motor comes on, the refrigerator light dims momentarily. Similarly, you can see the passenger compartment light dim when you start the engine of your car (although this may be due to resistance inside the battery itself). What is happening in these high-current situations is illustrated in Figure 21.7. The device represented by 3 has a very low resistance, and so when it is switched on, a large current flows. This increased current causes a larger drop in the wires represented by 1, reducing the voltage across the light bulb (which is 2 ), which then dims noticeably. Figure 21.7 Why do lights dim when a large appliance is switched on? The answer is that the large current the appliance motor draws causes a significant drop in the wires and reduces the voltage across the light. Check Your Understanding Can any arbitrary combination of resistors be broken down into series and parallel combinations? See if you can draw a circuit diagram of resistors that cannot be broken down into combinations of series and parallel. Solution No, there are many ways to connect resistors that are not combinations of series and parallel, including loops and junctions. In such cases Kirchhoff\u2019s rules, to be introduced in Kirchhoff\u2019s Rules, will allow you to analyze the circuit. Problem-Solving Strategies for Series and Parallel Resistors 1. Draw a clear circuit diagram, labeling all resistors and voltage sources. This step includes a list of the knowns for the problem, since they are labeled in your circuit diagram. 2. Identify exactly what needs to be determined in the problem (identify the unknowns). A written list is useful. 3. Determine whether resistors are in series, parallel, or a combination of both series and parallel. Examine the circuit diagram to make this assessment. Resistors are in series if the same current must pass sequentially through them. 4. Use the appropriate list of major features for series or parallel connections to solve for the unknowns. There is one list for series and another for parallel. If your problem has a combination of series and parallel, reduce it in steps by considering individual groups of series or parallel connections, as done in this module and the examples. Special note: When finding, the reciprocal must be taken with care. 5. Check to see whether the answers are reasonable and consistent. Units and numerical results must be reasonable. Total series resistance should be greater, whereas total parallel resistance should be smaller, for example. Power should be greater for the same devices in parallel compared with series, and so on. 21.2 Electromotive Force: Terminal Voltage By the end of this section, you will be able to: Learning Objectives 924 Chapter 21 | Circuits, Bioelectricity, and DC Instruments \u2022 Compare and contrast the voltage and the electromagnetic force of an electric power source. \u2022 Describe what happens to the terminal voltage, current, and power delivered to a load as internal resistance of the voltage source increases. \u2022 Explain why it is beneficial to use more than one voltage source connected in parallel. The information presented in this section supports the following AP\u00ae learning objectives and science practices: \u2022 5.", "B.9.7 The student is able to refine and analyze a scientific question for an experiment using Kirchhoff\u2019s loop rule for circuits that includes determination of internal resistance of the battery and analysis of a nonohmic resistor. (S.P. 4.1, 4.2, 5.1, 5.3) When you forget to turn off your car lights, they slowly dim as the battery runs down. Why don\u2019t they simply blink off when the battery\u2019s energy is gone? Their gradual dimming implies that battery output voltage decreases as the battery is depleted. Furthermore, if you connect an excessive number of 12-V lights in parallel to a car battery, they will be dim even when the battery is fresh and even if the wires to the lights have very low resistance. This implies that the battery\u2019s output voltage is reduced by the overload. The reason for the decrease in output voltage for depleted or overloaded batteries is that all voltage sources have two fundamental parts\u2014a source of electrical energy and an internal resistance. Let us examine both. Electromotive Force You can think of many different types of voltage sources. Batteries themselves come in many varieties. There are many types of mechanical/electrical generators, driven by many different energy sources, ranging from nuclear to wind. Solar cells create voltages directly from light, while thermoelectric devices create voltage from temperature differences. A few voltage sources are shown in Figure 21.8. All such devices create a potential difference and can supply current if connected to a resistance. On the small scale, the potential difference creates an electric field that exerts force on charges, causing current. We thus use the name electromotive force, abbreviated emf. Emf is not a force at all; it is a special type of potential difference. To be precise, the electromotive force (emf) is the potential difference of a source when no current is flowing. Units of emf are volts. Figure 21.8 A variety of voltage sources (clockwise from top left): the Brazos Wind Farm in Fluvanna, Texas (credit: Leaflet, Wikimedia Commons); the Krasnoyarsk Dam in Russia (credit: Alex Polezhaev); a solar farm (credit: U.S. Department of Energy); and a group of nickel metal hydride batteries (credit: Tiaa Monto). The voltage output of each depends on its construction and load, and equals emf only if there is no load. Electromotive force is directly related to the source of potential difference, such as the particular combination of chemicals in a battery. However, emf differs from the voltage output of the device when current flows. The voltage across the terminals of a battery, for example, is less than the emf when the battery supplies current, and it declines further as the battery is depleted or loaded down. However, if the device\u2019s output voltage can be measured without drawing current, then output voltage will equal emf (even for a very depleted battery). Internal Resistance As noted before, a 12-V truck battery is physically larger, contains more charge and energy, and can deliver a larger current than a 12-V motorcycle battery. Both are lead-acid batteries with identical emf, but, because of its size, the truck battery has a smaller internal resistance. Internal resistance is the inherent resistance to the flow of current within the source itself. Figure 21.9 is a schematic representation of the two fundamental parts of any voltage source. The emf (represented by a script E in the figure) and internal resistance are in series. The smaller the internal resistance for a given emf, the more current and the more power the source can supply. This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 21 | Circuits, Bioelectricity, and DC Instruments 925 Figure 21.9 Any voltage source (in this case, a carbon-zinc dry cell) has an emf related to its source of potential difference, and an internal resistance related to its construction. (Note that the script E stands for emf.). Also shown are the output terminals across which the terminal voltage is measured. Since = emf \u2212, terminal voltage equals emf only if there is no current flowing. The internal resistance can behave in complex ways. As noted, increases as a battery is depleted. But internal resistance may also depend on the magnitude and direction of the current through a voltage source, its temperature, and even its history. The internal resistance of rechargeable nickel-cadmium cells, for example, depends on how many times and how deeply they have been depleted. Things Great and Small: The Submicroscopic Origin of Battery Potential Various types of batteries are available, with emfs determined by the combination of chemicals involved. We can view this as a molecular reaction (what much of chemistry is about) that separates charge. The lead-acid battery used in", " cars and other vehicles is one of the most common types. A single cell (one of six) of this battery is seen in Figure 21.10. The cathode (positive) terminal of the cell is connected to a lead oxide plate, while the anode (negative) terminal is connected to a lead plate. Both plates are immersed in sulfuric acid, the electrolyte for the system. Figure 21.10 Artist\u2019s conception of a lead-acid cell. Chemical reactions in a lead-acid cell separate charge, sending negative charge to the anode, which is connected to the lead plates. The lead oxide plates are connected to the positive or cathode terminal of the cell. Sulfuric acid conducts the charge as well as participating in the chemical reaction. The details of the chemical reaction are left to the reader to pursue in a chemistry text, but their results at the molecular level help explain the potential created by the battery. Figure 21.11 shows the result of a single chemical reaction. Two electrons are placed on the anode, making it negative, provided that the cathode supplied two electrons. This leaves the cathode positively charged, because it has lost two electrons. In short, a separation of charge has been driven by a chemical reaction. Note that the reaction will not take place unless there is a complete circuit to allow two electrons to be supplied to the cathode. Under many circumstances, these electrons come from the anode, flow through a resistance, and return to the cathode. Note also that since the chemical reactions involve substances with resistance, it is not possible to create the emf without an internal resistance. 926 Chapter 21 | Circuits, Bioelectricity, and DC Instruments Figure 21.11 Artist\u2019s conception of two electrons being forced onto the anode of a cell and two electrons being removed from the cathode of the cell. The chemical reaction in a lead-acid battery places two electrons on the anode and removes two from the cathode. It requires a closed circuit to proceed, since the two electrons must be supplied to the cathode. Why are the chemicals able to produce a unique potential difference? Quantum mechanical descriptions of molecules, which take into account the types of atoms and numbers of electrons in them, are able to predict the energy states they can have and the energies of reactions between them. In the case of a lead-acid battery, an energy of 2 eV is given to each electron sent to the anode. Voltage is defined as the electrical potential energy divided by charge: = E. An electron volt is the energy given to a single electron by a voltage of 1 V. So the voltage here is 2 V, since 2 eV is given to each electron. It is the energy produced in each molecular reaction that produces the voltage. A different reaction produces a different energy and, hence, a different voltage. Terminal Voltage The voltage output of a device is measured across its terminals and, thus, is called its terminal voltage. Terminal voltage is given by = emf \u2212, (21.44) where is the internal resistance and is the current flowing at the time of the measurement. is positive if current flows away from the positive terminal, as shown in Figure 21.9. You can see that the larger the current, the smaller the terminal voltage. And it is likewise true that the larger the internal resistance, the smaller the terminal voltage. Suppose a load resistance load is connected to a voltage source, as in Figure 21.12. Since the resistances are in series, the total resistance in the circuit is load +. Thus the current is given by Ohm\u2019s law to be = emf load +. (21.45) Figure 21.12 Schematic of a voltage source and its load load. Since the internal resistance is in series with the load, it can significantly affect the terminal voltage and current delivered to the load. (Note that the script E stands for emf.) We see from this expression that the smaller the internal resistance, the greater the current the voltage source supplies to its load load. As batteries are depleted, increases. If becomes a significant fraction of the load resistance, then the current is significantly reduced, as the following example illustrates. This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 21 | Circuits, Bioelectricity, and DC Instruments 927 Example 21.4 Calculating Terminal Voltage, Power Dissipation, Current, and Resistance: Terminal Voltage and Load A certain battery has a 12.0-V emf and an internal resistance of 0.100 \u03a9. (a) Calculate its terminal voltage when connected to a 10.0- \u03a9 load. (b) What is the terminal voltage when connected to a 0.500- \u03a9 load? (c) What power does the 0.500- \u03a9 load dissipate? (d) If the internal resistance grows to 0.500 \u03a9, find the current, terminal voltage, and", " power dissipated by a 0.500- \u03a9 load. Strategy The analysis above gave an expression for current when internal resistance is taken into account. Once the current is found, the terminal voltage can be calculated using the equation = emf \u2212. Once current is found, the power dissipated by a resistor can also be found. Solution for (a) Entering the given values for the emf, load resistance, and internal resistance into the expression above yields = emf load + = 12.0 V 10.1 \u03a9 = 1.188 A. Enter the known values into the equation = emf \u2212 to get the terminal voltage: = emf \u2212 = 12.0 V \u2212 (1.188 A)(0.100 \u03a9) = 11.9 V. Discussion for (a) (21.46) (21.47) The terminal voltage here is only slightly lower than the emf, implying that 10.0 \u03a9 is a light load for this particular battery. Solution for (b) Similarly, with load = 0.500 \u03a9, the current is = emf load + = 12.0 V 0.600 \u03a9 = 20.0 A. = emf \u2212 = 12.0 V \u2212 (20.0 A)(0.100 \u03a9) = 10.0 V. The terminal voltage is now Discussion for (b) (21.48) (21.49) This terminal voltage exhibits a more significant reduction compared with emf, implying 0.500 \u03a9 is a heavy load for this battery. Solution for (c) The power dissipated by the 0.500 - \u03a9 load can be found using the formula = 2. Entering the known values gives load = 2load = (20.0 A)2(0.500 \u03a9) = 2.00\u00d7102 W. (21.50) Discussion for (c) Note that this power can also be obtained using the expressions 2 this case). Solution for (d) or, where is the terminal voltage (10.0 V in Here the internal resistance has increased, perhaps due to the depletion of the battery, to the point where it is as great as the load resistance. As before, we first find the current by entering the known values into the expression, yielding Now the terminal voltage is = emf load + = 12.0 V 1.00 \u03a9 = 12.0 A. = emf \u2212 = 12.0 V \u2212 (12.0 A)(0.500 \u03a9) = 6.00 V, and the power dissipated by the load is (21.51) (21.52) 928 Chapter 21 | Circuits, Bioelectricity, and DC Instruments load = 2load = (12.0 A)2(0.500 \u03a9 ) = 72.0 W. (21.53) Discussion for (d) We see that the increased internal resistance has significantly decreased terminal voltage, current, and power delivered to a load. Applying the Science Practices: Internal Resistance The internal resistance of a battery can be estimated using a simple activity. The circuit shown in the figure below includes a resistor R in series with a battery along with an ammeter and voltmeter to measure the current and voltage respectively. Figure 21.13 The currents and voltages measured for several R values are shown in the table below. Using the data given in the table and applying graphical analysis, determine the emf and internal resistance of the battery. Your response should clearly explain the method used to obtain the result. Table 21.1 Resistance Current (A) Voltage (V) R1 R2 R3 R4 3.53 2.07 1.46 1.13 4.24 4.97 5.27 5.43 Answer Plot the measured currents and voltages on a graph. The terminal voltage of a battery is equal to the emf of the battery minus the voltage drop across the internal resistance of the battery or V = emf \u2013 Ir. Using this linear relationship and the plotted graph, the internal resistance and emf of the battery can be found. The graph for this case is shown below. The equation is V = -0.50I + 6.0 and hence the internal resistance will be equal to 0.5 \u03a9 and emf will be equal to 6 V. Figure 21.14 Battery testers, such as those in Figure 21.15, use small load resistors to intentionally draw current to determine whether the terminal voltage drops below an acceptable level. They really test the internal resistance of the battery. If internal resistance is high, the battery is weak, as evidenced by its low terminal voltage. This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 21 | Circuits, Bioelectricity, and DC Instruments 929 Figure 21.15 These two battery testers measure terminal voltage under a load to determine the condition of a battery. The large device is being used by a U.S. Navy", " electronics technician to test large batteries aboard the aircraft carrier USS Nimitz and has a small resistance that can dissipate large amounts of power. (credit: U.S. Navy photo by Photographer\u2019s Mate Airman Jason A. Johnston) The small device is used on small batteries and has a digital display to indicate the acceptability of their terminal voltage. (credit: Keith Williamson) Some batteries can be recharged by passing a current through them in the direction opposite to the current they supply to a resistance. This is done routinely in cars and batteries for small electrical appliances and electronic devices, and is represented pictorially in Figure 21.16. The voltage output of the battery charger must be greater than the emf of the battery to reverse current through it. This will cause the terminal voltage of the battery to be greater than the emf, since = emf \u2212, and is now negative. Figure 21.16 A car battery charger reverses the normal direction of current through a battery, reversing its chemical reaction and replenishing its chemical potential. Multiple Voltage Sources There are two voltage sources when a battery charger is used. Voltage sources connected in series are relatively simple. When voltage sources are in series, their internal resistances add and their emfs add algebraically. (See Figure 21.17.) Series connections of voltage sources are common\u2014for example, in flashlights, toys, and other appliances. Usually, the cells are in series in order to produce a larger total emf. But if the cells oppose one another, such as when one is put into an appliance backward, the total emf is less, since it is the algebraic sum of the individual emfs. A battery is a multiple connection of voltaic cells, as shown in Figure 21.18. The disadvantage of series connections of cells is that their internal resistances add. One of the authors once owned a 1957 MGA that had two 6-V batteries in series, rather than a single 12-V battery. This arrangement produced a large internal resistance that caused him many problems in starting the engine. Figure 21.17 A series connection of two voltage sources. The emfs (each labeled with a script E) and internal resistances add, giving a total emf of emf1 + emf2 and a total internal resistance of 1 + 2. 930 Chapter 21 | Circuits, Bioelectricity, and DC Instruments Figure 21.18 Batteries are multiple connections of individual cells, as shown in this modern rendition of an old print. Single cells, such as AA or C cells, are commonly called batteries, although this is technically incorrect. If the series connection of two voltage sources is made into a complete circuit with the emfs in opposition, then a current of magnitude = emf1 \u2013 emf2 1 + 2 flows. See Figure 21.19, for example, which shows a circuit exactly analogous to the battery charger discussed above. If two voltage sources in series with emfs in the same sense are connected to a load load, as in Figure 21.20, then = emf1 + emf2 1 + 2 + load flows. Figure 21.19 These two voltage sources are connected in series with their emfs in opposition. Current flows in the direction of the greater emf and is limited to = emf1 \u2212 emf2 1 + 2 by the sum of the internal resistances. (Note that each emf is represented by script E in the figure.) A battery charger connected to a battery is an example of such a connection. The charger must have a larger emf than the battery to reverse current through it. Figure 21.20 This schematic represents a flashlight with two cells (voltage sources) and a single bulb (load resistance) in series. The current that flows is = emf1 + emf2 1 + 2 + load. (Note that each emf is represented by script E in the figure.) Take-Home Experiment: Flashlight Batteries Find a flashlight that uses several batteries and find new and old batteries. Based on the discussions in this module, predict the brightness of the flashlight when different combinations of batteries are used. Do your predictions match what you observe? Now place new batteries in the flashlight and leave the flashlight switched on for several hours. Is the flashlight still quite bright? Do the same with the old batteries. Is the flashlight as bright when left on for the same length of time with old and new batteries? What does this say for the case when you are limited in the number of available new batteries? Figure 21.21 shows two voltage sources with identical emfs in parallel and connected to a load resistance. In this simple case, the total emf is the same as the individual emfs. But the total internal resistance is reduced, since the internal resistances are in parallel. The parallel connection thus can produce a larger current. Here, = emf tot + load flows through the load, and tot is less than those of the individual batteries. For example, some diesel-powered cars", " use two 12-V batteries in parallel; they produce a total emf of 12 V but can deliver the larger current needed to start a diesel engine. This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 21 | Circuits, Bioelectricity, and DC Instruments 931 Figure 21.21 Two voltage sources with identical emfs (each labeled by script E) connected in parallel produce the same emf but have a smaller total internal resistance than the individual sources. Parallel combinations are often used to deliver more current. Here = flows emf tot + load through the load. Animals as Electrical Detectors A number of animals both produce and detect electrical signals. Fish, sharks, platypuses, and echidnas (spiny anteaters) all detect electric fields generated by nerve activity in prey. Electric eels produce their own emf through biological cells (electric organs) called electroplaques, which are arranged in both series and parallel as a set of batteries. Electroplaques are flat, disk-like cells; those of the electric eel have a voltage of 0.15 V across each one. These cells are usually located toward the head or tail of the animal, although in the case of the electric eel, they are found along the entire body. The electroplaques in the South American eel are arranged in 140 rows, with each row stretching horizontally along the body and containing 5,000 electroplaques. This can yield an emf of approximately 600 V, and a current of 1 A\u2014deadly. The mechanism for detection of external electric fields is similar to that for producing nerve signals in the cell through depolarization and repolarization\u2014the movement of ions across the cell membrane. Within the fish, weak electric fields in the water produce a current in a gel-filled canal that runs from the skin to sensing cells, producing a nerve signal. The Australian platypus, one of the very few mammals that lay eggs, can detect fields of 30 mV m, while sharks have been found to be able to sense a field in their snouts as small as 100 mV m (Figure 21.22). Electric eels use their own electric fields produced by the electroplaques to stun their prey or enemies. Figure 21.22 Sand tiger sharks (Carcharias taurus), like this one at the Minnesota Zoo, use electroreceptors in their snouts to locate prey. (credit: Jim Winstead, Flickr) Solar Cell Arrays Another example dealing with multiple voltage sources is that of combinations of solar cells\u2014wired in both series and parallel combinations to yield a desired voltage and current. Photovoltaic generation (PV), the conversion of sunlight directly into 932 Chapter 21 | Circuits, Bioelectricity, and DC Instruments electricity, is based upon the photoelectric effect, in which photons hitting the surface of a solar cell create an electric current in the cell. Most solar cells are made from pure silicon\u2014either as single-crystal silicon, or as a thin film of silicon deposited upon a glass or metal backing. Most single cells have a voltage output of about 0.5 V, while the current output is a function of the amount of sunlight upon the cell (the incident solar radiation\u2014the insolation). Under bright noon sunlight, a current of about 100 mA/cm2 of cell surface area is produced by typical single-crystal cells. Individual solar cells are connected electrically in modules to meet electrical-energy needs. They can be wired together in series or in parallel\u2014connected like the batteries discussed earlier. A solar-cell array or module usually consists of between 36 and 72 cells, with a power output of 50 W to 140 W. The output of the solar cells is direct current. For most uses in a home, AC is required, so a device called an inverter must be used to convert the DC to AC. Any extra output can then be passed on to the outside electrical grid for sale to the utility. Take-Home Experiment: Virtual Solar Cells One can assemble a \u201cvirtual\u201d solar cell array by using playing cards, or business or index cards, to represent a solar cell. Combinations of these cards in series and/or parallel can model the required array output. Assume each card has an output of 0.5 V and a current (under bright light) of 2 A. Using your cards, how would you arrange them to produce an output of 6 A at 3 V (18 W)? Suppose you were told that you needed only 18 W (but no required voltage). Would you need more cards to make this arrangement? 21.3 Kirchhoff\u2019s Rules By the end of this section, you will be able to: Learning Objectives \u2022 Analyze a complex circuit using Kirchhoff\u2019s rules, applying the conventions for determining the correct signs of various terms. The information presented in this section supports the following AP\u00ae learning objectives and science practices:", " \u2022 5.B.9.1 The student is able to construct or interpret a graph of the energy changes within an electrical circuit with only a single battery and resistors in series and/or in, at most, one parallel branch as an application of the conservation of energy (Kirchhoff\u2019s loop rule). (S.P. 1.1, 1.4) \u2022 5.B.9.2 The student is able to apply conservation of energy concepts to the design of an experiment that will demonstrate the validity of Kirchhoff\u2019s loop rule in a circuit with only a battery and resistors either in series or in, at most, one pair of parallel branches. (S.P. 4.2, 6.4, 7.2) \u2022 5.B.9.3 The student is able to apply conservation of energy (Kirchhoff\u2019s loop rule) in calculations involving the total electric potential difference for complete circuit loops with only a single battery and resistors in series and/or in, at most, one parallel branch. (S.P. 2.2, 6.4, 7.2) \u2022 5.B.9.4 The student is able to analyze experimental data including an analysis of experimental uncertainty that will demonstrate the validity of Kirchhoff\u2019s loop rule. (S.P. 5.1) \u2022 5.B.9.5 The student is able to use conservation of energy principles (Kirchhoff\u2019s loop rule) to describe and make predictions regarding electrical potential difference, charge, and current in steady-state circuits composed of various combinations of resistors and capacitors. (S.P. 6.4) \u2022 5.C.3.1 The student is able to apply conservation of electric charge (Kirchhoff\u2019s junction rule) to the comparison of electric current in various segments of an electrical circuit with a single battery and resistors in series and in, at most, one parallel branch and predict how those values would change if configurations of the circuit are changed. (S.P. 6.4, 7.2) \u2022 5.C.3.2 The student is able to design an investigation of an electrical circuit with one or more resistors in which evidence of conservation of electric charge can be collected and analyzed. (S.P. 4.1, 4.2, 5.1) \u2022 5.C.3.3 The student is able to use a description or schematic diagram of an electrical circuit to calculate unknown values of current in various segments or branches of the circuit. (S.P. 1.4, 2.2) \u2022 5.C.3.4 The student is able to predict or explain current values in series and parallel arrangements of resistors and other branching circuits using Kirchhoff\u2019s junction rule and relate the rule to the law of charge conservation. (S.P. 6.4, 7.2) \u2022 5.C.3.5 The student is able to determine missing values and direction of electric current in branches of a circuit with resistors and NO capacitors from values and directions of current in other branches of the circuit through appropriate selection of nodes and application of the junction rule. (S.P. 1.4, 2.2) Many complex circuits, such as the one in Figure 21.23, cannot be analyzed with the series-parallel techniques developed in Resistors in Series and Parallel and Electromotive Force: Terminal Voltage. There are, however, two circuit analysis rules that can be used to analyze any circuit, simple or complex. These rules are special cases of the laws of conservation of charge and conservation of energy. The rules are known as Kirchhoff\u2019s rules, after their inventor Gustav Kirchhoff (1824\u20131887). This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 21 | Circuits, Bioelectricity, and DC Instruments 933 Figure 21.23 This circuit cannot be reduced to a combination of series and parallel connections. Kirchhoff\u2019s rules, special applications of the laws of conservation of charge and energy, can be used to analyze it. (Note: The script E in the figure represents electromotive force, emf.) Kirchhoff\u2019s Rules \u2022 Kirchhoff\u2019s first rule\u2014the junction rule. The sum of all currents entering a junction must equal the sum of all currents leaving the junction. \u2022 Kirchhoff\u2019s second rule\u2014the loop rule. The algebraic sum of changes in potential around any closed circuit path (loop) must be zero. Explanations of the two rules will now be given, followed by problem-solving hints for applying Kirchhoff\u2019s rules, and a worked example that uses them. Kirchhoff\u2019s First Rule Kirchhoff\u2019s first rule (the junction rule) is an application of", " the conservation of charge to a junction; it is illustrated in Figure 21.24. Current is the flow of charge, and charge is conserved; thus, whatever charge flows into the junction must flow out. Kirchhoff\u2019s first rule requires that 1 = 2 + 3 (see figure). Equations like this can and will be used to analyze circuits and to solve circuit problems. Making Connections: Conservation Laws Kirchhoff\u2019s rules for circuit analysis are applications of conservation laws to circuits. The first rule is the application of conservation of charge, while the second rule is the application of conservation of energy. Conservation laws, even used in a specific application, such as circuit analysis, are so basic as to form the foundation of that application. Figure 21.24 The junction rule. The diagram shows an example of Kirchhoff\u2019s first rule where the sum of the currents into a junction equals the sum of the currents out of a junction. In this case, the current going into the junction splits and comes out as two currents, so that 1 = 2 + 3. Here 1 must be 11 A, since 2 is 7 A and 3 is 4 A. 934 Chapter 21 | Circuits, Bioelectricity, and DC Instruments Kirchhoff\u2019s Second Rule Kirchhoff\u2019s second rule (the loop rule) is an application of conservation of energy. The loop rule is stated in terms of potential,, rather than potential energy, but the two are related since PEelec =. Recall that emf is the potential difference of a source when no current is flowing. In a closed loop, whatever energy is supplied by emf must be transferred into other forms by devices in the loop, since there are no other ways in which energy can be transferred into or out of the circuit. Figure 21.25 illustrates the changes in potential in a simple series circuit loop. Kirchhoff\u2019s second rule requires emf \u2212 \u2212 1 \u2212 2 = 0. Rearranged, this is emf = + 1 + 2, which means the emf equals the sum of the (voltage) drops in the loop. Figure 21.25 The loop rule. An example of Kirchhoff\u2019s second rule where the sum of the changes in potential around a closed loop must be zero. (a) In this standard schematic of a simple series circuit, the emf supplies 18 V, which is reduced to zero by the resistances, with 1 V across the internal resistance, and 12 V and 5 V across the two load resistances, for a total of 18 V. (b) This perspective view represents the potential as something like a roller coaster, where charge is raised in potential by the emf and lowered by the resistances. (Note that the script E stands for emf.) Applying Kirchhoff\u2019s Rules By applying Kirchhoff\u2019s rules, we generate equations that allow us to find the unknowns in circuits. The unknowns may be currents, emfs, or resistances. Each time a rule is applied, an equation is produced. If there are as many independent equations as unknowns, then the problem can be solved. There are two decisions you must make when applying Kirchhoff\u2019s rules. These decisions determine the signs of various quantities in the equations you obtain from applying the rules. 1. When applying Kirchhoff\u2019s first rule, the junction rule, you must label the current in each branch and decide in what direction it is going. For example, in Figure 21.23, Figure 21.24, and Figure 21.25, currents are labeled 1, 2, 3, and, and arrows indicate their directions. There is no risk here, for if you choose the wrong direction, the current will be of the correct magnitude but negative. 2. When applying Kirchhoff\u2019s second rule, the loop rule, you must identify a closed loop and decide in which direction to go around it, clockwise or counterclockwise. For example, in Figure 21.25 the loop was traversed in the same direction as the current (clockwise). Again, there is no risk; going around the circuit in the opposite direction reverses the sign of every term in the equation, which is like multiplying both sides of the equation by \u20131. Figure 21.26 and the following points will help you get the plus or minus signs right when applying the loop rule. Note that the resistors and emfs are traversed by going from a to b. In many circuits, it will be necessary to construct more than one loop. In traversing each loop, one needs to be consistent for the sign of the change in potential. (See Example 21.5.) This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 21 | Circuits, Bioelectricity, and DC Instruments 935 Figure 21.26 Each of these resistors and voltage sources is traversed from", " a to b. The potential changes are shown beneath each element and are explained in the text. (Note that the script E stands for emf.) \u2022 When a resistor is traversed in the same direction as the current, the change in potential is \u2212. (See Figure 21.26.) \u2022 When a resistor is traversed in the direction opposite to the current, the change in potential is +. (See Figure 21.26.) \u2022 When an emf is traversed from \u2013 to + (the same direction it moves positive charge), the change in potential is +emf. (See Figure 21.26.) \u2022 When an emf is traversed from + to \u2013 (opposite to the direction it moves positive charge), the change in potential is \u2212 emf. (See Figure 21.26.) Example 21.5 Calculating Current: Using Kirchhoff\u2019s Rules Find the currents flowing in the circuit in Figure 21.27. Figure 21.27 This circuit is similar to that in Figure 21.23, but the resistances and emfs are specified. (Each emf is denoted by script E.) The currents in each branch are labeled and assumed to move in the directions shown. This example uses Kirchhoff\u2019s rules to find the currents. Strategy This circuit is sufficiently complex that the currents cannot be found using Ohm\u2019s law and the series-parallel techniques\u2014it is necessary to use Kirchhoff\u2019s rules. Currents have been labeled 1, 2, and 3 in the figure and assumptions have been made about their directions. Locations on the diagram have been labeled with letters a through h. In the solution we will apply the junction and loop rules, seeking three independent equations to allow us to solve for the three unknown currents. Solution We begin by applying Kirchhoff\u2019s first or junction rule at point a. This gives 1 = 2 + 3, (21.54) since 1 flows into the junction, while 2 and 3 flow out. Applying the junction rule at e produces exactly the same equation, so that no new information is obtained. This is a single equation with three unknowns\u2014three independent equations are needed, and so the loop rule must be applied. 936 Chapter 21 | Circuits, Bioelectricity, and DC Instruments Now we consider the loop abcdea. Going from a to b, we traverse 2 in the same (assumed) direction of the current 2, and so the change in potential is \u221222. Then going from b to c, we go from \u2013 to +, so that the change in potential is +emf1. Traversing the internal resistance 1 from c to d gives \u221221. Completing the loop by going from d to a again traverses a resistor in the same direction as its current, giving a change in potential of \u221211. The loop rule states that the changes in potential sum to zero. Thus, \u221222 + emf1 \u2212 21 \u2212 11 = \u22122(2 + 1) + emf1 \u2212 11 = 0. Substituting values from the circuit diagram for the resistances and emf, and canceling the ampere unit gives Now applying the loop rule to aefgha (we could have chosen abcdefgha as well) similarly gives + 11 + 33 + 32 \u2212 emf2= +1 1 + 3 3 + 2 \u2212 emf2 = 0. \u22123 2 + 18 \u2212 6 1 = 0. (21.55) (21.56) (21.57) Note that the signs are reversed compared with the other loop, because elements are traversed in the opposite direction. With values entered, this becomes These three equations are sufficient to solve for the three unknown currents. First, solve the second equation for 2 : + 6 1 + 2 3 \u2212 45 = 0. Now solve the third equation for 3 : 2 = 6 \u2212 2 1. 3 = 22.5 \u2212 3 1. Substituting these two new equations into the first one allows us to find a value for 1 : 1 = 2 + 3 = (6 \u2212 2 1) + (22.5 \u2212 3 1) = 28.5 \u2212 5 1. Combining terms gives 6 1 = 28.5, and 1 = 4.75 A. Substituting this value for 1 back into the fourth equation gives.50 2 = \u22123.50 A. The minus sign means 2 flows in the direction opposite to that assumed in Figure 21.27. Finally, substituting the value for 1 into the fifth equation gives 3 = 22.5\u22123 1 = 22.5 \u2212 14.25 3 = 8.25 A. (21.58) (21.59) (21.60) (21.61) (21.62) (21.63) (21.64) (21.65) (21.66) (21.67) Discussion Just as a check, we note that indeed 1 = 2 +", " 3. The results could also have been checked by entering all of the values into the equation for the abcdefgha loop. Problem-Solving Strategies for Kirchhoff\u2019s Rules 1. Make certain there is a clear circuit diagram on which you can label all known and unknown resistances, emfs, and currents. If a current is unknown, you must assign it a direction. This is necessary for determining the signs of potential changes. If you assign the direction incorrectly, the current will be found to have a negative value\u2014no harm done. 2. Apply the junction rule to any junction in the circuit. Each time the junction rule is applied, you should get an equation with a current that does not appear in a previous application\u2014if not, then the equation is redundant. 3. Apply the loop rule to as many loops as needed to solve for the unknowns in the problem. (There must be as many independent equations as unknowns.) To apply the loop rule, you must choose a direction to go around the loop. Then This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 21 | Circuits, Bioelectricity, and DC Instruments 937 carefully and consistently determine the signs of the potential changes for each element using the four bulleted points discussed above in conjunction with Figure 21.26. 4. Solve the simultaneous equations for the unknowns. This may involve many algebraic steps, requiring careful checking and rechecking. 5. Check to see whether the answers are reasonable and consistent. The numbers should be of the correct order of magnitude, neither exceedingly large nor vanishingly small. The signs should be reasonable\u2014for example, no resistance should be negative. Check to see that the values obtained satisfy the various equations obtained from applying the rules. The currents should satisfy the junction rule, for example. The material in this section is correct in theory. We should be able to verify it by making measurements of current and voltage. In fact, some of the devices used to make such measurements are straightforward applications of the principles covered so far and are explored in the next modules. As we shall see, a very basic, even profound, fact results\u2014making a measurement alters the quantity being measured. Check Your Understanding Can Kirchhoff\u2019s rules be applied to simple series and parallel circuits or are they restricted for use in more complicated circuits that are not combinations of series and parallel? Solution Kirchhoff's rules can be applied to any circuit since they are applications to circuits of two conservation laws. Conservation laws are the most broadly applicable principles in physics. It is usually mathematically simpler to use the rules for series and parallel in simpler circuits so we emphasize Kirchhoff\u2019s rules for use in more complicated situations. But the rules for series and parallel can be derived from Kirchhoff\u2019s rules. Moreover, Kirchhoff\u2019s rules can be expanded to devices other than resistors and emfs, such as capacitors, and are one of the basic analysis devices in circuit analysis. Making Connections: Parallel Resistors A simple circuit shown below \u2013 with two parallel resistors and a voltage source \u2013 is implemented in a laboratory experiment with \u025b= 6.00 \u00b1 0.02 V and R1 = 4.8 \u00b1 0.1 \u03a9 and R2 = 9.6 \u00b1 0.1 \u03a9. The values include an allowance for experimental uncertainties as they cannot be measured with perfect certainty. For example if you measure the value for a resistor a few times, you may get slightly different results. Hence values are expressed with some level of uncertainty. Figure 21.28 In the laboratory experiment the currents measured in the two resistors are I1 = 1.27 A and I2 = 0.62 A respectively. Let us examine these values using Kirchhoff\u2019s laws. For the two loops, E - I1R1 = 0 or I1 = E/R1 E - I2R2 = 0 or I2 = E/R2 Converting the given uncertainties for voltage and resistances into percentages, we get E = 6.00 V \u00b1 0.33% R1 = 4.8 \u03a9 \u00b1 2.08% R2 = 9.6 \u03a9 \u00b1 1.04% We now find the currents for the two loops. While the voltage is divided by the resistance to find the current, uncertainties in voltage and resistance are directly added to find the uncertainty in the current value. I1 = (6.00/4.8) \u00b1 (0.33%+2.08%) = 1.25 \u00b1 2.4% = 1.25 \u00b1 0.03 A I2 = (6.00/9.6) \u00b1 (0.33%+1.04%) = 0.63 \u00b1 1.4% = 0.63 \u00b1 0.01 A Finally you can check that the two measured values in this case are within the uncertainty ranges found for the currents. However", " there can also be additional experimental uncertainty in the measurements of currents. 938 Chapter 21 | Circuits, Bioelectricity, and DC Instruments 21.4 DC Voltmeters and Ammeters By the end of this section, you will be able to: Learning Objectives \u2022 Explain why a voltmeter must be connected in parallel with the circuit. \u2022 Draw a diagram showing an ammeter correctly connected in a circuit. \u2022 Describe how a galvanometer can be used as either a voltmeter or an ammeter. \u2022 Find the resistance that must be placed in series with a galvanometer to allow it to be used as a voltmeter with a given reading. \u2022 Explain why measuring the voltage or current in a circuit can never be exact. Voltmeters measure voltage, whereas ammeters measure current. Some of the meters in automobile dashboards, digital cameras, cell phones, and tuner-amplifiers are voltmeters or ammeters. (See Figure 21.29.) The internal construction of the simplest of these meters and how they are connected to the system they monitor give further insight into applications of series and parallel connections. Figure 21.29 The fuel and temperature gauges (far right and far left, respectively) in this 1996 Volkswagen are voltmeters that register the voltage output of \u201csender\u201d units, which are hopefully proportional to the amount of gasoline in the tank and the engine temperature. (credit: Christian Giersing) Voltmeters are connected in parallel with whatever device\u2019s voltage is to be measured. A parallel connection is used because objects in parallel experience the same potential difference. (See Figure 21.30, where the voltmeter is represented by the symbol V.) Ammeters are connected in series with whatever device\u2019s current is to be measured. A series connection is used because objects in series have the same current passing through them. (See Figure 21.31, where the ammeter is represented by the symbol A.) This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 21 | Circuits, Bioelectricity, and DC Instruments 939 Figure 21.30 (a) To measure potential differences in this series circuit, the voltmeter (V) is placed in parallel with the voltage source or either of the resistors. Note that terminal voltage is measured between points a and b. It is not possible to connect the voltmeter directly across the emf without including its internal resistance,. (b) A digital voltmeter in use. (credit: Messtechniker, Wikimedia Commons) Figure 21.31 An ammeter (A) is placed in series to measure current. All of the current in this circuit flows through the meter. The ammeter would have the same reading if located between points d and e or between points f and a as it does in the position shown. (Note that the script capital E stands for emf, and stands for the internal resistance of the source of potential difference.) Analog Meters: Galvanometers Analog meters have a needle that swivels to point at numbers on a scale, as opposed to digital meters, which have numerical readouts similar to a hand-held calculator. The heart of most analog meters is a device called a galvanometer, denoted by G. Current flow through a galvanometer, G, produces a proportional needle deflection. (This deflection is due to the force of a magnetic field upon a current-carrying wire.) The two crucial characteristics of a given galvanometer are its resistance and current sensitivity. Current sensitivity is the current that gives a full-scale deflection of the galvanometer\u2019s needle, the maximum current that the instrument can measure. For example, a galvanometer with a current sensitivity of 50 \u03bcA has a maximum deflection of its needle when 50 \u03bcA flows through it, reads half-scale when 25 \u03bcA flows through it, and so on. 940 Chapter 21 | Circuits, Bioelectricity, and DC Instruments If such a galvanometer has a 25- \u03a9 resistance, then a voltage of only = = full-scale reading. By connecting resistors to this galvanometer in different ways, you can use it as either a voltmeter or ammeter that can measure a broad range of voltages or currents. (25 \u03a9) = 1.25 mV produces a 50 \u03bcA Galvanometer as Voltmeter Figure 21.32 shows how a galvanometer can be used as a voltmeter by connecting it in series with a large resistance,. The value of the resistance is determined by the maximum voltage to be measured. Suppose you want 10 V to produce a fullscale deflection of a voltmeter containing a 25-\u03a9 galvanometer with a 50-\u03bcA sensitivity. Then 10 V applied to the meter must produce a current of 50 \u03bcA. The total resistance must be tot = + = = 10 V 50 \u03bcA = 200 k\u03a9, or = tot \u2212 = 200", " k\u03a9 \u2212 25 \u03a9 \u2248 200 k \u03a9. (21.68) (21.69) ( is so large that the galvanometer resistance,, is nearly negligible.) Note that 5 V applied to this voltmeter produces a halfscale deflection by producing a 25-\u03bcA current through the meter, and so the voltmeter\u2019s reading is proportional to voltage as desired. This voltmeter would not be useful for voltages less than about half a volt, because the meter deflection would be small and difficult to read accurately. For other voltage ranges, other resistances are placed in series with the galvanometer. Many meters have a choice of scales. That choice involves switching an appropriate resistance into series with the galvanometer. Figure 21.32 A large resistance placed in series with a galvanometer G produces a voltmeter, the full-scale deflection of which depends on the choice of. The larger the voltage to be measured, the larger must be. (Note that represents the internal resistance of the galvanometer.) Galvanometer as Ammeter The same galvanometer can also be made into an ammeter by placing it in parallel with a small resistance, often called the shunt resistance, as shown in Figure 21.33. Since the shunt resistance is small, most of the current passes through it, allowing an ammeter to measure currents much greater than those producing a full-scale deflection of the galvanometer. Suppose, for example, an ammeter is needed that gives a full-scale deflection for 1.0 A, and contains the same 25- \u03a9 galvanometer with its 50-\u03bcA sensitivity. Since and are in parallel, the voltage across them is the same. These drops are = G so that = G =. Solving for, and noting that G is 50 \u03bcA and is 0.999950 A, we have = G = (25 \u03a9 ) 50 \u03bcA 0.999950 A = 1.2510\u22123 \u03a9. (21.70) Figure 21.33 A small shunt resistance placed in parallel with a galvanometer G produces an ammeter, the full-scale deflection of which depends on the choice of. The larger the current to be measured, the smaller must be. Most of the current ( ) flowing through the meter is shunted through to protect the galvanometer. (Note that represents the internal resistance of the galvanometer.) Ammeters may also have multiple scales for greater flexibility in application. The various scales are achieved by switching various shunt resistances in parallel with the galvanometer\u2014the greater the maximum current to be measured, the smaller the shunt resistance must be. Taking Measurements Alters the Circuit When you use a voltmeter or ammeter, you are connecting another resistor to an existing circuit and, thus, altering the circuit. Ideally, voltmeters and ammeters do not appreciably affect the circuit, but it is instructive to examine the circumstances under which they do or do not interfere. This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 21 | Circuits, Bioelectricity, and DC Instruments 941 First, consider the voltmeter, which is always placed in parallel with the device being measured. Very little current flows through the voltmeter if its resistance is a few orders of magnitude greater than the device, and so the circuit is not appreciably affected. (See Figure 21.34(a).) (A large resistance in parallel with a small one has a combined resistance essentially equal to the small one.) If, however, the voltmeter\u2019s resistance is comparable to that of the device being measured, then the two in parallel have a smaller resistance, appreciably affecting the circuit. (See Figure 21.34(b).) The voltage across the device is not the same as when the voltmeter is out of the circuit. Figure 21.34 (a) A voltmeter having a resistance much larger than the device ( Voltmeter >> ) with which it is in parallel produces a parallel resistance essentially the same as the device and does not appreciably affect the circuit being measured. (b) Here the voltmeter has the same resistance as the device ( Voltmeter \u2245 ), so that the parallel resistance is half of what it is when the voltmeter is not connected. This is an example of a significant alteration of the circuit and is to be avoided. An ammeter is placed in series in the branch of the circuit being measured, so that its resistance adds to that branch. Normally, the ammeter\u2019s resistance is very small compared with the resistances of the devices in the circuit, and so the extra resistance is negligible. (See Figure 21.35(a).) However, if very small load resistances are involved, or if the ammeter is not as low in resistance as it should be, then the total series resistance is significantly greater, and the current in the branch being measured is reduced.", " (See Figure 21.35(b).) A practical problem can occur if the ammeter is connected incorrectly. If it was put in parallel with the resistor to measure the current in it, you could possibly damage the meter; the low resistance of the ammeter would allow most of the current in the circuit to go through the galvanometer, and this current would be larger since the effective resistance is smaller. Figure 21.35 (a) An ammeter normally has such a small resistance that the total series resistance in the branch being measured is not appreciably increased. The circuit is essentially unaltered compared with when the ammeter is absent. (b) Here the ammeter\u2019s resistance is the same as that of the branch, so that the total resistance is doubled and the current is half what it is without the ammeter. This significant alteration of the circuit is to be avoided. One solution to the problem of voltmeters and ammeters interfering with the circuits being measured is to use galvanometers with greater sensitivity. This allows construction of voltmeters with greater resistance and ammeters with smaller resistance than when less sensitive galvanometers are used. There are practical limits to galvanometer sensitivity, but it is possible to get analog meters that make measurements accurate to a few percent. Note that the inaccuracy comes from altering the circuit, not from a fault in the meter. Connections: Limits to Knowledge Making a measurement alters the system being measured in a manner that produces uncertainty in the measurement. For macroscopic systems, such as the circuits discussed in this module, the alteration can usually be made negligibly small, but it cannot be eliminated entirely. For submicroscopic systems, such as atoms, nuclei, and smaller particles, measurement alters the system in a manner that cannot be made arbitrarily small. This actually limits knowledge of the system\u2014even limiting what nature can know about itself. We shall see profound implications of this when the Heisenberg uncertainty principle is discussed in the modules on quantum mechanics. There is another measurement technique based on drawing no current at all and, hence, not altering the circuit at all. These are called null measurements and are the topic of Null Measurements. Digital meters that employ solid-state electronics and null measurements can attain accuracies of one part in 106. 950 Chapter 21 | Circuits, Bioelectricity, and DC Instruments heart defibrillator is slightly more complex than the one in Figure 21.42, to compensate for magnetic and AC effects that will be covered in Magnetism. Check Your Understanding When is the potential difference across a capacitor an emf? Solution Only when the current being drawn from or put into the capacitor is zero. Capacitors, like batteries, have internal resistance, so their output voltage is not an emf unless current is zero. This is difficult to measure in practice so we refer to a capacitor\u2019s voltage rather than its emf. But the source of potential difference in a capacitor is fundamental and it is an emf. PhET Explorations: Circuit Construction Kit (DC only) An electronics kit in your computer! Build circuits with resistors, light bulbs, batteries, and switches. Take measurements with the realistic ammeter and voltmeter. View the circuit as a schematic diagram, or switch to a life-like view. Figure 21.45 Circuit Construction Kit (DC only) (http://cnx.org/content/m55370/1.3/circuit-construction-kit-dc_en.jar) Glossary ammeter: an instrument that measures current analog meter: a measuring instrument that gives a readout in the form of a needle movement over a marked gauge bridge device: a device that forms a bridge between two branches of a circuit; some bridge devices are used to make null measurements in circuits capacitance: the maximum amount of electric potential energy that can be stored (or separated) for a given electric potential capacitor: an electrical component used to store energy by separating electric charge on two opposing plates conservation laws: require that energy and charge be conserved in a system current: the flow of charge through an electric circuit past a given point of measurement current sensitivity: the maximum current that a galvanometer can read digital meter: a measuring instrument that gives a readout in a digital form electromotive force (emf): the potential difference of a source of electricity when no current is flowing; measured in volts full-scale deflection: the maximum deflection of a galvanometer needle, also known as current sensitivity; a galvanometer with a full-scale deflection of 50 \u03bcA has a maximum deflection of its needle when 50 \u03bcA flows through it galvanometer: an analog measuring device, denoted by G, that measures current flow using a needle deflection caused by a magnetic field force acting upon a current-carrying wire internal resistance: the amount of resistance within the voltage source Joule\u2019s law: the relationship between potential electrical power, voltage, and resistance in an electrical circuit, given by: = junction rule:", " Kirchhoff\u2019s first rule, which applies the conservation of charge to a junction; current is the flow of charge; thus, whatever charge flows into the junction must flow out; the rule can be stated 1 = 2 + 3 Kirchhoff\u2019s rules: a set of two rules, based on conservation of charge and energy, governing current and changes in potential in an electric circuit loop rule: Kirchhoff\u2019s second rule, which states that in a closed loop, whatever energy is supplied by emf must be transferred into other forms by devices in the loop, since there are no other ways in which energy can be transferred into or out of This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 21 | Circuits, Bioelectricity, and DC Instruments 951 the circuit. Thus, the emf equals the sum of the (voltage) drops in the loop and can be stated: emf = + 1 + 2 null measurements: methods of measuring current and voltage more accurately by balancing the circuit so that no current flows through the measurement device ohmmeter: an instrument that applies a voltage to a resistance, measures the current, calculates the resistance using Ohm\u2019s law, and provides a readout of this calculated resistance Ohm\u2019s law: the relationship between current, voltage, and resistance within an electrical circuit: = parallel: the wiring of resistors or other components in an electrical circuit such that each component receives an equal voltage from the power source; often pictured in a ladder-shaped diagram, with each component on a rung of the ladder potential difference: the difference in electric potential between two points in an electric circuit, measured in volts potentiometer: a null measurement device for measuring potentials (voltages) RC circuit: a circuit that contains both a resistor and a capacitor resistance: causing a loss of electrical power in a circuit resistor: a component that provides resistance to the current flowing through an electrical circuit series: a sequence of resistors or other components wired into a circuit one after the other shunt resistance: a small resistance placed in parallel with a galvanometer G to produce an ammeter; the larger the current to be measured, the smaller must be; most of the current flowing through the meter is shunted through to protect the galvanometer terminal voltage: the voltage measured across the terminals of a source of potential difference voltage: the electrical potential energy per unit charge; electric pressure created by a power source, such as a battery voltage drop: the loss of electrical power as a current travels through a resistor, wire or other component voltmeter: an instrument that measures voltage Wheatstone bridge: a null measurement device for calculating resistance by balancing potential drops in a circuit Section Summary 21.1 Resistors in Series and Parallel \u2022 The total resistance of an electrical circuit with resistors wired in a series is the sum of the individual resistances: s = 1 + 2 + 3 +.... \u2022 Each resistor in a series circuit has the same amount of current flowing through it. \u2022 The voltage drop, or power dissipation, across each individual resistor in a series is different, and their combined total adds up to the power source input. \u2022 The total resistance of an electrical circuit with resistors wired in parallel is less than the lowest resistance of any of the components and can be determined using the formula +.... \u2022 Each resistor in a parallel circuit has the same full voltage of the source applied to it. \u2022 The current flowing through each resistor in a parallel circuit is different, depending on the resistance. \u2022 If a more complex connection of resistors is a combination of series and parallel, it can be reduced to a single equivalent resistance by identifying its various parts as series or parallel, reducing each to its equivalent, and continuing until a single resistance is eventually reached. 21.2 Electromotive Force: Terminal Voltage \u2022 All voltage sources have two fundamental parts\u2014a source of electrical energy that has a characteristic electromotive force (emf), and an internal resistance. \u2022 The emf is the potential difference of a source when no current is flowing. \u2022 The numerical value of the emf depends on the source of potential difference. \u2022 The internal resistance of a voltage source affects the output voltage when a current flows. \u2022 The voltage output of a device is called its terminal voltage and is given by = emf \u2212, where is the electric current and is positive when flowing away from the positive terminal of the voltage source. 952 Chapter 21 | Circuits, Bioelectricity, and DC Instruments \u2022 When multiple voltage sources are in series, their internal resistances add and their emfs add algebraically. \u2022 Solar cells can be wired in series or parallel to provide increased voltage or current, respectively. 21.3 Kirchhoff\u2019s Rules \u2022 Kirchhoff\u2019s rules can be used to analyze any circuit, simple or complex. \u2022 Kirchhoff\u2019s first rule\u2014the junction rule: The sum of all currents entering a junction must equal the sum", " of all currents leaving the junction. \u2022 Kirchhoff\u2019s second rule\u2014the loop rule: The algebraic sum of changes in potential around any closed circuit path (loop) must be zero. \u2022 The two rules are based, respectively, on the laws of conservation of charge and energy. \u2022 When calculating potential and current using Kirchhoff\u2019s rules, a set of conventions must be followed for determining the correct signs of various terms. \u2022 The simpler series and parallel rules are special cases of Kirchhoff\u2019s rules. 21.4 DC Voltmeters and Ammeters \u2022 Voltmeters measure voltage, and ammeters measure current. \u2022 A voltmeter is placed in parallel with the voltage source to receive full voltage and must have a large resistance to limit its effect on the circuit. \u2022 An ammeter is placed in series to get the full current flowing through a branch and must have a small resistance to limit its effect on the circuit. \u2022 Both can be based on the combination of a resistor and a galvanometer, a device that gives an analog reading of current. \u2022 Standard voltmeters and ammeters alter the circuit being measured and are thus limited in accuracy. 21.5 Null Measurements \u2022 Null measurement techniques achieve greater accuracy by balancing a circuit so that no current flows through the measuring device. \u2022 One such device, for determining voltage, is a potentiometer. \u2022 Another null measurement device, for determining resistance, is the Wheatstone bridge. \u2022 Other physical quantities can also be measured with null measurement techniques. 21.6 DC Circuits Containing Resistors and Capacitors \u2022 An circuit is one that has both a resistor and a capacitor. \u2022 The time constant for an circuit is =. \u2022 When an initially uncharged ( 0 = 0 at = 0 ) capacitor in series with a resistor is charged by a DC voltage source, the voltage rises, asymptotically approaching the emf of the voltage source; as a function of time, = emf(1 \u2212 \u2212 / )(charging). \u2022 Within the span of each time constant, the voltage rises by 0.632 of the remaining value, approaching the final voltage asymptotically. If a capacitor with an initial voltage 0 is discharged through a resistor starting at = 0, then its voltage decreases exponentially as given by In each time constant, the voltage falls by 0.368 of its remaining initial value, approaching zero asymptotically. = 0\u2212 / (discharging). \u2022 \u2022 Conceptual Questions 21.1 Resistors in Series and Parallel 1. A switch has a variable resistance that is nearly zero when closed and extremely large when open, and it is placed in series with the device it controls. Explain the effect the switch in Figure 21.46 has on current when open and when closed. This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 21 | Circuits, Bioelectricity, and DC Instruments 953 Figure 21.46 A switch is ordinarily in series with a resistance and voltage source. Ideally, the switch has nearly zero resistance when closed but has an extremely large resistance when open. (Note that in this diagram, the script E represents the voltage (or electromotive force) of the battery.) 2. What is the voltage across the open switch in Figure 21.46? 3. There is a voltage across an open switch, such as in Figure 21.46. Why, then, is the power dissipated by the open switch small? 4. Why is the power dissipated by a closed switch, such as in Figure 21.46, small? 5. A student in a physics lab mistakenly wired a light bulb, battery, and switch as shown in Figure 21.47. Explain why the bulb is on when the switch is open, and off when the switch is closed. (Do not try this\u2014it is hard on the battery!) Figure 21.47 A wiring mistake put this switch in parallel with the device represented by. (Note that in this diagram, the script E represents the voltage (or electromotive force) of the battery.) 6. Knowing that the severity of a shock depends on the magnitude of the current through your body, would you prefer to be in series or parallel with a resistance, such as the heating element of a toaster, if shocked by it? Explain. 7. Would your headlights dim when you start your car\u2019s engine if the wires in your automobile were superconductors? (Do not neglect the battery\u2019s internal resistance.) Explain. 8. Some strings of holiday lights are wired in series to save wiring costs. An old version utilized bulbs that break the electrical connection, like an open switch, when they burn out. If one such bulb burns out, what happens to the others? If such a string operates on 120 V and has 40 identical bulbs, what is the normal operating voltage of each? Newer versions use bulbs that short circuit, like a closed switch, when they burn", " out. If one such bulb burns out, what happens to the others? If such a string operates on 120 V and has 39 remaining identical bulbs, what is then the operating voltage of each? 9. If two household lightbulbs rated 60 W and 100 W are connected in series to household power, which will be brighter? Explain. 10. Suppose you are doing a physics lab that asks you to put a resistor into a circuit, but all the resistors supplied have a larger resistance than the requested value. How would you connect the available resistances to attempt to get the smaller value asked for? 11. Before World War II, some radios got power through a \u201cresistance cord\u201d that had a significant resistance. Such a resistance cord reduces the voltage to a desired level for the radio\u2019s tubes and the like, and it saves the expense of a transformer. Explain why resistance cords become warm and waste energy when the radio is on. 12. Some light bulbs have three power settings (not including zero), obtained from multiple filaments that are individually switched and wired in parallel. What is the minimum number of filaments needed for three power settings? 21.2 Electromotive Force: Terminal Voltage 13. Is every emf a potential difference? Is every potential difference an emf? Explain. 14. Explain which battery is doing the charging and which is being charged in Figure 21.48. 954 Chapter 21 | Circuits, Bioelectricity, and DC Instruments Figure 21.48 15. Given a battery, an assortment of resistors, and a variety of voltage and current measuring devices, describe how you would determine the internal resistance of the battery. 16. Two different 12-V automobile batteries on a store shelf are rated at 600 and 850 \u201ccold cranking amps.\u201d Which has the smallest internal resistance? 17. What are the advantages and disadvantages of connecting batteries in series? In parallel? 18. Semitractor trucks use four large 12-V batteries. The starter system requires 24 V, while normal operation of the truck\u2019s other electrical components utilizes 12 V. How could the four batteries be connected to produce 24 V? To produce 12 V? Why is 24 V better than 12 V for starting the truck\u2019s engine (a very heavy load)? 21.3 Kirchhoff\u2019s Rules 19. Can all of the currents going into the junction in Figure 21.49 be positive? Explain. Figure 21.49 20. Apply the junction rule to junction b in Figure 21.50. Is any new information gained by applying the junction rule at e? (In the figure, each emf is represented by script E.) Figure 21.50 21. (a) What is the potential difference going from point a to point b in Figure 21.50? (b) What is the potential difference going from c to b? (c) From e to g? (d) From e to d? 22. Apply the loop rule to loop afedcba in Figure 21.50. 23. Apply the loop rule to loops abgefa and cbgedc in Figure 21.50. 21.4 DC Voltmeters and Ammeters This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 21 | Circuits, Bioelectricity, and DC Instruments 955 24. Why should you not connect an ammeter directly across a voltage source as shown in Figure 21.51? (Note that script E in the figure stands for emf.) Figure 21.51 25. Suppose you are using a multimeter (one designed to measure a range of voltages, currents, and resistances) to measure current in a circuit and you inadvertently leave it in a voltmeter mode. What effect will the meter have on the circuit? What would happen if you were measuring voltage but accidentally put the meter in the ammeter mode? 26. Specify the points to which you could connect a voltmeter to measure the following potential differences in Figure 21.52: (a) the potential difference of the voltage source; (b) the potential difference across 1 ; (c) across 2 ; (d) across 3 ; (e) across 2 and 3. Note that there may be more than one answer to each part. Figure 21.52 27. To measure currents in Figure 21.52, you would replace a wire between two points with an ammeter. Specify the points between which you would place an ammeter to measure the following: (a) the total current; (b) the current flowing through 1 ; (c) through 2 ; (d) through 3. Note that there may be more than one answer to each part. 21.5 Null Measurements 28. Why can a null measurement be more accurate than one using standard voltmeters and ammeters? What factors limit the accuracy of null measurements? 29. If a potentiometer is used to measure cell emfs on the order of a", " few volts, why is it most accurate for the standard emfs to be the same order of magnitude and the resistances to be in the range of a few ohms? 21.6 DC Circuits Containing Resistors and Capacitors 30. Regarding the units involved in the relationship =, verify that the units of resistance times capacitance are time, that is, \u03a9 \u22c5 F = s. 31. The time constant in heart defibrillation is crucial to limiting the time the current flows. If the capacitance in the defibrillation unit is fixed, how would you manipulate resistance in the circuit to adjust the constant? Would an adjustment of the applied voltage also be needed to ensure that the current delivered has an appropriate value? 32. When making an ECG measurement, it is important to measure voltage variations over small time intervals. The time is limited by the constant of the circuit\u2014it is not possible to measure time variations shorter than. How would you manipulate and in the circuit to allow the necessary measurements? 33. Draw two graphs of charge versus time on a capacitor. Draw one for charging an initially uncharged capacitor in series with a resistor, as in the circuit in Figure 21.41, starting from t = 0. Draw the other for discharging a capacitor through a resistor, as in the circuit in Figure 21.42, starting at t = 0, with an initial charge 0. Show at least two intervals of. 34. When charging a capacitor, as discussed in conjunction with Figure 21.41, how long does it take for the voltage on the capacitor to reach emf? Is this a problem? 956 Chapter 21 | Circuits, Bioelectricity, and DC Instruments 35. When discharging a capacitor, as discussed in conjunction with Figure 21.42, how long does it take for the voltage on the capacitor to reach zero? Is this a problem? 36. Referring to Figure 21.41, draw a graph of potential difference across the resistor versus time, showing at least two intervals of. Also draw a graph of current versus time for this situation. 37. A long, inexpensive extension cord is connected from inside the house to a refrigerator outside. The refrigerator doesn\u2019t run as it should. What might be the problem? 38. In Figure 21.44, does the graph indicate the time constant is shorter for discharging than for charging? Would you expect ionized gas to have low resistance? How would you adjust to get a longer time between flashes? Would adjusting affect the discharge time? 39. An electronic apparatus may have large capacitors at high voltage in the power supply section, presenting a shock hazard even when the apparatus is switched off. A \u201cbleeder resistor\u201d is therefore placed across such a capacitor, as shown schematically in Figure 21.53, to bleed the charge from it after the apparatus is off. Why must the bleeder resistance be much greater than the effective resistance of the rest of the circuit? How does this affect the time constant for discharging the capacitor? Figure 21.53 A bleeder resistor bl discharges the capacitor in this electronic device once it is switched off. This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 21 | Circuits, Bioelectricity, and DC Instruments 957 Problems & Exercises 21.1 Resistors in Series and Parallel Note: Data taken from figures can be assumed to be accurate to three significant digits. 1. (a) What is the resistance of ten 275-\u03a9 resistors connected in series? (b) In parallel? 2. (a) What is the resistance of a 1.00\u00d7102 -\u03a9, a 2.50-k\u03a9, and a 4.00-k \u03a9 resistor connected in series? (b) In parallel? 3. What are the largest and smallest resistances you can obtain by connecting a 36.0-\u03a9, a 50.0-\u03a9, and a 700-\u03a9 resistor together? 4. An 1800-W toaster, a 1400-W electric frying pan, and a 75-W lamp are plugged into the same outlet in a 15-A, 120-V circuit. (The three devices are in parallel when plugged into the same socket.). (a) What current is drawn by each device? (b) Will this combination blow the 15-A fuse? 5. Your car\u2019s 30.0-W headlight and 2.40-kW starter are ordinarily connected in parallel in a 12.0-V system. What power would one headlight and the starter consume if connected in series to a 12.0-V battery? (Neglect any other resistance in the circuit and any change in resistance in the two devices.) 6. (a) Given a 48.0-V battery and 24.0-\u03a9 and 96.0-\u03a9 resistors, find the current and power for each when connected in", " series. (b) Repeat when the resistances are in parallel. 7. Referring to the example combining series and parallel circuits and Figure 21.6, calculate 3 in the following two different ways: (a) from the known values of and 2 ; (b) using Ohm\u2019s law for 3. In both parts explicitly show how you follow the steps in the Problem-Solving Strategies for Series and Parallel Resistors. 8. Referring to Figure 21.6: (a) Calculate 3 and note how it compares with 3 found in the first two example problems in this module. (b) Find the total power supplied by the source and compare it with the sum of the powers dissipated by the resistors. 9. Refer to Figure 21.7 and the discussion of lights dimming when a heavy appliance comes on. (a) Given the voltage source is 120 V, the wire resistance is 0.400 \u03a9, and the bulb is nominally 75.0 W, what power will the bulb dissipate if a total of 15.0 A passes through the wires when the motor comes on? Assume negligible change in bulb resistance. (b) What power is consumed by the motor? 10. A 240-kV power transmission line carrying 5.00\u00d7102 A is hung from grounded metal towers by ceramic insulators, each having a 1.00109 -\u03a9 resistance. Figure 21.54. (a) What is the resistance to ground of 100 of these insulators? (b) Calculate the power dissipated by 100 of them. (c) What fraction of the power carried by the line is this? Explicitly show how you follow the steps in the Problem-Solving Strategies for Series and Parallel Resistors. Figure 21.54 High-voltage (240-kV) transmission line carrying 5.00\u00d7102 A is hung from a grounded metal transmission tower. The row of ceramic insulators provide 1.00\u00d7109 \u03a9 of resistance each. 11. Show that if two resistors 1 and 2 are combined and one is much greater than the other ( 1 >>2 ): (a) Their series resistance is very nearly equal to the greater resistance 1. (b) Their parallel resistance is very nearly equal to smaller resistance 2. 12. Unreasonable Results Two resistors, one having a resistance of 145 \u03a9, are connected in parallel to produce a total resistance of 150 \u03a9. (a) What is the value of the second resistance? (b) What is unreasonable about this result? (c) Which assumptions are unreasonable or inconsistent? 13. Unreasonable Results Two resistors, one having a resistance of 900 k\u03a9, are connected in series to produce a total resistance of 0.500 M\u03a9. (a) What is the value of the second resistance? (b) What is unreasonable about this result? (c) Which assumptions are unreasonable or inconsistent? 21.2 Electromotive Force: Terminal Voltage 14. Standard automobile batteries have six lead-acid cells in series, creating a total emf of 12.0 V. What is the emf of an individual lead-acid cell? 15. Carbon-zinc dry cells (sometimes referred to as nonalkaline cells) have an emf of 1.54 V, and they are produced as single cells or in various combinations to form other voltages. (a) How many 1.54-V cells are needed to make the common 9-V battery used in many small electronic devices? (b) What is the actual emf of the approximately 9-V battery? (c) Discuss how internal resistance in the series connection of cells will affect the terminal voltage of this approximately 9-V battery. 16. What is the output voltage of a 3.0000-V lithium cell in a digital wristwatch that draws 0.300 mA, if the cell\u2019s internal resistance is 2.00 \u03a9? 17. (a) What is the terminal voltage of a large 1.54-V carbonzinc dry cell used in a physics lab to supply 2.00 A to a circuit, if the cell\u2019s internal resistance is 0.100 \u03a9? (b) How much 958 Chapter 21 | Circuits, Bioelectricity, and DC Instruments represent the situation. (b) If the internal resistance of the power supply is 2000 \u03a9, what is the current through his body? (c) What is the power dissipated in his body? (d) If the power supply is to be made safe by increasing its internal resistance, what should the internal resistance be for the maximum current in this situation to be 1.00 mA or less? (e) Will this modification compromise the effectiveness of the power supply for driving low-resistance devices? Explain your reasoning. 27. Electric fish generate current with biological cells called electroplaques, which are physiological emf devices. The electroplaques in the South American eel are arranged in 140 rows", ", each row stretching horizontally along the body and each containing 5000 electroplaques. Each electroplaque has an emf of 0.15 V and internal resistance of 0.25 \u03a9. If the water surrounding the fish has resistance of 800 \u03a9, how much current can the eel produce in water from near its head to near its tail? 28. Integrated Concepts A 12.0-V emf automobile battery has a terminal voltage of 16.0 V when being charged by a current of 10.0 A. (a) What is the battery\u2019s internal resistance? (b) What power is dissipated inside the battery? (c) At what rate (in \u00baC/min ) will its temperature increase if its mass is 20.0 kg and it has a specific heat of 0.300 kcal/kg \u22c5 \u00baC, assuming no heat escapes? 29. Unreasonable Results A 1.58-V alkaline cell with a 0.200-\u03a9 internal resistance is supplying 8.50 A to a load. (a) What is its terminal voltage? (b) What is the value of the load resistance? (c) What is unreasonable about these results? (d) Which assumptions are unreasonable or inconsistent? 30. Unreasonable Results (a) What is the internal resistance of a 1.54-V dry cell that supplies 1.00 W of power to a 15.0-\u03a9 bulb? (b) What is unreasonable about this result? (c) Which assumptions are unreasonable or inconsistent? 21.3 Kirchhoff\u2019s Rules 31. Apply the loop rule to loop abcdefgha in Figure 21.27. 32. Apply the loop rule to loop aedcba in Figure 21.27. 33. Verify the second equation in Example 21.5 by substituting the values found for the currents 1 and 2. 34. Verify the third equation in Example 21.5 by substituting the values found for the currents 1 and 3. 35. Apply the junction rule at point a in Figure 21.55. electrical power does the cell produce? (c) What power goes to its load? 18. What is the internal resistance of an automobile battery that has an emf of 12.0 V and a terminal voltage of 15.0 V while a current of 8.00 A is charging it? 19. (a) Find the terminal voltage of a 12.0-V motorcycle battery having a 0.600-\u03a9 internal resistance, if it is being charged by a current of 10.0 A. (b) What is the output voltage of the battery charger? 20. A car battery with a 12-V emf and an internal resistance of 0.050 \u03a9 is being charged with a current of 60 A. Note that in this process the battery is being charged. (a) What is the potential difference across its terminals? (b) At what rate is thermal energy being dissipated in the battery? (c) At what rate is electric energy being converted to chemical energy? (d) What are the answers to (a) and (b) when the battery is used to supply 60 A to the starter motor? 21. The hot resistance of a flashlight bulb is 2.30 \u03a9, and it is run by a 1.58-V alkaline cell having a 0.100-\u03a9 internal resistance. (a) What current flows? (b) Calculate the power supplied to the bulb using 2 bulb. (c) Is this power the same as calculated using 2 bulb? 22. The label on a portable radio recommends the use of rechargeable nickel-cadmium cells (nicads), although they have a 1.25-V emf while alkaline cells have a 1.58-V emf. The radio has a 3.20-\u03a9 resistance. (a) Draw a circuit diagram of the radio and its batteries. Now, calculate the power delivered to the radio. (b) When using Nicad cells each having an internal resistance of 0.0400 \u03a9. (c) When using alkaline cells each having an internal resistance of 0.200 \u03a9. (d) Does this difference seem significant, considering that the radio\u2019s effective resistance is lowered when its volume is turned up? 23. An automobile starter motor has an equivalent resistance of 0.0500 \u03a9 and is supplied by a 12.0-V battery with a 0.0100-\u03a9 internal resistance. (a) What is the current to the motor? (b) What voltage is applied to it? (c) What power is supplied to the motor? (d) Repeat these calculations for when the battery connections are corroded and add 0.0900 \u03a9 to the circuit. (Significant problems are caused by even small amounts of unwanted resistance in low-voltage, high-current applications.) 24. A child\u2019s electronic toy is supplied by three 1", ".58-V alkaline cells having internal resistances of 0.0200 \u03a9 in series with a 1.53-V carbon-zinc dry cell having a 0.100-\u03a9 internal resistance. The load resistance is 10.0 \u03a9. (a) Draw a circuit diagram of the toy and its batteries. (b) What current flows? (c) How much power is supplied to the load? (d) What is the internal resistance of the dry cell if it goes bad, resulting in only 0.500 W being supplied to the load? 25. (a) What is the internal resistance of a voltage source if its terminal voltage drops by 2.00 V when the current supplied increases by 5.00 A? (b) Can the emf of the voltage source be found with the information supplied? 26. A person with body resistance between his hands of 10.0 k \u03a9 accidentally grasps the terminals of a 20.0-kV power supply. (Do NOT do this!) (a) Draw a circuit diagram to This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 21 | Circuits, Bioelectricity, and DC Instruments 959 44. Find the resistance that must be placed in series with a 25.0-\u03a9 galvanometer having a 50.0-\u03bcA sensitivity (the same as the one discussed in the text) to allow it to be used as a voltmeter with a 0.100-V full-scale reading. 45. Find the resistance that must be placed in series with a 25.0-\u03a9 galvanometer having a 50.0-\u03bcA sensitivity (the same as the one discussed in the text) to allow it to be used as a voltmeter with a 3000-V full-scale reading. Include a circuit diagram with your solution. 46. Find the resistance that must be placed in parallel with a 25.0-\u03a9 galvanometer having a 50.0-\u03bcA sensitivity (the same as the one discussed in the text) to allow it to be used as an ammeter with a 10.0-A full-scale reading. Include a circuit diagram with your solution. 47. Find the resistance that must be placed in parallel with a 25.0-\u03a9 galvanometer having a 50.0-\u03bcA sensitivity (the same as the one discussed in the text) to allow it to be used as an ammeter with a 300-mA full-scale reading. 48. Find the resistance that must be placed in series with a 10.0-\u03a9 galvanometer having a 100-\u03bcA sensitivity to allow it to be used as a voltmeter with: (a) a 300-V full-scale reading, and (b) a 0.300-V full-scale reading. 49. Find the resistance that must be placed in parallel with a 10.0-\u03a9 galvanometer having a 100-\u03bcA sensitivity to allow it to be used as an ammeter with: (a) a 20.0-A full-scale reading, and (b) a 100-mA full-scale reading. 50. Suppose you measure the terminal voltage of a 1.585-V alkaline cell having an internal resistance of 0.100 \u03a9 by placing a 1.00-k \u03a9 voltmeter across its terminals. (See Figure 21.57.) (a) What current flows? (b) Find the terminal voltage. (c) To see how close the measured terminal voltage is to the emf, calculate their ratio. Figure 21.57 51. Suppose you measure the terminal voltage of a 3.200-V lithium cell having an internal resistance of 5.00 \u03a9 by placing a 1.00-k \u03a9 voltmeter across its terminals. (a) What current flows? (b) Find the terminal voltage. (c) To see how close the measured terminal voltage is to the emf, calculate their ratio. 52. A certain ammeter has a resistance of 5.0010\u22125 \u03a9 on its 3.00-A scale and contains a 10.0-\u03a9 galvanometer. What is the sensitivity of the galvanometer? 53. A 1.00-M\u03a9 voltmeter is placed in parallel with a 75.0-k \u03a9 resistor in a circuit. (a) Draw a circuit diagram of the connection. (b) What is the resistance of the combination? (c) If the voltage across the combination is kept the same as it was across the 75.0-k \u03a9 resistor alone, what is the percent increase in current? (d) If the current through the combination is kept the same as it was through the 75.0-k \u03a9 resistor alone, what is the percentage decrease in voltage? (e) Are the changes found in parts (c) and (d) significant? Discuss. Figure 21.55 36. Apply the loop rule", " to loop abcdefghija in Figure 21.55. 37. Apply the loop rule to loop akledcba in Figure 21.55. 38. Find the currents flowing in the circuit in Figure 21.55. Explicitly show how you follow the steps in the ProblemSolving Strategies for Series and Parallel Resistors. 39. Solve Example 21.5, but use loop abcdefgha instead of loop akledcba. Explicitly show how you follow the steps in the Problem-Solving Strategies for Series and Parallel Resistors. 40. Find the currents flowing in the circuit in Figure 21.50. 41. Unreasonable Results Consider the circuit in Figure 21.56, and suppose that the emfs are unknown and the currents are given to be 1 = 5.00 A, 2 = 3.0 A, and 3 = \u20132.00 A. (a) Could you find the emfs? (b) What is wrong with the assumptions? Figure 21.56 21.4 DC Voltmeters and Ammeters 42. What is the sensitivity of the galvanometer (that is, what current gives a full-scale deflection) inside a voltmeter that has a 1.00-M \u03a9 resistance on its 30.0-V scale? 43. What is the sensitivity of the galvanometer (that is, what current gives a full-scale deflection) inside a voltmeter that has a 25.0-k \u03a9 resistance on its 100-V scale? 960 Chapter 21 | Circuits, Bioelectricity, and DC Instruments 54. A 0.0200-\u03a9 ammeter is placed in series with a 10.00-\u03a9 resistor in a circuit. (a) Draw a circuit diagram of the connection. (b) Calculate the resistance of the combination. (c) If the voltage is kept the same across the combination as it was through the 10.00-\u03a9 resistor alone, what is the percent decrease in current? (d) If the current is kept the same through the combination as it was through the 10.00-\u03a9 resistor alone, what is the percent increase in voltage? (e) Are the changes found in parts (c) and (d) significant? Discuss. 55. Unreasonable Results Suppose you have a 40.0-\u03a9 galvanometer with a 25.0-\u03bcA sensitivity. (a) What resistance would you put in series with it to allow it to be used as a voltmeter that has a full-scale deflection for 0.500 mV? (b) What is unreasonable about this result? (c) Which assumptions are responsible? 56. Unreasonable Results (a) What resistance would you put in parallel with a 40.0-\u03a9 galvanometer having a 25.0-\u03bcA sensitivity to allow it to be used as an ammeter that has a full-scale deflection for 10.0-\u03bcA? (b) What is unreasonable about this result? (c) Which assumptions are responsible? 21.5 Null Measurements 57. What is the emfx of a cell being measured in a potentiometer, if the standard cell\u2019s emf is 12.0 V and the potentiometer balances for x = 5.000 \u03a9 and s = 2.500 \u03a9? 58. Calculate the emfx of a dry cell for which a potentiometer is balanced when x = 1.200 \u03a9, while an alkaline standard cell with an emf of 1.600 V requires s = 1.247 \u03a9 to balance the potentiometer. 59. When an unknown resistance x is placed in a Wheatstone bridge, it is possible to balance the bridge by adjusting 3 to be 2500 \u03a9. What is x if 2 1 = 0.625? 60. To what value must you adjust 3 to balance a Wheatstone bridge, if the unknown resistance x is 100 \u03a9, 1 is 50.0 \u03a9, and 2 is 175 \u03a9? 61. (a) What is the unknown emfx in a potentiometer that balances when x is 10.0 \u03a9, and balances when s is 15.0 \u03a9 for a standard 3.000-V emf? (b) The same emfx is placed in the same potentiometer, which now balances when s is 15.0 \u03a9 for a standard emf of 3.100 V. At what resistance x will the potentiometer balance? 62. Suppose you want to measure resistances in the range from 10.0 \u03a9 to 10.0 k\u03a9 using a Wheatstone bridge that = 2.000. Over what range should 3 be has 2 1 adjustable? This content is available for free at http://cnx.org/content/col11844/1.13 21.6 DC Circuits Containing Resistors and Capacitors 63. The timing device in an automobile\u2019s intermittent wiper", " system is based on an time constant and utilizes a 0.500-\u03bcF capacitor and a variable resistor. Over what range must be made to vary to achieve time constants from 2.00 to 15.0 s? 64. A heart pacemaker fires 72 times a minute, each time a 25.0-nF capacitor is charged (by a battery in series with a resistor) to 0.632 of its full voltage. What is the value of the resistance? 65. The duration of a photographic flash is related to an time constant, which is 0.100 \u03bcs for a certain camera. (a) If the resistance of the flash lamp is 0.0400 \u03a9 during discharge, what is the size of the capacitor supplying its energy? (b) What is the time constant for charging the capacitor, if the charging resistance is 800 k\u03a9? 66. A 2.00- and a 7.50-\u03bcF capacitor can be connected in series or parallel, as can a 25.0- and a 100-k\u03a9 resistor. Calculate the four time constants possible from connecting the resulting capacitance and resistance in series. 67. After two time constants, what percentage of the final voltage, emf, is on an initially uncharged capacitor, charged through a resistance? 68. A 500-\u03a9 resistor, an uncharged 1.50-\u03bcF capacitor, and a 6.16-V emf are connected in series. (a) What is the initial current? (b) What is the time constant? (c) What is the current after one time constant? (d) What is the voltage on the capacitor after one time constant? 69. A heart defibrillator being used on a patient has an time constant of 10.0 ms due to the resistance of the patient and the capacitance of the defibrillator. (a) If the defibrillator has an 8.00-\u03bcF capacitance, what is the resistance of the path through the patient? (You may neglect the capacitance of the patient and the resistance of the defibrillator.) (b) If the initial voltage is 12.0 kV, how long does it take to decline to 6.00\u00d7102 V? 70. An ECG monitor must have an time constant less than 1.00\u00d7102 \u03bcs to be able to measure variations in voltage over small time intervals. (a) If the resistance of the circuit (due mostly to that of the patient\u2019s chest) is 1.00 k\u03a9, what is the maximum capacitance of the circuit? (b) Would it be difficult in practice to limit the capacitance to less than the value found in (a)? 71. Figure 21.58 shows how a bleeder resistor is used to discharge a capacitor after an electronic device is shut off, allowing a person to work on the electronics with less risk of shock. (a) What is the time constant? (b) How long will it take to reduce the voltage on the capacitor to 0.250% (5% of 5%) of its full value once discharge begins? (c) If the capacitor is charged to a voltage 0 through a 100-\u03a9 resistance, calculate the time it takes to rise to 0.8650 (This is about two time constants.) Chapter 21 | Circuits, Bioelectricity, and DC Instruments 961 Consider a rechargeable lithium cell that is to be used to power a camcorder. Construct a problem in which you calculate the internal resistance of the cell during normal operation. Also, calculate the minimum voltage output of a battery charger to be used to recharge your lithium cell. Among the things to be considered are the emf and useful terminal voltage of a lithium cell and the current it should be able to supply to a camcorder. Figure 21.58 72. Using the exact exponential treatment, find how much time is required to discharge a 250-\u03bcF capacitor through a 500-\u03a9 resistor down to 1.00% of its original voltage. 73. Using the exact exponential treatment, find how much time is required to charge an initially uncharged 100-pF capacitor through a 75.0-M \u03a9 resistor to 90.0% of its final voltage. 74. Integrated Concepts If you wish to take a picture of a bullet traveling at 500 m/s, then a very brief flash of light produced by an discharge through a flash tube can limit blurring. Assuming 1.00 mm of motion during one constant is acceptable, and given that the flash is driven by a 600-\u03bcF capacitor, what is the resistance in the flash tube? 75. Integrated Concepts A flashing lamp in a Christmas earring is based on an discharge of a capacitor through its resistance. The effective duration of the flash is 0.250 s, during which it produces an average 0.500 W from an average 3.00 V. (a) What energy does it dissipate? (b) How much", " charge moves through the lamp? (c) Find the capacitance. (d) What is the resistance of the lamp? 76. Integrated Concepts A 160-\u03bcF capacitor charged to 450 V is discharged through a 31.2-k \u03a9 resistor. (a) Find the time constant. (b) Calculate the temperature increase of the resistor, given that its mass is 2.50 g and its specific heat is 1.67 kJ kg \u22c5 \u00baC, noting that most of the thermal energy is retained in the short time of the discharge. (c) Calculate the new resistance, assuming it is pure carbon. (d) Does this change in resistance seem significant? 77. Unreasonable Results (a) Calculate the capacitance needed to get an time constant of 1.00\u00d7103 s with a 0.100-\u03a9 resistor. (b) What is unreasonable about this result? (c) Which assumptions are responsible? 78. Construct Your Own Problem Consider a camera\u2019s flash unit. Construct a problem in which you calculate the size of the capacitor that stores energy for the flash lamp. Among the things to be considered are the voltage applied to the capacitor, the energy needed in the flash and the associated charge needed on the capacitor, the resistance of the flash lamp during discharge, and the desired time constant. 79. Construct Your Own Problem 962 Chapter 21 | Circuits, Bioelectricity, and DC Instruments Test Prep for AP\u00ae Courses 21.1 Resistors in Series and Parallel 1. Figure 21.59 The figure above shows a circuit containing two batteries and three identical resistors with resistance R. Which of the following changes to the circuit will result in an increase in the current at point P? Select two answers. a. Reversing the connections to the 14 V battery. b. Removing the 2 V battery and connecting the wires to close the left loop. c. Rearranging the resistors so all three are in series. d. Removing the branch containing resistor Z. 2. In a circuit, a parallel combination of six 1.6-k\u03a9 resistors is connected in series with a parallel combination of four 2.4-k\u03a9 resistors. If the source voltage is 24 V, what will be the percentage of total current in one of the 2.4-k\u03a9 resistors? a. 10% b. 12% c. 20% d. 25% 3. If the circuit in the previous question is modified by removing some of the 1.6 k\u03a9 resistors, the total current in the circuit is 24 mA. How many resistors were removed? a. 1 b. 2 c. 3 d. 4 4. Figure 21.60 Two resistors, with resistances R and 2R are connected to a voltage source as shown in this figure. If the power dissipated in R is 10 W, what is the power dissipated in 2R? a. 1 W b. 2.5 W c. 5 W d. 10 W 5. In a circuit, a parallel combination of two 20-\u03a9 and one 10-\u03a9 resistors is connected in series with a 4-\u03a9 resistor. The source voltage is 36 V. a. Find the resistor(s) with the maximum current. b. Find the resistor(s) with the maximum voltage drop. c. Find the power dissipated in each resistor and hence the total power dissipated in all the resistors. Also find the power output of the source. Are they equal or not? Justify your answer. d. Will the answers for questions (a) and (b) differ if a 3 \u03a9 resistor is added in series to the 4 \u03a9 resistor? If yes, repeat the question(s) for the new resistor combination. This content is available for free at http://cnx.org/content/col11844/1.13 e. If the values of all the resistors and the source voltage are doubled, what will be the effect on the current? 21.2 Electromotive Force: Terminal Voltage 6. Suppose there are two voltage sources \u2013 Sources A and B \u2013 with the same emfs but different internal resistances, i.e., the internal resistance of Source A is lower than Source B. If they both supply the same current in their circuits, which of the following statements is true? a. External resistance in Source A\u2019s circuit is more than Source B\u2019s circuit. b. External resistance in Source A\u2019s circuit is less than Source B\u2019s circuit. c. External resistance in Source A\u2019s circuit is the same as Source B\u2019s circuit. d. The relationship between external resistances in the two circuits can\u2019t be determined. 7. Calculate the internal resistance of a voltage source if the terminal voltage of the source increases by 1 V when the current supplied decreases by 4 A? Suppose this source is connected in series", " (in the same direction) to another source with a different voltage but same internal resistance. What will be the total internal resistance? How will the total internal resistance change if the sources are connected in the opposite direction? 21.3 Kirchhoff\u2019s Rules 8. An experiment was set up with the circuit diagram shown. Assume R1 = 10 \u03a9, R2 = R3 = 5 \u03a9, r = 0 \u03a9 and E = 6 V. Figure 21.61 a. One of the steps to examine the set-up is to test points with the same potential. Which of the following points can be tested? a. Points b, c and d. b. Points d, e and f. c. Points f, h and j. d. Points a, h and i. b. At which three points should the currents be measured so that Kirchhoff\u2019s junction rule can be directly confirmed? a. Points b, c and d. b. Points d, e and f. c. Points f, h and j. d. Points a, h and i. c. If the current in the branch with the voltage source is upward and currents in the other two branches are downward, i.e. Ia = Ii + Ic, identify which of the following can be true? Select two answers. a. b. c. d. Ii = Ij - If Ie = Ih - Ii Ic = Ij - Ia Id = Ih - Ij Chapter 21 | Circuits, Bioelectricity, and DC Instruments 963 d. The measurements reveal that the current through R1 is 0.5 A and R3 is 0.6 A. Based on your knowledge of Kirchoff\u2019s laws, confirm which of the following statements are true. a. The measured current for R1 is correct but for R3 is incorrect. b. The measured current for R3 is correct but for R1 is incorrect. c. Both the measured currents are correct. d. Both the measured currents are incorrect. e. The graph shown in the following figure is the energy dissipated at R1 as a function of time. Figure 21.62 Which of the following shows the graph for energy dissipated at R2 as a function of time? c. Figure 21.65 d. Figure 21.66 9. For this question, consider the circuit shown in the following figure. Figure 21.67 a. Assuming that none of the three currents (I1, I2, and I3) are equal to zero, which of the following statements is false? a. b. c. The current through R3 is equal to the current I3 = I1 + I2 at point a. I2 = I3 - I1 at point e. through R5. d. The current through R1 is equal to the current through R5. b. Which of the following statements is true? a. E1 + E2 + I1R1 - I2R2 + I1r1 - I2r2 + I1R5 = 0 b. - E1 + E2 + I1R1 - I2R2 + I1r1 - I2r2 - I1R5 = 0 c. E1 - E2 - I1R1 + I2R2 - I1r1 + I2r2 - I1R5 = 0 d. E1 + E2 - I1R1 + I2R2 - I1r1 + I2r2 + I1R5 = 0 c. If I1 = 5 A and I3 = -2 A, which of the following statements is false? a. The current through R1 will flow from a to b and will be equal to 5 A. b. The current through R3 will flow from a to j and will be equal to 2 A. c. The current through R5 will flow from d to e and will be equal to 5 A. d. None of the above. d. If I1 = 5 A and I3 = -2 A, I2 will be equal to a. 3 A -3 A b. c. 7 A -7 A d. 10. a. Figure 21.63 b. Figure 21.64 964 Chapter 21 | Circuits, Bioelectricity, and DC Instruments Figure 21.68 In an experiment this circuit is set up. Three ammeters are used to record the currents in the three vertical branches (with R1, R2, and E). The readings of the ammeters in the resistor branches (i.e. currents in R1 and R2) are 2 A and 3 A respectively. a. Find the equation obtained by applying Kirchhoff\u2019s loop rule in the loop involving R1 and R2. b. What will be the reading of the third ammeter (i.", "e. the branch with E)? If E were replaced by 3E, how would this reading change? If the original circuit is modified by adding another voltage source (as shown in the following circuit), find the readings of the three ammeters. c. Figure 21.69 11. 0.19 A. According to the results calculated in part (d) identify the resistor(s). Justify any difference in measured and calculated value. 21.6 DC Circuits Containing Resistors and Capacitors 12. A battery is connected to a resistor and an uncharged capacitor. The switch for the circuit is closed at t = 0 s. a. While the capacitor is being charged, which of the following is true? a. Current through and voltage across the resistor increase. b. Current through and voltage across the resistor decrease. c. Current through and voltage across the resistor first increase and then decrease. d. Current through and voltage across the resistor first decrease and then increase. b. When the capacitor is fully charged, which of the following is NOT zero? a. Current in the resistor. b. Voltage across the resistor. c. Current in the capacitor. d. None of the above. 13. An uncharged capacitor C is connected in series (with a switch) to a resistor R1 and a voltage source E. Assume E = 24 V, R1 = 1.2 k\u03a9 and C = 1 mF. a. What will be the current through the circuit as the switch is closed? Draw a circuit diagram and show the direction of current after the switch is closed. How long will it take for the capacitor to be 99% charged? b. After full charging, this capacitor is connected in series to another resistor, R2 = 1 k\u03a9. What will be the current in the circuit as soon as it\u2019s connected? Draw a circuit diagram and show the direction of current. How long will it take for the capacitor voltage to reach 3.24 V? Figure 21.70 In this circuit, assume the currents through R1, R2 and R3 are I1, I2 and I3 respectively and all are flowing in the clockwise direction. a. Find the equation obtained by applying Kirchhoff\u2019s junction rule at point A. b. Find the equations obtained by applying Kirchhoff\u2019s loop rule in the upper and lower loops. c. Assume R1 = R2 = 6 \u03a9, R3 = 12 \u03a9, r1 = r2 = 0 \u03a9, E1 = 6 V and E2 = 4 V. Calculate I1, I2 and I3. d. For the situation in which E2 is replaced by a closed switch, repeat parts (a) and (b). Using the values for R1, R2, R3, r1 and E1 from part (c) calculate the currents through the three resistors. e. For the circuit in part (d) calculate the output power of the voltage source and across all the resistors. Examine if energy is conserved in the circuit. f. A student implemented the circuit of part (d) in the lab and measured the current though one of the resistors as This content is available for free at http://cnx.org/content/col11844/1.13 Appendix A 1531 A ATOMIC MASSES Table A1 Atomic Masses Atomic Number 10 11 Name neutron Hydrogen Deuterium Tritium Helium Lithium Beryllium Boron Carbon Nitrogen Oxygen Fluorine Neon Sodium Atomic Mass Number, A Symbol 10 11 11 12 13 14 13 14 15 15 16 18 18 19 20 22 22 1H 2H or D 3H or T 3He 4He 6Li 7Li 7Be 9Be 10B 11B 11C 12C 13C 14C 13N 14N 15N 15O 16O 18O 18F 19F 20Ne 22Ne 22Na 3.016 050 3.016 030 4.002 603 6.015 121 7.016 003 7.016 928 9.012 182 10.012 937 11.009 305 11.011 432 12.000 000 13.003 355 14.003 241 13.005 738 14.003 074 15.000 108 15.003 065 15.994 915 17.999 160 18.000 937 Atomic Mass (u) 1.008 665 Percent Abundance or Decay Mode \u2212 Half-life, t1/2 10.37 min 1.007 825 99.985% 2.014 102 0.015% \u2212 12.33 y 1.38\u00d710\u22124% \u2248100% 7.5% 92.5% EC 100% 19.9% 80.1% EC, + 98.90% 1.10% \u2212 + 99.63% 0.37% EC, + 99.76% 0.200%", " EC, + 53.29 d 5730 y 9.96 min 122 s 1.83 h 2.602 y 18.998 403 100% 19.992 435 90.51% 21.991 383 21.994 434 9.22% + 1532 Appendix A Atomic Number, Z Name Atomic Mass Number, A Symbol Atomic Mass (u) Percent Abundance or Decay Mode Half-life, t1/2 12 13 14 Magnesium Aluminum Silicon 15 Phosphorus 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 Sulfur Chlorine Argon Potassium Calcium Scandium Titanium Vanadium Chromium Manganese Iron Cobalt Nickel Copper Zinc Gallium 23 24 24 27 28 31 31 32 32 35 35 37 40 39 40 40 45 48 51 52 55 56 59 60 58 60 63 65 64 66 69 23Na 24Na 24Mg 27Al 28Si 31Si 31P 32P 32S 35S 35Cl 37Cl 40Ar 39K 40K 40Ca 45Sc 48Ti 51V 52Cr 22.989 767 23.990 961 100% \u2212 23.985 042 78.99% 26.981 539 100% 14.96 h 27.976 927 92.23% 2.62h 30.975 362 30.973 762 31.973 907 \u2212 100% \u2212 31.972 070 95.02% 14.28 d 87.4 d 34.969 031 34.968 852 36.965 903 39.962 384 38.963 707 39.963 999 44.955 910 47.947 947 50.943 962 51.940 509 \u2212 75.77% 24.23% 99.60% 93.26% 100% 73.8% 99.75% 83.79% 100% 0.0117%, EC, \u2212 1.28\u00d7109 y 39.962 591 96.94% 55Mn 54.938 047 5.271 y 56Fe 59Co 60Co 58Ni 60Ni 63Cu 65Cu 64Zn 66Zn 69Ga 55.934 939 91.72% 58.933 198 59.933 819 57.935 346 59.930 788 62.939 598 64.927 793 63.929 145 65.926 034 68.925 580 100% \u2212 68.27% 26.10% 69.17% 30.83% 48.6% 27.9% 60.1% This content is available for free at http://cnx.org/content/col11844/1.13 Appendix A 1533 Atomic Number, Z Name Atomic Mass Number, A Symbol Atomic Mass (u) Percent Abundance or Decay Mode Half-life, t1/2 71.922 079 73.921 177 74.921 594 79.916 520 78.918 336 83.911 507 84.911 794 85.909 267 87.905 619 89.907 738 88.905 849 89.907 152 27.4% 36.5% 100% 49.7% 50.69% 57.0% 72.17% 9.86% 82.58% \u2212 100% \u2212 89.904 703 51.45% 92.906 377 100% 97.905 406 24.13% 97.907 215 101.904 348 102.905 500 105.903 478 106.905 092 108.904 757 113.903 357 114.903 880 \u2212 31.6% 100% 27.33% 51.84% 48.16% 28.73% 32 Germanium 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 Arsenic Selenium Bromine Krypton Rubidium Strontium Yttrium Zirconium Niobium Molybdenum Technetium Ruthenium Rhodium Palladium Silver Cadmium Indium Tin Antimony Tellurium Iodine Xenon 72 74 75 80 79 84 85 86 88 90 89 90 90 93 98 98 102 103 106 107 109 114 115 120 121 130 127 131 132 136 72Ge 74Ge 75As 80Se 79Br 84Kr 85Rb 86Sr 88Sr 90Sr 89Y 90Y 90Zr 93Nb 98Mo 98Tc 102Ru 103Rh 106Pd 107Ag 109Ag 114Cd 115In 120Sn 121Sb 130Te 127I 131I 132Xe 136Xe 95.7%, \u2212 4.4\u00d71014y 119.902 200 32.59% 120.903 821 57.3% 129.906 229 126.904 473 130.906 114 131.904 144", " 135.907 214 33.8%, \u2212 2.5\u00d71021y 100% \u2212 26.9% 8.9% 8.040 d 28.8 y 64.1 h 4.2\u00d7106y 1534 Atomic Number, Z 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 Name Cesium Barium Lanthanum Cerium Praseodymium Neodymium Promethium Samarium Europium Gadolinium Terbium Dysprosium Holmium Erbium Thulium Ytterbium Lutecium Hafnium Tantalum Tungsten Rhenium Osmium Iridium Platinum Gold Mercury Atomic Mass Number, A Symbol Atomic Mass (u) Percent Abundance or Decay Mode Half-life, t1/2 Appendix A 133 134 137 138 139 140 141 142 145 152 153 158 159 164 165 166 169 174 175 180 181 184 187 191 192 191 193 195 197 198 199 133Cs 134Cs 137Ba 138Ba 139La 140Ce 141Pr 142Nd 145Pm 152Sm 153Eu 158Gd 159Tb 164Dy 165Ho 166Er 132.905 429 133.906 696 136.905 812 137.905 232 138.906 346 139.905 433 140.907 647 100% EC, \u2212 11.23% 71.70% 99.91% 88.48% 100% 141.907 719 27.13% 144.912 743 151.919 729 152.921 225 EC, 26.7% 52.2% 157.924 099 24.84% 2.06 y 17.7 y 158.925 342 163.929 171 164.930 319 165.930 290 169Tm 168.934 212 174Yb 175Lu 180Hf 181Ta 173.938 859 174.940 770 179.946 545 180.947 992 184W 183.950 928 100% 28.2% 100% 33.6% 100% 31.8% 97.41% 35.10% 99.98% 30.67% 187Re 191Os 192Os 191Ir 193Ir 195Pt 197Au 198Au 199Hg 186.955 744 190.960 920 191.961 467 190.960 584 192.962 917 194.964 766 196.966 543 197.968 217 62.6%, \u2212 4.6\u00d71010y \u2212 41.0% 37.3% 62.7% 33.8% 100% \u2212 15.4 d 2.696 d 198.968 253 16.87% This content is available for free at http://cnx.org/content/col11844/1.13 Appendix A 1535 Atomic Number, Z Name Atomic Mass Number, A Symbol Atomic Mass (u) Percent Abundance or Decay Mode Half-life, t1/2 202Hg 201.970 617 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 Thallium Lead Bismuth Polonium Astatine Radon Francium Radium Actinium Thorium Protactinium Uranium Neptunium Plutonium Americium Curium Berkelium 202 205 206 207 208 210 211 212 209 211 210 218 222 223 226 227 228 232 231 233 235 236 238 239 239 239 243 245 247 205Tl 206Pb 207Pb 208Pb 210Pb 211Pb 212Pb 209Bi 211Bi 210Po 218At 222Rn 223Fr 226Ra 227Ac 228Th 232Th 231Pa 233U 235U 236U 238U 239U 239Np 239Pu 204.974 401 205.974 440 206.975 872 207.976 627 209.984 163 210.988 735 211.991 871 208.980 374 210.987 255 209.982 848 218.008 684 222.017 570 223.019 733 226.025 402 227.027 750 228.028 715 29.86% 70.48% 24.1% 22.1% 52.4% \u2212 \u2212 \u2212 100% \u2212 \u2212 \u2212 \u2212 22.3 y 36.1 min 10.64 h 2.14 min 138.38 d 1.6 s 3.82 d 21.8 min 1.60\u00d7103y 21.8 y 1.91 y 232.038 054 100%, 1.41\u00d71010y 231.035 880 233.039 628 235.043 924 0.720%, 236.045 562 238.050 784 99.2745%, 239.054 289 239.052 933 239.052 157 \u2212 \u2212 243Am 243.061 375 fission 245Cm 245.0", "65 483 247Bk 247.070 300 3.28\u00d7104y 1.59\u00d7103y 7.04\u00d7108y 2.34\u00d7107y 4.47\u00d7109y 23.5 min 2.355 d 2.41\u00d7104y 7.37\u00d7103y 8.50\u00d7103y 1.38\u00d7103y 1536 Appendix A Atomic Number, Z Name Atomic Mass Number, A Symbol Atomic Mass (u) Percent Abundance or Decay Mode Half-life, t1/2 98 99 100 101 102 103 104 105 106 107 108 109 Californium Einsteinium Fermium Mendelevium Nobelium Lawrencium Rutherfordium Dubnium Seaborgium Bohrium Hassium Meitnerium 249 254 253 255 255 257 261 262 263 262 264 266 249Cf 254Es 253Fm 255Md 255No 257Lr 261Rf 262Db 263Sg 262Bh 264Hs 266Mt 249.074 844 254.088 019 253.085 173 255.091 081 255.093 260 257.099 480 261.108 690 262.113 760 263.11 86 262.123 1 264.128 5 266.137 8 \u2212 EC, EC, EC, EC, fission fission 351 y 276 d 3.00 d 27 min 3.1 min 0.646 s 1.08 min 34 s 0.8 s 0.102 s 0.08 ms 3.4 ms This content is available for free at http://cnx.org/content/col11844/1.13 Appendix B 1537 B SELECTED RADIOACTIVE ISOTOPES Decay modes are, \u2212, + would + are roughly one-half the maxima. decay. IT is a transition from a metastable excited state. Energies for \u00b1, electron capture (EC) and isomeric transition (IT). EC results in the same daughter nucleus as decays are the maxima; average energies Table B1 Selected Radioactive Isotopes Isotope t 1/2 DecayMode(s) Energy(MeV) Percent \u03b3 -Ray Energy(MeV) Percent 100% 100% 100% 90% 1.27 100% 0.0186 0.156 1.20 0.55 1.71 0.167 0.710 1.31 0.827 100% 100% 100% 89% 87% 0.257 100% 3.69 28% 1.80 43% 0.273 0.466 0.318 45% 55% 100% 3H 14C 13N 22Na 32P 35S 36Cl 40K 43K 45Ca 51Cr 52Mn 52Fe 59Fe 60Co 65Zn 67Ga 12.33 y 5730 y 9.96 min 2.602 y 14.28 d 87.4 d 3.00\u00d7105y 1.28\u00d7109y 22.3 h 165 d 27.70 d 5.59d 8.27 h 44.6 d 5.271 y 244.1 d 78.3 h \u2212 \u2212 + + \u2212 \u2212 \u2212 \u2212 \u2212 \u2212 EC + + \u2212 s \u2212 EC EC 75Se 118.5 d EC s s s s s s 0.373 0.618 0.320 1.33 1.43 0.169 0.378 1.10 1.29 1.17 1.33 1.12 0.0933 0.185 0.300 others 0.121 0.136 0.265 87% 87% 10% 28% 28% 43% 43% 57% 43% 100% 100% 51% 70% 35% 19% 20% 65% 68% 1538 Isotope t 1/2 DecayMode(s) Energy(MeV) Percent \u03b3 -Ray Energy(MeV) Appendix B Percent 20% 9% 0.280 others 1.08 0.514 100% 0.142 0.392 0.159 0.364 others 0.0400 0.372 0.411 others 0.662 0.030 0.044 0.537 others 0.412 100% 100% \u2248100% 85% 35% 32% 25% 95% 25% 65% 24% \u2248100% 0.0733 100% 0.186 100% \u2248100% s numerous <0.400% s s 23% 77% 11% 15% 73% 0.050 23% numerous <0.250% 7.5\u00d710\u22125 0.013 0.052 73% 15% 10% 86Rb 18.8 d 85Sr 90Sr 90Y 99mTc 113mIn 123I 131I 64.8 d 28.8 y 64.1 h 6.02 h 99.5 min 13.0 h 8.040 d \u2212 s EC \u2212 \u2212 IT IT EC \u2212 s 129Cs 32.3 h EC 137Cs 30.17 y 140Ba 12.79 d 198", "Au 197Hg 210Po 226Ra 2.696 d 64.1 h 138.38 d 1.60\u00d7103y 235U 238U 7.038\u00d7108y 4.468\u00d7109y 237Np 2.14\u00d7106y 239Pu 2.41\u00d7104y \u2212 s \u2212 \u2212 EC s s s s 0.69 1.77 0.546 2.28 0.248 0.607 others 0.511 1.17 1.035 s s 9% 91% 100% 100% 7% 93% 95% 5% \u2248100% s 1.161 \u2248100% 100% 5% 95% 5.41 4.68 4.87 4.68 4.22 4.27 numerous 4.96 (max.) 5.19 5.23 5.24 This content is available for free at http://cnx.org/content/col11844/1.13 Appendix B 1539 Isotope t 1/2 DecayMode(s) Energy(MeV) Percent \u03b3 -Ray Energy(MeV) Percent 243Am 7.37\u00d7103y s Max. 5.44 s 5.37 5.32 others 88% 11% others 0.075 others 1540 Appendix B This content is available for free at http://cnx.org/content/col11844/1.13 Appendix C 1541 C USEFUL INFORMATION This appendix is broken into several tables. \u2022 Table C1, Important Constants \u2022 Table C2, Submicroscopic Masses \u2022 Table C3, Solar System Data \u2022 Table C4, Metric Prefixes for Powers of Ten and Their Symbols \u2022 Table C5, The Greek Alphabet \u2022 Table C6, SI units \u2022 Table C7, Selected British Units \u2022 Table C8, Other Units \u2022 Table C9, Useful Formulae Table C1 Important Constants [1] Symbol Meaning Best Value Approximate Value \u03c3 \u03b50 \u03bc0 Speed of light in vacuum Gravitational constant Avogadro\u2019s number Boltzmann\u2019s constant Gas constant StefanBoltzmann constant Coulomb force constant Charge on electron Permittivity of free space Permeability of free space Planck\u2019s constant 2.99792458 \u00d7 108 m / s 3.00 \u00d7 108 m / s 6.67384(80) \u00d7 10\u221211 N \u22c5 m2 / kg2 6.67 \u00d7 10\u221211 N \u22c5 m2 / kg2 6.02214129(27) \u00d7 1023 6.02 \u00d7 1023 1.3806488(13) \u00d7 10\u221223 J / K 1.38 \u00d7 10\u221223 J / K 8.3144621(75) J / mol \u22c5 K 8.31 J / mol \u22c5 K = 1.99 cal / mol \u22c5 K = 0.0821atm \u22c5 L / mol \u22c5 K 5.670373(21) \u00d7 10\u22128 W / m2 \u22c5 K 5.67 \u00d7 10\u22128 W / m2 \u22c5 K 8.987551788... \u00d7 109 N \u22c5 m2 / C2 8.99 \u00d7 109 N \u22c5 m2 / C2 \u22121.602176565(35) \u00d7 10\u221219 C \u22121.60 \u00d7 10\u221219 C 8.854187817... \u00d7 10\u221212 C2 / N \u22c5 m2 8.85 \u00d7 10\u221212 C2 / N \u22c5 m2 4\u03c0 \u00d7 10\u22127 T \u22c5 m / A 1.26 \u00d7 10\u22126 T \u22c5 m / A 6.62606957(29) \u00d7 10\u221234 J \u22c5 s 6.63 \u00d7 10\u221234 J \u22c5 s Table C2 Submicroscopic Masses [2] Symbol Meaning Best Value Approximate Value Electron mass 9.10938291(40)\u00d710\u221231kg 9.11\u00d710\u221231kg Proton mass 1.672621777(74)\u00d710\u221227kg 1.6726\u00d710\u221227kg 1. Stated values are according to the National Institute of Standards and Technology Reference on Constants, Units, and Uncertainty, www.physics.nist.gov/cuu (http://www.physics.nist.gov/cuu) (accessed May 18, 2012). Values in parentheses are the uncertainties in the last digits. Numbers without uncertainties are exact as defined. 2. Stated values are according to the National Institute of Standards and Technology Reference on Constants, Units, and Uncertainty, www.physics.nist.gov/cuu (http://www.physics.nist.gov/cuu) (accessed May 18, 2012). Values in parentheses are the uncertainties in the last digits. Numbers without", " uncertainties are exact as defined. 1542 Appendix C Symbol Meaning Best Value Approximate Value u Neutron mass 1.674927351(74)\u00d710\u221227kg 1.6749\u00d710\u221227kg Atomic mass unit 1.660538921(73)\u00d710\u221227kg 1.6605\u00d710\u221227kg Table C3 Solar System Data Sun mass average radius Earth-sun distance (average) Earth mass average radius orbital period Moon mass average radius orbital period (average) 1.99\u00d71030kg 6.96\u00d7108m 1.496\u00d71011m 5.9736\u00d71024kg 6.376\u00d7106m 3.16\u00d7107s 7.35\u00d71022kg 1.74\u00d7106m 2.36\u00d7106s Earth-moon distance (average) 3.84\u00d7108m Table C4 Metric Prefixes for Powers of Ten and Their Symbols Prefix Symbol Value Prefix Symbol Value tera giga mega kilo hecto deka T G M k h da 1012 109 106 103 102 101 deci centi milli micro nano pico \u2014 \u2014 100( = 1) femto d c m n p f 10\u22121 10\u22122 10\u22123 10\u22126 10\u22129 10\u221212 10\u221215 Table C5 The Greek Alphabet Alpha \u0391 Eta \u0397 Nu \u039d Tau \u03a4 Beta \u0392 Theta \u0398 Xi \u039e Upsilon \u03a5 Gamma \u0393 Iota \u0399 Omicron \u039f Phi \u03a6 Delta \u0394 Kappa \u039a Pi Epsilon \u0395 Lambda \u039b Rho \u03a0 Chi \u03a1 Psi \u03a7 \u03a8 Zeta \u0396 Mu \u039c Sigma \u03a3 Omega \u03a9 This content is available for free at http://cnx.org/content/col11844/1.13 Appendix C 1543 Table C6 SI Units Entity Abbreviation Name Fundamental units Length Mass Time Current Supplementary unit Angle Derived units Force Energy Power Pressure m kg s A rad meter kilogram second ampere radian N = kg \u22c5 m / s2 newton J = kg \u22c5 m2 / s2 joule W = J / s Pa = N / m2 watt pascal hertz volt farad coulomb Frequency Hz = 1 / s Electronic potential V = J / C Capacitance Charge Resistance \u03a9 = V / A ohm Magnetic field T = N / (A \u22c5 m) tesla Nuclear decay rate Bq = 1 / s becquerel Table C7 Selected British Units Length 1 inch (in.) = 2.54 cm (exactly) 1 foot (ft) = 0.3048 m 1 mile (mi) = 1.609 km Force 1 pound (lb) = 4.448 N Energy 1 British thermal unit (Btu) = 1.055\u00d7103 J Power 1 horsepower (hp) = 746 W Pressure 1 lb / in2 = 6.895\u00d7103 Pa Table C8 Other Units Length 1 light year (ly) = 9.46\u00d71015 m 1 astronomical unit (au) = 1.50\u00d71011 m 1 nautical mile = 1.852 km 1 angstrom(\u00c5) = 10\u221210 m Area 1 acre (ac) = 4.05\u00d7103 m2 1 square foot (ft2) = 9.29\u00d710\u22122 m2 1 barn () = 10\u221228 m2 Volume 1 liter () = 10\u22123 m3 1544 Appendix C 1 U.S. gallon (gal) = 3.785\u00d710\u22123 m3 Mass 1 solar mass = 1.99\u00d71030 kg Time Speed Angle 1 metric ton = 103 kg 1 atomic mass unit () = 1.6605\u00d710\u221227 kg 1 year () = 3.16\u00d7107 s 1 day () = 86400 s 1 mile per hour (mph) = 1.609 km / h 1 nautical mile per hour (naut) = 1.852 km / h 1 degree () = 1.745\u00d710\u22122 rad 1 minute of arc (') = 1 / 60 degree 1 second of arc ('') = 1 / 60 minute of arc 1 grad = 1.571\u00d710\u22122 rad Energy 1 kiloton TNT (kT) = 4.2\u00d71012 J 1 kilowatt hour (kW \u22c5 ) = 3.60\u00d7106 J 1 food calorie (kcal) = 4186 J 1 calorie (cal) = 4.186 J 1 electron volt (eV) = 1.60\u00d710\u221219 J Pressure 1 atmosphere (atm) = 1.013\u00d7105 Pa 1 millimeter of mercury (mm Hg) = 133.3 Pa 1 torricelli (torr) = 1 mm Hg = 133.3 Pa Nuclear decay rate 1 curie (Ci) = 3.70\u00d71010 Bq Table C", " horizontal axis capacitive reactance inductive reactance root mean square diffusion distance vertical axis This content is available for free at http://cnx.org/content/col11844/1.13 Appendix D 1555 Symbol Definition elastic modulus or Young's modulus atomic number (number of protons in a nucleus) impedance 1556 Appendix D This content is available for free at http://cnx.org/content/col11844/1.13 Answer Key 1557 ANSWER KEY Chapter 1 Problems & Exercises 1 a. 27.8 m/s b. 62.1 mph 3 s \u00d7 3600 s 1 hr 1.0 m s = 1.0 m = 3.6 km/h. \u00d7 1 km 1000 m 5 length: 377 ft ; 4.53\u00d7103 in. width: 280 ft ; 3.3\u00d7103 in. 7 8.847 km 9 (a) 1.3\u00d710\u22129 m (b) 40 km/My 11 2 kg 13 a. 85.5 to 94.5 km/h b. 53.1 to 58.7 mi/h 15 (a) 7.6\u00d7107 beats (b) 7.57\u00d7107 beats (c) 7.57\u00d7107 beats 17 a. 3 b. 3 c. 3 19 a) 2.2% (b) 59 to 61 km/h 21 80 \u00b1 3 beats/min 23 2.8 h 25 11 \u00b1 1 cm3 27 12.06 \u00b1 0.04 m2 1558 Answer Key 29 Sample answer: 2\u00d7109 heartbeats 31 Sample answer: 2\u00d71031 if an average human lifetime is taken to be about 70 years. 33 Sample answer: 50 atoms 35 Sample answers: (a) 1012 cells/hummingbird (b) 1016 cells/human Chapter 2 Problems & Exercises 1 (a) 7 m (b) 7 m (c) +7 m 3 (a) 13 m (b) 9 m (c) +9 m 5 (a) 3.0\u00d7104 m/s (b) 0 m/s 7 2\u00d7107 years 9 34.689 m/s = 124.88 km/h 11 (a) 40.0 km/h (b) 34.3 km/h, 25\u00ba S of E. (c) average speed = 3.20 km/h, - = 0. 13 384,000 km 15 (a) 6.61\u00d71015 rev/s (b) 0 m/s 16 4.29 m/s2 18 (a) 1.43 s (b) \u22122.50 m/s2 20 (a) 10.8 m/s This content is available for free at http://cnx.org/content/col11844/1.13 Answer Key (b) 1559 Figure 2.48. 21 38.9 m/s (about 87 miles per hour) 23 (a) 16.5 s (b) 13.5 s (c) \u22122.68 m/s2 25 (a) 20.0 m (b) \u22121.00 m/s (c) This result does not really make sense. If the runner starts at 9.00 m/s and decelerates at 2.00 m/s2, then she will have stopped after 4.50 s. If she continues to decelerate, she will be running backwards. 27 0.799 m 29 (a) 28.0 m/s (b) 50.9 s (c) 7.68 km to accelerate and 713 m to decelerate 31 (a) 51.4 m (b) 17.1 s 33 (a) \u221280.4 m/s2 (b) 9.33\u00d710\u22122 s 35 (a) 7.7 m/s (b) \u221215\u00d7102 m/s2. This is about 3 times the deceleration of the pilots, who were falling from thousands of meters high! 37 (a) 32.6 m/s2 (b) 162 m/s (c) > max, because the assumption of constant acceleration is not valid for a dragster. A dragster changes gears, and would have a greater acceleration in first gear than second gear than third gear, etc. The acceleration would be 1560 Answer Key greatest at the beginning, so it would not be accelerating at 32.6 m/s2 during the last few meters, but substantially less, and the final velocity would be less than 162 m/s. 39 104 s 40 (a) = 12.2 m/s ; = 4.07 m/s2 (b) = 11.2 m/s 41 (a) 1 = 6.28 m ; 1 = 10.1 m/s (b) 2 = 10.1 m ; 2 = 5.20 m/s (c) 3 = 11", ".5 m ; 3 = 0.300 m/s (d) 4 = 10.4 m ; 4 = \u22124.60 m/s 43 0 = 4.95 m/s 45 (a) = \u22129.80 m/s2 ; 0 = 13.0 m/s ; 0 = 0 m (b) = 0m/s. Unknown is distance to top of trajectory, where velocity is zero. Use equation 2 = 0 because it contains all known values except for, so we can solve for. Solving for gives 2 + 2( \u2212 0) 2 = 2( \u2212 00 m/s)2 \u2212 (13.0 m/s)2 \u22129.80 m/s2 2 = 8.62 m (2.100) Dolphins measure about 2 meters long and can jump several times their length out of the water, so this is a reasonable result. (c) 2.65 s 47 Figure 2.57. (a) 8.26 m (b) 0.717 s 49 1.91 s 51 (a) 94.0 m (b) 3.13 s This content is available for free at http://cnx.org/content/col11844/1.13 1561 = (11.7 \u2212 6.95)\u00d7103 m (40.0 \u2013 20.0) s = 238 m/s (2.114) Answer Key 53 (a) -70.0 m/s (downward) (b) 6.10 s 55 (a) 19.6 m (b) 18.5 m 57 (a) 305 m (b) 262 m, -29.2 m/s (c) 8.91 s 59 (a) 115 m/s (b) 5.0 m/s2 61 63 Figure 2.63. 65 (a) 6 m/s (b) 12 m/s (c) 3 m/s2 (d) 10 s Test Prep for AP\u00ae Courses 1 (a) 3 a. Use tape to mark off two distances on the track \u2014 one for cart A before the collision and one for the combined carts after the collision. Push cart A to give it an initial speed. Use a stopwatch to measure the time it takes for the cart(s) to cross the marked distances. The speeds are the distances divided by the times. If the measurement errors are of the same magnitude, they will have a greater effect after the collision. The speed of the combined carts will be less than the initial speed of cart A. As a result, these errors will be a greater percentage of the actual velocity value after the collision occurs. (Note: Other arguments could properly be made for \u2018more error before the collision' and error that \u2018equally affects both sets of measurement.') b. 5 1562 Answer Key The position vs. time graph should be represented with a positively sloped line whose slope steadily decreases to zero. The y-intercept of the graph may be any value. The line on the velocity vs. time graph should have a positive yintercept and a negative slope. Because the final velocity of the book is zero, the line should finish on the x-axis. The position vs. time graph should be represented with a negatively sloped line whose slope steadily decreases to zero. The y-intercept of the graph may be any value. The line on the velocity vs. time graph should have a negative yintercept and a positive slope. Because the final velocity of the book is zero, the line should finish on the x-axis.] 7 (c) Chapter 3 Problems & Exercises 1 (a) 480 m (b) 379 m, 18.4\u00b0 east of north 3 north component 3.21 km, east component 3.83 km 5 19.5 m, 4.65\u00b0 south of west 7 (a) 26.6 m, 65.1\u00b0 north of east (b) 26.6 m, 65.1\u00b0 south of west 9 52.9 m, 90.1\u00b0 with respect to the x-axis. 11 x-component 4.41 m/s y-component 5.07 m/s 13 (a) 1.56 km (b) 120 m east 15 North-component 87.0 km, east-component 87.0 km 17 30.8 m, 35.8 west of north 19 (a) 30.8 m, 54.2\u00ba south of west (b) 30.8 m, 54.2\u00ba north of east 21 18.4 km south, then 26.2 km west(b) 31.5 km at 45.0\u00ba south of west, then 5.56 km at 45.0\u00ba west of north 23 7.34 km, 63.5\u00ba south of east 25 = 1.30 m\u00d7102 = 30.9 m. 27 (a) 3.50 s (b) 28.6", " m/s (c) 34.3 m/s (d) 44.7 m/s, 50.2\u00b0 below horizontal 29 This content is available for free at http://cnx.org/content/col11844/1.13 Answer Key (a) 18.4\u00b0 (b) The arrow will go over the branch. 31 = 0 sin2\u03b80 For = 45\u00b0, = 0 1563 = 91.8 m for 0 = 30 m/s ; = 163 m for 0 = 40 m/s ; = 255 m for 0 = 50 m/s. 33 (a) 560 m/s (b) 8.00\u00d7103 m (c) 80.0 m. This error is not significant because it is only 1% of the answer in part (b). 35 1.50 m, assuming launch angle of 45\u00b0 37 = 6.1\u00b0 yes, the ball lands at 5.3 m from the net 39 (a) \u22120.486 m (b) The larger the muzzle velocity, the smaller the deviation in the vertical direction, because the time of flight would be smaller. Air resistance would have the effect of decreasing the time of flight, therefore increasing the vertical deviation. 41 4.23 m. No, the owl is not lucky; he misses the nest. 43 No, the maximum range (neglecting air resistance) is about 92 m. 45 15.0 m/s 47 (a) 24.2 m/s (b) The ball travels a total of 57.4 m with the brief gust of wind. 49 \u2212 0 = 0 = 0 \u2212 1 22 = (0 sin ) \u2212 1 22, so that = 2(0 sin ) \u2212 0 = 0 = (0 cos ) = and substituting for gives: = 0 cos 20 sin = 20 2 sin cos since 2 sin cos = sin 2\u03b8 the range is: = 0 2 sin 2\u03b8. 52 (a) 35.8 km, 45\u00ba south of east (b) 5.53 m/s, 45\u00ba south of east 1564 Answer Key (c) 56.1 km, 45\u00ba south of east 54 (a) 0.70 m/s faster (b) Second runner wins (c) 4.17 m 56 17.0 m/s, 22.1\u00ba 58 (a) 230 m/s, 8.0\u00ba south of west (b) The wind should make the plane travel slower and more to the south, which is what was calculated. 60 (a) 63.5 m/s (b) 29.6 m/s 62 6.68 m/s, 53.3\u00ba south of west 64 (a) average = 14.9km/s Mly (b) 20.2 billion years 66 1.72 m/s, 42.3\u00ba north of east Test Prep for AP\u00ae Courses 1 (d) 3 We would need to know the horizontal and vertical positions of each ball at several times. From that data, we could deduce the velocities over several time intervals and also the accelerations (both horizontal and vertical) for each ball over several time intervals. 5 The graph of the ball's vertical velocity over time should begin at 4.90 m/s during the time interval 0 - 0.1 sec (there should be a data point at t = 0.05 sec, v = 4.90 m/s). It should then have a slope of -9.8 m/s2, crossing through v = 0 at t = 0.55 sec and ending at v = -0.98 m/s at t = 0.65 sec. The graph of the ball's horizontal velocity would be a constant positive value, a flat horizontal line at some positive velocity from t = 0 until t = 0.7 sec. Chapter 4 Problems & Exercises 1 265 N 3 13.3 m/s2 7 (a) 12 m/s2. (b) The acceleration is not one-fourth of what it was with all rockets burning because the frictional force is still as large as it was with all rockets burning. 9 (a) The system is the child in the wagon plus the wagon. (b This content is available for free at http://cnx.org/content/col11844/1.13 Answer Key 1565 Figure 4.10. (c) = 0.130 m/s2 in the direction of the second child\u2019s push. (d) = 0.00 m/s2 11 (a) 3.68\u00d7103 N. This force is 5.00 times greater than his weight. (b) 3750 N; 11.3\u00ba above horizontal 13 1.5\u00d7103 N, 150 kg, 150 kg 15 Force on shell: 2.64\u00d7107 N Force exerted on ship = \u22122.64\u00d7107 N, by Newton\u2019s third law 17", " a. b. 0.11 m/s2 1.2\u00d7104 N 19 (a) 7.84\u00d710-4 N (b) 1.89\u00d710\u20133 N. This is 2.41 times the tension in the vertical strand. 21 Newton\u2019s second law applied in vertical direction gives = \u2212 2 sin = 0 = 2 sin =. 2 sin () () () 23 1566 Answer Key Figure 4.26. Using the free-body diagram: net = \u2212 \u2212 =, so that = \u2212 \u2212 = 1.250\u00d7107 N \u2212 4.50\u00d7106 \u2212 (5.00\u00d7105 kg)(9.80 m/s2) 5.00\u00d7105 kg = 6.20 m/s2. 25 1. Use Newton\u2019s laws of motion. Figure 4.26. 2. Given : = 4.00 = (4.00)(9.80 m/s2 ) = 39.2 m/s2 ; = 70.0 kg, Find:. 3. \u2211 =+ \u2212 =, so that = + = + = ( + ). = (70.0 kg)[(39.2 m/s2 ) + (9.80 m/s2)] down on the ground, but is up from the ground and makes him jump. = 3.43\u00d7103N. The force exerted by the high-jumper is actually 4. This result is reasonable, since it is quite possible for a person to exert a force of the magnitude of 103 N. This content is available for free at http://cnx.org/content/col11844/1.13 1567 Answer Key 27 (a) 4.41\u00d7105 N (b) 1.50\u00d7105 N 29 (a) 910 N (b) 1.11\u00d7103 N 31 = 0.139 m/s, = 12.4\u00ba north of east 33 1. Use Newton\u2019s laws since we are looking for forces. 2. Draw a free-body diagram: Figure 4.29. 3. The tension is given as = 25.0 N. Find app. Using Newton\u2019s laws gives: = 0, so that applied force is due to the y-components of the two tensions: = 2 sin = 2(25.0 N)sin The x-components of the tension cancel. \u2211 = 0. 15\u00ba = 12.9 N 4. This seems reasonable, since the applied tensions should be greater than the force applied to the tooth. 40 10.2 m/s2, 4.67\u00ba from vertical 42 1568 Answer Key Figure 4.35. 1 = 736 N 2 = 194 N 44 (a) 7.43 m/s (b) 2.97 m 46 (a) 4.20 m/s (b) 29.4 m/s2 (c) 4.31\u00d7103 N 48 (a) 47.1 m/s (b) 2.47\u00d7103 m/s2 (c) 6.18\u00d7103 N. The average force is 252 times the shell\u2019s weight. 52 (a) 1\u00d710\u221213 (b) 1\u00d710\u221211 54 102 Test Prep for AP\u00ae Courses 1 This content is available for free at http://cnx.org/content/col11844/1.13 Answer Key 1569 Figure 4.4. Car X is shown on the left, and Car Y is shown on the right. i. Car X takes longer to accelerate and does not spend any time traveling at top speed. Car Y accelerates over a shorter time and spends time going at top speed. So Car Y must cover the straightaways in a shorter time. Curves take the same time, so Car Y must overall take a shorter time. ii. The only difference in the calculations for the time of one segment of linear acceleration is the difference in distances. That shows that Car X takes longer to accelerate. The equation = corresponds to Car Y traveling for a time at 4 top speed. Substituting = 1 into the displacement equation in part (b) ii gives = 3 2 1. This shows that a car takes less time to reach its maximum speed when it accelerates over a shorter distance. Therefore, Car Y reaches its maximum speed more quickly, and spends more time at its maximum speed than Car X does, as argued in part (b) i. 3 A body cannot exert a force on itself. The hawk may accelerate as a result of several forces. The hawk may accelerate toward Earth as a result of the force due to gravity. The hawk may accelerate as a result of the additional force exerted on it by wind. The hawk may accelerate as a result of orienting its body to create less air resistance, thus increasing the net force forward. 5 (a) A soccer player, gravity, air, and friction commonly exert forces on a soccer ball being kicked. (", "b) Gravity and the surrounding water commonly exert forces on a dolphin jumping. (The dolphin moves its muscles to exert a force on the water. The water exerts an equal force on the dolphin, resulting in the dolphin\u2019s motion.) (c) Gravity and air exert forces on a parachutist drifting to Earth. 7 (c) 9 Figure 4.14. The diagram consists of a black dot in the center and two small red arrows pointing up (Fb) and down (Fg) and two long red arrows pointing right (Fc = 9.0 N) and left (Fw=13.0 N). In the diagram, Fg represents the force due to gravity on the balloon, and Fb represents the buoyant force. These two forces are equal in magnitude and opposite in direction. Fc represents the force of the current. Fw represents the force of the wind. The net force on the balloon will be \u2212 = 4.0 N and the balloon will accelerate in the direction the wind is blowing. 11 1570 Answer Key Since = /, the parachutist has a mass of 539 N/9.8 km/s2 = 55 kg. For the first 2 s, the parachutist accelerates at 9.8 m/s2. \u2022 2s = = 9.8 m s2 = 17.6m s Her speed after 2 s is 19.6 m/s. From 2 s to 10 s, the net force on the parachutist is 539 N \u2013 615 N, or 76 N upward. = = \u221276 N 55 kg = \u22121.4 m s2 Since = 0 +, = 17.6 m/s2 + ( \u2212 1.4 m/s2)(8s) = 6.5 m/s2. At 10 s, the parachutist is falling to Earth at 8.4 m/s. 13 The system includes the gardener and the wheelbarrow with its contents. The following forces are important to include: the weight of the wheelbarrow, the weight of the gardener, the normal force for the wheelbarrow and the gardener, the force of the gardener pushing against the ground and the equal force of the ground pushing back against the gardener, and any friction in the wheelbarrow\u2019s wheels. 15 The system undergoing acceleration is the two figure skaters together. Net force = 120 N \u2013 5.0 N = 115 N. Total mass = 40 kg + 50 kg = 90 kg. Using Newton\u2019s second law, we have that = = 115 N 90 kg = 1.28 m s2 The pair accelerates forward at 1.28 m/s2. 17 The force of tension must equal the force of gravity plus the force necessary to accelerate the mass. = can be used to calculate the first, and = can be used to calculate the second. For gravity: = = (120.0 kg)(9.8 m/s2) = 1205.4 N For acceleration: = = (120.0 kg)(1.3 m/s2) = 159.9 N The total force of tension in the cable is 1176 N + 156 N = 1332 N. 19 (b) This content is available for free at http://cnx.org/content/col11844/1.13 1571 Answer Key 21 Figure 4.24. The diagram has a black dot and three solid red arrows pointing away from the dot. Arrow Ft is long and pointing to the left and slightly down. Arrow Fw is also long and is a bit below a diagonal line halfway between pointing up and pointing to the right. A short arrow Fg is pointing down. Fg is the force on the kite due to gravity. Fw is the force exerted on the kite by the wind. Ft is the force of tension in the string holding the kite. It must balance the vector sum of the other two forces for the kite to float stationary in the air. 23 (b) 25 (d) 27 A free-body diagram would show a northward force of 64 N and a westward force of 38 N. The net force is equal to the sum of the two applied forces. It can be found using the Pythagorean theorem: net = 2 + = (38 N)2 + (64 N)2 = 74.4 N Since =, = 74.4 N 825 kg = 0.09 m/s2 The boulder will accelerate at 0.09 m/s2. 29 (b) 31 (b) 33 (d) Chapter 5 Problems & Exercises 1 5.00 N 4 (a) 588 N (b) 1.96 m/s2 Answer Key (5.30) (5.58) (5.59) 1572 6 (a) 3.29 m/s2 (b) 3.52 m/s2", " (c) 980 N; 945 N 10 1.83 m/s2 14 (a) 4.20 m/s2 (b) 2.74 m/s2 (c) \u20130.195 m/s2 16 (a) 1.03\u00d7106 N (b) 3.48\u00d7105 N 18 (a) 51.0 N (b) 0.720 m/s2 20 115 m/s; 414 km/hr 22 25 m/s; 9.9 m/s 24 2.9 26 28 0.76 kg/m \u22c5 s 29 [] = s [][] = kg \u22c5 m/s2 m \u22c5 m/s = kg m \u22c5 s 1.90\u00d710\u22123 cm 31 (a)1 mm (b) This does seem reasonable, since the lead does seem to shrink a little when you push on it. 33 (a)9 cm (b)This seems reasonable for nylon climbing rope, since it is not supposed to stretch that much. 35 8.59 mm 37 1.49\u00d710\u22127 m 39 (a) 3.99\u00d710\u22127 m (b) 9.67\u00d710\u22128 m 41 4\u00d7106 N/m2. This is about 36 atm, greater than a typical jar can withstand. 43 1.4 cm This content is available for free at http://cnx.org/content/col11844/1.13 Answer Key 1573 Test Prep for AP\u00ae Courses 1 (b) 3 (c) Chapter 6 Problems & Exercises 1 723 km 3 5\u00d7107 rotations 5 117 rad/s 7 76.2 rad/s 728 rpm 8 (a) 33.3 rad/s (b) 500 N (c) 40.8 m 10 12.9 rev/min 12 4\u00d71021 m 14 a) 3.47\u00d7104 m / s2, 3.55\u00d7103 b) 51.1 m / s 16 a) 31.4 rad/s b) 118 m/s c) 384 m/s d)The centripetal acceleration felt by Olympic skaters is 12 times larger than the acceleration due to gravity. That's quite a lot of acceleration in itself. The centripetal acceleration felt by Button's nose was 39.2 times larger than the acceleration due to gravity. It is no wonder that he ruptured small blood vessels in his spins. 18 a) 0.524 km/s b) 29.7 km/s 20 (a) 1.35\u00d7103 rpm (b) 8.47\u00d7103 m/s2 (c) 8.47\u00d710\u201312 N (d) 865 21 (a) 16.6 m/s Answer Key 1574 (b) 19.6 m / s2 (c) Figure 6.10. (d) 1.76\u00d7103 N or 3.00, that is, the normal force (upward) is three times her weight. (e) This answer seems reasonable, since she feels like she's being forced into the chair MUCH stronger than just by gravity. 22 a) 40.5 m / s2 b) 905 N c) The force in part (b) is very large. The acceleration in part (a) is too much, about 4 g. d) The speed of the swing is too large. At the given velocity at the bottom of the swing, there is enough kinetic energy to send the child all the way over the top, ignoring friction. 23 a) 483 N b) 17.4 N c) 2.24 times her weight, 0.0807 times her weight 25 4.14\u00ba 27 a) 24.6 m b) 36.6 m / s2 c) c = 3.73 This does not seem too large, but it is clear that bobsledders feel a lot of force on them going through sharply banked turns. 29 a) 2.56 rad/s b) 5.71\u00ba 30 a) 16.2 m/s b) 0.234 32 a) 1.84 b) A coefficient of friction this much greater than 1 is unreasonable. c) The assumed speed is too great for the tight curve. 33 a) 5.979\u00d71024 kg b) This is identical to the best value to three significant figures. This content is available for free at http://cnx.org/content/col11844/1.13 Answer Key 35 a) 1.62 m / s2 b) 3.75 m / s2 37 a) 3.42\u00d710\u20135 m / s2 b) 3.34\u00d710\u20135 m / s2 1575 The values are nearly identical. One would expect the gravitational force to be the same as the centripetal force at the core of the system. 39 a) 7.01\u00d7", "10\u20137 N b) 1.35\u00d710\u20136 N, 0.521 41 a) 1.66\u00d710\u201310 m / s2 b) 2.17\u00d7105 m/s 42 a) 2.94\u00d71017 kg b) 4.92\u00d710\u20138 of the Earth's mass. c) The mass of the mountain and its fraction of the Earth's mass are too great. d) The gravitational force assumed to be exerted by the mountain is too great. 44 1.98\u00d71030 kg 46 48 a) 7.4\u00d7103 m/s = 316 b) 1.05\u00d7103 m/s c) 2.86\u00d710\u22127 s d) 1.84\u00d7107 N e) 2.76\u00d7104 J 49 a) 5.08\u00d7103 km b) This radius is unreasonable because it is less than the radius of earth. c) The premise of a one-hour orbit is inconsistent with the known radius of the earth. Test Prep for AP\u00ae Courses 1 (a) 3 (b) Answer Key (7.8) (7.9) 3.00 J = 7.17\u00d710\u22124 kcal 3.14\u00d7103 J 1576 5 (b) Chapter 7 Problems & Exercises 1 3 (a) 5.92\u00d7105 J (b) \u22125.88\u00d7105 J (c) The net force is zero. 5 7 (a) \u2212700 J (b) 0 (c) 700 J (d) 38.6 N (e) 0 9 1 / 250 11 1.1\u00d71010 J 13 2.8\u00d7103 N 15 102 N 16 (a) 1.961016 J (b) The ratio of gravitational potential energy in the lake to the energy stored in the bomb is 0.52. That is, the energy stored in the lake is approximately half that in a 9-megaton fusion bomb. 18 (a) 1.8 J (b) 8.6 J 20 22 = 2 + 0 2 = 2(9.80 m/s2)( \u2212 0.180 m) + (2.00 m/s)2 = 0.687 m/s 7.81105 N/m (7.45) (7.60) 24 9.46 m/s 26 4104 molecules 27 Equating \u0394PEg and \u0394KE, we obtain = 2 + 0 29 (a) 25\u00d7106 years 2 = 2(9.80 m/s2)(20.0 m) + (15.0 m/s)2 = 24.8 m/s This content is available for free at http://cnx.org/content/col11844/1.13 1577 (7.81) Answer Key (b) This is much, much longer than human time scales. 210\u221210 30 32 (a) 40 (b) 8 million 34 $149 36 (a) 208 W (b) 141 s 38 (a) 3.20 s (b) 4.04 s 40 (a) 9.46107 J (b) 2.54 y 42 Identify knowns: = 950 kg, slope angle = 2.00\u00ba, = 3.00 m/s, = 600 N Identify unknowns: power of the car, force that car applies to road Solve for unknown: = = = = where is parallel to the incline and must oppose the resistive forces and the force of gravity: = + = 600 N + sin Insert this into the expression for power and solve: = = + sin 600 N + = 2.77\u00d7104 W 950 kg 9.80 m/s2 sin 2\u00ba (30.0 m/s) About 28 kW (or about 37 hp) is reasonable for a car to climb a gentle incline. 44 (a) 9.5 min (b) 69 flights of stairs 46 641 W, 0.860 hp 48 31 g 50 14.3% 52 (a) 3.21104 N (b) 2.35103 N (c) Ratio of net force to weight of person is 41.0 in part (a); 3.00 in part (b) 54 Answer Key 1578 (a) 108 kJ (b) 599 W 56 (a) 144 J (b) 288 W 58 (a) 2.501012 J (b) 2.52% (c) 1.4104 kg (14 metric tons) 60 (a) 294 N (b) 118 J (c) 49.0 W 62 (a) 0.500 m/s2 (b) 62.5 N (c) Assuming the acceleration of the swimmer decreases linearly with time over the 5.00 s interval, the frictional force must therefore be increasing linearly with time, since = \u2212. If the acceleration decreases linearly with time, the velocity will contain a term dependent on time squared", " ( 2 ). Therefore, the water resistance will not depend linearly on the velocity. 64 (a) 16.1\u00d7103 N (b) 3.22\u00d7105 J (c) 5.66 m/s (d) 4.00 kJ 66 (a) 4.65\u00d7103 kcal (b) 38.8 kcal/min (c) This power output is higher than the highest value on Table 7.5, which is about 35 kcal/min (corresponding to 2415 watts) for sprinting. (d) It would be impossible to maintain this power output for 2 hours (imagine sprinting for 2 hours!). 69 (a) 4.32 m/s (b) 3.47\u00d7103 N (c) 8.93 kW Test Prep for AP\u00ae Courses 1 (b) 3 (d) 5 (a) 7 This content is available for free at http://cnx.org/content/col11844/1.13 Answer Key 1579 The kinetic energy should change in the form of \u2013cos, with an initial value of 0 or slightly above, and ending at the same level. 9 Any force acting perpendicular will have no effect on kinetic energy. Obvious examples are gravity and the normal force, but others include wind directly from the side and rain or other precipitation falling straight down. 11 Note that the wind is pushing from behind and one side, so your KE will increase. The net force has components of 1400 N in the direction of travel and 212 N perpendicular to the direction of travel. So the net force is 1420 N at 8.5 degrees from the direction of travel. 13 Gravity has a component perpendicular to the cannon (and to displacement, so it is irrelevant) and has a component parallel to the cannon. The latter is equal to 9.8 N. Thus the net force in the direction of the displacement is 50 N \u2212 9.8 N, and the kinetic energy is 60 J. 15 The potato cannon (and many other projectile launchers) above is an option, with a force launching the projectile, friction, potentially gravity depending on the direction it is pointed, etc. A drag (or other) car accelerating is another possibility. 17 The kinetic energy of the rear wagon increases. The front wagon does not, until the rear wagon collides with it. The total system may be treated by its center of mass, halfway between the wagons, and its energy increases by the same amount as the sum of the two individual wagons. 19 (d) 21 0.049 J; 0.041 m, 0.25 m 23 20 m high, 20 m/s. 25 (a) 27 (d) 29 (c) 31 (b) 33 (c) 35 (c) 37 (c), (d) 39 (a) 41 (b) Chapter 8 Problems & Exercises 1 (a) 1.50\u00d7104 kg \u22c5 m/s (b) 625 to 1 (c) 6.66\u00d7102 kg \u22c5 m/s 3 (a) 8.00\u00d7104 m/s (b) 1.20\u00d7106 kg \u00b7 m/s 1580 Answer Key (c) Because the momentum of the airplane is 3 orders of magnitude smaller than of the ship, the ship will not recoil very much. The recoil would be \u22120.0100 m/s, which is probably not noticeable. 5 54 s 7 9.00\u00d7103 N 9 a) 2.40\u00d7103 N toward the leg b) The force on each hand would have the same magnitude as that found in part (a) (but in opposite directions by Newton\u2019s third law) because the change in momentum and the time interval are the same. 11 a) 800 kg \u22c5 m/s away from the wall b) 1.20 m/s away from the wall 13 (a) 1.50\u00d7106 N away from the dashboard (b) 1.00\u00d7105 N away from the dashboard 15 4.69\u00d7105 N in the boat\u2019s original direction of motion 17 2.10\u00d7103 N away from the wall 19 = 2 p = v \u21d2 2 = 22 \u21d2 2 \u21d2 2 = 1 2 = 2 2 22 = (8.35) 21 60.0 g 23 0.122 m/s 25 In a collision with an identical car, momentum is conserved. Afterwards f = 0 for both cars. The change in momentum will be the same as in the crash with the tree. However, the force on the body is not determined since the time is not known. A padded stop will reduce injurious force on body. 27 22.4 m/s in the same direction as the original motion 29 0.250 m/s 31 (a) 86.4 N perpendicularly away from the bumper (b) 0.389 J (c) 64.0% 33 (a) 8.06 m/s (b) -56", ".0 J (c)(i) 7.88 m/s; (ii) -223 J 35 (a) 0.163 m/s in the direction of motion of the more massive satellite This content is available for free at http://cnx.org/content/col11844/1.13 Answer Key 1581 (b) 81.6 J (c) 8.70\u00d710\u22122 m/s in the direction of motion of the less massive satellite, 81.5 J. Because there are no external forces, the velocity of the center of mass of the two-satellite system is unchanged by the collision. The two velocities calculated above are the velocity of the center of mass in each of the two different individual reference frames. The loss in KE is the same in both reference frames because the KE lost to internal forces (heat, friction, etc.) is the same regardless of the coordinate system chosen. 37 0.704 m/s \u20132.25 m/s 38 (a) 4.58 m/s away from the bullet (b) 31.5 J (c) \u20130.491 m/s (d) 3.38 J 40 (a) 1.02\u00d710\u22126 m/s (b) 5.63\u00d71020 J (almost all KE lost) (c) Recoil speed is 6.79\u00d710\u221217 m/s, energy lost is 6.25\u00d7109 J. The plume will not affect the momentum result because the plume is still part of the Moon system. The plume may affect the kinetic energy result because a significant part of the initial kinetic energy may be transferred to the kinetic energy of the plume particles. 42 24.8 m/s 44 (a) 4.00 kg (b) 210 J (c) The clown does work to throw the barbell, so the kinetic energy comes from the muscles of the clown. The muscles convert the chemical potential energy of ATP into kinetic energy. 45 (a) 3.00 m/s, 60\u00ba below -axis (b) Find speed of first puck after collision: 0 = \u20321 sin 30\u00ba\u2212\u20322 sin 60\u00ba \u21d2 \u20321 = 2 sin 60\u00ba sin 30\u00ba = 5.196 m/s KE = 1 KE = 1 21 2\u20321 2 = 18 J 2 + 1 2\u20322 KE KE\u2032 2 = 18 J = 1.00 Verify that ratio of initial to final KE equals one: 47 (a) \u22122.26 m/s (b) 7.63\u00d7103 J (c) The ground will exert a normal force to oppose recoil of the cannon in the vertical direction. The momentum in the vertical direction is transferred to the earth. The energy is transferred into the ground, making a dent where the cannon is. After long barrages, cannon have erratic aim because the ground is full of divots. 49 (a) 5.36\u00d7105 m/s at \u221229.5\u00ba (b) 7.52\u00d710\u221213 J 51 1582 Answer Key We are given that 1 = 2 \u2261. The given equations then become: and Square each equation to get 1 = 1 cos 1 + 2 cos 2 0 = \u20321 sin 1 + \u20322 sin 2. 1 0 2 = \u20321 = \u20321 2 cos2 1 + \u20322 2 sin2 1 + \u20322 2 cos2 2 + 2\u20321 \u20322 cos 1 cos 2 2 sin2 2 + 2\u20321 \u20322 sin 1 sin 2. 1 Add these two equations and simplify: 2 + \u20322 2 + \u20322 2 + \u20322 2 = \u20321 = \u20321 2 + 2\u20321 \u20322 cos 1 cos 2 + sin 1 sin 2 1 2 + 2\u20321 \u20322 2 2 + 2\u20321 \u20322 cos 1 cos cos. 1 + 2 + 1 2 cos 1 \u2212 2 \u2212 1 2 cos 1 + 2 (8.107) (8.108) (8.109) (8.110) Multiply the entire equation by 1 2 to recover the kinetic energy: 2 + 1 2 = 1 2\u20321 2\u20322 1 21 2 + \u20321 \u20322 cos 1 \u2212 2 (8.111) 53 39.2 m/s2 55 4.16\u00d7103 m/s 57 The force needed to give a small mass \u0394 an acceleration \u0394 is = \u0394\u0394. To accelerate this mass in the small time interval \u0394 at a speed e requires e = \u0394\u0394, so = e \u0394 in magnitude to the thrust force acting on the rocket, so thrust = e \u0394 Newton\u2019s second law to the rocket gives thrust \u2212 = \u21d2 = e and unburnt fuel. 60 2.63\u00d7103 kg 61 \u2212, where is the mass of the rocket, where all quantities are positive. Applying \u0394 \u0394 \u0394 \u0394.", " By Newton\u2019s third law, this force is equal (a) 0.421 m/s away from the ejected fluid. (b) 0.237 J. Test Prep for AP\u00ae Courses 1 (b) 3 (b) 5 (a) 7 (c) (based on calculation of = \u0394 \u0394 ) 9 (c) 11 (d) 13 (b) This content is available for free at http://cnx.org/content/col11844/1.13 Answer Key 1583 15 (d) 17 (b) 19 (c) 21 (b) 23 (c) 25 (b) 27 (a) 29 (c) 31 (b) 33 (a) 35 (b) 37 (a) 39 (a) 41 (d) 43 (c). Because of conservation of momentum, the final velocity of the combined mass must be 4.286 m/s. The initial kinetic energy is (0.5)(2.0)(15)2 = 225 J. The final kinetic energy is (0.5)(7.0)(4.286)2 = 64 J, so the difference is \u2212161 J. 45 (a) 47 (d) 49 (c) 51 (b) Chapter 9 Problems & Exercises 1 a) 46.8 N\u00b7m b) It does not matter at what height you push. The torque depends on only the magnitude of the force applied and the perpendicular distance of the force's application from the hinges. (Children don't have a tougher time opening a door because they push lower than adults, they have a tougher time because they don't push far enough from the hinges.) 3 23.3 N 5 Given: 1 = 26.0 kg, 2 = 32.0 kg, s = 12.0 kg, 1 = 1.60 m, s = 0.160 m, find (a) 2, (b) p (9.26) a) Since children are balancing: Answer Key (9.27) (9.28) (9.29) (9.30) 1584 So, solving for 2 gives: net cw = \u2013 net ccw \u21d2 1 1 + s s = 22 26.0 kg)(1.60 m) + (12.0 kg)(0.160 m) 32.0 kg b) Since the children are not moving: = 1.36 m So that net = 26.0 kg + 32.0 kg + 12.0 kg)(9.80 m / s2) = 686 N 6 wall = 1.43\u00d7103 N 8 a) 2.55\u00d7103 N, 16.3\u00ba to the left of vertical (i.e., toward the wall) b) 0.292 10 B = 2.12\u00d7104 N 12 a) 0.167, or about one-sixth of the weight is supported by the opposite shore. b) = 2.0\u00d7104 N, straight up. 14 a) 21.6 N b) 21.6 N 16 350 N directly upwards 19 25 50 N 21 a) MA = 18.5 b) i = 29.1 N c) 510 N downward 23 1.3\u00d7103 N 25 a) = 299 N b) 897 N upward 26 This content is available for free at http://cnx.org/content/col11844/1.13 Answer Key 1585 B = 470 N; 1 = 4.00 cm; a = 2.50 kg; 2 = 16.0 cm;b = 4.00 kg; 3 = 38.0 cm 16.0 cm 9.80 m / s2 4.0 cm 38.0 cm 9.80 m / s2 4.00 cm \u2013 1 \u2013 1 = 2.50 kg + 4.00 kg = 407 N 28 1.1\u00d7103 N = 190\u00ba ccw from positive axis 30 V = 97 N, = 59\u00ba 32 (a) 25 N downward (b) 75 N upward 33 (a) A = 2.21\u00d7103 N upward (b) B = 2.94\u00d7103 N downward 35 (a) teeth on bullet = 1.2\u00d7102 N upward (b) J = 84 N downward 37 (a) 147 N downward (b) 1680 N, 3.4 times her weight (c) 118 J (d) 49.0 W 39 a) 2 = 2.33 m b) The seesaw is 3.0 m long, and hence, there is only 1.50 m of board on the other side of the pivot. The second child is off the board. c) The position of the first child must be shortened, i.e. brought closer to the pivot. Test Prep for AP\u00ae Courses 1 (a) 3 Both objects are in equilibrium. However, they will respond differently if a force is applied to their sides. If the cone placed on its base is displaced to the", " side, its center of gravity will remain over its base and it will return to its original position. When the traffic cone placed on its tip is displaced to the side, its center of gravity will drift from its base, causing a torque that will accelerate it to the ground. 5 (d) 7 a. FL = 7350 N, FR = 2450 N b. As the car moves to the right side of the bridge, FL will decrease and FR will increase. (At exactly halfway across the bridge, FL and FR will both be 4900 N.) 9 1586 Answer Key The student should mention that the guiding principle behind simple machines is the second condition of equilibrium. Though the torque leaving a machine must be equivalent to torque entering a machine, the same requirement does not exist for forces. As a result, by decreasing the lever arm to the existing force, the size of the existing force will be increased. The mechanical advantage will be equivalent to the ratio of the forces exiting and entering the machine. 11 a. The force placed on your bicep muscle will be greater than the force placed on the dumbbell. The bicep muscle is closer to your elbow than the downward force placed on your hand from the dumbbell. Because the elbow is the pivot point of the system, this results in a decreased lever arm for the bicep. As a result, the force on the bicep must be greater than that placed on the dumbbell. (How much greater? The ratio between the bicep and dumbbell forces is equal to the inverted ratio of their distances from the elbow. If the dumbbell is ten times further from the elbow than the bicep, the force on the bicep will be 200 pounds!) b. The force placed on your bicep muscle will decrease. As the forearm lifts the dumbbell, it will get closer to the elbow. As a result, the torque placed on the arm from the weight will decrease and the countering torque created by the bicep muscle will do so as well. Chapter 10 Problems & Exercises 1 = 0.737 rev/s 3 (a) \u22120.26 rad/s2 (b) 27 rev 5 (a) 80 rad/s2 (b) 1.0 rev 7 (a) 45.7 s (b) 116 rev 9 a) 600 rad/s2 b) 450 rad/s c) 21.0 m/s 10 (a) 0.338 s (b) 0.0403 rev (c) 0.313 s 12 0.50 kg \u22c5 m2 14 (a) 50.4 N \u22c5 m (b) 17.1 rad/s2 (c) 17.0 rad/s2 16 3.96\u00d71018 s or 1.26\u00d71011 y This content is available for free at http://cnx.org/content/col11844/1.13 Answer Key 18 1587 2 = + 42 = 1 2 32 \u2212 1 42 = 1 122 Thus, = \u2212 1 19 (a) 2.0 ms (b) The time interval is too short. (c) The moment of inertia is much too small, by one to two orders of magnitude. A torque of 500 N \u22c5 m is reasonable. 20 (a) 17,500 rpm (b) This angular velocity is very high for a disk of this size and mass. The radial acceleration at the edge of the disk is > 50,000 gs. (c) Flywheel mass and radius should both be much greater, allowing for a lower spin rate (angular velocity). KErot = 434 J (10.104) 21 (a) 185 J (b) 0.0785 rev (c) = 9.81 N 23 (a) 2.57\u00d71029 J (b) KErot = 2.65\u00d71033 J 25 27 (a) 128 rad/s (b) 19.9 m 29 (a) 10.4 rad/s2 (b) net = 6.11 J 34 (a) 1.49 kJ (b) 2.52\u00d7104 N 36 (a) 2.66\u00d71040 kg \u22c5 m2/s (b) 7.07\u00d71033 kg \u22c5 m2/s The angular momentum of the Earth in its orbit around the Sun is 3.77\u00d7106 times larger than the angular momentum of the Earth around its axis. 38 22.5 kg \u22c5 m2/s 40 25.3 rpm 43 Answer Key 1588 (a) 0.156 rad/s (b) 1.17\u00d710\u22122 J (c) 0.188 kg \u22c5 m/s 45 (a) 3.13 rad/s (b) Initial KE = 438 J, final KE = 438 J 47 (a) 1.70 rad/s (b) Initial KE = 22.5 J,", " final KE = 2.04 J (c) 1.50 kg \u22c5 m/s 48 (a) 5.64\u00d71033 kg \u22c5 m2 /s (b) 1.39\u00d71022 N \u22c5 m (c) 2.17\u00d71015 N Test Prep for AP\u00ae Courses 1 (b) 3 (d) 5 (d) You are given a thin rod of length 1.0 m and mass 2.0 kg, a small lead weight of 0.50 kg, and a not-so-small lead weight of 1.0 kg. The rod has three holes, one in each end and one through the middle, which may either hold a pivot point or one of the small lead weights. 7 (a) 9 (c) 11 (a) 13 (a) 15 (b) 17 (c) 19 (b) 21 (b) 23 (c) 25 (d) 27 A door on hinges is a rotational system. When you push or pull on the door handle, the angular momentum of the system changes. If a weight is hung on the door handle, then pushing on the door with the same force will cause a different increase in angular momentum. If you push or pull near the hinges with the same force, the resulting angular momentum of the system will also be different. 29 Since the globe is stationary to start with, This content is available for free at http://cnx.org/content/col11844/1.13 1589 Answer Key = \u0394 \u0394 \u22c5 \u0394 = \u0394 By substituting, 120 N\u2022m \u2022 1.2 s = 144 N\u2022m\u2022s. The angular momentum of the globe after 1.2 s is 144 N\u2022m\u2022s. Chapter 11 Problems & Exercises 1 1.610 cm3 3 (a) 2.58 g (b) The volume of your body increases by the volume of air you inhale. The average density of your body decreases when you take a deep breath, because the density of air is substantially smaller than the average density of the body before you took the deep breath. 4 2.70 g/cm3 6 (a) 0.163 m (b) Equivalent to 19.4 gallons, which is reasonable 8 7.9\u00d7102 kg/m3 9 15.6 g/cm3 10 (a) 1018 kg/m3 (b) 2\u00d7104 m 11 3.59\u00d7106 Pa ; or 521 lb/in2 13 2.36\u00d7103 N 14 0.760 m 16 units = (m) = kg/m3 kg \u22c5 m/s2 m/s2 1/m2 = kg \u22c5 m2 / m3 \u22c5 s2 (11.30) 18 (a) 20.5 mm Hg = N/m2 (b) The range of pressures in the eye is 12\u201324 mm Hg, so the result in part (a) is within that range 20 1.09\u00d7103 N/m2 22 24.0 N 24 1590 Answer Key 2.55\u00d7107 Pa ; or 251 atm 26 5.76\u00d7103 N extra force 28 (a) = ii = oo \u21d2 o = i i o. Now, using equation: Finally = oo = i o i i i o = ii = i. (11.32) (11.33) In other words, the work output equals the work input. (b) If the system is not moving, friction would not play a role. With friction, we know there are losses, so that out = in \u2212 f ; therefore, the work output is less than the work input. In other words, with friction, you need to push harder on the input piston than was calculated for the nonfriction case. 29 Balloon: = 5.00 cm H2 O, g abs = 1.035\u00d7103 cm H2 O. Jar: = \u221250.0 mm Hg, g abs = 710 mm Hg. 31 4.08 m 33 \u0394 = 38.7 mm Hg, Leg blood pressure = 159 119. 35 22.4 cm2 36 91.7% 38 815 kg/m3 40 (a) 41.4 g (b) 41.4 cm3 (c) 1.09 g/cm3 42 (a) 39.5 g (b) 50 cm3 (c) 0.79 g/cm3 It is ethyl alcohol. This content is available for free at http://cnx.org/content/col11844/1.13 1591 Answer Key 44 8.21 N 46 (a) 960 kg/m3 (b) 6.34% She indeed floats more in seawater. 48 (a) 0.24 (b) 0.68 (c) Yes, the cork will float because obj < ethyl alcohol(0", ".678 g/cm3 < 0.79 g/cm3) 50 The difference is 0.006%. 52 net = fl \u2212 1fl 2 \u2212 1 = 2 \u2212 1 fl where fl = density of fluid. Therefore, net = (2 \u2212 1)fl = flfl = fl = fl where is fl the weight of the fluid displaced. 54 592 N/m2 56 2.23\u00d710\u22122 mm Hg 58 (a) 1.65\u00d710\u22123 m (b) 3.71\u00d710\u20134 m 60 6.32\u00d710\u22122 N/m Based on the values in table, the fluid is probably glycerin. 62 w = 14.6 N/m2 a = 4.46 N/m2 sw = 7.40 N/m2. Alcohol forms the most stable bubble, since the absolute pressure inside is closest to atmospheric pressure. 64 5.1\u00ba This is near the value of = 0\u00ba for most organic liquids. 66 \u22122.78 The ratio is negative because water is raised whereas mercury is lowered. Answer Key 1592 68 479 N 70 1.96 N 71 \u221263.0 cm H2 O 73 (a) 3.81\u00d7103 N/m2 (b) 28.7 mm Hg, which is sufficient to trigger micturition reflex 75 (a) 13.6 m water (b) 76.5 cm water 77 (a) 3.98\u00d7106 Pa (b) 2.1\u00d710\u22123 cm 79 (a) 2.97 cm (b) 3.39\u00d710\u22126 J (c) Work is done by the surface tension force through an effective distance / 2 to raise the column of water. 81 (a) 2.01\u00d7104 N (b) 1.17\u00d710\u22123 m (c) 2.56\u00d71010 N/m2 83 (a) 1.38\u00d7104 N (b) 2.81\u00d7107 N/m2 (c) 283 N 85 (a) 867 N (b) This is too much force to exert with a hand pump. (c) The assumed radius of the pump is too large; it would be nearly two inches in diameter\u2014too large for a pump or even a master cylinder. The pressure is reasonable for bicycle tires. Test Prep for AP\u00ae Courses 1 (e) 3 (a) 100 kg/m3 (b) 60% (c) yes; yes (76% will be submerged) (d) answers vary 5 (d) Chapter 12 Problems & Exercises 1 This content is available for free at http://cnx.org/content/col11844/1.13 1593 Answer Key 2.78 cm3 /s 3 27 cm/s 5 (a) 0.75 m/s (b) 0.13 m/s 7 (a) 40.0 cm2 (b) 5.09\u00d7107 9 (a) 22 h (b) 0.016 s 11 (a) 12.6 m/s (b) 0.0800 m3 /s (c) No, independent of density. 13 (a) 0.402 L/s (b) 0.584 cm 15 (a) 127 cm3 /s (b) 0.890 cm 17 = Force Area () units = N/m2 = N \u22c5 m/m3 = J/m3 = energy/volume 19 184 mm Hg 21 2.54\u00d7105 N 23 (a) 1.58\u00d7106 N/m2 (b) 163 m 25 (a) 9.56\u00d7108 W (b) 1.4 27 1.26 W 29 (a) 3.02\u00d710\u22123 N (b) 1.03\u00d710\u22123 31 1.60 cm3 /min 1594 33 8.7\u00d710\u221211 m3 /s 35 0.316 37 (a) 1.52 Answer Key (b) Turbulence will decrease the flow rate of the blood, which would require an even larger increase in the pressure difference, leading to higher blood pressure. 225 mPa \u22c5 s 0.138 Pa \u22c5 s, (12.98) (12.99) 39 41 or Olive oil. 43 (a) 1.62\u00d7104 N/m2 (b) 0.111 cm3 /s (c)10.6 cm 45 1.59 47 2.95\u00d7106 N/m2 (gauge pressure) 51 R = 1.99\u00d7102 < 2000 53 (a) nozzle: 1.27\u00d7105, not laminar (b) hose: 3.51\u00d7104, not laminar. 55 2.54 << 2000, laminar. 57 1.02 m/s 1.28\u00d710\u20132 L/s 59 (a) \u2265 13.0 m (b", ") 2.68\u00d710\u22126 N/m2 61 (a) 23.7 atm or 344 lb/in2 (b) The pressure is much too high. (c) The assumed flow rate is very high for a garden hose. (d) 5.27\u00d7106 > > 3000, turbulent, contrary to the assumption of laminar flow when using this equation. 62 1.41\u00d710\u22123 m 64 1.3\u00d7102 s 66 This content is available for free at http://cnx.org/content/col11844/1.13 1595 Answer Key 0.391 s Test Prep for AP\u00ae Courses 1 (c) 3 (a) 5 (a) 7 (a) 9 (d) Chapter 13 Problems & Exercises 1 102\u00baF 3 20.0\u00baC and 25.6\u00baC 5 9890\u00baF 7 (a) 22.2\u00baC \u0394(\u00baF) = 2 (\u00baF) \u2212 1(\u00baF) (b\u00baC) + 32.0\u00ba \u2212 5 = 9 5 2 (\u00baC) \u2212 1(\u00baC) 1 (\u00baC) + 32.0\u00ba \u0394(\u00baC) 9 169.98 m 11 5.4\u00d710\u22126 m 13 Because the area gets smaller, the price of the land DECREASES by ~17000. 15 = 0 + \u0394 = 0(1 + \u0394) (13.25) = (60.00 L) 1 + 950\u00d710\u22126 / \u00baC (35.0\u00baC \u2212 15.0\u00baC) = 61.1 L 17 (a) 9.35 mL (b) 7.56 mL 19 0.832 mm 21 We know how the length changes with temperature: \u0394 = 0\u0394. Also we know that the volume of a cube is related to its length by = 3, so the final volume is then = 0 + \u0394 = 0 + \u0394 3. Substituting for \u0394 gives = 0 + 0\u0394 3 = 0 3(1 + \u0394)3. Now, because \u0394 is small, we can use the binomial expansion: \u2248 0 3(1 + 3\u03b1\u0394T) = 0 3 + 3\u03b10 3\u0394. (13.26) (13.27) 1596 So writing the length terms in terms of volumes gives = 0 + \u0394 \u2248 0 + 3\u03b1 0\u0394 and so \u0394 = 0\u0394 \u2248 3\u03b1 0\u0394 or \u2248 3\u03b1. Answer Key (13.28) 22 1.62 atm 24 (a) 0.136 atm (b) 0.135 atm. The difference between this value and the value from part (a) is negligible. 26 (a) = (mol)(J/mol \u22c5 K)(K) = J (b) = (mol)(cal/mol \u22c5 K)(K) = cal (c) = (mol)(L \u22c5 atm/mol \u22c5 K)(K) = L \u22c5 atm = (m3)(N/m2) = N \u22c5 m = J 28 7.86\u00d710\u22122 mol 30 (a) 6.02\u00d7105 km3 (b) 6.02\u00d7108 km 32 \u221273.9\u00baC 34 (a) 9.14\u00d7106 N/m2 (b) 8.23\u00d7106 N/m2 (c) 2.16 K (d) No. The final temperature needed is much too low to be easily achieved for a large object. 36 41 km 38 (a) 3.7\u00d710\u221217 Pa (b) 6.0\u00d71017 m3 (c) 8.4\u00d7102 km 39 1.25\u00d7103 m/s 41 (a) 1.20\u00d710\u221219 J (b) 1.24\u00d710\u221217 J 43 458 K 45 1.95\u00d7107 K 47 6.09\u00d7105 m/s 49 This content is available for free at http://cnx.org/content/col11844/1.13 1597 Answer Key 7.89\u00d7104 Pa 51 (a) 1.99\u00d7105 Pa (b) 0.97 atm 53 3.12\u00d7104 Pa 55 78.3% 57 (a) 2.12\u00d7104 Pa (b) 1.06 % 59 (a) 8.80\u00d710\u22122 g (b) 6.30\u00d7103 Pa ; the two values are nearly identical. 61 82.3% 63 4.77\u00baC 65 38.3 m 67 B / Cu B / Cu circumstances. 69 (a) 4.41\u00d71010 mol/m3 = 1.02. The buoyant force supports nearly the exact same amount of force on the copper block in both (b) It\u2019s unreasonably large. (c)", " At high pressures such as these, the ideal gas law can no longer be applied. As a result, unreasonable answers come up when it is used. 71 (a) 7.03\u00d7108 m/s (b) The velocity is too high\u2014it\u2019s greater than the speed of light. (c) The assumption that hydrogen inside a supernova behaves as an idea gas is responsible, because of the great temperature and density in the core of a star. Furthermore, when a velocity greater than the speed of light is obtained, classical physics must be replaced by relativity, a subject not yet covered. Test Prep for AP\u00ae Courses 1 (a), (c) 3 (d) 5 (b) 7 (a) 7.29 \u00d7 10-21 J; (b) 352K or 79\u00baC Chapter 14 Problems & Exercises 5.02\u00d7108 J 3.07\u00d7103 J 0.171\u00baC Answer Key (14.18) (14.19) (14.20) 1598 1 3 5 7 10.8 9 617 W 11 35.9 kcal 13 (a) 591 kcal (b) 4.94\u00d7103 s 15 13.5 W 17 (a) 148 kcal (b) 0.418 s, 3.34 s, 4.19 s, 22.6 s, 0.456 s 19 33.0 g 20 (a) 9.67 L (b) Crude oil is less dense than water, so it floats on top of the water, thereby exposing it to the oxygen in the air, which it uses to burn. Also, if the water is under the oil, it is less efficient in absorbing the heat generated by the oil. 22 a) 319 kcal b) 2.00\u00baC 24 20.6\u00baC 26 4.38 kg 28 (a) 1.57\u00d7104 kcal (b) 18.3 kW \u22c5 h (c) 1.29\u00d7104 kcal 30 (a) 1.01\u00d7103 W (b) One 32 84.0 W 34 2.59 kg 36 (a) 39.7 W (b) 820 kcal 38 35 to 1, window to wall This content is available for free at http://cnx.org/content/col11844/1.13 1599 Answer Key 40 1.05\u00d7103 K 42 (a) 83 W (b) 24 times that of a double pane window. 44 20.0 W, 17.2% of 2400 kcal per day 45 10 m/s 47 85.7\u00baC 49 1.48 kg 51 2\u00d7104 MW 53 (a) 97.2 J (b) 29.2 W (c) 9.49 W (d) The total rate of heat loss would be 29.2 W + 9.49 W = 38.7 W. While sleeping, our body consumes 83 W of power, while sitting it consumes 120 to 210 W. Therefore, the total rate of heat loss from breathing will not be a major form of heat loss for this person. 55 \u221221.7 kW Note that the negative answer implies heat loss to the surroundings. 57 \u2212266 kW 59 \u221236.0 W 61 (a) 1.31% (b) 20.5% 63 (a) \u221215.0 kW (b) 4.2 cm 65 (a) 48.5\u00baC (b) A pure white object reflects more of the radiant energy that hits it, so a white tent would prevent more of the sunlight from heating up the inside of the tent, and the white tunic would prevent that heat which entered the tent from heating the rider. Therefore, with a white tent, the temperature would be lower than 48.5\u00baC, and the rate of radiant heat transferred to the rider would be less than 20.0 W. 67 (a) 3\u00d71017 J (b) 1\u00d71013 kg (c) When a large meteor hits the ocean, it causes great tidal waves, dissipating large amount of its energy in the form of kinetic energy of the water. 69 (a) 3.44\u00d7105 m3 /s 1600 Answer Key (b) This is equivalent to 12 million cubic feet of air per second. That is tremendous. This is too large to be dissipated by heating the air by only 5\u00baC. Many of these cooling towers use the circulation of cooler air over warmer water to increase the rate of evaporation. This would allow much smaller amounts of air necessary to remove such a large amount of heat because evaporation removes larger quantities of heat than was considered in part (a). 71 20.9 min 73 (a) 3.96\u00d710-2 g (b) 96.2 J (c) 16.0 W 75 (a) 1.102 (b) 2.79\u00d7104 J (c) 12.6 J. This will not cause a significant cooling of the air because it is much less than", " the energy found in part (b), which is the energy required to warm the air from 20.0\u00baC to 50.0\u00baC. 76 (a) 36\u00baC (b) Any temperature increase greater than about 3\u00baC would be unreasonably large. In this case the final temperature of the person would rise to 73\u00baC (163\u00baF). (c) The assumption of 95% heat retention is unreasonable. 78 (a) 1.46 kW (b) Very high power loss through a window. An electric heater of this power can keep an entire room warm. (c) The surface temperatures of the window do not differ by as great an amount as assumed. The inner surface will be warmer, and the outer surface will be cooler. Test Prep for AP\u00ae Courses 1 (c) 3 (a) 5 (b) 7 (a) 9 (d) Chapter 15 Problems & Exercises 1 1.6\u00d7109 J 3 -9.30108 J 5 (a) \u22121.0\u00d7104 J, or \u22122.39 kcal (b) 5.00% 7 (a) 122 W This content is available for free at http://cnx.org/content/col11844/1.13 Answer Key (b) 2.10\u00d7106 J 1601 (c) Work done by the motor is 1.61\u00d7107 J ;thus the motor produces 7.67 times the work done by the man 9 (a) 492 kJ (b) This amount of heat is consistent with the fact that you warm quickly when exercising. Since the body is inefficient, the excess heat produced must be dissipated through sweating, breathing, etc. 10 6.77\u00d7103 J 12 (a) = \u0394 = 1.76\u00d7105 J (b) = = 1.76\u00d7105 J. Yes, the answer is the same. 14 = 4.5\u00d7103 J 16 is not equal to the difference between the heat input and the heat output. 20 (a) 18.5 kJ (b) 54.1% 22 (a) 1.32 \u00d7 109 J (b) 4.68 \u00d7 109 J 24 (a) 3.80 \u00d7 109 J (b) 0.667 barrels 26 (a) 8.30 \u00d7 1012 J, which is 3.32% of 2.50 \u00d7 1014 J. (b) \u20138.30 \u00d7 1012 J, where the negative sign indicates a reduction in heat transfer to the environment. 28 403\u00baC 30 (a) 244\u00baC (b) 477\u00baC (c)Yes, since automobiles engines cannot get too hot without overheating, their efficiency is limited. 32 (a) (b) 1 = 1 \u2212 c,1 h,1 = 1 \u2212 543 K 723 K = 0.249 or 24.9% 2 = 1 \u2212 423 K 543 K = 0.221 or 22.1% 1602 (c) 1 = 1 \u2212 c,1 h,1 \u21d2 c,1 = h,1 1, \u2212, similarly, c,2 = h,2 1 \u2212 1 2 using h,2 = c,1 in above equation gives c,2 = h,,1 2 1 \u2212 overall 1 \u2212 1 overall = 1 \u2212 (1 \u2212 0.249)(1 \u2212 0.221) = 41.5% 2 Answer Key 1 \u2212 overall (d) overall = 1 \u2212 423 K 723 K = 0.415 or 41.5% 34 The heat transfer to the cold reservoir is c = h \u2212 = 25 kJ \u2212 12 kJ = 13 kJ, so the efficiency is = 1 \u2212 c h = 1 \u2212 13 kJ 25 kJ = 0.48. The Carnot efficiency is C = 1 \u2212 c h = 1 \u2212 300 K 600 K = 0.50. The actual efficiency is 96% of the Carnot efficiency, which is much higher than the best-ever achieved of about 70%, so her scheme is likely to be fraudulent. 36 (a) \u201356.3\u00baC (b) The temperature is too cold for the output of a steam engine (the local environment). It is below the freezing point of water. (c) The assumed efficiency is too high. 37 4.82 39 0.311 41 (a) 4.61 (b) 1.66\u00d7108 J or 3.97\u00d7104 kcal (c) To transfer 1.66\u00d7108 J, heat pump costs $1.00, natural gas costs $1.34. 43 27.6\u00baC 45 (a) 1.44\u00d7107 J (b) 40 cents (c) This cost seems quite realistic; it says that running an air conditioner all day would cost $9.59 (if it ran continuously). 47 (a) 9.78\u00d7104 J/K (b) In order to gain more energy, we must generate it from things within", " the house, like a heat pump, human bodies, and other appliances. As you know, we use a lot of energy to keep our houses warm in the winter because of the loss of heat to the outside. 49 8.01\u00d7105 J 51 (a) 1.04\u00d71031 J/K (b) 3.28\u00d71031 J 53 199 J/K 55 This content is available for free at http://cnx.org/content/col11844/1.13 Answer Key (a) 2.47\u00d71014 J (b) 1.60\u00d71014 J (c) 2.85\u00d71010 J/K (d) 8.29\u00d71012 J 1603 57 It should happen twice in every 1.27\u00d71030 s or once in every 6.35\u00d71029 s 1 y 6.35\u00d71029 s 365.25 d 1 h 3600 s 1 d 24 h 2.0\u00d71022 y = 59 (a) 3.0\u00d71029 (b) 24% 61 (a) -2.3810 \u2013 23 J/K (b) 5.6 times more likely (c) If you were betting on two heads and 8 tails, the odds of breaking even are 252 to 45, so on average you would break even. So, no, you wouldn't bet on odds of 252 to 45. Test Prep for AP\u00ae Courses 1 (d) 3 (a) 5 (b) 7 (c) 9 (d) 11 (a) 13 (c) 15 (b) Chapter 16 Problems & Exercises 1 (a) 1.23\u00d7103 N/m (b) 6.88 kg (c) 4.00 mm 3 (a) 889 N/m (b) 133 N 5 (a) 6.53\u00d7103 N/m Answer Key 1604 (b) Yes 7 16.7 ms 8 0.400 s / beats 9 400 Hz 10 12,500 Hz 11 1.50 kHz 12 (a) 93.8 m/s (b) 11.3\u00d7103 rev/min 13 2.37 N/m 15 0.389 kg 18 94.7 kg 21 1.94 s 22 6.21 cm 24 2.01 s 26 2.23 Hz 28 (a) 2.99541 s (b) Since the period is related to the square root of the acceleration of gravity, when the acceleration changes by 1% the period changes by (0.01)2 = 0.01% so it is necessary to have at least 4 digits after the decimal to see the changes. 30 (a) Period increases by a factor of 1.41 ( 2 ) (b) Period decreases to 97.5% of old period 32 Slow by a factor of 2.45 34 length must increase by 0.0116%. 35 (a) 1.99 Hz (b) 50.2 cm (c) 1.41 Hz, 0.710 m 36 (a) 3.95\u00d7106 N/m (b) 7.90\u00d7106 J 37 a). 0.266 m/s b). 3.00 J 39 This content is available for free at http://cnx.org/content/col11844/1.13 1605 (16.75) (16.76) (16.77) (16.78) (16.79) Answer Key \u00b1 3 2 42 384 J 44 (a). 0.123 m (b). \u22120.600 J (c). 0.300 J. The rest of the energy may go into heat caused by friction and other damping forces. = 9.26 d = 40.0 Hz w = 16.0 m/s = 700 m = 34.0 cm 46 (a) 5.00\u00d7105 J (b) 1.20\u00d7103 s 47 49 51 53 55 57 = 4 Hz 59 462 Hz, 4 Hz 61 (a) 3.33 m/s (b) 1.25 Hz 63 0.225 W 65 7.07 67 16.0 d 68 2.50 kW 70 3.38\u00d710\u20135 W/m2 Test Prep for AP\u00ae Courses 1 (d) 3 (b) 5 The frequency is given by = 1 = 50 30 = 1.66 Time period is: 1606 Answer Key = 1 = 1 1.66 = 0.6 s 7 (c) 9 The energy of the particle at the center of the oscillation is given by = 1 \u00d70.2 kg\u00d7(5 m\u00b7s\u22121)2 = 2.5 J 22 = 1 2 11 (b) 13 19.7 J 15 (c) 17 = 2 2 \u2212 2 = 50 N \u22c5 m\u22121 = 0.06 = 0.5kg = 50N \u22c5 m\u22121 2\u00d70.06\u00d79.8m \u22c5 s\u2212", "2 (0.2)2 \u2212 0.06\u00d70.5kg\u00d79.8m \u22c5 s\u22122) (50N \u22c5 m\u22121)2 2 = 1.698 m 19 The waves coming from a tuning fork are mechanical waves that are longitudinal in nature, whereas electromagnetic waves are transverse in nature. 21 The sound energy coming out of an instrument depends on its size. The sound waves produced are relative to the size of the musical instrument. A smaller instrument such as a tambourine will produce a high-pitched sound (higher frequency, shorter wavelength), whereas a larger instrument such as a drum will produce a deeper sound (lower frequency, longer wavelength). 23 2 m 25 The student explains the principle of superposition and then shows two waves adding up to form a bigger wave when a crest adds with a crest and a trough with another trough. Also the student shows a wave getting cancelled out when a crest meets a trough and vice versa. 27 The student must note that the shape of the wave remains the same and there is first an overlap and then receding of the waves. 29 (c) Chapter 17 Problems & Exercises 1 0.288 m 3 332 m/s 5 7 0.223 9 (a) 7.70 m w = (331 m/s) 273 K = (331 m/s) 293 K 273 K = 343 m/s (17.12) This content is available for free at http://cnx.org/content/col11844/1.13 Answer Key 1607 (b) This means that sonar is good for spotting and locating large objects, but it isn\u2019t able to resolve smaller objects, or detect the detailed shapes of objects. Objects like ships or large pieces of airplanes can be found by sonar, while smaller pieces must be found by other means. 11 (a) 18.0 ms, 17.1 ms (b) 5.00% (c) This uncertainty could definitely cause difficulties for the bat, if it didn\u2019t continue to use sound as it closed in on its prey. A 5% uncertainty could be the difference between catching the prey around the neck or around the chest, which means that it could miss grabbing its prey. (17.23) (17.24) (17.25) (17.36) 3.16\u00d710\u20134 W/m2 3.04\u00d710\u20134 W/m2 1.45\u00d710\u20133 J 3.79\u00d7103 Hz 12 14 16 106 dB 18 (a) 93 dB (b) 83 dB 20 (a) 50.1 (b) 5.01\u00d710\u20133 or 1 200 22 70.0 dB 24 100 26 28 28.2 dB 30 (a) 878 Hz (b) 735 Hz 32 34 (a) 12.9 m/s (b) 193 Hz 36 First eagle hears 4.23\u00d7103 Hz Second eagle hears 3.56\u00d7103 Hz 38 0.7 Hz 40 0.3 Hz, 0.2 Hz, 0.5 Hz 42 (a) 256 Hz (b) 512 Hz 44 180 Hz, 270 Hz, 360 Hz Answer Key 1608 46 1.56 m 48 (a) 0.334 m (b) 259 Hz 50 3.39 to 4.90 kHz 52 (a) 367 Hz (b) 1.07 kHz 54 (a) = 47.6 Hz, = 1, 3, 5,..., 419 (b) = 95.3 Hz, = 1, 2, 3,..., 210 55 1\u00d7106 km (17.49) 57 498.5 or 501.5 Hz 59 82 dB 61 approximately 48, 9, 0, \u20137, and 20 dB, respectively 63 (a) 23 dB (b) 70 dB 65 Five factors of 10 67 (a) 2\u00d710\u221210 W/m2 (b) 2\u00d710\u221213 W/m2 69 2.5 71 1.26 72 170 dB 74 103 dB 76 (a) 1.00 (b) 0.823 (c) Gel is used to facilitate the transmission of the ultrasound between the transducer and the patient\u2019s body. 78 (a) 77.0 \u03bcm (b) Effective penetration depth = 3.85 cm, which is enough to examine the eye. (c) 16.6 \u03bcm 80 (a) 5.78\u00d710\u20134 m (b) 2.67\u00d7106 Hz This content is available for free at http://cnx.org/content/col11844/1.13 Answer Key 82 1609 (a) w = 1540 m/s = = 1540 m/s 100\u00d7103 Hz = 0.0154 m < 3.50 m. Because the wavelength is much shorter than \u21d2 the distance in question, the wavelength is not the limiting factor. (b) 4.55 ms 84 974", " Hz (Note: extra digits were retained in order to show the difference.) Test Prep for AP\u00ae Courses 1 (b) 3 (e) 5 (c) 7 Answers vary. Students could include a sketch showing an increased amplitude when two waves occupy the same location. Students could also cite conceptual evidence such as sound waves passing through each other. 9 (d) 11 (c) 13 (a) 15 (c) 17 (b) 19 (a), (b) 21 (c) Chapter 18 Problems & Exercises 1 (a) 1.25\u00d71010 (b) 3.13\u00d71012 3 -600 C 5 1.03\u00d71012 7 9.09\u00d710\u221213 9 1.48\u00d7108 C 15 (a) = 1.00 cm = \u2212 \u221e (b) 2.12\u00d7105 N/C (c) one charge of + 1610 17 (a) 0.252 N to the left (b) = 6.07 cm 19 Answer Key (a)The electric field at the center of the square will be straight up, since and are positive and and are negative and all have the same magnitude. (b) 2.04\u00d7107 N/C (upward) 21 0.102 N in the \u2212 23 direction = 4.36\u00d7103 N/C 35.0\u00ba, below the horizontal. \u2192 (a) (b) No 25 (a) 0.263 N (b) If the charges are distributed over some area, there will be a concentration of charge along the side closest to the oppositely charged object. This effect will increase the net force. 27 The separation decreased by a factor of 5. 31 = |1 2| 2 = = 2 \u21d2 2 9.00\u00d7109 N \u22c5 m2/ C2 = 1.60\u00d710\u201319 m 2 1.67\u00d710\u201327 kg 2.00\u00d710\u20139 m = 3.45\u00d71016 m/s2 2 32 (a) 3.2 (b) If the distance increases by 3.2, then the force will decrease by a factor of 10 ; if the distance decreases by 3.2, then the force will increase by a factor of 10. Either way, the force changes by a factor of 10. 34 (a) 1.04\u00d710\u22129 C (b) This charge is approximately 1 nC, which is consistent with the magnitude of charge typical for static electricity 37 1.02\u00d710\u221211 39 a. 0.859 m beyond negative charge on line connecting two charges b. 0.109 m from lesser charge on line connecting two charges 42 8.75\u00d710\u22124 N 44 (a) 6.94\u00d710\u22128 C (b) 6.25 N/C 46 (a) 300 N/C (east) This content is available for free at http://cnx.org/content/col11844/1.13 Answer Key 1611 (b) 4.80\u00d710\u221217 N (east) 52 (a) 5.58\u00d710\u221211 N/C (b)the coulomb force is extraordinarily stronger than gravity 54 (a) \u22126.76\u00d7105 C (b) 2.63\u00d71013 m/s2 (upward) (c) 2.45\u00d710\u221218 kg 56 The charge 2 is 9 times greater than 1. Test Prep for AP\u00ae Courses 1 (b) 3 (c) 5 (a) 7 (b) 9 (a) -0.1 C, (b) 1.1 C, (c) Both charges will be equal to 1 C, law of conservation of charge, (d) 0.9 C 11 W is negative, X is positive, Y is negative, Z is neutral. 13 (c) 15 (c) 17 (b) 19 a) Ball 1 will have positive charge and Ball 2 will have negative charge. b) The negatively charged rod attracts positive charge of Ball 1. The electrons of Ball 1 are transferred to Ball 2, making it negatively charged. c) If Ball 2 is grounded while the rod is still there, it will lose its negative charge to the ground. d) Yes, Ball 1 will be positively charged and Ball 2 will be negatively charge. 21 (c) 23 decrease by 77.78%. 25 (a) 27 (d) 29 (a) 3.60\u00d71010 N, (b) It will become 1/4 of the original value; hence it will be equal to 8.99\u00d7109 N 31 (c) 33 (a) 35 (b) 37 (a) 350 N/C, (b) west, (c) 5.6\u00d710\u221217 N, (d) west. 39 (b) 1612 41 Answer Key (a) i) Field vectors near objects point toward negatively charged objects and away from positively charged objects. (a) ii)", " The vectors closest to R and T are about the same length and start at about the same distance. We have that / 2 = / 2, so the charge on R is about the same as the charge on T. The closest vectors around S are about the same length as those around R and T. The vectors near S start at about 6 units away, while vectors near R and T start at about 4 units. We have that / 2 = / 2, so / = 2/ 2 = 36/ 16 = 2.25, and so the charge on S is about twice that on R and T. (b) Figure 18.35. A vector diagram. (c) = \u2212 ( + )2 + 2 ()2 + ( \u2212 )2 (d) The statement is not true. The vector diagram shows field vectors in this region with nonzero length, and the vectors not shown have even greater lengths. The equation in part (c) shows that, when 0 < <, the denominator of the negative term is always greater than the denominator of the third term, but the numerator is the same. So the negative term always has a smaller magnitude than the third term and since the second term is positive the sum of the terms is always positive. Chapter 19 Problems & Exercises 1 42.8 4 1.00105 K 6 (a) 4104 W (b) A defibrillator does not cause serious burns because the skin conducts electricity well at high voltages, like those used in defibrillators. The gel used aids in the transfer of energy to the body, and the skin doesn\u2019t absorb the energy, but rather lets it pass through to the heart. 8 (a) 7.40103 C (b) 1.541020 electrons per second 9 3.89106 C 11 (a) 1.44\u00d71012 V (b) This voltage is very high. A 10.0 cm diameter sphere could never maintain this voltage; it would discharge. (c) An 8.00 C charge is more charge than can reasonably be accumulated on a sphere of that size. 15 This content is available for free at http://cnx.org/content/col11844/1.13 Answer Key (a) 3.00 kV (b) 750 V 1613 17 (a) No. The electric field strength between the plates is 2.5\u00d7106 V/m, which is lower than the breakdown strength for air ( 3.0\u00d7106 V/m ). (b) 1.7 mm 19 44.0 mV 21 15 kV 23 (a) 800 KeV (b) 25.0 km 24 144 V 26 (a) 1.80 km (b) A charge of 1 C is a very large amount of charge; a sphere of radius 1.80 km is not practical. 28 \u20132.22\u00d710 \u2013 13 C 30 (a) 3.31\u00d7106 V (b) 152 MeV 32 (a) 2.78\u00d710-7 C (b) 2.00\u00d710-10 C 35 (a) 2.96\u00d7109 m/s (b) This velocity is far too great. It is faster than the speed of light. (c) The assumption that the speed of the electron is far less than that of light and that the problem does not require a relativistic treatment produces an answer greater than the speed of light. 46 21.6 mC 48 80.0 mC 50 20.0 kV 52 667 pF 54 (a) 4.4 \u00b5F (b) 4.0\u00d710 \u2013 5 C 56 (a) 14.2 kV (b) The voltage is unreasonably large, more than 100 times the breakdown voltage of nylon. 1614 Answer Key (c) The assumed charge is unreasonably large and cannot be stored in a capacitor of these dimensions. 57 0.293 \u03bcF 59 3.08 \u00b5F in series combination, 13.0 \u00b5F in parallel combination 60 2.79 \u00b5F 62 (a) \u20133.00 \u00b5F (b) You cannot have a negative value of capacitance. (c) The assumption that the capacitors were hooked up in parallel, rather than in series, was incorrect. A parallel connection always produces a greater capacitance, while here a smaller capacitance was assumed. This could happen only if the capacitors are connected in series. 63 (a) 405 J (b) 90.0 mC 64 (a) 3.16 kV (b) 25.3 mC 66 (a) 1.42\u00d710\u22125 C, 6.38\u00d710\u22125 J (b) 8.46\u00d710\u22125 C, 3.81\u00d710\u22124 J 67 (a) 4.43\u00d710 \u2013 12 F (b) 452 V (c) 4.52\u00d710 \u2013 7 J 70 (a) 133 F", " (b) Such a capacitor would be too large to carry with a truck. The size of the capacitor would be enormous. (c) It is unreasonable to assume that a capacitor can store the amount of energy needed. Test Prep for AP\u00ae Courses 1 (a) 3 (b) 5 (c) 7 (a) 9 (b) 11 (b) 13 (a) 15 (c) 17 This content is available for free at http://cnx.org/content/col11844/1.13 Answer Key 1615 (b) 19 (a) 21 (d) 23 (d) 25 (a) 27 (b) 29 (c) 31 (d) 33 (a) 35 (c) 37 (c) 39 (b) 41 (a) 43 (d) Chapter 20 Problems & Exercises 1 0.278 mA 3 0.250 A 5 1.50ms 7 (a) 1.67k \u03a9 (b) If a 50 times larger resistance existed, keeping the current about the same, the power would be increased by a factor of about 50 (based on the equation = 2 ), causing much more energy to be transferred to the skin, which could cause serious burns. The gel used reduces the resistance, and therefore reduces the power transferred to the skin. 9 (a) 0.120 C (b) 7.501017 electrons 11 96.3 s 13 (a) 7.81 \u00d7 1014 He++ nuclei/s (b) 4.00 \u00d7 103 s (c) 7.71 \u00d7 108 s 15 \u22121.1310\u22124 m/s 1616 17 9.421013 electrons 18 0.833 A 20 7.3310\u22122 \u03a9 22 (a) 0.300 V (b) 1.50 V Answer Key (c) The voltage supplied to whatever appliance is being used is reduced because the total voltage drop from the wall to the final output of the appliance is fixed. Thus, if the voltage drop across the extension cord is large, the voltage drop across the appliance is significantly decreased, so the power output by the appliance can be significantly decreased, reducing the ability of the appliance to work properly. 24 0.104 \u03a9 26 2.810\u22122 m 28 1.1010\u22123 A 30 \u22125\u00baC to 45\u00baC 32 1.03 34 0.06% 36 \u221217\u00baC 38 (a) 4.7 \u03a9 (total) (b) 3.0% decrease 40 2.001012 W 44 (a) 1.50 W (b) 7.50 W = V2 V/A 46 V2 \u03a9 48 1 kW \u22c5 h= = AV = \u00d7103 J 1 s (1 h) 3600 s 1 h = 3.60\u00d7106 J 50 $438/y 52 $6.25 54 1.58 h 56 $3.94 billion/year 58 25.5 W 60 (a) 2.00109 J This content is available for free at http://cnx.org/content/col11844/1.13 1617 Answer Key (b) 769 kg 62 45.0 s 64 (a) 343 A (b) 2.17103 A (c) 1.10103 A 66 (a) 1.23\u00d7103 kg (b) 2.64\u00d7103 kg 69 (a) 2.08\u00d7105 A (b) 4.33\u00d7104 MW (c) The transmission lines dissipate more power than they are supposed to transmit. (d) A voltage of 480 V is unreasonably low for a transmission voltage. Long-distance transmission lines are kept at much higher voltages (often hundreds of kilovolts) to reduce power losses. 73 480 V 75 2.50 ms 77 (a) 4.00 kA (b) 16.0 MW (c) 16.0% 79 2.40 kW 81 (a) 4.0 (b) 0.50 (c) 4.0 83 (a) 1.39 ms (b) 4.17 ms (c) 8.33 ms 85 (a) 230 kW (b) 960 A 87 (a) 0.400 mA, no effect (b) 26.7 mA, muscular contraction for duration of the shock (can't let go) 89 1.20105 \u03a9 91 (a) 1.00 \u03a9 1618 (b) 14.4 kW 93 Temperature increases 860\u00ba C. It is very likely to be damaging. 95 80 beats/minute Test Prep for AP\u00ae Courses Answer Key 1 (a) 3 10 A 5 (a) 7 3.2 \u03a9, 2.19 A 9 (b), (d) 11 9.72 \u00d7 10\u22128 \u03a9\u00b7m 13 18 \u03a9 15 10:3 or 3.33 Chapter 21 Problems & Exercises 1 (a) 2.75 k \u03a9", " (b) 27.5 \u03a9 3 (a) 786 \u03a9 (b) 20.3 \u03a9 5 29.6 W 7 (a) 0.74 A (b) 0.742 A 9 (a) 60.8 W (b) 3.18 kW 11 (a >>2 (b, so that >>2. 13 This content is available for free at http://cnx.org/content/col11844/1.13 Answer Key (a) -400 k \u03a9 (b) Resistance cannot be negative. 1619 (c) Series resistance is said to be less than one of the resistors, but it must be greater than any of the resistors. 14 2.00 V 16 2.9994 V 18 0.375 \u03a9 21 (a) 0.658 A (b) 0.997 W (c) 0.997 W; yes 23 (a) 200 A (b) 10.0 V (c) 2.00 kW (d) 0.1000 \u03a9 80.0 A, 4.0 V, 320 W 25 (a) 0.400 \u03a9 (b) No, there is only one independent equation, so only can be found. 29 (a) \u20130.120 V (b) -1.4110\u22122 \u03a9 (c) Negative terminal voltage; negative load resistance. (d) The assumption that such a cell could provide 8.50 A is inconsistent with its internal resistance. \u221222 + emf1 \u2212 21 + 33 + 32 - emf2 = 0 3 = 1 + 2 emf2 - 22 - 22 + 15 + 11 - emf1 + 11 = 0 (21.69) (21.70) (21.71) 31 35 37 39 (a) I1 = 4.75 A (b) I2 = -3.5 A (c) I3 = 8.25 A 41 (a) No, you would get inconsistent equations to solve. (b) 1 \u2260 2 + 3. The assumed currents violate the junction rule. 42 30 44 1.98 k \u03a9 46 1.2510-4 \u03a9 (21.75) Answer Key 1620 48 (a) 3.00 M \u03a9 (b) 2.99 k \u03a9 50 (a) 1.58 mA (b) 1.5848 V (need four digits to see the difference) (c) 0.99990 (need five digits to see the difference from unity) 52 15.0 \u03bcA 54 (a) Figure 21.39. (b) 10.02 \u03a9 (c) 0.9980, or a 2.0\u00d710\u20131 percent decrease (d) 1.002, or a 2.0\u00d710\u20131 percent increase (e) Not significant. 56 (a) \u221266.7 \u03a9 (b) You can\u2019t have negative resistance. (c) It is unreasonable that G is greater than tot (see Figure 21.36). You cannot achieve a full-scale deflection using a current less than the sensitivity of the galvanometer. Range = 5.00 \u03a9 to 5.00 k \u03a9 (21.82) 57 24.0 V 59 1.56 k \u03a9 61 (a) 2.00 V (b) 9.68 \u03a9 62 63 range 4.00 to 30.0 M \u03a9 65 (a) 2.50 \u03bcF (b) 2.00 s 67 86.5% 69 (a) 1.25 k \u03a9 (b) 30.0 ms 71 (a) 20.0 s This content is available for free at http://cnx.org/content/col11844/1.13 1621 Answer Key (b) 120 s (c) 16.0 ms 73 1.73\u00d710\u22122 s 74 3.3310\u22123 \u03a9 76 (a) 4.99 s (b) 3.87\u00baC (c) 31.1 k \u03a9 (d) No Test Prep for AP\u00ae Courses 1 (a), (b) 3 (b) 5 (a) 4-\u03a9 resistor; (b) combination of 20-\u03a9, 20-\u03a9, and 10-\u03a9 resistors; (c) 20 W in each 20-\u03a9 resistor, 40 W in 10-\u03a9 resistor, 64 W in 4-\u03a9 resistor, total 144W total in resistors, output power is 144 W, yes they are equal (law of conservation of energy); (d) 4 \u03a9 and 3 \u03a9 for part (a) and no change for part (b); (e) no effect, it will remain the same. 7 0.25 \u03a9, 0.50 \u03a9, no change 9 a. (c) b. (c) c. (d) d. (d) 11 a.", " I1 + I3 = I2 b. E1 - I1R1 - I2R2 - I1r1 = 0; - E2 + I1R1 - I3R3 - I3r2 = 0 c. d. I1 = 8/15 A, I2 = 7/15 A and I3 = -1/15 A I1 = 2/5 A, I2 = 3/5 A and I3 = 1/5 A e. PE1 = 18/5 W and PR1 = 24/25 W, PR2 = 54/25 W, PR3 = 12/25 W. Yes, PE1 = PR1+ PR2 + PR3 f. R3, losses in the circuit 13 (a) 20 mA, Figure 21.44, 5.5 s; (b) 24 mA, Figure 21.35, 2 s Chapter 22 Problems & Exercises 1 (a) Left (West) (b) Into the page (c) Up (North) (d) No force (e) Right (East) (f) Down (South) 3 (a) East (right) 1622 (b) Into page (c) South (down) 5 (a) Into page (b) West (left) (c) Out of page Answer Key 7 7.50\u00d710\u22127 N perpendicular to both the magnetic field lines and the velocity 9 (a) 3.01\u00d710\u22125 T (b) This is slightly less then the magnetic field strength of 5\u00d710\u22125 T at the surface of the Earth, so it is consistent. 11 (a) 6.67\u00d710\u221210 C (taking the Earth\u2019s field to be 5.00\u00d710\u22125 T ) (b) Less than typical static, therefore difficult 12 4.27 m 14 (a) 0.261 T (b) This strength is definitely obtainable with today\u2019s technology. Magnetic field strengths of 0.500 T are obtainable with permanent magnets. 16 4.36\u00d710\u22124 m 18 (a) 3.00 kV/m (b) 30.0 V 20 0.173 m 22 7.50\u00d710\u22124 V 24 (a) 1.18 \u00d7 10 3 m/s (b) Once established, the Hall emf pushes charges one direction and the magnetic force acts in the opposite direction resulting in no net force on the charges. Therefore, no current flows in the direction of the Hall emf. This is the same as in a current-carrying conductor\u2014current does not flow in the direction of the Hall emf. 26 11.3 mV 28 1.16 \u03bcV 30 2.00 T 31 (a) west (left) (b) into page (c) north (up) (d) no force (e) east (right) (f) south (down) This content is available for free at http://cnx.org/content/col11844/1.13 Answer Key 33 (a) into page (b) west (left) (c) out of page 35 (a) 2.50 N (b) This is about half a pound of force per 100 m of wire, which is much less than the weight of the wire itself. Therefore, it does not cause any special concerns. 1623 37 1.80 T 39 (a) 30\u00ba (b) 4.80 N 41 (a) \u03c4 decreases by 5.00% if B decreases by 5.00% (b) 5.26% increase 43 10.0 A 45 A \u22c5 m2 \u22c5 T = A \u22c5 m2 47 3.48\u00d710\u221226 N \u22c5 m 49 (a) 0.666 N \u22c5 m west b) This is not a very significant torque, so practical use would be limited. Also, the current would need to be alternated to make the loop rotate (otherwise it would oscillate). 50 (a) 8.53 N, repulsive (b) This force is repulsive and therefore there is never a risk that the two wires will touch and short circuit. 52 400 A in the opposite direction 54 (a) 1.67\u00d710\u22123 N/m (b) 3.33\u00d710\u22123 N/m (c) Repulsive (d) No, these are very small forces 56 (a) Top wire: 2.65\u00d710\u22124 N/m s, 10.9\u00ba to left of up (b) Lower left wire: 3.61\u00d710\u22124 N/m, 13.9\u00ba down from right (c) Lower right wire: 3.46\u00d710\u22124 N/m, 30.0\u00ba down from left 58 (a) right-into page, left-out of page (b) right-out of page, left-into", " page (c) right-out of page, left-into page Answer Key 1624 60 (a) clockwise (b) clockwise as seen from the left (c) clockwise as seen from the right 61 1.01\u00d71013 T 63 (a) 4.80\u00d710\u22124 T (b) Zero (c) If the wires are not paired, the field is about 10 times stronger than Earth\u2019s magnetic field and so could severely disrupt the use of a compass. 65 39.8 A 67 (a) 3.14\u00d710\u22125 T (b) 0.314 T 69 7.55\u00d710\u22125 T, 23.4\u00ba 71 10.0 A 73 (a) 9.09\u00d710\u22127 N upward (b) 3.03\u00d710\u22125 m/s2 75 60.2 cm 77 (a) 1.02\u00d7103 N/m2 (b) Not a significant fraction of an atmosphere 79 17.0\u00d710\u22124%/\u00baC 81 18.3 MHz 83 (a) Straight up (b) 6.00\u00d710\u22124 N/m (c) 94.1 \u03bcm (d)2.47 \u03a9/m, 49.4 V/m 85 (a) 571 C (b) Impossible to have such a large separated charge on such a small object. (c) The 1.00-N force is much too great to be realistic in the Earth\u2019s field. 87 (a) 2.40\u00d7106 m/s (b) The speed is too high to be practical \u2264 1% speed of light This content is available for free at http://cnx.org/content/col11844/1.13 Answer Key 1625 (c) The assumption that you could reasonably generate such a voltage with a single wire in the Earth\u2019s field is unreasonable 89 (a) 25.0 kA (b) This current is unreasonably high. It implies a total power delivery in the line of 50.0x10^9 W, which is much too high for standard transmission lines. (c)100 meters is a long distance to obtain the required field strength. Also coaxial cables are used for transmission lines so that there is virtually no field for DC power lines, because of cancellation from opposing currents. The surveyor\u2019s concerns are not a problem for his magnetic field measurements. Test Prep for AP\u00ae Courses 1 (a) 3 (b) 5 (b) 7 (a) 9 (b) 11 (e) 13 (c) 15 (c) Chapter 23 Problems & Exercises 1 Zero 3 (a) CCW (b) CW (c) No current induced 5 (a) 1 CCW, 2 CCW, 3 CW (b) 1, 2, and 3 no current induced (c) 1 CW, 2 CW, 3 CCW 9 (a) 3.04 mV (b) As a lower limit on the ring, estimate R = 1.00 m\u03a9. The heat transferred will be 2.31 mJ. This is not a significant amount of heat. 11 0.157 V 13 proportional to 1 17 (a) 0.630 V (b) No, this is a very small emf. 19 2.22 m/s 25 1626 (a) 10.0 N (b) 2.81\u00d7108 J (c) 0.36 m/s Answer Key (d) For a week-long mission (168 hours), the change in velocity will be 60 m/s, or approximately 1%. In general, a decrease in velocity would cause the orbit to start spiraling inward because the velocity would no longer be sufficient to keep the circular orbit. The long-term consequences are that the shuttle would require a little more fuel to maintain the desired speed, otherwise the orbit would spiral slightly inward. 28 474 V 30 0.247 V 32 (a) 50 (b) yes 34 (a) 0.477 T (b) This field strength is small enough that it can be obtained using either a permanent magnet or an electromagnet. 36 (a) 5.89 V (b) At t=0 (c) 0.393 s (d) 0.785 s 38 (a) 1.92\u00d7106 rad/s (b) This angular velocity is unreasonably high, higher than can be obtained for any mechanical system. (c) The assumption that a voltage as great as 12.0 kV could be obtained is unreasonable. 39 (a) 12.00 \u03a9 (b) 1.67 A 41 72.0 V 43 0.100 \u03a9 44 (a) 30.0 (b) 9.75\u00d710\u22122 A 46 (a) 20.0 mA (b) 2.40 W (c) Yes, this amount of power is quite reasonable for a small appliance.", " 48 (a) 0.063 A (b) Greater input current needed. 50 (a) 2.2 (b) 0.45 This content is available for free at http://cnx.org/content/col11844/1.13 Answer Key (c) 0.20, or 20.0% 52 (a) 335 MV (b) way too high, well beyond the breakdown voltage of air over reasonable distances (c) input voltage is too high 1627 54 (a) 15.0 V (b) 75.0 A (c) yes 55 1.80 mH 57 3.60 V 61 (a) 31.3 kV (b) 125 kJ (c) 1.56 MW (d) No, it is not surprising since this power is very high. 63 (a) 1.39 mH (b) 3.33 V (c) Zero 65 60.0 mH 67 (a) 200 H (b) 5.00\u00baC 69 500 H 71 50.0 \u03a9 73 1.00\u00d710\u201318 s to 0.100 s 75 95.0% 77 (a) 24.6 ms (b) 26.7 ms (c) 9% difference, which is greater than the inherent uncertainty in the given parameters. 79 531 Hz 81 1.33 nF 83 (a) 2.55 A (b) 1.53 mA 85 63.7 \u00b5H 87 1628 (a) 21.2 mH (b) 8.00 \u03a9 89 (a) 3.18 mF (b) 16.7 \u03a9 92 Answer Key (a) 40.02 \u03a9 at 60.0 Hz, 193 \u03a9 at 10.0 kHz (b) At 60 Hz, with a capacitor, Z=531 \u03a9, over 13 times as high as without the capacitor. The capacitor makes a large difference at low frequencies. At 10 kHz, with a capacitor Z=190 \u03a9, about the same as without the capacitor. The capacitor has a smaller effect at high frequencies. 94 (a) 529 \u03a9 at 60.0 Hz, 185 \u03a9 at 10.0 kHz (b) These values are close to those obtained in Example 23.12 because at low frequency the capacitor dominates and at high frequency the inductor dominates. So in both cases the resistor makes little contribution to the total impedance. 96 9.30 nF to 101 nF 98 3.17 pF 100 (a) 1.31 \u03bcH (b) 1.66 pF 102 (a) 12.8 k\u03a9 (b) 1.31 k\u03a9 (c) 31.9 mA at 500 Hz, 312 mA at 7.50 kHz (d) 82.2 kHz (e) 0.408 A 104 (a) 0.159 (b) 80.9\u00ba (c) 26.4 W (d) 166 W 106 16.0 W Test Prep for AP\u00ae Courses 1 (c) This content is available for free at http://cnx.org/content/col11844/1.13 Answer Key 1629 Figure 23.6. 3 (c) 5 (a), (d) 7 (c) Chapter 24 Problems & Exercises 3 150 kV/m 6 (a) 33.3 cm (900 MHz) 11.7 cm (2560 MHz) (b) The microwave oven with the smaller wavelength would produce smaller hot spots in foods, corresponding to the one with the frequency 2560 MHz. 8 26.96 MHz 10 5.0\u00d71014 Hz 12 14 0.600 m 16 = = 3.00108 m/s 1.201015 Hz = 2.5010 \u2013 7 m () (a) = = 3.00108 m/s 110-10 m = 31018 Hz (b) X-rays 19 (a) 6.00\u00d7106 m (b) 4.33\u00d710\u22125 T 21 (a) 1.50 \u00d7 10 6 Hz, AM band (b) The resonance of currents on an antenna that is 1/4 their wavelength is analogous to the fundamental resonant mode of an air column closed at one end, since the tube also has a length equal to 1/4 the wavelength of the fundamental oscillation. 23 (a) 1.55\u00d71015 Hz (b) The shortest wavelength of visible light is 380 nm, so that 1630 Answer Key () visible UV = 380 nm 193 nm = 1.97. In other words, the UV radiation is 97% more accurate than the shortest wavelength of visible light, or almost twice as accurate! 25 3.90\u00d7108 m 27 (a) 1.50\u00d71011 m (b) 0.500 s (c) 66.7 ns 29 (a) \u22123.5102 W/m2 (b) 88% (c) 1.7 T 30 = =", " 2 0 0 2 () 3.00108 m/s 8.8510\u201312 C2 /N \u22c5 m2 (125 V/m)2 2 = 20.7 W/m2 32 (a) = = 2 = 0.25010\u22123 W 0.50010\u22123 m 2 = 318 W/m2 ave = 2 0 2\u03bc0 \u21d2 0 = 2\u03bc0 1 / 2 = 410\u22127 T \u22c5 m/A 2 318.3 W/m2 3.00108 m/s 1 / 2 = 1.6310\u22126 T 0 = 0 = 3.00108 m/s 1.63310\u22126 T = 4.90102 V/m (b) (c) 34 (a) 89.2 cm (b) 27.4 V/m 36 (a) 333 T (b) 1.331019 W/m2 (c) 13.3 kJ 38 This content is available for free at http://cnx.org/content/col11844/1.13 Answer Key (a) = = 4\u03c0 2 \u221d 1 2 1631 (b 40 13.5 pF 42 (a) 4.07 kW/m2 (b) 1.75 kV/m (c) 5.84 T (d) 2 min 19 s 44 (a) 5.00103 W/m2 (b) 3.88\u00d710\u22126 N (c) 5.18\u00d710\u221212 N 46 (a) = 0 (b) 7.50\u00d710\u221210 s (c) 1.00\u00d710\u22129 s 48 (a) 1.01\u00d7106 W/m2 (b) Much too great for an oven. (c) The assumed magnetic field is unreasonably large. 50 (a) 2.53\u00d710\u221220 H (b) L is much too small. (c) The wavelength is unreasonably small. Test Prep for AP\u00ae Courses 1 (b) 3 (a) 5 (d) 7 (d) 9 (d) 11 (a) Chapter 25 Problems & Exercises 1 1632 Answer Key Top 1.715 m from floor, bottom 0.825 m from floor. Height of mirror is 0.890 m, or precisely one-half the height of the person. 5 2.25\u00d7108 m/s in water 2.04\u00d7108 m/s in glycerine 7 1.490, polystyrene 9 1.28 s 11 1.03 ns 13 = 1.46, fused quartz 17 (a) 0.898 (b) Can\u2019t have < 1.00 since this would imply a speed greater than. (c) Refracted angle is too big relative to the angle of incidence. 19 (a) 5.00 (b) Speed of light too slow, since index is much greater than that of diamond. (c) Angle of refraction is unreasonable relative to the angle of incidence. 22 66.3\u00ba 24 > 1.414 26 1.50, benzene 29 46.5\u00ba, red; 46.0\u00ba, violet 31 (a) 0.043\u00ba (b) 1.33 m 33 71.3\u00ba 35 53.5\u00ba, red; 55.2\u00ba, violet 37 5.00 to 12.5 D 39 \u22120.222 m 41 (a) 3.43 m (b) 0.800 by 1.20 m 42 (a) \u22121.35 m (on the object side of the lens). (b) +10.0 (c) 5.00 cm This content is available for free at http://cnx.org/content/col11844/1.13 1633 Answer Key 43 44.4 cm 45 (a) 6.60 cm (b) \u20130.333 47 (a) +7.50 cm (b) 13.3 D (c) Much greater 49 (a) +6.67 (b) +20.0 (c) The magnification increases without limit (to infinity) as the object distance increases to the limit of the focal distance. 51 \u22120.933 mm 53 +0.667 m 55 (a) \u20131.5\u00d710\u20132 m (b) \u201366.7 D 57 +0.360 m (concave) 59 (a) +0.111 (b) -0.334 cm (behind \u201cmirror\u201d) (c) 0.752cm 61 63 6.82 kW/m2 = i o = \u2212 i o = \u2212 \u221225.61) Test Prep for AP\u00ae Courses 1 (c) 3 (c) 5 (a) 7 Since light bends toward the normal upon entering a medium with a higher index of refraction, the upper path is a more accurate representation of a light ray moving", " from A to B. 9 First, measure the angle of incidence and the angle of refraction for light entering the plastic from air. Since the two angles can be measured and the index of refraction of air is known, the student can solve for the index of refraction of the plastic. Next, measure the angle of incidence and the angle of refraction for light entering the gas from the plastic. Since the two angles can be measured and the index of refraction of the plastic is known, the student can solve for the index of refraction of the gas. 11 1634 Answer Key The speed of light in a medium is simply c/n, so the speed of light in water is 2.25 \u00d7 108 m/s. From Snell\u2019s law, the angle of incidence is 44\u00b0. 13 (d) 15 (a) 17 (a) 19 (b) Chapter 26 Problems & Exercises 1 52.0 D 3 (a) \u22120.233 mm (b) The size of the rods and the cones is smaller than the image height, so we can distinguish letters on a page. 5 (a) +62.5 D (b) \u20130.250 mm (c) \u20130.0800 mm 6 2.00 m 8 (a) \u00b10.45 D (b) The person was nearsighted because the patient was myopic and the power was reduced. 10 0.143 m 12 1.00 m 14 20.0 cm 16 \u20135.00 D 18 25.0 cm 20 \u20130.198 D 22 30.8 cm 24 \u20130.444 D 26 (a) 4.00 (b) 1600 28 (a) 0.501 cm (b) Eyepiece should be 204 cm behind the objective lens. 30 (a) +18.3 cm (on the eyepiece side of the objective lens) This content is available for free at http://cnx.org/content/col11844/1.13 Answer Key (b) -60.0 (c) -11.3 cm (on the objective side of the eyepiece) 1635 (d) +6.67 (e) -400 33 \u221240.0 35 \u22121.67 37 +10.0 cm 39 (a) 0.251 \u03bcm (b) Yes, this thickness implies that the shape of the cornea can be very finely controlled, producing normal distant vision in more than 90% of patients. Test Prep for AP\u00ae Courses 1 (a) 3 (c) 5 (a) 7 (b) 9 (d) 11 (c) Chapter 27 Problems & Exercises 1 1 / 1.333 = 0.750 3 1.49, Polystyrene 5 0.877 glass to water 6 0.516\u00ba 8 1.22\u00d710\u22126 m 10 600 nm 12 2.06\u00ba 14 1200 nm (not visible) 16 (a) 760 nm (b) 1520 nm 18 For small angles sin \u2212 tan \u2248 (in radians). For two adjacent fringes we have, 1636 and Subtracting these equations gives sin m = sin m + 1 = ( + 1) ( + 1) \u2212 sin m + 1 \u2212 sin tan = = Answer Key (27.11) (27.12) (27.13) 20 450 nm 21 5.97\u00ba 23 8.99\u00d7103 25 707 nm 27 (a) 11.8\u00ba, 12.5\u00ba, 14.1\u00ba, 19.2\u00ba (b) 24.2\u00ba, 25.7\u00ba, 29.1\u00ba, 41.0\u00ba (c) Decreasing the number of lines per centimeter by a factor of x means that the angle for the x\u2010order maximum is the same as the original angle for the first- order maximum. 29 589.1 nm and 589.6 nm 31 28.7\u00ba 33 43.2\u00ba 35 90.0\u00ba 37 (a) The longest wavelength is 333.3 nm, which is not visible. (b) 333 nm (UV) (c) 6.58\u00d7103 cm 39 1.13\u00d710\u22122 m 41 (a) 42.3 nm (b) Not a visible wavelength The number of slits in this diffraction grating is too large. Etching in integrated circuits can be done to a resolution of 50 nm, so slit separations of 400 nm are at the limit of what we can do today. This line spacing is too small to produce diffraction of light. 43 (a) 33.4\u00ba (b) No 45 (a) 1.35\u00d710\u22126 m This content is available for free at http://cnx.org/content/col11844/1.13 1637 Answer Key (b) 69.9\u00ba 47 750 nm 49 (a) 9.04\u00ba (b) 12 51 (a) 0.0150", "\u00ba (b) 0.262 mm (c) This distance is not easily measured by human eye, but under a microscope or magnifying glass it is quite easily measurable. 53 (a) 30.1\u00ba (b) 48.7\u00ba (c) No (d) 2\u03b81 = (2)(14.5\u00ba) = 29\u00ba 2 \u2212 1 = 30.05\u00ba\u221214.5\u00ba=15.56\u00ba. Thus, 29\u00ba \u2248 (2)(15.56\u00ba) = 31.1\u00ba. 55 23.6\u00ba and 53.1\u00ba 57 (a) 1.63\u00d710\u22124 rad (b) 326 ly 59 1.46\u00d710\u22125 rad 61 (a) 3.04\u00d710\u22127 rad (b) Diameter of 235 m 63 5.15 cm 65 (a) Yes. Should easily be able to discern. (b) The fact that it is just barely possible to discern that these are separate bodies indicates the severity of atmospheric aberrations. 70 532 nm (green) 72 83.9 nm 74 620 nm (orange) 76 380 nm 78 33.9 nm 80 4.42\u00d710\u22125 m 82 The oil film will appear black, since the reflected light is not in the visible part of the spectrum. 84 45.0\u00ba 86 Answer Key 1638 45.7 mW/m2 88 90.0% 90 0 92 48.8\u00ba 94 41.2\u00ba 96 (a) 1.92, not diamond (Zircon) (b) 55.2\u00ba 98 2 = 0.707 1 100 (a) 2.07\u00d710-2 \u00b0C/s (b) Yes, the polarizing filters get hot because they absorb some of the lost energy from the sunlight. Test Prep for AP\u00ae Courses 1 (b) 3 (b) and (c) 5 (b) 7 (b) 9 (b) 11 (d) 13 (b) 15 (d) 17 (b) Chapter 28 Problems & Exercises 1 (a) 1.0328 (b) 1.15 3 5.96\u00d710\u22128 s 5 0.800 7 0.140 9 (a) 0.745 (b) 0.99995 (to five digits to show effect) This content is available for free at http://cnx.org/content/col11844/1.13 1639 Answer Key 11 (a) 0.996 (b) cannot be less than 1. (c) Assumption that time is longer in moving ship is unreasonable. 12 48.6 m 14 (a) 1.387 km = 1.39 km (b) 0.433 km (c) = Thus, the distances in parts (a) and (b) are related when = 3.20. 16 (a) 4.303 y (to four digits to show any effect) (b) 0.1434 y (c) \u0394t = \u03b3\u0394t0 \u21d2 = \u0394t \u0394t0 = 4.303 y 0.1434 y = 30.0 Thus, the two times are related when 30.00. 18 (a) 0.250 (b) must be \u22651 (c) The Earth-bound observer must measure a shorter length, so it is unreasonable to assume a longer length. 20 (a) 0.909 (b) 0.400 22 0.198 24 a) 658 nm b) red c) / 9.92\u00d710\u22125 (negligible) 26 0.991 28 \u22120.696 30 0.01324 32 \u2032 =, so = 1 + ( 2) = 1 + ( / 2) = 1 + ( / ) = () = 34 a) 0.99947 1640 b) 1.2064\u00d71011 y c) 1.2058\u00d71011 y (all to sufficient digits to show effects) 35 4.09\u00d710\u201319 kg \u22c5 m/s 37 (a) 3.000000015\u00d71013 kg \u22c5 m/s. Answer Key (b) Ratio of relativistic to classical momenta equals 1.000000005 (extra digits to show small effects) 39 2.9957\u00d7108 m/s 41 (a) 1.121\u00d710\u20138 m/s (b) The small speed tells us that the mass of a proton is substantially smaller than that of even a tiny amount of macroscopic matter! 43 8.20\u00d710\u221214 J 0.512 MeV 45 2.3\u00d710\u221230 kg 47 (a) 1.11\u00d71027 kg (b) 5.56\u00d710\u22125 49 7.1\u00d710\u22123 kg 7.1\u00d710\u22123 The ratio is greater for hydrogen. 51 208 0.999988 53 6.92\u00d7105 J 1.54 55 (a) 0.9", "14 (b) The rest mass energy of an electron is 0.511 MeV, so the kinetic energy is approximately 150% of the rest mass energy. The electron should be traveling close to the speed of light. 57 90.0 MeV 59 This content is available for free at http://cnx.org/content/col11844/1.13 Answer Key 1641 (a) 2 \u2212 1 2 = 22 + 24 = 22 4 so that 2 4, and therefore 2 2 = ()2 2 2 = 2 \u2212 1 (b) yes 61 1.07\u00d7103 63 6.56\u00d710\u22128 kg 4.37\u00d710\u221210 65 0.314 0.99995 67 (a) 1.00 kg (b) This much mass would be measurable, but probably not observable just by looking because it is 0.01% of the total mass. 69 (a) 6.3\u00d71011 kg/s (b) 4.5\u00d71010 y (c) 4.44\u00d7109 kg (d) 0.32% Test Prep for AP\u00ae Courses 1 (a) 3 The relativistic Doppler effect takes into account the special relativity concept of time dilation and also does not require a medium of propagation to be used as a point of reference (light does not require a medium for propagation). 5 Relativistic kinetic energy is given as KErel = ( \u2212 1)2 where = 1 1 \u2212 2 2 Classical kinetic energy is given as KEclass = 1 22 At low velocities = 0, a binomial expansion and subsequent approximation of gives: = 1 + 12 22 or \u2212 1 = 12 22 Substituting \u2212 1 in the expression for KErel gives KErel = 12 22 2 = 1 22 = KEclass 1642 Answer Key Hence, relativistic kinetic energy becomes classical kinetic energy when \u226a. Chapter 29 Problems & Exercises 1 (a) 0.070 eV (b) 14 3 (a) 2.21\u00d71034 J (b) 2.26\u00d71034 (c) No 4 263 nm 6 3.69 eV 8 0.483 eV 10 2.25 eV 12 (a) 264 nm (b) Ultraviolet 14 1.95\u00d7106 m/s 16 (a) 4.02\u00d71015 /s (b) 0.256 mW 18 (a) \u20131.90 eV (b) Negative kinetic energy (c) That the electrons would be knocked free. 20 6.34\u00d710\u22129 eV, 1.01\u00d710\u221227 J 22 2.42\u00d71020 Hz 24 = 6.62607\u00d710\u221234 J \u22c5 s 2.99792\u00d7108 m/s 109 nm 1 m 1.00000 eV 1.60218\u00d710\u221219 J (29.22) = 1239.84 eV \u22c5 nm \u2248 1240 eV \u22c5 nm 26 (a) 0.0829 eV (b) 121 (c) 1.24 MeV (d) 1.24\u00d7105 28 This content is available for free at http://cnx.org/content/col11844/1.13 1643 Answer Key (a) 25.0\u00d7103 eV (b) 6.04\u00d71018 Hz 30 (a) 2.69 (b) 0.371 32 (a) 1.25\u00d71013 photons/s (b) 997 km 34 8.33\u00d71013 photons/s 36 181 km 38 (a) 1.66\u00d710\u221232 kg \u22c5 m/s (b) The wavelength of microwave photons is large, so the momentum they carry is very small. 40 (a) 13.3 \u03bcm (b) 9.38\u00d710-2 eV 42 (a) 2.65\u00d710\u221228 kg \u22c5 m/s (b) 291 m/s (c) electron 3.86\u00d710\u221226 J, photon 7.96\u00d710\u221220 J, ratio 2.06\u00d7106 44 (a) 1.32\u00d710\u221213 m (b) 9.39 MeV (c) 4.70\u00d710\u22122 MeV 46 = 2 and =, so = 2 = 2. (29.35) As the mass of particle approaches zero, its velocity will approach, so that the ratio of energy to momentum in this limit is lim\u21920 = 2 = (29.36) which is consistent with the equation for photon energy. 48 (a) 3.00\u00d7106 W (b) Headlights are way too bright. (c) Force is too large. 49 7.28\u00d710\u20134 m Answer Key 15.1 keV (29.42) 1644 51 6.62\u00d7107 m/s 53 1.32\u00d710\u201313 m 55 (a) 6.62\u00d7107 m", "/s (b) 22.9 MeV 57 59 (a) 5.29 fm (b) 4.70\u00d710\u221212 J (c) 29.4 MV 61 (a) 7.28\u00d71012 m/s (b) This is thousands of times the speed of light (an impossibility). (c) The assumption that the electron is non-relativistic is unreasonable at this wavelength. 62 (a) 57.9 m/s (b) 9.55\u00d710\u22129 eV (c) From Table 29.1, we see that typical molecular binding energies range from about 1eV to 10 eV, therefore the result in part (b) is approximately 9 orders of magnitude smaller than typical molecular binding energies. 64 29 nm, 290 times greater 66 1.10\u00d710\u221213 eV 68 3.3\u00d710\u221222 s 70 2.66\u00d710\u221246 kg 72 0.395 nm 74 (a) 1.3\u00d710\u221219 J (b) 2.1\u00d71023 (c) 1.4\u00d7102 s 76 (a) 3.35\u00d7105 J (b) 1.12\u00d710\u20133 kg \u22c5 m/s (c) 1.12\u00d710\u20133 m/s (d) 6.23\u00d710\u20137 J This content is available for free at http://cnx.org/content/col11844/1.13 Answer Key 78 (a) 1.06\u00d7103 (b) 5.33\u00d710\u221216 kg \u22c5 m/s (c) 1.24\u00d710\u221218 m 80 (a) 1.62\u00d7103 m/s 1645 (b) 4.42\u00d710\u221219 J for photon, 1.19\u00d710\u221224 J for electron, photon energy is 3.71\u00d7105 times greater (c) The light is easier to make because 450-nm light is blue light and therefore easy to make. Creating electrons with 7.43 \u03bceV of energy would not be difficult, but would require a vacuum. 81 (a) 2.30\u00d710\u22126 m (b) 3.20\u00d710\u221212 m 83 3.69\u00d710\u22124 \u00baC 85 (a) 2.00 kJ (b) 1.33\u00d710\u22125 kg \u22c5 m/s (c) 1.33\u00d710\u22125 N (d) yes Test Prep for AP\u00ae Courses 1 (b) 3 (c) 5 (b) 7 (c) 9 (c) 11 (a) 13 (a) 15 (c) 17 (d) 19 (d) Chapter 30 Problems & Exercises 1 1.84\u00d7103 3 50 km 1646 4 6\u00d71020 kg/m3 6 (a) 10.0 \u03bcm Answer Key (b) It isn\u2019t hard to make one of approximately this size. It would be harder to make it exactly 10.0 \u03bcm. i \u22c5 f)2 2 2 \u2212 f i i = 2, f = 1, so that = m 1.097\u00d7107 (2\u00d71)2 22 \u2212 12 = 1.22\u00d710\u22127 m = 122 nm, which is UV radiation6.626\u00d710\u221234 J\u00b7s)2 4 2(9.109\u00d710\u221231 kg)(8.988\u00d7109 N\u00b7m2 / C2)(1)(1.602\u00d710\u221219 C)2 = 0.529\u00d710\u221210 m 11 0.850 eV 13 2.12\u00d710\u201310 m 15 365 nm It is in the ultraviolet. 17 No overlap 365 nm 122 nm 19 7 21 (a) 2 (b) 54.4 eV = 2 23 2 2 velocity, giving: = 2, so that = so that = 2 2 2. From the equation = 2, we can substitute for the 2 2 4 2 = 2 B where B = 2 2 4\u03c02. 25 (a) 0.248\u00d710\u221210 m (b) 50.0 keV (c) The photon energy is simply the applied voltage times the electron charge, so the value of the voltage in volts is the same as the value of the energy in electron volts. 27 (a) 100\u00d7103 eV, 1.60\u00d710\u221214 J (b) 0.124\u00d710\u221210 m 29 (a) 8.00 keV This content is available for free at http://cnx.org/content/col11844/1.13 1647 Answer Key (b) 9.48 keV 30 (a) 1.96 eV (b) (1240 eV\u00b7nm) / (1.96 eV) = 633 nm (c) 60.0 nm 32 693 nm 34 (a) 590 nm (b) (1240 e", "V\u00b7nm) / (1.17 eV) = 1.06 \u03bcm 35 = 4, 3 are possible since < and \u2223 \u2223 \u2264. 37 = 4 \u21d2 = 3, 2, 1, 0 \u21d2 = \u00b13, \u00b1 2, \u00b1 1, 0 are possible. 39 (a) 1.49\u00d710\u221234 J \u22c5 s (b) 1.06\u00d710\u221234 J \u22c5 s 41 (a) 3.66\u00d710\u221234 J \u22c5 s (b) = 9.13\u00d710\u221235 J \u22c5 s (c) = 12 3 / 4 = 4 43 = 54.7\u00ba, 125.3\u00ba 44 (a) 32. (b) 2 in 6 in 10 in and 14 in, for a total of 32. 46 (a) 2 (b) 3 9 48 (b) \u2265 is violated, (c) cannot have 3 electrons in subshell since 3 > (2 + 1) = 2 (d) cannot have 7 electrons in subshell since 7 > (2 + 1) = 2(2 + 1) = 6 50 (a) The number of different values of overall factor of 2 since each can have equal to either +1 / 2 or \u22121 / 2 \u21d2 2(2 + 1). is \u00b1 \u00b1 ( \u2212 1),...,0 for each > 0 and one for = 0 \u21d2 (2 + 1). Also an (b) for each value of, you get 2(2 + 1) = 0, 1, 2,...,(\u20131) \u21d2 2 (2)(0) + 1 + (2)(1) + 1 +.... + (2)( \u2212 1) + 1 = 2 1 + 3 +... + (2 \u2212 3) + (2 \u2212 1) \u23df terms to see that the expression in the box is = 2 imagine taking ( \u2212 1) from the last term and adding it to first term 1648 Answer Key = 2 1 + (\u20131) + 3 +... + (2 \u2212 3) + (2 \u2212 1)( \u2212 1) = 2. Now take ( \u2212 3) from penultimate term and add to the second term 2[ + +... + + ] + 3 +.... + (2 \u2212 3) + = 22. \u23df terms 52 The electric force on the electron is up (toward the positively charged plate). The magnetic force is down (by the RHR). 54 401 nm 56 (a) 6.54\u00d710\u221216 kg (b) 5.54\u00d710\u22127 m 58 1.76\u00d71011 C/kg, which agrees with the known value of 1.759\u00d71011 C/kg to within the precision of the measurement 60 (a) 2.78 fm (b) 0.37 of the nuclear radius. 62 (a) 1.34\u00d71023 (b) 2.52 MW 64 (a) 6.42 eV (b) 7.27\u00d710\u221220 J/molecule (c) 0.454 eV, 14.1 times less than a single UV photon. Therefore, each photon will evaporate approximately 14 molecules of tissue. This gives the surgeon a rather precise method of removing corneal tissue from the surface of the eye. 66 91.18 nm to 91.22 nm 68 (a) 1.24\u00d71011 V (b) The voltage is extremely large compared with any practical value. (c) The assumption of such a short wavelength by this method is unreasonable. Test Prep for AP\u00ae Courses 1 (a), (d) 3 (a) 5 (a) 7 (b) 9 (a) 11 (d) 13 (d) 15 (a), (c) This content is available for free at http://cnx.org/content/col11844/1.13 Answer Key 1649 Chapter 31 Problems & Exercises 1 1.67\u00d7104 5 = = 3 \u21d2 = 1 3 1/3 = 2.3\u00d71017 kg 1000 kg/m3 = 61\u00d7103 m = 61 km 7 1.9 fm 9 (a) 4.6 fm (b) 0.61 to 1 11 85.4 to 1 13 12.4 GeV 15 19.3 to 1 17 19 21 23 25 27 29 3 H2 \u2192 2 1 \u00af 3 He1 + \u2212 + 50 25 \u2192 24 25 50 Cr26 + + + 7 Be3 + \u2212 \u2192 3 4 7 Li4 + 210 Po126 \u2192 82 84 206 Pb124 + 2 4He2 137 Cs82 \u2192 56 55 \u00af 137 Ba81 + \u2212 + 232 Th142 \u2192 88 90 228 Ra140 + 2 4He2 (31.17) (31.47) (31.48) (31.49) (31.50) (31.51) (31", ".52) (a) charge:(+1) + (\u22121) = 0 electron family number: (+1) + (\u22121b) 0.511 MeV (c) The two rays must travel in exactly opposite directions in order to conserve momentum, since initially there is zero momentum if the center of mass is initially at rest. 31 33 35 = ( + 1) \u2212 1; = ; efn : 0 = (+1) + (\u22121) - 1 = \u2212 1; = ; efn :(+1) = (+1) (31.53) (31.54) (a) 88 226 Ra138 \u2192 86 222 Rn136 + 2 4He2 (b) 4.87 MeV 37 1650 \u00af (a) n \u2192 p + \u2212 + (b) ) 0.783 MeV 39 1.82 MeV 41 (a) 4.274 MeV (b) 1.927\u00d710\u22125 Answer Key (c) Since U-238 is a slowly decaying substance, only a very small number of nuclei decay on human timescales; therefore, although those nuclei that decay lose a noticeable fraction of their mass, the change in the total mass of the sample is not detectable for a macroscopic sample. 43 (a) 8 15 O7 + \u2212 \u2192 7 15 N8 + (b) 2.754 MeV 44 57,300 y 46 (a) 0.988 Ci (b) The half-life of 226 Ra is now better known. 48 1.22\u00d7103 Bq 50 (a) 16.0 mg (b) 0.0114% 52 1.48\u00d71017 y 54 5.6\u00d7104 y 56 2.71 y 58 (a) 1.56 mg (b) 11.3 Ci 60 (a) 1.23\u00d710\u22123 (b) Only part of the emitted radiation goes in the direction of the detector. Only a fraction of that causes a response in the detector. Some of the emitted radiation (mostly particles) is observed within the source. Some is absorbed within the source, some is absorbed by the detector, and some does not penetrate the detector. 62 (a) 1.68\u00d710 \u2013 5 Ci (b) 8.65\u00d71010 J (c) $ 2.9\u00d7103 64 (a) 6.97\u00d71015 Bq This content is available for free at http://cnx.org/content/col11844/1.13 Answer Key (b) 6.24 kW (c) 5.67 kW 68 (a) 84.5 Ci 1651 (b) An extremely large activity, many orders of magnitude greater than permitted for home use. (c) The assumption of 1.00 \u03bcA is unreasonably large. Other methods can detect much smaller decay rates. 69 1.112 MeV, consistent with graph 71 7.848 MeV, consistent with graph 73 (a) 7.680 MeV, consistent with graph (b) 7.520 MeV, consistent with graph. Not significantly different from value for 12 C, but sufficiently lower to allow decay into another nuclide that is more tightly bound. 75 (a) 1.46\u00d710\u22128 u vs. 1.007825 u for 1 H (b) 0.000549 u (c) 2.66\u00d710\u22125 76 (a) \u20139.315 MeV (b) The negative binding energy implies an unbound system. (c) This assumption that it is two bound neutrons is incorrect. 78 22.8 cm 79 (a) 92 235 U143 \u2192 90 231 Th141 + 2 4 He2 (b) 4.679 MeV (c) 4.599 MeV 81 a) 2.4\u00d7108 u (b) The greatest known atomic masses are about 260. This result found in (a) is extremely large. (c) The assumed radius is much too large to be reasonable. 82 (a) \u20131.805 MeV (b) Negative energy implies energy input is necessary and the reaction cannot be spontaneous. (c) Although all conversation laws are obeyed, energy must be supplied, so the assumption of spontaneous decay is incorrect. Test Prep for AP\u00ae Courses 1 (c) 3 (a) 5 When 95 7 241 Am undergoes \u03b1 decay, it loses 2 neutrons and 2 protons. The resulting nucleus is therefore 93 237 Np. 1652 Answer Key During this process, the nucleus emits a particle with -1 charge. In order for the overall charge of the system to remain constant, the charge of the nucleus must therefore increase by +1. 9 a. No. Nucleon number is conserved (238 = 234 + 4), but the atomic number or charge is NOT conserved (92 \u2260 88+2). b. Yes. Nucleon number is conserved (223 = 209 + 14), and atomic number is", " conserved (88 = 82 + 6). c. Yes. Nucleon number is conserved (14 = 14), and charge is conserved if the electron\u2019s charge is properly counted (6 = 7 + (-1)). d. No. Nucleon number is not conserved (24 \u2260 23). The positron released counts as a charge to conserve charge, but it doesn\u2019t count as a nucleon. 11 This must be alpha decay since 4 nucleons (2 positive charges) are lost from the parent nucleus. The number remaining is found from: \u22120.693 N() = N0 1 2 = 3.4\u00d71017 \u2212(0.693)(0.035) 0.00173 N() = 4.1\u00d71011 nuclei Chapter 32 Problems & Exercises 1 5.701 MeV 3 99 Mo57 \u2192 43 42 5 1.43\u00d710\u22129 g 7 \u00af 99 Tc56 + \u2212 + (a) 6.958 MeV (b) 5.7\u00d710\u221210 g 8 (a) 100 mSv (b) 80 mSv (c) ~30 mSv 10 ~2 Gy 12 1.69 mm 14 1.24 MeV 16 7.44\u00d7108 18 4.92\u00d710\u20134 Sv 20 4.43 g 22 0.010 g 24 95% 26 This content is available for free at http://cnx.org/content/col11844/1.13 Answer Key 1653 (a) =1+1=2, =1+1=1+1, efn = 0 = \u22121 + 1 (b) =1+2=3, =1+1=2, efn=0=0 (c) =3+3=4+1+1, =2+2=2+1+1, efn=0=0 28 = (i \u2212 f)2 4 \u2212 4(1.007825) \u2212 4.002603 4 He = 1 H 2 = = 26.73 MeV (931.5 MeV) 30 3.12\u00d7105 kg (about 200 tons) 32 = (i \u2212 f)2 1 = (1.008665 + 3.016030 \u2212 4.002603)(931.5 MeV) = 20.58 MeV 2 = (1.008665 + 1.007825 \u2212 2.014102)(931.5 MeV) = 2.224 MeV 4 He is more tightly bound, since this reaction gives off more energy per nucleon. 34 1.19\u00d7104 kg 36 2\u2212 + 41 H \u2192 4 He + 7\u03b3 + 2 38 (a) =12+1=13, =6+1=7, efn = 0 = 0 (b) =13=13, =7=6+1, efn = 0 = \u22121 + 1 (c) =13 + 1=14, =6+1=7, efn = 0 = 0 (d) =14 + 1=15, =7+1=8, efn = 0 = 0 (e) =15=15, =8=7+1, efn = 0 = \u22121 + 1 (f) =15 + 1=12 + 4, =7+1=6 + 2, efn = 0 = 0 40 = 20.6 MeV 4 He = 5.68\u00d710-2 MeV 42 (a) 3\u00d7109 y (b) This is approximately half the lifetime of the Earth. 43 (a) 177.1 MeV (b) Because the gain of an external neutron yields about 6 MeV, which is the average BE/ for heavy nuclei. (c) = 1 + 238 = 96 + 140 + 1 + 1 + 1, = 92 = 38 + 53 efn = 0 = 0 45 (a) 180.6 MeV 1654 Answer Key (b) = 1 + 239 = 96 + 140 + 1 + 1 + 1 + 1, = 94 = 38 + 56 efn = 0 = 0 47 238 U + \u2192 239 U + 4.81 MeV 239 U \u2192 239 Np + \u2212 + 0.753 MeV Np \u2192 Pu + \u2212 + 0.211 MeV 49 (a) 2.57\u00d7103 MW (b) 8.03\u00d71019 fission/s (c) 991 kg 51 0.56 g 53 4.781 MeV 55 (a) Blast yields 2.1\u00d71012 J to 8.4\u00d71011 J, or 2.5 to 1, conventional to radiation enhanced. (b) Prompt radiation yields 6.3\u00d71011 J to 2.1\u00d71011 J, or 3 to 1, radiation enhanced to conventional. 57 (a) 1.1\u00d71025 fissions, 4.4 kg (b", ") 3.2\u00d71026 fusions, 2.7 kg (c) The nuclear fuel totals only 6 kg, so it is quite reasonable that some missiles carry 10 overheads. The mass of the fuel would only be 60 kg and therefore the mass of the 10 warheads, weighing about 10 times the nuclear fuel, would be only 1500 lbs. If the fuel for the missiles weighs 5 times the total weight of the warheads, the missile would weigh about 9000 lbs or 4.5 tons. This is not an unreasonable weight for a missile. 59 7\u00d7104 g 61 (a) 4.86\u00d7109 W (b) 11.0 y Test Prep for AP\u00ae Courses 1 (b) 3 (c) 5 (d) 7 (d) 9 (b) Chapter 33 Problems & Exercises 1 3\u00d710\u221239 s 3 This content is available for free at http://cnx.org/content/col11844/1.13 Answer Key 1655 1.99\u00d710\u221216 m (0.2 fm) 4 (a) 10\u221211 to 1, weak to EM (b) 1 to 1 6 (a) 2.09\u00d710\u22125 s (b) 4.77\u00d7104 Hz 8 78.0 cm 10 1.40\u00d7106 12 100 GeV 13 67.5 MeV 15 (a) 1\u00d71014 (b) 2\u00d71017 17 (a) 1671 MeV (b) = 1. = \u2212 1; \u2032 = \u2212 1; = 0; \u2032 = \u2212 1 + 1 = 0 \u00af \u2212 \u2192 \u2212+ + \u00af \u21d2 \u2212 antiparticle of +; of \u00af ; of (c) 19 (a) 3.9 eV (b) 2.9\u00d710\u22128 21 (a) The composition is the same as for a proton. (b) 3.3\u00d710\u221224 s (c) Strong (short lifetime) 23 a) \u0394++() = ) 1656 Answer Key Figure 33.20. 25 (a) +1 (b) = 1 = 1 + 0, = = 0 + ( \u2212 1), all lepton numbers are 0 before and after (c) () \u2192 () + ( ) 27 (ab) 277.9 MeV (c) 547.9 MeV 29 No. Charge = \u22121 is conserved. i 31 = 0 \u2260 f = 2 is not conserved. = 1 is conserved. (a)Yes. = \u22121 = 0 + ( \u2212 1), = 1 = 1 + 0, all lepton family numbers are 0 before and after, spontaneous since mass greater before reaction. (b) \u2192 + 33 (a) 216 (b) There are more baryons observed because we have the 6 antiquarks and various mixtures of quarks (as for the \u03c0-meson) as well. = \u22121, \u2212 1 3 35 \u03a9. 37 = 1, (a)803 MeV This content is available for free at http://cnx.org/content/col11844/1.13 Answer Key (b) 938.8 MeV (c) The annihilation energy of an extra electron is included in the total energy. 1657 39 \u00af 41 a)The antiproton - \u2192 0 + \u2212 b) 43 (a) 5\u00d71010 (b) 5\u00d7104 particles/m2 45 2.5\u00d710\u221217 m 47 (a) 33.9 MeV (b) Muon antineutrino 29.8 MeV, muon 4.1 MeV (kinetic energy) 49 (a) 7.2\u00d7105 kg (b) 7.2\u00d7102 m3 (c) 100 months Test Prep for AP\u00ae Courses 1 (d) 3 (d) 5 (b) 7 (a) 9 (c), though this comes from Einstein's special relativity 11 (a) 13 (d) 15 (b) 17 (b) Chapter 34 Problems & Exercises 1 3\u00d71041 kg 3 (a) 3\u00d71052 kg (b) 2\u00d71079 1658 (c) 4\u00d71088 5 0.30 Gly 7 (a) 2.0\u00d7105 km/s (b) 0.67 Answer Key 9 2.7\u00d7105 m/s 11 6\u00d710\u221211 (an overestimate, since some of the light from Andromeda is blocked by gas and dust within that galaxy) 13 (a) 2\u00d710\u22128 kg (b) 1\u00d71019 15 (a) 30km/s \u22c5 Mly (b) 15km/s \u22c5 Mly 17 960 rev/s 19 89.999773\u00ba (many digits are used to show the difference between 90\u00ba ) 22 23.6 km 24 (a) 2.95\u00d71012 m (b) 3.12\u00d710\u22124 ly", " 26 (a) 1\u00d71020 (b) 10 times greater 27 29 31 1.5\u00d71015 0.6 m\u22123 0.30 \u03a9 (34.6) (34.7) (34.8) This content is available for free at http://cnx.org/content/col11844/1.13 Index Index Symbols (peak) emf, 1030 RC circuit, 945 A aberration, 1176 aberrations, 1175 absolute pressure, 454, 478 absolute zero, 532, 565 AC current, 887, 901 AC voltage, 887, 901 acceleration, 43, 82, 146, 178 acceleration due to gravity, 68, 82 accommodation, 1155, 1176 Accuracy, 22 accuracy, 29 acoustic impedance, 755, 761 active transport, 518, 519 activity, 1400, 1410 adaptive optics, 1174, 1176 adhesive forces, 465, 478 adiabatic, 632 adiabatic process, 661 air resistance, 113, 128 alpha, 1379 alpha decay, 1391, 1410 alpha rays, 1410 Alternating current, 886 alternating current, 901 ammeter, 950 ammeters, 938 ampere, 869, 901 Ampere\u2019s law, 991, 999 amplitude, 681, 711, 1074, 1090, 1092 amplitude modulation, 1080 amplitude modulation (AM), 1092 analog meter, 950 Analog meters, 939 analytical method, 128 Analytical methods, 107 Anger camera, 1427, 1453 angular acceleration, 391, 425 angular magnification, 1172, 1176 angular momentum, 413, 425 angular momentum quantum number, 1353, 1364 angular velocity, 222, 249 antielectron, 1395, 1410 antimatter, 1394, 1410 antinode, 706, 711, 743, 761 approximation, 29 approximations, 27 arc length, 221, 249 Archimedes' principle, 459, 478 astigmatism, 1161, 1176 atom, 1318, 1364 atomic de-excitation, 1340, 1364 atomic excitation, 1340, 1364 atomic mass, 1386, 1410 atomic number, 1358, 1364, 1386, 1410 atomic spectra, 1279, 1306 Average Acceleration, 43 average acceleration, 84, 82 Average speed, 41 average speed, 82 Average velocity, 40 average velocity, 82 Avogadro\u2019s number, 546, 565 axions, 1518, 1524 axis of a polarizing filter, 1213, 1224 B B-field, 973, 999 back emf, 1031, 1056 banked curve, 249 banked curves, 230 barrier penetration, 1408, 1410 baryon number, 1478, 1492 Baryons, 1478 baryons, 1492 basal metabolic rate, 296, 301 beat frequency, 707, 711 becquerel, 1400, 1410 Bernoulli's equation, 496, 519 Bernoulli's principle, 497, 519 beta, 1379 beta decay, 1393, 1410 beta rays, 1410 Big Bang, 1504, 1524 binding energy, 1281, 1306, 1403, 1410 binding energy per nucleon, 1405, 1410 bioelectricity, 895, 901 Biot-Savart law, 991, 999 birefringent, 1221, 1224 Black holes, 1511 black holes, 1524 blackbodies, 1278 blackbody, 1306 blackbody radiation, 1278, 1306 Bohr radius, 1331, 1364 Boltzmann constant, 544, 565 boson, 1476, 1492 bottom, 1486, 1492 bow wake, 739, 761 break-even, 1442, 1453 breeder reactors, 1448, 1454 breeding, 1448, 1454 bremsstrahlung, 1287, 1306 Brewster\u2019s angle, 1216, 1224 Brewster\u2019s law, 1216, 1224 bridge device, 950 bridge devices, 944 Brownian motion, 1319, 1364 buoyant force, 458, 478 C capacitance, 843, 855, 945, 950 capacitive reactance, 1049, 1056 capacitor, 841, 855, 945, 950 capillary action, 470, 478 carbon-14 dating, 1398, 1410 Carnot cycle, 640, 661 Carnot efficiency, 640, 661 Carnot engine, 640, 661 carrier particle, 178 carrier particles, 177 carrier wave, 1080, 1092 cathode-ray tube, 1364 cathode-ray tubes, 1320 Celsius, 531 1659 Celsius scale, 565 center of gravity", ", 364, 380 center of mass, 237, 249 centrifugal force, 233, 249 centrifuge, 226 centripetal acceleration, 225, 249 centripetal force, 228, 249 change in angular velocity, 391, 425 change in entropy, 650, 661 change in momentum, 319, 343 Chaos, 1520 chaos, 1524 characteristic time constant, 1045, 1057 characteristic x rays, 1287, 1306 charm, 1486, 1492 chart of the nuclides, 1389, 1410 chemical energy, 287, 301 classical physics, 12, 29 Classical relativity, 125 classical relativity, 128 classical velocity addition, 1267 coefficient of linear expansion, 537, 565 coefficient of performance, 648, 661 coefficient of volume expansion, 539, 565 coherent, 1191, 1224 cohesive forces, 465, 478 Colliding beams, 1473 colliding beams, 1492 color, 1487, 1492 color constancy, 1164, 1176 commutative, 104, 129, 128 complexity, 1519, 1524 component (of a 2-d vector), 128 components, 106 compound microscope, 1165, 1176 Compton effect, 1291, 1306 Conduction, 590 conduction, 606 conductor, 782, 805 Conductors, 786 confocal microscopes, 1223, 1224 conservation laws, 933, 950 conservation of mechanical energy, 279, 301 conservation of momentum principle, 324, 343 Conservation of total, 1478 conservation of total baryon number, 1478, 1492 conservation of total electron family number, 1492 conservation of total, 1478 conservation of total muon family number, 1492 conservative force, 277, 301 constructive interference, 704, 711 constructive interference for a diffraction grating, 1197, 1224 constructive interference for a double slit, 1193, 1224 contact angle, 470, 478 Contrast, 1222 contrast, 1224 Convection, 590 convection, 607 1660 Index converging (or convex) lens, 1122 converging lens, 1142 converging mirror, 1142 conversion factor, 19, 29 Coriolis force, 234, 249 corner reflector, 1116, 1142 correspondence principle, 1277, 1306 cosmic microwave background, 1505, 1524 cosmological constant, 1516, 1525 cosmological red shift, 1504, 1525 Cosmology, 1502 cosmology, 1525 Coulomb force, 792, 805 Coulomb forces, 791 Coulomb interaction, 799, 805 Coulomb's law, 790, 805 critical angle, 1112, 1142 Critical damping, 694 critical damping, 711 critical density, 1516, 1525 critical mass, 1446, 1454 critical point, 557, 565 Critical pressure, 557 critical pressure, 565 critical temperature, 557, 565, 1521, 1525 criticality, 1447, 1454 curie, 1400, 1411 Curie temperature, 970, 999 current, 915, 950 Current sensitivity, 939 current sensitivity, 950 cyclical process, 636, 661 cyclotron, 1472, 1492 D Dalton\u2019s law of partial pressures, 560, 565 dark matter, 1515, 1525 daughter, 1411 daughters, 1390 de Broglie wavelength, 1296, 1306 decay, 1379, 1390, 1411 decay constant, 1398, 1411 decay equation, 1392, 1396, 1411 decay series, 1390, 1411 deceleration, 82 defibrillator, 853, 855 deformation, 203, 212, 675, 711 degree Celsius, 531, 565 degree Fahrenheit, 531, 565 Density, 441 density, 478 dependent variable, 75, 82 derived units, 16, 29 destructive interference, 704, 711 destructive interference for a double slit, 1193, 1225 destructive interference for a single slit, 1202, 1225 dew point, 561, 565 dialysis, 518, 519 diastolic pressure, 455, 478 Diastolic pressure, 474 dielectric, 846, 855 dielectric strength, 855 dielectric strengths, 846 diffraction, 1190, 1225 diffraction grating, 1196, 1225 Diffusion, 515 diffusion, 519 digital meter, 950 digital meters, 939 dipole, 799, 805 Direct current, 886 direct current, 901 direction, 102 direction (of a vector), 128 direction of magnetic field lines, 973, 999 direction of polarization, 1213, 1225 Dispersion, 1118 dispersion, 1142 displacement, 34, 82 Distance, 36 distance, 82 Distance traveled, 36 distance traveled, 82 diverging lens, 1124, 1142 diverging", " mirror, 1142 domains, 969, 999 Doppler effect, 736, 762 Doppler shift, 736, 762 Doppler-shifted ultrasound, 759, 762 Double-slit interference, 1329 double-slit interference, 1364 down, 1481, 1492 drag force, 198, 212 drift velocity, 872, 901 dynamic equilibrium, 358, 380 Dynamics, 141, 144, 179 dynamics, 178 E eddy current, 1024, 1057 efficiency, 290, 301 Elapsed time, 39 elapsed time, 83 elastic collision, 328, 343 elastic potential energy, 677, 711 electric and magnetic fields, 1090 electric charge, 777, 805 electric current, 869, 901 electric field, 786, 805, 1074, 1092 electric field lines, 805, 1092 Electric field lines, 1071 electric field strength, 793, 1092 electric fields, 794 electric generator, 1057 Electric generators, 1028 electric potential, 824, 855 electric power, 883, 901 Electrical energy, 287 electrical energy, 301 electrocardiogram (ECG), 900, 901 electromagnet, 999 electromagnetic force, 806 electromagnetic induction, 1018, 1057 electromagnetic spectrum, 1092 electromagnetic waves, 1073, 1090, 1092 Electromagnetism, 970 electromagnetism, 999 electromagnets, 970 electromotive force, 924 This content is available for free at http://cnx.org/content/col11844/1.13 electromotive force (emf), 950, 1092 electron, 806 Electron capture, 1396 electron capture, 1411 electron capture equation, 1396, 1411 electron family number, 1478, 1492 electron volt, 828, 856 electrons, 777 electron\u2019s antineutrino, 1394, 1411 electron\u2019s neutrino, 1396, 1411 electrostatic equilibrium, 786, 806 electrostatic force, 790, 806 electrostatic precipitators, 803, 806 Electrostatic repulsion, 783 electrostatic repulsion, 806 electrostatics, 800, 806 electroweak epoch, 1508, 1525 electroweak theory, 1488, 1493 emf, 934 emf induced in a generator coil, 1029, 1057 emissivity, 603, 607 endoscope, 1114 energies of hydrogen-like atoms, 1332, 1364 energy, 301 energy stored in an inductor, 1044, 1057 energy-level diagram, 1330, 1364 English units, 15, 29 entropy, 649, 661 equipotential line, 856 equipotential lines, 838 escape velocity, 1511, 1525 event horizon, 1511, 1525 external force, 146, 178 external forces, 144 External forces, 179 Extremely low frequency (ELF), 1079 extremely low frequency (ELF), 1092 eyepiece, 1166, 1176 F Fahrenheit, 531 Fahrenheit scale, 565 far point, 1158, 1176 Faraday cage, 788, 806 Faraday\u2019s law of induction, 1019, 1057 Farsightedness, 1158 farsightedness, 1176 fermion, 1476, 1493 ferromagnetic, 969, 999 Feynman diagram, 1470, 1493 Fiber optics, 1114 fiber optics, 1142 fictitious force, 233, 249 field, 806 fine structure, 1351, 1364 first law of thermodynamics, 621, 661 first postulate of special relativity, 1239, 1267 fission fragments, 1445, 1454 flat (zero curvature) universe, 1516, 1525 flavors, 1481, 1493 Flow rate, 490 flow rate, 519 fluid dynamics, 519 Index 1661 fluids, 440, 478 Fluorescence, 1340 fluorescence, 1364 focal length, 1122, 1142 focal point, 1122, 1142 Food irradiation, 1437 food irradiation, 1454 force, 144, 178 Force, 179 force constant, 676, 711 force field, 175, 176, 178, 792 fossil fuels, 298, 301 free charge, 806 free charges, 786 free electron, 806 free electrons, 782 free radicals, 1438, 1454 free-body diagram, 144, 179, 167, 178 free-fall, 68, 83, 149, 178 Frequency, 680 frequency, 712, 1074, 1092 frequency modulation, 1080 frequency modulation (FM), 1092 friction, 147, 178, 212, 301 Friction, 192, 282 full-scale deflection, 939, 950 fundamental, 744, 762 fundamental frequency, 706, 712 fundamental particle, 1480, 1493 fundamental units, 16, 29 G galvanometer, 939", ", 950 gamma, 1379 gamma camera, 1427, 1454 Gamma decay, 1397 gamma decay, 1411 gamma ray, 1088, 1092, 1306 Gamma rays, 1284 gamma rays, 1411 gauge boson, 1493 gauge bosons, 1476 gauge pressure, 454, 478 gauss, 975, 999 Geiger tube, 1383, 1411 general relativity, 1509, 1525 geometric optics, 1103, 1142 glaucoma, 475, 478 gluons, 1470, 1493 Gluons, 1489 grand unified theory, 1493 Grand Unified Theory (GUT), 1488 gravitational constant, 237 gravitational constant, G, 249 gravitational potential energy, 272, 301 Gravitational waves, 1512 gravitational waves, 1525 gray (Gy), 1429, 1454 greenhouse effect, 605, 607 grounded, 801, 806 grounding, 839, 856 GUT epoch, 1508, 1525 H Hadrons, 1476 hadrons, 1493 half-life, 1397, 1411 Hall effect, 982, 999 Hall emf, 982, 999 harmonics, 744, 762 head, 100 head (of a vector), 128 head-to-tail method, 100, 129, 128 Hearing, 724, 749 hearing, 762 heat, 576, 607 heat engine, 627, 661 heat of sublimation, 589, 607 heat pump, 661 heat pump's coefficient of performance, 646 Heisenberg uncertainty principle, 1302 Heisenberg\u2019s uncertainty principle, 1303, 1306 henry, 1041, 1057 hertz, 1092 Higgs boson, 1490, 1493 high dose, 1430, 1454 hologram, 1347, 1364 Holography, 1347 holography, 1364 Hooke's law, 203, 212 horizontally polarized, 1213, 1225 Hormesis, 1431 hormesis, 1454 horsepower, 293, 301 Hubble constant, 1504, 1525 hues, 1162, 1176 Human metabolism, 625 human metabolism, 661 Huygens\u2019s principle, 1188, 1225 Hydrogen spectrum wavelength, 1329 hydrogen spectrum wavelengths, 1364 hydrogen-like atom, 1331, 1364 hydrogen-spectrum wavelengths, 1328 hyperopia, 1158, 1176 I ideal angle, 249 ideal banking, 230, 249 ideal gas law, 544, 565 ideal speed, 249 Ignition, 1442 ignition, 1454 Image distance, 1128 impedance, 1051, 1057 impulse, 319, 343 Incoherent, 1191 incoherent, 1225 independent variable, 75, 83 index of refraction, 1108, 1142 inductance, 1040, 1057 induction, 783, 806, 1057 inductive reactance, 1047, 1057 inductor, 1042, 1057 inelastic collision, 332, 343 inertia, 146, 179 Inertia, 179 inertial confinement, 1442, 1454 inertial frame of reference, 165, 179, 1239, 1267 inflationary scenario, 1509, 1525 Infrared radiation, 1083 infrared radiation, 1306 infrared radiation (IR), 1092 Infrared radiation (IR), 1289 infrasound, 749, 762 ink jet printer, 802 ink-jet printer, 806 Instantaneous acceleration, 49 instantaneous acceleration, 83 Instantaneous speed, 41 instantaneous speed, 83 Instantaneous velocity, 40 instantaneous velocity, 83 insulator, 806 insulators, 782 intensity, 709, 712, 732, 762, 1091, 1092 intensity reflection coefficient, 756, 762 Interference microscopes, 1222 interference microscopes, 1225 internal energy, 622, 661 Internal kinetic energy, 328 internal kinetic energy, 343 internal resistance, 924, 950 intraocular pressure, 475, 478 intrinsic magnetic field, 1351, 1364 intrinsic spin, 1351, 1364 ionizing radiation, 1284, 1306, 1381, 1411 ionosphere, 787, 806 irreversible process, 635, 661 isobaric process, 628, 661 isochoric, 630 isochoric process, 661 isolated system, 325, 343 isothermal, 632 isothermal process, 661 isotopes, 1387, 1411 J joule, 265, 301 Joule\u2019s law, 916, 950 junction rule, 933, 950 K Kelvin, 532 Kelvin scale, 565 kilocalorie, 576, 607 kilogram, 16, 29 kilowatt-hour, 301 kilowatt-hours, 295 kinematics, 83, 114, 128 kinematics of rotational motion, 395, 425 kinetic energy, 268, 301 kinetic friction, 192, 212 Kirchhoff\u2019", "s rules, 932, 950 L Laminar, 504 laminar, 519 laser, 1343, 1364 laser printer, 806 Laser printers, 802 Laser vision correction, 1161 laser vision correction, 1176 latent heat coefficient, 607 latent heat coefficients, 585 law, 11, 29 law of conservation of angular momentum, 416, 425 law of conservation of charge, 780, 806 1662 Index law of conservation of energy, 287, 301 law of inertia, 146, 179, 179 law of reflection, 1142 law of refraction, 1110 Length contraction, 1248 length contraction, 1267 Lenz\u2019s law, 1019, 1057 leptons, 1476, 1493 linear accelerator, 1474, 1493 linear hypothesis, 1431, 1454 Linear momentum, 316 linear momentum, 343 liquid drop model, 1445, 1454 liter, 490, 519 longitudinal wave, 702, 712 loop rule, 934, 950 Lorentz force, 975, 999 loudness, 749, 762 low dose, 1430, 1454 M MACHOs, 1518, 1525 macrostate, 656, 661 magic numbers, 1389, 1411 magnetic confinement, 1442, 1454 magnetic damping, 1024, 1057 magnetic field, 973, 999, 1074, 1092 magnetic field lines, 973, 1000, 1093 Magnetic field lines, 1072 magnetic field strength, 1093 magnetic field strength (magnitude) produced by a long straight currentcarrying wire, 990, 1000 magnetic field strength at the center of a circular loop, 991, 1000 magnetic field strength inside a solenoid, 992, 1000 magnetic flux, 1018, 1057 magnetic force, 975, 1000 magnetic monopoles, 972, 1000 Magnetic resonance imaging (MRI), 998 magnetic resonance imaging (MRI), 1000 magnetized, 969, 1000 magnetocardiogram (MCG), 999, 1000 magnetoencephalogram (MEG), 999, 1000 magnification, 1128, 1143 magnitude, 102 magnitude (of a vector), 128 magnitude of kinetic friction, 212 magnitude of kinetic friction fk, 193 magnitude of static friction, 212 magnitude of static friction fs, 193 magnitude of the intrinsic (internal) spin angular momentum, 1355, 1364 mass, 146, 179 Mass, 179 mass number, 1386, 1411 massive compact halo objects, 1518 maximum field strength, 1090, 1093 Maxwell\u2019s equations, 991, 1000, 1071, 1093 mechanical advantage, 371, 380 mechanical energy, 279, 301, 856 Mechanical energy, 829 mechanical equivalent of heat, 577, 607 meson, 1468, 1493 Mesons, 1478 metabolic rate, 296, 301 metastable, 1342, 1364 meter, 16, 29, 1000 Meters, 988 method of adding percents, 24, 29 metric system, 17, 29 Michelson-Morley experiment, 1240, 1267 Microgravity, 242 microgravity, 249 microlensing, 1518, 1525 microshock sensitive, 894, 901 microstate, 656, 661 Microwaves, 1082, 1289 microwaves, 1093, 1306 micturition reflex, 477, 478 mirror, 1143 model, 11, 29, 41, 83 moderate dose, 1430, 1454 Modern physics, 14 modern physics, 29 mole, 546, 565 moment of inertia, 401, 401, 425 motion, 113, 128 motor, 1000 Motors, 986 muon family number, 1478, 1493 Mutual inductance, 1040 mutual inductance, 1057 myopia, 1158, 1176 N natural frequency, 698, 712 near point, 1158, 1176 Nearsightedness, 1158 nearsightedness, 1176 negatively curved, 1516, 1525 Nerve conduction, 895 nerve conduction, 901 net external force, 147, 179 net rate of heat transfer by radiation, 604, 607 net work, 267, 301 neutral equilibrium, 366, 380 neutralinos, 1518, 1525 neutrino, 1393, 1411 neutrino oscillations, 1518, 1525 neutron, 1386, 1411 Neutron stars, 1512 neutron stars, 1525 Neutron-induced fission, 1445 neutron-induced fission, 1454 newton, 149 Newton's universal law of gravitation, 236, 249 newton-meters, 265 Newton\u2019s first law of motion, 145, 179, 179 Newton\u2019s second law of motion, 146, 179 Newton\u2019s third law of motion, 153, 180, 179 node, 743, 762 Nodes, 705 This content is available for free at http://cnx.org/content/col11844/1.", "13 nodes, 712 non-inertial frame of reference, 233, 249 nonconservative force, 282, 301 normal force, 158, 179 north magnetic pole, 967, 1000 note, 762 notes, 749 Nuclear energy, 287 nuclear energy, 301 Nuclear fission, 1444 nuclear fission, 1454 Nuclear fusion, 1439 nuclear fusion, 1454 nuclear magnetic resonance (NMR), 998, 1000 nuclear radiation, 1379, 1411 nuclear reaction energy, 1392, 1411 nucleons, 1386, 1411 nucleus, 1411 nuclide, 1386, 1411 Null measurements, 942 null measurements, 951 numerical aperture, 1176 numerical aperture, 1167 O objective lens, 1166, 1177 ohm, 875, 901 Ohm's law, 874, 901 ohmic, 875, 901 ohmmeter, 951 ohmmeters, 943 Ohm\u2019s law, 915, 951 optically active, 1220, 1225 orbital angular momentum, 1350, 1364 orbital magnetic field, 1350, 1364 order, 1193, 1225 order of magnitude, 17, 29 oscillate, 712, 1093 Osmosis, 518 osmosis, 519 osmotic pressure, 518, 519 Otto cycle, 638, 661 over damping, 712 overdamped, 695 overtones, 706, 712, 744, 762 P parallel, 917, 951 parallel plate capacitor, 842, 856 parent, 1390, 1411 Partial pressure, 560 partial pressure, 565 particle physics, 1493 particle-wave duality, 1295, 1306 Particle-wave duality, 1304 Pascal's principle, 451 Pascal's Principle, 478 Pauli exclusion principle, 1358, 1365 peak emf, 1057 percent relative humidity, 563, 565 percent uncertainty, 24, 29 perfectly inelastic collision, 332, 343 period, 680, 712 periodic motion, 680, 712 permeability of free space, 990, 1000 perpendicular lever arm, 360, 380 Index 1663 phase angle, 1054, 1057 phase diagram, 565 phase diagrams, 558 phase-contrast microscope, 1223, 1225 phon, 750, 762 Phosphorescence, 1342 phosphorescence, 1365 photoconductor, 801, 806 photoelectric effect, 1280, 1306 photomultiplier, 1384, 1411 photon, 1280, 1290, 1306 photon energy, 1280, 1306 photon momentum, 1291, 1307 physical quantity, 15, 29 Physics, 8 physics, 29 pion, 1467, 1493 pit, 221, 249 pitch, 726, 749, 762 Planck\u2019s constant, 1278, 1307 planetary model of the atom, 1326, 1365 point charge, 792, 806 point masses, 336, 343 Poiseuille's law, 507, 519 Poiseuille's law for resistance, 506, 519 polar molecule, 799, 806, 847, 856 polarization, 783, 806, 1225 Polarization, 1213 polarization microscope, 1223, 1225 polarized, 786, 806, 1213, 1225 population inversion, 1343, 1365 position, 34, 83 positively curved, 1516, 1525 positron, 1396, 1411 positron decay, 1395, 1411 positron emission tomography (PET), 1427, 1454 potential difference, 824, 924, 951 potential difference (or voltage), 856 potential energy, 277, 279, 301 potential energy of a spring, 278, 301 potentiometer, 943, 951 power, 292, 301, 1123, 1143 power factor, 1054, 1057 precision, 23, 29 presbyopia, 1158, 1177 pressure, 444, 447, 478 Pressure, 451 probability distribution, 1300, 1307 projectile, 113, 128 Projectile motion, 113 projectile motion, 128 Proper length, 1248 proper length, 1267 Proper time, 1243 proper time, 1267 proton, 806 proton-proton cycle, 1440, 1454 protons, 777, 1386, 1411 PV diagram, 557, 565 Q quality factor, 1429, 1454 quantized, 1277, 1307 quantum chromodynamics, 1487, 1490, 1493 quantum electrodynamics, 1470, 1493 Quantum gravity, 1509, 1525 quantum mechanical tunneling, 1408, 1411 Quantum mechanics, 14 quantum mechanics, 29, 1277, 1307 quantum numbers, 1353, 1365 quark, 343, 1493 quarks, 327, 1480 quasars, 1512, 1525 R R factor, 607 factor,", " 594 rad, 1429, 1454 Radar, 1082 radar, 1093 radians, 221, 250 radiant energy, 287, 302 radiation, 590, 602, 607 radiation detector, 1383, 1411 radio waves, 1070, 1079, 1093 radioactive, 1379, 1411 Radioactive dating, 1398 radioactive dating, 1411 radioactivity, 1379, 1412 radiolytic products, 1438, 1454 radiopharmaceutical, 1425, 1454 radiotherapy, 1435, 1454 radius of a nucleus, 1387, 1412 radius of curvature, 221, 250 rainbow, 1143 range, 119, 128 range of radiation, 1381, 1412 rate of conductive heat transfer, 593, 607 rate of decay, 1400, 1412 ray, 1102, 1143 Ray tracing, 1125 Rayleigh criterion, 1204, 1225 RC circuit, 951 real image, 1127, 1143 reflected light is completely polarized, 1216 reflected light that is completely polarized, 1225 refraction, 1106, 1143 relative biological effectiveness, 1429 relative biological effectiveness (RBE), 1454 relative humidity, 561, 565 relative osmotic pressure, 518, 519 relative velocities, 125 relative velocity, 128 relativistic Doppler effects, 1269, 1267 Relativistic kinetic energy, 1263 relativistic kinetic energy, 1267 Relativistic momentum, 1257 relativistic momentum, 1267 relativistic velocity addition, 1253, 1267 Relativity, 14 relativity, 29, 125, 128, 1239, 1267 Renewable forms of energy, 298 renewable forms of energy, 302 resistance, 874, 901, 914, 951 resistivity, 878, 902 resistor, 914, 945, 951 resonance, 698, 712 resonant, 1075, 1093 resonant frequency, 1053, 1057 resonate, 698, 712 Rest energy, 1259 rest energy, 1267 rest mass, 1257, 1267 restoring force, 675, 712 resultant, 101, 128 resultant vector, 101, 128 retinex, 1177 retinex theory of color vision, 1165, 1177 retinexes, 1165 reverse dialysis, 518, 519 Reverse osmosis, 518 reverse osmosis, 519 reversible process, 633, 661 Reynolds number, 512, 519 right hand rule 1, 975 right hand rule 1 (RHR-1), 1000 right hand rule 2, 989 right hand rule 2 (RHR-2), 1000 right-hand rule, 423, 425 RLC circuit, 1093 rms current, 888, 902 rms voltage, 888, 902 rods and cones, 1162, 1177 roentgen equivalent man, 1429 roentgen equivalent man (rem), 1454 rotation angle, 221, 250 rotational inertia, 401, 425 rotational kinetic energy, 405, 425 Rydberg constant, 1328, 1333, 1365 S saturation, 561, 565 scalar, 37, 83, 106, 128, 830, 856 Schwarzschild radius, 1511, 1525 scientific method, 12, 29 scintillators, 1384, 1412 screening, 799, 806 second, 16, 29 second law of motion, 317, 344 second law of thermodynamics, 635, 636, 640, 661 second law of thermodynamics stated in terms of entropy, 661 second postulate of special relativity, 1240, 1268 Self-inductance, 1042 self-inductance, 1057 semipermeable, 517, 519, 895, 902 series, 915, 951 shear deformation, 209, 212 shell, 1359, 1365 shielding, 1432, 1454 shock hazard, 890, 902, 1036, 1057 short circuit, 891, 902 shunt resistance, 940, 951 SI unit of torque, 361 SI units, 15, 29 SI units of torque, 380 sievert, 1430, 1454 significant figures, 25, 29 simple circuit, 875, 902 Simple Harmonic Motion, 681 simple harmonic motion, 712 simple harmonic oscillator, 681, 712 1664 Index simple pendulum, 686, 712 simplified theory of color vision, 1163, 1177 single-photon-emission computed tomography (SPECT), 1454 single-photon-emission computed tomography(SPECT), 1427 slope, 75, 83 solenoid, 992, 1000 Solid-state radiation detectors, 1385 solid-state radiation detectors, 1412 sonic boom, 739, 762 sound, 724, 762 sound intensity level, 733, 762 sound pressure level,", " 735, 762 south magnetic pole, 967, 1000 space quantization, 1351, 1365 special relativity, 1268 special relativity., 1239 specific gravity, 461, 478 specific heat, 579, 607 speed of light, 1093 spin projection quantum number, 1355, 1365 spin quantum number, 1355, 1365 spontaneous symmetry breaking, 1509, 1525 stable equilibrium, 365, 380 Standard Model, 1490 standard model, 1493 standing wave, 705, 1075, 1093 static electricity, 806 static equilibrium, 358, 368, 380 static friction, 192, 212 statistical analysis, 658, 661 Stefan-Boltzmann law of radiation, 603, 607 step-down transformer, 1034, 1057 step-up transformer, 1034, 1057 Stimulated emission, 1343 stimulated emission, 1365 Stokes' law, 202, 212 strain, 208, 212 strange, 1481, 1493 strangeness, 1478, 1493 stress, 208, 212 sublimation, 558, 566, 607 Sublimation, 589 subshell, 1359, 1365 Superconductors, 1521, 1525 supercriticality, 1447, 1454 superforce, 1508, 1525 superposition, 704, 712 Superstring theory, 1491, 1515, 1525 superstring theory, 1493 surface tension, 465, 478 synchrotron, 1472, 1493 synchrotron radiation, 1472, 1493 system, 146, 179 systolic pressure, 455, 478 Systolic pressure, 474 T tagged, 1425, 1455 tail, 100, 128 tangential acceleration, 392, 425 tau family number, 1493 Television, 1081 Temperature, 530 temperature, 566 temperature coefficient of resistivity, 880, 902 tensile strength, 205, 212 tension, 161, 179 terminal speed, 514, 519 terminal voltage, 926, 951 tesla, 975, 1000 test charge, 792, 806 the second law of thermodynamics stated in terms of entropy, 652 theory, 11, 29 theory of quark confinement, 1487, 1493 therapeutic ratio, 1435, 1455 thermal agitation, 1082, 1093 thermal conductivity, 593, 607 thermal energy, 282, 287, 302, 552, 566 thermal equilibrium, 536, 566 thermal expansion, 537, 566 thermal hazard, 890, 902, 1036, 1057 Thermal stress, 541 thermal stress, 566 thin film interference, 1208, 1225 thin lens, 1125 thin lens equations, 1128 thought experiment, 1509, 1525 three-wire system, 1036, 1057 thrust, 154, 180, 179 timbre, 750, 762 time, 39, 83 Time dilation, 1242 time dilation, 1268 TOE epoch, 1508, 1525 tone, 750, 762 top, 1486, 1493 Torque, 360 torque, 380, 400, 425 Total energy, 1259 total energy, 1268 total internal reflection, 1112 trajectory, 113, 128 transformer, 1057 transformer equation, 1034, 1058 Transformers, 1032 transverse wave, 702, 712, 1075, 1093 triple point, 559, 566 Tunneling, 1409 tunneling, 1412 turbulence, 504, 519 TV, 1093 twin paradox, 1268 U ultra high frequency, 1081 ultra-high frequency (UHF), 1093 ultracentrifuge, 227, 250 ultrasound, 749, 762 Ultraviolet (UV) microscopes, 1222 ultraviolet (UV) microscopes, 1225 Ultraviolet radiation, 1287 ultraviolet radiation, 1307 ultraviolet radiation (UV), 1085, 1093 uncertainty, 23, 29 uncertainty in energy, 1303, 1307 uncertainty in momentum, 1301, 1307 This content is available for free at http://cnx.org/content/col11844/1.13 uncertainty in position, 1301, 1307 uncertainty in time, 1303, 1307 under damping, 712 underdamped, 695 uniform circular motion, 250 units, 15, 29 unpolarized, 1213, 1225 unstable equilibrium, 365, 380 up, 1481, 1493 useful work, 296, 302 V Van de Graaff, 1472, 1493 Van de Graaff generator, 806 Van de Graaff generators, 801 vapor, 559, 566 Vapor pressure, 560 vapor pressure, 566 vector, 37, 83, 99, 128, 806, 830, 856 vector addition, 122, 128, 796, 806 vectors, 97, 794 velocity, 122, 128 vertically polarized, 1213, 1225 very high frequency, 1081 very high frequency (VHF), 10", "93 virtual image, 1130, 1143 virtual particles, 1467, 1494 viscosity, 506, 519 viscous drag, 513, 519 Visible light, 1084 visible light, 1093, 1288, 1307 voltage, 824, 915, 951 voltage drop, 915, 951 voltmeter, 951 Voltmeters, 938 W watt, 292, 302 wave, 700, 712 wave velocity, 700, 712 wavelength, 701, 712, 1074, 1093 wavelength in a medium, 1187, 1225 weakly interacting massive particles, 1518 weight, 149, 179 Weight, 157 Wheatstone bridge, 944, 951 WIMPs, 1518, 1525 work, 263, 302 work-energy theorem, 268, 302, 406, 427, 425 X x ray, 1307 x rays, 1285, 1365 X rays, 1334 X-ray, 1088, 1093 x-ray diffraction, 1338, 1365 xerography, 801, 806 Y y-intercept, 75, 83 Z z-component of spin angular momentum, 1365 z-component of the angular momentum, 1365 Index 1665 -component of spin angular momentum, 1355 -component of the angular momentum, 1353 Zeeman effect, 1350, 1365 zeroth law of thermodynamics, 536, 566 zircon, 1143ent with greater accuracy than is possible with the means at our disposal. The reasons will be apparent after studying later paragraphs (page 15). 28.7 (c) Approximate Numbers In all measurement, you must first decide what will be the limits of accuracy of your work and then realize that a possible error exists in the result. In order to make this small, the possible error is not permitted to exceed one-half of the smallest unit employed in the operaThus, in the previous examples, tion. if 1/8 inch were the smallest unit on the measuring instrument used, a possible error of 1/16 inch more or less is found in the answer. Accordingly the length of the room in part (a) would be between 14 feet, 15/16 inch and 15 feet, 1/16 inch and the width of the door would be between 29 15/16 and 30 1/16 Similarly, if the present room inches. temperature is given as 21.5\u00b0C., and the thermometer is marked off into 0.1 degree units, the possible error will be.05 centigrade degrees and the actual reading may be between 21.45\u00b0C. and 21.55\u00b0C. The reading 21.5\u00b0C. represents the temperature as accurately as the means of measurement at our disposal will permit. Other examples follow: Measurement 55.3 cm. 719 ft. 19.0 ft. 0.0032 in. From 55.25 cm. Limits of Accuracy up to but not including 55.35 cm. a \u201c \u201c 718.5 ft. 18.95 ft. 0.00315 in. \u201c \u201c \u201c \u201c \u201c \u201c \u201c 719.5 ft. 19.05 ft. 0.00325 in. To allay the fear that naturally arises over the presence of this possible error, let us calculate what per cent of the whole it represents. The percentage error is possible error divided by the accepted number times Using the first measurement in the preceding table we get: 100. The percentage error =: From this, we see that approximate numbers contain an extremely small error. The more precise the measuring device is, the smaller the unit of measurement will be. The smaller the unit of measurement is, the smaller the possible error will be. (d) How to Deal with Approximate Numbers 55.3 - 55.25 55.3 - 55.35 X 100 or 55.3 = +_X100or- \u2014 X.oo 55.3.05.05 = -f.09% or -.09% X 100 1. Significant Digits When performing a measurement, only those numbers in the measurement that are certain, like the 15 feet, 30 inches, 21.5\u00b0C. are recorded. The digits in these numbers paragraphs, previous of 13 Chap. 2 MECHANICS are called the significant digits in the result, there being two in the first two numbers above, and three in the third. Mathematicians have laid down certain rules to guide us in working with them. They are as follows: 1. All the digits from 1 to 9 including any zeros between them or after them are significant digits. 2. The position of the decimal is disregarded in determining the number of significant digits. 3. Rounding off Approximate Numbers A number like the value of pi (tt), 3.14159, is correct to 6 significant digits. It may be made consistent with measurements having fewer than six significant digits by the process of", " rounding-off. This means dropping necessary number of digits off the end of a number and adding one to the last remaining digit if the next digit was five or more. the 4. Examples 3. The zeros preceding the first digit Rounding-off Accuracy are not significant. Significant Digits IN Numbers Number 604 60.4 6.04 0.604 0.0604 0.06004 604.0 60.40 Significant Digits 3 3 3 3 3 4 4 4 2. Representing Significant Digits in Large Numbers digit, Where a number like 600 is correct it should be to one significant written as 6 X 10^. To express it as 600 indicates digits. Similarly, the distance from the earth to the sun is correct to only two signiwritten ficant 93 X 10\u00ae rather than 93,000,000 miles. digits and should significant three be 3.14159 3.142 3.1 6 4 2 significant digits a a Calculations With Significant Digits How to carry out mathematical operations with approximate numbers is indicated by the following rules: 1. Addition and Subtraction In operations involving addition and subtraction of approximate numbers of which the least precise has N places of decimals, round off the other numbers where possible to N + 1 places and the answer to N places. 2. Multiplication and Division In operations involving multiplication and division of approximate numbers of which the least accurate has N significant digits, round off the others where possible to N + 1 digits and the answer to N digits. Examples 1. Add 2.0149 3.02864 1.239 1.97 14 Method Round off all to three decimals where possible. Round off answer to 2 decimals. Result 2.015 3.029 1.239 1.97 Total = 8.253 Proper answer = 8.25 MEASUREMENT Method Result 2. Subtract 21.347 As for addition from 32.5 Sec. 1:3 32.5 21.35 3. Multiply 2.1 by 2.56 and by 9.547 Round oflF all numbers to 3 significant digits. Keep 3 significant digits in all partial products. Round off answer to 2 significant digits. 4. Divide 96.568 As for multiplication by 7.02 Difference = 11.15 Proper answer = 11.2 2.1 X 2.56 = 5.376 Proper product = 5.38 5.38 X 9.55 = 51.3790 Proper answer = 51. 96.57 = 13.75 7.02 = 13.75 Quotient Proper answer =13.8 On page 13 we multiplied 7.54 by getting 28.7274 for an answer. 3.81, However, as the multiplier and multiplicand are correct to three significant digits only, 28.7 is the proper answer. You may wonder about the usefulness of the above material, but be assured that the method is used daily by scientists and mathematicians whose work has contributed so much to each modern development and invention. (e) Applying Approximate Numbers in Elementary Physics The diameter of a cylindrical solid is 2.50 cm. and it is 10.04 cm. long. (7r = 3.1416). Its mass is 280.76 gm. Calculate its density. Radius = 1.25 cm. Length = 10.04 cm. =3.1416 = 3.142 TT Area of the end of the cylinder = tt R^ = 3.142 X 1.25 X 1.25 = 4.910 sq. cm. Volume of the cylinder = area of end X length = 4.910 X 10.04 = 49.30 c.c. The mass given = 280.76 gm. The proper mass = 280.8 gm. 49.30 c.c. of solid weigh 280.8 gm. \u2022.. 1 1 1-^ c.c. of solid weighs r \u2022 u 280.8 49.30 = 5.695 gm. The density = 5.70 gm. per c.c. 15 Chap. 2 : 4 I MECHANICS QUESTIONS 1. 2. (a) Why Is measurement necessary? (b) Why was it necessary to establish standard units? (a) Name the two systems of measurement and name the fundamental units for each. (b) What advantages has the metric system over the British system of measurement? (c) Why is the British system still used in a few countries? 3. (a) What apparatus is commonly used in the laboratory for the measurement of (i) length (ii) mass (ili) time? (b) Suggest other pieces of apparatus that could be used to measure these fundamental units more ac- curately. 4. 5. 6. 7. 8. decimetres (a) State the number of millimetres, centimetres and metre. How many metres are there In 1 kilometre? (b)", " Using the above units construct (i) square measure (ii) tables of in 1 cubic measure. (c) Define: litre, millilitre. (a) Distinguish between the mass and weight of an object. (b) State the number of milligrams, centigrams, decigrams, and grams in 1 kilogram. (a) Why was the apparent motion of the sun adopted as the basis for reckoning time? (b) Define: mean solar day, second. (a) Distinguish between exact and approximate numbers. Give an example of each. (b) What gives rise to approximate numbers? Why? (a) What do you mean by possible error? 16 9. (b) How do you calculate percentage error? (a) What is meant by significant digits? (b) State the number of significant digits in each of: 32060, 36.060, 0.32060,.032060, 3 X 10^, 3.56 X 105. 10. (a) What is meant by rounding off a number? (b) Round off the following to two significant digits: 36.7, 34.32, 37.495. 11. (a) State the rules for carrying out the following mathematical operations with approximate numbers: (i) addition and subtraction, (ii) multiplication and division. (b) Do the following: (i) 10.3575 + 9.75-8.65248. (ii) 7.935 X 2.4248 2.3. B 1. (a) Express in cm.: 15.2 m., 38 mm., 6 m. 5 cm. 4 mm. (b) Express in sq. cm.: 3 sq. m., 236 sq. mm., 6 sq. m. 5 sq. cm. 44 sq. mm. (c) Express in c.c.: 2.5 cu. m., 2300 cu. mm., 6 cu. m. 50 c.c. 465 cu. mm. 2. (a) Determine the number of inches In 950 mm., 40 cm., 10 dm. (b) Determine the number of (i) cm. in 1 foot, (ii) km. in 1 mile. (c) Which is the greater distance, 100 yd. or 100 metres? Express the difference In (i) (d) An object is at the rate of 40 miles per hour. Calculate the rate in (i) ft. per sec. (ii) metres per sec. (iii) kilometres per hour. 3. A tank is 50 cm. long, 3 dm. wide, ft. (ii) cm. travelling and 150 mm. high. (a) Calculate the area of a cover MEASUREMENT. Sec. 1:4 in sq. cm. (ii) in sq. for the tank (i) dm. (b) Calculate the volume of the tank, and express In (i) c.c. (ii) cu. dm. (iii) litres. 4. (a) What mass of water will the above tank hold (1 c.c. of water weighs 1 gm.)? Express this mass in gm., mg., eg., dg., kg. (b) Calculate grams In an ounce (ii) kilograms in a number the of (i) ton? (c) Calculate your own weight in kilograms. 5. A beaker is 20 cm. high and has a diameter of 1 4 cm. Calculate its volume in (i) litres (ii) millilitres (iii) pints. 6 Using proper \"rounding-off\u201d tech- niques, make the following calculations: (a) Add 9.75+10.357+76.92 + 5.674. (b) Subtract (i) 10.357 (ii) 5.674 (iii) 9.75 from 76.92. (c) Multiply (i) 2.6X7.93X1.732. (ii) 77.5X1.4142X.0032. (iii) 46X23.55X0.25. (d) Divide (i) 154 by.1 1. (ii) 9.5 by 19.03. (iii) 134.5 by 15. 17 CHAPTER 3 DENSITY AND SPECIFIC GRAVITY 1 gm. per c.c. G.G.S, system of units is In the F.P.S. system, however, a cu. ft. of water is found to weigh 62.5 ib. (approx.), and hence the density of water in this system of units will be expressed as 62.5 lb. per cu. ft. Densiof various subties stances are given in the table on page 21. (in gm. per c", ".c.) Fig. 3:1 Relative Densities of A\u2014 Solids, B\u2014 Liquids. To find the corresponding densities in lb. per cu. ft. the numerical values must be multiplied by 62.5. I is e.g., iron \u201cheavier\u201d ; 5 MEANING OF DENSITY In ordinary conversation we often say that one substance is \u201cheavier\u201d than than another, aluminum. Obviously we cannot mean that any given piece of iron is heavier than every piece of aluminum, but rather that for pieces of equal size, the iron would be the heavier. In science we use the term density to express the physical difference implied in the above everyday statement, that is, we say that iron has a greater density than aluminum. Density is defined as the mass of a unit volume of a substance. The method of determining density is described in chapter 5, experiments 1, 2 and 3. numerical densities measure must always be accompanied by suitable units, e.g., gm. per c.c., lb. per cu. ft., etc., according to the units; in which the mass and volume of the substance have been measured. It should further be noted that the numerical value of the density of any given substance will depend on the system of Thus, since one gram of units used. water (at its maximum density) occupies a volume of one cubic centimetre (Sec. 1:2), the density of water in the stating the In 18 DENSITY AND SPECIFIC GRAVITY Sec. 1:6 Research Scientist Determining the Density of a Substance by Comparison with Standard Density Floats Suspended in a Solution of Known Density. Canadian Industries Ltd. We have previously defined mass as the quantity of matter in a body. According to modern theory, matter is comprised of molecules, the molecules of any given substance being identical Hence, to each other (Sec. differences in density between various substances (Fig. 3:1), are due to the relative masses of their molecules as spatial arrangement. well as Variations in the density of a given substance are due to changes which vary the closeness of packing of the molecules. Ill: 2). their to I of water is : 6 DENSITY OF WATER It is frequently said that the density 1 gm. per c.c. However, solids and gases, expand liquids, when heated and contract when cooled with no change whatever in mass. They therefore at different densities have like different temperatures. Careful experiments to show changes in density of water with changes in temperature can be carried out with the aid of the dilatometer shown in Fig. 3:2. If this instrument, filled with water. 19 Chap. 3 MECHANICS is placed in a water bath with a thermometer, and ice slowly added, the volume change can be observed on the Fig. 3:2 A Dilatometer. scale, for changes of temperature down to 0\u00b0C, The water contracts, i.e., density increases, until a temperature of its scale) to (not Volume is smallest which the volume is the temperature at which the density is the greatest, namely 4\u00b0C. We say that water has its maximum density at 4\u00b0C. It is at this temperature that 1 cubic centimetre of water has the mass of 1 gram. The changes mentioned above are shown graphically 3:3. (Most liquids show a gradual increase in goes down.) temperature density Fig. the in as at 0\u00b0G. This fact, together with the fact that there is a sudden expansion as water (evidenced by those freezes burst pipes in winter!) is of profound importance in nature. With wintry conditions the coldest layers of a pond or lake are those at the surface, and when these layers freeze the ice so formed remains on the surface because its density is less than that of the water beneath it. Without this unusual behaviour of water the pond would freeze solid from the bottom upwards, greatly to the detriment of all life and to aquatic life in particular. : 7 MEANING OF SPECIFIC GRAVITY I For many purposes, instead of density, it is found more convenient to use the density of a substance relative to that of water. The relationship so obtained is called the specific gravity of It may be calculated in the substance. the following way: g ^ _ density of the substance density of water Temperature Fig. 3:3 Graph to Show the Effect of Changes of Temperature on the Density of Water. o.o. \u2014 - \u201e mass of unit volume of the substance mass of unit volume of water g g _ mass of any volume of the substance mass of an equal volume of water reached. 4\u00b0G. is further cooling makes it expand, i.e., its density The temperature at decreases After that, again. Thus specific gravity is a ratio and no It is simply a numunits are required. 20 DENSITY AND SPECIFIC GRA", "VITY Sec. 1:8 7 ber stating how many times as heavy as water, bulk for bulk, the substance is. Further, it must be evident that the number giving the specific gravity of a substance will be the same whatever the units in which the masses are measured, (Chap, 5, Exp. 4). In the C.G.S. system the density of water is 1 gm. per mb, and thus in this system, density and specific gravity are numerically equal. The two tenns are not interchangeable, however, although frequently, but wrongly, so used. : 8 DENSITY OR SPECIFIC GRAVITY I OF VARIOUS SUBSTANCES or specific gravity The density of solids, liquids and gases is a property that helps us to identify them. Some of these you will have determined experimentally while others will be found in the accompanying table of specific gravi- ties. Table of Approximate Specific Gravities 10.5 11.4 13.6 19.3 21.5 8.4-8. 8. 7-8.9 2.6 0.6 0.8 1.74 2.70 7.15 7.30 7.85 8.90 Silver Lead Mercury Gold Platinum 7.0-7.7 7.1-7.7 Brass Bronze Sand Pine Oak 0.24 0.92 2.2 0.70 0.79 0.87 Metals Magnesium Aluminum Zinc Tin Iron (pure) Copper Alloys Steel Iron (cast)^ Miscellaneous Solids Cork Ice (0\u00b0C.) Salt Liquids Gasoline Alcohol Turpentine Gases (at S.T.P.) Hydrogen Helium Air Carbon tetrachloride Sea- water Cone, sulphuric acid 1.60 1.01-1.05 1.83 0.00009 0.00018 0.00129 Oxygen Carbon dioxide Chlorine 0.00143 0.00198 0.00322 Gases, being so light, generally have their densities expressed in grams Also, air or hydrogen is used as the standard, rather than water, for per litre, instead of grams per cubic centimetre. (preferably the latter) purposes of comparison. 21 Chap. 3 i:9 MECHANICS QUESTIONS A 1. (a) Define density and state what two measurements must be made in order to calculate the density of an 5. Find the mass of 20 cu. ft. of material whose density is 3.2 Ib./cu. ft. 6. Find the volume of an object whose mass Is 42.7 lb. and whose density is 2.10 object. Ib./cu. ft. 2. (b) Calculate the density of a piece of aluminum whose volume is 150 c.c. and whose mass is 405 gm. (a) What effect does an increase in temperature have on the density of most substances? Why? (b) In what respect may water be said to be an unusual liquid? (c) Explain why this behaviour of water is important to life. (d) Why does it a much longer and more severe period of cold weather to cause a layer of ice to form on deep bodies of water than to form on shallow bodies? unusual require 3. (a) Define specific gravity and state clearly what is needed to find the specific gravity of an object. (b) Calculate the specific gravity of a substance whose volume is 20 c.c. and whose mass is 1 60 gm. 4. (a) Distinguish between density and specific gravity. (b) What relationship exists between the specific gravity of a substance and its density? (c) What is the specific gravity of the aluminum in 1. (b)? 7. Find the specific gravity of a substance whose mass is 148.5 gm. and whose volume is 30.5 c.c. 8. The specific gravity of a substance is 1.85. What volume of It weighs 1 00 gm.? Find the mass of 0.5 litres of a liquid 9. whose specific gravity is 1.6. 10. Find the specific gravity of a substance whose mass Is 3.2 lb. and whose volume is.75 cu. ft. (Density of water is 62.5 lb. per cu. ft.) 11. The specific gravity of a substance is 2.7. What volume of It will weigh 1 00 lb.? 12. Find the mass of 3.2 cu. ft. of material whose specific gravity is 8.9. 13. An irregular object has a mass of 72.6 gm. On placing it in water in a graduated cylinder the level rises from 12.0 ml. to 21.5 ml. Calculate its density. 14. A specific gravity bottle weighed 24.20 gm. when empty, 67.81 gm. when filled with turpentine and 74.20 gm. when with distilled water", ". What is the filled specific gravity of the turpentine? 15. A flask weighs 8.8 gm. when empty, 33.6 gm. when filled with water and 28.6 gm. when filled the with specific gravity of the alcohol. alcohol. Find B 1. Calculate the density of a rectangular solid 25 cm. long, 15 cm. wide, 3 cm. thick, whose mass is 5625 gm. 2. Find the mass of 75 ml. of a liquid whose density is 0.70 gm./ml. 3. Find the volume of an object whose mass is 750 gm. and whose density is 2.80 gm./c.c. 4. Find the density of water if 1 5 cu. ft. weigh 937.5 lb. 16. The density of a salt solution is 1.20 gm. per ml. If a flask weighs 1 2.6 gm. when empty and 62.8 gm. when filled with water, how much will it weigh when filled with the solution? 17. The composition of brass is 75% copper and 25% zinc by volume. Calculate its density. 1 8. An alloy of tin and lead has a specific gravity of 10.6. Calculate the proportion of tin present in the alloy (a) by volume (b) by weight. 22 2 DENSITY AND SPECIFIC GRAVITY Sec. 1:9 19. A piece of wax whose real specific gravity is 0.96 has an apparent specific gravity of 0.92 owing to a bubble of air it. The volume of the being enclosed in whole is 10.0 c.c. Find the volume of the air enclosed, assuming the weight of the air to be negligible, 20. A piece of metal 1 1.2 cm. long, 4.5 cm. wide and 1 mm. thick, has a mass of 30.2 gm. Find its density. 21. Calculate the density of a cylinder 22. Calculate the density of a sphere whose radius is 1.4 cm. and whose mass is 100 gm. 23. A cube of ice whose side is 4 cm. is allowed to melt. The volume of the water is found to be 58.2 ml. Find the density of ice. 24. Four solutions of salt, of densities 1.12, 1.17, 1.19 and 1.20 gm. per c.c., proportion are 1:2:3:4 by volume. Find the density of the together mixed the in whose length is 5.2 cm., diameter is mm., and whose mass fs 58 gm. 1 mixture. 23 CHAPTER 4 BUOYANCY force will be apparent after a consideration of the forces that water exerts on an object immersed in it (Fig. 4 : 1). The water exerts a downward pressure upon the top surface of the object, and an upward pressure upon the bottom. Because the bottom of the object is deeper in the liquid than is the top, and because pressure increases with depth, the upward pressure upon the bottom surface will exceed the downward pressure upon the top surface. There will, therefore, be a net upward pressure upon It is this upward force that the block. accounts for the apparent lightness of an object when immersed in the water. The same principle applies to all bodies immersed in any liquid or gas. Archimedes (287-212 b.g.), a Greek mathematician and inventor, was the first to study the buoyancy of liquids and to enunciate an important principle connected therewith. The story is told that Hiero, king of Syracuse, had sent his jeweller a known mass of gold to be made into a new crown. When the crown was delivered and tested it was found to have the right weight, but there was a suspicion that some silver had been substituted for gold in the interior of the Consequently Archimedes was crown. commissioned to determine whether or not the crown was pure gold, at the same time being instructed not to mar the crown in any way. Archimedes puzzled over this problem at great length. All that he had to work I : 10 INTRODUCTION TO BUOYANCY We are all familiar with the fact that things seem lighter under water. Those of us who have helped to build a dock know that a large stone which can only be raised with difficulty when out of the water may be raised quite easily when Similarly, we find that under water. Fig. 4:1 Explanation of Buoyancy. an anchor becomes heavier on emerging from the water. In general then, bodies immersed in water (or in any fluid) appear to lose some of their weight. This is, of course, due to the fluid exerting a buoyant force or lift on them. The reason why fluids exert a buoyant 24 BUOYANCY Sec. 1:11 with was the well-known fact that an object is easier to lift when immersed in water than when on", " land. medes\u2019 Principle (Chap. 5, Exp. 5, 6) should be studied carefully in conjunction with the following statements: One day as he hopped into his bath, which had been filled brimful of water, he caused a considerable quantity to overflow onto the brick floor, and he was suddenly struck by the idea that the weight he lost on submerging was equal to that of the water which overflowed. He was so excited by this discovery, so the story goes, that he ran home \u201cEureka! Eureka!\u201d (\u201cI have found it! I have found it!\u201d) As you that Archimedes enunciated, see if ypu can suggest the experiment that Archimedes probably performed to determine whether* or not the crown was pure gold. unclothed shouting, principle study the 1:11 ARCHIMEDES' PRINCIPLE The experiment to demonstrate Archi- 1. When a body is immersed in a liquid it displaces some of the liquid to make room for itself. Note that the volume of the liquid displaced is equal to the volume of the object. 2. The body is buoyed up by the liquid and therefore seems to weigh less than it does in air. 3. This loss in weight is exactly equal the weight of the liquid dis- to placed (Fig. 4:2). Archimedes\u2019 Principle, therefore, may be stated as follows: An object, when immersed in a liquid, loses in apparent weight an amount equal to the weight of the liquid displaced. or an alternative statement is: The buoyant force of a fluid (liquid or gas) upon an object immersed in it, equal to the weight of the fluid displaced. is Examples 1. A rectangular piece of metal, 15 cm. long, 6 cm. wide and 3 cm. thick, weighs 2700 gm. in air. Find its weight when immersed in water. = 2700 gm. Weight of object in air = 15 X 6 X 3 = 270 c.c. Volume of object Volume of water displaced = 270 c.c. Weight of water displaced = 270 gm. (Density of water = 1 gm./cc.) Weight of object in water = 2700 \u2014 270 (Archimedes\u2019 Principle) = 2430 gm. 25 Chap. 4 MECHANICS. 2. Find the weight of the object in example 1, if it were immersed in carbon tetrachloride (S.G. = 1.60). Weight of object in air = 2700 gm. Volume of object = 15 X 6 X 3 = 270 c.c. Volume of carbon tetrachloride displaced = 270 c.c. 1 c.c. of carbon tetrachloride weighs 1.60 gm. (S.G. = 1.60) Weight of carbon tetrachloride displaced = 270 X 1.60 = 432 gm. Weight of object in carbon tetrachloride = 2700 \u2014 432 Archimedes\u2019 Principle is used for the accurate determination of the specific gravity of solids and liquids (Chap. 5, Exp. 7-9) (Archimedes\u2019 Principle) = 2268 gm. : 12 PRINCIPLE OF FLOTATION I Archimedes\u2019 Principle applies to floating bodies as well as to those which are submerged in a fluid. When a body is floating in a liquid it appears to lose all its weight, as can be shown by lowering into water a block of wood, suspended by a thread from the hook of a spring balance (Fig. 4:3). As the wood settles more deeply into the water the reading of the spring balance decreases until, when the wood is floating, it records zero weight. The entire weight of the wood is now being supported by the upthrust of the water, this being equal to the weight of water displaced. We thus have, for floating bodies, the Principle of Flotation, which states that the weight of a floating object is equal to the weight of the fluid (liquid or gas) it displaces when floating. A fairly heavy object will of course sink lower in the liquid in which it is floating than will a lighter In either case the submerged object. displaced an amount of portion has liquid equal to the weight of the floating object. This statement is embraced in Archimedes\u2019 Principle, i.e., a body which floats has lost its whole weight. (Remember loss is an apparent one, not that this real. The pull of the earth on the floating body is still the same.) Experimental proof for this Principle is supplied in Chap. 5, Exp. 10. Fig. 4:3 Principle of Flotation. 26 BUOYANCY Examples Sec. 1:13 1. An object loaded onto a flat barge 18 feet long and 10 feet wide, causes it to settle 2 inches deeper into the water. Calculate the weight of", " the object. Volume of water displaced =z 18 X 10 X 2/12 = 30 cu. ft. Weight of water displaced = 30 X 62.5 = 1875 lb. (Density of water = 62.5 Ib./cu. ft.).'. weight of object = 1875 lb. (Principle of Flotation) 2. A plastic tray 25 cm. long, 15 cm. wide is floating on water. A lead weight whose mass is 300 gm. when placed in it causes it to sink deeper into the water. Calculate the depth to which it sinks. Let the depth to which it sinks be x cm. Volume of water displaced = 25 X 15 X x = 375x c.c. Density of water = 1 gm./c.c. Weight of water displaced = 375 x gm. Weight of water displaced = weight of floating object. (Principle of Flotation) Weight of water displaced = 300 gm. 375 x = 300 - = ^\u201c =.8 375 Depth to which the tray sinks is.8 cm. : 13 HYDROMETERS AND THEIR USES I (a) Structure displaced its own weight of the liquid, the lighter the liquid, the deeper the 5, for the Exp. (Chap. instrument The Principle of Flotation finds application in the hydrometer, which is a convenient rapid determination of the specific gravity of 12). A liquids common type of hydrometer consists of a cylindrical stem, graduated or containing a paper scale, an expansion of the stem called the float, and a bulb weighted with mercury or lead shot, the ballast, to make it float upright (Fig. 4:4). The float increases the buoyancy of the hydrometer. 11, 0.80 0.85 * 0.90 Stem 0.95 Scale 1.00N / Float The liquid whose specific gravity is to be determined is poured into a tall jar. The hydrometer is gently lowered into the liquid until it floats freely. The specific gravity of the liquid is indicated by the number on the scale which is even with the surface liquid. Since the hydrometer sinks until it has of the Weighted Bulb Fig. 4:4 Structure of a Hydrometer. hydrometer will sink in it. the largest specific gravity readings are Therefore, 27 Chap. 4 MECHANICS long and having a cross-sectional area of 1 sq. cm. It is weighted at one end with a plug of lead. On one face is marked at the bottom of the scale, and the smallest at the top. Hydrometers used lighter than water have a for liquids large float, and scale gradations starting with a specific gravity of 1.000 at the bottom. Those used for liquids heavier than water have a small float, and scale gradations starting with 1.000 at the top. A universal hydrometer has 1.000 in the centre so that it may be used for Fig. 4:6 Principle of the Hydrometer. a centimetre scale, and the rod has been rendered impervious to water by dipping in hot paraffin. Float the rod in water and note the depth to which it sinks, for example, 15 cm. Hence it displaces 15 c.c. or 15 gm. of water, and thus the weight of the hydrometer is 15 gm. Now float it in some liquid whose specific gravity is to be determined, and note the level to which it sinks, for example, 20 cm. Hence the volume of the liquid displaced is 20 c.c. and the mass of an equal volume of water is 20 gm. But note, the weight of liquid displaced is equal to the weight of the hydrometer, namely, 15 gm. That is, the mass of liquid displaced is 15 gm. and the mass of an equal volume of water is 20 gm. therefore, that the specific It follows, gravity of the liquid is 15/20 or.75, and its density is.75 gm. per c.c. (c) Uses Fig, 4:5 Hydrometers of Different Types. A\u2014 for liquids less dense than water B\u2014 universal C\u2014 for liquids denser than water both kinds of liquids (Fig. 4:5). Since few scale readings can be accommodated above and below, such an instrument has a narrow range of usefulness. (b) Principle The principle of the hydrometer can best be illustrated by means of the simple hydrometer (Fig, 4:6). This consists of a straight wooden rod about 25 cm. Hydrometers, of course, are widely used beyond the walls of the physics laboratory for testing liquids. Industries using or preparing syrups, salt solutions, acids and heavy petroleum products, 28 BUOYANCY Sec. 1:14 is used for chemicals make extensive use of them. The storage-battery hydrometer (Fig. testing the specific 4:7) gravity", " of the acid in a battery, and hence for estimating the degree to which the battery is charged. Antifreeze hydrometers are used to measure the specific gravity of antifreeze and thereby indicate how low a temperature it can stand without freezing. Similarly lactometers are used in of milk for checking possible dilution, and alcoholometers the strength of beers and wines. estimating testing used are the for Courtesy of Exide Automotive Division Fig. 4:7 Storage Battery Hydrometer. : 14 OTHER APPLICATIONS OF I BUOYANCY Archimedes\u2019 Principle and the Principle of Flotation apply wherever fluids displaced, and the following exare The Plimsoll mark is painted amidships on the ship's hull to indicate safe-loading levels under different conditions (scale: 1 14 in. = 1 ft.). 29 Chap. 4 MECHANICS amples are important these principles are: a few which show how its being hollow, because, (a) Ships Even when its hull is made of iron (density 7.9 gm. per c.c.) a ship floats on water average density is less than that of water. When it is launched it will sink to the level at which the weight of the water it displaces is equal to the entire weight of the ship, and hence the meaning of the term \u201cdisplacement\u201d. A ship like the Queen Elizabeth, of 85,000 tons displacement, displaces 85,000 tons of sea-water when afloat. To ensure that it does not sink more deeply than is safe in the water, each ship has a safe-loading line, known as the Plimsoll line (Fig. 4:8) painted on its hull. Actually there are several such lines, to allow for regional These marks and seasonal variations. indicate the depth to which the ship may be safely loaded under the different conditions. (b) Submarines A submarine is a vessel with a cylindrically shaped enclosed hull fitted with ballast tanks at bow and stern and with smaller tanks on either side amidships. In order to submerge, the submarine allows water to enter these tanks until the total weight of the boat and ballast is nearly as great as that of the water it can displace. The submarine is now in \u201cdiving trim\u201d and, by a proper use of horizontal or diving rudders, it can submerge completely and maintain its depth below the surface. In order to resurface, it forces the water out of its ballast tanks with compressed air. This action reduces its total weight until, when the weight is less than the upthrust, the ship rises. (c) Floating Docks When the hull of a great liner is to be serviced and reconditioned, a floating dock is used to lift it out of the water. 30 This dock consists, essentially, of a large flat tank divided into several compartments which can be filled with water to Fig. 4:9 Floating Dry-dock. sink the dock to a sufficient depth for the ship to be drawn into it by tugs. The water is then blown out of the -tanks by compressed air, and the dock rises, lifting the ship with it (Fig. 4:9). (d) Balloons When we apply Archimedes\u2019 Principle to air, we see that a body in air will experience a force of buoyancy equal to the weight of air it displaces. A balloon will rise if the weight of the air displaced is greater than the weight of the balloon envelope and its attachments. It will continue to rise until it reaches a level of more rarified air where the weight of the air displaced is equal to that of the balloon. The first balloons, built in 1783, employed hot air for upthrust, having open bottoms with burning braziers slung underneath them, but by August of the same year the French scientist Charles had sent up the first balloon to be filled with the newly discovered gas, hydrogen. Two years later a trip was made from France to England in such a balloon, motive power being provided by oars. Modern balloons are made of a gastight silk fabric fitted with valves. Increased height is obtained by releasing water or sand ballast, while the descent BUOYANCY Sec. 1:14 is brought about by slowly releasing the gas from the envelope. Then, since atmospheric pressure decreases with height, the gas inside the balloon expands as it rises (in accordance with Boyle\u2019s law) and thus there is danger that the balloon For this will burst at great heights. reason, in balloon ascents to high altitudes, the envelope is not filled to capacity at ground level. The greatest height reached by a manned balloon, 72,395 feet (over I 3/2 miles) was made by Stevens and Anderson, two United States Army officers, at Rapid City, South Dakota, in 1935. Their trip was made in a tightly-sealed hollow metal sphere attached to", " a huge helium-filled balloon. (e) Weather or Sounding Balloons The main use of hydrogen balloons today is to collect information about the upper atmosphere for meteorological pur- poses. These sounding balloons, as they are called, expand as they rise, until they spring a leak or burst, when the meteorological instruments which they carry are parachuted to earth. Temperature, pressure and humidity reports at various altitudes are automatically sent to weather stations by radio transmitters. Analysis of the readings have shown that sounding balloons have reached heights of 25 miles or more. (f) Airships An airship is a balloon built on a light rigid framework, propelled by air-screws and steered by rudders. The design of this type of craft developed rapidly in the early decades of this century. Although long flights were successfully made in these ships, a series of disasters resulting from the ease with which they caught fire, has caused their further development to be abandoned. 31 Chap. 4 : 15 I A MECHANICS QUESTIONS 1. (a) State Archimedes\u2019 Principle. 7. air a piece of iron whose (b) In 1 00 c.c. has a mass of volume Is 890 gm. When this is immersed in water, calculate: (i) the buoyant force of the water on it. (ii) the weight of the iron in water. 2. Describe briefly how Archimedes\u2019 Principle is used to determine the specific gravity of (a) a solid denser than water (b) a liquid. 3. (a) State the Principle of Flotation. (b) Show that it is a modification of Archimedes\u2019 Principle. (c) When a piece of wood is floated in water in a graduated cylinder the level rises from 1 5.7 ml. to 1 8.3 ml. (i) Calculate the mass of the wood. (ii) If its specific gravity is 0.60, what is its volume? 4. Explain how Archimedes could have determined whether King Hiero\u2019s crown was pure gold or a mixture of gold and silver. 5. Explain how Archimedes\u2019 Principle or the Principle of Flotation applies to each of the following statements: (a) In landing a fish, you find that it seems to weigh more when it is pulled out of the water than it does beneath the water surface. (b) As a ship in harbour is being unloaded, it slowly rises higher in the water. In order to make a submarine (c) submerge, large tanks aboard it are filled with water. (d) The same ship with the same cargo will ride higher on the Atlantic Ocean than on the Great Lakes. (e) Plimsoll lines are used on ships. (a) What is hydrometer? (b) Why are the smaller numbers of purpose the of a 6. 32 the hydrometer scale near the top? (c) Compare the size of float required for denser and less dense liquids. Explain the difference. for hydrometers used State three methods used for finding the specific gravity of a liquid. Which of the three do you think Is the most accurate? Why? B 1. 15 c.c. material weigh of (a) If 45 gm. in air, find the weight when immersed in water. 2. (b) If 3 cu. ft. of a substance weigh 350 lb. in air, find the weight when immersed in water. (a) Iron has a density of 7.8 gm. per c.c. Find the weight of 10 c.c. of it when immersed in water. (b) A substance has a density of 1 87.5 lb. per cu. ft. Find the weight when 5 cu. ft. of it are immersed in water. 3. (a) If an object weighs 140 gm. in air and 1 1 5 gm. in water, what is the volume of the water displaced? (b) If an object weighs 140 lb. in air and 115 lb. in water, what is the volume of the water displaced? 4. A piece of silver weighs 65.1 gm. in air and 58.9 gm. in water. Find its specific gravity. 5. A piece of metal weighs 500 gm. in air and 430 gm. in water, (i) What is specific gravity? (Ii) What is its volume? Its 6. A 15 lb. weight weighs only 9 lb. in (ii) Find water, (i) Find its specific gravity, Its volume, (iii) Find its density. 7. A piece of metal weighing 1 20.4 gm. has a volume of 14.5 c.c. (i) What will it weigh In water? (Ii) Find also the density of the metal. 8. An object weighs 42.2 gm. in air, 29.4 gm.", " in water and 25.6 gm. in a liquid.. BUOYANCY Sec. 1 : 15 9 Calculate the specific gravity of the liquid. A metal weighs 56.3 gm. in air, 45.8 gm. when immersed in water and 48.6 gm. when immersed in a liquid. Calculate the specific gravity of (i) the metal (ii) the liquid. 10. A piece of metal weighs 138.8 gm. in air, 123.2 gm. in water and 125.7 gm. in a liquid. Find the specific gravity of the metal and of the liquid. 11. An object weighs 42.2 gm. in air and 29.4 gm. in water. How much will it weigh in a liquid of density 0.80 gm. per c.c.? 12. Was King Hiero\u2019s crown made of pure gold if in air it weighed 1500 gm., and when immersed in water it weighed 1 400 gm.? 13. A rectangular block of soap 8 cm. long and 6 cm. wide floats in water with 2.5 cm. of its thickness submerged. Calculate the mass of the soap. 14. A cube of wood, side 50 cm., floats in water with its base horizontal and 6 cm. of its height above the surface. Find its density. 15. A wooden raft 5 ft. long and 4 ft. wide floats in water. When a person steps on the raft it sinks 1.5 inches deeper into the water. Calculate the person\u2019s weight. 16. A cork of volume 60 c.c. and density 0.24 gm. per c.c. floats in a liquid of density 0.85 gm. per c.c. Find the least weight required to sink it. 17. What volume of lead of density 11.2 gm. per c.c. will be required to sink a piece of wood in water, the weight of the wood being 425 gm. and its volume 556 C.C.? 18. A block of wood of volume 100 c.c. floats in a liquid of specific gravity 1.2 with 75 c.c. immersed. Calculate the density of the wood. 19. A wooden hydrometer sinks in water to a depth of 1 8 cm. and in a liquid to a depth of 1 4 cm. What is the specific gravity of the liquid? 20. A hydrometer sinks in water to a depth of 1 5 cm. How far would it sink in a liquid whose specific gravity Is 0.80? 21. A hydrometer sinks to a depth of 1 2 cm. in a liquid whose specific gravity is 1.7. To what depth would it sink in water? 22. A piece of wood whose volume is 1 50 c.c. floats with of its volume submerged in water. Find its mass. 23. A piece of wood whose mass is 75.0 gm. floats in water with ^ of its volume above the surface. Find its volume. 24. A piece of cork of density 0.25 gm. per c.c. floats in a liquid of density 1.2 gm. per c.c. What proportion of the volume of the cork will be immersed? 25. An object floats in water with half its volume submerged. How much will be submerged when it floats in a liquid of specific gravity 1.5? 26. To what depth will a block of wood 20 cm. high and of density 0.63 gm. per c.c. sink in a liquid of density 0.90 gm. per c.c.? \u2022 33 CHAPTER 5 EXPERIMENTS ON MECHANICS INTRODUCTION Before proceeding with the following experiments the student should review the following techniques used in measurement. A. \u2014 Use of the ruler 1. Avoid using the ends of the ruler. 2. Place your eye directly above the point where the reading is to be taken to avoid the error due to parallax. B / Measuring Length with a Fig. 5:1 Ruler\u2014How to Avoid the Error Due to Parallax. Fig. 5:2 Measuring Volume with a Graduated Cylinder, B, \u2014 Use of the graduated cylinder Place your eye directly opposite the centre of the meniscus curve. Take the reading at this level. C. \u2014 Use of the balance 1. Clean and level the balance. 2. Support the beam of the balance on the knife-edge as demonstrated by the instructor. 3. Adjust for zero reading with all weights removed. Note that in all readings the pointer should swing an equal number of divisions on either side of the zero mark on the scale. Do not wait for the pointer to come to rest. 4. Place object to be weighed on the centre of the left pan of the balance. 5. Commencing with a weight that is definitely too heavy on the right-", " 34 EXPERIMENTS ON MECHANICS A B Fisher Scientific Co. Canadian laboratory Supplies Ltd. Fig. 5:3 The Balances A\u2014Triple-beam balance. B\u2014 Equal-arm balance and box of weights. hand pan of the equal-arm balance (or on the arm of the triple-beam balance), systematically reduce the weight until balance is attained. 6. Total the weights used. 7. Disengage the knife-edge and return all weights to their box or to their zero position. EXPERIMENT 1 To determine the density of a regular solid. (Ref. Sec. 1:5) Apparatus Rectangular solid, ruler (graduated in mm.), balance. 1. Carefully measure to the nearest millimetre the length, width and 35 Chap. 5 MECHANICS thickness of a rectangular solid.* Record your measurements, and calculate the volume of the object in cubic centimetres. 2. Determine the mass of the object and record it in grams. Observations = Length = Width Thickness = Mass =: cm. cm. cm. gm. Calculations 1. What is the volume of the object? 2. Determine the mass of one cubic centimetre. 3. What is the average of the results obtained by the class? Conclusion What is the density of this material? Questions 1. What is the correct value for the density of this material? (Table P-21). 2. Express the difference between the class average and the true value as a percentage of the true value. This is a measure of the experimental error. 3. Suggest sources of experimental error. * Note The density of other regular solids, such as a cylinder (Volume = 77 h), or a sphere (Volume = - tt 4 3 may be determined in a similar way (Fig. 5:4). EXPERIMENT 2 To determine the density of an irregular solid (rubber stopper), (Ref. Sec, 1:5) 36 EXPERIMENTS ON MECHANICS Apparatus Rubber stopper, thread, graduated cylinder, water, balance. Method 1. Determine the mass of the object and record it in grams. 2. Half fill a graduated cylinder with water. Note and record the Tie a thread to the object and carefully volume of the water. is completely immersed. Note and lower it into the water until it record the final volume of the water and object. Determine the volume of the object and record it in cubic centimetres. Observations Mass of object Initial volume of water Final volume of water and object.'. Volume of object'= = = rr gm. c.c. c.c. c.c. Conclusion What is the density of this material? Question Why should the object be weighed before its volume is determined? EXPERIMENT 3 To determine the density of a liquid by measurement, (Ref. Sec. 1:5) Apparatus Beaker, graduated cylinder, balance, liquid (water, alcohol, etc.). Method 1. Determine the mass of a clean dry beaker. 2. Add about 50 c.c.^of the liquid to the beaker, weigh, and hence determine the mass of the liquid used. 3. Pour the liquid into a graduated cylinder and determine its volume. Observations = Mass of beaker Mass of beaker plus liquid = =.'. Mass of liquid Volume of liquid gm. gm. gm. c.c. Calculations Determine the mass of one cubic centimetre. What is the average of the results obtained by the class? Conclusion What is the density of this liquid? 37 Chap. 5 MECHANICS Questions 1. What is the correct value for the density of this liquid? (Table p. 21) 2. Calculate the percentage error. 3. Suggest sources of experimental error. 4. If the balance at your disposal is suitable, a graduated cylinder, instead of a beaker, could be used in steps 1 and 2 in the above experiment. Why should this tend to reduce the experimental error? EXPERIMENT 4 To determine the specific gravity of a liquid by means of the specific gravity bottle. (Ref. Sec. 1:7) Apparatus Specific-gravity bottle, water, balance, liquid (carbon tetrachloride, alcohol, etc.) X Fig. 5:6 Method 1. Carefully clean and dry the bottle. Determine its mass. 2. Fill completely with the liquid whose specific gravity is to be Insert the stopper, wipe off any excess liquid that determined. exudes through the opening. Determine the mass of the bottle plus the liquid. 3. Pour out the liquid. Rinse out the bottle with water. Fill the bottle completely with water. Determine the mass of the bottle plus the water. 38 EXPERIMENTS ON MECHANICS Observations = Mass of specific-gravity bottle empty Mass of specific-gravity bottle full of liquid", " = = Mass of specific-gravity bottle full of water = =.\u2019. Mass of water Mass of liquid gm. gm. gm. gm. gm. Conclusion What is the specific gravity of the liquid? Questions 1. Define specific gravity. 2. Calculate the percentage error. 3. Suggest sources of experimental error. 4. What is the purpose of the hole through the centre of the stopper of the specific-gravity bottle? EXPERIMENT 5 To demonstrate Archimedes' Principle. (Ref. Sec. 1:11) Apparatus Balance, bucket and cylinder apparatus, beaker, water. Method 1. Hook the cylinder A on the bottom of the bucket B. Suspend them from the hook on the balance. Adjust the weights until the balance is in equilibrium. 2. Completely immerse the cylinder in a beaker of water. Be sure that the cylinder does not touch the bottom or sides of the beaker. Note the effect on the equilibrium. 3. Carefully add water to the bucket until it is completely full. Again note the effect on the equilibrium. 39 Chap. 5 MECHANICS Observations 1. What was observed when the cylinder was completely immersed in the water? 2. What was observed when the bucket was filled with water? Conclusions 1. Why was the equilibrium disturbed in step 2? 2. Why was the equilibrium restored in step 3? 3. State Archimedes\u2019 Principle. Questions 1. Why do objects apparently weigh less when immersed in a liquid? 2. What would be the effect of immersing the object in a denser liquid? EXPERIMENT 6 Alternative method to demonstrate Archimedes' Principle. (Ref. Sec. 1:11) Apparatus Balance, object (glass stopper), beaker, overflow can, catch bucket, water, several other liquids (alcohol, carbon tetrachloride, brine, etc.) Method 1. Suspend the object from the hook on the balance. Determine its mass in air. 40 EXPERIMENTS ON MECHANICS 2. Completely immerse the object in a beaker of water. Be sure that it does not touch the beaker. Weigh the object while immersed in water. 3. Weigh a dry empty catch bucket. 4. Completely fill an overflow can with water. Let any excess water flow freely from the spout and discard it. Do not disturb the overflow can. Place the catch bucket under the spout. Carefully lower the object into the water and catch all of the overflow in the bucket. Weigh the catch bucket and overflow water. 5. Repeat the above using other liquids and fill in the table below. Observations Liquid Used Water Alcohol Carbon TETRAC H LORIDE Weight of object in air Weight of object in liquid.'. Apparent loss of weight in liquid Weight of empty catch bucket Weight of bucket plus displaced liquid.*. Weight of displaced liquid Conclusions 1. How does the apparent loss of weight compare with the weight of liquid displaced? 2. State Archimedes\u2019 Principle. Questions 1. Why is a glass stopper an ideal solid to use in the above experiment? 2. What type of solid must be avoided? EXPERIMENT 7 To determine the specific gravity of a solid which is more dense than water using Archimedes' Principle. (Ref. Sec. 1:11) Apparatus Balance, beaker, water, thread, several solid objects more dense than water. Fig. 5:9 41 ). Chap. 5 MECHANICS Method 1. Suspend a solid object by a thread from the hook on the balance. Determine its weight in air. 2. Completely immerse the object in a beaker of water. Be sure that it Determine the weight of the object does not touch the beaker. when immersed in water. 3. Repeat the above weighings for other objects supplied and fill in the table below. Observations Object Weight in Air Weight in Water Apparent Loss IN Weight 1. 2. 3. Calculations Weight of object in air = Apparent loss in weight = Weight of water displaced = gm. gm. gm. (Archimedes\u2019 Principle) Specific gravity =.\u2019. Specific gravity = Weight of object in air Weight of equal volume of water Conclusion What is the specific gravity of the object? Questions 1. Calculate the percentage error. 2. Suggest sources of experimental error. 3. Explain how the weight of an equal volume of water was deter- mined. 4. State the density of the above objects in the metric system of units. 5. Discuss advantages and disadvantages of this method for finding density compared with the method used formerly (Exp. 2) EXPERIMENT 8 To determine the specific gravity of a liquid using Archimedes' Principle. (Ref. Sec. 1 : 1 1 Apparatus Balance, glass stopper, thread, several", " beakers, water, liquid (alcohol, carbon tetrachloride, etc.). Method 1. Suspend the object by a thread from the hook on the balance. Determine its weight in air. 42 EXPERIMENTS ON MECHANICS 2. Immerse the object in a beaker of water, being careful not to let it touch the beaker. Determine the weight of the object when immersed in water. 3. Rinse the object in a reserve supply of the liquid whose specific gravity is to be determined. Immerse the object in a beaker of this liquid and again weigh. 4. Repeat for other liquids supplied. Weight of Object in Air Weight of Object in Water Weight of Object in Liquid Observations Liquid Used Alcohol Carbon tetrachloride Calculations Mass of water displaced = Mass of liquid displaced = Calculate the specific gravity of this liquid. gm. gm. Conclusion What is the specific gravity of the liquid used? Questions 1. Calculate the percentage error. 2. Suggest sources of error. 3. Explain how the mass of water displaced or of liquid displaced was obtained in the above experiment. 4. Why was it correct to say that the volumes of water displaced and of liquid displaced were equal? EXPERIMENT 9 To demonstrate the Principle of Flotation. (Ref. Sec. 1:12) Apparatus Balance, paraffined wooden block, overflow can and catch bucket, water, other liquids. 43 Chap. 5 MECHANICS Method 1. Place the overflow can on the pan of the balance. Fill the overflow can to the spout with water (Exp. 6). Balance it. 2. Without adjusting the weights, and after placing the catch bucket under the spout of the overflow can, carefully lower the wooden block into the water. Let it float freely being careful not to let it touch the sides of the can. Note all changes that occur until the water ceases to flow. 3. Repeat this experiment using the other liquids provided. Observations Describe the changes in equilibrium that occurred. Conclusions 1. How does the mass of the floating block compare with the mass of liquid it displaces? 2. State the Principle of Flotation. Questions 1. Why was the block of wood used in the above experiment coated with a thin film of paraffin? 2. Why does a steel ship float? EXPERIMENT 10 To show the principle of the hydrometer, (Ref. Sec. 1:13) Apparatus Simple hydrometer (Fig. 4:6), two tall cylindrical vessels, water, other liquids. Method 1. Float the hydrometer in a cylinder of water, being careful that it does not touch the sides. Note the depth to which it sinks in the water. Calculate the mass of the hydrometer. 2. Rinse the hydrometer in a reserve supply of the liquid to be used. Then float the hydrometer in a cylinder of the liquid, again being careful not to let it touch the sides. Record the depth to which it sinks in the liquid. 3. Repeat part 2 for other liquids. Observations = 1. Depth to which hydrometer sinks in water 2. Depth to which hydrometer sinks in the liquid = Calculations Calculate the specific gravity of the liquid. 44. EXPERIMENTS ON MECHANICS Conclusions 1. What is the specific gravity of the liquid? 2. On what principle does the use of the hydrometer depend? Questions 1. How do you find the mass of the hydrometer? 2. How do you find the mass of the liquid displaced? 3. How is the depth that the hydrometer sinks related to the specific gravity of the liquid? 4. What is the use of the hydrometer? EXPERIMENT 11 To determine the specific gravity of a liquid using a hydrometer, (Ref. Sec. 1:13) Apparatus Several tall cylindrical vessels, several liquids (brine, alcohol, etc.), three hydrometers (one for heavy liquids, one for light liquids, one universal) Method 1. Float an appropriate hydrometer in a cylinder containing the liquid whose specific gravity is to be determined. Be careful not to let it touch the sides of the vessel. Determine the specific gravity by reading the hydrometer scale at the liquid surface level. 2. Repeat for other liquids. Note This hydrometer method can be used to check the specific gravities of liquids obtained in previous experiments. Observations Li^^uid Specific Gravity Conclusion State the specific gravity of each liquid used. Questions 1. Describe the construction of a hydrometer. 2. Why is the lower end of the hydrometer weighted (with mercury or lead shot)? 3. Is the flotation bulb on a hydrometer for low-density liquids larger or smaller than the bulb on a hydrometer for high-density liquids? Explain. 4. Why is the", " hydrometer scale graduated with the smallest readings at the top and the largest at the bottom? Explain. 45 UNIT II SOUND Describe the different ways in which the sound of this depth charge would be heard below and above the water, and why. star Newspaper Service CHAPTER 6 PRODUCTION AND TRANSMISSION OF SOUND 11:1 INTRODUCTION From earliest childhood our ears grow. accustomed to sounds about us : first the sound of our mother\u2019s voice, then the : sounds of home, of nature, and the busy world. Our consideration of their nature rarely goes further than calling the sounds we dislike noises, and some of the more pleasant ones music. These, however, are often subjective definitions, 1 as can be seen from the fact that a \u201chot-rodder\u201d may drive for miles in his unmuffled car and think the sound it is making is \u201cmusic\u201d, while he will hurriedly turn off the radio because of the \u201cnoise\u201d Beethoven is making. I I I [ I I Man, in fact, has been interested in inventing devices for making music and noises for much longer than in investigating the nature of sound. References to musical instruments in the Old Testament date back to 4000 b.c. Remember I the story of Joshua and the walls of! Jericho. Yet, although Aristotle and the early philosophers knew something of the physical nature of sound, it is only in the last four or five hundred years that a fuller understanding of it has been gained. [ j i The word sound has been used frequently already, but no effort has so far been made to define it. One definiit the sensation that results tion calls when the auditory nerve is stimulated, while another refers physical the to causes of this sensation, in terms of the three necessary agencies for any sound: a source, a medium and a receiver. It is this second definition of sound which will be our concern in this unit. 11:2 THE ORIGIN OF SOUND When we ring a bell, bow a string, or strike a tuning-fork, and bring each into contact with a light object such as a pith ball, the object moves away as if being struck regularly (Fig. 6:1). Some demonstrators may prefer to touch the sounding body to some water and note Fig. 6:1 Sounding Bodies Vibrate. the splash and waves set up. This is ample proof that the sounding body is vibrating, i.e., moving to and fro. It may be concluded that sound always 49 Chap. 6 SOUND originates in a rapidly vibrating body. of Only sufficiently rapid vibrations cause sound, but some vibrations, whether or not they emit sound, have other effects, such as the destruction buildings, bridges, or parts of moving machinery. To take one instance, the vibration set up by a body of troops marching would be sufficient to destroy some bridges and so troops marching across them have to break of vibratory motion is one of great importance. Obviously the study step. 11:3 A STUDY OF VIBRATORY MOTION (a) The Pendulum and Transverse Vibrations A simple device for demonstrating vibrations is the pendulum (Chap. 10, Exp. 1 ). This consists of a weight, called the bob, attached to the free end of a vertical cord, the other end of which is securely attached to a support (Fig. / \\ \\ \\ \\\u2014V- \\ \\ / / / / / / / / Simple Pendulum in the Mean Position Amplitude of Vibration y- Movement of the Bob during one Vibration (Cycle) Fig. 6:2 Transverse Vibrations. 6:2). When the weight is drawn aside and allowed to swing, it will be -\u201cseen the pendulum swings back \u201cand that forth in a regular manner and that it 50 have moves about the same distance on either side of the point of rest or mean position. The advantage of using a pendulum to study vibrations is that the action is slow enough to be noted in every detail. For example, if we draw the bob aside and let it move to and fro once pendulum will performed the Most objects one vibration or cycle. would vibrate too fast to allow this to be seen. The horizontal displacement that the bob undergoes on either side of its natural rest position is the amplitude of the vibration. This is the destructive element of vibratory motion that affects rigid and brittle components of buildings, bridges, machinery and the like. Moreover, we shall see later (Sec. 11:11) that it affects the loudness of sound. To the movement of the pendulum, which is obviously at right angles to the mean position, is ascribed the term transverse vibration. Although there are other kinds of vibrations, you will recognize this kind in some of the objects used to demonstrate the origin of sound. Counting the number of vibrations in a given time, say 30 seconds, enables us to", "e the same period. It is for this reason that the pendulum is used in For that physics as a timing device. reason also, it is the primary component of large clocks. Were you to experiment with pendulums of different lengths, the period of a short one would be less than It is for this reason that of a long one. that you shorten the pendulum of a frequently ; V PRODUCTION AND TRANSMISSION OF SOUND Sec. II:4 clock which loses time and lengthen it for one that gains. Another demonstration may be arranged using a long rnetal rod, clamped (b) LoyigitiuUnal Vibrations A coil spring with a weight attached is supported vertically from a strong support as in Fig. 6:3. When it is vibrated its pith centre, having a at in contact with one end as in Fig. 6:4. When the half of the rod farthest from the pith ball is stroked with a chamois shrill sound is coated with resin, ball a emitted. Meanwhile the pith ball is displaced in the direction of the axis of the rod, thus showing longitudinal vibrations, although these are too rapid to be seen in detail. These examples indiare cate characterized by motion back and forth along the length of the vibrating body. longitudinal vibrations that 11:4 MEDIA FOR THE TRANSMISSION OF SOUND The need for a material medium for the transmission of sound is studied in Fig. 6:3 Longitudinal Vibrations. so that the weight moves vertically by alternately stretching and compressing the spring, longitudinal vibrations result. Fortunately vibrations slow the are Fig. 6:4 Longitudinal Vibrations. enough to enable us to observe all the details as in the pendulum (Chap. 10, Exp. 2). Chap. 10, Exp. 3. There it will be found that as the air is removed from the belljar in which there is a vibrating bell (Fig. 6:5), the sound gradually becomes fainter. The bell is heard again when air is reintroduced into the bell- jar. Since Of THE UNIVERSITY Of ALBERTA 51 Chap. 6 SOUND from or towards the point of origin of the waves. Again, if part of our rope were chalk-marked, this part would be seen as a white line perpendicular to the length of the rope when the latter was vibrated at one end to set up a train of Electromagnetic waves waves along it. (Sec. IV : 38), which include light waves, are also transverse in character. Fig. 6:6 shows the displacements of the particles of a medium transmitting a transverse wave. Particles at the crests (B, F, etc.) of the wave are undergoing a maximum displacement upwards, those at the bottom of the troughs (D, H, etc.) a maximum displacement downwards. Some terms used in wave motion follow: Amplitude is the maximum displacement from the mean position (BBi or DDi, etc.). Phase. Particles at the same distance from their mean positions and which are moving in the same direction are said to be in the same phase. Thus particles P and Q are in the same phase, as also are particles B and F. On the other hand, particles B and D are completely out of phase. Wave-Length is the distance, (/) usually expressed in centimetres or inches, between two consecutive particles in the same phase. Thus the distance BF between two adjacent crests, or the distance DH from one trough to the next, gives the wave-length of the disturbance. This we could see the vibrating bell throughout the entire experiment but could not hear the sound when there was no air present, we conclude that sound, unlike light, cannot travel through a vacuum. Sounds can also be conveyed through liquids and most solids. Thus the noise of the engines of a submarine can be picked up by underwater microphones, and the sound of vibrating telegraph wires can be clearly heard by an ear pressed against a telegraph post. 11:5 WAVE MOTION (a) Transverse Waves The disturbances set up by a vibrating body are propagated in the form of waves in the medium (Chap. 10, Exp. 4). A wave may be defined as a disturbance of any kind which travels without change of form and without the medium moving bodily with it. Simple examples of waves are to be seen when one end of a taut rope is jerked or when a stone is thrown into a pond and makes ripples on the surface. Waves produced in this way are known as transverse waves, since the disturbances in the medium are perpendicular or transverse to the direction of propagation of the waves. The ripples on water can be seen to travel outwards from the centre of the disturbance, but a cork floating on the surface will execute an up-anddown motion without moving away 52 PRODUCTION AND TRANSMISSION OF SOUND Sec. II:", " 5 represents the distance that the motion has travelled during the execution of one com]3lete vibration. Period of a \\ ibratioii is the time of one complete vibration, or is the time taken by a particle in travelling from its mean position through the maximum displacement first in one direction and then in the other, finally returning to its mean position. Frequency (n) is the number of vibra- tions in one second. Wave-Train is a succession of waves caused by continuous vibration of the source. DISPLACEMENT OF END A. % Vibration Vi Vibration Ff V* Vibration n 1 Vibration DISTANCE TRAVELLED BY DISTURBANCE IN CORD AB g '/4 Wave-length Vi Wave-length Ti Wave-length 1 Wave-length Fig. 6:7 Proving that V = nl. Velocity (F) is the distance covered in a unit of time (a second). Since during one vibration the disturbance travels I cm. (Fig. 6:7), then during n vibrations, the disturbance travels nl cm. (n wave-lengths). Now, (one wave-length) if n the frequency is vibrations per second, the disturbance travels nl cm. in one second, i.e., the velocity is nl cm. per second. This gives us the wave formula found to be equally useful in all branches of physics. Velocity = frequency X wave-length or V = nl. R I -C.1 R 53 Chap. 6 SOUND audible in all directions, each wave-front, i.e., the leading edge of each wave, must be spherical. How the air responds to the vibrating bell may be illustrated with the aid of the apparatus shown in Fig. 6:10. The Fig. 6:10 Illustrating Longitudinal Wave Motion. steel balls are suspended so that they just touch. When the first one is drawn aside and allowed to hit the ball next in line, none moves except the one at the opIt flies out about as far as posite end. the first was drawn aside. Because steel is elastic, the impact is passed through Such waves (b) Longitudinal Waves Sound waves differ from those described above in that the particles of the medium are displaced from their mean positions backwards and forwards along the line of travel of the wave are known as motion. longitudinal waves, and may be illustrated by reference to Fig. 6:8. There we have a coil spring stretched between two supports (S, Si). A piece of cloth is tied near its centre. When several coils of the spring are squeezed together a compression, formed. When the coils are released, their elasticity causes them to return to their normal position. The momentum so produced causes them to move past this position, thereby forming a stretched region or rarefaction (R). or condensation (C) is A study of the jerking of the cloth to and fro along the length of the spring will give ample proof of what is happening. It will now be good practice to draw a longitudinal wave-train in a coil spring and label amplitude, particles in the same phase, and one wave-length. Longitudinal waves are always characterized by condensations and rarefactions. Rarefoction Sound Waves from a Vibrating Tuning-fork. the line as a compression wave or condensation followed by an expansion or rarefaction. This is similar to the way in which sound waves are transmitted through air. Ii:6 THE SUPERPOSITION OF WAVES (a) Interference Interesting effects are observed when Fig. Sound waves from a vibrating bell are 6:9. The transverse depicted in vibrations of the gong give rise to condensations and rarefactions alternately, i.e., longitudinal waves. As the sound is 54 PRODUCTION AND TRANSMISSION OF SOUND Sec. II: 6 two waves are simultaneously propagated through the same medium. The resulting displacement at any point of the medium is the algebraic sum of the displacements produced by the two separate waves. When these are in the same direction the effects are thus reinforced, and when In the opposed they are diminished. special case where the two waves are of the same frequency and amplitude, as in Fig. 6:11 (a), each wave will assist the other at all points when die two waves crests and phase, are! troughs of the two waves exactly coinciding throughout the medium. If, however, they are completely out of phase * as in Fig. 6: 11(b) the two waves are i in exact opposition at all points, the crests of one coinciding with the troughs of the other, and accordingly the result- exactly the in I ; j I ing effect in the medium is nil. These phenomena are referred to under the general heading of interference. Special cases of particular interest to us in our study of sound will be presented in Sec. 11:24. (b) Standing Waves A", " very important case of interference is seen when two trains of waves of the same frequency and amplitude travel in opposite directions through a medium, for example, original and reflected waves (Chap. 10, Exp. 5). To demonstrate this, attach a light flexible silk cord to one prong of a large tuning-fork. Pass the other end of the cord over a pulley and attach a weight to it. The tuningfork should be activated by an electromagnet to give continuous vibration. In (b) Two Waves In Opposite Phase They Nullify Each Other Fig. 6:11 Production of Standing Waves. 55 Chap. 6 SOUND Vi Period After (a) '/2 Period After (a) Vi Period After (o) ^ Period After (o) A combination of (a) (b) (c) (d) (e) Fig. 6:12 Superposition of Waves. 56 PRODUCTION AND TRANSMISSION OF SOUND Sec. II to place of the tuning-fork an electric bell with gong removed may be used, the cord being tied the clapper, or a special vibrator as shown in Fig. 6:11 (c) may be employed. For reasons to be discussed later, the length of the cord is adjusted until it takes the form shown in the figure. The hazy oval regions where displacement of the cord is greatest are called loops. The points of quiet tlie ends where reflection as well It might be occurs are called nodes. imagined that a loop would occur where the cord meets the vibrator but, in reality, the amplitude of the vibrator is so small compared to that of the cord that it must be considered to be a node. In any case some reflection occurs there, as depends The reason for adjusting the length of to be the cord is rather too difficult explained here. Flowever, this much can be said, that the wave-length of the disturbance tension (caused by the weight) and the frequency of the source. Since each loop is half a wave-length and there must be a whole number of loops in the cord, the length must be adjusted to accommodate a whole number of loops. the on An examination of Fig. 6:12 will help us to understand the phenomenon. The wave composed of dashes is proceeding to the right, the dotted one to the left. They have the same amplitude and wave-length. The solid line is the resultant of the former two. At the start let us assume that the waves are completely out of phase as in (a). We know that the result will be Onea line of undisturbed particles. quarter of a period later (b), each will have shifted one-quarter wave-length but in opposite directions. -The waves will be in phase now and will reinforce each the resultant other. has wider amplitude than either of the original waves. Diagrams (c), (d), and (e), may be explained as above except For this reason, that the phase is different after each quarter vibration. The combined effect is shown in (f). 4, 8, Examination of the diagrams reveals 10 which are that points 2, 6, one-half a wave-length apart are always at rest and hence constitute the nodes. Points 1, 3, 5, 7, 9 move from rest to a point of maximum displacement on one side then back through the point of rest to a point of maximum displacement on the other and return. These are the loops. Thus standing waves consist of nodes and loops. The distance between successive nodes or is one-half a wave-length. These waves will be useful in understanding vibrations in strings and air columns which will be presented in succeeding chapters. loops 11:7 REFLECTION OF SOUND WAVES Sound waves travelling through the air and striking a smooth hard surface undergo reflection, obeying the same laws of reflection as light waves. That is so can be demonstrated using this the apparatus shown in Fig. 6:13. Sound A B Fig. 6:13 Reflection of Sound. waves from a source S (a watch) are directed by a tube to a hard surface, AB (a drawing-board is suitable for the purpose). A receiver, R (the ear), is 57 Chap. 6 SOUND Canadian National Exhibition Band Shell. Canadian National Exhibition placed at the end of a second tube which, to detect the signals, must be inclined to AB at the same angle as the first In short, the angle of reflection tube. equals the angle of incidence (i). (r) Also, the incident sound (SO), the perpendicular (CO) and the reflected sound The (RO) are in screen CD acts as a shield to protect R from the sound waves transmitted directly from S. the same plane. A very interesting demonstration of the reflection of sound is found in the Museum of Science and Industry at Chicago. Two concave mirrors are arranged a long distance apart. A person standing at the focus of one may whisper softly and the words will be clearly audible to a person standing at", " the focus of the other. Echoes are due to reflection. They are produced as the result of a sound 58 PRODUCTION AND TRANSMISSION OF SOUND Sec. II: 8 all etc. will sides, signal directed towards a distant ob(e.g., a wall) being returned to stacle the listener, who thus hears a repetition of the sound a short period after it has been produced. Forests, clifTs, hillreflect sound and cause the formation of echoes. If there are a number of reflecting surfaces at different distances from the source, a series of repetitions of the sound signal, each following the other, will be received. These are known as multiple echoes, and they are heard, for example, when an Alpine horn is blown amidst a number of mountain peaks. The reflection of sounds may have unpleasant results in auditoria that are not properly constructed. The sounds echo and re-echo from the walls so that the effects of one sound have not died The away before the next is made. resulting jumble of sounds is known as reverberation. Further reference of sound will be made in the next section and in section 11:32. reflection to 11:8 THE VELOCITY OF SOUND (a) Methods of Measuring 1. By Direct Measurement The early experiments to determine the velocity of sound in air, made in the late seventeenth and early eighteenth centuries, were based on estimating the difference in time for the light and the of sound to travel to an observer from a cannon fired some distance away (Fig. 6:15). Since the speed of light is very great (186,000 miles per second), the the explosion is seen almost flash instantaneously. Hence the time elapsing between an observer seeing the flash and hearing the report of the explosion may be taken as the time needed for the sound to travel the measured distance between the cannon and observer, and this enables the speed of sound to be calculated. There are two main objections to these simple \u201cflash-bang\u201d experiments. If there is a wind the result will be greater or less than the true value according to whether the wind is blowing from the cannon to the observer or in the opposite direction. The other chief source of error is the \u201creflex time\u201d of the observer. Thus, if a stop-watch is used in such an experiment there is a difference between the response time in seeing the flash and the starting report and stopping it. This error varies with different observers, and with the same individual from time to time, and also even with the loudness of the sound. If the cannon is one mile from the obtake about 5 server seconds to travel the distance, and hence a personal error of 1/5 second will introduce an error of four per cent in the the watch, and hearing the sound will final result. 59 Chap. 6 SOUND The accepted values for the velocity of sound in air are tabulated below. It will be noted that for a change of 1 centigrade is a corresponding change in the velocity of sound in air of 2 feet per second or 0.6 metres per second. degree, there Temperature 0\u00b0C. 10\u00b0C. 20\u00b0C. Velocity Ft. per Sec. Velocity Metres PER Sec. 1089 1109 1129 332 338 344 The first accurate determination of the speed of sound was carried out by 1738. Two the French Academy in cannons were used, separated by a distance of about 18 miles to reduce the error due to the \u201creflex time\u201d. To eliminate the effect of wind, timings were taken (by means of pendulums) in both directions and the average used to calculate the speed of sound. The result obtained was 337 metres per sec. at 6\u00b0G., or 332 metres per sec. at 0\u00b0C. Later experiments, carried out in the same way, but using chronometers, accurate to one-tenth of a second, gave a mean result of 331 metres per sec. at 0\u00b0C. 2. By Echoes If a sound signal is directed to a distant wall or obstacle an echo will be received some time later, the sound wave having travelled twice the distance between the wall and the observer during This suggests a possible the period. method of measuring the velocity of sound. For clear echoes, however, the wall (or reflecting surface) should not be too far distant, and this involves the difficulty of measuring very practical short time-intervals. With a wall 100 yards distant, for example, the echo of a sound signal will be received only about one-half second later. This difficulty has been overcome by using a metronome or 60 an electrically controlled tapper to send out a sequence of sound signals at regular intervals. The distance of the metronome from the wall is adjusted until the echo of one click is heard simultaneously with the next click. The sound signal has clearly travelled to and from the wall during the interval between the clicks. This time-interval is accurately known, and so the velocity of", " sound can be It should be noted that since found. the sound signal has to travel back along its track this method automatically eliminates the effect of the wind. Example A metronome was 280 feet from the interval between wall and the time clicks was 1/2 second. Distance covered in 1 /2 sec. was 560 feet In 1/2 sec. sound has travelled 560 feet In 1 sec. sound has travelled 1/2 \u2018.. the velocity of sound ^ = 1120 feet per sec. 3. By Resonance In section 11:22, and experiment 10, chapter 10, we shall study resonance experimentally, and shall use the results in the formula V \u2014 nl to determine the velocity of sound. (h) Breaking the Sound Barrier In recent years breaking the sound barrier, i.e., flying at the speed of sound (742.5 miles per hour at 0\u00b0C.) and more, has become a great test both of the skill of operation and the construction of aircraft. When an advancing craft catches up with its own sound waves a giant barrier of compressed molecules of air must be penetrated, and this requires the power of jet-type engines and an extremely strong construction of the aircraft. This situation PRODUCTION AND TRANSMISSION OF SOUND Sec. II: 9 presents just one of the many practical problems which can only be dealt with because of our modern grasp of the nature of sound. The Velocity of Sound in Various Media Medium Velocity Gases feet per sec. metres per sec. Carbon Dioxide (0\u00b0C.) Oxygen (0\u00b0C.) Air (0\u00b0C.) Hydrogen (0\u00b0C.) Liquids Water (9\u00b0C.) Sea-Water (9\u00b0G.) Solids Brass Oak Glass Iron Aluminum 846 1041 1089 4165 4708 4756 11480 12620 16410 16410 16740 i, 11:9 QUESTIONS 258 317 332 1269.5 1435 1450 3500 3850 5000 5000 5104 A 1. Which of the following statements is correct? (a) All vibrating objects produce sound, (b) All sound is produced by vibrating objects. Explain your answer fully. 2. 3. between (a) Distinguish and longitudinal vibrations. (b) Define: complete vibration, ampli- transverse tude, frequency, period. (a) What is necessary for the transmission of sound? (b) Compard^the transmission of sound through the three states of matter and suggest a theoretical explanation for any differences. 4. (a) Distinguish between transverse and longitudinal waves. (b) Define: amplitude, wave-length, period, frequency. (c) By what fype of wave-motion is sound transmitted? What are the components of each complete sound wave? 5. 6. (d) What is the fundamental characteristic of wave-motion? Explain fully. (a) Establish the wave formula. (b) Calculate the velocity of sound in air if its frequency is 250 v.p.s. and its wave-length is 4.4 ft. (a) What is meant by the terms \"in phase\u201d and \"out phase\u201d as applied to any wave-motion? (b) How are standing waves produced? of (c) Define: node, loop. (d) Explain why the distance between two successive nodes is one-half a wave-length. 7. 8. (a) State the laws of reflection of sound waves. (b) Distinguish echoes and reverbera- tions. (a) In what three ways may the velocity of sound be determined? (b) What is the effect of a change of 61 Chap. 6 SOUND 7. At what temperature will the velocity of sound be (a) 1,1 19 ft. per sec. (b) 336 metres per sec.? 8. A thunder-clap is heard 5 seconds after the lightning-flash was seen. How far away was the flash if the temperature of the air were 1 5\u00b0C.? 9. When the temperature of the air is 15\u00b0C., calculate the wave-length in metric units of the sound from a tuning-fork having a frequency of 256 v.p.s. TO. When the temperature of the air is 25\u2019/2\u00b0C., and the wave-lengfh of a sound is 4.40 ft., calculate the frequency of the sound. TT. A signal of 128 v.p.s. has a wavelength of 279. cm. (a) Find the velocity of sound in air. (b) What would be the temperature of the air to the nearest degree centigrade? Express your answer in metric units. T2. The human ear is incapable of disindividual sounds unless they tinguishing are separated by a time interval of at least ]/(q sec. (a) Calculate the length of the shortest auditorium that would give a distinct echo,", " (b) What would be the effect if the auditorium were shorter? (Assume that the temperature of the air is 20\u00b0C.] T3. A 220 yd. dash over a straight course was timed at 23.2 sec. What would the time have been had the timer started the watch on hearing the sound instead of seeing the flash? (Temperature of air -20\u00b0C.) T4. Calculate the minimum speed of an aircraft in miles per hour which has broken sound the air = 5.5\u00b0C.) barrier. (Temperature of temperature on the velocity of sound In air? 9. The sound of a gun was heard 1 0 sec. after the flash was seen. If the distance to the gun was 1 1,500 ft., calculate the probable velocity of sound in air. Why is this merely a probable velocity? TO. Two and one-half seconds elapse between shouting across a river 1,375 ft. wide and hearing the opposite bank, (a) Find the velocity of sound, (b) Compare the accuracy of this velocity with that of question 5. from echo the B 1. The horizontal distance between the end points in the swing of a pendulum Is 7.5 cm. What is (a) the amplitude, (b) the distance covered by the bob in one com- plete vibration? 2. The pendulum in question 1 makes 45 complete vibrations in 30 seconds. Calculate the period of vibration. What would amplitude were be the doubled? period the If 3. Calculate the velocity of sound in air if the frequency of sound and its wave1 80 v.p.s. and 6.50 ft., length are (a) (b) 360 v.p.s. and 80.0 cm. respectively. if the 4. Calculate the frequency of a sound and are (a) 1,120 ft. per sec. and 3.50 ft., (b) 340 metres per sec. and 51.0 cm. wave-length velocity 5. What are the wave-lengths of the notes when (a) the frequency is 900 v.p.s. and velocity of sound 1,350 ft. per sec. and (b) the frequency is 625 v.p.s. and velocity of sound is 350 metres per sec.? 6. Calculate the velocity of sound in air in (a) feet per second and (b) metres per second at: 5\u00b0C., - 17\u00b0C., 23\u00b0C. 62 CHAPTER 7 CHARACTERISTICS OF MUSICAL SOUNDS 11:10 INTRODUCTION (a) How a Musical Sound Differs from a Noise As we have seen, people are quick to classify the sounds they hear, as either musical sounds or noises. Almost everyone finds noises, like the slamming of a door or the rumble of machinery unpleasant, and musical sounds like those or pleasant. of We can establish a more objective difference than this between noise and music. When an oscilloscope is used to compare the sound of machinery and that of a tuning-fork, a trace similar tuning-fork violin a Oscilloscope Tracing of Fig. 7:1 A\u2014 Noise B\u2014Musical Sound It is obvious to that in Fig. 7 : 1 results. that the musical sound is caused by rapid regular (periodic) vibrations, while the noise is the result of irregular (non- periodic) vibrations. (b) How Musical Sounds Differ from Each Other If one tuning-fork is struck lightly, and another more vigorously, the second will Sounds that differ emit a louder sound. in loudness differ in intensity. If the sounds from two vibrating tuning-forks with differing frequencies are compared, the fork with the greater frequency emits the \u201chigher\u201d sound. Sounds that differ Further, in \u201chighness\u201d differ in pitch. if the sounds of a vibrating tuning-fork and a vibrating string of the same frequency are compared, there is no difficulty in identifying the origin of each Sounds from different sources sound. may be distinguished because they differ in quality. Thus musical sounds differ from each other in intensity, pitch and quality. It is the purpose of this chapter to study these three characteristics fur- ther. 11:11 THE INTENSITY OF SOUND If a bell that is rung cannot be heard at a distance, we know that we may be able to make it heard by striking it harder. This is so because striking with greater force transfers more energy to the vibrating body, increasing what we call the sounds. This increases the amplitude (energy of vibration) and gives If the sound is it a greater intensity. still inaudible, the other obvious thing that can be done to make it heard is 63 Chap. 7 SOUND the shorten distance between the to It might seem source and the receiver. from this that there are just two factors affecting the intensity of sound, namely amplitude and distance, but the experience of undersea workers in caissons and diving-", "bells where the air is under great pressure, is that quite ordinary sounds are unexpectedly loud there. This greater intensity of the sound transmitted results from the fact that increased pressure on a gas crowds the molecules closer and increases the density of the medium. For the same reason, as is well known, sounds transmitted by solids and liquids are louder than when transmitted by air. Thus, a third factor affecting intensity of sound is density of medium. fles, while neighbouring buildings can be protected by double windows. The importance of such efforts to reduce Threshold of Painful Noise Airplane Engine Riveting Machine Heavy Traffic Motor Truck Ordinary Conversation Vacuum Cleaner Average Office Quiet Home, Quiet Conversation Rustle of Leaves Quiet Whisper \u2014;130 \u2014h20 -^110 \u2014 100 -^90 \u2014 80.^ -70|.( \u2014;60 Q \u2014 50 \u2014 40 -30 \u2014 20 \u2014 10 \u201c0 The particular intensity of sound are: laws governing the Threshold of Hearing 1. The intensity of sound varies directly as the square of the energy of vibration (amplitude) of the source. 2. The intensity of sound varies in- versely as the square of the distance of the receiver from the source. 3. The intensity of sound increases with an increase in the density of the transmitting medium. in vary intensity. (deci \u2014 1/10), The intensity of sound is measured in bels and decibels the decibel being the faintest sound that can be perceived by a normal human ear. Fig. 7:2 shows the amount of noise or \u201cnoise level\u201d in decibels in a few common locations, for, like musical sounds, Naturally, noises noises are inevitable wherever there is machinery, but unnecessary discomfort can be avoided by the noise level being measured and steps being taken to eliminate all unnecessary vibration. Noise levels around machinery can be reduced by the use of rubber mountings, mufflers and the like; walls and ceilings can be covered with sound-absorbent wallboard, and air-ducts with sound-absorbent baf- Fig. 7:2 Noise Levels. noise levels is indicated by the fact that temporary or permanent deafness and many other illnesses can result from long proximity to noisy machinery. 11:12 THE PITCH OF SOUND That musical sounds vary in highness or pitch was stated in a previous section. In chap. 10, exp. 6, Savart\u2019s toothed Fig. 7:3 Savart's Toothed Wheel. wheel ( Fig. 7:3) was used to show what determines pitch. The card held against the teeth of the rapidly rotating wheel received a sequence of taps, and a note, whose pitch increased with the speed of 64 CHARACTERISTICS OF MUSICAL SOUNDS Sec. 11:12 rotation, was heard. If the rate of rotation was doubled, the number of taps per second (the frequency) was doubled. The second note, whose frequency is twice as great, is said to be an octave higher than the first. Thus for a note of high pitch to be sounded, the body must be vibrating rapidly, whereas a slow rate of vibration produces a note of low pitch. In short, the pitch of sound depends on the frequency. The frequency of the sound produced may be determined by applying the formula: frequency of note produced ( v.p.s.) = number of teeth on Savart\u2019s wheel X speed of rotation revolutions per second. in The limits of frequency for notes of are from about 20 to pitch audible 20,000 v.p.s. for the average ear (Fig. 7:4). The range of frequencies used in music is from about 30 to 5,000 v.p.s., the keyboard of a piano extending from 27 to 3,500 v.p.s. A man\u2019s speaking voice embraces frequencies ranging between 100 and 150 v.p.s. approximately. Extreme limits of audibility (for very sensitive ears) may be taken as extending from 20 to 35,000 v.p.s. The lowest notes of a large organ are in the neighbourhood of this lower audible limit, while the squeak of a bat or the noise of a cricket are examples of frequencies in the region of the upper audible limit. Frequencies above this point, referred to as ultrasonic frequencies, are becoming of increasing importance (Sec. 11:33). sounds heard by the human ear wind instruments string instruments frequency limits of human hearing ^ Fig. 7:4 Frequency Limits of the Human Ear. Bell Telephone Company of Canada. 65 Chap. 7 SOUND 11:13 THE DOPPLER EFFECT When a car with its horn sounding approaches a pedestrian at high speed, the pitch of the sound appears to be higher than the true pitch which the driver hears (Fig. 7:5). After the car has passed, the pitch appears to be lower than the true pitch. Similar changes in pitch occur when the origin is stationary and", " the observer moves past it. To determine the cause, let us take the case of the origin approaching the In any second, a uniform observer. number of wave-fronts are sent out and they will be a uniform distance apart. However, because the origin is approaching the observer, there will be more than the usual number packed in a given space, i.e,, they will be closer together than before. As a result, more than the normal number will be received by the observer in one second. This will cause an apparent rise in the pitch of the sound. This phenomenon is called the doppler effect. Fig. 7:5 The Doppler Effect. 11:14 THE SONOMETER In stringed instruments, the stretched strings of steel, gut, or silk, are set in a state of transverse vibration by being struck, bowed, or picked as in the piano, violin and guitar respectively. Examination of a piano and violin will reveal that sounds of different pitches are obtained by the use of strings of different lengths, tensions, diameters and densities. To ensure that the sound is loud enough to be distinctly heard, the string is attached to a sounding-box or board. The natural frequency of this is not that of the string but the string will set it in vibration with forced vibrations will increase the volume of the sound. that The laboratory device embodying all these features is the sonometer shown in Fig. 7:6. It consists of a hollow wooden box (A) on which one or more strings (B) are stretched. Permanent (C) and movable (D) bridges and a means of varying the tension (E) are provided. CD AD B C 66 CHARACTERISTICS OF MUSICAL SOUNDS Sec. 11:15 \\Ve shall now proceed to use it to study the laws of vibrating strings. 11:15 THE LAWS OF VIBRATING STRINGS (a) The Relationship between Frequency and Letigth After doing experiment 7, chapter 10, results similar to those in the following table are found. Examination of trials 1, 4 and 5 will show that while the frequency of 256 is emitted by 34.4 cm. of string, v.p.s. twice that frequency is produced by onehalf that length and four times the frequency by one-quarter that length. Other possible results would be that three times the frequency is produced by one-third the length, five times the frequency by one-fifth, and so on. Thus we see, as in column 3 below, that the product of frequency times the length of vibrating string is constant within the limits of experimental error. Therefore, the frequency of the note produced by a vibrating string varies inversely as its length. This is the law of lengths. Results for Law of Lengths Constant Tension \u2014 1000 gm. Length of Wire Producing Unison Frequency of Fork Frequency X Length \u201c 1. 2. 256 v.p.s. 320 384 512 5. 1024 4. 3. \u201c \u201c \u201c 8806 8832 8832 8806 8806 34.4 cm. 27.6 cm. 23.0 cm. 17.2 cm. 8.6 cm. Example If 30 cm. of wire at a certain tension produces a note with a frequency of 256 v.p.s., what would be the frequency when the length is 40 cm.? Solution 1 The ratio of the new length to the old = \u2014 40 30 the frequency varies inversely as the length. the new frequency is \u2014 of the old. 40. the new frequency = 256 X \u2014 = 192 v.p.s. 30 40. Solution II the frequency varies inversely as the length, frequency X length is constant. 67 Chap. 7 SOUND the new frequency X the new length old length. the old frequency X the Let the new frequency be x a; X 40 = 30 X 256 ;c = 40 = 192 the new frequency =192 v.p.s. (b) Relationship between Frequency and Tension Adjust the length and tension of a string on a sonometer until it is in unison with a tuning-fork of frequency 256 v.p.s. Keeping a constant length of 25 cm., adjust the weights that stretch the string to get unison with a second, and then again to get unison with a third fork. Record the results as follows, and from them determine the relationship. it After comparing the results of trials 1 and 2, is noted that twice the frequency is caused by four times the tension. A comparison of trials 1 and 3 shows that 4 times the frequency is caused by 16 times the tension. Since the multiplier for the frequency is the square root of the multiplier for the it follows that the frequency tension, of the note emitted by a vibrating string varies directly as the square root of the tension. This is the law of tensions. Frequency of Fork 1. 128 v.p.s. 2. 256 v.p.s. 3. 512 v", ".p.s. Tension 385 gm. 1540 gm. 6160 gm. Example A string with tension of 2000 gm. produces a note with a frequency of 300 v.p.s. What would be the frequency of the note if the tension were 4500 gm.? Ratio of the new tension to the old = 4500 9, = 4 2000 \u2019. the frequency varies directly as the square root of the tension. the new frequency will be the new frequency = 300 X 4500 of the old 2000 ^ = 300 X - = 450 v.p.s. 4 2 (c) Relationship between Frequency and Diameter (d) Relationship between Frequency and Density By an experiment somewhat analagous to (a) we may determine the law of diameters, i.e., the frequency of the note emitted by a vibrating string varies inversely as the diameter. frequency of The law of densities states that the the note emitted by a vibrating string varies inversely as the square root of the density of the material. All four laws may be illustrated by 68 CHARACTERISTICS OF MUSICAL SOUNDS Sec. II; 16 the tuned, examination of a violin. To increase the frequency of a note, the violinist shortens the string with his finger. When a violin tension to is increase the frequency and decreased to It will also be decrease the frequency. noted that those strings with greatest densities and diameters the lowest notes. increased produce is 11:16 HOW A STRETCHED STRING VIBRATES We have just considered the simplest mode of vibration of a stretched string wherein the string vibrates as a whole. There is a loop in the centre with a node at each end (Sec. 11:6). When vibrating thus, the string is emitting the note of lowest frequency, the fundamental. However, the in other ways to produce notes of higher frequency (Chap. 10, Exp. 8). string may vibrate When the string vibrates in halves, the note produced has twice the frequency of the fundamental and one-half the wave-length. This is the first over- wave-lengths tone or.second harmonic. When vibratthirds, quarters and fifths, the ing in frequencies produced are three, four and five times that of the fundamental, and onethe are quarter and one-fifth of the fundamental. The notes are called the second overtone or third harmonic and so on. Fig. 7 : 7 shows some modes of vibration of a stretched string. one-third, fundamental when that We see that a string may vibrate in parts, and as a whole as well. When it is vibrating in parts, the frequency of the note is a multiple of that of the fundamental and the notes are called the harmonics or overtones of the string. Frequently these overtones accompany the is sounded, and give quality to the sound Very produced (Chap. 10, Exp. 9). few sources, on the other hand, produce the fundamental free of overtones. The tuning-fork is one that does, but even in it the overtones are present at the bevibration, vanishing as ginning of its It is this absence of overtime goes on. note FREQUENCY WAVE-LENGTH n I Fundamenfal l\\ & First Overtone A\\ N A\\ N N Second Overtone N A\\ A\\ A\\ N N N N Third Overtone N N 2N 3N 4N Fig. 7:7 Nodes and Loops in a Vibrating String. 'A I 'A I 'A I 69 Chap. 7 SOUND 8 Fig. 7:8 Oscilloscope Tracings of Tuning Forks Sounded (a) Singly A\u2014Tuning-fork B\u2014Organ Pipe (b) Pairs tones that makes the tuning-fork valuable in the study of sound, though at the same time it makes the note dull and uninteresting, for it is the overtones that make a note rich and interesting to the listener. 11:17 QUALITY OF SOUND If respectively. Let us analyse the vibration of a fundamental and its harmonics by means of a cathode-ray oscilloscope. the fundamental has a frequency of 128, the first two overtones have frequencies of 256 and 384, shows the results of sounding tuningforks of these frequencies singly and in groups. When sounded singly, we see differences in frequency and wave-length of the notes. When sounded in groups, we see complicated wave forms which represent the blending of the fundamental and one or more overtones. Fig. 7 : Next, with the aid of an oscilloscope, let us analyse notes from a tuning-fork and several other different sources having the same fundamental frequency. Fig. 7:9 shows several traces made in such a way. The regular trace of the is emitting tuning-fork indicates that it but a single tone. The complicated wave forms of the others indicate that", ": D\u2014Violin Fig. 7:9 Oscilloscope Tracings. 70 CHARACTERISTICS OF MUSICAL SOUNDS Sec. II : 18 1. They contain tones (overtones) in addition to the fundamental. 2. That some have more of these over- tones than others. 3. That in some, the overtones are more prominent than in others, We may conclude therefore that the quality of a musical note is dependent on the number and relative prominence of the overtones that occur along with the fundamental. is the quality of It the sound that enables us to distinguish notes of the same pitch and intensity from different sources. 7. QUESTIONS II : 18 1. A between (a) Distinguish sound and a noise. (b) What are the three distinguishing characteristics of musical sounds? musical a 2. (a) Define intensity of sound. (b) State three factors that affect the intensity of sound. Illustrate each with a suitable example. 3. (a) Define pitch. (b) Describe an experiment to illustrate upon what the pitch of sound depends. (c) Make up a set of observations and show how the frequency of a given note may be calculated. 4. (a) Describe the sound of a train whistle as the train moves rapidly away from you. (b) Explain this phenomenon fully. 5. Describe a sonometer and state the purpose of each of its parts. 6. (a) State four factors that affect the frequency of a vibrating string. the Law of Lengths. A (b) State stretched string 50 cm. long vibrates with a frequency of 1 50 v.p.s. What will be its frequency when the length Is (i) 10 cm., (ii) 75 cm.? (c) State the Law of Tensions. A stretched length vibrates with a frequency of 1 25 v.p.s. when its tension is 900 gm. What will be its frequency when the fixed string of tension is (i) increased to 2500 gm. (ii) decreased to 144 gm.? (a) Describe an experiment to show the modes of vibration of vibrating strings. (b) Define; fundamental, overtone. (c) What governs the quality of a note? B 1. A certain note has a frequency of the frequencies 480 v.p.s. (a) Determine of notes that are one, two, and four octaves above the given note. (b) Find the frequencies of notes that are one, three and five octaves below the note. 2. A toothed wheel with 40 teeth Is rotated at the rate of 360 revolutions per minute while a card is in contact with the teeth. Calculate the frequency of the note heard. 3. A toothed wheel having 66 teeth is rotated while in contact with a card. What will be the speed of rotation in revolutions per minute when the frequency of the note produced Is 352 v.p.s.? 4. How many teeth Savart\u2019s wheel have if the speed of rotation is 540 revolutions per minute and the frequency of the note produced Is 1 350 v.p.s.? will a 5. Savart\u2019s toothed wheels generally are arranged in sets of four on a common shaft with 12, 15, 18, 24 teeth respectively. 71 Chap. 7 SOUND. frequency of 540 v.p.s. What would be the weight the became 16000 gm., 1000 gm.? frequency stretching If 10 A piano string is 60.0 in. long. It vibrates at 260 v.p.s. A piano tuner changes the tension from 25.0 lb. to 36.0 lb. What will be the new frequency? n. A string 40.0 cm. long and having a tension of 1 600 gm. emits a note of frequency 1 28 v.p.s. Determine the tension of this string when it vibrates with a frequency of: 64 v.p.s., 1 60 v.p.s. 12. A string 36.0 in. long under a tension of 1 6.0 lb. vibrates with a frequency of 256 v.p.s. What is the vibration frequency if the length is increased to 54.0 in. and the tension is increased to 81.0 lb.? long 13. A string 100 cm. under a tension of 4900 gm. has a frequency of 280 v.p.s. What is the frequency if the 1 25 cm. and the length is increased to tension reduced to 2500 gm.? When rotating at the average rate of 21 Vz revolutions per second, they produce frequencies corresponding to C major chord at middle C on the piano, i.e., CEGC'. Determine the frequency of each note in the chord. 6. A stretched string 45.0 cm. long emits a note with a frequency of 300 v.p.s", ". What would be the frequency if length became 15.0 cm., 60.0 cm., 20.0 cm.? the 7. The A string of a violin vibrates at 440 v.p.s. The string is 40.0 cm. long from If the violinist moves his bridge to nut. finger so that only 30.0 cm. of the string vibrates, what will be the frequency of vibration? 8. A certain vibrating string 50.0 cm. long emits a note with a frequency of 320 v.p.s. Whaf length of string would vibrate with the following frequencies: 640 v.p.s., 200 v.p.s., 457 v.p.s.? 9. A string 30.0 cm. long stretched by a weight of 4000 gm. emits a note having a 72 CHAPTER 8 RESONANCE AND INTERFERENCE PHENOMENA I I, I! 11:19 THE MEANING OF RESONANCE As resonance is a new idea, we shall find out what it means using the apparatus shown in Fig. 8:1. This consists of several pendulums attached to a cord tied between two supports. When one is vibrated transversely, the motion will be transmitted the supporting cord. Any other pendulum through the rest to Fig. 8:1 Mechanical Illustration of Resonance. will vibrate erratically, starting, stopping, but never accomplishing the persistence of vibration referred to above. Now, when impulses from one body affect another having the same period of vibration, the second will begin to vibrate with increasing amplitude. If it is already in motion, the amplitude will become greatThis effect is known as resonance er. and will be very valuable in explaining the phenomena that follow. 11:20 RESONANCE IN AIR COLUMNS That air columns can be set in vibration and made to produce sounds of a definite pitch is well illustrated by such simple experiments as blowing across empty test-tubes of various lengths. Such air columns have a natural period of vibration depending on their length. If the fluctuations of pressure at the end of the column (caused by blowing) have the same period as that of the air column, resonance will occur. The column will be in a state of violent sympathetic vibration, and a strong note will be heard. In our study of air columns, we shall use tubes of uniform cross-section. If the tube is closed at one end, it is called a closed tube, while if it is open at both ends it is designated an open tube. of the same length and period of vibration will take up the vibratory motion and move with increasing amplitude. The others that have different periods (a) The Closed Tube After performing the experiment to demonstrate resonance in a closed tube (Chap. 10, Exp. 10), let us consider 73 Chap. 8 SOUND what occurred. When the sound that proceeds down the tube is reflected at the closed end the wave returns without change of phase. Thus a condensa- Condensation goes down and is reflected up. li Wave-length Fig. 8:2 Closed Tube in Resonance with Tuning-fork. phase occurs tion is reflected as a condensation and a rarefaction as a rarefaction. As the sound arrives at the open end a change as follows : when a in rarefaction reaches the open end some air is taken into the tube, and a condensation goes down the tube; when a condensation reaches the open end some air spills out of the tube and a rarefaction goes down the tube. Thus a condensation returns as a rarefaction, and a rarefaction as a condensation from the open end. half vibration, diagram (b), a rarefaction will go into the tube which tends to reinforce the rarefaction already proceeding downward. This process continues until the air in the tube is vibrating with such wide amplitude that it becomes the major source of the sound heard. It should be apparent that the sound travelled twice the length of the closed tube during one-half vibration, and therefore the length of the closed tube is equal to one-quarter of the wave-length. (b) The Open Tube Open tubes will also vibrate in resonance with sources of sound such as It has been found that tuning-forks. during one vibration the sound travels twice the length of the tube, and therefore the length of the open tube is equal to one-half of the wave-length. It will be evident, therefore, that the open tube that vibrates in resonance with a tuningfork of a certain frequency is twice the length of the closed tube. Closed Tubes Open Tubes in resonance with it. Fig. 8:2 shows a vibrating tuningfork held over a closed air column which For greater is simplicity we shall confine our attention to the movements of the lower prong, since movements of the upper one do not alter the final result. As the prong of the tuning-fork traces one-half a vibration", ", diagram (a), a condensation is sent down the tube and is reflected at the closed end as a condensation. When it reaches the open end it will be reflected down the tube as a rarefaction while the air that spills out forms a condensation the condensation pro- that duced above the prong of the fork. the length of the tube is such that the fork is about to execute the next one- reinforces If 74 Fig. 8:3 Modes of Vibration in Air Columns. 11:21 MODES OF VIBRATION IN CLOSED AND OPEN TUBES In vibrating air columns (Fig. 8:3), there will be nodes and loops ( Sec. 11:6). In the closed tube there will be RESONANCE AND INTERFERENCE PHENOMENA Sec. 11:22 a node at the closed end and a loop at the open end. The length of closed tube will be one-quarter of a wave-length when it is responding to its fundamental For the overtones, the distance tone. from node to loop must also be onequarter of a wave-length. After examination of the diagram, it will be evident that the closed tube can be in overtones whose resonance the frequencies are odd-number multiples of that of the fundamental. with In the open tube there will be a loop at the open ends and a node will occur in the middle when in resonance with the fundamental. The length of the tube will be one-half a wave-length. For overtones, the tube must be capable of containing several half wave-lengths. Figure 8:3 shows how this can be done and makes it clear that the open tube can be in resonance with all the over- tones. I I j [ V I II : 22 DETERMINING THE VELOCITY OF SOUND IN AIR BY RESONANCE IN AIR COLUMNS (a) Closed Tubes As explained in section II: 20(a), the condensation travelled twice the length of the tube during one-half a vibration of the prong of the tuning-fork. Hence, sound must travel four times the length of the tube during one vibration of the prong. Since the distance energy travels during one vibration is one wave-length, the wave-length of sound must be four times the length of the closed tube which is in resonance with the tuning-fork. In the true wave-length actual practice, must be augmented by.3 times the diameter of the tube (see note) but for purposes we may disregard our Therefore, wave-length (/) of sound = 4 X length of closed tube (L) giving resonance. it. Example A tuning-fork whose vibration frequency is 256 v.p.s. produces resonance with a closed tube 13.0 inches long. Calculate the velocity of sound in air.? n = 256 v.p.s. l:=4L = 4x222. 12 ft. (Sec. 11:5) T = 256 X =1109 4 X 13.0 12.'. Velocity =1109 feet per second Since the measurements are accurate to three significant digits, the proper answer is 1 1 1 X 10 feet per second ( Sec. 1:3). What would be the approximate temperature of the air in the above example? (b) Open Tubes In section II: 20(b) ; it was stated that sound travels twice the length of the open tube during one vibration. There- fore, the wave-length (/) of sound = 2 X the length of the open tube (L) (see note), A tuning-fork whose vibration frequency is 1024 v.p.s. produces resonance with a tube 17.2 cm. long. Calculate the velocity of sound in air. Example 75 Chap. 8 SOUND V=? n = 1024 v.p.s. l = 2L = 2 X 17.2 cm. \\'V^nl V = 1024 X 2 X 17.2 100 = 352.3 Velocity = 352.3 metres per second or 352 metres per second. What would be the approximate temperature of the air in the above example? Note In very accurate work a correction must be made for the change in pressure influencing the sound waves a short distance from the end of the tube. In a first course in physics, however, this factor need not be considered. For closed tubes, the end correction of.3 X the diameter of the tube must be added. For open tubes, you make the same correction for each end. The phenomenon is caused by the inertia of the air molecules. 11:23 SYMPATHETIC VIBRATIONS Two tuning-forks frequencies are mounted on hollow wooden boxes each open at one end (Fig. 8:4). identical of Fig. 8:4 Identical Tuning-forks to Illustrate Sympathetic Vibrations. The size of the air column is such that its natural period of vibration is the same", " as that of the fork. The boxes are placed a short distance apart with their open ends facing each other. When one fork is vibrated and then silenced shortly afterwards, a sound of the same pitch is still heard (Chap. 10, Exp. 11.) It is found to originate from the other fork. This response of one body to the sound 76 waves caused by the vibrations of another is called sympathetic vibrations. When a piece of plasticine is attached to the prongs of one fork its frequency is alIf the above experiment were tered. repeated this fork would not set up vibrations in the other. This shows that the two forks must have identical periods for sympathetic vibrations to occur. Vibrations from the fork cause forced vibrations in the box and in turn the air in the box vibrates in resonance with them. The vibrations pass through the air to the other box, through the same sequence of events as above, but in reverse order, causing the second fork to vibrate. 11:24 INTERFERENCE OF SOUND WAVES (a) Silent Points around a Tuning-Fork When a vibrating tuning-fork is held vertically and rotated near the ear alternate loud and faint sounds will be It will be heard (Chap. 10, Exp. 12). found that the faint sounds are obtained when the tuning-fork is held cornerwise to the ear. To understand this phenomenon, we should recall that in transverse RESONANCE AND INTERFERENCE PHENOMENA Sec. 11:24 vibration the prongs of the fork move together for half a vibration and apart for the next half. When they approach a compression each other (Fig. 8:5) R. tube (Chap. 10, Exp. 13). It consists of two U-shaped tubes that telescope in and out of each other (Fig. 8:6). One side has a speaking-tube, the other an listen. When a opening at which to sound is sent in and the length of the two paths is adjusted, faint sounds are heard in some positions and loud ones at If the two paths that the sound others. follows differ by one-half a wave-length, or an odd number of half wave-lengths, the two parts of the sound will arrive out of phase at the observer and a faint If the two paths sound will be heard. are equal, or differ by a whole number of wave-lengths, the two parts of the I j rarefaction (Ri) at either side. These r two waves spread out in all directions and, because they are in opposite phase, interfere with each other, producing silence when they meet at the corners (S). It is often noticed that the intensity of sound varies in different parts of an auditorium without any obvious cause. One possible reason for this is the interference of direct waves with reflected If these are out of phase they will produce a faint i sound as happened at the corners of the 1 waves from walls and ceiling. I I tuning-fork above. Another cause is discussed in Sec. 11:30. (b) The Herschel Divided Tube Fig. 8:6 The Herschel Divided Tube. I be can produced Interference by dividing a wave disturbance into two parts, conducting each along a separate path, and then blending the two. This'can be accomplished by the Herschel sound will arrive in phase at the observer and a loud sound will be heard. This phenomenon is used to find the wavelength of sound. If the frequency of the 77 Chap. 8 SOUND note is known, the velocity of sound can be calculated. (c) Beats If the prongs of one of two middle C tuning-forks are loaded with plasticine, the frequency of its vibration will be slightly lower than that of the other. If these forks are sounded together, a sound will be heard which periodically increases and decreases in intensity. The alterations in the loudness of the sound are called beats. Consider each of the waves sent out by the forks as transverse waves. By representing one with a solid line and the other by a dotted line, as in Fig. 8:7 (a), we see that they become progressively more out of step until one cancels the other. If the lines are continued they will eventually arrive in step, although one will be a wave-length in front of the other. Now, when waves are out of phase they interfere with each other with consequent reduction in amplitude of vibration or loudness of the sound. When completely out of phase will be no movement and no there sound. When completely in phase, there will be a greater amplitude and a louder sound. cases there will be a gradual increase or de- Between these extreme 78 RESONANCE AND INTERFERENCE PHENOMENA Sec. 11:25 crease in both amplitude and loudness. 8: 7(b) shows the result of such Fig. : interference. i I i I j j Fig. easily. 8; 7(c) Flaving studied beats with reference", " to transverse waves, we may understand their production in sound waves more shows two sound, waves of slightly different frequencies being produced simultaneously. Assuming that they begin in phase, two condensations or rarefactions will occur together, producing a loud sound or a beat. As 1 the two waves get out of phase, the sound will become fainter and the more out of phase they are, the fainter the be. When completely out of phase, as when a condensation from i one and a rarefaction from the other'occur together, silence results. As they \u2018 become progressively more in phase, the a! maximum, at which point a loud sound'sound will increases intensity reaches until it I'or beat will occur as before. A repetition of this sequence of events gives us noticed when two tuningforks of slightly different frequencies are effect the vibrated together. If the forks being used had a difference in frequency of one v.p.s., one beat per second would result. Similarly, a difference of two v.p.s. would produce two beats per second and so on. In general, the number of beats per second equals the difference between the frequencies of the two notes. This provides a convenient means of determining the frequency of a sound. (How could Moreover, musical inthis be done?) struments are tuned by listening for beats. The fewer the beats, the more nearly alike are the two frequencies. When unison is achieved, no beats may be discerned II : 25 QUESTIONS (a) What is meant by resonance? fully. 5. (a) How are beats produced? Explain (b) Explain resonance in closed tubes. 2. (a) A closed tube 1 2 in. long is in resonance with a tuning-fork whose is 300 v.p.s. vibration frequency Calculate (i) the wave-length of the sound (ii) the velocity of the sound in air. (b) What would be the length of a tube open at both ends that would be in resonance with the tuning-fork used in part (a)? (b) What determines the number of beats per second? 6. A person holds down the \"loud pedal\u201d of a piano and sings a note. Account for the humming sound heard. Why does more than one string respond? 7. Account for the sound produced by blowing across the top of an empty testtube. What would be the wave-length of such a note? 3. Explain and give examples of sym- pathetic vibrations. 8. Account for the rise in the pitch of sound heard as a cylinder is gradually 4. (a) What is the cause of interference in sound? (b) Explain (i) silent points around a tuning-fork, (ii) variations in the loud- ness of sound as Herschel\u2019s divided tube is elongated. filled with water. 9. Explain why a wavy sound is frequently heard when a tuning-fork mounted on a sounding-box, the open end of which faces a wall, is moved towards and away from the wall. Try it. 79 Chap. 8 SOUND B 1. (a) Calculate the wave-length of a note that gives resonance with (i) a closed tube 15 in. long, (ii) an open tube 6 In. long. (Disregard the cor- rection for diameter.) 5. A closed tube 4.0 ft. long responds to a frequency of 70 v.p.s. Find the temperature of the air. 6. A closed tube 40.0 cm. long responds the to a frequency of 220 v.p.s. temperature of the air. Find 7. A resonance box is to be made for (n = 440 v.p.s.). When the a tuning-fork velocity of sound in air is 330 metres per second, what would be the shortest length of box, closed at one end, that would resonate with it? 8. Two forks, having frequencies of 384 and 380 v.p.s. respectively, are sounded together. How many beats per second will be produced? 9. When a tuning-fork (n \u2014 4S0 v.p.s.) is sounded with another of slightly different pitch and there are 6 beats per second, what are the possible frequencies of the second fork? How would you determine whether its frequency would be higher or lower than the other? (b) Determine the velocity of sound (n = 220 results shown in tuning-fork gives v.p.s.) the air a in if question 1 (a). 2. Compare the frequencies to which a closed tube 1 2 in. long and an open tube of the same length will respond, the temperature of the air being 1 91/2\u00b0 C. 3. Find the frequency of a note that resonates with a closed tube 1 0.5 in. long, the temperature of the air being '\\5Vi\u00b0Q. 4. Find the length of closed", " tube that will respond to a frequency of 288 v.p.s., the temperature of the air being 25V2\u00b0C. Express the answer in both British and metric units. 80 CHAPTER 9 APPLICATIONS OF SOUND ing between these membranes sets them in vibration in a way similar to blowing between the strands of a stretched elastic band. The faster the air moves, the the intensity of the sound greater is produced. The particular quality of the sound depends on resonance in the cavities of the mouth (m) and (n). Variaare caused by muscles altering tions the size and shape of these cavities. In the mouth, the tongue measure, makes the major changes. large (t), in 11:27 THE EAR The ear is the most wonderful sound It con- receiver that can be imagined. the outer ear, the sists of three parts: middle ear, and the inner ear. Sounds are collected by the pinna (Fig. 9:2) and directed into the ear canal to the eardrum. The ear-drum consists of a thin (3/1000 inch thick) tightly stretched membrane that is set in vibration by the sound waves and serves as the gateway to the middle ear. the ear there Within middle are three bones named because of their shape, the hammer, the anvil and the stirrup. The hammer is in contact with the eardrum, the stirrup with the oval window leading to the inner ear, and the anvil connects the two so that the vibrations of the ear-drum are transmitted to the inner ear. The middle ear is joined to the throat by the eustachian tube, the purpose of which is to equalize the air pressure on either side of the ear-drum. ^! j I! : 26 THE VOICE Of all sources of sound the voice is the most wonderful. The vocal cords in the larynx (Fig. 9:1) are two (c) elastic membranes whose thickness, length and tension affect the pitch in response to the will of the person and in keeping with the maturity and sex of the individual. Air (a) from die lungs (1) pass- Chap. 9 SOUND This adjustment of pressure can be felt when motoring in hilly country. The inner ear contains a spirallyshaped organ, the cochlea, containing a fluid which is agitated when the oval window vibrates. Movement in this fluid will cause hair-like projections to vibrate, transmitting small nerve impulses through the auditory nerve to the brain. In addition to the cochlea, the inner ear contains another organ known as the semicircular canals, which is associated with posture and balance. 11:28 MUSICAL SCALES The story of the evolution of the existing musical scale is a long and interesting one. The scale which gives maximum pleasure to us is one in which the frequencies of the notes are in the simple ratios shown (page 83). This scale is known as the diatonic scale, which on the tonic sol-fa corresponds to the notes doh, ray, me, fah, soh, lah, te, doh, or more familiarly perhaps, C, D, E, F, G, A, B, C'. The first note on the scale is called the tonic, and the last note, of twice the frequency of the first, the octave. The number of the note (counting from the tonic) defines a musical interval on the scale; thus the interval from C to D is a second, that from C to E a third, and so on. These intervals correspond respectively to frequency ratios of 9/D _ 288\\ 8\\C ~ 256/\u2019 4\\C ~ 256 / 5/E _ The last row of figures gives the ratios 82 6 APPLICATIONS OF SOUND Sec. 11:28 The Diatonic Scale No. of note Notation Absolute frequencies (scientific pitch)* Frequency ratios 2 D 1 3 4 C 256 288 320 341.3 384 426.6 480 512 C 15 8 Interval ratios 9 8 10 9 16 15 9 8 10 9 9 8 16 15 *In science C represents a frequency of 256 v.p.s. but for concert work however, C is usually tuned to 261 v.p.s. I of the frequencies between successive These are known notes on the scale. as the interval ratios, of which it will be seen that there are three. These are 9 / 0 8VC ^ and 288 \\ = \u2014 ), \u2014 ( \u2014 10/E _320 9\\D \"~288 256/ 16/F 15\\E ~ 341.3\\ 320 } 1 etc. A note is sharpened when raised by an interval of 25 - e.g., Clt = \u2014 X 256 = 266.7 v.p.s. 25 24 24 A note is flattened when lowered by an interval of 24 \u2014, e.g., Db = \u2014 X 288 = 276.5 v", ".p.s. 24 25 25 9 10 The ratios - and \u2014 are called major 9 and minor tones respectively, those of \u2014 being called semitones. The intervals o 1 15 between major and minor tones are too large for musical requirements and accordingly intermediate notes are inserted, known as sharps and flats. These are the black notes on the piano, the one between C and D, for example, being C sharp (CJj:) or D flat (D[)). not exactly equal in Thus on the diatonic scale Cjf and Db frequency. are Further practical difficulties arise with the diatonic scale when changing from key to key\u2014a process known as modulation. Thus in the key of D (the scale obtained using D as the tonic) it will be seen that intervals obtained using the white notes of the piano occur in a different sequence (minor tone, semitone, major tone,..) from those obtained when using the key of C (major tone.. Fig. 9:3 The Piano Keyboard Showing Two Octaves of the Chromatic Scale. 83 Chap. 9 SOUND The Scale of Equal Temperament or the Chromatic Scale Number of Note Notation Absolute frequencies Interval ratios 1 G 3 2 C# D Db 4 D# Eb 5 E 6 F 7 FJt Gb 8 9 10 G G# A Ab 11 A# Bb 12 B 13 C' 256 271 287 304 323 342 362 384 406 431 456 483 512 1.059 1.059 1.059 etc. minor tone, semitone,...). In order to correct this in all the various keys the number of \u201cblack\u201d notes required would be very large indeed. This would make keyed instruments such as the piano and to play, although organ very difficult with the violin and the human voice, where a continuous range of frequencies is possible, there is no difficulty in producing the notes required. fixed The difficulty is overcome with instrupitch by adopting a ments of scale of equal temperament or the chroin which the matic scale differences in the intervals of the diatonic the octave being scale divided into twelve equal semitone intervals each of which equals are abolished, (Fig. 9:3) 1.059, e.g., \u2014 = C 256 =r 1.059. There is no difference between 0 % and Db, etc., on this scale and difficulties of modulation are overcome. It should be noted that a small amount of discord is inevitably present in instruments of fixed pitch, such as the piano and organ, which are tuned according to this scale. Thus reference to the tables will show that a chord of the three notes C, E, G, which is known as a major triad, does not have exactly the desired frequency ratio 4: 5: 6 as on the diatonic scale. However, in spite of this imperfection, this scale meets all the requirements admir- ably. 84 Photo by Everett Roseborou^h Ltd, APPLICATIONS OF SOUND Sec. 11:29 11:29 MUSICAL INSTRUMENTS (a) Stringed Instruments (Fig. 9:4). extensive list of these The reader will be able to suggest an All! hav'e a sounding-box or board over which one or more strings are stretched. This is made to vibrate at the same frequency as the vibrating strings to give greater'intensity of sound. The frequencies of the notes emitted by the strings are de, termined by their lengths, tensions, diameters, and densities. Some instruments, : e.g., the banjo-like group, have a fretboard to which the string is pressed, thus pre-determining the length required for a certain note. Those of the violin group have no frets and the performer : must rely on his ear to obtain the desired I note when he presses and vibrates the string. In the piano, there is a string of tension and All stringed in-!' density for each note. diameter, certain length, I j I j j struments produce notes with one or more overtones, their number depending on the manner of vibrating the string and the place where it is bowed, picked, or struck. (b) Wind Instruments These include the pipe and reed organs, the wood-winds and the brass instruments (Fig. 9:5). They involve a means of vibrating a resonant air column either of the open or closed variety. In some the air column is of fixed length and in others it may be varied by means of valves or a sliding telescoping device. The closed pipe or flute type of pipe of an organ is pictured in Fig. 9: 6(a). Compressed air (A) enters the space (C), is forced through the slit (S) and, on striking the lip (L), causes periodic variations in pressure. The length of the pipe is adjusted so that it reson", " beamed and on range. being detected are of such a nature as not to be confused with other vibrations in the water. 11:33 THE FUTURE OF SOUND the No one can foresee future of sound. The properties of sounds in the audible range, that is, 20-20,000 v.p.s., are well understood. However, there is much to be learned about the ultrasonic vibrations whose frequencies are from 20,000 to 500,000,000 v.p.s. The dog whistle (20,000 v.p.s.) and the squeak of a bat (30,000 v.p.s.) are at the lower limit of this group of vibrations. sea was used at Some present uses for ultrasonic vibrations follow. When used at an intensity of 160 decibels or more they have been used to remove the dust and soot from chimney gases. During World War II Sonar (Sound, Navigation, and Rangfor sounding, ing) locating submarines or other ships, and for underwater communication. Peacetime underwater uses include locating schools of fish and sunken ships. Another use is to cause molten metals to set more thus giving them finer grain quickly, structure and, thereby, greater strength. Conversely, it is used by large organizastructural tions materials such as concrete. Still another is in homogenizing milk. The physiological these high-frequency sound waves are only Research beginning to be understood. IS proceeding on measuring the body\u2019s a view to tolerance deriving possible curative values. Truly, the future of sound may be amazing! discover them, effects flaws with of to to in Sea Bed Fig. 9:14 Determining the Depth of the Sea. By means of a timingunder water. device the interval may be determined. This must be halved when finding the depth of the sea. (Why?) Knowing the velocity of sound in water and the time, the depth can be determined. The hydrophone is a special receiver for underwater work designed to respond to vibrations from one direction only In time of war, it serves a useful purpose in locating enemy submarines. 92 APPLICATIONS OF SOUND Sec. 11:34 QUESTIONS II : 34 1. 2. A (a) Describe the larynx. (b) How do we produce sound in the larynx? (c) Give varying reasons the for quality of different voices. (a) What are the three main divisions of the ear? Name the parts and purpose of each. (b) Describe how we hear. 3. (a) Define: tonic, octave, major tone, minor tone, semitone, major triad. (b) Distinguish between diatonic scale, and scale of equal temperament or chromatic scale. orchestra under the headings: (a) stringed (b) wood-wind (c) brass Give at least three examples of each. percussion. (d) 5. Describe how the acoustics of lecture halls may be improved. 6. Describe several ways of recording sounds. 7. In sounding a lake, the time lapse between producing a sound and hearing the echo is 0.75 sec. The velocity of sound in water is 4750 ft. per sec. Calculate the depth of the lake at that point. 8. (a) Distinguish ultrasonic and supersonic. between the terms (b) Give several uses for ultrasonic 4. Classify the instruments of a school vibrations. 93 CHAPTER 10 EXPERIMENTS ON SOUND EXPERIMENT 1 To study transverse vibrations. (Ref. Sec. II;3)j Apparatus A simple pendulum and support, stop-watch Fig. 10:1 Method 1. Attach the pendulum to the support. Draw the bob aside and let it swing freely. 2. With the bob at rest, mark its position by a chalk mark on the table. Draw the bob aside and measure the distance it travels to either side of the rest position. 3. While the bob is swinging measure the time required for 30 complete vibrations. Calculate (a) the number of vibrations per second and (b) the time required for one vibration. 4. Repeat part 3 using a greater and a smaller amplitude. 5. Repeat pai't 3 with a longer and a shorter pendulum. Observations 1. (a) In the simple pendulum, what is the direction of motion relative to the length of the vibrating object? (b) What is one complete vibration? 94 2. 3. EXPERIMENTS ON SOUND How do the distances that the bob swings to either side of the rest position compare in magnitude? Time for 30 VIBRATIONS The number OF VIBRATIONS PER SEC. The time for 1 VIBRATION The given pendulum The same with greater amplitude The same with smaller amplitude The pendulum made shorter The pendulum made longer Conclusions 1. (a) What", " type of vibratory motion is illustrated by the pendulum? (b) Define complete vibration. 2. Define amplitude of vibration. 3. Define (a) frequency of vibration, (b) period of vibration. 4. What efTect has changing the amplitude on the frequency and period of vibration? 5. What effect has changing the length of the pendulum on its fre- quency and period? Questions 1. Why do we call these vibrations transverse? 2. What changes take place in amplitude as the body is allowed to vibrate for a long time? What effect has this on the period or frequency of vibration? 3. Why is the period not dependent on the amplitude? 4. Why is the pendulum a suitable device for controlling a clock? EXPERIMENT 2 To study longitudinal vibrations, (Ref. Sec. 11:3) Apparatus A coil spring, weight, support, stop-watch. Method 1. Suspend the weight from the support by the coil spring. Draw the bob down and release it. Fig. 10:2 95 Chap. 10 SOUND 2. When the weight is at rest, mark its position by a chalk mark on some vertical object such as a ruler. Draw the weight down and note the distance that it travels above and below the rest position. 3. While the weight is moving measure the time required for 30 complete vibrations. Calculate (a) the number of vibrations per second and (b) the time required for one vibration. Observations 1. (a) In the coil spring what is the direction of motion relative to the length of the vibrating object? (b) What is one complete vibration? 2. How do the distances that the weight moves to either side of the rest position compare in magnitude? 3. (a) What is the number of vibrations in 30 seconds? (b) What is the number of vibrations in one second? (c) What is the length of time for one vibration? Conclusion 1. What type of vibratory motion is illustrated by the coil spring? 2. Give the meaning of the terms complete vibration, amplitude, fre- quency and period. Questions 1. Why do we describe these vibrations as longitudinal? 2. What would be the mode of vibration of a tuning-fork at a, b, c, d? Test your answers by exploring the fork with a pith ball. a b c d EXPERIMENT 3 To determine whether or not sound requires a material medium for its transmission, (Ref. Sec. II: 4) Apparatus Bell-in-vacuo (Fig. 6:5), exhaust pump, wax or vaseline, electric wires, two dry cells, switch. Method 1. Seal the bell-in-vacuo onto the pump plate with the wax and connect with the exhaust pump. Connect the bell, cells and switch. Close the switch. 2. Start the pump and gradually evacuate the jar. 3. Stop the pump and let the air slowly return. 96. EXPERIMENTS ON SOUND Observations What changes in loudness are observed? Conclusion What would you be led to conclude about the ability of sound to be transmitted in the absence of a material medium? Questions 1. Why was sound not entirely eliminated? 2. What changes in the apparatus would improve this experiment? 3. What effect does changing the density of the medium have on the transmission of sound? 4. Verify your answer to question 3 experimentally, using different media, e.g., wood, water and air, between your ear and a sounding object, e.g., a waterproof watch. 5. Does light require a material medium for its transmission? EXPERIMENT 4 To illustrate the different kinds and fundamental characteristics of wave motion. (Ref. Sec. II; 5) Apparatus A length of rubber tubing, a long coil spring, two rigid supports. (a) Method 1. Tie one end of the rubber tubing to one of the supports. Tie a piece of string to the tube, leaving one end dangling. Vibrate the free end of the tube up and down with the hand. Note the effect of the vibration on the tube. 2. Attach the coil spring between the two supports. Tie a piece of string Squeeze several coils together and release them. to it as in part ( 1 ) Note the movements of the string. 97 Chap. 10 SOUND Observations 1. (a) What is observed when the end of the tube is vibrated? (b) What is the direction of the motion of the particles of the tubing relative to the length? 2. (a) What is observed when the coils of the spring are released? (b) What is the direction of motion of the coils of the spring relative to the length? Conclusions 1. What are the two parts of transverse waves? Define and give examples of transverse wave-motions. 2. What are the two parts of a longitudinal wave? Define and", " give examples of longitudinal wave-motion. 3. Define; crest, trough, condensation, rarefaction, wave-train. 4. How is the energy from the source of disturbance transmitted through a medium? 5. What is the fundamental characteristic of wave-motion? Questions 1. What is the meaning of wave-length, the amplitude of the wave, the frequency of the wave, the period of the wave? 2. Establish the relationship between velocity, frequency and wave- length. EXPERIMENT 5 To show standing waves in a stretched cord. (Ref. Sec. 11:6) Apparatus Electric bell with gong removed, rigid stand and clamp, pulley and support, pan, with weights, length of light silk cord, batteries, connecting wire. Method Tie one end of the cord to the clapper of the bell and the other to the pan of weights. Assemble the apparatus as in diagram and close the circuit. Adjust the tension of the cord by changing the weights on the pan until the cord takes up a steady appearance. Observations 1. What effect has the clapper on the cord? 98 EXPERIMENTS ON SOUND 2. Wliat happens to this disturbance when it reaches the support at the distant end of the cordi\u2019 3. What is the appearance of tlie cord? Conclusions 1. What causes standing waves in a stretched cord? 2. Define; node, loop. Explain the cause of each. EXPERIMENT 6 To show what determines the pitch of sound. (Ref. Sec. 11:12) Apparatus Savart\u2019s toothed wheel, rotator, cardboard card. Method 1. Assemble the apparatus as in Fig. 7:3 and rotate the disc while holding the card against it. Note the pitch of the sound produced. 2. Repeat part 1 while rotating the disc at a slower rate. 3. Repeat part 1 while rotating the disc at a faster rate. Observations What is observed in each of the above parts? Conclusion What determines the pitch of sound? Questions 1. How would you determine the frequency of the note produced in the above experiment? 2. What is the frequency of the note an octave higher than another? 3. What kind of sound would have been produced by a disc with irregularly spaced teeth? Explain. /Vote Experiment 6 may be done using a perforated disc and a jet of compressed air. EXPERIMENT 7 To determine the law of lengths for vibrating strings. (Ref. Sec. 11:15) Apparatus Sonometer (Fig. 7:6), one steel string, movable bridge, tuning-forks with different frequencies such as 256, 320, 384, 512, 1024 v.p.s. Method 1. Using a string 100 cm. long, adjust its tension until the pitch of sound that it produces is the same as that of the fork whose frequency is 256 v.p.s. Record the frequency of this note and the length of string that produces it in the table below. 2. Without changing the tension, adjust the length of the string by 99 Chap. lO SOUND inserting the movable bridge beneath it. Determine the lengths of string that will produce notes of the same pitch as the other forks provided. Tabulate these frequencies and lengths. Observations 1. What length of string produces a sound of low frequency? 2. What change in frequency of sound occurs as the string is shortened? 3. Table of Results. Frequency OF Note Length of String Ratio of Frequencies Ratio of Lengths t Product of Frequency and Length of String 256 v.p.s. 100 320 v.p.s. 384 v.p.s. 512 v.p.s. 1024 v.p.s. Explanation 1. What is true of the ratio of the frequencies compared to the ratio of the lengths? 2. What is true of the product of the frequency of the note times the length of the string producing it? Conclusion State the law of lengths for vibrating strings.- Questions 1. In the above experiment, what length of string would have a frequency of 768 v.p.s., 128 v.p.s.? 2. In the experiment, what frequency would be produced by a string of length 150 cm., 20 cm.? 3. In part 1, what effect does changing the tension have on the fre- quency of the note produced? 4. List two factors that affect the frequency of a note produced by a vibrating string. EXPERIMENT 8 To illustrate the modes of vibration of vibrating strings, (Ref. Sec. II; 16) Apparatus Sonometer, one steel string, bow, several V-shaped paper riders. 100 EXPERIMENTS ON SOUND Method 1. Place three paper riders at equal intervals along the string. Bow the string at its centre and note the effect on the riders and the pitch of the note produced. Record your results in the table.", " 2. Place three riders on the string as before. Touch the string lightly at its centre and bow the string in the middle of one of the halves. Make the same observ^ations as before and tabulate them. 3. Repeat part 2 using five riders, touching the string one-third of its length from one end and bowing in the middle of this third. 4. Repeat using seven riders, touching the string one-quarter of its length from one end and bowing in the middle of this quarter. 5. Repeat using nine riders, touching the string one-fifth the length of the string from one end and bowing in the middle of this fifth. Observations Position of Damping Effect on Riders Diagram to Si-iow Mode OF Vibration of String Frequency of Note Produced 1. None 2. 1/2 3. 1/3 4. 1/4 5. 1/5 1 1 Conclusions 1. What is the manner of vibration of a vibrating string when producing its fundamental? Its various overtones? 2. What are the frequencies of the various overtones compared to that of the fundamental? 3. Define: fundamental, overtone. Question How do you account for differences in quality of the same note from various sources? EXPERIMENT 9 To study the effect of the superposition of waves on the quality of sound produced, (Ref. Sec. II; 17) Apparatus Four tuning-forks (n = 256, 320, 384, 512 v.p.s.), rubber mallet, hard mallet, bow, tuning-fork on resonance box. Method 1. (a) Vibrate each tuning-fork separately by striking it with the rubber mallet and note the quality and pitch of the sound produced by each. 101'Chap. 10 SOUND (b) Vibrate the forks, n = 256 v.p.s. and 320 v.p.s., simultaneously and note the quality of the note. (c) Repeat part 1 one, two or three of the others. using the fork, n = 256 v.p.s., and with (b) 2. (a) Bow the -tuning-fork on the resonance box and observe the quality of sound produced. (b) Vibrate the same fork by striking it with the hard mallet and again note the quality of sound produced. (c) Vibrate the same fork by bowing it and striking it with the hard mallet simultaneously. Note the quality of this sound. Observations What is observed in the above parts of the experiment? Conclusions 1. What determines the quality of sound? 2. Explain in terms of superposition of waves. EXPERIMENTIO To demonstrate resonance in a closed air column and to find the velocity of sound in air, (Ref. Sec. 11:20) > Apparatus Retort stand, clamp, cylinder, water, open tube about 15 inches long, several tuning-forks (256, 384, 512 v.p.s.). Method 1. (a) Fill a tall glass jar with water at room temperature to the three-quarter mark and place a smaller glass tube, open at both This tube should be clamped to a ends, in the water as shown. retort stand so that it may be raised, lowered or secured at will. Sound a tuning-fork {n = 256 v.p.s.) and hold it close to the open 102 EXPERIMENTS ON SOUND end of the tube. Raise or lower tlie tube until a position is found where the air column resounds most loudly. Measure the length of the air column. (b) Repeat (a) with the other tuning-forks. Observations 1. What was the room temperature? 2. What is observed when the tuning-fork is brought to the top of the tube before and after adjustment of the length? 3. Table of Results. Frequency of Fork Length of Closed Tube in Resonance 256 384 512 Explanation What is the cause of resonance? Calculations 1. What is the wave-length of the note produced by each tuning-fork? 2. Calculate the velocity of sound in air from the result obtained for each fork and average your answers. Conclusions 1. What is resonance? 2. What is the average velocity of sound in air? Questions 1. From the temperature recorded during the experiment, calculate the velocity of sound in air and compare it with the experimental value. 2. Determine the percentage error in your experimental result. 3. What is the relationship between the length of vibrating air column and the frequency of fork that gives resonance? EXPERIMENT 11 To study the production of sympathetic vibrations. 11:23) (Ref. Sec. Apparatus Two matched tuning-forks mounted on resonance boxes (Fig. 8:4), rubber mallet, wax or plasticine. Method 1. Place the tuning-forks close together with the open ends of the Strike", " one fork and silence it after a short boxes facing each other. time. Note what happens. 2. Strike the other fork and repeat part 1. 103 Chap. 10 SOUND 3. Load the prongs of one of the forks with wax or plasticine. Sound the forks separately. What is observed? 1 and 2 using the forks as they are now. What 4. Repeat parts happens? Observation State your observations for each part above. Explanation Account for your observations. Conclusions 1. What are sympathetic vibrations? 2. Explain their cause. Questions 1. What would occur in this experiment if one fork were an octave higher than the other? Try it, and explain. 2. Suggest other examples of sympathetic vibrations. EXPERIMENT 12 To illustrate interference of sound waves by means of a study of silent points round a vibrating tuning-fork. (Ref. Sec. II: 24(a) Apparatus A tuning-fork, rubber mallet, closed air column as in experiment 10. Method 1. Vibrate the tuning-fork and hold it over the closed air column. Adjust the length of the column until resonance results. 2. Now slowly rotate the vibrating tuning-fork while holding it above the air column and observe. Observations Note the changes in the intensity of the sound heard and the position of the tuning-fork for each change. Fig. 10:7 Explanation With the aid of a diagram account for these changes. Conclusion What is meant by interference of sound waves? 104 EXPERIMENTS ON SOUND EXPERIMENT 13 To illustrate the interference of sound waves by the use of Herschel's divided tube, (Ref. Sec. II; 24(b) Apparatus Tuning-fork, rubber mallet, Herschel\u2019s divided tube (Fig. 8:6), rubber hose and ear-trumpet. Method 1. Adjust Herschel\u2019s tube so that the two paths CAD and C B D are equal. Hold a vibrating tuning-fork in front of the opening C and listen at the ear-trumpet joined to D. 2. Gradually draw out A and note the positions of minimum and maximum loudness of sound. Measure the difference in lengths of the two paths for each such position. Tabulate your results. Observations 1. Describe carefully what is heard. 2. Table of Results. Intensity of Sound Difference in Lengths OF Paths (C A D-C B D) ( 1 ) Maximum (2) Minimum (3) Maximum (4) Minimum Explanation 1. Account for the variations in the intensity of sound heard. 2. How could the wave-length of the note be calculated? Calculation Calculate the wave-length of the note. Conclusion How does Herschel\u2019s divided tube illustrate interference of sound waves? Question Knowing the frequency of a sound, how would you determine the velocity of sound in air? 105 Chap. 10 SOUND EXPERIMENT 14 To study the production of beats. (Ref. Sec. II: 24(c) Apparatus Two matched tuning-forks on resonance boxes, rubber mallet, wax or plasticine. Method 1. Vibrate the two forks together and note the result. 2. Change the frequency of one slightly by putting a little wax or plasticine on its prongs (or rearranging the weights if your forks are of this type) and repeat part 1. 3. Make a still greater difference in the frequencies of the forks (how?) and repeat part 1. Observations What do you observe in each part? Conclusions 1. How are beats produced? 2. What governs the number of beats per second? 3. Suggest an explanation for beats. Question How does a musician use beats to tune his musical instrument? 106 UNIT HEAT This? in Role Important an Plays Effect Heat What Jet-Propulsion. by Air the Through Sent is Aircraft This CHAPTER 11 THE NATURE AND SOURCES OF HEAT III : 1 HISTORICAL BACKGROUND Perhaps the greatest single advance made by mankind was the discovery of The presence of how to make fire. ashes and charcoal in caves inhabited in very ancient, times, shows that the discovery may have taken place in the Stone Age or earlier. Man\u2019s mastery of fire has enabled him to protect himself from exposure to cold, to kill diseaseproducing organisms and to frighten off It has allowed him to smelt wild beasts. copper and iron from which new weafashioned. pons Without this discovery man might not have risen from barbarism. could and tools be friction between two dry History reveals the various fire-producing devices used. One was the igniting of dry moss or leaves by heat from the sticks. Another was the production of sparks when a piece of pyrite (fools\u2019 gold, a sulphide of iron) was struck a glancing blow with a sharp piece of flint (a hard, compact mass of silica,", " the mineral of which sand is composed). Both of these are a far cry from the use of such modern devices igniters etc., although ihe principle of the flint gas and still employed in as matches, electrical igniter is cigarette lighters. The many developments in the production of heat that have occurred since early times give ample evidence of man\u2019s creativeness. Equally interesting are the forward strides made in the understanding of the nature of heat and the laws that govern its use. Not only has man learned how to produce heat but he can now control it, retain it, measure it, transfer it from place to place and convert it into motive power. This study of heat is sufficiently large and important to warrant a special branch of physics, namely, \u201cthermodynamics\u201d. Ill : 2 THE NATURE OF HEAT (a) Caloric Theory Until well into the nineteenth cen- in most the spite people believed tury, Caloric Theory of Heat in of questionable proofs to support it. This theory insisted that all empty spaces in matter contained a fluid called caloric and that the warming or cooling of a body was due to the gain or loss in the amount of this fluid. The early Greeks speculated that heat was the rapid vibratory motion of the molecules of a body. Francis Bacon produced some promising experimental evidence to (1561-1626) this effect. In an 1798 American, Benjamin Thompson, who later became Count Rumford, made further investigations. While directing the boring of cannon, he became interested in the amount of 109 Chap. 11 HEAT heat produced in the process and decided to investigate the problem of the nature of heat. The adherents of the caloric theory argued that caloric came out when iron shavings were formed and that the shavings had more of it than the iron. To test this theory, he applied a blunt drill (to produce few shavings) to the cannon with the whole assembly immersed in a box filled with water. In a short time the assembly became warm and finally the water boiled. Because heat was long the mechanism continued to turn, Rumford concluded that anything which could be as heat had produced without limit, been, could not be a material substance (caloric). He reasoned that heat must be caused by a vibratory motion in the produced as as material. In 1799 Sir Humphrey Davy of England dealt the caloric theory the coup de grace by rubbing together two pieces of ice in a vacuum at a temperature The below the melting-point of caloric theory held that as ice contained no caloric it could not melt under these conditions. But melt it did and in doing so afforded yet more proof that heat must be a product of motion. ice. (h) The Kinetic Theory of Matter According to theory, matter is this composed of numerous, tiny, moving called molecules, each being particles separated from its neighbour by empty space. A molecule is defined as the smallest particle of a substance which can exist alone and possess the properties of that substance. Scientists are forced to accept such a theory because they know that gases, and to a lesser extent liquids and solids, diffuse, and that gases do not settle but maintain a uniform pressure on the walls of the container. All the evidence indicates that molecules of gases are separated by comparatively vast distances, e.g., water vapour molecules are no more than ten times as far apart as water molecules, and possess great freedom of movement. Molecules of liquids have freedom of movement but must be fairly close together, since liquids resist compression. In solids the molecules exhibit great cohesion as illustrated by their rigidity. Their closeness is shown by their resistance,to compression. Their movement is said to consist of vibration about certain positions in a prearranged pattern. (c) Heat\u2014A Form of Energy Energy is the ability to do work. All moving bodies can do work and therefore have energy. Heat, which is caused by the motion of the molecules of a body, is capable of doing work and therefore must be a form of energy. THE NATURE AND SOURCES OF HEAT Sec. 111:3 that This may be illustrated with the aid of the apparatus shown in Fig. 11:1. When the tube is heated an effect is imitates somewhat the observed motion of the molecules. The mercury vaporizes and the molecules of mercury vapour drive the bits of glass upward. The pieces move about erratically, colliding with mercury molecules and each Heat is causing the motion of other. the particles (molecules and glass) and hence heat must be a form of energy. Under other circumstances heat energy may be transformed into other forms of energy such as electricity, light, etc. if from the Law of Conservation of Energy. This fundamental law of nature states that energy can neither be created nor destroyed although it may be transformed into any of its many forms. In a moving automobile is practice, brought to a stop by its brakes, the disappears enei\u2018gy changes to heat energy in the brakes. From this and numerous other examples", ", it is evident that heat is produced at the expense of some other form of energy. A brief discussion of several sources of heat (Fig. 11:2) will follow. motion which of (a) Mechanical Action Ill : 3 SOURCES OF HEAT The origin of heat may be inferred In every mechanical operation, the output of work is always somewhat less (a) MECHANICAL (a) MECHANICAL (a) MECHANICAL Compression (c) ELECTRICAL Combustion Fig. 11:2 Sources of Heat. Ill Chap. 11 HEAT than the input of energy, the loss being equal to the amount of energy converted into heat. This is stated thus: input = output -|- heat. All such operations involve friction, because no surface is perfectly smooth. When one body slides or rolls over another, as when the hands are rubbed together, some of the energy devoted to the purpose is converted to heat. Frequently, percussion, which is the sudden stopping of a moving object when it collides with one at rest, e.g., hitting a nail, is employed in mechanienergy of cal motion is converted into molecular motion within both bodies which is mani- Here the operations. fested as heat. Some operations involve compression of gases. We know from operating a bicycle pump that some of the energy applied changes into heat. In the Diesel engine (Sec. 111:31), the heat produced by compression of the air in the cylinder is sufficient to ignite the fuel. (h) Chemical Change When a fuel is burned in air, new are formed and energy is materials released. Such a happening is called a chemical change. Every substance has its share of stored chemical energy, a form of potential energy, and when it burns the products of combustion generally possess less of it than do the original materials. The difference in the amount two represents changed into heat. For example: energies the (c) The Electric Current it This Whenever a conductor carries elecbecomes warmer. tricity, is because electrical energy encounters resistance to its flow in much the same way as water encounters resistance (friction) while flowing through a pipe. Just as the moving water loses energy as it overcomes resistance, so too does an electric current. The electrical energy lost becomes transformed into heat. Heatingelements clearly demonstrate this. (d) The Atom All matter in the universe is made up from about one hundred different kinds of elements. Elements are simple substances that have not been decomposed by ordinary chemical means. They are composed of atoms which are the building-blocks for the molecules of all sub- like those Some atoms, this energy. of stances. uranium or radium, are very large and complex and change into new atoms As they do so a small spontaneously. amount of mass changes into energy some of which becomes heat. There are more details of this process in chap. 32. Huge structures called atomic piles conContrary to popular trol belief, the energy from this source is not amazingly limitless large. For example, it has been estimated that one pound of uranium can produce three million times as much energy as one pound of coal or one pint of oil. Imagine how little uranium would be required to heat your home for one year! nevertheless, but, fuel + oxygen carbon dioxide + water + heat. The total energy of the fuel and of the oxygen equals the total energy of the carbon dioxide and of the water plus the energy which was converted into heat. 112 (e) The Sun Few of us realize the importance of the sun as a source of energy. We accept its daily warmth, and take for granted its energy stored in plant and animal products, in the water vapour of the air, in water and air currents, and in the THE NATURE AND SOURCES OF HEAT Sec. Ill: 3 Fig. 11:3 Trapping Solar Energy. fossil fuels\u2014oil, coal and gas. If the sun were to be suddenly extinguished and all the sources of energy at our disposal were tapped at one time, our accustomed temperature would be maintained for After that we would only three days. quickly freeze to death! In spite of the fact that the earth is only a tiny dot in space, it receives a million-trillion kilowatt hours (Sec. V:75) of energy per year, of which all but.05 per cent slips from our grasp. Green plants trap the major part of this percentage, as follows: carbon dioxide + water + light energy (in the presence of chlorophyll) carbohydrates + oxygen This process is known as photosynthesis. The sun\u2019s internal temperature, estimated at about 20 million degrees centi- grade, is maintained by a complicated process which is essentially the union of 4 atoms of hydrogen to form 1 atom of helium gas. Dr. Hans Bethe at The Bell Telephone Company of Canada. Fig. 11:4 The Solar Battery. 113 Chap. 11 HEAT Cornell University in 1938 showed that there is a decrease in the mass during the process and that", " this is converted to heat. For some time man has dreaded a world scarcity of fuel, knowing that at our present rate of consumption we shall be at that critical point in two or three centuries. Accordingly, research workers are constantly seeking ways and means of using solar energy. Some pin their hopes in part on utihzing an improved photosynthetic process, while others are investigating the use of light-sensitive 11:3, 11:4). As the chemicals (Figs. sun will yield its fabulous supply of energy at the present rate for an estimated 10 billion years, our future is assured, provided inexpensive ways and means of sunlight can be found. collecting that III : 4 QUESTIONS A 1. (a) Present an argument to show that heat is a form of energy. (b) Describe the changes in size and state that occur on intensely heating a piece of iron. Explain each by means of the kinetic theory of matter. 2. Explain each of the following: (a) A bullet is found to be warmer after hitting a target. when concentrated sulphuric acid is added to water. (d) A fuse burns out in an overloaded electrical circuit. (e) The origin of heat from the splitting of the atom. (f) The origin of energy from the sun. 3. State all the energy transformations that are involved in the sequence: sunlight, water-power, electricity, heat from a toaster. (b) Bearings frequently \"burn out\" when they run short of oil. (c) Considerable heat is produced 4. If the heat from 6 tons of coal will heat a home for one year, what mass of uranium (U235) will do the same thing? 114 CHAPTER 12 EXPANSION CAUSED BY HEAT (Fig. 12:2) consists of equal lengths of iron and brass welded together. On being heated, the bar bends with the III : 5 EXPANSION OF SOLIDS Almost all bodies expand on being heated and contract on being cooled. The ball and ring experiment (Chap. 15, Exp. 1), demonstrates the expansion and contraction of metals (Fig. 12:1). That different metals expand and contract by different amounts when heated or cooled through a given change of Fig. 12:2 Unequal Expansion of Solids. Since the brass on the outside. distance round the outside of a curved path is longer than round the inside, the brass has expanded more than the iron for a given change in temperature. On being cooled in a freezing mixture, it bends with the brass on the inside. For this reason, we say that brass contracts more than iron for a given temperature change. The expansion of metals, though small, exerts great force and is of particular concern to the engineer. For example, rivets are heated before being put into 115 temperature is shown by the \u201ccompound bar\u201d experiment (Chap. 15, Exp. 2). The compound bar or bimetallic strip Chap. 12 HEAT Fig. 12:3 Compensating tor Expansion, fa) Bridge fb) Balance Wheel of a Watch. is it the necessary provided To determine the clearances and allowances required, to know exactly the small changes in length, area and volume resulting from temperature changes. These can be calcuappropriate lated that coefficient of expansion is known. Coefficients of expansion are of two kinds, namely, linear and volume. The linear coefficient of expansion of a substance is a number representing the increase in length of a unit length of the substance when its temperature is raised one centigrade degree. The volume coefficient may be defined in a similar way. The Note (Circled) the effect o f atmos- pheric heat on the bolts in the struc- ture of this bridge. place so that they will contract and hold the parts securely in place when cool. Steel bridges are subject to considerable contraction and expansion and for this reason are provided with a small gap at the end while the end may rest on a steel roller supported by the abutment (Fig. 12:3a). For similar reasons, there are expansion bends in metal piping, and gaps between the of railway tracks; hydro and telephone wires are left slack and a slight clearance is provided between the piston and the cylinder walls of internal combustion engines to avoid \u201cseizing\u201d. sections 116 EXPANSION CAUSED BY HEAT Sec. III:6 experimental determination of these co- is beyond the scope of this efficients course in physics but an explanation of their meaning makes the various applications very much more easily understood. Coefficients of Linear Expansion OF Solids Substance Aluminum Copper Iron Nickel Platinum Silver Zinc Brass Invar Steel Glass (flint) (soda) Silica (fused) Pyrex* Coefficient 0.000025 0.000017 0.000012 0.000013 0.0000089 0.000019 0.000026 0.000019 0.0000009 0.000011 0.0000088 0.00000", "85 0.0000004 0.0000036 *Pyrex consists of 80 per cent silica and 20 per cent various oxides of metals, chiefly of boron. 111:6 APPLICATIONS OF EXPANSION OF METALS Some applications of expansion have been mentioned already and it is clear that the expansion of metals, though small, must always be taken into consideration. The errors in using metal surveying tapes have been largely overcome by the use of invar steel, a nickel-steel alloy containing 36 per cent of nickel and having a coefficient of linear expansion which is almost negligible. The same material is used for the pendulums of clocks to ensure almost constant length and accurate time-keeping. Watches are controlled by a metal balance-wheel (Fig. 12:3b) and hairoscillation of the spring, wheel being determined by its diameter. A rise in temperature would cause the diameter of the wheel to increase and. the time of consequently, the watch would lose time. To compensate for this defect the rim of the wheel is made in segments, each being a bimetallic strip of brass and steel with the more expansible metal on the outside. When the temperature rises, the segments curl inward, reducing the \u201cefthe wheel and fective of compensating for the troublesome increase in diameter that would otherwise occur. diameter\u201d strip, with The principle of the bimetallic strip finds other applications. One is the dial thermometer (Fig. 12:4a). The essential part of this instrument is a coiled bithe more again metallic expansible metal on the outside. One end of the coil is firmly attached to the case of the instrument and the other is connected to the pointer. As the coil winds or unwinds with a rise or fall in temperature, the movement of the free end is transmitted to the pointer moving over a scale graduated in degrees. Although not as accurate as other thermometers to be described in Sec. Ill: 8, is a robust instrument and has the it advantage of containing no liquid to vaporize or solidify. Such a device is also an essential part of the thermograph or Fig. 12:4 (a) The Dial Thermometer. 117 Chap, 12 HEAT fb) The Thermo- graph. Compound Bar continuous recording thermometer (Fig. 12:4b). Thermostats for automatically regulating temperature contain a bimetallic Such as the controlling feature. strip devices are pictured in Fig. 12:5. Fig. 12:6 shows a simple circuit controlled Thermostats control the temby one. perature of ovens, refrigerators, rooms, homes and hot-air furnaces. The par- Thermostats fa) Household Fig. 12:5 Type, (b) Type with Mercury Switch. Fig, 12:6 A Simple Thermostat. 118 EXPANSION CAUSED BY HEAT Sec. Ill: 8 ticular type used in hot-water furnaces is called an aquastat. Thermostats may be also used to activate fire-alarms and for a variety of other purposes, where the bending of the metal closes or opens a switch which controls of electricity to the device. In every case the supply involved. j I heat is Ill : 7 EXPANSION OF LIQUIDS I 1 I When liquids are heated, they too expand and, when cooled they contract (Fig. 12:7). The actual expansion or contraction is always somewhat greater i than that observed because of the ex- pansion and contraction of the container. With suitable apparatus, the expansion in volume per centigrade degree, i.e., the coefficient of expansion in volume, may alcoefficient than'be determined. Some liquids, T; cohol, show a larger like is It at others. different coefficient Most, except mercury, show a temdifferent peratures. The table shows a few of these coefficients including some coefficients of expansion for water at various temperathis property of a liquid tures. which often determines its suitability for certain therAgain allowance must be mometers. made for expansion wherever liquids are being heated in conhned spaces, such as in hot-water heating systems (Sec. 111:14). purposes, such in as Coefficients of Cubical Expansion OF Liquids Liquid Alcohol Glycerine Mercury Water 5-10\u00b0C. 10-20 20-40 40-60 60-80 80-100 Coefficient 0.00110 0.00053 0.00018 0.00005 0.00015 0.00030 0.00046 0.00059 0.00070 III : 8 THERMOMETRY (a) Temperatiire If a scale is attached to the long glass tube (Fig. 12:7) while performing the experiment described in Sec. Ill: 7, a form of thermometer registering an increase or decrease in the temperature will be constructed. The temperature of a body may be defined as that condition which determines the direction of heat flow between it and", " its surroundings. Thus, a body at a high temperature will give heat to cooler objects while a body at a low temperature will take in heat from warmer objects. This will proceed until all objects are at the same final temperature. (b) Thermometers Originally, man relied solely on his 119 Chap. 12 HEAT sense of touch to measure temperature. Obviously, judgments obtained in this way are not very precise. For example, a door-knob feels colder to the touch than the wooden door. Again, if one hand is placed in a beaker of hot water, and the other in a beaker of cold water, and then both hands are placed simultaneously in lukewarm water, the first hand will get the impression of coolness and the second that of warmth. Clearly, i'* B W Fig. 12:8 Filling a Mercury Ther- mometer. scale therefore, some means of measuring temperature that is more sensitive and more reliable than that provided by human sensations is needed for scientific purposes. We must have a precise, conof temperature and an sistent instrument for measuring it accurately. The evolution the modern thermometer is an interesting story. Students are advised to consult a good encyclopedia for the contributions of such men as Galileo, Viviani, Rey, Boulliau and others who have shared in its perfection. of 120 to register rapidly. The modern thermometer is constructed from a length of capillary tubing of uniform bore, sealed at one end by heating it in a flame. By gently blowing down the tube when it is hot, a small bulb is produced at B (Fig. 12:8). This should be very thin if the instrument \u2018is After the tube has cooled, a small funnel is attached to the open end A, and clean dry mercury is poured into it. Before the mercury will fill the bore it is necessary to heat and cool the bulb alternately to force the air past the metal. When the tube is full, it is heated to expel any remaining traces of air. The bulb is now placed in a bath of liquid which has a tempera15\u00b0 higher than the ture 10\u00b0 of to ther- temperature which the highest mometer will be required to register. Using the fine blow-pipe flame, the tube is sealed at a point just below the free surface of the mercury. On removing the thermometer from the the mercury contracts in the stem, leaving a vacuum in the space above it. bath, To graduate the thermometer, we choose two fixed temperatures which can be easily obtained, and mark the level of the mercury on the stem when each of these temperatures has been maintained for some little time. The temperatures chosen are the freezing- and boiling-points of pure water at standard atmospheric pressure (760 m.m. of mercury). The former is called the lower fixed point and is marked on the stem by making a groove in the glass with a file at the level of the mercury when it has been standing for some time in melting ice (Fig. 12:9a). To obtain the upper fixed point, the thermometer is placed in the apparatus shown in Fig. 12:9b where the bulb and stem are surrounded by steam. When the mercury level is stationary the upper fixed point is scratched on the stem. If the pressure is not standard, it is necessary EXPANSION CAUSED BY HEAT Sec. Ill: 8 i to apply a correction before making this mark. 100\u00b0C. Upper Fixed 212\u00b0F. Having determined the positions of the fixed points, we divide the distance n V Point f 100 divs. C. = 180 divs. F. ^.'. 1 div. C. = \u2014 div. F. 9 0\u00b0C. Lower Fixed k 32\u00b0F. Point CENTIGRADE SCALE FAHRENHEIT SCALE Fig. 12:10 Comparison of Temperature Scales. Daniel Gabriel Fahrenheit (16861736), a German instrument maker at Amsterdam, selected points 212\u00b0F. and 32\u00b0F. and constructed the scale that bears his name. In choosing fixed the >-55C.\u00b0 100\u00b0C. 45\u00b0C. 30\u00b0C. 15\u00b0C. \u00b0C or Actual Temperatures Changes in Temperatures or Fig. 12:11 The Comparison of \u00b0C and C\u00b0, 121 j grees. Then we test the thermometer I at various temperatures against a stan- dard instrument for accuracy. II - (c) Temperature Scales Two thermometer scales, the Fahren! heit and Centigrade or Celsius are in common use in English-speaking countries. The former is used in everyday practice while is used in science. In countries that are not English-speaking the centigrade thermometer is used for all purposes. latter the i, 'I I Chap. 12 HEAT those fixed points, he was influenced by the incorrect thought that 0\u00b0F. was the lowest temperature that could be reached. The centigrade scale introduced by the Swedish scientist Celsius, in 1742,", " had the fixed points of 100\u00b0C. and 0\u00b0C. boilingrepresenting, and freezing-points of pure water. respectively, -the The comparison of these two scales may be seen by reference to Fig. 12:10 and with its help we are able to convert a temperature on one scale to a corresponding temperature on the other. However, before we attempt any con- versions it should be stressed that \u00b0C. and \u00b0F. refer to actual temperatures whereas C.\u00b0 and F.\u00b0 refer -to a change of temperature anywhere on the scale. For example, difference between 15\u00b0G. and 15C.\u00b0 is shown in Fig. 12:11. the (d) Conversion of Temperatures both scales Since temperature are legal, it is important that we be able to convert a centigrade reading into the corresponding Fahrenheit reading, and vice versa. The following examples will show how this is done. Examples 1. Convert 20\u00b0C. to a Fahrenheit reading. 20\u00b0C. is 20C.\u00b0 above the freezing-point (0\u00b0C.) 100C.\u00b0 = ISOF.r 100 20C.\u00b0 = 20 X - = 36F.\u00b0 5 5 20\u00b0C. is 36F.\u00b0 above the freezing-point (32\u00b0F.) 20\u00b0C. = (32 + 36) = 68\u00b0F. 2. Convert 14\u00b0F. to a centigrade reading. 14\u00b0F. is 18F.\u00b0 below the freezing-point (32\u00b0F.) 180F.\u00b0 = 100C.\u00b0 lF.o=l^ = ^C.\" 180 9 18F.\u00b0 = 18 X - = 10C.\u00b0 9 14\u00b0F. is 10C.\u00b0 below the freezing-point (0\u00b0C.) 14\u00b0F. = (0 \u2014 10) = \u2014 10\u00b0C. The above conversions may be accomplished more conveniently by applying the following formula: not to attempt to use the formula until they have mastered the previous solu- tions. \u00b0C. =^(\u00b0F. -32) However, it is only by a study of the foregoing examples that the reasons for the various operations in the formula will be understood. Students are advised Ill ; 9 EXPANSION OF GASES You will remember from your earlier studies of science that gases expand on heating and contract on cooling (Fig. 12:12). In addition, you will recall that gases expand much more than liquids 122 EXPANSION CAUSED BY HEAT Sec. Ill: 9 and solids for a given change of temperature, i.e., they have a greater coIt may seem expansion. efficient of Fig. 12:12 Expansion ond Contraction of Gases. strange, but is nevertheless true, that all the same gases have This may be coefficient of expansion. exactly almost expressed thus: \u201cAt constant pressure, the volume of a given mass of gas Increases by 1/273 of its volume at 0\u00b0C. for each centigrade degree rise in temperature\u201d. It should be noted that, because gases are compressible, constant pressure must be relationship prescribed this for to hold. A special use of the coefficient of expansion of gases is in determining \u201cabIf we were provided with solute zero\u201d. a tube containing 273 c.c. of gas at 0\u00b0C. it would contain 263 c.c. at \u2014 10\u00b0C., 200 c.c. at \u2014 73\u00b0C., and theoretically, 0 c.c. at \u2014 273\u00b0C. We know, of course, that we cannot destroy matter in this way and the gas would have changed in that state before reaching this temperature. This temperature, \u2014 273\u00b0C., is called absolute zero, a temperature at which bodies all molecular motion having ceased. More accurately, absolute zero is \u2014 273.16\u00b0C. The lowest temperature so far recorded is.005\u00b0 above absolute zero. possess no heat whatever, Absolute zero is the lowest point on another temperature scale, the Absolute or Kelvin Temperature Scale, first proposed by Lord Kelvin, a great English scientist (1824-1907). This finds application in the calculation of the volumes of gases and will be used extensively for that purpose in your chemistry course. Fig. 12:13 shows the relationship between centigrade and Kelvin tempera- 273'^C. 546\u00b0K. I00\u00b0C. -- 373\u00b0K. 0\u00b0C. -- 273\u00b0K. -273\u2018\u2019C, J- 0\u00b0K. Fig. 12:13 Comparison of Centigrade and Kelvin Scales. tures. You will see that to obtain a Kelvin temperture 273 is added to the centigrade reading. 123 Chap. 12 HEAT A practical application of the effect of heat on the volume of a gas is shown in rise the pressure 12:12, the instrument is sealed so that the gas is maintained at constant volume and a rise in temperature causes a proportionate (Fig.", " This pressure change is read 12:14). directly in degrees. For low temperature work hydrogen or helium is used. Above 500\u00b0C. they would diffuse through the bulb and for this reason nitrogen is This device is used to used instead. calibrate thermometers. Ill : 10 THE EFFECT OF EXPANSION ON DENSITY Since changes in temperature cause changes in volume without affecting the mass, densities of substances vary with the temperature. When heat is applied, substances usually expand and a decrease Substances are said in density occurs. to be \u201clighter\u201d then. The opposite effect occurs when they are cooled. A few exceptions to this rule are known, the most outstanding being water. As was shown in Sec. 1:6, water contracts when its temperature rises from 0\u00b0C. to 4\u00b0C. and thus its density increases. This is known as the anomalous behaviour of water. Above and below these temperatures water behaves normally. in the gas thermometer. Instead of allowing the gas to escape as in Fig. Ill : 11 QUESTIONS 1. 2. A (a) A threaded metal cover on a glass sealer fits too tightly. How may it be released? Explain your method, (b) What error would be introduced a into surveyor\u2019s tape made of copper? What material is used to avoid such errors? Why? measurement by using a explain and (a) Describe happens when a bimetallic strip heated and then cooled. (b) What is purpose of the what is the 124 balance-wheel of a watch? How does it accomplish its purpose? (a) How is a centigrade thermometer scale calibrated? (b) Under what conditions is mercury preferable to alcohol as the liquid in a thermometer? Give reasons for your answers. (a) When would gases be used in thermometers? Where and when are such thermometers used? (b) What is absolute zero? (a) Define linear and cubical efficients of expansion. co- 3. 4. 5. EXPANSION CAUSED BY HEAT Sec. Ill: 11 (b) Why does increasing the temperature usually cause a decrease. in the density of a substance? B 1. Find the readings on the Fahrenheit thermometer corresponding tol5\u00b0C., 200\u00b0 C, -60\u00b0C, -273\u00b0C 2. Find the readings on the centigrade Fahrenheit thermometer scale when the reads: 100\u00b0, 350\u00b0, -220\u00b0, -50\u00b0. Fahrenheit and centigrade readings the same? (b) At the Fahrenheit reading double the centigrade reading? temperature what is 4 (a) (i) Express 57\u00b0C, \u2014 23\u00b0C as Kelvin temperatures. (ii) Convert 298\u00b0K., 237\u00b0K. to centigrade temperatures. (b) (i) Express 98\u00b0F., 0\u00b0F. as Kelvin temperatures. (ii) Convert 373\u00b0K., 0\u00b0K. to 3. (a) At what temperature are the Fahrenheit temperatures. 125 CHAPTER 13 TRANSFER OF HEAT is transmitted from molecule to molecule along the length of the bar until the far end becomes hot. Metals are generally good conductors of heat, some better than others. The differences in the conductivities of four different metals may be shown by performing experiment 3, chapter 15, using a conductometer similar to that shown in Fig. 13:1. The relative conductivi- ties of some common metals are shown (The figures used in the in the table. table indicate the number of heat units, calories, conducted in one second by a cube 1 cm. to the edge for each centigrade degree.) III; 12 HOW HEAT IS DISTRIBUTED In a previous chapter we studied the sources of heat energy. Here we shall learn how heat is conveyed from the source so that it may be made to go where it is required, or prevented from is not needed. When a going where it saucepan touches a hot stove it becomes warm: heat has travelled by conduction. The current of warm air above a hot radiator is carried upwards by convection. A fire-place sends out heat by radiation. Thus, the three methods of heat transfer are; conduction, convection and radiation. Ill : 13 CONDUCTION (a) Solids If one end of an iron bar is placed in a fire, the other end will soon become warm. The heat energy has been transferred along the bar by the process of conduction. The rate of vibration of the molecules at the hot end, and therefore their energy, has been greatly increased, and this results in the molecules in sucbar acquiring cessive increased energy by the chain of colliIn this manner heat sions that results. sections the of 126 TRANSFER OF HEAT Conductivities of Some Common Substances Alcohol Petroleum Oil (a) Metals Silver Copper Aluminum Brass Iron German Silver (b) Other Solids Porcelain Glass Cork Sawdust (c) Liqu", "ids Mercury Water.97.92.50.26.16.10.0025.0020.0001.0001.0148.00139 Canadian Pittsburgh Industries Ltd. One Kind of Window Construction. Fig. 13:2 Methods of Reducing Heat Loss. Sec. 111:13.000423.000355.0000568 ( d ) Gases Air Metals are used for many purposes transfer heat efficiently. they because Boilers^ hot-water tanks, cooking utensils and radiators frequently employ copper because of its superior conductivity and durability. However, aluminum may be is lighter, cheaper and used because it more easily shaped. Iron, though heavier Coarse Brick 1\u201d Air Space_ Pope, 1'll Sheathing 1 ] 'j'1 1 ^ Studding I f u \u201e'^ Rock Lath tM : Plaster 127 Chap. 13 HEAT and not as conductive as the others, is used for boilers because it can withstand the forces of high pressure better than the other metals. Certain other solid substances, such as china, plastic material, wood, bone, etc., are employed for their low conduci.e., their inability to pass heat tivity, from molecule to molecule readily. These substances are known as insulators. Further examples are cork used in the walls of refrigerators, rock wool, used in outer walls of homes and the wrapping of water tanks, wool for clothing and bedding, glass and porcelain food containers, sawdust in ice-houses, and many others. Some of these, because of the looseness pockets of which contribute immeasurably to their insulating properties, air being a most contain texture, their air efficient insulator. (h) Liquids When a test-tube full of water or other liquid is held at the bottom while the top is heated by a Bunsen burner is noted that while the (Fig. 13:3), it bottom remains cold the liquid above Fig. 13:3 Poor Conductivity of Liquids. may be boiling. liquids are poor conductors of heat. If the positions of the hand and burner This indicates that 128 are reversed, heat will be transferred quickly from bottom to top, but by convection currents (Sec. 111:14), not by conduction. To avoid confusion, it should be stressed that the rule about the poor conductivity of liquids does not include mercury, which, being a metal, is a good conductor. (c) Gases If the hand is held close to a Bunsenburner flame, the resulting burn is not as intense as when gripping a metal bar at the same distance from the flame. This demonstrates that air (or any gas) is a poor conductor of heat. When we recall that gases are composed of molecules that are very far apart and that heat conductivity depends on the actual contact between molecules, we understand why gases are poor conductors of heat. The above fact concerning gases has many practical applications. In part (a) of this section, reference was made to certain solid insulating materials with loose texture. Many of these, such as fur, wool, sawdust, rock wool, asbestos, snow, etc., depend on the poor conductivity of pockets of air trapped in them for a large insulating properties. Storm-windows, thermopane and the hollow construction of the outer walls of buildings (Fig. 13:2) likewise have insulating value because of the poor heat conductivity of the enclosed their part of air. Ill : 14 CONVECTION (a) Liquids (Chap. If a small crystal of potassium permanganate is dropped into a beaker of cold water heated gently by a Bunsen red burner 15, streaks will be observed as the crystal dissolves (Fig. 13:4). The streaks will rise, move just under the surface of the water for some distance, and fall. Some of the colour may be seen to return to 4A), Exp. TRANSFER OF HEAT Sec. 111:14 If we realize that its point of origin. different parts of the liquid in the beaker have different temperatures, then the streaming of the colour signifies that there are rising and falling currents in the water caused by these differences in temperature. These currents are known as convection currents and are the means by which the heat is circulated through water and liquids. The movement is established because of the expansion and consequent decrease in density of the water immediately above the source of heat. The mass of hot water is being pushed up continually and replaced by the surrounding denser water. Convection, then, is the transfer of heat in a substance by the actual, sometimes observable, motion of its parts. It should Fig. 13:4 Convection Currents in Liquids. Fig. 13:5 Applications of Convection Currents in Liquids. (a) Domestic Hot-water Supply (b) Hot-water Heating 129 Chap. 13 HEAT be clear from the nature of solids, in which the", " molecules occupy certain fixed positions, that convection is impossible, is possible in liquids and gases but it where the molecules have greater freedom of movement. The applications of convection currents in liquids are numerous but only a few will be mentioned. In nature, ocean drifts, which are produced by differences in the temperature of seawater, are an interesting example. The Gulf Stream is a well-known illustration. Others, such as domestic hot-water supply and hot-water heating are presented diagrammatically 13:5). Students should study these and explain how each operates. above (Fig. (h) Gases If a glass tube of large diameter is lowered over a burning candle (Fig. 13:6a), the flame will burn fitfully and then go out unless an air inlet is proIf some burning vided at the bottom. smoke-paper (blotting-paper soaked in concentrated potassium nitrate solution Hot is It heated, forcing the and dried) is held close to the bottom, the smoke will move into the tube and upward. is evident that the cool, dense air less This movement of dense air upward. the air caused by a difference in the temperatures of its various parts and the resulting difference in densities is called All gases show a convection current. this phenomenon. The apparatus used to illustrate convection currents in gases Fig. 13:6b shows the is quite varied. form used in Exp. 4B, Chap. 15. (air) The applications of convection curare commonplace rents in gases and not too intricate for the student to explain. A partial list includes ventilation, draught in a chimney, circulation of air in a refrigerator, hot-air heating and winds. Although all winds are caused by convection currents, we shall confine our attention to land- and sea-breezes (Fig. 13:7). During the day, the land warms faster than In consequence, land will be hotter than that above the water adjoining water. air above the the the (a) 130 Fig. 13:6 Convection Currents in Gases. TRANSFER OF HEAT Sec. 111:14 Night Fig. 13:7 On-shore and OfF-shore Breezes. (i.e., and a giant convection current, a seaConbreeze on-shore) versely, at night, the land will cool faster by radiation (Sec. 111:15) and the reverse in a land- situation will results. result breeze (i.e., off-shore). Hot-air heating systems (Fig. 13:8) depend upon convection currents for However, both the transfer of heat. bution of heat on cold windy days, when it is hard to heat the windward This situation is side of the building. largely corrected by the use of \u201cforcedair\u201d heating where a motor-driven fan This system accomplishes the transfer. has the further advantage that the air is \u201cconditioned\u201d, that is, dust is filtered out and the humidity is more efficiently Fig. 13:8 Hot-air Heating (a) Pipeless (b) Conduit Type (c) Forced-air. 131 Chap. 13 III : 15 RADIATION (a) Introduction HEAT tions, When you stand before a camp-fire, you are aware of its intense heat. Since the effect may be prevented by holding up a blanket between yourself and the fire, you will conclude that the energy travels in straight lines. Since the same thing happens on all sides of the fire, this energy must radiate in all direci.e., travel along the radii of a sphere with the fire at the centre. This kind of energy is a form of radiant energy and the method of transfer is It is only when this called radiation. energy strikes an object and is absorbed that it changes to heat energy. Transfer of energy by radiation is different from conduction and convection since the latter require a material medium, whereas radiation may proceed through a vacuum. For example, radiant energy from the sun traverses 93 X 10\u00ae miles of space, most of which is empty. Or again, energy may radiate from the filament to the glass envelope of an evacuated radio tube. Radiant energy is a wave-motion and has many properties in common with is the radiation light. The major heat effect comes from the infra-red radiations just beyond the red of the visible spectrum (Sec. IV: 38). Subsequent references to radiant energy in this section refer to these infra-red radiations. All bodies whose temperatures are above absolute zero (Sec. III:9) emit this kind of energy at the expense of the energy of motion of their atoms or molecules. The rate of emission and the wave-length of proportional to the temperature: the higher the temperature, the faster the rate and the shorter the wave-length. The waves are believed to be of the transverse variety which, according to one theory, are set up by a minute pulse", " of energy, called a quantum, from the source. These waves are a part of the great electromagnetic family of waves (Fig. 19:4) that includes visible light. X-rays, ultraviolet rays, radio waves, cosmic rays, all of which have a velocity of etc., 186,000 miles per sec. For their transmission, early physicists invented an imaginary, weightless, all-pervading medium called ether, but the Theory of Relativity proposed by Einstein denies its existence. The nature of the medium still remains a mystery. 132 Light Surfaces. TRANSFER OF HEAT Sec. 111:15 (h) The Emission of Radiant Energy (c) The A bsorption of Radiant In the introduction to this section, it was stated that radiant energy is released at the expense of motion of the molecules. It may well be asked whether or not all objects under the same conditions emit this form of energy. To find the answer, experiment 5, chapter 15, should be performed. For the purposes of discussion, let us take two cans (Fig. 13:9), one dark and dull on the outside, the other light and shiny, but identical in other respects. Energy As this objects absorb previously, was suggested heat results when radiant energy is absorbed. A critical thinker will want to know if energy different equally well. To answer this in part, experiment 6, chapter 15, should be performed. Another demonstration (Fig. 13:10) involves two thermometers, one darkened and dulled by the soot from a candle flame, the other left light and shiny, placed at equal distances on either side of a source of heat such as a Bunsen burner. The temperature of the one with the dark, dull surface rises more quickly than that with the light, shiny surface. We know that dark, dull surfaces absorb light without reflecting much of it and a light, shiny one reflects most of the light without absorbing much of it. In Comparing the Ability of Fig. 13:10 Dull Dark and Shiny Light Surfaces to Absorb Radiant Energy. Place a quantity of hot water and a thermometer in each and support them Although on identical insulating-bases. both are at the same temperature initially, the water in the dark, dull can cools more quickly than that in the light, shiny one. No matter how we perform such an experiment we always find that dark, dull surfaces are good emitters of radiant energy, while light, shiny ones are poor in this respect. It is admitted that other factors, such as starting temperature and area of surface also affect the rate but for the purpose of our discussion, these were kept constant. Can Pratt and Whitney Aircraft. Radial Engine of Airplane. Note cooling fins on cylinders. the same manner dark, dull surfaces are good absorbers and poor reflectors of radiant energy, while light, shiny sur- 133 Chap. 13 HEAT faces are poor absorbers and good reflectors. Knowing this we wear lightcoloured clothes in summer and dark ones in winter. (d) The Transmission of Radiant Energy are Certain materials \u201ctransparent\u201d or \u201copaque\u201d toward radiant energy just as some are toward light. As an example, ice does not transmit much radiant energy, while rock salt transmits almost all that falls upon it. Glass, on the other hand, transmits well the shorter wavelengths that originate from a high-temperature source like the sun but does not transmit the longer ones that originate from a low-temperature source such as the earth or a living object. This property of glass makes it greenhouses (part (f) below). useful in (e) Some Detectors of Radiant Energy The simplest device is the darkened air-thermometer, or thermoscope, where radiant energy is converted into molecular motion which manifests itself as a rise of temperature (Fig. 13:11a). The radiometer (Fig. 13:11b) consists of an almost completely evacuated glass bulb in which four light aluminum vanes are mounted so as to turn easily. One side of each vane is blackened while left shiny. When radiant the other is energy falls upon the vanes, the black surfaces become warmer than the others. Accordingly the few air molecules adjacent to the black sides will become heated and will move away from the vanes. The reaction of the vanes causes them to turn about their pivot. The more radiant energy that enters, the faster will the vanes turn. This instrument is very sensitive to small amounts of radiant energy. 134 Fig. 13:11 Some Detectors of Ra- diant Energy, (a) Thermoscope. (b) Radiometer. (f) Applications of Radiant Energy The vacuum or thermos bottle (Fig. 13:12) is a double- walled glass bottle, with a high vacuum between the walls, contained in a suitable protective carrying case. The inner glass walls facing each other are silvered. Liquids, whether hot or cold, will remain at very nearly", " the same temperature for several hours. The reason is that the bottle is so constructed that it is very difficult for heat to be transferred by any of the three methods described above. We shall consider the storing of a hot liquid here. Similar explanations obtain for a cold liquid. 1. The vacuum prevents the loss of heat by conduction owing to the lack of molecules present. The transfer of heat through the glass and the stopper is slow owing to the poor conducting property of each. 2. Convection from inside the bottle TRANSFER OF HEAT Sec. 111:15 reflected back inside. Thus, these devices act as heat traps for the energy from the sun. A further application, the screening action of the clouds, depends on the inability of water to transmit radiant upwards is prevented by the stopper. Loss by convection in the air space between the glass and the case is prevented by being closed at the top. 3. Heat loss by radiation is prevented by the silvered surfaces of the walls. These reflect back into the bottle any radiant energy that tends to escape. Greenhouses and cold frames (Fig. are heated by radiation. The 13:13) short wave-length radiant-heat energy from the sun is readily transmitted by the glass. This is absorbed by the plants, etc. within; as their temperatures soil, rise, they lose heat by radiation. Since this longer wave-length radiant energy is not transmitted by the glass it is largely Shiny Metal Cap Cork Stopper Double-Walled Glass Bottle Silvered Inside Vacuum Silvered Outside Metal Case Spring Felt Fig. 13:12 The Thermos Bottle, Fig. 13:13 A Greenhouse Acts as a \"Fleat Trap\". well. The moisture energy present in the atmosphere absorbs much of the sun\u2019s heat by day, thereby preventing the scorching of plant and animal life. At night the clouds provide a blanket which prevents the escape of radiation from the earth\u2019s surface, the temperature of which is largely maintained. On the other hand, in hot, dry, arid regions, the absence of water vapour results in extreme temperature changes, being very hot by day and very cold by night. We have included these few applications for their general appeal since they come within the realm of everyone\u2019s experience. However, many other applications are to be found both in nature and elsewhere. It is to be hoped that with this introduction to the subject, the student will be able to recognize others as he encounters them. 135 Chap. 13 HEAT Installation for radia heating ant in building. Anaconda American Brass Ltd. III : 16 QUESTIONS 1. Name three methods of heat transfer and explain how they are involved in heating water in a kettle over an electric (b) Should the bottom of a kettle be polished for economical heating? Explain. 5. Make a chart comparing conduction, convection and radiation, under the following headings (a) the media in which the transference takes place, (b) direction of the transference, (c) a brief comparison of the theories which explain how the transference occurs. 7. 6. Explain the action of a radiometer. (a) Make a labelled diagram of a thermos bottle. (b) Write a note to show how (i) conduction, (ii) convection, (iii) radiation are reduced to a minimum when a hot liquid is placed in the bottle. heating-coil. it, in fact, 2. (a) On a cold day, why does the metal door handle feel colder than the wooden door? Is colder? (b) Name three good conductors and three good Insulators of heat, and state the use for each. (a) What are convection Explain how they are produced. (b) Explain the production of an on- currents? 3. shore breeze. 4. (a) Why does more rapidly when dirty, than when clean? snow melt 136 CHAPTER 14 MEASUREMENT OF HEAT III; 17 WHY WE MEASURE HEAT We know that heat is a form of energy (Sec. Ill :2c) and that other forms of energy can be changed into heat, but why do we bother to measure it? Were we required to determine the efficiency of an electric heater, the energy yield when a gallon of gasoline, a pound of tablespoonful coal sugar or of is a Fig. 14:1 Distinction between Quan- tity of Heat and Temperature. burned, we should be able to measure the quantities of heat produced. Various fuels and foods are used widely because of their large energy content. Hi: 18 THE COMPARISON OF QUANTITY OF HEAT AND TEMPERATURE When two equal masses of water are heated by the same source for the same length of time, each will show the same If the experiment rise in temperature. is repeated with one mass larger than the other (Fig. 14:1), the smaller mass will show a greater rise in temperature. If two unequal masses of", " water are heated to the same temperature by the same source, the larger mass will require to be heated for a longer time. is evident that temperature and quantity of heat are entirely different and should never be confused. It Ill : 19 FACTORS THAT INFLUENCE THE QUANTITY OF HEAT We are all familiar with the fact that a basin of hot water may be cooled by the addition of cold water and that the final temperature of the mixture will be lower than that of the hot water and higher than that of the cold. We realize that the hot water becomes cool as ft gives heat to the cold water while the cold water becomes warm because it gains heat from the hot. This is the principle of heat exchange and it applies whenever substances at different temperatures are mixed (Sec. 111:22). 137 Chap. 14 HEAT : at each different equal masses of water, Let us mix two equal masses of water, with temperatures, the same at Since the warmer water temperature. gives rise to the higher final temperature (see example), the mass at the higher temperature obviously contains the greater quantity of the heat. quantity of heat contained in a body varies as its temperature. For example: When 100 gm. of water at 80 \u00b0C. are added to 100 gm. of water at 20\u00b0C., the final temperature is 50\u00b0 C. When 100 gm. of water at 40\u00b0C. are added to 100 gm. of water at 20\u00b0C., the final temperature is 30\u00b0C. Therefore, Let us mix two different masses of water, at the same temperature, each with equal masses of water, at the same temperature. As the larger mass gives rise to the higher final temperature (see example), it follows that the larger mass larger quantity of heat. contains the Thus, the quantity of heat contained in a body varies as its mass. For example: When 100 gm. of water at 80\u00b0 C. are added to 100 gm. of water at 20\u00b0C., the final temperature is 50\u00b0C. When 200 gm. of water at 80\u00b0C. are added to 100 gm. of water at 20\u00b0C., the final temperature is 60\u00b0 C. So far, we have dealt with quantities of water in the above examples, but what would be the effect of using one different substance along with water? us mix two equal For example, let masses, one of water and one of iron filings, at the same temperature, with two equal masses of water, also at the same temperature. The water will give rise to the higher final temperature (see example), because it contains more heat than the iron. Hence, the quantity of heat contained in a body depends upon the nature of the material of which it is composed. For example: When 100 gm. of water at 80\u00b0C. are added to 100 gm. of water at 20\u00b0C., the final temperature is 50\u00b0C. When 100 gm. of iron filings at 80\u00b0 C. are added to 100 gm. of water at 20\u00b0C., the final temperature is 26\u00b0C. Ill : 20 THE UNITS FOR MEASURING THE QUANTITY OF HEAT Because water is a common substance and its capacity for heat is so great, it is used as a reference material in defining the units for measuring the quanIn the metric system, the tity of heat. unit of quantity of heat is the calorie. A calorie is the quantity of heat gained or lost when the temperature of one gram of water rises or falls one centigrade degree. How many calories of heat are gained by 100 gms. of water when its Example temperature rises from 20\u00b0G. to 80\u00b0C.? Change in temperature = 80 \u2014 20 = 60C.\u00b0 Quantity of heat required to raise the temperature of 1 gm. of water 1C.\u00b0 = 1 cal. 100 gm. of water 1C.\u00b0 = 100 X 1 = 100 cal. 100 gm. of water 60C.\u00b0 = 100 X 1 X 60 = 6000 cal..*. Quantity of heat gained = 6000 calories. 138 MEASUREMENT OF HEAT Sec. 111:21 In the British system, the unit of quantity of heat is the British Thermal Unit. One B.T.U. is the quantity of heat gained or lost when the temperature of one pound of water rises or falls one Fahrenheit degree. Example How many B.T.U. are lost when 100 pounds of water cool from 170\u00b0F. to 100\u00b0F.? The change in temperature 170 \u2014 100 = 70F.\u00b0 Quantity of heat lost by: 1 lb. of water cooling 100 lb. of water cooling 100 lb. of water cooling 70F.\u00b0 = 100 X 1 X 70 = 7000 B.T.U. 1F.\u00b0 = 1 B.T.U. 1F.\u00b0 = 100 X 1 = 100 B.T.U. I j 1 1 Quantity of heat lost = 7000 B", ".T.U. Note 1. The calorie used when measuring the energy content of foods and fuels (sometimes called the kilogram calorie), is equivalent to 1000 of the calories above. 2. 1 B.T.U. is equivalent to 252 calories. Ill : 21 SPECIFIC HEAT To find the quantity of heat gained or lost by a substance other than water, we must take into account the nature of the substance as well as its mass and the change in its temperature. The calculation is done by multiplying the mass by the change in temperature by a quantity, related to the nature of the substance, called the specific heat. The specific heat of a substance is a number representing the quantity of heat gained or lost by a unit mass of substance when its temperature rises or falls one degree. In the metric system, this is the number of calories of heat gained or lost when the temperature of one gram of the substance rises or falls one centigrade degree. is the number In the British system, it of B.T.U. gained or lost when the temperature of one pound of the substance rises or falls one Fahrenheit degree. The Specific Heats of Some Common Substances Substance Specific Heat Substance Specific Heat Water Alcohol Ice Steam Aluminum 1.000 0.548 0.500 0.500 0.214 Iron Copper Silver Mercury Lead 0.110 0.092 0.056 0.033 0.031 139 : : Chap. 14 HEAT Example 1 How much heat is gained by 50 gm. of mercury when its temperature rises from 20\u00b0G. to 60\u00b0C.? Solution 1 Change in temperature = 60 \u2014 20 = 40C.\u00b0. Quantity of heat required to raise the temperature of 1C.\u00b0 =.033 cal. 1 gm. of mercury 1C.\u00b0 =.033 X 50 cal. 50 gm. of mercury 50 gm. of mercury 40C.\u00b0 =.033 X 50 X 40 = 66 cal. Quantity of heat gained = 66 calories. Solution 2 Change in temperature = 60 \u2014 20 = 40C.\u00b0 Quantity of heat gained = mass X change in temperature X specific heat = 50 X 40 X.033 = 66 cal. Quantity of heat gained = 66 calories. How much heat is lost by a piece of iron weighing 10 lb. when it cools from 150\u00b0F. to 70\u00b0F.? Example 2 Solution 1 Change in temperature = 150 \u2014 70 = 80F.\u00b0 Quantity of heat lost by 1 lb. of iron in cooling 1F.\u00b0 =.110 B.T.U. 1F.\u00b0 =.110 X 10 B.T.U. 10 lb. of iron in cooling 10 lb. of iron in cooling 80F.\u00b0 =.1 10 X 10 X 80 = 88 B.T.U. Quantity of heat lost = 88 B.T.U. Solution 2 Change in temperature = 150 \u2014 70 = 80F.\u00b0 Quantity of heat lost = mass X change in temperature X specific heat = 10 X 80 X.110 = 88 B.T.U. Quantity of heat lost = 88 B.T.U. 111:22 THE PRINCIPLE OF HEAT EXCHANGE IN MIXTURES As was explained in Sec. 111:19, whenever two substances different temperatures are mixed or in contact, heat passes from the warm one to the cool one until both have attained the at same temperature. Because heat is a form of energy and energy can neither be created nor destroyed (Sec. HI: 3), it follows that the quantity of heat lost by the warm body equals the quantity of heat gained by the cool one. This is the principle of heat exchange. Example 1 A piece of lead weighing 200 gm. and at a temperature of 100\u00b0C. is placed in water and the final temperature of the mixture is 25 \u00b0C. How much heat is transferred to the water!* 140 MEASUREMENT OF HEAT Sec. 111:22 M = 200 gm. Lead Change in temperature 100 - 25 = 75C.\u00b0 100\u00b0G. 25\u00b0G. Water Quantity of heat lost by the lead = mass X change in temperature X specific heat 200 X 75 X.031 = 465 cal. Heat lost by the lead = gained by the water.. Quantity of heat gained by the water = 465 cal. \u2018. Example 2 A mass of 200 gm. of mercury at 100\u00b0C. is mixed with an unknown mass of water at 20\u00b0 C. and the final temperature is 25 \u00b0C. Find the mass of the water used. Mercury M = 200 gm. S =.033 Change in temperature = 100 \u2014 25 = 75C.\u00b0 100\u00b0C. 25\u00b0G. 20\u00b0C. Quantity of heat lost by the mercury = mass X change in temperature X specific heat = 200 X 75 X.033 = 495 cal. Water M = gm. S= 1 Change in temperature 25 - 20 = 5G.\u00b0 Heat lost by mercury = heat gained by", " the water 495 = 5x Quantity of heat gained by the water = mass X change in temperature X specific heat = a: X 5 X 1 = 5x cal. Mass of water used 99 gm. 141 Chap. 14 HEAT Stirrer Bakelite Cover Polished Aluminum Inner Vessel Fibre Ring Polished Aluminum Outer Vessel (a, (b) Canadian Laboratory Supplies Ltd. Fig. 14:2 The Calorimeter. Ill : 23 THE CALORIMETER Students will recognize at once that an operation such as is outlined in ex- 142 Fig. possible. ample 2 above cannot be conducted without some heat exchange with the surroundings. An instrument called a calorimeter (literally, heat measurer) is designed to prevent as much of this as Its component parts are seen 14:2. The small metal can, in or inner vessel, of known mass and specific heat is smooth and shiny to avoid radiation of heat from within and absorption from without. a container during experiments and the heat that it absorbs may be calculated It is supported by a fibre ring readily. which is a poor conductor of heat. The outer metal can is smooth and shiny like the inner vessel to reduce radiation and absorption. The dead air space between the cans reduces loss of heat by Loss of heat through conconduction. vection is avoided by the cover. Since is made of wood, a poor conductor of heat, the cover also prevents loss of heat by conduction. The stirrer, of the acts as It it MEASUREMENT OF HEAT Sec. 111:23 same material as is used to permit uniform mixing of the the inner vessel, contents. The materials put into the inner vessel must be so chosen that they do not interact to generate heat. Further, they must not react with the metal. The final temperatures of the materials should not be far above or below room temperature, If the final temperature for best results. is above room temperature, it should be arranged that the initial temperature be the same number of degrees below room temperature and vice versa. An example of a problem involving a a calorimeter follows: Example A mass of 200 gm. of mercury at 113\u00b0C. is mixed with an unknown mass of water at 18\u00b0C. contained in a calorimeter, the inner can and stirrer of which has a mass of 100 gm. and a specific heat of.22. If the final temperature is 23 \u00b0C., find the mass of the water. Mercury Water Calorimeter Quantity of heat lost by the mercury =: mass X change in temperature X specific heat = 200 X 90 X.033 = 594 cal. Quantity of heat gained by the water = mass X change in temperature X specific heat = V X 5 X 1 = 5.V cal. Quantity of heat gained by the calorimeter = mass X change in temperature X specific heat = 100 X 5 X.22 = no cal. Heat lost by the mercury = heat gained by the water + heat gained by the calorimeter. 5v + 110 594 5v = 594 \u2014 no = 484 X = 96.8 Mass of water required = 97 gm. 143 Chap. 14 HEAT Another form of calorimeter, the bomb calorimeter (Fig. 14:3) is used in the determination of the energy content of foods and fuels. Some typical results follow: Calorific Values of Some Common Fuels Fuel B.T.U. per lb. Fuel B.T.U. per cu. ft. Gasoline Fuel Oil Alcohol Soft Goal Hard Goal Wood (average) 20,750 18,500 11,600 14,000 11,600 5,000 III : 24 FINDING THE SPECIFIC HEAT OF A METAL The method generally employed is known as the method of mixtures. The substance whose specific heat is to be determined, say a metal, is mixed with a material which absorbs its heat, say water. From the various observations within the calorimeter the specific heat of the metal can be calculated. A brief summary of the method and a model the Full solution follow. details of Propane Acetylene Natural Gas Goal Gas 2,450 1,450 1,000 300 method are to be found in experiment 7, chapter 15. is A known mass of copper shot heated to a known temperature in a water boiler (Fig. 15:5). The metal is transferred to a certain mass of water at a known temperature contained in the inner vessel of a calorimeter. The mass and specific heat of the calorimeter are known. The mixture is stirred until the highest constant temperature is obtained. A table of data and a model calculation follow. = 100 gm. = 200 gm. Mass of the calorimeter vessel Mass of the vessel and water Mass of the water = 200 \u2014 100 Mass of the copper shot 100 gm. = 200 gm. = 95 \u00b0G. Initial temperature of the copper Initial temperature of the water and vessel = 15\u00b0C. Specific heat", " of the water Specific heat of the calorimeter Final temperature of the mixture Let the specific heat of the copper = 1 =.22 = 25.5\u00b0C. = x 144 MEASUREMENT OF HEAT Sec. 111:25 Change in temperature zi: 95 - 25.5 = 69.5C.\u00b0 Quantity of heat lost by the copper = 200 X 69.5 X X = 13900x cal. Copper M = 200 gm. S = a: 95\u00b0C. Water Change in temperature = 25.5 - 15 \u2014 10.5C.\u00b0 Quantity of heat gained = 100 X 10.5 X 1 = 1050 cal. Quantity of heat gained by the calorimeter = 100 X 10.5 X.22 z= 231 cal. Heat lost by the copper = heat gained by the water + heat gained by the calorimeter. 13900x = 1050 + 231 = 1281 1281 X =\u2014 =.092 13900 the specific heat of the copper shot =.092 As in all experiments, some error is unavoidable. Some heat, not accounted for in our method, will be absorbed by other parts of the calorimeter, by the thermometer and a small amount will escape by the methods of heat transfer. With care these losses are quite small. Ill : 25 APPLICATIONS OF SPECIFIC HEAT Specific heats affect our lives more than we realize. Water has the highest specific heat of all common substances Substances with a low (Sec. 111:21). specific heat undergo a great rise in temperature when a given quantity of heat is absorbed. When cooled, those same substances undergo a large drop in temperature. On the other hand, water gains or loses a great quantity of heat without much change in temperature. The high specific heat of water makes it useful in the cooling system of an engine and in automobile hot-water In each case it absorbs large heating. quantities of heat and carries it to a 145 Chap. 14 HEAT radiator to be dissipated. Because it has a higher specific heat than land, water does not reach as high a temperature during the summer season or during the day. In the winter season, or at night, water will not cool to as low a temperature as land for the same reason. Thus temperatures over water or near it will always be more moderate than inland. For example the Niagara region has a more moderate climate because of the water round it. The Prairie Provinces, on the other hand, will experience extremes of temperature since there are no moderating influences. The daily differences in temperature referred to above are also responsible for land- and sea-breezes (Sec. 111:14) in coastal reAgriculturalists know well that gions. wet soils do not warm as rapidly in because this high spring specific heat of water. Dry, sandy soils warm up more quickly, produce crops earlier and frequently yield more than one crop in a season. dry, of as Metals, as a rule, have low specific 111:21), and this makes heats them ideal for cooking utensils. (Sec. Ill : 26 HEAT EXCHANGE DURING CHANGES OF STATE gases, states three either physical intermediate in All matter is found in the solid, liquid or gaseous state. These are the of matter. Each state of matter consists of moving molecules separated from each other by spaces that vary with the state, largest liquids and in smallest in solids. The rate of motion of the molecules is faster, and the amount of space between them larger, at higher temperatures. Each state may be converted into one of the others by the addition or removal of heat. In solids and liquids, there is a force of attraction between the molecules known as the force of cohesion which must be overcome by the absorption of heat energy before a liquid or gas state may result. Fig. 14:4 shows these changes in state diagrammatically. The heat exchange during melting (fusion) can be illustrated by stirring some chopped ice or snow with a thermometer while warming it very gently over a low flame. The temperature is 0\u00b0C. when we begin and does not rise Heat Added Fig. 14:4 Changes of State. 146 MEASUREMENT OF HEAT Sec. 111:26 a wide range of temperature. We should realize also that freezing occurs at the same temperature as melting and that the same quantity of heat is released during freezing as was absorbed during melting. Moreover, when the temperature of a mass of substance is kept at its freezing-point without any change in the quantity of heat, melting and freezing are both occurring at the same rate, i.e., equilibrium will exist between the ice and the water. It is only when heat is added or removed that one or other process predominates. To illustrate heat exchange during boiling, let us heat a quantity of water from 0\u00b0G. to 100\u00b0C., stirring constantly with a thermometer. It is found that the temperature does not rise above 100\u00b0C., although heat is being", " absorbed continually. The heat is being used to overcome the force of cohesion rather than the but, since until all the ice has melted. Heat is being temperature absorbed, does not rise, the heat is being used to melt the ice, that is, to overcome the force of cohesion between the molecules. The temperature at which the solid becomes a melting-point. Some substances, ice, sulphur or like salt, have a definite melting-point, while others, like glass, wax and tar, melt over liquid the is Pressure (a) High Pressure Raises Boiling Point. The Pressure Cooker. (b) Low Pressure Lowers Boiling Point. A Commercial Evaporator. (c) Graph Relating Pressure and Boiling Point. Fig. 14:5 The Effect of Changes of Pressure on the Boiling Point of Water. 147 Chap. 14 HEAT and to raise the temperature. The temperaat which the water is changing ture from liquid to vapour throughout the whole mass is called the boiling-point. evaporation Boiling distinguished by the fact that vapour forms throughout the body of the liquid in the former while occurring only at the Because boilingsurface in the latter. points vary with the atmospheric presthey are expressed sure with relation to standard atmospheric 14:5), (Fig. are (760 mm. of mercury). pressure It should be recognized that the boilingpoint is the same as the temperature at which condensation occurs, and the quantity of heat released when a unit mass of vapour condenses is the same as that absorbed during vaporization. Moreover, if the temperature of a mass of liquid is kept at the boiling-point with no change in heat, vaporization and condensation take place simultaneously at i.e., equilibrium exists between liquid and It is only when a change in vapour. the quantity of heat occurs that either process predominates. the same rate, quantity of the Ill : 27 DETERMINING THE HEAT OF FUSION OF ICE In the melting of ice, the quantity of heat required to convert a unit mass to water without a change in of ice temperature is called the heat of fusion of ice. In the metric system, the heat of fusion of ice is the quantity of heat in calories required to change one gram of ice at 0\u00b0C. to water at 0\u00b0C. To determine it, the method of mixtures is employed again. A brief summary of the method, a set of typical results and a sample calculation are presented below but the method in detail will be found in experiment 8, chapter 15. Small pieces of ice that have been dried with a cloth are allowed to melt with continuous stirring in a known mass of warm water of known temperature, contained in the weighed inner vessel of a calorimeter of known specific heat. When the ice is completely melted, the final temperature is recorded, and, after weighing the vessel and contents, the quantity of ice used may be calculated. Example = 1 00 gm. Mass of calorimeter vessel = 500 gm. Mass of calorimeter and warm water Mass of calorimeter, warm water and melted ice = 592 gm. Mass of warm water = 500 \u2014 100 = 400 gm. Mass of ice used = 592 \u2014 500 = 92 gm. = 32 \u00b0C. Initial temperature of water and vessel = 12\u00b0C. =.22 = 1 = x Final temperature of water and vessel Let the heat of fusion of ice Specific heat of water * Specific heat of the vessel Before we attempt the calculation, let us examine the heat exchange that occurs. Since the water and the vessel cooled they lost heat. This heat did two things. First it melted the ice at 0\u00b0C. to water at 0\u00b0C. and then it warmed this water to 12\u00b0C. 148 MEASUREMENT OF HEAT Sec. 111:27 Quantity of heat lost by the water = 400 X 20 X 1 = 8000 cal. Quantity of heat lost by the calorimeter = 100 X 20 X.22 = 440 cal. Ice Water M = 92 gm. Heat of F M = 92 gm. S= 1 No change in temperature Quantity of heat gained by the ice melting = 92 X X = 92^: cal. Quantity of heat gained Change in temperature by the ice water warming = 12 - 0 = 12C.\u00b0 = 92 X 12 X 1 = 1104 cal. Heat lost by the water + heat lost by the calorimeter = heat gained by the ice + heat gained by the ice-water. 8000 + 440 = 92x +1104 92^ = 8000 + 440- 1104 = 7336 92 the heat of fusion of ice = 79.7 cal. There is bound to be a small error in this experiment owing to the use of ice that was not entirely dry and to heat released by the thermometer and other parts of the equipment. With care a fairly accurate result may be obtained. The accepted value for the heat of fusion of ice is 80 calories per", " gram. In the British system, the heat of fusion of ice is 144 B.T.U. per pound. It is the quantity of heat required to change a pound of ice at 32\u00b0F. to water at 32\u00b0F. 149 Chap. 14 HEAT Some Typical Heats of Fusion and Melting-Points Substance M.-P. rc.) H. of F. ( Cal. per gm.) Ice Aluminum Copper 0 660 1083 80 77 42 Substance Lead Cast Iron Mercury M.-P. (^C.) 327 1230 -39 H. of F. ( Cal. per gm.) 6 5.5 3 III : 28 THE IMPORTANCE OF THE HEAT OF FUSION OF ICE When one gram of ice melts without a change in temperature, 80 calories of heat are absorbed. This is enough heat to raise the temperature of one gram of water at 20\u00b0C. to the boiling-point, or of 80 grams of water through one centigrade degree. Since such a large quantity of heat is required to melt ice, it is not difficult to understand why ice is useful in preserving food. There are important consequences of this large heat of fusion in nature. Icebergs float long distances before absorbing enough heat to melt. Large snowfalls melt slowly and disastrous floods are avoided. Since ice on lakes and streams melts slowly in spring, sudden extreme changes in temperature do not occur. On the other hand, the heat released when water freezes is useful. Tubs of water are set near fruits and vegetables in unheated basements when there is a chance of frost damage. As the temperature drops, the water freezes and the heat released prevents the fruits from freezing. The heat released when lakes and rivers freeze has a moderating effect and prevents extremes of temperature. This helps to make the climate of regions like Southern Ontario milder than regions more distant from large bodies of Furthermore, the weather genwater. erally becomes milder before or during a snowstorm because of the heat released when the water vapour changes to solid. 150 These examples and many more may be cited to show the usefulness of the heat of fusion of ice. Ill : 29 DETERMINING THE HEAT OF VAPORIZATION OF WATER the called In boiling water, the quantity of heat required to convert a unit mass of water at its boiling-point to steam without a change in temperature is heat of vaporization of water. In the metric system, the heat of vaporization is the quantity of heat required to change one gram of water at 100\u00b0C. to steam at 100\u00b0G. (atmospheric pressure being 760 mm. of mercury). Experiment 9, chapter 15, describes the method used to determine it. A brief summary of the method, some typical results and a set of calculations follow. required of heat The quantity to vaporize a unit mass of water at its boiling-point is the same as that released when a unit mass of steam condenses. Since the latter is more easily determined, the method involves using it rather than the former. Hence, steam from a boiler is passed through a steam trap to free it of water (It is now called \u201cdry\u201d or \u201clive\u201d steam). It is then conducted into a quantity of cool water of known mass and temperature contained in the weighed inner vessel of a calorimeter of known specific (Fig. 14:6). The water is continually stirred and, shortly, the final temperature is recorded. The vessel and contents are weighed to find out the weight of steam used. heat MEASUREMENT OF HEAT Sec. 111:29 Mass of inner vessel of calorimeter = 100 gm. = 500 gm. Mass of vessel and cool water Mass of vessel, water and condensed steam = 521.7 gm, Mass of cool water = 500 \u2014 100 = 400 gm. = 521.7 \u2014 500 = 21.7 gm. Mass of steam = 5\u00b0C. Initial temperature of water and vessel = 36\u00b0C. =.22 = 1 = X Let the heat of vaporization of water Specific heat of the vessel Specific heat of water Final temperature Let us analyse these results before proceeding. The heat absorbed by the water and the vessel originated not only from the condensation of the steam, but also from the cooling of the resulting water from 100\u00b0C. to 36\u00b0C. We must recognize these two sources of heat in our calculations. Steam Water formed by condensa- tion Cool Water Calorimeter M = 21.7 gm. HofV M = 21.7 gm. S = 1 No change in temperature Quantity of heat lost by the steam condensing = 21.7 X x^2\\Jx cal. Change in temperature = 100 \u2014 36 = 64C.\u00b0 Quantity of heat lost by the resulting water cool- ing = 21.7 X 64 X 1 = 1389 cal. 100\u00b0C. 36\u00b0G. 5'^C. M = 400 gm", ". S = 1 M = 100 gm. S =.22 Change in temperature = 36 \u2014 5 = 31C.\u00b0 Quantity of heat gained by the cool water = 400 X 31 X 1 = 12400 cal. Quantity of heat gained by the calorimeter = 100 X 31 X.22 = 682 cal. 151 Chap. 14 HEAT The heat lost by the steam + heat lost by the resulting water = the heat gained by the cool water + heat gained by the calorimeter. 21.7a: + 1389 = 12400 + 682 21.7a: = 12400 + 682 \u2014 1389 = 11693 11693 ^ =\u2014 = 538.8.\u2019. the calculated heat of vaporization of water is 539 calories. The accepted value is 540 calories. In the British system, the heat of vaporization of water is 972 B.T.U. per pound and is the quantity of heat required to change a pound of water at 212\u00b0F. to steam at 212\u00b0F. Heats of Vaporization and Boiling-Points of Various Useful Materials Substance Water Ethyl Alcohol Ethyl Ether Chloroform Ammonia Methyl Chloride Sulphur Dioxidfe Freon 12 B.-P. rc.) 100 78 35 61 \u2014 33 \u2014 24 \u2014 10 \u2014 30 H. of V. ( Cal. per gm.) 540 204 84 59 327 102 95 41 III : 30 APPLICATIONS OF HEAT OF VAPORIZATION (a) Water of heat The high vaporization of water is of great practical importance in nature. Evaporation of soil water is slow because of the vast quantity of heat required for this purpose. Thus extremes of drought and torrential that would attend excessive evaporation are avoided. rains Steam-heating, as used in most large ex- is an excellent buildings, public 152 Fig. 14:6 Determining the Heat of Vaporization of Water. ample of heat of vaporization at work. Water is boiled; the steam is conducted to radiators where it condenses and yields its heat of vaporization. The hot water now flows back to the boiler to be used again. This system is cheaper than hotwater heating (Sec. 111:14) as smaller radiators are required and the steam pipe itself often serves as a water return. Cooling is frequently accomplished by evaporation. The faster-moving molecules of the liquid escape first and the Further evaporation temperature falls.. MEASUREMENT OF HEAT Sec. 111:30 occurs only if more heat is absorbed from its surroundings. This is the principle involved in the use of perspiration to regulate body temperature, the sprinkling of city streets with water on hot summer evenings, and the alcohol rub to reduce fever. same principle. The chief requirement is a gas which can be liquefied without difficulty and which has a high heat of vaporization. Some gases of this kind are ammonia, sulphur dioxide, methyl chloride, and\u2014the most widely used\u2014-freon 1 2 ( dichlorodifluorom ethane ) (b) Refrigerators The refrigerator makes use of the a Fig. 14:8. The parts of a refrigerator are shown In it, a small electric in compressor which motor operates forces the gas into the coils of a conHere it condenses because of denser. the pressure and cooling (See Faraday\u2019s experiment below). A small fan is often provided for cooling purposes, though loss of heat by radiation is frequently sufficient. The liquid proceeds to the cooling-unit where it vaporizes, removing Graph Summarizing Heat Fig. 14:7 Exchanges for the Changes of State of Water. Fig. 14:8 The Operation of an Elec- tric Refrigerator. 153 Chap. 14 HEAT heat from the food compartment. The gas now returns to the compressor to Ice-making machines, be recirculated. silent automatic. The gas refrigerator (Fig. 14:9) involves no moving parts and is, conseThe and quently, sealed tubes contain water, ammonia and hydrogen. Ammonia, which is ordinarily very soluble, is expelled from the water by the heat of the flame. A combination of the pressure created and the simultaneous cooling by the air causes amIt now proceeds to monia to liquefy. absorbs the evaporates, cooling-unit, the water and is heat, ready to repeat the process. Hydrogen is used to prevent the formation of a vacuum which otherwise would be caused by the rapid dissolving of the 'ammonia. This avoids \u201cpounding\u201d. The refrigerator is simple in its principle of operation but its structure is somewhat more complicated than this explanation would lead one to believe. redissolves in Courtesy of Servel (Canada) Ltd. Fig. 14:9 The Operation of a Gas Refrigerator. quick-freeze units etc., all work on the same principle. (c) The Liquefaction", " of Gases learn more later Michael Faraday, about whom we (Sec. V:57), shall devised an ingenious method of liquefying gases. He filled a thick-walled glass tube of the type shown in Fig. 14:10 with chlorine gas. One end of the tube was surrounded by a freezing mixture of Fig. 14:10 (a) Illustrating Faraday's Method of Liquefying Gases, (b) Producing Cold Artificially. 154 MEASUREMENT OF HEAT Sec. Ill: 30 ice and salt. Heating the gas in the other end caused a rise in pressure while the gas liquefied in the cool end. In this way he was also able to liquefy several other gases. This principle is employed in the gas refrigerator above. methyl chloride and freon 12 are said to be easily liquefied at ordinary tem- peratures. When a compressed gas is allowed As an exto expand, cooling results. ample, the air escaping from an inflated oxygen, hydrogen, Faraday found that certain gases, for example, nitrogen, air and many others, could not be liquefied in this way. It was found that these gases had to be cooled to a certain temperature, called the critical temperabefore any amount of pressure ture, would liquefy them. The pressure required to liquefy the gas at this temperature is known as the critical pressure. It should be noted that when the operating temperature is lower than the critical temperature, the quired for liquefaction pressure reis very much lower than the critical pressure. not difficult to understand, then, why dioxide, ammonia. the sulphur gases It is Some Critical Temperatures AND Pressures Substance C.T. C.P. (\u00b0C.) (Atm.-^) 78 132 143 112 157 144 76 66 112 40 Sulphur Dioxide Chlorine Methyl Chloride Ammonia Freon 12 Carbon Dioxide Oxygen Air Nitrogen Hydrogen Helium *One atmosphere pressure = 760 m.m of mercury. - 119 \u2014 141 \u2014 147 - 240 - 268 73 50 34 37 13 31 2 ABC High Pressure D E F Low Pressure Fig. 14:11 The Production of Liquid Air. 155 Chap. 14 HEAT automobile tire feels cold. In liquefying a gas of low critical temperature, the gas is subjected to the cooling effect of an expanding cold gas around it, in order to reach the low temperature required. This principle is seen in opera- tion in the liquefaction of air (Fig, 14:11). The air is first compressed and then to of the through needle-valve circulated through cooling-coils remove the heat of compression. It passes on the liquefier where it expands, cools, and returns to the compressor by way of a pipe which surrounds the one that carAs these ried it processes are repeated continuously, the gas is soon cooled to its critical temperature and hence it liquefies. to the needle-valve. Liquid air at atmospheric pressure boils at between \u2014 182\u00b0G. and \u2014 194\u00b0C., the variation depending on the percentIt is so cold tage of nitrogen present. that it will boil when in contact with convert mercury (F.-P. ice and will \u2014 39\u00b0 C.) into a solid so hard that it can be used as a hammer. It is drawn off into open Dewar flasks of the same construction in which it may be kept for a short time. On standing, the nitrogen will boil off first leaving a pale blue liquid which is largely oxygen. \u201cthermos\u201d bottles as Liquid air is used in industry as a source for the very low temperature needed for causing metals to contract. It is a source of industrial nitrogen used for preparing nitrogen compounds. From it may be obtained oxygen used for welding, hospitals, aviation etc., together with the inert gases which are used in neon signs, argon bulbs and the like. Liquid carbon dioxide is prepared in a similar way. When this liquid is allowed to escape from a valve, its evaporation removes heat from the surroundings and from the escaping liquid converting it to \u201cdry\u201d ice. This solid carbon dioxide 156 a temperature of \u2014 80\u00b0 C. and has changes to a gas without melting (sublimation), hence the apt use of the word \u201cdry\u201d. As the heat absorbed in sublimation (87 cal. per gm.) is large and only a small amount of solid required for a considerable cooling effect, is used as a refrigerant in shipping. it It saves freight and keeps foods at such a low temperature that is no danger carbon dioxide is also used for fire-extinguishers. spoilage. Liquid there of is 111:31 THE RELATIONSHIP BETWEEN HEAT AND WORK Ancient civilization accomplished work, such as building the Pyramids, by utilizing slave-labour. In recent times various machines have been developed which convert heat into work, so releasing man from much drud", "gery. be converted Man has known since the time of that Rumford and Davy (Sec. Ill: 2) work could heat. into Later it was shown that heat can be made to do work. When a pound of coal burns, for example, enough heat (14,500 B.T.U.) is generated to raise a ton of material through 5,000 feet. In practice, only a fraction of the energy from fuel is available to do useful work in an engine because of losses through radiation, conduction, convecetc. The small fraction tion, actually available compared to the total possible amount of energy is called the follow efficiency. show the efficiency of various kinds of engines. The examples that friction Efficiencies of Heat Engines Locomotives Steam turbine, condensing steam- 6-8% engine 16-30% 22-28% Gasoline (automobiles, aircraft) Diesel (railway, marine, trucks, buses) 32-38% G.M Corp. Heat engines are of two types (a) the external combustion engine of which MEASUREMENT OF HEAT Sec. 111:31 Eccentric Live Steam from Boiler Slide Valve Steam Chest Crank Bearing II! i Flywheel Fig. 14:12 The Reciprocating Steam Engine. the steam-engine is an example, and the internal-combustion engine as (b) in automobiles and aircraft. In the former the fuel burns outside of the engine proper while in the burns inside the cylinder. latter it The Steam-Engine In the common steam- or \u201creciprocating\u201d-engine, the pressure exerted by live steam (dry and much hotter than pushes the piston back and 100\u00b0C.) forth in the cylinder. The back-andforth motion is made possible by an automatic slide-valve which first admits the steam at one end of the cylinder, thus driving the piston to the opposite end, and then admits the steam at this end, thus forcing the piston back to its This original motion rotates the fly-wheel which, being heavy, enables the engine to operate 14:12). position (Fig. smoothly. The Steam Turbine It consists of a specially designed paddlewheel against which steam at high pressure is directed through nozzles (Fig. Fig. 14:13 The Steam Turbine. (a) Principle of the Steam Turbine. This type of steam-engine is widely used in power-plants and large ships. 14:13). The force of the vapour rotates the paddle-wheel. are These engines 157 @ CONT\u00abOt \u00abO0M (?) STATION SERVICE TRANSfORMER \u00a9 ELECTRIC' GENERATOR \u00a9 CONDENSER \u00a9TURBINE \u00a9BOOSTER PUMP \u00a9 FEED PUMP \u00a9STEAM GENERATOR CONTROL \u00a9CONDENSATE PUMP \u00a9STATION SERVICE SWITCHBOARD @ HEATERS \u00a9COAL FEEDER @ TRIPPER \u00a9 SCALES @ PULVERIZER \u00a9 STEAM-GENERATOR @ COAL CONVEYOR \u00a9 STEAM LINE \u00a9 FORCED DRAFT FAN \u00a9 AIR INTAKE \u00ae IN OUCED DRAFT FAN \u00ae MECHANICAL OUST COLLECTOR \u00ae ELECTROSTATIC PRECIPITATOR \u00a9 CRANE Fig. 14:13 (b) How a Steam Generating Station Works. Ontario Hydro Richard L. Hearn Generating Station. Ontario Hydro 158 MEASUREMENT OF HEAT Sec. 111:31 Fig. 14:14 The Four Cycle Internal Combustion Engine. A Sectional View. possible. designed to use the pressure as efficiently They run at very high as speeds and are more efficient and smaller than ordinary steam-engines of the same capacity. The Internal-Combustion Engine If a compact, powerful, mobile source of power is required, the internal-com- bustion engine is the choice. This type of engine can be adapted to the use of any fuel that can be vaporized, such as Coal gasoline, gas is used in some engines, while the Diesel type employs cheap petroleum alcohol, and kerosene. oils. The Gasoline Engine The type generally used is the fourcycle engine, so named because the piston makes four strokes for each explosion of gas in the cylinder. Once started it will run automatically as long as the three necessities of fuel, compression and spark are met. Smoothness of operation is accomplished by the use of a heavy flywheel and of several cylinders which fire Students will have at different times. understanding engine after studying Fig. 14:14. Further treatment than this is beyond the scope of this text. difficulty little this in Chrysler Corporation A Cut-Away Photograph of a Modern Internal Combustion Engine. The Diesel Engine This engine operates like a four-stroke gasoline engine but is without carburetor or electrical ignition system. Air is 159 Chap. 14 HEAT to about one-sixteenth of forced into the cylinder and", " is compressed its volume and for that reason becomes hot. When oil is forced into this hot gas, it burns, without any need for a spark. The Jet Engine The last few years have seen the very rapid development of a new type of internal combustion engine, the Jet Engine. There are several types of such engines; the commonest, however, is the TurboJet. Air is scooped into the intake at the front of the engine. It is compressed, and consequently heated, by a compressor. This heated air is driven under high pressure into the combustion chamber where fuel is injected in and combustion occurs. The hot expanding gases stream away at a high velocity. A small portion of their energy is used to drive the turbine Most which operates the compressor. of the energy is in the stream of hot gases which is ejected from the rear of the engine. The force exerted by this jet creates an equal of hot gases (action) and opposite force (reaction) that drives the plane forward. Rockets are another type of modern reaction engine that operate very similarly to the jet engine described above. They differ in that they carry their own supply of oxygen to burn the fuel. As a result rockets can travel through outer space where there is no air. Inter-planetary travel, a thing long dreamed about, now seems to be becoming a real pos- sibility. III : 32 QUESTIONS A 1. (a) Distinguish between quantity of heat and temperature. (b) State the factors that govern the quantity of heat possessed by a body. 2. (a) Define: calorie, British Thermal Unit, specific heat. (b) What quantity of heat is needed to: (i) warm 25 gm. of water from 13\u00b0C. to 27\u00b0C.? (ii) heat 37 lb. of water from 68\u00b0F. to 212\u00b0F.? 3. of 25 copper 22\u00b0C. gm. from (iii) heat (S.H. =.092) 100\u00b0C.? (a) What is the exchange in mixtures? (b) How long does heat exchange continue between two substances in principle of heat to contact? (c) What is the purpose of a calorimeter? How does it fulfil its purpose? 160 4. (a) Describe an experiment to find the specific heat of a metal. (b) When a 200 gm. mass of metal at a temperature of 85\u00b0C. is immersed in 300 gm. of water at 30\u00b0C., the final temperature is 33\u00b0C. Calculate the specific heat of the metal. 5. (a) Distinguish between boiling and evaporation. (b) Define melting-point and boiling- point. 6. (a) Define heat of fusion of ice and state its numerical value in the metric and British systems. 7. (b) Describe how you would determine its value experimentally. (a) How much heat will be released when 50 gm. of water at 0\u00b0C. freeze to Ice at 0\u00b0C.? (b) How much heat will be absorbed in the melting of 20 gm. of ice at 0\u00b0C. to water at 0\u00b0C.? (c) How much heat will be required MEASUREMENT OF HEAT Sec. 111:32 to convert 80 gm. of ice at 0\u00b0C. to water at 25\u00b0C.? 8. (a) Define heat of vaporization of 9. 10. that from water. (b) Why is a burn from steam much more severe hot than water? (a) How much heat will be required to convert 50 gm. of water at 1 00\u00b0C. to steam at 1 00\u00b0C.? (b) How much heat will be released when 1 5 gm. of steam at 1 00\u00b0C. are condensed to water at 100\u00b0C.? (c) How much heat will be released when 35 gm. of steam at 100\u00b0C. are condensed to water and the water is cooled to 20\u00b0C.? 8. When 25 gm. of water at 100\u00b0C. are added to 50 gms. of water at 10\u00b0C., what is the final temperature? 9. When 200 gm. of metal at 100\u00b0C. are placed in 200 gm. of water at 1 5.0\u00b0C., final temperature becomes 23.0\u00b0C. the Calculate the specific heat of the metal. 10. A brass kilogram weight at a tem- of 90.0\u00b0C. is submerged perature in 440 gm. of water at 10.0\u00b0C. The final temperature is 24.0\u00b0C. Find the specific heat of the brass. 11. 49 gm. of water at 1 3\u00b0C. are contained in an aluminum calorimeter weighing 50 gm. If 35 gm. of glass at 87\u00b0C. are dropped Into the calorimeter the temperature becomes 21\u00b0C.", " Find the specific heat (a) Explain the principle of operation of the glass. of the electric refrigerator. (b) How is air liquefied? B 1. How much heat is required to raise the temperature of 2 kg. of water from 25\u00b0C. to 75\u00b0C.? 2. How many B.T.U. will be absorbed when 30 gallons of water in a hot-water tank are heated from 70\u00b0F. to 200\u00b0F.? (1 gallon of water weighs 10 lb.) 3. How much heat is lost when 1.3 kg. of water are cooled from 90\u00b0C. to 20\u00b0C.? 4. How much heat in will be released when 15 gallons of water cool from 1 65\u00b0F. to 1 25\u00b0F.? B.T.U. 5. How many calories of heat must be supplied to heat 200 gm. of cast iron from 20.0\u00b0C. to 80.0\u00b0C.? 6. How much heat does a silver spoon weighing 30.0 gm. absorb when placed in a cup of coffee that raises its temperature from 20.0\u00b0C. to 80.0\u00b0C.? 7. How many grams of water at 85. 0\u00b0C. must be added to 100 gm. of water at 10.0\u00b0C. to give a final temperature of 37.0\u00b0C.? 12. In an experiment, 500 gm. of lead at 100\u00b0C. are placed in 100 gm. of water at 14\u00b0C. contained in a copper calorimeter weighing 80 gm. The final temperature is 25\u00b0C. Find the specific heat of the lead. 13. What mass of iron at 90.0\u00b0C. when added to 200 gm. of water at 1 5.0\u00b0C. contained in a copper calorimeter weighing 100 gm. will give a final temperature of 25.0\u00b0C.? 14. When 400 gm. of silver at 100\u00b0C. was placed in water at 1 6.0\u00b0C. contained in an aluminum calorimeter weighing 40.0 gm. the final temperature was 24.0\u00b0C. What mass of water was used? 15. Calculate the final temperature when 120 gm. of iron at 100\u00b0C. are added to 400 gm. of water at 10\u00b0C. in a copper calorimeter having a mass of 80 gm. 16. A copper calorimeter weighing 65 gm. contains 30 gm. of turpentine at 15\u00b0C. When 45 gm. of Iron at 98\u00b0C. are placed temperature becomes 32\u00b0C. Calculate the specific heat of the turpentine. 17. How much ice af 0\u00b0C. can be melted the it, in by 1 kg. of water at 100\u00b0C.? 18. How much ice at 0\u00b0C. will be required 161 Chap. 14 HEAT to cool 1 kg. of drinking-water from 15\u00b0C. to 0\u00b0C.? ture is 27\u00b0C. Find the heat of vaporization of water. 19. What mass of ice at 0\u00b0C. will be required to cool 750 gm. of water from 35\u00b0C to 10\u00b0C? 20. When 5,0 gm. of ice at 0\u00b0C. are melted in 30 gm. of water at 25\u00b0C. the final temperature is 10\u00b0C. Find the heat of fusion. 21. A copper calorimeter weighing 55 gm. contains 90 gm. of water at 25\u00b0C. When 15 gm. of ice at 0\u00b0C. are melted in the water the resulting temperature is 1 1 \u00b0C. Find the heat of fusion of ice. 22. In an experiment 203 gm. of water at 40\u00b0C. are contained in a calorimeter vessel weighing 50 gm. having a specific heat of.22. After ice at 0\u00b0C. was melted in the water the temperature became 25\u00b0C. and the mass 233 gm. Find the heat of fusion of ice. If 15 gm. of ice at \u2014 20\u00b0C. are 23. melted in 50 gm. of water at 40\u00b0C. and the resulting temperature is 10\u00b0C., cal- culate the specific heat of ice. 24. How many grams of water can be freezing-point raised boilingpoint by the condensation of 5.0 gm. of steam? from to 25. To what temperature will 75 gm. of water at 25\u00b0C. be heated by the condensation of 3.0 gm. of steam? 26. When 6.6 gm. of steam at 1 00\u00b0C. are passed into 1 80 gm. of water at 6.0\u00b0C. contained calorimeter weighing 45 gm., the resulting tempera- aluminum an in 27. If 15 gm. of steam at 100\u00b0C. are added to 150", " gm. of water at 20\u00b0C. in a calorimeter (S.H. = 0.10) weighing 75 gm. the final temperature is 74\u00b0C. Calculate the heat of vaporization of water. 28. When 160 gm. of water at 7.7\u00b0C., contained in an aluminum calorimeter weigh1 26.2 gm., are heated by the condensation of 10.1 gm. of steam, the final temperature is 39.8\u00b0C. Find the heat of vaporization of water. ing the Find final 29. Find the resultant temperature when 8.0 gm. of steam at 100\u00b0C. are passed into a vessel of negligible mass containing 40 gm. of ice at 0\u00b0C. 30. temperature when 20 gm. of steam at 1 00\u00b0C. are passed into 240 gm. of water at 10\u00b0C. contained in an aluminum calorimeter weighing 150 gm. 31. What quantity of heat will convert at \u2014 1 6\u00b0C. to steam at 5 gm. of ice 100\u00b0C.? 32. When 45.0 gm. of iron at 95.0\u00b0C. are placed in a cavity in a block of ice at 0\u00b0C. and the temperature has dropped to 0\u00b0C., 6.0 gm. of ice are melted. Knowing the heat of fusion of ice, find the specific heat of iron. 33. An ice-water mixture weighing 200 gm. is contained in a calorimeter weighing 100 gm. (S.H. = 0.20). When 35 gm. of steam at 100\u00b0C. are added the temperature becomes 50\u00b0C. Calculate the mass of ice used. 162 CHAPTER 15 EXPERIMENTS IN HEAT INTRODUCTION Before commencing these experiments in heat, students should be familiar with the following. A. The Use of the Bunsen Burner 1. Structure of the Burner Examine a Bunsen burner and identify the gas inlet, the orifice, the air-inlet valve, the mixing tube. Make a labelled diagram of the burner. 2. Lighting the Burner (a) Close the air inlet, turn on the gas and ignite it. Gradually open the air inlet until you have the desired flame. If the flame \u201cstrikes back\u201d, i.e., burns at the bottom of the (b) mixing tube, turn off the gas and repeat (a) above. 163 Chap. 15 HEAT 3. Regulating the Size and Temperature (Colour) of the Flame (a) The size of the flame may be changed by increasing or decreasing the supply of gas. (b) When first low for most purposes. When the air inlet is gradually opened, the colour changes from yellow, through blue to nearly colourless, and the temperature increases until it is a maximum at the last stage. lit, the flame is yellow and its temperature is too 4. The Structure of the Flame Fig. 15:2 shows the various regions (cones) of a Bunsen flame. An object to be heated is held just above the turquoise cone. B. The Use of the Thermometer 1. Examine the instrument and find the centigrade scale. (Some ther- mometers have centigrade and Fahrenheit scales.) 2. Wait until the liquid has come to rest before taking a reading. 3. Adjust the thermometer so that you are able to view the top surface of the liquid at right angles in order to avoid the error of parallax. 4. Take readings to a fraction of a degree. EXPERIMENT 1 To study the expansion of solids, (Ref. Sec. III:5) Apparatus Bunsen burner, ball-and-ring apparatus, cold water. Method 1. Try to pass the ball through the ring when both are cold. 164 EXPERIMENTS IN HEAT 2. Heat the ball strongly and try to pass the ball through the ring again. 3. Cool the ball by placing it in cold water and again try to pass it through the ring. Observations What do you observe in the above steps? Conclusion What is the effect of heating and cooling on the volume of a solid? Questions 1. What would be the probable effect of heating the ring and trying to pass the heated ball through it? 2. What would be the probable effect of chilling the ring in a freezing mixture and trying to pass the ball through it? 3. By means of labelled diagrams, show how you would demonstrate the effect of heating and cooling on the volume of (a) a liquid, (b) a gas. EXPERIMENT 2 To compare the expansion of different metals when heated. (Ref. Sec. Ill: 5) Apparatus Bunsen burner, compound bar consisting of strips of copper and iron fastened together, cold water. Method 1. Heat the long straight compound bar in", " the Bunsen-burner flame and note any change. 2. Cool the bar in cold water and again note the change. Observations 1. Describe the changes that occurred in parts 1 and 2. 2. Which metal was on the outside of the curve? Explanation Account for the changes. Conclusion What is the relative amount of expansion and contraction that occurs when different metals undergo the same change in temperature? Question Make a labelled diagram to show how to use a compound bar as a thermostat to control an oil burner and thus regulate the temperature in a room. 165 Chap. 15 HEAT EXPERIMENT 3 To study the transfer of heat by conduction. (Ref. Sec. 111:13) Apparatus Bunsen burner, metal rod about 12 in. long with a wooden handle, a conductometer (Fig. 13:1), wax. Method 1. Place drops of wax at three-inch intervals along the length of the long metal rod. Hold it by the wooden handle and heat the end of the rod strongly in the flame of the burner. 2. Place drops of wax at equal intervals along the rods of the conductometer and heat the metals simultaneously at the point where the rods meet. Observations 1. (a) What happens to the wax in part 1? (b) Is there any change noted in the temperature of the wooden handle? 2. State the differences observed in each metal rod in part 2. Conclusions 1. State the meaning of the term conduction of heat. Explain how it occurs. 2. What have we learned about the heat conductivity of different metals? How does wood compare with metals in this regard? 3. List the metals studied in order of their relative heat conductivities. Questions 1. How do metals compare with other substances in heat conductivity? 2. What use is made of the conductivity of heat through metals? 3. Why are liquids and gases very poor conductors of heat? 4. List some materials that are good insulators and state where they are used for this purpose. EXPERIMENT 4 To study the transfer of heat by convection. (Ref. Sec. 111:14) A. IN LIQUIDS Apparatus A large beaker, retort stand, ring, Bunsen burner, cold water, potassium permanganate. Method Fill the beaker with water and place it on the ring attached to the stand. Make sure that the water is at rest. Drop a crystal of potassium 166 EXPERIMENTS IN HEAT permanganate into the water near the edge. Using the tip of the low Bunsen flame, heat the liquid beneath the crystal. Observations Describe all phenomena. Explanation Account for the changes observed. B. IN GASES Apparatus A candle, smoke-paper, convection apparatus (Fig. 13; 6b). Method Light the candle and place it beneath one of the chimneys. Close the Hold a piece of lighted smoke-paper above the other glass front. chimney. Observations State what occurs. Explanation Account for these results. Conclusion State the meaning of the term convection currents and explain how they occur. Question Show by means of a diagram how heat is transferred from the furnace to an upper room in a home heated by (a) a simple hot- water system (b) a simple hot-air system. EXPERIMENT 5 To compare the abilities of dull/ dark and shiny/ light surfaces to emit radiant energy. (Ref. Sec. Ill; 15(b) Apparatus A differential thermometer with both bulbs blackened with soot from a candle, metal vessel with one side blackened and the other polished, some boiling water. 167 Chap. 15 HEAT Method Mark the levels of the coloured liquid in the arms of the differential thermometer and then place the metal vessel full of boiling water midway between the bulbs. Note any change in the levels of the coloured liquid. Observations What is observed? Explanation Account for your observations. Conclusion What effect has the nature of the surface of an object on its ability to emit radiant energy? Questions 1. Why is this apparatus called a differential thermometer? 2. Why should tea-pots be shiny rather than dull? EXPERIMENT 6 To compare the abilities of dull, dark and shiny, light surfaces to absorb radiant energy. (Ref. Sec. Ill; 15(c) Apparatus A differential thermometer with one bulb shiny and one blackened, dull, dark metal vessel, supply of boiling water. Method Mark the levels of the coloured liquid in the differential thermometer and place the vessel filled with boiling water midway between the two bulbs. Note any changes in the levels of the liquid. Observations What is observed? Explanation Account for the observations. 168 EXPERIMENTS IN HEAT Conclusion What effect has the nature of the surface on its ability to absorb radiant energy? Questions 1. In experiment 5, why were the bulbs of the differential thermometer darkened? 2. In experiment 6", ", why is a dull, dark vessel used? 3. Why do people wear dark clothing in winter and light-coloured clothes in summer? 4. Examine a radiometer and make a labelled diagram of it. Note and explain what happens when a source of radiant energy is brought near it. EXPERIMENT 7 To determine the specific heat of a metal, (Ref. Sec. 111:24) Apparatus A quantity of copper (or lead) shot, balance and weights, flask, testtube, water, retort stand, ring, gauze, Bunsen burner, calorimeter, two thermometers. Method 1. Fill the test-tube three-quarters full of shot and carefully insert the Place the shot and its container in bulb of a thermometer into it. the flask and boil the water while carrying out parts 2, 3 and 4. Note the temperature of the shot to within a fraction of a degree after the mercury stops rising. 2. Find the mass of the inner vessel of the calorimeter and stirrer. Record the specific heat of the metal of which both are made. 3. Place about 100 ml. of cold tap-water whose temperature is slightly lower than room temperature in the inner vessel. Find the mass of the vessel and water, and determine the mass of the water. 169 Chap. 15 HEAT 4. Place the inner vessel and contents into the outer vessel. Cover with the lid. Stir the water and take its temperature. 5. Open the calorimeter, add the shot to the water, close it, and after stirring the mixture, again take its temperature. 6. Find the mass of the vessel, stirrer, and contents, and determine the mass of the shot. Observations = 1. Temperature of the shot 2. Mass of the inner vessel and stirrer == = 3. Mass of the inner vessel, stirrer and water = 4. Initial temperature of the vessel and water 5. Final temperature of the mixture of water and shot = 6. Mass of the vessel, stirrers and mixture =: = = = Specific heat of the vessel Specific heat of the water Let the specific heat of the shot x Calculations Study the worked example in Sec. 111:24 and calculate the specific heat of the metal from the observations recorded above. Conclusions 1. What is your experimental value? 2. What is the class average? Questions 1. What is the percentage error? 2. What are the sources of error in this experiment? 3. Why is it desirable to have the initial temperature a few degrees lower than room temperature? EXPERIMENT 8 To determine the heat of fusion of ice. (Ref. Sec. 111:27) Apparatus Quantity of ice, paper towels, thermometer, balance and weights, quantity of water at about 10C.\u00b0 warmer than room temperature, calorimeter. Method 1. Find the mass of the inner vessel and stirrer. 2. Place about 100 ml. of the warm water in the vessel and again find the mass. 3. Place the inner vessel in the outer one. Cover with the lid. Stir the water and take its temperature, estimating it to a fraction of a degree. 170 EXPERIMENTS IN HEAT 4. Wipe dry about 25 gm. of ice with the paper towels and quickly drop it into the warm water. Replace the cover and stir until the ice has melted completely. Record the lowest temperature reached by the water. 5. Find the mass of the vessel, stirrer, and contents, and determine the mass of the ice. Observations = 1. Mass of inner vessel and stirrer = 2. Mass of inner vessel, stirrer and water 3. Initial temperature of the water and vessel = 4. Final temperature of the mixture of original water and melted ice 5. Mass of vessel, stirrer and the mixture Specific heat of the vessel Specific heat of the water Initial temperature of the ice Let the heat of fusion of ice = =: = = = = x Calculations Study the worked example in Sec. 111:27 and calculate the heat of fusion of ice from the observations recorded above. Conclusions 1. What is the experimental value for the heat of fusion of ice? 2. What is the class average? Questions 1. What is the percentage error? 2. What are the sources of error in this experiment? 3. Why is it desirable to have the initial temperature a few degrees higher than room temperature? 4. Why do we dry the ice before placing it in the water? 171 Chap. 15 HEAT EXPERIMENT 9 To determine the heat of vaporization of water, 111:29) (Ref. Sec. Apparatus A calorimeter, quantity of water at about 15C.\u00b0 below room temperature, thermometer, Bunsen burner, retort stand, ring, gauze, steamboiler, steam-trap, rubber connectors (Fig. 14:6). Method Place the boiler containing water on the ring of the retort stand and heat it", ". 1. Find the mass of the inner vessel and stirrer. 2. Put about 100 ml. of the cold water into the inner vessel and find the combined mass. 3. Place the inner vessel in the outer one and put on the lid. Stir the water and take its temperature, estimating it to a fraction of a degree. 4. Connect the steam-trap with the boiler and conduct steam into the water. Stir constantly until its temperature has risen as much above room temperature as it was originally below it. Discontinue passing steam into the water and take the highest temperature reached by the water. 5. Find the mass of the inner vessel and contents. Observations 1. Mass of inner vessel and stirrer 2. Mass of inner vessel, stirrer and water 3. Initial temperature of the mixture 4. Final temperature of the mixture 5. Mass of vessel, stirrer and mixture Specific heat of the vessel Specific heat of the water Initial temperature of the steam Let the heat of vaporization of water - <2 - X Calculations Study the worked example in Sec. 111:29 and calculate the heat of vaporization of water from the observations recorded above. Conclusions 1. What is the experimental value for the heat of vaporization of water? 2. What is the class average? Questions 1. What is the percentage error? 2. What are the sources of error in this experiment? 3. How may the effect of the sources of error be minimized? 4. Why is it important to use the steam-trap? 172 UNIT IV LIGHT These Searchlights Create an Interesting Construction, Showing both a Converging and a Diverging Pencil of Light Rays. Wheeler Newspaper Syndicate CHAPTER 16 NATURE AND PROPAGATION OF LIGHT IV : 1 NATURE OF LIGHT (sight). Light is that agency which affects the eye and produces the sensation of \u201cseeThat branch of physics ing\u201d that covers all the phenomena pertainlight is called Optics (Greek, ing to ops\u2014eye). Some knowledge of light existed from very early times, though this was limited effects rather than to any fundamental understanding. primarily to As to the nature of light, the early Greeks believed it to consist of streams of minute particles of some sort. There was considerable debate as to whether these particles originated in the eye or Plato (428-348 in the object viewed. B.G.) and Euclid (about 300 b.c.) held to the idea that invisible feelers were emitted from the eye, and that the eye sees a body somewhat as the hand may feel it with a rod. The Pythagoreans, Aristotle (284-322 b.c.) in particular, opposed this view and taught that light consisted of minute particles projected into the eye from the object. Both these conflicting ideas were mere guesses and as such were worthless. However, in the eleventh century Alhazen, an Arabian evidence physicist, provided definite showing that the cause of vision proceeded from the object and not the eye. Even to-day, much mystery still surrounds the nature of light (Sec. IV; 7). In view of the fact that light can be produced from other forms of energy, e.g., heat energy, and that light can be transformed into other forms of energy, (Sec. V:82), we shall e.g., simply say that light is a form of energy. electricity IV: 2 SOURCES OF LIGHT Few objects give out light and these are termed luminous bodies. Most objects are non-luminous, becoming visible only when they reflect light from some outside source to our eyes (Fig. 16:1). Our main source of light is the sun. When we think how important the sun has always been in human affairs, it not surprising that in prehistoric times it became a prime object of worship. the ancient Egyptian History records Sun-god, Ra, and an ancient Persian god of light, Ahura-mazda. The terms \u201cray\u201d of light, and \u201cmazda\u201d lamps are derived from these names. is Many objects are rendered luminous by being heated to incandescence. This may be accomplished by mechanical means as shown when sparks are produced by friction between flint and steel in a gas-lighter; by resistance to an electric current in the thin wire used in electric-light bulbs; and by chemical action as in the burning of a fuel. Some objects are luminous at ordinary tem- 175 Chap. 16 LIGHT 16:1 Self-LumFig. inous and Non-Lum- inous Objects. Tigerstedt Studios, Calgary It such as frosted glass or oiled paper, is one that transmits some light, but in doing so distorts or scatters the light so that we cannot see clearly objects on the other side. Opaque substances, like wood, do not transmit light at all and hence we cannot see through them. is common knowledge that light travels in straight lines. Our", " inability to see around corners, the formation of shadows, and other examples point to this conclusion. This behaviour of light is termed rectilinear propagation. Knowing this, we represent a path of light by a straight line called a ray. (Note: the light that travels along this path is also called a ray.) The direction in which the light is travelling is indicated by arrow-heads placed on the rays. Several parallel rays form a beam of light. Rays of light proceeding towards a point form a converging pencil; when they spread out from a point they form a diverging pencil (Fig. 16:2). IV: 4 PIN-HOLE CAMERA The pin-hole camera is an interesting of electricity of phosphorus, for peratures, a stick example, and fluorescent bodies. Then there is the glow produced by the discharge certain gases, e.g., neon tubes. Another interesting example of \u201ccold light\u201d is that produced by fireflies, and by certain deepsea fish. Probably such light is produced by chemical means. through IV: 3 TRANSMISSION OF LIGHT fact that light Unlike sound, light does not require a material medium for its transmission. Evidence in support of this is supplied by travels through a the vacuum in coming to us from objects in space, and from the glowing filament of an evacuated tube. Further evidence was provided in experiment 4, chapter 10, where we could not hear the bell when the air was evacuated from the jar yet could still see it ringing. Various media diflfer in their ability to transmit light. Transparent objects such as air, glass and water transmit is easy to see light so readily that it through them. A translucent substance. 176 NATURE AND PROPAGATION OF LIGHT Sec. IV: 4 Glass bricks are used in the many construction modern of buildings. What are the advantages? Canadian Pittsburgh Industries Ltd. application of the rectilinear propagaIt consists of an opaque tion of light. box, having a small hole (pin-hole) in the middle of one end, and a translucent screen (piece of ground glass, or oiled paper) at the other. If a lighted candle is placed a little distance in front of the pin-hole, an inverted image of the candle will be seen on the translucent screen. \\ / / \\ (a) Fig. 16:2 Rectilinear Propagation of Light (a) Ray (b) Beam (c) Converging Pencil. (d) Diverging Pencil. 177. Chap. 16 LIGHT however, because of the small amount of light admitted through the pin-hole. IV : 5 SHADOWS AND ECLIPSES A shadow is the dark space behind an opaque object, an area from which light has been partially or completely excluded. An opaque object in front of a point source of light will cut off all the light, and a sharply defined shadow is produced. If the light comes from a larger source, the shadow will vary in intensity, the dark central portion of the shadow which receives no light from any part of the source being the umbra 16:4a), the lighter shadow sur(Fig. rounding the umbra which receives some light being the penumbra (Fig. 16:4b). An eclipse of the sun is an interesting shadow phenomenon caused when the moon comes between the sun and earth 16:5). A person located in the (Fig. moon\u2019s umbra will total If in the moon\u2019s eclipse of the sun. observe a This kind of image is formed as a result of very narrow diverging pencils of light from each point of the object passing through the pin-hole, and producing small patches of light, identical in shape to the pin-hole, on the screen. Fig. 16:3 The Pin-Hole Camera. or The resulting image is formed by a large number of these overlapping patches of light, producing an exact replica of the original object. Such an image is a real image. The image so obtained will have a somewhat blurred, out-of-focus appearance due to the circular edges of each patch of light not entirely overlapping. The larger the pin-hole the brighter will be the image on the screen, but the more blurred will be its edges. The image is inverted because the light rays cross at the pin-hole (Fig. 16:3). The size of the image is governed by the size of the object, its distance from the camera, and the distance of the image from the (Chap. 21, Exp. 1 ) pin-hole Size of Image Distance of Image Size of Object Distance of Object Hi _ Di 1h~1^ A consideration of the equiangular (or similar) triangles formed by the rays of should light as shown in 16:3 Fig. establish these relationships fairly readily. If we replace the translucent screen with a light-sensitive paper, or photographic plate, quite acceptable photographs of distant objects can be obtained. A very long exposure is necessary, 178 Fig", ". 16:4 Shadows (a) Using Point Source. (b) Using Large Source. eclipse of the sun. penumbra, he will be able to see part of the sun. The latter is called a parBecause the tial moon\u2019s orbit around the earth is slightly its distance from the earth Thus it occasionally happens elliptical varies. NATURE AND PROPAGATION OF LIGHT Sec. IV: 6 that the moon comes between the sun and the earth at a time when its umbra does not reach the surface of the earth. A person located on the earth below the tip of the moon\u2019s umbra would see a ring of the sun around the edge of the moon. Such an eclipse is called an annular very eclipse sun and occurs the of rarely. Eclipses of the moon also occur, and at fairly frequent intervals. The moon is a non-luminous body, and is seen only when sunlight is reflected from its surface to the earth. The full moon occurs when the moon is on the opposite side of the earth from the sun. At such a time the moon may pass through the earth\u2019s A partial shadow and be eclipse of the moon is caused when it eclipsed. (a) (b) Fig. 16:5 Eclipses. (a) Total and Partial Eclipse of the Sun. (b) Annular Eclipse of the Sun. (c) Eclipse of the Moon. is partly in the earth\u2019s umbra, and a total eclipse when it is completely in the earth\u2019s umbra. When in the earth\u2019s penumbra, the moon is not eclipsed, but only less bright as it receives, and hence reflects, less light from the sun. IV : 6 VELOCITY OF LIGHT It has long been suspected that light travels with a finite velocity, but early attempts to measure this velocity were Fig. 16:6 Velocity of Light Using the Moons of Jupiter. light too crude to be successful. The first reasonably accurate value was obtained by a young Danish astronomer Olaus Romer in 1676. He found that intervals between the successive eclipses of one of the moons of the planet Jupiter were longer when Earth was receding from Jupiter (going from Ei to E 2 ) and shorter when Earth was approaching (going from Eg to E^, Fig. 16:6). Romer ascribed the discrepancy to the time required for the to diameter of the earth\u2019s orbit. The timelag was found to be about 16.5 minutes or approximately 1000 seconds. Since the diameter of the earth\u2019s orbit is about 186,000,000 miles, the velocity of light is found to be 186,000 miles per second. The first determination of the velocity distances on the of earth was made in 1849 by A. H. Fizeau. His method was to pass a beam of light through one of the gaps in a toothed wheel, and reflect it back on its path from a mirror three or four miles away light over short across travel 179 Chap. 16 LIGHT : (Fig. 16:7). When the wheel was at rest, the return beam passed back through the same gap and was visible on the other side. When the wheel was rotated rapidly, a speed could be found at which Fizeau's ApFig. 16:7 paratus for Measuring Velocity of Scheme of Light. the return way was blocked by the next tooth. The time spent by the wheel in spinning through this small part of a revolution, is also the time required for the light to travel to the distant mirror and back again. Hence, knowing these facts, the velocity of light could be easily calculated. A better method was devised by J. L. Foucault in 1850, who used a rotating mirror instead of a toothed wheel. This method was used in more elaborate form by A. A. Michelson in 1926, with an eight-sided mirror and a considerably increased light path (Fig. 16:8). More recently, Michelson in collaboration with others, set up a mile-long evacuated tube with a mirror arrangement for causing a beam of light to traverse this path back and forth many times before being obAgain using a rotating-mirror served. method, they obtained a quite accurate value for the velocity of light in a vacuum which was found to be slightly higher than its velocity in air. The approximate values for the velo- city of light, C, in air are C = 300,000 kilometres per second or 3 X 10^\u00ae centimetres per second or C = 186,000 miles per second. The velocity of light is a most important physical determination, since it is the speed with which many forms of energy travel through space. It is interesting to note that the vast distances of space are measured in terms of the light-year. This is the distance travelled 180 NATURE AND PROPAGATION OF LIGHT Sec. IV: 7 by light in one", " year. Some of the more distant stars and nebulae are so remote from the earth that the light by which we now see them set out on its journey to the earth thousands of years ago. IV: 7 THEORIES OF LIGHT The first rational attempt to explain the propagation of light was made by Isaac Newton (1642-1728). His Sir the Emission or Corpuscular theory, Theory, postulated that light energy was conveyed through space by a swiftly moving stream of particles or corpuscles shot out from the luminous body. Most of the properties of light known at Newton\u2019s time such as the rectilinear propagation of light and the effects of reflection (Chap. 17) and of refraction (Chap. 18) were adequately accounted for by this theory. A rival postulate was put forward by Christian Huygens (1629-1695), the son of a Dutch diplomat and poet. He sought to explain the behaviour of light in terms of waves, and hence his theory is called the Wave Theory. Again, reflection and refraction were readily explicable in terms of this wave theory. Difficulties were encountered, however, when seeking to explain the rectilinear propagation of light, and also in the need for postulating the existence of a medium, the ether, completely filling space in which the waves could travel. As *a result, the wave theory remained undeveloped and Newton\u2019s corpuscular theory was generally accepted. nineteenth the century, Thomas Young (1773-1829), and A. J. Fresnel (1788-1827), provided valuable Early in experimental support for the wave theory of light. They were able to show that two beams of light could be made to interfere with and to reinforce each other, thereby producing alternate dark and bright lines. This could only be explained by \u201csuperposition of waves\u201d. At one position when in opposite phase, a crest with a trough, these waves produce a dark line. At another position when in the same phase, a crest with a crest, or a trough with a trough, they produce a bright line. (Compare with superposition of sound waves. Sec. 11:6.) Another line of experimental support in favour of the wave theory was to show that the velocity of light is smaller in the denser of two media. The corpuscular theory had predicted the exact reverse of this. Recent work seems to favour a combination of the corpuscular and wave theories in the explanation of many of the observed effects. This theory called the Quantum Theory, was first devised by Max Planck in 1901. According to this theory, light is emitted by the atoms of a luminous body in separate packets or bundles of energy called quanta or photons. Probably one or more of the electrons revolving about the nucleus of an atom ( Sec. V : 1 2 ) can be made to jump from one orbit or \u201cenergy level\u201d to another. As they do so, one or more quanta of energy, or photons are emitradiates from the ted. luminous body as electromagnetic waves (Sec. IV: 38). The energy content of a photon determines the length and frequency of the wave, and hence the colour of the light observed. energy This 181 Chap. 16 IV : 8 LIGHT QUESTIONS. (b) How many minutes are required for light to travel from the sun to the earth (93X10*^ miles)? (c) What is a light-year? 8 What contributions Newton, Huygens and Planck make to a theory of did light? 1. Calculate the distance of an object 12 ft. high whose image is 4 in. high in a pin-hole camera 10 in. long. 2. Calculate the size of the image of a tree 30 ft. high, 100 yards distant, in a pin-hole camera 8 in. long. 3. Calculate the height of a building 300 metres distant which produces an image 2.5 cm. high in a pin-hole camera 2.0 in. long. 4. How long does it take for light to 0'* travel from the moon to the earth (24 X 1 miles)? 5. Calculate the number of (a) miles, (b) kilometres in 1 light-year. 6. Sirius, the brightest star in the sky, is 9 light-years away. How far away is this In miles? 7. Our nearest neighbour among the stars, excepting the sun, is Proxima Centauri which is about 25X 1 0\u2019^ miles away. How long does it take for light from this star to reach us? 1. 2. 3. 4. 5. (a) Define optics. (b) What is the nature of light? \"luminous\u201d (a) Distinguish between and \"illuminated\u201d objects. (b) How may objects be rendered luminous? (a) What evidence have we to prove a that vacuum", "? through travel can light (b) Distinguish between opaque, and sub- transparent translucent stances. Give examples of each. (a) What Is meant by the \"rectilinear propagation of light\u201d? Discuss evidence in support of it. (b) Define: ray, beam, converging pencil, diverging pencil. (a) Describe the image obtained in a pin-hole camera. (b) Explain how it is produced. (c) Name three factors that govern its size. (d) The length of a pin-hole camera is 10 in. An object 6ft. high Is placed at a distance of 30 ft. from the pinhole. Calculate the size of the image produced. 6. (a) Define: umbra, penumbra. (b) Construct a labelled diagram to show how both total and partial eclipses of the sun are produced. (a) What is the velocity of light in air? 7. 182 CHAPTER 17 REFLECTION OF LIGHTMIRRORS IV: 9 THE LAWS OF REFLECTION In section IV:2, we learned that most are made visible when light objects falling on them is reflected back to our eyes. The rays of light that fall upon a body are called incident rays, while those that are sent back by the body are called reflected rays. A mirror is a smooth, highly polished surface, designed to reflect a maximum amount of light. Mirrors usually consist of pieces of glass silvered on one surface. Some are flat and are called plane mirrors, while others are called Fig. 17:1 The Optical Disc, for Dem- onstrating the Laws of Optics. 183 Chap. 17 LIGHT curved mirrors. However, any smooth surface, such as polished metal, polished wood, or still water, will serve as a mirror. to the reflecting surface and angles meeting it at the point of incidence is called the normal. Rotate the disc and thus cause the incident ray to strike the mirror at different angles. In each case note the direction of the reflected ray and compare the size of angle of incidence (the angle between the incident ray and the normal), with that of the angle of reflection (the angle between the reflected ray and the normal). In each case these two angles will be found to be equal, and the incident ray, the reflected ray and the normal will all be found to be in the plane of the disc. Hence we can state the two laws of reflection as follows: First Law: The angle of reflection althe angle of ways equals incidence or Z r = Z i Fig. 17:2 Reflection of Light by a Plane Mirror (a) Using Optical Disc. (b) Terms. Second Law: The incident ray, the normal, and the reflected ray all lie in the same plane. To study reflection of light and other is 17:1). optical phenomena an optical disc This consists of a used (Fig. circular flat disc, graduated in degrees, to which various pieces of optical equipment can be fastened by means of thumb-screws. The disc can be turned about a horizontal axis by means of a handle fastened to the back. Surroundis an opaque collar in ing the which is a window containing one or more horizontal slits. disc Mount a plane mirror at the centre of the disc, with the face of the mirror at right angles to the zero line marked on the disc (Fig. 17:2). Allow a ray of light from a lantern to pass through a single slit in the window of the optical falls upon the mirror. disc so that it Adjust so that the point of incidence coincides with the point where the zero line meets the mirror. This line at right 184 are laws quite These simple and straightforward. We apply them daily in games such as handball, tennis and basketball where the bounce of a ball Experiment 2, chapter 21, is is an alternative method for proving these two laws of reflection. utilized. IV : 10 REGULAR AND DIFFUSE REFLECTION a upon rays fall surface, such as When parallel a smooth reflecting plane mirror, they form the same angle of incidence with the surface, and in consequence they will be reflected as a beam of 17:3a). Such reflection is called regular reflection and often produces undesirable glare. For example, one finds it difficult to read from a glazed paper in parallel (Fig. rays sunlight. Diffuse or irregular reflection occurs REFLECTION OF LIGHT\u2014MIRRORS Sec. IV: 11 when light strikes a rough surface (Fig. 17:3b). Such a surface may be conto be composed of a large sidered number of tiny, flat surfaces that face Thus when parallel in all rays of light strike such a surface, the directions. individual rays are scattered or diffused due to their being reflected in different directions. To avoid glare it is often necessary to promote diffusion of light. Unglazed paper is used for newspapers, and for Smooth Surface Rough Surface Regular Reflection", " Irregular Reflection (a) (b) Fig. 17:3 Regular and Diffuse Reflection. Frequently the woodwork wall-papers. in our homes is left with a dull finish. These, and other devices, all cause dif- bulbs, and lamp-shades. Prism light glass and other types of roughened glass in windows serve the same purpose. IV : 1 1 IMAGES IN PLANE MIRRORS Mirrors have many and varied uses. Large plate-glass mirrors are frequently placed on the walls of our homes and of public rooms to give an impression of spaciousness. Rear-view mirrors are now compulsory in all automobiles. Mirrors are often used in scientific instruments to reflect light onto a scale, in projectors to intensify the light beam, in the view-finder of reflex cameras, in periscopes, and the like. In view of such it should be of real widespread uses, interest to us to study the images produced by plane mirrors (Fig. 17:4). On doing experiment 3, chapter 21, you will learn the following facts about images in plane mirrors: The Position of the Image. The image is as far behind the mirror as the object is in front, and a line joining the two right the passes through mirror at fuse reflection of light. angles. Diffusion, and hence the elimination of glare, is also obtained by transmission of light through frosted or opalescent The Characteristics of the Image. The image is the same size as the object. it only It is a virtual image, that is, Fig. 17:4 Images in a Plane Mirror (a) The Position and Characteristics of the Image. (b) Lateral Inversion. 185 Chap. 17 LIGHT appears to be there. No light emanates from it. Such an image cannot be proThe image is jected onto a screen. vertically erect, but laterally inverted. When we say \u201claterally.inverted\u201d we mean that the right and left sides are interchanged (Fig. 17:4b). Note that when the mirror is horizontal, the image is vertically as well as laterally inverted. Recall what you observe on looking into a still body of water to verify this fact. IV: 12 TO LOCATE IMAGES IN PLANE MIRRORS These phenomena observed experimentally can be shown to be true as a geometrical consequence of the laws of reflection, Sec. IV; 9. Let us consider the simplest possible case, that of a point object, O, the image of which, I, is viewed by an eye BD (Fig. 17:5). O is sending out light rays in all directions. Point in a Plane Mirror. lying between OAB and Only those OCD enter the pupil of the eye after being reflected from the mirror. The reflected rays AB and CD appear to be 186 coming from I, their point of inter- section. In order to locate the image of an object in a plane mirror geometrically, the following construction is necessary (Fig. 17:6). From each point of the to object, draw a perpendicular the mirror, and extend it an equal distance behind the mirror. Join the ends of all such lines and you will have an outline of the image of the object. Using the half arrow ( j ) enables you to indicate lateral inversion nicely. To show how the eye sees the image place a diagram of an eye on the same side of the mirror as the object. Draw a cone of rays from the tip of the image to just fill the pupil of the eye. Draw two light rays from the tip of the object to the two points on the mirror where the previous rays met the mirror. Put arrows on the Repeat this real light rays as shown. same procedure for each point on the object. Thus the eye is receiving light that appears to originate from the virtual image behind the mirror, but actually comes from the object. In all diagrams, use faint lines for construction, dotted lines for imaginary rays or virtual image, and solid lines for real rays, real image, object, etc.. REFLECTION OF LIGHT\u2014MIRRORS Sec. IV: 13 Fig. 17:7 Images in Parallel Mirrors. IV: 13 PLANE MIRROR SYSTEMS 1. Parallel Mirrors On looking into a mirror facing a parallel mirror on the opposite wall of a room, such as is used in barber shops, a very large number of images of the room can be seen stretching away almost endlessly. To discover how these are formed study Fig. 17:7 carefully. {ii, 12, h) A series of images representing the side of O facing mirror Mi are formed (/i, 1 2, 1.3,) and also a series of images representing the side of O facing mirror These multiple images '^^2 are due to the image formed in one mirror, acting as the object, which in turn forms an image in the second mirror, and so is absorbed at each reflection, each succeeding", " image is fainter than the one preceding. Since some light on. 2. Mirrors at Right Angles When two mirrors are placed at right angles to each other (Chap. 21, Exp. 4), three images of an object will be 17:8). Ii and h are observed images of O in mirrors Mi and M^. the mirror is the image of Ii (Fig. in Is M 2 produced (Mg produced is really an image of M 2 in Mi) and of I 2 in mirror Mi produced, these two images coinciding. How the these eye sees 1 3. Fig. 17:8 Images in Mirrors at Right Angles. images may be shown by a construction similar to that used previously. Mirrors Inclined at Sixty Degrees When mirrors are inclined, the number of images obtained depends upon the 187 Chap. 17 LIGHT angle between the mirrors. we can say: In general the number of images _ 360 ^ Z Inclination For example, using angles we obtained three images mirrors at If two mirrors are inclined at 60\u00b0, five images will be obtained 17:9). (Fig. right IV : 14 CURVED MIRRORS These form multiple images of differently shaped pieces of coloured glass placed between them. Many beautiful and fascinating designs can be produced at every turn of the instrument. Curved mirrors are frequently more suitable than plane mirrors for certain purposes. Curved mirrors are used as rear-vision mirrors, as shaving mirrors and as reflectors for car headlights. One important type of curved mirror is the spherical mirror whose reflecting surface is a portion of the surface of a sphere. If the inner surface is the reflecting surface, it is a concave mirror; if reflection occurs at the outer surface it is a convex mirror. We represent such mirrors in cross-section by the arc of a circle (Fig. 17:10). Silvered Surface Reflecting Surface Silvered Surface Reflecting Surface Fig. 17:9 Images in Mirrors inclined at 60\u00b0. Sir David Brewster, of Edinburgh, making use of the images produced by inclined mirrors, in 1819 invented the kaleidoscope. This consists of three mirrors set at angles of 60\u00b0 to each other. Spherical Mirrors. Fig. 17:10 (a) Concave or Converging. (b) Convex or Diverging. A few terms used in connection with curved mirrors need to be defined (Fig. 17:11): 188 REFLECTION OF LIGHT\u2014MIRRORS Sec. IV: 14 Fig. 17:12 Action of Curved Mirrors. (A) Concave. (B) Convex. Centre of Curvature, C, is the centre of the imaginary sphere from which the mirror was cut. that is, points through which reflected However, light rays are converged. there is only one principal focus as Vertex, F, is the mid point of the mir- defined. ror. Principal Axis, PV, is the line through the centre of curvature and the vertex of the mirror. Secondary Axis, CD, is any other line drawn through the centre of curvature to the mirror. Since all radii of a circle meet the circumference at right angles, and since all axes behave like radii since they pass through the centre of curvature, it follows, that a Normal to the surface of a curved mirror at the point of incidence is simply a secondary axis. to it) pass, close Principal Focus, F, is the point on the principal axis, through which rays, travelling parallel to the principal axis (and fairly after being reflected from the mirror. For a convex mirror, the principal focus is a virtual focus, and is that point from which such rays appear to diverge after reflection from the mirThis point will be found to be ror. midway between the centre of curvature and the vertex of the mirror. There are many other foci possible. Focal Length, FV, is the distance from the principal focus to the vertex of the mirror. The action of curved mirrors (Chap. 21, Exp. 5), may be demonstrated by means of the optical disc. Mount a concave mirror on the optical disc so that the 0\u00b0\u20140\u00b0 line becomes the prin- cipal axis of the mirror. Insert a metal shield with several parallel slits in the window of the opaque collar and shine light from a projection lantern through these slits onto the mirror (Fig. 17:12). You will note that the reflected rays conIf the disc is rotated so that the verge. incident rays are parallel to the principal axis it will be found that the reflected rays converge through a point. This point is the principal focus of the mirror. Similarly, if we insert a convex mirror in place of the concave one, and repeat the above, it will be found that the reflected rays diverge as though coming from a point behind the mirror. This virtual point locates the principal focus of the convex mirror", ". 189 Chap. 17 LIGHT IV: 15 IMAGES IN A CONCAVE MIRROR In chapter experiment 21 we 6, studied the characteristics and position of images formed by a concave mirror for difTerent positions of the object. As the object was moved closer to the mirthe image, which was real and ror, Rays Used to Locate the Fig. 17:13 Image, I, of an Object, O, in a Concave Mirror. inverted, gradually grew larger in size and moved farther from the mirror. When the object was located inside the principal focus of the mirror, the image became virtual and erect and was located behind the mirror. All of these results can be verified with fair accuracy by means of simple geometric diagrams. The principles used here are very similar to those used previously with plane mirrors. Each point on the object is sending out rays of light in straight lines in all directions. If we trace the paths of several of these rays from a point on the object, to the mirror, and then their rays back from the mirror, we will find that they intersect. The point of intersection is the point on the image that corresponds to the original point on the object from whence the light rays came. reflected Suitable rays for this purpose, as illustrated in Fig. 17:13, are: 190 1. A ray from the object parallel to the principal axis, which will be reflected through the principal focus. 2. A ray from the object through the centre of curvature which will be reflected back along the same path, since the incident ray strikes the mirror at right angles. 3. A ray from the object through the principal focus, which will be reflected parallel to the principal axis. 4. A ray from the object to the vertex of the mirror, which will be reflected so that the angle of incidence equals the angle of reflection. 5. Any other ray from the object to the mirror will be reflected so that Z f = Z r. In actual practice any two of these rays will suffice. The first two mentioned are the most easily drawn and hence are the most convenient to use. In order to verify the observations obtained in our experiment, it is necessary to make accurate scale drawings. To Locate the Image in a Fig. 17:14 Concave Mirror of an Object Placed Beyond F. For example, if the focal length of the concave mirror was 20 cm., the object 60 cm. from the mirror, and 12 cm. high, our construction would be as above (Fig. 17:14), using the scale 10 cm. = 1 cm. REFLECTION OF LIGHT\u2014MIRRORS Sec. IV; 17 Draw principal axis, PV. With centre C, and radius of curvature 4 cm. (twice focal length), draw an arc to represent the mirror. Locate the principal focus F, midway between C and V (2 cm. from mirror). Locate the object O, 6 cm. from the mirror. Draw the object (OOi) 1.2 cm. high and perpendicular to the axis. From the tip of the object (Oi) draw the two rays and their reflected rays as outlined on previous page. These rays intersect at /i, and hence locate the position of the tip of the (Rays from other points on the image. would corresponding object image points. sufficient.) Draw the image 11 1, perpendicular to PV. The distance of the image from the mirror and its size can be obtained by measuring accurately. Notice that when the object is beyond the image is between F and C, C, real, smaller than the object, and inverted. See how perfectly this diagram confirms your observations recorded in the table on page 242. produce However, one the is To Locate the Image in a Fig. 17:15 Concave Mirror of an Object Placed Between F and V. of for other positions The student should construct similar diagrams the object, and verify the other observations made. The special case, where a virtual image is obtained, is slightly more difficult. Study the following diagram carefully, and note the method is that identical to that given previously (Fig. 17:15). However, this time the reflected rays diverge, and therefore appear to come from a point behind the mirror, thereby creating the virtual image. IV : 16 IMAGES IN A CONVEX MIRROR One has often observed the images produced by the back of a spoon, the side of a tea-kettle or the shiny fender of a car. Such objects as these are all acting as convex mirrors. In experiment Fig. 17:16 To Locate the Image in a Convex Mirror. 7, chapter 21, we found that the images in a convex mirror were always behind virtual, smaller than the the mirror, object, and erect. The same method is used as with the concave mirror to verify this geometrically (Fig. 17:16). IV: 17 HOW THE EYE SEES THE IMAGE discussed As previously section IV", ": 12, the eye sees the image by means of rays which actually come from the object, but which appear to come from in Fig. 17:17 How the Eye Sees a Real Image in a Concave Mirror. 191 Chap. 17 LIGHT : the image. Fig. 17:17 and 17:18 show how this is accomplished. In each case the light rays start from the object, are reflected at the mirror, pass through a real image or appear to come from a Fig. 17:18 How the Eye Sees a Virtual Image In a Convex Mirror. virtual image, and enter the eye. In order to construct these diagrams just reverse the previous order. That is, join 7 to the outer edges of the pupil of the eye. This cone is projected back to the mirror or cuts the mirror. From there it is drawn back to the object.!V: 18 THE MIRROR FORMULAE In the previous sections we have been studying the images produced by curved It is possible by means of two mirrors. simple formulae to determine the location and characteristics of the image for various positions of the object. To do a consistent convention regarding so, signs must be followed. Values for distances of objects and real images are always positive; those for virtual images are negative. Similarly, the focal length of a concave mirror (which has a real is positive, that of a principal focus) convex mirror (with a virtual principal is negative. To state this confocus) cisely, the convention of signs is: \u201creal is positive, virtual is negative\u201d The two formulae, with worked examples to show how to use them, follow (a) Magnification Formula Height of Image Distance of Image Height of Object Distance of Object 77o Do Note the similarity of this formula with that obtained for the pin-hole camera (Sec. IV: 4). (b) Distance Formula Distance of Object Distance of Image _ 1 Focal Length 1. An object 2 in. tall is placed 15 in. from a concave mirror whose focal length is 5 in. Find the position and size of the image. Examples Ho= 2 in. Do = 15 in. f = 5 in. Di=? 1 1 _ 1 \u2014 + Do Dr 7 \u2014 + 1 1 1 T 1 15 15 1 _ 1 D,\u201c 5 :.2Di \u2014 15 :.Di =7.5 The image is located 7.5 in. from the mirror. 192 Ho\u2014 2 in. Do = 15 in. Dj \u2014 7.5 in. REFLECTION OF LIGHT\u2014MIRRORS. Hi ^ A \u2018 Ho~ Do. Hi _ 7.5 ~ 15'2?. \u2018 Sec. IV: 19 The image is 1 in. tall. Check these results by making an accurate scale construction to locate the image geometrically. Both position and size of image should agree closely with the above values obtained by calculation. 2. An object 5 cm. high is placed 30 cm. from a convex mirror whose focal length is 20 cm. Find the position, size, and nature of the image. Ho= 5 cm. Do = 30 cm. f = \u2014 20 cm. Di =?'Do^ Di f 30 1 _ ~~ \u2018 Di. \u2018 Di \u2014 20 1 1 20 30 _ -3 -2 \u201d 60 \u2014 5Di=:60 Di = \u2014 12 cm. Image is 12 cm. from mirror and is virtual (since sign is negative). Ho= 5 cm. Do = 30 cm. 12 cm. Di? \u2018 Ho Do \u2018 \u2019 5 ~ 30 Hi = 2 cm. Image is 2 cm. high. IV: 19 APPLICATIONS OF MIRRORS Reference has already been made in section IV : 1 1 to the uses of plane mirrors. Many of these uses, however, are more satisfactorily fulfilled by using curved mirrors. For example, a plane mirror is frequently used as a shaving mirror. However, a concave mirror so used produces an enlarged erect image when the face is held between the principal focus and the vertex of the mirThe enlarged image is often a ror. Similarly, a convex distinct advantage. mirror is frequently used as a rear vision mirror instead of a plane mirror. The erect, smaller image produced gives a wider field of view with the same size mirror. Spherical mirrors such as we have been describing have one serious defect, known as spherical aberration. In such mirrors, only those rays parallel to the principal axis and fairly near to it pass through the principal focus on being 193 Chap. 17 LIGHT are axis (Fig. 17:19). reflected from the mirror. Rays farther from the principal reflected through points some distance from the Conversely, if a focus source of light is placed at the principal focus, the reflected rays will not form a parallel beam, but the outer ones will be scattered and hence the light is weakened. Fig. 17:19 Spherical Aberration in Cur", "ved Mirrors. To overcome this parabolic rather than spherical mirrors are used defect, 17:20). A parabola is a section (Fig. from a cone obtained by cutting the cone Fig. 17:20 A Parabolic Mirror Used to Prevent Spherical Aberration. by a plane parallel to a line from the apex of the cone to any point on the circumference of the circular base. All rays which emanate from the principal focus of such a mirror, after reflection are parallel to the principal axis, no matter how great the aperture is. As a result parabolic mirrors are used as reflectors for searchlights, car headlights and in the reflecting telescope. IV : 20 QUESTIONS A 1. (a) Define: angle of incidence, angle of reflection. (b) State the two laws of reflection and indicate how you could test them. 5. 2. (a) Distinguish between diffuse and locate the image of an inclined arrow in a plane mirror. (b) Show how an eye sees the image. (a) How many images will be observed in 2 mirrors inclined at 45\u00b0? (b) Draw a diagram to locate these images. glare de- 6. (a) Define: principal axis, principal regular reflection. (b) How is undesirable creased? 3. (a) State the rule for determining the position of an image in a plane mirror. (b) Describe the Image. between (c) Distinguish real and virtual images. (a) Show by a diagram how you can 4. 194 focus, focal length. (b) By means of accurate construction show how the of (i) a concave mirror, and (ii) a convex mirror may be located. principal focus 7. (a) Under what conditions does a concave mirror form (i) a real image, (ii) a virtual image? REFLECTION OF LIGHT-MIRRORS Sec. IV: 20 (b) Describe the size and location of all the real images produced by a concave mirror. (c) When a concave mirror is used in shaving, where is the person\u2019s face in relation to the principal focus of the mirror? What kind of image is seen? 8. An object is located just beyond the centre of curvature of a concave mirror. (a) By means of a diagram locate its image. (b) Show how the eye sees the image. (c) State the characteristics of the image. 9. (a) State uses of plane, concave and convex mirrors. (b) How does a searchlight produce a narrow beam of light? B 1. A watch seen in a mirror seems to read (a) 2 o\u2019clock, (b) 6.15 o\u2019clock. What does it actually read? Explain. 2. Support a mirror vertically on a desk and stand a book in front of it to serve as a screen. Draw a triangle on a piece of paper and lay it flat between the book and the mirror. Watching the image in the mirror, but not the triangle itself, try to retrace the triangle. Why do you have difficulty? 3. What changes would you observe in your image as you walked toward (a) a plane mirror, (b) a concave mirror, (c) a convex mirror? the position grams locate of the image for each position of the object, (b) State the characteristics of each image. accurate 5. By means of an scale diagram locate the image produced by a convex mirror, whose focal length is 15 in., of an object 6 in. high and 2 ft. distant. State the characteristics of the image. 6. An object is 4 in. tall and its image is 6 in. tall when the object is placed 2 ft. from the mirror. How far is the image from the mirror? 7. A 6 in. pencil is ft. in front of a curved mirror. Find the length of the image if it is 8 in. from the mirror. 1 8. How tall is an object if it produces an image 3 in. tall located 7 in. behind a convex mirror when the object is located 1 0 ft. in front of the mirror? 9. An object 4 in. in front of a concave mirror produces an image 1 2 in. behind the mirror. Find the focal length of the mirror. 10. An object is 1 8 cm. in front of a concave mirror which has a focal length of 1 2 cm. How far is the image from the mirror? 11. An object 9 in. from a convex mirror produces an image 3 in. behind the mirror. Find the focal length of the mirror. 12. An object is 15 cm. from a convex mirror of focal length 20 cm. Calculate the image distance. 4. An object 15 cm. high is located (i) 75 cm. (ii) 60 cm. (iii) 45", " cm. (iv) 30 cm. (v) 20 cm. from a concave mirror whose focal length is 30 cm. 13. The focal length of a concave mirror is 12 in. How tall is the image of a 5 in. candle standing 15 in. from the mirror? 14. Repeat question 13 using a convex (a) By means of accurate scale dia- instead of a concave mirror. 195 CHAPTER 18 REFRACTION OF LIGHTLENSES phate (hypo) and a few drops of hydrochloric acid. These materials will render the water slightly turbid and make Incident Mirror Refraction of Light on PassFig. 18:2 ing from Air to Water or from Water to Air. visible a light beam projected through Shine a beam obliquely upon the it. surface of the water (Fig. 18:2). Part of this beam will be reflected at the surface of the water, and part will be If a refracted as it enters the water. normal (perpendicular) is placed at the point of incidence, it will be observed that the light beam is bent toward the normal. Similarly, if we shine the beam of light obliquely up through the water and into the air, it will be observed to bend away from the normal. Both effects can be shown simultaneously by placing a plane mirror on the bottom of the tank Note that no refraction (Fig. occurs if the light enters or leaves the water at right angles to the surface. 18:2). We may also show that a beam of light is refracted on passing obliquely IV; 21 MEANING OF REFRACTION Water frequently appears to be much shallower than it actually is. An oar or stick when only partly submerged appears to be bent upwards at the surface of the water (Fig. 18:1). These phenomena, and many others similar to them, are due to the bending of light rays as they pass obliquely from one medium into another of different optical Fig. 18:1 Partly Immersed Stick Ap- pears Bent. density (Sec. IV: 22). Such bending of the light rays is called refraction. Refraction of light as it passes from air into water, or from water into air, can be illustrated Fill a tank with water containing a small amount of fluorescein, or a little sodium thiosul- easily. 196 REFRACTION OF LIGHT-LENSES Sec. IV: 23 from air into glass, or from glass into Place a semicircular block of glass air. on the optical disc, so that its flat edge is bisected at right angles by the 0\u00b0-0\u00b0 Shine a ray axis of the optical disc. of light so that it strikes the surface at this central point where the axis crosses it. The axis thus is made the normal to the refracting surface. The light will Fig. 18:3 Refraction of Light on Pass- ing from Air to Glass. be refracted as shown in Fig. 18:3. Why is the light not refracted on leaving, or on entering the circular surface of the glass? From the preceding observations we may summarize the behaviour of light rays in passing from one medium into another of different optical density, as follows: 1. When a ray of light passes obliquely from one medium into another of greater optical density, it is refracted toward the normal. 2. When a ray of light passes obliquely from one medium into another of less optical density, it is refracted away from the normal. It follows from the above that when a ray of light enters a new medium at right angles to the surface, no refraction occurs. IV: 22 EXPLANATION OF REFRACTION We may ask what causes refraction of light. The following illustration should help us to understand it. Suppose the brakes of an automobile are improperly adjusted, those on the right wheels holding better than those on the left. When the brakes are applied, the car will swerve to the right, that is, to the side that is slowed up most. Similarly, the bending of the light beam, or refraction, is caused by the change in the velocity of light as it passes from one medium into another of different optical density. The greater the optical density of a substance the more slowly light will travel through it. According to the wave theory, light travels out from the source as spherical waves. When a wave-front enters an optically denser medium obliquely (Fig. 18:4), that part of it that enters the new medium first will be slowed, while the rest of it continues to advance at the same speed as before. Consequently, the wave-front swerves toward the normal as it enters the denser medium. Conversely, the wave-front would swerve away from the normal as it speeds up on passing into an optically less dense medium. No bending occurs if the light enters the new medium at right angles to the surface, for the entire wave", "-front would be slowed down or speeded up at the same instant. IV : 23 INDEX OF REFRACTION From Fig. 18:4 it is evident that the light is travelling the distance PR in one medium (air), while it travels the distance OQ in the other medium. Hence these distances must be proportional to 197 Chap. 18 LIGHT the velocities of the light in the two media. Also, it should be evident that the amount of refraction is governed by these In the study of refraction only acute angles are involved. When the angle is contained in a right-angled triangle, its sine is a constant quantity found by dividing the length of the side opposite A the angle by the length of the hypotenuse. For example, in A ABC, Fig. 18:4 Refraction of Wave Front on Entering an Optically Denser Medium. relative distances or these relative velocities, and would be constant for any two given media. This ratio is called the index of refraction, and may be defined as follows: Index of Refraction of a medium (yu,) Velocity of light in air Velocity of light in the medium V (air) V (medium) and AB Sin Z ACB = Sin Z CAB = \u2014 AC BC AC *To prove that: Index of Refraction (u) = Sin Z i Sin Ir PR OQ PR OQ V (air) ^ Sin Z i V {medium) PR/ OR Sin Ir OQ/OR P ~ Sin Z i TZ Sin Z ^ In experiment 8, chapter 21, by means of a simple geometric construction it shown that Index of Refraction is Sine of angle of incidence* Sine of angle of refraction The sine of an angle ( abbreviated sin Z ) is a property of an angle found useful in calculations in mathematics and science. The index of refraction is an important physical property of a transparent medium. An instrument, called a refractometer is used to measure' its value quickly and accurately. This enables scientists to identify substances and to If the index of recheck their purity. fraction of a substance is known the velocity of light may be determined for that substance. A substance with a large index of refraction is said to have a REFRACTION OF LIGHT-LENSES Sec. IV: 25 high optical density, because it permits light to travel through it at a relatively Such substances refract slow velocity. light to a greater extent than do those with a smaller index of refraction. The brilliance of diamonds, and other precious stones is largely due to this fact (Sec. IV: 26). The index of refraction of a substance varies with the colour of the light that passes through it. The rainbow and the beautiful colours obtained when light is refracted through cut glass and precious stones are due to Temperature, too, this has an effect on the index of refraction because it alters the optical density of The waviness observed a when light rising above a hot object is caused by the differing.refractive indices of various layers of hot and cold air. passes through air (Sec. IV: 37). substance. The Index of Refraction of Some Common Substances W ater Crown Glass Flint Glass 1.3 1.5 1.7 Quartz Zircon Diamond 1.5 1.9 2.4 IV : 24 REFRACTION THROUGH A GLASS PLATE When a ray of light passes obliquely through a glass plate with parallel sides (Chap. 21, Exp. 9), it is refracted both on entering and on leaving the glass. On entering, the light is slowed down and Fig. 18:5 Refraction Through a Glass Plate with Parallel Sides. is, therefore, refracted toward the normal. On emerging, speeds up again, and hence is refracted away from the normal. Since this second effect exactly counteracts the first the speeding up exactly compensates for the light the (i.e., Fig. 18:6 Why an Object Appears 'Closer When Viewed Through a Glass Plate. original slowing down), the emergent ray will be parallel to the incident ray but laterally displaced (Fig. 18:5). The amount of this lateral displacement depends on the refractive index of the glass, on the angle of incidence, and on the thickness of the glass. On looking' at an object through such a plate, it will be slightly displaced in position and will appear nearer to the eye than it is in fact (Fig. 18:6). The same thing occurs in water and this accounts for the diffi-culty in locating the exact position of an immersed object (Chap. 21, Exp. 10). IV: 25 REFRACTION THROUGH PRISMS A prism consists of a wedge-shaped portion of a refracting substance, bounded by two plane surfaces inclined at an angle to each other, this angle being called the refracting angle. When a ray of light enters such a prism it is slowed down, and therefore", " refracted toward the normal. On leaving, it speeds up, and therefore bends away from the normal. This can be shown by mounting a 60\u00b0 prism on an optical disc and shining a ray of monochromatic light, e.g., red, through it. From Fig. 18:7 it is seen 199. Chap. 18 LIGHT that both these refractions are toward the thick base of the prism. Hence the light is bent, or deviated, quite consid- A prism when the refracted ray passes through the prism parallel to the base. Finally, the amount of deviation depends upon the colour, or the wavelength, of the light used, which is why monochromatic light was used in these experiments. gives prisms one of their large areas of usefulness. More discussion on this follows in the chapter on colour, page 213. factor This latter Fig. 18:7 Deviation Through a Glass Prism. erably from its original path. The angle of deviation, D, is obtained by extending the incident and emergent rays to meet (Chap. 21, Exp. 11 ) The amount of deviation produced by a prism depends upon a number of facFirst, the material comprising the tors. prism\u2014^the greater its index of refrac- tion, the greater will be the deviation produced. Secondly, the shape of the prism\u2014^^the greater its refracting angle, the greater will be the angle of deviation. Thirdly, the angle of incidence at which the light meets the prism. This can be shown by rotating the optical disc and observing the amount of deviation for various angles of incidence. Minimum deviation is obtained from an equilateral IV : 26 TOTAL REFLECTION It is more usual to consider light passing from air into an optically denser medium; however, when it goes from the optically denser medium into air, or a less dense medium such as from water or glass into air, a peculiar phenomenon occurs. As the light speeds up on entering the air obliquely it bends away from the normal. As the angle of in the denser medium inincidence creases, we finally come to a position when the refracted ray just grazes the surface, that is, the angle of refraction equals 90\u00b0. This angle of incidence is angle. When the called the critical incident angle exceeds the critical angle the light is completely reflected, a phenomenon known as total reflection (Fig. 18:8). Using the semicircular block of glass on the optical disc as in section IV:21, shine a ray of light through it. Rotate the disc to increase the angle of Fig. 18:8 Total Reflection of Light (a) On Passing from Water to Air. (b) On Passing from Glass to Air. 200 REFRACTION OF LIGHT\u2014LENSES Sec. IV: 26 incidence and note the size when total This angle, the reflection first occurs. critical angle for glass, is about 42\u00b0. Fig, 18:9 An Illustration of Total Reflection. V An example of total reflection is seen when an empty test-tube is placed in a Canadian Industries Ltd. An Interesting Example of Light Being ''Bent\" in a \"Perspex\" Rod. beaker of water (Fig. 18:9). On looking down into the water the sides of Applications of Total ReFig. 18:10 flection Prisms (a) A Simple Periscope. (b) Field Glasses. 201 Chap. 18 LIGHT, is very Total reflection test-tube appear silvery. This is the due to the light striking the surface of the tube at an angle greater than the This light is critical angle for glass. totally reflected to the eye as shown. Similarly a sooted ball appears silvery on being lowered into a beaker of water. A layer of air entrapped by the coating of soot produces a water-air -boundary at which reflection occurs. a useful phenomenon. Total reflection prisms are usually right-angled prisms with wellpolished faces. Light enters such a prism at an angle of incidence of 45\u00b0, which is greater than the critical angle for glass (p. 201), and hence total reflection occurs. Such prisms are used in periscopes (Fig. 18:10a), range finders, field-glasses (Fig, 18:10b), and reflecting telescopes. They are much more efficient reflectors than mirrors, since mirrors reflect only about seventy per cent of the light they receive, whereas prisms reflect a much greater proportion. In addition, prisms are more robust, there is no silvering to tarnish and they give rise to a single well-defined image. The brilliancy of diamonds, brilliants, and cut-glass dishes is due to total reflection. The greater the index of refraction of a substance the smaller is its critical angle. Diamonds have a large index of refraction (Sec. IV: 23) and consequently have a small critical angle.", " The surfaces of the diamonds meet each other at such angles that much of the light entering a diamond is totally reflected a number of times internally, before eventually being refracted out. This lights up many surfaces, and gives the diamond its sparkle. Brilliants, and cut-glass articles are often made of leaded The addition of lead increases the index of refraction, and consequently the cut faces are able to cause much total reflection. glass. 202 IV : 27 ATMOSPHERIC REFRACTION Light travels faster in a vacuum than in air. Therefore, light reaching us from the sun and stars will slow up and be refracted as it enters the atmosphere of the earth. Since the atmosphere gradually becomes denser as the altitude decreases, the light will be refracted more and more as it passes through successive layers of denser air nearer the surface. Consequently, when light comes to us obliquely from the sun and stars these S' A' Fig. 18:11 Atmospheric Refraction (a) The Sun Low on the Horizon. (b) A Mirage. the distances increased bright objects appear higher than they really are (Fig. 18:11a). This effect is most pronounced at low altitudes because of the larger angles of incidence, and be travelled through the successive layers of air. As a result the sun appears to set several minutes after it has actually passed below the horizon. The enlarged and sometimes elliptical appearance of the sun and moon when near the horizon is due to the fact that rays from the to REFRACTION OF LIGHT\u2014LENSES Sec. IV: 28 lower edge are refracted more than those from the upper edge. A mirage is a well-known optical illusion caused by refraction and sometimes total reflection of light as it passes through layers of atmosphere of varying density (Fig. 18:11b). Objects in the distance may be raised above or depressed below their normal position and may be distorted into irregular fantastic shapes. The most commonly observed mirage is that of an apparent layer of water over a hot level sandy surface, or paved roadway. The mirage in these cases sky, produced by total reflection from layers of air near the ground. really an image of the is IV: 28 LENSES A lens is a piece of transparent refracting medium, usually glass, bounded by two spherical surfaces, or by a plane and a spherical surface. There are two main types: (a) Converging or Convex Lenses are thicker at the centre than at the outer edge. Such lenses always refract rays of light so as to converge them. They therefore collect light. CONVERGING or CONVEX DIVERGING or CONCAVE Fig. 18:12 Kinds of Lenses. (b) Diverging or Concave Lenses are thinner at the centre than at the These lenses cause outer edge. They to diverge. light rays therefore scatter light. Fig. 18:13 Comparison of Lenses and Prisms, (a) Action of a Converging Lens. (b) Action of a Diverging Lens. 203 Chap. 18 LIGHT These two types of lenses may be of varying shapes as shown in Fig. 18:12. Each shape is devised for a specific purpose. The action of a lens is similar to that o-f two prisms base to base (Fig. 18:13). As discussed in section IV: 25, the light is bent toward the base of the prism both on entering and leaving the prism. Similarly, the light is bent toward the thicker part of the lens both on entering and leaving it. These examples explain the converging action of a convex lens, and the diverging action of a concave lens. Fig. 18:14 Terms Pertaining to Lenses. The study of lenses makes use of a new vocabulary. The terms used are explained below (Fig. 18:14) : Centre of Curvature C, In most lenses there are two centres of curvature. They are the centres of the spherical surfaces that bound the lens. Principal Axis of a lens is the line passing through the centres of curvature of the two faces, or, in the case of a lens which has one face plane, it is the line passing through the centre of the curved face and curvature of which is normal to the plane face. Optical Centre O, is the point on the principal axis midway between the two surfaces of the lens. All distances along the principal axis are measured from this point. Principal Focus F, of a convex lens is that point on the principal axis to which a beam of light which is parallel to the principal axis converges 204 after refraction through the lens. (In a concave lens, the principal focus is a virtual point, and is that point on the principal axis from which a beam of light, which is parallel to the prindiverge on cipal being refracted through the lens.) axis, appears to Focal Plane is a surface that passes through the principal focus perpendicular to the principal axis of", " the lens. Focal Length is the distance of the principal focus from the optical centre of the lens. Note 1: In Fig. 18:15 (see below) a line has been drawn through the optical centre of the lens perpendicular to the principal axis. For simplicity, we may represent the entire refraction of light as occurring at this line. Actually, of course, a light ray will be refracted both Fig. 18:15 Represented and Actual Paths of Light Through a Lens. on entering and on leaving the lens. This actual path of the light through is shown by drawing a line the lens between the point where the incident ray enters the lens, and the point where the refracted ray leaves the lens. Note 2: A ray through the optical centre of a lens may be considered as passing straight through the lens. This lens is through which the light is travelling are almost parallel. Therefore, the light will be refracted on entering and on leaving because surfaces the the of REFRACTION OF LIGHT\u2014LENSES Sec. IV: 29 in such a way that the emergent ray will be parallel to, but laterally displaced from, the incident ray (Sec. IV: 24). We shall consider all our lenses to be very thin, so that the amount of lateral displacement is negligible. Increasing either or both of these increases -.the amount of bending of the light and so shortens the focal length. The powers of lenses used in most optical instruments are usually expressed in terms of their focal lengths. IV: 29 FOCAL LENGTH OF LENSES To determine the focal length of a lens the position of the principal focus In optometry it is more usual to deal with the power of a lens rather than its focal length. The unit of power is the dioptre, which is the power of a converging lens of focal length one metre (100 centimetres). The shorter the focal the greater the power of the length, lens, and accordingly the power can be the focal related length the by to formula: r,, P (dioptres) X 100 / (cm.) According to our convention of signs (Sec. IV: 18, and IV: 32), a convex lens with a real principal focus has a positive focal length and a 4- power; a concave lens with a virtual principal focus has a negative focal length and a \u2014 power. If a lens is used in a different medium, its focal length and therefore its power will change. This can be shown by placing a lens in a tank of turbid water (Fig. 18:17). Shine a parallel beam of Comparison of Focal Lengths Fig. 18:17 of Lens (a) In Air (b) Immersed in Water. 205 Concave Fig. 18:16 Principal Focus of a Convex Lens and a Concave Lens must first be determined and then its distance from the optical centre of the lens measured (Chap. 21, Exp. 12). This may be demonstrated by placing a convex lens on an optical disc. Shine a number of rays, parallel to the principal axis, through the disc and note the point through which they converge. This point is the principal focus (Fig. 18:16). The focal length depends upon the index of refraction of the lens and its thickness. Chap. 18 LIGHT : For example, a glass globe of water in sunlight could focus the sun\u2019s rays. If flammable material should be located at the principal focus of this lens a fire could easily result. Use is made of this very fact in some types of sunlight recorders used by weather bureaus. A sunlight recorder is a device used to determine the number of hours of bright sunshine. In Ontario, the meteorological department uses Campbell-Stokes Sunshine Recorder consists essentially of two parts 18:18). (Fig. the It ( 1 ) A glass sphere which brings the sun\u2019s rays to a focus. (2) An approximately spherical metal light through the water and the lens, and compare the focal length in water with that previously obtained in air. It will be found to be longer. This is to be expected as there is a smaller decrease in velocity of the light going from water to glass than when going from air to glass. Consequently there is less bending of the light, and therefore a longer focal length. A convex air lens in water would function as a diverging lens. ExIt can be constructed by plain why. cementing watch glasses together using a waterproof cement. Care should be observed in the use and location of spherical transparent objects. 206 ; REFRACTION OF LIGHT\u2014LENSES Sec. IV: 31 bowl carries cards which form a belt on which the sun burns a record. lens. This ray may be considered straight through the as lens (Sec. IV; 28). passing recorder. Real care must be observed in setting up There must be no obthe structions that would shield the recorder It must be placed from the sun\u2019s rays.", " table, and made perfectly on a rigid Instructions are always provided level. for adjusting for latitude, and for adjusting for the time meridian. The glass ball must be kept perfectly clean at all times. IV: 30 IMAGES IN CONVEX LENSES Previously we saw how images were ( Sec. IV formed in concave mirrors 15). Convex lenses form a very similar series of images by refracting the light that passes through them. In experiment 13, chapter 21 the method of studying the characteristics and position of the images formed is given. It will be found that as the object approaches the lens up to the focal plane the image formed on the opposite side of the lens is real, inverted, and gradually becomes larger in size as it moves farther from the lens. When the object is located inside the focal plane, the image becomes virtual, erect, and is located on the same side of the lens as the object. As in mirrors, so in lenses, it is possible to verify these results by means of simple geometric diagrams. To do so it is necessary to draw two rays from any point on object, and determine through what point they are focused by (Fig. 18:19). The two most the lens the suitable rays are; ( 1 ) A ray from the tip of the object parallel to the principal axis. This ray on passing through the lens is refracted principal through the focus., (2) A ray from the tip of the object through the optical centre of the Fig. 18:19 To Locate a Real Image in a Convex Lens. The point where these two refracted rays cross locates the tip of the image. A similar construction for other points on the object will locate the corresponding points on the image. Fig. 18:20 shows how to locate the virtual image obtained when the object is inside the focal plane of the lens. Fig. 18:20 To Locate a Virtual Image in a Convex Lens. The construction is identical to that just Because the refracted rays described. diverge from each other, it is necessary to produce them back to where they appear to meet. IV; 31 IMAGES IN CONCAVE LENSES By referring to experiment 14 chapter 21 it will be observed that concave lenses can form virtual images only. These are 207 Chap. 18 LIGHT always erect, smaller than the object, and located at less than the focal dis(Fig. 18:21). A tance from the lens close similarity will be noted between the images formed by concave lenses and those formed by convex mirrors (Sec. IV:16). Fig. 18:22 shows how the eye sees the Fig. 18:21 To Locate the Image of image produced by a lens. an Object in a Concave Lens. (b) Virtual Images. IV: 32 THE LENS FORMULAE The formulae we obtained for curved mirrors (Sec. IV: 18) also apply to lenses, that is, (a) Magnification Formula Height of Image Distance of Image FI eight of Object Distance of Object A fio~~ Do (b) Distance Formula Distance of Object Distance of Image Focal Length 1 + 1 = 1 D. D< f The same convention of signs must be followed, namely: real is posi- tive, virtual is negative. 208 REFRACTION OF LIGHT\u2014LENSES Sec. IV:32 Examples 1. An object 5 cm. tall is placed 30 cm. from a convex lens whose focal length is 10 cm. (a) By means of an accurate scale diagram locate the image, and state its characteristics. Fig. 18:23 Scale 5 cm. = ^ in. (b) By using the lens formulae, determine the position of the image, and its size. How could you tell from your answer whether it is real or virtual? Do \u2014 30 cm. Di \u2014? 10 cm. / Do \u2014 + 1, 30 Di 1 / 1 Di~ 10 1 1 D~ 10 1 30 _ 3 \u2014 1 ~ 30.'. Di=: 15 Note: Since the image distance i; + 15 cm., therefore the image is real..*. Imaae distance is 15 cm. Do = 30 cm. Di = 15 cm. Ho = 5 cm. H,=?..^_D, \u2019 Ho~ Do \u2018. Hi _ 15 ~ 30 Hi = 2.5 5 Height of image is 2.5 cm. 209 Chap. 18 LIGHT 2. A concave lens has a focal length of 4 in. An object 1 in. high is 12 in. from the lens. (a) Determine the position and size of the image. (b) Verify your answer by making an accurate scale diagram. / = \u2014 4 in. Do = ( \u2014 since / is virtual) 12 in...Ill D~ f.\u2022. 1 + '- = -1 12 Di 4. 1 _ 1 1 _ \u20143 \u20141.\\Di = -3 Image distance is", " 3 in. Note: Since the image distance is negative, therefore, the image is virtual. Hi =? Ho = I in. Di = 3 in. Do= 12 in...Hi _1^ \u2018 Ho~ Do \u2018 ^- 1.''T~ 12.\\Hi =. 25 Height of image is.25 in. IV: 33 APPLICATIONS OF LENSES part As lenses are an essential of almost every optical instrument, a discussion of their major applications is reserved until chapter 20 where several optical instruments are described. However, to relate specifically to the facts learned in the preceding sections we will mention a few simple uses here. The camera (Sec. IV; 44), (Sec. IV: 45), IV; 48) the telescope and the projection lantern (Sec. IV: 49) all contain a convex lens as an essential part of their construction. All these in- the eye (Sec. struments produce real, inverted images because the object viewed in each case is beyond the principal focus of the lens. (Sec. IV;46) In the magnifying glass and microscope (Sec. IV; 47) an enlarged, erect, virtual image is obtained because the object is inside the principal focus of the convex lens. Concave lenses are used along with convex lenses in many optical instruments to overcome certain defects, e.g., chromatic aberra(Sec. IV; 42), that would be apthe convex lens were used parent if tion alone. 210 REFRACTION OF LIGHT\u2014LENSES Sec. IV: 34 IV : 34 QUESTIONS 1. 2. 3. A (a) Define refraction of light. (b) Describe and explain fraction of light as it passes obliquely from one medium into another of different optical density. the re- (a) Define index of refraction. (b) Draw a diagram showing how you can see a coin lying on the bottom of a dish filled with water, though the coin would be hidden if the dish contained no water. (c) What is the velocity of light in quartz? (See table p. 1 99 for index of refraction of quartz). (d) The velocity of light in a diamond is 75,300 miles per second. What is its index of refraction? (a) How do you account for shimmering effect seen in the above a hot radiator? (b) How does the same principle account the twinkling the the for of air stars? 4. Place a thick glass plate on a line drawn on a piece of paper. View the line obliquely. Describe and explain what is observed. 5. aid of a diagram (a) With the explain why an oar appears bent when partly immersed in water and viewed obliquely. (b) Is the index of refraction greater for glass in which the velocity of light is 1 24,000 miles per second, or for water, in which the velocity is 1 40,000 miles per second? Why? 7. (a) Define critical angle. Illustrate your definition with a labelled diagram. (b) Would the be greater for water (Index of refrac- angle critical 8. 9. 1.33) or for glass (index of tion refraction 1.5)? Why? (a) Draw a diagram to show how a right-angled prism may be used to secure (i) one total internal reflection, (ii) two total internal reflections. (b) Explain why total reflection occurs in these two cases. (c) Why are total-reflection prisms preferable to mirrors in many optical instruments? (a) Compare the action of lenses to that of two prisms. (b) Define principal focus of a lens. (c) How can you determine experimentally the focal length of a lens? (d) Calculate the power of a lens whose focal length is 0.15 metres. 10. (a) What is the purpose of a sunlight recorder? (b) What approximate position relative to the glass sphere should the recording-belt occupy In the sunlight recorder? 11. (a) Distinguish between convex and concave lenses. (b) Summarize the types of images possible with both types of lenses and the conditions under which each is obtained. 12. An object is located at a point more than twice the focal length from a convex 6. (a) Define angle of deviation. lens. (b) State four factors that govern the amount of deviation produced by a prism. (c) Is the index of refraction of glass (a) By means of a diagram locate its image. (b) State the characteristics of the image. constant for all colours of light? 13. (a) What do we mean by the magni- Explain your answer. fication produced by a lens? 211 Chap. 18 LIGHT 9. (b) How does the magnification depend on image distance and object distance? B 1", ". What is the index of refraction of a liquid in which the speed of light is 1 55,000 miles per second? 2. The index of refraction of diamond is 2.47; that of window glass is 1.51. How much faster does light travel in the glass than in diamond? 3. Light surface strikes the of glass making an angle of incidence of (a) 60\u00b0, (b) 45\u00b0, (c) 30\u00b0. The index of refraction of 1.5. By means of accurate geoglass is metric diagrams draw the refracted ray for each case. Using a protractor, measure the angles of refraction. 4. By means of an accurate construction determine the size of the angle of incidence when the angle of deviation is a minimum in an equilateral crown-glass prism. Measure the angle of deviation. 5. In which material does light travel faster, one with a critical angle of 25\u00b0 or one with a critical angle of 30\u00b0? Explain, using appropriate diagrams. 6. An object 1 5 cm. high is located (i) 75 cm. (ii) 60 cm. (iii) 45 cm. (iv) 30 cm. (v) 20 cm. from a convex lens whose focal length is 30 cm. (a) By means of accurate scale diagrams locate of the image for each position of the object. (b) State the characteristics of each position the image. 7. By means of an accurate scale diagram locate the image produced by a concave lens whose focal length Is 1 5 in. of an object 6 in. high and 2 ft. distant. State the characteristics of the image. 8. A camera forms an image 8 cm. from the lens. If the object is 400 cm. away and 250 cm. tall, what is the height of the image? 212 The image of a tree in a miniature camera is 50 mm. from the lens and 30 mm. high. The tree is 1 5 metres away. How tall is the tree? 10. The image of an object 3 in. from a lens is formed 20 ft. from the lens. How many times is it magnified? 1 1. The image of an object 24 ft. from a lens is focused clearly on a screen 3 ft. from the lens. What is the focal length of the lens? 12. A convex lens forms a virtual image at a point 1 2 cm. from the lens. The object distance is 8 cm. Find the focal length of the lens. 13. A tree 1 00 ft. from a camera lens has its image very close to the principal focus of the lens. If the tree is 66 ft. tall and the image is 4 in. tall, what is the focal length of the lens? 14. A candle is 1 2 cm. from a convex lens of focal length 8 cm. What is the distance from the lens to the image of the candle? 15. A student uses a convex lens to look at an object held 4 cm. from the lens. If the focal length of the lens is 5 cm., how far is the image from the lens? What kind of image is it? 16. The image in a camera is 1 0 cm. high and 1 4 cm. from the lens. If the object is 100 cm. tall, what is the focal length of the lens? 17. A jeweller uses a converging lens of in. to examine a diamond. focal length 1 The virtual image is 10 in. from the lens. Find (a) the object distance, (b) the magnifi- cation. 18. When photographing a scene at a distance of 6 ft. from the lens, you find that the distance between the lens and the film is 6 in. (a) What is the focal length of the lens? (b) What is the actual size of a portion of a scene which occupies a space 3 in. X 5 in. on the film? CHAPTER 19 COLOUR IV: 35 INTRODUCTION TO COLOUR Imagine how drab and uninteresting the world would be if there were no pleasure and colour. stimulation we receive from the colours of nature\u2014the blue sky, the green grass, Think of the the beautiful flowers, the gorgeous hues of the sunrise and sunset. The use of colour in photography, in movies, and in book illustrations has added tremendously to our enjoyment of these things. How appealing and satisfying are some of the beautifully coloured mastei-pieces of art! Economically too, colour plays a very important role, as shown by the varied colours used in home decorating, clothing, advertising, and the like. For thousands of years men have known that colourless glass of certain shapes, as well as frost, diamonds and other crystals, produce light of many colours when illuminated by white light. Sir Isaac Newton Until the time of everyone supposed that or crystals produced the light by giving something to the light as it was reflected by, or transmitted through, them. It", " was he who, after thorough scientific investigation arrived at the true explanation of the nature of colour and the character of white light. glass the IV: 36 COMPOSITION OF WHITE LIGHT (a) Dispersion In 1666, Newton permitted a beam of sunlight, passing through a circular hole in a window blind, to fall on a triangular glass prism. He found that the light was refracted or deviated, from its original Instead of obtaining a simple path. image of the hole, he obtained a band of colours which he called a spectrum. Its colours were the same as those found in the rainbow\u2014 red, orange, yellow, green, blue and violet, with each colour merging imperceptibly into the ne.xt (Fig. Red \u2022Orange Yellow Green Blue Indigo Violet Fig. 19:1 Dispersion of White Light into its Spectrum. that Newton reasoned white 19:1). light must be composite, that is, made up of a combination of the above colours. The separation of the colours by the prism he called dispersion. A further study of dispersion is made in experiment 15, chapter 21. 213 Chap. 19 LIGHT.'As shown in Fig. 19:1 this dispersion occurs because the different colours are refracted different amounts by the prism and are deviated different amounts from their original direction. Red is always bent the least from its original direction and violet the most, with the other colours intermediate between these. Red light has the longest waves and violet the shortest, the wave-lengths of the other colours being between these two. the rays encounter less opposition to their passage through the glass prism and, therefore pass through more rapidly than do the violet rays. longer red Evidently, As we learned earlier (Sec. IV; 22), refraction is caused by a change in velocity of light on passing from one medium to another. In free space, or in air, light of all colours travels at the same rate, 186,000 miles per second. On entering an optically denser medium such as glass the light is slowed down, and on leaving this medium it speeds up to regain its original velocity in air. Since the different colours are bent different amounts by the prism, it follows that they must be velocities through the prism. Red light, bent the least, must slow down the least on entering the prism and therefore speeds up the least on leaving. In contrast, violet light must slow down the most on entering, and speed up the most on leaving the prism. travelling different at A question that immediately arises of course is, \u201cWhy does the prism have this different effect on the different colours?\u201d The answer relates back to our wave theory of light (Sec. IV: 7). The different colours of light result from different wave-lengths. Table of Wave-Lengths* (1 Angstrom (A) = 10'\u00ae cm. Infra-red above 7000 A Red Orange Yellow Green Blue ^ Visible Spectrum 6500A 6000A 5800A 5200A 4700A 4100A ^ Violet Ultra-violet below 4000 A [ * The wave-length shown for each colour is representative only. Each colour consists of wave-lengths that merge into those of the colours adjacent to it. For example, the wave-lengths of red lie between 6470 A and 7000 A. Colour bears the same relation to light that pitch does to sound. The pitch of a sound depends upon the number of vibrations per second that reach the ear Similarly the colour of (Sec. 11:12). light depends upon the number of vibrations per second that reach the eye. In light, however, since the frequency is so great, it is customary to describe the colour in terms of wave-lengths, rather than in terms of vibration frequency. Example Calculate the vibration frequency of red light. Velocity = 3X1 0^\u00ae cm. per sec ( Sec. IV : 6 ) Wave-length = 6500 A \u2014 6500 X 10\"\u00ae cm. Velocity = Frequency X Wave-length (Sec. 11:5) Frequency = Velocity Wave-length _ 3 X 1010 - 6500 X 10-8 =.46 X 1013 The vibration frequency of red light is.46 X lOi^ vibrations per second. 214 COLOUR Sec. IV: 36 Fig. 19:2 Recomposition of the Spectrum into White Light, (a) By Reversed Prisms, (b) By a Converging Lens, (c) By Newton's Disc. (b) Recomposition Newton further supported his theory concerning the composite nature of white light by showing that the colours of the spectrum could be recombined, giving white light (Chap. 21, Exp. 16). He first arranged two prisms with their refracting edges in opposite directions. On passing white light through these reversed prisms, white light was obtained through Fig. 19:2a shows that the first them. prism disperses the colours, while the second prism recombines them by reversing", " the original refraction and causing the light waves to be superimposed on each other. A similar effect can be secured by using a converging lens to catch the dispersed coloured light from If this light is brought to a a prism. focus on a screen a spot of white light will be obtained (Fig. 19:2b). If the screen is moved beyond the focus, a rethe original colours will be versal of obtained. Newton also prepared a colour disc on which were coloured sectors whose sizes and colours corresponded fairly closely to the coloured bands obtained in a pure spectrum of white light (Fig. 19:2c). If this disc is strongly illuminated and rapidly rotated it will appear white. This phenomenon is due to what is called \u2018the persistence of vision\u201d. Any visual impression on the retina of the eye persists for a short period of time after the It is on this cause has been removed. principle that movies are made to appear continuous. In reality, each picture is thrown on the screen for a fraction of a second, its image persisting in our vision until the next appears. Similarly, if the coloured disc is rotated rapidly enough, the impression produced by one colour persists, while impressions produced by all the other colours are received on the same portion of the retina. Thus all the colours of the spectrum will be superimposed on the retina, and will give the sensation of white light. 215 Chap. 19 LIGHT IV : 37 THE RAINBOW The rainbow is a spectrum of sunlight formed by water droplets. A ray of sunlight entering a drop of water is refracted at A (Fig. 19:3a), the violet rays being refracted more than the red rays. Fig. 19:3 The Rainbow (a) Refraction and Total Reflection in a Raindrop. (b) The Primary Bow. (c) Double Reflection to Produce the Sec- ondary Bow. The refracted light is totally reflected at B and is again refracted at C so that the different colours are dispersed. Each drop of water forms its own little specIn the actual bow which the trum. observer sees, the red rays come at an angle of 42\u00b0 from drops of water higher 216 (Fig. in the sky, and the violet rays come at an angle of 40\u00b0 from drops of water lower 19:3b). The other in the sky colours come from drops between these angles. The rainbow has the shape of a bow, since the eye of the observer is at the apex of a cone from which he sees the coloured rays refracted from drops, all of which must subtend approximately the same angle at the eye (between 40\u00b0 and 42\u00b0). Sometimes a larger, but fainter secondary bow is seen above the primary. The colours in it are reversed, the violet being on the outside. The light enters the lower part of the water drops, is refracted, and twice totally reflected before it leaves the drop (Fig. 19:3c). The light is refracted from the drops of water at angles of from 51\u00b0 to 54\u00b0. The double reflection not only reverses the colours, but also absorbs more light thus causing the secondary bow to be fainter than the primary bow. IV : 38 BEYOND THE VISIBLE SPECTRUM So far in our study of the spectrum we have considered only those radiations to which the eye is sensitive. Actually the spectrum of sunlight extends well beyond Sir William Herschel, its visible limits. in 1800, on placing the blackened bulb of a thermometer in the various parts of the spectrum, discovered that the heating effect observed at the red end was continued when the thermometer was placed well beyond the visible limit. He thus indicated the existence of a wide range of invisible radiation beyond the red end of the spectrum (Sec. 111:15). These infra-red radiations, as they are called, convey almost half of the sun\u2019s total outpouring of energy into space. They can penetrate mist, smoke and haze and hence are very suitable for distance photophotography, graphy in the dark, and detection of reconnaissance, ^ <c ^ III 3 | i 0) -c ^ ^ I <D \"D Q- c s D I c c O (D o O) S \u20224^ Q. O O \"D ^ ^-2 < fi-g O O O _C CO \u2014 CO u -tu C <13 1 \u00b0? CO Q. to LU O Frequency 10 H lo' lO' 10- lo\"- 10 lO'-l 10 lo'H *- 12 10 10 *- 10 10 * \u2014 lo' lo' lo' lo' *- 10 lo'- lo'- lo\u2019 10 10 COLOUR Sec. IV: 38 Electromagnetic Wave Spectrum Generation (Uncertain) Disintegration of atomic nuclei Sadden stopping of fast-moving electrons Radiated by very hot bodies and ionized gases Radiated from hot bodies Heat radiations from hot bodies Spark-", "gap discharge Oscillating three-element vacuum tube ^ Used in Radio Very long waves Coil rotating in a magnetic field Fig. 19:4 217 Chap. 19 LIGHT (a) American Optical Company (Canada) Ltd. value of a holiday away from the city. For ultra-violet treatment one must be exposed directly to the sun\u2019s rays, or use windows of special \u201cvita glass\u201d which is transparent to some ultra-violet rays. Proper precautions must be taken as overtreatment can produce severe sunburn, as well as damage to the retina of Ultra-violet lamps simply conthe eye. sist of carbon or mercury electrodes between which an electric current produces an arc. Such lamps are usually fitted with quartz lenses which are transparent to the ultra-violet rays. The presence of ultra-violet rays beyond the visible spectrum of white light may be shown by using a quartz prism in place of the glass prism in Fig. 19:1. If a sample of anthracene is placed beyond the violet it will be seen to fluoresce. In addition to the invisible radiations forgeries or substitutions. Under these conditions special photographic film sensitive to infra-red must be used. These rays also have therapeutic value such as the treatment of rheumatism and in other aches and pains. The year following Herschel\u2019s discovery of the infra-red, Ritter discovered, by its action in blackening certain silver salts, a band of invisible radiation beyond the violet, called the ultra-violet. Ultra-violet rays do have special photographic effects, and also make some substances such as vaseline and certain minerals \u201cfluoresce\u201d or give out \u201ccold light\u201d. They have a highly beneficial effect on health, accelerating the manufacture of vitamin D under the skin. These ultra-violet rays are easily absorbed by mist, smoke and ordinary glass. Consequently, one readily the recognizes 218 COLOUR Sec. IV: 39 discussed above in connection with the solar spectrum, there exist other invisible radiations such as radio waves, X-rays, and gamma rays (rays emitted by radioAll these different active substances). radiations are propagated as waves with the same velocity as light, the difference in their properties being due to their different wave-lengths. The relationship between these various electromagnetic is shown waves, diagrammatically in Fig. 19:4. as they are called, IV: 39 SPECTRUM ANALYSIS The spectroscope is an important instrument by means of which the spectra of luminous bodies can be produced and measured. It consists of a tube called the collimator (Fig. 19:5), at one end of which is a convex lens and at the other end an adjustable vertical slit. The length of the tube is equal to the focal length of the collimator lens, so that when the slit is illuminated by light from the source, S, under examination, a parallel beam can be directed onto a prism situated on a small turn-table at the centre of the instrument. The dispersed beams produced by the prism are received by a telescope focused for parlight and fitted with cross-wires. allel This telescope can be moved round a circular scale against which the deviations of the constituent parts of the spectrum formed can be measured. The this way of the light examination in from different luminous bodies reveals that each spectrum is characteristic of the source, and a study of these spectra yields much valuable information to the physicist and astronomer. Accordingly we shall briefly consider here some of the more important spectra and their significance (Fig. 19:6), Continuous Spectra: These consist of a number of coloured bands each shading off gradually into the next. They are produced by incandescent solids, e.g., arc-lamps, white-hot iron, etc. Incandescent gases and Line Spectra: vapours emit light which when analysed produces spectra consisting of a number of well - defined coloured lines or col oured images of the slit. Each element has its own characteristic line spectrum which provides a certain and accurate means of identifying the element (Exp. 17, Chap. 21). The presence of even a minute quantity of an element in a mixture can be detected by spectroscopic analysis, and it is interesting to note that the method has been the means of discovering new elements. If the spectrum of a substance contains lines which do not correspond with those of any known element, the obvious conclusion is that there is present an element as yet not known. In this way the elements caesium and rubidium were discovered by Bunsen. Line spectra can be produced by the Bunsen flame only in the case of subvaporized. stances Other substances must be vaporized, using an electric arc or a spark dis- which easily are Fig. 19:7 An Electrical Discharge Tube. The spectra of incandescent charge. gases are usually obtained by means of special tubes containing a sample of the gas", " at low pressure through which the discharge from an induction passed (Fig. 19:7). coil is Absorption Spectra: If a sodium flame is set up between the slit of a spectroscope and a high-power electric lamp (Fig. 19:8), a dark line will be observed in the continuous spectrum of the light from the relatively hotter lamp in a position corresponding to that of the 219 Chap. 19 LIGHT yellow line of the incandescent sodium vapour. With other vapours dark lines would be formed in positions corresponding to their line spectra. A spectrum Fig. 19:8 The Production of an Ab- sorption Spectrum. crossed by dark lines in this way is called an absorption spectrum, and it was found that all substances when interposed in the path of light originating from a higher temperature source absorb from it light of the same wave-length that they themselves emit. Most of the The Solar Spectrum is an absorption spectrum containing hundreds of dark lines carefully studied by Fraunhofer in 1814. These lines are due to selective absorption from the radiation emanating from the extremely hot core of the sun by the relatively cooler gases in the sun\u2019s atmosphere. lines have been found to correspond to lines in the spectra of elements present on the earth, and we have thus good grounds for believing that the chemical components of the sun and the earth are similar. A certain group of lines did not correspond with those of any known element. They were accordingly attributed to an element which was named helium, and which was ultimately discovered on the earth twenty-six years later. The constitution of the stars is determined in a similar way from an examination of their absorption lines. IV : 40 NATURE OF COLOUR (a) Colour We have that colour is a property of light waves (Sec. established already 220 It may be defined generally, IV; 36). as the response of vision to different wave-lengths of light. Colour is the alphabet in our visual language, through which we make interesting our description of things and ideas which otherwise would be dull and prosaic. Our present knowledge and use of colour is based largely on the sciences of psychology, chemistry, and physics. Psychology considers colour through the physical operation of the eye and the mental impression created by colour on the brain. As a result, the actual perception of colour is a highly personal experience which may be influenced by such factors as health, fatigue and response of the eye to colour (colourAccording to psychologists blindness). the average person recognizes four distinct primary colours, red, yellow, blue and green. Colours do have a strong influence on people. Tensions can be heightened or relaxed through a change in colour harmonies; a feeling of warmth or coldness can be obtained by colour schemes used; colour can be stimulating or restful, it can attract or it can repel. Chemistry deals with the production of materials which have the ability to absorb or reflect part of the light which falls upon them. These are chiefly dyes and pigments. The explanation for the colours of these materials is given in part (b) below. Physics deals with colour from the standpoint of differences in wave-lengths of the coloured lights, and the effects of superimposing these on one another. The additive theory of colour to explain (c) these effects is discussed in part below. These three colour theories are the basis for our modern usage of colour. Though separate, they are also interdependent. The range and intensity of the colours we see depend upon the quality and quantity of light, the nature of the. COLOUR Sec. IV: 40 surface visible, and our ability and interpret what meets our eyes. to see (b) Colour of Objects\u2014 The Subtractive Theory There are three colours, yellow, red, and blue, which in pigments are the source of all other colours. By mixing these three colours in the right proportion, all other colours may be obtained. No mixture of other colours will produce any of these three colours. For this reason yellow, red, and blue are called primary colours. A colour chart has been prepared (Fig. 19:9) in which these primary colours are placed the same distance apart along the edge of the circle. Mixing any two primary colours produces a third colour called a binary colour which is intermediate between those primaries, e.g., yellow and red produce orange. Various hues can be produced by mixing a binary colour with a primary colour used in making the binary, e.g., yellow and orange produce yellow orange. Certain colours seem to strengthen each other when they are seen together, and hence are said to be complementary to each other. In the chart the colours on the opposite sides of the circle are complementary. White is the total addition of colour, is of then, the result spectrum, and reflects and not, as often believed, the absence of colour. It is produced when a surface reflects all colours equally. Black, on the other hand, is perceived when a surface absorbs all colours and reflects none. Colour, partial absorption, and", " consequent subtraction, of a band of colour from the spectrum. The surface of an object will appear red when it is of such nature that it absorbs all but the red wave-lengths of red the wave-lengths (Chap. 21, Exp. 18). When we mix pigments, each one subtracts certain colours from white light, and the resulting colour depends upon the light that is not absorbed. For example, if we mix a blue pigment with a yellow one we get a green colour. The blue colours from subtracts or absorbs all white light, except green, blue and violet, while the yellow subtracts all except green, yellow and orange. Green is the only colour not subtracted by either pigment. For this reason the two pigments produce green. these Similarly, transparent objects are able certain colours to absorb or subtract from white light. If white light is passed through a piece of red glass, the glass absorbs all light except red, and a little orange, and we obtain red light (Fig. 19:10a). Objects that are transparent to only one colour are called colour filters. The combination of a yellow and a blue filter placed over a single white light source will result in a green light because all except the green wave-lengths of the spectrum have been absorbed a combination of three filters, each corresponding to a primary colour, will absorb all the colours and allow practically no light to pass (Fig. 19:10c). (Chap. 21, Exp. 18) 19:10b). Similarly (Fig. The light which falls upon an object can also determine its colour. If the light does not contain all the colours of 221 Chap. 19 LIGHT RED YELLOW BLUE (a) (b) Fig. 19:10 Colour Filters (The Subtractive Theory) (a) Transmission of Light by Different Coloured Filters. (b) Combination of Yellow and Blue. (c) Combination of Red, Yellow, and Blue. o the spectrum then the true colour of the object will not be observed. The light given off by a light bulb does not contain all the colours found in sunlight. For this reason an article of clothing may appear quite different in the sunlight than when seen under artificial light. (c) Coloured Lights\u2014 The Additive Theory Having produced different coloured lights by the use of filters, two or more of these may be added together to produce still other colours. The three colours, red, green and blue, which combine to produce white are called the three primary colours for lights. None of these primary colours can be produced by adding other colours of lights. However, any other colour can be pro- 222 19:11). duced by a proper combination of them For example, red and (Fig. blue light when combined produce violet; red and green light added together Fig. 19:11 Coloured Lights (The Ad- ditive Theory). COLOUR Sec. IV: 41 will produce yellow, and so on. The colour resulting from the simple combination of each pair of primary colours is complementary to the third primary; for example, violet is complementary to green, and yellow is complementary to If we superimpose these comblue. plementary coloured lights we again get white light. Contrast how colour is obtained in mixing pigments with how it is obtained in mixing coloured lights. In the former the resulting colour is the one not absorbed or subtracted by either pigment in the latter the (the subtractive effect) ; resulting colour is obtained by adding the effect of the individual colours (the additive effect). Both methods are widely used in industry. The mixing of pigments to produce our multi-coloured paints, the many vegetable and synthetic dyes for treating fabrics, etc., and the use of filters in photography, all are applications of the subtractive effect mentioned above. The additive effect is applied in the use of coloured spotlights to illuminate actors on stage, or participants in a carnival or pageant, as well as to produce changing and spectacular effects in the illumination of Niagara Falls and in some of the coloured advertising that is so common nowadays. textiles, Colour is of tremendous importance Its use in advertising has in industry. been mentioned above. Bright contrasting colours of paint on walls, floors, and machinery not only provide better visual conditions but also boost morale, speed up production, and affect safety re- cords. Colour codes are used in busi- ness for invoicing, in electrical indus- tries to denote polarity, voltage, resistance, and the like, and in shipping to encourage due precautions in the han- dling of dangerous articles. All, of course, are familiar with its use in traffic control. (d) Colour Vision are there theory, widely The way in which light is changed into the sensation of sight is not definitely The most known. accepted the Young-Helmholtz theory, of colour three states nerves in the normal eye. One of these sets of nerves responds only to blue light, another set to", " green light, and the third set to red light. The colour of an object is determined by the relative degree of stimulation of each type of nerve. sets A small proportion of men (about 3/2%) and women (about ^2%), do not have all three of these sets of colour nerves and hence they do not receive the same sensation of colour from an object as a person with all three sets of colour nerves. Such people are said to be colour blind. By this we do not mean that they do not see colour in objects, but that they do not see the same colours as a person with normal eyes. This defect through inherited is usually exists from birth; and there is no known cure for it. it mother; the pattern Because a colour-blind person sees a different colour pattern in a landscape seen by a person from the having normal vision, colour-blind observers are very efficient in detecting camouflage and are so used during wars. Some colour-blind persons are unable to distinguish reds and greens from each other and, consequently, have real difficulty with traffic signals. IV: 41 COLOUR PRINTING Most of the coloured pictures in books and magazines are made by using four separate printing-plates, each plate printing a different coloured ink on white paper\u2014yellow, red, blue and black, in that order. The plates are made from four photographic negatives of the same are made negatives subject. through These fine-meshed coloured filters. 223 Chap. 19 LIGHT the the creation of each stained with one of three primary colours (the black one is not the actually used in Fig. 19:12). The coloured picture in fine-meshed filter breaks the subject into numerous small dots on the negative. The filter determines what colours can pass through the camera and become recorded on the negative. Thus dots representing only certain colours are recorded on each negative. The first plate reproduces the dots of the first negative in yellow ink on white paper. The second plate reproduces those of the second negative in red ink on or between the yellow dots, etc. The final effect is one of individual and overlapping dots of colour. Where a dot of one colour overlaps a dot of another colour a new colour may be obtained. The eye blends these dots of the four original and the mixed colours to give the total effect. IV: 42 CHROMATIC ABERRATION in contact (Sec. IV: 28), Since the action of a convex lens is similar to that of two prisms with their bases dent rays, in addition to being deviated to form a convergent beam will be dispersed on passing through the lens. The various colours are brought to diffoci on the principal axis, the ferent focus for violet rays being nearer the inci- than red rays that for (Fig. lens 19:13a). In consequence it is impossible to obtain a well-defined image when white light is used, the image received on a screen having a red fringe when placed at A and a blue fringe when placed at B, there being no position be- (a) Cr. FI. V R F fb) Fig. 19:13 Chromatic Aberration. (a) Its Cause. (b) Its Correction\u2014An Achromatic Lens. tween A and B at which the image is This non-focusing of sharply in focus. light of different colours is known as chromatic aberration. By using two reversed prisms, one of crown glass and the other of flint glass which has a higher index of refraction, it is possible to obtain a non-dispersive combination which still has the power of deviating the light (Fig. 19:13b). is possible to combine a convex lens of crown glass with a weaker concave lens of flint glass so that the dispersion produced by the two lenses is equal and opposite. A lens combination of this type is called an achromatic (without colour) lens. These achromatic combinations are still converging and accordingly can be used as objective in field glasses, cameras, and other optical instruments. Similarly it telescopes, lenses 224 Red, Print- green- a Yellow, through 5. made filter; Four-Colour In Note; negative red-yellow print. Plates. Blue and Plate, a Red finished Red 2. through the Yellow, made filter; in the shades to negative and Addition violet-green Plate, a through Blue 4. made yellow; on negative colours in Used various is Plate the Block Notice a ing Red 3. combined. filter; Blue yellow and Plate, Yellow L IV : 43 COLOUR QUESTIONS Sec. IV: 43 1. (a) Define: spectrum, dispersion. (b) Why does a prism deviate violet light more than red light? (c) Calculate the vibration frequency of violet light (See table, p. 214, for wave-length of violet light). 2. Describe and explain three methods for recombining the spectral colours into white light. 3. Explain how ref", "raction produces a solar spectrum (rainbow). 4. Write a note on (a) infra-red rays, (b) ultra-violet rays, indicating, (i) how each differs from visible light. (ii) how each can be detected. (iii) how each is used. (a) What is the purpose of a spectroscope? 5. 8. (i) a continuous (b) Distinguish a line spectrum from an absorption spectrum. How is each produced? spectrum, (ii) (c) Why does red glass appear red, blue glass appear blue, and ordinary glass appear colourless, white light? State a general rule for the in colour of transparent objects. (d) What would be the colour of the objects in (b) when viewed through (i) a red-glass filter, (ii) a blue-glass filter? (e) Why is it unwise to buy coloured clothing by artificial light? (a) How does the additive theory of subtractive light differ from the theory? (b) In the additive theory what are (i) the primary colours, (ii) complementary colours? (c) If coloured spotlights are shone on a white screen, (i) what colour will be obtained if red and blue lights are used? (ii) what coloured used to produce yellow? lights should be 6. What is the physicist\u2019s explanation of (a) coloured light, (b) white light? 7. (a) In the subtractive theory of light what are (i) the primary colours, (ii) complementary colours? (b) Why does a rose appear red, grass green and a dandelion yellow when seen in sunlight? State a general rule for the colour of an opaque object. 9. How does the physicist account for the difference between mixing coloured lights and coloured pigments? 10. What is colour-blindness? How does the Helmholtz theory explain it? 11. Describe the process of colour print- ing. 12. (a) What is chromatic aberration? (b) What is an achromatic lens? 225 CHAPTER 20 OPTICAL INSTRUMENTS 1553, Porta, Baptista and camera. Johann Kepler, 1604, later added a lens and then a combination of lenses. From these simple beginnings have come many types of cameras ranging from the relatively inexpensive family-type cameras, through the more expensive press and movie cameras, up into the highly specialized cameras used in television and in aerial photography. (b) Structure All cameras are constructed on the same basic principles as were the first crude, wooden, box-cameras. The essential parts of a camera (Fig. 20:1) are: 1. A light-tight box or bellows. 2. A lens to form the image on the film. 3. A shutter (and diaphragm) to con- IV : 44 THE CAMERA The photographic camera is essentially an enclosed box or bellows chamber with an opening or aperture in the front through which light from an object The image is recorded on a passes. light-sensitive material at the back. (a) History The evolution of the camera is credited to men like Roger Bacon, 1267, and Leonardo da Vinci, 1519, who dedeveloped pin-hole scribed and the 226 OPTICAL INSTRUMENTS Sec. IV; 45 the amount of trol admitted to the film. light that is 4. A negative holder or carrier to hold films in position during exposure. 5. A view-finder or ground glass to determine picture area. By the addition of special accessories and other refinements, the utility and scope of the camera may be extended to meet almost any photographic requirement. (c) Focusing In order to focus objects at different distances from the camera, the lens is often attached to a bellows so that it may be moved nearer to the light-sensitive plate or film, or farther away. In section IV : 30 we learned that as the object is moved nearer to a converging lens, the real inverted image gradually moves farther away from the lens. This should indicate how to focus the camera. For distant objects the lens is moved so that the sensitive plate is located at its principal focus. For close objects, the lens is moved away from the sensitive film, for now the image is focused beyond the principal focus of the lens. Most cameras include a focusing scale on the lens mount or on the bed of the camera and many have built-in range finders to assist in focusing. (d) Light Control As has been indicated previously the film or plate in a camera contains materials which are sensitive to light. The plate is coated with an emulsion consisting of gelatine and a silver salt. Light decomposes this silver salt so that when \u201cdeveloped\u201d and \u201cfixed\u201d a negative is obtained consisting of a dark deposit of silver in places corresponding to the light parts of the original image, and light transparent portions", " in positions corresponding to the darker parts of the original image. One sees, therefore, that it is necessary to be able to control the amount of light that enters the camera. This is done in two ways : ( 1 ) by controlling the size of the aperture or opening through which the light enters; and (2) by regulating the length of time that the light is permitted to enter. The former is accomplished by using a variable diaphragm behind the lens which enables the aperture to be varied, and the latter by means of a shutter, which is a delicate mechanical device, often consisting of a number of overlapping thin metal blades, held in position by means of springs. On being released or tripped, when taking a picture, the shutter opens and closes for a measured time interval. When using a small aperture a longer exposure is required than with a large It is wise practice to use as small one. an aperture as possible as this permits using the central part of the lens only, and prevents errors due to inaccuracies near the edges of the lens. Also greater depth of focus is obtained, that is, objects relatively smaller and greater disstill be at tances from the camera will sufficiently in focus to look clear in the print. IV: 45 THE HUMAN EYE to be perceived. The eye is the sense organ enabling This, the most light wonderful of all optical instruments, can be compared in many respects to a camera (Fig. 20:2). (a) Structure and Action The eyeball is approximately spherical in shape and slightly less than one inch in diameter (Fig. 20:3). The wall of this sphere is composed of two major layers: ( 1 ) the outer covering, the sclerotic coat, substance forming the white of the eye. The front portion of this sclerotic coat forms a transparent, curved section called the tough opaque white a is 227 Chap. 20 LIGHT Fig. 20:2 Comparison of the Eye and the Camera. cornea that protects the eye and aids in refracting light. (2) The inner layer, the choroid coat, is black to prevent internal reflection and to protect the light-sensitive parts of the eye. is also largely nutritive in function. The aqueous and vitreous humours are jellylike materials filling the spaces within It the eyeball, which help to keep its spherical shape. Behind the cornea is a coloured diaphragm called the iris. This is really an Fig. 20:3 The Structure of the Eye. 228 extension of the choroid coat, and its colour is mainly due to the variable amounts of pigment in it. In the centre of the iris is a circular aperture, called the pupil, which appears black due to the black interior of the eye. The iris contains muscles, which dilate and contract the pupil adjusting the eye to different amounts of light. The pupil is dilated in dim light, and contracted in bright light. This action occurs involuntarily, i.e., without any control on our part, and hence is called the iris reflex. To note its action, stand in front of a mirror in a dark room for several minutes. Then turn on the light and note the immediate contraction of the pupil. Optometrists use drugs, such as atropine, to temporarily dilate the pupil and permit internal examination of the eye. Behind the pupil and iris is the crystalline lens. This is a transparent structure, made up of numerous concentric layers, increasing in density toward the middle. OPTICAL INSTRUMENTS Sec. IV: 45 it so that tension The curvature of its front surface is considerably less than that of the back Both of these features tend surface. to diminish distortion of the image. The lens is encased in an elastic capsule and is held in place by suspensory ligaments which are attached to the choroid coat. These ligaments hold the lens under a normally certain tends to be flattened and adjusted for distant vision. In the choroid layer near the suspensory ligaments are small ciliary muscles. The contraction of these muscles tension exerted by the suspensory ligaments on the lens which therefore becomes more convex and so adjusted for close vision. The adjustment of the lens to form a sharp image on the retina is called accommodation (Fig. 20:4). The normal eye can accommodate itself to objects ranging from infinity to a point about twenty-five ten inches from the eye. This latter point is known as the nearest point of distinct Objects closer to the eye than centimeters or relieve vision. tends the to \u2022 Distant Vision (b) Close Vision. this cannot be clearly focused by the unaided eye. their characteristic The retina is the innermost layer of the eye and is sensitive to light. The retina is composed of visual cells, nerve cells, and the fibres which connect these and conduct the nerve impulses to the brain. The visual cells are of two types, called rods and cones, so named because rod and cone of shapes", ". The rods are concerned with colourless and dim-light vision, while the cones handle colour and bright-light vision. The nerve cells integrate and relay the impulses which they receive, to the optic nerve, and through it to That part of the retina where the optic nerve enters the eye, is insensitive to light and is known as the Its existence can be readily blind spot. demonstrated as shown in Fig. 20:5. The most sensitive part of the retina is a small yellow spot located on the principal axis of the lens. An object being viewed is automatically focused on this brain. the area. X Fig. 20:5 The Blind Spot. Close one eye. Hold the book at arm's length. Stare at the X. Move the book toward you. At a certain distance the circle will disappear, for its image is formed on the blind spot. 229. Chap. 20 LIGHT (b) Defects of Vision and Their Correction In view of the delicate adjustments made by our eyes and the way we abuse them these days with excessive reading, watching movies, television, only natural that many abnormalities or defects develop. Here, however, we are only concerned with defects or errors caused by refraction of light. etc., it is is rest, parallel When an eye with no refractive error rays of light are at focused exactly on the retina. Many eyes, owing to abnormalities of the refracting mechanism, do not focus the light exactly on the retina. Such defects of the eye can in large measure be overcome with the (spectacles) of suitable lenses aid /. Near-sightedness to close Persons suffering from this defect are unable to focus on distant objects distinctly, although objects the eyes are clearly seen. This is due to the axis of the eyeball being too long or the crystalline lens being too powerful, so that rays from a distant object tend to be focused in front of the retina (Fig. This defect is largely heredi20:6a). also aggravated by too strenuous use of the eyes during the early years. Concave lenses placed in front of the eye will make the light more divergent and bring the image to a focus on the retina. but is tary, 2. Far-sightedness A far-sighted person cannot see clearly objects close to the eye although distant objects may be distinctly The defect is due to the axis of the eyeball being too short or the crystalline lens not sufficiently converging so that rays from a nearby object tend to converge seen. to a point behind the (Fig. It is the most common of re20:6b). fractive errors of the eyes and is present retina 230 to some degree in about two-thirds of all adults. Convex lenses placed in front of the eye will reduce the divergence of the rays which can then be focused on the retina. (c) Fig. 20:6 Defects of Vision. (a) Near-sightedness, Corrected by Con- cave Lens. (b) Far-sightedness, Corrected by Con- vex Lens. (c) Test for Astigmatism. the lines appear equally distinct to you? Do all 3. Astigmatism As a result in different planes are brought to different foci and in consequence one set of lines of this defect lines OPTICAL INSTRUMENTS Sec. IV: 46 (Fig. 20:6c). will be in sharp focus, while others inclined to them are blurred and indisFor example, an tinct astigmatic eye may clearly discern all the vertical lines of a building, while those in a horizontal plane are scarcely perceived. The main cause of this defect is the lack of sphericity of the cornea, in which the vertical section is frequently more curved than the horizontal. To overcome the defect cylindrical lenses are used to assist refraction in the plane of least curvature of the cornea. Astigmatism is present to varying degrees in most eyes. ing rays from the object without undue With the use of a short-focus strain. convex lens, however, it is possible to bring the object even closer, thereby enabling it to be seen still more distinctly without straining the eye. The object is placed inside the focus of the lens at such a position that the eye focuses the magnified virtual image formed at the nearest point of distinct vision. The path of the rays by which the eye sees the image is shown in Fig. 20:7. The magnifying power of such a lens the size of image on retina when is object is viewed through the lens divided 4. Presbyopia It loss This is a defect of age resulting from of accommodation brought the about by hardening of the crystalline lens, or of the ciliary muscles. is evidenced by both near and far objects being indistinct. Clear vision by the unaided eye is possible over only a very restricted range. Two pairs of spectacles are required by such people, one pair (convex) for reading, and another pair (concave) for distant", " vision. Both purposes may be served by a pair of bifocal lenses, the top part being used for distance, and the lower part for reading. IV : 46 THE MAGNIFYING GLASS The apparent size of an object depends on the angle it subtends at the eye. This angle is known as the visual angle of the object, and in order to see a small object clearly one instinctively brings it close to the eye, with consequent increase in its visual angle. With the unaided eye there is, however, a limit to which this process can be carried. This is reached when the object is at the nearest point of distinct vision (normally 25 cm. or 10 in.) at which position the object is seen most clearly. If brought within this natural limit the eye cannot accommodate itself to the widely diverg- The Action of a MagnifyFig. 20:7 ing Glass. The image is focused at the least distance of distinct vision. by size of image on retina when object is viewed directly at the least distance of distinct vision least distance of distinct vision focal length of the lens 10 in. 25 cm. In using a magnifying glass we focus the image at the least distance of distinct vision by moving the lens back and forth from the object. A near-sighted person might focus on an image 7 inches from the eye, for example, whereas a far-sighted person would change the lensobject distance so that the image might be 13 inches from the eye. If / 1 inch., the former person would only get a magnification of 7 X, whereas the farsighted person would get a magnification of 13 X. To save argument the average value for least distance of distinct vision has been set at 10 inches. 231 a as simple magnifying principal focus, F', of the eyepiece which acts glass. Through this lens the eye sees a large virtual image, pq, when the length of the tube is properly adjusted so that the image is sharply in focus at the nearest As with the point of magnifying glass, position of the the virtual image depends upon the accommodation of the observer\u2019s eye. distinct vision. The magnification produced by the compound microscope is obtained in two stages. Both lenses individually produce a magnified image, so that the total magnification is the product of the two. Microscopes for visual use can be constructed to give magnifications of 2000 and above, the higher magnifications being obtained by using objective lenses of very short focal length. Often such lenses are designed to employ a medium of cedar oil between the specimen and Microthe objective (Oil-Immersion This medium increases the scopes). light-gathering power and the resolving power of the objective by reducing reflection and scattering of light at the surface Single lenses of very short of the lens. images which are length give focal LIGHT Chap. 20 Note: Magnification = \u2014 := \u2014 (Sec. IV: 32) Ho and Di \u2014 least distance of distinct vision Do Hi Di r= 10 in. or 25 cm. and Do = focal length (/) approximately Magnification 10 in. ~T~ or 25 cm. / of short focal length, IV : 47 THE COMPOUND MICROSCOPE The compound microscope is very widely used to produce great magnifications of small objects. The instrument (Fig. 20:8a) consists mainly of a convex called the lens objective, O, and a second convex lens, called the eyepiece, E. These lenses are mounted in a tube of adjustable length. The object, PQ, to be viewed is placed just outside the principal focus, F, of the objective (Fig. 20:8b). This produces a real magnified image of the object at P'Q,', and this image falls inside the American Optical Company (Canada) Ltd. (a) Fig. 20:8 The Compound Microscope. 232 OPTICAL INSTRUMENTS Sec. IV:48 Virtual Image Fig. 20:9 The Refracting Astronomical Telescope. greatly distorted, and which have coloured edges. To avoid this the objectives of actual microscopes are usually compound IV:42). achromatic lenses (Sec. of Microscopes generally have a concave mirror, and frequently a condensing lens above it, to focus sufficient light on the object being examined. The inventhe microscope in 1590 by tion Zacharias Janssen, with the many improvements added since then, has contributed advances made in medicine and public health, as well as revealing many marvels of plant and animal life, and hidden secrets of the mineral kingdom. immensely the to IV: 48 THE TELESCOPE (a) Refracting Telescope The method by which the telescope works is similar to that of the microscope. The refracting telescope consists be the focal larger length of an arrangement of two convex lenses for viewing very distant objects (Chap. 21, Exp. 19), The objective should have a very great focal length, for the greater the will the image", " of an object at infinity. It should also be of large diameter to enable a maximum amount of light to be collected, thereby producing a brighter image. This lens will produce a real inverted image at its principal focus. This should be located just within the principal focus of the eyepiece through which the eye sees a magnified virtual image (Fig. 20:9). The magnifying power of the telescope is approximately equal to the focal length, F, of the objective divided by the focal length, /, of the eyepiece. Magnifying power = \u2014 Fig. 20:10 Two Forms of the Reflecting Telescope. 233 Chap. 20 LIGHT Palomar Mountain Observatory, California. Star Newspaper Service (h) Reflecting Telescope Reflecting telescopes are used for astronomical purposes. The objective is a parabolic mirror (Sec. IV; 19) which light from the distant object. collects Since the light is reflected back toward the object, a means of viewing the image with an eyepiece necessitates the use of an additional mirror or prism 20 : 10 ). (Fig. Large reflecting telescopes include the 74 inch diameter telescope at the David Dunlap Observatory at Richmond Hill, Ontario, and the 72 inch telescope at the Dominion Observatory near Victoria, British Columbia. Larger Astro-Physical telescopes of this type are located at Mount Wilson, California (100 in.) and at Palomar Mountain, California (200 in. ). This latter is the largest telescope in the world, having one million times the light-gathering power of the human eye. (c) Terrestrial Telescope The drawback to the astronomical telescope is that the image is inverted. For observing familiar terrestrial objects a telescope requires an image-erecting system. The simplest means of obtaining this is to insert a third convex lens to re-invert the image (Fig. 20:11). Such terrestrial telescopes are used by surveyors, by mariners and by military men. Final Image First Image 234 OPTICAL INSTRUMENTS Sec. IV; 50 Fig. 20:12 The Slide Projector. IV: 49 THE SLIDE PROJECTOR The purpose of the slide projector is to cast a magnified image of a slide or transparency upon a screen coloured some distance away. Fig. 20:12 shows the essential structure of such a projector. A study of the paths of the light rays shown in the diagram will serve as an excellent review of some of the work covered earlier in concave mirrors and converging lenses. For this instrument a very powerful light source is needed. This is placed at the centre of curvature of a concave reflector (mirror) M, so that many rays that tend to be lost, on hitting this mirror are reflected back on themselves thereby intensifying the light source. It is also placed at the principal focus of is this condenser a condensing lens system, C. As a result the diverging cone of rays from the light source on striking concentrated into a parallel beam of light and sent through the slide, O. The slide must be located outside the principal focus of the objective, L. This lens as a result produces a real, enlarged, and inverted image on the screen. The slide is always inserted upside down in the carrier, so that its image will be erect on the screen, /. The size of the image may be increased by increasing the distance between the projector and the screen. The objective is contained in a moveable mount, to permit focusing the image for different distances of projecOptically the motion-picture pro- tion. jector and slide projector are similar. 3. IV: 50 QUESTIONS 1. Compare lens camera and the human eye under the following headings: contrast the and (a) structure. (b) focusing. (c) light control. (a) an object close to the camera. (b) an object distant from the camera. (c) an object on a cloudy day. (d) an object on a sunny day. (e) a moving object. The focal length of a camera lens is 2. What adjustments would you make in your camera for taking pictures under the following conditions: 3.0 in. (a) How far from the film is the lens when a man standing 1 2 ft. from the 235 (a) Make a sketch of a compound microscope locating the two images, (b) If the first (objective) lens magnifies the object 8 times and the eyepiece magnifies the real image 20 times, how many times is the final image larger than the object? 8. Discuss (a) the refracting telescope, (b) the reflecting telescope and (c) the terrestrial telescope, as to: 9. (i) essential parts. (ii) how each works. (iii) types of images obtained. In a slide projector: (a) Where is the light source relative to (i) the concave reflector, (ii) the condensing lens? Why is it placed in this position? (b) Where is", " the slide relative to the principal focus of the objective lens? In what relative position is the slide inserted in the carrier? Why? (c) If the image formed by the projector is too large for the screen should you move the projector closer to the screen or farther away from it? Why? What further adjustment would be necessary? (d) The object is 16 mm. high and 50 mm. from the lens. If the screen is 6 metres away, what b the height of the image? Chap. 20 LIGHT camera is brought into focus upon the 7. film? (b) How tall will be the man\u2019s image on the unenlarged picture his if 4. iris reflex, height is 6 ft.? (a) What is meant by (i) (ii) lens accommodation? (b) Why do your eyes become more sensitive when you go into a darkened room? (c) By means of a diagram show how the lens of the eye changes when the object viewed is moved closer to the eye. 5. Describe, with the aid of diagrams, (a) near-sightedness, (b) far-sightedness, (c) astigmatism, under the following head- ings: (i) cause. (ii) effect. (iii) correction. 6. (a) What is the purpose of the magnifying glass? (b) Describe the image you see through a magnifying glass. (c) Calculate the magnifying power of such a glass having a focal length 1,5 in. of (i) 3.0 cm, (ii) (d) What would be the size of the image of a crystal 3 mm. long when viewed through each of the magnify- ing glasses in 6 (c)? 236 CHAPTER 21 EXPERIMENTS ON LIGHT EXPERIMENT 1 To study the characteristics of an image produced by a pin-hole camera. (Ref. Sec. IV: 4) Apparatus Pin-hole camera (Fig. 16:3), one large candle, one small candle. Method 1. With the adjustable section of the camera held stationary so that the ground-glass screen is a fixed distance from the pin-hole, observe the size and other characteristics of the image of the candle placed at some distance from the camera. Move the camera closer to the candle and note the size of the image. 2. Holding the camera a fixed distance from the candle, note the size of the image. Move the adjustable section thereby decreasing the distance of the screen, and hence of the image, from the pin-hole, and again note the size of the image. 3. With the adjustable section held stationary as in step 1, observe the size of the image produced by first the large, and then the small candle at a fixed distance from the camera. 4. If possible, enlarge the pin-hole and note the effect on the brightness and sharpness of the image. Observations Record carefully the observations for each of the previous steps. Conclusions 1. State the characteristics of the image. 2. What effect did each of the following have on the size of the image: (a) decreasing object distance? (b) decreasing image distance? (c) decreasing size of object? 3. Establish the formula: Hi Di Ho~~Do 4. On what does the brightness of the image and its sharpness of defini- tion depend? Explain fully. 237 Chap. 21 LIGHT Questions 1. Why do we get an image of the candle and not an image of the pin-hole? 2. What image would be obtained if the hole were increased in size until it has a diameter of 1 in. or more? Why? EXPERIMENT 2 To establish the laws of reflection. (Ref. Sec. IV: 9) Apparatus Plane mirror, pins, sheet of paper, tack board, ruler, protractor. Method A. Demonstration Experiment\u2014Optical Disc Method (Sec. IV: 9). B. Student Experiment^\u2014-Pin Method. 1. Draw a straight line AB (Fig. 21:1). Place the mirror perpen- dicular to the paper so that its back edge is along this line. 2. Place pins at P and Q about 2 in. apart so that the line PQ strikes the mirror obliquely. 3. Look into the mirror with one eye at the level of the paper and line up the images of P and Q. Insert pins R and S in line with these images. 4. Mark the position of all the pins. Remove the mirror and the pins. Draw the lines PQ and RS. Produce these lines until they meet at T. Through T draw TN perpendicular to AB. With a protractor measure angles PTN and STN. 5. Repeat, with other positions of the pins. Observations 1. Observation No. Angle of Incidence Z PTN Angle of Reflection Z STN 1. 2. 3. 238 EXPER", "IMENTS ON LIGHT 2. When the incident ray is on the plane of the paper where are the reflected ray and the normal found? Conclusions 1. State the two Laws of Reflection. 2. Define, and label on your diagram: incident ray, reflected ray, point of incidence, normal, angle of incidence, angle of reflection. Questions 1. If a polished piece of metal were used as the mirror in the above experiment, where would T be found? Why? 2. Add an observer\u2019s eye to the diagram and trace the path of the rays to the eye. EXPERIMENT 3 To study the position and characteristics of the image in a plane mirror, (Ref. Sec. IV: 11) Apparatus Plane mirror, two wooden pegs, paper, ruler, protractor. A Mb \u2014 izi-:. I o Fig. 21:2 Method 1. Draw a straight line AB in the middle of the sheet of paper. Place a plane mirror perpendicular to the paper, so that its reflecting surface is on AB. 2. Place the object (wooden peg) about 3 in. in front of the mirror. Mark its position O. 3. Look into the mirror to see the image of this object, and using the second wooden peg as a finder place it exactly where the image This can be done quite accurately by using the appears to be. method of parallax. That is, look into the mirror at the image, and over the top of the mirror at the finder. Adjust the position of the latter so that when you move your eye from side to side there is no relative movement between the finder and the image. Mark the position of the image I. Compare the size of the image to the size of the object. 239 Chap. 21 LIGHT 4. Remove the mirror. Join OI and let the line cut AB at M. Measure the lengths of OM and IM, and measure the sizes of the angles at M. 5. Repeat for other positions of the object. 6. Look at yourself in a mirror and raise your left hand. Note what you see. Observations 1. Observation Number Distance of Object OM Distance of Image IM Size of Angles AT M 1. 2. 3. 2. How does the size of an object and its image compare? 3. Which hand appeared to be raised in the mirror? What term applied to the image describes this phenomenon? 4. What kind of image is it? Conclusions 1. State your conclusion relating to the position of an image in a plane mirror. 2. Summarize the characteristics of such an image. 3. Define or explain; parallax, lateral inversion, virtual image. Questions 1. How could you locate geometrically the position of the image of an object in a plane mirror? Show this by a diagram. 2. Include in the diagram an eye, and show the path of the light rays whereby the eye sees the image. EXPERIMENT 4 To show the location of images produced by two plane mirrors at right angles. (Ref. Sec. IV: 13) Apparatus Two plane mirrors, four wooden pegs, paper, ruler, protractor. Method Repeat parts 1, 2 and 3 of experiment 3 using two plane mirrors placed at right angles to each other. Observations Describe the number and location of the images. Conclusion Construct a diagram to show how the images were obtained. 240 EXPERIMENTS ON LIGHT Questions 1. Add an observer\u2019s eye to the above diagram and trace the path of the rays to the eye. 2. Determine experimentally how many images are obtained when the mirrors are inclined at 60\u00b0, at 45\u00b0, etc. Use your results to verify the formula; Number of images = 360\u00b0 \u2014\u2014 : ; angle ol inclination 1 EXPERIMENT 5 To study the action of a concave mirror, and to determine its focal length (f), (Ref. Sec. IV: 14) Apparatus Optical bench, concave mirror, candle, screen. V D 1'U. Fig. 21:3 J U Method A. Demonstration Experiment\u2014Optical-Disc Method (Sec. IV: 14). B. Student Experiment\u2014Optical-Bench Method. 1. Mount the concave mirror and screen on the optical bench. Hold the mirror so that rays from the sun fall directly upon it. Failing that, hold it so that rays from a candle held at the opposite end of the room fall upon it. Focus the image on the screen. To do this move the screen until the image is clear and sharply defined. 2. Measure the distance of the image from the vertex of the mirror. Observations 1. What is observed when parallel incident rays are reflected from the mirror? 2. Describe the image obtained when the object is at infinity. 3. What is the distance of the image from the vertex of the mirror? Conclusions 1. What effect does a conc", "ave mirror have on light rays incident upon it? 2. Define; principal focus, focal length. 241 Chap. 21 LIGHT Questions 1. Why may a concave mirror be called a converging mirror? 2. Why did we use an infinitely distant object in the above experiment? EXPERIMENT 6 To study the images produced by a concave mirror, (Ref. Sec. IV: 15) Apparatus Optical bench, concave mirror of known focal length, candle, screen, finder. Method 1. Place the candle at more than twice the focal length (/) from the mirror. Locate the image on the screen and state its characteristics. 2. Repeat the above for two or three other positions of the candle as it is gradually moved nearer to the principal focus of the mirror. 3. Place the candle between the principal focus and the mirror. Look into the mirror to see the image. Note its characteristics and locate its position by the method of parallax using the finder. Observations Observation Number Position of Object Position of Image Characteristics of Image Kind Attitude Size 1. 2. 3. Conclusions 1. Where may an object be placed so that a real image of it is produced by a concave mirror? 2. Describe the changes in position and characteristics of the image as the object is gradually moved from infinity to the principal focus of a concave mirror. 3. Describe the position and characteristics of the image when the object is between the principal focus and the mirror. Question Make accurate scale diagrams to locate the images for the above positions of the object. EXPERIMENT 7 To study the images produced by a convex mirror, (Ref. Sec. IV: 16) 242 EXPERIMENTS ON LIGHT Apparatus Optical bench, convex mirror, candle, finder. Method Using the candle at three different positions from the mirror, locate the image each time by the method of parallax, and state its characteristics. Observations Observation Number Position of Object Position of Image Characteristics of Image Kind Attitude Size 1. 2. 3. Conclusions 1. What kind of image is produced by a convex mirror? 2. Describe the position and characteristics of the images produced by a convex mirror for all positions of the object. Questions 1. How would you find the focal length of a convex mirror? 2. Why may such a mirror be called a diverging mirror? 3. Construct a geometric diagram to locate the image of an object produced by a convex mirror. EXPERIMENT 8 To measure the refractive index of glass, (Ref. Sec. IV: 21, 23) Apparatus Rectangular block of plate glass with one pair of opposite edges polished, sheet of paper, tack board, pins, geometrical instruments. M 243 Chap. 21 LIGHT Method 1. Lay the sheet of paper on the tack board. Place the glass block on the sheet of paper. Outline its position carefully. 2. Stick a pin A upright in the paper, near the centre of one edge of the glass and touching it. 3. Stick a second pin B upright in the paper about 2 in. from A so that a line joining the two meets the glass obliquely. 4. Look through the block, line up the images of A and B and insert pin C against the opposite edge of the glass block so that it and the images of A and B appear in a straight line. Mark the position of each pin. 5. Remove the block and pins and draw BA, and AC. Draw the normal MN at A. With centre A describe a circle to cut AB at D and AC at E. From D and E draw perpendiculars DF and EG to the normal MN. 6. Measure the distances DF and EG and calculate the value of DF With a protractor measure the angles of incidence ( Z BAF) and of refraction ( Z CAG). Using a table of sines calculate sin Z i -r- sin Z r. EG. 7. Repeat for several other positions of the pins and arrange your results in the following observation table. Observations Obs. No. DF EG DF/EG Z i Z r sin Z i/sin Z r 1. 2. 3. 4. Average Average Conclusion Define: Index of Refraction. Questions 1. Compare your experimental value with the theoretical value (table p. 199). Calculate your percentage error. 2. Which way does light bend on entering obliquely a different medium of greater optical density? Why? EXPERIMENT 9 To trace the path of light through a glass plate with parallel sides. (Ref. Sec. IV: 24) 244 EXPERIMENTS ON LIGHT Apparatus Same apparatus as in experiment 8. Method 1. Lay the sheet of paper on the tack board. Place the glass block on the sheet of paper. Outline its position carefully. 2. Stick two pins A and B upright in the paper, about 2 in", ". apart, so that a line joining them (the incident ray) strikes the glass obliquely. 3. Look through the block, line up the images of A and B and insert pins C and D about 2 in. apart so that they and the images of A and B appear in a straight line. Mark the position of each pin. 4. Remove the block and pins. Join AB and produce it to meet the block at E. Join CD and produce it to meet the block at F. Join EF. On your diagram label the incident, refracted, and emergent rays. 5. Draw the normals at E and F. Measure the angle of incidence at E, and the angle of emergence at F. 6. Repeat the experiment for angles of incidence of different sizes. Observations 1. In what direction relative to the normal does the light ray bend on entering the glass at E and on leaving it at F? 2. Angle of incidence at F = Angle of emergence at F = 1. 2. 3. Conclusions 1. How does the angle of incidence compare with the angle of emer- gence? 245 Chap. 21 LIGHT 2. What relationship exists between the directions of the incident and emergent rays? 3. Describe the path of a light ray through a glass plate with parallel sides. Explanation Explain the observed bending of the light ray on entering and on leaving the glass. Question What factors affect the amount of lateral displacement of the emergent ray? EXPERIMENT 10 To illustrate the apparent change in depth due to refraction of light at a plane surface. (Ref. Sec. IV: 24) Apparatus Tall beaker, two pins, water. r_- Water Real Depth Apparent Depth ^Search Pin S Object Pin O Fig. 21:6 Method 1. Place a pin at the bottom of the beaker and pour in water to a depth of about 15 cm. This pin is the object, O. 2. Hold the search pin, S, horizontally outside the beaker. View from directly above and adjust the position of the search pin until there is no parallax between it and the image of the object pin. 3. Determine the distance of the object pin, and of the search pin from the surface of the water, thus finding the real and apparent depths. 4. Repeat the experiment using different depths of water. 5. Tabulate your observations and determine the average value for real depth apparent depth. 246 EXPERIMENTS ON LIGHT Real Depth Apparent Depth Real Depth Apparent Depth Observations Observation Number 1. 2. 3. 4. Conclusions 1. Construct a diagram to show the path of the light rays from the immersed object to an observer\u2019s eye. 2. Why is the apparent depth always less than the real depth? 3. Compare the value obtained by dividing the real depth by the appar- ent depth to the index of refraction of water (table, page 199). Questions 1. In step 3 of the method, why must the measurements be made on the outside of the beaker? 2. If the water in the above experiment were replaced with an optically denser liquid, e.g., carbon tetrachloride, what effect would this have on the apparent position of the object? Why? Try the experiment. EXPERIMENT 11 To show deviation produced by refraction through a prism. (Ref. Sec. IV: 25) Apparatus A 60\u00b0 prism, sheet of paper, tack board, four pins, geometrical instruments. A Fig. 21:7 Method A. Demonstration Experiment\u2014Optical-Disc Method (Sec. IV: 25). B. Student Experiment\u2014Pin Method. 247 Chap. 21 LIGHT 1. Lay the sheet of paper on the tack board. Place the prism on the paper. Outline its position carefully. Label its vertices A, B and C. 2. Stick two pins P and Q upright in the paper about 2 in. apart so that Q is touching the prism, and a line joining them (the incident ray) strikes the surface of the prism, AB, obliquely. 3. Look through the prism on the side AC. Line up the images of P and Q and insert pins R and S about 2 in. apart with R touching the prism, so that they and the images of P and Q appear in a straight line. Mark the position of each pin. 4. Remove the prism and the pins. Join PQ, QR and RS. Label the incident, refracted and emergent rays. 5. Extend PQ and SR to meet at D. Measure the size of Z D (angle of deviation). 6. Construct the normals at Q and R. Measure the angle of incidence and the angle of emergence. 7. Repeat on separate diagrams for three other angles of incidence, including the special case where the angle of incidence is equal to the angle of emergence. Observations 1. Which way does the light ray bend", " (a) on entering the prism? (b) on leaving the prism? Angle of Incidence Angle of Deviation Angle of Emergence 2. Obs. No. 1. 2. 3. 4. Conclusions 1. Describe the path of a light ray through a prism. 2. Describe how the angle of deviation varies as the angle of incidence is changed. 3. What is the relationship of the angle of incidence to the angle of emergence when the deviation is a minimum? Questions 1. Draw a diagram showing a ray of light passing through the prism at the position of minimum deviation. 2. What factors affect the amount of deviation produced by a prism? EXPERIMENT 12 To study the action of a convex and a concave lens, and to determine their focal lengths, (Ref. Sec. IV: 28, 29) 248 EXPERIMENTS ON LIGHT Apparatus Optical bench, convex lens, concave lens, candle, screen, finder. [ 'r'^ I I'' Fig. 21:8 Method A. Demonstration Experiment\u2014Optical-Disc Method (Sec. IV: 29). B. Student Experiment\u2014Optical-Bench Method. 1. Mount the convex lens and screen on the optical bench. Hold the lens so that rays of light from the sun, or from a distant candle, fall directly upon it. Focus the image on the screen. 2. Measure the distance of the image from the optical centre of the lens. 3. Repeat using the concave lens. In this case, in order to see the image, it is necessary to look into the lens toward the object. The position of the image can be determined by the method of parallax with the aid of the finder. Observations 1. What is observed when parallel incident rays are refracted through (a) the convex lens, (b) the concave lens? 2. Describe the image obtained when the object is at infinity using (a) the convex lens (b) the concave lens. 3. What is the distance of the image from the optical centre of the lens? Conclusions 1. What effect do (a) a convex lens, (b) a concave lens, have on light rays incident upon them? 2. Define: principal focus, focal length. Questions 1. What other name may be given to (a) a convex lens, (b) a concave lens? Why? 2. Why did we use an infinitely distant object in the above experiment? EXPERIMENT 13 To study the images produced by a convex lens, (Ref. Sec. IV: 30) 249 Chap. 21 LIGHT Apparatus Optical bench, convex lens of known focal length, candle, screen, finder. Method 1. Place the candle at more than twice the focal length (/) from the lens. Locate the image on the screen and state its characteristics. 2. Repeat the above for two or three other positions of the candle as it is gradually moved nearer to the principal focus of the lens. 3. Place the candle between the principal focus and the lens. Look into the lens to see the image. Note its characteristics and locate its position by the method of parallax using the finder. Observations Observation Number Position of Object Position of Image Characteristics of Image Kind Attitude Size 1. 2. 3. Conclusions 1. Where may an object be placed so that a real image of it is produced by a convex lens? 2. Describe the changes in position and characteristics of the image as the object is gradually moved from infinity to the principal focus of a convex lens. 3. Describe the position and characteristics of the image when the object is between the principal focus and the lens. Question Make accurate scale diagrams to locate the images for the above positions of the object. EXPERIMENT 14 To study the images produced by a concave lens. (Ref. Sec. IV: 31) Apparatus Optical bench, concave lens, candle, finder. Method Using the candle at three different positions from the lens, locate the image each time by the method of parallax, and state its characteristics. Observations Observation Number Position of Object Position of Image Characteristics of Image Kind Attitude Size 1. 2. 2 250 EXPERIMENTS ON LIGHT Conclusions 1. What kind of image is produced by a concave lens? 2. Describe the position and characteristics of the images produced by a concave lens for all positions of the object. Questions 1. How would you find the focal length of a concave lens? 2. Why may such a lens be called a diverging lens? 3. Construct a geometric diagram to locate the image of an object produced by a concave lens, EXPERIMENT 15 To study the dispersion of white light into the spectrum. (Ref. Sec. IV:36(a) Apparatus Projection lantern, single", " slit, converging-lens, prism, white cardboard screen. Method Direct a beam of light from the projection lantern on to the slit. Place the converging-lens in the light that diverges from the slit, and focus these rays on the screen at A. Place the prism in the path of the beam and note what happens to the light. Rotate the prism so that the emergent beam sweeps to and fro. Adjust the position of the prism until the emergent beam is as near as possible to A. The prism is now in the position of minimum deviation. Move the screen into the path of the beam and adjust its position until the light which falls on it is in clear focus. Observations 1. What happened to the beam of light on passing through the prism? 2. Name the colours of the spectrum that you see. 3. Which coloured rays are deviated least, and which most, by the prism? 251 Chap. 21 LIGHT Explanation 1. Why is light deviated on passing through the prism? 2. Why are different colours deviated different amounts? 3. Which is deviated the least? Why? Which the most? Why? Conclusions 1. What must white light consist of? 2. What is meant by dispersion of white light? Questions 1. What would be observed if a sensitive heat detector were placed just beyond the red end of the spectrum? Why? 2. What would be observed if a fluorescent material (e.g., anthracene), were placed just beyond the violet end of the spectrum? Why? EXPERIMENT 16 To recombine the colours of the spectrum into white light. (Ref. Sec. IV:36(b) Apparatus As in experiment 15, second prism identical to one used there, second converging lens, Newton\u2019s colour disc, rotating-machine. Method 1. With apparatus arranged as in experiment 15, place the second prism with its refracting edge in the opposite position to that of Note what is first. Move the screen to catch the image. the obtained. 2. With apparatus arranged as in experiment 15, place the second converging- lens in the spectrum. Move the lens until the image is brought to a focus on the screen. Again note what is observed. 3. Examine Newton\u2019s colour disc. Strongly illuminate it with white light from the projection lantern. Rapidly rotate it. Note the effect obtained. Place on the rotator. Observations Describe carefully what is observed in each part above. Include diagrams. Conclusion How may the colours of the spectrum be combined to form white light? Explanation 1. Why did the reversed prisms and the converging- lens re-form white light? 2. Why did Newton\u2019s disc appear white when rapidly rotated? 252 EXPERIMENTS ON LIGHT Question How could a concave mirror be used to recombine the colour of the spectrum? EXPERIMENT 17 To study the spectra of a few common elements, (Ref. Sec. IV: 39) Apparatus Direct-vision spectroscope, Bunsen burner with monochromatic flame attachment, sodium, calcium and lithium salts; electrical discharge tubes containing neon, nitrogen and hydrogen, induction coil, source of direct current. Method 1. Place a sample of the sodium salt in the attachment on the burner. Light the burner and heat the salt. Observe the flame through the spectroscope. Repeat for the other salts. 2. By means of the induction coil send a high-voltage discharge of electricity through the tube containing neon. Observe the discharge Repeat using the tubes containing the through the spectroscope. other gases. Observations Describe the spectra obtained and compare with those given in Fig. 19:6. Conclusion How is the spectrum of an element obtained? Questions 1. What is meant by spectrum analysis? 2. What are some of its advantages? EXPERIMENT 18 To study colour in natural objects (a) opaque (b) transparent, (Ref. Sec. IV: 40) Apparatus As in experiment 15, red, yellow and blue cardboard screens; red, yellow and blue glass filters. Method Produce a pure spectrum as outlined in experiment 15. 1. Replace the white screen by the coloured screens in turn. Note any change in the appearance of the spectrum each time. 253 Chap. 21 LIGHT 2. With the white screen in place, insert each of the coloured filters in the path of the beam. Note any change in the appearance of the spectrum each time. 3. Repeat 2 using both the yellow and blue filters in the path of the light beam. Note the colour of the light obtained. 4. Repeat 2 using all three filters (red, yellow and blue) in the path of the beam. Note what is obtained. Observations Describe carefully all that is observed. Represent your observations by means of simple diagrams. Explanation Account for each of the observations. Conclusion What is the cause of colour in (", "-\u2018v<:.ir.'^i.'I-,. -\u2022;-'/ /'b\u2019v^.a a.,r,.a'>aa'. \u2022' o, -..^.\u2022jv :i-,.,_ 'W,.yji Si. UNIT V MAGNETISM and ELECTRICITY Happens What Exactly Explain Fire. this CHAPTER 22 MAGNETISM V ; 1 INTRODUCTION TO MAGNETISM The Study of magnetism began with the observation of the power of attracLodestone tion of lodestone for iron. is the mineral magnetite, an oxide of iron, which was first discovered in the province of Magnesia (from which the name magnet was derived) Asia this mineral are Minor. Deposits of found in many other parts of the world including large areas of Canada and the United States. in The original discovery of magnetite is recorded in an ancient legend which tells of a Cretan shepherd whose irontipped crook was so strongly drawn to the earth that he dug to find the cause. Fables tell of magnetic rocks covered on their seaward sides with nails drawn from passing ships, and of iron statues supported in mid air. For centuries men have been intrigued by the phenomena of magnetic propIn the writings of Lucretius, a erties. Roman who died in 52 b.c., the following reference is found: \u201cAnd iron filings in the brazen bowls furiously; when underneath seethe was set the magnet stone.\u201d It was in the period of the Renaissance physician emphasized when William Gilbert, to Queen Elizabeth I of England, published his famous and hitherto unequalled book entitled De Magnete. In his book Gilbert of scientific method, and of arriving at conclusions only after careful experiment. He dedicated his work \u201cto you alone, true philosophers, ingenious minds, who not only in books but in things themselves look for knowledge.\u201d importance the De Magnete laid the foundation of magnetic science. It described in detail a wide variety of experiments in magnetism. The book proved, however, to be too much in advance of its time and was soon forgotten. For the next hundred and fifty years there was little progress, but in the latter part of the eighteenth century magnetism began to be recognized as the exact science that it has become today. 22:1). V : 2 SOME PROPERTIES OF MAGNETS When an iron bar is stroked with a natural magnet, the iron itself becomes capable of attracting other pieces of iron It has become an artificial or steel. magnet (Fig. this magnet If is brought near to a number of different materials in succession, it is found that only a few of these, called magnetic substances, are attracted. Among these are iron, nickel, cobalt and a few special alloys. The materials which are not attracted are said to be non-magnetic (Chap. 31, Exp. 1). 259 Chap. 22 MAGNETISM AND ELECTRICITY If a bar magnet is rolled in iron filings, the filings are seen to be attracted in large clumps around the ends of the magnet (Fig. These points at which the magnetic force is concentrated are called the poles of the magnet (Chap. 31, Exp. 2). 22:2). Fig. 22:1 Artificial Magnets (a) A Horseshoe Magnet. (b) A Bar Magnet. A magnet which is suspended freely from a non-magnetic stand will always come to rest with its axis along a line running approximately north-south. The same end of the magnet will always point towards the north. This is designated the north-seeking (N-pole) of the Magnetic Force Concentrated ^ near Ends'Clustered Fig. 22:2 Around the Poles of a Bar Magnet. Filings Iron magnet and the other end is the southseeking pole (S-pole) (Chap. 31, Exp. 3). If the N-pole of the suspended magnet is now approached by the N-pole of 260 another magnet, the poles will be observed to move farther apart (Fig. Fig. 22:3 Demonstrating the Law of Magnetism. of the This motion is caused by a 22:3). force, here one of repulsion. When the suspended magnet is S-pole approached by an N-pole of another magnet they will be attracted to each other. From such observations we may conclude that: like magnetic poles repel; unlike magnetic poles attract. This is called the Law of Magnetism (Chap. 31, Exp. 4). V:3 THE EARTH AS A MAGNET In the previous section it was shown that a suspended magnet always comes to rest along a north-south line. The magnet behaves in this way because the earth itself possesses a magnetic field as if it contained a huge bar magnet lying (Fig. 22:4). nearly parallel to its axis Why is the end of the magnet that points north labelled 6'? The magnetic north pole is", " located about 1400 miles south of the geographic north pole in the extreme north-central part of Canada. MAGNETISM Sec. V:3 Fig. 22:5 Demonstration Magnetic Compass. consists The modern magnetic compass (Fig. of a magnetized steel 22:5) needle suspended on a point so that it is free to move in a horizontal plane. The N-pole of the compass will point Fig. 22:6 Map of Canada Showing Points of Equal Declination, 1955. 261 Chap. 22 MAGNETISM AND ELECTRICITY always to the earth\u2019s magnetic north pole. The angle between the true north and the direction in which the compass points is called the angle of declination (Fig. 22:6). It is evident that the angle of declination will vary with position on the earth\u2019s surface. A navigator must determine its value from charts in order find true north from the compass to bearing at his position. The declination at Toronto, Ontario, was approximately 7\u00b0 west in 1955. This indicates that the angle between the lines drawn from Toronto to the geographic north pole and to the magnetic north pole is 7\u00b0, and the magnetic line is west of the line to the geographic pole. The magnetic north pole shifts its position slowly westward, and its exact location must be resurveyed frequently in order to correct charts for navigation purposes. V : 4 THE MAGNETIC FIELD compass-needle at a number of positions at varying distances from the magnet and observing the direction in which the needle points. The space surrounding a magnet in which its magnetic influence can be detected is known as the magnetic field of the magnet. in as the century, Magnetic fields were studied by Gilbut bert sixteenth Michael Faraday (1791-1867) showed that they could be best described and interpreted by means of magnetic maps made up of a series of lines known as lines of force. A line of force may be the path that would be defined travelled by a free N-pole if at liberty to move in a magnetic field. By specifying the movement of an N-pole in our definition we obtain the conventional direction for the lines of force, i.e., from the N-pole of the magnet, which would repel the free N-pole, to the S-pole of the magnet which would attract it. The effect of a magnet is noticeable over a considerable space surrounding it. This can be demonstrated by placing a An effective method of demonstrating the disposition of the lines of force about a magnet is described in chapter 31, Fig. 22:7 The Magnetic Field about (a) A Bar Magnet (b) Unlike Poles (c) Like Poles. 262 MAGNETISM Sec. V:6 experiment 5. Some examples of magnetic maps obtained in this way are shown in Fig. 22 : 7. The properties of lines of force as shown on these maps may be summarized as follows: the 1. Lines of force form closed curves, the magnet, N-pole of leaving proceeding through space to the S-pole, and completing their path through the The apparently loose ends magnet. shown in the diagrams are actually parts of complete lines linking the poles of the magnets. of a magnetized needle that is free to turn in a vertical plane, and a protractor to measure its inclination from the horizontal (Fig. 22:8). The plane of this needle must be adjusted so that it is parallel to the direction indicated by a compass. The instrument will then indicate the direction of the lines of force. At the magnetic poles the needle will repel each other, 2. Lines of force never cross but tend thus spreading to farther and farther apart as they leave the N-pole of the magnet. 3. Lines of force behave like stretched elastic bands, tending to contract and thus to shorten their paths. 4. The concentration of the lines of force determines the strength of the magnetic field at any point. Thus the field is strongest close to the poles of the magnet. The property of that allows magnetic lines of force to pass through it is called permeability. Substances such as soft iron and permalloy, an alloy of nickel and iron, which allow pass through them lines of readily are said to have a high permeability. For this reason these materials are used where a strong magnetic field is desirable as in the electromagnet (Sec. V:50) V:53). galvanometer substance force and (Sec. the to a V ; 5 THE ANGLE OF INCLINATION In section V : 3 the earth was described as possessing a magnetic field of its own. It is the horizontal component of this field that activates the magnetic compass. The exact direction of the magnetic lines of force may be determined by the use of a dipping needle. This consists Fig. 22:8 Dipping Needle. thus be 90\u00b0. stand vertically and the reading on its At the protractor will magnetic equator the needle will be horizontal and the angle of inclination or dip will be 0\u00b0. Other locations on the earth\u2019", "s surface will give readings between these extremes. At Toronto the angle of inclination is about 74\u00b0. V:6 INDUCED MAGNETISM An unmagnetized piece of soft iron If, however, will not attract iron filings. we bring it near the pole of a bar magnet, the iron will become a magnet and exhibit powers of attraction. When the bar magnet is removed the soft iron will lose its magnetic properties very rapidly. The iron is said to have been magnetized by induction. If the S-pole of the bar magnet approaches the soft iron it can be shown that the end of the iron 263 Chap. 22 MAGNETISM AND ELECTRICITY A Permanent MagUsed to net Being Separate Metal Sheets. Canadian General Electric nearest the magnet becomes an N-pole and the otlier end becomes an S-pole (Chap. 31, Exp. 6). If a bar of soft iron is held at the angle to the earth\u2019s surface that is indicated by a dipping needle and one end is tapped sharply with a hammer, it can be demonstrated that the iron has become a weak magnet (Chap. 31, Exp. 7 ). Thus we see that even the weak ships, hulls of etc. may of buildings, become magnetized by induction. The magnetic fields from these structures may seriously affect delicate measuring instruments such as watches and com- passes. In order to protect such instruments they are frequently enclosed in a magnetic shield of some highly permeable metal. The lines of force pass through the shielding material more readily than through the space enclosed by it (Chap. 31, Exp. 8), and so the instruments operate free from magnetic disturbance. The effect illustrated in Fig. 22:9. of magnetic shielding is Induced magnetism plays an important part in many of the devices that will be studied in later chapters of this unit. V:7 A THEORY OF MAGNETISM If a piece of iron wire magnetized as in experiment 9, chapter 31, is subjected to continued division and sub-division, each piece will be found to be a small magnet possessing an N-pole and an 10). When S-pole these pieces finally become too small for further division, we may imagine the process to be continued until finally the (Chap. 31, Exp. Fig. 22:9 Magnetic Shielding. Lines of Force Follow the Soft Iron Ring. Instrument I, in centre, is Unaffected by Magnetic Field. magnetic field of the earth is capable of causing magnetism by induction. It may be readily understood then, that magnetic substances such as the iron parts of machines, steel frameworks 264 MAGNETISM Sec. V:7 molecule of the iron is reached. Thus each molecule would possess its own Nand S-pole of equal strength. We may then reason that all magnetic substances are composed of molecules that are tiny magnets in themselves (Chap. 31, E.xp. 11 )-, Using this theory the unmagnetized state is accounted for by the random EJ m LEi] LXJ 0:1 LX] aa 03 m 03 aJLJJ rm ni] na ra DU [. i\u00a3i LL_j rm m aji Id (b) The Theory of Magnetism Fig. 22:10 (a) Unmagnetized State (b) Magnetized State. we can now account for the concentration of magnetic properties at the poles of the magnet. taken across The fact This molecular theory of magnetism provides a simple explanation of many magnetic phenomena. that there are the same numbers of unbalanced N- and S-poles at either end of a magnetized iron bar means that the poles of a bar magnet will be equally strong. A section the magnetized bar will expose an equal number of N- and S-poles showing that equal and opposite poles will be produced on breaking the magnet. Saturai.e., a magnetic substance tion effect, can be magnetized only up to a certain strength is clearly due to the fact that when all molecular magnets have been drawn into line there can be no further increase in magnetic strength. The loss of magnetism due to rough handling or heat is due to the disturbance of the molecules and a reversion to the random arrangement. of of N- arrangement distribution of molecular millions magnets in the bar (Fig. 22:10a). The haphazard and S-poles neutralizes their effect and no external magnetism is evident. When the bar is stroked with the N-pole of a magnet the S-poles of the molecular magnets are attracted in the direction of the stroking magnet and the molecules become aligned as shown in Fig. 22: 10b. At the end of the bar last touched by the N-pole of the magnet there will be many S-poles of the molecular magnets and thus it will act as the S-pole of our new magnet. The opposite end, having many molecular N-poles will be the new N-pole. In the central region of the bar the N- and", " S-poles of the molecules will again neutralize each other so that Fig. 22:11 Keepers (a) on Bar Magnets, (b) on a Horseshoe Magnet. The mutual repulsion of the similar molecular poles at the ends of a magnet tends to push the molecular magnets out 265 Chap. 22 MAGNETISM AND ELECTRICITY of alignment and so slowly to destroy its magnetism. To prevent this, bar magnets are usually stored in pairs with opposite poles adjacent and a piece of soft iron, called a keeper, placed across each end as shown in Fig. 22:11. The keepers become magnetized by induction and their molecules tend to hold those of the magnet in alignment. A horseshoe magnet requires one keeper as indicated. only A material which is readily magnetized but loses its magnetism rapidly, such as soft iron, is used to form temporary magnets used in electromagnets (Sec. V:50). The molecules of such substances are readily rearranged. Materials which have a more stable arrangement of molecules such as steel of aluminum, and alnico, an alloy nickel, and cobalt, are more difficult to magnetize, but tend to retain their magnetic properties much longer. Such substances are used to make the permanent magnets used in telephone receivers, galvanometers and other instruments. V:8 QUESTIONS A 1. (a) What is a natural magnet? (b) What is the common name of a natural magnet? 2. From the following list select those materials which are magnetic and those which nickel, aluminum, wood, rubber, iron, copper, lead, non-magnetic: glass, are cobalt, steel, zinc, tin. 3. 4. 5. 6. (a) State the Law of Magnetism. (b) Explain how, with the aid of a compass-needle, you would determine which end of a magnet is the N-pole. simplest form of (a) Describe the magnetic compass. (b) Explain why the N-pole of the compass points to the magnetic north pole of the earth. (c) Why doesn\u2019t a compass-needle point to the true north pole? Define angle of declination. (a) Define line of magnetic force. (b) Draw the magnetic field about a bar magnet. (c) State three fundamental characteristics of these lines of force. (a) Define angle of inclination or dip. (b) What would be the angle of inclination at (i) the magnetic north 266 pole, (ii) the magnetic south pole and (iii) the magnetic equator? 7. (a) What is induction? meant by magnetic (b) By means of a labelled diagram show how you could induce magnetism into a nail so that the head is an N-pole. 8. 10. 9. (a) Define magnetic permeability. (b) What is \"magnetic shielding\u201d? (c) Where and why would shielding be necessary? such (a) Describe how you could magnetize a steel needle. (b) How could a piece of steel wire be magnetized by a bar magnet without returning the magnet in a wide arc? (a) Outline the theory of magnetism, (b) Use this theory to explain: (i) that a magnet when repeatedly cut gives sections that are magnets also. (ii) polarity of a magnet. (iii) that there is no magnetic power in the centre of a magnet. (iv) that magnetism is lost on heating or jarring a magnet. (v) magnetic induction. (vi) magnetic saturation. MAGNETISM Sec. V:8 (vii) the difference between temporary, and permanent magnets. B 1. A rod of soft iron is held vertically and struck with a hammer. test (a) Describe how you would whether the rod is magnetized. (b) Which end should become an S-pole? (c) Why strike it with a hammer? 2. How could you use a compass-needle to distinguish between a magnet and a magnetic substance? 3. (a) How could the lines of magnetic force of a horseshoe magnet be traced? (b) Show by means of a diagram the pattern that you would expect to see. 4. A steel bar is repeatedly stroked from A to B with the N-pole of a bar magnet. A compass-needle is then placed near the end B of the steel bar. Make a diagram of the steel bar and compass-needle and label all poles. 5. How would you use a compassneedle to distinguish between three bars similar in shape and appearance if one is brass, the second is unmagnetized iron and the third is a magnet? 6. You are given two iron bars \u2014 one magnetized, the other not \u2014 and nothing else. Describe carefully how you could determine which is the magnet. 267 CHAPTER 23 ELECTROSTATICS 2500 years ago", ". However, similar discoveries have been made by everyone early in life. What child has not been fascinated to discover that on a cold dry day the comb which has just straightened his hair will now pick up bits of paper? Who has not been startled by a shock after scuffing across a carpet to touch another person or a metallic object? Materials which have gained the property of attracting small bits of paper and other light objects are said to have an electric charge (Chap. 31, Exp. 12). Substances that do not have this charge are said to Since many non-magnetic be neutral. materials are attracted, it is evident that electric charges differ from magnetism. It has been found by many experiments that whenever two different substances are rubbed together each will become this charge electrically charged. that builds up in gasoline tank trucks as they move along the highways. To prevent sparking, which could cause an explosion, a chain drags behind the truck to allow the charge to leak off gradually. Responsibility for many fires has been traced to charges of static electricity and precautions are necessary in many industrial plants to ensure that sparking does It is not occur. V : 1 1 KINDS OF ELECTRIC CHARGES It is a simple matter to prove that a hard rubber or ebonite rod rubbed with cat\u2019s fur or flannel, or a glass rod rubbed with silk, will become electrically V : 9 INTRODUCTION TO ELECTROSTATICS is attributed Greece, about 600 b.c. Every high-school student is familiar with many devices operated by electricity. However, this is electricity in motion, a comparatively recent scientific development. Early experiments in this field were confined to electricity at rest, or static electricity. The discovery of generally electricity to Thales of Miletus, one of the seven wise men of He learned that amber, a yellowish fossilized resin, when rubbed with a fabric would attract light objects. This knowledge an of for over two thousand years. science From the beginning of the seventeenth century electrical science began to grow, and it is from the early experiments of such men as Gilbert and Gray in England, du Fay in France, and Franklin in America that many of our fundamental concepts of electricity have arisen. The study of electrostatics has paved the way for advancement in current electricity. remained isolated fact V: 10 ELECTRIC CHARGES As mentioned above, the phenomenon of electric charges was first noted over 268 ELECTROSTATICS Sec. V:12 charged (Sec. V:10). If we charge an ebonite rod in tliis way, suspend it by a stirrup from an insulated stand, and approach it with another charged ebon- Xt-N. Fig. 23:1 Like Charges Repel. ite rod (Fig. 23:1), the suspended rod is repelled. If the suspended ebonite rod is approached by a charged glass rod, attraction will be observed (Chap. 31, Exp. 13). Thus we can see that these introduced the term negative to describe the charge produced on an ebonite rod rubbed with fur or wool, and positive for the charge on a glass rod rubbed with silk. A simple law of electrical charges may now be stated: like charges repel each other; unlike charges attract each other. V: 12 THE ELECTRON THEORY Our knowledge of the structure of matter has made great progress since Experimental the turn of the centuiy. work beyond the scope of this text has led to a concept of the atom far different from that held until 1897. Until that time the atom was thought to be the Since then smallest particle of matter. our concept of the structure of the atom has been constantly changing. For our is convenient to think of the purposes it atom as follows: 1. It may be pictured as a miniature solar system with a central relatively large mass, called the nucleus corresponding to our sun, and a number of smaller objects, called electrons travelling in their orbits about the nucleus much as the planets move around the sun. 2. The nucleus is composed of two kinds of particles: (a) Protons\u2014which are positively charged and are considered to have unit mass. (b) Neutrons\u2014which have no electric charge and have the same mass as the proton. 3. The electrons are negatively charged and are approximately as heavy as a proton. University of Toronto Benjamin Franklin are two charges. In different kinds electric 1750 Benjamin Franklin of 4. Atoms are neutral normally in charge. Therefore we may conclude that in each atom there must be as many electrons moving around the nucleus as there are protons within the nucleus. 269 Chap. 23 MAGNETISM AND ELECTRICITY Hydrogen Oxygen Neon Magnesium Chlorine Fig. 23:2 Diagrams of Several Atoms\u2014 Electrons (\u2022), Protons ( + ), Neutrons (n). 5. There are over one hundred different kinds of atoms known at the Scientists believe that present time. even more will be identified in the near future. The atoms", " of different elements differ from each other in the numbers of protons and neutrons contained in the nucleus and in the number and arrangements of electrons in their orbits. Some typical diagrams of atoms are shown in Fig. 23:2. An understanding of this theory of atomic structure enables us to explain why an object becomes charged by friction. Tremendous forces are required to separate a proton from the nucleus of an atom. However, it is relatively easy to cause an electron to leave an atom or to move to another atom. When an ebonite rod is rubbed with cat\u2019s fur, electrons are transferred from the fur to the ebonite and the rod becomes negaThe fur, having lost tively charged. electrons, becomes positively charged. Similarly, when a glass rod is rubbed with silk, electrons are transferred from the glass to the silk so that the glass becomes positive and the silk negative. We may now define a positive charge as a deficiency of electrons. A negative charge is a surplus of electrons. A neutral body has equal numbers of protons and electrons. 270 V:13 ELECTROSCOPES Before proceeding further in our discussion of electrostatics it is convenient to study the means by which electric charges may be detected and identified. Instruments used for this purpose are called electroscopes. (a) The Pith Ball Electroscope The simplest form of electroscope consists of two balls of dried elder pith, each about the size of a pea, suspended side by side from an insulated stand by silk threads (Fig. 23:3). When the pith balls are approached by an uncharged object no attraction or repulsion will be noted. When a charged ebonite rod is touched to the pith balls, they will be ELECTROSTATICS Sec. V:13 seen to be attracted, cling to the rod for a moment, and then be violently repelled from it. The pith balls have become negatively charged and they will spring apart due to the repulsion of like charges. The stronger the charge, the greater is the repulsion. When the charged pith balls are now approached by an object bearing the same kind of charge, they will be repelled from it. If the charge on the object is opposite to that placed on the pith balls, they will be attracted to it (Chap. 31, Exp. 14). ball to the rod which, being positively charged, has a deficiency of electrons. Thus, the pith ball, when charged by contact, receives the same charge as the charging object. (b) The Gold-Leaf Electroscope This instrument, shown in Fig. 23:5 is much more sensitive than that described above. It consists of a piece of These observations may be explained electron the readily by reference The to being rod, negatively theory. charged, has an excess of electrons. Because the rod is charged and the pith balls are not, they are attracted to it. As the pith balls touch the rod some electrons move from the rod to the electroscope. Thus the pith balls become negatively charged and are repelled. This is shown diagrammatically in Fig. 23:4. It is frequently convenient to use an electroscope with only one pith ball as shown for electrostatic experiments. events series of Similarly, the pith ball may be given a positive charge by touching it with a charged glass rod. In this case the electrons will be transferred from the pith Fig. 23:5 Gold-Leaf Electroscope. gold-leaf or thin aluminum-foil attached at its upper edge to a brass plate at the end of a brass rod. This assembly is enclosed in a metal case with glass windows for observation. The rod is in- 271 Chap. 23 MAGNETISM AND ELECTRICITY manner (Sec. V:40, V:77). Substances in which the electrons can not move insulators. A few readily common conductors and insulators are tabulated below. called are V : 15 INDUCED CHARGES Two metal spheres, A and B (Fig. 23:6), supported on insulated stands are placed in contact so that they form a ring causes sulated from the case by a of sulphur, wax, or rubber. The rod generally terminates at the upper end in a knob or flat disc to receive the charge. A negative charge on the electroscope may be obtained by touching the knob with a negatively charged rod (Chap. Electrons will be trans31, Exp. 15). ferred from the charged rod. to the knob, rod, plate and leaf of the electroscope. The similar charge on the plate and leaf leaf diverges. The extent of this divergence is an indication of the strength of the charge. If the knob is now approached by an object bearing the same kind of charge as that on the electroscope, there will be further divergence of the goldIf, however, the object bears the leaf. opposite charge, the leaf will be seen to fall. When touched by a positively charged object", ", electrons will leave the electroscope and it will be left with a positive charge. repulsion and the V : 14 CONDUCTORS AND INSULATORS In some materials, mostly metals, the loosely bound orbital electrons are so that they will leave the atom readily. In some other substances the electrons are so tightly bound to their nuclei that their movement is slight. Materials in which electrons can move readily are called conductors (Chap. 31, Exp. 16). Conductors may be solids, liquids, or gases, although liquids and gases conduct electricity in a slightly different single conductor. If a negatively charged ebonite rod is now brought near to A as described in chapter 31, experiment 17, and the surfaces of the spheres are examined for charge by means of a charged electroscope and a proof plane (a small metal disc at the end of an insulating-handle used to transfer charges to an electroscope), it will be found that the surface of A near the ebonite rod has acquired a positive charge. A negative charge will be found on the surface of B remote from the rod. If the ebonite rod is removed and the spheres Good Conductors metals graphite water solutions of acids, bases, and salts Poor Conductors dry wood paper alcohol kerosene pure water Good Insulators hard rubber paraffin wax sulphur mica porcelain dry air 272. ELECTROSTATICS Sec. V:17 will again examined, no charges be This is an example of electrofound. static induction. The charges discovered on the spheres while the charged rod was near are known as induced charges. They are due to the repulsion of electrons from sphere A, which thus becomes positively charged, to sphere B, which thus acquired a negative charge. Induced charges are possible only with conductors since there are no free electrons in insulators to allow the moveIf A and B are separment necessary. ated while the ebonite rod is in position, peiTnanent charges will remain on the spheres (positive on A, negative If the rod is now removed and on B) the spheres allowed to touch, it will be found that no charge remains on either sphere. We may conclude then, that induced charges are equal. still Exp. 18). The leaf diverges as shown in Fig. 23:7 because electrons are repelled to the plate and leaf. Keeping the charged rod near, touch the knob with a finger, that is, \u201cground\u201d it. The leaf will fall as electrons are repelled from it to \u201cground\u201d. Next, remove your finger, and then the charged rod. The leaf will diverge again. The remaining electrons are distributed over the electroscope which is left with a deficiency of electrons and thus a positive charge. In order to give the electroscope a negative charge the above procedure must be followed positively charged rod. This method is illustrated Students are advised to in Fig. 23:8. draw the diagrams and make the necessary explanations bearing in mind that \u201cground\u201d acts a extra source using of as a electrons. V:16 CHARGING BY INDUCTION V: 17 THE ELECTROPHORUS Approach, but do not touch, the knob of a gold-leaf electroscope with a negatively charged ebonite rod (Chap. 31, About 1775 Volta invented the simplest form of machine to produce electric charges by induction, called the electro3) Ground ( 4 ) (5) Fig. 23:7 Charging Gold-Leaf Electroscope Positively by Induction. (1) Uncharged electroscope. (2) Approach rod (\u2014 ). with charged ebonite (3) Touch with finger (ground) (4) Remove finger (5) Remove charged rod. 273 Chap. 23 MAGNETISM AND ELECTRICITY Fig. 23:8 Charging Gold-Leaf Electroscope Negatively by Induction. (1) Uncharged electroscope (2) Approach with charged glass rod ( + ) (3) Touch with finger (ground) (4) Remove finger (5) Remove charged rod It consists of a hard rubber or phorus. ebonite cake and a separate metal disc provided with an insulating handle. The ebonite cake is charged negatively by rubbing with fur. The metal disc is then placed on it and touched with a finger. If the disc is then removed by means of the insulating handle it will be found to possess a relatively strong positive charge. The disc has been charged by induction. In order to explain this we must that neither the disc nor the realize cake are perfectly smooth. Thus actual contact between them is made at very few points and we may assume that the two are separated by a thin air space over most of their areas. By induction a positive charge will appear on the under surface of the disc and a negative charge on its upper surface. When the Fig. 23:9 Charging an Electrophorus by Induction. (1) Ebonite", " rubbed and given neg- (3) Electrons repelled through finger to ative charge. ground. (2) Electrons repelled to top surface of metal disc. 274 (4) Metal disc positively charged. Ebonite cake still negatively charged. ELECTROSTATICS Sec. V:18 finger touches the upper surface, elec- are disc positive repelled from the trons to \u2018Aground\u201d and on removal of the finger the disc is left with a deficiency of electrons and a These changes are illustrated in Fig. 23:9. will be noted that very little of the charge is removed from the ebonite cake. The disc, therefore, can be charged repeatedly in the manner described above without recharging the ebonite. charge. It V:18 THE WIMSHURST MACHINE This is a well-known laboratory ma- metal sectors which serve both as inductors and carriers. Wire brushes, supported at the ends of two metal rods placed at right angles to one another across each plate, rub against the sectors as they pass towards collecting combs consisting of a number of sharp metal teeth directed towards the sectors on both front and back plates. These combs are placed at either end of the horizontal diameter of the plates and each is connected to a discharging knob supported above the machine. chine for the continuous production of The action of this machine is beyond electric charge by inductive separation and was first made by James WimsIt consists of two hurst about 1878. circular ebonite or shellacked glass plates mounted together geared close and as to rotate in opposite directions so (Fig. 23:10). On the outer surface of each plate near the rim are fixed an even number of equally spaced thin Fig. 23:10 Wimshurst Machine. the scope of this Interested students are advised to refer to a good text. encyclopedia. is interesting to note that with a 1 cm. air gap, 30,000 volts It are required to cause a spark discharge between the terminals. In spite of the high voltage there is little danger from the discharge because of the small quan- tity of electricity involved. Canadian Laboratory Supplies Ltd. 275 Chap. 23 MAGNETISM AND ELECTRICITY 1. In conductor (a) the leaves of the electroscope show the same divergence matter what part no is touched by the proof plane. 2. In conductor (b) the divergence proves to be a little greater from tests at the ends than at the centre. 3. In conductor (c) the divergence is much greater from tests at the more pointed parts of the conductor than elsewhere. From these tests we may conclude that the charge on a conductor tends to be concentrated at the more pointed parts of the surface. V : 20 LIGHTNING-RODS About the middle of the eighteenth century Benjamin Franklin demonstrated in his classic kite experiment that thunder-clouds carry electrical charges and that lightning is an electrical discharge clouds and objects on the earth. spark between clouds, or or If a metallic point is connected to one of the discharge knobs of a Wims- V: 19 DISTRIBUTION OF CHARGES The location of charges on a charged body may be demonstrated as in chap31, experiment 19A, using Biot's ter Spheres shown in Fig. 23:11. The apparatus consists of an insulated metal over which two tightly-fitting sphere can metal hemispheres be insulated placed to cover it completely. A charge is given to the sphere and then the tightly over it. hemispheres are fitted When the hemispheres are removed and tested using a proof plane and electroscope, each will be found to possess a charge, while the sphere is found to have lost its charge. The charge, then, must have passed to the hemispheres as they formed the outer surface of the sphere. Charges reside only on the outer surface of a charged conductor. To determine how the charge is distributed over the surface of a charged conductor we can charge several insulated conductors of different shapes as described in chapter 31, experiment 19B. On investigation with a proof plane and electroscope the following observations are made: 276 Fig. 23:12 Action of Points hurst machine, as shown in Fig. 23:12, the streaming of charge away from the point is sufficient to blow aside the ELECTROSTATICS Sec. V:21 flame of a small candle (Chap. 31, Exp. 20). The action of points is as if the point produces a self-discharging action, the charge on the conductor streaming away from it as an \u201celectric wind\u201d. Franklin made use of this action of points as he designed the lightning-rod to protect buildings against the dangers of lightning. As a charged cloud passes its ning with occurs over the earth it induces the opposite charge on objects below. If the charge becomes great enough a flash of lightaccompanying thunder, a shock-wave, caused by the sudden and violent expansion of the air heated by the discharge. The rods are", " the induced charge so designed escapes from the sharp points and so an accumulation of charge is prevented. If the lightning does strike the rods, a good that conductor leads the electricity safely to ground through a metal rod deeply buried in the damp ground near the building (Fig. 23:13). V : 21 QUESTIONS 1. 2. (a) What is the meaning of static electricity? (b) How is it produced? (c) How is it detected? (a) Name two kinds of static electricity. How is each produced? (b) How would you prove that there are two kinds? (c) State the law for electrical charges. 3. (a) Describe in point form our modern concept of the structure of an atom. (b) In terms of the electron theory, what is (i) a positively charged, (ii) a negatively charged, (iii) a neutral body? (c) Why does (i) ebonite become 5. (ii) glass become negative when rubbed with wool, when rubbed with silk? What is the electrical condition of the wool and silk afterward? positive 4. (a) What are the purposes of electroscopes? (b) Describe and explain what occurs when (i) a positively charged, (ii) a negatively charged object Is brought up to an uncharged pith ball electro- scope. (c) How would you use it to identify an unknown charge? the (a) Describe the gold-leaf electroscope and state the purpose of each part. structure of 277 Chap. 23 MAGNETISM AND ELECTRICITY 8. Describe, using a series of diagrams to illustrate your answer, how to charge a gold-leaf electroscope (a) negatively by induction. positively, (b) 9. Explain the following: (a) the metal disc of an electrophorus can be repeatedly charged without recharging the ebonite cake. (b) the purpose of the Wimshurst machine. (c) electrical charges reside on the 10. outside of a conductor. (d) lightning-rods are sharply pointed. (e) lightning-rods are connected by a good conductor to ground. A positively charged cloud hovers over a tall church spire. What will be the resulting electrical condition of the spire? What will happen if a large enough charge accumulates on the cloud? 6. 7. (b) Explain how to charge it positively, (ii) negatively by contact. (c) How would you use it to identify an unknown charge? (i) (a) What is a good conductor of electrical charges? Give examples. (b) What is an insulator of electricity? Give examples. (c) How could you use a gold-leaf electroscope to show that iron is a better conductor than wax? (a) What is meant by electrostatic induction? (b) You have a charged glass rod and two metal spheres mounted on insulated stands. Describe how you would charge one sphere negatively, and the other positively, by electro- static induction. (c) How would you verify that the spheres were oppositely charged? 278 1 ^. CHAPTER 24 ELECTRIC CURRENT V:22 INTRODUCTION In chapter 23 it was shown that electrons moved between the terminals of the Wimshurst machine and that lightning is actually a movement of electrons between earth and charged clouds. Such movement of electrons is called an electric current. In the eighteenth century Luigi Galvani, an Italian physiologist, discovered that a freshly dissected frog\u2019s leg showed violent muscular contractions when touched by two dissimilar metals. He erroneously attributed this property to the frog, and it remained for Alessandro Volta, an Italian professor of physics, to demonstrate a short time later that the frog\u2019s leg had acted only as a sensitive detector of electricity which had been produced by contact with the dissimilar metals. Volta showed that this electricity was identical with that produced by frictional means (Sec. V:10). From this beginning he constructed the first source of continuous electric current, the voltaic cell (Sec. V:23). From the time of of Volta\u2019s works in 1800, there has been continuous progress in the field of electricity. The sections that follow will publication help to give you an understanding of the fundamentals of electricity that mean so much in our modern way of life. V:23 THE VOLTAIC CELL Every high-school student is familiar with the tiny dry cell that is used to operate a flashlight and the larger storage battery used to start a car; these are just two modern forms of the voltaic cell. A voltaic cell consists of two dissimilar materials immersed in a solution of an acid, called the base, or salt, electrolyte. voltaic The simplest form of cell consists of a copper plate and a zinc plate placed in a vessel containing dilute (Chap. 31, Exp. 21). sulphuric acid As a result of the", " chemical action in the cell the copper plate becomes posiand plate negatively charged. If the plates are charged tively zinc the i\u2014O\u2014 1 Dilute Sulphuric Acid Flashlight Bulb 1 ^. 11111 1 1 11 1111111 'HWT'Copper N. 1 111 II Zinc - Fig. 24:1 Diagram of Simple Voltaic Cell. 279 Chap. 24 MAGNETISM AND ELECTRICITY then connected by a conductor (Fig. 24:1) the potential difference between them will cause electrons to flow from the zinc to the copper. The circuit is As completed through the electrolyte. the current flows, hydrogen is deposited on the copper plate and zinc sulphate dissolves in the electrolyte as the zinc is gradually used up. The potential difference between the plates is approximately 1.1 volts. This P.D. is independent of the size of the plates used, but In all varies with the materials used. such cells chemical energy is transformed into electrical energy. V : 24 DEFECTS OF THE VOLTAIC CELL 1. Polarization If a small flashlight bulb is connected in the circuit of a voltaic cell it will glow brightly for a short time and then fade gradually until no light is seen (Chap 31, Exp. 21). It is apparent that the current delivered by the cell has decreased. This effect is caused by po- METAL CAP, with washer to insulate the cap from the metal cover. EXPANSION SPACE, for expansion of cell contents during use. as the ZINC CAN serves as the negative plate and at the same time container for the cell. ELECTROLYTIC PASTE sal-ammoniac containing and zinc This reacts with the zinc making it electro-negative and producing hydrogen which moves toward the carbon rod. chloride. larization, which is the accumulation of bubbles of hydrogen on the positive plate. The hydrogen deters the acid from reaching the plate and serves as an insulator preventing further transfer of electrons. To restore a polarized cell the bubbles may be removed mechanically by wiping the positive plate, shaking it, or, chemically, by adding an oxidizing agent such as potassium dichromate to the solution. Such a chemical releases oxygen which unites with the hydrogen to form water. Substances added for depolarizing purpose called are this agents. 2. Local Action cell, the If impure or commercial zinc is used the plates are found to in deteriorate rapidly even when the cell is not in use. This is caused by small particles of such impurities as lead, iron, carbon, etc., occurring in the plates used. Many small voltaic cells are set up between these impurities and the zinc, and local action continues until the plate METAL COVER, which closes the cell tightly at the top, making it safe against bulging and breakage. ASPHALT INNER SEAL. CENTERING WASHER. DEPOLARIZING MIX. This contains carbon to provide conductivity, and manganese dioxide which reacts with the hydrogen forming water which keeps the contents moist. CARBON ROD, serves as the positive plate. It is powdered composed of bonded carbon particles together and baked at a very high temperature. COMPLETE CELI.. BOTTOM WASHER. 280 Fig. 24:2 Structure of a Dry Cell. ELECTRIC CURRENT Sec. V:26 is consumed. Thus zinc is used up while delivering no useful current outside the thinner and may eventually break open allowing the paste to dry out. Tliis in- cell. The use of pure zinc, of course, would be a simple remedy, but pure zinc is expensive. A more practical method of eliminating local action is to amalgamate the surface of the zinc plate with mercury. The zinc dissolves in tlie mercury and gradually comes to the surface where it reacts with the acid. The impurities are covered by the mercury and their contact with the electrolyte is prevented. V:25 THE DRY CELL Obviously, the voltaic cell described in section V : 23 has several disadvanThe dry cell tages eliminates many of these undesirable Its structure is shown in Fig. features. practical use. in 24:2. creases the internal resistance of the cell to such an extent that current will flow in the circuit. if any, little, The potential difTerence of a fresh dry cell is approximately 1.5 volts regardless of size. Large dry cells are able to produce current over longer periods than smaller cells. Dry cells of various sizes and shapes have been made to fill many needs. They are widely used for flashlights, portable radios, telephones, hearand many other ing devices used by the armed forces and industry. lanterns, aids, V:26 THE ELECTRIC CIRCUIT The flow of electrons in a wire is so similar to the flow of water in a pipe a comparison may be made as that As the cell is used the zinc becomes follows: The Water System The Electric Circuit", " 1. Water current is the flow of water through the system. 1. Electric current is the flow of electrons through the wires. 2. Water flows through a closed system of pipes between source and user. 3. The quantity of water in a system is measured in gallons or similar units. 4. The current of water is measured in gallons moving through the pipe in a unit of time. 5. Water current is reduced by the friction between the water and the pipes. 2. Electrons flow through a closed path of wires called a circuit. 3. The quantity of electricity is measured in coulombs. A coulomb is equal to approximately 6.28 billion billion electrons. 4. The electric current is measured in amperes. An ampere is the current when one coulomb passes a point in the circuit in 1 second. 5. Electric current is decreased by the resistance of the conductor. The unit of resistance is the ohm. 281 : Chap. 24 MAGNETISM AND ELECTRICITY The Water System 6. Water is driven through the pipes by pressure which is measured in pounds per square inch or related units. Pressure is provided by increasing the potential energy of the water by high tanks. The force so created pushes the water to its destination although the water pressure decreases as the length of pumping into it pipe increases. The Electric Circuit 6. Electrons move through a circuit because of a difference in electrical pressures or potential difference (P.D.) between points in the wire. This potential difference is measured in volts. The pressure difference is maintained by the use of batteries or cells which provide a source of electrons by means of the transformation of chemical energy. This potential difference is the electromotive force necessary to move the electrons through the wire. Resistance in the circuit causes a lowering of potential ence between points the conductor. This is referred to as a voltage drop. differ- in The units of electricity in common use and their symbols are sum- marized below for convenient reference (b) Current of electricity (a) Quantity of electricity (Q) \u2014-unit is the coulomb (/) \u2014 unit is the ampere {V ) \u2014 unit is the volt (R) \u2014 unit is the ohm Potential difference (d) Resistance (c) Smaller or larger units are obtained by using submultiples or multiples of these as: 1 milliampere = 1/1000 ampere 1 microampere = 1/1, 000, 000 ampere 1 millivolt = 1/1000 volt 1 megohm V : 27 ELECTRIC CURRENT AND ELECTRON FLOW Electric current has been defined as the flow of electrons through the wires of a circuit (Sec. V:26). From the beginning of electrical theory, the current has been considered to flow, as is the case with water, from the place of high potential to the place of lower potential. Early scientists considered the positive side of a source of electricity to be at 282 1 microvolt 1 microhm 1 kilovolt = 1,000,000 ohms etc. = 1/1,000,000 volt = 1/1,000,000 ohm = 1,000 volts a higher potential than the negative side. Thus, the convention that electric current flows from positive to negative came into common use. We now know that the negative side of a source has a surplus of electrons and the positive side a deficiency of a electrons fundamental law of nature, electrons move from where there are more to where there are fewer. Electrons, there- (Sec. V:12). Following fore, actually move from negative to (b) Parallel Circuit ELECTRIC CURRENT Sec. V:29 positive. In all work on electronics, including radio, it is necessary to refer to electron movement in order to understand the operation of modern devices. However, in most other electrical work, unless definitely specified, references have been confined in the past to the conventional current flow, i.e., positive to negative. In most modern writings, including this text, all references are to the electron flow. No difficulty need arise if the difference between current flow and electron flow is recognized. V : 28 TYPES OF CIRCUITS Two main types of connection are in common use in electric circuits and all others are merely combinations of these. (a) Series Circuit In this arrangement the current goes from one terminal of the source to one terminal of the first appliance. On passing through this, it flows to the next and the next until it reaches the other terminal of the source (Fig. 24:3). All In the parallel or shunt type of connection the conductor from the positive terminal of the source is connected to one terminal of each appliance and the negative terminal of the source is connected to the other terminal of each Electron Flow 4- Battery Appliance = \u00a9 (I) (\u00a3) (p Fig. 24:4 Parallel Circuit. appliance (Fig. 24:4). This is typical of wiring circuits used in your home. Each appliance receives current independently from the others", " and thus a break in one does not interfere with the operation of the rest. V:29 ARRANGEMENT OF CELLS (a) Series If the resistance in a circuit is so great that the P.D. of one cell is not sufficient to provide the required current, a number of cells may be connected in series. The positive terminal of the is connected to the negative of the second, the positive of the second cell first the current must pass through each appliance, causing a voltage drop (Sec. V:26) in the circuit. A break in any part of the circuit will stop all flow of current. Series circuits are frequently found in Christmas-tree lights, but are seldom used for other purposes. Fig. 24:5 Battery of Cells Connected in Series. 283 Chap. 24 MAGNETISM AND ELECTRICITY to the negative of the third, etc. Such an arrangrnent of two or more dry cells is called a battery. When cells are connected in series the total P.D. is equal to the sum of the voltages of the indiradio vidual B-battery, for example, is composed of thirty dry cells connected in series (Fig. cells. A forty-five volt 24:5). (b) Parallel If the resistance in a circuit is low and a large current is to be used, dry cells are connected in parallel. All the positive and negative terminals are connected as shown in Fig. 24:6. The P.D. of such a battery remains the same as that of a single cell. Since the current is provided equally by all cells the parallel arrangement gives a large current for a relatively long period of time. V : 30 SWITCHES Just as taps are inserted in a pipe line to stop the flow of water, so switches are used to interrupt the flow of electric current in a circuit. These are devices which break or complete the conducting path. Many types of switches are in common use, but only two will be described here. erally of the type shown in Fig. 24:7a. Wall switches are smaller variations of the same type. (b) Push Button Switch This familiar type is shown in Fig. 24:7b. It is used to control low-voltage Fig. 24:7 Switches (a) Main House Switch (b) Push Button Switch circuits in which the current is required for short intervals only. Electric doorbells, signal buzzers, etc. make use of this type of switch. at this their description Many newer types of switches have been devised for specific purposes, but the scope of this text does not allow The interested student will find information on mercury switches, automatic circuit breakers, and special switching arrangements such as those used to control a stairway light from either upstairs or downstairs in any good wiring handbook. time. V:31 USE OF SYMBOLS (a) Knife Switch Modifications of the common knife switch are used to control many household circuits. The main switch is gen- In order to represent the many pieces of equipment encountered in electrical drawings and circuit diagrams, a number of conventional symbols have been de- 284 ELECTRIC CURRENT Sec. V:31 vised. A few of the most common symbols are introduced at this point so that the student may use them in the work which follows. SOURCES OF \u2022potential DIFFERENCES + Cell \u2014wwvw Fixed Resistors WIRING I Battery (Cells in Series) RESISTORS AWWW- L T T T Battery (Cells in Parallel) METERS Wires Joined Wires Crossing Not Joined RADIO Voltmeter Ammeter Fig. 24:8 Common Electrical Symbols 285 Chap. 24 V : 32 MAGNETISM AND ELECTRICITY QUESTIONS 5. Define the following: electric circuit, ampere, coulomb, electrical resistance, potential difference. 6. (a) Name the two principal types of circuits and define each. electrical (b) State the advantages and disadvantages of lights each way. (c) Make a diagram showing three lights connected in each of these ways. connecting electric 7. (a) Distinguish between the arrangement of cells in series and in parallel. (b) What is the purpose of each of these arrangements? (c) What would be the difference of three dry cells connected potential (I) in series (ii) in parallel? 8. Make a diagram of a circuit containing a cell, a fixed resistor, two wires joined, two wires crossing but not joined, a fuse, a lamp, a switch. 1. 2. 3. (a) What is an electric current? (b) How is an electric current produced? (a) Make a diagram of a simple voltaic cell. Mark the direction of the electric current (electron flow). (b) What determines the voltage of such a cell? (c) What transformation of energy takes place in a voltaic cell? (a) What are defects in a volta", "ic cell? (b) How are the defects overcome? two the principal 4. (a) Make a labelled diagram of a dry cell. (b) State the purpose of each struc- ture or material labelled in part (a). (c) What is the voltage of a dry cell? (d) Why are different sizes? dry cells made in 286 ; CHAPTER 25 OHM\u2019S LAW AND RESISTANCE V:33 OHM'S lAW relationship The between current strength, potential difference, and resistance of the circuit is of the utmost importance in practical electricity (Chap. In such an experiment 31, Exp. 22). the potential difference across the revaried by connecting sistance different numbers of dry cells coil is Dry Cells ment is shown in Fig. 25:1. Let us consider some sample results of such an experiment No. of Cells Galvanometer deflection No. of Cells (divisions) Deflection 1 2 3 4 5 4.9 9.8 14.8 19.7 24.5 0.204 0.204 0.203 0.203 0.205 Within the limits of experimental error. Number of cells in the Galvanometer deflection And, since the P.D. across the resistance coil varies directly as the number of cells used, and the galvanometer deflection is proportional to the current through the coil, we have Potential Difference ~ a constant Current Strength or Current Strength is proportional to Potential Difference The strength current circuit. flowing through the coil is gauged by the deflection of a galvanometer needle ( Sec. V : 53 ). A circuit for this arrange- the of in the This relationship was first investigated by nineteenth early George Simon Ohm. The results of his work, published in 1826, are embodied in the law bearing his name. Ohm\u2019s Law may be stated as follows: century The current through a given conductor is proportional to the potential difference between its ends. It must, however, be emphasized that 287. : Chap. 25 MAGNETISM AND ELECTRICITY this law is true only if the temperature remains constant (Sec. V:34). The value of the constant derived above is known as its resistance {R) The unit of resistance is the ohm which may be defined on the basis of Ohm\u2019s Law as being the resistance of a conductor which permits a current of one ampere to flow when a potential difference of one volt is applied across it. Ohm\u2019s Law may be written, then, as Potential Difference \u201e = Resistance. or = Ohms Current Volts Amperes V I This formula may be rearranged to read V = IR or I = V/R All of these forms will be found useful in numerical work. Problem examples are given in sec- tion V:36. FACTORS AFFECTING >f:34 RESISTANCE The resistance of a conductor depends on four main factors; (a) Length\u2014The resistance of a con- ductor varies directly as its length i.e., twice the length gives twice the resistance; half the length gives half the resistance; etc. (b) Cross-section\u2014The resistance of a conductor varies inversely as the area of twice the cross-sectional area gives half the cross-section i.e., its resistance; half the cross-sectional area gives twice the resistance; etc. (c) Temperature\u2014In most metals an increase in temperature causes an inin some crease resistance, but in 288 substances such as glass, carbon and electrolytes an increase in temperature causes a decrease in resistance. (d) Material\u2014Materials differ widely in their resistances. Good conductors such as copper and aluminum have very low resistances. Poor conductors such as nichrome and manganin have much higher resistances. Good insulators, such as glass, mica, etc., have very high resistances. Relative Resistances of Some Common Materials (in ohm-cms at 20\u00b0C.) Aluminum Copper Iron Mercury Nichrome Platinum Silver Tungsten X 10-6 2.83 1.72 10.0 95.8 100 10.0 1.63 5.51 V : 35 RESISTORS IN SERIES AND IN PARALLEL The potential difference which causes the current to flow between the ends of a resistor is frequently called the voltage drop across the resistor. The value of a voltage drop (F) may be readily calculated by using Ohm\u2019s Law. (a) Resistors in Series If resistances of r^, and ohms are AVWV-t-AAAA/V Fig. 25:2 Resistors in Series. joined end to end as shown in Fig. 25:2, and a current of 1 amperes is passed OHM\u2019S LAW AND RESISTANCE Sec. V:36 through Ts- The potential difference beV volts and the total tween A and B effective resistance in the circuit is R ohms. through", " them, the voltage drop across V \u2014 IR) will be respectively: each Vi \u2014 hi, Vo z=z lr2. Vs = hs will be The total voltage drop Vi h Vo + Vs or I {ri + rg + rs). But V = IR, where R is the total re- sistance in the circuit. IR \u2014 I [ti + rg + Ts) and R \u2014 Ti r 2 h Ts (b) Resistors in Parallel By connecting resistors in parallel, \u2022 alternative paths are provided for the --I. Ij \u2014 \u2014, 12 \u2014 \u2014, 13 \u2014 V. V.. ri rg But I = ii is + is V rs + L + ri rg R And \u2014 = \u2014 + \u2014 + \u2014 Ts Ti rs rs R reciprocal of the total Thus we see that when resistors are in series, the total resistance of the circuit equals the sum of the individual resistances. When resistors are in parallel, resistance the equals the sum of the reciprocals of the individual resistances. Note The reciprocal of the resistance is known as the conductance of a conductor. The unit of conductance is the mho. Thus a wire with a resistance of 2 ohms will have a conductance of half a mho. current through the circuit (Fig. 25:3). The current, /, divides at A so that ii passes through r^, fg through rg, and is V : 36 PROBLEM EXAMPLES INVOLVING OHM'S LAW Example 1 Calculate the resistance of a light bulb which carries a current of 2 amperes when connected to the 110 volt circuit. T = 1 10 volts 1 = 2 amp. R =? / no. \u201e V ^'R = 55 Resistance of the bulb 55 ohms. Example 2 If two lamps Li, Lg, of resistance 30 ohms and 20 ohms are connected in the total resistance in the series in a 110 volt circuit, determine (a) 289 Chap. 25 MAGNETISM AND ELECTRICITY circuit (b) the current in the circuit (c) the voltage drop across each lamp. fi = 30 ohms t2 = 20 ohms (a) (b) ri = 2)0 ohms 72 = 20 ohms R =? v = no volts /? nr 50 ohms 1 =? / = 2.2 amp. 7i = 30 ohms rg rr 20 ohms 1*^ V =110 volts H'R = Ti + 72 (series connection) i? = 30 + 20 = 50 total resistance in circuit = 50 ohms I = \u2014 (Ohm\u2019s Law) R 7 = 112 = 22 50 current in the circuit = 2.2 amp. Vi \u2014 hi (Ohm\u2019s Law) vi = 2.2 X 30 = 66 voltage drop across Li = 66 volts V2 = Ir2.\\v2 = 2.2 X 20 = 44 voltage drop across = 44 volts. Example 3 If two lamps Li, Lg, of resistance 30 ohms and 20 ohms are connected in parallel in a 110 volt circuit, determine (a) the effective resistance of the circuit (b) the current through each lamp. 290 OHM\u2019S LAW AND RESISTANCE Sec. V:37 :a) Ti \u2014 30 ohms r* =20 ohms R =? (b) = 110 volts = 30 ohms = 20 ohms _ p \u2014? OF RESISTORS Ti 72 \u2014 = \u2014 + \u2014 (parallel connection) R 1 _ 1 \u201cso R=\\2 effective resistance of the circuit =12 ohms. 1 _ 5 _ 1 20 ~~60 ~ 12 V I j =z \u2014 (Ohm\u2019s Law) ri 110 II 11 CO 30 current through Li \u2014 3.7 amp. rz.\u2022./5 = \u2014 = 5.5 current through Lg = 5.5 amp. 20 be made from special constantan or manganin (alloys of copper and nickel), having an accuracy within 1 %. These are in general use laboratory work and measuring- for instruments. A number of coils of fixed resistance are frequently arranged together in a Resistors are used primarily to control current and potential difference in eleccircuits. They are constructed in tric many forms and sizes varying from those which have resistances of only a fraction of an ohm, as used in some measuring instruments, to those having resistances of many megohms as found in radio receivers. In general, however, resistors may be classified or variable. either fixed as (a) Fixed Resistors These may be constructed of short lengths of metal strip for very low resistances. Carbon or wire coils will provide higher resistances. Carbon resistors are frequently used in radios and other equipment in which some variation in resistance is permissible since such resistors may have an error tolerance of up to 20%. Wire-wound resistors may", " Fig. 25:4 Section of Resistance Box. Details of construction resistance box. are shown in Fig. 25:4. The ends of the 291 Chap. 25 MAGNETISM AND ELECTRICITY the two (Chap. 31, Exp. 23 and 24) which follow: (a) Voltmeter\u2014Ammeter Method The unknown resistance, R, is connected in the circuit shown in Fig. 25:6. The rheostat is adjusted until the amme(Sec. V:54) records any suitable ter coils are attached to brass blocks on the top of the box and the current passes through the blocks when the shortingplugs are placed in position. Pulling out the plug puts the resistor below it into the circuit. Such a box provides a large range of standardized resistors for laboratory use. To avoid errors in using the plugs must be a resistance box, Plugs and sockets must inserted firmly. be kept clean at all times. (h) Variable Resistors (Rheostats) Such resistors are usually continuously variable between certain limits. A common form of rheostat is shown in Fig. It consists of a number of turns 25:5. Fig. 25:6 for Determining Voltmeter-Ammeter Method the Value of an Unknown Resistance. value for the current passing through the resistor. The voltmeter (Sec. V:54) across the resistor indicates the potential difference between the ends of the resistor. From this information the resistance may be calculated by using Ohm\u2019s Law. V = LToUs R=? Example = J.\u2022./e = l^ = 4 3.'. The resistance is 4 ohms. (b) Substitution Method The unknown resistance may be connected in a circuit as shown in Fig. 25:7. The rheostat is then adjusted until a large deflection is shown on the ammeter, or galvanometer (Sec. V:53). The meter reading is noted and the un- Fig. 25:5 Rheostat. of wire wound on a porcelain tube. Above the coil is supported a metal bar along which moves a metal spring which can make contact at any turn of the coil. Connections are made to the end of the bar and to the end of the coil to enable the resistance in the circuit to be readily adjustable by moving the spring. Variable resistors such as this are of various sizes and shapes and are used in radio volume controls, for dimmer lights in theatres, starter boxes for heavy electric motors, etc. V:38 METHODS OF MEASURING RESISTANCE Although there are several methods the scope of of measuring resistance, this text permits the discussion of only 292 OHM\u2019S LAW AND RESISTANCE Sec. V:39 known resistance is replaced by a rePlugs are removed until sistance box. the meter reading is the same as before. The resistance of the box is now equal to the unknown resistance. WAW\\/\\ Rheostat Substitution Fig. 25:7 for Determining the Value of an Unknown Resistance. Method Battery Galvanometer O Ammeter or \u2014WWW Unknown Resistance or Resistance Box V : 39 QUESTIONS 1. 2. (a) State Ohm\u2019s Law and explain how it may be determined. (b) Calculate resistance the of a light bulb which carries a current of 5.0 amperes when connected in a 110 volt circuit. (a) In terms of Ohm's Law define: ohm, volt, ampere. (b) What weakness in these definitions is apparent? 3. (a) Explain electrical resistance. (b) List four factors affecting re- sistance. State the effect of each. 4. (a) What is meant by voltage drop across a resistor? (b) What is the effect on the total resistance of connecting a number of resistors (i) in series, (ii) in paral- lel? If two lamps of resistance 50 (c) ohms and 40 ohms are connected in series in a 110 volt circuit calculate (i) the total resistance of the lamps (ii) the current in the circuit (iii) the voltage drop across each lamp. (d) If two lamps of resistance 20 ohms and 40 ohms are connected in parallel in a 110 volt circuit, determine (i) the effective resistance of the circuit. (ii) the current through each lamp. 5. 6. (a) Describe a resistance box and a rheostat. State the purposes of each, (b) What is the effect of removing a plug from a resistance box? (a) Describe two methods for finding the resistance of a conductor. (b) In an electrical circuit, the ammeter reading is 5.5 amperes and the is 77 volts. potential What is the resistance? difference B 1. What is the resistance of a flashlight bulb which carries a current of 0.", "50 ampere when connected in series with a 6.0 volt battery? 2. The resistance of a conductor is 25 ohms and it carries a current of 8.5 amperes. What is the potential difference? 3. How much current does a 36 ohm resistance draw when operated on a 120 volt line? 4. (a) What current flows through a 99.5 ohm resistance connected in series with a 1 2 volt battery having an internal resistance of 0.50 ohm? (b) What is the voltage drop across the resistance and across the battery? (a) What is the electric toaster which passes a current of 5.0 amperes when connected in alio volt circuit? (b) What resistance must be placed resistance of an 5. 293 Chap. 25 MAGNETISM AND ELECTRICITY in series with the toaster to lower the current to 4.0 amperes? ference of 20 volts is applied between points A and B. 6. Eight lamps, each having a resistance of 330 ohms, are connected in parallel in the 110 volt circuit. Calculate (a) the effective resistance of the circuit. (b) the current in each lamp in the circuit. (c) the total current in the circuit. 7. A battery of resistance 0.10 ohm is connected in series with an ammeter of resistance 0.5 ohm. The ammeter registers 1 0 amperes. When a resistance is placed in series with the ammeter the current drops to 4 amperes. (a) Draw a diagram of the circuit described. (b) Find the voltage of the battery. (c) Find the value of the added resistance. 8. The difference of potential between the ends of a certain resistance coil when a current of 0.36 ampere is passed through is 1.2 volts. What resistance must be connected in parallel with the coil so that, with the same total current, the difference of potential will be only 1.0 volt? 6 ohm the total (a) Calculate resistance between A and B when the switch is closed. (b) Find the current in the 6 ohm resistance with switch open. (c) Find the current in the 1 2 ohm resistance with switch closed. 10. A cell with a potential difference of 2 volts and negligible resistance sends a current through two resistances of 6 ohms and 9 ohms connected in parallel. In series with the cell is a third resistance of 2 ohms. Calculate the current in the 6 ohm resis- tance. 11. A current of 4.0 amperes is passed through a resistance of 2.0 ohms in series with a parallel combination of 4.0 ohms and 6.0 ohms. Calculate (a) the current in each of the parallel resistances. 9. Three resistances are connected as shown in the diagram and a potential dif- (b) the potential difference across the whole circuit. 294 CHAPTER 26 CHEMICAL EFFECTS OF ELECTRIC CURRENT reason that one can receive a severe electric shock while standing in water or on a damp floor. V : 40 ELECTROLYSIS others All chemical compounds are composed of two or more elements. The elements are composed of atoms as described in Section V:12. Frequently when atoms unite, electrons are transferred from one kind of atom to the other. The atoms electrons become charged which lose positively and the negatively. These charged particles are called ions. The oppositely charged ions attract each other to form ion-pairs. However, the total number of positive and negative charges has not been altered, so that the ion-pairs formed are neutral. When substances formed in this way are dissolved in water, some of these particles and separate dissociate negative ions. Compounds which form ions in solution are capable of carrying electric current and are called electrolytes. Examples of good electrolytes are solutions of acids bases and salts. Many other substances do not ionize when in solution and so do not conduct electricity. These are classed as non-electrolytes. Sugar, alcohol and distilled water are examples of these. Ordinary water salts which usually contains dissolved make it a weak electrolyte. It is for this positive into the water, decomposition of Shortly after the discovery of current electricity by Galvani and the construction of the first voltaic cells, experimenters like Sir Humphrey Davy of England and Svante Arrhenius of Sweden investigated the decomposition of water by electricity, or as we speak of it now, the electrolysis of water. Davy showed that the in volume of hydrogen produced is double His most striking disthat of oxygen. coveries were the breaking apart by electricity of the alkalis, caustic soda and caustic potash. Arrhenius explained these phenomena by his theory of ionization which assumes that in solutions of electrolytes there is at least a partial dissociation of the dissolved substances into", " It was not until 1834 that separate ions. Faraday introduced the term anode for the positive plate and cathode for the negative plate cell (Fig. 26:1). The negatively charged ions which are attracted to the anode became known as anions, and the positively charged ions which are attracted to the cathode as cations. When compounds dissociate in solution, metals and hydrogen form cations while non-metals radicals become and in an electrolytic chemical most anions. A typical electrolytic cell consists of two conducting-plates or electrodes immersed in an electrolyte. The anode is 295 Chap. 26 MAGNETISM AND ELECTRICITY the negative connected to the positive terminal of a battery or generator and becomes positively charged while the cathode is conterminal and nected to becomes negatively charged. Anions will move to the anode where they give up their surplus electrons, are neutralized, and released as neutral atoms. These electrons flow from the anode, through the connecting wire to the battery terminal. At the same time cations move to the cathode where they receive electrons and became neutral atoms. Electrons move from the negative battery terminal to the cathode to replenish the to the positive terminal of the battery. The electrolyte is decomposed by the electric current. This process is known as electrolysis. V : 41 ELECTROLYSIS OF WATER Water may be decomposed readily in a simple electrolytic cell (Fig. 26:1), or in the HoflFman water voltameter (Fig. 26:2), by the passage of a direct current through it (Chap. 31, Exp. 25). Fig. 26:2 Hoffman Water Voltameter. A small amount of sulphuric acid (about 10% of the total volume) is added to the water to make it a better electrolyte. The anode and cathode are made of platinum which is unaffected by the electrolyte and is a good conductor of electricity. In solution the acid is highly dissociated into positive hydrogen ions (H\"^) and negative sulphate ions (S04\u201c\u201c). H2S04^2H^ + SO4-- The water is slightly dissociated into positive hydrogen ions and negative hydroxyl ions (OH\u201c) : H 2O + OH-. Fig. 26:1 Electrolytic Cell as Used in the Electrolysis of Water. supply. Thus we have electrons moving from the negative battery terminal to the cathode, through the electrolyte by way of the ions to the anode, and back 296. CHEMICAL EFFECTS OF ELECTRIC CURRENT Sec. V:42 When the voltage is applied to the electrodes the following reactions occur: within the electrolyte completes the circuit (Sec. V:40) At the Cathode 1. The positive hydrogen ions are attracted to the negative cathode. 2. Each hydrogen ion is neutralized by gaining an electron from the cathode and becomes a hydrogen atom: + le-\u00bbH\u00b0 3. The hydrogen atoms immediately combine in pairs to form molecules of hydrogen gas which escape as bubbles: H\u00b0 + H\u00b0 ^ H 2. At the Anode 1. The negative sulphate and hydroxyl ions are attracted to the positive anode. 2. The hydroxyl ions are discharged in preference to the sulphate ions. Each hydroxyl ion gives up an electron to the anode to become a neutral hydroxyl group. OH--le-^OH\u00b0 3. Pairs of hydroxyl groups then combine to form water and an atom of oxygen. pairs OH\u00b0 + OH\u00b0 ^ H 2O + 0 \u00b0 4. The oxygen atoms immediately comto form molecules of bine in oxygen \u2019gas which escape as bubbles. 0\u00b0 + 0\u00b0 ^ O2 As the ions are discharged more water dissociates into ions. The sulphuric acid remains in the electrolyte throughout the process and only the water is decomposed, producing two volumes of hydrogen for every one volume of oxygen. The electrons given up by the hydroxyl ions at the anode are propelled through the external circuit by the battery to the cathode where an equivalent number of electrons is taken up by the hydrogen ions. This constitutes the current in the external circuit. The movement of the ions to the oppositely charged electrodes V:42 ELECTROLYSIS OF COPPER SULPHATE SOLUTION If copper sulphate solution is used as the electrolyte and carbon rods as the anode and cathode, copper will be deposited at the cathode soon after the voltage is applied (Chap. 31, Exp. 26). the copper sulphate is highly dissociated into positive copper ions (Cu^^) and negative sulphate ions (SO4--). In solution Cu SO 4 Cu\"\" + SO 4 - - The water is slightly dissociated into positive hydrogen ions (H\"^) and negative hydroxyl ions (OH\u201c). H 2O + OHis applied to When", " the voltage electrodes the following reactions occur: the At the Cathode 1. The copper ions and the hydrogen ions are attracted. 2. The copper ions are discharged in preference to the hydrogen ions. Each copper ion takes two electrons from the cathode and becomes a neutral copper atom. Cu\"\" + 2e ^ Cu\u00b0 3. The copper atoms are deposited on the carbon rod. At the Anode 1. The sulphate ions and the hydroxyl ions are attracted. 2. The reactions are exactly the same as in the electrolysis of water (Sec. V:42), in which oxygen bubbles were formed. Commercial use has been made of such a process in the electroplating industry (Sec. V:45), in which copper is used as the anode and the object to be plated as the cathode. Also, in the purification of various metals, a bar of impure metal 297 Chap. 26 MAGNETISM AND ELECTRICITY is used as the anode, pure metal as the cathode and a solution of a salt of the same metal as the electrolyte. Only the pure metal is transferred from the anode to the cathode where bars of pure metal are obtained. The impurities are left in the electrolytic cell. V:43 LAWS OF ELECTROLYSIS From your experimental work on electrolysis it will be apparent that there is a close connection between the current strength used and the amount of material that is decomposed (Chap. 31, Exp. It will also be readily noted in 27). the previous examples that the greater the time that the current flows, the greater the amounts of water decomposed or copper deposited. Further, it is not difficult to show that a greater mass of silver than of copper will be deposited using the same current for the same length of time. These observations were made as long ago as 1834 by Faraday who formulated Faraday\u2019s Laws of Electrolysis as follows: 1. The amount of chemical change produced (i.e., the amount of any sub- 298 stance deposited) by an electric current is proportional to the quantity of electricity passed (Quantity = Current X Time.) 2. The amounts of different substances deposited by the same quantity of electricity are proportional to their equiva(The equivalent weight lent weights. of a substance is the weight of the substance deposited by 96,400 coulombs.) The number of grams of various elements liberated by one ampere in one electrochemical second equivalent of the element. called the is Electrochemical Equivalents Aluminum Chlorine Copper Hydrogen Magnesium Oxygen Potassium Silver Sodium Zinc 0.000093 0.000368 0.000329 0.0000105 0.000126 0.0000829 0.000405 0.001118 0.000238 0.000339 V ;44 THE COPPER VOLTAMETER The internationally accepted definition of the ampere is based on the deposition of metal at the cathode of an electrolytic cell or voltameter. Silver was selected as the standard, since the relatively large amount of silver deposited in a short time enables a greater degree of accuracy in weighing, and because silver is very resistant to oxidation. The international ampere is defined as that current which, when passed through a solution of silver nitrate in accordance with given specifiat the rate of cations, 0.001118 gm. per second. deposits silver While silver is the best substance for accurate results, a copper voltameter is current satisfactory strength in elementary work (Chap. 31, Exp. 28). In this experiment both anode determining for CHEMICAL EFFECTS OF ELECTRIC CURRENT Sec. V:45 and cathode should be copper and the electrolyte copper sulphate solution. The circuit is adjusted so that a small current flows. The cathode should be thoroughly cleaned with fine emery paper, washed and dried. It is then carefully weighed. Next the is connected, as circuit in Fig. 26:3, and the current is allowed to flow for exactly twenty minutes. The cathode is removed, dipped in alcohol or ether and allowed to dry by evaporaIt is then carefully reweighed. The this tion. current may be calculated as example. in Example Initial weisfht of cathode = 5.52 gm. = 5.71 gm. =0.19 gm. = 20 min. = 1200 sec..\u2018. Weight of copper deposited Time of current flow Electrochemical equivalent of copper = 0.000329 In 1200 sec. weight of copper deposited =0.19 gm..'. In 1 sec. weight of copper deposited = 0.19 = 0.000158 srm. Current flowing when 0.000329 gm. copper is deposited in 1 sec. = 1 amp. Current flowing when 1 gm. copper is deposited in 1 sec. = amp. 000329 1200 Current flowing when.000158 gm. copper is deposited in 1 sec", ". = 0.000158 X = 0.48 amp. 0.000329.'. The current strength is 0.48 amp. V:45 ELECTROPLATING Electroplating is an important industrial process designed to improve the appearance, or resistance increase to corrosion, of various metals. It consists of covering the metal with a thin layer silver, chromium, copper, etc., by of electrolysis. The article to be plated serves as the cathode and a solution of a compound of the metal to be deposited makes up the electrolyte (Chap. 31, Exp. 29). Before plating, the object must be thoroughly cleaned of rust, grease, etc. This is usually accomplished by treatment with abrasives followed by rinsing in strong acid, then strong alkali, then weak acid again, and finally distilled In order to obtain a fine longwater. lasting deposit a small current over a long time interval must be used. Special platings require modifications in the process. For example, silver plating is usually preceded by copper plating to give a suitable base. In copper plating better results are obtained if a little sulphuric acid is added to the electrolyte. Chromium plating directly on another metal leaves small openings like pin-holes in the surface so that corrosion may begin below the chromium layer. To avoid this the base metal is often lightly coated with nickel before the chromium is applied. Electroplating has also been applied to the printing industry in the reproduction of pages of type and illustrations, a process called electrotyping. An impression of the original is made in wax or plastic. This mould is covered with graphite or metallic powder to serve as a conductor 299 Chap. 26 MAGNETISM AND ELECTRICITY Industrial Electroplating. By Electrolysis These Metal Plates are Being Coated with a Thin Layer of Copper. International Nickel Co. of Canada and is then used as the cathode in a copper or nickel electrolytic cell. The metal deposited on it is stripped off, backed with a metal of low melting-point to strengthen it and is then ready for use. This reproduction can be used over and over again and is readily stored for later editions. V:46 THE LEAD-ACID STORAGE BATTERY When the simple voltaic cell (Sec. V;23), and the dry cell (Sec. V:25), become discharged they must be discarded and replaced by new cells. Cells of this type are classified as primary cells. The lead storage cell differs from these in that once its chemical energy 300 has been exhausted, the cell may be restored to its original condition by electrical means. called secondary cells. The familiar lead-acid storage battery used in the electrical system of many automobiles consists of six secondary cells connected in series Such cells are 12 volts. provide approximately to Chemical energy is stored in the active materials of the plates and in the electrolyte when current is passed through during charging. This chemical energy is transformed into electrical energy as the cells are discharged in providing current in a circuit (Chap. 31, Exp. 30). (a) Structure of a Cell Several positive plates, consisting of CHEMICAL EFFECTS OF ELECTRIC CURRENT Sec. V : 46 VENT PLUGS SEALED COVER ELEMENT PROTECTOR SEPARATOR NEGATIVE PLATE PLATE HARD RUBBER CONTAINER SEDIMENT SPACE Exide Automotive Division Fig. 26:4 Structure of a Lead-Acid Storage Battery. lead peroxide forced into a strong grid of lead alloy, are joined to each other by metal strips (Fig. 26:4). Between each of these, and at the ends, are negative electrodes of spongy lead. These are connected by metal strips and kept plates by from touching the positive wood separators. The plates are contained in a moulded, hard rubber case holding the dilute sulphuric acid which serves as the electrolyte. (b) Discharging the Cell As the cell is used to provide the E.M.F. which causes current to flow in a circuit, it becomes discharged. The sulphuric acid combines with the porous materials of the plates and the following chemical reaction occurs: negative positive diluted plates electrolyte lead + lead peroxide + sulphuric acid -\u00bb lead sulphate + water both plates plates electrolyte As the two plates become coated with the same material the potential difference decreases. Simultaneously, the removal of sulphuric acid from the electrolyte and the formation of water causes dilution of the acid in the cell. Thus, discharged cells are indicated by a lowering of the P.D. of the cell and a decrease in the specific gravity of the electrolyte. A fully charged cell should have a P.D. of approximately 2.2 volts, and 301 Chap. 26 MAGNETISM AND ELECTRICITY a specific gravity of 1.275 to 1.300. specific gravity falls to 1.185. (c) Recharging the Cell It is", " ready for recharging when the In order to recharge the cell direct current from a rectifier or some other source must be passed through the cell in the direction opposite to that of discharge. The chemical reactions in the cells are reversed: diluted negative both plates lead sulphate + water electrolyte positive plates plates \u2014\u00bb lead peroxide + lead + sulphuric acid electrolyte Thus, the plates are again made dissimilar and the P.D. increases to 2.2 volts. At the same time water is used up and sulphuric acid is produced by the reaction, so the specific gravity of the electrolyte increases. The charging and discharging reactions may be summarized as: negative positive electrolyte both plates diluted plates plates lead + lead peroxide + s (d) Care of Storage Batteries 1. Keep the battery clean and dry. 2. Do not allow the battery to dis- 3. Keep the charge below specific gravity 1.185. above the level of the wood separators by addition of distilled water. electrolyte 4. Do not allow a fully discharged cell to stand in this condition for more than a few days before charging. (e) Battery Rating The life of a battery is rated in ampere-hours, i.e., amperes X hours. For example, a battery rated at 100 ampere-hours would be capable of de- discharge iric acid ^ lead sulphate + water electrolyte charge livering 5 amperes for 20 hours, or 12.5 amperes for 8 hours, etc. In general, the greater the number of plates per cell, the greater the ampere-hour capacity of the 3. battery. (f) Uses of Storage Batteries Storage cells are frequently used to provide a source of E.M.F. for extended periods of time, as in some radio transmitters and receivers. They are used in the starting circuits of automobiles and aircraft, for auxiliary power supplies for telephones, trains, submarines and ships and for emergency lighting-systems. V : 47 Q U E S noNS A 1. (a) Define: ion, electrolyte, non-elec- trolyte, electrolytic cell, anode, cath- ode, anion, cation, electrolysis. (b) Outline Arrhenius\u2019 theory of ion- ization. 2. Explain how the electrolyte completes In the electrolysis of water describe: (a) the condition of the water and the sulphuric acid before the voltage is applied. (b) what happens in the electrolyte at the moment when the voltage is the circuit in electrolysis. applied. 302 CHEMICAL EFFECTS OF ELECTRIC CURRENT Sec. V:47 (c) the reactions that occur at (i) the cathode (ii) the anode. lysis of water by a current of 2.0 amperes flowing for 4 hours? 4. Explain the electrolysis of copper 4. What current would deposit 0.1 sulphate solution. gm. of silver in 8 hours? 5. Describe how electrolysis may be used in (a) silver plating (b) refining of copper 5. A constant current is passed through 20 copper voltameter a minutes and it is found that 0.500 gm. of copper is deposited on the cathode. exactly for 6. 7. (c) electrotyping. (a) State Faraday\u2019s Laws of Electro- lysis. (b) Define electrochemical equiva- lent. (a) Define an ampere in terms of the deposition of metal at the cathode of an electrolytic cell or voltameter. (b) Why Is acceptable Ohm\u2019s Law? (c) Why was silver selected as the international standard rather than more on definition based than that this copper? (a) Describe the construction of a 8. lead-acid storage cell. (b) What transformations of energy (i) charging, (il) distake place in charging a lead-acid cell? (c) Give word equations describing the reactions as a lead-acid cell is 9. (i) charged, (Ii) discharged. (d) How may the condition of charge of such a storage battery be determined? (a) Define ampere-hour. (b) What determines the amperehour capacity of a battery? (c) What precautions should be obstorage served caring for a in battery? B Calculate the strength of the current. 6. An ammeter connected in series with a silver voltameter reads 1.50 amperes. In 50 minutes the increase in weight of the cathode is 4.947 gm. What Is the error in the ammeter reading? 7. A copper voltameter and a water voltameter are connected in series with a direct current supply. In 25 minutes 0.09 gm. of copper is", " deposited on the cathode of the copper voltameter.. (a) What current is flowing in the circuit? (b) What mass of hydrogen is liberated in the water voltameter? 8. A current of 2 amperes is passed through a copper voltameter. Copper is deposited evenly on the cathode which has an area of 66 sq. cm. Find the thickness of the layer when the current has been flowing for 30 minutes. (Density of copper = 9.0 gm. per c.c.) 9. The anode of a copper voltameter is made of impure copper, and the particles of impurity detach themselves and fall to as cell bottom of the the electrolysis continues. A current of 2.0 amperes flows for 1 hour and 40 minutes. At the start the anode weighs 85.69 gm. and at the end 81.37 gm. Find the mass of copper dissolved from the anode and the mass of impurity released. 10 1. What weight of (a) copper (b) silver will be deposited in 3 hours by a current of 1 ampere? 2. What weight of (a) copper (b) silver will be deposited in 4 hours by a current of 2.0 amperes? 3. What weight of (a) oxygen (b) hydrogen would be liberated in the electro- series connected (a) If four 2 volt lead-acid cells, each of 0.1 ohm internal resistance, a are in conductor of 3.6 ohms resistance, what current would flow? (b) How long continue to flow if the battery has a capacity of 30 ampere-hours? would current with this 303 CHAPTER 27 MAGNETIC EFFECTS OF ELECTRIC CURRENT through a piece of cardboard, as shown in Fig. 27:1. When the switch is closed and iron filings are sprinkled lightly over the cardboard, gentle tapping will cause the filings to arrange themselves in concentric circles with the wire as centre. Thus the pattern of the lines of magnetic force becomes apparent. The direction of these lines of force may be determined Left Hand Rule to DeterFig. 27:2 mine the Direction of the Magnetic Field about a Conductor. by placing a number of small magnetic compasses on the cardboard. The N-poles of the compasses indicate that the lines go in a counter-clockwise direction about the wire (Chap. 31, Exp. 31). It is apparent, then, that if the direction of current is known, the direction of the lines of force about a conductor may be determined as follows: Grasp the conductor with the left hand with the thumb extended in the direction of the electron flow ( \u2014 to + ). The fingers V:48 ELECTROMAGNETIC EFFECT In 1820, Hans Christian Oersted, professor of Physics at the University of Copenhagen, made a discovery that has made possible most of the modern advancements in electrical knowledge. He observed that a wire carrying an electric current deflected a compass-needle. Further that whenever a current was passing through a conductor a magnetic field was set up around it. experiments established It is possible to investigate this magconductor passing field by a netic Fig. 27:1 Magnetic Field about a Conductor. 304 MAGNETIC EFFECTS OF ELECTRIC CURRENT Sec. V:49 When considering the magnetic field about a helix, it is convenient to represent it as in Fig. 27:5. The helix is shown in cross-section with the front half removed. The electron flow is up the back section of the turns and down the front. In the diagram, electron flow is indicated by a dot (.) representing the head of an arrow and a cross ( + ) representing the Fig. 27:6a shows the lines of magtail. netic force around a few turns of wire. The Left-Hand Rule will confirm the will now point in the direction in which the lines of magnetic force encircle the is This called statement the wire. Left-Hand Rule (Fig. 27:2). Conversely, this rule may be used to determine the direction of the electron flow if the direction of the lines of force is known. V : 49 MAGNETIC FIELD ABOUT A HELIX A single turn of wire (Fig. 27:3) is called a loop or coil. A series of such Fig. 27:3 Magnetic Field about a Loop Carrying Electric Current. loops in a wire (Fig. 27:4) is referred to as a helix or solenoid. If a conductor is formed into a single loop, the lines of Fig. 27:4 A Helix or Solenoid. magnetic force which surround the wire will pass through the centre of the loop as shown. Note that the magnetic field inside the loop will be more dense than outside since the lines of force are crowded into a smaller area. The Lines of Magnetic Force Fig. 27:6 Around the Individual Turns of", " Wire in a Helix as Shown at (a). Combine to Form the Resultant Field Shown at (b). Fig. 27:5 Cross-sectional View of a Helix with the Front Half Removed. The Direction of Electron Flow is Indicated. direction of these lines. Between the turns the forces are in opposite directions and so tend to cancel each other out. In the centre of the coils the forces act in the same direction and are very concentrated. On the outside of the helix the forces also reinforce each other to give us Fig. 27:6b. Note the similarity between this field represented in the 305 Chap. 27 MAGNETISM AND ELECTRICITY pattern and that formed by a bar magnet (Sec. V:4). The existence of field may be readily proved by inserting a helix, carrying current, in a slit in a piece of cardboard Iron filings 27:7). (Fig. will this Apparatus Used to DemonFig. 27:7 strate the Field of Force about a Helix Carrying Electric Current. the pattern of the magnetic indicate field (Chap. 31, Exp. 32). The polarity of the field may be determined by testing with a compass-needle. From this information a simple rule may be confirmed: Grasp the helix in the left hand so that the fingers extend in the direction the electrons are flowing around the turns. The extended thumb will then point towards the N-pole of the helix This may be called the (Fig. 27:8). Left-Hand Rule for the Helix or, to avoid confusion with other rules, simply the Helix Rule. V:50 ELECTROMAGNETS If a bar of iron is placed in the centre of a coil of wire through which a cur- 306 rent is flowing, the iron will become magnetized. The elementary magnets of the iron become aligned in such a direction as to reinforce the magnetic field of the coil. Thus the iron core greatly strengthens the magnetic field. Such an arrangement is known as an electromagnet. Soft iron makes the most satisfactory core, since a temporary magnet is produced. When the current in the is stopped the magnetism of the coil core is lost in a short time. An increase in the number of turns in the coil or an increase in the strength of the electric current flowing through it will result in a more powerful electromagnet (Chap. 31, Exp. 33). Electromagnets have been designed in a variety of shapes to meet many needs. The iron clad type is widely used where strong magnetic fields are desirable as Iron Core * ^ 1 \\ Electron Flow 1 ) ) ) ) ) )..--1 1 I. ^ ^ 2 r 1 Fig. 27:9 Iron-Clad Electromagnet. for a lifting magnet for scrap iron or in radio loudspeakers. The iron core not only passes through the centre, but almost completely surrounds it as shown in Fig. 27:9. The horseshoe electromagnet (Fig. 27:10) is used in such devices as earphones, telephone receivers, electric bells and electric buzzers. The electric bell (Fig. 27:11) consists of a gong, a horseshoe electromagnet, soft iron armature and contact screw arranged in a circuit as shown. When the switch is closed MAGNETIC EFFECTS OF ELECTRIC CURRENT Sec. V:50 loses The electromagnet broken. its magnetism and the armature is pulled back to the contact point by the spring. This completes the circuit and the action Thus, the hammer will is vibrate against the gong as long as the switch is depressed and a continuous ringing is produced. repeated. Electromagnets are being used more and more to protect circuits carrying large currents. Magnetic circuit breakers employ a switch which is opened by the magnet if current becomes too strong. A simple circuit is shown in Fig. 27:12. The student should trace the the Fig. 27:10 The Horseshoe Electromagnet. current flows in the winding ot the electromagnet, causing the armature to be attracted. As the armature moves away from the contact screw the circuit is Magnetic Circuit Breaker. Fig. 27:12 If current becomes too great, core is pulled upwards by magnetism of the coil, tripping the catch held by Si. pulls up the knife switch Spring S 2 breaking the circuit. path of the electrons and explain the action of the circuit breaker. The automobile generator cut-out relay is a magnetic switch which opens and closes the circuit between the genbattery erator (Sec. V:63) and the 307 Chap. 27 MAGNETISM AND ELECTRICITY An Electromagnet Used for Lifting Scrap Iron. steel Co. of Canada Ltd. (Sec. V:46). It serves to connect the generator to the battery when the generator is operating at charging speeds and to open the circuit when the generator stops or slows down to prevent back the through", " the generator. A simple circuit is shown in Fig. 27 : 13. discharging battery from The cut-out relay has two windings assembled on the same soft iron core. The shunt winding (dotted) is connected in parallel with the generator so that, when the generator starts to operate, the potential difference created causes a current to flow through the winding. This 308 produces a magnetic field strong enough to pull the armature toward the core, and the circuit is closed as the contact points meet. Current then flows from the and back generator through \u201cground\u201d, passing through the series winding in such a direction as to add armature down. the magnetism holding battery the the to to When the generator slows down, the voltage produced by the generator becomes less than the voltage. Thus, a current begins to flow in the reverse direction through the series winding but continues to flow in the same battery MAGNETIC EFFECTS OF ELECTRIC CURRENT Sec. V:52 the shunt winding. The direction in magnetic fields of the two windings are now in opposite directions and so tend to cancel each other. The armature is pulled upward by spring tension to open Fig. 27:14 A Simple Galvanoscope. parison of current strengths may be made. For weak currents the coils with many turns will be used and for strong currents the coils with few turns are satisfactory (Sec. V:50). This instrument has been largely replaced by more sensitive and more accurate current-measuring devices. Fig. 27:13 Automobile Generator Cut-Out Relay. V:52 THE MOTOR PRINCIPLE the contact points and the broken between the is battery and the circuit generator. V;51 THE GALVANOSCOPE The galvanoscope is a simple device used to determine the directions and comparative strengths of electric curIt consists of several coils having rents. varying numbers of turns of wire (Fig. 27:14). The current is made to pass over a compass-needle in one direction and under the needle in the opposite In this way the magnetic direction. effect of the current is magnified and the compass-needle will be deflected. By Left-Hand applying (Sec. V:48), the direction of current may be determined. By noting the amount of deflection of the needle a rough com- Rule the In experiment 34, chapter 31, a conductor AB was suspended between the poles of a strong permanent magnet as shown in Fig. 27:15. When a heavy current was sent through the conductor from a storage battery or similar source, the conductor was pushed aside. When the current was reversed, the conductor was thrust in the opposite direction. Thus we see that a conductor carrying current in a magnetic field is acted upon by forces which cause the conductor to move. The magnetic field between the poles of the permanent magnet may be represented as shown in Fig. 27: 16(a). The current in the conductor sets up a magnetic field as in Fig. 27: 16(b). A crosssection of the conductor is shown in Fig. 27: 16(c). When the conductor is in the permanent magnetic field, Fig. as in 309 Chap. 27 MAGNETISM AND ELECTRICITY 27: 16(d), the lines of force from both the conductor and the permanent magnet are in the same direction on the force are like stretched elastic bands (Sec. V:4), they try to straighten themselves and consequently, push the conductor to the left. Similarly, if the electron flow in the conductor is reversed, the motion will be in the opposite direction. If now conductor AB (Fig. 27:15), is replaced by a single loop ABCD (Fig. 27:17), section AB will be moved toward the left by the combined forces of the permanent magnetic field and the electromagnetic field, while section CD will be thrust toward the right. Thus, the loop will set itself so that the plane of the coil will become perpendicular to the direction of the lines of force of the permanent magnet. This behaviour may be more simply explained if we consider of a Conductor Fig. 27:15 Carrying Electric Current in a Mag- Action netic Field. right and in opposite directions on the left. Thus the lines tend to be crowded on the right side, while on the left they cancel each other. Since lines of magnetic Fig. 27:17 Action of a Loop CarryElectric Current in a Magnetic ing Field. the loop to be a form of helix. The Helix Rule (Sec. V:49), indicates that the loop, when carrying current, will have an N-pole and an S-pole. Following the Law of Magnetism (Sec. V:2), the N-pole of the loop will be attracted to the S-pole of the permanent magnet while the S-pole of the loop will face the N-pole of the magnet. V:53 THE D'ARSONVAL GALVANOMETER Fig. 27:16 The Motor Principle. The D\u2019", "Arsonval Galvanometer is a 310 MAGNETIC EFFECTS OF ELECTRIC CURRENT Sec. V;54 sensitive instrument designed to detect the presence of small direct currents of It also serves to determine electricity. Fig. 27:18 A Simple Galvanometer. the direction and compare the strengths of such currents. The simplest form of this type of gal- vanometer is shown in Fig. 27:18. It consists of a permanent horseshoe magnet and a coil of wire which is free to swing between the poles of the magnet. A stationary soft iron core is mounted inside the coil to concentrate the lines of force. Springs attached to the coil serve to conduct the electrons into and out of the coil and also to bring the coil to rest in a position where the plane of the coil faces away from the poles of the magnet, as shown in the diagram. As current passes through the coil it will turn, so that the N-pole of the coil is toward the S-pole of the magnet, and is toward the the S-pole of the N-pole of the magnet (Sec. V:52). This movement is restricted by the springs. The amount of turning is indicated by the attached pointer moving across a calibrated scale. The stronger the electric current through the coil, the greater the magnetic field produced, and so the coil higher the reading on the scale. This moving-coil type of meter is the basis of most standard electrical measuring instruments in common use. V : 54 AMMETERS AND VOLTMETERS Ammeters and voltmeters are instruments designed to measure electric current and potential difference respectively. They are essentially galvanometers of the moving-coil type (Sec. V:53), modified by the addition of suitable resisTheir scales are calibrated to tances. read directly in amperes or volts. (a) Ammeters The wire winding on the moving coil of an ammeter must be very fine and light to allow the necessary freedom of movement. Such a conductor cannot carry a large current without undue (Sec. V:71), and consequent heating melting of the wire. The movable coil of such an instrument is rarely allowed to carry more than 0.05 amperes. If the ammeter is to be used to mea- Fig. 27:19 Ammeter Connected in Series. 311 ) Chap. 27 MAGNETISM AND ELECTRICITY sure higher currents than this the current must be divided so that no more than 0.05 amperes will flow through the moving coil, and the rest of the current will be carried around the coil through a shunt connected in parallel with the instrument (Fig. 27:19). Such a shunt must have a very low resistance and be accurate over a wide range of temperatures. To measure the current in a circuit the ammeter must be connected in series. As a result, all the current, or a known fraction of the total current, will pass through the instrument coil. In order to avoid affecting the total current in the circuit, the resistance of the ammeter must be very low. If the ammeter were accidentally connected across the circuit, a high current would flow through it and the instrument would be \u201cburned out\u201d. To avoid this a fuse (Sec. V:72) should be connected in series with the ammeter. Example A galvanometer has a resistance of 5 ohms and gives a full-scale deflection with a current of 0.05 amperes. What value of shunt resistance must be used to convert it to an ammeter reading up to 5 amperes? Current to be carried by ammeter = 5 amp. Current to be carried by galvanometer = 0.05 amp. Current to be carried by shunt = 5 \u2014 0.05 \u2014 4.95 amp. Resistance of galvanometer = 5 ohms. Potential difference across galvanometer \u2014 0.05 X 5 = 0.25 volts {V = IR) Potential difference across shunt = 0.25 volts (parallel connection) Resistance of shunt to be used = ^ 4.95 0.05 ohms {R \u2014 \u2014 I Fig. 27:20 Voltmeter Connected in Parallel. 312 MAGNETIC EFFECTS OF ELECTRIC CURRENT Sec. V:55 (b) Voltmeters potential Since a voltmeter is used to measure difference between two the points in a circuit, it must be connected circuit between the in these points. Therefore, a voltmeter must have a very high resistance to avoid parallel with drawing a large current. As a result the total current in the circuit will not be affected significantly. A galvanometer may be converted to a voltmeter by the addition of a high resistance in series with the moving coil (Fig. 27:20). Example A galvanometer has a resistance of 5 ohms and gives a full-scale deflection with a current of 0.05 amperes. What value of series resistance must be used to convert it to a voltmeter", " reading up to 5 volts? Potential difference across voltmeter = 5 volts Current through voltmeter = 0.05 amp Resistance of voltmeter = 5 =100 ohms (R 0.05 Resistance of galvanometer = 5 ohms Series resistance to be added = 100\u2014 5 = 95 ohms. V:55 THE ELECTRIC MOTOR The electric motor is studied in experiment 35, chapter 31, using a St. Louis motor (Fig. 27:21). In order to understand its consider an electromagnet mounted on an axis so that it is free to rotate between the poles of a magnet (Fig. 27:22). If a current is passed through the electromagnet so as to produce an S-pole at the top, as operation shown, the S-pole will be attracted by the N-pole of the stationary field. Similarly the N-pole of the electromagnet will be attracted by the S-pole of the These forces cause stationary magnet. the electromagnet, which serves as the motor armature, to rotate. The momentum of this armature causes it to rotate so that the S-pole of the armature passes slightly beyond the N-pole of the magnet. Commutator Brushes Commutator Plates Armature \u2014 Field Magnet Fig. 27:21 St. Louis Motor. Central Scientific Co. of Canada Ltd. 313 Chap. 27 MAGNETISM AND ELECTRICITY ments of the commutator also rotate. Connection to the commutator segments is made by a pair of brushes, one leading the electrons in ( \u2014 ) and the other out ( + ). At exactly the correct instant, segment R makes contact with brush Bi causing the electrons to flow through the armature. One-half turn later, segment R makes contact with brush B 2 as segment S contacts brush Bi, etc. Thus the commutator serves to reverse the direction of electron flow with every 180\u00b0 rotation of the armature. The magnetic field in which the armature turns is usually provided by an Fig. 27:24 Electric Motor. The Commutator Reverses the Direction of Electron Flow Every 180\u00b0. electromagnet, since such a magnet may be made more powerful than a permanIts windings are called the ent type. field coils to difTerentiate them from the armature coil. The field coils may be in series or in parallel with the armature coil, depending on the motor characteristics desired. A more advanced text should be consulted if greater detail is required. The motor described here operates on direct current only, and thus is seldom used. However, it serves to illustrate the principle electric operating on of operation of including motors, those all alternating current. S-poIe Electric Motor, The ArFig. 27:22 mature is Free to Rotate Between the Poles of the Magnet. If the current in the armature windings could be reversed at this instant, the polarity of the armature would change and repulsion and attraction would carry the armature through another 180\u00b0 (Fig. Successive reversals of current 27:23). Electric Motor. The Poles Fig. 27:23 Reverse in this Position so Armature Continues to Rotate. and consequent changes of armature polarity cause the rotation to continue. Such a reversal of current can be caused of a commutator (Fig. by the 27:24). The armature wires are attached to two segments of good conducting material which are insulated from each other. As the armature rotates the seg- use 314 MAGNETIC EFFECTS OF ELECTRIC CURRENT Sec. V:56 V : 56 QUESTIONS A (b) State the Left-Hand Rule. 1. 2. (a) Draw diagrams showing the lines of magnetic force for (i) a straight wire (ii) a single turn of wire when carrying an electric current. Indicate clearly on each diagram the direction flow and of the electron of the magnetic field. (a) On a diagram of a solenoid indicate electron direction flow and the magnetic field rounding it. the sur- of the the (c) Indicate current in wires (1), (2) and (3) in which the direction of the lines of direction of force is shown. (d) Indicate the direction of the lines (5) and of force around wires (4), (6) carrying a current. (b) State the Helix Rule. (c) On the following diagrams mark the direction of the current and the polarity of the solenoids. 315 Chap. 27 MAGNETISM AND ELECTRICITY 3. (a) Why is soft iron, rather than steel, used as the core in most electromagnets? (b) How may the strength electromagnet be increased? of an 8. (d) How may a galvanometer be (i) an modified to convert it into ammeter (ii) a voltmeter? Make a labelled diagram of a simple D.C. motor and describe its opera", "- 4. (a) Make a fully labelled diagram of an electric bell connected in a tion. B circuit. (b) With the aid of this diagram describe the operation of an elec- tric bell. 5. 6. (a) What is the purpose of an automobile generator cut-out relay? (b) Describe its operation. (a) What is the purpose of a galvanoscope? (b) Describe and explain the operation of a galvanoscope. 7. (a) State the motor principle. (b) Make a labelled diagram of a simple moving-coil galvanometer and explain how its operation employs the motor principle. (c) What precaution should be obobserved when a galvanometer is connected in a circuit? 1. A galvanometer has a resistance of 1 0 ohms and gives a full-scale deflection with a current of 0.04 ampere. What value of shunt resistance must be used to convert it to an ammeter reading up to (a) 5 amperes (b) 50 amperes (c) 500 amperes. 2. A galvanometer has a resistance of 2 ohms. It gives a full-scale deflection with a current of 0.01 ampere. What value of shunt resistance must be used to convert it to an ammeter reading to 10 amperes? 3. What value of series resistance must be added to convert the galvanometer of question 1 to a voltmeter reading up to (c) 500 volts? (a) 5 volts 4. What value of resistance must be connected in series with the galvanometer (b) 50 volts of question 2 to convert it to a voltmeter reading up to 100 volts? 316 CHAPTER 28 ELECTROMAGNETIC INDUCTION V:57 THE STORY OF FARADAY Michael Faraday was one the greatest experimental scientists the world has ever known. He was born in England in 1791 and, after a very ordinary education, became an apprentice to a book- of pressed the great chemist. Sir Humphrey Davy, so much that at the age of twentyone he became his assistant at the Royal Institution. Within twelve years he was made a director of the Institution. From the time that Oersted demonstrated the magnetic effect of an electric current Faraday dreamed of the converse effect, the production of electricity from magnetism. After many unsuccessful experiments he finally produced a momentary current in a coil of wire wound on an iron ring by starting and stopping a current in another coil wound on the same ring (Fig. 28:1). With this small beginning Faraday began a series of experiments on electro- University of Toronto Michael Faraday binder. He read every scientific book which passed through his hands and quickly became well educated. He im- Fig. 28:1 Electricity from Magnetism. to magnetism that eventually led his invention of the dynamo. The services of Faraday were in great demand by firms but he refused great industrial wealth in order to devote his talents to scientific discovery. Honours of all kinds were showered upon him but he remained the humble scientist and experimenter to the end, devoting his time and 317 Chap. 28 MAGNETISM AND ELECTRICITY energy to discovery. When all else is forgotten, he will be remembered as the father of the electric current that serves our homes, offices, factories and communities. Some of his other discoveries, such as the methods of liquefying gases, making optical glass, of benzine, and the like, are not remembered so well even though of great importance. the isolation Although he was never one to seek publicity his name has become immortal because of his contributions to humanity through physics and chemistry. V:58 CAUSE OF AN INDUCED CURRENT It is a simple matter to demonstrate, as Faraday did, that, just as electrons in motion can set up a magnetic field, a magnetic field in motion can cause a Three such experiflow of electrons. ments are described in chapter 31, experiment 36. is i.e., no coil, (Fig. current If a galvanometer is connected to the terminals of a solenoid and a bar magnet or an electromagnet is held above the solenoid 28:2), obtained. When the magnet is thrust into the centre of the the strength of the magnetic field is increased, the galvanometer needle is momentarily deflected from its centre position. The galvanometer needle returns to the zero position when the magnet remains stationary in the solenoid, i.e., while the strength of the magnetic field remains constant. When the magnet is withdrawn quickly from the solenoid, the strength of the magnetic field decreases, and the galvanometer needle is again deflected, but this time in the opposite direction. It is apparent, then, that a flow of electrons originated each time the magnet was moved. This is called induced current, and the effect caused by the changing magnetic field around a conductor is called electromagnetic induction. 318 For an induced current to", " flow in the circuit there must, of course, be an electromotive in (Sec. V:26). This induced E.M.F. is caused by the changing magnetic field about the conductor. circuit force the Similarly, in the circuit used by Faraday (Fig. 28:1), when the switch in circuit containing the battery and the primary coil (C) was opened, the electron flow was stopped. Thus, the magnetic field about the primary coil deThis changing creased rapidly to zero. magnetic field the the secondary coil in the circuit and an E.M.F. was induced. When the switch was closed, current flowed in the primary circuit and as the magnetic field increased from zero to a maximum, an E.M.F. was induced again in the secondary circuit, causing a current to flow in the turns cut (5') of opposite direction. From these experiments we may conclude that an induced E.M.F. is caused by the changing magnetic field about the conductor. V:59 MAGNITUDE OF INDUCED E.M.F. In experiment 37, chapter 31, it will be observed that, if the magnet is slowly plunged into and withdrawn from the solenoid, there will be slight deflection of the galvanometer needle. If the action is repeated more and more rapidly, it will be seen that the amount of deflection is directly proportional to the speed of movement of the magnet. Thus, we may conclude that the strength of the induced current is proportional to the speed at which the magnet is moved or the E.M.F. induced in a circuit is proportional to the rate of change of the magnetic field cutting the conductor. If a magnet is thrust first into a coil of few turns and then, at the same speed, into a coil of many turns, and the deflection of a galvanometer in the ELECTROMAGNETIC INDUCTION Sec. V: 60 circuit is noted, it will be found that a stronger induced current flows in the coil having the larger number of turns. Therefore, it is evident that the E.M.F. induced in a circuit is proportional to the number of turns of the conductor cut by the varying magnetic field. N-pole at the end of the solenoid where the magnet enters. This N-pole repels the N-pole of the magnet and opposes the motion of the magnet. As the N-pole of When a powerful electromagnet with an iron core is substituted for the bar magnet in the experiment described above it will be observed that, with the rate of movement and the number of turns in the coil remaining constant, the deflection of the galvanometer needle is greatly increased. It is apparent, then, that the E.M.F. induced in a circuit is the proportional the changing magnetic field. strength of to V:60 DIRECTION OF AN INDUCED CURRENT The direction of an induced current is studied in experiment 38, chapter 31. Here, the direction of the current is noted when (a) the N-pole of a magnet is thrust into the coil; (b) the N-pole is Induced Current as the Bar Fig. 28:3 Magnet is Withdrawn from the Solenoid. the magnet is withdrawn from the coil (Fig. 28:3) the induced current is in such a direction as to produce an S-pole at the same end of the solenoid. The attraction of this S-pole of the solenoid for the N-pole of the magnet tends to oppose the withdrawal of the magnet. Similar results may be found when the electromagnet is used. These eflfects are summarized in Lenz\u2019s Law which states: The direction of an induced current is such that the magnetic field which it produces opposes the motion or change that induces the current. Induced Current as the Bar Fig. 28:2 Magnet is Plunged into the Solenoid. withdrawn; (c) the S-pole is inserted; (d) the S-pole is withdrawn. As the Npole enters the solenoid (Fig. 28:2) a current is induced which produces an Because of their importance, primary and secondary currents require special It was noted (Sec. V:58), treatment. that secondary currents are in opposite directions when the primary is opened compared to when it is and demonstrable secondary currents are in the same direction when the primary circuit is opened (at the \u201cbreak\u201d) and in the opposite direction when it the \u201cmake\u201d). The idea of opposing mag- primary is closed. closed that the (at It is 319 Chap. 28 MAGNETISM AND ELECTRICITY As it nears the position at right angles to the first, much cutting of the lines of force netic fields is carried out here in that the induced current, while flowing in the same direction as the primary, attempts to maintain the magnetic field that is Similarly, at the completion collapsing. of the primary circuit, the", " induced current is attempting to oppose a magnetic field which is being built up. It is further to be noted that the induced E.M.F. is greater at the \u201cbreak\u201d because the induced current tends to strengthen the field of force of the primary. V:61 THE EARTH INDUCTOR The earth inductor is a coil with a diameter of about 18 in., made of several hundred turns of fine insulated copper wire. When the two end leads of the coil are connected to a galvanometer and the rotated through 360\u00b0, the galvanometer needle will be seen to swing first to one side and then to the other (Chap. 31, Exp. 39A). This action continues as long as the inductor is rotated. inductor rapidly is In Fig. 28:4, as the coil rotates clockwise, A downward and B upward, little cutting of the lines of force occurs and a small E.M.F. will be developed at first. occurs, and the E.M.F. produced will be larger. Throughout this 90\u00b0 of rotation the induced E.M.F. has been built up from zero. It will be obvious that the E.M.F. will gradually decrease to zero in the next 90\u00b0. Throughout this rotation of 180\u00b0, the E.M.F. has gone from zero to a maximum and back to zero again (Fig. 28:5). The direction of the current has been constant. All this time DIRECTION OF CURRENT 320 ELECTROMAGNETIC INDUCTION Sec. V: 62 Fig. 28:6 Simple Alternating Current (A.C.) Generator. the magnetic field of the induced current has been opposing the motion causing it (Lenz\u2019s Law). Rotation through the next 180\u00b0 in which A will move up and B down will result in the same cycle of events as before except that the direction of the current will be reversed. The energy required to overcome the opposition of the fields of force is transformed into electrical energy. While the earth inductor itself is of no practical value it does permit a preliminary study of the generation of electricity by induction. This simple knowledge has made possible the development of all the many forms of electric generators which supply power for modern living. V:62 THE ALTERNATING-CURRENT GENERATOR As indicated in section V:61, mechanical energy may be used to produce electrical energy. A device used for this purpose is called a generator or dynamo. A simple generator (Fig. 28:6) consists of a coil of wire called an armature which is made to rotate between the poles of a magnet (Chap. 31, Exp. 39B). To facilitate connections to the rotating armature, slip rings {Ai, A 2 ) are connected to each end of the coil and rotate with it. Contact is made by metal or carbon brushes {Bi, B2 ) which connect the coil with the external circuit containing the lamp (L). As the coil rotates in the direction shown in the diagram, it cuts the magnetic field between the pole-pieces {S, N) of the magnet and a current is induced In position (1) the in the conductor. sides ab and cd of the conductor are momentarily moving parallel to the lines of magnetic force so there is no cutting and no induced E.M.F. Therefore no electrons will flow. In position (2) the coil is cutting an increasing number of lines of force so the induced E.M.F. is Electrons flow in the direcincreasing. tion abed so as to produce an N-pole and an S-pole in the coil which will oppose its motion (Lenz\u2019s Law). Electrons are now leaving the slip ring B, passing through the lamp, and returning to A. In position (3) the conductor is cutting the maximum number of lines of force and the induced E.M.F. and induced electron flow are at a maximum. As the 321 Chap. 28 MAGNETISM AND ELECTRICITY (4) coil reaches position the electron flow is decreasing and, in position (5) as the motion is parallel to the lines of force once more, it ceases entirely. In the electron flow builds up again (6) but this time in the direction dcba so that electrons leave slip ring A, pass through the lamp, and return through B. The current again reaches a maximum (8), and falls to in zero in position (9). Thus, during one complete rotation of the coil, the induced current starts at zero, increases to a maximum, falls to zero, increases to a maximum in the opposite direction, and again falls These (7), decreases in zero. to changes are summarized in Fig. 28:7. A current with such characteristics is said to be an alternating current (A.C.) and one cycle has been described above. Alternating current is provided by all hydro-electric installations. Those who have lived in, or visited, 25-cycle power", " areas will have noticed the rapid flickering of the lights. This is caused by the reversal of current flow occurring 25 times per second. When the number of reversals is increased to 60 times per second as in 60-cycle power, this flickering is not noticeable. Very few areas 25-cycle of power. the world use this still 322 ELECTROMAGNETIC INDUCTION Sec. V: 64 V:63 THE DIRECT-CURRENT GENERATOR When electrons flow continuously in one direction it is said to be a direct current. Such current is required for special installations such as electroplatIn order to obtain direct current ing. from an A.C. generator, a commutator must be used in place of the slip rings (Chap. 31, Exp. 39B). For a single-coil generator as described in section V:62, the commutator consists of a collectingring made of two segments or com- Brush trons are about to reverse direction. Thus the commutator bars change brushes just as the electron flow reverses so that the current flowing through the external circuit is always in the same direction (Fig. 28:9). The current from a single turn of wire is clearly a pulsating current. To eliminate this the armature of a commercial generator consists of many such coils each with its commutator arranged around the shaft (Fig. 28:10). Such an arrangement will give a more continuous flow of current, since some of the coils will always be cutting the magnetic lines of force. Such generators are used in autoremote mobiles, places, direct-current generating systems in schools, hospitals, etc., and for electroplating and electro-purification of metals. electrical systems in Segments Fig. 28:10 Armature of a Commer- cial D.C. Generator mutator bars insulated from each other. Each bar is connected to a terminal of the coil (Fig. 28:8). The brushes are so placed that they rest on the insulating material between the bars at the instant when the elec- V : 64 TRANSFORMERS Transformers are used to change the voltage in an A.C. circuit as required. A transformer consists of two separate coils of insulated wire, the primary and secondary coils, wound on the same soft iron core (Fig. 28:11). Alternating currents from a generator flow through the windings of the input or primary coil. As the current changes in intensity and direction, the magnetic field surrounding the coil changes as well. Thus lines 323 \u2014 Chap. 28 MAGNETISM AND ELECTRICITY of magnetic force of changing strength and direction are made to cut the turns of the output or secondary coil. An alternating E.M.F. is induced in the secondary coil. Since the magnitude of Fig. 28:11 Step-up Transformer. the induced E.M.F. is proportional to the number of turns of the conductor cut by the changing magnetic field, the strength of the induced voltage may be varied by using a different number of the primary and secondary turns windings. For any transformer in Output Voltage _ No. of turns on secondary coil Input Voltage No. of turns on primary coil A step-up transformer has more turns on the secondary coil than on the primary coil and so serves to increase or \u201cstep up\u201d the voltage. A step-down transformer has more turns on the primary coil than on the secondary and so serves to decrease or \u201cstep down\u201d the voltage. In section V : 75, it will be established that the power of an electric current = the number of volts X the number of Obviously the power of the amperes. secondary current must equal the power of the primary current (except for a very small decrease owing to conversion to heat). Thus, if a step-up transformer causes a voltage rise, there must be a corresponding fall in the current. Students will be familiar with the small step-down transformers used to operate electric trains, electric bells and electric janitors around the home. Some other uses of the transformer will be described in the sections that follow. Example A transformer is to be used to provide 6 volts to operate an electric If the primary coil has 2000 turns, how door-bell on a 120 volt circuit. many turns should be on the secondary coil? Output voltage ~ 6 volts Input voltage \u2014 120 volts Number of turns on primary = 2000 Number of turns on secondary = x Output voltage Number of turns on secondary coil Input voltage Number of turns on primary coil 6 X 120 2000 6.\u2022.x=r\u2014X 2000 =100.\u2019. there should be 100 turns on the secondary coil. 324 ELECTROMAGNETIC INDUCTION Sec. V:65 V:65 THE TELEPHONE The first telephone was invented by Alexander Graham Bell, a Scottish emigrant to the United States, 1875, and first used between Brantford and Bell\u2019s main interest was Paris, Ontario. he had deaf-mutes, but in many other interests and", " the telephone was one of the many products of his fertile mind. From his simple instru- teaching in Fig. 28:12 Telephone Transmitter. vibrate. When the vibrations are transmitted to the carbon granules, they are compressed to a vaiying degree. This, in turn, causes their electrical resistance to vary and fluctuations in the current strength are produced at the same freThis fluctuatquencies as the sounds. ing current is transmitted to the re- ceiver. into sound again. As we know so well, the purpose of the receiver is to convert these fluctuatThe ing currents receiver (Fig. 28:13) consists of a permanent magnet (M) having an electromagnet at the poles. As the fluctuating current caused by the transmitter passes through the coils {CC) of the electro- Diaphragm (D) Electromagnet (C) Star Newspaper Service Alexander Graham Bell. ment has been developed the worldthat we wide communication system know to-day. The essential parts are the transmitter and the receiver. The transmitter (Fig. 28:12) consists of a capsule containing carbon granules (C) connected the to diaphragm (D) on one side and to a carbon block (B) on the other. A current from a battery passes through the granules. metal called sheet thin a It responds to sound waves as follows: the condensations and rarefactions falling on the diaphragm cause it to Fig. 28:13 Telephone Receiver. 325 Chap. 28 MAGNETISM AND ELECTRICITY Transmitter Ground Fig. 28:14 Simple Telephone Circuit. magnet, the magnetic field is caused to change at the same frequency as the Thus the steel original sound waves. diaphragm {D) is caused to vibrate at this frequency and sound waves almost identical to those at the transmitter are reproduced at the receiver. If the transmitter and receiver circuits are directly linked, only weak signals This is because of the will be heard. very small changes in resistance in the carbon granules of the transmitter as compared to the total resistance of the circuit. The slight variations in current do not cause enough variation in the receiver electromagnet to give satisfactory reception. To overcome this, the transmitter is connected to the primary winding of a step-up transformer. The secondary coil is connected in the line circuit through which magnified effects are carried to both receivers as shown in Fig. 28:14. Refinements such as the bell signal, automatic dialing, relays, ultra high-frequency beam wireless, sheathed cables, coaxial cables, and others are employed in the modern system. However, the above still remains the basis for all telephone communication. V:66 SELF-INDUCTANCE It has been observed that when an is broken, as when a circuit electric in a house-lighting switch turned off, a spark or \u201celectric arc\u201d freThis is caused by selfquently occurs. inductance which is explained below. circuit is r RINGER CO IL j DIAL MECHANISM Bell Telephone Co. Canada The Dial Telephone. 326 When a coil of wire of many turns, wound about an iron core, is connected to a battery, the current that begins to flow causes a magnetic field about the wire. As an increasing number of lines of force cut the conductor, an E.M.F. ELECTROMAGNETIC INDUCTION Sec. V:67 will be induced in the coil itself causing a current to flow in a direction opposite to the flow of current in the coil (Lenz\u2019s Law). Thus the current will build uo more slowly than might be expected in such a circuit (Chap. 31, Exp. 40). When the connection to the battery is broken, the decreasing current causes a decreasing magnetic field about the wires. An E.M.F. causing a current that flows in the same direction as the main current will result. Since this is of relatively high voltage, a definite spark (arc) will be observed. In a circuit, an induced current that opposes any change in the flow of current through it caused by self-inductance is called a self-induced current. It is so similar to the inertia of matter that it has been termed electrois an important magnetic inertia. consideration in all alternating-current equipment design. It V : 67 THE INDUCTION COIL The induction coil is very like the transformer in construction, as both consist of primary and secondary coils of different numbers of turns of wire wound about a common soft iron core. Whereas the transformer can utilize the changing magnetic field caused by alternating current to step up the voltage, the same operation is impossible with a direct current for lack of changing magnetic field. However, it is possible to obtain high voltage from a low voltage direct current by the use of an induction coil. is accomplished by continually This interrupting the primary current, with the aid of a circuit breaker similar to (Sec. V:50). that in the electric bell", " This results in a changing magnetic field around the primary which induces a high voltage current in the secondary coil. This device was first invented by Page in the United States in 1836 and later reinvented by Ruhmkorff in Europe in 1851. The of coil (Fig. thick induction 28:15) consists of a primary coil T of a few insulated copper wire turns wound about a soft iron core B which is not solid but built up from a bundle of soft iron wires to prevent self-induction effects. Around the primary coil is a secondary coil C consisting of many turns of very fine copper wire wound in sections with parafifined-paper insulation between the layers of each section. The primary circuit is completed through Secondary Terminals a make-and-break device D. This consists of a stiff piece of spring steel with a soft iron armature on one side and a platinum (or tungsten) contact on the other. It is normally in contact with the point on the platinum (or tungsten) As the end of the adjusting screw. primary current flows, the core becomes magnetized and attracts the armature, thereby breaking the circuit. The magnetism of the core now collapses and the armature thus released is drawn back by the spring to make contact with the The primary circuit again screw. complete for the action to be repeated. As the magnetic field of the primary circuit builds up and collapses at each make and break, an alternating induced is 327 Chap. 28 MAGNETISM AND ELECTRICITY E.M.F. is set up in the secondary coil. This E.M.F. may be many tens of thousands of volts and can cause a spark several inches long to pass between the terminals of the secondary coil. Oneof each alternation has a much half greater E.M.F. than the other (Sec. V:60). As a result this current usually is viewed as a pulsating direct current. In actual practice the weaker of the two is practically eliminated by the use of a condenser. Induction coils can be used for operalthough modern ating X-ray tubes, tubes are mainly transformer operated. They are also employed in some forms of laboratory research where moderately high voltages are required, and in the ignition systems of automobiles. Note: Because of the very large selfinductance of the primary winding, the upbuilding and decay of the primary current are both delayed, and the slower changes in the magnetic field mean correspondingly reduced voltages the secondary coil. The self-induced forward E.M.F. in the primary circuit at the break causes arcing across the contact points which become considerably worn as a result. To prevent this a condenser E, formed from alternate sheets of tinfoil and paraffined-paper, is placed across the make and break, and the self-induced current at break surges into the con- in denser instead of arcing across the points, thereby producing a very rapid break. The condenser plates, however, are connected through the primary winding, and no sooner is the condenser charged than it discharges through the primary coil, producing a current in it in a direction opposite to the now upbuilding current from the battery. As was noted before (Sec. V:60), the induced current is weaker at the make than at the break. The condenser weakens it still further. Accordingly, the coil produces practically 328 a pulsating, E.M.F. than that of the primary. direct current of higher V:68 THE AUTOMOBILE IGNITION SYSTEM the into fitted plugs The flammable mixture of gasoline vapour and air used in internal combustion engines is ignited by a series of sparks passing across the electrodes of spark cylinder heads. These sparks, which are correctly timed for each of the cylinders, are produced by a type of induction coil, the primary of which is connected to the battery through a circuit breaker, while the secondary is connected through a \u201cdistributor\u201d to the electrodes of the spark plugs. The scheme of the ignition system, is shown in Fig. 28:16. Spark Plugs Fig. 28:16 Simple Automobile Ignition System. The circuit breaker is operated by a cam revolving at half speed and driven from the engine crankshaft. When the platinum points are separated, the resulting change of magnetic field of force produces an induced voltage in the secondary coil. A condenser is connected to prevent across contact points the time. 3. 4. ELECTROMAGNETIC INDUCTION Sec. V: 69 arcing (and consequent damage to the points) and to ensure a rapid break as In described in the previous section. order to withstand the high voltages (of the order of 10,000 volts), the leads from the secondary coil to the distributor and spark plugs are covered with thick in- sulation. V : 69 QUESTIONS A 1. Read a more complete story of Faraday in a good encyclopedia and write a brief account of his accomplish-", " 5. ments. 2. Describe briefly Faraday's method of producing induced current for the first (a) What is the cause of an induced current? (b) What factors affect the magnitude of the E.M.F. induced in a circuit? (c) State Lenz's Law. (d) On the following series of diagrams label of the current and indicate the polarity of the upper end of the solenoid. direction the 6. the inductor with regard to direction and magnitude as the coil is revolved through 360\u00b0 in the earth's field. (a) Make a diagram of a simple generator with a single coil of wire rotating between the permanent regard to current produced with magnitude and direction as the coil rotates through 360\u00b0. (b) How is the current taken from the rotating armature and fed into the poles of a magnet. Describe the external circuit? (c) Describe how a generator differs direct-current from an alter- nating-current generator. (a) Describe, using simple diagrams, an (ii) a current produced alternating-current generator by the (i) direct-current generator. (b) What is meant by (i) a single (ii) 60-cycle current (iii) pulcycle sating direct-current? 7. Explain: (i) the structure (ii) the action of 10. 8. (a) a step-up transformer (b) a step-down transformer. (a) Describe (i) a simple telephone (ii) a simple telephone transmitter receiver with regard to purpose, (a) Describe the structure of an earth inductor. (b) Describe the current produced in structure, action. (b) Describe how transmitter and receiver circuits are linked in order to keep up the strength of signals. 9. What is the effect of self-inductance when a circuit containing a coil of wire is (a) completed (b) broken? Describe a simple induction coil as to (a) purpose (b) structure (c) action (d) uses. 329 Chap. 28 MAGNETISM AND ELECTRICITY B 1. A transformer is required to provide 6 volts to operate an electric door-bell on a 120 volt circuit. If the primary coil has 1 600 turns, how many turns should be on the secondary? 2. A step-down transformer has 2000 turns on the primary and 200 turns on the secondary. If the primary voltage Is 22,000 volts, what is the secondary voltage? 3. An electric door-bell transformer has 960 turns on the primary and 80 turns on the secondary. When it is connected to the 110 volt circuit, what voltage is provided for the bell? 4. A generator produces at 2200 volts. This current alternating is supplied through a step-up transformer with a voltage ratio of 1:100 to the transmission lines. This is fed to a step-down transformer of voltage ratio 1000:1. (a) What is the final voltage? (b) Draw a simple diagram to show these stages. 5. A generator with an E.M.F. of 550 volts supplies energy to a 2200-volt line. has The transformer primary winding. How many turns must there be on its secondary? 1 00 turns on its 6. A transformer, used to operate a toy electric train on a 1 20-volt line, has taps on the secondary to give 1 2, 8, 6 and 4 volts. If the primary winding has 960 turns, how many turns must there be on the secondary at each tap? 330 CHAPTER 29 ELECTRICAL ENERGY V:70 THE PRODUCTION OF ELECTRICITY towards industrial Not many years ago Canada was known throughout the world as an agricultural country. A survey of our exports at the beginning of the twentieth century shows that farm products formed the basis of our economy. In the years preceding the Second World War a gradual swing greater exFrom those pansion could be noted. years to the present time the change has been almost phenomenal. Canada has become an industrial nation. Our exports now, while still including a large part of the world wheat supply and other agricultural commodities, have been expanded to take in countless manufactured farm articles machinery to paper, clothing, electrical devices, chemicals and products of mining and smelting. ranging heavy from Closely connected with these changes has been the marvellous development of Canada\u2019s hydro-electric power resources. The energy of falling water has been turn the huge turbines harnessed which rotate the armatures of large genSteam-generating erators plants have become common in which (Fig. 29:1). to the energy from coal is utilized. In recent years heat from an atomic pile is These being used in the same way. latter methods provide the alternating current that is so vital to our modern way of life. Millions of horsepower have been made available to turn the wheels of industry across the country and to provide the luxuries of cheap electricity in our homes. Men of fore", "sight have planned for new and greater feats in engineering as they divert the waterways, tunnel through mountains and improve the generators to provide Canada with power for the future. V:71 THE TRANSMISSION OF ELECTRICITY Every high-school student knows that electrical energy is readily transformed into heat energy. A current of electricity consists of a shunting of free electrons in the circuit. The energy of these electrons is converted into heat as they encounter opposition to their motion in the conIt is apparent that the greater ductor. the number of electrons moving and the greater the resistance of the conductor, the greater will be the amount of heat produced. In 1841, James Joule proved experimentally that the heat produced was proportional to the square of the current, to the resistance, and to the time the current flowed {HaPRt). It is evident then that the amount of current in a conductor is the most important single factor in heat losses dur- 331 Aerial View of Hydro-Electric Plant, Niagara Falls. Ontario Hydro 332 ELECTRICAL ENERGY Sec. V: 72 Ing the transmission of electricity. The power of an electric current is equal to the voltage times the current strength, i.e., P\u2014 VI (Sec. V;75). This is constant for a given current. Thus, if the voltage is made higher, the current becomes smaller and heat losses will be reduced. For this reason, alternating current produced at a potential difference of 2,200 to 12,000 volts is stepped up to between 50,000 and 220,000 volts for transmission. the the voltage is stepped down by transformers so that the pole voltage is usually about 2,200 volts. The familiar pole transformers step this down further to 220 and 110 volts for household circuits. This current is used to operate electric ranges, heaters, lights and various other appliances. sub-station At V:72 FUSES ^ In designing any electrical circuit care must be taken to ensure that no circuit will carry an overload of current for fear of overheating the conductor. Re- Ontario Hydro House Fuse Box. call that the heat developed is proporcurrent the square of tional the to the times resistance of the strength, conductor. The wiring, then, must be of such a type as to carry the intended current safely. To prevent fires or failure in the circuit through melting of the conductor, fuses are put in as safeIn homes, these are usually guards. placed in special boxes near where the current enters. Each fuse of a length of metal alloy with a low melting-point enclosed in a case to confine the heat (Fig. 29:2). In the plug-type fuse one end is transparent, to permit inspection. In the cartridge fuse no opportunity for inspection is provided and larger ones sometimes contain a powder to extin- consists 333 Fig. 29:2 Fuses (a) Plug Type (b) Cartridge Type. Chap. 29 MAGNETISM AND ELECTRICITY frustrating research ending in success. In the years preceding 1879 Edison had tried almost every kind of fibre in his to find a filament that would efforts carry the current and emit light. The guish the arc when the current becomes interrupted by the melting of the metal strip. Each fuse is marked with a number denoting the maximum current in amperes that it is designed to carry. One should be certain that the rating of the fuse is not greater than that of the ciris used. Any overload cuit in which it should melt the fuse wire instead of harming the rest of the circuit. V:73 THE INCANDESCENT LAMP In 1879 Thomas Edison exhibited the first incandescent lamp (Fig. 29:3) at his laboratory in New Jersey. The dethis lamp is a story of velopment of Fig. 29:4 A Carbon Filament Bulb. Canadian General Electric filament had to have high electrical resistance and a high melting-point; it could not be too brittle even at high temperatures; it had to glow brightly when the current passed through it; it had to be enclosed in an finally, evacuated glass chamber so that it would not burn or unite with oxygen from the Eventually, Edison discovered that air. a carbon filament made by heating bamboo fibres met these conditions and the carbon filament lamp was born (Fig. 29:4). Other problems had to be solved before the lamps could be used extensively. Many improvements were found, so that now the modern electric-light Thomas A. Edison of Canada Ltd. Fig. 29:3 The Incandescent Lamp. 334 ELECTRICAL ENERGY Sec. V:74 \\ \\ \\'resistance wire. This is an alloy of nickel (80%) and copper (20%) which can be heated to a high temperature without melting, and oxidizing very slowly even when red hot. In addition, it has a resistance of about sixty times that of copper and this reduces the costs. / f { \\ /1 1 I- fp [1 ^ lp ^ 1 Canadian General Electric Fig. 29:", "5 A Modern Tungsten Filament Bulb. bulb (Fig. 29:5) bears little resemblance to the original. Modern filaments are made of tungsten, giving greater service and whiter light. Instead of evacuated glass bulbs, we now have bulbs filled with an inactive mixture of nitrogen and argon which allows the filament to be heated to higher temperatures. Platinum lead-in wires passing through the glass have been replaced by inexpensive nickeliron alloys. Costs have gone down while efficiency has been raised to heights not dreamed of a few decades ago. V:74 OTHER APPLICATIONS OF THE HEATING EFFECT (a) Domestic Heating Appliances Nowadays, in many homes, space heaters, hot-water heaters, ranges, etc., use electricity. The appliances designed for these various purposes contain heating-elements usually made of \u201cnichrome\u201d the heaters, In electric heating-elements are generally supported on fireclay forms, and in some of the smaller heaters concave polished metal reflectors are used to concentrate the radiant heat into a beam. The electric range has nichrome heating-elements embedded in some suitable insulating material for protection. The elements often consist of two parts. When the switch is set at \u201chigh\u201d these are in parallel; when at WVVWWVW \u2022-VvWVVVWWS^ fo) (b) (c) Fig. 29:6 Regulation of Heat in an Electric Stove Element. \u201cmedium\u201d one element is cut out; when at \u201clow\u201d both elements are in series (Fig. 29:6). The walls and door of the oven are filled with heat-insulating material such as fibre glass to reduce the heat loss, the oven temperature being regulated by some form of thermostat. (Fig. 29:7) With electric irons the heating-elements are usually wound on insulating-sheets of mica situated near the \u201csole\u201d of the iron. These heatingelements are covered by a layer of asbestos and above this is a heavy iron plate firmly screwed to Electric kettles are also provided with nichrome heating-elements wound on mica and these, sandwiched between mica strips, place below the are In other kettles base the heating-element is of the immerof sion type and consists of a coil the base. firmly kettle. held the of in 335 Chap. 29 MAGNETISM AND ELECTRICITY slowly apart. This phenomenon was discovered by Davy in 1806, using two thousand simple cells and charcoal rods. These rods were later replaced by gas carbon and the intensity of the light produced by this means led to the extensive use of electric \u201carc lamps\u201d in street lighting towards the end of the nineteenth century. Arc lights are used as spotlights in theatres. nichrome wire contained in a metal tube from which it is insulated. (h) The Electric Arc If two carbon rods are joined across an electric supply of not less than 45 volts, and the ends of the carbons are touched together, an intensely brilliant \u201carc\u201d of white light appears between the ends of the rods on drawing them The heavy current passing on striking the arc produces intense heat between the carbon tips which are thus to incandescence. The arc-like raised flame consists of white-hot particles of carbon torn from one of the rods which thus steadily wears away. If used with a D.C. supply the positive carbon is consumed twice as quickly as the negative rod, and it is found that a \u201ccrater\u201d develops at the end of it. Almost all the light originates at this \u201cpositive crater,\u201d about of which is temperature the 3500\u00b0C. Electric Arc Welding. Air Reduction C.anada Ltd. 336 ELECTRICAL ENERGY Sec. V:75 ) (c) The Electric Furnace Some types of electric furnaces make use of the high temperatures produced by the arc discharge. These arc furnaces the manufacture of are employed in carborundum and calcium carbide\u2014car- bon rods being immersed in the constituent chemicals through which an arcdischarge is passed. Another use of the arc furnace is in the \u201cfixation\u201d of atmospheric nitrogen for the preparation of fertilizers. (d) Electric Welding The high temperature developed by the electric arc is also applied in electric ships\u2019 plates, Boiler welding. plates, etc., can be welded by connecting the plates to the negative of a D.C. supply, applying the positive side to the weldingrod where the weld is to be made. V:75 BUYING ELECTRICAL ENERGY Since electricity is able to do work, it is a form of energy. The faster work is done, the faster energy is utilized. The rate of doing work, i.e., the rate of utilizing energy, is the power. Electrical power is measured in", " watts. One watt is the power provided when a current of one ampere flows with a potential difference of one volt. Thus: Power (watts) = P.D. (volts) X Current (amps. or P = VI watts P ={IR)ir.'V = IR,Ohm'sLaw) or P = PR watts If the power of one watt is provided for one hour, the consumer must pay for 1 watt-hour of energy. Similarly, if one kilowatt (1000 watts) is used for one hour, the consumer must pay for 1 kilowatt-hour (k.w.h.) of energy. Thus electrical energy (k.w.h.) == power (k.w.) X time (hours). Note: Since 746 watts = 1 horsepower (h.p.) it follows that P or P = = VI 746 PR 746 h.p. h.p. Example Find the cost of operating an electric toaster for two hours if it draws 8 amperes on a 110 volt circuit. The electrical energy costs 4 cents per kilowatt-hour. I =z 8 amp. no volts / P = VI P=\\10 X 8 = 880 watts 880= \u2014r\u2014- =.880 k.w. 1000 Electric energy used =.88 X 2 = 1.76 k.w.h, 1 k.w.h. costs 4 cents 1.76 k.w.h. cost 4 X 1.76 = 7.04 cents. The cost is 7 cents. 337 0 5 (i) Chap. 29 V : 76 MAGNETISM AND ELECTRICITY QUESTIONS A 1. (a) Describe what happens to alternating current from the time it produced at the generator until it is used to operate the electric toaster in your home. is Explain why the (b) changed before electricity is transmitted across the country. voltage is 2. (a) What is the function of a fuse in an electric circuit? (b) Describe the structure and action of a simple fuse. 3. (a) Make a labelled diagram of an electric-light bulb. (b) Explain the function of each part labelled in (a). 4. List three appliances found in each of the following, in which the heating effect of your employed, current electric (a) is house (b) your school (c) industry. 5. Define: power, watt, kilowatt, kilo- watt-hour. B 1. At the power-house, electricity at a potential of 1 2,000 volts is generated. It is prepared for long-distance transmission by being applied to a step-up transformer in which the number of turns of wire in the primary and secondary are in the ratio of 1 to 2000. (a) What will be the potential difference in the secondary? (b) How will the current strength be altered? (c) How will this affect the loss of energy through heating of the con- ductor during transmission? 2. (a) A pole transformer steps down the line potential from 5500 volts to 1 1 0 volts. What is the ratio of the number of turns of wire in the primary and secondary of the transformer? 338 (b) What change will occur in the current strength (ii) the heating effect? 3. Determine the power of an electric kettle which amperes when plugged into a 110 volt requires a current of 1 electric outlet. 4. What current is drawn by a 550 watt toaster when it is plugged into the 1 1 volt line? 5. What voltage is applied to a 2970 watt heating-element that draws a current of 1 3.5 amperes? 6. What is the power (a) in horsepower (b) in watts (c) in kilowatts, of a motor which draws a current of 3.38 amperes on the 1 1 0 volt line? 7. What is the maximum power that can be used in a circuit containing a 15 ampere fuse in the 1 1 0 volt line? 8. If an electric toaster of 550 watts and an electric kettle of 1000 watts were plugged into the same wall outlet in your home, would the 15 ampere fuse in the circuit melt? Explain your answer. 9. Find the cost of operating an electric toaster for 3 hours if it draws 5 amperes on a 110 volt circuit. The electricity costs 3.5 cents per kilowatt-hour. 10. What is the cost to a storekeeper of leaving a 40 watt light bulb burning near his safe for 36 hours if electricity costs 3 cents per kilowatt-hour? 11. An electric stove element draws 5 amperes on the 220 volt circuit. is turned on for 4 hours, what is the cost of operating the stove at 2.2 cents per kilo- If it watt-hour? 12. A Vi h.p. electric motor In an oil furnace comes on for a", " period of 5 minutes 48 times each day on the average. What is the cost of the electricity for 30 days If it costs 2.5 cents per kilowatt-hour? 13 A boy on returning home from school. ELECTRICAL ENERGY Sec. V:76 at 4:30 p.m. turns on lights in the kitchen, basement, hall and bedroom. The kitchen has one 100 watt bulb, the basement two 60 watt bulbs, the hall one 40 watt bulb and the bedroom one 40 watt bulb and one 60 watt bulb. If these lights are left burning 1 0:00 p.m. what is the cost when electricity costs 3 cents per kilowatt-hour? until 339 CHAPTER 30 ELECTRONICS points of which are to the secondary of an induction coil, the discharge close together. The induction coil is turned on while the tube is being evacuated (Chap. 31, Exp, 42). At first a discharge discharge occurs between the V : 77 INTRODUCTION (a) When a lighted match, or the flame of a gas burner, is brought near the knob of a positively charged electroscope, the leaves will be seen to fall (Chap. 31, Exp. 41). The flame liberates electrons from the atoms of the gases causing positively charged ions. The electrons so set free are attracted to the electroscope and neutralize its charge. Had the charge on the instrument been negative, charge would have been lost positively charged gaseous ions. become them the to to its (b) An electrical discharge tube with an open side-arm (Fig. 30:1) is attached Fig. 30:1 Apparatus to Show Dis- charge of Electricity at Low Pressures. 340 points but, as the tube is partially evacuthe current ceases between the ated, points and begins between the tertube even though minals the they are farther apart. This discharge, though varying in appearance with the even degree of though the tube is evacuated as com- evacuation, continues inside pletely as possible. The current flows through the gas at reduced pressure because the gas becomes ionized. The electrons expelled from the atoms of gas are attracted to the anode where they re-enter the metallic part of the circuit. The positively charged gaseous ions are attracted to the cathode where they remove electrons from it. Where the electricity is passing through a vacuum, electrons are expelled from the cathode and proceed in straight lines directly to the anode. The behaviour of electricity as it passes through a gas or a vacuum is known as electronics. The vacuum-tube and the low pressure gas-filled tube have opened the way for a seemingly endless number of new The most applications familiar application is the radio, but this is only one of the many being used in our modern way of living. Electronics electricity. of ELECTRONICS Sec. V:78 ticularly near the anode end, shine with a bright green fluorescent light. The phenomena occurring at this stage of the discharge proved to be most important. If an object is placed some distance in front of the cathode a sharp shadow of the object is cast on the end of the tube Experiments suggest near the anode. is used to guide our ships and aircraft, to perforin calculations in minutes that would take a mathematician days, to time delicate processes, to operate radio, television, sound movies, and to perform amazing feats of mechanical control. Its future is likely to be as remarkable as its past\u2014a challenging field for keen young scientists. V:78 CATHODE RAYS The main characteristics of the discharge of electricity through gases may be shown by the same method as was used in section V: 77(b). At normal pressures the insulating properties of the is of the the gas are too great for a spark to occur, but as the pressure is reduced a current passed and irregular streamers of light traverse the tube between the electrodes each of which is surrounded by a luminous glow. With continued lowdischarge ering pressure broadens out to a steady luminous column which extends from the anode to This is almost as far as the cathode. known as the positive column and is separated from the cathode glow by a the Faraday dark dark space called space. The pressure in the gas at this stage is about 5 mm. of mercury. Further reduction in the pressure of the gas to about positive column to shrink and to break up into alternate light and dark patches known as striations. The Faraday dark space increases in size at this low pressure, and the cathode glow moves away from the cathode, leaving a dark space between it and the cathode called the Crookes\u2019 dark space (Fig. 30:2). When the pressure is about 0.1 mm. the positive column disappears altogether and negative glow becomes considerthe ably extended. At still lower pressures the 1 mm. causes negative disappears, glow the the Crookes\u2019 dark space filling tube, and the glass walls of the tube, par-", " the Gases, (a) At Reduced Pressures\u2014 P, Positive Column; F, Faraday Dark Space; N, Cathode Glow; C, Crookes' Dark Space, (b) At Very Low Pressures\u2014 Cathode Rays in Crookes' Tube. that the fluorescence of the glass is due to something being emitted from the in Germany, cathode, and Goldstein, gave the name of cathode rays to this emanation. Crookes, in England, showed that the rays were shot out at right angles from the cathode, caused a rise of temperature of bodies interposed in their path, and that the rays were also capable of producing a mechanical force. 341. Chap. 30 MAGNETISM AND ELECTRICITY He hazarded a guess that the rays consisted of a stream of particles, and it was later shown that the rays could be deflected by a magnetic field and that they carried a negative charge (Chap. 31, Exp. 43) It was Sir J. J. Thomson, in England, of however, who showed in a series masterly experiments, carried out at the close of the nineteenth century, that the rays consisted of tiny, identical, negatively electrified particles which we now call electrons. In one of his experiments, in which the rays were deflected by both magnetic and electric fields, he was able to calculate their velocity and also the ratio of the charge to the mass of the the particles. The magnitude of electronic charge had been found by other means, and so he was able to obtain a value for the mass of the electron which he showed to be about 1/1840 part of the mass of the hydrogen atom. Various forms of the cathode ray tube are in use. One of these, the X-ray tube will be described in the next section. The other, the television picture-tube will be referred to in the next chapter. V : 79 X-RAYS While experimenting with discharge tubes at very low pressure, Roentgen, in Germany, in a fluorescent screen some distance away glowed brightly even when the discharge tube was covered with black paper, and 1895, noted that A Chest X-Ray Machine. General Electric 342 ELECTRONICS Sec. V:79 objects placed between the tube and screen produced shadows. Roentgen attributed these effects to a new form of tube, radiation its unknown which, of nature, was called X-rays. emanating account from the on cyanide, which is the substance frequently used on X-ray viewing-screens. They affect photographic plates, produce chemical changes, and are capable of discharging electrified bodies since they air. the are X-rays ionize electromagnetic waves of the same nature as visible or ultra-violet rays but having a very much shorter wave-length. One of the most characteristic properties of the rays is their penetrating power. They through many substances can opaque to ordinary light waves, the degree of penetration varying inversely as the density of the substance. Roentgen himself noticed this power of penetration and observed that human flesh was more transparent to the rays than bone. The pass It X-rays, is now known that or Roentgen rays, are produced whenever cathode rays impinge on a solid object Fig. 30:3 shows (Chap. 31, Exp. 44). a modern form of X-ray tube in which the electrons, constituting the cathode stream, originate from a heated filament (Sec. V:80) and are focused by a concave cathode on to a tungsten or molybdenum target set in a block of copper called the anti-cathode. The face of the anti-cathode is set at an angle of 45\u00b0 to the axis of the tube, the X-rays coming off in the direction indicated. Most of the energy of the cathode stream is dissipated as heat, only about 1 per cent of the energy is converted into X-rays. X-Ray Photograph of a Human Hand. St. Michael\u2019s Hospital X-rays travel in straight lines, are not deflected by either magnetic or electric fields and produce fluorescence in suitable materials such as platinum-barium medical uses of X-rays in locating fractures and foreign bodies in the human system are their most important application. Another use of X-rays is the detec- 343 Chap. 30 MAGNETISM AND ELECTRICITY tion of flaws in metal castings, and research scientists find them a useful aid in many of their investigations. V : 80 THE ELECTRONIC TUBE Just as a tap acts as a valve to control the flow of water through the pipes of a house, so the electronic tube acts as a valve to control the flow of electrons in an electric circuit (Chap. 31, Exp. Electronic tubes are able to 45, 46). increase or decrease the amount of current flow, or even to start or stop it, and the action is immediate. They use praccircuit. No tically no power in attempt", " will be made in this text to do any more than to outline the structure and action of the simplest forms of these the tubes. (a) Structure Any electronic tube has four basic parts (Fig. 30:4) : 1. A glass metal envelope enclosing the tube serves to sustain the vacuum or low pressure within the tube; 2. A cathode, which is heated either directly by an electric current or indirectly by a heater element which carries the current, serves to give ofT the electrons when heated; 3. An anode serves to receive the electrons from the cathode; 4. A base which has four or eight terminals or prongs for connecting the tube into its circuit. Such a tube is called a diode. A triode has an additional part called the grid (part d). (b) Action When the tube is connected into a circuit, the circuit is actually broken. When the cathode is heated electrons Hot Filament Fig. 30:5 Action of a Diode Tube. If a positive charge is leave its surface. applied to the anode, these electrons will be attracted across the intervening space and thus a flow of electrons will occur (Fig. 30:5). The anode is frequently referred to as the \u2018\u2018plate\u201d and the current 344 ELECTRONICS Sec. V:80 passing through the plate is called the \u201cplate current\u201d. (c) Rectification If the anode is connected to an alternating-current circuit (Fig. 30:6) it will be alternately positiv^ely and negatively While the anode is positive charged. electrons will flow from the cathode to the anode and through the plate circuit. During those half cycles when the anode the negative, electrons about the is Fig. 30:6 Electrons are Attracted Only When Plate Rectifier Circuit. is Positive. (a) (b) 345 Chap. 30 MAGNETISM AND ELECTRICITY flow. current Thus the cathode will be repelled and no current passes will through the tube in one direction only, from cathode to anode, and a direct current flows in the plate circuit. The tube has been used to obtain direct current from an alternating-current supply. Such a process is called rectification. The tube used for the purpose is designated a rectifier tube. Half-wave rectification by a diode is illustrated in Fig. 30:7. (d) Amplification A triode tube has a fifth part called the grid (Fig. 30:8) which controls the current passing through the tube. The grid is a special wire or mesh between the cathode and the anode. Any change in grid voltage causes a proportional change in the amount of current flowing through the tube. Thus, if we have a very weak alternating current, too weak to be used directly, it would be applied to the grid as a signal voltage. As the grid voltage varies it causes an identical amplified variation in the plate current. The grid is given a negative charge just large enough that it never actually becomes positive when the signal is applied. Thus, no electrons are attracted to the grid itself so the tube operates at maximum efficiency. Fig. 30:9 illustrates this amplification action as a radio signal is impressed on the grid of the triode. 346 Triode Tube as an Amplifier. ELECTRONICS Sec. V:82 Sound Waves Microphone Aerial - Aerial Vacuum Tube Amplifiers Vacuum'\\ Tube Generator of High Frequency Waves'Modulated Carrier Waves Tuner / Detector / 00000(5 ^ // 7 \u2018 ( Tube Used as an Amplifier Vacuum Tube Used as a Modulator (a) TRANSMISSION Audio-frequency Amplifier Radio Frequency Amplifier (b) RECEPTION Fig. 30:10 Radio Transmission and Reception. V:81 RADIO TRANSMISSION AND RECEPTION The triode tube can be used as an oscillator which is merely a type of alternating-current generator capable of providing frequencies in the order of 150,000 to 1,500,000 cycles or more per As these variations in current second. are impressed on a transmitting antenna, waves, electromagnetic carrier called In the waves, are sent out into space. transmitting-studio a microphone transforms sound waves into a fluctuating This current is used to direct current. modulate, or cause variations in, the amplitude of the carrier waves from the oscillator. differences At the receiving-station the varying electromagnetic waves induce minute antenna. potential Certain of these, selected by the tuning are impressed on the grid of circuit, an amplifier where they cause correspondingly larger variations in the plate These current of the amplifier tube. the in currents pass to a detector circuit in which another electronic tube separates the audio frequency component of the wave (i.e., the part that can be converted to audible sound) from the radio These frequency of the carrier wave. audio frequency components are further amplified and used to operate the diaphragm of earphones or a magnetic speaker. A simple diagram illustrating", " the above principles is shown in Fig. Students desiring further infor30:10. mation on the principles and structure of radio circuits are referred to the Radio Ajnateurs\u2019 Handbook or specialized texts on radio. V:82 PHOTOELECTRICITY In 1888, investigation. Photoelectric cells or, as they are commonly called, \u2018\u2018electric eyes\u201d, are the result of almost one hundred years of Hallcareful wacks, a German scientist, found that negatively charged zinc plates lost their charges when exposed ultra-violet light coming from an arc-lamp (Fig. 30:11). Thus electrons were shown to be substances when light falls upon them. This effect may be demonstrated visually if a zinc to an electroscope plate (Chap. 31, Exp. 47). This is called photo-emission of electrons. attached released certain from to is A short time later it was discovered that the alkali metals, sodium, potassium and caesium are sensitive even to orlight. Modern emissive dinary visible type photoelectric cells use these metals and are similar in appearance to radio vacuum-tubes. A layer of light-sensitive metal coats the entire inner surface of the tube with the exception of one small area which admits light. A positively charged anode is located in the centre of the tube and serves to attract elec- 347 Chap. 30 MAGNETISM AND ELECTRICITY trons from all sides as they are released (Fig. 30:12). The rate of electron emission is proportional to the amount and nature of the light falling upon the metal. Such cells control circuits which are used to open and close store doors, operate cesses with the exception of sound reproIt consists of a metal supportduction. ing disc upon which is deposited a layer of light-sensitive material. Over this is placed a transparent metal grid which acts as an electrode and collector for the electrons set free by light. Students Positively Charged Anode Serves as Plate to Attract Emitted Electrons Light-Sensitive Cathode Gives Off Electrons when Light Hits It Electrons Escape from Surface Electric Bell Fig, 30:12 Photoelectric Cell. burglar alarms, control dangerous machinery, reproduce sound for movies, count articles on production lines, and for hundreds of other purposes contributing to the efficiency and safety of industrial processes (Fig. 30:13). The photronic cell is used almost universally to-day for all the above pro- 348 Fig. 30:13 Photoelectric to Operate an Electric Bell. Cell Used ELECTRONICS Sec. V:83 The Component Parts of a Light-Meter. Canadian General Electric interested in photography will recognize the light-meter as a simple application of the photronic cell. In these devices light energy is con- verted into electrical energy. This, in part, provides the answer to the scientists\u2019 dream of eventually tapping the huge resources of the sun itself. V : 83 QUESTIONS 1. (a) What is the meaning of the term electronics? (b) State two facts fundamental to electronics. thermionic (c) Distinguish emission and photo-emission of elec- between trons. 2. (a) What are cathode rays? (d) Describe cathode rays. (c) How are cathode rays produced? properties some of 3. (a) Explain how X-rays are produced. (b) Describe some properties of Xrays. (c) How are these properties associated with the uses of X-rays? 4. (a) Make a labelled diagram of a simple diode tube. State the purpose of each part. (b) How does a triode tube differ from a diode tube? (a) What is meant by rectification? (b) Explain why a diode tube may be used as a rectifier. (a) What is meant by amplification? (b) Explain why a triode tube may be used as an amplifier. 5. 6. 7. Describe briefly how radio signals are sent from one station to another. 8. (a) Describe the structure of a simple photoelectric cell (ii) (i) a photronic cell. (b) What transformation of energy takes place in each of the above cells? 349 CHAPTER 31 EXPERIMENTS ON MAGNETISM AND ELECTRICITY EXPERIMENT 1 To distinguish between magnetic and non-magnetic substances. (Ref. Sec. V:2) Apparatus Bar magnet, assortment of small articles made from iron, paper, glass, wood, copper, nickel, rubber, tin, silver, plastic, etc. Method Approach each of the articles in turn with the bar magnet. Tabulate your results. Observations State which objects were attracted and which were not attracted by the magnet. Conclusions 1. State which were magnetic substances and which were non-magnetic substances. 2. Define (a) magnetic substances, (b) non-magnetic substances. EXPERIM", "ENT 2 To determine if the magnetic strength is equal in all parts of the bar magnet. (Ref. Sec. V;2) Apparatus Bar magnet, iron-filings. Method Sprinkle iron-filings liberally on a piece of paper. Roll the bar magnet in the filings and observe. Clean the magnet carefully with a cloth at the end of the experiment and return the filings to their container. Observations 1. Are the filings attracted equally to all parts of the bar magnet? 2. Where do most of the filings collect? 350 EXPERIMENTS ON MAGNETISM AND ELECTRICITY 3. Where does the least number of filings collect? 4. Compare the number of filings collected at each end of the magnet. Conclusions 1. Where is the magnetic strength of a bar magnet concentrated? What are the these areas called? 2. Where is the least magnetic strength in a bar magnet? EXPERIMENT 3 To determine the difference between the poles of a bar magnet, (Ref. Sec. V:2) Apparatus Bar magnet, chalk, non-magnetic stand, brass support. Method 1. Mark one end of the magnet with chalk so that it may be readily identified. 2. Suspend the magnet horizontally from the non-magnetic stand by means of thread supporting a brass support. 3. Allow the magnet to swing freely until it comes to rest. Note the direction in which the marked end is pointing. 4. Reverse the ends of the magnet in the support and again let it swing freely until it comes to rest. Observations 1. In what direction did the marked end of the magnet point when it came to rest the first time? 2. In what direction did the marked end of the magnet point when it came to rest the second time? Conclusions 1. Along what directional line will a freely suspended bar magnet always come to rest? 351 Chap. 31 2. MAGNETISM AND ELECTRICITY Why do we call one end of a bar magnet the north-seeking pole and the other end the south-seeking pole? What abbreviations for these are in common use? EXPERIMENT 4 To investigate the effect of bringing like and unlike magnetic poles together, (Ref. Sec. V:2) Apparatus Two bar magnets of known polarity, non-magnetic stand, brass support (Fig. 22:3). Method 1. Suspend one bar magnet from the non-magnetic stand by a thread and a brass support so that it is free to move in a horizontal plane, or balance a bar magnet horizontally on the convex surface of a watch glass. 2. Approach the N-pole of the suspended magnet with the N-pole of the other magnet. 3. Repeat, using the N-pole of the suspended magnet with the S-pole of of the other. 4. Now approach the S-pole of the suspended magnet with each end of the other. 5. Tabulate your results. Observations Pole of Suspended Magnet Pole of Other Magnet Observation N-pole N-pole S-pole S-pole N-pole S-pole N-pole S-pole Conclusions 1. What is the effect of like magnetic poles on each other? 2. How do unlike magnetic poles affect each other? 3. The above two conclusions combined into one statement give us the Law of Magnetism. State this important law. Questions 1. If the polarity of the magnets to be used is not known, how could you determine this information? 2. Why is it correct to say that the north magnetic pole of the earth contains S-pole magnetism? 352 EXPERIMENTS ON MAGNETISM AND ELECTRICITY EXPERIMENT 5 To study the magnetic field about a bar magnet, (Ref. Sec. V:4) Apparatus Two bar magnets, iron-filings, sheet of paper. Method 1. Lay a bar magnet on the desk and place a book of approximately the same thickness as the magnet on either side of it. Place the paper over the magnet so that the edges of the paper are supported by the books (a magnet board may be used to support the magnet Sprinkle iron-filings evenly over and paper instead of the books). the surface of the paper while tapping it gently. 2. Repeat the above procedure using the two bar magnets placed with their N-poles about 2 in. apart. 3. Repeat part 2 with an N-pole and an S-pole about 2 in. apart. Observations 1. Where are the lines of filings most concentrated? 2. Where are the lines least concentrated? 3. What is the general shape of the lines formed? 4. Do any of the lines appear to cross each other? 5. Where do the lines seem to begin and end in the case of the single bar magnet? 6. Describe the lines formed when like poles are adjacent to each other. 7. Describe the lines formed when unlike poles are adjacent to each other. Conclusions 1. What", " causes the iron-filings to take up their positions about the bar magnet? 2. What evidence have you found that supports the theory that lines of force repel each other? 3. What property of lines of force explains the attraction of unlike poles? 4. What evidence have you found that would indicate that the field is strongest close to the poles of the magnet? Questions 1. Make a diagram of each of the patterns formed by: (a) A single bar magnet. (b) Two bar magnets with like poles adjacent. (c) Two bar magnets with unlike poles adjacent. 2. Indicate by means of a diagram what you would expect the field of force about a horseshoe magnet to look like. Verify your prediction by experiment. 353 Chap. 31 MAGNETISM AND ELECTRICITY EXPERIMENT 6 To study induced magnetism. (Ref. Sec. V:6) Apparatus Bar magnet, long iron nail, iron tacks, magnetic compass. Fig. 31:2 Method 1, Place one end of the nail near the tacks to test for magnetic power. Now hold one end of the bar magnet near one end of the nail. Allow the other end of the nail to come in contact with the tacks. Lift the magnet and the nail together. Then move the magnet away from the nail. 2. With the head of the nail near the N-pole of the magnet, approach the N-pole of the compass-needle with the point of the nail. Repeat this with the head of the nail near the S-pole of the magnet. Observations 1. Did the nail by itself possess magnetic powers? 2. When a pole of the magnet was close to the nail, what was the effect on the tacks? 3. What occurred when the magnet was moved away from the nail? 4. With the N-pole of the magnet near the head of the nail, how was the N-pole of the compass-needle affected? 5. What occurred when the S-pole of the magnet was used? Conclusions 1. Why did the nail act as a magnet when under the influence of the bar magnet? 2. Why did it not remain a magnet when the bar magnet was removed? 3. What polarity was induced in the end of the nail (a) nearer to (b) farther from the pole of the magnet used? Questions 1. Why do we specify an iron nail and iron tacks? 2. Why should a long nail be used? 354 EXPERIMENTS ON MAGNETISM AND ELECTRICITY EXPERIMENT 7 To determine if the earth's magnetic field can cause induced magnetism. (Ref. Sec. V:6) Apparatus Dipping-needle (Fig. 22:8), soft iron bar about 12 in. long, hammer, magnetic compass. Method 1. Determine the direction of the magnetic lines of force at your location on the earth\u2019s surface by means of the dipping-needle (Sec. V:5). 2. Test the iron bar for magnetic polarity by approaching the N-pole of the compass-needle with each end in turn. 3. Holding the iron bar at the angle indicated by the dipping-needle, strike the bar sharply several times with the hammer. 4. Again test the bar for magnetic polarity using the magnetic compass. Observations 1. What is the magnitude of the angle of inclination? 2. What was the result of testing each end of the iron bar with the compass in the first case? 3. After striking the bar what changes were observed when it was tested with the compass? Conclusions 1. Why was the test different in the second case than in the first? 2. From your observations what magnetic polarity must the north mag- netic pole of the earth possess? EXPERIMENT 8 To study magnetic shielding. (Ref. Sec. V:6) Apparatus Demonstration magnetic compass-needle, bar magnet, thin sheets of iron, glass, copper, wood. Method 1. Hold the bar magnet to one side of the compass-needle so that it will be attracted towards the magnet. 2. Move the point of the compass-needle to one side and release it, allowing it to begin swinging from side to side. Note the number of vibrations in 10 seconds and the time in which it finally comes to rest. 3. Insert the iron sheet between the compass-needle and the magnet and repeat the above procedures, taking care that all other conditions remain the same as before. 4. Repeat part 3 using other materials. 355 Chap. 31 MAGNETISM AND ELECTRICITY Observations 1. Number of v.p.s. without shield 2. Time to come to full stop without shield 3. Number of v.p.s. with iron shield 4. Time to come to full stop with iron shield. 5. Repeat the above observations when other materials are inserted. Conclusions What is magnetic shielding? Questions 1", ". Why is it necessary to shield a magnetic compass on board a ship? 2. How can such magnetic shielding be accomplished? EXPERIMENT 9 To magnetize a steel wire, (Ref. Sec. V:7) Apparatus Bar magnet, steel wire or thin knitting-needle, magnetic compass. Method 1. Test the ends of the wire for magnetism, using the compass. 2. Using the N-pole of the bar magnet stroke the wire from one end to the other several times, always in the same direction. Lift the magnet well away from the wire as you return it to the starting point for each stroke. Test each end of the wire for polarity, using the compass. 3. Repeat the above steps using the S-pole of the bar magnet to stroke the wire. Observations State observations for each part described above. 356 EXPERIMENTS ON MAGNETISM AND ELECTRICITY Conclusions 1. What indication did you have that the wire was not magnetized before it was stroked with the magnet? 2. What makes you believe that the wire has become a bar magnet during the experiment? 3. What \\vas the polarity of the end of the wire last touched by the pole of the bar magnet? Questions 1. Why were both ends of the wire tested for magnetic polarity? 2. Why was the bar magnet lifted away from the wire as it was returned at the end of each stroke? 3. Explain why the wire became magnetized. EXPERIMENT 10 To investigate some experimental evidence for the theory of magnetism. (Ref. Sec. V:7) Apparatus A magnetized steel wire, wire cutters, magnetic compass. Method 1. Determine the polarity of the steel wire. 2. Cut the wire in half and determine the polarity of each half. 3. Cut the pieces into smaller and smaller sections and determine the polarity of each piece. Observations By means of diagrams illustrate your observations. Conclusion From these observations what inference can be made about the basic structure of a magnet? EXPERIMENT 11 To investigate further experimental evidence for the theory of magnetism. (Ref. Sec. V:7) Apparatus Test-tube, iron-filings, stopper, bar magnet, magnetic compass. Method 1. Fill the test-tube with iron-filings and insert the stopper. Test the tube for magnetic polarity. 2. Stroke the test-tube from end to end several times using one pole of the bar magnet as was done in experiment 9. Test the tube again for magnetic polarity. 3. Shake the test-tube vigorously and again test it for polarity. 357 Chap. 31 MAGNETISM AND ELECTRICITY Observations 1. Did the tube of filings act as a magnet when first tested? 2. Was any movement of the filings noticed as the test-tube was stroked? 3. What was observed in the second polarity test? 4. How did the shaking afifect the magnetic properties? Conclusion How does this experiment support the theory of magnetism? Question Would there be any limit to the magnetic strength produced? Explain. EXPERIMENT 12 To produce static electricity by friction, (Ref. Sec. V:10) Apparatus Hard rubber or ebonite rod, glass rod, cat\u2019s fur, piece of silk, small scraps of papers, sawdust and other light objects. Method 1. Rub the ebonite rod with the cat\u2019s fur. Bring the rod near the objects. 2. Repeat the above procedure, rubbing the glass rod with the silk. Note Lucite rubbed with a sheet of rubber or polystyrene (the material used in transparent vegetable containers) is an excellent substitute for glass and silk. Observations Describe the behaviour of the objects as the rod approaches in each case. Conclusion What is the efTect of rubbing an ebonite rod with cat\u2019s fur or a glass rod with silk? 358 EXPERIMENTS ON MAGNETISM AND ELECTRICITY Questions 1. What transformation of energy is involved in producing static elec- tricity by friction? 2. What evidence is there that this new power of attraction is not magnetism? 3. Describe instances encountered static electricity. in your everyday life in which you have EXPERIMENT 13 To show that there are different kinds of electrical charges and to establish the law of electrical charges. (Ref. Sec. V:11) Apparatus Two ebonite rods, two glass cat\u2019s fur, silk. rods, insulated stand, thread, stirrup, Method 1. Charge an ebonite rod by rubbing it with cat\u2019s fur. Suspend the charged rod in the stirrup attached by a thread to the insulated stand. Approach the charged end of the ebonite rod with the other charged ebonite rod. 2. Approach the charged end of the ebonite rod with the charged end of a glass rod which", " has been rubbed with silk. 3. Replace the suspended ebonite rod by a charged glass rod and repeat the procedures outlined above. Observations What is observed in each of the above parts? Conclusions 1. Are there different kinds of electrical charges? Name each kind. 2. What simple law of electrical charges may now be stated? Question What standard method have we of producing each kind of charge in the laboratory? EXPERIMENT 14 To study the use of the pith-ball electroscope. (Ref. Sec. V : 13) Apparatus Pith-ball electroscope, ebonite rod, cat\u2019s fur, glass rod, silk. Method 1. Charge the ebonite rod and slowly bring it near the pith balls. Allow them to touch the rod, and observe. 359 Chap. 31 MAGNETISM AND ELECTRICITY 2. After charging the pith balls approach them slowly with (a) a negatively charged rod and (b) a positively charged rod. 3. Repeat the above procedures using a charged glass rod in part 1. Observations Describe what is observed in each part above. Conclusions 1. Explain each observation. 2. How may the pith-ball electroscope be used to (a) detect (b) identify an electrical charge? Questions 1. What is meant by \u201ccharging by contact\u201d? Explain this process. 2. Why is repulsion the only sure evidence that an object is charged? EXPERIMENT 15 To study the use of the gold-leaf electroscope. (Ref. Sec. V:13) Apparatus Gold-leaf electroscope, ebonite rod, cat\u2019s fur, glass rod, silk, stick of sealing-wax, piece of flannel. Method 1. Charge the ebonite rod and slowly bring it near the knob of the electroscope until it touches. Withdraw the charged ebonite rod. 2. Approach but do not touch the knob of the charged electroscope with (a) another negatively charged rod (b) a positively charged glass rod (c) a stick of sealing-wax that has been rubbed with flannel. 3. Remove the charge from the electroscope by touching the knob with a finger. 4. Repeat parts 1 and 2, touching the knob with a charged glass rod. Observations Describe the observations for each part above. Conclusions 1. What charge is given to the electroscope when it is \u201ccharged by contact\u201d with (a) a positively (b) a negatively charged rod? 2. What procedure can be used to identify an unknown charge on an object? EXPERIMENT 16 To distinguish between conductors and insulators. (Ref. Sec. V:14) 360 7 EXPERIMENTS ON MAGNETISM AND ELECTRICITY Apparatus Gold-leaf electroscope, ebonite rod, cat\u2019s room. fur, various objects in the Method 1. Place a negative charge on the electroscope by touching the knob with the charged ebonite rod. 2. Approach the knob of the charged electroscope with the charged ebonite rod. Note what happens. 3. Rub the charged rod on a water pipe and again approach the knob. Note what happens. 4. Recharge the rod after each test and repeat step 3, rubbing the charged rod on rubber, slate, wood, your hand, and dipping it in water. Observations Note the effect on the electroscope leaves in each case. Conclusions 1. List those materials which allowed the charge to escape. What are such materials called? 2. List those that did not permit the charge to escape. What are they called? EXPERIMENT 1 To produce induced electrical charges. (Ref. Sec. V:15) Apparatus Two metal spheres on insulated stands, proof-plane, ebonite rod, cat\u2019s fur, glass rod, silk, gold-leaf electroscope. Method 1. Charge the electroscope negatively by the contact method. 2. Test each sphere for the presence of a charge by bringing it near the knob of the charged electroscope. 3. Place the two spheres A and B in contact as shown in Fig. 23:6. Approach, but do not touch, sphere A with the end of a negatively charged ebonite rod. Remove B from A and then remove the rod. Again test each sphere as in part 2. 4. Touch the spheres to each other and test each for a charge as in part 2. 5. Repeat parts 3 and 4, using a positively charged rod. Observations State what was observed. Explanation Explain your observations in this experiment with reference to the electron theory. 361 9 Chap. 31 MAGNETISM AND ELECTRICITY Conclusions 1. In each case what charge was induced on (a) the sphere near the inducing charge (sphere.4), (b) the sphere distant from the inducing charge (sphere B)? 2. What do you", " conclude about the magnitude of these two induced charges? EXPERIMENT 18 To charge a gold-leaf electroscope by induction. (Ref. Sec. V:16) Apparatus Gold-leaf electroscope, ebonite rod, cat\u2019s fur, glass rod, silk. Method 1. Charge an ebonite rod negatively and approach, but do not touch, the knob of the electroscope with it. Keeping the charged rod near, touch the knob with a finger, i.e., \u201cground\u201d the knob. Remove your finger. Then remove the charged rod. 2. Repeat the procedure, using a positively charged glass rod. Observations Record the behaviour of the leaves in each step of the method by means of a series of diagrams. Explanation Using the electron theory, indicate on the diagrams what occurred in each of the above steps. Conclusions What charge can be induced on the electroscope with (a) a negatively charged rod, (b) a positively charged rod? Question Describe how to charge an electrophorus (Sec. V:17) by induction. Explain fully. EXPERIMENT 1 To investigate the distribution of electric charges on a charged object. (Sec. V;19) Apparatus Electrophorus, proof-plane, gold-leaf electroscope, Biot\u2019s spheres (Fig. 23:11), several insulated conductors of different shapes. Method A. Using Biot's Spheres 1. Charge the disc of the electrophorus and transfer the charge by con- tact to the insulated metal sphere. 362 EXPERIMENTS ON MAGNETISM AND ELECTRICITY 2. Test the insulated metal hemispheres for electric charge by means of a proof-plane and the electroscope. 3. Then place the hemispheres tightly around the charged sphere. Remove the hemispheres and using the proof-plane and electroscope again test them for charge. 4. Test the sphere for the presence of electric charge. Observations Describe clearly what occurred in the above steps. Explain fully. Conclusion Where do electric charges reside on a charged object? B. Using Conductors of Different Shapes 1. Using the disc of the electrophorus, charge the conductors by contact. 2. Touch various parts of each conductor with the proof-plane and bring it to a given distance from a charged electroscope each time. Note the amount of deflection in each case and compare the density of charge on the various parts of each conductor. (a) Fig. 31:5 Observations Describe clearly what occurred. Conclusion Where is there the greatest density of charge on the surface of an irregularly-shaped object? EXPERIMENT 20 To study the action of pointed conductors. (Ref. Sec. V:20) Apparatus Wimshurst machine, metal point, candle. Method Attach the metal point to one terminal of the Wimshurst machine. Turn the handle rapidly to generate an electric charge. Hold the lighted candle in front of the point when it has become highly charged. Observation What happens to the flame of the candle? 363 Chap. 31 MAGNETISM AND ELECTRICITY Explanation What causes this phenomenon? Conclusion What effect do points have on the charge residing on a conductor? Application Explain the action of lightning-rods. EXPERIMENT 2 1 To study the action of a voltaic cell. (Ref. Sec. V:23, 24) Apparatus Glass vessel, strips of copper and zinc metal, flash-light bulb, socket, water, potassium dichromate. Method 1. Slowly add 10 ml. of concentrated sulphuric acid to 200 ml. of water, with continuous stirring. Pour this solution (the electrolyte) into the glass vessel. Immerse the two plates in the electrolyte parallel to each other but not touching. Note what occurs. 2. Connect the plates by wires joined to the flash-light bulb. Note what happens at both plates and to the flash-light bulb. 3. Add 14 teaspoonful of potassium dichromate and stir until it dis- solves. Observations 1. Is there any change noted at both plates in part 1? 2. (a) What changes occur at the plates in part 2? (b) What happens to the light bulb in part 2? 3. What is the result of adding potassium dichromate? 4. What eventually happens to the electrodes as the action continues? Explanations 1. What is the cause of the changes at the zinc plate? 2. What must have been produced to make the light bulb glow? 3. Account for the change in the glowing that is observed. 4. What is the action of the potassium dichromate? 5. Why will the action of the cell eventually stop? Conclusions 1. How is electric current produced in a voltaic cell? 2. What is (a) local action (b) polarization? How may they be prevented? Questions Why is the", " voltaic cell as used in this experiment not in common use to-day? 364 EXPERIMENTS ON MAGNETISM AND ELECTRICITY 2. What name, describing its action, can be given to the potassium dichromate in this experiment? EXPERIMENT 22 To determine the relationship between potential difference fV)^ current strength (I), and resistance (R) of an electric circuit (Ohm's Law). (Ref. Sec. V;33) Apparatus Resistance coil*, a number of dry cells, galvanometer (Sec. V:53), knife switch. VVWWW^ ( G I I.X I Method Fig. 31:6 1. Connect one dry cell, the resistance coil, the galvanometer, and the knife switch in a series circuit as indicated in the diagram. Close the switch and note the deflection of the galvanometer. Open the switch. 2. Repeat the above procedure using two, three, four and then five dry cells in series. Record all your results in the table below. Observations Number of Cells Galvanometer Deflection (Divisions) No. OF Cells Galvanometer Deflection 1 2 3 4 5 Explanations 1. What is represented by number of cells? How could it be repre- sented using an electrical symbol? 2. What is represented by the galvanometer deflection? How could it be represented using an electrical symbol? 3. How do the ratios obtained by dividing the number of cells by the *The instructor should ensure that the resistance used is of a correct value to give good results with the galvanometer and potentials employed. 365 Chap. 31 MAGNETISM AND ELECTRICITY galvanometer deflections compare? sistance R. This ratio is called the re- Conclusions 1. State Ohm\u2019s Law. 2. Express it, using symbols. Questions 1. Name the units used for measuring potential difference, current strength, and resistance. 2. Define \u201cohm\u201d. EXPERIMENT 23 To determine the resistance of an unknown resistance by the voltmeter-ammeter method. (Ref. Sec. V:38) Apparatus Voltmeter, ammeter, rheostat, dry cells, unknown resistance. Method 1. Connect the voltmeter, ammeter, rheostat and dry cells in a circuit with the unknown resistance X as shown in the diagram. 2. Adjust the rheostat, R, until any suitable value of current flows. 3. Determine the value of the current in amperes by reading the Determine the potential difference in volts by reading ammeter. the voltmeter. 4. Calculate the value of X in ohms. Observations V=^ I = Calculations Determine the value of X using Ohm\u2019s Law. Conclusion What is the value of the unknown resistance? 366 EXPERIMENTS ON MAGNETISM AND ELECTRICITY Questions 1. Why may the setting of the rheostat be varied without affecting the final result? 2. Why is the rheostat used in this circuit? 3. Why is the voltmeter connected in resistance? parallel with the unknown EXPERIMENT 24 To determine the resistance of an unknown resistance by the substitution method. (Ref. Sec. V:38) Apparatus Ammeter, resistance box, rheostat, dry cell, unknown resistance. Method 1. Connect the ammeter, rheostat and dry cell in series with the unknown resistance. Adjust the rheostat to get a large deflection of the ammeter. 2. Remove the unknown resistance and substitute the resistance box with all plugs removed. Replace the plugs singly and in groups of two or more until you get the same ammeter reading as before. Observation What is the total resistance indicated by the unreplaced plugs? Conclusion What is the value of the unknown resistance? EXPERIMENT 25 To study electrolysis of water. (Ref. Sec. V.41) Apparatus Simple electrolytic cell, pair of platinum electrodes, electrolyte consisting of one part sulphuric acid slowly added to approximately ten parts of water with constant stirring, source of direct current, test-tubes, wood splints. Fig. 31:8 367. Chap. 31 MAGNETISM AND ELECTRICITY Method 1. Fill the electrolytic cell almost to the top with electrolyte. Completely fill the two test-tubes with the same liquid and invert one over each electrode as shown in the diagram. Connect the electrodes to the current source noting which is the cathode ( \u2014 ), and which is the anode ( + ) 2. Allow the current to flow until one test-tube is almost filled with gas. Measure and compare the lengths of the gas column in the two test-tubes. 3. Remove the test-tubes and immediately lower a burning splint into each in turn. Note the results. Observations 1. What is observed at the an", "ode and at the cathode as the current flows? 2. Compare the volume of gas produced at the anode with that at the cathode in a given time. 3. What was observed when the splint was lowered into each of the gases in part 3? Conclusions 1. What gas is produced at (a) the anode (b) the cathode? 2. What are the relative volumes of the two gases? 3. Define cathode, anode, electrolyte, electrolysis. Explanation Write a brief explanation of electrolysis of water (Sec. V:41). Include in your explanation a description of what the current consists of in (a) the external circuit (b) the electrolyte. Questions 1. Why was sulphuric acid added to the water in making the electrolyte? 2. Why were the electrodes made of platinum? 3. What would be the eflfect on the quantities of gases produced of (a) increasing the current strength, (b) increasing the length of time that it flows. The Hoffman water voltameter (Fig. 26:2) may be used in place of the above apparatus. EXPERIMENT 26 To study electrolysis of copper sulphate solution. (Ref. Sec. V:42) Apparatus Electrolytic cell, pair of carbon electrodes, solution of copper sulphate, source of direct current, test-tube, wood splints. 368 EXPERIMENTS ON MAGNETISM AND ELECTRICITY Method 1. Fill the electrolytic cell almost to the top with the copper sulphate solution. Connect the electrodes to the current source. Allow the current to flow for several minutes. Note what occurs at both electrodes. Note any change in the colour of the electrolyte. 2. Collect a test-tube full of the gas liberated at the anode as described in the previous experiment and test it with a glowing splint. examine the deposit at the cathode. Also, Observations Describe all observations. Conclusion What is obtained at (a) the anode (b) the cathode? Explanation Write a brief e.xplanation of the electrolysis of copper sulphate solution. EXPERIMENT 27 To discover the factors that affect the amounts of materials liberated at the electrodes during electrolysis. (Ref. Sec. V:43) Apparatus As in experiment 25, with ammeter and rheostat in series with the electrodes. Fig. 31:9 Method 1. (a) Set up the electrolytic cell as in experiment 25, and allow a current of known strength to flow for one minute. Measure the length of the oxygen column collected over the anode. (b) Repeat part 1 (a) allowing the current to flow for two minutes, three minutes, etc. 3G9 \u2014 Chap. 31 2. MAGNETISM AND ELECTRICITY Repeat part 1 the strength etc., of that previously used. (a) using a current of one-half the strength, twice Observations 1. Current Strength Constant Time Current Flows (min.) Length of Oxygen Column (cm.) 2. Time Constant Current Strength (amp.) Length of Oxygen Column (cm.) Conclusion What factors affect the amounts of materials liberated at the electrodes during electrolysis? EXPERIMENT 28 To determine the strength of an electric current using a copper voltameter. (Ref. Sec. V:44) Apparatus Copper voltameter, copper sulphate solution, ammeter, rheostat, switch, source of direct current, stop-watch, balance, alcohol, emery-paper. Method 1. Clean the cathode with the emery-paper. Wash it in water. Dip 4. it in alcohol and let it dry by evaporation. Weigh it accurately. 2. Arrange apparatus as in Fig. 26:3. Simultaneously close the switch and start the stop-watch. Take ammeter readings every minute and let the current flow for exactly 15 minutes. 3. Remove the cathode, being careful not to dislodge any of the deposit from it. Rinse in water, dip in alcohol, and allow it to dry as before. Weigh it accurately. Observations Ammeter Readings 1. Initial weight of cathode 1 = 2. Final weight of cathode Weight of copper deposited = = 3. Time of current flow gm. gm. gm. sec. Current strength (average of ammeter readings) = amp. Calculations 1. Determine the weight of copper deposited in 1 second. 370 EXPERIMENTS ON MAGNETISM AND ELECTRICITY 2. Knowing that the electrochemical equivalent of copper is 0.000329, calculate the current flowing in the circuit. Conclusions 1. What was the calculated current? 2. How docs it compare with the ammeter reading? 3. Define ampere, coulomb. Questions 1. What possible sources of error were there in this experiment? 2. W'hy", " would a silver voltameter be more accurate than a copper voltameter? EXPERIMENT 29 To study electroplating. (Ref. Sec. V:45) Apparatus Electrolytic cell, copper sulphate solution, copper anode, object to be plated (a key) as cathode, source of direct current, rheostat, emerypaper, carbon tetrachloride. Method Carefully clean the object with emery-paper, wash in carbon tetrachloride and wash in water. Arrange apparatus as in the diagram. Close the circuit. Adjust the rheostat so that only a small current flows. After several minutes, remove the electrodes and examine them. Observations What is observed? Conclusion What is meant by electroplating? 371 Chap. 31 MAGNETISM AND ELECTRICITY Explanation Write a brief explanation for copper plating. Questions 1. In order to electroplate with any metal, what must (a) the cathode (b) the anode (c) the electrolyte consist of? 2. (a) Why should the cathode be cleaned? (b) Why should a small current be used? 3. Why was there no change in the colour of the copper sulphate solution in the above experiment? EXPERIMENT 30 To illustrate the principle of the lead-acid storage cell, (Ref. Sec. V:46) Apparatus Two strips of lead metal, emery-paper, two dry cells, glass tumbler, solution of sulphuric acid (one part concentrated sulphuric acid added to ten parts water), flash-light bulb, socket, galvanometer with fuse. Sulphuric Acid Solution Method 1. Thoroughly clean the lead strips, using the emery-paper. Fill the tumbler about three-quarters full of sulphuric acid solution. Immerse the lead strips in the solution so that they do not touch each other, connect them in series with the dry cells and let the current flow for about 5 minutes. Momentarily insert the galvanometer into the circuit to determine the direction of the current. Note any changes observed at each electrode. 372 2. EXPERIMENTS ON MAGNETISM AND ELECTRICITY Remove the diy cells and insert a flash-light bulb in their place. Note what happens. Momentarily insert the galvanometer to determine the direction of the current again. Let the current continue to flow until the flash-light bulb goes out and note any changes in the plates. Observations Describe all observations. Explanation 1. (a) What type of cell do you have in part 1? (b) What energy transformation takes place in it? 2. (a) What type of cell do you have in part 2? (b) What energy transformation takes place in it? 3. Describe briefly the chemical changes involved during the above energy transformations. Conclusions 1. What does a lead-acid storage cell consist of? 2. Describe what takes place during discharging and recharging. Questions 1. How may a hydrometer be used to determine the condition of a storage battery? 2. Why must water be added to a storage battery periodically? Why should it be distilled water? EXPERIMENT 31 To investigate the magnetic field surrounding a conductor of electricity. (Ref. Sec. VMS) Apparatus Several dry cells, conductor, iron-filings, small magnetic compass, piece of cardboard, switch. Method 1. Pass the conductor vertically through the piece of cardboard and connect it with the dry cell and switch so that the electrons will flow upward in the conductor when the switch is closed. Sprinkle ironfilings on the cardboard and gently tap it. Note the results. 2. Explore the magnetic field around the conductor with the compass- needle and plot the direction of the needle on a diagram. 3. Grasp the wire with your left hand so that the Angers point in the direction in which the N-poles are deflected. Compare the direction of the electron flow and that in which your thumb points. 4. Reverse the direction of electron flow and repeat the procedure in 3 and 4. 373 Chap. 31 MAGNETISM AND ELECTRICITY Observations Describe all observations made above. Conclusions 1. Describe the magnetic field around a wire carrying electric current. 2. State the Left-Hand Rule. Question Describe the magnetic field obtained if the above conductor were coiled into a single loop. EXPERIMENT 32 To investigate the magnetic field surrounding a helix carrying electric current, (Ref. Sec. V:49) Apparatus Dry cell, helix, iron-filings, magnetic compass, piece of cardboard. Method 1. Set the helix into a slit in the cardboard and support it horizontally. Join it to the dry cell and switch. Close the switch and sprinkle ironfilings on the cardboard, tapping it gently as you do so. Describe what occurs. 2", ". Determine the polarity of the magnetic field about the helix, using the N-pole of the compass. 3. Grasp the helix with your left hand so that the fingers point in the direction of the electron flow. Toward which pole does your thumb point? 4. Reverse the direction of electron flow and repeat parts 2 and 3 above. 374 EXPERIMENTS ON MAGNETISM AND ELECTRICITY Observations Describe all the observations made above. Conclusions 1. Compare the magnetic field about a helix carrying a current, with that about a bar magnet. 2. State the Helix Rule. EXPERIMENT 33 To discover the factors affecting the strength of an electromagnet, (Ref. Sec. V:50) Apparatus Two di-y cells, about 5 ft. of copper wire, cylindrical piece of soft iron about 4 in. long and in. in diameter, compass-needle, small iron weights (tacks, etc.) switch. Method 1. Wind about 10 turns of wire into a loose coil around the piece of soft \u201ciron\u201d. Remove the iron core and join the coil to a dry cell and switch. Close the switch and test the coil for magnetism. Determine the number of iron weights that can be lifted. 2. Insert the soft iron core, close the switch and test as before, noting any difference in the strength of the magnetism. 3. Using twenty turns of magnet wire around the iron core repeat the observations made in parts 1 and 2. 4. Using the two dry cells in series, repeat part 3. Observations State the number of iron weights lifted in each of the above cases. Conclusion What factors affect the strength of an electromagnet? Explanation State why each of these factors increases the strength of an electromagnet. Questions The galvanoscope (Fig. 27:14) consists of several coils of wire with different numbers of turns, e.g., 1, 25, 100, in each coil. 1. Describe how this instrument may be used (a) to determine the to compare the strengths of different direction of a current (b) currents. 2. What is the purpose of the coils with different numbers of turns? 375 Chap. 31 MAGNETISM AND ELECTRICITY EXPERIMENT 34 To determine the effect of a magnetic field upon a conductor carrying electric current, (Ref. Sec. V:52) Apparatus Battery of dry cells, strong horseshoe magnet, switch. Method 1. Assemble the apparatus as in the diagram with the conductor suspended between the poles of the magnet at right angles to the Close the switch and note what direction of the lines of force. happens. 2. Reverse the direction of the current and repeat part 1. Observations Describe what is observed. Conclusion What is the effect of a magnetic field upon a conductor carrying an electric current? This is the motor principle. Explanation What causes the conductor to move in each case? 376 EXPERIMENTS ON MAGNETISM AND ELECTRICITY Questions 1. If the above conductor were coiled into a helix and freely suspended in a magnetic field, what would be the effect (a) When a small current is passed through it? (b) When a larger current is passed through it? (c) When the current is reversed? (d) When the field is reversed? Make labelled sketches to indicate polarities and direction of movement. 2. What energy transformation occurs? Application Describe the construction of the D\u2019Arsonval galvanometer. EXPERIMENT 35 To investigate the principle of operation of a simple directcurrent motor, (Ref. Sec. V:55) Apparatus St. Louis motor (Fig. 27:21), dry cells, rheostat, magnetic compass, switch. Method 1. Examine the motor carefully. Identify the field magnets, armature, Connect the brushes in series with the brushes and commutator. battery, rheostat and switch. Close the switch and give the armature a gentle push to start it rotating. 2. Remove the magnets. Close the switch and adjust the rheostat so that a small current passes through the armature winding. Determine the polarity of the armature. Note the position of the commutator segments and brushes. Slowly rotate the armature through 180\u00b0. Note the changes in the position of the commutator segments and brushes as you do so. Again determine the polarity of the armature. Replace the magnets. 3. (a) Vary the current by changing the rheostat setting. (b) Vary the magnetic field by changing the distance of the poles from the ends of the armature. What effect do each of these have on the speed of rotation of the armature? 4. (a) Reverse the direction of the current. (b) Reverse the direction of the magnetic field. What effect do each of these have on the direction of rotation", " of the armature? Observations Describe all the observations. 377 Chap. 31 MAGNETISM AND ELECTRICITY Conclusions 1. Why does the armature of a direct current motor rotate? What energy transformation takes place? 2. What governs the speed of rotation of the armature? 3. What governs the direction of rotation of the armature? Explanation 1. What is the purpose of the commutator in a direct-current motor? Explain its action. 2. Explain why the factors mentioned in conclusions 2 and 3 affect the rotation of the armature. Question How does this motor differ from commercial D.G. motors? EXPERIMENT 36 To determine the cause of an induced current, (Ref. Sec. V:58) Apparatus Galvanometer, two solenoids wound on hollow spools so that one with few turns of wire will slip entirely inside one with many turns, bar magnet, dry cell, switch. 1 mm T (a) Fig. 31:14 Method Connect the larger solenoid to the galvanometer. 1. Thrust the magnet into the centre of the solenoid. Keep it stationary for a few moments and then withdraw it quickly. 2. Connect the terminals of the smaller solenoid through a switch to a dry cell. Close the circuit. Thrust smaller solenoid into the centre 378 EXPERIMENTS ON MAGNETISM AND ELECTRICITY of the larger. withdraw it. Leave it stationary for a few moments and then 3. Place the smaller solenoid (the primary coil) inside the larger one After a few (the secondary coil), and close the primary circuit. moments open the circuit again. Observations Observe the galvanometer needle at each stage of the experiment. Conclusion State the cause of an induced current. EXPERIMENT 37 To determine what factors affect the magnitude of an induced electromotive force (E.M.FJ. (Ref. Sec. V:59) Apparatus Same as for experiment 36, iron core to fit in hollow core of primary coil. Method 1. With the larger solenoid attached to the galvanometer, insert the bar magnet into the hollow core first slowly and then rapidly. Compare the strengths of the induced E.M.F.\u2019s. 2. With the larger solenoid attached to the galvanometer plunge the bar magnet into its hollow core. Then attach the galvanometer to the solenoid with fewer turns of wire and plunge the bar magnet into it at the same speed as before. 3. Attach the larger solenoid to the galvanometer. Connect the smaller to the dry cell to make an electromagnet. solenoid (primary coil) Plunge the electromagnet into the hollow core of the secondary coil. Now insert the iron core into the centre of the primary coil and plunge the primary into the secondary. Observations State what is observed in each part above. Conclusion What factors affect the magnitude of an induced E.M.F.? Question Why were the induced currents produced at the \u201cmake\u201d and \u201cbreak\u201d of the primary circuit in experiment 36 much greater than those produced by other means? EXPERIMENT 38 To investigate the direction of an induced E.M.F. (Ref. Sec. V:61) (Lenz's Law), 379 Chap. 31 MAGNETISM AND ELECTRICITY Apparatus Solenoid with many turns, bar magnet, galvanometer, dry cell, high resistance. Method 1. Connect the dry cell through the high resistance to the galvanometer. Note in what direction the galvanometer needle is deflected when a current of known direction is passing through it. 2. Connect the solenoid to the galvanometer and note the direction of the needle deflection and thus determine the direction of the current when: (a) The N-pole of the magnet is thrust into the coil. (b) The N-pole is withdrawn. (c) The S-pole is inserted. (d) The S-pole is withdrawn. 3. Using the Helix Law, determine the polarity of the upper end of the solenoid for each part of 2. Observations Direction of Current Across Front of Solenoid Polarity of Upper End of Solenoid Operation N-pole inserted N-pole withdrawn S-pole inserted S-pole withdrawn Conclusions 1. What effect does the magnetic field produced have on the motion that is inducing the current? 2. State Lenz\u2019s Law. Questions 1. Make a series of diagrams to illustrate the above observations. Show the direction of motion of the magnet and the polarity produced. 2. Use Lenz\u2019s Law to explain the production of an induced E.M.F. at the make and break of a primary circuit (Experiment 36, part 3). EXPERIMENT 39 To demonstrate the production of an alternating current and to study the principle of the A.C\u00bb and D.C. generator", ". (Ref. Sec. V:61 to 63) Apparatus Earth inductor (Fig. 28:4), galvanometer, connecting wires, St. Louis motor with both A.C. and D.C. armatures ( Fig. 27:21). 380 EXPERIMENTS ON MAGNETISM AND ELECTRICITY A. Using the Earth Inductor Method Connect the two leads of the earth inductor to the terminals of the galvanometer. Rotate the coil rapidly through 360\u00b0 and note the action of the galvanometer needle. Observations State what is observed. Explanation 1. What is the cause of the induced E.M.F.? 2. Why does it vary in direction throughout the 360\u00b0 rotation? Illus- trate your answer by a series of diagrams. Conclusion What is produced by each complete rotation of a coil in a magnetic held? B. Using the St. Louis Motor Model Method 1. Connect the leads from the A.C. armature to the galvanometer. Spin the armature with the huger and note the action of the galvanometer needle. 2. Replace the A.C. armature with the D.C. armature. Spin the armature as before and again note the action of the galvanometer needle. Observations State what occurred. Explanation 1. Answer the questions found in the explanation in part A. 2. How is the induced alternating current carried to an external circuit? 3. How may this induced alternating current be changed to direct current? Name and describe the device used to do this. Conclusions 1. What is a generator? 2. What energy transformation occurs in the operation of a generator? 3. What is the difference between an A.C. and a D.C. generator? Questions 1. How could you produce 60 cycle A.C. with the model used in this experiment? 2. What modihcation in the structure of the apparatus would permit a reduction in the speed of rotation of the armature in question 1? 381 Chap. 31 MAGNETISM AND ELECTRICITY EXPERIMENT 40 To study self-inductance. (Ref. Sec. V:67) Apparatus Coil with many turns wound around a soft iron core, three dry cells, switch, 6-volt lamp, neon lamp and socket. Fig. 31:15 Method 1. Connect the coil in series with the dry cells and switch as shown in the diagram. Join the lamp in parallel with the coil. 2. Close the switch. Note the effect on the lamp. 3. Open the switch. Note the effect on the lamp. Observations Make note of the observations. Explanation With the aid of Lenz\u2019s Law account for the phenomena noted above. Conclusion What is meant by self-inductance? Question What causes the arc when a switch is opened? EXPERIMENT 41 To demonstrate the conductivity of a gas (An introduction to electronics). (Ref. Sec. V:77) Apparatus Gold-leaf electroscope, ebonite rod, cat\u2019s fur, glass rod and silk, Bunsen burner. Method 1. Charge the electroscope positively. Hold a flame near the knob of the electroscope. Note what happens. 2. Charge the electroscope negatively and repeat step 1. 382 EXPERIMENTS ON MAGNETISM AND ELECTRICITY Observations Describe what is observed in each part above. Explanation What effect did the flame have on the particles of gas causing it to become a conductor of electricity? Why were both the positively and negatively charged electroscopes discharged? Conclusion What is electronics? EXPERIMENT 42 To study the changes in the conductivity of a gas as its pressure is reduced. (Ref. Sec. V:77) Apparatus Electrical discharge tube with open side-arm (Fig. 30; 1), vacuum-pump, induction coil, battery. Method Attach the electrodes of the discharge tube to the terminals of the secondaiy of the induction coil. The discharge points of the induction coil should be close together. Turn on the induction coil and evacuate the tube as completely as possible with the vacuum-pump. Note what happens. Observations 1. Where did the electrical discharge occur before and after evacuation of the tube? 2. Describe the discharge in the tube at various stages of evacuation. Explanation Why does the discharge occur in the evacuated tube rather than between the discharge points? Conclusion What effect does reducing the pressure of a gas have on its electrical conductivity? Question. According to the electron theory, what constitutes the current of electricity (a) through a gas at ordinary or reduced pressures (b) through a vacuum? EXPERIMENT 43 To study some properties of cathode rays. (Ref. Sec. V:78) Apparatus Crookes\u2019 tube (Fig. 30:2b), induction coil, dry cells, bar magnet. 383. Chap. 31 MAGNETISM AND ELECTRICITY Method 1.", " Set up apparatus as in diagram. Darken the room. Complete the circuit and observe the appearance of the end of the tube remote from the cathode. 2. Erect the target or metal cross in the path of the rays and again observe as in part 1 3. Approach the sides of the tube with first the N-pole and then the S-pole of the bar magnet and observe as above. Observations Describe what is observed. Conclusion List three properties of cathode rays. Question What evidence is provided in part 3 above to prove that cathode rays are streams of electrons? EXPERIMENT 44 To study the production and nature of X-rays. (Ref. Sec. V:79) Apparatus X-ray tube (Fig. 30:3), induction coil, gold-leaf electroscope. Method 1. Join the secondary terminals of the induction coil to the anode and cathode of the X-ray tube. Charge the electroscope and place it with its knob near the target of the tube. Start the induction coil and observe the leaf of the electroscope. 2. Repeat the above procedure placing (a) a piece of cardboard (b) a piece of wood (c) a piece of lead plate between the knob of the charged electroscope and the X-ray tube. Observations State what is observed in each case above. Conclusions 1. How are X-rays produced? 2. Account for the above observations. EXPERIMENT 45 To study thermionic emission of electrons in a vacuum-tube, (Ref. Sec. V:80) Apparatus Diode tube (B), two switches (Kj and Kg), \u201cA\u201d battery, \u201c5\u201d battery, galvanometer (G). 384 \u2014 EXPERIMENTS ON MAGNETISM AND ELECTRICITY Method 1. Connect the circuit as shown in the diagram using the plate and filament voltages recommended for the tube available. 2. With Ki open (a) close Kz making the plate positive and watch the galvanometer for any deflection. (b) reverse the terminals of the \u2018\u20185\u201d battery so the plate is negative and repeat part (a). 3. With Kr closed\u2014 (a) close K 2 making the plate positive and note any galvanometer deflection. (b) reverse the terminals of the \u201cB\u201d battery so the plate is negative and repeat part (a). Observations Switch Plate Charge Galvanometer Deflection 2(a) 2(b) 3(a) 3(b) Explanation. 1. Explain the effect on the filament of closing switch Ki. 2. Account for the above observations. Conclusions What two conditions must be fulfilled so that electrons will flow through a vacuum-tube? 385 Chap. 31 MAGNETISM AND ELECTRICITY EXPERIMENT 46 To demonstrate rectification by a diode tube. (Ref. Sec. V:80) Apparatus Diode tube, two 45-volt \u201cB\u201d batteries, \u201cA\u201d battery, neon bulb. Method 1. Connect the two \u201cB\u201d batteries in series and join these to the neon bulb. Reverse the battery connections. 2. Connect the neon bulb to the 110-volt alternating-current outlet. 3. In the circuit shown in experiment 45 replace the \u201cB\u201d battery by the 110- volt alternating-current source, and the galvanometer by the neon bulb. Observations Describe the way in which the neon bulb glows in each step above. Explanation Account for each observation. Conclusions 1. What is meant by rectification? 2. Explain how rectification occurs in the diode tube. 3. Make a diagram to illustrate the direct current produced in the plate circuit on the diode tube in step 3. EXPERIMENT 47 To demonstrate photo-emission of electrons. (Ref. Sec. V;82) Apparatus Zinc plate, gold-leaf electroscope, arc-lamp. Method 1. Attach the knob of the electroscope to the zinc plate by a fine wire. Charge the zinc plate negatively. Allow the beam of light from the arc-lamp to fall upon the charged zinc plate. 2. Repeat part 1 with the zinc plate positively charged. Observations What is observed in each step above? Explanation Account for the observations. Conclusions What is meant by photo-emission of electrons? 386 MODERN DEVELOPMENTS IN PHYSICS This Electron Microscope Permits Viewing of Particles Smaller Than One 10-Millionth of an Inch in Any Diameter. It Provides Magnification 50 Times Greater Than Heretofore Possible\u2014 So Great That a Human Hair Would Take on the Dimensions of a Subway Tunnel. R.C.A. Victor Company, Ltd. CHAPTER 32 MODERN DEVELOPMENTS IN PHYSICS gets of research have become ordinary articles of commerce. This century has seen the \ufffd", "\ufffdhorseless carriage\u201d, originally a luxury, become a necessity for business and pleasure. The airplane in one generation has undergone an amazing development for peace and war. Electronics has given birth in turn to radio, radar and television. In this chapter, a few modern developThese are ments are to be described. THIS TWENTIETH CENTURY This has been called the age of Physics \u2014a time when apparent luxuries have become commonplace necessities or gad- 389 Chap. 32 MAGNETISM AND ELECTRICITY selected because the preceding chapters serve as a preparation for them and because they are topics that will be of considerable interest to many students. Cathode-Ray Oscilloscope Fig. The principle of the cathode-ray tube (Sec. V;78) is shown in 32:1. Electrons emitted by the hot cathode travel at high speed toward the (c) (a). Many pass through the anode opening in the anode and form the electron beam (b) which passes between the (di d 2 ) two sets of deflecting plates and impinge on the fluorescent screen (i). A spot of light appears on the electron beam the screen wherever strikes. The spot of light is made to travel by applying voltages to the deflecting plates, thus causing the electron beam to sweep from side to side, move up and down, or undergo both motions simultaneously. Each point on the screen struck by the beam continues to fluoresce for a short time after the beam has gone by and does not quite die out before the next passage of the beam revives it. This fact combined with persistence of vision RCA Victor Company, Ltd. Canadian Marconi Co. Fig. 32:2 Commercial Cathode Ray Oscilloscope and Tube. Fig. 32:3 The Image Orthicon. 390 MODERN DEVELOPMENTS IN PHYSICS of the human eye produces the illusion of a continuous line of light on the screen. Thus, it is easy to see that if one voltage is used to make the beam sweep from side to side at a known uniform rate, another voltage applied to the ver- Illus. Courtesy of Canadian Marconi Co. A Picture Tube deal deflection plates will make the spot write the autograph of the latter voltage on the screen, revealing in graphic form frequency, wave form and other its characteristics. The commercial cathode-ray oscilloscope and tube is pictured in Fig. 32:2. It combines in one circuit a cathode-ray tube, a sweep circuit, amplifiers, a power supply, all properly synchronized. This is probably the most widely used instrument in electronics, particularly in the testing and repair field. All kinds of sound\u2014speech, noise, music\u2014can be ana- lyzed by the use of this instrument. Television In radio, electromagnetic waves modulated by the original sound-effects are transmitted through space. In television, a picture controlled by the light from the scene is transmitted at the same time as the sound. The feeble energy from such waves controls electric currents and magnetic fields in such a way as to reproduce the original sound and scene. In the modern television camera, the essential component is an image orthicon. Fig. 32:3, not much bigger than a rolled-up magazine, but very complicated, sensitive and costly. It receives an optical image, /, of the scene on a thin metal coating, M, on the inside are surface tube. ejected from this screen in direct propor- Electrons the of Fluorescent Coating 391 Chap. 32 MAGNETISM AND ELECTRICITY Scene in Television Studio During a Broadcast. Canadian Broadcasting Corp. tion to the brightness of the light that falls on it. These collect on the target screen, T, nearby and produce an elec- Fig. 32:5 The Effect of the Earth's Cur- vature on Television Waves. tron image of the scene. This electron image is scanned by an electron beam, B, from the electron gun, G, in much 392 the same manner as you read a page in this book but much faster (covering 525 lines in 1/3 sec.). The beam is controlled as in the cathode-ray tube. The speed of the electrons in the beam is so regulated that some of the electrons are repelled in a return beam whose intensity varies with the concentration of electrons on the target screen and, consequently, with the brightness of light from the scene. This return beam constitutes a weak current which is amplified, sent to a transmitter where it modulates ultra high frequency carrier waves that are broadcast much as are radio waves. The main part of the receiving aptube, paratus called a kinescope (Fig. 32:4), that is another cathode-ray is MODERN DEVELOPMENTS IN PHYSICS essentially the same as in the cathoderay oscilloscope. The signal received by the antenna is amplified and then is Bell Telephone Company of Canada Fig. 32:6 A Relay Station. applied to the tube in a manner that controls the", " aid of two aerials each above the other. To recognize friendly aircraft the radar pulse triggers a special transmitter and consequently the return pulse will be given a special mark which is distinguished on the cathode-ray tube. Radar has numerous peace-time apIt is used to follow the flight plications. of aircraft flying over lonely or thinlypopulated areas like the far north. Aerial and water transport employ it at night and in fog to avoid striking obstacles. Highway traffic authorities find it valuable in tracking down speeding drivers. T he Electron Microscope It can be shown that the ability of a microscope to reveal the fine detail of an object (the resolving power) is limited by the wave-length of the light used. Consequently, the practical upper limit of magnifying power that can be obtained with optical microscopes\u2014using visible light\u2014 is about 2000 diameters. Now it is known that an electron stream behaves in a manner very similar to a light beam but with an effective wave-length some 10,000 times smaller, permitting examination of bacteria, viruses and very large molecules. Accordingly scientists, especially Dr. F. Burton and his associates at the University of Toronto in the 1930\u2019s explored the possibility of using electrons in microscopy. As a result of their researches the electron microscope, shown diagrammatically in Fig. 32:9, was developed. The electrons are produced by a heated cathode and are accelerated under a potential of some 50,000 volts applied to an anode with a central hole from which they emerge in a stream of swiftly moving electrons. The electron stream is converged and focused by magnetic and electric fields much as light is by glass lenses in an optical microscope. These \u201cmagnetic lenses\u201d consist of short solenoids entirely encased in a steel shield except for a narrow gap in the centre through which the electrons pass and where they become deflected. The arrangement of lenses is similar to that in the optical microscope but the object to be studied is very much thinner (1/100,- 395 ). Chap. 32 MAGNETISM AND ELECTRICITY ELECTRON MICROSCOPE OPTICAL MICROSCOPE ^SOURCE OF ILLUMINATION (Electrons) (Light) CONDENSER LENS (Magnetic) (Glass) SPECIMEN STAGE OBJEaiVE LENS (Magnetic) (Glass) PROJECTOR LENS (Magnetic) (Glass) IMAGE (Viewing Screen) (Eye Piece) \u2014 Fig. 32:9 A Comparison of Electron and Optical Microscopes. R.C.A. Victor Company Ltd. 000 cm. thick). The instrument must be evacuated to prevent deflection of the electron stream as it collides with air molecules. The varying penetrations of the electrons through the object produce a sort of shadow picture on the fluorescent viewing screen or photographic film. Magnifications of up to 100,000 diameters have been obtained. The electron microscope has proved to be a very valuable tool in the hands of the research worker. It has revealed new facts about particle shapes and sizes which affect such processes as the covering power of paint pigments, the wear- ing quality of rubber tires and the operation of chemical catalysts. It has Comparison of Some Magnifiers Object Diameter (Microns Magnification (Diameters) Fine Machine Work 25-100 Pond Life Fungi 10-25 Bacteria Structure of Bacteria Large Virus Colloidal Particles Small Viruses 1-2 0.25 0.10 0.05 0.01 Instrument Hand lens Low power optical Medium power optical 8 20 200 1000. High power optical 2000. Special optical microscope or electron microscope 4000. Electron microscope 20,000 Electron microscope Large Molecules 0.002 100,000 Electron microscope 396 MODERN DEVELOPMENTS IN PHYSICS salts themselves. erated by the This phenomenon was called radioactivity and was extensively investigated by Pierre and Marie Curie. Using pitch-blende, a natural ore rich in uranium compounds, they succeeded, after a long series of repeated crystallizations, in isolating two new elements, polonium and radium, which than radioactive more were y- Rays provided new information in such fields as metallurgy, chemistiy, ceramics, crystallography, plastics, textiles, biology and medicine. NUCLEAR PHYSICS defined. protons, orbits \u2014 were In Section V:12, the composition of the atom was presented without any There the comsupporting evidence. ponents \u2014 electrons, neutrons, This nucleus, knowledge of the atom came about by a process of gradual evolution. First was the discovery of the electron by J. Thomson as a result of his Sir J. (Sec. V;78). work on cathode rays studies on radioactivity contriLater, buted much to knowledge about the nucleus of the atom. Radioactivity Fig. 32:10 Radium Emanations. In 1896 the French physicist Becquerel noticed that salts of uranium affected a neighbouring", " photographic plate. He concluded that these salts must be emitting rays or particles spontaneously gen- uranium. Later work by the Curies and others has revealed, indeed, that quite a number of heavier elements are radioactive. the Fig. 32:11 The Uranium Disintegration Series. 397 Chap. 32 MAGNETISM AND ELECTRICITY in In the Fig. into that rays, 1903 three undertook the opposite Rutherford a careful examination of the nature of the radiation from radioactive substances. He showed, by applying a magnetic radiation field, (/?), and gamma (y) 32:10, could be resolved distinct groups of rays which he called alpha (a), beta the a-rays being deflected in one direction, the /?-rays direction, while the y-rays do not undergo deFurther experiments showed flection. that the a-rays were positively charged helium nuclei, the /?-rays, negative electrons, and that the y-rays were highly penetrating, short wave-length, energy waves. Rutherford came to the conclusion that the nuclei of radioactive substances are spontaneously disintegrating into the above rays and changing into new elements as they do so. Fig. 32:11 shows the sequence of changes through which uranium goes changes into lead. as it gradually The Structure of the Atom Rutherford found that a-particles a sheet positively would pass through a thin of metal, and concluded that there was considerable empty space in the atom. His concept of the atom was a miniature solar system consisting of electrons revolving about charged Later, while at McGill Uninucleus. versity, he was able to isolate the hydrogen ion or proton. Moseley\u2019s experiments, in England, with X-ray phenomena on various elements, indicated the presence of these protons in every nucleus and a different number in the nuclei of different elements. The number of protons in the nucleus of the atom is the atomic number or nuclear charge. Since the atom is electrically neutral, the Fig. 32:12 Structure of Some Atoms. (MN = Mass Number; AN = Atomic Number) 398 MODERN DEVELOPMENTS IN PHYSICS Fig. 32:13 Isotopes of Hydrogen. Tritium MN = 3 AN = 1 number of electrons revolving about the nucleus must equal the number of protons in the nucleus, and hence must also equal the atomic number. of the This Rutherford\u2019s atom served until the discovery by Sir James Chadwick in 1932 neutron. particle (which was earlier predicted by Rutherford) has the same mass as the proton It is now bebut possesses no charge. lieved that atomic nuclei are built up of protons and neutrons. The sum of the protons plus the neutrons make up what is called the mass number of the atom. In general the number of neutrons the mass in number minus the nuclear charge or atomic number. For example, the helium nucleus which has an atomic number of 2 and a mass number of 4 must contain 2 protons and 2 neutrons. the nucleus is equal to Finally, to complete the picture of the atom, Niels Bohr postulated that the electrons revolved in various orbits, or energy levels, about the nucleus. The Bohr Concept for a number of typical atoms is shown in Fig. 32:12. Isotopes About this time, it was discovered by Rutherford and Aston in England that although all atoms of an element have the same chemical properties some differed from others in mass. Such forms of any element are called isotopes. They postulated that all the isotopes of any one element must have the same number of protons and electrons and hence the same atomic number, but must differ in the number of neutrons in the nucleus and hence must differ in mass num- (Fig. 32:13). ber. Most elements consist of a number For example, hydrogen is of isotopes. thought to consist of a mixture of three Protium is the isotopes commonest isotope of hydrogen. Deuterium is the isotope of hydrogen that is found in heavy water. The nucleus of called is much used in experimental work in nuclear physics. Tritium is a very rare isotope of hydrogen that is of importance in the hydrogen bomb. deuterium, deuteron, a Artificial Transmutation The above interpretation of natural radioactivity suggested to Rutherford the possibility of bringing about the disinteThis was acgration of other nuclei. complished by Rutherford in 1919 while at McGill by using a-particles from a radioactive source to bombard nitrogen atoms. As a result of the collisions he found that some of the nitrogen atoms had a proton ejected from them, the a-particle remaining with the disrupted These atoms therefore had a nucleus. net increase of 1 proton and 2 neutrons and hence became oxygen atoms (actually a heavy isotope of oxygen) : 14 4 1 17 7 N 2 18 He-^P + O Thus was achieved the first transmutation of the elements\u2014the old alchemists\u2019", " dream becoming a reality! artificial Rutherford next considered the use of artificially energized particles as a more effective means of bringing about nuclear changes, using protons accelerated under a potential difference of about half a million volts. These high-speed protons were directed on to a lithium target. 399 Chap. 32 MAGNETISM AND ELECTRICITY some of the atoms of which were split up to form helium atoms: l/+ P^-^He + He 3 12 2 Other light elements were disintegrated by this means, but it was realized that even more energetic particles were required to attack the nuclei of heavier elements on account of the large forces of repulsion between the bombarding particles and the larger nuclear positive Particles having the necessary charge. energy for this purpose can be produced by means of the cyclotron (Fig. 32:14). This latter machine, first invented by the American physicists Lawrence and Livingston in 1930, is capable of rapidly accelerating charged particles such as protons and deuterons. Nuclear Fission The discovery of the neutron provided physicists with a most effective projectile with which to bombard atomic nuclei 400 because, having no charge, a neutron can approach the positive nucleus of an atom without any repulsive force. A common way of producing neutrons for to bombard beryllium this purpose is metal by a beam of fast-moving deu- Slow Neutron AN 92 MN 235 Beta Particles Uranium Excess Mass Changed into Energy Fig. 32:15 Nuclear Fission. terons from a cyclotron\u2014neutrons being knocked out of the beryllium nuclei. It was while bombarding uranium 235 (a light unstable isotope of uranium) with neutrons in 1939, that Hahn and Strass- MODERN DEVELOPMENTS IN PHYSICS mann discovered an entirely new foiTn of nuclear transformation, Nuclear Fission, in which the uranium nucleus was split into two nearly equal parts with the release of a large amount of energy (Fig. 32:15). Another interesting feature of the fission of uranium is that during the process further neutrons are emitted in relatively larger numbers than those absorbed. Thus from each uranium nucleus undergoing fission two or more neutrons are released. The neutrons so released can then cause fission in other uranium nuclei and so on causIf allowed to ing a chain reaction. proceed in this way, provided that there is a suitable amount of fissionable matter present, the whole of the material is transformed in a very short time, with the release of vast quantities of energy. This is the principle of the atomic bomb. It has been proven that the combined mass of the products is a little less than the mass of original material. Dr. Albert Einstein as early as 1905 using his famous equation, E =; MC^, predicted what we now know\u2014that the mass lost becomes transformed into the energy of the explosion. E represents the energy, M the loss in mass and C, a constant quantity equal to the speed of light (Sec. IV: 6). Thus we can see that a small amount of mass is transformed into a tremendous amount of energy. Nuclear Reactors If the neutrons from the fission of U 235 are absorbed or slowed down as they are produced, the chain reaction can be controlled and the energy released put to useful purposes. Graphite and certain other light substances which possess the property of absorbing neutrons, are used to control the rate of neutron emission in atomic fission and are called moderators. The assembly of uranium (or other fissile material) and graphite (in the form of blocks or rods) is known as an atomic pile or nuclear reactor (page 402), the rate of energy release being controlled by adjusting the length of the graphite rods inserted in the pile. The energy released is transfomied into heat which is used for various purposes. Energy is variously measured. The engineer may express it as B.T.U. per pound (Sec. 111:20). The nuclear physicist may express it in electron volts per atom. One B.T.U. equals 665 X 10^^ million electron volts. estimated 13,600 B.T.U. per pound, this is 4 electron volts per atom of carbon. By comparison one atom of uranium produces 200 million This means that one electron atom of uranium produces 50 million times as much energy as one atom of carbon or if you have equal masses of coal and uranium the amount of energy released from the uranium is 3 million times as great. There is little wonder, therefore, that the search for uranium goes on at a feverish pace. gives volts. that, coal It is if Atomic Fusion to In 1951 when man was congratulating himself on his conquest of the atom through turning fission his advana new and still more exciting tage, process was evolved by Dr. Edward Teller in the United States. You will recall Hans Bethe\u2019s theory of the origin of the sun\u2019s continuous heat by the union of four atoms of hydrogen to", " form an atom of helium accompanied by the conversion of a small amount of mass into energy (Sec. Dr. Teller employed this process using the isotopes of hydrogen\u2014protium, deuterium and tritium\u2014resulting in the production of a source of energy far greater than ever The hydrogen bomb realized was created which was fantastically devastating. A source of energy was born that was so amazing that it may well be the source of energy of the future. Ill: 3). before. 401 Chap. 32 MAGNETISM AND ELECTRICITY Atomic Knergy of Canada Ltd. Model of N.R.X. Reactor, Chalk River, Ontario. star Mewspaper Service U.S.S. Nautilus. The First Atomic Powered Submarine. 402 MODERN DEVELOPMENTS IN PHYSICS COOLING WATER TRAVELS Space does not permit dealing at any length with the many and varied uses of atomic energy. Suffice it to say that biological, medical, agricultural and industrial research are greatly benefiting from the developments made in this field as are some users of industrial and domestic power. are as yet beyond man's fondest dreams. The possibilities 403 ANSWERS Unit I\u2014Mechanics A Chapter 2, Section I : 4\u2014Page 16 4 (a) 1000 mm., 100 c.m., 10 dm., 1000 m. 10.000 dg., 1000 gm. 1, 3. 11.46 (ii) 8.3. 9 (b) 5, 5, 5, 5, 5 (b) 1,000,000 mg., 100,000 eg., 10 (b) 37., 34., 37. 11 (b) (i) B (a) A A B A 1520, 3.8, 605.4 cm.; (b) 1 2.3, 6,000,050.465 c.c. (c) 100 m. (i) 28.1 ft. (hi) 64.4 km. per hr. (ii) 22.5 cu. dm. (hi) 22.5 litres 225.000 dg., 22.5 kg.; (b) (i) 28.4 gm. (ii) 909. kg. (hi) 5.4 pints 102.70; (b) (ii) 0.35 (hi) 27 X 101; (j) (i) 14 x 102 (h) 0.50 (hi) 9.0 Chapter 3, Section I : 9\u2014Page 22 30,000, 2.36, 60,005.44 sq. cm.; (c) 2,500,000, 2 (a) 37.4, 15.7, 39.4 in.; (b) (i) 30.5 cm. (ii) 1.61 km.; (ii) 856 cm.; (d) (i) 58.7 R. per sec. (ii) 17.9 m. per sec. 3 (a) (i) 1500 sq. cm. (ii) 15. sq. dm.; (b) (i) 22,500 c.c. 4 (a) 22,500 gm., 22,500,000 mg., 2,250,000 eg., 5 (i) 3.1 litres (ii) 3080 ml. 36. (i) 66.56 (ii) 71.25 (hi) 67.17; (c) 6 (a) (i) 1 (b) 2.70 gm. per c.c. 3 (b) 8.0. B 1 5 gm. per c.c. lb. 6 20.3 cu. ft. cu. ft. 16 72.8 gm. or 6. gm. per c.c. 24 1.18 gm. per c.c. 2 52.5 or 53. gm. 3 268 c.c. 4 62.5 or 63. lb. per cu. ft. 7 4.87 12 1781.3 or 18 X lO^ lb. 17 8.5 gm. per c.c. 8 54.1 c.c. 9 800 or 8 X 10^ gm. 13 7.64 gm. per c.c. 18 (a) 19.5% (b) 13.4% 5 64. 10 0.068 11 0.59 15 0.80 14 0.8722 20 5.9 19 0.42 c.c. 23 0.9 gm. per c.c. 21 9.9 gm. per c.c. 22 8.7 gm. per c.c. Chapter 4, Section I : 15\u2014Page 32 1 (b) (i) 100 gm. (ii) 790 gm. 3 (c) (i) 2.6 gm. (ii) 4.3 c.c", " B 5 (i) 7.14 (ii) 70 c.c 2 (a) 68 gm.; (b) 625.0 lb. 4 10.5 1 (a) 30 gm.; (b) 162.5 lb. cu. ft. per cu. ft. 10 8.90, 0.840 or 1.6 X 102 lb. 20 18.8 or 19. cm. 26 14. cm. 25 1/3 11 32.0 gm. 7 (i) 105.9 gm. (ii) 8.30 gm. per c.c. 12 No 13 120 gm. 17 11.7 c.c. 22 100 gm. 16 36.6 or 37. gm. 21 20.4 or 20 cm. 3 (a) 25 ml. (b) 0.40 6 (i) 2.5 (ii) 0.096 cu. ft. (hi) 156 lb. 9 (i) 5.36 (ii) 0.73 15 156. 19 1.3 24 0.21 8 1.30 14 0.9 gm. per c.c. 18 0.90 gm. per c.c. 23 187.5 or 188. c.c. Unit II\u2014Sound A Chapter 6, Section II : 9\u2014Page 61 5(b) 1 100 ft. per sec. 9 1 150 ft. per sec. 10(a) 1 100 ft. per sec. 1 (a) 3.75 cm.; (b) 15 cm. (b) 288 m. per sec. 6 (a) 1099, 1055, 1135 ft. per sec.; (b) 335, 321.8, 345.8 m. per sec. (b) 6.7\u00b0C. 12 (a) 56.5 ft. 2.67,.67 sec. 4 (a) 320 v.p.s.; (b) 667 v.p.s. 3 (a) 1170 or 11.7 X 10^ ft. per sec.; 5 (a) 1.50 ft.; (b) 56.0 cm. 7 (a) 15\u00b0C.; 11 (a) 357 m.persec.; (b) 42\u00b0C. 14 750 m.p.h.\u2019 10 259 v.p.s. 13 22.6 sec. 8 5595 ft. 9 1.33 m. Chapter 7, Section II : 18\u2014Page 71 6 (b) (i) 750 v.p.s., (ii) 100 v.p.s.; (c) (i) 208.3 or 208 v.p.s. (ii) 50 v.p.s. 404 ANSWERS B (a) 1 4 150 teeth 8 25, 80, 35 cm. v.p.s. 13 160 v.p.s. 960, 1920, 7680 v.p.s.; (b) 240, 60, 15 v.p.s. 5 256, 320, 384, 512 v.p.s. 9 1080, 270 v.p.s. 2 240 v.p.s. 6 900, 225, 675 v.p.s. 10 312 v.p.s. 11 400, 2500 gm. 3 320 r.p.m. 7 587 v.p.s. 12 384 A Ch.\\pter 8, Section II : 25\u2014 Page 79 2 (a) (i) 48 in. (ii) 1 200 ft. per sec.; (b) 24 in. B 1 (a).302 m. (i) 5 ft. (ii) 5 15.5\u00b0C. A 7 1781 ft. Unit III\u2014Heat A 4 4 X 10-3 lb. ft.; (b) 1100 ft. per sec. 1 6 33.3\u00b0C. 7 18.8 cm. 2 1:2 8 4 beats per sec. 3 320 v.p.s. 4 11.9 in., 9 486 or 474 v.p.s. Chapter 9, Section II : 34\u2014Page 93 Chapter 11, Section III : 4\u2014Page 114 Chapter 12, Section III : 11\u2014 Page 124 2 37.8, 176.7, \u2014140, \u2014 45.6\u00b0C. 3 (a) \u201440\u00b0; (b) 4 (a) (i) 330, 250\u00b0 K. (ii) 25, -36\u00b0 C.; (b) (i) 309.7, 255.2, \u00b0K. (ii) 212, B 1 59, 392, \u201476, \u2014 459.4\u00b0F. 320\u00b0 \u2014 459.4\u00b0F. Chapter 14, Section III : 32\u2014Page 160 2 1600, 8400 cal. (i) A (b) B 350. cal., (ii) 5328 B.T", ".U., (iii) 179.4 cal. 4 (b) 0.087. 7 4000, 9 (a) 27,000 cal; (b) 8100 cal.; (c) 21,700 cal. 1 100,000 or 10 X 104 4 6000 or 60 X 10\u201c B.T.U. cal. 8 40\u00b0C. gm. 9 0.104 14 204.3 or 204 gm. 18 187.5 or 188 gm. 24 27 gm. 28 534 cal. 1.2 X 102 25 46.6 or 47\u00b0C. 29 40\u00b0C. 30 53\u00b0C. 10 0.0933. 15 12.8 or 13\u00b0C. 19 208.3 gm. 2 39,000 or 39 X lO^ B.T.U. 5 1320 or 13.2 X 10^ cal. 11 0.21 12 0.031 16 0.44 20 80 cal. 26 530 or 5.3 X 10^ cal. 31 3640 cal. 6 101 cal. 3 91,000 or 91 X lO^ 7 56.3 13 292.6 or 293 gm, 17 1250 or 12.5 X 10^ gm. 22 82 cal. 23 0.50 27 541 or 5.4 X 10^ cal. 33 121 or 32 0.112 or 0.11 21 78 cal. Unit IV\u2014Light A Chapter 16, Section IV : 8\u2014Page 182 5 (d) 2 in. 7 (b) 8.3 min. B 1 30 ft. 2 0.8 in. 3 148 or 1.5 X 10^ m. 4 1.3 sec. 5 5.87 X lO^^ miles, 9.46 X 1042 km. 6 52.8 X 1042 7 4 3 yg^rs. Chapter 17, Section IV : 20\u2014Page 194 A. 5 (a) 7. 405 A 2 (c) dioptres. B I 1.20 B (i) 10 o\u2019clock (ii) 5.45 o\u2019clock (v) \u2014 60 cm. 5 \u2014 9.23 in. 11 \u2014 4.5 in. 12\u2014 8.6 cm. 1 oc 13 20 in. 14 2.2 in. Chapter 18, Section IV ; 34\u2014 Page 211 ANSWERS 4 (a) 6 3 ft. (i) 50 cm. (ii) 60 cm. (iii) 90 cm. (iv) 7 4 in. 10 36 cm. 8 4.3 ft. 9 3 in. 124,000 miles per sec.; (d) 2.47 5 (b) glass 7 (b) water 9 (d) 6.7 2 479 X 102 3 (i) 35\u00b0 (ii) 28\u00b0 (iii) 20\u00b0 per sec. ZD = 37\u00b0 5 One with critical angle of 30\u00b0 7 \u2014 9.2 in. (v) \u2014 60 cm. (iii) 90 cm. (iv) II 32 in. 13 6 in. (i) 0.9 in. (ii) 11 X 18 (a) 5.5 in.; (b) 36 in. x 60 in. cc 12 24 cm. 6 (a) 8 5 cm. 14 24 cm. 4 Zi = 48.5\u00b0 (i) 50 cm. (ii) 60 cm. 10 80 X 17 9 9 metres 15 \u201420 cm. 16 140, 12.7 cm. Chapter 19, Section IV : 43\u2014 Page 225 1 (c) 7.3 X 1014 ^ p 3 (a) 3.1 in.; (b) 1.5 in. 7 (b) 160 X 9 (d) 192 cm. Unit V\u2014Magnetism and Electricity Chapter 20, Section IV : 50\u2014Page 235 6 (c) (i) 8.3 X (ii) 6.7 X (d) (i) 25 mm. (ii) 20 mm. Chapter 25, Section V : 39\u2014 Page 293 A (b) 22 volts 4 (c) 1 ohms (ii) 5.5, 2.8 amp. (i) 90 ohms (ii) 6 (b) 14 ohms. 1.2 amp. (iii) 60, 48 volts; (d) (i) 13 B 2 212.5 or 21 X 10 volts 1 12 ohms 4 (a) 0.12 amp.; (b) 11.9 or 12 volts and 0.06 volts 5 (a) 22 ohms; (b) 5.5 ohms 6 (a) 41.3 ohms; (b) 0.333 amp.; (c) 2.66 amps. 7 (b) 6 volts; (c) 0.9 ohm 8 16.7 or 17 ohms. 9 (a) 3.4 ohms; (b) 3.3 amp.; (c) Chapter", " 26, Section V : 47\u2014-Page 302 11 (a) 2.4, 1.6 amp.; (b) 18 volts. 1.7 amp. 10 0.2 amp. 3 3.3 amp. B (a) 3.55 gm.; (b) 12.1 gm. 4 0.003 amp. 8 0.002 cm. 5 1.26 amp. 9 3.95, 0.37 gm. Chapter 27, Section V : 56\u2014^Page 315 6 0.02 amp. 10(a) 2 amp.; (b) 15 hr. 2 (a) 9.5 gm.; (b) 32 gm. 3 (a) 2.4 gm. (b) 0.30 7 (a) 0.18 amp.; (b) 0.0028 1 gm. gm B 1 (a) 0.08 ohm; (b) 0.008 ohm; (c) 0.0008 ohm, 20.002 ohms 3 (a) 115 ohms; (b) 1240 ohms; (c) 12490 ohms 4 9998 ohms. Chapter 28, Section V : 69\u2014Page 329 B 4 220 volts 5 400 turns 6 (i) 96 turns 1 80 turns 2 2200 volts (ii) 64 turns (iii) 48 turns (iv) 32 turns. 3 9.2 volts B Chapter 29, Section V ; 76\u2014Page 338 1 (a) 24 X 10\u00ae volts; (b) lZ2000th; (c) greatly diminished 50 times (ii) 2500 times (ii) 372 watts (iii) h.p. 4 5 amp. 7 1650 watts 3 1650 watts 0.372 kw. 8 No. cents 11 9.7 cents 12 $1.12 13 6 cents, 406 5 220 volts 2 (a) 50:1; (b) (i) 6 (i) 0.498 10 4.3 9 6 cents INDEX Aberration, chromatic, 210, 224; spherical, 193 Abnormal expansion of water, 124 Absolute, temperature, 123; zero, 123, 132 Absorption of radiant energy, 133, 142 {Exp. 168) _ Absorption spectra, 219 Accommodation, lens, 229 Accuracy, degree of, 11; limits of, 13 Achromatic lens, 224, 233 Acid, 295, 299 Acoustics of buildings, 87-89 Action of points, 276 {Exp. 363) Addition, 14 Additive theory of colour, 222 Air, lens, 205, 206; liquefaction of, 155; pres- sure effect on the boiling-point, 148 Airships, 31 Alcohol, 295 Alcoholometer, 29 Alkali, base, 299; metals, 347 Alloys, 291; densities of, 21 Alnico magnet, 266 Alpha rays, 398 Alternating current, 322; 25-cycle, 322; 60- cycle, 322 Alternating current generator, 321 {Exp. 380) Amalgamate with mercury, 281 Amber, 268 Ammeter, 311, 312 Ammonia gas, 152, 154 Amperes, 281, 288, 298 Ampere-hours, 302 Amplihcation, 346 Amplifier, 346 Amplitude, modulation, 393; of vibration, 50, 64, 76, 78; of wave, 52 Angle, critical, 200; of declination, 262; of deviation, 200; of incidence, 184; of inclination, 263; of reflection, 184; of refraction, 196, 198; visual, 231 Angstrom, 2l4 Anions, 295 Annular eclipse, 179 Anode, 295 Antenna, 285, 347, 393 Anthracene, 218 Anti-cathode, 343 Anti-freeze hydrometer, 29 Aperture, 227 Approximate numbers, 13; rules for using, 14 Aquastat, 119 Aqueous humour, 228 Arc, electric, 326, 336, 382; furnace, 337; lamp, 336, 386; welding, 336, 337 Archimedes, 1, 3, 24, 25 Archimedes\u2019 principle, 25 {Exp. 39, 40); ap- plications of, 29-31 Area, measurement of, 10; of conductor and resistance, 288 Argon, 335 Aristotle, 1, 3, 49, 175 Armature, 321, 323, 377 Arrangement of cells, 283 Arrhenius, Svante, 295 Artificial, magnets, 260; transmutation of ele- ments, 399 Astigmatism, 230, 231 Astronomical telescope, 233, 234 {Exp. 254) Atmosphere, pressure of, 120, 148, 150 Atmospheric refraction, 202 Atom, 112, 132, 397; structure of, 269, 270, 398 ; Bohr concept, 399 Atomic, bomb, 401; energy, 3, 401; fusion, 401; number, 398; pile", ", 112, 331, 401 Atropine, 228 Attraction, electrical, 268 {Exp. 358) ; mag- netic, 260 {Exp. 352) Audibility, limits of, 65 Audio-frequency, 347 Auditoria, acoustics in, 59, 77, 88-89 Auditory nerve, 49, 82 Automobile, engine, 159; cooling system, 145; generator cut out relay, 307, 309; ignition system, 328 Axis, principal, 189, 204; secondary, 189 Bacon, Roger, 3, 226 Balance, 12, 34, 35 Balance wheel of watch, 116 Ball and ring apparatus, 115 {Exp. 164) Balloons, 30 Bar magnet, 260 {Exp. 350, 351) Base (Alkali), 295 Battery, 29, 279, 284; hydrometer, 29; rat- ing, 302 Beam, of light, 176; wireless, 326 Beats, 78, 79 {Exp. 106) Becquerel, Henri, 397 Bel, 64 Bell, Alexander Graham, 325 Bell-in-vacuo, 51 {Exp. 96) Bell-jar, 51 Bellows, camera, 226 Beryllium, 400 Beta ray, 398 Bethe, Dr. Hans, 113 Bifocal lenses, 231 Bimetallic strip, 115 Binary colours, 221 Binoculars, 201 Biot\u2019s spheres, 276 {Exp. 362) Black, 221 Blind spot, 229 Blip, radar, 394 407 1 ; INDEX Bohr, Niels, 399; concept of the atom, 399 Boiling-point, 148; effect of air pressure, 148; table of, 152 Bomb calorimeter, 144 Break and make, 379 Breaking the sound barrier, 60, 61 Breezes, land and sea, 130, 131 Bridges, expansion in, 116 Brilliancy, 202 British system of measurement, 9 British thermal unit, 139, 401 Brushes, electric, 314, 321, 323 Bucket and cylinder, 39 Bunsen burner, 163, 164 Buoyancy, 24 Buoyant force, 24 Burette, 12 Burton, Dr. F., 395 Caesium, 219, 347 Calculations with approximate numbers, 13, 14, 15 Calibrating a thermometer, 120, 124 Calipers, 12 Caloric theory, 109 Calorie, 138; food, 139 Calorihc values of fuels, 144 Calorimeter, 142; bomb, 142, 144 Camera, lens, 226, 227; pin-hole, 176-178 Camouflage, 223 Can, overflow, 40, 43 Canadian National Exhibition Band Shell, 58 Carbon dioxide, 156 Carbon resistors, 291; rods, 297; filament, 334 Carrier waves, 347, 393 Catch bucket, 40, 43 Cathode, 295 Cathode rays, 341, 397 {Exp. 383) Cathode ray oscilloscope, 390; tube, 342, 389- 393 Cations, 295 Cat\u2019s fur, 268 Cell, arrangement of, 283; dry, 279; electrolytic, 295, 296, 367, 369; primary, 300; secondary, 300; storage, 300, 372; voltaic, 279, 364 Celsius, 122; scale, 121 Centigrade scale, 121 Centigram, 10; centimetre, 9, 10 Centre of curvature, mirrors, 189; lens, 204 C. G. S. system of measurement, 9 Chadwick, Sir James, 399 Chain reaction, 401 Changes of state, 146 Characteristics of musical sounds, 63 Charges, electric, kinds of, 268, 269; distri- bution of, 276 Charging, by induction, 273 {Exp. 362) ; by contact, 271 {Exp. 359, 360) Churchill, Sir Winston, 394 Ciliary muscles, 229 Circuit, 281 ; breaker, 284, 327 Climate, 146, 150 Clock, 50, 117 Closed tube, 73, 74 {Exp. 102) Clouds, screening effect of, 135 Coal, 112-113, 156, 401 Coaxial cable, 326 Cobalt, 259 Coefficient of expansion, of gases, 123; of liquids, 119; of solids, 116 Cohesion, force of, 146 Coil, 305 Coil spring, 54 Cold frames, 135 Cold light, 176, 218 Collimator tube, 219 Colour, blindness, 223; chart, 221; disc, 215; filters, 221; pigments, 221, 223 {Exp. 253); importance of, 213, 223; nature of, 220222; printing, 223, 224; television, 394; theories of, 220, 222; uses of, 223; vision, 223 Coloured lights, 222 Commutator, 314, 322, 323, 377 Compass, magnetic, 261 Compensating for expansion, 116 Complementary colours, 221 Compound bar, 115, 165 Compound microscope, 232, 396 Compression of gases, ll2 Concave lens, 203; images in", ", 208 {Exp. 250) focal length, 205 {Exp. 248) Concave mirror, 188; images in, 190 {Exp. 241, 242) Concave reflector, 58, 235 Condensation, 54, 74, 75, 79, 325 Condenser, 328 Condensing lens, 233, 235 Conductance, 289 Conduction of heat, 126, 156; in gases, 128; in liquids, 128; in solids, 126, 134 {Exp. 166) Conductometer, 126, 166 Conductors, electric, 272, 288 {Exp. 361) ; dif- ferent shapes, 276 {Exp. 363) Conductors, of electricity, 272; of heat, 127 Cones, for vision, 229 Conservation of energy, 1 1 Constantan, 291 Continuous spectra, 219 Convection currents, 128-131 ; in liquids, 128130 {Exp. 166); in gases, 130, 135, 142 {Exp. 167) Convention of signs, 192, 205, 208 Converging (Convex) lens, 203; images in, 207 {Exp. 250) ; focal length, 205 {Exp. 248) Chemical, change, 112; compounds, 295; effects of electric current, 295; energy, 300 Converging pencil, 176 Convex mirror, 188; images in, 191 {Exp. Chemistry and colour, 220 Chlorine, 154 Choroid coat, 228 Chromatic, aberration, 210, 224; scale, 84 Chromium plating, 299 243) Cooling system of automobile, 145 Copper, 288; ions, 297; plate, 279; plating, 299 Copper sulphate, electrolysis, 297 {Exp. 370) 408 INDEX Copper voltameter, 298 {Exp. 370) Cornea, 228 Corpuscular theory- of light, 181 Corrosion, 299 Cosmic rays, 132 Coulombs, 281 Crest, 52 Critical, angle, 200; pressure, 155; tempera- ture, 155 Crookes, Sir William, 341 Crookes\u2019 tube, 383; dark space, 341 Crystalline lens, 228 Cubic measure, 10 Curie, Pierre and Marie, 397 Current, see electric alternating, direct, convection, Curved mirrors, 188 Cut glass, 202 Cut-out relay, automobile generator, 307-309 Cycle, 50; alternating current, 322 Cyclotron, 400 Cylinder, volume of, 36; automobile, 328 D\u2019Arsonval galvanometer, 310, 377 Davy, Sir Humphrey, 110, 156, 295, 317, 336 Decibel, 64, 92 Decimetre, 9 Declination, angle of, 262 Decomposition of water, 295, 296 Defects, of vision, 230 Degrees, 120; absolute (Kelvin), 123; centi- grade (Celsius), 121; Fahrenheit, 121 Demagnetization, 265 Density, 15; definition, 18 {Exp. 35, 36, 37); effect of expansion on, 124; in relation to convection, 129; maximum density of water, 19, 20; table of, 21 Depolarizing agents, 280 Depth, apparent change in {Exp. 246) Detector circuit, 347 Deuterium, 399 Deuterons, 399 Deviation, angle of, 200 {Exp. 247) Dewar flask, 156 Dial, telephone, 326; thermometer, 117 Diamonds, 199, 202 Diaphragm, camera, 226, 227; telephone, 325 Diatonic scale, 82, 83 Diesel engine, 156, 159, 160 Difference of potential, 280-282 Differential thermometer, 167 Diffusion of light, 184, 185 Dilatometer, 19, 20 Diode tube, 344, 345; as a rectifier, 345 {Exp. 386) ; thermionic emission, 344 {Exp. 384) Dioptre, 205 Dip, magnetic, 263 Dipping needle, 263 Direct current, 346; generator 323 380); motor, 313, 314 {Exp. 377) {Exp. Direction of electron flow, 374 Discharge, electrical, 341 ; through gases, 340, 341; tubes, 219 Discontinuous spectra, 219, 220 Discord, 84 Dispersion, of light, 213 {Exp. 251) Displacement, 30 Dissociation, 295, 296 Distance formula, lens, 208; mirror, 192 Distilled water, 295, 373 Distinct vision, least distance of, 229, 231 Distribution of charges, 276 {Exp. 362) Distributor, 328 Diverging lens, 203 Diverging pencil, 176 Division, 14, 15 Doppler\u2019s principle, 66 Drums, 87 Dry, cell, 279-281; dock, 30; ice, 156; steam, 151 Dunlap Observatory, 234 Dynamo, alternating current, 317, 321 Ear, 81, 82; frequency limits of hearing, 65 Earth inductor, 320 {Exp. 380) Earth\u2019", "s magnetic field, 260 Ebonite rod, 268 Echoes, 58, 59, 394 Eclipses, of moon, 179; of sun, 178, 179 Edison, 334 Efficiencies, of heat engines, 156; of electric heater, 137 Einstein, Dr. Albert, 2, 4, 132, 401 Elasticity, 54 Electric, arc, 326, 336; bell, 306, 307; conductors, 361; eye, 348; furnace, 337; generator, A.C., 321, D.C., 323; heating appliances, 335; insulators, 361; light bulb, 334; organ, 88; power, 337; symbols, 285; welding, 337 Electric charges, 268; induced, 272, 273, 277 {Exp. {Exp. distribution 362) ; kinds of, 269; law of, 269 {Exp. 359) 361); 276 of, Electric circuit, 281; types of, 283 Electric current, 279, 281; and electron flow, 282; alternating, 346, 347; direct, 346; chemical effects of, 295 {Exp. 367-373); energy, 331; induced, 318 {Exp. 378-380); magnetic effects of, 304 {Exp. 373, 374) ; measurement of, 281, 298 {Exp. 370) ; rectified, 345 {Exp. 386) Electric motor, 309, 310; St. Louis, 313, 377; {Exp. 376) Electrical discharge tube, 340 {Exp. 382, 383) Electrical energy, 300, 301, 337 Electricity, 1; heat from, 331, 333; production See Static of, 331; transmission of, 331. Electricity, Electric Current Electrochemical equivalent, 298 Electrodes, 295 Electrolysis, 295, 296 ; uses, 297 ; laws of, 298 {Exp. 369) ; of water, 295, 296 {Exp. 367) ; of copper sulphate solution, 297 {Exp. 368) Electrolyte, 279, 295 Electrolytic cell, 295, 296, 367 Electromagnet, 52, 263, 304, 306 {Exp. 375) Electromagnetic, induction, 317, 318; inertia, 327 Electromagnetic, waves, 132, 181, 219, 347; spectrum, 217 Electromagnetism, 317 Electromotive force, 282 {Exp. 379, 380) 409 1 INDEX Electron, flow, 282, 296, 297 ; microscope, 388, 395; theory, 269; volt, 401 Electrons, 269, 342, 397 ; thermionic emission of, 344 [Exp. 384) ; photo-emission of, 347 {Exp. 386) Electronics, 283, 340, 389 {Exp. 382) Electronic tube, 344 ; as amplifier, 346 ; as rectifier, 345 Electrophorus, 273, 274 {Exp. 362) Electroplating, 297, 299 {Exp. 371) Electrorefining, 297 Electroscopes, 270. See gold-leaf and pith-ball Electrostatic, induction, 272, 273 {Exp. 361) ; machine, 275 Electrostatics, 268 Electrotyping, 299 Elements, 270, 295; discovery of, 219; mean- ing of, 112; transmutation of, 399 Elementary (molecular) magnets, 265 Emergent ray, 200 Emission of radiant energy, 133 {Exp. 167) Emission theory of light, 181 E.M.F., 282; induced, 318 {Exp. 379, 380) Energy, 3, 4, 137; atomic, 401; chemical and electrical, 300, 331; conservation of. 111; equation, 401 ; from foods, 139, 144; light, 175; transformation of, 3 Energy levels (orbits), 181, 399 Engines, 156-160 Equilibrium, 147, 148 Equivalent weight, 298 Error, percentage, 13; possible, 13 Ether, 132 Evaporation, boiling and, 148 Exact numbers, 1 Expansion, 115, 119, 122-124 {Exp. 164-165); and density, 124; unusual for water, 19, 20 Eye, 227 Eyepiece, 232 Fahrenheit, D. G., 121 ; scale, 121 Faraday, Michael, 3, 262, 295, 298, 317 Faraday\u2019s, Laws, 298; {Exp. 369) ; dark space, 341 Farsightedness, 230 Field coils, 314 Field glasses, 201 Field, magnetic, 262; around conductor, 304 {Exp. 373) ; around helix, 305, 306 {Exp. 374) Filament, 334, 343, 344, 346 Filters, colour, 221, 394 {Exp. 253) Fire extinguisher, 156 Fission, atomic, 400 Fixed points on thermometer, 120 Fixed resistors, 291 Fizeau, A. H., 179 Flame of Bunsen burner, 164; producing dis- charge, 340 {Exp. 383) Flannel, 268 Floating docks, 30 Flotation, principle", " of, 26 {Exp. 43) Fluid, 26 Fluorescein, 196 Fluorescence, 176, 218 410 Focal 189 lens, 204, 205; mirror, length, {Exp. 241) Focal plane, 204 Focus, principal, of lens, 204; of mirror, 189 Foods, 137 ; energy from, 139, 144 Foot, 7, 9 Force of cohesion, 146, 147 Forced-air heating, 131 Forced vibrations, 66 Formulae for: electrical energy conversion of centigrade degrees to absolute (Kelvin), 123, centigrade degrees to Fahrenheit, 122; distance from velocity and time, energy 92; (Kilowatt-hours), equation 337; (Einstein), 401; frequency of sound wave, 53; heat from electricity, 331, 333; images in inclined mirrors, 188; index of refraction, 198; lens formulae, distance, magnification, 208 ; magnifying power, compound microscope, 231, telescope, 233 ; mirror formulae, distance, magnification, 192; Ohm\u2019s law, 287, 288; pin-hole camera, 178; power of a lens, 205; power of electricity, horsepower, 337, watts, 333, 337; resistances, in parallel, 289, in series, 289; specific gravity, 20; transformer, of wave, 53; volumes of 324; solids, 36; wave-length of sound, 75, 77 magnifying velocity glass, 232, Four-cycle engine, 159 F.P.S. system of measurement, 9 Franklin, Benjamin, 268, 269, 276 Fraunhofer lines, 220 Freezing, 146; mixture, 155; point, 120; of pond, 20 Frequency of vibration, 50, 64, 65, 92; determination of, 64, 65; of waves, 53; of various sources, 65; ultrasonic, 65, 92 Frequency modulation, 393 Friction, 156; electricity produced by, 268, 270 {Exp. 358) Fuels, 113, 137, 144 Fundamental, 69, 75 Furnace, electric, 337 Fuses, 333 Fusion, meaning of, 146; atomic, 401; heat of, 148 {Exp. 170) Future of sound, 92 Galileo, 2, 3, 120 Galvani, L., 279, 295 Galvanometer, D\u2019Arsonval, 310, 311 Galvanoscope, 309, 375 Gamma ray, 219, 398 Gas, 113; thermometer, 124 Gaseous ions, 340 Gases, thermal conductivity of, 128; convection in, 130 {Exp. 167) ; density of, 21; expansion of, 122-124; liquefaction of, 154; molecular motion of, 110; sound transmission in, 54; spectra of, 219 Generator, 308; A.C., 321 {Exp. 380); D.C., 323 {Exp. 380) Geographic North Pole, 260 Geometric construction of images, 186, 190, 204, 207 Gilbert, Dr. William, 2, 3, 4, 259, 262, 268 INDEX Glare, light, 185 Glass, insulator, 288; plate, 199 {Exp. 245); rod, 268 Gold-leaf electroscope. 271; charging by contact, 272; charging by induction, 273, 274 {Exp. 362) ; identifying charge, 272 {Exp. 360) Graduated cylinder, 12, 34 Gram, 10 Graphite, 299, 401 Greenhouse, 134, 135 Grid, 301, 344, 346, 348 Ground, 273, 274, 285 of, 128-131 Half-wave rectification, 346 Hammond organ, 88 Harmonic, 69 Headlights, car, 194 Hearn, R. L., Generating Station, 158 Heat, 1, 3; absorption of, 133 {Exp. 168); conductivity of, 126-128 {Exp. 166) ; con166) ; expanvection sion caused by, 122-124 {Exp. 164. \u201965) ; insulators, 127, 128, 142; measurement of, 137-162; nature of, 110; of fusion, 146-150 {Exp. 170); of vaporization, 150132-136 153 {Exp. 167); sources of. 111; specific heat, and 139-146 {Exp. temperature, 169) ; 119-121; thermometers, 119-122; transfer of, 126; theory of, 109, 110; and work, 156 Heat exchange, principle of, 137, 140; dur- radiation 172) ; {Exp. {Exp. 109, of, ing changes of state, 146 Heat from electric current, 331, 333 Heat from nuclear fission, 401 Heating appliances, electrical, 335 Heating systems, hot water, 145, 152; hot air, 130, 131 119, 129, 130, Heavy water, 399 Helium, 113, 124, 220, 398, 400 Helix, 305; rule, 306 {Exp.", " 375) Herschel, Sir W., 216 Herschel\u2019s divided tube, 77, 78 {Exp. 105) Hoffman water voltameter, 296 Horse-power, 337 Horseshoe, magnet, 260 ; electromagnet, 306, 307 Hot-air heating, 130, 131 Hot-water, heating, 119, 129, 130, 145, 152; supply, 129-130 frequency limits of Hues, 221 Human, ear, 82; hearing, 65; eye, 227 81, Humours, aqueous, vitreous, 228 Huygens, C., 181 Hydro-electric power, 331 Hydrogen, 113, 154, 280, 295, 399; bomb, 401; ion, 296, 398; isotopes, 399, 401 Hydrometer, 27-29 {Exp. 44, 45) Hydroxyl ion, 296, 297 Hypo (sodium thiosulphate), 196 Image, in camera, 226; real, 178, 190, 192, 207; virtual, 185, 186, 191, 192, 207; in {Exp. 250), convex, lenses, concave, 207 185 207 {Exp. 239), concave, 190 {Exp. 242), con{Exp. 243), inclined, 187 {Exp. vex, 191 240), in pin-hole camera, 178 {Exp. 237) in mirrors, plane, {Exp. 250) : Image orthicon, 391 Incandescence, 175 Incandescent lamp, 334 Inch, 7, 9, 10 Incident rays, 183 Inclination, angle of, 263 Index of refraction, 198, 224 {Exp. 243) Induced charges, 272, 273, 277 {Exp. 361) Induced current (E.M.F.) 318; cause of, 318 {Exp. 378); direction of, 319 {Exp. 380); self, 327 magnitude of, 318 {Exp. 379) ; {Exp. 382) Induced magnetism, 263, 264 {Exp. 354, 355) Induction, 263; charging electroscope by, 273, 274 {Exp. 362) Induction coil, 327 ; uses, 328, 383 Infra-red radiations, 132, 216 In parallel, 284, 312 Input, 112 In series, 283, 312 Instruments, measuring, 12; musical, 85-88 Insulators, of electricity, 272, 288 {Exp. 361) ; of heat, 128, 142; of sound, 64, 88, 89 Intensity of sound, 63, 64, 77, 78 Interference, 54, 55; of sound waves, 76-79 Internal combustion engine, 159; ignition sys- tem, 328 International ampere, 298 Invar, 117 Invisible radiation, 216, 218 Ionization, in gases, 340; theory of, 295 Ions, 295; gaseous, 340 Ion-pairs, 295 Iris, 228; reflex, 228 Iron, 259; electric, 335, 336; filings, 259, 260; soft, 263, 264 Irregular reflection, 185 Isotopes, 399 ; of hydrogen, 399 Jupiter, moons of, 179 Kaleidoscope, 188 Keeper, magnetic, 265, 266 Kelvin, Lord, 123 Kelvin (absolute) temperature, 123 Kettle, electric, 335 Kilo, gram, 10; metre, 9; volt, 282; watt, 337 Kilogram calorie, 139 Kilowatt-hour, 113, 337 Kinescope, 391 Kinetic theory, 110 Knife switch, 284 Knowledge, seven steps to, 4 Ice, dry, 156; heat of fusion of, 150 {Exp. 170) Ignition system, 328 Lactometer, 29 Lamp, arc, 336, 386; electric 334 Land breezes, 130, 146 Larynx, 81 411 ) INDEX Laterally, displaced, 199; inverted, 186 Law of: conservation of energy 111; electrolysis, 298; electrostatics, 269 {Exp. 359) ; of sound, 64; Lenz, 319 {Exp. intensity 380); magnetism, 260, 352; reflection, of sound, 58, of light, 184; refraction of light, 197; vibrating strings, 67-69 {Exp. 99) Lead, 397, 398 ; spongy, 301 Lead-acid storage battery, 300 {Exp. 372) Lead peroxide, 301 Left-hand rule, 305 {Exp. 374) Length, measurement of, 7, 8, 9; of conduc- tor and resistance, 288 Lenses, accommodation, 229; action of, 203 {Exp. 250); applications of, 210; crystalline, 228; focal length of, 205 {Exp. 249); formulae, 208 ; power of, 205 Lenz\u2019s Law, 319 {Exp. 380) Lifting electromagnet, 308 Light, 1, 175; colour of, 213-214; diffusion of, 185; dispersion of, 213", "; nature of, 175; reflection of, 183 {Exp. 238) ; refraction of, 196 {Exp. 243); sources of, 175; theories of, 181 ; transmission of, 96, 132, 176; velocity of, 179, 180 Light, energy, 349; meter, 349; microscope, 396; sensitive metals, 347; year, 180; waves, 394 Lightning, 276; rods, 276, 277 Limits of audibility, 65 Line spectra, 219 Lines of magnetic force, 262, 263 {Exp. 353) Linear measure, table of, 9, 10 Liquefaction, 146; of gases, 154 Liquid, buoyancy of, 24, 25, 26 ; heat conductivity of, 128, 166; convection in, 128-130 {Exp. 166); density of, 18, 21; expansion of, 19-20, 119; specific gravity of, 21 {Exp. 38, 42, 45) Liquid air, 155, 156 Litre, 10 Local action, 280 {Exp. 364) Lodestone, 259 Longitudinal vibrations, 51 {Exp. 95) Longitudinal waves, 54 {Exp. 97) Loop or coil, 305 Loops, 57, 69, 74, 86 {Exp. 99) Loudness of sound, 50, 63-64, 77, 78 Lower fixed point, 120 Lucite, 358 Luminous bodies, 175 Magnet, 260; kinds, 266 Magnetic, circuit breakers, 307; compass, 261; effects of electric current, 304, 305 {Exp. 373, 374); field, {Exp. 353); lenses, 395, 396; poles, 260 {Exp. 350-351) ; shield, 264 separator, 264; substances, 259 {Exp. 350) {Exp. 355) ; 262, 306 Magnetism, 1, 259; induced, 263 {Exp. 354, 355) laws of, 260, 352 {Exp. 352); terrestrial, 260-263; theory of, 264 {Exp. 357) Magnetite, 259 Magnetization {Exp. 356) Magnification formula, lens, 208; mirror, 192 Magnifying glass, 231 412 Magnifying power, 231, 232, 233, 395 Major triad, 84 Make and break, 379 Manganin, 288, 291 Mass, 10, 11, 19, 112, 114; number, 398, 399 Matter, states of, 146; and energy, 112, 114, 401 Maximum density of water, 20 Mean, position, 50; solar day, 11 Measurement, 7 ; accuracy of, 11; of heat, 137-162; of of resistance, 292; standard units of, 8; systems of, 7 of mass, length, 10; 9; Measuring devices, 7, 11, 12 Mechanics, 1 Media, for transmitting sound, 51, 52 {Exp. 96); light, 52, 96; radiant energy, 132 Megacycle, 394 Megohm, 282 Melting, definition, 146; point, 147 Meniscus, 34 Mercury, 118-120, 128, 281; switches, 284 Metals, heat conductivity of, 126 {Exp. 166) ; 21; expansion of, density of, {Exp. 165); plating of, 299 {Exp. 371); purification of, 297; specific heat of, 139, 144, 146 {Exp. 169) 115 Meteorological balloons, 31 Method, of science, 4; of mixtures, 144, 148 Metre, 9 Metric system, 9 Mho, 289 Mica, insulator, 288 Michelson, A. A., 180 Microampere, 282 Micrometer screw gauge, 12 Microscope, electron, 388, 395-397 ; optical, 232, 396 Microwaves, 393 Mile, 9 Mini, 10; litre, gram, ampere, 282; metre, 9; volt, 282 Minimum deviation, 200 Mirage, 202 Mirrors, 183; applications of, 193; curved, 184; formulae, {Exp. 240); parallel, 187; plane, 183 {Exp. 239) 192; inclined, 187 10; Mixtures, method of, 144, 148 Moderators, 401 Modes of vibration, in air columns, 74, 75; in strings, 69 {Exp. 100) Modulation, 347, 393 Molecules, 19, 112, 128, 132, 134, 152, 297; motion of, 110; kinetic theory of, 110 Molecular theory of magnetism, 265 {Exp. 357) Monochromatic light, 199; flame attachment, 253 Moseley, 398 Motor, principle, 309, 310 {Exp. 376, 377); St. Louis, 313; structure of, 313; direct current, 314 {Exp. 377 Movies, projector, 235; sound, 90, 91 Multiplication, 14, 15 Museum of Science and Industry, 58 Music, 49; analysis of, 391 INDEX Musical, interval, scales, 82-84;", " sounds. 63 82; instruments, 85-88; Nautilus (submarine), 402 Nearsightedness, 230 Negative, charge, 269-270; photographic, 227 Neon, tube, 176; lamp, 382, 386 Nerve cells, 229 Neutral body, 268, 270 Neutralization, 297 Neutron, 269, 397, 399 Newton, Sir Isaac, 2, 3, 181, 213 Newton\u2019s disc, 215 {Exp. 252) Niagara, power, 332; region, 146 Nichrome, 288, 335 Nickel, 259, 300 Nitrogen, 124, 156, 335, 399; fixation, 337 Node, 57, 69, 74, 99 Noise, 49, 63, 391 ; noise level, 64 Non-electrolyte, 295 Non-luminous, 175 Non-magnetic substances, 259 {Exp. 350) Non-periodic vibrations, 63 Normal, 184, 189 North magnetic pole, 260 N-pole (north-seeking pole), 260 Nuclear, charge, 398; fission, 400; physics, 1, 397 ; reactors, 112, 401, 402 Nucleus, 269, 397 Numbers, exact, 11; approximate, 13 Objective, optical, 232 Octave, 65, 82 Oersted, H. C., 304, 317 Ohm, G. S., 287 Ohm, 3, 281, 288 Ohm\u2019s Law, 287 {Exp. 365) Oil, 112, 113, 160 Oil-immersion microscope, 232 Opaque, 134, 176 Open tube, 74, 86 Optical, centre, 204; density, 197, 199; disc, 183, 184; instruments, 226-235; microscope, 396 Optics, 175 Orbit (or energy level), 269, 399 Organ, electric, 88; pipes, 85-87 Origin of sound, 49 Orthicon, 391 Oscillator, 347 Oscilloscope, sound tracings, 63, 70 Ounce, 10 Output, 112 Overflow can, 40, 43 Overtone, 69, 75, 86 Oxidizing agent, 280 Oxygen, 156, 295, 399 Parabola, 194 Parabolic mirrors, 194, 234 Parallax, method of, 34 Parallel, circuit, 283, 284; mirrors, 187 Partial eclipse, 178 Pencils, converging and diverging, 176 Pendulum, 50, 117 {Exp. 94); clock, 12 Penumbra, 178 Percentage error, 13 Percussion, 112 Period of vibration, 50; of wave, 53 Periodic vibrations, 63 Periscope, 201 Perm.alloy, 263 Permanent magnets, 266 Permeability (magnetic), 263 Persistence of vision, 215, 390, 393 Perspex, 201 Phase, 52, 76-78 Phonograph, 90 Phosphors, 394 Phosphorus, 176 Photoelectric cell, 347 Photoelectricity, 347 Photoemission of electrons, 347 {Exp. 386) Photography, 178, 216, 223, 227, 349 Photon, 181 Photosynthesis, 113, 114 Photronic cell, 348 Physical states of matter, 146 Physics, 1 ; definition, 3, 8 Piano, 85 Pictures, motion, 90, 91, 235 Pigments, 221, 223 Pin-hole camera, 176, 177; image in, 226 {Exp. 237) 178, Pint, 10 Pipe organ, 85-87 Pipeless furnace, 131 Pitch, 50, 63-65, 214 {Exp. 99) Pitchblende, 397 Pith-ball electroscope, 270; charging by contact, 271; identifying charge, 271 {Exp. 359) Planck, Max, 181 Plane mirrors, 185; images in, 185, 186 {Exp. 239) Plate, 344; current, 345 Platinum, 296 Plimsoll line, 29, 30 Point of incidence, 184 Points, action of, 276 {Exp. 363) Polarization, 280, 364 Poles, magnetic, 260 Polonium, 397 Polystyrene, 358 Pond, freezing of, 20 Positive charge, 269, 270 Possible error, 13 Potassium, 347; dichromate, 280, 364; nitrate, {Exp. 350-352) 130; permanganate, 128, 166 Potential, difference, 280-282; energy, 112 Pound, 10 Power, meaning, 337; of electric current, 324, 337; of lens, 205 Prefixes (metric), 9 Presbyopia, 231 Pressure, atmospheric, 120; cooker, 147; critical, 155; effect on boiling point, 147, 148; in water, 282 Primary, cells, 300; coil, 323, 378; currents, 318, 319, 327 ; colours, 220-222 Principal axis, of lens, 204; of mirror, 189 Principal focus, of lens, 204; of mirror, 189 Principle, {Exp. 39, 40) ; Doppler\u2019s, 66; of flotation, 26", " {Exp. 43); Archimedes\u2019, 25 413 7 INDEX of generator, 321, 323 {Exp. 380) ; of heat exchange, 137, 140 {Exp. 169) ; of hydrometer, 28 {Exp. 44) ; of lead-acid storage direct current cell, 300 {Exp. 372) ; motor, 309, 310 {Exp. 376, 377) of Printing, colour, 223, 224 Prisms, 199, 213; reversed, 215, 224 {Exp. 247) ; total reflection, 202 Projector, slide, 235 Proof plane, 272 Protium, 399 Proton, 269, 397, 398 Protractor, 263 Psychology and colour, 220 Pulsating current, 323 Pupil of eye, 228 Purification of metals, 297 Push button switch, 284 Pyrex, 1 1 Pyrite (fool\u2019s gold), 109 Quality of sound, 63, 70, 81 {Exp. 101) Quantity, of heat, 137-139; of electricity, 281, 298 Quanta, 132, 181 Quantum theory, 181 Quartz, prism, 218; lenses, 218 Queen Elizabeth, the, 30 Quick-freeze units, 154 Radar, 394; screen, 395 Radiant, energy, 132; heating, 136 Radiation, 142, 135, 131, 126, 156 {Exp. 167, 168) Radioactivity, 397 Radio, frequency, 347 ; transmission and recep- tion, 347; waves, 219 Radiometer, 134 Radium, 112, 397 Rainbow, 199, 216 Range, electric, 335; finder, 227 Rarefaction, 54, 77, 79, 325 Rays, alpha, 398; beta, 398; cathode, 341, 397; cosmic, 132; gamma, 219, 398; infrared, 152, 216; light, 175, 176; ultra-violet, 132, 347 Reactors, nuclear, frontispiece, 401-403 Real image, 178, 207 Rear-vision mirror, 193 Receiver, telephone, 325 Receiving station, radio, 347 Reciprocating steam engine, 157 Recomposition of white light, 215 {Exp. 252) Recorder, sunlight, 206 ; tape, 90 Recordings, 90 Rectification, 345 {Exp. 386) Rectifier, tube, 302, 346 ; circuit, 345 Rectilinear propagation, 176, 177 Reducing heat loss, 127 Refining of metals, 297 Reflected rays, 183 Reflecting telescope, 233, 234 Reflection of light, 183, diffuse, 184; laws of, 184 {Exp. 238) ; regular, 184; total, 200 Reflection of sound, 57, 73; laws of, 57 Reflex time, 59 414 Refracting angle, 199 Refracting telescope, 233 Refraction, of light, 196; atmospheric, 202; 197, 198 {Exp. 243) ; laws of, index of, 197 {Exp. 243); through glass plate, 199 {Exp. {Exp. IM) 245) ; through prisms, 199 Refractive index, 197, 198 {Exp. 243) Refractometer, 198 Refrigerator, electric, 153; gas, 154 Regular reflection, 184 Relativity, Theory of, 132 Relay station, T.V., 393 Reproduction of sound, 90-92 Resistance, electrical, 112, 281, 288; factors affecting, 288; measurement of, 292 {Exp. 365-367) ; table of, 288; unit of, 281, 288 Resistance box, 291 Resistors, 288; in series, 288; in parallel, 289; types of, 291, 292 Resolving power, 395 Resonance, 73; and interference, 73-80; and velocity of sound, 60, 75 {Exp. 102) Retina, 229 Reverberation, 59, 88 Rheostat, 292 Rod, unit of measurement, 9 Rods, for vision, 229 Roentgen, 342 ; rays, 343 Rotor of steam turbine, 158 Rounding-off numbers, 14 Rubidium, 219 Ruhmkorff, 327 Ruler, 12, 34 Rumford, Count, 109, 156 Rutherford, 398 Salts, 295 Saturation effect (magnetism), 265 Savart\u2019s toothed wheel, 64 {Exp. 99) Scales, musical, 82-84; diatonic, 82; of equal temperament, 84, chromatic, 84 Scales, temperature, 121-123; absolute (Kelvin), 123; centigrade, 121; Fahrenheit, 121 Science, aim of, 8 Scientific method, 4, 259 Sclerotic coat, 227 Sea breeze, 131, 146 Searchlight, 194 Second, 11 Secondary, axis, 189; cells, 300; coil, 323, 327; currents, 318, 319 Segments, commutator, 314 Self-inductance, 326 {Exp. 382) Self-induced current, 327 Semitone, 83 Series circuit,", " 283 Shadows, 178 Shaving mirror, 193 Shorting plugs, 292 Shunt, 312; connection, 283 Shutter, camera, 226, 227 Significant digits, 13, 14 Silent points about tuning fork, 76, 77 {Exp. 104), Silk, 268 ; 1 INDEX Silver, 298; plating, 299; voltameter, 370 Simple hydrometer, 28 Sine of angle, 198 Slide projector, 235 Slip rings, 321 Smoke paper, 130 Sodium, 347; thiosulphate (hypo), 196 Soft iron, 263, 264 Solar, battery, 113; energy, 113; spectrum, 220 Solenoid, 305, 378 Solidihcation, 146 Solids, conductivity of heat, 126 {Exp. 166); density of, 21 {Exp. 35, 36); expansion of, 115 {Exp. 164, 165) ; specific gravity of, 21 [Exp. 41) 92; future, insulation, Sonar, 92 Sonometer, 66 Sound, 1, 49; characteristics, 63-72; dehnition, 89; 49; 54-57 intensity, {Exp. 104, 105, 106); media for transmisorigin, 49; pitch, 63-65 sion, 51 {Exp. 96) ; {Exp. 99); quality, 63 [Exp. 101); reflection, 57; reproduction, 90-92; resonance, 60, 73 {Exp. 102); transmission, 54; velocity of, 53 {Exp. 102) interference 63 ; 88, 50, 64, of, Sound, barrier, 60; ranging, 91; track, 90; waves, 54, 325 Sounding, 92; balloons, 31 Sources, of heat. 111, 114; of sound, 49; of light, 175 South magnetic pole, 261 Spark, 326; discharge, 275; plugs, 328 Specific gravity, 20, 301 ; determination of, 27, 28 {Exp. 38, 41, 42) Specific-gravity bottle, 38 Specific heat, definition of, 139 {Exp. 169) Spectacles, 230 Spectra, 213, 214; kinds, 219 {Exp. 251) Spectroscope, 218, 219 Spectrum analysis, 219 {Exp. 253) Sphere, volume of, 36 Spherical, aberration, 193; mirrors, 188 S-pole, 260 Spongy lead, 301 Spot lights, 336 Square measure, 10 Standard pressure, 120 Standing waves, 55-57 {Exp. 98) States of matter, 146 Static, 394; electricity, 268 {Exp. 358) Steam, engine, 157; generating plants, coloured, 223 158, 331; heating, 152; turbine, 157; trap, 151 Step, -down transformer, 324; -up, 324 Steps to knowledge, 4 St. Louis Motor, 313 {Exp. 377, 380) Storage battery, 279; hydrometer, 29 Storage cell, 300; action in, 301, 302 {Exp. 372); structure, 300, 301; uses, 302 Stringed instruments, 85 Strings, vibrating, modes of, 69-70 {Exp. 100); laws of, 67-69 {Exp. 99) Sublimation, 146, 156 Submarine, 30, 402 Subtraction, 14, 15 Subtractive theory of colour, 221 Substitution method for resistance, 292, 293 {Exp. 367) Sugar, 295 Sulphate ions, 296 Sulphuric acid, 279, 301 Sun, 1 12, 175, 349 Sunlight recorder, 206 Sunset, 202 Superposition, of light waves, 181; of.sound waves, 54-57 {Exp. 101) Switches, 284 Symbols, electric, 285 Sympathetic vibrations, 76 {Exp. 103) of liquids, 119; coefficient Tables of; boiling-points, 152; calorific values of fuels, 144; coefficient of cubical expansion of linear expansion of solids, 117; critical pressures and temperatures, 155; efficiencies of heat engines, 156; electrical conductors and insulators, 272; electrical symbols, 285; electrochemical electromagnetic waves, 217; heat conductivities, 127; heats of fusion, 150; heats of vaporization, 152; indices of refraction, 199; linear measure, 9; mass, 10; prefixes, 9; resistances, 288; specific gravities, 21; specific heats, 139; units of electricity, 282; velocity of sound and temperature, of sound in various media, 61; volume, 10; wave-lengths of coloured lights, 214 equivalents, velocity 298; 60; Tape recorder, 90 Telephone, 325, 326 Telescopes, 233, 234 {Exp. 254) Television, 391 ; colour, 394; picture tube, 391 Temperature, meaning of, 119; absolute, 123; critical, 155; scales, 121; and quantity of heat, 137; and resistance, 288 Temporary magnets, 266 Terrestrial telescope, 234 Theories, atomic, 270; colour (additive and subtractive),", " 220-223; electron, 269; heat, 109; light (corpuscular, 181; magnetism, 264-265; molecular (kinetic), (Einstein), 132; wave, of 110; relativity heat, 132, of light, 181, 197, of sound, 54 Thermionic emission of electrons, 344 {Exp. 295; quantum), (Arrhenius), ionization emission, 384) Thermodynamics, 109 Thermograph, 117, 118 Thermometers, 1 19-122 ; calibrating, 120, 124; construction of, 120; dial, 117; how to use, 164 1 Thermos bottle, 134, 156 Thermoscope, 134 Thermostat, 118 Thomson, Sir J. J., 342, 397 Time, 8, Tonic, 82 Total eclipse, 178 Total reflection, 200; prisms, 201, 202 Traffic signals, 223 Transfer of heat, 126 Transformers, 323, 324 415 INDEX Translucent, 176 Transmission of, 126, 128, 132; light, 176; radiant energy, 132134; radio waves, 347; sound, 49-62 electricity, 331 ; heat, Transmitter, telephone, 325 Transmitting studio, 347 Transmutation, artificial, 399 Transparent, 134, 176 Transverse vibrations, 50, 78, 132 {Exp. 94, 95) Transverse waves, 52, 78, 132 {Exp. 97) Triad, major, 84 Triode tube, 344, 346; amplifier, 346; oscil- lator, 347 Tritium, 399 Troughs, 52 Tubes, closed and open, 73, 85-87 Tuning-fork, 55 {Exp. 96) ; interference of sound around, 76 {Exp. l04) Tuning musical instruments, 79 Tungsten, 335 Turbines, 331; steam, 157 U, see Uranium below Ultrasonic frequencies, 65, 92 Ultra-violet, lamps, 218; radiation, 218; light, 132, 347 Umbra, 178 Unison, 79 Units, electrical, 282; heat, 138; measure- ment, 8-11; power, 337; sound, 64 Universal hydrometer, 28 Upper fixed point, 120 Uranium, U, 112; U-235, 400; disintegration series, 397; salts, 397; and coal, 401 Vacuum, 52, 132, 154; bottle, 134; evaporator, 147; tube, 132, 340 {Exp. 384, 386); tunnel (light), 180 Valves, 344 Vaporization, 146; heat of, 150 {Exp. 172) Variable resistors, 292 Velocity, definition, 53; of light, 179, 198; of radio waves, 132; of sound, 59, 75 {Exp. 102 ) Ventilation, 130 Vernier calipers, 12 Vertex, 189 Vibrating strings, laws of, 67-69 {Exp. 99) ; mode of vibrations 69 {Exp. 100) Vibration, 325; definition, 49; forced, 66; frequency for light, 214; kinds of, 50, 51; air columns, 74; of longitudinal, 51; of strings, 67-69; sympathetic, 76; transverse, 50 View finder, camera, 227 Virtual image, 185, 186, 191, 207 Vision, defects of, 230; persistence of, 215, 390, 393 Visual, angle, 231; cells, 229 Vita glass, 218 Vitreous humour, 228 Vocal cords, 81 Voice, 81 416 Volt, 282, 288 Volta, A., 3, 273 Voltage-drop, 282, 283, 288 Voltaic cell, 279 {Exp. 364) Voltameter, water, 296, 368; copper, 298 {Exp. 370) Voltmeter, 311-313 Voltmeter-ammeter method for resistance, 292 {Exp. 366) Volume, 10, 36 Watch, stop, 12; balance wheel, 116-117 Water, 295; boiling point of, 120; conductivity of, 127, 128; convection in, 129; density of, 18-20; displacement, 30; electrolysis of, 295, 296; expansion of, 124; heating {Exp. system, 170) ; heat of vaporization, 150-153 {Exp. 172); refractive index, 199; specific heat, 145; sound transmitted by, 52; supply (hot water), 129-130; total reflection in, 200, 201; turbine, 157, 331; voltameter, 296 129; heat of fusion, 148 Water, heavy, 399 Watt, Sir R. W., 394 Watt, definition, 337 ; hour, 337 Wave, 52; audio-frequency, 347; carrier, 347, 394; electromagnetic, 132, 181, 219, 347; formula, 53, 60, 75, 77, 86; front, 54, 197; heat, 132; length, 52, 70, 77, 214", "; light, 181; modulated, 347, 391, 393; motion, 3, 52-54 pulsating, 345; radio, 217, 219; sound, 54; train, 53 {Exp. 97) ; Wave length, of electromagnetic waves, 217; of sound waves, 75 Waves, electromagnetic, 132; longitudinal, 54; standing, 55-57 {Exp. 98) ; superposition of, 54-57 {Exp. 101); transverse, 52; television, 391, 392, 393 Weather balloons, 31 Weight, 7, 11 Welding, electric arc, 337 Wheel, balance, of watch, 116 Whirl, electric, 276 White, the colour, 221 White light, composition, 213, 215 Wimshurst machine, 275 Wind instruments, 85-86 Winds, 130 Wire-wound resistors, 291 Work from electricity, 337; and heat, 156; rate of doing, 337 X-rays, 132, 219, 342; nature of, 343 {Exp. 384) ; photograph, 343 X-ray tube, 328, 343 ; chest unit, 342 Yard, definition, 7, 9 Yellow spot, 229 Young-Helmholtz theory, 223 Zero, absolute, 123, 132 Zinc, photo-emission of electrons, 347, 348 {Exp. 386) ; plate in voltaic cell, 279 THB 85 Date Due 1 i 1 s QC 23 E8G 1958 C-3 EUBANK HOWARD LAWRENCE 1910- BASIC PHYSICS FOR SECONDARY SCHOOLS 399032G3 CURR HIST -00002252B7M3- OC 23 ESS 1958 C. Eubank, Howard Lawrence, 1910- 3 Basic physics for secondary schoo 1 39903263 CURR HIST,X- CJ STUDENT ohool BASIC PHYSICS THE MACMILLAN COMPANY OF CANADA LIMITED checks (NOTE TO SELF: Add to this table as we go along with examples from each section.) Now you don\u2019t have to memorise this table but you should read it. The best thing to do is to refer to it every time you do a calculation. 1.9 Temperature We need to make a special mention of the units used to describe temperature. The unit of temperature listed in Table 1.1 is not the everyday unit we see and use. Normally the Celsius scale is used to describe temperature. As we all know, Celsius temperatures can be negative. This might suggest that any number is a valid temperature. In fact, the temperature of a gas is a measure of the average kinetic energy of the particles that make up the gas. As we lower the temperature so the motion of the particles is reduced until a point is reached 8 where all motion ceases. The temperature at which this occurs is called absolute zero. There is no physically possible temperature colder than this. In Celsius, absolute zero is at 273oC. Physicists have de\ufb02ned a new temperature scale called the Kelvin scale. According to this scale absolute zero is at 0K and negative temperatures are not allowed. The size of one unit kelvin is exactly the same as that of one unit Celsius. This means that a change in temperature of 1 degree kelvin is equal to a change in temperature of 1 degree Celsius| the scales just start in di\ufb01erent places. Think of two ladders with steps that are the same size but the bottom most step on the Celsius ladder is labelled -273, while the \ufb02rst step on the Kelvin ladder is labelled 0. There are still 100 steps between the points where water freezes and boils. \u00a1 water boils ---> |----| |----| |----| |----| |----| 102 Celsius 101 Celsius 100 Celsius Celsius 99 Celsius 98 |----| |----| |----| |----| |----| 375 Kelvin 374 Kelvin 373 Kelvin 372 Kelvin 371 Kelvin ice melts ---> |----| |----| |----| |----| |----| 2 1 0 -1 -2 Celsius Celsius Celsius Celsius Celsius...... |----| |----| |----| |----| |----| 275 Kelvin 274 Kelvin 273 Kelvin 272 Kelvin 271 Kelvin |----| |----| |----| |----| |----| -269 Celsius -270 Celsius -271 Celsius -272 Celsius -273 Celsius |----| |----| |----| |----| |----| 4 Kelvin 3 Kelvin 2 Kelvin 1 Kelvin 0 Kelvin absolute zero ---> (NOTE TO SELF: Come up with a decent picture of two ladders with the labels |water boiling and freezing|in the same place but with di\ufb01erent labelling on the steps!) This makes the conversion from kelvin to Celsius and back very easy. To convert from Celsius to kelvin add 273. To convert from kelvin to Celsius subtract 273. Representing the Kelvin temperature by TK and the Celsius temperature by ToC, TK =", " ToC + 273: (1.1) It is because this conversion is additive that a di\ufb01erence in temperature of 1 degree Celsius is equal to a di\ufb01erence of 1 kelvin. The majority of conversions between units are multiplicative. For example, to convert from metres to millimetres we multiply by 1000. Therefore a change of 1m is equal to a change of 1000mm. 1.10 Scienti\ufb02c Notation, Signi\ufb02cant Figures and Rounding (NOTE TO SELF: still to be written) 9 1.11 Conclusion In this chapter we have discussed the importance of units. We have discovered that there are many di\ufb01erent units to describe the same thing, although you should stick to SI units in your calculations. We have also discussed how to convert between di\ufb01erent units. This is a skill you must acquire. 10 Chapter 2 Waves and Wavelike Motion Waves occur frequently in nature. The most obvious examples are waves in water, on a dam, in the ocean, or in a bucket. We are most interested in the properties that waves have. All waves have the same properties so if we study waves in water then we can transfer our knowledge to predict how other examples of waves will behave. 2.1 What are waves? Waves are disturbances which propagate (move) through a medium 1. Waves can be viewed as a transfer energy rather than the movement of a particle. Particles form the medium through which waves propagate but they are not the wave. This will become clearer later. Lets consider one case of waves: water waves. Waves in water consist of moving peaks and troughs. A peak is a place where the water rises higher than when the water is still and a trough is a place where the water sinks lower than when the water is still. A single peak or trough we call a pulse. A wave consists of a train of pulses. So waves have peaks and troughs. This could be our \ufb02rst property for waves. The following diagram shows the peaks and troughs on a wave. Peaks Troughs In physics we try to be as quantitative as possible. If we look very carefully we notice that the height of the peaks above the level of the still water is the same as the depth of the troughs below the level of the still water. The size of the peaks and troughs is the same. 2.1.1 Characteristics of Waves : Amplitude The characteristic height of a peak and depth of a trough is called the amplitude of the wave. The vertical distance between the bottom of the trough and the top of the peak is twice the amplitude. We use symbols agreed upon by convention to label the characteristic quantities of 1Light is a special case, it exhibits wave-like properties but does not require a medium through which to propagate. 11 the waves. Normally the letter A is used for the amplitude of a wave. The units of amplitude are metres (m). 2 x Amplitude Worked Example 1 Amplitude Amplitude Question: (NOTE TO SELF: Make this a more exciting question) If the peak of a wave measures 2m above the still water mark in the harbour what is the amplitude of the wave? Answer: The de\ufb02nition of the amplitude is the height that the water rises to above when it is still. This is exactly what we were told, so the answer is that the amplitude is 2m. 2.1.2 Characteristics of Waves : Wavelength Look a little closer at the peaks and the troughs. The distance between two adjacent (next to each other) peaks is the same no matter which two adjacent peaks you choose. So there is a \ufb02xed distance between the peaks. Looking closer you\u2019ll notice that the distance between two adjacent troughs is the same no matter which two troughs you look at. But, more importantly, its is the same as the distance between the peaks. This distance which is a characteristic of the wave is called the wavelength. Waves have a characteristic wavelength. The symbol for the wavelength is \u201a. The units are metres (m). \u201a \u201a \u201a The wavelength is the distance between any two adjacent points which are in phase. Two points in phase are separate by an integer (0,1,2,3,...) number of complete wave cycles. They don\u2019t have to be peaks or trough but they must be separated by a complete number of waves. 2.1.3 Characteristics of Waves : Period Now imagine you are sitting next to a pond and you watch the waves going past you. First one peak, then a trough and then another peak. If you measure the time between two adjacent peaks you\u2019ll \ufb02nd that it is the same. Now if you measure the time between two adjacent troughs you\u2019ll 12 \ufb02nd that its always the same, no matter which two adjacent troughs you pick. The time you have", " been measuring is the time for one wavelength to pass by. We call this time the period and it is a characteristic of the wave. Waves have a characteristic time interval which we call the period of the wave and denote with the symbol T. It is the time it takes for any two adjacent points which are in phase to pass a \ufb02xed point. The units are seconds (s). 2.1.4 Characteristics of Waves : Frequency There is another way of characterising the time interval of a wave. We timed how long it takes for one wavelength to pass a \ufb02xed point to get the period. We could also turn this around and say how many waves go by in 1 second. We can easily determine this number, which we call the frequency and denote f. To determine the frequency, how many waves go by in 1s, we work out what fraction of a waves goes by in 1 second by dividing 1 second by the time it takes T. If a wave takes 1 2 a second to go by then in 1 second two waves must go by. 1 = 2. The unit of frequency is the Hz or s\u00a11. 1 2 Waves have a characteristic frequency. f = 1 T f T : frequency (Hz or s\u00a11) : period (s) 2.1.5 Characteristics of Waves : Speed Now if you are watching a wave go by you will notice that they move at a constant velocity. The speed is the distance you travel divided by the time you take to travel that distance. This is excellent because we know that the waves travel a distance \u201a in a time T. This means that we can determine the speed. v = \u201a T v \u201a T : speed (m:s\u00a11) : wavelength (m) : period (s) There are a number of relationships involving the various characteristic quantities of waves. A simple example of how this would be useful is how to determine the velocity when you have the frequency and the wavelength. We can take the above equation and substitute the relationship between frequency and period to produce an equation for speed of the form v = f \u201a v \u201a f : speed (m:s\u00a11) : wavelength (m) : frequency (Hz or s\u00a11) Is this correct? Remember a simple \ufb02rst check is to check the units! On the right hand side we have velocity which has units ms\u00a11. On the left hand side we have frequency which is 13 measured in s\u00a11 multiplied by wavelength which is measure in m. On the left hand side we have ms\u00a11 which is exactly what we want. 2.2 Two Types of Waves We agreed that a wave was a moving set of peaks and troughs and we used water as an example. Moving peaks and troughs, with all the characteristics we described, in any medium constitute a wave. It is possible to have waves where the peaks and troughs are perpendicular to the direction of motion, like in the case of water waves. These waves are called transverse waves. There is another type of wave. Called a longitudinal wave and it has the peaks and troughs in the same direction as the wave is moving. The question is how do we construct such a wave? An example of a longitudinal wave is a pressure wave moving through a gas. The peaks in this wave are places where the pressure reaches a peak and the troughs are places where the pressure is a minimum. In the picture below we show the random placement of the gas molecules in a tube. The piston at the end moves into the tube with a repetitive motion. Before the \ufb02rst piston stroke the pressure is the same throughout the tube. JLK When the piston moves in it compresses the gas molecules together at the end of the tube. If the piston stopped moving the gas molecules would all bang into each other and the pressure would increase in the tube but if it moves out again fast enough then pressure waves can be set up. UWV 465 When the piston moves out again before the molecules have time to bang around then the increase in pressure moves down the tube like a pulse (single peak). The piston moves out so fast that a pressure trough is created behind the peak. \u00d8\u0141 \u00e6 \u00c6 As this repeats we get waves of increased and decreased pressure moving down the tubes. We can describe these pulses of increased pressure (peaks in the pressure) and decreased pressure (troughs of pressure) by a sine or cosine graph. \u00b4\u02dc\u02c6 \u00a3\u00a5\u2044 \u0192\u00a4\u00a7 \u02c7\u2014 \u00a2\u00a1 14! \" # $ % & \u2019 ( ) * +,. / Incident ray There are a number of examples of each type of wave. Not all can be seen with the naked eye but all can be detected. 2.3 Properties of Waves We have discussed some of the simple characteristics of waves that we need to know. Now we", " can progress onto some more interesting and, perhaps, less intuitive properties of waves. 2.3.1 Properties of Waves : Reection When waves strike a barrier they are reected. This means that waves bounce o\ufb01 things. Sound waves bounce o\ufb01 walls, light waves bounce o\ufb01 mirrors, radar waves bounce o\ufb01 planes and it can explain how bats can y at night and avoid things as small as telephone wires. The property of reection is a very important and useful one. (NOTE TO SELF: Get an essay by an air tra\u2013c controller on radar) (NOTE TO SELF: Get an essay by on sonar usage for \ufb02shing or for submarines) When waves are reected, the process of reection has certain properties. If a wave hits an obstacle at a right angle to the surface (NOTE TO SELF: diagrams needed) then the wave is reected directly backwards. If the wave strikes the obstacle at some other angle then it is not reected directly backwards. The angle that the waves arrives at is the same as the angle that the reected waves leaves at. The angle that waves arrives at or is incident at equals the angle the waves leaves at or is reected at. Angle of incidence equals angle of reection i = r 15 (2.1) Incident ray i = r i r : angle of incidence : angle of reection i r In the optics chapter you will learn that light is a wave. This means that all the properties we have just learnt apply to light as well. Its very easy to demonstrate reection of light with a mirror. You can also easily show that angle of incidence equals angle of reection. If you look directly into a mirror your see yourself reected directly back but if you tilt the mirror slightly you can experiment with di\ufb01erent incident angles. Phase shift of reected wave When a wave is reected from a more dense medium it undergoes a phase shift. That means that the peaks and troughs are swapped around. The easiest way to demonstrate this is to tie a piece of string to something. Stretch the string out at and then ick the string once so a pulse moves down the string. When the pulse (a single peak in a wave) hits the barrier that the string is tide to it will be reected. The reected wave will look like a trough instead of a peak. This is because the pulse had undergone a phase change. The \ufb02xed end acts like an extremely dense medium. If the end of the string was not \ufb02xed, i.e. it could move up and down then the wave would still be reected but it would not undergo a phase shift. To draw a free end we draw it as a ring around a line. This signi\ufb02es that the end is free to move. 16 2.3.2 Properties of Waves : Refraction Sometimes waves move from one medium to another. The medium is the substance that is carrying the waves. In our \ufb02rst example this was the water. When the medium properties change it can a\ufb01ect the wave. Let us start with the simple case of a water wave moving from one depth to another. The speed of the wave depends on the depth 2. If the wave moves directly from the one medium to the other then we should look closely at the boundary. When a peak arrives at the boundary and moves across it must remain a peak on the other side of the boundary. This means that the peaks pass by at the same time intervals on either side of the boundary. The period and frequency remain the same! But we said the speed of the wave changes, which means that the distance it travels in one time interval is di\ufb01erent i.e. the wavelength has changed. Going from one medium to another the period or frequency does not change only the wave- length can change. Now if we consider a water wave moving at an angle of incidence not 90 degrees towards a change in medium then we immediately know that not the whole wavefront will arrive at once. So if a part of the wave arrives and slows down while the rest is still moving faster before it arrives the angle of the wavefront is going to change. This is known as refraction. When a wave bends or changes its direction when it goes from one medium to the next. If it slows down it turns towards the perpendicular. 2 17 Air Water If the wave speeds up in the new medium it turns away from the perpendicular to the medium surface. Air Water When you look at a stick that emerges from water it looks like it is bent. This is because the light from below the surface of the water bends when it leaves the water. Your eyes project the light back in a straight line and so the object looks like it is a di\ufb01erent place. 18 Air Water 2.3.3 Properties of Waves : Interference If two waves meet interesting things can happen. Waves", " are basically collective motion of particles. So when two waves meet they both try to impose their collective motion on the particles. This can have quite di\ufb01erent results. If two identical (same wavelength, amplitude and frequency) waves are both trying to form a peak then they are able to achieve the sum of their e\ufb01orts. The resulting motion will be a peak which has a height which is the sum of the heights of the two waves. If two waves are both trying to form a trough in the same place then a deeper trough is formed, the depth of which is the sum of the depths of the two waves. Now in this case the two waves have been trying to do the same thing and so add together constructively. This is called constructive interference. A=0.5 B=1 A+B=1.5 If one wave is trying to form a peak and the other is trying to form a trough then they are competing to do di\ufb01erent things. In this case they can cancel out. The amplitude of the resulting 19 wave will depend on the amplitudes of the two waves that are interfering. If the depth of the trough is the same as the height of the peak nothing will happen. If the height of the peak is bigger than the depth of the trough a smaller peak will appear and if the trough is deeper then a less deep trough will appear. This is destructive interference. A=0.5 B=1 B-A=0.5 2.3.4 Properties of Waves : Standing Waves When two waves move in opposite directions, through each other, interference takes place. If the two waves have the same frequency and wavelength then a speci\ufb02c type of constructive interference can occur: standing waves can form. Standing waves are disturbances which don\u2019t appear to move, they look like they stay in the same place even though the waves that from them are moving. Lets demonstrate exactly how this comes about. Imagine a long string with waves being sent down it from either end. The waves from both ends have the same amplitude, wavelength and frequency as you can see in the picture below: 1 0 -1 -5 -4 -3 -2 -1 0 1 2 3 4 5 To stop from getting confused between the two waves we\u2019ll draw the wave from the left with a dashed line and the one from the right with a solid line. As the waves move closer together when they touch both waves have an amplitude of zero: 20 1 0 -1 -5 -4 -3 -2 -1 0 1 2 3 4 5 If we wait for a short time the ends of the two waves move past each other and the waves overlap. Now we know what happens when two waves overlap, we add them together to get the resulting wave. 1 0 -1 -5 -4 -3 -2 -1 0 1 2 3 4 5 Now we know what happens when two waves overlap, we add them together to get the resulting wave. In this picture we show the two waves as dotted lines and the sum of the two in the overlap region is shown as a solid line: 1 0 -1 -5 -4 -3 -2 -1 0 1 2 3 4 5 The important thing to note in this case is that there are some points where the two waves always destructively interfere to zero. If we let the two waves move a little further we get the picture below: 1 0 -1 -5 -4 -3 -2 -1 0 1 2 3 4 5 Again we have to add the two waves together in the overlap region to see what the sum of the waves looks like. 1 0 -1 -5 -4 -3 -2 -1 0 1 2 3 4 5 In this case the two waves have moved half a cycle past each other but because they are out of phase they cancel out completely. The point at 0 will always be zero as the two waves move past each other. 21 When the waves have moved past each other so that they are overlapping for a large region the situation looks like a wave oscillating in place. If we focus on the range -4, 4 once the waves have moved over the whole region. To make it clearer the arrows at the top of the picture show peaks where maximum positive constructive interference is taking place. The arrows at the bottom of the picture show places where maximum negative interference is taking place. 1 0 -1 0 As time goes by the peaks become smaller and the troughs become shallower but they do not -3 -1 -2 -4 1 3 2 4 move. 1 0 -1 For an instant the entire region will look completely at. -4 -3 -2 -1 0 1 2 3 4 1 0 -1 The various points continue their motion in the same manner. -4 -3 -2 -1 0 1 2 3 4 1 0 -1 0 Eventually the picture looks like the complete reection through the x-axis of what we started -2 -1 -4 -3 4", " 3 2 1 with: 1 0 -1 0 Then all the points begin to move back. Each point on the line is oscillating up and down -1 -3 -4 -2 2 1 4 3 with a di\ufb01erent amplitude. 22 1 0 -1 -3 -2 -4 0 If we superimpose the two cases where the peaks were at a maximum and the case where the same waves were at a minimum we can see the lines that the points oscillate between. We call this the envelope of the standing wave as it contains all the oscillations of the individual points. A node is a place where the two waves cancel out completely as two waves destructively interfere in the same place. An anti-node is a place where the two waves constructively interfere. -1 1 4 3 2 Important: The distance between two anti-nodes is only 1 2 \u201a because it is the distance from a peak to a trough in one of the waves forming the standing wave. It is the same as the distance between two adjacent nodes. This will be important when we workout the allowed wavelengths in tubes later. We can take this further because half-way between any two anti-nodes is a node. Then the distance from the node to the anti-node is half the distance between two anti-nodes. This is half of half a wavelength which is one quarter of a wavelength, 1 4 \u201a. Anti-nodes To make the concept of the envelope clearer let us draw arrows describing the motion of points along the line. Nodes Every point in the medium containing a standing wave oscillates up and down and the amplitude of the oscillations depends on the location of the point. It is convenient to draw the envelope for the oscillations to describe the motion. We cannot draw the up and down arrows for every single point! Reection from a \ufb02xed end If waves are reected from a \ufb02xed end, for example tieing the end of a rope to a pole and then sending waves down it. The \ufb02xed end will always be a node. Remember: Waves reected from a \ufb02xed end undergo a phase shift. The wavelength, amplitude and speed of the wave cannot a\ufb01ect this, the \ufb02xed end is always a node. Reection from an open end If waves are reected from end, which is free to move, it is an anti-node. For example tieing the end of a rope to a ring, which can move up and down, around the pole. Remember: The waves sent down the string are reected but do not su\ufb01er a phase shift. 23 Wavelengths of standing waves with \ufb02xed and open ends There are many applications which make use of the properties of waves and the use of \ufb02xed and free ends. Most musical instruments rely on the basic picture that we have presented to create speci\ufb02c sounds, either through standing pressure waves or standing vibratory waves in strings. The key is to understand that a standing wave must be created in the medium that is oscillating. There are constraints as to what wavelengths can form standing waves in a medium. For example, if we consider a tube of gas it can have both ends open (Case 1) one end open and one end closed (Case 2) both ends closed (Case 3). \u2020 \u2020 \u2020 Each of these cases is slightly di\ufb01erent because the open or closed end determines whether a node or anti-node will form when a standing wave is created in the tube. These are the primary constraints when we determine the wavelengths of potential standing waves. These constraints must be met. In the diagram below you can see the three cases di\ufb01erent cases. It is possible to create standing wave with di\ufb01erent frequencies and wavelengths as long as the end criteria are met. Case 1 L Case 2 L Case 3 L The longer the wavelength the less the number of anti-nodes in the standing waves. We cannot have a standing wave with 0 no anti-nodes because then there would be no oscillations. We use n to number to anti-nodes. If all of the tubes have a length L and we know the end constraints we can workout the wavelenth, \u201a, for a speci\ufb02c number of anti-nodes. Lets workout the longest wavelength we can have in each tube, i.e. the case for n = 1. \u201a = 2L \u201a = 4L n = 1 Case 1: In the \ufb02rst tube both ends must be nodes so we can place one anti-node in the 2 \u201a and we also know this middle of the tube. We know the distance from one node to another is 1 distance is L. So we can equate the two and solve for the wavelength: 1 2 \u201a = L \u201a = 2L Case", " 2: In the second tube one ends must be a node and the other must be an anti-node. We are looking at the case with one anti-node we we are forced to have it at the end. We know the distance from one node to another is 1 2 \u201a but we only have half this distance contained in the tube. So : 1 2 ( 1 2 \u201a) = L \u201a = 4L 24 NB: If you ever calculate a longer wavelength for more nodes you have made a mistake. Remember to check if your answers make sense! Case 3: Here both ends are closed and so we must have two nodes so it is impossible to construct a case with only one node. Next we determine which wavelengths could be formed if we had two nodes. Remember that we are dividing the tube up into smaller and smaller segments by having more nodes so we expect the wavelengths to get shorter. \u201a = L \u201a = 4 3 L \u201a = 2L n = 2 Case 1: Both ends are open and so they must be anti-nodes. We can have two nodes inside the tube only if we have one anti-node contained inside the tube and one on each end. This means we have 3 anti-nodes in the tube. The distance between any two anti-nodes is half a wavelength. This means there is half wavelength between the left side and the middle and another half wavelength between the middle and the right side so there must be one wavelength inside the tube. The safest thing to do is workout how many half wavelengths there are and equate this to the length of the tube L and then solve for \u201a. Even though its very simple in this case we should practice our technique: 2( 1 2 \u201a) = L \u201a = L Case 2: We want to have two nodes inside the tube. The left end must be a node and the right end must be an anti-node. We can have one node inside the tube as drawn above. Again we can count the number of distances between adjacent nodes or anti-nodes. If we start from the left end we have one half wavelength between the end and the node inside the tube. The distance from the node inside the tube to the right end which is an anti-node is half of the distance to another node. So it is half of half a wavelength. Together these add up to the length of the tube \u201a) = Case 3: In this case both ends have to be nodes. This means that the length of the tube is one half wavelength: So we can equate the two and solve for the wavelength: 1 2 \u201a = L \u201a = 2L To see the complete pattern for all cases we need to check what the next step for case 3 is when we have an additional node. Below is the diagram for the case where n=3 25 Case 1: Both ends are open and so they must be anti-nodes. We can have three nodes inside the tube only if we have two anti-node contained inside the tube and one on each end. This means we have 4 anti-nodes in the tube. The distance between any two anti-nodes is half a wavelength. This means there is half wavelength between every adjacent pair of anti-nodes. We count how many gaps there are between adjacent anti-nodes to determine how many half wavelengths there are and equate this to the length of the tube L and then solve for \u201a. 3( 1 2 \u201a) = L \u201a = 2 3 L Case 2: We want to have three nodes inside the tube. The left end must be a node and the right end must be an anti-node, so there will be two nodes between the ends of the tube. Again we can count the number of distances between adjacent nodes or anti-nodes, together these add up to the length of the tube. Remember that the distance between the node and an adjacent anti-node is only half the distance between adjacent nodes. So starting from the left end we 3 nodes so 2 half wavelength intervals and then only a node to anti-node distance: 2( 1 2 \u201a) + \u201a) = Case 3: In this case both ends have to be nodes. With one node in between there are two sets of adjacent nodes. This means that the length of the tube consists of two half wavelength sections: 2( 1 2 \u201a) = L \u201a = L 2.3.5 Beats If the waves that are interfering are not identical then the waves form a modulated pattern with a changing amplitude. The peaks in amplitude are called beats. If you consider two sound waves interfering then you hear sudden beats in loudness or intensity of the sound. The simplest illustration is two draw two di\ufb01erent waves and then add them together. You can do this mathematically and draw them yourself to see the pattern that occurs. Here is wave 1: 26 Now we add this to another wave, wave 2: When the two waves are added (", "drawn in coloured dashed lines) you can see the resulting wave pattern: To make things clearer the resulting wave without the dashed lines is drawn below. Notice that the peaks are the same distance apart but the amplitude changes. If you look at the peaks they are modulated i.e. the peak amplitudes seem to oscillate with another wave pattern. This is what we mean by modulation. 2Amax 2Amin The maximum amplitude that the new wave gets to is the sum of the two waves just like for constructive interference. Where the waves reach a maximum it is constructive interference. The smallest amplitude is just the di\ufb01erence between the amplitudes of the two waves, exactly like in destructive interference. The beats have a frequency which is the di\ufb01erence between the frequency of the two waves that were added. This means that the beat frequency is given by fB = f1 j f2 j \u00a1 (2.2) j \u00a1 f2 fB = f1 j : beat frequency (Hz or s\u00a11) : frequency of wave 1 (Hz or s\u00a11) : frequency of wave 2 (Hz or s\u00a11) fB f1 f2 27 2.3.6 Properties of Waves : Di\ufb01raction One of the most interesting, and also very useful, properties of waves is di\ufb01raction. When a wave strikes a barrier with a hole only part of the wave can move through the hole. If the hole is similar in size to the wavelength of the wave di\ufb01ractions occurs. The waves that comes through the hole no longer looks like a straight wave front. It bends around the edges of the hole. If the hole is small enough it acts like a point source of circular waves. This bending around the edges of the hole is called di\ufb01raction. To illustrate this behaviour we start by with Huygen\u2019s principle. Huygen\u2019s Principle Huygen\u2019s principle states that each point on a wavefront acts like a point source or circular waves. The waves emitted from each point interfere to form another wavefront on which each point forms a point source. A long straight line of points emitting waves of the same frequency leads to a straight wave front moving away. To understand what this means lets think about a whole lot of peaks moving in the same direction. Each line represents a peak of a wave. If we choose three points on the next wave front in the direction of motion and make each of them emit waves isotropically (i.e. the same in all directions) we will get the sketch below: What we have drawn is the situation if those three points on the wave front were to emit waves of the same frequency as the moving wave fronts. Huygens principle says that every point on the wave front emits waves isotropically and that these waves interfere to form the next wave front. To see if this is possible we make more points emit waves isotropically to get the sketch below: 28 You can see that the lines from the circles (the peaks) start to overlap in straight lines. To make this clear we redraw the sketch with dashed lines showing the wavefronts which would form. Our wavefronts are not perfectly straight lines because we didn\u2019t draw circles from every point. If we had it would be hard to see clearly what is going on. It Huygen\u2019s principle is a method of analysis applied to problems of wave propagation. recognizes that each point of an advancing wave front is in fact the center of a fresh disturbance and the source of a new train of waves; and that the advancing wave as a whole may be regarded as the sum of all the secondary waves arising from points in the medium already traversed. This view of wave propagation helps better understand a variety of wave phenomena, such as di\ufb01raction. Wavefronts Moving Through an Opening Now if allow the wavefront to impinge on a barrier with a hole in it, then only the points on the wavefront that move into the hole can continue emitting forward moving waves - but because a lot of the wavefront have been removed the points on the edges of the hole emit waves that bend round the edges. 29 The wave front that impinges (strikes) the wall cannot continue moving forward. Only the points moving into the gap can. If you employ Huygens\u2019 principle you can see the e\ufb01ect is that the wavefronts are no longer straight lines. For example, if two rooms are connected by an open doorway and a sound is produced in a remote corner of one of them, a person in the other room will hear the sound as if it originated at the doorway. As far as the second room is concerned, the vibrating air in the doorway is the source of the sound. The same is true of light passing the edge of an obstacle, but this is not as easily observed because of the short wavelength of visible light", ". This means that when waves move through small holes they appear to bend around the sides because there aren\u2019t enough points on the wavefront to form another straight wavefront. This is bending round the sides we call di\ufb01raction. 2.3.7 Properties of Waves : Dispersion Dispersion is a property of waves where the speed of the wave through a medium depends on the frequency. So if two waves enter the same dispersive medium and have di\ufb01erent frequencies they will have di\ufb01erent speeds in that medium even if they both entered with the same speed. We will come back to this topic in optics. 2.4 Practical Applications of Waves: Sound Waves 2.4.1 Doppler Shift The Doppler shift is an e\ufb01ect which becomes apparent when the source of sound waves or the person hearing the sound waves is moving. In this case the frequency of the sounds waves can 30 be di\ufb01erent. This might seem strange but you have probably experienced the doppler e\ufb01ect in every day life. When would you notice it. The e\ufb01ect depends on whether the source of the sound is moving away from the listener or if it is moving towards the listener. If you stand at the side fo the road or train tracks then a car or train driving by will at \ufb02rst be moving towards you and then away. This would mean that we would experience the biggest change in the e\ufb01ect. We said that it e\ufb01ects the frequency of the sound so the sounds from the car or train would sound di\ufb01erent, have a di\ufb01erent frequency, when the car is coming towards you and when it is moving away from you. Why does the frequency of the sound change when the car is moving towards or away from you? Lets convince ourselves that it must change! Imagine a source of sound waves with constant frequency and amplitude. Just like each of the points on the wave front from the Huygen\u2019s principle section. Remember the sound waves are disturbances moving through the medium so if the source moves or stops after the sound has been emitted it can\u2019t a\ufb01ect the waves that have been emitted already. The Doppler shift happens when the source moves while emitting waves. So lets imagine we have the same source as above but now its moving to the right. It is emitting sound at a constant frequency and so the time between peaks of the sound waves will be constant but the position will have moved to the right. In the picture below we see that our sound source (the black circle) has emitted a peak which moves away at the same speed in all directions. The source is moving to the right so it catches up a little bit with the peak that is moving away to the right. 31 When the second peak is emitted it is from a point that has moved to the right. This means that the new or second peak is closer to the \ufb02rst peak on the right but further away from the \ufb02rst peak on the left. If the source continues moving at the same speed in the same direction (i.e. with the same velocity which you will learn more about later). then the distance between peaks on the right of the source is the constant. The distance between peaks on the left is also constant but they are di\ufb01erent on the left and right. This means that the time between peaks on the right is less so the frequency is higher. It is higher than on the left and higher than if the source were not moving at all. On the left hand side the peaks are further apart than on the right and further apart than if the source were at rest - this means the frequency is lower. So what happens when a car drives by on the freeway is that when it approaches you hear higher frequency sounds and then when it goes past you it is moving away so you hear a lower frequency. 2.4.2 Mach Cone Now we know that the waves move away from the source at the speed of sound. What happens if the source moves at the speed of sound? This means that the wave peaks on the right never get away from the source so each wave is emitted on top of the previous on on the right hand side like in the picture below. 32 If the source moves faster than the speed of sound a cone of wave fronts is created. This is called a Mach cone. Sometimes we use the speed of sound as a reference to describe the speed of the object. So if the object moves at exactly the speed of sound we can say that it moves at 1 times the speed of sound. If it moves at double the speed of sound then we can say that it moves at 2 times the speed of sound. A convention for this is to use Mach numbers, so moving at the speed of sound is Mach one (one times the speed of sound) and moving at twice the speed of sound is called", " Mach two (twice the speed of sound). 2.4.3 Ultra-Sound Ultrasound is sound with a frequency greater than the upper limit of human hearing, approximately 20 kilohertz. Some animals, such as dogs, dolphins, and bats, have an upper limit that is greater than that of the human ear and can hear ultrasound. Ultrasound has industrial and medical applications. Medical ultrasonography can visualise muscle and soft tissue, making them useful for scanning the organs, and obstetric ultrasonography is commonly used during pregnancy. Typical diagnostic ultrasound scanners operate in the frequency range of 2 to 13 megahertz. More powerful ultrasound sources may be used to generate local heating in biological tissue, with applications in physical therapy and cancer treatment. Focused ultrasound sources may be used to break up kidney stones. Ultrasonic cleaners, sometimes called supersonic cleaners, are used at frequencies from 20-40 kHz for jewellery, lenses and other optical parts, watches, dental instruments, surgical instruments and industrial parts. These cleaners consist of containers with a uid in which the object to be cleaned is placed. Ultrasonic waves are then sent into the uid. The main mechanism for cleaning action in an ultrasonic cleaner is actually the energy released from the collapse of millions of microscopic cavitation events occurring in the liquid of the cleaner. 33 Interesting Fact: Ultrasound generator/speaker systems are sold with claims that they frighten away rodents and insects, but there is no scienti\ufb02c evidence that the devices work; controlled tests have shown that rodents quickly learn that the speakers are harmless. Medical Ultrasonography Medical ultrasonography makes uses the fact that waves are partially reected when the medium in which they are moving changes density. If the density increases then the reected waves undergoes a phase shift exactly like the case where the waves in a string were reected from a \ufb02xed end. If the density decreases then the reected waves has the same phase exactly like the case where the waves in a string were reected from a free end. Combining these properties of waves with modern computing technology has allowed medical professionals to develop an imaging technology to help with many aspects of diagnosis. Typical ultrasound units have a hand-held probe (often called a scan head) that is placed directly on and moved over the patient: a water-based gel ensures good contact between the patient and scan head. Ultrasonic waves are emitted from the scan head and sent into the body of the patient. The scan head also acts a receiver for reected waves. From detailed knowledge of interference and reection an image of the internal organs can be constructed on a screen by a computer programmed to process the reected signals. Interesting Fact: Medical ultrasonography was invented in 1953 at Lund University by cardiologist Inge Edler and Carl Hellmuth Hertz, the son of Gustav Ludwig Hertz, who was a graduate student at the department for nuclear physics. Uses Ultrasonography is widely utilized in medicine, primarily in gastroenterology, cardiology, gynaecology and obstetrics, urology and endocrinology. It is possible to perform diagnosis or therapeutic procedures with the guidance of ultrasonography (for instance biopsies or drainage of uid collections). Strengths of ultrasound imaging It images muscle and soft tissue very well and is particularly useful for \ufb02nding the interfaces between solid and uid-\ufb02lled spaces. It renders \"live\" images, where the operator can dynamically select the most useful section for diagnosing and documenting changes, often enabling rapid diagnoses. It shows the structure as well as some aspects of the function of organs. \u2020 \u2020 \u2020 34 It has no known long-term side e\ufb01ects and rarely causes any discomfort to the patient. Equipment is widely available and comparatively exible; examinations can be performed at the bedside. Small, easily carried scanners are available. \u2020 \u2020 \u2020 Much cheaper than many other medical imaging technology. \u2020 Weaknesses of ultrasound imaging \u2020 \u2020 \u2020 Ultrasound has trouble penetrating bone and performs very poorly when there is air between the scan head and the organ of interest. For example, overlying gas in the gastrointestinal tract often makes ultrasound scanning of the pancreas di\u2013cult. Even in the absence of bone or air, the depth penetration of ultrasound is limited, making it di\u2013cult to image structures that are far removed from the body surface, especially in obese patients. The method is operator-dependent. A high level of skill and experience is needed to acquire good-quality images and make accurate diagnoses. Doppler ultrasonography Ultrasonography can be enhanced with Doppler measurements, which employ the Doppler e\ufb01ect to assess whether structures (usually blood) are moving towards or away from the probe. By calculating the frequency shift of a particular sample volume, e.g. within a jet of blood ow over a heart valve, its speed and direction", " can be determined and visualised. This is particularily useful in cardiovascular studies (ultrasonography of the vasculature and heart) and essential in many areas such as determining reverse blood ow in the liver vasculature in portal hypertension. The Doppler information is displayed graphically using spectral Doppler, or as an image using colour Doppler or power Doppler. It is often presented audibly using stereo speakers: this produces a very distinctive, despite synthetic, sound. 2.5 Important Equations and Quantities Frequency: f = 1 T : (2.3) Quantity Amplitude Period Wavelength Frequency Speed Symbol A T \u201a f v S.I. Units m s m or m:s\u00a11 Hz s\u00a11 Direction | | | | | Table 2.1: Units used in Waves 35 Speed: v = f \u201a \u201a T = 36 Chapter 3 Geometrical Optics In this chapter we will study geometrical optics. Optics is the branch of physics that describes the behavior and properties of light and the interaction of light with matter. By geometrical optics we mean that we will study all the optics that we can treat using geometrical analysis (geometry), i.e. lines, angles and circles. As we have seen in the previous chapters, light propagates as a wave. The waves travel in straight lines from the source. So we will consider light as a set of rays. The wavelike nature will become apparent when the waves go from one medium to another. Using rays and Snell\u2019s law to describe what happens when the light ray moves from one medium to another we can solve all the geometrical optics problems in this chapter. 3.1 Refraction re-looked We have seen that waves refract as they move from shallower to deeper water or vise versa, thus light also refracts as it moves between two mediums of di\ufb01erent densities. We may consider the parallel beams of light in Fig 1(a) as a set of wheels connected by a straight rod placed through their centres as in Fig1(b). As the wheels roll onto the grass (representing higher density), they begin to slow down, while those still on the tarmac move at a relatively faster speed. This shifts the direction of movtion towards the normal of the grasstarmac barrier. Likewise light moving from a medium of low density to a medium of high density (Fig 2) moves towards the normal, hence the angle of incidence (i) is greater than the angle of Refraction (r): i > r for d1 < d2 (NOTE TO SELF: Diagram of ray moving into more dense medium) (NOTE TO SELF: Discussion of Snell\u2019s Law) (NOTE TO SELF: Follow by discussion of how we can reverse rays) (NOTE TO SELF: Show its all the same) 37 where both angles are measured from the normal to the ray. I is the incident ray and R is the refracted ray. Rays are reversible 3.1.1 Apparent and Real Depth: If you submerge a straight stick in water and look at the stick where it passes through the surface of the water you will notice that it looks like the stick bends. If you remove you will see that the stick did not bend. The bending of the stick is a very common example of refraction. How can we explain this? We can start with a simple object under water. We can see things because light travels from the object into our eyes where it is absorbed and sent to the brain for processing. The human brain interprets the information it receives using the fact that light travels in straight lines. This is why the stick looks bent. The light did not travel in a straight line from the stick underwater to your eye. It was refracted at the surface. Your brain assumes the light travelled in a straight line and so it intrepets the information so that it thinks the stick is in a di\ufb01erent place. This phenomenon is easily explained using ray diagrams as in Fig??. The real light rays are represented with a solid line while dashed lines depict the virtual rays. The real light rays undergo refraction at the surface of the water hence move away from the normal. However the eye assumes that light rays travel in straight lines, thus we extend the refracted rays until they converge to a point. These are virtual rays as in reality the light was refracted and did not originate from that point. 38 air water (NOTE TO SELF: Insert an example with water - can be worked example!) We note that the image of is seen slightly higher and ahead of the object. Where would we see the object if it was submerged in a uid denser than water? (NOTE TO SELF: Insert an example with denser medium show change - can be worked example!) 3.1.2 Splitting of White Light How is a rainbow formed? White is a", " combination of all other colours of light. Each colour has a di\ufb01erent wavelength and is thus di\ufb01racted through di\ufb01erent angles. red yellow blue The splitting of white light into its component colours may be demonstrated using a triangular prism (Fig4). White light is incident on the prism. As the white light enters the glass its component colours are di\ufb01racted through di\ufb01erent angles. This separation is further expanded as the light rays leave the prism. Why? What colour is di\ufb01racted the most? If red has the longest wavelength and violet the shortest, what is the relation between refraction and wavelength? As the sun appears after a rainstorm, droplets of water still remain in the atmosphere. These act as tiny prisms that split the suns light up into its component colours forming a rainbow. 39 white light red yellow blue 3.1.3 Total Internal Reection Another useful application of refraction is the periscope. We know that as light move from higher to lower density mediums, light rays tend to be di\ufb01racted towards the normal. We also know that the angle of refraction is greater than the angle of incidence and increases as we increase the angle of incidence (Fig 5(a),(b)). At a certain angle of incidence, c, the refracted angle equals 90o (Fig 5(c)), this angle is called the critical angle. For any angle of incidence greater than c the light will be refracted back into the incident medium, such refraction is called Total Internal reection (Fig 5(d)). A periscope uses two 90o triangular prisms as total internal reectors. Light enters the periscope and is reected by the \ufb02rst prism down the chamber, where again the light is reected to the observer. This may be illustrated using a ray diagram as in (Fig 6). 3.2 Lenses Lenses are used in many aspects of technology ranging from contact lenses to projectors. We shall again use light rays to explain the properties on lenses. There are two types of lenses, namely: Converging: Lenses that cause parallel light rays to converge to a point (Fig 7). Diverging: Lenses that cause parallel light rays to diverge (Fig 8). Such deviation of light rays are caused by refraction. It is now a good time to introduce a few new de\ufb02nitions: Optical Centre: The centre of the lens Principal Axis: The line joining two centres of curvature of the surfaces of the lens. Focal Point: The point at which light rays, parallel to the principal axis, converge, or appear to converge, after passing through the lens. Focal Length: The distance between the optical centre and the focal point. 40 As seen in Fig 7 and 8, the focal point of a converging lens is real, while the focal point of a diverging lens is vitual. 3.2.1 Convex lenses Convex lenses are in general converging lenses. Hence they possess real focal points (with one exception, discussed later). Such focal points allow for real images to be formed. One may verify the above by placing an illuminated object on one side of a convex lens, ensuring that the distance between them is greater than the focal length (explained later). By placing a screen on the other side of the lens and varing the distance, one will acquire a sharp image of the object on the screen. The ray diagrams below illustrate the images formed when an object is placed a distance d from the optical centre of a convex lens with focal length F. d Image type Magni\ufb02cation Orientation Image Position (I) Figure >2F Real <1 Inverted F<I<2F 9(a) 2F Real 1 Inverted 2F 9(b) F<d<2F Real >1 Inverted >2F 9(c) F No Image - - - 9(d) <F Virtual >1 Original >d 9(e) 3.2.2 Concave Lenses Concave lenses disperse parallel rays of light and are hence diverging lenses. All images formed by concave lenses are virtual and placed between the lens and the object. Furthermore, the image retains its original orientation while it is smaller in size (Fig 10). B0 A0 B A 41 B0 A0 B A 3.2.3 Magni\ufb02cation By De\ufb02nition the magni\ufb02cation, m, is: m = (Height of Image) / (Height of Object) however, using similar triangles, we may prove that: m = (Distance of Image) / (Distance of Object) where both distances are measured from the optical centre. The above method allows us to accurately estimate the magni\ufb02cation of an image. This is commonly used in a compound microscope as discussed in the next section. Worked", " Example 2 A n object is placed 0.5m in front of a convex a lens. \u2020 \u2020 At what distance should a screen be placed in order to create an image with a magni\ufb02cation of 1.5? If the height of the image and object are equal, what is the focal length of the lens? Solutions: a)m = (Distance of Image) / (Distance of Object) therefore Distance of Image = (Distance of Object) x m Distance of Image = 0.5m x 1.5 = 0.75m b)This implies that the magni\ufb02cation m = 1. Therefore from the above table it is seen that d = I = 2F Therefore F = d/2 = 0.25m 42 3.2.4 Compound Microscope This type of microscope uses two convex lenses. The \ufb02rst creates a real magni\ufb02ed image of the object that is in turn used by the second lens to create the \ufb02nal image. This image is virtual and again enlarged. The \ufb02nal image, as seen in Fig 11, is virtual, enlarged and inverted. The lens, L1 that forms the real image is called the objective lens, while L2 is referred to as the eyepiece. B A A1 B1 3.2.5 The Human Eye The eye also contains a biconvex lens that is used to focus objects onto the retina of the eye. However, in some cases the lens maybe abnormal and cause defects in ones vision. Hyperopia (Long-Sightedness) This occurs when the image is focussed beyond the retina. Hyperopia is due to the eyeball being too short or the lens not being convex enough. A convex lens is used to correct this defect (Fig12). Miopia (Short-Sightedness) Images are focussed before the retina. The lens being too convex or the eyeball being too long causes this. A concave lens corrects this defect (Fig 13). Astigmatism When this occurs, one is able to focus in one plane (e.g. vertical) but not another. This again is due to a distortion in the lens and may be cured by using relevant lenses. 43 3.3 Introduction Light is at \ufb02rst, something we feel incredibly familiar with. It can make us feel warm, it allows us to see, allows mirrors and lenses to \u2018work\u2019, allows for... Under more careful study light exhibits many fascinating and wonderful properties. The study of light has led to many important and amazing scienti\ufb02c discoveries and much progress. For example \ufb02bre optics, lasers and CCD\u2019s play a huge role in modern technology. \u2020 \u2020 \u2020 \u2020 Light is a form of energy. This is demonstrated by Crookes Radiometer or Solar Cells Light travels in straight lines. This is demonstrated by an experiment involving a card with an hole looking at light source e.g. candle. Also the simple camera: candle black box one side pin hole other side grease proof paper. We see the candle upside down on the paper Light travels at a constant speed. The speed depends on the medium it is in. (substance like air or water)1. Nothing travels faster than the speed of light in a vacuum c = 2:99790 of the fundamental constants in physics. \u00a3 108. This is one Probably the most important use of light by the majority of living things on earth is that it allows them to see. If the light from an object enters our optical detectors/sensors i.e. our eyes!, we can see that object. In some cases the light originates at the object itself. Objects which give out light are said to be luminous objects e.g. a lighted candle/torch/bulb, the Sun, stars. The moon is not a luminous object! Why? Most objects however are not luminous, objects which do not give out their own light. We can see them because they reect light into our eyes. 3.4 Reection 3.4.1 Di\ufb01use reection (NOTE TO SELF: diag of light rays hitting a rough surface) (NOTE TO SELF: diag of light rays hitting a polished surface) Most objects do not have perfectly smooth surfaces. Because of this di\ufb01erent parts of the surface reect light rays in di\ufb01erent directions (angles). 3.4.2 Regular reection Mirrors and highly polished surfaces give regular reection. A mirror is a piece of glass with a thin layer of silver (aluminium is commonly used) on the rear surface Light is reected according to the laws of reection. 3.4.3 Laws of reection (NOTE TO SELF: diag of i N r rays striking a mirror de\ufb02", "ne i r N) Laws of reection: The angle of incidence is always equal to the angle of reection The incident ray, the normal and the reected ray are all in the same plane 1There are some substances where light moves so slowly you can walk faster than it moves - more on this later! 44 Note that the angle of incidence i is between the incident ray and the normal2 to the surface, not between the incident ray and the surface of the mirror. There are two forms of an image formed by reection: real and virtual. A real image is formed by the actual intersection of light rays. It is always inverted (upside down) and can be formed on a screen A virtual image is formed by the apparent intersection of light rays. It is always erect and cannot be formed on a screen. Images formed in plane mirrors are virtual images. Real images are formed by lenses (e.g. the image on a cinema screen) or by curved mirrors. 3.4.4 Lateral inversion Where some thing is back to front e.g. AMBULANCE (NOTE TO SELF: How a periscope works, why images are right!) 3.5 Curved Mirrors 3.5.1 Concave Mirrors (Converging Mirrors) (NOTE TO SELF: how to remember a con cave from a convex mirror) (NOTE TO SELF: diag of light rays striking a concave mirror base ray diag! de\ufb02ne things) P = pole of mirror F = focal point (focus) C = centre of curvature |CP| = radius of curvature (r) |FP| = focal length (f) r = 2f Rules for ray tracing: Rays of light that arrive parallel to principal axis leave the surface of the mirror through the focal point Rays of light that arrive through the focal point leave the surface of the mirror parallel to principal axis Rays of light that arrive through the centre of curvature leave the surface back through the centre of curvature (NOTE TO SELF: Maybe diag for each of these above!) In order to obtain the position, nature (real or virtual) and size of the image we need just apply the rules above. The details depend on the distance of the object from the mirror surface.. Object outside C diag Image - located between c and f - inverted (upside down) - diminished (reduced in size, smaller) - real Object at C diag Image - located at c - inverted (upside down) - same size - real Object between C and f diag Image - located outside c - inverted (upside down) - magni\ufb02ed (increased in size, larger) - real Object at f diag 2The normal is a line that is perpendicular to the surface i.e. the angle between the line and the surface is 90o 45 Image - located at in\ufb02nity Object between f and P diag Image - located behind the mirror - erect (right side up) - magni\ufb02ed (increased in size, larger) - virtual 3.5.2 Convex Mirrors diag base ray diag! de\ufb02ne things Image - located behind the mirror - erect (right side up) - diminished (reduced in size, smaller) - virtual We can also arrive at the position and nature of the image by calculation, using the following formula 1/u + 1/v = 1/f AND Magni\ufb02cation (=M) m = v/u (We will deal with the \u2018pro\ufb01\u2019 for these later) N.B. distance between f and P = f (focal length negative for convex mirror) distance between o and P = u (object distance) distance between i and P = v (image distance negative for virtual image) concave mirror f is positive convex mirror f is negative real image v is positive virtual image v is negative Lots of examples with numbers e.g. u = 10cm v = 20cm m=2 things are bigger if m > 1 (and -1) things are smaller if 0 > m > 1 (and -1) Uses of Convex mirrors image is always erect wide range of view \u2020 \u2020 They often used in shops, double decker busses, dangerous bends in roads, wing mirrors of cars Disadvantage False sense of distance (objects seem closer than they actually are) uses of concave mirrors They are usually used as make-up mirrors or shaving mirrors and in reecting telescopes Advantages if object is inside f the image is magni\ufb02ed \u2020 MAYBE MENTION \u2018Virtual object\u2019 in convex mirror for completeness 3.5.3 Refraction When light travels from one medium to another it changes direction, except when it is incident normally on the separating surface. The change of direction is caused by the change in the velocity of", " the light as it passes from one medium to the other. 46 4 3 2 1 0 -1 -2 -3 -4 D A C B -4 -5 -3 diag de\ufb02ning angles i, r and N Light is refracted according to the laws of refraction: -1 -2 2 1 0 3 4 5 3.5.4 Laws of Refraction Laws of Refraction: sin i /sin r is a constant for two given media (Snell\u2019s Law) The incident ray, the normal and the refracted ray are all in the same plane Sin i /sin r is known as the refractive index (n) EXP verify Snell\u2019s law EXP coin in an empty cup, move head till it disappears. Then \ufb02ll with water. Can you see it? diag real apparent depth Due to refraction a body which is at O beneath the water appears to be at I when seen from above. As a result of refraction the pool appears to be 1.5m instead of 2m. The relationship between the real depth and the apparent depth is determined by the refractive index of water, which is 4/3 diag of water depth in pool n = sin i /sin r =|AP|/|PI| / |AP|/|PO| |PO|/|PI| However the above diagram is greatly exaggerated in size. In practise |PO| |AO| and |PI| |AI| So n =|AO|/|AI| == real/apparent depth! More than two media.g. air glass water The refractive index of glass is 3/2 and the refractive index of water is 4/3. The refractive \u00a3 index from water to glass a n w= 4/3 => w n a =3/=4 \u00a3 3=2 = 9=8 \u00a3 47 3.5.5 Total Internal Reection This can only happen when light is travelling into a less dense medium Refrectionnumbers are n1 = 2 and n2 = 11 -2 -3 6 4 3 0 5 1 2 -2 -3 -7 -1 -4 -5 -8 -6 diag air glass critical angle i r When light travels from glass to air it is refracted away from the normal as in a above. As the angle of incidence is increased the angle of refraction eventually reaches 90o as in b If the angle of incidence is increased further beyond this value total internal reection occurs as in c The critical angle c is the angle of incidence corresponding to the angle of refraction of 90o From the above the refractive index from from air to glass n=sin90/sin c = 1/sin c Also sin c = 1 / n = 2/3 in the case of glass. Sin c = 0.6667 => critical angle for glass is 41o490 Total internal reection has some very useful properties eg diag prism for turning light 180o daig prism for turning light 90o The size of the \ufb02nal image in a pair of binoculars depends on the distance travelled by the light within the binoculars. By using two prisms this distance can be increased without increasing the length of the binoculars. Fibre Optics In the same Water can be directed from one place to another by con\ufb02ning it within a pipe. way light can be directed from one place to another by con\ufb02ning it within a single glass \ufb02bre. The light is kept within the \ufb02bre by total internal reection. The amount of light which can be carried by a single \ufb02bre is very small so it is usual to form a light tube tapping a few thousand \ufb02bres together. On great advantage of such a light tube is exibility; it can be ties in knots 48 B 7 8 and still function. However since total internal reection only occurs when light is going from a medium to a less dense medium, it is necessary ti coat each \ufb02bre with glass of a lower refractive index. Otherwise light would leak from one \ufb02bre at their points of contact. Light tube can be used to bring light from a lamp to an object, thus illuminating the object. A second light tube can then used to carry light from the illuminated object to an observer, thus enabling the object to be seen and photographed. The procedure has been used to photograph the digestive system the reproductive system and many other parts of the human body. In the case of the light tube carrying light from the object to the observer, it is vital that the individual \ufb02bres in the tube do not cross each other, otherwise the image will become garbled. Like radio waves, light waves are electromagnetic. (cf section em) However, their shorter wavelength and higher frequency means that a single light beam can carry far more telephone conservations at one time compared with a radio wave. In the", " case of long \ufb02bre cables it would be necessary to incorporate a device to boost the intensity of the light to make up for losses due to absorption. Nevertheless the system has great potential for the communication industry, including the possibility of transmitting pictures over long distances. As we said earlier, the reason why light bends when going from one medium to another is because of the change of velocity. This will be dealt with in more detail later. For the moment we consider only the implication for the refractive index. 1 n 2 = velocity medium 1 / velocity medium 2 3.5.6 Mirage Yet another e\ufb01ect of refraction is the mirage. The most common mirage occurs in warm weather when motorists see what appears to be a pool of water on the road close to the horizon. The explanation is this: when air is heated it expands and becomes less dense and when it cools it contracts and becomes more dense. In summer the ground is hot and the layer of air nearest the ground is hot. The layer above that is cooler, etc etc. A of light coming from the blue sky passes down through the layers of air which are getting progressively less dense. As it does so it is progressively bent, as shown in the diagram. As a result an image of the blue sky is seen on the road and is taken to be a pool of water! 3.6 The Electromagnetic Spectrum They are transverse waves. They can travel through empty space. They travel at the speed of light. Short wavelength Long wavelength Dangers Microwaves kill living cells Ultraviolet light causes skin cancer and can kill living cells Uses Radio waves are used in TV\u2019s and radios Microwaves are used to heat foods, they are also used to transmit signals to satellites and back to earth. Infrared light is used to asses heat loss from buildings, weather forecasting, locating survivors in earthquakes and in TV remote controls. Visible light is used in \ufb02bre optics. Ultraviolet light produces vitamins in the skin if it is in small doses. Suntan. X-Rays are used to look at bones inside your body, my body and everybody\u2019s body. g-rays are used to treat cancer. 49 3.7 Important Equations and Quantities Quantity Symbol Unit S.I. Units Direction Units or Table 3.1: Units used in Optics 50 Chapter 4 Vectors (NOTE TO SELF: SW: initially this chapter had a very mathematical approach. I have toned this down and tried to present in a logical way the techniques of vector manipulation after \ufb02rst exploring the mathematical properties of vectors. Most of the PGCE comments revolved around the ommision of the graphical techniques of vector addition (i.e. scale diagrams), incline questions and equilibrium of forces. I have addressed the \ufb02rst two; equilibrium of forces and the triangle law of three forces in equilibrium I think would be better o\ufb01 in the Forces Chapter. Also the Forces chapter should include some examples of incline planes. Inclines are introduced here merely as an example of components in action!) 4.1 PGCE Comments \u2020 \u2020 \u2020 Use a+b = c to illustrate that a = c-b under subtraction. Summarise properties of vectors at end. Mention that vector multiplication exists in two forms- neither introduced at school, but used implicitly in work and motor rules. 4.2 \u2018TO DO\u2019 LIST \u2020 \u2020 \u2020 \u2020 \u2020 Mention that the direction of a vector is measured from its tail. Also include explanations of the ways in which to specify direction (e.g. bearing, compass points, angle from vertical, etc.) Include exam-type questions involving owing rivers or aeroplanes in wind. These questions link displacement, velocity and time and also test the learners understanding of vectors combining to give a combined result (i.e. resultant). Mention that displacement at a point is the directed line from the start to that point. Rewrite section on velocity to make clear the distinction between average and instantaneous rates of change (velocity and speed). The \u00a2\u2019s in the equations imply we are calculating average quantities. Mention that we take the limit of a small time interval to give instantaneous quantities. Perhaps the example of a parabola with average gradient and gradient of 51 tangent can be used as an illustration. Else defer until chapter on Graphs and Equations of Motion. Instantaneous velocity: reading on the speedometer in a direction tangent to the path. Instantaneous speed is magnitude of instantaneous velocity but average speed is not equal to magnitude of average velocity. Average speed and average velocity are the total distance and resultant displacement over the time interval related to that part of the path. The example of the circular track uses these de\ufb02nitions and is an important illustration of the di\ufb01erences. Instantaneous calculated at a certain instant in time while average is calculated over an interval", ". Include relative velocity. Address PGCE comments above and comments in the text. \u2020 \u2020 4.3 Introduction \\A vector is \u2018something\u2019 that has both magnitude and direction.\" \\\u2018Things\u2019? What sorts of \u2018things\u2019?\" Any piece of information which contains a magnitude and a related direction can be a vector. A vector should tell you how much and which way. Consider a man driving his car east along a highway at 100 km=h. What we have given here is a vector{ the car\u2019s velocity. The car is moving at 100 km=h (this is the magnitude) and we know where it is going{ east (this is the direction). Thus, we know the speed and direction of the car. These two quantities, a magnitude and a direction, form a vector we call velocity. De\ufb02nition: A vector is a measurement which has both magnitude and direction. In physics magnitudes often have directions associated with them. If you push something it is not very useful knowing just how hard you pushed. A direction is needed too. Directions are extremely important, especially when dealing with situations more complicated than simple pushes and pulls. Di\ufb01erent people like to write vectors in di\ufb01erent ways. Anyway of writing a vector so that it has both magnitude and direction is valid. Are vectors physics? No, vectors themselves are not physics. Physics is just a description of the world around us. To describe something we need to use a language. The most common language used to describe physics is mathematics. Vectors form a very important part of the mathematical description of physics, so much so that it is absolutely essential to master the use of vectors. 4.3.1 Mathematical representation Numerous notations are commonly used to denote vectors. In this text, vectors will be denoted by symbols capped with an arrow. As an example, \u00a1!s, \u00a1!v and \u00a1!F are all vectors (they have both magnitude and direction). Sometimes just the magnitude of a vector is required. In this case, is another the arrow is ommitted. In other words, F denotes the magnitude of vector \u00a1!F. way of representing the size of a vector. \u00a1!F j j 52 4.3.2 Graphical representation Graphically vectors are drawn as arrows. An arrow has both a magnitude (how long it is) and a direction (the direction in which it points). For this reason, arrows are vectors. In order to draw a vector accurately we must specify a scale and include a reference direction in the diagram. A scale allows us to translate the length of the arrow into the vector\u2019s magnitude. For instance if one chose a scale of 1cm = 2N (1cm represents 2N ), a force of magnitude 20N would be represented as an arrow 10cm long. A reference direction may be a line representing a horizontal surface or the points of a compass. Worked Example 3 Drawing vectors Question: Using a scale of 1cm = 2m:s\u00a11 represent the following velocities: a) 6m:s\u00a11 north b) 16m:s\u00a11 east Answer: scale 1cm = 2m:s\u00a11 m c 3 a) b) W N S E 8cm 4.4 Some Examples of Vectors 4.4.1 Displacement Imagine you walked from your house to the shops along a winding path through the veld. Your route is shown in blue in Figure 12.3. Your sister also walked from the house to the shops, but she decided to walk along the pavements. Her path is shown in red and consisted of two straight stretches, one after the other. Although you took very di\ufb01erent routes, both you and your sister walked from the house to the shops. The overall e\ufb01ect was the same! Clearly the shortest path from your house to the shops is along the straight line between these two points. The length of this line and the direction from the start point (the house) to the end point (the shops) forms a very special vector known as displacement. Displacement is assigned the symbol \u00a1!s. De\ufb02nition: Displacement is de\ufb02ned as the magnitude and direction of the straight line joining one\u2019s starting point to one\u2019s \ufb02nal point. 53 Finish (Shop) D isplace m ent Start (House) Figure 4.1: Illustration of Displacement OR De\ufb02nition: Displacement is a vector with direction pointing from some initial (starting) point to some \ufb02nal (end) point and whose magnitude is the straight-line distance from the starting point to the end point. (NOTE TO SELF: choose one of the above) In this example both you and your sister had the same displacement. This is shown", " as the black arrow in Figure 12.3. Remember displacement is not concerned with the actual path taken. It is only concerned with your start and end points. It tells you the length of the straight-line path between your start and end points and the direction from start to \ufb02nish. The distance travelled is the length of the path followed and is a scalar (just a number). Note that the magnitude of the displacement need not be the same as the distance travelled. In this case the magnitude of your displacement would be considerably less than the actual length of the path you followed through the veld! 4.4.2 Velocity De\ufb02nition: Velocity is the rate of change of displacement with respect to time. The terms rate of change and with respect to are ones we will use often and it is important that you understand what they mean. Velocity describes how much displacement changes for a certain change in time. We usually denote a change in something with the symbol \u00a2 (the Greek letter Delta). You have probably seen this before in maths{ the gradient of a straight line is \u00a2y \u00a2x. The gradient is just how much y changes for a certain change in x. In other words it is just the rate of change of y with respect to x. This means that velocity must be \u00a1!v = \u00a2\u00a1!s \u00a2t = \u00a1!s f inal \u00a1 \u00a1!s initial tinitial tf inal \u00a1 : (4.1) 54 (NOTE TO SELF: This is actually average velocity. For instantaneous \u00a2\u2019s change to differentials. Explain that if \u00a2 is large then we have average velocity else for in\ufb02nitesimal time interval instantaneous!) What then is speed? Speed is how quickly something is moving. How is it di\ufb01erent from velocity? Speed is not a vector. It does not tell you which direction something is moving, only how fast. Speed is the magnitude of the velocity vector (NOTE TO SELF: instantaneous speed is the magnitude of the instantaneous velocity.... not true of averages!). Consider the following example to test your understanding of the di\ufb01erences between velocity and speed. Worked Example 4 Speed and Velocity Question: A man runs around a circular track of radius 100m. It takes him 120s to complete a revolution of the track. If he runs at constant speed calculate: 1. his speed, 2. his instantaneous velocity at point A, 3. his instantaneous velocity at point B, 4. his average velocity between points A and B, 5. his average velocity during a revolution. A 100m B W N S E Direction the man runs Answer: 1. To determine the man\u2019s speed we need to know the distance he travels and how long it takes. We know it takes 120s to complete one revolution of the track. Step 1 : First we \ufb02nd the distance the man travels What distance is one revolution of the track? We know the track is a circle and we know its radius, so we can determine the perimeter or distance around the circle. We start with the equation for the circumference of a circle C = 2\u2026r = 2\u2026(100m) = 628:3 m: So we know the distance the man covers in one revolution is 628:3 m. 55 Step 2 : Now we determine speed from the distance and time. We know that speed is distance covered per unit time. So if we divide the distance covered by the time it took we will know how much distance was covered for every unit of time. v = Distance travelled time taken = 628:3m 120s = 5:23 m:s\u00a11 2. Step 3 : Determine his instantaneous velocity at A Consider the point A in the diagram. A Direction the man runs We know which way the man is running around the track and we know his speed. His velocity at point A will be his speed (the magnitude of the velocity) plus his direction of motion (the direction of his velocity). The instant that he arrives at A he is moving as indicated in the diagram below. A So his velocity vector will be 5:23 m:s\u00a11 West. 3. Step 4 : Determine his instantaneous velocity at B Consider the point B in the diagram. B Direction the man runs 56 We know which way the man is running around the track and we know his speed. His velocity at point B will be his speed (the magnitude of the velocity) plus his direction of motion (the direction of his velocity). The instant that he arrives at B he is moving as indicated in the diagram below. B So his velocity vector will be 5:23 m:s\u00a11 South. 4. Step 5 : Now we determine his average velocity between A and B (NOTE TO SELF: add this here to further stress the di\ufb01erence between average and instantaneous velocities, as well as the di\ufb01erence", " between magnitude of average velocity and average speed!) 5. Step 6 : Now we calculate his average velocity over a complete revolution. The de\ufb02nition of average velocity is given earlier and requires that you know the total displacement and the total time. The total displacement for a revolution is given by the vector from the initial point to the \ufb02nal point. If the man runs in a circle then he ends where he started. This means the vector from his initial point to his \ufb02nal point has zero length. So a calculation of his average velocity follows: \u00a1!v = \u00a2\u00a1!s \u00a2t 0m 120s = 0 m:s\u00a11 = Remember displacement can be zero when even is distance not! 4.4.3 Acceleration De\ufb02nition: Acceleration is the rate of change of velocity with respect to time. Acceleration is also a vector. Remember that velocity was the rate of change of displacement with respect to time so we expect the velocity and acceleration equations to look very similar. In fact, \u00a1!a = \u00a2\u00a1!v \u00a2t = \u00a1!v f inal \u00a1 \u00a1!v initial tinitial tf inal \u00a1 (4.2) (NOTE TO SELF: average and instantaneous distiction again! expand further- what does it mean.) Acceleration will become very important later when we consider forces. 57 Using vectors is an important skill you MUST master! 4.5 Mathematical Properties of Vectors Vectors are mathematical objects and we will use them to describe physics in the language of mathematics. However, \ufb02rst we need to understand the mathematical properties of vectors (e.g. how they add and subtract). We will now use arrows representing displacements to illustrate the properties of vectors. Remember that displacement is just one example of a vector. We could just as well have decided to use forces to illustrate the properties of vectors. 4.5.1 Addition of Vectors If we de\ufb02ne a displacement vector as 2 steps in the forward direction and another as 3 steps in the forward direction then adding them together would mean moving a total of 5 steps in the forward direction. Graphically this can be seen by \ufb02rst following the \ufb02rst vector two steps forward and then following the second one three steps forward: 2 steps + 3 steps = = 5 steps We add the second vector at the end of the \ufb02rst vector, since this is where we now are after the \ufb02rst vector has acted. The vector from the tail of the \ufb02rst vector (the starting point) to the head of the last (the end point) is then the sum of the vectors. This is the tail-to-head method of vector addition. As you can convince yourself, the order in which you add vectors does not matter. In the example above, if you decided to \ufb02rst go 3 steps forward and then another 2 steps forward, the end result would still be 5 steps forward. The \ufb02nal answer when adding vectors is called the resultant. De\ufb02nition: The resultant of a number of vectors is the single vector whose e\ufb01ect is the same as the individual vectors acting together. In other words, the individual vectors can be replaced by the resultant{ the overall e\ufb01ect is the same. If vectors \u00a1!a and \u00a1!b have a resultant \u00a1!R, this can be represented mathematically as, \u00a1!R = \u00a1!a + \u00a1!b : Let us consider some more examples of vector addition using displacements. The arrows tell you how far to move and in what direction. Arrows to the right correspond to steps forward, while arrows to the left correspond to steps backward. Look at all of the examples below and check them. 1 step 1 step 1 step 1 step + + = = 2 steps 2 steps = = 2 steps 2 steps 58 Let us test the \ufb02rst one. It says one step forward and then another step forward is the same as an arrow twice as long{ two steps forward. It is possible that you end up back where you started. In this case the net result of what you have done is that you have gone nowhere (your start and end points are at the same place). In this case, your resultant displacement is a vector with length zero units. We use the symbol \u00a1!0 to denote such a vector: 1 step 1 step 1 step 1 step + + = = 1 step 1 step 1 step 1 step = \u00a1!0 = \u00a1!0 Check the following examples in the same way. Arrows up the page can be seen as steps left and arrows down the page as steps right. Try a couple to convince yourself! + = = + = = + = = \u00a1!0 + = = \u00a1", "!0 It is important to realise that the directions aren\u2019t special{ \u2018forward and backwards\u2019 or \u2018left and right\u2019 are treated in the same way. The same is true of any set of parallel directions: + = = + = = + = = \u00a1!0 + = = \u00a1!0 In the above examples the separate displacements were parallel to one another. However the same \u2018tail-to-head\u2019 technique of vector addition can be applied to vectors in any direction. + = = + = = 59 + = = Now you have discovered one use for vectors; describing resultant displacement{ how far and in what direction you have travelled after a series of movements. Although vector addition here has been demonstrated with displacements, all vectors behave in exactly the same way. Thus, if given a number of forces acting on a body you can use the same method to determine the resultant force acting on the body. We will return to vector addition in more detail later. 4.5.2 Subtraction of Vectors What does it mean to subtract a vector? Well this is really simple; if we have 5 apples and we subtract 3 apples, we have only 2 apples left. Now lets work in steps; if we take 5 steps forward and then subtract 3 steps forward we are left with only two steps forward: 5 steps - 3 steps 2 steps = What have we done? You originally took 5 steps forward but then you took 3 steps back. That backward displacement would be represented by an arrow pointing to the left (backwards) with length 3. The net result of adding these two vectors is 2 steps forward: 5 steps + 3 steps 2 steps = Thus, subtracting a vector from another is the same as adding a vector in the opposite direction (i.e. subtracting 3 steps forwards is the same as adding 3 steps backwards). This suggests that in this problem arrows to the right are positive and arrows to the left are negative. More generally, vectors in opposite directions di\ufb01er in sign (i.e. if we de\ufb02ne up as positive, then vectors acting down are negative). Thus, changing the sign of a vector simply reverses its direction: - = - = - = - = - = - = 60 In mathematical form, subtracting \u00a1!a from \u00a1!b gives a new vector \u00a1!c : \u00a1!c = \u00a1!b \u00a1 \u00a1!a = \u00a1!b + ( \u00a1\u00a1!a ) This clearly shows that subtracting vector \u00a1!a from \u00a1!b is the same as adding ( Look at the following examples of vector subtraction. \u00a1\u00a1!a ) to \u00a1!b. - = + = \u00a1!0 - = + = 4.5.3 Scalar Multiplication What happens when you multiply a vector by a scalar (an ordinary number)? Going back to normal multiplication we know that 2 2 is just 2 groups of 2 added together to give 4. We can adopt a similar approach to understand how vector multiplication works. \u00a3 2 x = + = 4.6 Techniques of Vector Addition Now that you have been acquainted with the mathematical properties of vectors, we return to vector addition in more detail. There are a number of techniques of vector addition. These techniques fall into two main categories- graphical and algebraic techniques. 4.6.1 Graphical Techniques Graphical techniques involve drawing accurate scale diagrams to denote individual vectors and their resultants. We next discuss the two primary graphical techniques, the tail-to-head technique and the paralelogram method. The Tail-to-head Method In describing the mathematical properties of vectors we used displacements and the tail-tohead graphical method of vector addition as an illustration. In the tail-to-head method of vector addition the following strategy is followed: \u2020 \u2020 Choose a scale and include a reference direction. Choose any of the vectors to be summed and draw it as an arrow in the correct direction and of the correct length{ remember to put an arrowhead on the end to denote its direction. 61 \u2020 \u2020 \u2020 Take the next vector and draw it as an arrow starting from the arrowhead of the \ufb02rst vector in the correct direction and of the correct length. Continue until you have drawn each vector{ each time starting from the head of the previous vector. In this way, the vectors to be added are drawn one after the other tail-to-head. The resultant is then the vector drawn from the tail of the \ufb02rst vector to the head of the last. Its magnitude can be determined from the length of its arrow using the scale. Its direction too can be determined from the scale diagram. Worked Example 5 Tail-to-Head Graphical Addition I Question: A ship leaves harbour H and sails 6km north to port A. From here the ship travels 12km east to port", " B, before sailing 5:5km south-west to port C. Determine the ship\u2019s resultant displacement using the tail-to-head technique of vector addition. Answer: Now, we are faced with a practical issue: in this problem the displacements are too large to draw them their actual length! Drawing a 2km long arrow would require a very big book. Just like cartographers (people who draw maps), we have to choose a scale. The choice of scale depends on the actual question{ you should choose a scale such that your vector diagram \ufb02ts the page. Before choosing a scale one should always draw a rough sketch of the problem. In a rough sketch one is interested in the approximate shape of the vector diagram. Step 1 : Draw a rough sketch of the situation Its easy to understand the problem if we \ufb02rst draw a quick sketch. N A 6km H W E S B 45o 5:5km 12km 62 C In a rough sketch one should include all of the information given in the problem. All of the magnitudes of the displacements are shown and a compass has been included as a reference direction. Step 2 : Next we choose a scale for our vector diagram It is clear from the rough sketch that choosing a scale where 1cm represents 1km (scale: 1cm = 1km) would be a good choice in this problem ){ the diagram will then take up a good fraction of an A4 page. We now start the accurate construction. Step 3 : Now we construct our scaled vector diagram Contruction Step 1: Starting at the harbour H we draw the \ufb02rst vector 6cm long in the direction north (remember in the diagram 1cm represents 1km): m c 6 A H W N S E 1cm = 1km Construction Step 2: Since the ship is now at port A we draw the second vector 12cm long starting from this point in the direction east: 63 A m c 6 H 12cm B W N S E 1cm = 1km 64 Construction Step 3: Since the ship is now at port B we draw the third vector 5:5cm long starting from this point in the direction south-west. A protractor is required to measure the angle of 45o. A m c 6 H 12cm B 45o C 5.5c m W N S 1cm = 1km E Construction Step 4: As a \ufb02nal step we draw the resultant displacement from the starting point (the harbour H) to the end point (port C). We use a ruler to measure the length of this arrow and a protractor to determine its direction A m c 6 H 12cm B 45o C 5.5c m W N S 1cm = 1km E 75:4o 8. 3 8 c m 65 Step 4 : Apply the scale conversion We now use the scale to convert the length of the resultant in the scale diagram to the actual displacement in the problem. Since we have chosen a scale of 1cm = 1km in this problem the resultant has a magnitude of 8:38 km. The direction can be speci\ufb02ed in terms of the angle measured either as 75:4o east of north or on a bearing of 75:4o. Step 5 : Quote the \ufb02nal answer The resultant displacement of the ship is 8:38 km on a bearing of 75:4o! Worked Example 6 Tail-to-Head Graphical Addition II Question: A man walks 40 m East, then 30 m North. a) What was the total distance he walked? b) What is his resultant displacement? N S E W m 0 3 40m Answer: Step 1 : a) Determine the distance that the man traveled In the \ufb02rst part of his journey he traveled 40 m and in the second part he traveled 30 m. This gives us a total distance traveled of 40 + 30 = 70 m. Step 2 : b) Determine his resultant displacement - start by drawing a rough sketch The man\u2019s resultant displacement is the vector from where he started to where he ended. It is the sum of his two separate displacements. We will use the tail-to-head method of accurate construction to \ufb02nd this vector. Here is our rough sketch: 66 N S E W m 0 3 R es ulta n t 40m Step 3 : Choose a suitable scale A scale of 1cm represents 5m (1cm = 5m) is a good choice here. Now we can begin the process of construction. Step 4 : Draw the \ufb02rst vector according to scale We draw the \ufb02rst displacement as an arrow 8cm long (according to the scale 8cm = 8 5m = 40m) in the direction east: \u00a3 W N S E 1cm = 5m 8cm Step 5 : Draw the second vector according to scale Starting from the head of the \ufb02rst vector we draw the second", " displacement as an 5m = 30m) in the direction north: arrow 6cm long (according to the scale 6cm = 6 \u00a3 67 W N S E 1cm = 5m m c 6 8cm Step 6 : Determine the resultant vector Now we connect the starting point to the end point and measure the length and direction of this arrow (the resultant) W N S E 1cm = 5m 36:9o m c 6 1 0 c m 8cm Step 7 : Apply the scale conversion Finally we use the scale to convert the length of the resultant in the scale diagram to the actual magnitude of the resultant displacement. According to the chosen scale 1cm = 5m. Therefore 10cm represents 50m. The resultant displacement is then 50m 36:9o north of east. 68 The Parallelogram Method When needing to \ufb02nd the resultant of two vectors another graphical technique can be applied- the parallelogram method. The following strategy is employed: \u2020 \u2020 \u2020 \u2020 \u2020 Choose a scale and a reference direction. Choose either of the vectors to be added and draw it as an arrow of the correct length in the correct direction. Draw the second vector as an arrow of the correct length in the correct direction from the tail of the \ufb02rst vector. Complete the parallelogram formed by these two vectors. The resultant is then the diagonal of the parallelogram. Its magnitude can be determined from the length of its arrow using the scale. Its direction too can be determined from the scale diagram. Worked Example 7 Parallelogram Method of Graphical Addition I Question: A force of F1 = 5N is applied to a block in a horizontal direction. A second force F2 = 4N is applied to the object at an angle of 30o above the horizontal. N 4 = F 2 30o F1 = 5N Determine the resultant force acting on the block using the parallelogram method of accurate construction. Answer: Step 1 : Firstly make a rough sketch of the vector diagram N 4 30o 5N 69 Step 2 : Choose a suitable scale In this problem a scale of 1cm = 0:5N would be appropriate, since then the vector diagram would take up a reasonable fraction of the page. We can now begin the accurate scale diagram. Step 3 : Draw the \ufb02rst scaled vector Let us draw F1 \ufb02rst. According to the scale it has length 10cm: 10cm 1cm = 0:5N Step 4 : Draw the second scaled vector Next we draw F2. According to the scale it has length 8cm. We make use of a protractor to draw this vector at 30o to the horizontal: c m 8 30o 10cm Step 5 : Determine the resultant vector Next we complete the parallelogram and draw the diagonal: 1cm = 0:5N c m 8 1 7 : 4 c m 13:3o 10cm 1cm = 0:5N Step 6 : Apply the scale conversion Finally we use the scale to convert the measured length into the actual magnitude. Since 1cm = 0:5N, 17:4cm represents 8:7N. Therefore the resultant force is 8:7N at 13:3o above the horizontal. 70 The parallelogram method is restricted to the addition of just two vectors. However, it is arguably the most intuitive way of adding two forces acting at a point. 4.6.2 Algebraic Addition and Subtraction of Vectors Vectors in a Straight Line Whenever you are faced with adding vectors acting in a straight line (i.e. some directed left and some right, or some acting up and others down) you can use a very simple algebraic technique: \u2020 \u2020 \u2020 Choose a positive direction. As an example, for situations involving displacements in the directions west and east, you might choose west as your positive direction. In that case, displacements east are negative. Next simply add (or subtract) the vectors with the appropriate signs. As a \ufb02nal step the direction of the resultant should be included in words (positive answers are in the positive direction, while negative resultants are in the negative direction). Let us consider a couple of examples. Worked Example 8 Adding vectors algebraically I Question: A tennis ball is rolled towards a wall which is 10m away to the right. If after striking the wall the ball rolls a further 2:5m along the ground to the left, calculate algebraically the ball\u2019s resultant displacement. (NOTE TO SELF: PGCE suggest a \u2018more real looking\u2019 diagram, followed by a diagram one would draw to solve the problem (like our existing one with the positive direction shown as an arrow)) Answer: Step 1 : Draw a rough sketch of the situation 10 m 2.5 m Wall Start Step 2 : Decide which method to use to calculate the resultant We know that the resultant displacement", " of the ball (\u00a1!s resultant) is equal to the sum of the ball\u2019s separate displacements (\u00a1!s 1 and \u00a1!s 2): \u00a1!s resultant = \u00a1!s 1 + \u00a1!s 2 Since the motion of the ball is in a straight line (i.e. the ball moves left and right), we can use the method of algebraic addition just explained. 71 Step 3 : Choose a positive direction Let\u2019s make to the right the positive direction. This means that to the left becomes the negative direction. Step 4 : Now de\ufb02ne our vectors algebraically With right positive: \u00a1!s 1 = +10:0m and \u00a1!s 2 = 2:5m \u00a1 Step 5 : Add the vectors Next we simply add the two displacements to give the resultant: \u00a1!s resultant = (+10m) + ( 2:5m) \u00a1 = (+7:5)m Step 6 : Quote the resultant Finally, in this case right means positive so: \u00a1!s resultant = 7:5m to the right Let us consider an example of vector subtraction. Worked Example 9 Subtracting vectors algebraically I Question: Suppose that a tennis ball is thrown horizontally towards a wall at 3m:s\u00a11 to the right. After striking the wall, the ball returns to the thrower at 2m:s\u00a11. Determine the change in velocity of the ball. Answer: Step 1 : Draw a sketch A quick sketch will help us understand the problem (NOTE TO SELF: Maybe a sketch here?) Step 2 : Decide which method to use to calculate the resultant Remember that velocity is a vector. The change in the velocity of the ball is equal to the di\ufb01erence between the ball\u2019s initial and \ufb02nal velocities: \u00a2\u00a1!v = \u00a1!v f inal \u00a1 \u00a1!v initial Since the ball moves along a straight line (i.e. left and right), we can use the algebraic technique of vector subtraction just discussed. Step 3 : Choose a positive direction Let\u2019s make to the right the positive direction. This means that to the left becomes the negative direction. 72 Step 4 : Now de\ufb02ne our vectors algebraically With right positive: \u00a1!v initial = +3m:s\u00a11 and \u00a1!v f inal = 2m:s\u00a11 \u00a1 Step 5 : Subtract the vectors Thus, the change in velocity of the ball is: \u00a2\u00a1!v = ( = ( 2m:s\u00a11) 5)m:s\u00a11 \u00a1 \u00a1 (+3m:s\u00a11) \u00a1 Step 6 : Quote the resultant Remember that in this case right means positive so: \u00a2\u00a1!v = 5m:s\u00a11 to the left Remember that the technique of addition and subtraction just discussed can only be applied to vectors acting along a straight line. A More General Algebraic technique In worked example 3 the tail to head method of accurate construction was used to determine the resultant displacement of a man who travelled \ufb02rst east and then north. However, the man\u2019s resultant can be calculated without drawing an accurate scale diagram. Let us revisit this example. Worked Example 10 An Algebraic solution to Worked Example 3 Question: A man walks 40 m East, then 30 m North. Calculate the man\u2019s resultant displacement. Answer: Step 1 : Draw a rough sketch As before, the rough sketch looks as follows: 73 N S E W m 0 3 R es ulta n t \ufb01 40m Step 2 : Determine the length of the resultant Note that the triangle formed by his separate displacement vectors and his resultant displacement vector is a right-angle triangle. We can thus use Pythogoras\u2019 theorem to determine the length of the resultant. If the length of the resultant vector is called s then: s2 = (40m)2 + (30m)2 s2 = 2500m2 s = 50m Step 3 : Determine the direction of the resultant Now we have the length of the resultant displacement vector but not yet its direction. To determine its direction we calculate the angle \ufb01 between the resultant displacement vector and East. We can do this using simple trigonometry: tan \ufb01 = tan \ufb01 = opposite adjacent 30 40 \ufb01 = arctan(0:75) \ufb01 = 36:9o Step 4 : Quote the resultant Our \ufb02nal answer is then: Resultant Displacement: 50 m at 36:9o North of East This is exactly the same answer we arrived at after drawing a scale diagram! In the previous example we were able to use simple trigonometry to calculate a man", "\u2019s resultant displacement. This was possible since the man\u2019s directions of motion were perpendicular (north 74 and east). Algebraic techniques, however, are not limited to cases where the vectors to be combined are along the same straight line or at right angles to one another. The following example illustrates this. Worked Example 11 Further example of vector addition by calculation Question: A man walks from point A to point B which is 12km away on a bearing of 45o. From point B the man walks a further 8km east to point C. Calculate the man\u2019s resultant displacement. Answer: Step 1 : Draw a rough sketch of the situation 8km C B 45o G 12k m F A 45o B ^AF = 45o since the man walks initially on a bearing of 45o. Then, A ^BG = B ^AF = 45o (alternate angles parallel lines). Both of these angles are included in the rough sketch. Step 2 : Calculate the length of the resultant The resultant is the vector AC. Since we know both the lengths of AB and BC and the included angle A ^BC, we can use the cosine rule: AC 2 = AB2 + BC 2 = (12)2 + (8)2 = 343:8 2 2 \u00a2 \u00a2 \u00a1 \u00a1 AC = 18:5 km BC cos(A ^BC) AB (12)(8) cos(135o) \u00a2 Step 3 : Determine the direction of the resultant Next we use the sine rule to determine the angle : 75 sin 8 = sin = sin 1350 18:5 8 \u00a3 sin 135o 18:5 = arcsin(0:3058) = 17:8o Thus, F ^AC = 62:8o. Step 4 : Quote the resultant Our \ufb02nal answer is then: Resultant Displacement: 18:5km on a bearing of 62:8o 4.7 Components of Vectors In the discussion of vector addition we saw that a number of vectors acting together can be combined to give a single vector (the resultant). In much the same way a single vector can be broken down into a number of vectors which when added give that original vector. These vectors which sum to the original are called components of the original vector. The process of breaking a vector into its components is called resolving into components. While summing a given set of vectors gives just one answer (the resultant), a single vector can be resolved into in\ufb02nitely many sets of components. In the diagrams below the same black vector is resolved into di\ufb01erent pairs of components. These components are shown in red. When added together the red vectors give the original black vector (i.e. the original vector is the resultant of its components). In practice it is most useful to resolve a vector into components which are at right angles to one another. Worked Example 12 Resolving a vector into components 76 Question: A motorist undergoes a displacement of 250km in a direction 30o north of east. Resolve this displacement into components in the directions north (\u00a1!s N ) and east (\u00a1!s E). Answer: Step 1 : Draw a rough sketch of the original vector W N S 30o E k m 0 5 2 Step 2 : Determine the vector component Next we resolve the displacement into its components north and east. Since these directions are orthogonal to one another, the components form a right-angled triangle with the original displacement as its hypotenuse 30o \u00a1!s E N \u00a1!s Notice how the two components acting together give the orginal vector as their resultant. Step 3 : Determine the lengths of the component vectors Now we can use trigonometry to calculate the magnitudes of the components of the original displacement: sN = 250 sin 30o = 125 km 77 and sE = 250 cos 30o = 216:5 km Remember sN and sE are the magnitudes of the components{ they are in the directions north and east respectively. (NOTE TO SELF: SW: alternatively these results can be arrived at by construction. Include?) 4.7.1 Block on an incline As a further example of components let us consider a block of mass m placed on a frictionless surface inclined at some angle to the horizontal. The block will obviously slide down the incline, but what causes this motion? The forces acting on the block are its weight mg and the normal force N exerted by the surface on the object. These two forces are shown in the diagram below. N W k? W mg Now the object\u2019s weight can be resolved into components parallel and perpendicular to the inclined surface. These components are shown as red arrows in the diagram above and are at right angles to each other. The components have been drawn acting from the same point. Applying the parallelogram method, the two components of the block\u2019s weight sum to the weight vector", ". To \ufb02nd the components in terms of the weight we can use trigonometry: Wk = mg sin W? = mg cos The component of the weight perpendicular to the slope W? exactly balances the normal force N exerted by the surface. The parallel component, however, Wk is unbalanced and causes the block to slide down the slope. 78 Figure 4.2: An example of two vectors being added to give a resultant 4.7.2 Vector addition using components In Fig 4.3 two vectors are added in a slightly di\ufb01erent way to the methods discussed so far. It might look a little like we are making more work for ourselves, but in the long run things will be easier and we will be less likely to go wrong. In Fig 4.3 the primary vectors we are adding are represented by solid lines and are the same vectors as those added in Fig 4.2 using the less complicated looking method. Each vector can be broken down into a component in the x-direction and one in the ydirection. These components are two vectors which when added give you the original vector as the resultant. Look at the red vector in \ufb02gure 4.3. If you add up the two red dotted ones in the x-direction and y-direction you get the same vector. For all three vectors we have shown their respective components as dotted lines in the same colour. But if we look carefully, addition of the x components of the two original vectors gives the x component of the resultant. The same applies to the y components. So if we just added all the components together we would get the same answer! This is another important property of vectors. Worked Example 13 Adding Vectors Using Components Question: Lets work through the example shown in Fig. 4.3 to determine the resultant. Answer: Step 1 : Decide how to tackle the problem The \ufb02rst thing we must realise is that the order that we add the vectors does not matter. Therefore, we can work through the vectors to be added in any order. 79 Figure 4.3: Components of vectors can be added as well as the vectors themselves Step 2 : Resolve the red vector into components Let us start with the bottom vector. If you are told that this vector has a length of 5:385 units and an angle of 21:8o to the horizontal then we can \ufb02nd its components. We do this by using known trigonometric ratios. First we \ufb02nd the vertical or y component: sin = y hypotenuse sin(21:8) = y 5:385 y = 5:385 sin(21:8 Secondly we \ufb02nd the horizontal or x component: x hypotenuse cos = cos(21:8) = x 5:385 x = 5:385 cos(21:8) x = 5 80 We now know the lengths of the sides of the triangle for which our vector is the hypotenuse. If you look at these sides we can assign them directions given by the dotted arrows. Then our original red vector is just the sum of the two dotted vectors (its components). When we try to \ufb02nd the \ufb02nal answer we can just add all the dotted vectors because they would add up to the two vectors we want to add. Step 3 : Now resolve the second vector into components. The green vector has a length of 5 units and a direction of 53.13 degrees to the horizontal so we can \ufb02nd its components. sin(53:13) = sin = y hypotenuse y 5 y = 5 sin(53:13) y = 4 5 4 3 cos(53:13) = cos = x hypotenuse x 5 x = 5 cos(53:13) x = 3 Step 4 : Determine the components of the resultant vector Now we have all the components. If we add all the x-components then we will have the x-component of the resultant vector. Similarly if we add all the y-components then we will have the y-component of the resultant vector. The x-components of the two vectors are 5 units right and then 3 units right. This gives us a \ufb02nal x-component of 8 units right. The y-components of the two vectors are 2 units up and then 4 units up. This gives us a \ufb02nal y-component of 6 units up. Step 5 : Determine the magnitude and direction of the resultant vector Now that we have the components of the resultant, we can use Pythagoras\u2019 theorem to determine the length of the resultant. Let us call the length of the hypotenuse l and we can calculate its value; 81 l2 = (6)2 + (8)2 l2 = 100 l = 10: 8 6 1 0 The resultant has length of 10 units so all we have to do is calculate its direction. We can specify the direction as the", " angle the vectors makes with a known direction. To do this you only need to visualise the vector as starting at the origin of a coordinate system. We have drawn this explicity below and the angle we will calculate is labeled \ufb01. 82 Using our known trigonometric ratios we can calculate the value of \ufb01; tan \ufb01 = 6 8 \ufb01 = arctan \ufb01 = 36:8o: 6 8 Step 6 : Quote the \ufb02nal answer Our \ufb02nal answer is a resultant of 10 units at 36:8o to the positive x-axis. 4.8 Do I really need to learn about vectors? Are they really useful? Vectors are essential to do physics. Absolutely essential. This is an important warning. If something is essential we had better stop for a moment and make sure we understand it properly. 4.9 Summary of Important Quantities, Equations and Con- cepts Vector A vector is a measurement which has both magnitude and direction. Displacement Displacement is a vector with direction pointing from some initial (starting) point to some \ufb02nal (end) point and whose magnitude is the straight-line distance from the starting point to the \ufb02nal point. 83 Distance The distance travelled is the length of your actual path. Velocity Velocity is the rate of change of displacement with respect to time. Acceleration Acceleration is the rate of change of velocity with respect to time. Resultant The resultant of a number of vectors is the single vector whose e\ufb01ect is the same as the individual vectors acting together. Quantity Displacement Velocity Distance Speed Acceleration Symbol \u00a1!s \u00a1!u,\u00a1!v d v \u00a1!a S.I. Units Direction m m:s\u00a11 m m:s\u00a11 m:s\u00a12 X X { { X Table 4.1: Summary of the symbols and units of the quantities used in Vectors 84 Chapter 5 Forces 5.1 \u2018TO DO\u2019 LIST introduce concept of a system for use in momentum- in NIII for instance, also concept of isolated system and external forces add incline plane examples generally \u2018bulk-up\u2019 the Newton\u2019s Laws sections \u2020 \u2020 \u2020 5.2 What is a force? The simplest answer is to say a \u2018push\u2019 or a \u2018pull\u2019. If the force is great enough to overcome friction the object being pushed or pulled will move. We could say a force is something that makes objects move. Actually forces give rise to accelerations! In fact, the acceleration of a body is directly proportional to the net force acting on it. The word net is important{ forces are vectors and what matters in any situation is the vector sum of all the forces acting on an object. The unit of force is the newton (symbol N ). It is named after Sir Isaac Newton, whose three laws you will learn about shortly. Interesting Fact: Force was \ufb02rst described by Archimedes. Archimedes of Syracuse (circa 287 BC - 212 BC), was a Greek mathematician, astronomer, philosopher, physicist and engineer. He was killed by a Roman soldier during the sack of the city, despite orders from the Roman general, Marcellus, that he was not to be harmed. 5.3 Force diagrams The resultant force acting on an object is the vector sum of the set of forces acting on that one object. It is very important to remember that all the forces must be acting on the same object. 85 The easiest way to determine this resultant force is to construct what we call a force diagram. In a force diagram we represent the object by a point and draw all the force vectors connected to that point as arrows. Remember from Chapter?? that we use the length of the arrow to indicate the vector\u2019s magnitude and the direction of the arrow to show which direction it acts in. The second step is to rearrange the force vectors so that it is easy to add them together and \ufb02nd the resultant force. Let us consider an example to get started: Two people push on a box from opposite sides with a force of 5N. 5N 5N When we draw the force diagram we represent the box by a dot. The two forces are represented by arrows, with their tails on the dot. Force Diagram: 5N 5N See how the arrows point in opposite directions and have the same magnitude (length). This means that they cancel out and there is no net force acting on the object. This result can be obtained algebraically too, since the two forces act along the same line. Firstly we choose a positive direction and then add the two vectors taking their directions into account. Consider to the right as the positive direction Fres = (+5N ) + ( = 0N \u00a1 5N ) As you work with more complex force diagrams, in which the forces", " do not exactly balance, 2N ). What does this mean? you may notice that sometimes you get a negative answer (e.g. Does it mean that we have something the opposite of force? No, all it means is that the force acts in the opposite direction to the one that you chose to be positive. You can choose the positive direction to be any way you want, but once you have chosen it you must keep it. \u00a1 Once a force diagram has been drawn the techniques of vector addition introduced in the previous chapter can be implemented. Depending on the situation you might choose to use a graphical technique such as the tail-to-head method or the parallelogram method, or else an algebraic approach to determine the resultant. Since force is a vector all of these methods apply! Always remember to check your signs Worked Example 14 Single Force on a block Question: A block on a frictionless at surface weighs 100N. A 75N force is applied to the block towards the right. What is the net force (or resultant force) on the block? Answer: 86 Step 1 : Firstly, draw a force diagram for the block Fnormal = 100N Fapplied = 75N Fweight = 100N Be careful not to forget the two forces perpendicular to the surface. Every object with mass is attracted to the centre of the earth with a force (the object\u2019s weight). However, if this were the only force acting on the block in the vertical direction then the block would fall through the table to the ground. This does not happen because the table exerts an upward force (the normal force) which exactly balances the object\u2019s weight. Step 2 : Answer Thus, the only unbalanced force is the applied force. This applied force is then the resultant force acting on the block. 5.4 Equilibrium of Forces At the beginning of this chapter it was mentioned that resultant forces cause objects to accelerate. If an object is stationary or moving at constant velocity then either, no forces are acting on the object, or \u2020 the forces acting on that object are exactly balanced. \u2020 A resultant force would cause a stationary object to start moving or an object moving at constant velocity to speed up or slow down. In other words, for stationary objects or objects moving with constant velocity, the resultant force acting on the object is zero. The object is said to be in equilibrium. If a resultant force acts on an object then that object can be brought into equilibrium by applying an additional force that exactly balances this resultant. Such a force is called the equilibrant and is equal in magnitude but opposite in direction to the original resultant force acting on the object. De\ufb02nition: The equilibrant of any number of forces is the single force required to produce equilibrium. 87 Objects at rest or moving with constant velocity are in equilibrium and have a zero resultant force Resultant of \u00a1!F1 and \u00a1!F2 \u00a1!F1 \u00a1!F2 \u00a1!F3 Equilibrant of \u00a1!F1 and \u00a1!F2 In the \ufb02gure the resultant of \u00a1!F1 and \u00a1!F2 is shown in red. The equilibrant of \u00a1!F1 and \u00a1!F2 is then the vector opposite in direction to this resultant with the same magnitude (i.e. \u00a1!F3). \u00a1!F1, \u00a1!F2 and \u00a1!F3 are in equilibrium \u00a1!F3 is the equilibrant of \u00a1!F1 and \u00a1!F2 \u2020 \u2020 \u00a1!F1 and \u00a1!F2 are kept in equilibrium by \u00a1!F3 \u2020 As an example of an object in equilibrium, consider an object held stationary by two ropes in the arrangement below: 50o 40o Rope 1 Rope 2 Let us draw a force diagram for the object. In the force diagram the object is drawn as a dot and all forces acting on the object are drawn in the correct directions starting from that dot. In this case, three forces are acting on the object. 88 50o \u00a1!T1 40o \u00a1!T2 \u00a1!W Each rope exerts a force on the object in the direction of the rope away from the object. These tension forces are represented by \u00a1!T1 and \u00a1!T2. Since the object has mass, it is attracted towards the centre of the earth. This weight is represented in the force diagram as \u00a1!W. Since the object is stationary, the resultant force acting on the object is zero. In other words the three force vectors drawn tail-to-head form a closed triangle: 40o \u00a1!T2 50o \u00a1!T1 \u00a1!W In general, when drawn tail-to-head the forces acting on an object in equilibrium form a closed \ufb02gure with the head of the last", " vector joining up with the tail of the \ufb02rst vector. When only three forces act on an object this closed \ufb02gure is a triangle. This leads to the triangle law for three forces in equilibrium: Triangle Law for Three Forces in Equilibrium: Three forces in equilibrium can be represented in magnitude and direction by the three sides of a triangle taken in order. Worked Example 15 89 Equilibrium Question: A car engine of weight 2000N is lifted by means of a chain and pulley system. In sketch A below, the engine is suspended by the chain, hanging stationary. In sketch B, the engine is pulled sideways by a mechanic, using a rope. The engine is held in such a position that the chain makes an angle of 30o with the vertical. In the questions that follow, the masses of the chain and the rope can be ignored. 30o CHAIN CHAIN ENGINE ENGINE ROPE Sketch A Sketch B i) Draw a force diagram representing the forces acting on the engine in sketch A. ii) Determine the tension in the chain in sketch A. iii) Draw a force diagram representing the forces acting on the engine in sketch B. ii) In sketch B determine the magnitude of the applied force and the tension in the chain. Answer: Step 1 : Force diagram for sketch A i) Just two forces are acting on the engine in sketch A: \u00a1!T chain \u00a1!W 90 Step 2 : Determine the tension in the chain ii) Since the engine in sketch A is stationary, the resultant force on the engine is zero. Thus the tension in the chain exactly balances the weight of the engine, Tchain = W = 2000N Step 3 : Force diagram for sketch B iii) Three forces are acting on the engine in sketch B: \u00a1!F applied 30o \u00a1!W \u00a1!T chain 30o Since the engine is at equilibrium (it is held stationary) the three forces drawn tailto-head form a closed triangle. Step 4 : Calculate the magnitude of the forces in sketch B iv) Since no method was speci\ufb02ed let us calculate the magnitudes algebraically. Since the triangle formed by the three forces is a right-angle triangle this is easily done: Fapplied W = tan 30o Fapplied = (2000) tan 30o and = 1155 N Tchain W Tchain = = 1 cos 30o 2000 cos 30o = 2309 N 5.5 Newton\u2019s Laws of Motion Our current laws of motion were discovered by Sir Isaac Newton. It is said that Sir Isaac Newton started to think about the nature of motion and gravitation after being struck on the head by a falling apple. Newton discovered 3 laws describing motion: 91 5.5.1 First Law De\ufb02nition: Every object will remain at rest or in uniform motion in a straight line unless it is made to change its state by the action of an external force. For example, a cement block isn\u2019t going to move unless you push it. A rocket in space is not going to speed up, slow down nor change direction unless the engines are switched on. Newton\u2019s First Law may seem rather surprising to you when you \ufb02rst meet it. If you roll a ball along a surface it always stops, but Newton\u2019s First Law says that an item will remain in uniform motion in a straight line unless a force acts upon it. It doesn\u2019t seem like there is any force acting on the ball once you let it go, but there is! In real life (unlike many physics problems) there is usually friction acting. Friction is the external force acting on the ball causing it to stop. Notice that Newton\u2019s First Law says that the force must be external. For example, you can\u2019t grab your belt and pull yourself up to the ceiling. Of course, you could get someone else to pull you up, but then that person would be applying an external force. Worked Example 16 Newton\u2019s First Law in action Question: Why do passengers get thrown to the side when the car they are driving in goes around a corner? Answer: Newton\u2019s First Law Step 1 : What happens before the car turns Before the car starts turning both you and the car are travelling at the same velocity. (picture A) Step 2 : What happens while the car turns The driver turns the wheels of the car, which then exert a force on the car and the car turns. This force acts on the car but not you, hence (by Newton\u2019s First Law) you continue moving with the same velocity. (picture B) Step 3 : Why passengers get thrown to the side If the passenger is wearing a seatbelt it will exert a force on the passenger until the passenger\u2019s velocity is the same as that of the car (picture C). Without a seatbelt the passenger may hit the side of the car or even the windscreen! 92", " A: Both the car and the person travelling at the same velocity B: The cars turns but not the person C: Both the car and the person are travelling at the same velocity again 5.5.2 Second Law De\ufb02nition: The resultant force acting on a body results in an acceleration which is in the same direction as the resultant force and is directly proportional to the magnitude of this force and inversely proportional to the mass of the object. In mathematical form this law states, \u00a1!F Res = m\u00a1!a (5.1) So if a resultant force \u00a1!F Res acts on an object with mass m it will result in an acceleration \u00a1!a. It makes sense that the direction of the acceleration is in the direction of the resultant force. If you push something away from you it doesn\u2019t move toward you unless of course there is another force acting on the object towards you! Worked Example 17 Newton\u2019s Second Law Question: A block of mass 10kg is accelerating at 2m:s\u00a12. What is the magnitude of the resultant force acting on the block? Answer: Step 1 : Decide what information has been supplied We are given the block\u2019s mass the block\u2019s acceleration \u2020 \u2020 all in the correct units. 93 Step 2 : Determine the force on the block We are asked to \ufb02nd the magnitude of the force applied to the block. Newton\u2019s Second Law tells us the relationship between acceleration and force for an object. Since we are only asked for the magnitude we do not need to worry about the directions of the vectors: FRes = ma 2m:s\u00a12 = 10kg = 20N \u00a3 Thus, there must be a resultant force of 20N acting on the box. Worked Example 18 Newton\u2019s Second Law 2 Question: A 12N force is applied in the positive x-direction to a block of mass 100mg resting on a frictionless at surface. What is the resulting acceleration of the block? Answer: Step 1 : Decide what information has been supplied We are given the block\u2019s mass the applied force \u2020 \u2020 but the mass is not in the correct units. Step 2 : Convert the mass into the correct units 100mg = 100 1000g = 1kg \u00a3 10\u00a13g = 0:1g 1 1000g 1 = 1kg \u00a3 1kg 1000g = 0:1g = 0:1g = 0:1g 1 1kg 1000g \u00a3 \u00a3 = 0:0001 kg Step 3 : Determine the direction of the acceleration We know that net force results in acceleration. Since there is no friction the applied force is the resultant or net force on the block (refer to the earlier example of the block pushed on the surface of the table). The block will then accelerate in the direction of this force according to Newton\u2019s Second Law. 94 Step 4 : Determine the magnitude of the acceleration FRes = ma 12N = (0:0001kg)a a = = 120000 12N 0:0001kg N kg kg:m s2:kg m s2 = 120000 = 120000 = 1:2 \u00a3 105 m:s\u00a12 From Newton\u2019s Second Law the direction of the acceleration is the same as that of the 105 m:s\u00a12 resultant force. The \ufb02nal result is then that the block accelerates at 1:2 in the positive x-direction. \u00a3 Weight and Mass You must have heard people saying \\My weight is 60kg\". This is actually incorrect because it is mass that is measured in kilograms. Weight is the force of gravity exerted by the earth on an object with mass: Fweight = mg (5.2) As such, weight is measured in newtons. If you compare this equation to Newton\u2019s Second Law you will see that it looks exactly the same with the a replaced by g. Thus, when weight is the only force acting on an object (i.e. when Fweight is the resultant force acting on the object) the object has an acceleration g. Such an object is said to be in free fall. The value for g is the same for all objects (i.e. it is independent of the objects mass): g = 9:8ms\u00a12 10ms\u00a12 (5.3) \u2026 You will learn how to calculate this value from the mass and radius of the earth in Chapter??. Actually the value of g varies slightly from place to place on the earth\u2019s surface. The reason that we often get confused between weight and mass, is that scales measure your weight (in newtons) and then display your mass using the equation above. (NOTE TO SELF: include example of skydiver- calculate the acceleration before and after opening parachute{ free-fall and then negative acceleration) Worked Example 19 Calculating the resultant and then the acceleration 95 Question", ": A block (mass 20kg) on a frictionless at surface has a 45N force applied to it in the positive x-direction. In addition a 25N force is applied in the negative x-direction. What is the resultant force acting on the block and the acceleration of the block? Answer: Step 1 : Decide what information has been supplied We are given the block\u2019s mass Force F1 = 45N in the positive x-direction Force F2 = 25N in the negative x-direction \u2020 \u2020 \u2020 all in the correct units. Step 2 : Figure out how to tackle the problem We are asked to determine what happens to the block. We know that net force results in an acceleration. We need to determine the net force acting on the block. Step 3 : Determine the magnitude and direction of the resultant force Since F1 and F2 act along the same straight line, we can apply the algebraic technique of vector addition discussed in the Vectors chapter to determine the resultant force. Choosing the positive x-direction as our positive direction: Positive x-direction is the positive direction: FRes = (+45N ) + ( = +20N = 20N in the positive x 25N ) \u00a1 direction \u00a1 where we remembered in the last step to include the direction of the resultant force in words. By Newton\u2019s Second Law the block will accelerate in the direction of this resultant force. Step 4 : Determine the magnitude of the acceleration FRes = ma 20N = (20kg)a a = = 1 20N 20kg N kg kg:m s2:kg = 1 m:s\u00a12 = 1 The \ufb02nal result is then that the block accelerates at 1 m:s\u00a12 in the positive x-direction (the same direction as the resultant force). 96 Worked Example 20 Block on incline Question: insert question here maybe with friction! N? W mg W k 5.5.3 Third Law De\ufb02nition: For every force or action there is an equal but opposite force or reaction. Newton\u2019s Third Law is easy to understand but it can get quite di\u2013cult to apply it. An important thing to realise is that the action and reaction forces can never act on the same object and hence cannot contribute to the same resultant. Worked Example 21 Identifying action-reaction pairs Question: Consider pushing a box on the surface of a rough table. 1. Draw a force diagram indicating all of the forces acting on the box. 2. Identify the reaction force for each of the forces acting on the box. Answer: 1. The following force diagram shows all of the forces acting on the box 97 Remember to draw the object in your force diagram as a dot Fnormal Fpush Ff riction Fweight 2. The following table lists each of forces acting on the box (the actions) together with their corresponding reactions: Action Fpush : Person pushing on the box Reaction Box pushing on the person Fweight : The earth attracting The box attracting the box (the weight) Fnormal : The box pushing on the table the earth The table pushing on the box Ff riction : The table acting on The box acting on the box the table Notice that to \ufb02nd the reaction force you need to switch around the object supplying the force and the object receiving the force. Be careful not to think that the normal force is the reaction partner to the weight of the box. The normal force balances the weight force but they are both forces acting on the box so they cannot possibly form an action-reaction pair. There is an important thing to realise which is related to Newton\u2019s Third Law. Think about dropping a stone o\ufb01 a cli\ufb01. It falls because the earth exerts a force on it (see Chapter??) and it doesn\u2019t seem like there are any other forces acting. So is Newton\u2019s Third Law wrong? No, the reactionary force to the weight of the stone is the force exerted by the stone on the earth. This is illustrated in detail in the next worked example. Worked Example 22 Newton\u2019s Third Law Question: A stone of mass 0:5kg is accelerating at 10 m:s\u00a12 towards the earth. 98 1. What is the force exerted by the earth on the stone? 2. What is the force exerted by the stone on the earth? 3. What is the acceleration of the earth, given that its mass is 5:97 Answer: 1. Step 1 : Decide what information has been supplied We are given 1027kg? \u00a3 the stone\u2019s mass the stone\u2019s acceleration (g) \u2020 \u2020 all in the correct units. Step 2 : The force applied by the earth on the stone This force is simply the weight of the stone. Step 3 : The magnitude of the force of the earth on the stone Applying Newton\u2019s Second", " Law, we can \ufb02nd the magnitude of this force, FRes = ma = 0:5kg = 5N \u00a3 10 m:s\u00a12 Therefore the earth applies a force of 5N towards the earth on the stone. 2. By Newton\u2019s Third Law the stone must exert an equal but opposite force on the earth. Hence the stone exerts a force of 5N towards the stone on the earth. 3. We have the force acting on the earth the earth\u2019s mass \u2020 \u2020 in the correct units Step 4 : To \ufb02nd the earth\u2019s acceleration we must apply Newton\u2019s Second Law to \ufb02nd the magnitude FRes = ma 5N = (5:97 a = 5:97 \u00a3 = 8:37521 1027kg)a \u00a3 5N 1027kg 10\u00a128 m:s\u00a12 \u00a3 The earth\u2019s acceleration is directed towards the stone (i.e. in the same direction as the force on the earth). The earth\u2019s acceleration is really tiny. This is why you don\u2019t notice the earth moving towards the stone, even though it does. 99 Interesting Fact: Newton \ufb02rst published these laws in Philosophiae Naturalis Principia Mathematica (1687) and used them to prove many results concerning the motion of physical objects. Only in 1916 were Newton\u2019s Laws superceded by Einstein\u2019s theory of relativity. The next two worked examples are quite long and involved but it is very important that you understand the discussion as they illustrate the importance of Newton\u2019s Laws. Worked Example 23 Rockets Question: How do rockets accelerate in space? Answer: \u2020 \u2020 \u2020 \u2020 Gas explodes inside the rocket. This exploding gas exerts a force on each side of the rocket (as shown in the picture below of the explosion chamber inside the rocket). Note that the forces shown in this picture are representative. With an explosion there will be forces in all directions. Due to the symmetry of the situation, all the forces exerted on the rocket are balanced by forces on the opposite side, except for the force opposite the open side. This force on the upper surface is unbalanced. This is therefore the resultant force acting on the rocket and it makes the rocket accelerate forwards. Systems and External Forces The concepts of a system and an external forces are very important in physics. A system is any collection of objects. If one draws an imaginary box around such a system then an external force is one that is applied by an object or person outside the box. Imagine for example a car pulling a trailer. 100 5.6 Examples of Forces Studied Later Most of physics revolves around forces. Although there are many di\ufb01erent forces we deal with them all in the same way. The methods to \ufb02nd resultants and acceleration do not depend on the type of force we are considering. At \ufb02rst glance, the number of di\ufb01erent forces may seem overwhelming - gravity, drag, electrical forces, friction and many others. However, physicists have found that all these forces can be classi\ufb02ed into four groups. These are gravitational forces, electromagnetic forces, strong nuclear force and weak nuclear force. Even better, all the forces that you will come across at school are either gravitational or electromagnetic. Doesn\u2019t that make life easy? 5.6.1 Newtonian Gravity Gravity is the attractive force between two objects due to the mass of the objects. When you throw a ball in the air, its mass and the earth\u2019s mass attract each other, which leads to a force between them. The ball falls back towards the earth, and the earth accelerates towards the ball. The movement of the earth toward the ball is, however, so small that you couldn\u2019t possibly measure it. 5.6.2 Electromagnetic Forces Almost all of the forces that we experience in everyday life are electromagnetic in origin. They have this unusual name because long ago people thought that electric forces and magnetic forces were di\ufb01erent things. After much work and experimentation, it has been realised that they are actually di\ufb01erent manifestations of the same underlying theory. The Electric Force If we have objects carrying electrical charge, which are not moving, then we are dealing with electrostatic forces (Coulomb\u2019s Law). This force is actually much stronger than gravity. This may seem strange, since gravity is obviously very powerful, and holding a balloon to the wall seems to be the most impressive thing electrostatic forces have done, but think about it: for gravity to be detectable, we need to have a very large mass nearby. But a balloon rubbed in someone\u2019s hair can stick to a wall with a force so strong that it overcomes the force of gravity|with just the charges in the balloon and the wall! Magnetic force The magnetic force is a di\ufb01erent manifestation", " of the electromagnetic force. It stems from the interaction between moving charges as opposed to the \ufb02xed charges involved in Coulomb\u2019s Law. Examples of the magnetic force in action include magnets, compasses, car engines, computer data storage and your hair standing on end. Magnets are also used in the wrecking industry to pick up cars and move them around sites. 101 Friction We all know that Newton\u2019s First Law states that an object moving without a force acting on it will keep moving. Then why does a box sliding on a table stop? The answer is friction. Friction arises from the interaction between the molecules on the bottom of a box with the molecules on a table. This interaction is electromagnetic in origin, hence friction is just another view of the electromagnetic force. The great part about school physics is that most of the time we are told to neglect friction but it is good to be aware that there is friction in the real world. Friction is also useful sometimes. If there was no friction and you tried to prop a ladder up against a wall, it would simply slide to the ground. Rock climbers use friction to maintain their grip on cli\ufb01s. Drag Force This is the force an object experiences while travelling through a medium. When something travels through the air it needs to displace air as it travels and because of this the air exerts a force on the object. This becomes an important force when you move fast and a lot of thought is taken to try and reduce the amount of drag force a sports car experiences. The drag force is very useful for parachutists. They jump from high altitudes and if there was no drag force, then they would continue accelerating all the way to the ground. Parachutes are wide because the more surface area you show, the greater the drag force and hence the slower you hit the ground. 5.7 Summary of Important Quantities, Equations and Con- cepts Equilibrium Objects at rest or moving with constant velocity are in equilibrium and have a zero resultant force. Equilibrant The equilibrant of any number of forces is the single force required to produce equilibrium. Triangle Law for Forces in Equilibrium Three forces in equilibrium can be represented in magnitude and direction by the three sides of a triangle taken in order. Newton\u2019s First Law Every object will remain at rest or in uniform motion in a straight line unless it is made to change its state by the action of an external force. Newton\u2019s Second Law The resultant force acting on a body results in an acceleration which is in the same direction as the resultant force and is directly proportional to the magnitude of this force and inversely proportional to the mass of the object. Quantity Mass Acceleration Force Symbol m \u00a1!a \u00a1!F S.I. Units kg m:s\u00a12 kg:m:s\u00a12 or N Direction | X X Table 5.1: Summary of the symbols and units of the quantities used in Force 102 Newton\u2019s Third Law For every force or action there is an equal but opposite force or reaction. 103 Chapter 6 Rectilinear Motion 6.1 What is rectilinear motion? Rectilinear motion means motion along a straight line. This is a useful topic to study for learning how to describe the movement of cars along a straight road or of trains along straight railway In this section you have only 2 directions to worry about: (1) along the direction of tracks. motion, and (2) opposite to the direction of motion. To illustrate this imagine a train heading east. Train W P N S E If it is accelerating away from the station platform (P), the direction of acceleration is the same as the direction of the train\u2019s velocity - east. If it is braking the direction of acceleration is opposite to the direction of its motion, i.e. west. 6.2 Speed and Velocity Let\u2019s take a moment to review our de\ufb02nitions of velocity and speed by looking at the worked example below: 104 Worked Example 24 Speed and Velocity C B m 0 4 A 30m Question: A cyclist moves from A through B to C in 10 seconds. Calculate both his speed and his velocity. Answer: Step 1 : Decide what information has been supplied The question explicitly gives the distance between A and B the distance between B and C the total time for the cyclist to go from A through B to C \u2020 \u2020 \u2020 all in the correct units! Step 2 : Determine the cyclist\u2019s speed His speed - a scalar - will be v = = s t 30m + 40m 10s = 7 m s Step 3 : First determine the cyclist\u2019s resultant displacement Since velocity is a vector we will \ufb02rst need to \ufb02nd the resultant displacement of the cyclist. His velocity will be \u00a1!v = \u00a1!s t The total displacement is the vector from A to C,", " and this is just the resultant of the two displacement vectors, ie. \u00a1!s = \u00a1!AC = \u00a1\u00a1!AB + \u00a1\u00a1!BC 105 Remember to check the units! Using the rule of Pythagoras: \u00a1!s = (30m)2 + (40m)2 = 50m in the direction f rom A to C q Step 4 : Now we can determine the average velocity from the displacement and the time ) \u00a1!v = 50m 10s m s = 5 in the direction f rom A to C For this cyclist, his velocity is not the same as his speed because there has been a change in the direction of his motion. If the cyclist traveled directly from A to C without passing through B his speed would be v = 50m 10s m s = 5 and his velocity would be \u00a1!v = 50m 10s m s = 5 in the direction f rom A to C In this case where the cyclist is not undergoing any change of direction (ie. he is traveling in a straight line) the magnitudes of the speed and the velocity are the same. This is the de\ufb02ning principle of rectilinear motion. Important: For motion along a straight line the magnitudes of speed and velocity are the same, and the magnitudes of the distance and displacement are the same. 6.3 Graphs In physics we often use graphs as important tools for picturing certain concepts. Below are some graphs that help us picture the concepts of displacement, velocity and acceleration. 6.3.1 Displacement-Time Graphs Below is a graph showing the displacement of the cyclist from A to C: 106 50 ) \u00a2\u00a1!s \u00a2t time (s) 10 This graphs shows us how, in 10 seconds time, the cyclist has moved from A to C. We know the gradient (slope) of a graph is de\ufb02ned as the change in y divided by the change in x, i.e \u00a2y \u00a2x. In this graph the gradient of the graph is just \u00a2\u00a1!s - and this is just the expression for velocity. \u00a2t Important: The slope of a displacement-time graph gives the velocity. The slope is the same all the way from A to C, so the cyclist\u2019s velocity is constant over the entire displacement he travels. In \ufb02gure 6.1 are examples of the displacement-time graphs you will encounter) time ) time ) time Figure 6.1: Some common displacement-time graphs: a) shows the graph for an object stationary over a period of time. The gradient is zero, so the object has zero velocity. b) shows the graph for an object moving at a constant velocity. You can see that the displacement is increasing as time goes on. The gradient, however, stays constant (remember: its the slope of a straight line), so the velocity is constant. Here the gradient is positive, so the object is moving in the direction we have de\ufb02ned as positive. c) shows the graph for an object moving at a constant acceleration. You can see that both the displacement and the velocity (gradient of the graph) increase with time. The gradient is increasing with time, thus the velocity is increasing with time and the object is accelerating. 107 6.3.2 Velocity-Time Graphs Look at the velocity-time graph below \u00a2\u00a1!v \u00a2t time This is the velocity-time graph of a cyclist traveling from A to B at a constant acceleration, i.e. with steadily increasing velocity. The gradient of this graph is just \u00a2\u00a1!v - and this is just the expression for acceleration. Because the slope is the same at all points on this graph, the acceleration of the cyclist is constant. \u00a2t Important: The slope of a velocity-time graph gives the acceleration. Not only can we get the acceleration of an object from its velocity-time graph, but we can also get some idea of the displacement traveled. Look at the graph below: 10 ) time (s) 5 This graph shows an object moving at a constant velocity of 10m=s for a duration of 5s. The area between the graph and the time axis (the (NOTE TO SELF: SHADED) area) of the above plot will give us the displacement of the object during this time. In this case we just need to calculate the area of a rectangle with width 5s and height 10m/s area of rectangle = height width \u00a3 = \u00a1!v = 10 t \u00a3 m s \u00a3 5s = 50m = \u00a1!s = displacement 108 So, here we\u2019ve shown that an object traveling at 10m/s for 5s has undergone a displacement of 50m. Important: The area between a velocity-time graph and the \u2018time\u2019 axis gives the displacement of the object. Here are a", " couple more velocity-time graphs to get used to) time y t i c o l e v b) time Figure 6.2: Some common velocity-time graphs: In \ufb02gure 6.2 are examples of the displacement-time graphs you may encounter. a) shows the graph for an object moving at a constant velocity over a period of time. The gradient is zero, so the object is not accelerating. b) shows the graph for an object which is decelerating. You can see that the velocity is decreasing with time. The gradient, however, stays constant (remember: its the slope of a straight line), so the acceleration is constant. Here the gradient is negative, so the object is accelerating in the opposite direction to its motion, hence it is decelerating. 6.3.3 Acceleration-Time Graphs In this chapter on rectilinear motion we will only deal with objects moving at a constant acceleration, thus all acceleration-time graphs will look like these two) time ) time Here is a description of the graphs below: a) shows the graph for an object which is either stationary or traveling at a constant velocity. Either way, the acceleration is zero over time. 109 b) shows the graph for an object moving at a constant acceleration. In this case the acceler- ation is positive - remember that it can also be negative. We can obtain the velocity of a particle at some given time from an acceleration time graph it is just given by the area between the graph and the time-axis. In the graph below, showing an object at a constant positive acceleration, the increase in velocity of the object after 2 seconds corresponds to the (NOTE TO SELF: shaded) portion time (s) t area of rectangle = \u00a1!a \u00a3 m = 5 s2 \u00a3 m s = 10 2s = \u00a1!v Its useful to remember the set of graphs below when working on problems. Figure 6.3 shows how displacement, velocity and time relate to each other. Given a displacement-time graph like the one on the left, we can plot the corresponding velocity-time graph by remembering that the slope of a displacement-time graph gives the velocity. Similarly, we can plot an acceleration-time graph from the gradient of the velocity-time graph. 110 Displacement Velocity Acceleration Figure 6.3: A Relationship Between Displacement, Velocity and Acceleration 6.3.4 Worked Examples Worked Example 25 Relating displacement-, velocity-, and acceleration-time graphs Question: Given the displacement-time graph below, draw the corresponding velocitytime and acceleration-time graphs, and then describe the motion of the object time (s) 4 6 Answer: Step 1 : Decide what information is supplied The question explicitly gives a displacement-time graph. Step 2 : Decide what is asked? 3 things are required: 1. Draw a velocity-time graph 2. Draw an acceleration-time graph 111 3. Describe the behaviour of the object Step 3 : Velocity-time graph - 0-2 seconds For the \ufb02rst 2 seconds we can see that the displacement remains constant - so the object is not moving, thus it has zero velocity during this time. We can reach this conclusion by another path too: remember that the gradient of a displacement-time graph is the velocity. For the \ufb02rst 2 seconds we can see that the displacement-time graph is a horizontal line, ie. it has a gradient of zero. Thus the velocity during this time is zero and the object is stationary. Step 4 : Velocity-time graph - 2-4 seconds For the next 2 seconds, displacement is increasing with time so the object is moving. Looking at the gradient of the displacement graph we can see that it is not constant. In fact, the slope is getting steeper (the gradient is increasing) as time goes on. Thus, remembering that the gradient of a displacement-time graph is the velocity, the velocity must be increasing with time during this phase. Step 5 : Velocity-time graph - 4-6 seconds For the \ufb02nal 2 seconds we see that displacement is still increasing with time, but this time the gradient is constant, so we know that the object is now travelling at a constant velocity, thus the velocity-time graph will be a horizontal line during this stage. So our velocity-time graph looks like this one below. Because we haven\u2019t been given any values on the vertical axis of the displacement-time graph, we cannot \ufb02gure out what the exact gradients are and hence what the values of the velocity are. In this type of question it is just important to show whether velocities are positive or negative, increasing, decreasing or constant time (s) 4 6 Once we have the velocity-time graph its much easier to get the acceleration-time graph as we know that the gradient of a velocity-time graph is the just the acceleration. Step 6 : Acceleration-time graph - 0-", "2 seconds For the \ufb02rst 2 seconds the velocity-time graph is horizontal at zero, thus it has a gradient of zero and there is no acceleration during this time. (This makes sense because we know from the displacement time graph that the object is stationary during this time, so it can\u2019t be accelerating). Step 7 : Acceleration-time graph - 2-4 seconds For the next 2 seconds the velocity-time graph has a positive gradient. This gradient its constant) throughout these 2 seconds so there must be a is not changing (i.e. 112 constant positive acceleration. Step 8 : Accleration-time graph - 4-6 seconds For the \ufb02nal 2 seconds the object is traveling with a constant velocity. During this time the gradient of the velocity-time graph is once again zero, and thus the object is not accelerating. The acceleration-time graph looks like this time (s) 4 6 Step 9 : A description of the object\u2019s motion A brief description of the motion of the object could read something like this: At t = 0s and object is stationary at some position and remains stationary until t = 2s when it begins accelerating. It accelerates in a positive direction for 2 seconds until t = 4s and then travels at a constant velocity for a further 2 seconds. Worked Example 26 Calculating distance from a velocity-time graph Question: The velocity-time graph of a car is plotted below. Calculate the displacement of the car has after 15 seconds. 113 2 5 time (s) 12 15 Answer: Step 1 : Decide how to tackle the problem We are asked to calculate the displacement of the car. All we need to remember here is that the area between the velocity-time graph and the time axis gives us the displacement. Step 2 : Determine the area under the velocity-time graph For t = 0s to t = 5s this is the triangle on the left: Area 4 h \u00a3 b = 1 2 1 2 = 10m 5s = 4m=s \u00a3 For t = 5s to t = 12s the displacement is equal to the area of the rectangle h Area\u2044 = w = 7s \u00a3 = 28m \u00a3 4m=s For t = 12s to t = 14s the displacement is equal to the area of the triangle above the time axis on the right Area 4 = = 1 2 1 2 h b \u00a3 2s \u00a3 4m=s = 4m For t = 14s to t = 15s the displacement is equal to the area of the triangle below the 114 time axis Area 4 = = 1 2 1 2 h b \u00a3 1s \u00a3 2m=s = 1m Step 3 : Determine the total displacement of the car Now the total displacement of the car is just the sum of all of these areas. HOWEVER, because in the last second (from t = 14s to t = 15s) the velocity of the car is negative, it means that the car was going in the opposite direction, i.e. back where it came from! So, to get the total displacement, we have to add the \ufb02rst 3 areas (those with positive displacements) and subtract the last one (because it signi\ufb02es a displacement in the opposite direction). \u00a1!s = 10 + 28 + 4 1 \u00a1 = 41m in the positive direction Worked Example 27 Velocity from a displacement-time graph Question: Given the diplacement-time graph below, 1. what is the velocity of the object during the \ufb02rst 4 seconds? 2. what is the velocity of the object from t = 4s to t = 7s? 2 ) m ( s 4 t (s) 7 Answer: Step 1 : The velocity during the \ufb02rst 4 seconds The velocity is given by the slope of a displacement-time graph. During the \ufb02rst 4 seconds, this is \u00a1!v = \u00a2s \u00a2t 2m 4s = 0:5m=s = 115 Step 2 : The velocity during the last 3 seconds For the last 3 seconds we can see that the displacement stays constant, and that the gradient is zero. Thus \u00a1!v = 0m=s Worked Example 28 From an acceleration-time graph to a velocity-time graph Question: Given the acceleration-time graph below, assume that the object starts from rest and draw its velocity-time graph2 2 4 t (s) 6 Answer: Once again attempt to draw the graph in time sections, i.e. \ufb02rst draw the velocity-time graph for the \ufb02rst 2 seconds, then for the next 2 seconds and so ons) 6 116 6.4 Equations of Motion This section is about solving problems relating to uniformly accelerated motion. We\u2019ll \ufb02rst introduce the variables and the equations, then we\u2019ll show you how to derive", " them, and after that we\u2019ll do a couple of examples. u = starting velocity (m/s) at t = 0 v = \ufb02nal velocity (m/s) at time t s = displacement (m) t = time (s) a = acceleration (m/s2) v = u + at s = (u + v) 2 t s = ut + at2 1 2 v2 = u2 + 2as (6.1) (6.2) (6.3) (6.4) Make sure you can rhyme these o\ufb01, they are very important! There are so many di\ufb01erent types of questions for these equations. Basically when you are answering a question like this: 1. Find out what values you have and write them down. 2. Figure out which equation you need. 3. Write it down!!! 4. Fill in all the values you have and get the answer. Interesting Fact: Galileo Galilei of Pisa, Italy, was the \ufb02rst to determined the correct mathematical law for acceleration: the total distance covered, starting from rest, is proportional to the square of the time. He also concluded that objects retain their velocity unless a force { often friction { acts upon them, refuting the accepted Aristotelian hypothesis that objects \"naturally\" slow down and stop unless a force acts upon them. This principle was incorporated into Newton\u2019s laws of motion (1st law). Equation 6.1 By the de\ufb02nition of acceleration a = \u00a2v t 117 where \u00a2v is the change in velocity, i.e. \u00a2v = v a = \u00a1 v u. Thus we have u \u00a1 t v = u + at Equation 6.2 In the previous section we saw that displacement can be calculated from the area between a velocity-time graph and the time-axis. For uniformly accelerated motion the most complicated velocity-time graph we can have is a straight line. Look at the graph below - it represents an object with a starting velocity of u, accelerating to a \ufb02nal velocity v over a total time t time (s) To calculate the \ufb02nal displacement we must calculate the area under the graph - this is just the area of the rectangle added to the area of the triangle. (NOTE TO SELF: SHADING) Area \u00a3 \u00a3 vt \u00a1 (v u) \u00a1 1 2 ut h Area\u2044 = w = t \u00a3 = ut \u00a3 u Displacement = Area\u2044 + Area 4 1 2 \u00a1 ut s = ut + vt 1 2 (u + v) 2 t = Equation 6.3 This equation is simply derived by eliminating the \ufb02nal velocity v in equation 6.2. Remembering from equation 6.1 that v = u + at 118 then equation 6.2 becomes s = = t u + u + at 2 2ut + at2 2 1 2 at2 = ut + Equation 6.4 This equation is just derived by eliminating the time variable in the above equation. From Equation 6.1 we know Substituting this into Equation 6.3 gives t = v u \u00a1 a s = u( u ) + v \u00a1 a u )2 \u00a1 a v2 v a( 1 2 1 2 v2 2a \u00a1 a( + u2 uv a \u00a1 a u2 uv a \u00a1 a 2u2 + v2 + u2 + = = 2as = \u00a1 v2 = u2 + 2as \u00a1 2uv + u2 a2 ) uv a + u2 2a (6.5) This gives us the \ufb02nal velocity in terms of the initial velocity, acceleration and displacement and is independent of the time variable. Worked Example 29 Question: A racing car has an initial velocity of 100m=s and it covers a displacement of 725m in 10s. Find its acceleration. Answer: Step 1 : Decide what information has been supplied We are given the quantities u, s and t - all in the correct units. We need to \ufb02nd a. Step 2 : Find an equation of motion relating the given information to the acceleration We can use equation 6.3 s = ut + at2 1 2 Step 3 : Rearrange the equation if needed We want to determine the acceleration so we rearrane equation 6.3 to put acceleration on the left of the equals sign: 2(s a = ut) \u00a1 t2 119 Step 4 : Do the calculation Substituting in the values of the known quantities this becomes 2(725m a = 100 m 10s) \u00a1 102s2 s \u00a2 = = 2( 275m) \u00a1 100s2 m 5:5 s2 \u00a1 Step 5 : Quote the \ufb02nal answer The racing car is", " accelerating at -5.5 m s2, or we could say it is decelerating at 5.5 m s2. Worked Example 30 Question: An object starts from rest, moves in a straight line with a constant acceleration and covers a distance of 64m in 4s. Calculate its acceleration its \ufb02nal velocity at what time the object had covered half the total distance what distance the object had covered in half the total time. \u2020 \u2020 \u2020 \u2020 Answer: Step 1 : Decide what information is supplied We are given the quantities u, s and t in the correct units. Step 2 : Acceleration: Find an equation to calculate the acceleration and rearrange To calculate the acceleration we can use equation 6.3. Step 3 : Rearrange to make a subject of the formula 2(s a = ut) \u00a1 t2 Step 4 : Do the calculation Substituting in the values of the known quantities this becomes 2(64m 0 m s 4s \u00a1 42s2 a = = 128m 16s2 m s2 = 8 Step 5 : Final velocity: Find an equation to calculate the \ufb02nal velocity We can use equation 6.1 - remember we now also know the acceleration of the object. v = u + at 120 Step 6 : Do the calculation v = 0 + (8 m s2 )(4s) m s m s = 32 Step 7 : Time at half the distance: Find an equation to relate the unknown and known quantities Here we have the quantities s, u and a so we do this in 2 parts, \ufb02rst using equation 6.4 to calculate the velocity at half the distance, i.e. 32m: v2 = u2 + 2as = (0m)2 + 2(8m=s2)(32m) = 512m2=s2 v = 22:6m=s Now we can use equation 6.2 to calculate the time at this distance: t = 2s u + v = (2)(32m) 0m=s + 22:6m=s = 2:8s Step 8 : Distance at half the time: Find an equation to relate the distance and time To calculate the distance the object has covered in half the time. Half the time is 2s. Thus we have u, a and t - all in the correct units. We can use equation 6.3 to get the distance: s = ut + at2 1 2 = (0m=s)(2s) + = 16m 1 2 (8 m s2 )(2s)2 Worked Example 31 Question: A ball is thrown vertically upwards with a velocity of 10m=s from the balcony of a tall building. The balcony is 15m above the ground and gravitational accleration is 10m=s2. Find a) the time required for the ball to hit the ground, and b) the velocity with which it hits the ground. Answer: Step 1 : Draw a rough sketch of the problem In most cases helps to make the problem easier to understand if we draw ourselves a picture like the one below: 121 balcony s1 v1 u1 u2 a1; a2 s2 v2 ground where the subscript 1 refers to the upward part of the ball\u2019s motion and the subscript 2 refers to the downward part of the ball\u2019s motion. Step 2 : Decide how to tackle the problem First the ball goes upwards with gravitational acceleration slowing it until it reaches its highest point - here its speed is 0m=s - then it begins descending with gravitational acceleration causing it to increase its speed on the way down. We can separate the motion into 2 stages: Stage 1 - the upward motion of the ball Stage 2 - the downward motion of the ball. We\u2019ll choose the upward direction as positive - this means that gravitation acceleraton is negative. We have used this and we\u2019ll begin by solving for all the variables of Stage 1. Step 3 : For Stage 1, decide what information is given We have these quantities: u1 = 10m=s v1 = 0m=s a1 = \u00a1 t1 =? s1 =? 10m=s2 Step 4 : Find the time for stage 1 Using equation 6.1 to \ufb02nd t1: v1 = u1 + a1t1 v1 u1 t1 = \u00a1 a1 0m=s = \u00a1 = 1s 10m=s \u00a1 10m=s2 Step 5 : Find the distance travelled during stage 1 122 We can \ufb02nd s1 by using equation 6.4 s1 = 1 = u2 v2 1 + 2a1s1 u2 v2 1 \u00a1 1 2a (0m=s)2 2( = = 5m \u00a1 (10m=s)2 \u00a1 10m=s2) Step 6 :", " For Stage 2, decide what information is supplied For Stage 2 we have the following quantities: u2 = 0m=s v2 =? a2 = \u00a1 t2 =? s2 = 15m \u00a1 10m=s2 5m = 20m \u00a1 Step 7 : Determine the velocity at the end of stage 2 We can determine the \ufb02nal velocity v2 using equation 6.4: 2 = u2 v2 2 + 2a2s2 = (0m=s)2 + 2( = 400(m=s)2 \u00a1 10m=s2)( 20m) \u00a1 v2 = 20m=s downwards Step 8 : Determind the time for stage 2 Now we can determine the time for Stage 2, t2, from equation 6.1: v2 = u2 + a2t2 v2 u2 t2 = \u00a1 a2 20m=s 0m=s = \u00a1 = 2s \u00a1 10m=s2 \u00a1 Step 9 : Quote the answers to the problem Finally, a) the time required for the stone to hit the ground is t = t1 + t2 = 1s + 2s = 3s b) the velocity with which it hits the ground is just v2 = 20m=s \u00a1 These questions do not have the working out in them, but they are all done in the manner described on the previous page. Question: A car starts o\ufb01 at 10 m/s and accelerates at 1 m/s2 for 10 seconds. What is it\u2019s \ufb02nal velocity? Answer: 20 m/s 123 Question: A car starts from rest, and accelerates at 1 m/s2 for 10 seconds. How far does it move? Answer: 50 m Question: A car is going 30 m/s and stops in 2 seconds. What is it\u2019s stopping distance for this speed? Answer: 30 m Question: A car going at 20 m/s stops in a distance of 20 m/s. 1. What is it\u2019s deceleration? 2. If the car is 1 Tonne (1000 Kg, or 1 Mg) how much force do the brakes exert? 124 6.5 Important Equations and Quantities Quantity Displacement Velocity Distance Speed Acceleration Units Symbol Unit \u00a1!s \u00a1!u,\u00a1!v s v \u00a1!a - Base S.I. Units m + direction m:s\u00a11 + direction m m:s\u00a11 m:s\u00a11 + direction Table 6.1: Units used in Rectilinear Motion 125 Chapter 7 Momentum 7.1 What is Momentum? Momentum is a physical quantity which is closely related to forces. We will learn about this connection a little later. Remarkably momentum is a conserved quantity. This makes momentum extremely useful in solving a great variety of real-world problems. Firstly we must consider the de\ufb02nition of momentum. De\ufb02nition: The momentum of an object is de\ufb02ned as its mass multiplied by its velocity. Mathematically, \u00a1!p = m\u00a1!v : momentum (kg:m:s\u00a11 + direction) \u00a1!p m : mass (kg) \u00a1!v : velocity (m:s\u00a11 + direction) Thus, momentum is a property of a moving object and is determined by its velocity and mass. A large truck travelling slowly can have the same momentum as a much smaller car travelling relatively fast. Note the arrows in the equation de\ufb02ning momentum{ momentum is a vector with the same direction as the velocity of the object. Since the direction of an object\u2019s momentum is given by the direction of its motion, one can calculate an object\u2019s momentum in two steps: Momentum is a vector with the same direction as the velocity. calculate the magnitude of the object\u2019s momentum using, \u2020 p = mv : magnitude of momentum (kg:m:s\u00a11) p m : mass (kg) v : magnitude of velocity (m:s\u00a11) 126 include in the \ufb02nal answer the direction of the object\u2019s motion \u2020 Worked Example 32 Calculating Momentum 1 Question: A ball of mass 3kg moves at 2m:s\u00a11 to the right. Calculate the ball\u2019s momentum. Answer: Step 1 : Decide what information has been supplied The question explicitly gives the ball\u2019s mass, and the ball\u2019s velocity \u2020 \u2020 in the correct units! Step 2 : Decide how to tackle the problem What is being asked? We are asked to calculate the ball\u2019s momentum. From the de\ufb02nition of momentum, \u00a1!p = m\u00a1!v ; we", " see that we need the mass and velocity of the ball, which we are given. Step 3 : Do the calculation We calculate the magnitude of the ball\u2019s momentum, p = mv = (3kg)(2m:s\u00a11) = 6 kg:m:s\u00a11: Step 4 : Quote the \ufb02nal answer We quote the answer with the direction of the ball\u2019s motion included, \u00a1!p = 6 kg:m:s\u00a11 to the right Worked Example 33 Calculating Momentum 2 Question: A ball of mass 500g is thrown at 2m:s\u00a11. Calculate the ball\u2019s momentum. Answer: Step 1 : Decide what information is supplied The question explicitly gives the ball\u2019s mass, and the magnitude of the ball\u2019s velocity \u2020 \u2020 but with the ball\u2019s mass in the incorrect units! 127 Remember to check the units! Remember to check the units! Step 2 : Decide how to tackle the problem What is being asked? We are asked to calculate the momentum which is de\ufb02ned as \u00a1!p = m\u00a1!v : Thus, we need the mass and velocity of the ball but we have only its mass and the magnitude of its velocity. In order to determine the velocity of the ball we need the direction of the ball\u2019s motion. If the problem does not give an explicit direction we are forced to be general. In a case like this we could say that the direction of the velocity is in the direction of motion of the ball. This might sound silly but the lack of information in the question has forced us to be, and we are certainly not wrong! The ball\u2019s velocity is then 2m:s\u00a11 in the direction of motion. Step 3 : Convert the mass to the correct units 1000g = 1kg 1 = 1kg 1000g 500g \u00a3 1 = 500g \u00a3 = 0:500kg 1kg 1000g Step 4 : Do the calculation Now, let us \ufb02nd the magnitude of the ball\u2019s momentum, p = mv = (0:500kg)(2m:s\u00a11) = 1 kg:m:s\u00a11 Step 5 : Quote the \ufb02nal answer Remember to include the direction of the momentum: \u00a1!p = 1 kg:m:s\u00a11 in the direction of motion of the ball Worked Example 34 Calculating the Momentum of the Moon Question: The moon is 384 400km away from the earth and orbits the earth in 1022kg1 what is the magnitude of its 27.3 days. If the moon has a mass of 7:35 momentum if we assume a circular orbit? Answer: Step 1 : Decide what information has been supplied The question explicitly gives \u00a3 1This is 1 81 of the mass of the earth 128 with mass in the correct units but all other quantities in the incorrect units. The units we require are Remember to check the units! the moon\u2019s mass, the distance to the moon, and the time for one orbit of the moon seconds (s) for time, and metres (m) for distance \u2020 \u2020 \u2020 \u2020 \u2020 Step 2 : Decide how to tackle the problem What is being asked? We are asked to calculate only the magnitude of the moon\u2019s momentum (i.e. we do not need to specify a direction). In order to do this we require the moon\u2019s mass and the magnitude of its velocity, since Step 3 : Find the speed or magnitude of the moon\u2019s velocity Speed is de\ufb02ned as, p = mv: speed = Distance time We are given the time the moon takes for one orbit but not how far it travels in that time. However, we can work this out from the distance to the moon and the fact that the moon\u2019s orbit is circular. Firstly let us convert the distance to the moon to the correct units, 1km = 1000m 1000m 1km 1 = 384 400km \u00a3 1000m 1km 1 = 384 400km \u00a3 = 384 400 000m 108 m = 3:844 \u00a3 Using the equation for the circumference, C, of a circle in terms of its radius, we can determine the distance travelled by the moon in one orbit: C = 2\u2026r = 2\u2026(3:844 = 2:42 \u00a3 109 m: 108 m) \u00a3 Next we must convert the orbit time, T, into the correct units. Using the fact that a day contains 24 hours, an hour consists of 60 minutes, and a minute is 60 seconds long, 1day = (24)(60)(60)seconds 1 = (24)(60)(60)s 1day 27:3days \u00a3 1 = 27:3days (24)(60)(60)s 1day = 2:36 106s \u00a3 129 Therefore, T =", " 2:36 106s: \u00a3 Combining the distance travelled by the moon in an orbit and the time taken by the moon to complete one orbit, we can determine the magnitude of the moon\u2019s velocity or speed, v = Distance time = C T = 1:02 103 m:s\u00a11: \u00a3 Step 4 : Finally calculate the momentum and quote the answer The magnitude of the moon\u2019s momentum is: p = mv = (7:35 = 7:50 \u00a3 \u00a3 1022kg)(1:02 \u00a3 1025 kg:m:s\u00a11: 103 m:s\u00a11) 7.2 The Momentum of a System In Chapter?? the concept of a system was introduced. The bodies that make up a system can have di\ufb01erent masses and can be moving with di\ufb01erent velocities. In other words they can have di\ufb01erent momenta. De\ufb02nition: The total momentum of a system is the sum of the momenta of each of the objects in the system. Since momentum is a vector, the techniques of vector addition discussed in Chapter?? must be used to calculate the total momentum of a system. Let us consider an example. Worked Example 35 Calculating the Total Momentum of a System Question: Two billiard balls roll towards each other. They each have a mass of 0:3kg. Ball 1 is moving at v1 = 1 m:s\u00a11 to the right, while ball 2 is moving at v2 = 0:8 m:s\u00a11 to the left. Calculate the total momentum of the system. Answer: Step 1 : Decide what information is supplied The question explicitly gives the mass of each ball, the velocity of ball 1, \u00a1!v1, and \u2020 \u2020 130 the velocity of ball 2, \u00a1!v2, \u2020 all in the correct units! Step 2 : Decide how to tackle the problem What is being asked? We are asked to calculate the total momentum of the system. In this example our system consists of two balls. To \ufb02nd the total momentum we must sum the momenta of the balls, Remember to check the units! \u00a1!p total = \u00a1!p1 + \u00a1!p2 Since ball 1 is moving to the right, its momentum is in this direction, while the second ball\u2019s momentum is directed towards the left. System m1 \u00a1!p 1 m2 \u00a1!p 2 Thus, we are required to \ufb02nd the sum of two vectors acting along the same straight line. The algebraic method of vector addition introduced in Chapter?? can thus be used. Step 3 : Choose a positive direction Let us choose right as the positive direction, then obviously left is negative. Step 4 : Calculate the momentum The total momentum of the system is then the sum of the two momenta taking the directions of the velocities into account. Ball 1 is travelling at 1 m:s\u00a11 to the right or +1 m:s\u00a11. Ball 2 is travelling at 0:8 m:s\u00a11 to the lef t or 0:8 m:s\u00a11. Thus, \u00a1 Right is the positive direction \u00a1!p total = m1\u00a1!v1 + m2\u00a1!v2 = (0:3kg)(+1 m:s\u00a11) + (0:3kg)( = (+0:3 kg:m:s\u00a11) + ( = +0:06 kg:m:s\u00a11 = 0:06 kg:m:s\u00a11 to the right \u00a1 \u00a1 0:24 kg:m:s\u00a11) 0:8 m:s\u00a11) In the last step the direction was added in words. Since the result in the second last line is positive, the total momentum of the system is in the positive direction (i.e. to the right). 7.3 Change in Momentum If either an object\u2019s mass or velocity changes then its momentum too will change. If an object has an initial velocity \u00a1!u and a \ufb02nal velocity \u00a1!v, then its change in momentum, \u00a2\u00a1!p, is 131 \u00a2\u00a1!p = \u00a1!p f inal \u00a1 \u00a1!p initial = m\u00a1!v \u00a1 m\u00a1!u Worked Example 36 Change in Momemtum Question: A rubber ball of mass 0:8kg is dropped and strikes the oor at a velocity of 6 m:s\u00a11. It bounces back with an initial velocity of 4 m:s\u00a11. Calculate the change in momentum of the rubber ball caused by the oor. Answer: Step 1 : Decide what information has been supplied The question explicitly gives the ball\ufffd", "\ufffds mass, the ball\u2019s initial velocity, and the ball\u2019s \ufb02nal velocity \u2020 \u2020 \u2020 all in the correct units. Do not be confused by the question referring to the ball bouncing back with an \\initial velocity of 4 m:s\u00a11\". The word \\initial\" is included here since the ball will obviously slow down with time and 4 m:s\u00a11 is the speed immediately after bouncing from the oor. Step 2 : Decide how to tackle the problem What is being asked? We are asked to calculate the change in momentum of the ball, \u00a2\u00a1!p = m\u00a1!v m\u00a1!u : \u00a1 We have everything we need to \ufb02nd \u00a2\u00a1!p. Since the initial momentum is directed downwards and the \ufb02nal momentum is in the upward direction, we can use the algebraic method of subtraction discussed in the vectors chapter. Step 3 : Choose a positive direction Let us choose down as the positive direction. Then substituting, Down is the positive direction Step 4 : Do the calculation and quote the answer Remember to check the units! \u00a2\u00a1!p = m\u00a1!v m\u00a1!u \u00a1 = (0:8kg)( = (0:8kg)( = = 8 kg:m:s\u00a11 up 8 kg:m:s\u00a11 \u00a1 \u00a1 \u00a1 4 m:s\u00a11) \u00a1 10 m:s\u00a11) (0:8kg)(+6 m:s\u00a11) where we remembered in the last step to include the direction of the change in momentum in words. 132 \u00a1!u 1 m1 \u00a1!u 2 m2 Figure 7.1: Before the collision. 7.4 What properties does momentum have? You may at this stage be wondering why there is a need for introducing momentum. Remarkably momentum is a conserved quantity. Within an isolated system the total momentum is constant. No matter what happens to the individual bodies within an isolated system, the total momentum of the system never changes! Since momentum is a vector, its conservation implies that both its magnitude and its direction remains the same. This Principle of Conservation of Linear Momentum is one of the most fundamental principles of physics and it alone justi\ufb02es the de\ufb02nition of momentum. Since momentum is related to the motion of objects, we can use its conservation to make predictions about what happens in collisions and explosions. If we bang two objects together, by conservation of momentum, the total momentum of the objects before the collision is equal to their total momentum after the collision. Momentum is conserved in isolated systems! Principle of Conservation of Linear Momentum: The total linear momentum of an isolated system is constant. or In an isolated system the total momentum before a collision (or explosion) is equal to the total momentum after the collision (or explosion). Let us consider a simple collision of two pool or billiard balls. Consider the \ufb02rst ball (mass m1) to have an initial velocity (\u00a1!u1). The second ball (mass m2) moves towards the \ufb02rst ball with an initial velocity \u00a1!u2. This situation is shown in Figure 7.1. If we add the momenta of each ball we get a total momentum for the system. This total momentum is then \u00a1!p total bef ore = m1\u00a1!u1 + m2\u00a1!u2; After the two balls collide and move away they each have a di\ufb01erent momentum. If we call the \ufb02nal velocity of ball 1 \u00a1!v1 and the \ufb02nal velocity of ball 2 \u00a1!v2 (see Figure 7.2), then the total momentum of the system after the collision is \u00a1!p total af ter = m1\u00a1!v1 + m2\u00a1!v2; This system of two balls is isolated since there are no external forces acting on the balls. Therefore, by the principle of conservation of linear momentum, the total momentum before the collision is equal to the total momentum after the collision. This gives the equation for the conservation of momentum in a collision of two objects, 133 \u00a1!v 1 m1 m2 \u00a1!v 2 Figure 7.2: After the collision. \u00a1!p total bef ore = \u00a1!p total af ter m1\u00a1!u1 + m2\u00a1!u2 = m1\u00a1!v1 + m2\u00a1!v2 m1 m2 : mass of object 1 (kg) : mass of object 2 (kg) \u00a1!u1 \u00a1!u2 \u00a1!v1 \u00a1!v2 : initial velocity of object 1 (m:s\u00a11 + direction) : initial", " velocity of object 2 (m:s\u00a11 + direction) : \ufb02nal velocity of object 1 (m:s\u00a11 + direction) : \ufb02nal velocity of object 2 (m:s\u00a11 + direction) This equation is always true- momentum is always conserved in collisions. The chapter \u2018Collisions and Explosions\u2019 (Chapter??) deals with applications of momentum conservation. 7.5 Impulse At the beginning of this chapter it was mentioned that momentum is closely related to force. We will now explain the nature of this connection. Consider an object of mass m moving with constant acceleration \u00a1!a. During a time \u00a2t the object\u2019s velocity changes from an initial velocity \u00a1!u to a \ufb02nal velocity \u00a1!v (refer to Figure 7.3). We know from Newton\u2019s First Law that there must be a resultant force \u00a1!F Res acting on the object. Starting from Newton\u2019s Second Law, Momentum is always conserved in collisions! \u00a1!F Res = m\u00a1!a = m( \u00a1!v ) \u00a1 \u00a1!u \u00a2t m\u00a1!u m\u00a1!v = \u00a1 \u00a2t = \u00a1!p f inal \u00a1 \u00a1!p initial \u00a2t since \u00a1!a = \u00a1!v \u00a1 \u00a1!u \u00a2t = \u00a2\u00a1!p \u00a2t This alternative form of Newton\u2019s Second Law is called the Law of Momentum. 134 \u00a1!F Res m \u00a1!u t = 0 t = \u00a2t \u00a1!v \u00a1!F Res m Figure 7.3: An object under the action of a resultant force. Law of Momentum: The applied resultant force acting on an object is equal to the rate of change of the object\u2019s momentum and this force is in the direction of the change in momentum. Mathematically, \u00a1!F Res = \u00a2\u00a1!p \u00a2t \u00a1!F Res \u00a2\u00a1!p \u00a2t : resultant force (N + direction) : change in momentum (kg:m:s\u00a11 + direction) : time over which \u00a1!F Res acts (s) Rearranging the Law of Momentum, The product \u00a1!F Res\u00a2t is called impulse, \u00a1!F Res\u00a2t = \u00a2\u00a1!p : Impulse \u00b7 \u00a1!F Res\u00a2t = \u00a2\u00a1!p From this equation we see, that for a given change in momentum, \u00a1!F Res\u00a2t is \ufb02xed. Thus, if FRes is reduced, \u00a2t must be increased (i.e. the resultant force must be applied for longer). Alternatively if \u00a2t is reduced (i.e. the resultant force is applied for a shorter period) then the resultant force must be increased to bring about the same change in momentum. Worked Example 37 Impulse and Change in momentum 135 Question: A 150 N resultant force acts on a 300 kg object. Calculate how long it takes this force to change the object\u2019s velocity from 2 m:s\u00a11 to the right to 6 m:s\u00a11 to the right. Answer: Step 1 : Decide what information is supplied The question explicitly gives the object\u2019s mass, the object\u2019s initial velocity, the object\u2019s \ufb02nal velocity, and the resultant force acting on the object \u2020 \u2020 \u2020 \u2020 all in the correct units! Step 2 : Decide how to tackle the problem What is being asked? We are asked to calculate the time taken \u00a2t to accelerate the object from the given initial velocity to \ufb02nal velocity. From the Law of Momentum, \u00a1!F Res\u00a2t = \u00a2\u00a1!p m\u00a1!u = m\u00a1!v \u00a1 \u00a1 \u00a1!u ): = m(\u00a1!v Thus we have everything we need to \ufb02nd \u00a2t! Step 3 : Choose a positive direction Although not explicitly stated, the resultant force acts to the right. This follows from the fact that the object\u2019s velocity increases in this direction. Let us then choose right as the positive direction. Step 4 : Do the calculation and quote the \ufb02nal answer Right is the positive direction Remember to check the units! \u00a1!F Res\u00a2t = m(\u00a1!v \u00a1 \u00a1!u ) (+150N )\u00a2t = (300kg)((+6 (+150N )\u00a2t = (300kg)(+4 m s m s (300kg)(+4 m s ) +150N ) \u00a2t = (+2 ) \u00a1 m s )) \u00a2t = 8", "s Worked Example 38 Calculating Impulse Question: A cricket ball weighing 156g is moving at 54 km:hr\u00a11 towards a batsman. It is hit by the batsman back towards the bowler at 36 km:hr\u00a11. Calculate i) the 136 Remember to check the units! ball\u2019s impulse, and ii) the average force exerted by the bat if the ball is in contact with the bat for 0:13s. Answer: Step 1 : Decide what information is supplied The question explicitly gives the ball\u2019s mass, the ball\u2019s initial velocity, the ball\u2019s \ufb02nal velocity, and the time of contact between bat and ball \u2020 \u2020 \u2020 \u2020 all except the time in the wrong units! Answer to (i): Step 2 : Decide how to tackle the problem What is being asked? We are asked to calculate the impulse Impulse = \u00a2\u00a1!p = \u00a1!F Res\u00a2t: Since we do not have the force exerted by the bat on the ball (\u00a1!F Res), we have to calculate the impulse from the change in momentum of the ball. Now, since \u00a2\u00a1!p = \u00a1!p f inal \u00a1 \u00a1!p initial = m\u00a1!v m\u00a1!u ; \u00a1 we need the ball\u2019s mass, initial velocity and \ufb02nal velocity, which we are given. Step 3 : Convert to S.I. units Firstly let us change units for the mass 1000g = 1kg 1 = 1kg 1000g 156g \u00a3 1 = 156g \u00a3 = 0:156kg 1kg 1000g Next we change units for the velocity 1km = 1000m 1000m 1km 1 = 3600s = 1hr 1 = 1hr 3600s 54 km hr \u00a3 1 \u00a3 1 = 54 = 15 km hr \u00a3 m s 1000m 1km \u00a3 1hr 3600s 137 36 km hr \u00a3 1 \u00a3 1 = 36 = 10 km hr \u00a3 m s 1000m 1km \u00a3 1hr 3600s Step 4 : Choose your convention Next we must choose a positive direction. Let us choose the direction from the batsman to the bowler as the postive direction. Then the initial velocity of the ball 15 m:s\u00a11, while the \ufb02nal velocity of the ball is \u00a1!v = +10 m:s\u00a11 is \u00a1!u = Step 5 : Calculate the momentum Now we calculate the change in momentum, \u00a1 Direction from batsman to bowler is the positive direction \u00a2\u00a1!p = \u00a1!p f inal \u00a1 \u00a1!p initial = m\u00a1!v m\u00a1!u \u00a1 \u00a1 \u00a1!u ) = m(\u00a1!v = (0:156kg)((+10 m:s\u00a11) = +3:9 kg:m:s\u00a11 = 3:9 kg:m:s\u00a11 in the direction from batsman to bowler 15 m:s\u00a11)) ( \u00a1 \u00a1 where we remembered in the last step to include the direction of the change in momentum in words. Step 6 : Determine the impulse Finally since impulse is just the change in momentum of the ball, Impulse = \u00a2\u00a1!p = 3:9 kg:m:s\u00a11 in the direction from batsman to bowler Answer to (ii): Step 7 : Determine what is being asked What is being asked? We are asked to calculate the average force exerted by the bat on the ball, \u00a1!F Res. Now, Impulse = \u00a1!F Res\u00a2t = \u00a2\u00a1!p : We are given \u00a2t and we have calculated the change in momentum or impulse of the ball in part (i)! Step 8 : Choose a convention Next we choose a positive direction. Let us choose the direction from the batsman to the bowler as the postive direction. Step 9 : Calculate the force Then substituting, Direction from batsman to bowler is the positive direction 138 \u00a1!F Res\u00a2t = Impulse \u00a1!F Res(0:13s) = +3:9 kg:m s +3:9 kg:m s 0:13s \u00a1!F Res = = +30 kg:m s2 = 30N in the direction from batsman to bowler where we remembered in the \ufb02nal step to include the direction of the force in words. 7.6 Summary of Important Quantities, Equations and Con- cepts Quantity Momentum Mass Velocity Change in momentum Force Impulse Units Symbol Unit \u00a1!p m \u00a1!u,\u00a1!v \u00a2\u00a1!p \u00a1!F J N - S.I. Units Direction kg:m:s", "\u00a11 kg m:s\u00a11 kg:m:s\u00a11 kg:m:s\u00a12 kg:m:s\u00a11 X | X X X X Table 7.1: Summary of the symbols and units of the quantities used in Momentum Momentum The momentum of an object is de\ufb02ned as its mass multiplied by its velocity. Momentum of a System The total momentum of a system is the sum of the momenta of each of the objects in the system. Principle of Conservation of Linear Momentum: \u2018The total linear momentum of an isolated system is constant\u2019 or \u2018In an isolated system the total momentum before a collision (or explosion) is equal to the total momentum after the collision (or explosion)\u2019. Law of Momentum: The applied resultant force acting on an object is equal to the rate of change of the object\u2019s momentum and this force is in the direction of the change in momentum. 139 Chapter 8 Work and Energy 8.1 What are Work and Energy? During this chapter you will discover that work and energy are very closely related: We consider the energy of an object as its capacity to do work and doing work as the process of transferring energy from one object or form to another. In other words, \u2020 \u2020 an object with lots of energy can do lots of work. when work is done, energy is lost by the object doing work and gained by the object on which the work is done. Lifting objects or throwing them requires that you do work on them. Even making electricity ow requires that something do work. Something must have energy and transfer it through doing work to make things happen. 8.2 Work To do work on an object, one must move the object by applying a force with at least a component in the direction of motion. The work done is given by W = Fks W : work done (N:m or J) Fk s : component of applied force parallel to motion (N ) : displacement of the object (m) It is very important to note that for work to be done there must be a component of the applied force in the direction of motion. Forces perpendicular to the direction of motion do no work. As with all physical quantities, work must have units. As follows from the de\ufb02nition, work is measured in N:m. The name given to this combination of S.I. units is the joule (J). De\ufb02nition: 1 joule is the work done when an object is moved 1m under the application of a force of 1N in the direction of motion. 140 The work done by an object can be positive or negative. Since force (Fk) and displacement (s) are both vectors, the result of the above equation depends on their directions: \u2020 \u2020 If Fk acts in the same direction as the motion then positive work is being done. In this case the object on which the force is applied gains energy. If the direction of motion and Fk are opposite, then negative work is being done. This means that energy is transferred in the opposite direction. For example, if you try to push a car uphill by applying a force up the slope and instead the car rolls down the hill you are doing negative work on the car. Alternatively, the car is doing positive work on you! Worked Example 39 Calculating Work Done I Question: If you push a box 20m forward by applying a force of 15N in the forward direction, what is the work you have done on the box? Answer: Step 1 : Analyse the question to determine what information is provided \u2020 \u2020 \u2020 The force applied is F = 15N. The distance moved is s = 20m. The applied force and distance moved are in the same direction. Therefore, Fk = 15N. These quantities are all in the correct units, so no unit conversions are required. Step 2 : Analyse the question to determine what is being asked \u2020 We are asked to \ufb02nd the work done on the box. We know from the de\ufb02nition that work done is W = Fks Step 3 : Next we substitute the values and calculate the work done W = Fks = (15N )(20m) = 300 N = 300 J m \u00a2 Remember that the answer must be positive as the applied force and the motion are in the same direction (forwards). In this case, you (the pusher) lose energy, while the box gains energy. 141 Worked Example 40 Calculating Work Done II Question: What is the work done by you on a car, if you try to push the car up a hill by applying a force of 40N directed up the slope, but it slides downhill 30cm? Answer: Step 1 : Analyse the question to determine what information is provided \u2020 \u2020 \u2020 The force applied is F = 40N The distance moved is s =", " 30cm. This is expressed in the wrong units so we must convert to the proper S.I. units (meters): s = 30cm = 30cm 1m 100cm \u00a2 = 0:3m The applied force and distance moved are in opposite directions. Therefore, if we take s = 0:3m, then Fk = 40N. \u00a1 Step 2 : Analyse the question to determine what is being asked \u2020 We are asked to \ufb02nd the work done on the car by you. We know that work done is W = Fks Step 3 : Substitute the values and calculate the work done Again we have the applied force and the distance moved so we can proceed with calculating the work done: W = Fks = ( 40N )(0:3m) = = \u00a1 12N 12 J \u00a1 \u00a1 m \u00a2 Note that the answer must be negative as the applied force and the motion are in opposite directions. In this case the car does work on the person trying to push. What happens when the applied force and the motion are not parallel? If there is an angle between the direction of motion and the applied force then to determine the work done we have to calculate the component of the applied force parallel to the direction of motion. Note that this means a force perpendicular to the direction of motion can do no work. Worked Example 41 Calculating Work Done III Question: Calculate the work done on a box, if it is pulled 5m along the ground by applying a force of F = 10N at an angle of 60o to the horizontal. 142 F 60o Answer: Step 1 : Analyse the question to determine what information is provided The force applied is F = 10N The distance moved is s = 5m along the ground The angle between the applied force and the motion is 60o \u2020 \u2020 \u2020 These quantities are in the correct units so we do not need to perform any unit conversions. Step 2 : Analyse the question to determine what is being asked We are asked to \ufb02nd the work done on the box. \u2020 Step 3 : Calculate the component of the applied force in the direction of motion Since the force and the motion are not in the same direction, we must \ufb02rst calculate the component of the force in the direction of the motion. Fk F 60o Fjj From the force diagram we see that the component of the applied force parallel to the ground is Fjj = F cos(60o) \u00a2 = 10N = 5 N cos(60o) \u00a2 Step 4 : Substitute and calculate the work done Now we can calculate the work done on the box: W = Fks = (5N )(5m) = 25 J Note that the answer is positive as the component of the force Fk is in the same direction as the motion. We will now discuss energy in greater detail. 143 8.3 Energy As we mentioned earlier, energy is the capacity to do work. When positive work is done on an object, the system doing the work loses energy. In fact, the energy lost by a system is exactly equal to the work done by the system. Like work (W ) the unit of energy (E) is the joule (J). This follows as work is just the transfer of energy. A very important property of our universe which was discovered around 1890 is that energy is conserved. Energy is never created nor destroyed, but merely transformed from one form to another. Energy conservation and the conservation of matter are the principles on which classical me- chanics is built. Energy is conserved! IN THE ABSENCE OF FRICTION When work is done on an object by a system: -the object gains energy equal to the work done by the system Work Done = Energy Transferred IN THE PRESENCE OF FRICTION When work is done by a system: -only some of the energy lost by the system is transferred into useful energy -the rest of the energy transferred is lost to friction Total Work Done = Useful Work Done + Work Done Against Friction 8.3.1 Types of Energy So what di\ufb01erent types of energy exist? Kinetic, mechanical, thermal, chemical, electrical, radiant, and atomic energy are just some of the types that exist. By the principle of conservation of energy, when work is done energy is merely transferred from one object to another and from one type of energy to another. 144 Kinetic Energy Kinetic energy is the energy of motion that an object has. Objects moving in straight lines possess translational kinetic energy, which we often abbreviate as Ek. The translational kinetic energy of an object is given by Ek = 1 2 mv2 : kinetic energy (J) Ek m : mass of object (kg) v : speed of the object (m:s\u00a11) Note the dependence of the kinetic energy on the speed of the object{ kinetic energy is related to motion. The faster an object", " is moving the greater its kinetic energy. Worked Example 42 Calculation of Kinetic Energy Question: If a rock has a mass of 1kg and is thrown at 5m=s, what is its kinetic energy? Answer: Step 1 : Analyse the question to determine what information is provided The mass of the rock m = 1kg The speed of the rock v = 5m=s \u2020 \u2020 These are both in the correct units so we do not have to worry about unit conversions. Step 2 : Analyse the question to determine what is being asked \u2020 We are asked to \ufb02nd the kinetic energy. From the de\ufb02nition we know that to work out Ek, we need to know the mass and the velocity of the object and we are given both of these values. Step 3 : Substitute and calculate the kinetic energy mv2 Ek = = 1 2 1 2 (1kg)(5 )2 m s m2 \u00a2 s2 kg = 12:5 = 12:5 J 145 To check that the units in the above example are in fact correct: kg m2 \u00a2 s2 m kg \u00a2 s2 \u00a2 \u00b6 = = J m = N m \u00a2 The units are indeed correct! Study hint: Checking units is an important cross-check and you should get into a habit of doing this. If you, for example, \ufb02nish an exam early then checking the units in your calculations is a very good idea. Worked Example 43 Mixing Units and Kinetic Energy Calculations 1 Question: If a car has a mass of 900kg and is driving at 60km=hr, what is its kinetic energy? Answer: Step 1 : Analyse the question to determine what information is provided \u2020 \u2020 The mass of the car m = 900kg The speed of the car v = 60km=hr. These are not the units we want so before we continue we must convert to m=s. We do this by multiplying by one: 60 km hr \u00a3 1 = 60 1000m 1km km hr \u00a3 m hr = 60000 Now we need to change from hours to seconds so we repeat our procedure: 60000 m hr \u00a3 1 = 60000 = 16:67 1hr 3600s m hr \u00a3 m s and so the speed in the units we want is v = 16:67m=s. Step 2 : Analyse the question to determine what is being asked We are asked to \ufb02nd the kinetic energy. \u2020 Step 3 : Substitute and calculate 146 We know we need the mass and the speed to work out Ek and we are given both of these quantities. We thus simply substitute them into the equation for Ek: Ek = = 1 2 1 2 mv2 (900kg)(16:67 m s )2 = 125 000 kgm2 s2 = 125 000 J Worked Example 44 Mixing Units and Kinetic Energy Calculations 2 Question: If a bullet has a mass of 150g and is shot at a muzzle velocity of 960m=s, what is its kinetic energy? Answer: Step 1 : Analyse the question to determine what information is provided \u2020 We are given the mass of the bullet m = 150g. This is not the unit we want mass to be in. We need to convert to kg. Again, we multiply by one: 150g \u00a2 1 = 150g \u00a2 = 0:15kg 1kg 1000g \u2020 We are given the muzzle velocity which is just how fast the bullet leaves the barrel and it is v = 960m=s. Step 2 : Analyse the question to determine what is being asked We are asked to \ufb02nd the kinetic energy. \u2020 Step 3 : Substitute and calculate We just substitute the mass and velocity (which are known) into the equation for Ek: Ek = = 1 2 1 2 mv2 (150kg)(960 m s )2 = 69 120 kgm2 s2 = 69 120 J 147 Potential Energy If you lift an object you have to do work on it. This means that energy is transferred to the object. But where is this energy? This energy is stored in the object and is called potential energy. The reason it is called potential energy is because if we let go of the object it would move. De\ufb02nition: Potential energy is the energy an object has due to its position or state. As an object raised above the ground falls, its potential energy is released and transformed into kinetic energy. The further it falls the faster it moves as more of the stored potential energy is transferred into kinetic energy. Remember, energy is never created nor destroyed, but merely transformed from one type to another. In this case potential energy is lost but an equal amount of kinetic energy is gained. In the example of a falling mass the potential energy is known as gravitational potential energy as it is the gravitational force exerted by the earth which causes the mass to accelerate towards the ground. The gravitational \ufb02", "eld of the earth is what does the work in this case. Another example is a rubber-band. In order to stretch a rubber-band we have to do work on it. This means we transfer energy to the rubber-band and it gains potential energy. This potential energy is called elastic potential energy. Once released, the rubber-band begins to move and elastic potential energy is transferred into kinetic energy. Gravitational Potential Energy As we have mentioned, when lifting an object it gains gravitational potential energy. One is free to de\ufb02ne any level as corresponding to zero gravitational potential energy. Objects above this level then possess positive potential energy, while those below it have negative potential energy. To avoid negative numbers in a problem, always choose the lowest level as the zero potential mark. The change in gravitational potential energy of an object is given by: \u00a2EP = mg\u00a2h \u00a2EP : Change in gravitational potential energy (J) m : mass of object (kg) g \u00a2h : acceleration due to gravity (m:s\u00a12) : change in height (m) When an object is lifted it gains gravitational potential energy, while it loses gravitational potential energy as it falls. Worked Example 45 Gravitational potential energy 148 Question: How much potential energy does a brick with a mass of 1kg gain if it is lifted 4m. Answer: Step 1 : Analyse the question to determine what information is provided The mass of the brick is m = 1kg The height lifted is \u00a2h = 4m \u2020 \u2020 These are in the correct units so we do not have to worry about unit conversions. Step 2 : Analyse the question to determine what is being asked We are asked to \ufb02nd the gain in potential energy of the object. \u2020 Step 3 : Identify the type of potential energy involved Since the block is being lifted we are dealing with gravitational potential energy. To work out \u00a2EP, we need to know the mass of the object and the height lifted. As both of these are given, we just substitute them into the equation for \u00a2EP. Step 4 : Substitute and calculate \u00a2EP = mg\u00a2h = (1kg) 10 = 40 kg \u2021 m2 \u00a2 s2 = 40 J (4m) m s2 \u00b7 8.4 Mechanical Energy and Energy Conservation Kinetic energy and potential energy are together referred to as mechanical energy. The total mechanical energy (U ) of an object is then the sum of its kinetic and potential energies: U = EP + EK 1 2 U = mgh + mv2 (8.1) Now, IN THE ABSENCE OF FRICTION Mechanical energy is conserved Ubef ore = Uaf ter This principle of conservation of mechanical energy can be a very powerful tool for solving physics problems. However, in the presence of friction some of the mechanical energy is lost: 149 IN THE PRESENCE OF FRICTION Mechanical energy is not conserved (The mechanical energy lost is equal to the work done against friction) \u00a2U = Ubef ore \u00a1 Uaf ter = Work Done Against Friction Worked Example 46 Using Mechanical Energy Conservation Question: A 2kg metal ball is suspended from a rope. If it is released from point A and swings down to the point B (the bottom of its arc) what is its velocity at point B? A 0.5m B Answer: Step 1 : Analyse the question to determine what information is provided The mass of the metal ball is m = 2kg The change in height going from point A to point B is h = 0:5m The ball is released from point A so the velocity at point A is zero (vA = 0m=s). \u2020 \u2020 \u2020 These are in the correct units so we do not have to worry about unit conversions. Step 2 : Analyse the question to determine what is being asked Find the velocity of the metal ball at point B. \u2020 Step 3 : Determine the Mechanical Energy at A and B To solve this problem we use conservation of mechanical energy as there is no friction. Since mechanical energy is conserved, UA = UB Therefore we need to know the mechanical energy of the ball at point A (UA) and at point B (UB). The mechanical energy at point A is UA = mghA + 1 2 m(vA)2 150 We already know m, g and vA, but what is hA? Note that if we let hB = 0 then hA = 0:5m as A is 0:5m above B. In problems you are always free to choose a line corresponding to h = 0. In this example the most obvious choice is to make point B correspond to h = 0. Now we have, UA = (2kg) 10 = 10 J \u2021 m s2 \u00b7 (0:5m) + (2kg)(0)2 1 2 As already stated UB = UA. Therefore UB = 10J", ", but using the de\ufb02nition of mechanical energy UB = mghB + 1 2 m(vB)2 = 1 2 m(vB)2 because hB = 0. This means that (2kg)(vB)2 10J = 1 2 (vB)2 = 10 J kg vB = p10 m s 8.5 Summary of Important Quantities, Equations and Con- cepts Quantity Work Kinetic Energy Potential Energy Mechanical Energy Units Symbol Unit W EK EP U J J J J S.I. Units N:m or kg:m2:s\u00a12 N:m or kg:m2:s\u00a12 N:m or kg:m2:s\u00a12 N:m or kg:m2:s\u00a12 Direction | | | | Table 8.1: Summary of the symbols and units of the quantities used in Energy Principle of Conservation of Energy: Energy is never created nor destroyed, but merely transformed from one form to another. Conservation of Mechanical Energy: In the absence of friction, the total mechanical energy of an object is conserved. 151 Essay 1 : Energy Author: Asogan Moodaly Asogan Moodaly received his Bachelor of Science degree (with honours) in Mechanical Engineering from the University of Natal, Durban in South Africa. For his \ufb02nal year design project he worked on a 3-axis \ufb02lament winding machine for composite (Glass re-enforced plastic in this case) piping. He worked in Vereeniging, Gauteng at Mine Support Products (a subsidiary of Dorbyl Heavy Engineering) as the design engineer once he graduated. He currently lives in the Vaal Triangle area and is working for Sasol Technology Engineering as a mechanical engineer, ensuring the safety and integrity of equipment installed during projects. Energy and electricity. Why the fuss? Disclaimer: The details furnished below are very basic and for illustration purposes only. Why do we need energy? Note that I use the word energy and not electricity. On a broad scale it stimulates economic growth, etc, etc but on a personal level it allows us to lead a comfortable lifestyle. e.g. Flick a switch and Heat for cooking Entertainment such as television and radio Heat for water and interior of house Ironing Electronic and electrical devices such as alarms, garage doors, etc. In a modern household this energy is provided in the form of electricity which is powered via fossil fuels or nuclear. How is electricity made? In a nutshell: By moving a magnet through or near a set of conducting coils. Linear Rotational Conducting wire typically copper Magnet Light bulb Magnet Iron core in coil increases electricity Most power stations produce steam through heat (nuclear reaction or burning fossil fuels), the steam drives a turbine which moves a magnet relative to a coil (the generator like the above but on a much larger scale i.e. bigger magnets, bigger coils, etc), which produces electricity that is transmitted via a power network to our homes. Gas \ufb02red plants burn gas directly in a gas turbine to produce the same desired relative motion between permanent magnet and coil. 152 Steam rushes through Turbine, causing the blades to spin (attached to the shaft so shaft spins) Generator Boiler Shaft Steam Water Magnet attached to shaft Heat Turbine blades Coils Coal Electricity transmitted through power cable Power Station Pylons support cables transformers houses transmission cables Coal, oil and gas are fossil fuels. Fossil fuels were created by decomposing organic (plant and animal) matter a long, long time ago and are typically found underground. Di\ufb01erent temperatures and pressures resulted in the organic matter transforming into coal, oil or gas. Why the fuss about fossil fuels? 1. Fossil fuel power is bad news in the long run. It pollutes and contributes to the greenhouse e\ufb01ect (global warming resulting in melting polar ice caps, oods, droughts, disease, etc). 2. Its not going to last forever. 3. Nuclear power is cleaner in terms of emissions but theres no proven way of disposing of the nuclear waste. Oh, and it wont last forever either! Renewable Energy As the name suggests renewable energy lasts forever. Solar (sun), wind, geothermal, wave, hydro and biomass (organic) are all sources of energy that will last until the sun eventually explodes many millions of years from now. Hopefully the human race will have moved from the earth by then! Generally the principal of renewable electricity generation is similar to fossil fuel electricity generation in that electricity is generated by moving a magnet relative to a conducting coil. What is di\ufb01erent is the way energy is supplied to cause that motion. The below are a few di\ufb01erent types of available renewable energy technologies. Solar There are di\ufb01erent types of solar electricity technologies, the main ones being solar thermal and photovoltaic. 153 Sun Heated oil stored in insulated tank for night use", " Transmission cables & pylon Steam pipes Steam Turbine Generator Mirrored trough reflector, reflects sunlight onto black pipes Black pipes carrying oil Heat exchanger transfers heat from hot oil to water Water becomes steam Water pipes Solar thermal uses the heat of the sun to produce electricity. Sun is concentrated using mirrors. This heat either creates steam which drives a turbine which in turn drives a generator (as per fossil fuel generation), or drives an air engine (engine that uses expanding air to obtain motion) that drives a generator. Photovoltaic panels convert sunlight directly into electricity. The bene\ufb02t of photovoltaic panels is that there are no moving parts, and is therefore relatively maintenance free. The downside is that its very expensive at this stage (17/06/2004). Solar Water Heaters could save up to 30% of the total electricity used in a house. Heater on roof Hot Water out Insulated geyser to store water Cold water in Wooden box with glass top provides insulation. Silicon gel should be used to make the box airtight. If air escapes, so does heat. copper Water (in pipes painted black) is heated by sunlight. Cold water sinks. Hot water rises into the geyser. Wind Wind turbines catch wind that spins the blades. The blades are connected to a shaft that spins because of the wind. This spinning shaft spins another shaft that turns a permanent magnet relative to conducting coils. Note that gears are used to convert the slow spinning of the 1st shaft to a faster spin on the 2nd shaft. The generator shaft needs to spin at the correct speed to produce the right amount and quality of electricity. Some generators are now being modi\ufb02ed to run at slower speeds. This saves money as gears are not needed. 154 Vessel Nacelle containing all moving parts Blades Detail of Nacelle Vessel (top view) to blades Shaft 1 (slow spin) Shaft 2 (Fast spin) Gears Generator Biomass Biomass is anything organic i.e. plant or animal matter. It can be used in the place of coal as per a normal coal \ufb02red plant and is renewable as long as the biomass e.g. wood; is handled in a sustainable manner. By sustainable I mean that suitable farming practices are used so that the land is not over farmed which will result in the soil becoming barren and nothing growing there again. Pipe carrying biogas for heat or power Processed biomass fertilizer Biogas bubbles Upside down floating container to catch biogas Input biomass Water forms a seal to keep air out Input biomass being digested Biomass can also be processed using anaerobic digestion to produce a gas that can be burned for heat or electricity. This biogas is made up of a number of other gases that are similar to those found in fossil fuel natural gas Except the amount of the gases are di\ufb01erent. E.g. Natural gas has about 94 Anaerobic digestion: Anaerobic means No air. Therefore anaerobic digestion means to digest in the absence of air. Bacteria that naturally exist in organic matter will convert organic matter to biogas and fertilizer when all the air is removed. Thousands of anaerobic digesters have been installed in rural India, Nepal and China in rural areas where cow dung, human waste and chicken litter (faeces) are all processed using anaerobic digestion to produce gas that can be burned in the home for cooking and heating. The leftover is used as fertilizer. Geothermal Energy In some places on earth, the earths crust is thinner than others. As a result the heat from the earths core escapes. The heat can be captured by converting water to steam, and using the 155 steam to drive a steam generator as discussed above. Hydroelectric power Water from a river is diverted to turn a water turbine to create electricity similar to the principles of steam generation. The water is returned to the river after driving the turbine. Pipe diverting water Transmission cables Generator house River Wave Energy Some wave energy generators work similarly to wind turbines except that underwater ocean currents turns the blades instead of wind; and of course most of the structure is under water! Ocean current blades Ocean floor Underwater pipe to shore carrying transmission cable Another concept uses the rising and falling of the tides to suck air in using a one way valve. As a result air becomes compressed in a chamber and the compressed air is let out to drive a turbine which in turn drives a generator One way valve allows air to be sucked in then shuts when air tries to get out. Generator house with turbine and generator Chamber with compressed air Cliff face Ocean floor As the water level falls and rises, air is sucked in and compressed 156 These are relatively new technologies. Liquid Fuels Liquid fuels are used mainly for transportation. Petrol and diesel are the most common liquid fuels and are obtained from oil. Sasol is the only company in the world that makes liquid fuels from coal; and will be one of the leading companies in the world to make liquid fuels from", " natural gas! The Sasol petro-chemical plants are based in Sasolburg on the border of the Free State and in Secunda in Mpumalanga. However, as discussed above coal, gas and oil are fossil fuels and are not renewable. Petrol and diesel are obtained from fossil fuels and therefore pollute and contribute to the green house e\ufb01ect (global warming). Alternatives Biodiesel Oil can be extracted from plants such as the soya bean, sunower and rapeseed by pressing it through a \ufb02lter. This oil if mixed correctly with either methanol or dry ethanol and Sodium Hydroxide will separate the plant oil into biodiesel, glycerol and fertilizer. The biodiesel can be used as produced in a conventional diesel engine with little or no mod- i\ufb02cations required. The glycerol can be re\ufb02ned a bit further for pharmaceutical companies to use, or can be used to make soap. Ethanol Corn, maize and sugar cane can be used to make ethanol as a fuel substitute for petrol. Its made by the same fermentation process used to make alcohol. Enzymes are often used to speed up the process. In ethanol from sugar cane production, the leftover bagasse (the \ufb02bre part of the sugar cane) can be burned in a biomass power station to produce electricity. Hydrogen Through the process of electrolysis electricity (hopefully clean, renewable electricity!) can split water into hydrogen and oxygen. The stored hydrogen can be used in a fuel cell to create electricity in a process that is opposite to electrolysis; to drive electric motors in a car. The hydrogen can also be burned directly in a modi\ufb02ed internal combustion engine. In both cases the waste product is water. 157 Essay 2 : Tiny, Violent Collisions Author: Thomas D. Gutierrez Tom Gutierrez received his Bachelor of Science and Master degrees in physics from San Jose State University in his home town of San Jose, California. As a Master\u2019s student he helped work on a laser spectrometer at NASA Ames Research Centre. The instrument measured the ratio of di\ufb01erent isotopes of carbon in CO2 gas and could be used for such diverse applications as medical diagnostics and space exploration. Later, he received his Ph.D. in physics from the University of California, Davis where he performed calculations for various reactions in high energy physics collisions. He currently lives in Berkeley, California where he studies proton-proton collisions seen at the STAR experiment at Brookhaven National Laboratory on Long Island, New York. High Energy Collisions Take an orange and expanded it to the size of the earth. The atoms of the earth-sized orange would themselves be about the size of regular oranges and would \ufb02ll the entire \\earth-orange\". Now, take an atom and expand it to the size of a football \ufb02eld. The nucleus of that atom would be about the size of a tiny seed in the middle of the \ufb02eld. From this analogy, you can see that atomic nuclei are very small objects by human standards. They are roughly 10\u00a115 meters in diameter { one-hundred thousand times smaller than a typical atom. These nuclei cannot be seen or studied via any conventional means such as the naked eye or microscopes. So how do scientists study the structure of very small objects like atomic nuclei? The simplest nucleus, that of hydrogen, is called the proton. Faced with the inability to isolate a single proton, open it up, and directly examine what is inside, scientists must resort to a brute-force and somewhat indirect means of exploration: high energy collisions. By colliding protons with other particles (such as other protons or electrons) at very high energies, one hopes to learn about what they are made of and how they work. The American physicist Richard Feynman once compared this process to slamming delecate watches together and \ufb02guring out how they work by only examining the broken debris. While this analogy may seem pessimistic, with su\u2013cent mathematical models and experimental precision, considerable information can be extracted from the debris of such high energy subatomic collisions. One can learn about both the nature of the forces at work and also about the sub-structure of such systems. The experiments are in the category of \\high energy physics\" (also known as \\subatomic\" physics). The primary tool of scienti\ufb02c exploration in these experiments is an extremely violent collision between two very, very small subatomic objects such as nuclei. As a general rule, the higher the energy of the collisions, the more detail of the original system you are able to resolve. These experiments are operated at laboratories such as CERN, SLAC, BNL, and Fermilab, just to name a few. The giant machines that perform the collisions are roughly the size of towns. For example, the RHIC collider at BNL is", " a ring about 1 km in diameter and can be seen from space. The newest machine currently being built, the LHC at CERN, is a ring 9 km in diameter! Let\u2019s examine the kinematics of such a collisions in some detail... 158 Chapter 9 Collisions and Explosions In most physics courses questions about collisions and explosions occur and to solve these we must use the ideas of momentum and energy; with a bit of mathematics of course! This section allows you to pull the momentum and energy ideas together easily with some speci\ufb02c problems. 9.1 Types of Collisions We will consider two types of collisions in this section Elastic collisions Inelastic collisions \u2020 \u2020 In both types of collision, total energy and total momentum is always conserved. Kinetic energy is conserved for elastic collisions, but not for inelastic collisions. 9.1.1 Elastic Collisions De\ufb02nition: An elastic collision is a collision where total momentum and total kinetic energy are both conserved. (NOTE TO SELF: this should be in an environment for de\ufb02nitions!!) This means that the total momentum and the total kinetic energy before an elastic collision is the same as after the collision. For these kinds of collisions, the kinetic energy is not changed into another type of energy. Before the Collision In the following diagram, two balls are rolling toward each other, about to collide \u00a1!p 1, K1 \u00a1!p 2, K2 159 Before the balls collide, the total momentum of the system is equal to all the individual momenta added together. The ball on the left has a momentum which we call \u00a1!p 1 and the ball on the right has a momentum which we call \u00a1!p 2, it means the total momentum before the collision is \u00a1!p Before = \u00a1!p 1 + \u00a1!p 2 (9.1) We calculate the total kinetic energy of the system in the same way. The ball on the left has a kinetic energy which we call K1 and the ball on the right has a kinetic energy which we call K2, it means that the total kinetic energy before the collision is KBefore = K1 + K2 (9.2) After the Collision The following diagram shows the balls after they collide \u00a1!p 3, K3 \u00a1!p 4, K4 After the balls collide and bounce o\ufb01 each other, they have new momenta and new kinetic energies. Like before, the total momentum of the system is equal to all the individual momenta added together. The ball on the left now has a momentum which we call \u00a1!p 3 and the ball on the right now has a momentum which we call \u00a1!p 4, it means the total momentum after the collision is \u00a1!p After = \u00a1!p 3 + \u00a1!p 4 The ball on the left now has a kinetic energy which we call K3 and the ball on the right now has a kinetic energy which we call K4, it means that the total kinetic energy after the collision is (9.3) KAfter = K3 + K4 (9.4) Since this is an elastic collision, the total momentum before the collision equals the total momentum after the collision and the total kinetic energy before the collision equals the total kinetic energy after the collision Before \u00a1!p Before \u00a1!p 1 + \u00a1!p 2 After = \u00a1!p After = \u00a1!p 3 + \u00a1!p 4 and KBefore K1 + K2 = KAfter = K3 + K4 (9.5) (9.6) Worked Example 47 An Elastic Collision 160 We will have a look at the collision between two pool balls. Ball 1 is at rest and ball 2 is moving towards it with a speed of 2 [m:s\u00a11]. The mass of each ball is 0.3 [Kg]. After the balls collide elastically, ball 2 comes to a stop and ball 1 moves o\ufb01. What is the \ufb02nal velocity of ball 1? Step 1 : Draw the \\before\" diagram Before the collision, ball 2 is moving; we will call it\u2019s momentum P2 and it\u2019s kinetic energy K2. Ball 1 is at rest, so it has zero kinetic energy and momentum. 2 1 \u00a1!p 2, K2 \u00a1!p 1 = 0, K1 = 0 Step 2 : Draw the \\after\" diagram After the collision, ball 2 is at rest but ball 1 has a momentum which we call P3 and a kinetic energy which we call K3. \u00a1!p 4 = 0, K4 = 0 \u00a1!p 3, K3 2 1 Because the collision is elastic, we can solve the problem using momentum conservation or kinetic energy conservation. We will do it both ways to show that the answer is the same", ", whichever method you use. Step 3 : Show the conservation of momentum We start by writing down that the momentum before the collision \u00a1!p Before is equal to the momentum after the collision \u00a1!p After Before After \u00a1!p Before = \u00a1!p After \u00a1!p 1 + \u00a1!p 2 = \u00a1!p 3 + \u00a1!p 4 0 + \u00a1!p 2 = \u00a1!p 3 + 0 \u00a1!p 2 = \u00a1!p 3 We know that momentum is just P = mv, and we know the masses of the balls, so we can rewrite the conservation of momentum in terms of the velocities of the balls (9.7) \u00a1!p 2 = \u00a1!p 3 m2v2 = m3v3 0:3v2 = 0:3v3 v2 = v3 So ball 1 exits with the velocity that ball 2 started with! v3 = 2[m:s\u00a11] 161 (9.8) (9.9) Step 4 : Show the conservation of kinetic energy We start by writing down that the kinetic energy before the collision KBefore is equal to the kinetic energy after the collision KAfter Before After KBefore = KAfter K1 + K2 = K3 + K4 0 + K2 = K3 + 0 K2 = K3 (9.10) We know that kinetic energy is just K = mv 2, and we know the masses of the balls, so we can rewrite the conservation of kinetic energy in terms of the velocities of the balls 2 K2 = K3 (9.11) = m3v2 3 2 m2v2 2 2 0:15v2 2 = 0:15v2 3 2 = v2 v2 3 v2 = v3 So ball 1 exits with the velocity that ball 2 started with, which agrees with the answer we got when we used the conservation of momentum. v3 = 2[m:s\u00a11] (9.12) Worked Example 48 Elastic Collision 2 Question: Now for a slightly more di\u2013cult example. We have 2 marbles. Marble 1 has mass 50 g and marble 2 has mass 100 g. I roll marble 2 along the ground towards marble 1 in the positive x-direction. Marble 1 is initially at rest and marble 2 has a velocity of 3 m:s\u00a11 in the positive x-direction. After they collide elastically, both marbles are moving. What is the \ufb02nal velocity of each marble? Answer: Step 1 : Put all the quantities into S.I. units So: m1 = 0:05kg and m2 = 0:1kg Step 2 : Draw a rough sketch of the situation Before the collision: 2 1 \u00a1!p2, Ek2 \u00a1!p1 = 0, Ek1 = 0 162 After the collision: 2 1 \u00a1!p3, Ek3 \u00a1!p4, Ek4 Step 3 : Decide which equations to use in the problem Since the collision is elastic, both momentum and kinetic energy are conserved in the collision. So: EkBef ore = EkAf ter and \u00a1\u00a1\u00a1\u00a1!pBef ore = \u00a1\u00a1\u00a1\u00a1!pAf ter There are two unknowns (\u00a1!v1 and \u00a1!v2) so we will need two equations to solve for them. We need to use both kinetic energy conservation and momentum conservation in this problem. Step 4 : Solve the \ufb02rst equation Let\u2019s start with energy conservation. Then: 1 2 2 m1\u00a1!u1 + 2 m1\u00a1!u1 2 EkBef ore = EkAf ter 1 2 + m2\u00a1!u2 1 2 = m1\u00a1!v1 m1\u00a1!v1 2 m2\u00a1!u2 = 2 2 + 1 2 + m2\u00a1!v2 2 m2\u00a1!v2 2 But \u00a1!u1=0, and solving for \u00a1!v2 2 : 2 2 \u00a1!v2 \u00a1!v2 2 \u00a1!v2 2 = \u00a1!u2 \u00a1 = (3)2 = 9 \u00a1 \u00a1 1 2 \u00a1!v1 2 m1 m2 \u00a1!v1 (0:05) (0:10) \u00a1!v1 2 2 (A) Step 5 : Solve the second equation Now we have simpli\ufb02ed as far as we can, we move onto momentum conservation: \u00a1\u00a1\u00a1\u00a1!pBef ore = \u00a1\u00a1\u00a1\u00a1!pAf ter m1\u00a1!u1 + m2\u00a1!u2 = m1\u00a1!v1 + m2\u00a1", "!v2 But \u00a1!u1=0, and solving for \u00a1!v1: m2\u00a1!u2 = m1\u00a1!v1 + m2\u00a1!v2 m2\u00a1!v2 m1\u00a1!v1 = m2\u00a1!u2 m2 \u00a1!v2 \u00a1!u2 m1 \u00a1!v1 = \u00a1 m2 m1 \u00a1!v1 = 2(3) \u00a1!v1 = 6 \u00a1 \u00a1 2\u00a1!v2 \u00a1 2\u00a1!v2 (B) 163 Step 6 : Substitute one equation into the other Now we can substitute (B) into (A) to solve for \u00a1!v2: 2 2 2 2 2 2 \u00a1!v2 \u00a1!v2 \u00a1!v2 \u00a1!v2 3\u00a1!v2 \u00a1!v2 = 9 = 9 = 9 \u00a1 \u00a1 \u00a1 2\u00a1!v2)2 2 1 2 \u00a1!v1 1 (6 2 1 2 \u00a1 18 + 12\u00a1!v2 (36 \u00a1 = 9 = \u00a1 = 4\u00a1!v2 \u00a1 9 + 12\u00a1!v2 3 \u00a1 24\u00a1!v2 + 4\u00a1!v2 2 ) 2 2\u00a1!v2 \u00a1 2 \u00a1!v2 (\u00a1!v2 \u00a1 \u00a1 3)(\u00a1!v2 4\u00a1!v2 + 3 = 0 1) = 0 \u00a1!v2 = 3 or \u00a1!v2 = 1 \u00a1 We were lucky in this question because we could factorise. If you can\u2019t factorise, then you can always solve using the formula for solving quadratic equations. Remember: b x = \u00a1 \u00a7 4ac pb2 2a \u00a1 So, just to check: \u00a1!v2 = 4 \u00a7 42 \u00a1 2(1) 4(1)(3) 4 \u00a1 \u00a7 12 \u00a1!v2 = p p16 2 p4 \u00a7 2 1 \u00a1!v2 = 3 or \u00a1!v2 = 1 same as before \u00a1!v2 = \u00a1!v2 = 2 \u00a7 4 Step 7 : Solve for and quote the \ufb02nal answers So \ufb02nally, substituting into equation (B) to get \u00a1!v1: \u00a1!v1 = 6 2\u00a1!v2 \u00a1 If \u00a1!v2 = 3 m:s\u00a11 then \u00a1!v1 = 6 \u00a1 2(3) = 0 m:s\u00a11 But, according to the question, marble 1 is moving after the collision. So \u00a1!v1 \u00a1!v2 = 3. Therefore: = 0 and \u00a1!v2 = 1 m:s\u00a11 in the positive x and \u00a1!v1 = 4 m:s\u00a11 in the positive x direction direction \u00a1 \u00a1 164 6 6 9.1.2 Inelastic Collisions De\ufb02nition: An inelastic collision is a collision in which total momentum is conserved but total kinetic energy is not conserved; the kinetic energy is transformed into other kinds of energy. So the total momentum before an inelastic collisions is the same as after the collision. But the total kinetic energy before and after the inelastic collision is di\ufb01erent. Of course this does not mean that total energy has not been conserved, rather the energy has been transformed into another type of energy. As a rule of thumb, inelastic collisions happen when the colliding objects are distorted in some way. Usually they change their shape. To modify the shape of an object requires energy and this is where the \\missing\" kinetic energy goes. A classic example of an inelastic collision is a car crash. The cars change shape and there is a noticeable change in the kinetic energy of the cars before and after the collision. This energy was used to bend the metal and deform the cars. Another example of an inelastic collision is shown in the following picture. Here an asteroid (the small circle) is moving through space towards the moon (big circle). Before the moon and the asteroid collide, the total momentum of the system is: \u00a1\u00a1\u00a1\u00a1!pBef ore = \u00a1!pm + \u00a1!pa (\u00a1!pm stands for \u00a1\u00a1\u00a1!pmoon and \u00a1!pa stands for \u00a1\u00a1\u00a1\u00a1\u00a1! pasteroid) and the total kinetic energy of the system is: EBef ore = Ekm + Eka \u00a1!pm, Ekm \u00a1!pa, Eka When the asteroid collides inelastically with the moon, its kinetic energy", " is transformed mostly into heat energy. If this heat energy is large enough, it can cause the asteroid and the area of the moon\u2019s surface that it hit, to melt into liquid rock! From the force of impact of the asteroid, the molten rock ows outwards to form a moon crater. After the collision, the total momentum of the system will be the same as before. But since this collision is inelastic, (and you can see that a change in the shape of objects has taken place!), 165 total kinetic energy is not the same as before the collision. \u00a1\u00a1\u00a1\u00a1!pAf ter, EkAf ter \u00a1\u00a1\u00a1\u00a1!pBef ore = \u00a1\u00a1\u00a1\u00a1!pAf ter \u00a1!pm + \u00a1!pa = \u00a1\u00a1\u00a1\u00a1!pAf ter but = EkAf ter = EkAf ter EkBef ore Ekm + Eka So: Worked Example 49 Inelastic Collision Question: Let\u2019s consider the collision of two cars. Car 1 is at rest and Car 2 is moving at a speed of 2m:s\u00a11 in the negative x-direction. Both cars each have a mass of 500kg. The cars collide inelastically and stick together. What is the resulting velocity of the resulting mass of metal? Answer: Step 1 : Draw a rough sketch of the situation Before the collision: Car 1 \u00a1!p1 = 0 Car 2 \u00a1!p2 After the collision: \u00a1\u00a1\u00a1\u00a1!pAf ter 166 6 6 Step 2 : Decide which equations to use in the problem We know the collision is inelastic and there was a de\ufb02nite change in shape of the objects involved in the collision - there were two objects to start and after the collision there was one big mass of metal! Therefore, we know that kinetic energy is not conserved in the collision but total momentum is conserved. So: EkBef ore = EkAf ter but pT Bef ore = \u00a1\u00a1\u00a1\u00a1!pAf ter \u00a1\u00a1\u00a1\u00a1\u00a1\u00a1! Step 3 : Solve for and quote the \ufb02nal velocity So we must use conservation of momentum to solve this problem. Take the negative x-direction to have a negative sign: pT Bef ore = \u00a1\u00a1\u00a1\u00a1!pAf ter \u00a1\u00a1\u00a1\u00a1\u00a1\u00a1! \u00a1!p1 + \u00a1!p2 = \u00a1\u00a1\u00a1\u00a1!pAf ter m1\u00a1!u1 + m2\u00a1!u2 = (m1 + m2)\u00a1!v 2) = (500 + 500)\u00a1!v 0 + 500( \u00a1 \u00a1 1000 = 1000\u00a1!v \u00a1!v = 1 m:s\u00a11 \u00a1 Therefore, \u00a1!v = 1 m:s\u00a11 in the negative x direction: \u00a1 9.2 Explosions When an object explodes, it breaks up into more than one piece and it therefore changes its shape. Explosions occur when energy is transformed from one kind e.g. chemical potential energy to another e.g. heat energy or kinetic energy extremely quickly. So, like in inelastic collisions, total kinetic energy is not conserved in explosions. But total momentum is always conserved. Thus if the momenta of some of the parts of the exploding object are measured, we can use momentum conservation to solve the problem! Interesting Fact: The Tunguska event was an aerial explosion that occurred near the Podkamennaya (Stony) Tunguska River in what is now Evenkia, Siberia, at 7:17 AM on June 30, 1908. The size of the blast was later estimated to be equivalent to between 10 and 15 million tons of regular explosive. It felled an estimated 60 million trees over 2,150 square kilometers. At around 7:15 AM, Tungus natives and Russian settlers in the hills northwest of Lake Baikal observed a huge \ufb02reball moving across the sky, nearly as bright as the Sun. A few minutes later, there was a ash that lit up half of the sky, followed by a shock wave that 167 6 knocked people o\ufb01 their feet and broke windows up to 650 km away (the same as the distance from Bloemfontein to Durban!). The explosion registered on seismic stations across Europe and Asia, and produced uctuations in atmospheric pressure strong enough to be detected in Britain. Over the next few weeks, night skies over Europe and western Russia glowed brightly enough for people to read by. Had the object responsible for the explosion hit the Earth a few hours later, it would have exploded over Europe instead of the sparsely-populated T", "unguska region, producing massive loss of human life. In the following picture, a closed can of baked beans is put on a stove or \ufb02re: BAKED BEANS \u00a1\u00a1!pcan; Ekcan Before the can heats up and explodes, the total momentum of the system is: \u00a1\u00a1\u00a1\u00a1!pBef ore = \u00a1\u00a1!pcan = 0 and the total kinetic energy of the system is: EkBef ore = Ekcan = 0 since the can isn\u2019t moving. Once the mixture of beans, juice and air inside the can reach a certain temperature, the pressure inside the can becomes so great that the can explodes! Beans and sharp pieces of metal can y out in all directions. Energy in the system has been transformed from heat energy into kinetic energy. \u00a1!p4; Ek4 \u00a1!p3; Ek3 \u00a1!p1; Ek1 \u00a1!p2; Ek2 168 After the explosion, the can is completely destroyed. But momentum is always conserved, so: \u00a1\u00a1\u00a1\u00a1!pBef ore = \u00a1\u00a1\u00a1\u00a1!pAf ter \u00a1\u00a1\u00a1\u00a1!pBef ore = \u00a1!p1 + \u00a1!p2 + \u00a1!p3 + \u00a1!p4 0 = \u00a1!p1 + \u00a1!p2 + \u00a1!p3 + \u00a1!p4 However, the kinetic energy of the system is not conserved. The can\u2019s shape was changed in the explosion. Before the explosion the can was not moving, but after the explosion, the pieces of metal and baked beans were moving when they were ying out in all directions! So: EkB 6 = EkA Safety tip: Never heat a closed can on a stove or \ufb02re! Always open the can or make a hole in the lid to allow the pressure inside and outside the can to remain equal. This will prevent the can from exploding! Worked Example 50 Explosions 1 Question: An object with mass mT ot = 10 kg is sitting at rest. Suddenly it explodes into two pieces. One piece has a mass of m1 = 5 kg and moves o\ufb01 in the negative x-direction at \u00a1!v1 = 3ms\u00a11. What is the velocity of the other piece? Answer: Step 1 : Draw a rough sketch of the situation Before the explosion, the object is at rest: \u00a1\u00a1!pT ot = 0, EkT ot = 0 After the explosion, the two pieces move o\ufb01: 1 2 \u00a1!p1, Ek1 \u00a1!p2, Ek2 Step 2 : Decide which equations to use in the problem Now we know that in an explosion, total kinetic energy is not conserved. There is a de\ufb02nite change in shape of the exploding object! But we can always use momentum conservation to solve the problem. So: \u00a1\u00a1\u00a1\u00a1!pBef ore = \u00a1\u00a1\u00a1\u00a1!pAf ter \u00a1\u00a1\u00a1\u00a1!pBef ore = \u00a1!p1 + \u00a1!p2 169 But the object was initially at rest so: 0 = \u00a1!p1 + \u00a1!p2 0 = m1\u00a1!v1 + m2\u00a1!v2 (A) Step 3 : Find the mass of the second piece Now we know that m1= 5 kg but we do not know what the mass of m2 is. However, we do know that: mT ot = m1 + m2 m2 = mT ot \u00a1 = 10 kg \u00a1 m1 5 kg = 5 kg Step 4 : Solve for and quote the velocity of the other piece Now we can substitute all the values we know into equation (A) and solve for \u00a1!v2. Let\u2019s choose the positive x-direction to have a postive sign and the negative x-direction to have a negative sign: (A) 0 = m1\u00a1!v1 + m2\u00a1!v2 0 = 5( 3) + 5\u00a1!v2 0 = \u00a1 5\u00a1!v2 = 15 \u00a1!v2 = +3 m:s\u00a11 \u00a1 15 + 5\u00a1!v2 Therefore, \u00a1!v2 = 3 m:s\u00a11 in the positive x direction: \u00a1 Worked Example 51 Explosions 2 Question: An object with mass mT ot =15 kg is sitting at rest. Suddenly it explodes into two pieces. One piece has a mass of m1 = 5000 g and moves o\ufb01 in the positive x-direction at v1 = 7ms\u00a11. What is the \ufb02nal", " velocity of the other piece? Answer: Step 1 : Draw a rough sketch of the situation Before the collision: \u00a1\u00a1!pT ot = 0, EkT ot = 0 170 After the collision: 2 1 \u00a1!p2, Ek2 \u00a1!p1, Ek1 Step 2 : Convert all units into S.I. units m1 = 5000 g m1 = 5 kg Step 3 : Decide which equations to use in the problem Now we know that in an explosion, total kinetic energy is not conserved. There is a de\ufb02nite change in shape of the exploding object! But we can always use momentum conservation to solve the problem. So: \u00a1\u00a1\u00a1\u00a1!pBef ore = \u00a1\u00a1\u00a1\u00a1!pAf ter \u00a1\u00a1\u00a1\u00a1!pBef ore = \u00a1!p1 + \u00a1!p2 But the object was initially at rest so: 0 = \u00a1!p1 + \u00a1!p2 0 = m1\u00a1!v1 + m2\u00a1!v2 (A) Step 4 : Determine the mass of the other piece Now we know that m1= 5 kg but we do not know what the mass of m2 is. However, we do know that: mT ot = m1 + m2 m2 = mT ot \u00a1 = 15 kg \u00a1 m1 5 kg = 10 kg Step 5 : Solve for and quote the \ufb02nal velocity of the other piece Now we can substitute all the values we know into equation (A) and solve for \u00a1!v2. Let\u2019s choose the positive x-direction to have a postive sign and the negative x-direction to have a negative sign: (A) 0 = m1\u00a1!v1 + m2\u00a1!v2 0 = 5(7) + 10(\u00a1!v2) 0 = 35 + 10(\u00a1!v2) 35 3:5 m:s\u00a11 \u00a1 \u00a1 10(\u00a1!v2) = \u00a1!v2 = Therefore, \u00a1!v2 = 3:5 m:s\u00a11 in the negative x direction: \u00a1 171 9.3 Explosions: Energy and Heat In explosions, you have seen that kinetic energy is not conserved. But remember that total energy is always conserved. Let\u2019s look at what happens to the energy in some more detail. If a given amount of energy is released in an explosion it is not necessarily all transformed into kinetic energy. Due to the deformation of the exploding object, often a large amount of the energy is used to break chemical bonds and heat up the pieces. Energy is conserved but some of it is transferred through non-conservative processes like It will be radiated into the heating. This just means that we cannot get the energy back. environment as heat energy but it is all still accounted for. Now we can start to mix the ideas of momentum conservation with energy transfer to make longer problems. These problems are not more complicated just longer. We will start o\ufb01 short and them combine the di\ufb01erent ideas later on. Long problems should be treated like a number of smaller problems. Focus on them one at a time. Worked Example 52 Energy Accounting 1 Question: An object with a mass of mt = 17 kg explodes into two pieces of mass m1 = 7 kg and m2 = 10 kg. m1 has a velocity of 9ms\u00a11 in the negative x-direction and m2 has a velocity of 6:3ms\u00a11 in the positive x-direction. If the explosion released a total energy of 2000 J, how much was used in a non-conservative way? Answer: Step 1 : Draw a rough sketch of the situation Before the collision: After the collision: \u00a1\u00a1!pT ot = 0, EkT ot = 0 E = 2000 J 2 1 \u00a1!p2, Ek2 \u00a1!p1, Ek1 Step 2 : (NOTE TO SELF: step is deprecated, use westep instead.) Determine what is being asked We are asked how much energy was used in a nonconservative fashion. This is the di\ufb01erence between how much energy was used in a conservative fashion and how much was used in total. We are lucky because we have everything we need to determine the kinetic energy of both pieces. The kinetic energy of the pieces is energy that was used in a conservative way. 172 Step 3 : Determine the total kinetic energy The sum of the kinetic energy for the two blocks is the total kinetic energy of the pieces. So: EkT ot = Ek1 + Ek2 2 + = m1\u00a1!v1 1 2 1 2 2 1 m2\u00a1!v2 2 1 2 = (7)(9)2 + (10)(6", ":3)2 = 283:5 + 198:45 EkT ot = 481:95 J Step 4 : Solve for and quote the \ufb02nal answer The total energy that was transformed into kinetic energy is 481:95 J. We know that 2000 J of energy were released in total. the question makes no statements about other types of energy so we can assume that the di\ufb01erence was lost in a non-conservative way. Thus the total energy lost in non-conservative work is: EkT ot = 2000 481:95 \u00a1 = 1518:05 J E \u00a1 Worked Example 53 Energy Accounting 2 Question: An object at rest, with mass mT ot = 4 kg, explodes into two pieces (m1, m2) with m1 = 2:3 kg. m1 has a velocity of 17ms\u00a11 in the negative x-direction. If the explosion released a total energy of 800 J, 1. What is the velocity of m2? 2. How much energy does it carry? 3. And how much energy was used in a non-conservative way? Answer: Step 1 : Draw a rough sketch of the situation Before the collision: After the collision: \u00a1\u00a1!pT ot = 0, EkT ot = 0 173 E = 800 J 1 \u00a1!p1; Ek1 2 \u00a1!p2; Ek2 Step 2 : N ow we know that in an explosion, total kinetic energy is not conserved. There is a de\ufb02nite change in shape of the exploding object! But we can always use momentum conservation to solve the problem. So: \u00a1\u00a1\u00a1\u00a1!pBef ore = \u00a1\u00a1\u00a1\u00a1!pAf ter \u00a1\u00a1\u00a1\u00a1!pBef ore = \u00a1!p1 + \u00a1!p2 But the object was initially at rest so: 0 = \u00a1!p1 + \u00a1!p2 0 = m1\u00a1!v1 + m2\u00a1!v2 (A) Step 3 : (NOTE TO SELF: step is deprecated, use westep instead.) Now we know that m1= 2.3 kg but we do not know what the mass of m2 is. However, we do know that: mT ot = m1 + m2 m2 = mT ot \u00a1 = 4 kg \u00a1 = 1:7 kg m1 2:3 kg Step 4 : (NOTE TO SELF: step is deprecated, use westep instead.) Now we can substitute all the values we know into equation (A) and solve for \u00a1!v2. Let\u2019s choose the positive x-direction to have a postive sign and the negative x-direction to have a negative sign: 0 = m1\u00a1!v1 + m2\u00a1!v2 0 = (2:3)( 17) + 1:7\u00a1!v2 (A) \u00a1 39:1 + 1:7\u00a1!v2 0 = 1:7\u00a1!v2 = 39:1 \u00a1 \u00a1!v2 = 23 m:s\u00a11 \u2020 \u00a1!v2 = 23 m:s\u00a11 in the positive x -direction Step 5 : (NOTE TO SELF: step is deprecated, use westep instead.) Now we need to calculate the energy that the second piece carries: Ek2 = = 2 m2\u00a1!v2 (1:7)(23)2 1 2 1 2 = 449:65 J 174 The kinetic energy of the second piece is Ek2 = 449:65 J \u2020 Step 6 : (NOTE TO SELF: step is deprecated, use westep instead.) Now the amount of energy used in a non-conservative way in the explosion, is the di\ufb01erence between the amount of energy released in the explosion and the total kinetic energy of the exploded pieces: We know that: E \u00a1 EkT ot = 800 EkT ot \u00a1 EkT ot = Ek1 + Ek2 = = 1 2 1 2 m1\u00a1!v1 2 + 449:65 (2:3)(17)2 + 449:65 = 332:35 + 449:65 = 782 J Step 7 : (NOTE TO SELF: step is deprecated, use westep instead.) So going back to: E \u00a1 EkT ot = 800 = 800 EkT ot 782 \u00a1 \u00a1 = 18 J 18 J of energy was used in a non-conservative way in the explosion \u2020 9.4 Important Equations and Quantities Quantity velocity momentum energy Symbol Unit | | J \u00a1!v \u00a1!p E Units m s kg:m s kg:m s2 2 S.I. Units or m:s\u00a11 or kg:m:s\u00a11 or", " kg:m2s\u00a12 Direction X X | Table 9.1: Units commonly used in Collisions and Explosions Momentum: Kinetic energy: \u00a1!p = m\u00a1!v Ek = 1 2 m\u00a1!v 2 175 (9.13) (9.14) Chapter 10 Newtonian Gravitation 10.1 Properties Gravity is a force and therefore must be described by a vector - so remember magnitude and direction. Gravity is a force that acts between any two objects with mass. To determine the magnitude of the force we use the following equation: F = Gm1m2 r2 This equation describes the force between two bodies, one of mass m1, the other of mass m2 (both have units of Kilogrammes, or Kg for short). The G is Newton\u2019s \u2018Gravitational Constant\u2019 10\u00a111 [Nm2kg\u00a12]) and r is the straight line distance between the two bodies in meters. (6:673 This means the bigger the masses, the greater the force between them. Simply put, big things matter big with gravity. The 1=r2 factor (or you may prefer to say r\u00a12) tells us that the distance between the two bodies plays a role as well. The closer two bodies are, the stronger the gravitational force between them is. We feel the gravitational attraction of the Earth most at the surface since that is the closest we can get to it, but if we were in outer-space, we would barely even know the Earth\u2019s gravity existed! (10.1) \u00a3 Remember that F = ma (10.2) which means that every object on the earth feels the same gravitational acceleration! That means whether you drop a pen or a book (from the same height), they will both take the same length of time to hit the ground... in fact they will be head to head for the entire fall if you drop them at the same time. We can show this easily by using the two equations above (10.1 and 10.2). The force between the Earth (which has the mass me) and an object of mass mo is F = Gmome r2 and the acceleration of an object of mass mo (in terms of the force acting on it) is So we substitute equation (10.3) into equation (10.4), and we \ufb02nd that ao = F mo ao = Gme r2 176 (10.3) (10.4) (10.5) Since it doesn\u2019t matter what mo is, this tells us that the acceleration on a body (due to the Earth\u2019s gravity) does not depend on the mass of the body. Thus all objects feel the same gravitational acceleration. The force on di\ufb01erent bodies will be di\ufb01erent but the acceleration will be the same. Due to the fact that this acceleration caused by gravity is the same on all objects we label it di\ufb01erently, instead of using a we use g which we call the gravitational acceleration. 10.2 Mass and Weight Weight is a force which is measured in Newtons, it is the force of gravity on an object. People are always asking other people \\What is your weight?\" when in fact they should be asking \\What is your mass?\". Mass is measured in Kilograms (Kg) and is the amount of matter in an object, it doesn\u2019t change unless you add or remove matter from the object (if you continue to study physics through to university level, you will \ufb02nd that Einstein\u2019s theory of relativity means that mass can change when you travel as fast as light does, but you don\u2019t need to worry about that right now). There are 1000g in 1Kg and 1000Kg in a Tonne. To change mass into weight we use Newton\u2019s 2nd Law which is F = Ma. The weight is the force and gravity the acceleration, it can be rewritten as: W is the Weight, measured in Newtons. M is the Mass, measured in Kg and g is the acceleration due to gravity, measured in m=s2 it is equal to 10 on the Earth. W = mg (10.6) 10.2.1 Examples 1. A bag of sugar has a mass of 1Kg, what is it\u2019s weight? (Acceleration due to gravity = 10m=s2) - Step 1: Always write out the equation, it helps you to understand the question, and you will get marks as well. W = M g (10.7) - Step 2: Fill in all the values you know. (remember to make sure the mass is in Kg and NOT in grams or Tonnes!) W = 1 10 \u00a3 W = 10 (10.8) - Step 3: Write out the answer remembering to include the units! You will lose marks", " if you don\u2019t W = 10Newtons (10.9) 2. A space-man has a mass of 90Kg, what is his weight (a) on the earth? (b) on the moon? (c) in outer space? (The acceleration due to gravity on the earth is 10m=s2, on the moon gravity is 1/6 of the gravity on earth). (a) W = M g 10 = 900 W = 90 \u00a3 W = 900Newtons 177 (10.10) (b) W = 90 W = M g 1=6 = 150 10 \u00a3 \u00a3 W = 150Newtons (10.11) (c) Weightless in outer space because g = 0. So now when somebody asks you your weight, you know to reply \\Anything!! But my mass is a di\ufb01erent matter!\" 10.3 Normal Forces If you put a book on a table it does not accelerate it just lies on the table. We know that gravity is acting on it with a force F = G mEmbook r2 (10.12) but if there is a net force there MUST be an acceleration and there isn\u2019t. This means that the gravitational force is being balanced by another force1. This force we call the normal force. It is the reaction force between the book and the table. It is equal to the force of gravity on the book. This is also the force we measure when we measure the weight of something. The most interesting and illustrative normal force question, that is often asked, has to do with a scale in a lift. Using Newton\u2019s third law we can solve these problems quite easily. When you stand on a scale to measure your weight you are pulled down by gravity. There is no acceleration downwards because there is a reaction force we call the normal force acting upwards on you. This is the force that the scale would measure. If the gravitational force were less then the reading on the scale would be less. Worked Example 54 Normal Forces 1 Question: A man weighing 100kg stands on a scale (measuring newtons). What is the reading on the scale? Answer: Step 1 : Decide what information is supplied We are given the mass of the man. We know the gravitational acceleration that acts on him - g = 10m=s2. Step 2 : Decide what equation to use to solve the problem The scale measures the normal force on the man. This is the force that balances gravity. We can use Newton\u2019s laws to solve the problem: Fr = Fg + FN (10.13) where Fr is the resultant force on the man. 1Newton\u2019s third law! 178 Step 3 : Firstly we determine the net force acting downwards on the man due to gravity Fg = mg m s2 = 100kg 9:8 = 980 \u00a3 kgm s2 = 980N downwards Step 4 : Now determine the normal force acting upwards on the man We now know the gravitational force downwards. We know that the sum of all the forces must equal the resultant acceleration times the mass. The overall resultant acceleration of the man on the scale is 0 - so Fr = 0. Fr = Fg + FN 0 = FN = 980N upwards 980N + FN \u00a1 Step 5 : Quote the \ufb02nal answer The normal force is then 980N upwards. It exactly balances the gravitational force downwards so there is no net force and no acceleration on the man. Now we are going to add things to exactly the same problem to show how things change slightly. We will now move to a lift moving at constant velocity. Remember if velocity is constant then acceleration is zero. Worked Example 55 Normal Forces 2 s. What is the reading on the scale? Question: A man weighing 100kg stands on a scale (measuring newtons) inside a lift moving downwards at 2 m Answer: Step 1 : Decide what information is supplied We are given the mass of the man and the acceleration of the lift. We know the gravitational acceleration that acts on him. Step 2 : Decide which equation to use to solve the problem Once again we can use Newton\u2019s laws. We know that the sum of all the forces must equal the resultant acceleration times the mass (This is the resultant force, Fr). Fr = Fg + FN (10.14) 179 Step 3 : Firstly we determine the net force acting downwards on the man due to gravity Fg = mg m s2 = 100kg 9:8 = 980 \u00a3 kgm s2 = 980N downwards Step 4 : Now determine the normal force acting upwards on the man The scale measures this normal force, so once we\u2019ve determined it we will know the reading on the scale. Because the lift is moving at constant velocity the overall resultant acceleration of the man on the scale is 0. If we write out the equation: Fr = Fg + FN 0 = FN = 980N upwards 980N + FN \u00a1 Step 5 : Quote", " the \ufb02nal answer The normal force is then 980N upwards. It exactly balances the gravitational force downwards so there is no net force and no acceleration on the man. In this second example we get exactly the same result because the net acceleration on the man was zero! If the lift is accelerating downwards things are slightly di\ufb01erent and now we will get a more interesting answer! Worked Example 56 Normal Forces 3 s2. What is the reading on the scale? Question: A man weighing 100kg stands on a scale (measuring newtons) inside a lift accelerating downwards at 2 m Answer: Step 1 : Decide what information is supplied We are given the mass of the man and his resultant acceleration - this is just the acceleration of the lift. We know the gravitational acceleration that acts on him. Step 2 : Decide which equation to use to solve the problem Once again we can use Newton\u2019s laws. We know that the sum of all the forces must equal the resultant acceleration times the mass (This is the resultant force, Fr). Fr = Fg + FN (10.15) 180 Step 3 : Firstly we determine the net force acting downwards on the man due to gravity, Fg Fg = mg m s2 = 100kg 9:8 = 980 \u00a3 kgm s2 = 980N downwards Step 4 : Now determine the normal force acting upwards on the man, FN We know that the sum of all the forces must equal the resultant acceleration times the mass. The overall resultant acceleration of the man on the scale is 2 m s2 downwards. If we write out the equation: 100kg 2) ( \u00a1 \u00a3 200 \u00a1 kgm s2 Fr = Fg + FN m s2 = = \u00a1 980N + FN 980N + FN \u00a1 200N = \u00a1 FN = 780N upwards 980N + FN \u00a1 Step 5 : Quote the \ufb02nal answer The normal force is then 780N upwards. It balances the gravitational force downwards just enough so that the man only accelerates downwards at 2 m s2. Worked Example 57 Normal Forces 4 s2. What is the reading on the scale? Question: A man weighing 100kg stands on a scale (measuring newtons) inside a lift accelerating upwards at 4 m Answer: Step 1 : Decide what information is supplied We are given the mass of the man and his resultant acceleration - this is just the acceleration of the lift. We know the gravitational acceleration that acts on him. Step 2 : Decide which equation to use to solve the problem Once again we can use Newton\u2019s laws. We know that the sum of all the forces must equal the resultant acceleration times the mass (This is the resultant force, Fr). Fr = Fg + FN (10.16) 181 Step 3 : Firstly we determine the net force acting downwards on the man due to gravity, Fg Fg = mg m s2 = 100kg 9:8 = 980 \u00a3 kgm s2 = 980N downwards Step 4 : Now determine the normal force upwards, FN We know that the sum of all the forces must equal the resultant acceleration times the mass. The overall resultant acceleration of the man on the scale is 2 m s2 downwards. if we write out the equation: 100kg (4) \u00a3 Fr = Fg + FN m s2 = = \u00a1 980N + FN 980N + FN \u00a1 400 kgm s2 400N = FN = 1380N upwards 980N + FN \u00a1 Step 5 : Quote the \ufb02nal answer The normal force is then 1380N upwards. then in addition applies su\u2013cient force to accelerate the man upwards at 4 m s2. It balances the gravitational force and 10.4 Comparative problems Here always work with multiplicative factors to \ufb02nd something new in terms of something old. Worked Example 58 Comparative Problem 1 Question:On Earth a man weighs 70kg. Now if the same man was instantaneously beamed to the planet Zirgon, which has the same size as the Earth but twice the mass, what would he weigh? (NOTE TO SELF: Vanessa: isn\u2019t this confusing weight and mass?) Answer: 182 Step 1 : We start with the situation on Earth mEm r2 Step 2 : Now we consider the situation on Zirgon W = mg = G WZ = mgZ = G mZm r2 Z (10.17) (10.18) Step 3 : Relation between conditions on Earth and Zirgon but we know that mZ = 2mE and we know that rZ = r so we could write the equation again and substitute these relationships in: Step 4 : Substitute WZ = mgZ = G (2mE)m (r)2 WZ = 2(G (mE)m (r)2 ) Step 5 : Relation between weight on Zirgon and Earth Step 6 : Quote the \ufb02", "nal answer so on Zirgon he weighs 140kg. WZ = 2(W ) (10.19) (10.20) (10.21) 10.4.1 Principles Write out \ufb02rst case Write out all relationships between variable from \ufb02rst and second case Write out second case Substitute all \ufb02rst case variables into second case Write second case in terms of \ufb02rst case \u2020 \u2020 \u2020 \u2020 \u2020 Interesting Fact: The acceleration due to gravity at the Earth\u2019s surface is, by convention, equal to 9.80665 ms\u00a12. (The actual value varies slightly over the surface of the Earth). This quantity is known as g. The following is a list of the gravitational accelerations (in multiples of g) at the surfaces of each of the planets in our solar system: Mercury Venus Earth Mars Jupiter Saturn Uranus Neptune Pluto 0.376 0.903 1 0.38 2.34 1.16 1.15 1.19 0.066 183 Note: The \"surface\" is taken to mean the cloud tops of the gas giants (Jupiter, Saturn, Uranus and Neptune) in the above table. Worked Example 59 Comparative Problem 2 Question: On Earth a man weighs 70kg. On the planet Beeble how much will he weigh if Beeble has mass half of that of the Earth and a radius one quarter that of the earth. Answer: Step 1 : Start with the situation on Earth Step 2 : Now consider the situation on Beeble W = mg = G mEm r2 WB = mgB = G mBm r2 B (10.22) (10.23) Step 3 : Relation between conditions on Earth and on Beeble We know that mB = 1 again and substitute these relationships in: Step 4 : Substitute 2 mE and we know that rB = 1 4 r so we could write the equation WB = mgB = G = mgB = G (mB)m (rB)2 ( 1 2 mE)m ( 1 4 r)2 = 8(G (mE)m (r)2 ) Step 5 : Relation between weight on Earth and weight on Beeble WB = 8(W ) (10.24) Step 6 : Quote the \ufb02nal answer So the man weighs 560kg on Beeble! 184 Interesting Fact: Did you know that the largest telescope in the Southern Hemisphere is the South African Large Telescope (SALT) which came online in 2005 outside Sutherland in the Karoo. 10.5 Falling bodies Objects on the earth fall because there is a gravitation force between them and the earth - which results in an acceleration - as we saw above. So if you hold something in front of you and let it go - it will fall. It falls because of an acceleration toward the centre of the earth which results from the gravitational force between the two. These bodies move in a straight line from the point where they start to the centre for the earth. This means we can reuse everything we learnt in rectilinear motion. the only thing that needs thinking about is the directions we are talking about. We need to choose either up or down as positive just like we had to choose a positive direction in standard rectilinear motion problems. this is the hardest part. If you can do rectilinear motion you can do falling body problems. Just remember the acceleration they feel is constant and because of gravity - but once you have chosen your directions you can forget that gravity has anything to do with the problem - all you have is a rectilinear motion problem with a constant acceleration!! 10.6 Terminal velocity Physics is all about being simple - all we do is look at the world around us and notice how it really works. It is the one thing everyone is quali\ufb02ed to do - we spend most of our time when we are really young experimenting to \ufb02nd out how things work. Take a book - wave it in the air - change the angle and direction. what happens of course there is resistance. di\ufb01erent angles make it greater - the faster the book moves the greater it is. The bigger the area of the book moving in the direction of motion the greater the force. So we know that air resistance exists! it is a force. So what happens when an object falls? of course there is air resistance - or drag as it is normally called. There is an approximate formula for the drag force as well. The important thing to realise is that when the drag force and the gravitational force are equal for a falling body there is no net force acting on it - which means no net acceleration. That does not mean it does not move - but it means that its speed does not change. It falls at a constant velocity! This velocity is called terminal velocity. 10.7 Drag force The actual force of air resistance is quite complicated. Experiment", " by moving a book through the air with the face of the book and then the side of the book forward, you will agree that the area of the book makes a di\ufb01erence as to how much you must work in order to move the book at the same speed in both cases. This is why racing cars are slim-lined in design, and not shaped like a big box! 185 Get a plastic container lid (or anything waterproof) swing it around in air and then try to swing it around under water. The density of the water is much larger than the air, making you have to work harder at swinging the lid in water. This is why boats and submarines are a lot slower than aeroplanes! So we know that density, area and speed all play a role in the drag force. The expression we use for drag force is 1 2 where C is a constant which depends on the object and uid interactions, \u2030 is the density, A is the area and v is the velocity. C\u2030Av2 (10.25) D = 10.8 Important Equations and Quantities Quantity mass velocity force energy Symbol Unit | | N J m \u00a1!v \u00a1!F E Units kg m s kg:m s2 kg:m s2 2 S.I. Units or | or m:s\u00a11 or kg:m:s\u00a12 or kg:m2:s\u00a12 Direction | X X | Table 10.1: Units used in Newtonian Gravitation 186 Chapter 11 Pressure 11.1 Important Equations and Quantities Quantity Symbol Unit S.I. Units Direction Units or Table 11.1: Units used in Pressure 187 Essay 3 : Pressure and Forces Author: Asogan Moodaly Asogan Moodaly received his Bachelor of Science degree (with honours) in Mechanical Engineering from the University of Natal, Durban in South Africa. For his \ufb02nal year design project he worked on a 3-axis \ufb02lament winding machine for composite (Glass re-enforced plastic in this case) piping. He worked in Vereeniging, Gauteng at Mine Support Products (a subsidiary of Dorbyl Heavy Engineering) as the design engineer once he graduated. He currently lives in the Vaal Triangle area and is working for Sasol Technology Engineering as a mechanical engineer, ensuring the safety and integrity of equipment installed during projects. Pressure and Forces In the mining industry, the roof (hangingwall) tends to drop as the face of the tunnel (stope) is excavated for rock containing gold. As one can imagine, a roof falling on one\u2019s head is not a nice prospect! Therefore the roof needs to be supported. Roof Face The roof is not one big uniform chunk of rock. Rather it is broken up into smaller chunks. It is assumed that the biggest chunk of rock in the roof has a mass of less than 20 000 kgs therefore each support has to be designed to resist a force related to that mass. The strength of the material (either wood or steel) making up the support is taken into account when working out the minimum required size and thickness of the parts to withstand the force of the roof. Roof Face Supports 188 Sometimes the design of the support is such that the support needs to withstand the rock mass without the force breaking the roof.. Therefore hydraulic supports (hydro = water) use the principles of force and pressure such that as a force is exerted on the support, the water pressure increases. A pressure relief valve then squirts out water when the pressure (and thus the force) gets too large. Imagine a very large, modi\ufb02ed doctor\u2019s syringe. Hydraulic support Doctors syringe Force Force Seal to trap water in tube Water filled tube Pressure relief valve Rubber seal to trap medicine in syringe Pressure Pressure In the petrochemical industry, there are many vessels and pipes that are under high pressures. A vessel is a containment unit (Imagine a pot without handles, that has the lid welded to the pot that would be a small vessel) where chemicals mix and react to form other chemicals, amongst other uses. End Product Chemical The end product chemicals are sold to companies that use these chemicals to make shampoo, dishwashing liquid, plastic containers, fertilizer, etc. Anyway, some of these chemical reactions require high temperatures and pressures in order to work. These pressures result in forces being applied to the insides of the vessels and pipes. Therefore the minimum thickness of the pipe and vessels walls must be determined using calculations, to withstand these forces. These calculations take into account the strength of the material (typically steel, plastic or composite), the diameter and of course the pressure inside the equipment. Let examine the concepts of force and pressure in further detail. 189 Chapter 12 Heat and Properties of Matter 12.1 Phases of matter 12.1.1 Density Matter is a substance which has mass and occupies space. The density of matter refers to how much mass is in a", " given volume. Said di\ufb01erently, you can imagine the density to be the amount of mass packed into a given volume. density = M ass V olume If we consider a bar of soap and a bar of steel with the same volume, the steel will have more mass because it has a greater density. The density is greater in steal because more atoms are closely packed in comparison to the soap. Although they are both the same size, the bar of steel will be \"heavier\" because it has more mass. Worked Example 60 Density of objects A bar of aluminum (Al) has dimensions 2cm x 3cm x 5cm with a mass of 81g. A bar of lead (Pb) has dimensions 3cm x 3cm x 5cm and a mass of 510.3g. Calculate the density of the aluminum and lead. Solution: First we calculate the volume of Al and Pb: For Aluminum: volume = 2cm \u2044 3cm For Lead: volume = 3cm \u2044 \u2044 volume = Length W idth Height \u2044 \u2044 5cm = 30cm3 3cm 5cm = 45cm3 \u2044 We can now calculate the densities using the mass and volume of each material. 30cm3 = 2:7g=cm3 For Aluminum: density = 81g For Lead: density = 510:3g 45cm3 = 11:34g=cm3 190 2cm 5cm 3cm 5cm 3cm 3cm Now that you know the density of aluminum and lead, which object would be bigger (larger volume): 1kg of Lead or 1kg of Aluminum. Solution: 1kg of aluminum will be much larger in volume than 1kg of lead. Aluminum has a smaller density so it will take a lot more of it to have a weight of 1kg. Lead is much more dense, so it will take less for it to weigh 1kg. The density of liquids and gases can be calculated the same way as in solids. If the mass and volume of a liquid is known, the density can be calculated. We can often determine which liquid has a greater density by mixing two liquids and seeing how they settle. The more dense liquid will fall towards the bottom, or \u2019sink\u2019. If you have ever added olive oil to water, you have seen it sits on the surface, or \u2019oats\u2019. This is because olive oil is less dense than water. Fog occurs when water vapor becomes more dense than air(\"a cloud that sinks in air\"). This principle can be used with solids and liquids. In fact, it is the density of an object that determines if it will oat or sink in water. Objects with densities greater than water will sink. Worked Example 61 Objects oating in water Ivory soap is famous for \"soap that oats\". If a 5cm x 3cm x 10cm bar of ivory soap weighs 1.35 Newtons, show that its density is less than water. Solution: First calculate the bars volume: volume = 3cm Now we must determine the mass of the bar based on its weight. We will use Newton\u2019s Second law (F = ma): 10cm = 150cm3 5cm \u2044 \u2044 W eight = mass gravity = ) \u2044 W eight = 9:8m=s2 M ass \u2044 191 M ass = 1:35N 9:8m=s2 = :138kg Using the mass and the volume we determine the density of the soap: density = 138g 150cm3 = :92g=cm3 Water has a density of 1g=cm3, therefore the soap is less dense than water, allowing it to oat. 12.2 Phases of matter Although phases are conceptually simple, they are hard to de\ufb02ne precisely. A good de\ufb02nition of a phase of a system is a region in the parameter space of the system\u2019s thermodynamic variables in which the free energy is analytic. Equivalently, two states of a system are in the same phase if they can be transformed into each other without abrupt changes in any of their thermodynamic properties. All the thermodynamic properties of a system { the entropy, heat capacity, magnetization, compressibility, and so forth { may be expressed in terms of the free energy and its derivatives. For example, the entropy is simply the \ufb02rst derivative of the free energy with temperature. As long as the free energy remains analytic, all the thermodynamic properties will be well-behaved. When a system goes from one phase to another, there will generally be a stage where the free energy is non-analytic. This is known as a phase transition. Familiar examples of phase transitions are melting (solid to liquid), freezing (liquid to solid), boiling (liquid to gas), and condensation (gas to liquid). Due to this non-analyt", "icity, the free energies on either side of the transition are two di\ufb01erent functions, so one or more thermodynamic properties will behave very di\ufb01erently after the transition. The property most commonly examined in this context is the heat capacity. During a transition, the heat capacity may become in\ufb02nite, jump abruptly to a di\ufb01erent value, or exhibit a \"kink\" or discontinuity in its derivative. In practice, each type of phase is distinguished by a handful of relevant thermodynamic properties. For example, the distinguishing feature of a solid is its rigidity; unlike a liquid or a gas, a solid does not easily change its shape. Liquids are distinct from gases because they have much lower compressibility: a gas in a large container \ufb02lls the container, whereas a liquid forms a puddle in the bottom. Many of the properties of solids, liquids, and gases are not distinct; for instance, it is not useful to compare their magnetic properties. On the other hand, the ferromagnetic phase of a magnetic material is distinguished from the paramagnetic phase by the presence of bulk magnetization without an applied magnetic \ufb02eld. To take another example, many substances can exist in a variety of solid phases each corresponding to a unique crystal structure. These varying crystal phases of the same substance are called polymorphs. Diamond and graphite are examples of polymorphs of carbon. Graphite is composed of layers of hexagonally arranged carbon atoms, in which each carbon atom is strongly bound to three neighboring atoms in the same layer and is weakly bound to atoms in the neighboring layers. By contrast in diamond each carbon atom is strongly bound to four neighboring carbon atoms in a cubic array. The unique crystal structures of graphite and diamond are responsible for the vastly di\ufb01erent properties of these two materials. Metastable phases 192 Metastable states may sometimes be considered as phases, although strictly speaking they aren\u2019t because they are unstable. For example, each polymorph of a given substance is usually only stable over a speci\ufb02c range of conditions. For example, diamond is only stable at extremely high pressures. Graphite is the stable form of carbon at normal atmospheric pressures. Although diamond is not stable at atmospheric pressures and should transform to graphite, we know that diamonds exist at these pressures. This is because at normal temperatures the transformation If we were to heat the diamond, the rate of from diamond to graphite is extremely slow. transformation would increase and the diamond would become graphite. However, at normal temperatures the diamond can persist for a very long time. Another important example of metastable polymorphs occurs in the processing of steel. Steels are often subjected to a variety of thermal treatments designed to produce various combinations of stable and metastable iron phases. In this way the steel properties, such as hardness and strength can be adjusted by controlling the relative amounts and crystal sizes of the various phases that form. Phase diagrams The di\ufb01erent phases of a system may be represented using a phase diagram. The axes of the diagrams are the relevant thermodynamic variables. For simple mechanical systems, we generally use the pressure and temperature. The following \ufb02gure shows a phase diagram for a typical material exhibiting solid, liquid and gaseous phases. The markings on the phase diagram show the points where the free energy is non-analytic. The open spaces, where the free energy is analytic, correspond to the phases. The phases are separated by lines of non-analyticity, where phase transitions occur, which are called phase boundaries. In the above diagram, the phase boundary between liquid and gas does not continue indefinitely. Instead, it terminates at a point on the phase diagram called the critical point. This reects the fact that, at extremely high temperatures and pressures, the liquid and gaseous phases become indistinguishable. In water, the critical point occurs at around 647 K (374 C or 705 F) and 22.064 MPa. The existence of the liquid-gas critical point reveals a slight ambiguity in our above de\ufb02nitions. When going from the liquid to the gaseous phase, one usually crosses the phase boundary, but it is possible to choose a path that never crosses the boundary by going to the right of the critical point. Thus, phases can sometimes blend continuously into each other. We should note, however, that this does not always happen. For example, it is impossible for the solid-liquid phase boundary to end in a critical point in the same way as the liquid-gas boundary, because the solid and liquid phases have di\ufb01erent symmetry. An interesting thing to note is that the solid-liquid phase boundary in the phase diagram of most substances, such as the one shown above, has a positive slope. This is due to the solid phase having a higher density than the liquid, so that increasing the pressure increases the melting temperature. However, in the phase diagram for", " water the solid-liquid phase boundary has a negative slope. This reects the fact that ice has a lower density than water, which is an unusual property for a material. 193 12.2.1 Solids, liquids, gasses 12.2.2 Pressure in uids 12.2.3 change of phase 12.3 Deformation of solids 12.3.1 strain, stress Stress () and strain (\u2020) is one of the most fundamental concepts used in the mechanics of materials. The concept can be easily illustrated by considering a solid, straight bar with a constant cross section throughout its length where a force is distributed evenly at the ends of the bar. This force puts a stress upon the bar. Like pressure, the stress is the force per unit area. In this case the area is the cross sectional area of the bar. stress = F orce Areacrosssection = ) = F A (A) Bar under compression (B) Bar under tension Figure 12.1: Illustration of Bar The bar in \ufb02gure 1a is said to be under compression. If the direction of the force (\u00a1!F ), were reversed, stretching the bar, it would be under tension (\ufb02g. 1b). Using intuition, you can imagine how the bar might change in shape under compression and tension. Under a compressive load, the bar will shorten and thicken. In contrast, a tensile load will lengthen the bar and make it thinner. Figure 12.2: Bar changes length under tensile stress For a bar with an original length L, the addition of a stress will result in change of length L and L we can now de\ufb02ne strain as the ratio between the two. That is, strain is L. With 4 de\ufb02ned as the fractional change in length of the bar: 4 Strain L 4 L \u00b7 12.3.2 Elastic and plastic behavior Material properties are often characterized by a stress versus strain graph (\ufb02gure x.xx). One way in which these graphs can be determined is by tensile testing. In this process, a machine 194 L L 4 Figure 12.3: Left end of bar is \ufb02xed as length changes | | + Figure 12.4: dashed line represents plastic recovery **incomplete** stretches a the material by constant amounts and the corresponding stress is measured and plotted. Typical solid metal bars will show a result like that of \ufb02gure x.xx. This is called a Type II response. Other materials may exibit di\ufb01erent responses. We will only concern ourself with Type II materials. The linear region of the graph is called the elastic region. By obtaining the slope of the linear region, it is easy to \ufb02nd the strain for a given stress, or vice-versa. This slope shows itself to be very useful in characterizing materials, so it is called the Modulus of Elasticity, or Young\u2019s Modulus: E = stress strain = F=A \u00a2L=L The elastic region has the unique property that allows the material to return to its original shape when the stress is removed. As the stress is removed it will follow line back to zero. One may think of stretching a spring and then letting it return to its original length. When a stress is applied in the linear region, the material is said to undergo elastic deformation. When a stress is applied that is in the non-linear region, the material will no longer return to its original shape. This is referred to as plastic deformation. If you have overstretched a spring you have seen that it no longer returns to its initial length; it has been plastically deformed. The stress where plastic behavior begins is called the yield strength (point A, \ufb02g x). When a material has plastically deformed it will still recover some of its shape (like an overstretched spring). When a stress in the non-linear region is removed, the stress strain graph will follow a line with a slope equal to the modulus of elasticity (see the dashed line in \ufb02gure x.xx). The plastically deformed material will now have a linear region that follows the dashed line. Greater stresses in the plastic region will eventually lead to fracture (the material breaks). The maximum stress the material can undergo before fracture is the ultimate strength. 195 | | + Figure 12.5: dashed line represents plastic recovery **incomplete** 12.4 Ideal gasses Author: G\u00b6erald Wigger G\u00b6erald Wigger started his Physics studies at ETH in Zuerich, Switzerland. He moved to Cape Town, South Africa, for his Bachelor of Science degree (with honours) in Physics from the University of Cape Town in 1998. Returned to Switzerland, he \ufb02nished his Diploma at ETH in 2000 and followed up with a PhD in the Solid State Physics group of Prof. Hans", "-Ruedi Ott at ETH. He graduated in the year 2004. Being awarded a Swiss fellowship, he moved to Stanford University where he is currently continuing his Physics research in the \ufb02eld of Materials with novel electronic properties. Any liquid or solid material, heated up above its boiling point, undergoes a transition into a gaseous state. For some materials such as aluminium, one has to heat up to three thousand degrees Celsius (\u2013C), whereas Helium is a gas already at -269 \u2013C. For more examples see Table 12.1. As we \ufb02nd very strong bonding between the atoms in a solid material, a gas consists of molecules which do interact very poorly. If one forgets about any electrostatic or intermolecular attractive forces between the molecules, one can assume that all collisions are perfectly elastic. One can visualize the gas as a collection of perfectly hard spheres which collide but which otherwise do not interact with each other. In such a gas, all the internal energy is in the form of kinetic energy and any change in internal energy is accompanied by a change in temperature. Such a gas is called an ideal gas. In order for a gas to be described as an ideal gas, the temperature should be raised far enough above the melting point. A few examples of ideal gases at room temperature are Helium, Argon and hydrogen. Despite the fact that there are only a few gases which can be accurately described as an ideal gas, the underlying theory is widely used in Physics because of its beauty and simplicity. A thermodynamic system may have a certain substance or material whose quantity can be expressed in mass or mols in an overall volume. These are extensive properties of the system. In the following we will be considering often intensive versus extensive quantities. A material\u2019s intensive property, is a quantity which does not depend on the size of the material, such as temperature, pressure or density. Extensive properties like volume, mass or number of atoms on 196 Material Aluminium Water Ethyl alcohol Methyl ether Nitrogen Helium Temperature in Celsius Temperature in Kelvin 2467 \u2013C 100 \u2013C 78.5 \u2013C -25 \u2013C -195.8 \u2013C -268.9 \u2013C 2740 K 373.15 K 351.6 K 248 K 77.3 K 4.2 K Table 12.1: Boiling points for various materials in degrees Celsius and in Kelvin quantity pressure p volume V unit Pa m3 molar volume vmol m3/mol temperature T mass M density \u2030 internal energy E K kg kg/m3 J intensive or extensive intensive extensive intensive intensive extensive intensive extensive Table 12.2: Intensive versus extensive properties of matter the other hand gets bigger the bigger the material is (see Table 12.2 for various intensive/extensive properties). If the substance is evenly distributed throughout the volume in question, then a value of volume per amount of substance may be used as an intensive property. For an example, for an amount called a mol, volume per mol is typically called molar volume. Also, a volume per mass for a speci\ufb02c substance may be called speci\ufb02c volume. In the case of an ideas gas, a simple equation of state relates the three intensive properties, temperature, pressure, and molar or speci\ufb02c volume. Hence, for a closed system containing an ideal gas, the state can be speci\ufb02ed by giving the values of any two of pressure, temperature, and molar volume. 12.4.1 Equation of state The ideal gas can be described with a single equation. However, in order to arrive there, we will be introducing three di\ufb01erent equations of state, which lead to the ideal gas law. The combination of these three laws leads to a complete picture of the ideal gas. 1661 - Robert Boyle used a U-tube and Mercury to develop a mathematical relationship between pressure and volume. To a good approximation, the pressure and volume of a \ufb02xed amount of gas at a constant temperature were related by V = constant p \u00a2 p V : pressure (P a) : Volume (m3) In other words, if we compress a given quantity of gas, the pressure will increase. And if we put it under pressure, the volume of the gas will decrease proportionally. 197 Figure 12.6: Pressure-Volume diagram for the ideal gas at constant temperature. Worked Example 62 compressed Helium gas A sample of Helium gas at 25\u2013C is compressed from 200 cm3 to 0.240 cm3. pressure is now 3.00 cm Hg. What was the original pressure of the Helium? Solution: It\u2019s always a good idea to write down the values of all known variables, indicating whether the values are for initial or \ufb02nal states. Boyle\u2019s Law problems are essentially special cases of the Ideal Gas Law: Initial: p1 =?; V1 = 200 cm3; Final", ": p2 = 3.00 cm Hg; V2 = 0.240 cm3; Since the number of molecules stays constant and the temperature is not changed along the process, so Its V1 = p2 p1 \u00a2 V2 \u00a2 hence p1 = p2 \u00a2 V2=V1 = 3:00cmHg 0:240cm3=200cm3 \u00a2 Setting in the values yields p1 = 3.60 Did you notice that the units for the pressure are in cm Hg? You may wish to convert this to a more common unit, such as millimeters of mercury, atmospheres, or pascals. 3.60 10mm/1 cm = 3.60 10\u00a12 mm Hg 10\u00a13 Hg \u00a2 10\u00a13 cm Hg. \u00a2 \u00a2 \u00a2 198 3.60 \u00a2 10\u00a13 Hg \u00a2 1 atm/76.0 cm Hg = 4.74 10\u00a15 atm \u00a2 One way to experience this is to dive under water. There is air in your middle ear, which is normally at one atmosphere of pressure to balance the air outside your ear drum. The water will put pressure on the ear drum, thereby compressing the air in your middle ear. Divers must push air into the ear through their Eustacean tubes to equalize this pressure. Worked Example 63 pressure in the ear of a diver How deep would you have to dive before the air in your middle ear would be compressed to 75% of its initial volume? Assume for the beginning that the temperature of the sea is constant as you dive. Solution: First we write down the pressure as a function of height h: where we take for p0 the atmospheric pressure at height h = 0, \u2030 is the density of water at 20 degrees Celsius 998.23 kg/m3, g = 9.81 ms\u00a12. p = p0 + \u2030 h g \u00a2 \u00a2 As the temperature is constant, it holds for both heights h Now solving for h using the fact that V0 = (p0 + \u2030gh) p0 \u00a2 Ve \u00a2 yields Ve=V0 = 0:75 h = (0:75 p0 \u2044 \u00a1 p0)=(\u2030g) Now, how far can the diver dive down before the membranes of his ear brake. Solution: As the result is negative, h determines the way he can dive down. h is given as roughly 2.6 m. In 1809, the French chemist Joseph-Louis Gay-Lussac investigated the relationship between the Pressure of a gas and its temperature. Keeping a constant volume, the pressure of a gas sample is directly proportional to the temperature. Attention, the temperature is measured in Kelvin! The mathematical statement is as follows: 199 p1=T1 = p2=T2 = constant p1;2 T1;2 : pressures (P a) : Temperatures (K) That means, that pressure divided by temperature is a constant. On the other hand, if we plot pressure versus temperature, the graph crosses 0 pressure for T = 0 K = -273.15 \u2013C as shown in the following \ufb02gure. That point is called the absolute Zero. That is where any motion of molecules, electrons or other particles stops. Figure 12.7: Pressure-temperature diagram for the ideal gas at constant volume. Worked Example 64 Gay-Lussac Suppose we have the following problem: A gas cylinder containing explosive hydrogen gas has a pressure of 50 atm at a temperature of 300 K. The cylinder can withstand a pressure of 500 atm before it bursts, causing a building-attening explosion. What is the maximum temperature the cylinder can withstand before bursting? Solution: Let\u2019s rewrite this, identifying the variables: A gas cylinder containing explosive hydrogen gas has a pressure of 50 atm (p1) at a temperature of 300 K (T1). The cylinder can withstand a pressure of 500 atm 200 (p2) before it bursts, causing a building-attening explosion. What is the maximum temperature the cylinder can withstand before bursting? Plugging in the known variables into the expression for the Gay-Lussac law yields we \ufb02nd the answer to be 3000 K. T2 = p2=p1 T1 = 500atm=50atm \u2044 300K = 3000K \u2044 The law of combining volumes was interpreted by the Italian chemist Amedeo Avogadro in 1811, using what was then known as the Avogadro hypothesis. We would now properly refer to it as Avogadro\u2019s law: Equal volumes of gases under the same conditions of temperature and pressure contain equal numbers of molecules. This can be understood in the following. As in an ideal gas, all molecules are considered to be tiny particles with no spatial extension", " which collide elastically with each other. So, the kind of gas is irrelevant. Avogadro found that at room temperature, in atmospheric pressure the 1023 molecules or atoms, occupies the volume of 22.4 volume of a mol of a substance, i.e. 6.022 \u00a2 l. Figure 12.8: Two di\ufb01erent gases occupying the same volume under the same circumstances. Combination of the three empirical gas laws, described in the preceding three sections leads to the Ideal Gas Law which is usually written as: p \u00a2 V = n R T \u00a2 \u00a2 : pressure (P a) : Volume (m3) : number of mols (mol) p V n R : gas konstant (J=molK) : temperature (K) T 201 where p = pressure, V = volume, n = number of mols, T = kelvin temperature and R the ideal gas constant. The ideal gas constant R in this equation is known as the universal gas constant. It arises from a combination of the proportionality constants in the three empirical gas laws. The universal gas constant has a value which depends only upon the units in which the pressure and volume are measured. The best available value of the universal gas constant is: 8.3143510 Another value which is sometimes convenient is 0.08206 dm3 atm/mol K. R is related to the molK or 8.3143510 kP adm molK J 3 Boltzmann-constant as: R = N0 kB \u00a2 where N0 is the number of molecules in a mol of a substance, i.e. 6.022 \u00a2 1.308 10\u00a123 J/K is valid for one single particle. \u00a2 This ideal gas equation is one of the most used equations in daily life, which we show in the (12.1) 1023 and kB is following problem set: Worked Example 65 ideal gas 1 A sample of 1.00 mol of oxygen at 50 \u2013C and 98.6 kPa occupies what volume? Solution: We solve the ideal gas equation for the volume V = nRT p and plug in the values n = 1, T = 273.15 + 50 K = 323.15 K and p = 98.6 \u00a2 yielding for the volume V = 0.0272 m3 = 27.2 dm3. 103 Pa, This equation is often used to determine the molecular masses from gas data. Worked Example 66 ideal gas 2 A liquid can be decomposed by electricity into two gases. In one experiment, one of the gases was collected. The sample had a mass of 1.090 g, a volume of 850 ml, a pressure of 746 torr, and a temperature of 25 \u2013C. Calculate its molecular mass. Solution: To calculate the molecular mass we need the number of grams and the number of mols. We can get the number of grams directly from the information in the question. We can calculate the mols from the rest of the information and the ideal gas equation. 202 V = 850mL = 0:850L = 0:850dm3 P = 746torr=760torr = 0:982atm T = 25:0\u2013C + 273:15 = 298:15K pV = nRT (0:982atm)(0:850L) = (n)(0:0821Latmmol 1K \u00a1 \u00a1 1)(298:15K) n = 0:0341mol molecular mass = g/mol = 1.090 g/ 0.0341 mol = 31.96 g/mol. The gas is oxygen. Or the equation can be comfortably used to design a gas temperature controller: Worked Example 67 ideal gas 3 In a gas thermometer, the pressure needed to \ufb02x the volume of 0.20 g of Helium at 0.50 L is 113.3 kPa. What is the temperature? Solution: We transform \ufb02rst need to \ufb02nd the number of mols for Helium. Helium consists of 2 protons and 2 neutrons in the core (see later) and therefore has a molar volume of 4 g/mol. Therefore, we \ufb02nd plugging this into the ideal gas equation and solving for the temperature T we \ufb02nd: n = 0:20g=4g=mol = 0:05mol T = pV nR = 113:3 103P a 0:5 10\u00a13m3 \u00a2 0:05mol \u00a2 \u00a2 8:314J=molK = 136:3K \u00a2 The temperature is 136 Kelvin. 12.4.2 Kinetic theory of gasses The results of several experiments can lead to a scienti\ufb02c law, which describes then all experiments performed. This is an empirical, that is based on experience only, approach to", " Physics. A law, however, only describes results; it does not explain why they have been obtained. Significantly stronger, a theory is a formulation which explains the results of experiments. A theory usually bases on postulates, that is a proposition that is accepted as true in order to provide a basis for logical reasoning. The most famous postulate in Physics is probably the one formulated by Walter Nernst which states that if one could reach absolute zero, all bodies would have the same entropy. 203 The kinetic-molecular theory of gases is a theory of great explanatory power. We shall see how it explains the ideal gas law, which includes the laws of Boyle and of Charles; Dalton\u2019s law of partial pressures; and the law of combining volumes. The kinetic-molecular theory of gases can be stated as four postulates: \u2020 \u2020 \u2020 \u2020 A gas consists of particles (atoms or molecules) in continuous, random motion. Gas molecules inuence each other only by collision; they exert no other forces on each other. All collisions between gas molecules are perfectly elastic; all kinetic energy is conserved. The average energy of translational motion of a gas particle is directly proportional to temperature. In addition to the postulates above, it is assumed that the volumes of the particles are negligible as compared to container volume. These postulates, which correspond to a physical model of a gas much like a group of billiard balls moving around on a billiard table, describe the behavior of an ideal gas. At room temperatures and pressures at or below normal atmospheric pressure, real gases seem to be accurately described by these postulates, and the consequences of this model correspond to the empirical gas laws in a quantitative way. We de\ufb02ne the average kinetic energy of translation Et of a particle in a gas as Et = 1=2 mv2 (12.2) \u00a2 where m is the mass of the particle with average velocity v. The forth postulate states that the average kinetic energy is a constant de\ufb02ning the temperature, i.e. we can formulate Et = 1=2 \u00a2 mv2 = c T \u00a2 (12.3) where the temperature T is given in Kelvin and c is a constant, which has the same value for all gases. As we have 3 di\ufb01erent directions of motion and each possible movement gives kBT, we \ufb02nd for the energy of a particle in a gas as Et = 1=2 \u00a2 mv2 = 3=2kBT = 3=2 R NA T (12.4) Hence, we can \ufb02nd an individual gas particle\u2019s speed rms = root mean square, which is the average square root of the speed of the individual particles (\ufb02nd u) r where Mmol is the molar mass, i.e. the mass of the particle m times the Avogadro number vrms = 3RT Mmol (12.5) NA. Worked Example 68 kinetic theory 1 204 Calculate the root-mean-square velocity of oxygen molecules at room temperature, 25 \u2013C. Solution: Using vrms = 3RT =Mmol ; the molar mass of molecular oxygen is 31.9998 g/mol; the molar gas constant has the value 8.3143 J/mol K, and the temperature is 298.15 K. Since the joule is the s\u00a12, the molar mass must be expressed as 0.0319998 kg/mol. The root-meankg \u00a2 square velocity is then given by: m2 \u00a2 p vrms = 3(8:3143)(298:15)=(0:0319998) = 482:1m=s A speed of 482.1 m/s is 1726 km/h, much faster than a jetliner can y and faster than most rie bullets. p The very high speed of gas molecules under normal room conditions would indicate that a gas molecule would travel across a room almost instantly. In fact, gas molecules do not do so. If a small sample of the very odorous (and poisonous!) gas hydrogen sul\ufb02de is released in one corner of a room, our noses will not detect it in another corner of the room for several minutes unless the air is vigorously stirred by a mechanical fan. The slow di\ufb01usion of gas molecules which are moving very quickly occurs because the gas molecules travel only short distances in straight lines before they are deected in a new direction by collision with other gas molecules. The distance any single molecule travels between collisions will vary from very short to very long distances, but the average distance that a molecule travels between collisions in a gas can be calculated. This distance is called the mean free path l of the gas molecules. If the rootmean-square velocity is divided by the mean free path of the gas molecules, the result will be the number of collisions", " one molecule undergoes per second. This number is called the collision frequency Z1 of the gas molecules. The postulates of the kinetic-molecular theory of gases permit the calculation of the mean free path of gas molecules. The gas molecules are visualized as small hard spheres. A sphere (d=2)2 and length vrms each of diameter d sweeps through a cylinder of cross-sectional area \u2026 second, colliding with all molecules in the cylinder. \u00a2 The radius of the end of the cylinder is d because two molecules will collide if their diameters overlap at all. This description of collisions with stationary gas molecules is not quite accurate, however, because the gas molecules are all moving relative to each other. Those relative velocities range between zero for two molecules moving in the same direction and 2vrms for a head-on collision. The average relative velocity is that of a collision at right angles, which is p2vrms. The total number of collisions per second per unit volume, Z1, is Z1 = \u2026d2p2vrms (12.6) This total number of collisions must now be divided by the number of molecules which are present per unit volume. The number of gas molecules present per unit volume is found by rearrangement of the ideal gas law to n=V = p=RT and use of Avogadro\u2019s number, n = N=NA; thus N=V = pNA=RT. This gives the mean free path of the gas molecules, l, as (urms=Z1)=(N=V ) = l = RT =\u2026d2pNAp2 (12.7) 205 According to this expression, the mean free path of the molecules should get longer as the temperature increases; as the pressure decreases; and as the size of the molecules decreases. Worked Example 69 mean free path Calculate the length of the mean free path of oxygen molecules at room temperature, 25 \u2013C, taking the molecular diameter of an oxygen molecule as 370 pm. Solution: Using the formula for mean free path given above and the value of the root-meansquare velocity urms, l = \u2026(370 \u00a2 (8:3143kgm2s\u00a12=Kmol)(298:15K) 10\u00a112m)2(101325kg=ms2)(6:0225 1023mol\u00a11)p2 \u00a2 ; 10\u00a18 m = 67 nm. so l = 6.7 \u00a2 The apparently slow di\ufb01usion of gas molecules takes place because the molecules travel only a very short distance before colliding. At room temperature and atmospheric pressure, oxygen 10\u00a112 m) = 180 molecular diameters between collisions. 10\u00a18 m)/(370 molecules travel only (6.7 \u00a2 \u00a2 The same thing can be pointed out using the collision frequency for a single molecule Z1, which is the root-mean-square velocity divided by the mean free path: Z1 = \u2026d2pNAp2 =RT = vrms=l (12.8) For oxygen at room temperature, each gas molecule collides with another every 0.13 nanosec10+9 collisions per second 10\u00a19 s), since the collision frequency is 7.2 onds (one nanosecond is 1.0 \u00a2 \u00a2 per molecule. For an ideal gas, the number of molecules per unit volume is given using pV = nRT and n = N=NA as N=V = NAp=RT (12.9) which for oxygen at 25 \u2013C would be (6.022 1023 mol\u00a11)(101325 kg/m s2) / (8.3143 kg m2/s2 \u00a2 1025 molecules/m3. The number of collisions between two molecules K mol)(298.15 K) or 2.46 \u00a2 in a volume, Z11, would then be the product of the number of collisions each molecule makes times the number of molecules there are, Z1N=V, except that this would count each collision twice (since two molecules are involved in each one collision). The correct equation must be Z11 = \u2026d2p2N 2 Ap2vrms 2R2T 2 (12.10) If the molecules present in the gas had di\ufb01erent masses they would also have di\ufb01erent speeds, so an average value of vrms would be using a weighted average of the molar masses; the partial pressures of the di\ufb01erent gases in the mixture would also be required. Although such calculations involve no new principles, they are beyond our scope. 206 12.4.3 Pressure of a gas In the kinetic-molecular theory of gases, pressure is the force exerted against the wall of a container by the continual collision of molecules against it. From Newton\u2019s second law of motion", ", the force exerted on a wall by a single gas molecule of mass m and velocity v colliding with it is: F = m a = m \u00a2 \u00a2v \u00a2t (12.11) In the above equation, the change in a quantity is indicated by the symbol \u00a2, that means by changing the time t by a fraction, we change the velocity v by some other minimal amount. It is assumed that the molecule rebounds elastically and no kinetic energy is lost in a perpendicular collision, so \u00a2v = v - (-v) = 2v (see \ufb02gure below). If the molecule is moving perpendicular to the wall it will strike the opposite parallel wall, rebound, and return to strike the original wall again. If the length of the container or distance between the two walls is the path length l, then the time between two successive collisions on the same wall is \u00a2t = 2l/v. The continuous force which the molecule moving perpendicular to the wall exerts is therefore Figure 12.9: Change in momentum as a particle hits a wall. F = m 2v 2l=v = mv2 l (12.12) The molecules in a sample of gas are not, of course, all moving perpendicularly to a wall, but the components of their actual movement can be considered to be along the three mutually perpendicular x, y, and z axes. If the number of molecules moving randomly, N, is large, then on the average one-third of them can be considered as exerting their force along each of the three perpendicular axes. The square of the average velocity along each axis, v2(x), v2(y), or v2(z), will be one-third of the square of the average total velocity v2: v2(x) = v2(y) = v2(z) = v2=3 (12.13) The average or mean of the square of the total velocity can replace the square of the perpen- dicular velocity, and so for a large number of molecules N, Since pressure is force per unit area, and the area of one side of a cubic container must be l2, the pressure p will be given by F=l2 as: F = (N=3) mv2 l (12.14) 207 This equation rearranges to p = (N=3) mv2 l3 (12.15) pV = N mv2=3 (12.16) \u00a2 because volume V is the cube of the length l. The form of the ideal gas law given above shows the pressure-volume product is directly proportional to the mean-square velocity of the gas molecules. If the velocity of the molecules is a function only of the temperature, and we shall see in the next section that this is so, the kinetic-molecular theory gives a quantitative explanation of Boyle\u2019s law. Worked Example 70 gas pressure A square box contains He (Helium) at 25 \u2013C. If the atoms are colliding with the walls 1022 times per second, calculate the force perpendicularly (at 90\u2013) at the rate of 4.0 \u00a2 (in Newtons) and the pressure exerted on the wall per mol of He given that the area of the wall is 100 cm2 and the speed of the atoms is 600 ms\u00a11. Solution: We use the equation 12.14 to calculate the force. mv2 l The fraction v=l is the collision frequency Z1 = 0.6679 s\u00a11. The product of N Z1 is the number of molecules impinging on the wall per second. This induces for the force: = (N=3)mv F = (N=3) v l \u00a2 F = (N=3)mv\u00bf = 6:022 1023=3 \u00a2 \u00a2 0:004g=mol 6:022 1023 \u00a2 \u00a2 600m=s \u00a2 0:6679s\u00a11 yielding for the force F = 0.534 N. The pressure is the force per area: p = F=A = 0:534N=0:01m2 = 53:4P a: The calculated force is 0.534 N and the resulting pressure is 53.4 Pa. 12.4.4 Kinetic energy of molecules In the following, we will make the connection between the kinetic theory and the ideal gas laws. We will \ufb02nd that the temperature is an important quantity which is the only intrinsic parameter entering in the kinetic energy of a gas. We will consider an ensemble of molecules in a gas, where the molecules will be regarded as rigid large particles. We therefore neglect any vibrations or rotations in the molecule. Hence, making this assumption, Physics for a molecular gas is the same as for a single atom gas. 208 The square of the velocity is sometimes di\u2013cult to conceive, but", " an alternative statement can be given in terms of kinetic energy. The kinetic energy Ek of a single particle of mass m moving at velocity v is mv2=2. For a large number of molecules N, the total kinetic energy Ek will depend on the mean-square velocity in the same way: mv2=2 = n The second form is on a molar basis, since n = N=NA and the molar mass M = mNA where M v2=2 Ek = N (12.17) \u00a2 \u00a2 NA is Avogadro\u2019s number 6.022 1023. The ideal gas law then appears in the form: \u00a2 (12.18) Compare pV = nM v2=2. This statement that the pressure-volume product of an ideal gas is directly proportional to the total kinetic energy of the gas is also a statement of Boyle\u2019s law, since the total kinetic energy of an ideal gas depends only upon the temperature. pV = 2Ek=3 Comparison of the ideal gas law, pV = nRT, with the kinetic-molecular theory expression pV = 2Ek=3 derived in the previous section shows that the total kinetic energy of a collection of gas molecules is directly proportional to the absolute temperature of the gas. Equating the pV term of both equations gives which rearranges to an explicit expression for temperature, Ek = 3=2nRT ; T = 2 3R Ek n = M v2 3R (12.19) (12.20) We see that temperature is a function only of the mean kinetic energy Ek, the mean molecular velocity v, and the mean molar mass M. Worked Example 71 mean velocity 1 Calculate the kinetic energy of 1 mol of nitrogen molecules at 300 K? Solution: Assume nitrogen behaves as an ideal gas, then Ek = 3=2 \u00a2 RT = (3=2)8:3145J=(molK) \u00a2 300K = 3742J=mol(or3:74kJ=mol) At 300 K, any gas that behaves like an ideal gas has the same energy per mol. As the absolute temperature decreases, the kinetic energy must decrease and thus the mean velocity of the molecules must decrease also. At T = 0, the absolute zero of temperature, all motion of gas molecules would cease and the pressure would then also be zero. No molecules would be moving. Experimentally, the absolute zero of temperature has never been attained, although modern experiments have extended to temperatures as low as 1 \u201eK. However, at low temperatures, the interactions between the particles becomes important and we enter a new regime of Quantum Mechanics, which considers molecules, single atoms or protons and electrons simultaneously as waves and as rigid particles. However, this would go too far. 209 Worked Example 72 mean velocity 2 If the translational rms. speed of the water vapor molecules (H2O) in air is 648 m/s, what is the translational rms speed of the carbon dioxide molecules (CO2) in the same air? Both gases are at the same temperature. And what is the temperature we measure? Solution: The molar mass of H2O is \u00a2 As the temperature is constant we can write \u00a2 MH2O = 2 1g=mol + 1 16g=mol = 18g=mol T = M v2 3R = 0:018kg=mol (648m=s)2 K \u00a2 \u00a2 3 8:314J=mol \u00a2 Now we calculate the molar mass of CO2 = 303:0K = 29:9\u2013C MCO2 = 2 \u00a2 16g=mol + 1 \u00a2 12g=mol = 44g=mol The rms velocity is again calculated with eq. 12.20 vCO2 = 3 R T \u00a2 MCO2 s 3 \u00a2 = s 8:314J=molK 303:0K 0:044kg=mol \u00a2 = 414:5m=s The experiment was performed at 29.9 \u2013C and the speed of the CO2-molecules is 414.5 m/s, that is much slower than the water molecules as they are much heavier. 12.5 Temperature Let us look back to the equation for the temperature of an ideal gas, T = 2 3R We can see that temperature is proportional to the average kinetic energy of a molecule in the gas. In other words temperature is a measure of how much energy is contained in an object { in hot things the atoms have a lot of kinetic energy, in cold things they have less. It may be surprising that \u2018hot\u2019 and \u2018cold\u2019 are really just words for how fast molecules or atoms are moving around, but it is true. (12.21) Ek n De\ufb02nition: Temperature is a measure of the average kinetic energy of the particles in a body. It should", " now be clear that heat is nothing more than energy on the move. It can be carried 210 ice water Figure 12.10: A heat ow diagram showing the heat owing from the warmer water into the cooler ice cube. by atoms, molecules or electromagnetic radiation but it is always just transport of energy. This is very important when we describe movement of heat as we will do in the following sections. \u2018Cold\u2019 is not a physical thing. It does not move from place to place, it is just the word for a lack of heat, just like dark is the word for an absence of light. 12.5.1 Thermal equilibrium Now that we have de\ufb02ned the temperature of an isolated object (usually referred to as a body) we need to consider how heat will move between bodies at di\ufb01erent temperatures. Let us take two bodies; A which has a \ufb02xed temperature and B whose temperature is allowed to change. If we allow heat to move between the two bodies we say they are in thermal contact. First let us consider what happens if B is cooler than A. Remember { we have \ufb02xed the temperture of A so we need only worry about the temperature of B changing. An example of such a situation is an ice cube being dropped into a large pan of boiling water on a \ufb02re. The water temperature is \ufb02xed i.e. does not change, because the \ufb02re keeps it constant. It should be obvious that the ice cube will heat up and melt. In physical terms we say that the heat is owing out of the (warmer) boiling water, into the (cooler) ice cube. This ow of heat into the ice cube causes it to warm up and melt. In fact the temperature of any cooler object in thermal contact with a warmer one will increase as heat from the warmer object ows into it. The reverse would be true if B were warmer than A. We can now picture putting a small amount of warm water in to a freezer. If we come back in an hour or so the water will have cooled down and possibly frozen. In physical terms we say that the heat is owing out of the (warmer) water, into the (cooler) air in the freezer. This ow of heat out of the ice cube in to the aircauses it to cool down and (eventually) freeze. Again, any warm object in thermal contact with a cooler one will cool down due to heat owing out of it. There is one special case which we have not yet discussed { what happens if A and B are at the same temperature? In this case B will neither warm up nor cool down, in fact, its temperature will remain constant. When two bodies are at the same temperature we say that they are in thermal equilibrium. Another way to express this is to say that two bodies are in thermal equilibrium if the particles within those bodies have the same average kinetic energies. 211 water air Figure 12.11: A heat ow diagram showing the heat owing from the warmer water into the cooler air in the fridge. Convince yourself that the last three paragraghs are correct before you continue. You should notice that heat always ows from the warmer object to the cooler object, never the other way around. Also, we never talk about coldness moving as it is not a real physical thing, only a lack of heat. Most importantly, it should be clear that the ow of heat between the two objects always attempts to bring them to the same temperature (or in other words, into thermal equilibrium). The logical conclusion of all this is that if two bodies are in thermal contact heat will ow from the hotter object to the cooler one until they are in thermal equilibrium (i.e. at the same temperature). We will see how to deal with this if the temperature of object A is not \ufb02xed in the section on heat capacities. 12.5.2 Temperature scales Temperature scales are often confusing and even university level students can be tricked into using the wrong one. For most purposes in physics we do not use the familiar celcius (often innaccurately called centigrade) scale but the closely related absolute (or kelvin) scale { why? Let us think about the celcius scale now that we have de\ufb02ned temperature as a measure of the average kinteic energy of the atoms or molecules in a body. A scale is a way of assigning a number to a physical quantity. Consider distance { using a ruler we can measure a distance and \ufb02nd its legnth. This legnth could be measured in metres, inches, or miles. The same is true of temperatures in that many di\ufb01erent scales exist to measure them. Table 12.3 shows a few of these scales. Just like a ruler the scales have two de\ufb02ned points which \ufb02x the scale (consider the values at the beginning and end of", " the ruler e.g. 0cm and 15cm). This is usually achieved by de\ufb02ning the temperature of some physical process, e.g. the freezing point of water. Armed with our knowledge of temperture we can see that Celcius\u2019s scale has a big problem { it allows us to have a negative temperature. 212 Scale Fahrenheit Symbol De\ufb02nition \u2013F Temperature at which an equal mixture of ice and salt melts = 0\u2013F Temperature of blood = 96\u2013F Celcius \u2013C Temperature at which water freezes = 0\u2013C Temperature at which water boils = 100\u2013C Kelvin K Absolute zero is 0 K Triple point of water is 273:16 K Table 12.3: The most important temperature scales. We found that temperature is a measure of the average kinetic energy of the particles in a body. Therefore, a negative temperature suggests that that the particles have negative kinetic energy. This can not be true as kinetic energy can only be positive. Kelvin addressed this problem by rede\ufb02ning the zero of the scale. He realised that the coldest temperature you could achieve would be when the particles in a body were not moving at all. There is no way to cool something further than this as there is no more kinetic energy to remove from the body. This temperature is called absolute zero. Kelvin chose his scale so that 0K was the same as absolute zero and chose the size of his degree to be the same as one degree in the celcius scale. Interesting Fact: Rankine did a similar thing to Kelvin but set his degree to be the same size as one degree fahrenheit. Unfortunately for him, almost everyone preferred Kelvin\u2019s absolute scale and the rankine scale is now hardly ever used! It turns out that the freezing point of water, 0\u2013C, is equal to 273:15 K. So, in order to convert from celcius to kelvin need to subtact 273.15. De\ufb02nition: T (K) = T (\u2013C) 273:15 \u00a1 12.5.3 Practical thermometers It is often important to be able to determine an object\u2019s temperature precisely. This can be a challenge at very high or low tempertures or in inaccesible places. Consider a scientist who wishes to know how hot the magma in a volcano is. They are not going to be able to just lower a thermometer in to the magma as it will just melt as it reaches the superheated rock. We will now look at some less extreme situations and show how a variety of thermometry techniques can be developed. Consider \ufb02rst the gas cylinder which we tried to explode in worked example 7 by heating it while sealed. We decided that we would need to heat it to around 3000K before it explodes. How can we check this experimentally? In a sealed gas cylinder the volume of the gas and the number of moles of gas remain constant as we heat, this is why we could use the Gay-Lussac law in example 7. The Gay-Lussac law tells us that pressure is directly proportional 213 to temperature for \ufb02xed volume and amount of gas. Therefore by mesuring the pressure in the cylinder (which can be done by \ufb02tting a pressure gauge to the top of it) we can indirectly work out the temperature. This is similar to familiar alcohol or mercury thermometers. In these we use the fact that expansion of a liquid as it is heated is approximately proportional to temperature so we can use this expansion to as a measure of temperature. In fact, any thermometer you can imagine uses some physical property which varies with temperature to measure it indirectly. 12.5.4 Speci\ufb02c heat capacity Conversion of macroscopic energy to microscopic kinetic energy thus tends to raise the temperature, while the reverse conversion lowers it. It is easy to show experimentally that the amount of heating needed to change the temperature of a body by some amount is proportional to the amount of matter in the body. Thus, it is natural to write \u00a2Q = M C\u00a2T (23.4) where M is the mass of material, \u00a2Q is the amount of energy transferred to the material, and \u00a2T is the change of the material\u2019s temperature. The quantity C is called the speci\ufb02c heat of the material in question and is the amount of energy needed to raise the temperature of a unit mass of material one degree in temperature. C varies with the type of material. Values for common materials are given in table 22.2. Table 22.2: Speci\ufb02c heats of common materials. Material C (J kg\u00a11 K\u00a11) brass 385 glass 669 ice 2092 steel 448 methyl alcohol 2510 glycerine 2427 water 4184 12.5.5 Speci\ufb02c latent heat It can be seen that the speci\ufffd", "\ufffd\ufffdc heat as de\ufb02ned above will be in\ufb02nitely large for a phase change, where heat is transferred without any change in temperature. Thus, it is much more useful to de\ufb02ne a quantity called latent heat, which is the amount of energy required to change the phase of a unit mass of a substance at the phase change temperature. 12.5.6 Internal energy In thermodynamics, the internal energy is the energy of a system due to its temperature. The statement of \ufb02rst law refers to thermodynamic cycles. Using the concept of internal energy it is possible to state the \ufb02rst law for a non-cyclic process. Since the \ufb02rst law is another way of stating the conservation of energy, the energy of the system is the sum of the heat and work input, i.e., E = Q + W. Here E represents the heat energy of the system along with the kinetic energy and the potential energy (E = U + K.E. + P.E.) and is called the total internal energy of the system. This is the statement of the \ufb02rst law for non-cyclic processes. For gases, the value of K.E. and P.E. is quite small, so the important internal energy function is U. In particular, since for an ideal gas the state can be speci\ufb02ed using two variables, the state variable u is given by, where v is the speci\ufb02c volume and t is the temperature. Thus, by de\ufb02nition,, where cv is the speci\ufb02c heat at constant volume. 214 Internal energy of an Ideal gas In the previous section, the internal energy of an ideal gas was shown to be a function of both the volume and temperature. Joule performed an experiment where a gas at high pressure inside a bath at the same temperature was allowed to expand into a larger volume. picture required In the above image, two vessels, labeled A and B, are immersed in an insulated tank containing water. A thermometer is used to measure the temperature of the water in the tank. The two vessels A and B are connected by a tube, the ow through which is controlled by a stop. Initially, A contains gas at high pressure, while B is nearly empty. The stop is removed so that the vessels are connected and the \ufb02nal temperature of the bath is noted. The temperature of the bath was unchanged at the and of the process, showing that the internal energy of an ideal gas was the function of temperature alone. Thus Joule\u2019s law is stated as = 0. 12.5.7 First law of thermodynamics We now address some questions of terminology. The use of the terms \\heat\" and \\quantity of heat\" to indicate the amount of microscopic kinetic energy inhabiting a body has long been out of favor due to their association with the discredited \\caloric\" theory of heat. Instead, we use the term internal energy to describe the amount of microscopic energy in a body. The word heat is most correctly used only as a verb, e. g., \\to heat the house\". Heat thus represents the transfer of internal energy from one body to another or conversion of some other form of energy to internal energy. Taking into account these de\ufb02nitions, we can express the idea of energy conservation in some material body by the equation \u00a2E = \u00a2Q \u00a1 \u00a2W (\ufb02rst law of thermodynamics) where \u00a2E is the change in internal energy resulting from the addition of heat \u00a2Q to the body and the work \u00a2W done by the body on the outside world. This equation expresses the \ufb02rst law of thermodynamics. Note that the sign conventions are inconsistent as to the direction of energy ow. However, these conventions result from thinking about heat engines, i. e., machines which take in heat and put out macroscopic work. Examples of heat engines are steam engines, coal and nuclear power plants, the engine in your automobile, and the engines on jet aircraft. 12.6 Important Equations and Quantities Quantity Symbol Unit S.I. Units Direction Units or Table 12.4: Units used in Electricity and Magnetism 215 Chapter 13 Electrostatics 13.1 What is Electrostatics? Electrostatics is the study of electric charge which is not moving i.e. is static. 13.2 Charge All objects surrounding us (including people!) contain large amounts of electric charge. Charge can be negative or positive and is measured in units called coulombs (C). Usually, objects contain the same amount of positive and negative charge so its e\ufb01ect is not noticeable and the object is called electrically neutral. However, if a small imbalance is created (i.e. there is a little bit more of one type of charge than the other on the object) then", " the object is said to be electrically charged. Some rather amusing examples of what happens when a person becomes charged are for example when you charge your hair by combing it with a plastic comb and it stands right up on end! Another example is when you walk fast over a nylon carpet and then touch a metal doorknob and give yourself a small shock (alternatively you can touch your friend and shock them!) Charge has 3 further important properties: \u2020 \u2020 \u2020 Charge is always conserved. Charge, just like energy, cannot be created or destroyed. Charge comes in discrete packets. The smallest unit of charge is that carried by one electron called the elementary charge, e, and by convention, it has a negative sign (e = 10\u00a119C). 1:6 \u00a1 \u00a3 Charged objects exert electrostatic forces on each other. Like charges repel and unlike charges attract each other. Interesting Fact: The word \u2018electron\u2019 comes from the Greek word for amber! The ancient Greeks observed that if you rubbed a piece of amber, you could use it to pick up bits of straw. (Attractive electrostatic force!) 216 You can easily test the fact that like charges repel and unlike charges attract by doing a very simple experiment. Take a glass rod and rub it with a piece of silk, then hang it from its middle with a piece string so that it is free to move. If you then bring another glass rod which you have also charged in the same way next to it, you will see the rod on the string turn away from the rod in your hand i.e. it is repelled. If, however, you take a plastic rod, rub it with a piece of fur and then bring it close to the rod on the string, you will see the rod on the string turn towards the rod in your hand i.e. it is attracted! ////////// ////////// + + + + + + - - - - What actually happens is that when you rub the glass with silk, tiny amounts of negative charge are transferred from the glass onto the silk, which causes the glass to have less negative charge than positive charge, making it positively charged. When you rub the plastic rod with the fur, you transfer tiny amounts of negative charge onto the rod and so it has more negative charge than positive charge on it, making it negatively charged. Conductors and Insulators Some materials allow charge carriers to move relatively freely through them (e.g. most metals, tap water, the human body) and these materials are called conductors. Other materials, which do not allow the charge carriers to move through them (e.g. plastic, glass), are called nonconductors or insulators. Aside: As mentioned above, the basic unit of charge, namely the elementary charge, e, is carried by the electron. In a conducting material (e.g. copper), when the atoms bond to form the material, some of the outermost, loosely bound electrons become detached from the individual atoms and so become free to move around. The charge carried by these electrons can move around in the material! In insulators, there are very few, if any, free electrons and so the charge cannot move around in the material. 217 If an excess of charge is put onto an insulator, it will stay where it is put and there will be a concentration of charge in that area on the object. However, if an excess of charge is put onto a conductor, the charges of like sign will repel each other and spread out over the surface of the object. When two conductors are made to touch, the total charge on them is shared between the two. If the two conductors are identical, then each conductor will be left with half of the total charge. 13.3 Electrostatic Force As we now know, charged objects exert a force on one another. If the charges are at rest then this force between them is known as the electrostatic force. An interesting characteristic of the electrostatic force is that it can be either attractive or repulsive, unlike the gravitational force which is only ever attractive. The relative charges on the two objects is what determines whether the force between the charged objects is attractive or repulsive. If the objects have opposite charges they attract each other, while if their charges are similar they repel each other (e.g. two metal balls which are negatively charged will repel each other, while a positively charged ball and negatively charged ball will attract one another). F F - - F F - + It is this force that determines the arrangement of charge on the surface of conductors. When we place a charge on a spherical conductor the repulsive forces between the individual like charges cause them to spread uniformly over the surface of the sphere. However, for conductors with non-regular shape there is a concentration of charge near the point or points of the object. +++++ + + + + + + +++++ ----- - - - - - - - - ---- This collection of charge can actually allow charge to leak", " o\ufb01 the conductor if the point is sharp enough. It is for this reason that buildings often have a lightning rod on the roof to remove any charge the building has collected. This minimises the possibility of the building being struck by lightning. This \\spreading out\" of charge would not occur if we were to place the charge on an insulator since charge cannot move in insulators. 13.3.1 Coulomb\u2019s Law The behaviour of the electrostatic force was studied in detail by Charles Coulomb around 1784. Through his observations he was able to show that the electrostatic force between two point-like 218 charges is inversely proportional to the square of the distance between the objects. He also discovered that the force is proportional to the product of the charges on the two objects. F / Q1Q2 r2 ; where Q1 is the charge on the one point-like object, Q2 is the charge on the second, and r is the distance between the two. The magnitude of the electrostatic force between two point-like charges is given by Coulomb\u2019s Law: F = k Q1Q2 r2 (13.1) and the proportionality constant k is called the electrostatic constant. We will use the value k = 8:99 109N \u00a2 \u00a3 m2=C2: The value of the electrostatic constant is known to a very high precision (9 decimal places). Not many physical constants are known to as high a degree of accuracy as k. Aside: Notice how similar Coulomb\u2019s Law is to the form of Newton\u2019s Universal Law of Gravitation between two point-like particles: FG = G m1m2 r2 ; where m1 and m2 are the masses of the two particles, r is the distance between them, and G is the gravitational constant. It is very interesting that Coulomb\u2019s Law has been shown to be correct no matter how small the distance, nor how large the charge: for example it still applies inside the atom (over distances smaller than 10\u00a110m). Let\u2019s run through a simple example of electrostatic forces. Worked Example 73 Coulomb\u2019s Law I 10\u00a19C Question: Two point-like charges carrying charges of +3 are 2m apart. Determine the magnitude of the force between them and state whether it is attractive or repulsive. Answer: Step 1 : (NOTE TO SELF: step is deprecated, use westep instead.) First draw the situation: +3 10\u00a19C and 10\u00a19C 10\u00a19C 5 \u00a1 \u00a3 \u00a3 5 \u00a1 \u00a3 \u00a3 \u00a2 \u00a2 2m Step 2 : (NOTE TO SELF: step is deprecated, use westep instead.) Is everything in the correct units? Yes, charges are in coulombs [C] and distances in meters [m]. 219 Step 3 : (NOTE TO SELF: step is deprecated, use westep instead.) Determine the magnitude of the force: Using Coulomb\u2019s Law we have F = k Q1Q2 r2 = (8:99 \u00a3 = 3:37 \u00a1 \u00a3 109N m2=C2) \u00a2 10\u00a18N (+3 \u00a3 10\u00a19C)( 5 \u00a1 (2m)2 \u00a3 10\u00a19C) Thus the magnitude of the force is 3:37 two point charges having opposite signs. Step 4 : (NOTE TO SELF: step is deprecated, use westep instead.) Is the force attractive or repulsive? Well, since the two charges are oppositely charged, the force is attractive. We can also conclude this from the fact that Coulomb\u2019s Law gives a negative value for the force. 10\u00a18N. The minus sign is a result of the \u00a3 Next is another example that demonstrates the di\ufb01erence in magnitude between the gravita- tional force and the electrostatic force. Worked Example 74 Coulomb\u2019s Law II Question: Determine the electrostatic force and gravitational force between two electrons 1\u201dAapart (i.e. the forces felt inside an atom) Answer: Step 1 : (NOTE TO SELF: step is deprecated, use westep instead.) \u00a3 First draw the situation: e 10\u00a119C 10\u00a119C 1:60 1:60 \u00a3 e \u00a1 \u00a1 1\u201dA Step 2 : (NOTE TO SELF: step is deprecated, use westep instead.) Get everything into S.I. units: The charge on an electron is 10\u00a131kg, and 1\u201dA=1 of an electron is 9:11 Step 3 : (NOTE TO SELF: step is deprecated, use westep instead.) Calculate the electrostatic force using Coulomb\u2019s Law: 10\u00a110m 1:60 \u00a3 \u00a3 \u00a1 \u00a3 10\u00a119C, the mass FE", " = k e e \u00a2 1\u201dA2 Q1Q2 r2 = k 109N \u00a3 \u00a2 10\u00a18N = (8:99 = 2:30 \u00a3 m2=C2) 1:60 ( \u00a1 \u00a3 10\u00a119C)( 1:60 \u00a1 (10\u00a110m)2 10\u00a119C) \u00a3 10\u00a18N. Hence the magnitude of the electrostatic force between the electrons is 2:30 (Note that the electrons carry like charge and from this we know the force must be repulsive. Another way to see this is that the force is positive and thus repulsive.) \u00a3 220 Step 4 : (NOTE TO SELF: step is deprecated, use westep instead.) Calculate the gravitational force: FE = G m1m2 r2 = G me \u00a2 me (1\u201dA)2 = (6:67 10\u00a111N \u00a2 \u00a3 m2=kg2) (9:11 \u00a3 10\u00a131C)(9:11 (10\u00a110m)2 \u00a3 10\u00a131kg) = 5:54 10\u00a151N \u00a3 The magnitude of the gravitational force between the electrons is 5:54 10\u00a151N \u00a3 Notice that the gravitational force between the electrons is much smaller than the electrostatic force. For this reason, the gravitational force is usually neglected when determining the force between two charged objects. We mentioned above that charge placed on a spherical conductor spreads evenly along the surface. As a result, if we are far enough from the charged sphere, electrostatically, it behaves as a point-like charge. Thus we can treat spherical conductors (e.g. metallic balls) as point-like charges, with all the charge acting at the centre. Worked Example 75 Coulomb\u2019s Law: Challenge Question Question: In the picture below, X is a small negatively charged sphere with a mass of 10kg. It is suspended from the roof by an insulating rope which makes an angle of 60o with the roof. Y is a small positively charged sphere which has the same magnitude of charge as X. Y is \ufb02xed to the wall by means of an insulating bracket. Assuming the system is in equilibrium, what is the magnitude of the charge on X? /////////// 60o 10kg X { Y + 50cm n n n n n Answer: How are we going to determine the charge on X? Well, if we know the force between X and Y we can use Coulomb\u2019s Law to determine their charges as we know the distance between them. So, \ufb02rstly, we need to determine the magnitude of the electrostatic force between X and Y. Step 1 : (NOTE TO SELF: step is deprecated, use westep instead.) Is everything in S.I. units? The distance between X and Y is 50cm = 0:5m, and the mass of X is 10kg. 221 Step 2 : (NOTE TO SELF: step is deprecated, use westep instead.) Draw the forces on X (with directions) and label. T : tension from the thread 60o FE: electrostatic force X Fg: gravitational force Step 3 : (NOTE TO SELF: step is deprecated, use westep instead.) Determine the magnitude of the electrostatic force (FE). Since nothing is moving (system is in equlibrium) the vertical and horizonal components of the forces must cancel. Thus FE = T sin(60o); Fg = T sin(60o): The only force we know is the gravitational force Fg = mg. Now we can calculate the magnitude of T from above: T = Fg sin(60o) = (10kg)(10m=s2) sin(60o) = 1155N: Which means that FE is: FE = T cos(60o) = 1154N cos(60o) = 577:5N \u00a2 Step 4 : (NOTE TO SELF: step is deprecated, use westep instead.) Now that we know the magnitude of the electrostatic force between X and Y, we can calculate their charges using Coulomb\u2019s Law. Don\u2019t forget that the magnitudes of the charges on X and Y are the same:. The magnitude of the electrostatic = j force is QX j QY j j FE = k j QX j j = = j = k Q2 X r2 QXQY r2 FEr2 k r 8:99 s (577:5N)(0:5m)2 109N \u00a3 10\u00a14C \u00a2 m2=C2 = 1:27 Thus the charge on X is 1:27 \u00a1 \u00a3 10\u00a14C \u00a3 222 13.4 Electric Fields We have learnt that objects that carry charge feel forces from all other charged objects. It is useful to determine what the e\ufffd", "\ufffd\ufffdect of a charge would be at every point surrounding it. To do this we need some sort of reference. We know that the force that one charge feels due to another depends on both charges (Q1 and Q2). How then can we talk about forces if we only have one charge? The solution to this dilemna is to introduce a test charge. We then determine the force that would be exerted on it if we placed it at a certain location. If we do this for every point surrounding a charge we know what would happen if we put a test charge at any location. This map of what would happen at any point we call a \ufb02eld map. It is a map of the electric It tells us how large the force on a test charge would be and in what \ufb02eld due to a charge. direction the force would be. Our map consists of the lines that tell us how the test charge would move if it were placed there. 13.4.1 Test Charge This is the key to mapping out an electric \ufb02eld. The equation for the force between two electric charges has been shown earlier and is: F = k Q1Q2 r2 : (13.2) If we want to map the \ufb02eld for Q1 then we need to know exactly what would happen if we put Q2 at every point around Q1. But this obviously depends on the value of Q2. This is a time when we need to agree on a convention. What should Q2 be when we make the map? By convention we choose Q2 = +1C. This means that if we want to work out the e\ufb01ects on any other charge we only have to multiply the result for the test charge by the magnitude of the new charge. The electric \ufb02eld strength is then just the force per unit of charge and has the same magnitude and direction as the force on our test charge but has di\ufb01erent units: E = k Q1 r2 (13.3) The electric \ufb02eld is the force per unit of charge and hence has units of newtons per coulomb [N/C]. So to get the force the electric \ufb02eld exerts we use: F = EQ (13.4) Notice we are just multiplying the electric \ufb02eld magnitude by the magnitude of the charge it is acting on. 13.4.2 What do \ufb02eld maps look like? The maps depend very much on the charge or charges that the map is being made for. We will start o\ufb01 with the simplest possible case. Take a single positive charge with no other charges around it. First, we will look at what e\ufb01ects it would have on a test charge at a number of points. 223 Positive Charge Acting on Test Charge At each point we calculate the force on a test charge, q, and represent this force by a vector. +Q We can see that at every point the positive test charge, q, would experience a force pushing it away from the charge, Q. This is because both charges are positive and so they repel. Also notice that at points further away the vectors are shorter. That is because the force is smaller if you are further away. If the charge were negative we would have the following result. Negative Charge Acting on Test Charge -Q Notice that it is almost identical to the positive charge case. This is important { the arrows are the same length because the magnitude of the charge is the same and so is the magnitude of the test charge. Thus the magnitude of the force is the same. The arrows point in the opposite direction because the charges now have opposite sign and so the test charge is attracted to the charge. Now, to make things simpler, we draw continuous lines showing the path that the test charge would travel. This means we don\u2019t have to work out the magnitude of the force at many di\ufb01erent points. 224 Electric Field Map due to a Positive Charge +Q Some important points to remember about electric \ufb02elds: There is an electric \ufb02eld at every point in space surrounding a charge. Field lines are merely a representation { they are not real. When we draw them, we just pick convenient places to indicate the \ufb02eld in space. Field lines always start at a right-angle (90o) to the charged object causing the \ufb02eld. Field lines never cross! \u2020 \u2020 \u2020 \u2020 13.4.3 Combined Charge Distributions We look at the \ufb02eld of a positive charge and a negative charge placed next to each other. The net resulting \ufb02eld would be the addition of the \ufb02elds from each of the charges. To start o\ufb01 with let us sketch the \ufb02eld maps for each of the charges as though", " it were in isolation. Electric Field of Negative and Positive Charge in Isolation +Q -Q 225 Notice that a test charge starting o\ufb01 directly between the two would be pushed away from the positive charge and pulled towards the negative charge in a straight line. The path it would follow would be a straight line between the charges. +Q -Q Now let\u2019s consider a test charge starting o\ufb01 a bit higher than directly between the charges. If it starts closer to the positive charge the force it feels from the positive charge is greater, but the negative charge does attract it, so it would move away from the positive charge with a tiny force attracting it towards the negative charge. As it gets further from the positive charge the force from the negative and positive charges change and they are equal in magnitude at equal distances from the charges. After that point the negative charge starts to exert a stronger force on the test charge. This means that the test charge moves towards the negative charge with only a small force away from the positive charge. +Q -Q Now we can \ufb02ll in the other lines quite easily using the same ideas. The resulting \ufb02eld map is: +Q -Q Two Like Charges I: The Positive Case For the case of two positive charges things look a little di\ufb01erent. We can\u2019t just turn the arrows around the way we did before. In this case the test charge is repelled by both charges. This tells us that a test charge will never cross half way because the force of repulsion from both charges will be equal in magnitude. 226 +Q +Q The \ufb02eld directly between the charges cancels out in the middle. The force has equal magnitude and opposite direction. Interesting things happen when we look at test charges that are not on a line directly between the two. +Q +Q We know that a charge the same distance below the middle will experience a force along a reected line, because the problem is symmetric (i.e. if we ipped vertically it would look the same). This is also true in the horizontal direction. So we use this fact to easily draw in the next four lines. +Q +Q 227 Working through a number of possible starting points for the test charge we can show the electric \ufb02eld map to be: +Q +Q Two Like Charges II: The Negative Case We can use the fact that the direction of the force is reversed for a test charge if you change the sign of the charge that is inuencing it. If we change to the case where both charges are negative we get the following result: -Q -Q 13.4.4 Parallel plates One very important example of electric \ufb02elds which is used extensively is the electric \ufb02eld between two charged parallel plates. In this situation the electric \ufb02eld is constant. This is used for many practical purposes and later we will explain how Millikan used it to measure the charge on the electron. 228 Field Map for Oppositely Charged Parallel Plates + + + + + + + + + - - - - - - - - - This means that the force that a test charge would feel at any point between the plates would be identical in magnitude and direction. The \ufb02elds on the edges exhibit fringe e\ufb01ects, i.e. they bulge outwards. This is because a test charge placed here would feel the e\ufb01ects of charges only on one side (either left or right depending on which side it is placed). Test charges placed in the middle experience the e\ufb01ects of charges on both sides so they balance the components in the horizontal direction. This isn\u2019t the case on the edges. The Force on a Test Charge between Oppositely Charged Parallel Plates + + + + + + + + + - - - - - - - - - 13.4.5 What about the Strength of the Electric Field? When we started making \ufb02eld maps we drew arrows to indicate the strength of the \ufb02eld and the direction. When we moved to lines you might have asked \\Did we forget about the \ufb02eld strength?\". We did not. Consider the case for a single positive charge again: 229 A m mg h B Figure 13.1: A mass under the inuence of a gravitational \ufb02eld. +Q Notice that as you move further away from the charge the \ufb02eld lines become more spread out. In \ufb02eld map diagrams the closer \ufb02eld lines are together the stronger the \ufb02eld. This brings us to an interesting case. What is the electric \ufb02eld like if the object that is charged has an irregular shape. 13.5 Electrical Potential 13.5.1 Work Done and Energy Transfer in a Field When a charged particle moves in an electric", " \ufb02eld work is done and energy transfers take place. This is exactly analogous to the case when a mass moves in a gravitational \ufb02eld such as that set up by any massive object. Work done by a \ufb02eld Gravitational Case A mass held at a height h above the ground has gravitational potential energy since, if released, it will fall under the action of the gravitational \ufb02eld. Once released, in the absence of friction, only the force of gravity acts on the mass and the mass accelerates in the direction of the force (towards the earth\u2019s centre). In this way, work is done by the \ufb02eld. When the mass 230 + + + + + + + A +Q QE s B - - - - - - - Figure 13.2: A charged particle under the inuence of an electric \ufb02eld. falls a distance h (from point A to B), the work done is, W = F s = mgh In falling, the mass loses gravitational potential energy and gains kinetic energy. The work done by the \ufb02eld is equal to the energy transferred, Energy is conserved! W = Gain in Ek = Loss in Ep (a falling mass) Electrical Case A charge in an electric \ufb02eld has electrical potential energy since, if released, it will move under the action of the electric \ufb02eld. When released, in the absence of friction, only the electric force acts on the charge and the charge accelerates in the direction of the force (for positive charges the force and acceleration are in the direction of the electric \ufb02eld, while negative charges experience a force and acceleration in the opposite direction to the electric \ufb02eld.) Consider a positive charge +Q placed in the uniform electric \ufb02eld between oppositely charged parallel plates. The positive charge will be repelled by the positive plate and attracted by the negative plate (i.e. it will move in the direction of the electric \ufb02eld lines). In this way, work is done by the \ufb02eld. In moving the charge a distance s in the electric \ufb02eld, the work done is, W = F s = QEs since E = F Q : In the process of moving, the charge loses electrical potential energy and gains kinetic energy. The work done by the \ufb02eld is equal to the energy transferred, W = Gain in Ek = Loss in Ep (charge moving under the inuence of an electric \ufb02eld) Work done by us Gravitational Case In order to return the mass m in Fig.13.1 to its original position (i.e. lift it a distance h from B back to A) we have to apply a force mg to balance the force of gravity. An amount of work mgh is done by the lifter. In the process, the mass gains gravitational potential energy, mgh = Gain in Ep (lifting a mass) 231 Electrical Case In order to return the charge in Fig.13.2 to its original position (i.e. from B back to A) we have to exert a force QE on the charge to balance the force exerted on it by the electric \ufb02eld. An amount of work QEs is done by us. In the process, the charge gains electrical potential energy, Energy is conserved! QEs = Gain in Ep (charge moved against an electric \ufb02eld) In summary, when an object moves under the inuence of a \ufb02eld, the \ufb02eld does work and potential energy is transferred into kinetic energy. Potential energy is lost, while kinetic energy is gained. When an object is moved against a \ufb02eld we have to do work and the object gains potential energy. Worked Example 76 Work done and energy transfers in a \ufb02eld Question: A charge of +5nC is moved a distance of 4 cm against a uniform electric \ufb02eld of magnitude 2 1012N:C\u00a11 from A to B 4cm +5nC A - - - - - - - (a) Calculate the work done in moving the charge from A to B. (b) The charge is now released and returns to A. Calculate the kinetic energy of the charge at A. Answer: (a) Step 1 : (NOTE TO SELF: step is deprecated, use westep instead.) We are given the values of the charge, the \ufb02eld and the distance the charge must move. All are in the correct units. Step 2 : (NOTE TO SELF: step is deprecated, use westep instead.) Since the charge is positive we have to do work to move it from A to B (since this is against the \ufb02eld). This work is given by, W = Q", "started at rest, the gain in kinetic energy is the \ufb02nal kinetic energy, Eat A k = 400 J 13.5.2 Electrical Potential Di\ufb01erence Consider a positive test charge +Q placed at A in the electric \ufb02eld of another positive point charge. + +Q A B The test charge moves towards B under the inuence of the electric \ufb02eld of the other charge. In the process the test charge loses electrical potential energy and gains kinetic energy. Thus, at A, the test charge has more potential energy than at B { A is said to have a higher electrical potential than B. The potential energy of a charge at a point in a \ufb02eld is de\ufb02ned as the work required to move that charge from in\ufb02nity to that point. The potential di\ufb01erence between two points in an electric \ufb02eld is de\ufb02ned as the work required to move a unit positive test charge from the point of lower potential to that of higher potential. If an amount of work W is required to move a charge Q from one point to another, then the potential di\ufb01erence between the two points is given by, V = W Q unit : J:C\u00a11 or V (the volt) 233 From this equation it follows that one volt is the potential di\ufb01erence between two points in an electric \ufb02eld if one joule of work is done in moving one coulomb of charge from the one point to the other. Worked Example 77 Potential di\ufb01erence Question: A positively charged object Q is placed as shown in the sketch. The potential di\ufb01erence between two points A and B is 4 10\u00a14 V. \u00a3 +Q A B (a) Calculate the change in electrical potential energy of a +2nC charge when it moves from A to B. (b) Which point, A or B, is at the higher electrical potential? Explain. (c) If this charge were replaced with another of charge -2nC, in what way would its change in energy be a\ufb01ected? Answer: (a) The electrical potential energy of the positive charge decreases as it moves from A to B since it is moving in the direction of the electric \ufb02eld produced by the object Q. This loss in potential energy is equal to the work done by the \ufb02eld, Loss in Electrical Potential Energy = W = V Q (Since V = W Q ) = (4 = 8 \u00a3 10\u00a14)(2 10\u00a113 J \u00a3 10\u00a19) \u00a3 (b) Point A is at the higher electrical potential since work is required by us to move a positive test charge from B to A. (c) If the charge is replaced by one of negative charge, the electrical potential energy of the charge will increase in moving from A to B (in this case we would have to do work on the charge). As an example consider the electric \ufb02eld between two oppositely charged parallel plates a distance d apart maintained at a potential di\ufb01erence V. 234 + + + + + + + P d O +Q - - - - - - - This electric \ufb02eld is uniform so that a charge placed anywhere between the plates will experience the same force. Consider a positive test charge Q placed at point O just o\ufb01 the surface of the negative plate. In order to move it towards the positive plate we have to apply a force QE. The work done in moving the charge from the negative to the positive plate is, W = F s = QEd; but from the de\ufb02nition of electrical potential, Equating these two expressions for the work done, W = V Q: QEd = V Q; E = V d : and so, rearranging, Worked Example 78 Parallel plates Question: Two charged parallel plates are at a distance of 180 mm from each other. The potential di\ufb01erence between them is 3600 V as shown in the diagram. + + + + + + + d = 180mm X Y V = 3 600V - - - - - - - 235 + + + + + + + d -Q Fup=QE Fdown=mg - - - - - - - Figure 13.3: An oil drop suspended between oppositely charged parallel plates. 10\u00a19 C, is (a) If a small oil drop of negligible mass, carrying a charge of +6:8 placed between the plates at point X, calculate the magnitude and direction of the electrostatic force exerted on the droplet. (b) If the droplet is now moved to point Y, would the force exerted on it be bigger, smaller or the same as in (a)? Answer: (a) Step 1 : (NOTE TO SELF: step is", " deprecated, use westep instead.) First \ufb02nd the electric \ufb02eld strength between the plates, \u00a3 E = = V d 3600 0:180 = 20000 N:C\u00a11 from the positive to the negative plate Step 2 : (NOTE TO SELF: step is deprecated, use westep instead.) Now the force exerted on the charge at X is, F = QE = (6:8 = 1:36 \u00a3 \u00a3 10\u00a19)(20000) 10\u00a14 down (b) Step 3 : (NOTE TO SELF: step is deprecated, use westep instead.) The same. Since the electric \ufb02eld strength is uniform, the force exerted on a charge is the same at all points between the plates. 13.5.3 Millikan\u2019s Oil-drop Experiment Robert Millikan measured the charge on an electron by studying the motion of charged oil drops between oppositely charged parallel plates. Consider one such negative drop between the plates in Fig.13.3. Since this drop is negative, the electric \ufb02eld exerts an upward force on the drop. In addition to this upward force, gravity exerts a downward force on the drop. Millikan adjusted 236 the electric \ufb02eld strength between the plates by varying the potential di\ufb01erence applied across the plates. In this way, Millikan was able to bring the drops to rest. At equlibrum, Since E = V d, and, therefore, Fup = Fdown QE = mg Q V d = mg; Q = mgd V Millikan found that all drops had charges which were multiples of 1:6 become charged by gaining or losing electrons, the charge on an electron must be The magnitude of the electron\u2019s charge is denoted by e, \u00a3 10\u00a119 C. Since objects 10\u00a119 C. 1:6 \u00a1 \u00a3 e = 1:6 10\u00a119 C \u00a3 Worked Example 79 Charge Question: A metal sphere carries a charge of +3:2 to lose to attain its charge? Answer: Since the sphere is positive it lost electrons in the process of charging (when an object loses negative charges it is left positive). In fact, it lost, 10\u00a18 C. How many electrons did it have \u00a3 10\u00a18 10\u00a119 = 2 \u00a3 3:2 1:6 \u00a3 \u00a3 1011 electrons Worked Example 80 Millikan oil-drop experiment Question: In a Millikan-type experiment a positively charged oil drop is placed between two horizontal plates, 20 mm apart, as shown. 237 - - - - - - - + + + + + + + The potential di\ufb01erence across the plates is 4000V. The drop has a mass of 1:2 10\u00a114kg and a charge of 8 (a) Draw the electric \ufb02eld pattern between the two plates. (b) Calculate: 10\u00a119C. \u00a3 \u00a3 1. the electric \ufb02eld intensity between the two plates. 2. the magnitude of the gravitational force acting on the drop. 3. the magnitude of the Coulomb force acting on the drop. (c) The drop is observed through a microscope. What will the drop be seen to do? Explain. (d) Without any further calculations, give two methods that could be used to make the drop remain in a \ufb02xed position. Answer: (a 4000 0:02 = 2 \u00a3 105 V:m\u00a11 up (b) 1. 2. Fgrav = mg = (1:2 = 1:2 10\u00a114)(10) 10\u00a113N \u00a3 \u00a3 238 3. FCoulomb = QE = (8 = 1:6 \u00a3 \u00a3 10\u00a119)(s 10\u00a113N 105) \u00a3 (c) Since Fup > Fdown, the drop accelerates upwards. (d) The Coulomb force can be decreased by decreasing the electric \ufb02eld strength between the plates. Since E = V d, this can be done either by increasing d or decreasing V. 13.6 Important Equations and Quantities Units Quantity charge force mass acceleration radial distance electric \ufb02eld work potential di\ufb01erence Unit Symbol q or Q C (Coulomb) N (Newton) | | | A:s kg:m s2 kg m s2 m N=C or V =m kg:m A:s3 2 kg:m s2 2 kg:m A:s3 \u00a1!F m \u00a1!a r \u00a1!E W V J V (Volt) S.I. Units or | or kg:m:s\u00a12 or | or m:s\u00a12 or | or kg:m:A\u00a11:s\u00a13 or kg:m2:s\u00a12 or kg:", "m2:A\u00a11:s\u00a13 Direction | X | X | X | | Table 13.1: Units used in Electrostatics 239 Chapter 14 Electricity Warning: We believe in experimenting and learning about physics at every opportunity, BUT playing with electricity can be EXTREMELY DANGEROUS! Do not try to build home made circuits without someone who knows if what you are doing is safe. Normal electrical outlets are dangerous. Treat electricity with respect in your everyday life. You will encounter electricity everyday for the rest of your life and to make sure you are able to make wise decisions we have included an entire chapter on electrical safety. Please read it - not only will it make you safer but it will show the applications of many of the ideas you will learn in this chapter. 14.1 Flow of Charge The normal motion of \"free\" electrons in a conductor has no particular direction or speed. However, electrons can be inuenced to move in a coordinated fashion through a conductive material. This motion of electrons is what we call electricity, or electric current. This is in contrast to static electricity, which is an unmoving accumulation of electric charge. Just like water owing through the emptiness of a pipe, electrons are able to move between the atoms of a conductor. The conductor may appear to be solid to our eyes, but any material composed of atoms is mostly empty space! The liquid-ow analogy is so \ufb02tting that the motion of electrons through a conductor is often referred to as a \"ow.\" As each electron moves through a conductor, it pushes on the one ahead of it. This push of one electron on another makes all of the electrons move together as a group. The motion of the each electron in a conductor may be very slow. However, the starting and stopping of electron ow through a conductor is virtually instantaneous from one end of it to the other. As an analogy consider a tube \ufb02lled end-to-end with marbles: Tube Marble Marble The tube is full of marbles, just as a conductor is full of free electrons. If a single marble is suddenly inserted into this full tube on the left-hand side, another marble will immediately try to exit the tube on the right. Even though each marble only traveled a short distance, the 240 transfer of motion through the tube is virtually instantaneous from the left end to the right end. The nearly instantaneous transfer of motion through the tube occurs no matter how long the tube is. With electricity, the overall e\ufb01ect from one end of a conductor to the other is e\ufb01ectively instantaneous. Each individual electron, though, travels through the conductor at a much slower pace. If we want electrons to ow in a certain direction to a certain place, we must provide the proper path for them to move. A path for electrons must be provided just as a plumber must install piping to get water to ow where he or she wants it to ow. Wires made of highly conductive metals such as copper or aluminum are used to form this path. This means that there can be electric current only where there exists a continuous path of conductive material (wire) providing a path for electrons. In the marble analogy, marbles can ow into the left-hand side of the tube only if the tube is open on the right-hand side for marbles to ow out. If the tube is blocked on the right-hand side, the marbles will just \"pile up\" inside the tube. Marble \"ow\" will not occur if the tube is blocked. The same holds true for electric current: the continuous ow of electrons requires there be an unbroken path. Let\u2019s look at a diagram to illustrate how this works: A thin, solid line (as shown above) is the conventional symbol for a continuous piece of wire. The wire is made of a conductive material, such as copper or aluminum. The wire\u2019s constituent atoms have many free electrons which can easily move through the wire. However, there will never be a continuous ow of electrons within this wire unless they have a place to come from and a place to go. Let\u2019s add an hypothetical electron \"Source\" and \"Destination:\" Electron Source Electron Destination Now, with the Electron Source pushing new electrons into the wire on the left-hand side, electron ow through the wire can occur (as indicated by the arrows pointing from left to right). However, the ow will be interrupted if the conductive path formed by the wire is broken: Electron Source no flow! no flow! (break) Electron Destination Air is an insulator that impedes the ow of electrons. An air gap separates the two pieces of wire, the path has now been broken, and electrons cannot ow from Source to Destination. This is like cutting a water pipe in two and capping o\ufb01 the broken ends of the pipe: water can\u2019t ow if there\u2019s no exit out of the pipe. If we", " were to take another piece of wire leading to the Destination and connect it with the wire leading to the Source, we would once again have a continuous path for electrons to ow. The two dots in the diagram indicate physical (metal-to-metal) contact between the wire pieces: Electron Source no flow! (break) Electron Destination 241 Now, we have continuity from the Source, to the newly-made connection, down, to the right, and up to the Destination. Please take note that the broken segment of wire on the right hand side has no electrons owing through it. This is because it is no longer part of a complete path from Source to Destination. It is interesting to note that no \"wear\" occurs within wires due to this electric current. This is in contrast to water-carrying pipes which are eventually corroded and worn by prolonged ows. Electrons do encounter some degree of friction as they move and this friction can generate heat in a conductor. This is a topic we\u2019ll discuss later. 14.2 Circuits In order for the Source-and-Destination scheme to work, both would have to have a huge reservoir of electrons in order to sustain a continuous ow! Using the marble-and-tube analogy, the source and destination buckets would have to be large reservoirs to contain enough marble capacity for a \"ow\" of marbles to be sustained. The answer to this paradox is found in the concept of a circuit: a never-ending looped pathway for electrons. If we take a wire, and loop it around so that it forms a continuous pathway, we have the means to support a uniform ow of electrons without having to resort to huge reservoirs. electrons can flow in a path without beginning or end, continuing forever! A marble-andhula-hoop \"circuit\" Each electron advancing clockwise in this circuit pushes on the one in front of it, which pushes on the one in front of it, and so on. A circuit is just like a hula-hoop \ufb02lled with marbles. Now, we have the capability of supporting a continuous ow of electrons inde\ufb02nitely without the need for reservoirs. All we need to maintain this ow is a continuous means of motivation for those electrons. This topic will be addressed in the next section of this chapter. It must be realized that continuity is just as important in a circuit as it is in a straight piece of wire. Just as in the example with the straight piece of wire between the electron Source and Destination, any break in this circuit will prevent electrons from owing through it: 242 no flow! continuous electron flow cannot occur anywhere in a \"broken\" circuit! (break) no flow! no flow! An important principle to realize here is that it doesn\u2019t matter where the break occurs. Any discontinuity in the circuit will prevent electron ow throughout the entire circuit. no flow! continuous electron flow cannot occur anywhere in a \"broken\" circuit! no flow! (break) no flow! A combination of batteries and conductors with other components is called an electric circuit or circuit. The word circuit implies that you return to your starting point and this is an important property of electric circuits. They must contain a closed loop before charge can ow. The simplest possible circuit is a battery with a single conductor. Now how do we form a closed loop with these two components? The battery has two terminals (connection points). One is called the positive terminal and one the negative. When we describe charge owing we consider charges moving from the positive terminal of the battery around the conductor and back into the battery at the negative terminal. As much charge ows out of the positive terminal as ows into the negative terminal. Because of this, there is no build up of charge in the battery. The battery does work on the charges causing them to move round the circuit. We have covered the topics of batteries and circuits but we need to draw these things to help keep the ideas clear in our minds. To do this we need to agree on how to draw things so that other people can understand what we are doing. We need a convention. 243 14.3 Voltage and current As was previously mentioned, we need more than just a continuous path (circuit) before a continuous ow of electrons will occur. We also need some means to push these electrons around the circuit. Just like marbles in a tube or water in a pipe, it takes some kind of inuencing force to initiate ow. With electrons, this force is the same force at work in static electricity: the force produced by an imbalance of electric charge. The electric charge di\ufb01erence serves to store a certain amount of energy. This energy is not unlike the energy stored in a high reservoir of water that has been pumped from a lower-level pond: Reservoir Energy stored Water flow Pump Pond The inuence of gravity on the water in the reservoir creates a force that attempts to move the water down to the lower level again. If a suitable pipe is run from", " the reservoir back to the pond, water will ow under the inuence of gravity down from the reservoir, through the pipe: 244 Reservoir Energy released Pond It takes energy to pump that water from the low-level pond to the high-level reservoir. The movement of water through the piping back down to its original level constitutes a releasing of energy stored from previous pumping. If the water is pumped to an even higher level, it will take even more energy to do so, and so more energy will be stored. The more energy is stored the more energy is released if the water is allowed to ow through a pipe back down again: 245 Energy stored Reservoir Pump Pond Reservoir Energy released More energy stored More energy released Pump Pond Electrons are not much di\ufb01erent. If we \"pump\" electrons away from their normal \"levels,\" we create a condition where a force exists as the electrons seek to re-establish their former positions. The force attracting electrons back to their original positions is analogous to the force gravity exerts on water in the reservoir. Just as gravity tries to draw water down to its former level, the force exerted on the electrons attracts them back to their former posisions. Just as the pumping of water to a higher level results in energy being stored, \"pumping\" electrons to create an electric charge imbalance results in a certain amount of energy being stored in that imbalance. Providing a way for water to ow back down from the heights of the 246 reservoir results in a release of that stored energy. Similarly, providing a way for electrons to ow back to their original \"levels\" results in a release of stored energy. When the electrons are poised in that static condition (just like water sitting still, high in a reservoir), the energy stored there is called potential energy. It is given that name because it has the possibility (potential) of release that has not been fully realized yet. This potential energy, stored in the form of an electric charge imbalance, can be expressed as a term called voltage. Technically, voltage is a measure of potential energy per unit charge of electrons, or something a physicist would call speci\ufb02c potential energy. De\ufb02ned in the context of static electricity, voltage is the measure of work required to move a unit charge from one location to another. This work is against the force which tries to keep electric charges balanced. In the context of electrical power sources, voltage is the amount of potential energy available (work to be done) per unit charge, to move electrons through a conductor. Voltage is an expression of potential energy. As such, it represents the possibility or potential for energy release as the electrons move from one \"level\" to another. Because of this, voltage is always referenced between two points. Consider the water reservoir analogy: Reservoir Drop Drop Location #1 Location #2 Because of the di\ufb01erence in the height of the drop, there\u2019s potential for much more energy to be released from the reservoir through the piping to location 2 than to location 1. The principle can be intuitively understood in dropping a rock: which results in a more violent impact, a rock dropped from a height of one foot, or the same rock dropped from a height of one kilometre? Obviously, the drop of greater height results in greater energy released (a more violent impact). We cannot assess the amount of stored energy in a water reservoir simply by measuring the volume of water. Similarly, we cannot predict the severity of a falling rock\u2019s impact simply from knowing the weight of the rock: in both cases we must also consider how far these masses will drop from their initial height. The amount of energy released by allowing a mass to drop is relative to the distance between its starting and ending points. Likewise, the potential energy available for moving electrons from one point to another is relative to those two points. Therefore, voltage is always expressed as a quantity between two points. Interestingly enough, the analogy of a mass potentially \"dropping\" from one height to another is such an apt model that voltage between two points is sometimes called a voltage drop. 247 Voltage can be generated in many ways. Chemical reactions, radiant energy, and the inuence of magnetism on conductors are a few ways in which voltage may be produced. Respective examples of these three sources of voltage are batteries, solar cells, and generators. For now, we won\u2019t go into detail as to how each of these voltage sources works. The important thing is that we understand how voltage sources can be applied to create electron ow in a circuit. Let\u2019s take the symbol for a chemical battery and build a circuit step by step: 1 - + 2 Battery Any source of voltage, including batteries, have two points for electrical contact. In this case, we have point 1 and point 2 in the above diagram. The horizontal lines of varying length indicate that this is a battery. The horizontal lines further indicate the direction which this battery\u2019s voltage will try to push electrons through a circuit. The", " horizontal lines in the battery symbol appear separated, and so make it appear as if the battery is unable to serve as a path for electrons to move. This is no cause for concern: in real life, those horizontal lines represent metallic plates immersed in a liquid or semi-solid material that not only conducts electrons, but also generates the voltage to push them along by interacting with the plates. Notice the little \"+\" and \"-\" signs to the immediate left of the battery symbol. The negative (-) end of the battery is always the end with the shortest dash, and the positive (+) end of the battery is always the end with the longest dash. By convention, electrons are said to be \"negatively\" charged, so the negative end of a battery is that end which tries to push electrons out of it. Likewise, the positive end is that end which tries to attract electrons. With the \"+\" and \"-\" ends of the battery not connected to anything, there will be voltage between those two points. However, there will be no ow of electrons through the battery, because there is no continuous path for the electrons to move. 248 Water analogy Reservoir Electric Battery No flow 1 - + 2 Battery No flow (once the reservoir has been completely filled) Pump Pond The same principle holds true for the water reservoir and pump analogy: without a return pipe back to the pond, stored energy in the reservoir cannot be released in the form of water ow. Once the reservoir is completely \ufb02lled up, no ow can occur, no matter how much pressure the pump may generate. There needs to be a complete path (circuit) for water to ow from the pond, to the reservoir, and back to the pond in order for continuous ow to occur. We can provide such a path for the battery by connecting a piece of wire from one end of the battery to the other. Forming a circuit with a loop of wire, we will initiate a continuous ow of electrons in a clockwise direction: 249 Electric Circuit Battery 1 - + 2 electron flow! Water analogy Reservoir water flow! water flow! Pump Pond So long as the battery continues to produce voltage and the continuity of the electrical path isn\u2019t broken, electrons will continue to ow in the circuit. Following the metaphor of water moving through a pipe, this continuous, uniform ow of electrons through the circuit is called a current. So long as the voltage source keeps \"pushing\" in the same direction, the electron ow will continue to move in the same direction in the circuit. This single-direction ow of electrons is called a Direct Current, or DC. Because electric current is composed of individual electrons owing in unison through a conductor, just like marbles through a tube or water through a pipe, the amount of ow throughout a single circuit will be the same at any point. If we were to monitor a cross-section of the wire in a single circuit, counting the electrons owing by, we would notice the exact same quantity per unit of time as in any other part of the circuit. The same quantity would be observed regardless 250 of conductor length or conductor diameter. If we break the circuit\u2019s continuity at any point, the electric current will cease in the entire loop. Futhermore, the full voltage produced by the battery will be manifested across the break, between the wire ends that used to be connected: 1 - + 2 no flow! Battery no flow! - (break) + voltage drop Notice the \"+\" and \"-\" signs drawn at the ends of the break in the circuit, and how they correspond to the \"+\" and \"-\" signs next to the battery\u2019s terminals. These markers indicate the direction that the voltage attempts to push electron ow. Remember that voltage is always relative between two points. Whether a point in a circuit gets labeled with a \"+\" or a \"-\" depends on the other point to which it is referenced. Take a look at the following circuit, where each corner of the loop is marked with a number for reference: 1 - + 4 no flow! 2 - Battery (break) + 3 no flow! With the circuit\u2019s continuity broken between points 2 and 3, voltage dropped between points 2 and 3 is \"-\" for point 2 and \"+\" for point 3. Now let\u2019s see what happens if we connect points 2 and 3 back together again, but place a break in the circuit between points 3 and 4: 251 1 - + 4 no flow! 2 Battery no flow! + - (break) 3 With the break between 3 and 4, the voltage drop between those two points is \"+\" for 4 and \"-\" for 3. Take special note of the fact that point 3\u2019s \"sign\" is opposite of that in the \ufb02rst example, where the break was between points 2 and 3 (where point 3 was labeled \"+\"). It is impossible for us to say that point 3 in this circuit will always be either \"+\" or \"-\", because the sign is not speci\ufb02c to a single point", ", but is always relative between two points! 14.4 Resistance The circuit in the previous section is not a very practical one. In fact, it can be quite dangerous to directly connect the poles of a voltage source together with a single piece of wire. This is because the magnitude of electric current may be very large in such a short circuit, and the release of energy very dramatic (usually in the form of heat). Usually, electric circuits are constructed in such a way as to make practical use of that released energy, in as safe a manner as possible. One practical and popular use of electric current is for the operation of electric lighting. The simplest form of electric lamp is a tiny metal \"\ufb02lament\" inside of a clear glass bulb, which glows white-hot (\"incandesces\") with heat energy when su\u2013cient electric current passes through it. Like the battery, it has two conductive connection points, one for electrons to enter and the other for electrons to exit. Connected to a source of voltage, an electric lamp circuit looks something like this: Battery - + electron flow electron flow Electric lamp (glowing) As the electrons work their way through the thin metal \ufb02lament of the lamp, they encounter more opposition to motion than they typically would in a thick piece of wire. This opposition to 252 electric current depends on the type of material, its cross-sectional area, and its temperature. It is technically known as resistance. It serves to limit the amount of current through the circuit with a given amount of voltage supplied by the battery. The \"short circuit\" where we had nothing but a wire joining one end of the voltage source (battery) to the other had not such a limiting resistance. Interesting Fact: Materials known as conductors have a low resistance, while insulators have a very high one. When electrons move against the opposition of resistance, \"friction\" is generated. Just like mechanical friction, the friction produced by electrons owing against a resistance manifests itself in the form of heat. The concentrated resistance of a lamp\u2019s \ufb02lament results in a relatively large amount of heat energy dissipated at that \ufb02lament. This heat energy is enough to cause the \ufb02lament to glow white-hot, producing light. The wires connecting the lamp to the battery hardly even get warm while conducting the same amount of current. This is because of their much lower resistance due to their larger cross-section. As in the case of the short circuit, if the continuity of the circuit is broken at any point, electron ow stops throughout the entire circuit. With a lamp in place, this means that it will stop glowing: no flow! - Battery - + no flow! + (break) voltage drop Electric lamp (not glowing) no flow! As before, with no ow of electrons the entire potential (voltage) of the battery is available across the break, waiting for the opportunity of a connection to bridge across that break and permit electron ow again. This condition is known as an open circuit, where a break in the continuity of the circuit prevents current throughout. All it takes is a single break in continuity to \"open\" a circuit. Once any breaks have been connected once again and the continuity of the circuit re-established, it is known as a closed circuit. What we see here is the basis for switching lamps on and o\ufb01 by switches. Because any break in a circuit\u2019s continuity results in current stopping throughout the entire circuit, we can use a device designed to intentionally break that continuity (called a switch). This switch can be mounted at any convenient location that we can run wires to. It controls the ow of electrons in the hole circuit: 253 switch Battery - + It doesn\u2019t matter how twisted or convoluted a route the wires take conducting current, so long as they form a complete, uninterrupted loop (circuit). This is how a switch mounted on the wall of a house can control a lamp that is mounted down a long hallway, or even in another room, far away from the switch. The switch itself is constructed of a pair of conductive contacts (usually made of some kind of metal) forced together by a mechanical lever actuator or pushbutton. When the contacts touch each other, electrons are able to ow from one to the other and the circuit\u2019s continuity is established; when the contacts are separated, electron ow from one to the other is prevented by the insulation of the air between, and the circuit\u2019s continuity is broken. In keeping with the \"open\" and \"closed\" terminology of circuits, a switch that is making contact from one connection terminal to the other provides continuity for electrons to ow through, and is called a closed switch. Conversely, a switch that is breaking continuity won\u2019t allow electrons to pass through and is called an open switch. This terminology is often confusing to the new student of electronics, because the words \"open\" and \"closed\" are commonly understood in the context of a door", ", where \"open\" is equated with free passage and \"closed\" with blockage. With electrical switches, these terms have opposite meaning: \"open\" means no ow while \"closed\" means free passage of electrons. 14.5 Voltage and current in a practical circuit Because it takes energy to force electrons to ow against the opposition of a resistance, there will be voltage manifested (or \"dropped\") between any points in a circuit with resistance between them. It is important to note that although the amount of current is uniform in a simple circuit, the amount of voltage between di\ufb01erent sets of points in a single circuit may vary considerably: 254 same rate of current... Battery 1 - + 4 2 3... at all points in this circuit (NOTE TO SELF: How do you actually measure currents and voltages? In the next two paragraphs there\u2019s a lot of ado what we \u2019see\u2019 at di\ufb01erent points of a circuit, but I don\u2019t see anything!) Take this circuit as an example. We labelled four points with the numbers 1, 2, 3, and 4. The amount of current conducted through the wire between points 1 and 2 is exactly the same as the amount of current conducted through the lamp (between points 2 and 3). This same quantity of current passes through the wire between points 3 and 4, and through the battery (between points 1 and 4). However, we will \ufb02nd the voltage appearing between any two of these points to be directly proportional to the resistance within the conductive path between those two points. In a normal lamp circuit, the resistance of a lamp will be much greater than the resistance of the connecting wires. So we should expect to see a substantial amount of voltage drop between points 2 and 3, and only a very small one between points 1 and 2, or between 3 and 4. The voltage drop between points 1 and 4, of course, will be the full voltage o\ufb01ered by the battery. This will be only slightly higher than the voltage drop across the lamp (between points 2 and 3). This, again, is analogous to the water reservoir system: 255 2 Reservoir 1 Waterwheel (energy released) 3 (energy stored) Pump 4 Pond Between points 2 and 3, where the falling water is releasing energy at the water-wheel, there is a di\ufb01erence of pressure between the two points. This reects the opposition to the ow of water through the water-wheel. From point 1 to point 2, or from point 3 to point 4, where water is owing freely through reservoirs with little opposition, there is little or no di\ufb01erence of pressure (no potential energy). However, the rate of water ow in this continuous system is the same everywhere (assuming the water levels in both pond and reservoir are unchanging): through the pump, through the water-wheel, and through all the pipes. So it is with simple electric circuits: the rate of electron ow is the same at every point in the circuit, although voltages may di\ufb01er between di\ufb01erent sets of points. 14.6 Direction of current ow in a circuit We know now that the moving charges in an electrical ciruit are the negatively chargend electrons. These electrons naturally ow from the negative pole of a battery to the positive pole. This form of symbology became known as electron ow notation: Electron flow notation + - Electric charge moves from the negative (surplus) side of the battery to the positive (deficiency) side. 256 However, for historical reasons the current ow in a circuit is conventionally denoted in the opposite direction. That is, it ows from the positive pole to the negative one. This became known as conventional ow notation: Conventional flow notation + - Electric charge moves from the positive (surplus) side of the battery to the negative (deficiency) side. In conventional ow notation, we show the motion of charge according to the (technically. This way the labels make sense, but the direction of charge ow is incorrect) labels of + and incorrect. \u00a1 Does it matter, really, how we designate charge ow in a circuit? Not really, so long as we\u2019re consistent in the use of our symbols. You may follow an imagined direction of current (conventional ow) or the actual (electron ow) with equal success insofar as circuit analysis is concerned. Concepts of voltage, current, resistance, continuity, and even mathematical treatments such as Ohm\u2019s Law (section 2 (NOTE TO SELF: make this reference dynamic)) and Kirchho\ufb01\u2019s Laws (section 6 (NOTE TO SELF: make this reference dynamic)) remain just as valid with either style of notation. Aside: Benjamin Franklin made a conjecture regarding the direction of charge ow when rubbing smooth wax with rough wool. By assuming that the observed charges ow from the wax to the wool, he set the precedent for electrical", " notation that exists to this day. Because Franklin assumed electric charge moved in the opposite direction that it actually does, electrons are said to have a negative charge, and so objects he called \"negative\" (representing a de\ufb02ciency of charge) actually have a surplus of electrons. By the time the true direction of electron ow was discovered, the nomenclature of \"positive\" and \"negative\" had already been so well established in the scienti\ufb02c community that no e\ufb01ort was made to change it. It would have made more sense to call electrons \"positive\" in referring to \"excess\" charge. You see, the terms \"positive\" and \"negative\" are human inventions, and as such have no absolute meaning beyond our own conventions of language and scienti\ufb02c description. Franklin could have just as easily referred to a surplus of charge as \"black\" and a de\ufb02ciency as \"white\", in which case scientists would speak of electrons having a \"white\" charge. However, because we tend to associate the word \"positive\" with \"surplus\" and \"negative\" with \"de\ufb02ciency,\" the standard label for electron charge does seem backward. As discussed above, many engineers decided to retain the old concept of electricity with \"positive\" referring to a surplus of charge, and label charge ow (current) accordingly. 257 14.7 How voltage, current, and resistance relate First lets recap some of the ideas we have learnt so far. We will need these to understand how voltage, current and resistance relate. An electric circuit is formed when a conductive path is created to allow free electrons to continuously move. This continuous movement of free electrons through the conductors of a circuit is called a current. It is often referred to in terms of \"ow,\" just like the ow of a liquid through a hollow pipe. The force motivating electrons to \"ow\" in a circuit is called voltage. Voltage is a speci\ufb02c measure of potential energy that is always relative between two points. When we speak of a certain amount of voltage being present in a circuit, we are referring to the measurement of how much potential energy exists to move electrons from one particular point in that circuit to another particular point. Without reference to two particular points, the term \"voltage\" has no meaning. Free electrons tend to move through conductors with some degree of friction, or opposition to motion. This opposition to motion is more properly called resistance. The amount of current in a circuit depends on the amount of voltage available to motivate the electrons (NOTE TO SELF: Motivated electrons?), and also the amount of resistance in the circuit to oppose electron ow. Just like voltage, resistance is a quantity relative between two points. For this reason, the quantities of voltage and resistance are often stated as being \"between\" or \"across\" two points in a circuit. To be able to make meaningful statements about these quantities in circuits, we need to be able to describe their quantities in the same way that we might quantify mass, temperature, volume, length, or any other kind of physical quantity. For mass we might use the units of \"pound\" or \"gram\". For temperature we might use degrees Fahrenheit or degrees Celsius. Here are the standard units of measurement for electrical current, voltage, and resistance: Quantity Current Voltage Resistance Charge Symbol I V (or E) R Q Unit of Measurement Abbreviation of Unit Ampere volt Ohm coulomb A V \u203a C The \"symbol\" given for each quantity is the standard alphabetical letter used to represent that quantity in an algebraic equation. Standardized letters like these are common in the disciplines of physics and engineering, and are internationally recognized. The \"unit abbreviation\" for each quantity represents the alphabetical symbol used as a shorthand notation for its particular unit of measurement. And, yes, that strange-looking \"horseshoe\" symbol is the capital Greek letter \u203a (called omega), just a character in a foreign alphabet (apologies to any Greek readers here). Aside: Each unit of measurement is named after a famous experimenter in electricity: The amp after the Frenchman Andre M. Ampere, the volt after the Italian Alessandro Volta, and the ohm after the German Georg Simon Ohm. The mathematical symbol for each quantity is meaningful as well. The \"R\" for resistance and the \"V \" for voltage are both self-explanatory. The \"I\" is thought to have been meant to represent \"Intensity\" (of electron ow). The other symbol for voltage, \"E\", stands for \"Electromotive force\". The symbols \"E\" and \"V \" are interchangeable for the most part, although some texts reserve \"E\" to represent voltage across a source (such as a battery or generator) and \"V \" to represent voltage across anything else. 258 One foundational unit of electrical measurement is the unit of the coulomb. It is a measure of", " electric charge proportional to the number of electrons in an imbalanced state. One coulomb of charge is rougly equal to the charge of 6,250,000,000,000,000,000 electrons. The symbol for electric charge quantity is the capital letter \"Q\", with the unit of coulombs abbreviated by the capital letter \"C\". It so happens that the unit for electron ow, the ampere, is equal to 1 coulomb of electrons passing by a given point in a circuit in 1 second of time. Cast in these terms, current is the rate of electric charge motion through a conductor. As stated before, voltage is the measure of potential energy per unit charge available to motivate electrons from one point to another. Before we can precisely de\ufb02ne what a \"volt\" is, we must understand how to measure this quantity we call \"potential energy\". The general metric unit for energy of any kind is the joule, equal to the amount of work performed by a force of 1 newton exerted through a motion of 1 meter (in the same direction). (NOTE TO SELF: Make a reference to the Mechanics chapter.) De\ufb02ned in these scienti\ufb02c terms, 1 volt is equal to 1 joule of electric potential energy per (divided by) 1 coulomb of charge. Thus, a 9 volt battery releases 9 joules of energy for every coulomb of electrons moved through a circuit. These units and symbols for electrical quantities will become very important to know as we begin to explore the relationships between them in circuits. The \ufb02rst, and perhaps most important, relationship between current, voltage, and resistance is called Ohm\u2019s Law. It states that the amount of electric current through a metal conductor in a circuit is directly proportional to the voltage impressed across it, for any given temperature. It can be expressed in the form of a simple equation, describing how voltage, current, and resistance interrelate: \u00a2 In this algebraic expression, voltage (V ) is equal to current (I) multiplied by resistance (R). V = I R : (14.1) Aside: Georg Simon Ohm published his law in his 1827 paper, The Galvanic Circuit Investigated Mathematically. Using algebra techniques, we can manipulate this equation into two variations, solving for I and for R, respectively: Let\u2019s see how these equations might work to help us analyze simple circuits14.2) electron flow Battery + - Electric lamp (glowing) electron flow 259 (NOTE TO SELF: replace E by V) In the above circuit, there is only one source of voltage (the battery, on the left) and only one source of resistance to current (the lamp, on the right). This makes it very easy to apply Ohm\u2019s Law. If we know the values of any two of the three quantities (voltage, current, and resistance) in this circuit, we can use Ohm\u2019s Law to determine the third. In this \ufb02rst example, we will calculate the amount of current (I) in a circuit, given values of voltage (V ) and resistance (R): Worked Example 81 Question: What is the amount of current (I) in this circuit? Battery E = 12 V + - I =??? I =??? Lamp R = 3 W (NOTE TO SELF: replace E by V) Answer: I = V R = 12 V 3 \u203a = 4 A : (14.3) In the second example, we will calculate the amount of resistance (R) in a circuit, given values of voltage (V ) and current (I): Worked Example 82 Question: What is the amount of resistance (R) o\ufb01ered by the lamp? 260 Battery E = 36 Lamp R =??? (NOTE TO SELF: replace E by V) Answer: R = V I = 36 V 4 A = 9 \u203a : (14.4) In the last example, we will calculate the amount of voltage supplied by a battery, given values of current (I) and resistance (R): Worked Example 83 Question: What is the amount of voltage provided by the battery? Battery E =??? + - I = 2 A I = 2 A Lamp R = 7 W (NOTE TO SELF: replace E by V) Answer: V = I \u00a2 R = (2 A) \u00a2 (7 \u203a) = 14 V : (14.5) 261 Ohm\u2019s Law is a very simple and useful tool for analyzing electric circuits. It is used so often in the study of electricity and electronics that it needs to be committed to memory by the serious student. All you need to do is commit V = I R to memory and derive the other two formulae from that when you need them! \u00a2 14.8 Voltmeters, ammeters, and ohmmeters As we have seen", " in previous sections, an electric circuit is made up of a number of di\ufb01erent components such as batteries and resistors. In electronics, there are many types of meters used to measure the properties of the individual components of an electric circuit. For example, one may be interested in measuring the amount of current owing through a circuit, or measure the voltage provided by a battery. In this section we will discuss the practical usage of voltmeters, ammeters, and ohmmeters. A voltmeter is an instrument for measuring the voltage between two points in an electric circuit. In analogy with a water circuit, a voltmeter is like a meter designed to measure pressure di\ufb01erence. Since one is interested in measuring the voltage between two points in a circuit, a voltmeter must be connected in parallel with the portion of the circuit on which the measurement is made: V The above illustration shows a voltmeter connected in parallel with a battery. One lead of the voltmeter is connected to one end of the battery and the other lead is connected to the opposite end. The voltmeter may also be used to measure the voltage across a resistor or any other component of a circuit that has a voltage drop. An ammeter is an instrument used to measure the ow of electric current in a circuit. Since one is interested in measuring the current owing through a circuit component, the ammeter must be connected in series with the measured circuit component: A An ohmmeter is an instrument for measuring electrical resistance. The basic ohmmeter can function much like an ammeter. The ohmmeter works by suppling a constant voltage to the resistor and measuring the current owing through it. The measured current is then converted into a corresponding resistance reading through Ohm\u2019s law. One cautionary detail needs to be 262 mentioned with regard to ohmmeters: they only function correctly when measuring resistance that is not being powered by a voltage or current source. In other words, you cannot measure the resistance of a component that is already connected to a circuit. The reason for this is simple: the ohmmeter\u2019s accurate indication depends only on its own source of voltage. The presence of any other voltage across the measured circuit component interferes with the ohmmeter\u2019s operation. The circuit diagram below shows an ohmmeter solely connected with a resistor: \u203a The table below summarizes the use of each measuring instrument that we discussed and the way it should be connected to a circuit component. Instrument Measured Quantity Proper Connection Voltmeter Ammeter Ohmmeter In Parallel In Series Only with Resistor Voltage Current Resistance 14.9 An analogy for Ohm\u2019s Law In our water-and-pipe analogy, Ohm\u2019s Law also exists. Think of a water pump that exerts pressure (voltage) to push water around a \"circuit\" (current) through a restriction (resistance). If the resistance to water ow stays the same and the pump pressure increases, the ow rate must also increase. \" V = \" I R If the pressure stays the same and the resistance increases (making it more di\u2013cult for the water to ow), then the ow rate must decrease: V = # I \" R If the ow rate stays the same while the resistance to ow decreases, the required pressure from the pump decreases: # V = I # R As odd as it may seem, the actual mathematical relationship between pressure, ow, and resistance is actually more complex for uids like water than it is for electrons. If you pursue further studies in physics, you will discover this for yourself. Thankfully for the electronics student, the mathematics of Ohm\u2019s Law is very simple. 263 14.10 Power in electric circuits In addition to voltage and current, there is another measure of free electron activity in a circuit: power. The concept of power was introduced in Chapter 8. Basically, it is a measure of how rapidly a standard amount of work is done. In electric circuits, power is a function of both voltage and current: P = IV: So power (P ) is exactly equal to current (I) multiplied by voltage (V ) and there is no extra constant of proportionality. The unit of measurement for power is the Watt (abbreviated W). Aside: You can verify for yourself that the eqution for power in an electric cicuit makes sense. Remember that voltage is the speci\ufb02c work (or potential energy) per unit charge, while current is the amount electric charge that ow though a conductor per time unit. So the product of those two qunatities is the oumount of work per time unit, which is exactly the power. It is important to realise that only the combination of a voltage drop and the ow of current corresponds to power. So, a circuit with high voltage and low current may be dissipating the same amount of power as a circuit with low voltage and high current. In an open circuit, where voltage is present between the terminals of the source and there", " is zero current, there is zero power dissipated, no matter how great that voltage may be. Since P = IV and I = 0, the power dissipated in any open circuit must be zero. 14.11 Calculating electric power We\u2019ve seen the formula for determining the power in an electric circuit: by multiplying the voltage in volts by the current in Amperes we arrive at an answer in watts.\" Let\u2019s apply this to a circuit example: Battery E = 18 V + - I =??? I =??? Lamp R = 3 W In the above circuit, we know we have a battery voltage of 18 Volts and a lamp resistance of 3 \u203a. Using Ohm\u2019s Law to determine current, we get: I = V R = 18V 3\u203a = 6A: 264 Now that we know the current, we can take that value and multiply it by the voltage to determine power: P = IV = (6A)(18V = 108W: Answer: the lamp is dissipating (releasing) 108 W of power, most likely in the form of both light and heat. Let\u2019s try taking that same circuit and increasing the battery voltage to see what happens: Battery E = 36 V + - I =??? I =??? Lamp R = 3 W Since the resistance stays the same, the current will increase when we increase the voltage: 36V 3\u203a Note that Ohm\u2019s Law is linear, so the current exactly doubles when we double the voltage. = 12A: V R I = = Now, let\u2019s calculate the power: P = IV = (12A)(36V) = 432W: Notice that the power has increased just as we might have suspected, but it increased quite a bit more than the current. Why is this? Because power is a function of voltage multiplied by current, and both voltage and current doubled from their previous values, the power will increase by a factor of 2 x 2, or 4: the ratio of the new power 432 W and the old power 108 W, is exactly 4. We could in fact have arrived at this result without the intermediate step of calculating the current. From we can expres power directly as a function of voltage: I = V R and P = IV The analogous relation between power and current is P = IIR = I 2R: 265 Interesting Fact: It was James Prescott Joule, not Georg Simon Ohm, who \ufb02rst discovered the mathematical relationship between power dissipation and current through a resistance. This discovery, published in 1841, followed the form of the last equation (P = I2R), and is properly known as Joule\u2019s Law. However, these power equations are so commonly associated with the Ohm\u2019s Law equations relating voltage, current, and resistance (V = IR ; I = V =R ; and R = V =I) that they are frequently credited to Ohm. 14.12 Resistors Because the relationship between voltage, current, and resistance in any circuit is so regular, we can reliably control any variable in a circuit simply by controlling the other two. Perhaps the easiest variable in any circuit to control is its resistance. This can be done by changing the material, size, and shape of its conductive components (remember how the thin metal \ufb02lament of a lamp created more electrical resistance than a thick wire?). Special components called resistors are made for the express purpose of creating a precise quantity of resistance for insertion into a circuit. They are typically constructed of metal wire or carbon, and engineered to maintain a stable resistance value over a wide range of environmental conditions. Unlike lamps, they do not produce light, but they do produce heat as electric power is dissipated by them in a working circuit. Typically, though, the purpose of a resistor is not to produce usable heat, but simply to provide a precise quantity of electrical resistance. The most common schematic symbol for a resistor is a zig-zag line: Resistor values in ohms are usually shown as an adjacent number, and if several resistors are present in a circuit, they will be labeled with a unique identi\ufb02er number such as R1, R2, R3, etc. As you can see, resistor symbols can be shown either horizontally or vertically: R1 150 This is resistor \"R1\" with a resistance value of 150 ohms. R2 25 This is resistor \"R2\" with a resistance value of 25 ohms. In keeping more with their physical appearance, an alternative schematic symbol for a resistor looks like a small, rectangular box: Resistors can also be shown to have varying rather than \ufb02xed resistances. This might be for the purpose of describing an actual physical device designed for the purpose of providing an adjustable resistance, or it could be to show some component that just happens to have an unstable resistance: 266 variable resistance... or... In fact, any time", " you see a component symbol drawn with a diagonal arrow through it, that component has a variable rather than a \ufb02xed value. This symbol \"modi\ufb02er\" (the diagonal arrow) is standard electronic symbol convention. In practice, resistors are not only rated in terms of their resistance in Aside: ohms, but also in term the amount of power they can dissipate in watts. Resistors dissipate heat as the electric currents through them overcome the \"friction\" of their resistance and can in fact become quite hot in actual applications. Most resistors found in small electronic devices such as portable radios are rated at 1/4 (0.25) watt or less. The power rating of any resistor is roughly proportional to its physical size. Also note how resistances (in ohms) have nothing to do with size! 14.13 Nonlinear conduction Ohm\u2019s Law is a powerful tool for analyzing electric circuits, but it has a practical limitation. In the application of Ohm\u2019s Law, we alwasy assume that the restistance does not change as a function of voltage and current. For most conductors, this is a reasonable approximation as long ads the temperature does not change too much. In a normal lightbulb, the resistance of the \ufb02lament wire will increase dramatically as it warms from room temperature to operating temperature. If we increase the supply voltage in a real lamp circuit, the resulting increase in current causes the \ufb02lament to increase in temperature, which increases its resistance. This e\ufb01ectively limits the increase in current. Consequently, voltage and current do not follow the simple equation I = V =R, with a constant R (of 3 \u203a ion our example). The lamp\u2019s \ufb02lament resistance does not remain stable for di\ufb01erent currents. The phenomenon of resistance changing with variations in temperature is one shared by almost all metals, of which most wires are made. For most applications, these changes in resistance are small enough to be ignored. In the application of metal lamp \ufb02laments, which increase a lot in temperature (up to about 1000oC, and starting from room temperature) the change is quite large. A more realistic analysis of a lamp circuit over several di\ufb01erent values of battery voltage would generate a plot of this shape: 267 I (current) E (voltage) The plot is no longer a straight line. It rises sharply on the left, as voltage increases from zero to a low level. As it progresses to the right we see the line attening out, the circuit requiring greater and greater increases in voltage to achieve equal increases in current. If we apply Ohm\u2019s Law to \ufb02nd the resistance of this lamp circuit with the voltage and current values plotted above, the calculated values will change with voltage or curreny. We could say that the resistance here is nonlinear, increasing with increasing current and voltage. The nonlinearity is caused by the e\ufb01ects of high temperature on the metal wire of the lamp \ufb02lament. 14.14 Circuit wiring So far, we\u2019ve been analyzing single-battery, single-resistor circuits with no regard for the connecting wires between the components, so long as a complete circuit is formed. Does the wire length or circuit \"shape\" matter to our calculations? Let\u2019s look at a couple of circuit con\ufb02gurations and \ufb02nd out: Battery 10 V 1 4 2 Resistor 5 W 3 Battery 10 V 1 4 2 Resistor 5 W 3 268 When we draw wires connecting points in a circuit, we usually assume those wires have negligible resistance. As such, they contribute no appreciable e\ufb01ect to the overall resistance of the circuit, and so the only resistance we have to contend with is the resistance in the components. In the above circuits, the only resistance comes from the 5 \u203a resistors, so that is all we will consider in our calculations. In real life, metal wires actually do have resistance (and so do power sources!), but those resistances are generally so much smaller than the resistance present in the other circuit components that they can be safely ignored. If connecting wire resistance is very little or none, we can regard the connected points in a circuit as being electrically common. That is, points 1 and 2 in the above circuits may be physically joined close together or far apart, and it doesn\u2019t matter for any voltage or resistance measurements relative to those points. The same goes for points 3 and 4. It is as if the ends of the resistor were attached directly across the terminals of the battery, so far as our Ohm\u2019s Law calculations and voltage measurements are concerned. This is useful to know, because it means you can re-draw a circuit diagram or re-wire a circuit, shortening or lengthening the wires as desired without appreciably impacting the circuit\u2019", "s function. All that matters is that the components attach to each other in the same sequence. It also means that voltage measurements between sets of \"electrically common\" points will be the same. That is, the voltage between points 1 and 4 (directly across the battery) will be the same as the voltage between points 2 and 3 (directly across the resistor). Take a close look at the following circuit, and try to determine which points are common to each other: 1 Battery 10 V 4 2 3 Resistor 5 W 6 5 Here, we only have 2 components excluding the wires: the battery and the resistor. Though the connecting wires take a convoluted path in forming a complete circuit, there are several electrically common points in the electrons\u2019 path. Points 1, 2, and 3 are all common to each other, because they\u2019re directly connected together by wire. The same goes for points 4, 5, and 6. The voltage between points 1 and 6 is 10 volts, coming straight from the battery. However, since points 5 and 4 are common to 6, and points 2 and 3 common to 1, that same 10 volts also exists between these other pairs of points: Between points 1 and 4 = 10 volts Between points 2 and 4 = 10 volts Between points 3 and 4 = 10 volts (directly across the resistor) Between points 1 and 5 = 10 volts Between points 2 and 5 = 10 volts Between points 3 and 5 = 10 volts Between points 1 and 6 = 10 volts (directly across the battery) Between points 2 and 6 = 10 volts Between points 3 and 6 = 10 volts Since electrically common points are connected together by (zero resistance) wire, there is no signi\ufb02cant voltage drop between them regardless of the amount of current conducted from one 269 to the next through that connecting wire. Thus, if we were to read voltages between common points, we should show (practically) zero: Between points 1 and 2 = 0 volts Between points 2 and 3 = 0 volts Between points 1 and 3 = 0 volts Between points 4 and 5 = 0 volts Between points 5 and 6 = 0 volts Between points 4 and 6 = 0 volts Points 1, 2, and 3 are electrically common Points 4, 5, and 6 are electrically common This makes sense mathematically, too. With a 10 volt battery and a 5 \u203a resistor, the circuit current will be 2 amps. With wire resistance being zero, the voltage drop across any continuous stretch of wire can be determined through Ohm\u2019s Law as such: E = I R E = (2 A)(0 W) E = 0 V It should be obvious that the calculated voltage drop across any uninterrupted length of wire in a circuit where wire is assumed to have zero resistance will always be zero, no matter what the magnitude of current, since zero multiplied by anything equals zero. Because common points in a circuit will exhibit the same relative voltage and resistance measurements, wires connecting common points are often labeled with the same designation. This is not to say that the terminal connection points are labeled the same, just the connecting wires. Take this circuit as an example: 1 wire #2 2 wire #2 Battery 10 V 4 wire #1 3 Resistor 5 W 6 5 wire #1 wire #1 Points 1, 2, and 3 are all common to each other, so the wire connecting point 1 to 2 is labeled the same (wire 2) as the wire connecting point 2 to 3 (wire 2). In a real circuit, the wire stretching from point 1 to 2 may not even be the same color or size as the wire connecting point 2 to 3, but they should bear the exact same label. The same goes for the wires connecting points 6, 5, and 4. Knowing that electrically common points have zero voltage drop between them is a valuable troubleshooting principle. If I measure for voltage between points in a circuit that are supposed to be common to each other, I should read zero. If, however, I read substantial voltage between those two points, then I know with certainty that they cannot be directly connected together. 270 If those points are supposed to be electrically common but they register otherwise, then I know that there is an \"open failure\" between those points. One \ufb02nal note: for most practical purposes, wire conductors can be assumed to possess zero resistance from end to end. In reality, however, there will always be some small amount of resistance encountered along the length of a wire, unless it\u2019s a superconducting wire. Knowing this, we need to bear in mind that the principles learned here about electrically common points are all valid to a large degree, but not to an absolute degree. That is, the rule that electrically common points are guaranteed to have zero voltage between them is more accurately stated as such: electrically common points will have very little voltage dropped between them. That small, virtually unavoidable trace of resistance found in any piece of connecting wire is bound to create a small voltage", " across the length of it as current is conducted through. So long as you understand that these rules are based upon ideal conditions, you won\u2019t be perplexed when you come across some condition appearing to be an exception to the rule. 14.15 Polarity of voltage drops We can trace the direction that electrons will ow in the same circuit by starting at the negative (-) terminal and following through to the positive (+) terminal of the battery, the only source of voltage in the circuit. From this we can see that the electrons are moving counter-clockwise, from point 6 to 5 to 4 to 3 to 2 to 1 and back to 6 again. As the current encounters the 5 \u203a resistance, voltage is dropped across the resistor\u2019s ends. The signs of this voltage drop is negative (-) at point 4 with respect to positive (+) at point 3. We can mark the polarity of the resistor\u2019s voltage drop with these negative and positive symbols, in accordance with the direction of current (whichever end of the resistor the current is entering is negative with respect to the end of the resistor it is exiting: Battery 10 V 1 + - 6 2 3 current current - + Resistor 5 W 4 5 We could make our table of voltages a little more complete by marking the polarity of the voltage for each pair of points in this circuit: Between points 1 (+) and 4 (-) = 10 volts Between points 2 (+) and 4 (-) = 10 volts Between points 3 (+) and 4 (-) = 10 volts Between points 1 (+) and 5 (-) = 10 volts Between points 2 (+) and 5 (-) = 10 volts Between points 3 (+) and 5 (-) = 10 volts Between points 1 (+) and 6 (-) = 10 volts Between points 2 (+) and 6 (-) = 10 volts Between points 3 (+) and 6 (-) = 10 volts 271 While it might seem a little silly to document polarity of voltage drop in this circuit, it is an important concept to master. It will be critically important in the analysis of more complex circuits involving multiple resistors and/or batteries. It should be understood that polarity has nothing to do with Ohm\u2019s Law: there will never be negative voltages, currents, or resistance entered into any Ohm\u2019s Law equations! There are other mathematical principles of electricity that do take polarity into account through the use of signs (+ or -), but not Ohm\u2019s Law. 14.16 What are \"series\" and \"parallel\" circuits? Circuits consisting of just one battery and one load resistance are very simple to analyze, but they are not often found in practical applications. Usually, we \ufb02nd circuits where more than two components are connected together. There are two basic ways in which to connect more than two circuit components: series and parallel. First, an example of a series circuit: Series R1 R3 1 + - 4 2 R2 3 Here, we have three resistors (labeled R1, R2, and R3), connected in a long chain from one terminal of the battery to the other. (It should be noted that the subscript labeling { those little numbers to the lower-right of the letter \"R\" { are unrelated to the resistor values in ohms. They serve only to identify one resistor from another.) The de\ufb02ning characteristic of a series circuit is that there is only one path for electrons to ow. In this circuit the electrons ow in a counter-clockwise direction, from point 4 to point 3 to point 2 to point 1 and back around to 4. Now, let\u2019s look at the other type of circuit, a parallel con\ufb02guration: 1 + - 8 Parallel 2 3 4 R1 R2 R3 7 6 5 Again, we have three resistors, but this time they form more than one continuous path for electrons to ow. There\u2019s one path from 8 to 7 to 2 to 1 and back to 8 again. There\u2019s another from 8 to 7 to 6 to 3 to 2 to 1 and back to 8 again. And then there\u2019s a third path from 8 to 7 to 272 6 to 5 to 4 to 3 to 2 to 1 and back to 8 again. Each individual path (through R1, R2, and R3) is called a branch. The de\ufb02ning characteristic of a parallel circuit is that all components are connected between the same set of electrically common points. Looking at the schematic diagram, we see that points 1, 2, 3, and 4 are all electrically common. So are points 8, 7, 6, and 5. Note that all resistors as well as the battery are connected between these two sets of points. And, of course, the complexity doesn\u2019t stop at simple series and parallel either! We can have circuits that are a combination of series and parallel, too: Series-parallel R1 1", " + - 6 2 5 R2 3 4 R3 In this circuit, we have two loops for electrons to ow through: one from 6 to 5 to 2 to 1 and back to 6 again, and another from 6 to 5 to 4 to 3 to 2 to 1 and back to 6 again. Notice how both current paths go through R1 (from point 2 to point 1). In this con\ufb02guration, we\u2019d say that R2 and R3 are in parallel with each other, while R1 is in series with the parallel combination of R2 and R3. This is just a preview of things to come. Don\u2019t worry! We\u2019ll explore all these circuit con\ufb02g- urations in detail, one at a time! The basic idea of a \"series\" connection is that components are connected end-to-end in a line to form a single path for electrons to ow: Series connection R1 R2 R3 R4 only one path for electrons to flow! The basic idea of a \"parallel\" connection, on the other hand, is that all components are connected across each other\u2019s leads. In a purely parallel circuit, there are never more than two sets of electrically common points, no matter how many components are connected. There are many paths for electrons to ow, but only one voltage across all components: 273 Parallel connection These points are electrically common R1 R2 R3 R4 These points are electrically common Series and parallel resistor con\ufb02gurations have very di\ufb01erent electrical properties. We\u2019ll ex- plore the properties of each con\ufb02guration in the sections to come. 14.17 Simple series circuits Let\u2019s start with a series circuit consisting of three resistors and a single battery: 9 V 1 + - 4 R1 3 kW 5 kW R3 2 10 kW R2 3 The \ufb02rst principle to understand about series circuits is that the amount of current is the same through any component in the circuit. This is because there is only one path for electrons to ow in a series circuit, and because free electrons ow through conductors like marbles in a tube, the rate of ow (marble speed) at any point in the circuit (tube) at any speci\ufb02c point in time must be equal. From the way that the 9 volt battery is arranged, we can tell that the electrons in this circuit will ow in a counter-clockwise direction, from point 4 to 3 to 2 to 1 and back to 4. However, we have one source of voltage and three resistances. How do we use Ohm\u2019s Law here? An important caveat to Ohm\u2019s Law is that all quantities (voltage, current, resistance, and power) must relate to each other in terms of the same two points in a circuit. For instance, with a single-battery, single-resistor circuit, we could easily calculate any quantity because they all applied to the same two points in the circuit: 274 kW 3 =I 9 volts 3 kW = 3 mA Since points 1 and 2 are connected together with wire of negligible resistance, as are points 3 and 4, we can say that point 1 is electrically common to point 2, and that point 3 is electrically common to point 4. Since we know we have 9 volts of electromotive force between points 1 and 4 (directly across the battery), and since point 2 is common to point 1 and point 3 common to point 4, we must also have 9 volts between points 2 and 3 (directly across the resistor). Therefore, we can apply Ohm\u2019s Law (I = E/R) to the current through the resistor, because we know the voltage (E) across the resistor and the resistance (R) of that resistor. All terms (E, I, R) apply to the same two points in the circuit, to that same resistor, so we can use the Ohm\u2019s Law formula with no reservation. However, in circuits containing more than one resistor, we must be careful in how we apply Ohm\u2019s Law. In the three-resistor example circuit below, we know that we have 9 volts between points 1 and 4, which is the amount of electromotive force trying to push electrons through the series combination of R1, R2, and R3. However, we cannot take the value of 9 volts and divide it by 3k, 10k or 5k \u203a to try to \ufb02nd a current value, because we don\u2019t know how much voltage is across any one of those resistors, individually. 9 V 1 + - 4 R1 3 kW 5 kW R3 2 10 kW R2 3 The \ufb02gure of 9 volts is a total quantity for the whole circuit, whereas the \ufb02gures of 3k, 10k,", " and 5k \u203a are individual quantities for individual resistors. If we were to plug a \ufb02gure for total voltage into an Ohm\u2019s Law equation with a \ufb02gure for individual resistance, the result would not relate accurately to any quantity in the real circuit. For R1, Ohm\u2019s Law will relate the amount of voltage across R1 with the current through R1, given R1\u2019s resistance, 3k\u203a: 275 IR1 = ER1 3 kW ER1 = IR1(3 kW) But, since we don\u2019t know the voltage across R1 (only the total voltage supplied by the battery across the three-resistor series combination) and we don\u2019t know the current through R1, we can\u2019t do any calculations with either formula. The same goes for R2 and R3: we can apply the Ohm\u2019s Law equations if and only if all terms are representative of their respective quantities between the same two points in the circuit. So what can we do? We know the voltage of the source (9 volts) applied across the series combination of R1, R2, and R3, and we know the resistances of each resistor, but since those quantities aren\u2019t in the same context, we can\u2019t use Ohm\u2019s Law to determine the circuit current. If only we knew what the total resistance was for the circuit: then we could calculate total current with our \ufb02gure for total voltage (I=E/R). This brings us to the second principle of series circuits: the total resistance of any series circuit is equal to the sum of the individual resistances. This should make intuitive sense: the more resistors in series that the electrons must ow through, the more di\u2013cult it will be for those electrons to ow. In the example problem, we had a 3 k\u203a, 10 k\u203a, and 5 k\u203a resistor in series, giving us a total resistance of 18 k\u203a: Rtotal = R1 + R2 + R3 + Rtotal = 3 kW 10 kW 5 kW + Rtotal = 18 kW In essence, we\u2019ve calculated the equivalent resistance of R1, R2, and R3 combined. Knowing this, we could re-draw the circuit with a single equivalent resistor representing the series combination of R1, R2, and R3: 9 V 1 + - 4 R1 + R2 + R3 = 18 kW Now we have all the necessary information to calculate circuit current, because we have the voltage between points 1 and 4 (9 volts) and the resistance between points 1 and 4 (18 k\u203a): Itotal= Etotal Rtotal Itotal = 9 volts 18 kW = 500 mA Knowing that current is equal through all components of a series circuit (and we just determined the current through the battery), we can go back to our original circuit schematic and note the current through each component: 276 9 V 1 + - 4 R1 3 kW I = 500 mA I = 500 mA 2 R2 10 kW R3 5 kW 3 Now that we know the amount of current through each resistor, we can use Ohm\u2019s Law to determine the voltage drop across each one (applying Ohm\u2019s Law in its proper context): ER1 = IR1 R1 ER2 = IR2 R2 ER3 = IR3 R3 ER1 = (500 mA)(3 kW) = 1.5 V ER2 = (500 mA)(10 kW) = 5 V ER3 = (500 mA)(5 kW) = 2.5 V Notice the voltage drops across each resistor, and how the sum of the voltage drops (1.5 + 5 + 2.5) is equal to the battery (supply) voltage: 9 volts. This is the third principle of series circuits: that the supply voltage is equal to the sum of the individual voltage drops. However, the method we just used to analyze this simple series circuit can be streamlined for better understanding. By using a table to list all voltages, currents, and resistances in the circuit, it becomes very easy to see which of those quantities can be properly related in any Ohm\u2019s Law equation: R1 R2 R3 Total E I R Volts Amps Ohms Ohm\u2019s Law Ohm\u2019s Law Ohm\u2019s Law Ohm\u2019s Law The rule with such a table is to apply Ohm\u2019s Law only to the values within each vertical column. For instance, ER1 only with IR1 and R1; ER2 only with IR2 and R2; etc. You begin your analysis by \ufb02lling in those elements of the table that are given to you from the beginning: R1 R2 R3 3k 10k 5k E I R Total 9 Volts Amps", " Ohms As you can see from the arrangement of the data, we can\u2019t apply the 9 volts of ET (total 277 voltage) to any of the resistances (R1, R2, or R3) in any Ohm\u2019s Law formula because they\u2019re in di\ufb01erent columns. The 9 volts of battery voltage is not applied directly across R1, R2, or R3. However, we can use our \"rules\" of series circuits to \ufb02ll in blank spots on a horizontal row. In this case, we can use the series rule of resistances to determine a total resistance from the sum of individual resistances: R1 R2 R3 Total 9 3k 10k 5k 18k E I R Volts Amps Ohms Rule of series circuits RT = R1 + R2 + R3 Now, with a value for total resistance inserted into the rightmost (\"Total\") column, we can apply Ohm\u2019s Law of I=E/R to total voltage and total resistance to arrive at a total current of 500 \u201eA: R1 R2 R3 3k 10k 5k E I R Volts Amps Ohms Total 9 500m 18k Ohm\u2019s Law Then, knowing that the current is shared equally by all components of a series circuit (another \"rule\" of series circuits), we can \ufb02ll in the currents for each resistor from the current \ufb02gure just calculated: R1 R2 R3 E I R 500m 3k 500m 10k 500m 5k Total 9 500m 18k Volts Amps Ohms Rule of series circuits IT = I1 = I2 = I3 Finally, we can use Ohm\u2019s Law to determine the voltage drop across each resistor, one column at a time: 278 R1 1.5 500m 3k E I R R2 5 500m 10k R3 2.5 500m 5k Total 9 500m 18k Volts Amps Ohms Ohm\u2019s Law Ohm\u2019s Law Ohm\u2019s Law 14.18 Simple parallel circuits Let\u2019s start with a parallel circuit consisting of three resistors and a single battery: 1 + - 9 V 2 3 4 R1 10 kW R2 2 kW R3 1 kW 8 7 6 5 The \ufb02rst principle to understand about parallel circuits is that the voltage is equal across all components in the circuit. This is because there are only two sets of electrically common points in a parallel circuit, and voltage measured between sets of common points must always be the same at any given time. Therefore, in the above circuit, the voltage across R1 is equal to the voltage across R2 which is equal to the voltage across R3 which is equal to the voltage across the battery. This equality of voltages can be represented in another table for our starting values: R1 9 10k E I R R2 9 2k R3 9 1k Total 9 Volts Amps Ohms Just as in the case of series circuits, the same caveat for Ohm\u2019s Law applies: values for voltage, current, and resistance must be in the same context in order for the calculations to work correctly. However, in the above example circuit, we can immediately apply Ohm\u2019s Law to each resistor to \ufb02nd its current because we know the voltage across each resistor (9 volts) and the resistance of each resistor: 279 IR1 = ER1 R1 IR2 = ER2 R2 IR3 = ER3 R3 IR1 = 9 V 10 kW = 0.9 mA IR2 = IR3 = E I R 9 V 2 kW 9 V 1 kW R1 9 0.9m 10k = 4.5 mA = 9 mA R2 9 4.5m 2k R3 9 9m 1k Total 9 Volts Amps Ohms Ohm\u2019s Law Ohm\u2019s Law Ohm\u2019s Law At this point we still don\u2019t know what the total current or total resistance for this parallel circuit is, so we can\u2019t apply Ohm\u2019s Law to the rightmost (\"Total\") column. However, if we think carefully about what is happening it should become apparent that the total current must equal the sum of all individual resistor (\"branch\") currents: 1 + - 9 V IR1 IT IT 2 3 4 IR2 IR3 R1 10 kW 2 kW R2 R3 1 kW 8 6 7 5 As the total current exits the negative (-) battery terminal at point 8 and travels through the circuit, some of the ow splits o\ufb01 at point 7 to go up through R1, some more splits o\ufb01 at point 6 to go up through R2, and the remainder goes up through R3. Like a river branching into several smaller streams, the combined", " ow rates of all streams must equal the ow rate of the whole river. The same thing is encountered where the currents through R1, R2, and R3 join to ow back to the positive terminal of the battery (+) toward point 1: the ow of electrons from point 2 to point 1 must equal the sum of the (branch) currents through R1, R2, and R3. This is the second principle of parallel circuits: the total circuit current is equal to the sum of the individual branch currents. Using this principle, we can \ufb02ll in the IT spot on our table with the sum of IR1, IR2, and IR3: 280 R1 9 0.9m 10k E I R R2 9 4.5m 2k R3 9 9m 1k Total 9 14.4m Volts Amps Ohms Rule of parallel circuits Itotal = I1 + I2 + I3 Finally, applying Ohm\u2019s Law to the rightmost (\"Total\") column, we can calculate the total circuit resistance: R1 9 0.9m 10k E I R R2 9 4.5m 2k R3 9 9m 1k Total 9 Volts 14.4m Amps 625 Ohms Rtotal = Etotal Itotal = 9 V 14.4 mA = 625 W Ohm\u2019s Law Please note something very important here. The total circuit resistance is only 625 \u203a: less than any one of the individual resistors. In the series circuit, where the total resistance was the sum of the individual resistances, the total was bound to be greater than any one of the resistors individually. Here in the parallel circuit, however, the opposite is true: we say that the individual resistances diminish rather than add to make the total. This principle completes our triad of \"rules\" for parallel circuits, just as series circuits were found to have three rules for voltage, current, and resistance. Mathematically, the relationship between total resistance and individual resistances in a parallel circuit looks like this: Rtotal = 1 1 R2 + 1 R3 1 R1 + The same basic form of equation works for any number of resistors connected together in parallel, just add as many 1/R terms on the denominator of the fraction as needed to accommodate all parallel resistors in the circuit. 14.19 Power calculations When calculating the power dissipation of resistive components, use any one of the three power equations to derive and answer from values of voltage, current, and/or resistance pertaining to each component: 281 Power equations P = IE P = E2 E R P = I2R This is easily managed by adding another row to our familiar table of voltages, currents, and resistances: E I R P R1 R2 R3 Total Volts Amps Ohms Watts Power for any particular table column can be found by the appropriate Ohm\u2019s Law equation (appropriate based on what \ufb02gures are present for E, I, and R in that column). An interesting rule for total power versus individual power is that it is additive for any con\ufb02guration of circuit: series, parallel, series/parallel, or otherwise. Power is a measure of rate of work, and since power dissipated must equal the total power applied by the source(s) (as per the Law of Conservation of Energy in physics), circuit con\ufb02guration has no e\ufb01ect on the mathematics. 14.20 Correct use of Ohm\u2019s Law When working through worked examples it is important to try to \ufb02gure out what you are doing correctly as well as what you are doing wrong. Make sure you don\u2019t stop doing the good things and try to correct the mistakes. Circuit questions form a large part of high school and early university courses and it is important to understand the concepts properly. One common mistake which students make we\u2019ll discuss here so you know to look out for it when you are working through examples and studying. When applying Ohm\u2019s Laws students often mix up the contexts of voltage, current, and resistance. This means a student might mistakenly use a value for I through one resistor and the value for V across another resistor or a set of connected resistors. Remember this important rule: The variables used in Ohm\u2019s Law equations must be common to the same two points in the circuit under consideration. This is especially important in series-parallel combination circuits where nearby components may have di\ufb01erent values for both voltage drop and current. When using Ohm\u2019s Law to calculate a variable for a single component: be sure the voltage you\u2019re using is solely across that single component and the current you\u2019re referencing is solely through that single component and \u2020 \u2020 the resistance you\u2019re referencing is solely for that single component. \u2020 When calculating a variable for a set of components in", " a circuit, be sure that the voltage, current, and resistance values are speci\ufb02c to that complete set of components only! 282 A good way to remember this is to pay close attention to the two points on either side of the component or set of components. Making sure that the voltage in question is across those two points, that the current in question is the electron ow from one of those points all the way to the other point, that the resistance in question is the equivalent of a single resistor between those two points, and that the power in question is the total power dissipated by all components between those two points. The \"table\" method presented for both series and parallel circuits in this chapter is a way to keep the components correct when using Ohm\u2019s Law. In a table like the one shown below, you are only allowed to apply an Ohm\u2019s Law equation for the values of a single vertical column at a time: R1 R2 R3 Total E I R P Volts Amps Ohms Watts Ohm\u2019s Law Ohm\u2019s Law Ohm\u2019s Law Ohm\u2019s Law Deriving values horizontally across columns is allowable as per the principles of series and parallel circuits: For series circuits: R1 R2 R3 Total E I R P Add Volts Equal Amps Add Add Ohms Watts Etotal = E1 + E2 + E3 Itotal = I1 = I2 = I3 Rtotal = R1 + R2 + R3 Ptotal = P1 + P2 + P3 283 For parallel circuits: R1 R2 R3 Total E I R P Equal Volts Add Diminish Add Amps Ohms Watts Etotal = E1 = E2 = E3 Itotal = I1 + I2 + I3 Rtotal = 1 1 R2 + 1 R3 1 R1 + Ptotal = P1 + P2 + P3 The \"table\" method helps to keep track of all relevant quantities. It also facilitates crosschecking of answers by making it easy to solve for the original unknown variables through other methods, or by working backwards to solve for the initially given values from your solutions. For example, if you have just solved for all unknown voltages, currents, and resistances in a circuit, you can check your work by adding a row at the bottom for power calculations on each resistor, seeing whether or not all the individual power values add up to the total power. If not, then you must have made a mistake somewhere! While this technique of \"cross-checking\" your work is nothing new, using the table to arrange all the data for the cross-check(s) results in a minimum of confusion. Aside: Although checking your work might not be fun when you have just worked hard on the problem the bene\ufb02ts are great. Coming back to a problem after a small break (trying another problem) often helps to \ufb02nd simple mistakes. If you have done all the work then \ufb02nding a simple mistake will be quick to \ufb02x because you know exactly what you need to do. Also if you start \ufb02nding mistakes while checking you\u2019ll build a mental list and \ufb02nd that you\u2019ll stop making them after a while. Do it and you\u2019ll \ufb02nd it will pay o\ufb01! 14.21 Conductor size The width of a conductor a\ufb01ects the ow of electrons through it. The broader the cross-sectional area (thickness or area of a sl) of the conductor, the more room for electrons to ow, and consequently, the easier it is for ow to occur (less resistance). Electrical wire is usually round in cross-section (although there are some unique exceptions to this rule), and comes in two basic varieties: solid and stranded. Solid copper wire is just as it sounds: a single, solid strand of copper the whole length of the wire. Stranded wire is composed of smaller strands of solid copper wire twisted together to form a single, larger conductor. The greatest bene\ufb02t of stranded wire is its mechanical exibility, being able to withstand repeated 284 bending and twisting much better than solid copper (which tends to fatigue and break after time). 14.22 Fuses Normally, the ampacity rating of a conductor is a circuit design limit never to be intentionally exceeded, but there is an application where ampacity exceedence is expected: in the case of fuses. A fuse is nothing more than a short length of wire designed to melt and separate in the event of excessive current. Fuses are always connected in series with the component(s) to be protected from overcurrent, so that when the fuse blows (opens) it will open the entire circuit and stop current through the component(s). A fuse connected in one branch of a parallel circuit, of course, would not a\ufb01ect current through", " any of the other branches. Normally, the thin piece of fuse wire is contained within a safety sheath to minimize hazards of arc blast if the wire burns open with violent force, as can happen in the case of severe overcurrents. In the case of small automotive fuses, the sheath is transparent so that the fusible element can be visually inspected. Residential wiring used to commonly employ screw-in fuses with glass bodies and a thin, narrow metal foil strip in the middle. 14.23 Important Equations and Quantities Quantity Symbol Unit S.I. Units Direction Units or Table 14.1: Units used in Electricity and Magnetism \u2020 \u2020 \u2020 \u2020 \u2020 \u2020 \u2020 \u2020 REVIEW: A circuit is an unbroken loop of conductive material that allows electrons to ow through continuously without beginning or end. If a circuit is \"broken,\" that means it\u2019s conductive elements no longer form a complete path, and continuous electron ow cannot occur in it. The location of a break in a circuit is irrelevant to its inability to sustain continuous electron ow. Any break, anywhere in a circuit prevents electron ow throughout the circuit. REVIEW: Electrons can be motivated to ow through a conductor by a the same force manifested in static electricity. Voltage is the measure of speci\ufb02c potential energy (potential energy per unit charge) between two locations. In layman\u2019s terms, it is the measure of \"push\" available to motivate electrons. Voltage, as an expression of potential energy, is always relative between two locations, or points. Sometimes it is called a voltage \"drop.\" 285 \u2020 \u2020 \u2020 \u2020 \u2020 \u2020 \u2020 \u2020 \u2020 \u2020 \u2020 \u2020 \u2020 \u2020 \u2020 \u2020 \u2020 \u2020 \u2020 \u2020 \u2020 When a voltage source is connected to a circuit, the voltage will cause a uniform ow of electrons through that circuit called a current. In a single (one loop) circuit, the amount current of current at any point is the same as the amount of current at any other point. If a circuit containing a voltage source is broken, the full voltage of that source will appear across the points of the break. The +/- orientation a voltage drop is called the polarity. It is also relative between two points. REVIEW: Resistance is the measure of opposition to electric current. A short circuit is an electric circuit o\ufb01ering little or no resistance to the ow of electrons. Short circuits are dangerous with high voltage power sources because the high currents encountered can cause large amounts of heat energy to be released. An open circuit is one where the continuity has been broken by an interruption in the path for electrons to ow. A closed circuit is one that is complete, with good continuity throughout. A device designed to open or close a circuit under controlled conditions is called a switch. The terms \"open\" and \"closed\" refer to switches as well as entire circuits. An open switch is one without continuity: electrons cannot ow through it. A closed switch is one that provides a direct (low resistance) path for electrons to ow through. REVIEW: Connecting wires in a circuit are assumed to have zero resistance unless otherwise stated. Wires in a circuit can be shortened or lengthened without impacting the circuit\u2019s function { all that matters is that the components are attached to one another in the same sequence. Points directly connected together in a circuit by zero resistance (wire) are considered to be electrically common. Electrically common points, with zero resistance between them, will have zero voltage dropped between them, regardless of the magnitude of current (ideally). The voltage or resistance readings referenced between sets of electrically common points will be the same. These rules apply to ideal conditions, where connecting wires are assumed to possess absolutely zero resistance. In real life this will probably not be the case, but wire resistances should be low enough so that the general principles stated here still hold. REVIEW: Power is additive in any con\ufb02guration of resistive circuit: PT otal = P1 + P2 +... Pn REVIEW: 286 \u2020 \u2020 \u2020 \u2020 When electrons ow through a conductor, a magnetic \ufb02eld will be produced around that conductor. The left-hand rule states that the magnetic ux lines produced by a current-carrying wire will be oriented the same direction as the curled \ufb02ngers of a person\u2019s left hand (in the \"hitchhiking\" position), with the thumb pointing in the direction of electron ow. The magnetic \ufb02eld force produced by a current-carrying wire can be greatly increased by shaping the wire into a coil instead of a straight line. If wound in a coil shape, the magnetic \ufb02eld will be oriented along the axis of the coil\u2019s length. The magnetic \ufb02eld force produced by an", " electromagnet (called the magnetomotive force, or mmf), is proportional to the product (multiplication) of the current through the electromagnet and the number of complete coil \"turns\" formed by the wire. 287 Chapter 15 Magnets and Electromagnetism sectionPermanent magnets Magnetism has been known to mankind for many thousands of years. Lodestone, a magnetized form of the iron oxide mineral magnetite which has the property of attracting iron objects, is referred to in old European and Asian historical records, around 800 BC in Europe and earlier in the East, around 2600 BC. The root of the English word magnet is the Greek word magnes, thought to be derived from Magnesia in Asia Minor, once an important source of lodestone. Lodestone was used as a navigational compass as it was found to orient itself in a north-south direction if left free to rotate by suspension on a string or on a oat in water. Interesting Fact: A compass is a navigational instrument for \ufb02nding directions. It consists of a magnetised pointer free to align itself accurately with Earth\u2019s magnetic \ufb02eld. A compass provides a known reference direction which is of great assistance in navigation. The cardinal points are north, south, east and west. A compass can be used in conjunction with a clock and a sextant to provide a very accurate navigation capability. This device greatly improved maritime trade by making travel safer and more e\u2013cient. A compass can be any magnetic device using a needle to indicate the direction of the magnetic north of a planet\u2019s magnetosphere. Any instrument with a magnetized bar or needle turning freely upon a pivot and pointing in a northerly and southerly direction can be considered a compass. Aside: In 1269, Frenchmen Peter Peregrinus and Pierre de Maricourt, using a compass and a lodestone, found that the magnetic force of the lodestone was di\ufb01erent at the opposite ends, which they de\ufb02ned to be the poles of the magnet. Like poles of magnets repel one another whilst unlike poles attract. These poles always occur in pairs. It is impossible to isolate a single pole. Breaking a piece of magnet in half results in two pieces, each with it\u2019s own pair of poles. 288 N magnet S... after breaking in half... N magnet S N magnet S The Earth itself is a magnet. Its magnetic poles are approximately aligned along the Earth\u2019s axis of rotation. The magnitude of forces between the poles of magnets follows an inverse square law; i. e. it varies inversely as the square of the distance of separation. Magnetic forces are a result of magnetic \ufb02elds. By placing a magnet underneath a piece of paper and sprinkling iron \ufb02lings on top one can map the magnetic \ufb02eld. The \ufb02lings align themselves parallel to the \ufb02eld. Magnetic \ufb02elds can be represented by magnetic \ufb02eld lines which are parallel to the magnetic \ufb02eld and whose spacing represents the relative strength of the magnetic \ufb02eld. The strength of the magnetic \ufb02eld is referred to as the magnetic ux. Magnetic \ufb02eld lines form closed loops. In a bar magnet magnetic \ufb02eld lines emerge at one pole and then curve around to the other pole with the rest of the loop being inside the magnet. magnetic field N magnet S As already said, opposite poles of a magnet attract each other and bringing them together results in their magnetic \ufb02eld lines converging. Like poles of a magnet repel each other and bringing them together results in their magnetic \ufb02eld lines diverging. 289 Ferromagnetism is a phenomenon exhibited by materials like iron, nickel or cobalt. These materials are known as permanent magnets. They always magnetize so as to be attracted to a magnet, regardless of which magnetic pole is brought toward the unmagnetized iron: N iron S N magnet S attraction Interesting Fact: The cause of Earth\u2019s magnetic \ufb02eld is not known for certain, but is possibly explained by the dynamo theory. The magnetic \ufb02eld extends several tens of thousands of kilometers into space. The \ufb02eld is approximately a magnetic dipole, with one pole near the geographic north pole and the other near the geographic south pole. An imaginary line joining the magnetic poles would be inclined by approximately 11.3 from the planet\u2019s axis of rotation. The location of the magnetic poles is not static but wanders as much as several kilometers a year. The two poles wander independently of each other and are not at exact opposite positions on the globe. The Earth\u2019s magnetic \ufb02eld reverses at intervals, ranging from tens of thousands to many millions of years, with an average interval of 250,000", " years. It is believed that this last occurred some 780,000 years ago. The mechanism responsible for geomagnetic reversals is not well understood. When the North reappears in the opposite direction, we would interpret this as a reversal, whereas turning o\ufb01 and returning in the same direction is called a geomagnetic excursion. At present, the overall geomagnetic \ufb02eld is becoming weaker at a rate which would, if it continues, cause the \ufb02eld to disappear, albeit temporarily, by about around 3000-4000 AD. The deterioration began roughly 150 years ago and has accelerated in the past several years. So far the strength of the earth\u2019s \ufb02eld has decreased by 10 to 15 percent. The ability of a ferromagnetic material tends to retain its magnetization after an external \ufb02eld is removed is called it\u2019s retentivity. 290 Paramagnetic materials are materials like aluminum or platinum which become magnetized in an external magnetic \ufb02eld in a similar way to ferromagnetic materials but lose their magnetism when the external magnetic \ufb02eld is removed. Diamagnetism is exhibited by materials like copper or bismuth which become magnetized in a magnetic \ufb02eld with a polarity opposite to the external magnetic \ufb02eld. Unlike iron, they are slightly repelled by a magnet S diamagnetic material N N magnet S repulsion The cause of Earth\u2019s magnetic \ufb02eld is not known for certain, but is possibly explained by the dynamo theory. The magnetic \ufb02eld extends several tens of thousands of kilometers into space. The \ufb02eld is approximately a magnetic dipole, with one pole near the geographic north pole and the other near the geographic south pole. An imaginary line joining the magnetic poles would be inclined by approximately 11.3 from the planet\u2019s axis of rotation. The location of the magnetic poles is not static but wanders as much as several kilometers a year. The two poles wander independently of each other and are not at exact opposite positions on the globe. Currently the south magnetic pole is further from the geographic south pole than than north magnetic pole is from the north geographic pole. The strength of the \ufb02eld at the Earth\u2019s surface at this time ranges from less than 30 microtesla (0.3 gauss) in an area including most of South America and South Africa to over 60 microtesla (0.6 gauss) around the magnetic poles in northern Canada and south of Australia, and in part of Siberia. The \ufb02eld is similar to that of a bar magnet, but this similarity is super\ufb02cial. The magnetic \ufb02eld of a bar magnet, or any other type of permanent magnet, is created by the coordinated motions of electrons (negatively charged particles) within iron atoms. The Earth\u2019s core, however, is hotter than 1043 K, the temperature at which the orientations of electron orbits within iron become randomized. Therefore the Earth\u2019s magnetic \ufb02eld is not caused by magnetised iron deposits, but mostly by electric currents (known as telluric currents). Another feature that distinguishes the Earth magnetically from a bar magnet is its magnetosphere. A magnetosphere is the region around an astronomical object, in which phenomena are dominated by its magnetic \ufb02eld. Earth is surrounded by a magnetosphere, as are the magnetized planets Jupiter, Saturn, Uranus and Neptune. Mercury is magnetized, but too weakly to trap plasma. Mars has patchy surface magnetization. The distant \ufb02eld of Earth is greatly modi\ufb02ed by the solar wind, a hot outow from the sun, consisting of solar ions (mainly hydrogen) moving at about 400 km/s. Earth\u2019s magnetic \ufb02eld forms an obstacle to the solar wind. The Earth\u2019s magnetic \ufb02eld reverses at intervals, ranging from tens of thousands to many It is believed that this last oc- millions of years, with an average interval of 250,000 years. 291 curred some 780,000 years ago. The mechanism responsible for geomagnetic reversals is not well understood. When the North reappears in the opposite direction, we would interpret this as a reversal, whereas turning o\ufb01 and returning in the same direction is called a geomagnetic excursion. At present, the overall geomagnetic \ufb02eld is becoming weaker at a rate which would, if it continues, cause the \ufb02eld to disappear, albeit temporarily, by about around 3000-4000 AD. The deterioration began roughly 150 years ago and has accelerated in the past several years. So far the strength of the earth\u2019s \ufb02eld has decreased by 10 to 15 percent. 15.1", " Electromagnetism The discovery of the relationship between magnetism and electricity was, like so many other scienti\ufb02c discoveries, stumbled upon almost by accident. The Danish physicist Hans Christian Oersted was lecturing one day in 1820 on the possibility of electricity and magnetism being related to one another, and in the process demonstrated it conclusively by experiment in front of his whole class! By passing an electric current through a metal wire suspended above a magnetic compass, Oersted was able to produce a de\ufb02nite motion of the compass needle in response to the current. What began as conjecture at the start of the class session was con\ufb02rmed as fact at the end. Needless to say, Oersted had to revise his lecture notes for future classes! His serendipitous discovery paved the way for a whole new branch of science: electromagnetics. Detailed experiments showed that the magnetic \ufb02eld produced by an electric current is always oriented perpendicular to the direction of ow. A simple method of showing this relationship is called the left-hand rule. Simply stated, the left-hand rule says that the magnetic ux lines produced by a current-carrying wire will be oriented the same direction as the curled \ufb02ngers of a person\u2019s left hand (in the \"hitchhiking\" position), with the thumb pointing in the direction of electron ow: The \"left-hand\" rule I I I The magnetic \ufb02eld encircles this straight piece of current-carrying wire, the magnetic ux I lines having no de\ufb02nite \"north\" or \"south\u2019 poles. (NOTE TO SELF: Need to add wires attracting or wires repelling) While the magnetic \ufb02eld surrounding a current-carrying wire is indeed interesting, it is quite weak for common amounts of current, able to deect a compass needle and not much more. To 292 create a stronger magnetic \ufb02eld force (and consequently, more \ufb02eld ux) with the same amount of electric current, we can wrap the wire into a coil shape, where the circling magnetic \ufb02elds around the wire will join to create a larger \ufb02eld with a de\ufb02nite magnetic (north and south) polarity: S N magnetic field The amount of magnetic \ufb02eld force generated by a coiled wire is proportional to the current through the wire multiplied by the number of \"turns\" or \"wraps\" of wire in the coil. This \ufb02eld force is called magnetomotive force (mmf), and is very much analogous to electromotive force (E) in an electric circuit. An electromagnet is a piece of wire intended to generate a magnetic \ufb02eld with the passage of electric current through it. Though all current-carrying conductors produce magnetic \ufb02elds, an electromagnet is usually constructed in such a way as to maximize the strength of the magnetic \ufb02eld it produces for a special purpose. Electromagnets \ufb02nd frequent application in research, industry, medical, and consumer products. As an electrically-controllable magnet, electromagnets \ufb02nd application in a wide variety of \"electromechanical\" devices: machines that e\ufb01ect mechanical force or motion through electrical power. Perhaps the most obvious example of such a machine is the electric motor. Relay Applying current through the coil causes the switch to close. Relays can be constructed to actuate multiple switch contacts, or operate them in \"reverse\" (energizing the coil will open the switch contact, and unpowering the coil will allow it to spring closed again). 293 Multiple-contact relay Relay with \"normallyclosed\" contact 15.2 Magnetic units of measurement If the burden of two systems of measurement for common quantities (English vs. metric) throws your mind into confusion, this is not the place for you! Due to an early lack of standardization in the science of magnetism, we have been plagued with no less than three complete systems of measurement for magnetic quantities. First, we need to become acquainted with the various quantities associated with magnetism. There are quite a few more quantities to be dealt with in magnetic systems than for electrical systems. With electricity, the basic quantities are Voltage (E), Current (I), Resistance (R), and Power (P). The \ufb02rst three are related to one another by Ohm\u2019s Law (E=IR ; I=E/R ; R=E/I), while Power is related to voltage, current, and resistance by Joule\u2019s Law (P=IE ; P=I2R ; P=E2/R). With magnetism, we have the following quantities to deal with: Magnetomotive Force { The quantity of", " magnetic \ufb02eld force, or \"push.\" Analogous to electric voltage (electromotive force). Field Flux { The quantity of total \ufb02eld e\ufb01ect, or \"substance\" of the \ufb02eld. Analogous to electric current. Field Intensity { The amount of \ufb02eld force (mmf) distributed over the length of the elec- tromagnet. Sometimes referred to as Magnetizing Force. Flux Density { The amount of magnetic \ufb02eld ux concentrated in a given area. Reluctance { The opposition to magnetic \ufb02eld ux through a given volume of space or material. Analogous to electrical resistance. Permeability { The speci\ufb02c measure of a material\u2019s acceptance of magnetic ux, analogous to the speci\ufb02c resistance of a conductive material (\u2030), except inverse (greater permeability means easier passage of magnetic ux, whereas greater speci\ufb02c resistance means more di\u2013cult passage of electric current)... But wait. the fun is just beginning! Not only do we have more quantities to keep track of with magnetism than with electricity, but we have several di\ufb01erent systems of unit measurement for each of these quantities. As with common quantities of length, weight, volume, and temperature, we have both English and metric systems. However, there is actually more than 294 one metric system of units, and multiple metric systems are used in magnetic \ufb02eld measurements! One is called the cgs, which stands for Centimeter-Gram-Second, denoting the root measures upon which the whole system is based. The other was originally known as the mks system, which stood for Meter-Kilogram-Second, which was later revised into another system, called rmks, standing for Rationalized Meter-Kilogram-Second. This ended up being adopted as an international standard and renamed SI (Systeme International). Quantity Symbol Unit of Measurement and abbreviation Field Force mmf Gilbert (Gb) Amp-turn Amp-turn CGS SI English Field Flux Field Intensity Flux Density Reluctance F H B \u00b4 Permeability m Maxwell (Mx) Weber (Wb) Line Oersted (Oe) Amp-turns per meter Amp-turns per inch Gauss (G) Tesla (T) Lines per square inch Gilberts per Maxwell Amp-turns per Weber Amp-turns per line Gauss per Oersted Tesla-meters per Amp-turn Lines per inch-Ampturn And yes, the \u201e symbol is really the same as the metric pre\ufb02x \"micro.\" I \ufb02nd this especially confusing, using the exact same alphabetical character to symbolize both a speci\ufb02c quantity and a general metric pre\ufb02x! As you might have guessed already, the relationship between \ufb02eld force, \ufb02eld ux, and reluctance is much the same as that between the electrical quantities of electromotive force (E), current (I), and resistance (R). This provides something akin to an Ohm\u2019s Law for magnetic circuits: A comparison of \"Ohm\u2019s Law\" for electric and magnetic circuits: E = IR Electrical mmf = F\u00b4 Magnetic And, given that permeability is inversely analogous to speci\ufb02c resistance, the equation for \ufb02nding the reluctance of a magnetic material is very similar to that for \ufb02nding the resistance of a conductor: A comparison of electrical and magnetic opposition: R = r l A Electrical \u00b4 = l mA Magnetic 295 In either case, a longer piece of material provides a greater opposition, all other factors being equal. Also, a larger cross-sectional area makes for less opposition, all other factors being equal. 15.3 Electromagnetic induction While Oersted\u2019s surprising discovery of electromagnetism paved the way for more practical applications of electricity, it was Michael Faraday who gave us the key to the practical generation of electricity: electromagnetic induction. Faraday discovered that a voltage would be generated across a length of wire if that wire was exposed to a perpendicular magnetic \ufb02eld ux of changing intensity. An easy way to create a magnetic \ufb02eld of changing intensity is to move a permanent magnet next to a wire or coil of wire. Remember: the magnetic \ufb02eld must increase or decrease in intensity perpendicular to the wire (so that the lines of ux \"cut across\" the conductor), or else no voltage will be induced: Electromagnetic induction current changes direction with change in magnet motion voltage changes polarity with change in magnet motion - + V + - N S magnet moved back and forth Faraday was able to mathematically relate the rate of change", " of the magnetic \ufb02eld ux with induced voltage (note the use of a lower-case letter \"e\" for voltage. This refers to instantaneous voltage, or voltage at a speci\ufb02c point in time, rather than a steady, stable voltage.): e = N dF dt Where, e = (Instantaneous) induced voltage in volts N = F = t = Number of turns in wire coil (straight wire = 1) Magnetic flux in Webers Time in seconds The \"d\" terms are standard calculus notation, representing rate-of-change of ux over time. 296 \"N\" stands for the number of turns, or wraps, in the wire coil (assuming that the wire is formed in the shape of a coil for maximum electromagnetic e\u2013ciency). This phenomenon is put into obvious practical use in the construction of electrical generators, which use mechanical power to move a magnetic \ufb02eld past coils of wire to generate voltage. However, this is by no means the only practical use for this principle. If we recall that the magnetic \ufb02eld produced by a current-carrying wire was always perpendicular to that wire, and that the ux intensity of that magnetic \ufb02eld varied with the amount of current through it, we can see that a wire is capable of inducing a voltage along its own length simply due to a change in current through it. This e\ufb01ect is called self-induction: a changing magnetic \ufb02eld produced by changes in current through a wire inducing voltage along the length of that same wire. If the magnetic \ufb02eld ux is enhanced by bending the wire into the shape of a coil, and/or wrapping that coil around a material of high permeability, this e\ufb01ect of self-induced voltage will be more intense. A device constructed to take advantage of this e\ufb01ect is called an inductor, and will be discussed in greater detail in the next chapter. A device speci\ufb02cally designed to produce the e\ufb01ect of mutual inductance between two or more coils is called a transformer. Because magnetically-induced voltage only happens when the magnetic \ufb02eld ux is changing in strength relative to the wire, mutual inductance between two coils can only happen with alternating (changing { AC) voltage, and not with direct (steady { DC) voltage. The only applications for mutual inductance in a DC system is where some means is available to switch power on and o\ufb01 to the coil (thus creating a pulsing DC voltage), the induced voltage peaking at every pulse. A very useful property of transformers is the ability to transform voltage and current levels If the according to a simple ratio, determined by the ratio of input and output coil turns. energized coil of a transformer is energized by an AC voltage, the amount of AC voltage induced in the unpowered coil will be equal to the input voltage multiplied by the ratio of output to input wire turns in the coils. Conversely, the current through the windings of the output coil compared to the input coil will follow the opposite ratio: if the voltage is increased from input coil to output coil, the current will be decreased by the same proportion. This action of the transformer is analogous to that of mechanical gear, belt sheave, or chain sprocket ratios: 297 Torque-reducing geartrain Large gear (many teeth) Small gear (few teeth) + + high torque, low speed low torque, high speed \"Step-down\" transformer high voltage AC voltage source many turns low voltage few turns Load high current low current A transformer designed to output more voltage than it takes in across the input coil is called a \"step-up\" transformer, while one designed to do the opposite is called a \"step-down,\" in reference to the transformation of voltage that takes place. The current through each respective coil, of course, follows the exact opposite proportion. 15.4 AC Most students of electricity begin their study with what is known as direct current (DC), which is electricity owing in a constant direction, and/or possessing a voltage with constant polarity. DC is the kind of electricity made by a battery (with de\ufb02nite positive and negative terminals), or the kind of charge generated by rubbing certain types of materials against each other. As useful and as easy to understand as DC is, it is not the only \"kind\" of electricity in use. Certain sources of electricity (most notably, rotary electro-mechanical generators) naturally produce voltages alternating in polarity, reversing positive and negative over time. Either as a voltage switching polarity or as a current switching direction back and forth, this \"kind\" of electricity is known as Alternating Current (AC): 298 DIRECT CURRENT (DC) ALTERNATING CURRENT (AC) I I I I Whereas the familiar battery symbol is used as a generic symbol for any DC voltage source", ", the circle with the wavy line inside is the generic symbol for any AC voltage source. One might wonder why anyone would bother with such a thing as AC. It is true that in some cases AC holds no practical advantage over DC. In applications where electricity is used to dissipate energy in the form of heat, the polarity or direction of current is irrelevant, so long as there is enough voltage and current to the load to produce the desired heat (power dissipation). However, with AC it is possible to build electric generators, motors and power distribution systems that are far more e\u2013cient than DC, and so we \ufb02nd AC used predominately across the world in high power applications. To explain the details of why this is so, a bit of background knowledge about AC is necessary. If a machine is constructed to rotate a magnetic \ufb02eld around a set of stationary wire coils with the turning of a shaft, AC voltage will be produced across the wire coils as that shaft is rotated, in accordance with Faraday\u2019s Law of electromagnetic induction. This is the basic operating principle of an AC generator, also known as an alternator : Alternator operation Step #1 Step #2 S N no current! Load Step #3 N S no current! Load N S + I - I Load Step #4 S N - I I Load + Notice how the polarity of the voltage across the wire coils reverses as the opposite poles of 299 the rotating magnet pass by. Connected to a load, this reversing voltage polarity will create a reversing current direction in the circuit. The faster the alternator\u2019s shaft is turned, the faster the magnet will spin, resulting in an alternating voltage and current that switches directions more often in a given amount of time. While DC generators work on the same general principle of electromagnetic induction, their construction is not as simple as their AC counterparts. With a DC generator, the coil of wire is mounted in the shaft where the magnet is on the AC alternator, and electrical connections are made to this spinning coil via stationary carbon \"brushes\" contacting copper strips on the rotating shaft. All this is necessary to switch the coil\u2019s changing output polarity to the external circuit so the external circuit sees a constant polarity: (DC) Generator operation Step #1 Step #2 N S SN Load Step #3 N S SN - + N S + Load Step #4 - + SN + N S - I N S - I Load Load The generator shown above will produce two pulses of voltage per revolution of the shaft, both pulses in the same direction (polarity). In order for a DC generator to produce constant voltage, rather than brief pulses of voltage once every 1/2 revolution, there are multiple sets of coils making intermittent contact with the brushes. The diagram shown above is a bit more simpli\ufb02ed than what you would see in real life. The problems involved with making and breaking electrical contact with a moving coil should be obvious (sparking and heat), especially if the shaft of the generator is revolving at high speed. If the atmosphere surrounding the machine contains ammable or explosive vapors, the practical problems of spark-producing brush contacts are even greater. An AC generator (alternator) does not require brushes and commutators to work, and so is immune to these problems experienced by DC generators. The bene\ufb02ts of AC over DC with regard to generator design is also reected in electric motors. While DC motors require the use of brushes to make electrical contact with moving coils of wire, AC motors do not. In fact, AC and DC motor designs are very similar to their 300 generator counterparts (identical for the sake of this tutorial), the AC motor being dependent upon the reversing magnetic \ufb02eld produced by alternating current through its stationary coils of wire to rotate the rotating magnet around on its shaft, and the DC motor being dependent on the brush contacts making and breaking connections to reverse current through the rotating coil every 1/2 rotation (180 degrees). So we know that AC generators and AC motors tend to be simpler than DC generators and DC motors. This relative simplicity translates into greater reliability and lower cost of manufacture. But what else is AC good for? Surely there must be more to it than design details of generators and motors! Indeed there is. There is an e\ufb01ect of electromagnetism known as mutual induction, whereby two or more coils of wire placed so that the changing magnetic \ufb02eld created by one induces a voltage in the other. If we have two mutually inductive coils and we energize one coil with AC, we will create an AC voltage in the other coil. When used as such, this device is known as a transformer : Transformer AC voltage source Induced AC voltage The fundamental signi\ufb02cance of a transformer is its ability to step voltage up or down from the powered coil to the unpowered coil. The AC voltage induced in the unpowered (\"secondary\") coil is", " equal to the AC voltage across the powered (\"primary\") coil multiplied by the ratio of secondary coil turns to primary coil turns. If the secondary coil is powering a load, the current through the secondary coil is just the opposite: primary coil current multiplied by the ratio of primary to secondary turns. This relationship has a very close mechanical analogy, using torque and speed to represent voltage and current, respectively: 301 Speed multiplication geartrain Large gear (many teeth) Small gear (few teeth) high torque low speed + + low torque high speed \"Step-down\" transformer high voltage AC voltage source many turns low voltage few turns Load high current low current If the winding ratio is reversed so that the primary coil has less turns than the secondary coil, the transformer \"steps up\" the voltage from the source level to a higher level at the load: Speed reduction geartrain Large gear (many teeth) Small gear (few teeth) low torque high speed + + high torque low speed \"Step-up\" transformer high voltage AC voltage source low voltage few turns high current many turns Load low current The transformer\u2019s ability to step AC voltage up or down with ease gives AC an advantage unmatched by DC in the realm of power distribution. When transmitting electrical power over long distances, it is far more e\u2013cient to do so with stepped-up voltages and stepped-down currents 302 (smaller-diameter wire with less resistive power losses), then step the voltage back down and the current back up for industry, business, or consumer use use. high voltage Power Plant Step-up low voltage... to other customers Step-down Home or Business low voltage Transformer technology has made long-range electric power distribution practical. Without the ability to e\u2013ciently step voltage up and down, it would be cost-prohibitive to construct power systems for anything but close-range (within a few miles at most) use. As useful as transformers are, they only work with AC, not DC. Because the phenomenon of mutual inductance relies on changing magnetic \ufb02elds, and direct current (DC) can only produce steady magnetic \ufb02elds, transformers simply will not work with direct current. Of course, direct current may be interrupted (pulsed) through the primary winding of a transformer to create a changing magnetic \ufb02eld (as is done in automotive ignition systems to produce high-voltage spark plug power from a low-voltage DC battery), but pulsed DC is not that di\ufb01erent from AC. Perhaps more than any other reason, this is why AC \ufb02nds such widespread application in power systems. If we were to follow the changing voltage produced by a coil in an alternator from any point on the sine wave graph to that point when the wave shape begins to repeat itself, we would have marked exactly one cycle of that wave. This is most easily shown by spanning the distance between identical peaks, but may be measured between any corresponding points on the graph. The degree marks on the horizontal axis of the graph represent the domain of the trigonometric sine function, and also the angular position of our simple two-pole alternator shaft as it rotates: one wave cycle 0 90 180 270 Alternator shaft position (degrees) 180 90 360 (0) one wave cycle 270 360 (0) Since the horizontal axis of this graph can mark the passage of time as well as shaft position in degrees, the dimension marked for one cycle is often measured in a unit of time, most often seconds or fractions of a second. When expressed as a measurement, this is often called the period of a wave. The period of a wave in degrees is always 360, but the amount of time one period occupies depends on the rate voltage oscillates back and forth. A more popular measure for describing the alternating rate of an AC voltage or current wave than period is the rate of that back-and-forth oscillation. This is called frequency. The modern 303 unit for frequency is the Hertz (abbreviated Hz), which represents the number of wave cycles completed during one second of time. In the United States of America, the standard power-line frequency is 60 Hz, meaning that the AC voltage oscillates at a rate of 60 complete back-andforth cycles every second. In Europe, where the power system frequency is 50 Hz, the AC voltage only completes 50 cycles every second. A radio station transmitter broadcasting at a frequency of 100 MHz generates an AC voltage oscillating at a rate of 100 million cycles every second. Prior to the canonization of the Hertz unit, frequency was simply expressed as \"cycles per second.\" Older meters and electronic equipment often bore frequency units of \"CPS\" (Cycles Per Second) instead of Hz. Many people believe the change from self-explanatory units like CPS to Hertz constitutes a step backward in clarity. A similar change occurred when the unit of \"Celsius\" replaced that of \"Centigrade\" for", " metric temperature measurement. The name Centigrade was based on a 100-count (\"Centi-\") scale (\"-grade\") representing the melting and boiling points of H2O, respectively. The name Celsius, on the other hand, gives no hint as to the unit\u2019s origin or meaning. Period and frequency are mathematical reciprocals of one another. That is to say, if a wave has a period of 10 seconds, its frequency will be 0.1 Hz, or 1/10 of a cycle per second: Frequency in Hertz = 1 Period in seconds An instrument called an oscilloscope is used to display a changing voltage over time on a graphical screen. You may be familiar with the appearance of an ECG or EKG (electrocardiograph) machine, used by physicians to graph the oscillations of a patient\u2019s heart over time. The ECG is a special-purpose oscilloscope expressly designed for medical use. General-purpose oscilloscopes have the ability to display voltage from virtually any voltage source, plotted as a graph with time as the independent variable. The relationship between period and frequency is very useful to know when displaying an AC voltage or current waveform on an oscilloscope screen. By measuring the period of the wave on the horizontal axis of the oscilloscope screen and reciprocating that time value (in seconds), you can determine the frequency in Hertz. OSCILLOSCOPE vertical Y DC GND AC V/div trigger timebase 1m X s/div DC GND AC 16 divisions @ 1ms/div = a period of 16 ms Frequency = 1 period = 1 16 ms = 62.5 Hz Voltage and current are by no means the only physical variables subject to variation over time. Much more common to our everyday experience is sound, which is nothing more than 304 the alternating compression and decompression (pressure waves) of air molecules, interpreted by our ears as a physical sensation. Because alternating current is a wave phenomenon, it shares many of the properties of other wave phenomena, like sound. For this reason, sound (especially structured music) provides an excellent analogy for relating AC concepts. In musical terms, frequency is equivalent to pitch. Low-pitch notes such as those produced by a tuba or bassoon consist of air molecule vibrations that are relatively slow (low frequency). High-pitch notes such as those produced by a ute or whistle consist of the same type of vibrations in the air, only vibrating at a much faster rate (higher frequency). Here is a table showing the actual frequencies for a range of common musical notes: Note A A sharp (or B flat) B C (middle) C sharp (or D flat) D D sharp (or E flat) E F F sharp (or G flat) G G sharp (or A flat) A A sharp (or B flat) B C Musical designation Frequency (in hertz) A1 A# or Bb B1 C C# or Db D D# or Eb E F F# or Gb G G# or Ab A A# or Bb B C1 220.00 233.08 246.94 261.63 277.18 293.66 311.13 329.63 349.23 369.99 392.00 415.30 440.00 466.16 493.88 523.25 Astute observers will notice that all notes on the table bearing the same letter designation are related by a frequency ratio of 2:1. For example, the \ufb02rst frequency shown (designated with the letter \"A\") is 220 Hz. The next highest \"A\" note has a frequency of 440 Hz { exactly twice as many sound wave cycles per second. The same 2:1 ratio holds true for the \ufb02rst A sharp (233.08 Hz) and the next A sharp (466.16 Hz), and for all note pairs found in the table. Audibly, two notes whose frequencies are exactly double each other sound remarkably similar. This similarity in sound is musically recognized, the shortest span on a musical scale separating such note pairs being called an octave. Following this rule, the next highest \"A\" note (one octave above 440 Hz) will be 880 Hz, the next lowest \"A\" (one octave below 220 Hz) will be 110 Hz. A view of a piano keyboard helps to put this scale into perspective: 305 C# Db D# Eb F# Gb G# Ab A# Bb C# Db D# Eb F# Gb G# Ab A# Bb C# Db D# Eb F# Gb G# Ab A# Bb one octave As you can see, one octave is equal to eight white keys\u2019 worth of distance on a piano keyboard. The familiar musical mnemonic (doe-ray-mee-fah-so-lah-tee-doe) { yes, the same pattern immortalized in the", " whimsical Rodgers and Hammerstein song sung in The Sound of Music { covers one octave from C to C. While electromechanical alternators and many other physical phenomena naturally produce sine waves, this is not the only kind of alternating wave in existence. Other \"waveforms\" of AC are commonly produced within electronic circuitry. Here are but a few sample waveforms and their common designations: Square wave Triangle wave one wave cycle one wave cycle Sawtooth wave These waveforms are by no means the only kinds of waveforms in existence. They\u2019re simply a few that are common enough to have been given distinct names. Even in circuits that are supposed to manifest \"pure\" sine, square, triangle, or sawtooth voltage/current waveforms, the real-life result is often a distorted version of the intended waveshape. Some waveforms are so complex that they defy classi\ufb02cation as a particular \"type\" (including waveforms associated with many kinds of musical instruments). Generally speaking, any waveshape bearing close resemblance to a perfect sine wave is termed sinusoidal, anything di\ufb01erent being labeled as nonsinusoidal. Being that the waveform of an AC voltage or current is crucial to its impact in a circuit, we need to be aware of the fact that AC waves come in a variety of shapes. 306 15.5 Measurements of AC magnitude So far we know that AC voltage alternates in polarity and AC current alternates in direction. We also know that AC can alternate in a variety of di\ufb01erent ways, and by tracing the alternation over time we can plot it as a \"waveform.\" We can measure the rate of alternation by measuring the time it takes for a wave to evolve before it repeats itself (the \"period\"), and express this as cycles per unit time, or \"frequency.\" In music, frequency is the same as pitch, which is the essential property distinguishing one note from another. However, we encounter a measurement problem if we try to express how large or small an AC quantity is. With DC, where quantities of voltage and current are generally stable, we have little trouble expressing how much voltage or current we have in any part of a circuit. But how do you grant a single measurement of magnitude to something that is constantly changing? One way to express the intensity, or magnitude (also called the amplitude), of an AC quantity is to measure its peak height on a waveform graph. This is known as the peak or crest value of an AC waveform: Peak Time Another way is to measure the total height between opposite peaks. This is known as the peak-to-peak (P-P) value of an AC waveform: Peak-to-Peak Time Unfortunately, either one of these expressions of waveform amplitude can be misleading when comparing two di\ufb01erent types of waves. For example, a square wave peaking at 10 volts is obviously a greater amount of voltage for a greater amount of time than a triangle wave peaking at 10 volts. The e\ufb01ects of these two AC voltages powering a load would be quite di\ufb01erent: 307 10 V Time 10 V (peak) 10 V (peak) more heat energy dissipated (same load resistance) less heat energy dissipated One way of expressing the amplitude of di\ufb01erent waveshapes in a more equivalent fashion is to mathematically average the values of all the points on a waveform\u2019s graph to a single, aggregate number. This amplitude measure is known simply as the average value of the waveform. If we average all the points on the waveform algebraically (that is, to consider their sign, either positive or negative), the average value for most waveforms is technically zero, because all the positive points cancel out all the negative points over a full cycle: + + + + + + + + + - - - - - - - - - True average value of all points (considering their signs) is zero! This, of course, will be true for any waveform having equal-area portions above and below the \"zero\" line of a plot. However, as a practical measure of a waveform\u2019s aggregate value, \"average\" is usually de\ufb02ned as the mathematical mean of all the points\u2019 absolute values over a cycle. In other words, we calculate the practical average value of the waveform by considering all points on the wave as positive quantities, as if the waveform looked like this: 308 + + + + + + + + + + ++ + + + + + + Practical average of points, all values assumed to be positive. Polarity-insensitive mechanical meter movements (meters designed to respond equally to the positive and negative half-cycles of an alternating voltage or current) register in proportion to the waveform\u2019s (practical) average value, because the inertia of the pointer against the tension of the spring naturally", " averages the force produced by the varying voltage/current values over time. Conversely, polarity-sensitive meter movements vibrate uselessly if exposed to AC voltage or current, their needles oscillating rapidly about the zero mark, indicating the true (algebraic) average value of zero for a symmetrical waveform. When the \"average\" value of a waveform is referenced in this text, it will be assumed that the \"practical\" de\ufb02nition of average is intended unless otherwise speci\ufb02ed. Another method of deriving an aggregate value for waveform amplitude is based on the waveform\u2019s ability to do useful work when applied to a load resistance. Unfortunately, an AC measurement based on work performed by a waveform is not the same as that waveform\u2019s \"average\" value, because the power dissipated by a given load (work performed per unit time) is not directly proportional to the magnitude of either the voltage or current impressed upon it. Rather, power is proportional to the square of the voltage or current applied to a resistance (P = E2/R, and P = I2R). Although the mathematics of such an amplitude measurement might not be straightforward, the utility of it is. Consider a bandsaw and a jigsaw, two pieces of modern woodworking equipment. Both types of saws cut with a thin, toothed, motor-powered metal blade to cut wood. But while the bandsaw uses a continuous motion of the blade to cut, the jigsaw uses a back-and-forth motion. The comparison of alternating current (AC) to direct current (DC) may be likened to the comparison of these two saw types: Bandsaw blade motion wood Jigsaw wood blade motion (analogous to DC) (analogous to AC) The problem of trying to describe the changing quantities of AC voltage or current in a single, aggregate measurement is also present in this saw analogy: how might we express the speed of a jigsaw blade? A bandsaw blade moves with a constant speed, similar to the way DC voltage pushes or DC current moves with a constant magnitude. A jigsaw blade, on the other hand, moves back and forth, its blade speed constantly changing. What is more, the back-and-forth motion of any two jigsaws may not be of the same type, depending on the mechanical design 309 of the saws. One jigsaw might move its blade with a sine-wave motion, while another with a triangle-wave motion. To rate a jigsaw based on its peak blade speed would be quite misleading when comparing one jigsaw to another (or a jigsaw with a bandsaw!). Despite the fact that these di\ufb01erent saws move their blades in di\ufb01erent manners, they are equal in one respect: they all cut wood, and a quantitative comparison of this common function can serve as a common basis for which to rate blade speed. Picture a jigsaw and bandsaw side-by-side, equipped with identical blades (same tooth pitch, angle, etc.), equally capable of cutting the same thickness of the same type of wood at the same rate. We might say that the two saws were equivalent or equal in their cutting capacity. Might this comparison be used to assign a \"bandsaw equivalent\" blade speed to the jigsaw\u2019s back-and-forth blade motion; to relate the wood-cutting e\ufb01ectiveness of one to the other? This is the general idea used to assign a \"DC equivalent\" measurement to any AC voltage or current: whatever magnitude of DC voltage or current would produce the same amount of heat energy dissipation through an equal resistance: 10 V RMS 10 V 5 A RMS 5 A RMS 5 A 5 A 2 W 50 W power dissipated Equal power dissipated through equal resistance loads 2 W 50 W power dissipated Suppose we were to wrap a coil of insulated wire around a loop of ferromagnetic material and energize this coil with an AC voltage source: iron core wire coil As an inductor, we would expect this iron-core coil to oppose the applied voltage with its inductive reactance, limiting current through the coil as predicted by the equations XL = 2\u2026fL and I=E/X (or I=E/Z). For the purposes of this example, though, we need to take a more 310 detailed look at the interactions of voltage, current, and magnetic ux in the device. Kirchho\ufb01\u2019s voltage law describes how the algebraic sum of all voltages in a loop must equal zero. In this example, we could apply this fundamental law of electricity to describe the respective voltages of the source and of the inductor coil. Here, as in any one-source, one-load circuit, the voltage dropped across the load must equal the voltage supplied by the source, assuming zero voltage dropped along the resistance of any connecting wires. In other words, the", " load (inductor coil) must produce an opposing voltage equal in magnitude to the source, in order that it may balance against the source voltage and produce an algebraic loop voltage sum of zero. From where does this opposing voltage arise? If the load were a resistor, the opposing voltage would originate from the \"friction\" of electrons owing through the resistance of the resistor. With a perfect inductor (no resistance in the coil wire), the opposing voltage comes from another mechanism: the reaction to a changing magnetic ux in the iron core. Michael Faraday discovered the mathematical relationship between magnetic ux (') and induced voltage with this equation: e = N dF dt Where, e = (Instantaneous) induced voltage in volts N = F = t = Number of turns in wire coil (straight wire = 1) Magnetic flux in Webers Time in seconds The instantaneous voltage (voltage dropped at any instant in time) across a wire coil is equal to the number of turns of that coil around the core (N) multiplied by the instantaneous rate-ofchange in magnetic ux (d'/dt) linking with the coil. Graphed, this shows itself as a set of sine waves (assuming a sinusoidal voltage source), the ux wave 90o lagging behind the voltage wave: e = voltage F = magnetic flux e F Magnetic ux through a ferromagnetic material is analogous to current through a conductor: it must be motivated by some force in order to occur. In electric circuits, this motivating force is voltage (a.k.a. electromotive force, or EMF). In magnetic \"circuits,\" this motivating force is magnetomotive force, or mmf. Magnetomotive force (mmf) and magnetic ux (') are related to each other by a property of magnetic materials known as reluctance (the latter quantity symbolized by a strange-looking letter \"R\"): 311 A comparison of \"Ohm\u2019s Law\" for electric and magnetic circuits: E = IR Electrical mmf = F\u00b4 Magnetic In our example, the mmf required to produce this changing magnetic ux (') must be supplied by a changing current through the coil. Magnetomotive force generated by an electromagnet coil is equal to the amount of current through that coil (in amps) multiplied by the number of turns of that coil around the core (the SI unit for mmf is the amp-turn). Because the mathematical relationship between magnetic ux and mmf is directly proportional, and because the mathematical relationship between mmf and current is also directly proportional (no rates-of-change present in either equation), the current through the coil will be in-phase with the ux wave: e = voltage F = magnetic flux i = coil current e F i This is why alternating current through an inductor lags the applied voltage waveform by 90o: because that is what is required to produce a changing magnetic ux whose rate-of-change produces an opposing voltage in-phase with the applied voltage. Due to its function in providing magnetizing force (mmf) for the core, this current is sometimes referred to as the magnetizing current. It should be mentioned that the current through an iron-core inductor is not perfectly sinusoidal (sine-wave shaped), due to the nonlinear B/H magnetization curve of iron. In fact, if the inductor is cheaply built, using as little iron as possible, the magnetic ux density might reach high levels (approaching saturation), resulting in a magnetizing current waveform that looks something like this: e = voltage F = magnetic flux i = coil current e F i When a ferromagnetic material approaches magnetic ux saturation, disproportionately greater levels of magnetic \ufb02eld force (mmf) are required to deliver equal increases in magnetic \ufb02eld ux ('). Because mmf is proportional to current through the magnetizing coil (mmf = NI, where \"N\" 312 is the number of turns of wire in the coil and \"I\" is the current through it), the large increases of mmf required to supply the needed increases in ux results in large increases in coil current. Thus, coil current increases dramatically at the peaks in order to maintain a ux waveform that isn\u2019t distorted, accounting for the bell-shaped half-cycles of the current waveform in the above plot. The situation is further complicated by energy losses within the iron core. The e\ufb01ects of hysteresis and eddy currents conspire to further distort and complicate the current waveform, making it even less sinusoidal and altering its phase to be lagging slightly less than 90o behind the applied voltage waveform. This coil current resulting from the sum total of all magnetic e\ufb01ects in the core (d'/dt magnetization plus hysteresis losses, eddy current losses, etc.) is called the exciting current. The distortion of an iron-", "core inductor\u2019s exciting current may be minimized if it is designed for and operated at very low ux densities. Generally speaking, this requires a core with large cross-sectional area, which tends to make the inductor bulky and expensive. For the sake of simplicity, though, we\u2019ll assume that our example core is far from saturation and free from all losses, resulting in a perfectly sinusoidal exciting current. As we\u2019ve seen already in the inductors chapter, having a current waveform 90o out of phase with the voltage waveform creates a condition where power is alternately absorbed and returned to the circuit by the inductor. If the inductor is perfect (no wire resistance, no magnetic core losses, etc.), it will dissipate zero power. Let us now consider the same inductor device, except this time with a second coil wrapped around the same iron core. The \ufb02rst coil will be labeled the primary coil, while the second will be labeled the secondary: iron core wire coil wire coil If this secondary coil experiences the same magnetic ux change as the primary (which it should, assuming perfect containment of the magnetic ux through the common core), and has the same number of turns around the core, a voltage of equal magnitude and phase to the applied voltage will be induced along its length. In the following graph, the induced voltage waveform is drawn slightly smaller than the source voltage waveform simply to distinguish one from the other: 313 ep = primary coil voltage es = secondary coil voltage F = magnetic flux ip = primary coil current ep es F ip 314 Chapter 16 Electronics (NOTE TO SELF: Mark: I have very little idea of how to make this ow, \ufb02t in or even how best to explain any of it. All the content in here is just trawled from other GFDL projects: www.wikipedia.com www.wikibooks.com The syllabus document has NO meaningful information on this stu\ufb01.) Electronics: 16.1 capacitive and inductive circuits 16.1.1 A capacitor (NOTE TO SELF: If we are going to talk of capacitive circuits we need a de\ufb02nition of capacitor.) A capacitor (historically known as a \"condenser\") is a device that stores energy in an electric \ufb02eld, by accumulating an internal imbalance of electric charge. It is made of two conductors separated by a dielectric (insulator). The problem of two parallel plates with a uniform electric \ufb02eld between them is a capacitor. When voltage exists one end of the capacitor is getting drained and the other end is getting \ufb02lled with charge. This is known as charging. Charging creates a charge imbalance between the two plates and creates a reverse voltage that stops the capacitor from charging. This is why when capacitors are \ufb02rst connected to voltage charge ows only to stop as the capacitor becomes charged. When a capacitor is charged current stops owing and it becomes an open circuit. It is as if the capacitor gained in\ufb02nite resistance. Just as the capacitor charges it can be discharged. 16.1.2 An inductor (NOTE TO SELF: If we are going to talk of inductive circuits we need a de\ufb02nition of a inductor) An inductor is a device which stores energy in a magnetic \ufb02eld. Inductors are formed of a coil of conductive material. When current ows through the wire it creates a magnetic \ufb02eld which exists inside the coil. When the current stops the magnetic \ufb02eld gets less, but we have learnt that a changing magnetic \ufb02eld induces a current in a wire. So when the current turns o\ufb01 the magnetic \ufb02eld decreases inducing another current in the wire. As the \ufb02eld decreases in strength so does the induced magnetic \ufb02eld. Normally they are made of copper wire, but not always (Example: aluminum wire, or spiral pattern etched on circuit board). The material around and within the coil a\ufb01ects its properties; 315 common types are air-core (only a coil of wire), iron-core, and ferrite core. Iron and ferrite types are more e\u2013cient because they conduct the magnetic \ufb02eld much better than air; of the two, ferrite is more e\u2013cient because stray electricity cannot ow through it. Interesting Fact: Some inductors have more than a core, which is just a rod the coil is formed about. Some are formed like transformers, using two E-shaped pieces facing each other, the wires wound about the central leg of the E\u2019s. The E\u2019s are made of laminated iron/steel or ferrite. Important qualities of an inductor There are several important properties for an inductor. * Current carrying capacity is determined by wire thickness.", " * Q, or quality, is determined by the uniformity of the windings, as well as the core material and how thoroughly it surrounds the coil. * Last but not least, the inductance of the coil. The inductance is determined by several factors. * coil shape: short and squat is best * core material * windings: winding in opposite directions will cancel out the inductance e\ufb01ect, and you will have only a resistor. 16.2 \ufb02lters and signal tuning (NOTE TO SELF: I think this relies on an understanding of second order ODEs and thats beyond the scope of the maths syllabus - we can put something high level but there is no way they\u2019ll understand it properly - surely we should teach as little phenomonology as possible - the waves chapter has a ton of it already) 16.3 active circuit elements, diode, LED and \ufb02eld e\ufb01ect transistor, operational ampli\ufb02er 16.3.1 Diode A diode functions as the electronic version of a one-way valve. By restricting the direction of movement of charge carriers, it allows an electric current to ow in one direction, but blocks it in the opposite direction. It is a one-way street for current. 316 Diode behavior is analogous to the behavior of a hydraulic device called a check valve. A check valve allows uid ow through it in one direction only: Check valves are essentially pressure-operated devices: they open and allow ow if the pressure across them is of the correct \"polarity\" to open the gate (in the analogy shown, greater uid pressure on the right than on the left). If the pressure is of the opposite \"polarity,\" the pressure di\ufb01erence across the check valve will close and hold the gate so that no ow occurs. Like check valves, diodes are essentially \"pressure-\" operated (voltage-operated) devices. The essential di\ufb01erence between forward-bias and reverse-bias is the polarity of the voltage dropped across the diode. Let\u2019s take a closer look at the simple battery-diode-lamp circuit shown earlier, this time investigating voltage drops across the various components: 317 When the diode is forward-biased and conducting current, there is a small voltage dropped across it, leaving most of the battery voltage dropped across the lamp. When the battery\u2019s polarity is reversed and the diode becomes reverse-biased, it drops all of the battery\u2019s voltage and leaves none for the lamp. If we consider the diode to be a sort of self-actuating switch (closed in the forward-bias mode and open in the reverse-bias mode), this behavior makes sense. The most substantial di\ufb01erence here is that the diode drops a lot more voltage when conducting than the average mechanical switch (0.7 volts versus tens of millivolts). This forward-bias voltage drop exhibited by the diode is due to the action of the depletion region formed by the P-N junction under the inuence of an applied voltage. When there is no voltage applied across a semiconductor diode, a thin depletion region exists around the region of the P-N junction, preventing current through it. The depletion region is for the most part devoid of available charge carriers and so acts as an insulator: 318 16.3.2 LED A light-emitting diode (LED) is a semiconductor device that emits light when charge ows in the correct direction through it. If you apply a voltage to force current to ow in the direction the LED allows it will light up. This notation of having two small arrows pointing away from the device is common to the schematic symbols of all light-emitting semiconductor devices. Conversely, if a device is lightactivated (meaning that incoming light stimulates it), then the symbol will have two small arrows pointing toward it. It is interesting to note, though, that LEDs are capable of acting as lightsensing devices: they will generate a small voltage when exposed to light, much like a solar cell on a small scale. This property can be gainfully applied in a variety of light-sensing circuits. The color depends on the semiconducting material used to construct the LED, and can be in the near-ultraviolet, visible or infrared part of the electromagnetic spectrum. Interesting Fact: Nick Holonyak Jr. (1928 ) of the University of Illinois at Urbana-Champaign developed the \ufb02rst practical visible-spectrum LED in 1962. 319 Physical function Because LEDs are made of di\ufb01erent chemical substances than normal rectifying diodes, their forward voltage drops will be di\ufb01erent. Typically, LEDs have much larger forward voltage drops than rectifying diodes, anywhere from about 1.6 volts to over 3 volts, depending on the color", ". Typical operating current for a standard-sized LED is around 20 mA. When operating an LED from a DC voltage source greater than the LED\u2019s forward voltage, a series-connected \"dropping\" resistor must be included to prevent full source voltage from damaging the LED. Consider this example circuit: With the LED dropping 1.6 volts, there will be 4.4 volts dropped across the resistor. Sizing the resistor for an LED current of 20 mA is as simple as taking its voltage drop (4.4 volts) and dividing by circuit current (20 mA), in accordance with Ohm\u2019s Law (R=E/I). This gives us a \ufb02gure of 220?. Calculating power dissipation for this resistor, we take its voltage drop and multiply by its current (P=IE), and end up with 88 mW, well within the rating of a 1/8 watt resistor. Higher battery voltages will require larger-value dropping resistors, and possibly higher-power rating resistors as well. Consider this example for a supply voltage of 24 volts: Here, the dropping resistor must be increased to a size of 1.12 k? in order to drop 22.4 volts at 20 mA so that the LED still receives only 1.6 volts. This also makes for a higher resistor power dissipation: 448 mW, nearly one-half a watt of power! Obviously, a resistor rated for 1/8 watt power dissipation or even 1/4 watt dissipation will overheat if used here. Dropping resistor values need not be precise for LED circuits. Suppose we were to use a 1 k? resistor instead of a 1.12 k? resistor in the circuit shown above. The result would be a slightly greater circuit current and LED voltage drop, resulting in a brighter light from the LED and slightly reduced service life. A dropping resistor with too much resistance (say, 1.5 k? instead of 1.12 k?) will result in less circuit current, less LED voltage, and a dimmer light. LEDs are quite tolerant of variation in applied power, so you need not strive for perfection in sizing the dropping resistor. Also because of their unique chemical makeup, LEDs have much, much lower peak-inverse voltage (PIV) ratings than ordinary rectifying diodes. A typical LED might only be rated at 5 volts in reverse-bias mode. Therefore, when using alternating current to power an LED, you should connect a protective rectifying diode in series with the LED to prevent reverse breakdown every other half-cycle: 320 Light emission The wavelength of the light emitted, and therefore its color, depends on the materials forming the pn junction. A normal diode, typically made of silicon or germanium, emits invisible far-infrared light (so it can\u2019t be seen), but the materials used for an LED have emit light corresponding to near-infrared, visible or near-ultraviolet frequencies. Considerations in use Unlike incandescent light bulbs, which can operate with either AC or DC, LEDs require a DC supply of the correct electrical polarity. When the voltage across the pn junction is in the correct direction, a signi\ufb02cant current ows and the device is said to be forward biased. The voltage across the LED in this case is \ufb02xed for a given LED and is proportional to the energy of the emitted photons. If the voltage is of the wrong polarity, the device is said to be reverse biased, very little current ows, and no light is emitted. Because the voltage versus current characteristics of an LED are much like any diode, they can be destroyed by connecting them to a voltage source much higher than their turn on voltage. The voltage drop across a forward biased LED increases as the amount of light emitted increases because of the optical power being radiated. One consequence is that LEDs of the same type can be readily operated in parallel. The turn-on voltage of an LED is a function of the color, a higher forward drop is associated with emitting higher energy (bluer) photons. The reverse voltage that most LEDs can sustain without damage is usually only a few volts. Some LED units contain two diodes, one in each direction and each a di\ufb01erent color (typically red and green), allowing two-color operation or a range of colors to be created by altering the percentage of time the voltage is in each polarity. LED materials LED development began with infrared and red devices made with gallium arsenide. Advances in materials science have made possible the production of devices with ever shorter wavelengths, producing light in a variety of colors. Conventional LEDs are made from a variety of inorganic minerals, producing the following colors: aluminium gallium arsenide (AlGaAs): red and infrared gallium arsenide/phosphide (GaAsP): red, orange-red, orange, and yellow gallium nitride (GaN): green", ", pure green (or emerald green), and blue gallium phosphide (GaP): red, yellow and green zinc selenide (ZnSe): blue \u2020 \u2020 \u2020 \u2020 \u2020 321 indium gallium nitride (InGaN): bluish-green and blue silicon carbide (SiC): blue diamond (C): ultraviolet \u2020 \u2020 \u2020 silicon (Si) - under development \u2020 (NOTE TO SELF: The above list is taken from public sources, but at least one LED given as blue does not produce blue light. (There is a good chance that almost none do, because of the higher frequency of blue.) This is a common problem in daily life due to the majority of mankind being ignorant of colour theory and conating blue with light blue with cyan, the latter often called \"sky blue\". A cyan LED may be distinguished from a blue LED in that adding a yellow phosphor to the output makes green, rather than white light. And often aqua is called blue-green when in actuality the latter is cyan, and light cyan-green would be aqua. What adds to the confusion is that cyan LEDs are enclosed in blue plastic. A great amount of work is needed to dispel these intuitive myths of colour mixing before accurate descriptions of physical phenomena and their production can happen. - This needs to be sorted out) Blue and white LEDs and Other colors Commercially viable blue LEDs based invented by Shuji Nakamura while working in Japan at Nichia Corporation in 1993 and became widely available in the late 1990s. They can be added to existing red and green LEDs to produce white light. Most \"white\" LEDs in production today use a 450nm 470nm blue GaN (gallium nitride) LED covered by a yellowish phosphor coating usually made of cerium doped yttrium aluminium garnet (YAG:Ce) crystals which have been powdered and bound in a type of viscous adhesive. The LED chip emits blue light, part of which is converted to yellow by the YAG:Ce. The single crystal form of YAG:Ce is actually considered a scintillator rather than a phosphor. Since yellow light stimulates the red and green receptors of the eye, the resulting mix of blue and yellow light gives the appearance of white. The newest method used to produce white light LEDs uses no phosphors at all and is based on homoepitaxially grown zinc selenide (ZnSe) on a ZnSe substrate which simultaneously emits blue light from its active region and yellow light from the substrate. Other colors Recent color developments include pink and purple. They consist of one or two phosphor layers over a blue LED chip. The \ufb02rst phosphor layer of a pink LED is a yellow glowing one, and the second phosphor layer is either red or orange glowing. Purple LEDs are blue LEDs with an orange glowing phosphor over the chip. Some pink LEDs have run into issues. For example, some are blue LEDs painted with uorescent paint or \ufb02ngernail polish that can wear o\ufb01, and some are white LEDs with a pink phosphor or dye that unfortunately fades after a short tme. Ultraviolet, blue, pure green, white, pink and purple LEDs are relatively expensive compared to the more common reds, oranges, greens, yellows and infrareds and are thus less commonly used in commercial applications. The semiconducting chip is encased in a solid plastic lens, which is much tougher than the glass envelope of a traditional light bulb or tube. The plastic may be colored, but this is only for cosmetic reasons and does not a\ufb01ect the color of the light emitted. 322 Operational parameters and e\u2013ciency Most typical LEDs are designed to operate with no more than 30-60 milliwatts of electrical power. It is projected that by 2005, 10-watt units will be available. These devices will produce about as much light as a common 50-watt incandescent bulb, and will facilitate use of LEDs for general illumination needs. Interesting Fact: In September 2003 a new type of blue LED was demonstrated by the company Cree, Inc. to have 35% e\u2013ciency at 20 mA. This produced a commercially packaged white light having 65 lumens per watt at 20 mA, becoming the brightest white LED commercially available at the time. Organic light-emitting diodes (OLEDs) If the emissive layer material of an LED is an organic compound, it is known as an Organic Light Emitting Diode (OLED). To function as a semiconductor, the organic emissive material must have conjugated pi bonds. The emissive material can be a small organic molecule in a crystalline phase, or a polymer. Polymer materials can be exible; such LEDs are known as PLEDs or FLEDs. Compared with regular LEDs, OLEDs are lighter and polymer LEDs", " can have the added bene\ufb02t of being exible. Some possible future applications of OLEDs could be: Light sources Wall decorations Luminous cloth \u2020 \u2020 \u2020 LED applications Here is a list of known applications for LEDs, some of which are further elaborated upon in the following text: in general, commonly used as information indicators in various types of embedded systems (many of which are listed below) thin, lightweight message displays, e.g. in public information signs (at airports and railway stations, among other places) status indicators, e.g. on/o\ufb01 lights on professional instruments and consumers audio/video equipment infrared LEDs in remote controls (for TVs, VCRs, etc) clusters in tra\u2013c signals, replacing ordinary bulbs behind colored glass car indicator lights and bicycle lighting; also for pedestrians to be seen by car tra\u2013c \u2020 \u2020 \u2020 \u2020 \u2020 \u2020 323 calculator and measurement instrument displays (seven segment displays), although now mostly replaced by LCDs red or yellow LEDs are used in indicator and [alpha]numeric displays in environments where night vision must be retained: aircraft cockpits, submarine and ship bridges, astronomy observatories, and in the \ufb02eld, e.g. night time animal watching and military \ufb02eld use red or yellow LEDs are also used in photographic darkrooms, for providing lighting which does not lead to unwanted exposure of the \ufb02lm illumination, e.g. ashlights (a.k.a. torches, UK), and backlights for LCD screens signaling/emergency beacons and strobes movement sensors, e.g. in mechanical and optical computer mice and trackballs \u2020 \u2020 \u2020 \u2020 \u2020 \u2020 in LED printers, e.g. high-end color printers \u2020 LEDs o\ufb01er bene\ufb02ts in terms of maintenance and safety. \u2020 \u2020 \u2020 \u2020 The typical working lifetime of a device, including the bulb, is ten years, which is much longer than the lifetimes of most other light sources. LEDs fail by dimming over time, rather than the abrupt burn-out of incandescent bulbs. LEDs give o\ufb01 less heat than incandescent light bulbs and are less fragile than uorescent lamps. Since an individual device is smaller than a centimetre in length, LED-based light sources used for illumination and outdoor signals are built using clusters of tens of devices. Because they are monochromatic, LED lights have great power advantages over white lights where a speci\ufb02c color is required. Unlike the white lights, the LED does not need a \ufb02lter that absorbs most of the emitted white light. Colored uorescent lights are made, but they are not widely available. LED lights are inherently colored, and are available in a wide range of colors. One of the most recently introduced colors is the emerald green (bluish green, about 500 nm) that meets the legal requirements for tra\u2013c signals and navigation lights. Interesting Fact: The largest LED display in the world is 36 metres high (118 feet), at Times Square, New York, U.S.A. There are applications that speci\ufb02cally require light that does not contain any blue component. Examples are photographic darkroom safe lights, illumination in laboratories where certain photo-sensitive chemicals are used, and situations where dark adaptation (night vision) must be preserved, such as cockpit and bridge illumination, observatories, etc. Yellow LED lights are a good choice to meet these special requirements because the human eye is more sensitive to yellow light. 324 16.3.3 Transistor The transistor is a solid state semiconductor device used for ampli\ufb02cation and switching, and has three terminals. The transistor itself does not amplify current though, which is a common misconception, but a small current or voltage applied to one terminal controls the current through the other two, hence the term transistor; a voltage- or current-controlled resistor. It is the key component in all modern electronics. In digital circuits, transistors are used as very fast electrical switches, and arrangements of transistors can function as logic gates, RAM-type memory and other devices. In analog circuits, transistors are essentially used as ampli\ufb02ers. Transistor was also the common name in the sixties for a transistor radio, a pocket-sized portable radio that utilized transistors (rather than vacuum tubes) as its active electronics. This is still one of the dictionary de\ufb02nitions of transistor. The only functional di\ufb01erence between a PNP transistor and an NPN transistor is the proper biasing (polarity) of the junctions when operating. For any given state of operation, the current directions and voltage polarities for each type of transistor are exactly opposite each other. Bipolar transistors work as current-controlled current regulators. In other words, they restrict the", " amount of current that can go through them according to a smaller, controlling current. The main current that is controlled goes from collector to emitter, or from emitter to collector, depending on the type of transistor it is (PNP or NPN, respectively). The small current that controls the main current goes from base to emitter, or from emitter to base, once again depending on the type of transistor it is (PNP or NPN, respectively). According to the confusing standards of semiconductor symbology, the arrow always points against the direction of electron ow: 325 Bipolar transistors are called bipolar because the main ow of electrons through them takes place in two types of semiconductor material: P and N, as the main current goes from emitter to collector (or visa-versa). In other words, two types of charge carriers { electrons and holes { comprise this main current through the transistor. As you can see, the controlling current and the controlled current always mesh together through the emitter wire, and their electrons always ow against the direction of the transistor\u2019s arrow. This is the \ufb02rst and foremost rule in the use of transistors: all currents must be going in the proper directions for the device to work as a current regulator. The small, controlling current is usually referred to simply as the base current because it is the only current that goes through the base wire of the transistor. Conversely, the large, controlled current is referred to as the collector current because it is the only current that goes through the collector wire. The emitter current is the sum of the base and collector currents, in compliance with Kirchho\ufb01\u2019s Current Law. If there is no current through the base of the transistor, it shuts o\ufb01 like an open switch and prevents current through the collector. If there is a base current, then the transistor turns on like a closed switch and allows a proportional amount of current through the collector. Collector current is primarily limited by the base current, regardless of the amount of voltage available to push it. The next section will explore in more detail the use of bipolar transistors as switching elements. Importance The transistor is considered by many to be one of the greatest discoveries or inventions in modern history, ranking with banking and the printing press. Key to the importance of the transistor in modern society is its ability to be produced in huge numbers using simple techniques, resulting in vanishingly small prices. Computer \"chips\" consist of millions of transistors and sell for rands, with per-transistor costs in the thousandths-of-cents. The low cost has meant that the transistor has become an almost universal tool for nonmechanical tasks. Whereas a common device, say a refrigerator, would have used a mechanical device for control, today it is often less expensive to simply use a few million transistors and the appropriate computer program to carry out the same task through \"brute force\". Today 326 transistors have replaced almost all electromechanical devices, most simple feedback systems, and appear in huge numbers in everything from computers to cars. Hand-in-hand with low cost has been the increasing move to \"digitizing\" all information. With transistorized computers o\ufb01ering the ability to quickly \ufb02nd (and sort) digital information, more and more e\ufb01ort was put into making all information digital. Today almost all media in modern society is delivered in digital form, converted and presented by computers. Common \"analog\" forms of information such as television or newspapers spend the vast majority of their time as digital information, being converted to analog only for a small portion of the time. Interesting Fact: The transistor was invented at Bell Laboratories in December 1947 (\ufb02rst demonstrated on December 23) by John Bardeen, Walter Houser Brattain, and William Bradford Shockley, who were awarded the Nobel Prize in physics in 1956. 16.3.4 The transistor as a switch Because a transistor\u2019s collector current is proportionally limited by its base current, it can be used as a sort of current-controlled switch. A relatively small ow of electrons sent through the base of the transistor has the ability to exert control over a much larger ow of electrons through the collector. Suppose we had a lamp that we wanted to turn on and o\ufb01 by means of a switch. Such a circuit would be extremely simple: For the sake of illustration, let\u2019s insert a transistor in place of the switch to show how it can control the ow of electrons through the lamp. Remember that the controlled current through a transistor must go between collector and emitter. Since it\u2019s the current through the lamp that we want to control, we must position the collector and emitter of our transistor where the two contacts of the switch are now. We must also make sure that the lamp\u2019s current will move against the direction of the emitter arrow symbol to ensure that the transistor\u2019s junction bias will be", " correct: 327 In this example I happened to choose an NPN transistor. A PNP transistor could also have been chosen for the job, and its application would look like this: The choice between NPN and PNP is really arbitrary. All that matters is that the proper current directions are maintained for the sake of correct junction biasing (electron ow going against the transistor symbol\u2019s arrow). Going back to the NPN transistor in our example circuit, we are faced with the need to add something more so that we can have base current. Without a connection to the base wire of the transistor, base current will be zero, and the transistor cannot turn on, resulting in a lamp that is always o\ufb01. Remember that for an NPN transistor, base current must consist of electrons owing from emitter to base (against the emitter arrow symbol, just like the lamp current). Perhaps the simplest thing to do would be to connect a switch between the base and collector wires of the transistor like this: If the switch is open, the base wire of the transistor will be left \"oating\" (not connected to anything) and there will be no current through it. In this state, the transistor is said to be cuto\ufb01. If the switch is closed, however, electrons will be able to ow from the emitter through to the base of the transistor, through the switch and up to the left side of the lamp, back to the positive side of the battery. This base current will enable a much larger ow of electrons from the emitter through to the collector, thus lighting up the lamp. In this state of maximum circuit current, the transistor is said to be saturated. 328 Of course, it may seem pointless to use a transistor in this capacity to control the lamp. After all, we\u2019re still using a switch in the circuit, aren\u2019t we? If we\u2019re still using a switch to control the lamp { if only indirectly { then what\u2019s the point of having a transistor to control the current? Why not just go back to our original circuit and use the switch directly to control the lamp current? There are a couple of points to be made here, actually. First is the fact that when used in this manner, the switch contacts need only handle what little base current is necessary to turn the transistor on, while the transistor itself handles the majority of the lamp\u2019s current. This may be an important advantage if the switch has a low current rating: a small switch may be used to control a relatively high-current load. Perhaps more importantly, though, is the fact that the current-controlling behavior of the transistor enables us to use something completely di\ufb01erent to turn the lamp on or o\ufb01. Consider this example, where a solar cell is used to control the transistor, which in turn controls the lamp: Or, we could use a thermocouple to provide the necessary base current to turn the transistor on: 329 Even a microphone of su\u2013cient voltage and current output could be used to turn the transistor on, provided its output is recti\ufb02ed from AC to DC so that the emitter-base PN junction within the transistor will always be forward-biased: The point should be quite apparent by now: any su\u2013cient source of DC current may be used to turn the transistor on, and that source of current need only be a fraction of the amount of current needed to energize the lamp. Here we see the transistor functioning not only as a switch, but as a true ampli\ufb02er: using a relatively low-power signal to control a relatively large amount of power. Please note that the actual power for lighting up the lamp comes from the battery to the right of the schematic. It is not as though the small signal current from the solar cell, thermocouple, or microphone is being magically transformed into a greater amount of power. Rather, those small power sources are simply controlling the battery\u2019s power to light up the lamp. Field-E\ufb01ect Transistor (FET) (NOTE TO SELF: Schematic can be found under GFDL on wikipedia) The schematic symbols for p- and n-channel MOSFETs. The symbols to the right include an extra terminal for the transistor body (allowing for a seldom-used channel bias) whereas in those to the left the body is implicitly connected to the source. The most common variety of \ufb02eld-e\ufb01ect transistors, the enhancement-mode MOSFET (metaloxide semiconductor \ufb02eld-e\ufb01ect transistor) consists of a unipolar conduction channel and a metal gate separated from the main conduction channel by a thin layer of (SiO2) glass. This is why an alternative name for the FET is \u2019unipolar transistor.\u2019 When a potential di\ufb01erence (of the proper polarity)", " is impressed across gate and source, charge carriers are introduced to the channel, making it conductive. The amount of this current can be modulated, or (nearly) completely turned o\ufb01, by varying the gate potential. Because the gate is insulated, no DC current ows to or from the gate electrode. This lack of a gate current and the ability of the MOSFET to act like a switch, allows particularly e\u2013cient digital circuits to be created, with very low power consumption at low frequencies. The power consumption increases markedly with frequency, because the capacitive loading of the FET control terminal takes more energy to slew at higher frequencies, in direct proportion to the frequency. Hence, MOSFETs have become the dominant technology used in computing hardware such as microprocessors and memory devices such as RAM. Bipolar transistors are more rugged and hence more useful for low-impedance loads and inductively reactive (e.g. motor) loads. Power MOSFETs become less conductive with increasing temperature and can therefore be applied in shunt, to increase current capacity, unlike the bipolar transistor, which has a negative 330 temperature coe\u2013cient of resistance, and is therefore prone to thermal runaway. The downside of this is that, while the power FET can protect itself from overheating by diminishing the current through it, high temperatures need to be avoided by using a larger heat sink than for an equivalent bipolar device. Macroscopic FET power transistors are actually composed of many little transistors. They are stacked (on-chip) to increase breakdown potential and paralleled to reduce Ron, i.e. allowing for more current, bussing the gates to provide a single control (gate) terminal. The depletion mode FET is a little di\ufb01erent. It uses a back-biased diode for the control terminal, which presents a capacitive load to the driving circuit in normal operation. With the gate tied to the source, a DFET is fully on. Changing the potential of a DFET (pulling an Nchannel gate downward, for example) will turn it o\ufb01, i.e. \u2019deplete\u2019 the channel (drain-source) of charge carriers. MOSFETs, formerly called IGFETs (for Insulated Gate Field-E\ufb01ect Transistor) can be depletion-mode, enhancement-mode, or mixed-mode, but are almost always enhancement mode in modern commercial practice. This means that, with the source and gate tied together (thus equipotential) the channel will be o\ufb01 (high impedance or non-conducting). The n-channel device (reverse for P-channel), like in the DFET, is turned on by raising the potential of the gate. Typically, the gate on a MOSFET will withstand +-20V, relative to the source terminal. If one were to raise the gate potential of an n-channel device without limiting the current to a few milliamps, one would destroy the gate diode, like any other small diode. Why do we typically think of n-channel devices as the default? In silicon devices, the ones that use majority carriers that are electrons, rather than holes, are slightly faster and can carry more current than their P-type counterparts. The same is true in GaAs devices. The FET is simpler in concept than the bipolar transistor and can be constructed from a wide range of materials. The most common use of MOSFET transistors today is the CMOS (complementary metallic oxide semiconductor) integrated circuit which is the basis for most digital electronic devices. These use a totem-pole arrangement where one transistor (either the pull-up or the pull-down) is on while the other is o\ufb01. Hence, there is no DC drain, except during the transition from one state to the other, which is very short. As mentioned, the gates are capacitive, and the charging and discharging of the gates each time a transistor switches states is the primary cause of power drain. The C in CMOS stands for \u2019complementary.\u2019 The pull-up is a P-channel device (using holes for the mobile carrier of charge) and the pull-down is N-channel (electron carriers). This allows busing of the control terminals, but limits the speed of the circuit to that of the slower P device (in silicon devices). The bipolar solutions to push-pull include \u2019cascode\u2019 using a current source for the load. Circuits that utilize both unipolar and bipolar transistors are called Bi-Fet. A recent development is called \u2019vertical P.\u2019 Formerly, BiFet chip users had to settle for relatively poor (horizontal) P-type FET devices. This is no longer the case and allows for quieter and faster analog circuits. A clever variant of the FET", " is the dual-gate device. This allows for two opportunities to turn the device o\ufb01, as opposed to the dual-base (bipolar) transistor which presents two opportunities to turn the device on. FETs can switch signals of either polarity, if their amplitude is signi\ufb02cantly less than the gate swing, as the devices (especially the parasitic diode-free DFET) are basically symmetrical. This means that FETs are the most suitable type for analog multiplexing. With this concept, one can construct a solid-state mixing board, for example. The power MOSFET has a \u2019parasitic diode\u2019 (back-biased) normally shunting the conduction channel that has half the current capacity of the conduction channel. Sometimes this is useful in driving dual-coil magnetic circuits (for spike protection), but in other cases it causes problems. 331 The high impedance of the FET gate makes it rather vulnerable to electrostatic damage, though this is not usually a problem after the device has been installed. A more recent device for power control is the insulated-gate bipolar transistor, or IGBT. This has a control structure akin to a MOSFET coupled with a bipolar-like main conduction channel. These have become quite popular. 16.4 principles of digital electronics logical gates, counting circuits 16.4.1 Electronic logic gates The simplest form of electronic logic is diode logic (DL). This allows AND and OR gates to be built, but not inverters, and so is an incomplete form of logic. To built a complete logic system, valves or transistors can be used. The simplest family of logic gates using bipolar transistors is called resistor-transistor logic, or RTL. Unlike diode logic gates, RTL gates can be cascaded inde\ufb02nitely to produce more complex logic functions. These gates were used in early integrated circuits. For higher speed, the resistors used in RTL were replaced by diodes, leading to diodetransistor logic, or DTL. It was then discovered that one transistor could do the job of two diodes in the space of one diode, so transistor-transistor logic, or TTL, was created. In some types of chip, to reduce size and power consumption still further, the bipolar transistors were replaced with complementary \ufb02eld-e\ufb01ect transistors (MOSFETs), resulting in complementary metal-oxide-semiconductor (CMOS) logic. For small-scale logic, designers now use prefabricated logic gates from families of devices such as the TTL 7400 series invented by Texas Instruments and the CMOS 4000 series invented by RCA, and their more recent descendants. These devices usually contain transistors with multiple emitters, used to implement the AND function, which are not available as separate components. Increasingly, these \ufb02xed-function logic gates are being replaced by programmable logic devices, which allow designers to pack a huge number of mixed logic gates into a single integrated circuit. Electronic logic gates di\ufb01er signi\ufb02cantly from their relay-and-switch equivalents. They are much faster, consume much less power, and are much smaller (all by a factor of a million or more in most cases). Also, there is a fundamental structural di\ufb01erence. The switch circuit creates a continuous metallic path for current to ow (in either direction) between its input and its output. The semiconductor logic gate, on the other hand, acts as a high-gain voltage ampli\ufb02er, which sinks a tiny current at its input and produces a low-impedance voltage at its output. It is not possible for current to ow between the output and the input of a semiconductor logic gate. Another important advantage of standardised semiconductor logic gates, such as the 7400 and 4000 families, is that they are cascadable. This means that the output of one gate can be wired to the inputs of one or several other gates, and so on ad in\ufb02nitum, enabling the construction of circuits of arbitrary complexity without requiring the designer to understand the internal workings of the gates. In practice, the output of one gate can only drive a \ufb02nite number of inputs to other gates, a number called the \u2019fanout limit\u2019, but this limit is rarely reached in the newer CMOS logic circuits, as compared to TTL circuits. Also, there is always a delay, called the \u2019propagation delay\u2019, from a change an input of a gate to the corresponding change in its output. When gates are cascaded, the total propagation delay is approximately the sum of the individual delays, an e\ufb01ect which can become a problem in high-speed circuits. The US symbol for an AND gate is: AND symbol and the IEC symbol is AND symbol", ". The US circuit symbol for an OR gate is: OR symbol and the IEC symbol is: OR symbol. 332 The US circuit symbol for a NOT gate is: NOT symbol and the IEC symbol is: NOT symbol. In electronics a NOT gate is more commonly called an inverter. The circle on the symbol is called a bubble, and is generally used in circuit diagrams to indicate an inverted input or output. The US circuit symbol for a NAND gate is: NAND symbol and the IEC symbol is: NAND symbol. The US circuit symbol for a NOR gate is: NOR symbol and the IEC symbol is: NOR symbol. In practice, the cheapest gate to manufacture is usually the NAND gate. Additionally, Charles Peirce showed that NAND gates alone (as well as NOR gates alone) can be used to reproduce all the other logic gates. Two more gates are the exclusive-OR or XOR function and its inverse, exclusive-NOR or XNOR. Exclusive-OR is true only when exactly one of its inputs is true. In practice, these gates are built from combinations of simpler logic gates. The US circuit symbol for an XOR gate is: XOR symbol and the IEC symbol is: XOR symbol. 16.5 Counting circuits An arithmetic and logical unit (ALU) adder provides the basic functionality of arithmetic operations within a computer, and is a signi\ufb02cant component of the arithmetic and logical unit. Adders are composed of half adders and full adders, which add two-bit binary pairs, and ripple carry adders and carry look ahead adders which do addition operations to a series of binary numbers. (NOTE TO SELF: Pictures on wikipedia under GFDL) 16.5.1 Half Adder A half adder is a logical circuit that performs an addition operation on two binary digits. The half adder produces a sum and a carry value which are both binary digits. Sum(s) = A xor B Cot(c) = A and B Half adder circuit diagram Half adder circuit diagram Following is the logic table for a half adder: A B Sum Cot 16.5.2 Full adder A full adder is a logical circuit that performs an addition operation on three binary digits. The full adder produces a sum and carry value, which are both binary digits. Sum = (A xor B) xor Cin Cot = (A nand B) nand (Cin nand (A xor B)) Full adder circuit diagram Full adder circuit diagram 333 A B Cin Sum Cot Quantity Symbol Unit S.I. Units Direction Units or Table 16.1: Units used in Electronics 334 Chapter 17 The Atom Atoms are the building blocks of matter. They are the basis of all the structures and organisms in the universe. The planets, the sun, grass and trees, the air we breathe, and people are all made up of atoms. 17.1 Models of the Atom 17.2 Structure of the Atom Atoms are very small and cannot be seen with the naked eye. They consist of two main parts: the positively charged nucleus at the centre and the negatively charged elementary particles called electrons which surround the nucleus in their orbitals. (Elementary particle means that the electron cannot be broken down to anything smaller and can be thought of as a point particle.) The nucleus of an atom is made up of a collection of positively charged protons and neutral particles called neutrons. interesting fact: the neutrons and protons are not elementary particles. They are actually made up of even smaller particles called quarks. Both protons and neutrons are made of three quarks each. There are all sorts of other particles composed of quarks which nuclear physicists study using huge detectors - you can \ufb02nd out more about this by reading the essay in Chapter??. (NOTE TO SELF: Insert diagram of atomic structure - see lab posters) Atoms are electrically neutral which means that they have the same number of negative electrons as positive protons. The number of protons in an atom is called the atomic number which is sometimes also called Z. (NOTE TO SELF: check A and Z) The atomic number is what distinguishes the di\ufb01erent chemical elements in the Periodic table from each other. In fact, the elements are listed on the Periodic table in order of their atomic numbers. For example, the \ufb02rst element, hydrogen (H), has one proton whereas the sixth element, carbon (C) has 6 protons. Atoms with the same number of protons (atomic number) share physical properties and show similar chemical behaviour. The number of neutrons plus protons in the nucleus is called the atomic mass of the atom. 335 17.3 Isotopes Two atoms are considered to be the same element if they have the same number of protons (atomic number). However, they do not have to have the same number of neutrons or overall", " atomic mass. Atoms which have the same number of protons but di\ufb01erent numbers of neutrons are called isotopes. For example, the hydrogen atom has one proton and no neutrons. Therefore its atomic number is Z=1 and atomic mass is A=1. If a neutron is added to the hydrogen nucleus, then a new atom is formed with atomic mass A=2 but atomic number is still Z=1. This atom is called deuterium and is an isotope of hydrogen. 17.4 Energy quantization and electron con\ufb02guration 17.5 Periodicity of ionization energy to support atom ar- rangement in Periodic Table 17.6 Successive ionisation energies to provide evidence for arrangement of electrons into core and valence [Brink and Jones sections: de Broglie - matter shows particle and wave characteristics, proved by Davisson and Germer. Shroedinger and Heisenberg developed this model into quantum mechanics] The nucleus (atomic nucleus) is the center of an atom. It is composed of one or more protons and usually some neutrons as well. The number of protons in an atom\u2019s nucleus is called the atomic number, and determines which element the atom is (for example hydrogen, carbon, oxygen, etc.). Though the positively charged protons exert a repulsive electromagnetic force on each other, the distances between nuclear particles are small enough that the strong interaction (which is stronger than the electromagnetic force but decreases more rapidly with distance) predominates. (The gravitational attraction is negligible, being a factor 1036 weaker than this electromagnetic repulsion.) The discovery of the electron was the \ufb02rst indication that the atom had internal structure. This structure was initially imagined according to the \"raisin cookie\" or \"plum pudding\" model, in which the small, negatively charged electrons were embedded in a large sphere containing all the positive charge. Ernest Rutherford and Marsden, however, discovered in 1911 that alpha particles from a radium source were sometimes scattered backwards from a gold foil, which led to the acceptance of a planetary model, in which the electrons orbited a tiny nucleus in the same way that the planets orbit the sun. Interesting Fact: The word atom is derived from the Greek atomos, indivisible, from a-, not, and tomos, a cut. An atom is the smallest portion into which a chemical element can be divided while still retaining its properties. Atoms are the basic constituents of molecules and ordinary matter. Atoms are composed of subatomic particles. Atoms are composed mostly of empty space, but also of smaller subatomic particles. At the center of the atom is a tiny positive nucleus composed of nucleons (protons and neutrons). The rest of the atom contains only the fairly exible electron shells. Usually atoms are electrically neutral with as many electrons as protons. 336 Atoms are generally classi\ufb02ed by their atomic number, which corresponds to the number of protons in the atom. For example, carbon atoms are those atoms containing 6 protons. All atoms with the same atomic number share a wide variety of physical properties and exhibit the same chemical behavior. The various kinds of atoms are listed in the Periodic table. Atoms having the same atomic number, but di\ufb01erent atomic masses (due to their di\ufb01erent numbers of neutrons), are called isotopes. The simplest atom is the hydrogen atom, having atomic number 1 and consisting of one proton and one electron. It has been the subject of much interest in science, particularly in the early development of quantum theory. In The chemical behavior of atoms is largely due to interactions between the electrons. particular the electrons in the outermost shell, called the valence electrons, have the greatest inuence on chemical behavior. Core electrons (those not in the outer shell) play a role, but it is usually in terms of a secondary e\ufb01ect due to screening of the positive charge in the atomic nucleus. There is a strong tendency for atoms to completely \ufb02ll (or empty) the outer electron shell, which in hydrogen and helium has space for two electrons, and in all other atoms has space for eight. This is achieved either by sharing electrons with neighboring atoms or by completely removing electrons from other atoms. When electrons are shared a covalent bond is formed between the two atoms. Covalent bonds are the strongest type of atomic bond. When one or more electrons are completely removed from one atom by another, ions are formed. Ions are atoms that possess a net charge due to an imbalance in the number of protons and electrons. The ion that stole the electron(s) is called an anion and is negatively charged. The atom that lost the electron(s) is called a cation and is positively charged. Cations and anions are attracted to each other due to coulombic forces between the positive and negative charges. This attraction is called ionic bonding and", " is weaker than covalent bonding. As mentioned above covalent bonding implies a state in which electrons are shared equally between atoms, while ionic bonding implies that the electrons are completely con\ufb02ned to the anion. Except for a limited number of extreme cases, neither of these pictures is completely accurate. In most cases of covalent bonding, the electron is unequally shared, spending more time around the more electronegative atom, resulting in the covalent bond having some ionic character. Similarly, in ionic bonding the electrons often spend a small fraction of time around the more electropositive atom, resulting in some covalent character for the ionic bond. [edit] Models of the atom * Democritus\u2019 shaped-atom model (for want of a better name) * The plum pudding model * Cubical atom * The Bohr model * The quantum mechanical model The Plum pudding model of the atom was made after the discovery of the electron but before the discovery of the proton or neutron. In it, the atom is envisioned as electrons surrounded by a soup of positive charge, like plums surrounded by pudding. This model was disproved by an experiment by Ernest Rutherford when he discovered the nucleus of the atom. The Bohr Model is a physical model that depicts the atom as a small positively charged nucleus with electrons in orbit at di\ufb01erent levels, similar in structure to the solar system. Because of its simplicity, the Bohr model is still commonly used and taught today. In the early part of the 20th century, experiments by Ernest Rutherford and others had established that atoms consisted of a small dense positively charged nucleus surrounded by orbiting negatively charged electrons. However classical physics at that time was unable to explain why the orbiting electrons did not spiral into the nucleus. The simplest possible atom is hydrogen, which consists of a nucleus and one orbiting electron. Since the nucleus is positive and the electron are oppositely charged they will attract one another by coulomb force, in much the same way that the sun attracts the earth by gravitational force. 337 However, if the electron orbits the nucleus in a classical orbit, it ought to emit electromagnetic radiation (light) according to well established theories of electromagnetism. If the orbiting electron emits light, it must lose energy and spiral into the nucleus, so why do atoms even exist? What\u2019s more, the spectra of atoms show that the orbiting electrons can emit light but only at certain frequencies. This made no sense at all to the scientists of the time. These di\u2013culties were resolved in 1913 by Niels Bohr who proposed that: * (1) The orbiting electrons existed in orbits that had discrete quantized energies. That is, not every orbit is possible but only certain speci\ufb02c ones. The exact energies of the allowed orbits depends on the atom in question. * (2) The laws of classical mechanics do not apply when electrons make the jump from one allowed orbit to another. * (3) When an electron makes a jump from one orbit to another the energy di\ufb01erence is carried o\ufb01 (or supplied) by a single quantum of light (called a photon) which has a frequency that directly depends on the energy di\ufb01erence between the two orbitals. f = E / h where f is the frequency of the photon, E the energy di\ufb01erence, and h is a constant of proportionality known as Planck\u2019s constant. De\ufb02ning we can write where? is the angular frequency of the photon. * (4) The allowed orbits depend on quantized (discrete) values of orbital angular momentum, L according to the equation Where n = 1,2,3, and is called the angular momentum quantum number. These assumptions explained many of the observations seen at the time, such as why spectra consist of discrete lines. Assumption 4) states that the lowest value of n is 1. This corresponds to a smallest possible radius (for the mathematics see Ohanian-principles of physics or any of the large, usually American, college introductory physics textbooks) of 0.0529 nm. This is known as the Bohr radius, and explains why atoms are stable. Once an electron is in the lowest orbit, it can go no further. It cannot emit any more light because it would need to go into a lower orbit, but it can\u2019t do that if it is already in the lowest allowed orbit. The Bohr model is sometimes known as the semiclassical model because although it does include some ideas of quantum mechanics it is not a full quantum mechanical description of the atom. Assumption 2) states that the laws of classical mechanics don\u2019t apply during a quantum jump but doesn\u2019t state what laws should replace classical mechanics. Assumption 4) states that angular momentum is quantised but does not explain why. In order to fully describe an atom we need to use the full theory of quantum", " mechanics, which was worked out by a number of people in the years following the Bohr model. This theory treats the electrons as waves, which create 3D standing wave patterns in the atom. (This is why quantum mechanics is sometimes called wave mechanics.) This theory considers that idea of electrons as being little billiard ball like particles that travel round in orbits as absurdly wrong; instead electrons form probability clouds. You might \ufb02nd the electron here with a certain probability; you might \ufb02nd it over there with a di\ufb01erent probability. However it is interesting to note that if you work out the average radius of an electron in the lowest possible energy state it turns out to be exactly equal to the Bohr radius (although it takes many more pages of mathematics to work it out). The full quantum mechanics theory is a beautiful theory that has been experimentally tested and found to be incredibly accurate, however it is mathematically much more advanced, and often using the much simpler Bohr model will get you the results with much less hassle. The thing to remember is that it is only a model, an aid to understanding. Atoms are not really little solar systems. * See also: Hydrogen atom, quantum mechanics, Schrdinger equation, Niels Bohr. * An interactive demonstration (http://webphysics.davidson.edu/faculty/dmb/hydrogen/) of the prob- 338 ability clouds of electron in Hydrogen atorm according to the full QM solution. 17.7 Bohr orbits Brink and Jones sections: Standing waves (quantisation). Atom seen as positive nucleus with vibrating electron waves surrounding it. Shrodinger\u2019s equation calucaltes the energy of these waves and their shape and position{ most probable region of movement of electrons called orbitals (talk about n=1,2 energy levels and spdf orbitals). 17.8 Heisenberg uncertainty Principle Quantum mechanics is a physical theory that describes the behavior of physical systems at short distances. Quantum mechanics provides a mathematical framework derived from a small set of basic principles capable of producing experimental predictions for three types of phenomena that classical mechanics and classical electrodynamics cannot account for: quantization, wave-particle duality, and quantum entanglement. The related terms quantum physics and quantum theory are sometimes used as synonyms of quantum mechanics, but also to denote a superset of theories, including pre-quantum mechanics old quantum theory, or, when the term quantum mechanics is used in a more restricted sense, to include theories like quantum \ufb02eld theory. Quantum mechanics is the underlying theory of many \ufb02elds of physics and chemistry, includ- ing condensed matter physics, quantum chemistry, and particle physics. 17.9 Pauli exclusion principle The Pauli exclusion principle is a quantum mechanical principle which states that no two identical fermions may occupy the same quantum state. Formulated by Wolfgang Pauli in 1925, it is also referred to as the \"exclusion principle\" or \"Pauli principle.\" The Pauli principle only applies to fermions, particles which form antisymmetric quantum states and have half-integer spin. Fermions include protons, neutrons, and electrons, the three types of elementary particles which constitute ordinary matter. The Pauli exclusion principle governs many of the distinctive characteristics of matter. Particles like the photon and graviton do not obey the Pauli exclusion principle, because they are bosons (i.e. they form symmetric quantum states and have integer spin) rather than fermions. The Pauli exclusion principle plays a role in a huge number of physical phenomena. One of the most important, and the one for which it was originally formulated, is the electron shell structure of atoms. An electrically neutral atom contains bound electrons equal in number to the protons in the nucleus. Since electrons are fermions, the Pauli exclusion principle forbids them from occupying the same quantum state. For example, consider a neutral helium atom, which has two bound electrons. Both of these electrons can occupy the lowest-energy (1s) states by acquiring opposite spin. This does not violate the Pauli principle because spin is part of the quantum state of the electron, so the two electrons are occupying di\ufb01erent quantum states. However, the spin can take only two di\ufb01erent In a lithium atom, which contains three bound electrons, the third values (or eigenvalues.) electron cannot \ufb02t into a 1s state, and has to occupy one of the higher-energy 2s states instead. Similarly, successive elements produce successively higher-energy shells. The chemical properties of an element largely depends on the number of electrons in the outermost shell, which gives rise to the periodic table of the elements. 339 The Pauli principle is also responsible for the large-scale stability of matter. Molecules cannot be pushed arbitrarily close together, because the bound electrons in each molecule are forbidden from entering the", " same state as the electrons in the other molecules - this is the reason for the repulsive r-12 term in the Lennard-Jones potential. The Pauli principle is the reason you do not fall through the oor. Astronomy provides the most spectacular demonstrations of this e\ufb01ect, in the form of white dwarf stars and neutron stars. In both types of objects, the usual atomic structures are disrupted by large gravitational forces, leaving the constituents supported only by a \"degeneracy pressure\" produced by the Pauli exclusion principle. This exotic form of matter is known as degenerate matter. In white dwarfs, the atoms are held apart by the degeneracy pressure of the electrons. In neutron stars, which exhibit even larger gravitational forces, the electrons have merged with the protons to form neutrons, which produce a larger degeneracy pressure. Another physical phenomenon for which the Pauli principle is responsible is ferromagnetism, in which the exclusion e\ufb01ect implies an exchange energy that induces neigboring electron spins to align (whereas classically they would anti-align). 17.10 Ionization Energy (\ufb02rst, second etc.) 17.11 Electron con\ufb02guration i.e. \ufb02lling the orbitals starting from 1s..... Aufbau principle unpaired and paired electrons Hund\u2019s rule: 1 e- in each orbital before pairing in p orbitals shorthand: 1s22s22p1 etc 17.12 Valency Capacity for bonding Covalent bonding is a form of chemical bonding characterized by the sharing of one or more pairs of electrons, by two atoms, in order to produce a mutual attraction; atoms tend to share electrons, so as to \ufb02ll their outer electron shells. Such bonds are always stronger than the intermolecular hydrogen bond and similar in strength or stronger than the ionic bond. Commonly covalent bond implies the sharing of just a single pair of electrons. The sharing of two pairs is called a double bond and three pairs is called a triple bond. Aromatic rings of atoms and other resonant structures are held together by covalent bonds that are intermediate between single and double. The triple bond is relatively rare in nature, and two atoms are not observed to bond more than triply. Covalent bonding most frequently occurs between atoms with similar electronegativities, where neither atom can provide su\u2013cient energy to completely remove an electron from the other atom. Covalent bonds are more common between non-metals, whereas ionic bonding is more common between two metal atoms or a metal and a non-metal atom. Covalent bonding tends to be stronger than other types of bonding, such as ionic bonding. In addition unlike ionic bonding, where ions are held together by a non-directional coulombic attraction, covalent bonds are highly directional. As a result, covalently bonded molecules tend to form in a relatively small number of characteristic shapes, exhibiting speci\ufb02c bonding angles. 340 17.13 341 Chapter 18 Modern Physics 18.1 Introduction to the idea of a quantum Imagine that a beam of light is actually made up of little \"packets\" or \"bundles\" of energy, called quanta. It\u2019s like looking at a crowd of people from above. At \ufb02rst, it seems as though they are one huge patch, without any spaces between them. You would never suspect that they were people. But as you move closer, you slowly begin to see that they are individuals, and when you get even closer, you may even recognize a few. Light seems like a continuous wave at \ufb02rst, but when we zoom in at the subatomic level, we notice that a beam of light actually consists of little \"packets\" of energy, or quanta. This idea introduces the concept of the quantum (particle) nature of light, which is demonstrated by the photoelectric e\ufb01ect. When a metal surface is illuminated with light, electrons can be emitted from the surface. This is known as the photoelectric e\ufb01ect. 18.2 The wave-particle duality The wave nature of light is demonstrated by di\ufb01raction, interference, and polarization of light; and the particle nature of light is demonstrated by the photoelectric e\ufb01ect. So light has both wave-like and particle-like properties, but only shows one or the other, depending on the kind of experiment we perform. A wave-type experiment shows the wave nature, and a particle-type experiment shows particle nature. When you\u2019re watching a cricketer on the \ufb02eld, you see only that side of his personality. So to you, he is just a good cricketer. You do not see his gol\ufb02ng side, for example. Only when he is playing golf, will that side be", " revealed to you. The same applies to light. Now, we consider light to behave not as a wave, but as particles. But what do we call a \u2019particle\u2019 of light? Photon : A photon is a quantum (energy packet) of light. Imagine a sheet of metal. On the surface, there are electrons that are waiting to be set free. If a photon comes along and strikes the surface of the metal, then it will give its entire energy packet to one electron. This means that the electron now has some energy, and it may escape (leave the surface) if this energy Ek is greater than the minimum energy required to free an electron Emin. Now, suppose the electron needs 5eV of kinetic energy to escape. And suppose this little 342 photon has just 2eV of energy in its energy packet. Then the electron will not leave the surface of the metal. But suppose the photon has 8eV of energy. This means that the electron will emerge with 3eV of kinetic energy. Note that this does not mean the photon can give 5eV of energy to one electron and 3eV to another. A photon will give all of its energy to just one electron. The minimum amount of energy needed for an electron to escape (electrons do not normally leave a metal whenever they please), is called the work function of the metal. In our example, the work function is 5eV. The work function has a di\ufb01erent value for each metal: 4.70eV for copper and 2.28eV for sodium. It is worth mentioning that the best conductors are those with the smallest work functions. The frequency of the radiation is very important, because if it is below a certain threshold value, no electrons will be emitted. Even if the intensity of the light is increased, and the light is allowed to fall on the surface for a long period of time, if the frequency of the radiation is below the threshold frequency, electrons will not be emitted. We therefore reason that E = h? Where E is the energy of the photon, h = 6:57X10\u00a134Js is Plancks constant, and? is the frequency of the radiation. This means that the kinetic energy acquired by the electron is equal to the energy of the photon minus the work function,?, i.e. Ek = h? -? The electrons emerge with a range of velocities from zero up to a maximum vmax. The maximum kinetic energy, (1/2)mvmax2, depends (linearly) on the frequency of the radiation, and is independent of its intensity. For incident radiation of a given frequency, the number of electrons emitted per unit time is proportional to the intensity of the radiation. Electron emission takes place from the instant the light shines on the surface, i.e. there is no detectable time delay. What are the uses of the photoelectric e\ufb01ect? For this work, Einstein received the Nobel prize in 1905. 18.3 Practical Applications of Waves: Electromagnetic Waves In physics, wave-particle duality holds that light and matter simultaneously exhibit properties of waves and of particles. This concept is a consequence of quantum mechanics. In 1905, Einstein reconciled Huygens\u2019 view with that of Newton; he explained the photoelectric e\ufb01ect (an e\ufb01ect in which light did not seem to act as a wave) by postulating the existence of photons, quanta of energy with particulate qualities. Einstein postulated that light\u2019s frequency,?, is related to the energy, E, of its photons: E = hf 10\u00a134Js). (18.1) where h is Planck\u2019s constant (6:626 \u00a3 In 1924, De Broglie claimed that all matter has a wave-like nature; he related wavelength,?, and momentum, p: \u201a = h p (18.2) This is a generalization of Einstein\u2019s equation above since the momentum of a photon is given by, p = ; (18.3) E c where c is the speed of light in vacuum, and? = c /?. 343 De Broglie\u2019s formula was con\ufb02rmed three years later by guiding a beam of electrons (which have rest mass) through a crystalline grid and observing the predicted interference patterns. Similar experiments have since been conducted with neutrons and protons. Authors of similar recent experiments with atoms and molecules claim that these larger particles also act like waves. This is still a contoversial subject because these experimenters have assumed arguments of waveparticle duality and have assumed the validity of deBroglie\u2019s equation in their argument. The Planck constant h is extremely small and that explains why we don\u2019t perceive a wave-like quality of everyday objects: their wavelengths are exceedingly small. The fact that matter can have very", " short wavelengths is exploited in electron microscopy. In quantum mechanics, the wave-particle duality is explained as follows: every system and particle is described by state functions which encode the probability distributions of all measurable variables. The position of the particle is one such variable. Before an observation is made the position of the particle is described in terms of probability waves which can interfere with each other. 344 Chapter 19 Inside atomic nucleus Amazingly enough, human mind that is kind of contained inside a couple of liters of human\u2019s brain, is able to deal with extremely large as well as extremely small objects such as the whole universe and its smallest building blocks. So, what are these building blocks? As we already know, the universe consists of galaxies, which consist of stars with planets moving around. The planets are made of molecules, which are bound groups (chemical compounds) of atoms. There are more than 1020 stars in the universe. Currently, scientists know over 12 million chemical compounds i.e. 12 million di\ufb01erent molecules. All this variety of molecules is made of only a hundred of di\ufb01erent atoms. For those who believe in beauty and harmony of nature, this number is still too large. They would expect to have just few di\ufb01erent things from which all other substances are made. In this chapter, we are going to \ufb02nd out what these elementary things are. 19.1 What the atom is made of The Greek word \ufb01\u00bf o\u201eo\u201d (atom) means indivisible. The discovery of the fact that an atom is actually a complex system and can be broken in pieces was the most important step and pivoting point in the development of modern physics. It was discovered (by Rutherford in 1911) that an atom consists of a positively charged nucleus and negative electrons moving around it. At \ufb02rst, people tried to visualize an atom as a microscopic analog of our solar system where planets move around the sun. This naive planetary model assumes that in the world of very small objects the Newton laws of classical mechanics are valid. This, however, is not the case. The microscopic world is governed by quantum mechanics which does not have such notion as trajectory. Instead, it describes the dynamics of particles in terms of quantum states that are characterized by probability distributions of various observable quantities. For example, an electron in the atom is not moving along a certain trajectory but rather along all imaginable trajectories with di\ufb01erent probabilities. If we were trying to catch this electron, after many such attempts we would discover that the electron can be found anywere around the nucleus, even very close to and very far from it. However, the probabilities of \ufb02nding the electron 345 P (r) RBohr r Figure 19.1: Probability density P (r) for \ufb02nding the electron at a distance r from the proton in the ground state of hydrogen atom. at di\ufb01erent distances from the nucleus would be di\ufb01erent. What is amazing: the most probable distance corresponds to the classical trajectory! You can visualize the electron inside an atom as moving around the nucleus chaotically and extremely fast so that for our \\mental eyes\" it forms a cloud. In some places this cloud is more dense while in other places more thin. The density of the cloud corresponds to the probability of \ufb02nding the electron in a particular place. Space distribution of this density (probability) is what we can calculate using quantum mechanics. Results of such calculation for hydrogen atom are shown in Fig. 19.1. As was mentioned above, the most probable distance (maximum of the curve) coincides with the Bohr radius. Quantum mechanical equation for any bound system (like an atom) can have solutions only at a discrete set of energies E1; E2; E3 : : :, etc. There are simply no solutions for the energies E in between these values, such as, for instance, E1 < E < E2. This is why a bound system of microscopic particles cannot have an arbitrary energy and can only be in one of the quantum states. Each of such states has certain energy and certain space con\ufb02guration, i.e. distribution of the probability. A bound quantum system can make transitions from one quantum state to another either spontaneously or as a result of interaction with other systems. The energy conservation law is one of the most fundamental and is valid in quantum world as well as in classical world. This means that any transition between the states with energies Ei and Ej is accompanied with either Ei \u00a1 emission or absorption of the energy \u00a2E = j. This is how an atom emits light. Ejj Electron is a very light particle. Its mass is negligible as compared to the total mass of the atom. For example, in the lightest of all atoms, hydrogen, the electron constitutes only 0.054% of the atomic mass. In the silicon atoms", " that are the main component of the rocks around us, all 14 electrons make up only 0.027% of the mass. Thus, when holding a heavy rock in your hand, you actually feel the collective weight of all the nuclei that are inside it. 346 19.2 Nucleus Is the nucleus a solid body? Is it an elementary building block of nature? No and no! Although it is very small, a nucleus consists of something even smaller. 19.2.1 Proton The only way to do experiments with such small objects as atoms and nuclei, is to collide them with each other and watch what happens. Perhaps you think that this is a barbaric way, like colliding a \\Mercedes\" and \\Toyota\" in order to learn what is under their bonnets. But with microscopic particles nothing else can be done. In the early 1920\u2019s Rutherford and other physicists made many experiments, changing one element into another by striking them with energetic helium nuclei. They noticed that all the time hydrogen nuclei were emitted in the process. It was apparent that the hydrogen nucleus played a fundamental role in nuclear structure and was a constituent part of all other nuclei. By the late 1920\u2019s physicists were regularly referring to hydrogen nucleus as proton. The term \\proton\" seems to have been coined by Rutherford, and \ufb02rst appears in print in 1920. 19.2.2 Neutron Thus it was established that atomic nuclei consist of protons. Number of protons in a nucleus is such that makes up its positive charge. This number, therefore, coincides with the atomic number of the element in the Mendeleev\u2019s Periodic table. This sounded nice and logical, but serious questions remained. Indeed, how can positively charged protons stay together in a nucleus? Repelling each other by electric force, they should y away in di\ufb01erent directions. Who keeps them together? Furthermore, the proton mass is not enough to account for the nuclear masses. For example, if the protons were the only particles in the nucleus, then a helium nucleus (atomic number 2) would have two protons and therefore only twice the mass of hydrogen. However, it actually is four times heavier than hydrogen. This suggests that it must be something else inside nuclei in addition to protons. These additional particles that kind of \\glue\" the protons and make up the nuclear mass, apparently, are electrically neutral. They were therefore called neutrons. Rutherford predicted the existence of the neutron in 1920. Twelve years later, in 1932, his assistant James Chadwick found it and measured its mass, which turned out to be almost the same but slightly larger than that of the proton. 19.2.3 Isotopes Thus, in the early 1930\u2019s it was \ufb02nally proved that atomic nucleus consists of two types of particles, the protons and neutrons. The protons are positively charged while the neutrons are electrically neutral. The proton charge is exactly equal but opposite to that of the electron. The masses of proton and neutron are almost the same, approximately 1836 and 1839 electron masses, respectively. 347 Apart from the electric charge, the proton and neutron have almost the same properties. This is why there is a common name for them: nucleon. Both the proton and neutron are nucleons, like a man and a woman are both humans. In physics literature, the proton is denoted by letter p and the neutron by n. Sometimes, when the di\ufb01erence between them is unimportant, it is used the letter N meaning nucleon (in the same sense as using the word \\person\" instead of man or woman). Chemical properties of an element are determined by the charge of its atomic nucleus, i.e. by the number of protons. This number is called the atomic number and is denoted by letter Z. The mass of an atom depends on how many nucleons its nucleus contains. The number of nucleons, i.e. total number of protons and neutrons, is called the atomic mass number and is denoted by letter A. Standard nuclear notation shows the chemical symbol, the mass number and the atomic number of the isotope. number of nucleons number of protons A ZX chemical symbol For example, the iron nucleus (26-th place in the Mendeleev\u2019s periodic table of the elements) with 26 protons and 30 neutrons is denoted as 56 26Fe ; where the total nuclear charge is Z = 26 and the mass number A = 56. The number of neutrons Z (here, it is used the same letter N, as for nucleon, but this is simply the di\ufb01erence N = A should not cause any confusion). Chemical symbol is inseparably linked with Z. This is why the lower index is sometimes omitted and you may encounter the simpli\ufb02ed notation like 56Fe.", " \u00a1 If we add or remove a few neutrons from a nucleus, the chemical properties of the atom remain the same because its charge is the same. This means that such atom should remain in the same place of the Periodic table. In Greek, \\same place\" reads \u00b6\u00b6o& \u00bf \u00b6o\u2026o& (isos topos). The nuclei, having the same number of protons, but di\ufb01erent number of neutrons, are called therefore isotopes. Di\ufb01erent isotopes of a given element have the same atomic number Z, but di\ufb01erent mass numbers A since they have di\ufb01erent numbers of neutrons N. Chemical properties of di\ufb01erent isotopes of an element are identical, but they will often have great di\ufb01erences in nuclear stability. For stable isotopes of the light elements, the number of neutrons will be almost equal to the number of protons, but for heavier elements, the number of neutrons is always greater than Z and the neutron excess tends to grow when Z increases. This is because neutrons are kind of glue that keeps repelling protons together. The greater the repelling charge, the more glue you need. 348 19.3 Nuclear force Since atomic nuclei are very stable, the protons and neutrons must be kept inside them by some force and this force must be rather strong. What is this force? All of modern particle physics was discovered in the e\ufb01ort to understand this force! Trying to answer this question, at the beginning of the XX-th century, physicists found that all they knew before, was inadequate. Actually, by that time they knew only gravitational and electromagnetic forces. It was clear that the forces holding nucleons were not electromagnetic. Indeed, the protons, being positively charged, repel each other and all nuclei would decay in a split of a second if some other forces would not hold them together. On the other hand, it was also clear that they were not gravitational, which would be too weak for the task. The simple conclusion was that nucleons are able to attract each other by yet unknown nuclear forces, which are stronger than the electromagnetic ones. Further studies proved that this hypothesis was correct. Nuclear force has rather unusual properties. Firstly, it is charge independent. This means 10\u00a113 cm, that in all pairs nn, pp, and np nuclear forces are the same. Secondly, at distances 100 times stronger than the electromagnetic the nuclear force is attractive and very strong, \u00bb repulsion. Thirdly, the nuclear force is of a very short range. If the nucleons move away from each other for more than few fermi (1 fm=10\u00a113 cm) the nuclear attraction practically disappears. Therefore the nuclear force looks like a \\strong man with very short hands\". \u00bb 19.4 Binding energy and nuclear masses 19.4.1 Binding energy When a system of particles is bound, you have to spend certain energy to disintegrate it, i.e. to separate the particles. The easiest way to do it is to strike the system with a moving particle that carries kinetic energy, like we can destroy a glass bottle with a bullet or a stone. If our bulletparticle moves too slow (i.e. does not have enough kinetic energy) it cannot disintegrate the system. On the other hand, if its kinetic energy is too high, the system is not only disintegrated but the separated particles acquire some kinetic energy, i.e. move away with some speed. There is an intermediate value of the energy which is just enough to destroy the system without giving its fragments any speed. This minimal energy needed to break up a bound system is called binding energy of this system. It is usually denoted by letter B. 19.4.2 Nuclear energy units The standart unit of energy, Joule, is too large to measure the energies associated with individual nuclei. This is why in nuclear physics it is more convenient to use a much smaller unit called Mega-electron-Volt (MeV). This is the amount of energy that an electron acquires after passing between two charged plates with the potential di\ufb01erence (voltage) of one million Volts. Sounds very huge, isn\u2019t it? But look at this relation and think again. In the units of MeV, most of the energies in nuclear world can be expressed by values with only few digits before decimal point and without ten to the power of something. For 1 MeV = 1:602 10\u00a113 J \u00a3 349 example, the binding energy of proton and neutron (which is the simplest nuclear system and is called deuteron) is Bpn = 2:225 MeV : The simplicity of the numbers is not the only advantage of using the unit MeV. Another, more important advantage, comes from the fact that most of experiments in nuclear physics are collision experiments, where particles are accelerated by electric \ufffd", "\ufffd\ufffdeld and collide with other particles. From the above value of Bpn, for instance, we immediately know that in order to break up deuterons, we need to bombard them with a ux of electrons accelerated through a voltage not less than 2.225 million Volts. No calculation is needed! On the other hand, if we know that a charged particle (with a unit charge) passes through a voltage, say, 5 million Volts, we can, without any calculation, say that it acqures the energy of 5 MeV. It is very convenient. Isn\u2019t it? 19.4.3 Mass defect Comparing the masses of atomic nuclei with the masses of the nucleons that constitute them, we encounter a surprising fact: Total mass of the nucleons is greater than mass of the nucleus! For example, for the deuteron we have md < mp + mn ; where md, mp, and mn are the masses of deuteron, proton, and neutron, respectively. The di\ufb01erence is rather small, but on the nuclear scale is noticeable since the mass of proton, for example, (mp + mn) \u00a1 md = 3:968 \u00a3 10\u00a130 kg ; mp = 1672:623 10\u00a130 kg \u00a3 is also very small. This phenomenon is called \\mass defect\". Where the mass disappears to, when nucleons are bound? To answer this question, we notice that the energy of a bound state is lower than the energy of free particles. Indeed, to liberate them from a bound complex, we have to give them some energy. Thinking in the opposite direction, we conclude that, when forming a bound state, the particles have to get rid of the energy excess, which is exactly equal to the binding energy. This is observed experimentally: When a proton captures a neutron to form a deuteron, the excess energy of 2.225 MeV is emitted via electromagnetic radiation. A logical conclusion from the above comes by itself: When proton and neutron are bounding, some part of their mass disappears together with the energy that is carried away by the radiation. And in the opposite process, when we break up the deutron, we give it the energy, some part of which makes up the lost mass. Albert Einstein came to the idea of the equivalence between the mass and energy long before any experimental evidences were found. In his theory of relativity, he showed that total energy E of a moving body with mass m is ; v2 c2 mc2 E = 1 \u00a1 r 350 (19.1) where v is its velocity and c the speed of light. Applying this equation to a non-moving body (v = 0), we conclude that it possesses the rest energy E0 = mc2 (19.2) simply because it has mass. As you will see, this very formula is the basis for making nuclear bombs and nuclear power stations! All the development of physics and chemistry, preceding the theory of relativity, was based on the assumption that the mass and energy of a closed system are conserving in all possible processes and they are conserved separately. In reality, it turned out that the conserving quantity is the mass-energy, Ekin + Epot + Erad + mc2 = const ; i.e. the sum of kinetic energy, potential energy, the energy of radiation, and the mass of the system. In chemical reactions the fraction of the mass that is transformed into other forms of energy (and vise versa), is so small that it is not detectable even in most precise measurements. In nuclear processes, however, the energy release is very often millions times higher and therefore is observable. You should not think that mutual transformations of mass and energy are the features of only nuclear and atomic processes. If you break up a piece of rubber or chewing gum, for example, in two parts, then the sum of masses of these parts will be slightly larger than the mass of the whole piece. Of course we will not be able to detect this \\mass defect\" with our scales. But we can calculate it, using the Einstein formula (19.2). For this, we would need to measure somehow the mechanical work A used to break up the whole piece (i.e. the amount of energy supplied to it). This can be done by measuring the force and displacement in the breaking process. Then, according to Eq. (19.2), the mass defect is \u00a2m = A c2 : To estimate possible e\ufb01ect, let us assume that we need to stretch a piece of rubber in 10 cm before it breaks, and the average force needed for this is 10 N (approximately 1 kg). Then and hence A = 10 N \u00a3 0:1 m = 1 J ; \u00a2m = 1 J (299792458 m=s)2 \u2026 10\u00a117 kg: 1:1 \u00a3 This is very small value for measuring with a", " scale, but huge as compared to typical masses of atoms and nuclei. 19.4.4 Nuclear masses Apparently, an individual nucleus cannot be put on a scale to measure its mass. Then how can nuclear masses be measured? This is done with the help of the devices called mass spectrometers. In them, a ux of identical nuclei, accelerated to a certain energy, is directed to a screen where it makes a visible mark. 351 Before striking the screen, this ux passes through magnetic \ufb02eld, which is perpendicular to velocity of the nuclei. As a result, the ux is deected to certain angle. The greater the mass, the smaller is the angle (because of inertia). Thus, measuring the displacement of the mark from the center of the screen, we can \ufb02nd the deection angle and then calculate the mass. Since mass and energy are equivalent, in nuclear physics it is customary to measure masses of all particles in the units of energy, namely, in MeV. Examples of masses of subatomic particles are given in Table 19.1. The values given in this table, are the energies to which the nuclear particle number of protons number of neutrons mass (MeV) e p n 2 1H 3 1H 3 2He 4 2He 7 3Li 9 4Be 12 6C 16 8O.511 938.272 939.566 1875.613 2808.920 2808.391 3727.378 6533.832 8392.748 11174.860 14895.077 238 92U 92 146 221695.831 Table 19.1: Masses of electron, nucleons, and some nuclei. masses are equivalent via the Einstein formula (19.2). \u00a3 There are several advantages of using the units of MeV to measure particle masses. First of all, like with nuclear energies, we avoid handling very small numbers that involve ten to the power of something. For example, if we were measuring masses in kg, the electron mass would 10\u00a131 kg. When masses are given in the equivalent energy units, it is very be me = 9:1093897 easy to calculate the mass defect. Indeed, adding the masses of proton and neutron, given in the second and third rows of Table 19.1, and subtracting the mass of 2 1H, we obtain the binding energy 2.225 MeV of the deuteron without further ado. One more advantage comes from particle physics. In collisions of very fast moving particles new particles (like electrons) can be created from vacuum, i.e. kinetic energy is directly transformed into mass. If the mass is expressed in the energy units, we know how much energy is needed to create this or that particle, without calculations. 352 19.5 Radioactivity As was said before, the nucleus experiences the intense struggle between the electric repulsion of protons and nuclear attraction of the nucleons to each other. It therefore should not be surprising that there are many nuclei that are unstable. They can spontaneously (i.e. without an external push) break in pieces. When the fragments reach the distances where the short range nuclear attraction disappears, they \ufb02ercely push each other away by the electric forces. Thus accelerated, they move in di\ufb01erent directions like small bullets making destruction on their way. This is an example of nuclear radioactivity but there are several other varieties of radioactive decay. 19.5.1 Discovery of radioactivity Nuclear radioactivity was discovered by Antoine Henri Becquerel in 1896. Following Wilhelm Roentgen who discovered the X-rays, Becquerel pursued his own investigations of these mysterious rays. The material Becquerel chose to work with contained uranium. He found that the crystals containing uranium and exposed to sunlight, made images on photographic plates even wrapped in black paper. He mistakingly concluded that the sun\u2019s energy was being absorbed by the uranium which then emitted X-rays. The truth was revealed thanks to bad weather. On the 26th and 27th of February 1896 the skies over Paris were overcast and the uranium crystals Becquerel intended to expose to the sun were returned to a drawer and put over (by chance) the photographic plates. On the \ufb02rst of March, Becquerel developed the plates and to his surprise, found that the images on them were clear and strong. Therefore the uranium emitted radiation without an external source of energy such as the sun. This was the \ufb02rst observation of the nuclear radioactivity. Later, Becquerel demonstrated that the uranium radiation was similar to the X-rays but, unlike them, could be deected by a magnetic \ufb02eld and therefore must consist of charged particles. For his discovery of radioactivity, Becquerel was awarded the 1903 Nobel Prize for physics. 19.5.2 Nuclear \ufb01, \ufb02, and rays Classical", " experiment that revealed complex content of the nuclear radiation, was done as follows. The radium crystals (another radioactive element) were put at the bottom of a narrow straight channel made in a thick piece of lead and open at one side. The lead absorbed everything except the particles moving along the channel. This device therefore produced a ux of particles moving in one direction like bullets from a machine gun. In front of the channel was a photoplate that could register the particles. Without the magnetic \ufb02eld, the image on the plate was in the form of one single dot. When the device was immersed into a perpendicular magnetic \ufb02eld, the ux of particles was split in three uxes, which was reected by three dots on the photographic plate. One of the three uxes was stright, while two others were deected in opposite directions. This showed that the initial ux contained positive, negative, and neutral particles. They were 353 named respectively the \ufb01, \ufb02, and particles. The \ufb01-rays were found to be the 4He nuclei, two protons and two neutrons bound together. They have weak penetrating ability, a few centimeters of air or a few sheets of paper can e\ufb01ectively block them. The \ufb02-rays proved to be electrons. They have a greater penetrating power than the \ufb01-particles and can penetrate 3 mm of aluminum. The -rays are not deected because they are high energy photons. They have the same nature as the radio waves, visible light, and the X-rays, but have much shorter wavelength and therefore are much more energetic. Among the three, the -rays have the greatest penetrating power being able to pass through several centimeters of lead and still be detected on the other side. 19.5.3 Danger of the ionizing radiation The \ufb01, \ufb02, and particles moving through matter, collide with atoms and knock out electrons from them, i.e. make positive ions out of the atoms. This is why these rays are called ionizing radiation. Apart from ionizing the atoms, this radiation destroys molecules. For humans and all other organisms, this is the most dangerous feature of the radiation. Imagine thousands of tiny tiny bullets passing through your body and making destruction on their way. Although people do not feel any pain when exposed to nuclear radiation, it harms the cells of the body and thus can make people sick or even kill them. Illness can strike people years after their exposure to nuclear radiation. For example, the ionizing particles can randomly modify the DNA (long organic molecules that store all the information on how a particular cell should function in the body). As a result, some cells with wrong DNA may become cancer cells. Fortunately, our body is able to repair some damages caused by radiation. Indeed, we are constantly bombarded by the radiation coming from the outer space as well as from the inner parts of our own planet and still survive. However, if the number of damages becomes too large, the body will not cope with them anymore. There are established norms and acceptable limits for the radiation that are considered safe for human body. If you are going to work in contact with radioactive materials or near them, make sure that the exposure dose is monitored and the limits are adhered to. You should understand that no costume can protect you from -rays! Only a thick wall of concrete or metal can stop them. The special costumes and masks that people wear when handling radioactive materials, protect them not from the rays but from contamination with that materials. Imagine if few specks of radioactive dirt stain your everyday clothes or if you inhale radioactive atoms. They will remain with you all the time and will shoot the \\bullets\" at you even when you are sleeping. In many cases, a very e\ufb01ective way of protecting yourself from the radiation is to keep certain distance. Radiation from nuclear sources is distributed equally in all directions. Therefore the number n of dangerous particles passing every second through a unit area (say 1 cm2) is the total number N of particles emitted during 1 second, divided by the surface of a sphere n = N 4\u2026r2 ; 354 where r is the distance at which we make the observation. From this simple formula, it is seen that the radiation intensity falls down with incresing distance quadratically. In other words, if you increase the distance by a factor of 2, your exposure to the radiation will be decreased by a factor of 4. 19.5.4 Decay law Unstable nuclei decay spontaneously. A given nucleus can decay next moment, next day or even next century. Nobody can predict when it is going to happen. Despite this seemingly chaotic and \\unscienti\ufb02c\" situation, there is a strict order in all this. Atomic nuclei, being microscopic objects, are ruled by quantum probabilistic laws. Although we cannot predict the exact moment of its decay, we can calculate the probability that a nucleus will decay", " within this or that time interval. Nuclei decay because of their internal dynamics and not because they become \\old\" or somehow \\rotten\". To illustrate this, let us imagine that yesterday morning we found that a certain nucleus was going to decay within 24 hours with the probability of 50%. However, this morning we found that it is still \\alive\". This fact does not mean that the decay probability for another 24 hours increased. Not at all! It remains the same, 50%, because the nucleus remains the same, nothing wrong happened to it. This can go on and on for centuries. Actually, we never deal with individual nuclei but rather with huge numbers of identical nuclei. For such collections (ensembles) of quantum objects, the probabilistic laws become statictical laws. Let us assume that in the above example we had 1 million identical nuclei instead of only one. Then by this morning only half of these nuclei would survive because the decay probability for 24 hours was 50%. Among the remaining 500000 nuclei, 250000 will decay by tomorrow morning, then after another 24 hours only 125000 will remain and so on. The number of unstable nuclei that are still \\alive\" continuously decreases with time according to the curve shown in Fig. 19.2. If initially, at time t = 0, their number is N0, then after certain time interval T1=2 only half of these nuclei will remain, namely, 1 2 N0. Another one half of the remaining half will decay during another such interval. So, after the time 2T1=2, we will have only one quarter of the initial amount, and so on. The time interval T1=2, during which one half of unstable nuclei decay, is called their half-life time. It is speci\ufb02c for each unstable nucleus and vary from a fraction of a second to thousands and millions of years. A few examples of such lifetimes are given in Table 19.2 19.5.5 Radioactive dating Examining the amounts of the decay products makes possible radioactive dating. The most famous is the Carbon dating, a variety of radioactive dating which is applicable only to matter which was once living and presumed to be in equilibrium with the atmosphere, taking in carbon dioxide from the air for photosynthesis. Cosmic ray protons blast nuclei in the upper atmosphere, producing neutrons which in turn bombard nitrogen, the major constituent of the atmosphere. This neutron bombardment produces the radioactive isotope 14 6C. The radioactive carbon-14 combines with oxygen to form 355 N (t) N0 1 2 N0 1 4 N0 1 8 N0 0 T1=2 2T1=2 3T1=2 4T1=2 t Figure 19.2: The time T1=2 during which one half of the initial amount of unstable particles decay, is called their half-life time. isotope T1=2 decay mode 214 84Po 89 36Kr 222 86Rn 90 38Sr 226 88Ra 14 6C 238 92U 115 49In 1:64 10\u00a14 s \u00a3 3:16 min 3.83 days 28:5 years 103 years 1:6 \u00a3 5:73 4:47 4:41 \u00a3 \u00a3 \u00a3 103 years 109 years 1014 years \ufb01; \ufb02\u00a1; \ufb01; \ufb02\u00a1 \ufb01; \ufb02\u00a1 \ufb01; \ufb02\u00a1 Table 19.2: Half-life times of several unstable isotopes. carbon dioxide and is incorporated into the cycle of living things. The isotope 14 6C decays (see Table 19.2) inside living bodies but is replenished from the air 356 and food. Therefore, while an organism is alive, the concentration of this isotope in the body remains constant. After death, the replenishment from the breath and food stops, but the isotopes that are in the dead body continue to decay. As a result the concentration of 14 6C in it gradually decreases according to the curve shown in Fig. 19.2. The time t = 0 on this Figure corresponds to the moment of death, and N0 is the equilibrium concentration of 14 6C in living organisms. Therefore, by measuring the radioactive emissions from once-living matter and comparing its activity with the equilibrium level of emissions from things living today, an estimation of the time elapsed can be made. For example, if the rate of the radioactive emissions from a piece of wood, caused by the decay of 14 6C, is one-half lower than from living trees, then we can conclude it is elapsed exactly one half-life-time that we are at the point t = T1=2 on the curve 19.2, i.e. period. According to the Table 19.2), this means that the tree, from which this piece of wood was made, was cut approximately 5730 years ago. This is how physicists help archaeologists", " to assign dates to various organic materials. 19.6 Nuclear reactions Those of you who studied chemistry, are familiar with the notion of chemical reaction, which, in essence, is just regrouping of atoms that constitute molecules. As a result, reagent chemical compounds are transformed into product compounds. In the world of nuclear particles, similar processes are possible. When nuclei are close to each other, nucleons from one nucleus can \\jump\" into another one. This happens because there are attractive and repulsive forces between the nucleons. The complicated interplay of these forces may cause their regrouping. As a result, the reagent particles are transformed into product particles. Such processes are called nuclear reactions. For example, when two isotopes 3 in such a way that the isotope 4 reactions, this process is denoted as 2He collide, the six nucleons constituting them, can rearrange 2He is formed and two protons are liberated. Similarly to chemical 2He + 3 3 2He \u00a1! 4 2He + p + p + 12:86 MeV : (19.3) The same as in chemical reactions, nuclear reactions can also be either exothermic (i.e. releasing energy) or endothermic (i.e. requiring an energy input). The above reaction releases 12.86 MeV of energy. This is because the total mass on the left hand side of Eq. (19.3) is in 12.86 MeV greater than the total mass of the products on the right hand side (you can check this using Table 19.1). Thus, when considering a particular nuclear reaction, we can always learn if it releases or absorbs energy. For this, we only need to compare total masses on the left and right hand sides of the equation. Now, you can understand why it is very convenient to express masses in the units of energy. Composing equations like (19.3), we should always check the superscripts and subscripts of the nuclei in order to have the same number of nucleons and the same charge on both sides of the equation. In the above example, we have six nucleons and the charge +4 in both the initial and \ufb02nal states of the reaction. To make the checking of nucleon number and charge conservation easier, sometimes the proton and neutron are denoted with superscripts and subscripts as well, 357 namely, 1 subscripts are the same on both sides of the equation. 1p and 1 0n. In this case, all we need is to check that sum of superscripts and sum of 19.7 Detectors How can we observe such tiny tiny things as protons and \ufb01-particles? There is no microscope that would be able to discern them. From the very beginning of the sub-atomic era, scientists have been working on the development of special instruments that are called particle detectors. These devices enable us either to register the mere fact that certain particle has passed through certain point in space or to observe the trace of its path (the trajectory). Actually, this is as good as watching the particle. Although the particle sizes are awfully small, when passing through some substances, they leave behind visible traces of tens of centimeters in length. By measuring the curvature of the trajectory of a particle deected in electric or magnetic \ufb02eld, a physicist can determine the charge and mass of the particle and thus can identify it. 19.7.1 Geiger counter The most familiar device for registering charged particles is the Geiger counter. It cannot tell you anything about the particle except the fact that it has passed through the counter. The counter consists of a thin metal cylinder \ufb02lled with gas. A wire electrode runs along the center of the tube and is kept at a high voltage ( 2000 V) relative to the cylinder. When a particle passes through the tube, it causes ionization of the gas atoms and thus an electric discharge between the cylinder and the wire. The electric pulse can be counted by a computer or made to produce a \\click\" in a loudspeaker. The number of counts per second tells us about intensity of the radiation. \u00bb 19.7.2 Fluorescent screen The very \ufb02rst detector was the uorescent screen. When a charged particle hits the screen, a human eye can discern a ash of light at the point of impact. In fact, we all use this kind of detectors every day when watching TV of looking at a computer (if it does not have an LCD screen of course). Indeed, the images on the screens of their electron-ray tubes are formed by the accelerated electrons. 19.7.3 Photo-emulsion Another type of particle detector, dating back to Becquerel, is the nuclear photographic emulsion. Passage of charged particles is recorded in the emulsion in the same way that ordinary black and white photographic \ufb02lm records a picture. The only di\ufb01erence is that nuclear photo", "emulsion is made rather thick in order to catch a signi\ufb02cant part of the particle path. After the developing, a permanent record of the charged particle trajectory is available. 19.7.4 Wilson\u2019s chamber In the \ufb02elds of sub-atomic physics and nuclear physics, Wilson\u2019s cloud chamber is the most fundamental device to observe the trajectories of particles. Its basic principle was discovered by C. T. R. Wilson in 1897, and it was put to the practical use in 1911. 358 The top and the side of the chamber are covered with round glasses of several centimeters in diameter. At the bottom of the chamber, a piston is placed. The air \ufb02lled in the chamber is saturated with vapor of water. When pulling down the piston quickly, the volume of the chamber would be expanded and the temperature goes down. As a result, the air inside would be supersaturated with the vapor. If a fast moving charged particle enters the chamber when it is in such a supersaturated state, the vapor of water would condense along the line of the ions generated by the particle, which is the path of the particle. Thus we can observe the trace, and also take a photograph. To make clear the trace, a light is sometimes illuminated from the side. When placing the cloud chamber in a magnetic \ufb02eld, we can obtain various informations about the charged particle by measuring the curvature of the trace and other data. The bubble chamber and the spark chamber have taken place of the cloud chamber which is nowadays used only for the educational purposes. Wilson\u2019s cloud chamber has however played a very important role in the history of physics. 19.7.5 Bubble chamber Bubble chamber is a particle detector of major importance during the initial years of high-energy physics. The bubble chamber has produced a wealth of physics from about 1955 well into the 1970s. It is based on the principle of bubble formation in a liquid heated above its boiling point, which is then suddenly expanded, starting boiling where passing charged particles have ionized the atoms of the liquid. The technique was honoured by the Nobel prize award to D. Glaser in 1960. Even today, bubble chamber photographs provide the aesthetically most appealing visualization of subnuclear collisions. 19.7.6 Spark chamber Spark chamber is a historic device using electric discharges over a gap between two electrodes with large potential di\ufb01erence, to render passing particles visible. Sparks occurred where the gas had been ionized. Most often, multiple short gaps were used, but wide-gap chambers with gaps up to 40 cm were also built. The spark chamber is still of great scienti\ufb02c value in that it remains relatively simple and cheap to build as well as enabling an observer to view the paths of charged particles. 19.8 Nuclear energy Nuclei can produce energy via two di\ufb01erent types of reactions, namely, \ufb02ssion and fusion reactions. Fission is a break up of a nucleus in two or more pieces (smaller nuclei). Fusion is the opposite process: Formation of a bigger nucleus from two small nuclei. A question may arise: How two opposite processes can both produce energy? Can we make an inexhaustible souce of energy by breaking up and then fusing the same nuclei? Of cousre not! The energy conservation law cannot be circumvented in no way. When speaking about fusion and \ufb02ssion, we speak about di\ufb01erent ranges of nuclei. Energy can only be released when either light nuclei fuse or heavy nuclei \ufb02ssion. To understand why this is so, let us recollect that for releasing energy the mass of initial nuclei must be greater than the mass of the products of a nuclear reaction. The mass di\ufb01erence is transformed into the released energy. And why the product nuclei can loose some mass as compared to the initial nuclei? Because they are more tightly bound, i.e. their binding energies 359 are lager. Fig. 19.3 shows the dependence of the binding energy B per nucleon on the number A of 9 MeV nucleons constituting a nucleus. As you see, the curve reaches the maximum value of per nucleon at around A 50. The nuclei with such number of nucleons cannot produce energy neither through fusion nor through \ufb02ssion. They are kind of \\ashes\" and cannot serve as a fuel. In contrast to them, very light nuclei, when fused with each other, make more tightly bound products as well as very heavy nuclei do when split up in lighter fragments. \u00bb \u00bb B=A, (MeV) 10 8 6 4 2 0 | {z } \"fusion % | {z } - \u02c6 \ufb02ssion | {z }. \u02c6 10 8 6 4 2 0 0 50 100 150 200 250 number of nucleons, A Figure 19.3", ": Binding energy per nucleon. In \ufb02ssion processes, which were discovered and used \ufb02rst, a heavy nucleus like, for example, uranium or plutonium, splits up in two fragments which are both positively charged. These fragments repel each other by an electric force and move apart at a high speed, distributing their kinetic energy in the surrounding material. In fusion reactions everything goes in the opposite direction. Very light nuclei, like hydrogen or helium isotopes, when approaching each other to a distance of a few fm (1 fm = 10\u00a113 cm), experience strong attraction which overpowers their Coulomb (that is electric) repulsion. As a result the two nuclei fuse into a single nucleus. They collapse with extremely high speeds towards each other. To form a stable nucleus they must get rid of the excessive energy. This energy is emitted by ejecting a neutron or a photon. 19.8.1 Nuclear reactors Since the discovery of radioactivity it was known that heavy nuclei release energy in the processes of spontaneous decay. This process, however, is rather slow and cannot be inuenced (speed up or slow down) by humans and therefore could not be e\ufb01ectively used for large-scale energy production. Nonetheless, it is ideal for feeding the devices that must work autonomously in remote 360 places for a long time and do not require much energy. For this, heat from the spontaneousdecays can be converted into electric power in a radioisotope thermoelectric generator. These generators have been used to power space probes and some lighthouses built by Russian engineers. Much more e\ufb01ective way of using nuclear energy is based on another type of nuclear decay which is considered next. Chain reaction The discovery that opened up the era of nuclear energy was made in 1939 by German physicists O. Hahn, L. Meitner, F Strassmann, and O. Frisch. They found that a uranium nucleus, after absorbing a neutron, splits into two fragments. This was not a spontaneous but induced \ufb02ssion n + 235 92U \u00a1! 54Xe + 94 140 38Sr + n + n + 185 MeV (19.4) \u00bb that released 185 MeV of energy as well as two neutrons which could cause similar reactions on surrounding nuclei. The fact that instead of one initial neutron, in the reaction (19.4) we obtain two neutrons, is crucial. This gives us the possibility to make the so-called chain reaction schematically shown in Fig. 19.4 Figure 19.4: Chain reaction on uranium nuclei. In such process, one neutron breaks one heavy nucleus, the two released neutrons break two more heavy nuclei and produce four neutrons which, in turn, can break another four nuclei and so on. This process develops extremely fast. In a split of a second a huge amount of energy can be released, which means explosion. In fact, this is how the so-called atomic bomb works. Can we control the development of the chain reaction? Yes we can! This is done in nuclear reactors that produce energy for our use. How can it be done? Critical mass First of all, if the piece of material containing \ufb02ssile nuclei is too small, some neutrons may reach its surface and escape without causing further \ufb02ssions. For each type of \ufb02ssile material there is therefore a minimal mass of a sample that can support explosive chain reaction. It is called the critical mass. For example, the critical mass of 235 92U is approximately 50 kg. If the mass is 361 below the critical value, nuclear explosion is not possible, but the energy is still released and the sample becomes hot. The closer mass is to its critical value, the more energy is released and more intensive is the neutron radiation from the sample. The criticality of a sample (i.e. its closeness to the critical state) can be reduced by changing its geometry (making its surface bigger) or by putting inside it some other material (boron or cadmium) that is able to absorb neutrons. On the other hand, the criticality can be increased by putting neutron reectors around the sample. These reectors work like mirrors from which the escaped neutrons bounce back into the sample. Thus, moving in and out the absorbing material and reectors, we can keep the sample close to the critical state. How a nuclear reactor works In a typical nuclear reactor, the fuel is not in one piece, but in the form of several hundred vertical rods, like a brush. Another system of rods that contain a neutron absorbing material (control rods) can move up and down in between the fuel rods. When totally in, the control rods absorb so many neutrons, that the reactor is shut down. To start the reactor, operator gradually moves the control rods up. In an emergency situation they are dropped down automatically. To collect", " the energy, water ows through the reactor core. It becomes extremely hot and goes to a steam generator. There, the heat passes to water in a secondary circuit that becomes steam for use outside the reactor enclosure for rotating turbines that generate electricity. Nuclear power in South Africa By 2004 South Africa had only one commercial nuclear reactor supplying power into the national grid. It works in Koeberg located 30 km north of Cape Town. A small research reactor was also operated at Pelindaba as part of the nuclear weapons program, but was dismantled. Koeberg Nuclear Power station is a uranium Pressurized Water Reactor (PWR). In such a reactor, the primary coolant loop is pressurised so the water does not boil, and heat exchangers, called steam generators, are used to transmit heat to a secondary coolant which is allowed to boil to produce steam. To remove as much heat as possible, the water temperature in the primary loop is allowed to rise up to about 300 \u2013C which requires the pressure of 150 atmospheres (to keep water from boiling). The Koeberg power station has the largest turbine generators in the southern hemisphere and produces 10000 MWh of electric energy. Construction of Koeberg began in 1976 and two of its Units were commissioned in 1984-1985. Since then, the plant has been in more or less continuous operation and there have been no serious incidents. \u00bb Eskom that operates this power station, may be the current technology leader. It is developing a new type of nuclear reactor, a modular pebble-bed reactor (PBMR). In contrast to traditional nuclear reactors, in this new type of reactors the fuel is not assembled in the form of rods. The uranium, thorium or plutonium fuels are in oxides (ceramic form) contained within spherical pebbles made of pyrolitic graphite. The pebbles, having a size of a tennis ball, are in a bin or can. An inert gas, helium, nitrogen or carbon dioxide, circulates through the spaces between the fuel pebbles. This carries heat away from the reactor. 362 Ideally, the heated gas is run directly through a turbine. However since the gas from the primary coolant can be made radioactive by the neutrons in the reactor, usually it is brought to a heat exchanger, where it heats another gas, or steam. The primary advantage of pebble-bed reactors is that they can be designed to be inherently safe. When a pebble-bed reactor gets hotter, the more rapid motion of the atoms in the fuel increases the probability of neutron capture by 238 92U isotopes through an e\ufb01ect known as Doppler broadening. This isotope does not split up after capturing a neutron. This reduces the number of neutrons available to cause 235 92U \ufb02ssion, reducing the power output by the reactor. This natural negative feedback places an inherent upper limit on the temperature of the fuel without any operator intervention. The reactor is cooled by an inert, \ufb02reproof gas, so it cannot have a steam explosion as a water reactor can. A pebble-bed reactor thus can have all of its supporting machinery fail, and the reactor will not crack, melt, explode or spew hazardous wastes. It simply goes up to a designed \"idle\" temperature, and stays there. In that state, the reactor vessel radiates heat, but the vessel and fuel spheres remain intact and undamaged. The machinery can be repaired or the fuel can be removed. A large advantage of the pebble bed reactor over a conventional water reactor is that they operate at higher temperatures. The reactor can directly heat uids for low pressure gas turbines. The high temperatures permit systems to get more mechanical energy from the same amount of thermal energy. Another advantage is that fuel pebbles for di\ufb01erent fuels might be used in the same basic design of reactor (though perhaps not at the same time). Proponents claim that some kinds of pebble-bed reactors should be able to use thorium, plutonium and natural unenriched Uranium, as well as the customary enriched uranium. One of the projects in progress is to develop pebbles and reactors that use the plutonium from surplus or expired nuclear explosives. On June 25, 2003, the South African Republic\u2019s Department of Environmental A\ufb01airs and Tourism approved ESKOM\u2019s prototype 110 MW pebble-bed modular reactor for Koeberg. Eskom also has approval for a pebble-bed fuel production plant in Pelindaba. The uranium for this fuel is to be imported from Russia. If the trial is successful, Eskom says it will build up to ten local PBMR plants on South Africa\u2019s seacoast. Eskom also wants to export up to 20 PBMR plants per year. The estimated export revenue is 8 billion rand a year, and could employ about 57000 people. 19.8.2 Fusion energy For a", " given mass of fuel, a fusion reaction like 1H + 3 2 1H \u00a1! 4 2He + n + 17:59 MeV : (19.5) yield several times more energy than a \ufb02ssion reaction. This is clear from the curve given in Fig. 19.3. Indeed, a change of the binding energy (per nucleon) is much more signi\ufb02cant for a fusion reaction than for a \ufb02ssion reaction. Fusion is, therefore, a much more powerful source of energy. For example, 10 g of Deuterium which can be extracted from 500 litres of water and 15 g of Tritium produced from 30 g of Lithium would give enough fuel for the lifetime electricity 363 needs of an average person in an industrialised country. But this is not the only reason why fusion attracted so much attention from physicists. Another, more fundamental, reason is that the fusion reactions were responsible for the synthesis of the initial amount of light elements at primordial times when the universe was created. Furthermore, the synthesis of nuclei continues inside the stars where the fusion reactions produce all the energy which reaches us in the form of light. Thermonuclear reactions If fusion is so advantageous, why is it not used instead of \ufb02ssion reactors? The problem is in the electric repulsion of the nuclei. Before the nuclei on the left hand side of Eq. (19.5) can fuse, 10\u00a113 cm. This is not an we have to bring them somehow close to each other to a distance of easy task! They both are positively charged and \\refuse\" to approach each other. \u00bb What we can do is to make a mixture of the atoms containing such nuclei and heat it up. At high temperatures the atoms move very fast. They \ufb02ercely collide and loose all the electrons. The mixture becomes plasma, i.e. a mixture of bare nuclei and free moving electrons. If the temperature is high enough, the colliding nuclei can overcome the electric repulsion and approach each other to a fusion distance. When the nuclei fuse, they release much more energy than was spent to heat up the plasma. Thus the initial energy \\investment\" pays o\ufb01. The typical temperature needed to ignite the reaction of the type (19.5) is extremely high. In fact, it is the same temperature that our sun has in its center, namely, 15 million degrees. This is why the reactions (19.3), (19.5), and the like are called thermonuclear reactions. \u00bb Human-made thermonuclear reactions The same as with \ufb02ssion reactions, the \ufb02rst application of thermonuclear reactions was in weapons, namely, in the hydrogen bomb, where fusion is ignited by the explosion of an ordinary (\ufb02ssion) plutonium bomb which heats up the fuel to solar temperatures. In an attempt to make a controllable fusion, people encounter the problem of holding the plasma. It is relatively easy to achieve a high temperature (with laser pulses, for example). But as soon as plasma touches the walls of the container, it immediately cools down. To keep it from touching the walls, various ingenious methods are tried, such as strong magnetic \ufb02eld and laser beams directed to plasma from all sides. In spite of all e\ufb01orts and ingenious tricks, all such attempts till now have failed. Most probably this straightforward approach to controllable fusion is doomed because one has to hold in hands a \\piece of burning sun\". Cold fusion To visualize the struggle of the nuclei approaching each other, imagine yourself pushing a metallic ball towards the top of a slope shown in Fig. 19.5. The more kinetic energy you give to the ball, the higher it can climb. Your purpose is to make it fall into the narrow well that is behind the barrier. 364 Coulomb barrier Ve\ufb01 \u00a1 \u00a1\u201c projectile x R Figure 19.5: E\ufb01ective nucleus{nucleus potential as a function of the separation between the nuclei. In fact, the curve in Fig. 19.5 shows the dependence of relative potential energy Ve\ufb01 between two nuclei on the distance R separating them. The deep narrow well corresponds to the strong short-range attraction, and the 1=R barrier represents the Coulomb (electric) repulsion. The nuclei need to overcome this barrier in order to \\touch\" each other and fuse, i.e. to fall into the narrow and deep potential well. One way to achieve this is to give them enough kinetic energy, which means to rise the temperature. However, there is another way based on the quantum laws. \u00bb As you remember, when discussing the motion of the electron inside an atom (see Sec. 19.1), we said that it formed a \\cloud\" of probability around the nucleus. The density of this cloud dimin", "ishes at very short and very long distances but never disappears completely. This means that we can \ufb02nd the electron even inside the nucleus though with a rather small probability. The nuclei moving towards each other, being microscopic objects, obey the quantum laws as well. The probability density for \ufb02nding one nucleus at a distance R from another one also forms a cloud. This density is non-zero even under the barrier and on the other side of the barrier. This means that, in contrast to classical objects, quantum particles, like nuclei, can penetrate through potential barriers even if they do not have enough energy to go over it! This is called the tunneling e\ufb01ect. The tunneling probability strongly depends on thickness of the barrier. Therefore, instead of lifting the nuclei against the barrier (which means rising the temperature), we can try to make the barrier itself thinner or to keep them close to the barrier for such a long time that even a low penetration probability would be realized. How can this be done? The idea is to put the nuclei we want to fuse, inside a molecule where they can stay close to each other for a long time. Furthermore, in a molecule, the Coulomb barrier becomes thinner because of electron screening. In this way fusion may proceed even at 365 room temperature. This idea of cold fusion was originally (in 1947) discussed by F. C. Frank and (in 1948) put forward by A. D. Sakharov, the \\father\" of Russian hydrogen bomb, who at the latest stages of his career was worldwide known as a prominent human rights activist and a winner of the Nobel Prize for Peace. When working on the bomb project, he initiated research into peaceful applications of nuclear energy and suggested the fusion of two hydrogen isotopes via the reaction (19.5) by forming a molecule of them where one of the electrons is replaced by a muon. The muon is an elementary particle (see Sec. 19.9), which has the same characteristics as an electron. The only di\ufb01erence between them is that the muon is 200 times heavier than the electron. In other words, a muon is a heavy electron. What will happen if we make a muonic atom of hydrogen, that is a bound state of a proton and a muon? Due to its large mass the muon would be very close to the proton and the size of such atom would be 200 times smaller than that of an ordinary atom. This is clearly seen from the formula for the atomic Bohr radius where the mass is in the denominator. RBohr = ~2 me2 ; Now, what happens if we make a muonic molecule? It will also be 200 times smaller than an ordinary molecule. The Coulomb barrier will be 200 times thinner and the nuclei 200 times closer to each other. This is just what we need! Speaking in terms of the e\ufb01ective nucleus{nucleus potential shown in Fig. 19.5, we can say that the muon modi\ufb02es this potential in such a way that a second minimum appears. Such a modi\ufb02ed potential is (schematically) shown in Fig. 19.6. Ve\ufb01 probability density \u00a1 \u00a1\u201c x R Figure 19.6: E\ufb01ective nucleus{nucleus potential (thick curve) for nuclei con\ufb02ned in a molecule. Thin curve shows the corresponding distribution of the probability for \ufb02nding the nuclei at a given distance from each other. The molecule is a bound state in the shallow but wide minimum of this potential. Most of 366 the time, the nuclei are at the distance corresponding to the maximum of the probability density distribution (shown by the thin curve). Observe that this density is not zero under the barrier (though is rather small) and even at R = 0. This means that the system can (with a small probability) jump from the shallow well into the deep well through the barrier, i.e. can tunnel and fuse. Unfortunately, the muon is not a stable particle. Its lifetime is only 10\u00a16 sec. This means that a muonic molecule cannot exist longer than 1 microsecond. As a matter of fact, from a quantum mechanical point of view, this is quite a long interval. \u00bb The quantum mechanical wave function (that describes the probability density) oscillates with a frequency which is proportional to the energy of the system. With a typical binding 1017 s\u00a11. This means that the particle energy of a muonic molecule of 300 eV this frequency is hits the barrier with this frequency and during 1 microsecond it makes 1011 attempts to jump 10\u00a17. Therefore, during through it. The calculations show that the penetration probability is 1 microsecond nuclei can penetrate through the barrier 10000 times and fusion can happen much faster than the decay rate of the muon. \u00bb \u00bb Cold fusion via the", " formation of muonic molecules was done in many laboratories, but unfortunately, it cannot solve the problem of energy production for our needs. The obstacle is the negative e\u2013ciency, i.e. to make muonic cold fusion we have to spend more energy than it produces. The reason is that muons do not exist like protons or electrons. We have to produce them in accelerators. This takes a lot of energy. Actually, the muon serves as a catalyst for the fusion reaction. After helping one pair of nuclei to fuse, the muon is liberated from the molecule and can form another molecule, and so on. It was estimated that the e\u2013ciency of the energy production would be positive only if each muon ignited at least 1000 fusion events. Experimentalists tried their best, but by now the record number is only 150 fusion events per muon. This is too few. The main reason why the muon does not catalyze more reactions is that it is eventually trapped by a 4He nucleus which is a by-product of fusion. Helium captures the muon into an atomic orbit with large binding energy, and it cannot escape. Nonetheless, the research in the \ufb02eld of cold fusion continues. There are some other ideas of how to keep nuclei close to each other. One of them is to put the nuclei inside a crystal. Another way out is to increase the penetration probability by using molecules with special properties, namely, those that have quantum states with almost the same energies as the excited states on the compound nucleus. Scientists try all possibilities since the energy demands of mankind grow continuously and therefore the stakes in this quest are high. 19.9 Elementary particles In our quest for the elementary building blocks of the universe, we delved inside atomic nucleus and found that it is composed of protons and neutrons. Are the three particles, e, p, and n, the blocks we are looking for? The answer is \\no\". Even before the structure of the atom was understood, Becquerel discovered the redioactivity (see Sec. 19.5.1) that afterwards puzzled physicists and forced them to look deeper, i.e. inside protons and neutrons. 367 19.9.1 \ufb02 decay Among the three types of radioactivity, the \ufb01 and rays were easily explained. The emission of \ufb01 particle is kind of \ufb02ssion reaction, when an initial nucleus spontaneously decays in two fragments one of which is the nucleus 4 2He (i.e. \ufb01 particle). The rays are just electromagnetic quanta emitted by a nuclear system when it transits from one quantum state to another (the same like an atom emits light). The \ufb02 rays posed the puzzle. On the one hand, they are just electrons and you may think that it looks simple. But on the other hand, they are not the electrons from the atomic shell. It was found that they come from inside the nucleus! After the \ufb02-decay, the charge of the nucleus increases in one unit, A Z (parent nucleus) \u00a1! A Z+1 (daughter nucleus) + e ; which is in accordance with the charge conservation law. There was another puzzle associated with the \ufb02 decay: The emitted electrons did not have a certain energy. Measuring their kinetic energies, you could \ufb02nd very fast and very slow electrons as well as the electrons with all intermediate speeds. How could identical parent nuclei, after loosing di\ufb01erent amount of energy, become identical daughter nuclei. May be energy is not conserving in the quantum world? The fact was so astonishing that even Niels Bohr put forward the idea of statistical nature of the energy conservation law. To explain the \ufb02rst puzzle, it was naively suggested that neutron is a bound state of proton and electron. At that time, physicists believed that if something is emitted from an object, this something must be present inside that object before the emission. They could not imagine that a particle could be created from vacuum. The naive (pe) model of the neutron contradicted the facts. Indeed, it was known already that the pe bound state is the hydrogen atom. Neutron is much smaller than the atom. Therefore, it would be unusually tight binding, and perhaps with something elese involved that keeps the size small. By the way, this \\something elese\" could also save the energy conservation law. In 1930, Wolfgang Pauli suggested that in addition to the electron, the \ufb02 decay involves another particle, \u201d, that is emitted along with the electron and carries away part of the energy. For example, 234 90Th \u00a1! 234 91Pa + e\u00a1 + \u201e\u201d : (19.6) This additional particle was called neutrino (in Italian the word \\neutrino\" means small neutron). The neutrino is electrically neutral, has extremely small mass", " (maybe even zero, which is still a question in 2004) and very weakly interacts with matter. This is why it was not detected experimentally till 1956. The \\bar\" over \u201d in Eq. (19.6) means that in this reaction actually the anti-neutrino is emitted (see the discussion on anti-particles further down in Sec. 19.9.2). 19.9.2 Particle physics In an attempt to explain the \ufb02 decay and to understand internal structure of the neutron a new branch of physics was born, the particle physics. The only way to explore the structure of subatomic particles is to strike them with other particles in order to knock out their \\constituent\" parts. The simple logic says: The more powerful the impact, the smaller parts can be knocked 368 out. At the beginning the only source of energetic particles to strike other particles were the cosmic rays. Earth is constantly bombarded by all sort of particles coming from the outer space. Atmosphere protects us from most of them, but many still reach the ground. Antiparticles In 1932, studying the cosmic rays with a bubble chamber, Carl Anderson made a photograph of two symmetrical tracks of charged particles. The measurements of the track curvatures showed that one track belonged to an electron and the other was made by a particle having the same mass and equal but positive charge. These particles were created when a cosmic quantum of a high energy collided with a nucleus. The discovered particle was called positron and denoted as e+ to distinguish it from the electron, which sometimes is denoted as e\u00a1. It was the \ufb02rst antiparticle discovered. Later, it was found that every particle has its \\mirror reection\", the antiparticle. To denote an antiparticle, it is used \\bar\" over a particle symbol. For example, \u201ep is the anti-proton, which has the same mass as an ordinary proton but a negative charge. When a particle collides with its \\mirror reection\", they annihilate, i.e. they burn out completely. In this collision, all their mass is transformed into electromagnetic energy in the form of quanta. For example, if an electron collides with a positron, the following reaction may take place e\u00a1 + e+ + ; \u00a1! (19.7) where two photons are needed to conserve the total momentum of the system. In principle, stable antimatter can exist. For example, the pair of \u201ep and e+ can form an atom of anti-hydrogen with exactly the same energy states as the ordinary hydrogen. Experimentally, atoms of anti-helium were obtained. The problem with them is that, surrounded by ordinary matter, they cannot live long. Colliding with ordinary atoms, they annihilate very fast. There are speculations that our universe should be symmetric with respect to particles and antiparticles. Indeed, why should preference be given to matter and not to anti-matter? This implies that somewhere very far, there must be equal amount of anti-matter, i.e. anti-universe. Can you imagine what happens if they meet? Muon, mesons, and the others In yet another cosmic-ray experiment a particle having the same properties as the electron but 207 times heavier, was discovered in 1935. It was given the name muon and the symbol \u201e. For \u00bb a long time it remained \\unnecessary\" particle in the picture of the world. Only the modern theories harmonically included the muon as a constituent part of matter (see Sec 19.9.3). The same inexhaustible cosmic rays revealed the \u2026 and K mesons in 1947. The \u2026 mesons (or simply pions) were theoretically predicted twelve years before by Yukawa, as the mediators of the strong forces between nucleons. The K mesons, however, were unexpected. Furthermore, they showed very strange behaviour. They were easily created only in pairs. The probability 369 of the inverse process (i.e. their decay) was 1013 times lower than the probability of their creation. It was suggested that these particles possess a new type of charge, the strangeness, which is conserving in the strong interactions. When a pair of such particles is created, one of them has strangeness +1 and the other 1, so the total strangeness remains zero. When decaying, they act individually and therefore the strangeness is not conserving. According to the suggestion, this is only possible through the weak interactions that are much weaker than the strong interactions (see Sec. 19.9.4) and thus the decay probability is much lower. \u00a1 The golden age of particle physics began in 1950-s with the advent of particle accelerators, the machines that produced beams of electrons or protons with high kinetic energy. Having such beams available, experimentalists can plan the experiment and repeat it, while with the cosmic rays they were at the mercy of", " chance. When the accelerators became the main tool of exploration, the particle physics acquired its second name, the high energy physics. During the last half a century, experimentalists discovered so many new particles (few of them are listed in Table 19.3) that it became obvious that they cannot all be elementary. When colliding with each other, they produce some other particles. Mutual transformations of the particles is their main property. family photon leptons hadrons Lifetime T1=2 (s) stable stable 2:2 10\u00a16 \u00a3 10\u00a113 stable stable stable 2:6 0:8 1:2 0:9 5:2 10\u00a18 10\u00a116 10\u00a18 10\u00a110 10\u00a18 \u00a3 \u00a3 \u00a3 \u00a3 \u00a3 10\u00a118 stable 900 particle photon electron muon tau electron neutrino muon neutrino tau neutrino pion pion kaon kaon kaon eta meson proton neutron lambda sigma sigma sigma omega symbol e\u00a1, e+ \u201e\u00a1, \u201e+ \u00bf \u00a1, \u00bf + \u201de \u201d\u201e \u201d\u00bf \u2026+, \u2026\u00a1 \u20260 K +, K \u00a1 K 0 S K 0 L \u00b70 p n \u20440 \u00a7+ \u00a70 \u00a7\u00a1 \u203a\u00a1, \u203a+ mass (MeV) 0 0.511 105.7 1777 0 \u00bb 0 \u00bb 0 \u00bb 139.6 135.0 493.7 497.7 497.7 548.8 938.3 939.6 1116 1189 1192 1197 1672 \u00a3 \u00a3 \u00a3 \u00a3 \u00a3 Table 19.3: Few representatives of di\ufb01erent particle families. 2:6 0:8 6 1:5 0:8 10\u00a110 10\u00a110 10\u00a120 10\u00a110 10\u00a110 Physicists faced the problem of particle classi\ufb02cation similar to the problems of classi\ufb02cation of animals, plants, and chemical elements. The \ufb02rst approach was very simple. The particles were leptons (light particles, like electron), mesons divided in four groups according to their mass: (intermediate mass, like pion), baryons (heavy particles, like proton or neutron), and hyperons 370 (very heavy particles). Then it was realized that it would be more logical to divide the particles in three families according to their ability to interact via weak, electromagnetic, and strong forces (in addition to that, all particles experience gravitational attraction towards each other). Except for the gravitational interaction, the photon ( quantum) participates only in electromagnetic interactions, the leptons take part in both weak and electromagnetic interactions, and hadrons are able to interact via all forces of nature (see Sec. 19.9.4). In addition to conservation of the strangeness, several other conservation laws were discovered. For example, number of leptons is conserving. This is why in the reaction (19.6) we have an electron (lepton number +1) and anti-neutrino (lepton number 1) in the \ufb02nal state. Similarly, the number of baryons is conserving in all reactions. \u00a1 The quest for the constituent parts of the neutron has led us to something unexpected. We found that there are several hundreds of di\ufb01erent particles that can be \\knocked out\" of the neutron but none of them are its parts. Actually, the neutron itself can be knocked out of some of them! What a mess! Further e\ufb01orts of experimentalists could not \ufb02nd an order, which was \ufb02nally discovered by theoreticians who introduced the notion of quarks. 19.9.3 Quarks and leptons While experimentalists seemed to be lost in the maze, the theoreticians groped for the way out. Using an extremely complicated mathematical technique, they managed to group the hadrons in such families which implied that all known (and yet unknown) hadrons are build of only six types of particles with fractional charges. The main credit for this (in the form of Nobel Prize) was given to M. Gell-Mann and G. Zweig. At \ufb02rst, they considered a subset of the hadrons and developed a theory with only three types of such truly elementary particles. When Murray Gell-Mann thought of the name for them, he came across the book \"Finnegan\u2019s Wake\" by James Joyce. The line \"Three quarks for Mister Mark...\" appeared in that fanciful book (in German, the word \\quark\" means cottage cheese). He needed a name for three particles and this was the answer. Thus the term quark was coined. Later, the theory was generalized to include all known particles, which required six types of", " quarks. Modern theories require also that the number of di\ufb01erent leptons should be the same as the number of di\ufb01erent quark types. According to these theories, the quarks and leptons are truly elementary, i.e. they do not have any internal structure and therefore are of a zero size (pointlike). Thus, the world is constructed of just twelve types of elementary building blocks that are given in Table 19.4. Amazingly enough, the electron that was discovered before all other particles, more than a century ago, turned out to be one of them! After Gell-Mann, who used a funny name (quark) for an elementary particle, the fundamental physics was ooded with such names. For example, the six quark types are called avors (for cottage cheese, this is appropriate indeed), the three di\ufb01erent states in which each quark can be, are called colors (red, green, blue), etc. Modern physics is so complicated and mathematical, that people working in it, need such kind of jokes to \\spice unsavoury dish with avors\". The funny names should not confuse anybody. Elementary particles do not have any smell, taste, or colour. These terms simply denote certain properties (similar to electric charge) that do not 371 family elementary particle symbol charge leptons quarks electron muon tau electron neutrino muon neutrino tau neutrino up down strange charmed top (truth) bottom (beauty) e\u00a1 \u201e\u00a1 \u00bf \u00a1 \u201de \u201d\u201e \u201d\u00bf 2=3 1=3 \u00a1 1=3 \u00a1 +2=3 +2=3 1=3 \u00a1 lepton number 1 baryon number /3 1/3 1/3 1/3 1/3 1/3 mass (MeV) 0.511 105.7 1777 \u00bb \u00bb 0 0 0 \u00bb 360 360 1500 540 174000 5000 Table 19.4: Elementary building blocks of the universe. exist in human world. Hadrons There are particles that are able to interact with each other by the so-called strong forces. An10\u00a115 m), other name for these forces is nuclear forces. They are very strong at short distances ( and very quickly vanish when the distance between the particles increases. All these particles are called hadrons. The protons and neutrons are examples of hadrons. \u00bb As you remember, we learned about the existence of huge variety of particles when trying to look inside a nucleon, more particularly, the neutron. So, what the neutron is made of? Can we get the answer at last, after learning about the quarks? Yes, we can. According to modern theories, all hadrons are composed of quarks. The quarks can be combined in groups of two or three. The bound states of two quarks are called mesons, and the bound complexes of three quarks are called baryons. No other numbers of quarks can form observable particles1. Nucleons are baryons and therefore consist of three quarks while the pion is a meson containing only two quarks, as schematically shown in Fig. 19.7. Comparing this \ufb02gure with Table 19.4, you can see why quarks have fractional charges. Counting the total charge of a hadron, you should not forget that anti-quarks have the opposite charges. The baryon number for an anti-quark also has the opposite sign (negative). This is why mesons actually consist of a quark and anti-quark in order to have total baryon number zero. 1Recently, experimentalists and theoreticians started to actively discuss the possibility of the existence of pentaquarks, exotic particles that are bound complexes of \ufb02ve quarks. 372 \u201d\u2022u \u201d\u2022u \u201e\u201a \u201e\u201a \u201d\u2022d \u201e\u201a proton \u201d\u2022d \u201e\u201a \u201d\u2022d \u201e\u201a \u201d\u2022u \u201d\u2022u \u201d\u2022\u201ed \u201e\u201a \u201e\u201a \u201e\u201a neutron \u2026+ meson Figure 19.7: Quark content of the proton, neutron, and \u2026+-meson. Particle reactions At the early stages of the particle physics development, in order to \ufb02nd the constituent parts of various particles, experimentalists simply collided them and watched the \\fragments\". However, this straightforward approach led to confusion. For example, the reaction between the \u2026 \u00a1 meson and proton, (19.8) would suggest (if naively interpreted) that either K 0 or \u20440 is a constituent part of the nucleon while the pion is incorporated into the other \\fragment\". On the other hand, the same collision can knock out di\ufffd", "\ufffderent \\fragments\" from the same proton. For example, \u00a1! \u2026\u00a1 + p K 0 + \u20440 ; which leads to an absurd suggestion that neutron is a constituent part of proton. \u2026\u00a1 + p \u00a1! \u20260 + n ; (19.9) The quark model explains all such \\puzzles\" nicely and logically. Similarly to chemical reactions that are just rearrangements of atoms, the particle reactions of the type (19.8) and (19.9) are just rearrangements of the quarks. The only di\ufb01erence is that, in contrast to chemistry where the number of atoms is not changing, the number of quarks before the collision is not necessarily equal to their number after the collision. This is because a quark from one colliding particle can annihilate with the corresponding antiquark from another particle. Moreover, if the collision is su\u2013ciently powerful, the quark-antiquark pairs can be created from vacuum. It is convenient to depict the particle transformations in the form of the so-called quark ow diagrams. On such diagrams, the quarks are represented by lines that may be visualized as the trajectories showing their movement from the left to the right. For example, the diagram given in Fig. (19.8), shows the quark rearrangement for the reaction (19.8). As you can see, when the pion collides with proton, its \u201eu quark annihilates with the u quark from the proton. At the same time, the s\u201es pair is created from the vacuum. Then, the \u201es quark binds with the d quark to form the strange meson K 0, while the s quark goes together with the ud pair as the strange baryon \u20440. The charge-exchange reaction (19.9) is a more simple rearrangement process shown in Fig. 19.9. You may wonder why the quark and antiquark of the same avor in the \u20260 meson do not annihilate. Yes they do, but not immediately. And due to this annihilation, the lifetime of \u2026 0 is 100 million times shorter than the lifetime of \u2026\u00a7 (see Table 19.3). 373 \u2026\u00a1 \u2030 p ( d \u201eu u u d d \u201es s u d K 0 \u20440 ) Figure 19.8: Quark-ow diagram for the reaction \u2026\u00a1 + p K 0 + \u20440. \u00a1! \u2026\u00a1 \u2030 p ( \u201eu d u u d \u201eu u d u d \u20260 n ) Figure 19.9: Quark-ow diagram for the reaction \u2026\u00a1 + p \u20260 + n. \u00a1! Despite its simplicity, the quark-ow diagram technique is very powerful method not only for explaining the observed reactions but also for predicting new reactions that have not yet been seen in experiments. Knowing the quark content of particles (which is available in modern Physics Handbooks), you can draw plenty of such diagrams that will describe possible particle transformations. The only rule is to keep the lines continuous. They can disappear or emerge only for a quark-antiquark pair of the same avor. However, the continuity of the quark lines is valid only for the processes caused by the strong interaction. Indeed, the \ufb02-decay of a free neutron (caused by the weak forces), as well as the \ufb02-decay of the nuclei, indicate that quarks can change avor. In particular, the \ufb02-decay (19.10) or (19.6) happens because the d quark transformes into the u quark, n \u00a1! p + e\u00a1 + \u201e\u201de ; (19.10) d \u00a1! u + e\u00a1 + \u201e\u201de ; (19.11) due to the weak interaction, as shown in Fig. 19.10 Quark con\ufb02nement At this point, it is very logical to ask if anybody observed an isolated quark. The answer is \\no\". Why? And how can one be so con\ufb02dent of the quark model when no one has ever seen an 374 \u201e\u201de e Figure 19.10: Quark-ow diagram for the \ufb02 decay of neutron. isolated quark? Basically, you can\u2019t see an isolated quark because the quark-quark attractive force does not let them go. In contrast to all other systems, the attraction between quarks grows with the distance separating them. It is like a rubber cord connecting two balls. When the balls are close to each other, the cord is not stretched and the balls do not feel any force. If, however, you try to separate the balls, the", " cord pulls them back. The more you stretch the cord, the stronger the force becomes (according to the Hook\u2019s law of elasticity). Of course, a real rubber cord would eventually break. This does not happen with the quark-quark force. It can grow to in\ufb02nity. This phenomenon is called the con\ufb02nement of quarks. Nonetheless, we are sure that the nucleon consists of three quarks having fractional charges. A hundred years ago Rutherford, by observing the scattering of charged particles from an atom, proved that its positive charge is concentrated in a small nucleus. Nowadays, similar experiments prove the existence of fractional point-like charges inside the nucleon. The quark model actually is much more complicated than the quark-ow diagrams. It is a consistent mathematical theory that explains a vast variety of experimental data. This is why nobody doubts that it reects the reality. 19.9.4 Forces of nature If asked how many types of forces exist, many people start counting on their \ufb02ngers, and when the count exceeds ten, they answer \\plenty of\". Indeed, there are gravitational forces, electrical, magnetic, elastic, frictional forces, and also forces of wind, of expanding steam, of contracting muscles, etc. If, however, we analyze the root causes of all these forces, we can reduce their number to just a few fundamental forces (or fundamental interactions, as physicists say). For example, the elastic force of a stretched rubber cord is due to the attraction between the molecules that the rubber is made of. Looking deeper, we \ufb02nd that the molecules attract each other because of the electromagnetic attraction between the electrons of one molecule and nuclei of the other. Similarly, if we depress a piece of rubber, it resists because the molecules refuse to approach each other too close due to the electric repulsion of the nuclei. Therefore the elasticity 375 of rubber has the electromagnetic origin. Any other force in the human world can be analyzed in the same manner. After doing this, we will \ufb02nd that all forces that we see around us (in the macroworld), are either of gravitational or electromagnetic nature. As we also know, in the microworld there are two other types of forces: The strong (nuclear) forces that act between all hadrons, and the weak forces that are responsible for changing the quark avors. Therefore, all interactions in the Universe are governed by only four fundamental forces: Strong, electromagnetic, weak and gravitational. These forces are very di\ufb01erent in strength and range. Their relative strengths are given in Table 19.5. The most strong is the nuclear interaction. The strength of the electromagnetic forces is one hundred times lower. The weak forces are nine orders of magnitude weaker than the nuclear forces, and the gravity is 38 orders of magnitude weaker! It is amazing that this subtle interaction governs the cosmic processes. The reason is that the gravitational forces are of long range and always attractive. There is no such thing as negative mass that would screen the gravitational \ufb02eld, like negative electrons screen the \ufb02eld of positive nuclei. Force Relative Strength Range Strong 1 Electromagnetic 0.0073 Short Long Weak Gravitational 10\u00a19 10\u00a138 Very Short Long Table 19.5: Four fundamental forces and their relative strengths. Towards the uni\ufb02ed force Physicists always try to simplify things. Since there are only four fundamental forces, it is tempting to ask \"If only four, then why not only one?\". Can it be that all interactions are just di\ufb01erent faces of one master force? The \ufb02rst who started the quest for uni\ufb02cation of forces was Einstein. After completing his general theory of relativity, he spent 30 years in unsuccessful attempts to unify the electromagnetic and gravity forces. At that time, it seemed logical because both of them were in\ufb02nite in range and obeyed the same inverse square law. Einstein failed because the uni\ufb02cation should be done on the basis of quantum laws, but he tried to do it using the classical concepts. Electro-weak uni\ufb02cation Now it is known that despite the similarities in form of the gravity and electromagnetic forces, the gravity will be the last to yield to uni\ufb02cation. The more implausible uni\ufb02cation of the electro- 376 magnetic and weak forces turned out to be the \ufb02rst successful step towards the uni\ufb02ed interaction. In 1979, the Nobel prize was awarded to Weinberg, Salam, and Glashow, who developed a uni\ufb02ed theory of electromagnetic and weak interactions. According to that theory, the electromagnetic and weak forces converge to one electro-weak interaction at very high collision energies. The theory also predicted the existence of heavy particles, the W and Z, with masses around", " 80000 MeV and 90000 MeV, respectively. These particles were discovered in 1983, which brought experimental veri\ufb02cation to the new theory. Grand uni\ufb02cation The next step was to try to combine the electro-weak theory with the theory of the strong interactions (i.e. quark theory) in a single theory. This work was called the grand uni\ufb02cation. Currently, physicists discuss versions of such theory that predicts the convergence of the three forces at aw1017 MeV. The quarks and leptons in this theory, are the uni\ufb02ed leptoquarks. fully high energies \u00bb The grand uni\ufb02cation is not that successful as the electro-weak theory. It has the problem of mathematical consistency and contradicts to at least one experiment. The matter is that it predicts the proton decay, that does not conserve both the baryon and lepton numbers, with the lifetime of The measurements show, however, that the lifetime of the proton is at least 1032 years. \u00bb 1029 years. p \u00a1! e+ + \u20260 ; Theory of everything Some people believe that the grand uni\ufb02cation has an inherent principal aw. According to them, one cannot unify the forces step by step (leaving the gravity out), and the correct way is to combine all four forces in the so-called theory of everything. There are few di\ufb01erent approaches to unifying everything. One of them suggests that all fundamental particles (quarks and leptons) are just vibrating modes of string loops in multidimensional space. The electron is a string vibrating one way, the up-quark is a string vibrating another way, and so on. The other approach introduces a new level of fundamental particles, the preons, that could be constituent parts of quarks and leptons. The quest goes on. Everyone agrees that constructing the theory of everything would in no way mean that biology, geology, chemistry, or even physics had been solved. The universe is so rich and complex that the discovery of the fundamental theory would not mean the end of science. The ultimate theory of everything would provide an unshakable pillar of coherence forever assuring us that the universe is a comprehensible place. 19.10 Origin of the universe Looking deep inside microscopic particles, physicists need to collide them with high kinetic energies. The smaller parts of matter they want to observe, the higher energy they need. This is why they build more and more powerful accelerators. However, the accelerators have natural limitations. Indeed, an accelerator cannot be bigger than the size of our planet. And even if we manage to build a circular accelerator around the whole earth (along the equator, for example), 377 it would not be able to reach the energy of mental interactions takes place. \u00bb 1017 MeV at which the grand uni\ufb02cation of funda- So, what are we to do? How can we test the theory of everything? Is it possible at all? 1017 MeV, should be looked for in the cosmos, Yes, it is! The astronomically high values, like \u00bb of course. Our journey towards extremely small objects eventually leads us to extremely large objects, like whole universe. Equations of Einstein\u2019s theory of relativity can describe the evolution of the universe. Physicists solved these equations back in time and found that the universe had its beginning. Approximately 15 billion years ago, it started from a zero size point that exploded and rapidly expanded to the present tremendous scale. At the \ufb02rst instants after the explosion, the matter was at such incredibly high density and temperature that all particles had kinetic energies even higher than 1017 MeV. This means that at the very beginning there was only one the uni\ufb02cation energy sigle force and no di\ufb01erence among fundamental particles. Everything was uni\ufb02ed and \\simple\". \u00bb You may ask \\So what? How can so distant past help us?\". In many ways! The development of the universe was governed by the fundamental forces. If our theories about them are correct, we should be able to reproduce (with calculations) how that development proceeded step by step. During the expansion, all the nuclei and atoms in the cosmos were created. The amounts of di\ufb01erent nuclei are not the same. Why? Their relative abundances were determined by the processes in the \ufb02rst moments after the explosion. Thus, comparing what follows from the theories with the observed abundances of chemical elements, we can judge validity of our theories. Nowadays, the most popular theory, describing the history of the universe, is the so{called Big-Bang model. The diagram given in Fig. 19.11, shows the sequence of events which led to the creation of matter in its present form. Nobody knows what was before the Big Bang and why", " it happened, but it is assumed that just after this enigmatic cataclysm, the universe was so dense and hot that all four forces of nature (strong, electromagnetic, weak, and gravitational) were indistinguishable and therefore gravity was governed by quantum laws, like the other three types of interactions. A complete theory of quantum gravity has not been constructed yet, and this very \ufb02rst \\epoch\" of our history remains as enigmatic as the Big Bang itself. The ideal \\democracy\" (equality) among the forces lasted only a small fraction of a second. 1032 K and the gravity separated. By the time t The other three forces, however, remained uni\ufb02ed into one universal interaction mediated by an extremely heavy particle, the so-called X boson, which could transform leptons into quarks and vice versa. 10\u00a143 sec the universe cooled down to \u00bb \u00bb \u00bb When at t 10\u00a135 sec most of the X bosons decayed, the quarks combined in trios and pairs 10\u00a110 sec, to form nucleons, mesons, and other hadrons. The only symmetry which lasted up to was between the electromagnetic and weak forces mediated by the Z and W particles. From 10\u00a110 sec) until the universe was about one the moment when this last symmetry was broken ( second old, neutrinos played the most signi\ufb02cant role by mediating the neutron-proton transmutations and therefore \ufb02xing their balance (neutron to proton ratio). \u00bb \u00bb Already in a few seconds after the Big Bang nuclear reactions started to occur. The protons 378??????? BIG BANG???? single uni\ufb02ed force gravitational force separated strong force separated weak force separated n+\u201d! p+e\u00a1, p+ \u201e\u201d! n+e+ p+n! 2H+, 2H+2H! 4He+ pp{chain today 1032 K 1028 K 1015 K 1010 K 109 K 107 K 2:9 K \u00bb? temperature 0 10 10 10 \u00a143sec \u00a135sec \u00a110sec 1 sec 10 sec 500 sec \u00bb? 15 \u00a3 109years time Figure 19.11: Schematic \\history\" of the universe. and neutrons combined very rapidly to form deuterium and then helium. During the very \ufb02rst seconds there were too many very energetic photons around which destroyed these nuclei immediately after their formation. Very soon, however, the continuing expansion of the universe changed the conditions in favour of these newly born nuclei. The density decreased and the photons could not destroy them that fast anymore. During a short period of cosmic history, between about 10 and 500 seconds, the entire universe behaved as a giant nuclear fusion reactor burning hydrogen. This burning took place via a chain of nuclear reactions, which is called the pp-chain because the \ufb02rst reaction in this sequence is the proton-proton collision leading to the formation of a deuteron. Nowadays, the same pp-chain is the main source of energy in our sun and other stars. But how do we know that the scenario was like this? In other words, how can we check the Big{Bang theory? Is it possible to prove something which happened 15 billion years ago and in such a short time? Yes, it is! The pp-chain fusion, pp-chain: p + p e\u00a1+p + p p + 2H 3He + 3He 3He + 4He 2H + e+ + \u201de 2H + \u201de 3He + 4He + p + p 7Be +!!!!! 379. & p + 7Be 8B 8Be\u2044 7Li + \u201de 8Be + 4He + 4He e\u00a1+7Be p + 7Li 8Be!!! 8B + 8Be\u2044 + e+ +\u201de 4He + 4He!!! is the key for such a proof. \u2030=\u2030p 6 1 \u00a12 \u00a14 \u00a16 \u00a18 10 10 10 10 \u00a110 \u00a112 10 10 helium deuterium - 10 102 103 104 t (sec) Figure 19.12: Mass fractions \u2030 (relative to hydrogen \u2030p) of primordial deuterium and 4He versus the time elapsed since the Big Bang. As soon as the nucleosynthesis started, the amount of deuterons, helium isotopes, and other light nuclei started to increase. This is shown in Fig. 19.12 for 2H and 4He. The temperature and the density, however, continued to decrease. After a few minutes the temperature dropped to such a level that the fusion practically stopped because the kinetic energy of the nuclei was not su\u2013cient to overcome the electric repulsion between nuclei anymore. Therefore the abundances of", " light elements in the cosmos were \ufb02xed (we call them the primordial abundances). Since then, they practically remain unchanged, like a photograph of the past events, and astronomers can measure them. Comparing the measurements with the predictions of the theory, we can check whether our assumptions about the \ufb02rst seconds of the universe are correct or not. Astronomy and the physics of microworld come to the same point from di\ufb01erent directions. The Big Bang theory is only one example of their common interest. Another example is related to the mass of neutrino. When Pauli suggested this tiny particle to explain the nuclear \ufb02-decay, it was considered as massless, like the photon. However, the experiments conducted recently, indicate that neutrinos may have small non-zero masses of just a few eV. In the world of elementary particles, this is extremely small mass, but it makes a huge difference in the cosmos. The universe continues to expand despite the fact that the gravitational forces pull everything back to each other. The estimates show, that the visible mass of all galaxies is not su\u2013cient to stop and reverse the expansion. The universe is \ufb02lled with a tremendous number of neutrinos. Even with few eV per neutrino, this amounts to a huge total mass of them, which is invisible but could reverse the expansion. Thus, the cooperation of astronomers and particle physicists has led to signi\ufb02cant advances in our understanding of the universe and its evolution. The quest goes on. A famous German 380 philosopher Friedrich Nietzsche once said that \\The most incomprehensible thing about this Universe is that it is comprehensible.\" 381 Appendix A GNU Free Documentation License 2000,2001,2002 Free Software Foundation, Inc. Version 1.2, November 2002 Copyright c 59 Temple Place, Suite 330, Boston, MA 02111-1307 USA Everyone is permitted to copy and distribute verbatim copies of this license document, but changing it is not allowed. PREAMBLE The purpose of this License is to make a manual, textbook, or other functional and useful document \\free\" in the sense of freedom: to assure everyone the e\ufb01ective freedom to copy and redistribute it, with or without modifying it, either commercially or non-commercially. Secondarily, this License preserves for the author and publisher a way to get credit for their work, while not being considered responsible for modi\ufb02cations made by others. 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If your document contains nontrivial examples of program code, we recommend releasing these examples in parallel under your choice of free software license, such as the GNU General Public License, to permit their use in free software. 388 meet ; with this point of intersection as a centre and with any line as a radius, describe a circle ; let L be the length of the arc of this circle intercepted between the two lines and let R be the length of the radius of the circle ; then by definition the ratio \u2014 is the numerical value of the angle. Thus, in the figure, AB and CD are the two lines lying in a plane ; 0 is their point of intersection; OP is the radius of the circle, i.e. R; PQ is the intercepted arc, i.e. L ; hence the PQ angle has the value \u2014 ^- This definition of the value for an A p, angle is adopted because it is known \u2014 R\u2014 \u2014 i from geometry that the ratio of the FIG. i.- Numerical^ value of the intercepted arc to the radius of its i8^ circle is the same for all\" values of the radius ; and therefore it is not necessary to specify the latter. If Nis the numerical value of the angle, this relation can be expressed N= =|, or L = RN; in words, the length of the intercepted arc equals the product of the length of the corresponding radius and the value of the angle. A unit angle is called a \" radian \" ; it is an angle such that the lengths of the intercepted arc and the radius are equal. In ordinary language angles are expressed in \"degrees,\" \"minutes,\" and \"seconds\" ; there being 60 seconds in a min- ute, 60 minutes in a degree, and 90 degrees in a right an<rl<\u00ab. INTRODUCTION a, e s u c f i R It is easy to find the relation between a radian and a degree ; e b a the arc intercepted by two lines making a right angle, where TT= 8.141\u00bbl approximately ; and therefore \u2014, or 1.5708, is the \u2014 value of a right angle in terms of radians. Hence, l.\")708 radians = 90\u00b0, or 1 radian = 57\u00b0.2958 = 57\u00b0 17' 45\".? / An angle has a sign as well as a numerical value. Thus, if OA be chosen as a fixed direction, there is a difference between the angles (AOB) and (y40(7), although", " they are numeric- ally equal. The latter corresponds \u00b0' to a rotation like that of the hands of a watch ; the former to a contrary rotation. One angle, it is immaterial which, is called plus ( + ); the other, minus (\u2014 ). When an angle becomes very small, the ratio of the value of its sine to its own value approaches unity. For, referring to the cut, the value of the angle between OA and OB is the ratio of the arc AB to the radius; the value of the sine of this angle is defined to be the ratio of CB to the radius; therefore, the ratio of the an^le t\u00bb\u00bb hs sine is the ratio of the an-, 1 />' t.> t '/>'. Fio 8 -In the limit when '^lis (M!ll;ils tn<> rati\u00b0 \u00b0f tne arc /;'/; to its ch'ord BCD; and, as the angle is made smaller and smaller, this ratio /t)-(AOB).. approaches unity, because in tin- limit an arc and its chord eqaaL 3. Vectors and vector quantities. \u2014 A vector is a limited port i\u00ab\u00bbll nf a MlMl-ht illir ill a definite diivrtinn. TllUS tin- INTRODUCTION straight lines AB and CD are vectors ; their lengths are the distances between A and B and between C and D ; and their directions are indicated by the arrows. Three ideas are involved: the direction of the line, the sense of this direction (i.e. a distinction is made between a line drawn to the right and one drawn to the left, etc.), and the length of the line. The position of the line is immaterial; so two vectors of the same length and in the same direction, wherever placed, are equal. A vector, then, is a straight line traced by a point moving from one position to another, as is indicated by the use of an \"arrow\" in the line. FIG. 4. \u2014 Two vectors. The process of \"addition of vectors\" is defined as follows: move one vector parallel to itself until one of its ends meets that end of the other which causes the arrows to indicate continuous ad- vance from the free end of one vector to that of the other, and then join the former free end to the latter by a straight line. The \"sum\" is therefore a vector. Thus AB and CD may be added in two ways : (1) move CD parallel to itself until C coincides with B, \u2014 the arrows now indicate continu- ous advance from A to D, \u2014 and join these points by a PIG. 5. \u2014 Three methods for tho_additton of the vectors AB and CD. tffTRODUi TION 27 straight line, tlius forming the vector AD: (2) move CD parallel to itself until I) coincides with A, \u2014 the arrows now indicate continuous advance from C to 1?, \u2014 and join these points by a straight line, thus forming the vector CB. It is evident from geometry that these two vectors are identical, having the same length and the same direction and sense. (If a parallelogram is formed, having the two vectors as adjacent sides, both starting from the same point. the diagonal is their sum.) This process is called \" geometrical addition v ; and it can obviously be extended to three and more vectors. The simplest case is evidently that when the two vectors are in the same straight line : if they are in the same sense, the numerical value of the sum is the ordinary arithmetical sum; while, if they are in opposite senses, it is their arithmetical difference. If, then, two vectors are in the same line and in the same sense, both may be called posi- A > B tive ; but if they have opposite senses, D < c we should call one positive ( + ) and the other negative ( \u2014 ) ; and their A D B,C geometrical sum equals in numerical FIO. B.-A.I \u2022 i\u00abni- value the algebraic sum in both cases, and has the direction of the two vectors. Its sense of direc- tion in the former case is that of both vectors; in the latter, that of the greater. Looking at this process of geometrical addition in a con- verse manner, it may be said that the vector j&? is the geo- metrical sum of AH ii\\\\<\\ ( 7>, where.1 // and f'/>aiv any two vectors such that, when added, their initial and final points are A and D: the vector AD is said to be \"resolved into components.\" The case when the two components are at right angles is the most important. Let* AD be any vector and (JT any straight line; drop 28 IN", "TRODUCTION perpendiculars AA' and DD' upon OP\\ A'D' is called the \"projection of AD upon OP.\" Draw through A a line par- allel to OP; it intersects DD' in B. Then the vector AD B,C FIG. 7. \u2014 Resolution of the vector AD into components. A' D' FIG. 8. \u2014 Projection of the vector AD upon the line OP. equals the geometrical sum of the vectors AB and BD. Let the lengths of AB, BD, and AD be 5, v, and A; then, by geometry, A2 = b2 -f v2 ; and, if JVis the angle (BAD), by the definitions of trigonometry : ^ = sine N, - = cosine N, - = tangent N, h n b or, as ordinarily written, v = h sin TV, b = h cos N, v = b tan N. So the projection of AD on OP equals the product of AD and cosine JV.* The vector ^..B is called \" the component in the direction OP of the vector AD\" (If AB and BD were not perpen- dicular, i.e. if (ABD) were not a right angle, the latter vector might be so resolved as to have a component in the direction OP ; and in that case the former would not be the only component of the vector^/) in this direction.) But, as just shown, AB = AD cos N. The general rule, then, for * In a similar manner if perpendicular lines are dropped upon a plane from the points forming the contour of any limited surface, the area inclosed by the feet of these lines is called the projection on this plane of the limited surface. If this surface is plane and has the area A, if the projected area is Aij and if the angle between the two planes (i.e. between lines perpendicular to them) is N, it is seen that A\\ = A cos N. INTRODUCTION 29 obtaining the numerical value of the component in a particu- lar direction of a given vector is to multiply the numerical vulue of the vector by the cosine of the angle between it and the specified direr' It is obvious from geometry that the component of any or AC along any line OP, i.e. Afi^ is the algebraic sum of the components along this same line of any two vec- tors whose geometrical sum is AC, e.g. AB and BC; viz., The geometrical addition of the two vectors AB and BC to produce the vector AC may be expressed in words as follows: In order to ob- tain the vector AC by joining a vector to AB, it is necessary to add the vector BC. Therefore, Itc is i In' \"geometrical difference\" between the _ I vector.1 <'and the vector 5 AT\" BT~ cT P AB. Thus, the nile for ^IQ \u201e._ ^ projection' of the vector ^T upon Sllht 1 -acting One Vector or e<lual8 the_sum ofjhe projections of the two from another is evident; thr difference is such a vector that, if added to the former, it L,rivcs thr latter. component* AB and BC. There are many physical quantities which require for their description a numerical value and a direction, e.g. velocity, force; they can be represented graphically by a vector, and are called \"vector quantities.\" L Average or mean values. \u2014 If there is a set of similar quantities, ar ay \u2022\u2022\u2022,\u00ab\u201e, their \"arithmetical mean \" is defined tobe-1 \u2014 ^\u2014 \u2014 \u2022 1 1 of ten happens, however, th // to the physical or mathemati. -al < -(.n. lit ions, the various quan- tities at. DO! \u00ab.t thr same importance. For instance, suppose thr mean age of a class of students is desired: let the age 30 INTRODUCTION of 2 members be 16, of 5 be 17, of 10 be 18, of 1 be 19 ; the ages are then 16, 17, 18, and 19, but the importance of 17 in the average is 5 times that of 19. In this case evidently the proper mode of finding the mean value is to multiply 16 by 2, 17 by 5, etc. ; add these products and divide by the num- ber in the class, i.e. 2 + 5 4- 10 -f- 1. So in the general case, if m1 is the importance of ar m2 that bf \u00ab2, etc., the mean value is w?, + ra2 + \u2022\u2022\u2022 + mn If a quantity assumes different values in succession, but in a continuous manner, e.g. the speed of a falling body, it is often necessary to find its", " proper- \u00ab\u00bbf matter: inertia, weight, and tin- one which includes the varied characteristics of size and shape. Kach of these will now l>e considered in!_;Tealer or less detail. As has been said before, the science of Mechanics is that branch of Physics which deals witli the inertia of matter. It is often divided into two parts. \u2022\u2022Kinematics\" and \u2022\u2022 Kinetic^\": the former is the science of motion considered apart from matter; that is, it treats of possible motion; the latter is -trictly the science of the inertia of matter. If there is no change of any kind in the motion, what we call \"rest \" beiiiLT ecial case of this, the science is called \"Statics\"; while if the motion is chan^in^, the science is called \"Dynamics.\" These two sciences are branches, then, of Kinetics. Statics can. however, be considered as a special limiting case of Dynamics; and this plan is adopted in the present book: 8O Mechanics will be treated under the two divisions. Kine- matics and Dynamics. In the former, the (jnestion as to the different possible kinds of motion will be discussed; in the latter, the physical conditions under which these types of motion occur. Kinematics is a geometrical science; dynamics, a physical one. \\\\YMii.\\ill l\u00bbc discussed under the more general head of.Station. This will be f\u00abill..\\\\,-d by several chapters on the properties of solids, liquids, and gases. AMES'S PHYSIC* W CHAPTER I KINEMATICS General Description. \u2014 Kinematics has been defined as the science of motion apart from matter ; that is, it is concerned with the study of the possible motions of the geometrical quantities : a point, a plane figure, and a solid figure. For the sake of illustration, many material bodies will be referred to ; but all the statements and theorems are meant to apply to figures, not to bodies, unless the contrary is expressly noted. If the motion of any actual body is observed (for instance, a stick thrown at random in the air, a moving baseball, the wheel of a moving wagon), it is seen that there are two types of motion involved : the object moves as a whole, and it also turns. These motions are independent of each other; one may occur without the other. In the up and down motion of an elevator, in the motion of a railway car on a straight track, etc., there is no turning ; in the motion of the fly- wheel of a stationary engine, in the opening or closing of a door, etc., we may say that the motion is one of turning only. The name \" translation \" is given to that kind of motion during which all lines in the figure remain parallel to their original positions ; further, all points of the figure move through paths that are geometrically identical. Thus, to describe completely any case of translation, all that is neces- sary is to describe the motion of any one point of the moving figure. The name \" rotation \" is given to that kind of motion dur- ing which each point of the figure moves in a circle ; e.g. a door as it opens or closes. The planes of these circles are 34 KIM-MAUL'S 35 parallel, and their centres all lie on a straight line which is called the --axis.\" If a plane section, perpendicular to the axi>, is taken through the rotating figure, all the lines of the figure in this plane have identically the same angular mo- tions: otherwise the figure would break up into parts. To \u2022 leseribe, therefore, motion of rotation at any instant, we must know two things: the position and motion of the axis and the angular motion of any line fixed in the figure with reference to any line fixed in space, provided the ^ lines lie in the same plane per- pendicular to the axis. Thus, con- sider the rotation of a figure like that of a grindstone ; its axis is the FIO. i2.-Rouuon of figure..inn,, of the axle. In the cut, ^^^^^ which represents a cross section by fixed in \u00bbpace, and />e, \u00bb line fixed a plane perpendicular to the axis, let PQ be a line fixed in the moving figure and AB be a line fixed in space; the position and motion of the figure at any instant are given by a knowledge of tin anurle l>etwern these lines and of its changes in value. It should l\u00bbe noted that this is a special case of rotation, because tin- axis does imt ci. position, as it does in general. To describe the most complicated motion, tin -i must consider it resolved into two parts, a translation and a rotation", ", and must discuss each separately. Translation In translation, as lias been already explained, it is neces- tn describe tin* motion of a point only. The simplest case of this i> motion almi<r a straight line; but the | general case, that of motion along a curved line, is not ditli- eult. To describe this nmtion the first thing that it is neces- to know is the position of the point at.my inMant with re nee to soi i figure. 36 MECHANICS Linear Displacement. \u2014 Let the path of the point with reference to some fixed figure be represented by the curve in the cut. Let 0 be its position at any instant, and P that at a later time. The vector OP is called the \"linear displacement\" of the point with reference to the fixed figure during this interval of time. This same vector might be the displacement for any motion that passed through 0 and P ; or, in other words, a point may pass from 0 to P by various paths. The displacement is then a vector quantity and may be resolved into components in as many ways as we choose ; con- FIG. is. -The vector OP versely, two or more displacements may is the linear displacement of \\>Q compounded by geometrical addition. P with reference to O.,\u201e,. The importance of mentioning the \"fixed figure of reference \" may be seen from an illustration : if a stone is dropped from the top of the mast of a moving steamer, it will fall at its foot ; the displacement with refer- ence to the steamer is a vertical line, while with reference to the earth it is an oblique one, being the geometrical sum of the vertical line and the displacement of the steamer. Linear Velocity. \u2014 If the interval of time taken for the displacement is extremely small, P is very close to 0; and, in the limit, the displacement OP coincides with the actual path along the curve, and has, in fact, the direction of the tangent to the curve at the point 0. If, as the displace- ment becomes very small, its length is represented by Aa?, and the corresponding interval of time by A\u00a3, the ratio \u2014 in the limit is called the \" linear velocity \" at the point 0 with reference to the fixed figure ; that is, it is the \" rate of change\" of the displacement. It is evidently a vector quantity for it is defined by its numerical value, which is that of \u2014 in the limit, and by the direction and \" sense \" of KINEMATICS 37 the displticement in the limit : viz., its direction is that of tht? tangent at 0 drawn from 0 to P, when P is close to 0. The numerical value of the linear velocity is called the k- linear speed\"; so that the velocity at any point is char- acterized by the value of the speed and by the direction and \"sense\" of the tangent to the path at that point. If the motion is uniform along a straight line, that is, if the velocity is constant (both in amount and in direction), the speed is equal numerically to the distance traversed in a unit of time; and if the motion is not uniform, the speed at any instant is the distance which the point would travel during the next unit of time if the motion were to remain uniform. The unit of linear speed on the C. G. S. system is the speed of \"1 cm. in 1 sec.\"; and the unit of linear velocity in a definite direction is the unit of speed in that direction. (This C. G. S. unit speed has not received a name; in fact, the only system of units in which there is a unit speed which has received a name is that based on the nautical mile \u2014 6080 ft. \u2014 as the unit length, and the hour as the unit time : the speed \" one nautical mile in one hour \" is called a \"Knot.\" The expression \"16 knots per hour\" is therefore incorrect; for a knot is a speed, not a length.) Since the linear velocity is a vector quantity, it can be resolved into components; and, conversely, two or more linear velocities may be compounded by geometrical addition. These statements are illustrated by many familiar facts: if a man \\\\alks across a moving railway carriage, his velocity with reference to the ground is com- pounded \u00ab>f that of the train and of that which he would have 38 MECHANICS if the train were at rest ; if a boat is rowed across a river, the actual velocity with reference to the earth is the geometrical sum of that of the water of the river and of that due to the oars ; the velocity of a raindrop with reference to the window pane of a moving carriage as it strikes it is the geometrical", " sum of a velocity equal but opposite to that of the carriage and of its own downward velocity at that instant ; if a man walks in a northeast direction with a speed of 8 cm. per second, his velocity may be represented by a vector PQ whose length is proportional to 8 and whose direction is northeast; and his velocity in a northern direction is given by the component, PR, in the direction north and south, whose numerical value is 8 cos 45\u00b0, or, more properly, s cos j- Similarly, one velocity may be subtracted from another, the difference being also a velocity. We will consider two illustra- tions: a body falling freely toward the earth and an extremely small particle mov- ing in a circle with a constant speed. In the first case, the velocity at any instant is represented by a vertical vector AB and FIG. is. - Rectilinear at some later instant by another vertical motion: AB and CD are -~=r f., the velocities at different vector CD ot greater length, because as BD l8 th6lr the body falls' its 8Peed increases. Call the length of AB sr and of CD \u00ab2. The change in velocity is the difference between these vectors ; that is, it is a vertical vector BD of length equal to *2 ~ *r In the second case in which the particle is moving in a circle let the constant speed be \u00ab; and let the direction of motion be that indicated by the arrows. When the particle is at the point A, its velocity has the direction of the tangent and the numerical value \u00ab; it can therefore be represented KIXEMATirs 30 by the vector PQ which is parallel to the tangent at A, and has a length proportional to \u00ab. Similarly, when the part irk- is at the point B, its velocity can be represented by the vector PS which is parallel to the tangent at B and whose length is equal to that of 7Vtv ( since the speed does not alter). The change in the velocity in the time taken for the particle to move from A to B is the difference between the vectors PS and PQ; that is, it is the vector s v. Linear Acceleration. - - To Fio.J_6. \u2014 Uniform motion In \u00bb circle. PQ and PS are the velocities of the point at A and B. return to the original problem, that of describing the general case of the motion of a point in a curved path, ire have defined the displacement and the velocity at any point, the latter being the rate of change of the former. The velocity may change, however, both in direction and in speed ; and its rate of change at any instant is called the \" linear acceleration \" at that instant ; that is, if At; is the change in the velocity dm MIL: the time At, the limiting value of \u2014 - is the acceleration. Moreover, since the change in the velocity, A>\\ is a vector quantity, so is also the accel- eration. Its numerical value is that of -- in the limit ; and At its direction is that of A'- in the limit. We may consider separately two cases; in one of which the din-rtion remains constant hut the speed changes, -while in the other th\u00ab speed remain^ constant but the direction changes. As an illustration \"f the former we may take the mot; : falling l\u00bbodv: and of the latter, the uniform 40 MECHANICS motion of a particle in a circle. These two cases have already been partially discussed. In the former motion let the change in speed from s1 to s2 take place in the interval of time T2 \u2014 T^\\ then the accel- when the interval eration has the numerical value T2 \u2014 Tl is taken infinitely small, and its direction is ver- tically down. If j:he acceleration is constant, it is, therefore, the change in the speed in a unit of time. In the latter case, that of uniform motion in a circle, let the interval of time during which the particle moves from A to B, and the velocity accordingly changes from PQ to PS, be taken extremely small, so that the length of the arc AB becomes minute also ; then, if this interval of time is called A\u00a3, the acceleration at the point A is the limiting value of fvector Q8\\ Call the lengths of the various straight lines in the diagram by the letters marking their terminal points : by geometry the triangles (SPQ)_*,in\\_(BOA) are similar, hence, -^ = ^-r ; but PQ equals s, the value of the constant speed ; OA equals the radius, r, of the circle; and in the limit, when A\u00a3 is infinitely small, the length of the chord AB equals the length of the arc AB, the value of which is s \u2022 A*, because s is the distance tra- vers", "a- Since the acceleration is constant, its numeri- cal value equals that of the change in speed in one second; and therefore, if the value of the speed at any point Pl and instant of time 2\\ is sr that at the point P2 which is reached at the instant T2 is given by the following equation, in which a is the value of the accel- Fl\u00b0- l9- ~ e ration : MT motion. sa-Sl = a(T^- T,), or s, = Sl + a(Ts- 7\\). (1) (It should be noted that, if the sense of the direction of a is that of the velocity, the speed increases in value with the time, e.g. a falling body ; whereas, if the sense of the two diivrtions are opposite, a and * have different signs, and the formula iH'coim-s s^ = s1 \u2014 a(T% \u2014 2j) if a is here the numerical value of the acceleration ; therefore the speed decreases, e.g. a body thrown vertically upward -or a body thrown along a slnTt of ice.) 1 1 tin- speed were constant, the distance traversed in t sec- onds would be the product of this by the value of the speed; but the latter is varying from instant to instant. Since the speed is, however, by assumption changing at a uniform rate, its mean value (see page 32) between the instants 7*, and 5P, is the average of \u00ab2 and sr i.e. **\"*\"**; and the distance passed over iu the interval of time (!Ta \u2014 71,) is &^il x (T, - r,) = s^T, - T,) +J a(r, - r,)'. If anv point 0 in the line of motion is chosen as a] reference or \"origin,\" and if the distances OI\\ ami OPt are \u2022 \u2022all. --I A, and /.,. ili.- displacement in the interval of time 7!,- T, /., - Lv That is, 44 MECHANICS L2- Ll = 9l(Tt - 7\\) + | a(T2 - r,)2, or ^\u2022=^i + \u00abi(r,-r1)+Ja(r,-rl)^. (2) (If the acceleration is an opposite sense to the speed, a must be given a negative value.) Thus, if the acceleration is known, and if the position and speed at any one instant are given, they can be predicted for any future instant. Con- versely, if any motion is found to obey either of these laws (for one is a consequence of the other), it is known that the acceleration is a constant. These two formulae assume their simplest form when we agree to measure time and distance from the instant and position in which the moving point is at rest. For instance, let the point be at rest at Pv i.e. s1 = 0 ; then we will choose 0 to coincide with Pr i.e. L^ = 0 ; and also choose this in- stant as the one from which to measure time. Hence the formulse become These are due to Galileo, and it was by showing that when a body moved down an inclined plane the displacement varied as the square of the time taken, that he convinced himself of the constancy of the acceleration parallel to the plane. Since this is constant, so is that for a body falling freely. In the general formula (2) it is seen that the displacement, L2 \u2014 Lv is made up of two parts : s^T2 \u2014 T^) is the distance the point would have gone if there had been no acceleration ; %a(T2\u2014 j^)2 is therefore the additional displacement owing to the acceleration. Another general formula may be obtained from equations (1) and (2) by eliminating (T^ \u2014 T^) from them : substitute in (2) the value of (T^- TJ obtained from (1,) viz. This gives a(L2-L1)=KV-V> (3) KINEMATICS Comparing equations (1) and (3), it is seen that the former defines a as a function of the interval of time, viz.,, while the latter expresses it as a function of the 1 displacement, viz., a == - 2 -. A i>\"lid moves with a constant acceleration in one direction with a constant velocity in a direction at r'ujht anjlt's to this. Let the constant accelera- tion have the numerical value a; and let distances in its direction, measured from the point at which the speed in this direction is zero, be called y. Then, by count in^ time from the instant when the point is in the position for which y=0, we have the relation", " shown further how to describe the angular motion around a fixed axis by means of two lines, one fixed in space, the other in the moving figure, but both lying in a plane perpendicular to the axis. Angular Displacement. \u2014 In the cut, as before, let AB be a line fixed in space and PQ a line fixed in the rotating figure ; let 0 be the point where the axis cuts the plane. As the figure turns around the axis, the angle between AB and PQ varies: It is called the \"an- gular displacement\" at any instant of PQ with refer- ence to AB. To describe --B this displacement more fully, however, we must know the position of the axis and the FIG. 25. \u2014 Angular displacement: 0 is trace,,. r \u2022 u i~ i_ J of axis, AB is a line fixed in space, PQ is a line \" S6nS6 in Which the body fixed in the figure. js turning; that is, whether, looking at the lines in the cut, PQ is rotating like the hands of a watch or in the opposite sense. These three ideas \u2014 the numerical value of the angle, the position of the axis, and the sense of the rotation \u2014 can all be represented by a vector placed so as to coincide with the axis; for its posi- tion indicates that of the axis, its length can represent the value of the angle through which the figure turns, and its direction can be made by some agreement to indicate the sense of the rotation. The connection between the direction of the vector and the kind of rotation which is usually adopted is as follows : if an ordinary right-handed screw partially in a board is placed so as to coincide with the K I \\EMATICS 53 \\ector and is turned in the sense of the rotation, the direc- tion which it moves lengthwise into or out of the board is that L,ri\\ -en the vector. Thus, in the cut, if AB is the vec- tor, and if an observer looks at a rotating figure in the direction from A toward B* the rotation is like that of the hands of a clock. If tin- arrow were in the opposite direction, the vector would indicate rotation in the opposite direction. This connection between the directions of the vector and the rotation is called the \"ri^ht-handed screw relation/' Fin. 26. \u2014 Rlffat- A Vector located 111 a definite position 111 this handed screw relation manner is called a \"rotor.\" It may be proved J^^J,** without dillicnlty that one can add two rotors geometrically : (1) if they lie in the same line, when the addition is therefore algebraic, because rotation in one sense is posit ive ( -f- ), and in the opposite is negative ( \u2014 ), and tin- two rotors are lines either in the same sense or in opposite senses; (2) if their axes lie in the same plane, i.e. if they meet at a point, and if the angular displacements to which they are proportional are infinitely small. < this pn.of reference maybe made to any treatise on Media such as that of Xiwet or Williamson.) Angular Velocity. - I he rate of change of the angular displace, in-lit around a given axis is called the \"angular velocity around that u numerical value of this velocity is called the \" angular speed \" ; 80 that in order to describe an B :ill.-ular vdodty three things um>t be Specified : the position <>( tin t\\ i^. the sense of the rota- *7.-AmruUr Speed. til,n,]\u201e. jmLrnl,lP gpeed. If the ilar motion is uniform, the \\alue of the angular speed is that of the angular displacement in a unit of time. In 64 MECHANICS the cut let, as before, AB and PQ be the two fixed lines of reference, one in space and one in the figure; but in this case let them be drawn through the axis at 0. Q is then any fixed point in the figure; call its distance from the axis, i.e. OQ, r. If the axis is fixed in space, the length of the arc described by Q is evidently equal, by the definition of the value of an angle, to the product of r and the value of the angular displacement. It follows, then, that the linear speed of Q at any instant, i.e. the rate at which Q passes along the arc of its circle, equals the product of r by the value of the angular speed of that instant, i.e. the rate at which the radius describes the angle. To return to the idea of angular velocity, it is evident that it can be represented by a rotor ; and since an angular velocity is the rate of change of an angle", ", i.e. is the limit of the ratio of the value of a small angle to that of a cor- responding small interval of time, it is proportional to an infinitely small angle, and therefore two rotors representing angular velocities may be added geometrically, if the two axes lie in the same plane. Illustrations of angular velocities are common ; a few may be described as follows : that of a flywheel is given by a line having a definite sense and length, coinciding with the central line of the axle ; that of a door or gate by a limited portion of a vertical line drawn so as to coincide with the central line of the hinges ; that of a cylindrical barrel rolling down an inclined plane by a line coinciding with the line of contact between the cylinder and the plane \u2014 in this case the axis is moving parallel to itself down the plane. As a hoop rolls along a floor, the rotor giving its angular velocity is a horizontal line perpendicular to its plane. If the hoop were at rest in an upright position, a sidewise push at the top would give it a rotation around a horizontal axis in its own plane ; therefore, if a sidewise push is given a rolling hoop at its top, the rotor of the resulting motion is the geometrical sum of the two separate rotors, and is in a horizontal plane but in a different direction from either \u2014 this explains why pushing a rolling hoop sidewise at the top changes the direction of its path. (The independent action of two or more forces is assumed again.) KIM-:.MATIC8 55 Angular Acceleration. \u2014 The rate of change of the angular velocity is called the angular acceleration. There are two special cases : in one, the direction of the axis remains fixed in space, while tin- angular speed varies, e.g. a door when it is opened <\u00bbr closed, a grindstone when it is set in motion or radually stopped, etc.; in the other, the angular speed remains constant and the direction of the axis changes, e.g. a spinning top whose axis is not vertical, a rolling hoop turn- ing a corner, etc. In the general case, of course, both the angular speed and the direction of the axis change. In tin- case of rotation around a fixed axis there is evidently a simple connection between the angular acceleration and the linear acceleration of any point of the figure. If A is the angular acceleration and a the linear acceleration of a point at a distance r from the axis, it follows at once from the defini- tion of the value of an angle that a = rA. (See page 47, where a similar formula for the velocities is proved.) There is also harmonic motion of rotation around a fixed. analogous to that of translation; it is illustrated by vibrations of the balance wheel of a watch, by those of a.) ^ clock's pendulum, etc. It is defined as follows: if the value of the angular displacement in a particular sense is called N^ harmonic motion of rotation is such that the angular accel- '.on equals \u2014 f?y, where c3 is a constant, depending upon the vibrating system. The amplitude is the maximum value of N; the period may be proved to be - id two harmonic c vibrations may differ in phase. It may be shown, fun that this definition is equivalent to saying that the angular displacement, equals A cos (ct \u2014 M ), where A is the ampli- tude: ~ \u2014. the period : and (ct \u2014 AT), I lie phase. C General Remarks. - The displacement,,f a point in any direction is independent of a displacement ll u at ri'^ht angles to ihix,,-.;/. a man I 'liuard does not 56 MECHANICS move to the east or to the west ; and, since it is possible to draw from any point three lines that are mutually perpen- dicular, like the three lines meeting at the corner of a room, a point, and therefore a solid figure, may have three indepen- dent directions of translation. Similarly, rotation around any axis is independent of rotation around an axis perpendicular to it; and therefore a solid figure has three independent modes of rotation. So, in general, a solid figure may be displaced in any one of six independent ways \u2014 three of translation and three of rotation; it is said to have six \" degrees of freedom.\" Freedom of motion may of course be hampered by various constraints ; thus a figure in a straight groove has only one degree of freedom of translation, a figure turning on a fixed axis has only one degree of freedom of rotation. It is always possible to produce a given displacement of a figure in several ways. Thus, if a figure like a wire, AB in the cut, is displaced by a rotation around 0 into the position A^v this same final position may be obtained by a translation from AB to A2B2 and", " bodies of ordinary size, when these are produced by the interaction of the bodies themselves. If two billiard balls meet, the velocities of both change, i.e. they are accelerated ; if a magnet and a small piece of iron are suspended by strings at the same level, each moves toward the other with an accel- eration ; if a man stands on a box which rests on a smooth floor, and jumps off sidewise, the box moves in a direction opposite to that iii which the man jumps; etc. It is a familiar fact, too, that if one of the moving bodies is much heavier than the other, its acceleration is much less; and, as the bodies are varied, there is apparently a connection between some property of the body and its acceleration. Fundamental Principles 1. Principle of Inertia. \u2014 In none of the cases described above is there an acceh-ration of one of th- without Hi-' re being at the same time an acceleration of the other. ve make, as the fundamental assumption in regard to iiat the acceleration of a body depends upon itt potition reference to neighboring bodies and upon their velocities. ii asure these velocities and accelerations, some suitable geometrical Hum re \u00ab,f reference must be selected.) There is no way of proving this assumption or the following ones; but all observation-.uv in iOOOrd with them. 60 MECHANICS 2. Principle of Independence of Action of Forces. \u2014 Again, we shall assume that, when a body is tunh'r the influence of several forces, the action of each one is independent of the actions of the others. 3. Principle of Action and Reaction. Definition of \" Mass.\" \u2014 Then, if we have an isolated system of two bodies, each will have a\" linear acceleration ; and, in order to speak defi- nitely, we shall consider the bodies as being so small that they occupy points. Such bodies are called \"particles.\" We assume that their accelerations are in the straight line joining the particles, but in opposite directions, and that the ratio of these accelerations is a constant quantity. We can, therefore, assign a number to each body such that, if m^ and w2 are these numbers, and al and #2 are the accelerations, m^ = \u2014m2a2. Similarly, if we have a third particle, we can assign a number to it by allowing it to \"act\" upon the first particle, using ml as its number, or upon the second one, using m2 for it. Experiments prove that the numbers thus obtained for the third particle are the same. Therefore, if we adopt an arbi- trary number for any one particle, the numbers obtained for all other particles are definite. These numbers are the values of what is called the \" mass \" of a body. (The system of masses in ordinary use will be described presently.) 4. Definition of \" Force.\" \u2014 When a particle of mass m has an acceleration a, the product ma is a vector with a definite value and direction ; and it is defined to be the value of what we have called the force. Thus, in the case of the two par- ticles, there are two equal and opposite forces ; and we say that the force of a particle A upon a particle B is equal numerically but opposite in direction to that of B upon A, or that \"action and reaction are equal and opposite.\" Since, then, F = ma, calling the value of force F, a= \u2014, or m when a given force acts upon a particle, the acceleration is in the direction of the force and its numerical value varies \u2022pi I>Y.;i inversely as the mass of the particle. That i>, if m is large, -mall : and conversely; so m measures the inertia of [Kirtirle with reference to translation. 5. Centre of Mass. \u2014 In practice we cannot obtain particles, for all material 1 todies occupy finite volumes ; and so we can- not apply the definition of a force directly. Further, under the action of a force a body, as a rule, has both linear and angular acceleration ; for instance, if a rod lying on a smooth table is struck at some point near the end by a ball rolling on the table the rod will move in the direction of the blow, and it will also rotate. Hut we can prove, as will be shown \u2022 \u2022ntly, that, if we assume tJmf <i / 1>ody may be \u00abw- red as built up of particles, there is a geometrical point e-iiineeted with tin- body that has the same linear accelera- tion when the body is acted on by a force, as if this a- on a particle whose mass equaled that of the body. This point is (ailed the \"centre of mass\" of the body. A few illustrations", "Unc* \u00abM^| fur rln,: 64 MECHANICS whose mass is -|, we must make two bodies of equal mass \u2014 as shown by producing the same elongation \u2014 which when suspended together will produce the same elongation as does the body of unit mass. Proceeding in this manner, we may obtain a u set \" of bodies all of whose masses are known in terms of the standard. Then to obtain the mass of any body, it may be suspended from the spring and that com- bination of bodies from this set determined which will produce the same elongation. (Other and more accurate methods are used in practice, as will be shown later in speaking of the \"chemical balance.\") In using a spiral spring for ordinary purposes a different method is followed from that just described. Experiments show that the elongation of a spiral spring is proportional to the stretching force, most approximately ; and a divided scale may be attached to the frame carrying the spring, the readings on which are proportional to the elongation of the spring. Then if a body whose mass is ml produces an elongation hv and one whose mass is unknown but which may be called m produces an elongation A, h^ : h = m^g : mg, since these last are the forces. Therefore, h^ : h = ml : m ; or m = ^-^ As a rule the instrument maker divides and marks the scale so that it gives the values of the masses directly on some known system ; that is, when a body of mass 1 is suspended, the pointer which marks the elongation stands at division 1 on the scale, etc. Mass and Weight. \u2014 It should be carefully noted that this method of comparing the masses of two bodies is in reality one which compares their weights; but, since g at any one locality is the same for all kinds and quantities of matter, the weight of a body is proportional to its mass. In other words, two bodies that have the same weight at any one point on the earth's surface also have equal masses. This 'l\\ DYNAMICS i;:, fact is not self-evident ; for the weight of a body is de- pendent upon tlie presence and proximity of tin- earth; its mass, upon the acceleration it would receive from a deli force anywhere in the univew. 'Hie former is a variable quantity, as we shall see. because it is different for the same body at different latitudes on the earth and at different heights above sea level ; while the latter is, to the be> our knowledge, an unvarying constant quantity for a given body. There is no more a jn-t'>ri reason tor believing that two 1 mdies which have equal masso have equal weights, than for believing that they have equal volumes, which is ob- viously not true in general. It is a question to be settled by experiment, and depends upon proving the constancy of i all kinds and amounts of matter at any one point on the earth's surface. This was first shown by (ialileo in 1590, who allowed two bodies of different weights to fall from the top of the Leaning Tower of l'i>a. and observed that \u2022 lied the ground at the same instant, thus proving that they had the same acceleration. (Galileo, however, had no conception of the property of mass, and performed hi* ex- periment with a different object.) Newton was Un- to real i/e the fact that a material body had a pi < iiieh we have called -mass\"; and he devised a most ingen experiment for the purpose of learning \\\\hether.-/ was the same for all bodies. His method depends upon the use of and will be described later. It ha- been used Bessel ; and all experiments pm\\e ih.it we may.7 as a constant at any one point. Conservation of Matter. l-Vw quantitiea can he measured with the exactness of the weight, and 1*61106 mass, of tic use of a \"chemical balance\" 80 railed. One,.(\" the questions ti: loans of it was as to the < stancy of mass of a body, o: when combined.. I rom \\\\hat has been said in the -ion. evident ti imh'rgo van AMI> 6 66 MECHANICS changes ; it may have its shape, its size, its temperature, etc., altered ; it may explode into fragments ; it may be \" electrified \" or \" magnetized \" ; it may be melted if it is a solid, or* evaporated if it is a liquid, or vice versa. Simi- larly, if two pieces of matter are brought together, they may stick to each other like putty and glass ; or they may unite to produce new substances, like a piece of coal burning, a process in which the oxygen", " in the air unites with the car- bon in coal to form a new gas, called \"carbon dioxide.\" In all these changes, however, there is, so far as we know, absolutely no alteration in the total mass of the body or bodies concerned. This fact is sometimes called the \"Principle of the Conservation of Matter \" ; or, more properly, the \" con- servation of mass.\" The Unit of Mass. \u2014 The standard body whose mass forms the basis of the accepted system of physical units is a piece of platinum which is kept in Paris, and which is called the Kilogramme des Archives. It was officially adopted in 1799, at the same time as the metre bar. Its mass is called a kilogram (Kg.) ; and a body whose mass is one thousandth of this is said to have a mass of 1 \"gram\" (g.)' A. mass of one tenth of a gram is a \" decigram \" ; one of one thou- sandth of a gram is a \" milligram,\" etc. When the kilogram was originally made it was designed to have a mass equal to that of 1000 cu. cm. of pure water at a temperature of 4\u00b0 C., because under those conditions the water is more dense than at any other temperature. (The temperature must be specified, because, as it changes, the volume of a given quantity of matter varies.) More exact experiments have, however, shown that this relation is not quite exact. In fact, the mass of 1000 cu. cm. of pure water at 4\u00b0 C. is about 999.96 g. In England and the United States the commercial Unit Mass is that of a piece of platinum kept at the Standards Office at Westminster, marked \" P. S. 1844 1 lb.,\" and called the \" Imperial Avoirdupois Pound.\" It has been determined by experiment that the number of grams in one pound is 453.5924277. This unit is subdivided in such a way that 16 ounces equal one pound. /\u00bb'.v.i M 67 C.G.S. System. \u2014 In all scientific work the units in terms of which lengths, intervals of time, and masses are expressed are the centimetre, the gram, and the mean solar second. This is called the \"C.G.S. system.\" Force F= ma The Unit of Force. \u2014 The unit of force on this C. G. S. em is that which corresponds to the product ma being unity ; that is, a force which produces in a body whose mass is 1 g. an acceleration of 1 cm. per second each second (or an acceleration 2 in a body where mass is J, etc.). This force died a \"dyne.\" A force of one million dynes, i.e. 106 dynes, is called a \"megadyne.\" As estimated by our muscles, a dyne is extremely small ; for, as we can find by experiment, the value of the acceleration of a falling body is not far from 980 on the C. G. S. system, so the weight of a body whose mass is a milli'jrnm i* the product of 0.001 and 980 or 0.980. Conse- quently, we feel approximately the force of 1 dyne when we hold a milli- gram \"weight\" in our hands. On the pound-foot-second system the unit force is one which gives in 1 sec. an acceleration of 1 ft. per second to a particle whose mass is 1 Ib. (This unit therefore equals 13,825 \u2022 approximately.) Other unit forces often used are the \"weight of a body whose mass is 1 g.,\" or of one whose mass is 1 Ib. : these units have the great disadvantage of being variable, owing to variation in \" g.\" Force Effects. \u2014 In defining the numerical value of a force, we made use of the idea of a particle and of the \"fo effect \" acceleration. But the production of acceleration is, as we have seen, not the only effect of a force. This is evi- dent frnin ili,' formula of definition itself, which has the form /= ma, if / ivjuvsi-nts the t e, m the mass of the par- ticle, and a the acceleration. I r \\\\\u00ab- li.tvo assumed that M aet independently; that is, they are vector quantities. 11 the fiM-mul;,. / may be the geometries 1 sum of several s. In |,,irti. -nlsr, if a = 0, /= 0 ; hut this does not mean necessarily that th\u00ab-re U n<\u00bb foroe; it may mean that there are two equal and opposite forces acting on the particle. 68 MECHANICS Thus, if a particle is suspended \" at rest \" at the end of a vertical wire, it has no acceleration, owing to the", " fact that the two forces acting on it, its weight and the tension in the wire, are equal and opposite. It may also happen that when a particle is under the action of two opposing forces, its acceleration is not zero ; in this case one force is greater than the other. So, in general, we may say that a second effect of a force is to overcome or neutralize, more or less completely, another force. (If we speak of matter when being accelerated as offering an opposing \" force of inertia \" equal to ma, we may say that the effect of a force is always to overcome another force.) This second effect of a force is illustrated in many ways : it requires a force to stretch a spring, to bend a stick, to twist a wire, to push a body over a rough table, to drive in a nail, etc. As the properties of matter are gradually better understood, we hope to explain all these effects in terms of the acceleration of particles of matter. We can do this in certain cases already, as we shall see later. (See Kinetic Theory of Matter, Chapter IV.) Similarly, as we have shown before, forces may be pro- duced in various ways. If a stretched spring or wire or cord is fastened to a body, it will be accelerated unless there is an opposing action ; if a moving body strikes another, each exerts a force on the other, etc. Measurement of a Force. \u2014 We make use of some one of these force effects in order to measure a force in practice. We know that a body which is supported at rest free from the earth must be acted upon by an upward force whose numerical value is mg, where m is its mass and g is the accel- eration of a body falling freely in a vacuum under the influ- ence of gravity. We have shown how to measure m, and methods will be described shortly for obtaining the value of g. Thus, we can observe how much a spring elongates under the stretching action of different bodies ; and, assuming that 69 the spring does not change during the operations, we may thus obtain readings on a scale attached to the spring, which correspond to known forces. This process is called ** calibra- tion \" of the spring. Then, to measure any force, we can observe how much it elongates the spring. (It -h to exert a known force in a definite direction upon a bodv, we can attach one end of a cord to the body and the other to a spiral spring, and then pull the farther end of the spring in the desired direction until it elongates the proper amount. ) < )ther elastic bodies maybe calibrated in a similar manner and used for the measurement of forces. Linear Momentum. \u2014 The general formula for the value of a force, F = ma, may be expressed in a different manner. Since the acceleration is the rate of change of the velo. the value of the force is the product of the mass by this rate of change ; and, since the mass is a constant for a given particle, the force equals the rate of change of the product mv, where v is the velocity. This product is called the \u2022\u2022 linear momentum\" of the particle ; so the force equals the rate of change of the linear momentum. If the acceleration is constant, i.e. if the speed is changing at a uniform rate but the direction of motion is unchanged, writing i'o and?'j as the velocities at the end and the be- ^inniiiLC of an interval of time t. the rate of elian^e, of the velocity is -*\u2014 \u2014 -\u2022 Hence, we may write F Ft = m(v9-vl). The product Ft is called the \"impulse\" If a sudden blow is struck the {.article, its momentum in the direction of the iWe.o will 1)0 changed; ami the amount of this change measures the impulse of the blow. If a particle is moving with a constant velocity, there is restdtanl force acting: but to produce e in the ;orce is required. A useful ill nila F=ma is furnished by suspending a heavy body 70 MECHANICS by a long cord and attaching a thread to it also, so that it may be drawn side- wise. If it is pulled slowly, a small force is required ; if it is jerked sud- denly, the force required may be so great that the thread breaks. Illustrations of Forces. \u2014 There are two types of acceleration, one when the speed changes and the direction does not, the other when the direction changes and the speed does not ; and correspond- ing to each of these is a definite type of force. In the one the force produces a change in speed ; in the other, a change in direction of motion. a. Rectilinear force. \u2014 Thus, if we ob- serve that a particle is moving in a straight line with varying", " force must be applied at right tingles to this ; and conversely, if a particle is moving in a curved path under the action of a certain force, and the force is removed, the particle will continue to move with a constant velocity in ths direction of the tangent to the curve at the point where it Avas at the instant the force ceased. If a particle is moving with constant speed in a curved path which 73 lies in a plane, the acceleration at any instant is toward the centre of curvature of the path and has for its numerical value \u2014, where \u00ab is the speed and r is the radius of curvature. ( Whatever the path is, a circle can always be drawn which will coincide with the curved path at any point, and its centre and radius are called the \"centre and radius of curvature\" at that point.) There must then be a force whose direction is perpendicular to the line of motion and whose numerical value is - s. in order to make the particle whose mass is m and speed * change its direction of motion and move in a circle of radius r. This force, F= \u2014, is called \" centrifugal force.\" The less the force, the less the change in direction, i.e. the greater the value of r\\ or, if the speed is increased, tin- radius must be also, unless the force is increased; and \\vhen a j \u00bbarticle is revolving in a circle, if the force toward the centre is diminished, or, if the speed is increased, tin- particle moves farther away from the centre, if such motion is possible, so as to have a larger radius of motion. This fact i.s SM],I, -times described by saying \"a particle re- volving around a centre tends to move as far a\\\\ay from it as possible.\" Thus, a wet mop may be freed from the water by revolving it rapidly; clothes may be dried by inclosing them in a perforated Cylinder which is made to rotate rapidly : etc. The force that is required to hold a particle in a circle \\ aries dhvctly as its ma>s ; and. if the force applied i than \" *~. the particle will move toward the centre. \\\\ hile, if it is less than this, the particle will move away from the centre. Therefore, if an emulsion of two liquids, one imnv dense than the othi M in milk, is put into a raj. idly rotat- ing cylinder, the heavier of the two \u2014 milk \u2014 will go to the \"inside wall, while the lighter \u2014 cream \u2014 will come closer to the,: 74 MECHANICS c. Pendulum. \u2014 If both the speed and the direction of motion are changing, the force must be oblique to the latter, i.e. it must have a component in the direction of motion and one at right angles to it. One particular case of this kind of force is furnished by the action of the earth on a \" simple pendulum,\" i.e. on a particle of matter suspended by a mass- less cord from a fixed support so that it is 'free to move in a vertical plane. When at rest, it hangs in a vertical direc- tion ; but, if disturbed slightly, it will make vibrations. (Of course a simple pendulum cannot be made, but we can ap- proximate to one by using a small but heavy bob and a very fine wire to support it.) Let OM be a vertical line through the point of support 0 ; let OP be the position of the pendulum at any instant (not necessarily at the end of the swing) ; let a circle be drawn with radius OP, thus indicating the path of the particle; draw through P two lines: one vertically down, to indicate the direction of the force of gravity ; and one tangent to the circle. Call the mass of the particle w, and the length of the pendulum I. There are two forces acting on the particle ; its weight mg vertically down, and the tension in the supporting cord acting along the line P 0. The latter has, however, no effect on the speed, serving merely to change the direction of motion ; similarly, only that component of mg which is along the tangent has any influence on the speed. This component has the value mg sin N, where N is the angle (MOP). Let us suppose the arc of vibration is made so small that its chord coincides with it; in this FIG. 82. \u2014 Simple pendulum. DYNAMICS 75 limit the sine of an angle equals the angle itself (see j and so the force at any instant along the tangent to the curve \u2014 which in the limit is nearly horizontal \u2014 has the value mgN. If the displacement of the particle along the arc measured from its lowest point is #, N= - ; and there- fore the force has the value mgj. Its direction is opposite to", " the displacement ; and therefore its value must be written /y. The acceleration is, then, \u2014 &x. It follows that t d the motion of a particle along its infinitesimal arc is harmonic ; and its period is 2 w\\\u2014 (See page 50.) 9 A method is thus offered for the measurement of g at any point, and therefore for the investigation of the question as to whether it is the same for all kinds of matter and for all amounts. That portion of the above formula which states that the period of a pendulum varies as the square root of its length was deduced by Galileo as early as 1638. Parallelogram and Triangle of Forces. \u2014 Another method of stating that tones act independently is to say that they can be added geometrically like vectors ; and accordingly this theorem is sometimes red to as the \"parallelogram of forces.\" For if tho two forces acting on a particle art; repre- \" & rented liv /'O -iml K\";' **\u2022-\". \" P\u00bbr\u00bblWo,rr\u00bbm of Forces \": UJ V'C\u00bbl sum \u00bbf r<J and PR U PS. I'll ill tin- rut, tli.-ir ft. \"Trianfle of Forew\": th\u00ab fwoMtrlttl tarn of FQ* U'coinrtrical sum is ^ *\"d \u00a76r\u00b0* < '.sen ted by PS, the diagonal of the parallelogram which has PR and PQ for two adjacent sides. It isrvidnit that it tlnj particle is acted on by three forces, 76 MECHANICS PR, PQ, and one equal but opposite to PS, their geometri- cal sum is zero ; for, adding these lines geometrically, a closed triangle is formed. Conversely, if a particle under the action of three forces has no acceleration, they will, if added geo- metrically, form a closed triangle. This theorem is called the \"triangle of forces.\" This principle was first stated, for a special case, however, by Stevin, as early as 1G05. JFio.84. \u2014 Composition of Forcesj_ OA + OB= OC\\_ Or, ~OA= OD+l>A ;_OB= OE +EB^_DA = - EB and OD=EC\\ there- fore OA+OB=OC. The composition of forces is illustrated by the motion of a boat that is being rowed across a river, there being two forces, one due to the oars, the other to the current ; and by many other similar motions. If a cord carrying at its two ends particles whose masses are ml and m2 is supported by two pulleys, and if a third particle of mass m3 is attached to it at some point between the pulleys, as shown in the cut, the system will come to rest under the action of gravity in some definite position. There are now three forces acting at the point O : m^g, vertically down; mlgl in the direction OA, because all that the pulley does is to change the direction of the force of the earth on the first particle ; w2gr, in the direction OB. If OA is a length proportional to the product m^, 77 and OB is a length proportional to m^, their geometrical sum OC will be a vertical line proportional to m^g. (Another mode of considering the geometrical sum of two vectors OA and OB is to see that, if OA is resolved into two vectors OD and ZM, where OD is along the diagonal OC and DA is perpendicular to it, and if OB is resolved in a similar manner into OE and EB, DA = - EB and OD + OE = OU.) Action and Reaction. \u2014 In the case of the interaction of two particles, the force exerted by one on the other is equal and opposite to that exerted by the latter on the former, and is in the straight line joining them. That is, if m1 and ro2 are the masses of the particles, and al and a., their accelera- tions owing to their mutual action, these accelerations are in tin- line joining the particles, and their numerical values are such that mla1 = \u2014 w2a2. This law may be expressed in terms of the linear momenta. It may be written m^ + m^ = 0 ; and this is equivalent to saying that the rate of change of the sum (^m1vl -f w2v2) is zero, where v1 and v2 are the com- ponents at any instant of the velocities of the two particles in the same direction along the line joining than. If the rate of change of a quantity is zero, the value of the quantity itself must remain unchanged. So we may write the formula in", " at any instant be called xr x^ x3, etc. If we assign to each of these distances as a measure of its u importance \" the value of the mass of the particle at that distance, i.e. m1 to xv m% to x2, etc., the mean distance of the system of particles from the plane is mlXl + m^ +... (gee 3Q } ml + m2 + -. This distance may be called x ; and, writing for the total mass of the system, m1 -+- mz + \u2022\u00ab\u00ab, M, we have or MX = m1x1 + But the particles are moving, and each one has a velocity whose component perpendicular to this plane may be called u with a proper suffix. Thus uv this component of the velocity of the first particle, is the rate of change of x1 ; etc. Owing to these motions x does not, in general, remain con- stant ; and calling u its rate of change, we have from the above definition of #, taking the rate of change of both terms of the equation, M- = m^ + m^ +... But the principle of the conservation of momentum states that this sum, m^ + m2u2 + \u2022\u2022\u2022, remains constant so long as there are no external actions ; therefore, M u is a constant or u is constant. Further, the total momentum of the system away from the fixed plane is Mu. In order to describe definitely the position of a particle, its distances from three fixed planes at right angles to each other must be given ; for instance, the position of the corner of a table in a room can be described by stating its height above the floor and its distances from two of the walls that make a corner. Thus, to describe the positions of the particles that are being considered, two other planes at right angles DYNAMICS 81 ich oilier and to the first must be chosen; and we have throe distances for each particle, which we may call x. each with its proper suffix. We can then define a point x, y, z, for which y and z have values similar to that of x\\ Mz = mft If v and w are the rates of change of y and z at any instant, they are also constant, provided there are no external forces. Therefore the geometrical sum of w, v, w* which is the actual linear velocity of the point a;,?/, z\", remains unchanged. (The product of this velocity by M* the total mass, is evidently equal to the total momentum of the system, the momenta of the separate particles being added geometrically.) The point defined by the distances ir, y, z is called the \" centre of mass \" of the system of particles ; and the state- i units just proved in regard to its velocity are seen to be equivalent to stating that \" the centre of mass of a system of particles free from all external actions moves with a constant velocity.\" This is another expression of the principle of the nervation of linear momentum. Tin- acceleration of the centre of mass is. in a perfectly similar manner, seen to be given by three equations, Ma = m,aj + m^a, + \u2022\u2022\u2022, two others, in which at is the rate of change of t*j, etc. l.ut ///,//j is the total force on the particle \\\\li\u00ab^e mas> i^ ///,. and if there are no external forces, this is due to the action of the.it her pai tides, resolved parallel to the direction o! etc. And, since all the internal forces neutralize each other. this sum. j/yij -^m^-\\ ----.must e<|iial /ero. 1 a is -. as was shown in the preceding paragraph. External Forces. \u2014 L\u00ab't us now suppose that the systen acted upon 1>\\ some externa. such as Lrra\\it\\. 1 dole will then he under the action \u00ab.f two sets <>t AMES'S piirnirs \u2014 0. 82 MECHANICS internal and external ; so that if a is the acceleration in some particular direction of a particle whose mass is w, it is due to the sum of the components in this direction of the forces acting on this particle due to the two sets. If, then, we form the sum m^ 4- wz2\u00ab2 + \"'\u00bb ^ *s ^ne sum \u00b0f tne compo- nents in this direction of all the forces, internal and external ; but the sum of the components of the internal forces is zero, and so only the sum of the components of the external forces need be considered. Call this sum X. Then m^a^m^a^ \u2014 equals X. We have just shown, however, that this sum equals Jfefa", ", where a is the component of the acceleration of the centre of mass in the chosen direction. That is, M a = X or a = \u2014. In words, the acceleration in any direc- tion of the centre of mass of a set of particles equals the sum of the components of the external forces in that direction divided by the total mass of the system. We thus see the exact agreement between the properties of a single particle with those of a set of particles, the centre of mass of the set playing the part of the point occupied by the single particle. j\u00a3- The General Problem of Dynamics. \u2014 So far we have con- sidered only the applications of forces to particles; but in nature, of course, material bodies are never in this form. The actual cases of forces are always those concerned with extended bodies; and it is evident that the effect of a force on an ordinary material body depends upon three things: its numerical value, its direction, and its point of application. Tims, as has been explained, if a blow is struck a body, the effect depends upon the point where the blow is applied as well as upon its numerical value and direction; and as a rule both translation and rotation are produced. We shall inves- tigate these two questions separately; that is, we shall deduce first the effect of a force in producing linear acceleration and then its effect in producing angular acceleration. DY.\\AMICS 83 Translation Translation of an Extended Body. \u2014 We can at once apply all our deduetions for a set of particles to a material body that lias a finite volume or to a system of such bodies, if we assume that we ean regard one as made up of particles. The centre of mass of such a body is defined by the same equations as for a set of particles; and its position can be calculated in many simple cases, e.g. a sphere, a uniform rod, etc. The linear momentum of such a body is the sum of the momenta of its parts, and therefore it equals the product of its total mass by the linear velocity of its centre of mass. The accel- eration of its centre of mass in any direction is equal to the sum of the components in this direction of all the external forces divided by the total mass ; and, therefore, the actual acceleration of the centre of mass is in the direction of the geometrical sum of the external forces and is numrric -ally tMjual t<> tin- value of this resultant force divided by that of the total mass. This is the great physical property of the centre of mass. In general language we may say that no matter what the shape or size of the body or where the exter- nal forces are applied, no matter how it turns or spins in its motion, the centre of mass follows the same path with the same velocity and acceleration that would be observed if these force* w. re transferred parallel to themselves and acted on a particle whose mass was that of the body. 1 '\u2022!\u00ab or two illustrations maybe given. If a parti' 1- i- tlr oMiijiit-ly in the air, it will describe the path of a parahola;.similarly. \\\\ li.-n a chair i- thrown in tin- air. or wln-n an acrobat jump* otV a >| i. turning \u2022'\u00bb >'>m\u00ab-rsault, tli.-ir \u2022\u2022\u2022\u2022ntrrs of maw describe paral I liMinKxhrll \u00ab'xpl\u00abMli's in mid air, the centre of mass \u2022 fragments describes a parabola as they gradually fall \u00ab\u00ab.\\\\anl the earth. \u2022.in.l..m Mow is struck a body, its centre of mass moves off in tin- tion <>f the blow. Translation of a System of Bodies. \u2014 It > ve a system of bodies, it may l>o considered, 90 far n is con- 84 MECHANICS cemed, as a system of particles coinciding with the centres of mass of the bodies and having their masses. This follows at once from what has just been said in regard to the properties of a single body. It is this fact that justifies us in using the illustrations we have given in discussing the laws concerning the motion of particles. It also justifies the language used in speaking of mass and its measurement. The properties of matter in rotation will be considered in the next Section. Illustrations of the calculation of the position of the centre of mass. It is easy to calculate the position of the centre of mass of any regular solid provided the matter is distributed uniformly throughout it, e.g. a cylindrical wire, a cube, a sphere, etc., and also of a system of bodies whose masses and the positions of whose centres of mass are known. 1. Uniform rod. \u2014 The centre of mass of a uniform rod is its middle", " point. For, consider the rod as made mnninnniniiufininiiiH 2 up of equal separate particles ; and _ x _ ^ let m1 and w2 be two which are at the ends. Take as the plane of reference FIG. ST. -centre of mass of a one perpendicular to the rod, and let uniform rod: m, and \u00bbit are mi- x^ and #2 be the distances of ml and 7W2 from the plane. Then, by defini- tion, their centre of mass is given by the equation : 2 But m1 = ra2 ; hence x \u2014 X^ _ x%, i.e. the centre of mass of these two particles is halfway between them. A similar statement is true for the other masses which make up the rod, always combining those which are equidistant from the two ends ; and therefore the centre of mass of the rod is this same point. Q.E.D. DYNAMICS 85 The centre of mass of a uniform sphere (or spherical shell ) Jso its eentre of ligure. 2. Uniform triangular board. \u2014 Draw the three medial lines Aa, Bb, Co, connecting the vertices with the middle points of the opposite sides. They meet in a point 0. Since the straight line Bb divides the tri- angle into two equal halves, the centre of mass must lie on it; for the triangle may be considered built up of a great number of strips parallel to the side AC, and as the centre of mass of each of W lies on the medial line Bb, the centre of mass of the entire triangle must lie on it also. Similarly, it must lie on Aa and Cc ; that is, it must be the point 0, their common point Fio. 88.\u2014 Centre of mass of a uniform triangular board. of intersection. 3. A uniform rod, mass 7\u00bb8 = 25, carrying two symmetrical bobs whose masses are ^=15, #^=20; the dimen- sions and distances being as indicated ill the CUt. The centre of mass of the rod itself is its middle point ; which is at a dis- tance 15 cm. from the ends. Take as Fi\u00ab,..T.\u00bb. Ontn-.if rn weighted bar. at its left of\u00bb the plane from which to measure \u00ab tances one perpemlieuhir to the rod Then m, = 15, r, = 5; mf = 20, rf = 20 ; 1113 = 25, ar, = 15; and tl - _ mlxl + m, + m, + m, 75 + 400 + 875 80 14.17. centre of mass must, then, lie HI.nee of 14.17 cm. tin plan, at the end of the rod; and since the bobs are 86 MECHANICS symmetrical, it must lie in the axis of the rod at that distance t'nun the end. masses m 4. A rigid framework lying in a plane ; two bodies, whose 20, 7W2 = 10, are connected by massless wires to a uniform rod whose mass 7H8 = 10; the dimensions being as shown in the cut. Take as the two planes of reference one perpendicular to the rod at its lower end, the other through the rod perpendicular to the two wires. Hence nij = 20, xl = 0, yl = 10, m2 = 10, x2 = 20, y2 = 5, FIG. 40. \u2014 Centre of mass of a rigid framework. So i = \\V = 7.5; y = W=6.25. That is, the centre of mass is a point at a distance 7.5 cm. from the plane perpendicular to the rod at its lower end; and a distance 6.25 cm. from the rod itself in a direction parallel to the wires ; therefore it is at the point 0 as shown. Rotation Introduction. \u2014 Let us now consider the rotation of a material body when it is acted on by a force. A simple case is that of a body pivoted on an axle, e.g. a door. If a push is given it, in general an angular acceleration will be produced ; but if the push is so directed that its line of action passes through the axis, it has no effect on the rotation. It is a fact easily observed that the effect increases as the direction of the line of action of the push is removed farther and farther from the axis. Moment of a Force. \u2014 To be definite, let the cut represent a cross section of the body so taken as to be perpendicular to the axis and to include the point of application of the DYNAMICS 87 force. Resolve the force into two, one parallel to the a the other in this plane section. Only the latter has any effect on the rotation. Let P be the intersection of the plane l", "f rw2r32 -f..., where rx is the radius of the path in which the particle who>e mavs is m^ is moving, etc. This sum is, of course, an arithmetical one. It is called the \"Moment of Inertia\" of the body with reference to the axis of rotation, and is ordinarily written /. Moments of inertia may be calculated for the regular solids provided the matter is uniformly distributed in them, and in many other special cases. Processes of the infinitesimal calculus are, however, required. A few illustrations may be given. Fora sphere of radius H and mass M^ the axis being a diameter, I\u2014 \\MJ&-, for a cylindrical cylinder of radius //and mass.)/. the axis being the axis of symmetry. I=^MR*. (If, however, the axis is a generating line of the surface, /= % ^f/22.) If the length of the cylinder U //. and if the axis p;i>s\u00ab-s ilir\u00ab\u00bbngli its middle point perpendicular t o its length, /= M -f 7 Equation of Motion for Rotation of a Rigid Body around a Fixed Axis. \u2014 \\Ve must next discuss the rotation of H body of definite si/.e and shape round a tixed axis under the action ot an,-xtenial moment. Such a b<.dy is called in dynamics auri_urid\" one. AS before, describe a cross section through the body perpendicular to the axis and including the point the force; and let the external force be re- solved int.. t\\\\,, components, one parallel to the axis, the other in the plane. L,-t the latter i\u00bbe represented in the cut '.M) MECHANICS by F ; let P be the trace of the axis ; and I the lever arm. The moment of the force is Fl. It should be expressly noted that this quantity has the same value wherever the point of application of the force is, provided only that the force keeps its line of action, i.e. provided I does not change. (Thus the force may be applied at Av or Av or Ay etc., in the line of action.) In discussing translation it was shown that the effect of a force on the motion of the centre of mass did not depend upon the position of the line of action nor of the point of application of the force, but. FIG. 44. -A rigid body. * under the action of a force simply, on its amount and direction; con- F. A^ A* A, are points sequently, the total effect of a force upon in the line of action of F. 1 J a rigid body depends upon its amount, its direction, and the position of its line of action with reference to the body, not upon the position in the line of action of its point of application. (If the body is made up of particles so connected as to form a figure of variable size and shape, e.g. an elastic body, I and I would change.) There are also internal forces between the particles; in any actual case there is friction between the material pivot on which the body turns and the body itself ; and, further, the pivot in general exerts a force on the body. The moment of this last force is zero, because its lever arm is zero ; and we shall assume for our present purposes that there is no friction. The moments of the internal forces neutralize each other, moreover, because the forces between any two particles have been assumed to be equal and opposite and in the line joining them, so the lever arms are equal, and there- fore the two moments are equal, but in opposite senses of rotation. The total moment around the axis is, then, that of the external force F\\ and calling, as before, Fl=L, we have DYNAMICS 91 the fundamental equation for the rotation of a rigid body around a tixed axis, ^ where A is the angular acceleration, L is the moment of the external force, and I is the moment of inertia of the body, the last two quantities being referred to the fixed axis. Illustration. \u2014 A simple but important illustration of this be a vertical formula is afforded by a pendulum. Let line through the point of support, and OP the position of the pendulum at any instant, so that the angular displacement is the angle (M OP), or N. If m is the mass of the bob of the pendulum, and r the length OP, the moment of inertia of the pendulum around an axis at 0 perpen- dieular to the plane of motion is mr2. Tl it- re are two forces acting on the pendu- lum hob: its weight, mg^ whose moment around the axis through 0 is mgr sin N, F,O. \u00ab.- simple Pendu- and the tension", " of the string, whose moment is zero. But this moment due to gravity is in an opposite direction to the displacement; hence, in the mr*. Therefore, formula A = \u2014, L = \u2014 m</r sin N and /= marsinff a.,_ ^ = 15 \u2014 = \u2014 -smiV. If the amplitude of vibration is made infinitesimal, we may replace sin ^V by JV; and A = \u2014 -N. Therefore, the motion is harmonic, and the Period is 27r\\/-, as was found before. v Composition of Moments. \u2014 It is evident, since forces act iii'leprmlentlv, that if t here are several external forces acting, A ifl the sum of tin- moments of the separate forces, proper attention brinir irivrn the ul^ehraie signs. 92 MECHANICS It is easily proved by geometry that if a rigid body is acted on by two forces that lie in a plane, the sum of ilic moments about any axis perpendicular to this plane equals the moment of their geometrical sum, provided its line of action passes through the point of intersection of the two forces. For, let the two forces have their lines of action in the lines OA and OB, and let them be represented in direc- tion and amount by the vectors OA and OB. Their geo- metrical sum is then 00, and its line of action is here represented as passing through 0, the intersection of OA and OB. Let P be the trace in this plane of the perpen- dicular axis ; draw the lines PO, PA, PB, and PC. The numerical value of the moment of the force OA about the axis through P is the product of the length of OA by the perpendicular dis- tance from P to this line ; that is, it equals twice the numerical value of the area of the triangle (POA). Similarly, the moment of OB is FIG. 46,-composition of the moments given by twice the area of the tri- of OB and OA about axis at P. -\\ /-T-\u00bb/^r>x j 1-1 c /T7T i angle (POB), and that of OO by twice the area of the triangle (P0(7). But the area of the latter triangle equals the sum of the areas of the three tri- angles (POB), (PBC), and (6>\u00a3<7); and the combined areas of these last two equals the area of the triangle (POA), because all three have bases of the same length, viz.; OA or BC, and the combined altitude of the first two equals that of the third. Therefore the area (P<9<7) equals the sum of the areas (POB) and (POA); and it follows at once that the moment of 00 equals the sum of the moments of OA and OB. (In the cut, these last two moments are in the same direction, viz., that corresponding to rotation counter- clockwise. If P is placed elsewhere, the two moments might be in opposite directions ; in which case they would DYXAMH S 0,3 be given opposite signs, and their algebraic sum must be taken. ) Conservation of Angular Momentum. \u2014 Returning to the general formula, it is seen that A = 0, if L= 0 ; that is, the angular velocity of a body turning on a fixed axis remains constant if either the moment of the external force is zero or if the sum of the moments of the external forces is zero. This is perfectly analogous to the case of translation when ^=0, and is illustrated by the rotation of a wheel whose frietion with its axle can be neglected. If the value of the angular velocity at any instant is written h. the product Ih is called the \"angular momentum\" of the ri-_rid body around the given axis; and the general law may rated by saying that the moment of the force around the equals the rate of change of the angular momentum about the same axis. So, if the total moment is zero, the angular momentum remains constant. If the rotating body is not rigid, the angular momentum is the sum where mr rr /^ apply to one particle of the body ; etc. The statement that this sum is constant when the external moment is zero is still true, however. Several illustrations are worth noting, if the angular velocity of all the particles is the same, the angular moment um may be expressed (wjfj2 -f nyf + \u2014)h\\ and now, if o\\viii\u00abjf to any internal cause the values etc., become smaller, the value of h must increase. This \\\\;is the case with the planets in their early history and is s<> \\\\ith the sun at present. There are forces acting <>n these bodies", ", but their moments about the axes of rotation : and the formula may be applied. As time goes on, the planets and <\\\\\\\\ have contracted owing to internal gravi- s ; and therefore, as proved above, their angular velocities have increased. A^ain, as an acrobat turns a :lt in the air. while at the same time he jumps over 94 MECHANICS an obstacle, his centre of mass describes the path of a parab- ola ; but he can increase his angular velocity by drawing in his arms and legs, thus diminishing his moment of inertia, because there is no moment due to the force of gravity. Illustrations of Rotation. \u2014 If a rigid body is turning on a fixed axis, a moment round the same axis will change the angular speed, either increasing or decreasing it, as is illus- trated by setting in motion a grindstone by means of a crank handle or in stopping one by means of a brake. If, however, the moment is around an axis at right angles to that of the existing angular velocity, the direction of this axis will be changed ; this is illustrated by the motion of a rolling hoop whose upper edge is pushed sidewise, as explained on page 54, or by the motion of a spinning top whose axis is inclined to the vertical. Principal Axes. \u2014 When a material body is rotating on a fixed axle there are in general certain forces and moments which the body exerts on the axle and which are borne by the bearings that hold the axle. If the axis does not pass through the centre of mass, there is a pull on the axle toward this point as it moves in a circle around the axis, due to the reaction of the centrifugal force. Its amount is Mrh?, if r is the radius of this circle, M the mass of the body, and h its angular speed. At any point of the moving body there are three directions, called \"the principal axes at that point,\" such that if the axis of rotation does not coincide with one of them there is a twist on the axle tending to make it turn. This push and twist must of course be withstood by the axle or its bearings. So, if the body is to turn freely, producing no forces or moments on the axle, the axis of rotation must pass through the centre of mass and must be one of the prin- cipal axes at this point. In other words, to make a body maintain its axis of rotation in a definite position and direc- tion other than one which is a principal axis at its centre of mass, a force or moment is required ; and, if no such force DYNAMICS 95 or moment is applied, the position or the direction of the axis of rotation will change. But, if a body is set spinning about a principal axis at its centre of mass, it will maintain its rotation unchanged in every respect, if no moment acts upon it. This last statement is illustrated by the throwing of a quoit, whose axis remains parallel to itself if it is set spinning in the proper way ; by the motion of the earth on its axis, which moves in space parallel to itself (omitting small perturbations and the effect due to the protuberances at the equator); by the motion of projectiles shot out by \"rifled\" guns; etc. Translation and Rotation It is interesting to arrange in parallel columns correspond- ing properties of translation and of rotation around a fixed axis. Translation of a Particle Rotation of a Rigid Body a. mass a. moment of inertia l>. force A. moment of force c. linear momentum c. angular momentum F = ma L = I A Forces act independently. Moments act independently. If F= 0, the linear momen- If L= 0, the angular mo- tuin remains constant. mentum remains constant. [f the direction of the force If the axis of the moment is perpendicular to that of the is perpendicular to that of the motion, th\u00ab- direction of the. motion, the direction of the latter is changed. latter is changed. Motion in General of a Material Body General Description. A material ln.d\\ will, in general, receive both linear and angular acceleration when acted upon by external forces; but these are independent <>f each other. The i. utre of DIMS of the body will re.ri \\ c a linear arc. 96 MECHANICS tion; and, as this point moves in space, the rotation will take place about it exactly as if it were a fixed point in the figure. Several illustrations have been given already ; viz., the motion of an acrobat, that of a chair thrown in the air, etc. If a rigid body is struck a blow at random, its centre of mass will move in the direction of the blow, and the body will rotate, in general ; but, if the line of action of the blow passes through the centre of mass, there is no rotation. Consequently, if", " two lines of action are found such that blows along them do not produce rotation, they must intersect at the centre of mass. Therefore, to discuss completely the most general problem in dynamics, all that is necessary is to know the laws of motion of translation and those of rotation about an axis passing through a fixed point. Resultant. \u2014 There are certain cases in which the resulting accelerations of a body under the action of several forces- might have been produced under the action of a single force; if such is the case, this force is called the \" resultant \" of the others. The action of a single force on a body is to accel- erate the centre of mass in its direction and to cause angular acceleration around an axis through the centre of mass at right angles to the plane including it and the line of action of the force. If the various forces acting in a body all lie in a plane, or if they are all parallel, it may be shown that they have a resultant, with the exception of one case, which will receive due attention. Non-parallel Forces. \u2014 Let the body be acted upon by two coplanar non-parallel forces Fl and F2. Their geomet- rical sum R may be found as usual ; and its effect in accel- erating the centre of mass equals the combined effects of Fl and Fz. But if R is to be the resultant, a position for its line of action must be found such that its moment shall equal the combined moments of F-^ and F^. It has been shown on It Y \\AM1CS 97 '.':! that this is the case if the line of action of R passes through the intersection of the lines of action of.Fj and Fv Therefore the obvious geometrical method of determining the resultant of two coplanar non-parallel forces is to take a plane section through the body so as to include the lines of action of the two forces, prolong these lines until they meet at a point, 0; lay <>ff from 0 along the lines of the forces, distances OA and OB propor- tional to their numerical values; and complete the parallelogram. The di- agonal OC represents in direction, <\u00bbpi\u00abnar forces /\\ \u00bbn,i amount, and position the resultant /2; its point of application may be anywhere along the line OC, e.g. at Q.,...,,,. \u201e Their resultant Is R. In a perfectly similar manner, by continuing the process, the resultant of any number of coplanar non-parallel forces may be determined. It M is the mass of the body; /, its moment of inertia about an axis through tin* centre of inertia perpendicular to the plane of the forces; and /. the perpendicular distance from the centre of inertia to the line of action of the re- sultant It \\ the linear acceleration of the centre of mass is /' /*/ \u2014 ; and the angular acceleration of the body is -\u2014. Equilibrium. \u2014 It is evident that if there are three forces :\u2022_!\u2022 MM the body, Fl% Fy and one e<pial and \u00bb>pp<> aloiiLT the same line of arti.m. they will neiitrali/e eaeh uthei- completely: there will Ix? no accelerat ion of.my kind. Such a state is called one of juilibriuin.\" The general law max then be stated that, if a body is in equilibrium under the n of three non-parallel forces, they must lie in a plane, their lin- [on must meet in a point, and any one must \\MI |*| i-in -i. H \u2014 7 98 MECHANICS be equal and opposite to the geometrical sum of the other two, \u2014 or, what is the same thing, if the three are added geometrically, they will form a closed triangle. (It was by considering the special case of equilibrium of a body on an inclined plane that Stevin was led to the statement of the parallelogram of forces. See page 76.) Parallel Forces. \u2014 If the two forces acting on the body are parallel, their resultant, so far as translation is concerned, must be a force parallel to them, whose numerical value is their algebraic sum. That is, if Fl and F2 are their numerical values, that of the resultant is Fl + F^ if the forces are in the same direc- n tion; but, if they are opposite directions, and i is the greater, the resultant is in the direction of this force, and has the value F% \u2014 Fr Further, in order to satisfy the requirements in regard to rotation, this resultant must have such a position thajb its moment around any axis equals the algebraic sum of the moments of Fl and F2 around the same axis. Describe a plane section", " through the body, including the parallel lines of action of Fl and FY The line of action of the resultant must also lie in this plane ; other- wise the resultant would have a moment about any axis Fm. 48. \u2014 Rijrid body under the action of two parallel forces 1\\ and Ft in the same direction. Their resultant is Ft + Ft. lying in it, which Fl and F2 do not. We shall consider first the case when the two forces are in the same direction. Imagine an axis perpendicular to their plane, and let its trace on the plane be 0. From 0 draw a line OA 0 perpendicular to the lines of the two forces ; if the DYNAMICS parallel force (^Fl -f P., ) is to be the resultant, it must be so placed that its moment around the axis through 0 equals the si 1111 of the moments of Fl and Fv Let its position be indi cated as shown in the cut, its intersection with the line OAC being at B. The condition that B must satisfy is that (F, + F2) OK = F}OA + FtOC, or FlAB = F3BC,i.e.j\u00a3 = %. This may also be expressed as follows: i.e. or These relations are independent of the position of 0, and therefore hold true for any axis. They determine uniquely the line of action of the resultant. If the forces are parallel but in opposite senses of direction, and if JPj is the greater, the resultant is F^\u2014 Fl and is in the direction of FV and it is so placed that its moment around any axis is equal to the difference in the moments of I'\\ and Fr The same formula*. as above apply, giving Fl in these a negative sign. As in the previous case, the acceleration of the centre of mass is in the direction of the resultant and has the value - ^ *; and the angular i celeration has the F 4- F Fto. tt.-Rlfftd body am). of two fttnUM fbTOM /', and value'l T *, where I is the perpendicular distan the centre of mass to the line of Action of the resultant, 100 and / is the moment of inertia about an axis through the centre of mass and perpendicular to the plane of the forces. This process may be continued so as to determine the resul- tant of any number of parallel forces. F.-F, \" Couples. \" \u2014 An ambiguity arises when the two parallel forces are equal but in opposite directions, i.e. when JF1= \u2014 Fr In this case there is no resultant. (On substituting these values in the previous solution, it is seen that, if there were a resultant, its value would be zero, and its line of application would be at an infinite distance.) Such a combination of two equal but opposite parallel forces is called of a \"couple.\" Fio. 50. \u2014 Rigid body under the action a \"couple.\" Their sum is zero, therefore the linear acceleration of the centre of mass is zero, i.e. it has a con- stant linear velocity ; but there is an angular acceleration around an axis through the centre of mass perpendicular to the plane of the couple. Describe a plane section through the body so as to include the two parallel forces, and consider any axis perpendicular to this plane. Let its trace be 0; and from it drop a per- pendicular OAB upon the lines of action of the forces. The sum of their moments, taken contrary to the direction of rotation of the hands of a watch, is F1OB\u2014F1OA = F1AB. This product is called the \"strength of the couple,\" and is evidently independent of the situation of the axis. There- fore, while the centre of mass of the body retains a constant velocity, the angular acceleration around an axis through the centre of mass and perpendicular to the plane of the couple equals its \"strength\" divided by the moment of inertia of the body about this axis. I>Y \\AMICS 101 Equilibrium. \u2014 If a body is under the action of three parallel forces that lie in a plane whose algebraic sum is zero, and the algebraic sum of whose moments around any axis is zero, there is neither linear nor angular acceleration: tin- bndv is in equilibrium. Conversely, if a body is in equilibrium under the action of three, or of any number o/, parallel forces, their algebraic sum is zero, and that of their moments around any axis is zero. A couple can evidently be balanced by another couple that is in the same plane, but that is not necessarily parallel to the first, of equal strength but of opposite sign. Centre of Gravity. \u2014 The most important illustration of parallel forces is furnished by gravity. A body, considered made up of particles, is acted upon by as many forces as there are particles; and these forces are", " parallel because their lines of action join the separate particles to the centre of the earth. Their resultant is their arithmetical sum, and is called the \"weight of the body \" ; its numerical value is the product of the total mass of the body by g, the acceleration of a falling body; and the position of its line of application mav he found as follows: Let m* and >//., l>e the masses of any the action of gravity. Their tun {.articles of the body at an in van- able distance apart, AS. The forces of gravity are vertical and their numerical values are mtf and wi^Mlirir resultant centre of gravity Is C. is Q^+rttg).?, so placed that (ml + m^gQR = (m^PR, wh6TC / ontd line, intersecting the lines of the three forces in the points P, R, Q. That is. (m^m^QR = mlP/f. This line of action \u00ab.f the resultant meets the line joining the particles in a point (7, such that, by similar triangles, AB : CB = PR : QR- 102 MECHANICS But PR : QR, as we have just seen, equals \u2014 J -- 2- There- OR ml m1 + ra2 fore, 41 = or G^S = _i_. 45. It is seen, then, that the point on the line joining the two particles through which their resultant weight passes, depends upon the masses of the particles and upon their distance apart, but is inde- pendent of the position or direction with reference to the earth of the line joining them ; it is therefore a fixed point ori this line. Similarly, in the general case, there is a point fixed, with reference to the particles, through which the line of action of the resultant weight passes, however the body is situated with reference to the earth. This is called the \" centre of gravity \" of the body ; and it is seen from the above equation that it coincides with the \" centre of mass.\" N.B. \u2014 The above proof of the existence of a centre of gravity and of its coincidence with the centre of mass depends upon the fact that \" g \" is a constant for all amounts and all kinds of matter. Equilibrium The state of equilibrium of a body has already been defined as that in which there is no acceleration, either linear or angular; and the obvious conditions are that both the sum of the components of the forces in any direction and the sum of the moments around any axis should be zero. (We may speak in the same way of the equilibrium of a system of bodies.) If the body is at rest with reference to any standard figure, \u2014 e.g. a book lying on a table is at rest with reference to the table, \u2014 the equilibrium is called \"statical\"; while, if the body is in motion, \u2014 unaccelerated, of course, \u2014 the equilibrium is called \" kinetic,\" e.g. a sphere rolling on a smooth horizontal table. There are several kinds of equi- librium, however, depending upon what changes in the motion of the body (or system of bodies) take place when a slight impulse or blow is given it. 103 Stable. \u2014 If the equilibrium of the body is such that, as a result of the impulse, it does not continue to move away from its former position, but makes oscillations about it, it is said to be \"stable.\" This is illustrated by practically all bodies in nature that are in equilibrium. An ordinary pendulum when at rest, a block when it rests on a table, a body hang- ing at rest from a spiral spring, etc., are illustrations of statical stable equilibrium. If an ellipsoidal body is set spinning around its longest or shortest axis, the motion is stable. Unstable. \u2014 But if the equilibrium is such that as a result of the impulse the body departs farther and farther from its former position, it is said to be \"unstable.\" This is illus- trated by a Wrd balanced in a vertical position on one corner, by a conical body balanced on its point, etc., or by an ellipsoidal body spinning around its intermediate axis. It is obvious that when a body is in unstable equilibrium Fio. M. - G la the centre of gravity ; If the vertical line fall* inside the base \u00a3<?, there Is equilibrium. w for impulses in some din \u2022< 'tions, it may be stable for others; and again a body may be stable for an extremely small impulse and unstable f..r a larger one, so that there are \u2022\u2022degrees of Stability.\" Thus a block shaped as shown in the rut and resting on a hori/.ontal support is in stable equi- librium, because the force of gravity acting vertically do\\\\n through the centre of Lrnmty is balanei", "-d b\\ an upward force due to the table. P.iit this last fOTOC CM pO88 through the 104 MECHANICS centre of gravity and therefore neutralize gravity completely only so long as the line of action of the force of gravity falls inside the area of contact between the block and the table. If the shape of the block is so changed that this line of action approaches the edge of this area, the stability becomes less and less \u2014 for an impulse in the proper direction. When the line of action of the weight falls outside the edge, this down- ward force forms a couple with the upward force due to the table, and the block will turn over. Neutral. \u2014 Another state of equilibrium is also recognized; namely, that in which, when an impulse is given the body, the change in motion produced remains permanent. This kind of equilibrium is called \" neutral,\" and is illustrated by a sphere or a cylinder lying on a smooth horizontal table, by a body pivoted on an axis passing through its centre of gravity, etc. It is evident that when the condition of a body in unstable equilibrium is disturbed, it passes over into either a stable or a neutral condition; and, as disturbances are always occur- ring in nature, the condition of unstable equilibrium can exist for only infinitesimal intervals of time. Principle of Stable Equilibrium. \u2014 Any disturbance of stability must produce a reaction which tends to restore the body or system to its previous condition ; and this principle can be applied to any stable condition, whether it is a purely mechanical one or not. Consider some illustrations of stability. (1) A body hanging suspended by a spiral spring is in stable equilibrium. If a blow downward is given it, the initial velocity will be decreased owing to the increased tension of the spring. Hence, if the tension of a stretched spiral spring is increased by any means, it will raise the suspended body. (2) An iron bar surrounded by some medium, e.g. water, at a constant temperature is in stable equilibrium ; for if its temperature is suddenly increased in any way, the tendency will be for it to return to the tern- DYNAMICS 105 perature of the surrounding medium. Now, when the tem- perature of an iron bar is increased, its length is increased ; but this act of increasing in length produces a tendency for tin- bar to return to its former temperature. That is, if an iron bar is stretched by mechanical means, its temperature will fall. (\u2022>) Just the opposite effect happens with a piece of rubber cord from which a weight is hanging. When its temperature is lowered, it elongates; consequently stretch- ing a rubber cord raises its temperature..*\u00bb{. Illustration of tlirvr kiml.s of \u00ab-<|iillit>riiiiii. Work and Energy Measurement of the Effect of a Force. \u2014 In the previous sections of Dynamics we have; considered, generally spcak- \u2022nly one property of a force, viz., the fact that it pro- duces a change in momentum. It was shown, however, on page 45 that when a particle is m\u00ab>\\in<r in a straight line with a constant ROOeforaf i\u00bbu. this.juaiitity could be expressed in two ways : where x, is the speed..|' the particle at th-- instant 7\\ when 106 MECHANICS it has reached a point at a distance x1 from a fixed point of reference in the line of motion ; and \u00ab2 is the speed at the instant T2 and position xv Multiplying each of these values - f_ of a by M, the mass of the moving particle, we have the force. Thus, X2 - X 1 \u2022> J ~ The former is the ordinary expression for the value of a constant force, stating that it equals the change in the linear momentum in a unit of time. The product/ (T2 \u2014 ^i) ^s called the impulse of the force ; so this formula expresses the fact that the impulse of the force equals the change in linear momentum of the particle. If the force varies, we must understand by f (T2 \u2014 T^ the sum of a series of terms each of which is a force multiplied by its time of action. Definition of Work and Kinetic Energy. \u2014 The latter for- mula, however, is a new expression. Suitable names have been given its terms : \\ ms2 is called the \" kinetic energy \" of the particle whose mass is m when its speed is s ; / (z2\u2014 x^) is called the \" work done by the force \" / in the distance x2 \u2014 xv provided the speed is increasing, or the \" work done against the force \" / if the speed is decreasing \u2014 thus this equation reads : either, \" the work done by the force in the distance x2 \u2014 xl equals the increase of the kinetic", " energy of the moving particle in that space \" ; or, \" the work done against the force/ in the distance xz \u2014 xl equals the decrease of the kinetic energy of the moving particle in that space.\" Several things should be noticed : 1. The distance #2 \u2014 x1 is measured in the line of action of the force ; if the line of motion makes an angle N with the line of the force, the work is the product of F cos N and O2-*i)- 2. Tho idea of work involves both force and motion in the direction of the force / no work is done unless there is motion ; 107 and this motion must be in tin- direction of the force. Tims, a pillar supporting a building does no work, neither does a horizontal table on which a ball rolls. 3. In the expression for the kinetic energy # is the speed, not the velocity ; in other words, kinetic energy does not depend upon the direction of motion. This is evident, be- cause to produce a change of direction (and no change of speed) the force must be at right angles to the direction of the motion, and therefore, by what has just been said, no work is done. 4. The same relation between work and change in kinetic energy holds true even if the force is not constant ; because we can in that case consider the force as constant for a short distance, during which the formula holds, thru assume another constant value for a short distance, etc. The total work done \u2014 that is, the sum of the amounts done by the separate forces \u2014 will then equal the total change in the kinetic energy. Illustrations. \u2014 Let us consider several cases of motion from the standpoint of both momentum and kinetic energy. When a ball is thrown, the momentum gained depends upon the impulse of the force, i.e. upon the time during which it acts; while the kinetic energy gained depends upon the work -lone by the force, i.e. upon the distance through which it aets. If a bullet is set in motion 1>\\ a compressed spring, {is in a toy gun, the spring in relaxing axerta an impulse ami also does \\\\ork. producing momentum and kirn-tic energy. As a bullet fired from a rille enters a wooden target, the distance it penetrates depends upon its kinetic energy ; the interval of time required to bring it to rest depends upon its momentum. The destm. ti\\e power of a moving l\u00bb\u00bb>'ly is due to its kinetic energy; its power. to its moment uin. Conservative Forces. \u2014 For the time being we shall limit ourselves to the consideration of what we m.i\\ call purely 108 Ml < IIANICS mechanical forces, e.g. those produced by gravity, by elastic bodies which are deformed, such as a bent bow, compressed spring, etc. In all these cases we make the assumption, which is based upon the results of experiments, that the work done by a force during a displacement from one point to another depends upon the initial and final positions, not upon the path followed ; such forces are called \" conservative \" for reasons that will soon appear. This may be illustrated by the force of gravity. Describe two horizontal planes PQ and P1Q1 at a distance h apart. Draw an oblique line AB, making an angle N with the ver- tical line AO\\ its length is Let a particle of mass cos^V m move under the action of gravity down BA ; at any point the force in a vertical direction is mg, and therefore the component along the line BA is mg cos N. The work done, then, between B and A is P, * < 1 mg cos N \u2022 \u2014 or man. cosN Similarly, if the particle in passing from B to A _ _A~ -Q had followed the path EQ FIG. 64. \u2014 PQ and I\\Ql are two horizontal planes at an(J (JJ^ ^}ie WOrk done would nave been the a distance apart, h. same ; for, in the path BO the motion is perpendicular to the force of gravity, and therefore no work is done ; while in the vertical motion OA the work done is mgh. Since any line curved or broken may be considered made up of straight lines, it is seen that when a particle of mass m falls from any point in one horizontal plane to any point in a lower parallel plane at a vertical distance A, the work done by gravity is mgh, and is independent of the path. This is a consequence of the fact that the force of gravity is vertical and is con- stant in amount at all points near the surface of the earth at any one locality. I)Y \\AMICS 109 Potential Energy. \u2014 Since, then, in all the cases that we are to consider at present the work done by forces depends upon only the initial and final points, we may write whore Fis", " a quantity whose value depends simply upon the point considered. Thus, our fundamental formula becomes or This means that during the motion the quantity F-f Jw*2 remains constant. Consider an illustration: a ball being set in motion by a compressed spring. The above formula states that as the speed of the ball increases, the value of V decreases; or, vice venta, if the ball is made to strike the spring and compress it. as the speed of the ball decreases, tin- value of V increases. Again, in the case of a falling body, as the speed increases, the value of V decreases; and, if tin- ball is thrown upward, as its speed decreases, the value of V increases. The quantity whose value is V is called \"potential energy\"; and it is seen by the above illustrations that when a spring is compressed, the potential energy increases; when a particle is raised vertically upward, the potential energy increases ; and. conversely, when the spring relaxes or the particle falls, the potential ener-\\ decreases. We say that the 4* com pressed spring has potential energy,\" and that the \"system of the particle and the earth has potential energy\"; or, in the latter case, more simply, \"the particle ; IMS potential energy\"; but these words are only a description of the expei i men ts just mentioned. In the case of the p article a -id t lie earth, the former has not changed its size, its shap IM, or any of its physical properties; it has therefore n,.t l.rrii clian'_:\u00ab'd nor has anything been added to or taken from it; but it> relation to the earth has been 110 MM'1IA\\ICS altered. The same is true of the particles of a compressed spring; their relative positions are changed. In a similar way, a twisted wire, a bent bow, a clock spring that has been wound up, etc., all have potential energy ; and, in gen- eral, a body or a system of bodies has potential energy if the particles composing it are in such a condition that a force is required to maintain it. The formula F(x% \u2014 x^) = F\"t \u2014 F^ gives a means of cal- culating only the change in the potential energy ; and so what is meant by \" the potential energy for a given position or condition \" is the work required to bring the system into that condition from some other one which is taken as the standard one. Thus, in dealing with gravity, it is customary to reckon from the surface of the earth ; and the potential energy of a particle of mass m at a vertical height h above the earth's surface is therefore mgh. In compressing a spring, the standard condition is that when the spring is entirely relaxed ; and since experiments show that the force which the spring exerts at any instant when compressed varies directly as the amount of compression, this force may be written ex, where x is the compression and c is a constant to indicate the proportionality ; but, as the spring is com- pressed more and more, the force varies, and therefore during the compression from 0 to #, the mean value of the force is \u00a3 ex (the average of 0 and ex \u2014 see page 32) ; and the potential energy of the compressed spring is the prod- uct of this mean value of the force by the distance, i.e. it is \\cx*. Conservation of Energy. \u2014 If a particle has kinetic energy, or if a system has potential energy, it is in a condition such that it can do work. A falling body can compress a spring or bend a board, thus overcoming a force ; or it may strike another body and change its speed, thus doing work also. Similarly, a bent bow may change the speed of an arrow or it may raise a body up from the earth. Two things should DYNAM1' 8 111 l)e noted : (1) There are two ways of doing work corre- sponding to the two types of forces referred to on page 68, namely, producing acceleration in a particle, in which case it gains kinetic energy ; or overcoming some opposing force, e.g. gravity, in which case the system on which the work is done gains potential energy. (2) If a particle or a system does work on another particle or a system, the latter gains energy and the former loses energy. The exact relation between this gain and loss is stated in the general formula V+ \\ m& = constant, which is true only for so-called \"con- servative\" forces. (See page 108.) This says that, if a tern has both kinetic and potential energies, the sum of the two remains constant; if one decreases, the other increases by an equal amount. This is a special case of the principle of the \"Conservation of Energy.\" (See page 115.) Tims, if one part of a system does", " work on the other, e.g. a com- pressed spring and a ball, a bent bow and arrow, one loses a certain amount of energy, the other gains it. Similarly, in the system made up of the earth and a falling body, the potential energy decreases by an amount equal to that by which the kinetic energy increases. If there are no changes in the potential energy of a system, the total kinetic energy does not change, the loss in one part equals the ^ain in another; an illustration is ^iven by the impact of two billiard balls. (All cases of impact between inelastic bodies are excluded t'r\u00abun considerat i\u00ab>n here, because the forces acting durin^'the impact of such bodies do not satisfy our assumption made above in regard to mechanical forces. As will be shown shortly, part of the energy in the case of impact of inelastic bodies disappears from view and is mani- to our senses in the production of heat effects, such as rise of temperature, etc.) Unit of Work and Energy. \u2014 The act of transfer of energy t'n.m one particle or one system to another involves what we called -work.\" Its numerical \\alue is. tVom iis deli- 112 MECHANICS nition, the product of the values of the force and the dis- placement in the direction of the force. Work, kinetic energy, and potential energy are, then, all similar quantities and are all measured in terms of the same unit. On the C. G. S. system, this is the work done by a force of one dyne in a displacement of one centimetre, or the energy of a particle whose mass is two grams moving with a speed of one centimetre per second ; it is called an \" erg \" ; but it is too small for practical purposes, and so 10,000,000, or 107, ergs is the unit in common use ; it is called a \" joule,\" in honor of the great English physicist who did so much to teach correct ideas in regard to energy. Another unit often used is the \"foot pound,\" or the work required to raise in a vertical direction a distance of one foot a body whose mass is one pound. This, then, equals approximately 1.356 joules, assuming g to equal 980. Motion in a Vertical Circle. \u2014 One case of transfer of energy deserves special notice ; it is that of a particle sus- pended by a massless cord and making vibrations in a vertical circle under the action of gravity. As it swings through its lowest point, it has its greatest kinetic energy and its least potential ; and, as it gradually rises, the former decreases and the latter increases, until at the end of its swing it has its energy entirely in the potential form. If the arc of vibration is extremely small, this particle is a simple circle under the action pendulum ; but even when the arc is large, we can deduce certain general laws. If the particle moves freely from Pl downward along the circle of radius r, starting from rest, its speed at the bottom A is such that s* = 2 (j QtA, where Ql is the projection of P1 on the vertical diameter. DYNAMICS 113 But, by geometry, we know that \\vlit-re Pj-A is the chord, and r is the radius. Hence, s = J\\ j y i 2 If tlie particle had moved from P2 down the circle, its 2 r'speed at the bottom would have been So the speed at the bottom of the path is directly propor- tional to the length of the chord of the arc through whieh it falls. Work and Energy in Motion of Rotation. \u2014 When work is done in producing angular acceleration of a rigid body about a fixed axis, somewhat different expressions for the work and the kinetic energy may be deduced, whieh are more useful. Let a plane section be taken through the body, perpendicular to the axis, and pass- ing through the point of application of the force, and let the force be resolved into one parallel to the axis and one parallel to this plane. In the cut let P be the trace of the pivot; F, the com- ponent of the force, whose point of appli- eation is O and whose line of action is OA ; and /M, the lever arm of length /: and call the length of the line PO, r. In the motion neither the component of the force parallel to the udi nor the force exerted by the pivot on of \u2022 rljri.l body pei-pen-.llr.ilar to ftxU at /\u00bb;/\u2022!\u2022 the component of the force in thl. \u00ab plane. '\u2022\u00ab. \u2014 Plane section the body does any work. ( \\\\'e may suppose that gravity does not act ; or, the axis may be", " considered as being ver- tical. ) The force /\u2022' may be resolved into two components, nmiot \u2014 8 114 MECHANICS urn- along OP which therefore does no work, the other per- pendicular to OP. Calling the angle (OP A) N, this last component has the value.Fcos.ZV; and, as the body turns, this component remaining always perpendicular to the line OP, the work done equals the product of.FcosJVand the length of the arc described by the point 0. This arc equals the product of the angle turned through and the length of the radius PO. Calling this angle M, the work is then rMFcos N; but r cos N = 7, the lever arm ; and Fl = L, the moment of the force F around the axis through P. Conse- quently the work equals the product of the moment of the force by the angular displacement. The kinetic energy of the rotating rigid body, which we may consider made up of particles, is the sum of the. energy of these taken separately. A particle of mass m1 at the distance rl from the axis has the linear -speed rji, if h is the angular speed of the body, and therefore the kinetic energy 2 mi(ri^)2 or J m^ffi. The sum of the energy of all' the particles is then \\(m^r^ + ra2r22 -+- --.)A2 or J/^2, where I is the moment of inertia about the given axis. (Notice the exact correspondence of the values for work and energy in translation and rotation.) Properties of Potential Energy. \u2014 The connection between potential energy and force deserves consideration. The formula of definition is F(x2 \u2014 x^) = }\\ \u2014 Vv This is equiva- lent to saying that, if by a displacement from a point 1 to a point 2, the potential energy decreases, i.e. if Vl>Vy F is positive, and is, therefore, in the direction from point 1 to point 2 ; so there is a force acting in the direction of the displacement, whose value is \u2014 -. This is illustrated by 2*2 - X\\ the force of gravity, that of a compressed spring, etc. Another interpretation is this : in a system left to itself under the action of its own forces, motions take place, if at all, in such a manner as to produce a decrease in the poten-. \". 1 1 tial energy. If a system is in equilibrium, the total force in any direction is zero, and therefore any slight displacement may be produced without there being any work done ; hence the potential energy remains constant during these displace- ments. This shows, then, that at a position of equilibrium, the potential energy has either a maximum or a minimum value. If the equilibrium is stable, it is not difficult to prove that the value of the potential energy is a minimum; as is illustrated by a pendulum being in stable equilibrium when it is hanging at rest at the bottom of its path. Power. \u2014 The rate at which work is done, that is \u2014 if this rate is uniform \u2014 the work done in a unit time, is called the \"activity\" or \"power.\" On the C. G. S. system the unit is \" one' erg per second \" ; but for practical purposes \" one joule per second \" is taken : this is called a M watt.\" in honor of the great Scottish engineer who made so many impn>\\r- mcnts in the steam engine. Since work equals the product \"f force by displacement or of moment of force by angular displacement. power equals the product of force by linear speed or of moment by angular speed. POW.M- is also often measured in terms of a unit called \"one horse power,\" which is defined to he :J:},000 foot pounds per minute. This equals 746 watts approximately. Other Forms of Energy Conservation of Energy. \u2014 There are many other forms forces than those which have been considered. S hese correspond to forms of potential energy, such as th< surface tension of liquids, forces due to electric charges ;o magnets, etc. ; others, h\u00ab\u00bbwe\\rr. d<> not Am.\u00bbn^ tin- latter the force of frietion is the most important. I force.anifest whenever twn pieces of matter in contact with each other move relatively: and in all the cases of motion i-sed the condit \u00ab'SSed as to assume the eomph-t.- lion. It is* a force 116 MECHANICS that always opposes the motion : and its numerical proper- ties will be discussed later. Let us consider several cases of friction and the immediate results. If two blocks of ice are rubbed together, some of the ice is melted ; if the bearing between a wheel and its axle is not well lubricated so as", " to avoid friction, there is a \"hot box,\" the bearings become hot and the parts expand; if a paddle is stirred rapidly in water, thus producing friction between different currents of the water, the temperature of the water rises and it will finally boil ; if an inelastic body, like a piece of lead or putty, is deformed, different layers move over one another, there is friction, and the temperature rises. These various changes \u2014 melting, boiling, rise in temperature, and expan- sion\u2014 are called \"heat effects,\" and will be discussed more fully in the next section of this book ; but what is of funda- mental importance here is to recognize that these effects are all produced when work is done against friction. It will be shown later how we can measure these effects numerically; and experiments show that their amount is proportional to the work thus done against friction. In doing this work, energy is lost by the body or system doing the work ; and so it is natural to assume that the heat effects are manifestations of the addition of energy to those parts of the bodies which are directly affected by the friction, namely, the most minute portions \u2014 in certain cases, the molecules. This assumption is completely supported by all experiments and observations. We have seen that in the transfer of energy in purely mechanical rases. Him- is no loss \u2014 \\\\-lui um- liodv loses, an- other gains; so \\ve extend this idoa t\u00ab> ;ill processes in nature, and state our belief that in no case is there any loss in energy. It may be present as energy of bodies of sensible size, of mole- cules or their parts, or of the ether. This statement is called the \"Principle of the Conservation of Energy.\" Nature of Potential Energy. \u2014 A few words more should be said in regard to what is meant by \" potential energy.\" As UYNAMICS 117 have used the expression, it describes a condition of a body with reference to other bodies or of the parts of a b\u00bb\u00bbdy witli ivtVivnce to each other, which is primarily concerned with the idea of force and its production. We cannot ex- plain it in terms of such simple quantities as intervals of \u2022e or time and mass. We understand, however, its ti formation into kinetic energy; and it is possible that it is the manifestation to our senses of the existence of the kinetic energy of portions of a medium which has inertia and which is intimately connected with ordinary matter, but which does not appeal directly to us. Friction External and Internal Friction. \u2014 As has been said in dis- cussing different methods of doing work, there is a force that opposes the relative motions of any two pieces of matter that are in contact; this is called the force of \"friction.\" liscussion does not properly belong to mechanics; but it is convenient to give it here. Distinction must be made between two kinds of friction, internal and external. The latter is illustrated when solid body is made to move in contact with another, or when layer of a fluid flows past another; e.g. a block of wood moving <>ver a tal\u00bble, currents of water produced in a vessel by stirring a paddle in it, currents in the air produced by blowing. In all these cases the relati\\. motion is soon stopped unless some force maintains it. Friction between moving layers of fluids is said to be due to \" viscosity.*' Internal friction is illustrated when a solid body is deformed in an\\ \\\\.iv, for in every case, to a greater or less extent, portions of tl body move over each other. The only case < tiou \\\\hieh will be considered now is that of one solid moving over another; the discussion of viscosity and of internal friction is defer 118 MECHANICS Sliding Friction. \u2014 The most important cases of friction between solid bodies are those when the two surfaces in con- tact are plane and when one body rolls on the plane surface of another. It will be seen that the explanation of the fric- tion in these two cases is quite different. Consider the motion of a rectangular block over a plane. Let AB be the section of the plane by the paper and CDEF be that of the block. Let a force whose value is F produce acceleration of the block parallel to the plane surface ; there is a force of friction opposing this, call its value F^\\ then the total force producing the acceleration is (F \u2014 F^), and if m TTf TTf is the mass of the moving block, its acceleration is ^. m If there is no acceleration, and therefore the speed remains constant, the applied force exactly balances the friction, F = Fl ; and we have thus an experimental method for determining the force of fric- Fio. 57. \u2014 Motion of a block along a rough tion between two given mate- plane. F is the force producing the", ", the power may be measured. There arc many modifications of this instrument, the most important one being the substitution of liquid friction for solid. A set of blades is fastened to the shaft, which is then inclosed in a cylinder containing water and so supported that it may be kept from turning by suitable levers. As the shaft revolves, currents are produced in the water and a moment is required to keep the outer cylinder from turning ; if L is the amount of this moment and h the angular speed of the shaft, the power it furnishes is Lh. (The energy goes into heat ellV< Machines General Principles. \u2014 A mechanism rnn>istin^ of rigid or iiiextensihle parts by means of which energy is transferred from one point to another is called a maehine. A force applied at one point overcomes another force applied at a different point. Thus, the lever funning a pump handle is a maehine, hecausc when work is dune by the man pumping at one end, the lever 'does work in raising the water at tin- other end. A pulley over which a cord passes is a machine, because if one end of the eord is attached to a body and the other is held by a man, the latin- l.\\ d-.in^ work pulling on tin- c.\u00bbrd may raise the body off the ground. In all machines there are parts which move over each other and therefor.- produce friction. ( '..nsc.jucnt ly, a 'nine never delivers as much energy as it receives; i ic hitter is spent in tion u\u00bbd is therefore : \" in heat effects. A -a in. if the foffM produces ieceleration of the working parts of the machine, part of the work done is thus spent in producing kinetic energy not in overcomiii ie ; while, if the velocity of the n parta of the machine is decreasing, they t! \u2022* do 122 MECHANICS work in helping to overcome the opposing force. Therefore, in the discussion that follows, the effect of friction will be neg- lected; and we shall assume that there is no acceleration. Con- sequently, the energy furnished the machine equals the work it does. Although this is the case, the force which the machine exerts need not be the same in amount or in direc- tion as that which is exerted on it ; for the distances througli which these forces are displaced need not be the same. In fact, the distinct object of a machine is to obtain in return for a given force a larger one in a suitable direction. The problem, then, is to determine the connection between the force that is doing work on the machine and the force that is being overcome by the machine. The ratio of the latter force to the former is called the mechanical advantage of the machine. There are two general methods for the dis- cussion of this problem. One is to consider the energy rela- tions ; that is, to express by equations the fact that the work done by one force equals that done against the other. The other method is to express the fact that the two forces are keeping the machine in a condition of equilibrium, because we have assumed that there is no acceleration. We shall discuss a few simple machines and deduce the ratio of the two forces. The Lever. \u2014 In its general form the lever consists of any rigid body capable of rotation around a fixed axis ; but most levers in actual use consist of straight rods. If a force p Jj is applied to this body at any point, FIO. 6i. - Equilibrium of a it will have a moment around the rigid body which is pivoted at ^^ ^ t p j where j js the lever P, when under the action of two 1 forces^ and F+ Principle of arm ; and, if there is motion, the work done by the force is the product of this moment by the angular displacement. If this moment does work by overcoming another force F% acting on the DYNAMICS body, whose lever arm is lv the work done is FJ2 multiplied by the angular displacement. But since the body is rigid, the angular displacements of all points are the same; and, therefore, since the force due to the reaction of the pivot does no work, the work done by Fl equals that done against Fv i.e. or I'This relation also expresses the fact that the body is in equilibrium under the action of two forces whose values are I-\\ and /-'2, and which are properly directed. Illustrations of the lever are given by a pump handle, a crowbar, a pair of scissors, a pair of tongs, nutcrackers, etc. The formula for a lever was given first by Archimedes in the simple case of a straight bar acted on by two weights. The more general 18 lirst solved by Leonardo da Vinci", ". The Pulley. \u2014 A pulley consists of a circular wheel which has a groove in its edge to hold a cord, and which turns on an axle supported by a framework called the H)lo< k.\" A pulley is used in two ways: in one its block is fastened to a firm support, and in the other it is kept from falling by being ried in the bi^ht of a cord passing round the whet-1. The former arrangement is called a \"fixed pulley\"; the latter, a \"free\" one. In the former case, if a force Fl is applied to one end of the cord and does the work Fio. 09. \u2014 A siin- pie form of pulley.! by producing the displace- ment xv it can overcome a force F% at the other end whose displace- ment is xy provided F^ = / FM.68.-Aflz0dpullry. But if the cord is inextensible, This is the same relation that follows if the pulley is in and IK-IK.- Fl=F 1:24 MECHANICS equilibrium under the two forces Fl and Fv Thus a pulley simply changes the direction of the force. As an illustration of a free pulley, let the arrangement be as shown in the cut : the cord, one end of which is fastened to a fixed support, passes under the pulley, and its two branches make an angle N with the vertical ; to the free end of the cord a force Fl is applied which balances a force F2 applied to the block vertically downward. Since a pulley sim- ply changes the direction of a force, there are two upward forces acting on the block owing to the two branches of the cord, which are equal in amount but inclined at equal angles N to the vertical on opposite sides. If the end of the cord is displaced in the direction of the force a distance x^ the work done is x^F^ ; but the vertical displacement of the block owing to this motion is x2, where x1 = 2 x% cos N. The work done against FIG. 64. \u2014 A free pulley. the force F% is F^ ; and therefore Flxl = or 2 F1 cos N= F2. The proof of this relation between xl and x2 is as follows: Let AOB be the original position of the cord ; and A CD its final position after the displacement ; so that A D = A O -f OB. Lay off AC equal to OB, thusjnaking CD = AO\\ and drop a perpendicular RE upon AD. Then the vertical displacement of 0, i.e.. r2, is A O cos N\\ and the displacement of B parallel to the line OB, i.e. xr is ED. But it is evident from the figure that CD = A O = CB ; and hence ED = CD + CB COS (BCE) = A 0 (1 + COS 2 N) = 2 A O COS2 N. FIG. 64a. \u2014 Geometrical treatment of the motion of a free pulley. B Is the end of the cord before displace- ment ; Z>, its position after. II. Mice, [Mica x = 40 cos N. a ;, = L'This is, again, the condition that the block should be in equilibrium under the three forces. If, as is usually the case, the two branches of the cord are 0, cos-ZV=l; and 2 Fl = Fv i.e. themselves vertical, Pulleys are combined in many ways; but all forms can be easily explained on the above principles. Thus let there be a free pulley and a fixed pulley with two wheels turning inde- pendently on the same axle ; let a cord, one end of which is fastened to the block of the free pulley, be run over them as shown; and let a vertical force act downward on the free pul- ley. Let the pulleys be so *mall I'Minpared with their ;>arl that t In- \u00ab. \u2014 Combination of pullovn: appcr flx\u00abd piilli-v li\u00bb\u00ab \u2022 -; |..\u00abrr |>ull various hranrhi's of the cord between the pull dlel. /\u2022. applied in the fn-e end of the \u00bb-\u00ab.rd and pro- duces a display-men! /-p the free pnll.-v \\\\ill rise a vertical distance xtt=}-\u00b1\\ and then ical force F~ can be overcome, wlm-r value is!-\\ 8 /' \u2022.. \u2022 /' - I, 126 MECHANICS This is evidently the condition that the free pulley should be in equilibrium under the action of Fl and Fv The formulte for pul", "leys were first deduced by Stevin about 1600. The Inclined Plane. \u2014 If a body of mass m1 is moved upward along an inclined plane whose angle of inclination is JV, the force m^g sin^V must be overcome. So, if a body of mass m1 lying on an inclined plane is joined by an inextensible cord, parallel to the plane \u2022 and passing over a pul- ley, to a body of mass 7W2 which hangs free, the weight of the latter is just sufficient to balance that of the former provided that m^g = m^g sin JV, or w2 = m^ sin N. FIG. 66. \u2014 Principle of the inclined plane. If the first body is displaced up the plane a distance xv the second will sink an equal distance xv Hence the two amounts of work, m^g x sin N and m2gxv are equal. That is, ml sin N= mv as before. The theory of the inclined plane was also given by Stevin. The Screw. \u2014 If a piece of paper the shape of a right-angle triangle (ABD) is cut out and wrapped around a circular cylinder, the edge AD being kept perpen- dicular to the axis of the cylinder, the hypot- enuse AB will trace a spiral line on the surface of the cylinder. If a groove is cut fol- i., i. -i. FIG. 67. \u2014 Principle of the screw. lowing this line, we \u2014 -2-TT-R- -*> have what is called a cylindrical \"screw.\" The portions of the cylinder between the grooves are called the \"threads\" of the screw ; and the distance from the edge of one thread DY\\ 1-27 to tin- corresponding edge i.f the next measured parallel to the axis of tin- cylinder is called the \"pitch\" of the screw. If a distance equal to the circumference of the cylinder is marked off on the line AD, beginning at ^4, and a perpen- dicular line erected, the length included between the base and the hypotenuse is evidently equal to the pitch of the screw. If R is the radius of the cylinder, tin- length of the circumference is 2irR\\ and the vertical distance on the triangular piece of paper which corresponds to this horizontal distance is given by the proportion, vertical distance : 2 irR = BD : AD; or, the vertical distance equals ZirR tan N. This, then, is the pitch. I f a spiral groove Is made on the inner surface of a hollow cylinder, we have what is called a \"nut\"; and by cutting the grooves of a screw and a nut at the same pit eh an suitable depths, the screw will lit inside the nut. Then, if the nut is held ; -crew can advance through it by a rotation on its axis: or, if the screw is lix.-d. the nut can only advance along it by a rotation round its axis. In either case, for one complete rotation, th<- ad\\anee equals the pitch of the screw. If a moment L is applied to the screw. : lifting JM*. turning in a!i\\.-d nut, and if by means! -w.n,i nut \"I (his a I'M overcome, which is so applied as to oppose the advance of the screw, the \\\\ d'.ne by the moment in on.- n.tati.m is the product of the moment by the an-h- L'TT. /.-. L'TT/.. and that done againgt the fon \u00ab\u2022/\u2022', / TrR tan N. If the im-mrm A is due to a 7\u00abTj applied to the circumference \u00ab>f th\u00ab n such a din \u00abti.. a as to have the lever ann ///, onsequently r\u00ab tan N, 128 or so that Fl = F., tan F, tan Screws are used in letter presses, cotton presses, lifting jacks, etc. It is thus seen that by these various machines a force may be magnified as much as desired, with a corresponding de- crease in the displacement ; that the direction of a force may be changed ; and that a moment may produce a force. Chemical Balance. \u2014 This is not a machine in the ordinary use of this word, but is an instrument involving the principle of the lever, which is used to determine when the weights of two bodies are equal. It consists essentially of a hori- zontal beam, carrying at its ends, by means of knife edges, two \" pans,\" arid sup- ported at its middle point by a knife edge which rests on a fixed support. The bodies whose weights are to be compared are placed one in each pan ; and by adding \" weights \" from a set, the balance may be brought to a", " state of equilibrium. The arrangement of the parts of the apparatus is such that the centre of gravity comes below the knife edge. For details of the instrument, reference may be made to Ames and Bliss, Manual of Experiments in Physics^ page 151. Fm. f.O. \u2014 Chemical balance. CHAPTER III GRAVITATION Law of Universal Gravitation. -- The property of matter that is railed inertia and forms the basis of dynamics was not recognized until Sir Isaac Newton stated the laws of motion in his great treatise, Philosophies Naturalis Principia Mathematica^ which was published in lb'87; but the prop- erty of weight was familiar to every philosopher. The laws tiling bodies were first stated in 1638 by Galileo, who iiowed that all bodies fall with the same acceleration neglecting the effect of the air. Previous to thK in 1609 and Ml*, Kepler had announced the laws that bear his name concerning the motion of the planets around the sun: and in searching for an explanation of these laws and of the in<>ti<\u00bbn <>f the moon around the earth. Newton was led to a nerali/.ation concerning matter. He thought that inasmuch as a body falls to the earth, and as there was no \u2022 n why such a phenomenon should be limited to the earth, then- might be forces acting between all portions of matter in tin- universe, and that the moon is held in its orbit and the planets in theirs by forces of the nine nature as that which draws an apple as it falls toward the earth. Other philosophers had had this idea 1 \u2022\u00ab fore, notably Hooke ; but no.me had expressed it clearly until Newton did M ii. ncipia. The concept ion occurred to him as early as 1666; but for various reasons he did not make it known until 1 He found that the observed facts would l>e explained if he assumed that this force obeyed the foil. \\v : Between two particles of matter whose masses are m, and m,and which I-IM-ICH \u2014 9 I-\"-' 130 MECHANICS are at a distance r apart, there is a force of attraction propor- tional to * 2. This can be expressed by writing /= Gr *2, where Gr is a constant of proportionality and is independent of the material of the particles or their distance apart. This law is in iiccord with all known observations and experiments. Illustrations. \u2014 We shall consider a few special cases; and, in discussing them shall make use of the fact first proved by Newton that, if the above law is true, a solid spherical body acts as if all the matter were concentrated into a particle at its centre, provided the body is homogeneous or can be re- garded as made up of spherical shells each of which is homo- geneous. The sun and planets and the various satellites may for present purposes be considered as satisfying these conditions. 1. Fatting bodies. \u2014 In the case of a body falling toward the earth, we may let ml be the mass of the falling particle, m2 be the mass of the earth, and r its radius. Then the force between them, acting on each, is as above, /= -- ^\u2014 -. Con- sidering the motion of the falling particle, we may write /= m^, where g takes the place of ~-^t and is therefore a constant. This is the ordinary formula for the weight of a body whose mass is m^. If Gr is independent of the kind of matter in the particle and its amount, g should be also, and to test this Newton performed some experiments with pendu- lums ; for, as has been shown in the discussion of the simple pendulum, the period of vibration is 2w'y-, where Z is the length of the pendulum. So, if g is different for different kinds of matter, it would be apparent if the periods of pendu- lums of different materials were determined. This question was investigated by Newton and later by Bessel; and all experiments agreed in showing that g is independent of the y GRAVITATION l-'.l kind of matter. (This fact is also shown by the experiments of Galileo on the two cannon halls falling from the Leaning To\\\\<-r ai Pisa; for, as they fell in the same time, they had tin- same acceleration.) Variations in \"$r.\"- \u2014 Since the earth is not a sphere, but is slightly flattened at the poles, the distance from its centre to the surface decreases as one proceeds from the equator to cither pole, and for this reason g increases. Again, owing to the fact that the earth is spinning rapidly on its axis, a cer- tain force is required to make any particle on its surface move in its circular path; consequently a portion of the force of gravitation is spent in accomplishing", " this, and the differ- ence between the two produces the acceleration of the falling body. Since the radius of the circle of motion of a particle is greatest at the equator, and therefore the centrifugal force. the weight of a body is least there, or g is least. Therefore, owing to both these causes,.; at diflVn-nt latitudes on the earth, increasing as the latitude is increased. Several formulae have been advanced to connect these two variable quantities: the most satisfactory of which gives as the value of g at any latitude I, g = 978(1 -I- 0.005310 sin8/)- The fact that a pendulum of a constant length had dif- ferent periods ut different points on the earth's surface was recognized as early us 167 Land it was explained 1>\\ Iluygens as due to differences in the centrifugal force at these points. -, as we now know, only part \u00ab>f the cause.) Of course^ varies also, o\\\\ in^ t\u00ab\u00bb 1 ises, such as the rness of a mountain, '^reat inequalities in the constitution of the surface of the earth, etc.; hut these variations are as a rule most minute. Methods of measuring g exactly will be discussed pre^-ntlv. 2. The motion of the moon. \u2014 The moon moves around the earth in if ne..rl\\ eireul.ir \\\\ith a p.!j.pni\\im.ii.-ly -11 days 8 hr.: and so there must IKS an acceleration toward the centre of its orbit, i.e. the centre of 132 the earth, equal to \u2014, where r is the radius of this orbit and 2 \u00ab is the linear speed of the moon; or, substituting for s its 4 rt J_ value in terms of the period T, s = \u2014?f, this acceleration is. So, since r and T are both known, it can be calculated. If this acceleration toward the earth is due to gravitation and if Newton's law is true, it can also be calculated in terms of the acceleration of a falling body at the surface of the earth, i.e. g. For, using again the general formula, in which ml is the mass of a particle, mz that of the earth, and r the dis- tance from the centre of the earth to the particle, F= m^ \u2022 and therefore the acceleration of the particle toward the r earth is \u2014^. Consequently, calling a the acceleration of the moon, rl the radius of the moon's orbit, and r2 the radius of ri rz the earth, a : g = \u2014 ^ : \u2014 ~ ; and thus, since #, rv and r2 are known, a can be calculated. Newton showed that the two values, one based upon direct observation, the other upon his law of gravitation, agreed. 3. The motions of the planets. \u2014 As a result of a laborious study of numerous observations on the motions of the planets around the sun, and after many futile trials, Kepler suc- ceeded in discovering three laws in regard to these motions, with which all observations are approximately in accord. These are : (1) The areas swept over by the straight line joining a planet to the sun are directly proportional to the time ; i.e. equal areas are described in equal intervals of time. (2) The orbit of a planet is an ellipse, having the sun at one of its foci. (3) The squares of the periods of different planets are proportional to the cubes of the major axes of their orbits. VITAT10N 133 Newton showed that these laws, and many slight variations from tin-in, were direct consequences of his law of gravita- tion. The first law follows because the force of gravitation, acting on a planet, is always directly toward a fixed point, vi/.., the centre of the sun, which in the statement of Kepler's laws is supposed not to move. (Forces like this which are directed toward a fixed point are called -central forces.\") The second and third laws follow because gravitation is a central force which varies inversely as the square of the distances between the bodies. There are of course many irregularities in the motions of the moon and of the planets because of the action of other portions of matter than the earth and the sun, of variations in their distances apart, of the departure of the earth from a spherical form, etc. ; but all these irregularities can be fully explained as consequences of this law of gravitation. I is the science of \u2022\u2022 <;ra\\ national Astronomy.\" 4. The \"Cavendish experiment.\" \u2014 Various experiments have been performed since the days of Newton to see whether the force of gravitation between bodies of ordinary size could be measured. The first of these was carried out by Caven- dish in 1797-8", ". His method was to place two bodies on the ends of a light rod which was suspended horizontally by a line vertical wire attached to its middle point, then to bring up near these suspended bodies two others so placed as by their force of gravitation to turn th\u00ab I thus twist the supporting win-. He observed an effect, and measured the >\u2022 exerted. This experiment lias been repeated often and in raiioUfl forms. (In one it was shown that the force d with the masses of the bodies and inversely as the square of the distance.) II.iv ing thus measured the force between two bodies of known mass at a known distance apart, and assuming New- s law to be true, one can at once calculate the value of a in the formula. It is 0.000000066570, or 6.6576 x 10-\u00bb, 134 MECHANICS on the C. G. S. system. This leads to a value for the mass and the average density of the earth. (By the value of the \" density \" of a homogeneous body is meant the ratio of the value of the mass of a certain portion of it to the value of the volume of this portion. Thus if D is the value of the density, and m and v are those of the mass and volume of any portion, D = \u2014 or m = Dv.) We have seen that in accord- ^ r* ance with Newton's law g = \u2014\u2014i where m is the mass and r is the radius of the earth. Thus m=^-'] and all the quan- tities in the second term are known. Further, if the earth can be considered as a sphere, its volume is | Trr3 ; and there- fore in terms of the average density m \u2014 | Trr*D, and accord- 6r ingly D = ^\u2014 and can be calculated. Its value is 5.5270 4 TrCrr on the C. G. S. system. As will be shown later, the value of the density of water at ordinary temperature does not differ far from 1 on this same system ; and so the density of the earth is about 5J times as great as that of water. The student should consult Mackenzie, The Laws of Gravi- tation, Scientific Memoir Series, New York, 1900. Centre of Gravity. \u2014 A few more things should be said in regard to gravitation as we observe it here on the surface of the earth. It is a force directed toward the centre of the earth approximately ; and therefore the forces acting upon the particles of a body are parallel to each other. Their re- sultant is called the weight of the body, and we have proved that there is a fixed point connected with the body through which this resultant always passes, however the body is turned. This point is called the \"centre of gravity.\" If a rigid body is pivoted so as to be free to turn around a horizontal axis, but is at rest with reference to the earth, a vertical line through the centre of gravity must intersect the axis ; otherwise the weight would have a moment about <;i:.i\\-ir.\\Tloy LS6 it, and the body would turn. The equilibrium is evidently stable if the axis is above the centre of gravity ; unstable, if it is below ; and neutral, if it passes through this point. The fact that the centre of gravity lies vertically below the of suspension when the equilibrium is stable furnishes a method for its experimental determination. If the body is suspended at a point, the centre of gravity must lie in vertical line through it ; and so, if the body is suspended in turn from two points, the centre of gravity must be the intersection of the two corresponding vertical lines. Compound Pendulum. \u2014 If a body is suspended free to turn about a horizontal axis, it is called a com- pound pendulum; and if it is set vibrating through an infinitesimal amplitude, it will have harmonic motion. Let the cut be a section through the centre of gravity of the body (7, and perpendicular to tin- axis at P. Call the length of the line and the angle it makes at any instant with a vertical line through N; this is then the angular displace- ment. The force mg acting vertically down through Q- has a moment about the axis equal to?////// sin X. which is in the opposite direction to the disp in. -nt ; tli' :ie angular acceleration is pendulum. PUoeMetkm ofarlffklbodyplTOtodby \u2022 flx\u00abdbortxoatalav 70.- to Uw axU r. ud toclnd Inff tbo ocotra of frmrlty. n N. wl, / the moment of in. : ihe body O. ;il... nt the a\\ix. If th.- amplitii'i ill. sin", " \\..in he re- plar.-d b).V. the angle itself; and th.- acceleration has the Y i,iis.'qii.-nlly the motion is hannome; and the peri!n-\u00b1-. The period maj be observed, and in pendnlui! ipes, m, A, and / may all be i: therei nay be determined. Calling th\u00ab p\u00ab 136 MECHANICS T, g = 4 7T2. A simple pendulum is a special case of a compound one, in which there is only a particle vibrating ; m h l * so / = mh2, and the period becomes 2 ir\\-. A simple pen- dulum cannot actually be constructed ; but one can be imagined of such a length that its period equals that of any compound pendulum. 9 By suspending the compound pendulum described above so as to vibrate in turn about different axes, all parallel to the original axis, one may be found on the opposite side of the centre of gravity, so placed that this point lies on a line perpendicular to the two axes and such that the period of vibration about it is the same as that about the original axis. This fact was deduced by Huygens as early as 1673. The distance between the two axes may be shown to be equal to the length of a simple pendulum having the same period as that of the compound one. Therefore, although it is impos- sible to construct a simple pendulum, a compound one may be made provided with two axes of vibration and so ad- justed that the periods of vibration about them are the same. If T is this period, and I is the distance apart of the axes, T= 2 ir^|-, or g = -j^-. Galileo, in 1583, first called attention to the apparent isoch- ronism of a pendulum ; and he made use of this fact in cer- tain observations. He also, in 1641, described a plan of using a pendulum to regulate a clock, and had a drawing of his invention made. This fact was not generally known, how- ever; and in 1656 Huygens independently invented a pen- dulum clock, which came into immediate use. Historical Sketch of Mechanics Although the main facts in regard to the historical devel- opment of mechanical principles have been stated in connec- GRAVITATION 1;>>7 tion \\vitli them, it may be well to give a brief review. Up to the time of Newton the fundamental property of matter was thought to be weight ; and the only forces considered were those produced by weight. Archimedes and da Vinci had stated the laws of the lever and Stevin had explained the equilibrium of a body on an inclined plane, making use of what we call the parallelogram of forces; the \"proofs\" in each case were made to rest upon certain assumptions which appealed to the philosopher as being fundamental and which could not be proved themselves. Galileo made a great step in advance, because he undertook the experimental study of \u2022 ///////////'\u2022a, and formulated certain statements in regard to the properties of matter in motion. He assumed that if a body were free from any force, it would continue to move in a straight line with a constant speed ; he showed that the acceleration of a falling body is constant ; and he deduced many well-known theorems; he further assumed what is equivalent to saying that a force acting on a body produces tion independently of the existing motion of the body. (ialileo's experiments on Mechanics were published in 1638. Newton's attention was attracted to a property of matter different from weight, and to other forces than that of weight, by his conception of the explanation of the motion of the planets. In his />/\u2022///'\u2022//*/\u00bb/. published in lt'\u00bb87, he proposed three laws of motion which are equivalent to the following: 1. A h\u00bb\u00bbdy left to itself will maintain its velocity constant. \u2022_'. It a l.od\\ i- BOted on by an external force, it will r< < an acceh -uch that f\u2014 nm \\ forces act independent 1\\,,!' each other. 3. Action and reaction are equal and opposite. se laws are in accord with the principles enunciated on. and the\\ have served for over two hundred years as the basis of all work in Mechanics. \\e\\\\toii t hus int induced the ideas of ni;i^s,iiid of the proper measure of an\\ force. I'.- fore NeWton'> ideas \\\\cl-e accepted, there \\\\as a dispute as to the value to be assigned \"quantity of motion of mar 138 MECHANICS Descartes maintained that the proper value was wv, where w was the weight of the body and v its velocity", ", while Leibnitz adopted wv*. It was shown by d'Alembert, in 1743, that both were correct ; mv measures the momentum or the impulse of the force, \u00a3rav2, the kinetic energy or the work done by the force. The greatest of Newton's contemporaries was Huygens, whose treatise, De Horologio Oscillator io, published in 1673, is equaled in importance only by the Principia. In this he discusses the motions of pendulums, simple and compound; the laws of centrifugal force, etc. He had no conception of mass and used less elementary assumptions than did Newton. He adopted in his work, as the fundamental property of matter in motion, wv2, and showed the great importance in questions of rotation of the quantity which we call the moment of inertia. Since the publication of the Principia progress in Mechanics has been (1) in a philosophical study of the nature of the postulates and definitions of Dynamics, (2) in deducing from Newton's laws other principles which are more useful for particular classes of motion. BOOKS OF REFERENCE MACH. Science of Mechanics. Chicago. 1893. This gives an interesting history of the progress of Mechanics, together with a critical study of the principles on which the science is based. POYNTING. The Mean Density of the Earth. London. 1894. This contains a full account of the experimental investigation of Newton's Law of Gravitation. ZIWET. Mechanics. New York. 1893. This is a more advanced text-book than the present one, but will be found most useful for purposes of reference. PERRY. Spinning Tops. London. 1890. This is a most interesting series of lectures on the mechanics of spin- ning bodies. WORTHINGTON. Dynamics of Rotation. London. 1892. This is a most useful book of reference for elementary students. CHAPTER IV PROPERTIES OF SIZE AND SHAPE OF MATTER Solids and Fluids ; Liquids and Gases. \u2014 The most obvious property of a material body is that it has a certain shape and size, both of which can be changed by suitable forces. As has been explained before, the name \"solid\" is given to a body that keeps its si/c and shape under all ordinary conditions; and the name \"fluid,\" to a body that yields to any force, however small, that acts in such a manner as to make one layer move over another. Fluids are divided into two classes, accordiug as they can form drops or not; the former are called \"liquids\"; the latter, \"gases.\" See Introduction, page 16. Elasticity. Viscosity, etc. \u2014 Some bodies when deformed slightly by a force will return to their previous condition after the force is removed; they are called \"elastic.\" Thus, glass, steel, ivory, etc., and all fluids are elastic. Certain solids, however, when deformed in a similar manner, remain so after the force ceases ; they are called \" inelastic \" or \" plastic.\" Such bodies continue to yield to a force as long as it is applied. In the deformation of all inelastic bodies tin-re is a sliding of portions of matter over each other, as when a piece of lead is bent or hammered; and consequently there is what has been called \"internal\" friction between these parts. There is a sliding of this kind whenever any actual solid, however elastic, is deformed, all hough its amount may be very small. This is shown by the fact that, if a body as elastic as a glass rod or a steel tuning fork is set in vibra- tion, the motion soon ceases, and the temperature of the body 140 MECHANICS is raised slightly. Similarly, when currents are produced in a liquid or a gas by stirring them in any way, the motion soon ceases, and the temperature is found to be increased. A fluid which offers great frictional opposition to the relative motion of its parts is said to be \" viscous,\" while those which flow easily are said to be \"limpid.\" As will be explained later, when a fluid is made to flow through a long tube, the quantity that escapes from the open end is independent in most cases of the material of which the tube is made. This proves that the fluid layers on the inner surface of the tube stick to it, and so the fluid actually flows through a tube of the same material as itself. Consequently, in this flow the velocity is zero at the surface of the tube and increases toward the axis ; and so layers of the fluid flow over each other. It requires work to accomplish this ; and the quan- tity of fluid escaping under a given force or \"head\" measures the viscosity of the fluid, being inversely proportional to it. Similarly, when a solid moves in a fluid, there is a", " layer of the fluid attached to it, which moves, then, over other layers of the fluid. Consequently, if a pendulum vibrates, or if a disk is supported by a vertical wire which twists to and fro around its axis, making torsional vibrations, the rate of decrease in the amplitude of the oscillations measures the viscosity of the surrounding fluid. In this way the viscosity of various fluids has been measured ; and it is found that it varies greatly for different fluids and with the temperature of any one fluid. Rise in temperature decreases the viscosity of a liquid, but increases that of a gas. Diffusion. \u2014 Whenever any two gases are brought together, they mix ; and after a short time the mixture is homogeneous. This process is called \"diffusion.\" If two liquids like water and alcohol are brought in contact, one will diffuse into the other ; and, even in other cases like mercury and water, where there is no apparent mingling, it may be proved that at the surface of contact there is a slight diffusion. OF -///\u2022:.i.v/> >//.!/'/\u2022; OF.\\/.i /\u2022//\u2022:/; 111 The most important investigation of the phenomena of diffusion was carried out by Graham (1850). He was led t\u00ab> divide substances into two classes \u2014 \"crystalloids\" and \"coll The former diffuse much more rapidly than the latter, and can as a rule be obtained in a crystalline form, while the latter are amorphous. The mineral acids and salts arc crystalloids; the gums, starch, and albumen are colloids. If the former are dissolved in water, the solutions have prop- erties most markedly different from the water ; while if the latter are dissolved in small amounts in water, they have little, if any, effect, in some cases the colloids being merely suspended in the water in a very finely divided state. If colloids are mixed with not too much water, they form jellies OF membranes ; and crystalloids are able to diffuse through many of these with almost as much ease as through pure water. This process is called \"osmosis,\" and one case of it will l)e discussed later. (This evidently offers a method for the separation of crystalloids from colloids if there is a mix ture of them. Osmosis was first observed by the Abbe Nollet in 174s, who used a piece of bladder as the membrane. Parchment paper is often used.) Similarly, gases can pass through a thin sheet of India rubber; the latter absorbs the gas on one side and gives it off on the other. Many gases can pass through metals with ease if the latter aiv red hot; thus, hydrogen Can pass through red-hot platinum, oxygen through red-hot silver, etc. If two soft solids like lead and gold are brought into c t.ict. experiment! show that alter the lapse of suilicient time there has been diffusion \u00ab,f,,ne into the other; and the same is believed to be true to a certain extent at the surface sepa- rating B -olids, or in fact any two portions of matter. I,' :\u2022\u2022 inured by placing two bodies in OOntad OV< r a known area and determining the (plan: of either which pass this surface in a given time and the 142 MECHANICS distances they permeate. There are great differences in the rates of diffusion of different bodies, and these rates vary with the temperature. Solution. \u2014 One of the most important phenomena deal- ing with the properties of matter is illustrated when some common salt is put into a basin of water : the salt as a solid disappears, it is said to be \"dissolved.\" The salt is called the \" dissolved substance \" or \" solute,\" and the water the \"solvent.\" Mixtures which are homogeneous and from which the constituent parts cannot be separated by mechani- cal processes, are called \"solutions.\" The formation of a solution is evidently closely connected with the process of diffusion. Similarly, we can have solutions of other solids in liquids, of solids in solids, of liquids in liquids, etc. In the case of salt dissolving in water, it is found that, if the temperature is kept constant and more and more salt is added, a condition is reached such that, if more is intro- duced, it does not dissolve, but remains as a solid pre- cipitate: the solution is said to be \"saturated.\" If the temperature is lowered, salt will be precipitated if some solid salt is already present; otherwise, this does not in general take place. If the liquid thus contains in solution more salt than would saturate it at a given temperature, it is said to be \"supersaturated \"; and its condition is unstable, for by adding a minute piece of salt, all the salt in solution in excess of that", " required to produce saturation is precipi- tated. Similar phenomena are observed in many other solu- tions, but not in all. It is found also that, as one substance dissolves in another, there are temperature changes ; thus, as common salt is dis- solved in water, the temperature of the water falls, while if sulphuric acid is dissolved in water, the temperature rises. These changes will be discussed later. Kinetic Theory of Matter. \u2014 It is impossible to explain these facts of diffusion without assuming that the minute /'/,'o /'/\u2022:/,'\u2022///\u2022> OF SIZE AXD SHAPE OF MATTER 1 1 '. portions of material bodies \u2014 their molecules \u2014 are endowed with motion of translation; while, if they are free to move and are moving, the general explanation of the phenomena is at once evident and needs no statement. To account for tiit- differences between solids, liquids, and gases, it is only necessary to assume different degrees of freedom of motion (\u2022I the molecules. Since solid bodies offer great opposition. in general, to changes in size and shape, we assume that in them the molecules are held together as if by a \" framework.\" so that they can vibrate, but cannot move from one part of the body to another unless the \"framework\" breaks down : this may happen with difficulty or with ease, thus causing the difference between elastic and inelastic bodies. The word \u2022\u2022framework\" is used to describe a condition, not an actual tiling; we mean simply that there are forces between tin- molecules which hold them together exactly as the frame- work of a building or bridge holds it together. In a liquid assume that the molecules are so free that they can and do move about limn one point to another, but yet that the forces are sufficiently strong to prevent them on the whole from getting far apart. Of course, if a molecule strikes tin- surface with suffici cut velocity it may escape, and thus evapo- ration is explained. In a gas we assume that the forces between the molecules are so minute that the freedom of motion is practically perfect; they can move freely from any point to another in the space open to them; and we think of tin; molecules as having rapid motion to and fro through this space. We shall show later h\u00bb>\\v simple it is to ;in in general terms all the properties of a gas as due to notion nf its moleeules. It is not kno\\\\n whether these M b.t ween the molecules which are so evident in the case of solids and liquids are due to gravitation or not, but ist possible. The phenomenon of viscosity is at once explained by this assumption of moving molecules; for, if one layer of a fin id 144 MECHANICS i> moving over another, molecules will pass between the layers, and each one that passes from the layer flowing more slowly into the other retards the latter ; while, if one moves from this layer into the former, it accelerates it. Thus, owing to the continual interchange, the two layers finally have no relative velocity; and, if one of them is at rest being in contact with a solid, all the fluid comes to rest. (This is a case where a force is explained in terms of the motion of particles of matter. See page 68.) Coefficients of Elasticity ; Hooke's Law. \u2014 When an elas- tic solid is subjected to a force, it will, in general, yield slightly and then come to rest, e.g. bending a bow, stretch- ing a wire. This means that the changes in the molecular forces which are called into action at any point by the defor- mation are sufficient to neutralize the action of the external force. There is thus at any point of a deformed elastic solid a change in the position of the molecules immediately around it and a corresponding \"force of restitution.\" When there is equilibrium between this internal force and the external one, the former may be determined from a knowledge of the latter. The elasticity of the body is measured by the ratio of this change in the internal force at any point, which is produced by the deformation of the matter near it, to 'the amount of this deformation. These internal forces between portions of the body are called \" stresses \" ; and their numeri- cal value is defined as follows : let the internal force between two portions of the body whose area of contact is A have the value F, then the limiting value of the ratio \u2014 as A is made smaller and smaller is that of the stress at that point. (If the stress is uniform, it equals the force per unit area.) Owing to the deformation the internal forces are changed ; and calling the change in the force A jP, the ratio \u2014 \u2014 is the stress corresponding to the deformation. The deformation F -A A F A. OP ^///-: AND SHAPE", " OF MATTKIi 1!.\"\u00bb which produces this stress is called the \"strain\"; and its numerical value is defined differently, depending upon the kind of deformation. Thus, if the volume of each minute portion of the body is changed, without there being at the same time a change in shape, let v be the value of the origi- nal volume of a minute portion of the body at any point, and Av be the change in this ; then the value of the ratio \u2014 v is defined to be that of the strain at that point. Similarly, if the length of a wire or rod is changed by stretching or compression, the strain is defined to be the ratio of the change in length to the original length. If the shape of a solid body is changed, the measure of the strain may be defined also, as will be shown later. It is found that, if the strain is small, the corresponding stress is proportional to it: this is called Hooke's law, and \\vas fust stated by Robert Hooke in 1676, in the form \"Ut tensio sic vis.\" Tin- ratio of the stress to the corresponding strain in any elastic body is called the \" coefficient of elasticity\" of that body with reference to the type of strain. Hooke's law. then, states that all coefficients of elasticity are constants for a given body ; or, in more common language, the amount of the deformation of an clastic body is proportional to the force applied. \"nee this proportionality between internal force of rexti- tution and displacement is t rue. and since one is in the oppo- site direction to the other, the elastic vibrations of any body must l.e harmonic; because, when in the course of its vibra- tions the body has a certain strain, i.,-. display-menu the elastic force of restitution is, in accordance with Hooke's proportional to it. and therefore the acceleration is I Again, if I lookers law is true, the elastic force corresponding to any displacement must!\u00bb\u00ab\u2022 directly propt.rt ionul to it, as has just been said; HO, if/ is the value of tin nd x that of the displacement/^ or, where c is a factor..f pro] tionality depending upon the nature and dimensions of the AMES'S MMHICI- ]o 146 MECHANICS strained body and the character of the strain. As the dis- placement, then, increases from 0 to some value xv the aver- age force of restitution is \u00a3 cxv and therefore the work done in producing the displacement x1 is the product of xl by \u00a3 cxv or \u00a3 cxf. A solid body can undergo two independent deformations : a change in shape and a change in volume ; and correspond- ing to these an elastic solid has two coefficients of elasticity. If the coefficient with reference to change in volume is large, the body is said to be nearly \"incompressible \" ; while if the one with reference to a change in shape is large, the body is said to be \" rigid.\" A fluid, 011 the other hand, has only one coeffi- cient of elasticity, that corresponding to a change in volume. Gases are very compressible ; liquids are not. These kinds of matter will be discussed separately. Density and Specific Gravity. \u2014 Before, however, proceed- ing to do this, a physical quantity should be defined which must be used often in the following pages. This is the \" density \" of a body. It is that property of a body which expresses its denseness, using this word in its ordinary mean- ing. If the body is homogeneous, and if m and v are the values of its mass and volume, the ratio \u2014 is defined to be v 931 the value of the density. But, if the body is heterogeneous, the density at any point is defined to be a quantity whose value equals that of the ratio in the limit, where Av is the volume of a small portion around the point and Aw is the value of its mass. On the C. G. S. system, the den- Ai> sity of pure water at 4\u00b0 C. is almost exactly one ; for the kilogram was so constructed that its mass almost perfectly equaled that of 1000 cu. cm. of pure water at 4\u00b0 C. ; i.e. the mass of this volume of water is most approximately 1000 g., and therefore the density of the water is as stated above..* Of >///\u2022; AM> >//.!/\u00bb/; OF MATTEL 1 17 The ratio of the density of a body to that of a standard both is called its \" specific gravity\" with reference to the latter. Thus, it water at 4\u00b0 C. is chosen as the standard sul \u00bbMaiice, its density is \u2014, where m is the mass of a volume", " v\\ and, if M is the mass of an equal volume of another substance, its (Unsity is \u2014, and its specific gravity is therefore \u2014 v m Thus \"specific gravity\" is independent of the choice of units in terms of which the mass and the volume are measured, but some standard body must be selected. In order to measure the density of any body \u2014 solid, liquid, or gaseous \u2014 one of two methods must be followed: eitlu-r its volume and mass must be measured directly, or its spe- cific gravity must be determined with reference to some body whose density is known. The details of these different pro- cesses may be found in various Laboratory Manuals. In the following table are given the values of the densities of some substances in ordinary use. DENSITIES SOLIDS Aluminium (cast).. 2.58 Brass (about)...8.5 Copper.... 8.92 ond.... - Glass (common).. 2.6 Mint). IcestO\u00b0C... 0.9107 I nm (wrought)... 7.86 Iron (gray cast)...7.1 Lead UJO Platinum.... 21.50 Silvr 10.5 Tii 759 /in.- 7.15 Ktl.yl alrnl.,.1 at 0\u00b0C Kthyl ether at 0\u00b0C. 0.791 l 1SJM Water at utli.-r t. mperaturea, see p. 148. 148 MECHANICS GASES AT 0\u00b0C. AND 76 CM. OF MERCURY Air (dry) Argon. Carbon dioxide Chlorine. 0.001293. 0.001700. 0.001977. 0.003133 Helium. Hydrogen Nitrogen Oxygen.. 0.00021. 0.0000895. 0.001257. 0.001429 WATER AT DIFFERENT TEMPERATURES 0\u00b0C.. 1\u00b0 2\u00b0 3\u00b0 4\u00b0 5\u00b0 6\u00b0 7\u00b0 8\u00b0 9\u00b0 10\u00b0 11\u00b0 12.999878. 0.999933. 0.999972. 0.999993. 1.000000. 0.999992. 0.999969. 0.999933. 0.999882. 0.999819. 0.999739. 0.999650. 0.999544. 0.999430. 0 999297 0.999154 16\u00b0 C... 0.999004 17\u00b0.... 0.998839 18\u00b0.... 0.998663 19\u00b0.... 0.998475. 0.998272 21\u00b0.... 0.998065. 0.997849 23\u00b0.... 0.997623 24\u00b0.... 0.997386. 0.99714 25\u00b0.... 0.99686 26\u00b0.... 0.99659 27\u00b0... 28\u00b0... 29\u00b0... 30\u00b0.... 0.99632. 0.99600. 0.99577 0.99547 2 \u00b0 2 0 \u00b0 2 \u00b0 5 31\u00b0 CHAPTER V SOLIDS General Description of the Strains of a Solid. \u2014 A solid body, being eharacteri/.ed 1>\\ a definite shape and size, can, as has been seen, be deformed in two independent ways; and, in general, under the action of forces both the size and shape are changed. These changes, if small enough, will disappear in the case of an elastic body when the force is removed; hut, if the force is too large, certain permanent effects are experienced. These will be described in one particular case, that of a vertical wire whose upper end is fastened to a fixed support and to whose lower end is attached a scale pan into which weights may be loaded. So long as the stretching force is not too great, the elongation varies directly as the load ; and, if this is removed, the wire returns to its original length. If the load is increased, however, a point is reached, known as the \"elastir limit.\" such that if the force exceeds this in value, the wire acquires a permanent elongation or \"set\" which does not disappear when the load is removed. It tin- load is increased still more, the elongation becomes greater; and at length a condition is reached sueh that. greater force is applied, the extension increases v< i \\ rapidly and the wire becomes plastic, because the amount of the ex- tension now varies with the time the load acts. This point is called the \"yield point/' If the load is increased beyond. the cross section of wire will eontraet until the \u2022\u2022 break- s reached. Changes similar to these go on when the shape of a body is altered by twisting it. In what follows we shall discuss", " mily those chants \\\\hich take place below II'.' 150 MECHANICS the elastic limit; so that we can consider the strain as proportional to the stress. Change of Volume. \u2014 In this case the strain is, as has been explained, \u2014, where Av is the change in volume of a portion v of the body whose volume originally was v ; and, if the cor- responding stress or force per unit area is Ap, the coefficient of elasticity is A# \u2022 \u2014 \u2022 Av In order to produce a change in size of the minute portions of a solid without changing their shape, it is necessary to immerse the solid in a liquid and then to compress the liquid. This is done by having the liquid inclosed in a stout trans- parent cylinder, one end of which is closed by a piston which can be screwed in or out. Such an instrument is called a \"piezometer.\" When the piston is pushed in, the liquid presses against the immersed solid and compresses it, the volume of each minute portion of the solid being decreased proportionally. To measure the change in volume of the solid, the latter is as a rule made in the shape of a rod ; two fine parallel lines are scratched on it, one near each end ; and by means of a comparator the distance apart of these lines is measured before and under the compression. If 1Q is the original length and I the length when the stress is increased by an amount Ap, experiments show that I \u2014 1Q = \u2014 eZ0Ajo, where c is a factor of proportionality and is extremely small. If Ap is measured, the value of c may be determined. If the body is homogeneous, we may assume that similar changes take place in a plane at right angles to the length of the rod. So, if a cube were subjected to the increase in stress A/?, its change in volume would be found as follows : the original volume VQ is Z03 ; when the stress is increased by Ajt?, the volume is Z8, or Z03(l - <?Ap)8, which equals Z08(l - 3 <?Ap), since c is so small that terms involving c2 and <P may be neglected ; hence, the change in volume is \u2014 3 Ifa&p or SOLIDS 151 - 3 v0<?Ap. So, in the original experiment the strain corre- sponding to the stress A/?, i.e. the ratio of the decrease in volume to the original volume, is \u2014 3 c Ap ; and the coeffi- cient of elasticity corresponding to a change in volume is tl irii \u2014 - \u2022 The stress Ajo is due to the change in the internal o c forces, but, since there is equilibrium between it and the forces due to the pressure of the liquid, it is determined by measuring the force per unit area with which the liquid presses on the solid, as can be done in a manner to be described presently. The result of subjecting a hollow solid body like a bottle to such a liquid pressure as this is to decrease its volume exactly as if there were no hollow spaces, provided the liquid has access to them ; for each minute portion of the body has its volume decreased exactly as if there were no ies. It is as if, before the compression, the solid were built up of large \"bricks\"; while, under compression, it is made up of the same number of smaller ones. This coefficient of elasticity is sometimes called the \" hulk modulus.\" Change of Shape. \u2014 An illustration of the < -han<je in shape large solid without any appreciable change in volume is given by the following experiment. Make a rectangular l.lo.-k of wood; to two opposite faces fasten two boards; and push these boards sidewise in their own planes, but in opposite directions. The shape will be changed as shown. tin- edges of the block being now <\u00bbl\u00bblique; and the angle veen the two positions of an edge, that is (BAB1)* is the measure of the strain. It the sidrwise force on \u2022 \u2022iihi-i I ><> nd is t\\ there is an equal but opposite force of restitution; and, it tin- an -a of the cross section at the block 1 the stress corresponding to the strain is \u2014 \u2022 This stress A i% acting across all the planes of the block that are parallel to the boards. The coefficient of elasticity for a change in shape is, then, the ratio <>f this ipiantity to the angle referred t<> j.\" 152 MECHANICS r 7i above. This coefficient is also called the \"coefficient of rigidity\"; and the particular kind of stress that arises when one layer is moved parallel to another is called a \"shearing stress,\" because it is like the force produced by a pair of shears. / // C B / 1 \\ L= /", " / / 7 When a rod or wire is twisted on its axis of figure, each minute por- tion of the body experi- ences a shearing stress, because different plane cross sections of the rod or wire are turned through different amounts ; and, therefore, this is an illus- tration of a pure change in shape. B'B C'< \\ \\ FIG. 71. \u2014 Change of shape or shearing strain. The deformation is produced by pushing the two boards in opposite directions. If one end of the rod or wire is held firm, and the other twisted by a moment L around the axis of figure, the angle through which the lower end will be turned is given by the following formula, which may be \"=\u00a7\u2022 deduced by the help of higher mathematics : where I is the length of the rod or wire, n is the coefficient of rigidity, and B is a constant for any one rod or wire depending upon its dimen- sions. For a circular cylinder of radius r, B = - ; and so, for a wire of circular cross section, 21L N = If the wire is maintained in this state of torsion by this moment, and if it is in equilibrium, there must be in each plane cross section two internal moments equal to L, due to the elasticity of the wire; one moment acts on one face at the cross section, the other, on the opposite i r n r 4 SOLIDS 153 and the two moments are equal and opposite in amount. It is exactly analogous to the case of a wire or cord under the action of a 4 force ; at each of its points there are two equal and opposite \u2022ns, each equal to the stretching force. If such a wire is twisted and then set free and allowed to make torsional vibrations, the moment tending to oppose the motion at any instant, due to the reaction of the. will be M the angle of torsion is N. Hence, if a disk whose moment of iiu-rtia around the axis is / is fastened to the free end of the wire, its angular acceleration at any instant of the vibration will be \u2014 or \u2014 wn \u2022 so the vibrations will be harmonic, and the period of one complete vibration will be The elasticity of a helical.spring, that is, a wire coiled nj\u00bb in a spiral, is due to the fact that, when it is ]ni lied out, the wire is twisted, not elongated. Actually there is a sliding of one layer over another when a wire is twisted, or, in general, when there is a shearing stress; but in an ideal solid there would be no such sliding, if the shape were changed; and in sueh bndies as steel, glass, ivory, etc., the slipping is very small. Young's Modulus. \u2014 When a rod or wire is elongated. <>r when a pillar is compressed, both the shape and si/.e <>f its minute portions are altei-.-d ; and so both the coefficients previously discussed are involved. Since, however, the case is such an important one in all practical work, a new coefficient is defined \\\\ hi.-h refers to the elasticity shown w lien ire or rod is stretched or compressed. Let I be the original length ; A/, its increase when there is an increase &F in the stretching force ; A* the area of the cross section. Since the 154 MECHANICS wire or rod comes to equilibrium, the change in the internal force equals A.F, too ; and so the stress is - - \u2022 The corre- sponding strain is \u2014 ; and the coefficient is - - -s- \u2014 \u2022 This I A. I \u2022 is called \" Young's Modulus \" ; and, writing its value E, A -A/ E=l'^F. This is found to be a constant for any one kind of matter, e.g. a definite kind of brass, iron, etc. ; and, if its value is known, the elongation of a particular wire or rod under the action of a given force can be calculated, because If a pillar supports a building, its length is less than if it carried no weight ; but, if the weight supported becomes too great, the pillar will buckle and give way. It may be shown by a more elaborate discussion that a circular pillar pressed between two supports, and having both its ends fixed, becomes unstable when the force of compression equals -71\" J, where r is the radius of the pillar and I is its length. If a beam is supported horizontally by resting on two horizontal knife edges placed near its ends, and if a weight is hung from its middle point, it will be bent. It is obvious that the lower side will be stretched, while the upper is shortened. There will there- fore be a layer of lines in the beam about halfway through, F.\u00ab. 73 -Standard form of metre", " of the containing vessel. When a fluid is not flowing, it is said to be at \"rest,\" although this does not imply that the molecules are not moving : it simply means that there is no motion of portions of the fluid over each other. We shall discuss in turn the two eruditions : that of rest and that of flowing. Fluids at Rest Thrust. \u2014 The fluid exerts a force against the walls of the 1 that eontains it; and, conversely, the wall reacts against the fluid. For instance, if a toy balloon is inflated, the _r;ts presses outward against the rubber envelope, and this presses inward, tending to compress the gas; a dam holding baek a river is pressed against by the water; a t containing water presses inward \\\\ith sufficient force to with- stand the outward force of the water, otherwise it bursts. Similar! \\, there ire forces against any foreign body im- 1 in tin- fluid. If the fluid is at rest, this force 167 158 MECHANICS between the fluid and the wall or immersed body is per- pendicular to the separating surface at each point ; for, if there were a component parallel to the surface, the fluid would flow. The total force acting on the surface is called the \"thrust.\" Fluid Pressure. \u2014 The properties at various points of a fluid are best described in terms of what is called the \" pres- sure.\" If any small portion of a fluid is considered as in- closed in a solid figure with plane faces, there is a stress across each plane face due to various causes; but, if the fluid is at rest, this force is perpendicular to the face, as just ex- plained. The limiting value of the ratio of this perpendicu- lar force to the area over which it acts is defined to be the value of the \" pressure \" at the point considered, in the direc- tion of the force. If the pressure is uniform, it equals, then, the force per unit area. We thus speak of the pressure at the bottom of a tank of water, etc. Pressure at a Point. \u2014 At any point of a fluid at rest the pressure has the same value in all directions ; for, consider a small portion of the fluid inclosed in a tetrahedron, ABCD, and express the con- dition that it shall be in equilibrium FIG 74 \u2014A portion of a un(ler the action of the forces on its faces. fluid inclosed in a tetrahedron, The SUm of the Components of these or triangular pyramid. <\u2022 \u2022 -i... forces in any direction must equal zero. Choose as this direction that of the line AB ; then the forces on the two faces BAD and ABO have no components parallel to this line, because they are perpendicular to it. Call the area of the face A CD, Av and the force acting on it Fl ; the area of the face BCD, Av and the force acting on it F2. The component of Fl parallel to AB is Fl multiplied by the cosine of the angle between AB and a line perpendicular to the plane ACD. Imagine a plane perpendicular to AB\\ the projection on this plane of the triangle ACD is another tri- angle whose area may be called A, and which is the same 169 as the projection on this same plane of tin- trian^. *l Therefore - - equals the cosine of the angle between the line AH and a line perpendicular to the plane ACD; ami component of the force Fl along AB is, accord ii. Similarly, the component of F2 along Afl is F^\u2014\\ and -M AI.since tliere is equilibrium, Fl \u2014- + F^ \u2014 = 0, or=-L = \u2014 \u00b1i. Al AI Al AI refore, in the limit, when the tetrahedron becomes intini- nal. tlie pressure in the direction of F1 is equal numeri- cally to that in the direction of F2. Hut these directions may be any two : and consequently the pressure at the point around which we have imagined the tetrahedron has the numerical value in all directions-. Work done when the Volume of a Fluid is Changed. \u2014 If a fluid is contained in a cylinder one end of which 1 bv \u2022 vable piston, the work done on the fluid in order to c.impivss it may he calculated at once. If the ssure of the fluid against the piston is uni- ; its value be p (if the pressure is not uniform, let j> be its mean value): then, if A MC area <,f th.- piston, the force that mu.st be over-.1. Let the piston be displaced in\\\\ itance r : the work done on the fluid /\u2022.I: the decrease in volume is./.I: ami..,';\" '. the work done in comp", ", another pressure at any point in a fluid due to the forces between the molecules. This we cannot measure ; but we can form an estimate of its value for a liquid by measuring the amount of work required to evaporate it, assuming that we can neglect this pressure for a gas.) Pascal's Law. \u2014 Let us, then, consider the properties at the centre of the earth of a fluid inclosed in some vessel which it fills; for instance, a cylinder closed with pistons of different areas. To maintain the fluid in a definite condi- tion, forces Fl and F2 must be applied to the pistons : \u2014 FI from without ; and there is, therefore, a corresponding pressure throughout the FIG. 78. \u2014 A fluid is inclosed in a vessel closed _. _ _., _., by two pistons of different areas. Neglecting mild. But, Since the fluid gravity, ^1=^, Pascal's Law. is at rest, this pressure must be the same at all points; for, if it were not, the fluid would flow from a point of high pressure to one of low, there being no force to counterbalance the difference in pressure. Therefore, the fluid pressure due to the reaction of the walls of the contain- ing vessel is the same at all points throughout the fluid. If this pressure is p, and if the area of one piston is Av the force necessary to keep it from moving outward is pA1 ; and, if the area of the other piston is Av the force acting on it is pAv Thus we have a \" machine \" by means of which a force F1 which equals p A1 can balance one F2 which equals pAz ; and so a small force may produce a great one. If the fluid is a gas, a great pressure, and therefore large forces, cannot be secured unless the volume is made very small; but if it is a liquid, the pressure may be made as great as desired without any marked decrease in volume. Thus, FLUIDS water or some other liquid as the fluid in a cylinder provided with two pistons, a small force acting on the smaller piston may produce a great force by means of the larger one. This is the principle of the \" hydrostatic press,\" which is shown in the cut. A pump forcing a small piston do\\vn produces an upward motion of the large piston; and thus a force is exerted as much greater than the original one at the pump as the area of the large piston is greater than that of the other. (Of course, since this press is used at the surface of the earth, there are additional pressures due 79. \u2014 Hydrostatic press. avity ; but they are nearly the same on the two pistons, and in any case their effect can be neglected when compared with that due to the pistons.) I n a piezometer (see page 150) the force per unit area on the immersed solid Imdy equals the pressure throughout the liquid: and, if the force acting on the piston and the area \"f the latter an* known, the value of the pressure equals their niiio. 1 law that tin- pressure is the same at all points in a fluid when not under 1 In- adi\u00ab\u00bbn of external forces was t stated byF. :\u2022 (felled M Pascal's Pressure due to Gravity. \u2014 The external force to which all fluids on the surface of the earth are subject is that of 164 MECHANICS gravity ; and since, when the fluid is at rest, the lower layers have to maintain the weight of the fluid above, there are dif- ferences in pressure at different points in a vertical line in the fluid. Thus imagine two horizontal planes at a distance // apart, and consider in them two portions having equal c D areas, of value A, one vertically above the other. Let their traces on the paper be AB and CD. The pressure over the plane CD due to gravity depends upon the fluid above it ; similarly for the pressure over the plane AB ; consequently the excess in the upward force across AB over that across CD equals the weight of the fluid contained in a cylinder planes in a fluid, at a whose cross section equals A, the area of either of the planes, and whose height is A, their distance apart. If d is the density of the fluid, this weight is dhAg, because hA is the volume and dhA the mass. Therefore the excess of pressure due to the vertical height h is dgh. It follows that the pressure at all points in the same horizontal plane of a fluid is the same ; because, if it were different, there would be a flowing of the fluid from the point of high pressure to that of low, as there would be no force to oppose the motion. Archimedes' Principle. \u2014 If a solid body (or a body with a solid envelope) is immersed in a fluid at rest, or if a liquid drop or bubble is immersed in a gas, it", " is acted on by the pressures against its surface ; and these produce a resultant force in a vertical direction upward, as can be easily seen. For if the solid body were replaced by a portion of the fluid having the same size and shape inclosed in a massless envel- ope, there would be identically the same surface pressures on this envelope as there were on the solid. But under these conditions the fluid in the envelope is at rest ; and this shows that the forces due to the surface pressures have a resultant FLl'llx 165 vertically upward whose amount equals the weight of tin- fluid in the envelope and whose line of act inn is through tin- centre of gravity of this portion of the fluid. Therefore, when a solid is immersed in a fluid, the surface forces due t<> the weight of the fluid combine to form a buoyant force which is equal in amount to the weight of the fluid displa he solid and whose line of action is vertically upward through the centre of gravity of the displaced fluid. (If the solid hody is homogeneous, its centre of gravity coincides with that of the fluid displaced ; otherwise, it will not in general.) This statement, having been first expressed in words by the great philosopher of Syracuse, is called \" Archi- medes' Principle.\" It is illustrated by floating balloons and soap bubbles; by all solid bodies here on the earth, whose apparent weight is therefore less than their true weight; by bodies immersed in water ; etc. I f the solid is denser than the fluid, the buoyant force is less than its own weight, and it sinks; while, if its density is less than that of the fluid, it will rise. In both cases the in takes place in such a manner as to make the potential energy decrease. Tin- j.rinri],!,- furnishes a method for the determination of the specific gravity of a solid body in terms of any liquid which does not dissolve \u2022i ise affect it If the weight of the solid is measured when Tee, its value is mg or dvg, when d is its density and v ito volume, due allowance being made for the bu<> If it is weighed again, hanging imineraed in the liquid, the difference in weight is is the density of tin- liquid. Thus the ratio of the on t to this loss in weight is ^, or the specific gravity of the solid. If at be the other is known, may Unit of Pressure. \u2014 The unit of pressure on the C. G. S. system dyne per square centimetre \"; hut, since pree- s are as a rule produced and measured by using columns :\u00abjuids. a more convenient practi.-il unit has been chosen, pressure due to a vertical height \u00ab\u2022\u2022 calculated once. 166 MECHANICS metre of mercury at the temperature of 0\u00b0 Centigrade, under the force of gravity which is observed at sea level at 45\u00b0 latitude.\" This unit is called a \"centimetre of mercury\"; and its value in terms of dynes per square centimetre may be calculated at once by substituting proper values in the formular p = dgh. On the C. G. S. system, the density of mercury at 0\u00b0 C. = 13.5950 ; and the value of g at sea level at 45\u00b0 latitude is 980.692. Consequently the pressure of one \" centimetre of mercury \" is the product of these two quan- tities ; that is, it is 13332.5 dynes per square centimetre. The pressure of 75 cm. of mercury is called a \"barie\";\" and its value in dynes per square centimetre is 75 x 13332.5, or almost exactly 106 dynes, i.e. a \"mega-dyne.\" The pres- sure of 76 cm. of mercury is called \" one^ atmosphere,\" because, as we shall see later, this is about the pressure of the atmosphere at sea level. Other units are \" one pound per square inch,\" \"one ton per square foot,\" etc., where \"one pound \" means the weight of one pound, etc. Atmospheric Pressure. \u2014 Owing to the smallness of the density of a gas, there are only slight variations in pressure at different points in a gas confined in a reservoir of any moderate dimensions. There are, however, marked differ- ences in the pressure of the surrounding atmosphere in which we live as one rises far above the earth's surface or goes up a mountain. This is owing to the large value of h in the above formula, which may in this manner be secured. Owing to the presence of the gases forming the atmosphere, there is a pressure exerted by it against every solid or liquid surface with which it is in contact. This is called the \"atmospheric pressure.\" It may be measured at any point on the earth's surface by balancing it against a column of some liquid of known density, as will be", " shown presently. The fact that the atmosphere exerts a pressure on solid and liquid surfaces was first clearly understood by Torricelli, a pupil of Galileo's, and by Pascal. Conditions at the surface FLUIDS!\u00bb;; nf the earth were in their minds comparable with those at tlu- bottom of an ocean of water, so far as fluid pressure was concerned. Torricelli devised the famous experiment, which bears his name, of filling with mercury a long glass tube \u2022 \u2022d at one end, inverting it, and placing the open end under the surface of mercury in an open vessel, care being taken to prevent the entrance of any air. The mercury column stands in the tube to a height of about 76 cm. (at the sea level) ; lu -ing held up in the tube by the downward pressure \u00ab>f the atmosphere on the mercury in the open vessel. The space above the mercury column is called a \"Torricellian vacuum \" ; and it is evident that the only matter present in it is mercury vapor, if the experiment has been carefully per- formed. This \u00bb \\JM rim, nt was performed for Torricelli by his fiu-inl Viviani in 1643. Pascal varied it by carrying a rometer,\" as this apparatus of Torricelli's is called, to ditlVrent heights and noting the change in the height of the column. Many experiments to show the effects of the atmos- phri -ir pressure were devised after the air pump was invented by Von Guericke (about 1657) and improved by Boyle. Liquid- are in general contained in open vessels which they only partially fill. The atmosphere presses against the free : ace of the liquid, exactly as if there were a piston press- down on it. Therefore in the case of liquids in open vessels the pressure due to the containing walls equals the atmospheric pressure, and, as said above, this pressure is the same at all points in the liquid. Fluids in Motion If there is a difference in pressure between two points in.1 tlnid. tin i, will be mot: 11 high to low pressure unless re vents it. We shall consider several cases. Uniform Tubes. \u2014 1 uents sh. within certain limits, thr.pi. in :luid that MM, -h a tube under 168 MECHANICS a difference of pressure at its two ends is independent of the material of the tube. This proves that what actually hap- pens is that a layer of the fluid sticks to the walls of the tube, and the escaping fluid thus moves through a tube made of the same material as itself. There is therefore friction be- tween the moving layers of the fluid, not between the tube and the fluid; and the quantity of fluid that escapes under definite conditions varies inversely as its viscosity. If the tube is horizontal, the pressure is uniform through- out it, so long as there is no flow; this is called the \"statical\" pressure. But as soon as the motion begins, the pressure falls throughout the tube. If a constant pres- sure is maintained at one end A \u2014 as in water or gas mains \u2014 and if the other end B is closed, there is a uniform pres- sure, as just said ; but if B is open so as to allow the fluid to escape, owing to the friction the pressure will decrease uniformly from A to B if the cross section of the tube is the same throughout ; and, if the tube is sufficiently long, the fluid will barely flow out, however great is the pressure at A. If the tube is bent, the flow is still further decreased. If the fluid is flowing uniformly, the quantity that passes through a cross section of the tube is the same at all points along the tube. If v is the average velocity of the fluid over any cross-section whose area is A, and d the density of the fluid, the quantity that passes in a unit of time is vAd. The manner in which the fall in pressure along a tube and the quantity of fluid flowing through it depend upon its length and cross section, has been found as the result of numerous experiments, and is expressed in various empirical formulae. Irregular Tubes. \u2014 If the fluid is flowing uniformly through a tube of irregular cross section, the quantity passing all cross sections in a unit of time is the same; and that passing any one cross section is, as shown above, vAd, where v is the mean velocity over the section, A is the area of the section, and d is the density of the fluid at that point. If the fluid FLUIDS is flowing slowly, there is no change in density at different points, and therefore there is the following relation between tin* velocities at different cross sections, VjAj = v^Ay in which t-j is the vdority at a point where the cross section Lfl ami j-2 is that at a", " point where the cross section is Ay '1 MS that, a< the cross section diminishes, the velocity in- creases, and vice versa. (Compare the flowing of a i into a lake and out through a narrow defile.) If tin- velocity iic'i'eases from one point to another in the direction of motion of the fluid, there must be a fall of pressure in this direction so as to produce this increase in the velocity. II' nee, if the velocity increases in one direction, the pres- sure must fall in this direction; and conversely. When the velocity is greatest the pressure is least, and where t la- the pressure is greatest. This is illustrated by the atomizer, the steam injector, the hall nozzle, etc. Another illustration of this principle is furnished by the \u2022\u2022 -\"ball. If a ball were in motion of translation only, < 1 be symmetrical about its Hue of motion. As >ved through the air, a layer of gas would stick to it; and, owing to friction between \u2022 and the surrounding air, currents would be produced h were symmetrical on all sides. But, if th*> 4 also as it moves, things are cliff- - advancing, as indicated in th. \u2022\u2022 direction of the arrow and be spinning on an to the plane of the pap<M clock I I on the si.l- mark-d A is equal to the turn of r, the he ball, and rA, where r In the ball and A..city of a point of the ball on the side marked B is th. wwn these. Consequently. th<> < - the air by friction are i greater on the former aide; and the relative* Telocity between the i nd the ball is!\u2022\u2022*\u2022*; therefore the /\u00bbrr\u00abruir in the air is greater. iff to this fact the ball is pushed sidewiae in the direct A to*. 170 MECHANICS Solid moving through a Fluid. \u2014 As a solid moves through the air, there are forces that oppose its motion ; and many experiments have been performed to determine the connec- tion between this force and the velocity of the moving body. Newton proposed the law that the force varies as the square of the velocity ; but the accepted relation to-day is due to Duchemin : R = av2 + bv8, for speeds below 1400 feet per second, where R is the force, v the velocity, and a and b are constants. The forces acting on a solid moving through a fluid are the same as those it would experience if it were at rest and the fluid flowed past it in the opposite direction. If a board is placed obliquely across a current of a fluid, the pres- sure will be greater at the edge which is \" up stream \" than at the other, because the stream strikes it directly and then flows down along the board. There is thus a moment which turns the board directly across the current. Similarly, as a board or a piece of paper moves through a fluid, for instance the air, it turns so as to move broad face forward. This is illustrated by a sheet of paper or a leaf falling in air, by a flat shell falling through water, etc. Liquids and Gases As has been said several times, both liquids and gases are fluids; but liquids are distinguished by the fact that when contained in an open vessel they have a free upper surface in contact with the air, or if left to themselves they form drops inclosed in spherical surfaces, while gases completely occupy any space open to them ; liquids are comparatively incompres- sible, while gases can be easily compressed. In discussing, then, the properties of a liquid as distinct from a gas, its surface of separation from other bodies and its incompres- sibility form the features to be studied; while the corre- sponding properties of a gas as distinct from a liquid are its power to expand so as to fill any space and its great compres- sibility. CHAPTER VII LIQUIDS Compressibility of Liquids. \u2014 For many years it was thought that liquids were absolutely incompressible; but later it was shown that all liquids could have their volumes changed by the application of sufficiently great pressure. This is done by tin- piezometer, which has already been described. The liquid whose compression is to be studied is placed in a glass bulb provided with a fine-bored stem; this stem is open and contains an index which moves up or down as the volume of the liquid in the hull) is altered; the bulb is placed in piezometer, which is filled with some transparent liquid like water. As the piston is screwed in, the pressure increases, and Uoth the bulb and its contents are compressed. The me may be determined if the force acting on the pis- ton and its area are known. Owin^ t<> the compression of bulb itself the index would rise", ", atmospheric pressure and gra\\ Written P, and it tin- \\ntieal depth of the 1 in quest! -n In-low tli.- Mil-far,- is //. th.- d. | the liquid being rf, the total pressure at the point is P + dgh. Thi> pressure is entirely independent of the shape or size of 174 MECHANICS the vessel, depending simply on the vertical distance between the point and the plane in which lies the upper surface. For, consider any point Q in a liquid at rest which is con- tained in a cylindrical vessel. The pressure is the same at all points in a horizontal plane through it, otherwise the liquid would flow, and the same is evidently true of a liquid contained in any other shaped vessel, e.g. a conical one. But the pressure at Q is P+dgh\\ and therefore all other points in the horizontal plane through it have the same pressure, wherever they are. (The pressure at a point under the slope of the side wall in the vessel shown in the cut owes its value directly to the weight of the column of liquid vertically above it and to the component downward of the reaction of the sloping side wall against the thrust of the liquid.) Fro. 84. \u2014Pressure at a point whose dis- tance below the free surface is h. The pressure at a point in the liquid increases uniformly as the point is taken lower and lower ; and therefore the mean pressure in the case when the vessel is a cylinder is the average of the pressure in the surface, P, and that at the bottom, P + dgH, where H is the vertical depth ; that is, it is p _j_ 1 dgH. The force on the bottom of the vessel, if it is horizontal and if its area is A, is (P-f dgH) A\\ and the thrust on the side wall, if it is rectangular and vertical, is the average pressure multiplied by the area of the wall in contact with the liquid. If this area is A, the thrust is then (P + J dgH) A. Its point of application is found from the consideration that it is the resultant of a great number of parallel forces whose values increase uniformly from the surface down. In the FIG. 85. \u2014 A vessel with oblique walls. case of a rectangular wall, this \" centre of pressure,\" as it is called, is at a distance of one third the depth of the liquid LIQUIDS 176 from the bottom. In the general case, when the wall is not rectangular, the total thrust is found by adding all the infinitesimal parallel forces which act on the elementary portions of the wall ; and the centre of pressure is found 1 > y expressing the fact that the moments of these infinitesimal forces around any axis must equal tin* moment of tin- thrust when applied at the centre of pressure. These operations require, however, the use of the infinitesimal calculus. If a liquid at rest is contained in a vessel that has several vertical branches of different shapes and sizes, its upper surface is at the same horizontal level in them all, provided they are not of such small bore as to cause capil- lary phenomena. This is evident from the fact that the pressure at all points in a horizontal plane AB through the body of the vessel is the same \u2014 otherwise the liquid would flow; and therefore the free surface must be at the A... FIG. M. \u2014 Same liquid* In conntctln* tubes. AB and CD are horizontal plan**. same height above this level in all tin- branches. Similarly, the pressure is the same at all points of the liquid in the branches that lie in the same horizontal plane; e.g. in the 10 CD. Liquids in Connecting Tubes. \u2014 If two liqui do not mix are |\u00bbla\u00ab ed in the same vessel, the denser will sink to the hot torn, because by so doing the potential energy becomes A heavier liquid may, however, rest upon a lighter one provided there is no jarring; but the equilibrium is unstable. 176 If a vertical U tube contains two liquids that do not mix, the levels of the upper surfaces of the two liquids in the two branches are not the same. The heavier of the two liquids will occupy the bottom of tlio tube and rise to a certain height in one of the arms, while the lighter one will stand to a greater height I in the other. The pressure at the fl surface of contact between the two I liquids is the same as at a point in '\u00ab! \u2014 *-- the heavier liquid at the same hori- zontal level, from what has been said in the previous paragraph ; so, FIG. ST.- TWO liquids of different if the upper surface of the heavier liquid is at a vertical distance h^ above this level, and that of the lighter at a distance 7/2, P + djt", "/Aj = P + d^gliy where d1 and d2 are the densities of the heavier and lighter liquids respectively. Therefore, density which do not mix..,.. d^h^dji^ and -^i = -,-?\u2022 This is, then, a method for the \u00ab2 \"i determination of the specific gravity of one liquid with ref- erence to another; and, if the density of either is known, that of the other may be at once calculated. Barometer. \u2014 The pressure of the atmosphere is, as a rule, measured by balancing it against a column of mercury. The apparatus consists of a long, wide tube, which is closed at one end and which contains a column of mercury, but no air or other gas (except mercury vapor). The tube is placed in a vertical position; and either its open end dips into a basin of mercury or the tube is bent into the shape of the letter J. The space above the mercury may be entirely freed from gases by different means. (One is to hold the tube, closed end down, fill it with mercury, cover the open end with the finger, invert it carefully, and place it upright in a basin containing mercury with the open end under the LIQUIDS 177 surf; i (\u2022\u00ab-.) The pressure on the surface open to the air holds tin mercury in the tube and is the same as that at a point at its level in the mercury in the tube; so, if the surface of the mercury in the tube stands at a height h above this outer u surface, the pressure due to the atmosphere equals dgh, where <1 is the density of the mercury at the temperature of the air, because there is no pressure on the top of the column. and t lie pressure at any point is that due to. ity alone, (h is observed to be about \u2022in. when the pressure is measured at sea level; and so the tube must have at : this length.) It should be noted that this height is independent of the shape or H section of the tube, provided only that it is so wide as to avoid capillary action. age 188. If a liquid other than mercury were used, it would stand at a height as much greater than this as the density of mercury is greater than its own. (Thus, since the density of mercury is about 13.6 times that of water, liquid would be forced up in a barom- eter by ordinary atmospheric pressure to a iit 1.').\u00bb> x 7\u00abi. MI. ) The pressure of the M ;_n\\eii by a barometer is dgh dynes per square centimetre on the ('.(J.S. system. Flo gg..^. however, ordinarilv expressed in \"cen- b*r<...... Mid dlphon forms. timetres of merciir\\. that is, in terms ol a unit pressure equal to that due to a vertical column of mercury at 0\u00b0 C., 1 cm. in height, when g has the value it possesses at sea level and!.\"> latitude. (See page 166.) Calling this value of the deiiMt\\..f mercury dQ, and that '. //4V this unit pressure equals d^g^ dynes per square S.i. it the observed height of the barometer is AMES'S PHYSICS\u2014 12 178 MECHANICS h cm., when the temperature is t\u00b0 C. and the latitude is Z, we may write the value of the pressure in dynes per square centimetre, P = dgh ; and its value in \" centimetres of mer- cury,\" therefore, is p = This is a height ; and so we speak of a barometric \" pressure of 76 cm.,\" meaning \" 76 cm. of mercury.\" The relation between g and g^ is known (see page 131) ; and that between d and dQ is also known. For, as will be shown presently, if t is the temperature on the Centigrade scale, d0 = d(\\ + 0.00018180- Therefore, if the temperature and latitude at the place of observation are known, the pressure in terms of centimetres of mercury may be at once calculated from the observed barometric height. In the formula, h is the height in centimetres. Sometimes, how- ever, the reading is made on a divided scale which is correct at 0\u00b0 C. ; and in this case the readings must be corrected in order to give A. If the scale is made of such a material that each centimetre increases in length an amount a cm. for each degree rise in temperature, two divisions which are 1 cm. apart at 0\u00b0 are a distance (l+a\"0 cm. apart at t\u00b0 C. Conse- quently, if the observed reading is H scale di- visions, the", " height is H (\\ + at) cm. ; and therefore h=H(\\+af). Open Manometer. \u2014 In a similar manner, the pressure in a gas inclosed in any vessel can be measured. Let a bent tube containing some liquid be joined to If there is a difference in pressure FIG. 89. \u2014 Open manometer. the vessel as shown. LIQUIDS 17l\u00bb l)ct \\veen the gas inside the vessel and the air outside, there will be a difference in level of the columns of the. liquid in tin- two arms. Call this difference A, and the density of the liquid d. Then, if the level is lower in the arm at i ached to the vessel, the pressure in the gas inside is P -f- dgh. (If the level is higher in this arm, the pressure if I' \u2014 dgh.) This instrument is called an \"open manoni- The pressure, as here expressed, is in dynes per square centimetre, if the C. G. S. system is used ; its value in centimetres of mercury can be deduced, as has just been ained for a barometer. Floating Bodies. \u2014 There is one application of Archimedes' principle to liquids that is of special interest. It is to the case of a body floating on the surface of a liquid. If a solid of less density than a liquid is immersed in it and allowed to move, it will rise to the surface, but will come to a position quilibrium when, as it floats, it displaces a volume of the liquid whose weight equals its own; for, under these con- ditions, tin- upward buoyant force due to the liquid equals tin- downward weight of the solid. The line of action of the fon IK r is vertically through the centre of gravity of the di-jilaccd liquid: that of the latter, vertically through the <\u00abntre of gravity of the floating body. Therefore, when there is equilibrium, these two centres of gravity must lie in thcsamr vertical line; otherwise there would he a moment which would make tin- body turn around a hori/ontal axis. This equilibrium is stable if, when the body is tipped slightly, the resulting moment is in such a direction as to turn it back a.^aiu : it is unstable if this moment, under similar < onditions, is such as to tip it over. Thus, a board :HLT on its side is in stable equilibrium : but, if made to float upright, its equilibrium is unstable. Osmosis and Osmotic Pressure. \u2014 As one substance dissolves in another, it breaks up into small particles which diffuse through the solvent. These particles by their presence affect 180 MECHANICS its molecular forces, as is shown by many facts. One of these may be mentioned here. It is found that certain solid bodies allow some liquids to pass through them, but not others (see page 141) ; and it is possible to make a membrane that will permit the molecules of a liquid to pass through perfectly free, but will not permit the passage of any dis- solved molecules. Such a membrane is called \" semi-perme- able.\" If now a solution is placed in a wide tube closed at one end with such a membrane, and is supported upright in a large vessel containing the pure liquid, which can pass through the membrane, it is observed that the levels of the liquid in the tube and in the outer vessel are not the same, as they would be if the membrane were absent; the height of the level in the tube is the greater. There is therefore on the two sides of the membrane a difference in hydrostatic pressure which is maintained by some force due to the difference in the conditions on these two sides. The molecules of the solvent can sure: inner tube contains pass freely through the membrane, and solution ; outer vessel, pure they Continue to do SO Until, when 60 ui- golvent. J librium is reached, the hydrostatic pres- sure prevents any further passage. (Of course molecules may still continue to pass through ; but, if they do, an equal number pass out.) There is therefore a difference between the pure solvent and the solution. If the density of the solvent is d, that of the solution does not differ much from this ; and if the difference in level of the two free surfaces is h, the hydrostatic pressure dgh measures this tendency of the pure solvent to pass through the membrane into the solution, that is, it measures the effect the dissolved sub- stance has upon the solvent in affecting its molecular forces. This passage of a liquid through a membrane or Fio. 90. \u2014 Osmotic pres-.,..,. LKiflDS 181 ml porous partition is called \"osmosis,\" as has been already stated ; and the above pressure, dgh, is said to measure the \"osmotic pressure\"", " of the solution. Experiments show that, as the solution is made more and more concentrated, the osmotic pressure increases. If m grams of a substance are dissolved in ml grams of a solvent, the ratio - - is called tin- *l concentration\"; and in certain simple cases the osmotic pressure varies directly as this, while in others it varies more rapidly. This law is the same as that for a gas ; viz., the pressure varies directly as the density of the gas, i.e. its concentration ; but if the gas changes its character by its molecules dissociating into parts, then the pressure varies more rapidly than the density. (See page 200.) We are thus led to believe that this abnormal osmotic pressure is due to a dissociation of the dissolved molecules into simpler parts. Liquids in Motion Efflux of Liquids. \u2014 If a liquid is contained in a large 1 which has thin walls, and if a small opening is made in cither the bottom or side, the liquid will escape. This motion is called \"efflux\" or \"effusion.\" The velocity of escape may be at once calculated, because, since the wall is assumed to be thin, there is no friction, and since the opening is small, we may neglect any motion of the liquid except that of the escaping stream. Thus the phenomenon is the same as if a drop of the liquid disappeared off the surface and reappeared lower down with a certain speed. ic opening is at a depth // below the surface, and if the speed of ctllux is x, each drop of mass m loses an amount of potential energy mgh and gains an amount of kinetic energy % mt*. Therefore these two quantities are equal, or * = V2 gh. This is, of course, the formula for a particle falling freely toward the earth; and therefore, if the jet 182 MECHANICS were turned upward, it would rise to the height of the level of the liquid in the vessel, were it not for the opposing action of the air. (This formula was first deduced by Torricelli.) The pressure in the liquid at the opening is P + dgh, while that on the outside is P ; so the difference in pressure caus- ing the flow is dgh. Calling this jo, the speed of efflux may be expressed in terms of it, viz., &= 2^-, or \u00ab=\u2022%/\u2014\u00a3\u2022 The direction of the jet depends, of course, upon the position of the opening ; and, unless this is on the bottom, the path of the jet is a parabola. a % a Other cases of motions of liquids will be discussed in Chapter IX. Capillarity and Surface Tension Fundamental Principle. \u2014 If a liquid is left to itself, free from external forces, it assumes the shape of a sphere ; and this is approximately the condition with falling drops of rain or of molten metal (like shot) and with soap bubbles. It is rigorously so if a small quantity of a liquid is immersed in another liquid of the same density with which it does not mix \u2014 Archimedes' principle. Of all solid geometrical fig- ures having the same volume the sphere has the least sur- face ; so this fundamental property of a liquid surface is that it becomes as small as it can. Thus the surface of a drop contracts until the resulting pressure in the liquid balances the contracting force ; it requires a force to blow a soap bubble, and, if one is left attached to the pipe and the lips are removed, it will contract. Again, if a glass plate is dipped in water (or any solid is dipped in any liquid that wets it), and is then raised slightly, the surface of the liquid near the plate is curved with the concavity upward. It has contracted from a rectangular shape, in doing which some of the liquid is raised above the horizontal surface ; and the liquid comes to rest when the weight of this elevated portion LIQUIDS 183 balances tin- contracting force of the surface. Similarly, if a glass plate is dipped in mercury (or any solid is dipped in any liquid that does not wet it), the surface of the liquid near the plate is curved s<> as to be convex. Since the liquid does not wet the plate, its surface continues around hel< >\\v the plate ; and, as it contracts, it rounds off the corners, thus leaving a free space which the force of gravity would cause the liquid to till were it not for the contracting force of the surface. There is equilib- rium, then, when these two forces balance each other. This phenomenon in the neighbor- Fio. 91. -Capillary action when a solid plate is dipped in a dipping ill a liquid liquid: (1) when the liquid wets the solid ; (2) when the liquid is most marked doe8not when the former is a tube with a small bore. If such a tube is dipped in a liquid that w", "ets it and is then raised slightly so as to leave a liquid surface on the inner and outer walls, the whole liquid surface includes that on the walls and that \"f the liquid in which the tube dips. So considering the liquid surface inside the tube, it has the appearance of the inside of the finger of A glove. Tin- liquid is then raised in the tube, owing to the contraction of the surface; ami equi- librium is reached when this emit ract MILT force is balanced by the effect of gravity on the raised portion of the liquid. Similarly, when a glass tube is dipped in mercury, the surface in t he t Illic is de[i| rssrd. Surface Tension. There is, therefore, a force produced by a liquid surface; and the simplest manner of defining it is to consider the force acting across a line of unit length in the surface. This is a molecular force and is evidently du \u2022 to the fa.-t that a molecule in ih- a liquid is in a 184 MECHANICS different condition from one in the interior. For a surface of a given liquid in contact with a definite medium, then, no matter whether its area is large or small, this force is a con- stant quantity ; and unless it is stated otherwise, the sur- rounding medium is always understood to be ordinary air. The force acting across a line of unit length of the surface of a given liquid in contact with a definite medium is called its surface tension with reference to the medium, and has the symbol T. A simple direct experiment showing the amount of this tension is to construct a rectangular frame of wire, one side of which is movable, and to make a film of liquid _\u2022 fill the area. (This may be done by dip- ping the frame for a moment into soapy water and then removing it.) It will be found that a force must be exerted on the movable wire to keep the film from contracting. Calling the length of the movable wire Z, this force equals 2 Tl, because the film has two surfaces. If under the action of the force the wire is moved so as to make the film larger, work is done. If the distance the wire moves is called x, this work is 2 Tlx ; but the increase in area on the two sides is 2 Ix, and therefore the work done per unit increase of area is T. In other words, to increase the surface of a liquid by one unit of area requires an amount of work equal to T\\ or, the potential energy of a surface of area A is numeric- ally equal to TA. As the surface of a liquid is increased, it is not stretched, as a rubber bag or toy balloon is, but new surface is formed by molecules coining up into the surface from the interior ; the new and the old surfaces being identically alike. (If in the experiment just described the movable wire be drawn out too far, or if a soap bubble be blown up too large, a point is reached when the two surfaces of the film come so close together that, if their area is FIG. 92. \u2014 A soap-film stretched on a wire frame, one side of which is mov- able. LIQUIDS 185 further increased, the interior molecules are no longer in the lition in which they are when the film is thicker; and all the properties of the film are changed.) Connection between Pressure and Surface Tension. \u2014 Let us n<>\\\\ examine imuv closely the illustrations of the contracting force of a liquid surface that were given above. A spher- ic al drop may be considered as made up of two halves touch- ing at an equatorial section ; they are held together by the ion in the surface, acting across the equator; and there is a reaction, as shown by the pressure in the liquid, acting over tin- equatorial section in which the two halves touch. If r is the radius of the sphere, the length of the equator is 2?rr, and the force of contraction due to the surface tension across it is therefore T2irr; the area of the equatorial section is Trr2, and, if p is tin- pressure in the drop, the force of reac- <\u00bbver this section is ptrr*. Since the drop is supposed to have ceased to contract, these two forces must be equal, or 2 T T - 77 /\u2022 : i.e. p = \u2014 \u2014. This pressure is felt, of course, throughout the drop, and is in addition to the pressures due to the atmosphere or to gravity. The formula states that, if a liquid whose surface tension is T has a portion of its sur- face (or all of it) curved in the form of a >phere of radius r, there is a pressure - - toward the centre of curvature, ami \u2022_' '/' i IK- surface is at rest, there must be an external pressure \u2022'", " is given by the same formula. This formula can be used to measure the surface tension of a liquid ; for 7&, d, g, and r can all be measured with accuracy. There are, however, other'methods which in some respects are more satisfactory. The values of the FIG. 94. -Capillary action be- surface tension of a few liquids in con- tween mercury and glass. ta(jt ^^ ^ jn dynes per centimetre, at about 30\u00b0 C., are given in the following table: Water. Mercury Alcohol TABLE 72.8 513. 22. Olive oil Turpentine Petroleum 34.6 28.8 29.7 Another mode of considering the curvature of a liquid surface near a solid wall is as follows : imagine the liquid to have a horizontal surface clear up to the wall; a particle of the liquid surface near the wall will be under the action of three forces (apart from gravity), viz. : 1, one owing to the molec- ular forces of the rest of the liquid ; it is represented in the cut by F\\ 2, one owing to the molecular forces between the liquid and the solid; provided the latter is wet by the former, it is represented by Fl ; 3, one owing to the forces between the upper medium \u2014 the air, in general \u2014 and the particle of liquid; this is neglected here. The resultant of these is represented by R; and therefore the surface of the liquid must be so curved as to be at right angles to its direction. FIG. 95. \u2014 Forces acting on a par- ticle of a liquid in its surface at a point near a solid wall. If the surface of the liquid is not spherical, the formula for the pres- sure may be shown tobe/>= T( 1 ), where Rl and R2 are the so- \\Rl R.2/ called principal radii of curvature. Thus, if the surface is a cylinder of LIQUIDS 189 T radius r, Rl = 0 and R = r ; so p = \u2014 \u2022 This formula may be proved di- rectly by considering a cylindrical drop instead of a spherical one as done on page 185. It may be shown further that, if a cylindrical por- i\u00abf liquid (as, for instance, a jet of water issuing under considerable pressure) has a small radius, it breaks up into spherical drops. The ex- act statement is that if / is the length of the cylinder and r its radius, the surface is unstable when I > 2 TIT. Effect of Surface Impurities. \u2014 If a liquid is homogeneous, the surface tension of all points of its surface must be same ; and therefore it is impossible to blow a bubble of such a liquid or to stretch a film of it on wire frames ; for it is evi- dent that, if a vertical film exists, the tension in the upper portions of the surface must be greater than in the 1< i use the former have to support the weight of the latter, and such inequalities in tension cannot exU in a homogene- ous surface. If bubbles or films are to be formed, thm. tin- liquid >urface must be made heterogeneous, e.g. soapy water be used. In these films the upper portions are observed to be purer than tin; lo\\\\\u00bb-r ones, showing that a pure film has a greater surface tension than a contaminated or hetero- geneous one. If iiiimit.' portions of camphor ar 1 upon a clfan surface of water, they are seen t...lari '\u2022 iW\\\\ar\u00abl \\\\i\\\\i a m. i H stopped by an extremely small trace of oil on the surface. A piece of camplmr is never exactly symmetrical and there- fore dissolves in the water more My at one point than at Jg. lluton the surface :.- makes < n elsewhere ; and so the face < drawn away 1-y th.- he opposit.i.IT-..J. it spreads the surface under the action of the surf art* tensions of the various re are three of these forces acting on any portion of the edge of the drop: the tension 7*, between the water and the air, T9 190 MECHANICS between the oil and the air, Tj between the oil and the water. The oil spreads because 7'3 is greater than the resultant of T2 and Tr This thin layer of oil will prevent the dissolving of the camphor, and its motions will cease. If a drop of alcohol is poured on a glass plate that is covered with a thin layer of water, the tension of the surface of the solution of alcohol is so much less than that of the pure water that the latter surface con- tracts, tearing apart the former and leaving the glass quite dry. The surface tension of a liquid varies with the tempera- ture, decreasing as", " the latter rises. This may be shown by many obvious experiments. Ripples ; Effect of Oil upon Waves. \u2014 It requires work to increase the area of a liquid surface, and so if the surface is increased slightly by some disturbance, there is a force of restitution, and waves will be propagated over the surface, which are quite distinct from those due to gravity. These are due to surface tension, and the crests come so close to- gether that they are called \" ripples.\" These may be seen if a fine wire in a vertical position and dipping in a liquid is moved sidewise. They will be discussed later. As the wind blows over a liquid surface it will soon magnify ripples into regular gravitational waves, and it is evident that the less surface there is exposed to this action of the wind so much the less is its effect. If a thin layer of oil is spread over a liquid surface, the wind blowing over it will tend to gather the oil in the same regions where it would heap up the water ; this excess of oil over one region produces a scarcity of it over others, and the surface tension in the latter is therefore greater than in the former, and so it opposes the action of the wind in causing the water to form waves. CHAPTER VIII GASES General Properties Dalton's Law. \u2014 The fundamental properties of a gas as (list in. i from a liquid are explained by assuming that its molecules are so far apart and have such freedom of motion that it distributes itself uniformly throughout a vessel of any shape and may be compressed with ease into a much small* -r 1 he aetnal space occupied by the particles of the gas must l>e \u00ab-\\tremely small, because if two different gases are inclosed in the same vessel, they mix uniformly, and each hehaves almost perfectly as if the other were al^ent : the pressure at any point (or on the walls) is the sum of the pressures \\\\hieh each gas by itself would produce. This is km.un as Dal ton's Law of (iases. ha\\in^ heen discovered by him in ISM-J; hut careful experiments show that it is not utely exact. Effusion. \u2014 If any gas is inclosed in a vessel that is sur- rounded l>y a d i fferent gas, and is allowed to escape into the ior all practical purposes as if it were escaping ;i vacuum, for each gas acts independently of the other, and the only difference enters from the fact that a longer time is required for a gas to diffuse into another gas than ; ilmte itself in an empty space. If the in the vessel f-.-m whieh the gas is escaping is small, and if \\alls are thin, the same formula applies as for the efllux i.piid. vi/.. f.y\u00b1\u00a3, where * is the sj d of the 191 I (I \"2 MECHANICS ing gas, d is its density, and p is its pressure. (JP is really the difference in pressure of the escaping gas, inside and out ; but if the outside space is very large compared with the vessel, the pressure outside may be neglected.) While the one gas is escaping, the surrounding gas is entering, and the two processes go on independently. This relation between the velocity of escape of a gas and its density sug- gested to Bunsen a method for determining the relative den- sities of two gases, because the quantity of gas that escapes in a given time can be measured with ease. If the openings through which the gas escapes are extremely fine, a different law is obeyed, which was investigated by Graham. Sensitive Flame. \u2014 If a gas escapes through a tube with a small opening under considerable pressure, it forms a \"jet,\" which preserves its identity for some distance without diffus- ing into the surrounding gas. The jet is inclosed, as it were, in an envelope which prevents the two gases from mingling. This envelope is made up of particles of the escaping gas which are given a rotation owing to their rapid motion over the particles of the surrounding gas. If this envelope can be broken into, the two gases mix and the jet loses its character. If a jet of illuminating gas is formed in this way and is lighted, it forms a tall, narrow flame nearly cylindrical ; but if a whistle is blown, or a bunch of keys rattled near by, the jet breaks down into an ordinary fan-shaped gas flame. This is owing to the disturbances sent out by the sounding body in the foim of short waves or pulses in the air, which disrupt the gaseous envelope of the flame. For this reason a jet like this is called a \"sensitive flame.\" It is used to study the many wave phenomena in connection witli Sound. Compressibility of Gases. \u2014 If a gas is inclosed in a cylin- der provided with", " nothing about the actual size or shape of a molecule ; but we can prove that, if we had inclosed in a vessel with rigid walls a great number of small, perfectly elastic spheres, moving at random but with great speed, this collection of particles would have many properties similar to those of a gas. For ease of calculation, let us assume that the vessel is a rectangular one, having edges of length #, 6, and c. At any instant a definite par- ticle has a certain velocity, but owing to impacts with other particles and with the walls, this changes frequently, and this is true of all the particles ; so, apparently, there is no regularity. But if things are in a steady state, there is a certain unvarying proportion of the particles \u2014 not the same individual ones, however \u2014 that have a given component velocity parallel to any one edge of the rectangle. Let N be the number of particles per unit volume that have the component velocity v parallel to the edge whose length is a; then the total number in the vessel that have this com- ponent velocity is Nabc. If each of these particles has the mass TW, the momentum of each parallel to the edge referred to is mv ; and therefore as each strikes the wall at the end, and its velocity is changed from v to \u2014 v, its momentum is changed from mv to \u2014 mv, or by an amount 2 mv. The time taken for a particle with the velocity v to pass from this wall across to the other end and return is - \u2014 (This is the time taken for the effect of the particle to be again felt at the wall, if, instead of moving over the whole distance and back again, it impinges on another particle and so hands on its momentum.) Therefore the number of impacts each par- ticle makes on the wall in a unit of time is - \u2014 Since at each impact of each particle there is a change in momentum \u2022> GASES 199. the change in momentum at the wall owing to one par- tiele during a unit of time is \u2014 \u2022 2mv or ; and, as there a are Nabc particles having this component velocity v, the change in momentum in a unit of time, owing to the particles whose component velocity is v, is Nabc \u2022 \u2014, or NmiPbc. This is, thru, the force on the end wall owing to the above particles; and since the area of this wall is fo, the pressure is NmiP. The total pressure is found by adding up similar terms for the particles having other velocities. The number of par- ticles per unit volume, however, which have the component velocity \u2014 v must also equal JVnn the principle of probabili- and so the number which have the same value v2 is 2N. Therefore, writing vv v2, etc., for different velocities and Nv NT etc., for corresponding numbers per unit volume, otal pressure is (2 N}v* + 2.Vsy2a + etc.) m. Further, the mean value of ^ 2 JVy^ -f 2 JV>,2 + etc. (See page 30.) But 2 JVj + 2 AJ + etc. is the totoJ number nf particles in a unit volume; call its value N and write for the mean value of v2, v2. Then the total pressure equals mN\\?. The tntal velocity V of any particle expressed in terms of its components parallel to three directions at ri^ht lc,s to each other, u, v, \u00ab\\ is given by F3 = u2 4- v2 + \u00ab*\u2022 l'>ut, since the nun ion of the particles is a perfectly random t he m. MII \\ alaes \u00ab -f //-, \u00bb\u2022-. and w2 must be equal ; so \\\\ i it - the mean value of F*, P^-Sv2: and the tin d : for the pivssiire owiii'^ to the {.articles 18 p = JmA^M. Since tVis the t..tal niii: particles in a unit volume. and since each particle has a mass m, the density, d, or the 200 MECHANICS mass in a unit volume, is 7HJV; hence the pressure may be written i This states that, if the mean kinetic energy of translation of the particles does not change, the pressure varies directly as the density. This is Boyle's law, assuming that the tem- perature of a gas corresponds to the mean kinetic energy of translation of its particles. If this formula can be applied to an actual gas, the mean squared velocity of its molecules may be at once calculated, because V2 = \u2014\u00a3, and the density of a gas at a certain pres- sure may be determined by experiment. The density of a gas varies with its temperature as well as with the pressure,", " and the manner in which any increase in the kinetic energy of one portion is distributed throughout the whole set ; and if we have two sets of such particles, we can explain the diffusion of one into the other. Further, we can calculate what the force of viscosity, the rate of distribution of kinetic energy, i.e. of conductivity of \"heat,\" and the rate of diffusion of such sets of spheres are, expressing these quantities in terms of the mass of a particle, its mean energy, its mean free path, its radius, and the number of spheres in a unit volume. Then, if we assume that an actual gas behaves approximately like a set of spheres, we can measure its pressure and density at a given temperature, its viscosity and conductivity for heat, and its rate of diffusion, and, by comparison with the me- chanical formulae deduced for a set of spheres, obtain approxi- mate values for the various properties of a gas molecule. A OASES of these may be mentioned. At a pressure of 76 cm. of men Miry and a temperature of about 20\u00b0 C. (i.e. 70\u00b0 F.), the in. an free path of a hydrogen molecule is 0.0000185 (in., and tin- number of impacts it makes in a second is 9480 million ; for oxygen, these figures are 0.0000099 cm. and 0 million ; for carbonic acid gas, 0.0000068 cm. and 5510 million. By various processes the dimensions of a molecule and the number in a unit volume may be approximately determined ; the \" radius \" of a molecule is found to be of the order of a ten-millionth of a millimetre, and the number in a cubic centimetre is of the order of 2 x 1019, i.e. twenty quintillions. Fourth State of Matter. \u2014 If a gas is inclosed in a glass bulb which can he gradually exhausted by means of an air pump, as will be explained later, the most evident change produced is the decrease in density and the consequent increase in length of the mean free path. (If the exhaus- tion is (anied so far that the pressure in the bulb is that of one thousandth of a centimetre of mercury, the mean free path is 7630 times as great as it is at the pressure of 1 cm., or about 1.5 cm. for hydrogen.) The properties of matter in this condition are quite different from those of ordinary gases; and for this reason the matter is now said to be in a urth State.\" Its chief properties were investigated by \\Villiam Crookes, and they will be described later when electrical phenomena are discussed. One purely mechanical property should, however, be mentioned here. It is illus-.straicd l.y the following experiment : a framework is made \u2022>tin'_r of t\\vo or more CM-OSS arms, which carry at each i small piece of mica blackened on one face and not on tin other: the plane of each mica vane is perpendicular to that of the cross arms, but includes the line of direction of the arm which carries it, and the blackened face of one vane is turned toward the polished face of the next one. This little wheel i> suspended in a Imlh in such a manner as to be 204 MKCHAMCS KIG. 100. \u2014 Crookc radiometer. free to turn alx>ut an axis perpendicular to the plane of the cross arms. If a hot body, like a burning match, is brought near the bulb, nothing noticeable happens if the gas inside the bulb is at ordinary pressure ; but, if the gas is exhausted to a few thousandths or hun- dredths of a centimetre of mercury, a stage is reached when the framework begins to rotate on its axis in the direction which it would move if the blackened face of each vane were repelled by the hot body. (If the gas is exhausted as completely as possible, this motion does not arise ; and at slightly higher pressures there are complications in the phe- nomena which need not be discussed here.) The explanation of this action is as follows: a blackened surface becomes hotter than a polished one when exposed to a source of heat, as is known to every one, and therefore those molecules in the gas that rebound from or leave the blackened face of a mica vane do so with a greater velocity than those that rebound from the polished faces. If the gas, however, is at ordinary pressure, the impacts are so numerous that the condition throughout the bulb is nearly uniform, and there is no mechanical action on the vanes ; but if the gas is so rarefied that the mean free path is a centimetre or more, the molecules leaving the blackened face give the vane a backward push which is not counterbalanced by that given the other face of the vane", ", and consequently the framework rotates in the direction described above. This instrument is called a \"radiometer\"; and its action was discovered by chance by Sir William Crookes in the course of a research which involved weighing small amounts in a vacuum. It has been modified recently so as to serve as an instrument for measuring certain quantities that are of importance in the discussion of heat phenomena. CHA1TKU IX HYDRAULIC MA( HIXKS: 1TMPS, ETC. Barker's Wheel. \u2014 This is n simple instrument, consisting of \u2022>[ lixed tube to one end of which is attached by a pivot a framework consisting of a number of cross tubes, whose ends are bent at right angles t<> their length and to that of tin- fixed tube. Connection is made through the pivot from tin- fixed tube into the mova- ble ones. If now a stream of. either liquid or,reed through thr li, tube and out the bent ends of the cross tubes, the Litter will rotate rapidly in a direct ion opposite to that in which the fluid escapes. This is then simply an illustration of the conservation of momentum.. he.-ls an- often seen connected with devices for 101. \u2014 Barker's wheel sprinkling lawn> with water. Turbines. \u2014 A turbine consists \u00ab>f a wheel that can rotate <>n an axis and that lias I'm- i; --s curved flanges SO I that as a lluid presses against them it exerts a uioinrnt annnid the a\\i*. ( ( 'ompare the action <\u00bbf the wind \u00ab'ii the fans of a windmill. ) In e, a water turbine is it the bottom of a deep well (or high cylindrical till).' ) \\\\ hi< 1 ihe wlie- under great pi* -T is admitted through the Ml 206 MECHANICS turbine near its axis, flows out along the flanges, and escapes at the edges, so that the wheel is set in rotation by the pressure. In a similar manner steam can be used to drive a turbine, as is done in the so-called \" turbine boats,\" in which there are several turbines fastened directly to the shaft of the boat. Hydraulic Ram. \u2014 In this instrument, which was invented by Montgolfier, in 1796, and is in such general use for forcing water from springs into tanks at a considerable elevation above them, the principle made use of is that a large quantity of water falling through a small distance may raise a small quantity through a great distance. A simple form is shown in the cut. The essential features of the machine are a large tube, down which the water flows and which is closed by the escape valve P opening inward, and by another valve Q opening outward into an air-tight reservoir called the \"air cham- ber.\" This contains some air; and into it enters for some distance the outlet pipe which carries the water to the tank. The escape valve P has a weight which exceeds slightly the upward force against it due to the water when there is no flowing \u2014 this upward force is the excess of that on the lower side of the valve over the downward force on its upper side. Therefore, at the beginning of the operation the valve drops ; as it does so, the water escapes, and, since in moving water the pressure is less than in water at rest, the downward force on the upper side of the valve, over which the water is flowing, is diminished, and the upward force is FIG. 102. \u2014 Hydraulic ram. P and Q are valves opening automatically. HYDRAULIC MACH1NX8 1 I'l'.MPs. ETC. 207 sutlii -lent to raise the valve and close it, thus stopping the flow; the valve therefore again drops owing to its weight, and the operation is repeated automatically. When the water is at rest at the beginning of the operation, the other valve Q is down, closing the opening; it remains so as long the eseape valve is open, for the pressure on its lower side is now small since the water is flowing. When the escape valve closes, there is an immediate increase in pressure throughout the whole tube, the outlet valve Q is pushed up and some water enters the air chamber ; then the valve drops as the pressure is thus relieved, and the operation is re- peated. As more and more water enters the air chamber, a time is reached when the level of the water covers the open end of the outlet pipe which connects the chamber with the tank ; after this time, as the water enters, the air trapped above it is compressed and has its pressure increased. Water is thus forced up the pipe into the tank. This operation is more or less continuous; for, as the water enters the air chamber rapidly, the air is compressed and some water flows up into the tank : and then, during the interval of", " time which passes before some more water enters, the compressed air ;ids and continues the flow. Siphon. \u2014 This consists of a large tube or pipe bent into the form of a U \u2022\u2022 it with its two arms of unequal h. This is placed in a vertical.on, with its shorter arm dipping belnw the surface of a liquid in a vessel, and its longer arm outside. l',\\ is of suction applied to the open end. the siphon is now tilled with liquid ; and. if left to itself, the liquid in the vessel will How nut through the siphon until its sur- comes below the end of the short arm. The explana- tion is evident if one considers the conditions that exist 208 MECI1AMCS when the siphon is full, at tin- instant before the flow starts. The pressure in the liquid in the longer arm at a level with the liquid surface in the vessel is, of course, that of the atmosphere P; so, if the open end of the siphon is at a depth h below this, the pressure in the liquid at this point is P -\\-dgh if its density is d. But the opposing pressure is simply that of the atmosphere P, and the difference of pressure dgh forces the liquid out. The shorter arm of the siphon must not be too long ; for if it is greater than the height to which the liquid would rise in a barometer, the pressure on the free surface of the liquid will not be sufficient to force the liquid up to the turn of the siphon. (If the tube is of fine bore, other actions than gravity and atmospheric pressure come into play.) Liquid Pumps. \u2014 These are instruments devised for the purpose of raising liquids from one tank or well into another at a greater height, or for forcing a liquid through a long pipe against friction. There are two types : the u lift pump \" and the \" force pump.\" The former consists of a cylinder in which fits an air-tight piston provided with a valve B opening upward, and whose lower end is closed by a valve A, also opening upward, where the pipe leading to the tank containing the liquid to be raised is attached. The vertical distance from the lower end of the cylinder to the level of the liquid in the well or tank must not exceed the barometric height of that liquid. Then if the piston is raised, some liquid is forced up through the lower valve into the cylinder by the pressure of the atmosphere on the surface of the liquid FIG. ID*. - Lift pump. A in the tank; if now the piston is brought and B are valves opening up-,, 111 ^11 ward8. to rest and then pushed down, the lower.\\fAfin\\Ks: valve drops juul the one in the piston is lifted, the liquid ;n_r through it from below the piston to the space above. When the piston is again 1, the liquid on top of lifted and may escape through a side outlet into a tank ; at the same time more liquid is being drawn up through the lower valve into the cylinder, and the process may be repeated indefinitely. In the force pump the piston 10 valve, and an air cham- ber, like that of the hydraulic, ram, is attached to one side of the cylinder. The explanation of its action is self-evident. This pump is as a rule placed near the surface of the liquid which is to be pumped, and the upper tank may be as high as is necessary. Ki... 105. \u2014 Force pump. Air Pumps, a. Mechanical. \u2014 These are instruments de- i either to force more and more ijas into a given space, 01 to withdraw as much gas as is desired from a closed sel ; in other words, to increase or to decrease the pressure inside the vessel. The former are called \"compression\" pumps: the latter, \"exhaust\" punn The simplest form of exhaust pump is illustrated on pa^c -in. Its mode of action is essentially that of the lift pump :dy described, the main point of dilTerence bein^ that in the latter the valves open and close automatically, while in the air pump they mu*t be operated h\\ mechanical means, the difference in pressure of the gas on the two sides of the \\al\\cs is nut snrtieient to move them. Such pumps as this are called \"mechanical\" ones. Other forms in general use are the Sprengel and the Geissler-Toepler. AMES'S PHYSICS \u2014 14 210 MECHANICS b. Sprengel pump. \u2014 The action of this purnp consists in having drops of mercury so fall as to trap the gas between them and thus carry it away. There is an elongated glass bulb, to the side of which is joined a long tube, as shown in the cut, whose lower end is connected by a rubber tube with a reservoir,", " so that the mercury may be thus forced Fro. 106. \u2014 Mechanical air pump. The vertical rod, A, is held by the moving piston'with sufficient friction to move it up or down until brought to rest by the conical ends entering their sockets ; then the piston slips along the rod. In the cut the piston is moving down. FIG. 107. \u2014 Sprengel air pump. slowly into the bulb. At the lower end of the bulb is joined a glass tube of narrow bore and at least 80 cm. long, and at the upper end of the bulb is a connection with the space to be exhausted. The tubes at the side and bottom of the bulb are so arranged that, as the drops of mercury break off and fall, they hit the opening of the lower tube and pass down it HYDRAULIC Jf.-l(7//.v/>: I'UMPS, ETC. 211 in the form of short cylinders. The space between these cylinders thus formed is occupied by small amounts of the \u2022 1 raw a in from the connected vessel; and so these drops act like a succession of small pistons forcing out the gas. The lower end of the long tube may dip into a basin of mer- cury, and the gas will bubble out at the surface, or it may be bent so as to form a \"trap.\" As the exhaustion continues, the mercury will rise in the long tube, and will finally stand at the barometric height when the vacuum is as complete as it can be made. c. O-eissler- Toepler pump. \u2014 In this pump there is a large bulb to which are joined two tubes, \u2014 one at the top, the other at the bottom. The lower one is at least 80 cm. long and is connected at its lower end to a large vessel of mercury by means of a long rubber tube. The upper tube is bent over into a vertical direction downward, and dips into a basin of mercury, or forms a trap. Around the large bulb there is a branch tube connecting the upper and lower tubes just as they leave the bulb; and into this branch is joined a long ver- tical tube leading to the vessel which is to VESSEL TO BE EXHAUSTED Fio. 108. \u2014 G\u00abtoftler- Toepler air pump. be exhausted. (This tube is replaced often by a short ver- tical one containing a glass valve.) I f the large vessel of mercury is now raised, as it can be owing to the flexible rubber tubing, the mercury will rise into the hnlh and the connecting tubes, shutting off connec- tion with the vessel to be exhausted, and will drive out all t!i\u00ab -.ras in the bulb through the tube in the top, so that it will bubble out through the mercury in the basin at its end. 1 1. now, the movable vessel of mercury is lowered, no air can enter through the tube at the top of the bulb, because it is \"sealed\" by the mm-nry in the basin, which \\\\ill rise in the 212 MECHANICS tube; but as soon as the mercury falls below the opening to the long vertical tube, the gas in the vessel to be exhausted will expand and fill the bulb and the connecting tubes. When the movable vessel of mercury is again raised, it drives out the gas in the bulb; and as the process continues, the exhaustion of the vessel proceeds rapidly. The tube leading from the top of the bulb around to the basin of mercury must be at least 80 cm. high, and the long vertical tube leading to the vessel to be exhausted must be still longer. BOOKS OF REFERENCE KIMBALL. The Physical Properties of Gases. Boston. 1890. BARUS. The Laws of Gases. New York. 1899. This contains Boyle's original paper and also Amagat's memoirs on the variations of a gas from Boyle's law. BOYS, Soap Bubbles. London. 1890. This is a description of many most interesting capillary phenomena. GREENHILL. Hydrostatics. London. 1894. A standard advanced text-book. TAIT. Properties of Matter. Edinburgh. 1885. A most interesting and useful text-book. RISTEEN. Molecules and the Molecular Theory of Matter. Boston. 1895. A popular, yet accurate, description of the kinetic theory of matter. HOLMAN. Matter, Energy, Force, and Work. New York. 1898. This is a philosophical discussion of the properties of matter. POYNTING AND THOMSON. Properties of Matter. London. 1902. This is an advanced text-book, and will be found most useful for reference. JONES. Elements of Physical Chemistry. New York. 1002. This contains an excellent description of the properties of solids, liquids, and gases. HEAT INTRODUCTION Tm: properties of matter that have been discussed in", " the previous pages are mass, weight, shape, size, elasticity, sure, etc. The mass of a body cannot be changed by any meehanical means, nor can its weight at any one point on tin- earth's surface; but the other properties may be changed at will. One of the simplest methods of doing this is to alter the temperature of the body; and this process will be ;>sed in the following pages. Molecular Energy. \u2014 We have proved, in the discussion of <li tins i<>n. of viscosity, and of the properties of gases, that matter consists of minute parts which are in motion, the extent of the freedom of this motion varying with the con- ditinn of the matter. In a solid these minute particles as a rule only make oscillations; while in thuds they can move :i one portion of space to another. Thus these partu -les have both kinetic and potential energy, \u2014 the former owing to their motion, the latter owing to the fact that work is required to hrin^ two mulreules into a certain <1. -finite posi- tion with reference to each other. The molecules themselves of parts, and these have energy in both the kinetic and tli.- potrntial forms. '1 'hi- internal energy of a body is quite apart from its energy owing to its motion as a whole, or to its push inn with e to the earth, and may be increased, as is obvious, loing WMI! i^ainsi forces that act in connection with the inuleeiile8,e.^. bv niiiLT friction or 1>\\ compressing a gas. 213 214 UK AT It may be varied in the case of an elastic body by setting it in vibration, or by sending waves or pulses through it ; for under these conditions the kinetic and potential energies of the molecules are altered. Thus a bell if struck by a hammer vibrates, and as a result waves are produced in the surround- ing air, the particles of which therefore are set in vibration. We may, therefore, consider two kinds of internal molecu- lar motions : one corresponding to a state of wave motion when all the particles or molecules are in similar vibrations ; the other, to a condition in which there is no regularity in the vibrations or motions. This last condition exists when a body is in its ordinary undisturbed state, and is altered when friction is overcome, when a gas is suddenly compressed, etc. The phenomena associated with variations in this in- ternal energy of bodies, owing to their irregular molecular motions, belong to that branch of Physics which is called \" Heat.\" CHAPTER X UK AT PIIKXOMKNA Preliminary Ideas. \u2014 In describing and discussing mechan- ical phenomena the sequence of ideas was somewhat as fol- lows: by means of our muscular sense we experience certain sensations which we associate with matter, and we are led t<> distinguish certain properties of matter which we consider as independent of each other for the time being, viz., mass, weight, elasticity, etc.; we study these at first by means of our own muscles, but later discover physical methods for the same purpose. By means of our other senses we can also in vestigate other properties of matter and the corresponding -.sot* nature.\" Everyone knows what is meant l>y tin- words \"hot\" and \"cold\"; and if a man dips his hand in turn into two basins of water, he is as a rule able to dis- uish between them by means of his temperature sense, and so can say one is hotter or colder than the other. We experience this sensation of hotness when we stand in the nine or 11. -ap a fire, when \\ve put our hands in a K of water on a stove, when we touch a body that has been rubbed violently against another, etc.; and we feel the sensation of coldness when we touch or stand near a block of ice, when we wet our hands and allow the water to evaporate, etc. If we expose inanimate objects to the same in ions, they undergo changes; and. in tact, in general all tin -ir physical properties with the exception of their mass and weight change. Tim-. M <\u00bbf iron is exposed to the Hun or put on a stove, its volume increases, its elasticity < -han^es, it feels hot 116 216 HEAT to our fingers. If a piece of ice is put in a basin on a stove, it changes its state, becoming a liquid ; this water gradually feels hotter and hotter to our fingers, and finally boils away in the form of a gas. If a gas is inclosed in a glass bulb and exposed to a flame, both its pressure and volume change, etc. All these changes could be produced equally well by friction. Similarly, if a piece of iron is put on a block of ice, its volume becomes", " smaller and it feels cold to our hands ; water can be frozen by making some of it evaporate rapidly, etc. These changes are called \"heat effects.\" Nature and Cause of Heat Effects. \u2014 If we investigate the conditions under which these changes occur, we see that in them all work is being done either on the minute portions of the body or by them. We shall consider one or two of these conditions in detail. When two solid bodies are rubbed together, the force of friction is overcome, and the changes produced, in general, are increase in hotness, in- crease in volume, etc. In friction, however, the force owes its origin to minute inequalities in the two surfaces, which are leveled off or altered as the work is done. Similarly, in all cases of fluid friction, the work done in maintaining the motion is clearly spent, as has been shown, in giving energy to the molecules or minute moving parts. An illus- tration of the heating effect produced by friction is furnished by meteors and \"shooting stars.\" As these pieces of matter enter the atmosphere of the earth, they are heated to incan- descence by friction against the air. Again, when a gas is compressed, it becomes hot and its pressure increases ; but, as we have shown on page 197, in this case work is done in increasing the kinetic energy of translation of the particles of the gas. If the gas is allowed to expand, doing work against some external force, it becomes cold and its pressure decreases ; but it has been shown that while this happens the molecules of the gas lose kinetic energy. In a flame, or any process of combustion, there UK AT rilKSOMENA -IT molecular changes going on which must necessarily in\\<>l\\v the energy of the molecules. It will be shown later that all bodies are emitting waves in the. ether; and, if those whose lengths art- comparable with the size of mole- cules are absorbed by bodies, they become hot, expand, etc. This is the explanation of the effect of exposure to sunlight, etc. Again, if we consider the effects themselves that are produced in bodies under the conditions described, it is seen that they correspond to what we would expect if they are due t\u00ab> work being done on or by molecules. When a body amis, the molecules are separated, and work is done against the molecular forces ; similarly, if a body contracts, these forces do work. When a solid is incited, the molecules iter freedom of motion, and therefore work must be done on them; similarly, when a liquid solidities, the molecules lose their freedom of motion, and in this change lose energy. As a gas becomes hot, owing to com- mon, its molecules gain kinetic energy of translation ; as the gas cools, owing to expansion against external forces, iis molecules lose kinetic energy of translation ; so that in tin- case of a gas, at least, changes in temperature correspond to changes in kinetic energy. Similarly, changes in elas- ticity, hardness, viscosity, etc., may be explained if we think of the energy,,f the molecules as being aii'-'cted. It is natural, therefore, to believe that all these changes are ;lie fact that t: I the molecules ha-* been increased by the result -rn;il forces or that it has been decreased by the molecules doin^ work against external s. Then two question I )\u00ab> th.se changes depend ply upon the Y/////////V of cilery that the bodies receive ip\u00ab>ii the manner in which the work is doi \\ml is the 'hat llie molecules receive exactly equal to the \\\\ork done on them'' The experiments that have been performed 111 0 ; hesc oi. \\\\ ill be described ', 218 HEAT but it may be stated here that there is every reason for believing that the changes depend upon the quantity of energy received by the molecules, not upon the manner in which it is delivered, and also that there is neither loss nor gain of energy in the process. Application of the Principle of the Conservation of Energy. \u2014 During most of these changes the volume of the body that undergoes them alters ; and, in this process, work is done on or by whatever external force or pressure is acting on the body maintaining its volume, and, in general, by the force of gravity also. Thus, if a pillar supporting a build- ing expands, the building is raised and work is done ; the air presses against the sides of the pillar, and this force is also overcome as it expands ; again, the centre of gravity of the pillar itself is raised, and thus more work is done against gravity. Consequently, when work is done against the molecular forces of a body so that it undergoes changes in temperature, in size, in state, etc., a definite amount of energy is given the body and this is spent in two ways : (1) in increasing its internal energy ;", " (2) in doing external work as just described. (During these changes the mole- cules are affected, and some of their kinetic energy may become potential or vice versa ; but in these internal changes there is neither loss nor gain of energy.) Similarly, when reverse changes take place and the.body becomes cool and contracts, the external forces of pressure and gravity do work on the body, the internal energy decreases, and the molecules of the body do work in such a manner as to give energy to external bodies; the amount of this last must therefore equal the sum of the work done on the body by the external forces and the amount of the decrease in the internal energy. Thus, when a piece of iron is placed in a basin of hot water, the latter loses a certain amount of inter- nal energy, and, since it contracts, the atmospheric pressure does work on it; and in return for these two supplies of in: AT ru }-:\\<>\\ii-;_\\ A 219 energy the internal energy of the iron increases, and as it mis it pushes back the atmosphere and so does work, i \\\\'e are neglecting purposely all losses of energy by radia- tion and conduction.) In any change, then, we may write the equation : quantity of energy received by work done on the molec-nlo = increase in internal energy + work done against ex- ternal forces, such as surface pressure and gravity, or the equivalent one : quantity of energy given up by the molecules doing work against external forces = decrease in internal energy -f work done on the body by external forces, such as surface pressure and gravity. Heat Energy. \u2014 The energy that is given a body when work is done on it against molecular forces is called \"heat energy,\" and the effects that bodies experience when they gain or lose this energy are called \"heat effects.\" In ordi- nary language the word \"heat\" is often used in place of heat energy, and we speak of k% adding heat to a body/ OK withdrawing it, of a \"source of heat.\" etc., where the mean- ing is obvious. As has been said, all the properties of a body except its mass and weight in general change when heat energy is added to it or taken from it. The most obvious of these changes are the following: 1. ('hange in hotness, as perceived by our temperature sense. _'. Change' in volume, if the external pressure is kept const 3. Change in pressure, if the volume is kept constant, in case the body is a fluid. I. Change in state, such as fusion, evaporation, etc. Chan-.- in electrical or magnetic properties, such as electrical conductivity, magnetic strength, etc. UK AT Heat Quantities. \u2014 In the discussion of heat effects two physical quantities enter that have not hitherto been described with exactness. In the first place, we must explain what is meant by the word \" temperature,\" which is used ordinarily as giving an idea of the hotness of a body, and must describe a method by which a numerical value may be assigned it. Further, in all heat effects we are concerned primarily with quantities of energy entering or leaving bodies; and some convenient unit must be denned in terms of which this energy may be measured. Changes in volume and pressure can be measured by means already described, and may be expressed in terms of the ordinary units \u2014 the cubic centi- metre and dynes per square centimetre or centimetres of mercury. In changes of state we have to consider alterations in volume, in elastic properties, etc.; but these require no new definitions or units. Temperature Preliminary Ideas in Regard to Temperature. \u2014 We have used, whenever convenient, the word \" temperature \" in its everyday meaning as a quantity describing the hotness of a body, and have said that a body which felt hot to us had a higher temperature than one which felt cold. This sensa- tion is due to some property of the molecules of the body, in virtue of which they affect our nerves. We saw in speak- ing of gases, page 197, that the mean kinetic energy of trans- lation of its molecules obeys the same laws as does its temperature ; or, in other words, the temperature or hotness of a gas is due to and is measured by the kine'tic energy of translation of its molecules. There are many reasons for believing that this is true also of solids and liquids ; and so, when the temperature of a body is raised, we believe that the kinetic energy of its molecules is increased. Temperature Scales ; Thermometers. \u2014 Evidently this prop- erty of a body cannot be measured, because it is impossible HEAT to conceive what is meant by a unit of hotness ; but we can assign it a numerical value. For when the temperature of a material body is changed, its physical properties all change; and so. instead of using our hands or bodies as instruments", " for investigating temperature, we can employ any material I unly and observe those properties which change as the tem- perature is changed. This body would then be called a *\u2022 tliLTinonu't Thus we might use a homogeneous metal rod and observe its length. If the rod had the same length ID immersed in two different baths of oil, we should say that their temperatures were the same with this thermome- ter ; whereas, it the length were greater when the rod was in one bath than when in the other, the former would be said to have the higher temperature. Kxperiments show that, if two bodies at different tempera- tures are placed together in such a manner that heat energy (an pass between them, e.g. if two liquids are stirred up together, if a solid is immersed in a fluid, etc., they finally come to the same temperature intermediate between their initial temperatures ; the body at the lower temperature must therefore gain energy, and the one at the higher tem- perature must lose it. (The former also loses energy in general, but it receives more than it loses, and so on the whole gains. Similarly, the latter in general gains energy, but its loss exceeds its gain.) Thus, from a physical point of view, the difference of temperature between two bodies determines whirh is to lose energy, or it determines the \u2022\u2022direction of the How of heat energy.\" The temperature of a h\u00ab>dy. then, is a property defining its thermal relations with neighboring bodies. The general method of assigning a number to the tem- perature [fl as follows: two stand, in 1 thermal conditions \u2022elected, and numbers are given them arbit rarily \u2014 let \u2022\u2022 lie f} and /., ; then some definite p of a definite. which can be ni.M>uied and \\\\hich changes with the \u20222'2-J. HEAT temperature, is selected ; its numerical value is determined in the two standard conditions and in the one to which a number is to be assigned \u2014 let these values be ar \u00ab2, and a ; finally, the number for the temperature is obtained from those arbitrarily given the standard conditions by simple proportion between the numbers thus obtained, viz., calling it \u00a3, t-tl:t2-tl = a-al:a2-al. Experiments show that the temperatures of a mixture of pure ice and water when in equilibrium, and of steam rising from boiling water, are constant and the same the world over, provided the external atmospheric pressure is the same ; that is, a definite metal rod always has the same length if it is put in a bath of water and ice no matter when or where it is done, a given quantity of mercury has the same volume, etc. For this reason, and because they are easily obtained and include the ordinary temperatures, these two thermal condi- tions when the external pressure is 76 cm. of mercury are selected as the standard ones ; and the numbers 0 and 100 are assigned them on the \" Centigrade \" or Celsius scale. The quantity agreed upon by physicists, the change of which is measured, is the pressure of a definite amount of hydrogen gas whose volume is kept constant. Let the values of the pressure of this gas at the standard temperatures be p0 and jt?100, and that at a temperature for which a number is desired p. Then, calling this number t, it is given by the equation : t - 0 : 100 - 0 = p - p, :Pm - Pot or t = 100 P-P*. Pun - Po This is the \" temperature on the constant volume hydrogen thermometer,\" using the Centigrade scale ; and, whenever hereafter the temperature of a body is referred to, its value on this instrument and scale is meant. This number is always ex- pressed as a certain number of \" degrees,\" and is written t\u00b0 C. HEAT PHENOMENA Several other thermometers are in more or less common use. In one the volume of a definite quantity of nitrogen (or of air), the pressure being kept constant, is the property measured as the temperature changes; in another it is the apparent volume (see page 234) of a quantity of mer- cury inclosed in a glass bulb with a fine stem that is measured. The former is called a \"constant pressure nitrogen (or air) thermometer\"; the latter, a \" mercury-in-glass thermometer \" ; in them both the number for the temperature on the Centigrade scale is given by t = 100 v ~ \"o, where the symbols have obvious meanings. Another form of instrument is one in which the elec- trical resistance of a piece of platinum wire is measured. If /,'\u201e, 7?100, and R are the values of this quantity at the tin. \u2022\u2022 temperatures, the number for the temperature is given by R-R. t = 100 Such a thermometer as this is called a", " \"platinum resist- ance\" in>trnment. The numbers obtained by these last three instruments for the temperature of the same body will not agree with each other, or with that obtained with the standard thermometer, except at 0\u00b0 and 100\u00b0; but careful com- parisons have been made by different observers of the readings on all four instruments at the same temp through wide limits; so that, if a number is obtained for the temperature of a body using a mercury thermometer, the value which would have been obtained if the standard hydrogen instrument had been used is known: this is tin- temperature of the body. (The cor- 'ii to I..- applied to the scale reading <\u00bbf an ordinary MI -y thermometer, such as is used in laboratories, is small.) other scales than the Centigrade are in use for non-, title purposes. In the Fahrenheit system the num- bers 32 and 212 are assigned the standard temperatures of melting ice and sl.-.im. and in the Keaumur the num- bers 0 and 80 are used. < Thu>. 1>0\u00b0C. = 68\u00b0F.= \\>> B 'O.-Mer- \u2022 The ordinary laboratory thermometer is a men-ury-in-glass ment. who.-- stem i, divided into numbered parts. The maker,.f the 224 HEAT instrument aims to have these numbers come at, such points that, \\\\licn the division corresponding to the top of tin- nn-ivury column is read, it gives the temperature on the mercury scale, i.e. the numbers correspond to equal increments in apparent volume. If the instrument is to be used for accurate measurements, however, the readings at the standard tem- peratures must be noted, and the volume of different portions of the stem must be measured, in order to determine exactly the error of each division as marked by the maker. Moreover, a glass thermometer is subject to an error due to two facts : a glass bulb whose temperature is raised from one value to another, and then lowered again to the former value, has a larger volume at the end than it had at the beginning ; and this increase is not permanent, but disappears gradually after the lapse of weeks or months. This is owing to the heterogeneous character of glass ; the molecular changes produced by raising the temperature persist after it is again lowered. Thus, if a glass thermometer reads 0\u00b0.02 C. when put in melting ice and is then, after being heated to 80\u00b0 or 90\u00b0, again put in melting ice, it may read \u2014 0\u00b0.01 C., showing that the volume of the glass bulb has increased. This is known as the \" depression of the zero point.\" If the thermometer is kept in ice for some months, the readings will gradually rise. There are numerous other defects in the mercury ther- mometer which must be carefully guarded against. To give a number to extremely low temperatures some substance should be used whose properties are the same in kind as at ordinary temperatures. Thus, mercury should not be used, for it solidifies at about \u2014 39\u00b0 C., and the changes in volume of the solid mer- cury cannot be compared with the similar changes of liquid mercury at ordinary temperatures. Hydrogen gas at a small pressure may be used, or a platinum resistance thermometer. Similarly, to give numbers to extremely high tem- peratures special precautions must be taken. The best Fro. 110. \u2014Rutherford's maximum and minimum thermometers. The former contains mercury, the hitter alcohol. in-: AT /\u2022///\u2022;vo.v/-;v.i 2-25 \u2022ical methods depend upon certain laws of radiation which will be issed later; but for standardizing purposes a hydrogen thermometer mu>t be used. There are many special types of thermometers devised for particular purposes. Among these it may be worth while to describe briefly one that registers the extreme temperatures which occur during ;i CHI tain tl of time, and one that is used by physicians for clinical purposes. Rutherford's \"maximum and mini- mum thermometers\" are two instruments, as shown in ut, which are supported with their stems horizontal : one contains alcohol, the other mercury. Just inside the \u2022 f the alcohol column is a small glass rod with rounded ends, as shown on an enlarged scale in a portion of the (lit ; so, as the temperature falls and the alcohol contracts, this rod is drawn back by the curved surface of the alcohol ; when the temperature rises sixain, it remains station- ary. Similarly, just outside the end of the mercury column is a small iron rod which is pushed forward by the mer- as it expands and remains stationary when the mer- contracts. Thus both the lowest and the highest temperature- readied are recorded. The glass index can b\u00ab- jarred bark in place again, and the iron one can be dra\\\\n hack by a magnet. (In Six's", " form of instrument these two thermometers are combined in one.) :iical thermometer is shown in the cut. Its stem has a tin.- bun- \\\\ith a sw. -llin^ at its lower end separating in the bulb containing the mercury : so that, as the tem|K*rature rises, the mercury does not reach the diviile.l stem until a temperature of 90\u00b0 F. is reached. At the swelling then- -triciimi in the tube; and if the ills when the mercury is above this, thecol- umt the mercury in the stem. Thus. if the instrument is inserted in the month ami tin- mer-! to a certain divi-ion. it will -.till stand at this K... 111.\u2014 I'Hnl- r\u00abl thermometer. point aft\u00abT the thermometer i* withdrawn. The mercury may be driven into the bulb by -liakiu- the instrument in the proper manner, \u2022rifiiijal fore\u00ab- is usually applied.) fari thermometer \\\\a> made by (ialileo, as early as 1598. It nded upon the expansion of air. and was not devised in such a!\"|-eudeiit of ban-met rie changes. I'.otilliean. in in-glass instrument. The credit of showing that i MY8IC8 \u2014 15 226 UK AT melting ice and boiling water furnish \"fixed points,\" and of proposing that they be adopted in a thermometric scale, belongs to Huygens (1665). Many improvements were made by Fahrenheit Units of Heat Energy. \u2014 In all heat effects, as has been said, we are concerned with the quantity of energy that must be added to or withdrawn from a body in order to produce a given change ; and so a convenient unit must be chosen, and suitable methods of measurement must be devised. The scientific unit for the expression of amounts of all kinds of energy, including therefore heat energy, is the erg or the joule, i.e. 107 ergs (see page 112); but heat effects are not as a rule produced by direct mechanical proc- esses in which the amount of work done can be measured by a dynamometer. The standard method of producing a heat effect in a body is to immerse it \u2014 if it is a solid \u2014 in a quantity of water at a different temperature ; the temperature of the water falls or rises because it loses or gains the heat energy that enters or leaves the body, allowance being made for external work and for the influence of surrounding bodies. The natural unit of heat energy is, then, either the amount required to raise the temperature of a unit mass of water through 1\u00b0, or one nth the amount required to raise its temperature through n\u00b0. (These two quantities of energy are not in general the same.) By stirring a paddle rapidly and continuously in a known quantity of water, the amount of work (measured in ergs) required to raise its temperature through a known number of degrees (on any scale) may be determined by a dynamometer; and so the value of the practical unit of heat energy may be expressed in ergs. Experiments show that the number of ergs required to raise the temperature of a definite quantity of water through 1\u00b0 is different for different temperatures, i.e. it is not the same when the temperature is raised from 5\u00b0 to 6\u00b0 as if the limits were 10\u00b0 and 11\u00b0, etc. ; but the difference is very small. The work required, however, to raise the in-: A -i temperature of a definite quantity of water from 0\u00b0 to 100\u00b0 C. most exactly 100 times as much as that required to raise it from 15\u00b0 C. to 16\u00b0 C. For this reason the practical unit of heat energy is defined to be the \"amount required to raise the temperature of 1 g. of water from 15\u00b0 to 16\u00b0 C.\"; this is called a \"gram calorie at 15\u00b0 C.,\" or, simply, the calorie. Its value in ergs, as determined by Rowland, Callendar and IJarnes, and others, is 4.187 x 107; that is, it is 4.187 joules. The great disadvantage in having as a \"heat unit\" one that depends upon a range of temperature (other than from 0\u00b0 to 100\u00b0 C.) lies in the difficulty of determining temperature accurately, and in the fact that so many arbitrary quantities and ideas enter into the definition of a temperature scale. If it were practicable, it would be much better to take as a heat unit the amount required to melt 1 g. of ice at 0\u00b0 C., or to produce some other change in state, because during these changes the temperature does not vary. Transfer of Heat Energy. \u2014 Before discussing the various heat effects in detail, a few words should be said in regard to the various methods by which heat energy is", " added to or taken from a body. These are three in number, and are illustrated in the following experiment : if one's hand is held above a heated stove, it feels hot. and at the same time one 'iiscious of an ascending current of air. Similarly, if a body is held in the upper portion of any fluid whose lower portimi is maintained at a hi^h temperature, it will receive heat energy from the ascending currents of heated tluid. This process is known as \"convection.\" If one end of a metal nnl is put into a lire, its temperature rises, and that of \u2022r neighboring portions of the rod also. In this process there is n<> actual displacement of the matter, and therefore there is no convert ion : hut the energy is handed on from niolreuh- to iii\"l,Tiile,l,,\\vn the rod. This is called ''con- duction, ii, if a l\u00bbod\\,^M! to the sun or is held at one side ot,t hot stove, its temperature iii general rises, it 228 HEAT is receiving heat energy \u2014 not, however, by convection or conduction. This process is called \"radiation,\" and will be shown later to consist in the absorption by the body of waves in the ether. All these pro- cesses will be described in detail in a later chapter. Convection and conduction cannot take place through a vacuum, and radiation is almost entirely prevented by having the sur- face of the body or vessel covered with a highly polished metallic layer. In his experi- ments on liquid air and hydrogen, Dewar has used a flask, called by his name, which con- sists of a double-walled glass vessel, the space between the walls being exhausted as completely as possible. Traces of mercury vapor are left in this space ; and at low temperatures this, freezes, forming a metallic surface over the glass walls. Fio. 112. \u2014 Dewar flask. CHAPTER XI CHANGES IN VOLUME AND PRESSURE Introduction. \u2014 The fact that all bodies, with the excep- tion of water below 4\u00b0 C. and one or two unimportant sub- stances, increase in volume when their temperature is raised, is most familiar to every one; and numerous measurements have been made of these changes. Naturally the amount of the increase in volume depends upon the external force act- ing, and upon whether this is constant or not. The mechan- ical force required to influence the expansion of a solid or a liquid is so great that ordinary changes in the atmospheric pressure have no measurable effect ; but this is not so in the case of a gas. It is therefore necessary, in studying those variations in volume of a body which accompany changes in temperature, to describe the external conditions, if one wishes to be definite. The condition which is always assumed, unless the contrary is stated, is that of constant external pressure. Solids Linear and Cubical Expansion. \u2014 In measuring the change in volume of a solid it is, as a rule, easier to measure the changes in length of certain linear dimensions of the body and from these to calculate the change in volume. It tin- body is isotropio, i.e. has the same properties in all direc- tions, this calculation is most simple. Imagine the body in the form of a cube, the length of whose edges at any one trm p. Mature tf is /, and at the temperature fa\u00b0 is /3, then tin- voluin.- ;it /, ii /,:! and at tj is lf\\ so the change in volume is I* \u2014 I*- It the body is not isotropic, but has no 230 HEAT different properties in different directions, three directions in it (which depend upon the arrangement of the molecules) may be determined, such that the changes along these are independent of one another \u2014 these are called \"axes.\" (In crystals they are the crystallographic axes.) If, then, a rectangular solid is made of this body with its edges parallel to these directions, and if lv mv n^ are the lengths of the edges at t^ and?2, m2, n2 their lengths at \u00a32\u00b0, the correspond- ing volumes are l^n^ and Z2w2w2; and the change in volume is?2W2W2 ~~ liminr The change in length of any straight line in the surface of a solid or of any straight edge may be measured by various means. The body is immersed in a bath of some fluid, whose temperature may be varied ; and the lengths are determined by a comparator of some kind. (Reference may be made to any laboratory manual.) Experiments show that, to a sufficient degree of accuracy, the change in length varies directly as the original length and as the change in temperature, but is different for bodies of", " which are shown as single black lines are of iron and the others of brass. It is seen that, as the temperature rises, the expansion of the iron rods lowers the pendulum bob, but that of the brass ones raises it. The linear expan- sion of brass is about 1\u00a3 times that of iron ; and so if the combined length of the two iron rods on each side and the middle one is 1* times that of the two brass ones on each side, there will be no change in length as the temperature varies. (A simpler method is used in the clock in the tower of the Houses of Parliament, London. The pendulum consists of an iron rod sur- rounded by a zinc cylinder, which in turn is surrounded by an iron one carrying the pendulum bob. The lower ends of the iron rod and the zinc cylinder are attached, and the upper ends of the two cylinders. Since the linear expansion of zinc is about 2\u00a3 times that of iron, the combined lengths of the iron rod and cylinder must equal 2i times that of the length of the zinc cylinder.) Graham's compensation pendulum consists of an iron rod whose lower end screws into a cast-iron cylindrical vessel partially filled with mercury. The quantity of mercury is so chosen that by its expansion combined with that of the iron rod and cylinder the centre of gravity of the whole does not move when the temperature is varied. FIG. 113. \u2014 Com- pensation \"grid- iron \" pendulum. The single black lines represent iron rods ; the double lines brass ones. Again, the period of a watch or of the ordinary spring clock is regu- lated by the vibrations of the balance wheel. This consists of a small Cll. 1 \\ '. ES IN VOL I M /\u2022; - 1 V It PRESSURE 233 wheel attached to a flat coiled spring, which coils and uncoils, making harmonic vibrations. If the temperature rises, two changes occur: the elasticity of the spring is decreased, and the wheel expands. Owing to both these effects, the period of the spring would be increased were it not for the peculiar <a^<^ construction of tin* rim of the wheel. As is shown in the cut, this consists of two (some- times tlnve) parts or sections. One end of >tened to a spoke, but the other is free. Near this latter end are screws or mov- able weights. Each section of the rim is double, consisting of iron on the inside and brass on the outside. Consequently, owing to the greater expansion of the brass, when the Fro. iu. \u2014 Balance wheel of a temperature is raised each section of the rim watch ; the Inner P01*00 of the rim is iron ; the outer la brass. is made more curved, and the weights are brought in nearer the centre. This decreases the moment of inertia, in i\u00abl thus tends to decrease the period of vibration. Therefore, by suit- ably adjusting tin- \\v rights or screws, the balance wheel may be made to r-'t.iin a constant period, however the temperature changes. ite recently Guillaume has discovered that an alloy of nickel and M\u00ab'*-l in the proportion of 36 parts of nickel to 64 parts of steel has a coefficient of linear expansion equal to 0.00000087; and so its expansion l\u00bbe neglected in all ordinary apparatus or work. The change in length of this alloy with change in temperature takes place very slowly if tli. Hue in t. inj>\u00ab-rature is small; for a rod made of it does not reach its full exj>.in>i<>ii for one or two months. Liquids Apparent and Absolute Expansion. \u2014 In measuring the change in volume \u2022\u2022!' ;i liquid, mviu^ to any cause, a ditli- culty is met which has been referred to heiWe ; namely, the fact that the liquid must!\u2022\u00ab\u2022 contained in a solid vessel, and whatever affects the volume of the former will also, iu general, alteri that of the latter. Thus, \\\\heu the contain- ing vessel is a hull) with a tube attached, if the liquid tills the hull. ;md pan,,f the tube, it is observed that, when the l>ull> is heated suddenly hy immersing it in a basin of hot i \u00ab>r in any other \\va\\, the top of the column of liquid 234 UK AT in the tube immediately sinks and then rises gradually, ascending finally higher than it was originally. This is owing to the fact that the first effect of the application of the hot bath is to raise the temperature of the bulb, and it therefore expands before the liquid inside is affected ; as soon, however, as its temperature is raised, it expands and rises in the tube. If the bulb is chilled, instead of heated, the reverse of these changes takes place ; the", " liquid first rises in the tube and then sinks, falling below its original position. It is evident, then, that the apparent change in volume of the liquid is less than the real change by an amount equal to the change in volume of the containing solid. So, if the coefficient of expansion of the solid is known, the direct method for determining the change in volume of a liquid that accompanies a change in temperature is to inclose the liquid in a bulb with a finely divided stem whose volumes are known, and to measure the volume of the liquid at any one temperature and the apparent change in volume when the temperature is altered. Let vl be the initial volume at the temperature t\u00b1, and v the apparent increase in volume when the temperature is raised to \u00a32\u00b0 ; further, let the coefficient of cubical expansion of the solid have the value b. Then the increase in volume of the solid is v1 [1 + b (\u00a32 \u2014 ^)] ; and hence the true change in volume of the liquid is v + vl [1. + 6(\u00a32 \u2014 tfj)]. Experiments show that fyor nearly all liquids \u2014 water is an exception, as will be explained below \u2014 the relation between the change in volume and that in tempera- ture is of the same form as for solids, viz., v2 \u2014 v1 = vj) (\u00a32 \u2014 ^), where b is the coefficient of cubical expansion of the liquid, referred to ^\u00b0. If the initial temperature is 0\u00b0 C., this becomes, as before, v = vQ(l + bQt~) ; and it is to be noted that b = Q \u2014, and therefore only if 50 is small can it 1 + bQt replace b. The coefficient of expansion is found to be different for CHANGES IN VOLUME AND PRESSURE 235 different liquids, as is shown in the following table. It should b\u00bb- noted that the expansion of liquids is, in general, m in -h greater than that of solids. Kthyl alcohol.. between 0\u00b0 and 80\u00b0 C. 0.00104 Kthyl ether... between - 15\u00b0 and 38\u00b0 C. 0.00215 (ilycerine... 0.000534 \u2022ury... between 0\u00b0 and 100\u00b0 C. 0.000182 Turpentine... between - 9\u00b0 and 106\u00b0 C. 0.00105 It is evident that, if the coefficient of expansion of a liquid is known, tfl of observations and measurements gives a method for '{\u2022\u2022termination of the coefficient of cubical expansion of the solid that contains the liquid. This method therefore can be used for all solids that can be formed into bulbs, e.g. glass or quartz. Mercury is the liquid that is, in general, used, because its expansion is known, and for r obvious reasons. Measurement of Coefficient of Expansion. \u2014 A better method, however, for the determination of the coefficient of expansion liquid, which does not involve a knowledge of that of solid vessel, was devised by Dulong and Petit and im- proved by Regnault. It depends upon the fact that the titfl at whieh two liquids of different densities stand when they balance each other in a (J-tube is independent of the material of the tube. If 7/t is the height of the column of LUC liquid of density dr and 7/.2 that of the column of the liquid whose density is dT -* == -2 (see page 176). But the formula for expansion, 2 l l>e replaced by v = *0(1 + /, 4 = ''0 + V). ii the density at \u00a3\u00b0, and </,, that at 0\u00b0; for tin- density of any ln\u00bbdy whose mass remains constant varies inversely as its volume (m = dv). Therefore, if the ratio of the densities of \" liquid at <\u00bb and /' is known, its coefficient of expansion can be deduced at once. 'I'!\"1 simplest i.\u00bbnu of the actual experiment is as follows : A tul..- ifl made in the form of a W- with a cross tube at the 236 HEAT top, as shown in the cut ; this last has a small opening -4 near its middle point and is kept horizontal ; there is a branch tube joined to the apparatus at B, which is con- nected with a reservoir K containing air that can be compressed to a pressure greater than that of the at- mosphere ; a quantity of the liquid whose expansion is to be studied is poured in, just sufficient to flow out of the opening A in the upper cross tube, thus insuring the con- dition that the upper level surfaces of the two columns are at the same height. These two separate portions of the liquid are then surrounded ==H: FIG. 115. \u2014 Apparatus for the determination of the", " coefficient of expansion of a liquid. by baths, one at 0\u00b0, the other at t\u00b0. When equilibrium is established, the difference in level of the two surfaces of one portion of the liquid is not the same as that of the other. Call this difference for the liquid at 0\u00b0, A0, and that for the one at t\u00b0, h. Then dQghQ = dyh, or \u00a5& = \u2014. Consequently, if hQ and h are measured, or d'may be calculated ; for, ^ = 1 + bQt..-. 1 + b0t = n, Expansion of Water. \u2014 When water is studied, it is found that as the temperature rises from 0\u00b0 C. to 4\u00b0C., it contracts; but above 4\u00b0 it expands. These changes in volume of water (or of any substance) which accompany changes in tempera- ture may be best shown graphically. Lay off two axes, one to indicate temperatures; the other, volumes of a definite CHANGE* l.\\ VOLUME.l.\\/> 237 quantity of the substance. For water the curve giving the connection between v and t is as shown in the cut. (If the coefficient of expansion is a con-.t, the curve is a straight line sloping upward from left to right.) This fact, that the density of water is a maximum at 4\u00b0 C., and that it decreases continuously from this temperature to 0\u00b0 and to 100\u00b0 is of great importance in nature ; for as the temperature of the water in a lake or a pond sinks in winter below 4\u00b0 and reaches 0\u00b0, the water at 0\u00b0 floats on top and does not sink to the bottom, as would be the case with other liquids; and consequently ice forms on the u]>]H-r surface, not at the bottom. TEMPERATURES CENTIGRADE 0 2 4 6 8 10 This jN\u00bbculiarity of water is explained by assuming that when ice melts the liquid formed is not at first made up of molecules that are all alike, but contains two kinds, \"ice molecules\" and \u00abT molecules.\" The former are assumed to be lighter than th\u00ab\u00ab latter; and, as the temperature rises, they are transformed gradually.\u2022han-,,1. Ki<;. 110. \u2014 Curve showing expan- sion of water as the temperature is he latter. Gases Expansion at Constant Pressure. -\u2014When the temperature of a gas is increased, its volume increases also, unless special autions are taken; but, if the external pressure varies \u2022luring tin- change in temperature, the change in volume is moditi. \u2022\u00ab!. \\Vt\u00bb shall then assume that during the change the rare is maintained constant. A gas is always inclosed in some solid vessel: and its expansion must a 1 \\\\ays be taken i i it \u00ab> account although its effect is small in mmpai -ison \\\\ith the - n of the gas. When this is done, experiments that the general formula for expansion is true. 238 HEAT This coefficient of expansion is constant for any one gas ; and, further, it is found by experiments to be practically the same for all gases. This is known as the \" Law of Gay- Lussac.\" The value of this constant is very nearly 0.0036600, or 2^-j, using the Centigrade scale. It is to be noted, then, that gases expand more than liquids. (The apparatus used by Gay-Lussac in his investigation on the expansions of gases is shown in the cut.) General Law for a Gas. \u2014 Having thus determined the relation between the volume and temperature of a definite quantity of gas when the pressure is kept constant, and \u00b0 ^ A -T-L I A. M ( ( '-& &? B i i o o FIG. 117. \u2014 Gay-Lussac's apparatus. \u2022 knowing \u2014 as stated in Boyle's law \u2014 the relation between volume and pressure when the temperature is kept constant, Ti\u00b0 we can combine the two and deduce an expression that will include them both. If we start with a given quantity of gas at 0\u00b0 and at a pressure p, it will have a definite volume which we can call VQ. If now the temperature is raised to \u00a3\u00b0, the pressure being kept constant, the resulting volume v = vQ(l. 4- \u00a30\u00a3). If the pressure is then changed to P, the temperature being kept constant, the volume will be altered and the new value F'must be such that PV=pv=pvQ(\\ + #0\u00a3). Thus, if the volume of a gas is measured at a temperature t\u00b0 CHANGES IN VOLUME AND PRESSURE 239 and pressure 1\\ the volume whieh this same quantity of gas would occupy if its temperature were made 0\u00b0 and its pres- sure y> ean be calculated; for ro-7 _p", " v (The \"standard conditions\" for a gas are a temperature of 0\u00b0 C., and a pressure of 76 cm. of mercury ; and when the \u2022\u2022density of a gas\" is referred to, it is always implied that the gas is in its standard condition, unless the contrary is stated.) This formula can be expressed in a simpler form, viz., PF or - 1 1 1 1 1 VQ is the volume of a given quantity of the gas at 0\u00b0 and at pressure JD, consequently the product pvQ is a constant for PV this gas by Boyle's law. Therefore, \u2022= - is a constant for \\ any one gas, but has a different value for different gases. This m a y - - = 7 a \u00b0o i \u2022 / * any one gas. The quantity \u2014 + e, which, using the Centi grade scale, approximately equals 273 + 1, is called the i + fyabsolute gas temperature\" on the Centigrade scale, and is written T. It is evident that, if different quantities of til*- same gas at the same triii)><-r.ttinv and pressure are taken, their volumes will vary directly as the masses. Thus, writing M for the mass of the gas, the complete law is 240 I1KAT The constant R can be found for any gas by measuring its_ density at a known temperature and pressure. Thus, if t\u00b0 C. is the temperature, p the pressure, and d the density, f -^ f\u00b0r hydrogen is, 4.14 x *o -j- bo 107; for oxygen, 0.259 x 107 ; for nitrogen, 0.296 x 107. If the same law were to hold for a gas as the pressure became smaller and smaller and finally vanished without the volume at the same time being made infinitely great, the temperature of this condition would be given by T = 0, or t\u00b0 C = - - = - 273\u00b0 C. approximately. Similarly, if the same law could be applied to a gas as its volume was made less and less and finally vanished, the pressure remaining finite, the temperature would be given by the same value, viz., \u2014 273\u00b0 C. approximately. Of course the above law for a gas does not apply to one if greatly compressed, and the pressure of a gas can become zero only by its volume being infinite ; and so the above deductions have no physical meaning. We shall see later, however, by other reasoning, that this temperature --, or C, 273\u00b0 \u2014 can reach. For that reason it is called the \"absolute zero.\" (See page 308.) universe lowest which marks body any our the in temperature Change of Pressure at Constant Volume. \u2014 Another fact is apparent from this formula. If the changes take place in such a manner that the initial and final volumes are the same, V= i>0, and therefore P = p(\\ + b0t). Therefore the pressure increases at the same rate with increase of tempera- ture when the volume is kept constant, as does the volume when the pressure is kept constant. (Actually, this is not exactly true of any gas ; but this discrepancy is a consequence of the fact that Boyle's law is not exact for an actual gas urn- is the coefficient of expansion a constant.) Laws of Gay-Lussac and Charles. \u2014 The fact that all gases have approximately the same coefficient for change of pressure when the volume is kept constant was discovered by the French physicist Charles ; while the corresponding one that all gases expand alike when the pressure is kept constant CHANGES IN VOU'Mi: AM) PEM88URM was discovered a few years later, ill 1802, by Gay-Lussac. se statements of fact are therefore called Charles's and -hussar's lav Other Forms of the Gas Law. \u2014 This formula for a gas, /' /'\u2014 HM'l\\ ma\\ be expressed differently and more simply if we assume the truth of Avogadro's hypothesis (see page I. If m is the mass of each molecule, and N is the num- ber in a unit volume, M= mNV\\ consequently, on substitut- ing in the formula, we have'P = RmNT, p RmT or N= - \u2014 -. But by Avogadro's hypothesis, if the pressure and temperature of two gases are the same, they have t In- sane number of molecules in each unit volume : that is, if P ami T air thr same for two gases, N is also. Therefore. fn\u00bbm the above formula, Rm must be the same for all gases; it is a constant of nature. Calling its value RQ, the formula becomes P = R0NT, which states that the pressure in a gas (or a mixture of gases) varies directly as the number of molecules in a unit volume and as the absolute temperature. This equation, in turn, may be expressed", " in a form which is nioiv us.-1'ul for practical purposes, l.rraiisr. of course, we have no accurate knowledge of the value of N. The \" molec- ular weight \" of a!_cas has hceii defined to l\u00bbe a number, char- acteristic of the Leas, which is proportional to the mass of one te molecules, and which is so chosen that the number for oxygen i> :\\~2. Thus if //' is the ///\u2022\u2022/,\u2022,\u2022///,//\u2022 //\u2022./;//// of a gas, each of whose molecules has the mass 1*1,10= <\u2022>//, when :it foi- all gases and has such a numerical value as to //\u2022 r.jiial :\\'2 for oxygen. A numb' unsofasuh- I06 iMjual to its molecular weight (e.g. 82 gram- Q| oxy- gen) is called uone gram-m\u00ab.l. \"one mol\" of that substance. Thus, if there are.V U of a gas in a unit vol- Miue. ['- I and M=N*ic = N'cm in the general formula: hrnr.e, P=RmcN'T= R AMES'S PHYSICS\u2014 16 242 HEAT The product R^c is a constant, the same for all gases ; call itR1. Hence P = R'N'T, or, the pressure varies directly as the number of mols in a unit volume. The value of R1 may be found by experiments on the densities of gases, because P is the pressure of a gas at temperature T when the density is such that there are N1 mols in a unit volume. It is found to be 8.28 x 107 on the C. G. S. system. Energy Relations during Expansion Mechanical Expansion. \u2014 When the dimensions of a body are increased, either by the addition of heat energy or by mechanical forces, there is a certain amount of external work done and there are alterations in the kinetic and potential energies of the molecules. In the case of the thermal effect this is evident from what has been already said ; change in relative position of the molecules, change in temperature, and external work occur together. Similarly, when a brass wire is stretched, its temperature falls ; when a gas is com- pressed, its temperature is raised, etc. We might have predicted that these changes in tempera- ture would be as just stated. (See page 104.) If the tem- perature of a brass wire should rise when it is stretched, it would be in unstable equilibrium. If the wire hangs verti- cally under the stretching force of a heavy body, and if a sudden downward blow is given this weight, it will move down, stretching the wire still more, then come to rest and move up, etc., making harmonic vibrations, showing that it was in stable equilibrium. But if, owing to this stretching, as the hanging body moves down, the temperature of the wire were to rise, it would lengthen ; and this increase in length would cause another rise in temperature, etc., so the equilibrium would be unstable. Therefore, only if the wire CHANGES l.\\ VOLUME AND PRESSURE 243 cools when stretched, is the equilibrium stable. The general law is that, if a body expands when its temperature is raised, its expansion by mechanical means will cool it, while mechan- ical compression will raise its temperature. The converse is true of those bodies which contract when their temperature is raised, e.g. water between 0\u00b0 and 4\u00b0 C. As the sun radiates energy, it contracts; and owing to this cause its parts are slowly coming closer together, and therefore energy is being liberated to make up for that loss. Thus, in ordinary language, the sun owes its heat to its slow contraction. Undoubtedly also, meteoric pieces of matter are falling into the sun, and their energy is thus also added to that of the sun. If the expanding body is a liquid or a gas, or if it is a solid immersed in a fluid, the amount of external work done equals the product of the increase in volume by the pressure. (See page 159.) The internal changes consist of change in kinetic energy and change in potential energy. The first of \u2022 \u2022is connected intimately with changes in temperature, as has been already shown. (See pages 197 and 220.) Changes in potential energy occur if there are molecular forces. The fact that these exist in solids and liquids and are large is evident from the obvious properties of these forms <\u00bbf matter, e.g. they retain definite volumes; but noth- ing is known in regard to the amount or cause of these forces. The case is different with a gas, for in it the forces are extremely small. (Nothing, however, can", " be said as to their ran Internal Work in a Gas. \u2014 This fact that the internal forces in a gas are extremely small was first shown by Gay-Lussac and latrr by Joule, working independently and also in col- laboration \\\\ith Thomson (now Lord Kelvin). The earlv exporinirnts \u00ab,f.Iniilr arr perhaps the simplest to consider. II apparatus consisted of l\\\\o strong metal cylinders con- bed b\\ ;, tube in which was a stopcock. In one of these cylinders quantities of the gas to be studied were compressed 244 II EAT until the pressure was as high as the apparatus would permit ; while from the other the gas was exhausted. The whole apparatus was then submerged in a tank containing water, and this was stirred until the temperature came to a steady state. Then the stopcock in the tube connecting the two cylinders was opened; and, when the gas had redistributed itself, occupying a greater volume and thus coming to a smaller pressure, but doing no external work, the temperature of the water in the tank after being well stirred was again observed. In no case was there any meas- urable change. If there had been molecular forces of attrac- tion between the molecules, it would have required work to separate them when the gas increased in volume, this energy would have necessarily come from the molecules themselves, which would have thus lost kinetic energy ; and therefore the temperature of the gas would have fallen. If there had been molecular forces of repulsion between the molecules, their potential energy would have been decreased by the increase in volume, their kinetic energy would then have increased an equal amount ; and therefore the temperature of the gas would have risen. Either of these changes in the temperature of the gas would have affected the temperature of the surrounding water ; and this would have been observed by the thermometer, if it Fio. 113. \u2014Joule's apparatus for study- Ing the free expansion of a gas: two cylinders, R and J\u00a3, are connected by a tube in which there is a valve ; the whole is immersed in a tank containing water. were sufficiently delicate. Consequently Joule's experiments prove that to the degree of sensitiveness of his thermometer, there are no forces between the molecules of a gas ; or, in other words, the internal energy of a gas is entirely kinetic. The later experiments of Joule and Thomson show, how- VBANQEb l.\\ VOLUME A.\\l> 246 ever, that gases do have measurable molecular fon-es, al- though they are extremely small, and that these in general are forces of attraction. (In hydrogen, at ordinary tempera- tures, the forces are repulsive ; but at very low temperatures they are attractive.) Expansion of a Gas when External Work is Done. \u2014 Joule modiiied hi> experiment by inclosing the two cylinders in sep- urate tanks <>t' water; and, when the expansion took place, he noted the temperatures of the water in the two tanks. He observed that the temperature of the one holding the cylinder in whieh there was the high -sure fell, while that of the other rose. The explanation is evident : the gas that stays behind in the high pressure cylinder after the expansion, has done work in giving kinetic energy to the escaping gas a% awhole, i.e. in producing a wind, consequently its temperature falls. This hlastof gas entering the other cylinder soon ceases, owing iscosity, and the kinetic energy of the moving gas passes into the energy of the molecules, and is apparent by a rise of temjM -rat u re. In other words, the gas in the first cylinder expands, doing work, and so its temperature falls; work is done upon the other gas in setting it flowing, and so its tem- perature rises after the motion is stopped by friction. 1 10. liy. \u2014 Jouie s second exporuueu Met that when a gas expands under such conditions as to do work its temp'-ratiin- falls in shown l\u00bby all\u00ab>\\\\in- (lain)- air inclosed in a vessel to expand -udd.-nly. I f t li.-n- an- nuclei in tin- air. drops of water will!\u00bb\u00ab> loosed around them, thus forming a visil.li- mist and showing that the air has been chilled. (See page 186.) This is the explanation in many cases of the formation of clouds in the air. Expansion of a Gas in General. \u2014 When a compressed gas ilowed to expand out through a fine opening into a space 246 HEAT where its pressure is less, changes in temperature take place owing to numerous causes. We shall consider two cases of practical importance. Let a cloth with fine meshes \u2014 a piece of cheese cloth or toweling \u2014 be folded over the nozzle from which the gas is escaping ; it will expand with violence through these", " numerous openings into the air, thus forming a wind ; the gas that does not have time to escape during any small interval thus does work on that portion which it blows out and so experiences a fall of temperature itself. This may be sufficient to liquefy or even solidify it ; and the cloth will in this last case be found to contain the solidified gas. (This is the ordinary way of obtaining carbon dioxide in the solid form.) Again, let the opening or openings through which the compressed gas escapes be so fine and tortuous that the outcoming gas has no kinetic energy as a whole, i.e. there is no wind, it does noi flow ; in this case its temperature is lower than when in the compressed con- dition (except with hydrogen at ordinary temperatures), as shown by Thomson and Joule, owing to the fact that there are minute molecular forces of attraction, and as the poten- tial energy of the expanded gas is increased, its kinetic energy, and therefore temperature, must be diminished. This fall in temperature varies directly as the difference in pressure of the gas in its two conditions, and so may be considerable. In any ordinary expansion of a gas from a small opening, both of these actions take place. The gas that is escaping at any instant has done work in pushing out the portion that was just before it, and so its temperature is lowered, quite apart from the influences of the molecular forces. (Some distance away from the opening, however, in the case of a jet or a blast, the temperature of the gas is increased, owing to the friction of the moving currents.) This method of securing a decrease in temperature by allowing a gas to expand through a fine nozzle is being used practically in recent machines for the liquefaction of gases. (See page 280.) CHAPTER XII CHANGES IN TEMPERATURE Energy Relations when the Temperature is Raised. \u2014 It is 1)\\ tin- change in temperature of bodies when exposed to some \"source of heat\" that our attention is directed to heat phenomena ; and it is in terms of changes of temperature that heat energy is measured, as we have already seen in the definition of the \"calorie.\" If, however, we wish to deter- mine lm\\v much energy goes to producing the rise in tempera- ture, it is necessary to ascertain how much is used in doing : nal work, and how much in overcoming the molecular forces. If there is a uniform pressure p over the body, and it its volume increases from v1 to vv the external work done against this pressure is p (v2 \u2014 Vj), and so may be calculated. It is only, however, in tin- case of a gas that anything is known <iuant it atively in regard to the molecular forces. 'Ill- '.hen the temperature of a solid or a liquid is raised, although tin- external work may be calculated if the external pressure is known, it is impossible to separate into its parts the energy that is spent in internal work ; and so do not know how much goes to producing rise in tem- r.ut tin; case of a gas is different. We can date the amount of external work done as well as meas- ure it. and we know further that the molecular forces are SO 1 that they can be neglected. ( '..us. -.jin -ntly when the temperature of a gas is raised owing to the addition of heat _ry, if we deduct the amount that is used in doing exter- nal work, the d nice is all spent in raising the tempera; i.e. in increasing the kinetic energy of the mole. ules. The Ml 248 HEAT amount of energy that must be added to the gas, then, in order to raise its temperature a definite number of degrees, depends upon the amount of external work done by the gas ; that is, upon the external conditions under which the gas is kept during the change in temperature. If this change is the same in different experiments, the energy that is spent in producing the increase in the kinetic energy of the mole- cules is the same whatever the external conditions ; and so the difference in the amounts of heat energy added must equal the difference in the amounts of the external work done. Special Case of a Gas. \u2014 For various reasons, partly prac- tical and partly theoretical, the change in temperature of a gas is considered, as a rule, under two different conditions : (1), when the volume is kept constant, and so no external work is done ; (2), when the pressure is maintained constant, so the external work equals the product of the values of this pressure and the change in volume. If Q1 is the heat energy added in one experiment when the volume is kept constant, Q2 that in another when the pressure is kept constant at a value p while the volume is increased from v1 to Vy, the change in temperature being the same in both' e^o", ", =,(.,-,>, where, of course, (?2 and Q1 are measured in mechanical units; i.e. in ergs or joules, since the product p(v2 \u2014 v^) is so ex- pressed. Reverse Changes. \u2014 When a body cools, it gives out heat energy ; but the external work done on it by the external forces is not necessarily the same as is done by the body against these forces when the body is heated and rises through the same range of temperature, because the external forces may have changed, or they may vary differently dur- ing the two processes. If, however, the pressure remains unchanged, or, when it is variable, if the series of changes is CHANGES 1\\ il-MI'l-HMTRE tly reversed during the two processes, the external work done by the body during the rise in temperature equals that \u2022 on the body during the opposite change. If this is the case, the heat energy added in the former process must equal that LTiven out during the latter; for at the end of the two diaii^rs tin* body is hack in its original condition, and so there is no change in its internal energy, and there has also been no gain or loss of energy owing to external work. This fact serves as the basis of all methods for the measurement of quantities of heat energy. Many simple experiments show that, to produce the same change in temperature in equal quantities of different bodies, requires different amounts of heat energy. One of these, due to Tyndall, is to raise to the same temperature in a bath of heated oil several spheres, made of different materials, but of the same weight and of the same size (natu- rally, some are hollow), and then to place them upon a cake of wax that melts at a low temperature. As the spheres cool down they melt <li tit-rent quantities of the wax, as is shown by their entering the wax to different dUtancf.s; and at the end of the process they all have tin* same tem|x>rature. This proves that in cooling over the same range of tem- perature different ] \u2022 \u2022\u2022\u2022 out different quantities of heat energy. - experiment does not pro\\.- this fact beyond question, l.e.-au- eomi'lir.ite.l 1,\\ differ.-nws in the conductivity of the different bodies and l.y the rnn-eqii'Mit fact that some give om more quickly than <>th. -i \u2022<. and so more of the wax is melted before it can conduct the energy away. There are other actions, too; l\u00bbut in reality the main effect is that descried above, and so the experiment is a proper one to illustrate the question referred to.) Measurement of Heat Energy. \u2014 Tin-re are three methods in if.-in-ral use I'm- tin- III\u00abM^II i viiHMit of the heal en- required to raiae the temperature of bodies, or of that given Ifhen a body cools : (1 \u00bb To put tin- body in a hath of r at a different temper, tt mv. and measure tin- change in perature of tin- lien equilibrium is reached. It the maSB of tin- \\vatrr is in ^rams. and it^ dian^c of tempera- ture i.s / ('.. //// calories enter or leave the water, if \\\\\u00ab- neg- bhe slig] oni in the value of tin caloric at different 250 HEAT temperatures. (2) To put the body in a bath of ice and water, so that a certain amount of ice is melted. It will be shown later that it requires 80 calories, very approximately, to melt 1 g. of ice at 0\u00b0 C. ; and so, if m grams are melted, the ice must receive 80 m calories. (3) To put the body in a bath of steam, so that a certain amount of this is condensed into water by contact with the cooler body. It will be shown later that 536 calories, approximately, are given out by 1 g. of steam when it condenses into water at 100\u00b0 C. So, if m grams are condensed by the body at this temperature, it must receive 536 m calories. \"Specific Heat.\" \u2014 The number of calories that must be added to a body whose mass is 1 g. in order to raise its temperature through one degree Centigrade, from t\u00b0 to (t + 1)\u00b0, is called the \"specific heat\" of that body at t\u00b0 C. It should be noted that the specific heat of a substance is a number that is independent, in a way, of the heat unit used or of the tempera- ture scale adopted. We could adopt as a heat unit that quantity of heat energy required to raise the", " temperature of a unit mass (on any system) of water from t\u00b0 to (t 4- 1)\u00b0 (on any scale) ; and define the specific heat of a substance at t\u00b0 as the number of these heat units that is required to raise the temperature of a unit mass of it from t\u00b0 to (/ + 1)\u00b0. Or, if we assume that the specific heat is the same at all temperatures, we may define the specific heat of a body as the ratio of the amount of heat energy required to raise the temperature of the body through any range of temperature to that required to raise the temperature of an equal mass of water through the same range of temperature. With solids and liquids, as has been explained, we assume that the pressure is that of the atmosphere ; but with gases we distinguish two special conditions, constant pressure or constant volume, and so have two corresponding specific heats. For most bodies the specific heat is practically con- stant for all temperatures that are far removed from the melting points; and so, if c is the specific heat of a body whose mass is m, and if the temperature is raised through <2 \u2014 \u00a3j degrees, the number of calories added is mc(t^ \u2014 ^). C7/.l.V(;/:> /.V 'IKMrKHATURB 251 In iron, boron, carbon, and a few other substances the specific heat varies to a marked degree at different temperatures ; and with them c in the above expression is the \"mean spe- cific heat \" for the range from tl to tv Measurement of Specific Heat. \u2014 Corresponding to the three methods of measurement of heat energy referred to above there are, then, three methods for the measurement of specific heat of a given substance. We assume in each case that there is no loss of heat by radiation, etc. 1. Method of Mixtures. If iw, = mass of body, wij = mass of water, /, = original temperature of the body, fa = original temperature of the water, ts = final temperature of equilibrium, ~\u00abift-W OF m2(<3-<,) The water is contained in some solid vessel, called a \" calo- rimeter \" ; and its temperature is affected by the changes in that of the water. If mz is its mass and c1 its specific heat, and if we assume that the changes in its temperature are the same as those of the water, the number of calories it receives is w8c'(/8\u2014 \u00a3a); and, since this energy comes from the foreign body hit rod u\u00ab \u2022\u00ab -d into the water, it is seen that, in the above formula t'.u- <?, wa must be increased by w8<?'. This quantity!lrd tlir M wuter equivalent\" of the calorimeter. mrtlind was fust used by Joseph Black about 1760; and it is the one most generally used at the present time. Tli.-ivaiv in an v objections to it, however. Chief among these are tin- ditVn -ulties of determining the water equivalent of the calorimeter and of avoiding losses by radiation, etc. Most of these are overcome in a modification of the apparatus due to l'rof\u00ab->sor \\Vah-nnan <>f Smith College. 252 HEAT 2. Method of Melting Ice. If w<! = mass of body, 7/i2 = mass of ice melted, t = initial temperature of the body, m^t = 80 w?2. This method was also first used by Joseph Black. The obvious difficulty in it is the measurement of the quantity of ice melted. This has been overcome most successfully in a form of apparatus due to Bunsen ; but great skill is required to use it properly. 3. Method of Condensation of Steam. If 7/jj = mass of the body, m2 = mass of steam condensed, /! = initial temperature of the body, t2 = final temperature of the body, which is never far from 100\u00b0 C., mlc(f2 ~ ll) ~ 536 m2' With this method there is a correction for the water equivalent of the calorimeter; and one of the chief difficulties is to measure accurately the quantity of steam condensed. The apparatus used was invented by Professor Joly of Dublin. Specific Heats of a Gas. \u2014 Any one of these methods can be used to measure the specific heat of a solid or a liquid ; but with gases a difficulty enters owing to their small density \u2014 the correction for the calorimeter is the larger part of the heat energy. To measure the specific heat of a gas at con- stant volume, the third method may be used if. the gas is compressed into a hollow sphere; but this is not very satis- factory. To measure the specific heat at a constant pressure the best method is to pass a", " large quantity slowly through a spiral tube which is immersed in a bath at a high tempera- ture, then through another spiral surrounded by water which is cooler and whose temperature is thus raised, and finally r/M.y ';/\u2022;> L\\ I i-'.Mi '/\u2022;/,'. i /T/;/-: 253 out into the air or into some large reservoir. If the gas is forced through very slowly, its pressure remains practically constant. There are two indirect methods by which the specific heat of a gas at constant volume may be determined : one depends upon a knowledge of the ratio of this to the specific heat at tant pressure ; the other, upon a knowledge of the differ- ence between these two quantities. It may be proved by higher mathematics that the ratio of the specific heat at constant pressure to that at constant volume equals the ratio of the adiabatic coefficient of elas- ticity to the one at constant temperature (see page 194). But this last ratio is a constant for any one gas, which may be determined by measuring the velocity of compressional waves in this gas, as will be proved in the next section of this book. (See page 337.) It will be shown there that this velocity is given by the following formula, V = where p is the pressure of the gas, d is its density, and c is the ratio of the two elasticities, and therefore of the two specific heats. It is not difficult, then, to determine <?, since v, p, and d can all be measured; and. if < \\, and Ov are the two ilic heats, one at constant piv>surc the other at con- 1 volume, Cp=cCv\\ so, if <: and ('/( arc known. Of can be calculate. 1. The difference (7p\u2014 (7, expressed in mechanical units, eqna Is. i what has been ^aid on page 248, the amount of external work done when 1 g. of the gas expands at constant pressure/?, the temperature rising through 1\u00b0C. This work eq luils p(y^\u2014 r{ ). wl id r} are the volumes of the gas at two temperatures dill'erin^ l.\\ 1\u00b0C. We may calculate this l.y ii>in-- the -_ras f. -nil n I i '' = Rm. Iii this case m= 1, 254 HEAT hence pvl = ET^ and pvz = RTV where Tz- Tl = 1. There- fore p(v2 -v^ = fi. So that Q, - Ov = R. If Cp and (7, are measured in calories, and if J is the value of a calorie in ergs, \u00ab/( Op \u2014 Cv ) =.#. The value of R may be found by ordi- nary methods of gas measurement, as has been shown; and, therefore, if the value at Cp is known, that of Cv may be calculated. Ratio of the Specific Heats. \u2014 We may obtain a better physical conception of c, the ratio of the two specific heats, by interpreting it in terms of the kinetic theory of gases; i.e. by considering the properties of a gas as identical with those of a set of elastic spheres. The internal energy of a gas may be regarded as entirely kinetic; but this energy is not entirely energy of motion of the molecules themselves. The phenomena of radiation show that as the temperature of a gas is raised, the energy of motion of the parts of the molecules is increased. Let us assume, following Clausius, that the entire kinetic energy of a molecule \u2014 that is, the energy of its parts and its own energy as a whole \u2014 is proportional to its energy of transla- tion. We can write it then \\ mbVz, where m is the mass of the molecule, V its velocity of translation, and b a factor, which has the value 1 if there is no motion inside the mole- cule, and otherwise is greater than 1. If ^Vis the number of molecules in a unit volume of the gas, the energy of the molecules in this volume is then \\mW^N. But wiN is the mass of the molecules in this volume; so the internal energy of a unit mass of the gas is \\ bVz. This can be expressed in terms of the temperature; for the pressure of a gas is given by the formula p = dRT, if d is the density; and on the kinetic theory p = ^dF2. Therefore V* = $RT\\ and the internal energy per unit mass is J bR T. So, if the tempera- ture is raised one degree, this energy is increased by an amount f bR. This quantity, then, is the specific heat of the gas at constant volume expressed in mechanical units ; or, in CHANGES", " IN TEMPERATURE 255 symbols, Cp = f bR. But Cp - CV=R-, hence Cp = Cv + -f 1). Hence the ratio of the specific heats __ ~~ 21 + The least value of b is 1 ; and therefore the greatest possible value of c is 1 + f or 1.67. For all other values of b, c is less than t Ins. | 6 It is a most striking fact that for certain gases, viz., mer-^ cury vapor, argon, helium, and a few others, the value of c found by experiments on the velocity of waves in them is 1.67, while for all other gases it is less than this, being about 1.41 for air, hydrogen, and oxygen, 1.26 for carbonic acid etc. Those gases for which c equals 1.67 are called by i-li'-mists \"monatomic\"; and, whatever value may be attached to the above assumptions, it is certain that a large value of c indicates an extremely simple construction of the molecule or a molecule whose internal energy is small; while a small value \u00ab\u00bbf /\u2022 indicates the contrary. A few values of specific heats are given in the following tabi AVERAGE SPECIFIC HEATS Alcohol.. 0\u00b0-40\u00b0C. 0.597 Aluminium. 0\u00b0-100\u00b0C. 0.2185 Brass.... 0.09 Copper.. 0\u00b0-100\u00b0C. <> (crown).. 0.161 In.n.. 0\u00b0-100\u00b0C. 0.11:1, 0\u00b0-100\u00b0C. 0.031 Mercury. Paraffin. Platinum Bihra Tin r, Water. Turpenti iii-. 20\u00b0-50\u00b0C.. 0\u00b0-100\u00b0C.. 0\u00b0-100\u00b0C.. 0\u00b0-100\u00b0C.. 0.o:W 0.0:^:5 0.467 1.00 Air.... j Chlorine. Carbon dioxide (JASKS. o.i\u00bb:J7.. 0.1'Jl.. O.'JO'J Of 0.171 \"iv.; RATIO. l.40i. 1.06. ; l.:to 256 HEAT GASES \u2014 Continued Cp Cv RATIO Helium Hydrogen.... 3.40.. 2.40 Mercury (vapor) Xitrogen 0.244 Oxygen 0.217.(17.41.41.41 Law of Dulong and Petit. \u2014 When the values of the spe-, cific heats of a great many substances are compared, a connec- tion becomes evident between them and the \" atomic weights \" of the substances. (For an explanation of this last quantity some book on chemistry should be consulted.) This was first noted by Dulong and Petit. It is found that the product of the value of the atomic weight of any solid substance and that of its specific heat is a quantity that is approximately the same for all substances, viz., 6.4, using the ordinary system of units; while the same constant for gases is 3.4. This means that the same amount of energy is required to raise the tem- perature of an atom, whatever solid substance it belongs to. This product is called the \" atomic heat.\" Naturally, this law of Dulong and Petit is only approximately true ; for, as has been said, the specific heat of a substance varies with the temperature, and it is impossible to know when different sub- stances are at temperatures such that their conditions are comparable. CHAPTER XIII CHANGE OF STATE Introductory. \u2014 The fact that heat energy enters or leaves a body when it changes its state is familiar to every one. In order to melt ice or boil water it must be exposed to some source of heat ; if water evaporates from one's hand or from the surface of a porous jar, tin* latter is chilled, showing that heat energy has been taken from it ; as steam condenses into water in steam coils or \" radiators,\" they are heated, showing they have received energy ; when an acid or salt is dissolved in water, its temperature is changed; when water freezes or dew is formed, the temperature of the surrounding air is raised slightly, etc. During these changes of state, not alone are there heat changes, but alteration in volume, and so external work is done ; and we shall see that the external conditions are of fundamental importance. We shall con- sider in detail a few of the most important cases of change of state : viz., fusion, evaporation, sublimation, solution, and chemical changes. Fusion Freezing and Melting Point. \u2014 \u2022\u2022 Kusion \" is the name \u00ab_ri\\ en tin- prOOeSfl in which a solid lm.lv in. -Its and becomes liquid ; the reverse process is called ki solid", "ification.\" It' a solid l>od\\ that can form crystals, e.g. ice, i^ exposed to a smir. heat, its temperature will rise until a point is reached when it begins to melt; and then, so long as there is any solid to melt, the temperature of the mixture of the solid and its liquid remain^ unchanged; hut when the solid is entirch i m -i. * \u2014 17 267 258 UK AT melted, the temperature again rises. Conversely, if the liquid thus formed is placed in such a condition as to lose heat energy, its temperature will fall until a point is reached at which some of the liquid solidifies ; then the temperature remains unchanged until the liquid entirely changes into the solid form ; and from then on the temperature falls again. This temperature which marks the transition from liquid to solid is the same as that which marks the reverse change. It is called the \"melting point,\" or the \"freezing point.\" If the solid and its liquid exist together in contact, and no energy is added or taken away from them, they will remain in equilibrium ; so the melting point may be described also as that temperature at which the solid and its liquid can exist together in equilibrium. Effect of Variations in the Pressure. \u2014 Experiments show that this temperature of equilibrium depends upon the external pressure, varying as it is changed. Thus, if ice and water are in equilibrium together, under ordinary atmospheric pressure and in a region where the temperature is therefore 0\u00b0 C., an increase in pressure will cause some of the ice to melt, showing that the melting point is lowered and that heat energy flows into the ice from the surrounding region ; a decrease in pressure will cause some of the water to freeze, showing that the melting point has been raised and heat energy flows out from the water into the surround- ing region. The explanation of this variation of the melting point with the external pressure depends upon the fact that when a solid melts its volume changes, in some cases increas- ing, in others decreasing. If ice melts, its size decreases ; that is, the volume of a certain mass of water in the solid form is greater than in the liquid. (Blocks of ice float on the surface of lakes.) Those metals which are used to form castings also expand when solidified ; but in general bodies expand. Thus, gold and silver coins are stamped, not cast, because they expand when melted. When ice melts, the CHAXQB \"/' -I ATE 259 change in volume is of the same kind as would be produced l\u00bbv an increase in the pressure by mechanical means. In other w.\u00bbr<ls, increase of pressure helps on the process of melting; so that if ice is being melted by the addition of heat energy, the temperature does not need to be so high in order to secure melting ; i.e. the melting point is lowered. If a hody expands on melting, an increase in pressure will, for similar reasons, raise the melting point. Regelation. \u2014 This pressure effect is small. In the case of ice and water, if the pressure is increased from one atmos- phere to two, that is, by 76 cm. of mercury, the change in the melting point is only 0\u00b0.0072 C.; and, consequently, ordinary barometric variations have no measurable effect on the melting point of ice. The fact, though, that the melting point is lowered by an increase of pressure is shown by many familiar illustrations. If two pieces of ice with sharp points are squeezed together, the pressure may be enormous because the area of contact may be extremely small ; so the melting point of the ice at the points where the pressure is great will be lowered and, if the surrounding bodies are at 0\u00b0 C., some of the ice will melt, and the result- ing water will be pressed out, so that it is at atmospheric pressure, and its freezing point is again 0\u00b0. But in order to melt this ice, heat energy must be taken from portions of the body near it, and their temperature is reduced below 0\u00b0 ; so the melted water, with a freezing point of 0\u00b0, is in contact with ice at a temperature below 0\u00b0, and it will there- fore immediately freeze again. This is the explanation of t he formation of snowballs ; and this action also plays a most important part in the motion of glaciers. The phenomenon is called \"regelation.\" n<>ti<m <>f :i in : is due to the fact that the ice \u2022\u2022 pressure of the edge of the skate, and so he is actually \u2022!iin layer of water. As the skate moves on, this water freezes again. Another illustration is the formation of so-called \" ground 260 in-: AT ice,\" which is ice formed at the bottom of streams where there are eddies. The ice crystals are whirled round in the current and stick against the bottom, owing to regelation ; then others stick to", " them, etc. Non-crystalline Substances. \u2014 Bodies that are not crystal- line, like waxes, plumbers' solder, etc., do not have a definite melting point ; but as they are exposed to a source of heat, their temperature rises continuously until they are entirely melted. They pass through a \" pasty \" condition ; and the temperature at which this begins is sometimes called the melting point. Similarly, as the liquid is cooled it begins to pass into the intermediate condition at a temperature called the freezing point. These two temperatures are not the same. These temperatures are affected by changes in pres- sure exactly in the same manner as those of crystalline bodies. Undercooling. \u2014 The transformation from the liquid into the solid condition does not always take place as described above ; for instance, if water is cooled gradually, its tem- perature will fall far below 0\u00b0 and yet there is no ice formed. Its condition is, however, most unstable, because if it is shaken or if a minute piece of ice is thrown in, the liquid will solidify immediately and the temperature will rise to 0\u00b0. This phenomenon of a liquid existing below its freezing point, as above described, is called \" undercooling \" ; it was discovered by Fahrenheit. \"Heat of Fusion. \" \u2014 The number of calories that must be added to a solid body at its melting point in order to make 1 g. of it melt, is called the \"heat of fusion\" at that temperature. (It should be noted that this number has the same value if we define the heat unit to be such a quan- tity of heat energy as will raise the temperature of a unit mass of water one degree Centigrade, and define heat of fusion as that number of these heat units which is required to melt a unit mass of the substance, quite regardless of the size of the unit mass.) This energy is spent in overcoming molecular forces and in doing external work if the body expands on r//.i.v<;A' OF 8TATM 261 melting; if it contracts, the external forces also do work in overcoming the molecular forces. The exact way in which this work is done cannot be determined ; but, it is evident tiiat, if the reverse process is carried out, the same amount of heat energy is given out by the body as is received during the direct one. Thus, the temperature of the air in a closed room may be kept from falling far below 0\u00b0, if a tub of water is {-lured in it; for, as the water freezes, a definite amount of heat energy is given off to the air. Again, by placing a pail of water under a fruit tree on a cold night, the fruit may be kept from being injured by the cold. The heat of fusion of a substance may be determined by various methods, such as are used for the measurement of specific heats. Thus, a known number of grams of ice may immersed (not allowed to float) in a vessel containing a known mass of water at a known temperature, and the fall in temperature may be noted. If rw, = mass of ice, ms = mass of water, including the watt-r equivalent of the calorimeter, /, = initial temperature of water, /t = final temperature of water, L = heat of fusion of the ice ; Effect of Dissolved Substances : Freezing Mixtures. - freezing point of a liquid is affected it' tin-re i> ;i I'oivi-rn sub- stance dissolved in it. In every case the free/in- point is lowered : ami the change is, within certain limits, propor- tional t<> the amount of dissolved substance iii a jjiven quan- tity of the solvent, for en-tain substances. With others, the change is abnormally great ; and it is to l>e noted that these -'instances arc those which have an abnormal osmotic pressure. (See page 1M, 262 HEAT If the temperature of a solution is lowered to its freezing point, the solid formed is that of the pure solvent, in general ; so that the solution becomes more concentrated. (In certain cases some of the dissolved substance is caught in the meshes of the solid solvent ; but this is a mechanical process, not a thermal one.) Then, in order to freeze out more of the pure solvent, the temperature must be lowered still further ; for, as said above, the freezing point of the solution falls as its concentration increases. A condition is finally reached with certain solutions such that the solution is saturated ; if now heat energy is withdrawn, some of the solvent separates out in the solid form, and at the same time some of the dissolved substance is precipitated; the temperature remains unchanged, and as more and more heat energy is withdrawn, equivalent amounts of solid solvent and dissolved substance separate out. This complex solid mixture is called the \" cryohydrate \" of the two", " parts. It is in equilibrium with the solution of the same concentration, as we have just seen, at a definite temperature ; so, if a cryohydrate is placed in a region at a higher temperature, it will melt. Thus, the cryohydrate of common salt and water has a composition of 23.8 parts by weight of salt to 76.2 parts of water, and its equilibrium temperature is \u2014 22\u00b0 C. ; so, if salt and ice are mixed thor- oughly and are at a temperature greater than this, the ice will melt and dissolve the salt. In this process the tempera- ture of the mixture and of surrounding bodies falls, because heat energy must be supplied both to melt the ice and to dissolve the salt. (See page 283.) If the salt and ice are in exactly right proportions, this process will cease when the temperature \u2014 22\u00b0 C. is reached. Such a mixture of two bodies as this is called a \" freezing mixture \" ; and the above description explains the use of salt and ice in lowering the temperature of surrounding bodies in \"freezers,\" and also the effect observed when salt is thrown on ice or snow. A freezing mixture of solid carbon dioxide and ordinary CHANGE OF STATE 268 sulphuric ether, known as Thirlorier's mixture, allows one to secure a temperature as low as \u2014 77\u00b0 C. The fusion constants of a few substances are given in the accompanying table : FUSION POINT HEAT or Fusion 1100\u00b0 C. Ice Iron 1400\u00b0-1600\u00b0C. Lead B2PC. -39\u00b0C. iry Sulphur Zinc 0\u00b0C. llf,\u00b0C. 415\u00b0 C. 80 23-33 5.86 2.82 9.37 28.1 Evaporation Boiling Point. \u2014 If a liquid stands in an open vessel ex- 1 to the air, it is observed that the quantity of liquid continually diminishes ; it is said to \" evaporate \" ; the sub- stance passes from the liquid to the gaseous condition. The gas rising from a liquid is called a \" vapor,\" and an exact di >tinet ion between gases and vapors will be made later. (See page 278.) This process of evaporation requires that energy should be constantly added to the liquid, as may be proved by direct experiment, or as is seen by the fact that the hand is chilled when any liquid evaporates from it. ie the process may be hastened by applying some int. -use source of heat to the liquid. If this is done, its tem- perature rises until a point IN reached \\\\hen bubbles of the : (mm in the liquid, rise to the surface, and break. When this stage of \"boiling,\" or \"ebullition,\" is reached, the temperature ceases to rise, and remains constant until all the liquid is boiled away. This temperature is known as the \" l><>iliii'_r point,\" and it is found to vary with the pressure of the m en the surface of the liquid. This is 264 HEAT what we should expect, because in order that the bubbles may form and rise to the surface, the pressure of the vapor in them must be at least as great as the pressure of the air on the surface, and the pressure of the vapor as it rises from the surface must equal this; so, if this pressure on the surface is increased, the liquid must be raised to a higher tempera- ture before it will boil. Similarly, if the external pressure is decreased, the boiling point is lowered. The process of boiling is one, then, of what may be called kinetic equilibrium, depending upon the equality of the pressure on the surface and that of the vapor as it rises from the surface. Saturated Vapor. \u2014 If the liquid is contained in a closed vessel, it is observed that after a time the evaporation appar- ently ceases. The vapor above the liquid is now said to be \"saturated.\" If the pressure and temperature of this vapor are noted, it is found that if the temperature is raised, the pressure increases, and more liquid is evaporated ; while, if the temperature is lowered, the pressure decreases, and some of the vapor condenses to form more liquid. If the tem- perature is kept constant, however, the pressure remains the same, entirely independent of whether there is a small or a large amount of liquid present. This condition may be called one of statical equilibrium. On the kinetic theory of matter it is easily explained. The molecules of the liquid may attain sufficient velocity to break through the surface, thus requiring work to be done upon them. Similarly, the molecules of the vapor may strike against the surface and become entangled, thus losing kinetic energy. So, if both these processes go on together, there will be equilibrium when the number of liquid molecules which escape in any interval of time equals the number of vapor molecules which are retained by the liquid surface in the same time. Sphe", "roidal State. \u2014 A simple illustration of the kinetic nature of evaporation is afforded by what is called the \" spheroidal state.\" If a small quantity of water is allowed CUAXGE OF STATE 265 t<\u00bb llow gently out of a tube or spoon on to a metal surface which is at a high temperature \u2014 far \"above 100\u00b0 C. \u2014 and which is slightly hollowed out so that the water will not run it is seen to collect in a flattened drop, which does not on the surface, and it now rapidly evaporates. The explanation is evident; for, owing to the rapid evaporation on the lower side occasioned by the heat energy received from the hot metal, the molecules are leaving tin- drop on this side with such velocity and in such quantity that their mechanical reaction holds up the drop. There is thus a layer of vapor between the drop and the hot plate. This spheroidal state can be noticed when water is spilled on a hot stove, and in fact the hotness of a stove or a flatiron is often tested in this manner by seeing if it can produce this state in small drops of water. (This condition of a drop is sometimes called \" Leidenfrost's Phenomenon,\" because tin- first observation of it U attributed to him, 1766.) Vapor Pressure. - One of the simplest :i<\u00bbds of obser \\inir the phenomena of satu- d vapor is to intro- duce a small quantity of the liquid to be erimented <\u00bbn above the mercury column in a barometer which! a deep basin. Some of :i 190. \u2014 Experiment* illtintrmtlnjr tl preMare of Mtnrated vapor : \u00bb umall u ll.ini.l. t.g. water, to Introduced there tl the liquid will evaporate, and equilibrium will lx\u00bb reached at a pressure d. -pending upon the temperature. This ma\\ 266 HEAT be varied at will by surrounding the tube with a bath of some liquid whose temperatures can be regulated. The pres- sure of this saturated vapor may be measured in terms of the atmospheric pressure by the ordinary law of hydrostat it- pressure. (If h is the difference in height of the mercury column in this tube and of that in a barometer, the pressuiv of the vapor is less than the atmospheric pressure by dgh, where d is the density of mercury.) If the temperature is kept constant and the ba- rometer tube slowly raised, some liquid will evaporate, but the pressure remains con- stant; similarly, if the tube is slowly pushed down, some of the vapor is condensed, but the pressure does not change. (If there were a gas above the mercury column in- stead of the vapor, its mass would not L00\u00b0 200\u00b0 Fm. 121. \u2014 Curve showing connection between the tem- perature and the pressure of saturated water vapor. change, but its pres- sure would decrease and increase during the above changes, obeying Boyle's law : pv = constant. But when there is a vapor in contact with its liquid, there is no change in pressure, but the mass of the vapor increases and decreases.) If, however, the temperature is increased, some liquid evaporates and the pressure increases ; and, if it is lowered, some vapor condenses, and the pressure decreases. (If CHANGE OF STATE 267 there were a gas above the mercury column, its mass would remain constant during the above changes in temperature, a n\u00ab I its pressure would change, but at a different rate from that of the saturated vapor.) There is thus seen to be a delinite pressure of the saturated vapor which corresponds to a definite temperature when there is equilibrium, and conversely; the corresponding values may be found by either the statical method or the kinetic one, in which the boiling point at different pressures is determined. The results may be expressed by a curve drawn with axes of temperature and pressure. This curve for water vapor is given. That the boiling point varies directly with the pressure is shown l>y the fact that the temperature of boiling water is much less than 100\u00b0 C. on a mountain top ; by the high tem- perature in steam boilers where the pressure is great, etc. From the description given above of the statical method, it is seen that there are two general methods available for con- dt nsiii'^ a vapor into a liquid : one is to lower the temperature ; the other is to decrease the volume. These will be discussed in full in a later section. If a vapor is not saturated, it obeys hiiyle's law quite closely ; and Dalton's law is also approximately exaet for the pressure produced by a mixture of vapors. A curve may be drawn that will >s this law of a vapor in con- tact with its liquid when the tem- i tu re is constant. Lay off axes \u00ab of pressure", " and volume; then, since a vapor keeps its pressure constant so long as the temperature is con- VOLUMES. the isothermal is a straight Fi\u00ab.i\u00ab. line parallel!\u2022\u2022 theaxis of volumes. Formation of Dew, Clouds, etc. \u2014 We have seen that, as the temporal UP- d, the corresponding vapor pressure of -a i u rated vapor becomes less ; that is, if there is a certain 268 HEAT amount of unsaturated vapor in a closed vessel, it will become saturated if the temperature is lowered sufficiently ; and then, if it is lowered still more, some of the vapor will condense. This is illustrated by the formation of dew, of clouds, etc. There is always a certain amount of moisture in the air, and the method of expressing it is as follows : we measure the temperature of the air and then by experiment find what temperature some solid body \u2014 like a metal can \u2014 must have in order to make moisture condense on it. This is called the \" dew-point.\" The vapor pressure corresponding to these two temperatures is found from tables or from the curve given on page 266 : one of these expresses the pressure that the water vapor in the air might have if it were saturated at the existing temperature ; the other gives the pressure that the water vapor in the air actually has. The ratio of the latter to the former gives what is called the \"humidity.\" If we assume the truth of Dalton's law, we can easily calculate the mass of a unit volume of ordinary damp air. This equals the sum of the masses of the air itself and of the water vapor in the space. The quan- tity of water vapor in a unit volume corresponding to various dew-points is given in tables ; e.g. if the dew-point is 10\u00b0 C., the mass per cubic metre is 9.3 g. The pressure of this vapor corresponding to 10\u00b0 C. is 0.914 cm. of mercury ; thus, if the barometric pressure is 76 cm., the pres- sure due to the air is 75.086 cm.; and, if the temperature of the air is known, e.g. let it be 20\u00b0 C., the mass of the air can be calculated, since the density of dry air at 0\u00b0 C. and 76 cm. pressure is known to be 0.00129. Thus, using the gas law, the density at 20\u00b0 C. and at 75.086 cm. pressure equals |Z?. 0.00129, or 0.00119. Therefore the mass of the air in a cubic metre is 1190 g. ; and the total mass of the cubic metre of damp air is 1190 + 9.3 = 1199.3 g. If the air were perfectly dry and at 20\u00b0 and 76 cm. pressure, its density would be |^ x 0.00129, or 0.00120 ; and so the mass of a cubic metre of it would be 1200 g. It is seen, then, that the mass of dry air is greater than that of damp. Boiling. \u2014 As already explained, the process of boiling consists in the formation of bubbles of the vapor in the interior of the liquid. Nuclei of some kind are required in order for CHANGE OF STATE these to form, such as sharp points or minute bubbles of some foreign gas, like air. As a liquid boils, the supply of such nuclei is used up, unless in some way it is renewed constantly, and it becomes more and more difficult for the liquid to boil. The temperature rises above the boiling point until the molec- ular forces are sufficient to form the bubble; there is a miniature explosion; and the temperature falls back to the boiling point. If the nuclei are removed from the liquid as completely as possible and if the walls of the containing vessel are smooth, the temperature of the liquid may be raided far above the boiling point; but this condition is of course unstable. If the liquid were entirely free from nuclei, it would never boil; but,! temperature were gradually raided, it would finally explode. Heat of Vaporization. \u2014 The num- ber of calorie.s that must be added to a liquid in order to make one i^ram of it evaporate, or boil at a \u2022 If finite temperature, is called the \u2022\u2022l!'-at of Vapori/.ation \" at that temperature. (This number does not depriid upon the size of the unit of mass if the heat unit is suitably defined. See page 260.) This eneiLfv is spent ill securing inter- nal changes corresponding to the it inn from liquid to vapor and also in doinur external work; this la>t can he calculated because the pressure must be kept con- stant, inasmuch as the temperature nilnntluii <>f thr li.'.i; \u2022 ofall.j.ii.l. /? Is \u00bb", " ring-burner for hcatinir tlio ll.|iii.t.-..ntnln.-d In tho v<\u2014 < ^undented In the colls In the lower veaseL is assumed not to vary. The same number of calories is also given out when one gram of a vapor condenses at that t< -\\\\\\- 270 HEAT perature. This physical quantity is measured ordinarily by the method of mixtures ; a known mass of vapor is condensed by making it enter a quantity of its liquid through a spiral tube, or in such a manner that it gives up the heat energy entirely to the liquid. A simple form of apparatus is shown in the cut. If 7/ij = the mass of vapor condensed, m2 = the mass of liquid in the calorimeter originally, including the correction for the calorimeter, *! = initial temperature of the liquid, t2 = final temperature of the liquid, ts = boiling point of the liquid at the given pressure, c = specific heat of the liquid, L = heat of vaporization, then m^L + mtc (t3 - t2) = m2c (t2 - ^), or L = -t 1* _ c (ts - *2). - m2c (t2 - tj). Fio. 124. \u2014 Two forms of cryophorus. Several illustrations have been given already of the changes in heat energy when a liquid evaporates or a vapor condenses, but one or two more may be described. In an experiment due originally to Boyle, a small quantity of water is placed under a bell jar attached to an air CIIAM.I: <>r >TATE 271 care being exercised to guard against any possible conduc- tion of heat to the water; as the air is exhausted, thus diminishing the pressure on the water and removing the vapor, the water evaporates so rapidly that the heat energy required is taken from the water left behind, and its temperature falls until it freezes. A somewhat similar experi- ment is one that involves the use of the *' cryophorus,\" an instrument invented l\u00bby Wolluston. This consists of two glass bulbs connected by a bent glass tube, as shown in the cut in two forms, a and b. There is sufficient water inside to half fill one of the bulbs. The experiment con- sists in placing the instrument a in a vertical position, with the curved portion uppermost, or the instrument b horizontal ; the water is poured into one bulb, which is carefully shielded against heat loss or gain, and the other is surrounded by a freezing mixture of salt and ice. After a short time the water will be found to be frozen, owing to rapid evapora- tion at its surface, which is caused by the continuous condensation and ing of the vapor in the other bulb. The action, then, is the same as if a substance \" cold \" were carried from one bulb to the other ; hence th\u00bb- name \u2022\u2022 rryophorus,\" which means \"carrier of cold.\" Some forms of apparatus for making artificial ice depend upon the same fact, that when a liquid evaporates, heat energy is required. In most cases the liquid which is evaporated is ammonia. This is placed in the space between the two walls of a double-walled vessel which contains water, and as the ammonia is evaporated, the water is frozen. Steam Engine. \u2014 As a further illustration of the proper- ol a vapor, the steam engine may be mentioned \u2014 a dia- gram of a simple form of which is given in the cut. The principle of its action is as follows: Steam is produced in a \"boiler\" under hi^h pressure, and therefore at high tem- perature; this steam is allowed at regular intervals to enter the \"cylinder.\" in which there is a movable piston, at the instants when the piston reaches one end of the cylinder, and in such a manner as to exert a pressure on this piston, pushing it away from this end. Steam continues to enter, and the pressure is that of the steam in the boiler as Imi^ as the connection is maintained; but after the supply of steam is cut off, the steam, as it expands, decreases in pres- sure. In the meantime the pressure in the cylinder on the other side of the piston has been made as small as possible 272 HEAT by one of three methods: (1) by opening it to the atmos- phere\u2014 this makes a kt non-condensing\" engine; (2) by join- ing it to a large vessel, which is kept as nearly exhausted as possible by means of pumps and as cool as possible by means of coils or jets of water \u2014 this makes a \"condens- ing \" engine ; (3) by join- ing it to another cylinder, exactly like the first one, only larger, and allowing the steam ejected from the first cylinder", " to work the piston FIG. 125. \u2014 Steam Engine. CHANGE OF 8TATM 273 in the second \u2014 this makes a \"double, triple, etc., expan- sion\" engine. When the piston reaches the other end of the cylinder, connection is made with the boiler, as above described, at this end, and the expanded steam is expelled, as the piston retraces its path, in one of the ways just mentioned. In the case of the non-condensing engine the steam escapes to the air and is lost; in that of the condensing engine it is condensed into water in the con- denser and pumped back into the boiler; in the expan- sion or compound engine it is used over again, expanding more and more until, finally, it is condensed in a condenser and pumped back to the boiler. These changes are made automatically by certain valves ; the \" sliding valve \" opens and closes the inlet pipes and also the exhaust pipe to the condenser. In all cases a certain quantity of saturated vapor is received at a definite pressure (and corresponding temperature) ; this expands, doing work on the piston, until a certain lower pressure is reached, then \u2014 in the condens- ini: engine \u2014 it is condensed to water at the pressure corre- sponding to the temperature of the condenser; this water is pumped into the boiler, its temperature is raised until it boils, and the process begins again. There is t hus a \" cycle \" of changes. While in the boiler, the water receives heat energy; and when water is lu-ing formed in the condenser, heat energy is given out by the steam. The steam does work in pushing on the piston, and work is done on it and on the water formed from it when the piston performs its reverse motion. If H is the heat energy received at the temperature of the boiler, and W is the net external work W H. done, the ratio -= is called the \"efficiency\" of the process. The amount of work done may he measured by a simple mechanical device. We have seen (page 161) that on a pressure- \\o] u in. diagram, a closed curve.describing the series of changes tlin>u<_:h whieh a tin id passes indicates by its area AMES'S PHYSICS \u2014 18 274 HEAT the total net work done. The curve giving the cycle of changes just described for the steam leaving the boiler, ex- panding, etc., must be somewhat as shown in the cut. A marks the instant when the water in the boiler begins to be changed into steam ; B, when this process is finished, the pressure and temperature remaining constant; this steam can be imagined as formed directly back of the piston and exert- ing a pressure on it; so the instant marked by B is that of the \" cut-off \"; the steam now expands, and both temperature and pressure fall; at (7, connection is made with the condenser, in which the pressure is that marked by the horizontal line ED\\ so the pressure falls to D, and the steam condenses to water as marked by E\\ this water is forced by a pump into the boiler, and its temperature and pressure both increase until the point A VOLUMES v is again reached. Looked at from another point of view, this curve indicates very approxi- mately the changes in volume and pressure of the space in the cylinder which is open to the steam as the piston moves to and fro : when the piston is close to one end, the volume is small, and as steam is admitted, the pressure rises from E to A ; then it remains constant as the piston is pushed out, thus increasing the volume, until the cut-off is reached at B ; then, as the piston continues to move forward, the volume increases and the pressure falls; O marks the end of the motion of the piston ; as it moves back, the volume decreases, but the pressure remains unchanged, being that of the condenser ; etc. These pressure and volume changes may be recorded automatically by the engine in several ways : a piece of paper may be fixed to a drum and a pencil moved over it, motion in one direction being secured < I1AXGE OF STA 1 1-: 275 Attaching the pehcil to a spring gage which measures the pressure in the cylinder and moves out and in as this i IK leases and decreases, and motion at right angles to this ln-ing obtained by last* -n ing the drum carry - the paper to a cord which is attached to the moving piston, and so lengthens or shortens as this moves, and thus makes the volume in the cylinder i nerease or decrease. These curves are called \"indi- cator diagrams,\" and are of the greatest assistance in studying the actual work- in LT of an engine. A form of indicator mechanism often Fio. 127. \u2014 Indicator mechanism", ". I is shown in the cut. The first successful steam engine was made by Newcomen in 1705; but all the great improvements, as shown in modern engines, are with one exception due to.James Watt. He invented the condenser in 1763; in succeeding years he con- ceived the idea of introducing the steam alternately on the two sides of the piston, and <>f cutting off the steam from the boiler so as to allow it to expand; In also invented the plan for surrounding the cylinder with a jacket into which steam \u00ab -ould be admitted, so as to keep the cylinder always hot. Hornblower was the first to construct a compound engine, i.e. one with two cylinders, so as to have double expansion. This was done in 1781. Effect of Dissolved Substances. \u2014 The vapor pressure of a liquid is alTeetrd it some substance is dissolved in it, being always lower \u00ab n t. mj>erature than that over the pure liquid. Then-ion- the boiling point of a solution is higher 276 // /\u2022;. I / than that of the pure liquid at a given pressure. These changes in the pressure or in the boiling point vary with the concentration of the solution. They are proportional to it in some cases, but in others the change is abnormally great. These last solutions are those for which there is an abnormal lowering of the freezing point and an abnormal osmotic pressure. When a solution evaporates or boils, the vapor formed is that of the pure solvent unless the dissolved sub- stance is volatile ; consequently, when the temperature of the boiling solution is to be measured, the thermometer must be immersed in the liquid, not in the vapor. The vaporization constants of a few liquids are contained in the accompanying table : BOILING POINT AT PRESSURE OF 76 CM. OF MERCURY HEAT OF VAPORIZATION 78\u00b0 C. - 80\u00b0 C. Alcohol (ethyl) Carbon dioxide Chloroform.... Cyanogen.... Ether (ethyl).... Hydrogen.... Mercury..... 357\u00b0 C. Oxygen - 184\u00b0 C. - 20\u00b0.7 C. 34\u00b0.6 C. - 238\u00b0 C. 61\u00b0.20 C. 444\u00b0.5 C. Water 100\u00b0 C. 209 58.5 72 at - 25\u00b0 90 103 at 0\u00b0 \u2014 \u2014 62 ___ 535.9 Isothermal for a Change in State from Vapor to Liquid. - The various physical properties of liquids and vapors can best be expressed and studied by graphical means. One method is to draw isothermals on a diagram whose axes correspond to pressure and volume. Let us imagine the vapor inclosed in a cylinder that has a movable piston. We can then represent by curves the various conditions of the vapor, as the piston is forced in gradually, the temperature CBAJfOM <>F STATE 277 kept constant. At first, as the volume is diminished, i In- pressure increases according tu Boyle's law, since the vapor is unsatu rated, as shown hy the curve AB. But finally, a pressure is ivaehed such that th\u00ab vapor is now saturated for the given temperature; and, if the volume is still further decreased, the pressure remains unchanged, as shown by the curve BC. As this decrease in volume goes on, more and more of the vapor condenses and sinks to the bottom of the cylinder. When all the vapor is condensed, the cylinder is full of liquid, and to produce any further decrease in volume a great increase in pressure is Fio. 128. \u2014 Isothermal for change from a vapor to a liquid. 450780 VOLUMES IN CC PER GRAM. Fio. 1\u00bb.\u2014 Isothermal* of carbonic ari.i (\u00ab>,). n hy the curve CD, which is nearly vertical. Cun ijx kind were first determined and studied by working with carbonic acid Lfas. in the years previ- 278 HEAT ous to 1869. The results of his investigation are given in the accompanying cut. It is seen that, as the isothermals at higher and higher temperature are drawn, they have the same general shape, but the horizontal portions become shorter, until a temperature is reached, the isothermal for which has no horizontal portion. This is called the \"critical\" temperature. For temperatures higher than it, the isothermals approximate more and more closely to those of a gas. The Critical Temperature. \u2014 The only points on the dia- gram for which the matter is in the form of a liquid are along the horizontal portions of the isothermals, where there is a free surface separating the liquid and its vapor, and along the continuations of these lines to the left, where the liquid completely fills the vessel. Consequently, whenever we see a liquid partially filling a vessel, it is represented on the diagram by a point on a horizontal portion of an iso- thermal. In", " other words, if a vapor is to be liquefied, it must be at a temperature whose isothermal has a horizontal portion, that is, it must be at a temperature lower than the critical one. Bearing this fact in mind, all gases, with the possible exception of helium, have been liquefied. The name \"vapor\" may then be restricted to a body in the gaseous condition which is at 'a temperature below the critical one ; while the name \" gas \" may be limited to temperatures above this. The exact point on the diagram, marked by A, where the critical isothermal has its point of inflection, that is, the point where an isothermal infinitely near the critical one, but below it, has a minute horizontal portion, is called the \"critical point.\" If we have matter in this condition filling a vessel, and if the temperature is lowered, even the least amount, the liquid will separate out and sink to the bottom, filling approximately half the space and leaving the rest full of vapor. At the critical point, then, the surface of separa- tion disappears, and the liquid and vapor dissolved in each CHAtfQM OF -/.I//-: 'J7'.' other make a homogeneous form of matter. The pressure corresponding to the critical point is called the \"critical pressure,\" and the volume of one gram of the substance in this critical condition is called the \"critical volume.\" In order to liquefy a gas, then, two steps are necessary ; the temperature must be lowered below the critical tempera- ture, and the volume must be decreased until the pressure is reached that corresponds to the state of saturation of the vapor for the temperature. After this, any further decrease in volume or temperature will cause the vapor to condense. The values of the critical temperatures for various gases are given in the accompanying table : CRITICAL TEMPERATUI:I - Alcohol... 243\u00b0.6C. Ammonia... 130\u00b0 C. Argon. - 120\u00b0 C. Carbon dioxide.. 30\u00b0.9 C. Chloroform 260\u00b0 C. Hydrogen... - 242\u00b0 C. Nitrogen. - 146\u00b0 C. Oxygen... - 119\u00b0 C. Sulphur dioxide.. 156\u00b0 C. Water. 365\u00b0 C. Liquefaction of Gases. \u2014 The critical temperatures of such gases as hydrogen, oxygen, nitrogen, etc., are seen to be extremely low ; so that special means must be adopted in order to liquefy them. There are three methods in use for the production of low temperature ; application of freezing mixtures, rapid evaporation of a liquid, expansion of a gas from lii^h to low pressure. (See page 246.) The standard method for liquefying gases is a combination of these to a certain degree. The gas to be liquefied is compressed by pumps to a high pressure and is cooled by a freezing mixture or l\u00bby evaporation of a liquid : it is then allowed to expand through a small opening, is again compressed ; and tin- process is repeated. This expanded \u2022 older than it was : iind 1). -fore iM-injr compressed again, as.is it expandfl through the opening, it i^> drawn i the mi IQ unexpand : hus chilling 280 HEAT it. As the process continues, the temperature of the com- pressed gas gets lower and lower, until finally, on expansion, the critical temperature is passed and drops of the liquefied gas fall to the bottom of a Dewar bulb. Two forms of apparatus are shown in the cut on page 281 ; one due to Linde, the other to Dewar. The most important parts of the former are the two-cylinder air compressor and the \"counter-current interchanger.\" This last consists of a triple spiral of three copper tubes wound one inside the other. The cycle of operation is performed in such a manner that compressed air at the temperature of the coil g and at about 200 Atm. flows through the inmost tube of the spiral from top to bottom, and passes out at the lower end through a valve a under a pressure of some 16 Atm., it returns upwards through the annular space between the inner and middle pipes, is again raised to a pressure of 200 Atm. by the smaller cylinder d of the compressor, and then begins the same cycle over. The larger cylinder e of the compressor pumps a small amount of air from the atmosphere into the suction pipe of the small cylinder, that is to say at 16 Atm. A similar quantity of air must therefore leave the cycle at another point so that the pressures may remain constant. This escape of air takes place at the lower end of the counter-current interchanger immediately after the discharge from 200 to 16 Atms. so that a controllable amount of air issues from 16 Atms. to 1 Atm. through a", " second valve &. Part of this air, when the apparatus is cooled down to the temperature of liquefaction, becomes liquefied and collects in a \"Dewar flask\" c. That part of the air issuing from the second valve which is not liquefied leaves the apparatus, es- caping through the space between the middle and outside pipes of the spiral into the atmosphere. An iron pipe in the form of a coil #, which is cooled down to a few degrees below zero by a freezing mixture of ice and chloride of calcium freezes out the small quantities of water vapor contained in the highly compressed air until only traces remain, and also cools the air. The properties of bodies at extremely low temperatures are entirely different from what they are at ordinary temperatures. The tempera- ture of liquid air when boiling at atmospheric pressure is \u2014 182\u00b0 C. ; under these conditions lead becomes elastic and iron and india rubber extremely brittle ; egg shells, leather, etc., become phosphorescent ; the electrical resistance of all pure metals decreases greatly; etc. By suitable methods, air, oxygen, and even hydrogen may be solidified. CUA\\>.i: OF 8TA /'/\u2022: 281 Apparatus of Llnde for the liquefaction of air and other gases. Apparatus of Dewar. A is a ryllmlrr containing the gas HIH|IT lilirh pr A U a Dewar flask containing a freezing mixture \u00bbuch aa solid carl etlMr ; C Is a second flask containing some liquid which Is made to evsponit- rnpidly. *.</., etl, - a thlrtl ring the g\u00ab* thr\u00bbui;li //.''. \u2022 Nrft|K- tliniinrh hjr the rod f. As the gas escni <lrawn out through t a-\u00ab It rises it c<>. 282 HEAT The melting point of hydrogen is estimated at \u2014 257\u00b0 C. ; and the max- imum density of liquid hydrogen is 0.086. By allowing a quantity of liquid air to evaporate slowly, and using the methods of fractional distillation, Ramsay discovered three new constituents of our atmos- phere, which he called Krypton, Xenon, and Neon. Continuity of Matter. \u2014 It is seen from the experiments of Andrews that it is possible to make a body pass from the state of vapor to that of liquid, and vice versa, by a series of continuous changes, because a path can be drawn from a point on the pressure-volume diagram where the body is in the form of a vapor to one where it is liquid, which does not pass through the region where the isothermals are horizontal, and where therefore the liquid and vapor exist separately, but in contact. The isothermal has two points of discon- tinuity, at the two ends of its horizontal portion, which correspond to molecular \u2022 rearrangements ; but a series of changes can be imagined, as above described, during which there will be no sudden molecular changes, but by which a vapor can be gradually and continuously changed into a liquid. This is ordinarily expressed by saying that matter is continuous from the liquid to the vapor state. Similarly, matter is continuous from the solid to the liquid state. Sublimation In many cases a solid body evaporates directly without passing through the liquid condition. This process is called \" sublimation \" ; and it is illustrated by camphor, arsenic, iodine, carbon, many metals, snow or ice, etc. It is found by experiment that, if this process takes place in a closed vessel, there will be equilibrium between the solid and the vapor when a definite pressure is reached, which depends upon the temperature. If the latter is increased, the equi- librium pressure is higher, and conversely. The reverse of this process of sublimation is seen in the formation of frost. It is found that all these substances which sublime under \u2022 ll^^<,l\u2022: or STA n-: ordinary conditions may be obtained in the liquid condition it' a suitable pressure ami temperature are applied. (This i> one step in Moissan's method of making artificial diamonds. ) The number of calories required to make one gram of a solid sublime at a definite temperature is called the \"Heat of Sublimation.\" It equals the sum of the Heats of Fusion and of Evaporation at that temperature, in accordance with the principle of the Conservation of Energy. Solution Heat of Solution. \u2014 When a body \u2014 solid, liquid, or gas \u2014 is \u2022 Ivi-d in a liquid, both being at the same temperature, tin-re is a change in temperature, showing that heat changes are involved. If the temperature falls, it shows that work is required to make the substance dissolve, if we assume that tin-re are no secondary molecular changes such as the forma- tion of ne\\v molecules or dissociation. The heat", " energy that is gained or lost is measured by the product of the mass of the solution, its specific heat and the increase or decrease of temperature. If the temperature rises, heat energy is said to be \" evolved \" ; if it falls, the energy is said to be \"absorbed.\" If one gram of the substance is dissolved in a eertain quantity of solvent, the heat energy thus involved will, in general, vary with the quantity of solvent ; but, by continually increasing this, it is found that after a certain point the heat changes are independent of the quantity of solvent. The heat energy gained or lost, when one gram of a substance is dissolved in such a large quantity of the solvent as this, is called the \" Heat of Solution \" of the substance. Some values are given in the following table, the solvent being water. A plus sign indicates a rise in temperature, or \"evolution of heat,\" and a mi mis si LTD, the opposite. Ammonia ffaa. +495.6 calories. Caustic potash. \u2022\u2022 ones. Ethyl alcohol. + 55.3 calories. Sodium Chloride. - 1 XL' calories. Sulphuric ;i<i<l. + 182.5 calories. Silver chloride. \u2014110. calories. IIKAT Effect of Rise in Temperature upon Solution. \u2014 We can from these facts predict whether raising the temperature of a solu- tion increases its solubility or not. Let us consider a saturated solution with an excess of dissolved substance precipitated ; and let it be one in which heat energy is absorbed when solution takes place. If now heat energy is added, the sol- vent will dissolve more of the substance ; and if heat energy is withdrawn, the solvent will precipitate some of its dissolved substance ; because in this last case, for instance, if the effect were to make the solution more soluble, this act of solution would withdraw some heat energy, etc., and the condition would be unstable. Therefore, with a solution of this kind, an increase in temperature makes it capable of dissolving more ; a decrease, less. Just the converse is true of those solutions in whose formation heat energy is evolved. Dissociation ; Ions. \u2014 The fact that certain solutions have an abnormally great osmotic pressure, and also have freezing and boiling points that differ abnormally from those of the pure solvent, can all be explained if it is assumed that in these solutions a certain proportion of the dissolved molecules are broken up into smaller parts. This has the immediate effect of increasing abnormally the number of moving particles due to the dissolved substance. If we assume, further, that these fragments of the molecules are electrically charged, we can explain the phenomena of electrolysis, as we shall see later ; and all solutions of this kind do conduct electric currents. The whole science of Physical Chemistry is based upon these two assumptions, and they may be regarded as justified by experiments. When such a solution is formed, some of the molecules dissociate into their parts since thereby the poten- tial energy is made less ; or, as we may express it, there is a force producing dissociation. The process ceases \u2014 or equi- librium is reached \u2014 when this \"solution pressure\" is balanced by the electrical forces that are called into action. The equilibrium is not one in which there is no further dissocia- tion, luit one in which for each molecule dissociated there is formed. The dissociated parts, called \"ions,\" are mov- ing t<> and fro in the solvent, and combinations and dissocia- tions are taking place continually but at the same rate. Chemical Reactions Heat of Combination. \u2014 In all chemical reactions there are molecular changes, and consequent heat changes. If the ;in^ bodies are gases, these changes depend to a marked degree upon the external conditions that are maintained ; for these determine the amount of external work. It is entirely immaterial, however, whether the change takes place in one or more stages. Thus, if one gram of carbon in the form of a diamond is converted into carbon monoxide (CO), -11<\u00bb calories are evolved: and if this is converted into carbon dioxide (CO2), 5720 more calories are involved \u2014 7860 in all. And, if the same amount of carbon is oxidized at once into carbon dioxide, the heat involved is the same. This is an illustration of the Conservation of Energy. A few illustrations of these heat changes may be given. : <j. of hydrogen L^as and 16 g. of oxygen, at 0\u00b0 C. and 76 cm. pressure, combine to form 18 g. of water at 0\u00b0, the heat energy evolved is 68,834 calories. If 65 g. of zinc are \u20221 ved in dilute sulphuric acid, 38,066 calories are evolved ; while if 63 g. of copper are dissolved in dilute sulphuric acfd, 12,500", " calories are absorbed. Dissociation. \u2014 One of the most interesting chemical changes is the dissoci.it inn of a '_r;ls i,,((, other gases. Thus some gases with complex molecules break down into others with simpler molecules when the temperature is raised to a high degree. In many cases it is observed that for a definite temperature equilibrium is reached at a definite pre- nd, if the temperature is increased. H Lfl the \u2022ndiiiLf pressure. This condition of equilibrium is one of continual recombination and d \"ii. CHAPTER XIV CONVECTION, CONDUCTION, AND RADIATION It has been shown that there are three general methods by which heat energy is added to or taken from a body : Con- vection, Conduction, and Radiation. Each will now be dis- cussed in turn. Convection When a vessel containing a liquid is placed on a hot stove, the upper layers of the liquid receive heat energy from the lower ones by the process known as \" convection \" ; the por- tions of the liquid in the lower layers have their temperature raised and therefore expand ; and, since their density is thus diminished, they are forced upward by gravity, the cooler upper portions sinking. Thus the temperature of the whole liquid is made uniform. The mechanics of the phenomenon is not difficult to understand, because the molecules of the hot upward-moving portions of the fluid communicate by their impacts some of their energy to the other molecules ; and thus the internal energy is distributed. It is evident that convection processes can occur only in fluids under the action of gravity, and when the heat energy is applied to its lower portions. It should be noted that the energy is gained by the portions at low temperature and is lost by those at a higher temperature. This method of distribution of heat energy is the one that forms the basis of the ordinary means of heating houses, \u2014 hot-air furnaces and stoves, hot-water systems, steam pipes, etc. Convection is of funda- mental importance in the economy of nature, as is explained in Physical Geography, in connection with winds, ocean currents, etc. The metal 286 CONVECTION, CONDUCTION, AND RADIAtlON 287 \u2022in of a tea kettle (or of a steam boiler) does not become unduly hot so long as there is water in it, localise, owing to convection, the cooler portions of the water are being continuously brought down to the >m. Conduction General Description. \u2014 When one end of a inetal rod, like a poker, is put into a fire, the temperature of this end rises, and in a short time that of the other portions not too far from the fire rises also. The temperature of the end in the fire is the highest; and that of the other points of the rod decreases gradually as one passes from the fire, until a point in the rod is reached that is at the temperature of the sur- rounding air. If a thin transverse portion or slab of the rod is considered after it has come to a steady state, it is evident that its side near the flame is at a higher temperature than the one away from it; the molecules at the former section have more energy of motion than those at the latter. As a consequence, the former molecules give up some of their energy to the molecules of the slab ; and its temperature would rise were it not for the fact that the slab is losing lit at energy by convection in the surrounding air (and by radiation, also, a process to be described presently), and that tin- molecules in the cooler end of the slab are themselves handing on energy to the other portions of the rod. This process by which molecules give up some of their energy to iguous molecules, there being no actual displacement as in convection, is called \"conduction.\" Thus, considering lah across the rod, we say that it gains heat energy at the hot face and loses it at the cooler one by conduction : and the difference between the \u00ab|uantities gained and lost must e<|iial that lost at the surface of the rod by convention 1 radiation), -iuce the rod i- in a steady state. It is important to note that the heat energy is conducted from the hotter portions of matter to the colder ones. When the >t in a steady state, e.g. immediately after one end is 288 BEAT put in the fire, part of the energy that enters the slab to raising its ti-inperature, to doing external work, etc. \" Conductivity \" for Heat. \u2014 If the rod is in a vacuum, there is very little energy lost, in general, from the surface, because there is now no convection ; and, when the bar is in a steady state the energy conducted in at one section of a slab equals that conducted out at the other. If t\u00b1 is the temperature of one section and t2 that of the other, if A", " is the area of each section and a the thickness of the slab, ex- periments show that the quantity of heat energy conducted through from the former section to the latter, if t\u00b1 is greater than \u00a32, is proportional to \u2014 ^\u2014, but is different for rods Sf \u00a3 \\^ of different material. This fact may be expressed by the following equation, in which Q is the quantity of heat energy conducted by the slab, Q = k\u2014 2A, where k is a factor of proportionality, which is different for different bodies. It is called the \" conductivity \" for heat. If the conductivity of one body is greater than that of another, it is said to \"conduct better.\" Thus silver conducts better than copper; copper, better than iron ; all metals, better than wood and other non-metals ; etc. The conductivity of any one body varies slightly with its temperature. If the conductivity of a fluid is to be determined, the upper surface must be made the hotter, so as to avoid con- vection. All liquids conduct poorly, with the exception* of fused metals ; and all gases conduct still worse. Thus loss of energy from a body by the processes of conduction and convection may be avoided by inclosing it in a quantity of eider down, feathers, or loose wool or felt; because these solids are poor conductors and motion of the air inclosed by them is prevented, as it is contained in small cavities. The best method of all, however, for avoiding these losses is to have the body inclosed by another and to have the space CONVECTION, CONDUCTION. AND HA />/ J77OJV 289 between completely exhausted of air. I>\\ using a Dewar tlask (sec page -'28), liquid air and hydrogen may be kept for hours in a room at ordinary temperatures. Illustrations. \u2014 The fact that metals conduct well is shown by count- less experiments. Thus, if a piece of wire gauze with fine meshes is lowered over a flame, e.g. one from ;i Bunsen burner, the latter burns below the gauze only ; because the molecules of the gas as they pass through the wire meshes lose so much of their heat energy by conduction to the outer portions of the gauze beyond the flame that the temperature of the gas as it rises through the -;ui/. \u2022 is lower than that at which if burns. However, the temperature of the gauze gradually rises, owing to the Maine, and as soon as the temperature of combustion is reached, t In- flame will Imrn on both sides of the gauze. Or, if the gas is turned on through the burner, but is not lighted, and the gauze is held close to the burner, the gas rising through the gauze may be ignited by a match, but the flame will not strike back below it. (This is the principle of the miner's safety lamp invented by Sir Humphry Davy.) Again, a bright luminous flame may be made smoky by bringing a large piece of metal close to it, so as to conduct off the heat energy and thus lower the temj>erature. The cracking of a tea cup or tumbler when hot water is poured into it is due to the sudden expansion of the inner surface before the outer one has time to be affected; this may often be prevented by putting a silver spoon (not a plated one) in the cup or tumbler and pouring in the water along it ; the silver is such a good conductor that it prevents the temperature from rising too high at once. Conduction in a Gas. \u2014 The process of conduction in a gas is evidently simply the redistribution of the kinetic energy of the particles. When the temperature is high at one point in the gas, the kinetic energy there is great; and SO, owiiiLT t() the increased velocity of these particles, this energy is communicated to the neighboring ones. On the assumption that a gas behaves like a set of elastic spin we can deduce a value for the conduct i\\ it \\ in terms of the mean free path, etc. (See page 202.) The conduetU ;i few bodies at 0\u00b0 C. are given in the following tahle, in which the heat unit is a calorie and the C.G.S. system is \\\\\u00ab-<\\: AMES'S PHYSICS \u2014 10 HEAT Silver 1.096 Copper 0.82 Aluminium 0.34 Zinc 0.307 Iron 0.16 Mercury 0.0148 Water 0.0012 Radiation Radiation as a Wave Motion. \u2014 When one's hand is ex- posed to sunlight, a sensation of hotness is perceived ; simi- larly, if a body is brought near a flame, \u2014 even when not above it, \u2014 its temperature rises, or if brought near a block of ice, its temperature falls. There is neither convection nor conduction", " involved in these changes of temperature, yet heat energy is being gained or lost. The process is called \"radiation.\" Boyle noted as early as the seventeenth century that it went on through a vacuum, and this fact is proved also by the heating action of the sun which we observe here on the earth. When we discuss, in Chapters XVI et seq., the phenomena of waves and show how wave motions may be detected, it will be proved that this process of radia- tion consists in the motion of waves in the ether, i.e. in the medium which occupies space when ordinary matter is removed and which permeates ordinary matter as water does a sponge or as air does the stream of motes revealed by a beam of sunlight entering a darkened room. Without going into details in regard to waves, several facts may be men- tioned which are familiar to every one from observations of waves on lakes or the ocean, or of waves along a rope. One is that in wave-motion we do not have the advance of matter, but the propagation of a certain disturbance or condition; each particle of matter makes oscillations about its centre of equilibrium, but does not move away from this as the wave itself advances ; and therefore by the \" velocity of waves \" is meant the distance this disturbance advances in a unit of CONVECTION, CONDUCTION, AND RADIATION 291 time. (Consider the waves produced in a long, stretched rope when one end is shaken sidewise.) Thus, in order to produce waves, there must be some centre of disturbance or vibration, and this centre is giving out energy, for it is evi- dent that a medium through which waves are passing has both kinetic and potential energy. Thus, as waves advance into a medium, energy is carried forward, owing to the action of the particles of the medium on each other. We say, then, that \"waves carry energy,\" although of course this energy is associated with the material particles. If wave motion ceases gradually as the waves enter a different medium (e.g. if a stretched rope is so arranged as to pass through some viscous liquid, waves sent along it will cease when they enter the liquid), this medium is said to \"absorb\" the waves; it gains the energy which the waves carry. Further, waves t different lengths, depending upon the nature of the dis- t iii-bin^ vibration ; to produce short waves, that is, waves in which the distance from crest to crest is short, requires very rapid vibrations; while long waves are due to slow vibra- tions. We know, too, that waves suffer reflection, as is seen when water waves strike a large pier with a solid wall. If several waves are passing through the same medium at the same time, the resulting motion is the geometrical sum of idividnal waves. Production of Radiation. \u2014 When ether waves are discussed, it will be shown that they have a velocity of 8 x 1010 cm. per second, or about 187,000 mi. per second, and that they are known to have lengths varying from -m-fc^ cm. up to man\\ kilometres, depending upon the frequency of the vibrating centre where they are produced. It will be shown presently that all portions of matter, whatever their temper- ature, are producing spontaneously waves in the ether, whose lengths are so small as to be comparable with the size <>f HIM], The exact mechanism of this is not kno\\Mi : but it is clear that there must be some mechanical connection 292 HEAT between the ether and the minute particles of matter, and that these last must be making exceedingly rapid vibrations. If the wave length is called I and the velocity of the waves v, the number of vibrations in a unit of time is -, because L during each vibration the waves advance a distance I ; and so, if there are n vibrations in a unit of time, the waves advance a distance nl in that time, or v = ril. If, then, there are waves in the ether whose length is y^ol7 cm'> the num- ber of vibrations per second is 3 x 1010 x 104 or 3 x 1014, i.e. 300 trillions. Consequently these vibrating particles are thought to be the infinitesimal parts of a molecule. Our conception, then, of the structure of ordinary matter is as follows : it consists of molecules which are moving to and fro, vibrating about centres of equilibrium in solids, or traveling from point to point in a fluid, and at the same time the parts of the molecule are vibrating and producing the waves in the ether \u2014 this is similar to the case of a man moving a ringing bell, for the bell moves as a whole and its parts are making vibrations. Ether waves are also produced during certain electrical changes, as will be shown later ; and they are short if the body experiencing this change is small, but long if the body is large. Measurement of Radiation. \u2014", " In order to study the nature of these waves in the ether and the connection between them and the material bodies which produce them, it is necessary to have some instrument which will detect their presence and measure the energy they carry. To do this some body must be found which absorbs the waves, for then some change which can be observed will be produced in it, depending upon its own properties and the length of the waves. Thus, if the waves are short, they may produce vibrations in the parti- cles of the molecules in accordance with a simple mechanical principle known as that of \"resonance.\" This is illustrated by a boy setting in motion another who sits in a swing ; this CONVECTION, <<).\\1>I < 770.V, AND RADIATION 203 has a natural period of vibration, and may receive a large amplitude if a series of pushes are given each time the swing passes through its lowest point, going in the same direction ; hut if the pushes are given at irregular intervals, one may neutralize another ; so the force applied must have the same period as the natural period of the swing. Similarly, if the waves of a certain period enter a material body, the particles <>t' whose molecules have a natural period the saint- as this, they will be set in vibration by the waves, and will therefore absorb them, gaining their energy. (If the waves have a much longer period, they may produce electrical changes in the body.) If the energy of the waves is absorbed by the particles of the molecules, further changes will occur, deter- mined by the nature of the molecules and its parts. Thus certain ether waves falling upon the eye produce changes which result in vision; again, if certain ether waves are absorbed by photographic films, molecular changes go on which may be detected later; but in general the energy gained by the particles of the molecules is diiVused among the molecules themselves, and is manifested by the appearance of heat effects. To measure the energy of the radiation, the absorbing instrument should absorb all the energy, and the corresponding change produced should be one which is in direct proportion to the energy added to the body. It is found (sec hfr-hiw) that polished bodies like bright metallic surfaces absorb very little energy, while i-on^li, blackened ones absorb nearly all the incident radiation, if the waves are not to\u00ab> long. So ordinary radiation produced in the \u2022 \u2022ili.T by material bodies is detected and measured by instru- m.-iits which are sensitive to heat changes, and whose sur- faces are covered with a layer of lampblack, or copper oxide or platinum black (finely divided platinum). It is worth while to describe a few of thrs,. indium, nts briefly. An \u2022 \u2022nlinary mercury ili.-riii..ni,.trr \\\\ ith its luilb l.larkmr.l has been used; i more sensitive instrument is a platinum ivsM;in<-\u00ab- thermometer, HEAT with its strip of platinum blackened; this is called a \"bolometer,\" and was invented by Professor Langley of the Smithsonian Institution. Another instrument is the thermocouple ; in some cases the electric cur- rent produced, as the temperature of one junction is raised, is measured by a galvanometer, while in other instruments, called \" radio-microm- eters,\" and invented by Professor Boys, the wire in which there is the junction, and which carries the current, is suspended between the poles of a magnet in such a manner that it rotates when there is a current in the wire. The most sensitive of all instruments, probably, is the \" radiometer,\" which is a modification of the instrument invented by Sir William Crookes and described on page 204 ; these alterations are in the main due to Professor E. F. Nichols, now of Columbia University. In its present form this instrument consists essentially of an exhausted glass bulb closed at one point by a window of fluorite, \u2014 which is particularly transparent to ether radiations of all lengths, while glass is not, \u2014 and containing a fine vertical quartz fibre which carries a horizontal arm; to each end of this is attached a thin piece of mica, polished on one side and blackened on the other. The blackened face of one mica disk comes opposite and parallel to the fluorite window; so, if radiation enters this, it falls upon the blackened face of the mica, whose temperature therefore rises and which then moves backward. This motion twists the quartz fibre; and when the torsional moment of reaction of the fibre equals the moment due to the \"repulsion\" of the blackened disk occasioned by its rise in temperature, everything comes to rest. The angle of deflection of the horizontal arm measures, then, the intensity of the radiation. Another class of instrument must also be used in order to describe radiation from any source; this is one which ana- lyzes it, and", " so distributes it that waves having different wave lengths proceed in different directions and may be studied separately. This process is called \" dispersion \" and is illus- trated by the action of a glass prism on the light from a lamp. Radiation Spectra; Energy Curves. \u2014 Using these instru- ments, certain facts have been established. All material bodies in the universe, so far as we know, are producing waves in the ether. Solid and liquid bodies emit waves of all wave lengths between certain limits, whereas gases emit trains of waves of definite wave lengths. The emission of a solid or liquid depends largely upon the condition of its < \u00bb\\VECT1ON, CONDUCTION, AND RADIATION 295 surface, other things being the same. A polished metallic sin face emits very little radiation, i.e. the energy of the radia- tion is small; whereas, rough or blackened surfaces emit a great deal. ( This is the reason why stoves, steam pipes, etc. are blackened.) Again, the amount of the radiation from a body depends largely upon its temperature. As this is raised, the energy carried by each train of waves of a definite wave length increases; but this increase is greater for the short than for the long waves. This fact can be represented by a graphical method. Let two axes be drawn, distances SHORT WAVE-LENGTHS LONG WAVE-LENGTHS Fio. 181. \u2014 Radiation from blackened copj.. r. along the hori/.ontal one to represent the wave lengths of the eonipMnent \\\\a\\cs, vertieal distances to represent quantities \"f energy carried by the individual trains of waves. Several eurves are given of the radiation from blackened copper at different temperatures. It is seen that these curves are in \u2022!-d with tin- statements made above. If we consider any individual wave length at the extreme ends of the curves, it marks evidently tin* limiting power of the measuring instru- ment used ; ;md therefore waves whose energy at one tem- perature of the l)od\\ is so small that they cannot be detected may be so intense at a higher temperature as to permit of 296 HEAT observation. For instance, the longest waves which affect the human eye in such a manner as to produce the sensation of light are those that cause the sensation we call red; therefore, if the temperature of a solid body, e.g. a piece of iron, is raised, the experiment being performed in a darkened room, the solid is invisible (except for stray re- flected light) until the temperature becomes so high that the energy of the waves whose length corresponds to \" red light \" is sufficiently intense to affect our eyes. (Actually, the fact must be taken into account that the human eye is not sensitive to all colors alike, and that if the light of any color is feeble, the eye perceives \" gray.\") The body is now \" red hot \" ; and as the temperature rises still higher, its color changes continually, and finally it appears white and is said to be \"white hot.\" Laws of Radiation. \u2014 Careful observations upon the radia- tion of a blackened body have shown a most intimate connection between the total quantity of energy emitted and the temperature. If Q is the energy emitted at a given temperature, and T the absolute temperature, i.e. T=t\u00b0C. +273, Q=cT^ in which c is a factor of propor- tionality. This statement is called \" Stefan's law,\" having been first proposed by him. Again, there is a connection observed between the temperature of a blackened body and the wave length of the train of waves which carries more energy than any other train, i.e. the wave length which corresponds to the highest point of an energy curve for that temperature, as shown on page 295. This relation is due to Weber ; and calling T the absolute temperature of the body and Lm the wave length just defined, TLm = \u00ab, where a is a constant. These two laws, which have been verified over wide ranges of temperature by most painstaking investigations, offer con- venient means of obtaining the temperature of bodies when they are so hot as to render it inconvenient to use ordinary CO.\\T7-:<T/M.\\. CQNDUi rfOJT, -i.v/; HAhiATioy 297 means. If we assume that the laws are true for temperatures which are higher than those for which they have been veri- fied, we may assign them numbers. (In this way, making the above assumption, the temperature of the sun is observed to be about 5700\u00b0 C.) Pre vest's Law of Exchanges. \u2014 It is thus seen that the radiation of any body is independent of other neighboring bodies, because it depends upon the vibrations of its own minute particles. So, if two bodies are associated", " in such a manner that one receives the radiation of the other, each radiates independently; and the temperature of either one will fall if it radiates more energy than it absorbs. This principle of the complete independence of the radiation of two bodies was first stated by Prevost (1792) in his \"Law \u2022 \u2022I' Kxchanges,\" which is equivalent to the above. Newton's Law of Cooling. \u2014 If a body is surrounded by an inclosure at a lower temperature, it loses more heat energy than it receives; and, if the radiation from the two bodies is that which is characteristic of a blackened body, this net loss may be expressed at once. Let Tj and T.2 l>e the absolute temperatures of the body and the inclosure: then the heat energy lost by the former diminished by that received from tin- inclosure is c(Tf \u2014 TJ). This equals and. if T{ is only slightly greater than 7!r it may be written f\\\\Tl \u2014 7!,). So the net loss in heat energy varies as the ditl ere nee in temperature between the body and the inel(,siire. This is called \"Newt\u00abm's law of cooling.\" and it is trin- of other bodies than \"black\" ones for small differ- ences in temperature. Absorption Reflection and Absorption. When radiation falls upon a l\"\u00abdy, SOUL- i> tboorbed, BOUM i* n-lle.-trd. and some is trans- mit ted. A body which allows waves of a certain wave length 298 HEAT to pass through it is said to be \"transparent\" to them/ but no body is perfectly transparent to any waves ; if it is suffi- ciently thick, it will absorb them. In a thin layer, however, a body may absorb certain waves completely and may trans- mit others comparatively freely. Thus, ordinary glass permits those waves to pass which affect our sense of sight, but either absorbs or reflects other waves which are shorter or longer. This is the explanation of the action of the glass roof of a greenhouse. The \" visible waves \" from the sun are transmitted through the glass and are then absorbed by the black earth or the green leaves. The tempera- ture of these is raised, \u2014 but not sufficiently to make them self-luminous, \u2014 and they radiate waves which are so long that they are reflected by the glass. Thus the energy which enters through the glass is trapped and stays inside; consequently the temperature is raised. A body which transmits comparatively freely those waves which carry the greatest amount of energy is called \" diathermous \" ; but the word is not often used, because of its indefinite character. If we wish to compare the properties of reflection and absorption, it is best to consider a body which is so thick as to transmit no waves. It is then at once evident that if a body absorbs well it must be a poor reflector, and conversely. Thus, a blackened surface absorbs well and reflects poorly ; while a polished metal reflects well and absorbs hardly at all. In order to secure what is called \"regular\" reflection, as from a mirror, not alone must the body be itself large in compari- son with the length of the waves, but its surface must be smooth to such an extent as to have no irregularities so large ; otherwise the different portions of 'the surface reflect the waves in different directions and so scatter them. Under the above conditions a body reflects at its surface waves of all lengths to a greater or less extent ; but in every case certain waves enter the body, although their intensity may be very small. Absorption. \u2014 Let us consider the process of absorption more closely. When ether waves fall upon a body, certain CONVECTION, CONDUCTION, AND RADIATION 299 partirK's in the molecules are set in more violent vibration bv ivsnnaiK-e, and thus the waves lose energy. la some bodies t IK-SI- vibrating particles emit waves immediately, without the temperature sensibly rising. This is the case with pieces of fluor spar, thin layers of kerosene oil, and with a few other bodies, as will be shown later under Fluores- cence and Phosphorescence. In other bodies the absorbed \u2022 \u2022ii'Tgy is distributed among the molecules and becomes ap- parent in heat effects. This absorption, where the energy goes into heat effects, is called \"body absorption.\" Many I xxlies absorb only waves of definite wave-length, and trans- mit others. 'I' bey are said to have \"selective\" absorption. Metals and substances that have strong selective absorp- tion reflect certain waves more intensely than others ; they are said to have \"selective\" reflection. Thus, the reason why gold appears yellow to our eyes is because, when viewed in ordinary white", " light, besides the waves that are reflected at the surface and that would make the gold appear white, thriv art- certain waves, of such a wave length as to produce the sensation of yellow, that are reflected more intensely than the others. Those waves which enter the gold are absorbed in the surface layer of molecules and produce heat effects. It is evident, then, that if white light is reflected again and again from a series of gold surfaces, in the end the only waves which will leave the last surface will be those which produce in our eyes the sensation of yellow. The waves which leave the last surface after a great number of reflections from the same material are called the \" residual \" ones. It one looks at a bundle of nonll.- in white light, the points being t ii mod toward the eye, they appear black, because the waves are reflected to and fro from needle to needle, but are continually getting weaker and weaker and being deflected down the needle ; thus no waves come back to the eye, and the points appear black. Finely divided silver and plati- num appear black for the same reason. Connection between Radiating and Absorbing Powers. \u2014 abruption i> due to resonance, it is simply a restate- oOO HEAT ment of this to say that a body absorbs to a marked extent waves of the same period as those which it has the power of emitting. But we can say more, if we consider the intensi- ties of the waves absorbed and emitted, and if we assume that there are no chemical or other molecular changes in the body. This excludes fluorescence, phosphorescence, etc. If several bodies at different temperatures could be inclosed inside a vessel which absolutely prevents any heat energy from entering or leaving, and which keeps a constant volume (so that no external work is done), there is every reason for believing that equilibrium would finally be reached, but not until the temperature of all the bodies inside was the same as that of the walls of the vessel. When this is the case, each body must be absorbing and turning into heat effects as much energy as it emits, provided there are no chemical or other energy changes, otherwise its temperature would change. That is, the absorbing power of a certain body at a definite temperature exactly equals its emissive power at that same temperature; where by absorbing power we refer to body absorption. (In general language, a body which absorbs well, in the sense of transforming radiant energy into heat energy, radiates well; e.g. a blackened surface.) We can imagine, moreover, a body in the vessel described above, which is entirely inclosed by some envelope which allows to pass through it waves of only one wave length ; therefore, when equilibrium is reached, the body inside must radiate as much energy in the form of waves of this wave length as it absorbs. Consequently, the amount of energy of a definite wave length which a body emits at a given temperature equals exactly the amount of energy in the form of waves of this same wave length which it absorbs at that temperature. In other words, the absorptive power of a body at a certain tempera- ture equals both in quantity and quality its emissive power at that same temperature. If at ordinary temperatures a body appears black when viewed in white light, it is owing to the CONVECTION, COMJUCTIO.\\. AM) i;Al>IATK)\\ 301 fact that it absorbs those waves which affect our sense of : : and. if raised to such a temperature as will enable a body to emit such short waves, it will emit them and so shine iitly in a darkened room. Similarly, if a body appears ivd at ordinary temperatures when. viewed in white light, it is because it absorbs all waves except those which produce in our eyes the sensation of red ; these are either transmitted or are reflected out from the interior by some small foreign particles. (Thus, a colored liquid appears perfectly black j.t by transmitted light, if it is entirely free from small solid particles; but, if a minute quantity of dust is stirred in it. it appears colored when viewed from any direction.) Then, if such a red body is heated until its temperature is suiliciently high, it will emit all the waves except those which < -spend to the sensation of red, and so, if viewed in a dark room, will appear bluish green. This law connecting radiation and absorption was fir>t stated by Balfour Stewart, but was discovered inde- pendently by Kirchhoft'. The latter, however, in expressing it, did so in a more mathematical form. He took as a stand- ard of absorption a hypothetical body, which is called a \"perfectly black body\" and which is defined to be such a body as will absorb and turn completely into heat effects all radiations which fall upon it. (Any non-reflecting body, if sufficiently thick, is such a body.) We can approximate to such bodies experiment", "ally by usin<j lampblack or platinum black as the surface layer. (It is seen, too, that the radiation inside an\\ hollow inclosure, provided it is rou^h, is that which is characteristic \u00ab.f a perfectly black body at thai perat ure, because after a sufficient number of irregular reflec- 18 any train of waves will he totally absorbed, SO the walls ic inclosure finally produce the effect of a black body.) It will be shown later that radiation may be produced and 1.1 m. ins than by raising or altering the tem- perature of the radiating h..d\\ ; but this law of Stewart and 302 HEAT Kirchhoff refers only to radiation that is due to the same cause which conditions the temperature of bodies, and to absorption that results in heat effects. Atmospheric Absorption. \u2014 One important case of absorp- tion of radiation is that of the solar rays incident upon the earth. As these pass through the earth's atmosphere, a cer- tain percentage of their energy is reflected by the floating particles and drops in the air, and also by the molecules of the air themselves ; and so does not reach the earth. There is also true body absorption as the waves traverse the atmos- phere. The energy that is not reflected or absorbed reaches the earth and is there almost completely absorbed and spent in producing heat effects. The earth itself is therefore also radiating energy, and this again passes through the atmos- phere and is partially absorbed. If there are clouds, this radiation from the earth itself is absorbed by them, and they radiate a certain proportion back toward the earth. Our appreciation of the temperature of the air in which we live depends largely upon the quantity of heat energy absorbed by the air, not so much upon the radiation received by us directly from the sun. Thus we see the reason why the temperature is lower upon mountain tops where the air is rare and so absorbs poorly than at sea level where the air is denser and absorbs more. CHAPTER XV THERMODYNAMICS Nature of Heat Effects. -- Throughout the previous chap- ters we have assumed that heat effects are due to work done against the molecular forces of a body and that for a definite amount of energy received the same effect is produced regardless of the source of the energy. These assumptions are justified by countless experiments, some direet and some indirect. Thus Joule, in a series of in- vestigations beginning about 1843 and lasting over forty : s, caused heat effects to be produced in many different ways ; compression of gases, friction of various kinds, con- duction of an electric current through a wire, chemical re- actions, etc. The only possible explanation of the results of these experiments is the assumption that heat effects are due to energy being given the molecules of the body and are proportional to the amount received. Thus, if a certain amount of hoat energy received from friction is used to boil some water or to melt some ice or to raise the temperature rater, we ean. by allowing this water to cool through tins temperature range, obtain an amount of energy which will melt an equal amount of ice, etc. \u2014 it being rem- that heat energy passes from high to low temperature. Again, when the energy received from any two source heat \u2014 for instance, a candle and the sun \u2014 is compared, if under any condition they raise the temperature of the same amount of water through the same range of temperature, 11 melt the same amount of ice, or boil the same water, et< \u2022. Therefore the production of heat m 304 HEAT effects depends upon the energy received, not upon the tem- perature or condition of the source of the energy. Mechanical Equivalent of Heat. \u2014 As stated before, the practical unit of heat energy is that required to raise the temperature of one gram of water from 15\u00b0 to 16\u00b0 C. The value of this in ergs, or the \" Mechanical Equivalent of Heat,\" has been determined by different observers. For many years it was thought that the specific heat of water was the same at all temperatures ; and in the early work no distinction was made between the amounts of energy required to raise the temperature of water between different degrees. Robert Mayer calculated the mechanical equivalent from experimental data secured by other observers on the heat energy required to raise the temperature of a gas at constant volume and at constant pressure. We have shown that, on the assumption of no internal forces, if J is the \" mechanical equivalent,\" J (<7p \u2014 \u2022#\u201e) = R for a gas ; and hence knowing 0^ Cv, and R for a gas, J may be calculated. Joule, and afterwards Rowland, used the method of turn- ing a paddle rapidly in water, and measuring the mechanical work, the quantity of water, and the rise in temperature. The mechanical work was measured by a simple dynamom- eter method (see page 121) ; and in Rowland's", " Carnot's engine; this has the great merit of being independent of the sub- stance used in the thermometer (or engine). Thomson's system of \" absolute \" thermometry, as it is therefore called, is equivalent to defining the ratio of the temperatures of two bodies as equal to the ratio of the quantities of heat energy received and given out by a Carnot engine working between these two temperatures. Thus, if in the above description of an ideal Carnot's engine the ratio of the tem- peratures on Thomson's scale of the hot and cold reservoirs T *i is written -=\u00b1t T H /2 //, 2 rI = 771' Hence the efficiency is 1 \u2014?. Since it is impossible to T \u2014 T have an efficiency greater than unity, there is a minimum value of trmpfi-ature, that for which T2 = 0 ; for, if T2 had a negative value, the efficiency would be greater than unity. This minimum temperature is called \"absolute zero.\" Thomson's definition of absolute temperature does not specify the \" size \" of a degree, but simply the ratio of two temperatures : we can choose the degree to suit ourselves. Let us agree to use the Centigrade scale, so that if T is the temperature of melting ice, T + 100 is that of boiling water. By a series of most ingenious experiments Thomson showed that this system of temperature agrees most approximately with that which we have been using, namely, that of a con- s tai it-pressure hydrogen thermometer on the Centigrade scale, if \\ve add to each temperature reading the reciprocal of the coefficient of expansion of the gas, i.e. 273 approximately. (This is what we have called on page 240 \"absolute gas temperature.\") Thus, if large quantities of boiling water and in\u00bb -It ing ice could be used as the reservoirs between which a Carnot engine worked, the quantities of heat energy received and given out, ITj and Hv would have such values 308 11KAT TT that their ratio ~~i almost exactly equaled. It can be x/2 2*1 & proved that, if the gas had no internal forces and obeyed Boyle's law exactly, the agreement between Thomson's ab- solute system of thermometry and that of a gas thermometer as above described would be perfect. Therefore, in the formula for a gas, pv = RmT, T is very nearly equal to Thomson's absolute temperature. The slight differences be- tween this system and that we have been using were deter- mined by Thomson and Joule ; and in all exact work in Heat this correction is made to the gas temperatures. There are several ways of determining the absolute tem- perature of a body and of calculating the value of absolute zero on the system of thermometry ordinarily used. The accepted value is -273\u00b0. 10 C. Historical Sketch of Heat Phenomena It was believed by the Greek philosophers that all the phenomena of a material body which we associate with the word \" heat,\" such as expansion, change in temperature, boiling, etc., were due to the addition to the body of a sub- stance; but to Newton and his immediate predecessors and associates it seemed clear that in some way they were due to motion of the parts of the body. Thus Boyle gave a correct explanation of the heat effects observed when a hammer strikes a nail. The materialistic theory of heat, however, was again proposed, and prevailed for nearly two hundred years. Its great defenders were Gassendi (1592- 1655), Euler (1707-1783), and Black (1728-1799). Even as late as 1856, in the eighth edition of the Encyclopaedia Britannica, this theory is offered as the accepted explana- tion of heat phenomena. One great reason why this theory was so universally accepted was because it was so analogous to the accepted explanation of combustion. Stahl (1660- 1734), who was professor at Halle, advanced the theory that Til Kit Unit VXAMICS 309 \u2022luring the process of combustion a material substance, called \"phlogiston,\" was given off; and this idea persisted until the \\\\ork of Lavoisier, about 1800, and even later. Fan- tastic theories in regard to the properties of phlogiston and of the substance heat (or \"caloric\") were of necessity brought forward in order to account for the observed facts. Joseph Black showed that when two bodies at different temperatures were brought together he could speak of '\u2022quantities of heat\" leaving or entering the bodies; and we owe to him our ideas and methods in regard to specific heats. Black considered two kinds of effects when \" heat \" was added to a body : if the temperature was raised, the heat was called \" sensible,\" and it was supposed to be free in the body; but if the temperature did not change, the heat was said to be \"latent,\" and it was supposed to form some kind of a", " compound with the molecules of the body changing their state. But the experiments of Rumford, in 1798, and of Davy, in 1799, convinced nearly every one that heat effects in a body were due primarily to the transmission of motion to its minute parts. Rum ford showed that in such a process as that of boring out a brass cannon, \" quantities of heat \" could be produced, limited only by the amount of work done. Similarly, Davy arranged an apparatus which caused one block of ice to rub violently against another, and showed that the quantity of ice melted varied directly with the work done. Tin- first one, however, to express clearly the belief that heat effects were due entirely to the addition of energy to tin- small parts of a body was Rolx-rt Mayer, in lsj-J. He followed hy Joule, in 1848, and later by I Irlmholt/., in 1M7. i;\\ the epoch-making research.- \u2022\u2022!' Joule, the prin- \u00ab -iplo of the conservation of energy \u2014 a phrase nf Rankine's -was soon extended so as to cover all heat phenomena. The fact that -radiation\" is a phenomenon due to wave motion in thr \u00ab th\u00ab i, of exactly the same nature as that win \u00bb-h 310 UEAT produces the sensation of light, has been established by a long line of investigators. William Herschel showed, in 1800, that there were rays in the solar spectrum invisible to the eye and yet having the power of affecting a ther- mometer. Herschel speaks of these rays as subject to the laws of reflection and refraction; and this fact was fully established by Melloni some thirty years later. It was proved in the following years that these rays could be dif- fracted, be made to interfere, be polarized, etc.; and that, in short, they were due to waves in the ether. BOOKS OF REFERENCE EDSER. Heat for Advanced Students. London. 1901. PRESTON. The Theory of Heat. London. 1894. An advanced text-book, containing extended descriptions of the experi- ments which form the basis of our knowledge of heat quantities. TYNDALL. Heat a Mode of Motion. London. 1887. In this the author describes a series of experiments illustrating the fact that heat effects are due to energy changes. BRACE. The Law of Radiation and Absorption. New York. 1901. This contains the original papers of Prevost, of Balfour Stewart, and of Kirchhoff. AMES. The Free Expansion of Gases. New York. 1898. This contains the papers of Gay-Lussac and Joule on the expansion of gases through small openings. RANDALL. The Expansion of Gases by Heat. New York. 1902. TAIT. Heat. London. 1884. MAXWELL. Theory of Heat. London. 1892. YOUMANS. The Correlation and Conservation of Forces. New York. 1876. This contains reprints of papers by Helmholtz, Mayer, and others on the Conservation of Energy. STEWART. The Conservation of Energy. New York. 1874. GRIFFITHS. Thermal Measurement of Energy. Cambridge. 1901. This gives a most interesting description of various methods for the measurement of heat energy. VIBRATIONS AND WAVES CHAPTER XVI WAVE MOTION General Description. One of the most important phe- nomena in nature, and one that is of most frequent occur- rence, is the transmission of energy from one point to another by what is called \" wave motion.\" This motion is illustrated in many ways: if a stone is dropped in a pond, waves are produced on its surface, which do work on any movable object which they meet; a vibrating bell produces waves in the air, which do work in a similar way, or by bending a suitably stretched membrane such as the drum of the human ear or the diaphragm of a telephone instrument ; if one end of a long, stretched rope is fastened to a movable object and if the other is given a sudden side wise or lentil i\\\\ ise motion, a disturbance will pass along the rope and will do work on the object; etc. In all these cases it is evident that there is a vibrating centre which produces motions in those portions of the surrounding medium immediately in contact with it; portions affect those next them, etc. The vibrations at th< centre of disturbance must be of such a nature as to produce in tin- medium a displacement or change which can be propagated l\u00bb\\ it. Thus, if the hand is moved through air or \\\\at.-r, <.r if a pendulum vibrate* in air or water, waves ai\u00ab not pn.duerd, because the fluid flows around the obstacle as it moves through it. and is", " not Q pressed; in order to Loot waves, the vibration must be so rapid or the motion so sudden that the fluid,t have time to flow, and is 312 VIBRATIONS AND WAVES therefore compressed on one side and expanded on the other. Again, any sidewise motion in a fluid of an object like a thin board, however sudden, would not produce waves, because in a fluid there is no elastic force of restitution when one layer is moved over another. In order that a medium should carry waves, there must be forces of restitution called into action when its parts are displaced ; for these forces are due to the action of the neighboring parts on each other ; and owing to the reaction of the displaced parts on those in contact with them the latter are displaced also, and so waves are produced and propagated. If these conditions as to the medium and the centre of disturbance are satisfied, and if the motion of this centre is a vibration, or a series of vibrations, the various portions of the surrounding medium will in turn be set in vibration. If the motion of the vibrating centre ceases, that in the medium will persist until forces of friction (or other causes) bring it to rest. In all these cases it is clear that the particles of the medium vibrate, but do not advance with the waves ; a certain \" condition \" moves out from the centre. \" Water Waves \" and Elastic Waves. \u2014 There are several kinds of forces of restitution that enable a medium to propagate waves. If the surface of a pond is disturbed at one point by the motion up and down of a stick dipping in the water, waves spread out, owing to the fact that the force of gravity tends to maintain a level surface. Again, if a vertical cord, like a fishing line, is moved sidewise through the surface of a pond (or, if a quietly flowing river flows past a stationary vertical cord), short waves called \" ripples \" may be observed on the side toward which the cord is mov- ing (or up stream in the other case). These waves are due to the force of restitution of surface tension. The waves just described occur on the surface of a liquid; but any portion of matter that is elastic can also carry waves through its interior. Thus, fluids can propagate com- pressional waves, that is, waves produced by having a centre WAVE MOTION 313 of disturbance where the fluid is compressed or expanded : as is illustrated by a bell vibrating in air or under the surface of a lake. Fluids cannot, however, propagate a distortional disturbance. An elastic solid body can carry both com- pivssiunal and distortional waves, as is illustrated by a long, stivtehed wire. If this is disturbed at some point by a transverse vibration, or if it is twisted back and forward, tin-re will be distortional waves; if it is pulled to and fro longitudinally, in the direction of its own length, at some point, there will be compressional waves. (This fact is also illustrated by the disturbances that we call earthquakes; for these are due to some great disturbance in the interior of the earth that produces both kinds of waves in the body of the earth. As we shall soon see, these two types of waves in a solid travel with different velocities ; and this fact is ob- served in all earthquakes.) Compressional waves are often called \"longitudinal,\" and distortional ones, \"transverse,\" \u2022bvions reasons. Polarization. \u2014 One distinction between longitudinal and transverse waves, other than the one which gives rise to the names, is worth noting. A longitudinal wave, in whieh the particles of the medium move to and fro along the line of advance of the waves, appears the same to the eye from what- ever side this line is viewed. But, since in a transverse wave the particles are vibrating in planes that are at right angles to the direction of propagation, it will as a rule appear different when viewed from different directions. Thus, if a t raus verse t rain of waves is produced in a long, stretched rope by n loving one end up and down in a vertical line, the rope will at any instant have a sinuous form when viewed from ide, but will appear straight when viewed from above. In this case all the particles are vibrating in straight lines through which a plane can be drawn; and such a train of '1 to be \" linearly \" or - plane polarized.\" Simi- larly, if the end of the rope is moved rapidly in a circle or in 314 VIBRATIONS AND WAVES an ellipse, all the particles of the rope will in turn move in circles or ellipses ; and the waves are said to be \" circularly \" or \" elliptically polarized.\" Thus only transverse waves can be polarized ; and, conversely, if in any wave motion we can detect polarization phenomena, we know that the waves must be transverse. Int", "ensity. \u2014 In all classes of waves it is at once evident that we are dealing with the propagation of energy. For wherever there is wave motion the moving parts have kinetic energy, and the existence of the waves presupposes forces of restitution, so when the parts are displaced there is potential energy. Therefore, as the waves spread out from a centre of disturbance, the medium into which they advance gains energy. If the cause of the waves is a temporary disturb- ance, any portion of the medium gains energy when the waves reach it and loses it again when the motion ceases, owing to the waves passing on. Thus a portion of the medium simply transmits the energy. It gains none perma- nently unless there are forces of friction when the particles of the medium move relatively to each other as the wave passes. (Of course if there are foreign bodies immersed in the medium carrying the waves, they may be set in vibration and may continue to vibrate after the wave passes ; in which case this portion of space \u2014 not the medium itself \u2014 has an increased amount of energy in it afterward.) If at any point in a medium a plane having a unit area be imagined described at right angles to the direction in which the waves are prop- agated, the amount of energy transmitted through it in a unit time is called the \" intensity \" of the waves at that point. (If the waves are varying, the exact definition is as follows : if E is the energy transmitted in time t through an area A, the intensity is the limiting value of the ratio \u2014 -, as t and A are taken smaller and smaller.) Detection of Waves. \u2014 The effect of the waves is perceived in two ways. If the medium is limited in one direction by \\\\ ATE MOTION 315 e object, with which it is connected in such a mannei that the vibration of this object would produce waves in the medium, this will be set in motion by the waves unless it is restrained by mechanical means. Thus, waves in the air set in motion the drum of the ear or the diaphragm of a tele- phone receiver. If the object is surrounded by the medium, l)ii t cannot move bodily with the waves, it may be set in vibration by them, owing to \" resonance,\" a process whieh will l)e described in detail later. Thus, waves produced in tin- air by one tuning fork may set in vibration another one, if the periods of vibration of the two are identical ; waves in the ether may set in vibration particles of matter, as described in the chapter on Radiation. Other Kinds of Waves. \u2014 So far we have spoken of mechan- ical waves only, that is, waves in matt-rial media or in the ether, in which the disturbances are displacements of par- ti, hs. lint we can have many other kinds of waves, de- pending upon the property of the medium that is varied. Tli us, if the temperature of one end of a metal rod is first d gradually, then lowered, raised again, etc., we have a vibration of temperature; and, if we observe the tempera- tun; of any point in the rod not too far away from this end, we shall find that its temperature also rises and falls. Since it takes time for the conduction of heat, the temperatures at rent points of the rod will not have their maximum values at the same time, so there is a wave of temperature in the rod. This is illustrated in the daily heating and cool- ing of the earth's surface as it is tunu-d t.. \\\\ard and a\\\\.i\\ ill. sun, also by the seasonal heating and cooling owiiiLr to th\u00ab- revolution of the earth around the sun. There are temperature waves, then, going down into the earth for a short distance. The daily wave is appreciable for a depth of about 2 or 3 ft. ; the annual one for about 50 i|Hrarup <>t th. * M it }i'-cru8t increases gradually but OOD- tn.'i..-. :\\- with the depth; the rate of increase varies greatly with the VIBRATION H AND WAVES geological conditions, but is on the average about 1\u00b0 C. for a depth of 28 m. This condition requires that heat energy should be continually flowing from the interior of the earth to the surface. From considera- tions based on this fact it is possible to make an estimate of the \" age of the earth \" ; that is, the interval of time since the earth was in a liquid condition. This is probably about 50,000,000 years. Similarly, if one end of a long electrical conductor, e.g. an ocean cable or a telephone wire, has its end suddenly joined to an electric batter}^ the effect is gradually felt along the conductor ; and, if the electric battery at the end is varied, there is an \" electric wave \" in the conductor", ". Again, if a metallic body is charged electrically, there are electric forces, so called, at points in space near by ; so, if this electric charge is varied, these forces will vary ; and, as they change, variations are produced at neighboring points. Therefore, when the electric charge on a body varies, that is, when there are \"electric oscillations,\" waves of electric force are produced in the surrounding medium. The medium which carries these waves has been proved to be the same as that which carries the waves that affect our sense of sight ; namely, the \"luminiferous ether.\" It has been proved, too, that, wherever there are variations in the electric force, there are also variations in the magnetic force ; so these waves are called \" electro-magnetic.\" (It should be borne in mind that if we look upon matter as the fundamental concept in nature, then as soon as we are able to explain electric and magnetic forces as in some way due to the motion of matter, we shall be able to describe electro-magnetic waves in the ether as dis- placements of material portions of the ether. But, if an elec- tric charge is the fundamental concept, as soon as we can explain the properties of matter as due to the motion of charges, we shall be able to describe all waves in matter in terms of electric forces.) These electro-magnetic waves may be detected by suitable means, as is shown by the various systems of \"wireless telegraphy.\" CHAPTER XVII HARMONIC AND COMPLEX VIBRATIONS \\Vi: shall now proceed to discuss in detail the two funda- mental features of waves : liist. the properties of the centre of disturbance; second, those of the medium through which the waves pass. The effects produced when waves in the air or in the ether are perceived by our senses of hearing or of vision will be considered later in the sections devoted to Sound and to Light. The Kinematics of Vibrations Simple Harmonic Vibration. \u2014 A disturbance may be peri- odic <>r not ; that is, it may after a definite period of time called the \"period\" be repeated identically, and again at the end of another period, etc.; or it may be irregular. Thus, the vibrations of a pendulum, of a tuning fork, of a violin string. \u00ab-tc., are periodic; while the motion of a piece of tin as it is \" crackled,\" of two stones when struck together, or of one's hand ; s it is moved to and fro at random, are not peri- The simplest case of periodic motion is that which is called \"simple harmonic.\" and which is discussed on page 48. The period of this has been defined above; and the numher \u00ab.f \\il.rati..us in a unit of time is called the u fre- quency\"; this is, of course, the reciprocal of the period, nplitude \" has been defined aa one half tin- length of the suing, or as the value of the maximum displacement. \u2022 harmonic motions having the same period and ampli- tude may yet differ in \"phase\"; that is. t h<- infant s at which they pass through their on ;ins ma\\ lie different. 317 318 VIBRATIONS AND WAVES Composition of Harmonic Motions. \u2014 1. In the Same Direc- tion. If a point is subjected to two harmonic motions, the resulting motion may be found by compounding the dis- placements geometrically at consecutive instants of time, or by simple algebraic processes. Thus, if the two harmonic motions have the same period and are in the same direc- tion, they may be represented by xl = A1 cos (nt \u2014 ax) and #2 = A2 cos (nt \u2014 a%) (see page 51) ; and the resultant mo- tion is x = xl + 2r2 = Al cos (nt \u2014 ctj) -f A2 cos (nt \u2014 a2). By ordinary trigonometrical formulae this takes the form x = A cos (nt \u2014 a), where A2 = Af + A22 + 2 AlAl cos (a, - a2), A, sin a, + A \u201e sin a9 and tan a = -^ \u2014 \u2022 A l cos ax + A 2 cos a2 This shows that the resulting motion has the same period as that of its two components, but a different amplitude and phase. Graphical Methods. \u2014 If the two component harmonic motions are in the same direction but have different periods, the algebraic formulae are much more complicated, but their resultant may always be found by a simple graphical method. FIG. 133. \u2014 Graphical representation of harmonic motion. Any harmonic motion may be represented by a curve drawn on a diagram whose axes are intervals of time and displacements. Thus,", " the latter is at Q1 ; and the geometrical sum is given by S1 ; etc. The resulting vibration is evidently a circle. FIG. 139. \u2014 Periods equal ; difference in phase 0, -,-,\u2014, IT. If the difference of phase is one eighth of a period, i.e. j, the resulting vibration is an ellipse. The curves are shown for a number of different differences in phase. If the amplitudes of the two vibrations are not the same, the geometrical methods are exactly similar. Lay off two lines, AB and CD, perpendicular to each other at their middle points; divide them into lengths that corre- spond to equal intervals of time ; at these points draw lines parallel to AB and CD. If the phase of vibra- tion is the same, the result- ing motion is in a straight line. If the difference in phase is one eighth of a FIG. 140. \u2014 Periods equal; amplitudes unequal; period, the vibration is in difference in phase one eighth of a period. an ellipse, as shown, etc. The simplest way, however, of compounding the vibrations is to eliminate t from the equations, and plot the resulting equation. Different curves will be obtained by giving a different values. The curves of Fig. 139 may be obtained by several physical processes. One is to use, as described on page 319, a pendulum AMt <-<>Mi>LEX VIBRATIONS 826 that can trace a path on a piece of glass ; but in this case the kept stationary and the pendulum is set swinging, not in a plane through its origin, but in a cone, as a result of a sidewise push given it when it is held out at the end of its swing. Another method, due to Lissajous, is to use two large tuning forks whose frequencies are the same, and to phice them, with their vibra- tion planes perpendicular to each other, in such a position that a pencil of light from a small source incident on the end of a prong of one fork is reflected to the end of a prong of the other and thence to a screen, or into the eye of the ver. This arrangement is.shown in the cut. If only one fork is vibrating, a straight line is seen; but if this fork is quiet and the other is vibrating, another straight line at right angles to the first is seen. These lines are caused by the rapid harmonic motions of the two forks. It now both forks are set vibrating, the path of li^ht seen is an ellipse. It' the forks are started again at NT I,.: Kio. 148. - LlM\u00abJou.' \u00abrrmn\u00ab\u00ab.ment of two tunlnjr millC tO rest, the shape of the ellipse will be different. in general, owing to the fact that their difference in phase i> not the same.1- 1.. This is on the assumption 326 VIBRATIONS AND WAVES that the frequencies of the two forks are exactly equal ; if they are not, the shape of the ellipse will change as one looks at it, showing that the difference of phase between the vibrations has changed. The reason for this is seen at once if one considers the two equations for the forks. If their periods are not quite the same, these may be written x = y = A2 cos (n^ - a), where n1 = n + bt and b is a small quantity. Therefore sub- stituting for Wj its value, y = A2 cos [n* \u2014 (a \u2014 &\u00a3)]. Comparing this with the equation for #, the difference in phase is seen to be a \u2014 bt\\ and this is different for different Periods in the ratio 1 : 2. Periods In the ratio 8 : 4. FIG. 143. \u2014 Lissajous' figures. HARMONIC AND COMPLEX VIBRATIONS values of the time. In words, if one vibration has a period slightly less than that of the other, it will gain on the latter, thus passing through various differences in phase; it will finally gain a whole vibration, when the cycle of changes will repeat itself. If the periods agree exactly, the par- ticular kind of ellipse seen will be a matter of chance, de- pending upon the manner of starting the forks, but it will persist until the motions of the forks die down owing to friction and to loss of energy caused by the production of waves. Lissajous' Figures. \u2014 Similar statements may be made in regard to compounding two harmonic vibrations of different periods, whose directions are at ri'_rht angles to each other. The curves obtained when the iiiencies bear certain simple relations to each other are given in the accompanying cuts. different curves in any Fl\u00b0- 144. -Geometrical method of com-", " i t- n> pounding two harmonic vibrations whoM out correspond to different ^rtot* \u00ab* in the ratio 8:4, who*. \u00bbm,.n todes are unequal, and whote phiiM are.litl.r.nt. olifferences in phase; and if, using Lissajous' method, the frequencies are not exactly in the proper ratio, one curve will gradually change into another. The number of cycles of changes in a unit of time depends, of course, upon the <li\\ \u2022\u2022]\u2022-_;,. nee of the frequencies \u00ab>f the |'..rkx t'rom, just- ment ; ami. if tliis numher is counted, the general si be curve noted, and the frequency of one fork km>\\\\n. tl-.it of the other may be deduced at once by a simple formula. NOTE. \u2014 Altli Hurli t hf se curves are generally called Lisaajous* figures, were first drawn and described by Nathaniel Bowditch of Salem, 1815, forty years before Lissajous did the same. 328 VIBRATIONS AND WAVES Energy of Vibrations Damped Vibrations; Dynamics of Vibrations. \u2014 If any actual harmonic motion is observed, it is seen that its ampli- tude slowly decreases; it is said to be \"damped.\" If this decrease is very gradual, there is no change in the period ; but if it is rapid, as for in- stance if a pendu- F.G. 145. - A damped vibration. lum has a plCCC of paper fastened to it, the period is sensibly increased. This decrease in ampli- tude is due to loss of energy by the vibrating body, generally by friction as it moves through the air or at the pivot. This is evident if we calculate the energy of the vibrating particle. If its motion is given by x = A cos (nt \u2014 a), its speed at any instant is 8 = An sin (nt \u2014 a) (see page 51) ; and so if its mass is m, its kinetic energy at this instant is \u00a3 mAW sin2 (nt \u2014 a). This varies at different instants of the vibration, but is always proportional to A2. (The mean value over one period may be proved by the infinitesimal cal- culus to be %mAW.) During the motion, as fast as kinetic energy is gained, potential energy is lost ; and so the mean total energy during' a vibration is twice the mean kinetic energy ; and this, from what has just been said, varies directly as the square of the amplitude. Therefore as the amplitude of vibration decreases, the particle loses energy. Forced Vibrations; Resonance. \u2014 Even though a body which can make vibrations like a simple pendulum has a definite period of its own, it may be given a different period, if it is attached to some other vibrating system ; thus, if the point of support of a pendulum is moved to and fro by a hand making harmonic motion, the motion of the pendulum is HARMONIC AND COMTLKX VIBRATIONS 329 i In.' resultant of two, one due to the force applied by the hand, the other its own natural motion. If this last is greatly damped by attaching a sheet of paper to the pendulum, it soon dies down ; and the iinal motion of the pendulum is that dui' to the harmonic force of the hand. This motion has the same period as that of the harmonic force; and is called a \u2022\u2022forced vibration,\" to distinguish it from the natural free vibration. The amplitude of this forced vibration depends to a great extent upon how closely the period of the force agrees with that of the natural vibration ; if they are exactly equal, the amplitude is very large. This condition is called Mmance,\" and is illustrated in many ways. The case of a child in a swing being set in motion by a series of pushes given at intervals agreeing exactly with the natural period of the swing has been mentioned already. In a similar manner a heavy church bell may be set swinging. If a tun- ing fork is vibrating near another one of the same frequency, tin- latter will be set in vihr.it ion. Many other simple mechanical cases of resonance are given in Rowland's Physi- cal Papers, page 28. If the ju-riod of the force varies slightly from the natural period of the vibrating body, the amplitude is not so great as when there is resonance ; and in most cases one can tell with considerable accuracy when the resonance is exact. The fad that a harmonic force produces in a system whose own vibrations are greatly damped a vibration whose period is the same as its own is of great importance. If the force is periodic, but complex, each of the component harmonic forces produces a corresponding harmonic vibration", " having its period ; but the phases and amplitudes of these component vibrations bear relations to each other that are not the same as for the component forces. Then-Tore, die resultant complex \\ibration is different from the complex force in \"form.\" Il Is Only a harmonic force that can - reproduce \". of OOUlVe \\\\ ith variations in the amplitude. CHAPTER XVIII VELOCITY OF WAVES OF DIFFERENT TYPES Wave Front. \u2014 When waves spread out from a centre of disturbance, a surface can be described that marks at any instant the points which the disturbance has reached. This is called the \"wave front.\" Thus, if a stone is dropped in a pond, or if a raindrop falls on a pool of water, the wave front is marked by an ever-expanding circle. If waves are pro- duced in air by a vibrating tuning fork, the wave front at some distance from the fork is very approximately a sphere. These are called \" spherical \" waves. The waves in the ether that reach us from a distant star, or from any distant terres- trial source of light which is small, have a spherical wave front ; but this sphere has such a large radius that the portion of the wave front that affects us is practically a plane; so we call these \"plane waves.\" Intensity of Spherical Waves. \u2014 If we consider a point source, which therefore produces spherical waves, we can easily calculate the relative intensities (see page 314) at different distances from the source. Let us describe two spherical surfaces of radii rl and r2 around the source ; their areas are 4<7rr12 and 4?rr22. So if the source emits in a unit of time an amount of energy equal to E, and if there is no absorption by the medium, the intensity at any point of the first surface Tfl TJ1 is 2 2' an^ ^at a^ any P0^ \u00b0f the other surface is 2- If the former intensity is called 7X and the latter J2, it is seen that l 1 '\u2022 2\"^V' 330 VI:L<>< ITY OF WAVES OF DIFFERENT TYPES 331 Or, in words, the intensity varies inversely as the square of the distance from the source. (If the source is of such a kind as to produce harmonic motions at all points in the surrounding medium, we see at once that the amplitude of tin- vibration at a point in the first spherical surface bears a ratio to that at a point in the second surface given by ri ri A1 : A% = \u2014 : \u2014 ; for we have shown on page 328 that the energy of a harmonic vibration is proportional to the square of the amplitude, i.e. /j : J2 = Af : A\u00a3 ; and therefore by the formula just deduced for the intensity, the one for the ampli- tude follows at once.) Velocity of Waves. \u2014 The rate at which the wave front ad van * < > is called the \"velocity\" of the waves. In the next article we shall deduce its value in certain simple cases in trims of the physical properties of the medium; but from general considerations it is evident that in an elastic medium the velocity will be increased if the elastic force of restitu- tion of the medium is increased, and will be decreased if the inertia of the medium is increased; and conversely. In fact, we can prove without difficulty that the velocity of compres- sional waves in a homogeneous fluid whose density is d and whose coefficient of elasticity is E^ is given by the formula Tli is formula is due to Newton, and is deduced in the Principia. WHY.- front of waves in the air is affected naturally by winds. if plane waves are advancing in a direction opposite to the \\vin.l. the iij>|\"T portions of the wave front will be more retarded than the lower, because, owinir to friction, the wiml m-ar the earth has a less velocity than hi^h.-r up. So the wave front will lean backward and will proceed up in the air away from the earth. Similarly, if the wind is l.lowini: in tin- direction of the advance of the waves, the wave front will lean forward. VIBRATIONS AND WAVES If a \u2022\u2022point source\" that is producing spherical waves has motion of translation apart from its vibration, the wave front may change. Thus, if the source is moving in a straight line with a constant velocity, it is evident that, so long as this velocity is less than that of the waves, the wave front remains spherical. If, however, the velocity of the FIG. 146. -Waves produced by a point source which source [g greater than is moving with a velocity greater than that of the waves \u00b0 in the medium. While the source moves", " from 0 to Ot, that of the WaVCS, the waves from O reach a distance OA. things are different. In the cut let 0, Or 02, Oy 04 be the position of the source at instants zero, \u00a3, 2 \u00a3, 3 \u00a3, 4 1 ; draw a sphere round 0 with Ji radius 4 tv, where v is the velocity of the waves ; one round Oj with a radius 3 tv ; etc. These spheres mark the distances the disturbances have spread out at the time 4 1 ; so, when the source is at 04, the wave front is a cone, whose section is shown by the cut. This fact is illustrated by the waves seen at the bow of a rapidly moving boat. A convenient model for the study of wave motion in an elastic medium is made by suspending a great number of lead balls by long strings at equal &w\u2122r(y^ intervals in a Straight Fl\u00b0- 147- \u2014 A baU and spring model for illustrating wave horizontal line, and joining them by spiral springs. If the end ball is pushed in quickly toward the next one, a compression is propagated down the line; if it is pulled away rapidly from the next motion. VEL(>< 11 Y <>!\u2022 \\VAVES OF />//\u2022'/\u2022'/\u2022:/,\u2022/\u2022; AT TYPES 333 ball, an expansion is propagated ; if it is given any periodic motion, a corresponding periodic train of waves is produced. The velocity of these disturbances is evidently increased by using stiffer springs or lighter balls, and diminished by usin!_r heavier balls or weaker springs. Reflection of Waves. \u2014 When waves in any medium meet an obstacle, there is, as a rule, reflection, as is illustrated by water waves being thrown back by a pier wall, by the phe- nomenon of echoes, by the use of a mirror in light, etc. We can deduce the exact conditions of reflection by considering t\\\\o l>all-an<l-spring models like the above, which have differ- ent velocities for compressional waves, and which are con- nected so as to form a continuous \"medium.\" Let the two sets of balls be arranged in the same straight line, the ones which carry waves with the less velocity on the left, as shown in tin- cut. We can call this set th\u00ab- --.slow\" one, the other the \"quick\" one. When a com- pression is prop- agate* 1 from Irft \u2022\u00bbg\u00abnn t IOXH1 f^^ noon k..-r,-r< k-m-rr. T>,T! rn > n n\"n \u2022i to ri\"ht. it trav- *'\"'\u2022 1K A \"\"\"1('l illustrating two media ; in the one on the left els with con- the velocity of waves Is less than in the one on the right stant velocity unt il the second set is reached. At the bound- arv the compression is carried ort more rapidly by the quick set of balls, and therefore the spring joining the two sets is not mm pressed as much as it would he if the velocity were not increased; the last ball of the slow set is thus pulled slightly toward the right, and so produces an extension of th\u00ab- spring hark <>! it; etc. Therefore a compression in the slow set produces a compression in the quiek set, and an expansion is reflected back along the slow set from the 334 VIBRATIONS AND WAVES boundary. If an expansion is propagated from left to right, there will be an expansion produced in the fast series of balls ; but at the boundary the first ball of this set yields more easily to the force of the spring pulling it toward the left than does the last ball of the slow set. Therefore the spring at the boundary is not extended as much as the others, and this will then produce a compres- sion in the springs of the slow set. An expansion, then, propagated along the slow set produces an expansion in the fast one, but a compression is reflected back along the slow one. If a series of waves consisting of alternate compressions and expansions, such as would be produced in this model by giving harmonic motion to the first ball of the slow set, meets a set of \" faster \" balls and springs, the reflected waves are of the same nature ; but whenever a com- pression reaches the boundary, an expansion is reflected, and conversely. In an exactly similar manner it may be shown that, if such a series of waves in a set of fast balls and springs is incident upon a boundary beyond which there is a slow set, a similar series of waves is reflected ; but in this case a com- pression produces a compression, and an expansion an expan- sion at the boundary. There is,a difference, then, in the reflection at the", " boundary between the fast and slow sets, depending upon the direction from which the incident waves come ; and this difference is equivalent to a substitution of a compression for an expansion, or vice versa. It should be particularly noted that if the velocities of the waves in the two media are the same, there is no reflection. (It is assumed that the waves are not damped, that is, that there is no absorption.) Therefore, in order to have reflec- tion of waves at a boundary separating two media, these must be such that the velocity of the waves is different in the two. (The \"rolling\" of thunder is an obvious illustra- tion of the reflection of air waves owing to the presence in the air of foreign bodies, namely clouds, or of regions in or M-.ir/-> Of i>in-i-:i;!-:.\\T TYPES 335 which the velocity is different.) Another obvious condition for securing reflection from un obstacle is that its area should hi- large compared with the length of the waves; otherwise they will puss around it. Velocity of Transverse Waves along a Flexible Stretched Cord. \u2014 It is not difficult to calculate the velocities of certain ses of waves. This is true of the propagation of trans- verse disturbances along a stretched but perfectly flexible cord in which the tension is constant. Imagine a tube, which is straight except for a circular portion near its middle, slipped over this cord and moved rapidly along it with a con- st ant velocity v. Let us consider the motion of the particles of the cord as the curved portion of the tube reaches it, and the forces which the tube exerts on the cord. As this curved P Q F i... 149. \u2014 A cord, over which has been slipped a bent tube, is stretched between Pand Q. portion of the tube reaches any particle of the cord, it gives the particle a motion which may be resolved into two com- ponents : uniform motion in a circle with constant speed v, and uniform motion parallel t<\u00bb that of the tube along the cord with constant speed v. The former motion will be con- si ( In -i \u2022\u00ab I presently. As the particle enters the curved portion, it is given, therefore, a momentum along the line of the cord, whirl i it keeps until it leaves the other end of the curved portion, when it is given an e<|iial momentum in the opposite (liivriinM and brought to rest. The two forces that pro- duce these changes in momentum are due to the tube; hut one balances the other exactly; so then- is no resultant action or reaction due to them. The only other acceleration is that occasion. <1 by the particle being made to move in a circle with constant speed. It' / is the length of any minute portion of the ronl. </ its mass per unit length, an<l r 336 VIBRATIONS AND WAVES tin- radius of the circular portion of the tube, the force which must be exerted on this portion of the cord to make 2 it move in the circle is dl - - There are three forces acting on it, the tension in the cord acting at its two ends and the reaction of the side of the tube against which it presses. Let us calculate the former. In the cut let A, B, O represent three consecutive points in the cord, drawn on an immense scale ; let the length of the arc ABO be Z, the portion of the cord which we are considering; let 0 be the centre of the circle drawn through these three points ; and T be the ten- sion in the cord. Owing to this tension B there are two forces acting on this por- tion of the cord, as shown in the cut. Calling the angle between AO and OB T (and also between OB and 00) N, the component of each of these forces along BO is T sin N, and their components perpendicular to BO balance each other. The resultant, then, of these two ten- Fio. 150. \u2014 A, B, C are i \u2014 i three consecutive points of sions is 2 T sin N\", and its direction is c\u00abrd.CUrV6d P\u00b0rti0n \u00b0f *\" along BO' N is an extremely small angle, and so sin N may be replaced by JV; but 2 N is the angle (^4.0(7), and this equals the length of the arc ABC, i.e. I, divided by r, the radius. So the resultant force due to the tension acting on the elementary portion of the cord considered is \u2014. If the tube is moving Tl with such a velocity that this equals the force required to make the portion I move in a circle, there is no reaction of the tube on the moving cord; that is, the tension in the cord is itself sufficient to maintain the cord in the given curv", "ature ; and the tube now exercises no force on the string whatever. In other words, the tube could now be removed entirely, and the \" hump \" in the cord would be propagated of itself VELOCITY OF IIM T/-> <>F l)irri-:i!l\\l IYPES 337 along the cord with the velocity of the tube before it was removed. The mathematical condition for this velocity, as given above, is that This si 10 us that the velocity of any hump, whatever its radius, is a constant quantity for a given cord under a definite tension ; therefore, any kind of a disturbance, whatever its nature, will be propagated with this same velocity. The fundamental conditions are that there is no force in the string except its tension and that this is constant. Velocity of Compressional Waves in a Fluid. \u2014 The velocity of compressions] waves in a fluid may also be calculated; but the process is not quite so simple as in the previous case. The student will find it given in Maxwell, Heat, 4th c<lit ion, page 223, and in many other text-books. The result \u2022 prove that, if E is the coefficient of elasticity of the tin id, and d its density, the velocity of the waves is given by * d Fri the formula V=\\\u2014 \u2022 Nothing need be said in explanation of d\\ but we have seen that the \"elasticity\" of a gas is an indefinite quantity, unless the condition is defined under which the change in volume occurs (see page 194). In the present case, the number of compressions and expansions in one second is so great that the change in volume of the fluid takes place adiabatically ; for there is not time for heat transfer to occur. Consequently in the above formula E is the adiabatic coefficient <>f elasticity. Its value for any gas has been given already (page 253). If c is the ratio of the two specific heats, and p the pressure of the gas, E equals the product cp. Therefore, for a gas, and this can be simplified by using the gas law p- RdT. Thus, V=^Tftft Then-fore, the velocity varies directly as the square mot of the absolute temperature; and AME*' -- 22 338 VIBRATIONS AND WAVES further, if F\", 72, and T are known for a gas, c may be calculated. Laplace was the first to see (1816) that the coefficient of elasticity in this formula was the adiabatic one. Newton, who was the first to derive and apply the formula, used the isothermal coefficient, whose value equals p, and thus made an error. Assuming the value of c for air to be 1.40, and substituting proper values of p and d, the velocity of air waves may be calculated at any temperature. The value of this velocity at 0\u00b0 C. is thus equal to 33,170 cm. per second, if the C. G. S. system is used. The fact that the velocity varies with the temperature is illustrated by the observations of arctic travelers who have noticed that the so-called \" velocity of sound \" is less at low temperatures. The velocity of waves in air is also seriously aifected by the presence of moisture, because the density of the air is changed. The velocity is seen to be independent of the nature of the disturbance propagated, and also of the pressure. When there is a violent explosion in the air, there are slight variations in the velocity near the centre of disturbance, owing to the fact that the value of the elasticity of the gas, as given above, is true for small variations in the pressure only. At some distance away from the centre, however, the velocity becomes normal. * a Similarly, the velocity in other gases may be calculated. The velocity of waves in liquids may also be deduced from the original formula V=\\\u2014 - For water at 8\u00b0 C. it is found, as stated before (page 172), that an increase in pressure of one atmosphere, i.e. of 76 cm. of mercury, decreases a unit volume by 0.000047 of its value. If the thermal effects may be neglected, \u201e 76 x 981 x 13.59, 7, therefore, V= 145,000 cm. per second. -\u00a3666617 \u2014 'and'/ = There are also experimental methods for the determination of these velocities in gases and liquids ; the details of which VEL\u00bb< ILY or WAVES <>r Dirri:nEM TTPMS 339 are described in larger text-books such as Poynting and Thomson, Sound. These methods may be divided into two classes : direct and indirect. In the former, a disturb- ance is produced at some point and the time taken for the waves to reach a point at a measured distance away is accu- rately measured", ". The disturbance may be the ringing of a bell, a mild explosion, etc. ; and the instant of arrival of the waves may be determined by the mechanical motion of a diaphragm or by the perception of the sound. In the indirect methods, the gas or the liquid is set in vibration by some periodic disturbance whose frequency is known : since this is due to resonance, the frequency of the vibration of the gas or liquid is known ; and, as will be shown presently, the velocity of waves in it may be at once calculated. Velocity of Waves in a Solid. \u2014 In solids a purely compres- sional wave cannot exist, because when there is a compression produced by two opposite forces there is at the same time a distortion. The velocity of longitudinal waves in a solid that extends in all directions is given by the formula when- /r is the coefficient of elasticity for a change in volume, and // the one for a chanty in shape. A train of waves can, however, be produced where there is only distortion, as when one end of a long wire is rapidly twisted to and fro. In this case, if n is the coefficient of rigidity for the solid, V\u2014 *\\pj- If longitudinal waves are produced in a wire or a rod by stroking it lengthwise with a resined or damp cloth, the ve- locity of the waves is given by V=^\u2014, where E is Young's modulus. ( S66 page I \u2022\"> I. ) Water Waves. \u2014 Tin- velocity of waves upon the surface of a liquid di-j.rmls upon many quantities, and we can do no 340 r//;/,M770ATs AMD WAVES more here than state certain facts in regard to liquids whose viscosity may be neglected. These statements involve the quantity known as the \" wave length,\" which in the case of waves on liquids may be defined to be the distance from crest to crest or trough to trough. If this quantity has the value Z, we have the following formulae for the velocity : If the liquid is deep, V\u2014\\\u2014, where g has its usual 2 7T meaning. If the liquid is shallow, V= Vfy/, where h is the \u2014, surface tension and d the density. The general formula for waves on liquids which are deep is Id 2* r y o V and it is clear that, if the waves are long, the first term is negligible, while, if they are short, the second one is. It is seen by calculation that in the case of water if Z>10 cm., the first term may be neglected, and if Z<0.3 cm., the second. For intermediate values of Z, the full expression must be used. There are many most interesting applications of these formulae. The fact that, if one sailing boat has a longer water-line than another, the latter is given a \" time allowance \" in a race, is due to an attempt to equalize the advantage of the longer boat ; for a boat moving through the water produces waves that are comparable in length with its own ; and as the boat is helped on by these waves, the longer boat is helped the more because the velocity of the waves it produces is greater. Again, as waves approach a shelving shore, if they are oblique to- the shore line, they will gradually turn so as to approach parallel to it, owing to the fact that in shallow water the waves are faster in the deeper portions than in the ones less so. The motion of the individual particle of a liquid as a wave passes over its surface is in general an elliptical path ; and the effect of the waves is felt only a short distance down VELOCITY <>F \\VAVES OF DIFFERENT TYPES 341 from the surface, as the amplitude of the vibration decreases rapidly \\\\ith the depth. This is not the place to discuss the velocity of temperature waves or of electric waves along wires. But it may be stated that in both these cases part of the energy of the waves is dissipated in heat effects throughout the media and in other ways, and as a consequence of this the waves die down and are not propagated as far as they otherwise would be. The FIG. 151. \u2014 A drawing1 of Lyman's wave model for water waves, showing the form of the wave, the motions of the indi- vidual particles, etc. waves are said to become \"attenuated.\" It may be proved also that long waves persist for a greater distance than short ones ; and this fact is of fundamental importance in tele- phone service, as will be shown later. CHAPTER XIX HARMONIC AND COMPLEX WAVES \u2014 \u00ab STATIONARY WAVES\" Trains of Waves and Pulses. \u2014 In discussing many of the properties of wave motion it is essential to distinguish two types of waves : one is produced by a sudden irregular", " dis- turbance, and may be called a \" pulse \" ; the other is pro- duced by a periodic disturbance, and is called a \" train of waves.\" Harmonic Waves : Wave Length, Wave Number, Amplitude. \u2014 The simplest type of a train of waves is one produced by a centre of disturbance whose motion is harmonic. This is called a \"train of harmonic waves,\" or a \"harmonic train.\" As a consequence of this disturbance, each particle of the medium will be set in harmonic motion, but the phase of the vibration varies from point to point at any one instant. (This may be illustrated on the ball-and-spring model.) The simplest mode of representing waves graphically is to choose two axes, one giving the distance the wave front advances in any direction, the other the displacement at any instant at points along a line drawn in this direction. A curve on this diagram gives the displacements at FIG. 152. \u2014 Distances along a line in the direction of prop-.,,,, agation. A harmonic train of waves. any instant Ot all the particles in the medium along a line in the direction of advance of the waves. 342 HARMONIC AND COMPLEX WAVES 343 I III! II I ILL LL ILL ILL I MM II i i in INI III I II III III Hill If the waves are due to a harmonic disturbance, the curve is as shown, where the line PQ indicates the displacement of the particle whose posi- \u00bb?????\u2022 tion in the undisturbed I M l medium is P. (If the waves are transverse, this line may be the actual displacement ; if they are longitudinal, it equals, or is proportional to, the displacement of P in the direction of the axis. In this latter case a displace- ment of P toward the right may be indicated by PQ being drawn ver- tically upward, as shown; and a displacement tow a n 1 the left by the line PQ being drawn downward.) As time goes on, the dis- placements all change, the waves advance ; and successive conditions in the medium arc illust rated i i i i 1 1 1 1 nun i i i ill Jn i i 1 1 rdt i i i i Mini 1 1 i 1 1 H\" i 1 1 ILL ILL M LLL I I III JJ_L ILL 11 in I i-. 1\"'L where three different curves are drawn for three different instants <>f time ; and it should be note.l that in drawing these it is aSSUinrd that there is no decrease in tin' amplitude. If the IT *iMMi\u00bb*fiit *\u2022\u00ab<\u2022\u00bb duplaoement of any particle P in the inr<liinii i-, i.Wrved, it is SITU that it is given in succession l>\\ /'',',. /'$,, and F.,.. IM.-A eompn\u00bb.Nl.>n\u00bbl train of waves id i nun MM n n m ILL U4 VIBRATIONS AND WAVES PQB; so it is making vibrations as previously described. The amplitude and the period of the vibration of any particle are called the amplitude and the period of the waves, and the frequency of vibration of the particle is called the \"wave number.\" The distance along the line of propagation from any point to the next one where at t la- same instant the motions are identical is called the \" wave length \" ; e.g. from P to R. (This may also be defined as tlip distance between two consecutive points in the direction FIG. 154. \u2014 Diagram showing three successive positions of the train of wavesj_the vibration of an individual particle of the medium P\\ and the wave length PR. of propagation at which the phase of vibration is the same.) In the time of one complete vibration of a particle the waves advance a distance equal to the wave length, and so in a unit of time the waves advance a distance equal to the product of the wave number and the wave length. Therefore, if V is the velocity of the waves, I the wave length, and ^V the wave number, V\u2014 Nl. Or, if T is the period of the waves, V= \u2014. (The velocity of an individual particle depends upon the instant we consider it, for it is making harmonic vibrations ; and so the velocity varies from zero to a maximum value, then decreases to zero, etc. It is clear that there is not the faintest connection between this varying velocity and the constant velocity of the waves.) //.l/M/o.y/r AM) ro.u/'/./;.v \\\\\\l\\ 345 Doppler's Principle. \u2014 If we speak of that portion of a train of waves whieh is a wave length long as a \" wave,\" we may say that, if the wave number is 2", "\\T, the source emits N waves in a unit of time ; and, in general, N waves pass any point in the medium in a unit of time. This is true if the vibrating source, the point in the medium, and the medium as a whole are not moving. It is interesting, however, to consider the two cases, when the source is moving and when the point in the medium where the waves are counted is moving, the medium not being in motion in either. If the vibrating source is at rest, and the point in the me- dium is moving toward it in a straight line, let N be the frequency of the source, V the velocity of the waves in the stationary medium, I their wave length, and v the velocity of the moving point. Owing to the motion of this point it would pass - waves in a time if the waves were stationary in the medium ; but since they are moving toward the point at such a rate that N waves would pass a fixed point in a unit of time, the total number that passes the moving point in that time is the sum JV+y. (But-ZV=-; and so this number jr. \\ may be written \u2014 j^-. J For similar reasons, if the point is moving away from tin- fixed source with a velocity v, the number of waves which pass it in a unit of time is N\u2014'/ T *\" \\ * If the point in the medium is fixed, and tin- vibrating source is approaching it in a straight line with the velocity v, the case is entirely different. If tin- source were n\u00ab>i the length \u00ab,f ;v wave would be I; but. when it is the wave fnmt advances a distance T from the in,i unit time, and in this same time the source advances a diMam-e v; so the X \\\\aves that have been 346 VIBRATIONS AND WAVES emitted in this time are crowded together in the interval of space V\u2014 v, and the length of a wave is now \u2014 ~^- The new wave number, or the number of these waves that pass a fixed point in a unit time, is the velocity V divided by this wave length, or \u2014 \u2014 N. Similarly, if the vibrating source v \u2014 v is receding from the fixed point with the velocity v, the new wave number is - \u2014 N. y V+v If v is small compared with FJ these last two expressions may be written fl + iW and (l - \u00a3\\flT, or N+ - and N\u2014 - ; so the formulae for this and the preceding case agree under this condition. It is thus seen that when the source of waves and the point under consideration are approaching each other, the wave number is apparently increased ; while, if they are receding from each other, it is apparently decreased. These formulae express what is called Doppler's Principle. It is illustrated in the case of a star approaching or receding from the earth, in a whistling locomotive approaching or receding from a station, etc. Attenuation of Waves. \u2014 The energy of a harmonic vibra- tion varies as the square of the amplitude, and hence the intensity of harmonic waves varies as the square of their amplitude. This amplitude decreases as the waves advance, owing to various causes. One case has been considered already on page 331, where it was proved that in spherical waves the amplitude varies inversely as the distance from the source. Further, there may be friction involved in the relative displacements of the particles of the medium, as is the case to a greater or less degree with all waves in all forms of matter ; or, motion may be given particles of foreign matter immersed in the medium (see page 311); in both of HARMONIC AND COMPLEX WAVES 347 which cast's the amplitude of the waves decreases and they lose energy. Similarly, if waves in one medium are inci- dent upon a boundary separating it from another in which the velocity is different (see page 333), waves are transmitted into the latter medium, and waves are also reflected back into the former. The combined intensity of these two trains of waves must equal that of the incident train ; and so the amplitude of the transmitted waves and that of the reflected waves must both be less than that of the incident waves. (An obvious law connects them.) Superposition of Waves; Complex Waves. \u2014 There can be two harmonic trains of waves of the same wave length and amplitude, but differing in phase at any instant, depending upon when or how their motion was begun. Similarly, we may have in any medium waves of different wave length, different amplitude, etc. ; and their combined action may be found by compounding them as was done for the vibrations of a particle, for it may be proved that this is allowable, pro- vided the individual displacements of the particles of the medium are small in comparison with", " the wave length. The best method of considering the superposition is a graphical one. Thus, two plane polarized transverse waves (see page 313) which are harmonic, and whose directions of vibrations are at right angles, may be compounded as shown in Lissa- jous' figures. In particular, two such waves having the same period would combine to produce an elliptically polar- ized train of waves, or a circularly polarized train if their amplitudes are equal and their difference in phase ^- Con- versely, an elliptically or circularly polarized train of waves may be resolved into two plane polarized trains of waves which are harmonic and whose vibrations are at right angles to each other. Two plane polarized transverse waves which are harmonic and whose directions of vibration an- the same, or two 348 VIBRATIONS AND WAVES tudinal harmonic trains of waves, may be compounded in the same manner as were two vibrations in the same direc- tion ; and any of the illustrations given on page 321 may be applied to the case of waves. Conversely, by Fourier's theorem, any complex train of waves may be resolved into trains of harmonic waves whose wave lengths are in the ratio of 1:2:3: etc. There are other modes of resolution, also, which often are more convenient. We see, then, that a harmonic vibration produces a har- monic train of waves ; a complex vibration, a complex train. A special case of this last is a non-periodic or a confused vibration ; it will produce a corresponding wave. If the disturbance is intense, but lasts only a short time, it produces in the medium what we have called a \"pulse.\" Its effect when it reaches a portion of the medium containing foreign matter is naturally different from that of a long train of harmonic waves ; because owing to these last there are peri- odic forces brought into action on the foreign particles, and resonance may follow. Distortion of Complex Waves. \u2014 As we have seen in speak- ing of the attenuation of waves owing to their decrease in amplitude, there are cases in which long waves are less affected than short ones. (See page 341.) In such cases, if a complex vibration is producing a train of complex waves, its harmonic components of long wave length will persist longer than those of short wave length ; and so the char- acter of the complex vibration at different points in the medium will vary. This phenomenon of the change in the \"form\" of the wave is called \"distortion.\" The further one is from the vibrating source, the more nearly does the vibration approach that of being simple harmonic. This fact is illustrated by ocean cables and by long-distance telephone wires. However complex the electrical disturbance at one end of a cable, that at the other is nearly, if not quite, har- monic. In using a telephone over a long distance the qual- 349 ity of the sound is entirely changed, only the graver notes l>cin\u00abr heard. The great merit of Pupin's new system of constructing cables and telephone lines is that it not alone il< -creases the attenuation, but also diminishes greatly the distortion by making the attenuation of all the waves, long and short, the same. Nodes and Loops \" Stationary Waves. \" \u2014 A most important effect of wave motion is illustrated by a simple experiment which may be performed with a long flexible cord; e.g. a long spiral spring or a long rubber tube, one of whose ends is fastened to a fixed support and the other is held in the hand. If the cord is stretched fairly tight and the free end moved sidewise with a rapid harmonic motion, waves will be produced in the cord which will be propagated up to the fixed end, and will there suffer reflection and be propagated back to the hand, etc. Consequently, at any instant there are in the cord two trains of waves traveling in opposite directions. If the frequency of the motion of the hand is exactly right, it will be observed that the cord ceases to have the appear- ance of being traversed by waves, and vibrates transversely in one or two or more portions or \" segments.\" That is, there are certain joints in the cord where there is very little, almost no, motion, which are called \"nodes\" ; and the cord in between these vibrates just like a short cord whose two ends are fastened. The points halfway het ween the nodes, where therefore the motion is greatest, are called \"loops.\" Tli is type of vibration is sometimes, but improperly, called \"stationary\" or \"standing\" Craves, In reality, the waves have disappeared in the production of a vibration. Such vibrations as this are extremely common; and may occur with waves of all kinds. The explanation i^ evident Consider a medium through which are passing in opposite directions two harmonic- in in- 350 VIBRATIONS AND WAVES of waves which", " are not suffering attenuation and whose wave lengths and amplitudes are equal. At any instant they may be represented by curves as in the cut. Since the actual motion in the medium is found by superimposing the two wave motions, there will be certain points, one of which is P2 in the cut, at which the displacement is zero, owing to one wave neutralizing the effect of the other. As the waves ad- vance\u2014 in opposite directions \u2014 they continue to neutralize x...../ \\. p FIG. 155. \u2014 Formation of nodes and loops by two trains of waves advancing in opposite direc- tions. PU P2, PSt etc., are nodes. each other at this point, as is seen from the cut; therefore this is a node. The importance of the conditions that the waves should not be damped and that the amplitudes of the two trains should be the same as well as their wave lengths is evident. Further, since in a train of waves at a distance of half a wave length from any point the displacement is exactly reversed, if two trains of waves neutralize each other at a point JP, they will also do so at points distant from P by half a wave length, a whole wave length, etc. Therefore, the distance apart of two nodes equals one half the wave length of either of the component trains of waves. In the case of the cord which was first considered, the fixed end is obviously a node ; and the one held in the hand is approximately one, as is evident if one observes the motion. If the long flexible cord is held suspended vertically from a balcony so that the lower end hangs free, a vibration of the same kind can be produced ; only in this case, since reflection takes place at a \" free end,\" this point is a loop. As the frequency of vibration of the hand is increased gradually, it is found that there are certain definite fre- \" STATION ABT \\\\.\\VES\" 351 quencies for which the cord separates into vibrating seg- ments ; ami the number of these segments increases, that is, the distance apart of the nodes decreases, as these critical frequencies increase. Similarly, if the tension of the cord is increased, the critical frequency must be increased also. The explanation of these facts is not difficult. Let us consider the first case, that of the cord with its end fixed ; and let the length of the cord be L, the wave length of the component waves be Z, the distance apart of two nodes be c?, the velocity of the waves in the cord be K, the frequency of the vibration be N, and the number of segments be n (necessarily a whole number). Then, we have the relations: c? = -, V \u2022= Nl, L = nd; and therefore, on substitution, N= \u2014. So if the tuision is kept constant, i.e. if V is constant, and if the length L of the cord is not varied, n varies directly as JV; 1 1 ml is, tin- frequency of the vibration must have a definite value, since n must be an integer, and if the number of seg- ments is increased from 1 to 2 to 3, etc., the frequency must be increased in an equal ratio. Again, if the tension in the cord is increased, the velocity V is increased ; and therefore, if // U tin- same, iV, the frequency, must be increased. Vibration of a Stretched Cord. \u2014 These vibrations in a stretched flexible cord are not always produced by the method described. If the cord is stretched between two fixed points, it may be set vibrating by using a violin bow, by plucking it with a tinker, by striking it a blow, etc. The vibrations are exactly like those just described. By lightly touching the middle point of the cord, so as to hold it nearly at rest, and bowing the cord or plucking it at a point distant from the end by quarter of the length of the cord, the string vibrates in two equal segments; if the point touched is at a distance from one end equal to one third the length of the \u2022 \u2022\u2022>! -d, it may be made to vibrate in three segments ; etc. The positions of the nodes and loops of any vibration of the cord 352 r //;/,-. i noNS AND WAVES may be determined experimentally by putting astride the vibrating cord, which is supposed to be horizontal, small paper saddles. They will be thrown off at the loops, but not at the nodes. (This method was used as early as 1677 by two Oxford scholars, William Noble and Thomas Pigott.) If the cord is bowed or plucked at random, the vibration is a complex one, which can be resolved into the simple compo- nents just described. (Naturally, the point of the cord which is plucked or bowed must be a loop ;", " and so only those component vibrations are present in the complex one which have a loop at this point.) We shall consider each of these simple modes of vibration n V more in detail. The general formula is N= \u2014 - ; and so, if V there is one segment, N\u2014 \u2022\u2014 ; if there are two segments, O pr *i g TT 2 L ^JL ^V\"=_; if there are three segments, N=\u2014 \u2014 ; etc. The frequencies are therefore in the ratio 1:2:3: etc. These ___ vibrations are, as we have ^5> said in speaking of Fou- rier's theorem (page 321), called \"harmonics.\" The first one, in which n = 1, is also called the \"funda- mental\"; and the others, \"upper partials.\" (It is seen that the frequency when there are two seg- ments is the same as if the cord were of half the length and had but one segment.) A few AFio.l56.-Vibrationsof stretched cords: (1) one <>UtS are given of these vibrating segment; (2) two vibrating segments ; tranSVCrSC vibratlOUS of (SUhree vibrating segments ; (4) superposition of \u2022\u2022 STATIONARY MM r/>\" 353 Tin- velocity of transverse waves in such a stretched flex- ible cord has been proved to be given by the formula, * d IT r=\\' \u2014, where T is the tension in the cord and d is the mass per unit length. Substituting this in the general for- mula, we have N= ^'^\\\u2014', which shows that if the tension of cord is increased, the frequency is increased ; if the length of the cord is increased, the frequency is decreased; etc. All of these facts are illustrated in various musical instru- 2.L * a ments, as will be noted later. Vibrations of a Column of Gas. \u2014 The same type of vibra- tion can be produced in a column of gas, such as we have in the case of an organ pipe, a flute, a horn, etc. There are many ways in which the vibrations may be produced: by blowing a blast of air over the sharp edge of an orifice open- ing into the column ; by making some solid body in contact with the gas at one end vibrate harmonically with a suitable frequency; etc. (The first is illustrated when one blows an ordinary whistle or when one blows over the end of a hollow key; the latter when one blows a horn by means of the vibrations of the lips or holds a tuning fork over the mouth of a bottle in which water is poured until there is resonance.) If a particle in a column of gas is at a node, that is, if its motion is a minimum, it must be in contact with the end wall or it must be held stationary by symmetrical conditions on its two sides, up and down the column. Thus, as the particles in the gas vibrate, there are the greatest fluctua- tions in pressure and density at the nodes. Similarly, at a loop there is the greatest motion, but tl <- least change in pressure and density. Tin- vibration..f a column of gas is illustrated in t he accompany MILT cut, in which the transverse lines indicate the positions of 1 1 -a us verse layers of gas. If the column <>f gas is closed by a solid partition, this point is a node; while if the en. I ix opm to thr air, so that AMES'S PHYSICS \u2014 28 354 VIBRATIONS AM) WAVES the pressure there cannot change greatly, this point is approx- imately a loop. (Actually, the loop is a short distance beyond the open end in the air outside. If the tube contain- I 1 FIG. 157. \u2014 Stationary vibration in a column of gas. Vertical lines represent positions of layers of gas. Curves represent by their vertical displacements the horizontal displacements of the layers of gas from their positions of equilibrium. Arrows represent the directions of motion of the layers of gas. ing the gas is a circular cylinder with a radius R, the loop is at a point beyond the end at a distance given approxi- mately by 0.57J2.) As in the case of a stretched cord, the column of air can vibrate in different ways, depending upon the number of seg- ments. We shall consider several special cases : \"STATION A It Y II.1 I'A'.s 855 1. A column of gas closed at both ends. \u2014 There is there- fore a node at each end. The simplest mode of vibration is wht'ii there is only one loop, which must then be at the mid- dle point : in this case the distance from node to node is", " the length of the column L. This must equal half the wave length of the component waves, J ; and, if 2 is the fre- quency of the vibration and V the velocity of the compres- sional waves in the gas, ^ = \u2014. Hence, L = A y *i JVj = ^-y. The next simplest mode of vibration is when the column of gas is divided into two segments by a node at its middle point. In this case the distance between two nodes V or is half the length of the column, \u2014. Hence, f Tin- analogy with the transverse vibrations of a cord stretched between two fixed points is complete. The possible vibrations make a complete harmonic series. 2. A column of gas open at both ends. \u2014 There is then a loop at each rnd. The simplest moil,- ut' vibration is wlu-n tin -re is only one node, which must be at the middle point. We saw in our general discus- sion that a loop came halfv 1..-I vreen two nodes; so the distance from loop to loop equals tl ii'Mlu to node. In this rasr. thr L N L N L N L '\u2022*. \u2014 Vibration* of A column of RM open \u00bbt both en.U : (1) ftimhuncnul ; (2) flrtt p*rtUl ; (.1) Mcond partial. 356 \\'li;ilATlONS AND WAVES distance from loop to loop is the length of the column, neglect- 2 2JV, ing the slight correction for the ends ; and so L = -1 = -\u2014, The next simplest case is when there is a loop at the middle point and a node at each of the points halfway from it to the ends. The distance from loop to loop is one half the length of the column. Hence, So this case is the same as the previous one ; the vibrations form a complete harmonic series. This kind of a column of gas is called an \" open \" one ; and, as in previous cases, if it is set vibrating by some random or indefinite means, the resulting motion is a complex vibra- tion equivalent to the addition of these simple ones. 3. A column of gas open at one end and closed at the other. \u2014 There is thus a loop at one end and a node at the other. The simplest mode of vibration is when there are no other nodes or loops. The distance from node to loop is one half the distance from node to node ; so in this case 2 L= J = \u2014 \u2014, v 2 2JV] orai-Jl The next simplest mode of vibration is when there is a node at a distance \u2014 from the open end, and a loop at a dis- 2 L 2 L tance \u2014 \u2022 The distance from node to node is then \u2014 ; and we have the relation =\u00a3 = * = -^-, or N2 = 2\u00b1- Simi- 9 T 7 V ^ V o 22 J\\/ 4 Jj larly, the next mode of vibration will give a frequency JV3 = \u2014 r ; etc. So in this case the vibrations have fre- quencies in the ratios 1:3:5: etc. ; a series of the odd har- monics only. It should be noted that the fundamental in \"8TAT10\\Ai;r 357 this column of gas has a frequency one half that of the fun- damental in the other two cases, when their lengths are the same. This kind of a column is called a \"stopped\" one. If it is set vibrating at random, its complex vibration is com- pounded of these simpler ones just described. Manometric Flame. \u2014 A simple method of studying the vibrations of the column of gas in a tube is to pierce openings at intervals along the tube and close them with flexible rubber diaphragms ; these are covered on the outside with small hemispherical caps into which there are two <>i -fil- ings: one permits the introduction of illuminating gas, the other is joined Fio. 169. \u2014Apparatus fur the comparison of UM vioraUng movement* of two sonorous tube*. to a pas burner with a fine opening. If the gas is ignited at this opening, it v. ill tlirk-T if tin-.li.ij.lir;i-.;iii '-nine.* at a node in the vibrating column of gas, because the fluctuations of pressure there will cause tli<- <lia|>hragm to vibrate and so affect the pressure on th<> illuminating gas. These tlui-t nations have the frequency of the vibration <>f th> column of gas, : ore, as a rule, very rapid. So the flickering of the 358 VIBRATIONS A\\D WAVES must be", " observed in a revolving mirror, which separates the images when one looks at the reflection of the flame in the mirror. This appa- ratus is called a manometric flame. In its present form it is due to Rudolph Konig. FIG. 160. \u2014 Attachment to organ pipe in order to obtain inanonietric flames. Longitudinal Vibrations of a Wire or Rod. \u2014 Similarly, a wire or a rod can be set in longitudinal vibrations by stroking or rubbing it lengthwise. If the wire is stretched between two fixed points at a distance L apart, there are nodes at these points ; and if V is the velocity of compressional waves in the wire, we have the following relation for the frequency of the fundamental vibration, FIG. 159 a. \u2014 Manometric dames. If a rod is clamped in the middle, this point is a node and the ends are loops. If L is the length of the rod and V the velocity of compressional waves in it, the frequency of the compressional vibrations is given as before by the formula \u00bb STATION AB\u00a5 \\\\AVEs\" 359 It may be proved by methods of the infinitesimal calculus, as has been already stated, that the velocity of these com- pressional waves in a wire or rod is given by the formula * d V = \\-, where E is Young's modulus of the wire and d is its densiu -. (This modulus E is found, however, by experi- ment not to have exactly the same value as the one found for the same material by the statical method as described on page 154.) Other longitudinal vibrations than the fundamentals can be produced in wires and rods; but their frequencies do not have simple ratios with that of the fundamental; so they are not harmonics, but upper partial*. Transverse Vibrations of a Rod. \u2014 A rod can also be set in Averse vibration. If it is not clamped at any point, but is supported in such a manner as to be perfectly free to vibrate, its fundamental mode is shown in the cut ; and its next X-x:-''' - --JiS: simplest mode of vibration is as '' ^ ^X shown. The ratio of the fre- quency of this vibration to that... N of the fundamental is 2.92; and,.,-J^V,...,'.>'C _____ '-jg^.., ill a similar manner the Other ^.,61.- Transverse vibrations of possible Vibrations do not have \u2022 rod: (I) fundamental; (2) first up- frequencies which hear simple relations to that of the fundamental or to each other; so are upper partials, but not harmonics. It the rod, instead of IMMULT straight, is bent into the shape of a long, narrow U- a|\u00abl has a projection attached at its middle point, as shown in the cut, it makes a \"tuning fork.\" In all actual forks the rod is not uniform, but is thickened at its base; so that the two nodes, which in the uniform rod are near i \\\\hen vibrating in its fundamental mode, are brought closer together. The point \\\\here the projeetin:.: ittached is still a loop, as may be shown l.y 360 VIBRATIONS AND WAVES the fork vibrating and then pressing the stem against a board; the latter will be set in vibration. A tuning fork is, in fact, generally attached to a hollow box, open at one end, and of such a length that the column of air inside has the same fre- quency as the fundamental of the fork. This is called a \" resonance box.\" An- other effect of giving the fork its peculiar shape is so to enfeeble the vibrations other than the fundamental FIG. 162. \u2014 Set of tuning forks on resonance that the latter is the Only one present when the fork is bowed or struck so as to be set in vibration. The frequency of a fork may be measured with great accuracy by comparing its period with that of a standard clock ; but for details of the methods of comparison, reference should be made to some larger text-book, such as Poynting and Thomson, Sound, Chapter III, or to some laboratory manual for advanced students. (The frequencies of two forks may be compared by Lissajous' figures, or by \" beats,\" if the frequencies are close together. See pages 327 and 412.).Vibrations of Metal Plates. \u2014 Thin metal plates cut in squares or circles can be made to vibrate transversely by clamping them at some point \u2014 generally the centre \u2014 and drawing a violin bow across their edges. When this is done, it is easy to show that there are certain lines along the plates where there is comparatively no motion ; these are called \"nodal lines.\" The simplest method of proving their existence is to scatter fine dry sand over the plates before they", " are set vibrating; when the vibrations begin, the sand collects along the nodal lines, being thrown there by the other vibrating parts of the plates. There are thus formed most beautiful, regular geometrical figures, which are \"STATIONARY WAVM&\" 361 called \"Chladni's figures,\" after the physicist who first sys- tematically investigated them. Their shape and complica- tions depend upon the point of support of the plate, the point where it is touched with the finger so as to make a node, the point of bowing, and the manner of bowing, which. to a certain degree, determines the number of vibrations, besides the fundamental, which are present. Similar nodal lines may be observed on stretched mem- branes, such as drumheads, when they are set in vibration in a proper manner. If some extremely light powder is used instead of the sand, with which to cover the plate, it is observed that it gathers, not at the nodal lines, but over those portions of the plate which are moving with the greatest amplitude. The reason for this was discovered by Faraday, who showed that small whirls were formed in the air near the vibrating portions of the plate, and that these carry the light powder off the nodal lines on to those portions. Vibrations of Bells. \u2014 Metal bells or bell-shaped objects, like glass tumblers or bowls, may be set in either transverse or longitudinal vibrations. The simplest mode of transverse vibration is shown in the cut. The position of the loops is fixed by the I MI i 1 1 1 where the clapper strikes. Many other vibrations than the fundamental are always present, but there is no simple relation between their frequen- It there ll an irregularity in the thickness of the rim at some point, this will produce comparatively little FIG. 168. - Ono mode of vibr*. effect if it comes at a node; but if it is at a loop, it will result in a change of the frequency. So a hell like this.-an vibrate in two ways, giving frequencies that are n..t very different : and if the bell is Struck a random l\u00bblo\\\\, both these vibrations will occur. This is the 362 V1BRATH).\\* AM) }\\'A\\rES cause of the \"beating\" of large church bells, as will be explained in a later chapter. (See page 412.) Other Illustrations of Stationary Waves. \u2014 The illustrations so far given of these vibrations due to the superposition of two trains of waves in opposite directions in a medium have been purely mechanical ones, the medium being either a solid or a fluid. But waves in the ether can produce these vibra- tions also, as is shown by the experiments of Wiener. A node in this case corresponds to a point where there is no motion in the ether (if we retain our mechanical concept of the ether) ; and a loop, to a point where there is the greatest motion. Rapid vibrations in the ether produce in the sur- rounding matter various effects, such as chemical changes which may be shown by photographic processes, fluores- cence, etc. ; and by all of these the existence of nodes and loops in the ether has been proved, when waves in it fall upon a mirror and suffer reflection. The distance apart of the nodes equals half the wave length of the incident waves. Electrical waves along wires and electromagnetic waves, produced by ordinary electric oscillations, cannot produce these vibrations with nodes and loops, because these waves are all \"damped\"; and owing to this fact the amplitudes of the reflected and incident waves are not equal. Measurement of the Velocity of Waves. \u2014 These facts in regard to the vibrations of stretched cords, of wires and rods, and of columns of gas (or, in fact, liquids also), lead at once to obvious methods of comparing the velocities of waves in different media. Thus, suppose two stretched cords have the same frequency of vibration when vibrating transversely in their fundamental modes ; then if L^ and L^ are their lengths, and Vl and F^ are the velocities of transverse waves along them, \u2022\u2022 BTA /vo.v.i/;)' HMFJ&S\" 363 Similarly, if the fundamental longitudinal vibrations of two \\viivs \u00ab'i rods of different material are the same, and if Zj and LI are their lengths, we have the same relation, I \\ : ] 2 = Ll : Ly for the velocities of compressional waves. Or. it t\\ 11 pipes contain different gases, and if their fundamental vibrations have the same frequency, the same formula applies for the velocities of waves in these gases. There an- two general methods for determining when two vibrations have the same frequency if the medium is a ma- il-rial one. As we shall see later, when two vibrating bodies that are not", " moving bodily are producing sounds, they have the same frequency if the \" pitches \" of the two sounds are the same; and this can be told with great exactness by a trained ear; or, if they differ slightly, the exact difference in frequency can be determined by the beats. Again, the vibrations may both be produced by resonance from a third vibration. Thus, if a vibrating tuning fork is held at the opening of a column of a gas whose length can be varied, resonance may be secured by noticing for what length the sound i> most reenforced ; and similarly with the column of a second gas. (A column of gas can vibrate, as has been shown, in many ways, breaking up into a different number of segments. So with a Lfiven tuning fork, resonance may be secured for several different lengths of the column of gas. The distance from node to node is the same, however, for all these different nodes of vibration ; and it is the quant it I. in the previous formula. If the frequency of the fork Nis known, and this length L is m.-asm-.-d, the velocity of tin- waves in the gas is obviously *2LN.) Again, stretched \u2022 \u2022ords or wires may have their lengths.-han^i-d until they are in resonance with a given tuning fork by the following mechanical method : stretch the cord or wire from two pegs or over two knife edges which are fastened to a large box open at its ends ; place loosely on the cord or wire a light saddle of paper ; set the stem of the vibrating tuning fork 364 VIBRATIONS AND WAVES on the box ; if there is resonance, the wire or cord will be set in vibration and the paper saddle will be thrown off. Kundt's Method. \u2014 We can also compare the velocity of congressional waves in a solid rod with their velocity in a gas or liquid by a method devised by Kundt. The gas is contained in a long tube, which is closed at one end by a tightly fitting piston and at the other by a very light one, which can move to and fro easily, but which nearly fits the tube. This last piston is attached rigidly to one end of the rod which is along the axis of the tube but projects beyond it, and which is clamped at its middle point. This rod is set in longitudinal vibration by stroking its free end with a damp cloth, and so the piston attached to its other end FIG. 164. \u2014 Kundt's apparatus for measuring the velocity of waves. vibrates and produces waves in the gas contained in the tube. The length of this column of gas is altered by means of the piston at its further end until it is vibrating with nodes and loops. The method of determining this condition is to sprinkle through the tube some light powder, such as is obtained from the finest cork dust ; when nodes and loops are formed, the powder collects in ridges across the bottom of the tube, leaving, however, the nodes perfectly bare. The frequency of the vibration of the rod is the same as that of the column of air, because the latter is \" forced \" by the former ; so, if L^ is the distance from loop to loop in the rod, that is, its length if we neglect the mass of the vibrating piston ; and if Lz is the distance from node to node in the column of air, the velocity of compressional waves in the rod is to the velocity of waves in the gas as Ll : Z2. (This method may also be used for a liquid instead of a gas by using suitable powder to mark the nodes.) 365 It should be observed that the piston on the end of the rod is at a loop for the rod, but a node for the gas; exactly as when the long elastic cord is set in vibration by the hand, as d---n il\u00bbed on page 350, \u2014 the motion at a loop in the solid rod is extremely small compared with that at a loop in the gas. The explanation of the formation of the transverse ridges of the pow- der depends upon the fact that, as the particles of gas away from the nodes vibrate back and forward between the particles of powder, their mean velocity depends upon the arrangement of the dust particles, and varies at different points, thus producing pressures in certain directions. (See page 169.) The full description of the process is long, and will not be given here. It may be found in Rayleigh's Theory of Sound, Vol. II, page 46. Therefore, if we know by direct experiment, or otherwise, the velocity of waves in air, we may by Kundt's method de- termine the velocity of compressional waves in any solid mate- rial out of which a rod can be made ; and then, by replacing the air in the tube by some other gas (or by a liquid), we may determine the velocity of waves in it. (This is the.standard", " method for all ^ascs which can be secured in a small quantity only, such as helium, argon, and the other new gas\u00ab The values of the velocity of compressional waves in a few substances is given in the following table : Air \u2022 \u2022 \u2022. 0\u00b0 C.... 33,140 cm. per second Hydrogen... 0\u00b0 C.... 128,600 cm. per second I Humiliating gas.. 0\u00b0 C.... 49,040 cm. per second Oxygen. 0\u00b0 C.... 31,7i>0 cm. per second...| (abx.lnt.-). 8\u00b0.4C.... 126,400 em. IMT. second :-nm... 73.4C.. i Water.. 8\u00b0 C.... 1 1.V)<H> em. i>er second Braes 350.000 cm. per second < \"pper... 20\u00b0 C.... 356,000 cm. per second 600,000 to 600,000 cm. per second. 20\u00b0 C.... f> 13.000 cm. IMT second tfin... 16\u00b0 C.... 130,400 cm. per second CHAPTER XX HUYGENS'S PRINCIPLE. REFLECTION AND REFRACTION Huygens's Principle. \u2014 One of the most important theorems in regard to wave motion is in part due to Huygens, and it is called by his name. In its most general form it is com- plicated, and can be demonstrated only by the aid of the infinitesimal calculus. We shall give certain special applica- tions of it ; and, although the statements to follow are not rigorous, they are sufficiently so for all present purposes. We cannot do better than to use Huygens's own language, as it appears in the translation of his TraitS de la Lumiere by Crew in The Wave Theory of Light (New York, 1900). Huygens's treatise was written in 1678, but was not pub- lished until 1690. \" In considering the propagation of waves, we must remem- ber that each particle of the medium through which the wave spreads does not communicate its motion only to that neigh- bor which lies in the straight line drawn from the luminous point, but shares it with all the particles which touch it and resist its motion. Each particle is thus to be considered as the centre of a wave. Thus, if DCF is a wave whose centre and origin is the luminous point A, a particle at B, inside the sphere DCF, will give rise to its own individual [secondary] wave, KCL, which will touch the wave DCF in the point (7, at the same instant in which the principal wave, origi- nating at A, reaches the position DCF. And it is clear that there will be only one point of the wave KCL which will touch the wave DCF, viz., the point which lies in the straight line from A drawn through B. In like manner, each of the 366 REFLECTION AND REFRACTION 867 other particles, bbbb, etc., lying within the sphere DCF, gives \\ n wave. The intensity of each of these waves may, however, be infinitesimal com- pared with that of I) OF, which is the resultant of all those parts of the other waves which are at a maxi- mum distance from the centre A. \" We see, moreover, that the wave DCF is determined by the extreme limit to which the motion has traveled from the point A within a certain interval of time. For there is no motion beyond this wave, whatever may have been produce (I inside by those parts of the secondary waves FIG. 165. which do not touch the sphere DCF.\" We shall also quote Huygens in his explanation of reflec- tion and n-fractioii. 1. Reflection of Plane Waves by a Plane Mirror. \u2014 \" Having explained the effects produced by light waves in a homogene- ous medium, we shall next consider what happens when they impinge upon other bodies. First of all we shall see how reflection is explained by these waves, and how the equality of angles fol- lows as a consequence. l.\u00ab-t AB represent a plane polished surface of some metal, glass, or other sub- stance, which, for the | Ft... 1M. ent, we shall consider as perfectly smooth (concern- in^ iiT(\"_nil;int i> - which are unavoidable, \\\\.- shall have some- thing to sav at the close of this demonstration); and let the liii..I/..', inclined to AB, represent a part of a light 368 VIBRATIONS AND \\YAVES wave whose centre is so far away that this", " part, AC, may be considered as a straight line. It may be mentioned here, once for all, that we shall limit our consideration to a single plane, viz., the plane of the figure, which passes through the centre of the spherical wave and cuts the plane AB at right angles. \" The region immediately about 0 on the wave A 0 will, after a certain interval of time, reach the point B in the plane AB, traveling along the straight line CB, which we may think of as drawn from the source of light, and hence drawn perpendicular to AC. Now, in this same interval of time, the region about A on the same wave is unable to transmit its entire motion beyond the plane AB ; it must, therefore, continue its motion on this side of the plane to a distance equal to CB, sending out a secondary spherical wave in the manner described above. This secondary wave is here rep- resented by the circle SNR, drawn with its centre at A and with its radius AN equal to CB. \" So, also, if we consider in turn the remaining parts, H, of the wave AC^ it will be seen that they not only reach the surface AB along the straight lines HK parallel to CB, but they will produce, at the centres, K, their own spherical waves in the transparent medium. These secondary waves are here represented by circles whose radii are equal to KM, that is, equal to the prolongations of HK to the straight line BGr, which is drawn parallel to AC. But, as is easily seen, all these circles have a common tangent in the straight line BN, viz., the same line which passes through B and is tangent to the first circle having A as centre and AN, equal to BC, as radius. \" This line BN (lying between B and the point N, the foot of the perpendicular let fall from A) is the envelope of all these circles, and marks the limit of the motion produced by the reflection of the wave AC. It is here that the motion is more intense than at any other point, because, as has been RE I '!.!\u2022:< T/o.Y l.\\/> UEFRACTION 369 explained. HX is tlie new position which tin? wave AC has assumed at the instant when the point C has reached B. For there is no other line which, like BN, is a common tan- gent to these circles... \" It is now evident that the angle of reflection is equal to the angle of incidence. For the right-angled triangles ABC and BXA have the side AB in common, and the side CB equal to the side NA, whence it follows that the angles oppo- site th.-se sides are equal, and hence also the angles CBA and NAB. But CB, perpendicular to CA, is the direction of the incident ray, while AN, perpendicular to the wave BN, has the direction of the reflected ray. These rays are, therefore, equally inclined to the plane AB. \" I remark, then, that the wave A C, so long as it is con- sidered merely a line, can produce no light. For a ray of light, however slender, must have a finite thickness in order to be visible. In order, therefore, to represent a wave whose path is along this ray, it is necessary to replace the straight line AC by a plane area... where the luminous point is supposed to be infinitely distant. From the preceding proof it is easily seen that each element of area on the wave [front], having reached the plane AB, will there give rise to its own secondary wave ; and when C reaches the point B, these will all have acommon tangent plane, vi/.., [a plane through BN~\\. This [plane] \\\\ill he cut... at right angles hy the same plane which thus cuts the [wave front at A C at right angles, i.e. the plane of incidence]. \"It is thus seen that the spherical second a r\\ waves can have no common tangent plane other than BN. In this plane will he located more of the reflected motion than in any other, and it will therefore receive the light transmitted from the \\\\.ive CH. ******* VM! -'\u00ab I II \u2022, - i 370 VIBRATIONS AND WAVES \"We must emphasize the fact that in our demonstration there is no need that the reflecting surface be considered a perfectly smooth plane, as has been assumed by all those who have attempted to explain the phenomena of reflection. All that is called for is a degree of smoothness such as would be produced by the particles of the reflecting medium being placed one near another. These particles are much larger than those of the ether, as will be shown later when we come to treat of the transparency and opacity of bodies. Since, now, the surface consists thus of particles assembled together, the ether particles being above and smaller, it is evident that", " one cannot demonstrate the equality of the angles of inci- dence and reflection from the time-worn analogy with that which happens when a ball is thrown against a wall. By our method, on the other hand, the fact is explained without difficulty. \" Take particles of mercury, for instance, for they are so small that we must think of the least visible portion of sur- face as containing millions, arranged like the grains in a heap of sand which one has smoothed out as much as possible ; this surface for our purpose is equal to polished glass. And, though such a surface may be always rough compared with ether particles, it is evident that the centres of all the second- ary waves of reflection which we have described above lie practically in one plane. Accordingly, a single tangent comes as near touching them all as is necessary for the pro- duction of light. And this is all that is required in our demonstration to explain the equality of angles without allowing the rest of the motion, reflected in various direc- tions, to produce any disturbing effect.\" The law of reflection in regard to the equality of the angles of incidence and reflection was known to the ancients, and was made use of by Euclid as early as 300 B.C. He also deduced some of the properties of concave mirrors. The law stating that the normals to the two waves and the surface lie Hl-:i-'LK< TIOH AND REFRACTION 371 in a plane was first given by the Arabian scholar Al Hazen, about 1000 A.I). 2. Refraction of Plane Waves at a Plane Surface. \u2014 Again quoting from Huygens : u In order to explain these phe- nomena on our theory, let the straight line AB, Fig. 10, represent the plane surface bounding a transparent body extending in a direction between 0 and N. \" By the use of the word plane we do not mean to imply a perfectly smooth surface, but merely such a one as was employed in treating of reflection, and for the same reason. Let the line AC represent a part of a light wave, whose source is so distant that this part may be treated as a straight line. The region (7, on the wave AQ will, after a certain interval of time, arrive at the plane AB, ;il<niLT the straight line CBi which we must think of as drawn from the source of light, and which will, therefore, intersect AC at right angles. But during this same interval of time, the region about A would have arrived at Q-, along the straight line AG. equal and parallel to CB\\ and, indeed, the whole of the wave AC would have reached the position Q-B* provided the transparent body were capable of transmitting waves as rapidly as the ether, lint suppose the rate of transmission is less rapid, say one third less. Thru the motion from the point A will extend into the transparent body to a distance which is only two thirds of CB. while producing its secondary spherical wave as described above. This wave is repr. ^. nted by the circle vV// mitre is at A and whose radius is equal to } CB. Fio. 167. It' \\\\e consider in like manner the. other points /STof the wave 372 VIBRATIONS AND WAVES AC, it will be seen that, during the same time which 0 cm- ploys in going to B, these points will not only have reached the surface AB, along the straight lines HK, parallel to OB, but they will have started secondary waves into the trans- parent body from the points Kas centres. These secondary waves are represented by circles, whose radii are respectively equal to two thirds of the distances KM ; that is, two thirds of the prolongations of HK to the straight line B G-. If the two transparent media had the same ability to transmit light, these radii would equal the whole lengths of the various lines KM. \" But all these circles have a common tangent in the line BN\\ viz., the same line which we drew from the point B tangent to the circle SNR first considered. For it is easy to see that all the other circles from B up to the point of contact N touch, in the same manner, the line BN, where N is also the foot of the perpendicular let fall from A upon BN. \" We may, therefore, say that BN is made up of small arcs of these circles, and that it marks the limits which the motion from the wave AC has reached in the transparent medium, and the region where this motion is much greater than anywhere else. And, furthermore, that this line, as already indicated, is the position assumed by the wave AC at the instant when the region C has reached the point B. For there is no other line below the plane AB, which, like BN, is a common tangent to all these secondary waves....", " \" If, now, using the same figure, we draw EAF normal to the plane AB at the point A, and draw DA at right angles to the wave AC, the incident ray of light will then be repre- sented by DA ; and AN, which is drawn perpendicular to BN, will be the refracted ray; for these rays are merely the straight lines along which the parts of the waves travel. \"From the foregoing, it is easy to deduce the principal law of refraction; viz., that the sine of the angle DAE always bears a constant ratio to the sine of the angle NAF, ui:rii:moN AND i;i-:n;.i< TION 373 whatever may be the direction of the incident ray, and that tin- ratio is the same as that which the speed of the waves in the medium on the side AE bears to their speed on the side AF. \" For, if we consider AB as the radius of a circle, the sine of the angle BAC is BQ and the sine of the angle ABN is AN. But the angles BAC and DAE are equal, for each is the complement of CAE. And the angle ABN is equal to NAF, since each is the complement of BAN. Hence the sine of the angle DAE is to the sine NAF as BO is to AN. But the ratio of BCto AN is the same as that of the speeds of light in the media on the side toward AE and the side inward AF, respectively; hence, also, the sine of the angle DAE bears to the sine of the angle NAF the same ratio as these two speeds of light.\" Since these speeds are properties of the media and not of the direction of the propagation of the waves, we have at once the law that the ratio of these sines is independ- ent of the angle of incidence. It is evidently different for different media, and will be shown to be different for waves of different wave length in the case of ether waves in a material medium. This law of refraction was first discovered experimentally by Snell (1591-1626). I! --fraction is a much more common phenomenon with ether waves than with air waves, and it. will be disenss.-d more fully in the section devoted t<> Light. CHAPTER XXI INTERFERENCE AND DIFFRACTION Interference Young's Experiments. \u2014 A most important phenomenon of wave motion, and one of particular interest historically because by means of its discovery Thomas Young in 1801 proved that light was due to waves, is wli.it is called inter- ference. In Young's own words: \"When two undulations, from different origins, coincide either perfectly or very nearly in direction, their joint effect is a combination of the motions belonging to each. \" Since every particle of the medium is affected by each undulation, wherever the directions coincide, the undulations can proceed no otherwise than by uniting their mo- tions, so that the joint motion may be the sum or difference of the separate motions, accordingly as similar or dissimilar parts of the undulations are ;. 16S. \u2014 Interference of waves on the surface of a liquid, which are sent out by two point sources. coincident.\" (This prin- ciple is illustrated in the cut, which represents the \" interference \" of two trains of water waves.) Two experiments may be described ; both are due to Young, and both may be performed easily with home-made apparatus. We shall describe them as if the waves to be 374 7JV TK /,\u2022/\u2022*/\u2022:/.' i:\\rf-:.i.v/> inrn;.ii IION 375 studied were light waves; but the same apparatus, suitably enlarged, would do equally well for air waves. Let there be trains of waves sent out by having some source placed near a long Marrow slit in an opaque screen. If the slit is sufficiently narrow, the disturbances will proceed out from the slit in all directions, making a train of waves with a cylindrical wave front. A second opaque screen with two X' if row slits, which are close together and both of which are parallel to the slit in the lirst screen, and at equal distances from it, is placed parallel to the latter. As the cylindrical waves reach these two slits, two cylindrical trains of waves are produced beyond the second screen. The importance of this arrangement lies in the fact that the two trains of waves thus produced are identical; that is, they have the same amplitude, the same wave length, and the same phase, because they arc produced by disturbances in the same wave front at the same distance from tin- lirst slit; so, if the original source of the waves changes its character in any way, the two cylindrical waves from the two slits both change in the same", " manner at the saint- instant. Then, if we consider the effect at any point in the space which is traversed l>y tin- two trains of waves, it is receiving disturb- ances from both waves, and the effect produced is the sum of two, one due to each train. Another mode of producing this result is to remove the screen with the two slits, and to place parallel to the slit in the first screen a narrow opaque object like a line wire or small needle. As will be shown in speaking of diffraetion (see page 880). disturbances are produced in the low. \\ l( tly as if there were a source of waves along the edge of the obstacle ; so, in the case of the wire or needle. the points in the shadow are receiving distnrhanees from two parallel line sources,,f waves along the two ed: These two disturban then, in this case also due to 1 \\\\ o '{\u2022'iff trains of Vfl 376 VIBRATIONS AND WAVES If these waves are light waves of a definite color, and if, from a point in the medium traversed by the two trains of waves, one looks in the direction of the two slits (or of the wire), or, better still, if a magnifying glass is used in front of the eye, a series of black and colored lines parallel to the slits (or wire) are seen. Similarly, if short sound waves are used, it is not difficult to prove by means of a sensitive flame (see page 192) that there are corresponding \" bands of silence and sound,\" meaning that at points along a line parallel to the slits there are disturbances in the air, while along a neighboring line there are not. B C A C FIG. 169. \u2014 Diagram of Young's Interference experiment. 0t and Ot are two sources of waves and ACi& a screen on which the two trains of waves are received. U The explanation is not difficult. Let us consider the effect at various points on a screen parallel to the plane of the two slits. Let Ol and 02 be the traces of the slits on the paper, and A C that of the last screen. Let B be a point halfway between the slits ; draw a line BA perpendicular to the screens, and let 0 be any other point on the screen. This point receives disturbances due to two trains of waves ; but the lengths of the paths from C to the two sources 01 and 02 are not the same. This is shown on a large scale in part of the diagram. i.\\TKi:ri-:nENCE AND DIFFRACTK>\\ 877 FKU n 0.2 draw O.J* perpendicular to the line 0-^C. Since OL a jid 0.2 are iii reality extremely close together compared with the other distances in the apparatus, O^P is the differ- ence in path from 01 and 02 to C. If it amounts to a wave length exactly, or to any integral number of wave lengths, tin- disturbances reach O in the same phase, and so the effect is great ; but, if this difference in path is exactly half a wave length, or any odd number of half wave lengths, the disturb- ances arrive at O in exactly opposite phases, and so there is no effect. At the point A the two paths are of equal length, so the effect is great ; and as points near it are considered, constantly receding from A in either direction, the effect decreases, becomes zero when the difference in path is half a wave length, increases to a maximum when this difference is a wave length, decreases to zero again; etc. The effect at all points on a line through C parallel to the two sources is evidently the same ; and so the screen is covered with a pat- tern of bands, or. as they are called, \"interference fringes,\" alternately dark and bright. The condition for a fringe where there is a maximum effect is, in terms of the figure, that where n is any whole number 0, 1, 2, 3, etc. ; and I is, as usual, the wave length. Similarly, the condition 1m a fringe of zero intensity is that O^P = (2 n+ 1 > _>- 2 These conditions may be expressed in t< -rms of the distance of ih\u00ab- friii'_r'- from tin- mitral one at A ; i.e. in terms of the distance AC. Call this distance x: the distance apart of th\u00ab- two screens, i.e. AB, a \\ thr di>taiir\u00ab- apart of the two sources 010a, 6; and tin- angle (ABC), N. It is to be re- mrmlirivd!l,;it this angle is always small, and that the two soiin-.-s", " are so close tOgethrr that the lim-s drawn from O to tin- points Or /f, and 03 make practically the same angle 378 VIBRATIONS AND WAVES with AB. Then, referring to the diagram, the angles (ABC) and (OjOjP) are equal, A0= AB tan (ABC), and O^P = 0X02 sin (^i^2-P); or, in symbols, or and hence 0*P = _ Therefore, if \u00a3 is small com- Va2 + z2 fo pared with a, 0XP = \u2014. \\i x gives the position of a fringe of maximum intensity, nl = \u2014, or x = \u2014 \u2014. The position of i i a al al the next similar fringe is given by xl = (n 4- 1) \u2014 - ; so the dis- tance apart of the two, or xl \u2014 x, is \u2014. This shows that all the bright fringes are equidistant. Similarly, if x gives the distance of a fringe where there is a zero effect, The distance of the next similar fringe is given by so their distance apart, or x1 \u2014 x, is \u2014, the same as for the fringes of maximum intensity. It is thus seen that the greater the wave length, the farther apart are the fringes ; and that, if their distance apart, that of the screens, and that of the sources are known, the wave length may be determined. This matter will be referred to later. It should be noted that in the formation of these interfer- ence fringes there is no destruction of energy ; it is simply distributed differently from what it would be if the screen t: A.\\h Dll-'Fl; ACTION 379 were receiving waves from* two sources which had no perma- nent phase relation. < rfertno* frtagM irtrttlntil by Young's method. 380 VIBRATIONS AND WAVES Other cases of interference will be described in the section devoted to Light, but all interference phenomena dealing with light can be reproduced with waves in the air. Diffraction FresnePs Principle. \u2014 It is a well-known fact that, if an opaque obstacle is interposed in a beam of light, a shadow will be cast on any suitably placed screen, which is more or less sharply defined, depending upon the smallness of the source of the light. This is sometimes expressed by saying that \"light travels in straight lines.\" In the case of sound, however, such an obstacle would not prevent a noise being heard behind it ; in other words, there is no sound shadow with such an obstacle. The explanation of the difference in the two cases was given by Fresnel, making use of Huygens's principle. It may be well to quote Fresnel's own words in Crew's translation. Fresnel's great memoir on Diffraction, from which these quotations are made, appeared in 1810. \"I shall now show how by the aid of these interference formulae and by the principle of Huygens alone it is possible to explain, and even to compute, all the phenomena of dif- fraction. This principle, which I consider as a rigorous deduction from the basal hypothesis, may be expressed thus: The vibrations at each point in the wave front may be considered as the sum of the elementary motions which at any one instant are sent to that point from all parts of this wave in any one of its previous * positions, each of these parts acting inde- pendently the one of the other. It follows from the principle of the superposition of small motions that the vibrations pro- duced at any point in an elastic fluid by several disturbances * I am here discussing only an infinite train of waves, or the most general vibration of a fluid. It is only in this sense that one can speak of two light waves annulling one another when they are half a wave length apart. The formulae of interference just given do not apply to the case of a single wave, not to mention the fact that such waves do not occur in nature. /A'/-/-;/,- /\u2022\u2022/\u2022;/,\u2022 /-;.v< B.i.v/> inrni ACTION 381 are equal to the resultant of all the disturbances reaching this point at the same instant from different centres of vibra- tion, whatever be their number, their respective positions, their nature, or the epoch of the different disturbances. This general principle must apply to all particular cases. I shall suppose that all of these disturbances, infinite in num- ber, are of the same kind, that they take place simulta- neously, that they are contiguous, and occur in the single plane or on a single spherieal surface.... I have thus reconstructed a primary wave out of partial [secondary] dis- turbances. We may, therefore, say that the vibrations", " at each point in the wave front can be looked upon as the resultant of all the secondary displacements which reach it at the same instant from all parts of this same wave in some previous position, each of these parts acting independently one of the other. \"If the intensity of the primary wave is uniform, it follows from theoretical as well as from all other considerations that this uniformity will be maintained throughout its path, pro- vided only that no part of the wave is intercepted or retarded with respect to its neighboring parts, because the resultant of the secondary displacements mentioned above will be the same at every point. But if a portion of the wave be stopped by the interposition of an opaque body, then the intensity of each point varies with its distance from the edge of tin- shadow, and these variations will be especially marked near the edge of the geometrical shadow.\" Rectilinear Propagation. \u2014 As a simple case, consider a train of plane waves advancing from left to right; let the paper be at ri.rht angles to them and let the trace on the the wave front at any instant be given in part by AB. The etl'eet at any later time at a point P in advance of the waves is determined, as already Stated, by deducing the effects there owing to the secondary waves from ea< -h point of the wave fn.nt. and adding these geometrically VIBRATIONS AND WAVES It is evident that the effect at P of the secondary waves from any point Q on the wave front depends upon the length of the line QP for two rea- sons : the decrease in amplitude of spherical waves varies inversely as the distance, and the phase of the disturbance as it reaches P varies with it. There- fore, if 0 is the foot of the perpendicular let fall upon the FIG. 171. \u2014 AB is a section of a plane wave advancing toward P. B wave front from P, i.e. its \"pole,\" as it is called, and if a circle with a radius equal to OQ be drawn on the wave front around 0, the secondary waves from each of the points on this circle will reach P with the same amplitude and in the same phase, because they start out with the same amplitude and in the same phase, and travel the same distance to reach P. But the directions of the displacements due to the sepa- rate secondary waves are not the same, and they must be added geometrically. (Since the phases are all the same, we have simply a case of vector addition.) Let us assume that the waves are longitudinal (the proof is similar, if they are transverse). Let Q1 and Q2 be two points at the end of a diameter of the circle round 0\\ and let the displacement at P due to the secondary waves from Q1 be represented by FIG. 172. \u2014 Diagram to represent the resultant action at P of all the secondary waves from points <?,, Qt, etc., on a circle around 0. i\\TKi;rEi;i-:\\f'E A.\\D DIFFRACTION 383 then that due to the secondary waves from Q2 is repre- sented by PA2. Their resultant is PA^ a displacement perpendicular to the plane wave front. Similarly, the resultant displacement due to all the secondary waves coming from points in the circle around 0, through Qr is proportional to PA. But calling the angle (OPQ^) the \" inclination \" of Qv and representing it by jY\", it is evident tliat PA = 2 PAl cos N; and therefore, as the inclination increases, the resultant displacement decreases. The task of compounding the effects at P, due to the secondary waves from all the points in the wave front, is not at first sight a simple one ; but Fresnel invented a most brilliant method. His plan is as follows : Calling the length <>f the line OP a, and the wave length of the waves /, describe around P as a centre a series of spherical surfaces of radii a, a -f- -, a -M, a H- \u2014, etc. ; these spheres will cut the wave 7 07 2 2 front in a series of concentric circles around 0, thus dividing the plane into a series of concentric zones or circular ri and the effect at P due to the secondary waves from each zone is considered as a whole. Fresnel's purpose in thus dividing the wave front into these particular zones was to secure the following condi- tion : If we consider the disturbance at P from any point in any /.one whose distance from P is 6, this will be exactly opposite in phase to disturbances reaching P at the same instant from certain points in the two emit i^mns /<>nes, tin- distances from which to P are b + - and ft \u2014 -; for. if two 2 2 waves differ by half a", ", in Fig. 176, the points P1 and B are the - represented by the same symbols in the previous cut. It is seen that the effect in the geometrical shadow is as if there were a source of waves at the point A, the edge '->!' the opaque obstacle. Fi... 177. \u2014 Photograph of diffraction bands near the geometrical shadow of a sharp edge. Naturally, everything depends upon the magnitude of the wave length of the waves. If this is small, as in light, the waves penetrate a very short distance into the ir.-omrtrieal shadow; or, better, tin- intensity of the waves decreases so rapidly inside the shadow that they can be percei\\<-d for only a short distance; and the region \u00ab\u00bbutsidi- the shadou in \\\\ Inch there are variations in the intensity, i.e. where there are \"bands,\" ii extremely limited. P.oth ()f these phenomena may be easily observed in the case of a shadow cast on a screen by any opaque obstacle if the source of light is small. 390 VIBRATIONS AND WAVES If we are dealing with air waves about 1 cm. long, all of the above phenomena, as described for light waves, are easily observed by the use of a sensitive flame ; and, if the waves are several metres long, \" sound shadows \" may be observed if the obstacle is sufficiently large, e.g. a mountain. But we see that, if the waves are long and the obstacle is of ordinary size, the former will penetrate a great distance in the shadow. This phenomenon of the peculiarities of a shadow produced by trains of waves incident upon an obstacle is called \" dif- fraction.\" It was first described in the case of light by the Italian priest Grimaldi in 1666 ; but its explanation was given by Fresnel. Diffraction through a Small Opening. \u2014 An important illus- tration of diffraction is afforded when a train of waves falls upon an opaque obstacle which has in it a single small open- ing. The general features of the phenomena may be deduced easily. Let the waves come from such a direction that their wave front is tangent to the plane of the opening. Draw the line OB perpendicular to this plane, and OP oblique to it ; we shall consider the effect at various points on a screen perpendicular to OB. The effect at B depends upon the number of zones that can be drawn in the small opening at 0; if it is even, the effect is zero; if it is odd, the effect is practically as great as if the whole obstacle were removed, viz., \u00a3 mv The effect at P in a similar manner depends upon the number of zones that can be drawn for it in the opening. The centre of the zones is the pole of P on the wave front, Fio. 178. \u2014Diffraction through a small opening 0. / \\ / /\u2022 /; /\u2022\u2022/;/; /\u2022; \\ i /\u2022;. i \\ i> DIFFRA GT/OJV* 391 and is therefore far from the opening ; and, since the edges of the zones get closer and closer as one recedes farther from the centre of the zones, P will have more zones \u2014 or rather portions of them \u2014 included by the opening than does B. So, if B has three zones in the opening, it will receive a maxi- mum effect; and, if P is sufficiently far away from B, it will have portions of four zones included, and so will receive a minimum effect ; farther out still, there will be a point for which there will be portions of five zones in the opening, and which accordingly receives a maximum effect; this is, however, much less than that at BI owing to the intimation of the zones and t<> the fact that only n:irr.-\\v -lit. \u2022 I a Mi, \u00bb.'!\u2022\u2022 'ton* of zones are intercepted by the opening; etc. Con- s' the opening 18 circular, the \"diffraction pattern \" mi tin- sen-en will cMiisist of concentric circular \"hands\" in which the intensity is alternately a maximum and minimum, hut which rapidlv fade in intensity as one recedes from their centre at B. It is also <-\\ ident from the same re, that 392 VIBRATIONS AND WAVES the smaller the opening, the farther apart are the bands; and. if the opening is sufficiently small, B will receive a maximum effect, and the first ring of minimum intensity recedes to an infinite distance ; so all the points on the screen receive dis- turbances from the opening just as if it were a minute source of waves (see page 375). This is a familiar fact, because if a small \"pin hole\" is made in a card- board or in any thin opaque screen, and a light of any kind is placed behind it, it appears bright when viewed", " from any direction on the far- ther side. (This is, of course, entirely apart from any light reflected from the edges or sides of the opening.) Similarly, if a room has a small open- ing in it, and if a tuning fork is vibrating in the room, it may be heard outside in all directions from the opening.. The treatment just given of diffraction past an edge and through a small opening must not be regarded as rigorous. It is based upon many assumptions, which have been pur- posely omitted, and which are not all justified. On the whole, however, the results are sufficiently accurate for our present purposes. Criteria of Wave Motion. \u2014 The existence of any of the phenomena that have been described in the foregoing arti- cles\u2014 vibrations with nodes and loops, diffraction, inter- ference \u2014 is proof positive of the presence of wave motion. This fact will be made use of immediately in describing and explaining the phenomena of Sound and Light. SOUND CHAPTER XXII ANALYSIS OF SOUND Fundamental Facts. \u2014 To one who has the sense of hearing tin- word --sound\" has a definite meaning, as describii.dn sensation, but it is impossible to define it so as to convey an idea to one who is born drat'. The cause of this Cation can always be traced to some body that is vibrat- ing rapidly. Thus, if we hear u sound from a tuning fork, a piano string, a metal bell, etc., it is a simple matter to prove that they are vibrating, and that, when tin- motion ceases, the sound does also. Again, it is a familiar fact that some time elapses between the instants when tin- vibration begins and the sound is heard, and between those when the former stops and the latter ceases; for there is a considerable inter- val of time between the instants when a distant gun is seen to be fired and when the sound of the report is heard, or when a mt steam whistle is seen to blow and when the noise is heard, etc. This proves that \\\\ a uses the sensation time for its transmission through space. The fact that the product ion of the sensation depends upon the pres- ence of a vmtt' rf'il medium between the vibrating bnd\\ and the ear may be proved by suspending the vibrating body in a space from which the air may be more an* 1 more exhausted ; as this is removed, leaving only the ether, the sound becomes less and less intense. Vibrating bodies will produce com- pre\u00bbioiiul waves in a surrounding fluid, provided the fre- quency of vibration is sufficiently great; and the fact that SOUND the sensation of sound is due to these waves may be proved most simply by allowing them to produce vibrations with nodes and loops, and showing that they cause sounds. Methods of doing this will be described in a few pages. We shall discuss first the characteristics of different sounds and the physical cause of these differences, then describe a few typical musical instruments and some acoustic phe- nomena, and finally give the physical explanation of harmony in musical compositions, with a brief description of musical scales. Noise and Musical Notes. \u2014 If we analyze our sensations of sounds, we are led at once to recognize two great classes which we call in ordinary language noises and musical notes. The latter have all the characteristics of periodic motion ; they are continuous and uniform in character, and are pleas- ant to the ear. The former are discontinuous, with abrupt changes, and are often extremely disagreeable to -the ear. Thus, the sounds due to a tuning fork, to a piano string, to the column of air in an organ pipe, etc., are musical notes. But the sounds heard when a piece of paper is torn, when a wagon rolls over cobblestones, when a slate pencil is sharpened, etc., are noises. We can study the nature of the vibrations of a body, as has been explained on page 319, and it is found that a musical note is always due to a periodic vibration ; a noise, to an extremely complex motion, consisting of differ- ent vibrations which differ slightly in frequency and which are rapidly damped. A confused vibration which causes a noise will produce a musical note, if the vibrating body is near a flight of steps ; for, when the pulse reaches the first step, a reflected pulse is produced; and in a similar manner others are produced when it reaches in turn the other steps. Therefore there will be in the air a series of reflected pulses at exactly equal inter- vals apart; and, as they reach the ear, a musical sound is heard. Thus, if one claps one's hands near a staircase, the noise is first heard, but it is followed immediately by a musical note. The same phenomenon occurs if u. noise is produced near a picket fence. ANALYSIS \"1 SOUND'395 Simple and Complex Notes; Quality. \u2014", " If we analyze our sensations of different musical notes. \\\\c (.ID >eparate them into two classes: one we call \"pure\" or \"simple\"; then: \"complex.\" Thus, the note produced by vibrating metal plates like cymbals is complex, while that due to a tuning fork <>r to a stopped organ pipe is pure. In fact, all notes, with the exception of these last, are more or less complex. If we examine the corresponding vibrations, it is found that a pure note is always due to a simple harmonic vibration, and a complex note to a complex vibration. Complex vibrations can be analyzed, in accordance with Fourier's theorem, into simple harmonic components who>e frequencies are in the ratio 1:2:3: etc., or into other com- ponents not so related. In a similar manner, if one listens attentively to a complex sound, various simple pure notes may be distinguished. (This statement that the human ear analy/.es mechanically a complex wa\\c into simpler com- ponents and hears the corresponding simple notes separately is known as Ohm's Law for Sound. ) If two complex notes differ, it is found that the corre- sponding complex vibrations differ; but t he converse state- ment that two different complex vibrations produce t\\\\o different complex notes is true only with one limitation. If the component parts of the two complex vibrations differ in their frequencies or in their amplitudes, the corresponding notes are different ; but the ditVerenees in phase between the coni| -nts may be different in the two complex vibrations or may be the same; they have no influence on the note. (This is seen to be in ace. n.l with ohm's law, as stated above.) Two different complex notes are said in \"quality.\" Thus, the complex notes produced by the vibrations of the column of air in an organ pipe, of a violin string, of a piano siring, of the column of air in a horn, etc., all differ in qualit) ; and it is by this property that we recoglii/e the 396 SOUND nature of the source of the sound. This quality depends upon the number and amplitudes of the other component vibrations besides the fundamental present in the complex vibrations ; it does not depend, however, as has been already said, upon the relative phases of these component vibra- tions. Pure notes are never used in music, because -they lack what may be called \"character,\" or individuality. Notes that are useful for musical compositions must be complex to a certain extent, seven or more components often being present. Analysis of Notes. \u2014 This process of analyzing a complex sound or a complex vibration into its harmonic components is greatly helped by the use of resonators. Helmholtz in his epoch-making work used those of the form shown in the cut. (One of their advan- tages is that when the inclosed air is set in vibration, the motion is simple harmonic.) He constructed a set of them, and accurately determined the frequencies of the vibrations of their inclosed volumes of air. Then by bringing them in turn near a vibrating body, he could tell by holding one of the ends of the resonator near the ear whether the par- ticular vibration that corresponded to that of the air in the resonator was present in the complex vibration of the body ; for, if it were, the corresponding sound would be intensified. In this manner it is a simple matter to detect the components. In the best and most recent work on analysis of sounds, phonographs are used, and the traces are magnified. FIG. 180. \u2014 A Helmholtz resonator. A\\AL}'SIS OF SOUND 397 Several other illustrations of resonance are worth mentioning. When a large seashell or a vase is lu-M near the ear, the roaring that is heard is due to the resonance of the inclosed air produced by certain sounds in tin- room. The sound may be varied by partly closing the opening of the sht'll or vase. The passage leading from the outside of the head into the eardrum forms a small resonator, and its action is often noticed when one is li>t\u00abMiing to an orchestra, by the strong resonance of certain very shrill sounds like the buzzing of insects. Helmholtz performed with his set of resonators the con- verse of the analysis of a complex sound; he produced one by HUM MS of the superposition of simple harmonic vibrations or of pure notes. He arranged in front of each resonator a tuning fork whose frequency of vibration was the same as that of the air in the resonator, and adjusted electro-magnets to these forks in such a manner that he could set them vibrating and maintain them in motion. He could also alter their amplitude. Then by making different forks vibrate he able to produce different complex sounds, and in fact to imitate the sounds", " characteristic of different instruments. In a complex sound, the component note corresponding to the fundamental vibration is called the \"fundamental\"; and other notes \"overtones.\" If the component vibrations form a harmonic series, the component notes are also called k* harmonics.\" Pitch and Loudness. \u2014 If we compare two simple notes, we recogni/e the fact that they may differ in two ways, in shrillness and in loudness. Thus, the notes of a piccolo are shriller than those from an nr-^an pipe; and any note of an in may keep the same shrillness and yet may vary in loudness. If we compare the < : ions, we tind that in every case if one note is shrill another, the frequency of its vibration, or rather the number \u2022 reaching the ear in a unit of time, is the greater. (This -st observed by G Kurt her, we find that ne note deoreaaei in Umd increases in amplitude, other things remaining unchanged. 398 We cannot measure the shrillness of a note, because we cannot imagine a unit of shrillness nor the idea of shrillness being made up of parts which can be compared. We can, however, give a number to the shrillness of a note by assign- ing it one equal to the frequency of the vibration, if the vibrating body and the observer are at rest relatively; or, more generally, we assign a number equal to the number of waves reaching the ear in one second. (See Doppler's Prin- ciple, page 345.) This number is called the \"pitch\" of the sound. Similarly, the pitch of a complex note is denned to be the pitch of its fundamental. The loudness of a sound, either pure or complex, varies as the intensity of the waves producing it. It is thus seen why, when a sounding body approaches the ear, the loudness of the sound heard increases. We can measure.the intensity of the waves, but we have no method of measuring the loudness, for this is a sensation, and not a physical quantity. Audibility of Waves. \u2014 In order to produce waves in the air, the frequency of the vibration must exceed a certain limit, as has been explained ; otherwise the air flows, but is not com- pressed. But all waves in the air do not affect our sense of hearing ; for this sense depends upon disturbances being conveyed to the nerve endings from the external air by a mechanism whose parts are set in motion by waves of certain wave lengths, and not by others. Thus, it is found that waves whose wave number is greater than 20,000 per second or less than 30 do not in general produce sounds ; but, of course, these limits are only approximate, and vary greatly with different individuals. In musical compositions as played by orchestras the maximum range of pitch is about from 40 to 4000. The Human Ear. \u2014 For a full description of the human ear reference should be made to some treatise on Physiology ; it is necessary to mention only a few details here. The ear consists of three parts : the external ear, which ends at the ear-.1 YALY818 OP SOtTJTD (Inini: the- middle car, which is connected with the throat and mouth I iy a tube, and in which there arc three little hones witli Hex il)lc connect ions, tli us making a mechanism joining the drum to a membrane which closes one opening into the third por- tion of ear; the inner ear, which is entirely inclosed in the hone of the skull, and which consists of several cavities tilled with a liquid. In one of these cavities there is a minute fibres of regularly decreasing length, with which the nerve endings of some of the auditory nerves are connected, and which are thought to play the part of reso- nators for musical notes. Other branches of the auditory nerve end in another cavity under conditions which have led several scientists to believe that their function was to respond to noises. In any case it is easy to trace a mechanical con- nection between the waves in the air and the nerve endings through the eardrum, the three bones, and the liquid in the inner ear. The student will find in the work of Helmholtz a full discussion of these various steps. CHAPTER XXIII MUSICAL INSTRUMENTS WE shall discuss only two types of musical instruments: stringed and wind instruments. The commonest stringed instruments are the piano, the violin, the violoncello, and the harp; and the commonest wind instruments are the organ, the flute, and the horn. Stringed Instruments. \u2014 For present purposes we may regard the vibrations of the strings in any stringed instru- ment as being identical with those of a perfectly flexible cord, although in reality musical strings are far from being per- fectly flexible, and their elasticity plays", "aking small changes a pipe may be \" tuned\" accurately. If the pipe is not of uniform cross section, but conical, there are marked differences produced \u2014 the position of the nodes and loops is affected ; as is shown by the difference between an organ note and one produced by a horn. The action of \"stops\" or \"pistons\" in horns is to vary the length of the vibrating column of air. If an opening is made in the side of the tube containing the column of gas, the vibrations must adjust themselves to this point being a loop ; and if two openings are made, their distance apart determines the length of a vibrating segment and therefore the frequency of the vibration. This explains the effect of making and closing openings in a tube as is done in flutes and similar instruments. Organ Pipes. \u2014 The column of air in a pipe or tube is set vibrating in several ways. In the ordinary organ pipe, a section of which is shown in the cut, there is a narrow pas- sage leading to the bellows or wind chest, through which a blast of air is directed against a sharp lip forming the upper edge of a narrow opening at the bottom of the pipe. This blast at the beginning of operations sends a disturbance up the tube, which is reflected at the upper end and returns. J/r>/'-.i/, INSTRUMENTS 403 When it reaches the bottom, its effect is to deflect the blast out of the opening in the tube. When this effect ceases, the' blast returns : and so there is an oscillation of the blast, which has a period determined by that of the stationary vibration of the column of air produced by the two trains of waves, the \u2022\u202211 direct and the reflected. This bottom of the tube is approximately a loop, because it is open to the air. The vibration of the air would soon cease owing to loss of energy by friction and by the production of waves, 8 it not for the sup- ply furnished periodic- ally by the blast. The MUM method of produc- Fio. 181. \u2014Section of an ordinary organ pipe. ing vibrations is used in flutes and whistles, the lungs or mouth In-ill^ the wind chest. Reed Pipe. \u2014In another form of organ pipe, known as tin- reed pipe, the wind chest is connected directly wit h an elongated l)o\\ ; and into this is inserted an inner tube, in which there is a rectangular opening closed by a strip of brass fastened at one end, called the \"reed.\" When pressure in the wind chest is sufficient, this spring door is pushed open, a 1.1 r passes, and the pressure being ':! 1\\. the Nprin^ returns and. o\\\\ in^ to its inertia. continues to vibrate in its own natural period. In this manner a series of puffs \u00ab.i delivered at intervals, 404 SOUND determined by the frequency of the metal spring. There is always attached to the pipe a resonance tube of some kind, the air in which is set in vibration owing to the intermittent puffs. Without this box the sound is most complex, but with it the note becomes fairly simple. The instrument is \" tuned \" by altering the length of the reed by a clamp. Horns. \u2014 In the case of horns, the vibration of the column of air is produced by means of the vibrations of the lips of the player. The column in a horn of fixed length can vibrate in only a limited number of ways ; and the lips must be stretched to exactly the right degree so that, when they are set vibrating by air from the mouth being forced through them, their frequency is one of those to which the horn responds. If one is playing a horn of variable length, like a trombone, and a definite note is being produced, a change in the length changes the frequency of the vibra- tion, and therefore requires a change in the frequency of the lips ; but this change is produced almost auto- matically, owing to the reaction of the column of air itself. The Siren. \u2014 There is another acoustic instru- ment which, although not, strictly speaking, a musical one, should be described. It is called the \" siren,\" because its action continues under water as well as above. In principle it is not unlike a reed pipe, inasmuch as it is designed PIG. 182. \u2014 Helinholtz's double siren. to deliver a number of M I > 1C A L INSTR UMENTS 4o;> pull's <>f air at iv^ular intervals, only with it this number can be varied at will and can be easily counted. As shown in,the cut, there is a wind chest, which is closed on its upper side by a thick circular plate perforated with a definite number of holes, at regular intervals, around a circle concentric with the plate itself. These passages in the", " form of instrument pictured are not perpendicular to the plate, but are inclined slightly, so that the axis of the passage con- Fio. 192 a. -The sidered as a vector has a component parallel to tlu of the circle through the holes at that point. Immediately over tins fixed plate is a movable on*. \\\\hi< -h can rotate on MI..m\u00abl which is identically like the fixed plate, except that its passages slope the opposite way. Therefore, if the movable plate is in su.-h a position that its openings are over those in the fixed plate, the air rushing out from the wind < -hest will have a momentum against the sloping sides of the passages in the former plate, and will set n m 406 SOUND rotation on its.axis. Each time the openings in the two plates coincide, a puff of air escapes ; and if there are n openings in each plate, there will be n puffs during each rotation. (This would be true, also, if there were n holes in one plate only and but one in the other ; if there are n holes in each, the intensity of the puffs is increased.) So, if the rate of rotation is ra turns per second, the number of puffs in a second is mn ; and this is therefore the pitch of the resulting sound. The speed of rotation may be altered at will by regulating the pressure of the air in the wind chest. The number of openings in the plates is easily observed; and the number of revolutions in any interval of time is determined by using a mechanical counter, such as are seen on steam engines. (A screw thread is cut on the shaft of the rotating plate, so that a worm wheel is turned ; this drives a train of cogwheels, which moves a hand over an indexed dial, like the face of a watch.) In other forms of this instrument, the passages in the plates are not slant- ing, and the movable plate is made to rotate by means of a mechanical or electric motor. The simple siren as just described has been modified in two ways. One is to make in the plates several concentric rows of openings, which contain different numbers, e.g. 8, 10, 12; and thus, if all these are opened at one time, a com- plex sound is heard whose component simple vibrations have frequencies in the ratio 8 : 10 : 12. Again, two sirens con- nected with the same bellows may be arranged one over the other with their movable plates on the same shaft and facing each other; this enables one to produce two sounds whose pitches have a known ratio, that is, are at a known \"inter- val\" apart. This instrument was invented by the German physicist Seebeck, and was improved by Cagniard de la Tour and more recently by Helmholtz. It is not, however, as much used now as formerly. MUS1LAL IX^ilil'MENTS 407 Phonograph. \u2014 Another acoustic instrument is the phono ;>h, which consists essentially of a hardened wax cylinder ;nst whose surface pre\u00bb\u00ab \u2022> a sharp point connected with a flexible membrane forming part of a mouthpiece. The cylinder is turned and advanced by clockwork; and, as >ounds are produced near the mouthpiece, the point makes \u2022nding indentations in the wax, which are faithful reproductions of the displacements of the vibrating body. :i. if another point attached to a membrane is made' by mechanical means to pass over these traces on a cylinder at the same rate as that at which the cylinder was turned originally, the membrane will be set in vibration, and its motion will therefore be very nearly the same as that of the one in the mouthpiece that caused the trace. These vibrations will then produce waves in the air which will affect the ear; and so the original sound is reproduced. The Human Voice. \u2014 The human voice is due to the vibra- tion of the vocal cords of the larynx and of the various mov- able parts of the mouth. Consonant sounds such as ft, e, etc., are produced by vibration of the lips, the tongue, etc.; while vowel sounds owe their origin directly to the larvnx. This consists of two stretched membranes which have free edges along ;i nearly straight line. Thoe, an beset in vibration \u2022lie air pressure in the Inn^s: and their frequency can be altered by voluntary changes in their tension. Owing to vibrations, which are, however, very \"damped.\" the air in the cavities of the mouth and throat is also set vibrating. When a definite vowel sound, such as <i/t. is produced, no matter what the pitch of its most prominent component, it urn! by analysis that there are present one or more com ponents of definite pitch. These are not partials of the fun- damental vibration of the", " exactness the dura- tion of reverberation in a hall when its dimensions and the materials used in its construction are known. The acoustic property of a hall depends upon other things than reverberation, for its shape may be such as to focus the iraYee at {.articular points, etc. \"Sounding boards\" which reflect the waves down upon the audience are often used. Another effect whirh must be taken into account is that due to ascending and descending currents of air: for wherever there are changes in the homogeneou of tho air, there are reflections of the waves. CHAPTER XXIV MUSICAL COMPOSITIONS Combinational Notes. \u2014 When two instruments are sound- ing at the same instant, there are several interesting phe- nomena besides the production of the two sounds. If the two instruments are setting in vibration directly the same portion of air, as when a double siren is used, or if two wind instruments are blown by the same wind chest, other sounds are heard than the two corresponding to the instruments. Thus, if n^ and n% are the frequencies of the two vibrations, other vibrations of frequencies, n^ -f- n2 and n^ \u2014 nv are pro- duced in the air. (The mathematical theory of this was given by Helmholtz.) The corresponding sounds are called \"com- binational,\" or \" summational \" and \" differential \" notes. Beats. \u2014 If the two vibrations have frequencies which are quite close together, a different phenomenon is observed. Thus, suppose the frequencies are n and n + m, where m is a small number \u2014 not necessarily an integer. Then, when the instruments are sounded at the same time, the loudness of the sound heard fluctuates; it rises and falls at regular in- tervals. If m = 4, these intervals are a quarter of a second ; or, in general, this is \u2014 th of a second. There are then said m to be 4 \"beats\" (or, in general, m \"beats\") per second. This \" beating \" is due obviously to the fact that as the two trains of waves traverse the same medium before they reach the ear, there will be points at regular intervals apart where the compression of one train will neutralize the expansion of the other, and, at points halfway between these, the com- 412 MUSICAL COMPOSITIONS 413 pressions of the two will coincide. Thus, if the waves have a velocity V, and if the difference in the wave numbers is an integer p, in a distance V in the direction of propagation of the waves, there are p points where the disturbances are almost, if not quite, neutralized and p points where the dis- turbance is abnormally great. This is shown in the figure, v/ \\v/ \\XAAx~ Fio. 184. \u2014 Diagram showing origin of beats. The two vibrations have frequencies whose ratio is 7 : 8. where n:n+p = 1 :8. Therefore, in the course of one sec- ond, as the waves enter the ear, it will happen p times that the sound almost vanishes and p times that it is abnormally loud. Thus, if we count the number of beats in a second, we can tell exactly the difference in the frequencies of the two vil (rations. (If m is not a whole number, a distance greater than V must be chosen in the above treatment. If 7\u00bb = 4.5, a distance 2 Fmay be chosen, in which there are then 9 points where the disturbances neutralize each other, etc.) If the vibrations are complex, beats may occur owing to the prox- imity of the frequencies of any two of the component vibra- tions, or to that of the frequencies of any of these and those of the combinational ones. (The explanation of beats was first given by Sauveur, about 1700.) It is recognized by every one thai beats produce a disagree- able sensation, and for the same reason that a tickling feather or a rapidly flashing li^ht do: namely, ouiiiLT t\u00ab\u00bb fatigue of the nerve* \\Vhen the b.-.us heroine so rapid that they cannot be individually re.-,.'_nii/ed, they cause a \"roughness\" in the sound whieh is unpleasant : hut the degree of di-.i- greeableness depends up<,n both the pitch \u00ab.f the sound and the number,,f. befttfl p--r second. Tins matter was fully investigated by Ilelmholt/.. 414 SOUND Harmony or Consonance. \u2014 It has been known to men of all nations since the earliest ages that there are certain com- binations of vibrating bodies which produce sounds pleasant to the ear, such as that heard when two or three stretched strings of definite lengths are vibrated at the same time.", " This fact has nothing to do with the state of civilization or of musical cultivation ; it is a property of the human ear. It was recognized by the Greeks before the days of Aristotle (probably as long ago as 500 B.C.) that, if two stretched strings of the same size and material and under the same tension, but of lengths in the ratio 1 : 2, were sounded together, the sound heard was agreeable ; and also that, if there were three similar strings whose lengths were in the ratios 4:5:6, the same was true. Mersenne, in 1636, showed that the frequencies of the vibrations of stretched strings varied, other things being equal, inversely as the lengths of the strings. So the problem of explaining the consonance of the sounds produced by the two or three strings as just described became this : why should two notes produced simultaneously by strings making vibrations of frequencies n and 2n, or three notes produced simultane- ously by strings making vibrations of frequencies 4 n, 5 n, 6 n, cause a pleasant sensation? Helmholtz's Explanation of Consonance and Discord. \u2014 The answer to these questions was first given by Helmlioltz. He showed that in every case of consonance when two or more notes are produced simultaneously, that is, a \" chord \" con- sisting of two or more notes, beats were nearly, if not en- tirely, absent ; and that in any other case, when two or more notes were sounded together, the degree of the discord could be predicted from calculations of the number of beats present and from a knowledge of the degree of their disagreeable- ness to the ear. Thus, when a note of pitch n is produced by a vibrating string, notes of pitch 2w, 3 ft, etc., may be heard in the complex sound ; and, when a note of pitch 2 n COJCPO&lTfO 415 is produced, others of pitch I //. b'n, etc., may be heard ; fur- ther, tin- combinational notes all have pitches n, 2n, etc.; so t lie re are no beats, and the two complex notes of pitches n ami 'In are therefore in harmony when produced by wind or stringed instruments. Hut, if the lower note had the pitch n + 5, its partials would have the pitches 2 n + 10, 3 n + 15, etc., and these would beat with the other note of pitch 2n and with its partials. Therefore, it t\\vo complex notes of pitch n + 5 and 2n are produced, they are discordant. It is evident that the same explanation applies to the har- mony of the chord consisting of the three notes of pitch 4n, 5n, and 6n. In general, two or more notes are consonant, or approximate to it. it' their pitches bear simple numerical \u2022 uch other, such as 1 : 2, 1 : 3, 2 : 3, 1 : \\. - If these relations can he expressed, however, only in terms of large numbers, the notes are discordant. On this fact is based the construction of all musical \"scales, that is, series of notes of different pitch, which are played in chords in musical compositions. In all music, however, another <|iiestion enters, namely, are two or n tes or chords played in rapid succession pleasant to the ear or n<>t / If they are, they are said to form a -melody.\" These questions belong to the Theory of Music and cannot he discussed here; but, in general, if :iant. they will also form a melod\\. Musical Scales. \u2014 Many musical scales have been devised and HN.-.I ; hut only two will be discussed here. All modem mii-i.-.il compositions are based upon the idea of sele< ; \u2022\u2022 note as a \" keynote,\" and using this as the starting point \u00ab.t\" the scale. I A i the pitch <>f the keynote, a note ;tch 'J X is. Billed its \"OOtftYQ \": and the inl\u00abT\\;d between tli. -in is called \"an octave.\" I \"interval\" l>etween notes is d< -lined to be equal to tin- ratio of their pit. In the \u2022,y scale, a definite numi- -'lave, whose,' im- ot diff. 416 SOUND widely, and which when played in chords do not produce too discordant sounds. Let their pitches be JV, 0, P, (?,\u2022\u2022\u2022, 2 N. Then, in the octave above this, that is, in the interval from '2Nto 4 N, the notes selected for the scale have the pitches '2N, 2 0, 2 P, 2 #,..., 4 JV; those in the", " octave above this, 4^, 4 0, 4P, 4 #, -, 8N; etc. The notes selected for the octave below the original one have the pitches J JV, JO, J P, J (?, \u2022\u2022\u2022, JVi etc. So, in defining a scale, it is necessary to choose only the pitches of the keynote and the notes in its octave. The Diatonic Scale. \u2014 The \" diatonic \" or \" natural \" scale consists of a series of notes whose pitches may be expressed as follows: If 24 n is the pitch of the keynote, the notes in its oc- tave have the pitches 24 n, 27 n, 30 n, 32 n, 36 n, 40 n, 45 n, 48 n. Thus it is seen that in the interval of an octave there are seven notes, counting only one of the end notes of the octave. If an instrument with strings of fixed lengths, like a piano, is to be constructed to play music written on this scale, some definite keynote must be chosen, and a string must be selected of such a length and size that under suitable tension it will give this note ; then other strings must be selected to pro- duce the other notes of the scale. But suppose a musical composer did not wish to use the same keynote for all his pieces ; suppose, for instance, that he wished to have as the key tone one whose pitch is 20 n. The diatonic scale of this is 20 n, gj x 20 n, fj x 20 n, fj x 20 n, Jf x 20 n,..., 40 n, or, 20 n, 22Jn, 25 n, 26 \u00a7 n, 30 n, \u2022\u2022\u2022, 40 n; and it is seen that, if the instrument described above is pro- vided with strings giving notes in the diatonic scale having 24 n as the keynote, many of the notes required for this composition cannot be produced ; for instance, 25 w, 26\u00a3 w, etc. For this reason, and also to make the intervals between two consecutive notes more nearly equal, the diatonic scale was altered by the introduction of five new notes in each octave : between 24 n and 27 n, 27 n and 30 n, 32 n and 36 n, MUSICAL COMPOSIT10 117 36 n and 40//. and 40 H and l~> n : and corresponding st i were introduced. But still the scale was unsatisfactory when musical compositions were written in different keys ; and so a final change was made, which solved this difficulty. hut introduced another less important one. The obvious advantage of the diatonic scale is that chords played on it are as nearly in harmony as is possible for a scale having seven notes in an octave, since the ratios defin- ing the scale are as simple as possible, vi/. - 1 : oil = 2: 3; 30:40 = 3:4; etc. But slight variations in the pitch of a note may occur in a chord without noticeable discord; and, in any case, people grow accustomed to a chord or to a piece of music and are not sensitive to its perfect harmony. The Equally Tempered Scale. \u2014 This fact was taken advan- tage of in the construction of the new scale. It is called the \"equally tempered scale,\" because the intervals between two consecutive notes are the same throughout the scale. Twelve notes are introduced in an octave; and calling tlie pitch of the keynote n and the constant interval <i. these notes and the octave have the pitches : //..i*n% a4n, \u2022\u2022\u2022, a^n. I Jut since d^n is the octave of n, a12 = 2 and a =^2, or 1.0595. The notes in the octave above and below this are formed as before by multiplying and dividing by 2; etc. It is evident, then, that if a piano or organ is made with strings or pipes corresponding to. the notes of this scale using any definite keynote, musical completion* written in any key whose note is any where in the scale iua\\ d on it. Standard Pitch. \u2014 The keynote adopted for musical scales \u2022 Stuttgart Congress in 1834 and later by the \"Society \\ : ts \" in Kn^Iand is one whose pitch is 2<U. In Midland. however, at the present time \"concert pitch \" is based upon a keynote whose pitch is 273. I i scientific purposes, tuning forks are used as standards of frequency or pitch; and they are generally made in sets following the diatonic scale. The most accurate forks in general use are those AMES'S MI THICK \u2014 27 418 SOUND made by the late Rudolph Koenig of Paris ; and he adopted as the fundamental frequency, or key tone, 256. As will be explained presently, suitable names and sym- bols have been", " given the notes in the various octaves of dif- ferent scales. For instance, \" Ut3\" is sometimes used as the name of the key tone above described. So it is seen that the pitch corresponding to a certain name or symbol in musical notation is not definite. Thus the note called Ut4 had a pitch less than 500 in the early part of the eighteenth cen- tury; in the days of Handel (1750) the note which had this same name and symbol in musical notation was one whose pitch was between 500 and 512 ; and in our days this same symbol is given a note whose frequency varies from 512 to 546, as we have seen above, for Ut4 = 2 Ut3. These facts are expressed by saying that there is a tendency as years go by for the pitch corresponding to a given symbol to rise. The octave of the equally tempered scale starting\" from the keynote 264 is made up, then, of the notes whose pitches are as follows: 264, 279.6, 296.3, 314.0, 332.6, 352.4, 373.3, 395.5, 419.0, 443.9, 470.3, 498.3, 528. If this note is used as the keynote of the diatonic or natural scale, the notes in the corresponding octave have the following pitches : 264, 297, 330, 352, 396, 440, 495, 528. It is thus seen how far apart the scales are at certain points. Violins, violoncellos, etc., do not have strings of invariable lengths, because the fingers of the player can alter them at will by \" stopping \" at any point ; and so one can play with them on the natural scale music written in any key, if the five additional notes are introduced, as previously explained. Musical Notation. \u2014 The seven notes in the octave of the diatonic scale are called <?, d, e, f, g,a,b\\ and different octaves are assigned different types or marks. Thus, the octave from 132 to 264 is written as above ; the one from 264 to 528, <?', d', etc.; the next one c\", d\", etc.; the one from 66 to 132, (7, D, etc. ; that from 33 to 66, Cv Dv etc. J/r>/<.\\L'Using the tempered scale and :M I as the keynote, the notes nearest these to whirh names ha\\c been given are railed by the Corresponding names; thus 'J'>1 ifl oalln.:!, d ; 332.6, e ; etr. Hie note between c and d, i.e. 279.6, iharp\" oru<f flat\"; that between <i and \u00ab,d \u2022harp or e flat : that between/ and g, f sharp or g flat; that be- tween y and a, g sharp or a flat; and that between a and 6, a sharp or 6 flat. <i sharp is written a$; a flat, at?; Strictly speaking, if a note is \"sharpened,\" its pitch is in- creased in the ratio 25: 24 ; thus, if the note has a pitch 264, its \"sharp\" i- _7.\">. Similarly, if it is flattened, its pitch is decreased in the ratio 24:25; thus the \"flat\" of 2(54 is \\\\. These in some cases differ appreciably from tin- notes which have received the names in the tempered s\u00ab Another system of notation is to call the notes making up an octave on the diatonic scale Ut, KY. Mi, Fa, Sol, La, Si, starting from the keynote, or a note which is a number of octaves above or below it ; and to distinguish the diflVrent octaves 1>\\ >u!\u00bberi|.ts. Thus, adopting *J\u00bb',J as the key tone, the octave from :'>3 to 66 is written Utj, Rer etc.; that from 66 to 132, Ut,,. I : etc. (In place of Ut, Do is now more often used. - BOOKS OF l:i i i \\M- THOMSON. S.MII.I. I. <.iid,,M. i is the best modem text-book and is a storehouse of facto. \\i l. On.S..MII.I. \\.-\\v Y,,,-k. 1*82. \\Mt li all the phenomena of S- \u2022\u2022resting description of numerous ezperimeuts dealing HMMH..II/. Sensations of Tone. (Translation by Ellw.) London. 1 966, IH contains a description of all the experiment* on \\\\ln.-h", " Hrlmholtx established his theory of harmony, and also a complete explanation of musical lustrum* Mt- ;iti<l scales. RAY i I vols. London. 1886. i * is a mathematical tren \u2022 1 1 contains \u2022 \\ i > < 1 1 > experimental knowledge of vibrations, waves, and musical notes. LIGHT CHAPTER XXV GENERAL PHENOMENA OF LIGHT To any one who has the sense of sight, the word \" light \" conveys a definite meaning, which, however, cannot be put into words ; but to a person who was born blind the word is unintelligible. The attention of all who have this sense of sight is attracted to many phenomena in nature, such as the colors of objects, the action of mirrors, the refraction produced by water, etc. ; and the study of these forms that branch of Physics called \"Light.\" It will be seen, as we go on, that we can subdivide this subject in certain definite ways. Fundamental Facts Light is Due to Ether Waves. \u2014 The statement has been made before several times that light is a sensation due to waves in a medium called the ether ; but a brief summary of the facts on which this is based may be given again. Thomas Young showed as early as 1802 that interference phenomena, such as described in Chapter XXI, could be observed with light ; Fresnel soon after performed numerous diffraction ex- periments ; and Wiener and others have obtained evidences of \"stationary waves.\" The experiments of Young and Fresnel may be repeated easily by any student. Thus it is established that light is a wave phenomenon. Again, Fresnel showed conclusively that these waves are transverse, because they admit of polarization (see page 313). The existence of a 420 GE.\\Ki;.lL I'l/l-:\\n.MENA OF LIGHT 1'Jl medium is proved by the fact of tin- wave motion; and, since we can see objects through spaces which are void of ordinary matter and through glass, water, etc., it is shown that this medium is one which fills all space known to us, even inside ordinary material bodies (see page 19). We often use the expressions \" light passes \" or travels, etc., meaning that the ether waves which produce light in our eyes pass or travel ; similarly, we speak of \"red light,\" etc., meaning those ether waves which produce in our eyes the sensation that we call \"red.\" We also speak of \"waves in air\" or in water, etc., meaning that the waves are in the ether, but that this medium is modified by the presence of air or water. \"Velocity of Light.\"\u2014 The fact that light travels with a measurable velocity was first shown by Roemer, a Danish astronomer, in 1676, from observations on the satellites of Jupiter; mid in later years experimental methods have been <levise\u00abl t<> measure this <juantity accurately. These will be discussed in a later chapter. The value of this velocity in the pure ether is not far from 3x 1(>10 cm. per second, or about 186,600 mi. per second. It was shown 1>\\ Foucault by direct experiment that the \" velocity in water \" is less than in air, \u2022 ing that the presence of minute particles oi water influ- ence the ether more than does that of similar \"particle* of air.\" Sources of Light. \u2014 As \"sources of light \" \\\\e make use in general of the sun, or of the sky; of electric discharges through rarefied gases; of solid bodies raised to a high tem- perature, such as the carbon rods in an electric \"arc li^ht.\" or the filament inside an ordinary electric \"glow lamp, gas-, lamp-, and candle-flames, for. in fact, the luminosity of any flame is due to the pretaDM in it of minute solid par- ticles which are raised to a hiurh temperature l\u00bby the buHtion of the gas, etc. These are all, in general, sources vhat we call \"white\" light. \\V owever, light of different colors, \u2014 red, green, yellow, etc., \u2014 by \u2022undiiiL: ilf ordinary source by a colored screen. Mich \\-2'2 LIGHT as a piece of colored glass; or we can in certain cases put salts in a flame and so color it \u2014 thus we can secure a brill- iant yellow light by putting common salt (NaCl) in a Bunsen flame, thus making what is called a \"sodium flame.\" Types of Waves. \u2014 If the source of light is very small, we have approximately spherical waves; while, if they come from a distant source like a star, they are plane. We can have two kinds of spherical waves, those which are expand- ing away from a source, and those which", " are contracting toward a point. It is a familiar fact that an ordinary reading lens, or magnifier, may produce an image of the sun upon some opaque screen \u2014 thus acting as a \" burn- ing glass \" ; this means that the plane waves reaching the lens from any point of the sun are changed 011 passing through the lens into spherical waves which converge to a point on the screen. This process of converging waves is exactly the reverse of that of waves diverging from a point source. Thus, if a spherical wave front is concave when considered from portions in the medium toward which it is advancing, it will -converge to a point ; if it is convex toward that side, it will diverge more and more. Fio. 185. \u2014 Dinprram illustrating stellar scintillation. This fact is illustrated in the familiar phenomenon of stellar scintillation. The waves coming from a star are naturally plane ; but if the atmosphere is disturbed by ascending currents of hot air, the wave front is no longer plane, owing to the fact that the velocity of light in cold air is different QKNSRAL PHSNOUBNA o/-' /./<,/// from that in hot. Thus the wave front at any instant may have a \"cor- rugated\" form, a> indicated in the cut. Therefore the light will be concentrated in certain points Av A* etc., the centres of the concave portions of the wave front ; and, as the heated portions of tin* air change their j these points move; so if the ry\u00ab- is at a ]><.iiit.-lj atone instant. th\u00ab> n<*xt it may be between two points,.1, ami. lo; etc. So the iiitrn^ity of the light will increase and decrease intermittently. The same phenomenon is observed in the \"shadow bands \" seen at times of total eclipses of the sun. Homogeneous Light and White Light. \u2014 In order to deter- mine tin- wave lengths of these \"light waves,\" it is simply necessary to use the interference method described on page 375. It is found that, if the source of light is white, the interference fringes or bands are all colored, with the excep- tion of the central one, which is white. If, however, the source is colored, the bands are alternately black and colored ; that is, in certain lines disturbances in the ether are entirely absent. In general, it will be observed that there are apparently two or more sets of fringes superimposed, each set having a definite color. It is possible, however, to secure such a colored source, that the bands are all of one color, separated by the dark fringes. (This is approximately the case with a sodium flame.) Under these conditions the light is said to be \"pure,\" or \"homogeneous.\" When the sourer is white, or when any ordinary colored source is used, we can analyze the complex interference pattern into series of fringes, each series having its own color and its own spacing. It was shown on page 378 that, if the waves have a definite wave length, the fringes are evenly spaced at a distance t which is proportional to the wave length. Therefore, when a source is hom-. it is emitting waves of a \u2022'.\u00ab\u2022 definite wave Imgih: and, in general, an ordinary source of light is emitting t rains of waves of different wave length-, 'fins, th\u00ab- interference apparatus \"disperses\" t lie complex from the MHirO6 into simpler trains,,-aeh ing a deiimi^ \\\\.i\\r Length, 424 LIGHT Connection between Wave Length and Color. \u2014 We can measure the wave length of the waves emitted by any homogeneous source, by using the formula deduced on page 378, viz., distance apart of fringes =\u2014, where I is the wave 6 length, a is the distance from the two sources to the screen, and b is the distance apart of the sources. In this way it is found that pure red light has a wave length greater than that of pure green ; and this in turn is greater than that of pure blue. The longest waves that affect our sense of sight produce the sensation of red and have a length of about 0.000077 cm. (i.e. 770/xft); while the shortest produce the sensation of violet and have a length of about 0.000039 cm. (i.e. 390 pp'). In between these limits there are all possible wave lengths, corresponding to which are all shades of color, ranging from the deepest red, through orange, yellow, green, blue, to the darkest violet. Waves shorter than 390 /i/i can be observed by photography ; and they have been obtained by Schumann as short as 100 /u/x. Waves longer than 770 pp may be studied, as described on page 292, by various means; and", " they have been observed as long as 25,000 pp, i.e. 0.025 cm. (Electrical oscillations produce waves in the ether, which are as a rule very long ; but, using minute conductors, waves as short as 0.6 cm. have been obtained. There is thus a gap between 0.6 cm. and 0.025 cm. which has not yet been investigated.) The fundamental facts are, then, that light is due to transverse waves in the ether ; that waves of different wave length produce different colors; that white light is, in general, due to a mixture of waves of all wave lengths which affect our sense of sight ; that the velocity of these waves is less in ether inclosed in matter than in the pure ether. (It will be shown on page 433 that, in the former case, waves of different wave length have different velocities, which is not the case in the pure ether.) mi: i OF LIGHT 425 General Properties of Light as Due to Wave Motion. \u2014 \\\\V described in Chapters XX and XXI certain general properties of wave motion which apply directly to light; viz., reflection, refraction, rectilinear propagation, interfer- ence, and diffraction. Each of these will be discussed more in detail later; but one or two points should be referred to here. (It should be remembered specially that we are now considering extremely short waves, viz., those whose wave length is not far from 500 pp.) Rays, Shadows. \u2014 It was shown that, if the waves are short, the disturbance at any point in an isotropic medium due to a train of waves depends directly upon the dis- turbances at previous instants along a straight line drawn backwards perpendicular to the wave front; this line is called a \"ray.\" If the rays are all parallel, we have plane waves, or a \" beam \" of light. If the waves are spherical, tin- rays are radii ; and, by considering only those rays which are close together, we have a \"cone\" or a \"pencil\" of light. (Although this theorem in regard to rays was proved only for a train of waves, it may be shown to hold true for a -pulse\" also.) It the waves meet an opaque obstacle, that is. on.- \\\\ hich does not transmit li^ht, a shadow is cast. If the source of light is a point, the shadow is that which we have called tin; \"geometrical\" shadow, if wo neglect diffraction phenomena, as We Shall f,,r the time beiO I f, however, the SOUrce is large, object^, representing the on \u2022 tureen rG\\y \u00bbn opaque,ik|, | tl;iMi(i MI. ;m i]lllmm;lted piece of paper, the shadow phenomena are evidently not quite so simple. 426 LiailT Thus, if the large source of light is represented by AB, and the opaque obstacle by CD, the shadow cast by the waves from A on a screen FG- is limited by EG-, and that cast by the waves from B by FH. Therefore the region EH on the screen receives no light ; and the region outside FCr receives light from all points of the source ; but the intermediate regions, EF and GH, receive light from only portions of the source. Therefore, the intensity of light on the screen fades away gradually toward the central region EH, where there is no light. The space back of the screen into which the waves do not penetrate is called the \"umbra,\" while the partly illuminated space surrounding it is called the \"penumbra.\" An illustration is afforded by solar eclipses, where the sun is the source, the moon is the opaque obstacle, and the earth the screen. In the dia- gram, S represents the sun, and M the moon. The umbra and penumbra are indicated by dark spaces. If the earth enters this region, the eclipse is total for all points on its sur- face which are inside the umbra, and par- tial for points outside this- but lying in the penumbra. If the earth just misses the umbra, it may happen that at certain points of its surface a ring of sunlight may be seen around the FIG. 187. \u2014 Diagram showing shadow cast in space by the moon, M, owing to the rays from the sun, S, edge of the moon ; this condition is called an \"annular\" eclipse. Another interesting case of rectilinear propagation is given by the formation of what are called \"pin-hole \" images. If a small hole is made in an opaque screen, any luminous object \u2014 e.g. a building in sunshine \u2014 situated on one side of it will produce on a screen on the other side an inverted image of itself, which is comparatively sharply denned. Thus if there is a small opening at", " O in the screen, and A is a point of an illuminated figure, there will be a cone of light from A passing through the opening. If this meets a screen, it will make at the point /* a bright './\u2022:\\/:/,M/. i'in-:.\\n.Mi-:\\.\\ LK.IIT i-JT spot, which will hsivi- tin- >ha|>\u00ab- of tli\u00ab- opening. If the opening is small and tin- t\\\\o.M-m-ns an- close together, the spot at B will be practically a point of light ; and hence, as each point of the illuminated figure pro- duces a point of light on the screen, there will be formed a well-defined image. As tin- images of the various points of the illuminated object overlap, the general appearance of the image is almost independent of the shape of the opening, if it is small. (The round or ellip- tical spotsof light which are v^ m._Pln_holeim*g<*. seen on the floors near cur- tained windows or under trees are images of the sun formed by minute openings in the curtains or leaves.) This is the principle of pin-hole photographic cameras, of the camera obscura, etc. Opacity; Transparency; Translucency. \u2014 A distinction is made between material media which are \" transparent \" and which are \"opaque.\" If an object when introduced between the eye and a source of light stops all the light, e.g. a board, a piece of tin, etc., we say that it is opaque, meaning that it does not transmit those ether waves which affect our <\u2022 of sight. An object may be opaque to some waves and not to others: thus, a piece of red glass is opaque to all visi- ble waves except those which produce the sensation of red. Again, a given material body may be opaque to waves which do not afieet our sense of sight, and may transmit all visible and there may be media with just the reverse prop-. Opacity is due to the fact that the waves which are incident upon the bodv do not pass through it: they are either reflected bn-k from it or are absorbed by it. Thus, a polished metallic surface is opaque largely owing to relleetioii, while a blackened surface is opaque owing to absorption. If we can look through an object and see sources of light 428 LIU / IT on the other side sharply defined, we say that it is \"trans- parent.\" Here, again, the body may be transparent to some waves and not to others. Waves, then, when incident upon a transparent body, will under ordinary conditions be trans- mitted by the ether in the body, maintaining a definite wave front. This is true only if the transparent body has surfaces which are \" smooth,\" in the sense that there are no inequal- ities comparable in size with the length of the waves. Thus, a window pane of glass, layers of water or alcohol, etc., are transparent. If the surface is rough, the waves suffer irreg- ular reflection. Again, we will see later that in certain cases the waves incident upon the surface of a smooth transparent body are entirely reflected. (This is called total reflection.') It must not be thought, however, that a bod}' which is not transparent is necessarily opaque ; a piece of opal glass or of oiled paper is not transparent, nor is it opaque. One cannot see objects through them, and yet they allow light to pass to a certain extent. Such bodies are said to be \" translucent.\" What happens is this. When waves from any source fall upon a translucent body, they are broken up and scattered by it in such a manner that each point of the body becomes a new and independent source of waves. So when such a body is held between the eye and a source of light, the waves which reach the eye come directly from the points of the body, not from the source ; and what the eye sees, then, is the surface of the body. A transparent body cannot be seen ; it is only owing to dirt on it that we are able to see a surface of water or a window pane. The explanation of translucency will be given in the following pages. Reflection and Refraction. \u2014 It has been shown that, if there are two media separated by a bounding surface, waves in one will be reflected at the boundary in general, if the velocity of the waves in the two media is different. The boundary surface must be large in comparison with the length of the waves, otherwise the waves will pass around the \"obstacle.\" 9XNMRAL /\u2022///\u2022: \\</.v/-;.\\.i or /,/'./// 429 If the surface is large, but possesses inequalities which are in it small in comparison with the", " wave length, each minute I M) it ion of the surface acts like a separate reflector, and so the waves are scattered. If, however, the surface is \u2022\u2022smooth,\" in the sense that any inequalities are extremely small, the reflection is regular, and the surface is called a '\u2022 mirror.\" This may have any shape, but in practice the only forms used are plane, spherical, cylindrical, and para- bolic surfaces. (These curved surfaces may, of course, be either concave or convex.) Similarly, in these cases there will In' waves entering the second medium with a definite wave front. As already stated, these are called \" refracted \" waves. If the medium is transparent, they continue indefi- nitely. Imt if it is absorbing, they soon die down. Diffuse Reflection. \u2014 When waves fall upon a roughened surface they are broken up and reflected irregularly, as just explained: they are said to be \"diffused.\" Each point of the surface now becomes a new source of waves. This dif- fusion of light is taking place from all natural objects, all walls, pieces of furniture, etc.; and it is owing to it that we are able to see any object. If a body reflects regularly, we do not see it, but the source of light reflected in it. Similarly, if a transparent body contains immersed in it a i numherof minute foreign particles, it diffuses the liijlit falling iip<\u00bbn it- In one such bodies the minute for- eign particles are opaque; whereas in another class they are transparent. l\u00bbut have irregular figures. In both cases the incident waves are broken up and sutler reflection to and fro from particle to particle ; and finally t lie disturbances emerge in the foil ii of spherical waves proceeding out from each point of the surface. (If the body is thin, these waves may emerge on l\u00bboth of its sides.) Diffusive reflection is thus due to one of three causes: etching '\"' roughness of the surface, the presence of opaque particles in a transparent medium. Of that of minute irregular 430 LHHIT transparent particles in such a medium. Ground glass, ordi- nary unglazed paper, etc., are illustrations of the first class of bodies; opal glass, celluloid films, etc., of the second; and foam, thin starch water dried on glass, etc., of the third. All these bodies which reflect light diffusively, diffuse it also when they transmit it ; that is, they are translucent. If these foreign particles in a transparent medium are very minute, it may happen that the shorter waves only are affected, while the longer ones will be transmitted through the body. In this case the body would appear bluish when viewed in reflected light. This is the explanation of the blue color of the sky, of fine smoke, of a hazy atmosphere, of blue eyes, etc. In each of these cases there are minute par- ticles existing as foreign bodies in a transparent medium, which diffuse the short waves, but allow the longer ones to pass. Regular Reflection and Refraction. \u2014 We gave in Chapter XX the treatment of reflection of plane waves at a plane sur- face, quoting from Huygens. But, as noted at the time, this method is not rigorous. In order to make it so, we must follow the same plan as did Fresnel in discussing rectilinear propagation ; namely, we must divide the wave front up into zones and deduce the effect B/ N \\ of the secondary waves, thus combining Huygens's principle of secondary waves and Young's princi- ple of interference. When this is done, we obtain Huygens's solution, and FIG. 189. -Reflection of plane waves by a plane also leam the effect of re- flection upon a ray. Thus, by Huygens's solution of the problem, plane waves, whose wave front at any instant is given in section by AB, inci- dent upon a plane surface MM, are reflected, and form the fXBAL I'llKMtMENA OF LK,11T 431 plane waves whose wa\\e front at a later time is given by T)C, where the angles (BAG) and (DC A) are equal. If draw a straight line OjO perpendicular to AB and meeting the reflecting surface at 0, and another OP, per- pendicular to the reflected wave front C/D, it is seen by geometry that the broken line OjOP is the shortest line that can be drawn from P to the wave front AB by way of the surface, and so Ol is the \"pole\" of P (see page 382) ; and. in drawing the zones around Or which are to be com- pounded in order to deduce the effect at P, it is seen that they are exactly", " the same as if the reflecting surface were* removed and the wave front had the position AB1 where the angles (B'AC) and (BAC) are equal. From this fact the ordinary laws of reflection follow at once. The disturbance at P is due directly to that at 0, and this in turn to that at Oj ; consequently, the ray Ol 0 is turned into the ray OP by reflection. If a normal ON is drawn to the surface at 0, the angles ( 01 ON) and (PON) are equal ; the former is called the -alible of incidence,\" the latter that of \"reflection.\" The plane including the lines 010 and ON is called the \"plane of incidence\"; it evidently includes also the re- flected ray OP. Reflection of li^ht may be demonstrated with ease by caus- ing a beam of sunlight, or a beam from a lantern, to fall upon an ordinary 1 MM king- glass ; for the path of the \\ beams may be seen if dust 0> ;iioke is distributed through the air.,; l.irl y, in the case of refraction of plane waves at a plane surface, FW. i90.-R\u00ab\u00abhwt\u00abonofPun\u00abwav\u00ab\u00abby\u00bbpton\u00ab Hu\\ -.rens's solution is that. >' ]J ifl the trace by the paper of the refracting surface, and AB that of the incident plane waves, tin., ted 432. LIGHT waves are plane and their trace is CD, where the angles (BAC) and (DC A) are connected by the relation sin ( BA C) _ velocity of waves in the upper medium sin (DC A) velocity of waves in the lower medium This ratio is a constant for the two media and for waves of a definite wave immU'r, and is, therefore, the same for all angles of incidence ; it is called the \" index of refraction \" of the lower medium, with reference to the upper for this wave number. This proof of Huygens's, however, is not rigorous ; but if Fresnel's method of treatment is followed, it may be made so. If we draw a line 0-fl perpendicular to the wave front of the incident waves at Ov and continue it by the line OP drawn perpendicular to the wave front of the refracted waves, it may be shown by geometry that the time taken for a dis- turbance to pass from 01 to P along this line is less than along any other line. (Huygens proved this.) Therefore 01 is the pole of P on the incident wave front ; and by constructing Fresnel's zones around Ov we are led at once to Huygens's solution. The disturbance at P is due directly to that at 0; and this, in turn, to that at 6^; so the incident ray Of) lias its direction changed at the surface into OP. Drawing a normal NON' to the refracting surface at 0, the angle (01ON^) is the angle of incidence; and the angle (PON'), the angle of refraction. These angles are equal respectively to (BAC) and (DC A). Therefore we can state the law for a ray: the Bine of the angle of incidence = ^ index Jf refraction. the sine of the angle of refraction Calling the angle of incidence i, that of refraction r, and the index of refraction n, this relation is smr QXNSRAL rui-:.\\<>Mi,\\A OF LIGHT 433 The other law of refraction is evident, viz., the refracted ray is in tin- plane of incidence. Refraction may be studied with ease by allowing a beam \u2022inlight to fall upon the surface of cloudy water in a tank. It will be observed that, if sources of light of different color arc used, the refraction is different; if the waves have the same angle of incidence, the angles of refraction are different. Thus the index of refraction varies with the wave number or color of the light. This is not easily shown in the experiment just described, unless the beam of light is made extremely narrow ; because the differences in the refraction are not great, but by means of a \"prism\" or a lens this phenomenon is m<>st apparent. If the incident light is white, each of the component trains of waves (see page 423) has its own index of refraction, and so the light is broken up or dispersed, forming a \"spectrum.\" In the case of ordinary transparent media such as glass, water, etc., the waves having the shorter wave lengths are refracted more than those having longer ones; i.e. blue light is refracted more than green, green more than red. This proves that in these media short waV6fl have a less velocity than do long ones. (In the pure ether all waves, so", " far as we kno\\v. have the same velocity.) This kind of medium is said to have ordinary or \" normal \" disper- sion. In other media, it may happen that some waves are refracted more than others which have a shorter wave length; they are said to have \"anomalous dispersion\" (see Chapter \\\\\\,. When waves are passing in the ether inclosed in anv material medium, such as water, the minute particles of this medium an- in motion also to a greater or less extent, owing to the waves; so, if there is a long train of waves, and if \\\\e consider any one point in space, the effect produced there by the read [on \u00ab\u00bbf the matter on each \u2022\u2022 wave\" as it passes it is different from what it would he if the matter were at rest; as, for instance, if a sudden \u2022\u2022 pulse \" came up to t his point and AMI *'i i iivsics \u2014 28 434 LIGHT passed. Since the velocity of a disturbance through the ether depends upon this reaction of the matter which incloses it, it is evident that the velocity of a train of waves is different from a pulse. Further, the method of Fresnel for consider- ing wave motion presupposes the existence of a train of waves. Thus the laws of refraction apply only to trains of waves. Geometrical Optics. \u2014 Other cases of reflection and refrac- tion will be considered in the following chapters : spherical waves upon a plane surface, plane waves upon a spherical surface, and spherical waves upon a spherical surface. There are two modes of procedure possible : one is to study the changes in the wave front produced by reflection or refraction ; the other is to study the changes in the direction of the rays, making use of the theorems just deduced ; for in the case of incidence upon a curved surface, we may con- sider the reflection or refraction of a ray at any point as due to an infinitesimal portion of the tangent plane of the surface at that point. The application of this latter method makes up what is called the science of \" Geometrical Optics.\" We shall use this in these chapters, but shall also outline in cer- tain cases the demonstrations in terms of waves. Real and Virtual Foci. \u2014 If spherical waves diverging from a point source are spherical also after reflection or refraction, we may have either of two conditions : the centre of the reflected or refracted waves may be in the medium in which the waves are advancing, i.e. the waves converge to a \"focus\"; or the centre of the waves may be in the other medium, i.e. the waves will diverge away from their centre. (Of course, in the former case, the waves, after converging to a point, will diverge again beyond it if no obstacle 'pre- vents.) The centre of the converging waves is called a \" real \" focus ; it is said to be a \" real image \" of the source or \" object.\" The centre of the diverging waves is called a \"virtual\" focus; it is said to be a \"virtual image.\" There GEXKHAL PHENOMENA of LIGHT L86 are cases, however, in which, even though the incident waves are spherical, the reflected <>r retracted waves are not. Homocentric and Astigmatic Pencils. \u2014 Similarly, from the standpoint of rays, if we consider any incident pencil of rays proceeding from a point source, it will, after reflection or i-i- fraction, form another pencil with its vertex in the medium into which the rays are advancing, if there is a real focus, or one with its vertex in the other medium, if the focus is vir- tual. But there are cases when, after reflection or refrac- tion, the rays do not form a cone. A pencil of rays which does form a cone is said to be \"homocentric\"; while one which does not is said to be \"astigmatic.\" In this latter case, as we shall sec, the rays of a homocentric pencil, after reflection or refraction, have as a focus (either real or virtual) not a point, but two short lines perpendicular to each other and a short distance apart; these are called \"focal lines.\" In describing the incidence of a pencil of rays it is simplest to give the direction of its central ray; so by speaking of \" normal incidence \" of a pencil we mean a case when the central ray of the small pencil is perpendicular to the surface at the point where the pencil meets it; and by \"oblique incidence\" is meant a case when the central ray of the pen- cil makes an an.^le different from zero with the normal to the surface at the point where this ray meets it. We shall see shortly that in all eases a pencil which is normal to a surface", " produces by reflection or refraction a homocentric pencil; and in nearly all cases an oblique pencil produces,in astigmatic one. Properties of a Focus from the Standpoint of Waves. \u2014 Since the locus of the points readied by the dis- turbances at any one instant, we may consider the existence of foci from a different standpoint. If \\\\ a ves di ver^injr from a point BOH 06 '-on verge after reflection or refraction to an- other point, we can draw various rays proceeding out from the former point and all meeting again at the latter. These 436 LIGHT rays have different paths ; but the time taken for the disturb- ances to pass along all of them must be the same. Thus, if Zj is the length of the portion of a ray in one medium in which the velocity of the waves is vv and if?2 is the length of its portion in a second medium in which the velocity is v2, the time taken for the propagation of the disturbance is v\\, or /Z, + Sl.0-' Bufc ^ is the index of refraction, n, of the second medium with reference to the first; so this time is (li + nl^)\u2014. The quantity (7X -f- nl2~) is called the \" \" of the ray. length optical is It dis- tance in the first medium which the waves would advance in the time taken for the actual propagation in the two media. Then we may state that the optical lengths of all rays from the point source to the focus are the same. We shall make use of this principle in discussing lenses. evidently equal the to Another fact in regard to foci should be emphasized. Re- flection and refraction are always produced by pieces of mat- ter of limited size, e.g. looking-glasses, prisms, lenses, etc. ; and so only a portion of the wave front undergoes the change. The effect at any point in the advance of the wave front must then be deduced by following Fresnel's method of combining the principles of Huygens and Young. The point at which the disturbance is greatest is the focus ; but this does not mean that there is no disturbance at other points. The effect at these latter points must in each case be calculated ; for some it is zero, and in no case does it approach in amount that at the focus. This fact is of great importance when we discuss the reflection or refraction of trains of waves from two points that are close together. CHAPTER XXVI PHOTOMETRY Intensity of Sources of Light ; Intensity of Illumination. \u2014 One of the most important practical questions hi regard to sources of light deals with their illuminating power. Of course only part of the energy radiated by these sources is in the form of waves of such lengths as to affect our sense of sight; and we speak of the \"luminous energy\" or the \u2022\u2022quantity of light\" that is radiated. The quantity of light emitted by a source in a unit of time measures its \"intensity.\" If the source is small compared with the dis- tance at which the light is received, the waves may be regarded as spherical. In other cases, for instance a white wall or screen, this is far from being true. In any case, however, if by means. of opaque screens with L L suitable openings in tin -in the radiation is limited to a \"beam,\" we can lind the relation between the quantities of li-_rht falling <>n sur- faces inclined at differ- Fio. 191. \u2014 Incidence of a beam of light upon \u00bbr. nMiijue screen. ent angles to the direction of the beam. Thus, as slio\\\\ n in the cut, let the beam be that luwsing through t\\\\\u00ab- equal and parallel openings AB and AVB^ of area A; and let the beam fall upon a screen whose normal makes the angle N with the direction \u00ab.f the heam. The illuminated surface on this screen, CZ), will have an ar< here 438 LIGHT BcosN=A. (See page 28.) If the screen is a dif- fusing one, this portion of it will appear bright when viewed from any direction. The \" intensity of its illumina- tion \" is defined to be the quantity of light received per unit area in a unit time. Therefore, if Q is the quantity of light carried by the beam in a unit of time, the intensity of illumination, 7, of a screen perpendicular to the beam is -\u00a3\u2022 That is, 1=-^- The intensity of illumination of the 0 oblique screen is \u00a3\u00ab Calling this Iv we have, then, If the source is small, we can, as has been said, consider the waves as spherical; and, if Q is the quantity of light emitted by the", " source in a unit of time, assuming that it radiates uniformly in all directions, the amount falling on a portion of the spherical surface of area A at a distance r is H tne illuminated surface is small and is oblique to the direction of propagation, let its area be B and let the angle made between a perpendicular to it and the direction of propagation be N\\ then the projection of this surface on a plane perpendicular to the direction of propagation has the area B cos N. So in the above formula if A is this projected area, A = B cos JV; and the light received in a unit time by the oblique surface whose area is B is \u2014 -- \u2014 \u2014 \u2022 The intensity of illumination of this oblique surface is then _Q_ cosjgV 47T r2 4?r r2 This formula offers at once a method for the comparison of the intensities of two sources of light. The general method is to illuminate a portion of a diffusing screen by one source and contiguous portions of the screen by the other, and then to vary the distances of the sources until the two portions of I'UOTOMETHY Jo'.' tin M-iven appear equally bright. When this is the case, the intensity of illumination must be the same for both portions; and, if the angles of inclination of the two sources are the i i Luminosity of Sources of Light. \u2014 If the source of light is not a point but an extended surface, like the surface of a white-hot metal or of a diffusing screen, let us consider the radiation of any small portion of the surface, whose area we may call A. If this were a point source, and it emitted a quantity of light Q in a unit of time, the quantity received in that time on a screen of unit area at right angles to the direction of propagation at a distance r, would be, as we have seen, \u2014^\u2014- Therefore the total amount of light actually 4 received per unit area by a screen, parallel to the luminous surface and at a distance r from it, varies inversely as r2, provided the area of the luminous surface and that of the portion of the illuminated screen considered are both small and face each other. That is, calling P the quantity of light received in a unit of time per unit area of the sen-en, P = \u2014, where c is a constant depending upon the properties of the luminous surface. Evidently c varies directly as the area of this surface ; so we may write c = LA, or P = \u2014, win -iv I, is a factor of proportionality. The quantity L is called tin- \u2022\u2022 intrinsic Ininiimsit y \" of tbe luminous\" surface prrjirmlienlar emission.\" It is a constant for the given luminous surface. The light received from a luminous surface by a parallel surface of unit area at a distance r is, as just shown, \u2014\u2022 Tin- brightness of a In in inoufl object as judged by our eyes <1- 1 M n,l u]>on the light received per unit area of the retina. Thus, if tin- light recei\\.--l by the eye, passing tli.-.liaphr;i-m f..nnrd by tin- pupil. i> <v'. ami if tin- an -.1 \"f tin- 440 LIGHT image of the luminous object formed on the retina is a, the brightness of the object is proportional to \u2014 \u2022 If the object is a surface parallel to the a eye and at a distance r, and if B is the area of the pupil, this light Q enter- ing in a unit of time is, from the above definition of L, \u2014 \u2014 \u2022 The image formed has the area a ; and it will be shown later in speaking of lenses that a is proportional inversely to r2; or a = \u2014, where A: is a constant de- pending upon the construction of the eye. Hence the brightness of the 1* luminous surface, - is \u2014 \u2014 divided by -, or \u2014 \u2014 \u2022 For a given a r2 r2 k luminous surface and a given eye, this is a constant quantity; or, in words, a luminous surface appears equally bright at all distances from the eye. Certain luminous objects, for instance a star, do not produce images on the retina, as do ordinary luminous bodies ; they give rise to a diffraction pattern determined by the size of the pupil. (See page 390.) Therefore the above statement does not hold for them. A telescope will increase the brightness of a star, because it introduces more of its light into the eye, but will not increase that of the sun or the moon. Experience shows that a luminous surface when viewed obliquely appears practically as bright as when viewed per- pendicularly. Thus, a luminous spherical solid appears to our eyes like a luminous disk of uniform brightness. If the area of", " the oblique surface is A, and if it is inclined at such an angle that a perpendicular to it makes the angle N with a line drawn to the eye, A cos N is the area of the projec- tion of this surface perpendicular to this line ; and a surface of this area placed parallel to the eye appears of the same brightness as the one of area A placed at the angle N. Let L1 be the intrinsic luminosity corresponding to the direction N; i.e. if the area of the surface is unity, the light received Tt per unit area at a distance r in this oblique direction is \u2014 \u2022 Then the light received by the eye from the oblique surface AU B is \u2014 \u2014 \u2014 ; whereas, if the parallel surface of area A cos N were used, the light received would be A\u2122*N' LB. But PHOTOMETRY 441 \u2022rience proves that these an- practically equal; so L' = L cosiV. This is called \" Lamhert's Law.\" The intrinsic luminosity in any direction of a small luminous surface is, in words, the quantity of light received per unit area by a screen perpendicular to this direction at a unit distance, divided by the area..\u00ab\u2022 luminous surface. So, if the screen is placed obliquely to this direction, making an angle NI with it ; and if its area is />'. the light received on it in a unit of time from a lu mi nous source of area A, making the angle N with this line referred to, and at a distance r, is AcosN - L- BcosN, A I B cos N cos Nt -I or - H As noted above, this statement is not absolutely true. We may regard L as a quantity which is not a constant factor but varies slightly with N. A method is thus evident for the comparison of the lumi- nosities of different sources of light. Each is surrounded by an opaque screen provided with a rectangular opening, or slit : the sources are so situated that a suitably placed diffusing screen receives light perpendicularly from these two open- ings; one portion of the screen receiving light from one source only, and contiguous portions receiving light from the other source only. The screen is now illuminated by li'^ht coming from the two rectangular openings as sources. Then 1>\\ some means the conditions are so altered that the hri'_rhtncss of one portion of the screen is diminished until that of the two portions appears equal. This diminution in intensity may be seemed in various ways: (1) remove one source to a greater distanc alter the widths of the rectangular slits. (:', ) interpose between the more intense sourer ;md the illuminated s.-r.-m ;m opaque disk which has certain i -it out, and cause this to revolve rapidly: by altering the si/e of the sector* the intensity may be diminished at will and in a known ratio, When the two por- tion reen are equally illuminated, they are receh 1 quantities of lj._rht per unit,n< -\\. I 'or simplicity let us assume that there is no interposing revolving disk. J*i 442 LIGHT Al be the area of one opening, L^ its luminosity, and rl its distance from the illuminated screen ; the light received by it per unit area per unit time is then \u2014 ^'\u2022 Similarly, giving A2, Ly and r2 corresponding meanings for the second source, the light received by the screen per unit area per unit time from the second source is \u2014 ^ Since, in the arrangement described above, these are equal, 121 = 2\"^2 ; and so L1 and Lz may be compared. 1 r% Standards of Light; Photometers. \u2014Various standard sources of light have been defined. One of these is the flame of a \"standard candle,\" which is a sperm candle weighing one sixth of a pound and burning 120 grains per hour. Another is the flame of a \" Hefner- Alteneck lamp \" (which burns amyl acetate), when the height of the flame is kept at 4 cm. Still another is a surface of platinum when it is raised to such a temperature that it is on the point of melting. An instrument for comparing the intensities or the lumi- nosities of two sources of light is called a \"photometer \"; and the special science involved in the study is called \" photom- etry.\" In Rumford's photometer, an opaque rod is placed close to the diffusing screen, so that two shadows are cast on it by the two sources ; the shadow cast by one receives light from the other only ; so, when the brightness of the two shadows is the same, the wished-for condition is obtained (care must be taken to have the angles of inclination the same). In Bunsen's photometer, a", " screen consisting of white unglazed paper, in the centre of which there is a small round or star-shaped grease spot, is placed between the two sources. Looking at this screen from either side, any portion is illu- minated by the transmitted light from the source on the other side and also by the reflected light from the source on that side. The screen is moved until the grease spot and the other portions appear equally bright when viewed from either PHOTOMETRY 443 side: and then tlu- above relation holds. For, let a be the proportion of light reflected by the unglazed paper, and b that reflected by the greased paper, and assume that there is no absorption. Then, if P1 is the quantity of light per unit area incident upon one side of the screen, and P2 that upon the other, the amount of light reflected per unit area by the unglazed portion on the former side is aPv and that received by transmission from the other side is (1 \u2014 a)P2 > similarly, the light reflected by the grease spot is bPr and that received by transmission is (1 \u2014 b)P2. Hence, when the two portions are equally bright, aPl + (1 - a)P, = bPl + (1 - &)/>\u201e or (a - 6) Pl = (a - 6) Py And therefore, since a does not equal 5, Pl = P2. The best photometer in use to-day is one designed by Lummer and Brodhun. For full details of these and other instruments reference should be made to some treatise on Photometry, such as Stine, Photometrical Measurements, or Palaz, Indus- trial Photometry. Naturally, the intensities of two sources of different color cannot be compared directly ; and, in general, if any two sources are to be compared, tin ir luminosities corresponding to each wave length should be investi- gated. This can be done by combining with a photometer a dispersing apparatus such as a prism. The complete apparatus is called a \" spectro- photometer,\" the simplest and most accurate form of which is one devised by Professor Brace of the University of Nebraska. CHAPTER XXVII REFLECTION WHAT is meant by regular reflection, and by a mirror, has already been explained ; and the law of reflection for a ray has been deduced (see page 431). We will now consider several special illustrations. /N, Plane Waves Incident upon a Plane Mirror. \u2014 This is the case already discussed on page 367, and needs no further treatment here. There is one illustration of it, however, which may be described. It is that of a plane mirror which is being rotated when plane waves are incident upon it. Let the trace of the mirror by the paper at any instant be MM, and let PO be any incident ray ; draw ON perpendicular to the mirror at 0; the re- flected ray OR will make with the normal an angle (RON) equal to the angle (PON). Let the mirror now be ro- tated about an axis through 0 perpendicular to the plane of the paper; that is, about an axis parallel to the inter- section of the plane wave At the end of a certain time front with the plane mirror. the mirror will have turned into the position indicated by MlMl ; so, iHXZVi is the position of the normal, the reflected ray will be ORl where the angles (P&ZVj) and (NflR^ are equal. The angle turned through by the reflected ray is 444 Fio. 192. \u2014 Rotating mirror: the incident ray iaPO. REFLECTION 445. This equals the difference between the angles (POR) and (POR^j that is, twice the difference between the angles (PON) and (PONJ, or twice the angle (NONJ. But this is the angle of rotation of the mirror ; so the reflected ray turns twice as fast as the mirror. This prin- ciple is made use of in many optical instruments : the sextant, which is used to measure the angle subtended at the eye of the observer by two distant points; the mirror attachment to a galvanometer ; etc. Spherical Waves Incident upon a Plane Mirror. \u2014 Let the sheet of paper be perpendicular to the plane of the mirror, and let the trace of the latter be MMr If 0 is the source of the waves, we may consider any two rays OP and OQ. l>v reflection they become PPj and QQ^ where the angles Q, (MPO) and (M^PP^ and (MQO) and (Jfi^Ci) are equal. If PPj is prolonged backwards, it will meet in the point (7 a line drawn from 0 perpendicular to the mirror. Let this line meet the mirror in the point A. Since the angles", " surface is only slightly curved, and that only a small portion of the mirror around M is used. For, since the line PC bisects the angle PO-.PO' = C or, putting Jlo = u, MO' = v, MC = r, PO'.PO' = u-r:r-v. If, however, the above conditions as to the curvature of the mirror and the closeness of P to M&TG satisfied, the distance PO nearly equals MO, and PO' nearly equals MO'. That is. Hence for definite values of u and r, that is, for waves from a definite point source 0 falling upon a concave mirror with the radius r, the value of v, which determines the position of the image 0', is independent of P. It is, in the case illustrated in the cut, a real focus of 0. Conversely, if Of is a source of rays, they will after reflection converge to 0. The two points are therefore called \"conjugate foci.\" The equation for v may be put in the form or REFLECTION 449 A simple geometrical method for determining 0' is as follows: Draw OOM through the centre of the mirror (7, draw OP to any point P near M, and OR through C parallel to it; draw a line PF so as to bisect OR at F\\ where this line intersects the line OM is the image 0' '. For, as has just been shown, O1 lies on the line OM and on PS, where the angles (\u00a3P(7) and (OPC) are equal ; and the intersec- tion of PS with CR may be proved to bisect it. CR is drawn parallel to OP, and F is its point of intersection with PAS'. The angles (RFP) and (FPO) are equal, and (RFP) equals the sum of (FPC) and (FCP) ; hence (FOP) equals the difference between (FPO) and (FPC), i.e. (OP0). But (CPO) and (FPC) are equal (angles of incidence and reflection); therefore (FCP) equals (.FP(7); the triangle (CFP) is isosceles ; and the sides JV and FP are equal. P is supposed to be close to M, and therefore to R ; and so FP {Tactically equals FR. Consequently FC equals FR. Q.K.D. A special case is when u is infi- nite; that is, when the incident waves are plane, with their wave normal parallel to OCM\\ hence v = -, or (X <:ts the line CM. This point is called the \"prill- : cipal focus\" on the line OM. Conversely, if a point source is at the middle point of CM, i.e. if u = r?, the reflected \\\\ aves will be plane and will liavr tln-ir wave normal parallel to CM. It is seen from the formula - + - = -, that, if u < -, v < 0. 112 T u v r 2 That [a, if the point source 0 is between the surf i the AMES'S PHYSICS \u2014 20 450 LIGHT principal focus, v is negative; and therefore 0' is on the opposite side of the surface. This means, then, that it is a virtual focus; and so the rays from 0, after reflection, diverge \"o7 away from the sur- face. It is thus shown that a normal pencil of rays gives rise on reflection to another such pencil, having FIG. 199. \u2014 Special case: 0 is nearer to the mirror M than either a real Or a virtual focus. If the pencil is oblique, this is not the case. Thus, if OPl and OP2 are any two oblique rays which are close together, after reflection they will cross at a point Fl off the axis ; and we can deduce at once the main phenomena for the whole oblique pencil of rays from 0, by rotating the plane fig- to its centre C. The image O is virtual. ure in the cut through a small angle around OM as an axis. The two rays OPl and OP2 will thus describe a cone having as its base a small rectangular IYM \\X c 0 area, of which PjP2 is one of the sides. Fio. 200. \u2014 Formation of focal lines at FI and Ft by an oblique pencil incident upon a concave spherical mirror. The point F^ will de- scribe a short straight line perpendicular to the plane of the paper. So all the rays making up this oblique pencil will, after reflection, pass through two small elongated areas which are practically short straight lines: one at F\u00b1 perpendicular to the plane of the paper; and REFLECTION 151 one at F^ in the plane of the paper at right", "/. Thin is called the \" principal focus.\" Thus a normal pencil gives rte to a homocentrio one; and it may be shown, by following the same method as was used for concave mirror.'-, that an oblique pencil produces an astigmatic one by reflection. 454 Plane Waves Incident upon a Parabolic Minor. \u2014 This is a mirror whose surface may be imagined described by rotating a parabola around its axis ; it is called a \" paraboloid of revo- lution.\" If the rays are all parallel to the axis, they will after reflection all converge to the focus, F, of the parabola. (To prove this requires a knowledge of the analytical prop- erties of the parabola.) In this case, then, there is no spherical aberration. Conversely, if a point source of light is placed at the focus -F, all the rays which are reflected by the surface proceed out parallel to the axis. This is the reason why such mirrors are used in search lights, the headlights of locomotives, etc. FIG. 206. \u2014 Parabolic mirrors. CHAPTER XXVIII REFRACTION Plane Waves Incident upon a Plane Surface. \u2014 This is the case already discussed on page 432. The incident plane waves give rise to refracted waves, which are plane and in such a direction that the incident and refracted portions of any ray and the normal to the surface at the point of incidence are all in the same plane, viz., the \"plane of incidence\"; and, if Ni and JV2 are the angles made with the normal by the two rays, sin is the same for all angles of FIG. 207. - Refraction of a r\u00bby. incidence. As already explained, this ratio, or the index of refraction of the second medium with reference to the Jirxt. as it is called, equals the ratio of the velocities of the waves in the two media. It should be noted that the waves are sup- posed to be homogeneous, i.e. to have a definite wave num- ber. Calling v1 and va these velocities, and writing n^ l for Conversely, the this index of refraction, na> t index of refraction of the first medium with reference to the sn second, n%l equals \u2022 Therefore, if v^v^ sin JV^sin NT and so ^ > N^ ; that is, the refracted ray is bent in closer to the normal than is the incident ray. This is the case illustrated in the cut. On the other hand, if V|<VP -ZV1<-ZV3; and tin- ivfrartrd ray is In-lit away t'nmi tin- normal. This 456 456 LIGHT may be illustrated by the cut, if the arrows indicating the directions of the rays are considered reversed. Total Reflection. \u2014 It is evident in this second class of refraction that, if the angle of incidence, Nv is sufficiently increased, the angle of refraction, Nv may finally equal 90\u00b0, i.e. ^. So the refracted ray just grazes the surface. When this occurs, sin JVj = 1 ; and, therefore, according to the for- mula, sin N2 = \u2014 \u2014 This angle of incidence is called the \" critical angle \" for the two media and for waves of a defi- nite wave number. If it can FIG. 2Ut>. \u2014 i'otal reflection. be measured, 7i2 j, or the corre- sponding index of refraction, may be at once calculated. If the angle of incidence exceeds this critical angle in value, there is no refracted ray, for the sine of an angle cannot exceed unity, and the ray suffers total reflection. The velocity of ether waves in water is less than in air, as is shown by direct experi- ment, or by the fact that a ray in air incident upon a plane surface of water is bent toward the normal. So this phenomenon of total reflection will be observed if rays are incident obliquely upon a surface of water from below. This condition may be secured if one holds a tumbler of water in such a manner that the eye looks up through the glass at the surface of water and turns so as to face an jllu- mmated object. If the direction in which one looks is suffi- ciently oblique to the surface, nothing is seen through it ; for it acts like a plane mirror. A piece of apparatus that is often used to change the direction of a beam of light, called a \" totally reflecting prism,\" consists of a glass \"1,0'REFRACTION 457 triangular prism whose cross section is an isosceles right-angle triangle. Light incident normally upon one of the smaller faces suffers total reflection at the hypotenuse face and emerges perpendicular to the third face. Index of Refraction.", " \u2014 As defined above, the value of the index of refraction of a substance depends upon its optical properties with reference to some other medium. The natural medium to which i7g prism.7'all indices of refraction should refer is the pure ether; and so, when the words \"the index of refrac- tion \" are used, this is implied. Thus, if VQ is the velocity of \\\\iives in the pure ether, and wlf0 the index of refraction of any substance, wx 0 = \u2014 ; similarly, w2 0 = \u2014 ) for a second substance, and therefore n2> 1 = -^- Experiments show that unity, nearly of air is very of refraction the index about 1.0003; and therefore the index of refraction of any solid <\u00bbr litpiid substance, such as glass or water, with reference to air is practically the same as with reference to the pure ether. In actual measurements of indices of refraction, it is customary to determine them with reference to air and then to in ahe the necessary correction. In the illustrations which f(\u00bblh\u00bb\\\\. // will indiejite the index of refraction of a substance witli reference to air. fact that air r - shown by the unsteadiness of objects \\V|HMI s\u00bb'\u00ab-n through airriiini; from a hot stovr or tirl.l : t In- ditY.'r.-nt J-T \u2022 \u00bbf the air, being at different t<-,, n-frart diffnvntly. and tin- rtT.-.-t i, thuaame as if a pane of glass whirh i> un. \\. u and \"streaky\" is moved up and down in front of t I'hr refraction of n ini|Nirtant. too. in all a^tron\u00ab'niic;il,>\\^, 'I'll.- fa.-t that the waves travel with diff.-r.-nt fdoottiM in layers of air at iliffi-r.-nt t.-nn-ratnn-i i-< shown by the fact that li^ht is rtjltcted l.v ^nr-h lavi \\|-lanation of mirage and similar phenomena depends upon this. 458 Lid I IT Special Cases. \u2014 The refracting matter is generally made into a figure with regular geometrical surfaces. There are three cases of special interest : (1) a \" plate,\" which is a figure bounded in part by two parallel planes ; (2) a \" prism,\" which is a figure bounded in part by two non-parallel planes ; (3) a \"spherical lens,'-' which is a figure bounded in part by two spherical surfaces, and which is symmetrical around the straight line joining their centres. We shall discuss briefly the path of a ray in passing through these various figures. 1. Plate. \u2014 Let the plate be placed with its parallel faces per- pendicular to the sheet of the paper, and consider a ray incident in this plane. This is illustrated in the cut. If JVj is the angle of incidence upon one plane face and N2 that of refraction, the angle of incidence of the ray upon the FIG. 210. \u2014 Refraction of a ray by a plate. other plane face, N2', must by the laws of geometry equal the angle N2 ; and the angle of refraction, or of emergence, out into the original medium, NI, must equal JVj ; for and, since N^ Therefore the emerging ray is parallel to the incident one, but is displaced sidewise an amount depending upon the angle of incidence, the thickness of the plate, and its index of refraction. It follows, then, that plane waves incident upon a plate emerge in the form of plane waves parallel to the incident waves. The case of spherical waves will be considered later. 2. Prism. \u2014 The straight line in which the two non-paral- lel surfaces meet (or would meet if prolonged) is called the REFRACTION 459 \"edge,\" and the angle between them is called the \"angle\" of the prism. Let the prism be placed with its edge perpen- dicular to the plane of the paper, and consider a ray incident in this plum-. This is illustrated in the cut. Call the angle of incidence upon the prism N-^ ; that of refraction, Nz ; that of incidence upon the second face of the prism, N% ; that of refraction out into the original medium, NJ ; that of the angle of the prism, A ; and that between the directions of the entering and the emerging rays, D. Fio. 211. \u2014 Refraction of a ray by a prism. It is evident from the geometry of the figure that the angle b\u00ab -tween the two", " normals drawn to the two surfaces equals the angle of the prism ; and that the following relations are trii.- : Further, JVj and NT and NJ and NJ are connected by the ivt'rart i\u00ab\u00bbn I'nrinula; and so the value of D, the \"de-. \" as it is called, can be deduced in terms of A, Nv and n. It \\\\a\\rs nf dilTrnmt wave numbers (or colors) are inci- dent upon a pn>in, it is observed that it deviates these waves 460 LIGHT to different degrees, thus showing that these waves have dif- ferent indices of refraction. If n is large, the ray is refracted more than if n is small ; and therefore the deviation is great, as is evident from the cut apart from the formula ; as a result, if white light enters the prism, it is dispersed into a spectrum of colors. (With glass or water the shorter waves, e.g. the \" blue ones,\" are deviated more than the longer ones, e.g. the \"red ones.\") (See Chapter XXX.) It may be seen by actual experiment, and it may be proved by methods of the infinitesimal calculus, that as the angle of incidence is varied gradually from normal to grazing inci- dence, i.e. from 0\u00b0 to 90\u00b0, the deviation gradually decreases, reaches a definite minimum value, and then increases ; and, further, that this minimum deviation is obtained when the angle of incidence, Nv equals that of emergence, N^ ; in other words, when the ray is symmetrical on the two sides of the prism. Call this angle of minimum deviation D; then, since N^_ = N^ N2 = NJ, and the two formulae above become It follows that, since we may write n = \u2022 A and D may both be measured with accuracy; and so n may be obtained. (Reference for details of the method should be made to some laboratory manual.) If homogeneous plane waves parallel to the edge of a prism are incident upon it, they will therefore emerge in the form of plane waves, but will be deviated through a certain angle. The case of spherical waves will be discussed later. ni: FRACTION 461 3. Spherical Lens. \u2014 Let the line joining the centres of the two spherical surfaces, the \"axis\" of the lens, as it is called, lie in the plane of the paper. The section of the lens will be one of the forms shown in the cut, for a plane surface is a Fio. 212. \u2014Different forms of lenses. special case of a spherical one. These are called \"double convex,\" \"plano-convex,\" \"concavo-convex,\" \"double con- cave,\" etc. Consider a ray incident upon the lens. It is refracted into the lens at one point and out at another, exactly as if there were minute plane surfaces tangent to the lens at these points. These minute planes constitute a prism ; and therefore a lens may be treated as a special case of a combination of a great number of pris- matic faces. (Other lenses than spherical ones are often used, especially in the construction of spec- tacles.) When Diane waves are Fio. 818. - Refraction of \u00bb r\u00bby by a tons. incident ii] inn ;i lens, its different rays meet the surface of the lens at different angles of incidence and on emerging they are refracted out at different angles; so nothing can be said immediately in regard to the nature of the trans- mitted waves. Plane waves are a special case of spherical ones ; and the theory of the action of a lens upon the latter will be given shortly. The values of tli\u00ab- indices \u00ab.f refraction of a few substances are given in the following table: LKillT INMCKS OF REFRACTION FOR YELLOW LIGHT OF WAVE LENGTH 01. SlTMTAXOX INDBX o\u00b0c. TEMPERATUBB 0\u00b0 pressure 0\u00b0 76 cm Air Oxygen Alcohol Carbon bisulphide.... Water Rock salt Flint glass Crown glass 1.000043 1.000L>9\u00b1> 1.000140 1 000297 1.000272 1 360 1 449.624.334 5441.651.524 5 6 \u00b0 \u00b0 2 1 Spherical Waves Incident upon a Plane Surface. \u2014 Let the surface be perpendicular to the plane of the paper, and let the point source 0 lie in this plane. Let OB be any ray incident upon the surface at a point B near the foot of the perpendicular OA dropped from 0 upon the surface. Let BO be the re- fracted ray, and \u00b0 4 2 Fw.314. \u2014 Formation of Image* of a point source 0 by a plane B01", " its prolonga- tion backward until \u2022nrfkoe: (1) when n>l ; (2) when n<l.,\u2022 \u2022> \u2022, j,-T it meets the normal OA. There are two cases to be con- sidered, depending upon whether the velocity of waves in the first medium is greater or less than that in the second. In either case let n be the index of refraction of the second medium with reference to the first. If the waves in the \u00b0 \u00b0 5 5 1 1 \u00b0 5 2 0\u00b0 REFRACTION 403 latter have a givati'r velocity than in tlie former, n > 1 ; if thi-ir velocity is greater in the former, w<l. These two cases are illustrated by the two cuts. In each the angle of incidence equals (BOA); and that of refraction j&OA. Therefore, nnce the sine of, and the sine of (^01^L)=' the index of 0 B l refraction, n = \u2014 \u2014. If B is extremely close to A, that is, OB if the ray OB is one of a normal pencil, we may replace the ratio Qi by -. So n = or O^A = n OA. It fol- lows, then, that Ol is a point at a fixed distance from the plane surface for all the rays of the normal pencil provided the waves are homogeneous, so that n is a constant. If n > 1, Oj is farther from the surface than 0 ; if n< 1, it is nearer the surface. In other words, a normal pencil from a point source 0 gives rise to a pencil of rays by refraction whose centre is Or the virtual image of 0. Conversely, if we imagine the directions of all the rays reversed, a normal pencil of rays in the second medium converging toward a point Oj in the first will actually meet at a point 0. A luminous object in one medium will thus give rise to a virtual image of itself. This image will not be of the same M/.r a^ the object; but their relative dimensions may be easily calculated. The treatment of this case of refraction of spherical waves at a plane surface l.y the method of waves is as follows: Let the source O be in the medium in which the waves hav. velocity, \\\\hich may therefor*'!\"\u2022 railed the \"slow medium,\" and let the waves meet tin- MI, face of the \"faster medium\" at A, as shown in -!.\"\u00bb. lierefore perpendicular to the surface. If the velocity..f the wav.-., in thr t\\\\\u00ab> media \\\\--r. \u2022 tin- same, the wave front would advance in a definite tim \u2022\u2022 t<> the ].<.siti,,n PQP* \u2022 but, since the new medium is the di>turl'an.-e t, :i,tead of Q, and the actual curve of the entering waves i- /'*,' /'. an arc of a circle whose centre n' is on the lin- O.4,but nearer the surface than O. O1 is then the virtual image of O. 464 LIGHT Consider also the case where the source 0 is in the \" faster medium \" ; the spherical waves meet the surface at A, the foot of the perpendicular from 0 upon the plane surface. The waves entering the other medium would have the form PQP' if the velocity were unchanged; but, since the medium is \"slower,\" the Case when n>l. Fio. 215. \u2014 Refraction of spherical waves by a plane surface. REFRACTION 465 actual curve is PQ'P', an arc of a circle whose centre is 0', a point on ()A, but farther away from the surface than O. (/ is, as before, the vir- tual image of O. The distance between the point source and its image may be calculated from the formula already deduced : O^A = nOA. For OOj = OA - O^A = OA (1 - n). Thus, if n > 1, the object appears to be farther from the surface than it really is ; while if n < 1, the opposite is true. This is illustrated when one looks at a stone lying at the bottom of a pond of water ; in this case 0 is at the bottom of the water, and 01 is nearer to the surface by a distance OA (1 \u2014 w), where OA is the actual depth of the water and n is the index of refraction of the air with reference to the water; i.e. it equals \u2014, where n' is the index of refraction of the water n 4 with reference to the air. (n' for water is approximately -; OA \\ so 00l is. J A method is thus offered", " for the measure- ment of w, since both OA and 00l can be measured. If the pencil of rays from 0 is oblique, it forms by refraction an astigmatic pencil. Thus, ^ Fw. 116. \u2014 Fnrin.itl..n of focal line* at A\\ and A', by the refraction of an oblique pencil at a plane Rurface. if OB and OS are two oblique rays, they form by refraction two rays which if prolonged backward cross at Fv -,\\ point off the perpendicular line OA. Therefore, the 217. \u2014 Caustic formed by reflection at a plane surface. AMES'S PHTSICS \u2014 30 466 LIGHT whole oblique pencil gives rise to two virtual focal lines: one at Fl perpendicular to the sheet of the paper : the other at JP2 along the line OA. Further, if we consider all the rays from 0 falling upon the surface, they give rise by refraction to a virtual caustic surface with a cusp at Ov the image of 0 for a normal pencil. Special Cases 1. Plate. \u2014 We shall consider the plate made of a material in which the waves have a less velocity than in the surround- ing medium, i.e. n>\\. Let 0 be the source of the spher- ical waves. The path of any ray is indicated by OB, BB', WC. 0' is the image of 0 in the first surface ; and 0\" is the image of 0' in the second one. So, if the pen- cil is a normal one, all the rays leaving 0 will diverge after emerging from the plate as if they came from FIG. 218. \u2014 Formation of an image of a point source O by refraction through a plate. 0\". Its position may be at once calculated. 2. Prism. \u2014 We shall consider the prism made of a material in which the waves have a less velocity than in the surround- ing medium, i.e. n>l. Let 0 be the source of spherical waves. The path of any ray is_indicated by OB, SW, WC. 0' is the image of 0 in the first surface ; and 0\" is the image of 0' in the second surface. It should be observed that, if the ray., Fro. 219. \u2014 Formation of an Image of a point IS One Of the normal pencil flource 0 by refraction through a pri8m. REFI! ACTION 467 for one surface it is oblique to the other when it is incident upon it. Therefore, in general, an incident pencil of rays will, after two refractions, emerge in the form of an astig- matic pencil; the rays will apparently come from two focal liiu's. It may be shown, however, that, if the pencil of rays is incident upon the prism at that angle which corresponds to minimum deviation, the virtual image from which the emerging rays apparently diverge is practically a point. So in this case the emerging pencil is homocentric. Owing to the difference in the index of refraction for waves of different wave number, it is evident that, if there is at 0 a source of several different trains of waves, each of them will have a different virtual image O\". The greater the value of n, i.e. the shorter the wave length, so much the more is O1 displaced from 0. The prism then disperses the light. 3. Spherical Lens. \u2014 It is evident that any ray which falls upon a lens suf- refraction \u2022\u00ab\u00bb, once on entering the lens ;ind a^ain on leaving it. We must therefore discuss, as a liminary to the treatment of the lens, the re f mo- tion of a ray at a spherical surface. There are two cases, which we shall consider separately: refraction at a concave surface and at a convex one. i. /'\u00ab. Fio. MO. -Formation of Images of a point nourre O by refrmction at \u2022 concave spheric*! surfcoe: (1) when n>l ; (2) when n<l. a. Concave Surface. \u2014 Let the section made by a plan, tli rough the oentre of the surface O and the point source 0, 468 LIGHT i.e. through the \"axis\" OA, be that shown in the cuts; let OB be any incident ray, and O'B the prolongation backward of the refracted ray ; and draw (7#, which is therefore normal at B. (The two cuts illustrate respectively the cases when the index of refraction of the second medium with reference to the first is greater and when it is less than one.) The angle of incidence is (OBC) = Nr The angle of refraction is (O'BQ) = N2. Then fay the well-known", " trigonometrical formula, which states that in any plane triangle the ratio of the sines of any two angles equals that of the lengths of the opposite sides, we have from the triangles (OBC) and (O'BC): sin ^ : sin (BOC) = OC : BC, sin N2 : sin (BO'C) = WC : BC. Therefore, since = n, and si\"^^> = ff (as is smBO'C) OB seen by dropping a perpendicular line from B upon OA). OC O'B O'c O B'If B is very close to A, that is, if the ray is one of a normal pencil and if the curvature of the surface issmall,we may replace the ratio by ; and so n = x - This OB OA 0 C OA formula shows that, if a normal pencil from 0 is incident upon the surface, it gives rise to a homocentric pencil with its centre at Of ; so 0' is the virtual image of 0. This can be expressed in a simpler mathematical form. Call the dis- Then tances OA, u ; V~A, v\\ CA,r. OC = u - r, O'C = v \u2014 r; and \u00bb = ^H\u00a3. v \u2014 r u Hence or > PEFR ACTION n- 469 If w < 1, it may happen, therefore, for a suitable value of 0 that v is negative. This means simply that Of is in on the opposite side of the surface from 0 ; for in this formula the positive direction is that defined by r, namely from 0 to A. b. Convex Surface. \u2014 The same treatment and the same formulae as those oc (y A above lead to the same result for a convex surface, viz., n = \u2014 \u2014 \u2022 \u2014 \u2014. L/C OA And since \u00ab= OA and r = CA, PC = OA + AC = OA - CA. Hence, as before, OC = u \u2014 r, and O'C = v \u2014 r ; and on substitution, we again have =, or - = - + r v u v u r If O is on the opposite side of the surface from C, u has a negative value. For instance, if in the cut the length of the line OA is x, the value of u is \u2014 x. If n' is the index of refraction of the first medium with reference to the second, n' = - ; and the general formula becomesn't; u n'r & u l=\u00a3, or! =!' Conversely, if a pencil of rays in the second medium is converging apparently toward a point C? in the first nn-dium, tlu-y will actually meet at the point 0. Fio. 221. \u2014 Formation of an Image of a point source 0 by \u2122*\u2122\"\u2122 \" \u2022 <*>\u00ab\"\u00ab* \"P\u00bb>w leal nurfa*, ease when n>l the pencil of rays is oblique, it gives rise to an astig- matic pencil ; and if all the incident rays are considered, the image is a caustic surface. This phenomenon is said to be \u2022 In* \u2022, as in other similar cases, to spherical aberration. We shall now return to the problem of the refraction pro- duced by a spherical lens. It is evident from what has just been shown that a normal pencil from any point on the axis 470 LIGHT of the lens will give rise to a homocentric pencil emerging from the lens after the two refractions ; the image of the source, though, may be either virtual or real. We shall deduce the formula for a double convex and for a double concave lens, and then show that, by a suitable agreement as to signs, one formula may be used for all lenses. We shall assume at first that the lens is so thin that a ray incident at any point of one surface emerges from the other surface at a point which is at the same distance from the axis as is the former. a. Double Convex Lens. \u2014 Consider a section through the axis. Let 0 be the point source. FIG. 222. \u2014 Formation of an image of a point source 0 by refraction through a double convex lens. Let PJ be the radius of the first spherical surface of the lens. Let r2 be the radius of the second spherical surface of the lens. Let n be the index of refraction of the lens with reference to the surrounding air. The formula for refraction at the first surface of the lens *>i ui ri is, then, \u2014 = h n ~~, where u* and v* are positive if 0 and its image lie on the right of the lens, because the centre of the first surface of the lens lies on this side. The refraction REFRACTION 471", " at the second surface is from the lens out into the air ; so the formula is \u2014 = \u2014 h \u2014 \u2014, where u2 and v2 are positive if the v2 \"2 r2 points to which they refer are to the left of the lens. But, if the lens is thin, v1 = \u2014 w2, for the rays incident upon the second surface are those diverging from the virtual image produced at the first ; but a quantity u or v which is positive with reference to one surface is negative with reference to the other, since their centres are on opposite sides of the lens. Therefore we have the two formulae : n _ 1 t n-1 \u00bb! tij r, 1 _ n | 1-n V9 t>! r II. -nee, I + l = -(n- ti, t;, In this formula, MX is the distance 0 lies on the axis to the right of the lens (referring to the cut) ; and v2 is the distance the image produced by the second surface of the lens lies to its left. Therefore, if we agree to call the distance the point source lies to the left of the lens w, and the distance the image lies to the right of the lens v, u = \u2014 ur v = \u2014 va. So the formula becomes Tin; quantity on the right-hand side of the equation is a con- staiit quantity t<u a -ivcn lens, and it is essentially positive if n>l, as it is in all ordinary cases, e.g. glass, quartz, etc. lenses surrounded by air. We write this quantity \u2014 ; and the formula then assumes the final form * 472 LIGHT b. Double Concave Lens. \u2014 In this case the formulae for refraction at the two surfaces are, as before, where u^ and v1 are positive if 0 and its image in the first surface lie on the left of the lens ; and Fio. 223. \u2014 Formation of an image of a point source 0 by refraction through a double concave lens. where uz and v2 are positive if the points to which they refer are on the right of the lens. Hence, or, introducing the same agreement as to signs as in the pre- vious case, u = uv v = v2 ; and hence u v f We can therefore use for both kinds \u2014 in fact, for all kinds \u2014 of lenses the one formula, 1,1 1 in which u and v have the meanings agreed upon above and / is positive for some lenses and negative for others. We RJCFBAi Y70.Y 473 shall discuss presently the properties of these two types of lenses. The formula for a lens may be deduced in another manner, possibly more instructive, if we assume the fact that a point source on the axis gives rise to a point image. The principle made use of in this second method is that the \" optical \" lengths of the paths from the point source to its image are the same along all rays. (See page 43G.) Fio. 224. \u2014 Refraction by a lens. Consider the case of a thin double convex lens, as shown in the cut on an enormously magnified scale. Let 0 and & be the source and its \", both on the axis ; so 0.4, AB, BO1 is the normal ray from 0. Let OC be any incident ray making a small angle with the axis. Since the I'M i- is assumed to lie thin, the portion of the ray within the lens, namely <'/>. is parallel to the axis; an.l th\u00ab- other portion is DO'. Then the \" optical \" lengths of the two rays are ~OA + nAB + Wf and ~UC + nCD+ D&. Since these are equal, UA + nJB + BV = UC + nCD -I- D&, or U&-OA +D0 -E& = n(AB - CD) = n(AE+ 7-70. to the. *ign* of the. lines, this may be written OC-UE + AE+D&-FO' + FB = n(A$ + FB), or <tE) + (S& - FCO = (n _ \\)(AE + FB). 474 LIGHT These three quantities may be expressed in a simple form provided the angles between the rays and the axis are small, as we have assumed to be the case. Thus, describing an arc of a circle RS around P as a centre, and drawing the two radii PR and PS, and the perpendicu- lar line RQ from R upon PS, we have the following formula : FIG. 226. \u2014 JiS is portion of a circle whose centre is P. Hence, if PR is large compared with RQ, we have, on expansion by the binomial theorem and neglecting small terms, 2 PR", " 2 PS 2 PR* PR-PQ = - Therefore, or. In the formula, as given above, for a lens, then, 2 PS 2 P R'O f f -S 1 CT? Consequently, \u00b1 + ^ = (\u201e - 1)(I + I). If the incident ray makes a sufficiently small angle with the axis, and if the curvatures of the two surfaces are small, we may replace OC by OA, i.e. u, and D<y by BO1, i.e. v ; and we have the general formula, as before, rj rzl f REFRACTION 475 We shall now discuss the two types of lenses referred to above : for one/ is positive ; for the other, negative. a. Lenses for which f is positive. \u2014 The general formula is A -f _ = _. Let us consider several special cases. u v f If the point source is removed farther and farther from the lens, but kept on the axis, u approaches an infinite value and the waves be- come plane as they reach the lens. When w = oo, it is seen from the for- mula that v =/; and, since / is positive, this means that there is a real imae'e F10-226- \u2014 Special case: the source 0 Is at an Infinite distance on at a distance / the axis. from the lens. This point is called a \" principal focus \" of the lens. We may express this fact in words by saying that plane waves advancing with their normal parallel to the axis are changed by the lens into converging spherical waves whose centre is at a distance / from the lens ; or, rays parallel to the axis are bent by the lens in such a manner as to converge to a point on the axis at a distance/ from the lens. Similarly, if w=/, it is seen from the formula that 0 = 00. In words this states that, if the point source is on the axis of the lens at a distance from it equal to /, the diverging spheri< -al waves are so converged by the lens as to become plane and to advance in a direction parallel to the axis; or, we may say that rays through a point on the axis at a dis- tance / from it are SO detleeted by the lens as to become parallel tn its axis. Conversely, we can have plane waves incident upon the lens from the other side, which will con- verge to a point at a distance/ from it. There are thus two \u2022I7\u00bbi ( LIGHT principal foci ; one on each side of the lens, and at the same distance / from it, if the lens is thin. This distance is called the \" focal length.\" Therefore, if we consider the point source as at an infinite distance from the lens, its image is the principal focus on the other side; and, if the source approaches the lens, the image recedes from it, until, when the source reaches the principal focus, the image is at an infinite distance. When the source is between the principal focus and the lens, i.e. when w</, it is seen from the formula that v < 0, so the image is on the same side of the lens as is the source, and is, therefore, virtual. When the source reaches the lens, i.e. when u = 0, v = 0 also ; so object and image coincide. If u has a negative value, the physical meaning is that rays are converging on the lens apparently toward a point on the other side, which may be called a \" virtual \" source ; and, in this case, as is seen from the formula, v is positive and less numerically than u ; but it should be noted that v does not exceed / in value so long as u is negative. Therefore, the converging rays are converged still more, and form a real image. It is thus seen that a lens for which / is positive always converges waves which fall upon it; for this reason it is called a \"converging lens.\" The ordinary form of a con- verging lens is double convex ; but any thin spherical lens thicker along the axis than elsewhere is a converging one. We may arrange in tabular form the facts proved above in regard to u and v : U = CO.... V = f \u00b0\u00b0>w>/.... *>>v>f U =f.... V = 00 />M>0.... \u00ab>>-l>>0 0>w>-cc... />y>0 It is a simple matter to find by geometrical methods the position of the image of an object, for we know the effect of UK FRACTION 177 the lens upon three of the rays from any point source : A ray parallel to the axis is deflected so as to pass through the principal focus on the other side of", " the lens ; a ray passing through the principal focus on the \" incident side \" of the lens will emerge on the other side parallel to the axis ; a ray meeting the lens at the point where it is intersected by the FIG. 887. \u2014 Formation of images by \u00bb converging lens., i.e. the \"centre of the lens,\" keeps its direction unal- tered, because at tliis point the two faces of the lens are par- allel, and it is equivalent to an infinitely thin plate. In showing graphically the formation of images l>y a lens, we shall represent the lens by a straight lino, lot- simplicity. Thus, let Fr F.,. and C!><\u2022 the principal foci and the centre <>f the lens, and P any point of the object; its image is atP'. Two cases are xlmwn : if /'is farther from the lens than I he principal fnens. the image is real: if /' is between the lens and the principal focus, the image is virtual. 478 LIGHT If the object is small, close to the axis, and perpendicular to it, as represented by OP, its image O'P' is also perpen- dicular to the axis. So 0' is the image of 0\\ and, since these points are on the axis, 00= u and C0' = v. Further, by similar triangles, P 0 : OP1 = u:v. The ratio of the length of the line O'P' to that of OP is called the \" linear magnifica- tion \" of the lens. It is evident from the geometry of the cut that Therefore, the magnification of the surface of any portion of C O O P O'P'the object perpendicular to the axis is ^-'O C A case of special interest is when the source, P, is placed in a plane perpendicular to the axis at the principal focus, or FIG. 228. \u2014 Special case : the point source P is in the focal plane of the converging lens. in the \"focal plane,\" as it is called. It is seen from the geometry of the cut that the two emerging rays which we can draw from known principles are parallel to the line join- ing P to the centre of the lens, (7; for the triangles (P-Fjtf), (PAC), and (ACF^) are all equal. Therefore, all the rays diverging from a point in the focal plane, which meet the lens, emerge parallel to the line joining this point to the centre of the lens ; or, in other words, spherical waves diverg- ing from such a point are made plane by the lens and proceed in the direction of the line referred to. Conversely, if we REFRACTION 479 these ravs reversed, parallel rays incident obliquely upon the lens. Imt. at a small angle, converge to a point in the focal plane where this plane is intersected by a line through the centre of the lens parallel to the beam of rays ; or, plane waves incident obliquely at a small angle upon a lens are brought to a focus at a point in the focal plane, as just defined. P, F,. Fi... 229.\u2014 Construction for the refraction of any ray PQ by a converging lens. This fact enables us to construct at once the refracted por- t ion of any incident ray. Thus, let PQ be a ray incident at @ upon the lens, whose principal foci and centre are Fv Fy and (7. Draw the focal plane through Fv and a line BC through 0 parallel to PQ\\ this meets the former in a point A. Draw the line QS, joining Q to A\\ this is the emerging ray. Kor, if there were two parallel rays PQ and BC, they would converge to A ; and therefore PQ must produce a ray which passes through A. Again, r, f. rring to Fig. 230, let there be two beams of parallel rays incident u]..,n the lens, one having the direction /'r. the other tiny \\\\ill be brought to a focus at Q and Qv in the focal plant-. (I hi- i- the case when an image of a ilistunt object is formed on a screen or photographic plate by a l<-n-. t'or \u2022\u2022arh point of the object sends out ;ippp. \\imal. \u2022!> parallel to each oth\u00ab-r \\\\licn they i-.-ach tlir lens.) F\u00ab D \"angular separation \" of the two beams, i.e. (/'CV,) tli- 'I f the focal length of th\u00ab>!-\u2022: increased. If this angle is small. is the focal length, the length of the line Q(^ <\u00abquals the | / l.y this", " nngl.-, for the an (PC/*,) an- 1 C<\"\\',) an- t-.jnal. If these two rays com*' from t\\\\o| 480 LIGHT on the edge of the distant object, its image will be bounded by the two points Q and Ql ; and the linear dimensions of this image will vary directly as the focal length of the lens. Therefore the area of the image will vary as the square of the focal length. If a lens having a long focal length is used, the size of the image is great; but, if a pho- tograph is to be taken, the time of exposure must be prolonged because FIG. 230. \u2014 Formation by a converging lens of an image of an object at an infinite distance. the energy is distributed over a large area. b. Lenses for which f is Negative. \u2014 The general formula is - -f- - = -^. If the source is at an infinite distance, u = oo, and therefore v=f; but since f is negative, v is also, and the image is a virtual one on the same side of the lens as the incident waves. This is called a \" principal focus \" ; and its distance, /, from the lens is called the \" focal length.\" Similarly, there is another F vx Fro. 231. \u2014 Special case: the point source is at an Infinite distance on the axis. principal focus on the axis on the opposite side of the lens, and at the same distance from it if the lens is thin. If u=f, i.e. if the rays are converging apparently toward the principal focus on the opposite side of the lens, it is seen from the formula that v = oo, i.e. the emerging rays are all parallel to the axis. We can, moreover, in a similar manner to that used in the previous case, discuss the relation between the positions of the object and image as the point source moves from -f- oo to \u2014 oo. It is seen at once that the effect /,'/\u2022;/\u2022\u2022/,'.!' T/o.Y 481 of the lens is to make the incident rays or waves diverge; for that reason lenses of this type are called u diverging lenses.\" All thin lenses which are thinnest at their centres are diverging. We can also deduce graphically the position of the image of any point source, because we know the effect of the lens upon three ravs : a ray parallel to the axis emerges in such a direction that if pro- longed backward it would ^ F Fio. 282. \u2014 Special case: the Incident rays are converging toward the principal focus on the farther side of the lens. meet the axis at the principal focus ; a ray pointed toward the principal focus on the other side of the lens emerges parallel to the axis; a ray through the centre of the lens retains its direction unchanged. A few cases will be drawn, the lens being represented as before by a straight line. F, O'Pia. 988.\u2014 Formation of an Image by a diverging lens. A real source P gives rise to a virtual image P' ; and a small object O/' perpendicular to the axis has an ini.r \u2022< 0 /' also perpendicular to the a\\ Thelinearmagnification prod viced by the lens, that is the ratio'> /' r r2 \u2014, equals as before - : and the surface magnification is -a. Vr a u* AMES'S PHYSICS \u2014 31 482 LIGHT A case of special interest is when the virtual source is in the focal plane. See Fig. 235. Let this point be P ; draw FIG. 234. \u2014 Formation of an image by a diverging lens. The \"virtual source\" P has a real image P', provided P is between the lens and the focal plane. two rays pointed toward it, one parallel to the axis, the other through the centre of the lens ; it is seen by geometry that the emerging rays are parallel, for the triangles (^(74), ((TAP), (CF2P) are all equal. Conversely, rays which are parallel to each other and inclined slightly to the axis diverge, Fio. 285. \u2014 Special case: the virtual point object P lies in the focal plane. after emerging from the lens, as if they proceeded from that point in the focal plane on the incident side where a line through the centre of the lens parallel to the rays meets the plane. This furnishes us with a method for the construction of the emerging portion of any ray. Let PQ be a ray meeting REFli ACTION 4K3 the lens at Q : draw through (7 a line CB parallel to PQ, ami intersecting the focal plane at Fliu AI draw through A the line AQSi QS is the continuation of the incident ray. For, if there", " were two parallel rays PQ and BC, they would diverge, on emerging, as if they came from A. F, \\S Fio. 286. \u2014 Construction for the refraction of any ray PQ by a diverging lens. Spherical Aberration, etc. \u2014 It may not be useless to state again the assumptions made in the above treatment of lenses: the lens is supposed to be thin; the object must be small and close to the axis; the pencils must be normal; the lens must have surfaces whose radii are large in comparison with its dimensions. If any of these conditions are violated, the laws cease to hold. Reference should be made to some special treatise such as Lummer, Photographic Optics, for a full discussion of the general subject. We have also assumed ttimughout that the waves were homogeneous, for n has been treated as a constant. Resolving Power of Lenses. \u2014 It has already been explained page 436) that when the waves from a point source fall upon a converging lens, the image of this source is the point to which liy far the greater amount of the energy is brought, but that owing to diffraction energy is relieved by other points also. In the case of a lens with a circular edge \u2014 such as most lenses have \u2014 the diffraction pattern consists of a bright area, whose brightest point is the geometrical image, 484 LIGHT as just explained, and this is surrounded by rings alter- nately dark and bright, if the waves are homogeneous; that is, the light gradually fades, then increases gradually, etc. So, if there are two point sources close together, the line joining which is perpendicular to the axis, there will be two diffraction patterns, which will overlap; and the resultant effect is due to their superposition. It is evident that if the two point sources are so close together that the centres of their diffraction patterns almost coincide, it may be impossi- ble to see these centres as separate bright points; but, if the centre of one pattern coincides with the first dark ring of the other, then the existence of two bright points may be recog- nized. There is thus a limiting value of the nearness of two point sources which can be perceived as such by the use of the lens. This quantity may be deduced by considering a simple case. Let AOB be the cross section of a converging lens by a plane through the axis, and let 0' be the image of 0, a point on the axis. 0' is the image of 0 because the disturb- ances along the 0' different rays from the latter reach the former in the same time, and so are in Fio.287.-Diagramtoillu8tratethere8olvlngpowerofalen8. the game phage Qf vibration; but, if a point near 0' is considered, the different disturbances from 0 reach there in different phases, because they pass over different \"optical\" paths; and it may happen that the difference in path is such that owing to the disturb- ances arriving there in opposite phases they annul each other's action. If this is the case, this point is on a dark ring. If there is another point source at P, where OP is perpendicu- lar to the axis, the disturbance produced by its waves at 0' REFRACTION 485 may be calculated in a similar manner. Draw PA and A&, PB and BO\\ these are the two lines from P to (7 whose difference in length is the greatest. If this difference amounts to a whole wave length of the waves, it may be assumed that the resultant action at O1 due to the disturbances from P which pass through one half the lens differs in phase by half a wave length from that due to those which pass through the other half of the lens; and therefore 0' will be on the first dark ring of the diffraction pattern due to P. From what was said above, when this is the case, P is as close to 0 as it is possible for it to be and yet to be seen separate from it. This condition may be expressed in an equation: (PB + B&) - (PA + ZO7) = /, where I is the wave length. But 0' is the image of 0\\ and, therefore, drawing 03 and OBy we have, since the optical lengths from 0 to O1 are equal, _ (OB + BO') = (OA+AOf). Subtracting this equation from p the previous one, we have ^ ^ (PB-OB) + (OA - PA)=l. These quantities may be ex- *\u00bb\u2022\u00ab\u00ab\u2022 -Portion of Fig. pressed in a simpler form. Draw- ing the diagram on a larger scale, and dropping a perpen- dicular 1>(J tmm P upon OA, it is seen by similar triangles _ l", "int since OP is extremely small, 0Q = &A -PA. It may be shown in a similar manner that 486 LIGHT Hence, substituting in the above formula, it follows that 2dP^ = l,oTdP = l -2fL = / \u00b0A. OA 2CA AB This marks the minimum value of OP which allows us to recognize the two independent point sources 0 and P. The 1 2 OA. 1 A.B, reciprocal of this, \u2014, or - \u2014 or - \u2014 -, is called the \" re- I (JA. I (JA solving power \" of the lens. There are two cases of special interest : one when we are considering two luminous points of an object near at hand ; the other when the two sources are at a great distance ; one deals with the power of a micro- scope to \" separate \" fine details of a structure ; the other, with the power of a telescope to separate the two components of a \"double star\" or to recognize the details of the surface of the moon or sun. The same formula applies to both, but it may be put in more convenient form. 1. Microscope. Call the angle (AOC), N; then ^4 = sin JV; hence (JA 1 = 2sinjy OP~ I Therefore, the larger the angle subtended at the object 0 by the lens, so much the greater is the resolving power, and therefore the lens should have a short focus, so that the point 0 may be close to the lens and yet may have a real image. Further, the shorter the wave length the greater this resolving power. Thus, fine details may be seen with a microscope if the object is illuminated with blue light, which are not distinguishable in red light. 2. Telescope. Since in this case the object is at a great distance, OA OP c/C equals OC, practically ; and \u2014 equals the angle subtended at the centre of the lens by the distant object. Call this angle M ; then from the formula OP = I Qj- it is seen that M = J-. This is the smallest angle which two objects can subtend and yet be seen as separate points of light in a telescope. \u2014 = - is called the \"angular resolving power.\" AB is the diameter of the lens; and we see the advantage then of using telescopes with large lenses (or mirrors) if one wishes to separate double stars, etc. RKi'n.\\< IION 487 (This elementary treatment of the resolving power of a lens is due to Lord Rayleigh. A rigid treatment leads, as he and others have sho\\\\ n, to the formula OP = 1.22 / QL). I it In using high-po\\u-r microscopes, it is customary to fill the space between the lens and the object with a liquid which has a large index of refraction \u2014 oil of cedar wood is generally used. (This oil has nearly the same index of refraction as the glass, so no light is lost by reflection at the surface of the glass; and, further, spherical aberration is avoided to a great extent.) Then the formula as deduced above must be modi- fied, because the optical paths of the rays from P and 0 include portions in a medium whose index of refraction, n, is different from that of air. Thus, (nPB + BO1) - (nPA + A&) = /, (nOB + BO\") - (nOA + AV) = 0. Hence, n (PB - OB) + n (OA - PA) = I', so the resolving power of the microsco]*' is increased in the ratio of w : 1. n N is called the \" numerical aperture \" of the lens.) The greatest value N can have is?, for which the sine is 1 ; so the least possible value of OP is -\u2014, or, very approximately, one half the wave length of \u2022 \u2014 the light used. No combination of lenses can therefore enable one to see separately two independent luminous points which are closer than thU together. Combination '/'>;> Thin Lenses. \u2014 L<-1 the two lenses be at a (listain. // apart; and let their focal lengths be fr fy The formula' I'm- tin- i\\v<> lenses are: 111 111 \u2014 +-=-, tod - + \u2014 =-i w, ', w, v, /s 488 LIGHT where uz \u2014 h \u2014 vr To find the focal length of the combina- tion on the right-hand side, with reference to Fig. 239, put MJ = oo. Hence, v1 =fr u2 = h \u2014 /x ; and f* or, Therefore, the focal length y2 = FIG. 289.\u2014 Formation of an image by a combination of two thin lenses. A special case", " of this is when the two lenses are placed close together, i.e. when h = 0. Then the focal length equals 0' ; or, calling it /, 1 = 1 + -i- /l-r/2 / /I /2 The reciprocal of the focal length of a lens (or of a combination of lenses) is called its \"power.\" The unit of power adopted by opticians is that of a lens whose focal length is one metre; it is called a \" diopter.\" To find the power of any lens, then, in diopters, its focal length in metres must be measured, and its reciprocal taken. A converging lens is called positive. Thick Lens. \u2014 If the lens is of such a thickness that it cannot be assumed to be \"thin,\" in the sense in which this word has been used above, the solution of the problem of re- fraction can be obtained, exactly as in the case of a thin lens, by considering the refraction of a pencil of rays at the two surfaces. But the distance from the first surface to the image produced by it no longer equals its distance REDACTION from the second surface ; that is, v1 does not equal wa numerically (see page 471). If t is the thickness of the lens, M2 = \u2014 (Vj + j); and, if this value is substituted in the equations, the iinul formula connecting the positions of the object and image may be deduced. Chromatic Aberration. \u2014 Attention has been called re- peatedly to the fact that the index of refraction of a sub- stance is different for trains of waves of different wave number, showing that the velocity of these different waves in the substance is different. (It should be remembered that the wave number of a train of waves does not change as it passes from one medium into another, e.g. air into glass, for the number of \"waves\" reaching any point in a given interval of time must equal the number that leaves it. But, calling N this wave number, and Fi, L, K, L, the velo- Fio. 240. \u2014 Chromatic aborrati-n. cities and wave lengths in the two media, l\\ = Nlr V2 = JV72 ; and since Vl is different from I',. /! is different from lv So, as the waves pass from one medium into another, their wave length changes.) In the case of ordinary dispersion the shorter waves are refracted more than the long ones ; or, in other words, the index of refraction varies in an inverse manner from the wave length. Methods for the study of tin- connection between these two quantities will be described in a later chapter. This fact that n varies with the. wave length is of great importance in dealing with the theory of lenses; for this quantity enters into the two fundamental formulae: and Linear Magnification =-. Therefore, if waves of different wave length are used to illu- minate a Ljiveli nl.j.-ct, or if tllC object itself emits SUci 490 LIGHT not alone will the corresponding images be at different dis- tances from the lens, but they will also be of different sizes. Since light waves of different wave length correspond to different colors, this phenomenon is called \" chromatic aberra- tion.\" For ordinary substances n is, as has just been said, greater for blue light than for green, etc. ; and so, if there are two trains of waves corresponding to these colors, / is less for the blue ones than for the green, and consequently the principal focus for the former is nearer the lens, and the magnification is less. If the object emits white light, there will be a series of colored images of different sizes. This fact is, naturally, most detrimental to the proper use of an optical instrument, for sharp clearly defined images are de- sired. As will be shown in the next paragraph, it is possible by a combination of two lenses of different material to remedy this defect to a certain extent by causing any two definite trains of waves of different wave length to be brought to the same focus, but waves of all wave lengths will not be simi- larly affected. So, when white light is used, if two of its components are thus brought to the same focus, the other components will have different foci \u2014 they form what is called a \"secondary spectrum.\" Thus chromatic aberration can be corrected only partially. The choice of the two trains of waves which shall be brought to the same focus is arbi- trary; if the instrument is to be used for visual purposes, two trains are chosen which affect the eye most intensely ; while, if it is to be used for photographic work, the two trains of waves chosen are those which act most intensely upon a photographic plate. Achromatism. \u2014 Since a lens can be considered as made up of prisms, we shall first", " show how one prism may be so chosen as to neutralize the dispersive action of another prism for two given trains of waves. In speaking of the effect of a prism, it was shown that the deviation produced depended upon the material and angle of the prism, the wave number of the REFRACTION 491 waves, and the angle of incidence. The difference in the de- viations for t\\v<\u00bb di tie rent trains <>f waves is called the \"dispersion\" of those waves; and by suitably choosing the material and angle of the prisms and the angles of incidence, it is evidently possible to secure the same dispersion for two dftiiiite trains of waves. When this is done, the dispersion of any two other trains is as a rule different ; for owing to the difference in the material of the prisms there is no connection Fio. 241. \u2014 Two different prisms of different materials may produce the same dispersion of two rays lt and /,. 1 \"-tween their dispersions in different parts of the spectrum. 'Hi us suppose that, when, white light is incident upon two prisms at angles /and /', as shown in the cut, a ray is dis- persed in such a manner that two of its components, of wave length Jj and lv have the same dispersion D. Let the angle hctween the emerging ray ^ and the normal to the second face of the prism be N and N' in the two prisms. Then if abeam of parallel rays of wave length Zj are ineident at an angle N' upon the second face of the second prism, they will emerge, alon<_r the direction of the incident ray in the cut, i.e. making the angle /' with the normal : and. if a beam of parallel rays of wave length l^ is incident upon the same face at an au^le (N1 + D), they will emerge in the same direction as the former rays. This condition may be secured by inverting the first prism and placing it with its edge parallel to that of the other pi-ism, l.iit M inclined that the angle of incidence upon it of the i. iv /, IV.. m the scc,,nd p] N\\ f..r. as is seen from geometry, the angle of incidence <\u00bbt the ray \u00a3s is, 492 LIGHT under these conditions, (N -f D). Therefore a ray incident upon the second prism at the angle I1 is dispersed by it ; and two of its components, ^ and Z2, fall upon the first prism, are refracted by it, and emerge parallel to each other, making an angle I with the normal to the last face of the prism. Therefore a beam of parallel rays of white light incident upon the second prism at an angle I' will be dispersed by the two prisms ; but all the rays of wave length l^ and \u00a32 will emerge parallel to each other in the direction defined by /. In practice it is found _ja that it is possible to choose the prisms so that, when combined as described above, FIG. 242. \u2014 One prism may neutralize the disper- their adjacent faC6S may be sion of two rays produced by another prism. n i i \u2022 j_i parallel ; but owing to the differences in the material and angles of the prisms the directions of the incident and emerging parallel rays are not necessarily the same. Therefore this double prism still deviates the two rays ^ and Z2, but does not disperse them. Such a prism is called an \" achromatic \" one ; or, either one of its component prisms is said to be \"achromatized.\" (It is evident also that two prisms may be compounded which will not deviate one particular train of waves, but will deviate the others, thus producing dispersion.) It follows at once that, by combining two lenses of different materials, one diverging, the other converging, a double lens may be secured which will deviate two rays of definite wave length, but not disperse them. Such a lens is shown in the cut. The converging lens is as a rule made of \"crown\" glass ; the diverging one, of \" flint \" glass. It is called an achromatic lens for the two definite trains of waves. FIG. 248. \u2014 Achromatic lens. FLINT QLA88 CROWN_GUSS REFRACTION 41 \u00bbo If the component lenses are thin and have focal lengths /! and /2, the focal length of the combination is given by \u2014 = \u2014 | Therefore, we can express the facts in regard / /i /2 to the achromatic combination by saying that / has the same value for the different values of n corresponding to l} ami /.,. In the case of this combination of thin lenses, not alone are the focal", " lengths the same for the two trains of \\\\a\\rs, but also the magnifications. This is not true if the combined lenses are thick. If a system of several lenses \u2014 not in contact with each other \u2014 is to be completely achro- matic for two definite trains of waves, each of its component lenses must be achromatic also. The first achromatic lens was made by John Dollond, a London optician, in 17.~>s. Previous to this, as early as 17*29, a method for achromatizing lenses had been discovered by Chester More Hall of Essex, England ; and he constructed several such lenses, but never pub- lished any account of his work. Newton and others had maintained that if one prism corrected the dis/n-rsinn of another, it would at the same time annul the deviatitm, and that therefore it would be impossible to make an achromatic lens. But when different kinds of glass were tested, it was found that the generalization drawn by Newton from his experiments on prisms of glass and water was not correct. Achromatic Combinations. \u2014 It may be proved that if two thin lenses of focal lengths /J and /2 of the same kind of glass are placed at a distance h apart, where h = \u2022*!, the combination has the same focal lengths for waves of all lengths. The distance of the principal focus -Jd h -f\\ -ft fn.m the end lens of focal length /* is, as we have seen, {*'~ ^ an\u00abl the distance of the oth\u00ab-i- principal focus from the other end is ft There are two special combinations \\\\i,i,-h an- <>f interest because of th-ii DM microscopes or telescopes. 1. / <'c?. \u2014 In this tli.-n- are two plano-convex lenses of the same kiml oi t!n\u00ab same fix-al l.-n-th placed \"ith th.-ir convex faces toward each other. The condition for achromatism is that 4U4 LIGHT their distance apart h = ~^=fv since fi=/2- Therefore, any scratches or dust particles on the surface of either lens would be seen if one were to look through the other with the eye accommodated for an infinitely distant object, as is the case when one is using a telescope. Therefore it is better to make the distance of the lenses apart slightly less, although by so doing the chromatic aberra- tion is not exactly corrected. Thus, if k = $/j, the distance of the prin- cipal focus from either end is Fio. 244. \u2014 A Ramsden eyepiece: the focal lengths of the two lenses are equal. In this case, any object placed at this distance from one of the lenses will be seen on looking through the other lens at apparently an infinite distance. In other words, this eye- piece has the focal properties of an ordinary converging lens of this focal length, but it is approximately achromatic. 2. Huygens eyepiece. \u2014 In this there are two plano-convex lenses, placed as shown in the cut, with focal lengths /, and /2, where /: \u2014 %f.,. In order to secure achromatism their distance apart should satis- fy the equation h = The distance of the principal focus from the first lens is given by /l(2/2~/2) _ 3/2 /2 2/,-/,-/\u00ab 2 - 2 This means that, if a ray is incident upon the first lens at such an angle as to be pointed toward a point between the lenses at a dis- tance \u00a3 from the first lens, it will emerge from the second lens parallel Fio. 245. \u2014 A Huygens eyepiece : the focal length of the first lens is three times that of the second. to the axis. Therefore, since the object must be virtual in order to have an image at infinity, this eyepiece cannot be used, like a converging lens, to magnify ordinary objects. CHAPTER XXIX OPTICAL INSTRUMENTS OPTICAL instruments, so called, are pieces of apparatus so designed as to make use of light waves in order to form images of luminous objects. In general they consist of com- binations of mirrors, lenses, and prisms; but some instruments do not have such parts. The number of these instruments is great, but in this chapter only the simplest and most gen- erally important ones will be described. Optical instruments may be divided roughly into three <lasses : \" pin-hole,\" reflecting, and refracting. In the first, no mirror or lens is used ; in the second, some form of mirror is the essential feature ; in the last, some prism or lens, or a combination of", " prisms or lenses. \" Pin-hole \" Instruments These have been described in Chapter XXV, and nothing further need be said here. They all depend upon the \"recti- ir propagation\" of light. The pin-hole camera and the \u2022\u2022 e.imera obscura\" are the only instruments of this class that of importance. Reflecting Instruments The action..I plane, spherical, and parabolic mirrors \\\\ as explained in Chapter \\\\YII: and their use as looking glasses, in searcl} lights, etc., was described. The great advantage of all forms of reflecting iiM rnmcnis over refract- ing om-s is that they are necessarily five from chromatic 4'.':, 4'.t\u00bbi LIGHT aberration. Spherical aberration, however, can be avoids 1 only by using normal pencils of light and mirrors with small curvatures. One important application of a concave spherical mirror was first made by Newton, namely, to form a \"telescope.\" When we look at a distant object, it subtends at the eye a comparatively small angle. Thus, if A and B are two points on opposite edges of the object, and if 0 marks the position of the eye, the linear angle subtended by the object is (AOB). We estimate the distance of an object from us by the angle which it subtends at our eyes ; the nearer it is, the larger is this angle. FIG. 246. -Diagram representing the effect of a The purpose of a telescope telescopejn increasing the angle between two rays jg to brinff a distant ob- AO&nlJBO.. ', ject apparently nearer, by changing the direction of a ray AO to Af 0, and a ray B 0 to B' 0 ; so that the angle subtended by the rays from A and B is now (A' OB'), an angle greater than (AOB). The \"power of the telescope\" is defined to be the ratio of the angles (A1 0&) and (A OB). In all telescopes either a concave mirror or a converging lens converges the rays so as to form a real image ; and this is viewed by a lens, or a combination of lenses, called the \"eyepiece.\" The simplest form of eyepiece is a converg- ing lens, but a Ramsden or Huygens eyepiece is ordinarily used. In doing this the eyepiece is so placed that the image to be viewed comes at its principal focus (or just inside it), and it forms a virtual magnified image at apparently an infinite distance. Reflecting Telescope. \u2014 In all forms of reflecting telescopes a concave spherical mirror is used to receive the rays from OPTICAL INSTRUMENTS 497 the distant object. The mirror is mounted in some frame- work or tube, so that it may be turned to point in any direction. In the cut, let C be the centre of this spher- ical surface, and I\\C and PC be lines drawn.from two opposite points ill the edge Of the F'\u00b0- 247- \u2014 Dia?rnin representing a reflectinp telescope. P remote object. All rays parallel to Pj(7are brought to a focus at a point Fv on PjC', halfway between C and the mirror; and all rays parallel to PC will be brought to a focus at a point F, on PC, halfway between C and the mirror (see page 449). Therefore, there will be a real image of the object in the plane FFr and I\\ are two points at an infinite distance. In order to see the magnified image of this image produced by the eyepiece, some plan must be adopted which will ren- der it unnecessary for the observer to stand in front of the mirror and so obstruct the view. Several have been tried : 1, the mirror is tipped slightly so that its centre lies outside the telescope tube, and the principal focus for parallel rays lies near the edge of this tul>e, and thus the image may be viewed \u2014 this arrangement was used by Herschel ; 2, a small plane mirror (or a totally reflecting prism ) is placed close to the principal focus, but between it and the mirror, and is so inclined as to turn the rays off sidewise through an opening in the side of the telescope tube; thus the image is formed outside or near the edge of the telescope \u2014 this was Newton V plan; 8, a small convj mirror is used in place of the plan, mirror, and is so turned as to reflect the rays back through a small opening at the centre of the concave mirror and form an image at iinur this arrangement is due to Casse- grain. 4. or. finally, a small concave mirror is placed beyond AMES'S PHYSICS \u2014 32 498 LIGHT the principal focus of the large mirror, and is turned so as to form", " an image in an opening in the large concave mirror (like Cassegrain's method) \u2014 this plan was proposed by Gregory. PIG. 248. \u2014Different forms of reflecting telescopes : (a) Herschel ; (b) Newton ; (c) Cassegrain ; (d) Gregory. The power of the telescope is found directly from simple geometrical considerations. As shown in Fig. 249, parallel rays from the distant point P are brought to a focus at F, OPTK.17. Of STRUM BA T8 halfway l>et \\\\<-ni (J and tin- mirror: parallel rays from the distant point Pl are brought to a focus at Fv The straight line at.1 represents the eyepiece, which is placed so that F and Fl lie in its focal plane; B is a point so chosen that AB = FA, and therefore F and B are the principal foci of the eyepiece. The rays from Pv after reflection, converge to Fr pass on through the eyepiece, and emerge as if they came from the virtual image of F-^ at an infinite distance in t In- direction^ A. Similarly, the reflected rays passing through F have a virtual image at an infinite distance on the lineal.. I>i:ii:r:im..f f..nn:iti..M \u00ab>f ima-c t.y mirn-r.in.] refore, the angle subtended by tin- distant object itself is ; while that subtended by the virtual inia!_re of this formed by the eyepiece is (!*/! /\u2022',>. lh.se angles may be compared by considering their tangents. The e<,nals (FCF^ and its tangent equals ^J. The FF tangent of (FAF\\ ) equals _J. These angles are, however, FA snOieiently small in general t<> allow us to replace their numerieal values 1\u00bb\\ those of their tangents; so the ratio of the angles (/'.I /, and (P^P) is, that is, it is the ratio i.f the f.u-al IniLTth of the mirror to that of the eyepiece. FA \\\\*f see. therefnre, the great ad \\, intake of having a teles< whose concave minor has a long focal length. 500 LIGHT The limit to the resolving power of a telescope is deter- mined, as we have seen on page 487, by the diameter of the mirror ; the larger it is, so much the greater is this power. A large mirror also gathers more light, and therefore enables fainter objects to be seen. Newton constructed his first reflecting telescope in 1668. It had a diameter of one inch, and magnified thirty or forty times. He later made a larger instrument. Previous to this, such instruments had been designed ; and the invention is attributed to Niccolo Zucchi (1586-1670) of Rome. Refracting Instruments All instruments containing lenses are subject to a certain amount of chromatic aberration, although this may be mini- mized by using only achromatic lenses and combinations. There is also always some spherical aberration ; but this may be avoided largely by using only normal pencils of light, and exposing only the central portions of the lenses. The number of the defects which are possible with a lens is so great that some special treatise on the subject should be consulted. Only a few of the simplest refracting instru- ments will be described here. Photographic Camera. \u2014 In this instrument a real image of an object, more or less remote, is formed on a plane photo- graphic plate. This image is produced by a converging lens or system of lenses. If the camera is used for taking photo- graphs of landscapes, the lens is ordinarily a single achroma- tic one, and a diaphragm with a circular opening of variable size is placed in front of it. If it is used for photographing buildings, special pains must be taken to avoid spherical aberration and the consequent distortion; two achromatic lenses symmetrical with reference to a plane halfway between them are used, the diaphragm being in this plane - this is called a \" rectilinear \" or \" orthoscopic \" lens. For portrait work, lenses of large diameter must be used in order to secure as much light as possible. OPTICAL IN STRUM E* TS 501 When photographs of distant objects are taken, the images are, in general, small ; but by combining a diverging lens with the converging system, a larger image may be secured. Sueh a compound lens system is called a \"teleobjective.\" Projection Lantern. \u2014 This instrument consists of a lamp, or a strong source of light, which by means of lenses illumi- nates a drawing or photograph on", " glass, or some object which is transparent in parts, and of a converging system of lenses which throws a real image of this illuminated object upon a suitable screen. The lens system between the light and the object \u2014 called the \"condenser\" \u2014 consists, in gen- \\ \\ Fie. SCO. \u2014 Projection lantern ; (.-1) arc light; (C) condenser; (5) slide to be Illuminated; (/\u2022') focusing lens. eral, of two plano-convex lenses with their curved surfaces in contact. Its function is to deflect down upon the object to be \" projected \" as much light as possible, so as to render it strongly luminous. Then by means of the \"focusing lens \" a real image of it is formed upon the screen. The focusing and condensing lenses must be achromatic, and the former must be corrected for spherical aberration also. Astronomical Telescope. \u2014 This instrument consists of a C()nver\u00abrin^ achromatic lens, called the \"object gla\u2014.\" \\\\liieh forms a real image of the distant object in its focal plane, and this is viewed by an eyepiece. (In the cut a Ramsden eye- piece is represented.) The eyepiece is so placed that the image formed by the object glass comes at or just inside its 502 LIGHT principal focus, so the object is seen apparently at infinity. The power of this telescope is, therefore, like that of the FIG. 251. \u2014 Astronomical telescope: O is source at an infinite distance; O' is image formed by object glass ; 0\" is virtual image formed at an infinite distance by the eyepiece. reflecting instrument (see page 499), equal to the ratio of the focal length of the object glass to that of the eyepiece. The resolving power of a telescope is, as we have seen, determined by the size of the object glass, as is also the quantity of light received. (See page 500.) The astronomical telescope forms, as just explained, a virtual image of the distant object which is inverted; that is, if the object is a tree, the image will have the tree pointed down instead of up. This is a disadvantage if the instrument is to be used for purposes that are not purely astronomical. Dutch Telescope (or Galileo's telescope). \u2014 This telescope is free from the disadvantage of the astronomical instrument FIG. 252. \u2014Dutch telescope. to which reference has just been made. It consists of a converging lens, forming the object glass, and a diverging <U>TK AL INSTRUMENTS 503 I- us so placed that the principal focus on the side next the observer coincides with the principal focus of the object glass. Thus, parallel rays from a point of the distant object are con- verged by the object glass toward a point behind the diverg- ing lens, which is in the focal plane of both lenses ; these rays are diverged again by the second lens and emerge parallel to a line drawn from the centre of the second lens to the point in the focal plane toward which the rays were converging. The image of the distant object is therefore a virtual one at an inlinite distance; but it is erect; that is, the image formed of a distant tree represents the tree in an upright position. The first telescope was probably constructed by Hans Lippershey of Middleburg in the Netherlands in 1608 ; and Galileo, upon hearing of the invention, but without knowing any details of the construction, made an exactly similar instrument in 1009. Kepler was the first to suggest the use of a convex lens for the eyepiece. Galileo immediately used his telescope for observing the heavenly bodies, and made many most im- portant discoveries. Previous to the construction of the reflecting telescope by Newton and the invention of the achromatic 1ms by Pollond, the only means of minimi/ing the color effects produced by telescopes was to use lenses of great focal length, which were clumsy and ott* \u00ab.1 many disadvantages. 1 1 _cens presented to the Royal Society of London a lens whose focal length was 123 ft. Microscope. \u2014 This instrument is one designed to -magnify\" a small ohjert, that is, to increase the apparent distance apart <>t any t \\vo of its point* wh'u-h are close together. a. Simple microscope. \u2014 As was said on page 494, in speak- iii'_r of eyepieces, a single converging lens or.t K'amsden eye- piece can be used as a microscope, the object being placed inside the principal focus. There is thus formed an virtual image of the object (see page 477). The dis- tance at which this image is formed is arbitrary: but it is generally chosen as about 2\"> om. (or 1\" in. > fn>ni the eye, because this U t: < <", "\u2022 at which j.copl,- with normal eyes hold an object in \u00ab>rder to see it most distinctly. 504 LIGHT b. Compound microscope. \u2014 The magnification secured by a single lens is not great ; and it is in general combined with another lens system, as shown in Fig. 253. The latter forms a real magnified image of the object, which is viewed and again magnified by the eyepiece. (In this instrument a Huygens eyepiece may be used, as is illustrated in the cut.) The lens system which is nearest the object is called the FIG. 258. \u2014 Compound microscope: OP is the object; 0\"P\", the virtual image formed by the eyepiece. \"objective.\" It consists of several lenses so chosen as to give an image as free as possible from chromatic and spher- ical aberration ; and it is so constructed as to give as much light as possible to the image and at the same time to have a large value of n sin ^V (see page 487), so that the resolving power is large. The magnifying power can be deduced by simple geometrical methods. The use of converging lenses as simple magnifiers was known to the ancients; but the compound microscope was probably invented by Zach- arias Joannides of Middleburg in the Netherlands and his father, some years before 1610. It was invented independently by others also, among OPT1< AL INSTRUMENTS 505 whom was Galileo. All the early microscopes had a concave lens for the eyepiece ; and Franciscus Fontana of Naples was the first to suggest the sul\u00bbtitution of a convex lens. Spectrometer. \u2014 A spectrometer is an instrument primarily <lfsigned to measure the angle of deviation produced in the direction of a beam of light by reflection or refraction. It consists essentially of four parts. There is a substan- tial metal base to which is rigidly connected an upright metal cylinder, whose axis is called the \"axis of the instrument.\" To this is attached a metal plate whose edge is divided into degrees, minutes, etc. Two bent metal arms are also attached to this cylin- der by means of collars, so tliat they can turn around it as an axle ; they carry metal tubes which lie in a plane perpendicular to the axis of the instrument and are thus movable in this plane. One of these tubes is a simple form of astronomical telescope; the other carries at one end an achromatic converging lens whusf focal length is that of the tube, and at the other end it is.-IMN.-.I by a metal cap in which there is a fine slit with straight parallel edges, which can be opened or closed \u2014 this is called a 4i enllimator.\" If the slit is illuminated, it serves as a s<>mvc <>f diverging waves which will emerge from the lens of the collimator in the f,.rm of plane waves. 506 LIGHT The telescope and collimator are turned until their axes intersect at the axis of the instrument ; and the object which is to produce the deviation of the light \u2014 a plane mirror or prism \u2014 is placed on a platform in the middle of the metal plate referred to above, having the normals to its reflecting or refracting faces in the plane of the axes of the collimator and telescope. If now by means of the former a beam of parallel rays falls upon the mirror or prism, they will be deviated, and their new direction may be found by turning the arm carrying the telescope until there is formed in the centre of the field of view an image of the illuminated slit. In order to determine this condition FIG. 255. -cross hairs exactly, it is customary to insert in the focal plane of the object glass a metal ring across which are stretched two fine silk fibres or spider lines, which are called the \"cross hairs.\" These are made to cross exactly at the centre of the tube. Since they are at the focal plane of the object glass, they will be seen through the eyepiece at the same apparent distance as the object which sends the parallel rays into the object glass. The positions of the collimator and telescope may be noted by means of the divided scale on the edge of the plate, if suitable pointers or verniers are attached to them. used in telescope. * With this instrument the laws of reflection and refraction may be verified ; the angle of a prism and the angle of minimum deviation produced by it for any train of waves may be measured; etc. Therefore, the index of refraction of any substance which can be made in a prism may be deter-. A + D sin\u2014 |- mined ; for n = ^ (See Ames and Bliss, Manual of sin- Experiments in Physics, pages 459-475.) OPTICAL M->T/,T.I//", "-;.\\ :,\u00ab\u00bb; If the laws of reflection are to be studied, ordinary white light may be used to illuminate the slit; but, if the phe- nomena of refraction are to be observed, special precautions must In- taken which will be described in the next chapter. Effect of Diaphragms. \u2014 One fact that must be taken into account in the description of all reflecting or refracting instruments is that the wave front of the waves is always limited by certain apertures or diaphragms. Thus, in the case of a telescope, the only portion of the wave front that enters the instrument has the size of the object glass. Again, if there are diaphragms in the tubes, they may limit the cone of rays which proceed from any point of the object and enter the instrument. Further, the only rays that are of practical use in the case of an instrument that is used visually are those which have such a direction as they leave the eyepiece as to enter the pupil of the eye. The effect of the presence of these various circular openings is felt in many ways. The brightness of the image, the resolving power of the instrument, the contrast of the background, the spherical aberration, etc., all depend upon their size and position. For details in regard to the matter, reference should In- made to some advanced text-book, such as Drude, Optics, or Luiuiuer, Photographic Optics. CHAPTER XXX DISPERSION IN speaking of chromatic aberration, achromatic lenses, etc., it was noted that the index of refraction of a given substance varies with the wave length of the light which suffers refraction, and that, in general, the shorter the wave length, the greater is the index of refraction. In this chapter the method of determining the connection between these two quantities will be discussed. Pure Spectrum. \u2014 Before any relation between index of refraction and wave length can be established, it is necessary to devise a method for securing homogeneous light of a FIG. 256. \u2014 Prism spectroscope. definite wave length. The method ordinarily adopted is to make use of the dispersive action of a prism. If waves of a definite index of refraction, emitted by a small source, fall upon one face of a prism, they suffer refraction and emerge 608 509 on the other side, diverging apparently from a virtual image of the source. They do not, therefore, of themselves come to a focus ; but, if a converging lens is introduced, the rays may be focused upon a screen. This real image is due to the virtual iina^e formed of the small source by the prism ; and it should be noted that the latter image is an astigmatic one, unless the pencil of incident rays meets the prism at the angle corresponding to minimum deviation. (See page 467.) The intensity of the effect can be increased if the point source is replaced by a series of such sources forming a line parallel to the edge of the prism. In general, a fine slit is made in an opaque solid, and the source of light, in the form of a flame, etc., is placed behind it. The image on the screen is in this case a narrow rectangle, practically a line, parallel to the slit. If the source is emitting several trains of waves of different wave length, there will then be formed as many images of tin- slit as there are separate trains of waves. The light is said to be \"analyzed\"; or a \"spectrum\" of the lii?ht is said to be formed. If the slit is very fine and the adjustment for minimum deviation as exact as possible the spectrum is said to be \"pure,\" because in this case each train of waves affects only a very narrow rectangle on the screen, and so there is only a small amount of overlapping of the images. (Since t}i\u00ab- angle of minimum deviation is different for waves of different wave lengths, it is best to make the adjustment of the prism for the mean wave length of the light which is Further, since the focusing lens \u2014 even if aohrom \u2014 has a different focus for different waves, the screen must be curved so as to obviate the error, or one must be satis- fied with a slightly i 114)11 re spectrum.) In this manner the nature of the light \u2014 or, speaking in a more general manner, of the ether waves \u2014 emitted by any source can ivestigated; and we shall discuss this matter in a later chapter. 510 LIGHT Resolving Power. \u2014 Owing to the fact that the waves which leave the slit have passed through several \"apertures\" on their way to the screen (or to the eye), and that their wave front has been limited by this means to the shape of these apertures, the final image is not a line. These apertures are the edges of the lenses, of the prism, and of", " any diaphragms there may be in the tubes. There is, therefore, a diffraction pattern produced, exactly as described for a lens on page 484. The image of the slit, provided the light is homogeneous, is a broadened line accompanied on each side by a series of alter- nately dark and light fringes, whose intensity is much less than that of the central line. The distance apart of these fringes varies with the FIG. 257. \u2014 Diagram to illustrate the resolving power of a prism. f,.,, width of the aper- ture. So, if the source of light is emitting two trains of waves of wave length I and I + AZ, there will be two overlapping diffraction patterns ; and the two trains of waves will produce two distinct images if the central line of one pattern is so far displaced by dispersion as to coincide with the first dark fringe of the other. If #j is the length of the shortest path of any of the rays through the prism, and \u00a32 is the length of the longest one, t2 \u2014 t\u00b1 is practically equal to the length of the base of the prism ; call its value b. If n and n + Aw are the indices of refraction of the two trains of waves, of wave length I and I + AZ, which are just resolved by the prism, it may be shown that An = Y \u2022 In other words, if the prism has a thick base 0 An is small, the image formed of the slit is extremely nar- row, and the resulting spectrum is pure ; but, if the prism is thin, the image formed is broad. D/> /\u2022/-:/,\u2022> ION 'A 1 Spectroscope. \u2014 The essential parts, then, of an instrument to In- used in order to form a pure speeti urn are a slit, a pi ism (or other means for securing dispersion), and a converging lens. This is called a \"spectroscope.\" In general, the arrange- ment is slightly different, a collimator and telescope heing used as described for a spectrometer on page 505. When the slit is illuminated, plane waves emerge from the lens; these enter the prism and are dispersed by it ; and, finally, as they leave the prism they enter the telescope which is focused for plane waves ; and a series of colored images of the slit is seen 1)\\ the observer looking in the eyepiece. Measurement of Dispersion. \u2014If the eyepiece is removed, and the images formed by the object glass are focused sharply on an opaque screen, and, if a fine slit, parallel to that in the collimator, is made in this screen, it will be illuminated by iiract i( -ally homogeneous waves, and will serve as a source K h waves. This entire instrument is called a \"mono- chromatic illuminator.\" By varying the position of this second slit (or by turning the prism) the wave length of the waves transmitted through it is altered..Methods will be discussed later which enable one to measure tin- wave length of any train of waves; and, granting that the wave lengths of the waves emitted by the source are known, we have thus a method for measuring the refractive index of any substance for waves of definite wave lengths. The substance to be studied is made in the form of a prism, and is placed on a spectrometer table. The entire instru- ment is turned until the slit of its eollimator eoinrides with the slit of the inonnehromatie illuminator. The angle of the prism and the angle of minimum deviation t<u different waves are m- md I mm a knowledge of these quantities the indi -inn may be calculated. Fraunhofer Lines. \u2014 It will be sho\\\\ n later that, when HUH- light, liy a spectroscope, it is found to OO1 trains of waves of varying lengths, forming \\\\hal a; 512 LIGHT first sight to be a \"continuous\" spectrum; that is, one in which waves of all wave lengths within certain limits are present. But, if examined with an instrument of fair resolv- ing power, the solar spectrum is seen to be distinguished by the absence of a great many trains of waves, as is shown by the presence of black lines across the bright colored spectrum. (The fact that this absence of the waves is manifested by the presence of lines is due, of course, to the use of a slit illumi- nated with sunlight as the source of the light in the spectro- scope.) These lines are called \" Fraunhofer lines,\" because they were first carefully studied by him. He measured \u2014 by methods to be discussed later \u2014 the wave lengths of the waves which correspond to these lines ; and to the most prominent of them he gave names, in the form of letters.", " (Thus the A line is at the limit of vision in the red end of the spectrum ; B and 0 are also in the red ; Dl and D2 are in the yellow, etc. ; K is at the limit of vision in the violet end of the spectrum.) It is customary, therefore, in studying the dispersion of any prism to illuminate the slit of the spectrom- eter with sunlight, and then to note the angle of minimum deviation that corresponds to the waves in the immediate neighborhood of the various strong Fraunhofer lines. Thus, nE means the index of refraction for waves whose wave length is that of the E line, etc. Dispersive Power. \u2014 The differences in the refractive index of various substances for different waves is shown in the fol- lowing table, in which the columns give the values for the A, B, C, etc., lines : A B C D E F 0 Water, 16\u00b0 C... Carbon bisulphide, 10\u00b0 Crown glass Flint glass. Rock salt, 17\u00b0.. 1.330 l.\u00ab16 1.528 1.678 1.537 1.630 1.681 1.539 1.332 1.626 1.531 1.583 1.640 1.334 1.635 1.534 1.587 1.644 1.338 1.661 1.540 1.597 1.653 1.537 1.502 1.549 1.546 1.606 1.661 // 1.344 1.708 1.561 1.614 1.568 DISPERSION 513 It is evident from this table that not alone is the index of refraction different for different substances, but that also the dispersion of any two rays, e.g. (nD \u2014 wc), may be the same for two substances and the dispersion of two other rays may differ ; thus for crown glass and rock salt nB \u2014 nA = 0.002, but for the former nn \u2014 n?= 0.011, while for the lat- ter it equals 0.015. Further, what is called the \"dispersive power \" is different. This is defined to be the ratio n\u00bb~nA. (In practice, the n in the denominator is taken as n^ or n^ or nE, etc., whichever happens to be best known ; for the error thus introduced is negligible.) Its value for the above substances is therefore : nD\u2014l Water 0.042 Flint glass.... 0.061 Carbon bisulphide.. O.H5 Rock salt.... 0.057 Crown glass... 0.043 The dispersive power of a substance is proportional to the length of the spectrum produced by a prism made of it, which has a small angle. For, and if A and therefore D are both small, we may write...5^, and \u00bb.-l >equently, the dispersive power equals \"~ \u2022D/i\u2014 I>A* the difference in the minimum deviations of waves of wave IfiiLTtlis A ami //. is practically the angle subtended at the prism by the length of the spectrum when the prism is in its position of minimum deviation. Therefore, prisms having':le but made of dinVivnt substances pro- duct (,f dillrivnt length and of dillVn-nt dispersion /)/>\u2014 1 PHYSICS \u2014 88 514 LIGHT for the same waves ; and the average deviation for the whole spectrum may be different also. It is owing to these facts in regard to dispersion that it is possible to construct an achromatic lens, or to combine two prisms in such a manner as to produce no deviation of a defi- nite train of waves, but to disperse the other waves. Direct-vision Spectroscope. \u2014 The spectroscope that has just been described suffers from the disadvantage that the collimator and telescope are not in the same direction, and therefore in some cases it is difficult to support the instru- ment rigidly. As has been just explained, however, it is a sim- ple matter to choose two prisms of different materials which, when combined with their edges parallel but turned in opposite -A direct vision spectroscope. directions? will not deviate OHO particular train of waves, but will the others. The disper- sion produced in this way by two prisms is not great ; but, if several are thus combined, it may be made as great as desired. If such a series of prisms is combined with a collimator and telescope, the apparatus is all along one straight line. Such an instrument is called a \" direct vision spectroscope.\" Normal and Anomalous Dispersion. \u2014 The table given above shows that in all the cases cited the index of refraction in- creases as the wave length decreases ; this is called \" normal", " dispersion.\" There are, however, many substances with which this is not true. As will be shown later, every sub- stance absorbs more or less completely ether waves of certain wave lengths; and some substances absorb only waves of wave lengths that lie within a narrow limit, e.g. only waves having wave lengths that differ but slightly. In such cases the index of refraction for waves whose wave lengths do not differ much from this wave length of the \"absorption DISPERSION 515 band \" does not follow the above law connecting it with the wave length. Such dispersion is called \" anomalous.\" There may, of course, be several such absorption bands, introduc- ing further complication. Dispersion Curves and Formulae. \u2014 The various facts in regard to dispersion, both normal and anomalous, can be best shown in a diagram whose axes are indices of refraction and wave lengths, or by a formula. Thus, for normal dis- 100 \u00ab00 300 tOO WO WO 700 800 800 1000 1100 1200 1300.1100 UOO 1000 1 'n.. 258. \u2014 Dispersion carve for quarts. persion, the connection between index of refraction and wave length is given by a curve like that shown in the cut, which is tin- \u2022\u2022 disjiersion curve\" for quartz. This maybe also expressed 1>\\ a formula, in which n is the index of i< traction and I the wave length, viz.: fin (lie neighborhood of an absorption band this formula ceases to he true. ) If then- is a single absorption kind whose wave length is J0, the dispersion curve has a general form as shown in tin- cut, which is the actual cur\\e for cyiinine as determined by 516 LIGHT Professor R. W. Wood. This substance has an absorption band at about wave-lengths 600 MJL. The formula which expresses the facts outside the absorp- tion band is n2 = A + \u2014 \u2014 \u2022 It is seen from this or from T* I \u2014 1Q the curve that for wave lengths not far from the absorption band, if 11>1Q>1Z, n1>n2; that is, the refractive index of the longer waves is the greater. So, if a prism is made of this substance, it will deviate the long waves near the ab- sorption band more than the short ones ; and the resulting S8iS3Sg\u00a7\u00a72gg\u00a78S!Sg 2.3'5>p'^>\u00ab oooSoo pc-t-V-t- WAVE LENGTHS IN fJ.fJL FIG. 2GO. \u2014 Dispersion curve of cyanine. spectrum will appear, in general terms, as if it were divided in two parts by the absorption band and as if these two halves were shifted toward and across each other. Whenever there is an absorption band, there is anomalous dispersion, and many substances show these phenomena in the visible spectrum ; such are the aniline dyes, the vapors of sodium and other metals, thin layers or films of metals, etc. It will be seen later that all these substances have other opti- cal phenomena which are closely connected with this fact. Rainbows. \u2014 An interesting illustration of dispersion is furnished by the phenomenon in nature called the \" rainbow.\" After a rain shower, DISPERSION 517 if tl.r sun is not far from the horizon, and the rain clouds have passed in such ;i.liivrtion that the observer is between them and the sun, a series of colored arcs or circles may be seen on looking away from the sun. These colored arcs are in the following order : violet on the inside shad- ing off to red on the outside, then a dark space, and another arc colored red on the inside and violet on the outer edge. They are all portions of circles whose centres coincide at a point lying in the prolongation of the line joining the sun with the eye of the observer. There are often also u supernumerary \" bows inside the primary one. G \u2014 - FIG. 261. \u2014 Rainbow : refraction oi a ray by a drop of water, single reflection. A complete explanation of these phenomena requires a consideration of the size of the drops, the nature of sun -light, the size of the sun, etc.; this was first given by Sir George Airy. An elementary, but imperfect, theory was given by Descartes ; an outline of which is as follows. Con- sider a raindrop as a sphere of water, and draw the paths of the rays inci- dent upon it from the sun. Certain rays will enter the drop, suffer reflection once and be refracted out, as shown in the cut, in which 0 is the centre of a drop, OS is the direction of the sun, and AB, BC, CD, DE", " represent the portions of a ray. The deviation of the ray is shown in the cut by the angle (GFE), where F is the intersection of the prolongations of the incident and emerging portions of the ray, AB and DE. By means of higher mathematics it may be shown that if there in a homo- geneous beam of rays, all parallel to OS, falling upon the drop, the angles of de- viation all exceed a definite value. In..\u2014Rainbow: JVI\u00bb angle of mini- mum deviation. N other words, there is a minimum value of the deviation, and this fixes a certain direction with reference to the line OS. If tins minimum angle is N, draw the line AB making such an angle with SA. Then of all the 518 LIGHT rays parallel to OS falling upon the drop, none are so deviated as to \u2022 \u2022iin-rge outside the angle (SAB)] some emerge in the directions ABr A Bp etc. This minimum angle is different for different wave lengths ; for violet it is about 140\u00b0, for red about 138\u00b0, etc. Therefore, if the observer looks up at the rain cloud in such a direction that his line of sight makes an angle of 180\u00b0 -140\u00b0, or 40\u00b0 (or less), with a line joining his eye to the sun, he will receive violet light from the rain- l / FIG. 268. \u2014 Kainbow. drops. If he looks up at an angle greater than 40\u00b0, he receives no violet light at all. Therefore, the observer will see a violet arc, each point of which subtends at his eye an angle of 40\u00b0 with the line drawn from the sun. Similarly, with the other colors, there is a red arc corresponding to the angle 180\u00b0\u2014 138\u00b0, or 42\u00b0, which is sharply denned on its outer edge, and to see which one must look higher up in the sky than was necessary in order to see the vio- let arc. But some of the rays from the sun suffer two reflections in the rain- drop, as shown in the cut. As before, we may show that there are FIG. 264. \u2014 Rainbow: refraction of a ray by a drop of water, double reflection. angles of minimum de- viation for the different colors, which give rise to a red bow at the angle 51\u00b0 and a violet one at 54\u00b0, these bows being sharply defined on their lower edges. CHAPTER XXXI INTERFERENCE OF LIGHT Interference Fringes. \u2014 The general phenomena of interfer- ence of waves were described in Chapter XXI, page 374, and the special ones dealing with light were discussed in Chap- ter XXV, page 420. It was shown that the simplest mode of illustrating interference was to place two identical sources of light close together and to allow them to illuminate a screen or to enter the eye directly. If the two sources are parallel slits, the interference pattern is a series of parallel colored fringes ; if homogeneous light is used, these are alternately bright and dark, and at regular intervals apart proportional to the wave length ; if white light is used, the fringes are a superposition of different sets, each of whirl i is due to a different color, thus proving that white light is equivalent to a superposition of waves of different wave length. Unless the two sources are identical, there is no permanent phase relation between the two sets of waves emitted l.y them, and so these cannot interfere. In demonstrating the interference fringes which are pro- duced by the two identical sources, a converging lens is always used, the arrangement being as shown in Fig. 265. Ol and 02 are the two sources ; L is the lens : M \\^ a sen-en placed at a distance fmm the lens equal t<> its focal length for parallel rays. The two sources are emitting niys in all directions; let OlAl and O^A^ be two parallel rays. After n-f motion by tin- 1ms, tln-\\ will meet at the point B in the screen, \\vhn. < I: a line drawn through the centre of tin- lens parallel to the incident rays. (See page IT1.'. > If 619 520 LIGHT the difference in length of the optical paths of these rays is half a wave length (or an odd number of half wave lengths), there will be complete inter- ference at B. This difference in path is found exactly, as on page 377, by drawing a line from 02 perpendicular to 0-^Ar For, if B were a source of waves, two of its rays would be BAfli and BA202i and the position of the wave front at any instant would be a plane perpendicular to A101 and A202. So, when the wave front reached 02, it would be at P on the line Aflv where 02P is perpendic- ular to A", "^O^ 111 Other words, the optical paths 02A2B and PA^B are equal in length ; and so the difference in length of the optical paths from 01 and 02 to B is O^P. b o M Fio. 265. \u2014 Young's Interference expert- By means of the lens, then, the interference fringes are focused onjthe screen. The central image is at 0, and the distance OB can be expressed at once in terms of the differ- ence in path of the rays from 01 and <92. Call this difference But the jingles (O&P) and (BOO) are equal ; and sin (BOO) O\\ D = OjP = equals \u2014 \u2014. Hence BO = EC sin (ECO) = BC sin (0i02P) = D-^- = -5- O\\ 0-2, If the focal length of the lens, 06\\ is called /, and if it is large compared with BO, we have the relation IM'Klil'KliENCE OF LIGHT 521 The condition for complete interference is that and for this value BO = (~ Therefore the distance between two dark fringes is There is thus offered a method for the measurement of the wave length of light, as the various quantities in the formula with the exception of I may all be measured directly. We shall now describe several methods of securing two identical sources. Young's Method. \u2014 This consists, as has been already de- scribed, in illuminating two parallel slits in an opaque screen by light from another slit parallel to them and at the same distance from each. The slits must be sufficiently narrow to dilVract the waves in all directions. (See page 392.) FresnePs Biprism. \u2014 This consists of a combination of two identical thin prisms which are placed base to base, as shown by the cross section in the cut. An opaque screen, with a slit in it, is so placed that the slit is parallel to the two edges of E Fra. 2\u00ab6. \u2014 Fresnern blprinm : two virtual Images 0, and Ot of the source 0 are produced. tin- biprism and at e<|iial distances fmm 1 linn. This slit, if illuminated, will emit waves, which will suffer refraction and deviation by the two hal\\\u00ab-s of the prism. If the slit is at 0, as sho\\\\ n in the cut, one half will form a virtual image at 0,, 522 LIGHT the other at 02. So the waves as they emerge from the biprism will come apparently from the two sources Ol and 02 ; and there are then two identical trains of waves, which will inter- fere, and may be focused on a screen by a lens, as described above. Therefore, if the distance 0^0^ is known (and it may easily be determined by experiment), the wave length of the light may be measured. Fresnel's Mirrors. \u2014 These are two plane mirrors which are carefully adjusted until their faces are slightly inclined to each other, but are in actual contact along a line. FIG. 267. \u2014 Fresnel's mirrors: two virtual images 0j and 02 of the source 0 WV are produced. A slit 0 is placed parallel to this line ; and therefore two virtual images of it, 01 and 02, are formed by the two mirrors. Let B be the line of contact of the two mirrors. Then a pencil of rays P^OB falling upon the first mirror will be reflected into the pencil Q^ OlBl ; and the pencil B OP2 falling upon the second mirror will be reflected into B202Q2. Therefore there will be a region included in the angle (^B^B,^) which is traversed by two trains of waves coming from identical sources. Lloyd's Mirror. - In this arrangement a slit is placed paral- lel to a plane mirror, at some distance from it, but only a slight distance above its plane. There will be a virtual image formed by the reflected rays ; and so any Pie. 268.\u2014 Lloyd's mirror: a virtual image 0, of the source 0, is formed by the mirror. INTERFI-:I;I-:\\<-E OF LIGUT 523 point above the mirror will receive waves directly from the slit, and also by reflection, apparently coming from the vir- tual linage of the slit. There will therefore be interference. (These two sources of waves are only approximately iden- tical, for one is the inverted image of the other.) Colors of Thin Plates. \u2014 The first interference phenomenon which was recognized as such and so explained is the pro- duction of the brilliant color effects by such thin films of transparent matter as soap bubbles, films of oil on water, layers of air between two pieces of nearly parallel glass, etc. These colors are due to the interference of the trains of", " waves which suffer reflection directly at one surface of the lil in with the waves which are refracted out from the film after one or more internal reflections. If we consider any point on the surface of a film which is receiving homoge- neous light from any point source, one ray from the latter is reflected at the point, and other rays emerge there which have entered the film at other points and have suffered reflection at the surfaces of the film. It is evident that these rays have had paths of different lengths ; and that also the ray directly reflected has suffered reflection when incident upon the surface of the film from the surround \\\\\\g medium, while the emerging rays have suffered reflection when inci- dent upon the surface of the Him fmm its interior. Owing to this last cause there is a difference in phase introduced, in addition to that caused by the difference in path, because the reflection in the one case is from a \"fast\" to a \"slow*' medium, and is the opposite in the other. (See page 834.) This additional difference in phase is equivalent, as was shown, to half a period of the vibration. If the total effect of all the rays at the point on the surface of the film is null owing to interference, this point will appear dark : while, if th. ravs do not destroy each other's action, the point will be bright. It is evident also that a film of such a thickness as to cause intc for waves of a definite wave length will 524 LIGHT not, in general, cause interference for other waves; so, if white light is used, a point where there is complete inter- ference for a definite train of waves will appear colored, owing to the fact that the other waves are not cut off ; and the effect is as if one color were completely removed from the constituent colors of white light. The trains of waves which interfere at the top surface of the film are not destroyed, for energy cannot be annihilated ; they are trans- mitted through the film and emerge on the lower side. Thus, when white light is incident upon a transparent film, some is reflected at the upper surface, the rest enters the film ; of this a certain amount is reflected once or more times, and is finally refracted out through the upper surface, while the rest is either directly, or after two or more reflections, refracted out through the lower surface. By far the greater amount of the light is transmitted, owing to the poor reflect- ing power of the film. There will then be colors visible if one looks at either of the two surfaces of the film ; but those seen by looking back at the second surface are much weaker than those at the other, owing to the presence of so much white light. If the film does not absorb the waves, the com- bined effects on its two sides are exactly equivalent to the incident waves; that is, they are \"complementary.\" We shall now consider in detail the case of a thin film with parallel faces, and we shall suppose that the film has a greater index of refraction than the surrounding medium, e.g. a film of water in air. Let the rays come from a homogeneous point source at such a distance compared with the area of the film that they may be regarded as parallel. Let 00 in the cut be an incident ray ; it undergoes reflection at O and gives rise to a ray OD in general. Other rays emerge by refraction at C\\ one of these is due to the incident ray PA. This is refracted into the ray AB, then reflected into BO, and finally refracted out. We can calculate the difference in optical path of these rays from the source to O. Draw UtTEBFEREN( /\u2022; LIGHT 525 AE perpendicular to the incident rays; then the phase at A.tiid K is the same because they are on the same wave front; draw FC perpendic- ular to AB, it repre- sents the refracted wave front; and there- fore the phase at C and F is the same. The difference in op- tical path of the two rays meeting at C is, then, n(FB + T*C), where n is the index of refraction of the film with reference to the surrounding medium. Calling the thickness of the film h, and the angle of refraction into the film r, it is seen from geometry that FB + BC= 2 h cos r. Therefore the difference in optical path is 2 nh cos r ; and FIG. 269.\u2014 Colors of thin plates. (The tram are omitted.) >itt*d rays'_ the corresponding time lag is, where v is the velocity There is also the of the waves in the outer medium. v.. 2 >///<\u2022\u00ab>\u2022>/\u2022. additional difference in phase of half a period, \u2014, due to the difference in the character of the reflection at B and at (7. So the total difference in", " phase, expressed in terms of time, between the two rays OCD and PABCD is 2 nh cos r, T This maybe expressed differently, by substituting for v its value \u2014, where I is the wave length in the outside medium : V1/\" r/2wAeosr 1\\ \\ / I'/ Another ra\\ which will have a compoip \u2022 Lfing at (7 is O 'A \\ incident at A' ^ etc. ; and it is n.\u00bbt dillicult to prove that 526 LIGHT the resultant emerging ray at C has the same amplitude as the reflected ray, but differs from it in phase by the amount given above. If this difference equals an odd number of half periods, the two rays interfere completely. This condi- tion is, then, that 2 nh cos r + 1 _ (2 m + 1) where m is any whole number 0, 1, 2, 3, etc. This reduces to 2 nh cos r = ml. Therefore if h = 0, or \u2014 - \u2014 or \u2014 - \u2014, etc., 2 n cos r n cos r the interference is complete. If, then, the film is wedge shaped, there will be inter- ference at points on the surface of the film corresponding to these critical thicknesses; and alternately dark and light bands across the film will be seen. The light used to illuminate films generally comes from an extended source, like a flame, which is not far from the film. In this case any point C on the surface of the film receives two rays from a point 0 of the source, viz., OCD and OABOD' ; and in general the directions of these rays after leaving 0 are not the same. Similarly, Q receives rays from other points of the source, which also have different directions after leaving it. wmwrn 'Mb So this point may be regarded as FIG. 270.- colors of thin plates: itself a point source ; and, if the eye localization of colors m the sur- fo focused on the surface of the film, the question as to whether C appears dark or not depends upon how exactly these pairs of rays from each point of the source neutralize each other. If the incidence is practically normal, the value of r in the formula is so small that cos r = 1 ; and the condition for complete interference is that 2 nh = ml, which is the same for all these face. INTERFERENCE OF LIGHT 527 pairs of rays. If the thickness of the film at the point C satisfies this condition, and if the eye is focused on the film, C will be a dark point. So, in general, when an extended source of light is used and the incidence is practically normal, the dark or light bands are to be seen by looking at the surface of the film. Another illustration of these formulae is given when the convex surface of a plano-convex glass lens is pressed closely against a glass plate. The film of air between the two pieces of glass may be regarded as made up of a great number of concentric rings, each ring having the same thickness at ^ \u2014 -~ \u2014 -\"\" all points. The film is then like that of a FIG. 271.\u2014 Apparatus for New. wedge one of whose surfaces is curved; and ton's rings. there is, of course, symmetry around the centre, or point of contact. Therefore if homogeneous light is incident nor- mally upon this film, there will be a central dark spot surrounded by alternately bright and dark rings. The dark rings are given by h = 0, \u2014 -, \u2014, etc., where h is the thickness of the film. These are known L' // 'J n as \" Newton's rings.\" h can be expressed in terms of the radius of the spherical surface of the lens and of the radius of the dark ring ; and a method is thus offered for the measurement of the wave length of light; or, if this is known, for the measurement of the radius of the spherical surface of the lens. If in any of the above experiments white light is used, there are no dark bands or rings \u2014 all are colored ; and the details can in each case be deduced from the general prin- ciples given on page 524. Interference over Long Paths.\u2014 In the cases so far treated, the inter- l rays have been considered to differ in path only by a small amount, 1'iit there are many interesting and important phenomena in which HUH condition is not fulfilled. Tims let 0 be a homogeneous point source of liLjht : /' \\te a plate of tran-par.-ut material; C be a converging lens; It be a screen placed in tin- />rincipal focus of the lens. Draw from C, the centre of the lens, a line CE at random. Then, all rays falling upon the lens parallel to this line \\\\ill", " lens be placed with its axis perpendicular to the plane of the grating, and let a screen be placed in its focal plane. In the cut, draw an arbitrary line OP from the centre of tin.- leii> to the screen; all the rays from the various open- ings of the grating which are parallel to this line will be brought to a focus at P. The difference in path of any two surh rays from the corresponding edges of two consecutive openings, or from two corresponding points in two consecu- tive openings, may be deduced at once. Call the distance from the edge of one opening to the corresponding edge of the next, i.e. the \"grating space,\" a; and the angle (P&A), jy. Then, the difference in path referred to is a sin N-, and, it this is a whole number of wave lengths, parallel rays from corresponding points in all the openings will coincide in phase at P\\ and it will be a bright point. (There is an exception to this, when rays from some of the points in any one opening interfere with those from other points in the same opening ; but this case need not be discussed here.) There will there- fore be a line of light through P parallel to the openings in the grating. The condition, then, that P should be bright is : where m = 0, 1, 2, 3, etc., and I is the wave length. Conse- quently, there is a series of bright lines determined by a sin jv\"0 = 0, a sin N{ = /, a sin Nt = 2 /, a sin N3 = 3 /, etc. ; that is, by riii JV9 = Ob \u00abnJ\\r, = 8inN9 =? sin JV8 = if, etc. The light along any line through a point P defined by these relations is bright ; thru for neighboring portions of the screen as one takes points farther ami farther away from P the light fades gradually away and vanishes. It rises to a maximum in another line detined l\u00bby the next value of N, etc. Then- Si.1 maximum!'\u00ab\u00bbr tin- dnvetion sin ^ = 0, or ^V0=0; that is along a line through the point A in the cut where 532 LIGHT the axis of the lens meets the screen; this is called the u central image \" ; the next maximum, at the angle Nv is called the \"first spectrum,\" etc. It is evident that there are maxima also on the other side of A, corresponding to negative values of N. The number m is said to give the \"order\" of the spectrum. If white light is used, instead of homogeneous light, each constituent train of waves has its own series of spectra : a central image, and spectra of different orders on its two sides. The spectra of the different colors overlap ; and the spectrum of any one order is not pure unless the individual FIG. 275. \u2014 Photograph of spectra produced by a grating, showing the different orders. spectra formed for any one train of waves are extremely narrow. We can easily determine how wide any one spec- trum \" line \" is by calculating the position of the point next it on either side where the intensity is zero. This condition involves complete interference at that point of all the rays reaching it from the grating. Let P be a point where there is a maximum, and let P1 be the nearest minimum on the side toward A. Draw the line OPV and call the angle (PjO^.), iVj ; then all rays parallel to OP1 are brought to a focus at.Pj. Let us suppose that there is an even number of openings in the grating ; if there is an odd number, we DIFFRACTION 583 may consider the last one by itself, and its effect in com- parison with that of the others may be neglected. If the number of openings is 2n, the condition that P should be the position of the mth spectrum may be expressed by saying that the difference in path of the two rays reach- ing it from corresponding points in the first opening and the middle one, equals //////. For this difference in path is na sin N* and it has been shown thai the con- dition for a maximum is that P F> A FIG. 276. \u2014 Diagram Illustrating the resolving power of a grating. a sin N= ml. If P1 is to be the nearest minimum, the difference in path between two rays from these same points must differ from this value for P by half a wave length; for, if this is true, the rays from the first and //th. the second and the (n+ l)th, etc., will interfere completely. The con- < N dition for a minimum at Pl is, then. //", "'/ sin NI = nml \u2014 -\u2022 illaily, for a minimum point on the other side of P, the vain.- \\\\onhl be nml + ^- ) The condition for a maxi- mum at P is, Ilrlier, :Jnn wi Mil.V - mill. / The entire number of openings in the grating is 2n, and each has the grating space a ; so 2 na is the width of the grating; and the formula shows that in order for N to be nearly e.pial to \\}. that is. toi />f to be very close to P, the 534 LIGHT width of the grating must be large. Under these conditions the spectrum \"lines\" are narrow. The distance from one spectrum line produced by a homo- geneous train of waves to the next one is determined by giv- ing m two consecutive values in the formula a sin N= ml. Thus, the fourth spectrum is at an angle -ZV4 whose sine satisfies the equation 4 / sin N. = \u2014 ; a and the corresponding formula for the fifth spectrum is Thus, sin N6 \u2014 sin N4 = -. This relation is general ; and it is seen, therefore, that in order to have consecutive spectra far apart the grating space a must be small. If white light is used, or waves from some complex source, the central image will receive light of all wave lengths ; and, in addition, a series of spectra will be produced on both sides of this. In order to have these spectra long, i.e. the disper- sion great, the grating space must be small ; and to have the spectra pure, i.e. the \"lines\" narrow, the grating must be wide. This condition for \" purity \" may be expressed differently. If there are two trains of waves of wave lengths I and I + AZ, which differ only slightly, their spectral images will, as a rule, overlap; but, if A\u00a3 is so large that the maximum of the waves of length / + AZ coincides with the minimum of the other waves, the two images or \" lines \" may be seen distinct from each other. The condition for a maximum of the forn id- waves is that N should be such that a sin N=m(l + A/); and we have just seen that the condition for the first minimum of waves of length / is that the angle NI should be such that na sin NI = nml + - > or that a sin Ni = ml+ \u2014 \u00bb 2n -, \\\\ lu-re 2 n is the number of grating spaces, cide, that is, if N=Nt If these two positions coin- I or This is the spectroscopic \"resolving power\" of the grating; and it is seen that, in order for A/ to be small, the number of grating spaces must be large and a high order of spectrum should be used. It should be noted that, even if the spectra formed of the components of white light are pure, the ones of different order are superimposed on each other. Thus, waves of wave length I have a spectrum line in the first order, at the same point where waves of wave length - have a line in the sec- ^ I ond order, and where waves of wave length ^ have a line in the third order, etc. It should be noted further that the above formulae are all independent of the material of the grating. If the plane waves are incident upon the grating at an angle /, different from 0\u00b0, the necessary modification in the formulae is deduced easily. Let O\\ and O-> be the edges of two consecutive grating openings; let PI 01 and PtOi be two..!' th\u00ab- im-idi-nt rays; and let <V7i and n..(i,!>\u00ab\u2022 th<> two rays from Oi and O2 which are diffracted in the direction defined by the angle N. v OtA perpendicular to PiO\\, and 09B perpendicular to 0\\Q\\. At the points A and Ot the rays are in the same phase ; therefore the difference^ in path <>f these rays when they are unit. -.1 l,\\- the converging lens is AO\\ + &J5. If the grating space O\\0t is called n, as before, this difference in path 1* writfr'i.V ) : an. I i!,.- condition for a spectrum line Pio. 877. \u2014 Oblique incidence upon a grating. K then, that a (sin /+ sin#) = ml. All the other formula* an- i lifi.-.l in a similar manner. If w\u00ab\u00ab mii*iili-r tli- s|.iTtniui lii", "APTER XXXIII DOUBLE REFRACTION General Phenomena. \u2014 It was observed by Bartholinus as early as ItJti'J that, when a luminous point was viewed through Iceland spar, two images were seen. With any ordinary transparent substance, such as glass or water, only one image is observed; so the phenomenon is called N \" double refraction.\" j \\ It was later discovered p! \\ A that many substances H ri;*\u00b0\\ ) had this property. If a doubly refracting (\u2014 substance is made in the form of a plate, any ray OA incident upon it from a point source 0 will there- fore give rise to two refracted rays AB1 and AB% ; and these will emerge as two parallel rays coming apparently from two virtual images Ol and Oy which are not in general in the line OPQ, drawn tVom O perpendicular to the sur- face; nor do. in general, the retracted rays lie in the plane of incidence. Since refraction in all eases is due to the fact that the Velocity of waves 18 not the same in all media, the explana- tion of double refract ii. ii in any i lium is that a centre of disturbance in such a medium produces waves whose wave M] front at an\\ instant OOnUStfl <>| t\\\\\u00bb> parts or \u2022\u2022 sheets\"; SO 542 LIGHT that, if a straight line is drawn out from this centre, there will be two points on it, at different distances from the centre, which mark the advance of the disturbance in that direction at any instant. In other words, waves spread out from the centre in such a manner that there are two distinct disturbances advancing with different velocities along any direction. In some doubly refracting substances there is, however, one direction in which these two disturbances have the same velocity, while in all others there are two such directions. This direction in a given body is not a, fixed line in the substance ; for the above statements are true for all lines in the body which have this direction ; it is called the direction of the \"optic axis.\" Those doubly refracting sub- stances which have only one such axis are called \"uniaxal\" ; the others, which have two axes, are called \" biaxal.\" All crystals which belong to the cubical system, so called, are single refracting; those that belong to the pyramidal or second system are doubly refracting and uniaxal ; the other crystals are doubly refracting and biaxal. Any ordinary iso- tropic transparent substance, such as glass, becomes doubly refracting if it is strained in one direction by pressure, by unequal annealing, etc. Uniaxal Substances. \u2014 In the case of uniaxal doubly re- fracting bodies, it is found that one of the rays obeys both of the ordinary laws of refraction, while the other in general obeys neither of them. The former is called the \" ordinary ray \" ; the latter, the \"extraordinary.\" In such a substance, then, a centre of disturbance gives rise to a spherical wave front, which accounts for the ordinary ray, and also to another wave front which advances with a different velocity, and which cannot be a sphere ; otherwise the extraordinary ray would obey the laws of ordinary refraction, but would have a different index of refraction from the ordinary ray. If a plate of Iceland spar is held between the eye and a bright object, two images of it are seen ; and if the plate is turned DOUBLE REFRACTION 543 around an axis perpendicular to its faces, one image will Ive ;ir<>mi<l the other. Huygens suggested that this ml wave front was the surface of an ellipsoid; and by his work and that of later investigators, this idea has been confirmed. Since the velocity of both disturbances is the same along the op- tic axis, these two surfaces, which make up at any instant the wave surface produced by a point source, the sphere and the ellipsoid, must be tangent to eacli other at the extremities of a diameter having the direction of the axis; and, since the phenomena are symmetrical around this axis, the ellip- soid must be one of revolution around this diameter. Tin- ellipsoid may lie inside the sphere or outside. Plane sec- tions of the two types of wave surface through the axis are ;i uniaxal doubly re- fracting substance. shown in the cut. It is a simple matter of geometry to draw the refracted rays, provided the wave surface is known. Thus, let a nniaxal substance be cut with a plane face making the angle N with the axis, and let plane waves be incident upon this surface in such a direction that the plain-", " \u00ab>t incidence is one that includes the normal to the surface at any point and the direction of the axis at that point sueh a plane is d a \u2022\u2022 principal section \" of the crystal. If the ineident wave front at any instant is represented in Fig. 283 by 00r waves will spread out from 0 in the douhly refracting sub- stance and will advance, as sj,,.un. f,,r a definite distance, while the disturbance from Ol reaches A. Following the method MI llnvgens, as explained in Chapter XX, the two wave fronts in the lower medium produced by the in. ident plme waves are obtained by drawing through a line per- 544 LIGHT pi'iidii-ular to the sheet of paper at.1 t\\vo planes which are tangent to the sphere and the ellipsoid of the wave surface around 0. The tan- gent plane to the sphere touches it at C, and OC, the radius, is perpen- dicular to it. The tangent plane to the ellipsoid touches it at D, but the per- pendicular from 0 upon the plane is OE, not OD. (If the plane of inci- dence were not a _Fm. 283. \u2014 Infraction of Iceland-spar. OB is the axis OC and OE are the normals to the refracted wave fronts OC and OD are the ordinary and extraordinary rays. principal section of the crystal, the point of tangency of the ellipsoid would not have been necessarily in the plane of the paper.) The point D marks that point on the wave front AED which the dis- turbance from 0 has reached at the instant of time considered. Therefore the line OD is the ray ; while the line OE is the direction of advance of the plane wave front AD, it is called the \"wave normal.\" It is seen that the \"ordinary ray,\" OC, obeys both the laws of refraction; while the \"extraordinary\" one, OD, does not, as a rule, obey either. The fact that the extraordinary ray has a direction different from the wave normal is an illustration of what was noted on page 386, that the ordinary case of the ray and the wave-normal being coincident was due to the velocity of the waves in the medium being the same in all directions. Separation of the Ordinary and Extraordinary Rays. - It is often important to separate the ordinary and extraor- dinary rays so that each may be used by itself. There are several methods for doing this : DOUBLM U1.FHACT10N 645 1. Prism. \u2014 If the uniaxal substance is made in the form of a prism and placed on a spectrometer table, the two beams of light produced by refraction of the beam from the col- limator will have different directions, and may be ob- served separately. If a glass prism of suitable angle and material is combined with a ORDIHARY RAY RAY EXTRAORDINARY prism of Iceland spar, the ordinary ray may be made to emerge parallel to its original direction, while the extraordinary ray will be deviated and the dispersion of white light produced by one prism may be largely neutralized. Such a compound prism is called \"achromatic.\" Fio. 284. \u2014 An achromatic Iceland spar prism. 2. Absorption. \u2014 In some uniaxal substances one of the 1 is absorbed much more easily than the other. Thus, a t/tiit plate of tourmaline transmits both rays; but a thicker / one absorbs the ordinary and transmits only the extraoriUnnrii. 3. Total reflection. \u2014 Since the two rays have different velocities in the crystal, it is evidently possible to select some isotropic sub- ice, the velocity of waves in which is inter- im-diate between these, and to have the ray incident upon a surface separating the latter substance from the doubly refracting one at such an alible that one <>f the rays u ill suffer total reflection and the other will not. Thus, in Iceland spar the extraordinary ray is refracted less than tin ordinary ; and Canada balsam, a transparent cement, has an intermediate veloc- | so if a piece of Iceland.spar 18 CUt into \u2022\u2022cos by an oblique ne. ti.m and these are cemented to by Canada balsam, we have a means of AMES'S PHYSICS \u2014 36 546 LIGHT separating the two rays. If light from any ordinary source, such as a flame, is incident upon such a compound plate, it will, on entering it, be broken up into two beams, ordinary and extraordinary, which will have different directions ; the former will suffer total reflection at the section of balsam, the latter will be transmitted and will emerge at the", " end of the plate. The sides of the plate are generally painted black ; so that the ordinary rays are all absorbed. This piece of apparatus was invented by William Nicol of Edinburgh, and is called a \"Nicol's prism,\" or a \"nicol.\" Biaxal Crystals. \u2014 In the case of biaxal crystals, neither of the rays, as a rule, obeys the ordinary laws of refraction. The form of the wave surface which was proposed by Fresnel is the true one so far as is known ; but it has not been veri- fied in all particulars. Its properties will be found discussed in any advanced book, such as Preston, Theory of Light, or Drude, Optics. CHAPTER XXXIV POLARIZATION Huygens's Experiment. \u2014 In his investigation of the doubly refracting properties of Iceland spar, Huygens no- ticed a remarkable fact concerning the two rays transmitted by a plate made of it. They both appeared to be like ordi- nary beams of light; they could be reflected and refracted; they affected the sense of sight; etc.; yet they were differ- ent in one respect, as was shown by an ingenious experiment. The simplest form of this is a slight modification of Huy- If8 original one. Two identical prisms are cut out of a piece of Iceland spar, so that the optic axis in each makes the same angle with the normal t<> the surface. These are made \"achro- matic\" (see page 545), and are mounted in tubes so that each can be turned around a line perpendicular to its faces. (This line may be called the \"axis of figure.\") If light from a small source falls upon one of these prisms, two pen- cils emerge : one, the ordinary : the other, the extraordinary. It now the second prism is placed parallel to the first, these two pencils emerging from the latin- \\\\ill fall upon the for- ii!\u2022\u2022!\u2022; and each will give rise to two pencils, one ordinary, th\u00ab other extraordinary; thus, four pencils in all will emerge, two ordinary and two extraordinary. This is true n. ral ; l.ut llnvgens observed that as the tube contain- ing the second |>rism Was turned around its axis of figure, there were f t mns, 90\u00b0 apart, dm inir a complete revo- lution \u00ab.f the tube, in which only two pencils emerged ; and in one of tlieqp positions the two emerging pencils coincide 647 548 LIGHT in direction. For an intermediate position between any two of these there are, as said, four pencils. They all appear equally bright for a position halfway between any two con- secutive ones of these four positions ; but, as the tube is turned, two of these grow feeble and vanish, while the other two grow more intense ; then, as the tube is turned farther, these two grow feebler and finally vanish, while two others appear and grow more intense. A rotation of 90\u00b0 is required to turn from one of these positions into the other. It is thus evident that the two pencils which emerge from the first prism are not like the incident light. Further, they are not like each other. For, calling the two pencils emerging from the first prism 0 and E ; and the two pencils produced in the second prism by the former pencil, 00 and OE; and those produced by the latter, EO and EE, we may state the above facts as follows: in general, 00, OE, EO, and EE are present ; as the second prism is turned, a position is reached for which only 00 and EE appear; as the rotation is con- tinued for 90\u00b0, these disappear, and OE and EO only are present; etc. Thus for one position of the second prism, one of the two pencils incident upon it gives rise to an ordi- nary pencil, while the other produces an extraordinary one ; and after a rotation of 90\u00b0 this condition is reversed. Huy- gens noticed that when the principal sections of the two prisms were parallel, 00 and EE were transmitted; but when these planes were perpendicular to each other, OE and EO were transmitted. The explanation of all these phenomena is simple if we consider the two rays 0 and E as plane polarized with their planes of vibration at right angles to each other. (See page 313.) The waves are then to be thought of as transverse, and the vibrations of any one beam are all in parallel straight lines in a plane at right angles to the direction of propaga- tion of the waves. The first prism breaks up the incident beam of light into two, whose vibrations are in directions POL A lilt. \\TION 549 perpendicular to each other. The experiment shows that e two", " directions are fixed in the prism ; so that, as it is turned on its axis of figure, the directions of the vibrations of the emerging beams turn also. In other words, at any point in the surface of the prism there are only two direc- tions in which vibrations are possible; and these are fixed in it with reference to the principal section at that point. Vibrations along one line make up the ordinary ray ; those along the other the extraordinary. Thus, let C be any point of the prism of Iceland spar; PCP be the position of the principal section through it ; and CA and CB be the two directions of /p possible vibrations, CA that of the I ordinary, CB that of the extraordi- t nary ray. When the two prisms are in whldi *\u2122 tbc two i \\a AM. L AM. \u2022 \u2022 \u2022 i ^ possible directions of vibration. so placed that their principal sections are parallel, the possible directions of vibrations in the two are parallel ; so that, calling the direction of the vibrations of th. ordinary ray in the second_one C^Ar and^that of the vibrations of the extraordinary C^B^ CA and ClAl areparal- B lei, and also fifl-and C\\Blt It is thus apparent why 00 and EE are transmitted. Again, when the second prism is rotated through 90\u00b0, CA and (^Bj are parallel, and and C*A.\\ so OE and EO are c,, \"^~\" * transmitted. If the second is turned so that the principal planes ''^A, are in. -lined to each other, let the ri\u00ab.f8T.-DtofTMnfliastrmtia\u00abiiay. line'',.'i make the angle N with lien a vihrat.o,, along CA having an amplitude CP will be resolved on entering the second prism into two Mluations; one along C\\Al having 550 LIGHT an amplitude OP cos N, the other along C^Bl with an amplitude OP sin N. That is, if the amplitude of 0, OP, is called A, we may write 0 = A, 00 = A cos N, OE = A sin N. Similarly, if there is a vibration along CB of ampli- tude A, we have E = A, EO = A sin N, EE=A cos JV. We must consider the light received from ordinary sources, such as the sun, flames, etc., as being made up of vibrations in all directions in the wave front, because it is not polarized in any way, and when these vibrations are analyzed by the first prism into two sets, along CA and CB, their intensities are equal; that is, their amplitudes are the same. Therefore the four pencils of light transmitted by the second prism have amplitudes given as above : 00 = A cos N; OE = AsmN; E0 = A sin N; EE = A cos N. For N= 0, 00 = A, OE = 0, EO = 0,EE = A. As N increases, 00 decreases, OE and EO increase, EE decreases. For N = 90\u00b0, 00 = Q,OE = A, EO = A, EE = 0. In this manner the phenomena observed by Huygens are all explained; but the hypothesis on which it is based, viz., that ether waves are transverse, was not advanced until the early part of the nineteenth century, when it was proposed independently by Young and Fresnel. The credit of explain- ing the various phenomena of polarization and of defending this hypothesis must be given the latter. Huygens recog- nized that the only way possible to account for his observa- tions was to assume a two-sided character for the ether waves ; but the only waves known to him were the longitudinal air waves which produce sounds, and the idea of transverse Waves does not seem to have occurred to him. POLARIZATION 551 Phase Differences. \u2014 These emerging rays have different phases, partly because 0 and E are not necessarily in the same phase at any instant, and also because they take dif- ferent times to traverse the plates, owing to their different velocities. Thus, if the incident waves are homogeneous, and if Vl is the velocity of the ordinary waves and V2 that of the extraordinary ones, and if h is the thickness of a plate, the difference in phase introduced by the plate, expressed in V>i V terms of time, between the two emerging rays is h ( Y The period of the rays is the same, since the light is assumed to be homogeneous ; and since an angle 2 TT corresponds to a time equal to a period T, the difference of phase of the two rays expressed as an angle is \u2014h(- - -\\ But calling the * V*", " I I ays /j and 1T Vl = \u2022\u00a3 and v2 = ^; wave lengths of the two rays u \u201e.,, ^ \u2014,a \u2014 so this quantity may be written 27rhf- \u2014 \u2014 j. (The ve- locity va depends, of course, upon the direction of the waves in the plate, and so is not a constant.) These formula; will be applied later to explain the colors observed in certain polarization phenomena. Polarization by Reflection. \u2014 From these experiments of lluygens, and from the explanation of them just given, mulling can be said in regard to the connection between the lion of the principal section at a point and that of the \u2022 liivctions of possible vibrations; but a discovery of Mains, which will bo described immediately, led to the proof that one \u2022 t these directions of vibration lies in the principal section. toning to the cut_on page 649, this means that POP coincides with either CA or OB. More recent experiments have shown that it coincides with CB.) Malus discovered by chance that the light reflected from a \u2022*j surface was plane pol. or less completely, ami lie observed that, if sunlight (or light from any onlin.uv 552 LIGHT source) is incident upon such a mirror at a definite angle, called the \"polarizing angle,\" practically all the reflected light is plane polarized, while the transmitted light contains waves that are also plane polarized, but with their vibrations at right angles to the former. Malus made his discovery in 1808 when looking through a plate of Iceland spar at the image of the sun reflected from the window panes of the Luxembourg Palace in Paris ; he noticed that the two trans- mitted pencils were of unequal intensity and that, for certain positions of the crystal, one image vanished while on rotation through 90\u00b0 the other one disappeared. This showed that the incident light on the crystal was plane polarized. If one looks through such a plate of Iceland spar at light reflected from a plane glass mirror at the polarizing angle, and slowly rotates the plate, it is observed that, when its principal section coincides with the plane of incidence, only the ordinary rays are transmitted ; while, if the principal section is at right angles to the plane of incidence, only the extraordinary rays are transmitted. For positions of the plate between these two, both rays are transmitted, but with different intensities except for the position halfway between. The direction of the vibration of the rays reflected from the plane mirror must by symmetry be either in the plane of incidence or at right angles to it, i.e. parallel to the plane of the mirror; and so the fact just described proves the statement made above in regard to the connection between the principal section of the Iceland spar plate and the possible directions of vibrations. (For many reasons it is believed that the vibrations in the rays polarized by reflection from a glass plate are parallel to its plane. This is in accord with the statement that the vibrations of the extraordinary rays are in the principal sec- tion. Therefore the vibrations transmitted by a Nicol's prism are in the principal section.) If the light is not reflected at the polarizing angle, only a portion of the reflected light is plane polarized ; the rest is POLARIZATION 563 like the incident light, made up of rays whose vibrations are in all directions in tin- wave front. At the polarizing angle more of the transmitted light is plane polarized than for any other angle ; but, as said above, these vibrations are at right angles to those of the reflected light, as may be shown by viewing it through an Iceland spar plate. Brewster's Law. \u2014 It is found by experiment that light reflected from a plane mirror of any transparent isotropic substance which shows ordinary normal dispersion, such as all kinds of glass, water, etc., may be plane polarized for certain definite polarizing angles. (This is not true of re- flection from metallic mirrors or from substances showing anom- alous dispersion.) A connection between the polarizing angle of any substance and its index of refraction was established by.vster. He found that at the polarizing angle the reflected and retracted rays were perpen- dicular to each other. If, in the Fio.m-Bwwst\u00abr'8Lawiiir<*\u00bbTdto cut, MM is the plane surface of msparent substance, whose index of refraction with ref< T- ence to the surrounding nu-d in m is n, and if BA* AD, and AC are the incident, tin- refracted, and the reflected rays respec- tively, experiments sliow, u just said, that if the angle (BA /') is the polari/mg angle, {CAD) is a right angle. Therefore the angle of refraction {DAP') equals (CAM) ; so, calling tin- angle of incidence N, (DAP')", " = *-N. The index of refraction n satisfies the emiat inn n s= 8in \\ _'; and hence, ut the polarizing angle, n = - -\u2014= tan N\\ or, the angle of cos A polari/.atinn -nUunrr ii such that its tangent equals the index of refraction. This is known as \" Brews ter's jy *\\r\\(DAP') 554 LIGHT Law.\" (More careful experiments have shown that there is no angle of incidence for which all the reflected light is plane polarized ; in reality there is always a small amount not so polarized, but at the polarizing angle this is small.). The value of the polarizing angle of pure water is 53\u00b0 11' ; of crown glass, about 57\u00b0 ; etc. Plane of Polarization. \u2014 The light which is plane polarized by reflection from a plane transparent surface is by definition said to be \" polarized in the plane of incidence \" ; or its \"plane of polarization\" is said to be that of the plane of incidence. (This definition is entirely independent of any conception of the directions of the vibrations of the rays in this reflected beam; but, accepting the statements made above in regard to these directions, it is seen that in a plane polarized beam the direction of vibration is in the wave front and at right angles to the plane of polarization.) Pile of Plates. \u2014 A means is obviously offered of securing plane polarized waves by reflecting ordinary sunlight or light from a flame, etc., from a glass or water surface at the polarizing angle. In general, only a small quantity of light is reflected owing to the poor reflecting power of glass or water; but the effect can be increased greatly in the case of glass by using several thin plates placed one on top of the other, thus forming a \"pile of plates.\" When light falls upon such a pile at the polarizing angle, the reflected light is plane polarized, but part of the refracted light is not ; this falls at the polarizing angle upon the surface where the top plate meets the next one, is partially reflected, and is refracted out so as to coincide in direction with the beam reflected from the plate, being polarized also like it. The light which enters the second plate is still not completely polarized, and when this meets the third plate, reflection again occurs ; etc. Thus both the light which is reflected and that which is transmitted by a pile of plates is practically completely plane polarized, but in planes at right angles to eacli other. POLARIZATION fi5f, If plant1 waves plane polari/.ed in the plane of incidence incident upon a i^lass surface at tlie polarizing angle, they will be entirely reflected; while, if they are polarized in a plane at right angles to this, they will be entirely transmitted. These and all the facts in repaid to polarization by reflection may be shown best by the use of a piece el apparatus invented by Norrenberg, which i own in the cut. It consists essentially, is seen, of two plane mirrors which may be turned about axes lying in their planes and also in such a manner that their nor- mals may lie in different planes. Plane Polarization. \u2014 We have, therefore, t\\\\o general methods for the production of plane polarized light; one is to use a crystal of Iceland spar, or other uniaxal crystal, and get rid of one of the rays by the means described on page 544 : the other is to use a pile of plates at the polarizing angle. Fi\u00bb.\u00aba.-N6rrenb\u00abrg'\u00ab apparatus. In practice a Nicol's prism is almost invariably used. Similarly, either one of these methods may be used to test whether a certain beam of light is plane polarized ; thus, if such a beam is incident upon a Nicol's prism, it will be en- tirely extinguished for some position of the latter, as it is slouly turned on its axis of liLnire; and when it is extin- guished, it is known that its vibrations are at right angles to principal section of the prism. In this manner it may be shown that, if one looks at the blue >k rection at right angles to a line joining tin- eye to the sun, the reflected sunlight \u2014 that which is scattered by the tine put icles (see page 480)\u2014 is plane polarized in the ie including the two directions just mentioned. We have thus another means of securing plane polari/.ed liijht. (It U predict from theory what the direction of the Nibi 111 tins case; and, by comparing this direction 556 LIGHT with the observed position of the plane of polarization, it is proved that the vibrations are at right angles to it. (See page 554.) Interference of Plane Polarized Waves. \u2014 Fresnel and Arago performed by means of two piles of plates a most ingenious experiment to", " determine whether the vibrations in the waves transmitted by them were exactly in planes at right angles to the direction of propagation of the light. If such is the case, by using independently the two piles with their planes of incidence at right angles to each other, two beams of light may be secured in which the vibrations are in the wave fronts but are perpendicular to each other. Two such trains of waves as this cannot \" interfere \" ; because, in order to have one train interfere with another, the vibrations of both must be in the same straight line. The experiment re- ferred to consisted in modifying Young's original interference one by introducing a pile of plates in front of each of the two slits. We shall quote from their own description, follow- ing Crew's translation in his Memoirs on the Wave Theory of Light : \" It has been known for a long time that if one cuts two very narrow slits close together in a thin screen and illuminates them by a single luminous point, there will be produced be- hind the screen a series of bright bands resulting from the meeting of the rays passing through the right-hand slit with those passing through the left. In order to polarize at right angles the rays passing through these two apertures,... we selected fifteen plates as clear as possible and superposed them. This pile was next cut in two by use of a sharp tool. So that now we had two piles of plates of almost exactly the same thickness, at least in those parts bordering on the line of bisection ; and this would be true even if the component plates had been perceptibly wedge shaped. The light trans- mitted by these plates was almost completely polarized when the angle of incidence was about thirty degrees. And it was I'ULAHI/.ATION 557 : ly at this angle of incidence that the plates were inclined when they were placed in front of the slits in the copper screen. \u2022\u2022 When the two planes of incidence were parallel, i.e. when the plates were inclined in the same direction; \u2014 up and down. for instance, \u2014 one could very distinctly see the interference hands produced by the two polarized pencils. In fact, they behave exactly as two rays of ordinary light. But if one of the piles be rotated about the incident ray until the two planes of incidence are at right angles to each other, the first pile, say, remaining inclined up and down while the second is inclined from right to left, then the two emergent pencils will be polarized at right angles to each other and will not, on meeting, produce any interference bands.... We must hide that rays of light polarized at right angles do not affect one another.\" Fresnel and Arago performed another most interesting experiment which is also described in the memoir from which we have just quoted. Its object was to determine the con- ditions under which interference may occur when two plane polarized waves with their vibrations parallel are brought together. It B incidentally explains the production pf \u00ab.f the beautiful colors which are seen when a thin plate made of a crystal or of any doubly refracting substance is interposed between two principal sections are \\ / X line so that one may look through Ixith at a sourer of white light. In any such plate, whether uninxnl or biaxal, there are only two possible direc- i whirli rations of the transmitted waves may have. Let these two directions at a point 0 of the plate be OA and OB\\ and let the vibrations of the waves 558 LIGHT upon itjrom the first nicol have the direction and the ampli- tude OP. This is resolved into two, a vibration along OA of amplitude OPV and one along OB of amplitude OPV where these lines are the components of OP in these directions; and these two trains of waves are transmitted by the plate. Let us consider the waves as homogeneous, so that they all have the same period. As they enter the plate they are in the same phase because they are components of the same vibration ; but they will emerge with a difference in phase, expressed in time by hi --- ), where h is the thickness of \\vl v2J the plate and v1 and v% are the velocities of the two waves. They will then fall upon the second nicol whose principal section may have the direction ON. The waves whose vibra- tions have the direction and amplitude OPl will be resolved into two, but the only one transmitted is that whose vibra- tions have the direction and amplitude OQV where this is the projection of OPl on ON. Similarly, the other train of waves whose vibrations are represented by OP2 will be re- solved into two, but the only one transmitted is that whose vibrations are represented by OQ^ the projection of OP2 on ON. Therefore, two trains of waves emerge from the second nicol, which are plane polarized with their vibrations in the same direction, and which", " have a difference in phase of h(\u2014\u2014 \u2014 ]. This expression may be simplified if the values v*i V of v1 and v2 in terms of the period and wave length are sub- stituted, viz., v1 = ^J, v2=-%i> hence tne difference in phase may be written Th(- - -). Expressed in terms of an angle, -\\h y If N! is the angle between OP and OA, 6Pl = OP cos Nv OP2 = OP sin NL. I POLARIZATION 559 And if NI is the angle between (Wand OA, OQl = OP, cos Nt = OP cos AT, cos N*, = 0Pa sin N* = OP sin ^ sin JV2. So if the original vibration as it leaves the first nicol is A cos nt, and if this difference of phase introduced by the plate is d, the resulting vibrations as they leave the second nicol may be written A cos JVj cos Nt cos nf, A sin AT, sin #2 cos (nt \u2014 d). If ON is perpendicular to OP, the nicols are said to be \"crossed,\" and no light is transmitted if there is no plate interposed between them. If this is the case, in the above formula ^=\u2014(90\u00b0 \u2014 ^); so cos JV2 = sin^Vr and sin N2 = \u2014 cos -ZVj ; and the two vibrations are A cos ^ sin N! cos nt and \u2014 A cos JVj sin N2 cos (nt \u2014 d ). This last vibration may be written A cos JVj sin Nt cos (n/ \u2014 d \u2014 TT). FIG. 291.\u2014 Special case of crossed nlcola. So the difference in phase is rf and the amplitudes are equal. If this quantity d -f- TT is equivalent to an odd numlMT <>f half periods, i.e. if d + TT = (2 m + 1) TT, or rf = 2 WITT, the interference will be complete: and thrst; homOgeneOQfl waves will bo extin- guished. (They suffer total reflection at the surface win i. the two halves of the last nicol are cemented together.) Other cases, in which the two nicols are not crossed, m ax- be found discussed in advanced text-books. So, if white light is used, certain waves will be absent in the transmitted light, ami it will he colored. Which par- ti< Milar waves are absent depends upon the thickness of the double refracting plate, as is evident from the formula. 560 LIGHT The first nicol, which polarizes the incident light, is called the \"polarizer\"; the second one is called the \" analyser.\" If there is no such plate between the crossed nicols, no light at all is transmitted ; but, if a plate is introduced, cer- tain waves appear, as just explained. The phenomenon is called \" depolarization \" ; and the experiment serves as an extremely delicate test of the double refraction of a substance. It at first sight appears as if, in the above experiment, the first nicol might be removed so that the light would fall directly upon the double refracting plate and then upon the second nicol, and there might still be interference ; for the plate would break up the light into two beams and introduce a dif- ference of phase between them before they were combined again by the second nicol. But the relations between the ampli- tudes and the phases in this case are not definite, because the light incident upon the plate is not polarized, but consists of vibrations in all directions ; and so there is no permanent interference. This fact is the fundamental one established by the experiments of Fresnel and Arago. Circular and Elliptical Polarization. \u2014 The fact that a plate of a doubly refracting substance breaks up an incident beam of plane polarized light into two such beams, polarized at right angles to each other and with a differ- ence in phase between them which varies directly with the thickness of the plate, renders it possible to secure a circular or an elliptical vibration. A beam of light whose vibrations are A of this character is said to be \"circu- FIG. 292. \u2014 Formation of cir- J J F larly\" or \"elliptically\" polarized. cularly and elliptically polarized Thus let OA and OB DG the tions of possible vibrations in the plate, and let OP be the direction of the principal section of the nicol through which the light is incident upon the plate. POL A UIZA T1ON If the amplitude of this plane polarized light is OP, those of the two beams transmitted by the plate are OP1 and OP2. Whatever the difference of phase between them introduced by the plate, these", " two vibrations will combine to form an elliptical one. (See page 324.) If this differ- ence in phase is equivalent to a quarter of a period, or to any odd number of quarter periods, the vibration is an ellipse whose axes coincide in direction with the lines OA and OB. If, in addition to this condition for the thickness of the plate being satisfied, the incident vibration OP bisects the angle between OA and OB, the amplitudes of the two trans- mitted beams will be the same ; and they will combine to form a circular vibration.'Such a plate is called a \"quarter wave plate\" ; and obviously plates of different thicknesses must be used for waves of different wave length or color. Fresnel's Rhomb. \u2014 If plane polarized light is totally re- flected from the surface of a transparent substance such as glass or water, it becomes, in general, elliptically polarized ; for, if the light is plane polarized in such a manner that the direction of the vibration is neither parallel to the surface nor in the plane of incidence, it is resolved by reflection into two plane polarized beams, one with its vibrations parallel to the surface, the other with its vibration in the plane of incidence. Their amplitudes are different, unless the direction of vibration in the incident beam bisects the an^le between two lines in the wave front, one parallel to the surface, the other in the plane of incidence. A change of phase is introduced by the reflection, which is not the same for the two beams; and the difference for the two depends upon the angle of incidence and the material at whose surface the reflection takes place. Fresnel m.ide a rhomb of such a particular kind of glass AMES'S PHYSICS \u2014 86 562 LIGHT and with such angles that when light was incident perpen- dicularly upon one of its end faces, it would suffer total reflection twice and emerge perpendicular to the opposite face with a difference of phase equivalent to a quarter of a period between the two plane polarized beams. By this means it is possible to obtain light circularly polarized, or elliptically polarized with one axis in the plane of incidence and the other at right angles to it. Detection of circularly or elliptically Polarized Light. - If circularly polarized light is incident upon a Fresnel's rhomb or upon a quarter wave plate, it will emerge plane polarized, for the effect of these pieces of apparatus is to in- troduce a difference of phase of a quartet* of a period between the two component plane polarized waves into which the incident waves are resolved. The existence of this plane polarized light may be detected by a nicol. If elliptically polarized light is passed through a Fresnel's rhomb or a quarter wave plate, it will, in general, emerge elliptically polarized ; but, if the plate is turned in its own plane, or the rhomb is turned around an axis perpendicular to the planes of its end faces, there will be four positions in one complete revolution for which this light will be plane polarized. This may be detected by a nicol. Other methods are described in advanced text-books. Ordinary Light. \u2014 The light which we receive from ordi- nary sources, such as diffused sunlight, etc., is not polarized in any manner ; yet it can be transformed into plane, circu- larly, or elliptically polarized light by methods which have been discussed. When ordinary light is passed through a doubly refracting substance, both the transmitted plane polar- ized beams are of equal intensity, and there is no permanent phase relation between them. This shows that we must consider ordinary light as due to transverse waves in which the vibration at any instant may be rectilinear or circular, etc.. but in which the vibration is continually changing its POLARIZATION 563 form. We may regard it, then, as due to an ever changing mixture of transverse rectilinear vibrations. Rotation of the Plane of Polarization. \u2014 There are certain substances which have a most remarkable property in regard to plane polarized light. If one of them is made in the form of a plate and a beam of homogeneous light plane polarized in a } (articular direction is transmitted through it, the emerg- ing light is plane polarized, but its plane of polarization has been rotated through a certain angle. Such substances are said to be \"optically active.\" Thus, if the incident light is produced by the use of a nicol, and a second nicol is \" crossed \" with it, no light passes before the \" active \" substance is intro- duced between them ; but after this is done, the second nicol must be turned on its axis of figure through a definite angle before the light is again extinguished. This angle varies directly with the thickness of the substance, and is different for waves of different wave lengths, being much _ri' ater for the short waves than for the long ones", ". This last phenomenon is called \" rotatory dispersion.\" There are two classes of these substances; one is made up of bodies which are naturally active, while the other contains bodies which are active only when they are under the influ- ence of a magnetic force. The former phenomenon was discovered by Hint ; the latter, by Faraday. A kinematic explanation of this rotation was given by Fresnel; but it is not necessary to state it here. It may be found in any advanced treatise. a. Naturally active bodies. \u2014 Examples of these bodies are quart/, when cut at ri^ht angles to its optic axis, an aqueniis solution of certain tartaric acids, of many of the irs, etc. In all these cases, if the plane polar i/ed li^ht is made to pass through a plate and then by means of a m is reflected back,i.;,nn, the plane of polarization is rotated in one direction and then in the opposite, so it emerges the second time polari/ed exactly as it was on incidence; 564 LIGHT it is as if one screwed a screw into a board and then unscrewed it. In certain bodies the plane of polarization is rotated in a right-handed direction, while in others it is turned in the opposite sense. Thus, if the light is emerging in a direction_perpendicularly up from the X paper, and if AB is the direction of the prin- cipal section of the second nicol, in the experi- ment described above, when it is so placed as to extinguish the light before the active FIG 294 - Rota- substance is introduced ; and, if after this tion of the plane of takes place, the nicol must be turned in the direction shown by the arrow in order to extinguish the light again, the rotation is said to be \"right- handed.\" If, on the other hand, the rotation of the nicol must be in the opposite direction, it is called \"left-handed.\" There are two varieties of quartz, left-handed and right- handed ; two varieties of active tartaric acids, etc. It was discovered by Pasteur that all optically active sub- stances were made up entirely or in part of certain crystals which had a \"hemihedral\" form ; of which there are for any substance two possible states. These two are symmetrical with reference to a plane, like the two hands of any individ- ual. Thus right-handed quartz has imbedded in it minute hemihedral crystals of one form ; while left-handed quartz has hemihedral crystals of the symmetrical form, etc. Crys- tals of tartaric acid are hemihedral; and when dissolved in water the molecules retain their asymmetric character. These facts are the basis of what is called \"stereochemistry,\" which is a branch of chemistry dealing with conceptions of the atomic arrangement in certain organic molecules. (See Richardson, Foundations of Stereochemistry, New York.) b. Magnetically active substances. \u2014 When any transpar- ent substance, such as glass, is placed in an intense magnetic field, it acquires the power of rotating the plane of polariza- POLARIZATION 565 tion ; but this rotation is different from that just described, because, if the rotated light is reflected back on its path, the plane is rotated still farther, in the same direction as before ; it is not turned back into its previous position. When the subject of electricity is discussed, it will be shown that if an electric current is passed through a wire wound in a helix, there is a strong magnetic field inside it, and that the direction of this field is reversed if that of the current is reversed. It is found by experiment that, if a piece of transparent matter is introduced in this helix, the direction of the rotation of the plane of polarization is that of the electric current in the helix. Metallic Reflection. \u2014 When plane polarized light falls upon a polished metal surface, it is reflected according to the ordinary laws; but the light is elliptically polarized, unless the incidence is normal. This is owing to the fact that the incident light is broken up into two plane polarized beams which have a difference in phase. If the metal surface is magnetic, that is, if it is made of iron, steel, nickel, etc., the character of the reflected light \u2014 the shape of the resulting ellipse \u2014 depends upon whether the metal is in its natural condition or is magnetized; if the latter is the case, the effect of the reflection also varies with the direction and the intensity of the magnetization. This IIMWII as the \" Kerr effect,\" and will l>e found fully described in any advanced text-book. CHAPTER XXXV VELOCITY OF LIGHT THE first experiments to determine whether, like sound, light traveled with a measurable velocity were performed by Galileo. They consisted in having one observer flash a", " light which was seen by a second one at a considerable dis- tance, who then flashed another light as quickly as possible, and this was seen and noted by the first observer. In any case there would necessarily be an interval between the instant when the first observer flashed his light and when he saw that flashed by the second one, owing to the time required to perform the manipulation, but, if time were re- quired for the passage of the light across the space between the two observers, this interval would vary directly with the distance between the observers. No such effect was observed. The interval of time between the events referred to was ap- parently independent of the distance apart of the observers, and was conditioned only by their quickness of motion and perception. It was therefore concluded that light traveled with an infinite velocity. Method of Roemer. \u2014 This opinion was maintained by every one until the year 1676, when Roemer, a Danish as- tronomer, then living in Paris, made certain observations on the eclipses of one of the satellites of Jupiter by the planet, which he interpreted as proving that the velocity of light was finite, but very great. As a satellite revolves around its planet \u2014 e.g. the moon around the earth \u2014 its motion is periodic, or may be assumed to be so to a very high degree of accuracy. So the interval of time which elapses between 566 7MLOCITT oi' LK.nr two consecutive instants of disappearance of a satellite of Jupiter behind the edge of the planet, when viewed from a point in space fixe<l with reference to the planet, must be a constant quantity. Roemer observed that this interval of time, as noted here on the earth, was not, however, the same when the earth in its motion around the sun was moving away from Jupiter, as when it was approaching it; it was longer in the former case than in the latter. Further, if this period of revolution of the satellite was noted when the earth was nearest Jupiter, and if a calculation was made, based on this, of the instant at which at the end of half a year, when the earth was farthest from Jupiter, an eelipse should take place, there was found to be a differ- ence of 996 sec. between this calculated instant and the observed one. Roemer saw that these facts could all be explained if the assumption were made, that it takes time for the propagation of light across space. As the earth is receding from Jupiter, light has to travel a greater distance at the instant of the second eclipse than at that of the first : and so the apparent period,,f the satellite is greater than it would l>e if the earth were at rest with reference to J uj liter. Just the reverse is true when the earth is approach- ing the planet. So, when the earth is farthest from Jupiter, light has to travel an additional dUtanee equal to the diameter lie earth's orbit ; and the interval of 996 SCC. observed, as stated above, between the calculated instant of an eclipse and the icc,,rd.-d one is the time required for light to pass ii stance; and so the velocity of light may be rmined. More recent observations have given 1002 sec. as the interval of tin)..'oenier; and the diameter of the earth*! orl.it may be taken as 2998 x 106 Km., as will be shown immediately; so the velocity of Iiurht in space given l.\\ this method is this quantity divided by 1003, or 2.984 x 1010 cm. per second. 568 LIGHT The angle subtended at the sun by the radius of the earth, N in the diagram, is called the \" solar parallax \" ; and it is a quantity whose value may be determined by astronomical observations with a high degree of accuracy. The accepted value of this constant at the present time is 8.79 sec. of arc. The radius of the earth is known, its value being 6378 Km. So the distance from the earth to the sun may be calculated. Referring to the cut, AE = 6.378 x 108 cm., But 27r in angular measure = 360 degrees of arc FIG. 295. \u2014 Solar parallax : S is the sun ; E is the earth. and N= 8.79 sec. of arc. = 360 x 60 x 60 sec. ; so 1 sec. = -, and 8.79 sec. =- 360x60x60\" 360x60x60 27TX8.79 Therefore, since N is so small, we may write AE = ES x N, or t\u2122 AE 6.378 x 108 x 360 x 60 x 60 27TX8.79 = 1.4966 x 1018cm. The diameter of the earth's orbit is, then, twice this, or 2.993 x 1018 cm.", " Method of Bradley. \u2014 Another method by which the veloc- ity of light could be determined was discovered by Bradley, the great English astronomer, in 1727. He had observed that if a fixed star was observed through a telescope, this instrument had to be pointed in slightly different directions at different times of the year ; so that, if the image of the star were kept on the cross hairs of the telescope during the whole year, the instrument had to be kept in continual motion in such a manner that its end described a small curve. The explana- tion given by Bradley was extremely simple. Consider a long tube closed at its two ends by caps in which there are two openings directly opposite each other. A particle entering at one open- ing, with a motion parallel to the axis Fio. 296. \u2014 Diagram repre- senting stellar aberration. 1 C B of the tube, will escape through the opening at the other end, if the tube is at rest. But, if the tube is moving at right VELOCITY OF LIGHT 569 angles to its length, the opening in the opposite end from tin- one at which the particle enters must be displaced in a direction opposite to the motion of the tube if the particle is to escape. The line of motion of the particle, with refer- ence to the tube, makes an angle with the axis of the tube whose tangent equals the ratio of the velocity of the tube to that of the particle; for, referring to the cut, if Vl is the velocity of the tube and V% that of the particle, and if t is the time taken for the particle to pass through the tube, 03= Vj and AB= F3*; so tan (BAG) = =. Simi- larly, if we consider light from a fixed star entering a tele- scope, if the instrument is at rest, the light will emerge directly from its farther end ; but, if the telescope is mov- ing at right angles to the direction from which the light is coming, it must be inclined forward in order to see the star, and the reason for this, according to Bradley, is because the \"path of the light\" is along the line AC. The angle through which the telescope has to be turned when it is pointed approximately perpendicular to the path of the earth in its orbit, is called the \"constant of aberration.\" (Its value according to recent astronomical observations is slightly less than 20.5 sec. of arc.) If this angle is determined by observations on the fixed stars, the velocity of liijht may be calculated, assuming that in the above formula Vl is the velocity of the earth in its orbit and \\\\ that of liurht, because the velocity of tin- earth in its orbit is known. The value thus deduced is 2.982 x 1010 cm. per second. This explanation of stellar aberration is insufficient to account I'm- all the facts. In the above formula T, w\u00ab>ul\u00abl be the velocity of light inside the telescope; therefore, if the telescope tube is filled with water, V^ is diminished in the ratio of the index of refraction of water to air, and so the aberration angle should l\u00bbe increased. This experiment was actually performed by Sir George Airy; 570 LIGHT and no change in the angle was observed. When we con- sider light as due to waves in the ether, and if we assume that the ether as a medium does not move as the earth travels through it in its orbital motion, the same formula, as given above, may be deduced for the aberration angle, only in it V% is the velocity of light in the pure ether, not when it is inside matter. Method of Fizeau. \u2014 The first method for measuring the velocity of light directly here on the earth, without making use of any astronomical data, was devised and applied by Fizeau in 1849. The principle is extremely simple. A source of light is placed in such a manner that it shines FIG. 297. \u2014 Fizeau's toothed wheel apparatus for measuring the velocity of light. through the space between two teeth of a cogwheel, which may be driven at a high speed. At some distance on the other side of this wheel is a mirror which reflects the light it receives back toward the source of light. If then the cogwheel is rotating so rapidly that the light which passes between any two teeth travels to the mirror and is reflected back at such an interval of time that the wheel has turned through a distance which brings a tooth to where an opening was, no light passes back through the wheel. If the speed of the wheel is increased, an opening will come where the previous one was, and the return light will pass through ; if the speed of the wheel is increased still more, the light will be again shut off; etc. If the speed of the wheel, and the VELOCITY OF L", "IGHT 571 distance over which the light travels are measured, the veloc- ity of light can be calculated. The actual arrangement of the apparatus is shown in the cut. W is the revolving toothed wheel ; P is the source of light ;.p is a plate of glass which partially reflects the light from P into the direction of the wheel ; L^ is a converging h-ii s which focuses the light from P at /, a point at the edge of the wheel ; Za is u converging lens so placed that / is its principal focus; Lz is another lens placed at a considerable distance from Lv but parallel to it, with its principal focus at the middle point of the concave mirror *S> whose centre of curvature is at the centre of the lens Ly The waves from P are thus converged at /, whence they diverge and are made plane by Z2 ; they are converged again by l/8, and are reflected back on their path, until they reach the glass plate p by which tliry an- in part transmitted; so an observer with an eye- piece 0 focused on / can determine when the reflected light passes through an opening and when it is shut off. 1 1 the number of revolutions of the wheel per second is n, its angular velocity is 2 TTH. If there are N teeth in the wlirrl, the angular distance from one edge of one tooth to the corresponding edge of the next one is -=\u2022; and so the 2 7T -Zv angular distance from the middle point of one opening to the ii ii< Idle point of a tooth is ~ The time required to turn tlm.uirj, this angle is ^ \u2022 - \u2014 =\u00bb- \u2014 \u2014 \u2022 If the distance from N 2 irn 2 nN the wheel to the distant minor is D (or what is practically thr sun,-, t he distance from Za to Z8), the path of the light is2Z>. So, if the angular speed STTTI is that \\\\hich corre- sponds to the first obscuration of ili\u00ab- litflit. it has traversed a distance 2 7) in the time - \u2014 \u2014 ; and the velocity of light '_' nN is, then, 2 D divided by - J\u2014, or * M fwAw 572 LIGHT In one of Fizeau's original experiments, the distance from La to L3 was 8.633 Kin. or 8.633 x 108 cm. ; the toothed wheel had 720 teeth ; and it was found that the first obscuration of the reflected light occurred when the wheel was making 12.6 revolutions per second. Therefore, in the above formula n = 12.6, N = 720, D = 8.633 x 1010, and so the velocity of light is determined to be 3.13 x 1010 cm. per second. The great experimental difficulty is to maintain a uniform speed of the wheel and to measure it accurately. This method has been used in more recent years by Cornu and Perrotin. In the work of the latter, the distance apart of the lenses L2 and L3 was about 12 Km. The results of these experiments is to give 2.99820 x 1010 cm. per second as the velocity of light in air. The object of using a concave mirror at S with its centre of curvature at the centre of Z-3, instead of a plane mirror, is apparent if it is remem- bered that the source of light at P is not a point, but is extended. So the waves from a point near P are converged at /, and are made plane by the lens L2 ; but their line of propagation is inclined slightly to that of the waves from P, and they are converged by the lens Ls upon a point of the mirror S, a short distance away from its middle point. If the mirror were plane, these waves would then be reflected off one side, and so would not return through the lens L3 ; but, since the mirror is concave, with its centre of curvature at the centre of L3, these incident waves are reflected back through the lens and finally reach the wheel. Therefore the reflected light is brighter than it would otherwise be. Method of Foucault. \u2014 Another method was suggested by Arago, but was first put in practical use by Foucault in 1850, and is always called by his name. (Fizeau also made some valuable suggestions in regard to it.) It consists of making use of a rotating plane mirror in the following manner : Referring to Fig. 298, there is a source of light at P ; a plate of glass at p ; a revolving mirror at m, whose axis of rotation is perpendicular to the plane of the paper ; a converging lens.L, which focuses upon a concave mirror S the light from P reflected at m ; the centre of", " curvature of this mirror S is at m. For a suitable position of the revolving mirror the light from the source P will be focused at the middle point of the concave mirror, and will be reflected back on its path until it reaches the glass plate, when it will be in part reflected and VELOCITY OF LIGHT 573 will form an image at a point P'. The revolving mirror is, of necessity, small, almost linear ; and so, in order to col- lect more light from the source P, the I;P\" mirror S is concave. (Another method, adopted later, was to place the lens L at a distance from m equal to its focal length ; and in this case the mirror S may be plane.) If, then, a reflected image Of P FIG. 298. \u2014 Foucault's revolving mirror apparatus for i is formed at P' when uring the veloclty of \"*ht- the revolving mirror is at rest, a displaced image P\" will be formed when the mirror is turning rapidly, because in the time taken for the waves to pass from m to S and back again, the mirror m will have turned a short distance, and so a reflected ray coming from S will have an angle of inci- dence upon m different from that which it would have if the minor had not moved. This ray, after reflection from m, \\vill have a different patli from the incident ray. If the mirror is making ti revolutions per second, its angu- lar velocity is 2 Trn, and that of the reflected ray is 4 Trn. (See page 445.) If the distance from m to S is D, if that from m to P is r, and if v is the velocity of light, - - is the time 2 D required for the light to pass from m to A^and back again; in this tim\u00ab. the mirror will have tnrne<l through an angle 27) o /> - \u2022 2 Trn, and the reflected ray through - -- 4 Trn ; therefore v the image of P will be displaced by a distance r times this. So, calling P*P\". :>74 LIGHT Foucault increased the effective distance D by having the light reflected several times back and forth between five mirrors before it was finally returned to the revolving mirror ; but in no case did he obtain a very large displacement P'P\". Michelson, however, by changing the arrangement of the apparatus was able to increase D to 600 m.; and even when the mirror was turning at the moderate speed of 200 revolutions per second, he obtained a displacement of 13 cm. This method was improved still more by Newcomb, who FIG. 298 a. \u2014 Michelson's modification of Foucault's apparatus. operated over a distance of 3721 m. The final result ob- tained for the velocity of light in air by this method was 2.999778 x 1010 cm. per second. The mean of all the best values for the velocity of light in the ether is 2.999880 x 1010 cm. per second, with a probable error of about 20 Km. It should be noted that the values of this velocity obtained directly by the methods of Fizeau and Foucault are for the velocity in air ; and, since the index of refraction of air is 1.00029, the velocity in the pure ether is greater in this ratio. The figure given above for the final value is corrected in this manner so as to apply to the pure ether. Velocity of Waves of Different Periods. \u2014 In the pure ether of interstellar space all ether waves, of whatever period, travel with the same velocity so far as is known, as is shown by the fact that the color of any one of Jupiter's satellites is VELOCITY OF LIGHT 575 the same to our eyes when we observe it as it goes into eclipse and as it emerges. If the short waves traveled faster than the longer ones, the satellite would appear red as it disappeared and blue as it reappeared ; and the converse would be true if the long waves traveled more rapidly. In ordinary transparent matter, however, not alone do all waves travel more slowly than in the pure ether, but the waves of different periods have different velocities. It is this which explains refraction and dispersion. Foucault showed by direct experiment that the velocity of light was less in water than in air, by placing a long tube of water in his apparatus immediately in front of the concave mirror. (Actually he used two concave mirrors, one for the light passing in air, the other for light passing in water, and thus olitaim-d two displaced images.) Michelson showed the same for water and for carbon bisulphide; and he also proved directly that in these substances red light travels more rapidly than blue. CHAPTER XXXVI RADIATION AND ABSORPTION SPECTRA Discovery by Newton of Nature of", " White Light. In the year 1672 Newton made the interesting discovery that when sunlight was admitted into a darkened room through a small opening and was allowed to traverse a glass prism, the trans- mitted light was no longer white, but consisted of beams of different colors, each color having a different refrangibility and therefore a direction differing slightly from that of its neighbors. He recognized as distinct colors violet, indigo, blue, green, yellow, orange, and red; but all other intermedi- ate shades were present also. He performed further the reverse experiment of combining these colors by means of a second prism, and produced white light again. He also showed that it was impossible by means of a second prism to break up any of these spectrum colors into parts. These observations prove that white light is due to a combination of simple elementary causes; and we know from Young's experiments that these are trains of waves of definite wave lengths, each train being characteristic of a definite color if it is perceived by the eye. These experiments of Newton form the basis of our expla- nation of the color of natural objects and of the science of spectrum analysis. Various bodies in the universe are emit- ting light (or, more generally, all bodies are emitting ether waves) ; all bodies reflect light (or ether waves) to a greater or less extent ; so, if we look at any object, the light (or ether waves) which we receive is due to various causes. We can analyze this radiation into its component parts by means 676 RADIATION AND ABSORPTION SPECTRA 577 of suitable dispersive apparatus, and can then detect these separate trains of waves by proper means. We shall con- sider in this chapter (1) methods of producing ether waves, especially those which appeal to our sense of sight ; (2) dif- ferent forms of dispersive apparatus and different modes of recognizing trains of waves of different wave length ; (3) the results of the examination of the radiations from different sources. Sources and Cause of Radiation Radiation owing to Temperature. \u2014 All substances in the universe are, so far as kumvn to us, emitting ether waves, owing to the vibrations of certain parts inside their mole- cules. If the body when placed in a darkened room can be seen by the eye, it is said to give off light. The ordinary method of making a body luminous is to raise its tempera- ture ; thus a body may be exposed to a hot flame, such as one from a Bun sen burner, or it may be placed in the poles of an electric arc light. (See page 665.) At such high tem- peratures, many bodies are vaporized, and their vapors are then at this temperature. The laws of radiation due to this cause have been discussed in Chapter XIV. Electro-luminescence. \u2014 Again, if an electric spark is made to pass between two metal points, they are vaporized ami the vapors are luminous ; not, however, owing entirely to the temperature being raised. (The same statement is true of the luminosity of the vapors in the electric arc ; it is only in part due to the temperature of the vapors.) Similarly, if a gas or vapor is inclosed in a hollow vessel, such as a glass bulb, and an electric discharge1 through it is produced 1>\\ any means, it becomes luminous. These cases of luminosity are said to be due to \"electro-luminescence.\" Chemical Luminescence. \u2014 In certain chemical reactions light is emitted ; for instance, when a piece of decayed wood slowly oxidizes, or when phosphorus is oxidized. These are illustrations of \"chemical luminescence.\" AMES'S PHYSICS \u2014 578 LIGHT Fluorescence and Phosphorescence. \u2014 There are many bodies which emit waves as a result of their having absorbed other ether waves, quite apart from any radiation due to tempera- ture alone. Some bodies emit these waves only while they are absorbing the other waves; while others continue to emit them even after the absorption ceases. All bodies of this kind are called \" fluorescent,\" and the phenomenon itself is called \" fluorescence \" ; while the second division of these bodies, as just described, are called \"phosphorescent,\" and the phenomenon is called \"phosphorescence.\" If a beam of light is passed into a fluorescent substance, certain trains of waves are absorbed and others are transmitted ; the energy of these absorbed waves is not spent in producing heat effects, but in emitting other ether waves, which proceed out in all directions. So this fluorescent light may best be seen by looking at the substance from one side. This phenomenon was first observed by Herschel and Brewster, but was first thoroughly investigated by the late Sir George Stokes. He showed that in all cases observed by him, the fluorescent light was of a wave length longer than that of the waves whose absorption caused the fluor- escence. This relation is", " not, however, true in all cases. Some common illustrations of fluorescence are the colors seen in certain forms of fluor spar (whence the name of the phe- nomenon); the color of canary glass \u2014 which is ordinary glass containing traces of certain salts of uranium ; the color of a decoction of the bark of chestnut trees ; the color of the surface layers of kerosene oil ; etc. Phosphorescence is exhibited by the sulphides of barium, calcium, strontium, etc., and by a great many ordinary sub- stances to a certain extent. Sometimes the light is emitted for only a minute fraction of a second ; but in other cases it continues for hours. Conclusion. \u2014 In many cases it is impossible to say exactly what is the cause of the luminosity ; and in nearly all there RADIATION AXD ABSORPTION -1'ECTRA 579 are several phenomena involved. We can, however, divide all cases of radiation into two classes : in one, the substance that is radiating does not change so long as its temperature is maintained constant; in the other, the substance does change even if its temperature is kept unchanged. In the tirst class of bodies, the radiation is a purely temperature effect; and to them Balfour Stewart's or KirchhofFs law \u2014 as it is more often called (see page 301) \u2014 and the other laws of radiation may be applied. This is not true of the bodies of the second class, in which molecular changes are going on. Spectroscopes Different Forms. \u2014 In order to study the radiation of any l><\u00bbdy, some method of dispersing it into a pure spectrum, an-1 some instrument which is sensitive to the various radia- tions, must be used. As we have seen, there are three ways in which dispersion may be secured: by the use of a prism. a grating, or some interference apparatus. Further, a slit (or small source of light) and a converging lens must be used. Thus we have prism, grating, and interference spectroscopes. The conditions as to the purity of the spectra formed by i nst n i in. -nts have been discussed in previous chapters. Precautions. \u2014 There is one obvious precaution which must be taken with these instruments; allowance must be made for aKsnrptimi <\u00bbf the ether waves produced in the apparatus itself. Thus, a glass prism can be used to study the spe.-tra of visible sources of liglit. luit not of those which emit only very short or very long waves, because glass absorbs them. A Ojiiart/ prism is ordinarily used for examining the spectra in.ed by very short waves; a prism of rock salt of tluorite, or of.silvite, those produced I iv lon\u00abr waves. AL: a reflecting grating does not reflect all waves equally; and air absorbs certain ves, eertan visible ones, and all the extremely >lmrt one* These absorption phenomena must be observed by preliminary investigations. 580 LIGHT Receiving Instruments. \u2014 In order to detect the radiations various means must be adopted, as has been already explained. For a limited range of wave lengths the eye may be used; for these and for shorter ones, photographic methods may be applied ; for longer waves some thermometric device is ordi- narily made use of, such as a bolometer, a thermopyle, a radi- ometer, etc. These instruments do not necessarily measure the intensity of the radiation, but by suitable calibration methods many of them may be used for this purpose. There are, of course, many other methods of observation. Scale PIG. 299. \u2014 Prism spectroscope. Scale Attachments. \u2014 Many spectroscopes are provided with divided scales so placed as to coincide with the spectra formed ; and in this manner the position of any definite train of waves may be recorded and thus described. Since by means of a grating the wave lengths of any radiation may be measured, it is a simple matter by using, in a preliminary nAblATI<>\\ A.\\D ABSORPTION SPECTRA f>81 experiment, certain radiations whose wave lengths are known, to calibrate this scale ; and then the instrument may be used to measure the wave lengths of other radiations. Different Kinds of Spectra Continuous and Discontinuous Spectra. \u2014 A distinction has been made already between continuous and discontinuous spectra. In the former all waves within certain limits are present; so there are no gaps in them. In the latter only certain isolated trains of waves are emitted, thus forming separate \"lines.\" Investigations show that all solids and liquids \u2014 with possibly a few exceptions \u2014 emit continuous spectra ; while all gases and vapors emit discontinuous ones. (This is obviously what one would expect to be the case from", " the kinetic theory of different forms of matter.) Gaseous Spectra. \u2014 The spectrum of a gas depends, natu- rally, upon the manner in which it is rendered luminous. So we have \"flame spectra,\" \"arc spectra,\" \"spark spectra.\" \" fluorescent spectra,\" etc. If, however, a gas is made lumi- nous in any definite manner, the waves it emits are definite and characteristic of the gas. Thus, different gases may be identified by their spectra ; and in many cases the discovery of new lines in the spectrum of a gas that was supposed to be pure has led t<> the identification of new elements. Absorption Spectra. \u2014 If the radiation from a solid or liquid falls upon any body, certain waves are absorbed ; and so only a portion of the incident waves are transmitted. The spectrum of this transmitted radiation is called the -orption spectrum \" of the body which produces the al KOI -j.t ion. This absorption takes place in many ways, as has l>cen already stated. In all, the absorption is due, in the main, to the resonance of the minute parts of the mole- cule or of the molecules themselves; and in the greater number of bodies the energy absorbed is distributed among the molecules of the body, and is manifest by heat effects. 582 LIGHT This is called \"body absorption.\" In other substances the energy absorbed in the interior is spent in emitting other waves of longer wave length, thus producing fluorescence. In certain bodies, the absorption takes place in a very thin surface layer ; but the larger portion of the energy incident upon the surface is reflected directly. This is the case with the metals and a few other bodies, and is therefore called \" metallic absorption.\" The law of Kirchhoff in regard to the equality of radiating and absorbing powers may be applied to a substance which exhibits body absorption only. Thus, if a substance absorbs certain trains of waves of definite wave lengths, it has the power of emitting them if rendered luminous by means of temperature alone (and, also, often if other means are used), and the intensities of absorption and of radiation are the same if the temperature of the substance is the same in the two cases. While if the temperature of either condition is decreased, so is the intensity of the effect. Thus, if a white- hot solid is placed behind a quantity of cooler gas or vapor, the absorption spectrum is a continuous one from which cer- tain isolated waves are absent ; and these are identical with those which the luminous gas would emit. The gas absorbs certain waves and transmits the others; it also radiates waves of the same wave length as those which it absorbs ; but the intensity of these radiations is so much less than that of those which are transmitted, that the spectrum is practically as if the gas did not radiate. If the gas or vapor is at a higher temperature than the white-hot solid, the spectrum will be that of the luminous gas with a continuous weak background ; i.e. it is a bright- line spectrum. If the gas or vapor and the solid are at the same temperature, the spectrum will be continuous. Solar and Stellar Spectra. \u2014 These facts are illustrated in the spectra of the sun and of the stars. The solar spectrum and certain stellar spectra are observed to be absorption 584 LIGHT ones ; while other stars produce bright-line emission spectra. The explanation of the latter is obvious : the stars producing them are surrounded by a luminous gaseous atmosphere, which is hotter than the interior portions. Similarly, in the case of the solar spectrum and other absorption stellar spectra, the explanation is that the interior portions are solid or liquid, and are at a higher temperature than the atmosphere of gases and vapors outside. These vapors are naturally those formed by the evaporation of the interior substances. The absorption spectrum in the case of the sun consists of the Fraunhofer lines ; and they can be identified almost completely with the emission spectra of the vapors of certain substances here on the earth ; and thus the con- stitution of the sun is known. The following are a few of the substances which are in this manner known to be in the sun: calcium, iron, hydrogen, oxygen, sodium, nickel, magnesium, cobalt, silicon, aluminium, carbon, copper, zinc, cadmium, silver, tin, lead, etc. Some of the absorption lines in the solar spectrum are due to* absorption by the atmosphere around the earth. Thus, certain groups of lines known as the \" A,\" \" B,\" \" a \" and \" B \" \" bands \" are due to absorption by the oxygen in the air, while numerous other lines are due to the presence of water vapor. A method for distinguishing between solar and terrestrial lines will be described presently. Similarly, by a study of the spectra of the stars, either emission or absorption, a great deal", " may be learned in regard to their constitution, and also motion, as will be shown immediately. The study of these and similar phenomena forms the science of Astrophysics. An excellent book to consult on this subject is Miss Clerke's Problems in Astro- physics, New York, 1903. In speaking of wave motion a certain general property, known as Doppler's principle, was described (see page 345). RADIATION AXD A I: >o /,\u2022/\u2022'/ /o.v sl'WTRA 585 It states that when a source of waves is approaching a point in space, the wave number at this point is increased, while the converse is true if the source is receding. In the case of ether waves that are being dispersed by a glass or quart/, prism or grating, this would be shown by a change in their refrangibility \u2014 an increase if the source is approach- ing the earth, a decrease if it is receding from it. There- fore, if a star is emitting certain trains of waves, their corresponding spectrum lines will all be shifted side wise by an amount depending upon the velocity of the star in the line of sight. If these lines, then, all apparently agree exactly with lines observed here on the earth in the labora- tory for any known vapor, except that they are all slightly displaced, the obvious explanation is that the star is moving in the line of sight; and its velocity in this direction may be deduced from the amount of the observed shift. (This statement is not absolutely correct, for shifts of the lines may sometimes be due to anomalous dispersion or to abnor- mal pressures in the atmospheres of the stars.) Similarly, if the image of the sun is focused by a lens upon the slit of a spectroscope, and it is so arranged that first one edge and then the other of the sun's ima^e is on the slit, the lines in the solar spectrum that are due to solar absorption will be shifted slightly, owing to the fact that one edge of tin- sun is receding from the earth while the other is approach- in LT it, because of the rotation of the sun. Hut those lines in the spectrum due to absorption in the earth's atmosphere will not be so displaced. student should consult Ames, Prismatic and Diffraction Spectra, for Fraunhofer's original memoirs, and Brace, The * of Radiation and Absorption, for the memoirs of Kin-h- hoff :.nd lJunseii. CHAPTER XXXVII EXPLANATION OF COLOR General Discussion. \u2014 The color of an object that is self- luminous depends upon the character of the light that it emits. If it radiates all the visible waves with suitable intensities, it will produce in a normal eye the sensation that we call \"white.\" (The case of defective eyes will be considered in the next chapter.) If the intensity of certain waves is abnormally great, the light appears colored, as is shown when \"red fire,\" a \"sodium flame,\" etc., are used. The color of most objects, however, is due to the fact that they are illuminated and either reflect or transmit light to the eye of an observer. It is obvious that the color of the object will depend fundamentally upon that of the illuminat- ing light ; but we are so accustomed to viewing objects in the white light produced by diffused sunlight, that in describ- ing the color of any object it is always assumed that white light is used with which to illuminate it. When we consider the color of an illuminated body, it is evident that it may be due to any one of several causes. It has been explained in the previous chapter that absorption of light may take place in many different ways, and corre- sponding to each of these there will be certain color phe- nomena. Again, we have seen how colors may be produced by any dispersive action, such as that of a prism, a grating, or an interference mechanism. These various cases will now be discussed briefly. 686 EXPLANATION OF COLOR Absorption Colors Body Absorption. \u2014 The most familiar kind of absorption is that shown when the incident light is absorbed in the interior of the body and the energy of the absorbed waves is spent in producing heat effects. The light that is trans- mitted appears colored, therefore, owing to the disappear- ance of certain trains of waves. If a single train of waves of wave length I is absorbed, which corresponds, therefore, to a definite color, the transmitted light will include all the other trains of waves, which will combine in the eye to pro- duce a definite color, called the \" complementary color \" of that of the waves which were absorbed. If the body absorbs two trains of waves, it may happen that the intensity of the absorption is not the same for both trains ; that is, it may n-.", "juire a greater thickness of the body to extinguish one color than is required for the other ; and it is thus apparent how such a body may appear of a different color as its thick- ness is varied. It is evident that if the absorbing substance is transparent for those waves which it does not absorb, it cannot itself be seen when viewed from the same end as the incident light, or from one side; but if owing to any cause the body diffuses tin- liirht which it does not absorb, then it will appear of the same color when viewed from any direction. Thus a tank ( -ontaininir colored water will appear practically black, except when the transmitted light is viewed, it there are no minute solid particles in suspension : luit if these are intro- duced, it appears colored from all points of view. two portions of matter having body absorption are so placed that the incident liurht tails upon one, and the trans- mit t\u00abl li'jlit is then incident upon the other, the color of the emerging li^ht is that due to the waves which are left after two absorpti The color of all leaves and flowers, of \u2022us clotlis, of paints, of bricks, etc., is due to body 588 LIGHT absorption.' If two paints are mixed, their color is, as just explained, that due to the absorption by both the paints; there is a double subtraction, as it were, from the incident light. The nature of the light that gives an object its color may be determined in two ways : one is to illuminate the object with white light and analyze by a spectroscope that light which is diffused ; the other is to form a continuous spec- trum on a white wall and move the object along this ; if it appears black for any position, it means that the color cor- responding to this position is absorbed, but if it transmits and so diffuses any particular color of the spectrum, it will in the corresponding position appear of this color. Fluorescence and Phosphorescence. \u2014 If the energy of the waves absorbed in the interior of a body is spent in produc- ing other waves, which are therefore radiated in all direc- tions, the phenomenon is called, as has been said, fluorescence. The fluorescent light consists in general of waves whose wave length is longer than that of the waves whose absorp- tion produces the fluorescence. In this case the color of the transmitted and the diffused light is not the same. It is evident that if the fluorescent body is thick, the waves which cause the fluorescence may be entirely absorbed in that portion of the body which is first traversed by the light; so that the fluorescence will occur in this portion only. In some cases this color is confined to almost the surface layers. If the emission of light continues after the incident light is intercepted, the phenomenon is called, as has been said, phosphorescence. Evidently there is some molecular trans- formation involved in this. Surface Color. \u2014 When polished metals and many of the aniline dyes in the solid form (e.g. a dried drop of red ink on paper) are viewed in white light, they have a peculiar appear- ance which is called \"metallic lustre.\" This is due to the EXPLANATION OF COLOR 589 fact that they reflect certain waves much more intensely than others, or, in other words, they have \" selective \" reflection. This process does not take place in the interior of the sub- stance, as in the case of a colored liquid, but at the surface. If a substance showing this metallic lustre, or surface color, is made in a film which is sufficiently thin, it will transmit certain waves. But the color by reflected light is not the same as by transmitted ; in some cases they are approximately complementary. These substances which exhibit surface color have anoma- lous dispersion and change plane polarized light into ellipti- cally polarized light by reflection. Scattering by Fine Particles. \u2014 If the light traverses a region where there are numerous minute particles, it may happen that they are of such a size as to scatter certain trains of waves, and to let pass unaffected all trains of longer wave length. The light so scattered is plane polarized if it is viewed at right angles to the incident beam. This scatter- ing is the explanation of the blue color of the sky, as has i already said, and of the color of sunset clouds, at least in part. The phenomenon also plays a most important part in (h'termining how much radiation (visible and invisible) reaches the earth from the sun. Dispersion It is not necessary to say anything here in regard to the dispersive action of prisms, gratings, etc., but a few illustra- tions may be given of colors due to it. Prismatic <lis|>\u00abTM(\u00bbn is slmun by rainlmws", ", halos around the sun and moon, dew- drops, diamonds when mutably cut, etc. Diffraction colors are seen when looking ;l( in,,ili,.r-\u00ab.f-p\u00ab-arl, at certain line feathers, at corome (the colored rings around tin- mmm \u00bb. tli rough fine-ni' ->bcd cl.,tb at a bright light, etc. Int< colors are shown by soap bubbles and other thin films of transparent matt< CHAPTER XXXVIII THE EYE AND COLOR SENSATION A TEXT-BOOK of Physics is not the proper place for a de- tailed description of the structure of the human eye or of the various theories which have been advanced to account for the sensation of color. Some treatise on Physiology or on Physi- ological Optics should be consulted. It is simply necessary to state here a few facts which are of physical importance. The Eye. \u2014 From an optical standpoint the eye consists of a converging lens which is provided in front with a diaphragm of adjustable diameter, the \" iris,\" and whose focal length can be changed at will to a cer- tain degree. (This power of accommodation is greatly decreased as one grows old.) This lens exhibits both spheri- cal and chromatic aberration, but not to a noticeable degree in general. The medium on one side the lens is the air, but on the other is a liquid filling the cavity of the eye. At the rear of this is the \" retina,\" upon which a normal eye forms an image of the object viewed. A \"near-sighted\" eye has its focus in front of the retina; while a \"far- sighted\" one has its back of it. In the former case, the image may be formed on the retina if a diverging lens is used in front of the eye ; in the latter, if a converging one is substituted. Fio. 801. \u2014 The human eye. 590 THE EYE AND COLOR SENSATION 591 Perception of Color. \u2014 The retina consists of a structure of minute parts which are intimately connected with the endings of the optic nerve. The exact mode of excitation of these nerve endings by the incident ether waves is not known. Certain portions of the retina, viz., those remote from its centre, play no part in color sensation ; for, when waves of all wave lengths lying within the limits of the visible spectrum are incident upon them, one is conscious of a sensation of gray only. This is true of all portions of the retina if the light is faint, with the exception of a small area, called the \" yellow spot,\" which gives color sensations only. This spot is slightly off the axis of the eye considered as a lens. There is also a minute area \u2014 called the \"blind point \"-\u2014near the cen- tre of the retina, where the optic nerve enters, at which no sen- sation of light is produced. Over the other central portions of tin: retina, light of all different colors may be perceived. Addition of Colors. \u2014 It has been known since the experi- ments of Newton that, in order to produce the sensation kk white,\" it was not necessary to have all the trains of waves in a continuous spectrum from violet to red. Corresponding to any color there is another such that if these two sensations are produced simultaneously in the eye, white is perceived. These two colors are called, as has been said, complementary. One way of producing these simultaneous sensations is to paint different sectors of a circular piece of cardboard with the two colors, and then to rotate it rapidly while it is illuminated with white light. Thus at consecutive minute intervals of time if one looks at the rotating disk, the eye receives first one impression and then the other; but since, if impressions reach the eye at intervals faster than about thirty or forty a second, a continuous effect is produced, the e\\ this case receives two simultaneous impressions. This is what may be called the addition of colors ; and it is evident that the mixing of paints, or the suht rael i\"ii of color, has HO connection with it. 592 LIGHT Similarly, a great variety of choices of three colors may be uuule which when added in suitable intensities will produce white light. Taking any three such colors and adding them in different intensities, any other color which is desired may be produced. This proves, then, that in order to account for the perception of colors of all kinds, it is simply necessary to assume that in the eye there are three sets of nerves corresponding respectively to these three colors. (Some writers claim that there are four such sets of nerves, while Hering has an entirely different theory. The statements made here are in accord with all ordinary facts, and embody a portion of the theory of color sensation that was advanced by Thomas Young and supported and extended by Helm- holtz.) The Young-Helmhol", "tz Theory of Color Sensation. \u2014 The choice of these three colors is to a certain extent arbitrary ; but it is limited by an investigation of different cases of color blindness. Many people are afflicted with an inability to recognize certain objects as colored which appear so to the normal eye ; to them they appear gray. Other colored objects appear to them to have colors different from those which would be ascribed to them by an observer whose eyes are normal. The simplest explanation is that in these cases one or more of the sets of nerves referred to above do not respond to stimuli. By an examination of a great many cases of color blindness the conclusion has been reached that the three fundamental color sensations are red (about wave length 671/Lt/^), green (505^), and blue (470 /^). The general explanation of color sensation may then be explained by assuming that there are these three sets of nerves which when excited produce these sensations respectively, red, green, and blue, and that when any train of waves of a definite wave length reaches the retina it stimulates all three sets of nerves, but to different degrees. Curves can be drawn as shown in the cut which give the TIIK EYE AND COLOR SENSATION 593 iv>ults of rxprrinu'iits on adding these three colors with varying intcnsi; Thus, considering the vertical line through F, the three curves are so dnnvn that if ml light of \\va\\v length 671 /JL/JL and of an intensity proportional to the length of the line F\\^ \u2022 Ided to green light of wave length 505/iyLt and of inten- sity proportional to \u2122LET ~ BLUE GREEN \u00a5ELLOW\" \"^ the length F2, and to \u2022 t i I TTO j j '-'\u2022 <' ii rvos of color sensation. blue light of wave length 41 Op/* and of intensity proportional to the length F3, the color perceived is blue corresponding to a wave length 486ft/i. Consequently, if we can assume that when waves of this wave length fall upon the retina they stimulate the \"red,\" \"green,\" and \"blue\" sets of nerves to degrees which are proportional to F 1, F 2, and F 3, the phenomena of color sensation have been explained. BOOKS OF REFERENCE ;:. Light for Adranoed Students. London. 1902. An excellent text-hook, contain in.: <!\u2022 \u2022-\u2022\u2022n|'tioiis of all the fundamental phenomena and references to all the recent investigations. PftBftTOW. Tin- Th.Miry of Liuht. Loii-lon. iM edition.!*!\u00bb:>. Tin- recognized book of reference for all the elementary phenomena of Light OptiOS. (Translation.) \\,-w York. 1902. I h. l....st mo. I. M -ii text-book ing the accepted theories and explanations of all optical phenoi LUMMKR. Photographic Optics. (Transit i-. ) LoDdOD. 1000. A text-book on geometrical Optics, with special reference to lenses. PUT8IC8 \u2014 38 MAGNETISM CHAPTER XXXIX PERMANENT AND INDUCED MAGNETIZATION Magnets. \u2014 A body which has the property of attracting pieces of iron is called a \"magnet\" ; that is, if a magnet is brought near a piece of iron there is a force between them which is shown by their approaching each other if either (or both) is free to move. Such bodies occur in nature, for one of the forms of iron ore which is not uncommon, a mix- ture of FeO and Fe2O3, is magnetic. It is, moreover, a simple matter to make any piece of iron or of ordinary steel into a magnet. There are two general methods for doing this : one depends upon a property of an electric current, the other, upon what is called magnetic induction. If an electric current is made to traverse a wire which is wound in the form of a spiral spring, or helix, the apparatus is called a \" solenoid \" ; and experiments show that, if a piece of iron or steel is placed inside this solenoid, it becomes a magnet. Or, if a piece of iron or steel is brought near, not necessarily in contact with, a magnet, it is made a magnet also. If a piece of iron is magnetized in this manner, ami the magnetizing agency is removed, the iron will lose its magnetism very easily, if it is jarred or subjected to an increase in temperature; but this is not true of the piece of steel \u2014 it remains a magnet under all ordinary conditions. All magnets in ordinary use are made therefore of steel, some kinds of which are much better than others. Much progress in this respect has been made in recent years. r", "i-:i;MANENT AND IMK' MAC \\ l-:i ItATWN 595 Magnets are usually made in the form of bars, rods, or elongated lozenge-shaped \"needles.\" Sometimes the bars or rods are bent into the shape of a U or of a horseshoe; and in this form they are called horseshoe magnets. Experiments show that long magnets are more permanent than short ones Fio. 808. \u2014Horseshoe magnet. and that they remain magnet- ized longer if their ends are joined by a piece of soft iron. Thus an iron bar, called the \"arma- ture,\" is always placed across the ends of a horseshoe magnet when it is not in use. Horseshoe mag- net with soft Iron Bar magnets with armatures. armature. FI.J. Magnetic and Diamagnetic Bodies. \u2014 It is found by ex- periment that a magnet can attract other kinds of matter than iron ; such as many forms of steel, nickel, cobalt, man- ganese, etc. These bodies are called \" magnetic,\" and any one of them can be made a magnet by the methods described above for iron or steel. Again, there are many other sub- stam-es which are repelled by a magnet; such as bismuth, antii i\\. ami /inc. These bodies are called \"diamagnetic.\" ulay made the most important observation that the \u2022 jiit-^tiuii as to whether a body is attracted or repelled by a magnet depends fundamentally upon tin- man-rial medium in winch the magnet an<l tin- body arc innm-rx \u00ab1. In the above dcti nit ions of magnetic ami diamagnetic bodies this medium is assumed to be the ordinary atmosphere. Faraday \u2022bowed that, while in OH\u20ac im-dium a l\u00bbod\\ mi^ht be attracted 1>\\ i m.i'jiirt. iii another it might be repelled. Thus tin- 596 MAG* I.TISM importance of the medium in the consideration of magnetic phenomena is shown. Poles. \u2014 If an iron or steel rod or \" needle \" is magnetized by means of a long solenoid, and if it is then removed and suspended by a fine thread or 011 a vertical pivot, so that it is free to rotate in a horizontal plane, it will turn and after a number of vibrations gradually come to rest in a direction which is approximately (or exactly) north and south. (This fact in regard to a mag- FIO. sos. -pivoted mag- netized bar or needle has been known for many centuries, and has been made use of by mariners and travelers.) The end which points toward the north is called the \" north pole \" of the magnet ; the other, the \"south pole.\" The direction in which it points is called \"magnetic north and south.\" If a magnet is suspended as just described, and another is brought near it, it may be shown that there is a force of attraction between a north and a south pole, but one of repul- sion between two north poles or between two south poles. \"Unlike poles attract, like ones repel.\" It is easily proved, further, that the greater the distance apart of the magnets, the less is the force. In order, then, to explain the reason why a magnet when freely suspended points in a north and south direction, all that it is necessary to assume is that the earth itself has the properties of a magnet. The particular magnetic properties that experiments show it to have will be described later in Chapter XLI. Magnetism a Molecular Property. \u2014 If a magnet is broken up into smaller pieces, each fragment, however minute, is found to be a magnet, with a north and a south pole. This leads one to believe that magnetism is a molecular property of all magnetic substances ; and all observations are in sup- port of this idea. Every property of a body except its mass and weight is changed when it is magnetized ; and con- \\ /:v/ AND L\\i>r<h'i> M.\\<; \\ ETIZATIOX 59 ely, any ehan^e that is known to afi'ect the moleeiii. a body will att'eet tin- magnetism of a magnet. Thus, when an iron rod is magnetized, its length, its volume, its elasticity, etc., are all changed; and when a magnet is hammered or twisted or heated, its magnetism is altered. as is shown by a change in the force which it exerts upon another magnet or upon a piece of iron at a fixed distance from it. Induction. \u2014 We make the assumption, then, that each molecule of a magnetic substance, e.g. of a piece of iron, or of nickel,", " etc., is a magnet; in other words, that each mole- cule of any one magnetic substance has a certain mass and other mechanical properties and is at the same time a magnet. When the substance is in its natural condition, we can assume that these molecular magnets are not arranged in any order, hut are distributed at random; so that, as far as external actions are concerned, each tiny magnet is neutral- ized by those around it. But if a magnet is brought near such a piece of magnetic sub- stance, each of the r*7vT \" becomes a sooth pole. la tter'8 molecular FIG. 806.-Magm-tlo induction: the end A of the Iron bar magnets is acted upon by a force due to the magnet; and the molecules are all turned, more or less completely, in an orderly and regular direction. Thus, if the magnetic substance is a rod or bar, and the magnet is in this form also and is brought near one en<l of the former, so that its north pole is nearest it. the nmlerules will turn so that their mmtli poles are ird the imrth pole <\u00bbf the magnet. Therefore the molec- ular magnets no longer neutralize each other; they now have an external a. tion, and. in fact, the bar which they con- stitute is now a magnet \\\\itl ;h pole toward the north polr of the magnet i/ini: magnet. The change produced in the ;he nioleeular magnets by the magnet is 598 MAGNETISM roughly indicated in the accompanying cut, where each mole- cule is represented by an elongated rectangle whose ends are shaded differently. (Of course, we do not assume that a molecule has actually the shape of a rectangle.) This explains, then, not alone why a piece of magnetic substance is magnetized by the magnet, but also why the two attract each other. The phenomenon is called \"magnetic FIG. 807. \u2014 Arrangement of molecules In an un magnetized and a magnetized iron bar. \u2022 nduction\"; or the former is said to be magnetized by \"in- duction.\" Strictly speaking, these names apply to the phe- nomenon only so long as the magnetizing magnet is kept in its position near the magnetic substance ; when the two are separated, the latter remains a magnet, although a weaker one, for a greater or less time, as described above ; but its magnetism is now spoken of not as induced, but as \" intrinsic \" or \"permanent.\" Similarly, when a rod is magnetized by the action of a solenoid, the magnetism is said to be induced ; etc. The reason why a long thin magnet is more permanent than a short one is clear, because in the latter the two ends are closer together and the molecular magnets at one end may disturb the direction of those at the other, and so produce demagnetization. The action of the armature of a PXSMANMNT AM) i\\i>r< I;D MA<;.\\ /\u2022;///. I y/o.v 599 horseshoe magnet is also easily explained : it keeps the molecular magnets at the ends from changing their positions. It is evident that, \\vh\u00ab-n a bar or rod is magnetized by the action of a magnet at one end, the molecular magnets in the former will be arranged in an orderly manner at the end near the magnet; but at its other end FIG. 80S. \u2014 Magnetization of an iron bar by induction. these minute magnets will not be so systematically distributed. Tha magnetizing of the bar or rod may be made more complete if two mag- nets are used, one at each end of the rod, and turned in opposite direc- tions as shown in the cut. The action may be made still more complete if the two magnets are plan \u2022\u00ab! on top of the rod at its middle point \u2014 oj >| \"\u00bbite poles being in contact \u2014 and are slightly inrlinrd to it, as shown in tin- cut (and if then the two magnets are <lra\\\\ti otf the bar lengthwise in opposite The process should be re- Fio.aw. \u2014 Processor mapneti/ peated several limes. If during any of these processes the rod is ham- mered or Jan..1. tin; magnetization is increased. Experience shows that it is impossible to magnetize a ba\u00bb more than up to a certain degree; ii is then said to be \"satu- rated.\" This fact and all those just described in regard tG methods of magnet i/.at ion are explained easily if it is assumed that the molecules of a magnetic substance are themselves ma '_n i.-i N. One can, in fad, make a model of a magnetic substance by placing on", " a board a great number nail magnets, all pivot.-d BO M t\u00ab\u00bb be five to turn around \\ei-tical axes; and the entire p..f magnetization can be imitated by placing this hoard inside a large solenoid OF l>rin^inur magnets n.-ar it. Temperature Effects. \u2014 It a \"permanent\" magnet has its temperature raised sufficiently high. it loses its magnetism, and becomes again.simply a magnetic substance. If, how- 600 MAGNETIC. M ever, when magnetism is being induced in a magnetic sub- stance, its temperature is raised, \u2014 not too high, \u2014 the induced magnetization is increased. These facts are at once explained on the molecular theory of magnetism. (Iron loses its magnetic properties at about 785\u00b0 C. ; nickel, at about 300\u00b0 C.) The Poles of a Magnet are of Equal Strength. \u2014 Several interesting facts are learned when one studies the force which a bar magnet exerts, either on another magnet or on a piece MAGNET \u00b0f magnetic sub- stance. One way FIG. 810. \u2014 Action of a large magnet upon of doing this quali- tatively is to place the magnet on a COMPASS NEEDLE,. a small pivoted magnetic needle. horizontal table, and to move near it a small pivoted magnetic needle \u2014 such as an ordinary pocket compass. If the latter is short, it may be pushed up close to the magnet ; and it is observed that when this is done, the small needle does not always point toward or away from one of two points, one in each end of the magnet, as it would if there were two centres of force in the magnet. On the contrary, the needle points nearly perpendicular to the FIG. 311. \u2014 Diagram representing arrangement of molecular magnets in a bar magnet. surface of the magnet at all points except near its middle ; and this is what we would expect from our knowledge of the nature of magnetization. The molecular magnets near the centre of the magnet are all more or less parallel to its axis ; but near the ends they are turned so that for one half I'KH. M. i. V/..V7- AND i\\i>r MM;M-:TIZATION 601 the magnet their north poles are in the surface, while for the other half their south poles are there. We must then regard each point of the surface of a magnet as a centre of force ; and the total force of a magnet on another magnet, or on a magnetic substance, is found by adding geometrically all the forces due to these magnetic poles in the surface. These may. however, be combined mathematically in certain cases in a manner that is instructive. Let a bar magnet be under the action of another magnet perpendicular to it at its middle point, whose south pole is turned toward the former, but which is so far away that all these minute surface forces R. Fio. 812. \u2014 Forces acting on a bar magnet owing to a distant magnet. may be considered parallel. (We can, if we wish to, con- sider the second magnet as so long that we need not take into account the forces due to its north end.) All the forces over the north half of the magnet may be added so as to form the resultant Rn : similarly, those over the south half may be added to form Rt. If the magnet is turned so that it is no longer perpendicular to the distant one, these forces still keep their value : and their direction is toward the latter. This condition is practi- cally the ease when a magnet is under the action of the earth, and if a magnet is suspended so as to be free, \u2014 not alone to turn, but also to move as a whole, e.g. if it is floated on a cork ivMiiiLr <>n the surface of a tank of water, \u2014 it is observed that the mag- net, if not originally in a north OF nth direct ion. turn* hut does not move i 18. \u2014 Mom.1.1 a maifiirt wh.-n pliwii! as a whole. ThU provei that the tun foroei /\u00bb', and //, are e\u00ab|ii.d in amount but opposite iii direction, forming a COUple. MAONET18M (See page 100.) If a bar magnet could be made whose molecular magnets were all parallel to its axis, there would In- magnetic forces only at its ends. So if in the actual magnet just described L is the distance apart of Rn and Rt measured along the magnet, and if this special bar magnet is made of the length Z, it might replace the former so far as action at a considerable distance is concerned. The forces on the two ends of the latter magnet due to the distant one are then equal in amount but opposite in direction. We are therefore led to make the assumption that each molecular magnet", " has two centres of forces, a minute distance apart, which are equal in amount but opposite in kind ; that is-, if one centre exerts a force of attraction on one end of any distant -Forces magnet, the other exerts on it a force of repul- between two small sion. Thus, if JV^ and N^S2 are ^wo molec- ular magnets, there are four forces acting on each, as shown, arid if S2 is at the same distance from both N! and Sr the two forces acting on it are equal in amount but are in nearly opposite directions. All observed facts in regard to magnetization are in accord with this assumption. We say that each molecular magnet has two poles whose \" strengths \" are equal but opposite ; or that they have equal but opposite \" quantities of magnetism \" or \" magnetic charges.\" The same statement in regard to magnetic charges must then be true of any magnet, however large or compli- cated, because it is made upof magnetic molecules, and therefore contains as much south as north magnetism. It is impossible to separate a north pole from an equal south pole and obtain them distinct from each other, because, when a magnetic mole- cule is decomposed into simpler parts, it ceases to be a magnet, and equal amounts of north and south magnetism vanish. Magnetic Field and Lines of Force. \u2014 When a small mag- net is placed near a large one, it is acted on by certain forces; I'Ki;MA\\i:\\T AM) 1XDUCED MAGNETIZATION 603 and, in general, a region in which a small magnet experiences forces is called u \" magnetic field.\" A simple mode of study- ing and describing the properties of such a field is to draw what are called \"lines of force.\" These are continuous lines such that any one indicates by its tangent at any point the direction in which a north pole would move if placed there. Thus, if P is any point on a line of force AB, a north pole, if placed there, would move in the direction of the curve at P, while a south pole would, of course, move in the opposite direction. (If the poles had no inertia, they would actually trace out the whole curve AB.) Therefore, if a small magnet is placed with its centre at P, there will be a moment acting on it which will make it turn and lie tangent to the line of force at P. Fie. 815.\u2014 Lines of mig- netic force, and the forces acting on a small magnet. This leads at once to two modes of actually drawing the lines of force in any li'-l.l. One is to move a small magnet from point to point in the fit-Id, noting at each its direction; thus, starting from any point P, the magnet should be moved a short distance in such a direction that in its n\u00ab'\\v |M.<iti<m it is tangent to the same curve to which it was tangent atP, etc.; a line can be drawn through these points. Then another starting I >oi nt is chosen and another line is drawn, etc. The other method is to place a horaontal sheet of paper or a glass plate in the field of force,.sprinkl.- tin.- inm filings Ov\u00ab-r it, ami jar the paper or plate until tin* filings assume definite positions. Under the action of the magnet i< : of the field each minute tiling Kccomcs magnetized, and therefore acts likr a small magnet; and under tin- action of the couple it turns and places ; along the lim- <>f force. Thus tin- action is just as if one had ma ny thousand minute magnetic needles in the field at one time. (Actually in neither of these cases do the forces acting on the small magnet form a couple, unless the inagn-'t which causes the field of force is at a great 'li-tan<.\u2022 ; 1,1.1 the resultant force of translation is as a rule minute.) Several Qlostrations of lines of magnetic force, as mapped by means of iron til ings, are shown in theMOOmptnyingOUte] 604 MAGNETISM around a single bar magnet, and between two magnetic poles, alike and unlike. The direction of a line of force is defined FIG. 816. \u2014 A bar magnet. m by its being that in which a north pole would move. It follows, therefore, that lines of force start from north poles of magnets and end on south poles. Fur- ther, two lines of force cannot cross. It is interesting to note the effect up- on a field of force of the introduction of a piece of magnetic substance as shown in Figs. 317 and 3170. From the first of these it is appar- ent that the lines of Fie. 816", " far as was possible with the instruments at his command. It was verified also by Gauss, and to a greater degree of accu- racy ; but our main reason for believing that the law is exact is that all of its consequences are found to be in accord with the varied facts of electrical engineering, into which enter so many questions connected with magnets and magnetic fields. In order to assign a number to the magnetic charge of any magnet, it is necessary to define a unit charge ; and in doing this it must be remembered that magnetic forces are different in different media. (See page 595.) Making use of the C. G. S. system of units, a \" unit magnetic charge \" is defined to be such a charge that, if placed at a distance of 1 cm. in air from another equal charge, the force between them is 1 dyne. Then, if a charge equal to m of these units is placed at a distance r cm. from a charge m1 in air, the force between them expressed in dynes is given by the equation f=\u2014^- In any other medium the force is proportional to this, and therefore, following the accepted system of symbols and writing as a factor of proportionality -, the force in any f* MAGNETIC FORCE AND INDUCTION 609 medium is/ = From what has just been said, the value of fj, for air is one on the C. G. S. system of units and using as a unit magnetic charge that defined above; natu- rally, if another unit charge were adopted, the value of this constant for air would be different. The factor /A is a quan- tity which is characteristic of any medium \u2014 it is not, how- ever, necessarily a constant. It is called the \" permeability,\" for reasons which will appear later. Intensity of Magnetic Field ; Magnetic Moment. \u2014 When a magnet is placed in a magnetic field, it is acted on by two forces, one at each pole, they being the resultants of all the forces acting over the surface, as explained on page 601. Let us assume the simplest case, viz., that the charges are entirely at the ends ; then, if the magnet is short, the forces at the two ends are equal in amount, although opposite in direction, because the two ends are at almost the same point in the field. The \"intensity\" of the field at any point is defined to be the value of the force which would be exerted (\u00bbn a unit north charge if placed at that point. Therefore if the inten- sity of the field is R, and if m is the (1 large on either pole of a short ni;i'_rn\u00ab-t which is placed in the field, there is a force Rm acting on each pole. If I is the length of the magnet and if it is in such a posi- ti\"M that it makes an angle N with the direction of the field, the per- pendicular distance between the Fio. MO.\u2014 Moment acting on a tor magnet when placed In a field of intensity R. two i'\"ives is I sin N i so the strength of the couple acting on the nii_rii.t is Rml sin N. If the magnet is pivoted AMES'S PHYSICS\u2014 89 f>10 MAGNETISM around an axis perpendicular to a plane which includes the magnet and the line of force at its middle point, it will turn under the action of this couple toward the direction of the field ; so the moment of this couple should be written \u2014 Rml sin N. The product ml is evidently a property of the magnet itself, and it has received the name \" magnetic moment of the magnet.\" In the general case of any kind of magnet, the magnetic moment around any axis is defined to be the maximum value of the moment of the forces acting on the magnet when it is placed in a field whose inten- sity is one, with this axis at right angles to the field. If a magnet is broken up into parts, these are found, as a rule, to have different magnetic moments ; and the value of the \"intensity of magnetization\" at any point of the magnet is defined to be that of the magnetic moment per unit volume around that point. Thus, if M is the magnetic moment of a portion whose volume is V, the ratio \u2014 in the limit, as Vis taken smaller and smaller, is the value of the intensity of magnetization. If the magnetic charges are entirely at the ends, the intensity is the same throughout the magnet ; so if this is a cylinder of length I and cross section A, and if the charges on each end are m, the mag- netic moment M is ml and the volume V is IA. So the V J\\. intensity of magnetization \u2014 = m-\\ or, it has the same value as that of the charge per unit area on the ends ; this is called the \" surface density", " so \u2014 may be determined. R Measurement of R or M. \u2014 By a combination of the two formula' for RM and \u2014 -, it is seen that R B>= \u00bb\u00abV r\u00abr\u00bbtanJV* and so both R and M may be measured. It /! i> known fur any one fu-ld, it has been explained how its value for any other field may be determined by means of a \\ ilir.it in-_r magnet whose period can be measured. Magnetic Tubes. - If one refers to the illustrations of lines of force given on page 604, it is evident that these lines are most crowded together at those places where the intensity of the tirlil is the greatest, and are the farthest apart at those \u00bb'.U MAGNETISM points where the intensity is the least. This suggests a systematic mode of drawing lines of force. We can describe a small closed curve at some point near the magnet, and can imagine lines of force drawn through each point of this curve ; these lines, if continued, will of course be found to start from a north pole of a magnet and end on a south pole ; so they thus form a hollow tube leading from one pole to the other, whose cross section is small near each end, but greater at a distance. If the initial small curve is taken of exactly the proper size, this tube will inclose at its two ends a unit magnetic charge. Such a tube is called a \"unit tube\"; and, if the magnet has a charge m at each end, tnere are m tubes leaving the north pole and returning to the south pole. It is evident that where the cross section of a tube is least, 'the intensity of the field is greatest ; and vice versa. Similarly, if A is the area of any small surface in the field at right angles to the force, and if there are JV tubes passing through this surface, the intensity of the field at that point is pro- portional to the limiting value of the ratio \u2014, as A is taken -ZV A smaller and smaller. In words, the intensity of the field at any point is proportional to the number of tubes per unit area at that point. Magnetic Induction. \u2014 As was explained on page 605, and as is apparent from the cuts on that page, the effect of intro- ducing a piece of iron or other magnetic material into a field of force in air is to cause the lines of force to change their direction and enter the iron. If the iron is in the form of a rod, and if its cross section is A, more tubes pass through it than did through the same area of air before the iron was substituted for it. If the original field of force in the air is uniform, so that the intensity is the same at all points, the lines of force are all parallel, and the tubes are all of the same cross section. If the intensity of the field is 72, the number of tubes per unit area is proportional to this. If, MA<; \\KTH- FOWK AM) IMK'CTION 615 now, a long iron rod is introduced parallel to the field, the number of tubes per unit area of its cross section is greatly increased; and it may be proved by methods of the infinitesi- mal calculus that the ratio of this number to the previous one equals the value of the quantity ft for iron, as defined on page 608. It is for this reason that- p is called the permea- bility. (For different kinds of iron, and for different con- ditions. p may have values as great as 2000.) It is thus seen that for any magnetic substance /A is greater than for air. The number of tubes per unit area in the iron (or other magnetic substance), when in a field of intensity 72, is, then, proportional to the product pR ; and this quantity has re- ceived the name of the \"magnetic induction\" at the point where Ii is the intensity of the field and p is the permeability. The fact that the tubes do not simply end on the iron rod, l>ut must be considered as passing through it, may be proved by certain phenomena of electric currents which will be dis- cussed in a later chapter, and also by the simple experiment of cutting the rod into two pieces by a transverse section and separating tin-in slightly ; the field of force in the crevasse is found to be much more intense than in the original fit-Id. When the magnet causing the field of force is removed, the iron remains magnet i/.\u00ab-d. and therefore some tubes still remain passing through it. So we are led to believe that in tin- case of an ordinary magnet these tubes do not end at its poles, but continue thron-h it, forming endless closed ri exactly as if <..... wen- to take a piece of rubber tubing and bring its two ends", " together. This conception of tul\u00bb M-tie induction is due to Faraday. Diamagnetic Bodies. \u2014 The case of a diamagnetic body ma\\ ited in a similar manner. Kxperiments.show that if such a hody is made in the form of a 1, n. and brought near a l> net, it is magnet i/ed 1>\\- induction, hut in a direction opposite t\u00ab\u00bb what it would he if it \\\\eiv iron. In the cut, where AB is the diamagnetic body, the end.1, 616 MAQNBT18M F,o. ^.-Action nearest the north pole of the magnet, is thus found to be a north pole, while B is a south one. There is repulsion, therefore, between the magnet and the diamagnetic body. jf the ^ Q\u00a3 force are drawn, they are found to be as shown in the cut, showing that, instead of being crowded together in the diamagnetic body, they avoid it. There is therefore a smaller number of tubes in the bar than there was in the original field before the bar was intro- DIAMAGNET.C BAR bod,: A Fio. 828. \u2014 Effect of introducing in a uniform magnetic field: (1) a sphere of magnetic sub- stance ; (2) a sphere of diamagnetic substance. duced. This proves that the value of the permeability JJL for a diamagnetic substance is less than for air. Energy Relations. \u2014 Since all motions in a system of bodies take place in such a manner that the potential energy of the system becomes less (see page 114), it must be possible to explain from this point of view the attraction and repulsion observed with magnetic and diamagnetic substances. If we consider the formula /= \u2014 -\u00a3 for the forces between mag- netic charges, it is evident that there are changes in the potential energy whenever magnets are moved with refer- ence to each other ; because work is done in overcoming the forces, or by the forces. The magnets themselves are not changed; and so we are led to believe that the energy is MAGNETIC FORCE AND INDUCTION 617 located in the surrounding medium where the magnetic field exists. It follows from the formula that / is small if //. is huL,re, or, in words, the forces are small if the permeability of the medium is large ; and consequently in such a medium the energy per unit volume is also small, since small amounts of work are involved in any changes, other things being equal. (Magnetic forces can be felt through a vacuum, and so the energy of a magnetic field is, in the case of any material medium, both in the ether and in the matter.) Attraction and Repulsion. \u2014 Therefore, if a piece of iron \u2014 for which ft is greater than for air \u2014 is introduced into a field of force in air near a magnet, the energy in the space occu- pied by the iron is less than when it was occupied by air; and the decrease in the energy is greater if the field of force is intense than if it is feeble. In other words, if a piece of iron is moved up gradually toward a magnet, the potential energy of the field becomes less and less ; therefore, if a mag- net and a piece of iron are left to themselves, there is a force of attraction between them, and they will approach each other. Similarly, a magnet will attract a piece of any magnetic sub- stance in air. Conversely, and for obvious reasons, a magnet will repel a piece of any diamagnetic substance in air. In general, if /z for any substance is greater than for the sur- rounding medium, it is attracted by a magnet; while, if it is less than for the medium, there is repulsion. The obser- vations described on page 595 are therefore explained. An exactly analogous phenomenon in mechanics is afforded by the motions of a block of stone and a block of wood wln-n immersed in a tank of water: the fonwr will be attracted by the earth and will sink; tin- latt.-r will be repelled and will rise. The explanation in both canes is that the motion takes place in such a direction as to make the potential energy of the system less. The stone sinks because it is heavier than the water; and therefore by replacing an equal volume of water closer to the earth, the potential energy of gravitation is decreased. The wood rises because it is lighter than the water; ami. th<-i<!\u2022\u00bb\u2022. it it moves up iin.l \\\\at.-r rrj.Isu-.\"* it. tin- potential energy is again decreased. CHAPTER XLI MAGNETISM OF THE EARTH Magnetic Elements. \u2014 The fact that there is a magnetic field of force on the surface of the earth is proved by the ob- servations on the", " motion of a suspended magnet, which were referred to on page 596. If a bar magnet or a magnetic needle is suspended in such a manner that it can turn freely in all direc- tions, it will finally come to rest in a position such that its axis is inclined with reference to a horizontal plane and lies in a vertical plane which nearly, if not quite, coincides with the geographical meridian at the point of suspension. This vertical plane is called 'the \" magnetic meridian \" at the point ; and the angle it makes with the geographical meridian is called the \" magnetic declina- tion \" or the \" variation.\" The angle which the axis of the needle makes with the hori- zontal plane is called the \"magnetic inclina- tion \" or udip.\" The earth's magnetic field at any point is then completely denned by its intensity, the declination and the inclination. These three quantities are called the \"magnetic elements.\" FIG. 324. \u2014 A mag- netic needle suspended free to turn in any di- rection. The dip can be measured by observing the angle which the needle makes with the horizontal plane. The variation is most easily determined by mounting the needle so that it is free to turn about a vertical pivot, and noting the angle it makes with a true north-and-south line, which may be 618 MAVXETISM OF THE AM/,' 777 619 1 orated by astronomical methods. It is convenient for most purposes of measurement to consider the earth's magnetic force as resolved into two com- ponents, one horizontal, the other vertical. Thus, if OD is the direc- tion of the field of force at 0, and if OA and OB are horizontal and vertical lines through 0 in the same plane as 0/>, the angle (AOD) is the dip ; and calling it N and the intensity of the force R, the horizontal com- ponent is R cos N, and the vertical one R sin N. The former can be meas- ured with great accuracy by the method described on pages 610-613. Therefore, since the angle of dip, N, can be measured directly, the value of R may be deduced. Further, if the ratio of the horizontal and vertical components can be meas- ured, the dip may be calculated ; for, calling these H and V* \"26. \u2014 Diagram rep- resenting the horizontal and vertical components of the earth's magnetic force. \\, A dip circl.-. Y=-~. Variations in the Elements. \u2014 Observations show that the values of all three of these elements at any one point are con- tinually (hanging. So far as is known, these changes are pei i(,di(, that is, for instance, the dip makes a pendulum-like oscillation during the twenty-four hours, increasing slowly, t hen decreasing, etc. ; further, the mean value for any one day iges slightly the next day, and so on, having an oscilla- tion whose period is a year; and, again, the mean value for any one year is not the same for the next year, but changes slightly; but the period of this change is not known, for 620 MAGNETISM since regular observations began to be taken \u2014 about the year 1540 \u2014 this oscillation in the mean annual value of the dip has not been completed. Similar statements may be made in regard to the other two magnetic elements ; there are daily, yearly, and secular changes, so called. (There are other periodic changes than these, but they are the most important.) It often happens that there FIG. 827. \u2014 Chart showing secular change In IG.. \u2014 r w cr cg. -, \u2022, j the earth's magnetism, from observations made 1S a Sudden and Unexpected at London ; the black line indicates both the in- disturbance of the magnetic clination and the declination. elements of a magnitude far greater than the regular changes ; this constitutes a \" mag- netic storm.\" The explanation of such phenomena is not known ; but observations have shown that they occur most frequently when the spots on the sun and when aurorse in our atmosphere are most numerous. \u201e.,,.. Magnetic Maps. \u2014 The magnetic field over the earth's sur- face may best be described by drawing on a map of the earth certain lines which indicate the values of the elements at any one epoch. Thus lines can be drawn such that at each point of the earth's surface, through whose position on the map any one line passes, the value of the declination (or variation) is the same. Such lines are called \"isog- onals,\" and are of the greatest possible assistance to mariners and surveyors. They are shown in the cut for the year 1900, and each one is marked with a certain number, e.g. 5\u00b0, which indicates the value of the", " variation for all points on that line. These lines run approximately north and south ; and it should be observed that for two lines the variation is zero, i.e. at points on them a mag- netic needle points true north and south: they are called \"agonic\" lines. One of these is approximately a great MAGNETISM OF THE EARTH 621 circle of the earth ; the other lies in northern Asia, and is called the \"Siberian oval.\" Again, lines can be drawn which indicate in a similar man- ner the inclination or dip ; they are called \" isoclinals,\" and are approximately parallels of latitude. The line of zero dip is called an \"aclinic\" line, or the \"magnetic equator.\" There are two points in the earth's surface where the dip is 90\u00b0; these are often called the \"magnetic poles.\" The lines for the year 1900 are shown in the cut. Other lines, giving other information, may also be drawn ; but they need not be described here. Conclusion. \u2014 The explanation of the magnetic action of the earth is not known. It has been proved, however, that it is due almost entirely to causes which are within the earth itself. Certain of the periodic changes are occasioned, how- ever, by external causes, such as electric currents in the atmosphere. Historical Sketch of Magnetism The property which the lodestone possesses, of attracting iron, was known centuries before the beginning of the Christian Era, because it is mentioned by Thales, who lived from the year 640 to M\u00bb; i..<. The Greeks and the Unmans were acquainted with the fact that the intervention of other bodies, like brass, does not destroy magnetic effects. That like poles repel and unlike attract, and that a lodestone possesses the power to eniniminicate polarity to inert iron, were known at least as rarly as the twelfth century. The compass was in daily use in Kumpo also as early as this, l>ut the disenvrries nf magnetic declination and its variation t'mm place to place were made by Columbus in Hart man is reputed to have di.senvn-ed the dip in 1644. He obtained a value of 9\u00b0 when- he should have obtained 624 MAGNETISM 70\u00b0. This fact was not published, and Norman, in 1576, independently discovered it in London, obtaining a value of 71\u00b0 50'. Norman was probably the first to suggest that the source of attraction is in the earth, and not in the heavens as gen- erally supposed. He also showed that the earth's field is simply directive and produces no motion of translation, by floating a needle on water. The variations in the magnetism of the earth were discovered by Gellibrand in 1636. The first systematic treatise on magnetism was William Gilbert's De Magnete. It was published in 1600, and con- tains a complete account of what was known as magnetism up to that time, as well as a great number of new ideas and experiments which are due to Gilbert himself. Gilbert was the first to recognize the difference between temporary and permanent magnets ; to detect the effect of a change in temperature ; to show that the fragments of a magnet are themselves magnets; to observe the effect of hammering, etc. ; to make use of the idea of lines of force, although in an imperfect manner. The fact that iron was not the only magnetic substance was shown by Brandt, who proved in 1733 that cobalt was magnetic. The diamagnetism of bis- muth was recognized by Brugmans in 1778, but the first systematic study of the subject was made by Faraday in 1845. It was he also who made the great discovery that the forces of attraction and repulsion depend fundamentally upon the surrounding medium. ELECTROSTATICS CHAPTER XLII 1 IXDA.MKNTAL PIIKNOMEXA Introduction. \u2014 It is observed if a piece of silk is rubbed against a glass rod and is then separated from it that both now have the power of attracting small fragments of paper, of metal foil, of thread, etc., toward those portions of their surfaces which had been in contact, and that, further, if either one of them is suspended so as to be free to move, it may be attracted by the other. The silk and the glass are said to be \" electrified,\" to have on them \" electrical charges \" or \" charges of electricity,\" or, more simply, to be u charged.\" The same phenomena may be observed with any two por- tions.)f different kinds of matter; but with certain kinds the forces of attraction are manifested, not alone by those portions of the surface where they were in contact with the other body, but also over all their surface. This is true of metals, for instance. So, if one end of a long metal wire is charged, the forces are evident over all its", " length; the wire is said to \"conduct\" the charge, and it and similar bodies are called \"conductors.\" l'\u00bb\\ suitable means the charges on the other end of the wire may be removed; but, if the charge on the first end is continually renewed, charges will appear again at the former one, etc. This may be called, then, an electric \"current,\" and while it is going on, many interesting phenomena occur both in the wire and ide it. We are thus led to divide the subject of Elec- AMBS'S PHYSICS \u2014 40 ''25 626 ELECTROSTA TICS tricity into two parts : one deals with electrical phenomena when the charges are at rest, it is called \" Electrostatics \" ; the other with the phenomena of electric currents, it is called \" Electrodynamics.\" We shall begin with the former. Electrical Charges. \u2014 As said above, when two portions of different kinds of matter are rubbed together and then separated, they are charged, and can produce forces which they could not when in an uncharged or neutral condition. The act of friction is not essential ; all that is necessary is that the two pieces of matter should be brought closely in contact. We distinguish, too, as stated above, between \"conductors\" and \" non-conductors.\" The following bodies are the commonest illustrations of conductors : all metals, either solid or liquid; water containing in solution almost any salt or acid ; the human body ; the earth. The follow- ing are illustrations of non-conductors : glass, silk, paper, cloths, dry wood, porcelain, rubber, sulphur. In order to produce any appreciable charge, therefore, in a conductor, it must not come in contact with the hand, but must be \" insulated \" by holding it in a piece of paper or cloth. Energy of Charges. \u2014 The fact that forces are exhibited near charged bodies and that therefore work can be done by producing motion, proves that there must be energy associ- ated with charges. This is evident also, because, as stated above, when two bodies are charged by rubbing them against each other and then separating them, one attracts the other, and this proves that in order to separate them work was required. In other words, work is necessary in order to have electrical charges. This energy which is associated with charges is not in the bodies themselves: it is in the medium which surrounds them wherever the electrical forces may be felt, that is, throughout the \"electric field.\" This fact is proved by the phenomena of electric sparks. It is known to every one that if the electric charges are too intense, sparks take place in the medium (e.g. ordinary / \u2022'[ \u2022 \\ DA M /\u2022; \\ T. 1 /. I'll h'\\<>M i:\\. 1 627 all(l these are due to the breaking down of the material structure of the medium. It' a spark passes through a sheet of paper or a pane of glass, a hole is made in it : it the spark is in air, the molecules of its gases are broken into parts. This proves that the medium must have been greatly >t rained just before the sparks passed ; and, if it was strained, it must have possessed potential energy. Electric forces may be shown in a vacuum; and therefore the seat of the energy of electric charges is in both the surrounding ether and the material medium immersed in it. The importance of the nature of this medium in all electrical phenomena is thus established. r Positive and Negative Charges. \u2014 If two rods of the same kind of L,rlass are charged by means of a piece of silk, and if \u2022 >nr is suspended horizontally in a paper sling so that it is free t<\u00bb turn, it may be seen, \"ii bringing the charged por- tion of the other rod near it, that one repels the other. Whereas, if the piece of silk \\vhieh was used to charge the glass rods is brought near, there is attraction. Simi- larly, if other charged bodies are brought near the sus- P'-nded glass rod, some repel it and the others attract it. All those charged bodi\u00ab \u2022> whieh repel it are said to be \" positively \" charged ; while those which attract it are said to be \"negatively' d. Thil unonntfl definition of positive cr plus ( 4- ) and negative or minus ( \u2014 ) ehurijes. Thus the experiim-nt s jn that 628 ELECTROSTATICS glass rubbed with silk is charged positively ; and that the silk is charged negatively. Similarly, in all cases, experi- ments show that when any two bodies are brought in con- tact and then separated, they are charged oppositely. If different charged bodies are suspended in turn, it is observed that it is a general law that a positive charge attracts a negative", " one but repels another positive charge, and that a negative charge repels another negative one. \"Like charges repel; unlike ones attract.\" It is found, further, that the force becomes less as the distance apart of the charges which are acting is increased. A body which is charged positively when rubbed with some definite body may be charged negatively when rubbed with another one. And, further, the character of the charge received by a body often depends upon the condition of its surface, whether it is smooth or scratched, etc. Thus, glass is charged positively by a piece of silk, but negatively by a piece of flannel ; and smooth glass may be charged positively, while, if it is rough, it may be charged negatively. By a careful study of the character of the charges produced on different bodies when rubbed with other ones, it is found that it is possible to arrange all bodies in a series, A, B, (7, etc., such that if B is rubbed with A it is negatively charged, whereas if it is rubbed with C it is charged positively. Such an arrangement is called the \"electrostatic series.\" A few of its terms are : cat's fur, flannel, glass, cotton, silk, wood, the metals, rubber, sealing wax, resin, sulphur. Conductors. \u2014 We say that a body is charged at any point if electrical forces are exhibited when a small piece of matter is brought near that point. If the charged body is a con- ductor, there are no forces shown in its interior ; if it is a hollow solid, \u2014 like a hollow ball, \u2014 there are no forces in the interior region ; in other words, if a conducting body is charged, the charge is entirely on its surface. This phenomenon may be considered as due to the repul- l--l-\\l>.\\Ml-:\\TAL PHENOMENA 629 sion of a charge by a similar charge ; the charges distribute themselves as far apart as possible, and, since a conductor allows charges to flow, they will all be on the surface. (This is true only after the charges have come to rest ; it does not hold when there are currents.) This fact may be proved by direct experiment in many ways. Faraday made a metal box large enough to allow him to enter it and carry with him his instruments ; and he showed that, however the box was charged, there were no effects inside after the charges came to rest. (Similarly, he showed that whatever electrical charges or changes he produced inside, there was no elec- trical force outside. The explanation of this will be given later.) Lines of Force. \u2014 The region around charged bodies in which electrical forces may be shown is called the \" electric field \" ; and a \" line of electric force \" is a line in the field such that at each of its points its tangent is the direction in which a minute body charged positively would move if left to itself. (A negatively charged body would, of course, move in the opposite direction.) If a line of force is con- tinued, it will be found, therefore, to start from a positive charge and to end on a negative one. Two lines of force < iiiniot cross, for that would mean that at the point of inter- section a charged body would move in two directions. 'I 'here are no lines of force inside a conducting body; they all end at its surface. In I i^. :>:51, lines of force are drawn for several special cases. It is seen that the phenomena of attraction and re- pulsion and the distribution of the lines themselves may be described by saying that lines in the same direction repel each other, and that there is a tension in the lines tending to m.ike them contract. It may not be unnecessary to state the obvious fact that these lines have no physical existence, but are merely geometrical constructions. The lines of force may be mapped by a method exactly 030 ELECTROSTA TICS Two unlike charges, the positive one four times Two similar charges, one four tiim.\u00ab as great as the negative one. as great as the other. FIG. 381. \u2014 Lines of electrostatic force. FUNDAMI-:M.\\L ni I:\\OMENA 631 similar to that described for a magnetic field. It will be shown in the next paragraph that when any piece of matter is put in an electric field it becomes electrically charged, some portions with plus electricity, others with minus. If it is an elongated body, its two ends become charged oppo- sitely : and so, if it is short and is suspended or pivoted at its middle point, it will turn and set itself along the line of force. It plays, then, the same part in an electric field as does the short magnet in the magnetic case. Induction. \u2014 If an uncharged body is insulated from the earth and is brought near a charged body which is", " also in- sulated, the former will exhibit electric forces. On the side nearer the charged body, it will apparently be charged with the opposite kind of elec- tricity to that of the latter ; while on its more remote side it will be charged with the >ame kind. Thus, if in the FIO. 882. \u2014 An electrified bo<ly is brought near cut the charged body has a positive charge, the end A of the other body which was originally uncharged will exhibit the properties of a negative charge, while the end B will have those of a positive charge. If th<- charged body is withdrawn, these charges on the other disappear. The phenomenon is called \"electrostatic induc- tion\" ; and tin; charges are railed \u2022\u2022 induced \" ones. If the uncharged hody which is brought near the chared one is a piece of a non-coin! net or, \u2014 <\u2022.//. of L^lass, of sulphur,. \u2014 its molecules are affected by the electric fours and the whole body is strained, if is said to he -polari/ed.\" The trie charges are distributed \u00ab\u2022 \\actly as are the magnetic in tin- case of,i pi. MM- <.f imn which is brought near a magnet. ( See page 597.) There is, in fact, almost complete analogy between the case of a piece of non-conductor put in an electric tidd and a piece,,f magnetic substance put in a magnetic field. There may be one ditVerence; after the 632 ELECTROSTATICS latter is removed from the field, it remains a magnet, while in the former case the electric charges in general disappear. Cases have, however, been observed in which the electric charges remained; and so the analogy in these cases is exact. I A FIG. 888. \u2014 Induced charges on a conductor. JL If the uncharged body which is brought near the charged one is a conductor, it becomes charged, oppo- sitely on its two ends or faces, as described above. There is an essential difference, however, between this case and that of a piece of non-conductor, owing to the fact that lines of force do not pass through a conductor and, there- fore, end on its surface, while they can and do pass through a non-conductor. This difference will be explained more fully in the next chapter. These induced charges on a conductor are caused by the attraction of the charge on the charged body for an unlike charge and its repulsion of a similar charge ; it being borne in mind that these forces are due to the fact that when unlike charges approach each other, or when like charges recede from each other, the potential energy in the medium becomes less. Thus, if a conductor is joined to the earth by a con- PIG. 884. \u2014 Charging a conductor by induction. ducting wire, and if a positively charged body is brought near it, a positive charge is repelled to the earth and the rr.\\i>.\\Mi-;\\ i AL 633 conductor itself has a negative charge ; if now the conduct- ing wire is removed, the conductor retains its charge. The distribution of the lines of force is shown in the cut. This process of charging a conductor is known as \"charging by induction.\" Fio. 884 a. \u2014 Charging a conductor by induction. Experiments show that, if a charged body has points on its sur- face, the electric force in the air is greatest near them ; and, in fact, if such a charged conductor is carried into a darkened room, faint sparks will be seen at the points. The charges are passing off to the particles of dust and to other small portions of matter in the surrounding air. These thus become charged with the same kind of electricity as that on the body, and are, there- fore, repelled by the latter, forming a current in the air, or a wind. This is often sufficient to be felt by the hand or to blow out a candle flame. If, then, a pointed conductor is brought near a \u00ab haiLced body, so that its points are to wanl the latter, which may be either a conductor or a_ non-conductor, t In- latter will induce charges on the for- mer ; and those on the points turned toward the charged body will escape, Fw. 886. -Action of,K,int\u00bb by induction. be drawn to the latter, and \"discharge \" it by neutrali/in^ the charges on it; the other induced charges, which are like tlmse on the body originally charged, will remain on the eondiu -tor. The final action, therefore, is as if the obi 634 ELECTRO 8T A TICS were bodily transferred to the pointed conductor. This action of such a pointed conductor,", " or a \"comb,\" is made use of in many electrical machines. (Its importance was first recognized by Benjamin Franklin (1747). It is the reason why lightning rods are always made with sharp points.) Electroscopes. \u2014 It may be well to explain at this point one or two simple instruments which are used in the study of electric phenomena. One of the most useful of these is the \"gold-leaf electroscope,\" which consists essentially of two vertical slender strips of thin gold foil connected at their upper ends to a metal rod which is attached to a metal plate or ball. The gold leaves are, as a rule, inclosed in a glass bottle so as to prevent any action of draughts of air. If the plate or ball is given a charge, this will spread over the leaves, and since they are now charged alike, they will repel each other, and will diverge. The angle of divergence will vary with the intensity of the force of repul- sion. Further, if a charged body is simply brought near the plate (or ball), charges will FIG. 886.\u2014 Gold-leaf electroscope. be induced on the leaves and they will diverge. In most gold-leaf electroscopes there are thin strips of tin foil fastened to the walls of the glass vessel and attached to the metal base of the instrument, so that if the gold leaves are diverged too far they will not communicate their charge to the non-conducting glass walls, but to the conducting strip, which will carry the charges to the outside of the instrument. Another simple instrument is the \"pith-ball electroscope.\" It consists of a small pith ball covered with a thin layer of metal foil and supported from a vertical metal rod by a fine wire or other conductor. If the rod is charged, it will trans- fer some of its charge to the pith ball, which will be repelled. The angle its supporting wire makes with the rod is a meas- FUMtAM K.\\TAL I'll K.\\< >M KNA (J35 urc of the force. It is obvious that a single gold leaf could be used in place of the pith hall, or that two pith halls could be used in place of the two gold leaves in the former instrument. Electrical Machines. \u2014 As we have seen, electrical charges may be produced by two independent methods: by friction or contact between two different bodies, and by induc- tion on a conductor. Corresponding to these are two types of machines for producing (barges continuously. a. Friction Machine. \u2014 There are various forms of these so-called friction machines ; but a description of the one shown in the cut will apply to all. There is a large glass plate pivoted on an axle, which is clasped at one point by two metal clips lined with leather; so that. as the wheel is turned, the glass becomes charged positively and the clamps negatively. The charges are removed from Fi... M, the latter by joining them to the earth, and from the former by the use of a point. -d conductor or \"comb.\" A positive 036 ELECTROSTATICS charge is thus accumulated on the large conductor which is joined to the comb. b. Induction Machine. \u2014 The simplest form of instrument for producing charges by induction is the so-called \"elec- trophorus,\" which was invented by Volta about 1775. It consists of a thick plate A, of some non-conducting sub- stance such as glass or hard rubber, which rests in a metal base B ; and of a loose metal cap (7, provided with an insu- lating handle D. In using the instrument, the cap is removed and the upper surface of A is charged by friction with a piece of -flan- nel or cat's fur ; let it be assumed that it is thus charged negatively. This charge will induce a plus charge on the upper surface of the metal base B, and the induced minus charge flows off to the earth. (The function of this in- duced charge on B is by its attraction for the charge on A to prevent the latter from escaping or leaking.) The metal cap 0 is now lowered on A. Actually, it touches it at the most in only a few points and so does not receive any appreciable charge from A directly. But the charge on A induces a positive charge on the lower side of C and a nega- tive one on the upper side. This cover 0 is now touched with the finger or otherwise connected to the earth ; so the negative charge is removed, and only the positive one re- mains. Connection with the earth is now broken, and if the cap is lifted by its handle, it will carry with it its positive charge. This charge may be transferred to some conductor ; and the cap being discharged may be replaced on the plate A, which still retains its charge. So the process may be repeated indefinitely. FIG.", " 889.\u2014 Electrophorus. Machines have been made by which these various steps are M i-;.\\ i. i L /'// I-:.\\\u00bbM I:.\\A 037 carried out automatically. One of these is shown in the cut. The explanation of its action is simple, but is so long that it need not be given here. It may be found in almost Fio. 840. \u2014Induction electrical machine. any special treatise on Electricity, such as dimming, Elec- tricity ; Perkins, J5Y\u00ab///W/v <tu<l Magnetism; S. P. Thomp- son, Elementary Lessons in Electricity and Magnetism, etc. CHAPTER XLIII ELECTRIC FORCE; MEASUREMENT OF ELECTRIC QUANTITIES Quantities of Electricity. \u2014 The fact that an electric charge is a quantity to which a numerical value may be assigned is suggested by many experiments. If a hollow metal vessel, like a can, is placed on top of a gold-leaf electro- scope, and if a charged body is lowered into it by means of a silk fibre (or other non-conductor), the leaves diverge owing to induction ; but the amount of the divergence is found to depend upon the charge lowered, not upon its position inside the vessel. Further, if another similar charge is lowered into the vessel, the leaves diverge still more, but the amount of this does not change if the two charged bodies are brought in contact, or even if they touch the walls of the vessel. We are thus led to speak of the \"quantity\" or \"amount\" of charge, or of electricity. It should be noted that the last experiment described shows that the total quantity of electric- ity on two or more conductors is the same before and after they touch. Thus we speak of the \" Conservation of Elec- tricity.\" Equal Quantities of + and -- always produced. \u2014 If two uncharged bodies are lowered into the vessel, e.g. a piece of silk and a piece of glass, the leaves do not diverge, even when these two bodies are touched or rubbed together and then separated. But, if one of them is now removed, the leaves do diverge, showing that the two bodies were charged, but with exactly equal amounts of opposite kinds of electric- ity. Similarly, if an insulated conductor is lowered into 638 ELECTRIC FORCE the can in which there is already a charged body, there is no change in the di\\ er^vnce of tlie leaves, thus proving that the two induced charges are of exactly equal amounts of opposite kind. Thus it can be stated as a general law that whenever a charge of any kind is produced, an equal charge of the opposite kind also appears. Faraday's \"Ice-pail Experiment. \" Dielectrics. \u2014 An in- teresting experiment in this connection is one due to Faraday. If a charged body is lowered by means of a silk cord into the interior of a nearly closed hollow conducting v< which is joined by a wire to an electroscope, the leaves of the latter will diverge ; but, as said above, the amount of the divergence does not change as the charged body is moved about inside the vessel, or even if the two touch. aday in his original experiment used a metal \"ice-pail \" as the vessel.) If the. charged body is a conduc- tor, it will lose its cha when it touches the metal vessel, because the charge will all go to the outside of the hollow conductor. (See page 628.) Hut since the divergence of the leaves is not affected, there can hare been no change in the ce*p*11 e*p6r\"nent charge on this ronihietor. The explanation is that when the charged body is lowered into the hollow vessel, it induces an equal amount of electricity on the inner wall of the vessel of a kind opposite to its own \u2014 and therefore also an equal amount of the same kind as its o\\\\-n on the outside of the wall of the vessel; so, when the charged conductor touches the inner wall of the vessel, equal amounts of plus and minus charges pass to the outer surface, and there is con- sequently no change in the external conditions. It is \"I, 640 ELECT If <>* T. I TICS served, further, that when the charged body is lowered into the hollow vessel, the divergence of the gold leaves, is the same whatever non-conducting medium is used to fill the vessel or is present in it: air, oil, sulphur, etc. Thus, electri- cal effects are transmitted through these various substances; and for that reason Faraday called them \"dielectrics.\" (Actually there is a distinction between the idea of a dielectric and that of a non-conductor or insulator, but it need not be emphasized here.) Law", " of Force. \u2014 The force between two charged bodies is found to depend upon the amounts of their charges, their distance apart, and the nature of the surrounding medium or dielectric. So far as distance is concerned, Cavendish proposed the relation that the force varied inversely as the square of the distance. He showed by most ingenious mathe- matical reasoning that, if this were true, the charge on a spherical conductor must be entirely on its outer surface, even if there were bodies in its interior which were joined with it ; and he then proved by direct experiment that the charge was entirely on the outer surface. (This was pre- vious to 1773.) This same suggestion as to the law of force was made independently by Coulomb (1785) ; and he verified it by direct experiment, placing charges at different distances apart. A unit electrical charge may be defined in a manner simi- lar to that used for a magnetic charge. On the C. G. S. system of units a unit charge is defined to be such a one that, if it is at a distance of 1 cm. in air from an equal '\u2022- charge, the force is 1 dyne. This is called the \"C. G. S. Electrostatic Unit.\" Then, if a charge whose value is e is at a distance of r cm. from - a charge e^ in air, the force in dynes between them is \u2014\u2022 But it is found that the force depends upon the surrounding medium ; and this is expressed by writing the value of the force in ELECTRIC FORCE 641 dynes as /= *,. where K is a quantity which is character- istic (and constant) for any one dielectric. It is called the '\u2022dielectric constant.\" Using the system of units defined ahove, the value of ITfor air is one. Its value for all other dielectrics (with the exception of a few gases) is greater than for air, as may be proved by directly measuring forces in different media. Tubes of Induction. \u2014 The \"intensity\" of an electric field of force at any point is defined to be the value of the force \\vhieh would act on a unit positive charge if placed at that point. Kxactly as in the case of a magnetic field, too, tubes hounded by lines of electric force can be drawn; and if they are of the proper size they will end on unit plus and minus charges. They are called \" tubes of induction.\" Since there is no force inside a closed conductor, even if it is charged, these tubes must end on its surface, not traverse it. They do, however, pass through a dielectric, as is shown by i day's experiment described on page 639. In this the tubes all start fn>ni either the charge which is introduced or from the inner wall of the vessel, and end on the other charge. We can, in fact, define a real charge as one which origi- nates tubes of induction. Thus, in the cases of induction described on page 6tfl tin- tubes from the charged body pass into and through the dielectric bodyj'but they end on the iiictiiiLT on,-, and an e.jual number leave the other end. Thus, the charges on the latter body are /v/// ; while in the former case they are only <//'/\u00bb//v///, the forces manifested bein'_r di; iv be sliown by methods of the inlinitcsimal calculus, to the fact that the tubes are passing from one dielectric into another. The number of unit tubes per square centimetre at ri^ht angles to the Held at any point is, as in the case of magnet- \u2014 41 ism, prop.M-ii.mal to the intensity of the field at that point. 642 ELECTROSTATICS Explanation of Attraction and Repulsion. \u2014 The explanation of electric attraction and repulsion may be given in terms of energy exactly as was done for mag- netic forces ; the constant K taking the place of /*. Since K is less for air than for all other dielectrics, a small piece of any such dielectric is more permeable for tubes than is air, and is attracted by a charged body if air is the surrounding medium. Further, since there is no force and therefore \u2122 energy inside a conducting body, a small piece of a conductor is attracted even more than a piece of non-conductor. The action of charges on each other has already been discussed. Electric Potential. \u2014 The properties of electric charges and the condition for their being in equilibrium may be expressed in a different manner. When a charge is moved in an elec- tric field, work is done, either at the expense of the energy of the field or against the forces of the field. Thus, if a plus charge moves in the direction of the", " field of force (or a minus charge in the opposite direction), the field loses energy and the charged body gains kinetic energy as it moves ; if a plus charge is moved against a field of force (or a minus charge in the opposite direction), work is done by some outside agency i and the energy of the field is increased. We have a mechanical analogy in moving a body toward or away from the earth ; if it is raised, work is done against gravitational forces and the potential energy is increased ; if it falls, it gains kinetic energy at the expense of the potential energy. We may define the potential energy of a body of unit mass with reference to the earth when placed at any point as the \" potential at that point,\" or we may say that the potential at a point is the work required to raise a body of unit masft from the earth to it ; and we may describe /;/./,' ntic FORCE 643 the gravitational forces in terms of this quantity. The potential is evidently constant at all points of any horizontal plane; and the higher the plane is from the earth, so much the greater is the potential. Such a surface of constant potential is called an \" equipotential \" one. Evidently there is no change in energy as a body moves along such a surface ; but the change as a body of unit mass is moved from a point Pl in an equipo- tential surface, the -\u00a3 V, potential of whose points is Vv to a point P2 in a second equipotential sur- face whose potential I.2 - J r and FIQ m _ p. and F> ^ two horlzonul planes is entirely independ- ent of the path of the motion. (See page 108.) The line of action of the force at any point is perpendicular to the equipotential surface through that point ; because, if it were inclined to it, there would be a component force in the surface, and work would be required to move a body against it, \\\\hich is contrary to tin- idea <>t' an equipotential surface. The direction of the force is, obviously, from points of high to those of low potential. Similarly, in the case of electrical phenomena, we may >se the earth as our standard body, since it is a con- ductor, and is so large that its electrical condition ma\\ In- regarded as permanent, and may define the \"electric pot en - tial \" at any point in an electric field as the work required to a unit plus charge from the earth to that point. I not necessary to specify any particular point on tin- earth, because tin- potentials of all points of a conducting body are the same, if tin' charges are at rest. If this were not so, it would require work to carry a charge from one point to another in the conductor; this would presuppose that there was an elec- 644 EL EC TROSTA TICS trical force in a conductor ; and, as we know, this is not true. So, since the earth may be regarded on the whole as a con- ductor, all points of its surface are at the same potential, whose numerical value is zero in accordance with the definition of potential given above. (This does not imply a zero amount of anything; for potential is not a quantity which can be measured. See page 11. We give it a number, just as we give temperature a number. 0\u00b0 temperature does not mean a zero amount of anything, but indicates a temperature which serves as the starting point of a thermometer scale. So the potential of the earth is 0, because the earth is the body of reference. Actually the earth is not a good conductor, and there may be local differences of potential.) Similarly, the potential at any point far removed from the electric charges, that is, at \" infinity,\" is zero, because no work would be re- quired to move a unit charge from such point to one on the farther side of the earth, where by definition the potential is zero. We can draw equipotential sur- faces in the field of force ; the lines of force are at right angles to them ; and the direction of the lines is from high to low potential. Thus, if the field is due to a charged spherical conductor, whose complementary charge is at a great no. 844. -Lines of force and equi- distance, everything is symmet- potenttal surfaces around a charged rical with reference to its Centl'6 ; spherical conductor. \u201e the equipotential surfaces are con- centric spheres, and the lines of force are portions of radii starting from the spherical conductor. If the charge on the conductor is plus, the potential at points near it is higher than that at those more distant ; if the charge is negative, just the reverse is true. Thus, if a plus charge is put at any point, the potential of all points near ELECTRIC FORCE 1>", "\\ is raised ; while the contrary is the case if the charge is negative. Induction. \u2014 We can thus explain the appearance of in- duced charges on a conductor. Let a positively charged body be brought near an insulated uncharged conductor AB. All points of this must be at the same potential since it is a conductor ; but if tlu' conductor were absent, the potential at a point A near the charged body would be higher than at a point B which is more remote ; consequently, if the poten- tials at A and B are to be the same, a negative charge must appear at A so as to lower its potential, and a positive charge at 5 so as to raise its potential. Or, again, since when the conductor is absent, the potential at A is greater than at B^ the electric force is in the direction from A to B\\ and, when the conductor is introduced, a plus charge moves in tin- direction of the force toward B, and a negative charge moves in the opposite direction toward A. T I A FIG. 845. -Electrostatic Induction. Further, if the conductor is joined to the earth by a wire, its potential must be zero; but 'under the influence of the FIG. 846.- Effect upon line* of force and c.|ii1poti>ntUt Mirftrr*,,f introducing a uphoric*! dttctor in the field and HIM. joining It to the earth. charged body alone it would be some positive amount ; there- fore, in order to lower it to zero, a negative charge must appear on it. (The same explanation can be given of the inducing action of a negative charge.) Distribution of Charges. \u2014 The fact that the potentials at all points of a conductor on which the charges are at rest are the same is a consequence, as was shown above, of the fact that there are no forces in a closed conductor. This may be expressed in a different manner : the charges on a conductor are all on the surface, and they are so distributed that the inten- sity at any point inside is zero, or, what is the same thing, that the potential at all points is the same. Thus, consider a closed FI\u00ab. 347. -Diagram illustrating the conductor of any shape, and let P fact that the force inside a closed con- be any point in its interior and Qv ductor is zero. r\\ r\\ i \u2022 \u00b1 f \u2022 L Qv Qp etc., be any points of its surface. The charge at Ql is at the distance rl from P ; that at Qy at the distance rv etc. So, if Av A2, Ay etc., are small areas at Qv Qv $3, etc. ; and if dv d%, c?8, etc., are the values of the surface density of the charges at these points, i.e. the charge per unit area, the intensity at P, or the force acting on a unit plus charge, if placed there, is the geomet- rical sum of 1 21, 2 22, 3 23, etc. This sum must be zero. ri rz rz By considerations of this kind it may be proved that the sur- face density at a point is greater than over a plane surface. (See page 633.) Sparks. \u2014 One of the commonest phenomena associated with electric charges is that of sparks. They are occasioned, as has been explained, by the mechanical rupture of the mate- rial medium in an electric field ; and they prove the existence of a great strain due to the electric forces. The intensity of the field at any point may be expressed in terms of the poten- ELECTIUC FORCE 047 tiul. If F\"and V + AF\"are the potentials at two neighboring points at a distance apart Aa?, the electric force is in the direction from the second point to the first ; and if R is its numerical value R&x = &V, because each member of the equation expresses the work required to move a unit charge from one point to the other. Since R is actually in tin* direction in which V decreases, the exact formula is R&x = AF \u2014 AF; or 72=\u2014\u2014. Consequently, if the intensity is great, there must be a great fall of potential in a small dis- tance. Thus, if the difference of potential between two con- ductors is high, there is danger of a spark passing between them, and a connection may be found by experiment between the potential difference and the spark length in any dielectric under definite conditions. A limit is therefore fixed by the electric properties or \" strength \" of the air for the value to which the potential of a conductor may be raised; for, if it is exceeded, a spark will pass to the earth or to particles of foreign matter in the air. When a spark passes between two conductors,", " its path through the air is an excellent conductor ; and therefore both bodies are brought to the same potential. The pot. n- tial of one is raised by the passage to it of a certain amount of positive electricity, or by the \\\\ ithdrawal from it of a cer- tain amount of negative electricity; and that of the other is lowered by the opposite process. (See Electric Current*. page 663.) The luminous character of a spark or dischar^1 in any gas is due to the luminosity produced by the electrical changes \\\\-hich accompany the disruption of the molecules and the conduct ion of the current. Capacity of a Conductor. \u2014 If we consider an insulated eondiirt.ir 1.;. i ii space, it is eviden t that if it is charged pnMt ivdv. it \\\\,11 itself liave a posit ivi- potent ial. and that if it* ( -hai-irc is increased, so is its potential, because a greater amount of \\\\\u00ab.ik would be required to bring up to the 648 ELECT HOST A TICS conductor a unit charge from the earth. If the charge is doubled, so is this amount of work, and therefore so is the potential, etc. We may express this fact in a formula, writing e for the charge, V for the potential of the con- ductor, and 0 as a factor of proportionality, viz., e = OV. This quantity O is called the \" capacity \" of the isolated conductor ; it may be defined, as is seen from the formula, as equal in value to that charge which would raise the poten- tial of the conductor by a unit amount. It is evident from general considerations that 0 must depend upon the shape and dimensions of the conductor, and upon the dielectric constant of the surrounding dielectric. If air is the dielectric and if the conductor is charged with a quantity, e, the potential VA is, by definition, the work required to carry to it a unit plus charge from the earth; while if the dielectric has the value jfiT, the forces are dimin- ished JT-fold (since the electric forces vary inversely as 7f ), and the potential of the conductor, F^, or the work now re- quired to bring up a unit plus charge, is less than VA in the ratio 1 : K\\ or, VA=KVK. So if CA and CK are the capaci- ties in the two cases, e = CAVA=CKVK\\ and hence OK=KQA. In words, the capacity of a conductor varies directly as the dielectric constant of the surrounding medium. The connection between the capacity of a conductor and its shape and size may be deduced in certain simple cases by means of the infinitesimal calculus. Thus, it is known for a sphere, an ellipsoid, a cylinder, etc. The capacity of a sphere of radius a in air is numerically equal to a ; and, there- fore, in any other medium it is Ka. If a charge is distributed over two spherical conductors of radii rl and r2, which are in contact, their potentials are the same, but their charges are different. If the dielectric is air, we have, writing el and e2 for the charges and V for the common potential, e\\ = r\\V, e2 =?'2F; and the surface densities of the charges on the two spheres are. * a and 2 2> (/\" we assume the distribution over each to be uniform. Calling these d\\ and ELECTRIC FORCE 649 rfj = F orrf,:rf,= I:I. This in- 4?rr2 rt r2 dicates that if n > r2, </i < </2- So, if the curvature is great, the surface density is great. A point on the surface of a sphere may be compared roughly with a small sphere attached to it ; and so we see why the sur- face density of the charge on a point is so great. Energy of a Charge. \u2014 The energy of a charged conductor is located in the surrounding dielectric ; but its numerical value can be expressed in terms of the charge, e, and the potential, F, of the conductor itself. We can imagine the conductor as being originally uncharged, and the process of charging as consisting in the bringing up to it from the earth a series of minute charges. In this manner the charge grad- ually increases from 0 to its final value, e ; and the potential rises from 0 to F\". Since the potential at an instant varies directly as the charge at that moment, the mean value of the potential during the process is \\V\\ that is, the work done in charging the conductor is the same as if the whole charge, \u00ab, were brought", " up against this potential of JF\", in- stead of the small amounts having been brought up against the continually increasing potential. By definition, the po- tential is the work required to move a unit plus charge from the earth up to the conductor, and so tin- work required to move the charge e is the product of e and the potential. In the present case, then, the \\\\.uk don.- is the product of e and JFi or \\eV. This is the value of the energy of the field. It should!M\u00bb noted that in this mode of considering the charg- ing of the conductor, an equal charge, \u2014 e, of the opposite kind, is left on the earth, whose potential is /ero. Similarly, the energy due to a charge \u2014 e whose potential -%cV. (This does not mean that there is such a thing as negative energy ; for if a charge \u2014 e is by itself in space, its potential T has a negative value. \u00bb S... in general, if there are two conductors with equal and opposite eharges, + e and \u2014 e, at potentials /'..and \\'r the energy in the di- electric surrounding them I'., l\\i 650 ELECTROSTATICS The energy of an isolated conductor can be expressed in another form, which is often useful. Calling it W, the formula is W = \\eV\\ but e = CV, so we may write W- \\?- or W= J C V\\ The capacity is a constant C for a given conductor in a given dielectric, and is independent of the charge or its potential ; so these formulae show that the energy is inde- pendent of the sign of the charge, depending simply upon its numerical value. Mechanical and Thermal Analogies. \u2014 An analogy may be drawn between electrostatic potential and fluid pressure, which is useful. A fluid, either gas or liquid, always flows from points of high to those of low pressure : a positive elec- tric charge moves from points of high to those of low poten- tial. When a gas is compressed by a pump into a vessel of any kind, the pressure continues to increase until a point is reached at which the vessel breaks or the gas leaks, and this maximum pressure does not depend upon the size of the vessel, but upon its strength, etc. ; when the charge of a conductor is increased, the potential rises, and a condition is finally reached when a spark passes or the charge leaks off, but this maximum potential is determined by the \" strength \" of the surrounding dielectric, not by the size or capacity of the conductor. Similarly, heat energy always flows from bodies at high temperature to those at low ; and, if a small flame is main- tained at as high a temperature as a large one, it is just as useful. Condensers ; Capacity. \u2014 Owing to the liability of a charged conductor to lose its charge if its potential is high, a method has been devised by which the conductor may keep its charge unaltered, but may have its potential lowered. When this is done, its charge may be increased before there is again danger of its escaping. The apparatus is called a \"con- denser.\" The general principle, then, is to make use of any processes which will decrease the potential of a charged conductor. ELECTRIC FORCE 651 If the conductor is a plate and is charged positively, the charge will he distributed the same on its two sides, if it is isolated; but if, as shown in the second figure in the cut, another conducting plate is brought near it, minus and plus charges will be induced on this, and an additional amount of the plus charge on the first plate will be attracted around to the face opposite the second one. As a result, the poten- tial of the first plate is lowered ; because if a unit plus charge is brought up to it from the earth, less work is required than before, owing to the action of the induced negative charge on the second plate. The induced plus charge on this plate serves to keep the potential hi<_rh ; and if it is removed by join- in- the plate to the earth, the potential is lowered still more. This potential is now the work required to carry a unit plus charge across from the second plate, whose potential is zero, to the first one; and it may be decreased n if a dielectric, such as glass, is substituted for the air between the plates; for if K is increased, the force required to move a charge is decreased. Therefore, in the end practically all the charge on the first plate is on the face toward the second one, and there is an equal amount ot electricity of the opposite kind on that lace of the second plate which is toward the first one; and the potential of the iir>t plate is greatly below that which it was originally. Fio. 848. \u2014", " Different steps in the construction of a condenser. \"EARTH If the connection with the earth is removed, and the whole apparatus is moved elsewhere, possibly near some other charged bodies, the potentials of the two plates will change, but f/i.-ir difference remains constant, because it equals the \\\\ork re.|niivd to move a unit positive charge from one plate 652 ELECTROSTATICS to the other; and we may assume that the two plates are so close together that this work depends simply upon the charges on them. If, then, -f- e and \u2014 e are the charges on the two plates, and V^ and Vl are their potentials, F^ \u2014 V\\ is a constant so long as e does not change. If it varies, so does V<i \u2014 V\\\\ and one is proportional to the other. We may therefore write e \u2014 G(V^\u2014 Fi), where 0 is called the \"capacity of the condenser.\" It is evident that this quantity is a constant for a given combination of two conductors of definite size and shape separated by a definite dielectric of a definite thickness ; and it has, of course, no connection with the similar constant for a single isolated conductor. The numerical value of the capacity may be calculated for many simple cases. A condenser consists essentially of two similarly shaped conductors placed close together and sepa- rated by a dielectric, such as glass, mica, etc. The com- monest forms are those in which the conducting plates, or \"armatures,\" are parallel plates, concentric spheres, or coaxial cylinders. A few facts in regard to the capacity of these condensers are evident from the formula of definition : The capacity must vary directly as the dielec- tric constant of the two dielectrics, because for a given value of e, V^ \u2014 V\\ varies inversely as K\\ the capacity must vary directly as the area of the armatures, because for a given \"value of V% \u2014 F^, e varies directly as this area ; the capacity must increase as the armatures are made to approach each other, because for a given value of e, V%\u2014VV varies in- versely as the distance apart of the armatures. Exact calculations by means of the calculus show that the capacity, on the C. G.S. system, of two parallel plates of area A and at a distance d apart is (7= ; that the capacity of two con- wd Kr r centric spheres of radii rl and r2 is O= \u2014 \u2014 ; and that the r 1 KLi:< ilUC FORCE capacity of two coaxial cylinders of radii rl and rv per unit length, is C = 2 log 5 This formula for two parallel plates holds, of course, for those por- tions of the plates only which are not near the edges ; for it is only over 19. \u2014 Lines of electrostatic force in the case of a plate condenser, the dielectric being air. those portions that the conditions are uniform. This is shown in the cut, which represents the lines of force. Therefore if such a condenser is to be used for purposes of measurement, a device is employed, invented i. \\\\hii-li consists in having a 1 one plate near its centre cut out from the rest of the plate and supported independently of it. The rl this disc are uniformly distributed, and the formula given above applies to its capac- |.!.-it- -f it One form of con- denser most often used for qualitative experiments is the so-called whos. iuii.-r mil outer surfaces, except for t li. ii of a glass bottle \"\"^^^\u2022\u2022\u2022T\"\"' 654 ELECTROSTATICS upper portions, are coated with tin foil; the bottle is closed by a wooden stopper through which passes a brass rod tipped with a knob and ending below in a chain which makes contact with the inner coating. The capacity of such a condenser may be calculated fairly closely from the formula for two parallel plates. Discharge of Condensers. \u2014 If a conductor carrying a knob is attached to each armature of a condenser, and if these two knobs are gradually pushed toward each other by means of some insulating rod, a limiting distance will be reached at which a spark will pass between them. This distance varies with the difference of potential V^\u2014V^ the greater it is, the longer the sparking distance. When the spark takes place, the potentials of the two armatures become the same, and since the plates had equal and opposite charges, the final charge on both plates is zero : the condenser is said to be \"discharged.\" Experiments show that there are two types of discharge. In one, the charge on each plate becomes gradually less, the potentials gradually approach the same value ; and in the end the", " in parallel. condensers are said to be \"in parallel.\" Whereas, if Ql is joined to P2, #2 to P8, etc., they are said to be \"in series.\" Let the condensers all have the same capacity and all be charged alike before they are connected ; then their differ- ences in potential are all equal, but the potentials of any two plates, e.g. Pl the be and P, need same. the potentials of P1 and Qv V, and U2 those of P2 and Qr etc. Then in all cases not be and \u2022I.\u2014 Three condensers joined in series. Let FJ- tf1= F2- Z72 = etc.'1 1 the condensers are joined in parallel, Vl = V^ = Fg = etc., and Ul = U^ = Us= etc. ; so, it is exactly as if the condenser were made up of two large plates, one, Pv Pv P8, etc., the other, Qv Q2, Qy etc. The difference of potential 2, is Fj \u2014 Ur and the total charge on either \" plate \" is ne^ \\\\ here n is the number of condensers connected and e is the charge on each plate; so the capacity is increased n tim\u00ab \u2022>. Thus, joining in parallel gives an increased quantity^ but does not change the difference in potential. 1 1 the condensers are joi m-d in series, J7j = Vv U^ = V# etc. ; t there are n condensers, V\u00b1 \u2014 Un^n(Vl\u2014 U-^). Thus if P! and Qn are connected so as to discharge the condenser, the difference of potential is increased n-fold; but the quantity of electricity discharged is the same as for a single con- d. Miser. Since the distance between two conductors at which a spark will lak\u00ab- place is increased if their difference AMES1* PHT81C8 - 42 658 ELECT It OSTA TIC 8 of potential is incrt-usi'd, joining condensers in series in- creases their sparking distance. (When two or more con- densers are joined in series, the minus charge on Ql does not combine with the equal plus charge on P2, etc., until Pl is joined to Qn. Before this, the minus charge on Ql is held in place by the attraction of the plus charge on Pv etc.) Electrometers. \u2014 Before we can explain how the various electric quantities are measured, it is necessary to describe an instrument which enables us to measure differences in Principle of quadrant electrometer. potential. Such an instrument is called an \"electrometer.\" There are many forms which may be used to measure the ratio of two differ- ences in potential; so that, if one is known, the other may be calcu- lated. The best of these is the Km. 856. \u2014 Thomson's quadrant -l-'tP.meter; one of the quadrants is removed so as to show the \" needle.\" \" quadrant electrometer,\" which was invented by Lord Kelvin, then William Thomson. It consists, as shown in the drawing, of a cylindrical metal box which is divided by two trans- verse cuts into four \" quadrants,\" and of a horizontal metal \" needle \" shaped like a solid figure eight, which is sus- pended by a fibre. The pairs of diagonally opposite quad- ELECTRIC FORCE 659 rants are connected by wires, and the needle is raised to a high potential by some electrical machine. If the difference of potential of two plates of a condenser is to be measured, each is joined to a pair of quadrants ; and the needle, which takes a symmetrical position with reference to the quadrants when they are not at different potentials, will now move so as to enter one pair, until it is brought to rest by the torsion in the fibre. The needle forms with the two plates of a quadrant a condenser, and the motion takes place in such a direction as to make as small as possible the energy of the i ondeiisers it makes with the four quadrants. It may be pn\u00bbved by methods of the infinitesimal calculus that the angle through which the needle turns varies directly as the di Hen- nee of potential of its two sets of quadrants. Thus, two differences of potential may be compared by measuring the corresponding deflections of the needle. In order, however, to measure any one difference of poten- tial, a different instrument must be used. This is the \"absolute electrometer,\" which was also invented by Lord Kelvin. As shown in the cut, it consists of a paral- lel plate condenser, with a disc cut out of the upper plate as", " described on page 653. In practice this is suspended from one arm of ilance. The two phltCS of the condenser are joined tO the tWO OOndUOtOn WHOM Fl<1 ***\u2022 -Thomson'* original form of n1 difference of potential d.-sired: the plates are ihns charge. 1 with opposite kinds of electricity, and the!'<.!\u2022.\u2022<\u2022,,f attraction on the movable disc may he n. putting vreightfl in the balance pan. It the area of this i the distance.ip.u-t of the plates, d, the difference of potential, f^_ V^ the dielectric con- 660 ELECTROSTATICS stant, JT, the force of attraction on the disc is given by the formula (V.-V^AK B\u00ab\u00abP zL If the plates are, as usual, in air, K= 1, and ( F^\u2014 F1)2=\u2014 F, rf, and A can all be measured ; and so V^\u2014V^ is known. Measurement of Electric Quantities. \u2014 The four electrical quantities that have to be measured are quantity, potential, capacity, and dielectric constant. We have just shown how differences in potential may be measured ; and, if the poten- tial of a conductor is to be measured, it may be joined to one plate of an electrometer, while the other plate is connected with the earth. The capacity of a sphere or of a simple form of condenser may be calculated from a knowledge of its dimensions, as explained on page 652. But there is a simple method, due to Cavendish, for determining when the capacity of two con- densers is the same ; and so, if the capacity of any condenser is desired, it may be compared by this method with a con- denser whose capacity may the capacities of two condensers. B, known, e.g. a parallel plate FIO. 867.\u2014 cavendish's method of comparing condenser the distance apart of whose plates can be varied. The method is as follows : let Av B1 and Av J?2 be the two condensers ; charge them by joining A1 and A% to some electrical machine, while Bl and B% are joined to the earth ; then disconnect B2 from the earth, and A1 and A2 from the machine and from each other ; join Bl to A2 by a conductor, and A1 to the earth (or to B^). If the capacities are equal, an electroscope in contact with the wire joining Bl to A2 will show no effect when Al is earthed ; for let O1 and (72 be the two capacities, and let V be the potential given Al and Az by the machine ; the charge on Al is then + 01 V, on Bl is \u2014 Ol F, on A2 is + C2 V, and on B is ELECTRIC FORCE 661 - C'2 r. \\VlKMi Bl and A2 are joined, the two charges, \u2014 6\\T and -f-(72F, do not combine until Al is joined to the earth. Then they do, and the final charge, which is distributed over Bv AT and the wire joining them, is V^C^\u2014 C^)\\ and this will utYi'L-t an electroscope unless Ol=Cy This method also permits one to measure K for any di- electric, and was so used by Cavendish. The capacity of the second condenser may be measured when air is the dielectric and again when glass, or sulphur, etc., is substituted. The ratio of the latter capacity to the former is the value of K. In order to measure a charge, the accepted method is to place it on a condenser whose capacity is known and to meas- ure the resulting potential. Then, since e = C ( V^ \u2014 Fi), the value of e is known. There may be a difficulty in making the charge pass to the condenser, but the method described on page 639 may always be used. This is to put the charge inside a conducting vessel which is nearly closed; an equal charge will appear on the outside and this may be measuivd. An ingenious method was devised by Lord Kelvin for the measurement of the potential at any point in the atmosphere. Let A be the point of a conductor which is joined to an electrometer, and let some means be adopted to have a continuous current of small conducting particles leave it. Let B be such a particle. Then if the potential of A is higher than that of points in the air near it, a plus charge will he induced on B and a minus one on A ; B will carry this charge off as it moves away; and the process is repeated as the stream of particles is maintained. Finally, the potential", " of A will be lowered by this accumulation of negative charges until it is the same as that of the surround- ing air. Similarly, if the potential of A is lower than that of the air near by, it will be raised until it is the same. There- fore, when the potential of A ceases to change, it gives the potential <>! tin- air at that point, and may be measured by the electrometer. One means of causing a pointed conductor to t ; 1 5 '2 ELECTROS! A TICS oil particles is to use a small flame, because a burning gas is a good conductor. Another method is to have as the conductor A a vessel of water ending below in a small fun- ik-1, so that drops of water are continually forming and break- ing away. In this manner many interesting facts in regard to atmospheric electricity have been learned; one at least should be noted ; the potential of the lower layers of air is as a rule always higher than that of the earth, and its value is continually changing. Strains Due to Electrification. \u2014 The fact that the main phenomenon of electrification consists in a strain of the dielectric is shown, as has been said before, by the formation of sparks, and in many other ways also. One of the most direct proofs is furnished by what is known as the \" residual charges \" of a glass condenser. If one is charged to a high potential and then discharged, a second discharge may be obtained after the lapse of a short time; then a third may be obtained, etc., each one being feebler than the preceding one. These are said to be due to residual charges. They depend upon the fact that glass is non-homogeneous ; for they cannot be obtained with a homogeneous dielectric. Their explanation is as follows : When the condenser is first charged, the glass is mechanically strained, and when it is discharged, certain parts of the glass lose their strain and, owing to inertia, are strained again in the opposite manner, while other portions of the glass do not relax completely; these two portions, however, balance each other for the moment, and there is no resultant strain ; as time goes on, however, these strains, not being maintained by any force, gradually relax, but not to the same degree, so there is again a resultant strain ; this causes the second discharge when the armatures are joined, etc. If it is remembered that there is no field of force inside a conductor, so that such a body cannot maintain a strain, all the phenomena of induction, etc., may be at once explained. ELECTRODYNAMICS CHAPTER XLIV PRODUCTION OF ELECTRIC CURRENTS Definition of Terms. \u2014 The simplest case of an electric current is furnished by the steady discharge of a condenser. (See page 647.) In this, two plates having a difference of potential are joined by a conducting wire; and, as a result of the change, the charges of the two plates disappear. It is noticed further that the temperature of the wire is raised, and certain magnetic effects are produced in the region around the wire. All these phenomena constitute the elec- tric current. \\Ve speak of the current as beni- in the wire; Imt this is only a mode of speaking. As the discharge begins, the plus charge on the plate of higher potential decreases, and so does the minus charge on the plate of lower potential : if by some means these charges may be maintained constant by adding continually the nec- essary <|iiantitics of plus and minus charges, the potentials of the plates will remain unchanged: and the current i> said to be \"steady.\" The phenomenon in the Conducting wire, which constitutes the current, consists, as will be shown in the next chapter, of a motion of a stream of positively charged particles in the direction from hi^h to low potential in the wire, and of a stream of negatively charged particles in the opposite direction. lly \u2022/\u2022_///////\"// the former direction is called that of the current. If t\\ is the cjuantity of plus elec- tricity that passes through the cross section of the wire at 688 664 ELECTRODYNAMICS any point in a unit of time, and i2 is the quantity of minus electricity that passes at the same time in the opposite direction, the quantity i\\ + e'2 is called the \" strength of the current\" or, more often, \"the current.\" If the current is steady, the quantities ij and izt pass in an interval of time t ; and (tj + iz)t is called the \" quantity of the current.\" If the current is not steady, and if in any interval of time the quan- tities of plus and minus charges that pass are e1 and ez, the quantity of current is (^ 4-02). (Thus, in the discharge of", " the condenser whose plates are charged with + e and \u2014 e, the quantity of the current is e, because the plates will be dis- charged if e\u00b1 + 02 = e. If H- e passes from one plate to the other ; or if \u2014 e passes in the reverse way ; or if 4- ^ e and \u2014 $ e pass in opposite ways ; etc., the plates are discharged.) In order, then, to produce a current in a conducting wire it is necessary to have a difference in potential between any two of its points. This difference is called the \"electro- motive force\" (E.M.F.) between the two points. Work done by a Current. \u2014 The passage of a current evi- dently involves the idea of work. If V^ \u2014 V\\ is the differ- ence of potential between two points in a wire, and if (il + i'2) is the current strength, the quantity of positive electricity il moves from a point of high potential, Vv to one of low, V-p and therefore the electric forces do the amount of work i\\(y<i\u2014 Fi) ; and similarly, owing to the motion of a quan- tity of negative electricity in the opposite direction, the same forces do an amount of work ^'2( V2 \u2014 V\\). So the total amount of work done in a unit of time by the electric forces is (t'i + *2) (^2\u2014 V\\) 5 or, calling the current strength i and the difference in potential, E^ it is iE\\ and the work done in an interval of time, \u00a3, if the current is steady, is iEt. Or, in general, if e is the quantity of current, the work is eE. Ordinarily this work is spent in raising the temperature of the conductor which carries the current ; and the necessary amount of energy is furnished by whatever produces the cur- PRODUCTION OF ELECTRIC CURRENTS 665 rent. (The heat produced in the conductor may be meas- mvil if it is in the form of a wire by coiling it in a calorimeter of water. See chapter XII. If the C.G.S. system of units is used in defining the unit quantity of electricity, the product iEt is a certain number of ergs; and so the heat produced must be expressed in ergs.) Heating Effect of a Current. \u2014 This heating effect of a current is, of course, greatest where there is the greatest amount of work done ; that is, where the electromotive force, or drop in potential, is the greatest. This is illustrated in various forms of electric lights, in the electric furnace, in electric heaters, etc. 1 Fw. 868. \u2014The electric arc between two oarb< < The arc light, as used for illuminating purposes, consists of two rurbon rods \\\\lnrh ;uv connected to some source of a 666 ELECT ROD YNAMICS current, and which are so controlled by automatic mechanism that when no current is flowing they are in contact, and then as soon as the current begins they are slightly separated. The two rods when loosely in contact offer great opposition to the current, so the temperature rises at the points of con- tact ; this makes the surrounding gas a conductor, and now the rods are drawn apart. The current passes off one rod to the gas, and from this to the other rod. There is great resistance to the current passing off or on a solid ; and the temperature of the tips of the rod is raised to a \" white heat,\" if the current is sufficient. This produces the light. Experiments show that more heat is produced at the end of the rod from which the current proceeds to the gas than at the other; this is called the \"positive pole.\" In the ordinary \"incandescent light\" there is a glass bulb into which enter two platinum wires connected inside by a fine filament of carbonized wood fibre, and from which the gas has been exhausted as completely as possible. A current is made to flow through the filament, and its temperature is raised to white heat. It does not burn up, because there is no oxygen left inside the bulb. Fro. 859.\u2014 Incan descent lamp. In the Nernst lamp there is a small filament whose constitution is a commercial secret, which ends in two metal wires ; this filament is not a conductor unless its temperature iy high, and even then under the action of a current in one direction it decomposes and breaks down. Therefore the process of using the lamp is first to raise the temperature of the filament until it becomes conducting, and then to have it traversed by a current whose direction is reversed at short intervals. If this is done, the filament gives out a brilliant light; and, as it does not oxidize, it may be used in the open air. PRODUCTION OF ELECTRIC CURRENTS 667 In an electric", " furnace use is made of the high temperature of the arc ; and the carbon rods are inclosed in a space whose walls are non-conductors for heat, and in which the pressure of the i^as may be increased. Direction of Current. \u2014 In order to determine by experi- ment the direction of a current it is necessary to ascertain which of the two conductors between which the current flows has the higher potential. The simplest mode of doing this is one invented by Volta. The two conductors 0 whose potentials are F^ and Fi are joined by wires to the two plates of a condenser, A2 and Ar if ra>rr the plate A% becomes A8|A, charged positively, and Ar negatively ; be- cause lines of force pass across from A% 1 Those Charges Fio. 860. \u2014 Method of determining direction of an elt-rtric on the two faces *' current. Volta's condensing electroscope. nearest each other ; but if the wire leading to A l is broken, and the plate A% is then removed, the negative charge on A1 will spread over the whole plate and may be detected and studied. In Volta's arrangement the plate Al was the top plate of a gold-leaf electroscope, and A., was a similar plate coated with a thin layer of shellac and carried by a \u2022_rlass handle. Therefore in this apparatus, after the wire leading to Al is broken and A.2 is then removed, the nega- tive charge will spread over the plate and the, gold leaves, which will then diverge. If now a glass rod which has been rnhhed \\\\ith silk is brought near the electroscope, it will induce a positive charge on the leaves, which will in part neutralize their negative charges, and so they will collapse. It', on the other hand. J \"J, < Fj, the gold leaves will be- come charged positively, and a charged glass rod will cause ELECTRODYNAMICS them to diverge still farther. In this manner, then, it may be determined whether F2 > Fr or F^< V^\\ if the former is the case, and if a wire is made to join the two conductors, the direction of the current is from the one at potential V% to the one at potential Vl ; in the contrary case, the direction of the current is opposite to this. Detection of a Current. \u2014 When an electric current is flow- ing in a conductor, its temperature rises, as explained above, owing to the work done by the electrical forces against the molecular forces of the conductor. But this fact does not lead to a simple direct means of ob- serving a current, because with a feeble current the change in tem- perature is small. The magnetic action of a current offers, however, an extremely simple and direct method of detecting and even measuring a current. It was discovered by Oer- Apparatus of Oer- sted> a Danish physicist, in 1819-1820, that a wire carrying a current had a magnetic field around it. We shall take up this question more in detail in a later chapter ; but one or two facts may be stated here. sted for studying the action of an electric current upon a magnet. FIG. 361. If a magnetic needle is pivoted so as to be free to turn about a vertical axis, it will assume a north-and-south posi- tion, and now if a conductor carrying a current is placed parallel to it, but above it, the needle is deflected ; if the current is reversed, so is the deflection. Similarly, if the current is parallel to the needle, but below it, it is deflected ; but the direction Of the deflection is Opposite FlG. 862. -Section of a simple galvanoscope. PRODUCTION OF ELECTRIC CL'RREXT* to what it would be if the current were above the needle. Hence, it follows that if the conductor carrying the current is made in a loop lying in the magnetic meridian and inclos- ing the magnet, the deflection will be increased ; and if many loops are used, forming a flat coil, the deflection will be still greater. This constitutes a \"gal- vanoscope.\" Another mode of increasing the deflection still more, and at the same time of avoiding, to a large extent, any disturbances of the magnet due FIG. 868. -An astatic combination to other actions than those of the current in the coils, is to attach rigidly to it another magnet of equal magnetic moment, but turned so that its axis is in an opposite direction. Thus, a north pole of one comes opposite to the south pole of the other. If, now, one of -<j magnets is inclosed in the coil and the other is either above or below it, the deflec- tive force of the current is in the same", " direction on both magnets ; but the action of any other magnetic field is almost entirely prevented. Such a combination of mag- nets as this is called an Fio. 864. \u2014Section of \u2022 palvanoscope with MUttc needle. \"astatic needle,\" because if tin -ir magnetic moments were exactly equal, and if their netic axes were exactly parallel, the system would not be under a directive force due to the earth, and would remain stationary in any position. Actually these conditions are not satisfied ; and the earth has an action, but it is extremely small. So, by bnnun,ILr another magnet in-ar the astatic needle, it may be made to take any position that 670 ELECTRODYNAMICS is desired, and the action of the earth may be neutralized as completely as is desired. By thus using a 'control magnet,\" then, the coils to carry the current may be kept in any position which is convenient, and the astatic needle may be made to lie in their plane, while the field of force due to the earth and the control magnet may be very' small. This last is shown by the period of the magnet becoming very long when it is set in FIG. 866. \u2014 Galvanoscope. I \u00a3 vibration, for T \u2014 ZTT^ \u2014 - (see page 611) ; and so, if R is small, T is large. The field of magnetic force near any current may be studied by the use of iron filings or of a small magnetic needle, as* was described on page 603. It is found that the lines of magnetic force form closed curves around the cur- rent; the directions of the current and the lines of force being connected by the right-handed-screw law. Thus, if LINES OF FORCE A'CURRENT Fio. 866. \u2014 Diagram Illustrating connection between the direction of a current and that of the lines of magnetic force. AB is any portion of a conductor carrying a current from A to B, the lines of magnetic force near it are as shown ; or, if the total electric circuit is considered, the lines of force pass through it from one side, and return outside. Thus, a current and any one of its lines of magnetic force form two closed links threading each other, like two links of a chain. If a current, then, is passed through the coil of a PRODUCTION OF ELECTRIC CURRENTS 671 galvanoscope, the magnetic needle will be deflected; and if the current is reversed in direction, so is the deflection of the magnet. If the current in the coil is in the direction of the motion of the hands of a watch as one looks at the coil from one end, the magnetic force is directed away from the observer, so that a north pole is forced away from, and a south pole is forced toward the observer. By means of such an instrument one can determine, then, the direction of a current, and can roughly estimate its strength. Tangent Galvanometer. \u2014 If the coil of the galvanoscope is a circular one, that is, if the cylinder on which the wire is wound has a circular cross section, the intensity of the magnetic field at the centre of the coil may be proved (see page 711) to vary directly as the current strength and inversely as the radius of the cylinder referred to. Thus if i is the current strength and the radius, the intensity of the magnetic force is pro- portional to -; it also varies directly with the number of a turns of wire in the coil ; if this is n and if the turns of wire are so close together as practically to coincide, the intensity may be written /= c\u2014, where c is a factor of proportionality. a The numerical value of c depends, of course, on the units chosen for the magnet ie and electric charges. If the coil is placed in tin- magnetic meridian and a current is passed around ic magnet (not an astatic one) sus- pended at its centre is under the action of two opposing couples, one due to the magnetic field of the earth, and the other to that of the current in the coil; and it comes to rest when these balance each other. If M is the magnetic moment of the ma-net. // the l,,,rj- ELECTItOD YNAMICS zontal component of the intensity of the earth's field, and N the angle that the magnet makes with the magnetic meridian when it comes to rest, the moment due to the earth's force is JIM sin N; and that due to the electric current is /M\"cos N. Since these must bal- ance each other, or FIG. 368.\u2014 Tangent galvanometer. Therefore the strength of the current is meas- ured by the tangent of the angle of deflection ; that is, the strengths of two currents vary directly as the tangents of the angles of deflection. Such an", " instrument as this is called a \" tangent galvanometer. \" Electro-magnetic Unit Current. \u2014 The current strength is defined in terms of the quantities of charge which pass a cross section ; but actually these quantities cannot easily be measured directly. So, it is more convenient to define a \" unit current \" in terms of its magnetic properties, and then from this to deduce the value of a new unit quantity. Thus, a \" unit current,\" or one of unit strength, is defined to be such a current that if flowing in a galvanometer coil of one turn whose radius is 1 cm., the intensity of the magnetic field at its centre equals 2 IT dynes. (The reasons for this choice of unit current will appear later.) Thus, using the same symbols as in the above formula, /= c \u2014, this may be expressed by saying that, when i \u2014 1, n = 1, and a = 1, PRODUCTION OF ELK'TUH' CURRENTS 673 /'= -2 TT : so this < It-tin it ion of a unit current is equivalent to putting c = - TT in the formula. This unit current is called th. \u2022\u2022(.(..S. electro-magnetic\" unit, because its definition depends upon the intensity of a magnetic field ; that is, upon the force acting upon a unit magnetic charge. Then in a undent galvanometer the formula becomes.) _^ or, t = H-- tan N. Ha '2 -mi - - is called quantity it (r, i= - - tan N. IT and G may be measured, and the \"galvanometer constant \" ; TT N observed ; so the strength of a current may be measured. Tin- C. <i. S. electro-magnetic unit quantity of electricity is, then, the quantity carried past any cross section of the conductor in one second by a unit electro-magnetic current. Thru- must, of course, be a constant relation between this quantity and tin* ('. (i. 8. electrostatic unit quantity as de- fined on page 640 ; and experiments prove that one C. G. S. electro-magnetic unit charge equals 3 x 1010 C. G. S. electro- static units of charge. (It should he noted that this ratio of th\u00ab- units equals the velocity of litfht.) The work ivqnhvil to carry a unit electro-magnetic quan- tity of charge from one point to another is the difference of potential between them exprened on the C. G. 8. electro- magnetic system. Kxperimeiits sho\\\\ that the numbers which would be required to BZpI68fl ordinary potential differences on this system are so large that they are inconvenient. Consequently other units arc used in practice. Thus a utial dilYcreiK t^on the C. G. S. electro-magnetic system of I'\" \"100,000, or 108, is ealled one -volt/' Measurement of Quantity of Current. \u2014 It is often necessary to n 'ie t.-tal <piant it y of current which llows in a short time, <>.//. \\\\ : 1 1 -<\u2022 harmed. In this case the current is not constant, and further the time of flow is SO AMEfl'8 PHYSIC'- \u2022 7 i ELECTRODYNAMICS short that it is not possible to secure a permanent deflection of the galvanometer needle. If such a current is passed through a galvanometer, the needle will be acted on by an impulse and will have a certain \" fling \" ; that is, it will be deflected from its north and south position and will proceed to make a number of oscillations before coming to rest again in its former position. If the period of vibration of the needle is so long and the time of passage of the current so short that we may consider the current as over before the needle is deflected an appreciable amount, the maximum angle of deflection measures the quantity of current. The exact formula may be deduced without difficulty. If e is the quantity of the current on the C. G. S. electro-magnetic system; JV, the angle of fling from a north and south direction ; 6r, the galvanometer constant ; H, the horizontal component of the earth's magnetic force ; T, the period of vibration of the magnet, An instrument specially designed to measure quantities of current, as distinguished from current strengths, is called a \" ballistic \" galvanometer. Measurement of Electro-motive Force. \u2014 Since an electro- motive force is a difference of potential, it may be measured by any electrometer. (See page 658.) But, in general, other methods are adopted. One is to join the two points which are at different potentials to a condenser of known capacity, and then to discharge it through a", " ballistic galva- nometer. If E is the difference of ^potential and 0 the capacity of the condenser, the quantity of current measured will be CE. (The value of 0 on the electro-magnetic system must be used, if E is to be measured; but, if two electro- motive forces are to be compared, it is not necessary to know the value of (7.) PRODI < PlOJi \"F l-'.l.KVTRIC CURRENTS 675 Another method of comparing differences of potentials depends upon the fact, which will be discussed more fully later, that in the case of a steady current its strength is directly proportional to the E. M. F. producing it; but, if the E. M. K. is applied at the ends of a long, fine wire, the current is small, while if the wire is short or thick, the current is large. In the former case there is said to be a great \" re- sistance\"; in the latter, a small one. Thus, if a P \u2014 \u00bb steady current is flowing through the conductor 0000 PQ, and the value of the R difference Of potential be- FIO.SW.\u2014 Diagram Illustrating a method of i tween two points A and B is desired, these points may be connected by wires to a gal- vanometer, #, through coils of wire, R, which are so long and so fine that they offer such an opposition to the passage of a current that practically none flows from A around through Q- to -B, and so no difference is made in the conditions at A and B. If, however, the galvanometer is sufficiently sensi- tive, it will measure this minute current; and its strength is directly proportional to the difference of potential between A and B. Other methods may be found described in labora- tory manuals. Steady Current. \u2014 If by any means, mechanical, chemical, thermal, etc., it is possible to maintain a constant difference of potential between two conductors, a steady current may be produced by connecting these two conductors by a wire or other conductor. There are at least four methods by which this constant diffen -nee in potential may be produced. 1 in electrical machine such as described on page 685 is turn.-.! at a uniform rate, it may be used to furnish a steady current. It' \\\\\\ nt metals, sin-h as y.inr and copper, a re partl\\ iniinei ><-d in soiur li.pfid conductor other than a fused metal, Midi as a solution of sulphuric acid in 676 EL ECTROD YNA MICK water, it is found that the rods are at different potentials. If a closed metallic circuit is made by joining several wires of different material in series, and if the junctions of the different wires are at different temperatures, a current is pro- duced in the circuit. Again, if a closed circuit of some wire is moved about in a magnetic field in such a manner that the field of force through the circuit varies, a current arises; and, if this change in the field continues at a uniform rate, the current is steady : this constitutes a \" dynamo. \" Primary Cells. \u2014 Experiments show that, when a solid con- ductor is immersed, partly or completely, in a liquid conductor other than a fused metal, there is a difference of potential between them, which is characteristic of the two conductors. So, if two solid conductors dip in the same liquid, they will be at different potentials ; and, if they are joined outside the liquid by a wire, a current will flow in it. This fact was first observed by Volta (1800), who used zinc, copper, and dilute sulphuric acid in this manner. This is said to be a \"Voltaic cell.\" It is a question of experiment to determine which of the solid conductors has the higher potential. In the case of the voltaic cell, the copper rod is at a higher potential than the acid, and the acid is higher than the zinc ; so the current in the connecting wire outside is from the copper to the zinc. It is observed that, as the current continues to flow, the zinc gradually dissolves away and bubbles of hydrogen gas collect on the copper rod or break loose from it and rise to the surface. It is observed, further, that there is a current also through the dilute acid, and that its direction is from the zinc to the copper. Thus the current flows in a circuit; FIG. 870.\u2014 Voltaic cell. PRODUCTION OF ELECTRIC CUliliK\\T.< 677 outside the liquid, from copper to zinc ; inside the liquid, from zinc to copper. Since the direction of a current is itl ways from a point of high potential to one of low, it is thus evident that", " at the boundary separating the zinc and the acid there must be some mechanism which raises the poten- tial: so that the points on the zinc must have the lowest potential in the whole circuit, and contiguous points in the dilute aeid must have the highest potential. This phenom- enon is evidently connected closely with the dissolving of the zinc in the acid. If pieces of zinc are placed in dilute sulphuric acid in a tumbler or beaker, it is noted that the zinc dissolves, that hydrogen gas is evolved, and that the temperature of the acid is raised. This proves that, when zinc dissolves, energy is liberate! 1 : in the simple chemical experiment this energy is spent in producing heat effects; in the voltaic cell it is spent iii raising the potential of points in the acid, and this maintains the current and so heats the conductors, etc. At the surface of the copper, where the current enters it from the acid, work is required to raise the potential of the plus charges from that of the acid to that of the copper, and to lower that of the negative charges which are going in the opposite direction. This difference of potential at the sur- face is due to the evolution there of the hydrogen gas. The mechanism of the current through the acid and at the /ine and copper rods will l\u00bbe discussed in the next chapter. The two solid conductors which dip in the liquid are called \u2022\u2022poles\"; t he one which 18 at the higher potential is called the positive \"ne, while the other is called the negative one. The latter is always dissolving as the current flows; so if it contains any metallic impurities, e.g. if the zinc has particles of iron in it. there will be local currents from the zinc to the acid, th.n to i1, iron, and thence to the zinc, etc. These currents have no external action ; and so should be pre- vented, it possible, because the zinc consumed in producing 678 them is \\\\;i>tr<l. This can be done in many cases by rubbing mercury over the xinc rod before it is immersed in the liquid, and thus making a surface of mercury amalgam with the metal, which is practically uniform. As the current flows in a voltaic cell, hydrogen bubbles collect over the copper pole, and thus hinder the action of the cell. Various devices have been invented in order to prevent this. The most successful is due to Daniell. He made a cell, which bears his name, consisting of a porous cup \u2014 such as unglazed porcelain \u2014 inside a larger vessel ; the cup contains a saturated aqueous solution of copper sul- phate, and the outer vessel, dilute sulphuric acid ; the zinc rod dips in the latter, the copper rod in the former. When the two rods are joined outside by a wire, the current flows from the copper to the zinc. As it flows, the zinc dissolves as before, but now copper is deposited out of the copper sul- phate solution on the copper rod. Consequently there is no change in the nature of the surface of the latter. This cell of Daniell is a typical \"two-fluid cell.\" Other cells, both one and two fluid, can be made by using other metals than zinc and copper, and other liquids than sulphuric acid. They are called \" primary cells \" in distinc- tion to \" secondary \" ones, which will be described presently. Cells may be joined \"in parallel\" or \"in series.\" Thus if Ol and Z1 are the posi- tive and negative poles of one cell, <72 and Z2 those of the second, etc., the cells are said to be in series if Ol and Zv 02 and Zy etc., are con- nected by wires ; while if Ov Cv <73, etc., and Zv Four cells joined tn parallel. 2' 3' Vt0m 87i. nected, the cells are said PRODUCTION OF F/./.r//;/r i 679 t\u00ab> he in parallel. If the cells are all of one kind, let E be tin- difference of potential between the two poles of each; then it n cells are joined in series, the difference in potential between C\" and Zn is nE. Whereas, if they are joined in parallel, the only effect is to make what is practically one (dl \\viili poles n times as large; this does not affect the difference of potential between the poles. A mechanical analogy of a simple voltaic cell is furnished by a pump or paddle wheel working in a horizontal tube connecting two tall vertical pipes containing some liquid, such as water. If the pump is open, the liquid will stand at the same level in the two vertical pipes; but, when the pump or wheel is set in action, the liquid will be forced through so as", " to stand higher on one side. A difference of pressure on the two sides of the pump or wheel is thus produced ; and, if sufficient, it will stop the action of the latter. If now a connecting tube between the upper portions of the pipes is opened, the liquid will flow from the one at the foot of \\\\hich the pressure is the higher over into the other, and a continuous current will be produced. This pij.j- in which tin- pressure against the pump or wheel is the greater corresponds to the o<>pp\u00ab -r rod in the voltaic c\u00ab-ll ; tin- other pip.- to tin- xinc; and the pump or wheel to the energy furnished by the dissolving zinc. taic cell. Fio. 872.-Model rep- resenting action of vol- Thermoelectricity. \u2014 If a closed circuit of linear conduct- ors, like wires, includes at least two different substances, tin-re is in general an electric current produced in the circuit if the junctions nf the.se substances are kept at different temperatures. Thus, if two \\vires I and II make up a cir- cuit having junctions at.1 and B, there will be, in general, a current if the temperatures <.f A and //are not the same. The direction and strength,,f the current depend Upon the t \\vo substances and upon the difference in temperature. It 'iind by experiment that, beginning with a condition when A and //are at the same temperature, if that of A is 680 EL EC Tit OD YNA MICS kept unchanged and that of B is continuously increased, the current will be in a definite direction and will gradually increase, while if the temperature of B is decreased, the current will be in the opposite direction and will gradually increase ; as the temperature of B is made to differ more and more from that of A in one direction, \u2014 in certain cases when it is higher, in others when it is cuit made up of two con- lower, \u2014 there comes a point when the doctors i and //, having current begins to decrease, and finally junctions at A and B. \u00b0 J one at which the current ceases; while if the difference in temperature is increased still more, a current is produced, but it is in the opposite direction to that which it was before, and as the change in the tem- perature of B continues, this reverse current increases in strength. If tA is the temperature of A during the experi- ment, and if \u00a3/ is that of B at which the current ceases, FIG. 378. -A closed cir- experiments show that their mean A^'r is a constant quan- tity for any two substances : it is known as their \" neutral temperature\"; and tj is called the \"temperature of inver- Zi sion,\" corresponding to tA. In order to explain these thermocurrents, as they are called, it is necessary to assume that at any cross section in the conductors where two different substances come in con- tact there is a difference of potential. If P and Q are two different substances meeting over a surface, the fundamental experi- ments of electrification show that, when they are separated, one is charged with plus, the other with minus electricity. This proves that when they are in contact there is some electric force \u2014 due to the difference in the elec- tric properties of the molecules \u2014 acting at the surface of con- FIG. 874. \u2014 Junction of two con- ductors P and Q. PRODI < won or I-:LK<-TIUC CURRENTS 681 t.ut, and resulting in a separation of the plus from the minus charges. Let us suppose that P is the substance which is charged positively; then the direction of the force producing this charge must be across the surface of contact from Q to P. As a result of the plus charge on P and the minus charge on Q, the potential of P is higher than that of Q ; so that if P ami Q are conductors, and if they are joined by some wire, a current would tend to flow, owing to this fact, from P through the wire to Q. This difference of potential at the surface of contact would be maintained by the molecular forces. Calling this difference of potential E, we may say that there is a \" contact electro-motive force \" E at the boundary. The proof of the existence of this E. M. F. across the surface of contact is afforded if P and Q are conductors, and if an electric current is forced by some source, such as a voltaic cell, across this surface, first in the direction from P to (), then in the opposite direction. It is found that in the former case the temperature of the junction rises; in the latter it falls. If the current i flows for an interval of time t", " from P to Qi the electricity is passing from high to low potential, and so the external electric forces do the work itE at the junction ; mid this energy appears in the form of heat effects. If, however, the current is in the opposite direction, the elec- tricity is having its potential raised at the junction, and so the work itE must be done on the electric forces at the expense of the energy of the molecules at the junction ; and there- fore its temperature falls. (Or, we may say that in the f on un- case work is done against the molecular forces which produce the electrical M-paraii\u00ab>n : \\vhih- in the latter, these forces do work themselves in helping on the current.) These forces at tin- surface <\u2022!' miitact of two substances are called \"Pel- tier elect i-M-im.tive forces,\" having been first discovered by him. They can be measured by putting a junction in a rimeter \"t water, ;m\u00abl measuring the heat produced, the rui-rent, and the time. Direct experiments prove that they 682 ELECTROJ) Y.\\A MICS vary in amount with the temperature. Thus, in the thermo- couple described above there are two such forces, at A and B\\ and, if the temperatures at these points are different, these forces are unequal. But there are other similar forces in each conductor between A and B, if the temperatures of these points are different. For, consider either of these conductors, the two ends of which are at different temperatures ; if a section is taken across the wire at any intermediate point, the tempera- tures on its two sides differ slightly, and so the condition of the molecules which are in contact across this section is different on the two sides. Therefore, we might expect an electro-motive force at each point in the conductor. This was proved by Lord Kelvin \u2014 then Sir William Thomson \u2014 by the following experiment : let an electric current be forced through a wire of some definite material whose ends are kept at a higher temperature than its middle point, and let the temperatures be noted at two intermediate points, one in each half, which are such that their temperatures are 100o Oo 10Qo the same when no cur- rent flows; it is observed p J^ Q FIG. 875.\u2014 Diagram representing Thomson's that, when the Current is flowing, the tempera- ture at one of these points rises, while that at the other falls. Thus if the wire is PQR, let the current be from P to R, and the temperature of P and R be higher than that of Q\\ and let A1 and Az be the two points whose temperatures are the same before the current begins. The molecular forces at Al producing the E. M. F. due to the temperature effect just described are either in the direction from Q toward P, i.e. from a cold point to a hot one, or from P toward Q. If the former is true, as the current flows from P to Q, the temperature at Al rises ; while if the latter is true, the temperature at A1 falls. Similarly, the temperature at Az either falls or rises ; but, if the temperature at Al A^ R PROi>r< y/o.v of t-:i.i-:< TR1C CURRENTS G83. if the molecular forces are in the direction from a cold point to a hot one, the current at A1 is in a direction opposite to that of these forces, while at A^ the forces and the current are in the same direction, and so the temperature at A^ falls. These forces in a wire which is homogeneous except for differences in temperature are called \"Thomson electro-motive forces.\" In a simple circuit made up of two w i res there are then these forces at each point of both. The electric current produced in a circuit made up of dif- ferent substances whose junctions are at different tempera- tures is due to the Thomson and Peltier electro-motive forces. These currents were discovered by Seebeck in 1821, but their explanation was not known for muny years. It is evident that, if a sensitive method is known for the detection of an electric current, a means is offered for detecting differences in temperature between two points ; for the junctions of a thermocouple may be placed at them. The sensitiveness of the instru- ment may be increased by joining in series several pairs of tli\u00ab- two conductors, as shown in the cut. If the alternate junctions are kept at one temperature, and the other junctions are kept at a different one, the current will be increased; and so a less difference in temperature mav l\u00bbe detected. Such an ins! ruim-nt is called a \" thermopile.\" A cut of an actual", " in>truuicnt is shown. Fio. 876. \u2014 A thermopile. CHAPTER XLV MECHANISM OF THE CURRENT Electrolysis. \u2014 It is found by experiment that many liquids are conductors, while others are not. A metal in a liquid condition is a conductor, and its properties are exactly like those of the solid conductor. There are, however, certain liquid conductors such that, when a current is made to traverse them, there is an evolution of matter at the points where the current enters and leaves. The liquid must be held in some vessel and two metal rods or wires connected with some source of electric current \u2014 such as a series of cells \u2014 must dip into it. The conductor, or \"elec- trode,\" at which the cur- rent enters the liquid is called the \"anode\"; that at which it leaves, the \"cathode.\" Thus the potential of the former is higher than that of the latter; and the direction of the current in the liquid is from the anode to the cathode. The matter that is liberated at the anode and cathode may bubble off in the form of a gas, it may combine chem- ically with the metal rods themselves, or it may simply form a solid deposit on them. Liquid conductors which have this property are called \"electrolytes\"; and the process of conduction in them is called \" electrolysis.\" A careful study of the nature of electrolytes has shown that in every case they are solutions, e.g. common salt or sulphuric acid in water; and 684 MECHANISM OF THE CURRENT 685 that the solutions are of the kind which exhibit an abnormal osmotic pressure, an abnormal depression of the freezing point, and an abnormal elevation of the boiling point. (See pages 2(H and -J76.) Faraday's Laws. \u2014 A careful study has been made of the character of the substances which are evolved at the anode and the cathode in different electrolytes. It is found that hydrogen and all metals are liberated at the cathode, while oxygen, chlorine, iodine, etc., are liberated at the anode. Further, the amounts of the substances evolved under dif- ferent conditions were systematically studied by Faraday. As a result of his investigation, he was able to describe all of his observations in two simple laws which bear his name. Before stating these, however, it is necessary to define a chemical term which was used much more commonly in former days than now, and yet which is convenient. Ex- p\u00ab i iments have established the fact that a molecule of any chemical compound consists of a certain number of smaller parts, called atoms, the atoms of any element being alike in all respects. Thus, a molecule of steam consists of two atoms of hydrogen and one of oxygen, so its symbol is HaO; a mole- cule of sulphuric acid may be expressed by the symbol H2SO4 ; one of copper sulphate by CuSO4 ; one of hydrochloric acid IIC1; etc. A molecule of hydrogen gas has the symbol II., : \"iic of oxygen gas, Oa; etc. The kk molecular weight\" iy definite compound has been already defined to be a number which is proportional to the weight of one of its molecules; and a method lias l>een described for the determi- nation of this (piantity in certain cases. Other methods arc kn\u00ab.\\vn. Similarly, the \u2022\u2022 atomic weight \" of any element number proportional to the weight of one of its atoms. Thus, the molecular weight of oxygen is 82; so its atomic weight is lr, :. \u2022 It is seen from the above illustrations of the com] -.f molecules that in some cases one atom, in others two, of hydrogen are contained in a molecule. Thus, ill hydrogen gas and hydrochloric acid, one atom of hydrogen combines with an atom of hydrogen or one of chlorine respec- tively; in steam and in sulphuric acid, two atoms of hydro- gen combine with one of oxygen or with the \" radical \" (S()4); etc. The number of hydrogen atoms which is re- quired to form a stable molecule with the atom of a substance or with a certain \"radical\" (or group of atoms), is called the \" valence \" of that substance or of that radical. Thus, the valence of hydrogen and of chlorine is one; that of oxygen and of SO4 is two ; etc. Experiments show that if any molecule is regarded as made up of two parts, the valences of the two parts are the same. Thus, since a mole- cule of copper sulphate is CuSO4, the valence of copper is two, as is shown also by the fact that the saturated oxide of copper is CuO. (An atom may have a different valence in different compounds", " ; but only one of these is in general a stable mole- cule.) The ratio of the atomic weight of an element, or of the sum of the atomic weights of the atoms in a radical, to its valence is called its \" chemical equivalent.\" We can now state Faraday's two laws : 1. The quantity or mass of a substance liberated from any electrolyte at either the anode or the cathode is directly proportional to the quantity of the current that passes. 2. The masses of different substances liberated at the anode and cathode in any electrolyte by the passage of the same quantity of current are directly proportional to their chemical equivalents. If the current flows through several electrolytes arranged in series, let ml and m^ be the masses of the substances liberated at the anode and the cathode in the first electrolyte, and cl and <?/ their chemical equivalents ; mv w2', <?2, and c2f be similar quantities for the second electrolyte, etc. Fara- day's first law states that any m varies directly as the quan- tity of current ; so, if the current is steady, and if i is its strength, m is directly proportional to the product of i by \u00a3, MXCHAX18M <>r THE CURRENT 687 tlu- interval of time taken to liberate law states that m^. //^': wa: mz' : etc. = Voltameters. - - The first law offers a convenient method for the mass m. The second : etc. tin- comparison of the strength of two different currents. An elec- trolyte is placed in series with the two currents in turn, and the (plan titles of matter liberated in < It-finite intervals of time at either anode or cathode are measured. If the strength of one current is called ir and if the mass it liber- - in an interval of time tl is mv raj is proportional to tfa. So, if the strength of the other current is called /2. and if it liberates a mass wa of the same substance in time tv 7H2 is proportional to t.,<\\, : \u00bb\u00bbr ml:mz = tlil:t^r Hence /,:/, = 5j.?b. /, t% i.il instruments have been designed for ihjs ]nirix>ae of coniparin.i,' cunvnt -lli>: tli. -van- call.-.l * voltaui.-t. : () of the commonest form- i> -ho\\vn in tin- rut. It is rall\u00abMl a \"water volta- meter\": an. I in it the electrolyte is dilute Hul j.l i uric acid, and the anode and cathode are sheets of platinum inserted at the lower ends of HM- t\\\\ \u2022 {\u00bb long U tube. The substances li are oxygen gas at the anode, and hydrogen gas at the catlx.d.-: tlu>se collect in tl.. U|.J..-i :,,,,!\u201e, ilistrum.-i.l,,ay be measured. (Tl,.- mid. II.- ti.U-. as Fiu. 873. -A water volUm\u00abt\u00abr. 688 EL KCTR 01) YXA MICS in the cut, is open to the air at the top, and the quantity of liquid in it may be varied so as to make the pressure on either gas equal to that of the atmosphere, when its volume is measured. From a knowledge of its pressure, volume, and temperature, and of its density as given in tables its mass may be calculated.) In another form of instrument the electrolyte is a solution of copper sul- phate, and the anode and cathode are both sheets of copper ; the quantity of copper deposited on the cathode is deter- mined by weighing it before and after the current passes. This is called a \"copper voltameter.\" In the standard instrument the electrolyte is a solution of silver nitrate in water, of a definite concentration; the anode is a plate of silver and the cathode which receives a deposit of silver is the platinum bowl which holds the silver nitrate; this bowl is weighed before and after the current passes, and suitable precautions are taken to prevent any mechanical currents in the liquid, for it is found that they, for purely chemical reasons, affect slightly the quantity deposited. FIG. 879. \u2014 A simple form of silver voltameter. Electro-chemical Equivalent. \u2014 By means of Faraday's second law we can calculate the relative amounts of differ- ent substances which are liberated by the same current in the same time. The quantity, i.e. the mass, of any substance which is set free at either anode or cathode as", " a unit C. G. S. electro-magnetic quantity of electricity passes is called the u electro-chemical equivalent\" of that substance. It is a matter of experiment to determine its value for any one substance ; but, being known for this, its value for any other is given at once by the second law. Careful experiments show that the electro-chemical equivalent of any substance apparently varies with the kind of voltameter used ; but the variations are undoubtedly due to secondary reactions or causes. If the voltameter is so constructed as to avoid MECIIAyiSM OF THE CURRENT 689 these, the elect ro-eheniical equivalent of silver is found to be 0.011175 g. The chemical equivalent of silver is 107.93; so calling the electro-chemical equivalent of any other substance M and its chemical equivalent <?, we have the relation <\u00bb.\u00ab ill 175: w = 107.93 :c, _ 0.011175 xc= c Thus, for hydrogen, in = 0.00010354; for copper, m = 107.93 9658* 0.003211:.\".!; for zinc, m = 0.0033857 ; etc. Since m is the quantity of any substance liberated by a unit electro-magnetic quantity of electricity, the quantity liberated by 9658 such units is a number of grams equal to c, the chemical equivalent. Ions. \u2014 The explanation of electrolysis which was advanced 1 \u00bb y Faraday was that in any electrolyte there are present cer- tain charged particles, some with +, others with \u2014 charges, and that these particles are driven by the electric force in one direction or the other. Positively charged particles will move in the direction of the force, that is toward the cathode; while those negatively charged will move in the opposite di reet ion, toward the anode. These charged particles Fara- day called \" ions \"; and those which move toward the cathode he called \"cations,\" while those which move toward the anode he called \"anions.\" Thus all cations are positively charged; all anions, negatively. The electric current in the electrolyte consists, then, from this point of view, of the passage in opposite directions of these two sets of charged hodies. When they reach the electrodes, they give np their charts and in some manner cause the liberation of uncharged Allies or ordinary matter. Hydrogen and all metals are liberated at the cathode; therefore we must consider a hy- drogen ion or any metallic ion as being positively charged. Similarly, we must consider an oxygen or a ehlorine ion as being negatively charged. \\MI *'\u00ab. i Hvsics \u2014 44 690 ELECTRODYNAMICS Since the electrolyte itself is not electrically charged, any volume of it must contain as much positive electricity on its cations as negative on its anions. The current consists of the passage across any cross section of the electrolyte of these ions ; if the current strength is i, and if ^ is the positive charge carried on the cations, and z'2 is the negative charge carried on the anions, i = z\\ + i2. In the main body of the electrolyte the positively and negatively charged ions balance each other ; but in the immediate neighborhood of the cath- ode, positively charged ions carrying a charge ^ come up in a unit of time, and negatively charged ions carrying a charge iz leave in the same time ; so this space gains in this time cations carrying charges (i^ + i%) which are not balanced by anions ; and these give up their charges to the cathode. Similarly, at the anode, anions carrying a charge i2 come up, and cations carrying a charge i^ leave, thus causing a con- centration, as it were, of anions carrying a charge (^ -h ^'2) which are not balanced by cations. In other words, the current of strength i enters the electrolyte owing to the fact that anions carrying a charge i are liberated at the anode ; and it leaves the electrolyte owing to the fact that cations carrying a charge i are liberated at the cathode. This fact may also be expressed by saying that the same quantity of electricity \u2014 not regarding its sign \u2014 is carried on that number of ions of any substance, whose mass equals its electro-chemical equivalent. This may be described differently : When a unit quantity of electricity passes, tj + \u00ab2 = 1 ; if ml and m2 are the electro-chemical equivalents of the two sets of ions, m1 and m2 grams are liberated at the two elec- trodes in this interval of time. The liberation of the m1 grams at the cathode is due, as has been shown, in part to the bringing up of the cations carrying the charge il and in part to the withdrawal in the opposite direction", " of the anions carrying the charge ia ; so the effect is the same as if there were no anions, but only cations, carrying a unit charge. In Ml-:< HAM-.M OF ////,' Cl'RRENT 601 other words, a mass of cations equal to //^ carries a unit elect n '-magnetic charge ; and similarly a mass of anions equal to 7H.2 carries a unit cliarge. 'I'll us, in the case of hydrogen, the electro-chemical equiva- lent equals \u00a73*53; and therefore this number of grams of hydrogen ions carries a unit charge, or one gram of hydrogen ions carries a charge equal to 9658. Similarly, a number of grams of ions of any substance equal to its chemical equivalent carries a charge equal to 9658. (The ratio, then. of the charge carried on a hydrogen ion to its mass equals ^g^, or, approximately, 1 x 10\"4.) Faraday showed that both of his experimental laws could be explained if it were assumed that the ions of any one substance were all alike, and that the charge on any ion was proportional to the valence of the substance. For, if these assumptions are true, it is evident that the quantity of elec- tricity carried through the liquid must vary directly as the mass liberated at either electrode ; this is the first law. Further, if the same quantity is being carried by two dif- ferent sets of ions, the masses of these substances liberated, if the charges carried by all the ions are equal, are propor- tional to their atomic weights; whereas, if each ion of one set carries twice the charge carried by each ion of the other set, \u2014 that is, if the valence of the former set is twice that of the latter, \u2014 only one half as many of the ions of the former are involved in the current as of the latter, and so the ratio lie mass of the former to that of the latter is equal to that of their chemical equivalents; this is the second law. On this assumption, the charge carried on an ion whose valence is one is the smallest charge involved in electrolysis. It is called an \"atom of electricity.\" Nature of Ions. - The question as to the nature of the ions is a most important one. As was said above, all electrolytes are solutions which show abnormal osmotic pressures, depres- ^ of the free/in^ point, etO.J and it is shown in ; 692 ELECTROD YNAM1CS on Physical Chemistry that these various abnormal phe- nomena can all be explained if it is assumed that in these solutions a certain proportion of the dissolved molecules are dissociated into simpler parts. It is natural to expect that, if a molecule is broken up into parts, they should be electric- ally charged, so that equal amounts of positive and negative electricity are produced. If this is the case, it is seen at once that the charge on any atom or radical is proportional to its valence. Thus, if a molecule of hydrochloric acid, HC1, breaks up into two parts, H and Cl, and if one is charged positively, the other will have an equal amount of negative electricity ; and the valences of hydrogen and chlorine are the same. Similarly, if a molecule of sulphuric acid, H2SO4, dissociates into three parts, H, H, and (SO4), the two hydro- gen atoms will be charged alike, and therefore the radical (SO4) will have an opposite charge equal numerically to twice that on a hydrogen atom ; and the valence of (SO4) is two. We assume, then, that when an electrolytic solution is made, a certain proportion of the dissolved molecules are dissociated into simpler parts, and that these parts are elec- trically charged, some positively, some negatively, so that the total charge is zero. (This condition of dissociation is not to be thought of as a static one, but as dynamic ; mole- cules are constantly dissociating, and others are being formed by combinations of the parts, which are moving about in the solution ; but at any temperature and concentration a certain definite proportion of the molecules are in a state of dissocia- tion.) These charged fragments of molecules form the ions when two electrodes at different potentials are lowered into the solution ; the positively charged ions move toward the cathode during their intervals of existence, before they com- bine with other ions and form electrically neutral molecules ; the negatively charged ones move toward the anode. The fact should be emphasized that the ions are produced in the act of solution, not by any action of the electric current : MECHANISM OF THE CURRENT 693 the current merely liberates the matter at the electrodes. It should also be emphasized that an ion is an electrically charged atom or radical, and is not", " a molecule ; and that the properties of matter as we observe them, e.g. gases, liquids,. are the properties of molecules or groups of molecules. Tli us, there is no connection between the general properties of a hydrogen ion and those of a hydrogen molecule. Again, as an ion moves through a solution, it is extremely probable that it carries with it a certain number of molecules, and that the number associated with a negative ion is not the same as that associated with a positive ion. If this is true, the effec- tive mass of an ion is much greater than its actual mass, considered merely as a fragment of a molecule. The student should consult Jones, The Modern Theory of Solution, New York, 1899, for the original memoirs of Van't Hoff, Arrhenius, and others. The question as to whether an ion is positive or negative is settled by observing whether it is a cation or an anion. Thus an ion of hydrogen or of any metal is positive ; while one of oxygen, or chlorine, etc., is negative. Again, since the same masses of any substance are liberated by the same quantity of electricity, regardless of the nature of the elec- trolyte, e.g. when hydrogen is liberated from dilute sulphuric acid, or nitric acid, or hydrochloric acid, etc., it is proved that an ion of any one substance always has the same charge in an electrolyte, no matter to what molecule it owes its in. Thus a hydrogen ion in a liquid always has a defi- nite plus charge ; etc. \\\\V shall now consider in detail one or t wo cases of elec- trolysis. If sulphuric acid, HaSO4, is dissolved in water, let us consider the ions as being II. II where the first two are charged with equal amounts of positive electricity, which we may call + \u00ab, and the last has a charge \u2014 2 e. Under the action of the electrical force the hydrogen ions move in the direction of the cathode, they combine with SO4 894 ions, other molecules dissociate, etc.; but, as the current flows, hydrogen ions continuously come up to the cathode, give up their charges, combine with other hydrogen atoms to form molecules of hydrogen gas which is liberated. The (SO4) ions in a similar manner migrate toward the anode ; but, since a molecule of (SO4) radicals cannot exist under present conditions of temperature, pressure, etc., when these ions reach the anode and give up their charges, there is a reaction with the molecules of the water near the anode, which takes place according to the following formula : S04 + H20 = H2S04 + O. Consequently for each (SO4) radical an oxygen atom appears, and these oxygen atoms form molecules of oxygen gas which may bubble off at the surface of the electrolyte or may act chemically upon the anode and oxidize it. Again, let the electrolyte be a solution of copper sulphate, CuSO4, in water; and let both the electrodes be copper plates. A molecule of copper sulphate dissociates into two ions, Cu and (SO4); the former is charged positively, the latter negatively. When the copper ions reach the cathode, they form molecules and are deposited on it. When the (SO4) ions reach the anode, they react upon the copper mole- cules of the plate in such a manner that the copper goes into solution. Since metal ions are charged positively, the copper dissolves in the form of positive ions ; so the current is carried off the anode by them. These + ions serve, then, to balance the \u2014 (SO4) ions which are being continually brought up to the anode. This obviously offers a method for \"copper plating\" an object. Its surface must be so prepared that it is a conductor and that copper will adhere to it; and it then must be used as a cathode in an electrolytic bath of copper sulphate, the anode being a plate of copper. Similarly, in an aqueous solution of silver nitrate, AgNO3, the ions are Ag and (NO3); the former is the cation, the OF V7//-; < ri;i;i-:.\\T 695 latter tin- union. If ;i plate of silver is the anode, it dis- solves, and silver is deposited on the cathode. This offers a method of silver plating. In tin; last two illustrations it is seen that the mechanism by which tin- current enters the electrolyte from the anode consists in the copper or the silver plates dissolving; this is done by the positively charged ions of copper or of silver leaving the plates and entering the liquid. In the water voltameter, where the anode and cathode are platinum plates, the case is not quite so simple. The negative", " ions of oxygen are formed at the anode by the reaction of SO4 upon the water molecules ; and in some manner positive charges pass from the anode to certain of these ions, first neutralizing their negative charges and then giving them positive ones ; and, after this takes place, a positive oxygen ion combines with a negative one and forms a molecule of oxygen gas, whieh bubbles off at the surface. At the cathode, the mech- anism is similar. Thus, in the case of copper sulphate, the positive copper ions reach the cathode ; under the electric force negative charges pass from this upon certain of the copper ions, making them negative ; then a negative copper 01 M hinrs with a positive one to form a copper molecule which is deposited on the cathode. Polarization. \u2014 If two platinum electrodes dip in a solution of sulphuric arid, and if a \\.-ry small elect ro-inotive force is applied to these electrodes, a current will flow, but will i cease. This is owing to the fact that under the aetion of t he electrical force the positively charged hydrogen ions collect at thr rathodr, and t hr negatively charged anioiis at the anode : |Q that, if theiv is not sullicieiit force to make the ssary charges pass from the electrodes to the ions and then to form the molecules, these charged particles will lower the potential at the anode and raise it at the cathode until the. 1 1 iv electrical force in t IK; electrolyte, and the current stops. As t he applied elect n \u2022 mot ive force 696 ELECTRODYNAMICS is gradually increased, a value is reached which will cause the evolution of the gases at the electrodes ; and the current will now continue to flow. The same description applies in general to any case of electrolysis ; a definite E. M. F. must be applied before electrolysis begins; but its value is dif- ferent for different electrolytes. This may be calculated, however, from a knowledge of the heats of combination of the compounds which are separated by the electrolysis and of their electro-chemical equivalents. Thus, experiments prove that, when 18 g. of water are formed by the combination of 2 g. of hydrogen and 16 g. of oxygen, 68,800 calories of heat energy are evolved; and, therefore, when 18 g. of water are broken up into 2 g. of hydrogen and 16 g. of oxygen, an amount of work equal to the mechanical equivalent of 68,800 calories must be done, i.e. 68,800 x 4.2 x 107 ergs. As a quantity of currents equal to e is passed through the electrolyte consisting of H2SO4, the quantity of hydrogen evolved is me, where m is the electro-chemical equivalent of hydrogen ; and an \"equivalent\" amount of oxygen is liberated at the anode. Since it requires 68,800 x 4.2 x 107 ergs to liberate 2 g. of hydrogen, that required to liberate me g. is me 68>80\u00b0 * 4>2 X 1QT. If E is the E. M. F. applied to the electrodes, which just causes the electrolysis, the work required to pass a quantity of electricity e between the electrodes is Ee. Therefore, since this work is spent in liberating the hydrogen and oxygen, \u2014 neglecting the work done in heating the electrolyte, and assuming that there is no other source of energy, \u2014 we have the equation \u201e me 68,800 x 4.2 x 107 \u2014 -' or E = m 68,800 x 2.1 x 107. For hydrogen, m = 1.035 x 10~4, and therefore E = 688 x 1.036 x 2.1 x 10* = 1.5 x 108 = 1.5 volts. The E. M. F. of a Daniell cell is approxi- mately 1.1 volts; so at least two Daniell cells are required to decompose water. Calculation of the E. M. F. of a Primary Cell. \u2014 We may consider the mechanism of a voltaic or of a Daniell cell from this standpoint of ions. In the former, the cause of the current is the solution of the zinc in the acid, that is, the MECHANISM OF THE CURRENT 697 parsing off of positive zinc ions into the liquid. The chem- ical action is the solution of the zinc in the acid and the evolution of hydrogen at the copper pole ; and experiments have proved that when 65.4 g. of zinc are dissolved in dilute sulphuric acid, 38,066 calories are evolved. The electro- chemical equivalent of zinc is 0.00338. So when a quantity of current e passes off the zinc rod into the acid, the mass of zinc dissolved is 0.00338 e ; and", " the energy liberated is 38.066 x2x 10' If E is the difference of potential between the copper and the zinc electrodes, it follows that ~,-, _ 38,066 x 4.2 x 107 x 0.00338 or E = 38,066x4.2x107x0.00338 = ^ x 10- = 0.83 volt* 65.4 In this calculation we neglect the loss of energy of heating tin- liquid, and we assume that the only source of energy is that furnished by the solution of the zinc. In a similar manner, when a current is produced by a Daniell cell, /inc dissolves at the zinc plate and copper is deposited at the copper electrode. When 63.6 g. of copper dissolve in sulphuric acid, 12,500 calories are evolved ; and the electro-chemical equivalent of copper is 0.00329. So the E. M. F. of the cell is found by calculation to be 1.1 volts, making the same assumptions as before. Storage Cell. \u2014 In certain cases of electrolysis the anode and eith'ide are so modified l>y the passage of the current, and the < >nsequent liberation of matter at them, that they be used afterward t\u00ab. form a coll for the production of a current. Therefore, if the battery of cells or the dynamo which was producing the current through the electrolyte is removed, and if the electrodes are joined by a wire, a cur- rent will flow through it. This action does not continue 698 EL ECTI! 0 1) YNA MICS indefinitely, for in producing a current, those modifications which were the result of the electrolysis are reversed, and the electrodes return to an inactive condition. If the bat- tery of cells or the dynamo is again used to send a cur- rent through the electrolyte, the process may be repeated. Such a combination of elec- trodes and electrolyte is called a \"storage cell\" or a \"secondary cell\"; and when it is in a condition to produce a current itself, it is said to be \" charged \" \u2014 but it is evident that this expression has no connec- tion with what has been called a \" charge \" in pre- vious chapters. FIG. 880. -The ordinary form of storage cell. The commonest form Qf storage cell has dilute sulphuric acid as the electrolyte, and as the anode and cathode two lead grids whose interstices are filled with a paste of lead sulphate. When such a cell is charged, it has an E. M. F. at first of about 2.1 volts; but this falls to about 1.8 volts as the current flows. Conduction of Electricity through Gases. \u2014 In speaking in a previous chapter of the passage of sparks through air or any gas, occasion was taken to state that the discharge con- sists in the disruption of molecules into simpler parts, and that the path of a spark is an excellent electric conductor. We are thus led to believe that a gas becomes a conductor owing to the presence in it of small parts of molecules, which we may again call ions, although there is no reason for believing that these ions are the same as those in evi- dence in the electrolysis of a liquid. The charged particles v/.s.u OF nn: < t HHENT 699 may In- the same, hut the in -\u2022>< iciat ed with thiMii are diileivMi ( >ee page 693). All the known facts in regard to condiu -lion through a gas are in support of this idea of tin- ionic nature of the proce The method of determining whether a gas is a conductor or not is, of course, to immerse in it two electrodes, at a small distance apart, and to observe whether a current flows when a difference of potential is produced between these plates by some cell or combination of cells. It is found that a pure dry gas is an extremely poor conductor ; but it may be made conducting by many means. A few will be mentioned. By passing a spark through any portion of the gas, all other neighboring portions have their conductivity increased. In many cases of complex gases, a sufficient rise in temperature makes them conducting. If the ultra-violet waves from a source of light pass through a gas, it becomes a conductor. There are many solids, e.g. alkaline earths and metals, which when illuminated with ultra-violet light make the gas near them conducting. This is called \" photo- electric\" action. Further, if metals (or carbon) are made very hot, they make the surrounding gas conducting. If the cathode rays <\u00bbr the X-rays (see below) traverse a gas, it i> made cm id noting. Certain bodies, known as \"radio- active \" substances, have", " the power of emitting radiations which make the i,ras through which they pass a conductor; these will be discussed more fully below. In all these cases the tid tn l>e \u2022\u2022 iolii/ed.\" Spark and Arc Discharge. \u2014 If two electrodes immersed in a gas at not t ; a distance apart are raised to a Milli- cient difference <>f potential, a spark will pass between them, i. The character of the dUchar^e and the utial difference required depend upon the nature and the condition <\u00bbf the ^;,s: its pressure, its temperature, its purity, etc. If the two electrode^ are lu-oiiLrlit sulVicieiit ly Close together and ti iidiic.tOP, 700 ELECTRODYNAMICS so that a large current passes, the discharge is called an \"arc,\" as is illustrated by the ordinary arc lights in the streets. In both the ordinary spark and the arc the elec- trodes are gradually vaporized, and their vapors, as well as the gas itself, are rendered luminous. The phenomena of the discharge are so varied that some special treatise on the subject should be consulted. The best is J. J. Thomson, Conduction of Electricity through G-ases, London, 1903. We shall describe here only one or two cases of special interest. Vacuum Tube Discharge. \u2014 If the gas is inclosed in a glass bulb into which two metallic electrodes enter, and if the pressure is gradually lowered by means of an air pump, the FIG. 381. \u2014Vacuum discharge tube. character of the discharge changes in a most marked man- ner. When the pressure is about that of 1 mm. of mer- cury, it is observed that at the cathode there is a velvety light covering it entirely or in part, which is separated by a dark space from a luminous region, and that this is separated by a second dark space from a luminous striated column extending to the anode, the end of which \u2014 if it is a wire \u2014 has a bright spot of light. The first dark space near the cathode is called the \" Crookes dark space \" ; the second one, the \" Faraday dark space \" ; the region separating them, the \" negative glow \"; and the striated portion near the anode, the \"positive column.\" As the bulb is more and more exhausted, the Faraday dark space extends farther down '01 or I III-:'URRENT toward the anode, and another phenomenon becomes most prominent. There is a radiation of something from the cathode, proceeding in straight lines perpendicular to its surface, quite independ- ent of the position of the anode. This radia- tion produces a faint luminescence of the traces of gas left in the bulb ; and so the path of the rays through the bulb may be seen. They are called the \"cathode rays.\" Where they strike the walls of the hull), it is made lumin- ous and its temperature a. 6. Fio. 882. \u2014 Two forms of vacuum tube discharge: a, moderate vacuum ; b, high vacuum. rises ; the color of the light which is thus produced depends upon the material of the bulb, but ordinary glass emits a greenish yellow light. (If certain other solids are intro- duced in the bulb in the path of the rays, e.g. coral, they emit characteristic colors.) The fact that the rays proceed in straight lines is proved by introducing in the tube solid bodies which are opaque to the rays; for it is observed that they cast sharp shadows on the walls of the tube. The radiation passes directly through metallic films, if sufficiently thin. The path of the rays in the tube is deflected by a strong electric field, provided the vacuum is p< \u2022! feet enough ; and the deflection is in such a direction as to l.-ad one to believe that the radiation OOnifotl of n. -Datively _M-d particles. In fact, all experiments lead to this con- \u2022ItiiT.l t.y r:ith'\u00bblr rny*. 702 ELECTRODYNAMICS elusion. If the rays are made to enter a hollow cylinder which is connected to an electrometer, it is seen that the cylinder is receiving a negative charge ; and the fact that, when the rays strike a solid its temperature is raised, is ex- plained by assuming that the rays consist of material parti- cles. Again, as will be shown in a few pages, a charged particle in rapid motion has the same action on a magnet as does an electric current ; and, since a wire carrying a current may be made to move under the action of a magnet (which is simply the reverse of the fact that a current can move a magnet), so a particle in rapid motion may have its", " direction altered if it is charged ; and experiments show that the cath- ode rays may be deflected by a magnet in exactly the man- ner which one would expect if they were negatively charged particles in rapid motion. The velocity of these rays may be measured in many ways ; and it is found to depend upon several conditions : difference of potential between the elec- trodes, pressure of the gas, etc. ; its value is not far from one tenth of the velocity of light. The masses of these particles and their electrical charges may also be measured with a fair degree of accuracy ; and it is believed that the charge of any particle is the same as that on a hydrogen ion in ordinary electrolysis, while its mass is approximately one thousandth of that of a hydrogen atom. So far as experiments can prove, these particles which constitute the cathode rays are the same no matter what gas is put in the tube. If the cathode is in the form of a metal plate with many small openings in it, it is observed that in addition to the cathode rays which are emitted from one side there are rays proceeding in the opposite direction apparently through the holes in the cathode. These are called \" canal rays,\" and have been proved to consist of positively charged particles, moving much more slowly than the cathode rays. Their charges are probably the same as those of the latter rays ; but their masses are comparable with those of an atom. MECHANISM oF THK (Tltl:i-:\\T 703 Tin- gas which is traversed by cither the cathmle or the canal rays is ionized; that is, becomes a conductor. This was proved for the former rays by Lenard, who constructed a glass bulb in such a manner that a portion of the wall which was struck by the rays had an opening in it which \\\\as covered by a thin layer of aluminium. Surrounding this bulb was another which could be exhausted if desired. So, as the cathode rays passed through the aluminium win- dow and entered the outer bulb, their action could be studied. Pto. 884.-AnX-\u00abytube. The aluminium itself emits cathode r,(\\s, as a result of being struck by the interior cathode rays, in addition to t: mittiiiLT some ; and the total radiation from the \\vindo\\\\ is called \"the Lenard r. This includes also* t\\ pe of radi- ation cut irel\\ distinct from cathode rays. first sho\\\\ n d that whenever cathode rays strike lid BQrfaoe, \u2022 radiation is produced which lias most in- -tiir_: properties, and to which he gavr the name X-rays. Thesr raya pro eed in -n-ai^ht lines from their sniirce; they arc not reflected, ivt'ra. \u2022 ij thej arc not 704 ELECTRODYNAMICS affected by an electric or magnetic field ; they cannot be polarized ; they affect a photographic plate ; they cause many bodies to become luminous ; they pass with ease through many bodies opaque to light, e.g. wood, aluminium, the human flesh, etc. ; they are absorbed by other bodies much more than is light, e.g. glass ; they ionize a gas. It is believed that they are transverse pulses in the ether, in dis- tinction to being trains of waves ; and all their properties may be explained on this assumption. If we consider what we would expect to happen when a moving charge is suddenly brought to rest, as the cathode rays are by the solid which they strike, we can understand how these pulses are pro- duced, and how irregular they are. Rontgen's original papers and Stokes' memoirs giving an explanation of the X-rays are reprinted in Barker, Rontgen Rays, Scientific Memoir Series, New York, 1899. Radioactive Bodies. \u2014 Certain substances of high atomic weight, namely uranium, thorium, radium, and one or two others, possess, in common with their compounds, such as thorium nitrate, etc., the most remarkable property of emit- ting spontaneously a radiation which can ionize a gas ; they are said to be \"radioactive.\" This radiation is complex, consisting of what are called \" a rays,\" which are positively charged particles analogous to canal rays, and \"/3 rays,\" which are negatively charged particles analogous to cathode rays. Radium also emits rays which are like X-rays in many respects; they are called \"7 rays.\" This radiation is accompanied by changes inside the mole- cules of the substances which emit it ; and in some cases the products of the molecular changes are gaseous. Thus, if thorium nitrate is dissolved in water and ammonia is added, a precipitate is formed, which may be separated by filtration. The precipitate is not radioactive at first but gradually", " becomes so, in fact returning to the same condition as the original thorium nitrate. The filtrate, on the other hand, is 3/AY7M-V/.S.V <>r mi-: CUKRKM 705 mo>t radioactive, but loses this property in time; as it does this, it L,rives off a ^a>emis emanation, which is radioactive. This emanation is at lirst uncharged, but by losing its nega- tive charges it becomes positively charged and may be attracted to any negatively charged body. It undergoes further changes; and during each, radiations are emitted. The emanation of radium finally decomposes into helium gas. During these processes, further, heat energy is evolved, and the temperature is raised several degrees Centigrade. Electrons. \u2014 When a gas is ionized, either by the action of X-rays or by the rapid motion of minute positively or nega- tively charged particles through it, all the negative ions are found to be alike in all respects and to be the same as the eathode rays. It is therefore believed that, as such charged particles or as pulses pass through a gas, they break off from its atoms these negative ions. A theory of the constitution of an atom has been based on this idea. An atom is thought to contain within it a great many minute negatively charged particles which are making rapid revolutions \u2014 not unlike the constitution of the solar system, the other portions of the atom making up the positive charge. Then ioni/ation would consist in causing one or more of these negative par- ticles to leave the atom. These particles while inside the at oiii \u2014 and when outside also, provided they have no other mat. -rial particles elin^in^ to them \u2014 are called \"electrons.\" Their vibrations inside the atom irive rise to waves in the ether. So tar as is known a moving piece of uncharged matter does not affect the ether; but, it' charged. it does; and. if it> motion has a* \u2022 //. waves are produced. It is known from theoretical considerations that when a charged particle U in motion, its kinetic energy is greater than it would he if it were not charged; and, therefore, an electric charge in motion by itself \u2014 quite apart from matter would have kinetic energy; that is, would have : The (jiiestion thm arisi.1 not the inertia of matter really AMKft'8 PI1T8IC8 \u2014 46 706 ELECTRODYNAMICS due to the motion of electric charges in its minute parts? In other words, is not a moving charge the fundamental fact in nature? This question is fully discussed, and a most in- teresting description of the general properties of electrons is given, in a series of papers in the London Electrician during 1902-1903, by Sir Oliver Lodge. Conduction in a Solid. \u2014 It has been proved in what has gone before that conduction in an electrolyte and in a gas consists in the actual motion of charged particles, called ions. In the case of a solid conductor the question as to the mechanism of the conduction of a current is not so simple, owing primarily to the fact that the particles of a solid have so little freedom of motion and can only vibrate. But there is every reason for believing that in a solid also the process of conduction is by means of ions. The existence of free electrons moving about inside the solid conductor from atom to atom may be proved by many experiments ; and the evidence in favor of this explanation of conduction is accumulating continually. Convection Currents. \u2014 It was Faraday who first conceived the idea that the essential feature of an electric current was the rapid motion of an electric charge ; but the first to prove by direct experiment that such a charge in motion had the same magnetic action as an ordinary current produced by a voltaic cell was the late Professor Rowland. He charged a circular metallic disc and caused it to rotate rapidly on an axle perpendicular to its faces; he observed that when a magnetic needle was brought near the disc, it was deflected exactly as if electric currents were flowing in circles in the disc. He was able to prove that, within the range of veloc- ities used, a charge e moving with a velocity v is equivalent to a current whose strength is ev. It is probable that this is true, even if v is very great, much greater than it is possible to attain by any mechanical means. A current due to a moving charge is called a \"convection current.\" CHAPTER XLVI MAGNETIC ACTION OF A CURRENT General Description. \u2014 In a previous chapter a general description of the magnetic field due to an electric current was given ; and it was seen that a conductor carrying a cur- rent is surrounded by a field of magnetic force, such that the lines of force form closed curves around the current. The", " relation between the direction of the current and that of the lines of force is given by the right-handed-screw law. The magnetic field, then, due to a circuit carrying a current is the same as if a great many minute magnets, of the same length and strength, are taken and placed side by side so that their north poles are all turned one way and their south poles in the opposite direetion, thus forming what is called a \"magnetic shell.\" having the same contour as is made by the conductor earn inur the current. For. lines of force would proceed out from the north poles of the slu-11 and all return tn the south poles. We < an thus speak <>f the M north face \" (.!' a eiivuit irivnt and of its \"south face.\" Again, if a wire \u2014 or other eoiid net \u00ab.r \u2014 is wound in the form of a helix, and if a current is 708 ELEL'TRODYSAMICS passed through it, thus forming a \" solenoid,\" the magnetic field enters one end and emerges from the other exactly as if it were a bar magnet. Electro-magnets. \u2014 If, then, a bar of iron or of any mag- netic substance is inserted in a solenoid, it is magnetized because each little molecular magnet turns and places itself along a line of force. A bar of iron wrapped with a helix of insulated wire is called an \"electro-magnet.\" The bar is FIG. 886. \u2014 A powerful electro-magnet arranged to show magnetic or diamagnetic action on suspended sphere. usually made in the shape of a horseshoe, or of the general form shown in the cut. These are used in a countless number of instruments, such as call bells, telegraph instru- ments, etc. The first electro-magnet was made by Sturgeon (1825) ; but the idea of wrapping several layers of wire, like thread on a spool, is due to Joseph Henry. By this means the intensity of a magnet may be greatly increased. Electro-magnetic Forces. \u2014 Since a solenoid is equivalent to a bar magnet, it will, if suspended free to move, turn and MA'.M-TK X OF A CURRENT plarr itM-lt in the magnetic meridian; and, if another solen- oid or a liar magnet is brought near it, it will be seen that 7'\".' like i H.Irs irjH'l and unlike ones attract. These motions, pro- duced owing to the magnetic field of a current, are said to be diit to \u2022\u2022 electro-magnetic forces\" ; and we can describe them in another manner which is perhaps more definite. Since the south end of a solenoid is the one which tubes of force enter, and since it is the one which is attracted by the north pole of a magnet or a solenoid from which tubes of force proceed, it is evident that by the act of approaching one another, more tubes of force enter the south end of the solenoid than before. Other cases of attraction and repulsion may be con- sidered in a similar manner ; and it is easily seen that they may all be described by saying that motions of conductors \u2022 any ing currents take place in such a manner that as many tubes of force as possible emerge through their north ends or faces. This is equivalent to saying that the electro-magnetic forces are in such a direction as to produce these motions. If a loosely wound spiral spring is sus- pended in a vertical position with its lower end just dipping in a cup of mercury, and an electric current is passed through it, the separate turns of the coil will attract each other, because by coming closer together more tubes of force pass through them, instead of escaping out from th\u00bb- sides; as th\u00ab- >]>ring thus contracts, the electrical connection at the bottom i- lu-okt-n. and so the force of attraction vanish*-* : tin- >j>rinij tli* ii drops, connection is again made, the..t Hows, etc. The action is increased \u00ab'rting an iron rod inside the coil. Plo. 8W._Kli0tro.IIIIIfBrtte tllf-e. UOD of parallel current*. These phenomena of electro-magnetic forces were discov- 1 by Ampere, and lie invented many most beautiful experi- 710 ELECTRODYNAMICS 999.9 - 9999 merits to illustrate them. They may be found described in many text-books. He also proposed a theory of magnetism based upon his observations. He advanced the hypothesis that in each molecule of a magnetic substance there is an electric current, flowing in a fixed chan- nel; and therefore if a bar of su", "cli a substance is brought into a magnetic field, \u2014 due either to a magnet or to a solenoid, \u2014 the molecules will all turn so as to include as many tubes of magnetic force as possible. The bar will be saturated magnet. Ampere's theory when the molecules have so arranged themselves that their currents are parallel to each other and to the ends of the bar ; and under these conditions the lines of force due to these currents will emerge at one end and return into the other, exactly like a solenoid. If two parallel wires or rods are placed in a horizontal plane and are joined by a fixed wire BO, containing a cell, and by a movable wire PP', which can roll on the wires, FIG. 888. \u2014 South pole of * ELECTRO MAGNETIC FORCE Fio. 890. \u2014 Relation between directions of current, magnetic Fio. 889. \u2014 Electro-magnetic force : magnetic field is upward fleltl> and electro-magnetic force, through circuit. Each is perpendicular to the other two. this movable wire will be set in motion if there is a mag- netic field through the space between the parallel wires, because by so moving a change is made in the number of tubes of force which pass through the circuit (PP1 OB). If MA<.\\I-:TI<- ACT1OB or.1 i-rniiKNT 711 the current is in the direction shown in the cut, the upper face is tin- north one; and, if the magnetic field of force is in tin- direction shown, tin- cross wire PP' will move trd the right, so that the circuit incloses more tubes coming out from its north face. This law of force is given by the diagram which describes the connection between t la- directions of the magnetic force, the electric current, and tlu- electro-magnetic force. Magnetic Force Due to a Current. \u2014 Since the magnetic lines of force due to a current form closed curves around it, it would require a certain amount of work to carry a unit north pole around such a line of force in a direction opposite to that of the line itself; and it is evident that this work must vary directly as the strength of the current. It may be shown by experimental methods that the amount of work required to carry a unit pole around any closed curve encir- cling the current is the same for all paths. Thus, calling this work IT. we may write W= d, where c is a factor of proportionality depending on the system of units adopted for the measurement of the current. The C. G. S. electro- magnetic system is based on the definition of such a unit current as will make this factor equal to 4 TT. (The reason for this choice depends upon the connection between a cur- ivut and a magnetic shell, and need not be explained here. It may 1..- found in any advanced text-honk.) If the unit pole is carried around the circuit /// times, the work done is evidently ITT////. One consequent- of this definition of a unit current is that the intensity of the magnetic field at the centre of a circular coil of radius a, of n turns, carrying a current of strength /'. \u2014. (See page 678.) If the condun m of \\\\ helix, the magnetic fOTOe in-ide il..!i it. may he deduced at once. Tin- work done in carrying a unit pole around a closed path tf/W \\\\n in the cut, where ab is parallel 712 ELECTRODYNAMICS to the axis of the helix and inside it, be and da are perpen- dicular to this axis and cd is outside the helix, equals the product of the intensity of the magnetic field inside and the length of the path ab, be- cause the lines of force are along the axis, and so no work is required to tra- verse the portions da and be, and the force outside is so small that it may be neglected if the helix is long. So, calling the inten- sity of the field inside R and the distance ab x, the work is Rx. But, if there are N turns of the helix per unit length, there are xN turns in the length x ; and since each of these carries the current t, the work done in threading them by a unit pole is, in accordance with the above definition of a unit current, 4 irN x i. So, Rx = 4 TrN x i, or, R = 4 irNi ; a most important formula.. \u2014 Magnetic force inside a long solenoid. If there is a rod of iron, of permeability ^, filling the solenoid, the number of tubes of induction per unit area passing through the iron is proportional to pR. (See page 615.) Since p for iron is large, this means that", " the number of tubes of induction passing through the solenoid is greatly increased by inserting in it a rod of iron. These additional tubes are due to the magnetization of the iron by the current. Since each of these tubes of induction passes N times through a circuit carrying the current i in a unit distance, it is evident that the magnetic action of the solenoid is propor- tional to that of a single turn of wire carrying a current equal to N*i. Energy of a Current. \u2014 The fact that, when a current is flowing in a conductor, forces may be experienced in the surrounding medium, proves that there is a certain amount of energy in this medium due to the current. This energy is MAGNETIC ACTION OF A iTRRENT 713 not in the form of a strain \u2014 no sparks are observed in the medium, etc. So it is natural to think of the energy as being kinetic- in its nature; and this idea will become more evident in a later chapter. What will be shown is this : as a current is first started, e.g. by joining the two poles of a primary cell, the i urrent does not rise to its full strength instantly, for part of the energy furnished by the source of the current is spent in producing those motions in the surrounding medium which constitute the magnetic field, and it is not until these motions are established that all the energy of the source of the current goes into forcing the current through the con- ductor, and so heating it. Similarly, if the source of the current is suddenly removed, the current does not instantly cease, because the energy in the medium disappears gradu- ally, being spent in maintaining the current for a short time. Compare the case of a railway train starting from rest; it does not In its full speed instantly because the energy furnished by the loco- motive is used in producing kinetic energy ; but, after the desired speed in reached, the only work done is against the friction of the wheels, the resistance of the air, etc. Similarly, when the train is to be stopped, the power furnished l\u00bby the locomotive is shut off, but motion continues until the kinetic energy of the train is used up in ov\u00ab -n \u2022oiuin^ friction. The analogy with an electric current is not particularly good, because in the latter caae the kinetic energy is not in the conductor, but in tin- medium, while the work done in producing the heat is spent in the conductor. From what was said above in regard to solenoids, it is evident that th<5 energy of their currents is proportional to vhere N is the number of turns per unit length. There are many electrical instruments and machines whose artiMiis depend ap<m electro-magnetic forces: and a few will The Electric Bell. \u2014 This consists \u00ab,f a gong Q- ; an elect n> magnet /.iluatiii^ clapper of spring brass which has a small knob If with \\\\ hich to st like the gong, and a strip of iron ihe polo of the electro-magnet; a stiff piece 714 ELECTRODYNAMICS of brass A> which is attached to the clapper, and presses against a lixed metal stud O. One end of the wire around the electro-magnet is joined to the clapper, while the other FIG. 392. \u2014 Electric bell and push button. end is connected to a binding post at D. When the bell is in use, D and 0 are joined to some primary cell B ; a contact key P being introduced in the circuit. When the key is pressed so as to make contact, the current flows through the Fi<;. :!!\u00bb:;. Tin; relay. electro-magnet, attracting the clapper and thus ringing the bell ; but as the clapper moves toward the magnet, contact is broken at (7, and the current ceases ; then, owing to its MAGNETIC A(T1<>.\\ oF.! rri;i:i-:NT 715 elasticity, the clapper vibrates back, making contact at (7, and the current again flows; etc. Relay. \u2014 A relay consists of an electro-magnet in front of which is an iron plate called the armature, carried by a metal rod pivoted at its base. The wire around the electro-magnet may be connected to a primary cell or a battery of cells at a distance, with a key in circuit. So, when the key is pressed, a current will flow around the magnet. Even if the current is extremely feeble, the armature will be attracted ; and by means of suitable contact points a sec- ond cell may be closed through any circuit; and thus any elect n> magnetic effect may be produced,. Vi:i|.l. k.'\\.'111.1 -oini.liT. such us rin'_mi", "L,r a bell. etc. NVheii the key is broken, the current ceases, and the armature is drawn Lack from the electro-raagiK'i\u00bbiral spring attached to its rear. It* a second eh-- \\\\itlianannatuivisintroduced in the circuit <>f the second \u00bb -ell, the sound made by the 716 ELECTRODYNAMICS armature as it clicks against the electro-magnet may be clearly heard. This is the principle of the ordinary telegraph system, different letters being distinguished by different combinations of \" dots and dashes \" ; that is, short and long intervals of time between consecutive clicks of the \" sounder.\" Duplex Telegraphy. \u2014 In the \" Duplex \" system of telegra- phy it is possible to receive and send messages from a station at the same time. The instrument consists essentially of a receiving instrument E, such as an electro-magnet or a galvanoscope, around which are wound two coils of wire in opposite directions ; so, if equal currents are passed through both coils, no effect is pro- duced, while, if a current passes through one coil alone, there is an effect. One of these coils is con- nected through the \"line wire \" to the distant sta- tion, while the other is joined through several coils of wire of adjust- able lengths, R, to the earth. The arrangement of the cell and the key K is as shown in the cut. The coils R are so adjusted that when the key K is pressed and makes connection with the cell jB, equal currents pass around the receiving instrument E, and there is no effect ; but a current passes over the line wire to the distant station. When a current is received over the line wire, it passes to the earth, entirely regardless of the position of the key K, thus affecting the instrument. (If the key K is pressed down when a current is being received from the distant station, we may regard the current in the line wire as FIG. 895. \u2014 Diagram of one form of duplex telegraph instrument. if.-i<..Y /\u2022:/\u2022/<\u2022 ACTION OF A rr/,'/;/v\\vr 717 neutralized. Init part of the current from U passes around E through R to the earth. Telephone. \u2014 The ordinary Bell telephone receiver consists steel magnet M, around one end of which is wound a coil of wire, and in front of which is a thin iron plate railed the dia- phragm. This is at- tra.-t ed toward the magnet, but is kept from motion as a whole by the frame. If a current is passed through the coil, it will either strengthen or weaken the steel magnet, drpniding upon its direction; and so the diaphragm is either attracted or repelled. Thus, if the current fluctuates, the diaphragm vibrates.,,, K i... 396.\u2014 The Bell telephone. Microphone or Transmitter. \u2014 This consists of a thin metal diaphragm whose edges are held but whieh can vibrate like a drumhead ; against its centre presses a carbon \" button,\" which is held firmly by suitable metal supports. A cell is connected to the carbon button and to its supports. There is poor electrical contact between these; and it varies in its conducting power as the pressure of the diaphragm against the carbon button varies. If the pressure is increased, more current flows; if it is decreased, less current. So, if the diaphragm vi- brates, the current fluctuates in st rength. It is thus evidrnt that, if a telephone I1]1 is included in the mi\u00ab roplion. FIO. 8\u00bb7. \u2014 nimrnun of BUk\u00ab vibrations of tin- mi.TMpli. -m- diaphragm produce corresponding \\ M. rations of the telephone diaphragm. In this manner. the MT produced b\\ the human voice will cause \\ MS of 718 EL ECTUOD YXAMICS the telephone diaphragm, which will in turn send out sound waves ; and thus sounds may be said to be transmitted. D'Arsonval Galvanometer. \u2014 This consists of a permanent steel horseshoe magnet, between whose poles a coil 0 is supported by means of a vertical wire. The wire in this coil is continu- ous from A to B, two fixed binding screws. When no current is passing, the coil is held so that its plane is parallel to the line joining the two poles ; but if a current is transmitted through the coil by means of A and B, it will turn so as to include as many of the tubes of force of the magnet as possible. It will be brought to rest by the torsion of the wire ; and so its deflections meas- u", " of these conditions, it satisfies all. A \" variable \" current is one that is not steady. In order to produce a steady current one may use a source of constant K. M. F., sin -h as a I Smell's cell, or a thermocouple whose junctions are maintained at constant temperatures, or a dynamo \u2014 as will be described in the next chapter. Uniformity of Current. \u2014 One of the most important properties of a steady current is that its strength is uni- form throughout the circuit ; that is. if the circuit includes conductor* of diflVrent material, of different sizes, etc., the strength of the current is the same in them all. This may!>\u2022\u2022 shown by proving that the magnetic or the hea1 \u2022 ii of the current is the same for all portions of the lit. A L: i in, if the current were not the same at all points, there would be accumulations of charges at certain points; and, as th.'sr innv.e>c<l, they could be detected; but such is not the case. mlailv. if at an\\ point of the circuit, it branches SO as i 721 722 EL ECTItOD YNAM1CS to form two or more parallel conductors, the strength of the current in the single conductor must equal the sum of the strengths of the currents in the branches. This may be expressed in a formula, K,W, x. \u201e\u201e is the current in any FIG. 402. \u2014 A divided circuit. J conductor at a branch point, the direction of the current being called positive if it is toward the point, the summation of all the currents at that point is zero, or, in symbols, 24 = 0. \u2022 \u2022 \u00b0- Ohm's Law. \u2014 We have seen that a current flows between two points of a conductor only if there is a difference in potential between them ; so that we may in a way regard the E. M. F. as the cause of the current, and it is not un- natural, judging from analogy with the flow of heat in a bar owing to difference in temperature between two points, to advance the hypothesis that the current strength in a con- ductor varies directly as the E. M. F. between two points, provided the current is steady. (We are considering the case where there is no cell or other source of E. M. F. introduced in the conductor between the two points.) That is, if A and B are any two points in a circuit in which is flow- ing a steady current of strength i, and if E is the E. M. F. between FIG. 408. \u2014 Diagram to illustrate A i T\u00bb.LI i A i \u2022 \u2022 1 1 Ohm's law. A and B, the hypothesis is that E = Ri, where R is a constant depending upon the nature of the conductor between A and B, but not upon the values of E or i. This hypothesis has been found to be true, so far as experiments can decide ; E has been varied by intro- ducing more cells or a dynamo, and the resulting current or STXADT m;i;i-:yT8 723 has been measured. It is called Ohm's law, having been proposed by Georg Ohm in the year 1826. Resistance. \u2014 It is evident from the formula that if R is lar^e. / i> >mall, provided E remains constant; while, if // small, i is u'reat. For tliis reason R is called the \"resistant. \" of the conductor between A and B. This law can also be E 1 H H written i = \u2014- ; and, if for \u2014 the symbol 0 is substituted, the formula becomes i = CE. For obvious reasons O is called the u conductanee \" of the conductor between A and B. If the conductor between A and B is a uniform wire, it is evident that the K. M. F. between A and B is exactly twice what it is between A and a point halfway to B. Therefore, since the current is uniform, the value of R for the conductor l>et ween.1 and Jt must be twice that for half the length. So, in general. the value of R for any portion of a uniform con- ductor of constant cross section varies directly as the length of this portion. It, while a constant E. M. F. is maintained between the two points A and B, a second conductor is introduced be- tween them, identical with the first one, each will carry a current i = \u2014, and so the current is doubled or the total E H resistance is halved. The same would be true if, instead of using two conductors, one of twice the cross section were in trod need. So, in general, the resistance of a conductor", " \\aries inversely as its cross section. Direct experiments show that if the same E. M 1 applied at the ends of conductors of the same length and M section, but of different matt rials, the resulting cur- rent is different. This and the two previous statements may be expressed in a formula. p a where R is the i i uniform conductor of length / ~- t i 724 ELECTRO!) YNA AIICS and of constant cross section a, and c is a constant for a con- ductor of any one material, but differs for different ones. This quantity c is called the \" specific resistance \" of a sub- stance, or its \"resistivity.\" Similarly, the conductance C = \u2014 = --, and may be written C = k-, where k = -. This R c I I c constant k is called the \" conductivity \" of a substance. Illustrations of Ohm's Law. \u2014 1. Con- ductors in series. Let the circuit consist of several conductors in series, and let the resistances of the portions A^AV A2 A^ ^.3^.4, be Rv R2, Rs ; further, let the potentials at the points A^ A2, A%, etc., be Vv Vv V& etc. Then applying Ohm's law to the separate sections, FIG. 404. \u2014 Conductors in series. Hence, I = The total resistance between A1 and A\u00b1 is by definition i = 1 \"7 \u2014 - ; and it is seen that its value is R1 + R2 + Ry In general, then, the total resistance of a number of conductors in series equals the sum of the resistances of the separate parts. (The fact that R varies as I, the length of a conductor, is a special case of this.) 2. Conductors in parallel. \u2014 Let the circuit branch at any point A into two or more conductors which meet again at B; let the resistance of these branches be Rv R2, R& etc. ; and let the currents FIG. 405.\u2014 Conductors in parallel. LAWS \u00ab/\u2022 STEADY friiliBNTS flowing in each be t\\. i.2. iy etc. The total current is / = /! -f /2 4- i3 -I- \u2022\u2022\u2022 ; and the total resistance between A and y _ YR B is by definition \u2014 \u2014. Applying Ohm's law to the various branches, we have *1 D D'8 D ; etc. and therefore, calling the total resistance R R RZ R This may be expressed more simply in terms of conductances, for 0 = \u2014 ; hence, C= C + Ca+ C + \u2014 \u2022 or. in a branched circuit the total conductance equals the sum of the conductances of the branches. (The fact that R varies inversely as the cross section of a conductor is a special case of this.) It should be noted, further, that the ratio of the currents in any two branches equals the inverse ratio of the resistances of these branches. Thus, 1. 1 3. Wheatstone bridge. \u2014 This is a particular arrangement of \u2022onductors ; four form a < iivuit ABCD, and two connect the diagonal points A and <\\ and B an<l />. This network of con- ductors is used in many experi- mental methods ; l.ut only one will be descrihe.1 here. In this a cell is introduced in one of the diagonal branches, say AC. and 726 ELECTRODYNAMICS a galvanoscope in the other. If the arrangement is such that A is joined to the positive pole of the cell, the potential of A is higher than that of (7; and the potentials of B and D are both less than that of A and greater than that of C. So it must be possible to find a point D in the branch ADC whose potential equals that of any given point B in the branch ABC. If this is the case in the actual arrangement, no current will flow across from B to D, and the current flowing from A to B will equal that from B to C ; and that flowing from A to D will equal that from D to C. Call the poten- tials, at A, B, C, and D, VA, VB, V& and VD (it should be noted that VB = F^>); the resistances of AB, BO, AD, and DO, RV R2, Ry, and R\u00b1 ; and the currents in AB and BO, FIG. 407. \u2014 Arrangement of Wheatstone's bridge for the comparison of resistances. -F, e'j ; and in AD and DO, ^'2. Then But F\u00bb= Vn\\ hence 4? = or This formula evidently offers a method for the comparison of resistances. For", " strength t is E If1 \u2014 = \u2014 '-\u00a3. Therefore, if the numerical value of the current K 10 /fcj is in l>e deduced from Ohm's law, using ohms and volts in which to express resistances and electro-motive forces, a unit current must be defined whose value is one tenth that of a C.G. S. electro-magnetic unit current. Such a current is called an -ampere.\" For if, in the above experiment, the cur- rent strength is i, ampere^ i = -1 = \u2014 * \u2014 - ; or il = 1. The V 1 V <piant ity nf current carried by a current of tj ampere flow- ing fort seconds is called (tj*) \"coulombs\"; and the capacity of a condenser, which when charged with,> coulombs is V 10 Rl 10 /i'j volts, is said to be -^ \"farads.\" A capacity of one millionth '1 is called a \"micro-farad.\" The definitions just given of the ohm and the volt are imt. strictly speaking, correct. As may be easily understood, the amplest way of defining a unit oi nee is to select some standard conductor and call its resistance the unit : hut. <iner there are grent advantages in using the 0. G. 8. sys- \u00ab\u00bbr (|iiantities which may he expressed in t. \u2022: it by ; lain number of factors 10, the h\u00ab-st manner of defining a unit is to select some condm tor whose resistance, when de- t'lmiiM'd as accurately as possible in termi of the (/. G. S. system, is simph --\\ pressed in terms..t n. and then to adopt the resistance of this conductor as the unit. Thus, the ohm i d. -lined to he t ipial to the resistance of a column of mereui\\ 730 KlJX'TltoDY \\AMirs at 0\u00b0 C., of uniform cross section, of length 106.3 cm. and having the mass 14.4521 g. (This column, then, has a cross section of almost exactly 1 sq. mm., accepting the usual value for the density of mercury at 0\u00b0 C.) The resistance of this column of mercury is equal to 109 C. (i. S. electro-magnetic units, to within the limits of accuracy of our present experimental methods. The volt is defined to be the E. M. F. which, steadily applied to a conductor whose resistance is 1 ohm, will pro- duce a current of 1 ampere. It is therefore practically equiv- alent to 108 C. G. S. electro-magnetic units. The E. M. F. of a certain cell, known as the \" Clark cell,\" which can be made in a definite manner, is found by careful experiments to be 1.4322 volts at 15\u00b0 C. The E. M. F. of the \" cadmium cell,\" which is another standard cell, is found to be 1.0186 at 20\u00b0 C. (The E. M. F. of the latter cell changes with the temperature much less than that of the former.) Heating Effect. \u2014 It was shown on page 664 that the heat energy developed in a conductor carrying a current of strength i in a period of time t was Hit, where E is the dif- ference of potential at the ends of the conductor considered. As was also noted, the number expressing this quantity -of heat is in ergs, if the C. G. S. electro-magnetic system is used. If the current is steady, this quantity may be expressed in other ways, for E \u2014 iR. So, writing W '= Eit, we have W= izRt = \u2014~. It is seen, then, that the heating effect R is independent of the direction of the current, because tin- square of the current enters the formula, and it has the samr value for either a plus or a minus sign. If the current is z\\ amperes and the resistance is R1 ohms, i = ^, and R = R1 109; so W= i?Rj 107 ergs. But 107 ergs equal 1 joule, and the power of a machine which does an amount of work of 1 joule per second is said to be 1 watt. LA II - ; AI>Y CURRENTS 731 I currents are measured in ampeivs and re.xistanees in ohms, tin- power of the current is i?Rl watts. Measurement of Resistance Absolutely. \u2014 By means of a Wheatstone bridge one may determine the ratio of the resist- ances of two conductors,", " but it does not furnish a method tor the measurement of a resistance directly. The formula just deduced for the heating effect of a current does, however, suggest a method. If a coil of wire is immersed in a calorimeter containing water, the heat produced in a given time by passing a current through it may be measured in calories, and since the mechanical equivalent of heat is known, the value in ergs may be deduced. The current strength may be measured by a galvanometer and, since thus TF, t, and t are known, the value of R may be determined ; for Ttr R = \u2014. (There are other methods whicli are more accurate.) Temperature Effect. \u2014 Experiments show that the resistance of a given eoml urt<>r varies with its temperature. This is what might be expected, be- no.\u2014 < mwwuring the hmtiaf effect of \u2022 current. \u2022\u2022 it was shown that the resista ne.- < \u2022!' any uniform con- \u2022 I net or of length / and of < ross section a could be expressed // = r, where c was a constant, characteristic of the con- a doctor. Hut. if the temperature of the conductor is altered, the motion and the distance apart of its molecules are ted: and BO ii is no longer the same snl\u00bbstaueo. It is found that U the temperature i> inen-asrd the resistance of all solid conductors \u2014 with one or two excepti 732 ELECTRODYNAMICS increases, while that of most liquid conductors decreases. (The effect in this last case is complicated by the fact that the extent of the dissociation produced in the act of solution varies with the temperature.) As the temperature of a pure solid is lowered toward absolute zero, its resistance almost vanishes. This change in resistance of a conductor with change in temperature offers at once a method of making a \"resistance thermometer \" ; and, in fact, a \" platinum thermometer \" con- sisting of a coil of platinum wire whose resistance can be measured is at the present time the most satisfactory ther- mometer in use for accurate work. Similarly, the same phenomenon is made use of in the \" bolometer,\" an instru- ment for the detection and the measurement of radiation in the form of ether waves. A strip of platinum is covered with lampblack, so that it absorbs as completely as possible all radiation that falls upon it, and is made to form one arm of a Wheatstone bridge. As ether waves are incident upon it, its resistance changes, and the amount of the change measures the intensity of the radiation. CHAPTER XLVIII INDUCED CURRENTS Tin: discovery by Oersted of the fact that an electric current produced a magnetic field, and the subsequent dis- covery of methods for making a bar of iron a magnet by means of a current, led many investigators to seek for means \u00ab \u00bbt' producing an electric current by means of a magnet. The method of doing this was discovered independently by Joseph Hi-nryin America and Michael Faraday in England about 1831. Experiments of Henry. \u2014 Henry's experiments were the \u2022 \u2022arlier. H\u00ab observed that, if a circuit in which there was a UUUUUUi Fio. 411. -A solenoid, Ultutntlog Henry'* fl\u00abt \u00abp\u00abrtm\u00abot rv \u00ab.f rells was lu\u00ab.k m at any point, there was a f spark; and, further, if the l>n-ak was made by means of the 788 734 EL ECTR ()1)Y. \\A.M /r.s hands, so that the circuit was completed by the arms and body, a shock was felt. He noticed, too, that both the effects were increased by increasing the length of the conductor and by coiling it up into a helix. There is thus an \" extra- current \" on breaking a circuit, in addition to the one due to the battery; and Henry's experiments prove that this cur- rent varies as the magnetic field of the original current ; for, if the conductor forms a helix, the magnetic field is much greater than if the conductor forms simply an approximately circular circuit. A few years later Henry observed that if a wire were wound around the soft iron armature of a horseshoe electro-magnet and if the current were suddenly broken, or if the armature were suddenly removed from the magnet, a shock would be felt, if the two ends of the wire were held in the hands ; or, if these ends were joined to a galvanometer, a sudden deflection of the needle would be produced, but the needle would return to its original posi- tion. The same effects are produced if LJI V FIG. 412.\u2014 Diagram rep- resenting Henry's second the current is again made or if", " the arma- ture, when separated from the magnet, is brought close to the magnet, but the current in the galva- nometer is in the opposite direction. The quantity of the current in the galvanometer, or the shock received by the arm, varies with the number of turns of wire on the arma- ture ; and the shock varies with the suddenness of the motion of the armature ; the current also varies with the material of the conductor, while the shock does not. It is evident that these \"induced\" currents, as they are called, are due to the change in the number of tubes of magnetic induc- tion which pass through the coil of wire wound on the armature. INDUCED CUi;i;i 73f> Experiments of Faraday. \u2014 Faraday's experiments were somewhat different, lie had two separate coils of \\\\irc wound on the same iron ring, one coil being joined to a cell, the other to a galvanometer ; and he observed that, if he broke the current or made it again, there was a sudden fling of the needle, but that the current was only a transient one. Here, again, the induced current is due evidently to the change in the number of tubes of magnetic induction through the circuit which is GAIVANOMFTCT joined to the galva- nometer. Faraday then showed by a series of most brilliant experi- ments that if the num- ber of tubes of magnetic induction inclosed by any closed conducting circuit is varied in any manner, e.g. by bring- ing up or removing a magnet or another cir- cuit carrying a current, there is an induced cur- ivnt, whose strength varies directly as the change in the number of tubes of magnetic induction and as the r.ite of this change, and also depends upon the material of the cir- cuit. If iron is inside the circuit, it is magnetized bv the current ; and thus the induction is changed. (It \\\\a>, in Fio. 418. - Faraday '\u00bb double coll in his first experiment. iiUr to this Study of induced currents that Farada\\ Was h-d to his Conception of tubes of induction and to the id.-. i of these tubes being continuous through a magnet. See page 61 M.in\\ years later Faraday rediaoovered tin- phenomena,,f the e\\t ra current on I,: i^'ed his apparatus as sho\\\\ n in Fi^. II I, where K i^ a cell, C is a helix 736 ELECTRODYNAMICS or electro-magnet, and A and B are the two ends of a broken wire in parallel with the helix. He observed that, if A and \u00a3 are held in the hands and the electrical current is broken at E^ a shock is felt. Similarly, if A and B are joined to a gal- vanometer, there is a sudden fling of the needle when the circuit is broken. Just before the current is broken, there is a magnetic field through the helix ; but, when the circuit is broken at E, there is still a closed circuit around the helix and through ^ 5 'dnd the magnetic field in thi\u00bb ^W de- creases, since the cell is out of circuit, and so there is no E. M. F. to maintain the current. Owing to the change in the number of tubes of induction in this cir- cuit there is the extra current. Law of Induced E. M. F. \u2014 All of the facts discovered by Henry and Faraday in regard to the strength of induced currents may be expressed by saying that, when the number of tubes of magnetic induction inclosed by a closed conduct- ing circuit is varied, there is an induced E. M. F. in this circuit whose value is proportional directly to the change in this number, and inversely to the time taken for the change. If there are n turns of the wire, as in a helix or coil, the tubes pass through each, and the induced E. M. F. is n times as great as if there was but one turn. Thus, calling the change in the number of tubes of magnetic induction A^V, and the time taken for this change A\u00a3, the induced E. M. F. during this change equals c-r\u2014, where c is a factor of pro- portionality. It may be proved by methods of the infinitesi- mal calculus that using the C. G. S. electro-magnetic system to express the E. M. F. and the C. G. S. definition of a unit pole as the source of a unit magnetic tube, this factor c has A7V the numerical value 1. Thus, we may write E = -\u2014-. INDUCED CURRENTS 737 If A* is the resistance of tin- circuit, t", "lic strength of the induced current is ', \u00ab>r i = \u2014 \u2014 \u2014. Hence the quantity of /i R A\u00a3 the < -in-rent in time A\u00a3, or i&t, is \u2022--\u2022\u2022 The current strength varies as the change is made, but the total induced quantity, or the summation of?Af during the entire change, equals the suiuinatioii of ~JT\\ that is, it equals the total change in -ZV divided by A*. It must be remembered that if the wire is coiled up, this quantity N varies directly as the number of turns. Thus the induced I-;. M. I-', varies with the sudden- ness of the change, while the induced quantity does not; the latter depends upon the resistance of the circuit, and upon the total change in the number of tubes; i.e. upon the field of magnetic force, and upon the area and number of turns of the coils. These facts are shown by Henry's and Faraday's experiments; because the shock received by one's arms is conditioned by the induced E. M. F., while the fling of the Lcalvanometer needle measures the induced quantity of the current. It must he particularly noted that there is an induced cur- rent in the ciivuit only so long as there is a change in the number of tubes of magnetic induction through the circuit. The direction of this current was investigated by both Henry and Faraday. Their conclusions may he expressed by saying that, if the change in tin- magnetic field through the eireuit is an ////\u2022/\u2022\u2022 <(*\u2022' in the number of tulx's of induction. the induced current is in such a direction as by its Own inai,'- netir lield to decrease the number : or, if the change in the lield is a decrease in the number \u00ab.f t ul\u00bb.-s. the induced current h a direction as to incre<i*> tin- number. In general. then, the induced em-rent produced by any change in the magnetic field throu-h i h a direction as to tend to neutralize this change. ( If this \\\\.-re n..t true, an increase in AMBS'8 FIIT\u00abICfl \u2014 47 738 ELECTRODYNAMICS the magnetic field would induce a current in such a direction as to increase the field still more ; this second increase would produce a second induced current in the same direction, etc. ; so conditions would be unstable.) If there is already a cur- rent flowing in the circuit, the induced current is superim- posed upon it, either increasing or decreasing it. Special Cases. \u2014 A few simple cases will be considered ; if a current is flowing in a circuit, and if a bar magnet is made to approach it or to recede from it, the direction of the induced current may be at once predicted. If the north pole of the magnet is nearest the south face of the circuit, some tubes due to the magnet pass out of the north face of the circuit. So, if the magnet is brought nearer the current, more tubes will pass through it, and the induced current will be in such a direction as to oppose this change; i.e. it will be in a direction opposite to that of the original current. Thus the current in the circuit is decreased as long as the magnet is approaching. This means, expressed in other lan- FIG. 415. \u2014 Diagram to illustrate induced guage, that WOrk is required to move the magnet, and since this is done by the current, only part of the energy of the cell is available for forcing the current around the circuit. Con- versely, if the magnet is withdrawn, the field of force through the circuit is decreased, and the induced current is in the same direction as is the original current. This means that work is being done by whatever agency moves the magnet ; and this work appears as an increased current. The case when the bar magnet is turned with its south face toward the south face of the circuit may be treated in a similar manner. Earth Inductor. \u2014 If a coil of wire is arranged so as to turn on an axis parallel to its plane faces, it may be so placed < r ///;/:. \\ ro 739 that this axis is vertical; and then, if the face of the coil is perpendicular t<> the magnetic meridian, it will include a tit-Id of magnetic force due to the earth. If the coil is turned into the magnetic meridian, t licit- is no field through the coil : and, if it is turned 90\u00b0 farther, the original field of force will pass through it, hut in the opposite direc- tion with reference to the coil. So it is just the saun- as if the coil had remained ionary and the field of force had changed from N to 0 to -N. The total change, then,", " is 2 N. If A is the area of the face of the coil, if there are n turns of wire in the coil, and if // is the horizontal component of the earth's magnetic Held, N= nAH. So, if the terminals of this coil are joined to a hallistic ^;i\\ \\ anometer, the (juantity of cur- rent measured when the coil just described is suddenly turned through 180\u00b0, from a position perpendicular to the R magnetic meridian, equals \" \u2014, where R is the resistance \u2022in- circuit. Similarly, if the coil is so turned that its axis of revolution is in the magnetic meridian and its faces an- hori/nntal. X=nAV* where V is the vertical component of the earth's magnetic force; and if the coil is turned on the axis through 180\u00b0, the (piantity of the induced current is -\u2022If V '\u2014= \u2014 Therefore the ratio of these two quantities equa 1 which is the tangent of the angle of dip. (Seepage 619.) Such an in^t i inimt ix.-ailed an \"earih inductor.\" It was invented by the great German physicist, Weber. 740 ELECT ROD YN AM n.s Induction Coil. \u2014 A case of special interest is one studied by both Henry and Faraday : a coil of wire is wound on a spool or cylinder of such a size that it will slip inside another Fm. 417. \u2014 Methods for production of induced currents. cylinder, on which is wound another coil. Call the first coil J., the second B. If the terminals of A are joined to a cell, and those of B are connected to a ballistic galvanometer, no 711 current will flow in the hitter until tin- current in.4 is varied; but. it' this is <h>nc. a current is induced. If there is a rod of soft iron inside A. the induced currents are greatly increased when the current is made or broken; because when the cur- rent is flowing, this rod is magnetized, and so the magnetic field is increased. If a magnet is brought near this iron rod or is taken away, there are also induced currents. Similarly, if the terminals of B are held in the hand, a shock is felt when the current in A is made or broken ; and this shock is increased by inserting a piece of iron in A. The shock on breaking the circuit is greater than on making, because the time taken for the change in the magnetic field is less in the former ease, and so the induced E. M. F. is greater. The induced nuantitii is the same in both cases. If a thin copper (or conducting ) cylindrical tube is interposed between 1 /A it prevents almost completely the shock felt at the terminals of B, but does not affect the quantity of current as shown by a ballistic galvanometer. The reason is that, as the current in.1 is changed, electro-motive forces are in- duced both in B and in the copper tube; and so, as the (in rent in the tube changes, it also induces an E. M. K. in J9; these two induced electro-motiye forces are in opposite directions and so the resultant effect is small. If the current in A is large, and there are enough turns of wire on B, an E. M.!\u2022'. may be induced in B. when the current in.1 i- broken, snllicicnt t o spark across considerable distances in case the terminals of B are separated. This ui.il. rdinarj \u2022\u2022induction c\u00bb.il,\" as shown in Fig. 418. I [| a mechanical arrangement for automatically break- md making the circuit in A. \\\\hose principle is evident. The iron core of >iidi a coil is alwa\\s made of iron wires insulated from each other and is not a solid rod, because as the current in.1 d. induced currents would be pro- duced in i iron rod in circles around the axis of the rod, and thoe arc piv\\ciitc\u00abl by the division of the rod into 742 ELECT R OD YXA MICK wires. (These currents produced in a solid core are called \"eddy,\" or Foucault currents.) A condenser is always in- troduced in the battery circuit in parallel with the \" primary \" coil A. One of its chief functions is to prevent sparking FIG. 418. \u2014 Diagram of induction coil. at the points where the circuit is broken ; it does this by diverting the extra current in the primary from the two points where the circuit is broken into the two plates of the condenser. (In other words, to produce a spark, a definite potential difference is re- quired, depending upon the distance ; and the differ-", " ence of potential of the two plates of the condenser does not rise sufficiently high to allow a spark to pass, provided the capacity is great; for Vl\u2014V^ = ^> See page 652.) Thus, if the extra current on breaking the primary circuit is prevented, the change in the field through the \" secondary \" circuit B is very sudden, and the induced E. M. F. is intense. FKJ. 418 \u00ab. \u2014 Induction coll.., -,,., i\\i>i < /\u2022:/> CUBES* 743 When the current in the primary is sixain made, the change in the magnetic field is comparatively slow, and so the induced E. M. F. in the secondary is not great Self and Mutual Induction. \u2014 1. Self-induction. If a cur- rent is Mowing in a circuit, it has a field of force of its own : if / is the current strength, the number of tubes of magnetic induction which thread this current is proportional to it and may he written LL where A is a constant for the given con- ductor and for a given medium surrounding it. L is called the \"coefficient of self-induction\" or the \"inductance.\" It is evident that L -varies directly as /A, the permeability of the medium ; for the magnetic induction equals p times the mag- netic force, and the latter depends simply upon the current and the shape of the conducting circuit. (Thus the effect of introducing an iron rod into the circuit is explained. In the case of iron it must be remembered that /u, is not a constant, lor it depends upon the intensity of magnetization. So L is not a constant unless the medium is kept the same.) 1 Hither, L must increase as the area of the circuit in- creases, because the circuit will include more tubes. In tin- case of a solenoid which has N turns per unit length, the magnetic force inside has been shown to be 4 irNi ; therefore, it I is the length of the solenoid, each tube of force passes through the current Nl times; and, it.1 is the area of the cross section of the solenoid, L = 4 7rjV-/M it' the medium is air, and equal> 4 Tr^N^lA in general. Therefore, if the em-rent is varied in any \\\\a\\...;/. by alter- ing the K. M. V. of the cell, there will be an induced I-;. M. V. whose value equals the rate of cha ii-'..l A/: and the greater /. i, so moch the greater is the indaoed K..M.K. The induced quantity of \u00ab urrent equals the total change in the number of tubes of induction divided l>\\ the resistance of the circuit. If the applied K. M. V. is 1 1 1 ; i 1. so as to tend to increase the current, and thu^ Increase the li\u00ab-M. l he induced '\u2022nrr. -MI must he in the opposite direction ; and as a c<- 744 ELECTRODYNAMICS quence the current does not rise instantly to the value corre- sponding to the applied E. M. F. Similarly, if the applied K. M. F. is decreased, the induced current is in the direction of the original current ; and so this does not decrease instantly to its final steady value. These facts may be expressed in a formula by writing the induced E. M. K. =, where the minus sign means that if AJVis positive, the induced E. M. F. is negative, while if AiV is negative, it is positive. Particular cases of these changes are -when the circuit is suddenly broken and when it is suddenly made, e.g. by removing one of the electrodes from the cell and by then plunging it in. In the former case, the current does not instantly fall to zero ; there is the extra current, as shown by the spark, etc., as observed by Henry and Faraday. In the latter case, the current does not rise instantly to its fixed value. The time taken for these changes evidently varies directly as L ; so that L measures what may be called the 44 inertia of the current.\" When the circuit is broken, the energy of the magnetic field is no longer maintained by the cell, and it returns into the conductor, continuing the current until all the energy is consumed in heating the conductor. Then the current ceases. Similarly, when the circuit is closed, part of the energy fur- nished by the cell is spent in producing the magnetic field, and only a portion of it is available for producing the cur- rent in the conductor. It is not until the magnetic field is established, then, that all the energy supplied", " by the cell goes into maintaining the current. As a consequence it takes time to produce a steady current. These intervals of time required for a current to come practically to rest when a circuit is broken, or to be produced, are, as a rule, extremely short, a few milliontlis of a second; but if the circuit has ti large value of L, e.g. if it is in the form of a D rr/;/;/-;.vrs 71-\". solenoid inclosing a rod of iron, the time may be as great second. The energy of the current, i.e. of the magnetic field due to it, is thus in the surrounding medium. 2. Mutual induction. \u2014 If a circuit carrying a current is near another circuit also carrying a current, some of the tubes of force due to each current will pass through the other riivuit. Thus if the currents in the two circuits are tj and /2. the number of tubes due to the first current which pass through the circuit carrying the second one may be written Mfa. (It must be noted that if the second circuit has n turns, the tubes pass through its current n times.) Simi- larly, the number of tubes due to the second current that pass through the first may be written Mj,v It may be proved by the infinitesimal calculus that Afj = Mv and that this quantity is a constant for the two circuits, depending upon their shape, size, number of turns, and relative positions, and also upon the permeability of the surrounding medium ; it :lled the coefficient of \"mutual induction,\" or the \"mu- tual inductance.\" (The unit of induction is called the \u2022\u2022 Henry.\" It is the induction which a coil has if it is of such a size and shape that a variation in it of a current at the rate of one ampere per second produces an induced E. M. 1. of one volt.) Tin- value of M may be calculated provi<l\u00bbl </// th,- till,-* due to one current pass thron<//t (ill the coils of the other l> This condition may be secured practically in two wa'1'I' IM. ing one helix inclose the other and making them nearly as possible of the same length and < TOSS section, and then 111! n><ll|rlll'_r,, lnllLr,.,\u201e!,,(' jm,, ; iniliiig the t\\v<> helices on a rin^ of iron. ;is shown in the cut : for iron lias such a great |>erme;il\u00bbilit v,((mp;uvd with air....... Fie. 419.\u2014 Dteffnun of ttmntibnntr that all the til!'ir. illy May, colU wound on. 746 ELECTRODYNAMICS in the iron and do not escape into the air. The former piece of apparatus constitutes an \" induction coil \" and has already been described ; the latter is called a \" transformer \" and is used for many practical com- mercial purposes. FIG. 419 a. \u2014 Section of a com- In the case of an induction coil, the intensity of the magnetic field at a point inside the primary far from the ends owing to a current iv is 4 TrNfa, where -ZV^ is the number of turns per unit length. If there are N2 turns per unit length in the secondary, there are N2l turns in a length I ; and if A is the area of the cross section of the primary, 4 irfjbNlN,2lAil tubes of induction thread these coils ; so for a length I of the induction coil \u2014 near its middle part \u2014 M = 4 jr^^N^A. It is thus seen how the induced E. M. F. in the secondary is increased by introducing an iron core \u2014 so as to have p large, and also by having N^ and N^ large. mercial transformer. Transformer. \u2014 In the case of the transformer the case is somewhat different, owing to the fact that it is assumed that none of the tubes escape into the air, and an exact calculation can be made of the effect of all the coils, not simply of those near the middle. If n^ is the total number of turns in the primary, and n2 that in the secondary, the coefficient of self-induction of the primary is proportional to nf ; and the coefficient of mutual induction of the two coils is proportional to n$iv So, if the current in the primary suddenly ceases, or if it is reversed in direction, the E. M. F. induced in the primary is proportional to nf, and that induced in the sec- ondary is proportional to n^ny As will be shown in", " armature makes ra revolutions per minute, the E. M. F. will be reversed nm times each minute. On the armature shaft there are two conducting rings, which are insulated from each other, and to which are joined the terminals of the wire wound on the armature. On them rest the two brushes joined to the external circuit ; so an alternating E. M. F. is applied to it. The pole pieces of the dynamo are magnetized by a separate direct-current dynamo. Oscillatory Discharge of a Condenser. \u2014 It has been stated in the description of the discharge of a condenser that under certain conditions it is oscillatory. The reasons for this may now be discussed more fully. When the condenser is charged, the surrounding medium has a certain amount of electrostatic energy. If its two plates are joined by a conductor, a cur- rent will flow in it, and thus, if the resistance is small, so that the energy is not in the main spent in heating the con- ductor, a considerable amount of it will be consumed in pro- ducing a magnetic field. When all the electrostatic energy is thus exhausted, the electro-magnetic energy will flow back into the conductor, continuing the current in the same direc- tion, and thus charging the condenser plates again, but in the opposite manner to their original charges. Finally, all the energy which has not gone into heating the conductor or to producing waves in the ether is again in the form of electrostatic energy ; and the process is repeated in the oppo- site direction ; etc. After a certain number of oscillations the energy is all exhausted, and everything comes to rest. It is evident that the greater the self-induction of the con- ductor joining the condenser plates, so much the more energy goes into the magnetic field for a given current. IMil'CED CUR If I A useful mechanical analogy is furnished by the vibrations spiral spring carrying a heavy body. When the sj st stretched, the energy is entirely poten- tial ; then when the body passes through its position of equilibrium, the energy is entirely kinetic; as the body moves np, the spring '\u2022d, the strain being op| what it was originally, and the energ again potential; etc. Further, the greater the mass of the moving body, so much the greater is the kinetic energy for a given velocity. As the vibrations continue, the amplitudes decrease gradually. owin^r t\" l< of energy by friction, and finally the motion ceases. The period of the motion is evi- dently increased if the mass is increased, and is decreased if the stiffness of the sprii increased. In the electrical oscillation, then, electro- static energy corresponds to potential energy. and electro-magnetic to kinetic. The potential Fio. 4\u00bb4.-Vibrmtlnir \u2022plrd.print. charged condenser is and for a given charge is greater if C is small ; so - corre- C spends to the stiffness of the spring. The self-induction of the elertrical conduct.. r corresponds to the ii; the vibrating matter. By means of the infinitesimal calculus it maybe shown that if the resistance of the conduct' small, the period of oscillation of tl irge is given the formula T l-rrVEG; so, if L is large, the period is great; and. t f is large, the period is small, other things _T equal. \u2014 48 AMKft'8 754 ELECTROD YNAMICS Electrical Waves along Wires. \u2014 This mechanical analogy is illustrated still further by the phenomena of electrical waves along wires. If the potential of one end of a long conductor is first raised, then lowered, raised again, etc., or, in other words, if it varies periodically, it is found that the potential at all points along the conductor is not the same at any one instant, and that there is a train of waves of elec- trical potential passing down it. This is the process by which messages are transmitted across submarine cables, and by which telephone messages are sent. The velocity of these waves can be measured, and the wave length also. Experi- ence and theory both agree in showing that the velocity of short waves along wires is the same as the velocity of light in the free ether, viz., 3 x 1010 cm. per second. These waves may be compared with ordinary mechanical waves along a stretched string or rope. The self-induction of a unit length of the conductor corresponds to the mass per unit length of the string ; the reciprocal of the capacity per unit length of the conductor, to the stiffness of the string ; the resistance of the conductor, to the internal friction of the string. Waves of all lengths decrease in amplitude as they advance along the conductor ; but the long waves decrease least ; so as a complex train of waves advances, it", " becomes more and more simple, because its shorter components vanish. If the self-induction of the conductor is increased sufficiently, not alone is the attenuation of all waves decreased, but it is the same for waves of all wave lengths ; thus there is no distor- tion of the waves. In Pupin's system of constructing tele- phone lines, this condition is secured by inserting in the line, at regular intervals of every few miles, a helix whose self- induction is large. By this means it is possible to telephone with distinctness over intervals of a thousand miles or more. Electrical Waves in the Ether. \u2014 Electrical oscillations pro- duce waves in the surrounding ether, as has been already i.\\in < i:i> < n;i;i-:\\ i s stated. These waves are identical in their properties with those which produce the sensation of light; and there are many ways by which they may be detected. They will pro- duce oscillations in other conductors : and these may be made evident by sparks or by the heating effects. Again, if a n umber of small metallic particles are put loosely together, hey offer a great resistance to a current : l\u00bbnt, when these long ether waves fall upon them, they cohere in such a manner as to have comparatively Mnall resistance. It the particles are jarred slightly, after the v as, their resistance again increases. Such an apparatus is.ailed a \"coherer.\" It is evident that by introducing wires int- two ends of a coherer, and putting it in circuit with a battery and a Lralvanoscope, or a telegraph sounder, one can observe the passage of these \" electric waves.\" There are many m\u00ab >i \u2022\u2022\u2022 methods, descriptions of which may be found in advanced text-books. The papers of Henry and Faraday, on the subject of induced currents, have been reprinted in the Seiei. liemoin - Vols. Xi and Xll, New York, CHAPTER XLIX OTHER ELECTRICAL PHENOMENA THERE are several phenomena, showing the connection between electricity and light, which should be mentioned. Faraday Effect. \u2014 The fact discovered by Faraday that, if a beam of plane polarized light is passed through a strong magnetic field, parallel to the lines of force, the plane of po- larization is rotated has already been discussed. (See page 564.) The direction of rotation follows the right-handed- screw law ; so that, if the field of force is inside a solenoid, the direction of rotation is that of the current flowing in the coils. This rotation leads one to believe that associated with a line of magnetic force there is a rotational motion in the ether ; so that any vibration in the ether at right angles to the line of force will have its direction changed. The energy of the magnetic field of force should then be considered as due to this motion ; and it is thus seen why it is kinetic. Hall Effect. \u2014 This rotational action of a magnetic 'field is shown also in what is called the \" Hall effect,\" a phenome- non discovered in 1879 by E. H. Hall, now of Harvard Uni- versity. If an electric current flows through a thin metallic film, it will so distribute itself that corresponding to any point on one edge of the film there is another on the op- posite edge which has the same potential. Let A and B, in the cut, be two corresponding points ; if they are joined by a wire which includes a galvanometer, no current will flow in it. If now this film is placed between the poles of a mag- net, so that a magnetic field is produced perpendicular to the current sheet, a current will flow from A to B through the 766 KLECTRICAL 757 galvanometer, showing that they are no longer at the same potential. This proves that the lines of flow of the m. Fio. 488. \u2014 Lines of flow In a thin metal Htrip. Dotted lines are lines of constant potential have been rotated by the magnetic field. \\\\\\ moving one terminal of the wire slightly along the edge of the film, another point may be found fur which there is again no current. Before the magnetic field is produced, the lines of tlo the current are as shown in the cut ; and lines of constant ;itial, at right angles to these, are r A also shown by dotted lines. After tin- field is produ these lines are all rotated, as shown. The direction of the rotation is different in different films, and is reversed if FM. 416. - Dtacrmn ahowlor Hall field or the cu s reversed. \u2022i may be described in a different way. Before Mtroduced into the magnetic field, the lin.,,f the electric force may be represented by ~FQ\\ after the fiel produced it U Lfiven by FQV if the r< the", ". Boys, rmdlomlcrometer, 184, 719. Mr:,.-,'. M.,, -t,,,,.!,..!..,!.,-!. r. Hi BradleyVmetiiod of determining veJodty of!.\u2022\u00ab at friction, 190. \u2022\u2022: \u2022i it, MT, \u2022\"!. ilorto, Ml,. Ml CUortaMtar, 151... 719. ktBsM \\t, ;v:; Atomic WtljM, Mt MB \"f radlatkw, 801 m Capillan \u00ab\u00bb\u00bbH - hi \u2022 i i 9 - : 762 INDEX Cathode rays, 701. Cation. 68t. ( austic, by reflection, 451 ; by refraction, 465. Cavendish, gravitation experiment, 183 ; lu\\v <>f electric force, 64U ; measurement of capacity, 660. Cells, primary, 676; secondary, 698. Centigrade scale of temperature, 222. Centimetre. '-'I. Centre, of gravity, 99, 134; of mass, 61, 80; of pressure, 174. Centrifugal force, 72. Charge, electrical, 638 ; magnetic, 607. Charles' law, 240. Chemical equivalent, 686 ; electro-, 688. Chladni's figures, 861. Chromatic aberration, 489. Circle, uniform motion in a, 40, 47. Clark cell, standard of E. M. F., 730. Clinical thermometer, 225. Clock, pendulum, 186. Clouds, formation of, 267. Coefficient of expansion, 280, 234, 287. Collimator, 505. Colloids, 141. Color, 586. Color blindness, 592. Color sensation, 592. Colors, absorption, 587 ; complementary, 524, 587 ; connection between wave number and, 424 ; mixtures of, 587 ; of thin plates, 528 ; polarization, 557 ; surface, 588. Combination, chemical, heat of, 285 ; of lenses, 487 ; of notes, 412. Combustion, 285. Commutator of dynamo and motor, 748. Compensated pendulum, 232. Complex pendulum, 322. Complex vibrations, 320. Complex waves, 347, 754. Composition, of a uniform acceleration and a uniform velocity, 45 ; of displacements, 36 ; of forces, 60, 75 ; of harmonic vibrations, 318, 822; of moments, 91 ; of velocities, 87, 53. Compound microscope, 504. Compound pendulum, 185. Compressed glass, double refraction in, 542. Compressibility, of a gas, 192 ; of a liquid, 171 ; of a solid, 150. Concave grating, 537. Concave mirror, 447. Condensation of vapors, 279. Condensers, electric, 650 ; capacity of, 652, 660 ; discharge of, 654, 752 ; energy of, 656 ; sec- ondary discharge of, 662. Conduction, electric, 668 ; of heat, 227, 287. Conservation, of electricity, 638; of energy. 110, 115, 218, 810; of linear momentum, 78; of mass, 65. Conservative forces, 107. Consonance, 414. Contact, difference of potential, 681 ; electrifi- cation, 625. Continuity of matter, 282. Convection, electric currents, 706 ; of heat, 227, 286. Converging lenses, 470, 475. Convex mirror, 452. Cooling, Newton's law of, 297. Coplanar forces, 96. Cords, vibrations of stretched, 851, 400 ; vocal, 407. Coulomb (unit of electricity), 729. Coulomb's law of electrostatic force, 640. Coulomb's law of magnetic force, 608. Couple, 100 ; theruio-, 294, G7'J. Critical angle, l.\">r,. Critical temperature, state, etc., 278. Crookcs, fourth state of matter, 208. Crookes, radiometer, 204. Cross hairs of telescope, 506. Cryohydrates, '^t>2. Cryopnorus, 270. Crystalloids, 141. Crystals, biaxal and uniaxal, 542; expansion of, 281. Current, electric, 663. d'Alembert, laws of mechanics, 188. Dalton's law of mixtures of gases, 190, 195. Damping of vibrations, 828. Daniell's cell, CMN d'Arsonval galvanometer, 718. da Vinci, lever, 123, 187. Davy's experiment on the nature of heat, 309.", ". 226, 804 ; of evaporation, 269 ; of fusion, 260 ; of solution, 283 ; sources of, 215 ; specific, 250 ; transfer of, 227, 286. Heat effects, 216, 303. Heat energy, 219, 226. Heating effect of electric currents, 665, 730. Helmholtz, analysis of sounds, 396 ; explanation Henry, electrical waves, 656 ; induced electric of harmony, 414. currents, 738. Henry, the, unit of inductance, 745. Herschel, fluorescence, 578 ; infra-red radiation, 310. measurement of, 613. HerschePs telescope, 497. Homocentric pencils, 435. Homogeneous waves, 428. Hooke's law, 145. Horizontal intensity of earth's magnetic field, Horn blower, steam engine, 275. Horns, 858, 404. Horse power, 115. Humidity of air, 268. Huygens, clock, 186, 138 ; experiment in double refraction, 547 ; eyepiece, 494 ; fixed points of temperature, 226; impact experiments, 156; principle, 866; reversible pendulum, 136; theory of reflection and refraction, 307, 871 ; variation in \"g,\" 181 ; wave surface for Iceland spar, 548. Hydraulic ram, 206. Hydrostatic press, 163. Ice, lowering of melting point of, by pressure, 258. Ice calorimeter, 250, 252. 789. Iceland spar, 541. linage.-, real and virtual, 435. Impact, 7*. l.V>. Impulse, C.'.t, 106. Incandescent electric lamp, 666. Incidence, angle and plane of, 869, 481. Inclination, earth's magnetism, 618, 620, 622, Inclined plane, 42, 119, 126. Independence o!' forces, principle of, 60. Index of refraction, 482, 455, 457 ; measure- ment of. 4;.C., 4\u00bbSO, 465, 506. Indicator diagram, nil, 274. Induced electric currents, 738. Induction, electro-magnetic, 707 ; electrostatic, Induction coil. 741. Inertia, 18, 14, 15; moment of, 88; principle 631, 645 ; magnetic, 597, 614. of, 59. Infra-red radiation, 293, 424. Insulators, 626. Intensity, of electric field, 641 ; of light, 487 ; of magnetic field, 609 ; of magnetization, 610 ; of sound, 397 ; of waves, 814, 828, 330. Interference of light waves, 374, 424, 519 ; of sound waves, 374 ; of waves on surface of liquid, 374 ; over long paths, 527. Internal energy, 218, \u2022-'\u00bb:;. Internal work done when a gas expands, 248, 248, 253. Interval, musical, 416. Inverse square, law of, 180, 608, 640. Inversion, thermoelectric, 680. lonization of gases, 699. Ions, 689, 692. Isoclinal lines, 621, 628. Isogonal lines, 620, 622. Isothermals, 195, 267, 276. Jar, Leyden, 653. Jets of liquid, 189. Joly, steam calorimeter, 252. Joule, determination of mechanical equivalent, 303, 804, 810 ; internal work in a gas, '243. Joule, the unit of energy, 112. Jupiter, occultation of satellites, 421, 5G6, 574. Kaleidoscope, 446. Kathode (see Cathode). Kelvin, absolute scale of temperature, 307; electrometers, 658; expansion of gases through porous plugs, 243 ; thermoelectric- ity, 682. Kepler's laws, 132. Kerr effect, 565. Kinetic theory, of evaporation, 143, 264; of gases, 197, 241, 254; of matter, 142; of vis- KirChhofFs law of radiation and absorption, 301, cosity, 148, 202. 579. Knot, 37. Koenip, manometric flame, 858. Kundt, measurement of velocity of sound, 864. M& Lamp, arc, 665; incandescent, 666; Nernst, Langley, bolometer, 294, 732. Laplace, velocity of sound, 888. Latent heat, 809. Leibnitz, laws of mechanics, 188. Lenard rays, 708. Length, units of, 21. INDEX Lenses, 461, 467 ; achromatic, 498 ; combination ^7; converging", "; complex, 322; compound, 185 ; magnetic, 610 ; reversible, 136; simple, 74, 91. Penumbra, 426. Period of vibration, 48, 817 ; measurement of, 322. (jnograph, 396, 407. Permeability, magnetic, 609. Phase of simple harmonic motion, 49. \u2014 logiston, 309. Phosphorescence, 299, 578, 588. Photo-electric action, 699. Photographic lens, 500. Photometer, 442. Photometry, 437. Physical quantities, 10 et seq. Physics, 7, 9 ; divisions of, 10. Piezometer, 150. Pigment colors, 587. Pile of plates, 554. Pin-hole images, 426, 495. Pitch, of a screw, 127 ; of sounds, 397 ; stand- ard, 418. Plane of incidence, 431 ; of polarization, 554. Plasticity, 17. Plate, refraction through, 458, 466. Plates, colors of thin, 523 ; vibrations of metal, 360. Platinum thermometer, 223. Points, effect of, on electric charge, 633. Polarization, angle of, 552 ; by double refrac- tion, 547; by reflection, 551, 554; colors due to, 557 ; plane of, 554 ; rotation of, 563, 750. Polarized waves, circularly and elliptically, 818, 560, 562; interference of, 556; plane, 313, 555. Polarizer, 560. Porosity, 18. Potential, electric, 642. Potential energy, 109, 117; and force, 114. Pound, \u00ab'\u2022('\u2022. Power, 115. Practical system of electrical units, 728. Pressure, 158; atmospheric, 166, 176; centre of, 174 ; unit of, 165 ; in a bubble or drop, 185 ; in a gas, 158, 198. Pressure in a gas, measurement of, 178, 196. Pressure in liquids, 158; due to cohesion, 162; due to gravity, 163; due to surface-tension, 185. Prevost's theory of exchanges, 297. Primary colors,\" fi'.i''. Primary electric cell, 676; energy of, 697. Principal axes of rotation, 94. Principal section of a crystal, 548. Prism, angle, edge, face, 458, 466; resolving power of, 510. Projectile, 45. Projection lantern, 501. Projection of a line or area, 28. Pulley, 128. Pulses, 342, 348, 394, 409, 425, 704. Pump, air, -joy ; water, 208. Quadrant electrometer, 658. Quality of sound, 895. Quantity of heat, 226, 809. Quarter- wave-plate, 561. Radian, 24. Radiant energy, 290. Radiation, 290 ; and absorption, 298, 576. Radioactive bodies, 7'H. Radiometer, Crookes', 204, 294. Rainbow, 516. Ram, hydraulic, 206. Ramsden eyepiece, 494. Ratio of specific heatsof a gas, 253, 254, 337. Ray of light, 881, 425. Reaumur temperature scale, 223. Rectilinear motion, 43, 70. Rectilinear propagation of light, 381, 425. Reed pipe, 403. Reflection, 383, 367 ; angle of, 369, 431 ; of,'8b <. ether waves, 867, 429, 430. Reflection of light, concave surface, 447 ; con-. vex surfacju.452 ; diffuse, 429; fine particles, regular, 430 ; total, 428, 456. Reflection of sound, 367, 408; diffuse, \u2022Ir- regular. 430 ; total, 428, 456. Refraction, angle of, 373, 432; double, 4f><> ; in- dex of, 432, 455, 457, 460, 506. Refraction of light, concave surface, 467 ; con- vex surface, It'i!) ; plane surface, 371, 432, 455, 462 ; plate, 458, 466 ; prism, 458, 466. Refraction of sound, 408. Relay, 715. Residual electric charges, 662. Resistance, electric, 723, 726, 781 ; standards of, 729. Resolution offerees, 75. Resolving power, of grating, 535; of lens, 483 ; of prism, 510. Resonance, 292, 328. Resonator, Helmholtz, 396. Restitution, coefficient of, 144. Resultant, non-parallel forces, 96; parallel forces, 98. Reversible engine, 300. Reynolds and Moorby, mechanical", "11) (2.12) to get an expression for displacement in terms of initial and \ufb01nal velocities. We can get an even more useful relationship by eliminating the \ufb01nal velocity. If we use Eq. (2.8) to substitute for the \ufb01nal velocity, \u2206x = 1 2 (v0 + (v0 + at))t \u2206x = v0t + 1 2 at2. (2.13) This expression is often handy because it does not contain the \ufb01nal velocity. In many cases, we have information about the start of motion, but we rarely have information about the end of it (that\u2019s usually what we are trying to predict). We can get another useful relationship by solving Eq. (2.8) for time, t = v \u2212 v0 a, (2.14) and substituting into Eq. (2.12), v \u2212 v0 a \u2206x = \u2206x = (v0 + v) 1 2 v2 \u2212 v2 0 2a v2 = v2 0 + 2a\u2206x. This equation is very handy if information about time is not given in the problem; it still allows us to calculate the \ufb01nal velocity without knowing how long the object was accelerating. 19 Example: Car Chase A car traveling at a constant speed of 24.0 m/s passes a trooped hidden behind a billboard. One second after the speeding car passes the billboard, the trooper sets o\ufb00 in chase with a constant acceleration of 3.00 m/s2. (a) How long does it take the trooper to overtake the speeding car? (b) How fast is the trooper going at that time? Solution: We \ufb01rst need to be clear about how we are measuring time and position in this problem. Let\u2019s measure them both relative to the trooper\u2019s motion. That is, x = 0 at the billboard where the trooper starts moving and t = 0 when the trooper starts moving. This means that the speeding car passed the trooper at t = \u22121 s. (a) Now we need to think about what the problem is asking us to \ufb01nd. It asks for a time (\u201chow long\u201d) at which the trooper overtakes the speeding car. At the exact moment that the trooper overtakes the car, their positions on our axis will be equal, i.e. xt = xc. So, we want to \ufb01nd the time at which the positions are equal. We have just derived an equation which gives position as a function of time for constant acceleration, \u2206x = v0t + 1 2 at2. First check that we can use this equation; are both objects undergoing constant acceleration? Yes, the problem says the trooper has a constant acceleration and the car has constant velocity (a constant acceleration of 0). So let\u2019s \ufb01gure out the position of the trooper. The trooper\u2019s initial position is x0t = 0, his initial velocity is v0t = 0, and his acceleration is a = 3.00 m/s2, so we have xt = 1 2 at2. Now we do the same for the car. It\u2019s initial position is x0c = 24.0 m (it had a 1 s head start), it\u2019s initial velocity is v0c = 24.0 m/s, and it\u2019s acceleration is ac = 0, so we have xc \u2212 x0c = v0ct xc = x0c + v0ct. To \ufb01nd the time at which positions are the same, we set the two expressions equal and solve for time, xt = xc at2 = x0c + v0ct 1 2 at2 \u2212 v0ct \u2212 x0c = 0. 1 2 You get a quadratic expression for t, so you will need to use the quadratic formula (it\u2019s in your textbook), t = t = v0c \u00b1 24.0 m/s \u00b1 2 a (\u2212x0c) 0c \u2212 4 1 v2 2 a 2 1 (24.0 m/s)2 \u2212 4 1 2 1 2 \u00b7 3.00 m/s2 2 \u00b7 3.00 m/s2 (\u221224.0 m) t = 16.9 s. There is also a negative root, but since we know that the trooper could not overtake the car before it passed him, this root is not physically meaningful. 20 (b) The trooper\u2019s speed at that time is a fairly straightforward problem. We have an equation that gives us \ufb01nal velocity if we know the acceleration, time and initial velocity (and we know all those now). v = v0 + at v = (0) +", " 3.00 m/s2(16.9 s) v = 50.7 m/s. As mentioned earlier, constant acceleration is particularly useful because objects moving due to the gravitational pull of an object will move with a constant acceleration if air resistance is neglected. Objects moving under the in\ufb02uence of gravity and without air resistance are said to be in free fall. Note that this does not mean that the object necessarily started from rest (v0 = 0). Objects can be moving upward, like when you throw a ball, and still be considered free-falling. The magnitude of the free-fall acceleration is denoted by g and has a value of 9.80 m/s2 on Earth (although it varies slightly depending on latitude). The direction of g is always towards the large object creating the gravitational pull. Because g is a constant, we can use all of the equations we just derived for constant acceleration. Example: Rookie Throw A ball is thrown from the top of a building with an initial velocity of 20.0 m/s straight upward, at an initial height of 50.0 m above the ground. The ball just misses the edge of the roof on its way down. Determine (a) the time needed for the ball to reach its maximum height, (b) the maximum height, (c) the time needed for the ball to return to the height from which it was thrown and the velocity of the ball at that instant, (d) the time needed for the ball to reach the ground. Neglect air drag. Solution: You can refer to the \ufb01gure in your textbook (Fig. 2.20) to get a visual idea of what\u2019s going on in the problem. We are expressly given an initial velocity and an initial height in the problem. There are a few other pieces of information that are not expressly stated in the problem. The ball is in free fall, so we know the acceleration, g = 9.80 m/s2, and we know that the velocity of the ball at it\u2019s maximum height will be zero. Let\u2019s set up our coordinate system so that y = 0 corresponds to the top of the building; this means that the bottom of the building is at -50.0 m. (a) The \ufb01rst part of the problem is concerned with the upward motion of the ball. We are given an initial velocity, we know acceleration and a \ufb01nal velocity, and we want time. We have an equation that contains all these quantities, v = v0 + at v \u2212 v0 a t = t = 0 \u2212 20.0 m/s \u22129.80 m/s2 t = 2.04 s. 21 (b) Now we want to know the maximum height reached by the ball. We have all the information from before and we now have the time at which we reach this maximum height, so we can use y = y0 + v0t + 1 2 at2 ymax = (0) + (20.0 m/s)(2.04 s) + ymax = 20.4 m. 1 2 (\u22129.80 m/s2)(2.04 s)2 (c) Now the ball begins to move downward and we want to know how long it takes to get back to it\u2019s initial height. We can use the same distance equation, but now use the fact that yf = 0 to solve for time. y = y0 + v0t + 1 2 at2 v0t + t(v0 + 1 2 1 2 at2 = 0 at) = 0. We will have two roots, t = 0 because the ball starts at that height, and the one we really want, v0 + 1 2 at = 0 t = \u2212 t = \u2212 2v0 a 2(20.0 m/s) \u22129.80 m/s2 t = 4.08 s. Note that this is twice the time it takes to get to the maximum height. The velocity of the ball at that time is v = v0 + at v = 20.0 m/s + (\u22129.80 m/s2)(4.08 s) v = \u221220.0 m/s. (d) Now our \ufb01nal position is yf = \u221250.0 m, but this is essentially the same problem we just solved. y = y0 + v0t + 1 2 at2 1 2 at2 + v0t \u2212 yf = 0. This leads to a quadratic equation, so it\u2019s a little more e\ufb00ort to solve, t = t = v0 \u00b1 20.0 m/s \u00b1 2 a (\u2212yf ) 0 \u2212 4 1 v2 2 a 2 1 (20.0 m/s).80 m/s2 2 \u00b7 9.80 m/s2 (\u2212(\u221250.", "0 m)) t = 5.83 s. 22 Chapter 3 Vectors and Two-Dimensional Motion We live in a three-dimensional world and the objects around us move in that three-dimensional space. While some simple examples of motion can be described in one dimension, extend what we\u2019ve learned about motion to more than one dimension will allow us to study many more systems. 3.1 Vector properties Displacement, velocity and acceleration are all vector quantities. That is, they have a magnitude and a direction. In one dimension, there were two options for the direction, left or right, and we could represent the direction of displacement, velocity, or acceleration simply by using a negative or positive sign. When we move to two dimensions, there are many more options for the direction of the vector and we will need to be more formal with the mathematics of vectors. Vectors are represented graphically as arrows with the length of the arrow representing the magnitude of the vector and the direction of the arrow giving the direction of the vector. This graphical representation might help you understand the basic arithmetic of vectors. Mathematically, we denote a vector, A, with an arrow over the variable name. If we use the variable name without the arrow, this means we are referring to the magnitude of the vector. Let\u2019s set up a coordinate system at the starting end of the vector (see Fig. 3.1). Then the vector points to a speci\ufb01c location in that space. We know that we can give the coordinates of that location using either Cartesian, (x, y), or polar, (r, \u03b8) coordinates. The polar coordinates of this point are quite straightforward; r is the length (magnitude) of the vector and \u03b8 is the angle between the vector and the x-axis. We can also determine the x and y coordinates by \ufb01nding the projections of the vector on the x- and y-axes. The projection of a vector A along the x-axis is called the x-component and is represented by Ax. The projection of A along the y-axis is called the y-component and is represented by Ay. We can \ufb01nd the x and y components by converting the polar coordinates to rectangular coordinates, Ax = A cos \u03b8 Ay = A sin \u03b8. (3.1) Note that we can go from the vector\u2019s x- and y-components back to the polar representation using the Pythagorean theorem and the de\ufb01nition of the tangent: x + A2 A2 = A2 y Ay Ax. tan \u03b8 = 23 (3.2) Figure 3.1: The projections of a vector on the x- and y-axes. Projections give the x- and y components of a vector. Equality of vectors Two vectors are equal only if they have the same magnitude and direction (two arrows are the same only if they are the same length and point in the same direction). This means that you can move a vector around in space as long as you don\u2019t change the length or direction of the vector. Adding vectors Just as when you are adding scalar quantities, you must ensure that the vectors you are trying to add have the same units. Vectors can be added either geometrically (graphically) or algebraically. To graphically add two vectors, A and B, draw the vectors (using the same scale for both) head to tail on a piece of paper. The resultant vector R = A + B is the vector drawn from the unmatched tail to the unmatched head (see Fig. 3.2). You can lay out many vectors head to tail to \ufb01nd the result of them all. While graphical addition of vectors is useful for visualizing the addition process. You will most often be adding the vectors algebraically. To add vectors algebraically, resolve the vectors into their x- and y- components. All the x-components are added to get the x-component of the resultant vector. All the y-components are added to get the y-component of the resultant vector. Never add the the x-components to y-components. You can get the magnitude and direction of the resultant by converting from the Cartesian representation (x and y components) using the equations presented earlier. Negative of a vector The negative of a vector, A, is de\ufb01ned as the vector that gives zero when added to A. If you think about the graphical addition of vectors, this means that the negative of A must have the same magnitude as A, but opposite direction (180\u25e6 di\ufb00erence). Subtracting vectors Subtraction is simply the addition of a negative quantity. Since we have now de\ufb01ned the negative of a vector, we can \ufb01gure out how to subtract. Remember that the negative of a vector points in the opposite direction,", " so for subtraction we \ufb02ip the direction of the vector to be subtracted, but otherwise use the same graphical method as for addition. Graphically \ufb02ipping a vector corresponds to algebraically changing the sign of both the x- and y- components of the vector, so we can use the same algebraic method for subtracting vectors as we do for adding vectors. 24 AAxAy\u03b8 Figure 3.2: Graphical addition of vectors. To graphically add vectors, the vectors are drawn head to tail. The resultant vector is the vector that connects the two \u201cloose ends\u201d, drawn from tail to head. Example: Take a Hike A hiker begins a trip by \ufb01rst walking 25.0 km 45.0\u25e6 south of east from her base camp. On the second day, she walks 40.0 km in a direction 60.0\u25e6 north of east, at which point she discovers a forest ranger\u2019s tower. (a) Determine the components of the hiker\u2019s displacements in the \ufb01rst and second days. (b) Determine the components of the hiker\u2019s total displacement for the trip. (c) Find the magnitude and direction of the displacement from base camp. Solution: (a) Let\u2019s set the origin of our coordinate system at the camp and have the x-axis pointing east and the y-axis pointing north. On the \ufb01rst day, the hiker\u2019s displacement, let\u2019s call it A has a magnitude of 25.0 km with a direction \u03b8 = \u221245\u25e6 \u2014 it\u2019s negative because she is moving south of the x-axis. We can use Eqs. 3.1 to \ufb01nd the x- and y-components, Ax = A cos \u03b8 = (25 km) cos(\u221245\u25e6) = 17.7 km Ay = A sin \u03b8 = (25 km) sin(\u221245\u25e6) = \u221217.7 km. On the second day, her displacement, let\u2019s call it B, has a magnitude of 40.0 km with a direction \u03b8 = 60.0\u25e6 \u2014 positive this time because it is north of east. The x- and y-components of this displacement are Bx = B cos \u03b8 = (40 km) cos(60\u25e6) = 20.0 km By = B sin \u03b8 = (40 km) sin(60\u25e6) = \u221234.6 km. (b) The total displacement is the vector sum of A and B. We\u2019ve just found the components of A and B, so we can \ufb01nd the components of the total displacement, Rx = Ax + Bx = 17.7 km + 20.0 km = 37.7 km Ry = Ay + By = \u221217.7 km + 34.6 km = 16.9 km. 25 (c) We just found the components of the total displacement, all we need to do is convert them to a magnitude and direction. The magnitude is found using the Pythagorean theorem, x + R2 y R2 = R2 R = (37.7 km)2 + (16.9 km)2 R = 41.3 km. The direction is found using the tangent function, tan \u03b8 = tan \u03b8 = Ry Rx 16.9 km 37.7 km 16.9 km 37.7 km \u03b8 = tan\u22121 \u03b8 = 24.1\u25e6. Don\u2019t forget to check that the angle is in the correct quadrant. components are both positive, so we are in the \ufb01rst quadrant and the angle is correct. In this case, the x- and y- 3.1.1 Displacement, velocity and acceleration in two dimensions In one dimension, we de\ufb01ned the displacement as the di\ufb00erence between the initial and \ufb01nal positions of an object. The position in that case was determined by a single coordinate. We now want to de\ufb01ne displacement in two dimensions where the position of the object is given by two coordinates. Let\u2019s call the initial position of the object ri and the \ufb01nal position of the object rf, where r is a position vector that goes from the origin to the position of the object. The displacement is de\ufb01ned as the vector di\ufb00erence between the initial and \ufb01nal position vectors, \u2206r = rf \u2212 ri. (3.3) With this generalized de\ufb01nition of displacement, we can also generalize the average velocity and average acceleration of an object, vav = aav = \u2206r \u2206t \u2206v \u2206t. Finally, we can also generalize the", " instantaneous velocity and instantaneous acceleration v = lim \u2206t\u21920 a = lim \u2206t\u21920 \u2206r \u2206t \u2206v \u2206t. (3.4) (3.5) (3.6) (3.7) If you look carefully at these de\ufb01nitions, you will see that x-component of acceleration is determined by the x-component of velocity which is determined by the x-component of the displacement. The same is true for the y-components of these quantities. This means that the horizontal and vertical components can be treated independently of each other. 26 Example: The swimmer The current in a river is 1.0 m/s. A woman swims across the river to the opposite bank and back. She can swim 2.0 m/s in still water and the river is 300 m wide. She swims perpendicular to the current so she ends up downstream from where she started. Find the time for the round trip Solution: Since the woman swims perpendicular to the current let\u2019s de\ufb01ne the y-axis as parallel to the river. We can treat the x and y motion independently. We are only interested in the motion in the x-direction (across the river) since this will determine how long the trip takes. She is swimming at a constant velocity of 2.0 m/s, so the time to travel a distance of 300 m is vx = t = t = \u2206x t \u2206x vx 300 m 2.0 m/s t = 150 s. It will take the same amount of time for her to travel back, so the round trip takes 300 s. Note that the current pushing the woman down the river is completely irrelevant here because it a\ufb00ects her motion in the y-direction (along the river) and this is independent of her motion in the x-direction. 3.2 Motion in two dimensions We have previously studied motion of objects moving in a straight line (one dimension). We will now extend our study to two dimensions. We know that if we break motion up into x and y components, that the motion in the two directions is independent, so that motion in the horizontal direction does not a\ufb00ect motion in the vertical direction and vice versa. This is particularly important when studying something called projectile motion which is the motion of any object thrown in some way. If we throw a ball with some horizontal initial velocity, its motion can be studied by breaking it up into the horizontal and vertical motions. In the vertical direction, the object undergoes acceleration due to gravity just as in free fall. In the horizontal direction, there is no acceleration and the velocity remains constant. The resulting two-dimensional motion is the combination of the two components. 27 Chapter 4 Laws of motion So far, we\u2019ve studied motion by describing what happens without being concerned about what causes the motion. Now, we will start to examine the causes of motion and we will learn the rules that govern changes in motion. 4.1 Newton\u2019s \ufb01rst law Isaac Newton developed the laws of motion in the 1600s when he started thinking about why objects close to the Earth tended to fall to Earth unless something was holding them up in some way. His ideas on motion are summed up in three laws that are based on the idea of forces. You probably have an intuitive sense of a force from everyday life. When you push or pull on an object, you are applying an external force on the object. These are examples of contact forces, forces which are caused by one object being in contact with another object. There are also forces that arise without contact of two objects. While this may seem strange (Newton was also uncomfortable with the idea of action-at-adistance), you are very familiar with one such force. Gravity causes all objects near Earth to fall towards the Earth even though the Earth is not touching the object. This is an example of a \ufb01eld force, so called because scientists use the idea of a force \ufb01eld emanating from an object to explain how it might a\ufb00ect the motion of objects that it hasn\u2019t touched. Essentially a force is something that can change the state of motion of an object. Note that force (contact or \ufb01eld) is a vector \u2014 it has both a magnitude and direction. If a force is something that can change the state of motion of an object, will objects move without a force? Obviously, if an object is at rest (not moving), it will just sit there forever unless something pushes or pulls it. Suppose now that the object is given a quick push. It will start moving because of the force that has been applied, but what happens after the initial push? In most cases, the object will start to slow down because there is friction between it and the object on which it moves. But suppose we could eliminate the friction, which is a force and changes", " the motion of the object? If we completely eliminate friction, then the object would continue moving without speeding up or slowing down. So yes, objects will move without the presence of a force, but with a very speci\ufb01c type of motion. This is the essence of Newton\u2019s \ufb01rst law, \u201cAn object moves with a velocity that is constant in magnitude and direction unless a non-zero net force acts on it.\u201d The net force is the vector sum of all the forces acting on the object. So this law can be used in two ways. If we know that there is no net force on the object, then we know that it will continue moving with a constant (possibly zero) velocity. Alternatively, if an object is moving with a constant (possibly zero) velocity, then the net force acting on it must be zero. This law is based on the notion of inertia which is the tendency of an object to continue its state of motion in the absence of a force. The law is sometimes stated as \u201ca body in motion will stay in motion and a body at rest will stay at rest unless acted upon by an outside force.\u201d This is closely related to the idea of 28 mass, which measures an object\u2019s resistance to changes in its velocity due to a force. If the same force acts on two objects with di\ufb00erent masses, the object with the smaller mass will experience a bigger change in its velocity than the more massive object. 4.2 Newton\u2019s second law In the absence of a force, an object will keep doing whatever it was doing. What happens if a force acts on the object? Clearly it\u2019s velocity will change in some way. If you push on an object, it will accelerate (change its velocity). If you push harder, it will accelerate faster. So the magnitude of the force is related to the acceleration. In fact, the force is proportional to the acceleration; so if you push twice as hard the acceleration will be twice as large. What other things might a\ufb00ect the acceleration? One quantity that we\u2019ve already discussed is the mass. The same force applied to objects with di\ufb00erent masses will result in di\ufb00erent accelerations. In this case the relationship is inversely proportional \u2014 if the mass is twice as large, the acceleration is halved. Newton put both of these observations together into his second law \u201cThe acceleration a of an object is directly proportional to the net force acting on it and inversely proportional to its mass.\u201d We can write this statement more compactly using mathematics a = F m, (4.1) where a is the acceleration (vector) of the object, F is the vector sum of of all the forces acting on the object and m is the mass of the object. To actually use this equation, we break it up into its x and y (and maybe z) components Fx = max Fy = may. (4.2) Note that the unit of force in the SI system is the newton where 1 N = 1 kg \u00b7 m/s2. 29 Example: Horses pulling a barge Two horses are pulling a barge with mass 2.0 \u00d7 103 kg along a canal. The cable connected to the \ufb01rst horse makes an angle of \u03b81 = 30.0\u25e6 with respect to the direction of the canal, while the cable connected to the second horse makes an angle of \u03b82 = \u221245.0\u25e6. Find the initial acceleration of the barge, starting at rest, if each horse exerts a force of magnitude 6.00 \u00d7 102 N on the barge. Ignore forces of resistance on the barge. Solution: We\u2019ve been given the mass of an object and the forces acting on it and we\u2019re asked to \ufb01nd accelerations. So we want to use Newton\u2019s second law to try to \ufb01nd the acceleration. Let\u2019s de\ufb01ne our coordinate system with the x-axis lying along the canal (so the angles are measured relative to the x-axis). Now we can break down the forces into x and y components, F1x = F cos \u03b81 = (6.00 \u00d7 102 N) cos(30.0\u25e6) = 5.2 \u00d7 102 N F1y = F sin \u03b81 = (6.00 \u00d7 102 N) sin(30.0\u25e6) = 3.00 \u00d7 102 N F2x = F cos \u03b82 = (6.00 \u00d7 102 N) cos(\u221245.0\u25e6) = 4.24 \u00d7 102 N F2y = F sin \u03b82 = (6.00 \u00d7 102 N) sin(\u221245.0\u25e6) = \u22124.24 \u00d7 102 N. Newton\u2019s second law tells us to \ufb01", "nd the net force in both the x and y directions, Fx = F1x + F2x = 5.2 \u00d7 102 N + 4.24 \u00d7 102 N = 9.44 \u00d7 102 N Fy = F1y + F2y = 3.00 \u00d7 102 N \u2212 4.24 \u00d7 102 N = \u22121.24 \u00d7 102 N. Now Newton\u2019s second law says that the net force is related to the acceleration, ax = ay = Fx m Fy m = = 9.44 \u00d7 102 N 2.0 \u00d7 103 kg \u22121.24 \u00d7 102 N 2.0 \u00d7 103 kg = 0.472 m/s2 = \u22120.062 m/s2. We have the x and y components of the acceleration, so we can \ufb01nd the magnitude and acceleration, a = x + a2 a2 y = 0.476 m/s2 \u03b8 = tan\u22121 ay ax = \u22127.46\u25e6. 4.2.1 Weight The weight of an object is not the same as its mass. The weight of an object is the magnitude of the gravitational force acting on an object, so on Earth the weight is Fw = mg. Because g is the same everywhere on Earth, we can use an object\u2019s weight to determine its mass and so the two terms are used interchangeably in everyday language. While mass is a fundamental property of an object and will not change if you move the object to another location, its weight can change depending on the object\u2019s location. 4.3 Newton\u2019s third law Newton\u2019s third law is perhaps the least intuitive of the three laws of motion. According to Newton, all forces come in pairs, 30 \u201cIf object 1and object 2 interact, the force F12 exerted by object 1 on object 2 is equal in magnitude but opposite in direction to the force F21 exerted by object 2 on object 1\u201d The force exerted by object 1 on object 2 is sometimes called the action force and the force exerted on object 2 by object 1 is called the reaction force. The law is sometimes stated as \u201cevery action has an equal and opposite reaction\u201d. Basically anytime two objects interact in some way, there will be two forces, one acting on each object. When you walk, your foot pushes on the \ufb02oor and the \ufb02oor pushes back on you. When you lean against a wall, the wall pushes back on you. Every time an object falls towards Earth because Earth\u2019s gravity is pulling it, the object also pulls Earth towards it. This may seem strange, but the Earth is so much more massive than other objects that its acceleration due to this force is negligible. One consequence of this law is a force called the normal force. Every object on Earth is being pulled towards the center of the Earth by gravity. Most objects are not moving downward because they are sitting on some surface and this surface is pushing up on the object. The upwards force is the normal force, so named because it is always perpendicular to the surface; this mean it does not always point straight up even though it is a reaction force to the pull of gravity. Example: Standing on a crate (a) A 38 kg crate rests on a horizontal \ufb02oor, and a 63 kg person is standing on the crate. Determine the magnitude of the normal force that the crate exerts on the person. (b) Determine the magnitude of the normal force that the \ufb02oor exerts on the crate. Solution: (a) Let\u2019s consider the forces acting on the person only. There is, of course a downward force due to gravity Fg = mpg. There is also a normal force from the crate pushing up on the person. The man is not moving so these two forces must be in equilibrium (the net force must be zero), \u2212Fg + Fcp = 0 Fcp = mpg. (b) Now let\u2019s look at the forces acting on the crate. Gravity acts on the crate, Fg = mcg, and the \ufb02oor pushes up on the crate through the normal force, Ff c. The person standing on the crate also pushes down on the crate with a \u201cnormal\u201d force that is equal in magnitude and opposite to the force of the crate pushing on the person, Fpc = \u2212Fcp. The crate is not moving,so these forces must be in equilibrium, Ff c \u2212 Fg \u2212 Fpc = 0 Ff c = mcg + mpg. When we are using Newton\u2019s laws to solve problems, we use several assumptions. We assume that each object is a point mass, or that they are particles without any spatial extent (0-dimensional objects). This means we don\u2019t have to worry about rotation of the objects. If strings or ropes are part of the problem, we assume that their mass is negligible and that any", " tension in the rope is the same at all points on the rope. When solving force problems, it is useful to draw a free-body diagram. The free body diagram is a drawing of all the forces acting on a particular object. It is very important to only draw the forces acting on the object, any force that the object exerts on its surroundings is not included in the free-body diagram. This diagram helps to isolate the forces of interest for our object and can then be used to apply Newton\u2019s laws. 31 Example: Tra\ufb03c light A tra\ufb03c light weighing 1.00 \u00d7 102 N hangs from a vertical cable tied to two other cables that are fastened to a support. The upper cables make angles of 37\u25e6 and 53\u25e6 with the horizontal (see Fig. 4.14 in your textbook). Find the tension in each of the three cables. Solution: We start by drawing free-body diagrams. First for the tra\ufb03c light which has gravity acting downward (the weight, W and tension, T3 from the rope pulling it upward. Because both forces are in the y direction only, this leads to a single equation, T3 \u2212 W = 0. Although we now know the value of T3, this does not tell us anything about the tension in the other two ropes. We can also consider the forces acting on the knot. The knot has three ropes pulling on it: tension downwards from T3, tension up to the left from T1 and tension up to the right from T2. Because forces are vectors, we need to break everything into x and y components and apply Newton\u2019s second law along each axis. Note that the knot is in equilibrium, so there is no acceleration in either direction, x \u2212 direction : y \u2212 direction : T2 cos(53\u25e6) \u2212 T1 cos(37\u25e6) = 0 T1 sin(37\u25e6) + T2 sin(53\u25e6) \u2212 T3 = 0. We know T3 from the \ufb01rst equation, so we are left T1 and T2 as unknowns. Luckily, we have two equations for our two unknowns. so we can solve one equation, T2 = T1 cos(37\u25e6) cos(53\u25e6) and substitute into the other equation, T1 sin(37\u25e6) + T1 cos(37\u25e6) cos(53\u25e6) sin(53\u25e6) \u2212 W = 0 T1(sin(37\u25e6) + cos(37\u25e6) cos(53\u25e6) sin(53\u25e6)) = W T1 = sin(37\u25e6) + cos(37\u25e6) W cos(53\u25e6) sin(53\u25e6) T1 = 60.1 N. We can now \ufb01nd T2 as well, T2 = (60.1 N) T2 = 79.9 N. cos(37\u25e6) cos(53\u25e6) 4.4 Friction An object moving on a surface or through some medium encounters resistance as it moves. This resistance is called friction. Friction is an essential force as it allows us to hold objects, drive a car and walk. The strength of the frictional force depends on whether an object is stationary or moving. You\u2019ve undoubtedly had the experience of trying to push a large heavy object; it takes more e\ufb00ort to get the object moving than to keep it moving once it\u2019s going. When an object is stationary, the frictional force is called the force 32 of static friction and when the object is moving, the frictional force is called the force of kinetic friction. Friction points opposite to the direction of motion in the kinetic case or the direction of impending motion in the static case. The force of static friction is larger than the force of kinetic friction. It has been shown experimentally that both kinetic and static friction are proportional to the normal force. The only way this can be true is if the two forces have di\ufb00erent constants of proportionality. For static friction, we have fs \u2264 \u00b5sN (4.3) where \u00b5s is the coe\ufb03cient of static friction. This is an inequality because the force of static friction can take on smaller values if less force is needed to hold an object in place. You will almost always use the \u2018=\u2019 sign in problems. For the force of kinetic friction, we have where \u00b5k is the coe\ufb03cient of kinetic friction. \u00b5k will almost always be less than \u00b5s. Note that the friction coe\ufb03cients do not have any units. fk = \u00b5kN (4.4) 33 Example: Block on a ramp Suppose a block with a mass of 2.50 kg is resting on a ramp. If the coe\ufb03cient of static friction between the block and ramp is", " 0.350, what maximum angle can the ramp make with the horizontal before the block starts to slip down? Solution: This is a problem involving forces, so we will need a free body diagram. There are three forces acting on the block: gravity pulls straight down, the normal force acts perpendicular to the surface, and the force of friction acts upwards along the ramp because the block would like to slide down the ramp. We will choose a tilted coordinate system such that the x-axis runs along the ramp. If we do this, the only force that needs to be broken into components is the force of gravity (friction pulls entirely along the x axis and the normal force pulls entirely along the y axis). We can now use Newton\u2019s second law for both axes to get two equations, Fx = 0 mg sin \u03b8 \u2212 f = 0 Fy = 0 N \u2212 mg cos \u03b8 = 0. Remember that the force of friction is related to the normal force, f = \u00b5sN. I have used the equal sign here because the static force of friction will be largest (and therefore equal) just before the object starts to slip. Let\u2019s use one equation to solve for N, and substitute into the other equation, N = mg cos \u03b8 mg sin \u03b8 \u2212 \u00b5smg cos \u03b8 = 0 tan \u03b8 = \u00b5s \u03b8 = 19.3\u25e6. The angle at which the block will slide depends only on the coe\ufb03cient of static friction between the ramp and block. 34 NWxyf Example: Two blocks A block of mass m = 5.00 kg rides on top of a second block of mass M = 10.0 kg. A person attaches a string to the bottom block and pulls the system horizontally across a frictionless surface. Friction between the two blocks keeps the 5.0 kg block from slipping o\ufb00. If the coe\ufb03cient of static friction is 0.305, (a) what maximum force can be exerted by the string on the 10.0 kg block without causing the 5.0 kg block to slip? (b) What is the acceleration? Solution: This is a problem involving forces, so we will need a free body diagram. There are \ufb01ve forces acting on block M : gravity pulls straight down, the normal force from the \ufb02oor pushes up, the normal force from block m pushes down, the tension from the rope, and friction from block m opposes the motion (so f points to the left). There are three forces acting on block m: the normal force from block M pushing up, gravity pulling down and the force of friction that opposes the potential motion of block m (so f points to the right). (a) Now let\u2019s use Newton\u2019s second law on both objects. First block M, and for the second block, Fx = M a T \u2212 f = M a Fy = 0 N2 \u2212 N1 \u2212 M g = 0, Fx = ma f = ma Fy = 0 N1 \u2212 mg = 0, Remember that the force of friction is related to the normal force, f = \u00b5sN1. I have used the equal sign here because the static force of friction will be largest (and therefore equal) just before the object starts to slip. Now we want to \ufb01nd T, N1 = mg \u00b5sN1 = ma a = \u00b5sg T \u2212 f = M a T \u2212 \u00b5smg = M \u00b5sg T = (m + M )\u00b5sg 35 T = 51.5 N (b) We already have an expression for the acceleration a = \u00b5sg a = 3.43 m/s2. NNTffM2m1W1W2N1 Chapter 5 Work and Energy 5.1 Work You probably de\ufb01ne work as something that expends some of your energy. Typing up a paper, or writing out a homework assignment is considered work as are more physically demanding tasks such as building furniture or moving heavy objects. In physics, work has a very speci\ufb01c de\ufb01nition that involves motion and forces. Work is done by a force only if that force causes a net displacement of the object. There are two key points here: a force is only responsible for motion if it is in the same direction as that motion, so forces that are perpendicular to motion do not result in work, and there must be a net displacement or no work has been done. The mathematical de\ufb01nition of work done on an object is W = F \u2206x cos \u03b8 (5.1) where F is the force applied to the object, \u2206x is the displacement, \u03b8 is the angle between the force and the direction of motion, and W is the work. Work is measured in units of joules (joule ). Note that if the force is doubled, work is doubled or if the", " object is displaced twice as far, then work is also doubled. Work is a scalar quantity \u2014 it has a magnitude, but no direction. Work can, however, be positive or negative; it\u2019s negative when the applied force is opposite to the direction of motion. 36 Example: Work on a block A block of mass m = 2.50 kg is pushed a distance d = 2.20 m along a frictionless horizontal table by a constant applied force of magnitude F = 16.0 N directed at an angle \u03b8 = 25.0\u25e6 below the horizontal. Determine the work done by (a) the applied force, (b) the normal force exerted by the table, (c) the force of gravity, and (d) the net force on the block. Solution: (a) The applied force has both horizontal and vertical components, but because the motion is entirely horizontal, only the horizontal component contributes to the force. W = F d cos \u03b8 W = (16.0 N)(2.20 m) cos(\u221225.0\u25e6) W = 32 J. (b) The normal force is perpendicular to the motion, so it does not do any work. (c) The force of gravity is also perpendicular to the motion, so it also does no work. (d) We could calculate the net force by vectorially adding the normal force, the applied force and the force of gravity and then \ufb01nd the work done by that force. Or we can save ourselves some e\ufb00ort by remembering that only forces along the direction of motion contribute to the work. The only force that has a horizontal component is the applied force, and we found the work due to that force in part (a). 5.2 Kinetic energy Energy is an indirectly observed quantity that measures an object\u2019s capacity to do work. Energy comes in many di\ufb00erent forms and can easily change from one form to another, but the total amount of energy in the universe (or in an isolated system) stays the same. That means that energy is a conserved quantity. The concept of energy provides an alternative formulation for Newton\u2019s laws. An object\u2019s energy determines it\u2019s potential to do work and the work it can do is related to the net force exerted by the object. If the force is constant, the acceleration is also constant and we can use kinematics equations, namely, Wnet = Fnet\u2206x = ma\u2206x. v2 \u2212 V 2 0 = 2a\u2206x a\u2206x = v2 \u2212 v2 0 2. We can substitute this into the work equation, Wnet = m v2 \u2212 v2 0 2 Wnet = 1 2 mv2 \u2212 1 2 mv2 0. (5.2) This equation tells us that the net work done on an object leads to a change in a quantity of the form 1 2 mv2. This term is called the kinetic energy of the object and it is the energy of the motion of the object. The equation tells us that any net work done on an object leads to a change in its kinetic energy and for this reason, the equation is known as the work-energy theorem. It\u2019s important to realize that this is just an alternative formulation of Newton\u2019s second law \u2014 two di\ufb00erent ways of looking at the same process. Both Newton\u2019s second law and the work-energy theorem tell us that interacting with an object in a certain way (by applying a force in Newton\u2019s view, or doing work in the 37 energy view) will lead to changes in the object\u2019s velocity. Why do we need two ways to describe the same process? It is sometimes more convenient to use one formulation over the other when solving problems. The energy formulation uses scalars rather than vectors, which can be easier for calculations, but it is sometimes hard to determine the net work done on an object without considering forces. Example: Stopping a ship A large cruise ship of mass 6.50 \u00d7 107 kg has a speed of 12.0 m/s at some instant. (a) What is the ship\u2019s kinetic energy at this time? (b) How much work is required to stop it? (c) What is the magnitude of the constant force required to stop it as it undergoes a displacement of 2.5 km? Solution: (a) We know the ship\u2019s initial speed and we know that the ship\u2019s kinetic energy is determined by its speed, mv2 K0 = 1 2 1 2 K0 = 4.7 \u00d7 109 J. K0 = (6.50 \u00d7 107 kg)(12.0 m/s)2 (b) We want the ship\u2019s \ufb01nal velocity (and also its kinetic energy) to be 0. The work-energy theorem tells us how much work is required to change an", " object\u2019s kinetic energy, W = Kf \u2212 K0 W = \u22124.7 \u00d7 109 J. (c) Remember that work is done by a force acting on an object that travels some distance. The force slowing down the ship in this case is the drag or friction force of the water. Remember that friction is always opposite to the direction of motion so \u03b8 = 180\u25e6. W = F \u2206x cos \u03b8 F = F = W \u2206x cos \u03b8 \u22124.7 \u00d7 109 J 2500 m cos(180\u25e6) F = 1.9 \u00d7 106 N. 5.2.1 Conservative and nonconservative forces Forces can be broken up into two types: conservative and non-conservative. Conservative forces are forces where you can easily get back the energy you put into system. Gravity is one example of a conservative force. If you lift a book, you will be doing work against gravity to raise that book. As you lower the book, the book is now doing work on you (the normal force still points the same way, but the motion is in the opposite direction, so work is negative) and you will recover all the energy you put into the book to lift it. A nonconservative force converts energy of objects into heat or sound \u2014 forms of energy that are hard to convert back to motion. Friction is one example of a nonconservative force \u2014 you can\u2019t recapture the energy lost to friction simply by moving the object back to where it started (like we did when we lowered the textbook). The proper physics de\ufb01nition of a conservative force is \u201cA force is conservative if the work it does moving an object between two points is the same no matter what path is taken.\u201d 38 This is based on the idea that, for conservative forces, we can get back the energy we put in simply by moving the object back to its starting point. We can re-write the work energy theorem to speci\ufb01cally separate these two types of forces, Wnc + Wc = \u2206KE, (5.3) where we\u2019ve separated the work done on the object into two parts: the work done by conservative forces and the work done by nonconservative forces. 5.3 Gravitational potential energy Conservative forces have the nice property that they essentially \u201cstore\u201d energy based on their position. When you lift a book, you\u2019ve done some work on that book and put energy into the book. You can get that energy back by lowering the book, or you can convert that energy to something else (like kinetic energy) by letting go of the book. The book is said to have potential energy because it now has the potential to do work on another object. Let\u2019s \ufb01gure out how much work is done by gravity as a book of mass m falls from yi to yf. Remember that the formula for work is W = F \u2206x cos \u03b8. The force of gravity is Fg = \u2212mg, the displacement is \u2206x = yf \u2212 yi, and the angle between them is \u03b8 = 0. So we have, Wg = \u2212mg(yf \u2212 yi). (5.4) Assuming we have no other conservative forces, we can explicitly put the e\ufb00ect of gravity into the work-energy theorem, Wnc + Wg = \u2206KE Wnc = \u2206KE + mg(yf \u2212 yi) Wnc = \u2206KE + \u2206P E. (5.5) Note that when you are using the work-energy theorem, it does not matter what you choose as your reference point for measuring the height of an object. It is only changes in height (and therefore changes in gravitational potential energy) that matter and the the actual value of the potential energy at an one point. In the absence of any nonconservative forces (which will be the case in most of your homework problems), we have 0 = \u2206KE + \u2206P E KEi + P Ei = KEf + P Ef. (5.6) This is a conservation law \u2014 it tells us that the total amount of kinetic energy and potential energy (sometimes called mechanical energy) for a particular system stays the same all the time. The amount of kinetic energy and potential energy might change as the object moves, but if you add the energies together, you will always get the same number. A ball sitting on the top of a hill has lots of gravitational potential energy and no kinetic energy. As it starts to roll down, it\u2019s potential energy decreases, but it\u2019s kinetic energy increases. When it gets to the bottom of the hill, it has no more potential energy, but lots of kinetic energy. So the amount of each type of energy changes, but the total will always be the same. 39 Example: Platform diver A diver of mass m drops from a board 10.0", " m above the water\u2019s surface. Neglect air resistance. (a) Find his speed 5.0 m above the water\u2019s surface. (b) Find his speed as he hits the water. Solution: (a) When solving problems dealing with gravitational potential energy, we need to set a reference point (it doesn\u2019t matter where the reference point is, we just need to be consistent for all measurements). Let\u2019s choose the bottom of the diving board as y = 0. We are told that the diver \u201cdrops\u201d from the board, so v0 = 0 which means that his kinetic energy is KEi = 0. The initial position of the diver is at the top of the board where his gravitational potential energy is P Ei = mgyi. The conservation of mechanical energy tells us, KEi + P Ei = KEf + P Ef 0 + mgyi = mv2 f + mgyf 1 2 2g(yi \u2212 yf ) vf = vf = 2(9.8 m/s2)(10.0 m \u2212 5.0 m) vf = 9.90 m/s. (b) We use the same procedure as for part (a), but with a di\ufb00erent end point where yf = 0. KEi + P Ei = KEf + P Ef 0 + mgyi = 1 2 f + 0 mv2 vf = 2gyi vf = 2(9.8 m/s2)(10.0 m) vf = 14.0 m/s. 40 Example: Waterslides Der Stuka is a waterslide at Six Flags in Dallas named for the German dive bombers of World War II. It is 21.9 m high. (a) Determine the speed of a 60.0 kg woman at the bottom of such a slide, assuming no friction is present. (b) If the woman is clocked at 18.0 m/s at the bottom of the slide, \ufb01nd the work done on the woman by friction. Solution: (a) Let\u2019s choose the bottom of the slide as y = 0. We can assume that the woman starts from rest, so v0 = 0 which means that her kinetic energy is KEi = 0 at the top of the slide. The initial position of the woman is at the top of the board where her gravitational potential energy is P Ei = mgyi. We are interested in the \ufb01nal position where her height is yf = 0 and so her gravitational potential energy is P Ef = 0. The conservation of mechanical energy tells us, KEi + P Ei = KEf + P Ef 0 + mgyi = 1 2 f + 0 mv2 vf = 2gyi vf = 2(9.8 m/s2)(21.9 m) vf = 20.7 m/s. (b) In this case we have to use the full work-energy theorem, Wnc = KEf \u2212 KEi + P Ef \u2212 P Ei f \u2212 0 + 0 \u2212 mgyi mv2 Wnc = 1 2 1 2 Wnc = \u22123.16 \u00d7 103 J. Wnc = (60.0 kg)(18.0 m/s)2 \u2212 (60.0 kg)(9.8 m/s2)(21.9 m) Note that the work done by friction is negative because the woman is losing her energy to friction. 5.4 Spring potential energy When you compress or stretch a string, you have to apply a force and therefore do some work on the spring. When you move the spring back to its original position, that energy is given back to you. Like gravity, the spring force is a conservative force \u2014 any energy you put into the spring when it is stretched or compressed is returned when the spring moves back to its original position. Springs exert a force on an object when they are stretched or compressed and the more you stretch or compress the spring, the larger the force trying to return the spring to its original position. So the force exerted by a spring is proportional to the displacement, Fs = \u2212k\u2206x, (5.7) where k is a proportionality constant called the spring constant (units of newtons per meter). This constant is di\ufb00erent for each spring. This equation is often called Hooke\u2019s law after Robert Hooke who discovered the relationship. In the case of a spring, we measure the displacement from the equilibrium position of the spring. That is x = 0 is the point at which the spring is neither compressed nor stretched. The spring force is sometimes called a restoring force because it tries to return the spring to equilibrium. Calculating the work done by a spring is not as straightforward as calculating the work done by gravity because the size of the spring force changes as the displacement changes", " (remember the force of gravity is the same no matter the height of the object). The work done by the spring force when an object moves from 41 xi to xf is Ws = \u2212( 1 2 kx2 f \u2212 1 2 kx2 i ). (5.8) We can include this in the work-energy theorem (if there is a spring involved in our system) as part of the work done by conservative forces, Wnc = \u2206KE + \u2206P Eg + \u2206P Es, (5.9) where the potential energy of a spring is 1 2 kx2. Example: Block on a spring A block with mass of 5.00 kg is attached to a horizontal spring with spring constant k = 4.00 \u00d7 102 N/m. The surface the block rests upon is frictionless. If the block is pulled out to xi = 0.05 m and released, (a) \ufb01nd the speed of the block when it \ufb01rst reaches the equilibrium point. (b) Find the speed when x = 0.025 m, and (c) repeat part (a) if friction acts on the block with coe\ufb03cient \u00b5k = 0.150. Solution: (a) In the \ufb01rst part of the problem there is no friction, so there aren\u2019t any nonconservative forces and the work-energy theorem can be written as KEi + P Egi + P Esi = KEf + P Egf + P Esf. We can simplify this a little more by realizing that all the action takes place at the same height, so there are no changes in gravitational potential energy and we can remove that from the equation, KEi + P Esi = KEf + P Esf. The problem states that the block is \u201creleased\u201d at a certain point \u2014 this means that the initial velocity is vi = 0. So the initial kinetic energy is also 0. The \ufb01nal position of the object is at the equilibrium point (xf = 0), so the spring potential energy at this point is also 0. 0 + 1 2 kx2 i = mv2 f + 0 1 2 vf = vf = kx2 i m (4.00 \u00d7 102 N/m)(0.05 m)2 5.00 kg vf = 0.45 m/s. (b) This time we\u2019re asked for the velocity at a non-equilibrium point. The method we use is still the same, we just have a non-zero \ufb01nal potential energy, 0 + 1 2 kx2 i = 1 2 mv2 f + k(x2 kx2 f 1 2 i \u2212 x2 f ) m vf = vf = (4.00 \u00d7 102 N/m)[(0.05 m)2 \u2212 (0.025 m)2] 5.00 kg vf = 0.39 m/s. 42 (c) Now we add friction, so we will need to consider the energy lost to this nonconservative force. First we need to \ufb01nd the magnitude of the frictional force, so we need a free-body diagram of the block. The block has four forces acting on it: gravity pulling down, the normal force pushing up, the spring force pulling towards equilibrium, and the force of friction pulling away from equilibrium. There is no acceleration in the y direction, so we must have Fy = 0 N \u2212 mg = 0 N = mg. We know that the frictional force is related to the normal force, The work done by friction then is fk = \u00b5kN fk = \u00b5kmg. Wf = fk\u2206x cos \u03b8 Wf = \u2212\u00b5kmgxi. The work-energy theorem now has to include the nonconservative work, Wnc = \u2206KE + \u2206P Es \u2212\u00b5kmgxi = mv2 f \u2212 1 2 kx2 i 1 2 vf = vf = kx2 i m \u2212 2\u00b5kgxi (4.00 \u00d7 102 N/m)(0.05 m)2 5.00 kg \u2212 2(0.150)(9.8 m/s2)(0.05 m) vf = 0.230 m/s. 43 Example: Circus acrobat A 50.0 kg circus acrobat drops from a height of 2.0 m straight down onto a springboard with a force constant of 8.00 \u00d7 103N/m By what maximum distance does she compress the spring? Solution: There aren\u2019t any nonconservative forces in this problem, but spring potential, gravitational potential and kinetic energy all play a role. In this problem, we need to be very clear about how we are measuring distances because there are two displacements that are relevant: her change in height and the compression of the spring", ". Let\u2019s set y = 0 to be the point of maximum spring compression and let\u2019s call the distance that the spring compresses d. We will call the acrobat\u2019s height above the uncompressed springboard h. Now let\u2019s use the work-energy theorem, KEi + P Egi + P Esi = KEf + P Egf + P Esf 0 + mg(h + d kd2 \u2212 mgd \u2212 mgh = 0. 1 2 kd2 We have a quadratic equation and will need to use the quadratic formula, d = d = mg \u00b1 m2g2 + 2kmgh k mg k 1 \u00b1 1 + 2kh mg d = 0.56 m (\u22120.44). 5.4.1 Power Power is the rate at which energy is transferred from one object to another. Remember that work is the amount of energy transferred from one object to another, so the average power will be the amount of work done over some period of time, P =. (5.10) W \u2206t The unit of power is the Watt (W) or joule/second (J/s). We can write the power in another form by using the de\ufb01nition of work W = F \u2206x cos \u03b8, P = F \u2206x cos \u03b8 \u2206t P = F v cos \u03b8. We can generalize this equation (if we use calculus) to get an equation for the instantaneous power, P = F v cos \u03b8, where P and v are the instantaneous power and velocity rather than the average power and velocity. (5.11) (5.12) 44 Example: Shamu Killer whales are able to accelerate up to 30 mph in a matter of seconds. Neglecting the drag force of water, calculate the average power a killer whale with mass 8.00 \u00d7 103 kg would need to generate to reach a speed of 12.0 m/s in 6.00 s. Solution: To \ufb01nd the power needed by the whale, we need to \ufb01gure out how much work the whale has to do to reach a speed of 12.0 m/s. We do not know the magnitude of the force needed to generate this acceleration, so we can\u2019t use the de\ufb01nition of work. (We also can\u2019t \ufb01nd the acceleration using kinematics because we can\u2019t assume constant acceleration). The other option is to use the work-energy theorem, Wnet = \u2206KE mv2 f \u2212 0 Wnet = 1 2 1 2 Wnet = 5.76 \u00d7 105 J. Wnet = (8.00 \u00d7 103 kg)(12.0 m/s)2 We know the elapsed time, so we can use the equation de\ufb01ning average power, P = P = W \u2206t 5.76 \u00d7 105 J 6.00 s P = 9.6 \u00d7 104 W. You might be wondering why we did not use the equation P = F v cos \u03b8. That equation uses the average velocity and we are given the \ufb01nal velocity. We can\u2019t \ufb01nd the average velocity without assuming constant acceleration, which we can\u2019t do in this case. 45 Chapter 6 Momentum and Collisions 6.1 Momentum and impulse You probably have an intuitive de\ufb01nition of momentum. Objects with a large amount of momentum are hard to stop, i.e. a larger force is required to stop an object with lots of momentum. In physics, of course, we like to have precise de\ufb01nitions for these concepts, so we de\ufb01ne the linear momentum as p = mv. (6.1) The linear momentum is proportional to both mass and velocity. That is the more massive an object, the more momentum it has and the faster an object moves the more momentum it has. The unit of momentum is kilogram meter per second (kg \u00b7 m/s). Notice that momentum is a vector that points in the same direction as an object\u2019s velocity. As usual, when dealing with vectors, you will break up momentum into its x and y components, px = mvx py = mvy. The momentum is related to the kinetic energy of an object, KE = 1 2 mv2 = (mv)2 2m = p2 2m. (6.2) The concept of momentum is closely tied to the idea of inertia and force. A force is required to change the momentum of an object. We can actually restate Newton\u2019s second law in terms of momentum, Fnet = ma = m \u2206v \u2206t = \u2206p \u2206t, (6.3) which tells us that the change in momentum over some time is equal to the net force. Two implications", " arise from this equation. First, if there is no net force, the momentum does not change. Second, to change an object\u2019s momentum you need to continuously apply a force over some time period (however small). We call the change in momentum of an object the impulse, I = \u2206p = Fnet\u2206t. (6.4) This is called the impulse-momentum theorem. Most forces vary over time, making it di\ufb03cult to use the idea of impulse without calculus. We can, however, replace a time-varying force with an average force which is a constant force that delivers the same impulse in the time \u2206t as the real time-varying force. In this case, \u2206p = Fav\u2206t. 46 (6.5) Example: Car crash In a crash test, a car of mass 1.50 \u00d7 103 kg collides with a wall and rebounds. The initial and \ufb01nal velocities are vi = \u221215.0 m/s and vf = 2.60 m/s. If the collision lasts for 0.150 s, \ufb01nd (a) the impulse delivered to the car due to the collision and (b) the size and direction of the average force exerted on the car. Solution: The impulse is the change in momentum, I = pf \u2212 pi I = m(vf \u2212 vi) I = (1.50 \u00d7 103 kg)(2.60 m/s \u2212 (\u221215.0 m/s)) I = 2.64 \u00d7 104 kg \u00b7 m/s. We know that the impulse is related to the average force, Fav = I \u2206t 2.64 \u00d7 104 kg \u00b7 m/s 0.150 s Fav = 1.76 \u00d7 105 N. Fav = 6.2 Conservation of momentum If the two objects are isolated, then we can Let\u2019s think about what happens when two objects collide. consider them as a single system. While the two objects will exert forces on each other during the collision, there are no net external forces acting on the system as a whole. If there are no net forces, then the total momentum of the system stays the same throughout the collision process. This is known as the conservation of momentum. Suppose a particle with mass m1 is travelling with velocity v1i towards a particle with mass m2 which is travelling towards the \ufb01rst particle with velocity v2i. These two particles will eventually collide. After they collide, the \ufb01rst particle moves away from the second with a velocity v1f and the second particle moves away with velocity v2f. While they are colliding, there will be a contact force between them. The force from particle 2 on particle 1 will change the momentum of particle 1, F21\u2206t = m1v1f \u2212 m1v1i, and similarly, the force from particle 1 on particle 2 will change the momentum of particle 2, F12\u2206t = m2v2f \u2212 m2v2i. (6.6) (6.7) By Newton\u2019s third law, we know that the contact forces are an action/reaction pair and so they must be equal in magnitude and opposite in direction, F21\u2206t = \u2212 F12\u2206t m1v1f \u2212 m1v1i = \u2212(m2v2f \u2212 m2v2i) m1v1i + m2v2i = m2v2f + m1v1f. (6.8) This equation tells us that if we add the momenta of the particles before the collision, that will be the same as the sum of all momenta after the collision. 47 Example: Littering \ufb01sherman A 75 kg \ufb01sherman in a 125 kg boat throws a package of mass 15 kg horizontally with a speed of 4.5 m/s. Neglecting water resistance, and assuming the boat is at rest before the package is thrown, \ufb01nd the velocity of the boat after the package is thrown. Solution: We will treat the \ufb01sherman/boat as a single object for the purpose of this problem; the package will be treated as a separate object. Everything is at rest before the package is thrown, so there is no initial momentum. After the package is thrown, the boat/\ufb01sherman will have some recoil velocity, 0 = mbVb + mpvp vb = \u2212 vb = \u2212 mpvp mb (15 kg)(4.5 m/s) 200 kg vb = \u22120.38 m/s. 6.2.1 Collisions While momentum is conserved during a collision, kinetic energy is not necessarily conserved. It is important to stress that total energy is always conserved, but certain types of energy", ", like kinetic energy are not always conserved. During a collision, energy is often lost to friction, sound, heat or deformation of the objects. We can classify collisions into several di\ufb00erent types: Elastic collision: An elastic collision is a collision in which both momentum and kinetic energy are con- served. Inelastic collision: An inelastic collision is a collision in which momentum is conserved, but kinetic energy is not. Perfectly inelastic collision: A perfectly inelastic collision is a collision in which momentum is conserved, kinetic energy is not conserved and the two objects stick together and move with the same velocity after the collision. 48 Example: Truck versus compact A pickup truck with mass 1.80 \u00d7 103 kg is travelling eastbound at 15.0 m/s, while a compact car with mass 9.00 \u00d7 102 kg is travelling westbound at -15.0 m/s. The vehicles collide head-on, becoming entangled. (a) Find the speed of the entangled vehicles after the collision. (b) Find the change in the velocity of each vehicle. (c) Find the change in the kinetic energy of the system consisting of both vehicles. Solution: (a) This problem involves a collision, so we should use conservation of momentum. Before the collision, we know the velocities of both the truck and compact. After the collision, they have the same velocity. mtvt + mcvc = (mt + mc)v mtvt + mcvc mt + mc v = v = (1.80 \u00d7 103 kg)(15.0 m/s) \u2212 (9.00 \u00d7 102 kg)(15.0 m/s) 1.80 \u00d7 103 kg + 9.00 \u00d7 102 kg v = 5.0 m/s. (b) The change in velocity of the truck is \u2206v = v \u2212 vt = 5.0 m/s \u2212 15.0 m/s = \u221210 m/s. The change in velocity of the car is \u2206v = v \u2212 vc = 5.0 m/s + 15.0 m/s = 20 m/s. (c) The change in kinetic energy is \u2206KE = KEf \u2212 KEi 1 2 \u2206KE = \u22122.7 \u00d7 105 J. \u2206KE = (mt + mc)v2 \u2212 1 2 mtv2 t \u2212 1 2 mcv2 c 49 Example: Billiard balls Two billiard balls of identical mass move toward each other. Assume that the collision between them is perfectly elastic. If the initial velocities of the balls are 30.0 cm/s and -20.0 cm/s, what are the velocities of the balls after the collision? Assume friction and rotation are unimportant. Solution: This problem involves a collision, so we should use conservation of momentum. Before the collision, we know the velocities of both balls. After the collision, both velocities are unknown. mv1i + mv2i = mv1f + mv2f v1i + v2i = v1f + v2f v1i \u2212 v1f = v2f \u2212 v2i. We only have one equation with two unknowns, so we will need to \ufb01nd another equation. We are told that the collision is perfectly elastic, so we know that kinetic energy is conserved, 1 2 mv2 mv2 mv2 1 1f + 2i = 2 1f + v2 2i = v2 2f 2f \u2212 v2 1f = v2 2i (v1i \u2212 v1f )(v1i + v1f ) = (v2f \u2212 v2i)(v2f + v2i). 1 1i + 2 v2 1i + v2 1i \u2212 v2 v2 mv2 2f 1 2 Let\u2019s divide the two equations, (v1i \u2212 v1f )(v1i + v1f ) v1i \u2212 v1f = (v2f \u2212 v2i)(v2f + v2i) v2f \u2212 v2i v1i + v1f = v2f + v2i. This equation is easier to deal with than the one with squared velocities, so let\u2019s solve it for v1f and substitute into the conservation of momentum equation v1f = v2f + v2i \u2212 v1i v1i \u2212 v2f \u2212 v2i + v1i = v2f \u2212 v2i v2f = v1i v2f = 30.0 cm/s v1f = v1i + v2i \u2212 v1i v1f = v2i v1f = \u221220.0 cm/s. The balls have swapped their vel", "ocities as if they had passed through each other. The linear equation derived in the above example v1i \u2212 v2i = \u2212(v1f \u2212 v2f ) (6.9) is actually true for any elastic collision, even if the masses are not equal. Instead of using the conservation of kinetic energy, which has quadratic terms and is hard to use, you can use the above linear equation for an elastic collision. 50 6.2.2 Collisions in two dimensions Most collisions are not head-on collisions with both objects travelling along a single line before and after the collision. Although most collisions occur in three dimensions, we will limit ourselves to two-dimensional collisions. When doing two-dimensional momentum problems, we break up momentum into x and y components, just as we did for Newton\u2019s second law. The conservation of momentum tells us that momentum will be conserved in the x direction and in the y direction. That is,we now have two equations m1v1ix + m2v2ix = m1v1f x + m2v2f x m1v1iy + m2v2iy = m1v1f y + m2v2f y. (6.10) There are now three subscripts on the velocity: one telling us which object, one telling us which direction, and one telling us whether it was before or after the collision. Example 6.8: Collision at an intersection A car with mass 1.50 \u00d7 103 kg travelling east at a speed of 25.0 m/s collides at an intersection with a 2.50 \u00d7 103 kg van travelling north at a speed of 20.0 m/s. Find the magnitude and direction of the velocity of the wreckage after the collision, assuming that the vehicles undergo a perfectly inelastic collision and assuming friction between the vehicles and the road can be neglected. Solution: Before the collision, the car has a velocity only in the x direction and the van has a velocity only in the y direction. After the collision, the velocity of the combined object will have components in both directions, mcvc + 0 = (mc + mv)vf x 0 + mvvv = (mc + mv)vf y. We can use these equations to \ufb01nd the x and y components of the \ufb01nal velocity vf x = vf y = mc mc + mv mv mc + mv vc = vv = 1.50 \u00d7 103 kg 1.50 \u00d7 103 kg + 2.50 \u00d7 103 kg 2.50 \u00d7 103 kg 1.50 \u00d7 103 kg + 2.50 \u00d7 103 kg (25.0 m/s) = 9.38 m/s (20.0 m/s) = 12.5 m/s, from which we can \ufb01nd the magnitude and direction f x + v2 v2 f y v = v = (9.38 m/s)2 + (12.5 m/s)2 v = 15.6 m/s. vf y vf x 12.5 m/s 9.38 m/s tan \u03b8 = tan \u03b8 = \u03b8 = 53\u25e6. 51 Example P54: Colliding blocks Consider a frictionless track as shown in Figure P6.54 of your textbook. A block of mass m1 = 5.00 kg is released from height h = 5.00 m. It makes a head-on elastic collision with a block of mass m2 = 10.0 kg that is initially at rest. Calculate the maximum height to which m1 rises after the collision. Solution: Let\u2019s think about what happens in this problem. The block m1 starts from some height with no initial velocity and will travel to the bottom of the ramp, gaining speed as it falls. At the bottom of the ramp, it will hit block 2 and transfer some of its speed to block 2. We want to know if block 1 will travel backward after the collision and, if so, how far back up the ramp it travels. Let\u2019s consider each step separately. First, the block falls from a height h. How fast is it moving at the bottom of the ramp? This is an energy conservation problem since all of the block\u2019s initial potential energy is converted to kinetic energy (no friction), KEi + P Ei = KEf + P Ef 0 + mgh = 1 2 f + 0 m1v2 vf = 2gh vf = 2(9.8 m/s2)(5.00 m) vf = 9.9 m/s. Now we can look at the collision. Block 1 has initial velocity v1i = 9.9 m/s and an unknown \ufb01nal velocity v1f. The second block is initially at rest v2i = 0 and has some unknown \ufb01nal velocity v", "2f. Use conservation of momentum because this is a collision m1v1i + 0 = m1v1f + m2v2f. We have one equation with two unknowns, but we also know that the collision is elastic will give us a second equation. We\u2019ll use the linear equation derived earlier instead of the quadratic equation to make the calculation easier, v1i \u2212 0 = \u2212(v1f \u2212 v2f ). We\u2019re actually not too interested in what block 2 does, so let\u2019s solve for its \ufb01nal velocity \ufb01rst and substitute into the other equation to \ufb01nd v1f v2f = v1i + v1f m1v1i = m1v1f + m2(v1i + v1f ) v1f (m1 + m2) = v1i(m1 \u2212 m2) v1f = v1f = v1i m1 \u2212 m2 m1 + m2 5.00 kg \u2212 10.0 kg 5.00 kg + 10.0 kg v1f = \u22123.3 m/s. (9.9 m/s) Finally, we do the reverse of what we did in the \ufb01rst part to see how high the block will go KEi + P Ei = KEf + P Ef 1 2 1f + 0 = 0 + mgh m1v2 h = 52 h = v2 1f 2g (3.3 m/s)2 2(9.8 m/s2) h = 0.56 m. Chapter 7 Rotational Motion All the motion that we have studied to this point is linear motion. All the objects travelled in a straight line (or a series of straight lines). Objects do not always move in a straight line, they often rotate or move in circles. Luckily, many of the concepts you have learned for linear motion have analogues in rotational motion. 7.1 Angular displacement, speed and acceleration When describing linear motion, the important quantities are displacement \u2206x, velocity v and acceleration, a. For rotational motion we use the angular displacement \u2206\u03b8, angular velocity \u03c9, and angular acceleration \u03b1. For linear motion, the displacement measured the change in linear position of the object. For rotational motion, we want to measure the net change in angle as the object moves around the circle. You are used to measuring angles in degrees, but a more natural unit for measuring angles is the radian. Remember that the circumference of a circle of radius r is s = 2\u03c0r. Rearranging this equation a little gives s/r = 2\u03c0. This quantity is dimensionless, but it tells us that the displacement around any circle is 2\u03c0. A displacement around half the circle is \u03c0; a quarter circle is \u03c0/2. This forms the basis of the unit of radians. Note that we can convert from degrees to radians using the relation 180\u25e6 = \u03c0. Angular quantities in physics must be expressed in radians, so be sure to set your calculators to radian mode when doing rotation problems. We can \ufb01nd the angle travelled by an object rotating at a distance r through an arc length s through \u03b8 = s r You might wonder why we bother de\ufb01ning a new type of displacement if we can just measure the arc length and use our linear displacement equations. Consider a solid object, called a rigid body, like a rotating compact disc where the entire object rotates as a unit. As the object rotates, all points on the CD will have the same angular displacement, but the arc length travelled by points on the disc will vary depending on the radius. Using the notion of angular displacement, we can easily describe the motion of the entire CD, but using linear displacement makes it di\ufb03cult to treat the CD as a single object. (7.1). Let\u2019s properly de\ufb01ne the angular properties. Suppose an object starts at an angle \u03b8i and ends at an angle \u03b8f after some time \u2206t. The angular displacement is determined by the initial and \ufb01nal angles, Note that for a rigid body the angular displacement is the same for all points on the object. The unit for angular displacement is the radian (rad). The average angular velocity of an object is the angular displacement divided by the time, \u2206\u03b8 = \u03b8f \u2212 \u03b8i. (7.2) \u03c9 = \u2206\u03b8 \u2206t. 53 (7.3) For a rigid body, again, all points will have the same angular velocity. The units of angular velocity are radian per second (rad/s). We will use the term angular speed when we are not concerned with the direction, but just using the magnitude of the velocity. A positive angular", " speed denotes counterclockwise rotation and a negative angular speed denotes clockwise rotation. Angular velocity is a vector and the direction is speci\ufb01ed by the right-hand rule. Take your right hand, curl your \ufb01ngers in the direction of the motion and your thumb will give the direction of the vector. This rule speci\ufb01es the rotation axis of the spinning object. Example: Spinning wheel A wheel has a radius of 2.0 m. (a) How far does a point on the circumference travel if the wheel is rotated through an angle of 30 rad. (b) If this occurs in 2 s, what is the average angular speed of the wheel? Solution: (a) We are \ufb01rst asked to \ufb01nd the arc length travelled by a point on the edge of the wheel when the wheel rotates through 30 rad. We use the equation relating the arc length to the angle, s = r\u03b8 s = (2.0 m)(30 rad) s = 60 m. (b) Angular speed is the angular displacement divided by time, \u03c9 = \u03c9 = \u2206\u03b8 \u2206t 30 rad 2 s \u03c9 = 15 rad/s. We de\ufb01ne the instantaneous angular speed by taking the limit (as we did for velocity), We use this instantaneous angular speed to de\ufb01ne the average angular acceleration, \u03c9 = lim \u2206t\u21920 \u2206\u03b8 \u2206t. \u03b1 = \u2206\u03c9 \u2206t. (7.4) (7.5) The units of angular acceleration are radian per second squared (rad/s2). A rigid body will have the same angular acceleration at all points on the body. The direction of angular acceleration is in the same direction as angular velocity if the object is accelerating, otherwise it is in the opposite direction. 7.1.1 Constant angular acceleration For linear motion, we developed a number of useful equations when we could assume constant linear acceleration. We can derive similar equations using the angular quantities under the assumption of constant angular acceleration. Linear motion Rotational motion 2 v = vf +vi vf = vi + at \u2206x = vit + 1 f = v2 v2 2 at2 i + 2a\u2206x \u03c9 = \u03c9f +\u03c9i \u03c9f = \u03c9i + \u03b1t 2 \u2206\u03b8 = \u03c9it + 1 f = \u03c92 \u03c92 2 \u03b1t2 i + 2\u03b1\u2206\u03b8 54 Example 7.2: Spinning wheel II A wheel rotates with a constant angular acceleration of 3.5 rad/s2. If the angular speed of the wheel is 2.00 rad/s at t = 0, (a) through what angle does the wheel rotate between t = 0 and t = 2.00 s? Give your answer in radians and revolutions. (b) What is the angular speed of the wheel at 2.00 s? (c) What angular displacement (in revolutions) results while the angular speed of part (b) doubles? Solution: (a) We are told that the angular acceleration is constant, so we can use the equations above. We are given angular acceleration, initial angular speed and a time and we want to \ufb01nd angular displacement, \u2206\u03b8 = \u03c9it + 1 2 \u03b1t2 \u2206\u03b8 = (2.00 rad/s)(2.00 s) + \u2206\u03b8 = 11.0 rad. 1 2 (3.5 rad/s2)(2.00 s)2 To convert this to revolutions, we remember that one revolution is 2\u03c0 rad, 11.0 rad/2\u03c0 = 1.75. (b) We know want the \ufb01nal angular speed, \u03c9f = \u03c9i + \u03b1t \u03c9f = (2.00 rad/s) + (3.5 rad/s2)(2.00 s) \u03c9f = 9.00 rad/s. (c) In this case, we have an initial and \ufb01nal angular speed and we want displacement, f = \u03c92 \u03c92 i + 2\u03b1\u2206\u03b8 f \u2212 \u03c92 \u03c92 i 2\u03b1 \u2206\u03b8 = \u2206\u03b8 = (18.00 rad/s)2 \u2212 (9.00 rad/s)2 2(3.5 rad/s2) \u2206\u03b8 = 34.7 rad = 5.52. 7.1.2 Relations between angular and linear quantities Remember that we could relate the distance travelled along an arced path to the angular displacement, \u2206s = r\u2206\u03b8. We can use this relationship to \ufb01nd relationships between other angular and linear quantities. For example, we can divide both sides of this equation by time to get the relationship between angular and linear velocity", ", = r \u2206\u03b8 \u2206s \u2206t \u2206t vt = r\u03c9. (7.6) I\u2019ve used the subscript t here to denote the tangential velocity. The instantaneous velocity of the rotating object is always tangent to the circle. That is, if the object were no longer forced to rotate, it would continue in a straight line tangent to the circle, as dictated by Newton\u2019s \ufb01rst law. So at every point on the path of a rotating object, the velocity is tangent to the path. 55 We can similarly derive a relationship between the angular and linear accelerations, where again, we use the subscript t to denote the tangential acceleration of the object. In this case, the distinction is very important because, as we shall see shortly, there is a second type of acceleration that is important during rotational motion. at = r\u03b1, (7.7) Example 7.3: Germy compact disc A compact disc rotates from rest up to an angular speed of 31.4 rad/s in a time of 0.892 s. (a) What is the angular acceleration of the disc, assuming the angular acceleration is uniform? (b) Through what angle does the disc turn while coming up to speed? (c) If the radius of the disc is 4.45 cm, \ufb01nd the tangential speed of a microbe riding on the rim of the disc when t = 0.892 s. (d) What is the magnitude of the tangential acceleration of the microbe at the given time? Solution: (a) We are told to assume uniform (constant) angular acceleration, so we can use the equations from the previous section. \u03c9f = \u03c9i + \u03b1t \u03c9f \u2212 \u03c9i t \u03b1 = \u03b1 = 31.4 rad/s 0.892 s \u03b1 = 35.2 rad/s2. (b) Now we want to \ufb01nd the angular displacement, f = \u03c92 \u03c92 i + 2\u03b1\u2206\u03b8 f \u2212 \u03c92 \u03c92 i 2\u03b1 \u2206\u03b8 = \u2206\u03b8 = (31.4 rad/s)2 2(35.2 rad/s2) \u2206\u03b8 = 14.0 rad. (c) Now we use the relationship relating angular velocity and tangential velocity, vt = r\u03c9 vt = (0.0445 m)(31.4 rad/s) vt = 1.4 m/s. (d) This time we need to relate the angular acceleration and the tangential acceleration, at = r\u03b1 at = (0.0445 m)(35.2 rad/s2) at = 1.57 m/s2. 7.2 Centripetal acceleration We just discussed the relationship between the tangential acceleration and the angular acceleration. The tangential acceleration of an object is determined by changes in how fast an object is spinning. There is 56 another type of acceleration that is present in all rotational motion, even when the rate of rotation is not changing. Recall that Newton\u2019s \ufb01rst law tells us that objects will continue to move in a straight line unless acted upon by an outside force. In order for an object to rotate, it must be continuously pulled from the straight line path that it wants to take. The direction of motion of the object is constantly changing, which means that there is some kind of acceleration. This acceleration is called the centripetal acceleration. This acceleration is towards the center of rotation and is responsible for changing the direction of motion. The tangential acceleration is responsible for changing the speed of rotation. The centripetal acceleration is given by ac = v2 r, where v is the tangential velocity. We can re-write this formula in terms of the angular speed, ac = (r\u03c9)2 r = r\u03c92. (7.8) (7.9) The total acceleration consists of both the tangential and the centripetal acceleration. Since the two are always perpendicular, the magnitude of the total acceleration is given by a = c + a2 a2 t. (7.10) 57 Example 7.5: At the racetrack A race car accelerates uniformly from a speed of 40.0 m/s to a speed of 60.0 m/s in 5.00 s while travelling counterclockwise around a circular track of radius 4.00 \u00d7 102 m. When the car reaches a speed of 50.0 m/s, \ufb01nd (a) the magnitude of the car\u2019s centripetal acceleration, (b) the angular speed, (c) the magnitude of the tangential acceleration, and (d) the magnitude of the total acceleration. Solution: (a) The centripetal acceleration can be determined from the tangential speed, ac = ac", " = v2 r (50.0 m/s)2 4.00 \u00d7 102 m ac = 6.25 m/s2. (b) The angular speed is related to the tangential speed, \u03c9 = v r \u03c9 = 50.0 m/s 4.00 \u00d7 102 m \u03c9 = 0.125 rad/s. (c) The acceleration is uniform, so we can use the constant acceleration equations, at = at = vf \u2212 vi \u2206t 60.0 m/s \u2212 40.0 m/s 5.00 s at = 4.00 m/s2. (d) The total acceleration is found from the centripetal and tangential accelerations a2 t + a2 c a = a = (4.00 m/s2)2 + (6.25 m/s2)2 a = 7.42 m/s2. Centripetal force Since there is an acceleration that is directed towards the center, there must be some force pulling objects towards the center of rotation. This is often called the centripetal force, but this is not actually a new force. The centripetal force is one of the forces you are already familiar with (friction, normal force, tension, gravity) that happens to be pulling an object towards the center of rotation. In the case of planets orbiting the sun, the centripetal force is gravity. In the case of a yo-yo being swung in a circle, the centripetal force is tension. When solving rotational motion problems, it is often useful to set up a coordinate system based on the radial and tangential directions of motion. In this case, the net force in the radial direction is the centripetal 58 force and gives rise to the centripetal acceleration, Fc = mac = m v2 r, (7.11) where we have used the formula for centripetal acceleration. If the centripetal force were to disappear, the spinning object would continue travelling in a straight line tangent to the circle. Centrifugal force Many people refer to the centrifugal force when discussing rotational motion. The centrifugal force is a \ufb01ctitious force; it does not exist. When we experience rotational motion, we often \ufb02ee like we are being pushed out from the center of rotation, but this feeling comes about because your body would like to continue in a straight line, tangent to the circle, and you have to exert a centripetal force to continue the rotation. Example 7.6: Car in a turn A car travels at a constant speed of 13.4 m/s on a level circular turn of radius 50.0 m. What minimum coe\ufb03cient of static friction between the tires and the roadway will allow the car to make the circular turn without sliding? Solution: There are three forces acting on the car: gravity, the normal force and friction. There is a frictional force pointing opposite to the direction of motion of the car, but we are not concerned with that kinetic frictional force. There is another frictional force, static friction in this case, due to the circular path of the car. Due to inertia, the car would like to move tangent to the circle, but friction prevents it from doing so. In this example, friction is the centripetal force. fs = m \u00b5sN = m \u00b5smg = m v2 r v2 r v2 r \u00b5s = \u00b5s = v2 gr (13.4 m/s)2 (50.0 m)(9.8 m/s2) \u00b5s = 0.366. 59 Example 7.8: Riding the tracks A roller coaster car moves around a frictionless circular loop of radius R. (a) What speed must the car have so that it will just make it over the top without any assistance from the track? (b) What speed will the car subsequently have at the bottom of the loop? (c) What will be the normal force on a passenger at the bottom of the loop if the loop has a radius of 10.0 m? Solution: (a) There are two forces acting on the car: gravity and the normal force, both acting downwards when the car is at the top of the loop. Since we want the car to make it through the loop without the assistance of the track, we set the normal force to 0. mg + N = m Fy = mac v2 t R v2 t mg = m R vt = gR. (b) We can use conservation of energy to \ufb01nd the car\u2019s speed at the bottom. At the top, the car has both potential and kinetic energy; at the bottom it only has kinetic energy, KEi + P Ei = KEf + P Ef 1 2 mv2 b + 0 mv2 t + mg(", "2R) = 1 2 1 2 5 v2 b = 2 vb = 5gR. gR (c) At the bottom of the loop the normal force and the force of gravity are in opposite directions, N \u2212 mg = m Fy = mac v2 b R 5gR R N = 6mg. N = m + mg 7.3 Gravitation One of the reasons we are interested in circular motion is because we know that astronomical bodies move in (roughly) circular orbits. The centripetal force that causes this rotation is the gravitational force \ufb01rst written by Isaac Newton. His law of universal gravitation states that any two objects will exert an attractive force because of their mass. If two particles with masses m1 and m2 are separated by a distance r, a gravitational force F acts along a line joining them, with magnitude given by where G = 6.673 \u00d7 10\u221211 kg\u22121 \u00b7 m3 \u00b7 s\u22122 is a constant of proportionality called the constant of universal gravitation. F = G m1m2 r2 (7.12) 60 It is important to notice that both objects feel an attractive force; gravitational forces form an action-reaction pair. Example 7.10: Ceres An astronaut standing on the surface of Ceres, the largest asteroid, drops a rock from a height of 10.0 m. It takes 8.06 s to hit the ground. (a) Calculate the acceleration of gravity on Ceres. (b) Find the mass of Ceres, given that the radius of Ceres is RC = 5.1 \u00d7 102 km. (c) Calculate the gravitational acceleration 50.0 km from the surface of Ceres. Solution: (a) We can use kinematics to \ufb01nd the acceleration, \u2206y = v0t + 1 2 at2 a = a = 2\u2206y t2 2(10.0 m) (8.06 s)2 a = 0.308 m/s2. (b) This acceleration is caused by the force of gravity pulling on the rock, mM R2 C aR2 C G ma = G M = M = (0.308 m/s2)(5.1 \u00d7 102 km)2 6.673 \u00d7 10\u221211 kg\u22121 \u00b7 m3 \u00b7 s\u22122 M = 1.2 \u00d7 1021 kg. (c) From the previous part, we have ma = G a = G mM R2 M R2 a = (6.673 \u00d7 10\u221211 kg\u22121 \u00b7 m3 \u00b7 s\u22122) 1.2 \u00d7 1021 kg (5.1 \u00d7 102 km + 50 km)2 a = 0.255 m/s2. 7.3.1 Kepler\u2019s Laws Before Isaac Newton discovered the law of gravity, another astronomer, Johannes Kepler discovered laws that helped him describe the motion of the planets in our solar system. Kepler\u2019s three laws state 1. All planets move in elliptical orbits with the Sun at one of the focal points. 2. A line drawn from the Sun to any planet sweeps out equal areas in equal time intervals, 3. The square of the orbital period of any planet is proportional to the cube of the average distance from the planet to the Sun. It turns out that these laws are a consequence of Newton\u2019s gravitational force. 61 First Law It turns out that any object moving under the in\ufb02uence of an inverse-square force (force varies as 1/r2) will move in an elliptical orbit. Ellipses are curves drawn such that the sum of the distances from any point on the curve to the two foci is constant. For planets in our solar system, the Sun is always at one focus, so a planets distance from the Sun will vary as the planet orbits. Second Law Kepler\u2019s second law has an interesting consequence. Since the planet is closer to the Sun at some points of its orbit than at other points, in order to sweep out equal area in equal time, it must change its speed as it moves around the Sun. The planet will move more slowly when it is far from the Sun and more quickly when it is close to the Sun. Third Law We can derive Kepler\u2019s third law from the law of gravity. Suppose that an object (Mp) is moving in a circular orbit around an object (Ms) with a constant velocity (Is this possible given the second law?). We know that the object undergoes centripetal acceleration and that the force causing this acceleration is the gravitational force, Mpac = Mpv2 r = GMpMs r2. The speed of the object is simply the circumference divided by the time required for one revolution (period) Substituting, we \ufb01nd v = 2\u03c0r T. Mp(4\u03c02r2 rT 2 = GMpMs r2 T 2 = 4\u03c02 GMs r3. (", "7.13) (7.14) 62 Example 7.13: Geosynchronous orbit From a telecommunications point of view, it\u2019s advantageous for satellites to remain at the same location relative to a location on Earth. This can occur only if the satellite\u2019s orbital period is the same as the Earth\u2019s period of rotation, 24.0 h. (a) At what distance from the center of the Earth can this geosynchronous orbit be found? (b) What\u2019s the orbital speed of the satellite? Solution: (a) We can use Kepler\u2019s third law to \ufb01nd the radius of the orbit, T 2 = 4\u03c02 r3 GMs GME 4\u03c02 r = 3 T 2 (6.673 \u00d7 10\u221211 kg\u22121 \u00b7 m3 \u00b7 s\u22122)(5.98 \u00d7 1024 kg (86400 s)2 r = 3 4(3.14)2 r = 4.23 \u00d7 107 m. (b) The speed is simply the circumference divided by the period, v = 2\u03c0r T 2(3.14)(4.23 \u00d7 107 m) 86400 s v = 3.08 \u00d7 103 m/s. v = 7.4 Torque When we studied forces, we treated all objects as point masses and assumed that it didn\u2019t matter where on the object a force was applied. The force would simply move linearly in response to the applied force. This is an extreme simpli\ufb01cation of reality, it actually does matter at which point on the object the force is applied. Forces applied near the edges of an extended object will tend to rotate the object rather than move it forward. If you lay your textbook on the table and push it with a force applied near the center of the book, it will slide forward. If you push the book with a force applied near the edge, it will rotate rather than move forward. Remember that forces cause an acceleration. Similarly, we de\ufb01ne something called a torque which causes an angular acceleration. Forces and torques are not completely independent \u2014 forces cause torques, but torques also depend on where the force is applied and on the angle at which the force is applied. Let F be a force acting on an object, and let r be a position vector from the point of rotation to the point of application of the force. The magnitude of the torque \u03c4 exerted by the force F is \u03c4 = rF sin \u03b8 (7.15) where r is the length of the position vector, F is the magnitude of the force and \u03b8 is the angle between r and F. The unit of torque is newton-meter. Torque is a vector with the direction given by the right hand rule. Point your \ufb01ngers in the direction of r and curl them toward the direction of F. Your thumb will point in the direction of the torque. This will be perpendicular to the plane that contains both r and F. When your \ufb01ngers point in the direction of the torque, your \ufb01ngers will curl in the direction of rotation that the torque will cause. 63 Example 8.2: Swinging door (a) A man applies a force of F = 3.00 \u00d7 102 N at an angle of 60.0\u25e6 to the door 2.00 m from the hinges. Find the torque on the door, choosing the position of the hinges as the axis of rotation. (b) Suppose a wedge is placed at 1.50 m from the hinges on the other side of the door. What minimum force must the wedge exert so that the force applied in part (a) won\u2019t open the door? Solution: (a) We use the above equation to calculate the torque on the door \u03c4 = F r sin \u03b8 \u03c4 = (F = 3.00 \u00d7 102 N)(2.00 m) sin(60.0\u25e6) \u03c4 = 520 N \u00b7 m. The direction of torque is out of the board/page (towards top of door). (b) We don\u2019t want the door to rotate, so there must be no net torque on the door. We need to identify all the torques/forces acting on the door. There are three forces acting on the door: the applied force, the force of the wedge and the force of the hinges. Although the hinges apply a force to the door, they do not exert a torque because they act at the point of rotation. \u03c4hinges + \u03c4wedge + \u03c4 = 0 0 + Fwedgerwedge sin(\u221290\u25e6) + \u03c4 = 0 Fwedge = Fwedge = \u03c4 rwedge sin(\u221290\u25e6) 520 N \u00b7 m (1.50 m) sin(\u221290\u25e6) Fwedge = 347 N. 7.4.1 Equilibrium An object that is in equilibrium must satisfy two conditions, 1. The", " net force must be zero, i.e. no linear acceleration ( F = 0). 2. The net torque must be zero, i.e. no angular acceleration ( \u03c4 = 0). This does not mean that the object is not moving or rotating, it can be moving at a constant velocity or rotating at a constant angular speed. If an object is in equilibrium, we can choose the axis of rotation so we should choose one which makes the calculation convenient. An axis where at least one torque is zero makes calculations easier. 64 Example 8.3: Balancing act A woman of mass m = 55.0 kg sits of the left side of a seesaw \u2014 a plank of length L = 4.00 m, pivoted in the middle. (a) First compute the torques on the seesaw about an axis that passes through the pivot point. Where should a man of mass M = 75.0 kg sit if the system is to be balanced? (b) Find the normal force exerted by the pivot if the plank has a mass of mp = 12.0 kg. (c) Repeat part (a), but this time compute the torques about an axis through the left end of the plank. Solution: (a) We need to identify all the forces acting on the plank and their point of application. The woman and the man will push down on the plank with a force equal to their respective weights. We know the woman\u2019s distance from the center of the plank, but the man\u2019s distance is unknown. Gravity also acts on the plank itself, and will pull down on the center of the plank as if all the plank\u2019s mass was concentrated at that one point. Finally, there is a normal force that pushes upwards from the pivot point. We want the seesaw to be in equilibrium, so the sum of all torques about some axis of rotation (the pivot point in this case) must be zero, \u03c4 = 0 \u03c4N + \u03c4g + \u03c4w + \u03c4m = 0 0 + 0 + mg L 2 \u2212 M gx = 0 x = x = mL 2M (55.0 kg)(4.00 m) 2(75.0 kg) x = 1.47 m. (b) We now want to \ufb01nd the normal force. We can use the \ufb01rst condition of equilibrium to do this, F = 0 \u2212M g \u2212 mg \u2212 mpg + N = 0 N = (M + m + mp)g N = (75.0 kg + 55.0 kg + 12.0 kg)(9.8 m/s2) N = 1.39 \u00d7 103 N. (c) We will now use the left end of the plank (where the woman sits) as the axis of rotation, \u03c4 = 0 \u03c4N + \u03c4g + \u03c4w + \u03c4m = 0 \u2212N L 2 + mpg L 2 + 0 + M g(N \u2212 mpg \u2212 M g) L 2 M g (M g + mg + mpg \u2212 mpg \u2212 M g) L 2 M g mL 2M x = 1.47 m. 65 7.4.2 Center of gravity As we saw in the previous problem, we treat gravity as if it acts on a single point of an extended body. This point is called the center of gravity. Suppose we have an object with some arbitrary shape. We can treat this object as if it is divided into very small pieces of weights m1g, m2g... at locations (x1,y1), (x2,y2).... Each piece contributes some torque about the axis of rotation due to its weight. For example, \u03c41 = m1gx1 and so forth. The center of gravity is the point where we apply a single force of magnitude F = i mig which has the same e\ufb00ect on the rotation of the object as all the individual little pieces. F xcg = ( i mig)xcg = xcg = i \u03c4i migxi i i mixi i mi. (7.16) This gives us the x coordinate of the center of gravity. We can \ufb01nd the y and z coordinates in a similar fashion, ycg = zcg = i miyi i mi i mizi i mi. (7.17) If an object is symmetric, the center of gravity will lie on the axis of symmetry, so it is sometimes possible to guess where the center of gravity is for such objects (like we did for the plank in the example problem). Example 8.4: Center of gravity Three objects are located on the x-axis as follows: a 5.00 kg mass sits at x = \u22120.500 m, a 2.00 kg mass sits at the origin, and a 4.00 kg mass sits at x = 1.00 m. Find the center of gravity. (b) How does the answer", " change if the object on the left is displaced upward by 1.00 m and the object on the right is displaced downward by 0.50 m? Solution: (a) We simply apply the formula for center of gravity that we just derived, xcg = xcg = i mixi i mi (5.00 kg)(\u22120.500 m) + (2.00 kg)(0) + (4.00 kg)(1.00 m) 5.00 kg + 2.00 kg + 4.00 kg xcg = 0.136 m. (b) The x coordinate of the center of gravity will not change since the masses have not been moved along the x-axis. We will, however, have to consider the y axis now, ycg = ycg = i miyi i mi (5.00 kg)(1.00 m) + (2.00 kg)(0) + (4.00 kg)(\u22120.500 m) 5.00 kg + 2.00 kg + 4.00 kg ycg = 0.273 m. 66 7.5 Torque and angular acceleration When an object is subjected to a torque, it undergoes an angular acceleration. We can derive a law similar to Newton\u2019s second law for the e\ufb00ect of a torque. Suppose we have an object of mass m connected to a very light rod of length r. The rod is pivoted about the end opposite the mass and its movement is con\ufb01ned to a horizontal frictionless table. Suppose a tangential force Ft acts on the mass. This will cause a tangential acceleration, Multiplying both sides of the equation by r, Ft = mat. Ftr = mrat, and substituting for the at = r\u03b1 for the tangential acceleration gives The left side is simply the torque, Ftr = mr2\u03b1. \u03c4 = (mr2)\u03b1. (7.18) (7.19) This tells us that the torque is proportional to the angular acceleration. The constant of proportionality is mr2 and is called the moment of inertia of the object. Moment of inertia has units of kg \u00b7 m2 and is denoted by I. So we can write, \u03c4 = I\u03b1. (7.20) This is the rotational analog of Newton\u2019s second law. 7.5.1 Moment of inertia The formula for moment of inertia that we just derived is true for a single point mass only. For extended objects, the moment of inertia will be di\ufb00erent. Consider a solid object rotating about its axis. We can break this object up into many little pieces like we did to \ufb01nd the center of gravity. The net torque on the object will be the sum of the torques caused by all the small pieces, i \u03c4i = ( mir2 i )\u03b1. i The moment of inertia of the whole object then is I = mir2 i. i (7.21) (7.22) We can \ufb01nd the moment of inertia of any object or any collection of objects by adding the moments of inertia of its constituents. Notice that the moment of inertia depends not just on the mass of an object, but on how the mass is distributed within the object. Importantly, it matters how the mass is distributed relative to the axis of rotation. 67 Table 7.1: Moments of inertia for various rigid objects of uniform composition Object Hoop or cylindrical shell Solid sphere Solid cylinder or disk Thin spherical shell Long thin rod Long thin rod Axis of Rotation Moment of inertia center center center center center end I = M R2 I = 2 5 M R2 I = 1 2 M R2 I = 2 3 M R2 I = 1 12 M L2 I = 1 3 M L2 Example 8.9: Baton twirler In an e\ufb00ort to be the star of the halftime show, a majorette twirls an unusual baton made up of four spheres fastened to the ends of very light rods. Each rod is 1.0 m long. Two of the spheres have a mass of 0.20 kg and the other two spheres have a mass of 0.30 kg. Spheres of equal masses are placed across from each other. (a) Find the moment of inertia of the baton through the point where the rods cross. (b) The majorette tries spinning her strange baton about the rod holding the 0.2 kg spheres. Calculate the moment of inertia of the baton about this axis. Solution: (a) When the baton is spinning around the point where the rods cross, all four spheres contribute to the moment of inertia. We can treat the spheres as point masses since their radius is small compared to the length of the rods. I = mir2 i i I = 2m1r2 + 2m2r2 I = 2(0.20 kg)(", "0.5 m)2 + 2(0.30 kg)(0.5 m)2 I = 0.25 kg \u00b7 m2. (b) In this case only the 0.3 kg spheres contribute to the moment of inertia because the 0.2 kg spheres lie along the axis of rotation (so r = 0). I = mir2 i i I = 2m2r2 I = (0.30 kg)(0.5 m)2 I = 0.15 kg \u00b7 m2. The moment of inertia for solid extended objects can be calculated using calculus. The moment of inertia for some common objects is given in Table 7.1. Note that the assumption is that the mass is distributed uniformly throughout these objects. Parallel axis theorem A useful property of the moment of inertia is that it is fairly easy to calculate the moment of inertia about an axis parallel to the axis through the center of gravity of the object. This result is called the parallel axis theorem and is as follows, Iz = Icm + M d2, (7.23) 68 where Icm is the moment of inertia of the object rotating about the center of mass, M is the mass of the object and d is the distance between the two parallel axes. 69 Example 8.11: Falling bucket A solid uniform frictionless cylindrical reel of mass M = 3.00 kg and radius R = 0.400 m is used to draw water from a well. A bucket of mass m = 2.00 kg is attached to a cord that is wrapped around the cylinder. (a) Find the tension T in the cord and acceleration a of the bucket. (b) If the bucket starts from rest at the top of the well and falls for 3.00 s before hitting the water, how far does it fall? Solution: (a) We will need free body diagrams for both the wheel and the bucket. The bucket has two forces acting on it: tension pulling up and gravity pulling down. Note that we don\u2019t care where these forces act on the bucket because this object is not rotating. The cylinder has three forces acting on it: gravity acting at the center and pulling down, a normal force (from the bar holding the cylinder) also acting at the center and pushing up, and tension acting at a distance R and pulling down. We know that the bucket is accelerating and the cylinder has an angular acceleration. We can use Newton\u2019s second law on the bucket, F = ma mg \u2212 T = ma. We can use the rotational analog of Newton\u2019s second law on the cylinder. In this case, we don\u2019t get to choose the point of rotation because the object is rotating about a speci\ufb01c axis, \u03c4 = I\u03b1 R2\u03b1 M R\u03b1. Gravity and the normal force don\u2019t contribute to the torque because they act at the axis of rotation. We now have two equations and three unknowns (a, \u03b1 and T ). We will need one more equation to solve this problem. Remember that the tangential acceleration is related to the angular acceleration of a rotating object. In this case, the rope is causing the tangential acceleration of the cylinder and we know that the acceleration of the rope is the same as that of the bucket, Using this relationship, we \ufb01nd a = R\u03b1. T = 1 2 M a, which we can use to substitute into the \ufb01rst equation, M a = ma mg \u2212 a(m + 1 2 1 2 M ) = mg a = 2 M ) mg (m + 1 (2.00 kg)(9.8 m/s2) 2.00 kg + 1 2 (3.00 kg) a = 5.60 m/s2. a = Now we can also \ufb01nd the tension 70 (3.00 kg)(5.60 m/s2) T = 8.4 N. (b) This is a kinematics problem, \u2206y = v0t + 1 2 at2 \u2206y = 1 2 (5.60 m/s2)(3.0 s)2 \u2206y = 25.2 m. 7.6 Rotational kinetic energy Recall that an object moving through space has kinetic energy. Similarly, a rotating object will have rotational kinetic energy. Consider the mass connected to a light rod rotating on a horizontal frictionless table. The kinetic energy of the mass is KE = mv2. 1 2 We know that the velocity is related to the angular speed, KE = 1 2 m(r\u03c9)2 = 1 2 (mr2)\u03c92 = 1 2 I\u03c92. (7.24) Notice that again the equations for translational (linear) kinetic energy and rotational kinetic energy are quite similar with moment of inertia replacing mass and angular speed replacing linear velocity. In the case of the rotating mass on a rod, either expression can be used to describe it\u2019s energy because it only undergo", "es rotational motion. There are cases, however when both expressions are used such as when balls or wheels are rolling. In this case, there is rotation about the center of mass while the center of mass itself is moving through space. The translational kinetic energy refers to the energy of the center of mass\u2019 motion while the rotational kinetic energy refers to the energy of the rotation. This new type of energy needs to be included in the work-energy theorem, Wnc = \u2206KEt + \u2206KEr + \u2206P E. (7.25) 71 Example 8.12: Ball on an incline A ball of mass M and radius R starts from rest at a height of 2.00 m and rolls down a 30\u25e6 slope. What is the linear speed of the ball when it leaves the incline? Assume that the ball rolls without slipping. Solution: We can use energy to solve this problem. Let\u2019s consider the energy at the top and bottom of the ramp, remembering that the ball rolls (rotates) down the ramp, KEti + KEri + P Eg = KEtf + KErf + P E 0 + 0 + M gh = 1 2 M v2 + 1 2 I\u03c92 + 0. Remember that there is a relationship between the translational velocity and the angular speed. Note that this relationship will only hold if the object \u201crolls without slipping.\u201d If the ball slips then the center of mass moves while the object is not rotating and the relationship does not hold. 2 5 1 2 M R2 2 v R M gh = gh = v = v = 1 5 v2 M v2 + 1 2 1 v2 + 2 10 7 10 7 gh (9.8 m/s2)(2.00 m) v = 5.29 m/s. The velocity is smaller than the velocity of a block sliding down the incline because some of the gravitational potential energy goes into rotational kinetic energy. 7.7 Angular momentum When we apply a torque to an object, we change its angular acceleration \u2014 and we have an equation relating the two. Just as we re-wrote Newton\u2019s second law in terms of momentum, we can re-write the rotational equivalent in terms of angular momentum. \u03c4 = I\u03b1 = I \u2206\u03c9 \u2206t = I\u2206\u03c9 \u2206t = \u2206L \u2206t, (7.26) where we have de\ufb01ned L = I\u03c9 as the angular momentum of an object. If there is no net torque, then the total angular momentum of a system does not change, Li = Lf. The law of conservation of angular momentum explains why \ufb01gure skaters spin faster when they bring their arms closer to their bodies. As their arms move in, their moment of inertia decreases, so their angular speed must increase to compensate. Note that angular momentum is a vector with the direction determined by the direction of angular velocity. Changes in direction of angular momentum (changes in the direction of the axis of rotation) are also subject to conservation of momentum. If the axis of rotation changes, there must be a change in the angular momentum of some other part of the system to compensate. 72 Example 8.14: Merry-go-round A merry-go-round modeled as a disk of mass M = 1.00 \u00d7 102 kg and radius R = 2.00 m is rotating in a horizontal plane about a frictionless vertical axle. (a) After a student with mass m = 60.0 kg jumps on the rim of the merry-go-round, the system\u2019s angular speed decreases to 2.00 rad/s. If the student walks slowly from the edge toward the center, \ufb01nd the angular speed of the system when she reaches a point 5.00 m from the center. (b) Find the change in the system\u2019s rotational kinetic energy caused by her movement to r = 0.500 m. (c) Find the work done on the student as she walks to r = 0.500 m. Solution: (a) There are two parts to the moment of inertia of the system, the moment of inertia of the disk and the moment of inertia of the person. It is the moment of inertia of the student that changes as she walks towards the center \u2014 the moment of inertia of the disk remains the same, \u03c9f = Li = Lf (Id + Ipi)\u03c9i = (Id + Ipf )\u03c9f (Id + Ipi)\u03c9i Id + Ipf ( 1 2 M R2 + mR2)\u03c9i 1 2 M R2 + mr2 ( 1 2 (1.00 \u00d7 102 kg)(2.00 m)2 + (60.0 kg)(2.00 m)2)(2.00 rad/s) 1 2 (1.00 \u00d7 102 kg)(0.500 m)2 \ufffd", "\ufffdf = \u03c9f = \u03c9f = 4.09 rad/s. (b) The change in rotational kinetic energy is \u2206KEr = ( \u2206KEr = 1 2 1 2 \u2206KEr = 920 J. (( \u2212 1 2 1 2 1 2 (Id + Ipf )\u03c92 f \u2212 1 2 (Id + Ipi)\u03c92 i (1.00 \u00d7 102 kg)(0.500 m)2 + (60.0 kg)(2.00 m)2)(4.09 rad/s)2 (1.00 \u00d7 102 kg)(2.00 m)2 + (60.0 kg)(2.00 m)2)(2.00 rad/s))2 (c) When calculating the work done by the student we need to use the change in kinetic energy of the student only, \u2206KEr = \u2206KEr = 1 2 1 2 Ipf \u03c92 f \u2212 1 2 Ipi\u03c92 i (60.0 kg)(2.00 m)2(4.09 rad/s)2 \u2212 1 2 (60.0 kg)(2.00 m)2)(2.00 rad/s))2 \u2206KEr = \u2212355 J. 73 Chapter 8 Vibrations and Waves We have now studied linear motion and circular motion. There is one more very important type of motion that arises in many aspects of physics. Periodic motion, such as waves or vibrations underlies sound and light and many other physical phenomena. 8.1 Return of springs One simple type of periodic motion is an object attached to a spring. Remember that the force of a spring is given by Hooke\u2019s law, F = \u2212kx. (8.1) This force is sometimes called a restoring force because it likes to pull the object back to the equilibrium position. The negative sign ensures that the force is pulling opposite to the direction of displacement. Suppose we pull the object so that the spring is stretched and let go. The spring force will cause an acceleration back towards the equilibrium position. The object will pick up speed as it moves back towards equilibrium and will overshoot the equilibrium position. Once it passes the equilibrium position the object starts to compress the spring and the force changes direction. The force now decelerates the object, eventually causing the object to stop. When the object stops, the spring is compressed and the force still points towards the equilibrium. So the object will accelerate towards the center again. In this way and object will move back and forth endlessly. This is an example of simple harmonic motion. Simple harmonic motion occurs when the net force along the direction of motion obeys Hooke\u2019s Law. Not all periodic motion is simple harmonic motion. Two people tossing a ball back and forth is not simple harmonic motion even though it is periodic. The force causing the motion of the ball is not of the form of Hooke\u2019s Law, so it cannot be simple harmonic motion. The acceleration of an object undergoing simple harmonic motion can be found using Newton\u2019s second law, F = ma \u2212kx = ma k m a = \u2212 x. (8.2) 8.1.1 Energy of simple harmonic motion Let\u2019s consider the energy of an object attached to a spring. Suppose that we pull the object and stretch the spring then release it. Just before the object is released, the spring is at it\u2019s maximum stretch. This is called the amplitude. The energy at this point is E = kA2, (8.3) 1 2 74 where A is the amplitude. Now we release the spring and as it moves it picks up speed. The object now has both potential energy and kinetic energy, so E = 1 2 kx2 + 1 2 mv2. We can use this to \ufb01nd the velocity at any position, 1 2 kA2 = 1 2 1 2 mv2 kx2 + k m v = \u00b1 (A2 \u2212 x2). (8.4) (8.5) The \u00b1 appears because of the square root. The usual convention is that if the object moves to the right, the velocity is positive; if it moves to the left, it is negative. 8.1.2 Connecting simple harmonic motion and circular motion When the object on the spring moves back and forth, it\u2019s similar to an object moving with constant angular velocity around a circle. The object moving around the circle will come back to its original position at regular time intervals, just like the mass on a spring. Remember that the period of a rotating object is T = 2\u03c0r v. (8.6) For the rotating object, r is the size of the spatial displacement and corresponds to the amplitude of the mass on a spring, T = 2\u03c0A v. The velocity of the mass after it has travelled a distance A can be found from Eq. (8.5) by setting x = 0", ", Now we can put this into the equation for the period, v = A k m. T = 2\u03c0 m k. (8.7) This represents the time it takes for a mass on a spring to return to its starting position. A larger mass gives a longer period, while a larger spring constant (sti\ufb00er spring) gives a shorter period. The frequency is the inverse of the period f = 1 T = 1 2\u03c0 k m. (8.8) The units of frequency are cycles per second or Hz. This is related the angular frequency (which is in radians per second), k m. (8.9) \u03c9 = 2\u03c0f = 75 Example 13.5: Shock absorbers A 1.3 \u00d7 103 kg car is constructed on a frame supported by four springs. Each spring has a spring constant of 2.00 \u00d7 104 N/m. If two people riding in the car have a combined mass of 1.6 \u00d7 102 kg, \ufb01nd the frequency of vibration of the car when it is driven over a pothole. Find also the period and the angular frequency. Assume the weight is evenly distributed. Solution: First we need to \ufb01nd the total mass, mt = mc + mp = 1.3 \u00d7 103 kg + 1.6 \u00d7 102 kg = 1.46 \u00d7 103 kg Each spring will hold up one quarter of the total mass. The frequency is f = f = 1 2\u03c0 1 2\u03c0 k m 2.00 \u00d7 104 N/m 365 kg f = 1.18 Hz. The period is the inverse of frequency and the angular frequency is T = 1 f = 1 1.18 Hz = 0.847 s, \u03c9 = 2\u03c0f = 2\u03c0(1.18 Hz) = 7.41 rad/s. 8.2 Position, velocity and acceleration Suppose a mass is moving on a circle with constant angular velocity. If we look at it\u2019s x position as it moves around the circle, we see that it oscillates somewhat like a mass on a spring. The x position of the mass is given by We know that the mass is moving with constant angular speed so x = A cos \u03b8. x = A cos(\u03c9t) = A cos(2\u03c0f t). (8.10) (8.11) This equation describes the position of an object undergoing simple harmonic motion as a function of time. We can substitute this into Eq. (8.5) v = \u00b1 (A2 \u2212 x2) k m k v = \u00b1 m v = A\u03c91 \u2212 cos2(2\u03c0f t) v = A\u03c9 sin(2\u03c0f t). (A2 \u2212 (A cos(2\u03c0f t))2 (8.12) The velocity also oscillates, but it is 90\u25e6 out of phase with the displacement. When the displacement is a maximum or minimum, velocity is zero and vice versa. The maximum value (amplitude) of velocity is A\u03c9 76 (when sin(\u03c0f t) = 1). We can also derive an expression for the acceleration a = \u2212 k m x a = \u2212A\u03c92 cos \u03c9(2\u03c0f t). (8.13) The acceleration is also sinusoidal and 180\u25e6 out of phase with the displacement. When the displacement is a maximum, acceleration is a minimum and vice versa. The maximum acceleration (amplitude) is A\u03c92. 77 Example 13.6: Vibrating system (a) Find the amplitude, frequency, and period of motion for an object vibrating at the end of a horizontal spring if the equation for its position as a function of time is x = (0.250 m) cos \u03c0 8.00 t. (b) Find the maximum magnitude of the velocity and acceleration. (c) What are the position, velocity, and acceleration of the object after 1.00 s has elapsed? Solution: (a) Compare the given function to the standard function for simple harmonic motion and we can just read o\ufb00 the amplitude, and the frequency x = A cos(2\u03c0f t) A = 0.250 m, 2\u03c0f = f = \u03c0 8.00 1 16 = 0.0625 Hz. The period is the inverse of frequency (b) The maximum velocity is T = 1 f = 1 0.0625 Hz = 16 s. vmax = A\u03c9 vmax = (0.250 m)(2\u03c0)(0.0625 Hz) vmax = 0.098 m/s, and the maximum acceleration is amax = A\u03c92 amax = (0.250 m)(2\u03c0)2(0.0625 Hz)2 amax = 0.039 m/s2. (c) We simply substitute into the given equation and for the velocity and for the acceleration x = (0.250 m) cos \u03c0 8", ".00 = 0.231 m, v = \u2212(0.098 m) cos \u03c0 8.00 = \u22120.038 m/s, a = \u2212(0.039 m) cos \u03c0 8.00 = \u22120.036 m/s2. 78 8.3 Motion of a pendulum Another type of periodic motion that you may have observed is that of a pendulum swinging back and forth. To determine if it is simple harmonic motion, we need to \ufb01gure out whether there is a Hooke\u2019s law type force causing the pendulum to move. There are two forces acting on the pendulum: the force of gravity pulls down and tension pulls towards the center of rotation. If the mass is pulled away from the equilibrium position, then the force trying to pull it back towards equilibrium is F = \u2212mg sin \u03b8, (8.14) where \u03b8 gives the angular displacement of the pendulum. We know that the linear displacement is s = L\u03b8 where L is the length of the pendulum (radius of the circle on which the pendulum moves). So the force pulling along the path towards equilibrium is F = \u2212mg sin. s L (8.15) This does not look like Hooke\u2019s law, so in general the motion of a pendulum is not simple harmonic. At small angles, however, the sine of an angle is approximately the same as the angle itself (as long as it\u2019s measured in radians). So for small angles, we can write F = \u2212mg s L = \u2212 mg L s. (8.16) Now the equation looks like Hooke\u2019s law. The force is propotional to the linear displacement with the \u201cspring constant\u201d given by k = mg/L. Remember, however, that this is only valid for small angles. Recall that the angular frequency for an object undergoing simple harmonic motion is \u03c9 = k m. We have an expression for k for a pendulum, so we can substitute, \u03c9 = mg/L m = g L. From that we can \ufb01nd the frequency and the period g L f = T = 1 2\u03c0 \u03c9 = 1 f = 2\u03c0 1 2\u03c0 L g. (8.17) (8.18) (8.19) Note that the period depends only on the length of the pendulum and not on its mass or the amplitude of the motion. 79 Example 13.7: Measuring g Using a small pendulum of length 0.171 m, a geophysicist counts 72.0 complete swings in a time of 6.00 s. What is the value of g in this location? Solution: We \ufb01rst need to determine the period of oscillation, T = T = time # of oscillations 6.00 s 72.0 T = 0.833 s. We can use this to \ufb01nd g, T = 2\u03c0 L g g = 4\u03c02 L T 2 g = 4\u03c02 0.171 m (0.833 s)2 g = 9.73 m/s2. 8.4 Damped oscillations So far we have assumed that the objects will continue oscillating forever. In the real world energy losses due to friction will cause the oscillating object to slow down. In this case the motion is said to be damped. If we consider the mass on the spring, we know that the mass will oscillate for some time but that the amplitude will decrease over time. This scenario is an underdamped oscillation. Suppose now we put the mass on a spring into a liquid. If the liquid is thick enough it will prevent the oscillations and simply allow the spring to come back to its equilibrium position. If the object returns to equilibrium rapidly without oscillating, then the motion is critically damped. If the object returns to equilibrium slowly, the motion is overdamped. 8.5 Waves A wave is typically thought of as a disturbance moving through a medium. When a wave passes through a medium, the individual components of that medium oscillate about some equilibrium point, but they do not move with the wave. Imagine a leaf \ufb02oating in a pond. You throw a pebble into the pond near the leaf. This creates a wave in the water. When the wave reaches the leaf, it causes the leaf to bo up and down, but it does not carry the leaf with it. The leaf was temporarily disturbed, but once the wave passes it goes back to its original state. 8.5.1 Types of waves Suppose you \ufb01x one end of a string to a wall and you hold the other end. If you quickly move your hand up and down, you will create a wave (in this case a pulse) that travels down the string. This is a traveling wave. In the case of the pulse on the string, the individual bits of string move up and", " down as the pulse goes through. They do not move in the direction of pulse. When the disturbed medium moves perpendicular to the direction of the wave, the wave is called a transverse wave. A longitudinal wave occurs when the disturbed medium oscillates along the direction of travel of the wave. A good example of this is to alternately stretch 80 and compress a spring. The stretched and compressed regions will move down the spring with each coil oscillating along the direction of the wave. While the coils oscillate along this direction they still do not actually move with the wave. It turns out that each point in the medium undergoes simple harmonic motion as the wave passes through. 8.5.2 Velocity of a wave The frequency of a wave is determined by the frequency of the individual oscillating points. The amplitude of the wave is the amplitude of the oscillations. The wavelength of a wave is the distance between two successive points that behave identically (peak to peak, for example). The wavelength is denoted by \u03bb. From these quantities we can determine the speed of the wave. The wave speed is the speed at which a particular part of the wave (like the peak) travels through the medium. Remember that speed is the displacement over time, v = \u2206x \u2206t. We know that the wave moves a distance of one wavelength in the time it takes for one point on the wave to move through a single cycle (it\u2019s period), v = \u03bb T = f \u03bb. (8.20) 8.5.3 Interference of waves One interesting aspect of waves is how they interact with other waves. Two travelling waves will pass right through each other when they meet. When you throw two pebbles into the water near each other they will each create waves rippling from the point of entry. When those two waves meet they don\u2019t destroy each other. Each wave comes out of the interaction undisturbed. At the point(s) where the two waves meet, they interact with each other in a process called interference. At these points, the motion of the points in space is determined by the principle of superposition: When two or more travelling waves encounter each other while moving through a medium, the resultant wave is found by adding together the displacements of the individual waves point by point. If the peaks and troughs of two waves occur at the same place at the same time, the waves are in phase and the resulting interference is constructive interference. If the peak of one wave occurs at the same time and place as the trough of another wave then the waves are inverted and the resulting interference is destructive interference. In this case the waves completely cancel each other in the region where they interact (they will re-appear once they pass through each other). 8.5.4 Re\ufb02ection of waves Waves cannot travel in a particular medium inde\ufb01nitely. Eventually the waves will reach a boundary. When the waves reach the boundary, some of the wave will be re\ufb02ected and some of the wave will be transmitted. The re\ufb02ected wave can sometimes be inverted with respect to the incoming wave. Consider a wave on a string that approaches a wall where the string is \ufb01xed. The string will pull on the wall as the wave hits. The wall will exert an equal and opposite force on the string (Newton\u2019s third law), pulling the string in the opposite direction. This causes the re\ufb02ected wave to be inverted. If the end of the string is free to move, the re\ufb02ected wave will have the same orientation as the original wave. 81 8.6 Sound waves Sound waves are longitudinal waves that are caused by vibrating objects. When an object vibrates, it pushes the air near it causing alternating compression and stretching of the spacing between molecules (density) in the air. This vibration is picked up by our ears and is interpreted by our brains as sound. In a sound wave, the air molecules oscillate along the direction of travel of the wave (think of the longitudinal wave on a spring). Sound waves can have a range of frequencies. The audible waves have frequencies between 20 and 20000 Hz. Infrasonic waves have frequencies below the audible range while ultrasonic waves have frequencies above the audible range. 8.6.1 Energy and intensity of sound waves Sound waves are created because a vibrating object pushes air molecules. The vibrating object exerts a force on the air and so is doing work on the air. The sound wave carries that energy away from the vibrating object. For waves, we don\u2019t typically measure the total energy in the wave, but instead measure the \ufb02ow of energy. The average intensity I of a wave on a given surface is de\ufb01ned as the rate at which energy \ufb02ows through the surface, divided by the surface area, I = 1 A \u2206E\u2206t, (8.21) where", " the direction of energy \ufb02ow is perpendicular to the surface. The rate of energy transfer is power, so we can also write this as. (8.22) The units of intensity are W/m2. I = P A The faintest sounds a human ear can hear have an intensity of about 1 \u00d7 10\u221212 W/m2. This is the threshold of hearing. The loudest sounds the ear can tolerate have an intensity of about 1 W/m2. This is the threshold of pain. You\u2019ll notice that the intensities that a human ear can detect vary over a very wide range. The quietest sounds don\u2019t seem to us to be 1\u00d71012 times quieter than the loudest sounds because our brains us an approximately logarithmic scale to determine loudness. This is measured by the intensity level de\ufb01ned by \u03b2 = 10 log, (8.23) I I0 where the constant I0 = 1 \u00d7 10\u221212 W/m2 is the reference intensity. \u03b2 is measured in decibels (dB). 82 Example 14.2: Noisy grinding machine A noisy grinding machine in a factory produces a sound intensity of 1.0 \u00d7 10\u22125 W/m2. Calculate (a) the decibel level of this machine and (b) the new intensity level when a second, identical machine is added to the factory. (c) A certain number of additional machines are put into operation alongside these two. The resulting decibel level is 77.0 dB. Find the sound intensity. Solution: (a) For a single grinder \u03b2 = 10 log I I0 \u03b2 = 10 log 1.0 \u00d7 10\u22125 W/m2 1.0 \u00d7 10\u221212 W/m2 \u03b2 = 70.0 dB. (b) With a second grinder the total intensity is 2.0 \u00d7 10\u22125 W/m2. The decibel level is \u03b2 = 10 log I I0 \u03b2 = 10 log 2.0 \u00d7 10\u22125 W/m2 1.0 \u00d7 10\u221212 W/m2 \u03b2 = 73.0 dB. (c) In this case we are given the decibel level and want the intensity \u03b2 = 10 log I I0 77 dB = 10 log I 1.0 \u00d7 10\u221212 W/m2 I 1.0 \u00d7 10\u221212 W/m2 I = 5.01 \u00d7 10\u22125 W/m2. 107.7 = There are \ufb01ve machines in all. Many sound waves can be thought of as coming from a point source. A point source is small compared to the waves and emits waves symmetrically. The waves emitted by a point source are spherical waves; they spread in a uniform sphere. Suppose that the average power emitted by the source is Pav. Then the intensity at a distance r is I = Pav A = Pav 4\u03c0r2. The average power always remains the same, no matter the distance so we can write Since the average power is the same, we get I1 = I2 = Pav 4\u03c0r2 1 Pav 4\u03c0r2 2. I1 I2 = r2 2 r2 1. 83 (8.24) (8.25) 8.6.2 The doppler e\ufb00ect When a moving object is making a sound, the frequency of the sound changes as as the object moves towards or away from the observer. Think of a train blowing its whistle \u2014 the whistle changes in pitch as the train approaches and as it moves away. This is known as the Doppler e\ufb00ect. Suppose that a source is moving through the air with velocity vs towards a stationary observer. Since the source is moving towards the observer, in the same direction as the sound wave, the waves emitted by the source get \u201csquished\u201d. The wavelength measured by the observer is shorter than the actual wavelength emitted by the source. During a single vibration, which lasts a time T (the period), the source moves a distance vsT = vs/fs. The wavelength detected by the observer is shortened by this amount, The frequency heard by the observer is \u03bbo = \u03bbs \u2212 vs fs. Rearranging, we get fo = v \u03bbo = v \u03bbs \u2212 vs fs = v \u2212 vs fs. v fs fo = fs v v \u2212 vs. (8.26) (8.27) The observed frequency increases when the source move towards the observer and decreases when it moves away (vs becomes negative). We can do a similar analysis for the case when the source is stationary and the observer is moving. In fact, both source and observer can be moving and this is covered by the general equation fo = fs v + vo v \u2212 vs. (8.28) The sign convention is that velocities are positive when source and observer move towards each other and negative when source and observer move away from each other.", " Example 14.4: Train whistle A train moving at a speed of 40.0 m/s sounds its whistle, which has a frequency of 5.00 \u00d7 102 Hz. Determine the frequency heard by a stationary observer as the train approaches the observer. Solution: The velocity of sound is 345 m/s. We use the equation for the doppler e\ufb00ect, keeping in mind that the train is approaching the observer, fo = fs v + vo v \u2212 vs fo = (500 Hz) fo = 566 Hz. 331 m/s 331 m/s \u2212 40 m/s 8.7 Standing waves Suppose we connect one end of a string to a stationary clamp and the other end to a vibrating object. The vibrating object will move down to the end of the string and will be re\ufb02ected. The re\ufb02ected wave will interact with the wave originating from the object and the two waves will combine according to the principle If the string vibrates at exactly the right frequency the wave appears to stand still so of superposition. 84 it is called a standing wave. A node occurs when the two travelling waves have the same amplitude but opposite displacement, so the net displacement is zero. Halfway between two nodes there will be an antinode where the string vibrates with the largest amplitude. Note that the distance between two nodes is half the wavelength dN N = \u03bb/2. Suppose we \ufb01x both ends of the string, then both ends must be nodes. We can then pluck the string so that we get a single antinode over the length of the string. This is the fundamental or \ufb01rst harmonic and we have half a wavelength on the string. Alternatively, we could set up our wave so that there is another node in the middle of the string. This is the second harmonic and we now have a full wavelength on the string. In fact there are many node/antinode patterns we can set up on the string always keeping in mind that the ends must be nodes. In general, we can set up waves whose wavelengths satisfy the condition \u03bbn = 2L n, (8.29) where n is the harmonic of the wave and can be any positive integer. The frequency of the harmonic is fn = v \u03bbn = vn 2L. (8.30) The velocity of a wave on a string depends on the tension in the string F and on the mass density of the string \u00b5, F \u00b5. v = This allows us to write fn = F \u00b5. n 2L This series of frequencies forms a harmonic series. Example 14.8: Harmonics of a stretched wire (8.31) (8.32) (a) Find the frequencies of the fundamental and second harmonics of a steel wire 1.00 m long with a mass per unit length of 2.00 \u00d7 10\u22123 kg/m and under a tension of 80.0 N. (b) Find the wavelengths of the sound waves created by the vibrating wire. Assume the speed of sound is 345 m/s. Solution: (a) We use the formula for frequency with n = 1 and n = 2 for the fundamental and second harmonics, f1 = f1 = F \u00b5 F \u00b5 1 2L 2 2L = = 1 2(1.00 m) 80.0 N 2.00 \u00d7 10\u22123 kg/m = 100 Hz 1 (1.00 m) 80.0 N 2.00 \u00d7 10\u22123 kg/m = 200 Hz. (b) The frequency of the vibrating string will be transferred to the air. The wave will then move at the speed of sound, \u03bb1 = \u03bb2 = v f1 v f2 = = 345 m/s 100 Hz 345 m/s 200 Hz = 3.45 m = 1.73 m 85 Standing waves can also be set up with sound waves in a pipe. Even if the end of the pipe is open, some of the sound wave will be re\ufb02ected back into the pipe by the edges of the pipe. The re\ufb02ected wave will interfere with the original wave and, if the frequency is right, a standing wave can be established. For pipes, the possible standing wave frequencies will depend on whether one end of the pipe is closed or if both ends are open. If both ends are open, then there must be antinodes at either end of the pipe. The \ufb01rst harmonic will have a single node in middle. If the length of the pipe is L, then the wavelength is \u03bb1 = 2L. The second harmonic will have two nodes and a wavelength of \u03bb2 = L. In general, the wavelength will be \u03bbn = 2L/n. The frequency will be fn = v \u03bbn nv 2L =, (8.33) where v is the speed of sound. For a pipe that is", " closed at one end, there must be a node at the closed end and an antinode at the open end. In this case, the \ufb01rst harmonic has only a quarter wavelength inside the pipe, \u03bb1 = 4L. The next possible harmonic will have one node inside the pipe, but not at the center. In this case, the wavelength is \u03bb3 = 4L/3. There are no even harmonics in a pipe with one end closed; only the odd harmonics are possible. In general, the wavelength is \u03bbn = 4L/n where n is odd integers only. The frequency is fn = nv 4L. (8.34) 8.8 Beats Standing waves are created by interference between two waves of the same frequency, More often, waves of di\ufb00erent frequencies will interfere with each other. When waves of di\ufb00erent frequencies interfere, then they will alternately go in and out of phase causing periods of constructive interference followed by periods of destructive interference. If you are listening to two sound waves of di\ufb00erent frequencies, you will hear a sound that alternates between loud and soft. This loud/soft pattern is known as beats and is also wave. The frequency of the beats is determined by the frquency di\ufb00erence of the two waves fb = |f2 \u2212 f1|. (8.35) 86 Example: Out of tune pipes Two pipes of equal length are each open at one end. Each has a fundamental frequency of 480 Hz when the speed of sound is 347 m/s. In one pipe the air temperature is increased so that the speed of sound is now 350 m/s. If the two pipes are sounded together, what beat frequency results? Solution: We will need to \ufb01nd the new fundamental frequency of the second pipe. In order to do this, we need to know the length of the pipe. We know the frequency of the unheated pipe, so we can \ufb01nd the length, f1 = L = L = v 2L v 2f 347 m/s 2(480 Hz L = 0.36 m. Now we can \ufb01nd the new fundamental frequency of the second pipe, f1 = f1 = v 2L 350 m/s 2(0.36 m f1 = 486 Hz. The frequency of beats is the di\ufb00erence between those two frequencies, fb = |f1 \u2212 f2| fb = |486 Hz \u2212 480 Hz| fb = 6 Hz. 87 Chapter 9 Solids and Fluids 9.1 States of matter The matter you interact with every day is typically classi\ufb01ed as being in one of three states: solid, liquid or gas. There is also a fourth state of matter that you will not ordinarily encounter, plasma. Matter consists of molecules, which are groups of atoms. The properties of particular states of matter are determined by how molecules of a substance interact. At low temperatures, most substances are solid. Macroscopically, solids have a de\ufb01nite volume and shape. In a solid, the molecules are held in (relatively) \ufb01xed positions relative to each other. There are usually electrical bonds between molecules that make it di\ufb03cult for molecules to move away from each other. As temperature increases, substances change from solid to liquid. A liquid has a de\ufb01nite volume, but no \ufb01xed shape. In a liquid, the molecules are weakly bound to each other and so can move within the substance with some freedom. Further increases in temperature completely break the bonds between molecules, changing the liquid into a gas. A gas has no de\ufb01nite shape and no de\ufb01nite volume. The molecules of a gas are far from each other (relative to the size of the molecules) and very rarely interact. This means that the gas can expand to \ufb01ll an volume. At extremely high temperatures (such as those encountered inside stars) the molecules and atoms of the substance are torn apart. Positive and negatively charged particles are free to move around within the substance creating long-range electrical and magnetic forces. This is a plasma. 9.1.1 Characterizing matter Even though substances may be in the same state and will have some broad general characteristics in common, they are by no means identical. For example, two equal masses of di\ufb00erent substances may not take up the same volume. This property is the density of a substance and is de\ufb01ned as the mass divided by its volume, \u03c1 =. (9.1) M V The SI unit for density is kg/m3. The density of a liquid or solid varies slightly with temperature and pressure. The density of a gas is very sensitive to temperature and pressure. Liquids are generally, but not always, less dense than solids. G", "asses are about 1000 times less dense than liquids or solids. It is sometimes convenient to standardize density by comparing it to some standard. The speci\ufb01c gravity of a substance is the ratio of its density to the density of water at 4 \u25e6C, which is 1.0 \u00d7 103 kg/m3. 88 9.2 Deformation of solids While a solid tends to have a de\ufb01nite shape, the shape can often be altered with the application of a force. While a strong enough force will permanently alter the shape, often when the force is removed, the substance will return to its original shape. This is called elastic behaviour. Di\ufb00erent substances have di\ufb00erent elastic properties, so we will need some way to quantify or characterize this. The stress on a material is the force per unit area that is causing some deformation. The strain is a measure of the amount of deformation in the material. For small stresses, stress is proportional to the strain with the constant of proportionality depending on the properties of the material. The proportionality constant is called the elastic modulus. The equation for the elasticity of a substance is similar to Hooke\u2019s law, F = \u2212k\u2206x and the elastic modulus can be thought of as a spring constant. It is a measure of the sti\ufb00ness of a material. A substance with a large elastic modulus is hard to deform. 9.2.1 Young\u2019s modulus Suppose we have a long bar of cross-sectional A and length L0 that is clamped at one end. When we apply an external force F along the bar, we can change the length of the bar. At this new length, the external force is balanced by internal forces that resist the stretch. The bar is said to be stressed. The tensile stress is the magnitude of the external force divided by the cross-sectional area. The tensile strain is the ratio of the change in length to the original length. Since we know that stress and strain are proportional, we have In this equation the proportionality constant is called Young\u2019s modulus. We can re-write this equation as F A = Y \u2206L L0. (9.2) F = Y A L0 \u2206L, so that it looks like Hooke\u2019s law with a spring constant of k = Y A/L0. The Young\u2019s modulus depends on whether the material is being stretched or compressed. Many materials are easier to stretch than compress. The elastic response is also not quite linear and substances have an elastic limit. The elastic limit is the point at which the stress is no longer proportional to the strain. If stretched (compressed) beyond this limit, substances will not return to their original shape once the force is released. The ultimate strength is the largest stress that the substance can endure and any force beyond that reaches the substance\u2019s breaking point. 9.2.2 Shear modulus Suppose we have a rectangular block of some substance. One side of the rectangle is held in a \ufb01xed position. A force is applied to the other side, parallel to the side (think of sliding the cover of a book that is sitting on a table). This is called shear stress. The shear stress is the ratio of the magnitude of the applied force to the area of the face being sheared. The shear strain is the ratio of the horizontal distance moved to the height of the object. F A \u2206x h = S, (9.3) where S is the shear modulus of the substance. In this case, the \u201cspring constant\u201d is k = SA/h. A substance with a large shear modulus is di\ufb03cult to bend. 89 9.2.3 Bulk modulus Suppose we have a block of some substance and we squeeze it uniformaly with a perpendicular force from all sides. This type of squeezing is common when a substance is immersed in a \ufb02uid. The volume stress is the ratio of the change in the magnitude of the applied force to the surface area. The volume strain is the ratio of the change in volume to the original volume. \u2206F A = \u2212B \u2206V V, (9.4) where B is the bulk modulus. The negative sign appears so that B is positive. An increase in the external force (more squeezing) results in a decrease of the volume. Materials with a large bulk modulus are di\ufb03cult to compress. Example: Shear stress on the spine Between each pair of vertebrae of the spine is a disc of cartilage of thickness 0.5 cm. Assume the disc has a radius of 0.04 m. The shear modulus of cartilage is 1 \u00d7 107 N/m2. A shear force of 10 N is applied to one end of the disc while the", " other end is held \ufb01xed. (a) What is the resulting shear strain? (b) How far has one end of the disc moved with respect to the other end? Solution: (a) The shear strain is caused by the shear force, strain = strain = F AS 10 N \u03c0(0.04 m)2(1 \u00d7 107 N/m2) strain = 1.99 \u00d7 10\u22124. (b) A shear strain is de\ufb01ned as the displacement over the height, strain = \u2206x h \u2206x = h \u00d7 strain \u2206x = (0.5 cm)(1.99 \u00d7 10\u22124) \u2206x = 0.99 \u00b5m. 9.3 Pressure and \ufb02uids While a force can deform or break a solid, forces applied to a \ufb02uid have a di\ufb00erent result. When a \ufb02uid is at rest, all parts of the \ufb02uid are in static equilibrium. This means that the forces are balanced for every point in the \ufb02uid. If there was some kind of a force imbalance at one point, then that part of the \ufb02uid would move. When discussing \ufb02uids, we often don\u2019t consider a force directly, but rather use the pressure which is the force per unit area, since \ufb02uids (and the forces that act on them) tend to be extended over some region of space. Mathematically, the pressure is given by the formula, P = F A. (9.5) The units of pressure are newton per meter2 or pascal (Pa). Suppose we have some \ufb02uid sitting in equilibrium in a large container. Consider the forces acting on a piece of the \ufb02uid extending from y1 to y2 (y = 0 is the top of the \ufb02uid) and having a cross-sectional area A. There are three forces acting on this piece of \ufb02uid: the force of gravity (M g), the force caused by the 90 pressure of the \ufb02uid above this piece pushing down (P1A), and the force caused by the pressure from the \ufb02uid below pushing up (P2A). This piece of \ufb02uid is not moving, so the forces must balance, P2A \u2212 P1A \u2212 M g = 0. (9.6) We can \ufb01nd the mass of the water from the density M = \u03c1V = \u03c1A(y1 \u2212 y2). Substituting into our equation, we get P2 = P1 + \u03c1g(y1 \u2212 y2). (9.7) You\u2019ll notice that the force of the \ufb02uid pushing upward is larger than the force of the \ufb02uid pushing down (the di\ufb00erence being the weight of the \ufb02uid we\u2019re considering). For a liquid near the surface of the earth exposed to the earth\u2019s atmosphere, this equation can give us the pressure at any depth h, P = P0 + \u03c1gh, (9.8) where P0 = 1.013 \u00d7 105 Pa is the atmospheric pressure at sea level. The atmospheric pressure arises because air is also a \ufb02uid and the large column of air over the surface of any point on earth will exert a downward pressure on the earth. This equation also suggests that if you change the pressure at the surface of a \ufb02uid, then the change is transmitted to every point in the \ufb02uid. This is known as Pascal\u2019s principle. The change in pressure will also be transmitted to the containers enclosing the \ufb02uid. Example: Oil and Water In a huge oil tanker, salt water has \ufb02ooded an oil tank to a depth h2 = 5.00 m. On top of the water is a layer of oil h1 = 8.00 m deep. The oil has a density of 700 kg/m3 and salt water has a density of 1025 kg/m3. Find the pressure at the bottom of the tank. Solution: The surface of the oil is exposed to air, so the pressure at that point will be atmospheric pressure. We can \ufb01nd the pressure at the bottom of the oil layer using our equation, P1 = P0 + \u03c1oilgh1 P1 = 1.01 \u00d7 105 Pa + (700 kg/m3)(9.8 m/s2)(8.00 m) P1 = 1.56 \u00d7 105 Pa. This is the pressure at the surface of the water layer. At the bottom of that layer, the pressure is, P2 = P1 + \u03c1watergh2 P2 = 1.56 \u00d7 105 Pa + (1025 kg/m3)(9.8 m/s2)(", "5.00 m) P2 = 2.06 \u00d7 105 Pa. 9.4 Buoyant forces The idea of buoyancy was discovered by the Greek mathematician Archimedes and is known as Archimedes\u2019 principle: Any object completely or partially submerged in a \ufb02uid is buoyed up by a force with magnitude equal to the weight of the \ufb02uid displaced by the object. Basically, buoyancy is the pressure di\ufb00erence between the \ufb02uid below and above an object. We know that the pressure from \ufb02uid below is larger than pressure from the \ufb02uid above, so the object will feel lighter if we try to lift it (since the \ufb02uid is helping us to move the object upward). 91 Suppose we replace the little piece of \ufb02uid in our container with a piece of lead of the same volume. The pressure above and below the lead will not change \u2014 their di\ufb00erence is still equal to the mass of the \ufb02uid. The lead is denser than water, so the piece of lead is heavier and the downward force of gravity is now larger. Since the forces are no longer in equilibrium, the lead will sink. The buoyant force is due to pressure di\ufb00erences in the surrounding \ufb02uid and will not change if a new substance is introduced. The buoyant force is given by, B = \u03c1\ufb02uidV\ufb02uidg, (9.9) where V\ufb02uid is the volume of \ufb02uid displaced by the object. 9.4.1 Fully submerged object For a fully submerged object, the buoyant force pushes upwards while the force of gravity pulls the object downwards. M g \u2212 B = M a \u03c1\ufb02uidV\ufb02uidg \u2212 \u03c1objectVobjectg = \u03c1objectVobjecta a = (\u03c1\ufb02uid \u2212 \u03c1object) g \u03c1object. (9.10) The acceleration will be positive (upwards) if the density of the \ufb02uid is larger than the density of the object. It will be negative (downwards) if the density of the object is larger than the density of the \ufb02uid. 9.4.2 Partially submerged object In this case, the object is in equilibrium since it is \ufb02oating in the \ufb02uid and not moving either up or down. This means that the forces must be in equilibrium. B = M g \u03c1\ufb02uidV\ufb02uidg = \u03c1objectVobjectg \u03c1object \u03c1\ufb02uid = V\ufb02uid Vobject. (9.11). 92 Example 9.8: Weighing a crown A bargain hunter purchases a \u201cgold\u201d crown at a \ufb02ea market. After she gets home, she hangs it from a scale and \ufb01nds its weight to be 7.48 N. She then weighs the crown while it is immersed in water and now the scale reads 6.86 N. Is the crown made of pure gold? Solution: To determine whether the crown is actually made of gold, we need to \ufb01nd the density of the crown. For this, we will need both the volume and mass of the crown. When the crown is weighed in the air, we have, When the crown is in water, the buoyant force needs to be included, Tair \u2212 mg = 0. Given these two equations, then, we must have that Twater + B \u2212 mg = 0. B = Tair \u2212 Twater B = 7.48 N \u2212 6.86 N B = 0.980 N. The buoyant force is equal to the weight of the water displaced, B = \u03c1waterVobjg Vobj = Vobj = B \u03c1waterg 0.980 N (1.0 \u00d7 103 kg/m3)(9.8 m/s2) Vobj = 1.0 \u00d7 10\u22124 m3. We can easily get the mass from the \ufb01rst equation, m = Tair g 7.48 N 9.8 m/s2 m = 0.800 kg. m = The density of the crown is \u03c1crown = \u03c1crown = m Vobj 0.800 kg 1.0 \u00d7 10\u22124 m3 \u03c1crown = 8.0 \u00d7 103 kg/m3. The density of gold is 19.3 \u00d7 103 kg/m3, so this crown is de\ufb01nitely not solid gold. 93 9.5 Fluids in motion When \ufb02uids move, there are two broad categories for the type of motion. Laminar or streamline motion occurs when every particle that passes a particular point moves along the same smooth path followed by previous particles passing that point. The path itself is called a streamline.", " During laminar motion, di\ufb00erent streamlines will not cross. When the motion of the \ufb02uid becomes irregular, or turbulent, the streamlines disappear and neighbouring particles can end up moving in very di\ufb00erent directions. In turbulent \ufb02ow, you tend to see eddy currents (little whirlpools) and other non-linear patterns. We will only study laminar motion in this course \u2014 turbulent motion is very complicated \u2014 and we will only consider the motion of an ideal \ufb02uid. The ideal \ufb02uid has the following properties: \u2022 The \ufb02uid is non-viscous, which means there is no internal friction between adjacent particles. (The viscosity of a \ufb02uid is a measure of the amount of internal friction.) \u2022 The \ufb02uid is incompressible, which means the density is constant. \u2022 The \ufb02uid motion is steady, meaning that the velocity and pressure at each point does not change in time. \u2022 The \ufb02uid moves without turbulence. This condition means that the particles have no rotational motion and no angular velocity \u2014 they only move in straight lines. 9.5.1 Equation of continuity Suppose a \ufb02uid \ufb02ows in a pipe whose cross-sectional area increases from left to right, going from A1 at one end to A2 at the other end. Suppose the \ufb02uid enters the pipe with a velocity v1. The \ufb02uid entering the pipe moves a distance \u2206x1 = v1\u2206t in a time \u2206t. The mass of water contained in this region is \u2206M1 = \u03c1waterA1\u2206x1 = \u03c1waterA1v1\u2206t. We can write a similar equation for the mass \ufb02owing out of the other end of the pipe, \u2206M2 = \u03c1waterA2v2\u2206t. Since mass is conserved (the \ufb02uid is incompressible), we must have the same amount of mass going in as is coming out, \u2206M1 = \u2206M2, or A1v1 = A2v2. (9.12) This equation is known as the equation of continuity. It tells us that \ufb02uid will speed up or slow down as the area through which they \ufb02ow changes. Fluid \ufb02ows faster through a pipe of small area than through a pipe of large area. The product Av is also known as the \ufb02ow rate. 94 Example 9.12: Garden hose A water hose 2.5 cm in diameter is used by a gardener to \ufb01ll a 30.0 L bucket. The gardener notices that it takes 1.0 min to \ufb01ll the bucket. A nozzle with an opening of cross-sectional are 0.500 cm2 is then attached to the hose. The nozzle is held so the water is projected horizontally from a point 1.0 m above the ground. Over what horizontal distance will the water be projected? Solution: We \ufb01rst need to determine the \ufb02ow rate for the water in the absence of the nozzle. We can \ufb01gure this out from how long it takes to \ufb01ll the bucket, A1v1 = 1000 cm3 1 L A1v1 = 5.0 \u00d7 10\u22124 m3/s. 30 L 1.00 min 1 m 100 cm 3 1 min 60 s The \ufb02ow rate remains constant when the new nozzle is attached, v2 = A1v1 = A2v2 A1v1 A2 5.0 \u00d7 10\u22124 m3/s 0.5 \u00d7 10\u22124 m2 v2 = v2 = 10.0 m/s. This is the initial horizontal velocity of the water. Once the water leaves the hose, it is a projectile undergoing acceleration in the vertical direction (but not horizontally). We can \ufb01nd how long it takes for the water to hit the ground, 1 2 gt2 y = v0yt \u2212 2y g t = t = 2(1.0 m) 9.8 m/s2 t = 0.452 s. Now we can \ufb01nd the horizontal distance travelled, x = v0xt x = (10 m/s)(0.452 s) x = 4.52 m. 9.6 Bernoulli\u2019s equation Suppose that the pipe with varying diameter is now angled upwards. Let\u2019s consider the work done on the \ufb02uid in the pipe. The \ufb02uid at the lower end is pushed by the \ufb02uid behind it. The \ufb02uid at the upper end is pushed by the \ufb02uid in front of it. The net work done on the \ufb02uid in the pipe", " then is W = F1\u2206x1 \u2212 F2\u2206x2 W = P1A1\u2206x1 \u2212 P2A2\u2206x2 W = P1V \u2212 P2V. 95 (9.13) The work done on the \ufb02uid can do one of two things: it can change the kinetic energy of the \ufb02uid, and it can change the gravitational potential energy, W = \u2206KE + \u2206P E. Combining these two equations we have, Now we can put in expressions for the kinetic and potential energies, P1V \u2212 P2V = \u2206KE + \u2206P E. We can re-write this as P1 \u2212 P2 = 1 2 \u03c1v2 2 \u2212 1 2 \u03c1v2 1 + \u03c1gy2 \u2212 \u03c1gy1. P1 + 1 2 \u03c1v2 1 + \u03c1gy1 = P2 + 1 2 \u03c1v2 2 + \u03c1gy2. (9.14) (9.15) (9.16) (9.17) This is known as Bernoulli\u2019s equation. Note that this is not really a new concept, it is just the conservation of energy applied to a \ufb02uid. This equation is only true, however, for laminar \ufb02ow. One consequence of this equation is that faster moving \ufb02uids exert less pressure than slowly moving \ufb02uids. We can see this by considering the pipe with a varying diameter when it is horizontally level. In this case, Bernoulli\u2019s equation simpli\ufb01es to P1 + 1 2 \u03c1v2 1 = P2 + 1 2 \u03c1v2 2. We know that the \ufb02uid moves faster in the narrow region, so the pressure in that region must be lower than in the wide region in order for the two sides of the equation to balance. Example 9.13: Shootout A nearsighted sheri\ufb00 \ufb01res at a cattle rustler with his trust six-shooter. Fortunately for the rustler, the bullet misses him and penetrates the town water tank, causing a leak. If the top of the tank is open to the atmosphere, determine the speed at which the water leaves the hole when the water level is 0.500 m above the hole. Solution: We can use Bernoulli\u2019s equation to \ufb01nd the velocity. Let\u2019s choose the \ufb01rst point to be the top of the tank and the second point will be the hole. The pressure at both of these points is just P0, the standard atmospheric pressure. We assume that the water level drops very slowly, so the velocity of the \ufb02uid at the top of the tank is zero. Putting these into Bernoulli\u2019s equation, P1 + 1 2 \u03c1v2 1 + \u03c1gy1 = P2 + 1 2 \u03c1v2 2 + \u03c1gy2 1 2 \u03c1gy1 = \u03c1v2 2 + \u03c1gy2 v = 2g(y1 \u2212 y2) v = 2(9.8 m/s2)(0.500 m) v = 3.13 m/s. 96 Chapter 10 Thermal physics There is a form of energy that we have to date neglected to consider in our description of objects. This is primarily because this form of energy is not typically involved in macroscopic motion. All objects have thermal energy, a type of energy which we intuitively detect as the object being hot or cold. 10.1 Temperature Determining whether an object is hot or cold is rather inexact and we would like to \ufb01nd a more quantitative way of measuring thermal energy. We say that two objects are in thermal contact if energy (particularly thermal energy) can be exchanged between them. Two objects are in thermal equilibrium if they are in thermal contact but there is no net exchange of energy between them. The zeroth law of thermodynamics (law of equilibrium) states: If objects A and B are separately in thermal equilibrium with a third object C, then A and B are in thermal equilibrium with each other. This law allows us to use a thermometer to compare the thermal energies of two objects. Suppose we want to compare the thermal energies of two objects A and B, we could put them into thermal contact and try to detect the direction of energy \ufb02ow. Suppose, however, that we cannot put the two objects into thermal contact directly. We can then use a third object, C, to compare the thermal energies of A and B. We \ufb01rst put the thermometer into thermal contact with A until it reaches thermal equilibrium at which point we read the thermometer. We then put the thermometer into thermal contact with B until it reaches thermal equilibrium and read the temperature again. If the two readings are the", " same then A and B are also in thermal equilibrium. This property allows us to de\ufb01ne temperature \u2014 if two objects are in thermal equilibrium, then they have the same temperature. The thermometer used to measure thermal energy must have some physical property that changes with temperature. Most thermometers use the fact that substances (solids, gasses, liquids) expand as temperature increases. This physical change can usually be measured visually allowing us to put a number on the temperature. We must \ufb01rst, however, calibrate the thermometer. That is, we must agree on a measurement scale for temperature. The Celsius temperature scale is de\ufb01ned by measuring the freezing point of water which is set to be 0 \u25e6C. and the boiling point of water which is set to be 100 \u25e6C. The scale most commonly used in the US is the Fahrenheit scale. On this scale, the freezing point is at 32\u25e6F and the boiling point is at 212\u25e6F. We can convert between the two using the formula The temperature scale most often used by scientists is the Kelvin scale. One of the problems with both the Celsius and Fahrenheit scales is that the freezing point and boiling point of water depend not only on the TF = 9 5 TC + 32. (10.1) 97 temperature but on the pressure. Scientists removed the pressure dependence by observing that the pressure of all substances goes to zero at a temperature of \u2212273.15 \u25e6C. This temperature is known as absolute zero and is de\ufb01ned to be 0 K. The second point used to de\ufb01ne the Kelvin scale is the triple point of water. This is the temperature and pressure at which water, water vapour, and ice exist in equilibrium. This point occurs at 0.01 \u25e6C and 4.58 mm of mercury. This temperature is de\ufb01ned to be 273.16 K. This means that the unit size of both the celsius and kelvin scales are the same. We convert between the two using, TC = TK \u2212 273.15. (10.2) 10.2 Thermal expansion Most substances increase in volume as their temperature (thermal energy) increases. The thermal energy of an object is actually a measure of the average velocity of the constituent atoms. As temperature increases, atoms move faster. In solids and liquids, these atoms cannot actually leave the substance, so their vibrational motion increases leading to an increased separation between atoms. Macroscopically, we see this as an increase in volume. If the expansion is small compared to the object\u2019s original size, the expansion in one dimension is approximately linear with temperature, where \u2206L is the change in length (not volume), L is the original length of the object, and \u03b1 is the coe\ufb03cient of linear expansion for a particular substance. \u2206L = \u03b1L0\u2206T, (10.3) Example 10.3: Expansion of a railroad track (a) A steel railroad track has a length of 30.0 m when the temperature is 0 \u25e6C. What is the length on a hot day when the temperature is 40.0 \u25e6C? (b) What is the stress caused by this expansion? Solution: (a) The change in length due to the temperature change, \u2206L = \u03b1L0\u2206T \u2206L = (11 \u00d7 10\u22126 /\u25e6C)(30.0 m)(40.0 \u25e6C) \u2206L = 0.013 m. So the new length is 30.013 m. (b) The railroad undergoes a linear expansion, so this is a tensile strain, = Y \u2206L L = (2.0 \u00d7 1011 Pa = 8.7 \u00d7 107 Pa. F A F A F A 0.013 m 30.0 m Since their is a linear expansion of objects with temperature, there must also be a change in their area and volume. Suppose we have a square of material with a length of L0. Each dimension of the square will undergo linear expansion and the new area is A = L2 A = (L0 + \u03b1L0\u2206T )(L0 + \u03b1L0\u2206T ) A = L2 0\u03b1\u2206T + (\u03b1L0\u2206T )2. 0 + 2L2 98 The last term in that equation will be very small, so we will ignore it, We can re-write this so that it looks like the linear expansion equation, A = A0 + 2\u03b1A0\u2206T. \u2206A = 2\u03b1A0\u2206T. (10.4) (10.5) We de\ufb01ne a new coe\ufb03cient \u03b3 = 2\u03b1 as the coe\ufb03cient of area expansion. We can perform the same type of derivation and show that the increase in volume of a substance is given by", " \u2206V = \u03b2V0\u2206T, (10.6) where \u03b2 is the coe\ufb03cient of volume expansion and is given by \u03b2 = 3\u03b1. 10.3 Ideal gas law The e\ufb00ect of temperature change on a gas is somewhat more complex than in solids and liquids. A gas will expand to \ufb01ll a particular container no matter what the temperature. What will change instead as the temperature increases is the pressure. There is usually a fairly complex relationship between the pressure, volume and temperature of gasses, but for an ideal gas, we can derive a simple relationship. An ideal gas is a gas that is maintained at low density or pressure. In an ideal gas, particles of the gas are so far apart that they rarely interact and so we can assume there are no forces acting on any of the particles and no collisions take place. Each particle of the gas moves randomly. Since gases contain large numbers of particles, we usually count the number of particles in moles where one mole is 6.02\u00d71023 gas particles. The number 6.02\u00d71023 is known as Avogadro\u2019s number and is denoted by NA. Avogadro\u2019s number was chosen so that the mass in grams of one mole of an element is numerically the same as the atomic mass units of the element. Carbon 12 has an atomic mass of 12 amu, so one mole of carbon 12 weighs 12 g. For an ideal gas, the relationship between pressure, volume, and temperature is P V = nRT, (10.7) where n is the number of moles of the substance, and R is the universal gas constant with a value of R = 8.31 J/mol \u00b7 K. The ideal gas law tells us that the pressure is linearly proportional to temperature and inversely proportional to the volume. As temperature increases, pressure increases. As volume increases, pressure decreases. 99 Example 10.6: Expanding gas An ideal gas at 20.0 \u25e6C and a pressure of 1.50 \u00d7 105 Pa is in a container having a volume of 1.0 L. (a) Determine the number of moles of gas in the container. (b) The gas pushes against a piston, expanding to twice its original volume, while the pressure falls to atmospheric pressure. Find the \ufb01nal temperature. Solution: (a) We need to be careful that all quantities are in SI units. We will need to convert the temperature to kelvins: T = 20 + 273 = 293 K. And we need to convert the volume to m3: V = 1.0 L = 1.0 \u00d7 10\u22123 m3. Now we can go ahead and plug the values into the gas law, n = P V = nRT P V RT (1.50 \u00d7 105 Pa)(1.0 \u00d7 10\u22123 m3) (8.31 J/mol \u00b7 K)(293 K) n = n = 6.16 \u00d7 10\u22122 mol. (b) We can \ufb01nd the new temperature from the gas law, T = T = P V nR (1.01 \u00d7 105 Pa)(2.0 \u00d7 10\u22123 m3) (6.16 \u00d7 10\u22122 mol)(8.31 J/mol \u00b7 K) T = 395 K. 100ed by classical physics. One reason for this is that they are small enough to travel at great speeds, near the speed of light. Figure 1.5 Using a scanning tunneling microscope (STM), scientists can see the individual atoms that compose this sheet of gold. (Erwinrossen) At particle colliders (Figure 1.6), such as the Large Hadron Collider on the France-Swiss border, particle physicists can make subatomic particles travel at very high speeds within a 27 kilometers (17 miles) long superconducting tunnel. They can then study the properties of the particles at high speeds, as well as collide them with each other to see how they exchange energy. This has led to many intriguing discoveries such as the Higgs-Boson particle, which gives matter the property of mass, and antimatter, which causes a huge energy release when it comes in contact with matter. Figure 1.6 Particle colliders such as the Large Hadron Collider in Switzerland or Fermilab in the United States (pictured here), have long tunnels that allows subatomic particles to be accelerated to near light speed. (Andrius.v ) Physicists are currently trying to unify the two theories of modern physics, relativity and quantum mechanics, into a single, comprehensive theory called relativistic quantum mechanics. Relating the behavior of subatomic particles to gravity, time, and space will allow us to explain how the universe works in a much more comprehensive way. Application of Physics You need not be a scientist to use physics. On the contrary, knowledge of physics is useful in everyday situations as well as", " in nonscientific professions. For example, physics can help you understand why you shouldn\u2019t put metal in the microwave (Figure 1.7), why a black car radiator helps remove heat in a car engine, and why a white roof helps keep the inside of a house cool. The operation of a car\u2019s ignition system, as well as the transmission of electrical signals through our nervous system, are much easier to understand when you think about them in terms of the basic physics of electricity. Figure 1.7 Why can't you put metal in the microwave? Microwaves are high-energy radiation that increases the movement of electrons in Access for free at openstax.org. 1.1 \u2022 Physics: Definitions and Applications 11 metal. These moving electrons can create an electrical current, causing sparking that can lead to a fire. (= MoneyBlogNewz) Physics is the foundation of many important scientific disciplines. For example, chemistry deals with the interactions of atoms and molecules. Not surprisingly, chemistry is rooted in atomic and molecular physics. Most branches of engineering are also applied physics. In architecture, physics is at the heart of determining structural stability, acoustics, heating, lighting, and cooling for buildings. Parts of geology, the study of nonliving parts of Earth, rely heavily on physics; including radioactive dating, earthquake analysis, and heat transfer across Earth\u2019s surface. Indeed, some disciplines, such as biophysics and geophysics, are hybrids of physics and other disciplines. Physics also describes the chemical processes that power the human body. Physics is involved in medical diagnostics, such as xrays, magnetic resonance imaging (MRI), and ultrasonic blood flow measurements (Figure 1.8). Medical therapy Physics also has many applications in biology, the study of life. For example, physics describes how cells can protect themselves using their cell walls and cell membranes (Figure 1.9). Medical therapy sometimes directly involves physics, such as in using X-rays to diagnose health conditions. Physics can also explain what we perceive with our senses, such as how the ears detect sound or the eye detects color. Figure 1.8 Magnetic resonance imaging (MRI) uses electromagnetic waves to yield an image of the brain, which doctors can use to find diseased regions. (Rashmi Chawla, Daniel Smith, and Paul E. Marik) Figure 1.9 Physics, chemistry, and biology help describe the properties of cell walls in plant cells, such as the onion cells seen here. (Umberto Salvagnin) BOUNDLESS PHYSICS The Physics of Landing on a Comet On November 12, 2014, the European Space Agency\u2019s Rosetta spacecraft (shown in Figure 1.10) became the first ever to reach and orbit a comet. Shortly after, Rosetta\u2019s rover, Philae, landed on the comet, representing the first time humans have ever landed a space probe on a comet. 12 Chapter 1 \u2022 What is Physics? Figure 1.10 The Rosetta spacecraft, with its large and revolutionary solar panels, carried the Philae lander to a comet. The lander then detached and landed on the comet\u2019s surface. (European Space Agency) After traveling 6.4 billion kilometers starting from its launch on Earth, Rosetta landed on the comet 67P/ChuryumovGerasimenko, which is only 4 kilometers wide. Physics was needed to successfully plot the course to reach such a small, distant, and rapidly moving target. Rosetta\u2019s path to the comet was not straight forward. The probe first had to travel to Mars so that Mars\u2019s gravity could accelerate it and divert it in the exact direction of the comet. This was not the first time humans used gravity to power our spaceships. Voyager 2, a space probe launched in 1977, used the gravity of Saturn to slingshotover to Uranus and Neptune (illustrated in Figure 1.11), providing the first pictures ever taken of these planets. Now, almost 40 years after its launch, Voyager 2 is at the very edge of our solar system and is about to enter interstellar space. Its sister ship, Voyager 1 (illustrated in Figure 1.11), which was also launched in 1977, is already there. To listen to the sounds of interstellar space or see images that have been transmitted back from the Voyager I or to learn more about the Voyager mission, visit the Voyager\u2019s Mission website (https://openstax.org/l/28voyager). Figure 1.11 a) Voyager 2, launched in 1977, used the gravity of Saturn to slingshot over to Uranus and Neptune. NASA b) A rendering of Voyager 1, the first space probe to ever leave our solar system and enter interstellar space. NASA Both Voyagers have electrical power generators based on the decay of radioisotopes. These generators have served them for almost 40 years. Rosetta, on the other hand, is solar-powered. In fact, Rosetta became", " the first space probe to travel beyond the asteroid belt by relying only on solar cells for power generation. At 800 million kilometers from the sun, Rosetta receives sunlight that is only 4 percent as strong as on Earth. In addition, it is very cold in space. Therefore, a lot of physics went into developing Rosetta\u2019s low-intensity low-temperature solar cells. In this sense, the Rosetta project nicely shows the huge range of topics encompassed by physics: from modeling the movement of gigantic planets over huge distances within our solar systems, to learning how to generate electric power from low-intensity light. Physics is, by far, the broadest field of science. GRASP CHECK What characteristics of the solar system would have to be known or calculated in order to send a probe to a distant planet, such as Jupiter? Access for free at openstax.org. 1.1 \u2022 Physics: Definitions and Applications 13 a. b. c. d. the effects due to the light from the distant stars the effects due to the air in the solar system the effects due to the gravity from the other planets the effects due to the cosmic microwave background radiation In summary, physics studies many of the most basic aspects of science. A knowledge of physics is, therefore, necessary to understand all other sciences. This is because physics explains the most basic ways in which our universe works. However, it is not necessary to formally study all applications of physics. A knowledge of the basic laws of physics will be most useful to you, so that you can use them to solve some everyday problems. In this way, the study of physics can improve your problem-solving skills. Check Your Understanding 1. Which of the following is notan essential feature of scientific explanations? a. They must be subject to testing. b. They strictly pertain to the physical world. c. Their validity is judged based on objective observations. d. Once supported by observation, they can be viewed as a fact. 2. Which of the following does notrepresent a question that can be answered by science? a. How much energy is released in a given nuclear chain reaction? b. Can a nuclear chain reaction be controlled? c. Should uncontrolled nuclear reactions be used for military applications? d. What is the half-life of a waste product of a nuclear reaction? 3. What are the three conditions under which classical physics provides an excellent description of our universe? a. b. c. d. 1. Matter is moving at speeds less than about 1 percent of the speed of light 2. Objects dealt with must be large enough to be seen with the naked eye. 3. Strong electromagnetic fields are involved. 1. Matter is moving at speeds less than about 1 percent of the speed of light. 2. Objects dealt with must be large enough to be seen with the naked eye. 3. Only weak gravitational fields are involved. 1. Matter is moving at great speeds, comparable to the speed of light. 2. Objects dealt with are large enough to be seen with the naked eye. 3. Strong gravitational fields are involved. 1. Matter is moving at great speeds, comparable to the speed of light. 2. Objects are just large enough to be visible through the most powerful telescope. 3. Only weak gravitational fields are involved. 4. Why is the Greek word for nature appropriate in describing the field of physics? a. Physics is a natural science that studies life and living organism on habitable planets like Earth. b. Physics is a natural science that studies the laws and principles of our universe. c. Physics is a physical science that studies the composition, structure, and changes of matter in our universe. d. Physics is a social science that studies the social behavior of living beings on habitable planets like Earth. 5. Which aspect of the universe is studied by quantum mechanics? a. objects at the galactic level b. objects at the classical level c. objects at the subatomic level d. objects at all levels, from subatomic to galactic 14 Chapter 1 \u2022 What is Physics? 1.2 The Scientific Methods Section Learning Objectives By the end of this section, you will be able to do the following: \u2022 Explain how the methods of science are used to make scientific discoveries \u2022 Define a scientific model and describe examples of physical and mathematical models used in physics \u2022 Compare and contrast hypothesis, theory, and law Section Key Terms experiment hypothesis model observation principle scientific law scientific methods theory universal Scientific Methods Scientists often plan and carry out investigations to answer questions about the universe around us. Such laws are intrinsic to the universe, meaning that humans did not create them and cannot change them. We can only discover and understand them. Their discovery is a very human endeavor, with all the elements of mystery, imagination, struggle, triumph, and disappointment inherent in any creative effort. The cornerstone of discovering natural laws is observation. Science must describe the universe as it is, not as we imagine or wish it to be. We all are curious to some extent. We look around, make generalizations, and try to understand", " what we see. For example, we look up and wonder whether one type of cloud signals an oncoming storm. As we become serious about exploring nature, we become more organized and formal in collecting and analyzing data. We attempt greater precision, perform controlled experiments (if we can), and write down ideas about how data may be organized. We then formulate models, theories, and laws based on the data we have collected, and communicate those results with others. This, in a nutshell, describes the scientific method that scientists employ to decide scientific issues on the basis of evidence from observation and experiment. An investigation often begins with a scientist making an observation. The scientist observes a pattern or trend within the natural world. Observation may generate questions that the scientist wishes to answer. Next, the scientist may perform some research about the topic and devise a hypothesis. A hypothesis is a testable statement that describes how something in the natural world works. In essence, a hypothesis is an educated guess that explains something about an observation. Scientists may test the hypothesis by performing an experiment. During an experiment, the scientist collects data that will help them learn about the phenomenon they are studying. Then the scientists analyze the results of the experiment (that is, the data), often using statistical, mathematical, and/or graphical methods. From the data analysis, they draw conclusions. They may conclude that their experiment either supports or rejects their hypothesis. If the hypothesis is supported, the scientist usually goes on to test another hypothesis related to the first. If their hypothesis is rejected, they will often then test a new and different hypothesis in their effort to learn more about whatever they are studying. Scientific processes can be applied to many situations. Let\u2019s say that you try to turn on your car, but it will not start. You have just made an observation! You ask yourself, \"Why won\u2019t my car start?\" You can now use scientific processes to answer this question. First, you generate a hypothesis such as, \"The car won\u2019t start because it has no gasoline in the gas tank.\" To test this hypothesis, you put gasoline in the car and try to start it again. If the car starts, then your hypothesis is supported by the experiment. If the car does not start, then your hypothesis is rejected. You will then need to think up a new hypothesis to test such as, \"My car won\u2019t start because the fuel pump is broken.\" Hopefully, your investigations lead you to discover why the car won\u2019t start and enable you to fix it. Modeling A model is a representation of something that is often too difficult (or impossible) to study directly. Models can take the form of physical models, equations, computer programs, or simulations\u2014computer graphics/animations. Models are tools that are especially useful in modern physics because they let us visualize phenomena that we normally cannot observe with our senses, such as very small objects or objects that move at high speeds. For example, we can understand the structure of an atom using models, despite the fact that no one has ever seen an atom with their own eyes. Models are always approximate, so they are simpler to consider than the real situation; the more complete a model is, the more complicated it must be. Models put the Access for free at openstax.org. 1.2 \u2022 The Scientific Methods 15 intangible or the extremely complex into human terms that we can visualize, discuss, and hypothesize about. Scientific models are constructed based on the results of previous experiments. Even still, models often only describe a phenomenon partially or in a few limited situations. Some phenomena are so complex that they may be impossible to model them in their entirety, even using computers. An example is the electron cloud model of the atom in which electrons are moving around the atom\u2019s center in distinct clouds (Figure 1.12), that represent the likelihood of finding an electron in different places. This model helps us to visualize the structure of an atom. However, it does not show us exactly where an electron will be within its cloud at any one particular time. Figure 1.12 The electron cloud model of the atom predicts the geometry and shape of areas where different electrons may be found in an atom. However, it cannot indicate exactly where an electron will be at any one time. As mentioned previously, physicists use a variety of models including equations, physical models, computer simulations, etc. For example, three-dimensional models are often commonly used in chemistry and physics to model molecules. Properties other than appearance or location are usually modelled using mathematics, where functions are used to show how these properties relate to one another. Processes such as the formation of a star or the planets, can also be modelled using computer simulations. Once a simulation is correctly programmed based on actual experimental data, the simulation can allow us to view processes that happened in the past or happen too quickly or slowly for us to observe directly. In addition, scientists can also run virtual experiments using computer-based models. In a model of planet formation, for example, the", " scientist could alter the amount or type of rocks present in space and see how it affects planet formation. Scientists use models and experimental results to construct explanations of observations or design solutions to problems. For example, one way to make a car more fuel efficient is to reduce the friction or drag caused by air flowing around the moving car. This can be done by designing the body shape of the car to be more aerodynamic, such as by using rounded corners instead of sharp ones. Engineers can then construct physical models of the car body, place them in a wind tunnel, and examine the flow of air around the model. This can also be done mathematically in a computer simulation. The air flow pattern can be analyzed for regions smooth air flow and for eddies that indicate drag. The model of the car body may have to be altered slightly to produce the smoothest pattern of air flow (i.e., the least drag). The pattern with the least drag may be the solution to increasing fuel efficiency of the car. This solution might then be incorporated into the car design. Snap Lab Using Models and the Scientific Processes Be sure to secure loose items before opening the window or door. In this activity, you will learn about scientific models by making a model of how air flows through your classroom or a room in your house. \u2022 One room with at least one window or door that can be opened \u2022 Piece of single-ply tissue paper 1. Work with a group of four, as directed by your teacher. Close all of the windows and doors in the room you are working in. Your teacher may assign you a specific window or door to study. 16 Chapter 1 \u2022 What is Physics? 2. Before opening any windows or doors, draw a to-scale diagram of your room. First, measure the length and width of your room using the tape measure. Then, transform the measurement using a scale that could fit on your paper, such as 5 centimeters = 1 meter. 3. Your teacher will assign you a specific window or door to study air flow. On your diagram, add arrows showing your hypothesis (before opening any windows or doors) of how air will flow through the room when your assigned window or door is opened. Use pencil so that you can easily make changes to your diagram. 4. On your diagram, mark four locations where you would like to test air flow in your room. To test for airflow, hold a strip of single ply tissue paper between the thumb and index finger. Note the direction that the paper moves when exposed to the airflow. Then, for each location, predict which way the paper will move if your air flow diagram is correct. 5. Now, each member of your group will stand in one of the four selected areas. Each member will test the airflow Agree upon an approximate height at which everyone will hold their papers. 6. When you teacher tells you to, open your assigned window and/or door. Each person should note the direction that their paper points immediately after the window or door was opened. Record your results on your diagram. 7. Did the airflow test data support or refute the hypothetical model of air flow shown in your diagram? Why or why not? Correct your model based on your experimental evidence. 8. With your group, discuss how accurate your model is. What limitations did it have? Write down the limitations that your group agreed upon. GRASP CHECK Your diagram is a model, based on experimental evidence, of how air flows through the room. Could you use your model to predict how air would flow through a new window or door placed in a different location in the classroom? Make a new diagram that predicts the room\u2019s airflow with the addition of a new window or door. Add a short explanation that describes how. a. Yes, you could use your model to predict air flow through a new window. The earlier experiment of air flow would help you model the system more accurately. b. Yes, you could use your model to predict air flow through a new window. The earlier experiment of air flow is not useful for modeling the new system. c. No, you cannot model a system to predict the air flow through a new window. The earlier experiment of air flow would help you model the system more accurately. d. No, you cannot model a system to predict the air flow through a new window. The earlier experiment of air flow is not useful for modeling the new system. Scientific Laws and Theories A scientific law is a description of a pattern in nature that is true in all circumstances that have been studied. That is, physical laws are meant to be universal, meaning that they apply throughout the known universe. Laws are often also concise, whereas theories are more complicated. A law can be expressed in the form of a single sentence or mathematical equation. For example, Newton\u2019s second law of motion, which relates the motion of an object to the force applied (F), the mass of the object (m), and the object\u2019s acceleration (a), is simply stated using the equation Scientific ideas and explanations", " that are true in many, but not all situations in the universe are usually called principles. An example is Pascal\u2019s principle, which explains properties of liquids, but not solids or gases. However, the distinction between laws and principles is sometimes not carefully made in science. A theory is an explanation for patterns in nature that is supported by much scientific evidence and verified multiple times by multiple researchers. While many people confuse theories with educated guesses or hypotheses, theories have withstood more rigorous testing and verification than hypotheses. As a closing idea about scientific processes, we want to point out that scientific laws and theories, even those that have been supported by experiments for centuries, can still be changed by new discoveries. This is especially true when new technologies emerge that allow us to observe things that were formerly unobservable. Imagine how viewing previously invisible objects with a Access for free at openstax.org. 1.2 \u2022 The Scientific Methods 17 microscope or viewing Earth for the first time from space may have instantly changed our scientific theories and laws! What discoveries still await us in the future? The constant retesting and perfecting of our scientific laws and theories allows our knowledge of nature to progress. For this reason, many scientists are reluctant to say that their studies proveanything. By saying supportinstead of prove, it keeps the door open for future discoveries, even if they won\u2019t occur for centuries or even millennia. Check Your Understanding 6. Explain why scientists sometimes use a model rather than trying to analyze the behavior of the real system. a. Models are simpler to analyze. b. Models give more accurate results. c. Models provide more reliable predictions. d. Models do not require any computer calculations. 7. Describe the difference between a question, generated through observation, and a hypothesis. a. They are the same. b. A hypothesis has been thoroughly tested and found to be true. c. A hypothesis is a tentative assumption based on what is already known. d. A hypothesis is a broad explanation firmly supported by evidence. 8. What is a scientific model and how is it useful? a. A scientific model is a representation of something that can be easily studied directly. It is useful for studying things that can be easily analyzed by humans. b. A scientific model is a representation of something that is often too difficult to study directly. It is useful for studying a complex system or systems that humans cannot observe directly. c. A scientific model is a representation of scientific equipment. It is useful for studying working principles of scientific equipment. d. A scientific model is a representation of a laboratory where experiments are performed. It is useful for studying requirements needed inside the laboratory. 9. Which of the following statements is correct about the hypothesis? a. The hypothesis must be validated by scientific experiments. b. The hypothesis must not include any physical quantity. c. The hypothesis must be a short and concise statement. d. The hypothesis must apply to all the situations in the universe. 10. What is a scientific theory? a. A scientific theory is an explanation of natural phenomena that is supported by evidence. b. A scientific theory is an explanation of natural phenomena without the support of evidence. c. A scientific theory is an educated guess about the natural phenomena occurring in nature. d. A scientific theory is an uneducated guess about natural phenomena occurring in nature. 11. Compare and contrast a hypothesis and a scientific theory. a. A hypothesis is an explanation of the natural world with experimental support, while a scientific theory is an educated guess about a natural phenomenon. b. A hypothesis is an educated guess about natural phenomenon, while a scientific theory is an explanation of natural world with experimental support. c. A hypothesis is experimental evidence of a natural phenomenon, while a scientific theory is an explanation of the natural world with experimental support. d. A hypothesis is an explanation of the natural world with experimental support, while a scientific theory is experimental evidence of a natural phenomenon. 18 Chapter 1 \u2022 What is Physics? 1.3 The Language of Physics: Physical Quantities and Units Section Learning Objectives By the end of this section, you will be able to do the following: \u2022 Associate physical quantities with their International System of Units (SI)and perform conversions among SI units using scientific notation \u2022 Relate measurement uncertainty to significant figures and apply the rules for using significant figures in calculations \u2022 Correctly create, label, and identify relationships in graphs using mathematical relationships (e.g., slope, y-intercept, inverse, quadratic and logarithmic) Section Key Terms accuracy ampere constant conversion factor dependent variable derived units English units exponential relationship fundamental physical units independent variable inverse relationship inversely proportional kilogram linear relationship logarithmic (log) scale log-log plot meter method of adding percents order of magnitude precision quadratic relationship scientific notation second semi-log plot SI units significant figures slope uncertainty variable y-intercept The Role of Units Physicists, like other scientists, make observations and ask basic questions. For example, how big is an object? How much mass does", " it have? How far did it travel? To answer these questions, they make measurements with various instruments (e.g., meter stick, balance, stopwatch, etc.). The measurements of physical quantities are expressed in terms of units, which are standardized values. For example, the length of a race, which is a physical quantity, can be expressed in meters (for sprinters) or kilometers (for long distance runners). Without standardized units, it would be extremely difficult for scientists to express and compare measured values in a meaningful way (Figure 1.13). Figure 1.13 Distances given in unknown units are maddeningly useless. All physical quantities in the International System of Units (SI) are expressed in terms of combinations of seven fundamental Access for free at openstax.org. 1.3 \u2022 The Language of Physics: Physical Quantities and Units 19 physical units, which are units for: length, mass, time, electric current, temperature, amount of a substance, and luminous intensity. SI Units: Fundamental and Derived Units There are two major systems of units used in the world: SI units (acronym for the French Le Syste\u0300me International d\u2019Unite\u0301s, also known as the metric system), and English units (also known as the imperial system). English units were historically used in nations once ruled by the British Empire. Today, the United States is the only country that still uses English units extensively. Virtually every other country in the world now uses the metric system, which is the standard system agreed upon by scientists and mathematicians. Some physical quantities are more fundamental than others. In physics, there are seven fundamental physical quantities that are measured in base or physical fundamental units: length, mass, time, electric current temperature, amount of substance, and luminous intensity. Units for other physical quantities (such as force, speed, and electric charge) described by mathematically combining these seven base units. In this course, we will mainly use five of these: length, mass, time, electric current and temperature. The units in which they are measured are the meter, kilogram, second, ampere, kelvin, mole, and candela (Table 1.1). All other units are made by mathematically combining the fundamental units. These are called derived units. Quantity Name Symbol Length Meter m Mass Time Kilogram kg Second Electric current Ampere Temperature Kelvin s a k Amount of substance Mole mol Luminous intensity Candela cd Table 1.1 SI Base Units The Meter The SI unit for length is the meter (m). The definition of the meter has changed over time to become more accurate and precise. The meter was first defined in 1791 as 1/10,000,000 of the distance from the equator to the North Pole. This measurement was improved in 1889 by redefining the meter to be the distance between two engraved lines on a platinum-iridium bar. (The bar is now housed at the International Bureau of Weights and Meaures, near Paris). By 1960, some distances could be measured more precisely by comparing them to wavelengths of light. The meter was redefined as 1,650,763.73 wavelengths of orange light emitted by krypton atoms. In 1983, the meter was given its present definition as the distance light travels in a vacuum in 1/ 299,792,458 of a second (Figure 1.14). Figure 1.14 The meter is defined to be the distance light travels in 1/299,792,458 of a second through a vacuum. Distance traveled is speed multiplied by time. 20 Chapter 1 \u2022 What is Physics? The Kilogram The SI unit for mass is the kilogram (kg). It is defined to be the mass of a platinum-iridium cylinder, housed at the International Bureau of Weights and Measures near Paris. Exact replicas of the standard kilogram cylinder are kept in numerous locations throughout the world, such as the National Institute of Standards and Technology in Gaithersburg, Maryland. The determination of all other masses can be done by comparing them with one of these standard kilograms. The Second The SI unit for time, the second (s) also has a long history. For many years it was defined as 1/86,400 of an average solar day. However, the average solar day is actually very gradually getting longer due to gradual slowing of Earth\u2019s rotation. Accuracy in the fundamental units is essential, since all other measurements are derived from them. Therefore, a new standard was adopted to define the second in terms of a non-varying, or constant, physical phenomenon. One constant phenomenon is the very steady vibration of Cesium atoms, which can be observed and counted. This vibration forms the basis of the cesium atomic clock. In 1967, the second was redefined as the time required for 9,192,631,770 Cesium atom vibrations (Figure 1.15). Figure 1.15 An atomic clock such as this one uses the vibrations of c", "esium atoms to keep time to a precision of one microsecond per year. The fundamental unit of time, the second, is based on such clocks. This image is looking down from the top of an atomic clock. (Steve Jurvetson/Flickr) The Ampere Electric current is measured in the ampere (A), named after Andre Ampere. You have probably heard of amperes, or amps, when people discuss electrical currents or electrical devices. Understanding an ampere requires a basic understanding of electricity and magnetism, something that will be explored in depth in later chapters of this book. Basically, two parallel wires with an electric current running through them will produce an attractive force on each other. One ampere is defined as the amount of 10\u20137 newton per meter of separation between the two wires (the electric current that will produce an attractive force of 2.7 newton is the derived unit of force). Kelvins The SI unit of temperature is the kelvin (or kelvins, but not degrees kelvin). This scale is named after physicist William Thomson, Lord Kelvin, who was the first to call for an absolute temperature scale. The Kelvin scale is based on absolute zero. This is the point at which all thermal energy has been removed from all atoms or molecules in a system. This temperature, 0 K, is equal to \u2212273.15 \u00b0C and \u2212459.67 \u00b0F. Conveniently, the Kelvin scale actually changes in the same way as the Celsius scale. For example, the freezing point (0 \u00b0C) and boiling points of water (100 \u00b0C) are 100 degrees apart on the Celsius scale. These two temperatures are also 100 kelvins apart (freezing point = 273.15 K; boiling point = 373.15 K). Metric Prefixes Physical objects or phenomena may vary widely. For example, the size of objects varies from something very small (like an atom) Access for free at openstax.org. 1.3 \u2022 The Language of Physics: Physical Quantities and Units 21 to something very large (like a star). Yet the standard metric unit of length is the meter. So, the metric system includes many prefixes that can be attached to a unit. Each prefix is based on factors of 10 (10, 100, 1,000, etc., as well as 0.1, 0.01, 0.001, etc.). Table 1.2 gives the metric prefixes and symbols used to denote the different various factors of 10 in the metric system. Prefix Symbol Value[1] Example Name Example Symbol Example Value Example Description exa peta tera giga mega kilo hector E P T G M k h deka da 1018 1015 1012 109 106 103 102 101 ____ ____ 100 (=1) deci centi d c milli m micro nano pico femto atto \u00b5 n p f a 10\u20131 10\u20132 10\u20133 10\u20136 10\u20139 10\u201312 10\u201315 10\u201318 Exameter Em Petasecond Ps Terawatt TW Gigahertz GHz Megacurie MCi Kilometer Hectoliter Dekagram Deciliter Centimeter km hL dag dL Cm Millimeter Mm Micrometer \u00b5m Nanogram Picofarad Ng pF 1018 m 1015 s 1012 W 109 Hz 106 Ci 103 m 102 L 101 g 10\u20131 L 10\u20132 m 10\u20133 m 10\u20136 m 10\u20139 g 10\u201312 F Distance light travels in a century 30 million years Powerful laser output A microwave frequency High radioactivity About 6/10 mile 26 gallons Teaspoon of butter Less than half a soda Fingertip thickness Flea at its shoulder Detail in microscope Small speck of dust Small capacitor in radio Femtometer Fm 10\u201315 m Size of a proton Attosecond as 10\u201318 s Time light takes to cross an atom Table 1.2 Metric Prefixes for Powers of 10 and Their Symbols Note\u2014Some examples are approximate. [1]See Appendix A for a discussion of powers of 10. The metric system is convenient because conversions between metric units can be done simply by moving the decimal place of a number. This is because the metric prefixes are sequential powers of 10. There are 100 centimeters in a meter, 1000 meters in a kilometer, and so on. In nonmetric systems, such as U.S. customary units, the relationships are less simple\u2014there are 12 inches in a foot, 5,280 feet in a mile, 4 quarts in a gallon, and so on. Another advantage of the metric system is that the same unit can be used over extremely large ranges of values simply by switching to the most-appropriate metric prefix. For example, distances in meters are suitable for building construction, but kilometers are used to describe road construction. Therefore, with the metric system, there is no need to invent new", " units when measuring very small or very large objects\u2014you just have to move the decimal 22 Chapter 1 \u2022 What is Physics? point (and use the appropriate prefix). Known Ranges of Length, Mass, and Time Table 1.3 lists known lengths, masses, and time measurements. You can see that scientists use a range of measurement units. This wide range demonstrates the vastness and complexity of the universe, as well as the breadth of phenomena physicists study. As you examine this table, note how the metric system allows us to discuss and compare an enormous range of phenomena, using one system of measurement (Figure 1.16 and Figure 1.17). Length (m) Phenomenon Measured 10\u201318 Present experimental limit to smallest observable detail Mass (Kg) 10\u201330 Phenomenon Measured[1] Mass of an electron (9.11 10\u201331 kg) 10\u201315 Diameter of a proton 10\u201327 Mass of a hydrogen atom 10\u201327 kg) (1.67 Time (s) 10\u201323 10\u201322 1014 Diameter of a uranium nucleus 10\u201315 Mass of a bacterium 10\u201315 10\u201310 Diameter of a hydrogen atom 10\u20135 Mass of a mosquito 10\u201313 Phenomenon Measured[1] Time for light to cross a proton Mean life of an extremely unstable nucleus Time for one oscillation of a visible light Time for one vibration of an atom in a solid Time for one oscillation of an FM radio wave 10\u20138 10\u20136 10\u20133 1 102 104 107 1011 1016 1021 Thickness of membranes in cell of living organism 10\u20132 Mass of a hummingbird 10\u20138 Wavelength of visible light Size of a grain of sand Height of a 4-year-old child Length of a football field Greatest ocean depth Diameter of Earth Distance from Earth to the sun Distance traveled by light in 1 year (a light year) 1 102 103 108 1012 1015 1023 1025 Diameter of the Milky Way Galaxy 1030 Mass of a liter of water (about a quart) 10\u20133 Duration of a nerve impulse 1 105 107 109 Time for one heartbeat One day (8.64 104 s) One year (3.16 107 s) About half the life expectancy of a human 1011 Recorded history 1017 Age of Earth 1018 Age of the universe Mass of a person Mass of a car Mass of a large ship Mass of a large iceberg Mass of the nucleus of a comet Mass of the moon (7.35 1022 kg) Mass of Earth (5.97 kg) 1024 Mass of the Sun (1.99 kg) 1024 Table 1.3 Approximate Values of Length, Mass, and Time [1] More precise values are in parentheses. Access for free at openstax.org. 1.3 \u2022 The Language of Physics: Physical Quantities and Units 23 Length (m) 1022 1026 Phenomenon Measured Mass (Kg) Phenomenon Measured[1] Time (s) Phenomenon Measured[1] Distance from Earth to the nearest large galaxy (Andromeda) Distance from Earth to the edges of the known universe 1042 1053 Mass of the Milky Way galaxy (current upper limit) Mass of the known universe (current upper limit) Table 1.3 Approximate Values of Length, Mass, and Time [1] More precise values are in parentheses. Figure 1.16 Tiny phytoplankton float among crystals of ice in the Antarctic Sea. They range from a few micrometers to as much as 2 millimeters in length. (Prof. Gordon T. Taylor, Stony Brook University; NOAA Corps Collections) Figure 1.17 Galaxies collide 2.4 billion light years away from Earth. The tremendous range of observable phenomena in nature challenges the imagination. (NASA/CXC/UVic./A. Mahdavi et al. Optical/lensing: CFHT/UVic./H. Hoekstra et al.) Using Scientific Notation with Physical Measurements Scientific notation is a way of writing numbers that are too large or small to be conveniently written as a decimal. For example, consider the number 840,000,000,000,000. It\u2019s a rather large number to write out. The scientific notation for this number is 8.40 1014. Scientific notation follows this general format In this format xis the value of the measurement with all placeholder zeros removed. In the example above, xis 8.4. The xis multiplied by a factor, 10y, which indicates the number of placeholder zeros in the measurement. Placeholder zeros are those at the end of a number that is 10 or greater, and at the beginning of a decimal number that is less than 1. In the example above, the factor is 1014. This tells you that you should move the decimal point 14 positions to the right, filling in placeholder zeros as you go. In this case", ", moving the decimal point 14 places creates only 13 placeholder zeros, indicating that the actual measurement value is 840,000,000,000,000. 24 Chapter 1 \u2022 What is Physics? Numbers that are fractions can be indicated by scientific notation as well. Consider the number 0.0000045. Its scientific notation is 4.5 10\u20136. Its scientific notation has the same format Here, xis 4.5. However, the value of yin the 10yfactor is negative, which indicates that the measurement is a fraction of 1. 10\u20136, the decimal point would be Therefore, we move the decimal place to the left, for a negative y. In our example of 4.5 moved to the left six times to yield the original number, which would be 0.0000045. The term order of magnitude refers to the power of 10 when numbers are expressed in scientific notation. Quantities that have the same power of 10 when expressed in scientific notation, or come close to it, are said to be of the same order of magnitude. For example, the number 800 can be written as 8 same value for y. Therefore, 800 and 450 are of the same order of magnitude. Similarly, 101 and 99 would be regarded as the same order of magnitude, 102. Order of magnitude can be thought of as a ballpark estimate for the scale of a value. The diameter of an atom is on the order of 10\u22129 m, while the diameter of the sun is on the order of 109 m. These two values are 18 orders of magnitude apart. 102, and the number 450 can be written as 4.5 102. Both numbers have the Scientists make frequent use of scientific notation because of the vast range of physical measurements possible in the universe, such as the distance from Earth to the moon (Figure 1.18), or to the nearest star. Figure 1.18 The distance from Earth to the moon may seem immense, but it is just a tiny fraction of the distance from Earth to our closest neighboring star. (NASA) Unit Conversion and Dimensional Analysis It is often necessary to convert from one type of unit to another. For example, if you are reading a European cookbook in the United States, some quantities may be expressed in liters and you need to convert them to cups. A Canadian tourist driving through the United States might want to convert miles to kilometers, to have a sense of how far away his next destination is. A doctor in the United States might convert a patient\u2019s weight in pounds to kilograms. Let\u2019s consider a simple example of how to convert units within the metric system. How can we want to convert 1 hour to seconds? Next, we need to determine a conversion factor relating meters to kilometers. A conversion factor is a ratio expressing how many of one unit are equal to another unit. A conversion factor is simply a fraction which equals 1. You can multiply any number by 1 and get the same value. When you multiply a number by a conversion factor, you are simply multiplying it by one. For example, the following are conversion factors: (1 foot)/(12 inches) = 1 to convert inches to feet, (1 meter)/(100 centimeters) = 1 to convert centimeters to meters, (1 minute)/(60 seconds) = 1 to convert seconds to minutes. In this case, we know that there are 1,000 meters in 1 kilometer. Now we can set up our unit conversion. We will write the units that we have and then multiply them by the conversion factor (1 km/1,000m) = 1, so we are simply multiplying 80m by 1: When there is a unit in the original number, and a unit in the denominator (bottom) of the conversion factor, the units cancel. In this case, hours and minutes cancel and the value in seconds remains. You can use this method to convert between any types of unit, including between the U.S. customary system and metric system. Notice also that, although you can multiply and divide units algebraically, you cannot add or subtract different units. An expression like 10 km + 5 kgmakes no sense. Even adding two lengths in different units, such as 10 km + 20 mdoes not make sense. You express both lengths in the same unit. See Appendix C for a more complete list of conversion factors. 1.1 Access for free at openstax.org. 1.3 \u2022 The Language of Physics: Physical Quantities and Units 25 WORKED EXAMPLE Unit Conversions: A Short Drive Home Suppose that you drive the 10.0 km from your university to home in 20.0 min. Calculate your average speed (a) in kilometers per hour (km/h) and (b) in meters per second (m/s). (Note\u2014Average speed is distance traveled divided by time of travel.) Strategy First we calculate the average speed using the given units. Then we can get the average speed into the desired units by picking the correct conversion factor and multiplying by it.", " The correct conversion factor is the one that cancels the unwanted unit and leaves the desired unit in its place. Solution for (a) 1. Calculate average speed. Average speed is distance traveled divided by time of travel. (Take this definition as a given for now\u2014average speed and other motion concepts will be covered in a later module.) In equation form, 2. Substitute the given values for distance and time. 3. Convert km/min to km/h: multiply by the conversion factor that will cancel minutes and leave hours. That conversion factor is. Thus, Discussion for (a) To check your answer, consider the following: 1. Be sure that you have properly cancelled the units in the unit conversion. If you have written the unit conversion factor upside down, the units will not cancel properly in the equation. If you accidentally get the ratio upside down, then the units will not cancel; rather, they will give you the wrong units as follows which are obviously not the desired units of km/h. 2. Check that the units of the final answer are the desired units. The problem asked us to solve for average speed in units of km/h and we have indeed obtained these units. 3. Check the significant figures. Because each of the values given in the problem has three significant figures, the answer should also have three significant figures. The answer 30.0 km/h does indeed have three significant figures, so this is appropriate. Note that the significant figures in the conversion factor are not relevant because an hour is definedto be 60 min, so the precision of the conversion factor is perfect. 4. Next, check whether the answer is reasonable. Let us consider some information from the problem\u2014if you travel 10 km in a third of an hour (20 min), you would travel three times that far in an hour. The answer does seem reasonable. Solution (b) There are several ways to convert the average speed into meters per second. 1. Start with the answer to (a) and convert km/h to m/s. Two conversion factors are needed\u2014one to convert hours to seconds, and another to convert kilometers to meters. 2. Multiplying by these yields 26 Chapter 1 \u2022 What is Physics? Discussion for (b) If we had started with 0.500 km/min, we would have needed different conversion factors, but the answer would have been the same: 8.33 m/s. You may have noted that the answers in the worked example just covered were given to three digits. Why? When do you need to be concerned about the number of digits in something you calculate? Why not write down all the digits your calculator produces? WORKED EXAMPLE Using Physics to Evaluate Promotional Materials A commemorative coin that is 2\u2033 in diameter is advertised to be plated with 15 mg of gold. If the density of gold is 19.3 g/cc, and the amount of gold around the edge of the coin can be ignored, what is the thickness of the gold on the top and bottom faces of the coin? Strategy To solve this problem, the volume of the gold needs to be determined using the gold\u2019s mass and density. Half of that volume is distributed on each face of the coin, and, for each face, the gold can be represented as a cylinder that is 2\u2033 in diameter with a height equal to the thickness. Use the volume formula for a cylinder to determine the thickness. Solution The mass of the gold is given by the formula where and Vis the volume. Solving for the volume gives If tis the thickness, the volume corresponding to half the gold is where the 1\u2033 radius has been converted to cm. Solving for the thickness gives Discussion The amount of gold used is stated to be 15 mg, which is equivalent to a thickness of about 0.00019 mm. The mass figure may make the amount of gold sound larger, both because the number is much bigger (15 versus 0.00019), and because people may have a more intuitive feel for how much a millimeter is than for how much a milligram is. A simple analysis of this sort can clarify the significance of claims made by advertisers. Accuracy, Precision and Significant Figures Science is based on experimentation that requires good measurements. The validity of a measurement can be described in terms of its accuracy and its precision (see Figure 1.19 and Figure 1.20). Accuracy is how close a measurement is to the correct value for that measurement. For example, let us say that you are measuring the length of standard piece of printer paper. The packaging in which you purchased the paper states that it is 11 inches long, and suppose this stated value is correct. You measure the length of the paper three times and obtain the following measurements: 11.1 inches, 11.2 inches, and 10.9 inches. These measurements are quite accurate because they are very close to the correct value of 11.0 inches. In contrast, if you had obtained a measurement of 12 inches, your measurement would not be very", " accurate. This is why measuring instruments are calibrated based on a known measurement. If the instrument consistently returns the correct value of the known measurement, it is safe for use in finding unknown values. Access for free at openstax.org. 1.3 \u2022 The Language of Physics: Physical Quantities and Units 27 Figure 1.19 A double-pan mechanical balance is used to compare different masses. Usually an object with unknown mass is placed in one pan and objects of known mass are placed in the other pan. When the bar that connects the two pans is horizontal, then the masses in both pans are equal. The known masses are typically metal cylinders of standard mass such as 1 gram, 10 grams, and 100 grams. (Serge Melki) Figure 1.20 Whereas a mechanical balance may only read the mass of an object to the nearest tenth of a gram, some digital scales can measure the mass of an object up to the nearest thousandth of a gram. As in other measuring devices, the precision of a scale is limited to the last measured figures. This is the hundredths place in the scale pictured here. (Splarka, Wikimedia Commons) Precision states how well repeated measurements of something generate the same or similar results. Therefore, the precision of measurements refers to how close together the measurements are when you measure the same thing several times. One way to analyze the precision of measurements would be to determine the range, or difference between the lowest and the highest measured values. In the case of the printer paper measurements, the lowest value was 10.9 inches and the highest value was 11.2 inches. Thus, the measured values deviated from each other by, at most, 0.3 inches. These measurements were reasonably precise because they varied by only a fraction of an inch. However, if the measured values had been 10.9 inches, 11.1 inches, and 11.9 inches, then the measurements would not be very precise because there is a lot of variation from one measurement to another. The measurements in the paper example are both accurate and precise, but in some cases, measurements are accurate but not precise, or they are precise but not accurate. Let us consider a GPS system that is attempting to locate the position of a restaurant in a city. Think of the restaurant location as existing at the center of a bull\u2019s-eye target. Then think of each GPS attempt to locate the restaurant as a black dot on the bull\u2019s eye. In Figure 1.21, you can see that the GPS measurements are spread far apart from each other, but they are all relatively close to the actual location of the restaurant at the center of the target. This indicates a low precision, high accuracy measuring system. However, in Figure 1.22, the GPS measurements are concentrated quite closely to one another, but they are far away from the target location. This indicates a high precision, low accuracy measuring system. Finally, in Figure 1.23, the GPS is both precise and accurate, allowing the restaurant to be located. 28 Chapter 1 \u2022 What is Physics? Figure 1.21 A GPS system attempts to locate a restaurant at the center of the bull\u2019s-eye. The black dots represent each attempt to pinpoint the location of the restaurant. The dots are spread out quite far apart from one another, indicating low precision, but they are each rather close to the actual location of the restaurant, indicating high accuracy. (Dark Evil) Figure 1.22 In this figure, the dots are concentrated close to one another, indicating high precision, but they are rather far away from the actual location of the restaurant, indicating low accuracy. (Dark Evil) Figure 1.23 In this figure, the dots are concentrated close to one another, indicating high precision, but they are rather far away from the actual location of the restaurant, indicating low accuracy. (Dark Evil) Uncertainty The accuracy and precision of a measuring system determine the uncertainty of its measurements. Uncertainty is a way to describe how much your measured value deviates from the actual value that the object has. If your measurements are not very accurate or precise, then the uncertainty of your values will be very high. In more general terms, uncertainty can be thought of as a disclaimer for your measured values. For example, if someone asked you to provide the mileage on your car, you might say that it is 45,000 miles, plus or minus 500 miles. The plus or minus amount is the uncertainty in your value. That is, you are indicating that the actual mileage of your car might be as low as 44,500 miles or as high as 45,500 miles, or anywhere in between. All measurements contain some amount of uncertainty. In our example of measuring the length of the paper, we might say that the length of the paper is 11 inches plus or minus 0.2 inches or 11.0 \u00b1 0.2 inches. The uncertainty in a Access for free at openstax.org. 1.3 \u2022 The Language of Physics: Physical Quantities and", " Units 29 measurement, A, is often denoted as \u03b4A(\"delta A\"), The factors contributing to uncertainty in a measurement include the following: 1. Limitations of the measuring device 2. The skill of the person making the measurement Irregularities in the object being measured 3. 4. Any other factors that affect the outcome (highly dependent on the situation) In the printer paper example uncertainty could be caused by: the fact that the smallest division on the ruler is 0.1 inches, the person using the ruler has bad eyesight, or uncertainty caused by the paper cutting machine (e.g., one side of the paper is slightly longer than the other.) It is good practice to carefully consider all possible sources of uncertainty in a measurement and reduce or eliminate them, Percent Uncertainty One method of expressing uncertainty is as a percent of the measured value. If a measurement, A, is expressed with uncertainty, \u03b4A, the percent uncertainty is 1.2 WORKED EXAMPLE Calculating Percent Uncertainty: A Bag of Apples A grocery store sells 5-lb bags of apples. You purchase four bags over the course of a month and weigh the apples each time. You obtain the following measurements: \u2022 Week 1 weight: \u2022 Week 2 weight: \u2022 Week 3 weight: \u2022 Week 4 weight: You determine that the weight of the 5 lb bag has an uncertainty of \u00b10.4 lb. What is the percent uncertainty of the bag\u2019s weight? Strategy First, observe that the expected value of the bag\u2019s weight, following equation to determine the percent uncertainty of the weight, is 5 lb. The uncertainty in this value,, is 0.4 lb. We can use the Solution Plug the known values into the equation Discussion We can conclude that the weight of the apple bag is 5 lb \u00b1 8 percent. Consider how this percent uncertainty would change if the bag of apples were half as heavy, but the uncertainty in the weight remained the same. Hint for future calculations: when calculating percent uncertainty, always remember that you must multiply the fraction by 100 percent. If you do not do this, you will have a decimal quantity, not a percent value. Uncertainty in Calculations There is an uncertainty in anything calculated from measured quantities. For example, the area of a floor calculated from measurements of its length and width has an uncertainty because the both the length and width have uncertainties. How big is the uncertainty in something you calculate by multiplication or division? If the measurements in the calculation have small uncertainties (a few percent or less), then the method of adding percents can be used. This method says that the percent uncertainty in a quantity calculated by multiplication or division is the sum of the percent uncertainties in the items used to make the calculation. For example, if a floor has a length of 4.00 m and a width of 3.00 m, with uncertainties of 2 percent and 1 30 Chapter 1 \u2022 What is Physics? percent, respectively, then the area of the floor is 12.0 m2 and has an uncertainty of 3 percent (expressed as an area this is 0.36 m2, which we round to 0.4 m2 since the area of the floor is given to a tenth of a square meter). For a quick demonstration of the accuracy, precision, and uncertainty of measurements based upon the units of measurement, try this simulation (http://openstax.org/l/28precision). You will have the opportunity to measure the length and weight of a desk, using milli- versus centi- units. Which do you think will provide greater accuracy, precision and uncertainty when measuring the desk and the notepad in the simulation? Consider how the nature of the hypothesis or research question might influence how precise of a measuring tool you need to collect data. Precision of Measuring Tools and Significant Figures An important factor in the accuracy and precision of measurements is the precision of the measuring tool. In general, a precise measuring tool is one that can measure values in very small increments. For example, consider measuring the thickness of a coin. A standard ruler can measure thickness to the nearest millimeter, while a micrometer can measure the thickness to the nearest 0.005 millimeter. The micrometer is a more precise measuring tool because it can measure extremely small differences in thickness. The more precise the measuring tool, the more precise and accurate the measurements can be. When we express measured values, we can only list as many digits as we initially measured with our measuring tool (such as the rulers shown in Figure 1.24). For example, if you use a standard ruler to measure the length of a stick, you may measure it with a decimeter ruler as 3.6 cm. You could not express this value as 3.65 cm because your measuring tool was not precise enough to measure a hundredth of a centimeter. It should be noted that the last digit in a measured value has been estimated in some way by the person performing the measurement. For example, the person measuring", " the length of a stick with a ruler notices that the stick length seems to be somewhere in between 36 mm and 37 mm. He or she must estimate the value of the last digit. The rule is that the last digit written down in a measurement is the first digit with some uncertainty. For example, the last measured value 36.5 mm has three digits, or three significant figures. The number of significant figures in a measurement indicates the precision of the measuring tool. The more precise a measuring tool is, the greater the number of significant figures it can report. Figure 1.24 Three metric rulers are shown. The first ruler is in decimeters and can measure point three decimeters. The second ruler is in centimeters long and can measure three point six centimeters. The last ruler is in millimeters and can measure thirty-six point five millimeters. Zeros Special consideration is given to zeros when counting significant figures. For example, the zeros in 0.053 are not significant because they are only placeholders that locate the decimal point. There are two significant figures in 0.053\u2014the 5 and the 3. However, if the zero occurs between other significant figures, the zeros are significant. For example, both zeros in 10.053 are significant, as these zeros were actually measured. Therefore, the 10.053 placeholder has five significant figures. The zeros in 1300 may or may not be significant, depending on the style of writing numbers. They could mean the number is known to the last zero, or the zeros could be placeholders. So 1300 could have two, three, or four significant figures. To avoid this ambiguity, Access for free at openstax.org. 1.3 \u2022 The Language of Physics: Physical Quantities and Units 31 write 1300 in scientific notation as 1.3 \u00d7 103. Only significant figures are given in the xfactor for a number in scientific notation (in the form ). Therefore, we know that 1 and 3 are the only significant digits in this number. In summary, zeros are significant except when they serve only as placeholders. Table 1.4 provides examples of the number of significant figures in various numbers. Number Significant Figures Rationale 4 4 3 6 3 7 4 4 1.657 0.4578 0.000458 2000.56 45,600 15895000 5.457 1013 6.520 10\u201323 Table 1.4 There are no zeros and all non-zero numbers are always significant. The first zero is only a placeholder for the decimal point. The first four zeros are placeholders needed to report the data to the ten-thoudsandths place. The three zeros are significant here because they occur between other significant figures. With no underlines or scientific notation, we assume that the last two zeros are placeholders and are not significant. The two underlined zeros are significant, while the last zero is not, as it is not underlined. In scientific notation, all numbers reported in front of the multiplication sign are significant In scientific notation, all numbers reported in front of the multiplication sign are significant, including zeros. Significant Figures in Calculations When combining measurements with different degrees of accuracy and precision, the number of significant digits in the final answer can be no greater than the number of significant digits in the least precise measured value. There are two different rules, one for multiplication and division and another rule for addition and subtraction, as discussed below. 1. For multiplication and division: The answer should have the same number of significant figures as the starting value with. Let us see the fewest significant figures. For example, the area of a circle can be calculated from its radius using how many significant figures the area will have if the radius has only two significant figures, for example, r= 2.0 m. Then, using a calculator that keeps eight significant figures, you would get But because the radius has only two significant figures, the area calculated is meaningful only to two significant figures or even though the value of is meaningful to at least eight digits. 2. For addition and subtraction: The answer should have the same number places (e.g. tens place, ones place, tenths place, etc.) as the least-precise starting value. Suppose that you buy 7.56 kg of potatoes in a grocery store as measured with a scale having a precision of 0.01 kg. Then you drop off 6.052 kg of potatoes at your laboratory as measured by a scale with a precision of 0.001 kg. Finally, you go home and add 13.7 kg of potatoes as measured by a bathroom scale with a precision of 0.1 kg. How many kilograms of potatoes do you now have, and how many significant figures are appropriate in the answer? The mass is found by simple addition and subtraction: 32 Chapter 1 \u2022 What is Physics? The least precise measurement is 13.7 kg. This measurement is expressed to the 0.1 decimal place, so our final answer must also", " be expressed to the 0.1 decimal place. Thus, the answer should be rounded to the tenths place, giving 15.2 kg. The same is true for non-decimal numbers. For example, We cannot report the decimal places in the answer because 2has no decimal places that would be significant. Therefore, we can only report to the ones place. It is a good idea to keep extra significant figures while calculating, and to round off to the correct number of significant figures only in the final answers. The reason is that small errors from rounding while calculating can sometimes produce significant errors in the final answer. As an example, try calculating to obtain a final answer to only two significant figures. Keeping all significant during the calculation gives 48. Rounding to two significant figures in the middle of the calculation changes it to which is way off. You would similarly avoid rounding in the middle of the calculation in counting and in doing accounting, where many small numbers need to be added and subtracted accurately to give possibly much larger final numbers. Significant Figures in this Text In this textbook, most numbers are assumed to have three significant figures. Furthermore, consistent numbers of significant figures are used in all worked examples. You will note that an answer given to three digits is based on input good to at least three digits. If the input has fewer significant figures, the answer will also have fewer significant figures. Care is also taken that the number of significant figures is reasonable for the situation posed. In some topics, such as optics, more than three significant figures will be used. Finally, if a number is exact, such as the 2in the formula, significant figures in a calculation., it does not affect the number of WORKED EXAMPLE Approximating Vast Numbers: a Trillion Dollars The U.S. federal deficit in the 2008 fiscal year was a little greater than $10 trillion. Most of us do not have any concept of how much even one trillion actually is. Suppose that you were given a trillion dollars in $100 bills. If you made 100-bill stacks, like that shown in Figure 1.25, and used them to evenly cover a football field (between the end zones), make an approximation of how high the money pile would become. (We will use feet/inches rather than meters here because football fields are measured in yards.) One of your friends says 3 in., while another says 10 ft. What do you think? Access for free at openstax.org. 1.3 \u2022 The Language of Physics: Physical Quantities and Units 33 Figure 1.25 A bank stack contains one hundred $100 bills, and is worth $10,000. How many bank stacks make up a trillion dollars? (Andrew Magill) Strategy When you imagine the situation, you probably envision thousands of small stacks of 100 wrapped $100 bills, such as you might see in movies or at a bank. Since this is an easy-to-approximate quantity, let us start there. We can find the volume of a stack of 100 bills, find out how many stacks make up one trillion dollars, and then set this volume equal to the area of the football field multiplied by the unknown height. Solution 1. Calculate the volume of a stack of 100 bills. The dimensions of a single bill are approximately 3 in. by 6 in. A stack of 100 of these is about 0.5 in. thick. So the total volume of a stack of 100 bills is 2. Calculate the number of stacks. Note that a trillion dollars is equal to, and a stack of one-hundred bills is equal to or. The number of stacks you will have is 1.3 3. Calculate the area of a football field in square inches. The area of a football field is, which gives. Because we are working in inches, we need to convert square yards to square inches This conversion gives us calculations.) for the area of the field. (Note that we are using only one significant figure in these 4. Calculate the total volume of the bills. The volume of all the $100-bill stacks is 5. Calculate the height. To determine the height of the bills, use the following equation 34 Chapter 1 \u2022 What is Physics? The height of the money will be about 100 in. high. Converting this value to feet gives Discussion The final approximate value is much higher than the early estimate of 3 in., but the other early estimate of 10 ft (120 in.) was roughly correct. How did the approximation measure up to your first guess? What can this exercise tell you in terms of rough guesstimatesversus carefully calculated approximations? In the example above, the final approximate value is much higher than the first friend\u2019s early estimate of 3 in. However, the other friend\u2019s early estimate of 10 ft. (120 in.) was roughly correct. How did the approximation measure up to your first guess? What can this exercise suggest about the value of rough guesstimatesversus carefully calculated approxim", "ations? Graphing in Physics Most results in science are presented in scientific journal articles using graphs. Graphs present data in a way that is easy to visualize for humans in general, especially someone unfamiliar with what is being studied. They are also useful for presenting large amounts of data or data with complicated trends in an easily-readable way. One commonly-used graph in physics and other sciences is the line graph, probably because it is the best graph for showing how one quantity changes in response to the other. Let\u2019s build a line graph based on the data in Table 1.5, which shows the measured distance that a train travels from its station versus time. Our two variables, or things that change along the graph, are time in minutes, and distance from the station, in kilometers. Remember that measured data may not have perfect accuracy. Time (min) Distance from Station (km) 0 24 36 60 84 97 116 140 0 10 20 30 40 50 60 70 Table 1.5 1. Draw the two axes. The horizontal axis, or x-axis, shows the independent variable, which is the variable that is controlled or manipulated. The vertical axis, or y-axis, shows the dependent variable, the non-manipulated variable that changes with (or is dependent on) the value of the independent variable. In the data above, time is the independent variable and should be plotted on the x-axis. Distance from the station is the dependent variable and should be plotted on the y-axis. Access for free at openstax.org. 1.3 \u2022 The Language of Physics: Physical Quantities and Units 35 2. Label each axes on the graph with the name of each variable, followed by the symbol for its units in parentheses. Be sure to leave room so that you can number each axis. In this example, use Time (min)as the label for the x-axis. 3. Next, you must determine the best scale to use for numbering each axis. Because the time values on the x-axis are taken every 10 minutes, we could easily number the x-axis from 0 to 70 minutes with a tick mark every 10 minutes. Likewise, the y-axis scale should start low enough and continue high enough to include all of the distance from stationvalues. A scale from 0 km to 160 km should suffice, perhaps with a tick mark every 10 km. In general, you want to pick a scale for both axes that 1) shows all of your data, and 2) makes it easy to identify trends in your data. If you make your scale too large, it will be harder to see how your data change. Likewise, the smaller and more fine you make your scale, the more space you will need to make the graph. The number of significant figures in the axis values should be coarser than the number of significant figures in the measurements. 4. Now that your axes are ready, you can begin plotting your data. For the first data point, count along the x-axis until you find the 10 min tick mark. Then, count up from that point to the 10 km tick mark on the y-axis, and approximate where 22 km is along the y-axis. Place a dot at this location. Repeat for the other six data points (Figure 1.26). Figure 1.26 The graph of the train\u2019s distance from the station versus time from the exercise above. 5. Add a title to the top of the graph to state what the graph is describing, such as the y-axis parameter vs. the x-axis parameter. In the graph shown here, the title is train motion. It could also be titled distance of the train from the station vs. time. 6. Finally, with data points now on the graph, you should draw a trend line (Figure 1.27). The trend line represents the dependence you think the graph represents, so that the person who looks at your graph can see how close it is to the real data. In the present case, since the data points look like they ought to fall on a straight line, you would draw a straight line as the trend line. Draw it to come closest to all the points. Real data may have some inaccuracies, and the plotted points may not all fall on the trend line. In some cases, none of the data points fall exactly on the trend line. 36 Chapter 1 \u2022 What is Physics? Figure 1.27 The completed graph with the trend line included. Analyzing a Graph Using Its Equation One way to get a quick snapshot of a dataset is to look at the equation of its trend line. If the graph produces a straight line, the equation of the trend line takes the form The bin the equation is the y-intercept while the min the equation is the slope. The y-intercept tells you at what yvalue the line intersects the y-axis. In the case of the graph above, the y-intercept occurs at 0, at the very beginning of the graph", ". The y-intercept, therefore, lets you know immediately where on the y-axis the plot line begins. The min the equation is the slope. This value describes how much the line on the graph moves up or down on the y-axis along the line\u2019s length. The slope is found using the following equation In order to solve this equation, you need to pick two points on the line (preferably far apart on the line so the slope you calculate describes the line accurately). The quantities Y2 and Y1 represent the y-values from the two points on the line (not data points) that you picked, while X2 and X1 represent the two x-values of the those points. What can the slope value tell you about the graph? The slope of a perfectly horizontal line will equal zero, while the slope of a perfectly vertical line will be undefined because you cannot divide by zero. A positive slope indicates that the line moves up the y-axis as the x-value increases while a negative slope means that the line moves down the y-axis. The more negative or positive the slope is, the steeper the line moves up or down, respectively. The slope of our graph in Figure 1.26 is calculated below based on the two endpoints of the line Equation of line: Because the xaxis is time in minutes, we would actually be more likely to use the time tas the independent (x-axis) variable and write the equation as Access for free at openstax.org. 1.3 \u2022 The Language of Physics: Physical Quantities and Units 37 only applies to linear relationships, or ones that produce a straight line. Another common type of line The formula in physics is the quadratic relationship, which occurs when one of the variables is squared. One quadratic relationship in physics is the relation between the speed of an object its centripetal acceleration, which is used to determine the force needed to keep an object moving in a circle. Another common relationship in physics is the inverse relationship, in which one variable decreases whenever the other variable increases. An example in physics is Coulomb\u2019s law. As the distance between two charged objects increases, the electrical force between the two charged objects decreases. Inverse proportionality, such the relation between xand yin the equation 1.4 for some number k, is one particular kind of inverse relationship. A third commonly-seen relationship is the exponential relationship, in which a change in the independent variable produces a proportional change in the dependent variable. As the value of the dependent variable gets larger, its rate of growth also increases. For example, bacteria often reproduce at an exponential rate when grown under ideal conditions. As each generation passes, there are more and more bacteria to reproduce. As a result, the growth rate of the bacterial population increases every generation (Figure 1.28). Figure 1.28 Examples of (a) linear, (b) quadratic, (c) inverse, and (d) exponential relationship graphs. Using Logarithmic Scales in Graphing Sometimes a variable can have a very large range of values. This presents a problem when you\u2019re trying to figure out the best scale to use for your graph\u2019s axes. One option is to use a logarithmic (log) scale. In a logarithmic scale, the value each mark labels 38 Chapter 1 \u2022 What is Physics? is the previous mark\u2019s value multiplied by some constant. For a log base 10 scale, each mark labels a value that is 10 times the value of the mark before it. Therefore, a base 10 logarithmic scale would be numbered: 0, 10, 100, 1,000, etc. You can see how the logarithmic scale covers a much larger range of values than the corresponding linear scale, in which the marks would label the values 0, 10, 20, 30, and so on. If you use a logarithmic scale on one axis of the graph and a linear scale on the other axis, you are using a semi-log plot. The Richter scale, which measures the strength of earthquakes, uses a semi-log plot. The degree of ground movement is plotted on a logarithmic scale against the assigned intensity level of the earthquake, which ranges linearly from 1-10 (Figure 1.29 (a)). If a graph has both axes in a logarithmic scale, then it is referred to as a log-log plot. The relationship between the wavelength and frequency of electromagnetic radiation such as light is usually shown as a log-log plot (Figure 1.29 (b)). Log-log plots are also commonly used to describe exponential functions, such as radioactive decay. Figure 1.29 (a) The Richter scale uses a log base 10 scale on its y-axis(microns of amplified maximum ground motion). (b) The relationship between the frequency and wavelength of electromagnetic radiation can be plotted as a straight", " line if a log-log plot is used. Virtual Physics Graphing Lines In this simulation you will examine how changing the slope and y-intercept of an equation changes the appearance of a plotted line. Select slope-intercept form and drag the blue circles along the line to change the line\u2019s characteristics. Then, play the line game and see if you can determine the slope or y-intercept of a given line. Click to view content (https://phet.colorado.edu/sims/html/graphing-lines/latest/graphing-lines_en.html) GRASP CHECK How would the following changes affect a line that is neither horizontal nor vertical and has a positive slope? 1. 2. increase the slope but keeping the y-intercept constant increase the y-intercept but keeping the slope constant a. Increasing the slope will cause the line to rotate clockwise around the y-intercept. Increasing the y-intercept will cause the line to move vertically up on the graph without changing the line\u2019s slope. Increasing the slope will cause the line to rotate counter-clockwise around the y-intercept. Increasing the y-intercept will cause the line to move vertically up on the graph without changing the line\u2019s slope. Increasing the slope will cause the line to rotate clockwise around the y-intercept. Increasing the y-intercept will cause the line to move horizontally right on the graph without changing the line\u2019s slope. b. c. Access for free at openstax.org. 1.3 \u2022 The Language of Physics: Physical Quantities and Units 39 d. Increasing the slope will cause the line to rotate counter-clockwise around the y-intercept. Increasing the y-intercept will cause the line to move horizontally right on the graph without changing the line\u2019s slope. Check Your Understanding 12. Identify some advantages of metric units. a. Conversion between units is easier in metric units. b. Comparison of physical quantities is easy in metric units. c. Metric units are more modern than English units. d. Metric units are based on powers of 2. 13. The length of an American football field is, excluding the end zones. How long is the field in meters? Round to the. nearest a. b. c. d. 14. The speed limit on some interstate highways is roughly. How many miles per hour is this if is about? a. 0.1 mi/h b. 27.8 mi/h c. 62 mi/h 160 mi/h d. 15. Briefly describe the target patterns for accuracy and precision and explain the differences between the two. a. Precision states how much repeated measurements generate the same or closely similar results, while accuracy states how close a measurement is to the true value of the measurement. b. Precision states how close a measurement is to the true value of the measurement, while accuracy states how much repeated measurements generate the same or closely similar result. c. Precision and accuracy are the same thing. They state how much repeated measurements generate the same or closely similar results. d. Precision and accuracy are the same thing. They state how close a measurement is to the true value of the measurement. 40 Chapter 1 \u2022 Key Terms KEY TERMS accuracy how close a measurement is to the correct value for that measurement ampere the SI unit for electrical current atom smallest and most basic units of matter classical physics physics, as it developed from the Renaissance to the end of the nineteenth century constant a quantity that does not change conversion factor a ratio expressing how many of one unit are equal to another unit century to the present, involving the theories of relativity and quantum mechanics observation step where a scientist observes a pattern or trend within the natural world order of magnitude the size of a quantity in terms of its power of 10 when expressed in scientific notation physics science aimed at describing the fundamental aspects of our universe\u2014energy, matter, space, motion, and time dependent variable the vertical, or y-axis, variable, which precision how well repeated measurements generate the changes with (or is dependent on) the value of the independent variable same or closely similar results principle description of nature that is true in many, but not derived units units that are derived by combining the all situations fundamental physical units quadratic relationship relation between variables that can English units (also known as the customary or imperial system) system of measurement used in the United States; includes units of measurement such as feet, gallons, degrees Fahrenheit, and pounds experiment process involved with testing a hypothesis exponential relationship relation between variables in which a constant change in the independent variable is accompanied by change in the dependent variable that is proportional to the value it already had be expressed in the form produces a curved line when graphed, which quantum mechanics major theory of modern physics which describes the properties and nature of atoms and their subatomic particles science the study or knowledge of how the physical world operates, based on objective evidence determined through observation and", " experimentation scientific law pattern in nature that is true in all fundamental physical units the seven fundamental circumstances studied thus far physical units in the SI system of units are length, mass, time, electric current, temperature, amount of a substance, and luminous intensity hypothesis testable statement that describes how something in the natural world works independent variable the horizontal, or x-axis, variable, which is not influence by the second variable on the graph, the dependent variable inverse proportionality a relation between two variables where k expressible by an equation of the form stays constant when xand ychange; the special form of inverse relationship that satisfies this equation inverse relationship any relation between variables where one variable decreases as the other variable increases kilogram the SI unit for mass, abbreviated (kg) linear relationships relation between variables that produce a straight line when graphed log-log plot a plot that uses a logarithmic scale in both axes logarithmic scale a graphing scale in which each tick on an axis is the previous tick multiplied by some value the SI unit for length, abbreviated (m) meter method of adding percents calculating the percent uncertainty of a quantity in multiplication or division by adding the percent uncertainties in the quantities being added or divided model system that is analogous to the real system of interest in essential ways but more easily analyzed modern physics physics as developed from the twentieth Access for free at openstax.org. scientific methods techniques and processes used in the constructing and testing of scientific hypotheses, laws, and theories, and in deciding issues on the basis of experiment and observation scientific notation way of writing numbers that are too large or small to be conveniently written in simple decimal form; the measurement is multiplied by a power of 10, which indicates the number of placeholder zeros in the measurement SI units second the SI unit for time, abbreviated (s) semi-log plot A plot that uses a logarithmic scale on one axis of the graph and a linear scale on the other axis. International System of Units (SI); the international system of units that scientists in most countries have agreed to use; includes units such as meters, liters, and grams; also known as the metric system significant figures when writing a number, the digits, or number of digits, that express the precision of a measuring tool used to measure the number slope the ratio of the change of a graph on the yaxis to the change along the x-axis, the value of min the equation of a line, theory explanation of patterns in nature that is supported by much scientific evidence and verified multiple times by various groups of researchers theory of relativity theory constructed by Albert Einstein which describes how space, time and energy are different for different observers in relative motion uncertainty a quantitative measure of how much measured values deviate from a standard or expected value universal applies throughout the known universe y-intercept the point where a plot line intersects the y-axis Chapter 1 \u2022 Section Summary 41 SECTION SUMMARY 1.1 Physics: Definitions and Applications \u2022 Physics is the most fundamental of the sciences, concerning itself with energy, matter, space and time, and their interactions. \u2022 Modern physics involves the theory of relativity, which describes how time, space and gravity are not constant in our universe can be different for different observers, and quantum mechanics, which describes the behavior of subatomic particles. \u2022 Physics is the basis for all other sciences, such as chemistry, biology and geology, because physics describes the fundamental way in which the universe functions. 1.2 The Scientific Methods \u2022 Science seeks to discover and describe the underlying order and simplicity in nature. \u2022 The processes of science include observation, hypothesis, experiment, and conclusion. \u2022 Theories are scientific explanations that are supported by a large body experimental results. \u2022 Scientific laws are concise descriptions of the universe that are universally true. 1.3 The Language of Physics: Physical Quantities and Units \u2022 Physical quantities are a characteristic or property of an KEY EQUATIONS 1.3 The Language of Physics: Physical Quantities and Units slope intercept form quadratic formula CHAPTER REVIEW Concept Items 1.1 Physics: Definitions and Applications 1. Which statement best compares and contrasts the aims and topics of natural philosophy had versus physics? object that can be measured or calculated from other measurements. \u2022 The four fundamental units we will use in this textbook are the meter (for length), the kilogram (for mass), the second (for time), and the ampere (for electric current). These units are part of the metric system, which uses powers of 10 to relate quantities over the vast ranges encountered in nature. \u2022 Unit conversions involve changing a value expressed in one type of unit to another type of unit. This is done by using conversion factors, which are ratios relating equal quantities of different units. \u2022 Accuracy of a measured value refers to how close a measurement is to the correct value. The uncertainty in a measurement is an estimate of the amount by which the measurement result may differ from this value. \u2022 Precision of measured values refers to how close the agreement is between repeated measurements. \u2022 Significant figures express the", " precision of a measuring tool. \u2022 When multiplying or dividing measured values, the final answer can contain only as many significant figures as the least precise value. \u2022 When adding or subtracting measured values, the final answer cannot contain more decimal places than the least precise value. positive exponential formula negative exponential formula a. Natural philosophy included all aspects of nature including physics. b. Natural philosophy included all aspects of nature excluding physics. c. Natural philosophy and physics are different. d. Natural philosophy and physics are essentially the 42 Chapter 1 \u2022 Chapter Review same thing. 2. Which of the following is not an underlying assumption essential to scientific understanding? a. Characteristics of the physical universe can be perceived and objectively measured by human beings. b. Explanations of natural phenomena can be established with absolute certainty. c. Fundamental physical processes dictate how characteristics of the physical universe evolve. d. The fundamental processes of nature operate the same way everywhere and at all times. 3. Which of the following questions regarding a strain of genetically modified rice is not one that can be answered by science? a. How does the yield of the genetically modified rice b. compare with that of existing rice? Is the genetically modified rice more resistant to infestation than existing rice? c. How does the nutritional value of the genetically modified rice compare to that of existing rice? d. Should the genetically modified rice be grown commercially and sold in the marketplace? 4. What conditions imply that we can use classical physics without considering special relativity or quantum mechanics? a. 1. matter is moving at speeds of less than roughly 1 percent the speed of light, 2. objects are large enough to be seen with the 3. naked eye, and there is the involvement of a strong gravitational field. b. 1. matter is moving at speeds greater than roughly 1 percent the speed of light, 2. objects are large enough to be seen with the 3. naked eye, and there is the involvement of a strong gravitational field. c. 1. matter is moving at speeds of less than roughly 1 percent the speed of light, 2. objects are too small to be seen with the naked 3. eye, and there is the involvement of only a weak gravitational field. 5. How could physics be useful in weather prediction? a. Physics helps in predicting how burning fossil fuel releases pollutants. b. Physics helps in predicting dynamics and movement of weather phenomena. c. Physics helps in predicting the motion of tectonic plates. d. Physics helps in predicting how the flowing water affects Earth\u2019s surface. 6. How do physical therapists use physics while on the job? Explain. a. Physical therapists do not require knowledge of physics because their job is mainly therapy and not physics. b. Physical therapists do not require knowledge of physics because their job is more social in nature and unscientific. c. Physical therapists require knowledge of physics know about muscle contraction and release of energy. d. Physical therapists require knowledge of physics to know about chemical reactions inside the body and make decisions accordingly. 7. What is meant when a physical law is said to be universal? a. The law can explain everything in the universe. b. The law is applicable to all physical phenomena. c. The law applies everywhere in the universe. d. The law is the most basic one and all laws are derived from it. 8. What subfield of physics could describe small objects traveling at high speeds or experiencing a strong gravitational field? a. general theory of relativity b. classical physics c. quantum relativity d. special theory of relativity 9. Why is Einstein\u2019s theory of relativity considered part of modern physics, as opposed to classical physics? a. Because it was considered less outstanding than the classics of physics, such as classical mechanics. b. Because it was popular physics enjoyed by average people today, instead of physics studied by the elite. c. Because the theory deals with very slow-moving objects and weak gravitational fields. d. Because it was among the new 19th-century d. 1. matter is moving at speeds of less than roughly 1 discoveries that changed physics. percent the speed of light, 2. objects are large enough to be seen with the 1.2 The Scientific Methods 3. naked eye, and there is the involvement of a weak gravitational field. 10. Describe the difference between an observation and a hypothesis. Access for free at openstax.org. Chapter 1 \u2022 Chapter Review 43 a. An observation is seeing what happens; a hypothesis is a testable, educated guess. b. An observation is a hypothesis that has been confirmed. c. Hypotheses and observations are independent of each other. d. Hypotheses are conclusions based on some observations. 11. Describe how modeling is useful in studying the structure of the atom. a. Modeling replaces the real system by something similar but easier to examine. b. Modeling replaces the real system by something more interesting to examine. c. Modeling replaces the real system by something with more realistic properties. a result. d. The dependent and independent variables are fixed by a", " convention and hence they are the same. 15. What could you conclude about these two lines? 1. Line A has a slope of 2. Line B has a slope of a. Line A is a decreasing line while line B is an increasing line, with line A being much steeper than line B. b. Line A is a decreasing line while line B is an increasing line, with line B being much steeper than line A. c. Line B is a decreasing line while line A is an increasing line, with line A being much steeper than line B. d. Modeling includes more details than are present in d. Line B is a decreasing line while line A is an the real system. 12. How strongly is a hypothesis supported by evidence compared to a theory? a. A theory is supported by little evidence, if any, at first, while a hypothesis is supported by a large amount of available evidence. b. A hypothesis is supported by little evidence, if any, at first. A theory is supported by a large amount of available evidence. c. A hypothesis is supported by little evidence, if any, at first. A theory does not need any experiments in support. d. A theory is supported by little evidence, if any, at first. A hypothesis does not need any experiments in support. 1.3 The Language of Physics: Physical Quantities and Units 13. Which of the following does not contribute to the uncertainty? a. b. c. d. other factors that affect the outcome (depending on the limitations of the measuring device the skill of the person making the measurement the regularities in the object being measured the situation) 14. How does the independent variable in a graph differ from the dependent variable? a. The dependent variable varies linearly with the independent variable. b. The dependent variable depends on the scale of the axis chosen while independent variable does not. c. The independent variable is directly manipulated or controlled by the person doing the experiment, while dependent variable is the one that changes as increasing line, with line B being much steeper than line A. 16. Velocity, or speed, is measured using the following where vis velocity, dis the distance formula: travelled, and tis the time the object took to travel the distance. If the velocity-time data are plotted on a graph, which variable will be on which axis? Why? a. Time would be on the x-axis and velocity on the yaxis, because time is an independent variable and velocity is a dependent variable. b. Velocity would be on the x-axis and time on the yaxis, because time is the independent variable and velocity is the dependent variable. c. Time would be on the x-axis and velocity on the yaxis, because time is a dependent variable and velocity is a independent variable. d. Velocity would be on x-axis and time on the y-axis, because time is a dependent variable and velocity is a independent variable. 17. The uncertainty of a triple-beam balance is. What is the percent uncertainty in a measurement of? a. b. c. d. 18. What is the definition of uncertainty? a. Uncertainty is the number of assumptions made prior to the measurement of a physical quantity. b. Uncertainty is a measure of error in a measurement due to the use of a non-calibrated instrument. c. Uncertainty is a measure of deviation of the measured value from the standard value. d. Uncertainty is a measure of error in measurement 44 Chapter 1 \u2022 Chapter Review due to external factors like air friction and temperature. Critical Thinking Items 1.1 Physics: Definitions and Applications 19. In what sense does Einstein\u2019s theory of relativity illustrate that physics describes fundamental aspects of our universe? a. It describes how speed affects different observers\u2019 measurements of time and space. It describes how different parts of the universe are far apart and do not affect each other. It describes how people think of other people\u2019s views from their own frame of reference. It describes how a frame of reference is necessary to describe position or motion. b. c. d. 20. Can classical physics be used to accurately describe a satellite moving at a speed of 7500 m/s? Explain why or why not. a. No, because the satellite is moving at a speed much smaller than the speed of the light and is not in a strong gravitational field. b. No, because the satellite is moving at a speed much smaller than the speed of the light and is in a strong gravitational field. c. Yes, because the satellite is moving at a speed much smaller than the speed of the light and it is not in a strong gravitational field. d. Yes, because the satellite is moving at a speed much smaller than the speed of the light and is in a strong gravitational field. 21. What would be some ways in which physics was involved in building the features of the room you are in right now? a. Physics is involved in structural strength, not age", " very much yourself? a. by traveling at a speed equal to the speed of light b. by traveling at a speed faster than the speed of light c. by traveling at a speed much slower than the speed of light d. by traveling at a speed slightly slower than the speed of light 1.2 The Scientific Methods 24. You notice that the water level flowing in a stream near your house increases when it rains and the water turns brown. Which of these are the best hypothesis to explain why the water turns brown. Assume you have all of the means to test the contents of the stream water. a. The water in the stream turns brown because molecular forces between water molecules are stronger than mud molecules b. The water in the stream turns brown because of the breakage of a weak chemical bond with the hydrogen atom in the water molecule. c. The water in the stream turns brown because it picks up dirt from the bank as the water level increases when it rains. d. The water in the stream turns brown because the density of the water increases with increase in water level. 25. Light travels as waves at an approximate speed of 300,000,000 m/s (186,000 mi/s). Designers of devices that use mirrors and lenses model the traveling light by straight lines, or light rays. Describe why it would be useful to model the light as rays of light instead of describing them accurately as electromagnetic waves. a. A model can be constructed in such a way that the dimensions, etc., of the room. speed of light decreases. b. Physics is involved in the air composition inside the b. Studying a model makes it easier to analyze the room. path that the light follows. c. Physics is involved in the desk arrangement inside c. Studying a model will help us to visualize why light the room. travels at such great speed. d. Physics is involved in the behavior of living beings d. Modeling cannot be used to study traveling light as inside the room. our eyes cannot track the motion of light. 22. What theory of modern physics describes the interrelationships between space, time, speed, and gravity? a. atomic theory b. nuclear physics c. quantum mechanics d. general relativity 23. According to Einstein\u2019s theory of relativity, how could you effectively travel many years into Earth\u2019s future, but 26. A friend says that he doesn\u2019t trust scientific explanations because they are just theories, which are basically educated guesses. What could you say to convince him that scientific theories are different from the everyday use of the word theory? a. A theory is a scientific explanation that has been repeatedly tested and supported by many experiments. b. A theory is a hypothesis that has been tested and Access for free at openstax.org. supported by some experiments. c. A theory is a set of educated guesses, but at least one of the guesses remain true in each experiment. d. A theory is a set of scientific explanations that has at least one experiment in support of it. 27. Give an example of a hypothesis that cannot be tested experimentally. a. The structure of any part of the broccoli is similar to the whole structure of the broccoli. b. Ghosts are the souls of people who have died. c. The average speed of air molecules increases with temperature. d. A vegetarian is less likely to be affected by night blindness. 28. Would it be possible to scientifically prove that a supreme being exists or not? Briefly explain your answer. a. It can be proved scientifically because it is a testable hypothesis. It cannot be proved scientifically because it is not a testable hypothesis. It can be proved scientifically because it is not a testable hypothesis. It cannot be proved scientifically because it is a testable hypothesis. b. c. d. 1.3 The Language of Physics: Physical Quantities and Units 29. A marathon runner completes a course in,, and. There is an uncertainty of in the distance traveled and an uncertainty of elapsed time. in the 1. Calculate the percent uncertainty in the distance. 2. Calculate the uncertainty in the elapsed time. 3. What is the average speed in meters per second? 4. What is the uncertainty in the average speed? a. b.,,,,,, Problems 1.3 The Language of Physics: Physical Quantities and Units 34. A commemorative coin that sells for $40 is advertised to be plated with 15 mg of gold. Suppose gold is worth about $1,300 per ounce. Which of the following best represents the value of the gold in the coin? a. $0.33 b. $0.69 Chapter 1 \u2022 Chapter Review 45 c. d.,,,,,, 30. A car engine moves a piston with a circular cross section of diameter a distance of to compress the gas in the cylinder. By what amount did the gas decrease in volume in cubic centimeters? Find the uncertainty in this volume. a. b. c. d. 31.", " What would be the slope for a line passing through the two points below? Point 1: (1, 0.1) Point 2: (7, 26.8) a. b. c. d. 32. The sides of a small rectangular box are measured and long and high. Calculate its volume and uncertainty in cubic centimeters. Assume the measuring device is accurate to a. b. c. d.. 33. Calculate the approximate number of atoms in a bacterium. Assume that the average mass of an atom in the bacterium is ten times the mass of a hydrogen atom. (Hint\u2014The mass of a hydrogen atom is on the order of 10\u221227 kg and the mass of a bacterium is on the order of 10\u221215 kg.) a. b. c. d. 1010 atoms 1011 atoms 1012 atoms 1013 atoms c. $3.30 d. $6.90 35. If a marathon runner runs in another direction and in one direction, in a third direction, how much distance did the runner run? Be sure to report your answer using the proper number of significant figures. a. b. c. 46 Chapter 1 \u2022 Test Prep d. 37. The length and width of a rectangular room are 36. The speed limit on some interstate highways is roughly. What is this in meters per second? How many, miles per hour is this? a. b. c. d.,,, Performance Task 1.3 The Language of Physics: Physical Quantities and Units 38. a. Create a new system of units to describe something that interests you. Your unit should be described using at least two subunits. For example, you can decide to measure the quality of songs using a new unit called song awesomeness. Song awesomeness TEST PREP Multiple Choice 1.1 Physics: Definitions and Applications 39. Modern physics could best be described as the combination of which theories? a. quantum mechanics and Einstein\u2019s theory of relativity b. quantum mechanics and classical physics c. Newton\u2019s laws of motion and classical physics d. Newton\u2019s laws of motion and Einstein\u2019s theory of relativity 40. Which of the following could be studied accurately using classical physics? a. b. c. d. the strength of gravity within a black hole the motion of a plane through the sky the collisions of subatomic particles the effect of gravity on the passage of time 41. Which of the following best describes why knowledge of physics is necessary to understand all other sciences? a. Physics explains how energy passes from one object to another. b. Physics explains how gravity works. c. Physics explains the motion of objects that can be seen with the naked eye. d. Physics explains the fundamental aspects of the universe. 42. What does radiation therapy, used to treat cancer patients, have to do with physics? a. Understanding how cells reproduce is mainly about Access for free at openstax.org. by measured to be. Calculate the area of the room and its uncertainty in square meters. a. b. c. d. is measured by: the number of songs downloaded and the number of times the song was used in movies. b. Create an equation that shows how to calculate your unit. Then, using your equation, create a sample dataset that you could graph. Are your two subunits related linearly, quadratically, or inversely? physics. b. Predictions of the side effects from the radiation therapy are based on physics. c. The devices used for generating some kinds of radiation are based on principles of physics. d. Predictions of the life expectancy of patients receiving radiation therapy are based on physics. 1.2 The Scientific Methods 43. The free-electron model of metals explains some of the important behaviors of metals by assuming the metal\u2019s electrons move freely through the metal without repelling one another. In what sense is the free-electron theory based on a model? a. Its use requires constructing replicas of the metal wire in the lab. It involves analyzing an imaginary system simpler than the real wire it resembles. It examines a model, or ideal, behavior that other metals should imitate. It attempts to examine the metal in a very realistic, or model, way. b. c. d. 44. A scientist wishes to study the motion of about 1,000 molecules of gas in a container by modeling them as tiny billiard balls bouncing randomly off one another. Which of the following is needed to calculate and store data on their detailed motion? a. a group of hypotheses that cannot be practically tested in real life b. a computer that can store and perform calculations on large data sets c. a large amount of experimental results on the molecules and their motion d. a collection of hypotheses that have not yet been tested regarding the molecules 45. When a large body of experimental evidence supports a hypothesis, what may the hypothesis eventually be considered? a. observation insight b. conclusion c. law d. 46. While watching", " some ants outside of your house, you notice that the worker ants gather in a specific area on your lawn. Which of the following is a testable hypothesis that attempts to explain why the ants gather in that specific area on the lawn. a. The worker thought it was a nice location. b. because ants may have to find a spot for the queen to lay eggs c. because there may be some food particles lying there d. because the worker ants are supposed to group together at a place. 1.3 The Language of Physics: Physical Quantities and Units 47. Which of the following would describe a length that is of a meter? a. b. kilometers megameters Short Answer 1.1 Physics: Definitions and Applications 51. Describe the aims of physics. a. Physics aims to explain the fundamental aspects of our universe and how these aspects interact with one another. b. Physics aims to explain the biological aspects of our universe and how these aspects interact with one another. c. Physics aims to explain the composition, structure and changes in matter occurring in the universe. d. Physics aims to explain the social behavior of living beings in the universe. 52. Define the fields of magnetism and electricity and state how are they are related. a. Magnetism describes the attractive force between a Chapter 1 \u2022 Test Prep 47 c. d. millimeters micrometers 48. Suppose that a bathroom scale reads a person\u2019s mass as 65 kg with a 3 percent uncertainty. What is the uncertainty in their mass in kilograms? a. a. 2 kg b. b. 98 kg c. d. d. 0 c. 5 kg 49. Which of the following best describes a variable? a. a trend that shows an exponential relationship b. something whose value can change over multiple measurements c. a measure of how much a plot line changes along d. the y-axis something that remains constant over multiple measurements 50. A high school track coach has just purchased a new stopwatch that has an uncertainty of \u00b10.05 s. Runners on the team regularly clock 100-m sprints in 12.49 s to 15.01 s. At the school\u2019s last track meet, the first-place sprinter came in at 12.04 s and the second-place sprinter came in at 12.07 s. Will the coach\u2019s new stopwatch be helpful in timing the sprint team? Why or why not? a. No, the uncertainty in the stopwatch is too large to effectively differentiate between the sprint times. b. No, the uncertainty in the stopwatch is too small to effectively differentiate between the sprint times. c. Yes, the uncertainty in the stopwatch is too large to effectively differentiate between the sprint times. d. Yes, the uncertainty in the stopwatch is too small to effectively differentiate between the sprint times. magnetized object and a metal like iron. Electricity involves the study of electric charges and their movements. Magnetism is not related to the electricity. b. Magnetism describes the attractive force between a magnetized object and a metal like iron. Electricity involves the study of electric charges and their movements. Magnetism is produced by a flow electrical charges. c. Magnetism involves the study of electric charges and their movements. Electricity describes the attractive force between a magnetized object and a metal. Magnetism is not related to the electricity. d. Magnetism involves the study of electric charges and their movements. Electricity describes the attractive force between a magnetized object and a metal. Magnetism is produced by the flow electrical charges. 48 Chapter 1 \u2022 Test Prep 53. Describe what two topics physicists are trying to unify with relativistic quantum mechanics. How will this unification create a greater understanding of our universe? a. Relativistic quantum mechanics unifies quantum mechanics with Einstein\u2019s theory of relativity. The unified theory creates a greater understanding of our universe because it can explain objects of all sizes and masses. b. Relativistic quantum mechanics unifies classical mechanics with Einstein\u2019s theory of relativity. The unified theory creates a greater understanding of our universe because it can explain objects of all sizes and masses. c. Relativistic quantum mechanics unifies quantum mechanics with Einstein\u2019s theory of relativity. The unified theory creates a greater understanding of our universe because it is unable to explain objects of all sizes and masses. d. Relativistic quantum mechanics unifies classical mechanics with the Einstein\u2019s theory of relativity. The unified theory creates a greater understanding of our universe because it is unable to explain objects of all sizes and masses. a. An understanding of force, pressure, heat, electricity, etc., which all involve physics, will help me design a sound and energy-efficient house. b. An understanding of the air composition, chemical composition of matter, etc., which all involves physics, will help me design a sound and energyefficient house. c. An understanding of material cost and economic factors involving physics will help me design a sound and energy-", "efficient house. d. An understanding of geographical location and social environment which involves physics will help me design a sound and energy-efficient house. 57. What aspects of physics would a chemist likely study in trying to discover a new chemical reaction? a. Physics is involved in understanding whether the reactants and products dissolve in water. b. Physics is involved in understanding the amount of energy released or required in a chemical reaction. c. Physics is involved in what the products of the reaction will be. d. Physics is involved in understanding the types of ions produced in a chemical reaction. 54. The findings of studies in quantum mechanics have 1.2 The Scientific Methods been described as strange or weird compared to those of classical physics. Explain why this would be so. a. It is because the phenomena it explains are outside the normal range of human experience which deals with much larger objects. It is because the phenomena it explains can be perceived easily, namely, ordinary-sized objects. It is because the phenomena it explains are outside the normal range of human experience, namely, the very large and the very fast objects. It is because the phenomena it explains can be perceived easily, namely, the very large and the very fast objects. b. c. d. 55. How could knowledge of physics help you find a faster way to drive from your house to your school? a. Physics can explain the traffic on a particular street and help us know about the traffic in advance. b. Physics can explain about the ongoing construction of roads on a particular street and help us know about delays in the traffic in advance. c. Physics can explain distances, speed limits on a particular street and help us categorize faster routes. d. Physics can explain the closing of a particular street and help us categorize faster routes. 56. How could knowledge of physics help you build a sound and energy-efficient house? Access for free at openstax.org. 58. You notice that it takes more force to get a large box to start sliding across the floor than it takes to get the box sliding faster once it is already moving. Create a testable hypothesis that attempts to explain this observation. a. The floor has greater distortions of space-time for moving the sliding box faster than for the box at rest. b. The floor has greater distortions of space-time for the box at rest than for the sliding box. c. The resistance between the floor and the box is less when the box is sliding then when the box is at rest. d. The floor dislikes having objects move across it and therefore holds the box rigidly in place until it cannot resist the force. 59. Design an experiment that will test the following hypothesis: driving on a gravel road causes greater damage to a car than driving on a dirt road. a. To test the hypothesis, compare the damage to the car by driving it on a smooth road and a gravel road. b. To test the hypothesis, compare the damage to the car by driving it on a smooth road and a dirt road. c. To test the hypothesis, compare the damage to the car by driving it on a gravel road and the dirt road. d. This is not a testable hypothesis. 60. How is a physical model, such as a spherical mass held in place by springs, used to represent an atom vibrating in a solid, similar to a computer-based model, such as that predicting how gravity affects the orbits of the planets? a. Both a physical model and a computer-based model should be built around a hypothesis and could be able to test the hypothesis. b. Both a physical model and a computer-based model should be built around a hypothesis but they cannot be used to test the hypothesis. c. Both a physical model and a computer-based model should be built around the results of scientific studies and could be used to make predictions about the system under study. d. Both a physical model and a computer-based model should be built around the results of scientific studies but cannot be used to make predictions about the system under study. 61. Explain the advantages and disadvantages of using a model to predict a life-or-death situation, such as whether or not an asteroid will strike Earth. a. The advantage of using a model is that it provides predictions quickly, but the disadvantage of using a model is that it could make erroneous predictions. b. The advantage of using a model is that it provides accurate predictions, but the disadvantage of using a model is that it takes a long time to make predictions. c. The advantage of using a model is that it provides predictions quickly without any error. There are no disadvantages of using a scientific model. d. The disadvantage of using models is that it takes longer time to make predictions and the predictions are inaccurate. There are no advantages to using a scientific model. 62. A friend tells you that a scientific law cannot be changed. State whether or not your friend is correct and then briefly explain your answer. a. Correct, because laws are theories that have been proved true. b.", " Correct, because theories are laws that have been c. d. proved true. Incorrect, because a law is changed if new evidence contradicts it. Incorrect, because a law is changed when a theory contradicts it. 63. How does a scientific law compare to a local law, such as that governing parking at your school, in terms of whether or not laws can be changed, and how universal a law is? a. A local law applies only in a specific area, but a Chapter 1 \u2022 Test Prep 49 scientific law is applicable throughout the universe. Both the local law and the scientific law can change. b. A local law applies only in a specific area, but a scientific law is applicable throughout the universe. A local law can change, but a scientific law cannot be changed. c. A local law applies throughout the universe but a scientific law is applicable only in a specific area. Both the local and the scientific law can change. d. A local law applies throughout the universe, but a scientific law is applicable only in a specific area. A local law can change, but a scientific law cannot be changed. 64. Can the validity of a model be limited, or must it be universally valid? How does this compare to the required validity of a theory or a law? a. Models, theories and laws must be universally valid. b. Models, theories, and laws have only limited validity. c. Models have limited validity while theories and laws are universally valid. d. Models and theories have limited validity while laws are universally valid. 1.3 The Language of Physics: Physical Quantities and Units 65. The speed of sound is measured at on a certain? Report your answer in day. What is this in scientific notation. a. b. c. d. 66. Describe the main difference between the metric system and the U.S. Customary System. a. In the metric system, unit changes are based on powers of 10, while in the U.S. customary system, each unit conversion has unrelated conversion factors. In the metric system, each unit conversion has unrelated conversion factors, while in the U.S. customary system, unit changes are based on powers of 10. In the metric system, unit changes are based on powers of 2, while in the U.S. customary system, each unit conversion has unrelated conversion factors. In the metric system, each unit conversion has unrelated conversion factors, while in the U.S. customary system, unit changes are based on b. c. d. 50 Chapter 1 \u2022 Test Prep powers of 2. 67. An infant\u2019s pulse rate is measured to be. What is the percent uncertainty in this measurement? a. b. c. d. 68. Explain how the uncertainty of a measurement relates to the accuracy and precision of the measuring device. Include the definitions of accuracy and precision in your answer. a. A decrease in the precision of a measurement increases the uncertainty of the measurement, while a decrease in accuracy does not. b. A decrease in either the precision or accuracy of a measurement increases the uncertainty of the measurement. c. An increase in either the precision or accuracy of a measurement will increase the uncertainty of that measurement. d. An increase in the accuracy of a measurement will increase the uncertainty of that measurement, while an increase in precision will not. 69. Describe all of the characteristics that can be and determined about a straight line with a slope of a y-intercept of on a graph. a. Based on the information, the line has a negative slope. Because its y-intercept is 50 and its slope is negative, this line gradually rises on the graph as the x-value increases. b. Based on the information, the line has a negative slope. Because its y-intercept is 50 and its slope is negative, this line gradually moves downward on Extended Response 1.2 The Scientific Methods 71. You wish to perform an experiment on the stopping distance of your new car. Create a specific experiment to measure the distance. Be sure to specifically state how you will set up and take data during your experiment. a. Drive the car at exactly 50 mph and then press harder on the accelerator pedal until the velocity reaches the speed 60 mph and record the distance this takes. b. Drive the car at exactly 50 mph and then apply the brakes until it stops and record the distance this takes. c. Drive the car at exactly 50 mph and then apply the brakes until it stops and record the time it takes. Access for free at openstax.org. the graph as the x-value increases. c. Based on the information, the line has a positive slope. Because its y-intercept is 50 and its slope is positive, this line gradually rises on the graph as the x-value increases. d. Based on the information, the line has a positive slope. Because its y-intercept is 50 and its slope is positive, this line gradually moves downward on the graph as the x-value increases. 70. The graph shows", " the temperature change over time of a heated cup of water. What is the slope of the graph between the time period 2 min and 5 min? a. \u201315 \u00baC/min b. \u20130.07 \u00baC/min c. 0.07 \u00baC/min 15 \u00baC/min d. d. Drive the car at exactly 50 mph and then apply the accelerator until it reaches the speed of 60 mph and record the time it takes. 72. You wish to make a model showing how traffic flows around your city or local area. Describe the steps you would take to construct your model as well as some hypotheses that your model could test and the model\u2019s limitations in terms of what could not be tested. a. 1. Testable hypotheses like the gravitational pull on each vehicle while in motion and the average speed of vehicles is 40 mph 2. Non-testable hypotheses like the average number of vehicles passing is 935 per day and carbon emission from each of the moving vehicle b. 1. Testable hypotheses like the average number of vehicles passing is 935 per day and the average speed of vehicles is 40 mph 2. Non-testable hypotheses like the gravitational pull on each vehicle while in motion and the carbon emission from each of the moving vehicle c. 1. Testable hypotheses like the average number of vehicles passing is 935 per day and the carbon emission from each of the moving vehicle 2. Non-testable hypotheses like the gravitational pull on each vehicle while in motion and the average speed of the vehicles is 40 mph d. 1. Testable hypotheses like the average number of vehicles passing is 935 per day and the gravitational pull on each vehicle while in motion 2. Non-testable hypotheses like the average speed of vehicles is 40 mph and the carbon emission from each of the moving vehicle 73. What would play the most important role in leading to an experiment in the scientific world becoming a scientific law? a. Further testing would need to show it is a universally followed rule. b. The observation would have to be described in a Chapter 1 \u2022 Test Prep 51 published scientific article. c. The experiment would have to be repeated once or twice. d. The observer would need to be a well-known scientist whose authority was accepted. 1.3 The Language of Physics: Physical Quantities and Units at this speed? What is its speed in 74. Tectonic plates are large segments of the Earth\u2019s crust that move slowly. Suppose that one such plate has an average speed of. What distance does it move in kilometers per million years? Report all of your answers using scientific notation. a. b. c. d. 75. At x = 3, a function f(x) has a positive value, with a positive slope that is decreasing in magnitude with increasing x. Which option could correspond to f(x)? a. b. c. d. 52 Chapter 1 \u2022 Test Prep Access for free at openstax.org. CHAPTER 2 Motion in One Dimension Figure 2.1 Shanghai Maglev. At this rate, a train traveling from Boston to Washington, DC, a distance of 439 miles, could make the trip in under an hour and a half. Presently, the fastest train on this route takes over six hours to cover this distance. (Alex Needham, Public Domain) Chapter Outline 2.1 Relative Motion, Distance, and Displacement 2.2 Speed and Velocity 2.3 Position vs. Time Graphs 2.4 Velocity vs. Time Graphs Unless you have flown in an airplane, you have probably never traveled faster than 150 mph. Can you imagine INTRODUCTION traveling in a train like the one shown in Figure 2.1 that goes over 300 mph? Despite the high speed, the people riding in this train may not notice that they are moving at all unless they look out the window! This is because motion, even motion at 300 mph, is relative to the observer. In this chapter, you will learn why it is important to identify a reference frame in order to clearly describe motion. For now, the motion you describe will be one-dimensional. Within this context, you will learn the difference between distance and displacement as well as the difference between speed and velocity. Then you will look at some graphing and problem-solving techniques. 54 Chapter 2 \u2022 Motion in One Dimension 2.1 Relative Motion, Distance, and Displacement Section Learning Objectives By the end of this section, you will be able to do the following: \u2022 Describe motion in different reference frames \u2022 Define distance and displacement, and distinguish between the two \u2022 Solve problems involving distance and displacement Section Key Terms displacement distance kinematics magnitude position reference frame scalar vector Defining Motion Our study of physics opens with kinematics\u2014the study of motion without considering its causes. Objects are in motion everywhere you look. Everything from a tennis game to a space-probe flyby of the planet Neptune involves motion. When you are resting, your heart moves", " blood through your veins. Even in inanimate objects, atoms are always moving. How do you know something is moving? The location of an object at any particular time is its position. More precisely, you need to specify its position relative to a convenient reference frame. Earth is often used as a reference frame, and we often describe the position of an object as it relates to stationary objects in that reference frame. For example, a rocket launch would be described in terms of the position of the rocket with respect to Earth as a whole, while a professor\u2019s position could be described in terms of where she is in relation to the nearby white board. In other cases, we use reference frames that are not stationary but are in motion relative to Earth. To describe the position of a person in an airplane, for example, we use the airplane, not Earth, as the reference frame. (See Figure 2.2.) Thus, you can only know how fast and in what direction an object's position is changing against a background of something else that is either not moving or moving with a known speed and direction. The reference frame is the coordinate system from which the positions of objects are described. Figure 2.2 Are clouds a useful reference frame for airplane passengers? Why or why not? (Paul Brennan, Public Domain) Your classroom can be used as a reference frame. In the classroom, the walls are not moving. Your motion as you walk to the door, can be measured against the stationary background of the classroom walls. You can also tell if other things in the classroom are moving, such as your classmates entering the classroom or a book falling off a desk. You can also tell in what direction something is moving in the classroom. You might say, \u201cThe teacher is moving toward the door.\u201d Your reference frame allows you to determine not only that something is moving but also the direction of motion. You could also serve as a reference frame for others\u2019 movement. If you remained seated as your classmates left the room, you would measure their movement away from your stationary location. If you and your classmates left the room together, then your perspective of their motion would be change. You, as the reference frame, would be moving in the same direction as your other moving classmates. As you will learn in the Snap Lab, your description of motion can be quite different when viewed from different reference frames. Access for free at openstax.org. 2.1 \u2022 Relative Motion, Distance, and Displacement 55 Snap Lab Looking at Motion from Two Reference Frames In this activity you will look at motion from two reference frames. Which reference frame is correct? \u2022 Choose an open location with lots of space to spread out so there is less chance of tripping or falling due to a collision and/or loose basketballs. \u2022 1 basketball Procedure 1. Work with a partner. Stand a couple of meters away from your partner. Have your partner turn to the side so that you are looking at your partner\u2019s profile. Have your partner begin bouncing the basketball while standing in place. Describe the motion of the ball. 2. Next, have your partner again bounce the ball, but this time your partner should walk forward with the bouncing ball. You will remain stationary. Describe the ball's motion. 3. Again have your partner walk forward with the bouncing ball. This time, you should move alongside your partner while continuing to view your partner\u2019s profile. Describe the ball's motion. 4. Switch places with your partner, and repeat Steps 1\u20133. GRASP CHECK How do the different reference frames affect how you describe the motion of the ball? a. The motion of the ball is independent of the reference frame and is same for different reference frames. b. The motion of the ball is independent of the reference frame and is different for different reference frames. c. The motion of the ball is dependent on the reference frame and is same for different reference frames. d. The motion of the ball is dependent on the reference frames and is different for different reference frames. LINKS TO PHYSICS History: Galileo's Ship Figure 2.3 Galileo Galilei (1564\u20131642) studied motion and developed the concept of a reference frame. (Domenico Tintoretto) The idea that a description of motion depends on the reference frame of the observer has been known for hundreds of years. The 17th-century astronomer Galileo Galilei (Figure 2.3) was one of the first scientists to explore this idea. Galileo suggested the following thought experiment: Imagine a windowless ship moving at a constant speed and direction along a perfectly calm sea. Is there a way that a person inside the ship can determine whether the ship is moving? You can extend this thought experiment 56 Chapter 2 \u2022 Motion in One Dimension by also imagining a person standing on the shore. How can a person on the shore determine whether the ship is moving? Galileo came to an amazing conclusion. Only by looking at each other can a person in the ship or", " a person on shore describe the motion of one relative to the other. In addition, their descriptions of motion would be identical. A person inside the ship would describe the person on the land as moving past the ship. The person on shore would describe the ship and the person inside it as moving past. Galileo realized that observers moving at a constant speed and direction relative to each other describe motion in the same way. Galileo had discovered that a description of motion is only meaningful if you specify a reference frame. GRASP CHECK Imagine standing on a platform watching a train pass by. According to Galileo\u2019s conclusions, how would your description of motion and the description of motion by a person riding on the train compare? a. b. c. d. I would see the train as moving past me, and a person on the train would see me as stationary. I would see the train as moving past me, and a person on the train would see me as moving past the train. I would see the train as stationary, and a person on the train would see me as moving past the train. I would see the train as stationary, and a person on the train would also see me as stationary. Distance vs. Displacement As we study the motion of objects, we must first be able to describe the object\u2019s position. Before your parent drives you to school, the car is sitting in your driveway. Your driveway is the starting position for the car. When you reach your high school, the car has changed position. Its new position is your school. Figure 2.4 Your total change in position is measured from your house to your school. Physicists use variables to represent terms. We will use d to represent car\u2019s position. We will use a subscript to differentiate between the initial position, d0, and the final position, df. In addition, vectors, which we will discuss later, will be in bold or will have an arrow above the variable. Scalars will be italicized. TIPS FOR SUCCESS In some books, x or s is used instead of d to describe position. In d0, said d naught, the subscript 0 stands for initial. When we begin to talk about two-dimensional motion, sometimes other subscripts will be used to describe horizontal position, dx, or vertical position, dy. So, you might see references to d0x and dfy. Now imagine driving from your house to a friend's house located several kilometers away. How far would you drive? The distance an object moves is the length of the path between its initial position and its final position. The distance you drive to your friend's house depends on your path. As shown in Figure 2.5, distance is different from the length of a straight line between two points. The distance you drive to your friend's house is probably longer than the straight line between the two houses. Access for free at openstax.org. 2.1 \u2022 Relative Motion, Distance, and Displacement 57 Figure 2.5 A short line separates the starting and ending points of this motion, but the distance along the path of motion is considerably longer. We often want to be more precise when we talk about position. The description of an object\u2019s motion often includes more than just the distance it moves. For instance, if it is a five kilometer drive to school, the distance traveled is 5 kilometers. After dropping you off at school and driving back home, your parent will have traveled a total distance of 10 kilometers. The car and your parent will end up in the same starting position in space. The net change in position of an object is its displacement, or The Greek letter delta,, means change in. Figure 2.6 The total distance that your car travels is 10 km, but the total displacement is 0. Snap Lab Distance vs. Displacement In this activity you will compare distance and displacement. Which term is more useful when making measurements? 1 recorded song available on a portable device 1 tape measure 3 pieces of masking tape \u2022 \u2022 \u2022 \u2022 A room (like a gym) with a wall that is large and clear enough for all pairs of students to walk back and forth without running into each other. Procedure 1. One student from each pair should stand with their back to the longest wall in the classroom. Students should stand at least 0.5 meters away from each other. Mark this starting point with a piece of masking tape. 2. The second student from each pair should stand facing their partner, about two to three meters away. Mark this point 58 Chapter 2 \u2022 Motion in One Dimension with a second piece of masking tape. 3. Student pairs line up at the starting point along the wall. 4. The teacher turns on the music. Each pair walks back and forth from the wall to the second marked point until the music stops playing. Keep count of the number of times you walk across the floor. 5. When the music stops, mark your ending position with the third piece of mask", "ing tape. 6. Measure from your starting, initial position to your ending, final position. 7. Measure the length of your path from the starting position to the second marked position. Multiply this measurement by the total number of times you walked across the floor. Then add this number to your measurement from step 6. 8. Compare the two measurements from steps 6 and 7. GRASP CHECK 1. Which measurement is your total distance traveled? 2. Which measurement is your displacement? 3. When might you want to use one over the other? a. Measurement of the total length of your path from the starting position to the final position gives the distance traveled, and the measurement from your initial position to your final position is the displacement. Use distance to describe the total path between starting and ending points,and use displacement to describe the shortest path between starting and ending points. b. Measurement of the total length of your path from the starting position to the final position is distance traveled, and the measurement from your initial position to your final position is displacement. Use distance to describe the shortest path between starting and ending points, and use displacement to describe the total path between starting and ending points. c. Measurement from your initial position to your final position is distance traveled, and the measurement of the total length of your path from the starting position to the final position is displacement. Use distance to describe the total path between starting and ending points, and use displacement to describe the shortest path between starting and ending points. d. Measurement from your initial position to your final position is distance traveled, and the measurement of the total length of your path from the starting position to the final position is displacement. Use distance to describe the shortest path between starting and ending points, and use displacement to describe the total path between starting and ending points. If you are describing only your drive to school, then the distance traveled and the displacement are the same\u20145 kilometers. When you are describing the entire round trip, distance and displacement are different. When you describe distance, you only include the magnitude, the size or amount, of the distance traveled. However, when you describe the displacement, you take into account both the magnitude of the change in position and the direction of movement. In our previous example, the car travels a total of 10 kilometers, but it drives five of those kilometers forward toward school and five of those kilometers back in the opposite direction. If we ascribe the forward direction a positive (+) and the opposite direction a negative (\u2013), then the two quantities will cancel each other out when added together. A quantity, such as distance, that has magnitude (i.e., how big or how much) but does not take into account direction is called a scalar. A quantity, such as displacement, that has both magnitude and direction is called a vector. WATCH PHYSICS Vectors & Scalars This video (http://openstax.org/l/28vectorscalar) introduces and differentiates between vectors and scalars. It also introduces quantities that we will be working with during the study of kinematics. Click to view content (https://www.khanacademy.org/embed_video?v=ihNZlp7iUHE) Access for free at openstax.org. GRASP CHECK 2.1 \u2022 Relative Motion, Distance, and Displacement 59 How does this video (https://www.khanacademy.org/science/ap-physics-1/ap-one-dimensional-motion/ap-physicsfoundations/v/introduction-to-vectors-and-scalars) help you understand the difference between distance and displacement? Describe the differences between vectors and scalars using physical quantities as examples. a. It explains that distance is a vector and direction is important, whereas displacement is a scalar and it has no direction attached to it. It explains that distance is a scalar and direction is important, whereas displacement is a vector and it has no direction attached to it. It explains that distance is a scalar and it has no direction attached to it, whereas displacement is a vector and direction is important. It explains that both distance and displacement are scalar and no directions are attached to them. b. c. d. Displacement Problems Hopefully you now understand the conceptual difference between distance and displacement. Understanding concepts is half the battle in physics. The other half is math. A stumbling block to new physics students is trying to wade through the math of physics while also trying to understand the associated concepts. This struggle may lead to misconceptions and answers that make no sense. Once the concept is mastered, the math is far less confusing. So let\u2019s review and see if we can make sense of displacement in terms of numbers and equations. You can calculate an object's displacement by subtracting its original position, d0, from its final position df. In math terms that means If the final position is the", " same as the initial position, then. To assign numbers and/or direction to these quantities, we need to define an axis with a positive and a negative direction. We also need to define an origin, or O. In Figure 2.6, the axis is in a straight line with home at zero and school in the positive direction. If we left home and drove the opposite way from school, motion would have been in the negative direction. We would have assigned it a negative value. In the round-trip drive, df and d0 were both at zero kilometers. In the one way trip to school, df was at 5 kilometers and d0 was at zero km. So, was 5 kilometers. TIPS FOR SUCCESS You may place your origin wherever you would like. You have to make sure that you calculate all distances consistently from your zero and you define one direction as positive and the other as negative. Therefore, it makes sense to choose the easiest axis, direction, and zero. In the example above, we took home to be zero because it allowed us to avoid having to interpret a solution with a negative sign. WORKED EXAMPLE Calculating Distance and Displacement A cyclist rides 3 km west and then turns around and rides 2 km east. (a) What is her displacement? (b) What distance does she ride? (c) What is the magnitude of her displacement? 60 Chapter 2 \u2022 Motion in One Dimension Strategy To solve this problem, we need to find the difference between the final position and the initial position while taking care to note the direction on the axis. The final position is the sum of the two displacements, and. Solution a. Displacement: The rider\u2019s displacement is b. Distance: The distance traveled is 3 km + 2 km = 5 km. c. The magnitude of the displacement is 1 km.. Discussion The displacement is negative because we chose east to be positive and west to be negative. We could also have described the displacement as 1 km west. When calculating displacement, the direction mattered, but when calculating distance, the direction did not matter. The problem would work the same way if the problem were in the north\u2013south or y-direction. TIPS FOR SUCCESS Physicists like to use standard units so it is easier to compare notes. The standard units for calculations are called SIunits (International System of Units). SI units are based on the metric system. The SI unit for displacement is the meter (m), but sometimes you will see a problem with kilometers, miles, feet, or other units of length. If one unit in a problem is an SI unit and another is not, you will need to convert all of your quantities to the same system before you can carry out the calculation. Practice Problems 1. On an axis in which moving from right to left is positive, what is the displacement and distance of a student who walks 32 m to the right and then 17 m to the left? a. Displacement is -15 m and distance is -49 m. b. Displacement is -15 m and distance is 49 m. c. Displacement is 15 m and distance is -49 m. d. Displacement is 15 m and distance is 49 m. 2. Tiana jogs 1.5 km along a straight path and then turns and jogs 2.4 km in the opposite direction. She then turns back and jogs 0.7 km in the original direction. Let Tiana\u2019s original direction be the positive direction. What are the displacement and distance she jogged? a. Displacement is 4.6 km,and distance is -0.2 km. b. Displacement is -0.2 km, and distance is 4.6 km. c. Displacement is 4.6 km, and distance is +0.2 km. d. Displacement is +0.2 km, and distance is 4.6 km. WORK IN PHYSICS Mars Probe Explosion Figure 2.7 The Mars Climate Orbiter disaster illustrates the importance of using the correct calculations in physics. (NASA) Access for free at openstax.org. 2.1 \u2022 Relative Motion, Distance, and Displacement 61 Physicists make calculations all the time, but they do not always get the right answers. In 1998, NASA, the National Aeronautics and Space Administration, launched the Mars Climate Orbiter, shown in Figure 2.7, a $125-million-dollar satellite designed to monitor the Martian atmosphere. It was supposed to orbit the planet and take readings from a safe distance. The American scientists made calculations in English units (feet, inches, pounds, etc.) and forgot to convert their answers to the standard metric SI units. This was a very costly mistake. Instead of orbiting the planet as planned, the Mars Climate Orbiter ended up flying into the Martian atmosphere. The probe disintegrated. It was one of the biggest embarrassments in NASA\u2019s history", ". GRASP CHECK In 1999 the Mars Climate Orbiter crashed because calculation were performed in English units instead of SI units. At one point the orbiter was just 187,000 feet above the surface, which was too close to stay in orbit. What was the height of the orbiter at this time in kilometers? (Assume 1 meter equals 3.281 feet.) 16 km a. 18 km b. 57 km c. d. 614 km Check Your Understanding 3. What does it mean when motion is described as relative? a. b. c. d. It means that motion of any object is described relative to the motion of Earth. It means that motion of any object is described relative to the motion of any other object. It means that motion is independent of the frame of reference. It means that motion depends on the frame of reference selected. 4. If you and a friend are standing side-by-side watching a soccer game, would you both view the motion from the same reference frame? a. Yes, we would both view the motion from the same reference point because both of us are at rest in Earth\u2019s frame of reference. b. Yes, we would both view the motion from the same reference point because both of us are observing the motion from two points on the same straight line. c. No, we would both view the motion from different reference points because motion is viewed from two different points; the reference frames are similar but not the same. d. No, we would both view the motion from different reference points because response times may be different; so, the motion observed by both of us would be different. 5. What is the difference between distance and displacement? a. Distance has both magnitude and direction, while displacement has magnitude but no direction. b. Distance has magnitude but no direction, while displacement has both magnitude and direction. c. Distance has magnitude but no direction, while displacement has only direction. d. There is no difference. Both distance and displacement have magnitude and direction. 6. Which situation correctly identifies a race car\u2019s distance traveled and the magnitude of displacement during a one-lap car race? a. The perimeter of the race track is the distance, and the shortest distance between the start line and the finish line is the magnitude of displacement. b. The perimeter of the race track is the magnitude of displacement, and the shortest distance between the start and finish line is the distance. c. The perimeter of the race track is both the distance and magnitude of displacement. d. The shortest distance between the start line and the finish line is both the distance and magnitude of displacement. 7. Why is it important to specify a reference frame when describing motion? a. Because Earth is continuously in motion; an object at rest on Earth will be in motion when viewed from outer space. b. Because the position of a moving object can be defined only when there is a fixed reference frame. 62 Chapter 2 \u2022 Motion in One Dimension c. Because motion is a relative term; it appears differently when viewed from different reference frames. d. Because motion is always described in Earth\u2019s frame of reference; if another frame is used, it has to be specified with each situation. 2.2 Speed and Velocity Section Learning Objectives By the end of this section, you will be able to do the following: \u2022 Calculate the average speed of an object \u2022 Relate displacement and average velocity Section Key Terms average speed average velocity instantaneous speed instantaneous velocity speed velocity Speed There is more to motion than distance and displacement. Questions such as, \u201cHow long does a foot race take?\u201d and \u201cWhat was the runner\u2019s speed?\u201d cannot be answered without an understanding of other concepts. In this section we will look at time, speed, and velocity to expand our understanding of motion. A description of how fast or slow an object moves is its speed. Speed is the rate at which an object changes its location. Like distance, speed is a scalar because it has a magnitude but not a direction. Because speed is a rate, it depends on the time interval of motion. You can calculate the elapsed time or the change in time, and the beginning time, of motion as the difference between the ending time The SI unit of time is the second (s), and the SI unit of speed is meters per second (m/s), but sometimes kilometers per hour (km/h), miles per hour (mph) or other units of speed are used. When you describe an object's speed, you often describe the average over a time period. Average speed, vavg, is the distance traveled divided by the time during which the motion occurs. You can, of course, rearrange the equation to solve for either distance or time Suppose, for example, a car travels 150 kilometers in 3.2 hours. Its average speed for the trip is A car's speed would likely increase and decrease many times over a 3.2 hour trip. Its speed at a specific instant in time,", " however, is its instantaneous speed. A car's speedometer describes its instantaneous speed. Access for free at openstax.org. 2.2 \u2022 Speed and Velocity 63 Figure 2.8 During a 30-minute round trip to the store, the total distance traveled is 6 km. The average speed is 12 km/h. The displacement for the round trip is zero, because there was no net change in position. WORKED EXAMPLE Calculating Average Speed A marble rolls 5.2 m in 1.8 s. What was the marble's average speed? Strategy We know the distance the marble travels, 5.2 m, and the time interval, 1.8 s. We can use these values in the average speed equation. Solution Discussion Average speed is a scalar, so we do not include direction in the answer. We can check the reasonableness of the answer by estimating: 5 meters divided by 2 seconds is 2.5 m/s. Since 2.5 m/s is close to 2.9 m/s, the answer is reasonable. This is about the speed of a brisk walk, so it also makes sense. Practice Problems 8. A pitcher throws a baseball from the pitcher\u2019s mound to home plate in 0.46 s. The distance is 18.4 m. What was the average speed of the baseball? a. 40 m/s b. - 40 m/s c. 0.03 m/s d. 8.5 m/s 9. Cassie walked to her friend\u2019s house with an average speed of 1.40 m/s. The distance between the houses is 205 m. How long did the trip take her? a. 146 s b. 0.01 s c. 2.50 min d. 287 s Velocity The vector version of speed is velocity. Velocity describes the speed and direction of an object. As with speed, it is useful to describe either the average velocity over a time period or the velocity at a specific moment. Average velocity is displacement divided by the time over which the displacement occurs. 64 Chapter 2 \u2022 Motion in One Dimension Velocity, like speed, has SI units of meters per second (m/s), but because it is a vector, you must also include a direction. Furthermore, the variable v for velocity is bold because it is a vector, which is in contrast to the variable vfor speed which is italicized because it is a scalar quantity. TIPS FOR SUCCESS It is important to keep in mind that the average speed is not the same thing as the average velocity without its direction. Like we saw with displacement and distance in the last section, changes in direction over a time interval have a bigger effect on speed and velocity. Suppose a passenger moved toward the back of a plane with an average velocity of \u20134 m/s. We cannot tell from the average velocity whether the passenger stopped momentarily or backed up before he got to the back of the plane. To get more details, we must consider smaller segments of the trip over smaller time intervals such as those shown in Figure 2.9. If you consider infinitesimally small intervals, you can define instantaneous velocity, which is the velocity at a specific instant in time. Instantaneous velocity and average velocity are the same if the velocity is constant. Figure 2.9 The diagram shows a more detailed record of an airplane passenger heading toward the back of the plane, showing smaller segments of his trip. Earlier, you have read that distance traveled can be different than the magnitude of displacement. In the same way, speed can be different than the magnitude of velocity. For example, you drive to a store and return home in half an hour. If your car\u2019s odometer shows the total distance traveled was 6 km, then your average speed was 12 km/h. Your average velocity, however, was zero because your displacement for the round trip is zero. WATCH PHYSICS Calculating Average Velocity or Speed This video (http://openstax.org/l/28avgvelocity) reviews vectors and scalars and describes how to calculate average velocity and average speed when you know displacement and change in time. The video also reviews how to convert km/h to m/s. Click to view content (https://www.khanacademy.org/embed_video?v=MAS6mBRZZXA) GRASP CHECK Which of the following fully describes a vector and a scalar quantity and correctly provides an example of each? Access for free at openstax.org. 2.2 \u2022 Speed and Velocity 65 a. A scalar quantity is fully described by its magnitude, while a vector needs both magnitude and direction to fully describe it. Displacement is an example of a scalar quantity and time is an example of a vector quantity. b. A scalar quantity is fully described by its magnitude, while a vector needs both magnitude and direction to fully describe it. Time is an example of a scalar quantity", " and displacement is an example of a vector quantity. c. A scalar quantity is fully described by its magnitude and direction, while a vector needs only magnitude to fully describe it. Displacement is an example of a scalar quantity and time is an example of a vector quantity. d. A scalar quantity is fully described by its magnitude and direction, while a vector needs only magnitude to fully describe it. Time is an example of a scalar quantity and displacement is an example of a vector quantity. WORKED EXAMPLE Calculating Average Velocity A student has a displacement of 304 m north in 180 s. What was the student's average velocity? Strategy We know that the displacement is 304 m north and the time is 180 s. We can use the formula for average velocity to solve the problem. Solution 2.1 Discussion Since average velocity is a vector quantity, you must include direction as well as magnitude in the answer. Notice, however, that the direction can be omitted until the end to avoid cluttering the problem. Pay attention to the significant figures in the problem. The distance 304 m has three significant figures, but the time interval 180 s has only two, so the quotient should have only two significant figures. TIPS FOR SUCCESS Note the way scalars and vectors are represented. In this book d represents distance and displacement. Similarly, v represents speed, and v represents velocity. A variable that is not bold indicates a scalar quantity, and a bold variable indicates a vector quantity. Vectors are sometimes represented by small arrows above the variable. WORKED EXAMPLE Solving for Displacement when Average Velocity and Time are Known Layla jogs with an average velocity of 2.4 m/s east. What is her displacement after 46 seconds? Strategy We know that Layla's average velocity is 2.4 m/s east, and the time interval is 46 seconds. We can rearrange the average velocity formula to solve for the displacement. Solution 2.2 Discussion The answer is about 110 m east, which is a reasonable displacement for slightly less than a minute of jogging. A calculator shows the answer as 110.4 m. We chose to write the answer using scientific notation because we wanted to make it clear that we only 66 Chapter 2 \u2022 Motion in One Dimension used two significant figures. TIPS FOR SUCCESS Dimensional analysis is a good way to determine whether you solved a problem correctly. Write the calculation using only units to be sure they match on opposite sides of the equal mark. In the worked example, you have m = (m/s)(s). Since seconds is in the denominator for the average velocity and in the numerator for the time, the unit cancels out leaving only m and, of course, m = m. WORKED EXAMPLE Solving for Time when Displacement and Average Velocity are Known Phillip walks along a straight path from his house to his school. How long will it take him to get to school if he walks 428 m west with an average velocity of 1.7 m/s west? Strategy We know that Phillip's displacement is 428 m west, and his average velocity is 1.7 m/s west. We can calculate the time required for the trip by rearranging the average velocity equation. Solution 2.3 Discussion Here again we had to use scientific notation because the answer could only have two significant figures. Since time is a scalar, the answer includes only a magnitude and not a direction. Practice Problems 10. A trucker drives along a straight highway for 0.25 h with a displacement of 16 km south. What is the trucker\u2019s average velocity? a. 4 km/h north b. 4 km/h south c. 64 km/h north d. 64 km/h south 11. A bird flies with an average velocity of 7.5 m/s east from one branch to another in 2.4 s. It then pauses before flying with an average velocity of 6.8 m/s east for 3.5 s to another branch. What is the bird\u2019s total displacement from its starting point? a. 42 m west b. 6 m west c. 6 m east d. 42 m east Virtual Physics The Walking Man In this simulation you will put your cursor on the man and move him first in one direction and then in the opposite direction. Keep the Introductiontab active. You can use the Chartstab after you learn about graphing motion later in this chapter. Carefully watch the sign of the numbers in the position and velocity boxes. Ignore the acceleration box for now. See if you can make the man\u2019s position positive while the velocity is negative. Then see if you can do the opposite. Access for free at openstax.org. 2.3 \u2022 Position vs. Time Graphs 67 Click to view content (https://archive.cnx.org/specials/e2ca52af-8c6b-450e-ac2f-9300", "b38e8739/moving-man/) GRASP CHECK Which situation correctly describes when the moving man\u2019s position was negative but his velocity was positive? a. Man moving toward 0 from left of 0 b. Man moving toward 0 from right of 0 c. Man moving away from 0 from left of 0 d. Man moving away from 0 from right of 0 Check Your Understanding 12. Two runners travel along the same straight path. They start at the same time, and they end at the same time, but at the halfway mark, they have different instantaneous velocities. Is it possible for them to have the same average velocity for the trip? a. Yes, because average velocity depends on the net or total displacement. b. Yes, because average velocity depends on the total distance traveled. c. No, because the velocities of both runners must remain the exactly same throughout the journey. d. No, because the instantaneous velocities of the runners must remain same midway but can be different elsewhere. 13. If you divide the total distance traveled on a car trip (as determined by the odometer) by the time for the trip, are you calculating the average speed or the magnitude of the average velocity, and under what circumstances are these two quantities the same? a. Average speed. Both are the same when the car is traveling at a constant speed and changing direction. b. Average speed. Both are the same when the speed is constant and the car does not change its direction. c. Magnitude of average velocity. Both are same when the car is traveling at a constant speed. d. Magnitude of average velocity. Both are same when the car does not change its direction. 14. Is it possible for average velocity to be negative? a. Yes, in cases when the net displacement is negative. b. Yes, if the body keeps changing its direction during motion. c. No, average velocity describes only magnitude and not the direction of motion. d. No, average velocity describes only the magnitude in the positive direction of motion. 2.3 Position vs. Time Graphs Section Learning Objectives By the end of this section, you will be able to do the following: \u2022 Explain the meaning of slope in position vs. time graphs \u2022 Solve problems using position vs. time graphs Section Key Terms dependent variable independent variable tangent Graphing Position as a Function of Time A graph, like a picture, is worth a thousand words. Graphs not only contain numerical information, they also reveal relationships between physical quantities. In this section, we will investigate kinematics by analyzing graphs of position over time. Graphs in this text have perpendicular axes, one horizontal and the other vertical. When two physical quantities are plotted against each other, the horizontal axis is usually considered the independent variable, and the vertical axis is the dependent variable. In algebra, you would have referred to the horizontal axis as the x-axis and the vertical axis as the y-axis. As in Figure 2.10, a straight-line graph has the general form. 68 Chapter 2 \u2022 Motion in One Dimension Here mis the slope, defined as the rise divided by the run (as seen in the figure) of the straight line. The letter bis the y-intercept which is the point at which the line crosses the vertical, y-axis. In terms of a physical situation in the real world, these quantities will take on a specific significance, as we will see below. (Figure 2.10.) Figure 2.10 The diagram shows a straight-line graph. The equation for the straight line is yequals mx+ b. In physics, time is usually the independent variable. Other quantities, such as displacement, are said to depend upon it. A graph of position versus time, therefore, would have position on the vertical axis (dependent variable) and time on the horizontal axis (independent variable). In this case, to what would the slope and y-intercept refer? Let\u2019s look back at our original example when studying distance and displacement. The drive to school was 5 km from home. Let\u2019s assume it took 10 minutes to make the drive and that your parent was driving at a constant velocity the whole time. The position versus time graph for this section of the trip would look like that shown in Figure 2.11. Figure 2.11 A graph of position versus time for the drive to school is shown. What would the graph look like if we added the return trip? As we said before, d0 = 0 because we call home our Oand start calculating from there. In Figure 2.11, the line starts at d = 0, as well. This is the bin our equation for a straight line. Our initial position in a position versus time graph is always the place where the graph crosses the x-axis at t= 0. What is the slope? The riseis the change in position, (i.e., displacement) and the runis the change in time. This relationship", " can also be written This relationship was how we defined average velocity. Therefore, the slope in a d versus tgraph, is the average velocity. TIPS FOR SUCCESS Sometimes, as is the case where we graph both the trip to school and the return trip, the behavior of the graph looks different during different time intervals. If the graph looks like a series of straight lines, then you can calculate the average velocity for each time interval by looking at the slope. If you then want to calculate the average velocity for the entire trip, you can do a 2.4 Access for free at openstax.org. weighted average. Let\u2019s look at another example. Figure 2.12 shows a graph of position versus time for a jet-powered car on a very flat dry lake bed in Nevada. 2.3 \u2022 Position vs. Time Graphs 69 Figure 2.12 The diagram shows a graph of position versus time for a jet-powered car on the Bonneville Salt Flats. Using the relationship between dependent and independent variables, we see that the slope in the graph in Figure 2.12 is average velocity, vavg and the intercept is displacement at time zero\u2014that is, d0. Substituting these symbols into y= mx+ bgives or 2.5 2.6 Thus a graph of position versus time gives a general relationship among displacement, velocity, and time, as well as giving detailed numerical information about a specific situation. From the figure we can see that the car has a position of 400 m at t= 0 s, 650 m at t= 1.0 s, and so on. And we can learn about the object\u2019s velocity, as well. Snap Lab Graphing Motion In this activity, you will release a ball down a ramp and graph the ball\u2019s displacement vs. time. \u2022 Choose an open location with lots of space to spread out so there is less chance for tripping or falling due to rolling balls. 1 ball \u2022 \u2022 1 board \u2022 2 or 3 books 1 stopwatch \u2022 \u2022 1 tape measure \u2022 6 pieces of masking tape 1 piece of graph paper \u2022 1 pencil \u2022 Procedure 1. Build a ramp by placing one end of the board on top of the stack of books. Adjust location, as necessary, until there is no obstacle along the straight line path from the bottom of the ramp until at least the next 3 m. 2. Mark distances of 0.5 m, 1.0 m, 1.5 m, 2.0 m, 2.5 m, and 3.0 m from the bottom of the ramp. Write the distances on the tape. 70 Chapter 2 \u2022 Motion in One Dimension 3. Have one person take the role of the experimenter. This person will release the ball from the top of the ramp. If the ball does not reach the 3.0 m mark, then increase the incline of the ramp by adding another book. Repeat this Step as necessary. 4. Have the experimenter release the ball. Have a second person, the timer, begin timing the trial once the ball reaches the bottom of the ramp and stop the timing once the ball reaches 0.5 m. Have a third person, the recorder, record the time in a data table. 5. Repeat Step 4, stopping the times at the distances of 1.0 m, 1.5 m, 2.0 m, 2.5 m, and 3.0 m from the bottom of the ramp. 6. Use your measurements of time and the displacement to make a position vs. time graph of the ball\u2019s motion. 7. Repeat Steps 4 through 6, with different people taking on the roles of experimenter, timer, and recorder. Do you get the same measurement values regardless of who releases the ball, measures the time, or records the result? Discuss possible causes of discrepancies, if any. GRASP CHECK True or False: The average speed of the ball will be less than the average velocity of the ball. a. True b. False Solving Problems Using Position vs. Time Graphs So how do we use graphs to solve for things we want to know like velocity? WORKED EXAMPLE Using Position\u2013Time Graph to Calculate Average Velocity: Jet Car Find the average velocity of the car whose position is graphed in Figure 1.13. Strategy The slope of a graph of dvs. tis average velocity, since slope equals rise over run. 2.7 Since the slope is constant here, any two points on the graph can be used to find the slope. Solution 1. Choose two points on the line. In this case, we choose the points labeled on the graph: (6.4 s, 2000 m) and (0.50 s, 525 m). (Note, however, that you could choose any two points.) 2. Substitute the d and tvalues of the chosen points into the equation. Remember in calculating change (\u0394) we always use final value minus initial value", ". 2.8 Discussion This is an impressively high land speed (900 km/h, or about 560 mi/h): much greater than the typical highway speed limit of 27 m/s or 96 km/h, but considerably shy of the record of 343 m/s or 1,234 km/h, set in 1997. But what if the graph of the position is more complicated than a straight line? What if the object speeds up or turns around and goes backward? Can we figure out anything about its velocity from a graph of that kind of motion? Let\u2019s take another look at the jet-powered car. The graph in Figure 2.13 shows its motion as it is getting up to speed after starting at rest. Time starts at zero for this motion (as if measured with a stopwatch), and the displacement and velocity are initially 200 m and 15 m/s, respectively. Access for free at openstax.org. 2.3 \u2022 Position vs. Time Graphs 71 Figure 2.13 The diagram shows a graph of the position of a jet-powered car during the time span when it is speeding up. The slope of a distance versus time graph is velocity. This is shown at two points. Instantaneous velocity at any point is the slope of the tangent at that point. Figure 2.14 A U.S. Air Force jet car speeds down a track. (Matt Trostle, Flickr) The graph of position versus time in Figure 2.13 is a curve rather than a straight line. The slope of the curve becomes steeper as time progresses, showing that the velocity is increasing over time. The slope at any point on a position-versus-time graph is the instantaneous velocity at that point. It is found by drawing a straight line tangent to the curve at the point of interest and taking the slope of this straight line. Tangent lines are shown for two points in Figure 2.13. The average velocity is the net displacement divided by the time traveled. WORKED EXAMPLE Using Position\u2013Time Graph to Calculate Average Velocity: Jet Car, Take Two Calculate the instantaneous velocity of the jet car at a time of 25 s by finding the slope of the tangent line at point Q in Figure 2.13. Strategy The slope of a curve at a point is equal to the slope of a straight line tangent to the curve at that point. Solution 1. Find the tangent line to the curve at 2. Determine the endpoints of the tangent. These correspond to a position of 1,300 m at time 19 s and a position of 3120 m at. time 32 s. 72 Chapter 2 \u2022 Motion in One Dimension 3. Plug these endpoints into the equation to solve for the slope, v. 2.9 Discussion The entire graph of v versus tcan be obtained in this fashion. Practice Problems 15. Calculate the average velocity of the object shown in the graph below over the whole time interval. a. 0.25 m/s b. 0.31 m/s c. 3.2 m/s d. 4.00 m/s 16. True or False: By taking the slope of the curve in the graph you can verify that the velocity of the jet car is at. a. True b. False Check Your Understanding 17. Which of the following information about motion can be determined by looking at a position vs. time graph that is a straight line? Access for free at openstax.org. 2.4 \u2022 Velocity vs. Time Graphs 73 a. frame of reference b. average acceleration c. velocity d. direction of force applied 18. True or False: The position vs time graph of an object that is speeding up is a straight line. a. True b. False 2.4 Velocity vs. Time Graphs Section Learning Objectives By the end of this section, you will be able to do the following: \u2022 Explain the meaning of slope and area in velocity vs. time graphs \u2022 Solve problems using velocity vs. time graphs Section Key Terms acceleration Graphing Velocity as a Function of Time Earlier, we examined graphs of position versus time. Now, we are going to build on that information as we look at graphs of velocity vs. time. Velocity is the rate of change of displacement. Acceleration is the rate of change of velocity; we will discuss acceleration more in another chapter. These concepts are all very interrelated. Virtual Physics Maze Game In this simulation you will use a vector diagram to manipulate a ball into a certain location without hitting a wall. You can manipulate the ball directly with position or by changing its velocity. Explore how these factors change the motion. If you would like, you can put it on the asetting, as well. This is acceleration, which measures the rate of change of velocity. We will explore acceleration in more detail later, but it might be interesting to take a look at it here. Click to view content (https://archive.cnx.org/specials/", "30e37034-2fbd-11e5-83a2-03be60006ece/maze-game/) GRASP CHECK Click to view content (https://archive.cnx.org/specials/30e37034-2fbd-11e5-83a2-03be60006ece/maze-game/#sim-mazegame) a. The ball can be easily manipulated with displacement because the arena is a position space. b. The ball can be easily manipulated with velocity because the arena is a position space. c. The ball can be easily manipulated with displacement because the arena is a velocity space. d. The ball can be easily manipulated with velocity because the arena is a velocity space. What can we learn about motion by looking at velocity vs. time graphs? Let\u2019s return to our drive to school, and look at a graph of position versus time as shown in Figure 2.15. 74 Chapter 2 \u2022 Motion in One Dimension Figure 2.15 A graph of position versus time for the drive to and from school is shown. We assumed for our original calculation that your parent drove with a constant velocity to and from school. We now know that the car could not have gone from rest to a constant velocity without speeding up. So the actual graph would be curved on either end, but let\u2019s make the same approximation as we did then, anyway. TIPS FOR SUCCESS It is common in physics, especially at the early learning stages, for certain things to be neglected, as we see here. This is because it makes the concept clearer or the calculation easier. Practicing physicists use these kinds of short-cuts, as well. It works out because usually the thing being neglectedis small enough that it does not significantly affect the answer. In the earlier example, the amount of time it takes the car to speed up and reach its cruising velocity is very small compared to the total time traveled. Looking at this graph, and given what we learned, we can see that there are two distinct periods to the car\u2019s motion\u2014the way to school and the way back. The average velocity for the drive to school is 0.5 km/minute. We can see that the average velocity for the drive back is \u20130.5 km/minute. If we plot the data showing velocity versus time, we get another graph (Figure 2.16): Figure 2.16 Graph of velocity versus time for the drive to and from school. We can learn a few things. First, we can derive a v versus tgraph from a d versus tgraph. Second, if we have a straight-line position\u2013time graph that is positively or negatively sloped, it will yield a horizontal velocity graph. There are a few other interesting things to note. Just as we could use a position vs. time graph to determine velocity, we can use a velocity vs. time graph to determine position. We know that v = d/t. If we use a little algebra to re-arrange the equation, we see that d = v t. In Figure 2.16, we have velocity on the y-axis and time along the x-axis. Let\u2019s take just the first half of the motion. We get 0.5 km/ minute If we calculate the same for the return trip, we get \u20135 km. If we add them together, we see that the net displacement for the 10 minutes. The units for minutescancel each other, and we get 5 km, which is the displacement for the trip to school. Access for free at openstax.org. 2.4 \u2022 Velocity vs. Time Graphs 75 whole trip is 0 km, which it should be because we started and ended at the same place. TIPS FOR SUCCESS You can treat units just like you treat numbers, so a km/km=1 (or, we say, it cancels out). This is good because it can tell us whether or not we have calculated everything with the correct units. For instance, if we end up with m \u00d7 s for velocity instead of m/s, we know that something has gone wrong, and we need to check our math. This process is called dimensional analysis, and it is one of the best ways to check if your math makes sense in physics. The area under a velocity curve represents the displacement. The velocity curve also tells us whether the car is speeding up. In our earlier example, we stated that the velocity was constant. So, the car is not speeding up. Graphically, you can see that the slope of these two lines is 0. This slope tells us that the car is not speeding up, or accelerating. We will do more with this information in a later chapter. For now, just remember that the area under the graph and the slope are the two important parts of the graph. Just like we could define a linear equation for the motion in a position vs. time graph", ", we can also define one for a velocity vs. time graph. As we said, the slope equals the acceleration, a. And in this graph, the y-intercept is v0. Thus,. But what if the velocity is not constant? Let\u2019s look back at our jet-car example. At the beginning of the motion, as the car is speeding up, we saw that its position is a curve, as shown in Figure 2.17. Figure 2.17 A graph is shown of the position of a jet-powered car during the time span when it is speeding up. The slope of a d vs. t graph is velocity. This is shown at two points. Instantaneous velocity at any point is the slope of the tangent at that point. You do not have to do this, but you could, theoretically, take the instantaneous velocity at each point on this graph. If you did, you would get Figure 2.18, which is just a straight line with a positive slope. 76 Chapter 2 \u2022 Motion in One Dimension Figure 2.18 The graph shows the velocity of a jet-powered car during the time span when it is speeding up. Again, if we take the slope of the velocity vs. time graph, we get the acceleration, the rate of change of the velocity. And, if we take the area under the slope, we get back to the displacement. Solving Problems using Velocity\u2013Time Graphs Most velocity vs. time graphs will be straight lines. When this is the case, our calculations are fairly simple. WORKED EXAMPLE Using Velocity Graph to Calculate Some Stuff: Jet Car Use this figure to (a) find the displacement of the jet car over the time shown (b) calculate the rate of change (acceleration) of the velocity. (c) give the instantaneous velocity at 5 s, and (d) calculate the average velocity over the interval shown. Strategy a. The displacement is given by finding the area under the line in the velocity vs. time graph. b. The acceleration is given by finding the slope of the velocity graph. c. The instantaneous velocity can just be read off of the graph. d. To find the average velocity, recall that Solution a. 1. Analyze the shape of the area to be calculated. In this case, the area is made up of a rectangle between 0 and 20 m/s stretching to 30 s. The area of a rectangle is length width. Therefore, the area of this piece is 600 m. 2. Above that is a triangle whose base is 30 s and height is 140 m/s. The area of a triangle is 0.5 length width. The area of this piece, therefore, is 2,100 m. 3. Add them together to get a net displacement of 2,700 m. b. 1. Take two points on the velocity line. Say, t= 5 s and t= 25 s. At t= 5 s, the value of v = 40 m/s. At t= 25 s, v = 140 m/s. 2. Find the slope. c. The instantaneous velocity at t= 5 s, as we found in part (b) is just 40 m/s. 1. Find the net displacement, which we found in part (a) was 2,700 m. d. 2. Find the total time which for this case is 30 s. 3. Divide 2,700 m/30 s = 90 m/s. Access for free at openstax.org. 2.4 \u2022 Velocity vs. Time Graphs 77 Discussion The average velocity we calculated here makes sense if we look at the graph. 100m/s falls about halfway across the graph and since it is a straight line, we would expect about half the velocity to be above and half below. TIPS FOR SUCCESS You can have negative position, velocity, and acceleration on a graph that describes the way the object is moving. You should never see a graph with negative time on an axis. Why? Most of the velocity vs. time graphs we will look at will be simple to interpret. Occasionally, we will look at curved graphs of velocity vs. time. More often, these curved graphs occur when something is speeding up, often from rest. Let\u2019s look back at a more realistic velocity vs. time graph of the jet car\u2019s motion that takes this speeding upstage into account. Figure 2.19 The graph shows a more accurate graph of the velocity of a jet-powered car during the time span when it is speeding up. WORKED EXAMPLE Using Curvy Velocity Graph to Calculate Some Stuff: jet car, Take Two Use Figure 2.19 to (a) find the approximate displacement of the jet car over the time shown, (b) calculate the instantaneous acceleration at t= 30 s, (c) find the instantaneous velocity at 30 s, and (d) calculate the approximate average velocity over the interval shown. Strategy a. Because this", " graph is an undefined curve, we have to estimate shapes over smaller intervals in order to find the areas. b. Like when we were working with a curved displacement graph, we will need to take a tangent line at the instant we are interested and use that to calculate the instantaneous acceleration. c. The instantaneous velocity can still be read off of the graph. d. We will find the average velocity the same way we did in the previous example. Solution a. 1. This problem is more complicated than the last example. To get a good estimate, we should probably break the curve into four sections. 0 \u2192 10 s, 10 \u2192 20 s, 20 \u2192 40 s, and 40 \u2192 70 s. 2. Calculate the bottom rectangle (common to all pieces). 165 m/s 3. Estimate a triangle at the top, and calculate the area for each section. Section 1 = 225 m; section 2 = 100 m + 450 m = 70 s = 11,550 m. 550 m; section 3 = 150 m + 1,300 m = 1,450 m; section 4 = 2,550 m. 4. Add them together to get a net displacement of 16,325 m. b. Using the tangent line given, we find that the slope is 1 m/s2. 78 Chapter 2 \u2022 Motion in One Dimension c. The instantaneous velocity at t= 30 s, is 240 m/s. d. 1. Find the net displacement, which we found in part (a), was 16,325 m. 2. Find the total time, which for this case is 70 s. 3. Divide Discussion This is a much more complicated process than the first problem. If we were to use these estimates to come up with the average velocity over just the first 30 s we would get about 191 m/s. By approximating that curve with a line, we get an average velocity of 202.5 m/s. Depending on our purposes and how precise an answer we need, sometimes calling a curve a straight line is a worthwhile approximation. Practice Problems 19. Figure 2.20 Consider the velocity vs. time graph shown below of a person in an elevator. Suppose the elevator is initially at rest. It then speeds up for 3 seconds, maintains that velocity for 15 seconds, then slows down for 5 seconds until it stops. Find the instantaneous velocity at t= 10 s and t= 23 s. a. b. c. d. Instantaneous velocity at t= 10 s and t= 23 s are 0 m/s and 0 m/s. Instantaneous velocity at t= 10 s and t= 23 s are 0 m/s and 3 m/s. Instantaneous velocity at t= 10 s and t= 23 s are 3 m/s and 0 m/s. Instantaneous velocity at t= 10 s and t= 23 s are 3 m/s and 1.5 m/s. Access for free at openstax.org. 20. 2.4 \u2022 Velocity vs. Time Graphs 79 Figure 2.21 Calculate the net displacement and the average velocity of the elevator over the time interval shown. a. Net displacement is 45 m and average velocity is 2.10 m/s. b. Net displacement is 45 m and average velocity is 2.28 m/s. c. Net displacement is 57 m and average velocity is 2.66 m/s. d. Net displacement is 57 m and average velocity is 2.48 m/s. Snap Lab Graphing Motion, Take Two In this activity, you will graph a moving ball\u2019s velocity vs. time. \u2022 your graph from the earlier Graphing Motion Snap Lab! \u2022 \u2022 1 piece of graph paper 1 pencil Procedure 1. Take your graph from the earlier Graphing Motion Snap Lab! and use it to create a graph of velocity vs. time. 2. Use your graph to calculate the displacement. GRASP CHECK Describe the graph and explain what it means in terms of velocity and acceleration. a. The graph shows a horizontal line indicating that the ball moved with a constant velocity, that is, it was not accelerating. b. The graph shows a horizontal line indicating that the ball moved with a constant velocity, that is, it was accelerating. c. The graph shows a horizontal line indicating that the ball moved with a variable velocity, that is, it was not accelerating. d. The graph shows a horizontal line indicating that the ball moved with a variable velocity, that is, it was accelerating. Check Your Understanding 21. What information could you obtain by looking at a velocity vs. time graph? a. acceleration b. direction of motion c. reference frame of the motion 80 Chapter 2 \u2022 Motion in One Dimension d. shortest path 22. How would you use a position vs. time graph to construct a velocity vs. time graph and vice versa? a. Slope of position vs. time curve is used to construct velocity vs. time curve, and slope of velocity vs. time curve", " is used to construct position vs. time curve. b. Slope of position vs. time curve is used to construct velocity vs. time curve, and area of velocity vs. time curve is used to construct position vs. time curve. c. Area of position vs. time curve is used to construct velocity vs. time curve, and slope of velocity vs. time curve is used to construct position vs. time curve. d. Area of position/time curve is used to construct velocity vs. time curve, and area of velocity vs. time curve is used to construct position vs. time curve. Access for free at openstax.org. KEY TERMS acceleration the rate at which velocity changes average speed distance traveled divided by time during which motion occurs average velocity displacement divided by time over which displacement occurs dependent variable the variable that changes as the independent variable changes displacement fixed axis the change in position of an object against a distance the length of the path actually traveled between an initial and a final position independent variable the variable, usually along the horizontal axis of a graph, that does not change based on human or experimental action; in physics this is usually SECTION SUMMARY 2.1 Relative Motion, Distance, and Displacement \u2022 A description of motion depends on the reference frame from which it is described. \u2022 The distance an object moves is the length of the path along which it moves. \u2022 Displacement is the difference in the initial and final positions of an object. 2.2 Speed and Velocity \u2022 Average speed is a scalar quantity that describes distance traveled divided by the time during which the motion occurs. \u2022 Velocity is a vector quantity that describes the speed and direction of an object. \u2022 Average velocity is displacement over the time period during which the displacement occurs. If the velocity is constant, then average velocity and instantaneous KEY EQUATIONS 2.1 Relative Motion, Distance, and Displacement Displacement 2.2 Speed and Velocity Average speed Average velocity Chapter 2 \u2022 Key Terms 81 time instantaneous speed speed at a specific instant in time instantaneous velocity velocity at a specific instant in time kinematics causes the study of motion without considering its magnitude size or amount position the location of an object at any particular time reference frame a coordinate system from which the positions of objects are described scalar a quantity that has magnitude but no direction speed rate at which an object changes its location tangent a line that touches another at exactly one point vector a quantity that has both magnitude and direction velocity the speed and direction of an object velocity are the same. 2.3 Position vs. Time Graphs \u2022 Graphs can be used to analyze motion. \u2022 The slope of a position vs. time graph is the velocity. \u2022 For a straight line graph of position, the slope is the average velocity. \u2022 To obtain the instantaneous velocity at a given moment for a curved graph, find the tangent line at that point and take its slope. 2.4 Velocity vs. Time Graphs \u2022 The slope of a velocity vs. time graph is the acceleration. \u2022 The area under a velocity vs. time curve is the displacement. \u2022 Average velocity can be found in a velocity vs. time graph by taking the weighted average of all the velocities. 2.3 Position vs. Time Graphs Displacement. 2.4 Velocity vs. Time Graphs Velocity Acceleration 82 Chapter 2 \u2022 Chapter Review CHAPTER REVIEW Concept Items 2.1 Relative Motion, Distance, and Displacement 1. Can one-dimensional motion have zero distance but a nonzero displacement? What about zero displacement but a nonzero distance? a. One-dimensional motion can have zero distance with a nonzero displacement. Displacement has both magnitude and direction, and it can also have zero displacement with nonzero distance because distance has only magnitude. b. One-dimensional motion can have zero distance with a nonzero displacement. Displacement has both magnitude and direction, but it cannot have zero displacement with nonzero distance because distance has only magnitude. c. One-dimensional motion cannot have zero distance with a nonzero displacement. Displacement has both magnitude and direction, but it can have zero displacement with nonzero distance because distance has only magnitude and any motion will be the distance it moves. d. One-dimensional motion cannot have zero distance with a nonzero displacement. Displacement has both magnitude and direction, and it cannot have zero displacement with nonzero distance because distance has only magnitude. 2. In which example would you be correct in describing an object in motion while your friend would also be correct in describing that same object as being at rest? a. You are driving a car toward the east and your friend drives past you in the opposite direction with the same speed. In your frame of reference, you will be in motion. In your friend\u2019s frame of reference, you will be at rest. b. You are driving a car toward the east and your friend is standing at the bus stop. In your frame of reference, you will be in motion. In your friend\u2019s frame of", " reference, you will be at rest. c. You are driving a car toward the east and your friend is standing at the bus stop. In your frame of reference, your friend will be moving toward the west. In your friend\u2019s frame of reference, he will be at rest. d. You are driving a car toward the east and your friend is standing at the bus stop. In your frame of reference, your friend will be moving toward the east. In your friend\u2019s frame of reference, he will be at rest. Access for free at openstax.org. 3. What does your car\u2019s odometer record? a. displacement b. distance c. both distance and displacement d. the sum of distance and displacement 2.2 Speed and Velocity 4. In the definition of velocity, what physical quantity is changing over time? a. speed b. distance c. magnitude of displacement d. position vector 5. Which of the following best describes the relationship between instantaneous velocity and instantaneous speed? a. Both instantaneous speed and instantaneous b. velocity are the same, even when there is a change in direction. Instantaneous speed and instantaneous velocity cannot be the same even if there is no change in direction of motion. c. Magnitude of instantaneous velocity is equal to instantaneous speed. d. Magnitude of instantaneous velocity is always greater than instantaneous speed. 2.3 Position vs. Time Graphs 6. Use the graph to describe what the runner\u2019s motion looks like. How are average velocity for only the first four seconds and instantaneous velocity related? What is the runner's net displacement over the time shown? a. The net displacement is 12 m and the average velocity is equal to the instantaneous velocity. b. The net displacement is 12 m and the average velocity is two times the instantaneous velocity. c. The net displacement is 10 + 12 = 22 m and the average velocity is equal to the instantaneous velocity. d. The net displacement is 10 + 12 = 22 m and the average velocity is two times the instantaneous velocity. 7. A position vs. time graph of a frog swimming across a. If each section lasts pond has two distinct straight-line sections. The slope of. The slope of the second section the first section is is, then what is the frog\u2019s total average velocity? a. b. c. d. 2.4 Velocity vs. Time Graphs 8. A graph of velocity vs. time of a ship coming into a harbor is shown. Chapter 2 \u2022 Chapter Review 83 Describe the acceleration of the ship based on the graph. a. The ship is moving in the forward direction at a steady rate. Then it accelerates in the forward direction and then decelerates. b. The ship is moving in the forward direction at a steady rate. Then it turns around and starts decelerating, while traveling in the reverse direction. It then accelerates, but slowly. c. The ship is moving in the forward direction at a steady rate. Then it decelerates in the forward direction, and then continues to slow down in the forward direction, but with more deceleration. d. The ship is moving in the forward direction at a steady rate. Then it decelerates in the forward direction, and then continues to slow down in the forward direction, but with less deceleration. Critical Thinking Items 2.1 Relative Motion, Distance, and Displacement 9. Boat A and Boat B are traveling at a constant speed in opposite directions when they pass each other. If a person in each boat describes motion based on the boat\u2019s own reference frame, will the description by a person in Boat A of Boat B\u2019s motion be the same as the description by a person in Boat B of Boat A\u2019s motion? a. Yes, both persons will describe the same motion because motion is independent of the frame of reference. b. Yes, both persons will describe the same motion because they will perceive the other as moving in the backward direction. c. No, the motion described by each of them will be different because motion is a relative term. d. No, the motion described by each of them will be different because the motion perceived by each will be opposite to each other. 10. Passenger A sits inside a moving train and throws a ball vertically upward. How would the motion of the ball be described by a fellow train passenger B and an observer C who is standing on the platform outside the train? a. Passenger B sees that the ball has vertical, but no horizontal, motion. Observer C sees the ball has vertical as well as horizontal motion. b. Passenger B sees the ball has vertical as well as horizontal motion. Observer C sees the ball has the vertical, but no horizontal, motion. c. Passenger B sees the ball has horizontal but no vertical motion. Observer C sees the ball has vertical as well as horizontal motion. d. Passenger B sees the ball has vertical as well as horizontal motion. Observer C sees the ball has horizontalbut no vertical motion", ". 2.2 Speed and Velocity 11. Is it possible to determine a car\u2019s instantaneous velocity from just the speedometer reading? a. No, it reflects speed but not the direction. b. No, it reflects the average speed of the car. c. Yes, it sometimes reflects instantaneous velocity of the car. d. Yes, it always reflects the instantaneous velocity of the car. 12. Terri, Aaron, and Jamal all walked along straight paths. 84 Chapter 2 \u2022 Chapter Review Terri walked 3.95 km north in 48 min. Aaron walked 2.65 km west in 31 min. Jamal walked 6.50 km south in 81 min. Which of the following correctly ranks the three boys in order from lowest to highest average speed? Jamal, Terri, Aaron a. Jamal, Aaron, Terri b. c. Terri, Jamal, Aaron d. Aaron, Terri, Jamal 13. Rhianna and Logan start at the same point and walk due Identify the time (ta, tb, tc, td, or te) at which at which the instantaneous velocity is greatest, the time at which it is zero, and the time at which it is negative. 2.4 Velocity vs. Time Graphs 15. Identify the time, or times, at which the instantaneous velocity is greatest, and the time, or times, at which it is negative. A sketch of velocity vs. time derived from the figure will aid in arriving at the correct answers. north. Rhianna walks with an average velocity Logan walks three times the distance in twice the time as Rhianna. Which of the following expresses Logan\u2019s average velocity in terms of a. Logan\u2019s average velocity =?.. b. Logan\u2019s average velocity = c. Logan\u2019s average velocity = d. Logan\u2019s average velocity =... 2.3 Position vs. Time Graphs 14. Explain how you can use the graph of position vs. time to describe the change in velocity over time. a. The instantaneous velocity is greatest at f, and it is negative at d, h, I, j, and k. b. The instantaneous velocity is greatest at e, and it is negative at a, b, and f. c. The instantaneous velocity is greatest at f, and it is negative at d, h, I, j, and k d. The instantaneous velocity is greatest at d, and it is negative at a, b, and f. Problems 2.1 Relative Motion, Distance, and Displacement Up is the positive direction. What are the total displacement of the ball and the total distance traveled by the ball? a. The displacement is equal to -4 m and the distance 16. In a coordinate system in which the direction to the is equal to 4 m. right is positive, what are the distance and displacement of a person who walks to the right, and then a. Distance is b. Distance is c. Distance is d. Distance is to the left, to the left? and displacement is and displacement is and displacement is and displacement is.... 17. Billy drops a ball from a height of 1 m. The ball bounces back to a height of 0.8 m, then bounces again to a height of 0.5 m, and bounces once more to a height of 0.2 m. Access for free at openstax.org. b. The displacement is equal to 4 m and the distance is equal to 1 m. c. The displacement is equal to 4 m and the distance is equal to 1 m. d. The displacement is equal to -1 m and the distance is equal to 4 m. 2.2 Speed and Velocity 18. You sit in a car that is moving at an average speed of 86.4 km/h. During the 3.3 s that you glance out the window, how far has the car traveled? 7.27 m a. 79 m b. c. 285 km 1026 m d. 2.3 Position vs. Time Graphs 19. Using the graph, what is the average velocity for the whole Chapter 2 \u2022 Chapter Review 85 moments in time. What is the minimum number of data points you would need to estimate the average acceleration of the object? a. 1 b. 2 3 c. d. 4 10 seconds? 22. Which option best describes the average acceleration from 40 to 70 s? a. The total average velocity is 0 m/s. b. The total average velocity is 1.2 m/s. c. The total average velocity is 1.5 m/s. d. The total average velocity is 3.0 m/s. 20. A train starts from rest and speeds up for 15 minutes until it reaches a constant velocity of 100 miles/hour. It stays at this speed for half an hour. Then it slows down for another 15 minutes until it is still. Which of the following correctly describes the position vs time graph of the train", "\u2019s journey? a. The first 15 minutes is a curve that is concave upward, the middle portion is a straight line with slope 100 miles/hour, and the last portion is a concave downward curve. b. The first 15 minutes is a curve that is concave downward, the middle portion is a straight line with slope 100 miles/hour, and the last portion is a concave upward curve. c. The first 15 minutes is a curve that is concave upward, the middle portion is a straight line with slope zero, and the last portion is a concave downward curve. d. The first 15 minutes is a curve that is concave downward, the middle portion is a straight line with slope zero, and the last portion is a concave upward curve. 2.4 Velocity vs. Time Graphs 21. You are characterizing the motion of an object by measuring the location of the object at discrete a. b. c. d. It is negative and smaller in magnitude than the initial acceleration. It is negative and larger in magnitude than the initial acceleration. It is positive and smaller in magnitude than the initial acceleration. It is positive and larger in magnitude than the initial acceleration. 23. The graph shows velocity vs. time. Calculate the net displacement using seven different divisions. Calculate it again using two divisions: 0 \u2192 40 s 86 Chapter 2 \u2022 Test Prep and 40 \u2192 70 s. Compare. Using both, calculate the average velocity. a. Displacement and average velocity using seven divisions are 14,312.5 m and 204.5 m/s while with two divisions are 15,500 m and 221.4 m/s respectively. b. Displacement and average velocity using seven divisions are 15,500 m and 221.4 m/s while with two divisions are 14,312.5 m and 204.5 m/s respectively. c. Displacement and average velocity using seven divisions are 15,500 m and 204.5 m/s while with two divisions are 14,312.5 m and 221.4 m/s respectively. d. Displacement and average velocity using seven divisions are 14,312.5 m and 221.4 m/s while with two divisions are 15,500 m and 204.5 m/s respectively. Performance Task 2.4 Velocity vs. Time Graphs 24. The National Mall in Washington, DC, is a national park containing most of the highly treasured memorials and museums of the United States. However, the National Mall also hosts many events and concerts. The map in shows the area for a benefit concert during which the president will speak. The concert stage is near the Lincoln Memorial. The seats and standing room for the crowd will stretch from the stage east to near the Washington Monument, as shown on the map. You are planning the logistics for the concert. Use the map scale to measure any distances needed to make the calculations below. The park has three new long-distance speakers. They would like to use these speakers to broadcast the concert audio to other parts of the National Mall. The speakers can project sound up to 0.35 miles away but they must be connected to one of the power supplies within the concert area. What is the minimum amount of wire needed for each speaker, in miles, in order to project the audio to the following areas? Assume that wire cannot be placed over buildings or any memorials. Part A. The center of the Jefferson Memorial using power supply 1 (This will involve an elevated wire that can travel over water.) Part B. The center of the Ellipse using power supply 3 (This wire cannot travel over water.) Part C. The president\u2019s motorcade will be traveling to the concert from the White House. To avoid concert traffic, the motorcade travels from the White House west down E Street and then turns south on 23rd Street to reach the Lincoln memorial. What minimum speed, in miles per hour to the nearest tenth, would the motorcade have to travel to make the trip in 5 minutes? Part D. The president could also simply fly from the White House to the Lincoln Memorial using the presidential helicopter, Marine 1. How long would it take Marine 1, traveling slowly at 30 mph, to travel from directly above the White House landing zone (LZ) to directly above the Lincoln Memorial LZ? Disregard liftoff and landing times and report the travel time in minutes to the nearest minute. TEST PREP Multiple Choice 2.1 Relative Motion, Distance, and Displacement 25. Why should you specify a reference frame when describing motion? a. a description of motion depends on the reference frame Access for free at openstax.org. b. motion appears the same in all reference frames c. reference frames affect the motion of an object d. you can see motion better from certain reference frames 26. Which of the following is true for the displacement of an object? a. It is always equal to the distance the object moved b. c. d.", " between its initial and final positions. It is both the straight line distance the object moved as well as the direction of its motion. It is the direction the object moved between its initial and final positions. It is the straight line distance the object moved between its initial and final positions. 27. If a biker rides west for 50 miles from his starting position, then turns and bikes back east 80 miles. What is his net displacement? a. b. c. d. Cannot be determined from the information given 130 miles 30 miles east 30 miles west 28. Suppose a train is moving along a track. Is there a single, correct reference frame from which to describe the train\u2019s motion? a. Yes, there is a single, correct frame of reference Chapter 2 \u2022 Test Prep 87 d. 1,500 m south 32. A bicyclist covers the first leg of a journey that is long in, at a speed of,, at a in. If his average speed is equal to the, then which of the following is and and the second leg of speed of average of true? a. b. c. d. 33. A car is moving on a straight road at a constant speed in a single direction. Which of the following statements is true? a. Average velocity is zero. b. The magnitude of average velocity is equal to the average speed. c. The magnitude of average velocity is greater than because motion is a relative term. the average speed. b. Yes, there is a single, correct frame of reference d. The magnitude of average velocity is less than the which is in terms of Earth\u2019s position. average speed. c. No, there is not a single, correct frame of reference because motion is a relative term. 2.3 Position vs. Time Graphs d. No, there is not a single, correct frame of reference because motion is independent of frame of reference. 29. If a space shuttle orbits Earth once, what is the shuttle\u2019s distance traveled and displacement? a. Distance and displacement both are zero. b. Distance is circumference of the circular orbit while displacement is zero. c. Distance is zero while the displacement is circumference of the circular orbit. d. Distance and displacement both are equal to circumference of the circular orbit. 2.2 Speed and Velocity 34. What is the slope of a straight line graph of position vs. time? a. Velocity b. Displacement c. Distance d. Acceleration 35. Using the graph, what is the runner\u2019s velocity from 4 to 10 s? 30. Four bicyclists travel different distances and times along a straight path. Which cyclist traveled with the greatest average speed? a. Cyclist 1 travels b. Cyclist 2 travels c. Cyclist 3 travels d. Cyclist 4 travels in in in in.... 31. A car travels with an average velocity of 23 m/s for 82 s. Which of the following could NOT have been the car's displacement? a. b. c. 1,700 m east 1,900 m west 1,600 m north a. \u20133 m/s b. 0 m/s c. d. 1.2 m/s 3 m/s 88 Chapter 2 \u2022 Test Prep 2.4 Velocity vs. Time Graphs 36. What does the area under a velocity vs. time graph line represent? a. acceleration b. displacement c. distance d. instantaneous velocity 37. An object is moving along a straight path with constant Short Answer 2.1 Relative Motion, Distance, and Displacement 38. While standing on a sidewalk facing the road, you see a bicyclist passing by toward your right. In the reference frame of the bicyclist, in which direction are you moving? a. b. c. d. in the same direction of motion as the bicyclist in the direction opposite the motion of the bicyclist stationary with respect to the bicyclist in the direction of velocity of the bicyclist 39. Maud sends her bowling ball straight down the center of the lane, getting a strike. The ball is brought back to the holder mechanically. What are the ball\u2019s net displacement and distance traveled? a. Displacement of the ball is twice the length of the lane, while the distance is zero. b. Displacement of the ball is zero, while the distance is twice the length of the lane. c. Both the displacement and distance for the ball are equal to zero. d. Both the displacement and distance for the ball are twice the length of the lane. 40. A fly buzzes four and a half times around Kit Yan\u2019s head. The fly ends up on the opposite side from where it started. If the diameter of his head is the total distance the fly travels and its total displacement? a. The distance is with a displacement of, what is zero. b. The distance is c. The distance is with a displacement of zero. with a displacement of. d. The distance is with a displacement of.", "eed of 3v m/s, how Access for free at openstax.org. and. acceleration. A velocity vs. time graph starts at ends at, stretching over a time-span of What is the object\u2019s net displacement? a. b. c. d. cannot be determined from the information given long is the return trip home? t/6 a. t/3 b. 3t c. d. 6t 42. What can you infer from the statement, Velocity of an object is zero? a. Object is in linear motion with constant velocity. b. Object is moving at a constant speed. c. Object is either at rest or it returns to the initial point. d. Object is moving in a straight line without changing its direction. 43. An object has an average speed of 7.4 km/h. Which of the following describes two ways you could increase the average speed of the object to 14.8 km/h? a. Reduce the distance that the object travels by half, keeping the time constant, or keep the distance constant and double the time. b. Double the distance that the object travels, keeping the time constant, or keep the distance constant and reduce the time by half. c. Reduce the distance that the object travels to onefourth, keeping the time constant, or keep the distance constant and increase the time by fourfold. Increase the distance by fourfold, keeping the time constant, or keep the distance constant and reduce the time by one-fourth. d. 44. Swimming one lap in a pool is defined as going across a pool and back again. If a swimmer swims 3 laps in 9 minutes, how can his average velocity be zero? a. His average velocity is zero because his total distance is zero. b. His average velocity is zero because his total displacement is zero. c. His average velocity is zero because the number of laps completed is an odd number. d. His average velocity is zero because the velocity of each successive lap is equal and opposite. 2.3 Position vs. Time Graphs 45. A hockey puck is shot down the arena in a straight line. Assume it does not slow until it is stopped by an opposing player who sends it back in the direction it came. The players are 20 m apart and it takes 1 s for the puck to go there and back. Which of the following describes the graph of the displacement over time? Consider the initial direction of the puck to be positive. a. The graph is an upward opening V. b. The graph is a downward opening V. c. The graph is an upward opening U. d. The graph is downward opening U. 46. A defensive player kicks a soccer ball 20 m back to her own goalie. It stops just as it reaches her. She sends it back to the player. Without knowing the time it takes, draw a rough sketch of the displacement over time. Does this graph look similar to the graph of the hockey puck from the previous question? a. Yes, the graph is similar to the graph of the hockey puck. b. No, the graph is not similar to the graph of the hockey puck. c. The graphs cannot be compared without knowing the time the soccer ball was rolling. 47. What are the net displacement, total distance traveled, and total average velocity in the previous two problems? a. net displacement = 0 m, total distance = 20 m, total average velocity = 20 m/s b. net displacement = 0 m, total distance = 40 m, total average velocity = 20 m/s c. net displacement = 0 m, total distance = 20 m, total average velocity = 0 m/s d. net displacement = 0 m, total distance = 40 m, total average velocity = 0 m/s 48. A bee flies straight at someone and then back to its hive along the same path. Assuming it takes no time for the bee to speed up or slow down, except at the moment it changes direction, how would the graph of position vs time look? Consider the initial direction to be positive. a. The graph will look like a downward opening V shape. b. The graph will look like an upward opening V shape. c. The graph will look like a downward opening parabola. Chapter 2 \u2022 Test Prep 89 a. b. c. d. It is a straight line with negative slope. It is a straight line with positive slope. It is a horizontal line at some negative value. It is a horizontal line at some positive value. 50. Which statement correctly describes the object\u2019s speed, as well as what a graph of acceleration vs. time would look like? a. The object is not speeding up, and the acceleration vs. time graph is a horizontal line at some negative value. b. The object is not speeding up, and the acceleration vs. time graph is a horizontal line at some positive value. c. The object is speeding up, and the acceleration vs. time graph is a horizontal line", " at some negative value. d. The object is speeding up, and the acceleration vs. time graph is a horizontal line at some positive value. d. The graph will look like an upward opening 51. Calculate that object\u2019s net displacement over the time parabola. shown. 2.4 Velocity vs. Time Graphs 49. What would the velocity vs. time graph of the object whose position is shown in the graph look like? 90 Chapter 2 \u2022 Test Prep a. 540 m b. 2,520 m c. 2,790 m 5,040 m d. a. 18 m/s b. 84 m/s c. 93 m/s 168 m/s d. 52. What is the object\u2019s average velocity? Extended Response 2.1 Relative Motion, Distance, and Displacement 53. Find the distance traveled from the starting point for each path. Which path has the maximum distance? a. The distance for Path A is 6 m, Path B is 4 m, Path C is 12 m and for Path D is 7 m. The net displacement for Path A is 7 m, Path B is \u20134m, Path C is 8 m and for Path D is \u20135m. Path C has maximum distance and it is equal to 12 meters. b. The distance for Path A is 6 m, Path B is 4 m, Path C is 8 m and for Path D is 7 m. The net displacement for Path A is 6 m, Path B is \u20134m, Path C is 12 m and for Path D is \u20135 m. Path A has maximum distance and it is equal to 6 meters. c. The distance for Path A is 6 m, Path B is 4 m, Path C is 12 m and for Path D is 7 m. The net displacement for Path A is 6 m, Path B is \u20134 m, Path C is 8 m and for Path D is \u20135 m. Path C has maximum distance Access for free at openstax.org. and it is equal to 12 meters. d. The distance for Path A is 6 m, Path B is \u20134 m, Path C is 12 m and for Path D is \u20135 m. The net displacement for Path A is 7 m, Path B is 4 m, Path C is 8 m and for Path D is 7 m. Path A has maximum distance and it is equal to 6 m. 54. Alan starts from his home and walks 1.3 km east to the library. He walks an additional 0.68 km east to a music store. From there, he walks 1.1 km north to a friend\u2019s house and an additional 0.42 km north to a grocery store before he finally returns home along the same path. What is his final displacement and total distance traveled? a. Displacement is 0 km and distance is 7 km. b. Displacement is 0 km and distance is 3.5 km. c. Displacement is 7 km towards west and distance is 7 km. d. Displacement is 3.5 km towards east and distance is 3.5 km. 2.2 Speed and Velocity 55. Two runners start at the same point and jog at a constant speed along a straight path. Runner A starts at time t = 0 s, and Runner B starts at time t = 2.5 s. The runners both reach a distance 64 m from the starting point at time t = 25 s. If the runners continue at the same speeds, how far from the starting point will each be at time t = 45 s? a. Runner A will be m away and Runner B will be m away from the starting point. b. Runner A will be will be c. Runner A will be will be d. Runner A will be will be m away and runner B m away from the starting point. away and Runner B away from the starting point. away and Runner B away from the starting point. 56. A father and his daughter go to the bus stop that is located 75 m from their front door. The father walks in a straight line while his daughter runs along a varied path. Despite the different paths, they both end up at the bus stop at the same time. The father\u2019s average speed is 2.2 m/s, and his daughter\u2019s average speed is 3.5 m/s. (a) How long does it take the father and daughter to reach the bus stop? (b) What was the daughter\u2019s total distance traveled? (c) If the daughter maintained her same average speed and traveled in a straight line like her father, how far beyond the bus stop would she have traveled? a. b. c. d. (a) 21.43 s (b) 75 m (c) 0 m (a) 21.43 s (b) 119 m (c) 44 m (a) 34 s (b) 75 m (c) 0 m (a)", " 34 s (b) 119 m (c) 44 m 2.3 Position vs. Time Graphs 57. What kind of motion would create a position graph like the one shown? a. uniform motion b. any motion that accelerates c. motion that stops and then starts d. motion that has constant velocity 58. What is the average velocity for the whole time period shown in the graph? Chapter 2 \u2022 Test Prep 91 a. b. c. d. 2.4 Velocity vs. Time Graphs 59. Consider the motion of the object whose velocity is charted in the graph. During which points is the object slowing down and speeding up? a. It is slowing down between dand e. It is speeding up between aand dand eand h It is slowing down between aand dand eand h. It is speeding up between dand eand then after i. It is slowing down between dand eand then after h. It is speeding up between aand dand eand h. It is slowing down between aand dand eand h. It is speeding up between dand eand then after i. b. c. d. 60. Divide the graph into approximate sections, and use those sections to graph the velocity vs. time of the object. 92 Chapter 2 \u2022 Test Prep Then calculate the acceleration during each section, and calculate the approximate average velocity. a. Acceleration is zero and average velocity is 1.25 m/s. b. Acceleration is constant with some positive value and average velocity is 1.25 m/s. c. Acceleration is zero and average velocity is 0.25 m/s. d. Acceleration is constant with some positive value and average velocity is 0.25 m/s. Access for free at openstax.org. CHAPTER 3 Acceleration Figure 3.1 A plane slows down as it comes in for landing in St. Maarten. Its acceleration is in the opposite direction of its velocity. (Steve Conry, Flickr) Chapter Outline 3.1 Acceleration 3.2 Representing Acceleration with Equations and Graphs You may have heard the term accelerator, referring to the gas pedal in a car. When the gas pedal is pushed INTRODUCTION down, the flow of gasoline to the engine increases, which increases the car\u2019s velocity. Pushing on the gas pedal results in acceleration because the velocity of the car increases, and acceleration is defined as a change in velocity. You need two quantities to define velocity: a speed and a direction. Changing either of these quantities, or both together, changes the velocity. You may be surprised to learn that pushing on the brake pedal or turning the steering wheel also causes acceleration. The first reduces the speedand so changes the velocity, and the second changes the directionand also changes the velocity. In fact, any change in velocity\u2014whether positive, negative, directional, or any combination of these\u2014is called an acceleration in physics. The plane in the picture is said to be accelerating because its velocity is decreasing as it prepares to land. To begin our study of acceleration, we need to have a clear understanding of what acceleration means. 3.1 Acceleration Section Learning Objectives By the end of this section, you will be able to do the following: \u2022 Explain acceleration and determine the direction and magnitude of acceleration in one dimension \u2022 Analyze motion in one dimension using kinematic equations and graphic representations 94 Chapter 3 \u2022 Acceleration Section Key Terms average acceleration instantaneous acceleration negative acceleration Defining Acceleration Throughout this chapter we will use the following terms: time, displacement, velocity, and acceleration. Recall that each of these terms has a designated variable and SI unit of measurement as follows: \u2022 Time: t, measured in seconds (s) \u2022 Displacement: \u0394d, measured in meters (m) \u2022 Velocity: v, measured in meters per second (m/s) \u2022 Acceleration: a, measured in meters per second per second (m/s2, also called meters per second squared) \u2022 Also note the following: \u25e6 \u0394 means change in \u25e6 The subscript 0 refers to an initial value; sometimes subscript i is instead used to refer to initial value. \u25e6 The subscript f refers to final value \u25e6 A bar over a symbol, such as, means average Acceleration is the change in velocity divided by a period of time during which the change occurs. The SI units of velocity are m/s and the SI units for time are s, so the SI units for acceleration are m/s2. Average acceleration is given by Average acceleration is distinguished from instantaneous acceleration, which is acceleration at a specific instant in time. The magnitude of acceleration is often not constant over time. For example, runners in a race accelerate at a greater rate in the first second of a race than during the following seconds. You do not need to know all the instantaneous accelerations at all times to calculate average acceleration. All you need to know is the change in velocity (i.e.,", " the final velocity minus the initial velocity) and the change in time (i.e., the final time minus the initial time), as shown in the formula. Note that the average acceleration can be positive, negative, or zero. A negative acceleration is simply an acceleration in the negative direction. Keep in mind that although acceleration points in the same direction as the changein velocity, it is not always in the direction of the velocity itself. When an object slows down, its acceleration is opposite to the direction of its velocity. In everyday language, this is called deceleration; but in physics, it is acceleration\u2014whose direction happens to be opposite that of the velocity. For now, let us assume that motion to the right along the x-axis is positiveand motion to the left is negative. Figure 3.2 shows a car with positive acceleration in (a) and negative acceleration in (b). The arrows represent vectors showing both direction and magnitude of velocity and acceleration. Figure 3.2 The car is speeding up in (a) and slowing down in (b). Access for free at openstax.org. 3.1 \u2022 Acceleration 95 Velocity and acceleration are both vector quantities. Recall that vectors have both magnitude and direction. An object traveling at a constant velocity\u2014therefore having no acceleration\u2014does accelerate if it changes direction. So, turning the steering wheel of a moving car makes the car accelerate because the velocity changes direction. Virtual Physics The Moving Man With this animation in, you can produce both variations of acceleration and velocity shown in Figure 3.2, plus a few more variations. Vary the velocity and acceleration by sliding the red and green markers along the scales. Keeping the velocity marker near zero will make the effect of acceleration more obvious. Try changing acceleration from positive to negative while the man is moving. We will come back to this animation and look at the Chartsview when we study graphical representation of motion. Click to view content (https://archive.cnx.org/specials/e2ca52af-8c6b-450e-ac2f-9300b38e8739/moving-man/) GRASP CHECK Figure 3.3 Which part, (a) or (b), is represented when the velocity vector is on the positive side of the scale and the acceleration vector is set on the negative side of the scale? What does the car\u2019s motion look like for the given scenario? a. Part (a). The car is slowing down because the acceleration and the velocity vectors are acting in the opposite direction. b. Part (a). The car is speeding up because the acceleration and the velocity vectors are acting in the same direction. c. Part (b). The car is slowing down because the acceleration and velocity vectors are acting in the opposite directions. d. Part (b). The car is speeding up because the acceleration and the velocity vectors are acting in the same direction. Calculating Average Acceleration Look back at the equation for average acceleration. You can see that the calculation of average acceleration involves three values: change in time, (\u0394t); change in velocity, (\u0394v); and acceleration (a). Change in time is often stated as a time interval, and change in velocity can often be calculated by subtracting the initial velocity from the final velocity. Average acceleration is then simply change in velocity divided by change in time. Before you begin calculating, be sure that all distances and times have been converted to meters and seconds. Look at these examples of acceleration of a subway train. 96 Chapter 3 \u2022 Acceleration WORKED EXAMPLE An Accelerating Subway Train A subway train accelerates from rest to 30.0 km/h in 20.0 s. What is the average acceleration during that time interval? Strategy Start by making a simple sketch. Figure 3.4 This problem involves four steps: 1. Convert to units of meters and seconds. 2. Determine the change in velocity. 3. Determine the change in time. 4. Use these values to calculate the average acceleration. Solution 1. Identify the knowns. Be sure to read the problem for given information, which may not looklike numbers. When the problem states that the train starts from rest, you can write down that the initial velocity is 0 m/s. Therefore, v0 = 0; vf = 30.0 km/h; and \u0394t= 20.0 s. 2. Convert the units. 3. Calculate change in velocity, velocity is to the right. 3.1 where the plus sign means the change in 4. We know \u0394t, so all we have to do is insert the known values into the formula for average acceleration. Discussion The plus sign in the answer means that acceleration is to the right. This is a reasonable conclusion because the train starts from rest and ends up with a velocity directed to the right (i.e., positive). So, acceleration is in the same direction as the changein velocity, as it should be. 3.2", " WORKED EXAMPLE An Accelerating Subway Train Now, suppose that at the end of its trip, the train slows to a stop in 8.00 s from a speed of 30.0 km/h. What is its average acceleration during this time? Strategy Again, make a simple sketch. Access for free at openstax.org. 3.1 \u2022 Acceleration 97 Figure 3.5 In this case, the train is decelerating and its acceleration is negative because it is pointing to the left. As in the previous example, we must find the change in velocity and change in time, then solve for acceleration. Identify the knowns: v0 = 30.0 km/h; vf = 0; and \u0394t= 8.00 s. Solution 1. 2. Convert the units. From the first problem, we know that 30.0 km/h = 8.333 m/s. 3. Calculate change in velocity, change in velocity points to the left. where the minus sign means that the 4. We know \u0394t= 8.00 s, so all we have to do is insert the known values into the equation for average acceleration. 3.3 Discussion The minus sign indicates that acceleration is to the left. This is reasonable because the train initially has a positive velocity in this problem, and a negative acceleration would reduce the velocity. Again, acceleration is in the same direction as the changein velocity, which is negative in this case. This acceleration can be called a deceleration because it has a direction opposite to the velocity. TIPS FOR SUCCESS \u2022 \u2022 It is easier to get plus and minus signs correct if you always assume that motion is away from zero and toward positive values on the x-axis. This way valways starts off being positive and points to the right. If speed is increasing, then acceleration is positive and also points to the right. If speed is decreasing, then acceleration is negative and points to the left. It is a good idea to carry two extra significant figures from step-to-step when making calculations. Do not round off with each step. When you arrive at the final answer, apply the rules of significant figures for the operations you carried out and round to the correct number of digits. Sometimes this will make your answer slightly more accurate. Practice Problems 1. A cheetah can accelerate from rest to a speed of in. What is its acceleration? a. b. c. d. 2. A women backs her car out of her garage with an acceleration of. How long does it take her to reach a speed of? a. b. c. d. 98 Chapter 3 \u2022 Acceleration WATCH PHYSICS Acceleration This video shows the basic calculation of acceleration and some useful unit conversions. Click to view content (https://www.khanacademy.org/embed_video?v=FOkQszg1-j8) GRASP CHECK Why is acceleration a vector quantity? a. b. c. d. It is a vector quantity because it has magnitude as well as direction. It is a vector quantity because it has magnitude but no direction. It is a vector quantity because it is calculated from distance and time. It is a vector quantity because it is calculated from speed and time. GRASP CHECK What will be the change in velocity each second if acceleration is 10 m/s/s? a. An acceleration of b. An acceleration of c. An acceleration of d. An acceleration of means that every second, the velocity increases by means that every second, the velocity decreases by means that every means that every, the velocity increases by, the velocity decreases by.... Snap Lab Measure the Acceleration of a Bicycle on a Slope In this lab you will take measurements to determine if the acceleration of a moving bicycle is constant. If the acceleration is constant, then the following relationships hold:, then and If You will work in pairs to measure and record data for a bicycle coasting down an incline on a smooth, gentle slope. The data will consist of distances traveled and elapsed times. \u2022 Find an open area to minimize the risk of injury during this lab. stopwatch \u2022 \u2022 measuring tape \u2022 bicycle 1. Find a gentle, paved slope, such as an incline on a bike path. The more gentle the slope, the more accurate your data will likely be. 2. Mark uniform distances along the slope, such as 5 m, 10 m, etc. 3. Determine the following roles: the bike rider, the timer, and the recorder. The recorder should create a data table to collect the distance and time data. 4. Have the rider at the starting point at rest on the bike. When the timer calls Start, the timer starts the stopwatch and the rider begins coasting down the slope on the bike without pedaling. 5. Have the timer call out the elapsed times as the bike passes each marked point. The recorder should record the times in the data table. It", " may be necessary to repeat the process to practice roles and make necessary adjustments. 6. Once acceptable data has been recorded, switch roles. Repeat Steps 3\u20135 to collect a second set of data. 7. Switch roles again to collect a third set of data. 8. Calculate average acceleration for each set of distance-time data. If your result for is not the same for different pairs of \u0394vand \u0394t, then acceleration is not constant. Interpret your results. 9. Access for free at openstax.org. 3.2 \u2022 Representing Acceleration with Equations and Graphs 99 GRASP CHECK If you graph the average velocity (y-axis) vs. the elapsed time (x-axis), what would the graph look like if acceleration is uniform? a. a horizontal line on the graph b. a diagonal line on the graph c. an upward-facing parabola on the graph d. a downward-facing parabola on the graph Check Your Understanding 3. What are three ways an object can accelerate? a. By speeding up, maintaining constant velocity, or changing direction b. By speeding up, slowing down, or changing direction c. By maintaining constant velocity, slowing down, or changing direction d. By speeding up, slowing down, or maintaining constant velocity 4. What is the difference between average acceleration and instantaneous acceleration? a. Average acceleration is the change in displacement divided by the elapsed time; instantaneous acceleration is the acceleration at a given point in time. b. Average acceleration is acceleration at a given point in time; instantaneous acceleration is the change in displacement divided by the elapsed time. c. Average acceleration is the change in velocity divided by the elapsed time; instantaneous acceleration is acceleration at a given point in time. d. Average acceleration is acceleration at a given point in time; instantaneous acceleration is the change in velocity divided by the elapsed time. 5. What is the rate of change of velocity called? a. Time b. Displacement c. Velocity d. Acceleration 3.2 Representing Acceleration with Equations and Graphs Section Learning Objectives By the end of this section, you will be able to do the following: \u2022 Explain the kinematic equations related to acceleration and illustrate them with graphs \u2022 Apply the kinematic equations and related graphs to problems involving acceleration Section Key Terms acceleration due to gravity kinematic equations uniform acceleration How the Kinematic Equations are Related to Acceleration We are studying concepts related to motion: time, displacement, velocity, and especially acceleration. We are only concerned with motion in one dimension. The kinematic equations apply to conditions of constant acceleration and show how these concepts are related. Constant acceleration is acceleration that does not change over time. The first kinematic equation relates displacement d, average velocity, and time t. 3.4 The initial displacement is often 0, in which case the equation can be written as 100 Chapter 3 \u2022 Acceleration This equation says that average velocity is displacement per unit time. We will express velocity in meters per second. If we graph displacement versus time, as in Figure 3.6, the slope will be the velocity. Whenever a rate, such as velocity, is represented graphically, time is usually taken to be the independent variable and is plotted along the xaxis. Figure 3.6 The slope of displacement versus time is velocity. The second kinematic equation, another expression for average velocity divided by two. is simply the initial velocity plus the final velocity Now we come to our main focus of this chapter; namely, the kinematic equations that describe motion with constant acceleration. In the third kinematic equation, acceleration is the rate at which velocity increases, so velocity at any point equals initial velocity plus acceleration multiplied by time 3.6 Note that this third kinematic equation does not have displacement in it. Therefore, if you do not know the displacement and are not trying to solve for a displacement, this equation might be a good one to use. The third kinematic equation is also represented by the graph in Figure 3.7. 3.5 The fourth kinematic equation shows how displacement is related to acceleration Figure 3.7 The slope of velocity versus time is acceleration. When starting at the origin, and, when starting from rest,, in which case the equation can be written as 3.7 Access for free at openstax.org. 3.2 \u2022 Representing Acceleration with Equations and Graphs 101 This equation tells us that, for constant acceleration, the slope of a plot of 2dversus t2 is acceleration, as shown in Figure 3.8. Figure 3.8 When acceleration is constant, the slope of 2dversus t2 gives the acceleration. The fifth kinematic equation relates velocity, acceleration, and displacement 3.8 This equation is useful for when we do not know, or do not need to know, the time. When starting from rest, the fifth equation simplifies to According to this equation, a graph of velocity squared versus twice the", " displacement will have a slope equal to acceleration. Figure 3.9 Note that, in reality, knowns and unknowns will vary. Sometimes you will want to rearrange a kinematic equation so that the knowns are the values on the axes and the unknown is the slope. Sometimes the intercept will not be at the origin (0,0). This will happen when v0 or d0 is not zero. This will be the case when the object of interest is already in motion, or the motion begins at some point other than at the origin of the coordinate system. Virtual Physics The Moving Man (Part 2) Look at the Moving Man simulation again and this time use the Chartsview. Again, vary the velocity and acceleration by sliding the red and green markers along the scales. Keeping the velocity marker near zero will make the effect of acceleration more obvious. Observe how the graphs of position, velocity, and acceleration vary with time. Note which are linear plots and which are not. Click to view content (https://archive.cnx.org/specials/e2ca52af-8c6b-450e-ac2f-9300b38e8739/moving-man/) GRASP CHECK On a velocity versus time plot, what does the slope represent? a. Acceleration 102 Chapter 3 \u2022 Acceleration b. Displacement c. Distance covered d. Instantaneous velocity GRASP CHECK On a position versus time plot, what does the slope represent? a. Acceleration b. Displacement c. Distance covered d. Instantaneous velocity The kinematic equations are applicable when you have constant acceleration., or, or 1. 2. 3. 4. 5. when d0 = 0 when v0 = 0, or, or when d0 = 0 and v0 = 0 when d0 = 0 and v0 = 0 Applying Kinematic Equations to Situations of Constant Acceleration Problem-solving skills are essential to success in a science and life in general. The ability to apply broad physical principles, which are often represented by equations, to specific situations is a very powerful form of knowledge. It is much more powerful than memorizing a list of facts. Analytical skills and problem-solving abilities can be applied to new situations, whereas a list of facts cannot be made long enough to contain every possible circumstance. Essential analytical skills will be developed by solving problems in this text and will be useful for understanding physics and science in general throughout your life. Problem-Solving Steps While no single step-by-step method works for every problem, the following general procedures facilitate problem solving and make the answers more meaningful. A certain amount of creativity and insight are required as well. 1. Examine the situation to determine which physical principles are involved. It is vital to draw a simple sketch at the outset. Decide which direction is positive and note that on your sketch. 2. Identify the knowns: Make a list of what information is given or can be inferred from the problem statement.Remember, not all given information will be in the form of a number with units in the problem. If something starts from rest, we know the initial velocity is zero. If something stops, we know the final velocity is zero. 3. Identify the unknowns: Decide exactly what needs to be determined in the problem. 4. Find an equation or set of equations that can help you solve the problem.Your list of knowns and unknowns can help here. For example, if time is not needed or not given, then the fifth kinematic equation, which does not include time, could be useful. 5. Insert the knowns along with their units into the appropriate equation and obtain numerical solutions complete with units.This step produces the numerical answer; it also provides a check on units that can help you find errors. If the units of the answer are incorrect, then an error has been made. 6. Check the answer to see if it is reasonable: Does it make sense?This final step is extremely important because the goal of physics is to accurately describe nature. To see if the answer is reasonable, check its magnitude, its sign, and its units. Are the significant figures correct? Summary of Problem Solving \u2022 Determine the knowns and unknowns. \u2022 Find an equation that expresses the unknown in terms of the knowns. More than one unknown means more than one equation is needed. \u2022 Solve the equation or equations. Access for free at openstax.org. 3.2 \u2022 Representing Acceleration with Equations and Graphs 103 \u2022 Be sure units and significant figures are correct. \u2022 Check whether the answer is reasonable. FUN IN PHYSICS Drag Racing Figure 3.10 Smoke rises from the tires of a dragster at the beginning of a drag race. (Lt. Col. William Thurmond. Photo courtesy of U.S. Army.) The object of the sport of drag racing is acceleration. Period! The races take", " place from a standing start on a straight onequarter-mile (402 m) track. Usually two cars race side by side, and the winner is the driver who gets the car past the quarter-mile point first. At the finish line, the cars may be going more than 300 miles per hour (134 m/s). The driver then deploys a parachute to bring the car to a stop because it is unsafe to brake at such high speeds. The cars, called dragsters, are capable of accelerating at 26 m/s2. By comparison, a typical sports car that is available to the general public can accelerate at about 5 m/s2. Several measurements are taken during each drag race: \u2022 Reaction time is the time between the starting signal and when the front of the car crosses the starting line. \u2022 Elapsed time is the time from when the vehicle crosses the starting line to when it crosses the finish line. The record is a little over 3 s. \u2022 Speed is the average speed during the last 20 m before the finish line. The record is a little under 400 mph. The video shows a race between two dragsters powered by jet engines. The actual race lasts about four seconds and is near the end of the video (https://openstax.org/l/28dragsters). GRASP CHECK A dragster crosses the finish line with a velocity of start to finish, what was its average velocity for the race? a. b. c. d.. Assuming the vehicle maintained a constant acceleration from WORKED EXAMPLE Acceleration of a Dragster The time it takes for a dragster to cross the finish line is unknown. The dragster accelerates from rest at 26 m/s2 for a quarter mile (0.250 mi). What is the final velocity of the dragster? Strategy The equation displacement, without involving the time. is ideally suited to this task because it gives the velocity from acceleration and 104 Chapter 3 \u2022 Acceleration Solution 1. Convert miles to meters. 3.9 2. 3. Identify the known values. We know that v0 = 0 since the dragster starts from rest, and we know that the distance traveled, d\u2212 d0 is 402 m. Finally, the acceleration is constant at a= 26.0 m/s2. Insert the knowns into the equation and solve for v. 3.10 Taking the square root gives us Discussion 145 m/s is about 522 km/hour or about 324 mi/h, but even this breakneck speed is short of the record for the quarter mile. Also, note that a square root has two values. We took the positive value because we know that the velocity must be in the same direction as the acceleration for the answer to make physical sense. An examination of the equation physical quantities: can produce further insights into the general relationships among \u2022 The final velocity depends on the magnitude of the acceleration and the distance over which it applies. \u2022 For a given acceleration, a car that is going twice as fast does not stop in twice the distance\u2014it goes much further before it stops. This is why, for example, we have reduced speed zones near schools. Practice Problems 6. Dragsters can reach a top speed of 145 m/s in only 4.45 s. Calculate the average acceleration for such a dragster. a. \u221232.6 m/s2 b. 0 m/s2 c. d. 32.6 m/s2 145 m/s2 7. An Olympic-class sprinter starts a race with an acceleration of 4.50 m/s2. Assuming she can maintain that acceleration, what is her speed 2.40 s later? a. 4.50 m/s 10.8 m/s b. c. 19.6 m/s d. 44.1 m/s Constant Acceleration In many cases, acceleration is not uniform because the force acting on the accelerating object is not constant over time. A situation that gives constant acceleration is the acceleration of falling objects. When air resistance is not a factor, all objects near Earth\u2019s surface fall with an acceleration of about 9.80 m/s2. Although this value decreases slightly with increasing altitude, it may be assumed to be essentially constant. The value of 9.80 m/s2 is labeled gand is referred to as acceleration due to gravity. Gravity is the force that causes nonsupported objects to accelerate downward\u2014or, more precisely, toward the center of Earth. The magnitude of this force is called the weight of the object and is given by mgwhere mis the mass of the object (in kg). In places other than on Earth, such as the Moon or on other planets, gis not 9.80 m/s2, but takes on other values. When using gfor the acceleration ain a kinematic equation, it is usually given a negative sign because the acceleration due to gravity is downward. Access for free at openstax.org. 3", ".2 \u2022 Representing Acceleration with Equations and Graphs 105 WORK IN PHYSICS Effects of Rapid Acceleration Figure 3.11 Astronauts train using G Force Simulators. (NASA) When in a vehicle that accelerates rapidly, you experience a force on your entire body that accelerates your body. You feel this force in automobiles and slightly more on amusement park rides. For example, when you ride in a car that turns, the car applies a force on your body to make you accelerate in the direction in which the car is turning. If enough force is applied, you will accelerate at 9.80 m/s2. This is the same as the acceleration due to gravity, so this force is called one G. One G is the force required to accelerate an object at the acceleration due to gravity at Earth\u2019s surface. Thus, one G for a paper cup is much less than one G for an elephant, because the elephant is much more massive and requires a greater force to make it accelerate at 9.80 m/s2. For a person, a G of about 4 is so strong that his or her face will distort as the bones accelerate forward through the loose flesh. Other symptoms at extremely high Gs include changes in vision, loss of consciousness, and even death. The space shuttle produces about 3 Gs during takeoff and reentry. Some roller coasters and dragsters produce forces of around 4 Gs for their occupants. A fighter jet can produce up to 12 Gs during a sharp turn. Astronauts and fighter pilots must undergo G-force training in simulators. The video (https://www.youtube.com/ watch?v=n-8QHOUWECU) shows the experience of several people undergoing this training. People, such as astronauts, who work with G forces must also be trained to experience zero G\u2014also called free fall or weightlessness\u2014which can cause queasiness. NASA has an aircraft that allows it occupants to experience about 25 s of free fall. The aircraft is nicknamed the Vomit Comet. GRASP CHECK A common way to describe acceleration is to express it in multiples of g, Earth's gravitational acceleration. If a dragster accelerates at a rate of 39.2 m/s2, how many g's does the driver experience? a. 1.5 g b. 4.0 g c. 10.5 g d. 24.5 g WORKED EXAMPLE Falling Objects A person standing on the edge of a high cliff throws a rock straight up with an initial velocity v0 of 13 m/s. (a) Calculate the position and velocity of the rock at 1.00, 2.00, and 3.00 seconds after it is thrown. Ignore the effect of air resistance. Strategy Sketch the initial velocity and acceleration vectors and the axes. 106 Chapter 3 \u2022 Acceleration Figure 3.12 Initial conditions for rock thrown straight up. List the knowns: time t= 1.00 s, 2.00 s, and 3.00 s; initial velocity v0 = 13 m/s; acceleration a= g= \u20139.80 m/s2; and position y0 = 0 m List the unknowns: y1, y2, and y3; v1, v2, and v3\u2014where 1, 2, 3 refer to times 1.00 s, 2.00 s, and 3.00 s Choose the equations. 3.11 3.12 These equations describe the unknowns in terms of knowns only. Solution Discussion The first two positive values for y show that the rock is still above the edge of the cliff, and the third negative y value shows that it has passed the starting point and is below the cliff. Remember that we set upto be positive. Any position with a positive value is above the cliff, and any velocity with a positive value is an upward velocity. The first value for v is positive, so the rock is still on the way up. The second and third values for v are negative, so the rock is on its way down. (b) Make graphs of position versus time, velocity versus time, and acceleration versus time. Use increments of 0.5 s in your graphs. Strategy Time is customarily plotted on the x-axis because it is the independent variable. Position versus time will not be linear, so calculate points for 0.50 s, 1.50 s, and 2.50 s. This will give a curve closer to the true, smooth shape. Solution The three graphs are shown in Figure 3.13. Access for free at openstax.org. 3.2 \u2022 Representing Acceleration with Equations and Graphs 107 Figure 3.13 Discussion \u2022 yvs. tdoes notrepresent the two-dimensional parabolic path of a trajectory. The path of the rock is straight up and straight down. The slope of a line tangent to the curve at any point on the curve equals the velocity at that point", "\u2014i.e., the instantaneous velocity. 108 Chapter 3 \u2022 Acceleration \u2022 Note that the vvs. tline crosses the vertical axis at the initial velocity and crosses the horizontal axis at the time when the rock changes direction and begins to fall back to Earth. This plot is linear because acceleration is constant. \u2022 The avs. tplot also shows that acceleration is constant; that is, it does not change with time. Practice Problems 8. A cliff diver pushes off horizontally from a cliff and lands in the ocean later. How fast was he going when he entered the water? a. b. c. d. 9. A girl drops a pebble from a high cliff into a lake far below. She sees the splash of the pebble hitting the water later. How fast was the pebble going when it hit the water? a. b. c. d. Check Your Understanding 10. Identify the four variables found in the kinematic equations. a. Displacement, Force, Mass, and Time b. Acceleration, Displacement, Time, and Velocity c. Final Velocity, Force, Initial Velocity, and Mass d. Acceleration, Final Velocity, Force, and Initial Velocity 11. Which of the following steps is always required to solve a kinematics problem? a. Find the force acting on the body. b. Find the acceleration of a body. c. Find the initial velocity of a body. d. Find a suitable kinematic equation and then solve for the unknown quantity. 12. Which of the following provides a correct answer for a problem that can be solved using the kinematic equations? for for. The body\u2019s final velocity is. The body\u2019s final velocity is.. a. A body starts from rest and accelerates at b. A body starts from rest and accelerates at c. A body with a mass of d. A body with a mass of is acted upon by a force of is acted upon by a force of. The acceleration of the body is. The acceleration of the body is.. Access for free at openstax.org. Chapter 3 \u2022 Key Terms 109 KEY TERMS acceleration due to gravity acceleration of an object that is subject only to the force of gravity; near Earth\u2019s surface this acceleration is 9.80 m/s2 average acceleration change in velocity divided by the time interval over which it changed instantaneous acceleration rate of change of velocity at a specific instant in time kinematic equations the five equations that describe motion in terms of time, displacement, velocity, and acceleration constant acceleration acceleration that does not change negative acceleration acceleration in the negative direction with respect to time SECTION SUMMARY 3.1 Acceleration \u2022 Acceleration is the rate of change of velocity and may be negative or positive. \u2022 Average acceleration is expressed in m/s2 and, in one dimension, can be calculated using 3.2 Representing Acceleration with Equations and Graphs \u2022 The kinematic equations show how time, displacement, KEY EQUATIONS 3.1 Acceleration Average acceleration \u2022 velocity, and acceleration are related for objects in motion. In general, kinematic problems can be solved by identifying the kinematic equation that expresses the unknown in terms of the knowns. \u2022 Displacement, velocity, and acceleration may be displayed graphically versus time. Average velocity Velocity, or when v0 = 0 3.2 Representing Acceleration with Equations and Graphs Average velocity, or when d0 = 0 Displacement when d0 = 0 and v0 = 0 Acceleration when d0 = 0 and v0 = 0, or, or CHAPTER REVIEW Concept Items 3.1 Acceleration 1. How can you use the definition of acceleration to explain the units in which acceleration is measured? a. Acceleration is the rate of change of velocity. Therefore, its unit is m/s2. b. Acceleration is the rate of change of displacement. Therefore, its unit is m/s. c. Acceleration is the rate of change of velocity. Therefore, its unit is m2/s. d. Acceleration is the rate of change of displacement. Therefore, its unit is m2/s. 2. What are the SI units of acceleration? a. b. c. d. 3. Which of the following statements explains why a racecar going around a curve is accelerating, even if the speed is constant? a. The car is accelerating because the magnitude as well as the direction of velocity is changing. b. The car is accelerating because the magnitude of velocity is changing. c. The car is accelerating because the direction of velocity is changing. 110 Chapter 3 \u2022 Chapter Review d. The car is accelerating because neither the magnitude nor the direction of velocity is changing. 3.2 Representing Acceleration with Equations and Graphs 4. A student calculated the final velocity of a train that decelerated from 30.5 m/s and got an answer of \u221243.34 m/", " s. Which of the following might indicate that he made a mistake in his calculation? a. The sign of the final velocity is wrong. b. The magnitude of the answer is too small. c. There are too few significant digits in the answer. d. The units in the initial velocity are incorrect. 5. Create your own kinematics problem. Then, create a flow chart showing the steps someone would need to take to solve the problem. a. Acceleration b. Distance c. Displacement d. Force 6. Which kinematic equation would you use to find the after she jumps from a plane velocity of a skydiver and before she opens her parachute? Assume the positive direction is downward. a. b. c. d. Critical Thinking Items 3.1 Acceleration 7. Imagine that a car is traveling from your left to your right at a constant velocity. Which two actions could the driver take that may be represented as (a) a velocity vector and an acceleration vector both pointing to the right and then (b) changing so the velocity vector points to the right and the acceleration vector points to the left? a. (a) Push down on the accelerator and then (b) push down again on the accelerator a second time. (a) Push down on the accelerator and then (b) push down on the brakes. (a) Push down on the brakes and then (b) push down on the brakes a second time. (a) Push down on the brakes and then (b) push down on the accelerator. b. c. d. 8. A motorcycle moving at a constant velocity suddenly accelerates at a rate of to a speed of in. What was the initial speed of the motorcycle? a. b. c. d. 3.2 Representing Acceleration with Equations and Graphs 9. A student is asked to solve a problem: An object falls from a height for 2.0 s, at which point it is still 60 m above the ground. What will be the velocity of the object when it hits the ground? Which of the following provides the correct order of kinematic equations that can be used to solve the problem? a. First use then use Access for free at openstax.org. b. First use then use c. First use then use d. First use then use 10. Skydivers are affected by acceleration due to gravity and by air resistance. Eventually, they reach a speed where the force of gravity is almost equal to the force of air resistance. As they approach that point, their acceleration decreases in magnitude to near zero. Part A. Describe the shape of the graph of the magnitude of the acceleration versus time for a falling skydiver. Part B. Describe the shape of the graph of the magnitude of the velocity versus time for a falling skydiver. Part C. Describe the shape of the graph of the magnitude of the displacement versus time for a falling skydiver. a. Part A. Begins with a nonzero y-intercept with a downward slope that levels off at zero; Part B. Begins at zero with an upward slope that decreases in magnitude until the curve levels off; Part C. Begins at zero with an upward slope that increases in magnitude until it becomes a positive constant b. Part A. Begins with a nonzero y-intercept with an upward slope that levels off at zero; Part B. Begins at zero with an upward slope that decreases in magnitude until the curve levels off; Part C. Begins at zero with an upward slope that increases in magnitude until it becomes a positive constant c. Part A. Begins with a nonzero y-intercept with a downward slope that levels off at zero; Part B. Begins at zero with a downward slope that Chapter 3 \u2022 Chapter Review 111 decreases in magnitude until the curve levels off; Part C. Begins at zero with an upward slope that increases in magnitude until it becomes a positive constant d. Part A. Begins with a nonzero y-intercept with an upward slope that levels off at zero; Part B. Begins at zero with a downward slope that decreases in magnitude until the curve levels off; Part C. Begins at zero with an upward slope that increases in magnitude until it becomes a positive constant 11. Which graph in the previous problem has a positive slope? a. Displacement versus time only b. Acceleration versus time and velocity versus time c. Velocity versus time and displacement versus time d. Acceleration versus time and displacement versus time Problems 3.1 Acceleration 12. The driver of a sports car traveling at 10.0 m/s steps down hard on the accelerator for 5.0 s and the velocity increases to 30.0 m/s. What was the average acceleration of the car during the 5.0 s time interval? a. \u20131.0 \u00d7 102 m/s2 b. \u20134.0 m/s2 c. 4.0 m/s2 d. 1.0 \u00d7 102 m/s2", " 13. A girl rolls a basketball across a basketball court. The ball slowly decelerates at a rate of \u22120.20 m/s2. If the initial velocity was 2.0 m/s and the ball rolled to a stop at 5.0 sec after 12:00 p.m., at what time did she start the ball rolling? a. 0.1 seconds before noon b. 0.1 seconds after noon 5 seconds before noon c. 5 seconds after noon d. Performance Task 3.2 Representing Acceleration with Equations and Graphs 16. Design an experiment to measure displacement and elapsed time. Use the data to calculate final velocity, average velocity, acceleration, and acceleration. Materials \u2022 a small marble or ball bearing \u2022 a garden hose \u2022 a measuring tape \u2022 a stopwatch or stopwatch software download \u2022 a sloping driveway or lawn as long as the garden 3.2 Representing Acceleration with Equations and Graphs 14. A swan on a lake gets airborne by flapping its wings and running on top of the water. If the swan must reach a velocity of 6.00 m/s to take off and it accelerates from rest at an average rate of 0.350 m/s2, how far will it travel before becoming airborne? a. \u22128.60 m b. 8.60 m c. \u221251.4 m d. 51.4 m 15. A swimmer bounces straight up from a diving board and falls feet first into a pool. She starts with a velocity of above the pool. and her takeoff point is How long are her feet in the air? a. b. c. d. e. 1.28 s hose with a level area beyond (a) How would you use the garden hose, stopwatch, marble, measuring tape, and slope to measure displacement and elapsed time? Hint\u2014The marble is the accelerating object, and the length of the hose is total displacement. (b) How would you use the displacement and time data to calculate velocity, average velocity, and acceleration? Which kinematic equations would you use? (c) How would you use the materials, the measured and calculated data, and the flat area below the slope to determine the negative acceleration? What would you measure, and which kinematic equation would you use? 112 Chapter 3 \u2022 Test Prep TEST PREP Multiple Choice 3.1 Acceleration 17. Which variable represents displacement? a. a b. d c. t d. v 18. If a velocity increases from 0 to 20 m/s in 10 s, what is the average acceleration? a. 0.5 m/s2 b. 2 m/s2 10 m/s2 c. 30 m/s2 d. 3.2 Representing Acceleration with Equations and Graphs 19. For the motion of a falling object, which graphs are Short Answer 3.1 Acceleration 21. True or False\u2014The vector for a negative acceleration points in the opposite direction when compared to the vector for a positive acceleration. a. True b. False 22. If a car decelerates from to in, what is a. b. c. d.? -5 m/s -1 m/s 1 m/s 5 m/s 23. How is the vector arrow representing an acceleration of magnitude 3 m/s2 different from the vector arrow representing a negative acceleration of magnitude 3 m/ s2? a. They point in the same direction. b. They are perpendicular, forming a 90\u00b0 angle between each other. c. They point in opposite directions. d. They are perpendicular, forming a 270\u00b0 angle between each other. 24. How long does it take to accelerate from 8.0 m/s to 20.0 m/s at a rate of acceleration of 3.0 m/s2? a. 0.25 s b. 4.0 s c. 9.33 s Access for free at openstax.org. straight lines? a. Acceleration versus time only b. Displacement versus time only c. Displacement versus time and acceleration versus time d. Velocity versus time and acceleration versus time 20. A bullet in a gun is accelerated from the firing chamber to the end of the barrel at an average rate of 6.30\u00d7105 m/ s2 for 8.10\u00d710\u22124 s. What is the bullet\u2019s final velocity when it leaves the barrel, commonly known as the muzzle velocity? a. b. c. d. 7.79 m/s 51.0 m/s 510 m/s 1020 m/s d. 36 s 3.2 Representing Acceleration with Equations and Graphs 25. If a plot of displacement versus time is linear, what can be said about the acceleration? a. Acceleration is 0. b. Acceleration is a non-zero constant. c. Acceleration is positive. d. Acceleration is negative. 26. True or False", ": \u2014The image shows a velocity vs. time graph for a jet car. If you take the slope at any point on the graph, the jet car\u2019s acceleration will be 5.0 m/s2. a. True b. False 27. When plotted on the blank plots, which answer choice would show the motion of an object that has uniformly accelerated from 2 m/s to 8 m/s in 3 s? Chapter 3 \u2022 Test Prep 113 the plot on the right shows a line from (0,2) to (3,2). d. The plot on the left shows a line from (0,8) to (3,2) while the plot on the right shows a line from (0,3) to (3,3). 28. When is a plot of velocity versus time a straight line and a. The plot on the left shows a line from (0,2) to (3,8) while the plot on the right shows a line from (0,2) to (3,2). b. The plot on the left shows a line from (0,2) to (3,8) while the plot on the right shows a line from (0,3) to (3,3). c. The plot on the left shows a line from (0,8) to (3,2) while b. c. d. when is it a curved line? a. It is a straight line when acceleration is changing and is a curved line when acceleration is constant. It is a straight line when acceleration is constant and is a curved line when acceleration is changing. It is a straight line when velocity is constant and is a curved line when velocity is changing. It is a straight line when velocity is changing and is a curved line when velocity is constant. Extended Response 3.1 Acceleration 29. A test car carrying a crash test dummy accelerates from and then crashes into a brick wall. Describe to the direction of the initial acceleration vector and compare the initial acceleration vector\u2019s magnitude with respect to the acceleration magnitude at the moment of the crash. a. The direction of the initial acceleration vector will point towards the wall, and its magnitude will be less than the acceleration vector of the crash. b. The direction of the initial acceleration vector will point away from the wall, and its magnitude will be less than the vector of the crash. c. The direction of the initial acceleration vector will point towards the wall, and its magnitude will be more than the acceleration vector of the crash. d. The direction of the initial acceleration vector will point away from the wall, and its magnitude will be more than the acceleration vector of the crash. 30. A car accelerates from rest at a stop sign at a rate of 3.0 m/s2 to a speed of 21.0 m/s, and then immediately begins to decelerate to a stop at the next stop sign at a rate of 4.0 m/s2. How long did it take the car to travel from the first stop sign to the second stop sign? Show your work. a. b. c. d. 1.7 seconds 5.3 seconds 7.0 seconds 12 seconds 3.2 Representing Acceleration with Equations and Graphs 31. True or False: Consider an object moving with constant acceleration. The plot of displacement versus time for such motion is a curved line while the plot of displacement versus time squared is a straight line. a. True b. False 32. You throw a ball straight up with an initial velocity of 15.0 m/s. It passes a tree branch on the way up at a height of 7.00 m. How much additional time will pass before the ball passes the tree branch on the way back down? a. 0.574 s b. 0.956 s 1.53 s c. 1.91 s d. 114 Chapter 3 \u2022 Test Prep Access for free at openstax.org. CHAPTER 4 Forces and Newton\u2019s Laws of Motion Figure 4.1 Newton\u2019s laws of motion describe the motion of the dolphin\u2019s path. (Credit: Jin Jang) Chapter Outline 4.1 Force 4.2 Newton's First Law of Motion: Inertia 4.3 Newton's Second Law of Motion 4.4 Newton's Third Law of Motion Isaac Newton (1642\u20131727) was a natural philosopher; a great thinker who combined science and philosophy to INTRODUCTION try to explain the workings of nature on Earth and in the universe. His laws of motion were just one part of the monumental work that has made him legendary. The development of Newton\u2019s laws marks the transition from the Renaissance period of history to the modern era. This transition was characterized by a revolutionary change in the way people thought about the physical universe. Drawing upon earlier work by scientists Galileo Galilei and Johannes Kepler, Newton\u2019s laws of motion allowed motion on Earth and in space to be predicted mathematically", ". In this chapter you will learn about force as well as Newton\u2019s first, second, and third laws of motion. 116 Chapter 4 \u2022 Forces and Newton\u2019s Laws of Motion 4.1 Force Section Learning Objectives By the end of this section, you will be able to do the following: \u2022 Differentiate between force, net force, and dynamics \u2022 Draw a free-body diagram Section Key Terms dynamics external force force free-body diagram net external force net force Defining Force and Dynamics Force is the cause of motion, and motion draws our attention. Motion itself can be beautiful, such as a dolphin jumping out of the water, the flight of a bird, or the orbit of a satellite. The study of motion is called kinematics, but kinematics describes only the way objects move\u2014their velocity and their acceleration. Dynamics considers the forces that affect the motion of moving objects and systems. Newton\u2019s laws of motion are the foundation of dynamics. These laws describe the way objects speed up, slow down, stay in motion, and interact with other objects. They are also universal laws: they apply everywhere on Earth as well as in space. A force pushes or pulls an object. The object being moved by a force could be an inanimate object, a table, or an animate object, a person. The pushing or pulling may be done by a person, or even the gravitational pull of Earth. Forces have different magnitudes and directions; this means that some forces are stronger than others and can act in different directions. For example, a cannon exerts a strong force on the cannonball that is launched into the air. In contrast, a mosquito landing on your arm exerts only a small force on your arm. When multiple forces act on an object, the forces combine. Adding together all of the forces acting on an object gives the total force, or net force. An external force is a force that acts on an object within the system from outsidethe system. This type of force is different than an internal force, which acts between two objects that are both within the system. The net external force combines these two definitions; it is the total combined external force. We discuss further details about net force, external force, and net external force in the coming sections. In mathematical terms, two forces acting in opposite directions have opposite signs(positive or negative). By convention, the negative sign is assigned to any movement to the left or downward. If two forces pushing in opposite directions are added together, the larger force will be somewhat canceled out by the smaller force pushing in the opposite direction. It is important to be consistent with your chosen coordinate system within a problem; for example, if negative values are assigned to the downward direction for velocity, then distance, force, and acceleration should also be designated as being negative in the downward direction. Free-Body Diagrams and Examples of Forces For our first example of force, consider an object hanging from a rope. This example gives us the opportunity to introduce a useful tool known as a free-body diagram. A free-body diagram represents the object being acted upon\u2014that is, the free body\u2014as a single point. Only the forces acting onthe body (that is, external forces) are shown and are represented by vectors (which are drawn as arrows). These forces are the only ones shown because only external forces acting on the body affect its motion. We can ignore any internal forces within the body because they cancel each other out, as explained in the section on Newton\u2019s third law of motion. Free-body diagrams are very useful for analyzing forces acting on an object. Access for free at openstax.org. 4.1 \u2022 Force 117 Figure 4.2 An object of mass, m, is held up by the force of tension. Figure 4.2 shows the force of tension in the rope acting in the upward direction, opposite the force of gravity. The forces are indicated in the free-body diagram by an arrow pointing up, representing tension, and another arrow pointing down, representing gravity. In a free-body diagram, the lengths of the arrows show the relative magnitude (or strength) of the forces. Because forces are vectors, they add just like other vectors. Notice that the two arrows have equal lengths in Figure 4.2, which means that the forces of tension and weight are of equal magnitude. Because these forces of equal magnitude act in opposite directions, they are perfectly balanced, so they add together to give a net force of zero. Not all forces are as noticeable as when you push or pull on an object. Some forces act without physical contact, such as the pull of a magnet (in the case of magnetic force) or the gravitational pull of Earth (in the case of gravitational force). In the next three sections discussing Newton\u2019s laws of motion, we will learn about three specific types of forces: friction, the normal force, and the gravitational force. To analyze situations involving forces, we will create free-body diagrams to organize", " the framework of the mathematics for each individual situation. TIPS FOR SUCCESS Correctly drawing and labeling a free-body diagram is an important first step for solving a problem. It will help you visualize the problem and correctly apply the mathematics to solve the problem. Check Your Understanding 1. What is kinematics? a. Kinematics is the study of motion. b. Kinematics is the study of the cause of motion. c. Kinematics is the study of dimensions. d. Kinematics is the study of atomic structures. 2. Do two bodies have to be in physical contact to exert a force upon one another? 118 Chapter 4 \u2022 Forces and Newton\u2019s Laws of Motion a. No, the gravitational force is a field force and does not require physical contact to exert a force. b. No, the gravitational force is a contact force and does not require physical contact to exert a force. c. Yes, the gravitational force is a field force and requires physical contact to exert a force. d. Yes, the gravitational force is a contact force and requires physical contact to exert a force. 3. What kind of physical quantity is force? a. Force is a scalar quantity. b. Force is a vector quantity. c. Force is both a vector quantity and a scalar quantity. d. Force is neither a vector nor a scalar quantity. 4. Which forces can be represented in a free-body diagram? a. Internal forces b. External forces c. Both internal and external forces d. A body that is not influenced by any force 4.2 Newton's First Law of Motion: Inertia Section Learning Objectives By the end of this section, you will be able to do the following: \u2022 Describe Newton\u2019s first law and friction, and \u2022 Discuss the relationship between mass and inertia. Section Key Terms friction inertia law of inertia mass Newton\u2019s first law of motion system Newton\u2019s First Law and Friction Newton\u2019s first law of motion states the following: 1. A body at rest tends to remain at rest. 2. A body in motion tends to remain in motion at a constant velocity unless acted on by a net external force. (Recall that constant velocitymeans that the body moves in a straight line and at a constant speed.) At first glance, this law may seem to contradict your everyday experience. You have probably noticed that a moving object will usually slow down and stop unless some effort is made to keep it moving. The key to understanding why, for example, a sliding box slows down (seemingly on its own) is to first understand that a net external force acts on the box to make the box slow down. Without this net external force, the box would continue to slide at a constant velocity (as stated in Newton\u2019s first law of motion). What force acts on the box to slow it down? This force is called friction. Friction is an external force that acts opposite to the direction of motion (see Figure 4.3). Think of friction as a resistance to motion that slows things down. Consider an air hockey table. When the air is turned off, the puck slides only a short distance before friction slows it to a stop. However, when the air is turned on, it lifts the puck slightly, so the puck experiences very little friction as it moves over the surface. With friction almost eliminated, the puck glides along with very little change in speed. On a frictionless surface, the puck would experience no net external force (ignoring air resistance, which is also a form of friction). Additionally, if we know enough about friction, we can accurately predict how quickly objects will slow down. Now let\u2019s think about another example. A man pushes a box across a floor at constant velocity by applying a force of +50 N. (The positive sign indicates that, by convention, the direction of motion is to the right.) What is the force of friction that opposes the motion? The force of friction must be \u221250 N. Why? According to Newton\u2019s first law of motion, any object moving at constant velocity has no net external force acting upon it, which means that the sum of the forces acting on the object must be zero. The mathematical way to say that no net external force acts on an object is N of force, then the force of friction must be \u221250 N for the two forces to add up to zero (that is, for the two forces to canceleach So if the man applies +50 or Access for free at openstax.org. other). Whenever you encounter the phrase at constant velocity, Newton\u2019s first law tells you that the net external force is zero. 4.2 \u2022 Newton's First Law of Motion: Inertia 119 Figure 4.3 For a box sliding across a floor, friction acts in the direction opposite to the velocity. The force of friction depends on two factors: the coefficient of friction and the normal force. For any two", " surfaces that are in contact with one another, the coefficient of friction is a constant that depends on the nature of the surfaces. The normal force is the force exerted by a surface that pushes on an object in response to gravity pulling the object down. In equation form, the force of friction is 4.1 where \u03bcis the coefficient of friction and N is the normal force. (The coefficient of friction is discussed in more detail in another chapter, and the normal force is discussed in more detail in the section Newton's Third Law of Motion.) Recall from the section on Force that a net external force acts from outside on the object of interest. A more precise definition is that it acts on the system of interest. A system is one or more objects that you choose to study. It is important to define the system at the beginning of a problem to figure out which forces are external and need to be considered, and which are internal and can be ignored. For example, in Figure 4.4 (a), two children push a third child in a wagon at a constant velocity. The system of interest is the wagon plus the small child, as shown in part (b) of the figure. The two children behind the wagon exert external forces on this system (F1, F2). Friction facting at the axles of the wheels and at the surface where the wheels touch the ground two other external forces acting on the system. Two more external forces act on the system: the weight W of the system pulling down and the normal force N of the ground pushing up. Notice that the wagon is not accelerating vertically, so Newton\u2019s first law tells us that the normal force balances the weight. Because the wagon is moving forward at a constant velocity, the force of friction must have the same strength as the sum of the forces applied by the two children. 120 Chapter 4 \u2022 Forces and Newton\u2019s Laws of Motion Figure 4.4 (a) The wagon and rider form a systemthat is acted on by external forces. (b) The two children pushing the wagon and child provide two external forces. Friction acting at the wheel axles and on the surface of the tires where they touch the ground provide an external force that act against the direction of motion. The weight W and the normal force N from the ground are two more external forces acting on the system. All external forces are represented in the figure by arrows. All of the external forces acting on the system add together, but because the wagon moves at a constant velocity, all of the forces must add up to zero. Mass and Inertia Inertia is the tendency for an object at rest to remain at rest, or for a moving object to remain in motion in a straight line with constant speed. This key property of objects was first described by Galileo. Later, Newton incorporated the concept of inertia into his first law, which is often referred to as the law of inertia. As we know from experience, some objects have more inertia than others. For example, changing the motion of a large truck is more difficult than changing the motion of a toy truck. In fact, the inertia of an object is proportional to the mass of the object. Mass is a measure of the amount of matter (or stuff) in an object. The quantity or amount of matter in an object is determined by the number and types of atoms the object contains. Unlike weight (which changes if the gravitational force changes), mass does not depend on gravity. The mass of an object is the same on Earth, in orbit, or on the surface of the moon. In practice, it is very difficult to count and identify all of the atoms and molecules in an object, so mass is usually not determined this way. Instead, the mass of an object is determined by comparing it with the standard kilogram. Mass is therefore expressed in kilograms. TIPS FOR SUCCESS In everyday language, people often use the terms weightand massinterchangeably\u2014but this is not correct. Weight is actually a force. (We cover this topic in more detail in the section Newton's Second Law of Motion.) WATCH PHYSICS Newton\u2019s First Law of Motion This video contrasts the way we thought about motion and force in the time before Galileo\u2019s concept of inertia and Newton\u2019s first law of motion with the way we understand force and motion now. Click to view content (https://www.khanacademy.org/embed_video?v=5-ZFOhHQS68) Access for free at openstax.org. 4.2 \u2022 Newton's First Law of Motion: Inertia 121 GRASP CHECK Before we understood that objects have a tendency to maintain their velocity in a straight line unless acted upon by a net force, people thought that objects had a tendency to stop on their own. This happened because a specific force was not yet understood. What was that force? a. Gravitational force b. Electrostatic force c. Nuclear force d. Frictional force Virtual Physics Forces", " and Motion\u2014Basics In this simulation, you will first explore net force by placing blue people on the left side of a tug-of-war rope and red people on the right side of the rope (by clicking people and dragging them with your mouse). Experiment with changing the number and size of people on each side to see how it affects the outcome of the match and the net force. Hit the \"Go!\" button to start the match, and the \u201creset all\u201d button to start over. Next, click on the Friction tab. Try selecting different objects for the person to push. Slide the applied forcebutton to the right to apply force to the right, and to the left to apply force to the left. The force will continue to be applied as long as you hold down the button. See the arrow representing friction change in magnitude and direction, depending on how much force you apply. Try increasing or decreasing the friction force to see how this change affects the motion. Click to view content (https://phet.colorado.edu/sims/html/forces-and-motion-basics/latest/forces-and-motionbasics_en.html) GRASP CHECK Click on the tab for the Acceleration Lab and check the Sum of Forcesoption. Push the box to the right and then release. Notice which direction the sum of forces arrow points after the person stops pushing the box and lets it continue moving to the right on its own. At this point, in which direction is the net force, the sum of forces, pointing? Why? a. The net force acts to the right because the applied external force acted to the right. b. The net force acts to the left because the applied external force acted to the left. c. The net force acts to the right because the frictional force acts to the right. d. The net force acts to the left because the frictional force acts to the left. Check Your Understanding 5. What does Newton\u2019s first law state? a. A body at rest tends to remain at rest and a body in motion tends to remain in motion at a constant acceleration unless acted on by a net external force. b. A body at rest tends to remain at rest and a body in motion tends to remain in motion at a constant velocity unless acted on by a net external force. c. The rate of change of momentum of a body is directly proportional to the external force applied to the body. d. The rate of change of momentum of a body is inversely proportional to the external force applied to the body. 6. According to Newton\u2019s first law, a body in motion tends to remain in motion at a constant velocity. However, when you slide an object across a surface, the object eventually slows down and stops. Why? a. The object experiences a frictional force exerted by the surface, which opposes its motion. b. The object experiences the gravitational force exerted by Earth, which opposes its motion c. The object experiences an internal force exerted by the body itself, which opposes its motion. d. The object experiences a pseudo-force from the body in motion, which opposes its motion. 122 Chapter 4 \u2022 Forces and Newton\u2019s Laws of Motion 7. What is inertia? a. b. c. d. Inertia is an object\u2019s tendency to maintain its mass. Inertia is an object\u2019s tendency to remain at rest. Inertia is an object\u2019s tendency to remain in motion Inertia is an object\u2019s tendency to remain at rest or, if moving, to remain in motion. 8. What is mass? What does it depend on? a. Mass is the weight of an object, and it depends on the gravitational force acting on the object. b. Mass is the weight of an object, and it depends on the number and types of atoms in the object. c. Mass is the quantity of matter contained in an object, and it depends on the gravitational force acting on the object. d. Mass is the quantity of matter contained in an object, and it depends on the number and types of atoms in the object. 4.3 Newton's Second Law of Motion Section Learning Objectives By the end of this section, you will be able to do the following: \u2022 Describe Newton\u2019s second law, both verbally and mathematically \u2022 Use Newton\u2019s second law to solve problems Section Key Terms freefall Newton\u2019s second law of motion weight Describing Newton\u2019s Second Law of Motion Newton\u2019s first law considered bodies at rest or bodies in motion at a constant velocity. The other state of motion to consider is when an object is moving with a changing velocity, which means a change in the speed and/or the direction of motion. This type of motion is addressed by Newton\u2019s second law of motion, which states how force causes changes in motion. Newton\u2019s second law of motion is used to calculate what", " happens in situations involving forces and motion, and it shows the mathematical relationship between force, mass, and acceleration. Mathematically, the second law is most often written as 4.2 where Fnet (or \u2211F) is the net external force, mis the mass of the system, and a is the acceleration. Note that Fnet and \u2211F are the same because the net external force is the sum of all the external forces acting on the system. First, what do we mean by a change in motion? A change in motion is simply a change in velocity: the speed of an object can become slower or faster, the direction in which the object is moving can change, or both of these variables may change. A change in velocity means, by definition, that an acceleration has occurred. Newton\u2019s first law says that only a nonzero net external force can cause a change in motion, so a net external force must cause an acceleration. Note that acceleration can refer to slowing down or to speeding up. Acceleration can also refer to a change in the direction of motion with no change in speed, because acceleration is the change in velocity divided by the time it takes for that change to occur, andvelocity is defined by speed and direction. From the equation we see that force is directly proportional to both mass and acceleration, which makes sense. To accelerate two objects from rest to the same velocity, you would expect more force to be required to accelerate the more massive object. Likewise, for two objects of the same mass, applying a greater force to one would accelerate it to a greater velocity. Now, let\u2019s rearrange Newton\u2019s second law to solve for acceleration. We get In this form, we can see that acceleration is directly proportional to force, which we write as 4.3 4.4 where the symbol means proportional to. This proportionality mathematically states what we just said in words: acceleration is directly proportional to the net external Access for free at openstax.org. 4.3 \u2022 Newton's Second Law of Motion 123 force. When two variables are directly proportional to each other, then if one variable doubles, the other variable must double. Likewise, if one variable is reduced by half, the other variable must also be reduced by half. In general, when one variable is multiplied by a number, the other variable is also multiplied by the same number. It seems reasonable that the acceleration of a system should be directly proportional to and in the same direction as the net external force acting on the system. An object experiences greater acceleration when acted on by a greater force. It is also clear from the equation that acceleration is inversely proportional to mass, which we write as 4.5 Inversely proportionalmeans that if one variable is multiplied by a number, the other variable must be dividedby the same number. Now, it also seems reasonable that acceleration should be inversely proportional to the mass of the system. In other words, the larger the mass (the inertia), the smaller the acceleration produced by a given force. This relationship is illustrated in Figure 4.5, which shows that a given net external force applied to a basketball produces a much greater acceleration than when applied to a car. Figure 4.5 The same force exerted on systems of different masses produces different accelerations. (a) A boy pushes a basketball to make a pass. The effect of gravity on the ball is ignored. (b) The same boy pushing with identical force on a stalled car produces a far smaller acceleration (friction is negligible). Note that the free-body diagrams for the ball and for the car are identical, which allows us to compare the two situations. Applying Newton\u2019s Second Law Before putting Newton\u2019s second law into action, it is important to consider units. The equation units of force in terms of the three basic units of mass, length, and time (recall that acceleration has units of length divided by time squared). The SI unit of force is called the newton (abbreviated N) and is the force needed to accelerate a 1-kg system at the rate of 1 m/s2. That is, because is used to define the we have One of the most important applications of Newton\u2019s second law is to calculate weight (also known as the gravitational force), which is usually represented mathematically as W. When people talk about gravity, they don\u2019t always realize that it is an acceleration. When an object is dropped, it accelerates toward the center of Earth. Newton\u2019s second law states that the net external force acting on an object is responsible for the acceleration of the object. If air resistance is negligible, the net external force on a falling object is only the gravitational force (i.e., the weight of the object). Weight can be represented by a vector because it has a direction. Down is defined as the direction in which gravity pulls, so weight is normally considered a downward force. By using Newton\u2019s second law, we can figure", " out the equation for weight. 4.6 Consider an object with mass mfalling toward Earth. It experiences only the force of gravity (i.e., the gravitational force or weight), which is represented by W. Newton\u2019s second law states that the gravitational force, we have Substituting these two expressions into Newton\u2019s second law gives We know that the acceleration of an object due to gravity is g, so we have Because the only force acting on the object is 124 Chapter 4 \u2022 Forces and Newton\u2019s Laws of Motion This is the equation for weight\u2014the gravitational force on a mass m. On Earth, now the direction of the weight) of a 1.0-kg object on Earth is 4.7 so the weight (disregarding for 4.8 Although most of the world uses newtons as the unit of force, in the United States the most familiar unit of force is the pound (lb), where 1 N = 0.225 lb. Recall that although gravity acts downward, it can be assigned a positive or negative value, depending on what the positive direction is in your chosen coordinate system. Be sure to take this into consideration when solving problems with weight. When the downward direction is taken to be negative, as is often the case, acceleration due to gravity becomes g = \u22129.8 m/s2. When the net external force on an object is its weight, we say that it is in freefall. In this case, the only force acting on the object is the force of gravity. On the surface of Earth, when objects fall downward toward Earth, they are never truly in freefall because there is always some upward force due to air resistance that acts on the object (and there is also the buoyancy force of air, which is similar to the buoyancy force in water that keeps boats afloat). Gravity varies slightly over the surface of Earth, so the weight of an object depends very slightly on its location on Earth. Weight varies dramatically away from Earth\u2019s surface. On the moon, for example, the acceleration due to gravity is only 1.67 m/s2. Because weight depends on the force of gravity, a 1.0-kg mass weighs 9.8 N on Earth and only about 1.7 N on the moon. It is important to remember that weight and mass are very different, although they are closely related. Mass is the quantity of matter (how much stuff) in an object and does not vary, but weight is the gravitational force on an object and is proportional to the force of gravity. It is easy to confuse the two, because our experience is confined to Earth, and the weight of an object is essentially the same no matter where you are on Earth. Adding to the confusion, the terms mass and weight are often used interchangeably in everyday language; for example, our medical records often show our weight in kilograms, but never in the correct unit of newtons. Snap Lab Mass and Weight In this activity, you will use a scale to investigate mass and weight. \u2022 \u2022 1 bathroom scale 1 table 1. What do bathroom scales measure? 2. When you stand on a bathroom scale, what happens to the scale? It depresses slightly. The scale contains springs that compress in proportion to your weight\u2014similar to rubber bands expanding when pulled. 3. The springs provide a measure of your weight (provided you are not accelerating). This is a force in newtons (or pounds). In most countries, the measurement is now divided by 9.80 to give a reading in kilograms, which is a of mass. The scale detects weight but is calibrated to display mass. If you went to the moon and stood on your scale, would it detect the same massas it did on Earth? 4. GRASP CHECK While standing on a bathroom scale, push down on a table next to you. What happens to the reading? Why? a. The reading increases because part of your weight is applied to the table and the table exerts a matching force on you that acts in the direction of your weight. b. The reading increases because part of your weight is applied to the table and the table exerts a matching force on you that acts in the direction opposite to your weight. c. The reading decreases because part of your weight is applied to the table and the table exerts a matching force on you that acts in the direction of your weight. d. The reading decreases because part of your weight is applied to the table and the table exerts a matching force on Access for free at openstax.org. you that acts in the direction opposite to your weight. 4.3 \u2022 Newton's Second Law of Motion 125 TIPS FOR SUCCESS Only net external forceimpacts the acceleration of an object. If more than one force acts on an object and you calculate the acceleration by using only one of these forces, you will not get the correct acceleration for that object. WATCH PHYSICS Newton\u2019s Second Law of Motion This video reviews Newton\u2019", "s second law of motion and how net external force and acceleration relate to one another and to mass. It also covers units of force, mass, and acceleration, and reviews a worked-out example. Click to view content (https://www.khanacademy.org/embed_video?v=ou9YMWlJgkE) GRASP CHECK True or False\u2014If you want to reduce the acceleration of an object to half its original value, then you would need to reduce the net external force by half. a. True b. False WORKED EXAMPLE What Acceleration Can a Person Produce when Pushing a Lawn Mower? Suppose that the net external force (push minus friction) exerted on a lawn mower is 51 N parallel to the ground. The mass of the mower is 240 kg. What is its acceleration? Figure 4.6 Strategy Because Fnet and mare given, the acceleration can be calculated directly from Newton\u2019s second law: Fnet = ma. Solution Solving Newton\u2019s second law for the acceleration, we find that the magnitude of the acceleration, a, is given values for net external force and mass gives Entering the 4.9 126 Chapter 4 \u2022 Forces and Newton\u2019s Laws of Motion Inserting the units for N yields 4.10 Discussion The acceleration is in the same direction as the net external force, which is parallel to the ground and to the right. There is no information given in this example about the individual external forces acting on the system, but we can say something about their relative magnitudes. For example, the force exerted by the person pushing the mower must be greater than the friction opposing the motion, because we are given that the net external force is in the direction in which the person pushes. Also, the vertical forces must cancel if there is no acceleration in the vertical direction (the mower is moving only horizontally). The acceleration found is reasonable for a person pushing a mower; the mower\u2019s speed must increase by 0.21 m/s every second, which is possible. The time during which the mower accelerates would not be very long because the person\u2019s top speed would soon be reached. At this point, the person could push a little less hard, because he only has to overcome friction. WORKED EXAMPLE What Rocket Thrust Accelerates This Sled? Prior to manned space flights, rocket sleds were used to test aircraft, missile equipment, and physiological effects on humans at high accelerations. Rocket sleds consisted of a platform mounted on one or two rails and propelled by several rockets. Calculate the magnitude of force exerted by each rocket, called its thrust, T, for the four-rocket propulsion system shown below. The sled\u2019s initial acceleration is the mass of the system is 2,100 kg, and the force of friction opposing the motion is 650 N. Figure 4.7 Strategy The system of interest is the rocket sled. Although forces act vertically on the system, they must cancel because the system does not accelerate vertically. This leaves us with only horizontal forces to consider. We\u2019ll assign the direction to the right as the positive direction. See the free-body diagram in Figure 4.8. Solution We start with Newton\u2019s second law and look for ways to find the thrust T of the engines. Because all forces and acceleration are along a line, we need only consider the magnitudes of these quantities in the calculations. We begin with 4.11 is the net external force in the horizontal direction. We can see from Figure 4.8 that the engine thrusts are in the where same direction (which we call the positive direction), whereas friction opposes the thrust. In equation form, the net external force is Access for free at openstax.org. Newton\u2019s second law tells us that Fnet= ma, so we get After a little algebra, we solve for the total thrust 4T: which means that the individual thrust is Inserting the known values yields 4.3 \u2022 Newton's Second Law of Motion 127 4.12 4.13 4.14 4.15 4.16 Discussion The numbers are quite large, so the result might surprise you. Experiments such as this were performed in the early 1960s to test the limits of human endurance and to test the apparatus designed to protect fighter pilots in emergency ejections. Speeds of 1000 km/h were obtained, with accelerations of 45 g. (Recall that g, the acceleration due to gravity, is of 45 g is ) Living subjects are no longer used, and land speeds of 10,000 km/h have now been obtained with rocket sleds. In this example, as in the preceding example, the system of interest is clear. We will see in later examples that choosing the system of interest is crucial\u2014and that the choice is not always obvious. which is approximately An acceleration Practice Problems 9. If 1 N is equal to 0.225 lb,", " how many pounds is 5 N of force? a. 0.045 lb b. 1.125 lb c. 2.025 lb 5.000 lb d. 10. How much force needs to be applied to a 5-kg object for it to accelerate at 20 m/s2? a. b. c. d. 1 N 10 N 100 N 1,000 N Check Your Understanding 11. What is the mathematical statement for Newton\u2019s second law of motion? a. F = ma b. F = 2ma c. d. F = ma2 12. Newton\u2019s second law describes the relationship between which quantities? a. Force, mass, and time b. Force, mass, and displacement c. Force, mass, and velocity d. Force, mass, and acceleration 13. What is acceleration? a. Acceleration is the rate at which displacement changes. b. Acceleration is the rate at which force changes. c. Acceleration is the rate at which velocity changes. 128 Chapter 4 \u2022 Forces and Newton\u2019s Laws of Motion d. Acceleration is the rate at which mass changes. 4.4 Newton's Third Law of Motion Section Learning Objectives By the end of this section, you will be able to do the following: \u2022 Describe Newton\u2019s third law, both verbally and mathematically \u2022 Use Newton\u2019s third law to solve problems Section Key Terms Newton\u2019s third law of motion normal force tension thrust Describing Newton\u2019s Third Law of Motion If you have ever stubbed your toe, you have noticed that although your toe initiates the impact, the surface that you stub it on exerts a force back on your toe. Although the first thought that crosses your mind is probably \u201couch, that hurt\u201d rather than \u201cthis is a great example of Newton\u2019s third law,\u201d both statements are true. This is exactly what happens whenever one object exerts a force on another\u2014each object experiences a force that is the same strength as the force acting on the other object but that acts in the opposite direction. Everyday experiences, such as stubbing a toe or throwing a ball, are all perfect examples of Newton\u2019s third law in action. Newton\u2019s third law of motion states that whenever a first object exerts a force on a second object, the first object experiences a force equal in magnitude but opposite in direction to the force that it exerts. Newton\u2019s third law of motion tells us that forces always occur in pairs, and one object cannot exert a force on another without experiencing the same strength force in return. We sometimes refer to these force pairs as action-reactionpairs, where the force exerted is the action, and the force experienced in return is the reaction (although which is which depends on your point of view). Newton\u2019s third law is useful for figuring out which forces are external to a system. Recall that identifying external forces is important when setting up a problem, because the external forces must be added together to find the net force. We can see Newton\u2019s third law at work by looking at how people move about. Consider a swimmer pushing off from the side of a pool, as illustrated in Figure 4.8. She pushes against the pool wall with her feet and accelerates in the direction opposite to her push. The wall has thus exerted on the swimmer a force of equal magnitude but in the direction opposite that of her push. You might think that two forces of equal magnitude but that act in opposite directions would cancel, but they do not because they act on different systems. In this case, there are two different systems that we could choose to investigate: the swimmer or the wall. If we choose the swimmer to be the system of interest, as in the figure, then motion. Because acceleration is in the same direction as the net external force, the swimmer moves in the direction of is an external force on the swimmer and affects her Because the swimmer is our system (or object of interest) and not the wall, we do not need to consider the force because it originates fromthe swimmer rather than acting onthe swimmer. Therefore, does not Note that the swimmer pushes in the direction directly affect the motion of the system and does not cancel opposite to the direction in which she wants to move. Access for free at openstax.org. 4.4 \u2022 Newton's Third Law of Motion 129 Figure 4.8 When the swimmer exerts a force on the wall, she accelerates in the direction opposite to that of her push. This means that the net external force on her is in the direction opposite to This opposition is the result of Newton\u2019s third law of motion, which dictates that the wall exerts a force on the swimmer that is equal in magnitude but that acts in the direction opposite to the force that the swimmer exerts on the wall. Other examples of Newton\u2019s third law are easy to", " find. As a teacher paces in front of a whiteboard, he exerts a force backward on the floor. The floor exerts a reaction force in the forward direction on the teacher that causes him to accelerate forward. Similarly, a car accelerates because the ground pushes forward on the car's wheels in reaction to the car's wheels pushing backward on the ground. You can see evidence of the wheels pushing backward when tires spin on a gravel road and throw rocks backward. Another example is the force of a baseball as it makes contact with the bat. Helicopters create lift by pushing air down, creating an upward reaction force. Birds fly by exerting force on air in the direction opposite that in which they wish to fly. For example, the wings of a bird force air downward and backward in order to get lift and move forward. An octopus propels itself forward in the water by ejecting water backward through a funnel in its body, which is similar to how a jet ski is propelled. In these examples, the octopus or jet ski push the water backward, and the water, in turn, pushes the octopus or jet ski forward. Applying Newton\u2019s Third Law Forces are classified and given names based on their source, how they are transmitted, or their effects. In previous sections, we discussed the forces called push, weight, and friction. In this section, applying Newton\u2019s third law of motion will allow us to explore three more forces: the normal force, tension, and thrust. However, because we haven\u2019t yet covered vectors in depth, we\u2019ll only consider one-dimensional situations in this chapter. Another chapter will consider forces acting in two dimensions. The gravitational force (or weight) acts on objects at all times and everywhere on Earth. We know from Newton\u2019s second law that a net force produces an acceleration; so, why is everything not in a constant state of freefall toward the center of Earth? The answer is the normal force. The normal force is the force that a surface applies to an object to support the weight of that object; it acts perpendicular to the surface upon which the object rests. If an object on a flat surface is not accelerating, the net external force is zero, and the normal force has the same magnitude as the weight of the system but acts in the opposite direction. In equation form, we write that Note that this equation is only true for a horizontal surface. The word tensioncomes from the Latin word meaning to stretch. Tension is the force along the length of a flexible connector, such as a string, rope, chain, or cable. Regardless of the type of connector attached to the object of interest, one must remember that the connector can only pull (or exert tension) in the direction parallelto its length. Tension is a pull that acts parallel to the connector, and that acts in opposite directions at the two ends of the connector. This is possible because a flexible connector is simply a long series of action-reaction forces, except at the two ends where outside objects provide one member of the actionreaction forces. Consider a person holding a mass on a rope, as shown in Figure 4.9. 4.17 130 Chapter 4 \u2022 Forces and Newton\u2019s Laws of Motion Figure 4.9 When a perfectly flexible connector (one requiring no force to bend it) such as a rope transmits a force T, this force must be parallel to the length of the rope, as shown. The pull that such a flexible connector exerts is a tension. Note that the rope pulls with equal magnitude force but in opposite directions to the hand and to the mass (neglecting the weight of the rope). This is an example of Newton\u2019s third law. The rope is the medium that transmits forces of equal magnitude between the two objects but that act in opposite directions. Tension in the rope must equal the weight of the supported mass, as we can prove by using Newton\u2019s second law. If the 5.00 kg mass in the figure is stationary, then its acceleration is zero, so weight W and the tension T supplied by the rope. Summing the external forces to find the net force, we obtain The only external forces acting on the mass are its 4.18 where T and W are the magnitudes of the tension and weight, respectively, and their signs indicate direction, with up being positive. By substituting mg for Fnet and rearranging the equation, the tension equals the weight of the supported mass, just as you would expect For a 5.00-kg mass (neglecting the mass of the rope), we see that 4.19 4.20 Another example of Newton\u2019s third law in action is thrust. Rockets move forward by expelling gas backward at a high velocity. This means that the rocket exerts a large force backward on the gas in the rocket combustion chamber, and the gas, in turn, exerts a large force forward on the rocket in response. This reaction force", " is called thrust. TIPS FOR SUCCESS A common misconception is that rockets propel themselves by pushing on the ground or on the air behind them. They actually work better in a vacuum, where they can expel exhaust gases more easily. LINKS TO PHYSICS Math: Problem-Solving Strategy for Newton\u2019s Laws of Motion The basics of problem solving, presented earlier in this text, are followed here with specific strategies for applying Newton\u2019s laws of motion. These techniques also reinforce concepts that are useful in many other areas of physics. First, identify the physical principles involved. If the problem involves forces, then Newton\u2019s laws of motion are involved, and it Access for free at openstax.org. is important to draw a careful sketch of the situation. An example of a sketch is shown in Figure 4.10. Next, as in Figure 4.10, use vectors to represent all forces. Label the forces carefully, and make sure that their lengths are proportional to the magnitude of the forces and that the arrows point in the direction in which the forces act. 4.4 \u2022 Newton's Third Law of Motion 131 Figure 4.10 (a) A sketch of Tarzan hanging motionless from a vine. (b) Arrows are used to represent all forces. T is the tension exerted on Tarzan by the vine, is the force exerted on the vine by Tarzan, and W is Tarzan\u2019s weight (i.e., the force exerted on Tarzan by Earth\u2019s gravity). All other forces, such as a nudge of a breeze, are assumed to be negligible. (c) Suppose we are given Tarzan\u2019s mass and asked to find the tension in the vine. We define the system of interest as shown and draw a free-body diagram, as shown in (d). is no longer shown because it does not act on the system of interest; rather, acts on the outside world. (d) The free-body diagram shows only the external forces acting on Tarzan. For these to sum to zero, we must have Next, make a list of knowns and unknowns and assign variable names to the quantities given in the problem. Figure out which variables need to be calculated; these are the unknowns. Now carefully define the system: which objects are of interest for the problem. This decision is important, because Newton\u2019s second law involves only external forces. Once the system is identified, it\u2019s possible to see which forces are external and which are internal (see Figure 4.10). If the system acts on an object outside the system, then you know that the outside object exerts a force of equal magnitude but in the opposite direction on the system. A diagram showing the system of interest and all the external forces acting on it is called a free-body diagram. Only external forces are shown on free-body diagrams, not acceleration or velocity. Figure 4.10 shows a free-body diagram for the system of interest. After drawing a free-body diagram, apply Newton\u2019s second law to solve the problem. This is done in Figure 4.10 for the case of Tarzan hanging from a vine. When external forces are clearly identified in the free-body diagram, translate the forces into equation form and solve for the unknowns. Note that forces acting in opposite directions have opposite signs. By convention, forces acting downward or to the left are usually negative. GRASP CHECK If a problem has more than one system of interest, more than one free-body diagram is required to describe the external forces acting on the different systems. a. True b. False 132 Chapter 4 \u2022 Forces and Newton\u2019s Laws of Motion WATCH PHYSICS Newton\u2019s Third Law of Motion This video explains Newton\u2019s third law of motion through examples involving push, normal force, and thrust (the force that propels a rocket or a jet). Click to view content (https://www.openstax.org/l/astronaut) GRASP CHECK If the astronaut in the video wanted to move upward, in which direction should he throw the object? Why? a. He should throw the object upward because according to Newton\u2019s third law, the object will then exert a force on him in the same direction (i.e., upward). b. He should throw the object upward because according to Newton\u2019s third law, the object will then exert a force on him in the opposite direction (i.e., downward). c. He should throw the object downward because according to Newton\u2019s third law, the object will then exert a force on him in the opposite direction (i.e., upward). d. He should throw the object downward because according to Newton\u2019s third law, the object will then exert a force on him in the same direction (i.e., downward). WORKED EXAMPLE An Accelerating Subway Train A physics teacher pushes a cart of demonstration equipment", " to a classroom, as in Figure 4.11. Her mass is 65.0 kg, the cart\u2019s mass is 12.0 kg, and the equipment\u2019s mass is 7.0 kg. To push the cart forward, the teacher\u2019s foot applies a force of 150 N in the opposite direction (backward) on the floor. Calculate the acceleration produced by the teacher. The force of friction, which opposes the motion, is 24.0 N. Figure 4.11 Strategy Because they accelerate together, we define the system to be the teacher, the cart, and the equipment. The teacher pushes backward with a force system. Because all motion is horizontal, we can assume that no net force acts in the vertical direction, and the problem becomes one dimensional. As noted in the figure, the friction f opposes the motion and therefore acts opposite the direction of of 150 N. According to Newton\u2019s third law, the floor exerts a forward force of 150 N on the Access for free at openstax.org. 4.4 \u2022 Newton's Third Law of Motion 133 We should not include the forces because these are exerted bythe system, not onthe system. We find the net external force by adding together the external forces acting on the system (see the free-body diagram in the figure) and then use Newton\u2019s second law to find the acceleration., or, Solution Newton\u2019s second law is The net external force on the system is the sum of the external forces: the force of the floor acting on the teacher, cart, and equipment (in the horizontal direction) and the force of friction. Because friction acts in the opposite direction, we assign it a negative value. Thus, for the net force, we obtain 4.21 The mass of the system is the sum of the mass of the teacher, cart, and equipment. Insert these values of net F and minto Newton\u2019s second law to obtain the acceleration of the system. 4.22 4.23 4.24 4.25 Discussion None of the forces between components of the system, such as between the teacher\u2019s hands and the cart, contribute to the net external force because they are internal to the system. Another way to look at this is to note that the forces between components of a system cancel because they are equal in magnitude and opposite in direction. For example, the force exerted by the teacher on the cart is of equal magnitude but in the opposite direction of the force exerted by the cart on the teacher. In this case, both forces act on the same system, so they cancel. Defining the system was crucial to solving this problem. Practice Problems 14. What is the equation for the normal force for a body with mass mthat is at rest on a horizontal surface? a. N = m b. N = mg c. N = mv d. N = g 15. An object with mass mis at rest on the floor. What is the magnitude and direction of the normal force acting on it? a. N = mvin upward direction b. N = mgin upward direction c. N = mvin downward direction d. N = mgin downward direction Check Your Understanding 16. What is Newton\u2019s third law of motion? a. Whenever a first body exerts a force on a second body, the first body experiences a force that is twice the magnitude and acts in the direction of the applied force. b. Whenever a first body exerts a force on a second body, the first body experiences a force that is equal in magnitude and acts in the direction of the applied force. c. Whenever a first body exerts a force on a second body, the first body experiences a force that is twice the magnitude but acts in the direction opposite the direction of the applied force. d. Whenever a first body exerts a force on a second body, the first body experiences a force that is equal in magnitude but 134 Chapter 4 \u2022 Forces and Newton\u2019s Laws of Motion acts in the direction opposite the direction of the applied force. 17. Considering Newton\u2019s third law, why don\u2019t two equal and opposite forces cancel out each other? a. Because the two forces act in the same direction b. Because the two forces have different magnitudes c. Because the two forces act on different systems d. Because the two forces act in perpendicular directions Access for free at openstax.org. Chapter 4 \u2022 Key Terms 135 KEY TERMS dynamics the study of how forces affect the motion of objects and systems external force a force acting on an object or system that originates outside of the object or system force a push or pull on an object with a specific magnitude and direction; can be represented by vectors; can be expressed as a multiple of a standard force free-body diagram a diagram showing all external forces acting on a body Newton\u2019s second law of motion the net external force, on an object is proportional to and in the same direction as the acceleration of the object", ", a, and also proportional to the object\u2019s mass, m; defined mathematically as or Newton\u2019s third law of motion when one body exerts a force on a second body, the first body experiences a force that is equal in magnitude and opposite in direction to the force that it exerts freefall a situation in which the only force acting on an normal force the force that a surface applies to an object; object is the force of gravity friction an external force that acts in the direction opposite to the direction of motion inertia the tendency of an object at rest to remain at rest, or for a moving object to remain in motion in a straight line and at a constant speed law of inertia Newton\u2019s first law of motion: a body at rest remains at rest or, if in motion, remains in motion at a constant speed in a straight line, unless acted on by a net external force; also known as the law of inertia mass the quantity of matter in a substance; measured in kilograms net external force the sum of all external forces acting on an object or system net force the sum of all forces acting on an object or system Newton\u2019s first law of motion a body at rest remains at rest or, if in motion, remains in motion at a constant speed in a straight line, unless acted on by a net external force; also known as the law of inertia SECTION SUMMARY 4.1 Force \u2022 Dynamics is the study of how forces affect the motion of objects and systems. \u2022 Force is a push or pull that can be defined in terms of various standards. It is a vector and so has both magnitude and direction. \u2022 External forces are any forces outside of a body that act on the body. A free-body diagram is a drawing of all external forces acting on a body. 4.2 Newton's First Law of Motion: Inertia \u2022 Newton\u2019s first law states that a body at rest remains at rest or, if moving, remains in motion in a straight line at a constant speed, unless acted on by a net external force. This law is also known as the law of inertia. Inertia is the tendency of an object at rest to remain at rest or, if moving, to remain in motion at constant velocity. Inertia is related to an object\u2019s mass. \u2022 acts perpendicular and away from the surface with which the object is in contact system one or more objects of interest for which only the forces acting on them from the outside are considered, but not the forces acting between them or inside them tension a pulling force that acts along a connecting medium, especially a stretched flexible connector, such as a rope or cable; when a rope supports the weight of an object, the force exerted on the object by the rope is called tension thrust a force that pushes an object forward in response to the backward ejection of mass by the object; rockets and airplanes are pushed forward by a thrust reaction force in response to ejecting gases backward weight the force of gravity, W, acting on an object of mass m; defined mathematically as W = mg, where g is the magnitude and direction of the acceleration due to gravity \u2022 Friction is a force that opposes motion and causes an object or system to slow down. \u2022 Mass is the quantity of matter in a substance. 4.3 Newton's Second Law of Motion \u2022 Acceleration is a change in velocity, meaning a change in speed, direction, or both. \u2022 An external force acts on a system from outside the system, as opposed to internal forces, which act between components within the system. \u2022 Newton\u2019s second law of motion states that the acceleration of a system is directly proportional to and in the same direction as the net external force acting on the system, and inversely proportional to the system\u2019s mass. In equation form, Newton\u2019s second law of motion is \u2022 or or. This is sometimes written as \u2022 The weight of an object of mass mis the force of gravity that acts on it. From Newton\u2019s second law, weight is 136 Chapter 4 \u2022 Key Equations \u2022 given by If the only force acting on an object is its weight, then the object is in freefall. 4.4 Newton's Third Law of Motion \u2022 Newton\u2019s third law of motion states that when one body exerts a force on a second body, the first body experiences a force that is equal in magnitude and opposite in direction to the force that it exerts. \u2022 When an object rests on a surface, the surface applies a force on the object that opposes the weight of the object. KEY EQUATIONS 4.2 Newton's First Law of Motion: Inertia Newton's first law of motion or This force acts perpendicular to the surface and is called the normal force. \u2022 The pulling force that acts along a stretched flexible connector, such as a rope or cable, is called tension. When a rope supports the weight of an object at rest, the tension in the rope is equal to the weight of the object.", " \u2022 Thrust is a force that pushes an object forward in response to the backward ejection of mass by the object. Rockets and airplanes are pushed forward by thrust. Newton\u2019s second law of motion to solve weight 4.4 Newton's Third Law of Motion 4.3 Newton's Second Law of Motion Newton\u2019s second law of motion or normal force for a nonaccelerating horizontal surface tension for an object at rest Newton\u2019s second law of motion to solve acceleration CHAPTER REVIEW Concept Items 4.1 Force 1. What is dynamics? a. Dynamics is the study of internal forces. b. Dynamics is the study of forces and their effect on motion. c. Dynamics describes the motion of points, bodies, and systems without consideration of the cause of motion. d. Dynamics describes the effect of forces on each other. 2. Two forces acting on an object are perpendicular to one another. How would you draw these in a free-body diagram? a. The two force arrows will be drawn at a right angle to one another. b. The two force arrows will be pointing in opposite directions. c. The two force arrows will be at a 45\u00b0 angle to one another. Access for free at openstax.org. d. The two force arrows will be at a 180\u00b0 angle to one another. 3. A free-body diagram shows the forces acting on an object. How is that object represented in the diagram? a. A single point b. A square box c. A unit circle d. The object as it is 4.2 Newton's First Law of Motion: Inertia 4. A ball rolls along the ground, moving from north to south. What direction is the frictional force that acts on the ball? a. North to south b. South to north c. West to east d. East to west 5. The tires you choose to drive over icy roads will create more friction with the road than your summer tires. Give another example where more friction is desirable. a. Children\u2019s slide b. Air hockey table Ice-skating rink c. Jogging track d. 6. How do you express, mathematically, that no external force is acting on a body? a. Fnet = \u22121 b. Fnet = 0 c. Fnet = 1 d. Fnet = \u221e 4.3 Newton's Second Law of Motion 7. What does it mean for two quantities to be inversely proportional to each other? a. When one variable increases, the other variable decreases by a greater amount. b. When one variable increases, the other variable also increases. c. When one variable increases, the other variable decreases by the same factor. d. When one variable increases, the other variable also increases by the same factor. Chapter 4 \u2022 Chapter Review 137 8. True or False: Newton's second law can be interpreted based on Newton's first law. a. True b. False 4.4 Newton's Third Law of Motion 9. Which forces cause changes in the motion of a system? a. internal forces b. external forces c. both internal and external forces d. neither internal nor external forces 10. True or False\u2014Newton\u2019s third law applies to the external forces acting on a system of interest. a. True b. False 11. A ball is dropped and hits the floor. What is the direction of the force exerted by the floor on the ball? a. Upward b. Downward c. Right d. Left Critical Thinking Items direction. 4.1 Force 12. Only two forces are acting on an object: force A to the left and force B to the right. If force B is greater than force A, in which direction will the object move? a. To the right b. To the left c. Upward d. The object does not move 13. In a free-body diagram, the arrows representing tension and weight have the same length but point away from one another. What does this indicate? a. They are equal in magnitude and act in the same direction. c. The magnitude is xand points in the downward direction. d. The magnitude is 2xand points in the downward direction. 15. Three forces, A, B, and C, are acting on the same object with magnitudes a, b, and c, respectively. Force A acts to the right, force B acts to the left, and force C acts downward. What is a necessary condition for the object to move straight down? a. The magnitude of force A must be greater than the magnitude of force B, so a > b. b. The magnitude of force A must be equal to the magnitude of force B, so a = b. c. The magnitude of force A must be greater than the b. They are equal in magnitude and act in opposite magnitude of force C, so A > C. directions. c. They are unequal in magnitude and act in the same d. The magnitude of force C must be greater than the", " magnitude of forces A or B, so A < C > B. direction. d. They are unequal in magnitude and act in opposite 4.2 Newton's First Law of Motion: Inertia directions. 14. An object is at rest. Two forces, X and Y, are acting on it. Force X has a magnitude of xand acts in the downward direction. What is the magnitude and direction of Y? a. The magnitude is xand points in the upward direction. 16. Two people push a cart on a horizontal surface by applying forces F1 and F2 in the same direction. Is the magnitude of the net force acting on the cart, Fnet, equal to, greater than, or less than F1 + F2? Why? a. Fnet < F1 + F2 because the net force will not include the frictional force. b. The magnitude is 2xand points in the upward b. Fnet = F1 + F2 because the net force will not include 138 Chapter 4 \u2022 Chapter Review the frictional force in acceleration and mass. c. Fnet < F1 + F2 because the net force will include the component of frictional force 4.4 Newton's Third Law of Motion d. Fnet = F1 + F2 because the net force will include the frictional force 17. True or False: A book placed on a balance scale is balanced by a standard 1-kg iron weight placed on the opposite side of the balance. If these objects are taken to the moon and a similar exercise is performed, the balance is still level because gravity is uniform on the moon\u2019s surface as it is on Earth\u2019s surface. a. True b. False 4.3 Newton's Second Law of Motion 18. From the equation for Newton\u2019s second law, we see that Fnet is directly proportional to a and that the constant of proportionality is m. What does this mean in a practical sense? a. An increase in applied force will cause an increase in acceleration if the mass is constant. b. An increase in applied force will cause a decrease in acceleration if the mass is constant. c. An increase in applied force will cause an increase in acceleration, even if the mass varies. d. An increase in applied force will cause an increase 19. True or False: A person accelerates while walking on the ground by exerting force. The ground in turn exerts force F2 on the person. F1 and F2 are equal in magnitude but act in opposite directions. The person is able to walk because the two forces act on the different systems and the net force acting on the person is nonzero. a. True b. False 20. A helicopter pushes air down, which, in turn, pushes the helicopter up. Which force affects the helicopter\u2019s motion? Why? a. Air pushing upward affects the helicopter\u2019s motion because it is an internal force that acts on the helicopter. b. Air pushing upward affects the helicopter\u2019s motion because it is an external force that acts on the helicopter. c. The downward force applied by the blades of the helicopter affects its motion because it is an internal force that acts on the helicopter. d. The downward force applied by the blades of the helicopter affects its motion because it is an external force that acts on the helicopter. Problems 4.3 Newton's Second Law of Motion 55.0 kg c. d. 91.9 kg on Earth. What is its 4.4 Newton's Third Law of Motion 21. An object has a mass of weight on the moon? a. b. c. d. 22. A bathroom scale shows your mass as 55 kg. What will it read on the moon? a. 9.4 kg 10.5 kg b. Performance Task 4.4 Newton's Third Law of Motion 24. A car weighs 2,000 kg. It moves along a road by applying a force on the road with a parallel component of 560 N. There are two passengers in the car, each weighing 55 kg. If the magnitude of the force of friction Access for free at openstax.org. 23. A person pushes an object of mass 5.0 kg along the floor by applying a force. If the object experiences a friction force of 10 N and accelerates at 18 m/s2, what is the magnitude of the force exerted by the person? a. \u221290 N b. \u221280 N c. 90 N 100 N d. experienced by the car is 45 N, what is the acceleration of the car? a. 0.244 m/s2 b. 0.265 m/s2 c. 4.00 m/s2 d. 4.10 m/s2 TEST PREP Multiple Choice 4.1 Force 25. Which of the following is a physical quantity that can be described by dynamics but not by kinematics? a. Velocity b. Acceleration c. Force 26. Which of the following is used to represent an object", " in a free-body diagram? a. A point b. A line c. A vector 4.2 Newton's First Law of Motion: Inertia 27. What kind of force is friction? a. External force b. Internal force c. Net force 28. What is another name for Newton\u2019s first law? a. Law of infinite motion b. Law of inertia c. Law of friction 29. True or False\u2014A rocket is launched into space and escapes Earth\u2019s gravitational pull. It will continue to travel in a straight line until it is acted on by another force. a. True b. False 30. A 2,000-kg car is sitting at rest in a parking lot. A bike and rider with a total mass of 60 kg are traveling along a road at 10 km/h. Which system has more inertia? Why? a. The car has more inertia, as its mass is greater than the mass of the bike. b. The bike has more inertia, as its mass is greater than the mass of the car. c. The car has more inertia, as its mass is less than the mass of the bike. d. The bike has more inertia, as its mass is less than the mass of the car. 4.3 Newton's Second Law of Motion 31. In the equation for Newton\u2019s second law, what does Fnet stand for? Internal force a. b. Net external force c. Frictional force 32. What is the SI unit of force? Chapter 4 \u2022 Test Prep 139 a. Kg b. dyn c. N 33. What is the net external force on an object in freefall on Earth if you were to neglect the effect of air? a. The net force is zero. b. The net force is upward with magnitude mg. c. The net force is downward with magnitude mg. d. The net force is downward with magnitude 9.8 N. 34. Two people push a 2,000-kg car to get it started. An acceleration of at least 5.0 m/s2 is required to start the car. Assuming both people apply the same magnitude force, how much force will each need to apply if friction between the car and the road is 300 N? a. 4850 N 5150 N b. c. 97000 N 10300 N d. 4.4 Newton's Third Law of Motion 35. One object exerts a force of magnitude F1 on another object and experiences a force of magnitude F2 in return. What is true for F1 and F2? a. F1 > F2 b. F1 < F2 c. F1 = F2 36. A weight is suspended with a rope and hangs freely. In what direction is the tension on the rope? a. parallel to the rope b. perpendicular to the rope 37. A person weighing 55 kg walks by applying 160 N of force on the ground, while pushing a 10-kg object. If the person accelerates at 2 m/s2, what is the force of friction experienced by the system consisting of the person and the object? 30 N a. b. 50 N c. 270 N d. 290 N 38. A 65-kg swimmer pushes on the pool wall and accelerates at 6 m/s2. The friction experienced by the swimmer is 100 N. What is the magnitude of the force that the swimmer applies on the wall? a. \u2212490 N b. \u2212290 N c. 290 N d. 490 N 140 Chapter 4 \u2022 Test Prep Short Answer 4.1 Force 39. True or False\u2014An external force is defined as a force generated outside the system of interest that acts on an object inside the system. a. True b. False 40. By convention, which sign is assigned to an object moving downward? a. A positive sign ( b. A negative sign ( c. Either a positive or negative sign ( d. No sign is assigned ) ) ) 41. A body is pushed downward by a force of 5 units and upward by a force of 2 units. How would you draw a free-body diagram to represent this? a. Two force vectors acting at a point, both pointing up with lengths of 5 units and 2 units b. Two force vectors acting at a point, both pointing down with lengths of 5 units and 2 units c. Two force vectors acting at a point, one pointing up with a length of 5 units and the other pointing down with a length of 2 units d. Two force vectors acting at a point, one pointing down with a length of 5 units and the other pointing up with a length of 2 units 42. A body is pushed eastward by a force of four units and southward by a force of three units. How would you draw a free-body diagram to represent this? a. Two force vectors acting at a point, one pointing left with a length of 4 units and the other pointing down with a length of 3 units b. Two force vectors acting", " at a point, one pointing left with a length of 4 units and the other pointing up with a length of 3 units c. Two force vectors acting at a point, one pointing right with a length of 4 units and the other pointing down with a length of 3 units d. Two force vectors acting at a point, one pointing right with a length of 4 units and the other pointing up with a length of 3 units 4.2 Newton's First Law of Motion: Inertia 43. A body with mass m is pushed along a horizontal surface by a force F and is opposed by a frictional force f. How would you draw a free-body diagram to represent this situation? a. A dot with an arrow pointing right, labeled F, and an arrow pointing left, labeled f, that is of equal length or shorter than F Access for free at openstax.org. b. A dot with an arrow pointing right, labeled F, and an arrow pointing right, labeled f, that is of equal length or shorter than F c. A dot with an arrow pointing right, labeled F, and a smaller arrow pointing up, labeled f, that is of equal length or longer than F d. A dot with an arrow pointing right, labeled F, and a smaller arrow pointing down, labeled f, that is of equal length or longer than F 44. Two objects rest on a uniform surface. A person pushes both with equal force. If the first object starts to move faster than the second, what can be said about their masses? a. The mass of the first object is less than that of the second object. b. The mass of the first object is equal to the mass of the second object. c. The mass of the first object is greater than that of the second object. d. No inference can be made because mass and force are not related to each other. 45. Two similar boxes rest on a table. One is empty and the other is filled with pebbles. Without opening or lifting either, how can you tell which box is full? Why? a. By applying an internal force; whichever box accelerates faster is lighter and so must be empty b. By applying an internal force; whichever box accelerates faster is heavier and so the other box must be empty c. By applying an external force; whichever box accelerates faster is lighter and so must be empty d. By applying an external force; whichever box accelerates faster is heavier and so the other box must be empty 46. True or False\u2014An external force is required to set a stationary object in motion in outer space away from all gravitational influences and atmospheric friction. a. True b. False 4.3 Newton's Second Law of Motion 47. A steadily rolling ball is pushed in the direction from east to west, which causes the ball to move faster in the same direction. What is the direction of the acceleration? a. North to south b. South to north c. East to west d. West to east 48. A ball travels from north to south at 60 km/h. After being hit by a bat, it travels from west to east at 60 km/ Chapter 4 \u2022 Test Prep 141 h. Is there a change in velocity? a. Yes, because velocity is a scalar. b. Yes, because velocity is a vector. c. No, because velocity is a scalar. d. No, because velocity is a vector 49. What is the weight of a 5-kg object on Earth and on the moon? a. On Earth the weight is 1.67 N, and on the moon the weight is 1.67 N. b. On Earth the weight is 5 N, and on the moon the weight is 5 N. b. F c. 2F 30F d. 52. A fish pushes water backward with its fins. How does this propel the fish forward? a. The water exerts an internal force on the fish in the opposite direction, pushing the fish forward. b. The water exerts an external force on the fish in the opposite direction, pushing the fish forward. c. The water exerts an internal force on the fish in the same direction, pushing the fish forward. c. On Earth the weight is 49 N, and on the moon the d. The water exerts an external force on the fish in the weight is 8.35 N. same direction, pushing the fish forward. d. On Earth the weight is 8.35 N, and on the moon the weight is 49 N. 50. An object weighs 294 N on Earth. What is its weight on the moon? 50.1 N a. b. 30.0 N c. 249 N 1461 N d. 4.4 Newton's Third Law of Motion 51. A large truck with mass 30 m crashes into a small sedan with mass m. If the truck exerts a force F on the sedan, what force will the sedan exert on the truck? a. Extended Response 4.1 Force 55. True or False", "\u2014When two unequal forces act on a body, the body will not move in the direction of the weaker force. a. True b. False 56. In the figure given, what is Frestore? What is its magnitude? a. Frestore is the force exerted by the hand on the spring, and it pulls to the right. b. Frestore is the force exerted by the spring on the hand, and it pulls to the left. 53. True or False\u2014Tension is the result of opposite forces in a connector, such as a string, rope, chain or cable, that pulls each point of the connector apart in the direction parallel to the length of the connector. At the ends of the connector, the tension pulls toward the center of the connector. a. True b. False 54. True or False\u2014Normal reaction is the force that opposes the force of gravity and acts in the direction of the force of gravity. a. True b. False c. Frestore is the force exerted by the hand on the spring, and it pulls to the left. d. Frestore is the force exerted by the spring on the hand, and it pulls to the right. 4.2 Newton's First Law of Motion: Inertia 57. Two people apply the same force to throw two identical balls in the air. Will the balls necessarily travel the same distance? Why or why not? a. No, the balls will not necessarily travel the same distance because the gravitational force acting on them is different. b. No, the balls will not necessarily travel the same distance because the angle at which they are thrown may differ. c. Yes, the balls will travel the same distance because the gravitational force acting on them is the same. d. Yes, the balls will travel the same distance because the angle at which they are thrown may differ. 58. A person pushes a box from left to right and then lets the box slide freely across the floor. The box slows down as it slides across the floor. When the box is sliding 142 Chapter 4 \u2022 Test Prep freely, what is the direction of the net external force? a. The net external force acts from left to right. b. The net external force acts from right to left. c. The net external force acts upward. d. The net external force acts downward. 4.3 Newton's Second Law of Motion 59. A 55-kg lady stands on a bathroom scale inside an elevator. The scale reads 70 kg. What do you know about the motion of the elevator? a. The elevator must be accelerating upward. b. The elevator must be accelerating downward. c. The elevator must be moving upward with a constant velocity. d. The elevator must be moving downward with a constant velocity. 60. True or False\u2014A skydiver initially accelerates in his jump. Later, he achieves a state of constant velocity called terminal velocity. Does this mean the skydiver becomes weightless? a. Yes b. No 4.4 Newton's Third Law of Motion 61. How do rockets propel themselves in space? a. Rockets expel gas in the forward direction at high velocity, and the gas, which provides an internal force, pushes the rockets forward. b. Rockets expel gas in the forward direction at high velocity, and the gas, which provides an external force, pushes the rockets forward. c. Rockets expel gas in the backward direction at high velocity, and the gas, which is an internal force, pushes the rockets forward. d. Rockets expel gas in the backward direction at high velocity, and the gas, which provides an external force, pushes the rockets forward. 62. Are rockets more efficient in Earth\u2019s atmosphere or in outer space? Why? a. Rockets are more efficient in Earth\u2019s atmosphere than in outer space because the air in Earth\u2019s atmosphere helps to provide thrust for the rocket, and Earth has more air friction than outer space. b. Rockets are more efficient in Earth\u2019s atmosphere than in outer space because the air in Earth\u2019s atmosphere helps to provide thrust to the rocket, and Earth has less air friction than the outer space. c. Rockets are more efficient in outer space than in Earth\u2019s atmosphere because the air in Earth\u2019s atmosphere does not provide thrust but does create more air friction than in outer space. d. Rockets are more efficient in outer space than in Earth\u2019s atmosphere because the air in Earth\u2019s atmosphere does not provide thrust but does create less air friction than in outer space. Access for free at openstax.org. CHAPTER 5 Motion in Two Dimensions Figure 5.1 Billiard balls on a pool table are in motion after being hit with a cue stick. (Popperipopp, Wikimedia Commons) Chapter Outline 5.1 Vector Addition and Subtraction: Graphical Methods 5.2 Vector Addition and Subtraction: Analytical Methods 5.3 Projectile Motion 5.4 Inclined Planes 5.", "5 Simple Harmonic Motion In Chapter 2, we learned to distinguish between vectors and scalars; the difference being that a vector has INTRODUCTION magnitude and direction, whereas a scalar has only magnitude. We learned how to deal with vectors in physics by working straightforward one-dimensional vector problems, which may be treated mathematically in the same as scalars. In this chapter, we\u2019ll use vectors to expand our understanding of forces and motion into two dimensions. Most real-world physics problems (such as with the game of pool pictured here) are, after all, either two- or three-dimensional problems and physics is most useful when applied to real physical scenarios. We start by learning the practical skills of graphically adding and subtracting vectors (by using drawings) and analytically (with math). Once we\u2019re able to work with two-dimensional vectors, we apply these skills to problems of projectile motion, inclined planes, and harmonic motion. 144 Chapter 5 \u2022 Motion in Two Dimensions 5.1 Vector Addition and Subtraction: Graphical Methods Section Learning Objectives By the end of this section, you will be able to do the following: \u2022 Describe the graphical method of vector addition and subtraction \u2022 Use the graphical method of vector addition and subtraction to solve physics problems Section Key Terms graphical method head (of a vector) head-to-tail method resultant resultant vector tail vector addition vector subtraction The Graphical Method of Vector Addition and Subtraction Recall that a vector is a quantity that has magnitude and direction. For example, displacement, velocity, acceleration, and force are all vectors. In one-dimensional or straight-line motion, the direction of a vector can be given simply by a plus or minus sign. Motion that is forward, to the right, or upward is usually considered to be positive(+); and motion that is backward, to the left, or downward is usually considered to be negative(\u2212). In two dimensions, a vector describes motion in two perpendicular directions, such as vertical and horizontal. For vertical and horizontal motion, each vector is made up of vertical and horizontal components. In a one-dimensional problem, one of the components simply has a value of zero. For two-dimensional vectors, we work with vectors by using a frame of reference such as a coordinate system. Just as with one-dimensional vectors, we graphically represent vectors with an arrow having a length proportional to the vector\u2019s magnitude and pointing in the direction that the vector points. Figure 5.2 shows a graphical representation of a vector; the total displacement for a person walking in a city. The person first walks nine blocks east and then five blocks north. Her total displacement does not match her path to her final destination. The displacement simply connects her starting point with her ending point using a straight line, which is the shortest distance. We use the notation that a boldface symbol, such as D, stands for a vector. Its magnitude is represented by the symbol in italics, D, and its direction is given by an angle represented by the symbol Note that her displacement would be the same if she had begun by first walking five blocks north and then walking nine blocks east. TIPS FOR SUCCESS In this text, we represent a vector with a boldface variable. For example, we represent a force with the vector F, which has both magnitude and direction. The magnitude of the vector is represented by the variable in italics, F, and the direction of the variable is given by the angle Figure 5.2 A person walks nine blocks east and five blocks north. The displacement is 10.3 blocks at an angle north of east. The head-to-tail method is a graphical way to add vectors. The tail of the vector is the starting point of the vector, and the head (or tip) of a vector is the pointed end of the arrow. The following steps describe how to use the head-to-tail method for graphical vector addition. Access for free at openstax.org. 1. Let the x-axis represent the east-west direction. Using a ruler and protractor, draw an arrow to represent the first vector (nine blocks to the east), as shown in Figure 5.3(a). 5.1 \u2022 Vector Addition and Subtraction: Graphical Methods 145 Figure 5.3 The diagram shows a vector with a magnitude of nine units and a direction of 0\u00b0. 2. Let the y-axis represent the north-south direction. Draw an arrow to represent the second vector (five blocks to the north). Place the tail of the second vector at the head of the first vector, as shown in Figure 5.4(b). Figure 5.4 A vertical vector is added. 3. If there are more than two vectors, continue to add the vectors head-to-tail as described in step 2. In this example, we have only two vectors, so we have finished placing arrows tip to tail. 4. Draw an arrow from the tail", " of the first vector to the head of the last vector, as shown in Figure 5.5(c). This is the resultant, or the sum, of the vectors. 146 Chapter 5 \u2022 Motion in Two Dimensions Figure 5.5 The diagram shows the resultant vector, a ruler, and protractor. 5. To find the magnitude of the resultant, measure its length with a ruler. When we deal with vectors analytically in the next section, the magnitude will be calculated by using the Pythagorean theorem. 6. To find the direction of the resultant, use a protractor to measure the angle it makes with the reference direction (in this case, the x-axis). When we deal with vectors analytically in the next section, the direction will be calculated by using trigonometry to find the angle. WATCH PHYSICS Visualizing Vector Addition Examples This video shows four graphical representations of vector addition and matches them to the correct vector addition formula. Click to view content (https://openstax.org/l/02addvector) GRASP CHECK There are two vectors and. The head of vector touches the tail of vector. The addition of vectors and gives a resultant vector. Can the addition of these two vectors can be represented by the following two equations? ; a. Yes, if we add the same two vectors in a different order it will still give the same resultant vector. b. No, the resultant vector will change if we add the same vectors in a different order. Vector subtraction is done in the same way as vector addition with one small change. We add the first vector to the negative of the vector that needs to be subtracted. A negative vector has the same magnitude as the original vector, but points in the opposite direction (as shown in Figure 5.6). Subtracting the vector B from the vector A, which is written as A \u2212 B, is the same as A + (\u2212B). Since it does not matter in what order vectors are added, A \u2212 B is also equal to (\u2212B) + A. This is true for scalars as well as vectors. For example, 5 \u2013 2 = 5 + (\u22122) = (\u22122) + 5. Access for free at openstax.org. 5.1 \u2022 Vector Addition and Subtraction: Graphical Methods 147 Figure 5.6 The diagram shows a vector, B, and the negative of this vector, \u2013B. Global angles are calculated in the counterclockwise direction. The clockwise direction is considered negative. For example, an from the positive x-axis. angle of south of west is the same as the global angle which can also be expressed as Using the Graphical Method of Vector Addition and Subtraction to Solve Physics Problems Now that we have the skills to work with vectors in two dimensions, we can apply vector addition to graphically determine the resultant vector, which represents the total force. Consider an example of force involving two ice skaters pushing a third as seen in Figure 5.7. Figure 5.7 Part (a) shows an overhead view of two ice skaters pushing on a third. Forces are vectors and add like vectors, so the total force on the third skater is in the direction shown. In part (b), we see a free-body diagram representing the forces acting on the third skater. In problems where variables such as force are already known, the forces can be represented by making the length of the vectors proportional to the magnitudes of the forces. For this, you need to create a scale. For example, each centimeter of vector length could represent 50 N worth of force. Once you have the initial vectors drawn to scale, you can then use the head-to-tail method to draw the resultant vector. The length of the resultant can then be measured and converted back to the original units using the scale you created. You can tell by looking at the vectors in the free-body diagram in Figure 5.7 that the two skaters are pushing on the third skater with equal-magnitude forces, since the length of their force vectors are the same. Note, however, that the forces are not equal because they act in different directions. If, for example, each force had a magnitude of 400 N, then we would find the magnitude of the total external force acting on the third skater by finding the magnitude of the resultant vector. Since the forces act at a right angle to one another, we can use the Pythagorean theorem. For a triangle with sides a, b, and c, the Pythagorean theorem tells us that Applying this theorem to the triangle made by F1, F2, and Ftot in Figure 5.7, we get 148 Chapter 5 \u2022 Motion in Two Dimensions or Note that, if the vectors were not at a right angle to each other Pythagorean theorem to find the magnitude of the resultant vector. Another scenario where adding two-dimensional vectors is necessary is for velocity, where the", " direction may not be purely east-west or north-south, but some combination of these two directions. In the next section, we cover how to solve this type of problem analytically. For now let\u2019s consider the problem graphically. to one another), we would not be able to use the WORKED EXAMPLE Adding Vectors Graphically by Using the Head-to-Tail Method: A Woman Takes a Walk Use the graphical technique for adding vectors to find the total displacement of a person who walks the following three paths north of (displacements) on a flat field. First, he walks 25 m in a direction east. Finally, he turns and walks 32 m in a direction Strategy Graphically represent each displacement vector with an arrow, labeling the first A, the second B, and the third C. Make the lengths proportional to the distance of the given displacement and orient the arrows as specified relative to an east-west line. Use the head-to-tail method outlined above to determine the magnitude and direction of the resultant displacement, which we\u2019ll call R. north of east. Then, he walks 23 m heading south of east. Solution (1) Draw the three displacement vectors, creating a convenient scale (such as 1 cm of vector length on paper equals 1 m in the problem), as shown in Figure 5.8. (2) Place the vectors head to tail, making sure not to change their magnitude or direction, as shown in Figure 5.9. Figure 5.8 The three displacement vectors are drawn first. (3) Draw the resultant vector R from the tail of the first vector to the head of the last vector, as shown in Figure 5.10. Figure 5.9 Next, the vectors are placed head to tail. Access for free at openstax.org. 5.1 \u2022 Vector Addition and Subtraction: Graphical Methods 149 Figure 5.10 The resultant vector is drawn. (4) Use a ruler to measure the magnitude of R, remembering to convert back to the units of meters using the scale. Use a protractor to measure the direction of R. While the direction of the vector can be specified in many ways, the easiest way is to measure the angle between the vector and the nearest horizontal or vertical axis. Since R is south of the eastward pointing axis (the x-axis), we flip the protractor upside down and measure the angle between the eastward axis and the vector, as illustrated in Figure 5.11. Figure 5.11 A ruler is used to measure the magnitude of R, and a protractor is used to measure the direction of R. In this case, the total displacement R has a magnitude of 50 m and points this vector can be expressed as south of east. Using its magnitude and direction, and 5.1 5.2 Discussion The head-to-tail graphical method of vector addition works for any number of vectors. It is also important to note that it does not matter in what order the vectors are added. Changing the order does not change the resultant. For example, we could add the vectors as shown in Figure 5.12, and we would still get the same solution. 150 Chapter 5 \u2022 Motion in Two Dimensions Figure 5.12 Vectors can be added in any order to get the same result. WORKED EXAMPLE Subtracting Vectors Graphically: A Woman Sailing a Boat A woman sailing a boat at night is following directions to a dock. The instructions read to first sail 27.5 m in a direction west of north). If the north of east from her current location, and then travel 30.0 m in a direction woman makes a mistake and travels in the oppositedirection for the second leg of the trip, where will she end up? The two legs of the woman\u2019s trip are illustrated in Figure 5.13. north of east (or Figure 5.13 In the diagram, the first leg of the trip is represented by vector A and the second leg is represented by vector B. Strategy We can represent the first leg of the trip with a vector A, and the second leg of the trip that she was supposed totake with a vector B. Since the woman mistakenly travels in the opposite direction for the second leg of the journey, the vector for second leg of the trip she actuallytakes is \u2212B. Therefore, she will end up at a location A + (\u2212B), or A \u2212 B. Note that \u2212B has the same south of east, as illustrated in Figure 5.14. magnitude as B (30.0 m), but is in the opposite direction, Figure 5.14 Vector \u2013B represents traveling in the opposite direction of vector B. We use graphical vector addition to find where the woman arrives A + (\u2212B). Access for free at openstax.org. 5.1 \u2022 Vector Addition and Subtraction: Graphical Methods 151 Solution (1) To determine the location at which the woman arrives by accident, draw", " vectors A and \u2212B. (2) Place the vectors head to tail. (3) Draw the resultant vector R. (4) Use a ruler and protractor to measure the magnitude and direction of R. These steps are demonstrated in Figure 5.15. Figure 5.15 The vectors are placed head to tail. In this case and 5.3 5.4 Discussion Because subtraction of a vector is the same as addition of the same vector with the opposite direction, the graphical method for subtracting vectors works the same as for adding vectors. WORKED EXAMPLE Adding Velocities: A Boat on a River A boat attempts to travel straight across a river at a speed of 3.8 m/s. The river current flows at a speed vriver of 6.1 m/s to the right. What is the total velocity and direction of the boat? You can represent each meter per second of velocity as one centimeter of vector length in your drawing. Strategy We start by choosing a coordinate system with its x-axis parallel to the velocity of the river. Because the boat is directed straight toward the other shore, its velocity is perpendicular to the velocity of the river. We draw the two vectors, vboat and vriver, as shown in Figure 5.16. Using the head-to-tail method, we draw the resulting total velocity vector from the tail of vboat to the head of vriver. 152 Chapter 5 \u2022 Motion in Two Dimensions Figure 5.16 A boat attempts to travel across a river. What is the total velocity and direction of the boat? Solution By using a ruler, we find that the length of the resultant vector is 7.2 cm, which means that the magnitude of the total velocity is By using a protractor to measure the angle, we find Discussion If the velocity of the boat and river were equal, then the direction of the total velocity would have been 45\u00b0. However, since the velocity of the river is greater than that of the boat, the direction is less than 45\u00b0 with respect to the shore, or xaxis. 5.5 Practice Problems 1. Vector, having magnitude, pointing south of east and vector having magnitude, pointing north of east are added. What is the magnitude of the resultant vector? a. b. c. d. 2. A person walks north of west for and east of south for. What is the magnitude of his displacement? a. b. c. d. Virtual Physics Vector Addition In this simulation (https://archive.cnx.org/specials/d218bf9b-e50e-4d50-9a6c-b3db4dad0816/vector-addition/), you will experiment with adding vectors graphically. Click and drag the red vectors from the Grab One basket onto the graph in the middle of the screen. These red vectors can be rotated, stretched, or repositioned by clicking and dragging with your mouse. Check the Show Sum box to display the resultant vector (in green), which is the sum of all of the red vectors placed on the Access for free at openstax.org. 5.2 \u2022 Vector Addition and Subtraction: Analytical Methods 153 graph. To remove a red vector, drag it to the trash or click the Clear All button if you wish to start over. Notice that, if you click on any of the vectors, the component, and Ryis its vertical component. You can check the resultant by lining up the vectors so that the head of the first vector touches the tail of the second. Continue until all of the vectors are aligned together head-to-tail. You will see that the resultant magnitude and angle is the same as the arrow drawn from the tail of the first vector to the head of the last vector. Rearrange the vectors in any order head-to-tail and compare. The resultant will always be the same. is its direction with respect to the positive x-axis, Rx is its horizontal is its magnitude, Click to view content (https://archive.cnx.org/specials/d218bf9b-e50e-4d50-9a6c-b3db4dad0816/vector-addition/) GRASP CHECK True or False\u2014The more long, red vectors you put on the graph, rotated in any direction, the greater the magnitude of the resultant green vector. a. True b. False Check Your Understanding 3. While there is no single correct choice for the sign of axes, which of the following are conventionally considered positive? a. backward and to the left b. backward and to the right forward and to the right c. forward and to the left d. 4. True or False\u2014A person walks 2 blocks east and 5 blocks north. Another person walks 5 blocks north and then two blocks east. The displacement of the first person will be more than the displacement of the second person. a. True b. False 5.2", " Vector Addition and Subtraction: Analytical Methods Section Learning Objectives By the end of this section, you will be able to do the following: \u2022 Define components of vectors \u2022 Describe the analytical method of vector addition and subtraction \u2022 Use the analytical method of vector addition and subtraction to solve problems Section Key Terms analytical method component (of a two-dimensional vector) Components of Vectors For the analytical method of vector addition and subtraction, we use some simple geometry and trigonometry, instead of using a ruler and protractor as we did for graphical methods. However, the graphical method will still come in handy to visualize the problem by drawing vectors using the head-to-tail method. The analytical method is more accurate than the graphical method, which is limited by the precision of the drawing. For a refresher on the definitions of the sine, cosine, and tangent of an angle, see Figure 5.17. 154 Chapter 5 \u2022 Motion in Two Dimensions Figure 5.17 For a right triangle, the sine, cosine, and tangent of \u03b8are defined in terms of the adjacent side, the opposite side, or the hypotenuse. In this figure, xis the adjacent side, yis the opposite side, and his the hypotenuse. Since, by definition, length of yby using, we can find the length xif we know hand by using. Similarly, we can find the. These trigonometric relationships are useful for adding vectors. When a vector acts in more than one dimension, it is useful to break it down into its x and y components. For a two-dimensional vector, a component is a piece of a vector that points in either the x- or y-direction. Every 2-d vector can be expressed as a sum of its x and y components. For example, given a vector like produce it. In this example, common situation in physics and happens to be the least complicated situation trigonometrically., add to form a right triangle, meaning that the angle between them is 90 degrees. This is a in Figure 5.18, we may want to find what two perpendicular vectors, and and Figure 5.18 The vector, with its tail at the origin of an x- y-coordinate system, is shown together with its x- and y-components, and These vectors form a right triangle. and triangle. are defined to be the components of along the x- and y-axes. The three vectors,,, and, form a right If the vector and y-components, we use the following relationships for a right triangle: is known, then its magnitude (its length) and its angle (its direction) are known. To find and, its x- and is the magnitude of A in the x-direction, where resultant with respect to the x-axis, as shown in Figure 5.19. is the magnitude of A in the y-direction, and is the angle of the Access for free at openstax.org. 5.2 \u2022 Vector Addition and Subtraction: Analytical Methods 155 Figure 5.19 The magnitudes of the vector components and can be related to the resultant vector and the angle with trigonometric identities. Here we see that and Suppose, for example, that Figure 5.20. is the vector representing the total displacement of the person walking in a city, as illustrated in Figure 5.20 We can use the relationships and to determine the magnitude of the horizontal and vertical component vectors in this example. Then A = 10.3 blocks and, so that This magnitude indicates that the walker has traveled 9 blocks to the east\u2014in other words, a 9-block eastward displacement. Similarly, 5.6 5.7 indicating that the walker has traveled 5 blocks to the north\u2014a 5-block northward displacement. 156 Chapter 5 \u2022 Motion in Two Dimensions Analytical Method of Vector Addition and Subtraction Calculating a resultant vector (or vector addition) is the reverse of breaking the resultant down into its components. If the perpendicular components definition, are known, then we can find analytically. How do we do this? Since, by of a vector and we solve for to find the direction of the resultant. Since this is a right triangle, the Pythagorean theorem (x2 + y2 = h2) for finding the hypotenuse applies. In this case, it becomes Solving for A gives In summary, to find the magnitude in Figure 5.21, we use the following relationships: and direction of a vector from its perpendicular components and, as illustrated Figure 5.21 The magnitude and direction of the resultant vector can be determined once the horizontal components and have been determined. Sometimes, the vectors added are not perfectly perpendicular to one another. An example of this is the case below, where the vectors are added to produce the resultant as illustrated in Figure 5.22. and Access for free at openstax.org", ". 5.2 \u2022 Vector Addition and Subtraction: Analytical Methods 157 Figure 5.22 Vectors and are two legs of a walk, and is the resultant or total displacement. You can use analytical methods to determine the magnitude and direction of. represent two legs of a walk (two displacements), then If and up at the tip of x-direction and then in the y-direction. Those paths are the x- and y-components of the resultant,. There are many ways to arrive at the same point. The person could have walked straight ahead first in the is the total displacement. The person taking the walk ends and If we know and, we can find and using the equations and. 1. Draw in the xand ycomponents of each vector (including the resultant) with a dashed line. Use the equations and angle of by vector A). to find the components. In Figure 5.23, these components are, and with its own x-axis (which is slightly above the x-axis used Vector makes an,, with the x-axis, and vector makes and angle of Figure 5.23 To add vectors and first determine the horizontal and vertical components of each vector. These are the dotted vectors shown in the image. 2. Find the xcomponent of the resultant by adding the xcomponent of the vectors and find the ycomponent of the resultant (as illustrated in Figure 5.24) by adding the ycomponent of the vectors. 158 Chapter 5 \u2022 Motion in Two Dimensions Figure 5.24 The vectors and add to give the magnitude of the resultant vector in the horizontal direction, Similarly, the vectors and add to give the magnitude of the resultant vector in the vertical direction, Now that we know the components of we can find its magnitude and direction. 3. To get the magnitude of the resultant R, use the Pythagorean theorem. 4. To get the direction of the resultant WATCH PHYSICS Classifying Vectors and Quantities Example This video contrasts and compares three vectors in terms of their magnitudes, positions, and directions. Click to view content (https://www.youtube.com/embed/Yp0EhcVBxNU) GRASP CHECK. Vector points in the northwest. If the vectors,, and were added points to the northeast. Vector points to the, and, have the same magnitude of. Vector Three vectors,, southwest exactly opposite to vector together, what would be the magnitude of the resultant vector? Why? a. b. c. d.. All of them will cancel each other out.. Two of them will cancel each other out. units. All of them will add together to give the resultant.. Two of them will add together to give the resultant. TIPS FOR SUCCESS In the video, the vectors were represented with an arrow above them rather than in bold. This is a common notation in math classes. Access for free at openstax.org. 5.2 \u2022 Vector Addition and Subtraction: Analytical Methods 159 Using the Analytical Method of Vector Addition and Subtraction to Solve Problems Figure 5.25 uses the analytical method to add vectors. WORKED EXAMPLE An Accelerating Subway Train Add the vector the y-axis is along the north\u2013south directions. A person first walks vector The person then walks in a direction to the vector shown in Figure 5.25, using the steps above. The x-axis is along the east\u2013west direction, and north of east, represented by in a direction north of east, represented by vector Figure 5.25 You can use analytical models to add vectors. Strategy The components of We will solve for these components and then add them in the x-direction and y-direction to find the resultant. along the x- and y-axes represent walking due east and due north to get to the same ending point. and Solution First, we find the components of and along the x- and y-axes. From the problem, we know that =, and. We find the x-components by using, which gives and Similarly, the y-components are found using and The x- and y-components of the resultant are and 160 Chapter 5 \u2022 Motion in Two Dimensions Now we can find the magnitude of the resultant by using the Pythagorean theorem 5.8 so that Finally, we find the direction of the resultant This is Discussion This example shows vector addition using the analytical method. Vector subtraction using the analytical method is very similar. It is just the addition of a negative vector. That is, components of. Therefore, the x- and y-components of the resultant. The components of \u2013 are the negatives of the are and and the rest of the method outlined above is identical to that for addition. Practice Problems 5. What is the magnitude of a vector whose x-component is 4 cm and whose y-component is 3 cm? 1 cm a. 5 cm b. c. 7 cm d", ". 25 cm 6. What is the magnitude of a vector that makes an angle of 30\u00b0 to the horizontal and whose x-component is 3 units? a. 2.61 units 3.00 units b. c. 3.46 units d. 6.00 units Access for free at openstax.org. 5.2 \u2022 Vector Addition and Subtraction: Analytical Methods 161 LINKS TO PHYSICS Atmospheric Science Figure 5.26 This picture shows Bert Foord during a television Weather Forecast from the Meteorological Office in 1963. (BBC TV) Atmospheric science is a physical science, meaning that it is a science based heavily on physics. Atmospheric science includes meteorology (the study of weather) and climatology (the study of climate). Climate is basically the average weather over a longer time scale. Weather changes quickly over time, whereas the climate changes more gradually. The movement of air, water and heat is vitally important to climatology and meteorology. Since motion is such a major factor in weather and climate, this field uses vectors for much of its math. Vectors are used to represent currents in the ocean, wind velocity and forces acting on a parcel of air. You have probably seen a weather map using vectors to show the strength (magnitude) and direction of the wind. Vectors used in atmospheric science are often three-dimensional. We won\u2019t cover three-dimensional motion in this text, but to go from two-dimensions to three-dimensions, you simply add a third vector component. Three-dimensional motion is represented as a combination of x-, y- and zcomponents, where zis the altitude. Vector calculus combines vector math with calculus, and is often used to find the rates of change in temperature, pressure or wind speed over time or distance. This is useful information, since atmospheric motion is driven by changes in pressure or temperature. The greater the variation in pressure over a given distance, the stronger the wind to try to correct that imbalance. Cold air tends to be more dense and therefore has higher pressure than warm air. Higher pressure air rushes into a region of lower pressure and gets deflected by the spinning of the Earth, and friction slows the wind at Earth\u2019s surface. Finding how wind changes over distance and multiplying vectors lets meteorologists, like the one shown in Figure 5.26, figure out how much rotation (spin) there is in the atmosphere at any given time and location. This is an important tool for tornado prediction. Conditions with greater rotation are more likely to produce tornadoes. GRASP CHECK Why are vectors used so frequently in atmospheric science? a. Vectors have magnitude as well as direction and can be quickly solved through scalar algebraic operations. b. Vectors have magnitude but no direction, so it becomes easy to express physical quantities involved in the atmospheric science. c. Vectors can be solved very accurately through geometry, which helps to make better predictions in atmospheric science. d. Vectors have magnitude as well as direction and are used in equations that describe the three dimensional motion of the atmosphere. Check Your Understanding 7. Between the analytical and graphical methods of vector additions, which is more accurate? Why? a. The analytical method is less accurate than the graphical method, because the former involves geometry and 162 Chapter 5 \u2022 Motion in Two Dimensions trigonometry. b. The analytical method is more accurate than the graphical method, because the latter involves some extensive calculations. c. The analytical method is less accurate than the graphical method, because the former includes drawing all figures to the right scale. d. The analytical method is more accurate than the graphical method, because the latter is limited by the precision of the drawing. 8. What is a component of a two dimensional vector? a. A component is a piece of a vector that points in either the xor ydirection. b. A component is a piece of a vector that has half of the magnitude of the original vector. c. A component is a piece of a vector that points in the direction opposite to the original vector. d. A component is a piece of a vector that points in the same direction as original vector but with double of its magnitude. 9. How can we determine the global angle (measured counter-clockwise from positive ) if we know and? a. b. c. d. 10. How can we determine the magnitude of a vector if we know the magnitudes of its components? a. b. c. d. 5.3 Projectile Motion Section Learning Objectives By the end of this section, you will be able to do the following: \u2022 Describe the properties of projectile motion \u2022 Apply kinematic equations and vectors to solve problems involving projectile motion Section Key Terms air resistance maximum height (of a projectile) projectile projectile motion range trajectory Properties of Projectile Motion Projectile motion is the motion of an object thrown (projected) into the air. After the initial force that launches the object, it only experiences", " the force of gravity. The object is called a projectile, and its path is called its trajectory. As an object travels through the air, it encounters a frictional force that slows its motion called air resistance. Air resistance does significantly alter trajectory motion, but due to the difficulty in calculation, it is ignored in introductory physics. The most important concept in projectile motion is that horizontal and vertical motions areindependent, meaning that they don\u2019t influence one another. Figure 5.27 compares a cannonball in free fall (in blue) to a cannonball launched horizontally in projectile motion (in red). You can see that the cannonball in free fall falls at the same rate as the cannonball in projectile motion. Keep in mind that if the cannon launched the ball with any vertical component to the velocity, the vertical displacements would not line up perfectly. Access for free at openstax.org. Since vertical and horizontal motions are independent, we can analyze them separately, along perpendicular axes. To do this, we separate projectile motion into the two components of its motion, one along the horizontal axis and the other along the vertical. 5.3 \u2022 Projectile Motion 163 Figure 5.27 The diagram shows the projectile motion of a cannonball shot at a horizontal angle versus one dropped with no horizontal velocity. Note that both cannonballs have the same vertical position over time. We\u2019ll call the horizontal axis the x-axis and the vertical axis the y-axis. For notation, d is the total displacement, and x and y are its components along the horizontal and vertical axes. The magnitudes of these vectors are xand y, as illustrated in Figure 5.28. Figure 5.28 A boy kicks a ball at angle \u03b8, and it is displaced a distance of s along its trajectory. As usual, we use velocity, acceleration, and displacement to describe motion. We must also find the components of these variables along the x- and y-axes. The components of acceleration are then very simple ay = \u2013g = \u20139.80 m/s2. Note that this definition defines the upwards direction as positive. Because gravity is vertical, ax = 0. Both accelerations are constant, so we can use the kinematic equations. For review, the kinematic equations from a previous chapter are summarized in Table 5.1. (when ) (when ) Table 5.1 Summary of Kinematic Equations (constant a) Where x is position, x0 is initial position, v is velocity, vavg is average velocity, tis time and a is acceleration. 164 Chapter 5 \u2022 Motion in Two Dimensions Solve Problems Involving Projectile Motion The following steps are used to analyze projectile motion: 1. Separate the motion into horizontal and vertical components along the x- and y-axes. These axes are perpendicular, so and are used. The magnitudes of the displacement along x- and y-axes are called and The magnitudes of the components of the velocity are and, where is the magnitude of the velocity and is its direction. Initial values are denoted with a subscript 0. 2. Treat the motion as two independent one-dimensional motions, one horizontal and the other vertical. The kinematic equations for horizontal and vertical motion take the following forms Vertical motion (assuming positive is up ) 3. Solve for the unknowns in the two separate motions (one horizontal and one vertical). Note that the only common variable between the motions is time. The problem solving procedures here are the same as for one-dimensional kinematics.. We can use the analytical method of vector 4. Recombine the two motions to find the total displacement and velocity addition, which uses displacement and velocity. and to find the magnitude and direction of the total is the direction of the displacement, and is the direction of the velocity. (See Figure 5.29 Access for free at openstax.org. 5.3 \u2022 Projectile Motion 165 Figure 5.29 (a) We analyze two-dimensional projectile motion by breaking it into two independent one-dimensional motions along the vertical and horizontal axes. (b) The horizontal motion is simple, because and is thus constant. (c) The velocity in the vertical direction begins to decrease as the object rises; at its highest point, the vertical velocity is zero. As the object falls towards the Earth again, the vertical velocity increases again in magnitude but points in the opposite direction to the initial vertical velocity. (d) The x- and y-motions are recombined to give the total velocity at any given point on the trajectory. TIPS FOR SUCCESS For problems of projectile motion, it is important to set up a coordinate system. The first step is to choose an initial position for and. Usually, it is simplest to set the initial position of the object so that and. 166 Chapter 5 \u2022 Motion in Two Dimensions WATCH PHYSICS Projectile at an Angle This video presents an example of finding the displacement (or range) of a projectile launched", " at an angle. It also reviews basic trigonometry for finding the sine, cosine and tangent of an angle. Click to view content (https://www.khanacademy.org/embed_video?v=ZZ39o1rAZWY) GRASP CHECK Assume the ground is uniformly level. If the horizontal component a projectile's velocity is doubled, but the vertical component is unchanged, what is the effect on the time of flight? a. The time to reach the ground would remain the same since the vertical component is unchanged. b. The time to reach the ground would remain the same since the vertical component of the velocity also gets doubled. c. The time to reach the ground would be halved since the horizontal component of the velocity is doubled. d. The time to reach the ground would be doubled since the horizontal component of the velocity is doubled. WORKED EXAMPLE A Fireworks Projectile Explodes High and Away During a fireworks display like the one illustrated in Figure 5.30, a shell is shot into the air with an initial speed of 70.0 m/s at an angle of 75\u00b0 above the horizontal. The fuse is timed to ignite the shell just as it reaches its highest point above the ground. (a) Calculate the height at which the shell explodes. (b) How much time passed between the launch of the shell and the explosion? (c) What is the horizontal displacement of the shell when it explodes? Figure 5.30 The diagram shows the trajectory of a fireworks shell. Strategy The motion can be broken into horizontal and vertical motions in which be zero and solve for the maximum height. and. We can then define and to Solution for (a) By height we mean the altitude or vertical position above the starting point. The highest point in any trajectory, the maximum height, is reached when ; this is the moment when the vertical velocity switches from positive (upwards) to negative (downwards). Since we know the initial velocity, initial position, and the value of vy when the firework reaches its maximum height, we use the following equation to find Access for free at openstax.org. Because and are both zero, the equation simplifies to Solving for gives Now we must find initial velocity of 70.0 m/s, and is the initial angle. Thus,, the component of the initial velocity in the y-direction. It is given by 5.3 \u2022 Projectile Motion 167, where is the and is so that Discussion for (a) Since up is positive, the initial velocity and maximum height are positive, but the acceleration due to gravity is negative. The maximum height depends only on the vertical component of the initial velocity. The numbers in this example are reasonable for large fireworks displays, the shells of which do reach such heights before exploding. Solution for (b) There is more than one way to solve for the time to the highest point. In this case, the easiest method is to use is zero, this equation reduces to. Because Note that the final vertical velocity,, at the highest point is zero. Therefore, Discussion for (b) This time is also reasonable for large fireworks. When you are able to see the launch of fireworks, you will notice several seconds pass before the shell explodes. Another way of finding the time is by using equation for., and solving the quadratic Solution for (c) Because air resistance is negligible, velocity multiplied by time as given by and the horizontal velocity is constant. The horizontal displacement is horizontal, where is equal to zero where is the x-component of the velocity, which is given by Now, The time for both motions is the same, and so is Discussion for (c) The horizontal motion is a constant velocity in the absence of air resistance. The horizontal displacement found here could be useful in keeping the fireworks fragments from falling on spectators. Once the shell explodes, air resistance has a major effect, and many fragments will land directly below, while some of the fragments may now have a velocity in the \u2013x direction due to the 168 Chapter 5 \u2022 Motion in Two Dimensions forces of the explosion. The expression we found for while solving part (a) of the previous problem works for any projectile motion problem where air resistance is negligible. Call the maximum height ; then, This equation defines the maximum height of a projectile. The maximum height depends only on the vertical component of the initial velocity. WORKED EXAMPLE Calculating Projectile Motion: Hot Rock Projectile Suppose a large rock is ejected from a volcano, as illustrated in Figure 5.31, with a speed of the horizontal. The rock strikes the side of the volcano at an altitude 20.0 m lower than its starting point. (a) Calculate the time it takes the rock to follow this path. and at an angle above Figure 5.31 The diagram shows the projectile motion of a large rock from a volcano. Strategy Breaking this two-dimensional motion into two independent one-dimensional motions will allow us to solve for", " the time. The time a projectile is in the air depends only on its vertical motion. Solution While the rock is in the air, it rises and then falls to a final position 20.0 m lower than its starting altitude. We can find the time for this by using If we take the initial position vertical component of the initial velocity, found from to be zero, then the final position is Now the initial vertical velocity is the 5.9 Substituting known values yields Rearranging terms gives a quadratic equation in This expression is a quadratic equation of the form \u201320.0. Its solutions are given by the quadratic formula, where the constants are a= 4.90, b= \u201314.3, and c= Access for free at openstax.org. 5.3 \u2022 Projectile Motion 169 This equation yields two solutions t= 3.96 and t= \u20131.03. You may verify these solutions as an exercise. The time is t = 3.96 s or \u20131.03 s. The negative value of time implies an event before the start of motion, so we discard it. Therefore, Discussion The time for projectile motion is completely determined by the vertical motion. So any projectile that has an initial vertical velocity of and lands 20.0 m below its starting altitude will spend 3.96 s in the air. Practice Problems 11. If an object is thrown horizontally, travels with an average x-component of its velocity equal to, and does not hit the ground, what will be the x-component of the displacement after a. b. c. d.? 12. If a ball is thrown straight up with an initial velocity of upward, what is the maximum height it will reach? a. b. c. d. The fact that vertical and horizontal motions are independent of each other lets us predict the range of a projectile. The range is the horizontal distance R traveled by a projectile on level ground, as illustrated in Figure 5.32. Throughout history, people have been interested in finding the range of projectiles for practical purposes, such as aiming cannons. 170 Chapter 5 \u2022 Motion in Two Dimensions Figure 5.32 Trajectories of projectiles on level ground. (a) The greater the initial speed, the greater the range for a given initial angle. (b) The effect of initial angle on the range of a projectile with a given initial speed. Note that any combination of trajectories that add to 90 degrees will have the same range in the absence of air resistance, although the maximum heights of those paths are different. How does the initial velocity of a projectile affect its range? Obviously, the greater the initial speed shown in the figure above. The initial angle range of a projectile on level groundis also has a dramatic effect on the range. When air resistance is negligible, the, the greater the range, as is the initial speed and where apply to problems where the initial and final y position are different, or to cases where the object is launched perfectly horizontally. is the initial angle relative to the horizontal. It is important to note that the range doesn\u2019t Virtual Physics Projectile Motion In this simulation you will learn about projectile motion by blasting objects out of a cannon. You can choose between objects such as a tank shell, a golf ball or even a Buick. Experiment with changing the angle, initial speed, and mass, and adding in air resistance. Make a game out of this simulation by trying to hit the target. Click to view content (https://archive.cnx.org/specials/317dbd00-8e61-4065-b3eb-f2b80db9b7ed/projectile-motion/) Access for free at openstax.org. 5.4 \u2022 Inclined Planes 171 GRASP CHECK If a projectile is launched on level ground, what launch angle maximizes the range of the projectile? a. b. c. d. Check Your Understanding 13. What is projectile motion? a. Projectile motion is the motion of an object projected into the air, which moves under the influence of gravity. b. Projectile motion is the motion of an object projected into the air which moves independently of gravity. c. Projectile motion is the motion of an object projected vertically upward into the air which moves under the influence of gravity. d. Projectile motion is the motion of an object projected horizontally into the air which moves independently of gravity. 14. What is the force experienced by a projectile after the initial force that launched it into the air in the absence of air resistance? a. The nuclear force b. The gravitational force c. The electromagnetic force d. The contact force 5.4 Inclined Planes Section Learning Objectives By the end of this section, you will be able to do the following: \u2022 Distinguish between static friction and kinetic friction \u2022 Solve problems involving inclined planes Section Key Terms kinetic friction static friction Static Friction and Kinetic Friction Recall from the previous chapter that", " friction is a force that opposes motion, and is around us all the time. Friction allows us to move, which you have discovered if you have ever tried to walk on ice. There are different types of friction\u2014kinetic and static. Kinetic friction acts on an object in motion, while static friction acts on an object or system at rest. The maximum static friction is usually greater than the kinetic friction between the objects. Imagine, for example, trying to slide a heavy crate across a concrete floor. You may push harder and harder on the crate and not move it at all. This means that the static friction responds to what you do\u2014it increases to be equal to and in the opposite direction of your push. But if you finally push hard enough, the crate seems to slip suddenly and starts to move. Once in motion, it is easier to keep it in motion than it was to get it started because the kinetic friction force is less than the static friction force. If you were to add mass to the crate, (for example, by placing a box on top of it) you would need to push even harder to get it started and also to keep it moving. If, on the other hand, you oiled the concrete you would find it easier to get the crate started and keep it going. Figure 5.33 shows how friction occurs at the interface between two objects. Magnifying these surfaces shows that they are rough on the microscopic level. So when you push to get an object moving (in this case, a crate), you must raise the object until it can skip along with just the tips of the surface hitting, break off the points, or do both. The harder the surfaces are pushed together (such as if another box is placed on the crate), the more force is needed to move them. 172 Chapter 5 \u2022 Motion in Two Dimensions Figure 5.33 Frictional forces, such as f, always oppose motion or attempted motion between objects in contact. Friction arises in part because of the roughness of the surfaces in contact, as seen in the expanded view. The magnitude of the frictional force has two forms: one for static friction, the other for kinetic friction. When there is no motion between the objects, the magnitude of static friction fs is is the coefficient of static friction and N is the magnitude of the normal force. Recall that the normal force opposes the where force of gravity and acts perpendicular to the surface in this example, but not always. Since the symbol means less than or equal to, this equation says that static friction can have a maximum value of That is, Static friction is a responsive force that increases to be equal and opposite to whatever force is exerted, up to its maximum limit. Once the applied force exceeds fs(max), the object will move. Once an object is moving, the magnitude of kinetic friction fk is given by where is the coefficient of kinetic friction. Friction varies from surface to surface because different substances are rougher than others. Table 5.2 compares values of static and kinetic friction for different surfaces. The coefficient of the friction depends on the two surfaces that are in contact. System Static Friction Kinetic Friction Rubber on dry concrete Rubber on wet concrete Wood on wood 1.0 0.7 0.5 Waxed wood on wet snow 0.14 Metal on wood Steel on steel (dry) Steel on steel (oiled) Teflon on steel 0.5 0.6 0.05 0.04 Bone lubricated by synovial fluid 0.016 Shoes on wood 0.9 Table 5.2 Coefficients of Static and Kinetic Friction 0.7 0.5 0.3 0.1 0.3 0.3 0.03 0.04 0.015 0.7 Access for free at openstax.org. 5.4 \u2022 Inclined Planes 173 System Static Friction Kinetic Friction Shoes on ice Ice on ice Steel on ice 0.1 0.1 0.4 0.05 0.03 0.02 Table 5.2 Coefficients of Static and Kinetic Friction Since the direction of friction is always opposite to the direction of motion, friction runs parallel to the surface between objects and perpendicular to the normal force. For example, if the crate you try to push (with a force parallel to the floor) has a mass of 100 kg, then the normal force would be equal to its weight perpendicular to the floor. If the coefficient of static friction is 0.45, you would have to exert a force parallel to the floor greater than to move the crate. Once there is motion, friction is less and the coefficient of kinetic friction might be 0.30, so that a force of only 290 N would keep it moving at a constant speed. If the floor were lubricated, both coefficients would be much smaller than they would be without lubrication. The coefficient of friction is unitless and is a number usually between 0 and 1.0. Working with Inclined Planes We discussed previously that when an object rests on a", " horizontal surface, there is a normal force supporting it equal in magnitude to its weight. Up until now, we dealt only with normal force in one dimension, with gravity and normal force acting perpendicular to the surface in opposing directions (gravity downward, and normal force upward). Now that you have the skills to work with forces in two dimensions, we can explore what happens to weight and the normal force on a tilted surface such as an inclined plane. For inclined plane problems, it is easier breaking down the forces into their components if we rotate the coordinate system, as illustrated in Figure 5.34. The first step when setting up the problem is to break down the force of weight into components. Figure 5.34 The diagram shows perpendicular and horizontal components of weight on an inclined plane. When an object rests on an incline that makes an angle with the horizontal, the force of gravity acting on the object is divided into two components: A force acting perpendicular to the plane, perpendicular force of weight, acting parallel to the plane, the object, so it acts upward along the plane., is typically equal in magnitude and opposite in direction to the normal force, The force, opposes the motion of, causes the object to accelerate down the incline. The force of friction,, and a force acting parallel to the plane,. The 174 Chapter 5 \u2022 Motion in Two Dimensions It is important to be careful when resolving the weight of the object into components. If the angle of the incline is at an angle to the horizontal, then the magnitudes of the weight components are Instead of memorizing these equations, it is helpful to be able to determine them from reason. To do this, draw the right triangle formed by the three weight vectors. Notice that the angle of the incline is the same as the angle formed between and. Knowing this property, you can use trigonometry to determine the magnitude of the weight components WATCH PHYSICS Inclined Plane Force Components This video (https://www.khanacademy.org/embed_video?v=TC23wD34C7k) shows how the weight of an object on an inclined plane is broken down into components perpendicular and parallel to the surface of the plane. It explains the geometry for finding the angle in more detail. GRASP CHECK Click to view content (https://www.youtube.com/embed/TC23wD34C7k) This video shows how the weight of an object on an inclined plane is broken down into components perpendicular and parallel to the surface of the plane. It explains the geometry for finding the angle in more detail. When the surface is flat, you could say that one of the components of the gravitational force is zero; Which one? As the angle of the incline gets larger, what happens to the magnitudes of the perpendicular and parallel components of gravitational force? a. When the angle is zero, the parallel component is zero and the perpendicular component is at a maximum. As the angle increases, the parallel component decreases and the perpendicular component increases. This is because the cosine of the angle shrinks while the sine of the angle increases. b. When the angle is zero, the parallel component is zero and the perpendicular component is at a maximum. As the angle increases, the parallel component decreases and the perpendicular component increases. This is because the cosine of the angle increases while the sine of the angle shrinks. c. When the angle is zero, the parallel component is zero and the perpendicular component is at a maximum. As the angle increases, the parallel component increases and the perpendicular component decreases. This is because the cosine of the angle shrinks while the sine of the angle increases. d. When the angle is zero, the parallel component is zero and the perpendicular component is at a maximum. As the angle increases, the parallel component increases and the perpendicular component decreases. This is because the cosine of the angle increases while the sine of the angle shrinks. TIPS FOR SUCCESS Normal force is represented by the variable This should not be confused with the symbol for the newton, which is also represented by the letter N. It is important to tell apart these symbols, especially since the units for normal force ( to be newtons (N). For example, the normal force, difference is that normal force is a vector, while the newton is simply a unit. Be careful not to confuse these letters in your calculations!, that the floor exerts on a chair might be One important ) happen To review, the process for solving inclined plane problems is as follows: Access for free at openstax.org. 5.4 \u2022 Inclined Planes 175 1. Draw a sketch of the problem. 2. 3. Draw a free-body diagram (which is a sketch showing all of the forces acting on an object) with the coordinate system Identify known and unknown quantities, and identify the system of interest. rotated at the same angle as the inclined plane. Resolve the vectors into horizontal and vertical components and", " draw them on the free-body diagram. 4. Write Newton\u2019s second law in the horizontal and vertical directions and add the forces acting on the object. If the object does not accelerate in a particular direction (for example, the x-direction) then Fnet x= 0. If the object does accelerate in that direction, Fnet x= ma. 5. Check your answer. Is the answer reasonable? Are the units correct? WORKED EXAMPLE Finding the Coefficient of Kinetic Friction on an Inclined Plane A skier, illustrated in Figure 5.35(a), with a mass of 62 kg is sliding down a snowy slope at an angle of 25 degrees. Find the coefficient of kinetic friction for the skier if friction is known to be 45.0 N. Figure 5.35 Use the diagram to help find the coefficient of kinetic friction for the skier. Strategy. Therefore, The magnitude of kinetic friction was given as 45.0 N. Kinetic friction is related to the normal force N as we can find the coefficient of kinetic friction by first finding the normal force of the skier on a slope. The normal force is always perpendicular to the surface, and since there is no motion perpendicular to the surface, the normal force should equal the component of the skier\u2019s weight perpendicular to the slope. That is, Substituting this into our expression for kinetic friction, we get which can now be solved for the coefficient of kinetic friction \u03bck. Solution Solving for gives Substituting known values on the right-hand side of the equation, Discussion This result is a little smaller than the coefficient listed in Table 5.2 for waxed wood on snow, but it is still reasonable since values 176 Chapter 5 \u2022 Motion in Two Dimensions of the coefficients of friction can vary greatly. In situations like this, where an object of mass mslides down a slope that makes an angle \u03b8with the horizontal, friction is given by WORKED EXAMPLE Weight on an Incline, a Two-Dimensional Problem The skier\u2019s mass, including equipment, is 60.0 kg. (See Figure 5.36(b).) (a) What is her acceleration if friction is negligible? (b) What is her acceleration if the frictional force is 45.0 N? Figure 5.36 Now use the diagram to help find the skier's acceleration if friction is negligible and if the frictional force is 45.0 N. Strategy The most convenient coordinate system for motion on an incline is one that has one coordinate parallel to the slope and one perpendicular to the slope. Remember that motions along perpendicular axes are independent. We use the symbol perpendicular, and to mean parallel. to mean, and in the free-body diagram. The only external forces acting on the system are the skier\u2019s weight, friction, and the normal force exerted by the ski slope, labeled, direction of either axis, so we must break it down into components along the chosen axes. We define weight parallel to the slope and two separate problems of forces parallel to the slope and forces perpendicular to the slope. to be the component of the component of weight perpendicular to the slope. Once this is done, we can consider the is always perpendicular to the slope and is parallel to it. But is not in the Solution The magnitude of the component of the weight parallel to the slope is the component of the weight perpendicular to the slope is, and the magnitude of (a) Neglecting friction: Since the acceleration is parallel to the slope, we only need to consider forces parallel to the slope. Forces perpendicular to the slope add to zero, since there is no acceleration in that direction. The forces parallel to the slope are the amount of the skier\u2019s weight parallel to the slope acceleration parallel to the slope is and friction. Assuming no friction, by Newton\u2019s second law the Where the net force parallel to the slope, so that is the acceleration. (b) Including friction: Here we now have a given value for friction, and we know its direction is parallel to the slope and it opposes motion between surfaces in contact. So the net external force is now Access for free at openstax.org. 5.4 \u2022 Inclined Planes 177 and substituting this into Newton\u2019s second law, gives We substitute known values to get or which is the acceleration parallel to the incline when there is 45 N opposing friction. Discussion Since friction always opposes motion between surfaces, the acceleration is smaller when there is friction than when there is not. Practice Problems 15. When an object sits on an inclined plane that makes an angle \u03b8with the horizontal, what is the expression for the component of the objects weight force that is parallel to the incline? a. b. c. d. rests on a plane inclined from horizontal. What is the component of the weight force that 16. An object with a mass of is parallel to the incline? a. b. c. d", ". Snap Lab Friction at an Angle: Sliding a Coin An object will slide down an inclined plane at a constant velocity if the net force on the object is zero. We can use this fact to measure the coefficient of kinetic friction between two objects. As shown in the first Worked Example, the kinetic friction on a slope opposite directions, so when they have equal magnitude, the acceleration is zero. Writing these out, and the component of the weight down the slope is equal to. These forces act in Solving for, since we find that \u2022 \u2022 \u2022 1 coin 1 book 1 protractor 1. Put a coin flat on a book and tilt it until the coin slides at a constant velocity down the book. You might need to tap the book lightly to get the coin to move. 5.10 178 Chapter 5 \u2022 Motion in Two Dimensions 2. Measure the angle of tilt relative to the horizontal and find. GRASP CHECK True or False\u2014If only the angles of two vectors are known, we can find the angle of their resultant addition vector. a. True b. False Check Your Understanding 17. What is friction? a. Friction is an internal force that opposes the relative motion of an object. b. Friction is an internal force that accelerates an object\u2019s relative motion. c. Friction is an external force that opposes the relative motion of an object. d. Friction is an external force that increases the velocity of the relative motion of an object. 18. What are the two varieties of friction? What does each one act upon? a. Kinetic and static friction both act on an object in motion. b. Kinetic friction acts on an object in motion, while static friction acts on an object at rest. c. Kinetic friction acts on an object at rest, while static friction acts on an object in motion. d. Kinetic and static friction both act on an object at rest. 19. Between static and kinetic friction between two surfaces, which has a greater value? Why? a. The kinetic friction has a greater value because the friction between the two surfaces is more when the two surfaces are in relative motion. b. The static friction has a greater value because the friction between the two surfaces is more when the two surfaces are in relative motion. c. The kinetic friction has a greater value because the friction between the two surfaces is less when the two surfaces are in relative motion. d. The static friction has a greater value because the friction between the two surfaces is less when the two surfaces are in relative motion. 5.5 Simple Harmonic Motion Section Learning Objectives By the end of this section, you will be able to do the following: \u2022 Describe Hooke\u2019s law and Simple Harmonic Motion \u2022 Describe periodic motion, oscillations, amplitude, frequency, and period \u2022 Solve problems in simple harmonic motion involving springs and pendulums Section Key Terms amplitude deformation equilibrium position frequency Hooke\u2019s law oscillate period periodic motion restoring force simple harmonic motion simple pendulum Hooke\u2019s Law and Simple Harmonic Motion Imagine a car parked against a wall. If a bulldozer pushes the car into the wall, the car will not move but it will noticeably change shape. A change in shape due to the application of a force is a deformation. Even very small forces are known to cause some deformation. For small deformations, two important things can happen. First, unlike the car and bulldozer example, the object returns to its original shape when the force is removed. Second, the size of the deformation is proportional to the force. This Access for free at openstax.org. 5.5 \u2022 Simple Harmonic Motion 179 second property is known as Hooke\u2019s law. In equation form, Hooke\u2019s law is where x is the amount of deformation (the change in length, for example) produced by the restoring force F, and k is a constant that depends on the shape and composition of the object. The restoring force is the force that brings the object back to its equilibrium position; the minus sign is there because the restoring force acts in the direction opposite to the displacement. Note that the restoring force is proportional to the deformation x. The deformation can also be thought of as a displacement from equilibrium. It is a change in position due to a force. In the absence of force, the object would rest at its equilibrium position. The force constant k is related to the stiffness of a system. The larger the force constant, the stiffer the system. A stiffer system is more difficult to deform and requires a greater restoring force. The units of k are newtons per meter (N/m). One of the most common uses of Hooke\u2019s law is solving problems involving springs and pendulums, which we will cover at the end of this section. Oscillations and Periodic Motion What do an ocean buoy, a child in a swing, a guitar, and the beating of", " hearts all have in common? They all oscillate. That is, they move back and forth between two points, like the ruler illustrated in Figure 5.37. All oscillations involve force. For example, you push a child in a swing to get the motion started. Figure 5.37 A ruler is displaced from its equilibrium position. Newton\u2019s first law implies that an object oscillating back and forth is experiencing forces. Without force, the object would move in a straight line at a constant speed rather than oscillate. Consider, for example, plucking a plastic ruler to the left as shown in Figure 5.38. The deformation of the ruler creates a force in the opposite direction, known as a restoring force. Once released, the restoring force causes the ruler to move back toward its stable equilibrium position, where the net force on it is zero. However, by the time the ruler gets there, it gains momentum and continues to move to the right, producing the opposite deformation. It is then forced to the left, back through equilibrium, and the process is repeated until it gradually loses all of its energy. The simplest oscillations occur when the restoring force is directly proportional to displacement. Recall that Hooke\u2019s law describes this situation with the equation F = \u2212kx. Therefore, Hooke\u2019s law describes and applies to the simplest case of oscillation, known as simple harmonic motion. Figure 5.38 (a) The plastic ruler has been released, and the restoring force is returning the ruler to its equilibrium position. (b) The net force is zero at the equilibrium position, but the ruler has momentum and continues to move to the right. (c) The restoring force is in the opposite direction. It stops the ruler and moves it back toward equilibrium again. (d) Now the ruler has momentum to the left. (e) In the absence of damping (caused by frictional forces), the ruler reaches its original position. From there, the motion will repeat itself. 180 Chapter 5 \u2022 Motion in Two Dimensions When you pluck a guitar string, the resulting sound has a steady tone and lasts a long time. Each vibration of the string takes the same time as the previous one. Periodic motion is a motion that repeats itself at regular time intervals, such as with an object bobbing up and down on a spring or a pendulum swinging back and forth. The time to complete one oscillation (a complete cycle of motion) remains constant and is called the period T. Its units are usually seconds. Frequency fis the number of oscillations per unit time. The SI unit for frequency is the hertz (Hz), defined as the number of oscillations per second. The relationship between frequency and period is As you can see from the equation, frequency and period are different ways of expressing the same concept. For example, if you get a paycheck twice a month, you could say that the frequency of payment is two per month, or that the period between checks is half a month. If there is no friction to slow it down, then an object in simple motion will oscillate forever with equal displacement on either side of the equilibrium position. The equilibrium position is where the object would naturally rest in the absence of force. The maximum displacement from equilibrium is called the amplitude X. The units for amplitude and displacement are the same, but depend on the type of oscillation. For the object on the spring, shown in Figure 5.39, the units of amplitude and displacement are meters. Figure 5.39 An object attached to a spring sliding on a frictionless surface is a simple harmonic oscillator. When displaced from equilibrium, the object performs simple harmonic motion that has an amplitude X and a period T. The object\u2019s maximum speed occurs as it passes through equilibrium. The stiffer the spring is, the smaller the period T. The greater the mass of the object is, the greater the period T. The mass mand the force constant k are the onlyfactors that affect the period and frequency of simple harmonic motion. The period of a simple harmonic oscillator is given by and, because f= 1/T, the frequency of a simple harmonic oscillator is Access for free at openstax.org. 5.5 \u2022 Simple Harmonic Motion 181 WATCH PHYSICS Introduction to Harmonic Motion This video shows how to graph the displacement of a spring in the x-direction over time, based on the period. Watch the first 10 minutes of the video (you can stop when the narrator begins to cover calculus). Click to view content (https://www.khanacademy.org/embed_video?v=Nk2q-_jkJVs) GRASP CHECK If the amplitude of the displacement of a spring were larger, how would this affect the graph of displacement over time? What would happen to the graph if the period was longer? a. Larger amplitude would result in taller peaks and troughs and a longer period would result in greater separation in time", " between peaks. b. Larger amplitude would result in smaller peaks and troughs and a longer period would result in greater distance between peaks. c. Larger amplitude would result in taller peaks and troughs and a longer period would result in shorter distance between peaks. d. Larger amplitude would result in smaller peaks and troughs and a longer period would result in shorter distance between peaks. Solving Spring and Pendulum Problems with Simple Harmonic Motion Before solving problems with springs and pendulums, it is important to first get an understanding of how a pendulum works. Figure 5.40 provides a useful illustration of a simple pendulum. Figure 5.40 A simple pendulum has a small-diameter bob and a string that has a very small mass but is strong enough not to stretch. The linear displacement from equilibrium is s, the length of the arc. Also shown are the forces on the bob, which result in a net force of \u2212mgsin\u03b8 toward the equilibrium position\u2014that is, a restoring force. Everyday examples of pendulums include old-fashioned clocks, a child\u2019s swing, or the sinker on a fishing line. For small displacements of less than 15 degrees, a pendulum experiences simple harmonic oscillation, meaning that its restoring force is directly proportional to its displacement. A pendulum in simple harmonic motion is called a simple pendulum. A pendulum has an object with a small mass, also known as the pendulum bob, which hangs from a light wire or string. The equilibrium position for a pendulum is where the angle is zero (that is, when the pendulum is hanging straight down). It makes sense that without any force applied, this is where the pendulum bob would rest. The displacement of the pendulum bob is the arc length s. The weight mg has components mg cos along the string and mg sin tangent to the arc. Tension in the string exactly cancels the component mg cos parallel to the string. This leaves a net restoring force back toward the equilibrium position that runs tangent to the arc and equals \u2212mg sin. 182 Chapter 5 \u2022 Motion in Two Dimensions For a simple pendulum, The period is The only things that affect the period of a simple pendulum are its length and the acceleration due to gravity. The period is completely independent of other factors, such as mass or amplitude. However, note that Tdoes depend on g. This means that if we know the length of a pendulum, we can actually use it to measure gravity! This will come in useful in Figure 5.40. TIPS FOR SUCCESS Tension is represented by the variable T, and period is represented by the variable T. It is important not to confuse the two, since tension is a force and period is a length of time. WORKED EXAMPLE Measuring Acceleration due to Gravity: The Period of a Pendulum What is the acceleration due to gravity in a region where a simple pendulum having a length 75.000 cm has a period of 1.7357 s? Strategy We are asked to find g given the period Tand the length Lof a pendulum. We can solve for g, assuming that the angle of deflection is less than 15 degrees. Recall that when the angle of deflection is less than 15 degrees, the pendulum is considered to be in simple harmonic motion, allowing us to use this equation. Solution 1. Square and solve for g. 2. Substitute known values into the new equation. 3. Calculate to find g. Discussion This method for determining g can be very accurate. This is why length and period are given to five digits in this example. WORKED EXAMPLE Hooke\u2019s Law: How Stiff Are Car Springs? What is the force constant for the suspension system of a car, like that shown in Figure 5.41, that settles 1.20 cm when an 80.0-kg person gets in? Access for free at openstax.org. 5.5 \u2022 Simple Harmonic Motion 183 Figure 5.41 A car in a parking lot. (exfordy, Flickr) Strategy Consider the car to be in its equilibrium position x = 0 before the person gets in. The car then settles down 1.20 cm, which means it is displaced to a position x = \u22121.20\u00d710\u22122 m. At that point, the springs supply a restoring force F equal to the person\u2019s weight w = mg = (80.0 kg)(9.80 m/s2) = 784 N. We take this force to be F in Hooke\u2019s law. Knowing F and x, we can then solve for the force constant k. Solution Solve Hooke\u2019s law, F = \u2212kx, for k. Substitute known values and solve for k. Discussion Note that F and x have opposite signs because they are in opposite directions\u2014the restoring force is up, and the displacement is down. Also, note that", " the car would oscillate up and down when the person got in, if it were not for the shock absorbers. Bouncing cars are a sure sign of bad shock absorbers. Practice Problems applied to a spring causes it to be displaced by. What is the force constant of the spring? 20. A force of a. b. c. d. 21. What is the force constant for the suspension system of a car that settles when a person gets in? a. b. 184 Chapter 5 \u2022 Motion in Two Dimensions c. d. Snap Lab Finding Gravity Using a Simple Pendulum Use a simple pendulum to find the acceleration due to gravity g in your home or classroom. \u2022 \u2022 \u2022 1 string 1 stopwatch 1 small dense object 1. Cut a piece of a string or dental floss so that it is about 1 m long. 2. Attach a small object of high density to the end of the string (for example, a metal nut or a car key). 3. Starting at an angle of less than 10 degrees, allow the pendulum to swing and measure the pendulum\u2019s period for 10 oscillations using a stopwatch. 4. Calculate g. GRASP CHECK How accurate is this measurement for g? How might it be improved? a. Accuracy for value of gwill increase with an increase in the mass of a dense object. b. Accuracy for the value of gwill increase with increase in the length of the pendulum. c. The value of gwill be more accurate if the angle of deflection is more than 15\u00b0. d. The value of gwill be more accurate if it maintains simple harmonic motion. Check Your Understanding 22. What is deformation? a. Deformation is the magnitude of the restoring force. b. Deformation is the change in shape due to the application of force. c. Deformation is the maximum force that can be applied on a spring. d. Deformation is regaining the original shape upon the removal of an external force. 23. According to Hooke\u2019s law, what is deformation proportional to? a. Force b. Velocity c. Displacement d. Force constant 24. What are oscillations? a. Motion resulting in small displacements b. Motion which repeats itself periodically c. Periodic, repetitive motion between two points d. motion that is the opposite to the direction of the restoring force 25. True or False\u2014Oscillations can occur without force. a. True b. False Access for free at openstax.org. Chapter 5 \u2022 Key Terms 185 KEY TERMS air resistance a frictional force that slows the motion of objects as they travel through the air; when solving basic physics problems, air resistance is assumed to be zero maximum height (of a projectile) the highest altitude, or maximum displacement in the vertical position reached in the path of a projectile amplitude the maximum displacement from the oscillate moving back and forth regularly between two equilibrium position of an object oscillating around the equilibrium position analytical method the method of determining the magnitude and direction of a resultant vector using the Pythagorean theorem and trigonometric identities component (of a 2-dimensional vector) a piece of a vector that points in either the vertical or the horizontal direction; every 2-d vector can be expressed as a sum of two vertical and horizontal vector components points period time it takes to complete one oscillation periodic motion motion that repeats itself at regular time intervals projectile an object that travels through the air and experiences only acceleration due to gravity projectile motion the motion of an object that is subject only to the acceleration of gravity range the maximum horizontal distance that a projectile deformation displacement from equilibrium, or change in travels shape due to the application of force restoring force force acting in opposition to the force equilibrium position where an object would naturally rest caused by a deformation in the absence of force frequency number of events per unit of time graphical method drawing vectors on a graph to add them using the head-to-tail method head (of a vector) the end point of a vector; the location of the sum of the a collection of vectors resultant resultant vector simple harmonic motion the oscillatory motion in a the vector sum of two or more vectors system where the net force can be described by Hooke\u2019s law the vector\u2019s arrow; also referred to as the tip simple pendulum an object with a small mass suspended head-to-tail method a method of adding vectors in which from a light wire or string the tail of each vector is placed at the head of the previous vector Hooke\u2019s law proportional relationship between the force F static friction a force that opposes the motion of two systems that are in contact and are not moving relative to one another on a material and the deformation it causes, tail the starting point of a vector; the point opposite to the kinetic friction a force that opposes the motion of two systems that are in contact and moving relative to one another SECTION SUMMARY 5.1 Vector Addition and Subtraction: Graphical Methods \u2022", " The graphical method of adding vectors and involves drawing vectors on a graph and adding them by using the head-to-tail method. The resultant vector is defined such that A + B = R. The magnitude and direction of protractor. are then determined with a ruler and \u2022 The graphical method of subtracting vectors A and B involves adding the opposite of vector B, which is defined as \u2212B. In this case, Next, use the head-totail method as for vector addition to obtain the resultant vector. \u2022 Addition of vectors is independent of the order in which they are added; A + B = B + A. \u2022 The head-to-tail method of adding vectors involves head or tip of the arrow trajectory the path of a projectile through the air vector addition adding together two or more vectors drawing the first vector on a graph and then placing the tail of each subsequent vector at the head of the previous vector. The resultant vector is then drawn from the tail of the first vector to the head of the final vector. \u2022 Variables in physics problems, such as force or velocity, can be represented with vectors by making the length of the vector proportional to the magnitude of the force or velocity. \u2022 Problems involving displacement, force, or velocity may be solved graphically by measuring the resultant vector\u2019s magnitude with a ruler and measuring the direction with a protractor. 5.2 Vector Addition and Subtraction: Analytical Methods \u2022 The analytical method of vector addition and subtraction uses the Pythagorean theorem and trigonometric identities to determine the magnitude 186 Chapter 5 \u2022 Key Equations and direction of a resultant vector. \u2022 The steps to add vectors method are as follows: and using the analytical 1. Determine the coordinate system for the vectors. Then, determine the horizontal and vertical components of each vector using the equations and 2. Add the horizontal and vertical components of each vector to determine the components resultant vector, and of the and 3. Use the Pythagorean theorem to determine the magnitude,, of the resultant vector \u2022 To solve projectile problems: choose a coordinate system; analyze the motion in the vertical and horizontal direction separately; then, recombine the horizontal and vertical components using vector addition equations. 5.4 Inclined Planes \u2022 Friction is a contact force between systems that opposes the motion or attempted motion between them. Simple friction is proportional to the normal force N pushing the systems together. A normal force is always perpendicular to the contact surface between systems. Friction depends on both of the materials involved. \u2022 \u00b5s is the coefficient of static friction, which depends on both of the materials. \u2022 \u00b5k is the coefficient of kinetic friction, which also depends on both materials. \u2022 When objects rest on an inclined plane that makes an angle with the horizontal surface, the weight of the object can be broken into components that act perpendicular the plane. and parallel ( ) to the surface of 4. Use a trigonometric identity to determine the direction, \u2022 An oscillation is a back and forth motion of an object 5.5 Simple Harmonic Motion, of 5.3 Projectile Motion \u2022 Projectile motion is the motion of an object through the air that is subject only to the acceleration of gravity. \u2022 Projectile motion in the horizontal and vertical directions are independent of one another. \u2022 The maximum height of an projectile is the highest altitude, or maximum displacement in the vertical position reached in the path of a projectile. \u2022 The range is the maximum horizontal distance traveled by a projectile. KEY EQUATIONS 5.2 Vector Addition and Subtraction: Analytical Methods resultant magnitude resultant direction x-component of a vector A (when an angle is given relative to the horizontal) Access for free at openstax.org. between two points of deformation. \u2022 An oscillation may create a wave, which is a disturbance that propagates from where it was created. \u2022 The simplest type of oscillations are related to systems that can be described by Hooke\u2019s law. \u2022 Periodic motion is a repetitious oscillation. \u2022 The time for one oscillation is the period T. \u2022 The number of oscillations per unit time is the frequency \u2022 A mass msuspended by a wire of length Lis a simple pendulum and undergoes simple harmonic motion for amplitudes less than about 15 degrees. y-component of a vector A (when an angle is given relative to the horizontal) addition of vectors 5.3 Projectile Motion angle of displacement velocity angle of velocity maximum height range Chapter 5 \u2022 Chapter Review 187 perpendicular component of weight on an inclined plane parallel component of weight on an inclined plane 5.4 Inclined Planes Hooke\u2019s law 5.5 Simple Harmonic Motion force of static friction force of kinetic friction period in simple harmonic motion frequency in simple harmonic motion period of a simple pendulum CHAPTER REVIEW Concept Items 5.1 Vector Addition and Subtraction: Graphical Methods d. 5.2 Vector Addition and Subtraction: Analytical Methods 1. There is a vector, with magnitude 5", " units pointing 4. What is the angle between the x and y components of a, with magnitude 3 units, towards west and vector pointing towards south. Using vector addition, calculate the magnitude of the resultant vector. a. 4.0 b. 5.8 c. 6.3 d. 8.0 2. If you draw two vectors using the head-to-tail method, how can you then draw the resultant vector? a. By joining the head of the first vector to the head of the last vector? a. b. c. d. 5. Two vectors are equal in magnitude and opposite in direction. What is the magnitude of their resultant vector? a. The magnitude of the resultant vector will be zero. b. The magnitude of resultant vector will be twice the magnitude of the original vector. b. By joining the head of the first vector with the tail of c. The magnitude of resultant vector will be same as the last magnitude of the original vector. c. By joining the tail of the first vector to the head of d. The magnitude of resultant vector will be half the the last magnitude of the original vector. d. By joining the tail of the first vector with the tail of the last 3. What is the global angle of south of west? a. b. c. 6. How can we express the x and y-components of a vector, and direction, global angle in terms of its magnitude,? a. b. c. 188 Chapter 5 \u2022 Chapter Review d. 7. True or False\u2014Every 2-D vector can be expressed as the c. d. 12. What equation gives the magnitude of kinetic friction? a. b. c. d. 5.5 Simple Harmonic Motion 13. Why is there a negative sign in the equation for Hooke\u2019s law? a. The negative sign indicates that displacement decreases with increasing force. b. The negative sign indicates that the direction of the applied force is opposite to that of displacement. c. The negative sign indicates that the direction of the restoring force is opposite to that of displacement. d. The negative sign indicates that the force constant must be negative. 14. With reference to simple harmonic motion, what is the equilibrium position? a. The position where velocity is the minimum b. The position where the displacement is maximum c. The position where the restoring force is the maximum d. The position where the object rests in the absence of force 15. What is Hooke\u2019s law? a. Restoring force is directly proportional to the displacement from the mean position and acts in the the opposite direction of the displacement. b. Restoring force is directly proportional to the displacement from the mean position and acts in the same direction as the displacement. c. Restoring force is directly proportional to the square of the displacement from the mean position and acts in the opposite direction of the displacement. d. Restoring force is directly proportional to the square of the displacement from the mean position and acts in the same direction as the displacement. displacement will be the same as it would have been had he followed directions correctly. a. True b. False product of its x and y-components. a. True b. False 5.3 Projectile Motion 8. Horizontal and vertical motions of a projectile are independent of each other. What is meant by this? a. Any object in projectile motion falls at the same rate as an object in freefall, regardless of its horizontal velocity. b. All objects in projectile motion fall at different rates, regardless of their initial horizontal velocities. c. Any object in projectile motion falls at the same rate as its initial vertical velocity, regardless of its initial horizontal velocity. d. All objects in projectile motion fall at different rates and the rate of fall of the object is independent of the initial velocity. 9. Using the conventional choice for positive and negative axes described in the text, what is the y-component of the acceleration of an object experiencing projectile motion? a. b. c. d. 5.4 Inclined Planes 10. True or False\u2014Kinetic friction is less than the limiting static friction because once an object is moving, there are fewer points of contact, and the friction is reduced. For this reason, more force is needed to start moving an object than to keep it in motion. a. True b. False 11. When there is no motion between objects, what is the relationship between the magnitude of the static friction and the normal force? a. b. Critical Thinking Items 5.1 Vector Addition and Subtraction: Graphical Methods 16. True or False\u2014A person is following a set of directions. He has to walk 2 km east and then 1 km north. He takes a wrong turn and walks in the opposite direction for the second leg of the trip. The magnitude of his total Access for free at openstax.org. 5.2 Vector Addition and Subtraction: Analytical Methods 17. What is the magnitude of a vector whose x-", "component and whose angle is? is a. b. c. d. 18. Vectors and are equal in magnitude and opposite in direction. Does have the same direction as vector or? a. b. Chapter 5 \u2022 Chapter Review 189 force pushes against it and it starts moving when is applied to it. What can be said about the coefficient of kinetic friction between the box and the floor? a. b. c. d. 22. The component of the weight parallel to an inclined with the horizontal is plane of an object resting on an incline that makes an angle of the object\u2019s mass? a. b. c. d.. What is 5.3 Projectile Motion 5.5 Simple Harmonic Motion 19. Two identical items, object 1 and object 2, are dropped building. Object 1 is dropped, while object 2 is thrown. from the top of a with an initial velocity of straight downward with an initial velocity of What is the difference in time, in seconds rounded to the nearest tenth, between when the two objects hit the ground? after object 2. a. Object 1 will hit the ground after object 2. b. Object 1 will hit the ground c. Object 1 will hit the ground at the same time as object 2. d. Object 1 will hit the ground after object 2. 20. An object is launched into the air. If the y-component of its acceleration is 9.8 m/s2, which direction is defined as positive? a. Vertically upward in the coordinate system b. Vertically downward in the coordinate system c. Horizontally to the right side of the coordinate system d. Horizontally to the left side of the coordinate system 5.4 Inclined Planes 21. A box weighing is at rest on the floor. A person Problems 5.1 Vector Addition and Subtraction: Graphical Methods 25. A person attempts to cross a river in a straight line by navigating a boat at. If the river flows at from his left to right, what would be the magnitude of the boat\u2019s resultant velocity? In what direction would the boat go, relative to the straight line 23. Two springs are attached to two hooks. Spring A has a greater force constant than spring B. Equal weights are suspended from both. Which of the following statements is true? a. Spring A will have more extension than spring B. b. Spring B will have more extension than spring A. c. Both springs will have equal extension. d. Both springs are equally stiff. 24. Two simple harmonic oscillators are constructed by attaching similar objects to two different springs. The force constant of the spring on the left is that of the spring on the right is force is applied to both, which of the following statements is true? a. The spring on the left will oscillate faster than and. If the same spring on the right. b. The spring on the right will oscillate faster than the spring on the left. c. Both the springs will oscillate at the same rate. d. The rate of oscillation is independent of the force constant. across it? a. The resultant velocity of the boat will be The boat will go toward his right at an angle of to a line drawn across the river. b. The resultant velocity of the boat will be The boat will go toward his left at an angle of to a line drawn across the river. c. The resultant velocity of the boat will be The boat will go toward his right at an angle of... 190 Chapter 5 \u2022 Chapter Review to a line drawn across the river. d. The resultant velocity of the boat will be. The boat will go toward his left at an angle of to a line drawn across the river. 26. A river flows in a direction from south west to north east. A boat captain wants to cross at a velocity of this river to reach a point on the opposite shore due east of the boat\u2019s current position. The boat moves at. Which direction should it head towards if the resultant velocity is a. b. c. d. It should head in a direction It should head in a direction It should head in a direction It should head in a direction east of south. south of east. east of south. south of east.? 5.2 Vector Addition and Subtraction: Analytical Methods 30. A person wants to fire a water balloon cannon such that it hits a target 100 m away. If the cannon can only be launched at 45\u00b0 above the horizontal, what should be the initial speed at which it is launched? 31.3 m/s a. 37.2 m/s b. c. 980.0 m/s 1,385.9 m/s d. 5.4 Inclined Planes 31. A coin is sliding down an inclined plane at constant to the horizontal, velocity. If the angle of the plane is what is the coefficient of kinetic friction? a. b. c. d. 27. A person walks 10.", "0 m north and then 2.00 m east. 32. A skier with a mass of 55 kg is skiing down a snowy slope Solving analytically, what is the resultant displacement of the person? a. b. c. d. = 10.2 m, \u03b8 = 78.7\u00ba east of north = 10.2 m, \u03b8 = 78.7\u00ba north of east = 12.0 m, \u03b8 = 78.7\u00ba east of north = 12.0 m, \u03b8 = 78.7\u00ba north of east that has an incline of 30\u00b0. Find the coefficient of kinetic friction for the skier if friction is known to be 25 N. a. b. c. d. 28. A person walks north of west for and 5.5 Simple Harmonic Motion south of west for. What is the magnitude 33. What is the time period of a long pendulum on of his displacement? Solve analytically. a. b. c. d. earth? a. b. c. d. 34. A simple harmonic oscillator has time period. If the, what is the force constant of mass of the system is the spring used? a. b. c. d. period 2 seconds). 5.3 Projectile Motion 29. A water balloon cannon is fired at at an angle of above the horizontal. How far away will it fall? a. b. c. d. Performance Task 5.5 Simple Harmonic Motion 35. Construct a seconds pendulum (pendulum with time Access for free at openstax.org. TEST PREP Multiple Choice 5.1 Vector Addition and Subtraction: Graphical Methods 36. True or False\u2014We can use Pythagorean theorem to calculate the length of the resultant vector obtained from the addition of two vectors which are at right angles to each other. a. True b. False 37. True or False\u2014The direction of the resultant vector depends on both the magnitude and direction of added vectors. a. True b. False with a headwind blowing. What is the resultant velocity 38. A plane flies north at from the north at of the plane? a. b. c. d. north south north south 39. Two hikers take different routes to reach the same spot. southeast, then turns and goes The first one goes at south of east. The second hiker goes south. How far and in which direction must the second hiker travel now, in order to reach the first hiker's location destination? a. b. c. d. east south east south 5.2 Vector Addition and Subtraction: Analytical Methods 40. When will the x-component of a vector with angle be greater than its y-component? a. b. c. d. 41. The resultant vector of the addition of vectors and. The magnitudes of is are, respectively. Which of the following is true? a., and,,, and b. c. Chapter 5 \u2022 Test Prep 191 d. 42. What is the dimensionality of vectors used in the study of atmospheric sciences? a. One-dimensional b. Two-dimensional c. Three-dimensional 5.3 Projectile Motion 43. After a projectile is launched in the air, in which direction does it experience constant, non-zero acceleration, ignoring air resistance? a. The x direction b. The y direction c. Both the x and y directions d. Neither direction 44. Which is true when the height of a projectile is at its maximum? a. b. c. 45. A ball is thrown in the air at an angle of 40\u00b0. If the maximum height it reaches is 10 m, what must be its initial speed? a. 17.46 m/s b. 21.78 m/s 304.92 m/s c. d. 474.37 m/s 46. A large rock is ejected from a volcano with a speed of above the horizontal. The and at an angle rock strikes the side of the volcano at an altitude of lower than its starting point. Calculate the horizontal displacement of the rock. a. b. c. d. 5.4 Inclined Planes 47. For objects of identical masses but made of different materials, which of the following experiences the most static friction? a. Shoes on ice b. Metal on wood c. Teflon on steel 48. If an object sits on an inclined plane and no other object makes contact with the object, what is typically equal in magnitude to the component of the weight perpendicular to the plane? 192 Chapter 5 \u2022 Test Prep a. The normal force b. The total weight c. The parallel force of weight 49. A 5 kg box is at rest on the floor. The coefficient of static friction between the box and the floor is 0.4. A horizontal force of 50 N is applied to the box. Will it move? a. No, because the applied force is less than the maximum limiting static friction.", " b. No, because the applied force is more than the maximum limiting static friction. c. Yes, because the applied force is less than the maximum limiting static friction. d. Yes, because the applied force is more than the maximum limiting static friction. 50. A skier with a mass of 67 kg is skiing down a snowy slope with an incline of 37\u00b0. Find the friction if the coefficient of kinetic friction is 0.07. a. 27.66 N 34.70 N b. 36.71 N c. d. 45.96 N 5.5 Simple Harmonic Motion 51. A change in which of the following is an example of deformation? a. Velocity b. Length c. Mass d. Weight 52. The units of amplitude are the same as those for which of the following measurements? a. Speed b. Displacement c. Acceleration d. Force 53. Up to approximately what angle is simple harmonic motion a good model for a pendulum? a. b. c. d. 54. How would simple harmonic motion be different in the absence of friction? a. Oscillation will not happen in the absence of friction. b. Oscillation will continue forever in the absence of friction. c. Oscillation will have changing amplitude in the absence of friction. d. Oscillation will cease after a certain amount of time in the absence of friction. 55. What mass needs to be attached to a spring with a force in order to make a simple harmonic constant of oscillator oscillate with a time period of a. b. c. d.? Short Answer 5.1 Vector Addition and Subtraction: Graphical Methods 56. Find for the following vectors: the addition of these vectors? a. Zero b. Six c. Eight d. Twelve 108 cm, a. 108 cm, b. c. 206 cm, d. 206 cm, 57. Find for the following vectors: 108 cm, a. b. 108 cm, c. 232 cm, d. 232 cm, 59. Two people pull on ropes tied to a trolley, each applying 44 N of force. The angle the ropes form with each other is 39.5\u00b0. What is the magnitude of the net force exerted on the trolley? a. 0.0 N b. 79.6 N c. 82.8 N d. 88.0 N 5.2 Vector Addition and Subtraction: Analytical Methods 60. True or False\u2014A vector can form the shape of a right 58. Consider six vectors of 2 cm each, joined from head to tail making a hexagon. What would be the magnitude of angle triangle with its x and y components. a. True Access for free at openstax.org. b. False 61. True or False\u2014All vectors have positive x and y a. True b. False Chapter 5 \u2022 Test Prep 193 69. For what angle of a projectile is its range equal to zero?. What is in terms of and a. b. c. d. or or or or 5.4 Inclined Planes 70. What are the units of the coefficient of friction?. What is in terms of and a. b. c. d. unitless 71. Two surfaces in contact are moving slowly past each components. a. True b. False 62. Consider? a. b. c. d. 63. Consider? a. b. c. d. 64. When a three dimensional vector is used in the study of atmospheric sciences, what is z? a. Altitude b. Heat c. Temperature d. Wind speed 65. Which method is not an application of vector calculus? a. To find the rate of change in atmospheric temperature b. To study changes in wind speed and direction c. To predict changes in atmospheric pressure d. To measure changes in average rainfall 5.3 Projectile Motion 66. How can you express the velocity, terms of its initial velocity, time,? a. b. c. d. 67. In the equation for the maximum height of a projectile, what does stand for? Initial velocity in the x direction a. b. Initial velocity in the y direction c. Final velocity in the x direction d. Final velocity in the y direction other. As the relative speed between the two surfaces in contact increases, what happens to the magnitude of their coefficient of kinetic friction? a. It increases with the increase in the relative motion. It decreases with the increase in the relative motion. It remains constant and is independent of the relative motion. b. c. 72. When will an object slide down an inclined plane at constant velocity? a. When the magnitude of the component of the weight along the slope is equal to the magnitude of the frictional force. b. When the magnitude of the component of the weight along the slope is greater than the magnitude of the frictional force. c. When the magnitude of the component of the weight perpendicular to the slope is less than the magnitude of the fr", "ictional force. 73. A box is sitting on an inclined plane. At what angle of incline is the perpendicular component of the box's weight at its maximum? a. b. c. d. 5.5 Simple Harmonic Motion 74. What is the term used for changes in shape due to the, of a projectile in, acceleration,, and d. When the magnitude of the component of the weight perpendicular to the slope is equal to the magnitude of the frictional force. 68. True or False\u2014Range is defined as the maximum vertical distance travelled by a projectile. application of force? a. Amplitude 194 Chapter 5 \u2022 Test Prep b. Deformation c. Displacement d. Restoring force 75. What is the restoring force? a. The normal force on the surface of an object b. The weight of a mass attached to an object c. Force which is applied to deform an object from its original shape d. Force which brings an object back to its equilibrium position 76. For a given oscillator, what are the factors that affect its period and frequency? a. Mass only b. Force constant only c. Applied force and mass d. Mass and force constant 77. For an object in simple harmonic motion, when does the Extended Response 5.1 Vector Addition and Subtraction: Graphical Methods 80. True or False\u2014For vectors the order of addition is important. a. True b. False 81. Consider five vectors a, b, c, d,and e.Is it true or false that their addition always results in a vector with a greater magnitude than if only two of the vectors were added? a. True b. False 5.2 Vector Addition and Subtraction: Analytical Methods 82. For what angle of a vector is it possible that its magnitude will be equal to its y-component? a. b. c. d. 83. True or False\u2014If only the angles of two vectors are known, we can find the angle of their resultant addition vector. a. True b. False 84. True or false\u2014We can find the magnitude and direction of the resultant vector if we know the angles of two vectors and the magnitude of one. Access for free at openstax.org. maximum speed occur? a. At the extreme positions b. At the equilibrium position c. At the moment when the applied force is removed d. Midway between the extreme and equilibrium positions 78. What is the equilibrium position of a pendulum? a. When the tension in the string is zero b. When the pendulum is hanging straight down c. When the tension in the string is maximum d. When the weight of the mass attached is minimum 79. If a pendulum is displaced by an angle \u03b8, what is the net restoring force it experiences? a. mgsin\u03b8 b. mgcos\u03b8 c. \u2013mgsin\u03b8 d. \u2013mgcos\u03b8 a. True b. False 5.3 Projectile Motion 85. Ignoring drag, what is the x-component of the acceleration of a projectile? Why? a. The x-component of the acceleration of a projectile is because acceleration of a projectile is due to gravity, which acts in the y direction. b. The x component of the acceleration of a projectile is because acceleration of a projectile is due to gravity, which acts in the x direction. c. The x-component of the acceleration of a projectile is because acceleration of a projectile is due to gravity, which acts in the x direction. d. The x-component of the acceleration of a projectile is because acceleration of a projectile is due to gravity, which acts in the y direction. 86. What is the optimum angle at which a projectile should be launched in order to cover the maximum distance? a. b. c. d. 5.4 Inclined Planes 87. True or False\u2014Friction varies from surface to surface because different substances have different degrees of roughness or smoothness. a. True b. False 88. As the angle of the incline gets larger, what happens to the magnitudes of the perpendicular and parallel components of gravitational force? a. Both the perpendicular and the parallel component d. The force constant kis related to the friction in the system: The larger the force constant, the lower the friction in the system. Chapter 5 \u2022 Test Prep 195 will decrease. b. The perpendicular component will decrease and the parallel component will increase. c. The perpendicular component will increase and the parallel component will decrease. d. Both the perpendicular and the parallel component will increase. 5.5 Simple Harmonic Motion 89. What physical characteristic of a system is its force constant related to? a. The force constant kis related to the stiffness of a system: The larger the force constant, the stiffer the system. b. The force constant kis related to the stiffness of a system: The larger the force constant, the looser the system. c. The force constant kis related to the friction in the", " system: The larger the force constant, the greater the friction in the system. 90. How or why does a pendulum oscillate? a. A pendulum oscillates due to applied force. b. A pendulum oscillates due to the elastic nature of the string. c. A pendulum oscillates due to restoring force arising from gravity. d. A pendulum oscillates due to restoring force arising from tension in the string. 91. If a pendulum from earth is taken to the moon, will its frequency increase or decrease? Why? a. It will increase because on the Moon is less than on Earth. It will decrease because on the Moon is less than b. c. d. on Earth. It will increase because on the Moon is greater than on Earth. It will decrease because on the Moon is greater than on Earth. 196 Chapter 5 \u2022 Test Prep Access for free at openstax.org. CHAPTER 6 Circular and Rotational Motion Figure 6.1 This Australian Grand Prix Formula 1 race car moves in a circular path as it makes the turn. Its wheels also spin rapidly. The same physical principles are involved in both of these motions. (Richard Munckton). Chapter Outline 6.1 Angle of Rotation and Angular Velocity 6.2 Uniform Circular Motion 6.3 Rotational Motion You may recall learning about various aspects of motion along a straight line: kinematics (where we learned INTRODUCTION about displacement, velocity, and acceleration), projectile motion (a special case of two-dimensional kinematics), force, and Newton\u2019s laws of motion. In some ways, this chapter is a continuation of Newton\u2019s laws of motion. Recall that Newton\u2019s first law tells us that objects move along a straight line at constant speed unless a net external force acts on them. Therefore, if an object moves along a circular path, such as the car in the photo, it must be experiencing an external force. In this chapter, we explore both circular motion and rotational motion. 198 Chapter 6 \u2022 Circular and Rotational Motion 6.1 Angle of Rotation and Angular Velocity Section Learning Objectives By the end of this section, you will be able to do the following: \u2022 Describe the angle of rotation and relate it to its linear counterpart \u2022 Describe angular velocity and relate it to its linear counterpart \u2022 Solve problems involving angle of rotation and angular velocity Section Key Terms angle of rotation angular velocity arc length circular motion radius of curvature rotational motion spin tangential velocity Angle of Rotation What exactly do we mean by circular motionor rotation? Rotational motion is the circular motion of an object about an axis of rotation. We will discuss specifically circular motion and spin. Circular motion is when an object moves in a circular path. Examples of circular motion include a race car speeding around a circular curve, a toy attached to a string swinging in a circle around your head, or the circular loop-the-loopon a roller coaster. Spin is rotation about an axis that goes through the center of mass of the object, such as Earth rotating on its axis, a wheel turning on its axle, the spin of a tornado on its path of destruction, or a figure skater spinning during a performance at the Olympics. Sometimes, objects will be spinning while in circular motion, like the Earth spinning on its axis while revolving around the Sun, but we will focus on these two motions separately. When solving problems involving rotational motion, we use variables that are similar to linear variables (distance, velocity, acceleration, and force) but take into account the curvature or rotation of the motion. Here, we define the angle of rotation, which is the angular equivalence of distance; and angular velocity, which is the angular equivalence of linear velocity. When objects rotate about some axis\u2014for example, when the CD in Figure 6.2 rotates about its center\u2014each point in the object follows a circular path. Figure 6.2 All points on a CD travel in circular paths. The pits (dots) along a line from the center to the edge all move through the same angle in time. The arc length,, is the distance traveled along a circular path. The radius of curvature, r, is the radius of the circular path. Both are shown in Figure 6.3. Access for free at openstax.org. 6.1 \u2022 Angle of Rotation and Angular Velocity 199 Figure 6.3 The radius (r) of a circle is rotated through an angle. The arc length,, is the distance covered along the circumference. Consider a line from the center of the CD to its edge. In a given time, each pit(used to record information) on this line moves through the same angle. The angle of rotation is the amount of rotation and is the angular analog of distance. The angle of rotation is the arc length divided by the radius of curvature. The angle of rotation is often measured by using a unit called the radian. (Radians", " are actually dimensionless, because a radian is defined as the ratio of two distances, radius and arc length.) A revolution is one complete rotation, where every point on the radians (or 360 degrees), and therefore has an angle of rotation circle returns to its original position. One revolution covers of revolutions, and degrees using the relationship radians, and an arc length that is the same as the circumference of the circle. We can convert between radians, 1 revolution = rad = 360\u00b0. See Table 6.1 for the conversion of degrees to radians for some common angles. Degree Measures Radian Measures 6.1 Table 6.1 Commonly Used Angles in Terms of Degrees and Radians Angular Velocity How fast is an object rotating? We can answer this question by using the concept of angular velocity. Consider first the angular speed is the rate at which the angle of rotation changes. In equation form, the angular speed is 6.2 which means that an angular rotation given time, it has a greater angular speed. The units for angular speed are radians per second (rad/s). occurs in a time,. If an object rotates through a greater angle of rotation in a Now let\u2019s consider the direction of the angular speed, which means we now must call it the angular velocity. The direction of the 200 Chapter 6 \u2022 Circular and Rotational Motion angular velocity is along the axis of rotation. For an object rotating clockwise, the angular velocity points away from you along the axis of rotation. For an object rotating counterclockwise, the angular velocity points toward you along the axis of rotation. Angular velocity (\u03c9) is the angular version of linear velocity v. Tangential velocity is the instantaneous linear velocity of an object in rotational motion. To get the precise relationship between angular velocity and tangential velocity, consider again a pit on the rotating CD. This pit moves through an arc length so its tangential speedis in a shorttime From the definition of the angle of rotation,, we see that. Substituting this into the expression for vgives 6.3 says that the tangential speed vis proportional to the distance rfrom the center of rotation. Consequently, The equation tangential speed is greater for a point on the outer edge of the CD (with larger r) than for a point closer to the center of the CD (with smaller r). This makes sense because a point farther out from the center has to cover a longer arc length in the same amount of time as a point closer to the center. Note that both points will still have the same angular speed, regardless of their distance from the center of rotation. See Figure 6.4. Figure 6.4 Points 1 and 2 rotate through the same angle ( ), but point 2 moves through a greater arc length ( ) because it is farther from the center of rotation. means large vbecause Now, consider another example: the tire of a moving car (see Figure 6.5). The faster the tire spins, the faster the car moves\u2014large, will produce a greater linear (tangential) velocity, v, for the car. This is because a larger radius means a longer arc length must contact the road, so the car must move farther in the same amount of time.. Similarly, a larger-radius tire rotating at the same angular velocity, Access for free at openstax.org. 6.1 \u2022 Angle of Rotation and Angular Velocity 201 Figure 6.5 A car moving at a velocity, v, to the right has a tire rotating with angular velocity. The speed of the tread of the tire relative to the axle is v, the same as if the car were jacked up and the wheels spinning without touching the road. Directly below the axle, where the tire touches the road, the tire tread moves backward with respect to the axle with tangential velocity, where ris the tire radius. Because the road is stationary with respect to this point of the tire, the car must move forward at the linear velocity v. A larger angular velocity for the tire means a greater linear velocity for the car. However, there are cases where linear velocity and tangential velocity are not equivalent, such as a car spinning its tires on ice. In this case, the linear velocity will be less than the tangential velocity. Due to the lack of friction under the tires of a car on ice, the arc length through which the tire treads move is greater than the linear distance through which the car moves. It\u2019s similar to running on a treadmill or pedaling a stationary bike; you are literally going nowhere fast. TIPS FOR SUCCESS Angular velocity \u03c9 and tangential velocity v are vectors, so we must include magnitude and direction. The direction of the angular velocity is along the axis of rotation, and points away from you for an object rotating clockwise, and toward you for an object rotating counterclockwise. In mathematics this is described by the right-hand rule. Tangential velocity is usually described as up", ", down, left, right, north, south, east, or west, as shown in Figure 6.6. Figure 6.6 As the fly on the edge of an old-fashioned vinyl record moves in a circle, its instantaneous velocity is always at a tangent to the circle. The direction of the angular velocity is into the page this case. 202 Chapter 6 \u2022 Circular and Rotational Motion WATCH PHYSICS Relationship between Angular Velocity and Speed This video reviews the definition and units of angular velocity and relates it to linear speed. It also shows how to convert between revolutions and radians. Click to view content (https://www.youtube.com/embed/zAx61CO5mDw) GRASP CHECK For an object traveling in a circular path at a constant angular speed, would the linear speed of the object change if the radius of the path increases? a. Yes, because tangential speed is independent of the radius. b. Yes, because tangential speed depends on the radius. c. No, because tangential speed is independent of the radius. d. No, because tangential speed depends on the radius. Solving Problems Involving Angle of Rotation and Angular Velocity Snap Lab Measuring Angular Speed In this activity, you will create and measure uniform circular motion and then contrast it with circular motions with different radii. \u2022 One string (1 m long) \u2022 One object (two-hole rubber stopper) to tie to the end \u2022 One timer Procedure 1. Tie an object to the end of a string. 2. Swing the object around in a horizontal circle above your head (swing from your wrist). It is important that the circle be horizontal! 3. Maintain the object at uniform speed as it swings. 4. Measure the angular speed of the object in this manner. Measure the time it takes in seconds for the object to travel 10 revolutions. Divide that time by 10 to get the angular speed in revolutions per second, which you can convert to radians per second. 5. What is the approximate linear speed of the object? 6. Move your hand up the string so that the length of the string is 90 cm. Repeat steps 2\u20135. 7. Move your hand up the string so that its length is 80 cm. Repeat steps 2\u20135. 8. Move your hand up the string so that its length is 70 cm. Repeat steps 2\u20135. 9. Move your hand up the string so that its length is 60 cm. Repeat steps 2\u20135 10. Move your hand up the string so that its length is 50 cm. Repeat steps 2\u20135 11. Make graphs of angular speed vs. radius (i.e. string length) and linear speed vs. radius. Describe what each graph looks like. GRASP CHECK If you swing an object slowly, it may rotate at less than one revolution per second. What would be the revolutions per second for an object that makes one revolution in five seconds? What would be its angular speed in radians per second?. The angular speed of the object would be a. The object would spin at. b. The object would spin at. The angular speed of the object would be. Access for free at openstax.org. c. The object would spin at d. The object would spin at. The angular speed of the object would be. The angular speed of the object would be.. 6.1 \u2022 Angle of Rotation and Angular Velocity 203 Now that we have an understanding of the concepts of angle of rotation and angular velocity, we\u2019ll apply them to the real-world situations of a clock tower and a spinning tire. WORKED EXAMPLE Angle of rotation at a Clock Tower The clock on a clock tower has a radius of 1.0 m. (a) What angle of rotation does the hour hand of the clock travel through when it moves from 12 p.m. to 3 p.m.? (b) What\u2019s the arc length along the outermost edge of the clock between the hour hand at these two times? Strategy We can figure out the angle of rotation by multiplying a full revolution ( hour hand in going from 12 to 3. Once we have the angle of rotation, we can solve for the arc length by rearranging the equation radians) by the fraction of the 12 hours covered by the since the radius is given. Solution to (a) In going from 12 to 3, the hour hand covers 1/4 of the 12 hours needed to make a complete revolution. Therefore, the angle between the hour hand at 12 and at 3 is (i.e., 90 degrees). Solution to (b) Rearranging the equation we get Inserting the known values gives an arc length of 6.4 6.5 6.6 Discussion We were able to drop the radians from the final solution to part (b) because radians are actually dimensionless. This is because the radian is defined as the ratio of two distances (radius and arc length).", " Thus, the formula gives an answer in units of meters, as expected for an arc length. WORKED EXAMPLE How Fast Does a Car Tire Spin? Calculate the angular speed of a 0.300 m radius car tire when the car travels at 15.0 m/s (about 54 km/h). See Figure 6.5. Strategy In this case, the speed of the tire tread with respect to the tire axle is the same as the speed of the car with respect to the road, so we have v= 15.0 m/s. The radius of the tire is r= 0.300 m. Since we know vand r, we can rearrange the equation, to get and find the angular speed. Solution To find the angular speed, we use the relationship:. 204 Chapter 6 \u2022 Circular and Rotational Motion Inserting the known quantities gives Discussion When we cancel units in the above calculation, we get 50.0/s (i.e., 50.0 per second, which is usually written as 50.0 s\u22121). But the angular speed must have units of rad/s. Because radians are dimensionless, we can insert them into the answer for the angular speed because we know that the motion is circular. Also note that, if an earth mover with much larger tires, say 1.20 m in radius, were moving at the same speed of 15.0 m/s, its tires would rotate more slowly. They would have an angular speed of 6.8 6.7 Practice Problems 1. What is the angle in degrees between the hour hand and the minute hand of a clock showing 9:00 a.m.? a. 0\u00b0 b. 90\u00b0 180\u00b0 c. 360\u00b0 d. 2. What is the approximate value of the arc length between the hour hand and the minute hand of a clock showing 10:00 a.m if the radius of the clock is 0.2 m? a. 0.1 m b. 0.2 m c. 0.3 m d. 0.6 m Check Your Understanding 3. What is circular motion? a. Circular motion is the motion of an object when it follows a linear path. b. Circular motion is the motion of an object when it follows a zigzag path. c. Circular motion is the motion of an object when it follows a circular path. d. Circular motion is the movement of an object along the circumference of a circle or rotation along a circular path. 4. What is meant by radius of curvature when describing rotational motion? a. The radius of curvature is the radius of a circular path. b. The radius of curvature is the diameter of a circular path. c. The radius of curvature is the circumference of a circular path. d. The radius of curvature is the area of a circular path. 5. What is angular velocity? a. Angular velocity is the rate of change of the diameter of the circular path. b. Angular velocity is the rate of change of the angle subtended by the circular path. c. Angular velocity is the rate of change of the area of the circular path. d. Angular velocity is the rate of change of the radius of the circular path. 6. What equation defines angular velocity,? Take that is the radius of curvature, is the angle, and is time. a. b. c. d. 7. Identify three examples of an object in circular motion. Access for free at openstax.org. 6.2 \u2022 Uniform Circular Motion 205 a. an artificial satellite orbiting the Earth, a race car moving in the circular race track, and a top spinning on its axis b. an artificial satellite orbiting the Earth, a race car moving in the circular race track, and a ball tied to a string being swung in a circle around a person's head c. Earth spinning on its own axis, a race car moving in the circular race track, and a ball tied to a string being swung in a circle around a person's head d. Earth spinning on its own axis, blades of a working ceiling fan, and a top spinning on its own axis 8. What is the relative orientation of the radius and tangential velocity vectors of an object in uniform circular motion? a. Tangential velocity vector is always parallel to the radius of the circular path along which the object moves. b. Tangential velocity vector is always perpendicular to the radius of the circular path along which the object moves. c. Tangential velocity vector is always at an acute angle to the radius of the circular path along which the object moves. d. Tangential velocity vector is always at an obtuse angle to the radius of the circular path along which the object moves. 6.2 Uniform Circular Motion Section Learning Objectives By the end of this section, you will be able to do the following: \u2022 Describe centripetal acceleration and relate it to linear acceleration \u2022 Describe centripetal force and", " relate it to linear force \u2022 Solve problems involving centripetal acceleration and centripetal force Section Key Terms centrifugal force centripetal acceleration centripetal force uniform circular motion Centripetal Acceleration In the previous section, we defined circular motion. The simplest case of circular motion is uniform circular motion, where an object travels a circular path at a constant speed. Note that, unlike speed, the linear velocity of an object in circular motion is constantly changing because it is always changing direction. We know from kinematics that acceleration is a change in velocity, either in magnitude or in direction or both. Therefore, an object undergoing uniform circular motion is always accelerating, even though the magnitude of its velocity is constant. You experience this acceleration yourself every time you ride in a car while it turns a corner. If you hold the steering wheel steady during the turn and move at a constant speed, you are executing uniform circular motion. What you notice is a feeling of sliding (or being flung, depending on the speed) away from the center of the turn. This isn\u2019t an actual force that is acting on you\u2014it only happens because your body wants to continue moving in a straight line (as per Newton\u2019s first law) whereas the car is turning off this straight-line path. Inside the car it appears as if you are forced away from the center of the turn. This fictitious force is known as the centrifugal force. The sharper the curve and the greater your speed, the more noticeable this effect becomes. Figure 6.7 shows an object moving in a circular path at constant speed. The direction of the instantaneous tangential velocity is shown at two points along the path. Acceleration is in the direction of the change in velocity; in this case it points roughly toward the center of rotation. (The center of rotation is at the center of the circular path). If we imagine smaller, then the acceleration would point exactlytoward the center of rotation, but this case is hard to draw. We call the acceleration of an object moving in uniform circular motion the centripetal acceleration ac because centripetal means center seeking. becoming smaller and 206 Chapter 6 \u2022 Circular and Rotational Motion Figure 6.7 The directions of the velocity of an object at two different points are shown, and the change in velocity is seen to point approximately toward the center of curvature (see small inset). For an extremely small value of, points exactly toward the center of the circle (but this is hard to draw). Because, the acceleration is also toward the center, so ac is called centripetal acceleration. Now that we know that the direction of centripetal acceleration is toward the center of rotation, let\u2019s discuss the magnitude of centripetal acceleration. For an object traveling at speed vin a circular path with radius r, the magnitude of centripetal acceleration is Centripetal acceleration is greater at high speeds and in sharp curves (smaller radius), as you may have noticed when driving a car, because the car actually pushes you toward the center of the turn. But it is a bit surprising that ac is proportional to the speed squared. This means, for example, that the acceleration is four times greater when you take a curve at 100 km/h than at 50 km/h. We can also express ac in terms of the magnitude of angular velocity. Substituting into the equation above, we get. Therefore, the magnitude of centripetal acceleration in terms of the magnitude of angular velocity is 6.9 TIPS FOR SUCCESS The equation expressed in the form ac = r\u03c92 is useful for solving problems where you know the angular velocity rather than the tangential velocity. Virtual Physics Ladybug Motion in 2D In this simulation, you experiment with the position, velocity, and acceleration of a ladybug in circular and elliptical motion. Switch the type of motion from linear to circular and observe the velocity and acceleration vectors. Next, try elliptical motion and notice how the velocity and acceleration vectors differ from those in circular motion. Click to view content (https://archive.cnx.org/specials/317a2b1e-2fbd-11e5-99b5-e38ffb545fe6/ladybug-motion/) Access for free at openstax.org. 6.2 \u2022 Uniform Circular Motion 207 GRASP CHECK In uniform circular motion, what is the angle between the acceleration and the velocity? What type of acceleration does a body experience in the uniform circular motion? a. The angle between acceleration and velocity is 0\u00b0, and the body experiences linear acceleration. b. The angle between acceleration and velocity is 0\u00b0, and the body experiences centripetal acceleration. c. The angle between acceleration and velocity is 90\u00b0, and the body experiences linear acceleration. d. The angle between acceleration and velocity is 90\u00b0, and the body experiences centripetal acceleration. Centripet", "al Force Because an object in uniform circular motion undergoes constant acceleration (by changing direction), we know from Newton\u2019s second law of motion that there must be a constant net external force acting on the object. Any force or combination of forces can cause a centripetal acceleration. Just a few examples are the tension in the rope on a tether ball, the force of Earth\u2019s gravity on the Moon, the friction between a road and the tires of a car as it goes around a curve, or the normal force of a roller coaster track on the cart during a loop-the-loop. Any net force causing uniform circular motion is called a centripetal force. The direction of a centripetal force is toward the center of rotation, the same as for centripetal acceleration. According to Newton\u2019s second law of motion, a net force causes the acceleration of mass according to Fnet = ma. For uniform circular motion, the acceleration is centripetal acceleration: a = ac. Therefore, the magnitude of centripetal force, Fc, is. By using the two different forms of the equation for the magnitude of centripetal acceleration, get two expressions involving the magnitude of the centripetal force Fc. The first expression is in terms of tangential speed, the second is in terms of angular speed:, we and and. Both forms of the equation depend on mass, velocity, and the radius of the circular path. You may use whichever expression for centripetal force is more convenient. Newton\u2019s second law also states that the object will accelerate in the same direction as the net force. By definition, the centripetal force is directed towards the center of rotation, so the object will also accelerate towards the center. A straight line drawn from the circular path to the center of the circle will always be perpendicular to the tangential velocity. Note that, if you solve the first expression for r, you get From this expression, we see that, for a given mass and velocity, a large centripetal force causes a small radius of curvature\u2014that is, a tight curve. 208 Chapter 6 \u2022 Circular and Rotational Motion Figure 6.8 In this figure, the frictional force fserves as the centripetal force Fc. Centripetal force is perpendicular to tangential velocity and causes uniform circular motion. The larger the centripetal force Fc, the smaller is the radius of curvature rand the sharper is the curve. The lower curve has the same velocity v, but a larger centripetal force Fc produces a smaller radius. WATCH PHYSICS Centripetal Force and Acceleration Intuition This video explains why a centripetal force creates centripetal acceleration and uniform circular motion. It also covers the difference between speed and velocity and shows examples of uniform circular motion. Click to view content (https://www.youtube.com/embed/vZOk8NnjILg) GRASP CHECK Imagine that you are swinging a yoyo in a vertical clockwise circle in front of you, perpendicular to the direction you are facing. Now, imagine that the string breaks just as the yoyo reaches its bottommost position, nearest the floor. Which of the following describes the path of the yoyo after the string breaks? a. The yoyo will fly upward in the direction of the centripetal force. b. The yoyo will fly downward in the direction of the centripetal force. Access for free at openstax.org. 6.2 \u2022 Uniform Circular Motion 209 c. The yoyo will fly to the left in the direction of the tangential velocity. d. The yoyo will fly to the right in the direction of the tangential velocity. Solving Centripetal Acceleration and Centripetal Force Problems To get a feel for the typical magnitudes of centripetal acceleration, we\u2019ll do a lab estimating the centripetal acceleration of a tennis racket and then, in our first Worked Example, compare the centripetal acceleration of a car rounding a curve to gravitational acceleration. For the second Worked Example, we\u2019ll calculate the force required to make a car round a curve. Snap Lab Estimating Centripetal Acceleration In this activity, you will measure the swing of a golf club or tennis racket to estimate the centripetal acceleration of the end of the club or racket. You may choose to do this in slow motion. Recall that the equation for centripetal acceleration is or. \u2022 One tennis racket or golf club \u2022 One timer \u2022 One ruler or tape measure Procedure 1. Work with a partner. Stand a safe distance away from your partner as he or she swings the golf club or tennis racket. 2. Describe the motion of the swing\u2014is this uniform circular motion? Why or why", " not? 3. Try to get the swing as close to uniform circular motion as possible. What adjustments did your partner need to make? 4. Measure the radius of curvature. What did you physically measure? 5. By using the timer, find either the linear or angular velocity, depending on which equation you decide to use. 6. What is the approximate centripetal acceleration based on these measurements? How accurate do you think they are? Why? How might you and your partner make these measurements more accurate? GRASP CHECK Was it more useful to use the equation or in this activity? Why? a. It should be simpler to use because measuring angular velocity through observation would be easier. b. c. It should be simpler to use because measuring tangential velocity through observation would be easier. It should be simpler to use because measuring angular velocity through observation would be difficult. d. It should be simpler to use because measuring tangential velocity through observation would be difficult. WORKED EXAMPLE Comparing Centripetal Acceleration of a Car Rounding a Curve with Acceleration Due to Gravity A car follows a curve of radius 500 m at a speed of 25.0 m/s (about 90 km/h). What is the magnitude of the car\u2019s centripetal acceleration? Compare the centripetal acceleration for this fairly gentle curve taken at highway speed with acceleration due to gravity (g). 210 Chapter 6 \u2022 Circular and Rotational Motion Strategy Because linear rather than angular speed is given, it is most convenient to use the expression the centripetal acceleration. to find the magnitude of Solution Entering the given values of v= 25.0 m/s and r= 500 m into the expression for ac gives Discussion To compare this with the acceleration due to gravity (g= 9.80 m/s2), we take the ratio. Therefore,, which means that the centripetal acceleration is about one tenth the acceleration due to gravity. WORKED EXAMPLE Frictional Force on Car Tires Rounding a Curve a. Calculate the centripetal force exerted on a 900 kg car that rounds a 600-m-radius curve on horizontal ground at 25.0 m/s. b. Static friction prevents the car from slipping. Find the magnitude of the frictional force between the tires and the road that allows the car to round the curve without sliding off in a straight line. Access for free at openstax.org. 6.2 \u2022 Uniform Circular Motion 211 Strategy and Solution for (a) We know that. Therefore, Strategy and Solution for (b) The image above shows the forces acting on the car while rounding the curve. In this diagram, the car is traveling into the page as shown and is turning to the left. Friction acts toward the left, accelerating the car toward the center of the curve. Because friction is the only horizontal force acting on the car, it provides all of the centripetal force in this case. Therefore, the force of friction is the centripetal force in this situation and points toward the center of the curve. Discussion Since we found the force of friction in part (b), we could also solve for the coefficient of friction, since. Practice Problems 9. What is the centripetal acceleration of an object with speed going along a path of radius? a. b. c. d. 10. Calculate the centripetal acceleration of an object following a path with a radius of a curvature of 0.2 m and at an angular velocity of 5 rad/s. a. b. c. d. 1 m/s 5 m/s 1 m/s2 5 m/s2 Check Your Understanding 11. What is uniform circular motion? 212 Chapter 6 \u2022 Circular and Rotational Motion a. Uniform circular motion is when an object accelerates on a circular path at a constantly increasing velocity. b. Uniform circular motion is when an object travels on a circular path at a variable acceleration. c. Uniform circular motion is when an object travels on a circular path at a constant speed. d. Uniform circular motion is when an object travels on a circular path at a variable speed. 12. What is centripetal acceleration? a. The acceleration of an object moving in a circular path and directed radially toward the center of the circular orbit b. The acceleration of an object moving in a circular path and directed tangentially along the circular path c. The acceleration of an object moving in a linear path and directed in the direction of motion of the object d. The acceleration of an object moving in a linear path and directed in the direction opposite to the motion of the object 13. Is there a net force acting on an object in uniform circular motion? a. Yes, the object is accelerating, so a net force must be acting on it. b. Yes, because there is no acceleration. c. No, because there is acceleration. d. No, because there is no acceleration. 14", ". Identify two examples of forces that can cause centripetal acceleration. a. The force of Earth\u2019s gravity on the moon and the normal force b. The force of Earth\u2019s gravity on the moon and the tension in the rope on an orbiting tetherball c. The normal force and the force of friction acting on a moving car d. The normal force and the tension in the rope on a tetherball 6.3 Rotational Motion Section Learning Objectives By the end of this section, you will be able to do the following: \u2022 Describe rotational kinematic variables and equations and relate them to their linear counterparts \u2022 Describe torque and lever arm \u2022 Solve problems involving torque and rotational kinematics Section Key Terms angular acceleration kinematics of rotational motion lever arm tangential acceleration torque Rotational Kinematics In the section on uniform circular motion, we discussed motion in a circle at constant speed and, therefore, constant angular velocity. However, there are times when angular velocity is not constant\u2014rotational motion can speed up, slow down, or reverse directions. Angular velocity is not constant when a spinning skater pulls in her arms, when a child pushes a merry-go-round to make it rotate, or when a CD slows to a halt when switched off. In all these cases, angular acceleration occurs because the angular velocity rate of change of angular velocity. In equation form, angular acceleration is changes. The faster the change occurs, the greater is the angular acceleration. Angular acceleration is the increases, then is the change in angular velocity and is the change in time. The units of angular acceleration are (rad/s)/s, or rad/ where s2. If is negative. Keep in mind that, by convention, counterclockwise is the positive direction and clockwise is the negative direction. For example, the skater in Figure 6.9 is rotating counterclockwise as seen from above, so her angular velocity is positive. Acceleration would be negative, for example, when an object that is rotating counterclockwise slows down. It would be positive when an object that is rotating counterclockwise speeds up. decreases, then is positive. If Access for free at openstax.org. 6.3 \u2022 Rotational Motion 213 Figure 6.9 A figure skater spins in the counterclockwise direction, so her angular velocity is normally considered to be positive. (Luu, Wikimedia Commons) The relationship between the magnitudes of tangential acceleration, a, and angular acceleration, 6.10 These equations mean that the magnitudes of tangential acceleration and angular acceleration are directly proportional to each other. The greater the angular acceleration, the larger the change in tangential acceleration, and vice versa. For example, consider riders in their pods on a Ferris wheel at rest. A Ferris wheel with greater angular acceleration will give the riders greater tangential acceleration because, as the Ferris wheel increases its rate of spinning, it also increases its tangential velocity. Note that the radius of the spinning object also matters. For example, for a given angular acceleration, a smaller Ferris wheel leads to a smaller tangential acceleration for the riders. TIPS FOR SUCCESS Tangential acceleration is sometimes denoted at. It is a linear acceleration in a direction tangent to the circle at the point of interest in circular or rotational motion. Remember that tangential acceleration is parallel to the tangential velocity (either in the same direction or in the opposite direction.) Centripetal acceleration is always perpendicular to the tangential velocity. So far, we have defined three rotational variables: Table 6.2 shows how they are related.,, and. These are the angular versions of the linear variables x, v, and a. Rotational Linear Relationship x Table 6.2 Rotational and Linear Variables 214 Chapter 6 \u2022 Circular and Rotational Motion Rotational Linear Relationship v a Table 6.2 Rotational and Linear Variables, and We can now begin to see how rotational quantities like, that starts at rest has a large angular acceleration for a fairly long time, it ends up spinning rapidly and rotates through many revolutions. Putting this in terms of the variables, if the wheel\u2019s angular acceleration the final angular velocity undergoes a large linear acceleration, then it has a large final velocity and will have traveled a large distance. and angle of rotation are large. In the case of linear motion, if an object starts at rest and are related to each other. For example, if a motorcycle wheel is large for a long period of time t, then The kinematics of rotational motion describes the relationships between the angle of rotation, angular velocity, angular acceleration, and time. It only describesmotion\u2014it does not include any forces or masses that may affect rotation (these are part of dynamics). Recall the kinematics equation for linear motion: (constant a). As in linear kinematics, we assume a is constant, which means that angular acceleration The equation for the k", "inematics relationship between,, and tis is also a constant, because. is the initial angular velocity. Notice that the equation is identical to the linear version, except with angular analogs of where the linear variables. In fact, all of the linear kinematics equations have rotational analogs, which are given in Table 6.3. These equations can be used to solve rotational or linear kinematics problem in which a and are constant. Rotational Linear constant, a constant, a constant, a Table 6.3 Equations for Rotational Kinematics In these equations, and are initial values, is zero, and the average angular velocity and average velocity are 6.11 Access for free at openstax.org. FUN IN PHYSICS Storm Chasing 6.3 \u2022 Rotational Motion 215 Figure 6.10 Tornadoes descend from clouds in funnel-like shapes that spin violently. (Daphne Zaras, U.S. National Oceanic and Atmospheric Administration) Storm chasers tend to fall into one of three groups: Amateurs chasing tornadoes as a hobby, atmospheric scientists gathering data for research, weather watchers for news media, or scientists having fun under the guise of work. Storm chasing is a dangerous pastime because tornadoes can change course rapidly with little warning. Since storm chasers follow in the wake of the destruction left by tornadoes, changing flat tires due to debris left on the highway is common. The most active part of the world for tornadoes, called tornado alley, is in the central United States, between the Rocky Mountains and Appalachian Mountains. Tornadoes are perfect examples of rotational motion in action in nature. They come out of severe thunderstorms called supercells, which have a column of air rotating around a horizontal axis, usually about four miles across. The difference in wind speeds between the strong cold winds higher up in the atmosphere in the jet stream and weaker winds traveling north from the Gulf of Mexico causes the column of rotating air to shift so that it spins around a vertical axis, creating a tornado. Tornadoes produce wind speeds as high as 500 km/h (approximately 300 miles/h), particularly at the bottom where the funnel is narrowest because the rate of rotation increases as the radius decreases. They blow houses away as if they were made of paper and have been known to pierce tree trunks with pieces of straw. GRASP CHECK What is the physics term for the eye of the storm? Why would winds be weaker at the eye of the tornado than at its outermost edge? a. The eye of the storm is the center of rotation. Winds are weaker at the eye of a tornado because tangential velocity is directly proportional to radius of curvature. b. The eye of the storm is the center of rotation. Winds are weaker at the eye of a tornado because tangential velocity is inversely proportional to radius of curvature. c. The eye of the storm is the center of rotation. Winds are weaker at the eye of a tornado because tangential velocity is directly proportional to the square of the radius of curvature. d. The eye of the storm is the center of rotation. Winds are weaker at the eye of a tornado because tangential velocity is inversely proportional to the square of the radius of curvature. Torque If you have ever spun a bike wheel or pushed a merry-go-round, you know that force is needed to change angular velocity. The farther the force is applied from the pivot point (or fulcrum), the greater the angular acceleration. For example, a door opens slowly if you push too close to its hinge, but opens easily if you push far from the hinges. Furthermore, we know that the more 216 Chapter 6 \u2022 Circular and Rotational Motion massive the door is, the more slowly it opens; this is because angular acceleration is inversely proportional to mass. These relationships are very similar to the relationships between force, mass, and acceleration from Newton\u2019s second law of motion. Since we have already covered the angular versions of distance, velocity and time, you may wonder what the angular version of force is, and how it relates to linear force. The angular version of force is torque, which is the turning effectiveness of a force. See Figure 6.11. The equation for the magnitude of torque is where ris the magnitude of the lever arm, F is the magnitude of the linear force, and is the angle between the lever arm and the force. The lever arm is the vector from the point of rotation (pivot point or fulcrum) to the location where force is applied. Since the magnitude of the lever arm is a distance, its units are in meters, and torque has units of N\u22c5m. Torque is a vector quantity and has the same direction as the angular acceleration that it produces. Figure 6.11 A man pushes a merry-go-round at its edge and perpendicular to the lever arm to achieve maximum torque. Applying a stronger torque will produce a greater angular acceleration.", " For example, the harder the man pushes the merry-goround in Figure 6.11, the faster it accelerates. Furthermore, the more massive the merry-go-round is, the slower it accelerates for the same torque. If the man wants to maximize the effect of his force on the merry-go-round, he should push as far from the center as possible to get the largest lever arm and, therefore, the greatest torque and angular acceleration. Torque is also maximized when the force is applied perpendicular to the lever arm. Solving Rotational Kinematics and Torque Problems Just as linear forces can balance to produce zero net force and no linear acceleration, the same is true of rotational motion. When two torques of equal magnitude act in opposing directions, there is no net torque and no angular acceleration, as you can see in the following video. If zero net torque acts on a system spinning at a constant angular velocity, the system will continue to spin at the same angular velocity. WATCH PHYSICS Introduction to Torque This video (https://www.khanacademy.org/science/physics/torque-angular-momentum/torque-tutorial/v/introduction-totorque) defines torque in terms of moment arm (which is the same as lever arm). It also covers a problem with forces acting in opposing directions about a pivot point. (At this stage, you can ignore Sal\u2019s references to work and mechanical advantage.) GRASP CHECK Click to view content (https://www.openstax.org/l/28torque) If the net torque acting on the ruler from the example was positive instead of zero, what would this say about the angular Access for free at openstax.org. 6.3 \u2022 Rotational Motion 217 acceleration? What would happen to the ruler over time? a. The ruler is in a state of rotational equilibrium so it will not rotate about its center of mass. Thus, the angular acceleration will be zero. b. The ruler is not in a state of rotational equilibrium so it will not rotate about its center of mass. Thus, the angular acceleration will be zero. c. The ruler is not in a state of rotational equilibrium so it will rotate about its center of mass. Thus, the angular acceleration will be non-zero. d. The ruler is in a state of rotational equilibrium so it will rotate about its center of mass. Thus, the angular acceleration will be non-zero. Now let\u2019s look at examples applying rotational kinematics to a fishing reel and the concept of torque to a merry-go-round. WORKED EXAMPLE Calculating the Time for a Fishing Reel to Stop Spinning A deep-sea fisherman uses a fishing rod with a reel of radius 4.50 cm. A big fish takes the bait and swims away from the boat, pulling the fishing line from his fishing reel. As the fishing line unwinds from the reel, the reel spins at an angular velocity of 220 rad/s. The fisherman applies a brake to the spinning reel, creating an angular acceleration of \u2212300 rad/s2. How long does it take the reel to come to a stop? Strategy We are asked to find the time tfor the reel to come to a stop. The magnitude of the initial angular velocity is rad/s, rad/s2, and the magnitude of the final angular velocity. The signed magnitude of the angular acceleration is where the minus sign indicates that it acts in the direction opposite to the angular velocity. Looking at the rotational kinematic equations, we see all quantities but tare known in the equation problem., making it the easiest equation to use for this Solution The equation to use is. We solve the equation algebraically for t, and then insert the known values. 6.12 Discussion The time to stop the reel is fairly small because the acceleration is fairly large. Fishing lines sometimes snap because of the forces involved, and fishermen often let the fish swim for a while before applying brakes on the reel. A tired fish will be slower, requiring a smaller acceleration and therefore a smaller force. 218 Chapter 6 \u2022 Circular and Rotational Motion WORKED EXAMPLE Calculating the Torque on a Merry-Go-Round Consider the man pushing the playground merry-go-round in Figure 6.11. He exerts a force of 250 N at the edge of the merrygo-round and perpendicular to the radius, which is 1.50 m. How much torque does he produce? Assume that friction acting on the merry-go-round is negligible. Strategy To find the torque, note that the applied force is perpendicular to the radius and that friction is negligible. Solution 6.13 Discussion The man maximizes the torque by applying force perpendicular to the lever arm, so that maximizes his torque by pushing at the outer edge of the merry-go-round, so that he gets the largest-", "possible lever arm. and. The man also Practice Problems 15. How much torque does a person produce if he applies a force away from the pivot point, perpendicularly to the lever arm? a. b. c. d. 16. An object\u2019s angular velocity changes from 3 rad/s clockwise to 8 rad/s clockwise in 5 s. What is its angular acceleration? a. 0.6 rad/s2 1.6 rad/s2 b. 1 rad/s2 c. 5 rad/s2 d. Check Your Understanding 17. What is angular acceleration? a. Angular acceleration is the rate of change of the angular displacement. b. Angular acceleration is the rate of change of the angular velocity. c. Angular acceleration is the rate of change of the linear displacement. d. Angular acceleration is the rate of change of the linear velocity. 18. What is the equation for angular acceleration, \u03b1? Assume \u03b8is the angle, \u03c9is the angular velocity, and tis time. a. b. c. d. 19. Which of the following best describes torque? It is the rotational equivalent of a force. It is the force that affects linear motion. It is the rotational equivalent of acceleration. It is the acceleration that affects linear motion. a. b. c. d. 20. What is the equation for torque? Access for free at openstax.org. a. b. c. d. 6.3 \u2022 Rotational Motion 219 220 Chapter 6 \u2022 Key Terms KEY TERMS angle of rotation the ratio of the arc length to the radius of time curvature of a circular path lever arm the distance between the point of rotation (pivot angular acceleration the rate of change of angular velocity point) and the location where force is applied with time radius of curvature the distance between the center of a angular velocity ( ) the rate of change in the angular circular path and the path position of an object following a circular path rotational motion the circular motion of an object about an arc length ( ) the distance traveled by an object along a axis of rotation circular path spin rotation about an axis that goes through the center of centrifugal force a fictitious force that acts in the direction mass of the object opposite the centripetal acceleration tangential acceleration the acceleration in a direction centripetal acceleration the acceleration of an object moving in a circle, directed toward the center of the circle centripetal force any force causing uniform circular tangent to the circular path of motion and in the same direction or opposite direction as the tangential velocity tangential velocity the instantaneous linear velocity of an motion object in circular or rotational motion circular motion the motion of an object along a circular torque the effectiveness of a force to change the rotational path speed of an object kinematics of rotational motion the relationships between rotation angle, angular velocity, angular acceleration, and uniform circular motion the motion of an object in a circular path at constant speed that always points toward the center of rotation, perpendicular to the linear velocity, in the same direction as the net force, and in the direction opposite that of the radius vector. \u2022 The standard unit for centripetal acceleration is m/s2. \u2022 Centripetal force Fc is any net force causing uniform circular motion. 6.3 Rotational Motion \u2022 Kinematics is the description of motion. \u2022 The kinematics of rotational motion describes the relationships between rotation angle, angular velocity, angular acceleration, and time. \u2022 Torque is the effectiveness of a force to change the rotational speed of an object. Torque is the rotational analog of force. \u2022 The lever arm is the distance between the point of rotation (pivot point) and the location where force is applied. \u2022 Torque is maximized by applying force perpendicular to the lever arm and at a point as far as possible from the pivot point or fulcrum. If torque is zero, angular acceleration is zero. SECTION SUMMARY 6.1 Angle of Rotation and Angular Velocity \u2022 Circular motion is motion in a circular path. \u2022 The angle of rotation is defined as the ratio of the arc length to the radius of curvature. \u2022 The arc length is the distance traveled along a circular path and ris the radius of curvature of the circular path. \u2022 The angle of rotation (rad), where \u2022 Angular velocity where a rotation is measured in units of radians revolution. is the rate of change of an angle, occurs in a time. \u2022 The units of angular velocity are radians per second (rad/s). \u2022 Tangential speed vand angular speed are related by, and tangential velocity has units of m/s. \u2022 The direction of angular velocity is along the axis of rotation, toward (away) from you for clockwise (counterclockwise) motion. 6.2 Uniform Circular Motion \u2022 Centripetal acceleration ac is the acceleration experienced while in uniform circular motion. \u2022 Centripetal acceleration force is a center-seekingforce", " Access for free at openstax.org. KEY EQUATIONS 6.1 Angle of Rotation and Angular Velocity 6.3 Rotational Motion Chapter 6 \u2022 Key Equations 221 Angle of rotation Angular speed: Tangential speed: 6.2 Uniform Circular Motion Centripetal acceleration Centripetal force or,, Angular acceleration Rotational kinematic equations Tangential (linear) acceleration Torque,,, CHAPTER REVIEW Concept Items 6.2 Uniform Circular Motion 6.1 Angle of Rotation and Angular Velocity 4. What is the equation for centripetal acceleration in terms 1. One revolution is equal to how many radians? Degrees? a. b. c. d. 2. What is tangential velocity? a. Tangential velocity is the average linear velocity of an object in a circular motion. b. Tangential velocity is the instantaneous linear velocity of an object undergoing rotational motion. c. Tangential velocity is the average angular velocity of an object in a circular motion. d. Tangential velocity is the instantaneous angular velocity of an object in a circular motion. 3. What kind of motion is called spin? a. Spin is rotational motion of an object about an axis parallel to the axis of the object. of angular velocity and the radius? a. b. c. d. 5. How can you express centripetal force in terms of centripetal acceleration? a. b. c. d. 6. What is meant by the word centripetal? a. b. c. d. center-seeking center-avoiding central force central acceleration b. Spin is translational motion of an object about an 6.3 Rotational Motion axis parallel to the axis of the object. c. Spin is the rotational motion of an object about its center of mass. d. Spin is translational motion of an object about its own axis. 7. Conventionally, for which direction of rotation of an object is angular acceleration considered positive? a. b. c. d. the positive xdirection of the coordinate system the negative xdirection of the coordinate system the counterclockwise direction the clockwise direction 222 Chapter 6 \u2022 Chapter Review 8. When you push a door closer to the hinges, why does it 9. When is angular acceleration negative? open more slowly? a. It opens slowly, because the lever arm is shorter so the torque is large. It opens slowly because the lever arm is longer so the torque is large. It opens slowly, because the lever arm is shorter so the torque is less. It opens slowly, because the lever arm is longer so the torque is less. b. c. d. Critical Thinking Items 6.1 Angle of Rotation and Angular Velocity 10. When the radius of the circular path of rotational motion increases, what happens to the arc length for a given angle of rotation? a. The arc length is directly proportional to the radius of the circular path, and it increases with the radius. b. The arc length is inversely proportional to the radius of the circular path, and it decreases with the radius. c. The arc length is directly proportional to the radius of the circular path, and it decreases with the radius. d. The arc length is inversely proportional to the radius of the circular path, and it increases with the radius. 11. Consider a CD spinning clockwise. What is the sum of the instantaneous velocities of two points on both ends of its diameter? a. b. c. d. 6.2 Uniform Circular Motion 12. What are the directions of the velocity and acceleration of an object in uniform circular motion? a. Velocity is tangential, and acceleration is radially outward. b. Velocity is tangential, and acceleration is radially inward. c. Velocity is radially outward, and acceleration is tangential. d. Velocity is radially inward, and acceleration is tangential. 13. Suppose you have an object tied to a rope and are rotating it over your head in uniform circular motion. If Access for free at openstax.org. displacement and is negative when a. Angular acceleration is the rate of change of the increases. b. Angular acceleration is the rate of change of the decreases. displacement and is negative when c. Angular acceleration is the rate of change of angular velocity and is negative when increases. d. Angular acceleration is the rate of change of angular velocity and is negative when decreases. you increase the length of the rope, would you have to apply more or less force to maintain the same speed? a. More force is required, because the force is inversely proportional to the radius of the circular orbit. b. More force is required because the force is directly proportional to the radius of the circular orbit. c. Less force is required because the force is inversely proportional to the radius of the circular orbit. d. Less force is required because the force is directly proportional to the radius of the circular orbit. 6.3 Rotational Motion 14. Consider two spinning", " tops with different radii. Both have the same linear instantaneous velocities at their edges. Which top has a higher angular velocity? a. the top with the smaller radius because the radius of curvature is inversely proportional to the angular velocity the top with the smaller radius because the radius of curvature is directly proportional to the angular velocity the top with the larger radius because the radius of curvature is inversely proportional to the angular velocity b. c. d. The top with the larger radius because the radius of curvature is directly proportional to the angular velocity 15. A person tries to lift a stone by using a lever. If the lever arm is constant and the mass of the stone increases, what is true of the torque necessary to lift it? a. It increases, because the torque is directly proportional to the mass of the body. It increases because the torque is inversely proportional to the mass of the body. It decreases because the torque is directly proportional to the mass of the body. It decreases, because the torque is inversely proportional to the mass of the body. b. c. d. Chapter 6 \u2022 Test Prep 223 Problems d. 13, 333 N 6.1 Angle of Rotation and Angular Velocity 16. What is the angle of rotation (in degrees) between two hands of a clock, if the radius of the clock is the arc length separating the two hands is a. b. c. d. and? 17. A clock has radius of. The outermost point on its minute hand travels along the edge. What is its tangential speed? a. b. c. d. 6.2 Uniform Circular Motion 18. What is the centripetal force exerted on a 1,600 kg car that rounds a 100 m radius curve at 12 m/s? 192 N a. b. 1, 111 N c. 2, 300 N Performance Task 6.3 Rotational Motion 22. Design a lever arm capable of lifting a 0.5 kg object such as a stone. The force for lifting should be provided by TEST PREP Multiple Choice 6.1 Angle of Rotation and Angular Velocity 23. What is 1 radian approximately in degrees? 57.3\u00b0 360\u00b0 a. b. c. \u03c0\u00b0 d. 2\u03c0\u00b0 24. If the following objects are spinning at the same angular velocities, the edge of which one would have the highest speed? a. Mini CD b. Regular CD c. Vinyl record 25. What are possible units for tangential velocity? a. b. 19. Find the frictional force between the tires and the road that allows a 1,000 kg car traveling at 30 m/s to round a 20 m radius curve. a. 22 N b. 667 N c. d. 45, 000 N 1, 500 N 6.3 Rotational Motion 20. An object\u2019s angular acceleration is 36 rad/s2. If it were initially spinning with a velocity of 6.0 m/s, what would its angular velocity be after 5.0 s? 186 rad/s a. 190 rad/s2 b. c. \u2212174 rad/s d. \u2212174 rad/s2 21. When a fan is switched on, it undergoes an angular acceleration of 150 rad/s2. How long will it take to achieve its maximum angular velocity of 50 rad/s? a. \u22120.3 s b. 0.3 s 3.0 s c. placing coins on the other end of the lever. How many coins would you need? What happens if you shorten or lengthen the lever arm? What does this say about torque? in radians? c. 26. What is a. b. c. d. 27. For a given object, what happens to the arc length as the angle of rotation increases? a. The arc length is directly proportional to the angle of rotation, so it increases with the angle of rotation. b. The arc length is inversely proportional to the angle of rotation, so it decreases with the angle of rotation. c. The arc length is directly proportional to the angle of rotation, so it decreases with the angle of rotation. 224 Chapter 6 \u2022 Test Prep d. The arc length is inversely proportional to the angle of rotation, so it increases with the angle of rotation. 6.2 Uniform Circular Motion 28. Which of these quantities is constant in uniform circular motion? a. Speed b. Velocity c. Acceleration d. Displacement 29. Which of these quantities impact centripetal force? a. Mass and speed only b. Mass and radius only c. Speed and radius only d. Mass, speed, and radius all impact centripetal force 30. An increase in the magnitude of which of these quantities causes a reduction in centripetal force? a. Mass b. Radius of curvature c. Speed 31. What happens to centripetal acceleration as the radius of curvature decreases and the speed is constant,", " and why? a. It increases, because the centripetal acceleration is inversely proportional to the radius of the curvature. It increases, because the centripetal acceleration is directly proportional to the radius of curvature. It decreases, because the centripetal acceleration is inversely proportional to the radius of the curvature. It decreases, because the centripetal acceleration is directly proportional to the radius of the curvature. b. c. d. 32. Why do we experience more sideways acceleration while driving around sharper curves? Short Answer 6.1 Angle of Rotation and Angular Velocity 37. What is the rotational analog of linear velocity? a. Angular displacement b. Angular velocity c. Angular acceleration d. Angular momentum 38. What is the rotational analog of distance? a. Rotational angle b. Torque c. Angular velocity d. Angular momentum Access for free at openstax.org. a. Centripetal acceleration is inversely proportional to the radius of curvature, so it increases as the radius of curvature decreases. b. Centripetal acceleration is directly proportional to the radius of curvature, so it decreases as the radius of curvature decreases. c. Centripetal acceleration is directly proportional to the radius of curvature, so it decreases as the radius of curvature increases. d. Centripetal acceleration is directly proportional to the radius of curvature, so it increases as the radius of curvature increases. 6.3 Rotational Motion 33. Which of these quantities is not described by the kinematics of rotational motion? a. Rotation angle b. Angular acceleration c. Centripetal force d. Angular velocity 34. In the equation, what is F? a. Linear force b. Centripetal force c. Angular force 35. What happens when two torques act equally in opposite directions? a. Angular velocity is zero. b. Angular acceleration is zero. 36. What is the mathematical relationship between angular and linear accelerations? a. b. c. d. 39. What is the equation that relates the linear speed of a point on a rotating object with the object's angular quantities? a. b. c. d. 40. As the angular velocity of an object increases, what happens to the linear velocity of a point on that object? It increases, because linear velocity is directly a. proportional to angular velocity. It increases, because linear velocity is inversely proportional to angular velocity. b. Chapter 6 \u2022 Test Prep 225 c. d. It decreases because linear velocity is directly proportional to angular velocity. It decreases because linear velocity is inversely proportional to angular velocity. a. b. c. d. 41. What is angular speed in terms of tangential speed and 47. What are the standard units for centripetal force? the radius? a. b. c. d. 42. Why are radians dimensionless? a. Radians are dimensionless, because they are defined as a ratio of distances. They are defined as the ratio of the arc length to the radius of the circle. b. Radians are dimensionless because they are defined as a ratio of distances. They are defined as the ratio of the area to the radius of the circle. c. Radians are dimensionless because they are defined as multiplication of distance. They are defined as the multiplication of the arc length to the radius of the circle. d. Radians are dimensionless because they are defined as multiplication of distance. They are defined as the multiplication of the area to the radius of the circle. 6.2 Uniform Circular Motion 43. What type of quantity is centripetal acceleration? a. Scalar quantity; centripetal acceleration has magnitude only but no direction b. Scalar quantity; centripetal acceleration has magnitude as well as direction c. Vector quantity; centripetal acceleration has magnitude only but no direction d. Vector quantity; centripetal acceleration has magnitude as well as direction 44. What are the standard units for centripetal acceleration? a. m/s b. c. d. 45. What is the angle formed between the vectors of tangential velocity and centripetal force? a. b. c. d. 46. What is the angle formed between the vectors of centripetal acceleration and centripetal force? a. m b. m/s c. m/s2 d. newtons 48. As the mass of an object in uniform circular motion increases, what happens to the centripetal force required to keep it moving at the same speed? a. It increases, because the centripetal force is directly proportional to the mass of the rotating body. It increases, because the centripetal force is inversely proportional to the mass of the rotating body. It decreases, because the centripetal force is directly proportional to the mass of the rotating body. It decreases, because the centripetal force", " is inversely proportional to the mass of the rotating body. b. c. d. 6.3 Rotational Motion 49. The relationships between which variables are described by the kinematics of rotational motion? a. The kinematics of rotational motion describes the relationships between rotation angle, angular velocity, and angular acceleration. b. The kinematics of rotational motion describes the relationships between rotation angle, angular velocity, angular acceleration, and angular momentum. c. The kinematics of rotational motion describes the relationships between rotation angle, angular velocity, angular acceleration, and time. d. The kinematics of rotational motion describes the relationships between rotation angle, angular velocity, angular acceleration, torque, and time. 50. What is the kinematics relationship between,, and? a. b. c. d. 51. What kind of quantity is torque? a. Scalar b. Vector 226 Chapter 6 \u2022 Test Prep c. Dimensionless d. Fundamental quantity 52. If a linear force is applied to a lever arm farther away from the pivot point, what happens to the resultant torque? a. b. c. d. It decreases. It increases. It remains the same. It changes the direction. 53. How can the same force applied to a lever produce different torques? a. By applying the force at different points of the lever Extended Response 6.1 Angle of Rotation and Angular Velocity 54. Consider two pits on a CD, one close to the center and one close to the outer edge. When the CD makes one full rotation, which pit would have gone through a greater angle of rotation? Which one would have covered a greater arc length? a. The one close to the center would go through the greater angle of rotation. The one near the outer edge would trace a greater arc length. b. The one close to the center would go through the greater angle of rotation. The one near the center would trace a greater arc length. c. Both would go through the same angle of rotation. The one near the outer edge would trace a greater arc length. d. Both would go through the same angle of rotation. The one near the center would trace a greater arc length. 55. Consider two pits on a CD, one close to the center and one close to the outer edge. For a given angular velocity of the CD, which pit has a higher angular velocity? Which has a higher tangential velocity? a. The point near the center would have the greater angular velocity and the point near the outer edge would have the higher linear velocity. arm along the length of the lever or by changing the angle between the lever arm and the applied force. b. By applying the force at the same point of the lever arm along the length of the lever or by changing the angle between the lever arm and the applied force. c. By applying the force at different points of the lever arm along the length of the lever or by maintaining the same angle between the lever arm and the applied force. d. By applying the force at the same point of the lever arm along the length of the lever or by maintaining the same angle between the lever arm and the applied force. the same? a. It increases because tangential velocity is directly proportional to the radius. It increases because tangential velocity is inversely proportional to the radius. It decreases because tangential velocity is directly proportional to the radius. It decreases because tangential velocity is inversely proportional to the radius. b. c. d. 6.2 Uniform Circular Motion 57. Is an object in uniform circular motion accelerating? Why or why not? a. Yes, because the velocity is not constant. b. No, because the velocity is not constant. c. Yes, because the velocity is constant. d. No, because the velocity is constant. 58. An object is in uniform circular motion. Suppose the centripetal force was removed. In which direction would the object now travel? a. b. In the direction of the centripetal force In the direction opposite to the direction of the centripetal force In the direction of the tangential velocity In the direction opposite to the direction of the tangential velocity c. d. b. The point near the edge would have the greater 59. An object undergoes uniform circular motion. If the angular velocity and the point near the center would have the higher linear velocity. c. Both have the same angular velocity and the point near the outer edge would have the higher linear velocity. d. Both have the same angular velocity and the point near the center would have the higher linear velocity. 56. What happens to tangential velocity as the radius of an object increases provided the angular velocity remains Access for free at openstax.org. radius of curvature and mass of the object are constant, what is the centripetal force proportional to? a. b. c. d. 6.3 Rotational Motion 60. Why do tornadoes produce more wind speed at the Chapter 6 \u2022", " Test Prep 227 bottom of the funnel? a. Wind speed is greater at the bottom because rate of rotation increases as the radius increases. b. The force should be applied perpendicularly to the lever arm as far as possible from the pivot point. c. The force should be applied parallel to the lever arm b. Wind speed is greater at the bottom because rate of as far as possible from the pivot point. rotation increases as the radius decreases. d. The force should be applied parallel to the lever arm c. Wind speed is greater at the bottom because rate of as close as possible from the pivot point. rotation decreases as the radius increases. d. Wind speed is greater at the bottom because rate of rotation decreases as the radius increases. 61. How can you maximize the torque applied to a given lever arm without applying more force? a. The force should be applied perpendicularly to the lever arm as close as possible from the pivot point. 62. When will an object continue spinning at the same angular velocity? a. When net torque acting on it is zero b. When net torque acting on it is non zero c. When angular acceleration is positive d. When angular acceleration is negative 228 Chapter 6 \u2022 Test Prep Access for free at openstax.org. CHAPTER 7 Newton's Law of Gravitation Figure 7.1 Johannes Kepler (left) showed how the planets move, and Isaac Newton (right) discovered that gravitational force caused them to move that way. ((left) unknown, Public Domain; (right) Sir Godfrey Kneller, Public Domain) Chapter Outline 7.1 Kepler's Laws of Planetary Motion 7.2 Newton's Law of Universal Gravitation and Einstein's Theory of General Relativity INTRODUCTION What do a falling apple and the orbit of the moon have in common? You will learn in this chapter that each is caused by gravitational force. The motion of all celestial objects, in fact, is determined by the gravitational force, which depends on their mass and separation. Johannes Kepler discovered three laws of planetary motion that all orbiting planets and moons follow. Years later, Isaac Newton found these laws useful in developing his law of universal gravitation. This law relates gravitational force to the masses of objects and the distance between them. Many years later still, Albert Einstein showed there was a little more to the gravitation story when he published his theory of general relativity. 7.1 Kepler's Laws of Planetary Motion Section Learning Objectives By the end of this section, you will be able to do the following: \u2022 Explain Kepler\u2019s three laws of planetary motion \u2022 Apply Kepler\u2019s laws to calculate characteristics of orbits 230 Chapter 7 \u2022 Newton's Law of Gravitation Section Key Terms aphelion Copernican model eccentricity Kepler\u2019s laws of planetary motion perihelion Ptolemaic model Concepts Related to Kepler\u2019s Laws of Planetary Motion Examples of orbits abound. Hundreds of artificial satellites orbit Earth together with thousands of pieces of debris. The moon\u2019s orbit around Earth has intrigued humans from time immemorial. The orbits of planets, asteroids, meteors, and comets around the sun are no less interesting. If we look farther, we see almost unimaginable numbers of stars, galaxies, and other celestial objects orbiting one another and interacting through gravity. All these motions are governed by gravitational force. The orbital motions of objects in our own solar system are simple enough to describe with a few fairly simple laws. The orbits of planets and moons satisfy the following two conditions: \u2022 The mass of the orbiting object, m, is small compared to the mass of the object it orbits, M. \u2022 The system is isolated from other massive objects. Based on the motion of the planets about the sun, Kepler devised a set of three classical laws, called Kepler\u2019s laws of planetary motion, that describe the orbits of all bodies satisfying these two conditions: 1. The orbit of each planet around the sun is an ellipse with the sun at one focus. 2. Each planet moves so that an imaginary line drawn from the sun to the planet sweeps out equal areas in equal times. 3. The ratio of the squares of the periods of any two planets about the sun is equal to the ratio of the cubes of their average distances from the sun. These descriptive laws are named for the German astronomer Johannes Kepler (1571\u20131630). He devised them after careful study (over some 20 years) of a large amount of meticulously recorded observations of planetary motion done by Tycho Brahe (1546\u20131601). Such careful collection and detailed recording of methods and data are hallmarks of good science. Data constitute the evidence from which new interpretations and meanings can be constructed. Let\u2019s look closer at each of these laws. Kepler\u2019s First Law The orbit of each planet about the sun is an ellipse with the sun at one focus, as shown in Figure 7.2. The planet\u2019s closest approach to the sun is called aphelion and its farthest distance from the sun is called per", "ihelion. Access for free at openstax.org. 7.1 \u2022 Kepler's Laws of Planetary Motion 231 Figure 7.2 (a) An ellipse is a closed curve such that the sum of the distances from a point on the curve to the two foci (f1 and f2) is constant. (b) For any closed orbit, mfollows an elliptical path with Mat one focus. (c) The aphelion (ra) is the closest distance between the planet and the sun, while the perihelion (rp) is the farthest distance from the sun. If you know the aphelion (ra) and perihelion (rp) distances, then you can calculate the semi-major axis (a) and semi-minor axis (b). Figure 7.3 You can draw an ellipse as shown by putting a pin at each focus, and then placing a loop of string around a pen and the pins and tracing a line on the paper. 232 Chapter 7 \u2022 Newton's Law of Gravitation Kepler\u2019s Second Law Each planet moves so that an imaginary line drawn from the sun to the planet sweeps out equal areas in equal times, as shown in Figure 7.4. Figure 7.4 The shaded regions have equal areas. The time for mto go from A to B is the same as the time to go from C to D and from E to F. The mass mmoves fastest when it is closest to M. Kepler\u2019s second law was originally devised for planets orbiting the sun, but it has broader validity. TIPS FOR SUCCESS Note that while, for historical reasons, Kepler\u2019s laws are stated for planets orbiting the sun, they are actually valid for all bodies satisfying the two previously stated conditions. Kepler\u2019s Third Law The ratio of the periods squared of any two planets around the sun is equal to the ratio of their average distances from the sun cubed. In equation form, this is where Tis the period (time for one orbit) and ris the average distance (also called orbital radius). This equation is valid only for comparing two small masses orbiting a single large mass. Most importantly, this is only a descriptive equation; it gives no information about the cause of the equality. LINKS TO PHYSICS History: Ptolemy vs. Copernicus Before the discoveries of Kepler, Copernicus, Galileo, Newton, and others, the solar system was thought to revolve around Earth as shown in Figure 7.5 (a). This is called the Ptolemaic model, named for the Greek philosopher Ptolemy who lived in the second century AD. The Ptolemaic model is characterized by a list of facts for the motions of planets, with no explanation of cause and effect. There tended to be a different rule for each heavenly body and a general lack of simplicity. Figure 7.5 (b) represents the modern or Copernican model. In this model, a small set of rules and a single underlying force explain not only all planetary motion in the solar system, but also all other situations involving gravity. The breadth and simplicity of the laws of physics are compelling. Access for free at openstax.org. 7.1 \u2022 Kepler's Laws of Planetary Motion 233 Figure 7.5 (a) The Ptolemaic model of the universe has Earth at the center with the moon, the planets, the sun, and the stars revolving about it in complex circular paths. This geocentric (Earth-centered) model, which can be made progressively more accurate by adding more circles, is purely descriptive, containing no hints about the causes of these motions. (b) The Copernican heliocentric (sun-centered) model is a simpler and more accurate model. Nicolaus Copernicus (1473\u20131543) first had the idea that the planets circle the sun, in about 1514. It took him almost 20 years to work out the mathematical details for his model. He waited another 10 years or so to publish his work. It is thought he hesitated because he was afraid people would make fun of his theory. Actually, the reaction of many people was more one of fear and anger. Many people felt the Copernican model threatened their basic belief system. About 100 years later, the astronomer Galileo was put under house arrest for providing evidence that planets, including Earth, orbited the sun. In all, it took almost 300 years for everyone to admit that Copernicus had been right all along. GRASP CHECK Explain why Earth does actually appear to be the center of the solar system. a. Earth appears to be the center of the solar system because Earth is at the center of the universe, and everything revolves around it in a circular orbit. b. Earth appears to be the center of the solar system because, in the reference frame of Earth, the sun, moon, and planets all appear to move", " across the sky as if they were circling Earth. c. Earth appears to be at the center of the solar system because Earth is at the center of the solar system and all the heavenly bodies revolve around it. d. Earth appears to be at the center of the solar system because Earth is located at one of the foci of the elliptical orbit of the sun, moon, and other planets. Virtual Physics Acceleration This simulation allows you to create your own solar system so that you can see how changing distances and masses determines the orbits of planets. Click Helpfor instructions. Click to view content (https://archive.cnx.org/specials/ee816dff-0b5f-4e6f-8250-f9fb9e39d716/my-solar-system/) GRASP CHECK When the central object is off center, how does the speed of the orbiting object vary? a. The orbiting object moves fastest when it is closest to the central object and slowest when it is farthest away. b. The orbiting object moves slowest when it is closest to the central object and fastest when it is farthest away. 234 Chapter 7 \u2022 Newton's Law of Gravitation c. The orbiting object moves with the same speed at every point on the circumference of the elliptical orbit. d. There is no relationship between the speed of the object and the location of the planet on the circumference of the orbit. Calculations Related to Kepler\u2019s Laws of Planetary Motion Kepler\u2019s First Law Refer back to Figure 7.2 (a). Notice which distances are constant. The foci are fixed, so distance of an ellipse states that the sum of the distances perimeter of triangle distances of objects in a system that includes one object orbiting another. is also constant. These two facts taken together mean that the must also be constant. Knowledge of these constants will help you determine positions and is a constant. The definition Kepler\u2019s Second Law Refer back to Figure 7.4. The second law says that the segments have equal area and that it takes equal time to sweep through each segment. That is, the time it takes to travel from A to B equals the time it takes to travel from C to D, and so forth. Velocity v equals distance ddivided by time t:, so distance divided by velocity is also a constant. For example, if we know the average velocity of Earth on June 21 and December 21, we can compare the distance Earth travels on those days.. Then, The degree of elongation of an elliptical orbit is called its eccentricity (e). Eccentricity is calculated by dividing the distance f from the center of an ellipse to one of the foci by half the long axis a. When, the ellipse is a circle. The area of an ellipse is given by sweeps out in a given period of time, you can calculate the fraction of the year that has elapsed., where bis half the short axis. If you know the axes of Earth\u2019s orbit and the area Earth 7.1 WORKED EXAMPLE Kepler\u2019s First Law At its closest approach, a moon comes within 200,000 km of the planet it orbits. At that point, the moon is 300,000 km from the other focus of its orbit, f2. The planet is focus f1 of the moon\u2019s elliptical orbit. How far is the moon from the planet when it is 260,000 km from f2? Strategy Show and label the ellipse that is the orbit in your solution. Picture the triangle f1mf2 collapsed along the major axis and add up the lengths of the three sides. Find the length of the unknown side of the triangle when the moon is 260,000 km from f2. Solution Perimeter of Discussion The perimeter of triangle f1mf2 must be constant because the distance between the foci does not change and Kepler\u2019s first law says the orbit is an ellipse. For any ellipse, the sum of the two sides of the triangle, which are f1mand mf2, is constant. WORKED EXAMPLE Kepler\u2019s Second Law Figure 7.6 shows the major and minor axes of an ellipse. The semi-major and semi-minor axes are half of these, respectively. Access for free at openstax.org. 7.1 \u2022 Kepler's Laws of Planetary Motion 235 Figure 7.6 The major axis is the length of the ellipse, and the minor axis is the width of the ellipse. The semi-major axis is half the major axis, and the semi-minor axis is half the minor axis. Earth\u2019s orbit is slightly elliptical, with a semi-major axis of 152 million km and a semi-minor axis of 147 million km. If Earth\u2019s period is 365.26 days, what area does an Earth", "-to-sun line sweep past in one day? Strategy Each day, Earth sweeps past an equal-sized area, so we divide the total area by the number of days in a year to find the area swept past in one day. For total area use a year (i.e., its period).. Calculate A, the area inside Earth\u2019s orbit and divide by the number of days in Solution 7.2 The area swept out in one day is thus. Discussion The answer is based on Kepler\u2019s law, which states that a line from a planet to the sun sweeps out equal areas in equal times. Kepler\u2019s Third Law Kepler\u2019s third law states that the ratio of the squares of the periods of any two planets (T1, T2) is equal to the ratio of the cubes of their average orbital distance from the sun (r1, r2). Mathematically, this is represented by From this equation, it follows that the ratio r3/T2 is the same for all planets in the solar system. Later we will see how the work of Newton leads to a value for this constant. WORKED EXAMPLE Kepler\u2019s Third Law Given that the moon orbits Earth each 27.3 days and that it is an average distance of calculate the period of an artificial satellite orbiting at an average altitude of 1,500 km above Earth\u2019s surface. Strategy The period, or time for one orbit, is related to the radius of the orbit by Kepler\u2019s third law, given in mathematical form by from the center of Earth, 236 Chapter 7 \u2022 Newton's Law of Gravitation. Let us use the subscript 1 for the moon and the subscript 2 for the satellite. We are asked to find T2. The given information tells us that the orbital radius of the moon is, and that the period of the moon is. The height of the artificial satellite above Earth\u2019s surface is given, so to get the distance r2 from the center of. Now all Earth we must add the height to the radius of Earth (6380 km). This gives quantities are known, so T2 can be found. Solution To solve for T2, we cross-multiply and take the square root, yielding 7.3 Discussion This is a reasonable period for a satellite in a fairly low orbit. It is interesting that any satellite at this altitude will complete one orbit in the same amount of time. Practice Problems 1. A planet with no axial tilt is located in another solar system. It circles its sun in a very elliptical orbit so that the temperature varies greatly throughout the year. If the year there has 612 days and the inhabitants celebrate the coldest day on day 1 of their calendar, when is the warmest day? a. Day 1 b. Day 153 c. Day 306 d. Day 459 2. A geosynchronous Earth satellite is one that has an orbital period of precisely 1 day. Such orbits are useful for communication and weather observation because the satellite remains above the same point on Earth (provided it orbits in the equatorial plane in the same direction as Earth\u2019s rotation). The ratio for the moon is. Calculate the radius of the orbit of such a satellite. a. b. c. d. Check Your Understanding 3. Are Kepler\u2019s laws purely descriptive, or do they contain causal information? a. Kepler\u2019s laws are purely descriptive. b. Kepler\u2019s laws are purely causal. c. Kepler\u2019s laws are descriptive as well as causal. d. Kepler\u2019s laws are neither descriptive nor causal. 4. True or false\u2014According to Kepler\u2019s laws of planetary motion, a satellite increases its speed as it approaches its parent body and decreases its speed as it moves away from the parent body. a. True b. False 5. Identify the locations of the foci of an elliptical orbit. a. One focus is the parent body, and the other is located at the opposite end of the ellipse, at the same distance from the center as the parent body. b. One focus is the parent body, and the other is located at the opposite end of the ellipse, at half the distance from the center as the parent body. Access for free at openstax.org. 7.2 \u2022 Newton's Law of Universal Gravitation and Einstein's Theory of General Relativity 237 c. One focus is the parent body and the other is located outside of the elliptical orbit, on the line on which is the semi- major axis of the ellipse. d. One focus is on the line containing the semi-major axis of the ellipse, and the other is located anywhere on the elliptical orbit of the satellite. 7.2 Newton's Law of Universal Gravitation and Einstein's Theory of General Relativity Section Learning Objectives By the end of this section, you will be able to do the following: \u2022 Explain Newton\u2019s law of", " universal gravitation and compare it to Einstein\u2019s theory of general relativity \u2022 Perform calculations using Newton\u2019s law of universal gravitation Section Key Terms Einstein\u2019s theory of general relativity gravitational constant Newton\u2019s universal law of gravitation Concepts Related to Newton\u2019s Law of Universal Gravitation Sir Isaac Newton was the first scientist to precisely define the gravitational force, and to show that it could explain both falling bodies and astronomical motions. See Figure 7.7. But Newton was not the first to suspect that the same force caused both our weight and the motion of planets. His forerunner, Galileo Galilei, had contended that falling bodies and planetary motions had the same cause. Some of Newton\u2019s contemporaries, such as Robert Hooke, Christopher Wren, and Edmund Halley, had also made some progress toward understanding gravitation. But Newton was the first to propose an exact mathematical form and to use that form to show that the motion of heavenly bodies should be conic sections\u2014circles, ellipses, parabolas, and hyperbolas. This theoretical prediction was a major triumph. It had been known for some time that moons, planets, and comets follow such paths, but no one had been able to propose an explanation of the mechanism that caused them to follow these paths and not others. Figure 7.7 The popular legend that Newton suddenly discovered the law of universal gravitation when an apple fell from a tree and hit him on the head has an element of truth in it. A more probable account is that he was walking through an orchard and wondered why all the apples fell in the same direction with the same acceleration. Great importance is attached to it because Newton\u2019s universal law of gravitation and his laws of motion answered very old questions about nature and gave tremendous support to the notion of underlying simplicity and unity in nature. Scientists still expect underlying simplicity to emerge from their ongoing inquiries into nature. The gravitational force is relatively simple. It is always attractive, and it depends only on the masses involved and the distance 238 Chapter 7 \u2022 Newton's Law of Gravitation between them. Expressed in modern language, Newton\u2019s universal law of gravitation states that every object in the universe attracts every other object with a force that is directed along a line joining them. The force is directly proportional to the product of their masses and inversely proportional to the square of the distance between them. This attraction is illustrated by Figure 7.8. Figure 7.8 Gravitational attraction is along a line joining the centers of mass (CM) of the two bodies. The magnitude of the force on each body is the same, consistent with Newton\u2019s third law (action-reaction). For two bodies having masses mand Mwith a distance rbetween their centers of mass, the equation for Newton\u2019s universal law of gravitation is where F is the magnitude of the gravitational force and Gis a proportionality factor called the gravitational constant. Gis a universal constant, meaning that it is thought to be the same everywhere in the universe. It has been measured experimentally to be. If a person has a mass of 60.0 kg, what would be the force of gravitational attraction on him at Earth\u2019s surface? Gis given above, Earth\u2019s mass Mis 5.97 \u00d7 1024 kg, and the radius rof Earth is 6.38 \u00d7 106 m. Putting these values into Newton\u2019s universal law of gravitation gives We can check this result with the relationship: You may remember that g, the acceleration due to gravity, is another important constant related to gravity. By substituting g for a in the equation for Newton\u2019s second law of motion we get gravitation gives. Combining this with the equation for universal Cancelling the mass mon both sides of the equation and filling in the values for the gravitational constant and mass and radius of the Earth, gives the value of g,which may look familiar. This is a good point to recall the difference between mass and weight. Mass is the amount of matter in an object; weight is the Access for free at openstax.org. 7.2 \u2022 Newton's Law of Universal Gravitation and Einstein's Theory of General Relativity 239 force of attraction between the mass within two objects. Weight can change because gis different on every moon and planet. An object\u2019s mass mdoes not change but its weight mg can. Virtual Physics Gravity and Orbits Move the sun, Earth, moon and space station in this simulation to see how it affects their gravitational forces and orbital paths. Visualize the sizes and distances between different heavenly bodies. Turn off gravity to see what would happen without it! Click to view content (https://archive.cnx.org/specials/a14085c8-96b8-4d04-bb5a-56d9ccbe6e69/gravity-and-orbits/) GRASP CHECK Why doesn\u2019t the Moon travel in a smooth", " circle around the Sun? a. The Moon is not affected by the gravitational field of the Sun. b. The Moon is not affected by the gravitational field of the Earth. c. The Moon is affected by the gravitational fields of both the Earth and the Sun, which are always additive. d. The moon is affected by the gravitational fields of both the Earth and the Sun, which are sometimes additive and sometimes opposite. Snap Lab Take-Home Experiment: Falling Objects In this activity you will study the effects of mass and air resistance on the acceleration of falling objects. Make predictions (hypotheses) about the outcome of this experiment. Write them down to compare later with results. \u2022 Four sheets of -inch paper Procedure \u2022 Take four identical pieces of paper. \u25e6 Crumple one up into a small ball. \u25e6 Leave one uncrumpled. \u25e6 Take the other two and crumple them up together, so that they make a ball of exactly twice the mass of the other crumpled ball. \u25e6 Now compare which ball of paper lands first when dropped simultaneously from the same height. 1. Compare crumpled one-paper ball with crumpled two-paper ball. 2. Compare crumpled one-paper ball with uncrumpled paper. GRASP CHECK Why do some objects fall faster than others near the surface of the earth if all mass is attracted equally by the force of gravity? a. Some objects fall faster because of air resistance, which acts in the direction of the motion of the object and exerts more force on objects with less surface area. b. Some objects fall faster because of air resistance, which acts in the direction opposite the motion of the object and exerts more force on objects with less surface area. c. Some objects fall faster because of air resistance, which acts in the direction of motion of the object and exerts more force on objects with more surface area. d. Some objects fall faster because of air resistance, which acts in the direction opposite the motion of the object and exerts more force on objects with more surface area. It is possible to derive Kepler\u2019s third law from Newton\u2019s law of universal gravitation. Applying Newton\u2019s second law of motion to 240 Chapter 7 \u2022 Newton's Law of Gravitation angular motion gives an expression for centripetal force, which can be equated to the expression for force in the universal gravitation equation. This expression can be manipulated to produce the equation for Kepler\u2019s third law. We saw earlier that the expression r3/T2is a constant for satellites orbiting the same massive object. The derivation of Kepler\u2019s third law from Newton\u2019s law of universal gravitation and Newton\u2019s second law of motion yields that constant: where Mis the mass of the central body about which the satellites orbit (for example, the sun in our solar system). The usefulness of this equation will be seen later. The universal gravitational constant Gis determined experimentally. This definition was first done accurately in 1798 by English scientist Henry Cavendish (1731\u20131810), more than 100 years after Newton published his universal law of gravitation. The measurement of Gis very basic and important because it determines the strength of one of the four forces in nature. Cavendish\u2019s experiment was very difficult because he measured the tiny gravitational attraction between two ordinary-sized masses (tens of kilograms at most) by using an apparatus like that in Figure 7.9. Remarkably, his value for Gdiffers by less than 1% from the modern value. Figure 7.9 Cavendish used an apparatus like this to measure the gravitational attraction between two suspended spheres (m) and two spheres on a stand (M) by observing the amount of torsion (twisting) created in the fiber. The distance between the masses can be varied to check the dependence of the force on distance. Modern experiments of this type continue to explore gravity. Einstein\u2019s Theory of General Relativity Einstein\u2019s theory of general relativity explained some interesting properties of gravity not covered by Newton\u2019s theory. Einstein based his theory on the postulate that acceleration and gravity have the same effect and cannot be distinguished from each other. He concluded that light must fall in both a gravitational field and in an accelerating reference frame. Figure 7.10 shows this effect (greatly exaggerated) in an accelerating elevator. In Figure 7.10(a), the elevator accelerates upward in zero gravity. In Figure 7.10(b), the room is not accelerating but is subject to gravity. The effect on light is the same: it \u201cfalls\u201d downward in both situations. The person in the elevator cannot tell whether the elevator is accelerating in zero gravity or is stationary and subject to gravity. Thus, gravity affects the path of light, even though we think of gravity as acting between masses, while photons are massless. Access for free at openstax.org. 7.2 \u2022 Newton's Law of", " Universal Gravitation and Einstein's Theory of General Relativity 241 Figure 7.10 (a) A beam of light emerges from a flashlight in an upward-accelerating elevator. Since the elevator moves up during the time the light takes to reach the wall, the beam strikes lower than it would if the elevator were not accelerated. (b) Gravity must have the same effect on light, since it is not possible to tell whether the elevator is accelerating upward or is stationary and acted upon by gravity. Einstein\u2019s theory of general relativity got its first verification in 1919 when starlight passing near the sun was observed during a solar eclipse. (See Figure 7.11.) During an eclipse, the sky is darkened and we can briefly see stars. Those on a line of sight nearest the sun should have a shift in their apparent positions. Not only was this shift observed, but it agreed with Einstein\u2019s predictions well within experimental uncertainties. This discovery created a scientific and public sensation. Einstein was now a folk hero as well as a very great scientist. The bending of light by matter is equivalent to a bending of space itself, with light following the curve. This is another radical change in our concept of space and time. It is also another connection that any particle with mass orenergy (e.g., massless photons) is affected by gravity. Figure 7.11 This schematic shows how light passing near a massive body like the sun is curved toward it. The light that reaches the Earth then seems to be coming from different locations than the known positions of the originating stars. Not only was this effect observed, but the amount of bending was precisely what Einstein predicted in his general theory of relativity. To summarize the two views of gravity, Newton envisioned gravity as a tug of war along the line connecting any two objects in the universe. In contrast, Einstein envisioned gravity as a bending of space-time by mass. 242 Chapter 7 \u2022 Newton's Law of Gravitation BOUNDLESS PHYSICS NASA gravity probe B NASA\u2019s Gravity Probe B (GP-B) mission has confirmed two key predictions derived from Albert Einstein\u2019s general theory of relativity. The probe, shown in Figure 7.12 was launched in 2004. It carried four ultra-precise gyroscopes designed to measure two effects hypothesized by Einstein\u2019s theory: \u2022 The geodetic effect, which is the warping of space and time by the gravitational field of a massive body (in this case, Earth) \u2022 The frame-dragging effect, which is the amount by which a spinning object pulls space and time with it as it rotates Figure 7.12 Artist concept of Gravity Probe B spacecraft in orbit around the Earth. (credit: NASA/MSFC) Both effects were measured with unprecedented precision. This was done by pointing the gyroscopes at a single star while orbiting Earth in a polar orbit. As predicted by relativity theory, the gyroscopes experienced very small, but measureable, changes in the direction of their spin caused by the pull of Earth\u2019s gravity. The principle investigator suggested imagining Earth spinning in honey. As Earth rotates it drags space and time with it as it would a surrounding sea of honey. GRASP CHECK According to the general theory of relativity, a gravitational field bends light. What does this have to do with time and space? a. Gravity has no effect on the space-time continuum, and gravity only affects the motion of light. b. The space-time continuum is distorted by gravity, and gravity has no effect on the motion of light. c. Gravity has no effect on either the space-time continuum or on the motion of light. d. The space-time continuum is distorted by gravity, and gravity affects the motion of light. Calculations Based on Newton\u2019s Law of Universal Gravitation TIPS FOR SUCCESS When performing calculations using the equations in this chapter, use units of kilograms for mass, meters for distances, newtons for force, and seconds for time. The mass of an object is constant, but its weight varies with the strength of the gravitational field. This means the value of g varies from place to place in the universe. The relationship between force, mass, and acceleration from the second law of motion can be written in terms of g. In this case, the force is the weight of the object, which is caused by the gravitational attraction of the planet or moon on which the object is located. We can use this expression to compare weights of an object on different moons and planets. Access for free at openstax.org. 7.2 \u2022 Newton's Law of Universal Gravitation and Einstein's Theory of General Relativity 243 WATCH PHYSICS Mass and Weight Clarification This video shows the mathematical basis of the relationship between mass and weight. The distinction between mass and weight are clearly explained. The mathematical relationship between mass and weight are shown mathematically in terms of the equation for Newton\u2019s law of universal gravitation and in terms of his second law of motion", ". Click to view content (https://www.khanacademy.org/embed_video?v=IuBoeDihLUc) GRASP CHECK Would you have the same mass on the moon as you do on Earth? Would you have the same weight? a. You would weigh more on the moon than on Earth because gravity on the moon is stronger than gravity on Earth. b. You would weigh less on the moon than on Earth because gravity on the moon is weaker than gravity on Earth. c. You would weigh less on the moon than on Earth because gravity on the moon is stronger than gravity on Earth. d. You would weigh more on the moon than on Earth because gravity on the moon is weaker than gravity on Earth. Two equations involving the gravitational constant, G, are often useful. The first is Newton\u2019s equation,. Several of the values in this equation are either constants or easily obtainable. F is often the weight of an object on the surface of a large object with mass M, which is usually known. The mass of the smaller object, m, is often known, and Gis a universal constant with the same value anywhere in the universe. This equation can be used to solve problems involving an object on or orbiting Earth or other massive celestial object. Sometimes it is helpful to equate the right-hand side of the equation to mg and cancel the mon both sides. The equation is also useful for problems involving objects in orbit. Note that there is no need to know the mass of the object. Often, we know the radius ror the period Tand want to find the other. If these are both known, we can use the equation to calculate the mass of a planet or star. WATCH PHYSICS Mass and Weight Clarification This video demonstrates calculations involving Newton\u2019s universal law of gravitation. Click to view content (https://www.khanacademy.org/embed_video?v=391txUI76gM) GRASP CHECK and. are both the acceleration due to gravity Identify the constants a. b. c. d. and is acceleration due to gravity on Earth and is the gravitational constant and and are both the universal gravitational constant. is the universal gravitational constant. is the acceleration due to gravity on Earth. WORKED EXAMPLE Change in g The value of g on the planet Mars is 3.71 m/s2. If you have a mass of 60.0 kg on Earth, what would be your mass on Mars? What would be your weight on Mars? Strategy Weight equals acceleration due to gravity times mass: on Mars gMand weight on Mars WM.. An object\u2019s mass is constant. Call acceleration due to gravity 244 Chapter 7 \u2022 Newton's Law of Gravitation Solution Mass on Mars would be the same, 60 kg. Discussion The value of g on any planet depends on the mass of the planet and the distance from its center. If the material below the surface varies from point to point, the value of g will also vary slightly. 7.4 WORKED EXAMPLE Earth\u2019s g at the Moon Find the acceleration due to Earth\u2019s gravity at the distance of the moon. Express the force of gravity in terms of g. Combine with the equation for universal gravitation. Solution Cancel mand substitute. 7.5 7.6 7.7 Discussion The value of g for the moon is 1.62 m/s2. Comparing this value to the answer, we see that Earth\u2019s gravitational influence on an object on the moon\u2019s surface would be insignificant. 7.8 Practice Problems 6. What is the mass of a person who weighs? a. b. c. d. 7. Calculate Earth\u2019s mass given that the acceleration due to gravity at the North Pole is and the radius of the Earth is from pole to center. a. b. c. d. Check Your Understanding 8. Some of Newton\u2019s predecessors and contemporaries also studied gravity and proposed theories. What important advance did Newton make in the study of gravity that the other scientists had failed to do? a. He gave an exact mathematical form for the theory. Access for free at openstax.org. 7.2 \u2022 Newton's Law of Universal Gravitation and Einstein's Theory of General Relativity 245 b. He added a correction term to a previously existing formula. c. Newton found the value of the universal gravitational constant. d. Newton showed that gravitational force is always attractive. 9. State the law of universal gravitation in words only. a. Gravitational force between two objects is directly proportional to the sum of the squares of their masses and inversely proportional to the square of the distance between them. b. Gravitational force between two objects is directly proportional to the product of their masses and inversely proportional to the square of the distance between them. c. Gravitational force between two objects is directly proportional to the sum of the squares of their masses and", " inversely proportional to the distance between them. d. Gravitational force between two objects is directly proportional to the product of their masses and inversely proportional to the distance between them. 10. Newton\u2019s law of universal gravitation explains the paths of what? a. A charged particle b. A ball rolling on a plane surface c. A planet moving around the sun d. A stone tied to a string and whirled at constant speed in a horizontal circle 246 Chapter 7 \u2022 Key Terms KEY TERMS aphelion closest distance between a planet and the sun (called apoapsis for other celestial bodies) Copernican model the model of the solar system where the sun is at the center of the solar system and all the planets orbit around it; this is also called the heliocentric model eccentricity a measure of the separation of the foci of an ellipse Johannes Kepler that describe the properties of all orbiting satellites Newton\u2019s universal law of gravitation states that gravitational force between two objects is directly proportional to the product of their masses and inversely proportional to the square of the distance between them. perihelion farthest distance between a planet and the sun Einstein\u2019s theory of general relativity the theory that (called periapsis for other celestial bodies) gravitational force results from the bending of spacetime by an object\u2019s mass gravitational constant the proportionality constant in Newton\u2019s law of universal gravitation Kepler\u2019s laws of planetary motion three laws derived by SECTION SUMMARY 7.1 Kepler's Laws of Planetary Motion \u2022 All satellites follow elliptical orbits. \u2022 The line from the satellite to the parent body sweeps out equal areas in equal time. \u2022 The radius cubed divided by the period squared is a Ptolemaic model the model of the solar system where Earth is at the center of the solar system and the sun and all the planets orbit around it; this is also called the geocentric model 7.2 Newton's Law of Universal Gravitation and Einstein's Theory of General Relativity \u2022 Newton\u2019s law of universal gravitation provides a mathematical basis for gravitational force and Kepler\u2019s laws of planetary motion. constant for all satellites orbiting the same parent body. \u2022 Einstein\u2019s theory of general relativity shows that KEY EQUATIONS 7.1 Kepler's Laws of Planetary Motion Kepler\u2019s third law eccentricity area of an ellipse semi-major axis of an ellipse semi-minor axis of an ellipse gravitational fields change the path of light and warp space and time. \u2022 An object\u2019s mass is constant, but its weight changes when acceleration due to gravity, g, changes. 7.2 Newton's Law of Universal Gravitation and Einstein's Theory of General Relativity Newton\u2019s second law of motion Newton\u2019s universal law of gravitation acceleration due to gravity constant for satellites orbiting the same massive object CHAPTER REVIEW Concept Items 7.1 Kepler's Laws of Planetary Motion is different from other ellipses. a. The foci of a circle are at the same point and are located at the center of the circle. 1. A circle is a special case of an ellipse. Explain how a circle b. The foci of a circle are at the same point and are Access for free at openstax.org. located at the circumference of the circle. c. The foci of a circle are at the same point and are located outside of the circle. d. The foci of a circle are at the same point and are located anywhere on the diameter, except on its midpoint. 2. Comets have very elongated elliptical orbits with the sun at one focus. Using Kepler's Law, explain why a comet travels much faster near the sun than it does at the other end of the orbit. a. Because the satellite sweeps out equal areas in equal times b. Because the satellite sweeps out unequal areas in Chapter 7 \u2022 Chapter Review 247 illustration of this is any description of the feeling of constant velocity in a situation where no outside frame of reference is considered. c. Gravity and acceleration have the same effect and cannot be distinguished from each other. An acceptable illustration of this is any description of the feeling of acceleration in a situation where no outside frame of reference is considered. d. Gravity and acceleration have different effects and can be distinguished from each other. An acceptable illustration of this is any description of the feeling of acceleration in a situation where no outside frame of reference is considered. equal times 6. Titan, with a radius of, is the largest c. Because the satellite is at the other focus of the moon of the planet Saturn. If the mass of Titan is ellipse d. Because the square of the period of the satellite is proportional to the cube of its average distance from the sun 3. True or False\u2014A planet-satellite system must be isolated from other massive objects to follow Kepler\u2019s laws of planetary motion. a. True b. False 4. Explain why", " the string, pins, and pencil method works for drawing an ellipse. a. The string, pins, and pencil method works because the length of the two sides of the triangle remains constant as you are drawing the ellipse. b. The string, pins, and pencil method works because the area of the triangle remains constant as you are drawing the ellipse. c. The string, pins, and pencil method works because the perimeter of the triangle remains constant as you are drawing the ellipse. d. The string, pins, and pencil method works because the volume of the triangle remains constant as you are drawing the ellipse. 7.2 Newton's Law of Universal Gravitation and Einstein's Theory of General Relativity 5. Describe the postulate on which Einstein based the theory of general relativity and describe an everyday experience that illustrates this postulate. a. Gravity and velocity have the same effect and cannot be distinguished from each other. An acceptable illustration of this is any description of the feeling of constant velocity in a situation where no outside frame of reference is considered. b. Gravity and velocity have different effects and can be distinguished from each other. An acceptable, what is the acceleration due to gravity on the surface of this moon? a. b. c. d. 7. Saturn\u2019s moon Titan has an orbital period of 15.9 days. If Saturn has a mass of 5.68\u00d71023 kg, what is the average distance from Titan to the center of Saturn? 1.22\u00d7106 m a. b. 4.26\u00d7107 m 5.25\u00d7104 km c. d. 4.26\u00d71010 km 8. Explain why doubling the mass of an object doubles its weight, but doubling its distance from the center of Earth reduces its weight fourfold. a. The weight is two times the gravitational force between the object and Earth. b. The weight is half the gravitational force between the object and Earth. c. The weight is equal to the gravitational force between the object and Earth, and the gravitational force is inversely proportional to the distance squared between the object and Earth. d. The weight is directly proportional to the square of the gravitational force between the object and Earth. 9. Explain why a star on the other side of the Sun might appear to be in a location that is not its true location. It can be explained by using the concept of a. atmospheric refraction. It can be explained by using the concept of the special theory of relativity. It can be explained by using the concept of the general theory of relativity. It can be explained by using the concept of light d. b. c. 248 Chapter 7 \u2022 Chapter Review scattering in the atmosphere. 10. The Cavendish experiment marked a milestone in the study of gravity. Part A. What important value did the experiment determine? Part B. Why was this so difficult in terms of the masses used in the apparatus and the strength of the gravitational force? a. Part A. The experiment measured the acceleration due to gravity, g. Part B. Gravity is a very weak force but despite this limitation, Cavendish was able to measure the attraction between very massive objects. b. Part A. The experiment measured the gravitational Critical Thinking Items 7.1 Kepler's Laws of Planetary Motion 11. In the figure, the time it takes for the planet to go from A to B, C to D, and E to F is the same. constant, G. Part B. Gravity is a very weak force but, despite this limitation, Cavendish was able to measure the attraction between very massive objects. c. Part A. The experiment measured the acceleration due to gravity, g. Part B. Gravity is a very weak force but despite this limitation, Cavendish was able to measure the attraction between less massive objects. d. Part A. The experiment measured the gravitational constant, G. Part B. Gravity is a very weak force but despite this limitation, Cavendish was able to measure the attraction between less massive objects. a. Area X < Area Y; the speed is greater for area X. b. Area X > Area Y; the speed is greater for area Y. c. Area X = Area Y; the speed is greater for area X. d. Area X = Area Y; the speed is greater for area Y. 7.2 Newton's Law of Universal Gravitation and Einstein's Theory of General Relativity 14. Rhea, with a radius of 7.63\u00d7105 m, is the second-largest moon of the planet Saturn. If the mass of Rhea is 2.31\u00d71021 kg, what is the acceleration due to gravity on the surface of this moon? a. 2.65\u00d710\u22121 m/s b. 2.02\u00d7105 m/s c. 2.65\u00d710\u22121 m/s2 d. 2.02\u00d7105 m/s2 15. Earth has a mass of 5.971\u00d71024 kg and a radius of 6.371\u00d7106", " m. Use the data to check the value of the gravitational constant. a. it matches the value of the gravitational constant G. it matches the value of the gravitational constant G. it matches the value of the b. c. gravitational constant G. Compare the areas A1, A2, and A3 in terms of size. a. A1 \u2260 A2 \u2260 A3 b. A1 = A2 = A3 c. A1 = A2 > A3 d. A1 > A2 = A3 12. A moon orbits a planet in an elliptical orbit. The foci of the ellipse are 50, 000 km apart. The closest approach of the moon to the planet is 400, 000 km. What is the length of the major axis of the orbit? a. 400, 000 km b. 450, 000, km c. 800, 000 km d. 850, 000 km 13. In this figure, if f1 represents the parent body, which set of statements holds true? Access for free at openstax.org. d. it matches the value of the gravitational constant G. 16. The orbit of the planet Mercury has a period of 88.0 days and an average radius of 5.791\u00d71010 m. What is the mass of the sun? Problems 7.1 Kepler's Laws of Planetary Motion 17. The closest Earth comes to the sun is 1.47\u00d7108 km, and Earth\u2019s farthest distance from the sun is 1.52\u00d7108 km. What is the area inside Earth\u2019s orbit? a. 2.23\u00d71016 km2 b. 6.79\u00d71016 km2 7.02\u00d71016 km2 c. 7.26\u00d71016 km2 d. Performance Task 7.2 Newton's Law of Universal Gravitation and Einstein's Theory of General Relativity 19. Design an experiment to test whether magnetic force is inversely proportional to the square of distance. Gravitational, magnetic, and electrical fields all act at a distance, but do they all follow the inverse square law? One difference in the forces related to these fields is that gravity is only attractive, but the other two can repel as well. In general, the inverse square law says that force F equals a constant Cdivided by the distance between objects, d, squared: Incorporate these materials into your design:. TEST PREP Multiple Choice 7.1 Kepler's Laws of Planetary Motion 20. A planet of mass m circles a sun of mass M. Which distance changes throughout the planet\u2019s orbit? a. b. c. d. 21. The focal point of the elliptical orbit of a moon is from the center of the orbit. If the, what is the length of the eccentricity of the orbit is semi-major axis? a. b. c. d. Chapter 7 \u2022 Test Prep 249 3.43\u00d71019 kg a. 1.99\u00d71030 kg b. c. 2.56\u00d71029 kg 1.48\u00d71040 kg d. 18. Earth is 1.496\u00d7108 km from the sun, and Neptune is 4.490\u00d7109 km from the sun. What best represents the number of Earth years it takes for Neptune to complete one orbit around the sun? 10 years a. 30 years b. 160 years c. d. 900 years \u2022 Two strong, permanent bar magnets \u2022 A spring scale that can measure small forces \u2022 A short ruler calibrated in millimeters Use the magnets to study the relationship between attractive force and distance. a. What will be the independent variable? b. What will be the dependent variable? c. How will you measure each of these variables? If you plot the independent variable versus the d. dependent variable and the inverse square law is upheld, will the plot be a straight line? Explain. e. Which plot would be a straight line if the inverse square law were upheld? 22. An artificial satellite orbits the Earth at a distance of 1.45\u00d7104 km from Earth\u2019s center. The moon orbits the Earth at a distance of 3.84\u00d7105 km once every 27.3 days. How long does it take the satellite to orbit the Earth? a. 0.200 days b. 3.07 days c. 243 days 3721 days d. 23. Earth is 1.496\u00d7108 km from the sun, and Venus is 1.08\u00d7108 km from the sun. One day on Venus is 243 Earth days long. What best represents the number of Venusian days in a Venusian year? a. 0.78 days b. 0.92 days 1.08 days c. 1.21 days d. 250 Chapter 7 \u2022 Test Prep 7.2 Newton's Law of Universal Gravitation and Einstein's Theory of General Relativity 24. What did the Cavendish experiment measure? a. The mass of Earth b. The gravitational constant c. Acceleration due to gravity d. The eccentricity of Earth\u2019s", " orbit 25. You have a mass of and you have just landed on one of the moons of Jupiter where you have a weight of, on. What is the acceleration due to gravity, the moon you are visiting? a. b. c. d. 26. A person is in an elevator that suddenly begins to descend. The person knows, intuitively, that the feeling of suddenly becoming lighter is because the elevator is accelerating downward. What other change would Short Answer 7.1 Kepler's Laws of Planetary Motion 27. Explain how the masses of a satellite and its parent body must compare in order to apply Kepler\u2019s laws of planetary motion. a. The mass of the parent body must be much less than that of the satellite. b. The mass of the parent body must be much greater than that of the satellite. c. The mass of the parent body must be equal to the mass of the satellite. d. There is no specific relationship between the masses for applying Kepler\u2019s laws of planetary motion. 28. Hyperion is a moon of the planet Saturn. Its orbit has an eccentricity of and a semi-major axis of. How far is the center of the orbit from the center of Saturn? a. b. c. d. 29. The orbits of satellites are elliptical. Define an ellipse. a. An ellipse is an open curve wherein the sum of the distance from the foci to any point on the curve is constant. b. An ellipse is a closed curve wherein the sum of the distance from the foci to any point on the curve is constant. Access for free at openstax.org. produce the same feeling? How does this demonstrate Einstein\u2019s postulate on which he based the theory of general relativity? a. It would feel the same if the force of gravity suddenly became weaker. This illustrates Einstein\u2019s postulates that gravity and acceleration are indistinguishable. It would feel the same if the force of gravity suddenly became stronger. This illustrates Einstein\u2019s postulates that gravity and acceleration are indistinguishable. It would feel the same if the force of gravity suddenly became weaker. This illustrates Einstein\u2019s postulates that gravity and acceleration are distinguishable. It would feel the same if the force of gravity suddenly became stronger. This illustrates Einstein\u2019s postulates that gravity and acceleration are distinguishable. b. c. d. c. An ellipse is an open curve wherein the distances from the two foci to any point on the curve are equal. d. An ellipse is a closed curve wherein the distances from the two foci to any point on the curve are equal. 30. Mars has two moons, Deimos and Phobos. The orbit of. The average radius of the and an average Deimos has a period of radius of. According to orbit of Phobos is Kepler\u2019s third law of planetary motion, what is the period of Phobos? a. b. c. d. 7.2 Newton's Law of Universal Gravitation and Einstein's Theory of General Relativity 31. Newton\u2019s third law of motion says that, for every action force, there is a reaction force equal in magnitude but that acts in the opposite direction. Apply this law to gravitational forces acting between the Washington Monument and Earth. a. The monument is attracted to Earth with a force equal to its weight, and Earth is attracted to the monument with a force equal to Earth\u2019s weight. The situation can be represented with two force vectors of unequal magnitude and pointing in the same direction. b. The monument is attracted to Earth with a force equal to its weight, and Earth is attracted to the monument with a force equal to Earth\u2019s weight. The situation can be represented with two force vectors of unequal magnitude but pointing in opposite directions. c. The monument is attracted to Earth with a force equal to its weight, and Earth is attracted to the monument with an equal force. The situation can be represented with two force vectors of equal magnitude and pointing in the same direction. d. The monument is attracted to Earth with a force equal to its weight, and Earth is attracted to the monument with an equal force. The situation can be represented with two force vectors of equal magnitude but pointing in opposite directions. 32. True or false\u2014Gravitational force is the attraction of the mass of one object to the mass of another. Light, either Extended Response 7.1 Kepler's Laws of Planetary Motion 35. The orbit of Halley\u2019's Comet has an eccentricity of 0.967 and stretches to the edge of the solar system. Part A. Describe the shape of the comet\u2019s orbit. Part B. Compare the distance traveled per day when it is near the sun to the distance traveled per day when it is at the edge of the solar system. Part C. Describe variations in the comet's speed as it completes an orbit. Explain the variations in terms of Kepler's second law of planetary", " motion. a. Part A. The orbit is circular, with the sun at the center. Part B. The comet travels much farther when it is near the sun than when it is at the edge of the solar system. Part C. The comet decelerates as it approaches the sun and accelerates as it leaves the sun. b. Part A. The orbit is circular, with the sun at the center. Part B. The comet travels much farther when it is near the sun than when it is at the edge of the solar system. Part C. The comet accelerates as it approaches the sun and decelerates as it leaves the sun. c. Part A. The orbit is very elongated, with the sun near one end. Part B. The comet travels much farther when it is near the sun than when it is at the edge of the solar system. Part C. The comet decelerates as it approaches the sun and accelerates as it moves away from the sun. 36. For convenience, astronomers often use astronomical units (AU) to measure distances within the solar system. One AU equals the average distance from Earth to the Chapter 7 \u2022 Test Prep 251 as a particle or a wave, has no rest mass. Despite this fact gravity bends a beam of light. a. True b. False 33. The average radius of Earth is. What is Earth\u2019s mass? a. b. c. d. 34. What is the gravitational force between two apart? people sitting a. b. c. d. sun. Halley\u2019s Comet returns once every 75.3 years. What is the average radius of the orbit of Halley\u2019s Comet in AU? a. 0.002 AU b. 0.056 AU c. 17.8 AU d. 653 AU 7.2 Newton's Law of Universal Gravitation and Einstein's Theory of General Relativity 37. It took scientists a long time to arrive at the understanding of gravity as explained by Galileo and Newton. They were hindered by two ideas that seemed like common sense but were serious misconceptions. First was the fact that heavier things fall faster than light things. Second, it was believed impossible that forces could act at a distance. Explain why these ideas persisted and why they prevented advances. a. Heavier things fall faster than light things if they have less surface area and greater mass density. In the Renaissance and before, forces that acted at a distance were considered impossible, so people were skeptical about scientific theories that invoked such forces. b. Heavier things fall faster than light things because they have greater surface area and less mass density. In the Renaissance and before, forces that act at a distance were considered impossible, so people were skeptical about scientific theories that invoked such forces. c. Heavier things fall faster than light things because they have less surface area and greater mass density. In the Renaissance and before, forces that 252 Chapter 7 \u2022 Test Prep act at a distance were considered impossible, so people were quick to accept scientific theories that invoked such forces. d. Heavier things fall faster than light things because they have larger surface area and less mass density. In the Renaissance and before, forces that act at a distance were considered impossible because of people\u2019s faith in scientific theories. 38. The masses of Earth and the moon are 5.97\u00d71024 kg and 7.35\u00d71022 kg, respectively. The distance from Earth to the moon is 3.80\u00d7105 km. At what point between the Earth and the moon are the opposing gravitational forces equal? (Use subscripts e and m to represent Earth and moon.) a. b. c. d. 3.42\u00d7105 km from the center of Earth 3.80\u00d7105 km from the center of Earth 3.42\u00d7106 km from the center of Earth 3.10\u00d7107 km from the center of Earth Access for free at openstax.org. CHAPTER 8 Momentum Figure 8.1 NFC defensive backs Ronde Barber and Roy Williams along with linebacker Jeremiah Trotter gang tackle AFC running back LaDainian Tomlinson during the 2006 Pro Bowl in Hawaii. (United States Marine Corps) Chapter Outline 8.1 Linear Momentum, Force, and Impulse 8.2 Conservation of Momentum 8.3 Elastic and Inelastic Collisions We know from everyday use of the word momentumthat it is a tendency to continue on course in the same INTRODUCTION direction. Newscasters speak of sports teams or politicians gaining, losing, or maintaining the momentum to win. As we learned when studying about inertia, which is Newton's first law of motion, every object or system has inertia\u2014that is, a tendency for an object in motion to remain in motion or an object at rest to remain at rest. Mass is a useful variable that lets us quantify inertia. Momentum is mass in motion. Momentum is important because it is conserved in isolated systems; this fact is convenient for solving problems where objects collide. The magnitude of", " momentum grows with greater mass and/or speed. For example, look at the football players in the photograph (Figure 8.1). They collide and fall to the ground. During their collisions, momentum will play a large part. In this chapter, we will learn about momentum, the different types of collisions, and how to use momentum equations to solve collision problems. 254 Chapter 8 \u2022 Momentum 8.1 Linear Momentum, Force, and Impulse Section Learning Objectives By the end of this section, you will be able to do the following: \u2022 Describe momentum, what can change momentum, impulse, and the impulse-momentum theorem \u2022 Describe Newton\u2019s second law in terms of momentum \u2022 Solve problems using the impulse-momentum theorem Section Key Terms change in momentum impulse impulse\u2013momentum theorem linear momentum Momentum, Impulse, and the Impulse-Momentum Theorem Linear momentum is the product of a system\u2019s mass and its velocity. In equation form, linear momentum p is You can see from the equation that momentum is directly proportional to the object\u2019s mass (m) and velocity (v). Therefore, the greater an object\u2019s mass or the greater its velocity, the greater its momentum. A large, fast-moving object has greater momentum than a smaller, slower object. Momentum is a vector and has the same direction as velocity v. Since mass is a scalar, when velocity is in a negative direction (i.e., opposite the direction of motion), the momentum will also be in a negative direction; and when velocity is in a positive direction, momentum will likewise be in a positive direction. The SI unit for momentum is kg m/s. Momentum is so important for understanding motion that it was called the quantity of motionby physicists such as Newton. Force influences momentum, and we can rearrange Newton\u2019s second law of motion to show the relationship between force and momentum. Recall our study of Newton\u2019s second law of motion (Fnet = ma). Newton actually stated his second law of motion in terms of momentum: The net external force equals the change in momentum of a system divided by the time over which it changes. The change in momentum is the difference between the final and initial values of momentum. In equation form, this law is where Fnet is the net external force, is the change in momentum, and is the change in time. We can solve for by rearranging the equation to be is known as impulse and this equation is known as the impulse-momentum theorem. From the equation, we see that the impulse equals the average net external force multiplied by the time this force acts. It is equal to the change in momentum. The effect of a force on an object depends on how long it acts, as well as the strength of the force.Impulse is a useful concept because it quantifies the effect of a force. A very large force acting for a short time can have a great effect on the momentum of an object, such as the force of a racket hitting a tennis ball. A small force could cause the same change in momentum, but it would have to act for a much longer time. Newton\u2019s Second Law in Terms of Momentum When Newton\u2019s second law is expressed in terms of momentum, it can be used for solving problems where mass varies, since. In the more traditional form of the law that you are used to working with, mass is assumed to be constant. In fact, this traditional form is a special case of the law, where mass is constant. is actually derived from the equation: Access for free at openstax.org. 8.1 \u2022 Linear Momentum, Force, and Impulse 255 For the sake of understanding the relationship between Newton\u2019s second law in its two forms, let\u2019s recreate the derivation of from by substituting the definitions of acceleration and momentum. The change in momentum is given by If the mass of the system is constant, then By substituting for, Newton\u2019s second law of motion becomes for a constant mass. Because we can substitute to get the familiar equation when the mass of the system is constant. TIPS FOR SUCCESS We just showed how would not apply would be a moving rocket that burns enough fuel to significantly change the mass of the rocket. In this case, you would need to use Newton\u2019s second law expressed in terms of momentum to account for the changing mass. applies only when the mass of the system is constant. An example of when this formula Snap Lab Hand Movement and Impulse In this activity you will experiment with different types of hand motions to gain an intuitive understanding of the relationship between force, time, and impulse. \u2022 one ball \u2022 one tub filled with water Procedure: 1. Try catching a ball while givingwith the ball, pulling your hands toward your body. 2. Next, try catching a ball while keeping your hands still. 3. Hit water in a tub with your full", " palm. Your full palm represents a swimmer doing a belly flop. 4. After the water has settled, hit the water again by diving your hand with your fingers first into the water. Your diving hand represents a swimmer doing a dive. 5. Explain what happens in each case and why. GRASP CHECK What are some other examples of motions that impulse affects? 256 Chapter 8 \u2022 Momentum a. a football player colliding with another, or a car moving at a constant velocity b. a car moving at a constant velocity, or an object moving in the projectile motion c. a car moving at a constant velocity, or a racket hitting a ball d. a football player colliding with another, or a racket hitting a ball LINKS TO PHYSICS Engineering: Saving Lives Using the Concept of Impulse Cars during the past several decades have gotten much safer. Seat belts play a major role in automobile safety by preventing people from flying into the windshield in the event of a crash. Other safety features, such as airbags, are less visible or obvious, but are also effective at making auto crashes less deadly (see Figure 8.2). Many of these safety features make use of the concept of impulse from physics. Recall that impulse is the net force multiplied by the duration of time of the impact. This was expressed mathematically as. Figure 8.2 Vehicles have safety features like airbags and seat belts installed. Airbags allow the net force on the occupants in the car to act over a much longer time when there is a sudden stop. The momentum change is the same for an occupant whether an airbag is deployed or not. But the force that brings the occupant to a stop will be much less if it acts over a larger time. By rearranging the equation for impulse to solve for force see how increasing padded dashboard increases the time over which the force of impact acts, thereby reducing the force of impact. you can stays the same will decrease Fnet. This is another example of an inverse relationship. Similarly, a while Cars today have many plastic components. One advantage of plastics is their lighter weight, which results in better gas mileage. Another advantage is that a car will crumple in a collision, especially in the event of a head-on collision. A longer collision time means the force on the occupants of the car will be less. Deaths during car races decreased dramatically when the rigid frames of racing cars were replaced with parts that could crumple or collapse in the event of an accident. GRASP CHECK You may have heard the advice to bend your knees when jumping. In this example, a friend dares you to jump off of a park bench onto the ground without bending your knees. You, of course, refuse. Explain to your friend why this would be a foolish thing. Show it using the impulse-momentum theorem. a. Bending your knees increases the time of the impact, thus decreasing the force. b. Bending your knees decreases the time of the impact, thus decreasing the force. c. Bending your knees increases the time of the impact, thus increasing the force. d. Bending your knees decreases the time of the impact, thus increasing the force. Access for free at openstax.org. 8.1 \u2022 Linear Momentum, Force, and Impulse 257 Solving Problems Using the Impulse-Momentum Theorem WORKED EXAMPLE Calculating Momentum: A Football Player and a Football (a) Calculate the momentum of a 110 kg football player running at 8 m/s. (b) Compare the player\u2019s momentum with the momentum of a 0.410 kg football thrown hard at a speed of 25 m/s. Strategy No information is given about the direction of the football player or the football, so we can calculate only the magnitude of the momentum, p. (A symbol in italics represents magnitude.) In both parts of this example, the magnitude of momentum can be calculated directly from the definition of momentum: Solution for (a) To find the player\u2019s momentum, substitute the known values for the player\u2019s mass and speed into the equation. Solution for (b) To find the ball\u2019s momentum, substitute the known values for the ball\u2019s mass and speed into the equation. The ratio of the player\u2019s momentum to the ball\u2019s momentum is Discussion Although the ball has greater velocity, the player has a much greater mass. Therefore, the momentum of the player is about 86 times greater than the momentum of the football. WORKED EXAMPLE Calculating Force: Venus Williams\u2019 Racquet During the 2007 French Open, Venus Williams (Figure 8.3) hit the fastest recorded serve in a premier women\u2019s match, reaching a speed of 58 m/s (209 km/h). What was the average force exerted on the 0.057 kg tennis ball by Williams\u2019 racquet? Assume that the ball\u2019s speed just after impact was 58 m/", "s, the horizontal velocity before impact is negligible, and that the ball remained in contact with the racquet for 5 ms (milliseconds). Figure 8.3 Venus Williams playing in the 2013 US Open (Edwin Martinez, Flickr) Strategy Recall that Newton\u2019s second law stated in terms of momentum is 258 Chapter 8 \u2022 Momentum As noted above, when mass is constant, the change in momentum is given by where vf is the final velocity and vi is the initial velocity. In this example, the velocity just after impact and the change in time are given, so after we solve for to find the force., we can use Solution To determine the change in momentum, substitute the values for mass and the initial and final velocities into the equation above. Now we can find the magnitude of the net external force using 8.1 8.2 Discussion This quantity was the average force exerted by Venus Williams\u2019 racquet on the tennis ball during its brief impact. This problem could also be solved by first finding the acceleration and then using Fnet = ma, but we would have had to do one more step. In this case, using momentum was a shortcut. Practice Problems 1. What is the momentum of a bowling ball with mass and velocity? a. b. c. d. 2. What will be the change in momentum caused by a net force of acting on an object for seconds? a. b. c. d. Check Your Understanding 3. What is linear momentum? a. b. c. d. the sum of a system\u2019s mass and its velocity the ratio of a system\u2019s mass to its velocity the product of a system\u2019s mass and its velocity the product of a system\u2019s moment of inertia and its velocity 4. If an object\u2019s mass is constant, what is its momentum proportional to? a. b. c. d. Its velocity Its weight Its displacement Its moment of inertia 5. What is the equation for Newton\u2019s second law of motion, in terms of mass, velocity, and time, when the mass of the system is Access for free at openstax.org. 8.2 \u2022 Conservation of Momentum 259 constant? a. b. c. d. 6. Give an example of a system whose mass is not constant. a. A spinning top b. A baseball flying through the air c. A rocket launched from Earth d. A block sliding on a frictionless inclined plane 8.2 Conservation of Momentum Section Learning Objectives By the end of this section, you will be able to do the following: \u2022 Describe the law of conservation of momentum verbally and mathematically Section Key Terms angular momentum isolated system law of conservation of momentum Conservation of Momentum It is important we realize that momentum is conserved during collisions, explosions, and other events involving objects in motion. To say that a quantity is conserved means that it is constant throughout the event. In the case of conservation of momentum, the total momentum in the system remains the same before and after the collision. You may have noticed that momentum was notconserved in some of the examples previously presented in this chapter. where forces acting on the objects produced large changes in momentum. Why is this? The systems of interest considered in those problems were not inclusive enough. If the systems were expanded to include more objects, then momentum would in fact be conserved in those sample problems. It is always possible to find a larger system where momentum is conserved, even though momentum changes for individual objects within the system. For example, if a football player runs into the goalpost in the end zone, a force will cause him to bounce backward. His momentum is obviously greatly changed, and considering only the football player, we would find that momentum is not conserved. However, the system can be expanded to contain the entire Earth. Surprisingly, Earth also recoils\u2014conserving momentum\u2014because of the force applied to it through the goalpost. The effect on Earth is not noticeable because it is so much more massive than the player, but the effect is real. Next, consider what happens if the masses of two colliding objects are more similar than the masses of a football player and Earth\u2014in the example shown in Figure 8.4 of one car bumping into another. Both cars are coasting in the same direction when the lead car, labeled m2, is bumped by the trailing car, labeled m1. The only unbalanced force on each car is the force of the collision, assuming that the effects due to friction are negligible. Car m1 slows down as a result of the collision, losing some momentum, while car m2 speeds up and gains some momentum. If we choose the system to include both cars and assume that friction is negligible, then the momentum of the two-car system should remain constant. Now we will prove that the total momentum of the two-car system does in fact remain constant, and is therefore conserved. 260 Chapter 8 \u2022 Momentum Figure 8.", "4 Car of mass m1 moving with a velocity of v1 bumps into another car of mass m2 and velocity v2. As a result, the first car slows down to a velocity of v\u20321 and the second speeds up to a velocity of v\u20322. The momentum of each car is changed, but the total momentum ptot of the two cars is the same before and after the collision if you assume friction is negligible. Using the impulse-momentum theorem, the change in momentum of car 1 is given by where F1 is the force on car 1 due to car 2, and is the time the force acts, or the duration of the collision. Similarly, the change in momentum of car 2 is duration of the collision is the same for both cars. We know from Newton\u2019s third law of motion that F2 = \u2013F1, and so. where F2 is the force on car 2 due to car 1, and we assume the Therefore, the changes in momentum are equal and opposite, and. Because the changes in momentum add to zero, the total momentum of the two-car system is constant. That is, where p\u20321 and p\u20322 are the momenta of cars 1 and 2 after the collision. This result that momentum is conserved is true not only for this example involving the two cars, but for any system where the net external force is zero, which is known as an isolated system. The law of conservation of momentum states that for an isolated system with any number of objects in it, the total momentum is conserved. In equation form, the law of conservation of momentum for an isolated system is written as or where ptot is the total momentum, or the sum of the momenta of the individual objects in the system at a given time, and p\u2032tot is the total momentum some time later. The conservation of momentum principle can be applied to systems as diverse as a comet striking the Earth or a gas containing huge numbers of atoms and molecules. Conservation of momentum appears to be violated only when the net external force is not zero. But another larger system can always be considered in which momentum is conserved by simply including the source of the external force. For example, in the collision of two cars considered above, the two-car system conserves momentum while each one-car system does not. Access for free at openstax.org. 8.2 \u2022 Conservation of Momentum 261 TIPS FOR SUCCESS Momenta is the plural form of the word momentum. One object is said to have momentum, but two or more objects are said to have momenta. FUN IN PHYSICS Angular Momentum in Figure Skating So far we have covered linear momentum, which describes the inertia of objects traveling in a straight line. But we know that many objects in nature have a curved or circular path. Just as linear motion has linear momentum to describe its tendency to move forward, circular motion has the equivalent angular momentum to describe how rotational motion is carried forward. This is similar to how torque is analogous to force, angular acceleration is analogous to translational acceleration, and mr2 is analogous to mass or inertia. You may recall learning that the quantity mr2 is called the rotational inertia or moment of inertia of a point mass mat a distance rfrom the center of rotation. We already know the equation for linear momentum, p = mv. Since angular momentum is analogous to linear momentum, the moment of inertia (I) is analogous to mass, and angular velocity is analogous to linear velocity, it makes sense that angular momentum (L) is defined as Angular momentum is conserved when the net external torque ( external force is zero. ) is zero, just as linear momentum is conserved when the net Figure skaters take advantage of the conservation of angular momentum, likely without even realizing it. In Figure 8.5, a figure skater is executing a spin. The net torque on her is very close to zero, because there is relatively little friction between her skates and the ice, and because the friction is exerted very close to the pivot point. Both F and rare small, and so is negligibly small. Figure 8.5 (a) An ice skater is spinning on the tip of her skate with her arms extended. In the next image, (b), her rate of spin increases greatly when she pulls in her arms. Consequently, she can spin for quite some time. She can do something else, too. She can increase her rate of spin by pulling her arms and legs in. Why does pulling her arms and legs in increase her rate of spin? The answer is that her angular momentum is constant, so that L = L\u2032. Expressing this equation in terms of the moment of inertia, where the primed quantities refer to conditions after she has pulled in her arms and reduced her moment of inertia. Because I\u2032 is smaller, the angular velocity must increase to keep the angular momentum constant. This allows her to spin much faster without exerting any extra torque. A", " video (http://openstax.org/l/28figureskater) is also available that shows a real figure skater executing a spin. It discusses the physics of spins in figure skating. 262 Chapter 8 \u2022 Momentum GRASP CHECK Based on the equation L = I\u03c9, how would you expect the moment of inertia of an object to affect angular momentum? How would angular velocity affect angular momentum? a. Large moment of inertia implies large angular momentum, and large angular velocity implies large angular momentum. b. Large moment of inertia implies small angular momentum, and large angular velocity implies small angular momentum. c. Large moment of inertia implies large angular momentum, and large angular velocity implies small angular momentum. d. Large moment of inertia implies small angular momentum, and large angular velocity implies large angular momentum. Check Your Understanding 7. When is momentum said to be conserved? a. When momentum is changing during an event b. When momentum is increasing during an event c. When momentum is decreasing during an event d. When momentum is constant throughout an event 8. A ball is hit by a racket and its momentum changes. How is momentum conserved in this case? a. Momentum of the system can never be conserved in this case. b. Momentum of the system is conserved if the momentum of the racket is not considered. c. Momentum of the system is conserved if the momentum of the racket is also considered. d. Momentum of the system is conserved if the momenta of the racket and the player are also considered. 9. State the law of conservation of momentum. a. Momentum is conserved for an isolated system with any number of objects in it. b. Momentum is conserved for an isolated system with an even number of objects in it. c. Momentum is conserved for an interacting system with any number of objects in it. d. Momentum is conserved for an interacting system with an even number of objects in it. 8.3 Elastic and Inelastic Collisions Section Learning Objectives By the end of this section, you will be able to do the following: \u2022 Distinguish between elastic and inelastic collisions \u2022 Solve collision problems by applying the law of conservation of momentum Section Key Terms elastic collision inelastic collision point masses recoil Elastic and Inelastic Collisions When objects collide, they can either stick together or bounce off one another, remaining separate. In this section, we\u2019ll cover these two different types of collisions, first in one dimension and then in two dimensions. In an elastic collision, the objects separate after impact and don\u2019t lose any of their kinetic energy. Kinetic energy is the energy of motion and is covered in detail elsewhere. The law of conservation of momentum is very useful here, and it can be used whenever the net external force on a system is zero. Figure 8.6 shows an elastic collision where momentum is conserved. Access for free at openstax.org. 8.3 \u2022 Elastic and Inelastic Collisions 263 Figure 8.6 The diagram shows a one-dimensional elastic collision between two objects. An animation of an elastic collision between balls can be seen by watching this video (http://openstax.org/l/28elasticball). It replicates the elastic collisions between balls of varying masses. Perfectly elastic collisions can happen only with subatomic particles. Everyday observable examples of perfectly elastic collisions don\u2019t exist\u2014some kinetic energy is always lost, as it is converted into heat transfer due to friction. However, collisions between everyday objects are almost perfectly elastic when they occur with objects and surfaces that are nearly frictionless, such as with two steel blocks on ice. Now, to solve problems involving one-dimensional elastic collisions between two objects, we can use the equation for conservation of momentum. First, the equation for conservation of momentum for two objects in a one-dimensional collision is Substituting the definition of momentum p = mv for each initial and final momentum, we get where the primes (') indicate values after the collision; In some texts, you may see ifor initial (before collision) and ffor final (after collision). The equation assumes that the mass of each object does not change during the collision. WATCH PHYSICS Momentum: Ice Skater Throws a Ball This video covers an elastic collision problem in which we find the recoilvelocityof an ice skater who throws a ball straight forward. To clarify, Sal is using the equation Click to view content (https://www.khanacademy.org/embed_video?v=vPkkCOlGND4). GRASP CHECK The resultant vector of the addition of vectors respectively. Which of the following is true? a. b. and is. The magnitudes of,, and are,, and, 264 Chapter 8 \u2022 Momentum c. d. Now, let us turn to the second type of collision. An inelastic collision is", " one in which objects stick together after impact, and kinetic energy is notconserved. This lack of conservation means that the forces between colliding objects may convert kinetic energy to other forms of energy, such as potential energy or thermal energy. The concepts of energy are discussed more thoroughly elsewhere. For inelastic collisions, kinetic energy may be lost in the form of heat. Figure 8.7 shows an example of an inelastic collision. Two objects that have equal masses head toward each other at equal speeds and then stick together. The two objects come to rest after sticking together, conserving momentum but not kinetic energy after they collide. Some of the energy of motion gets converted to thermal energy, or heat. Figure 8.7 A one-dimensional inelastic collision between two objects. Momentum is conserved, but kinetic energy is not conserved. (a) Two objects of equal mass initially head directly toward each other at the same speed. (b) The objects stick together, creating a perfectly inelastic collision. In the case shown in this figure, the combined objects stop; This is not true for all inelastic collisions. Since the two objects stick together after colliding, they move together at the same speed. This lets us simplify the conservation of momentum equation from to for inelastic collisions, where v\u2032 is the final velocity for both objects as they are stuck together, either in motion or at rest. WATCH PHYSICS Introduction to Momentum This video reviews the definitions of momentum and impulse. It also covers an example of using conservation of momentum to solve a problem involving an inelastic collision between a car with constant velocity and a stationary truck. Note that Sal accidentally gives the unit for impulse as Joules; it is actually N s or k gm/s. Click to view content (https://www.khanacademy.org/embed_video?v=XFhntPxow0U) GRASP CHECK How would the final velocity of the car-plus-truck system change if the truck had some initial velocity moving in the same direction as the car? What if the truck were moving in the opposite direction of the car initially? Why? a. If the truck was initially moving in the same direction as the car, the final velocity would be greater. If the truck was initially moving in the opposite direction of the car, the final velocity would be smaller. If the truck was initially moving in the same direction as the car, the final velocity would be smaller. If the truck was initially moving in the opposite direction of the car, the final velocity would be greater. b. c. The direction in which the truck was initially moving would not matter. If the truck was initially moving in either Access for free at openstax.org. direction, the final velocity would be smaller. d. The direction in which the truck was initially moving would not matter. If the truck was initially moving in either direction, the final velocity would be greater. 8.3 \u2022 Elastic and Inelastic Collisions 265 Snap Lab Ice Cubes and Elastic Collisions In this activity, you will observe an elastic collision by sliding an ice cube into another ice cube on a smooth surface, so that a negligible amount of energy is converted to heat. \u2022 Several ice cubes (The ice must be in the form of cubes.) \u2022 A smooth surface Procedure 1. Find a few ice cubes that are about the same size and a smooth kitchen tabletop or a table with a glass top. 2. Place the ice cubes on the surface several centimeters away from each other. 3. Flick one ice cube toward a stationary ice cube and observe the path and velocities of the ice cubes after the collision. Try to avoid edge-on collisions and collisions with rotating ice cubes. 4. Explain the speeds and directions of the ice cubes using momentum. GRASP CHECK Was the collision elastic or inelastic? a. perfectly elastic b. perfectly inelastic c. Nearly perfect elastic d. Nearly perfect inelastic TIPS FOR SUCCESS Here\u2019s a trick for remembering which collisions are elastic and which are inelastic: Elastic is a bouncy material, so when objects bounceoff one another in the collision and separate, it is an elastic collision. When they don\u2019t, the collision is inelastic. Solving Collision Problems The Khan Academy videos referenced in this section show examples of elastic and inelastic collisions in one dimension. In onedimensional collisions, the incoming and outgoing velocities are all along the same line. But what about collisions, such as those between billiard balls, in which objects scatter to the side? These are two-dimensional collisions, and just as we did with twodimensional forces, we will solve these problems by first choosing a coordinate system and separating the motion into its xand y components. One complication with two-dimensional collisions is that the objects might rotate before or after their collision. For example, if two ice skaters hook", " arms as they pass each other, they will spin in circles. We will not consider such rotation until later, and so for now, we arrange things so that no rotation is possible. To avoid rotation, we consider only the scattering of point masses\u2014that is, structureless particles that cannot rotate or spin. We start by assuming that Fnet = 0, so that momentum p is conserved. The simplest collision is one in which one of the particles is initially at rest. The best choice for a coordinate system is one with an axis parallel to the velocity of the incoming particle, as shown in Figure 8.8. Because momentum is conserved, the components of momentum along the x- and y-axes, displayed as px and py, will also be conserved. With the chosen coordinate system, pyis initially zero and pxis the momentum of the incoming particle. 266 Chapter 8 \u2022 Momentum Figure 8.8 A two-dimensional collision with the coordinate system chosen so that m2 is initially at rest and v1 is parallel to the x-axis. Now, we will take the conservation of momentum equation, p1 + p2 = p\u20321 + p\u20322 and break it into its xand ycomponents. Along the x-axis, the equation for conservation of momentum is In terms of masses and velocities, this equation is But because particle 2 is initially at rest, this equation becomes 8.3 8.4 The components of the velocities along the x-axis have the form v cos \u03b8. Because particle 1 initially moves along the x-axis, we find v1x= v1. Conservation of momentum along the x-axis gives the equation where and are as shown in Figure 8.8. Along the y-axis, the equation for conservation of momentum is or But v1yis zero, because particle 1 initially moves along the x-axis. Because particle 2 is initially at rest, v2yis also zero. The equation for conservation of momentum along the y-axis becomes 8.5 8.6 8.7 The components of the velocities along the y-axis have the form v sin. Therefore, conservation of momentum along the y-axis gives the following equation: Virtual Physics Collision Lab In this simulation, you will investigate collisions on an air hockey table. Place checkmarks next to the momentum vectors Access for free at openstax.org. 8.3 \u2022 Elastic and Inelastic Collisions 267 and momenta diagram options. Experiment with changing the masses of the balls and the initial speed of ball 1. How does this affect the momentum of each ball? What about the total momentum? Next, experiment with changing the elasticity of the collision. You will notice that collisions have varying degrees of elasticity, ranging from perfectly elastic to perfectly inelastic. Click to view content (https://archive.cnx.org/specials/2c7acb3c-2fbd-11e5-b2d9-e7f92291703c/collision-lab/) GRASP CHECK If you wanted to maximize the velocity of ball 2 after impact, how would you change the settings for the masses of the balls, the initial speed of ball 1, and the elasticity setting? Why? Hint\u2014Placing a checkmark next to the velocity vectors and removing the momentum vectors will help you visualize the velocity of ball 2, and pressing the More Data button will let you take readings. a. Maximize the mass of ball 1 and initial speed of ball 1; minimize the mass of ball 2; and set elasticity to 50 percent. b. Maximize the mass of ball 2 and initial speed of ball 1; minimize the mass of ball 1; and set elasticity to 100 percent. c. Maximize the mass of ball 1 and initial speed of ball 1; minimize the mass of ball 2; and set elasticity to 100 percent. d. Maximize the mass of ball 2 and initial speed of ball 1; minimize the mass of ball 1; and set elasticity to 50 percent. WORKED EXAMPLE Calculating Velocity: Inelastic Collision of a Puck and a Goalie Find the recoil velocity of a 70 kg ice hockey goalie who catches a 0.150-kg hockey puck slapped at him at a velocity of 35 m/s. Assume that the goalie is at rest before catching the puck, and friction between the ice and the puck-goalie system is negligible (see Figure 8.9). Figure 8.9 An ice hockey goalie catches a hockey puck and recoils backward in an inelastic collision. Strategy Momentum is conserved because the net external force on the puck-goalie system is zero. Therefore, we can use conservation of momentum to find the final velocity of the puck and goalie system. Note that the initial velocity of the goalie is zero and that the final velocity of the puck and goalie are the same. Solution For", " an inelastic collision, conservation of momentum is where v\u2032 is the velocity of both the goalie and the puck after impact. Because the goalie is initially at rest, we know v2 = 0. This simplifies the equation to 8.8 8.9 Solving for v\u2032 yields 268 Chapter 8 \u2022 Momentum Entering known values in this equation, we get 8.10 8.11 Discussion This recoil velocity is small and in the same direction as the puck\u2019s original velocity. WORKED EXAMPLE Calculating Final Velocity: Elastic Collision of Two Carts Two hard, steel carts collide head-on and then ricochet off each other in opposite directions on a frictionless surface (see Figure 8.10). Cart 1 has a mass of 0.350 kg and an initial velocity of 2 m/s. Cart 2 has a mass of 0.500 kg and an initial velocity of \u22120.500 m/s. After the collision, cart 1 recoils with a velocity of \u22124 m/s. What is the final velocity of cart 2? Figure 8.10 Two carts collide with each other in an elastic collision. Strategy Since the track is frictionless, Fnet = 0 and we can use conservation of momentum to find the final velocity of cart 2. Solution As before, the equation for conservation of momentum for a one-dimensional elastic collision in a two-object system is The only unknown in this equation is v\u20322. Solving for v\u20322 and substituting known values into the previous equation yields 8.12 8.13 Discussion The final velocity of cart 2 is large and positive, meaning that it is moving to the right after the collision. Access for free at openstax.org. 8.3 \u2022 Elastic and Inelastic Collisions 269 WORKED EXAMPLE Calculating Final Velocity in a Two-Dimensional Collision Suppose the following experiment is performed (Figure 8.11). An object of mass 0.250 kg (m1) is slid on a frictionless surface into a dark room, where it strikes an initially stationary object of mass 0.400 kg (m2). The 0.250 kg object emerges from the room at an angle of 45\u00ba with its incoming direction. The speed of the 0.250 kg object is originally 2 m/s and is 1.50 m/s after the collision. Calculate the magnitude and direction of the velocity (v\u20322 and ) of the 0.400 kg object after the collision. Figure 8.11 The incoming object of mass m1 is scattered by an initially stationary object. Only the stationary object\u2019s mass m2 is known. By measuring the angle and speed at which the object of mass m1 emerges from the room, it is possible to calculate the magnitude and direction of the initially stationary object\u2019s velocity after the collision. Strategy Momentum is conserved because the surface is frictionless. We chose the coordinate system so that the initial velocity is parallel to the x-axis, and conservation of momentum along the x- and y-axes applies. Everything is known in these equations except v\u20322 and \u03b82, which we need to find. We can find two unknowns because we have two independent equations\u2014the equations describing the conservation of momentum in the xand ydirections. Solution First, we\u2019ll solve both conservation of momentum equations ( ) for v\u20322 sin. For conservation of momentum along x-axis, let\u2019s substitute sin comes from rearranging the definition of the trigonometric identity tan = sin /cos for cos /tan Solving for v\u20322 sin yields For conservation of momentum along y-axis, solving for v\u20322 sin yields and so that terms may cancel out later on. This. This gives us 8.14 8.15 8.16 270 Chapter 8 \u2022 Momentum Since both equations equal v\u20322 sin, we can set them equal to one another, yielding Solving this equation for tan, we get Entering known values into the previous equation gives Therefore, Since angles are defined as positive in the counterclockwise direction, m2 is scattered to the right. We\u2019ll use the conservation of momentum along the y-axis equation to solve for v\u20322. Entering known values into this equation gives Therefore, 8.17 8.18 8.19 8.20 8.21 8.22 8.23 Discussion Either equation for the x- or y-axis could have been used to solve for v\u20322, but the equation for the y-axis is easier because it has fewer terms. Practice Problems 10. In an elastic collision, an object with momentum collides with another object moving to the right that has a. After the collision, both objects are still moving to the right, but the first object\u2019s momentum. What is the final momentum of the second object? momentum changes to a. b. c. d. 11. In an elastic collision, an object with momentum 25 kg \ufffd", "\ufffd\ufffd m/s collides with another that has a momentum 35 kg \u22c5 m/s. The first object\u2019s momentum changes to 10 kg \u22c5 m/s. What is the final momentum of the second object? 10 kg \u22c5 m/s a. b. 20 kg \u22c5 m/s 35 kg \u22c5 m/s c. 50 kg \u22c5 m/s d. Check Your Understanding 12. What is an elastic collision? a. An elastic collision is one in which the objects after impact are deformed permanently. b. An elastic collision is one in which the objects after impact lose some of their internal kinetic energy. Access for free at openstax.org. 8.3 \u2022 Elastic and Inelastic Collisions 271 c. An elastic collision is one in which the objects after impact do not lose any of their internal kinetic energy. d. An elastic collision is one in which the objects after impact become stuck together and move with a common velocity. 13. Are perfectly elastic collisions possible? a. Perfectly elastic collisions are not possible. b. Perfectly elastic collisions are possible only with subatomic particles. c. Perfectly elastic collisions are possible only when the objects stick together after impact. d. Perfectly elastic collisions are possible if the objects and surfaces are nearly frictionless. 14. What is the equation for conservation of momentum for two objects in a one-dimensional collision? a. p1 + p1\u2032 = p2 + p2\u2032 b. p1 + p2 = p1\u2032 + p2\u2032 c. p1 \u2212 p2 = p1\u2032 \u2212 p2\u2032 d. p1 + p2 + p1\u2032 + p2\u2032 = 0 272 Chapter 8 \u2022 Key Terms KEY TERMS angular momentum the product of the moment of inertia and angular velocity change in momentum the difference between the final and initial values of momentum; the mass times the change in velocity elastic collision collision in which objects separate after after impact and kinetic energy is not conserved isolated system system in which the net external force is zero law of conservation of momentum when the net external force is zero, the total momentum of the system is conserved or constant impact and kinetic energy is conserved linear momentum the product of a system's mass and impulse average net external force multiplied by the time the force acts; equal to the change in momentum impulse\u2013momentum theorem the impulse, or change in momentum, is the product of the net external force and the time over which the force acts inelastic collision collision in which objects stick together SECTION SUMMARY 8.1 Linear Momentum, Force, and Impulse \u2022 Linear momentum, often referenced as momentumfor short, is defined as the product of a system\u2019s mass multiplied by its velocity, p = mv. \u2022 The SI unit for momentum is kg m/s. \u2022 Newton\u2019s second law of motion in terms of momentum states that the net external force equals the change in momentum of a system divided by the time over which it changes,. \u2022 Impulse is the average net external force multiplied by the time this force acts, and impulse equals the change in momentum,. \u2022 Forces are usually not constant over a period of time, so we use the average of the force over the time it acts. 8.2 Conservation of Momentum velocity point masses spin structureless particles that cannot rotate or recoil backward movement of an object caused by the transfer of momentum from another object in a collision In an isolated system, the net external force is zero. \u2022 \u2022 Conservation of momentum applies only when the net external force is zero, within the defined system. 8.3 Elastic and Inelastic Collisions \u2022 If objects separate after impact, the collision is elastic; If they stick together, the collision is inelastic. \u2022 Kinetic energy is conserved in an elastic collision, but not in an inelastic collision. \u2022 The approach to two-dimensional collisions is to choose a convenient coordinate system and break the motion into components along perpendicular axes. Choose a coordinate system with the x-axis parallel to the velocity of the incoming particle. \u2022 Two-dimensional collisions of point masses, where mass 2 is initially at rest, conserve momentum along the initial direction of mass 1, or the x-axis, and along the direction perpendicular to the initial direction, or the y-axis. \u2022 The law of conservation of momentum is written ptot = \u2022 Point masses are structureless particles that cannot constant or ptot = p\u2032tot (isolated system), where ptot is the initial total momentum and p\u2032tot is the total momentum some time later. spin. KEY EQUATIONS 8.1 Linear Momentum, Force, and Impulse Newton\u2019s second law in terms of momentum impulse impulse\u2013momentum theorem linear momentum Access for free at openstax.org. 8.2 Conservation of Momentum law of conservation of momentum ptot = constant, or ptot = p\u2032tot Chapter 8", " \u2022 Chapter Review 273 conservation of momentum along x-axis for 2D collisions conservation of momentum along y-axis for 2D collisions conservation of momentum for two objects p1 + p2 = constant, or p1 + p2 = p\u20321 + p\u20322 angular momentum L = I 8.3 Elastic and Inelastic Collisions conservation of momentum in an elastic collision conservation of momentum in an inelastic collision CHAPTER REVIEW Concept Items 8.2 Conservation of Momentum 8.1 Linear Momentum, Force, and Impulse 5. What is angular momentum? 1. What is impulse? a. Change in velocity b. Change in momentum c. Rate of change of velocity d. Rate of change of momentum 2. In which equation of Newton\u2019s second law is mass assumed to be constant? a. b. c. d. 3. What is the SI unit of momentum? a. b. c. d. 4. What is the equation for linear momentum? a. b. c. d. a. The sum of moment of inertia and angular velocity b. The ratio of moment of inertia to angular velocity c. The product of moment of inertia and angular velocity d. Half the product of moment of inertia and square of angular velocity 6. What is an isolated system? a. A system in which the net internal force is zero b. A system in which the net external force is zero c. A system in which the net internal force is a nonzero constant d. A system in which the net external force is a nonzero constant 8.3 Elastic and Inelastic Collisions 7. In the equation p1 + p2 = p'1 + p'2 for the collision of two objects, what is the assumption made regarding the friction acting on the objects? a. Friction is zero. b. Friction is nearly zero. c. Friction acts constantly. d. Friction before and after the impact remains the same. 8. What is an inelastic collision? 274 Chapter 8 \u2022 Chapter Review a. when objects stick together after impact, and their c. when objects stick together after impact, and always internal energy is not conserved come to rest instantaneously after collision b. when objects stick together after impact, and their d. when objects stick together after impact, and their internal energy is conserved internal energy increases Critical Thinking Items 8.1 Linear Momentum, Force, and Impulse 9. Consider two objects of the same mass. If a force of acts on the first for a duration of and on the, which of the following other for a duration of statements is true? a. The first object will acquire more momentum. b. The second object will acquire more momentum. c. Both objects will acquire the same momentum. d. Neither object will experience a change in momentum. 10. Cars these days have parts that can crumple or collapse in the event of an accident. How does this help protect the passengers? a. It reduces injury to the passengers by increasing the time of impact. It reduces injury to the passengers by decreasing the time of impact. It reduces injury to the passengers by increasing the change in momentum. It reduces injury to the passengers by decreasing the change in momentum. b. c. d. 11. How much force would be needed to cause a 17 kg \u22c5 m/s change in the momentum of an object, if the force acted for 5 seconds? 3.4 N a. b. 12 N c. 22 N d. 85 N 8.2 Conservation of Momentum 12. A billiards ball rolling on the table has momentum p1. It hits another stationary ball, which then starts rolling. Considering friction to be negligible, what will happen to the momentum of the first ball? Problems 8.1 Linear Momentum, Force, and Impulse is applied to an object for, and it, what could be the mass 16. If a force of changes its velocity by of the object? a. b. Access for free at openstax.org. a. b. c. d. It will decrease. It will increase. It will become zero. It will remain the same. 13. A ball rolling on the floor with momentum p1 collides with a stationary ball and sets it in motion. The momentum of the first ball becomes p'1, and that of the second becomes p'2. Compare the magnitudes of p1 and p'2. a. Momenta p1 and p'2 are the same in magnitude. b. The sum of the magnitudes of p1 and p'2 is zero. c. The magnitude of p1 is greater than that of p'2. d. The magnitude of p'2 is greater than that of p1. 14. Two cars are moving in the same direction. One car with momentum p1 collides with another, which has momentum p2. Their momenta become p'1 and p'2 respectively. Considering frictional losses,", " compare (p'1 + p'2 ) with (p1 + p2). a. The value of (p'1 + p'2 ) is zero. b. The values of (p1 + p2) and (p'1 + p'2 ) are equal. c. The value of (p1 + p2) will be greater than (p'1 + p'2 ). d. The value of (p'1 + p'2 ) will be greater than (p1 + p2). 8.3 Elastic and Inelastic Collisions 15. Two people, who have the same mass, throw two different objects at the same velocity. If the first object is heavier than the second, compare the velocities gained by the two people as a result of recoil. a. The first person will gain more velocity as a result of recoil. b. The second person will gain more velocity as a result of recoil. c. Both people will gain the same velocity as a result of recoil. d. The velocity of both people will be zero as a result of recoil. c. d. 17. For how long should a force of 130 N be applied to an object of mass 50 kg to change its speed from 20 m/s to 60 m/s? a. 0.031 s b. 0.065 s 15.4 s c. Chapter 8 \u2022 Test Prep 275 d. 40 s d. 50.0 m/s 8.3 Elastic and Inelastic Collisions 18. If a man with mass 70 kg, standing still, throws an object with mass 5 kg at 50 m/s, what will be the recoil velocity of the man, assuming he is standing on a frictionless surface? a. \u22123.6 m/s b. 0 m/s c. 3.6 m/s 19. Find the recoil velocity of a ice hockey goalie who hockey puck slapped at him at a. Assume that the goalie is at rest catches a velocity of before catching the puck, and friction between the ice and the puck-goalie system is negligible. a. b. c. d. Performance Task 8.3 Elastic and Inelastic Collisions 20. You will need the following: \u2022 balls of different weights \u2022 a ruler or wooden strip \u2022 some books \u2022 a paper cup Make an inclined plane by resting one end of a ruler on a stack of books. Place a paper cup on the other end. Roll TEST PREP Multiple Choice 8.1 Linear Momentum, Force, and Impulse 21. What kind of quantity is momentum? a. Scalar b. Vector 22. When does the net force on an object increase? a. When \u0394p decreases b. When \u0394tincreases c. When \u0394tdecreases 23. In the equation \u0394p = m(vf \u2212 vi), which quantity is considered to be constant? a. Initial velocity b. Final velocity c. Mass d. Momentum 24. For how long should a force of be applied to change the momentum of an object by a. b. c. d. 8.2 Conservation of Momentum 25. In the equation L = I\u03c9, what is I? a ball from the top of the ruler so that it hits the paper cup. Measure the displacement of the paper cup due to the collision. Now use increasingly heavier balls for this activity and see how that affects the displacement of the cup. Plot a graph of mass vs. displacement. Now repeat the same activity, but this time, instead of using different balls, change the incline of the ruler by varying the height of the stack of books. This will give you different velocities of the ball. See how this affects the displacement of the paper cup. a. Linear momentum b. Angular momentum c. Torque d. Moment of inertia 26. Give an example of an isolated system. a. A cyclist moving along a rough road b. A figure skater gliding in a straight line on an ice rink c. A baseball player hitting a home run d. A man drawing water from a well 8.3 Elastic and Inelastic Collisions 27. In which type of collision is kinetic energy conserved? a. Elastic b. Inelastic 28. In physics, what are structureless particles that cannot? rotate or spin called? a. Elastic particles b. Point masses c. Rigid masses 29. Two objects having equal masses and velocities collide with each other and come to a rest. What type of a collision is this and why? a. Elastic collision, because internal kinetic energy is conserved 276 Chapter 8 \u2022 Test Prep b. Inelastic collision, because internal kinetic energy is not conserved c. Elastic collision, because internal kinetic energy is d. not conserved Inelastic collision, because internal kinetic energy is conserved 30. Two objects having equal masses and velocities collide with each other and come to a rest. Is momentum conserved in this case", "? a. Yes b. No Short Answer 8.1 Linear Momentum, Force, and Impulse 31. If an object\u2019s velocity is constant, what is its momentum proportional to? a. b. c. d. Its shape Its mass Its length Its breadth 32. If both mass and velocity of an object are constant, what can you tell about its impulse? a. b. c. d. Its impulse would be constant. Its impulse would be zero. Its impulse would be increasing. Its impulse would be decreasing. 33. When the momentum of an object increases with respect to time, what is true of the net force acting on it? a. It is zero, because the net force is equal to the rate of change of the momentum. It is zero, because the net force is equal to the product of the momentum and the time interval. It is nonzero, because the net force is equal to the rate of change of the momentum. It is nonzero, because the net force is equal to the product of the momentum and the time interval. b. c. d. 34. How can you express impulse in terms of mass and velocity when neither of those are constant? a. b. c. d. 35. How can you express impulse in terms of mass and initial and final velocities? a. b. c. d. 36. Why do we use average force while solving momentum problems? How is net force related to the momentum of the object? a. Forces are usually constant over a period of time, Access for free at openstax.org. and net force acting on the object is equal to the rate of change of the momentum. b. Forces are usually not constant over a period of time, and net force acting on the object is equal to the product of the momentum and the time interval. c. Forces are usually constant over a period of time, and net force acting on the object is equal to the product of the momentum and the time interval. d. Forces are usually not constant over a period of time, and net force acting on the object is equal to the rate of change of the momentum. 8.2 Conservation of Momentum 37. Under what condition(s) is the angular momentum of a system conserved? a. When net torque is zero b. When net torque is not zero c. When moment of inertia is constant d. When both moment of inertia and angular momentum are constant 38. If the moment of inertia of an isolated system increases, what happens to its angular velocity? a. b. c. d. It increases. It decreases. It stays constant. It becomes zero. 39. If both the moment of inertia and the angular velocity of a system increase, what must be true of the force acting on the system? a. Force is zero. b. Force is not zero. c. Force is constant. d. Force is decreasing. 8.3 Elastic and Inelastic Collisions 40. Two objects collide with each other and come to a rest. How can you use the equation of conservation of momentum to describe this situation? a. m1v1 + m2v2 = 0 b. m1v1 \u2212 m2v2 = 0 c. m1v1 + m2v2 = m1v1\u2032 d. m1v1 + m2v2 = m1v2 41. What is the difference between momentum and impulse? a. Momentum is the sum of mass and velocity. Impulse is the change in momentum. b. Momentum is the sum of mass and velocity. Impulse is the rate of change in momentum. c. Momentum is the product of mass and velocity. Impulse is the change in momentum. d. Momentum is the product of mass and velocity. Impulse is the rate of change in momentum. 42. What is the equation for conservation of momentum along the x-axis for 2D collisions in terms of mass and velocity, where one of the particles is initially at rest? Chapter 8 \u2022 Test Prep 277 a. m1v1 = m1v1\u2032cos \u03b81 b. m1v1 = m1v1\u2032cos \u03b81 + m2v2\u2032cos \u03b82 c. m1v1 = m1v1\u2032cos \u03b81 \u2212 m2v2\u2032cos \u03b82 d. m1v1 = m1v1\u2032sin \u03b81 + m2v2\u2032sin \u03b82 43. What is the equation for conservation of momentum along the y-axis for 2D collisions in terms of mass and velocity, where one of the particles is initially at rest? a. 0 = m1v1\u2032sin \u03b81 b. 0 = m1v1\u2032sin \u03b81 + m2v2\u2032sin \u03b82 c. 0 = m1v1\u2032sin \u03b81 \u2212 m2v2\u2032sin \u03b82 d. 0", " = m1v1\u2032cos \u03b81 + m2v2\u2032cos \u03b82 Extended Response 8.2 Conservation of Momentum 8.1 Linear Momentum, Force, and Impulse 47. Why does a figure skater spin faster if he pulls his arms 44. Can a lighter object have more momentum than a heavier one? How? a. No, because momentum is independent of the velocity of the object. b. No, because momentum is independent of the mass of the object. c. Yes, if the lighter object\u2019s velocity is considerably high. d. Yes, if the lighter object\u2019s velocity is considerably low. 45. Why does it hurt less when you fall on a softer surface? a. The softer surface increases the duration of the impact, thereby reducing the effect of the force. b. The softer surface decreases the duration of the impact, thereby reducing the effect of the force. c. The softer surface increases the duration of the and legs in? a. Due to an increase in moment of inertia b. Due to an increase in angular momentum c. Due to conservation of linear momentum d. Due to conservation of angular momentum 8.3 Elastic and Inelastic Collisions 48. A driver sees another car approaching him from behind. He fears it is going to collide with his car. Should he speed up or slow down in order to reduce damage? a. He should speed up. b. He should slow down. c. He should speed up and then slow down just before the collision. d. He should slow down and then speed up just before the collision. impact, thereby increasing the effect of the force. 49. What approach would you use to solve problems d. The softer surface decreases the duration of the impact, thereby increasing the effect of the force. involving 2D collisions? a. Break the momenta into components and then 46. Can we use the equation when the mass is constant? a. No, because the given equation is applicable for the variable mass only. b. No, because the given equation is not applicable for the constant mass. c. Yes, and the resultant equation is F = mv d. Yes, and the resultant equation is F = ma choose a coordinate system. b. Choose a coordinate system and then break the momenta into components. c. Find the total momenta in the x and y directions, and then equate them to solve for the unknown. d. Find the sum of the momenta in the x and y directions, and then equate it to zero to solve for the unknown. 278 Chapter 8 \u2022 Test Prep Access for free at openstax.org. CHAPTER 9 Work, Energy, and Simple Machines Figure 9.1 People on a roller coaster experience thrills caused by changes in types of energy. (Jonrev, Wikimedia Commons) Chapter Outline 9.1 Work, Power, and the Work\u2013Energy Theorem 9.2 Mechanical Energy and Conservation of Energy 9.3 Simple Machines Roller coasters have provided thrills for daring riders around the world since the nineteenth century. INTRODUCTION Inventors of roller coasters used simple physics to build the earliest examples using railroad tracks on mountainsides and old mines. Modern roller coaster designers use the same basic laws of physics to create the latest amusement park favorites. Physics principles are used to engineer the machines that do the work to lift a roller coaster car up its first big incline before it is set loose to roll. Engineers also have to understand the changes in the car\u2019s energy that keep it speeding over hills, through twists, turns, and even loops. What exactly is energy? How can changes in force, energy, and simple machines move objects like roller coaster cars? How can machines help us do work? In this chapter, you will discover the answer to this question and many more, as you learn about 280 Chapter 9 \u2022 Work, Energy, and Simple Machines work, energy, and simple machines. 9.1 Work, Power, and the Work\u2013Energy Theorem Section Learning Objectives By the end of this section, you will be able to do the following: \u2022 Describe and apply the work\u2013energy theorem \u2022 Describe and calculate work and power Section Key Terms energy gravitational potential energy joule kinetic energy mechanical energy potential energy power watt work work\u2013energy theorem The Work\u2013Energy Theorem In physics, the term work has a very specific definition. Work is application of force, the direction that the force is applied. Work, W, is described by the equation, to move an object over a distance, d, in Some things that we typically consider to be work are not work in the scientific sense of the term. Let\u2019s consider a few examples. Think about why each of the following statements is true. \u2022 Homework is notwork. \u2022 Lifting a rock upwards off the ground iswork. \u2022 Carrying a rock in a straight path across the lawn at a constant speed is notwork. The first two examples are fairly simple. Hom", "ework is not work because objects are not being moved over a distance. Lifting a rock up off the ground is work because the rock is moving in the direction that force is applied. The last example is less obvious. Recall from the laws of motion that force is notrequired to move an object at constant velocity. Therefore, while some force may be applied to keep the rock up off the ground, no net force is applied to keep the rock moving forward at constant velocity. Work and energy are closely related. When you do work to move an object, you change the object\u2019s energy. You (or an object) also expend energy to do work. In fact, energy can be defined as the ability to do work. Energy can take a variety of different forms, and one form of energy can transform to another. In this chapter we will be concerned with mechanical energy, which comes in two forms: kinetic energy and potential energy. \u2022 Kinetic energy is also called energy of motion. A moving object has kinetic energy. \u2022 Potential energy, sometimes called stored energy, comes in several forms. Gravitational potential energy is the stored energy an object has as a result of its position above Earth\u2019s surface (or another object in space). A roller coaster car at the top of a hill has gravitational potential energy. Let\u2019s examine how doing work on an object changes the object\u2019s energy. If we apply force to lift a rock off the ground, we increase the rock\u2019s potential energy, PE. If we drop the rock, the force of gravity increases the rock\u2019s kinetic energy as the rock moves downward until it hits the ground. The force we exert to lift the rock is equal to its weight, w, which is equal to its mass, m, multiplied by acceleration due to gravity, g. The work we do on the rock equals the force we exert multiplied by the distance, d, that we lift the rock. The work we do on the rock also equals the rock\u2019s gain in gravitational potential energy, PEe. Kinetic energy depends on the mass of an object and its velocity, v. Access for free at openstax.org. 9.1 \u2022 Work, Power, and the Work\u2013Energy Theorem 281 When we drop the rock the force of gravity causes the rock to fall, giving the rock kinetic energy. When work done on an object increases only its kinetic energy, then the net work equals the change in the value of the quantity. This is a statement of the work\u2013energy theorem, which is expressed mathematically as The subscripts 2 and 1 indicate the final and initial velocity, respectively. This theorem was proposed and successfully tested by James Joule, shown in Figure 9.2. Does the name Joule sound familiar? The joule (J) is the metric unit of measurement for both work and energy. The measurement of work and energy with the same unit reinforces the idea that work and energy are related and can be converted into one another. 1.0 J = 1.0 N\u2219m, the units of force multiplied by distance. 1.0 N = 1.0 k\u2219m/s2, so 1.0 J = 1.0 k\u2219m2/s2. Analyzing the units of the term (1/2)mv2 will produce the same units for joules. Figure 9.2 The joule is named after physicist James Joule (1818\u20131889). (C. H. Jeens, Wikimedia Commons) WATCH PHYSICS Work and Energy This video explains the work energy theorem and discusses how work done on an object increases the object\u2019s KE. Click to view content (https://www.khanacademy.org/embed_video?v=2WS1sG9fhOk) GRASP CHECK True or false\u2014The energy increase of an object acted on only by a gravitational force is equal to the product of the object's weight and the distance the object falls. a. True b. False Calculations Involving Work and Power In applications that involve work, we are often interested in how fast the work is done. For example, in roller coaster design, the amount of time it takes to lift a roller coaster car to the top of the first hill is an important consideration. Taking a half hour on the ascent will surely irritate riders and decrease ticket sales. Let\u2019s take a look at how to calculate the time it takes to do work. Recall that a rate can be used to describe a quantity, such as work, over a period of time. Power is the rate at which work is done. In this case, rate means per unit of time. Power is calculated by dividing the work done by the time it took to do the work. Let\u2019s consider an example that can help illustrate the differences among work, force, and power. Suppose the woman in Figure 9.3 lifting the TV", " with a pulley gets the TV to the fourth floor in two minutes, and the man carrying the TV up the stairs takes five 282 Chapter 9 \u2022 Work, Energy, and Simple Machines minutes to arrive at the same place. They have done the same amount of work mass over the same vertical distance, which requires the same amount of upward force. However, the woman using the pulley has generated more power. This is because she did the work in a shorter amount of time, so the denominator of the power formula, t, is smaller. (For simplicity\u2019s sake, we will leave aside for now the fact that the man climbing the stairs has also done work on himself.) on the TV, because they have moved the same Figure 9.3 No matter how you move a TV to the fourth floor, the amount of work performed and the potential energy gain are the same. Power can be expressed in units of watts (W). This unit can be used to measure power related to any form of energy or work. You have most likely heard the term used in relation to electrical devices, especially light bulbs. Multiplying power by time gives the amount of energy. Electricity is sold in kilowatt-hours because that equals the amount of electrical energy consumed. The watt unit was named after James Watt (1736\u20131819) (see Figure 9.4). He was a Scottish engineer and inventor who discovered how to coax more power out of steam engines. Figure 9.4 Is James Watt thinking about watts? (Carl Frederik von Breda, Wikimedia Commons) Access for free at openstax.org. 9.1 \u2022 Work, Power, and the Work\u2013Energy Theorem 283 LINKS TO PHYSICS Watt\u2019s Steam Engine James Watt did not invent the steam engine, but by the time he was finished tinkering with it, it was more useful. The first steam engines were not only inefficient, they only produced a back and forth, or reciprocal, motion. This was natural because pistons move in and out as the pressure in the chamber changes. This limitation was okay for simple tasks like pumping water or mashing potatoes, but did not work so well for moving a train. Watt was able build a steam engine that converted reciprocal motion to circular motion. With that one innovation, the industrial revolution was off and running. The world would never be the same. One of Watt's steam engines is shown in Figure 9.5. The video that follows the figure explains the importance of the steam engine in the industrial revolution. Figure 9.5 A late version of the Watt steam engine. (Nehemiah Hawkins, Wikimedia Commons) WATCH PHYSICS Watt's Role in the Industrial Revolution This video demonstrates how the watts that resulted from Watt's inventions helped make the industrial revolution possible and allowed England to enter a new historical era. Click to view content (https://www.youtube.com/embed/zhL5DCizj5c) GRASP CHECK Which form of mechanical energy does the steam engine generate? a. Potential energy b. Kinetic energy c. Nuclear energy d. Solar energy Before proceeding, be sure you understand the distinctions among force, work, energy, and power. Force exerted on an object over a distance does work. Work can increase energy, and energy can do work. Power is the rate at which work is done. WORKED EXAMPLE Applying the Work\u2013Energy Theorem An ice skater with a mass of 50 kg is gliding across the ice at a speed of 8 m/s when her friend comes up from behind and gives her a push, causing her speed to increase to 12 m/s. How much work did the friend do on the skater? 284 Chapter 9 \u2022 Work, Energy, and Simple Machines Strategy The work\u2013energy theorem can be applied to the problem. Write the equation for the theorem and simplify it if possible. Solution Identify the variables. m= 50 kg, Substitute. 9.1 9.2 Discussion Work done on an object or system increases its energy. In this case, the increase is to the skater\u2019s kinetic energy. It follows that the increase in energy must be the difference in KE before and after the push. TIPS FOR SUCCESS This problem illustrates a general technique for approaching problems that require you to apply formulas: Identify the unknown and the known variables, express the unknown variables in terms of the known variables, and then enter all the known values. Practice Problems 1. How much work is done when a weightlifter lifts a barbell from the floor to a height of? a. b. c. d. 2. Identify which of the following actions generates more power. Show your work. \u2022 \u2022 carrying a carrying a TV to the second floor in or watermelon to the second floor in? a. Carrying a TV generates more power than carrying a watermelon to the same height because power is defined as work done times the time interval. b. Carrying a TV generates", " more power than carrying a watermelon to the same height because power is defined as the ratio of work done to the time interval. c. Carrying a watermelon generates more power than carrying a TV to the same height because power is defined as work done times the time interval. d. Carrying a watermelon generates more power than carrying a TV to the same height because power is defined as the ratio of work done and the time interval. Check Your Understanding 3. Identify two properties that are expressed in units of joules. a. work and force b. energy and weight c. work and energy d. weight and force Access for free at openstax.org. 9.2 \u2022 Mechanical Energy and Conservation of Energy 285 4. When a coconut falls from a tree, work Wis done on it as it falls to the beach. This work is described by the equation 9.3 Identify the quantities F, d, m, v1, and v2 in this event. a. Fis the force of gravity, which is equal to the weight of the coconut, dis the distance the nut falls, mis the mass of the earth, v1 is the initial velocity, and v2 is the velocity with which it hits the beach. b. Fis the force of gravity, which is equal to the weight of the coconut, dis the distance the nut falls, mis the mass of the coconut, v1 is the initial velocity, and v2 is the velocity with which it hits the beach. c. Fis the force of gravity, which is equal to the weight of the coconut, dis the distance the nut falls, mis the mass of the earth, v1 is the velocity with which it hits the beach, and v2 is the initial velocity. d. Fis the force of gravity, which is equal to the weight of the coconut, dis the distance the nut falls, mis the mass of the coconut, v1 is the velocity with which it hits the beach, and v2 is the initial velocity. 9.2 Mechanical Energy and Conservation of Energy Section Learning Objectives By the end of this section, you will be able to do the following: \u2022 Explain the law of conservation of energy in terms of kinetic and potential energy \u2022 Perform calculations related to kinetic and potential energy. Apply the law of conservation of energy Section Key Terms law of conservation of energy Mechanical Energy and Conservation of Energy We saw earlier that mechanical energy can be either potential or kinetic. In this section we will see how energy is transformed from one of these forms to the other. We will also see that, in a closed system, the sum of these forms of energy remains constant. Quite a bit of potential energy is gained by a roller coaster car and its passengers when they are raised to the top of the first hill. Remember that the potentialpart of the term means that energy has been stored and can be used at another time. You will see that this stored energy can either be used to do work or can be transformed into kinetic energy. For example, when an object that has gravitational potential energy falls, its energy is converted to kinetic energy. Remember that both work and energy are expressed in joules. Refer back to. The amount of work required to raise the TV from point A to point B is equal to the amount of gravitational potential energy the TV gains from its height above the ground. This is generally true for any object raised above the ground. If all the work done on an object is used to raise the object above the ground, the amount work equals the object\u2019s gain in gravitational potential energy. However, note that because of the work done by friction, these energy\u2013work transformations are never perfect. Friction causes the loss of some useful energy. In the discussions to follow, we will use the approximation that transformations are frictionless. Now, let\u2019s look at the roller coaster in Figure 9.6. Work was done on the roller coaster to get it to the top of the first rise; at this point, the roller coaster has gravitational potential energy. It is moving slowly, so it also has a small amount of kinetic energy. As the car descends the first slope, its PEis converted to KE. At the low point much of the original PEhas been transformed to KE, and speed is at a maximum. As the car moves up the next slope, some of the KEis transformed back into PEand the car slows down. 286 Chapter 9 \u2022 Work, Energy, and Simple Machines Figure 9.6 During this roller coaster ride, there are conversions between potential and kinetic energy. Virtual Physics Energy Skate Park Basics This simulation shows how kinetic and potential energy are related, in a scenario similar to the roller coaster. Observe the changes in KEand PEby clicking on the bar graph boxes. Also try the three differently shaped skate parks. Drag the skater to the track to start the animation. Click to view content (http://phet.colorado.edu/sims/html/energy-skate", "-park-basics/latest/energy-skate-parkbasics_en.html) GRASP CHECK This simulation (http://phet.colorado.edu/en/simulation/energy-skate-park-basics (http://phet.colorado.edu/en/ simulation/energy-skate-park-basics) ) shows how kinetic and potential energy are related, in a scenario similar to the roller coaster. Observe the changes in KE and PE by clicking on the bar graph boxes. Also try the three differently shaped skate parks. Drag the skater to the track to start the animation. The bar graphs show how KE and PE are transformed back and forth. Which statement best explains what happens to the mechanical energy of the system as speed is increasing? a. The mechanical energy of the system increases, provided there is no loss of energy due to friction. The energy would transform to kinetic energy when the speed is increasing. b. The mechanical energy of the system remains constant provided there is no loss of energy due to friction. The energy would transform to kinetic energy when the speed is increasing. c. The mechanical energy of the system increases provided there is no loss of energy due to friction. The energy would transform to potential energy when the speed is increasing. d. The mechanical energy of the system remains constant provided there is no loss of energy due to friction. The energy would transform to potential energy when the speed is increasing. On an actual roller coaster, there are many ups and downs, and each of these is accompanied by transitions between kinetic and potential energy. Assume that no energy is lost to friction. At any point in the ride, the total mechanical energy is the same, and it is equal to the energy the car had at the top of the first rise. This is a result of the law of conservation of energy, which says that, in a closed system, total energy is conserved\u2014that is, it is constant. Using subscripts 1 and 2 to represent initial and final energy, this law is expressed as Either side equals the total mechanical energy. The phrase in aclosed systemmeans we are assuming no energy is lost to the surroundings due to friction and air resistance. If we are making calculations on dense falling objects, this is a good assumption. For the roller coaster, this assumption introduces some inaccuracy to the calculation. Access for free at openstax.org. 9.2 \u2022 Mechanical Energy and Conservation of Energy 287 Calculations involving Mechanical Energy and Conservation of Energy TIPS FOR SUCCESS When calculating work or energy, use units of meters for distance, newtons for force, kilograms for mass, and seconds for time. This will assure that the result is expressed in joules. WATCH PHYSICS Conservation of Energy This video discusses conversion of PEto KEand conservation of energy. The scenario is very similar to the roller coaster and the skate park. It is also a good explanation of the energy changes studied in the snap lab. Click to view content (https://www.khanacademy.org/embed_video?v=kw_4Loo1HR4) GRASP CHECK Did you expect the speed at the bottom of the slope to be the same as when the object fell straight down? Which statement best explains why this is not exactly the case in real-life situations? a. The speed was the same in the scenario in the animation because the object was sliding on the ice, where there is large amount of friction. In real life, much of the mechanical energy is lost as heat caused by friction. b. The speed was the same in the scenario in the animation because the object was sliding on the ice, where there is small amount of friction. In real life, much of the mechanical energy is lost as heat caused by friction. c. The speed was the same in the scenario in the animation because the object was sliding on the ice, where there is large amount of friction. In real life, no mechanical energy is lost due to conservation of the mechanical energy. d. The speed was the same in the scenario in the animation because the object was sliding on the ice, where there is small amount of friction. In real life, no mechanical energy is lost due to conservation of the mechanical energy. WORKED EXAMPLE Applying the Law of Conservation of Energy A 10 kg rock falls from a 20 m cliff. What is the kinetic and potential energy when the rock has fallen 10 m? Strategy Choose the equation. 9.4 9.5 9.6 9.7 List the knowns. m= 10 kg, v1 = 0, g = 9.80 h1 = 20 m, h2 = 10 m Identify the unknowns. KE2 and PE2 Substitute the known values into the equation and solve for the unknown variables. 288 Chapter 9 \u2022 Work, Energy, and Simple Machines Solution 9.8 9.9 Discussion Alternatively, conservation of energy equation could be solved for v2", " and KE2 could be calculated. Note that mcould also be eliminated. TIPS FOR SUCCESS Note that we can solve many problems involving conversion between KEand PEwithout knowing the mass of the object in question. This is because kinetic and potential energy are both proportional to the mass of the object. In a situation where KE= PE, we know that mgh= (1/2)mv2. Dividing both sides by mand rearranging, we have the relationship 2gh= v2. Practice Problems 5. A child slides down a playground slide. If the slide is 3 m high and the child weighs 300 N, how much potential energy does ) the child have at the top of the slide? (Round gto a. 0 J 100 J b. c. 300 J d. 900 J 6. A 0.2 kg apple on an apple tree has a potential energy of 10 J. It falls to the ground, converting all of its PE to kinetic energy. What is the velocity of the apple just before it hits the ground? a. 0 m/s b. 2 m/s 10 m/s c. 50 m/s d. Snap Lab Converting Potential Energy to Kinetic Energy In this activity, you will calculate the potential energy of an object and predict the object\u2019s speed when all that potential energy has been converted to kinetic energy. You will then check your prediction. You will be dropping objects from a height. Be sure to stay a safe distance from the edge. Don\u2019t lean over the railing too far. Make sure that you do not drop objects into an area where people or vehicles pass by. Make sure that dropping objects will not cause damage. You will need the following: Materials for each pair of students: \u2022 Four marbles (or similar small, dense objects) \u2022 Stopwatch Materials for class: \u2022 Metric measuring tape long enough to measure the chosen height \u2022 A scale Instructions Access for free at openstax.org. 9.2 \u2022 Mechanical Energy and Conservation of Energy 289 Procedure 1. Work with a partner. Find and record the mass of four small, dense objects per group. 2. Choose a location where the objects can be safely dropped from a height of at least 15 meters. A bridge over water with a safe pedestrian walkway will work well. 3. Measure the distance the object will fall. 4. Calculate the potential energy of the object before you drop it using PE= mgh= (9.80)mh. 5. Predict the kinetic energy and velocity of the object when it lands using PE= KEand so, 6. One partner drops the object while the other measures the time it takes to fall. 7. Take turns being the dropper and the timer until you have made four measurements. 8. Average your drop multiplied by and calculate the velocity of the object when it landed using v = at= gt= (9.80)t. 9. Compare your results to your prediction. GRASP CHECK Galileo\u2019s experiments proved that, contrary to popular belief, heavy objects do not fall faster than light objects. How do the equations you used support this fact? a. Heavy objects do not fall faster than the light objects because while conserving the mechanical energy of the system, the mass term gets cancelled and the velocity is independent of the mass. In real life, the variation in the velocity of the different objects is observed because of the non-zero air resistance. b. Heavy objects do not fall faster than the light objects because while conserving the mechanical energy of the system, the mass term does not get cancelled and the velocity is dependent on the mass. In real life, the variation in the velocity of the different objects is observed because of the non-zero air resistance. c. Heavy objects do not fall faster than the light objects because while conserving the mechanical energy the system, the mass term gets cancelled and the velocity is independent of the mass. In real life, the variation in the velocity of the different objects is observed because of zero air resistance. d. Heavy objects do not fall faster than the light objects because while conserving the mechanical energy of the system, the mass term does not get cancelled and the velocity is dependent on the mass. In real life, the variation in the velocity of the different objects is observed because of zero air resistance. Check Your Understanding 7. Describe the transformation between forms of mechanical energy that is happening to a falling skydiver before his parachute opens. a. Kinetic energy is being transformed into potential energy. b. Potential energy is being transformed into kinetic energy. c. Work is being transformed into kinetic energy. d. Kinetic energy is being transformed into work. 8. True or false\u2014If a rock is thrown into the air, the increase in the height would increase the rock\u2019s kinetic energy, and then the increase in the velocity as it falls to the ground would increase its potential energy. a. True b. False 9. Identify", " equivalent terms for stored energyand energy of motion. a. Stored energy is potential energy, and energy of motion is kinetic energy. b. Energy of motion is potential energy, and stored energy is kinetic energy. c. Stored energy is the potential as well as the kinetic energy of the system. d. Energy of motion is the potential as well as the kinetic energy of the system. 290 Chapter 9 \u2022 Work, Energy, and Simple Machines 9.3 Simple Machines Section Learning Objectives By the end of this section, you will be able to do the following: \u2022 Describe simple and complex machines \u2022 Calculate mechanical advantage and efficiency of simple and complex machines Section Key Terms complex machine efficiency output ideal mechanical advantage inclined plane input work lever mechanical advantage output work pulley screw simple machine wedge wheel and axle Simple Machines Simple machines make work easier, but they do not decrease the amount of work you have to do. Why can\u2019t simple machines change the amount of work that you do? Recall that in closed systems the total amount of energy is conserved. A machine cannot increase the amount of energy you put into it. So, why is a simple machine useful? Although it cannot change the amount of work you do, a simple machine can change the amount of force you must apply to an object, and the distance over which you apply the force. In most cases, a simple machine is used to reduce the amount of force you must exert to do work. The down side is that you must exert the force over a greater distance, because the product of force and distance, fd, (which equals work) does not change. Let\u2019s examine how this works in practice. In Figure 9.7(a), the worker uses a type of lever to exert a small force over a large distance, while the pry bar pulls up on the nail with a large force over a small distance. Figure 9.7(b) shows the how a lever works mathematically. The effort force, applied at Fe, lifts the load (the resistance force) which is pushing down at Fr. The triangular pivot is called the fulcrum; the part of the lever between the fulcrum and Feis the effort arm, Le; and the part to the left is the resistance arm, Lr. The mechanical advantage is a number that tells us how many times a simple machine multiplies the effort force. The ideal mechanical advantage, IMA, is the mechanical advantage of a perfect machine with no loss of useful work caused by friction between moving parts. The equation for IMAis shown in Figure 9.7(b). Figure 9.7 (a) A pry bar is a type of lever. (b) The ideal mechanical advantage equals the length of the effort arm divided by the length of the resistance arm of a lever. In general, the IMA= the resistance force, Fr, divided by the effort force, Fe. IMAalso equals the distance over which the effort is applied, de, divided by the distance the load travels, dr. Getting back to conservation of energy, for any simple machine, the work put into the machine, Wi, equals the work the machine puts out, Wo. Combining this with the information in the paragraphs above, we can write Access for free at openstax.org. 9.3 \u2022 Simple Machines 291 The equations show how a simple machine can output the same amount of work while reducing the amount of effort force by increasing the distance over which the effort force is applied. WATCH PHYSICS Introduction to Mechanical Advantage This video shows how to calculate the IMAof a lever by three different methods: (1) from effort force and resistance force; (2) from the lengths of the lever arms, and; (3) from the distance over which the force is applied and the distance the load moves. Click to view content (https://www.youtube.com/embed/pfzJ-z5Ij48) GRASP CHECK Two children of different weights are riding a seesaw. How do they position themselves with respect to the pivot point (the fulcrum) so that they are balanced? a. The heavier child sits closer to the fulcrum. b. The heavier child sits farther from the fulcrum. c. Both children sit at equal distance from the fulcrum. d. Since both have different weights, they will never be in balance. Some levers exert a large force to a short effort arm. This results in a smaller force acting over a greater distance at the end of the resistance arm. Examples of this type of lever are baseball bats, hammers, and golf clubs. In another type of lever, the fulcrum is at the end of the lever and the load is in the middle, as in the design of a wheelbarrow. The simple machine shown in Figure 9.8 is called a wheel and axle. It is actually a form of lever. The difference is that the effort arm can rotate in a complete circle", " around the fulcrum, which is the center of the axle. Force applied to the outside of the wheel causes a greater force to be applied to the rope that is wrapped around the axle. As shown in the figure, the ideal mechanical advantage is calculated by dividing the radius of the wheel by the radius of the axle. Any crank-operated device is an example of a wheel and axle. Figure 9.8 Force applied to a wheel exerts a force on its axle. An inclined plane and a wedge are two forms of the same simple machine. A wedge is simply two inclined planes back to back. Figure 9.9 shows the simple formulas for calculating the IMAs of these machines. All sloping, paved surfaces for walking or driving are inclined planes. Knives and axe heads are examples of wedges. 292 Chapter 9 \u2022 Work, Energy, and Simple Machines Figure 9.9 An inclined plane is shown on the left, and a wedge is shown on the right. The screw shown in Figure 9.10 is actually a lever attached to a circular inclined plane. Wood screws (of course) are also examples of screws. The lever part of these screws is a screw driver. In the formula for IMA, the distance between screw threads is called pitchand has the symbol P. Figure 9.10 The screw shown here is used to lift very heavy objects, like the corner of a car or a house a short distance. Figure 9.11 shows three different pulley systems. Of all simple machines, mechanical advantage is easiest to calculate for pulleys. Simply count the number of ropes supporting the load. That is the IMA. Once again we have to exert force over a longer distance to multiply force. To raise a load 1 meter with a pulley system you have to pull Nmeters of rope. Pulley systems are often used to raise flags and window blinds and are part of the mechanism of construction cranes. Figure 9.11 Three pulley systems are shown here. Access for free at openstax.org. 9.3 \u2022 Simple Machines 293 WATCH PHYSICS Mechanical Advantage of Inclined Planes and Pulleys The first part of this video shows how to calculate the IMAof pulley systems. The last part shows how to calculate the IMAof an inclined plane. Click to view content (https://www.khanacademy.org/embed_video?v=vSsK7Rfa3yA) GRASP CHECK How could you use a pulley system to lift a light load to great height? a. Reduce the radius of the pulley. b. Increase the number of pulleys. c. Decrease the number of ropes supporting the load. Increase the number of ropes supporting the load. d. A complex machine is a combination of two or more simple machines. The wire cutters in Figure 9.12 combine two levers and two wedges. Bicycles include wheel and axles, levers, screws, and pulleys. Cars and other vehicles are combinations of many machines. Figure 9.12 Wire cutters are a common complex machine. Calculating Mechanical Advantage and Efficiency of Simple Machines In general, the IMA= the resistance force, Fr, divided by the effort force, Fe. IMAalso equals the distance over which the effort is applied, de, divided by the distance the load travels, dr. Refer back to the discussions of each simple machine for the specific equations for the IMAfor each type of machine. No simple or complex machines have the actual mechanical advantages calculated by the IMAequations. In real life, some of the applied work always ends up as wasted heat due to friction between moving parts. Both the input work (Wi) and output work (Wo) are the result of a force, F, acting over a distance, d. The efficiency output of a machine is simply the output work divided by the input work, and is usually multiplied by 100 so that it is expressed as a percent. Look back at the pictures of the simple machines and think about which would have the highest efficiency. Efficiency is related to friction, and friction depends on the smoothness of surfaces and on the area of the surfaces in contact. How would lubrication affect the efficiency of a simple machine? WORKED EXAMPLE Efficiency of a Lever The input force of 11 N acting on the effort arm of a lever moves 0.4 m, which lifts a 40 N weight resting on the resistance arm a 294 Chapter 9 \u2022 Work, Energy, and Simple Machines distance of 0.1 m. What is the efficiency of the machine? Strategy State the equation for efficiency of a simple machine, are the product Fd. Solution = (11)(0.4) = 4.4 J and = (40)(0.1) = 4.0 J, then and calculate Woand Wi. Both work values Discussion Efficiency in real machines will always be less than 100 percent because of work that is converted to unavailable heat by friction and air resistance. Wo", "and Wican always be calculated as a force multiplied by a distance, although these quantities are not always as obvious as they are in the case of a lever. Practice Problems 10. What is the IMA of an inclined plane that is long and high? a. b. c. d. 11. If a pulley system can lift a 200N load with an effort force of 52 N and has an efficiency of almost 100 percent, how many ropes are supporting the load? 1 rope is required because the actual mechanical advantage is 0.26. a. b. 1 rope is required because the actual mechanical advantage is 3.80. c. 4 ropes are required because the actual mechanical advantage is 0.26. d. 4 ropes are required because the actual mechanical advantage is 3.80. Check Your Understanding 12. True or false\u2014The efficiency of a simple machine is always less than 100 percent because some small fraction of the input work is always converted to heat energy due to friction. a. True b. False 13. The circular handle of a faucet is attached to a rod that opens and closes a valve when the handle is turned. If the rod has a diameter of 1 cm and the IMA of the machine is 6, what is the radius of the handle? A. 0.08 cm B. 0.17 cm C. 3.0 cm D. 6.0 cm Access for free at openstax.org. Chapter 9 \u2022 Key Terms 295 KEY TERMS complex machine a machine that combines two or more output work output force multiplied by the distance over simple machines which it acts efficiency output work divided by input work energy the ability to do work gravitational potential energy energy acquired by doing potential energy stored energy power pulley a simple machine consisting of a rope that passes the rate at which work is done work against gravity ideal mechanical advantage the mechanical advantage of an idealized machine that loses no energy to friction inclined plane a simple machine consisting of a slope input work effort force multiplied by the distance over which it is applied over one or more grooved wheels screw a simple machine consisting of a spiral inclined plane simple machine a machine that makes work easier by changing the amount or direction of force required to move an object joule the metric unit for work and energy; equal to 1 watt the metric unit of power; equivalent to joules per newton meter (N\u2219m) second kinetic energy energy of motion law of conservation of energy states that energy is neither wedge a simple machine consisting of two back-to-back inclined planes created nor destroyed wheel and axle a simple machine consisting of a rod fixed lever a simple machine consisting of a rigid arm that pivots to the center of a wheel on a fulcrum mechanical advantage the number of times the input force is multiplied mechanical energy kinetic or potential energy work force multiplied by distance work\u2013energy theorem states that the net work done on a system equals the change in kinetic energy SECTION SUMMARY 9.1 Work, Power, and the Work\u2013Energy Theorem \u2022 Doing work on a system or object changes its energy. \u2022 The work\u2013energy theorem states that an amount of work that changes the velocity of an object is equal to the change in kinetic energy of that object.The work\u2013energy theorem states that an amount of work that changes the velocity of an object is equal to the change in kinetic energy of that object. \u2022 Power is the rate at which work is done. 9.2 Mechanical Energy and Conservation of Energy \u2022 Mechanical energy may be either kinetic (energy of KEY EQUATIONS 9.1 Work, Power, and the Work\u2013Energy Theorem equation for work force work equivalencies motion) or potential (stored energy). \u2022 Doing work on an object or system changes its energy. \u2022 Total energy in a closed, isolated system is constant. 9.3 Simple Machines \u2022 The six types of simple machines make work easier by changing the fdterm so that force is reduced at the expense of increased distance. \u2022 The ratio of output force to input force is a machine\u2019s mechanical advantage. \u2022 Combinations of two or more simple machines are called complex machines. \u2022 The ratio of output work to input work is a machine\u2019s efficiency. kinetic energy work\u2013energy theorem power 296 Chapter 9 \u2022 Chapter Review 9.2 Mechanical Energy and Conservation of Energy conservation of energy 9.3 Simple Machines ideal mechanical advantage (general) ideal mechanical advantage (lever) ideal mechanical advantage (wheel and axle) CHAPTER REVIEW Concept Items 9.1 Work, Power, and the Work\u2013Energy Theorem 1. Is it possible for the sum of kinetic energy and potential energy of an object to change without work having been done on the object? Explain. a. No, because the work-energy theorem states that work done on an object is equal to the change in kinetic energy, and change in KE requires a change in velocity. It is assumed that mass is constant. b. No, because the work-energy theorem states that work done on an object is equal to the sum of kinetic energy,", " and the change in KE requires a change in displacement. It is assumed that mass is constant. c. Yes, because the work-energy theorem states that work done on an object is equal to the change in kinetic energy, and change in KE requires a change in velocity. It is assumed that mass is constant. d. Yes, because the work-energy theorem states that work done on an object is equal to the sum of kinetic energy, and the change in KE requires a change in displacement. It is assumed that mass is constant. 2. Define work for one-dimensional motion. a. Work is defined as the ratio of the force over the distance. b. Work is defined as the sum of the force and the distance. c. Work is defined as the square of the force over the Access for free at openstax.org. ideal mechanical advantage (inclined plane) ideal mechanical advantage (wedge) ideal mechanical advantage (pulley) ideal mechanical advantage (screw) input work output work efficiency output distance. d. Work is defined as the product of the force and the distance. 3. A book with a mass of 0.30 kg falls 2 m from a shelf to the floor. This event is described by the work\u2013energy theorem: Explain why this is enough information to calculate the speed with which the book hits the floor. a. The mass of the book, m, and distance, d, are stated. Fis the weight of the book mg. v1 is the initial velocity and v2 is the final velocity. The final velocity is the only unknown quantity. b. The mass of the book, m, and distance, d, are stated. Fis the weight of the book mg. v1 is the final velocity and v2 is the initial velocity. The final velocity is the only unknown quantity. c. The mass of the book, m, and distance, d, are stated. Fis the weight of the book mg. v1 is the initial velocity and v2 is the final velocity. The final velocity and the initial velocities are the only unknown quantities. d. The mass of the book, m, and distance, d, are stated. Fis the weight of the book mg. v1 is the final velocity and v2 is the initial velocity. The final velocity and the initial velocities are the only unknown quantities. Chapter 9 \u2022 Chapter Review 297 9.2 Mechanical Energy and Conservation of Energy 4. Describe the changes in KE and PE of a person jumping up and down on a trampoline. a. While going up, the person\u2019s KE would change to PE. While coming down, the person\u2019s PE would change to KE. b. While going up, the person\u2019s PE would change to KE. While coming down, the person\u2019s KE would change to PE. c. While going up, the person\u2019s KE would not change, but while coming down, the person\u2019s PE would change to KE. d. While going up, the person\u2019s PE would change to KE, but while coming down, the person\u2019s KE would not change. 6. The starting line of a cross country foot race is at the bottom of a hill. Which form(s) of mechanical energy of the runners will change when the starting gun is fired? a. Kinetic energy only b. Potential energy only c. Both kinetic and potential energy d. Neither kinetic nor potential energy 9.3 Simple Machines 7. How does a simple machine make work easier? a. b. c. d. It reduces the input force and the output force. It reduces the input force and increases the output force. It increases the input force and reduces the output force. It increases the input force and the output force. 5. You know the height from which an object is dropped. 8. Which type of simple machine is a knife? Which equation could you use to calculate the velocity as the object hits the ground? a. b. c. d. a. A ramp b. A wedge c. A pulley d. A screw Critical Thinking Items 9.1 Work, Power, and the Work\u2013Energy Theorem 9. Which activity requires a person to exert force on an object that causes the object to move but does not change the kinetic or potential energy of the object? a. Moving an object to a greater height with acceleration b. Moving an object to a greater height without acceleration c. Carrying an object with acceleration at the same height 9.2 Mechanical Energy and Conservation of Energy 11. True or false\u2014A cyclist coasts down one hill and up another hill until she comes to a stop. The point at which the bicycle stops is lower than the point at which it started coasting because part of the original potential energy has been converted to a quantity of heat and this makes the tires of the bicycle warm. a. True b. False 9.3 Simple Machines d. Carrying", " an object without acceleration at the same 12. We think of levers being used to decrease effort force. height 10. Which statement explains how it is possible to carry books to school without changing the kinetic or potential energy of the books or doing any work? a. By moving the book without acceleration and keeping the height of the book constant b. By moving the book with acceleration and keeping the height of the book constant c. By moving the book without acceleration and changing the height of the book d. By moving the book with acceleration and changing the height of the book Which of the following describes a lever that requires a large effort force which causes a smaller force to act over a large distance and explains how it works? a. Anything that is swung by a handle, such as a hammer or racket. Force is applied near the fulcrum over a short distance, which makes the other end move rapidly over a long distance. b. Anything that is swung by a handle, such as a hammer or racket. Force is applied far from the fulcrum over a large distance, which makes the other end move rapidly over a long distance. c. A lever used to lift a heavy stone. Force is applied near the fulcrum over a short distance, which 298 Chapter 9 \u2022 Chapter Review makes the other end lift a heavy object easily. d. A lever used to lift a heavy stone. Force is applied far from the fulcrum over a large distance, which makes the other end lift a heavy object easily 13. A baseball bat is a lever. Which of the following explains how a baseball bat differs from a lever like a pry bar? a. In a baseball bat, effort force is smaller and is applied over a large distance, while the resistance force is smaller and is applied over a long distance. Problems 9.1 Work, Power, and the Work\u2013Energy Theorem 14. A baseball player exerts a force of on a ball for a as he throws it. If the ball has a mass, what is its velocity as it leaves his hand? distance of of a. b. c. d. 15. A boy pushes his little sister on a sled. The sled accelerates from 0 to 3.2 m/s. If the combined mass of his sister and the sled is 40.0 kg and 18 W of power were generated, how long did the boy push the sled? a. 205 s b. 128 s c. 23 s 11 s d. 9.2 Mechanical Energy and Conservation of Energy 16. What is the kinetic energy of a bullet traveling? at a velocity of a. b. c. d. 17. A marble rolling across a flat, hard surface at rolls and no energy is up a ramp. Assuming that lost to friction, what will be the vertical height of the marble when it comes to a stop before rolling back down? Ignore effects due to the rotational kinetic energy. a. b. c. d. 18. The potential energy stored in a compressed spring is Access for free at openstax.org. b. c. d. In a baseball bat, effort force is smaller and is applied over a large distance, while the resistance force is smaller and is applied over a short distance. In a baseball bat, effort force is larger and is applied over a short distance, while the resistance force is smaller and is applied over a long distance. In a baseball bat, effort force is larger and is applied over a short distance, while the resistance force is smaller and is applied over a short distance., where kis the force constant and xis the distance the spring is compressed from the equilibrium position. Four experimental setups described below can be used to determine the force constant of a spring. Which one(s) require measurement of the fewest number of variables to determine k? Assume the acceleration due to gravity is known. I. An object is propelled vertically by a compressed spring. II. An object is propelled horizontally on a frictionless surface by a compressed spring. III. An object is statically suspended from a spring. IV. An object suspended from a spring is set into oscillatory motion. a. b. c. d. I only III only I and II only III and IV only 9.3 Simple Machines 19. A man is using a wedge to split a block of wood by hitting the wedge with a hammer. This drives the wedge into the wood creating a crack in the wood. When he hits the wedge with a force of 400 N it travels 4 cm into the wood. This caused the wedge to exert a force of 1,400 N sideways increasing the width of the crack by 1 cm. What is the efficiency of the wedge? a. 0.875 percent b. 0.14 c. 0.751 d. 87.5 percent 20. An electrician grips the handles of a wire cutter, like the one shown, 10 cm from the pivot and places a wire between the jaws 2 cm from the pivot. If the cutter blades are 2 cm wide and 0.", "3 cm thick, what is the overall IMA of this complex machine? Performance Task 9.3 Simple Machines 21. Conservation of Energy and Energy Transfer; Cause and Effect; and S&EP, Planning and Carrying Out Investigations Plan an investigation to measure the mechanical advantage of simple machines and compare to the IMA of the machine. Also measure the efficiency of each machine studied. Design an investigation to make these measurements for these simple machines: lever, inclined plane, wheel and axle and a pulley system. In addition to these machines, include a spring scale, a tape measure, and a weight with a loop on top that can be attached to the hook on the spring scale. A spring scale is shown in the image. Chapter 9 \u2022 Test Prep 299 a. b. c. d. 1.34 1.53 33.3 33.5 LEVER: Beginning with the lever, explain how you would measure input force, output force, effort arm, and resistance arm. Also explain how you would find the distance the load travels and the distance over which the effort force is applied. Explain how you would use this data to determine IMAand efficiency. INCLINED PLANE: Make measurements to determine IMAand efficiency of an inclined plane. Explain how you would use the data to calculate these values. Which property do you already know? Note that there are no effort and resistance arm measurements, but there are height and length measurements. WHEEL AND AXLE: Again, you will need two force measurements and four distance measurements. Explain how you would use these to calculate IMAand efficiency. SCREW: You will need two force measurements, two distance traveled measurements, and two length measurements. You may describe a screw like the one shown in Figure 9.10 or you could use a screw and screw driver. (Measurements would be easier for the former). Explain how you would use these to calculate IMAand efficiency. PULLEY SYSTEM: Explain how you would determine the IMAand efficiency of the four-pulley system shown in Figure 9.11. Why do you only need two distance measurements for this machine? Design a table that compares the efficiency of the five simple machines. Make predictions as to the most and least efficient machines. A spring scale measures weight, not mass. TEST PREP Multiple Choice 9.1 Work, Power, and the Work\u2013Energy Theorem 22. Which expression represents power? a. b. c. d. 23. The work\u2013energy theorem states that the change in the kinetic energy of an object is equal to what? a. The work done on the object b. The force applied to the object c. The loss of the object\u2019s potential energy d. The object\u2019s total mechanical energy minus its kinetic energy 24. A runner at the start of a race generates 250 W of power as he accelerates to 5 m/s. If the runner has a mass of 60 kg, how long did it take him to reach that speed? a. 0.33 s b. 0.83 s c. d. 1.2 s 3.0 s 300 Chapter 9 \u2022 Test Prep 25. A car\u2019s engine generates 100,000 W of power as it exerts a force of 10,000 N. How long does it take the car to travel 100 m? a. 0.001 s b. 0.01 s 10 s c. 1,000 s d. 9.2 Mechanical Energy and Conservation of Energy 26. Why is this expression for kinetic energy incorrect? is missing.. a. The constant b. The term should not be squared. c. The expression should be divided by. d. The energy lost to friction has not been subtracted. 27. What is the kinetic energy of a object moving at? a. b. c. d. 28. Which statement best describes the PE-KE transformations for a javelin, starting from the instant the javelin leaves the thrower's hand until it hits the ground. a. Initial PE is transformed to KE until the javelin reaches the high point of its arc. On the way back down, KE is transformed into PE. At every point in the flight, mechanical energy is being transformed into heat energy. Initial KE is transformed to PE until the javelin reaches the high point of its arc. On the way back down, PE is transformed into KE. At every point in the flight, mechanical energy is being transformed into heat energy. Initial PE is transformed to KE until the javelin reaches the high point of its arc. On the way back down, there is no transformation of mechanical energy. At every point in the flight, mechanical energy is being transformed into heat energy. Initial KE is transformed to PE until the javelin reaches the high point of its arc. On the way back down, there is no transformation of mechanical energy. At every point in the flight, mechanical energy is being transformed into heat energy. b. c. d. 29.", " At the beginning of a roller coaster ride, the roller coaster car has an initial energy mostly in the form of PE. Which statement explains why the fastest speeds of the car will be at the lowest points in the ride? a. At the bottom of the slope kinetic energy is at its Access for free at openstax.org. maximum value and potential energy is at its minimum value. b. At the bottom of the slope potential energy is at its maximum value and kinetic energy is at its minimum value. c. At the bottom of the slope both kinetic and potential energy reach their maximum values d. At the bottom of the slope both kinetic and potential energy reach their minimum values. 9.3 Simple Machines 30. A large radius divided by a small radius is the expression used to calculate the IMA of what? a. A screw b. A pulley c. A wheel and axle d. An inclined plane. 31. What is the IMA of a wedge that is long and thick? a. b. c. d. 32. Which statement correctly describes the simple machines, like the crank in the image, that make up an Archimedes screw and the forces it applies? a. The crank is a wedge in which the IMA is the length of the tube divided by the radius of the tube. The applied force is the effort force and the weight of the water is the resistance force. b. The crank is an inclined plane in which the IMA is the length of the tube divided by the radius of the tube. The applied force is the effort force and the weight of the water is the resistance force. c. The crank is a wheel and axle. The effort force of the crank becomes the resistance force of the screw. d. The crank is a wheel and axle. The resistance force of the crank becomes the effort force of the screw. 33. Refer to the pulley system on right in the image. Assume this pulley system is an ideal machine. How hard would you have to pull on the rope to lift a 120 N Chapter 9 \u2022 Test Prep 301 load? How many meters of rope would you have to pull out of the system to lift the load 1 m? a. 480 N 4 m b. 480 N c. d. m 30 N 4 m 30 N m Short Answer 9.1 Work, Power, and the Work\u2013Energy Theorem 34. Describe two ways in which doing work on an object can increase its mechanical energy. a. Raising an object to a higher elevation does work as it increases its PE; increasing the speed of an object does work as it increases its KE. b. Raising an object to a higher elevation does work as it increases its KE; increasing the speed of an object does work as it increases its PE. c. Raising an object to a higher elevation does work as it increases its PE; decreasing the speed of an object does work as it increases its KE. d. Raising an object to a higher elevation does work as it increases its KE; decreasing the speed of an object does work as it increases its PE. 35. True or false\u2014While riding a bicycle up a gentle hill, it is fairly easy to increase your potential energy, but to increase your kinetic energy would make you feel exhausted. a. True b. False 36. Which statement best explains why running on a track with constant speed at 3 m/s is not work, but climbing a mountain at 1 m/s is work? a. At constant speed, change in the kinetic energy is zero but climbing a mountain produces change in the potential energy. b. At constant speed, change in the potential energy is zero, but climbing a mountain produces change in the kinetic energy. c. At constant speed, change in the kinetic energy is finite, but climbing a mountain produces no change in the potential energy. d. At constant speed, change in the potential energy is finite, but climbing a mountain produces no change in the kinetic energy. 37. You start at the top of a hill on a bicycle and coast to the bottom without applying the brakes. By the time you reach the bottom of the hill, work has been done on you and your bicycle, according to the equation: If is the mass of you and your bike, what are a. and? is your speed at the top of the hill, and is your speed at the bottom. b. c. d. is your speed at the bottom of the hill, and is your speed at the top. is your displacement at the top of the hill, and is your displacement at the bottom. is your displacement at the bottom of the hill, and is your displacement at the top. 9.2 Mechanical Energy and Conservation of Energy 38. True or false\u2014The formula for gravitational potential energy can be used to explain why joules, J, are equivalent to kg \u00d7 mg2 / s2. Show your work. a. True b. False 39. Which statement best explains why accelerating a car from a. Because", " kinetic energy is directly proportional to quadruples its kinetic energy? to the square of the velocity. b. Because kinetic energy is inversely proportional to the square of the velocity. c. Because kinetic energy is directly proportional to the fourth power of the velocity. d. Because kinetic energy is inversely proportional to 302 Chapter 9 \u2022 Test Prep the fourth power of the velocity. plane. 40. A coin falling through a vacuum loses no energy to friction, and yet, after it hits the ground, it has lost all its potential and kinetic energy. Which statement best explains why the law of conservation of energy is still valid in this case? a. When the coin hits the ground, the ground gains potential energy that quickly changes to thermal energy. b. When the coin hits the ground, the ground gains kinetic energy that quickly changes to thermal energy. c. When the coin hits the ground, the ground gains thermal energy that quickly changes to kinetic energy. d. When the coin hits the ground, the ground gains thermal energy that quickly changes to potential energy. 41. True or false\u2014A marble rolls down a slope from height h1 and up another slope to height h2, where (h2 < h1). The difference mg(h1 \u2013 h2) is equal to the heat lost due to the friction. a. True b. False 9.3 Simple Machines 42. Why would you expect the lever shown in the top image to have a greater efficiency than the inclined plane shown in the bottom image? b. The effort arm is shorter in case of the inclined plane. c. The area of contact is greater in case of the inclined plane. 43. Why is the wheel on a wheelbarrow not a simple machine in the same sense as the simple machine in the image? a. The wheel on the wheelbarrow has no fulcrum. b. The center of the axle is not the fulcrum for the wheels of a wheelbarrow. c. The wheelbarrow differs in the way in which load is attached to the axle. d. The wheelbarrow has less resistance force than a wheel and axle design. 44. A worker pulls down on one end of the rope of a pulley system with a force of 75 N to raise a hay bale tied to the other end of the rope. If she pulls the rope down 2.0 m and the bale raises 1.0 m, what else would you have to know to calculate the efficiency of the pulley system? a. b. c. d. the weight of the worker the weight of the hay bale the radius of the pulley the height of the pulley from ground 45. True or false\u2014A boy pushed a box with a weight of 300 N up a ramp. He said that, because the ramp was 1.0 m high and 3.0 m long, he must have been pushing with force of exactly 100 N. a. True b. False a. The resistance arm is shorter in case of the inclined Extended Response 9.1 Work, Power, and the Work\u2013Energy Theorem 46. Work can be negative as well as positive because an object or system can do work on its surroundings as well as have work done on it. Which of the following statements describes: a situation in which an object does work on its surroundings by decreasing its velocity and a situation in which an object can do work on its surroundings by decreasing its altitude? a. A gasoline engine burns less fuel at a slower speed. Solar cells capture sunlight to generate electricity. Access for free at openstax.org. Chapter 9 \u2022 Test Prep 303 b. A hybrid car charges its batteries as it decelerates. Falling water turns a turbine to generate electricity. c. Airplane flaps use air resistance to slow down for landing. Rising steam turns a turbine to generate electricity. d. An electric train requires less electrical energy as it decelerates. A parachute captures air to slow a skydiver\u2019s fall. 47. A boy is pulling a girl in a child\u2019s wagon at a constant speed. He begins to pull harder, which increases the speed of the wagon. Which of the following describes two ways you could calculate the change in energy of the wagon and girl if you had all the information you needed? a. Calculate work done from the force and the velocity. Calculate work done from the change in the potential energy of the system. b. Calculate work done from the force and the displacement. Calculate work done from the change in the potential energy of the system. c. Calculate work done from the force and the velocity. Calculate work done from the change in the kinetic energy of the system. be 12 m/s. Due to the lack of air friction, there would be complete transformation of the potential energy into the kinetic energy as the rock hits the moon\u2019s surface. d. The velocity of the rock as it hits the ground would be 12 m/s. Due to the lack of air friction", ", there would be incomplete transformation of the potential energy into the kinetic energy as the rock hits the moon\u2019s surface. 49. A boulder rolls from the top of a mountain, travels across a valley below, and rolls part way up the ridge on the opposite side. Describe all the energy transformations taking place during these events and identify when they happen. a. As the boulder rolls down the mountainside, KE is converted to PE. As the boulder rolls up the opposite slope, PE is converted to KE. The boulder rolls only partway up the ridge because some of the PE has been converted to thermal energy due to friction. b. As the boulder rolls down the mountainside, KE is converted to PE. As the boulder rolls up the opposite slope, KE is converted to PE. The boulder rolls only partway up the ridge because some of the PE has been converted to thermal energy due to friction. d. Calculate work done from the force and the c. As the boulder rolls down the mountainside, PE is displacement. Calculate work done from the change in the kinetic energy of the system. 9.2 Mechanical Energy and Conservation of Energy 48. Acceleration due to gravity on the moon is 1.6 m/s2 or about 16% of the value of gon Earth. If an astronaut on the moon threw a moon rock to a height of 7.8 m, what would be its velocity as it struck the moon\u2019s surface? How would the fact that the moon has no atmosphere affect the velocity of the falling moon rock? Explain your answer. a. The velocity of the rock as it hits the ground would be 5.0 m/s. Due to the lack of air friction, there would be complete transformation of the potential energy into the kinetic energy as the rock hits the moon\u2019s surface. b. The velocity of the rock as it hits the ground would be 5.0 m/s. Due to the lack of air friction, there would be incomplete transformation of the potential energy into the kinetic energy as the rock hits the moon\u2019s surface. c. The velocity of the rock as it hits the ground would converted to KE. As the boulder rolls up the opposite slope, PE is converted to KE. The boulder rolls only partway up the ridge because some of the PE has been converted to thermal energy due to friction. d. As the boulder rolls down the mountainside, PE is converted to KE. As the boulder rolls up the opposite slope, KE is converted to PE. The boulder rolls only partway up the ridge because some of the PE has been converted to thermal energy due to friction. 9.3 Simple Machines 50. To dig a hole, one holds the handles together and thrusts the blades of a posthole digger, like the one in the image, into the ground. Next, the handles are pulled apart, which squeezes the dirt between them, making it possible to remove the dirt from the hole. This complex machine is composed of two pairs of two different simple machines. Identify and describe the parts that are simple machines and explain how you would find the IMA of each type of simple machine. a. Each handle and its attached blade is a lever with the 304 Chapter 9 \u2022 Test Prep fulcrum at the hinge. Each blade is a wedge. The IMA of a lever would be the length of the handle divided by the length of the blade. The IMA of the wedges would be the length of the blade divided by its width. b. Each handle and its attached to blade is a lever with the fulcrum at the end. Each blade is a wedge. The IMA of a lever would be the length of the handle divided by the length of the blade. The IMA of the wedges would be the length of the blade divided by its width. c. Each handle and its attached blade is a lever with the fulcrum at the hinge. Each blade is a wedge. The IMA of a lever would be the length of the handle multiplied by the length of the blade. The IMA of the wedges would be the length of the blade multiplied by its width. d. Each handle and its attached blade is a lever with the fulcrum at the end. Each blade is a wedge. The IMA of a lever would be the length of the handle multiplied by the length of the blade. The IMA of the wedges would be the length of the blade multiplied by its width. 51. A wooden crate is pulled up a ramp that is 1.0 m high and 6.0 m long. The crate is attached to a rope that is wound around an axle with a radius of 0.020 m. The axle is turned by a 0.20 m long handle. What is the overall IMA of the complex machine? A. 6 B. 10 16 C. D. 60 Access for free at openstax.org. CHAPTER 10 Special Relativity Figure 10.1 Special relativity explains why", " travel to other star systems, such as these in the Orion Nebula, is unlikely using our current level of technology. (s58y, Flickr) Chapter Outline 10.1 Postulates of Special Relativity 10.2 Consequences of Special Relativity Have you ever dreamed of traveling to other planets in faraway star systems? The trip might seem possible by INTRODUCTION traveling fast enough, but you will read in this chapter why it is not. In 1905, Albert Einstein developed the theory of special relativity. Einstein developed the theory to help explain inconsistencies between the equations describing electromagnetism and Newtonian mechanics, and to explain why the ether did not exist. This theory explains the limit on an object\u2019s speed among other implications. Relativity is the study of how different observers moving with respect to one another measure the same events. Galileo and Newton developed the first correct version of classical relativity. Einstein developed the modern theory of relativity. Modern relativity is divided into two parts. Special relativity deals with observers moving at constant velocity. General relativity deals with observers moving at constant acceleration. Einstein\u2019s theories of relativity made revolutionary predictions. Most importantly, his predictions have been verified by experiments. In this chapter, you learn how experiments and puzzling contradictions in existing theories led to the development of the theory of special relativity. You will also learn the simple postulates on which the theory was based; a postulate is a statement that is assumed to be true for the purposes of reasoning in a scientific or mathematic argument. 10.1 Postulates of Special Relativity Section Learning Objectives By the end of this section, you will be able to do the following: \u2022 Describe the experiments and scientific problems that led Albert Einstein to develop the special theory of relativity \u2022 Understand the postulates on which the special theory of relativity was based 306 Chapter 10 \u2022 Special Relativity Section Key Terms ether frame of reference inertial reference frame general relativity postulate relativity simultaneity special relativity Scientific Experiments and Problems Relativity is not new. Way back around the year 1600, Galileo explained that motion is relative. Wherever you happen to be, it seems like you are at a fixed point and that everything moves with respect to you. Everyone else feels the same way. Motion is always measured with respect to a fixed point. This is called establishing a frame of reference. But the choice of the point is arbitrary, and all frames of reference are equally valid. A passenger in a moving car is not moving with respect to the driver, but they are both moving from the point of view of a person on the sidewalk waiting for a bus. They are moving even faster as seen by a person in a car coming toward them. It is all relative. TIPS FOR SUCCESS A frame of reference is not a complicated concept. It is just something you decide is a fixed point or group of connected points. It is completely up to you. For example, when you look up at celestial objects in the sky, you choose the earth as your frame of reference, and the sun, moon, etc., seem to move across the sky. Light is involved in the discussion of relativity because theories related to electromagnetism are inconsistent with Galileo\u2019s and Newton\u2019s explanation of relativity. The true nature of light was a hot topic of discussion and controversy in the late 19th century. At the time, it was not generally believed that light could travel across empty space. It was known to travel as waves, and all other types of energy that propagated as waves needed to travel though a material medium. It was believed that space was filled with an invisible medium that light waves traveled through. This imaginary (as it turned out) material was called the ether (also spelled aether). It was thought that everything moved through this mysterious fluid. In other words, ether was the one fixed frame of reference. The Michelson\u2013Morley experiment proved it was not. In 1887, Albert Michelson and Edward Morley designed the interferometer shown in Figure 10.2 to measure the speed of Earth through the ether. A light beam is split into two perpendicular paths and then recombined. Recombining the waves produces an inference pattern, with a bright fringe at the locations where the two waves arrive in phase; that is, with the crests of both waves arriving together and the troughs arriving together. A dark fringe appears where the crest of one wave coincides with a trough of the other, so that the two cancel. If Earth is traveling through the ether as it orbits the sun, the peaks in one arm would take longer than in the other to reach the same location. The places where the two waves arrive in phase would change, and the interference pattern would shift. But, using the interferometer, there was no shift seen! This result led to two conclusions: that there is no ether and that the speed of light is the same regardless of the relative motion of source and observer. The Michelson\u2013Morley investigation has been called the most famous failed experiment in history. Access", " for free at openstax.org. 10.1 \u2022 Postulates of Special Relativity 307 Figure 10.2 This is a diagram of the instrument used in the Michelson\u2013Morley experiment. To see what Michelson and Morley expected to find when they measured the speed of light in two directions, watch this animation (http://openstax.org/l/28MMexperiment). In the video, two people swimming in a lake are represented as an analogy to light beams leaving Earth as it moves through the ether (if there were any ether). The swimmers swim away from and back to a platform that is moving through the water. The swimmers swim in different directions with respect to the motion of the platform. Even though they swim equal distances at the same speed, the motion of the platform causes them to arrive at different times. Einstein\u2019s Postulates The results described above left physicists with some puzzling and unsettling questions such as, why doesn\u2019t light emitted by a fast-moving object travel faster than light from a street lamp? A radical new theory was needed, and Albert Einstein, shown in Figure 10.3, was about to become everyone\u2019s favorite genius. Einstein began with two simple postulates based on the two things we have discussed so far in this chapter. 1. The laws of physics are the same in all inertial reference frames. 2. The speed of light is the same in all inertial reference frames and is not affected by the speed of its source. Figure 10.3 Albert Einstein (1879\u20131955) developed modern relativity and also made fundamental contributions to the foundations of quantum mechanics. (The Library of Congress) The speed of light is given the symbol cand is equal to exactly 299,792,458 m/s. This is the speed of light in vacuum; that is, in the absence of air. For most purposes, we round this number off to The term inertial reference frame simply 308 Chapter 10 \u2022 Special Relativity refers to a frame of reference where all objects follow Newton\u2019s first law of motion: Objects at rest remain at rest, and objects in motion remain in motion at a constant velocity in a straight line, unless acted upon by an external force. The inside of a car moving along a road at constant velocity and the inside of a stationary house are inertial reference frames. WATCH PHYSICS The Speed of Light This lecture on light summarizes the most important facts about the speed of light. If you are interested, you can watch the whole video, but the parts relevant to this chapter are found between 3:25 and 5:10, which you find by running your cursor along the bottom of the video. Click to view content (https://www.youtube.com/embed/rLNM8zI4Q_M) GRASP CHECK An airliner traveling at 200 m/s emits light from the front of the plane. Which statement describes the speed of the light? a. b. c. d. It travels at a speed of c+ 200 m/s. It travels at a speed of c\u2013 200 m/s. It travels at a speed c, like all light. It travels at a speed slightly less than c. Snap Lab Measure the Speed of Light In this experiment, you will measure the speed of light using a microwave oven and a slice of bread. The waves generated by a microwave oven are not part of the visible spectrum, but they are still electromagnetic radiation, so they travel at the speed of light. If we know the wavelength, \u03bb, and frequency, f, of a wave, we can calculate its speed, v, using the equation v = \u03bbf. You can measure the wavelength. You will find the frequency on a label on the back of a microwave oven. The wave in a microwave is a standing wave with areas of high and low intensity. The high intensity sections are one-half wavelength apart. \u2022 High temperature: Very hot temperatures are encountered in this lab. These can cause burns. \u2022 a microwave oven \u2022 one slice of plain white bread \u2022 a centimeter ruler \u2022 a calculator 1. Work with a partner. 2. Turn off the revolving feature of the microwave oven or remove the wheels under the microwave dish that make it turn. It is important that the dish does not turn. 3. Place the slice of bread on the dish, set the microwave on high, close the door, run the microwave for about 15 seconds. 4. A row of brown or black marks should appear on the bread. Stop the microwave as soon as they appear. Measure the distance between two adjacent burn marks and multiply the result by 2. This is the wavelength. 5. The frequency of the waves is written on the back of the microwave. Look for something like \u201c2,450 MHz.\u201d Hz is the unit hertz, which means per second. The M represents mega, which stands for million, so multiply the number by 106. 6. Express the wavelength in meters and multiply", " it times the frequency. If you did everything correctly, you will get a number very close to the speed of light. Do not eat the bread. It is a general laboratory safety rule never to eat anything in the lab. GRASP CHECK How does your measured value of the speed of light compare to the accepted value (% error)? Access for free at openstax.org. 10.1 \u2022 Postulates of Special Relativity 309 a. The measured value of speed will be equal to c. b. The measured value of speed will be slightly less than c. c. The measured value of speed will be slightly greater than c. d. The measured value of speed will depend on the frequency of the microwave. Einstein\u2019s postulates were carefully chosen, and they both seemed very likely to be true. Einstein proceeded despite realizing that these two ideas taken together and applied to extreme conditions led to results that contradict Newtonian mechanics. He just took the ball and ran with it. In the traditional view, velocities are additive. If you are running at 3 m/s and you throw a ball forward at a speed of 10 m/s, the ball should have a net speed of 13 m/s. However, according to relativity theory, the speed of a moving light source is not added to the speed of the emitted light. In addition, Einstein\u2019s theory shows that if you were moving forward relative to Earth at nearly c(the speed of light) and could throw a ball forward at c, an observer at rest on the earth would not see the ball moving at nearly twice the speed of light. The observer would see it moving at a speed that is still less than c. This result conforms to both of Einstein\u2019s postulates: The speed of light has a fixed maximum and neither reference frame is privileged. Consider how we measure elapsed time. If we use a stopwatch, for example, how do we know when to start and stop the watch? One method is to use the arrival of light from the event, such as observing a light turn green to start a drag race. The timing will be more accurate if some sort of electronic detection is used, avoiding human reaction times and other complications. Now suppose we use this method to measure the time interval between two flashes of light produced by flash lamps on a moving train. (See Figure 10.4) 310 Chapter 10 \u2022 Special Relativity Figure 10.4 Light arriving to observer A as seen by two different observers. A woman (observer A) is seated in the center of a rail car, with two flash lamps at opposite sides equidistant from her. Multiple light rays that are emitted from the flash lamps move towards observer A, as shown with arrows. A velocity vector arrow for the rail car is shown towards the right. A man (observer B) standing on the platform is facing the woman and also observes the flashes of light. Observer A moves with the lamps on the rail car as the rail car moves towards the right of observer B. Observer B receives the light flashes simultaneously, and sees the bulbs as both having flashed at the same time. However, he sees observer A receive the flash from the right first. Because the pulse from the right reaches her first, in her frame of reference she sees the bulbs as not having flashed simultaneously. Here, a relative velocity between observers affects whether two events at well-separated locations are observed to be simultaneous. Simultaneity, or whether different events occur at the same instant, depends on the frame of reference of the observer. Remember that velocity equals distance divided by time, so t = d/v. If velocity appears to be different, then duration of time appears to be different. This illustrates the power of clear thinking. We might have guessed incorrectly that, if light is emitted simultaneously, then two observers halfway between the sources would see the flashes simultaneously. But careful analysis shows this not to be the case. Einstein was brilliant at this type of thought experiment (in German, Gedankenexperiment). He very carefully considered how an observation is made and disregarded what might seem obvious. The validity of thought experiments, of course, is determined by actual observation. The genius of Einstein is evidenced by the fact that experiments have repeatedly confirmed his theory of relativity. No experiments after that of Michelson and Morley were able to detect any ether medium. We will describe later how experiments also confirmed other predictions of special relativity, such as the distance between two objects and the time interval of two events being different for two observers moving with respect to each other. In summary: Two events are defined to be simultaneous if an observer measures them as occurring at the same time (such as by receiving light from the events). Two events are not necessarily simultaneous to all observers. Access for free at openstax.org. 10.1 \u2022 Postulates of Special Relativity 311 The discrepancies between Newtonian mechanics and relativity theory illustrate an important point about how science advances. Einstein\u2019s theory did not replace Newton\u2019s but", " rather extended it. It is not unusual that a new theory must be developed to account for new information. In most cases, the new theory is built on the foundation of older theory. It is rare that old theories are completely replaced. In this chapter, you will learn about the theory of special relativity, but, as mentioned in the introduction, Einstein developed two relativity theories: special and general. Table 10.1 summarizes the differences between the two theories. Special Relativity General Relativity Published in 1905 Final form published in 1916 A theory of space-time A theory of gravity Applies to observers moving at constant speed Applies to observers that are accelerating Most useful in the field of nuclear physics Most useful in the field of astrophysics Accepted quickly and put to practical use by nuclear physicists and quantum chemists Largely ignored until 1960 when new mathematical techniques made the theory more accessible and astronomers found some important applications Also note that the theory of general relativity includes the theory of special relativity. Table 10.1 Comparing Special Relativity and General Relativity WORKED EXAMPLE Calculating the Time it Takes Light to Travel a Given Distance The sun is 1.50 \u00d7 108 km from Earth. How long does it take light to travel from the sun to Earth in minutes and seconds? Strategy Identify knowns. Identify unknowns. Time Find the equation that relates knowns and unknowns. Be sure to use consistent units. Solution 10.1 10.2 Discussion The answer is written as 5.00 \u00d7 102 rather than 500 in order to show that there are three significant figures. When astronomers witness an event on the sun, such as a sunspot, it actually happened minutes earlier. Compare 8 light minutesto the distance to stars, which are light yearsaway. Any events on other stars happened years ago. 312 Chapter 10 \u2022 Special Relativity Practice Problems 1. Light travels through 1.00 m of water in 4.42\u00d710-9 s. What is the speed of light in water? a. 4.42\u00d710-9 m/s b. 4.42\u00d7109 m/s c. 2.26\u00d7108 m/s d. 226\u00d7108 m/s 2. An astronaut on the moon receives a message from mission control on Earth. The signal is sent by a form of electromagnetic radiation and takes 1.28 s to travel the distance between Earth and the moon. What is the distance from Earth to the moon? a. 2.34\u00d7105 km b. 2.34\u00d7108 km 3.84\u00d7105 km c. 3.84\u00d7108 km d. Check Your Understanding 3. Explain what is meant by a frame of reference. a. A frame of reference is a graph plotted between distance and time. b. A frame of reference is a graph plotted between speed and time. c. A frame of reference is the velocity of an object through empty space without regard to its surroundings. d. A frame of reference is an arbitrarily fixed point with respect to which motion of other points is measured. 4. Two people swim away from a raft that is floating downstream. One swims upstream and returns, and the other swims across the current and back. If this scenario represents the Michelson\u2013Morley experiment, what do (i) the water, (ii) the swimmers, and (iii) the raft represent? the ether rays of light Earth a. rays of light the ether Earth b. c. the ether Earth rays of light d. Earth rays of light the ether 5. If Michelson and Morley had observed the interference pattern shift in their interferometer, what would that have indicated? a. The speed of light is the same in all frames of reference. b. The speed of light depends on the motion relative to the ether. c. The speed of light changes upon reflection from a surface. d. The speed of light in vacuum is less than 3.00\u00d7108 m/s. 6. If you designate a point as being fixed and use that point to measure the motion of surrounding objects, what is the point called? a. An origin b. A frame of reference c. A moving frame d. A coordinate system 10.2 Consequences of Special Relativity Section Learning Objectives By the end of this section, you will be able to do the following: \u2022 Describe the relativistic effects seen in time dilation, length contraction, and conservation of relativistic momentum \u2022 Explain and perform calculations involving mass-energy equivalence Section Key Terms binding energy length contraction mass defect time dilation Access for free at openstax.org. 10.2 \u2022 Consequences of Special Relativity 313 proper length relativistic relativistic momentum relativistic energy relativistic factor rest mass Relativistic Effects on Time, Distance, and Momentum Consideration of the measurement of elapsed time and simultaneity leads to an important relativistic effect. Time dilation is the phenomenon of time passing more slowly for an observer who is moving relative to", " another observer. For example, suppose an astronaut measures the time it takes for light to travel from the light source, cross her ship, bounce off a mirror, and return. (See Figure 10.5.) How does the elapsed time the astronaut measures compare with the elapsed time measured for the same event by a person on the earth? Asking this question (another thought experiment) produces a profound result. We find that the elapsed time for a process depends on who is measuring it. In this case, the time measured by the astronaut is smaller than the time measured by the earth bound observer. The passage of time is different for the two observers because the distance the light travels in the astronaut\u2019s frame is smaller than in the earth bound frame. Light travels at the same speed in each frame, and so it will take longer to travel the greater distance in the earth bound frame. Figure 10.5 (a) An astronaut measures the time for light to cross her ship using an electronic timer. Light travels a distance in the astronaut\u2019s frame. (b) A person on the earth sees the light follow the longer path and take a longer time The relationship between \u0394tand \u0394to is given by where is the relativistic factor given by and vand care the speeds of the moving observer and light, respectively. 314 Chapter 10 \u2022 Special Relativity TIPS FOR SUCCESS Try putting some values for vinto the expression for the relativistic factor ( a difference and when is so close to 1 that it can be ignored. Try 225 m/s, the speed of an airliner; 2.98 \u00d7 104 m/s, the speed of Earth in its orbit; and 2.990 \u00d7 108 m/s, the speed of a particle in an accelerator. ). Observe at which speeds this factor will make Notice that when the velocity vis small compared to the speed of light c, then v/cbecomes small, and becomes close to 1. When this happens, time measurements are the same in both frames of reference. Relativistic effects, meaning those that have to do with special relativity, usually become significant when speeds become comparable to the speed of light. This is seen to be the case for time dilation. You may have seen science fiction movies in which space travelers return to Earth after a long trip to find that the planet and everyone on it has aged much more than they have. This type of scenario is a based on a thought experiment, known as the twin paradox, which imagines a pair of twins, one of whom goes on a trip into space while the other stays home. When the space traveler returns, she finds her twin has aged much more than she. This happens because the traveling twin has been in two frames of reference, one leaving Earth and one returning. Time dilation has been confirmed by comparing the time recorded by an atomic clock sent into orbit to the time recorded by a clock that remained on Earth. GPS satellites must also be adjusted to compensate for time dilation in order to give accurate positioning. Have you ever driven on a road, like that shown in Figure 10.6, that seems like it goes on forever? If you look ahead, you might say you have about 10 km left to go. Another traveler might say the road ahead looks like it is about 15 km long. If you both measured the road, however, you would agree. Traveling at everyday speeds, the distance you both measure would be the same. You will read in this section, however, that this is not true at relativistic speeds. Close to the speed of light, distances measured are not the same when measured by different observers moving with respect to one other. Figure 10.6 People might describe distances differently, but at relativistic speeds, the distances really are different. (Corey Leopold, Flickr) One thing all observers agree upon is their relative speed. When one observer is traveling away from another, they both see the other receding at the same speed, regardless of whose frame of reference is chosen. Remember that speed equals distance divided by time: v = d/t. If the observers experience a difference in elapsed time, they must also observe a difference in distance traversed. This is because the ratio d/tmust be the same for both observers. The shortening of distance experienced by an observer moving with respect to the points whose distance apart is measured is called length contraction. Proper length, L0, is the distance between two points measured in the reference frame where the observer and the points are at rest. The observer in motion with respect to the points measures L. These two lengths are related by the equation Because is the same expression used in the time dilation equation above, the equation becomes Access for free at openstax.org. 10.2 \u2022 Consequences of Special Relativity 315 To see how length contraction is seen by a moving observer, go to this simulation (http://openstax.org/l/28simultaneity). Here you can also", " see that simultaneity, time dilation, and length contraction are interrelated phenomena. This link is to a simulation that illustrates the relativity of simultaneous events. In classical physics, momentum is a simple product of mass and velocity. When special relativity is taken into account, objects that have mass have a speed limit. What effect do you think mass and velocity have on the momentum of objects moving at relativistic speeds; i.e., speeds close to the speed of light? Momentum is one of the most important concepts in physics. The broadest form of Newton\u2019s second law is stated in terms of momentum. Momentum is conserved in classical mechanics whenever the net external force on a system is zero. This makes momentum conservation a fundamental tool for analyzing collisions. We will see that momentum has the same importance in modern physics. Relativistic momentum is conserved, and much of what we know about subatomic structure comes from the analysis of collisions of accelerator-produced relativistic particles. One of the postulates of special relativity states that the laws of physics are the same in all inertial frames. Does the law of conservation of momentum survive this requirement at high velocities? The answer is yes, provided that the momentum is defined as follows. Relativistic momentum, p, is classical momentum multiplied by the relativistic factor is the rest mass of the object (that is, the mass measured at rest, without any as before, is the relativistic factor. We use the mass of the object as measured at rest because we cannot where to an observer, and determine its mass while it is moving. factor involved), is its velocity relative 10.3 for velocity here to distinguish it from relative velocity between observers. Only one observer is being Note that we use considered here. With defined in this way, Again we see that the relativistic quantity becomes virtually the same as the classical at low velocities. That is, relativistic momentum is conserved whenever the net external force is zero, just as in classical physics. is very nearly equal to 1 at low velocities. at low velocities, because becomes the classical Relativistic momentum has the same intuitive feel as classical momentum. It is greatest for large masses moving at high velocities. Because of the factor approaching infinity as speed of light. If it did, its momentum would become infinite, which is an unreasonable value. however, relativistic momentum behaves differently from classical momentum by (See Figure 10.7.) This is another indication that an object with mass cannot reach the approaches Figure 10.7 Relativistic momentum approaches infinity as the velocity of an object approaches the speed of light. Relativistic momentum is defined in such a way that the conservation of momentum will hold in all inertial frames. Whenever the net external force on a system is zero, relativistic momentum is conserved, just as is the case for classical momentum. This 316 Chapter 10 \u2022 Special Relativity has been verified in numerous experiments. Mass-Energy Equivalence Let us summarize the calculation of relativistic effects on objects moving at speeds near the speed of light. In each case we will need to calculate the relativistic factor, given by where v and care as defined earlier. We use u as the velocity of a particle or an object in one frame of reference, and v for the velocity of one frame of reference with respect to another. Time Dilation Elapsed time on a moving object, on the moving object when it is taken to be the frame or reference. as seen by a stationary observer is given by Length Contraction Length measured by a person at rest with respect to a moving object, L, is given by where is the time observed where L0 is the length measured on the moving object. Relativistic Momentum Momentum, p, of an object of mass, m, traveling at relativistic speeds is given by object as seen by a stationary observer. where u is velocity of a moving Relativistic Energy The original source of all the energy we use is the conversion of mass into energy. Most of this energy is generated by nuclear reactions in the sun and radiated to Earth in the form of electromagnetic radiation, where it is then transformed into all the forms with which we are familiar. The remaining energy from nuclear reactions is produced in nuclear power plants and in Earth\u2019s interior. In each of these cases, the source of the energy is the conversion of a small amount of mass into a large amount of energy. These sources are shown in Figure 10.8. Figure 10.8 The sun (a) and the Susquehanna Steam Electric Station (b) both convert mass into energy. ((a) NASA/Goddard Space Flight Center, Scientific Visualization Studio; (b) U.S. government) The first postulate of relativity states that the laws of physics are the same in all inertial frames. Einstein showed that the law of conservation of energy is valid relativistically,", " if we define energy to include a relativistic factor. The result of his analysis is that a particle or object of mass mmoving at velocity u has relativistic energy given by This is the expression for the total energy of an object of mass mat any speed u and includes both kinetic and potential energy. Look back at the equation for and you will see that it is equal to 1 when u is 0; that is, when an object is at rest. Then the rest Access for free at openstax.org. 10.2 \u2022 Consequences of Special Relativity 317 energy, E0, is simply This is the correct form of Einstein\u2019s famous equation. This equation is very useful to nuclear physicists because it can be used to calculate the energy released by a nuclear reaction. This is done simply by subtracting the mass of the products of such a reaction from the mass of the reactants. The difference is the min Here is a simple example: A positron is a type of antimatter that is just like an electron, except that it has a positive charge. When a positron and an electron collide, their masses are completely annihilated and converted to energy in the form of gamma rays. Because both particles have a rest mass of 9.11 \u00d7 10\u201331 kg, we multiply the mc2 term by 2. So the energy of the gamma rays is 10.4 where we have the expression for the joule (J) in terms of its SI base units of kg, m, and s. In general, the nuclei of stable isotopes have less mass then their constituent subatomic particles. The energy equivalent of this difference is called the binding energy of the nucleus. This energy is released during the formation of the isotope from its constituent particles because the product is more stable than the reactants. Expressed as mass, it is called the mass defect. For example, a helium nucleus is made of two neutrons and two protons and has a mass of 4.0003 atomic mass units (u). The sum of the masses of two protons and two neutrons is 4.0330 u. The mass defect then is 0.0327 u. Converted to kg, the mass defect is 5.0442 \u00d7 10\u201330 kg. Multiplying this mass times c2 gives a binding energy of 4.540 \u00d7 10\u201312 J. This does not sound like much because it is only one atom. If you were to make one gram of helium out of neutrons and protons, it would release 683,000,000,000 J. By comparison, burning one gram of coal releases about 24 J. BOUNDLESS PHYSICS The RHIC Collider Figure 10.9 shows the Brookhaven National Laboratory in Upton, NY. The circular structure houses a particle accelerator called the RHIC, which stands for Relativistic Heavy Ion Collider. The heavy ions in the name are gold nuclei that have been stripped of their electrons. Streams of ions are accelerated in several stages before entering the big ring seen in the figure. Here, they are accelerated to their final speed, which is about 99.7 percent the speed of light. Such high speeds are called relativistic. All the relativistic phenomena we have been discussing in this chapter are very pronounced in this case. At this speed = 12.9, so that relativistic time dilates by a factor of about 13, and relativistic length contracts by the same factor. Figure 10.9 Brookhaven National Laboratory. The circular structure houses the RHIC. (energy.gov, Wikimedia Commons) Two ion beams circle the 2.4-mile long track around the big ring in opposite directions. The paths can then be made to cross, thereby causing ions to collide. The collision event is very short-lived but amazingly intense. The temperatures and pressures produced are greater than those in the hottest suns. At 4 trillion degrees Celsius, this is the hottest material ever created in a 318 Chapter 10 \u2022 Special Relativity laboratory But what is the point of creating such an extreme event? Under these conditions, the neutrons and protons that make up the gold nuclei are smashed apart into their components, which are called quarks and gluons. The goal is to recreate the conditions that theorists believe existed at the very beginning of the universe. It is thought that, at that time, matter was a sort of soup of quarks and gluons. When things cooled down after the initial bang, these particles condensed to form protons and neutrons. Some of the results have been surprising and unexpected. It was thought the quark-gluon soup would resemble a gas or plasma. Instead, it behaves more like a liquid. It has been called a perfectliquid because it has virtually no viscosity, meaning that it has no resistance to flow. GRASP CHECK Calculate the relativistic factor \u03b3, for a particle traveling at 99.7 percent of the speed of light. a", ". 0.08 b. 0.71 1.41 c. 12.9 d. WORKED EXAMPLE The Speed of Light One night you are out looking up at the stars and an extraterrestrial spaceship flashes across the sky. The ship is 50 meters long and is travelling at 95 percent of the speed of light. What would the ship\u2019s length be when measured from your earthbound frame of reference? Strategy List the knowns and unknowns. Knowns: proper length of the ship, L0= 50 m; velocity, v, = 0.95c Unknowns: observed length of the ship accounting for relativistic length contraction, L. Choose the relevant equation. Solution Discussion Calculations of sure to also square the decimal representing the percentage before subtracting from 1. Note that the aliens will still see the length as L0 because they are moving with the frame of reference that is the ship. can usually be simplified in this way when vis expressed as a percentage of cbecause the c2 terms cancel. Be Practice Problems 7. Calculate the relativistic factor, \u03b3, for an object traveling at 2.00\u00d7108 m/s. a. 0.74 b. 0.83 c. d. 1.2 1.34 8. The distance between two points, called the proper length, L0, is 1.00 km. An observer in motion with respect to the frame of Access for free at openstax.org. 10.2 \u2022 Consequences of Special Relativity 319 reference of the two points measures 0.800 km, which is L. What is the relative speed of the frame of reference with respect to the observer? 1.80\u00d7108 m/s a. b. 2.34\u00d7108 m/s 3.84\u00d7108 m/s c. 5.00\u00d7108 m/s d. 9. Consider the nuclear fission reaction. If a neutron has a rest mass of 1.009u, has rest mass of 136.907u, and has a rest mass of 96.937u, what is the value of Ein has a rest mass of 235.044u, joules? a. b. c. d. J J J J Solution The correct answer is (b). The mass deficit in the reaction is Converting that mass to kg and applying to find the energy equivalent of the mass deficit gives or 0.191u. 10. Consider the nuclear fusion reaction. If has a rest mass of 2.014u, has a rest mass of 3.016u, and has a rest mass of 1.008u, what is the value of Ein joules? a. b. c. d. J J J J Solution The correct answer is (a). The mass deficit in the reaction is mass to kg and applying to find the energy equivalent of the mass deficit gives, or 0.004u. Converting that Check Your Understanding 11. Describe time dilation and state under what conditions it becomes significant. a. When the speed of one frame of reference past another reaches the speed of light, a time interval between two events at the same location in one frame appears longer when measured from the second frame. b. When the speed of one frame of reference past another becomes comparable to the speed of light, a time interval between two events at the same location in one frame appears longer when measured from the second frame. c. When the speed of one frame of reference past another reaches the speed of light, a time interval between two events at the same location in one frame appears shorter when measured from the second frame. d. When the speed of one frame of reference past another becomes comparable to the speed of light, a time interval between two events at the same location in one frame appears shorter when measured from the second frame. 12. The equation used to calculate relativistic momentum is p= \u03b3 \u00b7 m \u00b7 u. Define the terms to the right of the equal sign and state how mand uare measured. a. \u03b3is the relativistic factor, mis the rest mass measured when the object is at rest in the frame of reference, and uis the velocity of the frame. b. \u03b3is the relativistic factor, mis the rest mass measured when the object is at rest in the frame of reference, and uis the velocity relative to an observer. c. \u03b3is the relativistic factor, mis the relativistic mass measured when the object is moving in the frame of reference, and uis the velocity of the frame. 320 Chapter 10 \u2022 Special Relativity d. \u03b3is the relativistic factor, mis the relativistic mass measured when the object is moving in the frame of reference, and uis the velocity relative to an observer. 13. Describe length contraction and state when it occurs. a. When the speed of an object becomes the speed of light, its length appears to shorten when viewed by a stationary observer", ". b. When the speed of an object approaches the speed of light, its length appears to shorten when viewed by a stationary observer. c. When the speed of an object becomes the speed of light, its length appears to increase when viewed by a stationary observer. d. When the speed of an object approaches the speed of light, its length appears to increase when viewed by a stationary observer. Access for free at openstax.org. Chapter 10 \u2022 Key Terms 321 KEY TERMS binding energy the energy equivalent of the difference between the mass of a nucleus and the masses of its nucleons effects that become significant only when an object is to be moving close enough to the speed of light for significantly greater than 1 ether scientists once believed there was a medium that carried light waves; eventually, experiments proved that ether does not exist relativistic energy the total energy of a moving object or which includes both its rest energy particle mc2 and its kinetic energy frame of reference the point or collection of points relativistic factor, where u is the velocity of a arbitrarily chosen, which motion is measured in relation to general relativity the theory proposed to explain gravity and acceleration inertial reference frame a frame of reference where all objects follow Newton\u2019s first law of motion length contraction the shortening of an object as seen by an observer who is moving relative to the frame of reference of the object mass defect the difference between the mass of a nucleus and the masses of its nucleons postulate a statement that is assumed to be true for the purposes of reasoning in a scientific or mathematic argument proper length the length of an object within its own frame of reference, as opposed to the length observed by an observer moving relative to that frame of reference relativistic having to do with modern relativity, such as the SECTION SUMMARY 10.1 Postulates of Special Relativity \u2022 One postulate of special relativity theory is that the laws of physics are the same in all inertial frames of reference. \u2022 The other postulate is that the speed of light in a vacuum is the same in all inertial frames. \u2022 Einstein showed that simultaneity, or lack of it, depends on the frame of reference of the observer. KEY EQUATIONS 10.1 Postulates of Special Relativity speed of light constant value for the speed of light moving object and cis the speed of light relativistic momentum p = \u03b3mu, where is the relativistic factor, mis rest mass of an object, and u is the velocity relative to an observer relativity the explanation of how objects move relative to one another rest mass the mass of an object that is motionless with respect to its frame of reference simultaneity the property of events that occur at the same time special relativity the theory proposed to explain the consequences of requiring the speed of light and the laws of physics to be the same in all inertial frames time dilation the contraction of time as seen by an observer in a frame of reference that is moving relative to the observer 10.2 Consequences of Special Relativity \u2022 Time dilates, length contracts, and momentum increases as an object approaches the speed of light. \u2022 Energy and mass are interchangeable, according to the relationship E = mc2. The laws of conservation of mass and energy are combined into the law of conservation of mass-energy. 10.2 Consequences of Special Relativity elapsed time relativistic factor length contraction relativistic momentum 322 Chapter 10 \u2022 Chapter Review relativistic energy rest energy CHAPTER REVIEW Concept Items 10.1 Postulates of Special Relativity 1. Why was it once believed that light must travel through a medium and could not propagate across empty space? a. The longitudinal nature of light waves implies this. b. Light shows the phenomenon of diffraction. c. The speed of light is the maximum possible speed. d. All other wave energy needs a medium to travel. 2. Describe the relative motion of Earth and the sun: 1. 2. if Earth is taken as the inertial frame of reference and if the sun is taken as the inertial frame of reference. 1. Earth is at rest and the sun orbits Earth. a. 2. The sun is at rest and Earth orbits the sun. b. c. d. 1. The sun is at rest and Earth orbits the sun. 2. Earth is at rest and the sun orbits Earth. 1. The sun is at rest and Earth orbits the sun. 2. The sun is at rest and Earth orbits the sun. 1. Earth is at rest and the sun orbits Earth. 2. Earth is at rest and the sun orbits Earth. 10.2 Consequences of Special Relativity 3. A particle (a free electron) is speeding around the track Critical Thinking Items 10.1 Postulates of Special Relativity 6. Explain how the two postulates of Einstein\u2019s theory of special relativity, when taken together, could lead to a situation that seems to contradict the mechanics and laws of motion as described by Newton. a. In Newtonian mechanics, velocities are multiplicative but the", " speed of a moving light source cannot be multiplied to the speed of light because, according to special relativity, the speed of light is the maximum speed possible. In Newtonian mechanics, velocities are additive but the speed of a moving light source cannot be added to the speed of light because the speed of light is the maximum speed possible. b. c. An object that is at rest in one frame of reference may appear to be in motion in another frame of reference, while in Newtonian mechanics such a situation is not possible. Access for free at openstax.org. in a cyclotron, rapidly gaining speed. How will the particle\u2019s momentum change as its speed approaches the speed of light? Explain. a. The particle\u2019s momentum will rapidly decrease. b. The particle\u2019s momentum will rapidly increase. c. The particle\u2019s momentum will remain constant. d. The particle\u2019s momentum will approach zero. 4. An astronaut goes on a long space voyage at near the speed of light. When she returns home, how will her age compare to the age of her twin who stayed on Earth? a. Both of them will be the same age. b. This is a paradox and hence the ages cannot be compared. c. The age of the twin who traveled will be less than the age of her twin. d. The age of the twin who traveled will be greater than the age of her twin. 5. A comet reaches its greatest speed as it travels near the sun. True or false\u2014 Relativistic effects make the comet\u2019s tail look longer to an observer on Earth. a. True b. False d. The postulates of Einstein\u2019s theory of special relativity do not contradict any situation that Newtonian mechanics explains. 7. It takes light to travel from the sun to the planet Venus. How far is Venus from the sun? a. b. c. d. 8. In 2003, Earth and Mars were the closest they had been in 50,000 years. The two planets were aligned so that Earth was between Mars and the sun. At that time it took light from the sun 500 s to reach Earth and 687 s to get to Mars. What was the distance from Mars to Earth? 5.6\u00d7107 km a. 5.6\u00d71010 km b. c. 6.2\u00d7106 km d. 6.2\u00d71012 km 9. Describe two ways in which light differs from all other forms of wave energy. a. 1. Light travels as a longitudinal wave. 2. Light travels through a medium that fills up the empty space in the universe. b. 1. Light travels as a transverse wave. 2. Light travels through a medium that fills up the empty space in the universe. c. 1. Light travels at the maximum possible speed in the universe. 2. Light travels through a medium that fills up the empty space in the universe. d. 1. Light travels at the maximum possible speed in the universe. 2. Light does not require any material medium to travel. 10. Use the postulates of the special relativity theory to explain why the speed of light emitted from a fastmoving light source cannot exceed 3.00\u00d7108 m/s. a. The speed of light is maximum in the frame of reference of the moving object. b. The speed of light is minimum in the frame of reference of the moving object. c. The speed of light is the same in all frames of reference, including in the rest frame of its source. d. Light always travels in a vacuum with a speed less than 3.00\u00d7108 m/s, regardless of the speed of the Problems 10.2 Consequences of Special Relativity 13. Deuterium (2 H) is an isotope of hydrogen that has one proton and one neutron in its nucleus. The binding energy of deuterium is 3.56\u00d710-13 J. What is the mass defect of deuterium? 3.20\u00d710-4 kg a. 1.68\u00d710-6 kg b. 1.19\u00d710-21 kg c. 3.96\u00d710-30 kg d. 14. The sun orbits the center of the galaxy at a speed of 2.3\u00d7105 m/s. The diameter of the sun is 1.391684\u00d7109 m. An observer is in a frame of reference that is stationary with respect to the center of the galaxy. True or false\u2014The sun is moving fast enough for the observer to notice length contraction of the sun\u2019s diameter. a. True b. False 15. Consider the nuclear fission reaction Chapter 10 \u2022 Chapter Review 323 source. 10.2 Consequences of Special Relativity 11. Halley\u2019s Comet comes near Earth every 75 years as it travels around its 22 billion km orbit at a speed of up to 700, 000 m/s. If it were possible to put a clock on the comet", " and read it each time the comet passed, which part of special relativity theory could be tested? What would be the expected result? Explain. a. It would test time dilation. The clock would appear to be slightly slower. It would test time dilation. The clock would appear to be slightly faster. It would test length contraction. The length of the orbit would appear to be shortened from Earth\u2019s frame of reference. It would test length contraction. The length of the orbit would appear to be shortened from the comet\u2019s frame of reference. b. c. d. 12. The nucleus of the isotope fluorine-18 (18 F) has mass defect of 2.44\u00d710-28 kg. What is the binding energy of 18F? a. 2.2\u00d710-11 J 7.3\u00d710-20 J b. c. 2.2\u00d710-20 J d. 2.4\u00d710-28 J has a rest mass of 1.009u, has a rest mass of. If a neutron has rest mass of 143.923u, and 235.044u, has a rest mass of 88.918u, what is the value of Ein joules? a. b. c. d. J J J J 16. Consider the nuclear fusion reaction. If has a rest mass of has a rest has a rest mass of 3.016u, 2.014u, mass of 4.003u, and a neutron has a rest mass of 1.009u, what is the value of Ein joules? a. b. c. d. J J J J 324 Chapter 10 \u2022 Test Prep Performance Task 10.2 Consequences of Special Relativity 17. People are fascinated by the possibility of traveling across the universe to discover intelligent life on other planets. To do this, we would have to travel enormous distances. Suppose we could somehow travel at up to 90 percent of the speed of light. The closest star is Alpha Centauri, which is 4.37 light years away. (A light year is the distance light travels in one year.) TEST PREP Multiple Choice 10.1 Postulates of Special Relativity 18. What was the purpose of the Michelson\u2013Morley experiment? a. To determine the exact speed of light b. To analyze the electromagnetic spectrum c. To establish that Earth is the true frame of reference a. How long, from the point of view of people on Earth, would it take a space ship to travel to Alpha Centauri and back at 0.9c? b. How much would the astronauts on the spaceship have aged by the time they got back to Earth? c. Discuss the problems related to travel to stars that are 20 or 30 light years away. Assume travel speeds near the speed of light. been in 50,000 years. People looking up saw Mars as a very bright red light on the horizon. If Mars was 2.06\u00d7108 km from the sun, how long did the reflected light people saw take to travel from the sun to Earth? a. b. c. d. 14 min and 33 s 12 min and 15 s 11 min and 27 s 3 min and 7 s d. To learn how the ether affected the propagation of 10.2 Consequences of Special Relativity light 19. What is the speed of light in a vacuum to three 23. What does this expression represent: significant figures? a. b. c. d. 20. How far does light travel in? a. b. c. d. a. b. c. d. time dilation relativistic factor relativistic energy length contraction 24. What is the rest energy, E0, of an object with a mass of 1.00 g? 3.00\u00d7105 J a. 3.00\u00d71011 J b. c. 9.00\u00d71013 J d. 9.00\u00d71016 J 21. Describe what is meant by the sentence, \u201cSimultaneity is 25. The fuel rods in a nuclear reactor must be replaced from not absolute.\u201d a. Events may appear simultaneous in all frames of reference. b. Events may not appear simultaneous in all frames of reference. c. The speed of light is not the same in all frames of reference. d. The laws of physics may be different in different inertial frames of reference. 22. In 2003, Earth and Mars were aligned so that Earth was between Mars and the sun. Earth and Mars were 5.6\u00d7107 km from each other, which was the closest they had time to time because so much of the radioactive material has reacted that they can no longer produce energy. How would the mass of the spent fuel rods compare to their mass when they were new? Explain your answer. a. The mass of the spent fuel rods would decrease. b. The mass of the spent fuel rods would increase. c. The mass of the spent fuel rods would remain the same. d.", " The mass of the spent fuel rods would become close to zero. Access for free at openstax.org. Chapter 10 \u2022 Test Prep 325 Short Answer 10.2 Consequences of Special Relativity 10.1 Postulates of Special Relativity 30. What is the relationship between the binding energy 26. What is the postulate having to do with the speed of light on which the theory of special relativity is based? a. The speed of light remains the same in all inertial frames of reference. b. The speed of light depends on the speed of the source emitting the light. c. The speed of light changes with change in medium through which it travels. d. The speed of light does not change with change in medium through which it travels. 27. What is the postulate having to do with reference frames on which the theory of special relativity is based? a. The frame of reference chosen is arbitrary as long as it is inertial. and the mass defect of an atomic nucleus? a. The binding energy is the energy equivalent of the mass defect, as given by E0 = mc. b. The binding energy is the energy equivalent of the mass defect, as given by E0 = mc2. c. The binding energy is the energy equivalent of the mass defect, as given by d. The binding energy is the energy equivalent of the mass defect, as given by 31. True or false\u2014It is possible to just use the relationships F= maand E= Fdto show that both sides of the equation E0 = mc2 have the same units. a. True b. False b. The frame of reference is chosen to have constant 32. Explain why the special theory of relativity caused the nonzero acceleration. c. The frame of reference is chosen in such a way that the object under observation is at rest. d. The frame of reference is chosen in such a way that the object under observation is moving with a constant speed. 28. If you look out the window of a moving car at houses going past, you sense that you are moving. What have you chosen as your frame of reference? the car a. the sun b. c. a house 29. Why did Michelson and Morley orient light beams at right angles to each other? a. To observe the particle nature of light b. To observe the effect of the passing ether on the speed of light c. To obtain a diffraction pattern by combination of light d. To obtain a constant path difference for interference of light Extended Response 10.1 Postulates of Special Relativity 34. Explain how Einstein\u2019s conclusion that nothing can travel faster than the speed of light contradicts an older concept about the speed of an object propelled from another, already moving, object. a. The older concept is that speeds are subtractive. For example, if a person throws a ball while running, the speed of the ball relative to the ground is the law of conservation of energy to be modified. a. The law of conservation of energy is not valid in relativistic mechanics. b. The law of conservation of energy has to be modified because of time dilation. c. The law of conservation of energy has to be modified because of length contraction. d. The law of conservation of energy has to be modified because of mass-energy equivalence. 33. The sun loses about 4 \u00d7 109 kg of mass every second. Explain in terms of special relativity why this is happening. a. The sun loses mass because of its high temperature. b. The sun loses mass because it is continuously releasing energy. c. The Sun loses mass because the diameter of the sun is contracted. d. The sun loses mass because the speed of the sun is very high and close to the speed of light. speed at which the person was running minus the speed of the throw. A relativistic example is when light is emitted from car headlights, it moves faster than the speed of light emitted from a stationary source. b. The older concept is that speeds are additive. For example, if a person throws a ball while running, the speed of the ball relative to the ground is the speed at which the person was running plus the speed of the throw. A relativistic example is when light is emitted from car headlights, it moves no 326 Chapter 10 \u2022 Test Prep faster than the speed of light emitted from a stationary source. The car's speed does not affect the speed of light. c. The older concept is that speeds are multiplicative. For example, if a person throws a ball while running, the speed of the ball relative to the ground is the speed at which the person was running multiplied by the speed of the throw. A relativistic example is when light is emitted from car headlights, it moves no faster than the speed of light emitted from a stationary source. The car's speed does not affect the speed of light. d. The older concept is that speeds are frame independent. For example, if a person throws a ball while running, the speed of the ball", " relative to the ground has nothing to do with the speed at which the person was running. A relativistic example is when light is emitted from car headlights, it moves no faster than the speed of light emitted from a stationary source. The car's speed does not affect the speed of light. 35. A rowboat is drifting downstream. One person swims 20 m toward the shore and back, and another, leaving at the same time, swims upstream 20 m and back to the boat. The swimmer who swam toward the shore gets back first. Explain how this outcome is similar to the outcome expected in the Michelson\u2013Morley experiment. a. The rowboat represents Earth, the swimmers are beams of light, and the water is acting as the ether. Light going against the current of the ether would get back later because, by then, Earth would have moved on. b. The rowboat represents the beam of light, the swimmers are the ether, and water is acting as Earth. Light going against the current of the ether would get back later because, by then, Earth would have moved on. c. The rowboat represents the ether, the swimmers are ray of light, and the water is acting as the earth. Light going against the current of the ether would get back later because, by then, Earth would have moved on. d. The rowboat represents the Earth, the swimmers are the ether, and the water is acting as the rays of light. Light going against the current of the ether would get back later because, by then, Earth would have moved on. 10.2 Consequences of Special Relativity 36. A helium-4 nucleus is made up of two neutrons and two protons. The binding energy of helium-4 is 4.53\u00d710-12 J. What is the difference in the mass of this helium nucleus and the sum of the masses of two neutrons and two protons? Which weighs more, the nucleus or its constituents? a. b. c. d. 1.51\u00d710-20 kg; the constituents weigh more 5.03\u00d710-29 kg; the constituents weigh more 1.51\u00d710-29 kg; the nucleus weighs more 5.03\u00d710-29 kg; the nucleus weighs more 37. Use the equation for length contraction to explain the relationship between the length of an object perceived by a stationary observer who sees the object as moving, and the proper length of the object as measured in the frame of reference where it is at rest. a. As the speed vof an object moving with respect to a stationary observer approaches c, the length perceived by the observer approaches zero. For other speeds, the length perceived is always less than the proper length. b. As the speed vof an object moving with respect to a stationary observer approaches c, the length perceived by the observer approaches zero. For other speeds, the length perceived is always greater than the proper length. c. As the speed vof an object moving with respect to a stationary observer approaches c, the length perceived by the observer approaches infinity. For other speeds, the length perceived is always less than the proper length. d. As the speed vof an object moving with respect to a stationary observer approaches c, the length perceived by the observer approaches infinity. For other speeds, the length perceived is always greater than the proper length. Access for free at openstax.org. CHAPTER 11 Thermal Energy, Heat, and Work Figure 11.1 The welder\u2019s gloves and helmet protect the welder from the electric arc, which transfers enough thermal energy to melt the rod, spray sparks, and emit high-energy electromagnetic radiation that can burn the retina of an unprotected eye. The thermal energy can be felt on exposed skin a few meters away, and its light can be seen for kilometers (Kevin S. O\u2019Brien, U.S. Navy) Chapter Outline 11.1 Temperature and Thermal Energy 11.2 Heat, Specific Heat, and Heat Transfer 11.3 Phase Change and Latent Heat Heat is something familiar to all of us. We feel the warmth of the summer sun, the hot vapor rising up out of INTRODUCTION a cup of hot cocoa, and the cooling effect of our sweat. When we feel warmth, it means that heat is transferring energy toour bodies; when we feel cold, that means heat is transferring energy away fromour bodies. Heat transfer is the movement of thermal energy from one place or material to another, and is caused by temperature differences. For example, much of our weather is caused by Earth evening out the temperature across the planet through wind and violent storms, which are driven by heat transferring energy away from the equator towards the cold poles. In this chapter, we\u2019ll explore the precise meaning of heat, how it relates to temperature as well as to other forms of energy, and its connection to work. 11.1 Temperature and Thermal Energy Section Learning Objectives By the end of this", " section, you will be able to do the following: \u2022 Explain that temperature is a measure of internal kinetic energy \u2022 Interconvert temperatures between Celsius, Kelvin, and Fahrenheit scales 328 Chapter 11 \u2022 Thermal Energy, Heat, and Work Section Key Terms absolute zero Celsius scale degree Celsius thermal energy degree Fahrenheit Fahrenheit scale heat kelvin (K) Kelvin scale temperature Temperature What is temperature? It\u2019s one of those concepts so ingrained in our everyday lives that, although we know what it means intuitively, it can be hard to define. It is tempting to say that temperature measures heat, but this is not strictly true. Heat is the transfer of energy due to a temperature difference. Temperature is defined in terms of the instrument we use to tell us how hot or cold an object is, based on a mechanism and scale invented by people. Temperature is literally defined as what we measure on a thermometer. Heat is often confused with temperature. For example, we may say that the heat was unbearable, when we actually mean that the temperature was high. This is because we are sensitive to the flow of energy by heat, rather than the temperature. Since heat, like work, transfers energy, it has the SI unit of joule (J). Atoms and molecules are constantly in motion, bouncing off one another in random directions. Recall that kinetic energy is the energy of motion, and that it increases in proportion to velocity squared. Without going into mathematical detail, we can say that thermal energy\u2014the energy associated with heat\u2014is the average kinetic energy of the particles (molecules or atoms) in a substance. Faster moving molecules have greater kinetic energies, and so the substance has greater thermal energy, and thus a higher temperature. The total internal energy of a system is the sum of the kinetic and potential energies of its atoms and molecules. Thermal energy is one of the subcategories of internal energy, as is chemical energy. To measure temperature, some scale must be used as a standard of measurement. The three most commonly used temperature scales are the Fahrenheit, Celsius, and Kelvin scales. Both the Fahrenheit scale and Celsius scale are relative temperature scales, meaning that they are made around a reference point. For example, the Celsius scale uses the freezing point of water as its reference point; all measurements are either lower than the freezing point of water by a given number of degrees (and have a negative sign), or higher than the freezing point of water by a given number of degrees (and have a positive sign). The boiling point of water is 100 for the Celsius scale, and its unit is the degree Celsius ). On the Fahrenheit scale, the freezing point of water is at 32. The unit of temperature on ). Note that the difference in degrees between the freezing and boiling points is greater this scale is the degree Fahrenheit for the Fahrenheit scale than for the Celsius scale. Therefore, a temperature difference of one degree Celsius is greater than a temperature difference of one degree Fahrenheit. Since 100 Celsius degrees span the same range as 180 Fahrenheit degrees, one degree on the Celsius scale is 1.8 times larger than one degree on the Fahrenheit scale (because, and the boiling point is at 212 ). This relationship can be used to convert between temperatures in Fahrenheit and Celsius (see Figure 11.2). Access for free at openstax.org. 11.1 \u2022 Temperature and Thermal Energy 329 Figure 11.2 Relationships between the Fahrenheit, Celsius, and Kelvin temperature scales, rounded to the nearest degree. The relative sizes of the scales are also shown. The Kelvin scale is the temperature scale that is commonly used in science because it is an absolute temperature scale. This means that the theoretically lowest-possible temperature is assigned the value of zero. Zero degrees on the Kelvin scale is known as absolute zero; it is theoretically the point at which there is no molecular motion to produce thermal energy. On the original Kelvin scale first created by Lord Kelvin, all temperatures have positive values, making it useful for scientific work. The official temperature unit on this scale is the kelvin, which is abbreviated as K. The freezing point of water is 273.15 K, and the boiling point of water is 373.15 K. Although absolute zero is possible in theory, it cannot be reached in practice. The lowest temperature ever created and measured K, at Helsinki University of Technology in Finland. In comparison, the coldest during a laboratory experiment was recorded temperature for a place on Earth\u2019s surface was 183 K (\u201389 \u00b0C ), at Vostok, Antarctica, and the coldest known place (outside the lab) in the universe is the Boomerang Nebula, with a temperature of 1 K. Luckily, most of us humans will never have to experience such extremes. The average normal body temperature is 98.6 ). to 111 ranging from 75 to 44 (24 (37.0 ), but people have been known to survive with body temperatures WATCH PHYSICS Comparing Celsius and Fahrenheit Temperature Scales This video shows how the Fahrenheit and Celsius temperature scales compare to one another. Click to view content (https://", "www.openstax.org/l/02celfahtemp) GRASP CHECK Even without the number labels on the thermometer, you could tell which side is marked Fahrenheit and which is Celsius by how the degree marks are spaced. Why? a. The separation between two consecutive divisions on the Fahrenheit scale is greater than a similar separation on the Celsius scale, because each degree Fahrenheit is equal to degrees Celsius. b. The separation between two consecutive divisions on the Fahrenheit scale is smaller than the similar separation on the Celsius scale, because each degree Celsius is equal to degrees Fahrenheit. c. The separation between two consecutive divisions on the Fahrenheit scale is greater than a similar separation on the Celsius scale, because each degree Fahrenheit is equal to degrees Celsius. d. The separation between two consecutive divisions on the Fahrenheit scale is smaller than a similar separation on the Celsius scale, because each degree Celsius is equal to degrees Fahrenheit. 330 Chapter 11 \u2022 Thermal Energy, Heat, and Work Converting Between Celsius, Kelvin, and Fahrenheit Scales While the Fahrenheit scale is still the most commonly used scale in the United States, the majority of the world uses Celsius, and scientists prefer Kelvin. It\u2019s often necessary to convert between these scales. For instance, if the TV meteorologist gave the local weather report in kelvins, there would likely be some confused viewers! Table 11.1 gives the equations for conversion between the three temperature scales. To Convert From\u2026 Use This Equation Celsius to Fahrenheit Fahrenheit to Celsius Celsius to Kelvin Kelvin to Celsius Fahrenheit to Kelvin Kelvin to Fahrenheit Table 11.1 Temperature Conversions WORKED EXAMPLE Room temperatureis generally defined to be 25 (a) What is room temperature in (b) What is it in K? STRATEGY To answer these questions, all we need to do is choose the correct conversion equations and plug in the known values. Solution for (a) 1. Choose the right equation. To convert from to, use the equation 2. Plug the known value into the equation and solve. Solution for (b) 1. Choose the right equation. To convert from to K, use the equation 2. Plug the known value into the equation and solve. 11.1 11.2 11.3 11.4 Discussion Living in the United States, you are likely to have more of a sense of what the temperature feels like if it\u2019s described as 77 as 25 (or 298 K, for that matter). than Access for free at openstax.org. WORKED EXAMPLE 11.1 \u2022 Temperature and Thermal Energy 331 Converting Between Temperature Scales: The Reaumur Scale The Reaumur scale is a temperature scale that was used widely in Europe in the 18th and 19th centuries. On the Reaumur If \u201croom temperature\u201d is 25 temperature scale, the freezing point of water is 0 the Celsius scale, what is it on the Reaumur scale? STRATEGY To answer this question, we must compare the Reaumur scale to the Celsius scale. The difference between the freezing point and boiling point of water on the Reaumur scale is 80 scales start at 0 for freezing, so we can create a simple formula to convert between temperatures on the two scales. and the boiling temperature is 80. On the Celsius scale, it is 100. Therefore, 100. Both on Solution 1. Derive a formula to convert from one scale to the other. 2. Plug the known value into the equation and solve. 11.5 11.6 Discussion As this example shows, relative temperature scales are somewhat arbitrary. If you wanted, you could create your own temperature scale! Practice Problems 1. What is 12.0 \u00b0C in kelvins? a. 112.0 K b. 273.2 K c. 12.0 K d. 285.2 K 2. What is 32.0 \u00b0C in degrees Fahrenheit? a. 57.6 \u00b0F b. 25.6 \u00b0F c. 305.2 \u00b0F d. 89.6 \u00b0F TIPS FOR SUCCESS Sometimes it is not so easy to guess the temperature of the air accurately. Why is this? Factors such as humidity and wind speed affect how hot or cold we feel. Wind removes thermal energy from our bodies at a faster rate than usual, making us feel colder than we otherwise would; on a cold day, you may have heard the TV weather person refer to the wind chill. On humid summer days, people tend to feel hotter because sweat doesn\u2019t evaporate from the skin as efficiently as it does on dry days, when the evaporation of sweat cools us off. Check Your Understanding 3. What is thermal energy? a. The thermal energy is the average potential energy of the particles in a system. b. The thermal energy is the total sum of the potential energies of the particles in a system. c. The thermal energy is the average kinetic energy of the particles due to the interaction among the particles in a system. d. The thermal energy is the", " average kinetic energy of the particles in a system. 4. What is used to measure temperature? 332 Chapter 11 \u2022 Thermal Energy, Heat, and Work a. a galvanometer b. a manometer c. a thermometer d. a voltmeter 11.2 Heat, Specific Heat, and Heat Transfer Section Learning Objectives By the end of this section, you will be able to do the following: \u2022 Explain heat, heat capacity, and specific heat \u2022 Distinguish between conduction, convection, and radiation \u2022 Solve problems involving specific heat and heat transfer Section Key Terms conduction convection heat capacity radiation specific heat Heat Transfer, Specific Heat, and Heat Capacity We learned in the previous section that temperature is proportional to the average kinetic energy of atoms and molecules in a substance, and that the average internal kinetic energy of a substance is higher when the substance\u2019s temperature is higher. If two objects at different temperatures are brought in contact with each other, energy is transferred from the hotter object (that is, the object with the greater temperature) to the colder (lower temperature) object, until both objects are at the same temperature. There is no net heat transfer once the temperatures are equal because the amount of heat transferred from one object to the other is the same as the amount of heat returned. One of the major effects of heat transfer is temperature change: Heating increases the temperature while cooling decreases it. Experiments show that the heat transferred to or from a substance depends on three factors\u2014the change in the substance\u2019s temperature, the mass of the substance, and certain physical properties related to the phase of the substance. The equation for heat transfer Qis 11.7 where mis the mass of the substance and \u0394Tis the change in its temperature, in units of Celsius or Kelvin. The symbol cstands for specific heat, and depends on the material and phase. The specific heat is the amount of heat necessary to change the temperature of 1.00 kg of mass by 1.00 \u00baC. The specific heat cis a property of the substance; its SI unit is J/(kg K) or J/(kg The temperature change ( closely related to the concept of heat capacity. Heat capacity is the amount of heat necessary to change the temperature of a, where mis mass and cis specific heat. Note that heat substance by 1.00 capacity is the same as specific heat, but without any dependence on mass. Consequently, two objects made up of the same material but with different masses will have different heat capacities. This is because the heat capacity is a property of an object, but specific heat is a property of anyobject made of the same material. ) is the same in units of kelvins and degrees Celsius (but not degrees Fahrenheit). Specific heat is. In equation form, heat capacity Cis ). Values of specific heat must be looked up in tables, because there is no simple way to calculate them. Table 11.2 gives the values of specific heat for a few substances as a handy reference. We see from this table that the specific heat of water is five times that of glass, which means that it takes five times as much heat to raise the temperature of 1 kg of water than to raise the temperature of 1 kg of glass by the same number of degrees. Substances Specific Heat (c) Solids Aluminum J/(kg ) 900 Table 11.2 Specific Heats of Various Substances. Access for free at openstax.org. 11.2 \u2022 Heat, Specific Heat, and Heat Transfer 333 Substances Specific Heat (c) Asbestos Concrete, granite (average) Copper Glass Gold Human body (average) Ice (average) Iron, steel Lead Silver Wood Liquids Benzene Ethanol Glycerin Mercury Water Gases (at 1 atm constant pressure) Air (dry) Ammonia Carbon dioxide Nitrogen Oxygen Steam 800 840 387 840 129 3500 2090 452 128 235 1700 1740 2450 2410 139 4186 1015 2190 833 1040 913 2020 Table 11.2 Specific Heats of Various Substances. 334 Chapter 11 \u2022 Thermal Energy, Heat, and Work Snap Lab Temperature Change of Land and Water What heats faster, land or water? You will answer this question by taking measurements to study differences in specific heat capacity. \u2022 Open flame\u2014Tie back all loose hair and clothing before igniting an open flame. Follow all of your teacher's instructions on how to ignite the flame. Never leave an open flame unattended. Know the location of fire safety equipment in the laboratory. \u2022 Sand or soil \u2022 Water \u2022 Oven or heat lamp \u2022 Two small jars \u2022 Two thermometers Instructions Procedure 1. Place equal masses of dry sand (or soil) and water at the same temperature into two small jars. (The average density of soil or sand is about 1.6 times that of water, so you can get equal masses by using 50 percent more water by volume.) 2. Heat both substances (using an oven or a heat lamp) for the same amount of time", ". 3. Record the final temperatures of the two masses. 4. Now bring both jars to the same temperature by heating for a longer period of time. 5. Remove the jars from the heat source and measure their temperature every 5 minutes for about 30 minutes. GRASP CHECK Did it take longer to heat the water or the sand/soil to the same temperature? Which sample took longer to cool? What does this experiment tell us about how the specific heat of water compared to the specific heat of land? a. The sand/soil will take longer to heat as well as to cool. This tells us that the specific heat of land is greater than that of water. b. The sand/soil will take longer to heat as well as to cool. This tells us that the specific heat of water is greater than that of land. c. The water will take longer to heat as well as to cool. This tells us that the specific heat of land is greater than that of water. d. The water will take longer to heat as well as to cool. This tells us that the specific heat of water is greater than that of land. Conduction, Convection, and Radiation Whenever there is a temperature difference, heat transfer occurs. Heat transfer may happen rapidly, such as through a cooking pan, or slowly, such as through the walls of an insulated cooler. There are three different heat transfer methods: conduction, convection, and radiation. At times, all three may happen simultaneously. See Figure 11.3. Access for free at openstax.org. 11.2 \u2022 Heat, Specific Heat, and Heat Transfer 335 Figure 11.3 In a fireplace, heat transfer occurs by all three methods: conduction, convection, and radiation. Radiation is responsible for most of the heat transferred into the room. Heat transfer also occurs through conduction into the room, but at a much slower rate. Heat transfer by convection also occurs through cold air entering the room around windows and hot air leaving the room by rising up the chimney. Conduction is heat transfer through direct physical contact. Heat transferred between the electric burner of a stove and the bottom of a pan is transferred by conduction. Sometimes, we try to control the conduction of heat to make ourselves more comfortable. Since the rate of heat transfer is different for different materials, we choose fabrics, such as a thick wool sweater, that slow down the transfer of heat away from our bodies in winter. As you walk barefoot across the living room carpet, your feet feel relatively comfortable\u2026until you step onto the kitchen\u2019s tile floor. Since the carpet and tile floor are both at the same temperature, why does one feel colder than the other? This is explained by different rates of heat transfer: The tile material removes heat from your skin at a greater rate than the carpeting, which makes it feelcolder. Some materials simply conduct thermal energy faster than others. In general, metals (like copper, aluminum, gold, and silver) are good heat conductors, whereas materials like wood, plastic, and rubber are poor heat conductors. Figure 11.4 shows particles (either atoms or molecules) in two bodies at different temperatures. The (average) kinetic energy of a particle in the hot body is higher than in the colder body. If two particles collide, energy transfers from the particle with greater kinetic energy to the particle with less kinetic energy. When two bodies are in contact, many particle collisions occur, resulting in a net flux of heat from the higher-temperature body to the lower-temperature body. The heat flux depends on the temperature difference water.. Therefore, you will get a more severe burn from boiling water than from hot tap Figure 11.4 The particles in two bodies at different temperatures have different average kinetic energies. Collisions occurring at the contact surface tend to transfer energy from high-temperature regions to low-temperature regions. In this illustration, a particle in the lower- temperature region (right side) has low kinetic energy before collision, but its kinetic energy increases after colliding with the contact 336 Chapter 11 \u2022 Thermal Energy, Heat, and Work surface. In contrast, a particle in the higher-temperature region (left side) has more kinetic energy before collision, but its energy decreases after colliding with the contact surface. Convection is heat transfer by the movement of a fluid. This type of heat transfer happens, for example, in a pot boiling on the stove, or in thunderstorms, where hot air rises up to the base of the clouds. TIPS FOR SUCCESS In everyday language, the term fluidis usually taken to mean liquid. For example, when you are sick and the doctor tells you to \u201cpush fluids,\u201d that only means to drink more beverages\u2014not to breath more air. However, in physics, fluid means a liquid or a gas. Fluids move differently than solid material, and even have their own branch of physics, known as fluid dynamics, that studies", " how they move. As the temperature of fluids increase, they expand and become less dense. For example, Figure 11.4 could represent the wall of a balloon with different temperature gases inside the balloon than outside in the environment. The hotter and thus faster moving gas particles inside the balloon strike the surface with more force than the cooler air outside, causing the balloon to expand. This decrease in density relative to its environment creates buoyancy (the tendency to rise). Convection is driven by buoyancy\u2014hot air rises because it is less dense than the surrounding air. Sometimes, we control the temperature of our homes or ourselves by controlling air movement. Sealing leaks around doors with weather stripping keeps out the cold wind in winter. The house in Figure 11.5 and the pot of water on the stove in Figure 11.6 are both examples of convection and buoyancy by human design. Ocean currents and large-scale atmospheric circulation transfer energy from one part of the globe to another, and are examples of natural convection. Figure 11.5 Air heated by the so-called gravity furnace expands and rises, forming a convective loop that transfers energy to other parts of the room. As the air is cooled at the ceiling and outside walls, it contracts, eventually becoming denser than room air and sinking to the floor. A properly designed heating system like this one, which uses natural convection, can be quite efficient in uniformly heating a home. Figure 11.6 Convection plays an important role in heat transfer inside this pot of water. Once conducted to the inside fluid, heat transfer to other parts of the pot is mostly by convection. The hotter water expands, decreases in density, and rises to transfer heat to other regions of the water, while colder water sinks to the bottom. This process repeats as long as there is water in the pot. Radiation is a form of heat transfer that occurs when electromagnetic radiation is emitted or absorbed. Electromagnetic Access for free at openstax.org. 11.2 \u2022 Heat, Specific Heat, and Heat Transfer 337 radiation includes radio waves, microwaves, infrared radiation, visible light, ultraviolet radiation, X-rays, and gamma rays, all of which have different wavelengths and amounts of energy (shorter wavelengths have higher frequency and more energy). You can feel the heat transfer from a fire and from the sun. Similarly, you can sometimes tell that the oven is hot without touching its door or looking inside\u2014it may just warm you as you walk by. Another example is thermal radiation from the human body; people are constantly emitting infrared radiation, which is not visible to the human eye, but is felt as heat. Radiation is the only method of heat transfer where no medium is required, meaning that the heat doesn\u2019t need to come into direct contact with or be transported by any matter. The space between Earth and the sun is largely empty, without any possibility of heat transfer by convection or conduction. Instead, heat is transferred by radiation, and Earth is warmed as it absorbs electromagnetic radiation emitted by the sun. Figure 11.7 Most of the heat transfer from this fire to the observers is through infrared radiation. The visible light transfers relatively little thermal energy. Since skin is very sensitive to infrared radiation, you can sense the presence of a fire without looking at it directly. (Daniel X. O\u2019Neil) All objects absorb and emit electromagnetic radiation (see Figure 11.7). The rate of heat transfer by radiation depends mainly on the color of the object. Black is the most effective absorber and radiator, and white is the least effective. People living in hot climates generally avoid wearing black clothing, for instance. Similarly, black asphalt in a parking lot will be hotter than adjacent patches of grass on a summer day, because black absorbs better than green. The reverse is also true\u2014black radiates better than green. On a clear summer night, the black asphalt will be colder than the green patch of grass, because black radiates energy faster than green. In contrast, white is a poor absorber and also a poor radiator. A white object reflects nearly all radiation, like a mirror. Virtual Physics Energy Forms and Changes Click to view content (http://www.openstax.org/l/28energyForms) In this animation, you will explore heat transfer with different materials. Experiment with heating and cooling the iron, brick, and water. This is done by dragging and dropping the object onto the pedestal and then holding the lever either to Heat or Cool. Drag a thermometer beside each object to measure its temperature\u2014you can watch how quickly it heats or cools in real time. Now let\u2019s try transferring heat between objects. Heat the brick and then place it in the cool water. Now heat the brick again, but then place it on top of the iron. What do you notice? Selecting the fast forward option lets you speed up the heat transfers, to save time. GRASP CHECK Compare how quickly the different materials are heated or cooled. Based on these results", ", what material do you think has the greatest specific heat? Why? Which has the smallest specific heat? Can you think of a real-world situation where you would want to use an object with large specific heat? a. Water will take the longest, and iron will take the shortest time to heat, as well as to cool. Objects with greater specific heat would be desirable for insulation. For instance, woolen clothes with large specific heat would prevent heat loss from the body. 338 Chapter 11 \u2022 Thermal Energy, Heat, and Work b. Water will take the shortest, and iron will take the longest time to heat, as well as to cool. Objects with greater specific heat would be desirable for insulation. For instance, woolen clothes with large specific heat would prevent heat loss from the body. c. Brick will take shortest and iron will take longest time to heat up as well as to cool down. Objects with greater specific heat would be desirable for insulation. For instance, woolen clothes with large specific heat would prevent heat loss from the body. d. Water will take shortest and brick will take longest time to heat up as well as to cool down. Objects with greater specific heat would be desirable for insulation. For instance, woolen clothes with large specific heat would prevent heat loss from the body. Solving Heat Transfer Problems WORKED EXAMPLE Calculating the Required Heat: Heating Water in an Aluminum Pan A 0.500 kg aluminum pan on a stove is used to heat 0.250 L of water from 20.0 What percentage of the heat is used to raise the temperature of (b) the pan and (c) the water? STRATEGY The pan and the water are always at the same temperature. When you put the pan on the stove, the temperature of the water and the pan is increased by the same amount. We use the equation for heat transfer for the given temperature change and masses of water and aluminum. The specific heat values for water and aluminum are given in the previous table.. (a) How much heat is required? to 80.0 Solution to (a) Because the water is in thermal contact with the aluminum, the pan and the water are at the same temperature. 1. Calculate the temperature difference. 11.8 2. Calculate the mass of water using the relationship between density, mass, and volume. Density is mass per unit volume, or. Rearranging this equation, solve for the mass of water. 3. Calculate the heat transferred to the water. Use the specific heat of water in the previous table. 4. Calculate the heat transferred to the aluminum. Use the specific heat for aluminum in the previous table. 5. Find the total transferred heat. Solution to (b) The percentage of heat going into heating the pan is Solution to (c) The percentage of heat going into heating the water is 11.9 11.10 11.11 11.12 11.13 11.14 Discussion In this example, most of the total heat transferred is used to heat the water, even though the pan has twice as much mass. This is Access for free at openstax.org. 11.2 \u2022 Heat, Specific Heat, and Heat Transfer 339 because the specific heat of water is over four times greater than the specific heat of aluminum. Therefore, it takes a bit more than twice as much heat to achieve the given temperature change for the water than for the aluminum pan. Water can absorb a tremendous amount of energy with very little resulting temperature change. This property of water allows for life on Earth because it stabilizes temperatures. Other planets are less habitable because wild temperature swings make for a harsh environment. You may have noticed that climates closer to large bodies of water, such as oceans, are milder than climates landlocked in the middle of a large continent. This is due to the climate-moderating effect of water\u2019s large heat capacity\u2014water stores large amounts of heat during hot weather and releases heat gradually when it\u2019s cold outside. WORKED EXAMPLE Calculating Temperature Increase: Truck Brakes Overheat on Downhill Runs When a truck headed downhill brakes, the brakes must do work to convert the gravitational potential energy of the truck to internal energy of the brakes. This conversion prevents the gravitational potential energy from being converted into kinetic energy of the truck, and keeps the truck from speeding up and losing control. The increased internal energy of the brakes raises their temperature. When the hill is especially steep, the temperature increase may happen too quickly and cause the brakes to overheat. Calculate the temperature increase of 100 kg of brake material with an average specific heat of 800 J/kg truck descending 75.0 m (in vertical displacement) at a constant speed. from a 10,000 kg STRATEGY We first calculate the gravitational potential energy (Mgh) of the truck, and then find the temperature increase produced in the brakes. Solution 1. Calculate the change in gravitational potential energy as the truck goes downhill. 2. Calculate the temperature change from the heat", " transferred by rearranging the equation to solve for 11.15 11.16 where mis the mass of the brake material (not the entire truck). Insert the values Q= 7.35\u00d7106 J (since the heat transfer is equal to the change in gravitational potential energy), m 100 kg, and c 800 J/kg to find 11.17 Discussion This temperature is close to the boiling point of water. If the truck had been traveling for some time, then just before the descent, the brake temperature would likely be higher than the ambient temperature. The temperature increase in the descent would likely raise the temperature of the brake material above the boiling point of water, which would be hard on the brakes. This is why truck drivers sometimes use a different technique for called \u201cengine braking\u201d to avoid burning their brakes during steep descents. Engine braking is using the slowing forces of an engine in low gear rather than brakes to slow down. 340 Chapter 11 \u2022 Thermal Energy, Heat, and Work Practice Problems 5. How much heat does it take to raise the temperature of 10.0 kg of water by 1.0 \u00b0C? a. 84 J b. 42 J c. 84 kJ d. 42 kJ 6. Calculate the change in temperature of 1.0 kg of water that is initially at room temperature if 3.0 kJ of heat is added. 358 \u00b0C a. 716 \u00b0C b. c. 0.36 \u00b0C d. 0.72 \u00b0C Check Your Understanding 7. What causes heat transfer? a. The mass difference between two objects causes heat transfer. b. The density difference between two objects causes heat transfer. c. The temperature difference between two systems causes heat transfer. d. The pressure difference between two objects causes heat transfer. 8. When two bodies of different temperatures are in contact, what is the overall direction of heat transfer? a. The overall direction of heat transfer is from the higher-temperature object to the lower-temperature object. b. The overall direction of heat transfer is from the lower-temperature object to the higher-temperature object. c. The direction of heat transfer is first from the lower-temperature object to the higher-temperature object, then back again to the lower-temperature object, and so-forth, until the objects are in thermal equilibrium. d. The direction of heat transfer is first from the higher-temperature object to the lower-temperature object, then back again to the higher-temperature object, and so-forth, until the objects are in thermal equilibrium. 9. What are the different methods of heat transfer? conduction, radiation, and reflection conduction, reflection, and convection convection, radiation, and reflection conduction, radiation, and convection a. b. c. d. 10. True or false\u2014Conduction and convection cannot happen simultaneously a. True b. False 11.3 Phase Change and Latent Heat Section Learning Objectives By the end of this section, you will be able to do the following: \u2022 Explain changes in heat during changes of state, and describe latent heats of fusion and vaporization \u2022 Solve problems involving thermal energy changes when heating and cooling substances with phase changes Section Key Terms condensation freezing latent heat sublimation latent heat of fusion latent heat of vaporization melting vaporization phase change phase diagram plasma Access for free at openstax.org. 11.3 \u2022 Phase Change and Latent Heat 341 Phase Changes So far, we have learned that adding thermal energy by heat increases the temperature of a substance. But surprisingly, there are situations where adding energy does not change the temperature of a substance at all! Instead, the additional thermal energy acts to loosen bonds between molecules or atoms and causes a phase change. Because this energy enters or leaves a system during a phase change without causing a temperature change in the system, it is known as latent heat (latent means hidden). The three phases of matter that you frequently encounter are solid, liquid and gas (see Figure 11.8). Solid has the least energetic state; atoms in solids are in close contact, with forces between them that allow the particles to vibrate but not change position with neighboring particles. (These forces can be thought of as springs that can be stretched or compressed, but not easily broken.) Liquid has a more energetic state, in which particles can slide smoothly past one another and change neighbors, although they are still held together by their mutual attraction. Gas has a more energetic state than liquid, in which particles are broken free of their bonds. Particles in gases are separated by distances that are large compared with the size of the particles. The most energetic state of all is plasma. Although you may not have heard much about plasma, it is actually the most common state of matter in the universe\u2014stars are made up of plasma, as is lightning. The plasma state is reached by heating a gas to the point where particles are pulled apart, separating the electrons from the rest of the particle. This produces an ion", "ized gas that is a combination of the negatively charged free electrons and positively charged ions, known as plasma. Figure 11.8 (a) Particles in a solid always have the same neighbors, held close by forces represented here by springs. These particles are essentially in contact with one another. A rock is an example of a solid. This rock retains its shape because of the forces holding its atoms or molecules together. (b) Particles in a liquid are also in close contact but can slide over one another. Forces between them strongly resist attempts to push them closer together and also hold them in close contact. Water is an example of a liquid. Water can flow, but it also remains in an open container because of the forces between its molecules. (c) Particles in a gas are separated by distances that are considerably larger than the size of the particles themselves, and they move about freely. A gas must be held in a closed container to prevent it from moving out into its surroundings. (d) The atmosphere is ionized in the extreme heat of a lightning strike. During a phase change, matter changes from one phase to another, either through the addition of energy by heat and the transition to a more energetic state, or from the removal of energy by heat and the transition to a less energetic state. Phase changes to a more energetic state include the following: \u2022 Melting\u2014Solid to liquid \u2022 Vaporization\u2014Liquid to gas (included boiling and evaporation) \u2022 Sublimation\u2014Solid to gas Phase changes to a less energetic state are as follows: \u2022 Condensation\u2014Gas to liquid \u2022 Freezing\u2014Liquid to solid Energy is required to melt a solid because the bonds between the particles in the solid must be broken. Since the energy involved in a phase changes is used to break bonds, there is no increase in the kinetic energies of the particles, and therefore no rise in temperature. Similarly, energy is needed to vaporize a liquid to overcome the attractive forces between particles in the liquid. There is no temperature change until a phase change is completed. The temperature of a cup of soda and ice that is initially at 0 stays at 0 until all of the ice has melted. In the reverse of these processes\u2014freezing and condensation\u2014energy is released 342 Chapter 11 \u2022 Thermal Energy, Heat, and Work from the latent heat (see Figure 11.9). Figure 11.9 (a) Energy is required to partially overcome the attractive forces between particles in a solid to form a liquid. That same energy must be removed for freezing to take place. (b) Particles are separated by large distances when changing from liquid to vapor, requiring significant energy to overcome molecular attraction. The same energy must be removed for condensation to take place. There is no temperature change until a phase change is completed. (c) Enough energy is added that the liquid state is skipped over completely as a substance undergoes sublimation. The heat, Q, required to change the phase of a sample of mass mis (for melting/freezing), (for vaporization/condensation), is the latent heat of fusion, and where needed to cause a phase change between solid and liquid. The latent heat of vaporization is the amount of heat needed to cause a is the latent heat of vaporization. The latent heat of fusion is the amount of heat Access for free at openstax.org. 11.3 \u2022 Phase Change and Latent Heat 343 phase change between liquid and gas. strength of intermolecular forces, and both have standard units of J/kg. See Table 11.3 for values of substances. are coefficients that vary from substance to substance, depending on the of different and and Substance Melting Point ( ) Lf (kJ/kg) Boiling Point ( ) Lv (kJ/kg) Helium \u2012269.7 Hydrogen \u2012259.3 Nitrogen \u2012210.0 Oxygen \u2012218.8 Ethanol \u2012114 Ammonia \u201278 Mercury \u201238.9 Water 0.00 Sulfur Lead 119 327 Antimony 631 Aluminum 660 Silver Gold Copper 961 1063 1083 Uranium 1133 Tungsten 3410 5.23 58.6 25.5 13.8 104 332 11.8 334 38.1 24.5 165 380 88.3 64.5 134 84 184 \u2012268.9 \u2012252.9 \u2012195.8 \u2012183.0 78.3 \u201233.4 357 100.0 444.6 1750 1440 2520 2193 2660 2595 3900 5900 20.9 452 201 213 854 1370 272 2256 326 871 561 11400 2336 1578 5069 1900 4810 Table 11.3 Latent Heats of Fusion and Vaporization, along with Melting and Boiling Points Let\u2019s consider the example of adding heat to ice to examine its transitions through all three phases\u2014solid to liquid to gas. A phase diagram indicating the temperature", " changes of water as energy is added is shown in Figure 11.10. The ice starts out at \u221220, and its temperature rises linearly, absorbing heat at a constant rate until it reaches 0 Once at this temperature, the ice gradually melts, absorbing 334 kJ/kg. The temperature remains constant at 0 melted, the temperature of the liquid water rises, absorbing heat at a new constant rate. At 100 the temperature again remains constant while the water absorbs 2256 kJ/kg during this phase change. When all the liquid has become steam, the temperature rises again at a constant rate. during this phase change. Once all the ice has, the water begins to boil and 344 Chapter 11 \u2022 Thermal Energy, Heat, and Work Figure 11.10 A graph of temperature versus added energy. The system is constructed so that no vapor forms while ice warms to become liquid water, and so when vaporization occurs, the vapor remains in the system. The long stretches of constant temperature values at 0 and 100 reflect the large latent heats of melting and vaporization, respectively. We have seen that vaporization requires heat transfer to a substance from its surroundings. Condensation is the reverse process, where heat in transferred away froma substance toits surroundings. This release of latent heat increases the temperature of the surroundings. Energy must be removed from the condensing particles to make a vapor condense. This is why condensation occurs on cold surfaces: the heat transfers energy away from the warm vapor to the cold surface. The energy is exactly the same as that required to cause the phase change in the other direction, from liquid to vapor, and so it can be calculated from. Latent heat is also released into the environment when a liquid freezes, and can be calculated from. FUN IN PHYSICS Making Ice Cream Figure 11.11 With the proper ingredients, some ice and a couple of plastic bags, you could make your own ice cream in five minutes. (ElinorD, Wikimedia Commons) Ice cream is certainly easy enough to buy at the supermarket, but for the hardcore ice cream enthusiast, that may not be satisfying enough. Going through the process of making your own ice cream lets you invent your own flavors and marvel at the physics firsthand (Figure 11.11). The first step to making homemade ice cream is to mix heavy cream, whole milk, sugar, and your flavor of choice; it could be as Access for free at openstax.org. 11.3 \u2022 Phase Change and Latent Heat 345 simple as cocoa powder or vanilla extract, or as fancy as pomegranates or pistachios. The next step is to pour the mixture into a container that is deep enough that you will be able to churn the mixture without it spilling over, and that is also freezer-safe. After placing it in the freezer, the ice cream has to be stirred vigorously every 45 minutes for four to five hours. This slows the freezing process and prevents the ice cream from turning into a solid block of ice. Most people prefer a soft creamy texture instead of one giant popsicle. As it freezes, the cream undergoes a phase change from liquid to solid. By now, we\u2019re experienced enough to know that this means that the cream must experience a loss of heat. Where does that heat go? Due to the temperature difference between the freezer and the ice cream mixture, heat transfers thermal energy from the ice cream to the air in the freezer. Once the temperature in the freezer rises enough, the freezer is cooled by pumping excess heat outside into the kitchen. A faster way to make ice cream is to chill it by placing the mixture in a plastic bag, surrounded by another plastic bag half full of ice. (You can also add a teaspoon of salt to the outer bag to lower the temperature of the ice/salt mixture.) Shaking the bag for five minutes churns the ice cream while cooling it evenly. In this case, the heat transfers energy out of the ice cream mixture and into the ice during the phase change. This video (http://www.openstax.org/l/28icecream) gives a demonstration of how to make home-made ice cream using ice and plastic bags. GRASP CHECK Why does the ice bag method work so much faster than the freezer method for making ice cream? a. Ice has a smaller specific heat than the surrounding air in a freezer. Hence, it absorbs more energy from the ice-cream mixture. Ice has a smaller specific heat than the surrounding air in a freezer. Hence, it absorbs less energy from the ice-cream mixture. Ice has a greater specific heat than the surrounding air in a freezer. Hence, it absorbs more energy from the ice-cream mixture. Ice has a greater specific heat than the surrounding air in a freezer. Hence, it absorbs less energy from the ice-cream mixture. b. c. d. Solving Thermal Energy Problems with Phase Changes WORKED EXAMPLE Calculating Heat Required for a Phase Change Calculate a) how much energy is needed to", " melt 1.000 kg of ice at 0 vaporize 1.000 kg of water at 100 STRATEGY FOR (A) Using the equation for the heat required for melting, and the value of the latent heat of fusion of water from the previous table, we can solve for part (a). (freezing point), and b) how much energy is required to (boiling point). Solution to (a) The energy to melt 1.000 kg of ice is STRATEGY FOR (B) To solve part (b), we use the equation for heat required for vaporization, along with the latent heat of vaporization of water from the previous table. 11.18 Solution to (b) The energy to vaporize 1.000 kg of liquid water is 11.19 346 Chapter 11 \u2022 Thermal Energy, Heat, and Work Discussion The amount of energy need to melt a kilogram of ice (334 kJ) is the same amount of energy needed to raise the temperature of 1.000 kg of liquid water from 0 energy associated with temperature changes. It also demonstrates that the amount of energy needed for vaporization is even greater.. This example shows that the energy for a phase change is enormous compared to to 79.8 WORKED EXAMPLE and with a mass of Calculating Final Temperature from Phase Change: Cooling Soda with Ice Cubes Ice cubes are used to chill a soda at 20 cubes is 0.018 kg. Assume that the soda is kept in a foam container so that heat loss can be ignored, and that the soda has the same specific heat as water. Find the final temperature when all of the ice has melted. STRATEGY The ice cubes are at the melting temperature of 0 occurs in two steps: first, the phase change occurs and solid (ice) transforms into liquid water at the melting temperature; then,, so more heat is transferred from the soda to this water until the temperature of this water rises. Melting yields water at 0 they are the same temperature. Since the amount of heat leaving the soda is the same as the amount of heat transferred to the ice.. Heat is transferred from the soda to the ice for melting. Melting of ice and the total mass of the ice. The ice is at 0 11.20 The heat transferred to the ice goes partly toward the phase change (melting), and partly toward raising the temperature after melting. Recall from the last section that the relationship between heat and temperature change is temperature change is. The total heat transferred to the ice is therefore. For the ice, the Since the soda doesn\u2019t change phase, but only temperature, the heat given off by the soda is Since, 11.21 11.22 11.23 Bringing all terms involving to the left-hand-side of the equation, and all other terms to the right-hand-side, we can solve for. Substituting the known quantities 11.24 11.25 Discussion This example shows the enormous energies involved during a phase change. The mass of the ice is about 7 percent the mass of the soda, yet it causes a noticeable change in the soda\u2019s temperature. TIPS FOR SUCCESS If the ice were not already at the freezing point, we would also have to factor in how much energy would go into raising its temperature up to 0 often below 0, before the phase change occurs. This would be a realistic scenario, because the temperature of ice is. Access for free at openstax.org. Practice Problems 11. How much energy is needed to melt 2.00 kg of ice at 0 \u00b0C? 11.3 \u2022 Phase Change and Latent Heat 347 334 kJ a. 336 kJ b. c. 167 kJ d. 668 kJ 12. If a. b. c. d. of energy is just enough to melt of a substance, what is the substance\u2019s latent heat of fusion? Check Your Understanding 13. What is latent heat? a. b. c. d. It is the heat that must transfer energy to or from a system in order to cause a mass change with a slight change in the temperature of the system. It is the heat that must transfer energy to or from a system in order to cause a mass change without a temperature change in the system. It is the heat that must transfer energy to or from a system in order to cause a phase change with a slight change in the temperature of the system. It is the heat that must transfer energy to or from a system in order to cause a phase change without a temperature change in the system. 14. In which phases of matter are molecules capable of changing their positions? a. gas, liquid, solid liquid, plasma, solid b. c. liquid, gas, plasma d. plasma, gas, solid 348 Chapter 11 \u2022 Key Terms KEY TERMS absolute zero lowest possible temperature; the temperature at which all molecular motion ceases Kelvin scale temperature scale in which 0 K is the lowest possible temperature, representing absolute zero Celsius scale temperature scale in which the freezing point", " latent heat heat related to the phase change of a substance of water is 0 at 1 atm of pressure and the boiling point of water is 100 condensation phase change from gas to liquid conduction heat transfer through stationary matter by physical contact convection heat transfer by the movement of fluid degree Celsius unit on the Celsius temperature scale degree Fahrenheit unit on the Fahrenheit temperature scale Fahrenheit scale temperature scale in which the freezing and the boiling point of water is point of water is 32 212 freezing phase change from liquid to solid heat transfer of thermal (or internal) energy due to a temperature difference heat capacity amount of heat necessary to change the rather than a change of temperature latent heat of fusion amount of heat needed to cause a phase change between solid and liquid latent heat of vaporization amount of heat needed to cause a phase change between liquid and gas melting phase change from solid to liquid phase change transition between solid, liquid, or gas states of a substance plasma ionized gas that is a combination of the negatively charged free electrons and positively charged ions radiation energy transferred by electromagnetic waves specific heat amount of heat necessary to change the temperature of 1.00 kg of a substance by 1.00 sublimation phase change from solid to gas temperature quantity measured by a thermometer thermal energy average random kinetic energy of a temperature of a substance by 1.00 molecule or an atom Kelvin unit on the Kelvin temperature scale; note that it is vaporization phase change from liquid to gas never referred to in terms of \u201cdegrees\u201d Kelvin SECTION SUMMARY 11.1 Temperature and Thermal Energy \u2022 Temperature is the quantity measured by a thermometer. \u2022 Temperature is related to the average kinetic energy of atoms and molecules in a system. \u2022 Absolute zero is the temperature at which there is no molecular motion. \u2022 There are three main temperature scales: Celsius, Fahrenheit, and Kelvin. \u2022 Temperatures on one scale can be converted into temperatures on another scale. 11.2 Heat, Specific Heat, and Heat Transfer \u2022 Heat is thermal (internal) energy transferred due to a temperature difference. \u2022 The transfer of heat Qthat leads to a change temperature of a body with mass m is where cis the specific heat of the material. \u2022 Heat is transferred by three different methods: in the, conduction, convection, and radiation. \u2022 Heat conduction is the transfer of heat between two objects in direct contact with each other. \u2022 Convection is heat transfer by the movement of mass. \u2022 Radiation is heat transfer by electromagnetic waves. 11.3 Phase Change and Latent Heat \u2022 Most substances have four distinct phases: solid, liquid, gas, and plasma. \u2022 Gas is the most energetic state and solid is the least. \u2022 During a phase change, a substance undergoes transition to a higher energy state when heat is added, or to a lower energy state when heat is removed. \u2022 Heat is added to a substance during melting and vaporization. \u2022 Latent heat is released by a substance during condensation and freezing. \u2022 Phase changes occur at fixed temperatures called boiling and freezing (or melting) points for a given substance. Access for free at openstax.org. KEY EQUATIONS 11.1 Temperature and Thermal Energy 11.2 Heat, Specific Heat, and Heat Transfer Chapter 11 \u2022 Key Equations 349 Celsius to Fahrenheit conversion Fahrenheit to Celsius conversion Celsius to Kelvin conversion Kelvin to Celsius conversion Fahrenheit to Kelvin conversion Kelvin to Fahrenheit conversion heat transfer density 11.3 Phase Change and Latent Heat heat transfer for melting/freezing phase change heat transfer for vaporization/ condensation phase change CHAPTER REVIEW Concept Items 11.1 Temperature and Thermal Energy 1. A glass of water has a temperature of 31 degrees Celsius. solid liquid What state of matter is it in? a. b. c. gas d. plasma 2. What is the difference between thermal energy and internal energy? a. The thermal energy of the system is the average kinetic energy of the system\u2019s constituent particles due to their motion. The total internal energy of the system is the sum of the kinetic energies and the potential energies of its constituent particles. b. The thermal energy of the system is the average potential energy of the system\u2019s constituent particles due to their motion. The total internal energy of the system is the sum of the kinetic energies and the potential energies of its constituent particles. c. The thermal energy of the system is the average kinetic energy of the system\u2019s constituent particles due to their motion. The total internal energy of the system is the sum of the kinetic energies of its constituent particles. d. The thermal energy of the system is the average potential energy of the systems\u2019 constituent particles due to their motion. The total internal energy of the system is the sum of the kinetic energies of its constituent particles. 3. What does the Celsius scale use as a reference point? a. The boiling point of mercury b. The boiling point of wax c. The freezing point of water d. The freezing point of wax 11.2 Heat, Specific Heat, and Heat Transfer 4. What are the SI units of specific heat? a", ". b. c. d. 5. What is radiation? a. The transfer of energy through emission and absorption of the electromagnetic waves is known as radiation. b. The transfer of energy without any direct physical 350 Chapter 11 \u2022 Chapter Review contact between any two substances. c. The transfer of energy through direct physical contact between any two substances. d. The transfer of energy by means of the motion of fluids at different temperatures and with different densities. 11.3 Phase Change and Latent Heat 6. Why is there no change in temperature during a phase change, even if energy is absorbed by the system? a. The energy is used to break bonds between particles, and so does not increase the potential energy of the system\u2019s particles. b. The energy is used to break bonds between particles, Critical Thinking Items 11.1 Temperature and Thermal Energy 8. The temperature of two equal quantities of water needs to be raised - the first container by degrees Celsius and the second by degrees Fahrenheit. Which one would require more heat? a. The heat required by the first container is more than the second because each degree Celsius is equal to degrees Fahrenheit. b. The heat required by the first container is less than the second because each degree Fahrenheit is equal to degrees Celsius. c. The heat required by the first container is more than the second because each degree Celsius is equal to degrees Fahrenheit. d. The heat required by the first container is less than the second because each degree Fahrenheit is equal to degrees Celsius. 9. What is 100.00 \u00b0C in kelvins? a. 212.00 K b. 100.00 K c. 473.15 K 373.15 K d. 11.2 Heat, Specific Heat, and Heat Transfer 10. The value of specific heat is the same whether the units are J/kg\u22c5K or J/kg\u22c5\u00baC. How? a. Temperature difference is dependent on the chosen temperature scale. b. Temperature change is different in units of kelvins and degrees Celsius. c. Reading of temperatures in kelvins and degree Celsius are the same. Access for free at openstax.org. and so increases the potential energy of the system\u2019s particles. c. The energy is used to break bonds between particles, and so does not increase the kinetic energy of the system\u2019s particles. d. The energy is used to break bonds between particles, and so increases the kinetic energy of the system\u2019s particles. 7. In which two phases of matter do atoms and molecules have the most distance between them? a. gas and solid b. gas and liquid c. gas and plasma d. liquid and plasma d. The temperature change is the same in units of kelvins and degrees Celsius. 11. If the thermal energy of a perfectly black object is increased by conduction, will the object remain black in appearance? Why or why not? a. No, the energy of the radiation increases as the temperature increases, and the radiation becomes visible at certain temperatures. b. Yes, the energy of the radiation decreases as the temperature increases, and the radiation remains invisible at those energies. c. No, the energy of the radiation decreases as the temperature increases, until the frequencies of the radiation are the same as those of visible light. d. Yes, as the temperature increases, and the energy is transferred from the object by other mechanisms besides radiation, so that the energy of the radiation does not increase. 12. What is the specific heat of a substance that requires 5.00 kJ of heat to raise the temperature of 3.00 kg by 5.00 \u00b0F? 3.33\u00d7103 J/kg \u22c5\u00b0 C a. b. 6.00\u00d7103 J/kg \u22c5\u00b0 C 3.33\u00d7102 J/kg \u22c5 \u00b0 C c. d. 6.00\u00d7102 J/kg \u22c5 \u00b0 C 11.3 Phase Change and Latent Heat 13. Assume 1.0 kg of ice at 0 \u00b0C starts to melt. It absorbs 300 kJ of energy by heat. What is the temperature of the water afterwards? a. 10 \u00b0C b. 20 \u00b0C c. 5 \u00b0C d. 0 \u00b0C Problems 11.1 Temperature and Thermal Energy 14. What is 35.0 \u00b0F in kelvins? 1.67 K a. 35.0 K b. c. -271.5 K d. 274.8 K 15. Design a temperature scale where the freezing point of water is 0 degrees and its boiling point is 70 degrees. What would be the room temperature on this scale? a. If room temperature is 25.0 \u00b0C, the temperature on the new scale will be 17.5 \u00b0. If room temperature is 25.0 \u00b0C, the temperature on the new scale will be 25.0\u00b0. If the room temperature is 25.0 \u00b0C, the temperature on the new scale will be 35.7\u00b0.", " If the room temperature is 25.0 \u00b0C, the temperature on the new scale will be 50.0\u00b0. b. c. d. 11.2 Heat, Specific Heat, and Heat Transfer 16. A certain quantity of water is given 4.0 kJ of heat. This raises its temperature by 30.0 \u00b0F. What is the mass of the water in grams? 5.7 g a. 570 g b. Performance Task 11.3 Phase Change and Latent Heat 20. You have been tasked with designing a baking pan that will bake batter the fastest. There are four materials available for you to test. \u2022 Four pans of similar design, consisting of aluminum, iron (steel), copper, and glass \u2022 Oven or similar heating source \u2022 Device for measuring high temperatures \u2022 Balance for measuring mass Instructions Procedure 1. Design a safe experiment to test the specific heat of each material (i.e., no extreme temperatures TEST PREP Multiple Choice 11.1 Temperature and Thermal Energy 21. The temperature difference of is the same as Chapter 11 \u2022 Test Prep 351 c. d. 5700 g 57 g 17. 5290 J of heat is given to 0.500 kg water at 15.00 \u00b0C. What will its final temperature be? a. 15.25\u00b0 C 12.47 \u00b0 C b. c. 40.3\u00b0 C 17.53\u00b0 C d. 11.3 Phase Change and Latent Heat 18. How much energy would it take to heat 1.00 kg of ice at 0 \u00b0C to water at 15.0 \u00b0C? a. 271 kJ b. 334 kJ c. 62.8 kJ 397 kJ d. 19. Ice cubes are used to chill a soda with a mass msoda = 0.300 kg at 15.0 \u00b0C. The ice is at 0 \u00b0C, and the total mass of the ice cubes is 0.020 kg. Assume that the soda is kept in a foam container so that heat loss can be ignored, and that the soda has the same specific heat as water. Find the final temperature when all ice has melted. a. 19.02 \u00b0C b. 90.3 \u00b0C c. 0.11 \u00b0C d. 9.03 \u00b0C should be used) 2. Write down the materials needed for your experiment and the procedure you will follow. Make sure that you include every detail, so that the experiment can be repeated by others. 3. Carry out the experiment and record any data collected. 4. Review your results and make a recommendation as to which metal should be used for the pan. a. What physical quantities do you need to measure to determine the specific heats for the different materials? b. How does the glass differ from the metals in terms of thermal properties? c. What are your sources of error? a. b. c. d. degree Celsius degree Fahrenheit degrees Celsius degrees Fahrenheit 352 Chapter 11 \u2022 Test Prep 22. What is the preferred temperature scale used in celsius fahrenheit scientific laboratories? a. b. c. kelvin d. rankine 11.2 Heat, Specific Heat, and Heat Transfer 23. Which phase of water has the largest specific heat? solid liquid a. b. c. gas 24. What kind of heat transfer requires no medium? a. b. c. d. conduction convection reflection radiation 25. Which of these substances has the greatest specific heat? a. copper b. mercury c. aluminum d. wood 26. Give an example of heat transfer through convection. a. The energy emitted from the filament of an electric bulb b. The energy coming from the sun c. A pan on a hot burner d. Water boiling in a pot 11.3 Phase Change and Latent Heat 27. What are the SI units of latent heat? Short Answer 11.1 Temperature and Thermal Energy 31. What is absolute zeroon the Fahrenheit scale? a. 0 \u00b0F 32 \u00b0F b. -273.15 \u00b0F c. -459.67 \u00b0F d. 32. What is absolute zeroon the Celsius scale? a. 0 \u00b0C b. 273.15 \u00b0C c. d. -459.67 \u00b0C -273.15 \u00b0C 33. A planet\u2019s atmospheric pressure is such that water there boils at a lower temperature than it does at sea level on Access for free at openstax.org. a. b. c. d. 28. Which substance has the largest latent heat of fusion? a. gold b. water c. mercury tungsten d. 29. In which phase changes does matter undergo a transition to a more energetic state? a. freezing and vaporization b. melting and sublimation c. melting and vaporization d. melting and freezing 30. A room has a window made from thin glass. The room is colder than the air outside. There is some condensation on the glass window. On which side of the glass would the condensation most likely be found?", " a. Condensation is on the outside of the glass when the cool, dry air outside the room comes in contact with the cold pane of glass. b. Condensation is on the outside of the glass when the warm, moist air outside the room comes in contact with the cold pane of glass. c. Condensation is on the inside of the glass when the cool, dry air inside the room comes in contact with the cold pane of glass. d. Condensation is on the inside of the glass when the warm, moist air inside the room comes in contact with the cold pane of glass. Earth. If a Celsius scale is derived on this planet, will it be the same as that on Earth? a. The Celsius scale derived on the planet will be the same as that on Earth, because the Celsius scale is independent of the freezing and boiling points of water. b. The Celsius scale derived on that planet will not be the same as that on Earth, because the Celsius scale is dependent and derived by using the freezing and boiling points of water. c. The Celsius scale derived on the planet will be the same as that on Earth, because the Celsius scale is an absolute temperature scale based on molecular motion, which is independent of pressure. d. The Celsius scale derived on the planet will not be the same as that on Earth, but the Fahrenheit scale Chapter 11 \u2022 Test Prep 353 will be the same, because its reference temperatures are not based on the freezing and boiling points of water. b. 63 \u00b0C c. d. 6.3 \u00b0C 1.8\u00d710-2 \u00b0C 34. What is the difference between the freezing point and 40. Aluminum has a specific heat of 900 J/kg\u00b7\u00baC. How much boiling point of water on the Reaumur scale? a. The boiling point of water is 80\u00b0 on the Reaumur scale. b. Reaumur scale is less than 120\u00b0. c. 100\u00b0 d. 80\u00b0 energy would it take to change the temperature of 2 kg aluminum by 3 \u00baC? a. 1.3 kJ b. 0.60 kJ 54 kJ c. 5.4 kJ d. 11.2 Heat, Specific Heat, and Heat Transfer 35. In the specific heat equation what does cstand for? a. Total heat b. Specific heat c. Specific temperature d. Specific mass 36. Specific heat may be measured in J/kg \u00b7 K, J/kg \u00b7 \u00b0C. What other units can it be measured in? a. kg/kcal \u00b7 \u00b0C b. kcal \u00b7 \u00b0C/kg c. kg \u00b7 \u00b0C/kcal d. kcal/kg \u00b7 \u00b0C 37. What is buoyancy? a. Buoyancy is a downward force exerted by a solid that opposes the weight of an object. b. Buoyancy is a downward force exerted by a fluid that opposes the weight of an immersed object. c. Buoyancy is an upward force exerted by a solid that opposes the weight an object. d. Buoyancy is an upward force exerted by a fluid that opposes the weight of an immersed object. 38. Give an example of convection found in nature. a. heat transfer through metallic rod b. heat transfer from the sun to Earth c. heat transfer through ocean currents d. heat emitted by a light bulb into its environment 39. Calculate the temperature change in a substance with specific heat 735 J/kg \u00b7 \u00b0C when 14 kJ of heat is given to a 3.0-kg sample of that substance. a. 57 \u00b0C 11.3 Phase Change and Latent Heat 41. Upon what does the required amount of heat removed to freeze a sample of a substance depend? a. The mass of the substance and its latent heat of vaporization b. The mass of the substance and its latent heat of fusion c. The mass of the substance and its latent heat of sublimation d. The mass of the substance only 42. What do latent heats, Lf and Lv, depend on? a. Lf and Lv depend on the forces between the particles in the substance. b. Lf and Lv depend on the mass of the substance. c. Lf and Lv depend on the volume of the substance. d. Lf and Lv depend on the temperature of the substance. 43. How much energy is required to melt 7.00 kg a block of aluminum that is at its melting point? (Latent heat of fusion of aluminum is 380 kJ/kg.) 54.3 kJ a. b. 2.66 kJ c. 0.0184 kJ d. 2.66\u00d7103 kJ 44. A 3.00 kg sample of a substance is at its boiling point. If 5,360 kJ of energy are enough to boil away the entire substance, what is its latent heat of vaporization? a. 2,685 kJ/kg b. 3,580 kJ/kg c. 895 kJ/", "kg d. 1,790 kJ/kg Extended Response 11.1 Temperature and Thermal Energy 45. What is the meaning of absolute zero? a. It is the temperature at which the internal energy of the system is maximum, because the speed of its b. constituent particles increases to maximum at this point. It is the temperature at which the internal energy of the system is maximum, because the speed of its constituent particles decreases to zero at this point. 354 Chapter 11 \u2022 Test Prep c. d. It is the temperature at which the internal energy of the system approaches zero, because the speed of its constituent particles increases to a maximum at this point. It is that temperature at which the internal energy of the system approaches zero, because the speed of its constituent particles decreases to zero at this point. 46. Why does it feel hotter on more humid days, even though there is no difference in temperature? a. On hot, dry days, the evaporation of the sweat from the skin cools the body, whereas on humid days the concentration of water in the atmosphere is lower, which reduces the evaporation rate from the skin\u2019s surface. b. On hot, dry days, the evaporation of the sweat from the skin cools the body, whereas on humid days the concentration of water in the atmosphere is higher, which reduces the evaporation rate from the skin\u2019s surface. c. On hot, dry days, the evaporation of the sweat from the skin cools the body, whereas on humid days the concentration of water in the atmosphere is lower, which increases the evaporation rate from the skin\u2019s surface. d. On hot, dry days, the evaporation of the sweat from the skin cools the body, whereas on humid days the concentration of water in the atmosphere is higher, which increases the evaporation rate from the skin\u2019s surface. 11.2 Heat, Specific Heat, and Heat Transfer 47. A hot piece of metal needs to be cooled. If you were to put the metal in ice or in cold water, such that the ice did not melt and the temperature of either changed by the same amount, which would reduce the metal\u2019s temperature more? Why? a. Water would reduce the metal\u2019s temperature more, because water has a greater specific heat than ice. b. Water would reduce the metal\u2019s temperature more, because water has a smaller specific heat than ice. Ice would reduce the metal\u2019s temperature more, because ice has a smaller specific heat than water. c. d. Ice would reduce the metal\u2019s temperature more, because ice has a greater specific heat than water. 48. On a summer night, why does a black object seem colder than a white one? a. The black object radiates energy faster than the white one, and hence reaches a lower temperature in less time. b. The black object radiates energy slower than the white one, and hence reaches a lower temperature in less time. c. The black object absorbs energy faster than the white one, and hence reaches a lower temperature in less time. d. The black object absorbs energy slower than the white one, and hence reaches a lower temperature in less time. 49. Calculate the difference in heat required to raise the temperatures of 1.00 kg of gold and 1.00 kg of aluminum by 1.00 \u00b0C. (The specific heat of aluminum equals 900 J/kg \u00b7 \u00b0C; the specific heat of gold equals 129 J/ kg \u00b7 \u00b0C.) 771 J a. b. 129 J c. 90 J d. 900 J 11.3 Phase Change and Latent Heat 50. True or false\u2014You have an ice cube floating in a glass of water with a thin thread resting across the cube. If you cover the ice cube and thread with a layer of salt, they will stick together, so that you are able to lift the icecube when you pick up the thread. a. True b. False 51. Suppose the energy required to freeze 0.250 kg of water were added to the same mass of water at an initial temperature of 1.0 \u00b0C. What would be the final temperature of the water? -69.8 \u00b0C a. 79.8 \u00b0C b. c. -78.8 \u00b0C d. 80.8 \u00b0C Access for free at openstax.org. CHAPTER 12 Thermodynamics Figure 12.1 A steam engine uses energy transfer by heat to do work. (Modification of work by Gerald Friedrich, Pixabay) Chapter Outline 12.1 Zeroth Law of Thermodynamics: Thermal Equilibrium 12.2 First law of Thermodynamics: Thermal Energy and Work 12.3 Second Law of Thermodynamics: Entropy 12.4 Applications of Thermodynamics: Heat Engines, Heat Pumps, and Refrigerators Energy can be transferred to or from a system, either through a temperature difference between it and INTRODUCT", "ION another system (i.e., by heat) or by exerting a force through a distance (work). In these ways, energy can be converted into other forms of energy in other systems. For example, a car engine burns fuel for heat transfer into a gas. Work is done by the gas as it exerts a force through a distance by pushing a piston outward. This work converts the energy into a variety of other forms\u2014into an increase in the car\u2019s kinetic or gravitational potential energy; into electrical energy to run the spark plugs, radio, and lights; and back into stored energy in the car\u2019s battery. But most of the thermal energy transferred by heat from the fuel burning in the engine does not do work on the gas. Instead, much of this energy is released into the surroundings at lower temperature (i.e., lost through heat), which is quite inefficient. Car engines are only about 25 to 30 percent efficient. This inefficiency leads to increased fuel costs, so there is great interest in improving fuel efficiency. However, it is common knowledge that modern gasoline engines cannot be made much more efficient. The same is true about the conversion to electrical energy in large power stations, whether they are coal, oil, natural gas, or nuclear powered. Why is this the case? The answer lies in the nature of heat. Basic physical laws govern how heat transfer for doing work takes place and limit the 356 Chapter 12 \u2022 Thermodynamics maximum possible efficiency of the process. This chapter will explore these laws as well their applications to everyday machines. These topics are part of thermodynamics\u2014the study of heat and its relationship to doing work. 12.1 Zeroth Law of Thermodynamics: Thermal Equilibrium Section Learning Objectives By the end of this section, you will be able to do the following: \u2022 Explain the zeroth law of thermodynamics Section Key Terms thermal equilibrium zeroth law of thermodynamics We learned in the previous chapter that when two objects (or systems) are in contact with one another, heat will transfer thermal energy from the object at higher temperature to the one at lower temperature until they both reach the same temperature. The objects are then in thermal equilibrium, and no further temperature changes will occur if they are isolated from other systems. The systems interact and change because their temperatures are different, and the changes stop once their temperatures are the same. Thermal equilibrium is established when two bodies are in thermal contactwith each other\u2014meaning heat transfer (i.e., the transfer of energy by heat) can occur between them. If two systems cannot freely exchange energy, they will not reach thermal equilibrium. (It is fortunate that empty space stands between Earth and the sun, because a state of thermal equilibrium with the sun would be too toasty for life on this planet!) If two systems, A and B, are in thermal equilibrium with each another, and B is in thermal equilibrium with a third system, C, then A is also in thermal equilibrium with C. This statement may seem obvious, because all three have the same temperature, but it is basic to thermodynamics. It is called the zeroth law of thermodynamics. TIPS FOR SUCCESS The zeroth law of thermodynamics is very similar to the transitive property of equality in mathematics: If a = b and b = c, then a = c. You may be wondering at this point, why the wacky name? Shouldn\u2019t this be called the firstlaw of thermodynamics rather than the zeroth? The explanation is that this law was discovered after the first and second laws of thermodynamics but is so fundamental that scientists decided it should logically come first. As an example of the zeroth law in action, consider newborn babies in neonatal intensive-care units in hospitals. Prematurely born or sick newborns are placed in special incubators. These babies have very little covering while in the incubators, so to an observer, they look as though they may not be warm enough. However, inside the incubator, the temperature of the air, the cot, and the baby are all the same\u2014that is, they are in thermal equilibrium. The ambient temperature is just high enough to keep the baby safe and comfortable. WORK IN PHYSICS Thermodynamics Engineer Thermodynamics engineers apply the principles of thermodynamics to mechanical systems so as to create or test products that rely on the interactions between heat, work, pressure, temperature, and volume. This type of work typically takes place in the aerospace industry, chemical manufacturing companies, industrial manufacturing plants, power plants (Figure 12.2), engine manufacturers, or electronics companies. Access for free at openstax.org. 12.1 \u2022 Zeroth Law of Thermodynamics: Thermal Equilibrium 357 Figure 12.2 An engineer makes a site visit to the Baghdad South power plant. The need for energy creates quite a bit of demand for thermodynamics engineers, because both traditional energy companies and alternative (green) energy startups rely on interactions between heat and work and so require", " the expertise of thermodynamics engineers. Traditional energy companies use mainly nuclear energy and energy from burning fossil fuels, such as coal. Alternative energy is finding new ways to harness renewable and, often, more readily available energy sources, such as solar, water, wind, and bio-energy. A thermodynamics engineer in the energy industry can find the most efficient way to turn the burning of a biofuel or fossil fuel into energy, store that energy for times when it\u2019s needed most, or figure out how to best deliver that energy from where it\u2019s produced to where it\u2019s used: in homes, factories, and businesses. Additionally, he or she might also design pollution-control equipment to remove harmful pollutants from the smoke produced as a by-product of burning fuel. For example, a thermodynamics engineer may develop a way to remove mercury from burning coal in a coal-fired power plant. Thermodynamics engineering is an expanding field, where employment opportunities are expected to grow by as much as 27 percent between 2012 and 2022, according to the U.S. Bureau of Labor Statistics. To become a thermodynamics engineer, you must have a college degree in chemical engineering, mechanical engineering, environmental engineering, aerospace engineering, civil engineering, or biological engineering (depending on which type of career you wish to pursue), with coursework in physics and physical chemistry that focuses on thermodynamics. GRASP CHECK What would be an example of something a thermodynamics engineer would do in the aeronautics industry? a. Test the fuel efficiency of a jet engine b. Test the functioning of landing gear c. Test the functioning of a lift control device d. Test the autopilot functions Check Your Understanding 1. What is thermal equilibrium? a. When two objects in contact with each other are at the same pressure, they are said to be in thermal equilibrium. b. When two objects in contact with each other are at different temperatures, they are said to be in thermal equilibrium. c. When two objects in contact with each other are at the same temperature, they are said to be in thermal equilibrium. d. When two objects not in contact with each other are at the same pressure, they are said to be in thermal equilibrium. 2. What is the zeroth law of thermodynamics? 358 Chapter 12 \u2022 Thermodynamics a. Energy can neither be created nor destroyed in a chemical reaction. b. If two systems, A and B, are in thermal equilibrium with each another, and B is in thermal equilibrium with a third system, C, then A is also in thermal equilibrium with C. c. Entropy of any isolated system not in thermal equilibrium always increases. d. Entropy of a system approaches a constant value as temperature approaches absolute zero. 12.2 First law of Thermodynamics: Thermal Energy and Work Section Learning Objectives By the end of this section, you will be able to do the following: \u2022 Describe how pressure, volume, and temperature relate to one another and to work, based on the ideal gas law \u2022 Describe pressure\u2013volume work \u2022 Describe the first law of thermodynamics verbally and mathematically \u2022 Solve problems involving the first law of thermodynamics Section Key Terms Boltzmann constant first law of thermodynamics ideal gas law internal energy pressure Pressure, Volume, Temperature, and the Ideal Gas Law Before covering the first law of thermodynamics, it is first important to understand the relationship between pressure, volume, and temperature. Pressure, P, is defined as where Fis a force applied to an area, A, that is perpendicular to the force. Depending on the area over which it is exerted, a given force can have a significantly different effect, as shown in Figure 12.3. 12.1 Figure 12.3 (a) Although the person being poked with the finger might be irritated, the force has little lasting effect. (b) In contrast, the same force applied to an area the size of the sharp end of a needle is great enough to break the skin. The SI unit for pressure is the pascal, where Pressure is defined for all states of matter but is particularly important when discussing fluids (such as air). You have probably heard the word pressurebeing used in relation to blood (high or low blood pressure) and in relation to the weather (high- and low-pressure weather systems). These are only two of many examples of pressures in fluids. The relationship between the pressure, volume, and temperature for an ideal gas is given by the ideal gas law. A gas is considered ideal at low pressure and fairly high temperature, and forces between its component particles can be ignored. The ideal gas law states that 12.2 where Pis the pressure of a gas, Vis the volume it occupies, Nis the number of particles (atoms or molecules) in the gas, and Tis Access for free at openstax.org. 12.2 \u2022 First law of Thermodynamics: Thermal Energy and Work 359 its absolute temperature. The constant kis called the Boltzmann constant and has", " the value purposes of this chapter, we will not go into calculations using the ideal gas law. Instead, it is important for us to notice from the equation that the following are true for a given mass of gas: For the \u2022 When volume is constant, pressure is directly proportional to temperature. \u2022 When temperature is constant, pressure is inversely proportional to volume. \u2022 When pressure is constant, volume is directly proportional to temperature. This last point describes thermal expansion\u2014the change in size or volume of a given mass with temperature. What is the underlying cause of thermal expansion? An increase in temperature means that there\u2019s an increase in the kinetic energy of the individual atoms. Gases are especially affected by thermal expansion, although liquids expand to a lesser extent with similar increases in temperature, and even solids have minor expansions at higher temperatures. This is why railroad tracks and bridges have expansion joints that allow them to freely expand and contract with temperature changes. To get some idea of how pressure, temperature, and volume of a gas are related to one another, consider what happens when you pump air into a deflated tire. The tire\u2019s volume first increases in direct proportion to the amount of air injected, without much increase in the tire pressure. Once the tire has expanded to nearly its full size, the walls limit volume expansion. If you continue to pump air into tire (which now has a nearly constant volume), the pressure increases with increasing temperature (see Figure 12.4). Figure 12.4 (a) When air is pumped into a deflated tire, its volume first increases without much increase in pressure. (b) When the tire is filled to a certain point, the tire walls resist further expansion, and the pressure increases as more air is added. (c) Once the tire is inflated fully, its pressure increases with temperature. Pressure\u2013Volume Work Pressure\u2013volume workis the work that is done by the compression or expansion of a fluid. Whenever there is a change in volume and external pressure remains constant, pressure\u2013volume work is taking place. During a compression, a decrease in volume increases the internal pressure of a system as work is done onthe system. During an expansion (Figure 12.5), an increase in volume decreases the internal pressure of a system as the system doeswork. 360 Chapter 12 \u2022 Thermodynamics Figure 12.5 An expansion of a gas requires energy transfer to keep the pressure constant. Because pressure is constant, the work done is. Recall that the formula for work is in terms of pressure. We can rearrange the definition of pressure, to get an expression for force Substituting this expression for force into the definition of work, we get Because area multiplied by displacement is the change in volume, pressure\u2013volume work is, the mathematical expression for 12.3 12.4 12.5 Just as we say that work is force acting over a distance, for fluids, we can say that work is the pressure acting through the change in volume. For pressure\u2013volume work, pressure is analogous to force, and volume is analogous to distance in the traditional definition of work. WATCH PHYSICS Work from Expansion This video describes work from expansion (or pressure\u2013volume work). Sal combines the equations and to get. Click to view content (https://www.openstax.org/l/28expansionWork) GRASP CHECK If the volume of a system increases while pressure remains constant, is the value of work done by the system Wpositive or negative? Will this increase or decrease the internal energy of the system? a. Positive; internal energy will decrease b. Positive; internal energy will increase c. Negative; internal energy will decrease d. Negative; internal energy will increase The First Law of Thermodynamics Heat (Q) and work (W) are the two ways to add or remove energy from a system. The processes are very different. Heat is driven Access for free at openstax.org. 12.2 \u2022 First law of Thermodynamics: Thermal Energy and Work 361 by temperature differences, while work involves a force exerted through a distance. Nevertheless, heat and work can produce identical results. For example, both can cause a temperature increase. Heat transfers energy into a system, such as when the sun warms the air in a bicycle tire and increases the air\u2019s temperature. Similarly, work can be done on the system, as when the bicyclist pumps air into the tire. Once the temperature increase has occurred, it is impossible to tell whether it was caused by heat or work. Heat and work are both energy in transit\u2014neither is stored as such in a system. However, both can change the internal energy, U, of a system. Internal energy is the sum of the kinetic and potential energies of a system\u2019s atoms and molecules. It can be divided into many subcategories, such as thermal and chemical energy, and depends only on the state of a system (that is, P, V, and T), not on how the energy enters or leaves the system.", " In order to understand the relationship between heat, work, and internal energy, we use the first law of thermodynamics. The first law of thermodynamics applies the conservation of energyprinciple to systems where heat and work are the methods of transferring energy into and out of the systems. It can also be used to describe how energy transferred by heat is converted and transferred again by work. TIPS FOR SUCCESS Recall that the principle of conservation of energy states that energy cannot be created or destroyed, but it can be altered from one form to another. The first law of thermodynamics states that the change in internal energy of a closed system equals the net heat transfer intothe system minus the net work done bythe system. In equation form, the first law of thermodynamics is 12.6 is the change in internal energy, U, of the system. As shown in Figure 12.6, Qis the net heat transferred into the Here, system\u2014that is, Qis the sum of all heat transfers into and out of the system. Wis the net work done by the system\u2014that is, Wis the sum of all work done on or by the system. By convention, if Qis positive, then there is a net heat transfer into the system; if Wis positive, then there is net work done by the system. So positive Qadds energy to the system by heat, and positive Wtakes energy from the system by work. Note that if heat transfers more energy into the system than that which is done by work, the difference is stored as internal energy. Figure 12.6 The first law of thermodynamics is the conservation of energyprinciple stated for a system, where heat and work are the methods of transferring energy to and from a system. Qrepresents the net heat transfer\u2014it is the sum of all transfers of energy by heat into and out of the system. Qis positive for net heat transfer intothe system. is the work done bythe system, and is the work done on the system. Wis the total work done on or bythe system. Wis positive when more work is done bythe system than onit. The change in the internal energy of the system,, is related to heat and work by the first law of thermodynamics: It follows also that negative Qindicates that energy is transferred awayfrom the system by heat and so decreases the system\u2019s internal energy, whereas negative Wis work done onthe system, which increases the internal energy. WATCH PHYSICS First Law of Thermodynamics/Internal Energy This video explains the first law of thermodynamics, conservation of energy, and internal energy. It goes over an example of energy transforming between kinetic energy, potential energy, and heat transfer due to air resistance. Click to view content (https://www.openstax.org/l/28FirstThermo) 362 Chapter 12 \u2022 Thermodynamics GRASP CHECK Consider the example of tossing a ball when there\u2019s air resistance. As air resistance increases, what would you expect to happen to the final velocity and final kinetic energy of the ball? Why? a. Both will decrease. Energy is transferred to the air by heat due to air resistance. b. Both will increase. Energy is transferred from the air to the ball due to air resistance. c. Final velocity will increase, but final kinetic energy will decrease. Energy is transferred by heat to the air from the ball through air resistance. d. Final velocity will decrease, but final kinetic energy will increase. Energy is transferred by heat from the air to the ball through air resistance. WATCH PHYSICS More on Internal Energy This video goes into further detail, explaining internal energy and how to use the equation the equation system., where Wis the work done onthe system, whereas we use Wto represent work done bythe Note that Sal uses Click to view content (https://www.openstax.org/l/28IntrnEnergy) GRASP CHECK are taken away by heat from the system, and the system does If system? a. b. c. d. of work, what is the change in internal energy of the LINKS TO PHYSICS Biology: Biological Thermodynamics We often think about thermodynamics as being useful for inventing or testing machinery, such as engines or steam turbines. However, thermodynamics also applies to living systems, such as our own bodies. This forms the basis of the biological thermodynamics (Figure 12.7). Figure 12.7 (a) The first law of thermodynamics applies to metabolism. Heat transferred out of the body (Q) and work done by the body (W) remove internal energy, whereas food intake replaces it. (Food intake may be considered work done on the body.) (b) Plants convert part of Access for free at openstax.org. 12.2 \u2022 First law of Thermodynamics: Thermal Energy and Work 363 the radiant energy in sunlight into stored chemical energy, a process called photosynthesis. Life itself depends on the", " biological transfer of energy. Through photosynthesis, plants absorb solar energy from the sun and use this energy to convert carbon dioxide and water into glucose and oxygen. Photosynthesis takes in one form of energy\u2014light\u2014and converts it into another form\u2014chemical potential energy (glucose and other carbohydrates). Human metabolismis the conversion of food into energy given off by heat, work done by the body\u2019s cells, and stored fat. Metabolism is an interesting example of the first law of thermodynamics in action. Eating increases the internal energy of the body by adding chemical potential energy; this is an unromantic view of a good burrito. The body metabolizes all the food we consume. Basically, metabolism is an oxidation process in which the chemical potential energy of food is released. This implies that food input is in the form of work. Exercise helps you lose weight, because it provides energy transfer from your body by both heat and work and raises your metabolic rate even when you are at rest. Biological thermodynamics also involves the study of transductions between cells and living organisms. Transductionis a process where genetic material\u2014DNA\u2014is transferred from one cell to another. This often occurs during a viral infection (e.g., influenza) and is how the virus spreads, namely, by transferring its genetic material to an increasing number of previously healthy cells. Once enough cells become infected, you begin to feel the effects of the virus (flu symptoms\u2014muscle weakness, coughing, and congestion). Energy is transferred along with the genetic material and so obeys the first law of thermodynamics. Energy is transferred\u2014not created or destroyed\u2014in the process. When work is done on a cell or heat transfers energy to a cell, the cell\u2019s internal energy increases. When a cell does work or loses heat, its internal energy decreases. If the amount of work done by a cell is the same as the amount of energy transferred in by heat, or the amount of work performed on a cell matches the amount of energy transferred out by heat, there will be no net change in internal energy. GRASP CHECK Based on what you know about heat transfer and the first law of thermodynamics, do you need to eat more or less to maintain a constant weight in colder weather? Explain why. a. more; as more energy is lost by the body in colder weather, the need to eat increases so as to maintain a constant weight b. more; eating more food means accumulating more fat, which will insulate the body from colder weather and will reduce c. d. the energy loss less; as less energy is lost by the body in colder weather, the need to eat decreases so as to maintain a constant weight less; eating less food means accumulating less fat, so less energy will be required to burn the fat, and, as a result, weight will remain constant Solving Problems Involving the First Law of Thermodynamics WORKED EXAMPLE Calculating Change in Internal Energy Suppose 40.00 J of energy is transferred by heat to a system, while the system does 10.00 J of work. Later, heat transfers 25.00 J out of the system, while 4.00 J is done by work on the system. What is the net change in the system\u2019s internal energy? STRATEGY You must first calculate the net heat and net work. Then, using the first law of thermodynamics, change in internal energy. find the Solution The net heat is the transfer into the system by heat minus the transfer out of the system by heat, or The total work is the work done by the system minus the work done on the system, or 12.7 12.8 364 Chapter 12 \u2022 Thermodynamics The change in internal energy is given by the first law of thermodynamics. Discussion A different way to solve this problem is to find the change in internal energy for each of the two steps separately and then add the two changes to get the total change in internal energy. This approach would look as follows: For 40.00 J of heat in and 10.00 J of work out, the change in internal energy is 12.9 For 25.00 J of heat out and 4.00 J of work in, the change in internal energy is The total change is 12.10 12.11 12.12 No matter whether you look at the overall process or break it into steps, the change in internal energy is the same. WORKED EXAMPLE Calculating Change in Internal Energy: The Same Change in Uis Produced by Two Different Processes What is the change in the internal energy of a system when a total of 150.00 J is transferred by heat from the system and 159.00 J is done by work on the system? STRATEGY The net heat and work are already given, so simply use these values in the equation Solution Here, the net heat and total work are given directly as so that 12.13 Access for free at openstax.org. Discussion 12.2 \u2022 First law", " of Thermodynamics: Thermal Energy and Work 365 Figure 12.8 Two different processes produce the same change in a system. (a) A total of 15.00 J of heat transfer occurs into the system, while work takes out a total of 6.00 J. The change in internal energy is \u0394U = Q \u2013 W = 9.00 J. (b) Heat transfer removes 150.00 J from the system while work puts 159.00 J into it, producing an increase of 9.00 J in internal energy. If the system starts out in the same state in (a) and (b), it will end up in the same final state in either case\u2014its final state is related to internal energy, not how that energy was acquired. A very different process in this second worked example produces the same 9.00 J change in internal energy as in the first worked example. Note that the change in the system in both parts is related to system ends up in the samestate in both problems. Note that, as usual, in Figure 12.8 above, and and not to the individual Q\u2019s or W\u2019s involved. The is work done onthe system. is work done bythe system, Practice Problems 3. What is the pressure-volume work done by a system if a pressure of causes a change in volume of? a. b. c. d. 4. What is the net heat out of the system when is transferred by heat into the system and is transferred out of it? a. b. c. d. 366 Chapter 12 \u2022 Thermodynamics Check Your Understanding 5. What is pressure? a. Pressure is force divided by length. b. Pressure is force divided by area. c. Pressure is force divided by volume. d. Pressure is force divided by mass. 6. What is the SI unit for pressure? a. pascal, or N/m3 coulomb b. c. newton d. pascal, or N/m2 7. What is pressure-volume work? a. b. c. d. It is the work that is done by the compression or expansion of a fluid. It is the work that is done by a force on an object to produce a certain displacement. It is the work that is done by the surface molecules of a fluid. It is the work that is done by the high-energy molecules of a fluid. 8. When is pressure-volume work said to be done ON a system? a. When there is an increase in both volume and internal pressure. b. When there is a decrease in both volume and internal pressure. c. When there is a decrease in volume and an increase in internal pressure. d. When there is an increase in volume and a decrease in internal pressure. 9. What are the ways to add energy to or remove energy from a system? a. Transferring energy by heat is the only way to add energy to or remove energy from a system. b. Doing compression work is the only way to add energy to or remove energy from a system. c. Doing expansion work is the only way to add energy to or remove energy from a system. d. Transferring energy by heat or by doing work are the ways to add energy to or remove energy from a system. 10. What is internal energy? a. b. c. d. It is the sum of the kinetic energies of a system\u2019s atoms and molecules. It is the sum of the potential energies of a system\u2019s atoms and molecules. It is the sum of the kinetic and potential energies of a system\u2019s atoms and molecules. It is the difference between the magnitudes of the kinetic and potential energies of a system\u2019s atoms and molecules. 12.3 Second Law of Thermodynamics: Entropy Section Learning Objectives By the end of this section, you will be able to do the following: \u2022 Describe entropy \u2022 Describe the second law of thermodynamics \u2022 Solve problems involving the second law of thermodynamics Section Key Terms entropy second law of thermodynamics Entropy Recall from the chapter introduction that it is not even theoretically possible for engines to be 100 percent efficient. This phenomenon is explained by the second law of thermodynamics, which relies on a concept known as entropy. Entropy is a measure of the disorder of a system. Entropy also describes how much energy is notavailable to do work. The more disordered a system and higher the entropy, the less of a system's energy is available to do work. Access for free at openstax.org. 12.3 \u2022 Second Law of Thermodynamics: Entropy 367 Although all forms of energy can be used to do work, it is not possible to use the entire available energy for work. Consequently, not all energy transferred by heat can be converted into work, and some of it is lost in the form of waste heat\u2014that is, heat that does not go toward doing work. The unavailability of energy is important in", " thermodynamics; in fact, the field originated from efforts to convert heat to work, as is done by engines. The equation for the change in entropy,, is where Qis the heat that transfers energy during a process, and Tis the absolute temperature at which the process takes place. Qis positive for energy transferred intothe system by heat and negative for energy transferred out ofthe system by heat. In SI, entropy is expressed in units of joules per kelvin (J/K). If temperature changes during the process, then it is usually a good approximation (for small changes in temperature) to take Tto be the average temperature in order to avoid trickier math (calculus). TIPS FOR SUCCESS Absolute temperature is the temperature measured in Kelvins. The Kelvin scale is an absolute temperature scale that is measured in terms of the number of degrees above absolute zero. All temperatures are therefore positive. Using temperatures from another, nonabsolute scale, such as Fahrenheit or Celsius, will give the wrong answer. Second Law of Thermodynamics Have you ever played the card game 52 pickup? If so, you have been on the receiving end of a practical joke and, in the process, learned a valuable lesson about the nature of the universe as described by the second law of thermodynamics. In the game of 52 pickup, the prankster tosses an entire deck of playing cards onto the floor, and you get to pick them up. In the process of picking up the cards, you may have noticed that the amount of work required to restore the cards to an orderly state in the deck is much greater than the amount of work required to toss the cards and create the disorder. The second law of thermodynamics states that the total entropy of a system either increases or remains constant in any spontaneous process; it never decreases.An important implication of this law is that heat transfers energy spontaneously from higher- to lower-temperature objects, but never spontaneously in the reverse direction. This is because entropy increases for heat transfer of energy from hot to cold (Figure 12.9). Because the change in entropy is Q/T, there is a larger change in at lower temperatures (smaller T). The decrease in entropy of the hot (larger T) object is therefore less than the increase in entropy of the cold (smaller T) object, producing an overall increase in entropy for the system. Figure 12.9 The ice in this drink is slowly melting. Eventually, the components of the liquid will reach thermal equilibrium, as predicted by the second law of thermodynamics\u2014that is, after heat transfers energy from the warmer liquid to the colder ice. (Jon Sullivan, PDPhoto.org) Another way of thinking about this is that it is impossible for any process to have, as its sole result, heat transferring energy from a cooler to a hotter object. Heat cannot transfer energy spontaneously from colder to hotter, because the entropy of the 368 Chapter 12 \u2022 Thermodynamics overall system would decrease. Suppose we mix equal masses of water that are originally at two different temperatures, say will be water at an intermediate temperature of has become unavailable to do work, and the system has become less orderly. Let us think about each of these results.. The result. Three outcomes have resulted: entropy has increased, some energy and First, why has entropy increased? Mixing the two bodies of water has the same effect as the heat transfer of energy from the higher-temperature substance to the lower-temperature substance. The mixing decreases the entropy of the hotter water but increases the entropy of the colder water by a greater amount, producing an overall increase in entropy. Second, once the two masses of water are mixed, there is no more temperature difference left to drive energy transfer by heat and therefore to do work. The energy is still in the water, but it is now unavailableto do work. Third, the mixture is less orderly, or to use another term, less structured. Rather than having two masses at different temperatures and with different distributions of molecular speeds, we now have a single mass with a broad distribution of molecular speeds, the average of which yields an intermediate temperature. These three results\u2014entropy, unavailability of energy, and disorder\u2014not only are related but are, in fact, essentially equivalent. Heat transfer of energy from hot to cold is related to the tendency in nature for systems to become disordered and for less energy to be available for use as work. Based on this law, what cannot happen? A cold object in contact with a hot one never spontaneously transfers energy by heat to the hot object, getting colder while the hot object gets hotter. Nor does a hot, stationary automobile ever spontaneously cool off and start moving. Another example is the expansion of a puff of gas introduced into one corner of a vacuum chamber. The gas expands to fill the chamber, but it never regroups on its own in the corner. The random motion of the gas molecules could take them all back to the corner, but this is never observed to happen (Figure 12.10). Access", " for free at openstax.org. 12.3 \u2022 Second Law of Thermodynamics: Entropy 369 Figure 12.10 Examples of one-way processes in nature. (a) Heat transfer occurs spontaneously from hot to cold, but not from cold to hot. (b) The brakes of this car convert its kinetic energy to increase their internal energy (temperature), and heat transfers this energy to the environment. The reverse process is impossible. (c) The burst of gas released into this vacuum chamber quickly expands to uniformly fill every part of the chamber. The random motions of the gas molecules will prevent them from returning altogether to the corner. We've explained that heat never transfers energy spontaneously from a colder to a hotter object. The key word here is spontaneously. If we do workon a system, it ispossible to transfer energy by heat from a colder to hotter object. We'll learn more about this in the next section, covering refrigerators as one of the applications of the laws of thermodynamics. Sometimes people misunderstand the second law of thermodynamics, thinking that based on this law, it is impossible for entropy to decrease at any particular location. But, it actually ispossible for the entropy of one partof the universe to decrease, as long as the total change in entropy of the universe increases. In equation form, we can write this as Based on this equation, we see that can be negative as long as is positive and greater in magnitude. How is it possible for the entropy of a system to decrease? Energy transfer is necessary. If you pick up marbles that are scattered about the room and put them into a cup, your work has decreased the entropy of that system. If you gather iron ore from the ground and convert it into steel and build a bridge, your work has decreased the entropy of that system. Energy coming from the sun can decrease the entropy of local systems on Earth\u2014that is, universe increases by a greater amount\u2014that is, although you made the system of the bridge and steel more structured, you did so at the expense of the universe. Altogether, the entropy of the universe is increased by the disorder created by digging up the ore and converting it to steel. Therefore, is positive and greater in magnitude. In the case of the iron ore, is negative. But the overall entropy of the rest of the 12.14 370 Chapter 12 \u2022 Thermodynamics and the second law of thermodynamics is notviolated. Every time a plant stores some solar energy in the form of chemical potential energy, or an updraft of warm air lifts a soaring bird, Earth experiences local decreases in entropy as it uses part of the energy transfer from the sun into deep space to do work. There is a large total increase in entropy resulting from this massive energy transfer. A small part of this energy transfer by heat is stored in structured systems on Earth, resulting in much smaller, local decreases in entropy. Solving Problems Involving the Second Law of Thermodynamics Entropy is related not only to the unavailability of energy to do work; it is also a measure of disorder. For example, in the case of a melting block of ice, a highly structured and orderly system of water molecules changes into a disorderly liquid, in which molecules have no fixed positions (Figure 12.11). There is a large increase in entropy for this process, as we'll see in the following worked example. Figure 12.11 These ice floes melt during the Arctic summer. Some of them refreeze in the winter, but the second law of thermodynamics predicts that it would be extremely unlikely for the water molecules contained in these particular floes to reform in the distinctive alligator- like shape they possessed when this picture was taken in the summer of 2009. (Patrick Kelley, U.S. Coast Guard, U.S. Geological Survey) WORKED EXAMPLE Entropy Associated with Disorder Find the increase in entropy of 1.00 kg of ice that is originally at STRATEGY The change in entropy can be calculated from the definition of and melts to form water at. once we find the energy, Q, needed to melt the ice. Solution The change in entropy is defined as Here, Qis the heat necessary to melt 1.00 kg of ice and is given by where mis the mass and is the latent heat of fusion. for water, so Because Qis the amount of energy heat adds to the ice, its value is positive, and Tis the melting temperature of ice, So the change in entropy is 12.15 12.16 12.17 12.18 Access for free at openstax.org. Discussion 12.3 \u2022 Second Law of Thermodynamics: Entropy 371 Figure 12.12 When ice melts, it becomes more disordered and less structured. The systematic arrangement of molecules in a crystal structure is replaced by a more random and less orderly movement of molecules without fixed locations or orientations. Its entropy increases because heat transfer occurs into it. Entropy is a measure of disorder. The change in entropy is positive", ", because heat transfers energy intothe ice to cause the phase change. This is a significant increase in entropy, because it takes place at a relatively low temperature. It is accompanied by an increase in the disorder of the water molecules. Practice Problems are added by heat to water at, what is the change in entropy? 11. If a. b. c. d. 12. What is the increase in entropy when of ice at melt to form water at? a. b. c. d. Check Your Understanding 13. What is entropy? a. Entropy is a measure of the potential energy of a system. b. Entropy is a measure of the net work done by a system. c. Entropy is a measure of the disorder of a system. d. Entropy is a measure of the heat transfer of energy into a system. 14. Which forms of energy can be used to do work? a. Only work is able to do work. b. Only heat is able to do work. c. Only internal energy is able to do work. d. Heat, work, and internal energy are all able to do work. 15. What is the statement for the second law of thermodynamics? a. All the spontaneous processes result in decreased total entropy of a system. b. All the spontaneous processes result in increased total entropy of a system. c. All the spontaneous processes result in decreased or constant total entropy of a system. d. All the spontaneous processes result in increased or constant total entropy of a system. 16. For heat transferring energy from a high to a low temperature, what usually happens to the entropy of the whole system? It decreases. It must remain constant. a. b. c. The entropy of the system cannot be predicted without specific values for the temperatures. 372 Chapter 12 \u2022 Thermodynamics d. It increases. 12.4 Applications of Thermodynamics: Heat Engines, Heat Pumps, and Refrigerators Section Learning Objectives By the end of this section, you will be able to do the following: \u2022 Explain how heat engines, heat pumps, and refrigerators work in terms of the laws of thermodynamics \u2022 Describe thermal efficiency \u2022 Solve problems involving thermal efficiency Section Key Terms cyclical process heat engine heat pump thermal efficiency Heat Engines, Heat Pumps, and Refrigerators In this section, we\u2019ll explore how heat engines, heat pumps, and refrigerators operate in terms of the laws of thermodynamics. One of the most important things we can do with heat is to use it to do work for us. A heat engine does exactly this\u2014it makes use of the properties of thermodynamics to transform heat into work. Gasoline and diesel engines, jet engines, and steam turbines that generate electricity are all examples of heat engines. Figure 12.13 illustrates one of the ways in which heat transfers energy to do work. Fuel combustion releases chemical energy that heat transfers throughout the gas in a cylinder. This increases the gas temperature, which in turn increases the pressure of the gas and, therefore, the force it exerts on a movable piston. The gas does work on the outside world, as this force moves the piston through some distance. Thus, heat transfer of energy to the gas in the cylinder results in work being done. Access for free at openstax.org. 12.4 \u2022 Applications of Thermodynamics: Heat Engines, Heat Pumps, and Refrigerators 373 Figure 12.13 (a) Heat transfer to the gas in a cylinder increases the internal energy of the gas, creating higher pressure and temperature. (b) The force exerted on the movable cylinder does work as the gas expands. Gas pressure and temperature decrease during expansion, indicating that the gas\u2019s internal energy has decreased as it does work. (c) Heat transfer of energy to the environment further reduces pressure in the gas, so that the piston can more easily return to its starting position. To repeat this process, the piston needs to be returned to its starting point. Heat now transfers energy from the gas to the surroundings, so that the gas\u2019s pressure decreases, and a force is exerted by the surroundings to push the piston back through some distance. A cyclical process brings a system, such as the gas in a cylinder, back to its original state at the end of every cycle. All heat engines use cyclical processes., from the high-temperature object (or hot reservoir), whereas heat transfers unused energy, Heat engines do work by using part of the energy transferred by heat from some source. As shown in Figure 12.14, heat transfers energy, temperature object (or cold reservoir), and the work done by the engine is W. In physics, a reservoiris defined as an infinitely large mass that can take in or put out an unlimited amount of heat, depending upon the needs of the system. The temperature of the hot reservoir is and the temperature of the cold reservoir is, into the low-. 374 Chapter 12 \u2022 Thermodynamics Figure 12.14 (a", ") Heat transfers energy spontaneously from a hot object to a cold one, as is consistent with the second law of thermodynamics. (b) A heat engine, represented here by a circle, uses part of the energy transferred by heat to do work. The hot and cold objects are called the hot and cold reservoirs. Qh is the heat out of the hot reservoir, Wis the work output, and Qc is the unused heat into the cold reservoir. As noted, a cyclical process brings the system back to its original condition at the end of every cycle. Such a system\u2019s internal energy, U, is the same at the beginning and end of every cycle\u2014that is,. The first law of thermodynamics states that where Qis the netheat transfer during the cycle, and Wis the network done by the system. The net heat transfer is the energy transferred in by heat from the hot reservoir minus the amount that is transferred out to the cold reservoir ( ). Because there is no change in internal energy for a complete cycle ( ), we have so that Therefore, the net work done by the system equals the net heat into the system, or for a cyclical process. 12.19 12.20 12.21 Because the hot reservoir is heated externally, which is an energy-intensive process, it is important that the work be done as efficiently as possible. In fact, we want Wto equal Unfortunately, this is impossible. According to the second law of thermodynamics, heat engines cannot have perfect conversion of heat into work. Recall that entropy is a measure of the disorder of a system, which is also how much energy is unavailable to do work. The second law of thermodynamics requires that the total entropy of a system either increases or remains constant in that cannot be used for work. The amount of heat rejected to the cold any process. Therefore, there is a minimum amount of reservoir,, the smaller the value of depends upon the efficiency of the heat engine. The smaller the increase in entropy,, and for there to be no heat to the environment (that is, )., and the more heat energy is available to do work. Heat pumps, air conditioners, and refrigerators utilize heat transfer of energy from low to high temperatures, which is the opposite of what heat engines do. Heat transfers energy into a hot one. This requires work input, W, which produces a transfer of energy by heat. Therefore, the total heat transfer to the hot reservoir is from a cold reservoir and delivers energy 12.22 The purpose of a heat pump is to transfer energy by heat to a warm environment, such as a home in the winter. The great advantage of using a heat pump to keep your home warm rather than just burning fuel in a fireplace or furnace is that a heat pump supplies You only pay for W, and you get an additional heat transfer of much energy is transferred to the heated space as is used to run the heat pump. When you burn fuel to keep warm, you pay for all of it. The disadvantage to a heat pump is that the work input (required by the second law of thermodynamics) is sometimes comes from the outside air, even at a temperature below freezing, to the indoor space. from the outside at no cost. In many cases, at least twice as. Heat Access for free at openstax.org. 12.4 \u2022 Applications of Thermodynamics: Heat Engines, Heat Pumps, and Refrigerators 375 more expensive than simply burning fuel, especially if the work is provided by electrical energy. The basic components of a heat pump are shown in Figure 12.15. A working fluid, such as a refrigerant, is used. In the outdoor coils (the evaporator), heat enters the working fluid from the cold outdoor air, turning it into a gas. Figure 12.15 A simple heat pump has four basic components: (1) an evaporator, (2) a compressor, (3) a condenser, and (4) an expansion valve. In the heating mode, heat transfers to the working fluid in the evaporator (1) from the colder, outdoor air, turning it into a gas. The electrically driven compressor (2) increases the temperature and pressure of the gas and forces it into the condenser coils (3) inside the heated space. Because the temperature of the gas is higher than the temperature in the room, heat transfers energy from the gas to the room as the gas condenses into a liquid. The working fluid is then cooled as it flows back through an expansion valve (4) to the outdoor evaporator coils. The electrically driven compressor (work input W) raises the temperature and pressure of the gas and forces it into the condenser coils that are inside the heated space. Because the temperature of the gas is higher than the temperature inside the room, heat transfers energy to the room, and the gas condenses into a liquid. The liquid then flows back through an expansion (pressure-reducing) valve. The liquid, having been cooled through expansion, returns", " to the outdoor evaporator coils to resume the cycle. The quality of a heat pump is judged by how much energy is transferred by heat into the warm space ( much input work (W) is required. ) compared with how Figure 12.16 Heat pumps, air conditioners, and refrigerators are heat engines operated backward. Almost every home contains a refrigerator. Most people don\u2019t realize that they are also sharing their homes with a heat pump. Air conditioners and refrigerators are designed to cool substances by transferring energy by heat to a warmer one, where heat is given up. In the case of a refrigerator, heat is moved out of the inside of the fridge into the out of a cool environment 376 Chapter 12 \u2022 Thermodynamics surrounding room. For an air conditioner, heat is transferred outdoors from inside a home. Heat pumps are also often used in a reverse setting to cool rooms in the summer. As with heat pumps, work input is required for heat transfer of energy from cold to hot. The quality of air conditioners and refrigerators is judged by how much energy is removed by heat W, is required. So, what is considered the energy benefit in a heat pump, is considered waste heat in a refrigerator. from a cold environment, compared with how much work, Thermal Efficiency In the conversion of energy into work, we are always faced with the problem of getting less out than we put in. The problem is that, in all processes, there is some heat that. A way to quantify how efficiently a machine runs is through a quantity called thermal efficiency. that transfers energy to the environment\u2014and usually a very significant amount at We define thermal efficiency, Eff, to be the ratio of useful energy output to the energy input (or, in other words, the ratio of what we get to what we spend). The efficiency of a heat engine is the output of net work, W, divided by heat-transferred energy, into the engine; that is, An efficiency of 1, or 100 percent, would be possible only if there were no heat to the environment ( ). TIPS FOR SUCCESS All values of heat ( plus or minus sign. For example, and ) are positive; there is no such thing as negative heat. The directionof heat is indicated by a is out of the system, so it is preceded by a minus sign in the equation for net heat. 12.23 Solving Thermal Efficiency Problems WORKED EXAMPLE Daily Work Done by a Coal-Fired Power Station and Its Efficiency A coal-fired power station is a huge heat engine. It uses heat to transfer energy from burning coal to do work to turn turbines, which are used then to generate electricity. In a single day, a large coal power station transfers burning coal and transfers What is the efficiency of the power station? STRATEGY We can use water is boiled under pressure to form high-temperature steam, which is used to run steam turbine-generators and then condensed back to water to start the cycle again. by heat from by heat into the environment. (a) What is the work done by the power station? (b) to find the work output, W, assuming a cyclical process is used in the power station. In this process, 12.24 12.25, because is given, and work, W, was calculated in the first part of this Solution Work output is given by Substituting the given values, STRATEGY The efficiency can be calculated with example. Solution Efficiency is given by Access for free at openstax.org. 12.4 \u2022 Applications of Thermodynamics: Heat Engines, Heat Pumps, and Refrigerators 377 The work, W, is found to be, and is given ( ), so the efficiency is 12.26 12.27 Discussion The efficiency found is close to the usual value of 42 percent for coal-burning power stations. It means that fully 59.2 percent of the energy is transferred by heat to the environment, which usually results in warming lakes, rivers, or the ocean near the power station and is implicated in a warming planet generally. While the laws of thermodynamics limit the efficiency of such plants\u2014including plants fired by nuclear fuel, oil, and natural gas\u2014the energy transferred by heat to the environment could be, and sometimes is, used for heating homes or for industrial processes. Practice Problems 17. A heat engine is given by heat and releases by heat to the environment. What is the amount of work done by the system? a. b. c. d. 18. A heat engine takes in 6.0 kJ from heat and produces waste heat of 4.8 kJ. What is its efficiency? a. 25 percent b. 2.50 percent c. 2.00 percent d. 20 percent Check Your Understanding 19. What is a heat engine? a. A heat engine converts mechanical energy into thermal energy. b. A heat engine converts thermal energy into mechanical energy. c. A heat engine converts thermal energy into electrical energy. d. A", " heat engine converts electrical energy into thermal energy. 20. Give an example of a heat engine. a. A generator b. A battery c. A water pump d. A car engine 21. What is thermal efficiency? a. Thermal efficiency is the ratio of work input to the energy input. b. Thermal efficiency is the ratio of work output to the energy input. c. Thermal efficiency is the ratio of work input to the energy output. d. Thermal efficiency is the ratio of work output to the energy output. 22. What is the mathematical expression for thermal efficiency? a. b. c. d. 378 Chapter 12 \u2022 Key Terms KEY TERMS Boltzmann constant constant with the value k= 1.38\u00d710\u221223 of the gas J/K, which is used in the ideal gas law cyclical process process in which a system is brought back internal energy sum of the kinetic and potential energies of a system\u2019s constituent particles (atoms or molecules) to its original state at the end of every cycle pressure force per unit area perpendicular to the force, entropy measurement of a system's disorder and how much energy is not available to do work in a system states that the change in first law of thermodynamics internal energy of a system equals the net energy transfer by heat intothe system minus the net work done bythe system over which the force acts second law of thermodynamics states that the total entropy of a system either increases or remains constant in any spontaneous process; it never decreases thermal efficiency ratio of useful energy output to the energy input heat engine machine that uses energy transfer by heat to thermal equilibrium condition in which heat no longer do work heat pump machine that generates the heat transfer of energy from cold to hot transfers energy between two objects that are in contact; the two objects have the same temperature zeroth law of thermodynamics states that if two objects ideal gas law physical law that relates the pressure and volume of a gas to the number of gas molecules or atoms, or number of moles of gas, and the absolute temperature are in thermal equilibrium, and a third object is in thermal equilibrium with one of those objects, it is also in thermal equilibrium with the other object SECTION SUMMARY 12.1 Zeroth Law of Thermodynamics: Thermal Equilibrium \u2022 Systems are in thermal equilibrium when they have the same temperature. \u2022 Thermal equilibrium occurs when two bodies are in contact with each other and can freely exchange energy. \u2022 The zeroth law of thermodynamics states that when two systems, A and B, are in thermal equilibrium with each other, and B is in thermal equilibrium with a third system, C, then A is also in thermal equilibrium with C. 12.2 First law of Thermodynamics: Thermal Energy and Work \u2022 Pressure is the force per unit area over which the force is applied perpendicular to the area. \u2022 Thermal expansion is the increase, or decrease, of the size (length, area, or volume) of a body due to a change in temperature. \u2022 The ideal gas law relates the pressure and volume of a gas to the number of gas particles (atoms or molecules) and the absolute temperature of the gas. \u2022 Heat and work are the two distinct methods of energy transfer. \u2022 Heat is energy transferred solely due to a temperature difference. \u2022 The first law of thermodynamics is given as, where is the change in internal energy of a system, Qis the net energy transfer into the system by heat (the sum of all transfers by heat into and out of the system), and Wis the net work done by the Access for free at openstax.org. system (the sum of all energy transfers by work out of or into the system). \u2022 Both Qand Wrepresent energy in transit; only represents an independent quantity of energy capable of being stored. \u2022 The internal energy Uof a system depends only on the state of the system, and not how it reached that state. 12.3 Second Law of Thermodynamics: Entropy \u2022 Entropy is a measure of a system's disorder: the greater the disorder, the larger the entropy. \u2022 Entropy is also the reduced availability of energy to do work. \u2022 The second law of thermodynamics states that, for any spontaneous process, the total entropy of a system either increases or remains constant; it never decreases. \u2022 Heat transfers energy spontaneously from higher- to lower-temperature bodies, but never spontaneously in the reverse direction. 12.4 Applications of Thermodynamics: Heat Engines, Heat Pumps, and Refrigerators \u2022 Heat engines use the heat transfer of energy to do work. \u2022 Cyclical processes are processes that return to their original state at the end of every cycle. \u2022 The thermal efficiency of a heat engine is the ratio of work output divided by the amount of energy input. \u2022 The amount of work a heat engine can do is determined by the net heat transfer of energy during a cycle; more waste heat leads to less work output. \u2022 Heat pumps draw energy by heat from cold outside air and use it to heat an interior room.", " \u2022 A refrigerator is a type of heat pump; it takes energy KEY EQUATIONS 12.2 First law of Thermodynamics: Thermal Energy and Work Chapter 12 \u2022 Key Equations 379 from the warm air from the inside compartment and transfers it to warmer exterior air. 12.3 Second Law of Thermodynamics: Entropy ideal gas law change in entropy first law of thermodynamics pressure pressure\u2013volume work CHAPTER REVIEW Concept Items 12.1 Zeroth Law of Thermodynamics: Thermal Equilibrium 1. When are two bodies in thermal equilibrium? a. When they are in thermal contact and are at different pressures b. When they are not in thermal contact but are at the same pressure c. When they are not in thermal contact but are at different temperatures d. When they are in thermal contact and are at the same temperature 2. What is thermal contact? a. Two objects are said to be in thermal contact when they are in contact with each other in such a way that the transfer of energy by heat can occur between them. b. Two objects are said to be in thermal contact when they are in contact with each other in such a way that the transfer of energy by mass can occur between them. c. Two objects are said to be in thermal contact when they neither lose nor gain energy by heat. There is no transfer of energy between them. d. Two objects are said to be in thermal contact when they only gain energy by heat. There is transfer of energy between them. 3. To which mathematical property is the zeroth law of 12.4 Applications of Thermodynamics: Heat Engines, Heat Pumps, and Refrigerators thermal efficiency of a heat engine work output for a cyclical process thermodynamics similar? a. Associative property b. Commutative property c. Distributive property d. Transitive property 12.2 First law of Thermodynamics: Thermal Energy and Work 4. Why does thermal expansion occur? a. An increase in temperature causes intermolecular distances to increase. b. An increase in temperature causes intermolecular distances to decrease. c. An increase in temperature causes an increase in the work done on the system. d. An increase in temperature causes an increase in the work done by the system. 5. How does pressure-volume work relate to heat and internal energy of a system? a. The energy added to a system by heat minus the change in the internal energy of that system is equal to the pressure-volume work done by the system. b. The sum of the energy released by a system by heat and the change in the internal energy of that system is equal to the pressure-volume work done by the system. c. The product of the energy added to a system by heat and the change in the internal energy of that system 380 Chapter 12 \u2022 Chapter Review d. is equal to the pressure-volume work done by the system. If the energy added to a system by heat is divided by the change in the internal energy of that system, the quotient is equal to the pressure-volume work done by the system. 6. On what does internal energy depend? a. The path of energy changes in the system b. The state of the system c. The size of the system d. The shape of the system 7. The first law of thermodynamics helps us understand the relationships among which three quantities? a. Heat, work, and internal energy b. Heat, work, and external energy c. Heat, work, and enthalpy d. Heat, work, and entropy 12.3 Second Law of Thermodynamics: Entropy 8. Air freshener is sprayed from a bottle. The molecules spread throughout the room and cannot make their way back into the bottle. Why is this the case? a. The entropy of the molecules increases. b. The entropy of the molecules decreases. c. The heat content (enthalpy, or total energy available 12.4 Applications of Thermodynamics: Heat Engines, Heat Pumps, and Refrigerators 10. What is the quality by which air conditioners are judged? a. The amount of energy generated by heat from a hot environment, compared with the required work input b. The amount of energy transferred by heat from a cold environment, compared with the required work input c. The amount of energy transferred by heat from a hot environment, compared with the required work output d. The amount of energy transferred by heat from a cold environment, compared with the required work output 11. Why is the efficiency of a heat engine never 100 percent? a. Some energy is always gained by heat from the environment. b. Some energy is always lost by heat to the environment. c. Work output is always greater than energy input. d. Work output is infinite for any energy input. 12. What is a cyclic process? a. A process in which the system returns to its original for heat) of the molecules increases. state at the end of the cycle d. The heat content (enthalpy, or total", " energy available b. A process in which the system does not return to its for heat) of the molecules decreases. 9. Give an example of entropy as experienced in everyday rotation of Earth formation of a solar eclipse life. a. b. c. filling a car tire with air d. motion of a pendulum bob original state at the end of the cycle c. A process in which the system follows the same path for every cycle d. A process in which the system follows a different path for every cycle Critical Thinking Items 12.1 Zeroth Law of Thermodynamics: Thermal Equilibrium 13. What are the necessary conditions for energy transfer by heat to occur between two bodies through the process of conduction? a. They should be at the same temperature, and they should be in thermal contact. b. They should be at the same temperature, and they should not be in thermal contact. c. They should be at different temperatures, and they should be in thermal contact. should not be in thermal contact. 14. Oil is heated in a pan on a hot plate. The pan is in thermal equilibrium with the hot plate and also with the oil. The temperature of the hot plate is 150 \u00b0C. What is the temperature of the oil? a. b. c. d. 160 \u00b0C 150 \u00b0C 140 \u00b0C 130 \u00b0C 12.2 First law of Thermodynamics: Thermal Energy and Work d. They should be at different temperatures, and they 15. When an inflated balloon experiences a decrease in size, Access for free at openstax.org. the air pressure inside the balloon remains nearly constant. If there is no transfer of energy by heat to or from the balloon, what physical change takes place in the balloon? a. The average kinetic energy of the gas particles decreases, so the balloon becomes colder. b. The average kinetic energy of the gas particles increases, so the balloon becomes hotter. c. The average potential energy of the gas particles decreases, so the balloon becomes colder. d. The average potential energy of the gas particles increases, so the balloon becomes hotter. 16. When heat adds energy to a system, what is likely to happen to the pressure and volume of the system? a. Pressure and volume may both decrease with added energy. b. Pressure and volume may both increase with added energy. c. Pressure must increase with added energy, while volume must remain constant. d. Volume must decrease with added energy, while pressure must remain constant. 17. If more energy is transferred into the system by net heat as compared to the net work done by the system, what happens to the difference in energy? a. b. c. d. It is transferred back to its surroundings. It is stored in the system as internal energy. It is stored in the system as potential energy. It is stored in the system as entropy. 18. Air is pumped into a car tire, causing its temperature to increase. In another tire, the temperature increase is due to exposure to the sun. Is it possible to tell what caused the temperature increase in each tire? Explain your answer. a. No, because it is a chemical change, and the cause of that change does not matter; the final state of both systems are the same. b. Although the final state of each system is identical, the source is different in each case. c. No, because the changes in energy for both systems are the same, and the cause of that change does not matter; the state of each system is identical. d. Yes, the changes in the energy for both systems are the same, but the causes of that change are different, so the states of each system are not identical. 19. How does the transfer of energy from the sun help plants? a. Plants absorb solar energy from the sun and utilize it during the fertilization process. b. Plants absorb solar energy from the sun and utilize Chapter 12 \u2022 Chapter Review 381 it during the process of photosynthesis to turn it into plant matter. c. Plants absorb solar energy from the sun and utilize it to increase the temperature inside them. d. Plants absorb solar energy from the sun and utilize it during the shedding of their leaves and fruits. 12.3 Second Law of Thermodynamics: Entropy 20. If an engine were constructed to perform such that there would be no losses due to friction, what would be its efficiency? a. b. c. d. It would be 0 percent. It would be less than 100 percent. It would be 100 percent. It would be greater than 100 percent. 21. Entropy never decreases in a spontaneous process. Give an example to support this statement. a. The transfer of energy by heat from colder bodies to hotter bodies is a spontaneous process in which the entropy of the system of bodies increases. b. The melting of an ice cube placed in a room causes an increase in the entropy of the room. c. The dissolution of salt in water is a spontaneous process in which the entropy of the system increases. d. A plant uses", " energy from the sun and converts it into sugar molecules by the process of photosynthesis. 12.4 Applications of Thermodynamics: Heat Engines, Heat Pumps, and Refrigerators 22. What is the advantage of a heat pump as opposed to burning fuel (as in a fireplace) for keeping warm? a. A heat pump supplies energy by heat from the cold, outside air. b. A heat pump supplies energy generated by the work done. c. A heat pump supplies energy by heat from the cold, outside air and also from the energy generated by the work done. d. A heat pump supplies energy not by heat from the cold, outside air, nor from the energy generated by the work done, but from more accessible sources. 23. What is thermal efficiency of an engine? Can it ever be 100 percent? Why or why not? a. Thermal efficiency is the ratio of the output (work) to the input (heat). It is always 100 percent. b. Thermal efficiency is the ratio of the output (heat) 382 Chapter 12 \u2022 Chapter Review to the input (work). It is always 100 percent. c. Thermal efficiency is the ratio of the output (heat) to the input (work). It is never 100 percent. environment b. When mass transferred to the environment is zero c. When mass transferred to the environment is at a d. Thermal efficiency is the ratio of the output (work) maximum to the input (heat). It is never 100 percent. d. When no energy is transferred by heat to the 24. When would 100 percent thermal efficiency be possible? a. When all energy is transferred by heat to the environment Problems 12.2 First law of Thermodynamics: Thermal Energy and Work 25. Some amount of energy is transferred by heat into a, while. What is the system. The net work done by the system is the increase in its internal energy is amount of net heat? a. b. c. d. 26. Eighty joules are added by heat to a system, while it are added by heat to the of work. What is the change in of work. Later, does system, and it does the system\u2019s internal energy? a. b. c. d. Performance Task 12.4 Applications of Thermodynamics: Heat Engines, Heat Pumps, and Refrigerators 29. You have been tasked to design and construct a thermometer that works on the principle of thermal expansion. There are four materials available for you to test, each of which will find use under different sets of conditions and temperature ranges: Materials \u2022 Four sample materials with similar mass or volume: copper, steel, water, and alcohol (ethanol or isopropanol) \u2022 Oven or similar heating source \u2022 Instrument (e.g., meter ruler, Vernier calipers, or micrometer) for measuring changes in dimension \u2022 Balance for measuring mass Procedure 1. Design a safe experiment to analyze the thermal Access for free at openstax.org. 12.4 Applications of Thermodynamics: Heat Engines, Heat Pumps, and Refrigerators 27. A coal power station functions at 40.0 percent efficiency. What is the amount of work it does if it takes in 1.20\u00d71012 J by heat? 3\u00d71010 J a. b. 4.8\u00d71011 J 3\u00d71012 J c. d. 4.8\u00d71013 J 28. A heat engine functions with 70.7 percent thermal efficiency and consumes 12.0 kJ from heat daily. If its efficiency were raised to 75.0 percent, how much energy from heat would be saved daily, while providing the same output? a. \u221210.8 kJ b. \u22121.08 kJ c. 0.7 kJ 7 kJ d. expansion properties of each material. 2. Write down the materials needed for your experiment and the procedure you will follow. Make sure that you include every detail so that the experiment can be repeated by others. 3. Select an appropriate material to measure temperature over a predecided temperature range, and give reasons for your choice. 4. Calibrate your instrument to measure temperature changes accurately. a. Which physical quantities are affected by temperature change and thermal expansion? b. How do such properties as specific heat and thermal conductivity affect the use of each material as a thermometer? c. Does a change of phase take place for any of the tested materials over the temperature range to be examined? d. What are your independent and dependent variables for this series of tests? Which variables need to be controlled in the experiment? e. What are your sources of error? f. Can all the tested materials be used effectively in the same ranges of temperature? Which applications might be suitable for one or more of the tested substances but not the others? Chapter 12 \u2022 Test Prep 383 TEST PREP Multiple Choice 12.1 Zeroth Law of Thermodynamics: Thermal Equilibrium 30. Which law of thermodynamics", " describes thermal equilibrium? a. zeroth b. first c. d. second third 31. Name any two industries in which the principles of thermodynamics are used. a. aerospace and information technology (IT) industries industrial manufacturing and aerospace b. c. mining and textile industries d. mining and agriculture industries 12.2 First law of Thermodynamics: Thermal Energy and Work 32. What is the value of the Boltzmann constant? a. b. c. d. 33. Which of the following involves work done BY a system? increasing internal energy compression a. b. c. expansion cooling d. 34. Which principle does the first law of thermodynamics state? a. b. c. d. the ideal gas law the transitive property of equality the law of conservation of energy the principle of thermal equilibrium a. A real gas behaves like an ideal gas at high temperature and low pressure. b. A real gas behaves like an ideal gas at high temperature and high pressure. c. A real gas behaves like an ideal gas at low temperature and low pressure. d. A real gas behaves like an ideal gas at low temperature and high pressure. 12.3 Second Law of Thermodynamics: Entropy 37. In an engine, what is the unused energy converted into? internal energy a. b. pressure c. work d. heat 38. It is natural for systems in the universe to _____ spontaneously. a. become disordered b. become ordered c. produce heat d. do work 39. If is and is, what is the change in entropy? a. b. c. d. 40. Why does entropy increase during a spontaneous process? a. Entropy increases because energy always transfers spontaneously from a dispersed state to a concentrated state. b. Entropy increases because energy always transfers spontaneously from a concentrated state to a dispersed state. c. Entropy increases because pressure always 35. What is the change in internal energy of a system when increases spontaneously. and? a. b. c. d. 36. When does a real gas behave like an ideal gas? d. Entropy increases because temperature of any system always increases spontaneously. 41. A system consists of ice melting in a glass of water. What happens to the entropy of this system? a. The entropy of the ice decreases, while the entropy of the water cannot be predicted without more 384 Chapter 12 \u2022 Test Prep specific information. b. The entropy of the system remains constant. c. The entropy of the system decreases. d. The entropy of the system increases. 12.4 Applications of Thermodynamics: Heat Engines, Heat Pumps, and Refrigerators 42. Which equation represents the net work done by a system in a cyclic process? a. b. c. d. 43. Which of these quantities needs to be zero for efficiency to be 100 percent? a. \u0394U b. W c. Qh d. Qc 44. Which of the following always has the greatest value in a system having 80 percent thermal efficiency? a. \u0394U Short Answer 12.1 Zeroth Law of Thermodynamics: Thermal Equilibrium 47. What does greenenergy development entail? a. Green energy involves finding new ways to harness clean and renewable alternative energy sources. b. Green energy involves finding new ways to conserve alternative energy sources. b. W c. Qh d. Qc 45. In the equation Q= Qh \u2212 Qc, what does the negative sign indicate? a. Heat transfer of energy is always negative. b. Heat transfer can only occur in one direction. c. Heat is directed into the system from the surroundings outside the system. d. Heat is directed out of the system. 46. What is the purpose of a heat pump? a. A heat pump uses work to transfer energy by heat from a colder environment to a warmer environment. b. A heat pump uses work to transfer energy by heat from a warmer environment to a colder environment. c. A heat pump does work by using heat to convey energy from a colder environment to a warmer environment. d. A heat pump does work by using heat to convey energy from a warmer environment to a colder environment. volume, which variable relates to pressure, and what is that relation? a. Temperature; inverse proportionality b. Temperature, direct proportionality to square root c. Temperature; direct proportionality d. Temperature; direct proportionality to square c. Green energy involves decreasing the efficiency of 50. When is volume directly proportional to temperature? nonrenewable energy resources. d. Green energy involves finding new ways to harness nonrenewable energy resources. 48. Why are the sun and Earth not in thermal equilibrium? a. The mass of the sun is much greater than the mass of Earth. b. There is a vast amount of empty space between the sun and Earth. c. The diameter of the sun is much greater than the diameter of Earth. d. The sun is in thermal contact with Earth. 12.2 First law of Thermodynamics: Thermal", " Energy and Work 49. If a fixed quantity of an ideal gas is held at a constant a. when the pressure of the gas is variable b. when the pressure of the gas is constant c. when the mass of the gas is variable d. when the mass of the gas is constant 51. For fluids, what can work be defined as? a. pressure acting over the change in depth b. pressure acting over the change in temperature c. d. pressure acting over the change in volume temperature acting over the change in volume, what does 52. In the equation indicate? a. b. c. d. the work done on the system the work done by the system the heat into the system the heat out of the system Access for free at openstax.org. Chapter 12 \u2022 Test Prep 385 53. By convention, if Qis positive, what is the direction in which heat transfers energy with regard to the system? a. The direction of the heat transfer of energy depends on the changes in W, regardless of the sign of Q. a. Entropy depends on the change of phase of a system, but not on any other state conditions. b. Entropy does not depend on how the final state is reached from the initial state. c. Entropy is least when the path between the initial b. The direction of Qcannot be determined from just state and the final state is the shortest. the sign of Q. d. Entropy is least when the path between the initial c. The direction of net heat transfer of energy will be state and the final state is the longest. the work was done 61. What is the change in entropy caused by melting 5.00 kg out of the system. d. The direction of net heat transfer of energy will be into the system. 54. What is net transfer of energy by heat? a. b. c. d. It is the sum of all energy transfers by heat into the system. It is the product of all energy transfers by heat into the system. It is the sum of all energy transfers by heat into and out of the system. It is the product of all energy transfers by heat into and out of the system. 55. Three hundred ten joules of heat enter a system, after which the system does of work. What is the change in its internal energy? Would this amount change if the energy transferred by heat were added after the work was done instead of before? a. ; this would change if heat added energy after the work was done ; this would change if heat added energy after ; this would not change even if heat added energy after the work was done ; this would not change even if heat added energy after the work was done 56. Ten joules are transferred by heat into a system,. What is the change in the followed by another system\u2019s internal energy? What would be the difference in this change if the system at once? a. ; the change in internal energy would be same of energy were added by heat to even if the heat added the energy at once ; the change in internal energy would be same even if the heat added the energy at once ; the change in internal energy would be more if the heat added the energy at once ; the change in internal energy would be more b. c. d. if the heat added the energy at once 12.3 Second Law of Thermodynamics: Entropy 57. How does the entropy of a system depend on how the system reaches a given state? b. c. d. 58. Which sort of thermal energy do molecules in a solid possess? a. electric potential energy b. gravitational potential energy c. translational kinetic energy d. vibrational kinetic energy 59. A cold object in contact with a hot one never spontaneously transfers energy by heat to the hot object. Which law describes this phenomenon? the first law of thermodynamics a. the second law of thermodynamics b. the third law of thermodynamics c. the zeroth law of thermodynamics d. 60. How is it possible for us to transfer energy by heat from cold objects to hot ones? a. by doing work on the system b. by having work done by the system c. by increasing the specific heat of the cold body d. by increasing the specific heat of the hot body of ice at 0 \u00b0C? a. 0 J/K b. 6.11\u00d7103 J/K c. 6.11\u00d7104 J/K d. \u221eJ/K 62. What is the amount of heat required to cause a change in the entropy of a system at? of a. b. c. d. 12.4 Applications of Thermodynamics: Heat Engines, Heat Pumps, and Refrigerators 63. In a refrigerator, what is the function of an evaporator? a. The evaporator converts gaseous refrigerant into liquid. b. The evaporator converts solid refrigerant into liquid. c. The evaporator converts solid refrigerant into gas. 386 Chapter 12", " \u2022 Test Prep d. The evaporator converts liquid refrigerant into gas. 64. Which component of an air conditioner converts gas into liquid? a. b. c. d. the condenser the compressor the evaporator the thermostat 65. What is one example for which calculating thermal efficiency is of interest? a. A wind turbine b. An electric pump c. A bicycle d. A car engine 66. How is the efficiency of a refrigerator or heat pump expressed? a. Extended Response 12.1 Zeroth Law of Thermodynamics: Thermal Equilibrium 69. What is the meaning of efficiency in terms of a car engine? a. An engine\u2019s efficiency equals the sum of useful energy (work) and the input energy. b. An engine\u2019s efficiency equals the proportion of useful energy (work) to the input energy. c. An engine\u2019s efficiency equals the product of useful energy (work) and the input energy. d. An engine\u2019s efficiency equals the difference between the useful energy (work) and the input energy. 12.2 First law of Thermodynamics: Thermal Energy and Work 70. Why does a bridge have expansion joints? a. because the bridge expands and contracts with the change in temperature b. because the bridge expands and contracts with the change in motion of objects moving on the bridge c. because the bridge expands and contracts with the change in total load on the bridge d. because the bridge expands and contracts with the change in magnitude of wind blowing 71. Under which conditions will the work done by the gas in a system increase? a. It will increase when a large amount of energy is added to the system, and that energy causes an increase in the gas\u2019s volume, its pressure, or both. Access for free at openstax.org. b. c. d. 67. How can you mathematically express thermal efficiency in terms of and? a. b. c. d. 68. How can you calculate percentage efficiency? a. percentage efficiency b. percentage efficiency c. percentage efficiency d. percentage efficiency b. c. d. It will increase when a large amount of energy is extracted from the system, and that energy causes an increase in the gas\u2019s volume, its pressure, or both. It will increase when a large amount of energy is added to the system, and that energy causes a decrease in the gas\u2019s volume, its pressure, or both. It will increase when a large amount of energy is extracted from the system, and that energy causes a decrease in the gas\u2019s volume, its pressure, or both. 72. How does energy transfer by heat aid in body metabolism? a. The energy is given to the body through the work done by the body (W) and through the intake of food, which may also be considered as the work done on the body. The transfer of energy out of the body is by heat (\u2212Q). b. The energy given to the body is by the intake of food, which may also be considered as the work done on the body. The transfer of energy out of the body is by heat (\u2212Q) and the work done by the body (W). c. The energy given to the body is by the transfer of energy by heat (Q) into the body, which may also be considered as the work done on the body. The transfer of energy out of the body is the work done by the body (W). d. The energy given to the body is by the transfer of energy by heat (Q) inside the body. The transfer of energy out of the body is by the intake of food and the work done by the body (W). 73. Two distinct systems have the same amount of stored internal energy. Five hundred joules are added by heat to the first system, and 300 J are added by heat to the second system. What will be the change in internal energy of the first system if it does 200 J of work? How much work will the second system have to do in order to have the same internal energy? a. b. c. d. 700 J; 0 J 300 J; 300 J 700 J; 300 J 300 J; 0 J 12.3 Second Law of Thermodynamics: Entropy 74. Why is it not possible to convert all available energy into work? a. Due to the entropy of a system, some energy is always unavailable for work. b. Due to the entropy of a system, some energy is always available for work. c. Due to the decrease in internal energy of a system, some energy is always made unavailable for work. d. Due to the increase in internal energy of a system, some energy is always made unavailable for work. 75. Why does entropy increase when ice melts into water? a. Melting converts the highly ordered solid structure into a disorderly liquid, thereby increasing entropy. b. Melting converts the highly ordered liquid into a disorderly solid structure, thereby increasing entropy", ". c. Melting converts the highly ordered solid structure into a disorderly solid structure, thereby increasing entropy. d. Melting converts the highly ordered liquid into a disorderly liquid, thereby increasing entropy. 76. Why is change in entropy lower for higher temperatures? a. Increase in the disorder in the substance is low for high temperature. Chapter 12 \u2022 Test Prep 387 b. Increase in the disorder in the substance is high for high temperature. c. Decrease in the disorder in the substance is low for high temperature. d. Decrease in the disorder in the substance is high for high temperature. 12.4 Applications of Thermodynamics: Heat Engines, Heat Pumps, and Refrigerators 77. In the equation W= Qh \u2212 Qc, if the value of Qc were equal to zero, what would it signify? a. The efficiency of the engine is 75 percent. b. The efficiency of the engine is 25 percent. c. The efficiency of the engine is 0 percent. d. The efficiency of the engine is 100 percent. 78. Can the value of thermal efficiency be greater than 1? Why or why not? a. No, according to the first law of thermodynamics, energy output can never be more than the energy input. b. No, according to the second law of thermodynamics, energy output can never be more than the energy input. c. Yes, according to the first law of thermodynamics, energy output can be more than the energy input. d. Yes, according to the second law of thermodynamics, energy output can be more than the energy input. 79. A coal power station transfers 3.0\u00d71012 J by heat from burning coal and transfers 1.5\u00d71012 J by heat into the environment. What is the efficiency of the power station? a. 0.33 b. 0.5 c. 0.66 1 d. 388 Chapter 12 \u2022 Test Prep Access for free at openstax.org. CHAPTER 13 Waves and Their Properties Figure 13.1 Waves in the ocean behave similarly to all other types of waves. (Steve Jurveston, Flickr) Chapter Outline 13.1 Types of Waves 13.2 Wave Properties: Speed, Amplitude, Frequency, and Period 13.3 Wave Interaction: Superposition and Interference Recall from the chapter on Motion in Two Dimensions that oscillations\u2014the back-and-forth movement INTRODUCTION between two points\u2014involve force and energy. Some oscillations create waves, such as the sound waves created by plucking a guitar string. Other examples of waves include earthquakes and visible light. Even subatomic particles, such as electrons, can behave like waves. You can make water waves in a swimming pool by slapping the water with your hand. Some of these waves, such as water waves, are visible; others, such as sound waves, are not. But every wave is a disturbance that moves from its source and carries energy. In this chapter, we will learn about the different types of waves, their properties, and how they interact with one another. 390 Chapter 13 \u2022 Waves and Their Properties 13.1 Types of Waves Section Learning Objectives By the end of this section, you will be able to do the following: \u2022 Define mechanical waves and medium, and relate the two \u2022 Distinguish a pulse wave from a periodic wave \u2022 Distinguish a longitudinal wave from a transverse wave and give examples of such waves Section Key Terms Mechanical Waves longitudinal wave mechanical wave medium wave periodic wave pulse wave transverse wave What do we mean when we say something is a wave? A wave is a disturbance that travels or propagatesfrom the place where it was created. Waves transfer energy from one place to another, but they do not necessarily transfer any mass. Light, sound, and waves in the ocean are common examples of waves. Sound and water waves are mechanical waves; meaning, they require a medium to travel through. The medium may be a solid, a liquid, or a gas, and the speed of the wave depends on the material properties of the medium through which it is traveling. However, light is not a mechanical wave; it can travel through a vacuum such as the empty parts of outer space. A familiar wave that you can easily imagine is the water wave. For water waves, the disturbance is in the surface of the water, an example of which is the disturbance created by a rock thrown into a pond or by a swimmer splashing the water surface repeatedly. For sound waves, the disturbance is caused by a change in air pressure, an example of which is when the oscillating cone inside a speaker creates a disturbance. For earthquakes, there are several types of disturbances, which include the disturbance of Earth\u2019s surface itself and the pressure disturbances under the surface. Even radio waves are most easily understood using an analogy with water waves. Because water waves are common and visible, visualizing water waves may help you in studying other types of waves, especially those that are not visible. Water waves have characteristics common to", " all waves, such as amplitude, period, frequency, and energy, which we will discuss in the next section. Pulse Waves and Periodic Waves If you drop a pebble into the water, only a few waves may be generated before the disturbance dies down, whereas in a wave pool, the waves are continuous. A pulse wave is a sudden disturbance in which only one wave or a few waves are generated, such as in the example of the pebble. Thunder and explosions also create pulse waves. A periodic wave repeats the same oscillation for several cycles, such as in the case of the wave pool, and is associated with simple harmonic motion. Each particle in the medium experiences simple harmonic motion in periodic waves by moving back and forth periodically through the same positions. Consider the simplified water wave in Figure 13.2. This wave is an up-and-down disturbance of the water surface, characterized by a sine wave pattern. The uppermost position is called the crestand the lowest is the trough. It causes a seagull to move up and down in simple harmonic motion as the wave crests and troughs pass under the bird. Figure 13.2 An idealized ocean wave passes under a seagull that bobs up and down in simple harmonic motion. Access for free at openstax.org. 13.1 \u2022 Types of Waves 391 Longitudinal Waves and Transverse Waves Mechanical waves are categorized by their type of motion and fall into any of two categories: transverse or longitudinal. Note that both transverse and longitudinal waves can be periodic. A transverse wave propagates so that the disturbance is perpendicular to the direction of propagation. An example of a transverse wave is shown in Figure 13.3, where a woman moves a toy spring up and down, generating waves that propagate away from herself in the horizontal direction while disturbing the toy spring in the vertical direction. Figure 13.3 In this example of a transverse wave, the wave propagates horizontally and the disturbance in the toy spring is in the vertical direction. In contrast, in a longitudinal wave, the disturbance is parallel to the direction of propagation. Figure 13.4 shows an example of a longitudinal wave, where the woman now creates a disturbance in the horizontal direction\u2014which is the same direction as the wave propagation\u2014by stretching and then compressing the toy spring. Figure 13.4 In this example of a longitudinal wave, the wave propagates horizontally and the disturbance in the toy spring is also in the horizontal direction. TIPS FOR SUCCESS Longitudinal waves are sometimes called compression wavesor compressional waves, and transverse waves are sometimes called shear waves. Waves may be transverse, longitudinal, or a combination of the two. The waves on the strings of musical instruments are transverse (as shown in Figure 13.5), and so are electromagnetic waves, such as visible light. Sound waves in air and water are longitudinal. Their disturbances are periodic variations in pressure that are transmitted in fluids. 392 Chapter 13 \u2022 Waves and Their Properties Figure 13.5 The wave on a guitar string is transverse. However, the sound wave coming out of a speaker rattles a sheet of paper in a direction that shows that such sound wave is longitudinal. Sound in solids can be both longitudinal and transverse. Essentially, water waves are also a combination of transverse and longitudinal components, although the simplified water wave illustrated in Figure 13.2 does not show the longitudinal motion of the bird. Earthquake waves under Earth\u2019s surface have both longitudinal and transverse components as well. The longitudinal waves in an earthquake are called pressure or P-waves, and the transverse waves are called shear or S-waves. These components have important individual characteristics; for example, they propagate at different speeds. Earthquakes also have surface waves that are similar to surface waves on water. WATCH PHYSICS Introduction to Waves This video explains wave propagation in terms of momentum using an example of a wave moving along a rope. It also covers the differences between transverse and longitudinal waves, and between pulse and periodic waves. Click to view content (https://openstax.org/l/02introtowaves) GRASP CHECK In a longitudinal sound wave, after a compression wave moves through a region, the density of molecules briefly decreases. Why is this? a. After a compression wave, some molecules move forward temporarily. b. After a compression wave, some molecules move backward temporarily. c. After a compression wave, some molecules move upward temporarily. d. After a compression wave, some molecules move downward temporarily. FUN IN PHYSICS The Physics of Surfing Many people enjoy surfing in the ocean. For some surfers, the bigger the wave, the better. In one area off the coast of central California, waves can reach heights of up to 50 feet in certain times of the year (Figure 13.6). Access for free at openstax.org. 13.1 \u2022 Types of Waves 393 Figure 13.6 A surfer negotiates a steep take", "-off on a winter day in California while his friend watches. (Ljsurf, Wikimedia Commons) How do waves reach such extreme heights? Other than unusual causes, such as when earthquakes produce tsunami waves, most huge waves are caused simply by interactions between the wind and the surface of the water. The wind pushes up against the surface of the water and transfers energy to the water in the process. The stronger the wind, the more energy transferred. As waves start to form, a larger surface area becomes in contact with the wind, and even more energy is transferred from the wind to the water, thus creating higher waves. Intense storms create the fastest winds, kicking up massive waves that travel out from the origin of the storm. Longer-lasting storms and those storms that affect a larger area of the ocean create the biggest waves since they transfer more energy. The cycle of the tides from the Moon\u2019s gravitational pull also plays a small role in creating waves. Actual ocean waves are more complicated than the idealized model of the simple transverse wave with a perfect sinusoidal shape. Ocean waves are examples of orbital progressive waves, where water particles at the surface follow a circular path from the crest to the trough of the passing wave, then cycle back again to their original position. This cycle repeats with each passing wave. As waves reach shore, the water depth decreases and the energy of the wave is compressed into a smaller volume. This creates higher waves\u2014an effect known as shoaling. Since the water particles along the surface move from the crest to the trough, surfers hitch a ride on the cascading water, gliding along the surface. If ocean waves work exactly like the idealized transverse waves, surfing would be much less exciting as it would simply involve standing on a board that bobs up and down in place, just like the seagull in the previous figure. Additional information and illustrations about the scientific principles behind surfing can be found in the \u201cUsing Science to Surf Better!\u201d (http://www.openstax.org/l/28Surf) video. GRASP CHECK If we lived in a parallel universe where ocean waves were longitudinal, what would a surfer\u2019s motion look like? a. The surfer would move side-to-side/back-and-forth vertically with no horizontal motion. b. The surfer would forward and backward horizontally with no vertical motion. Check Your Understanding 1. What is a wave? a. A wave is a force that propagates from the place where it was created. b. A wave is a disturbance that propagates from the place where it was created. c. A wave is matter that provides volume to an object. d. A wave is matter that provides mass to an object. 2. Do all waves require a medium to travel? Explain. a. No, electromagnetic waves do not require any medium to propagate. b. No, mechanical waves do not require any medium to propagate. c. Yes, both mechanical and electromagnetic waves require a medium to propagate. d. Yes, all transverse waves require a medium to travel. 3. What is a pulse wave? a. A pulse wave is a sudden disturbance with only one wave generated. b. A pulse wave is a sudden disturbance with only one or a few waves generated. 394 Chapter 13 \u2022 Waves and Their Properties c. A pulse wave is a gradual disturbance with only one or a few waves generated. d. A pulse wave is a gradual disturbance with only one wave generated. 4. Is the following statement true or false? A pebble dropped in water is an example of a pulse wave. a. False b. True 5. What are the categories of mechanical waves based on the type of motion? a. Both transverse and longitudinal waves b. Only longitudinal waves c. Only transverse waves d. Only surface waves 6. In which direction do the particles of the medium oscillate in a transverse wave? a. Perpendicular to the direction of propagation of the transverse wave b. Parallel to the direction of propagation of the transverse wave 13.2 Wave Properties: Speed, Amplitude, Frequency, and Period Section Learning Objectives By the end of this section, you will be able to do the following: \u2022 Define amplitude, frequency, period, wavelength, and velocity of a wave \u2022 Relate wave frequency, period, wavelength, and velocity \u2022 Solve problems involving wave properties Section Key Terms wavelength wave velocity Wave Variables In the chapter on motion in two dimensions, we defined the following variables to describe harmonic motion: \u2022 Amplitude\u2014maximum displacement from the equilibrium position of an object oscillating around such equilibrium position \u2022 Frequency\u2014number of events per unit of time \u2022 Period\u2014time it takes to complete one oscillation For waves, these variables have the same basic meaning. However, it is helpful to word the definitions in a more specific way that applies directly to waves: \u2022 Amplitude\u2014distance between the resting position and the maximum displacement of", " the wave \u2022 Frequency\u2014number of waves passing by a specific point per second \u2022 Period\u2014time it takes for one wave cycle to complete In addition to amplitude, frequency, and period, their wavelength and wave velocity also characterize waves. The wavelength is the distance between adjacent identical parts of a wave, parallel to the direction of propagation. The wave velocity is the speed at which the disturbance moves. TIPS FOR SUCCESS Wave velocity is sometimes also called the propagation velocityor propagation speedbecause the disturbance propagates from one location to another. Consider the periodic water wave in Figure 13.7. Its wavelength is the distance from crest to crest or from trough to trough. The wavelength can also be thought of as the distance a wave has traveled after one complete cycle\u2014or one period. The time for one complete up-and-down motion is the simple water wave\u2019s period T. In the figure, the wave itself moves to the right with a wave Access for free at openstax.org. 13.2 \u2022 Wave Properties: Speed, Amplitude, Frequency, and Period 395 velocity vw. Its amplitude Xis the distance between the resting position and the maximum displacement\u2014either the crest or the trough\u2014of the wave. It is important to note that this movement of the wave is actually the disturbancemoving to the right, not the water itself; otherwise, the bird would move to the right. Instead, the seagull bobs up and down in place as waves pass underneath, traveling a total distance of 2Xin one cycle. However, as mentioned in the text feature on surfing, actual ocean waves are more complex than this simplified example. Figure 13.7 The wave has a wavelength \u03bb, which is the distance between adjacent identical parts of the wave. The up-and-down disturbance of the surface propagates parallel to the surface at a speed vw. WATCH PHYSICS Amplitude, Period, Frequency, and Wavelength of Periodic Waves This video is a continuation of the video \u201cIntroduction to Waves\u201d from the \"Types of Waves\" section. It discusses the properties of a periodic wave: amplitude, period, frequency, wavelength, and wave velocity. Click to view content (https://www.openstax.org/l/28wavepro) TIPS FOR SUCCESS The crest of a wave is sometimes also called the peak. GRASP CHECK If you are on a boat in the trough of a wave on the ocean, and the wave amplitude is position? a. b. c. d., what is the wave height from your The Relationship between Wave Frequency, Period, Wavelength, and Velocity Since wave frequency is the number of waves per second, and the period is essentially the number of seconds per wave, the relationship between frequency and period is or 13.1 13.2 just as in the case of harmonic motion of an object. We can see from this relationship that a higher frequency means a shorter period. Recall that the unit for frequency is hertz (Hz), and that 1 Hz is one cycle\u2014or one wave\u2014per second. The speed of propagation vw is the distance the wave travels in a given time, which is one wavelength in a time of one period. In 396 Chapter 13 \u2022 Waves and Their Properties equation form, it is written as or 13.3 13.4 From this relationship, we see that in a medium where vw is constant, the higher the frequency, the smaller the wavelength. See Figure 13.8. Figure 13.8 Because they travel at the same speed in a given medium, low-frequency sounds must have a greater wavelength than high- frequency sounds. Here, the lower-frequency sounds are emitted by the large speaker, called a woofer, while the higher-frequency sounds are emitted by the small speaker, called a tweeter. These fundamental relationships hold true for all types of waves. As an example, for water waves, vw is the speed of a surface wave; for sound, vw is the speed of sound; and for visible light, vw is the speed of light. The amplitude Xis completely independent of the speed of propagation vw and depends only on the amount of energy in the wave. Snap Lab Waves in a Bowl In this lab, you will take measurements to determine how the amplitude and the period of waves are affected by the transfer of energy from a cork dropped into the water. The cork initially has some potential energy when it is held above the water\u2014the greater the height, the higher the potential energy. When it is dropped, such potential energy is converted to kinetic energy as the cork falls. When the cork hits the water, that energy travels through the water in waves. \u2022 Large bowl or basin \u2022 Water \u2022 Cork (or ping pong ball) \u2022 Stopwatch \u2022 Measuring tape Instructions Procedure 1. Fill a large bowl or basin with water and wait for the water to settle so there are no ripples. 2. Gently drop a cork into the middle", " of the bowl. 3. Estimate the wavelength and the period of oscillation of the water wave that propagates away from the cork. You can estimate the period by counting the number of ripples from the center to the edge of the bowl while your partner times it. This information, combined with the bowl measurement, will give you the wavelength when the correct formula is used. 4. Remove the cork from the bowl and wait for the water to settle again. 5. Gently drop the cork at a height that is different from the first drop. 6. Repeat Steps 3 to 5 to collect a second and third set of data, dropping the cork from different heights and recording the resulting wavelengths and periods. Access for free at openstax.org. 13.2 \u2022 Wave Properties: Speed, Amplitude, Frequency, and Period 397 7. Interpret your results. GRASP CHECK A cork is dropped into a pool of water creating waves. Does the wavelength depend upon the height above the water from which the cork is dropped? a. No, only the amplitude is affected. b. Yes, the wavelength is affected. LINKS TO PHYSICS Geology: Physics of Seismic Waves Figure 13.9 The destructive effect of an earthquake is a palpable evidence of the energy carried in the earthquake waves. The Richter scale rating of earthquakes is related to both their amplitude and the energy they carry. (Petty Officer 2nd Class Candice Villarreal, U.S. Navy) Geologists rely heavily on physics to study earthquakes since earthquakes involve several types of wave disturbances, including disturbance of Earth\u2019s surface and pressure disturbances under the surface. Surface earthquake waves are similar to surface waves on water. The waves under Earth\u2019s surface have both longitudinal and transverse components. The longitudinal waves in an earthquake are called pressure waves (P-waves) and the transverse waves are called shear waves (S-waves). These two types of waves propagate at different speeds, and the speed at which they travel depends on the rigidity of the medium through which they are traveling. During earthquakes, the speed of P-waves in granite is significantly higher than the speed of S-waves. Both components of earthquakes travel more slowly in less rigid materials, such as sediments. P-waves have speeds of 4 to 7 km/s, and S-waves have speeds of 2 to 5 km/s, but both are faster in more rigid materials. The P-wave gets progressively farther ahead of the S-wave as they travel through Earth\u2019s crust. For that reason, the time difference between the P- and S-waves is used to determine the distance to their source, the epicenter of the earthquake. We know from seismic waves produced by earthquakes that parts of the interior of Earth are liquid. Shear or transverse waves cannot travel through a liquid and are not transmitted through Earth\u2019s core. In contrast, compression or longitudinal waves can pass through a liquid and they do go through the core. All waves carry energy, and the energy of earthquake waves is easy to observe based on the amount of damage left behind after the ground has stopped moving. Earthquakes can shake whole cities to the ground, performing the work of thousands of wrecking balls. The amount of energy in a wave is related to its amplitude. Large-amplitude earthquakes produce large ground displacements and greater damage. As earthquake waves spread out, their amplitude decreases, so there is less damage the farther they get from the source. GRASP CHECK What is the relationship between the propagation speed, frequency, and wavelength of the S-waves in an earthquake? a. The relationship between the propagation speed, frequency, and wavelength is b. The relationship between the propagation speed, frequency, and wavelength is c. The relationship between the propagation speed, frequency, and wavelength is 398 Chapter 13 \u2022 Waves and Their Properties d. The relationship between the propagation speed, frequency, and wavelength is Virtual Physics Wave on a String Click to view content (http://www.openstax.org/l/28wavestring) In this animation, watch how a string vibrates in slow motion by choosing the Slow Motion setting. Select the No End and Manual options, and wiggle the end of the string to make waves yourself. Then switch to the Oscillate setting to generate waves automatically. Adjust the frequency and the amplitude of the oscillations to see what happens. Then experiment with adjusting the damping and the tension. GRASP CHECK Which of the settings\u2014amplitude, frequency, damping, or tension\u2014changes the amplitude of the wave as it propagates? What does it do to the amplitude? a. Frequency; it decreases the amplitude of the wave as it propagates. b. Frequency; it increases the amplitude of the wave as it propagates. c. Damping; it decreases the amplitude of the wave as it propagates. d. Damping; it increases the amplitude of", " the wave as it propagates. Solving Wave Problems WORKED EXAMPLE Calculate the Velocity of Wave Propagation: Gull in the Ocean Calculate the wave velocity of the ocean wave in the previous figure if the distance between wave crests is 10.0 m and the time for a seagull to bob up and down is 5.00 s. STRATEGY The values for the wavelength can use are given and we are asked to find Therefore, we to find the wave velocity. and the period Solution Enter the known values into Discussion This slow speed seems reasonable for an ocean wave. Note that in the figure, the wave moves to the right at this speed, which is different from the varying speed at which the seagull bobs up and down. 13.5 WORKED EXAMPLE Calculate the Period and the Wave Velocity of a Toy Spring The woman in creates two waves every second by shaking the toy spring up and down. (a)What is the period of each wave? (b) If each wave travels 0.9 meters after one complete wave cycle, what is the velocity of wave propagation? STRATEGY FOR (A) To find the period, we solve for, given the value of the frequency Access for free at openstax.org. 13.2 \u2022 Wave Properties: Speed, Amplitude, Frequency, and Period 399 Solution for (a) Enter the known value into 13.6 STRATEGY FOR (B) Since one definition of wavelength is the distance a wave has traveled after one complete cycle\u2014or one period\u2014the values for the wavelength as well as the frequency are given. Therefore, we can use to find the wave velocity. Solution for (b) Enter the known values into Discussion We could have also used the equation to solve for the wave velocity since we already know the value of the period from our calculation in part (a), and we would come up with the same answer. Practice Problems 7. The frequency of a wave is 10 Hz. What is its period? a. The period of the wave is 100 s. b. The period of the wave is 10 s. c. The period of the wave is 0.01 s. d. The period of the wave is 0.1 s. 8. What is the velocity of a wave whose wavelength is 2 m and whose frequency is 5 Hz? a. 20 m/s b. 2.5 m/s c. 0.4 m/s 10 m/s d. Check Your Understanding 9. What is the amplitude of a wave? a. A quarter of the total height of the wave b. Half of the total height of the wave c. Two times the total height of the wave d. Four times the total height of the wave 10. What is meant by the wavelength of a wave? a. The wavelength is the distance between adjacent identical parts of a wave, parallel to the direction of propagation. b. The wavelength is the distance between adjacent identical parts of a wave, perpendicular to the direction of propagation. c. The wavelength is the distance between a crest and the adjacent trough of a wave, parallel to the direction of propagation. d. The wavelength is the distance between a crest and the adjacent trough of a wave, perpendicular to the direction of propagation. 11. How can you mathematically express wave frequency in terms of wave period? a. b. c. d. 12. When is the wavelength directly proportional to the period of a wave? 400 Chapter 13 \u2022 Waves and Their Properties a. When the velocity of the wave is halved b. When the velocity of the wave is constant c. When the velocity of the wave is doubled d. When the velocity of the wave is tripled 13.3 Wave Interaction: Superposition and Interference Section Learning Objectives By the end of this section, you will be able to do the following: \u2022 Describe superposition of waves \u2022 Describe interference of waves and distinguish between constructive and destructive interference of waves \u2022 Describe the characteristics of standing waves \u2022 Distinguish reflection from refraction of waves Section Key Terms antinode constructive interference destructive interference inversion nodes reflection refraction standing wave superposition Superposition of Waves Most waves do not look very simple. They look more like the waves in Figure 13.10, rather than the simple water wave considered in the previous sections, which has a perfect sinusoidal shape. Figure 13.10 These waves result from the superposition of several waves from different sources, producing a complex pattern. (Waterborough, Wikimedia Commons) Most waves appear complex because they result from two or more simple waves that combine as they come together at the same place at the same time\u2014a phenomenon called superposition. Waves superimpose by adding their disturbances; each disturbance corresponds to a force, and all the forces add. If the disturbances are along the same line, then the resulting wave is a simple addition of the disturbances of the individual waves, that is, their amplitudes add. Wave Interference The two special cases of super", "position that produce the simplest results are pure constructive interference and pure destructive interference. Pure constructive interference occurs when two identical waves arrive at the same point exactly in phase. When waves are exactly in phase, the crests of the two waves are precisely aligned, as are the troughs. Refer to Figure 13.11. Because the disturbances add, the pure constructive interference of two waves with the same amplitude produces a wave that has twice the amplitude of the two individual waves, but has the same wavelength. Access for free at openstax.org. 13.3 \u2022 Wave Interaction: Superposition and Interference 401 Figure 13.11 The pure constructive interference of two identical waves produces a wave with twice the amplitude but the same wavelength. Figure 13.12 shows two identical waves that arrive exactly outof phase\u2014that is, precisely aligned crest to trough\u2014producing pure destructive interference. Because the disturbances are in opposite directions for this superposition, the resulting amplitude is zero for pure destructive interference; that is, the waves completely cancel out each other. Figure 13.12 The pure destructive interference of two identical waves produces zero amplitude, or complete cancellation. While pure constructive interference and pure destructive interference can occur, they are not very common because they require precisely aligned identical waves. The superposition of most waves that we see in nature produces a combination of constructive and destructive interferences. Waves that are not results of pure constructive or destructive interference can vary from place to place and time to time. The sound from a stereo, for example, can be loud in one spot and soft in another. The varying loudness means that the sound waves add partially constructively and partially destructively at different locations. A stereo has at least two speakers that create sound waves, and waves can reflect from walls. All these waves superimpose. An example of sounds that vary over time from constructive to destructive is found in the combined whine of jet engines heard by a stationary passenger. The volume of the combined sound can fluctuate up and down as the sound from the two engines varies in time from constructive to destructive. The two previous examples considered waves that are similar\u2014both stereo speakers generate sound waves with the same amplitude and wavelength, as do the jet engines. But what happens when two waves that are not similar, that is, having different amplitudes and wavelengths, are superimposed? An example of the superposition of two dissimilar waves is shown in Figure 13.13. Here again, the disturbances add and subtract, but they produce an even more complicated-looking wave. The resultant wave from the combined disturbances of two dissimilar waves looks much different than the idealized sinusoidal shape of a periodic wave. 402 Chapter 13 \u2022 Waves and Their Properties Figure 13.13 The superposition of nonidentical waves exhibits both constructive and destructive interferences. Virtual Physics Wave Interference Click to view content (http://www.openstax.org/l/28interference) In this simulation, make waves with a dripping faucet, an audio speaker, or a laser by switching between the water, sound, and light tabs. Contrast and compare how the different types of waves behave. Try rotating the view from top to side to make observations. Then experiment with adding a second source or a pair of slits to create an interference pattern. GRASP CHECK In the water tab, compare the waves generated by one drip versus two drips. What happens to the amplitude of the waves when there are two drips? Is this constructive or destructive interference? Why would this be the case? a. The amplitude of the water waves remains same because of the destructive interference as the drips of water hit the surface at the same time. b. The amplitude of the water waves is canceled because of the destructive interference as the drips of water hit the surface at the same time. c. The amplitude of water waves remains same because of the constructive interference as the drips of water hit the surface at the same time. d. The amplitude of water waves doubles because of the constructive interference as the drips of water hit the surface at the same time. Standing Waves Sometimes waves do not seem to move and they appear to just stand in place, vibrating. Such waves are called standing waves and are formed by the superposition of two or more waves moving in opposite directions. The waves move through each other with their disturbances adding as they go by. If the two waves have the same amplitude and wavelength, then they alternate between constructive and destructive interference. Standing waves created by the superposition of two identical waves moving in opposite directions are illustrated in Figure 13.14. Figure 13.14 A standing wave is created by the superposition of two identical waves moving in opposite directions. The oscillations are at fixed locations in space and result from alternating constructive and destructive interferences. As an example, standing waves can be seen on the surface of a glass of milk in a refrigerator. The vibrations from the refrigerator motor create waves on the milk that oscillate up and down but do not seem to move across the surface. The two", " waves that Access for free at openstax.org. 13.3 \u2022 Wave Interaction: Superposition and Interference 403 produce standing waves may be due to the reflections from the side of the glass. Earthquakes can create standing waves and cause constructive and destructive interferences. As the earthquake waves travel along the surface of Earth and reflect off denser rocks, constructive interference occurs at certain points. As a result, areas closer to the epicenter are not damaged while areas farther from the epicenter are damaged. Standing waves are also found on the strings of musical instruments and are due to reflections of waves from the ends of the string. Figure 13.15 and Figure 13.16 show three standing waves that can be created on a string that is fixed at both ends. When the wave reaches the fixed end, it has nowhere else to go but back where it came from, causing the reflection. The nodes are the points where the string does not move; more generally, the nodes are the points where the wave disturbance is zero in a standing wave. The fixed ends of strings must be nodes, too, because the string cannot move there. The antinode is the location of maximum amplitude in standing waves. The standing waves on a string have a frequency that is of the disturbance on the string. The wavelength is determined by the distance between related to the propagation speed the points where the string is fixed in place. Figure 13.15 The figure shows a string oscillating with its maximum disturbance as the antinode. Figure 13.16 The figure shows a string oscillating with multiple nodes. Reflection and Refraction of Waves As we saw in the case of standing waves on the strings of a musical instrument, reflection is the change in direction of a wave when it bounces off a barrier, such as a fixed end. When the wave hits the fixed end, it changes direction, returning to its source. As it is reflected, the wave experiences an inversion, which means that it flips vertically. If a wave hits the fixed end with a crest, it will return as a trough, and vice versa (Henderson 2015). Refer to Figure 13.17. Figure 13.17 A wave is inverted after reflection from a fixed end. TIPS FOR SUCCESS If the end is not fixed, it is said to be a free end, and no inversion occurs. When the end is loosely attached, it reflects without 404 Chapter 13 \u2022 Waves and Their Properties inversion, and when the end is not attached to anything, it does not reflect at all. You may have noticed this while changing the settings from Fixed End to Loose End to No End in the Waves on a String PhET simulation. Rather than encountering a fixed end or barrier, waves sometimes pass from one medium into another, for instance, from air into water. Different types of media have different properties, such as density or depth, that affect how a wave travels through them. At the boundary between media, waves experience refraction\u2014they change their path of propagation. As the wave bends, it also changes its speed and wavelength upon entering the new medium. Refer to Figure 13.18. Figure 13.18 A wave refracts as it enters a different medium. For example, water waves traveling from the deep end to the shallow end of a swimming pool experience refraction. They bend in a path closer to perpendicular to the surface of the water, propagate slower, and decrease in wavelength as they enter shallower water. Check Your Understanding 13. What is the superposition of waves? a. When a single wave splits into two different waves at a point b. When two waves combine at the same place at the same time 14. How do waves superimpose on one another? a. By adding their frequencies b. By adding their wavelengths c. By adding their disturbances d. By adding their speeds 15. What is interference of waves? a. b. c. d. Interference is a superposition of two waves to form a resultant wave with higher or lower frequency. Interference is a superposition of two waves to form a wave of larger or smaller amplitude. Interference is a superposition of two waves to form a resultant wave with higher or lower velocity. Interference is a superposition of two waves to form a resultant wave with longer or shorter wavelength. 16. Is the following statement true or false? The two types of interference are constructive and destructive interferences. a. True b. False 17. What are standing waves? a. Waves that appear to remain in one place and do not seem to move b. Waves that seem to move along a trajectory 18. How are standing waves formed? a. Standing waves are formed by the superposition of two or more waves moving in opposite directions. b. Standing waves are formed by the superposition of two or more waves moving in the same direction. c. Standing waves are formed by the superposition of two or more waves moving in perpendicular directions. d. Standing waves are formed by the superposition of two or more waves", " moving in any arbitrary directions. 19. What is the reflection of a wave? a. The reflection of a wave is the change in amplitude of a wave when it bounces off a barrier. b. The reflection of a wave is the change in frequency of a wave when it bounces off a barrier. c. The reflection of a wave is the change in velocity of a wave when it bounces off a barrier. d. The reflection of a wave is the change in direction of a wave when it bounces off a barrier. Access for free at openstax.org. 13.3 \u2022 Wave Interaction: Superposition and Interference 405 20. What is inversion of a wave? a. b. c. d. Inversion occurs when a wave reflects off a fixed end, and the wave amplitude changes sign. Inversion occurs when a wave reflects off a loose end, and the wave amplitude changes sign. Inversion occurs when a wave reflects off a fixed end without the wave amplitude changing sign. Inversion occurs when a wave reflects off a loose end without the wave amplitude changing sign. 406 Chapter 13 \u2022 Key Terms KEY TERMS antinode location of maximum amplitude in standing motion waves pulse wave sudden disturbance with only one wave or a few constructive interference when two waves arrive at the waves generated same point exactly in phase; that is, the crests of the two waves are precisely aligned, as are the troughs reflection change in direction of a wave at a boundary or fixed end destructive interference when two identical waves arrive at refraction bending of a wave as it passes from one medium the same point exactly out of phase that is precisely aligned crest to trough inversion vertical flipping of a wave after reflection from a fixed end longitudinal wave wave in which the disturbance is parallel to another medium with a different density standing wave wave made by the superposition of two waves of the same amplitude and wavelength moving in opposite directions and which appears to vibrate in place superposition phenomenon that occurs when two or more to the direction of propagation waves arrive at the same point mechanical wave wave that requires a medium through which it can travel transverse wave wave in which the disturbance is perpendicular to the direction of propagation medium solid, liquid, or gas material through which a wave disturbance that moves from its source and carries wave propagates energy nodes points where the string does not move; more generally, points where the wave disturbance is zero in a standing wave periodic wave wave that repeats the same oscillation for several cycles and is associated with simple harmonic SECTION SUMMARY 13.1 Types of Waves \u2022 A wave is a disturbance that moves from the point of creation and carries energy but not mass. \u2022 Mechanical waves must travel through a medium. \u2022 Sound waves, water waves, and earthquake waves are all examples of mechanical waves. \u2022 Light is not a mechanical wave since it can travel through a vacuum. \u2022 A periodic wave is a wave that repeats for several cycles, whereas a pulse wave has only one crest or a few crests and is associated with a sudden disturbance. \u2022 Periodic waves are associated with simple harmonic motion. \u2022 A transverse wave has a disturbance perpendicular to its direction of propagation, whereas a longitudinal wave has a disturbance parallel to its direction of propagation. 13.2 Wave Properties: Speed, Amplitude, Frequency, and Period \u2022 A wave is a disturbance that moves from the point of creation at a wave velocity vw. \u2022 A wave has a wavelength, which is the distance between adjacent identical parts of the wave. \u2022 The wave velocity and the wavelength are related to the wave\u2019s frequency and period by or Access for free at openstax.org. wave velocity speed at which the disturbance moves; also called the propagation velocity or propagation speed wavelength distance between adjacent identical parts of a wave \u2022 The time for one complete wave cycle is the period T. \u2022 The number of waves per unit time is the frequency \u0192. \u2022 The wave frequency and the period are inversely related to one another. 13.3 Wave Interaction: Superposition and Interference \u2022 Superposition is the combination of two waves at the same location. \u2022 Constructive interference occurs when two identical waves are superimposed exactly in phase. \u2022 Destructive interference occurs when two identical waves are superimposed exactly out of phase. \u2022 A standing wave is a wave produced by the superposition of two waves. It varies in amplitude but does not propagate. \u2022 The nodes are the points where there is no motion in standing waves. \u2022 An antinode is the location of maximum amplitude of a standing wave. Inversion occurs when a wave reflects from a fixed end. \u2022 Reflection causes a wave to change direction. \u2022 \u2022 Refraction causes a wave\u2019s path to bend and occurs when a wave passes from one medium into another medium with a different density. KEY EQUATIONS 13.2 Wave Properties: Speed, Amplitude, Frequency, and Period wave velocity or CHAPTER REVIEW Concept Items 13.1 Types of Waves 1. Do water waves push water from one place to another? Explain. a. No, water waves", " transfer only energy from one place to another. b. Yes, water waves transfer water from one place to another. 2. With reference to waves, what is a trough? a. b. c. d. the lowermost position of a wave the uppermost position of a wave the final position of a wave the initial position of the wave 3. Give an example of longitudinal waves. light waves a. b. water waves in a lake sound waves in air c. seismic waves in Earth\u2019s surface d. 4. What does the speed of a mechanical wave depend on? a. b. c. d. the properties of the material through which it travels the shape of the material through which it travels the size of the material through which it travels the color of the material through which it travels 13.2 Wave Properties: Speed, Amplitude, Frequency, and Period 5. Which characteristic of a transverse wave is measured along the direction of propagation? a. The amplitude of a transverse wave is measured along the direction of propagation. b. The amplitude and the wavelength of a transverse wave are measured along the direction of propagation. c. The wavelength of a transverse wave is measured along the direction of propagation. d. The displacement of the particles of the medium in a transverse wave is measured along the direction of propagation. 6. Which kind of seismic waves cannot travel through Chapter 13 \u2022 Key Equations 407 period frequency liquid? a. b. P-waves c. d. S-waves compressional waves longitudinal waves 7. What is the period of a wave? a. b. c. d. the time that a wave takes to complete a half cycle the time that a wave takes to complete one cycle the time that a wave takes to complete two cycles the time that a wave takes to complete four cycles 8. When the period of a wave increases, what happens to its frequency? a. b. c. Its frequency decreases. Its frequency increases. Its frequency remains the same. 13.3 Wave Interaction: Superposition and Interference 9. Is this statement true or false? The amplitudes of waves add up only if they are propagating in the same line. a. True b. False 10. Why is sound from a stereo louder in one part of the room and softer in another? a. Sound is louder in parts of the room where the density is greatest. Sound is softer in parts of the room where density is smallest. b. Sound is louder in parts of the room where the density is smallest. Sound is softer in parts of the room where density is greatest. c. Sound is louder in parts of the room where constructive interference occurs and softer in parts where destructive interference occurs. d. Sound is louder in parts of the room where destructive interference occurs and softer in parts where constructive interference occurs. 11. In standing waves on a string, what does the frequency depend on? a. The frequency depends on the propagation speed and the density of the string. 408 Chapter 13 \u2022 Chapter Review b. The frequency depends on the propagation speed a. Refraction is the phenomenon in which waves and the length of the string. c. The frequency depends on the density and the length of the string. d. The frequency depends on the propagation speed, the density, and the length of the string. 12. Is the following statement true or false? Refraction is useful in fiber optic cables for transmitting signals. a. False b. True 13. What is refraction? Critical Thinking Items 13.1 Types of Waves 14. Give an example of a wave that propagates only through a solid. a. Light wave b. Sound wave c. Seismic wave d. Surface wave 15. Can mechanical waves be periodic waves? a. No, mechanical waves cannot be periodic waves. b. Yes, mechanical waves can be periodic. 16. In a sound wave, which parameter of the medium varies with every cycle? a. The density of the medium varies with every cycle. b. The mass of the medium varies with every cycle. c. The resistivity of the medium varies with every cycle. d. The volume of the medium varies with every cycle. 17. What is a transverse wave in an earthquake called? a. L-wave b. P-wave c. S-wave d. R-wave 13.2 Wave Properties: Speed, Amplitude, Frequency, and Period 18. If the horizontal distance, that is, the distance in the direction of propagation, between a crest and the adjacent trough of a sine wave is 1 m, what is the wavelength of the wave? a. 0.5 m b. 1 m c. 2 m d. 4 m 19. How is the distance to the epicenter of an earthquake determined? Access for free at openstax.org. change their path of propagation at the interface of two media with different densities. b. Refraction is the phenomenon in which waves change their path of propagation at the interface of two media with the same density.", " c. Refraction is the phenomenon in which waves become non-periodic at the boundary of two media with different densities. d. Refraction is the phenomenon in which waves become non-periodic at the boundary of two media with the same density. a. The wavelength difference between P-waves and S- waves is used to measure the distance to the epicenter. b. The time difference between P-waves and S-waves is used to measure the distance to the epicenter. c. The frequency difference between P-waves and Swaves is used to measure the distance to the epicenter. d. The phase difference between P-waves and S-waves is used to measure the distance to the epicenter. 20. Two identical waves superimpose in pure constructive interference. What is the height of the resultant wave if the amplitude of each of the waves is 1 m? a. 1 m b. 2 m 3 m c. d. 4 m 13.3 Wave Interaction: Superposition and Interference 21. Two identical waves with an amplitude superimpose in a way that pure constructive interference occurs. What is the amplitude of the resultant wave? a. b. c. d. 22. In which kind of wave is the amplitude at each point constant? a. Seismic waves b. Pulse wave c. Standing waves d. Electromagnetic waves 23. Which property of a medium causes refraction? a. Conductivity b. Opacity c. Ductility d. Density 24. What is added together when two waves superimpose? a. Amplitudes Problems 13.2 Wave Properties: Speed, Amplitude, Frequency, and Period 25. If a seagull sitting in water bobs up and down once every 2 seconds and the distance between two crests of the water wave is 3 m, what is the velocity of the wave? 1.5 m/s a. 3 m/s b. c. 6 m/s Performance Task 13.3 Wave Interaction: Superposition and Interference 27. Ocean waves repeatedly crash against beaches and coasts. Their energy can lead to erosion and collapse of land. Scientists and engineers need to study how waves interact with beaches in order to assess threats to coastal communities and construct breakwater systems. In this task, you will construct a wave tank and fill it with water. Simulate a beach by placing sand at one end. Create waves by moving a piece of wood or plastic up and down in the water. Measure or estimate the TEST PREP Multiple Choice 13.1 Types of Waves 28. What kind of waves are sound waves? a. Mechanical waves b. Electromagnetic waves Chapter 13 \u2022 Test Prep 409 b. Wavelengths c. Velocities d. 12 m/s 26. A boat in the trough of a wave takes 3 seconds to reach the highest point of the wave. The velocity of the wave is 5 m/s. What is its wavelength? a. 0.83 m 15 m b. 30 m c. 180 m d. wavelength, period, frequency, and amplitude of the wave, and observe the effect of the wave on the sand. Produce waves of different amplitudes and frequencies, and record your observations each time. Use mathematical representations to demonstrate the relationships between different wave properties. Change the position of the sand to create a steeper beach and record your observations. Give a qualitative analysis of the effects of the waves on the beach. What kind of wave causes the most damage? At what height, wavelength, and frequency do waves break? How does the steepness of the beach affect the waves? that creates a wave. It refers to the wavelength of the wave. It refers to the speed of the wave. c. d. 13.2 Wave Properties: Speed, Amplitude, Frequency, and Period 29. What kind of a wave does a tuning fork create? 32. Which of these is not a characteristic of a wave? a. Pulse wave b. Periodic wave c. Electromagnetic wave 30. What kind of waves are electromagnetic waves? a. amplitude b. period c. mass d. velocity a. Longitudinal waves b. Transverse waves c. Mechanical waves d. P-waves a. 31. With reference to waves, what is a disturbance? It refers to the resistance produced by some particles of a material. It refers to an oscillation produced by some energy b. 33. If you are in a boat at a resting position, how much will your height change when you are hit by the peak of a wave with a height of 2 m? a. 0 m 1 m b. c. 2 m d. 4 m 34. What is the period of a wave with a frequency of 0.5 Hz? 410 Chapter 13 \u2022 Test Prep a. 0.5 s 1 s b. c. 2 s 3 s d. 35. What is the relation between the amplitude of a wave and its speed? a. The amplitude of a wave is independent of its speed.", " b. The amplitude of a wave is directly proportional to its speed. c. The amplitude of a wave is directly proportional to the square of the inverse of its speed. d. The amplitude of a wave is directly proportional to the inverse of its speed. 36. What does the speed of seismic waves depend on? a. The speed of seismic waves depends on the size of the medium. b. The speed of seismic waves depends on the shape of the medium. c. The speed of seismic waves depends on the rigidity of the medium. 13.3 Wave Interaction: Superposition and Interference 37. What is added together when two waves superimpose? a. amplitudes b. wavelengths c. velocities 38. Pure constructive interference occurs between two waves when they have the same _____. Short Answer 13.1 Types of Waves frequency and are in phase frequency and are out of phase a. b. c. amplitude and are in phase d. amplitude and are out of phase 39. What kind(s) of interference can occur between two identical waves moving in opposite directions? a. Constructive interference only b. Destructive interference only c. Both constructive and destructive interference d. Neither constructive nor destructive interference 40. What term refers to the bending of light at the junction interference of two media? a. b. diffraction scattering c. refraction d. 41. Which parameter of a wave gets affected after superposition? a. wavelength b. direction c. amplitude frequency d. 42. When do the amplitudes of two waves get added? a. When their amplitudes are the same b. When their amplitudes are different c. When they propagate in perpendicular directions d. When they are propagating along the same line in opposite directions d. The cone of a speaker vibrates to create small changes in the resistance of the air. 43. Give an example of a non-mechanical wave. a. A radio wave is an example of a nonmechanical wave. b. A sound wave is an example of a nonmechanical wave. 45. What kind of wave is thunder? a. Transverse wave b. Pulse wave c. R-wave d. P-wave c. A surface wave is an example of a nonmechanical 46. Are all ocean waves perfectly sinusoidal? wave. d. A seismic wave is an example of a nonmechanical wave. 44. How is sound produced by an electronic speaker? a. The cone of a speaker vibrates to create small changes in the temperature of the air. b. The cone of a speaker vibrates to create small changes in the pressure of the air. c. The cone of a speaker vibrates to create small changes in the volume of the air. a. No, all ocean waves are not perfectly sinusoidal. b. Yes, all ocean waves are perfectly sinusoidal. 47. What are orbital progressive waves? a. Waves that force the particles of the medium to follow a linear path from the crest to the trough b. Waves that force the particles of the medium to follow a circular path from the crest to the trough c. Waves that force the particles of the medium to follow a zigzag path from the crest to the trough d. Waves that force the particles of the medium to Access for free at openstax.org. follow a random path from the crest to the trough 48. Give an example of orbital progressive waves. a. light waves b. ocean waves sound waves c. seismic waves d. 13.2 Wave Properties: Speed, Amplitude, Frequency, and Period 49. What is the relation between the amplitude and height of a transverse wave? a. The height of a wave is half of its amplitude. b. The height of a wave is equal to its amplitude. c. The height of a wave is twice its amplitude. d. The height of a wave is four times its amplitude. 50. If the amplitude of a water wave is 0.2 m and its frequency is 2 Hz, how much distance would a bird sitting on the water\u2019s surface move with every wave? How many times will it do this every second? a. The bird will go up and down a distance of 0.4 m. It will do this twice per second. b. The bird will go up and down a distance of 0.2 m. It will do this twice per second. c. The bird will go up and down a distance of 0.4 m. It will do this once per second. d. The bird will go up and down a distance of 0.2 m. It will do this once per second. 51. What is the relation between the amplitude and the frequency of a wave? a. The amplitude and the frequency of a wave are independent of each other. b. The amplitude and the frequency of a wave are equal. c. The amplitude decreases with an increase in the frequency of a wave. d. The amplitude increases with an increase in", " the frequency of a wave. 52. What is the relation between a wave\u2019s energy and its amplitude? a. There is no relation between the energy and the amplitude of a wave. b. The magnitude of the energy is equal to the magnitude of the amplitude of a wave. c. The energy of a wave increases with an increase in the amplitude of the wave. d. The energy of a wave decreases with an increase in the amplitude of a wave. 53. A wave travels every 2 cycles. What is its wavelength? a. Chapter 13 \u2022 Test Prep 411 b. c. d. 54. A water wave propagates in a river at 6 m/s. If the river moves in the opposite direction at 3 m/s, what is the effective velocity of the wave? a. 3 m/s b. 6 m/s c. 9 m/s 18 m/s d. 13.3 Wave Interaction: Superposition and Interference 55. Is this statement true or false? Spherical waves can superimpose. a. True b. False 56. Is this statement true or false? Waves can superimpose if their frequencies are different. a. True b. False 57. When does pure destructive interference occur? a. When two waves with equal frequencies that are perfectly in phase and propagating along the same line superimpose. b. When two waves with unequal frequencies that are perfectly in phase and propagating along the same line superimpose. c. When two waves with unequal frequencies that are perfectly out of phase and propagating along the same line superimpose. d. When two waves with equal frequencies that are perfectly out of phase and propagating along the same line superimpose. 58. Is this statement true or false? The amplitude of one wave is affected by the amplitude of another wave only when they are precisely aligned. a. True b. False 59. Why does a standing wave form on a guitar string? a. due to superposition with the reflected waves from the ends of the string b. due to superposition with the reflected waves from the walls of the room c. due to superposition with waves generated from the body of the guitar 60. Is the following statement true or false? A standing wave is a superposition of two identical waves that are in phase and propagating in the same direction. 412 Chapter 13 \u2022 Test Prep a. True b. False 61. Why do water waves traveling from the deep end to the shallow end of a swimming pool experience refraction? a. Because the pressure of water at the two ends of the c. Because the density of water at the two ends of the pool is same d. Because the density of water at the two ends of the pool is different 62. Is the statement true or false? Waves propagate faster in pool is same b. Because the pressures of water at the two ends of the pool are different a less dense medium if the stiffness is the same. a. True b. False Extended Response 13.1 Types of Waves 63. Why can light travel through outer space while sound cannot? a. Sound waves are mechanical waves and require a medium to propagate. Light waves can travel through a vacuum. b. Sound waves are electromagnetic waves and require a medium to propagate. Light waves can travel through a vacuum. c. Light waves are mechanical waves and do not require a medium to propagate; sound waves require a medium to propagate. d. Light waves are longitudinal waves and do not require a medium to propagate; sound waves require a medium to propagate. 64. Do periodic waves require a medium to travel through? a. No, the requirement of a medium for propagation does not depend on whether the waves are pulse waves or periodic waves. b. Yes, the requirement of a medium for propagation depends on whether the waves are pulse waves or periodic waves. 65. How is the propagation of sound in solids different from that in air? a. Sound waves in solids are transverse, whereas in and moves with the wave in its direction. b. The gull experiences mostly side-to-side motion but does not move with the wave in its direction. c. The gull experiences mostly up-and-down motion and moves with the wave in its direction. d. The gull experiences mostly up-and-down motion but does not move in the direction of the wave. 67. Why does a good-quality speaker have a woofer and a tweeter? a. In a good-quality speaker, sounds with high frequencies or short wavelengths are reproduced accurately by woofers, while sounds with low frequencies or long wavelengths are reproduced accurately by tweeters. b. Sounds with high frequencies or short wavelengths are reproduced more accurately by tweeters, while sounds with low frequencies or long wavelengths are reproduced more accurately by woofers. 68. The time difference between a 2 km/s S-wave and a 6 km/s P-wave recorded at a certain point is 10 seconds. How far is the epicenter of the earthquake from that", " point? a. b. c. d. 15 m 30 m 15 km 30 km air, they are longitudinal. b. Sound waves in solids are longitudinal, whereas in 13.3 Wave Interaction: Superposition and Interference air, they are transverse. c. Sound waves in solids can be both longitudinal and transverse, whereas in air, they are longitudinal. d. Sound waves in solids are longitudinal, whereas in air, they can be both longitudinal and transverse. 13.2 Wave Properties: Speed, Amplitude, Frequency, and Period 66. A seagull is sitting in the water surface and a simple water wave passes under it. What sort of motion does the gull experience? Why? a. The gull experiences mostly side-to-side motion 69. Why do water waves sometimes appear like a complex criss-cross pattern? a. The crests and the troughs of waves traveling in the same direction combine to form a criss-cross pattern. b. The crests and the troughs of waves traveling in different directions combine to form a criss-cross pattern. 70. What happens when two dissimilar waves interfere? a. pure constructive interference b. pure destructive interference c. a combination of constructive and destructive Access for free at openstax.org. interference 71. Occasionally, during earthquakes, areas near the epicenter are not damaged while those farther away are damaged. Why could this occur? a. Destructive interference results in waves with greater amplitudes being formed in places farther away from the epicenter. b. Constructive interference results in waves with greater amplitudes being formed in places farther away from the epicenter. c. The standing waves of great amplitudes are formed in places farther away from the epicenter. d. The pulse waves of great amplitude are formed in places farther away from the epicenter. 72. Why does an object appear to be distorted when you Chapter 13 \u2022 Test Prep 413 view it through a glass of water? a. The glass and the water reflect the light in different directions. Hence, the object appears to be distorted. b. The glass and the water absorb the light by different amounts. Hence, the object appears to be distorted. c. Water, air, and glass are media with different densities. Light rays refract and bend when they pass from one medium to another. Hence, the object appears to be distorted. d. The glass and the water disperse the light into its components. Hence, the object appears to be distorted. 414 Chapter 13 \u2022 Test Prep Access for free at openstax.org. CHAPTER 14 Sound Figure 14.1 This tree fell some time ago. When it fell, particles in the air were disturbed by the energy of the tree hitting the ground. This disturbance of matter, which our ears have evolved to detect, is called sound. (B.A. Bowen Photography) Chapter Outline 14.1 Speed of Sound, Frequency, and Wavelength 14.2 Sound Intensity and Sound Level 14.3 Doppler Effect and Sonic Booms 14.4 Sound Interference and Resonance INTRODUCTION If a tree falls in a forest (see Figure 14.1) and no one is there to hear it, does it make a sound? The answer to this old philosophical question depends on how you define sound. If sound only exists when someone is around to perceive it, then the falling tree produced no sound. However, in physics, we know that colliding objects can disturb the air, water or other matter surrounding them. As a result of the collision, the surrounding particles of matter began vibrating in a wave-like fashion. This is a sound wave. Consequently, if a tree collided with another object in space, no one would hear it, because no sound would be produced. This is because, in space, there is no air, water or other matter to be disturbed and produce sound waves. In this chapter, we\u2019ll learn more about the wave properties of sound, and explore hearing, as well as some special uses for sound. 416 Chapter 14 \u2022 Sound 14.1 Speed of Sound, Frequency, and Wavelength Section Learning Objectives By the end of this section, you will be able to do the following: \u2022 Relate the characteristics of waves to properties of sound waves \u2022 Describe the speed of sound and how it changes in various media \u2022 Relate the speed of sound to frequency and wavelength of a sound wave Section Key Terms rarefaction sound Properties of Sound Waves Sound is a wave. More specifically, sound is defined to be a disturbance of matter that is transmitted from its source outward. A disturbance is anything that is moved from its state of equilibrium. Some sound waves can be characterized as periodic waves, which means that the atoms that make up the matter experience simple harmonic motion. A vibrating string produces a sound wave as illustrated in Figure 14.2, Figure 14.3, and Figure 14.4. As the string oscillates back and forth, part of", " the string\u2019s energy goes into compressing and expanding the surrounding air. This creates slightly higher and lower pressures. The higher pressure... regions are compressions, and the low pressure regions are rarefactions. The pressure disturbance moves through the air as longitudinal waves with the same frequency as the string. Some of the energy is lost in the form of thermal energy transferred to the air. You may recall from the chapter on waves that areas of compression and rarefaction in longitudinal waves (such as sound) are analogous to crests and troughs in transverse waves. Figure 14.2 A vibrating string moving to the right compresses the air in front of it and expands the air behind it. Figure 14.3 As the string moves to the left, it creates another compression and rarefaction as the particles on the right move away from the string. Access for free at openstax.org. 14.1 \u2022 Speed of Sound, Frequency, and Wavelength 417 Figure 14.4 After many vibrations, there is a series of compressions and rarefactions that have been transmitted from the string as a sound wave. The graph shows gauge pressure (Pgauge) versus distance xfrom the source. Gauge pressure is the pressure relative to atmospheric pressure; it is positive for pressures above atmospheric pressure, and negative for pressures below it. For ordinary, everyday sounds, pressures vary only slightly from average atmospheric pressure. The amplitude of a sound wave decreases with distance from its source, because the energy of the wave is spread over a larger and larger area. But some of the energy is also absorbed by objects, such as the eardrum in Figure 14.5, and some of the energy is converted to thermal energy in the air. Figure 14.4 shows a graph of gauge pressure versus distance from the vibrating string. From this figure, you can see that the compression of a longitudinal wave is analogous to the peak of a transverse wave, and the rarefaction of a longitudinal wave is analogous to the trough of a transverse wave. Just as a transverse wave alternates between peaks and troughs, a longitudinal wave alternates between compression and rarefaction. Figure 14.5 Sound wave compressions and rarefactions travel up the ear canal and force the eardrum to vibrate. There is a net force on the eardrum, since the sound wave pressures differ from the atmospheric pressure found behind the eardrum. A complicated mechanism converts the vibrations to nerve impulses, which are then interpreted by the brain. The Speed of Sound The speed of sound varies greatly depending upon the medium it is traveling through. The speed of sound in a medium is determined by a combination of the medium\u2019s rigidity (or compressibility in gases) and its density. The more rigid (or less compressible) the medium, the faster the speed of sound. The greater the density of a medium, the slower the speed of sound. The speed of sound in air is low, because air is compressible. Because liquids and solids are relatively rigid and very difficult to compress, the speed of sound in such media is generally greater than in gases. Table 14.1 shows the speed of sound in various media. Since temperature affects density, the speed of sound varies with the temperature of the medium through which it\u2019s traveling to some extent, especially for gases. Medium vw (m/s) Gases at 0 \u00b0C Air Carbon dioxide Oxygen Helium 331 259 316 965 Hydrogen 1290 Liquids at 20 \u00b0C Ethanol Mercury Water, fresh Sea water 1160 1450 1480 1540 Human tissue 1540 Solids (longitudinal or bulk) Vulcanized rubber 54 Polyethylene 920 Marble Glass, Pyrex Lead Aluminum Steel 3810 5640 1960 5120 5960 Table 14.1 Speed of Sound in Various Media 418 Chapter 14 \u2022 Sound Access for free at openstax.org. The Relationship Between the Speed of Sound and the Frequency and Wavelength of a Sound Wave 14.1 \u2022 Speed of Sound, Frequency, and Wavelength 419 Figure 14.6 When fireworks explode in the sky, the light energy is perceived before the sound energy. Sound travels more slowly than light does. (Dominic Alves, Flickr) Sound, like all waves, travels at certain speeds through different media and has the properties of frequency and wavelength. Sound travels much slower than light\u2014you can observe this while watching a fireworks display (see Figure 14.6), since the flash of an explosion is seen before its sound is heard. The relationship between the speed of sound, its frequency, and wavelength is the same as for all waves: where vis the speed of sound (in units of m/s), fis its frequency (in units of hertz), and is its wavelength (in units of meters). Recall that wavelength is defined as the distance between adjacent identical parts of a wave. The wavelength of a sound, therefore, is the distance between adjacent identical parts of a sound wave. Just as the distance", " between adjacent crests in a transverse wave is one wavelength, the distance between adjacent compressions in a sound wave is also one wavelength, as shown in Figure 14.7. The frequency of a sound wave is the same as that of the source. For example, a tuning fork vibrating at a given frequency would produce sound waves that oscillate at that same frequency. The frequency of a sound is the number of waves that pass a point per unit time. 14.1 Figure 14.7 A sound wave emanates from a source vibrating at a frequency f, propagates at v, and has a wavelength. One of the more important properties of sound is that its speed is nearly independent of frequency. If this were not the case, and high-frequency sounds traveled faster, for example, then the farther you were from a band in a football stadium, the more the sound from the low-pitch instruments would lag behind the high-pitch ones. But the music from all instruments arrives in cadence independent of distance, and so all frequencies must travel at nearly the same speed. Recall that between fand is inverse: The higher the frequency, the shorter the wavelength of a sound wave., and in a given medium under fixed temperature and humidity, vis constant. Therefore, the relationship The speed of sound can change when sound travels from one medium to another. However, the frequency usually remains the same because it is like a driven oscillation and maintains the frequency of the original source. If vchanges and fremains the same, then the wavelength must change. Since frequency., the higher the speed of a sound, the greater its wavelength for a given Virtual Physics Sound Click to view content (https://www.openstax.org/l/28sound) 420 Chapter 14 \u2022 Sound This simulation lets you see sound waves. Adjust the frequency or amplitude (volume) and you can see and hear how the wave changes. Move the listener around and hear what she hears. Switch to the Two Source Interference tab or the Interference by Reflection tab to experiment with interference and reflection. TIPS FOR SUCCESS Make sure to have audio enabled and set to Listener rather than Speaker, or else the sound will not vary as you move the listener around. GRASP CHECK In the first tab, Listen to a Single Source, move the listener as far away from the speaker as possible, and then change the frequency of the sound wave. You may have noticed that there is a delay between the time when you change the setting and the time when you hear the sound get lower or higher in pitch. Why is this? a. Because, intensity of the sound wave changes with the frequency. b. Because, the speed of the sound wave changes when the frequency is changed. c. Because, loudness of the sound wave takes time to adjust after a change in frequency. d. Because it takes time for sound to reach the listener, so the listener perceives the new frequency of sound wave after a delay. Is there a difference in the amount of delay depending on whether you make the frequency higher or lower? Why? a. Yes, the speed of propagation depends only on the frequency of the wave. b. Yes, the speed of propagation depends upon the wavelength of the wave, and wavelength changes as the frequency changes. c. No, the speed of propagation depends only on the wavelength of the wave. d. No, the speed of propagation is constant in a given medium; only the wavelength changes as the frequency changes. Snap Lab Voice as a Sound Wave In this lab you will observe the effects of blowing and speaking into a piece of paper in order to compare and contrast different sound waves. \u2022 \u2022 \u2022 sheet of paper tape table Instructions Procedure 1. Suspend a sheet of paper so that the top edge of the paper is fixed and the bottom edge is free to move. You could tape the top edge of the paper to the edge of a table, for example. 2. Gently blow air near the edge of the bottom of the sheet and note how the sheet moves. 3. Speak softly and then louder such that the sounds hit the edge of the bottom of the paper, and note how the sheet moves. Interpret the results. 4. GRASP CHECK Which sound wave property increases when you are speaking more loudly than softly? a. amplitude of the wave frequency of the wave b. speed of the wave c. Access for free at openstax.org. 14.1 \u2022 Speed of Sound, Frequency, and Wavelength 421 d. wavelength of the wave WORKED EXAMPLE What Are the Wavelengths of Audible Sounds? Calculate the wavelengths of sounds at the extremes of the audible range, 20 and 20,000 Hz, in conditions where sound travels at 348.7 m/s. STRATEGY To find wavelength from frequency, we can use. Solution (1) Identify the knowns. The values for vand fare given. (2) Solve the relationship between speed, frequency", " and wavelength for. (3) Enter the speed and the minimum frequency to give the maximum wavelength. (4) Enter the speed and the maximum frequency to give the minimum wavelength. 14.2 14.3 14.4 Discussion Because the product of fmultiplied by be, and vice versa. Note that you can also easily rearrange the same formula to find frequency or velocity. equals a constant velocity in unchanging conditions, the smaller fis, the larger must Practice Problems 1. What is the speed of a sound wave with frequency and wavelength? a. b. c. d. 2. Dogs can hear frequencies of up to. What is the wavelength of a sound wave with this frequency traveling in air at? a. b. c. d. 422 Chapter 14 \u2022 Sound LINKS TO PHYSICS Echolocation Figure 14.8 A bat uses sound echoes to find its way about and to catch prey. The time for the echo to return is directly proportional to the distance. Echolocation is the use of reflected sound waves to locate and identify objects. It is used by animals such as bats, dolphins and whales, and is also imitated by humans in SONAR\u2014Sound Navigation and Ranging\u2014and echolocation technology. Bats, dolphins and whales use echolocation to navigate and find food in their environment. They locate an object (or obstacle) by emitting a sound and then sensing the reflected sound waves. Since the speed of sound in air is constant, the time it takes for the sound to travel to the object and back gives the animal a sense of the distance between itself and the object. This is called ranging. Figure 14.8 shows a bat using echolocation to sense distances. Echolocating animals identify an object by comparing the relative intensity of the sound waves returning to each ear to figure out the angle at which the sound waves were reflected. This gives information about the direction, size and shape of the object. Since there is a slight distance in position between the two ears of an animal, the sound may return to one of the ears with a bit of a delay, which also provides information about the position of the object. For example, if a bear is directly to the right of a bat, the echo will return to the bat\u2019s left ear later than to its right ear. If, however, the bear is directly ahead of the bat, the echo would return to both ears at the same time. For an animal without a sense of sight such as a bat, it is important to know whereother animals are as well as whatthey are; their survival depends on it. Principles of echolocation have been used to develop a variety of useful sensing technologies. SONAR, is used by submarines to detect objects underwater and measure water depth. Unlike animal echolocation, which relies on only one transmitter (a mouth) and two receivers (ears), manmade SONAR uses many transmitters and beams to get a more accurate reading of the environment. Radar technologies use the echo of radio waves to locate clouds and storm systems in weather forecasting, and to locate aircraft for air traffic control. Some new cars use echolocation technology to sense obstacles around the car, and warn the driver who may be about to hit something (or even to automatically parallel park). Echolocation technologies and training systems are being developed to help visually impaired people navigate their everyday environments. GRASP CHECK If a predator is directly to the left of a bat, how will the bat know? a. The echo would return to the left ear first. b. The echo would return to the right ear first. Check Your Understanding 3. What is a rarefaction? a. Rarefaction is the high-pressure region created in a medium when a longitudinal wave passes through it. b. Rarefaction is the low-pressure region created in a medium when a longitudinal wave passes through it. c. Rarefaction is the highest point of amplitude of a sound wave. d. Rarefaction is the lowest point of amplitude of a sound wave. 4. What sort of motion do the particles of a medium experience when a sound wave passes through it? a. Simple harmonic motion Access for free at openstax.org. 14.2 \u2022 Sound Intensity and Sound Level 423 b. Circular motion c. Random motion d. Translational motion 5. What does the speed of sound depend on? a. The wavelength of the wave b. The size of the medium c. The frequency of the wave d. The properties of the medium 6. What property of a gas would affect the speed of sound traveling through it? a. The volume of the gas b. The flammability of the gas c. The mass of the gas d. The compressibility of the gas 14.2 Sound Intensity and Sound Level Section Learning Objectives By the end of this section, you will be able to do the following: \u2022 Relate amplitude of a", " wave to loudness and energy of a sound wave \u2022 Describe the decibel scale for measuring sound intensity \u2022 Solve problems involving the intensity of a sound wave \u2022 Describe how humans produce and hear sounds Section Key Terms amplitude decibel hearing loudness pitch sound intensity sound intensity level Amplitude, Loudness and Energy of a Sound Wave Figure 14.9 Noise on crowded roadways like this one in Delhi makes it hard to hear others unless they shout. (Lingaraj G J, Flickr) In a quiet forest, you can sometimes hear a single leaf fall to the ground. But in a traffic jam filled with honking cars, you may have to shout just so the person next to you can hear Figure 14.9.The loudness of a sound is related to how energetically its source is vibrating. In cartoons showing a screaming person, the cartoonist often shows an open mouth with a vibrating uvula (the hanging tissue at the back of the mouth) to represent a loud sound coming from the throat. Figure 14.10 shows such a cartoon depiction of a bird loudly expressing its opinion. A useful quantity for describing the loudness of sounds is called sound intensity. In general, the intensity of a wave is the power per unit area carried by the wave. Power is the rate at which energy is transferred by the wave. In equation form, intensity Iis where Pis the power through an area A. The SI unit for Iis W/m2. The intensity of a sound depends upon its pressure amplitude. 14.5 424 Chapter 14 \u2022 Sound The relationship between the intensity of a sound wave and its pressure amplitude (or pressure variation \u0394p) is 14.6 where \u03c1is the density of the material in which the sound wave travels, in units of kg/m3, and vis the speed of sound in the medium, in units of m/s. Pressure amplitude has units of pascals (Pa) or N/m2. Note that \u0394pis half the difference between the maximum and minimum pressure in the sound wave. We can see from the equation that the intensity of a sound is proportional to its amplitude squared. The pressure variation is proportional to the amplitude of the oscillation, and so Ivaries as (\u0394p)2. This relationship is consistent with the fact that the sound wave is produced by some vibration; the greater its pressure amplitude, the more the air is compressed during the vibration. Because the power of a sound wave is the rate at which energy is transferred, the energy of a sound wave is also proportional to its amplitude squared. TIPS FOR SUCCESS Pressure is usually denoted by capital P, but we are using a lowercase pfor pressure in this case to distinguish it from power Pabove. Figure 14.10 Graphs of the pressures in two sound waves of different intensities. The more intense sound is produced by a source that has larger-amplitude oscillations and has greater pressure maxima and minima. Because pressures are higher in the greater-intensity sound, it can exert larger forces on the objects it encounters. The Decibel Scale You may have noticed that when people talk about the loudness of a sound, they describe it in units of decibels rather than watts per meter squared. While sound intensity (in W/m2) is the SI unit, the sound intensity level in decibels (dB) is more relevant for how humans perceive sounds. The way our ears perceive sound can be more accurately described by the logarithm of the intensity of a sound rather than the intensity of a sound directly. The sound intensity level \u03b2is defined to be 14.7 where Iis sound intensity in watts per meter squared, and I0 = 10\u201312 W/m2 is a reference intensity. I0 is chosen as the reference point because it is the lowest intensity of sound a person with normal hearing can perceive. The decibel level of a sound having an intensity of 10\u201312 W/m2 is \u03b2= 0 dB, because log10 1 = 0. That is, the threshold of human hearing is 0 decibels. Each factor of 10 in intensity corresponds to 10 dB. For example, a 90 dB sound compared with a 60 dB sound is 30 dB greater, or three factors of 10 (that is, 103 times) as intense. Another example is that if one sound is 107 as intense as another, it is 70 dB higher. Since \u03b2is defined in terms of a ratio, it is unit-less. The unit called decibel (dB) is used to indicate that this ratio is multiplied by 10. The sound intensity level is not the same as sound intensity\u2014it tells you the levelof the sound relative to a reference intensity rather than the actual intensity. Access for free at openstax.org. 14.2 \u2022 Sound Intensity and Sound Level 425 Snap Lab Feeling Sound In this lab, you will play music with a heavy beat to literally feel the vibrations and explore what happens when the volume", " changes. \u2022 CD player or portable electronic device connected to speakers \u2022 \u2022 a lightweight table rock or rap music CD or mp3 Procedure 1. Place the speakers on a light table, and start playing the CD or mp3. 2. Place your hand gently on the table next to the speakers. 3. 4. Increase the volume and note the level when the table just begins to vibrate as the music plays. Increase the reading on the volume control until it doubles. What has happened to the vibrations? GRASP CHECK Do you think that when you double the volume of a sound wave you are doubling the sound intensity level (in dB) or the sound intensity (in a. The sound intensity in b. The sound intensity level in c. The sound intensity in d. The sound intensity level in, because it is a closer measure of how humans perceive sound. because it is a closer measure of how humans perceive sound. because it is the only unit to express the intensity of sound. because it is the only unit to express the intensity of sound. )? Why? Solving Sound Wave Intensity Problems WORKED EXAMPLE Calculating Sound Intensity Levels: Sound Waves Calculate the sound intensity level in decibels for a sound wave traveling in air at 0 \u00baC and having a pressure amplitude of 0.656 Pa. STRATEGY We are given \u0394p, so we can calculate Iusing the equation. Using I, we can calculate \u03b2straight from its definition in. Solution (1) Identify knowns: Sound travels at 331 m/s in air at 0 \u00b0C. Air has a density of 1.29 kg/m3 at atmospheric pressure and 0\u00baC. (2) Enter these values and the pressure amplitude into. (3) Enter the value for Iand the known value for I0 into decibels.. Calculate to find the sound intensity level in 426 Chapter 14 \u2022 Sound Discussion This 87.0 dB sound has an intensity five times as great as an 80 dB sound. So a factor of five in intensity corresponds to a difference of 7 dB in sound intensity level. This value is true for any intensities differing by a factor of five. WORKED EXAMPLE Change Intensity Levels of a Sound: What Happens to the Decibel Level? Show that if one sound is twice as intense as another, it has a sound level about 3 dB higher. STRATEGY You are given that the ratio of two intensities is 2 to 1, and are then asked to find the difference in their sound levels in decibels. You can solve this problem using of the properties of logarithms. Solution (1) Identify knowns: The ratio of the two intensities is 2 to 1, or: We want to show that the difference in sound levels is about 3 dB. That is, we want to show Note that (2) Use the definition of \u03b2to get Therefore, 14.8 14.9 14.10 Discussion This means that the two sound intensity levels differ by 3.01 dB, or about 3 dB, as advertised. Note that because only the ratio I2/I1 is given (and not the actual intensities), this result is true for any intensities that differ by a factor of two. For example, a 56.0 dB sound is twice as intense as a 53.0 dB sound, a 97.0 dB sound is half as intense as a 100 dB sound, and so on. Practice Problems 7. Calculate the intensity of a wave if the power transferred is 10 W and the area through which the wave is transferred is 5 square meters. a. 200 W / m2 50 W / m2 b. c. 0.5 W / m2 d. 2 W / m2 8. Calculate the sound intensity for a sound wave traveling in air at and having a pressure amplitude of. a. b. c. d. Hearing and Voice People create sounds by pushing air up through their lungs and through elastic folds in the throat called vocal cords. These folds Access for free at openstax.org. 14.2 \u2022 Sound Intensity and Sound Level 427 open and close rhythmically, creating a pressure buildup. As air travels up and past the vocal cords, it causes them to vibrate. This vibration escapes the mouth along with puffs of air as sound. A voice changes in pitch when the muscles of the larynx relax or tighten, changing the tension on the vocal chords. A voice becomes louder when air flow from the lungs increases, making the amplitude of the sound pressure wave greater. Hearing is the perception of sound. It can give us plenty of information\u2014such as pitch, loudness, and direction. Humans can normally hear frequencies ranging from approximately 20 to 20,000 Hz. Other animals have hearing ranges different from that of humans. Dogs can hear sounds as high as 45,000 Hz, whereas bats and dolphins can hear up to 110,000 Hz sounds. You may have noticed that dogs respond to", " the sound of a dog whistle which produces sound out of the range of human hearing. Sounds below 20 Hz are called infrasound, whereas those above 20,000 Hz are ultrasound. The perception of frequency is called pitch, and the perception of intensity is called loudness. The way we hear involves some interesting physics. The sound wave that hits our ear is a pressure wave. The ear converts sound waves into electrical nerve impulses, similar to a microphone. Figure 14.11 shows the anatomy of the ear with its division into three parts: the outer ear or ear canal; the middle ear, which runs from the eardrum to the cochlea; and the inner ear, which is the cochlea itself. The body part normally referred to as the ear is technically called the pinna. Figure 14.11 The illustration shows the anatomy of the human ear. The outer ear, or ear canal, carries sound to the eardrum protected inside of the ear. The middle ear converts sound into mechanical vibrations and applies these vibrations to the cochlea. The lever system of the middle ear takes the force exerted on the eardrum by sound pressure variations, amplifies it and transmits it to the inner ear via the oval window. Two muscles in the middle ear protect the inner ear from very intense sounds. They react to intense sound in a few milliseconds and reduce the force transmitted to the cochlea. This protective reaction can also be triggered by your own voice, so that humming during a fireworks display, for example, can reduce noise damage. Figure 14.12 shows the middle and inner ear in greater detail. As the middle ear bones vibrate, they vibrate the cochlea, which contains fluid. This creates pressure waves in the fluid that cause the tectorial membrane to vibrate. The motion of the tectorial membrane stimulates tiny cilia on specialized cells called hair cells. These hair cells, and their attached neurons, transform the motion of the tectorial membrane into electrical signals that are sent to the brain. The tectorial membrane vibrates at different positions based on the frequency of the incoming sound. This allows us to detect the pitch of sound. Additional processing in the brain also allows us to determine which direction the sound is coming from (based on comparison of the sound\u2019s arrival time and intensity between our two ears). 428 Chapter 14 \u2022 Sound Figure 14.12 The inner ear, or cochlea, is a coiled tube about 3 mm in diameter and 3 cm in length when uncoiled. As the stapes vibrates against the oval window, it creates pressure waves that travel through fluid in the cochlea. These waves vibrate the tectorial membrane, which bends the cilia and stimulates nerves in the organ of Corti. These nerves then send information about the sound to the brain. FUN IN PHYSICS Musical Instruments Figure 14.13 Playing music, also known as \u201crocking out\u201d, involves creating vibrations using musical instruments. (John Norton) Yet another way that people make sounds is through playing musical instruments (see the previous figure). Recall that the perception of frequency is called pitch. You may have noticed that the pitch range produced by an instrument tends to depend upon its size. Small instruments, such as a piccolo, typically make high-pitch sounds, while larger instruments, such as a tuba, typically make low-pitch sounds. High-pitch means small wavelength, and the size of a musical instrument is directly related to the wavelengths of sound it produces. So a small instrument creates short-wavelength sounds, just as a large instrument creates long-wavelength sounds. Most of us have excellent relative pitch, which means that we can tell whether one sound has a different frequency from another. We can usually distinguish one sound from another if the frequencies of the two sounds differ by as little as 1 Hz. For example, 500.0 and 501.5 Hz are noticeably different. Musical notes are particular sounds that can be produced by most instruments, and are the building blocks of a song. In Western music, musical notes have particular names, such as A-sharp, C, or E-flat. Some people can identify musical notes just by listening to them. This rare ability is called perfect, or absolute, pitch. When a violin plays middle C, there is no mistaking it for a piano playing the same note. The reason is that each instrument produces a distinctive set of frequencies and intensities. We call our perception of these combinations of frequencies and intensities the timbreof the sound. It is more difficult to quantify timbre than loudness or pitch. Timbre is more subjective. Evocative adjectives such as dull, brilliant, warm, cold, pure, and rich are used to describe the timbre of a sound rather than Access for free at openstax.org. 14.2 \u2022 Sound Intensity and Sound Level 429 quantities with units, which makes for a difficult topic to dissect", " with physics. So the consideration of timbre takes us into the realm of perceptual psychology, where higher-level processes in the brain are dominant. This is also true for other perceptions of sound, such as music and noise. But as a teenager, you are likely already aware that one person\u2019s music may be another person\u2019s noise. GRASP CHECK If you turn up the volume of your stereo, will the pitch change? Why or why not? a. No, because pitch does not depend on intensity. b. Yes, because pitch is directly related to intensity. Check Your Understanding 9. What is sound intensity? a. b. c. d. Intensity is the energy per unit area carried by a wave. Intensity is the energy per unit volume carried by a wave. Intensity is the power per unit area carried by a wave. Intensity is the power per unit volume carried by a wave. 10. How is power defined with reference to a sound wave? a. Power is the rate at which energy is transferred by a sound wave. b. Power is the rate at which mass is transferred by a sound wave. c. Power is the rate at which amplitude of a sound wave changes. d. Power is the rate at which wavelength of a sound wave changes. 11. What word or phrase is used to describe the loudness of sound? frequency or oscillation intensity level or decibel timbre a. b. c. d. pitch 12. What is the mathematical expression for sound intensity level? a. b. c. d. 13. What is the range frequencies that humans are capable of hearing? a. b. c. d. to to to 14. How do humans change the pitch of their voice? a. Relaxing or tightening their glottis b. Relaxing or tightening their uvula c. Relaxing or tightening their tongue d. Relaxing or tightening their larynx References Nave, R. Vocal sound production\u2014HyperPhysics. Retrieved from http://hyperphysics.phy-astr.gsu.edu/hbase/music/voice.html 430 Chapter 14 \u2022 Sound 14.3 Doppler Effect and Sonic Booms Section Learning Objectives By the end of this section, you will be able to do the following: \u2022 Describe the Doppler effect of sound waves \u2022 Explain a sonic boom \u2022 Calculate the frequency shift of sound from a moving object by the Doppler shift formula, and calculate the speed of an object by the Doppler shift formula Section Key Terms Doppler effect sonic boom The Doppler Effect of Sound Waves The Doppler effect is a change in the observed pitch of a sound, due to relative motion between the source and the observer. An example of the Doppler effect due to the motion of a source occurs when you are standing still, and the sound of a siren coming from an ambulance shifts from high-pitch to low-pitch as it passes by. The closer the ambulance is to you, the more sudden the shift. The faster the ambulance moves, the greater the shift. We also hear this shift in frequency for passing race cars, airplanes, and trains. An example of the Doppler effect with a stationary source and moving observer is if you ride a train past a stationary warning bell, you will hear the bell\u2019s frequency shift from high to low as you pass by. What causes the Doppler effect? Let\u2019s compare three different scenarios: Sound waves emitted by a stationary source (Figure 14.14), sound waves emitted by a moving source (Figure 14.15), and sound waves emitted by a stationary source but heard by moving observers (Figure 14.16). In each case, the sound spreads out from the point where it was emitted. If the source and observers are stationary, then observers on either side see the same wavelength and frequency as emitted by the source. But if the source is moving and continues to emit sound as it travels, then the air compressions (crests) become closer together in the direction in which it\u2019s traveling and farther apart in the direction it\u2019s traveling away from. Therefore, the wavelength is shorter in the direction the source is moving (on the right in Figure 14.15), and longer in the opposite direction (on the left in Figure 14.15). Finally, if the observers move, as in Figure 14.16, the frequency at which they receive the compressions changes. The observer moving toward the source receives them at a higher frequency (and therefore shorter wavelength), and the person moving away from the source receives them at a lower frequency (and therefore longer wavelength). Figure 14.14 Sounds emitted by a source spread out in spherical waves. Because the source, observers, and air are stationary, the wavelength and frequency are the same in all directions and to all observers. Figure 14.15 Sounds emitted by a source moving to the right spread out from the points at which they", " were emitted. The wavelength is Access for free at openstax.org. reduced and, consequently, the frequency is increased in the direction of motion, so that the observer on the right hears a higher-pitch sound. The opposite is true for the observer on the left, where the wavelength is increased and the frequency is reduced. 14.3 \u2022 Doppler Effect and Sonic Booms 431 Figure 14.16 The same effect is produced when the observers move relative to the source. Motion toward the source increases frequency as the observer on the right passes through more wave crests than she would if stationary. Motion away from the source decreases frequency as the observer on the left passes through fewer wave crests than he would if stationary. We know that wavelength and frequency are related by medium and has the same speed vin that medium whether the source is moving or not. Therefore, fmultiplied by constant. Because the observer on the right in Figure 14.15 receives a shorter wavelength, the frequency she perceives must be higher. Similarly, the observer on the left receives a longer wavelength and therefore perceives a lower frequency. where vis the fixed speed of sound. The sound moves in a is a The same thing happens in Figure 14.16. A higher frequency is perceived by the observer moving toward the source, and a lower frequency is perceived by an observer moving away from the source. In general, then, relative motion of source and observer toward one another increases the perceived frequency. Relative motion apart decreases the perceived frequency. The greater the relative speed is, the greater the effect. WATCH PHYSICS Introduction to the Doppler Effect This video explains the Doppler effect visually. Click to view content (https://www.openstax.org/l/28doppler) GRASP CHECK If you are standing on the sidewalk facing the street and an ambulance drives by with its siren blaring, at what point will the frequency that you observe most closely match the actual frequency of the siren? a. when it is coming toward you b. when it is going away from you c. when it is in front of you For a stationary observer and a moving source of sound, the frequency (fobs) of sound perceived by the observer is 14.11 where fs is the frequency of sound from a source, vs is the speed of the source along a line joining the source and observer, and vw is the speed of sound. The minus sign is used for motion toward the observer and the plus sign for motion away from the observer. TIPS FOR SUCCESS Rather than just memorizing rules, which are easy to forget, it is better to think about the rules of an equation intuitively. Using a minus sign in will decrease the denominator and increase the observed frequency, which is consistent with the expected outcome of the Doppler effect when the source is moving toward the observer. Using a plus sign will increase the denominator and decrease the observed frequency, consistent with what you would expect for the source 432 Chapter 14 \u2022 Sound moving away from the observer. This may be more helpful to keep in mind rather than memorizing the fact that \u201cthe minus sign is used for motion toward the observer and the plus sign for motion away from the observer.\u201d Note that the greater the speed of the source, the greater the Doppler effect. Similarly, for a stationary source and moving observer, the frequency perceived by the observer fobs is given by 14.12 where vobs is the speed of the observer along a line joining the source and observer. Here the plus sign is for motion toward the source, and the minus sign is for motion away from the source. Sonic Booms What happens to the sound produced by a moving source, such as a jet airplane, that approaches or even exceeds the speed of sound? Suppose a jet airplane is coming nearly straight at you, emitting a sound of frequency fs. The greater the plane\u2019s speed, vs, the greater the Doppler shift and the greater the value of fobs. Now, as vs approaches the speed of sound, vw, fobs approaches infinity, because the denominator in approaches zero. This result means that at the speed of sound, in front of the source, each wave is superimposed on the previous one because the source moves forward at the speed of sound. The observer gets them all at the same instant, and so the frequency is theoretically infinite. If the source exceeds the speed of sound, no sound is received by the observer until the source has passed, so that the sounds from the source when it was approaching are stacked up with those from it when receding, creating a sonic boom. A sonic boom is a constructive interference of sound created by an object moving faster than sound. An aircraft creates two sonic booms, one from its nose and one from its tail (see Figure 14.17). During television coverage of space shuttle landings, two distinct booms could often be heard. These were separated by exactly the time it would", " take the shuttle to pass by a point. Observers on the ground often do not observe the aircraft creating the sonic boom, because it has passed by before the shock wave reaches them. If the aircraft flies close by at low altitude, pressures in the sonic boom can be destructive enough to break windows. Because of this, supersonic flights are banned over populated areas of the United States. Figure 14.17 Two sonic booms, created by the nose and tail of an aircraft, are observed on the ground after the plane has passed by. Solving Problems Using the Doppler Shift Formula WATCH PHYSICS Doppler Effect Formula for Observed Frequency This video explains the Doppler effect formula for cases when the source is moving toward the observer. Click to view content (https://www.openstax.org/l/28dopplerform) GRASP CHECK Let\u2019s say that you have a rare phobia where you are afraid of the Doppler effect. If you see an ambulance coming your way, what would be the best strategy to minimize the Doppler effect and soothe your Doppleraphobia? Access for free at openstax.org. 14.3 \u2022 Doppler Effect and Sonic Booms 433 a. Stop moving and become stationary till it passes by. b. Run toward the ambulance. c. Run alongside the ambulance. WATCH PHYSICS Doppler Effect Formula When Source is Moving Away This video explains the Doppler effect formula for cases when the source is moving away from the observer. Click to view content (https://www.openstax.org/l/28doppleraway) GRASP CHECK Sal uses two different formulas for the Doppler effect-one for when the source is moving toward the observer and another for when the source is moving away. However, in this textbook we use only one formula. Explain. a. The combined formula that can be used is, Use ( source is moving away from the observer. b. The combined formula that can be used is, ) when the source is moving toward the observer and ( ) when the source is moving away from the ) when the. Use ( observer and ( ) when the source is moving toward the observer. c. The combined formula that can be used is,. Use ( ) when the source is moving toward the observer and ( ) when the source is moving away from the observer. d. The combined formula that can be used is,. Use ( ) when the source is moving away from the observer and ( ) when the source is moving toward the observer. WORKED EXAMPLE Calculate Doppler Shift: A Train Horn Suppose a train that has a 150 Hz horn is moving at 35 m/s in still air on a day when the speed of sound is 340 m/s. What frequencies are observed by a stationary person at the side of the tracks as the train approaches and after it passes? Strategy To find the observed frequency, must be used because the source is moving. The minus sign is used for the approaching train, and the plus sign for the receding train. Solution (1) Enter known values into to calculate the frequency observed by a stationary person as the train approaches: (2) Use the same equation but with the plus sign to find the frequency heard by a stationary person as the train recedes. Discussion The numbers calculated are valid when the train is far enough away that the motion is nearly along the line joining the train and the observer. In both cases, the shift is significant and easily noticed. Note that the shift is approximately 20 Hz for motion toward and approximately 10 Hz for motion away. The shifts are not symmetric. 434 Chapter 14 \u2022 Sound Practice Problems 15. What is the observed frequency when the source having frequency is moving towards the observer at a speed of and the speed of sound is? a. b. c. d. 16. A train is moving away from you at a speed of. If you are standing still and hear the whistle at a frequency of, what is the actual frequency of the produced whistle? (Assume speed of sound to be.) a. b. c. d. Check Your Understanding 17. What is the Doppler effect? a. The Doppler effect is a change in the observed speed of a sound due to the relative motion between the source and the observer. b. The Doppler effect is a change in the observed frequency of a sound due to the relative motion between the source and the observer. c. The Doppler effect is a change in the observed intensity of a sound due to the relative motion between the source and the observer. d. The Doppler effect is a change in the observed timbre of a sound, due to the relative motion between the source and the observer. 18. Give an example of the Doppler effect caused by motion of the source. a. The sound of a vehicle horn shifts from low-pitch to high-", "pitch as we move towards it. b. The sound of a vehicle horn shifts from low-pitch to high-pitch as we move away from it. c. The sound of a vehicle horn shifts from low-pitch to high-pitch as it passes by. d. The sound of a vehicle horn shifts from high-pitch to low-pitch as it passes by. 19. What is a sonic boom? a. b. c. d. It is a destructive interference of sound created by an object moving faster than sound. It is a constructive interference of sound created by an object moving faster than sound. It is a destructive interference of sound created by an object moving slower than sound. It is a constructive interference of sound created by an object moving slower than sound. 20. What is the relation between speed of source and value of observed frequency when the source is moving towards the observer? a. They are independent of each other. b. The greater the speed, the greater the value of observed frequency. c. The greater the speed, the smaller the value of observed frequency. d. The speed of the sound is directly proportional to the square of the frequency observed. 14.4 Sound Interference and Resonance Section Learning Objectives By the end of this section, you will be able to do the following: \u2022 Describe resonance and beats \u2022 Define fundamental frequency and harmonic series \u2022 Contrast an open-pipe and closed-pipe resonator \u2022 Solve problems involving harmonic series and beat frequency Access for free at openstax.org. 14.4 \u2022 Sound Interference and Resonance 435 Section Key Terms beat beat frequency damping fundamental harmonics natural frequency overtones resonance resonate Resonance and Beats Sit in front of a piano sometime and sing a loud brief note at it while pushing down on the sustain pedal. It will sing the same note back at you\u2014the strings that have the same frequencies as your voice, are resonating in response to the forces from the sound waves that you sent to them. This is a good example of the fact that objects\u2014in this case, piano strings\u2014can be forced to oscillate but oscillate best at their natural frequency. A driving force (such as your voice in the example) puts energy into a system at a certain frequency, which is not necessarily the same as the natural frequency of the system. Over time the energy dissipates, and the amplitude gradually reduces to zero- this is called damping. The natural frequency is the frequency at which a system would oscillate if there were no driving and no damping force. The phenomenon of driving a system with a frequency equal to its natural frequency is called resonance, and a system being driven at its natural frequency is said to resonate. Most of us have played with toys where an object bobs up and down on an elastic band, something like the paddle ball suspended from a finger in Figure 14.18. At first you hold your finger steady, and the ball bounces up and down with a small amount of damping. If you move your finger up and down slowly, the ball will follow along without bouncing much on its own. As you increase the frequency at which you move your finger up and down, the ball will respond by oscillating with increasing amplitude. When you drive the ball at its natural frequency, the ball\u2019s oscillations increase in amplitude with each oscillation for as long as you drive it. As the driving frequency gets progressively higher than the resonant or natural frequency, the amplitude of the oscillations becomes smaller, until the oscillations nearly disappear and your finger simply moves up and down with little effect on the ball. Figure 14.18 The paddle ball on its rubber band moves in response to the finger supporting it. If the finger moves with the natural frequency of the ball on the rubber band, then a resonance is achieved, and the amplitude of the ball\u2019s oscillations increases dramatically. At higher and lower driving frequencies, energy is transferred to the ball less efficiently, and it responds with lower-amplitude oscillations. Another example is that when you tune a radio, you adjust its resonant frequency so that it oscillates only at the desired station\u2019s broadcast (driving) frequency. Also, a child on a swing is driven (pushed) by a parent at the swing\u2019s natural frequency to reach the maximum amplitude (height). In all of these cases, the efficiency of energy transfer from the driving force into the oscillator is best at resonance. 436 Chapter 14 \u2022 Sound Figure 14.19 Some types of headphones use the phenomena of constructive and destructive interference to cancel out outside noises. All sound resonances are due to constructive and destructive interference. Only the resonant frequencies interfere constructively to form standing waves, while others interfere destructively and are absent. From the toot made by blowing over a bottle to the recognizability of a great singer\u2019s voice, resonance and standing waves play a vital role in sound. Interference happens to all types of waves,", " including sound waves. In fact, one way to support that something is a waveis to observe interference effects. Figure 14.19 shows a set of headphones that employs a clever use of sound interference to cancel noise. To get destructive interference, a fast electronic analysis is performed, and a second sound is introduced with its maxima and minima exactly reversed from the incoming noise. In addition to resonance, superposition of waves can also create beats. Beats are produced by the superposition of two waves with slightly different frequencies but the same amplitude. The waves alternate in time between constructive interference and destructive interference, giving the resultant wave an amplitude that varies over time. (See the resultant wave in Figure 14.20). This wave fluctuates in amplitude, or beats, with a frequency called the beat frequency. The equation for beat frequency is where f1 and f2 are the frequencies of the two original waves. If the two frequencies of sound waves are similar, then what we hear is an average frequency that gets louder and softer at the beat frequency. TIPS FOR SUCCESS Don\u2019t confuse the beat frequency with the regular frequency of a wave resulting from superposition. While the beat frequency is given by the formula above, and describes the frequency of the beats, the actual frequency of the wave resulting from superposition is the average of the frequencies of the two original waves. 14.13 Figure 14.20 Beats are produced by the superposition of two waves of slightly different frequencies but identical amplitudes. The waves alternate in time between constructive interference and destructive interference, giving the resulting wave a time-varying amplitude. Virtual Physics Wave Interference Click to view content (https://www.openstax.org/l/28interference) Access for free at openstax.org. 14.4 \u2022 Sound Interference and Resonance 437 For this activity, switch to the Sound tab. Turn on the Sound option, and experiment with changing the frequency and amplitude, and adding in a second speaker and a barrier. GRASP CHECK According to the graph, what happens to the amplitude of pressure over time. What is this phenomenon called, and what causes it? a. The amplitude decreases over time. This phenomenon is called damping. It is caused by the dissipation of energy. b. The amplitude increases over time. This phenomenon is called feedback. It is caused by the gathering of energy. c. The amplitude oscillates over time. This phenomenon is called echoing. It is caused by fluctuations in energy. Fundamental Frequency and Harmonics Suppose we hold a tuning fork near the end of a tube that is closed at the other end, as shown in Figure 14.21, Figure 14.22, and Figure 14.23. If the tuning fork has just the right frequency, the air column in the tube resonates loudly, but at most frequencies it vibrates very little. This means that the air column has only certain natural frequencies. The figures show how a resonance at the lowest of these natural frequencies is formed. A disturbance travels down the tube at the speed of sound and bounces off the closed end. If the tube is just the right length, the reflected sound arrives back at the tuning fork exactly half a cycle later, and it interferes constructively with the continuing sound produced by the tuning fork. The incoming and reflected sounds form a standing wave in the tube as shown. Figure 14.21 Resonance of air in a tube closed at one end, caused by a tuning fork. A disturbance moves down the tube. Figure 14.22 Resonance of air in a tube closed at one end, caused by a tuning fork. The disturbance reflects from the closed end of the tube. Figure 14.23 Resonance of air in a tube closed at one end, caused by a tuning fork. If the length of the tube Lis just right, the disturbance gets back to the tuning fork half a cycle later and interferes constructively with the continuing sound from the tuning fork. This interference forms a standing wave, and the air column resonates. The standing wave formed in the tube has its maximum air displacement (an antinode) at the open end, and no displacement (a 438 Chapter 14 \u2022 Sound node) at the closed end. Recall from the last chapter on waves that motion is unconstrained at the antinode, and halted at the node. The distance from a node to an antinode is one-fourth of a wavelength, and this equals the length of the tube; therefore,. This same resonance can be produced by a vibration introduced at or near the closed end of the tube, as shown in Figure 14.24. Figure 14.24 The same standing wave is created in the tube by a vibration introduced near its closed end. Since maximum air displacements are possible at the open end and none at the closed end, there are other, shorter wavelengths that can resonate in the tube see Figure 14.25). Here the standing wave has three-fourths of its wavelength in the tube, or, so", " that. There is a whole series of shorter-wavelength and higher-frequency sounds that resonate in the tube. We use specific terms for the resonances in any system. The lowest resonant frequency is called the fundamental, while all higher resonant frequencies are called overtones. All resonant frequencies are multiples of the fundamental, and are called harmonics. The fundamental is the first harmonic, the first overtone is the second harmonic, and so on. Figure 14.26 shows the fundamental and the first three overtones (the first four harmonics) in a tube closed at one end. Figure 14.25 Another resonance for a tube closed at one end. This has maximum air displacements at the open end, and none at the closed end. The wavelength is shorter, with three-fourths equaling the length of the tube, so that. This higher-frequency vibration is the first overtone. Figure 14.26 The fundamental and three lowest overtones for a tube closed at one end. All have maximum air displacements at the open end and none at the closed end. The fundamental and overtones can be present at the same time in a variety of combinations. For example, the note middle C on a trumpet sounds very different from middle C on a clarinet, even though both instruments are basically modified versions of a Access for free at openstax.org. 14.4 \u2022 Sound Interference and Resonance 439 tube closed at one end. The fundamental frequency is the same (and usually the most intense), but the overtones and their mix of intensities are different. This mix is what gives musical instruments (and human voices) their distinctive characteristics, whether they have air columns, strings, or drumheads. In fact, much of our speech is determined by shaping the cavity formed by the throat and mouth and positioning the tongue to adjust the fundamental and combination of overtones. Open-Pipe and Closed-Pipe Resonators The resonant frequencies of a tube closed at one end (known as a closed-pipe resonator) are where f1 is the fundamental, f3 is the first overtone, and so on. Note that the resonant frequencies depend on the speed of sound vand on the length of the tube L. Another type of tube is one that is openat both ends (known as an open-pipe resonator). Examples are some organ pipes, flutes, and oboes. The air columns in tubes open at both ends have maximum air displacements at both ends. (See Figure 14.27). Standing waves form as shown. Figure 14.27 The resonant frequencies of a tube open at both ends are shown, including the fundamental and the first three overtones. In all cases the maximum air displacements occur at both ends of the tube, giving it different natural frequencies than a tube closed at one end. The resonant frequencies of an open-pipe resonator are where f1 is the fundamental, f2 is the first overtone, f3 is the second overtone, and so on. Note that a tube open at both ends has a fundamental frequency twice what it would have if closed at one end. It also has a different spectrum of overtones than a tube closed at one end. So if you had two tubes with the same fundamental frequency but one was open at both ends and the other was closed at one end, they would sound different when played because they have different overtones. Middle C, for example, would sound richer played on an open tube since it has more overtones. An open-pipe resonator has more overtones than a closed-pipe resonator because it has even multiples of the fundamental as well as odd, whereas a closed tube has only odd multiples. In this section we have covered resonance and standing waves for wind instruments, but vibrating strings on stringed instruments also resonate and have fundamentals and overtones similar to those for wind instruments. Solving Problems Involving Harmonic Series and Beat Frequency WORKED EXAMPLE Finding the Length of a Tube for a Closed-Pipe Resonator If sound travels through the air at a speed of 344 m/s, what should be the length of a tube closed at one end to have a fundamental frequency of 128 Hz? 440 Chapter 14 \u2022 Sound Strategy The length Lcan be found by rearranging the equation. Solution (1) Identify knowns. \u2022 The fundamental frequency is 128 Hz. \u2022 The speed of sound is 344 m/s. (2) Use to find the fundamental frequency (n= 1). (3) Solve this equation for length. (4) Enter the values of the speed of sound and frequency into the expression for L. 14.14 14.15 14.16 Discussion Many wind instruments are modified tubes that have finger holes, valves, and other devices for changing the length of the resonating air column and therefore, the frequency of the note played. Horns producing very low frequencies, such as tubas, require tubes so long that they are coiled into", " loops. WORKED EXAMPLE Finding the Third Overtone in an Open-Pipe Resonator If a tube that\u2019s open at both ends has a fundamental frequency of 120 Hz, what is the frequency of its third overtone? Strategy Since we already know the value of the fundamental frequency (n = 1), we can solve for the third overtone (n = 4) using the equation. Solution Since fundamental frequency (n = 1) is and 14.17 14.18 Discussion To solve this problem, it wasn\u2019t necessary to know the length of the tube or the speed of the air because of the relationship between the fundamental and the third overtone. This example was of an open-pipe resonator; note that for a closed-pipe resonator, the third overtone has a value of n = 7 (not n = 4). WORKED EXAMPLE Using Beat Frequency to Tune a Piano Piano tuners use beats routinely in their work. When comparing a note with a tuning fork, they listen for beats and adjust the string until the beats go away (to zero frequency). If a piano tuner hears two beats per second, and the tuning fork has a Access for free at openstax.org. 14.4 \u2022 Sound Interference and Resonance 441 frequency of 256 Hz, what are the possible frequencies of the piano? Strategy Since we already know that the beat frequency fBis 2, and one of the frequencies (let\u2019s say f2) is 256 Hz, we can use the equation to solve for the frequency of the piano f1. Solution Since, we know that either or. Solving for f1, Substituting in values, So, 14.19 14.20 14.21 Discussion The piano tuner might not initially be able to tell simply by listening whether the frequency of the piano is too high or too low and must tune it by trial and error, making an adjustment and then testing it again. If there are even more beats after the adjustment, then the tuner knows that he went in the wrong direction. Practice Problems 21. Two sound waves have frequencies and. What is the beat frequency produced by their superposition? a. b. c. d. 22. What is the length of a pipe closed at one end with fundamental frequency? (Assume the speed of sound in air is.) a. b. c. d. Check Your Understanding 23. What is damping? a. Over time the energy increases and the amplitude gradually reduces to zero. This is called damping. b. Over time the energy dissipates and the amplitude gradually increases. This is called damping. c. Over time the energy increases and the amplitude gradually increases. This is called damping. d. Over time the energy dissipates and the amplitude gradually reduces to zero. This is called damping. 24. What is resonance? When can you say that the system is resonating? a. The phenomenon of driving a system with a frequency equal to its natural frequency is called resonance, and a system being driven at its natural frequency is said to resonate. b. The phenomenon of driving a system with a frequency higher than its natural frequency is called resonance, and a system being driven at its natural frequency does not resonate. c. The phenomenon of driving a system with a frequency equal to its natural frequency is called resonance, and a system being driven at its natural frequency does not resonate. d. The phenomenon of driving a system with a frequency higher than its natural frequency is called resonance, and a system being driven at its natural frequency is said to resonate. 442 Chapter 14 \u2022 Sound 25. In the tuning fork and tube experiment, in case a standing wave is formed, at what point on the tube is the maximum disturbance from the tuning fork observed? Recall that the tube has one open end and one closed end. a. At the midpoint of the tube b. Both ends of the tube c. At the closed end of the tube d. At the open end of the tube 26. In the tuning fork and tube experiment, when will the air column produce the loudest sound? a. b. c. d. If the tuning fork vibrates at a frequency twice that of the natural frequency of the air column. If the tuning fork vibrates at a frequency lower than the natural frequency of the air column. If the tuning fork vibrates at a frequency higher than the natural frequency of the air column. If the tuning fork vibrates at a frequency equal to the natural frequency of the air column. 27. What is a closed-pipe resonator? a. A pipe or cylindrical air column closed at both ends b. A pipe with an antinode at the closed end c. A pipe with a node at the open end d. A pipe or cylindrical air column closed at one end 28. Give two examples of open-pipe resonators. a. piano, violin b. drum, tabla c. d. rlectric guitar, acoustic", " guitar flute, oboe Access for free at openstax.org. KEY TERMS amplitude the amount that matter is disrupted during a sound wave, as measured by the difference in height between the crests and troughs of the sound wave. beat a phenomenon produced by the superposition of two waves with slightly different frequencies but the same amplitude beat frequency the frequency of the amplitude fluctuations of a wave damping the reduction in amplitude over time as the energy of an oscillation dissipates decibel a unit used to describe sound intensity levels Doppler effect an alteration in the observed frequency of a sound due to relative motion between the source and the observer fundamental harmonics the term used to refer to the fundamental and the lowest-frequency resonance its overtones hearing the perception of sound SECTION SUMMARY 14.1 Speed of Sound, Frequency, and Wavelength \u2022 Sound is one type of wave. \u2022 Sound is a disturbance of matter that is transmitted from its source outward in the form of longitudinal waves. \u2022 The relationship of the speed of sound v, its frequency f, and its wavelength is given by same relationship given for all waves., which is the \u2022 The speed of sound depends upon the medium through \u2022 which the sound wave is travelling. In a given medium at a specific temperature (or density), the speed of sound vis the same for all frequencies and wavelengths. 14.2 Sound Intensity and Sound Level \u2022 The intensity of a sound is proportional to its amplitude squared. \u2022 The energy of a sound wave is also proportional to its amplitude squared. \u2022 Sound intensity level in decibels (dB) is more relevant for how humans perceive sounds than sound intensity (in W/m2), even though sound intensity is the SI unit. \u2022 Sound intensity level is not the same as sound intensity\u2014it tells you the levelof the sound relative to a reference intensity rather than the actual intensity. \u2022 Hearing is the perception of sound and involves that transformation of sound waves into vibrations of parts within the ear. These vibrations are then transformed Chapter 14 \u2022 Key Terms 443 loudness the perception of sound intensity natural frequency the frequency at which a system would oscillate if there were no driving and no damping forces overtones all resonant frequencies higher than the fundamental pitch the perception of the frequency of a sound rarefaction a low-pressure region in a sound wave resonance the phenomenon of driving a system with a frequency equal to the system's natural frequency resonate to drive a system at its natural frequency sonic boom a constructive interference of sound created by an object moving faster than sound sound a disturbance of matter that is transmitted from its source outward by longitudinal waves sound intensity the power per unit area carried by a sound wave sound intensity level the level of sound relative to a fixed standard related to human hearing into neural signals that are interpreted by the brain. \u2022 People create sounds by pushing air up through their lungs and through elastic folds in the throat called vocal cords. 14.3 Doppler Effect and Sonic Booms \u2022 The Doppler effect is a shift in the observed frequency of a sound due to motion of either the source or the observer. \u2022 The observed frequency is greater than the actual source\u2019s frequency when the source and the observer are moving closer together, either by the source moving toward the observer or the observer moving toward the source. \u2022 A sonic boom is constructive interference of sound created by an object moving faster than sound. 14.4 Sound Interference and Resonance \u2022 A system\u2019s natural frequency is the frequency at which the system will oscillate if not affected by driving or damping forces. \u2022 A periodic force driving a harmonic oscillator at its natural frequency produces resonance. The system is said to resonate. \u2022 Beats occur when waves of slightly different frequencies \u2022 are superimposed. In air columns, the lowest-frequency resonance is called the fundamental, whereas all higher resonant frequencies are called overtones. Collectively, they are 444 Chapter 14 \u2022 Key Equations called harmonics. \u2022 The resonant frequencies of a tube closed at one end are, where f1is the fundamental and L is the length of the tube. \u2022 The resonant frequencies of a tube open at both ends are KEY EQUATIONS 14.1 Speed of Sound, Frequency, and Wavelength speed of sound 14.2 Sound Intensity and Sound Level intensity sound intensity sound intensity level CHAPTER REVIEW Concept Items 14.1 Speed of Sound, Frequency, and Wavelength 1. What is the amplitude of a sound wave perceived by the loudness human ear? a. b. pitch c. d. intensity timbre 2. The compressibility of air and hydrogen is almost the same. Which factor is the reason that sound travels faster in hydrogen than in air? a. Hydrogen is more dense than air. b. Hydrogen is less dense than air. c. Hydrogen atoms are heavier than air molecules. d. Hydrogen atoms are lighter than air molecules. 14.3 Doppler Effect and Sonic Booms Doppler effect observed frequency (moving source) Doppler effect observed frequency", " (moving observer) 14.4 Sound Interference and Resonance beat frequency resonant frequencies of a closed-pipe resonator resonant frequencies of an open-pipe resonator a. b. c. d. 4. How does the \"decibel\" get its name? a. The meaning of deci is \u201chundred\u201d and the number of decibels is one-hundredth of the logarithm to base 10 of the ratio of two sound intensities. b. The meaning of deci is \"ten\" and the number of decibels is one-tenth of the logarithm to base 10 of the ratio of two sound intensities. c. The meaning of deci is \u201cone-hundredth\u201d and the number of decibels is hundred times the logarithm to base 10 of the ratio of two sound intensities. d. The meaning of deci is \u201cone-tenth\u201d and the number of decibels is ten times the logarithm to base 10 of the ratio of two sound intensities. 14.2 Sound Intensity and Sound Level 5. What is \u201ctimbre\u201d of sound? a. Timbre is the quality of the sound that distinguishes 3. What is the mathematical relationship between intensity, it from other sound power, and area? Access for free at openstax.org. Chapter 14 \u2022 Chapter Review 445 b. Timbre is the loudness of the sound that distinguishes it from other sound. c. Timbre is the pitch of the sound that distinguishes it from other sound. down is greater than the amplitude of the yo-yo b. when the amplitude of the finger moving up and down is less than the amplitude of the yo-yo c. when the frequency of the finger moving up and d. Timbre is the wavelength of the sound that down is equal to the resonant frequency of the yo-yo distinguishes it from other sound. d. when the frequency of the finger moving up and 14.3 Doppler Effect and Sonic Booms 6. Two sources of sound producing the same frequency are moving towards you at different speeds. Which one would sound more high-pitched? the one moving slower a. the one moving faster b. 7. When the speed of the source matches the speed of sound, what happens to the amplitude of the sound wave? Why? a. It approaches zero. This is because all wave crests are superimposed on one another through constructive interference. It approaches infinity. This is because all wave crests are superimposed on one another through constructive interference. It approaches zero, because all wave crests are superimposed on one another through destructive interference. It approaches infinity, because all wave crests are superimposed on one another through destructive interference. b. c. d. 8. What is the mathematical expression for the frequency perceived by the observer in the case of a stationary observer and a moving source? a. b. c. d. 14.4 Sound Interference and Resonance 9. When does a yo-yo travel the farthest from the finger? a. when the amplitude of the finger moving up and down is different from the resonant frequency of the yo-yo 10. What is the difference between harmonics and overtones? a. Harmonics are all multiples of the fundamental frequency. The first overtone is actually the first harmonic. b. Harmonics are all multiples of the fundamental frequency. The first overtone is actually the second harmonic. c. Harmonics are all multiples of the fundamental frequency. The second overtone is actually the first harmonic. d. Harmonics are all multiples of the fundamental frequency. The third overtone is actually the second harmonic. 11. What kind of waves form in pipe resonators? a. damped waves b. propagating waves c. high-frequency waves d. standing waves 12. What is the natural frequency of a system? a. The natural frequency is the frequency at which a system oscillates when it undergoes forced vibration. b. The natural frequency is the frequency at which a system oscillates when it undergoes damped oscillation. c. The natural frequency is the frequency at which a system oscillates when it undergoes free vibration without a driving force or damping. d. The natural frequency is the frequency at which a system oscillates when it undergoes forced vibration with damping. Critical Thinking Items d. It remains constant. 14.1 Speed of Sound, Frequency, and Wavelength 13. What can be said about the frequency of a monotonous sound? a. b. c. It decreases with time. It decreases with distance. It increases with distance. 14. A scientist notices that a sound travels faster through a solid material than through the air. Which of the following can explain this? a. Solid materials are denser than air. b. Solid materials are less dense than air. c. A solid is more rigid", " than air. d. A solid is easier to compress than air. 446 Chapter 14 \u2022 Chapter Review 14.2 Sound Intensity and Sound Level 15. Which property of the wave is related to its intensity? How? a. The frequency of the wave is related to the intensity of the sound. The larger-frequency oscillations indicate greater pressure maxima and minima, and the pressure is higher in greater-intensity sound. b. The wavelength of the wave is related to the intensity of the sound. The longer-wavelength oscillations indicate greater pressure maxima and minima, and the pressure is higher in greaterintensity sound. c. The amplitude of the wave is related to the intensity of the sound. The larger-amplitude oscillations indicate greater pressure maxima and minima, and the pressure is higher in greater-intensity sound. d. The speed of the wave is related to the intensity of the sound. The higher-speed oscillations indicate greater pressure maxima and minima, and the pressure is higher in greater-intensity sound. 16. Why is decibel (dB) used to describe loudness of sound? a. Because, human ears have an inverse response to the amplitude of sound. b. Because, human ears have an inverse response to the intensity of sound. c. Because, the way our ears perceive sound can be more accurately described by the amplitude of a sound rather than the intensity of a sound directly. d. Because, the way our ears perceive sound can be more accurately described by the logarithm of the intensity of a sound rather than the intensity of a sound directly. 17. How can humming while shooting a gun reduce ear damage? a. Humming can trigger those two muscles in the outer ear that react to intense sound produced while shooting and reduce the force transmitted to the cochlea. b. Humming can trigger those three muscles in the outer ear that react to intense sound produced while shooting and reduce the force transmitted to the cochlea. c. Humming can trigger those two muscles in the middle ear that react to intense sound produced while shooting and reduce the force transmitted to the cochlea. d. Humming can trigger those three muscles in the middle ear that react to intense sound produced while shooting and reduce the force transmitted to the cochlea. 18. A particular sound, S1, has an intensity times that of Access for free at openstax.org. another sound, S2. What is the difference in sound intensity levels measured in decibels? a. b. c. d. 14.3 Doppler Effect and Sonic Booms 19. When the source of sound is moving through the air, does the speed of sound change with respect to a stationary person standing nearby? a. Yes b. No 20. Why is no sound heard by the observer when an object approaches him at a speed faster than that of sound? a. If the source exceeds the speed of sound, then destructive interference occurs and no sound is heard by the observer when an object approaches him. If the source exceeds the speed of sound, the frequency of sound produced is beyond the audible range of sound. If the source exceeds the speed of sound, all the sound waves produced approach minimum intensity and no sound is heard by the observer when an object approaches him. If the source exceeds the speed of sound, all the sound waves produced are behind the source. Hence, the observer hears the sound only after the source has passed. b. c. d. 21. Does the Doppler effect occur when the source and observer are both moving towards each other? If so, how would this affect the perceived frequency? a. Yes, the perceived frequency will be even lower in this case than if only one of the two were moving. b. No, the Doppler effect occurs only when an observer is moving towards a source. c. No, the Doppler effect occurs only when a source is moving towards an observer. d. Yes, the perceived frequency will be even higher in this case than if only one of the two were moving. 14.4 Sound Interference and Resonance 22. When does the amplitude of an oscillating system become maximum? a. When two sound waves interfere destructively. b. When the driving force produces a transverse wave in the system. c. When the driving force of the oscillator to the oscillating system is at a maximum amplitude. d. When the frequency of the oscillator equals the natural frequency of the oscillating system. 23. How can a standing wave be formed with the help of a tuning fork and a closed-end tube of appropriate length? a. If the tube is just the right length, the reflected sound arrives back at the tuning fork exactly half a cycle later, and it interferes constructively with the continuing sound produced by the tuning fork. If the tube is just the right length, the reflected sound arrives back at the tuning fork exactly half a cycle later, and it interfe", "res destructively with the continuing sound produced by the tuning fork. If the tube is just the right length, the reflected sound arrives back at the tuning fork exactly one b. c. Problems 14.1 Speed of Sound, Frequency, and Wavelength 25. A bat produces a sound at and wavelength. What is the speed of the sound? a. b. c. d. 26. A sound wave with frequency of is traveling. By how much will its wavelength through air at change when it enters aluminum? a. b. c. d. 14.2 Sound Intensity and Sound Level 27. Calculate the sound intensity for a sound wave traveling through air at 15\u00b0 C and having a pressure amplitude of 0.80 Pa. (Hint\u2014Speed of sound in air at 15\u00b0 C is 340 m/s.) a. 9.6\u00d710\u22123 W / m2 7.7\u00d710\u22123 W / m2 b. c. 9.6\u00d710\u22124 W / m2 7.7\u00d710\u22124 W / m2 d. Chapter 14 \u2022 Chapter Review 447 d. full cycle later, and it interferes constructively with the continuing sound produced by the tuning fork. If the tube is just the right length, the reflected sound arrives back at the tuning fork exactly one full cycle later, and it interferes destructively with the continuing sound produced by the tuning fork. 24. A tube open at both ends has a fundamental frequency. What will the frequency be if one end is of closed? a. b. c. d. 14.3 Doppler Effect and Sonic Booms 29. An ambulance is moving away from you. You are standing still and you hear its siren at a frequency of. You know that the actual frequency of the siren. What is the speed of the ambulance?.) is (Assume the speed of sound to be a. b. c. d. 30. An ambulance passes you at a speed of. If its siren has a frequency of, what is difference in the frequencies you perceive before and after it passes you? (Assume the speed of sound in air is a. b. c. d..) 14.4 Sound Interference and Resonance 31. What is the length of an open-pipe resonator with a? (Assume the.) fundamental frequency of speed of sound is a. b. c. d. 28. The sound level in dB of a sound traveling through air at is. Calculate its pressure amplitude. 32. An open-pipe resonator has a fundamental frequency of. By how much would its length have to be a. b. c. d. changed to get a fundamental frequency of (Assume the speed of sound is a. b. c. d..)? 448 Chapter 14 \u2022 Test Prep Performance Task 14.4 Sound Interference and Resonance 33. Design and make an open air resonator capable of playing at least three different pitches (frequencies) of sound using a selection of bamboo of varying widths and lengths, which can be obtained at a local hardware store. Choose a piece of bamboo for creating a musical TEST PREP Multiple Choice 14.1 Speed of Sound, Frequency, and Wavelength 34. What properties does a loud, shrill whistle have? a. high amplitude, high frequency b. high amplitude, low frequency low amplitude, high frequency c. low amplitude, low frequency d. 35. What is the speed of sound in fresh water at degrees Celsius? a. b. c. d. 36. A tuning fork oscillates at a frequency of, creating sound waves. How many waves will reach the eardrum of a person near that fork in seconds? a. b. c. d. 37. Why does the amplitude of a sound wave decrease with distance from its source? a. The amplitude of a sound wave decreases with distance from its source, because the frequency of the sound wave decreases. b. The amplitude of a sound wave decreases with distance from its source, because the speed of the sound wave decreases. c. The amplitude of a sound wave decreases with distance from its source, because the wavelength of the sound wave increases. d. The amplitude of a sound wave decreases with distance from its source, because the energy of the wave is spread over a larger and larger area. 38. Does the elasticity of the medium affect the speed of sound? How? a. No, there is no relationship that exists between the speed of sound and elasticity of the medium. Access for free at openstax.org. pipe. Calculate the length required for a certain frequency to resonate and then mark the locations where holes should be placed in the pipe to achieve their desired pitches. Use a simple hand drill or ask your wood shop department for help drilling holes. Use tuning forks to test and calibrate your instrument. Demonstrate your pipe for the class. b. Yes. When particles are more easily compressed in a medium, sound does", " not travel as quickly through the medium. c. Yes. When the particles in a medium do not compress much, sound does not travel as quickly through the medium. d. No, the elasticity of a medium affects frequency and wavelength, not wave speed. 14.2 Sound Intensity and Sound Level 39. Which of the following terms is a useful quantity to intensity frequency describe the loudness of a sound? a. b. c. pitch d. wavelength 40. What is the unit of sound intensity level? a. decibels b. hertz c. watts 41. If a particular sound S1 is times more intense than another sound S2, then what is the difference in sound intensity levels in dB for these two sounds? a. b. c. 42. By what minimum amount should frequencies vary for humans to be able to distinguish two separate sounds? a. b. c. d. 43. Why is I0chosen as the reference for sound intensity? a. Because, it is the highest intensity of sound a person with normal hearing can perceive at a frequency of 100 Hz. b. Because, it is the lowest intensity of sound a person with normal hearing can perceive at a frequency of 100 Hz. c. Because, it is the highest intensity of sound a person with normal hearing can perceive at a frequency of 1000 Hz. d. Because, it is the lowest intensity of sound a person with normal hearing can perceive at a frequency of 1000 Hz. 14.3 Doppler Effect and Sonic Booms 44. In which of the following situations is the Doppler effect absent? a. The source and the observer are moving towards each other. b. The observer is moving toward the source. c. The source is moving away from the observer. d. Neither the source nor the observer is moving relative to one another. 45. What does the occurrence of the sonic boom depend on? speed of the source a. b. frequency of source c. amplitude of source d. distance of observer from the source 46. What is the observed frequency when the observer is? The source moving away from the source at frequency is. and the speed of sound is a. b. c. d. 47. How will your perceived frequency change if the source is moving towards you? Short Answer 14.1 Speed of Sound, Frequency, and Wavelength 52. What component of a longitudinal sound wave is analogous to a trough of a transverse wave? a. b. c. node d. antinode compression rarefaction 53. What is the frequency of a sound wave as perceived by the human ear? timbre a. loudness b. intensity c. d. pitch 54. What properties of a solid determine the speed of sound traveling through it? Chapter 14 \u2022 Test Prep 449 a. The frequency will become lower. b. The frequency will become higher. 14.4 Sound Interference and Resonance 48. Observation of which phenomenon can be considered interference proof that something is a wave? a. b. noise c. d. reflection conduction 49. Which of the resonant frequencies has the greatest amplitude? a. The first harmonic b. The second harmonic c. The first overtone d. The second overtone 50. What is the fundamental frequency of an open-pipe resonator? a. b. c. d.? 51. What is the beat frequency produced by the superposition of two waves with frequencies and a. b. c. d. a. mass and density b. rigidity and density c. volume and density d. shape and rigidity 55. Does the density of a medium affect the speed of sound? a. No b. Yes 56. Does a bat make use of the properties of sound waves to locate its prey? a. No b. Yes 57. Do the properties of a sound wave change when it travels from one medium to another? a. No b. Yes 450 Chapter 14 \u2022 Test Prep 14.2 Sound Intensity and Sound Level meter squared. 58. When a passing driver has his stereo turned up, you 61. Why is the reference intensity cannot even hear what the person next to you is saying. Why is this so? a. The sound from the passing car\u2019s stereo has a higher amplitude and hence higher intensity compared to the intensity of the sound coming from the person next to you. The higher intensity corresponds to greater loudness, so the first sound dominates the second. b. The sound from the passing car\u2019s stereo has a higher amplitude and hence lower intensity compared to the intensity of the sound coming from the person next to you. The lower intensity corresponds to greater loudness, so the first sound dominates the second. c. The sound from the passing car\u2019s stereo has a higher frequency and hence higher intensity compared to the intensity of the sound coming from the person next to you. The higher frequency corresponds to greater loudness so the first sound dominates the second. d. The sound from the passing car\u2019s stereo has", " a lower frequency and hence higher intensity compared to the intensity of the sound coming from the person next to you. The lower frequency corresponds to greater loudness, so the first sound dominates the second. 59. For a constant area, what is the relationship between intensity of a sound wave and power? a. The intensity is inversely proportional to the power transmitted by the wave, for a constant area. b. The intensity is inversely proportional to the square of the power transmitted by the wave, for a constant area. c. The intensity is directly proportional to the square of the power transmitted by the wave, for a constant area. d. The intensity is directly proportional to the power transmitted by the wave, for a constant area.? decibels, a. The upper limit of human hearing is, i.e.. For. b. The lower threshold of human hearing is decibels, i.e.. For, c. The upper limit of human hearing is decibels, i.e.. For, d. The lower threshold of human hearing is decibels, i.e.,. For, 62. Given that the sound intensity level of a particular wave, what will be the sound intensity for that wave? is a. b. c. d. 63. For a sound wave with intensity, calculate the pressure amplitude given that the sound. travels through air at a. b. c. d. 64. Which nerve carries auditory information to the brain? a. buccal nerve b. peroneal nerve c. cochlear nerve d. mandibular nerve 65. Why do some smaller instruments, such as piccolos, produce higher-pitched sounds than larger instruments, such as tubas? a. Smaller instruments produce sounds with shorter wavelengths, and thus higher frequencies. b. Smaller instruments produce longer wavelength, and thus higher amplitude, sounds. c. Smaller instruments produce lower amplitude, and thus longer wavelength sounds. 60. What does stand for in the equation d. Smaller instruments produce higher amplitude,? What is its unit? and thus lower frequency, sounds. a. Yes, is the sound intensity in watts per meter squared in the equation,. 14.3 Doppler Effect and Sonic Booms is the sound illuminance and its unit is lumen 66. How will your perceived frequency change if you move b. c. per meter squared. is the sound intensity and its unit is watts per meter cubed. d. is the sound intensity and its unit is watts per Access for free at openstax.org. away from a stationary source of sound? a. The frequency will become lower. b. The frequency will be doubled. c. The frequency will be tripled. d. The frequency will become higher. 67. True or false\u2014The Doppler effect also occurs with waves other than sound waves. a. False b. True 68. A source of sound is moving towards you. How will what you hear change if the speed of the source increases? a. The sound will become more high-pitched. b. The sound will become more low-pitched. c. The pitch of the sound will not change. 69. Do sonic booms continue to be created when an object is traveling at supersonic speeds? a. No, a sonic boom is created only when the source exceeds the speed of sound. b. Yes, sonic booms continue to be created when an Chapter 14 \u2022 Test Prep 451 a. Human speech is produced by shaping the cavity formed by the throat and mouth, the vibration of vocal cords, and using the tongue to adjust the fundamental frequency and combination of overtones. b. Human speech is produced by shaping the cavity formed by the throat and mouth into a closed pipe and using tongue to adjust the fundamental frequency and combination of overtones. c. Human speech is produced only by the vibrations of the tongue. d. Human speech is produced by elongating the vocal cords. 75. What is the possible number of nodes and antinodes along one full wavelength of a standing wave? a. nodes and antinodes or antinodes and object is traveling at supersonic speeds. nodes. 70. Suppose you are driving at a speed of and you.? hear the sound of a bell at a frequency of What is the actual frequency of the bell if the speed of sound is a. b. c. d. 71. What is the frequency of a stationary sound source if you hear it at 1200.0 Hz while moving towards it at a speed of 50.0 m/s? (Assume speed of sound to be 331 m/s.) a. b. c. d. 1410 Hz 1380 Hz 1020 Hz 1042 Hz 14.4 Sound Interference and Resonance 72. What is the actual frequency of the wave produced as a result of superposition of two waves? a. It is the average of the frequencies of the two original waves that were superimposed. It", " is the difference between the frequencies of the two original waves that were superimposed. It is the product of the frequencies of the two original waves that were superimposed. It is the sum of the frequencies of the two original waves that were superimposed. b. c. d. 73. Can beats be produced through a phenomenon different from resonance? How? a. No, beats can be produced only by resonance. b. Yes, beats can be produced by superimposition of any two waves having slightly different frequencies. 74. How is human speech produced? b. c. nodes and antinodes or antinodes and nodes. nodes and antinodes or antinodes and nodes. d. nodes and antinodes or antinodes and nodes. 76. In a pipe resonator, which frequency will be the least second overtone frequency intense of those given below? a. b. first overtone frequency fundamental frequency c. third overtone frequency d. 77. A flute is an open-pipe resonator. If a flute is long, what is the longest wavelength it can produce? a. b. c. d. 78. What is the frequency of the second overtone of a? closed-pipe resonator with a length of (Assume the speed of sound is a. b. c. d..) when the speed of sound is 79. An open-pipe resonator has a fundamental frequency of. What will its fundamental frequency be when the speed of sound is a. b. c. d.? 452 Chapter 14 \u2022 Test Prep Extended Response eardrum. 14.1 Speed of Sound, Frequency, and Wavelength 80. How is a human able to hear sounds? a. Sound waves cause the eardrum to vibrate. A complicated mechanism converts the vibrations to nerve impulses, which are perceived by the person as sound. b. Sound waves cause the ear canal to vibrate. A complicated mechanism converts the vibrations to nerve impulses, which are perceived by the person as sound. c. Sound waves transfer electrical impulses to the eardrum. A complicated mechanism converts the electrical impulses to sound. d. Sound waves transfer mechanical vibrations to the ear canal, and the eardrum converts them to electrical impulses. 81. Why does sound travel faster in iron than in air even though iron is denser than air? a. The density of iron is greater than that of air. However, the rigidity of iron is much greater than that of air. Hence, sound travels faster in it. b. The density of iron is greater than that of air. However, the rigidity of iron is much less than that of air. Hence, sound travels faster in it. c. The density of iron is greater than that of air. However, the rigidity of iron is equal to that of air. Hence, sound travels faster in it. d. The mass of iron is much less than that of air and the rigidity of iron is much greater than that of air. Hence, sound travels faster in it. 82. Is the speed of sound dependent on its frequency? a. No b. Yes 14.2 Sound Intensity and Sound Level 83. Why is the sound from a tire burst louder than that from a finger snap? a. The sound from the tire burst has higher pressure amplitudes, hence it can exert smaller force on the eardrum. b. The sound from the tire burst has lower pressure amplitudes, hence it can exert smaller force on the eardrum. c. The sound from the tire burst has lower pressure amplitudes, hence it can exert larger force on the ear drum. d. The sound from the tire burst has higher pressure amplitudes, hence it can exert larger force on the Access for free at openstax.org. 84. Sound A is times more intense than Sound B. What will be the difference in decibels in their sound intensity levels? a. b. c. d. 85. The ratio of the pressure amplitudes of two sound waves. What will be the is traveling through water at difference in their sound intensity levels in dB? a. b. c. d. 86. Which of the following most closely models how sound is produced by the vocal cords? a. A person plucks a string. b. A person blows over the mouth of a half-filled glass bottle. c. A person strikes a hammer against a hard surface. d. A person blows through a small slit in a wide, stretched rubber band. 14.3 Doppler Effect and Sonic Booms 87. True or false\u2014The Doppler effect occurs only when the sound source is moving. a. False b. True 88. True or false\u2014The observed frequency becomes infinite when the source is moving at the speed of sound. a. False b. True 89. You are driving alongside a train. You hear its horn at a pitch that is lower than the actual frequency. What should you do to match the speed of the train", "? Why? a. In order to match the speed of the train, one would need to increase or decrease the speed of his/her car because a lower pitch means that either the train (the source) is moving away or that you (the observer) are moving away. In order to match the speed of the train, one would need to drive at a constant speed because a lower pitch means that the train and the car are at the same speed. b. 14.4 Sound Interference and Resonance 90. How are the beat frequency and the regular frequency of a wave resulting from superposition of two waves different? a. Beat frequency is the sum of two frequencies and regular frequency is the difference between frequencies of two original waves. overtone so resonance will occur. c. The frequency formed is a harmonic and third overtone so resonance will occur. b. Beat frequency is the difference between the d. The frequency formed is a harmonic and fourth Chapter 14 \u2022 Test Prep 453 constituent frequencies, but the regular frequency is the average of the frequencies of the two original waves. c. Beat frequency is the sum of two frequencies and regular frequency is the average of frequencies of two original waves. d. Beat frequency is the average of frequencies of two original waves and regular frequency is the sum of two original frequencies. 91. In the tuning fork and tube experiment, if resonance is is the length of the tube formed for, where and is the wavelength of the sound wave, can resonance also be formed for a wavelength Why? a. The frequency formed is a harmonic and first? overtone so resonance will occur. b. The frequency formed is a harmonic and second overtone so resonance will occur. 92. True or false\u2014An open-pipe resonator has more overtones than a closed-pipe resonator. a. False b. True 93. A flute has finger holes for changing the length of the resonating air column, and therefore, the frequency of the note played. How far apart are two holes that, when closed, play two frequencies that are apart, if the first hole is the flute? a. b. c. d. away from the mouthpiece of 454 Chapter 14 \u2022 Test Prep Access for free at openstax.org. CHAPTER 15 Light Figure 15.1 Human eyes detect these orange sea goldiefish swimming over a coral reef in the blue waters of the Gulf of Eilat, in the Red Sea, using visible light. (credit: David Darom, Wikimedia Commons) Chapter Outline 15.1 The Electromagnetic Spectrum 15.2 The Behavior of Electromagnetic Radiation INTRODUCTION The beauty of a coral reef, the warm radiance of sunshine, the sting of sunburn, the X-ray revealing a broken bone, even microwave popcorn\u2014all are brought to us by electromagnetic waves. The list of the various types of electromagnetic waves, ranging from radio transmission waves to nuclear gamma-ray (\u03b3-ray) emissions, is interesting in itself. Even more intriguing is that all of these different phenomena are manifestations of the same thing\u2014electromagnetic waves (see Figure 15.1). What are electromagnetic waves? How are they created, and how do they travel? How can we understand their widely varying properties? What is the relationship between electric and magnetic effects? These and other questions will be explored. 15.1 The Electromagnetic Spectrum Section Learning Objectives By the end of this section, you will be able to do the following: \u2022 Define the electromagnetic spectrum, and describe it in terms of frequencies and wavelengths \u2022 Describe and explain the differences and similarities of each section of the electromagnetic spectrum and the applications of radiation from those sections Section Key Terms electric field electromagnetic radiation (EMR) magnetic field Maxwell\u2019s equations 456 Chapter 15 \u2022 Light The Electromagnetic Spectrum We generally take light for granted, but it is a truly amazing and mysterious form of energy. Think about it: Light travels to Earth across millions of kilometers of empty space. When it reaches us, it interacts with matter in various ways to generate almost all the energy needed to support life, provide heat, and cause weather patterns. Light is a form of electromagnetic radiation (EMR). The term lightusually refers to visible light, but this is not the only form of EMR. As we will see, visible light occupies a narrow band in a broad range of types of electromagnetic radiation. Electromagnetic radiation is generated by a moving electric charge, that is, by an electric current. As you will see when you study electricity, an electric current generates both an electric field, E, and a magnetic field, B. These fields are perpendicular to each other. When the moving charge oscillates, as in an alternating current, an EM wave is propagated. Figure 15.2 shows how an electromagnetic wave moves away from the source\u2014indicated by the ~ symbol. WATCH PHYSICS Electromagnetic Waves and the Electromagnetic Spectrum This video, link below, is closely related to the following figure. If", " you have questions about EM wave properties, the EM spectrum, how waves propagate, or definitions of any of the related terms, the answers can be found in this video (http://www.openstax.org/l/28EMWaves). Click to view content (https://www.openstax.org/l/28EMWaves) GRASP CHECK In an electromagnetic wave, how are the magnetic field, the electric field, and the direction of propagation oriented to each other? a. All three are parallel to each other and are along the x-axis. b. All three are mutually perpendicular to each other. c. The electric field and magnetic fields are parallel to each other and perpendicular to the direction of propagation. d. The magnetic field and direction of propagation are parallel to each other along the y-axis and perpendicular to the electric field. Virtual Physics Radio Waves and Electromagnetic Fields Click to view content (https://www.openstax.org/l/28Radiowaves) This simulation demonstrates wave propagation. The EM wave is propagated from the broadcast tower on the left, just as in Figure 15.2. You can make the wave yourself or allow the animation to send it. When the wave reaches the antenna on the right, it causes an oscillating current. This is how radio and television signals are transmitted and received. GRASP CHECK Where do radio waves fall on the electromagnetic spectrum? a. Radio waves have the same wavelengths as visible light. b. Radio waves fall on the high-frequency side of visible light. c. Radio waves fall on the short-wavelength side of visible light. d. Radio waves fall on the low-frequency side of visible light. Access for free at openstax.org. 15.1 \u2022 The Electromagnetic Spectrum 457 Figure 15.2 A part of the electromagnetic wave sent out from an oscillating charge at one instant in time. The electric and magnetic fields (E and B) are in phase, and they are perpendicular to each other and to the direction of propagation. For clarity, the waves are shown only along one direction, but they propagate out in other directions too. From your study of sound waves, recall these features that apply to all types of waves: \u2022 Wavelength\u2014The distance between two wave crests or two wave troughs, expressed in various metric measures of distance \u2022 Frequency\u2014The number of wave crests that pass a point per second, expressed in hertz (Hz or s\u20131) \u2022 Amplitude: The height of the crest above the null point As mentioned, electromagnetic radiation takes several forms. These forms are characterized by a range of frequencies. Because frequency is inversely proportional to wavelength, any form of EMR can also be represented by its range of wavelengths. Figure 15.3 shows the frequency and wavelength ranges of various types of EMR. With how many of these types are you familiar? Figure 15.3 The electromagnetic spectrum, showing the major categories of electromagnetic waves. The range of frequencies and wavelengths is remarkable. The dividing line between some categories is distinct, whereas other categories overlap. Take a few minutes to study the positions of the various types of radiation on the EM spectrum, above. Sometimes all radiation with frequencies lower than those of visible light are referred to as infrared (IR) radiation. This includes radio waves, which overlap with the frequencies used for media broadcasts of TV and radio signals. The microwave radiation that you see on the diagram is the same radiation that is used in a microwave oven. What we feel as radiant heat is also a form of low-frequency EMR. All the high-frequency radiation to the right of visible light is sometimes referred to as ultraviolet (UV) radiation. This includes X-rays and gamma (\u03b3) rays. The narrow band that is visible light extends from lower-frequency red light to higher-frequency violet light, thus the terms are infrared(below red) and ultraviolet(beyond violet). BOUNDLESS PHYSICS Maxwell\u2019s Equations The Scottish physicist James Clerk Maxwell (1831\u20131879) is regarded widely to have been the greatest theoretical physicist of the 458 Chapter 15 \u2022 Light nineteenth century. Although he died young, Maxwell not only formulated a complete electromagnetic theory, represented by Maxwell\u2019s equations, he also developed the kinetic theory of gases, and made significant contributions to the understanding of color vision and the nature of Saturn\u2019s rings. Maxwell brought together all the work that had been done by brilliant physicists, such as \u00d8rsted, Coulomb, Ampere, Gauss, and Faraday, and added his own insights to develop the overarching theory of electromagnetism. Maxwell\u2019s equations are paraphrased here in words because their mathematical content is beyond the level of this text. However, the equations illustrate how apparently simple mathematical statements can elegantly unite and express a multitude of concepts\u2014why mathematics is the language of science. Maxwell\u2019s Equations 1. Electric field lines originate on", " positive charges and terminate on negative charges. The electric field is defined as the force per unit charge on a test charge, and the strength of the force is related to the electric constant, \u03b50. 2. Magnetic field lines are continuous, having no beginning or end. No magnetic monopoles are known to exist. The strength of the magnetic force is related to the magnetic constant, \u03bc0. 3. A changing magnetic field induces an electromotive force (emf) and, hence, an electric field. The direction of the emf opposes the change, changing direction of the magnetic field. 4. Magnetic fields are generated by moving charges or by changing electric fields. Maxwell\u2019s complete theory shows that electric and magnetic forces are not separate, but different manifestations of the same thing\u2014the electromagnetic force. This classical unification of forces is one motivation for current attempts to unify the four basic forces in nature\u2014the gravitational, electromagnetic, strong nuclear, and weak nuclear forces. The weak nuclear and electromagnetic forces have been unified, and further unification with the strong nuclear force is expected; but, the unification of the gravitational force with the other three has proven to be a real head-scratcher. One final accomplishment of Maxwell was his development in 1855 of a process that could produce color photographic images. In 1861, he and photographer Thomas Sutton worked together on this process. The color image was achieved by projecting red, blue, and green light through black-and-white photographs of a tartan ribbon, each photo itself exposed in different-colored light. The final image was projected onto a screen (see Figure 15.4). Figure 15.4 Maxwell and Sutton\u2019s photograph of a colored ribbon. This was the first durable color photograph. The plaid tartan of the Scots made a colorfulphotographic subject. GRASP CHECK Describe electromagnetic force as explained by Maxwell\u2019s equations. a. According to Maxwell\u2019s equations, electromagnetic force gives rise to electric force and magnetic force. b. According to Maxwell\u2019s equations, electric force and magnetic force are different manifestations of electromagnetic force. c. According to Maxwell\u2019s equations, electric force is the cause of electromagnetic force. d. According to Maxwell\u2019s equations, magnetic force is the cause of electromagnetic force. Characteristics of Electromagnetic Radiation All the EM waves mentioned above are basically the same form of radiation. They can all travel across empty space, and they all Access for free at openstax.org. 15.1 \u2022 The Electromagnetic Spectrum 459 travel at the speed of light in a vacuum. The basic difference between types of radiation is their differing frequencies. Each frequency has an associated wavelength. As frequency increases across the spectrum, wavelength decreases. Energy also increases with frequency. Because of this, higher frequencies penetrate matter more readily. Some of the properties and uses of the various EM spectrum bands are listed in Table 15.1. Types of EM Waves Radio and TV Production Applications Life Sciences Aspect Issues Accelerating charges Communications, remote controls MRI Requires controls for band use Microwaves Accelerating charges & thermal agitation Communications, microwave ovens, radar Deep heating Cell phone use Infrared Thermal agitation & electronic transitions Thermal imaging, heating Visible Light Thermal agitation & electronic transitions All pervasive Greenhouse effect Absorption by atmosphere Photosynthesis, human vision Ultraviolet Thermal agitation & electronic transitions Sterilization, slowing abnormal growth of cells Vitamin D production Ozone depletion, causes cell damage X-rays Gamma Rays Inner electronic transitions & fast collisions Medical, security Medical diagnosis, cancer therapy Causes cell damage Nuclear decay Nuclear medicine, security Medical diagnosis, cancer therapy Causes cell damage, radiation damage Table 15.1 Electromagnetic Waves This table shows how each type of EM radiation is produced, how it is applied, as well as environmental and health issues associated with it. The narrow band of visible light is a combination of the colors of the rainbow. Figure 15.5 shows the section of the EM spectrum that includes visible light. The frequencies corresponding to these wavelengths are at the red end to at the violet end. This is a very narrow range, considering that the EM spectrum spans about 20 orders of magnitude. Figure 15.5 A small part of the electromagnetic spectrum that includes its visible components. The divisions between infrared, visible, and ultraviolet are not perfectly distinct, nor are the divisions between the seven rainbow colors TIPS FOR SUCCESS Wavelengths of visible light are often given in nanometers, nm. One nm equals wavelength of about 600 nm, or m. m. For example, yellow light has a 460 Chapter 15 \u2022 Light As a child, you probably learned the color wheel, shown on the left in Figure 15.6. It helps if you know what color results when you mix different colors of paint together. Mixing two of the primary pigmentcolors\u2014magenta, yellow, or cyan\u2014together results in a secondary color. For example, mixing cyan and yellow makes green. This is called subtractivecolor mixing. Mixing different colors of light", "together is quite different. The diagram on the right shows additivecolor mixing. In this case, the primary colors are red, green, and blue, and the secondary colors are cyan, magenta, and yellow. Mixing pigments and mixing light are different because materials absorb light by a different set of rules than does the perception of light by the eye. Notice that, when all colors are subtracted, the result is no color, or black. When all colors are added, the result is white light. We see the reverse of this when white sunlight is separated into the visible spectrum by a prism or by raindrops when a rainbow appears in the sky. Figure 15.6 Mixing colored pigments follows the subtractive color wheel, and mixing colored light follows the additive color wheel. Virtual Physics Color Vision Click to view content (https://www.openstax.org/l/28Colorvision) This video demonstrates additive color and color filters. Try all the settings except Photons. GRASP CHECK Explain why only light from a blue bulb passes through the blue filter. a. A blue filter absorbs blue light. b. A blue filter reflects blue light. c. A blue filter absorbs all visible light other than blue light. d. A blue filter reflects all of the other colors of light and absorbs blue light. LINKS TO PHYSICS Animal Color Perception The physics of color perception has interesting links to zoology. Other animals have very different views of the world than humans, especially with respect to which colors can be seen. Color is detected by cells in the eye called cones. Humans have three cones that are sensitive to three different ranges of electromagnetic wavelengths. They are called red, blue, and green cones, although these colors do not correspond exactly to the centers of the three ranges. The ranges of wavelengths that each cone detects are red, 500 to 700 nm; green, 450 to 630 nm; and blue, 400 to 500 nm. Most primates also have three kinds of cones and see the world much as we do. Most mammals other than primates only have two cones and have a less colorful view of things. Dogs, for example see blue and yellow, but are color blind to red and green. You might think that simplerspecies, such as fish and insects, would have less sophisticated vision, but this is not the case. Many birds, reptiles, amphibians, and insects have four or five different cones in their eyes. These species don\u2019t have a wider range of perceived colors, but they see more hues, or combinations of colors. Also, some animals, such as bees or rattlesnakes, see a Access for free at openstax.org. 15.1 \u2022 The Electromagnetic Spectrum 461 range of colors that is as broad as ours, but shifted into the ultraviolet or infrared. These differences in color perception are generally adaptations that help the animals survive. Colorful tropical birds and fish display some colors that are too subtle for us to see. These colors are believed to play a role in the species mating rituals. Figure 15.7 shows the colors visible and the color range of vision in humans, bees, and dogs. Figure 15.7 Humans, bees, and dogs see colors differently. Dogs see fewer colors than humans, and bees see a different range of colors. GRASP CHECK The belief that bulls are enraged by seeing the color red is a misconception. What did you read in this Links to Physics that shows why this belief is incorrect? a. Bulls are color-blind to every color in the spectrum of colors. b. Bulls are color-blind to the blue colors in the spectrum of colors. c. Bulls are color-blind to the red colors in the spectrum of colors. d. Bulls are color-blind to the green colors in the spectrum of colors. Humans have found uses for every part of the electromagnetic spectrum. We will take a look at the uses of each range of frequencies, beginning with visible light. Most of our uses of visible light are obvious; without it our interaction with our surroundings would be much different. We might forget that nearly all of our food depends on the photosynthesis process in plants, and that the energy for this process comes from the visible part of the spectrum. Without photosynthesis, we would also have almost no oxygen in the atmosphere. The low-frequency, infrared region of the spectrum has many applications in media broadcasting. Television, radio, cell phone, and remote-control devices all broadcast and/or receive signals with these wavelengths. AM and FM radio signals are both lowfrequency radiation. They are in different regions of the spectrum, but that is not their basic difference. AM and FM are abbreviations for amplitude modulationand frequency modulation. Information in AM signals has the form of changes in amplitudeof the radio waves; information in FM signals has the form of changes in wave frequency. Another application of long-wavelength radiation is found in microwave ovens. These appliances cook or warm food by irradiating it with EM radiation in the microwave", " frequency range. Most kitchen microwaves use a frequency of Hz. These waves have the right amount of energy to cause polar molecules, such as water, to rotate faster. Polar molecules are those that have a partial charge separation. The rotational energy of these molecules is given up to surrounding matter as heat. The first microwave ovens were called Radarangesbecause they were based on radar technology developed during World War II. Radar uses radiation with wavelengths similar to those of microwaves to detect the location and speed of distant objects, such as airplanes, weather formations, and motor vehicles. Radar information is obtained by receiving and analyzing the echoes of microwaves reflected by an object. The speed of the object can be measured using the Doppler shift of the returning waves. This is the same effect you learned about when you studied sound waves. Like sound waves, EM waves are shifted to higher frequencies by an object moving toward an observer, and to lower frequencies by an object moving away from the observer. Astronomers use this same Doppler effect to measure the speed at which distant galaxies are moving away from us. In this case, the shift in frequency is called the red shift, because visible frequencies are shifted toward the lower-frequency, red end of the spectrum. 462 Chapter 15 \u2022 Light Exposure to any radiation with frequencies greater than those of visible light carries some health hazards. All types of radiation in this range are known to cause cell damage. The danger is related to the high energy and penetrating ability of these EM waves. The likelihood of being harmed by any of this radiation depends largely on the amount of exposure. Most people try to reduce exposure to UV radiation from sunlight by using sunscreen and protective clothing. Physicians still use X-rays to diagnose medical problems, but the intensity of the radiation used is extremely low. Figure 15.8 shows an X-ray image of a patient\u2019s chest cavity. One medical-imaging technique that involves no danger of exposure is magnetic resonance imaging (MRI). MRI is an important imaging and research tool in medicine, producing highly detailed two- and three-dimensional images. Radio waves are broadcast, absorbed, and reemitted in a resonance process that is sensitive to the density of nuclei, usually hydrogen nuclei\u2014protons. Figure 15.8 This shadow X-ray image shows many interesting features, such as artificial heart valves, a pacemaker, and wires used to close the sternum. (credit: P.P. Urone) Check Your Understanding 1. Identify the fields produced by a moving charged particle. a. Both an electric field and a magnetic field will be produced. b. Neither a magnetic field nor an electric field will be produced. c. A magnetic field, but no electric field will be produced. d. Only the electric field, but no magnetic field will be produced. 2. X-rays carry more energy than visible light. Compare the frequencies and wavelengths of these two types of EM radiation. a. Visible light has higher frequencies and shorter wavelengths than X-rays. b. Visible light has lower frequencies and shorter wavelengths than X-rays. c. Visible light has higher frequencies and longer wavelengths than X-rays. d. Visible light has lower frequencies and longer wavelengths than X-rays. 3. How does wavelength change as frequency increases across the EM spectrum? a. The wavelength increases. b. The wavelength first increases and then decreases. c. The wavelength first decreases and then increases. d. The wavelength decreases. 4. Why are X-rays used in imaging of broken bones, rather than radio waves? a. X-rays have higher penetrating energy than radio waves. b. X-rays have lower penetrating energy than radio waves. c. X-rays have a lower frequency range than radio waves. d. X-rays have longer wavelengths than radio waves. 5. Identify the fields that make up an electromagnetic wave. a. both an electric field and a magnetic field b. neither a magnetic field nor an electric field Access for free at openstax.org. 15.2 \u2022 The Behavior of Electromagnetic Radiation 463 c. only a magnetic field, but no electric field d. only an electric field, but no magnetic field 15.2 The Behavior of Electromagnetic Radiation Section Learning Objectives By the end of this section, you will be able to do the following: \u2022 Describe the behavior of electromagnetic radiation \u2022 Solve quantitative problems involving the behavior of electromagnetic radiation Section Key Terms illuminance interference lumens luminous flux lux polarized light Types of Electromagnetic Wave Behavior In a vacuum, all electromagnetic radiation travels at the same incredible speed of 3.00 \u00d7 108 m/s, which is equal to 671 million miles per hour. This is one of the fundamental physical constants. It is referred to as the speed of light and is given the symbol c. The space between celestial bodies is a near vacuum, so the light we see from the Sun, stars, and other planets has traveled here at the speed of light", ". Keep in mind that all EM radiation travels at this speed. All the different wavelengths of radiation that leave the Sun make the trip to Earth in the same amount of time. That trip takes 8.3 minutes. Light from the nearest star, besides the Sun, takes 4.2 years to reach Earth, and light from the nearest galaxy\u2014a dwarf galaxy that orbits the Milky Way\u2014travels 25,000 years on its way to Earth. You can see why we call very long distances astronomical. When light travels through a physical medium, its speed is always less than the speed of light. For example, light travels in water at three-fourths the value of c. In air, light has a speed that is just slightly slower than in empty space: 99.97 percent of c. Diamond slows light down to just 41 percent of c. When light changes speeds at a boundary between media, it also changes direction. The greater the difference in speeds, the more the path of light bends. In other chapters, we look at this bending, called refraction, in greater detail. We introduce refraction here to help explain a phenomenon called thin-film interference. Have you ever wondered about the rainbow colors you often see on soap bubbles, oil slicks, and compact discs? This occurs when light is both refracted by and reflected from a very thin film. The diagram shows the path of light through such a thin film. The symbols n1, n2, and n3 indicate that light travels at different speeds in each of the three materials. Learn more about this topic in the chapter on diffraction and interference. Figure 15.9 shows the result of thin film interference on the surface of soap bubbles. Because ray 2 travels a greater distance, the two rays become out of phase. That is, the crests of the two emerging waves are no longer moving together. This causes interference, which reinforces the intensity of the wavelengths of light that create the bands of color. The color bands are separated because each color has a different wavelength. Also, the thickness of the film is not uniform, and different thicknesses cause colors of different wavelengths to interfere in different places. Note that the film must be very, very thin\u2014somewhere in the vicinity of the wavelengths of visible light. 464 Chapter 15 \u2022 Light Figure 15.9 Light striking a thin film is partially reflected (ray 1) and partially refracted at the top surface. The refracted ray is partially reflected at the bottom surface and emerges as ray 2. These rays will interfere in a way that depends on the thickness of the film and the indices of refraction of the various media. You have probably experienced how polarized sunglasses reduce glare from the surface of water or snow. The effect is caused by the wave nature of light. Looking back at, we see that the electric field moves in only one direction perpendicular to the direction of propagation. Light from most sources vibrates in all directions perpendicular to propagation. Light with an electric field that vibrates in only one direction is called polarized. A diagram of polarized light would look like. Polarized glasses are an example of a polarizing filter. These glasses absorb most of the horizontal light waves and transmit the vertical waves. This cuts down glare, which is caused by horizontal waves. Figure 15.10 shows how waves traveling along a rope can be used as a model of how a polarizing filter works. The oscillations in one rope are in a vertical plane and are said to be vertically polarized. Those in the other rope are in a horizontal plane and are horizontally polarized. If a vertical slit is placed on the first rope, the waves pass through. However, a vertical slit blocks the horizontally polarized waves. For EM waves, the direction of the electric field oscillation is analogous to the disturbances on the ropes. Figure 15.10 The transverse oscillations in one rope are in a vertical plane, and those in the other rope are in a horizontal plane. The first is said to be vertically polarized, and the other is said to be horizontally polarized. Vertical slits pass vertically polarized waves and block horizontally polarized waves. Light can also be polarized by reflection. Most of the light reflected from water, glass, or any highly reflective surface is polarized horizontally. Figure 15.11 shows the effect of a polarizing lens on light reflected from the surface of water. Access for free at openstax.org. 15.2 \u2022 The Behavior of Electromagnetic Radiation 465 Figure 15.11 These two photographs of a river show the effect of a polarizing filter in reducing glare in light reflected from the surface of water. Part (b) of this figure was taken with a polarizing filter and part (a) was taken without. As a result, the reflection of clouds and sky observed in part (a) is not observed in part (b). Polarizing sunglasses are particularly useful on snow and water. WATCH PHYSICS Polarization of Light, Linear and Circular This video explains the polarization of light in great detail. Before viewing the video, look back at the", " drawing of an electromagnetic wave from the previous section. Try to visualize the two-dimensional drawing in three dimensions. Click to view content (https://www.openstax.org/l/28Polarization) GRASP CHECK How do polarized glasses reduce glare reflected from the ocean? a. They block horizontally polarized and vertically polarized light. b. They are transparent to horizontally polarized and vertically polarized light. c. They block horizontally polarized rays and are transparent to vertically polarized rays. d. They are transparent to horizontally polarized light and block vertically polarized light. Snap Lab Polarized Glasses \u2022 EYE SAFETY\u2014Looking at the Sun directly can cause permanent eye damage. Avoid looking directly at the Sun. \u2022 \u2022 two pairs of polarized sunglasses OR two lenses from one pair of polarized sunglasses Procedure 1. Look through both or either polarized lens and record your observations. 2. Hold the lenses, one in front of the other. Hold one lens stationary while you slowly rotate the other lens. Record your observations, including the relative angles of the lenses when you make each observation. 3. Find a reflective surface on which the Sun is shining. It could be water, glass, a mirror, or any other similar smooth surface. The results will be more dramatic if the sunlight strikes the surface at a sharp angle. 4. Observe the appearance of the surface with your naked eye and through one of the polarized lenses. 5. Observe any changes as you slowly rotate the lens, and note the angles at which you see changes. 466 Chapter 15 \u2022 Light GRASP CHECK If you buy sunglasses in a store, how can you be sure that they are polarized? a. When one pair of sunglasses is placed in front of another and rotated in the plane of the body, the light passing through the sunglasses will be blocked at two positions due to refraction of light. b. When one pair of sunglasses is placed in front of another and rotated in the plane of the body, the light passing through the sunglasses will be blocked at two positions due to reflection of light. c. When one pair of sunglasses is placed in front of another and rotated in the plane of the body, the light passing through the sunglasses will be blocked at two positions due to the polarization of light. d. When one pair of sunglasses is placed in front of another and rotated in the plane of the body, the light passing through the sunglasses will be blocked at two positions due to the bending of light waves. Quantitative Treatment of Electromagnetic Waves We can use the speed of light, c, to carry out several simple but interesting calculations. If we know the distance to a celestial object, we can calculate how long it takes its light to reach us. Of course, we can also make the reverse calculation if we know the time it takes for the light to travel to us. For an object at a very great distance from Earth, it takes many years for its light to reach us. This means that we are looking at the object as it existed in the distant past. The object may, in fact, no longer exist. Very large distances in the universe are measured in light years. One light year is the distance that light travels in one year, which is kilometers or miles (\u2026and 1012 is a trillion!). A useful equation involving cis where fis frequency in Hz, and is wavelength in meters. WORKED EXAMPLE Frequency and Wavelength Calculation For example, you can calculate the frequency of yellow light with a wavelength of STRATEGY Rearrange the equation to solve for frequency. Solution Substitute the values for the speed of light and wavelength into the equation. m. 15.2 15.1 15.3 Discussion Manipulating exponents of 10 in a fraction can be tricky. Be sure you keep track of the + and \u2013 exponents correctly. Checking back to the diagram of the electromagnetic spectrum in the previous section shows that 1014 is a reasonable order of magnitude for the frequency of yellow light. The frequency of a wave is proportional to the energy the wave carries. The actual proportionality constant will be discussed in a later chapter. Since frequency is inversely proportional to wavelength, we also know that wavelength is inversely proportional to energy. Keep these relationships in mind as general rules. The rate at which light is radiated from a source is called luminous flux, P, and it is measured in lumens (lm). Energy-saving light bulbs, which provide more luminous flux for a given use of electricity, are now available. One of these bulbs is called a compact fluorescent lamp; another is an LED(light-emitting diode) bulb. If you wanted to replace an old incandescent bulb with an energy saving bulb, you would want the new bulb to have the same brightness as the old one. To compare bulbs accurately, you would need to compare the lumens each one puts out. Comparing wattage\u2014that is, the electric power used\u2014would be Access for free at openstax.org. 15.2 \u2022", " The Behavior of Electromagnetic Radiation 467 misleading. Both wattage and lumens are stated on the packaging. The luminous flux of a bulb might be 2,000 lm. That accounts for all the light radiated in all directions. However, what we really need to know is how much light falls on an object, such as a book, at a specific distance. The number of lumens per square meter is called illuminance, and is given in units of lux (lx). Picture a light bulb in the middle of a sphere with a 1-m radius. The total surface of the sphere equals 4\u03c0r2 m2. The illuminance then is given by What happens if the radius of the sphere is increased 2 m? The illuminance is now only one-fourth as great, because the r2 term in the denominator is 4 instead of 1. Figure 15.12 shows how illuminance decreases with the inverse square of the distance. 15.4 Figure 15.12 The diagram shows why the illuminance varies inversely with the square of the distance from a source of light. WORKED EXAMPLE Calculating Illuminance A woman puts a new bulb in a floor lamp beside an easy chair. If the luminous flux of the bulb is rated at 2,000 lm, what is the illuminance on a book held 2.00 m from the bulb? STRATEGY Choose the equation and list the knowns. Equation: P= 2,000 lm \u03c0 = 3.14 r= 2.00 m Solution Substitute the known values into the equation. Discussion Try some other distances to illustrate how greatly light fades with distance from its source. For example, at 3 m the illuminance is only 17.7 lux. Parents often scold children for reading in light that is too dim. Instead of shouting, \u201cYou\u2019ll ruin your eyes!\u201d it might be better to explain the inverse square law of illuminance to the child. 468 Chapter 15 \u2022 Light Practice Problems 6. Red light has a wavelength of 7.0 \u00d7 10\u22127 m and a frequency of 4.3 \u00d7 1014 Hz. Use these values to calculate the speed of light in a vacuum. a. b. c. d. 3 \u00d7 1020 m/s 3 \u00d7 1015 m/s 3 \u00d7 1014 m/s 3 \u00d7 108 m/s 7. A light bulb has a luminous flux of 942 lumens. What is the illuminance on a surface from the bulb when it is lit? a. b. c. d. Check Your Understanding 8. Give an example of a place where light travels at the speed of 3.00 \u00d7 108 m/s. a. outer space b. water c. Earth\u2019s atmosphere d. quartz glass 9. Explain in terms of distances and the speed of light why it is currently very unlikely that humans will visit planets that circle stars other than our Sun. a. The spacecrafts used for travel are very heavy and thus very slow. b. Spacecrafts do not have a constant source of energy to run them. c. If a spacecraft could attain a maximum speed equal to that of light, it would still be too slow to cover astronomical distances. d. Spacecrafts can attain a maximum speed equal to that of light, but it is difficult to locate planets around stars. Access for free at openstax.org. Chapter 15 \u2022 Key Terms 469 KEY TERMS electric field a field that tells us the force per unit charge at all locations in space around a charge distribution electromagnetic radiation (EMR) radiant energy that consists of oscillating electric and magnetic fields lux unit of measure for illuminance magnetic field the directional lines around a magnetic material that indicates the direction and magnitude of the magnetic force illuminance number of lumens per square meter, given in Maxwell\u2019s equations equations that describe the units of lux (lx) interference increased or decreased light intensity caused by the phase differences between waves lumens unit of measure for luminous flux luminous flux rate at which light is radiated from a source interrelationship between electric and magnetic fields, and how these fields combine to form electromagnetic radiation polarized light light whose electric field component vibrates in a specific plane SECTION SUMMARY 15.1 The Electromagnetic Spectrum \u2022 The electromagnetic spectrum is made up of a broad range of frequencies of electromagnetic radiation. \u2022 All frequencies of EM radiation travel at the same speed in a vacuum and consist of an electric field and a magnetic field. The types of EM radiation have different frequencies and wavelengths, and different energies and penetrating ability. 15.2 The Behavior of Electromagnetic Radiation \u2022 EM radiation travels at different speeds in different media, produces colors on thin films, and can be polarized to oscillate in only one direction. \u2022 Calculations can be based on the relationship among the speed, frequency, and wavelength of light, and on the relationship among luminous flux,", " illuminance, and distance. KEY EQUATIONS 15.2 The Behavior of Electromagnetic Radiation frequency and wavelength illuminance CHAPTER REVIEW Concept Items 15.1 The Electromagnetic Spectrum 1. Use the concepts on which Maxwell\u2019s equations are based to explain why a compass needle is deflected when the compass is brought near a wire that is carrying an electric current. a. The charges in the compass needle and the charges in the electric current have interacting electric fields, causing the needle to deflect. b. The electric field from the moving charges in the current interacts with the magnetic field of the compass needle, causing the needle to deflect. c. The magnetic field from the moving charges in the current interacts with the electric field of the compass needle, causing the needle to deflect. d. The moving charges in the current produce a magnetic field that interacts with the compass needle\u2019s magnetic field, causing the needle to deflect. 2. Consider these colors of light: yellow, blue, and red. Part A. Put these light waves in order according to wavelength, from shortest wavelength to longest wavelength. Part B. Put these light waves in order according to frequency, from lowest frequency to highest frequency. a. wavelength: blue, yellow, red frequency: blue, yellow, red b. wavelength: red, yellow, blue frequency: red, yellow, blue c. wavelength: red, yellow, blue frequency: blue, yellow, red d. wavelength: blue, yellow, red frequency: red, yellow, blue 3. Describe the location of gamma rays on the electromagnetic spectrum. 470 Chapter 15 \u2022 Chapter Review a. At the high-frequency and long-wavelength end of the spectrum b. At the high-frequency and short-wavelength end of thickness of the wall of a soap bubble? Explain your answer. a. The thickness of the bubble wall is ten times that of the spectrum the wavelength of light. c. At the low-frequency and long-wavelength end of b. The thickness of the bubble wall is similar to that of the spectrum the wavelength of light. d. At the low-frequency and short-wavelength end of c. The thickness of the bubble wall is half the the spectrum wavelength of light. 4. In which region of the electromagnetic spectrum would you find radiation that is invisible to the human eye and has low energy? a. Long-wavelength and high-frequency region b. Long-wavelength and low-frequency region c. Short-wavelength and high-frequency region d. Short-wavelength and low-frequency region 15.2 The Behavior of Electromagnetic Radiation 5. Light travels at different speeds in different media. Put these media in order, from the slowest light speed to the fastest light speed: air, diamond, vacuum, water. a. diamond, water, air, vacuum b. vacuum, diamond, air, water c. diamond, air, water, vacuum d. air, diamond, water, vacuum 6. Visible light has wavelengths in the range of about 400 to 800 nm. What does this indicate about the approximate Critical Thinking Items 15.1 The Electromagnetic Spectrum 8. Standing in front of a fire, we can sense both its heat and its light. How are the light and heat radiated by the fire the same, and how are they different? a. Both travel as waves, but only light waves are a form of electromagnetic radiation. b. Heat and light are both forms of electromagnetic radiation, but light waves have higher frequencies. c. Heat and light are both forms of electromagnetic radiation, but heat waves have higher frequencies. d. Heat and light are both forms of electromagnetic radiation, but light waves have higher wavelengths. 9. Light shines on a picture of the subtractive color wheel. The light is a mixture of red, blue, and green light. Part A\u2014Which part of the color wheel will look blue? Explain in terms of absorbed and reflected light. Part B\u2014Which part of the color wheel will look yellow? Explain in terms of absorbed and reflected light. a. A. The yellow section of the wheel will look blue because it will reflect blue light and absorb red Access for free at openstax.org. d. The thickness of the bubble wall equals the cube of the wavelength of light. 7. Bright sunlight is reflected from an icy pond. You look at the glare of the reflected light through polarized glasses. When you take the glasses off, rotate them 90\u00b0, and look through one of the lenses again, the light you see becomes brighter. Explain why the light you see changes. a. The glass blocks horizontally polarized light, and the light reflected from the icy pond is, in part, polarized horizontally. b. The glass blocks vertically polarized light, and the light reflected from the icy pond is, in part, polarized vertically. c. The glass allows horizontally polarized light to pass, and the light reflected from the icy pond is, in part, polarized vertically. d. The glass allows horizontally", " polarized light to pass, and the light reflected from the icy pond is, in part, polarized horizontally. and green. B. The blue section of the wheel will look yellow because it will reflect red and green light and absorb blue. b. A. The blue section of the wheel will look blue because it will absorb blue light and reflect red and green. B. The yellow section of the wheel will look yellow because it will absorb red and green light and reflect blue. c. A. The yellow section of the wheel will look blue because it will absorb blue light and reflect red and green. B. The blue section of the wheel will look yellow because it will absorb red and green light and reflect blue. d. A. The blue section of the wheel will look blue because it will reflect blue light and absorb red and green. B. The yellow section of the wheel will look yellow because it will reflect red and green light and absorb blue. 10. Part A. When you stand in front of an open fire, you can sense light waves with your eyes. You sense another type of electromagnetic radiation as heat. What is this other type of radiation? Part B. How is this other type of radiation different front light waves? a. A. X-rays B. The X-rays have higher frequencies and shorter wavelengths than the light waves. b. A. X-rays B. The X-rays have lower frequencies and longer wavelengths than the light waves. c. A. infrared rays B. The infrared rays have higher frequencies and shorter wavelengths than the light waves. d. A. infrared rays B. The infrared rays have lower frequencies and longer wavelengths than the light waves. 11. Overexposure to this range of EM radiation is dangerous, and yet it is used by doctors to diagnose medical problems. Part A\u2014Identify the type of radiation. Part B\u2014Locate the position of this radiation on the EM spectrum by comparing its frequency and wavelength to visible light. Part C\u2014Explain why this radiation is both dangerous and therapeutic in terms of its energy, based on your answer to Part B. a. A. X-rays B. X-rays have shorter wavelengths (1 \u00d7 10\u20138 \u2013 5 \u00d7 10\u201312 m) and higher frequencies (3 \u00d7 1016 \u2013 6 \u00d7 1019 Hz) than visible light (7.5 \u00d7 10\u20137 \u2013 4.0 \u00d7 10\u20137 m; 4.0 \u00d7 1014 \u2013 7.5 \u00d7 1014 Hz). C. X-rays have low energies because of their high frequencies, and so can penetrate matter to greater depths. b. A. X-rays B. X-rays have shorter wavelengths (1 \u00d7 10\u20138 \u2013 5 \u00d7 10\u201312 m) and higher frequencies (3 \u00d7 1010 \u2013 6 \u00d7 1013 Hz) than visible light (7.5 \u00d7 10\u20137 \u2013 4.0 \u00d7 10\u20137 m; 4.0 \u00d7 1014 \u2013 7.5 \u00d7 1014 Hz). C. X-rays have low energies because of their low frequencies, and so can penetrate matter to greater depths. c. A. X-rays B. X-rays have longer wavelengths (1 \u00d7 10\u20136 \u2013 5 \u00d7 10\u20137 m) and higher frequencies (3 \u00d7 1015 \u2013 6 \u00d7 1015 Hz) than visible light (7.5 \u00d7 10\u20137 \u2013 4.0 \u00d7 10\u20137 m; 4.0 \u00d7 1014 \u2013 7.5 \u00d7 1014 Hz). C. X-rays have high energies because of their high Chapter 15 \u2022 Chapter Review 471 frequencies, and therefore can penetrate matter to greater depths. d. A. X-rays B. X-rays have shorter wavelengths (1 \u00d7 10\u20138 \u2013 5 \u00d7 10\u201312 m) and higher frequencies (3 \u00d7 1016 \u2013 6 \u00d7 1019 Hz) than visible light (7.5 \u00d7 10\u20137 \u2013 4.0 \u00d7 10\u20137 m; 4.0 \u00d7 1014 \u2013 7.5 \u00d7 1014 Hz). C. X-rays have high energies because of their high frequencies, and so can penetrate matter to greater depths. 15.2 The Behavior of Electromagnetic Radiation 12. Explain how thin-film interference occurs. Discuss in terms of the meaning of interference and the pathways of light waves. a. For a particular thickness of film, light of a given wavelength that reflects from the outer and inner film surfaces is completely in phase, and so undergoes constructive interference. b. For a particular thickness of film, light of a given wavelength that reflects from the outer and inner surfaces is completely in phase, and so undergoes destructive interference. c. For a particular thickness of film, light of a given wavelength that reflects from the outer and inner film surfaces is completely out of phase, and so undergoes constructive interference. d. For a particular thickness of film, light of a given wavelength that reflects from the outer and inner film surfaces is completely out of phase, and so undergoes no interference.", " 13. When you move a rope up and down, waves are created. If the waves pass through a slot, they will be affected differently, depending on the orientation of the slot. Using the rope waves and the slot as a model, explain how polarizing glasses affect light waves. a. If the wave\u2014electric field\u2014is vertical and slit\u2014polarizing molecules in the glass\u2014is horizontal, the wave will pass. If the wave\u2014electric field\u2014 is vertical and slit\u2014polarizing molecules in the glass\u2014is vertical, the wave will not pass. If the wave\u2014electric field\u2014is horizontal and slit\u2014polarizing molecules in the glass\u2014is horizontal, the wave will pass. If the wave\u2014electric field\u2014is horizontal and slit\u2014polarizing molecules in the glass\u2014is horizontal, the wave will not pass. b. c. d. 472 Chapter 15 \u2022 Test Prep Problems 15.2 The Behavior of Electromagnetic Radiation 14. Visible light has a range of wavelengths from about 400 nm to 800 nm. What is the range of frequencies for visible light? a. b. c. d. 3.75 \u00d7 106 Hz to 7.50 \u00d7 106 Hz 3.75 Hz to 7.50 Hz 3.75 \u00d7 10\u22127 Hz to 7.50 \u00d7 10\u22127 Hz 3.75 \u00d7 1014 Hz to 7.50 \u00d7 1014 Hz Performance Task 15.2 The Behavior of Electromagnetic Radiation 16. Design an experiment to observe the phenomenon of thin-film interference. Observe colors of visible light, and relate each color to its corresponding wavelength. Comparison with the magnitudes of visible light wavelength will give an appreciation of just how very thin a thin film is. Thin-film interference has a number of practical applications, such as anti-reflection coatings and optical filters. Thin films used in filters can be designed to reflect or transmit specific wavelengths of light. This is done by depositing a film one molecular layer at a time from a vapor, thus allowing the thickness of the film to be exactly controlled. \u2022 EYE SAFETY\u2014Chemicals in this lab are poisonous if ingested. If chemicals are ingested, inform your teacher immediately. \u2022 FUMES\u2014Certain chemicals or chemical reactions in this lab create a vapor that is harmful if inhaled. Follow your teacher's instructions for the use of fume hoods and other safety apparatus designed to prevent fume inhalation. Never smell or otherwise breath in any chemicals or vapors in the lab. \u2022 FLAMMABLE\u2014Chemicals in this lab are highly flammable and can ignite, especially if exposed to a spark or open flame. Follow your teacher's TEST PREP Multiple Choice 15.1 The Electromagnetic Spectrum 17. Which type of EM radiation has the shortest wavelengths? a. gamma rays b. c. blue light d. microwaves infrared waves Access for free at openstax.org. 15. Light travels through the wall of a soap bubble that is 600 nm thick and is reflected from the inner surface back into the air. Assume the bubble wall is mostly water and that light travels in water at 75 percent of the speed of light in vacuum. How many seconds behind will the light reflected from the inner surface arrive compared to the light that was reflected from the outer surface? a. 4.0 \u00d7 10\u20138 s 5.3 \u00d7 10\u20136 s b. c. 2.65 \u00d7 10\u201315 s 5.3 \u00d7 10\u201315 s d. instructions carefully on how to handle flammable chemicals. Do not expose any chemical to a flame or other heat source unless specifically instructed by your teacher. \u2022 HAND WASHING\u2014Some materials may be hazardous if in extended contact with the skin. Be sure to wash your hands with soap after handling and disposing of these materials during the lab. \u2022 WASTE\u2014Some things in this lab are hazardous and need to be disposed of properly. Follow your teacher's instructions for disposal of all items. \u2022 A large flat tray with raised sides, such as a baking tray \u2022 Small volumes of motor oil, lighter fluid or a penetrating oil of the type used to loosen rusty bolts, and cooking oil \u2022 Water \u2022 A camera a. Thin-film interference causes colors to appear on the surface of a thin transparent layer. Do you expect to see a pattern to the colors? b. How could you make a permanent record of your observations? c. What data would you need to look up to help explain any patterns that you see? d. What could explain colors failing to appear under some conditions? 18. Which form of EM radiation has the most penetrating ability? a. red light b. microwaves c. gamma rays d. infrared radiation 19. Why are high-frequency gamma rays more dangerous to humans than visible light? a. Gamma rays have a lower frequency range than visible light. 22. What is the wavelength of red light with a frequency of Chapter 15 \u2022 Test Prep 473 b. Gamma rays have a longer wavelength range", " than visible light. c. Gamma rays have greater energy than visible light for penetrating matter. d. Gamma rays have less energy than visible light for penetrating matter. 20. A dog would have a hard time stalking and catching a red bird hiding in a field of green grass. Explain this in terms of cone cells and color perception. a. Dogs are red-green color-blind because they can see only blue and yellow through two kinds of cone cells present in their eyes. b. Dogs are only red color-blind because they can see only blue and yellow through two kinds of cones cells present in their eyes. c. Dogs are only green color-blind because they can see only blue and yellow through two kinds of cones cells present in their eyes. d. Dogs are color-blind because they have only rods and no cone cells present in their eyes. 15.2 The Behavior of Electromagnetic Radiation 21. To compare the brightness of light bulbs for sale in a frequency store, you should look on the labels to see how they are rated in terms of ____. a. b. watts c. amps d. lumens 4.00 \u00d7 1014 Hz? a. 2.50 \u00d7 1014 m b. 4.00 \u00d7 1015 m c. 2.50 \u00d7 106 m d. 4.00 \u00d7 10-7 m 23. What is the distance of one light year in kilometers? a. 2.59 \u00d7 1010 km 1.58 \u00d7 1011 km b. c. 2.63 \u00d7 109 km d. 9.46 \u00d7 1012 km 24. How does the illuminance of light change when the distance from the light source is tripled? Cite the relevant equation and explain how it supports your answer. a. if distance is tripled, then the illuminance increases by 19 times. b. c. d. if distance is tripled, then the illuminance decreases by 13 times. then the illuminance decreases by 9 times. if distance is tripled, if distance tripled, then the illuminance increases by 3 times. 25. A light bulb has an illuminance of 19.9 lx at a distance of 2 m. What is the luminous flux of the bulb? 500 lm a. b. 320 lm c. 250 lm d. 1,000 lm Short Answer 15.1 The Electromagnetic Spectrum 26. Describe one way in which heat waves\u2014infrared radiation\u2014are different from sound waves. a. Sound waves are transverse waves, whereas heat waves\u2014infrared radiation\u2014are longitudinal waves. b. Sound waves have shorter wavelengths than heat waves. c. Sound waves require a medium, whereas heat waves\u2014infrared radiation\u2014do not. d. Sound waves have higher frequencies than heat waves. 27. Describe the electric and magnetic fields that make up an electromagnetic wave in terms of their orientation relative to each other and their phases. a. They are perpendicular to and out of phase with each other. b. They are perpendicular to and in phase with each other. c. They are parallel to and out of phase with each other. d. They are parallel to and in phase with each other. 28. Explain how X-radiation can be harmful and how it can be a useful diagnostic tool. a. Overexposure to X-rays can cause HIV, though normal levels of X-rays can be used for sterilizing needles. b. Overexposure to X-rays can cause cancer, though in limited doses X-rays can be used for imaging internal body parts. c. Overexposure to X-rays causes diabetes, though normal levels of X-rays can be used for imaging internal body parts. d. Overexposure to X-rays causes cancer, though normal levels of X-rays can be used for reducing 474 Chapter 15 \u2022 Test Prep cholesterol in the blood. 31. What is it about the nature of light reflected from snow 29. Explain how sunlight is the original source of the energy in the food we eat. a. Sunlight is converted into chemical energy by plants; this energy is released when we digest food. b. Sunlight is converted into chemical energy by animals; this energy is released when we digest food. c. Sunlight is converted into chemical energy by fish; this energy is released when we digest food. that causes skiers to wear polarized sunglasses? a. The reflected light is polarized in the vertical direction. b. The reflected light is polarized in the horizontal direction. c. The reflected light has less intensity than the incident light. d. The reflected light has triple the intensity of the incident light. d. Sunlight is converted into chemical energy by 32. How many lumens are radiated from a candle which has humans; this energy is released when we digest food. 15.2 The Behavior of Electromagnetic Radiation 30. Describe what happens to the path of light when the light slows down as it passes from one medium to another? a", ". The path of the light remains the same. b. The path of the light becomes circular. c. The path of the light becomes curved. d. The path of the light changes. an illuminance of 3.98 lx at a distance of 2.00 m? a. 400 lm 100 lm b. c. 50 lm d. 200 lm 33. Saturn is 1.43\u00d71012 m from the Sun. How many minutes does it take the Sun\u2019s light to reach Saturn? a. b. c. d. 7.94 \u00d7 109 minutes 3.4 \u00d7 104 minutes 3.4 \u00d7 10\u20136 minutes 79.4 minutes Extended Response C. ultraviolet radiation 15.1 The Electromagnetic Spectrum 35. A mixture of red and green light is shone on each of the 34. A frequency of red light has a wavelength of 700 nm. Part A\u2014Compare the wavelength and frequency of violet light to red light. Part B\u2014Identify a type of radiation that has lower frequencies than red light. Part C\u2014Identify a type of radiation that has shorter wavelengths than violet light. a. A. Violet light has a lower frequency and longer wavelength than red light. B. ultraviolet radiation infrared radiation C. b. A. Violet light has a lower frequency and longer wavelength than red light. B. infrared radiation C. ultraviolet radiation c. A. Violet light has a higher frequency and shorter wavelength than red light. B. ultraviolet radiation infrared radiation C. d. A. Violet light has a higher frequency and shorter wavelength than red light. infrared radiation B. Access for free at openstax.org. subtractive colors. Part A\u2014Which of these colors of light are reflected from magenta? Part B\u2014Which of these colors of light are reflected from yellow? Part C\u2014Which these colors of light are reflected from cyan? a. Part A. red and green Part B. green Part C. red b. Part A. red and green Part B. red Part C. green c. Part A. green Part B. red and green Part C. red d. Part A. red Part B. red and green Part C. green 15.2 The Behavior of Electromagnetic Radiation 36. Explain why we see the colorful effects of thin-film interference on the surface of soap bubbles and oil slicks, but not on the surface of a window pane or clear plastic bag. a. The thickness of a window pane or plastic bag is more than the wavelength of light, and interference occurs for thicknesses smaller than the wavelength of light. b. The thickness of a window pane or plastic bag is less than the wavelength of light, and interference occurs for thicknesses similar to the wavelength of light. c. The thickness of a window pane or plastic bag is more than the wavelength of light, and interference occurs for thicknesses similar to the wavelength of light. d. The thickness of a window pane or plastic bag is Chapter 15 \u2022 Test Prep 475 less than the wavelength of light, and interference occurs for thicknesses larger than the wavelength of light. 37. The Occupational Safety and Health Administration (OSHA) recommends an illuminance of for desktop lighting. An office space has lighting hung. above desktop level that provides only To what height would the lighting fixtures have to be lowered to provide a. b. c. d. on desktops? 476 Chapter 15 \u2022 Test Prep Access for free at openstax.org. CHAPTER 16 Mirrors and Lenses Figure 16.1 Flat, smooth surfaces reflect light to form mirror images. (credit: NASA Goddard Photo and Video, via Flickr) Chapter Outline 16.1 Reflection 16.2 Refraction 16.3 Lenses INTRODUCTION \u201cIn another moment Alice was through the glass, and had jumped lightly down into the Looking-glass room.\u201d \u2014Through the Looking Glass by Lewis Carol Through the Looking Glasstells of the adventures of Alice after she steps from the real world, through a mirror, and into the virtual world. In this chapter we examine the optical meanings of real and virtual, as well as other concepts that make up the field of optics. The light from this page or screen is formed into an image by the lens of your eyes, much as the lens of the camera that made the photograph at the beginning of this chapter. Mirrors, like lenses, can also form images, which in turn are captured by your eyes. Optics is the branch of physics that deals with the behavior of visible light and other electromagnetic waves. For now, we concentrate on the propagation of light and its interaction with matter. It is convenient to divide optics into two major parts based on the size of objects that light encounters. When light interacts with an object that is several times as large as the light\u2019s wavelength, its observable behavior is similar to a ray; it does not display its 478 Chapter 16 \u2022 Mirrors and Lenses wave characteristics prominently. We call this part of optics geometric", " optics. This chapter focuses on situations for which geometric optics is suited. 16.1 Reflection Section Learning Objectives By the end of this section, you will be able to do the following: \u2022 Explain reflection from mirrors, describe image formation as a consequence of reflection from mirrors, apply ray diagrams to predict and interpret image and object locations, and describe applications of mirrors \u2022 Perform calculations based on the law of reflection and the equations for curved mirrors Section Key Terms angle of incidence angle of reflection central axis concave mirror convex mirror diffused focal length focal point geometric optics law of reflection law of refraction ray real image specular virtual image Characteristics of Mirrors There are three ways, as shown in Figure 16.2, in which light can travel from a source to another location. It can come directly from the source through empty space, such as from the Sun to Earth. Light can travel to an object through various media, such as air and glass. Light can also arrive at an object after being reflected, such as by a mirror. In all these cases, light is modeled as traveling in a straight line, called a ray. Light may change direction when it encounters the surface of a different material (such as a mirror) or when it passes from one material to another (such as when passing from air into glass). It then continues in a straight line\u2014that is, as a ray. The word raycomes from mathematics. Here it means a straight line that originates from some point. It is acceptable to visualize light rays as laser rays (or even science fiction depictions of ray guns). Figure 16.2 Three methods for light to travel from a source to another location are shown. (a) Light reaches the upper atmosphere of Earth by traveling through empty space directly from the source (the Sun). (b) This light can reach a person in one of two ways. It can travel through a medium, such as air or glass, and typically travels from one medium to another. It can also reflect from an object, such as a mirror. Because light moves in straight lines, that is, as rays, and changes directions when it interacts with matter, it can be described through geometry and trigonometry. This part of optics, described by straight lines and angles, is therefore called geometric optics. There are two laws that govern how light changes direction when it interacts with matter: the law of reflection, for situations in which light bounces off matter; and the law of refraction, for situations in which light passes through matter. In this section, we consider the geometric optics of reflection. Whenever we look into a mirror or squint at sunlight glinting from a lake, we are seeing a reflection. How does the reflected light travel from the object to your eyes? The law of reflection states: The angle of reflection,, equals the angle of incidence, Access for free at openstax.org. 16.1 \u2022 Reflection 479.This law governs the behavior of all waves when they interact with a smooth surface, and therefore describe the behavior of light waves as well. The reflection of light is simplified when light is treated as a ray. This concept is illustrated in Figure 16.3, which also shows how the angles are measured relative to the line perpendicular to the surface at the point where the light ray strikes it. This perpendicular line is also called the normal line, or just the normal. Light reflected in this way is referred to as specular (from the Latin word for mirror: speculum). We expect to see reflections from smooth surfaces, but Figure 16.4, illustrates how a rough surface reflects light. Because the light is reflected from different parts of the surface at different angles, the rays go in many different directions, so the reflected light is diffused. Diffused light allows you to read a printed page from almost any angle because some of the rays go in different directions. Many objects, such as people, clothing, leaves, and walls, have rough surfaces and can be seen from many angles. A mirror, on the other hand, has a smooth surface and reflects light at specific angles. Figure 16.3 The law of reflection states that the angle of reflection, \u03b8r, equals the angle of incidence, \u03b8i. The angles are measured relative to the line perpendicular to the surface at the point where the ray strikes the surface. The incident and reflected rays, along with the normal, lie in the same plane. Figure 16.4 Light is diffused when it reflects from a rough surface. Here, many parallel rays are incident, but they are reflected at many different angles because the surface is rough. When we see ourselves in a mirror, it appears that our image is actually behind the mirror. We see the light coming from a direction determined by the law of reflection. The angles are such that our image is exactly the same distance behind the mirror, di, as the distance we stand away from the mirror, do. Although these mirror images make objects appear to be where they cannot be (such as behind a solid", " wall), the images are not figments of our imagination. Mirror images can be photographed and videotaped by instruments and look just as they do to our eyes, which are themselves optical instruments. An image in a mirror is said to be a virtual image, as opposed to a real image. A virtual image is formed when light rays appear to diverge from a point without actually doing so. Figure 16.5 helps illustrate how a flat mirror forms an image. Two rays are shown emerging from the same point, striking the mirror, and reflecting into the observer\u2019s eye. The rays can diverge slightly, and both still enter the eye. If the rays are extrapolated backward, they seem to originate from a common point behind the mirror, allowing us to locate the image. The paths of the reflected rays into the eye are the same as if they had come directly from that point behind the mirror. Using the law of reflection\u2014the angle of reflection equals the angle of incidence\u2014we can see that the image and object are the same distance from the mirror. This is a virtual image, as defined earlier. 480 Chapter 16 \u2022 Mirrors and Lenses Figure 16.5 When two sets of rays from common points on an object are reflected by a flat mirror into the eye of an observer, the reflected rays seem to originate from behind the mirror, which determines the position of the virtual image. FUN IN PHYSICS Mirror Mazes Figure 16.6 is a chase scene from an old silent film called The Circus, starring Charlie Chaplin. The chase scene takes place in a mirror maze. You may have seen such a maze at an amusement park or carnival. Finding your way through the maze can be very difficult. Keep in mind that only one image in the picture is real\u2014the others are virtual. Figure 16.6 Charlie Chaplin is in a mirror maze. Which image is real? One of the earliest uses of mirrors for creating the illusion of space is seen in the Palace of Versailles, the former home of French royalty. Construction of the Hall of Mirrors (Figure 16.7) began in 1678. It is still one of the most popular tourist attractions at Versailles. Figure 16.7 Tourists love to wander in the Hall of Mirrors at the Palace of Versailles. (credit: Michal Osmenda, Flickr) GRASP CHECK Only one Charlie in this image (Figure 16.8) is real. The others are all virtual images of him. Can you tell which is real? Hint\u2014His hat is tilted to one side. Access for free at openstax.org. 16.1 \u2022 Reflection 481 Figure 16.8 a. The virtual images have their hats tilted to the right. b. The virtual images have their hats tilted to the left. c. The real images have their hats tilted to the right. d. The real images have their hats tilted to the left. WATCH PHYSICS Virtual Image This video explains the creation of virtual images in a mirror. It shows the location and orientation of the images using ray diagrams, and relates the perception to the human eye. Click to view content (https://openstax.org/l/28Virtualimage) Compare the distance of an object from a mirror to the apparent distance of its virtual image behind the mirror. a. The distances of the image and the object from the mirror are the same. b. The distances of the image and the object from the mirror are always different. c. The image is formed at infinity if the object is placed near the mirror. d. The image is formed near the mirror if the object is placed at infinity. Some mirrors are curved instead of flat. A mirror that curves inward is called a concave mirror, whereas one that curves outward is called a convex mirror. Pick up a well-polished metal spoon and you can see an example of each type of curvature. The side of the spoon that holds the food is a concave mirror; the back of the spoon is a convex mirror. Observe your image on both sides of the spoon. TIPS FOR SUCCESS You can remember the difference between concave and convex by thinking, Concave means caved in. Ray diagrams can be used to find the point where reflected rays converge or appear to converge, or the point from which rays appear to diverge. This is called the focal point, F. The distance from F to the mirror along the central axis (the line perpendicular to the center of the mirror\u2019s surface) is called the focal length, f. Figure 16.9 shows the focal points of concave and convex mirrors. 482 Chapter 16 \u2022 Mirrors and Lenses Figure 16.9 (a, b) The focal length for the concave mirror in (a), formed by converging rays, is in front of the mirror, and has a positive value. The focal length for the convex mirror in (b), formed by diverging rays,", " appears to be behind the mirror, and has a negative value. Images formed by a concave mirror vary, depending on which side of the focal point the object is placed. For any object placed on the far side of the focal point with respect to the mirror, the rays converge in front of the mirror to form a real image, which can be projected onto a surface, such as a screen or sheet of paper However, for an object located inside the focal point with respect to the concave mirror, the image is virtual. For a convex mirror the image is always virtual\u2014that is, it appears to be behind the mirror. The ray diagrams in Figure 16.10 show how to determine the nature of the image formed by concave and convex mirrors. Figure 16.10 (a) The image of an object placed outside the focal point of a concave mirror is inverted and real. (b) The image of an object Access for free at openstax.org. placed inside the focal point of a concave mirror is erect and virtual. (c) The image of an object formed by a convex mirror is erect and virtual. The information in Figure 16.10 is summarized in Table 16.1. Type of Mirror Object to Mirror Distance, do Image Characteristics 16.1 \u2022 Reflection 483 Concave Concave Convex Real and inverted Virtual and erect Virtual and erect Table 16.1 Curved Mirror Images This table details the type and orientation of images formed by concave and convex mirrors. Snap Lab Concave and Convex Mirrors \u2022 Silver spoon and silver polish, or a new spoon made of any shiny metal Instructions Procedure 1. Choose any small object with a top and a bottom, such as a short nail or tack, or a coin, such as a quarter. Observe the object\u2019s reflection on the back of the spoon. 2. Observe the reflection of the object on the front (bowl side) of the spoon when held away from the spoon at a distance of several inches. 3. Observe the image while slowly moving the small object toward the bowl of the spoon. Continue until the object is all the way inside the bowl of the spoon. 4. You should see one point where the object disappears and then reappears. This is the focal point. WATCH PHYSICS Parabolic Mirrors and Real Images This video uses ray diagrams to show the special feature of parabolic mirrors that makes them ideal for either projecting light energy in parallel rays, with the source being at the focal point of the parabola, or for collecting at the focal point light energy from a distant source. Click to view content (https://www.openstax.org/l/28Parabolic) Explain why using a parabolic mirror for a car headlight throws much more light on the highway than a flat mirror. a. The rays do not polarize after reflection. b. The rays are dispersed after reflection. c. The rays are polarized after reflection. d. The rays become parallel after reflection. You should be able to notice everyday applications of curved mirrors. One common example is the use of security mirrors in stores, as shown in Figure 16.11. 484 Chapter 16 \u2022 Mirrors and Lenses Figure 16.11 Security mirrors are convex, producing a smaller, upright image. Because the image is smaller, a larger area is imaged compared with what would be observed for a flat mirror; hence, security is improved. (credit: Laura D\u2019Alessandro, Flickr) Some telescopes also use curved mirrors and no lenses (except in the eyepieces) both to magnify images and to change the path of light. Figure 16.12 shows a Schmidt-Cassegrain telescope. This design uses a spherical primary concave mirror and a convex secondary mirror. The image is projected onto the focal plane by light passing through the perforated primary mirror. The effective focal length of such a telescope is the focal length of the primary mirror multiplied by the magnification of the secondary mirror. The result is a telescope with a focal length much greater than the length of the telescope itself. Figure 16.12 This diagram shows the design of a Schmidt\u2013Cassegrain telescope. A parabolic concave mirror has the very useful property that all light from a distant source, on reflection by the mirror surface, is directed to the focal point. Likewise, a light source placed at the focal point directs all the light it emits in parallel lines away from the mirror. This case is illustrated by the ray diagram in Figure 16.13. The light source in a car headlight, for example, is located at the focal point of a parabolic mirror. Figure 16.13 The bulb in this ray diagram of a car headlight is located at the focal point of a parabolic mirror. Parabolic mirrors are also used to collect sunlight and direct it to a focal point, where it is transformed into heat, which in turn can be used to generate electricity. This application is shown", " in Figure 16.14. Figure 16.14 Parabolic trough collectors are used to generate electricity in southern California. (credit: kjkolb, Wikimedia Commons) Access for free at openstax.org. 16.1 \u2022 Reflection 485 Using a concave mirror, you look at the reflection of a faraway object. The image size changes if you move the object closer to the mirror. Why does the image disappear entirely when the object is at the mirror's focal point? a. The height of the image became infinite. b. The height of the object became zero. c. The intensity of intersecting light rays became zero. d. The intensity of intersecting light rays increased. The Application of the Curved Mirror Equations Curved mirrors and the images they create involve a fairly small number of variables: the mirror\u2019s radius of curvature, R; the focal length, f; the distances of the object and image from the mirror, doand di, respectively; and the heights of the object and image, hoand hi, respectively. The signs of these values indicate whether the image is inverted, erect (upright), real, or virtual. We now look at the equations that relate these variables and apply them to everyday problems. Figure 16.15 shows the meanings of most of the variables we will use for calculations involving curved mirrors. The basic equation that describes both lenses and mirrors is the lens/mirror equation Figure 16.15 Look for the variables, do, di, ho, hi,and fin this figure. This equation can be rearranged several ways. For example, it may be written to solve for focal length. Magnification, m, is the ratio of the size of the image, hi, to the size of the object, ho. The value of mcan be calculated in two ways. This relationship can be written to solve for any of the variables involved. For example, the height of the image is given by We saved the simplest equation for last. The radius of curvature of a curved mirror, R, is simply twice the focal length. We can learn important information from the algebraic sign of the result of a calculation using the previous equations: \u2022 A negative diindicates a virtual image; a positive value indicates a real image \u2022 A negative hiindicates an inverted image; a positive value indicates an erect image \u2022 For concave mirrors, fis positive; for convex mirrors, fis negative Now let\u2019s apply these equations to solve some problems. 486 Chapter 16 \u2022 Mirrors and Lenses WORKED EXAMPLE Calculating Focal Length A person standing 6.0 m from a convex security mirror forms a virtual image that appears to be 1.0 m behind the mirror. What is the focal length of the mirror? STRATEGY The person is the object, so do= 6.0 m. We know that, for this situation, dois positive. The image is virtual, so the value for the image distance is negative, so di= \u20131.0 m. Now, use the appropriate version of the lens/mirror equation to solve for focal length by substituting the known values. Solution Discussion The negative result is expected for a convex mirror. This indicates the focal point is behind the mirror. WORKED EXAMPLE Calculating Object Distance Electric room heaters use a concave mirror to reflect infrared (IR) radiation from hot coils. Note that IR radiation follows the same law of reflection as visible light. Given that the mirror has a radius of curvature of 50.0 cm and produces an image of the coils 3.00 m in front of the mirror, where are the coils with respect to the mirror? STRATEGY We are told that the concave mirror projects a real image of the coils at an image distance di= 3.00 m. The coils are the object, and we are asked to find their location\u2014that is, to find the object distance do. We are also given the radius of curvature of the mirror, so that its focal length is f= R/2 = 25.0 cm (a positive value, because the mirror is concave, or converging). We can use the lens/mirror equation to solve this problem. Solution Because diand fare known, the lens/mirror equation can be used to find do. Rearranging to solve for do, we have Entering the known quantities gives us 16.1 16.2 16.3 Discussion Note that the object (the coil filament) is farther from the mirror than the mirror\u2019s focal length. This is a case 1image (do > fand f positive), consistent with the fact that a real image is formed. You get the most concentrated thermal energy directly in front of the mirror and 3.00 m away from it. In general, this is not desirable because it could cause burns. Usually, you want the rays to emerge parallel, and this is accomplished by having the filament at the focal point of", " the mirror. Note that the filament here is not much farther from the mirror than the focal length, and that the image produced is considerably farther away. Access for free at openstax.org. 16.2 \u2022 Refraction 487 Practice Problems 1. A concave mirror has a radius of curvature of. What is the focal length of the mirror? a. b. c. d. 2. What is the focal length of a makeup mirror that produces a magnification of 1.50 when a person\u2019s face is 12.0 cm away? Construct a ray diagram using paper, a pencil and a ruler to confirm your calculation. a. \u201336.0 cm b. \u20137.20 cm c. d. 7.20 cm 36.0 cm Check Your Understanding 3. How does the object distance, do, compare with the focal length, f, for a concave mirror that produces an image that is real and inverted? a. do > f, where do and f are object distance and focal length, respectively. b. do < f, where do and f are object distance and focal length, respectively. c. do = f, where do and f are object distance and focal length, respectively. d. do = 0, where do is the object distance. 4. Use the law of reflection to explain why it is not a good idea to polish a mirror with sandpaper. a. The surface becomes smooth, and a smooth surface produces a sharp image. b. The surface becomes irregular, and an irregular surface produces a sharp image. c. The surface becomes smooth, and a smooth surface transmits light, but does not reflect it. d. The surface becomes irregular, and an irregular surface produces a blurred image. 5. An object is placed in front of a concave mirror at a distance that is greater than the focal length of the mirror. Will the image produced by the mirror be real or virtual? Will it be erect or inverted? a. b. c. d. It is real and erect. It is real and inverted. It is virtual and inverted. It is virtual and erect. 16.2 Refraction Section Learning Objectives By the end of this section, you will be able to do the following: \u2022 Explain refraction at media boundaries, predict the path of light after passing through a boundary (Snell\u2019s law), describe the index of refraction of materials, explain total internal reflection, and describe applications of refraction and total internal reflection \u2022 Perform calculations based on the law of refraction, Snell\u2019s law, and the conditions for total internal reflection Section Key Terms angle of refraction corner reflector critical angle dispersion incident ray index of refraction refracted ray Snell\u2019s law total internal reflection The Law of Refraction You may have noticed some odd optical phenomena when looking into a fish tank. For example, you may see the same fish appear to be in two different places (Figure 16.16). This is because light coming to you from the fish changes direction when it 488 Chapter 16 \u2022 Mirrors and Lenses leaves the tank and, in this case, light rays traveling along two different paths both reach our eyes. The changing of a light ray\u2019s direction (loosely called bending) when it passes a boundary between materials of different composition, or between layers in single material where there are changes in temperature and density, is called refraction. Refraction is responsible for a tremendous range of optical phenomena, from the action of lenses to voice transmission through optical fibers. Figure 16.16 Looking at the fish tank as shown, we can see the same fish in two different locations, because light changes directions when it passes from water to air. In this case, light rays traveling on two different paths change direction as they travel from water to air, and so reach the observer. Consequently, the fish appears to be in two different places. This bending of light is called refractionand is responsible for many optical phenomena. Why does light change direction when passing from one material (medium) to another? It is because light changes speed when going from one material to another. This behavior is typical of all waves and is especially easy to apply to light because light waves have very small wavelengths, and so they can be treated as rays. Before we study the law of refraction, it is useful to discuss the speed of light and how it varies between different media. The speed of light is now known to great precision. In fact, the speed of light in a vacuum, c, is so important, and is so precisely known, that it is accepted as one of the basic physical quantities, and has the fixed value where the approximate value of 3.00 matter is less than it is in a vacuum, because light interacts with atoms in a material. The speed of light depends strongly on the type of material, given that its interaction with different atoms, crystal lattices, and other substructures varies. We define the index of refraction, n,", " of a material to be 108 m/s is used whenever three-digit precision is sufficient. The speed of light through 16.4 where vis the observed speed of light in the material. Because the speed of light is always less than cin matter and equals conly in a vacuum, the index of refraction (plural: indices of refraction) is always greater than or equal to one. Table 16.2 lists the indices of refraction in various common materials. Medium n Gases at 0 \u00b0C and 1 atm Table 16.2 Indices of Refraction The table lists the indices of refraction for various materials that are transparent to light. Note, that light travels the slowest in the materials with the greatest indices of refraction. Access for free at openstax.org. 16.2 \u2022 Refraction 489 Medium n Air 1.000293 Carbon dioxide 1.00045 Hydrogen Oxygen Liquids at 20 \u00b0C Benzene Carbon disulfide 1.000139 1.000271 1.501 1.628 Carbon tetrachloride 1.461 Ethanol Glycerin Water, fresh Solids at 20 \u00b0C Diamond Fluorite Glass, crown Glass, flint Ice at 0 \u00b0C Plexiglas Polystyrene 1.361 1.473 1.333 2.419 1.434 1.52 1.66 1.309 1.51 1.49 Quartz, crystalline 1.544 Quartz, fused Sodium chloride 1.458 1.544 Table 16.2 Indices of Refraction The table lists the indices of refraction for various materials that are transparent to light. Note, that light travels the slowest in the materials with the greatest indices of refraction. 490 Chapter 16 \u2022 Mirrors and Lenses Medium n Zircon 1.923 Table 16.2 Indices of Refraction The table lists the indices of refraction for various materials that are transparent to light. Note, that light travels the slowest in the materials with the greatest indices of refraction. Figure 16.17 provides an analogy for and a description of how a ray of light changes direction when it passes from one medium to another. As in the previous section, the angles are measured relative to a perpendicular to the surface at the point where the light ray crosses it. The change in direction of the light ray depends on how the speed of light changes. The change in the speed of light is related to the indices of refraction of the media involved. In the situations shown in Figure 16.17, medium 2 has a greater index of refraction than medium 1. This difference in index of refraction means that the speed of light is less in medium 2 than in medium 1. Note that, in Figure 16.17(a), the path of the ray moves closer to the perpendicular when the ray slows down. Conversely, in Figure 16.17(b), the path of the ray moves away from the perpendicular when the ray speeds up. The path is exactly reversible. In both cases, you can imagine what happens by thinking about pushing a lawn mower from a footpath onto grass, and vice versa. Going from the footpath to grass, the right front wheel is slowed and pulled to the side as shown. This is the same change in direction for light when it goes from a fast medium to a slow one. When going from the grass to the footpath, the left front wheel moves faster than the others, and the mower changes direction as shown. This, too, is the same change in direction as light going from slow to fast. Figure 16.17 The change in direction of a light ray depends on how the speed of light changes when it crosses from one medium to another. For the situations shown here, the speed of light is greater in medium 1 than in medium 2. (a) A ray of light moves closer to the perpendicular when it slows down. This is analogous to what happens when a lawnmower goes from a footpath (medium 1) to grass (medium 2). (b) A ray of light moves away from the perpendicular when it speeds up. This is analogous to what happens when a lawnmower goes from grass (medium 2) to the footpath (medium 1). The paths are exactly reversible. Snap Lab Bent Pencil A classic observation of refraction occurs when a pencil is placed in a glass filled halfway with water. Do this and observe the shape of the pencil when you look at it sideways through air, glass, and water. \u2022 A full-length pencil \u2022 A glass half full of water Instructions Procedure 1. Place the pencil in the glass of water. 2. Observe the pencil from the side. Access for free at openstax.org. 16.2 \u2022 Refraction 491 3. Explain your observations. Virtual Physics Bending Light Click to view content (https://www.openstax.org/l/28Bendinglight) The Bending Light simulation in allows you to show light refracting as it crosses", " the boundaries between various media (download animation first to view). It also shows the reflected ray. You can move the protractor to the point where the light meets the boundary and measure the angle of incidence, the angle of refraction, and the angle of reflection. You can also insert a prism into the beam to view the spreading, or dispersion, of white light into colors, as discussed later in this section. Use the ray option at the upper left. A light ray moving upward strikes a horizontal boundary at an acute angle relative to the perpendicular and enters the medium above the boundary. What must be true for the light to bend away from the perpendicular? a. The medium below the boundary must have a greater index of refraction than the medium above. b. The medium below the boundary must have a lower index of refraction than the medium above. c. The medium below the boundary must have an index of refraction of zero. d. The medium above the boundary must have an infinite index of refraction. The amount that a light ray changes direction depends both on the incident angle and the amount that the speed changes. For a ray at a given incident angle, a large change in speed causes a large change in direction, and thus a large change in the angle of refraction. The exact mathematical relationship is the law of refraction, or Snell\u2019s law, which is stated in equation form as In terms of speeds, Snell\u2019s law becomes Here, n1 and n2 are the indices of refraction for media 1 and 2, respectively, and \u03b81 and \u03b82 are the angles between the rays and the perpendicular in the respective media 1 and 2, as shown in Figure 16.17. The incoming ray is called the incident rayand the outgoing ray is called the refracted ray. The associated angles are called the angle of incidenceand the angle of refraction. Later, we apply Snell\u2019s law to some practical situations. Dispersion is defined as the spreading of white light into the wavelengths of which it is composed. This happens because the index of refraction varies slightly with wavelength. Figure 16.18 shows how a prism disperses white light into the colors of the rainbow. 16.5 492 Chapter 16 \u2022 Mirrors and Lenses Figure 16.18 (a) A pure wavelength of light ( ) falls onto a prism and is refracted at both surfaces. (b) White light is dispersed by the prism (spread of light exaggerated). Because the index of refraction varies with wavelength, the angles of refraction vary with wavelength. A sequence of red to violet is produced, because the index of refraction increases steadily with decreasing wavelength. Rainbows are produced by a combination of refraction and reflection. You may have noticed that you see a rainbow only when you turn your back to the Sun. Light enters a drop of water and is reflected from the back of the drop, as shown in Figure 16.19. The light is refracted both as it enters and as it leaves the drop. Because the index of refraction of water varies with wavelength, the light is dispersed and a rainbow is observed. Figure 16.19 Part of the light falling on this water drop enters and is reflected from the back of the drop. This light is refracted and dispersed both as it enters and as it leaves the drop. WATCH PHYSICS Dispersion This video explains how refraction disperses white light into its composite colors. Click to view content (https://www.openstax.org/l/28Raindrop) Which colors of the rainbow bend most when refracted? a. Colors with a longer wavelength and higher frequency bend most when refracted. b. Colors with a shorter wavelength and higher frequency bend most when refracted. c. Colors with a shorter wavelength and lower frequency bend most when refracted. d. Colors with a longer wavelength and a lower frequency bend most when refracted. Access for free at openstax.org. 16.2 \u2022 Refraction 493 A good-quality mirror reflects more than 90 percent of the light that falls on it; the mirror absorbs the rest. But, it would be useful to have a mirror that reflects all the light that falls on it. Interestingly, we can produce total reflection using an aspect of refraction. Consider what happens when a ray of light strikes the surface between two materials, such as is shown in Figure 16.20(a). Part of the light crosses the boundary and is refracted; the rest is reflected. If, as shown in the figure, the index of refraction for the second medium is less than the first, the ray bends away from the perpendicular. Because n1 > n2, the angle of refraction is greater than the angle of incidence\u2014that is, increased. This causes The critical angle, refraction of 90\u00b0. That is, as shown in Figure 16.20(c), then all the light is reflected back into medium 1, a condition called total internal reflection.,", " which produces an angle of, is greater than the critical angle,, for a combination of two materials is defined to be the incident angle, is the incident angle for which =90\u00b0. If the incident angle, to increase as well. The largest the angle of refraction,. Now, imagine what happens as the incident angle is, can be is 90\u00b0, as shown in Figure 16.20(b). > Figure 16.20 (a) A ray of light crosses a boundary where the speed of light increases and the index of refraction decreases\u2014that is, n2 < n1. The refracted ray bends away from the perpendicular. (b) The critical angle,, is the one for which the angle of refraction is 90\u00b0. (c) Total internal reflection occurs when the incident angle is greater than the critical angle. Recall that Snell\u2019s law states the relationship between angles and indices of refraction. It is given by 16.6 When the incident angle equals the critical angle ( Snell\u2019s law in this case becomes = ), the angle of refraction is 90\u00b0 ( = 90\u00b0). Noting that sin 90\u00b0 = 1, 494 Chapter 16 \u2022 Mirrors and Lenses The critical angle,, for a given combination of materials is thus for n1 > n2. Total internal reflection occurs for any incident angle greater than the critical angle, medium has an index of refraction less than the first. Note that the previous equation is written for a light ray that travels in medium 1 and reflects from medium 2, as shown in Figure 16.20., and it can only occur when the second There are several important applications of total internal reflection. Total internal reflection, coupled with a large index of refraction, explains why diamonds sparkle more than other materials. The critical angle for a diamond-to-air surface is only 24.4\u00b0; so, when light enters a diamond, it has trouble getting back out (Figure 16.21). Although light freely enters the diamond at different angles, it can exit only if it makes an angle less than 24.4\u00b0 with the normal to a given surface. Facets on diamonds are specifically intended to make this unlikely, so that the light can exit only in certain places. Diamonds with very few impurities are very clear, so the light makes many internal reflections and is concentrated at the few places it can exit\u2014hence the sparkle. Figure 16.21 Light cannot escape a diamond easily because its critical angle with air is so small. Most reflections are total and the facets are placed so that light can exit only in particular ways, thus concentrating the light and making the diamond sparkle. A light ray that strikes an object that consists of two mutually perpendicular reflecting surfaces is reflected back exactly parallel to the direction from which it came. This parallel reflection is true whenever the reflecting surfaces are perpendicular, and it is independent of the angle of incidence. Such an object is called a corner reflectorbecause the light bounces from its inside corner. Many inexpensive reflector buttons on bicycles, cars, and warning signs have corner reflectors designed to return light in the direction from which it originates. Corner reflectors are perfectly efficient when the conditions for total internal reflection are satisfied. With common materials, it is easy to obtain a critical angle that is less than 45\u00b0. One use of these perfect mirrorsis in binoculars, as shown in Figure 16.22. Another application is for periscopes used in submarines. Figure 16.22 These binoculars use corner reflectors with total internal reflection to get light to the observer\u2019s eyes. Fiber optics are one common application of total internal reflection. In communications, fiber optics are used to transmit telephone, internet, and cable TV signals, and they use the transmission of light down fibers of plastic or glass. Because the Access for free at openstax.org. fibers are thin, light entering one is likely to strike the inside surface at an angle greater than the critical angle and, thus, be totally reflected (Figure 16.23). The index of refraction outside the fiber must be smaller than inside, a condition that is satisfied easily by coating the outside of the fiber with a material that has an appropriate refractive index. In fact, most fibers have a varying refractive index to allow more light to be guided along the fiber through total internal reflection. Rays are reflected around corners as shown in the figure, making the fibers into tiny light pipes. 16.2 \u2022 Refraction 495 Figure 16.23 (a) Fibers in bundles are clad by a material that has a lower index of refraction than the core to ensure total internal reflection, even when fibers are in contact with one another. A single fiber with its cladding is shown. (b) Light entering a thin fiber may strike the inside surface at large, or grazing, angles, and is completely reflected if these angles exceed the critical angle. Such rays continue down the fiber, even following it around corners, because the angles of", " reflection and incidence remain large. LINKS TO PHYSICS Medicine: Endoscopes A medical device called an endoscopeis shown in Figure 16.24. Figure 16.24 Endoscopes, such as the one drawn here, send light down a flexible fiber optic tube, which sends images back to a doctor in charge of performing a medical procedure. The word endoscope means looking inside. Doctors use endoscopes to look inside hollow organs in the human body and inside body cavities. These devices are used to diagnose internal physical problems. Images may be transmitted to an eyepiece or sent to a video screen. Another channel is sometimes included to allow the use of small surgical instruments. Such surgical procedures include collecting biopsies for later testing, and removing polyps and other growths. Identify the process that allows light and images to travel through a tube that is not straight. a. The process is refraction of light. b. The process is dispersion of light. c. The process is total internal reflection of light. d. The process is polarization of light. Calculations with the Law of Refraction The calculation problems that follow require application of the following equations: 16.8 496 Chapter 16 \u2022 Mirrors and Lenses and These are the equations for refractive index, the mathematical statement of the law of refraction (Snell\u2019s law), and the equation for the critical angle. WATCH PHYSICS Snell\u2019s Law Example 1 This video leads you through calculations based on the application of the equation that represents Snell\u2019s law. Click to view content (https://www.openstax.org/l/28Snellslaw) Which two types of variables are included in Snell\u2019s law? a. The two types of variables are density of a material and the angle made by the light ray with the normal. b. The two types of variables are density of a material and the thickness of a material. c. The two types of variables are refractive index and thickness of each material. d. The two types of variables are refractive index of a material and the angle made by a light ray with the normal. WORKED EXAMPLE Calculating Index of Refraction from Speed Calculate the index of refraction for a solid medium in which the speed of light is 2.012 substance, based on the previous table of indicies of refraction. STRATEGY We know the speed of light, c, is 3.00 of refraction, n. 108 m/s, and we are given v. We can simply plug these values into the equation for index 108 m/s, and identify the most likely Solution This value matches that of polystyrene exactly, according to the table of indices of refraction (Table 16.2). Discussion The three-digit approximation for cis used, which in this case is all that is needed. Many values in the table are only given to three significant figures. Note that the units for speed cancel to yield a dimensionless answer, which is correct. 16.9 WORKED EXAMPLE Calculating Index of Refraction from Angles Suppose you have an unknown, clear solid substance immersed in water and you wish to identify it by finding its index of refraction. You arrange to have a beam of light enter it at an angle of 45.00\u00b0, and you observe the angle of refraction to be 40.30\u00b0. What are the index of refraction of the substance and its likely identity? STRATEGY We must use the mathematical expression for the law of refraction to solve this problem because we are given angle data, not speed data. The subscripts 1 and 2 refer to values for water and the unknown, respectively, where 1 represents the medium from which the 16.10 Access for free at openstax.org. light is coming and 2 is the new medium it is entering. We are given the angle values, and the table of indicies of refraction gives us nfor water as 1.333. All we have to do before solving the problem is rearrange the equation 16.2 \u2022 Refraction 497 Solution 16.11 16.12 The best match from Table 16.2 is fused quartz, with n= 1.458. Discussion Note the relative sizes of the variables involved. For example, a larger angle has a larger sine value. This checks out for the two indicates the ray has bent towardnormal. This result is to angles involved. Note that the smaller value of compared with be expected if the unknown substance has a greater nvalue than that of water. The result shows that this is the case. WORKED EXAMPLE Calculating Critical Angle Verify that the critical angle for light going from water to air is 48.6\u00b0. (See Table 16.2, the table of indices of refraction.) STRATEGY First, choose the equation for critical angle Then, look up the nvalues for water, n1, and air, n2. Find the value", " of and it compare with the given angle of 48.6\u00b0.. Last, find the angle that has a sine equal to this value Solution For water, n1 = 1.333; for air, n2 = 1.0003. So, 16.13 16.14 Discussion Remember, when we try to find a critical angle, we look for the angle at which light can no longer escape past a medium boundary by refraction. It is logical, then, to think of subscript 1 as referring to the medium the light is trying to leave, and subscript 2 as where it is trying (unsuccessfully) to go. So water is 1 and air is 2. Practice Problems 6. The refractive index of ethanol is 1.36. What is the speed of light in ethanol? a. 2.25\u00d7108 m/s b. 2.21\u00d7107 m/s c. 2.25\u00d7109 m/s d. 2.21\u00d7108 m/s 7. The refractive index of air is and the refractive index of crystalline quartz is. What is the critical angle for a ray of light going from crystalline quartz into air? a. b. c. d. 498 Chapter 16 \u2022 Mirrors and Lenses Check Your Understanding 8. Which law is expressed by the equation? a. This is Ohm\u2019s law. b. This is Wien\u2019s displacement law. c. This is Snell\u2019s law. d. This is Newton\u2019s law. 9. Explain why the index of refraction is always greater than or equal to one. a. The formula for index of refraction,, of a material is where, so is always greater than one. b. The formula for index of refraction,, of a material is where, so is always greater than one. c. The formula for index of refraction,, of a material is than one. d. The formula for refractive index,, of a material is, so is always greater than one. 10. Write an equation that expresses the law of refraction. a. b. c. d. where,, so is always greater where 16.3 Lenses Section Learning Objectives By the end of this section, you will be able to do the following: \u2022 Describe and predict image formation and magnification as a consequence of refraction through convex and concave lenses, use ray diagrams to confirm image formation, and discuss how these properties of lenses determine their applications \u2022 Explain how the human eye works in terms of geometric optics \u2022 Perform calculations, based on the thin-lens equation, to determine image and object distances, focal length, and image magnification, and use these calculations to confirm values determined from ray diagrams Section Key Terms aberration chromatic aberration concave lens converging lens convex lens diverging lens eyepiece objective ocular parfocal Characteristics of Lenses Lenses are found in a huge array of optical instruments, ranging from a simple magnifying glass to the eye to a camera\u2019s zoom lens. In this section, we use the law of refraction to explore the properties of lenses and how they form images. Some of what we learned in the earlier discussion of curved mirrors also applies to the study of lenses. Concave, convex, focal point F, and focal length fhave the same meanings as before, except each measurement is made from the center of the lens instead of the surface of the mirror. The convex lens shown in Figure 16.25 has been shaped so that all light rays that enter it parallel to its central axis cross one another at a single point on the opposite side of the lens. The central axis, or axis, is defined to be a line normal to the lens at its center. Such a lens is called a converging lens because of the converging effect it has on light rays. An expanded view of the path of one ray through the lens is shown in Figure 16.25 to illustrate how the ray changes Access for free at openstax.org. direction both as it enters and as it leaves the lens. Because the index of refraction of the lens is greater than that of air, the ray moves toward the perpendicular as it enters and away from the perpendicular as it leaves. (This is in accordance with the law of refraction.) As a result of the shape of the lens, light is thus bent toward the axis at both surfaces. 16.3 \u2022 Lenses 499 Figure 16.25 Rays of light entering a convex, or converging, lens parallel to its axis converge at its focal point, F. Ray 2 lies on the axis of the lens. The distance from the center of the lens to the focal point is the focal length, \u0192, of the lens. An expanded view of the path taken by ray 1 shows the perpendiculars and the angles of incidence and refraction at both surfaces. Note that rays from a light source placed at the focal point of a", " converging lens emerge parallel from the other side of the lens. You may have heard of the trick of using a converging lens to focus rays of sunlight to a point. Such a concentration of light energy can produce enough heat to ignite paper. Figure 16.26 shows a concave lens and the effect it has on rays of light that enter it parallel to its axis (the path taken by ray 2 in the figure is the axis of the lens). The concave lens is a diverging lens because it causes the light rays to bend away (diverge) from its axis. In this case, the lens has been shaped so all light rays entering it parallel to its axis appear to originate from the same point, F, defined to be the focal point of a diverging lens. The distance from the center of the lens to the focal point is again called the focal length, or \u201c\u0192,\u201d of the lens. Note that the focal length of a diverging lens is defined to be negative. An expanded view of the path of one ray through the lens is shown in Figure 16.26 to illustrate how the shape of the lens, together with the law of refraction, causes the ray to follow its particular path and diverge. Figure 16.26 Rays of light enter a concave, or diverging, lens parallel to its axis diverge and thus appear to originate from its focal point, F. The dashed lines are not rays; they indicate the directions from which the rays appear to come. The focal length, \u0192, of a diverging lens is negative. An expanded view of the path taken by ray 1 shows the perpendiculars and the angles of incidence and refraction at both surfaces. The power, P, of a lens is very easy to calculate. It is simply the reciprocal of the focal length, expressed in meters 16.15 The units of power are diopters, D, which are expressed in reciprocal meters. If the focal length is negative, as it is for the diverging lens in Figure 16.26, then the power is also negative. In some circumstances, a lens forms an image at an obvious location, such as when a movie projector casts an image onto a screen. In other cases, the image location is less obvious. Where, for example, is the image formed by eyeglasses? We use ray 500 Chapter 16 \u2022 Mirrors and Lenses tracing for thin lenses to illustrate how they form images, and we develop equations to describe the image-formation quantitatively. These are the rules for ray tracing: 1. A ray entering a converging lens parallel to its axis passes through the focal point, F, of the lens on the other side 2. A ray entering a diverging lens parallel to its axis seems to come from the focal point, F, on the side of the entering ray 3. A ray passing through the center of either a converging or a diverging lens does not change direction 4. A ray entering a converging lens through its focal point exits parallel to its axis 5. A ray that enters a diverging lens by heading toward the focal point on the opposite side exits parallel to the axis Consider an object some distance away from a converging lens, as shown in Figure 16.27. To find the location and size of the image formed, we trace the paths of select light rays originating from one point on the object. In this example, the originating point is the top of a woman\u2019s head. Figure 16.27 shows three rays from the top of the object that can be traced using the raytracing rules just listed. Rays leave this point traveling in many directions, but we concentrate on only a few, which have paths that are easy to trace. The first ray is one that enters the lens parallel to its axis and passes through the focal point on the other side (rule 1). The second ray passes through the center of the lens without changing direction (rule 3). The third ray passes through the nearer focal point on its way into the lens and leaves the lens parallel to its axis (rule 4). All rays that come from the same point on the top of the person\u2019s head are refracted in such a way as to cross at the same point on the other side of the lens. The image of the top of the person\u2019s head is located at this point. Rays from another point on the object, such as the belt buckle, also cross at another common point, forming a complete image, as shown. Although three rays are traced in Figure 16.27, only two are necessary to locate the image. It is best to trace rays for which there are simple ray-tracing rules. Before applying ray tracing to other situations, let us consider the example shown in Figure 16.27 in more detail. Access for free at openstax.org. 16.3 \u2022 Lenses 501 Figure 16.27 Ray tracing is used to locate the image formed by a lens. Rays originating from the same point on the object", " are traced. The three chosen rays each follow one of the rules for ray tracing, so their paths are easy to determine. The image is located at the point where the rays cross. In this case, a real image\u2014one that can be projected on a screen\u2014is formed. The image formed in Figure 16.27 is a real image\u2014meaning, it can be projected. That is, light rays from one point on the object actually cross at the location of the image and can be projected onto a screen, a piece of film, or the retina of an eye. In Figure 16.27, the object distance, do, is greater than f.Now we consider a ray diagram for a convex lens where do<f, and another diagram for a concave lens. 502 Chapter 16 \u2022 Mirrors and Lenses Virtual Physics Geometric Optics Click to view content (https://www.openstax.org/l/28Geometric) This animation shows you how the image formed by a convex lens changes as you change object distance, curvature radius, refractive index, and diameter of the lens. To begin, choose Principal Rays in the upper left menu and then try varying some of the parameters indicated at the upper center. Show Help supplies a few helpful labels., change with an increasing radius of curvature? How does change with an increasing How does the focal length, refractive index? a. The focal length increases in both cases: when the radius of curvature and the refractive index increase. b. The focal length decreases in both cases: when the radius of curvature and the refractive index increase. c. The focal length increases when the radius of curvature increases; it decreases when the refractive index increases. d. The focal length decreases when the radius of curvature increases; it increases in when the refractive index increases. Type Formed When Image Type di M Case 1 fpositive, do> f Real Positive Negative m>, <, or = \u20121 Case 2 fpositive, do< f Virtual Negative Positive m> 1 Case 3 fnegative Virtual Negative Positive m< 1 Table 16.3 Three Types of Images Formed by Lenses The examples in Figure 16.27 and Figure 16.28 represent the three possible cases\u2014case 1, case 2, and case 3\u2014summarized in Table 16.3. In the table, mis magnification; the other symbols have the same meaning as they did for curved mirrors. Figure 16.28 (a) The image is virtual and larger than the object. (b) The image is virtual and smaller than the object. Access for free at openstax.org. 16.3 \u2022 Lenses 503 Snap Lab Focal Length \u2022 Temperature extremes\u2014Very hot or very cold temperatures are encountered in this lab that can cause burns. Use protective mitts, eyewear, and clothing when handling very hot or very cold objects. Notify your teacher immediately of any burns. \u2022 EYE SAFETY\u2014Looking at the Sun directly can cause permanent eye damage. Do not look at the Sun through any lens. \u2022 Several lenses \u2022 A sheet of white paper \u2022 A ruler or tape measure Instructions Procedure 1. Find several lenses and determine whether they are converging or diverging. In general, those that are thicker near the edges are diverging and those that are thicker near the center are converging. 2. On a bright, sunny day take the converging lenses outside and try focusing the sunlight onto a sheet of white paper. 3. Determine the focal lengths of the lenses. Have one partner slowly move the lens toward and away from the paper until you find the distance at which the light spot is at its brightest. Have the other partner measure the distance from the lens to the bright spot. Be careful, because the paper may start to burn, depending on the type of lens. True or false\u2014The bright spot that appears in focus on the paper is an image of the Sun. a. True b. False Image formation by lenses can also be calculated from simple equations. We learn how these calculations are carried out near the end of this section. Some common applications of lenses with which we are all familiar are magnifying glasses, eyeglasses, cameras, microscopes, and telescopes. We take a look at the latter two examples, which are the most complex. We have already seen the design of a telescope that uses only mirrors in. Figure 16.29 shows the design of a telescope that uses two lenses. Part (a) of the figure shows the design of the telescope used by Galileo. It produces an upright image, which is more convenient for many applications. Part (b) shows an arrangement of lenses used in many astronomical telescopes. This design produces an inverted image, which is less of a problem when viewing celestial objects. 504 Chapter 16 \u2022 Mirrors and Lenses Figure 16.29 (a) Galileo made telescopes with a convex objective and a concave eyepiece. They produce an upright image and are used in spy", "glasses. (b) Most simple telescopes have two convex lenses. The objective forms a case 1 image, which is the object for the eyepiece. The eyepiece forms a case 2 final image that is magnified. Figure 16.30 shows the path of light through a typical microscope. Microscopes were first developed during the early 1600s by eyeglass makers in the Netherlands and Denmark. The simplest compound microscope is constructed from two convex lenses, as shown schematically in Figure 16.30. The first lens is called the objective lens; it has typical magnification values from 5 to 100. In standard microscopes, the objectives are mounted such that when you switch between them, the sample remains in focus. Objectives arranged in this way are described as parfocal. The second lens, the eyepiece, also referred to as the ocular, has several lenses that slide inside a cylindrical barrel. The focusing ability is provided by the movement of both the objective lens and the eyepiece. The purpose of a microscope is to magnify small objects, and both lenses contribute to the final magnification. In addition, the final enlarged image is produced in a location far enough from the observer to be viewed easily because the eye cannot focus on objects or images that are too close. Access for free at openstax.org. 16.3 \u2022 Lenses 505 Figure 16.30 A compound microscope composed of two lenses, an objective and an eyepiece. The objective forms a case 1 image that is larger than the object. This first image is the object for the eyepiece. The eyepiece forms a case 2 final image that is magnified even further. Real lenses behave somewhat differently from how they are modeled using rays diagrams or the thin-lens equations. Real lenses produce aberrations. An aberration is a distortion in an image. There are a variety of aberrations that result from lens size, material, thickness, and the position of the object. One common type of aberration is chromatic aberration, which is related to color. Because the index of refraction of lenses depends on color, or wavelength, images are produced at different places and with different magnifications for different colors. The law of reflection is independent of wavelength, so mirrors do not have this problem. This result is another advantage for the use of mirrors in optical systems such as telescopes. Figure 16.31(a) shows chromatic aberration for a single convex lens, and its partial correction with a two-lens system. The index of refraction of the lens increases with decreasing wavelength, so violet rays are refracted more than red rays, and are thus focused closer to the lens. The diverging lens corrects this in part, although it is usually not possible to do so completely. Lenses made of different materials and with different dispersions may be used. For example, an achromatic doublet consisting of a converging lens made of crown glass in contact with a diverging lens made of flint glass can reduce chromatic aberration dramatically (Figure 16.31(b)). Figure 16.31 (a) Chromatic aberration is caused by the dependence of a lens\u2019s index of refraction on color (wavelength). The lens is more powerful for violet (V) than for red (R), producing images with different colors, locations, and magnifications. (b) Multiple-lens systems can correct chromatic aberrations in part, but they may require lenses of different materials and add to the expense of optical systems such as cameras. 506 Chapter 16 \u2022 Mirrors and Lenses Physics of the Eye The eye is perhaps the most interesting of all optical instruments. It is remarkable in how it forms images and in the richness of detail and color they eye can detect. However, our eyes commonly need some correction to reach what is called normalvision, but should be called idealvision instead. Image formation by our eyes and common vision correction are easy to analyze using geometric optics. Figure 16.32 shows the basic anatomy of the eye. The cornea and lens form a system that, to a good approximation, acts as a single thin lens. For clear vision, a real image must be projected onto the light-sensitive retina, which lies at a fixed distance from the lens. The lens of the eye adjusts its power to produce an image on the retina for objects at different distances. The center of the image falls on the fovea, which has the greatest density of light receptors and the greatest acuity (sharpness) in the visual field. There are no receptors at the place where the optic nerve meets the eye, which is called the blind spot. An image falling on this spot cannot be seen. The variable opening (or pupil) of the eye along with chemical adaptation allows the eye to detect light intensities from the lowest observable to 1010 times greater (without damage). Ten orders of magnitude is an incredible range of detection. Our eyes perform a vast number of", "ly while traveling from air into the cornea. The lens provides the remaining magnification needed to produce an image on the retina. The cornea and lens can be treated as a single thin lens, although the light rays pass through several layers of material (such as the cornea, aqueous humor, several layers in the lens, and vitreous humor), changing direction at each interface. The image formed is much like the one produced by a single convex lens. This result is a case 1 image. Images formed in the eye are inverted, but the brain inverts them once more to make them seem upright. Material Index of Refraction Water Air Cornea Aqueous humor 1.33 1.00 1.38 1.34 *The index of refraction varies throughout the lens and is greatest at its center. Table 16.4 Refractive Indices Relevant to the Eye Access for free at openstax.org. 16.3 \u2022 Lenses 507 Material Index of Refraction Lens 1.41 average* Vitreous humor 1.34 *The index of refraction varies throughout the lens and is greatest at its center. Table 16.4 Refractive Indices Relevant to the Eye Figure 16.33 An image is formed on the retina, with light rays converging most at the cornea and on entering and exiting the lens. Rays from the top and bottom of the object are traced and produce an inverted real image on the retina. The distance to the object is drawn smaller than scale. As noted, the image must fall precisely on the retina to produce clear vision\u2014that is, the image distance, di, must equal the lensto-retina distance. Because the lens-to-retina distance does not change, di must be the same for objects at all distances. The eye manages to vary the distance by varying the power (and focal length) of the lens to accommodate for objects at various distances. In Figure 16.33, you can see the small ciliary muscles above and below the lens that change the shape of the lens and, thus, the focal length. The need for some type of vision correction is very common. Common vision defects are easy to understand, and some are simple to correct. Figure 16.34 illustrates two common vision defects. Nearsightedness, or myopia, is the inability to see distant objects clearly while close objects are in focus. The nearsighted eye overconvergesthe nearly parallel rays from a distant object, and the rays cross in front of the retina. More divergent rays from a close object are converged on the retina, producing a clear image. Farsightedness, or hyperopia, is the inability to see close objects clearly whereas distant objects may be in focus. A farsighted eye does not converge rays from a close object sufficiently to make the rays meet on the retina. Less divergent rays from a distant object can be converged for a clear image. Figure 16.34 (a) The nearsighted (myopic) eye converges rays from a distant object in front of the retina; thus, they are diverging when they strike the retina, and produce a blurry image. This divergence can be caused by the lens of the eye being too powerful (in other words, too short a focal length) or the length of the eye being too great. (b) The farsighted (hyperopic) eye is unable to converge the rays from a close object by the time they strike the retina and produce... blurry close vision. This poor convergence can be caused by insufficient power (in 508 Chapter 16 \u2022 Mirrors and Lenses other words, too long a focal length) in the lens or by the eye being too short. Because the nearsighted eye overconverges light rays, the correction for nearsightedness involves placing a diverging spectacle lens in front of the eye. This lens reduces the power of an eye that has too short a focal length (Figure 16.35(a)). Because the farsighted eye underconvergeslight rays, the correction for farsightedness is to place a converging spectacle lens in front of the eye. This lens increases the power of an eye that has too long a focal length (Figure 16.35(b)). Figure 16.35 (a) Correction of nearsightedness requires a diverging lens that compensates for the overconvergence by the eye. The diverging lens produces an image closer to the eye than the object so that the nearsighted person can see it clearly. (b) Correction of farsightedness uses a converging lens that compensates for the underconvergence by the eye. The converging lens produces an image farther from the eye than the object so that the farsighted person can see it clearly. In both (a) and (b), the rays that meet at the retina represent corrected vision, and the other rays represent blurred vision without corrective lenses. Calculations Using Lens Equ", "ations As promised, there are no new equations to memorize. We can use equations already presented for solving problems involving curved mirrors. Careful analysis allows you to apply these equations to lenses. Here are the equations you need where Pis power, expressed in reciprocal meters (m\u20131) rather than diopters (D), and fis focal length, expressed in meters (m). You also need where, as before, do and di are object distance and image distance, respectively. Remember, this equation is usually more useful if rearranged to solve for one of the variables. For example, The equations for magnification, m, are also the same as for mirrors where hi and ho are the image height and object height, respectively. Remember, also, that a negative di value indicates a virtual image and a negative hi value indicates an inverted image. These are the steps to follow when solving a lens problem: \u2022 Step 1. Examine the situation to determine that image formation by a lens is involved. \u2022 Step 2. Determine whether ray tracing, the thin-lens equations, or both should be used. A sketch is very helpful even if ray tracing is not specifically required by the problem. Write useful symbols and values on the sketch. \u2022 Step 3. Identify exactly what needs to be determined in the problem (identify the unknowns). \u2022 Step 4. Make a list of what is given or can be inferred from the problem as stated (identify the knowns). It is helpful to determine whether the situation involves a case 1, 2, or 3 image. Although these are just names for types of images, they have certain characteristics (given in Table 16.3) that can be of great use in solving problems. Access for free at openstax.org. 16.3 \u2022 Lenses 509 \u2022 Step 5. If ray tracing is required, use the ray-tracing rules listed earlier in this section. \u2022 Step 6. Most quantitative problems require the use of the thin-lens equations. These equations are solved in the usual manner by substituting knowns and solving for unknowns. Several worked examples were included earlier and can serve as guides. \u2022 Step 7. Check whether the answer is reasonable. Does it make sense?If you identified the type of image (case 1, 2, or 3) correctly, you should assess whether your answer is consistent with the type of image, magnification, and so on. All problems will be solved by one or more of the equations just presented, with ray tracing used only for general analysis of the problem. The steps then simplify to the following: Identify the unknown. Identify the knowns. 1. 2. 3. Choose an equation, plug in the knowns, and solve for the unknown. Here are some worked examples: WORKED EXAMPLE The Power of a Magnifying Glass Strategy The Sun is so far away that its rays are nearly parallel when they reach Earth. The magnifying glass is a convex (or converging) lens, focusing the nearly parallel rays of sunlight. Thus, the focal length of the lens is the distance from the lens to the spot, and its power, in diopters (D), is the inverse of this distance (in reciprocal meters). Solution The focal length of the lens is the distance from the center of the lens to the spot, which we know to be 8.00 cm. Thus, 16.16 To find the power of the lens, we must first convert the focal length to meters; then, we substitute this value into the equation for power. Discussion This result demonstrates a relatively powerful lens. Remember that the power of a lens in diopters should not be confused with the familiar concept of power in watts. 16.17 WORKED EXAMPLE Image Formation by a Convex Lens A clear glass light bulb is placed 0.75 m from a convex lens with a 0.50 m focal length, as shown in Figure 16.36. Use ray tracing to get an approximate location for the image. Then, use the mirror/lens equations to calculate (a) the location of the image and (b) its magnification. Verify that ray tracing and the thin-lens and magnification equations produce consistent results. Figure 16.36 A light bulb placed 0.75 m from a lens with a 0.50 m focal length produces a real image on a poster board, as discussed in the previous example. Ray tracing predicts the image location and size. 510 Chapter 16 \u2022 Mirrors and Lenses Strategy Because the object is placed farther away from a converging lens than the focal length of the lens, this situation is analogous to the one illustrated in the previous figure of a series of drawings showing a woman standing to the left of a lens. Ray tracing to scale should produce similar results for di. Numerical solutions for di and mcan be obtained using the thin-lens and magnification equations, noting that do = 0.75 m and f= 0.50 m. Solution The ray tracing to scale", " in Figure 16.36 shows two rays from a point on the bulb\u2019s filament crossing about 1.50 m on the far side of the lens. Thus, the image distance, di, is about 1.50 m. Similarly, the image height based on ray tracing is greater than the object height by about a factor of two, and the image is inverted. Thus, mis about \u20132. The minus sign indicates the image is inverted. The lens equation can be rearranged to solve for di from the given information. Now, we use to find m. 16.18 16.19 Discussion Note that the minus sign causes the magnification to be negative when the image is inverted. Ray tracing and the use of the lens equation produce consistent results. The thin-lens equation gives the most precise results, and is limited only by the accuracy of the given information. Ray tracing is limited by the accuracy with which you draw, but it is highly useful both conceptually and visually. WORKED EXAMPLE Image Formation by a Concave Lens Suppose an object, such as a book page, is held 6.50 cm from a concave lens with a focal length of \u201310.0 cm. Such a lens could be used in eyeglasses to correct pronounced nearsightedness. What magnification is produced? Strategy This example is identical to the preceding one, except that the focal length is negative for a concave or diverging lens. The method of solution is therefore the same, but the results are different in important ways. Solution 16.20 Now the magnification equation can be used to find the magnification, m, because both di and do are known. Entering their values gives 16.21 Discussion A number of results in this example are true of all case 3 images. Magnification is positive (as calculated), meaning the image is upright. The magnification is also less than one, meaning the image is smaller than the object\u2014in this case, a little more than half its size. The image distance is negative, meaning the image is on the same side of the lens as the object. The image is virtual. The image is closer to the lens than the object, because the image distance is smaller in magnitude than the object distance. The location of the image is not obvious when you look through a concave lens. In fact, because the image is smaller than the object, you may think it is farther away; however, the image is closer than the object\u2014a fact that is useful in correcting nearsightedness. Access for free at openstax.org. 16.3 \u2022 Lenses 511 WATCH PHYSICS The Lens Equation and Problem Solving The video shows calculations for both concave and convex lenses. It also explains real versus virtual images, erect versus inverted images, and the significance of negative and positive signs for the involved variables. Click to view content (https://www.openstax.org/l/28Lenses) If a lens has a magnification of a. The image is erect and is half as tall as the object. b. The image is erect and twice as tall as the object. c. The image is inverted and is half as tall as the object. d. The image is inverted and is twice as tall as the object., how does the image compare with the object in height and orientation? Practice Problems 11. A lens has a focal length of. What is the power of the lens? a. The power of the lens is b. The power of the lens is c. The power of the lens is d. The power of the lens is.... 12. If a lens produces a 5.00 -cm tall image of an 8.00 -cm -high object when placed 10.0 cm from the lens, what is the apparent image distance? Construct a ray diagram using paper, a pencil, and a ruler to confirm your calculation. a. \u22123.12 cm b. \u22126.25 cm 3.12 cm c. d. 6.25 cm Check Your Understanding 13. A lens has a magnification that is negative. What is the orientation of the image? a. Negative magnification means the image is erect and real. b. Negative magnification means the image is erect and virtual. c. Negative magnification means the image is inverted and virtual. d. Negative magnification means the image is inverted and real. 14. Which part of the eye controls the amount of light that enters? a. b. c. d. the pupil the iris the cornea the retina 15. An object is placed between the focal point and a convex lens. Describe the image that is formed in terms of its orientation, and whether the image is real or virtual. a. The image is real and erect. b. The image is real and inverted. c. The image is virtual and erect. d. The image is virtual and inverted. 16. A farsighted person buys a pair of glasses to correct her farsightedness. Describe", " the main symptom of farsightedness and the type of lens that corrects it. a. Farsighted people cannot focus on objects that are far away, but they can see nearby objects easily. A convex lens is used to correct this. b. Farsighted people cannot focus on objects that are close up, but they can see far-off objects easily. A concave lens is used to correct this. 512 Chapter 16 \u2022 Mirrors and Lenses c. Farsighted people cannot focus on objects that are close up, but they can see distant objects easily. A convex lens is used to correct this. d. Farsighted people cannot focus on objects that are either close up or far away. A concave lens is used to correct this. Access for free at openstax.org. Chapter 16 \u2022 Key Terms 513 KEY TERMS aberration a distortion in an image produced by a lens angle of incidence the angle, with respect to the normal, at which a ray meets a boundary between media or a reflective surface focal point the point at which rays converge or appear to converge incident ray the incoming ray toward a medium boundary or a reflective surface angle of reflection the angle, with respect to the normal, at index of refraction the speed of light in a vacuum divided which a ray leaves a reflective surface by the speed of light in a given material angle of refraction the angle between the normal and the law of reflection the law that indicates the angle of refracted ray reflection equals the angle of incidence central axis a line perpendicular to the center of a lens or law of refraction the law that describes the relationship mirror extending in both directions chromatic aberration an aberration related to color concave lens a lens that causes light rays to diverge from the central axis concave mirror a mirror with a reflective side that is curved inward converging lens a convex lens convex lens a lens that causes light rays to converge toward the central axis between refractive indices of materials on both sides of a boundary and the change in the path of light crossing the boundary, as given by the equation n1 sin = n2 sin ray light traveling in a straight line real image an optical image formed when light rays converge and pass through the image, producing an image that can be projected onto a screen refracted ray the light ray after it has been refracted Snell\u2019s law the law of refraction expressed mathematically convex mirror a mirror with a reflective side that is curved as outward critical angle an incident angle that produces an angle of refraction of 90\u00b0 dispersion separation of white light into its component wavelengths diverging lens a concave lens focal length the distance from the focal point to the mirror SECTION SUMMARY 16.1 Reflection \u2022 The angle of reflection equals the angle of incidence. \u2022 Plane mirrors and convex mirrors reflect virtual, erect images. Concave mirrors reflect light to form real, inverted images or virtual, erect images, depending on the location of the object. Image distance, height, and other characteristics can be calculated using the lens/mirror equation and the magnification equation. \u2022 16.2 Refraction \u2022 The index of refraction for a material is given by the speed of light in a vacuum divided by the speed of light in that material. \u2022 Snell\u2019s law states the relationship between indices of KEY EQUATIONS 16.1 Reflection lens/mirror equation (reciprocal version) total internal reflection reflection of light traveling through a medium with a large refractive index at a boundary of a medium with a low refractive index under conditions such that refraction cannot occur virtual image the point from which light rays appear to diverge without actually doing so refraction, the incident angle, and the angle of refraction. \u2022 The critical angle,, determines whether total internal refraction can take place, and can be calculated according to. 16.3 Lenses \u2022 The characteristics of images formed by concave and convex lenses can be predicted using ray tracing. Characteristics include real versus virtual, inverted versus upright, and size. \u2022 The human eye and corrective lenses can be explained using geometric optics. \u2022 Characteristics of images formed by lenses can be calculated using the mirror/lens equation. lens/mirror equation (solved version) 514 Chapter 16 \u2022 Chapter Review magnification equation critical angle radius/focal length equation R= 2f 16.2 Refraction index of Refraction Snell\u2019s law >Snell\u2019s law in terms of speed CHAPTER REVIEW Concept Items 16.1 Reflection 1. Part A. Can you see a virtual image? Part B. Can you photograph one? Explain your answers. a. A. yes; B. No, an image from a flat mirror cannot be photographed. b. A. no; B. Yes, an image from a flat mirror can be photographed. c. A. yes; B. Yes, an image from a flat mirror can be photographed. d. A. no; B.", " No, an image from a flat mirror cannot be photographed. 2. State the law of reflection. a. b. c. d., where is the angle of incidence., where is the angle of incidence., where is the angle of incidence., where is the angle of reflection and is the angle of reflection and is the angle of reflection and is the angle of reflection. 16.2 Refraction 3. Does light change direction toward or away from the normal when it goes from air to water? Explain. a. The light bends away from the normal because the index of refraction of water is greater than that of air. b. The light bends away from the normal because the index of refraction of air is greater than that of water. Access for free at openstax.org. 16.3 Lenses power and focal length mirror/lens (or thin-lens) equation rearranged mirror/lens equation magnification equation c. The light bends toward the normal because the index of refraction of water is greater than that of air. d. The light bends toward the normal because the index of refraction of air is greater than that of water. 16.3 Lenses 4. An object is positioned in front of a lens with its base resting on the principal axis. Describe two rays that could be traced from the top of the object and through the lens that would locate the top of an image. a. A ray perpendicular to the axis and a ray through the center of the lens b. A ray parallel to the axis and a ray that does not pass through the center of the lens c. A ray parallel to the axis and a ray through the center of the lens d. A ray parallel to the axis and a ray that does not pass through the focal point 5. A person timing the moonrise looks at her watch and then at the rising moon. Describe what happened inside her eyes that allowed her to see her watch clearly one second and then see the moon clearly. a. The shape of the lens was changed by the sclera, and thus its focal length was also changed, so that each of the images focused on the retina. b. The shape of the lens was changed by the choroid, and thus its focal length was also changed, so that each of the images focused on the retina. c. The shape of the lens was changed by the iris, and thus its focal length was also changed, so that each of the images focused on the retina. d. The shape of the lens was changed by the muscles, and thus its focal length was also changed, so that each of the images focused on the retina. 6. For a concave lens, if the image distance, di, is negative, where does the image appear to be with respect to the object? a. The image always appears on the same side of the Critical Thinking Items 16.1 Reflection 7. Why are diverging mirrors often used for rear-view mirrors in vehicles? What is the main disadvantage of using such a mirror compared with a flat one? a. It gives a wide range of view. The image appears to be closer than the actual object. It gives a narrow range of view. The image appears to be farther than the actual object. It gives a narrow range of view. The image appears to be closer than the actual object. It gives a wide range of view. The image appears to be farther than the actual object. b. c. d. 16.2 Refraction 8. A high-quality diamond may be quite clear and colorless, transmitting all visible wavelengths with little absorption. Explain how it can sparkle with flashes of brilliant color when illuminated by white light. a. Diamond and air have a small difference in their refractive indices that results in a very small critical angle. The light that enters a diamond may exit at only a few points, and these points sparkle because many rays have been directed toward them. b. Diamond and air have a small difference in their refractive indices that results in a very large critical angle. The light that enters a diamond may exit at only a few points, and these points sparkle because many rays have been directed toward them. c. Diamond has a high index of refraction with respect to air, which results in a very small critical angle. The light that enters a diamond may exit at only a few points, and these points sparkle because many rays have been directed toward them. d. Diamond has a high index of refraction with respect to air, which results in a very large critical angle. The light that enters a diamond may exit at only a few points, and these points sparkle because many rays have been directed toward them. 9. The most common type of mirage is an illusion in which light from far-away objects is reflected by a pool of water that is not really there. Mirages are generally observed in Chapter 16 \u2022 Chapter Review 515 lens. b. The image appears on the opposite side of", " the lens. c. The image appears on the opposite side of the lens only if the object distance is greater than the focal length. d. The image appears on the same side of the lens only if the object distance is less than the focal length. deserts, where there is a hot layer of air near the ground. Given that the refractive index of air is less for air at higher temperatures, explain how mirages can be formed. a. The hot layer of air near the ground is lighter than the cooler air above it, but the difference in refractive index is small, which results in a large critical angle. The light rays coming from the horizon strike the hot air at large angles, so they are reflected as they would be from water. b. The hot layer of air near the ground is lighter than the cooler air above it, and the difference in refractive index is large, which results in a large critical angle. The light rays coming from the horizon strike the hot air at large angles, so they are reflected as they would be from water. c. The hot layer of air near the ground is lighter than the cooler air above it, but the difference in refractive index is small, which results in a small critical angle. The light rays coming from the horizon strike the hot air at large angles, so they are reflected as they would be from water. d. The hot layer of air near the ground is lighter than the cooler air above it, and the difference in the refractive index is large, which results in a small critical angle. The light rays coming from the horizontal strike the hot air at large angles, so they are reflected as they would be from water. 16.3 Lenses 10. When you focus a camera, you adjust the distance of the lens from the film. If the camera lens acts like a thin lens, why can it not be kept at a fixed distance from the film for both near and distant objects? a. To focus on a distant object, you need to increase the image distance. b. To focus on a distant object, you need to increase the focal length of the lens. c. To focus on a distant object, you need to decrease the focal length of the lens. d. To focus on a distant object, you may need to increase or decrease the focal length of the lens. 11. Part A\u2014How do the refractive indices of the cornea, 516 Chapter 16 \u2022 Chapter Review aqueous humor, and the lens of the eye compare with the refractive index of air? Part B\u2014How do the comparisons in part A explain how images are focused on the retina? a. (A) The cornea, aqueous humor, and lens of the eye have smaller refractive indices than air. (B) Rays entering the eye are refracted away from the central axis, which causes them to meet at the focal point on the retina. (A) The cornea, aqueous humor, and lens of the eye have greater refractive indices than air. (B) Rays entering the eye are refracted away from b. Problems 16.1 Reflection 12. Some telephoto cameras use a mirror rather than a lens. What radius of curvature is needed for a concave mirror to replace a 0.800 -m focal-length telephoto lens? a. 0.400 m 1.60 m b. c. 4.00 m 16.0 m d. 13. What is the focal length of a makeup mirror that produces a magnification of 2.00 when a person\u2019s face is 8.00 cm away? a. \u201316 cm b. \u20135.3 cm c. d. 5.3 cm 16 cm c. d. the central axis, which causes them to meet at the focal point on the retina. (A) The cornea, aqueous humor, and lens of the eye have smaller refractive indices than air. (B) Rays entering the eye are refracted toward the central axis, which causes them to meet at the focal point on the retina. (A) The cornea, aqueous humor, and lens of the eye have greater refractive indices than air. (B) Rays entering the eye are refracted toward the central axis, which causes them to meet at the focal point on the retina. Calculate the amount the ray is displaced by the glass (\u0394x), given that the incident angle is 40.0\u00b0 and the glass is 1.00 cm thick. a. 0.839 cm b. 0.619 cm c. 0.466 cm d. 0.373 cm 16.2 Refraction 14. An optical fiber uses flint glass (n= 1.66) clad with crown 16.3 Lenses glass (n= 1.52). What is the critical angle? a. 33.2\u00b0 b. 23.7\u00b0 c. 0.92 rad 1.16 rad d. 15. Suppose this figure represents a", " ray of light going from air (n= 1.0003) through crown glass (n= 1.52) into water, similar to a beam of light going into a fish tank. 16. A camera\u2019s zoom lens has an adjustable focal length ranging from 80.0 to 200 mm. What is its range of powers? a. The lowest power is 0.05 D and the highest power is 0.125 D. b. The lowest power is 0.08 D and the highest power is 0.20 D. c. The lowest power is 5.00 D and the highest power is 12.5 D. d. The lowest power is 80 D and the highest power is 200 D. 17. Suppose a telephoto lens with a focal length of 200 mm is being used to photograph mountains 10.0 km away. (a) Where is the image? (b) What is the height of the image of a 1,000-m-high cliff on one of the mountains? a. (a) The image is 0.200 m on the same side of the lens. (b) The height of the image is \u2013 2.00 cm. (a) The image is 0.200 m on the opposite side of the Access for free at openstax.org. b. Chapter 16 \u2022 Test Prep 517 c. lens. (b) The height of the image is \u2013 2.00 cm. (a) The image is 0.200 m on the opposite side of the lens. (b) The height of the image is +2.00 cm. d. (a) The image is 0.100 m on the same side of the lens. (b) The height of the image is +2.00 cm. Performance Task 16.3 Lenses 18. In this performance task, you will investigate the lens- like properties of a clear bottle. \u2022 a water bottle or glass with a round cross-section and smooth, vertical sides \u2022 enough water to fill the bottle \u2022 a meter stick or tape measure \u2022 a bright light source with a small bulb, such as a pen light \u2022 a small bright object, such as a silver spoon. Instructions TEST PREP Multiple Choice 16.1 Reflection 19. In geometric optics, a straight line emerging from a point is called a (an) ________. a. b. c. d. object distance ray focal point image 20. An image of a 2.0 -cm object reflected from a mirror is 5.0 cm tall. What is the magnification of the mirror? a. 0.4 b. 2.5 3 c. 10 d. 21. Can a virtual image be projected onto a screen with additional lenses or mirrors? Explain your answer. a. Yes, the rays actually meet behind the lens or mirror. b. No, the image is formed by rays that converge to a point in front of the mirror or lens. Procedure 1. Look through a clear glass or plastic bottle and describe what you see. 2. Next, fill the bottle with water and describe what you see. 3. Use the water bottle as a lens to produce the image of a bright object. 4. Estimate the focal length of the water bottle lens. a. How can you find the focal length of the lens using the light and a blank wall? b. How can you find the focal length of the lens using the bright object? c. Why did the water change the lens properties of the bottle? b. c. d. the refractive index the speed of light in a vacuum the speed of light in a transparent material 23. What is the term for the minimum angle at which a light ray is reflected back into a material and cannot pass into the surrounding medium? critical angle a. b. incident angle c. angle of refraction d. angle of reflection 24. Consider these indices of refraction: glass: 1.52, air: 1.0003, water: 1.333. Put these materials in order from the one in which the speed of light is fastest to the one in which it is slowest. a. The speed of light in water > the speed of light in air > the speed of light in glass. b. The speed of light in glass > the speed of light in water > the speed of light in air. c. The speed of light in air > the speed of light in water > the speed of light in glass. d. The speed of light in glass > the speed of light in air c. Yes, any image that can be seen can be manipulated > the speed of light in water. so that it can be projected onto a screen. d. No, the image can only be perceived as being behind the lens or mirror. 16.2 Refraction 22. What does crepresent in the equation? a. the critical angle 25. Explain why an object in water always appears to be at a depth that is more shallow than it actually is. a. Because of the", " refraction of light, the light coming from the object bends toward the normal at the interface of water and air. This causes the object to appear at a location that is above the actual position of the object. Hence, the image appears to 518 Chapter 16 \u2022 Test Prep be at a depth that is more shallow than the actual depth. b. Because of the refraction of light, the light coming from the object bends away from the normal at the interface of water and air. This causes the object to appear at a location that is above the actual position of the object. Hence, the image appears to be at a depth that is more shallow than the actual depth. c. Because of the refraction of light, the light coming from the object bends toward the normal at the interface of water and air. This causes the object to appear at a location that is below the actual position of the object. Hence, the image appears to be at a depth that is more shallow than the actual depth. d. Because of the refraction of light, the light coming from the object bends away from the normal at the interface of water and air. This causes the object to appear at a location that is below the actual position of the object. Hence, the image appears to be at a depth that is more shallow than the actual depth. 16.3 Lenses 26. For a given lens, what is the height of the image divided by the height of the object ( ) equal to? a. power focal length b. c. magnification d. radius of curvature Short Answer 16.1 Reflection 30. Distinguish between reflection and refraction in terms of how a light ray changes when it meets the interface between two media. a. Reflected light penetrates the surface whereas refracted light is bentas it travels from one medium to the other. b. Reflected light penetrates the surface whereas refracted light travels along a curved path. c. Reflected light bouncesfrom the surface whereas refracted light travels along a curved path. d. Reflected light bouncesfrom the surface whereas refracted light is bentas it travels from one medium to the other. 31. Sometimes light may be both reflected and refracted as it meets the surface of a different medium. Identify a material with a surface that when light travels through Access for free at openstax.org. 27. Which part of the eye has the greatest density of light receptors? a. b. c. d. the lens the fovea the optic nerve the vitreous humor 28. What is the power of a lens with a focal length of 10 cm? a. b. c. d. 10 m\u20131, or 10 D 10 cm\u20131, or 10 D 10 m, or 10 D 10 cm, or 10 D 29. Describe the cause of chromatic aberration. a. Chromatic aberration results from the dependence of the frequency of light on the refractive index, which causes dispersion of different colors of light by a lens so that each color has a different focal point. b. Chromatic aberration results from the dispersion of different wavelengths of light by a curved mirror so that each color has a different focal point. c. Chromatic aberration results from the dependence of the reflection angle at a spherical mirror\u2019s surface on the distance of light rays from the principal axis so that different colors have different focal points. d. Chromatic aberration results from the dependence of the wavelength of light on the refractive index, which causes dispersion of different colors of light by a lens so that each color has a different focal point. the air it is both reflected and refracted. Explain how this is possible. a. Light passing through air is partially reflected and refracted when it meets a glass surface. It is reflected because glass has a smooth surface; it is refracted while passing into the transparent glass. b. Light passing through air is partially reflected and refracted when it meets a glass surface. It is reflected because glass has a rough surface, and it is refracted while passing into the opaque glass. c. Light passing through air is partially reflected and refracted when it meets a glass surface. It is reflected because glass has a smooth surface; it is refracted while passing into the opaque glass. d. Light passing through air is partially reflected and refracted when it meets a glass surface. It is reflected because glass has a rough surface; it is refracted while passing into the transparent glass. 32. A concave mirror has a focal length of 5.00 cm. What is the image distance of an object placed 7.00 cm from the center of the mirror? a. \u221217.5 cm b. \u22122.92 cm c. 2.92 cm 17.5 cm d. 33. An 8.0 -cm tall object is placed 6.0 cm from a concave mirror with a magnification of \u20132.0. What are the image height and the image distance? a. hi = \u2013", " 16 cm, di = \u2013 12 cm b. hi = \u2013 16 cm, di = 12 cm c. hi = 16 cm, di = \u2013 12 cm d. hi = 16 cm, di = 12 cm 16.2 Refraction 34. At what minimum angle does total internal reflection of toward light occur if it travels from water ice? a. b. c. d. 35. Water floats on a liquid called carbon tetrachloride. The two liquids do not mix. A light ray passing from water into carbon tetrachloride has an incident angle of 45.0\u00b0 and an angle of refraction of 40.1\u00b0. If the index of refraction of water is 1.33, what is the index of refraction of carbon tetrachloride? a. b. c. d. 1.60 1.49 1.21 1.46 36. Describe what happens to a light ray when it is refracted. Include in your explanation comparison of angles, comparison of refractive indices, and the term normal. a. When a ray of light goes from one medium to another medium with a different refractive index, the ray changes its path as a result of interference. The angle between the ray and the normal (the line perpendicular to the surfaces of the two media) is greater in the medium with the greater refractive index. b. When a ray of light goes from one medium to another medium with a different refractive index, the ray changes its path as a result of refraction. The angle between the ray and the normal (the line perpendicular to the surfaces of the two media) is less in the medium with the greater refractive index. Chapter 16 \u2022 Test Prep 519 another medium with a different refractive index, the ray does not change its path. The angle between the ray and the normal (the line parallel to the surfaces of the two media) is the same in both media. d. When a ray of light goes from one medium to another medium with a different refractive index, the ray changes its path as a result of refraction. The angle between the ray and the normal (the line perpendicular to the surfaces of the two media) is less in the medium with the lower refractive index. 16.3 Lenses 37. What are two equivalent terms for a lens that always causes light rays to bend away from the principal axis? a. a diverging lens or a convex lens b. a diverging lens or a concave lens c. a converging lens or a concave lens d. a converging lens or a convex lens 38. Define the term virtual image. a. A virtual image is an image that cannot be projected onto a screen. b. A virtual image is an image that can be projected onto a screen. c. A virtual image is an image that is formed on the opposite side of the lens from where the object is placed. d. A virtual image is an image that is always bigger than the object. 39. Compare nearsightedness (myopia) and farsightedness (hyperopia) in terms of focal point. a. The eyes of a nearsighted person have focal points beyond the retina. A farsighted person has eyes with focal points between the lens and the retina. b. A nearsighted person has eyes with focal points between the lens and the retina. A farsighted person has eyes with focal points beyond the retina. c. A nearsighted person has eyes with focal points between the lens and the choroid. A farsighted person has eyes with focal points beyond the choroid. d. A nearsighted person has eyes with focal points between the lens and the retina. A farsighted person has eyes with focal points on the retina. 40. Explain how a converging lens corrects farsightedness. a. A converging lens disperses the rays so they focus on the retina. b. A converging lens bends the rays closer together so they do not focus on the retina. c. A converging lens bends the rays closer together so c. When a ray of light goes from one medium to they focus on the retina. 520 Chapter 16 \u2022 Test Prep d. A converging lens disperses the rays so they do not 42. What is the magnification of a lens if it produces a focus on the retina. 41. Solve the equation for in such a way that it is not expressed as a reciprocal. 12-cm-high image of a 4 -cm -high object? The image is virtual and erect. a. b. a. b. c. d. c. d. Extended Response 16.1 Reflection 43. The diagram shows a lightbulb between two mirrors. One mirror produces a beam of light with parallel rays; the other keeps light from escaping without being put into the beam. What angle does the ray make from the instructor\u2019s face with the normal to the water (n=", " 1.33) at the point where the ray enters? Assume n= 1.00 for air. a. 68\u00b0 b. 25\u00b0 19\u00b0 c. 34\u00b0 d. 46. Describe total internal reflection. Include a definition of the critical angle and how it is related to total internal reflection. Also, compare the indices of refraction of the interior material and the surrounding material. a. When the interior material has a smaller index of refraction than the surrounding material, the incident ray may approach the boundary at an angle (called the critical angle) such that the refraction angle is 90\u00b0. The refracted ray cannot leave the interior, so it is reflected back inside and total internal reflection occurs. b. When the interior material has a smaller index of refraction than the surrounding material, the incident ray may approach the boundary at an angle (called the critical angle) such that the refraction angle is less than 90\u00b0. The refracted ray cannot leave the interior, so it is reflected back inside and total internal reflection occurs. c. When the interior material has the same index of refraction as the surrounding material, the incident ray approaches the boundary at an angle (called the critical angle) such that the refraction Where is the light source in relation to the focal point or radius of curvature of each mirror? Explain your answer. a. The bulb is at the center of curvature of the small mirror and at the focal point of the large mirror. b. The bulb is at the focal point of the small mirror and at the focal point of the large mirror. c. The bulb is at the center of curvature of the small mirror and at the center of curvature of the large mirror. d. The bulb is at the focal point of the small mirror and at the center of curvature of the large mirror. in front of a mirror that has. What is the radius of curvature 44. An object is placed a magnification of of the mirror? a. b. c. d. 16.2 Refraction 45. A scuba diver training in a pool looks at his instructor, as shown in this figure. The angle between the ray in the water and the normal to the water is 25\u00b0. Access for free at openstax.org. angle is less than 90\u00b0. The refracted ray cannot leave the interior, so it is reflected back inside and total internal reflection occurs. d. When the interior material has a greater index of refraction than the surrounding material, the incident ray may approach the boundary at an angle (called the critical angle) such that the refraction angle is 90\u00b0. The refracted ray cannot leave the interior, so it is reflected back inside and total internal reflection occurs. 16.3 Lenses 47. The muscles that change the shape of the lens in the eyes have become weak, causing vision problems for a person. In particular, the muscles cannot pull hard enough on the edges of the lens to make it less convex. Part A\u2014What condition does inability cause? Part B\u2014Where are images focused with respect to the retina? Part C\u2014Which type of lens corrects this person\u2019s problem? Explain. a. Part A\u2014This condition causes hyperopia. Part B\u2014Images are focused between the lens and the retina. Chapter 16 \u2022 Test Prep 521 Part C\u2014A converging lens gathers the rays slightly so they focus onto the retina. b. Part A\u2014This condition causes myopia. Part B\u2014Images are focused between the lens and the retina. Part C\u2014A converging lens gathers the rays slightly so they focus onto the retina. c. Part A\u2014This condition causes hyperopia. Part B\u2014Images are focused between the lens and the retina. Part C\u2014A diverging lens spreads the rays slightly so they focus onto the retina. d. Part A\u2014This condition causes myopia. Part B\u2014Images are focused between the lens and the retina. Part C\u2014A diverging lens spreads the rays slightly so they focus onto the retina. 48. If the lens-to-retina distance is, what is the power of the eye when viewing an object a. b. c. d. away? 522 Chapter 16 \u2022 Test Prep Access for free at openstax.org. CHAPTER 17 Diffraction and Interference Figure 17.1 The colors reflected by this compact disc vary with angle and are not caused by pigments. Colors such as these are direct evidence of the wave character of light. (credit: Reggie Mathalone) Chapter Outline 17.1 Understanding Diffraction and Interference 17.2 Applications of Diffraction, Interference, and Coherence Examine a compact disc under white light, noting the colors observed and their locations on the disc. Using INTRODUCTION the CD, explore the spectra of a few light sources, such as a candle flame, an incandescent bulb, and fluorescent light. If you have ever looked at the reds, blues, and greens in a", " sunlit soap bubble and wondered how straw-colored soapy water could produce them, you have hit upon one of the many phenomena that can only be explained by the wave character of light. That and other interesting phenomena, such as the dispersion of white light into a rainbow of colors when passed through a narrow slit, cannot be explained fully by geometric optics. In such cases, light interacts with small objects and exhibits its wave characteristics. The topic of this chapter is the branch of optics that considers the behavior of light when it exhibits wave characteristics. 17.1 Understanding Diffraction and Interference Section Learning Objectives By the end of this section, you will be able to do the following: \u2022 Explain wave behavior of light, including diffraction and interference, including the role of constructive and destructive interference in Young\u2019s single-slit and double-slit experiments \u2022 Perform calculations involving diffraction and interference, in particular the wavelength of light using data from a two-slit interference pattern Section Key Terms diffraction Huygens\u2019s principle monochromatic wavefront 524 Chapter 17 \u2022 Diffraction and Interference Diffraction and Interference We know that visible light is the type of electromagnetic wave to which our eyes responds. As we have seen previously, light obeys the equation m/s is the speed of light in vacuum, fis the frequency of the electromagnetic wave in Hz (or s\u20131), and where is its wavelength in m. The range of visible wavelengths is approximately 380 to 750 nm. As is true for all waves, light travels in straight lines and acts like a ray when it interacts with objects several times as large as its wavelength. However, when it interacts with smaller objects, it displays its wave characteristics prominently. Interference is the identifying behavior of a wave. In Figure 17.2, both the ray and wave characteristics of light can be seen. The laser beam emitted by the observatory represents ray behavior, as it travels in a straight line. Passing a pure, one-wavelength beam through vertical slits with a width close to the wavelength of the beam reveals the wave character of light. Here we see the beam spreading out horizontally into a pattern of bright and dark regions that are caused by systematic constructive and destructive interference. As it is characteristic of wave behavior, interference is observed for water waves, sound waves, and light waves. Figure 17.2 (a) The light beam emitted by a laser at the Paranal Observatory (part of the European Southern Observatory in Chile) acts like a ray, traveling in a straight line. (credit: Yuri Beletsky, European Southern Observatory) (b) A laser beam passing through a grid of vertical slits produces an interference pattern\u2014characteristic of a wave. (credit: Shim\u2019on and Slava Rybka, Wikimedia Commons) That interference is a characteristic of energy propagation by waves is demonstrated more convincingly by water waves. Figure 17.3 shows water waves passing through gaps between some rocks. You can easily see that the gaps are similar in width to the wavelength of the waves and that this causes an interference pattern as the waves pass beyond the gaps. A cross-section across the waves in the foreground would show the crests and troughs characteristic of an interference pattern. Figure 17.3 Incoming waves (at the top of the picture) pass through the gaps in the rocks and create an interference pattern (in the foreground). Light has wave characteristics in various media as well as in a vacuum. When light goes from a vacuum to some medium, such as water, its speed and wavelength change, but its frequency, f, remains the same. The speed of light in a medium is where nis its index of refraction. If you divide both sides of the equation is the wavelength in a medium, and by n, you get. Therefore,, where, Access for free at openstax.org. 17.1 \u2022 Understanding Diffraction and Interference 525 where is the wavelength in vacuum and nis the medium\u2019s index of refraction. It follows that the wavelength of light is smaller in any medium than it is in vacuum. In water, for example, which has n= 1.333, the range of visible wavelengths is (380 nm)/1.333 to (760 nm)/1.333, or not, since colors are associated with frequency. 285\u2013570 nm. Although wavelengths change while traveling from one medium to another, colors do The Dutch scientist Christiaan Huygens (1629\u20131695) developed a useful technique for determining in detail how and where waves propagate. He used wavefronts, which are the points on a wave\u2019s surface that share the same, constant phase (such as all the points that make up the crest of a water wave). Huygens\u2019s principle states, \u201cEvery point on a wavefront is a source of wavelets that spread out in the forward direction at the same speed as the wave itself. The new wavefront is", " a line tangent to all of the wavelets.\u201d Figure 17.4 shows how Huygens\u2019s principle is applied. A wavefront is the long edge that moves; for example, the crest or the trough. Each point on the wavefront emits a semicircular wave that moves at the propagation speed v. These are drawn later at a time, t, so that they have moved a distance. The new wavefront is a line tangent to the wavelets and is where the wave is located at time t. Huygens\u2019s principle works for all types of waves, including water waves, sound waves, and light waves. It will be useful not only in describing how light waves propagate, but also in how they interfere. Figure 17.4 Huygens\u2019s principle applied to a straight wavefront. Each point on the wavefront emits a semicircular wavelet that moves a distance. The new wavefront is a line tangent to the wavelets. What happens when a wave passes through an opening, such as light shining through an open door into a dark room? For light, you expect to see a sharp shadow of the doorway on the floor of the room, and you expect no light to bend around corners into other parts of the room. When sound passes through a door, you hear it everywhere in the room and, thus, you understand that sound spreads out when passing through such an opening. What is the difference between the behavior of sound waves and light waves in this case? The answer is that the wavelengths that make up the light are very short, so that the light acts like a ray. Sound has wavelengths on the order of the size of the door, and so it bends around corners. If light passes through smaller openings, often called slits, you can use Huygens\u2019s principle to show that light bends as sound does (see Figure 17.5). The bending of a wave around the edges of an opening or an obstacle is called diffraction. Diffraction is a 526 Chapter 17 \u2022 Diffraction and Interference wave characteristic that occurs for all types of waves. If diffraction is observed for a phenomenon, it is evidence that the phenomenon is produced by waves. Thus, the horizontal diffraction of the laser beam after it passes through slits in Figure 17.2 is evidence that light has the properties of a wave. Figure 17.5 Huygens\u2019s principle applied to a straight wavefront striking an opening. The edges of the wavefront bend after passing through the opening, a process called diffraction. The amount of bending is more extreme for a small opening, consistent with the fact that wave characteristics are most noticeable for interactions with objects about the same size as the wavelength. Once again, water waves present a familiar example of a wave phenomenon that is easy to observe and understand, as shown in Figure 17.6. Figure 17.6 Ocean waves pass through an opening in a reef, resulting in a diffraction pattern. Diffraction occurs because the opening is similar in width to the wavelength of the waves. WATCH PHYSICS Single-Slit Interference This video works through the math needed to predict diffraction patterns that are caused by single-slit interference. Click to view content (https://www.openstax.org/l/28slit) Which values of mdenote the location of destructive interference in a single-slit diffraction pattern? a. whole integers, excluding zero b. whole integers c. d. real numbers excluding zero real numbers The fact that Huygens\u2019s principle worked was not considered enough evidence to prove that light is a wave. People were also reluctant to accept light\u2019s wave nature because it contradicted the ideas of Isaac Newton, who was still held in high esteem. The acceptance of the wave character of light came after 1801, when the English physicist and physician Thomas Young (1773\u20131829) did his now-classic double-slit experiment (see Figure 17.7). Access for free at openstax.org. 17.1 \u2022 Understanding Diffraction and Interference 527 Figure 17.7 Young\u2019s double-slit experiment. Here, light of a single wavelength passes through a pair of vertical slits and produces a diffraction pattern on the screen\u2014numerous vertical light and dark lines that are spread out horizontally. Without diffraction and interference, the light would simply make two lines on the screen. When light passes through narrow slits, it is diffracted into semicircular waves, as shown in Figure 17.8 (a). Pure constructive interference occurs where the waves line up crest to crest or trough to trough. Pure destructive interference occurs where they line up crest to trough. The light must fall on a screen and be scattered into our eyes for the pattern to be visible. An analogous pattern for water waves is shown in Figure 17.8 (b). Note that regions of constructive and destructive interference move", " out from the slits at well-defined angles to the original beam. Those angles depend on wavelength and the distance between the slits, as you will see below. Figure 17.8 Double slits produce two sources of waves that interfere. (a) Light spreads out (diffracts) from each slit, because the slits are narrow. The waves overlap and interfere constructively (bright lines) and destructively (dark regions). You can only see the effect if the light falls onto a screen and is scattered into your eyes. (b) The double-slit interference pattern for water waves is nearly identical to that for light. Wave action is greatest in regions of constructive interference and least in regions of destructive interference. (c) When light that has passed through double slits falls on a screen, we see a pattern such as this. Virtual Physics Wave Interference Click to view content (https://www.openstax.org/l/28interference) This simulation demonstrates most of the wave phenomena discussed in this section. First, observe interference between two sources of electromagnetic radiation without adding slits. See how water waves, sound, and light all show interference patterns. Stay with light waves and use only one source. Create diffraction patterns with one slit and then with two. You may have to adjust slit width to see the pattern. 528 Chapter 17 \u2022 Diffraction and Interference Visually compare the slit width to the wavelength. When do you get the best-defined diffraction pattern? a. when the slit width is larger than the wavelength b. when the slit width is smaller than the wavelength c. when the slit width is comparable to the wavelength d. when the slit width is infinite Calculations Involving Diffraction and Interference The fact that the wavelength of light of one color, or monochromatic light, can be calculated from its two-slit diffraction pattern in Young\u2019s experiments supports the conclusion that light has wave properties. To understand the basis of such calculations, consider how two waves travel from the slits to the screen. Each slit is a different distance from a given point on the screen. Thus different numbers of wavelengths fit into each path. Waves start out from the slits in phase (crest to crest), but they will end up out of phase (crest to trough) at the screen if the paths differ in length by half a wavelength, interfering destructively. If the paths differ by a whole wavelength, then the waves arrive in phase (crest to crest) at the screen, interfering constructively. More, then generally, if the paths taken by the two waves differ by any half-integral number of wavelengths destructive interference occurs. Similarly, if the paths taken by the two waves differ by any integral number of wavelengths, then constructive interference occurs. Figure 17.9 shows how to determine the path-length difference for waves traveling from two slits to a common point on a screen. If the screen is a large distance away compared with the distance between the slits, then the angle between the path and a line from the slits perpendicular to the screen (see the figure) is nearly the same for each path. That approximation and simple trigonometry show the length difference,, where dis the distance between the slits,, to be To obtain constructive interference for a double slit, the path-length difference must be an integral multiple of the wavelength, or Similarly, to obtain destructive interference for a double slit, the path-length difference must be a half-integral multiple of the wavelength, or The number mis the orderof the interference. For example, m= 4 is fourth-order interference. Figure 17.9 The paths from each slit to a common point on the screen differ by an amount, assuming the distance to the screen is much greater than the distance between the slits (not to scale here). Figure 17.10 shows how the intensity of the bands of constructive interference decreases with increasing angle. Access for free at openstax.org. 17.1 \u2022 Understanding Diffraction and Interference 529 Figure 17.10 The interference pattern for a double slit has an intensity that falls off with angle. The photograph shows multiple bright and dark lines, or fringes, formed by light passing through a double slit. Light passing through a single slit forms a diffraction pattern somewhat different from that formed by double slits. Figure 17.11 shows a single-slit diffraction pattern. Note that the central maximum is larger than those on either side, and that the intensity decreases rapidly on either side. Figure 17.11 (a) Single-slit diffraction pattern. Monochromatic light passing through a single slit produces a central maximum and many smaller and dimmer maxima on either side. The central maximum is six times higher than shown. (b) The drawing shows the bright central maximum and dimmer and thinner maxima on either side. (c) The location of the minima are shown in terms of and D. The analysis of single-", "slit diffraction is illustrated in Figure 17.12. Assuming the screen is very far away compared with the size of the slit, rays heading toward a common destination are nearly parallel. That approximation allows a series of trigonometric operations that result in the equations for the minima produced by destructive interference. or relative to the original direction of the beam, each ray travels a different distance to the screen, and they can arrive in or farther than the ray from the top edge of the slit, they arrive out When rays travel straight ahead, they remain in phase and a central maximum is obtained. However, when rays travel at an angle out of phase. Thus, a ray from the center travels a distance of phase, and they interfere destructively. Similarly, for every ray between the top and the center of the slit, there is a ray between the center and the bottom of the slit that travels a distance interferes destructively. Symmetrically, there will be another minimum at the same angle below the direct ray. farther to the common point on the screen, and so 530 Chapter 17 \u2022 Diffraction and Interference Figure 17.12 Equations for a single-slit diffraction pattern, where \u03bbis the wavelength of light, Dis the slit width, is the angle between a line from the slit to a minimum and a line perpendicular to the screen, Lis the distance from the slit to the screen, yis the distance from the center of the pattern to the minimum, and mis a nonzero integer indicating the order of the minimum. Below we summarize the equations needed for the calculations to follow. The speed of light in a vacuum, c, the wavelength of the light,, and its frequency, f, are related as follows. The wavelength of light in a medium,, compared to its wavelength in a vacuum,, is given by To calculate the positions of constructive interference for a double slit, the path-length difference must be an integral multiple, m, of the wavelength. 17.1 where dis the distance between the slits and is the angle between a line from the slits to the maximum and a line perpendicular to the barrier in which the slits are located. To calculate the positions of destructive interference for a double slit, the path-length difference must be a half-integral multiple of the wavelength: For a single-slit diffraction pattern, the width of the slit, D, the distance of the first (m= 1) destructive interference minimum, y, the distance from the slit to the screen, L, and the wavelength,, are given by Also, for single-slit diffraction, is the angle between a line from the slit to the minimum and a line perpendicular to the screen, and mis the order of where the minimum. WORKED EXAMPLE Two-Slit Interference Suppose you pass light from a He-Ne laser through two slits separated by 0.0100 mm, and you find that the third bright line on a screen is formed at an angle of 10.95\u00ba relative to the incident beam. What is the wavelength of the light? STRATEGY The third bright line is due to third-order constructive interference, which means that m= 3. You are given d= 0.0100 mm and = 10.95\u00ba. The wavelength can thus be found using the equation for constructive interference. Access for free at openstax.org. 17.1 \u2022 Understanding Diffraction and Interference 531 Solution The equation is. Solving for the wavelength,, gives Substituting known values yields 17.2 17.3 Discussion To three digits, 633 nm is the wavelength of light emitted by the common He-Ne laser. Not by coincidence, this red color is similar to that emitted by neon lights. More important, however, is the fact that interference patterns can be used to measure wavelength. Young did that for visible wavelengths. His analytical technique is still widely used to measure electromagnetic spectra. For a given order, the angle for constructive interference increases with, so spectra (measurements of intensity versus wavelength) can be obtained. WORKED EXAMPLE Single-Slit Diffraction Visible light of wavelength 550 nm falls on a single slit and produces its second diffraction minimum at an angle of 45.0\u00b0 relative to the incident direction of the light. What is the width of the slit? STRATEGY From the given information, and assuming the screen is far away from the slit, you can use the equation D. to find Solution Quantities given are = 550 nm, m= 2, and gives = 45.0\u00b0. Solving the equation for Dand substituting known values Discussion You see that the slit is narrow (it is only a few times greater than the wavelength of light). That is consistent with the fact that light must interact with an object comparable in size to its wavelength in order to exhibit significant wave effects, such as this single-slit diffraction pattern. Practice Problems 1. Monochromatic light from a laser", " passes through two slits separated by. The third bright line on a screen is 17.4 relative to the incident beam. What is the wavelength of the light? formed at an angle of a. b. c. d. 2. What is the width of a single slit through which 610-nm orange light passes to form a first diffraction minimum at an angle of 30.0\u00b0? a. 0.863 \u00b5m b. 0.704 \u00b5m c. 0.610 \u00b5m d. 1.22 \u00b5m 532 Chapter 17 \u2022 Diffraction and Interference Check Your Understanding 3. Which aspect of a beam of monochromatic light changes when it passes from a vacuum into water, and how does it change? a. The wavelength first decreases and then increases. b. The wavelength first increases and then decreases. c. The wavelength increases. d. The wavelength decreases. 4. Go outside in the sunlight and observe your shadow. It has fuzzy edges, even if you do not. Is this a diffraction effect? Explain. a. This is a diffraction effect. Your whole body acts as the origin for a new wavefront. b. This is a diffraction effect. Every point on the edge of your shadow acts as the origin for a new wavefront. c. This is a refraction effect. Your whole body acts as the origin for a new wavefront. d. This is a refraction effect. Every point on the edge of your shadow acts as the origin for a new wavefront. 5. Which aspect of monochromatic green light changes when it passes from a vacuum into diamond, and how does it change? a. The wavelength first decreases and then increases. b. The wavelength first increases and then decreases. c. The wavelength increases. d. The wavelength decreases. 17.2 Applications of Diffraction, Interference, and Coherence Section Learning Objectives By the end of this section, you will be able to do the following: \u2022 Explain behaviors of waves, including reflection, refraction, diffraction, interference, and coherence, and describe applications based on these behaviors \u2022 Perform calculations related to applications based on wave properties of light Section Key Terms differential interference contrast (DIC) diffraction grating iridescence laser monochromator Rayleigh criterion resolution Wave-Based Applications of Light In 1917, Albert Einstein was thinking about photons and excited atoms. He considered an atom excited by a certain amount of energy and what would happen if that atom were hit by a photon with the same amount of energy. He suggested that the atom would emit a photon with that amount of energy, and it would be accompanied by the original photon. The exciting part is that you would have twophotons with the same energy andthey would be in phase. Those photons could go on to hit other excited atoms, and soon you would have a stream of in-phase photons. Such a light stream is said to be coherent. Some four decades later, Einstein\u2019s idea found application in a process called, light amplification by stimulated emission of radiation. Take the first letters of all the words (except byand \u201cof\u201d) and write them in order. You get the word laser(see (a)), which is the name of the device that produces such a beam of light. Laser beams are directional, very intense, and narrow (only about 0.5 mm in diameter). These properties lead to a number of applications in industry and medicine. The following are just a few examples: \u2022 This chapter began with a picture of a compact disc (see ). Those audio and data-storage devices began replacing cassette tapes during the 1990s. CDs are read by interpreting variations in reflections of a laser beam from the surface. \u2022 Some barcode scanners use a laser beam. \u2022 Lasers are used in industry to cut steel and other metals. \u2022 Lasers are bounced off reflectors that astronauts left on the Moon. The time it takes for the light to make the round trip can be used to make precise calculations of the Earth-Moon distance. \u2022 Laser beams are used to produce holograms. The name hologram means entire picture(from the Greek holo-, as in Access for free at openstax.org. 17.2 \u2022 Applications of Diffraction, Interference, and Coherence 533 holistic), because the image is three-dimensional. A viewer can move around the image and see it from different perspectives. Holograms take advantage of the wave properties of light, as opposed to traditional photography which is based on geometric optics. A holographic image is produced by constructive and destructive interference of a split laser beam. \u2022 One of the advantages of using a laser as a surgical tool is that it is accompanied by very little bleeding. \u2022 Laser eye surgery has improved the vision of many people, without the need for corrective lenses. A laser beam is used to change the shape of the lens of the eye, thus changing its focal length. Virtual Physics Lasers Click to view content (https://www", ".openstax.org/l/28lasers) This animation allows you to examine the workings of a laser. First view the picture of a real laser. Change the energy of the incoming photons, and see if you can match it to an excitation level that will produce pairs of coherent photons. Change the excitation level and try to match it to the incoming photon energy. In the animation there is only one excited atom. Is that the case for a real laser? Explain. a. No, a laser would have two excited atoms. b. No, a laser would have several million excited atoms. c. Yes, a laser would have only one excited atom. d. No, a laser would have on the order of 1023 excited atoms. An interesting thing happens if you pass light through a large number of evenly-spaced parallel slits. Such an arrangement of slits is called a diffraction grating. An interference pattern is created that is very similar to the one formed by double-slit diffraction (see and ). A diffraction grating can be manufactured by scratching glass with a sharp tool to form a number of precisely positioned parallel lines, which act like slits. Diffraction gratings work both for transmission of light, as in Figure 17.13, and for reflection of light, as on the butterfly wings or the Australian opal shown in Figure 17.14, or the CD pictured in the opening illustration of this chapter. In addition to their use as novelty items, diffraction gratings are commonly used for spectroscopic dispersion and analysis of light. What makes them particularly useful is the fact that they form a sharper pattern than do double slits. That is, their bright regions are narrower and brighter, while their dark regions are darker. Figure 17.15 shows idealized graphs demonstrating the sharper pattern. Natural diffraction gratings occur in the feathers of certain birds. Tiny, fingerlike structures in regular patterns act as reflection gratings, producing constructive interference that gives the feathers colors not solely due to their pigmentation. The effect is called iridescence. Figure 17.13 A diffraction grating consists of a large number of evenly-spaced parallel slits. (a) Light passing through the grating is diffracted in a pattern similar to a double slit, with bright regions at various angles. (b) The pattern obtained for white light incident on a grating. The central maximum is white, and the higher-order maxima disperse white light into a rainbow of colors. 534 Chapter 17 \u2022 Diffraction and Interference Figure 17.14 (a) This Australian opal and (b) the butterfly wings have rows of reflectors that act like reflection gratings, reflecting different colors at different angles. (credit: (a) Opals-On-Black.com, via Flickr (b) whologwhy, Flickr) Figure 17.15 Idealized graphs of the intensity of light passing through a double slit (a) and a diffraction grating (b) for monochromatic light. Maxima can be produced at the same angles, but those for the diffraction grating are narrower, and hence sharper. The maxima become narrower and the regions between become darker as the number of slits is increased. Snap Lab Diffraction Grating \u2022 A CD (compact disc) or DVD \u2022 A measuring tape \u2022 Sunlight near a white wall Instructions Procedure 1. Hold the CD in direct sunlight near the wall, and move it around until a circular rainbow pattern appears on the wall. 2. Measure the distance from the CD to the wall and the distance from the center of the circular pattern to a color in the rainbow. Use those two distances to calculate. Find.. 3. Look up the wavelength of the color you chose. That is 4. Solve 5. Compare your answer to the usual spacing between CD tracks, which is 1,600 nm (1.6 \u03bcm). for d. How do you know what number to use for m? a. Count the rainbow rings preceding the chosen color. Access for free at openstax.org. 17.2 \u2022 Applications of Diffraction, Interference, and Coherence 535 b. Calculate mfrom the frequency of the light of the chosen color. c. Calculate mfrom the wavelength of the light of the chosen color. d. The value of mis fixed for every color. FUN IN PHYSICS CD Players Can you see the grooves on a CD or DVD (see Figure 17.16)? You may think you can because you know they are there, but they are extremely narrow\u20141,600 in a millimeter. Because the width of the grooves is similar to wavelengths of visible light, they form a diffraction grating. That is why you see rainbows on a CD. The colors are attractive, but they are incidental to the functions of storing and retrieving audio and other data. Figure 17.16 For its size, this CD holds a surprising amount", " of information. Likewise, the CD player it is in houses a surprising number of electronic devices. The grooves are actually one continuous groove that spirals outward from the center. Data are recorded in the grooves as binary code (zeroes and ones) in small pits. Information in the pits is detected by a laser that tracks along the groove. It gets even more complicated: The speed of rotation must be varied as the laser tracks toward the circumference so that the linear speed along the groove remains constant. There is also an error correction mechanism to prevent the laser beam from getting off track. A diffraction grating is used to create the first two maxima on either side of the track. If those maxima are not the same distance from the track, an error is indicated and then corrected. The pits are reflective because they have been coated with a thin layer of aluminum. That allows the laser beam to be reflected back and directed toward a photodiode detector. The signal can then be processed and converted to the audio we hear. The longest wavelength of visible light is about 780 nm. How does that compare to the distance between CD grooves? a. The grooves are about 3 times the longest wavelength of visible light. b. The grooves are about 2 times the longest wavelength of visible light. c. The grooves are about 2 times the shortest wavelength of visible light. d. The grooves are about 3 times the shortest wavelength of visible light. LINKS TO PHYSICS Biology: DIC Microscopy If you were completely transparent, it would be hard to recognize you from your photograph. The same problem arises when using a traditional microscope to view or photograph small transparent objects such as cells and microbes. Microscopes using differential interference contrast (DIC) solve the problem by making it possible to view microscopic objects with enhanced contrast, as shown in Figure 17.17. 536 Chapter 17 \u2022 Diffraction and Interference Figure 17.17 This aquatic organism was photographed with a DIC microscope. (credit: Public Library of Science) A DIC microscope separates a polarized light source into two beams polarized at right angles to each other and coherent with each other, that is, in phase. After passing through the sample, the beams are recombined and realigned so they have the same plane of polarization. They then create an interference pattern caused by the differences in their optical path and the refractive indices of the parts of the sample they passed through. The result is an image with contrast and shadowing that could not be observed with traditional optics. Where are diffraction gratings used? Diffraction gratings are key components of monochromators\u2014devices that separate the various wavelengths of incoming light and allow a beam with only a specific wavelength to pass through. Monochromators are used, for example, in optical imaging of particular wavelengths from biological or medical samples. A diffraction grating can be chosen to specifically analyze a wavelength of light emitted by molecules in diseased cells in a biopsy sample, or to help excite strategic molecules in the sample with a selected frequency of light. Another important use is in optical fiber technologies where fibers are designed to provide optimum performance at specific wavelengths. A range of diffraction gratings is available for selecting specific wavelengths for such use. Diffraction gratings are used in spectroscopes to separate a light source into its component wavelengths. When a material is heated to incandescence, it gives off wavelengths of light characteristic of the chemical makeup of the material. A pure substance will produce a spectrum that is unique, thus allowing identification of the substance. Spectroscopes are also used to measure wavelengths both shorter and longer than visible light. Such instruments have become especially useful to astronomers and chemists. Figure 17.18 shows a diagram of a spectroscope. Figure 17.18 The diagram shows the function of a diffraction grating in a spectroscope. Light diffracts as it moves through space, bending around obstacles and interfering constructively and destructively. While diffraction allows light to be used as a spectroscopic tool, it also limits the detail we can obtain in images. Figure 17.19 (a) shows the effect of passing light through a small circular aperture. Instead of a bright spot with sharp edges, a spot with a fuzzy edge surrounded by circles of light is obtained. This pattern is caused by diffraction similar to that produced by a single slit. Light from different parts of the circular aperture interferes constructively and destructively. The effect is most noticeable when the aperture is small, but the effect is there for large apertures, too. Access for free at openstax.org. 17.2 \u2022 Applications of Diffraction, Interference, and Coherence 537 Figure 17.19 (a) Monochromatic light passed through a small circular aperture produces this diffraction pattern. (b) Two point light sources that are close to one another produce overlapping images because of diffraction. (c) If they are closer together, they cannot be resolved, that", " is, distinguished. How does diffraction affect the detail that can be observed when light passes through an aperture? Figure 17.19 (b) shows the diffraction pattern produced by two point light sources that are close to one another. The pattern is similar to that for a single point source, and it is just barely possible to tell that there are two light sources rather than one. If they are closer together, as in Figure 17.19 (c), you cannot distinguish them, thus limiting the detail, or resolution, you can obtain. That limit is an inescapable consequence of the wave nature of light. There are many situations in which diffraction limits the resolution. The acuity of vision is limited because light passes through the pupil, the circular aperture of the eye. Be aware that the diffraction-like spreading of light is due to the limited diameter of a light beam, not the interaction with an aperture. Thus light passing through a lens with a diameter of Dshows the diffraction effect and spreads, blurring the image, just as light passing through an aperture of diameter Ddoes. Diffraction limits the resolution of any system having a lens or mirror. Telescopes are also limited by diffraction, because of the finite diameter, D, of their primary mirror. Why are diffraction gratings used in spectroscopes rather than just two slits? a. The bands produced by diffraction gratings are dimmer but sharper than the bands produced by two slits. b. The bands produced by diffraction gratings are brighter, though less sharp, than the bands produced by two slits. c. The bands produced by diffraction gratings are brighter and sharper than the bands produced by two slits. d. The bands produced by diffraction gratings are dimmer and less sharp, but more widely dispersed, than the bands produced by two slits. Calculations Involving Diffraction Gratings and Resolution Early in the chapter, it was mentioned that when light passes from one medium to another, its speed and wavelength change, but its frequency remains constant. The equation, is related to the wavelength in a vacuum, shows how to the wavelength in a given medium,, and the refractive index, n, of the medium. The equation is useful for calculating the change in wavelength of a monochromatic laser beam in various media. The analysis of a diffraction grating is very similar to that for a double slit. As you know from the discussion of double slits in Young\u2019s double-slit experiment, light is diffracted by, and spreads out after passing through, each slit. Rays travel at an angle relative to the incident direction. Each ray travels a different distance to a common point on a screen far away. The rays start in phase, and they can be in or out of phase when they reach a screen, depending on the difference in the path lengths traveled. Each ray travels a distance that differs by equals an integral number of wavelengths, the rays all arrive in phase, and constructive interference (a maximum) is obtained. Thus, the condition necessary to obtain constructive interference for a diffraction grating is from that of its neighbor, where dis the distance between slits. If where dis the distance between slits in the grating, is the wavelength of the light, and mis the order of the maximum. Note that this is exactly the same equation as for two slits separated by d. However, the slits are usually closer in diffraction gratings than in double slits, producing fewer maxima at larger angles. 538 Chapter 17 \u2022 Diffraction and Interference WATCH PHYSICS Diffraction Grating This video (https://www.openstax.org/l/28diffraction) explains the geometry behind the diffraction pattern produced by a diffraction grating. Click to view content (https://www.openstax.org/l/28diffraction) The equation that gives the points of constructive interference produced by a diffraction grating is equation look familiar? a. b. c. d. It is the same as the equation for destructive interference for a double-slit diffraction pattern. It is the same as the equation for constructive interference for a double-slit diffraction pattern. It is the same as the equation for constructive interference for a single-slit diffraction pattern. It is the same as the equation for destructive interference for a single-slit diffraction pattern.. Why does that Just what is the resolution limit of an aperture or lens? To answer that question, consider the diffraction pattern for a circular aperture, which, similar to the diffraction pattern of light passing through a slit, has a central maximum that is wider and brighter than the maxima surrounding it (see Figure 17.19 (a)). It can be shown that, for a circular aperture of diameter D, the first minimum in the diffraction pattern occurs at wavelength of light, which is the case for most optical instruments. The accepted criterion for determining the diff", "raction limit to resolution based on diffraction was developed by Lord Rayleigh in the 19th century. The Rayleigh criterion for the diffraction limit to resolution states that two images are just resolvable when the center of the diffraction pattern of one is directly over the first minimum of the diffraction pattern of the other. See Figure 17.20 (b). The first minimum is at an angle of, so that two point objects are just resolvable if they are separated by the angle, provided that the aperture is large compared with the is the wavelength of the light (or other electromagnetic radiation) and Dis the diameter of the aperture, lens, mirror, where etc., with which the two objects are observed. In the expression above, has units of radians. Figure 17.20 (a) Graph of intensity of the diffraction pattern for a circular aperture. Note that, similar to a single slit, the central maximum is wider and brighter than those to the sides. (b) Two point objects produce overlapping diffraction patterns. Shown here is the Rayleigh criterion for their being just resolvable. The central maximum of one pattern lies on the first minimum of the other. Access for free at openstax.org. 17.2 \u2022 Applications of Diffraction, Interference, and Coherence 539 Snap Lab Resolution \u2022 A sheet of white paper \u2022 A black pen or pencil \u2022 A measuring tape Instructions Procedure 1. Draw two lines several mm apart on a white sheet of paper. 2. Move away from the sheet as it is held upright, and measure the distance at which you can just distinguish (resolve) the lines as separate. 3. Use to calculate Dthe diameter of your pupil. Use the distance between the lines and the maximum distance at which they were resolved to calculate. Use the average wavelength for visible light as the value for. 4. Compare your answer to the average pupil diameter of 3 mm. Describe resolution in terms of minima and maxima of diffraction patterns. a. The limit for resolution is when the minimum of the pattern for one of the lines is directly over the first minimum of the pattern for the other line. b. The limit for resolution is when the maximum of the pattern for one of the lines is directly over the first minimum of the pattern for the other line. c. The limit for resolution is when the maximum of the pattern for one of the lines is directly over the second minimum of the pattern for the other line. d. The limit for resolution is when the minimum of the pattern for one of the lines is directly over the second maximum of the pattern for the other line. WORKED EXAMPLE Change of Wavelength A monochromatic laser beam of green light with a wavelength of 550 nm passes from air to water. The refractive index of water is 1.33. What will be the wavelength of the light after it enters the water? STRATEGY You can assume that the refractive index of air is the same as that of light in a vacuum because they are so close. You then have all the information you need to solve for. Solution 17.5 Discussion The refractive index of air is 1.0003, so the approximation holds for three significant figures. You would not see the light change color, however. Color is determined by frequency, not wavelength. WORKED EXAMPLE Diffraction Grating A diffraction grating has 2000 lines per centimeter. At what angle will the first-order maximum form for green light with a wavelength of 520 nm? STRATEGY You are given enough information to calculate d, and you are given the values of and m. You will have to find the arcsin of a 540 Chapter 17 \u2022 Diffraction and Interference number to find. Solution First find d. Rearrange the equation for constructive interference conditions for a diffraction grating, and substitute the known values. 17.6 Discussion This angle seems reasonable for the first maximum. Recall that the meaning of sin\u20121 (or arcsin) is the angle with a sine that is (the unknown). Remember that the value of will not be greater than 1 for any value of. WORKED EXAMPLE Resolution What is the minimum angular spread of a 633-nm-wavelength He-Ne laser beam that is originally 1.00 mm in diameter? STRATEGY The diameter of the beam is the same as if it were coming through an aperture of that size, so D= 1.00 mm. You are given, and you must solve for. Solution 17.7 Discussion The conversion factor for radians to degrees is 1.000 radian = 57.3\u00b0. The spread is very small and would not be noticeable over short distances. The angle represents the angular separation of the central maximum and the first minimum. Practice Problems 6. A beam of yellow light has a wavelength of 600 nm in a vacuum and a wavelength of 397 nm in Plexiglas. What is the refractive index of Plexiglas? a. 1.51", " b. 2.61 3.02 c. 3.77 d. 7. What is the angle between two just-resolved points of light for a 3.00 mm diameter pupil, assuming an average wavelength of 550 nm? a. 224 rad 183 rad b. 1.83 \u00d7 10\u20134 rad c. d. 2.24 \u00d7 10\u20134 rad Check Your Understanding 8. How is an interference pattern formed by a diffraction grating different from the pattern formed by a double slit? Access for free at openstax.org. 17.2 \u2022 Applications of Diffraction, Interference, and Coherence 541 a. The pattern is colorful. b. The pattern is faded. c. The pattern is sharper. d. The pattern is curved. 9. A beam of light always spreads out. Why can a beam not be produced with parallel rays to prevent spreading? a. Light is always polarized. b. Light is always reflected. c. Light is always refracted. d. Light is always diffracted. 10. Compare interference patterns formed by a double slit and by a diffraction grating in terms of brightness and narrowness of bands. a. The pattern formed has broader and brighter bands. b. The pattern formed has broader and duller bands. c. The pattern formed has narrower and duller bands. d. The pattern formed has narrower and brighter bands. 11. Describe the slits in a diffraction grating in terms of number and spacing, as compared to a two-slit diffraction setup. a. The slits in a diffraction grating are broader, with space between them that is greater than the separation of the two slits in two-slit diffraction. b. The slits in a diffraction grating are broader, with space between them that is the same as the separation of the two slits in two-slit diffraction. c. The slits in a diffraction grating are narrower, with space between them that is the same as the separation of the two slits in two-slit diffraction. d. The slits in a diffraction grating are narrower, with space between them that is greater than the separation of the two slits in two-slit diffraction. 542 Chapter 17 \u2022 Key Terms KEY TERMS differential interference contrast (DIC) separating a tangent to all of the wavelets. polarized light source into two beams polarized at right angles to each other and coherent with each other then, after passing through the sample, recombining and realigning the beams so they have the same plane of polarization, and then creating an interference pattern caused by the differences in their optical path and the refractive indices of the parts of the sample they passed through; the result is an image with contrast and shadowing that could not be observed with traditional optics diffraction bending of a wave around the edges of an opening or an obstacle diffraction grating many of evenly spaced slits having dimensions such that they produce an interference pattern Huygens\u2019s principle Every point on a wavefront is a source of wavelets that spread out in the forward direction at the same speed as the wave itself; the new wavefront is a line SECTION SUMMARY 17.1 Understanding Diffraction and Interference iridescence the effect that occurs when tiny, fingerlike structures in regular patterns act as reflection gratings, producing constructive interference that gives feathers colors not solely due to their pigmentation laser acronym for a device that produces light amplification by stimulated emission of radiation monochromatic one color monochromator device that separates the various wavelengths of incoming light and allows a beam with only a specific wavelength to pass through Rayleigh criterion two images are just resolvable when the center of the diffraction pattern of one is directly over the first minimum of the diffraction pattern of the other resolution degree to which two images can be distinguished from one another, which is limited by diffraction wavefront points on a wave surface that all share an identical, constant phase 17.2 Applications of Diffraction, Interference, and Coherence \u2022 The wavelength of light varies with the refractive index \u2022 The focused, coherent radiation emitted by lasers has of the medium. many uses in medicine and industry. \u2022 Slits produce a diffraction pattern if their width and \u2022 Characteristics of diffraction patterns produced with separation are similar to the wavelength of light passing through them. Interference bands of a single-slit diffraction pattern can be predicted. Interference bands of a double-slit diffraction pattern can be predicted. \u2022 \u2022 KEY EQUATIONS 17.1 Understanding Diffraction and Interference speed of light, frequency, and wavelength change of wavelength with index of refraction two-slit constructive interference, for m = 0, 1, \u22121, 2, \u22122, \u2026 diffraction gratings can be determined. \u2022 Diffraction gratings have been incorporated in many instruments, including microscopes and spectrometers. \u2022 Resolution has a limit that can", " be predicted. two-slit destructive interference, for m= 0, 1, \u22121, 2, \u22122, \u2026 one-slit, first-order destructive interference; wavelength related to dimensions one-slit destructive interference Access for free at openstax.org. Chapter 17 \u2022 Chapter Review 543 17.2 Applications of Diffraction, Interference, and Coherence wavelength change with change in medium diffraction grating constructive interference resolution CHAPTER REVIEW Concept Items 17.1 Understanding Diffraction and Interference 1. Which behavior of light is indicated by an interference pattern? a. ray behavior b. particle behavior c. d. wave behavior corpuscular behavior 2. Which behavior of light is indicated by diffraction? a. wave behavior b. particle behavior ray behavior c. corpuscular behavior d. 17.2 Applications of Diffraction, Interference, and Coherence 3. There is a principle related to resolution that is expressed by this equation. 17.8 What is that principle stated in full? Critical Thinking Items 17.1 Understanding Diffraction and Interference 6. Describe a situation in which bodies of water and a line of rocks could create a diffraction pattern similar to light passing through double slits. Include the arrangement of the rocks, the positions of the bodies of water, and the location of the diffraction pattern. Note the dimensions that are necessary for the production of the pattern. a. When waves from a small body of water pass through two widely separated openings and enter a larger body of water, a diffraction pattern is produced that is similar tothe diffraction pattern formed by light passing through two slits. The width of each opening is larger than the size of the wavelength of the waves. b. When waves from a large body of water pass 4. A principle related to resolution states, \u201cTwo images are just resolved when the center of the diffraction pattern of one is directly over the first minimum of the diffraction pattern of the other.\u201d Write the equation that expresses that principle. a. b. c. d. 5. Which statement completes this resolution? Two images are just resolved when \u2014 a. The center of the diffraction pattern of one image is directly over the central maximum of the diffraction pattern of the other. b. The center of the diffraction pattern of one image is directly over the central minimum of the diffraction pattern of the other c. The center of the diffraction pattern of one image is directly over the first minimum of the diffraction pattern of the other d. The center of the diffraction pattern of one is directly over the first maximum of the diffraction pattern of the other through two narrow openings and enter a smaller body of water, a diffraction pattern is produced that is similar to the diffraction pattern formed by light passing through two slits. The widths and separation of the openings are similar to the size of the wavelength of the waves. c. When waves from a small body of water pass through two wide openings and enter a larger body of water, a diffraction pattern is produced that is similar tothe diffraction pattern formed by light passing through two slits. The separation between the openings is similar to the size of the wavelength of the waves. d. When waves from a large body of water pass through two wide openings and enter a smaller body of water, a diffraction pattern is produced that is similar to the diffraction pattern formed by light passing through two slits. The widths and 544 Chapter 17 \u2022 Chapter Review separation of the openings are larger than the size of the wavelength of the waves. 17.2 Applications of Diffraction, Interference, and Coherence 7. For what type of electromagnetic radiation would a grating with spacing greater than 800 nm be useful as a spectroscopic tool? a. It can be used to analyze spectra only in the infrared portion of the spectrum. It can be used to analyze spectra in the entire visible portion of the electromagnetic spectrum. b. Problems 17.1 Understanding Diffraction and Interference 9. What is the distance between two slits that produce a diffraction pattern with the first minimum at an angle of 45.0\u00b0 when 410-nm violet light passes through the slits? a. 2,030 nm b. 1,450 nm c. 410 nm d. 290 nm 10. A breakwater at the entrance to a harbor consists of a rock barrier with a 50.0 \u2212 m -wide opening. Ocean waves with a 20.0-m wavelength approach the opening straight on. At what angle to the incident direction are the boats inside the harbor most protected against wave action? 11.5\u00b0 a. 7.46\u00b0 b. c. 5.74\u00b0 d. 23.6\u00b0 Performance Task 17.2 Applications of Diffraction, Interference, and Coherence 13. In this performance task you will create one- and two- slit diffraction and observe the interference patterns that result. \u2022 A utility knife (a knife with a razor blade-like cutting edge)", " \u2022 Aluminum foil \u2022 A straight edge \u2022 A strong, small light source or a laser pointer \u2022 A tape measure \u2022 A white wall Access for free at openstax.org. c. d. It can only be used to analyze spectra in the short\u2010wavelength visible. It can only be used to analyze spectra in the short\u2010wavelength visible and ultraviolet. 8. A beam of green light has a wavelength of in a in Plexiglas. What vacuum and a wavelength of is the refractive index of Plexiglas? a. b. c. d. 17.2 Applications of Diffraction, Interference, and Coherence 11. A 500-nm beam of light passing through a diffraction grating creates its second band of constructive interference at an angle of 1.50\u00b0. How far apart are the slits in the grating? 38,200 nm a. b. 19,100 nm c. 667 nm 333 nm d. 12. The range of the visible-light spectrum is 380 nm to 780 nm. What is the maximum number of lines per centimeter a diffraction grating can have and produce a complete first-order spectrum for visible light? a. 26,300 lines/cm b. 13,200 lines/cm c. 6,410 lines/cm d. 12,820 lines/cm Procedure 1. Cut a piece of aluminum foil about 15 cm \u00d7 15 cm. 2. Use the utility knife and the straight edge to cut a straight slit several cm long in the center of the foil square. 3. With the room darkened, one partner shines the light through the slit and toward the wall. The other partner observes the pattern on the wall. The partner with the light changes the distance from the foil to the wall and the distance from the light to the foil. 4. When the sharpest, brightest pattern possible is obtained, the partner who is not holding the foil and light makes measurements. 5. Measure the perpendicular (shortest) distance from the slit to the wall, the distance from the center of the pattern to several of the dark bands, and the distance from the slit to the same dark bands. 6. Carefully make a second slit parallel to the first slit and 1 mm or less away. 7. Repeat steps 2 through 5, only this time measure the distances to bright bands. NOTE\u2014In your calculations, use 580 nm for if you used white light. If you used a colored laser pointer, look up the wavelength of the color. You from its tangent may find it easier to calculate TEST PREP Multiple Choice 17.1 Understanding Diffraction and Interference 14. Which remains unchanged when a monochromatic beam of light passes from air into water? a. b. c. d. the speed of the light the direction of the beam the frequency of the light the wavelength of the light 15. Two slits are separated by a distance of 3500 nm. If light with a wavelength of 500 nm passes through the slits and produces an interference pattern, the m = ________ order minimum appears at an angle of 30.0\u00b0. a. 0 b. 1 c. 2 3 d. 16. In the sunlight, the shadow of a building has fuzzy edges even if the building does not. Is this a refraction effect? Explain. a. Yes, this is a refraction effect, where every point on the building acts as the origin for a new wavefront. b. Yes, this is a refraction effect, where the whole building acts as the origin for a new wavefront. c. No, this is a diffraction effect, where every point on the edge of the building\u2019s shadow acts as the origin for a new wavefront. d. No, this is a diffraction effect, where the whole building acts as the origin for a new wavefront. 17.2 Applications of Diffraction, Interference, and Coherence 17. Two images are just resolved when the center of the diffraction pattern of one is directly over ________ of the diffraction pattern of the other. a. the center Chapter 17 \u2022 Test Prep 545 rather than from its sine. a. Which experiment gave the most distinct pattern\u2014one or two slits? b. What was the width of the single slit? Compare the calculated distance with the measured distance. c. What was the distance between the two slits? Compare the calculated distance with the measured distance. b. c. d. the first minimum the first maximum the last maximum 18. Two point sources of light are just resolvable as. What is the they pass through a small hole. The angle to the first minimum of one source is diameter of the hole? a. b. c. d. 19. Will a beam of light shining through a 1-mm hole behave any differently than a beam of light that is 1 mm wide as it leaves its source? Explain.? a. Yes, the beam passing through the hole will spread out as it travels,", " because it is diffracted by the edges of the hole, whereas the 1 -mm beam, which encounters no diffracting obstacle, will not spread out. b. Yes, the beam passing through the hole will be made more parallelby passing through the hole, and so will not spread out as it travels, whereas the unaltered wavefronts of the 1-mm beam will cause the beam to spread out as it travels. c. No, both beams will remain the same width as they travel, and they will not spread out. d. No, both beams will spread out as they travel. 20. A laser pointer emits a coherent beam of parallel light rays. Does the light from such a source spread out at all? Explain. a. Yes, every point on a wavefront is not a source of wavelets, which prevent the spreading of light waves. b. No, every point on a wavefront is not a source of wavelets, so that the beam behaves as a bundles of rays that travel in their initial direction. c. No, every point on a wavefront is a source of 546 Chapter 17 \u2022 Test Prep wavelets, which keep the beam from spreading. d. Yes, every point on a wavefront is a source of wavelets, which cause the beam to spread out steadily as it moves forward. Short Answer 17.1 Understanding Diffraction and Interference 21. Light passing through double slits creates a diffraction pattern. How would the spacing of the bands in the pattern change if the slits were closer together? a. The bands would be closer together. b. The bands would spread farther apart. c. The bands would remain stationary. d. The bands would fade and eventually disappear. 22. A beam of light passes through a single slit to create a diffraction pattern. How will the spacing of the bands in the pattern change if the width of the slit is increased? a. The width of the spaces between the bands will remain the same. spaced parallel lines that produces an interference pattern that is similar to but sharper and better dispersed than that of a double slit. b. A diffraction grating is a large collection of randomly spaced parallel lines that produces an interference pattern that is similar to but less sharp or well-dispersed as that of a double slit. c. A diffraction grating is a large collection of randomly spaced intersecting lines that produces an interference pattern that is similar to but sharper and better dispersed than that of a double slit. d. A diffraction grating is a large collection of evenly spaced intersecting lines that produces an interference pattern that is similar to but less sharp or well-dispersed as that of a double slit. b. The width of the spaces between the bands will 26. Suppose pure-wavelength light falls on a diffraction increase. c. The width of the spaces between the bands will decrease. d. The width of the spaces between the bands will first decrease and then increase. grating. What happens to the interference pattern if the same light falls on a grating that has more lines per centimeter? a. The bands will spread farther from the central maximum. b. The bands will come closer to the central 23. What is the wavelength of light falling on double slits maximum. if the third-order maximum is at c. The bands will not spread farther from the first maximum. d. The bands will come closer to the first maximum. 27. How many lines per centimeter are there on a diffraction grating that gives a first-order maximum for 473 nm blue light at an angle of 25.0\u00b0? a. b. c. 851 lines/cm d. 8,934 lines/cm 529,000 lines/cm 50,000 lines/cm 28. What is the distance between lines on a diffraction grating that produces a second-order maximum for 760-nm red light at an angle of 60.0\u00b0? a. 2.28 \u00d7 104 nm 3.29 \u00d7 102 nm b. c. 2.53 \u00d7 101 nm 1.76 \u00d7 103 nm d.? separated by an angle of a. b. c. d. 24. What is the longest wavelength of light passing through a single slit of width 1.20 \u03bcm for which there is a firstorder minimum? 1.04 \u00b5m a. b. 0.849 \u00b5m c. 0.600 \u00b5m d. 2.40 \u00b5m 17.2 Applications of Diffraction, Interference, and Coherence 25. Describe a diffraction grating and the interference pattern it produces. a. A diffraction grating is a large collection of evenly Access for free at openstax.org. Chapter 17 \u2022 Test Prep 547 grating. a. All three interference pattern produce identical bands. b. A double slit produces the sharpest and most distinct bands. c. A single slit produces the sharpest and most distinct bands. d. The diffraction grating produces", " the sharpest and most distinct bands. 32. An electric current through hydrogen gas produces several distinct wavelengths of visible light. The light is projected onto a diffraction grating having per centimeter. What are the wavelengths of the hydrogen spectrum if the light forms first-order maxima at angles of a., and?,, lines b. c. d. Extended Response 17.1 Understanding Diffraction and Interference 29. Suppose you use a double slit to perform Young\u2019s double-slit experiment in air, and then repeat the experiment with the same double slit in water. Does the color of the light change? Do the angles to the same parts of the interference pattern get larger or smaller? Explain. a. No, the color is determined by frequency. The magnitude of the angle decreases. b. No, the color is determined by wavelength. The magnitude of the angle decreases. c. Yes, the color is determined by frequency. The magnitude of the angle increases. d. Yes, the color is determined by wavelength. The magnitude of the angle increases. 30. A double slit is located at a distance xfrom a screen, with the distance along the screen from the center given by y. When the distance dbetween the slits is relatively large, there will be numerous bright bands. For small angles (where sin\u03b8= \u03b8, with \u03b8in radians), what is the distance between fringes? a. b. c. d. 17.2 Applications of Diffraction, Interference, and Coherence 31. Compare the interference patterns of single-slit diffraction, double-slit diffraction, and a diffraction 548 Chapter 17 \u2022 Test Prep Access for free at openstax.org. CHAPTER 18 Static Electricity Figure 18.1 This child\u2019s hair contains an imbalance of electrical charge (commonly called static electricity), which causes it to stand on end. The sliding motion stripped electrons away from the child\u2019s body, leaving him with an excess of positive charges, which repel each other along each strand of hair. (credit: Ken Bosma, Wikimedia Commons) Chapter Outline 18.1 Electrical Charges, Conservation of Charge, and Transfer of Charge 18.2 Coulomb's law 18.3 Electric Field 18.4 Electric Potential 18.5 Capacitors and Dielectrics You may have been introduced to static electricity like the child sliding down the slide in the opening INTRODUCTION photograph (Figure 18.1). The zapthat he is likely to receive if he touches a playmate or parent tends to bring home the lesson. But static electricity is more than just fun and games\u2014it is put to use in many industries. The forces between electrically charged particles are used in technologies such as printers, pollution filters, and spray guns used for painting cars and trucks. Static electricityis the study of phenomena that involve an imbalance of electrical charge. Although creating this imbalance typically requires moving charge around, once the imbalance is created, it often remains static for a long time. The study of charge in motion is called electromagnetismand will be covered in a later chapter. What is electrical charge, how is it associated 550 Chapter 18 \u2022 Static Electricity with objects, and what forces does it create? These are just some of the questions that this chapter addresses. 18.1 Electrical Charges, Conservation of Charge, and Transfer of Charge Section Learning Objectives By the end of this section, you will be able to do the following: \u2022 Describe positive and negative electric charges \u2022 Use conservation of charge to calculate quantities of charge transferred between objects \u2022 Characterize materials as conductors or insulators based on their electrical properties \u2022 Describe electric polarization and charging by induction Section Key Terms conduction conductor electron induction insulator law of conservation of charge polarization proton Electric Charge You may know someone who has an electricpersonality, which usually means that other people are attracted to this person. This saying is based on electric charge, which is a property of matter that causes objects to attract or repel each other. Electric charge comes in two varieties, which we call positiveand negative.Like charges repel each other, and unlike charges attract each other. Thus, two positive charges repel each other, as do two negative charges. A positive charge and a negative charge attract each other. How do we know there are two types of electric charge? When various materials are rubbed together in controlled ways, certain combinations of materials always result in a net charge of one type on one material and a net charge of the opposite type on the other material. By convention, we call one type of charge positive and the other type negative. For example, when glass is rubbed with silk, the glass becomes positively charged and the silk negatively charged. Because the glass and silk have opposite charges, they attract one another like clothes that have rubbed together in a dryer. Two glass rods rubbed with silk in this manner will repel one another, because each rod has positive charge on it. Similarly, two silk cloth", "s rubbed in this manner will repel each other, because both cloths have negative charge. Figure 18.2 shows how these simple materials can be used to explore the nature of the force between charges. Figure 18.2 A glass rod becomes positively charged when rubbed with silk, whereas the silk becomes negatively charged. (a) The glass rod is attracted to the silk, because their charges are opposite. (b) Two similarly charged glass rods repel. (c) Two similarly charged silk cloths repel. It took scientists a long time to discover what lay behind these two types of charges. The word electricitself comes from the Greek word elektronfor amber, because the ancient Greeks noticed that amber, when rubbed by fur, attracts dry straw. Almost 2,000 years later, the English physicist William Gilbert proposed a model that explained the effect of electric charge as being due to a mysterious electrical fluid that would pass from one object to another. This model was debated for several hundred years, but it was finally put to rest in 1897 by the work of the English physicist J. J. Thomson and French physicist Jean Perrin. Along with many others, Thomson and Perrin were studying the mysterious cathode raysthat were known at the time to consist of particles smaller than the smallest atom. Perrin showed that cathode rays actually carried negative electrical charge. Later, Thomson\u2019s work led him to declare, \u201cI can see no escape from the conclusion that [cathode rays] are charges of negative Access for free at openstax.org. 18.1 \u2022 Electrical Charges, Conservation of Charge, and Transfer of Charge 551 electricity carried by particles of matter.\u201d It took several years of further experiments to confirm Thomson\u2019s interpretation of the experiments, but science had in fact discovered the particle that carries the fundamental unit of negative electrical charge. We now know this particle as the electron. Atoms, however, were known to be electrically neutral, which means that they carry the same amount of positive and negative charge, so their net charge is zero. Because electrons are negative, some other part of the atom must contain positive charge. Thomson put forth what is called the plum pudding model, in which he described atoms as being made of thousands of electrons swimming around in a nebulous mass of positive charge, as shown by the left-side image of Figure 18.3. His student, Ernest Rutherford, originally believed that this model was correct and used it (along with other models) to try to understand the results of his experiments bombarding gold foils with alphaparticles (i.e., helium atoms stripped of their electrons). The results, however, did not confirm Thomson\u2019s model but rather destroyed it! Rutherford found that most of the space occupied by the gold atoms was actually empty and that almost all of the matter of each atom was concentrated into a tiny, extremely dense nucleus, as shown by the right-side image of Figure 18.3. The atomic nucleus was later found to contain particles called protons, each of which carries a unit of positive electric charge.1 Figure 18.3 The left drawing shows Thompson\u2019s plum-pudding model, in which the electrons swim around in a nebulous mass of positive charge. The right drawing shows Rutherford\u2019s model, in which the electrons orbit around a tiny, massive nucleus. Note that the size of the nucleus is vastly exaggerated in this drawing. Were it drawn to scale with respect to the size of the electron orbits, the nucleus would not be visible to the naked eye in this drawing. Also, as far as science can currently detect, electrons are point particles, which means that they have no size at all! Protons and electrons are thus the fundamental particles that carry electric charge. Each proton carries one unit of positive charge, and each electron carries one unit of negative charge. To the best precision that modern technology can provide, the charge carried by a proton is exactlythe opposite of that carried by an electron. The SI unit for electric charge is the coulomb (abbreviated as \u201cC\u201d), which is named after the French physicist Charles Augustin de Coulomb, who studied the force between charged objects. The proton carries protons required to make +1.00 C is and the electron carries. The number nof The same number of electrons is required to make \u22121.00 C of electric charge. The fundamental unit of charge is often represented as e. Thus, the charge on a proton is e, and the charge on an electron is \u2212e. Mathematically, 18.1 LINKS TO PHYSICS Measuring the Fundamental Electric Charge The American physicist Robert Millikan (1868\u20131953) and his student Harvey Fletcher (1884\u20131981) were the first to make a relatively accurate measurement of the fundamental unit of charge on the electron. They designed what is now a classic 1Protons were later found to contain sub particles called quarks, which have fractional", " electric charge. But that is another story that we leave for subsequent physics courses. 552 Chapter 18 \u2022 Static Electricity experiment performed by students. The Millikan oil-drop experiment is shown in Figure 18.4. The experiment involves some concepts that will be introduced later, but the basic idea is that a fine oil mist is sprayed between two plates that can be charged with a known amount of opposite charge. Some oil drops accumulate some excess negative charge when being sprayed and are attracted to the positive charge of the upper plate and repelled by the negative charge on the lower plate. By tuning the charge on these plates until the weight of the oil drop is balanced by the electric forces, the net charge on the oil drop can be determined quite precisely. Figure 18.4 The oil-drop experiment involved spraying a fine mist of oil between two metal plates charged with opposite charges. By knowing the mass of the oil droplets and adjusting the electric charge on the plates, the charge on the oil drops can be determined with precision. Millikan and Fletcher found that the drops would accumulate charge in discrete units of about within 1 percent of the modern value of times greater than the possible error Millikan reported for his results! which is Although this difference may seem quite small, it is actually five Because the charge on the electron is a fundamental constant of nature, determining its precise value is very important for all of science. This created pressure on Millikan and others after him that reveals some equally important aspects of human nature. First, Millikan took sole credit for the experiment and was awarded the 1923 Nobel Prize in physics for this work, although his student Harvey Fletcher apparently contributed in significant ways to the work. Just before his death in 1981, Fletcher divulged that Millikan coerced him to give Millikan sole credit for the work, in exchange for which Millikan promoted Fletcher\u2019s career at Bell Labs. Another great scientist, Richard Feynman, points out that many scientists who measured the fundamental charge after Millikan were reluctant to report values that differed much from Millikan\u2019s value. History shows that later measurements slowly crept up from Millikan\u2019s value until settling on the modern value. Why did they not immediately find the error and correct the value, asks Feynman. Apparently, having found a value higher than the much-respected value found by Millikan, scientists would look for possible mistakes that might lower their value to make it agree better with Millikan\u2019s value. This reveals the important psychological weight carried by preconceived notions and shows how hard it is to refute them. Scientists, however devoted to logic and data they may be, are apparently just as vulnerable to this aspect of human nature as everyone else. The lesson here is that, although it is good to be skeptical of new results, you should not discount them just because they do not agree with conventional wisdom. If your reasoning is sound and your data are reliable, the conclusion demanded by the data must be seriously considered, even if that conclusion disagrees with the commonly accepted truth. GRASP CHECK Suppose that Millikan observed an oil drop carrying three fundamental units of charge. What would be the net charge on this oil drop? a. \u22124.81 \u00d7 10\u221219 C b. \u22121.602 \u00d7 10\u221219 C 1.602 \u00d7 10\u221219 C c. d. 4.81 \u00d7 10\u221219 C Access for free at openstax.org. 18.1 \u2022 Electrical Charges, Conservation of Charge, and Transfer of Charge 553 Snap Lab Like and Unlike Charges This activity investigates the repulsion and attraction caused by static electrical charge. \u2022 Adhesive tape \u2022 Nonconducting surface, such as a plastic table or chair Instructions Procedure for Part (a) 1. Prepare two pieces of tape about 4 cm long. To make a handle, double over about 0.5 cm at one end so that the sticky side sticks together. 2. Attach the pieces of tape side by side onto a nonmetallic surface, such as a tabletop or the seat of a chair, as shown in Figure 18.5(a). 3. Peel off both pieces of tape and hang them downward, holding them by the handles, as shown in Figure 18.5(b). If the tape bends upward and sticks to your hand, try using a shorter piece of tape, or simply shake the tape so that it no longer sticks to your hand. 4. Now slowly bring the two pieces of tape together, as shown in Figure 18.5(c). What happens? Figure 18.5 Procedure for Part (b) 5. Stick one piece of tape on the nonmetallic surface, and stick the second piece of tape on top of the first piece, as shown in Figure 18.6(a). 6. Slowly peel off the two pieces by pulling on the handle of the bottom piece. 7. Gently stroke your finger along the top of the second piece of tape (i.e., the nonsticky side), as shown in Figure 18.6(b). 8.", " Peel the two pieces of tape apart by pulling on their handles, as shown in Figure 18.6(c). 9. Slowly bring the two pieces of tape together. What happens? Figure 18.6 GRASP CHECK In step 4, why did the two pieces of tape repel each other? In step 9, why did they attract each other? a. Like charges attract, while unlike charges repel each other. b. Like charges repel, while unlike charges attract each other. c. Tapes having positive charge repel, while tapes having negative charge attract each other. d. Tapes having negative charge repel, while tapes having positive charge attract each other. 554 Chapter 18 \u2022 Static Electricity Conservation of Charge Because the fundamental positive and negative units of charge are carried on protons and electrons, we would expect that the total charge cannot change in any system that we define. In other words, although we might be able to move charge around, we cannot create or destroy it. This should be true provided that we do not create or destroy protons or electrons in our system. In the twentieth century, however, scientists learned how to create and destroy electrons and protons, but they found that charge is still conserved. Many experiments and solid theoretical arguments have elevated this idea to the status of a law. The law of conservation of charge says that electrical charge cannot be created or destroyed. The law of conservation of charge is very useful. It tells us that the net charge in a system is the same before and after any interaction within the system. Of course, we must ensure that no external charge enters the system during the interaction and that no internal charge leaves the system. Mathematically, conservation of charge can be expressed as 18.2 where is the net charge of the system before the interaction, and is the net charge after the interaction. WORKED EXAMPLE What is the missing charge? Figure 18.7 shows two spheres that initially have +4 C and +8 C of charge. After an interaction (which could simply be that they touch each other), the blue sphere has +10 C of charge, and the red sphere has an unknown quantity of charge. Use the law of conservation of charge to find the final charge on the red sphere. Strategy The net initial charge of the system is. The net final charge of the system is is the final charge on the red sphere. Conservation of charge tells us that we can solve for., where, so 18.3 Solution Equating and and solving for gives The red sphere has +2 C of charge. Figure 18.7 Two spheres, one blue and one red, initially have +4 C and +8 C of charge, respectively. After the two spheres interact, the blue sphere has a charge of +10 C. The law of conservation of charge allows us to find the final charge on the red sphere. Discussion Like all conservation laws, conservation of charge is an accounting scheme that helps us keep track of electric charge. Practice Problems 1. Which equation describes conservation of charge? a. qinitial = qfinal = constant b. qinitial = qfinal = 0 c. qinitial \u2212 qfinal = 0 Access for free at openstax.org. 18.1 \u2022 Electrical Charges, Conservation of Charge, and Transfer of Charge 555 d. qinitial/qfinal = constant 2. An isolated system contains two objects with charges and. If object 1 loses half of its charge, what is the final charge on object 2? a. b. c. d. Conductors and Insulators Materials can be classified depending on whether they allow charge to move. If charge can easily move through a material, such as metals, then these materials are called conductors. This means that charge can be conducted (i.e., move) through the material rather easily. If charge cannot move through a material, such as rubber, then this material is called an insulator. Most materials are insulators. Their atoms and molecules hold on more tightly to their electrons, so it is difficult for electrons to move between atoms. However, it is not impossible. With enough energy, it is possible to force electrons to move through an insulator. However, the insulator is often physically destroyed in the process. In metals, the outer electrons are loosely bound to their atoms, so not much energy is required to make electrons move through metal. Such metals as copper, silver, and aluminum are good conductors. Insulating materials include plastics, glass, ceramics, and wood. The conductivity of some materials is intermediate between conductors and insulators. These are called semiconductors. They can be made conductive under the right conditions, which can involve temperature, the purity of the material, and the force applied to push electrons through them. Because we can control whether semiconductors are conductors or insulators, these materials are used extensively in computer chips. The most commonly used semiconductor is silicon. Figure 18.8 shows various materials arranged according to their ability to conduct electrons. Figure 18", ".8 Materials can be arranged according to their ability to conduct electric charge. The slashes on the arrow mean that there is a very large gap in conducting ability between conductors, semiconductors, and insulators, but the drawing is compressed to fit on the page. The numbers below the materials give their resistivityin \u03a9\u2022m (which you will learn about below). The resistivity is a measure of how hard it is to make charge move through a given material. What happens if an excess negative charge is placed on a conducting object? Because like charges repel each other, they will push against each other until they are as far apart as they can get. Because the charge can move in a conductor, it moves to the outer surfaces of the object. Figure 18.9(a) shows schematically how an excess negative charge spreads itself evenly over the outer surface of a metal sphere. What happens if the same is done with an insulating object? The electrons still repel each other, but they are not able to move, because the material is an insulator. Thus, the excess charge stays put and does not distribute itself over the object. Figure 18.9(b) shows this situation. 556 Chapter 18 \u2022 Static Electricity Figure 18.9 (a) A conducting sphere with excess negative charge (i.e., electrons). The electrons repel each other and spread out to cover the outer surface of the sphere. (b) An insulating sphere with excess negative charge. The electrons cannot move, so they remain in their original positions. Transfer and Separation of Charge Most objects we deal with are electrically neutral, which means that they have the same amount of positive and negative charge. However, transferring negative charge from one object to another is fairly easy to do. When negative charge is transferred from one object to another, an excess of positive charge is left behind. How do we know that the negative charge is the mobile charge? The positive charge is carried by the proton, which is stuck firmly in the nucleus of atoms, and the atoms are stuck in place in solid materials. Electrons, which carry the negative charge, are much easier to remove from their atoms or molecules and can therefore be transferred more easily. Electric charge can be transferred in several manners. One of the simplest ways to transfer charge is charging by contact, in which the surfaces of two objects made of different materials are placed in close contact. If one of the materials holds electrons more tightly than the other, then it takes some electrons with it when the materials are separated. Rubbing two surfaces together increases the transfer of electrons, because it creates a closer contact between the materials. It also serves to present freshmaterial with a full supply of electrons to the other material. Thus, when you walk across a carpet on a dry day, your shoes rub against the carpet, and some electrons are removed from the carpet by your shoes. The result is that you have an excess of negative charge on your shoes. When you then touch a doorknob, some of your excess of electrons transfer to the neutral doorknob, creating a small spark. Touching the doorknob with your hand demonstrates a second way to transfer electric charge, which is charging by conduction. This transfer happens because like charges repel, and so the excess electrons that you picked up from the carpet want to be as far away from each other as possible. Some of them move to the doorknob, where they will distribute themselves over the outer surface of the metal. Another example of charging by conduction is shown in the top row of Figure 18.10. A metal sphere with 100 excess electrons touches a metal sphere with 50 excess electrons, so 25 electrons from the first sphere transfer to the second sphere. Each sphere finishes with 75 excess electrons. The same reasoning applies to the transfer of positive charge. However, because positive charge essentially cannot move in solids, it is transferred by moving negative charge in the opposite direction. For example, consider the bottom row of Figure 18.10. The first metal sphere has 100 excess protons and touches a metal sphere with 50 excess protons, so the second sphere transfers 25 electrons to the first sphere. These 25 extra electrons will electrically cancel 25 protons so that the first metal sphere is left with 75 excess protons. This is shown in the bottom row of Figure 18.10. The second metal sphere lost 25 electrons so it has 25 more excess protons, for a total of 75 excess protons. The end result is the same if we consider that the first ball transferred a net positive charge equal to that of 25 protons to the first ball. Access for free at openstax.org. 18.1 \u2022 Electrical Charges, Conservation of Charge, and Transfer of Charge 557 Figure 18.10 In the top row, a metal sphere with 100 excess electrons transfers 25 electrons to a metal sphere with an excess of 50 electrons. After the transfer, both spheres have 75 excess electrons. In the bottom row, a metal sphere with 100 excess prot", "ons receives 25 electrons from a ball with 50 excess protons. After the transfer, both spheres have 75 excess protons. In this discussion, you may wonder how the excess electrons originally got from your shoes to your hand to create the spark when you touched the doorknob. The answer is that noelectrons actually traveled from your shoes to your hands. Instead, because like charges repel each other, the excess electrons on your shoe simply pushed away some of the electrons in your feet. The electrons thus dislodged from your feet moved up into your leg and in turn pushed away some electrons in your leg. This process continued through your whole body until a distribution of excess electrons covered the extremities of your body. Thus your head, your hands, the tip of your nose, and so forth all received their doses of excess electrons that had been pushed out of their normal positions. All this was the result of electrons being pushed out of your feet by the excess electrons on your shoes. This type of charge separation is called polarization. As soon as the excess electrons leave your shoes (by rubbing off onto the floor or being carried away in humid air), the distribution of electrons in your body returns to normal. Every part of your body is again electrically neutral (i.e., zero excess charge). The phenomenon of polarization is seen in. The child has accumulated excess positive charge by sliding on the slide. This excess charge repels itself and so becomes distributed over the extremities of the child\u2019s body, notably in his hair. As a result, the hair stands on end, because the excess negative charge on each strand repels the excess positive charge on neighboring strands. Polarization can be used to charge objects. Consider the two metallic spheres shown in Figure 18.11. The spheres are electrically neutral, so they carry the same amounts of positive and negative charge. In the top picture (Figure 18.11(a)), the two spheres are touching, and the positive and negative charge is evenly distributed over the two spheres. We then approach a glass rod that carries an excess positive charge, which can be done by rubbing the glass rod with silk, as shown in Figure 18.11(b). Because opposite charges attract each other, the negative charge is attracted to the glass rod, leaving an excess positive charge on the opposite side of the right sphere. This is an example of charging by induction, whereby a charge is created by approaching a charged object with a second object to create an unbalanced charge in the second object. If we then separate the two spheres, as shown in Figure 18.11(c), the excess charge is stuck on each sphere. The left sphere now has an excess negative charge, and the right sphere has an excess positive charge. Finally, in the bottom picture, the rod is removed, and the opposite charges attract each other, so they move as close together as they can get. 558 Chapter 18 \u2022 Static Electricity Figure 18.11 (a) Two neutral conducting spheres are touching each other, so the charge is evenly spread over both spheres. (b) A positively charged rod approaches, which attracts negative charges, leaving excess positive charge on the right sphere. (c) The spheres are separated. Each sphere now carries an equal magnitude of excess charge. (d) When the positively charged rod is removed, the excess negative charge on the left sphere is attracted to the excess positive charge on the right sphere. FUN IN PHYSICS Create a Spark in a Science Fair Van de Graaff generators are devices that are used not only for serious physics research but also for demonstrating the physics of static electricity at science fairs and in classrooms. Because they deliver relatively little electric current, they can be made safe for use in such environments. The first such generator was built by Robert Van de Graaff in 1931 for use in nuclear physics research. Figure 18.12 shows a simplified sketch of a Van de Graaff generator. Van de Graaff generators use smooth and pointed surfaces and conductors and insulators to generate large static charges. In the version shown in Figure 18.12, electrons are \u201csprayed\u201d from the tips of the lower comb onto a moving belt, which is made of an insulating material like, such as rubber. This technique of charging the belt is akin to charging your shoes with electrons by walking across a carpet. The belt raises the charges up to the upper comb, where they transfer again, akin to your touching the doorknob and transferring your charge to it. Because like charges repel, the excess electrons all rush to the outer surface of the globe, which is made of metal (a conductor). Thus, the comb itself never accumulates too much charge, because any charge it gains is quickly depleted by the charge moving to the outer surface of the globe. Access for free at openstax.org. 18.1 \u2022 Electrical Charges, Conservation of Charge, and Transfer of Charge 559 Figure 18.12 Van de Graaff generators transfer electrons onto a metallic sphere, where the electrons distribute themselves uniformly over the", " outer surface. Van de Graaff generators are used to demonstrate many interesting effects caused by static electricity. By touching the globe, a person gains excess charge, so his or her hair stands on end, as shown in Figure 18.13. You can also create mini lightning bolts by moving a neutral conductor toward the globe. Another favorite is to pile up aluminum muffin tins on top of the uncharged globe, then turn on the generator. Being made of conducting material, the tins accumulate excess charge. They then repel each other and fly off the globe one by one. A quick Internet search will show many examples of what you can do with a Van de Graaff generator. Figure 18.13 The man touching the Van de Graaff generator has excess charge, which spreads over his hair and repels hair strands from his neighbors. (credit: Jon \u201cShakataGaNai\u201d Davis) GRASP CHECK Why don\u2019t the electrons stay on the rubber belt when they reach the upper comb? a. The upper comb has no excess electrons, and the excess electrons in the rubber belt get transferred to the comb by contact. b. The upper comb has no excess electrons, and the excess electrons in the rubber belt get transferred to the comb by conduction. c. The upper comb has excess electrons, and the excess electrons in the rubber belt get transferred to the comb by conduction. d. The upper comb has excess electrons, and the excess electrons in the rubber belt get transferred to the comb by contact. Virtual Physics Balloons and Static Electricity Click to view content (http://www.openstax.org/l/28balloons) 560 Chapter 18 \u2022 Static Electricity This simulation allows you to observe negative charge accumulating on a balloon as you rub it against a sweater. You can then observe how two charged balloons interact and how they cause polarization in a wall. GRASP CHECK Click the reset button, and start with two balloons. Charge a first balloon by rubbing it on the sweater, and then move it toward the second balloon. Why does the second balloon not move? a. The second balloon has an equal number of positive and negative charges. b. The second balloon has more positive charges than negative charges. c. The second balloon has more negative charges than positive charges. d. The second balloon is positively charged and has polarization. Snap Lab Polarizing Tap Water This lab will demonstrate how water molecules can easily be polarized. \u2022 Plastic object of small dimensions, such as comb or plastic stirrer \u2022 Source of tap water Instructions Procedure 1. Thoroughly rub the plastic object with a dry cloth. 2. Open the faucet just enough to let a smooth filament of water run from the tap. 3. Move an edge of the charged plastic object toward the filament of running water. What do you observe? What happens when the plastic object touches the water filament? Can you explain your observations? GRASP CHECK Why does the water curve around the charged object? a. The charged object induces uniform positive charge on the water molecules. b. The charged object induces uniform negative charge on the water molecules. c. The charged object attracts the polarized water molecules and ions that are dissolved in the water. d. The charged object depolarizes the water molecules and the ions dissolved in the water. WORKED EXAMPLE Charging Ink Droplets Electrically neutral ink droplets in an ink-jet printer pass through an electron beam created by an electron gun, as shown in Figure 18.14. Some electrons are captured by the ink droplet, so that it becomes charged. After passing through the electron. How many electrons are captured by the ink droplet? beam, the net charge of the ink droplet is Access for free at openstax.org. 18.1 \u2022 Electrical Charges, Conservation of Charge, and Transfer of Charge 561 Figure 18.14 Electrons from an electron guncharge a passing ink droplet. STRATEGY A single electron carries a charge of of a single electron will give the number of electrons captured by the ink droplet.. Dividing the net charge of the ink droplet by the charge Solution The number nof electrons captured by the ink droplet are 18.4 Discussion This is almost a billion electrons! It seems like a lot, but it is quite small compared to the number of atoms in an ink droplet, which number about Thus, each extra electron is shared between about atoms. Practice Problems 3. How many protons are needed to make 1 nC of charge? 1 nC = 10\u22129 C 1.6 \u00d7 10\u221228 a. 1.6 \u00d7 10\u221210 b. 3 \u00d7 109 c. d. 6 \u00d7 109 4. In a physics lab, you charge up three metal spheres, two with and one with. When you bring all three spheres together so that they all touch one another, what is the total charge on the three spheres? a. b. c. d. Check Your Understanding 5", ". How many types of electric charge exist? a. one type two types b. three types c. four types d. 6. Which are the two main electrical classifications of materials based on how easily charges can move through them? 562 Chapter 18 \u2022 Static Electricity a. b. c. d. conductor and insulator semiconductor and insulator conductor and superconductor conductor and semiconductor 7. True or false\u2014A polarized material must have a nonzero net electric charge. a. b. true false 8. Describe the force between two positive point charges that interact. a. The force is attractive and acts along the line joining the two point charges. b. The force is attractive and acts tangential to the line joining the two point charges. c. The force is repulsive and acts along the line joining the two point charges. d. The force is repulsive and acts tangential to the line joining the two point charges. 9. How does a conductor differ from an insulator? a. Electric charges move easily in an insulator but not in a conducting material. b. Electric charges move easily in a conductor but not in an insulator. c. A conductor has a large number of electrons. d. More charges are in an insulator than in a conductor. 10. True or false\u2014Charging an object by polarization requires touching it with an object carrying excess charge. a. b. true false 18.2 Coulomb's law Section Learning Objectives By the end of this section, you will be able to do the following: \u2022 Describe Coulomb\u2019s law verbally and mathematically \u2022 Solve problems involving Coulomb\u2019s law Section Key Terms Coulomb\u2019s law inverse-square law More than 100 years before Thomson and Rutherford discovered the fundamental particles that carry positive and negative electric charges, the French scientist Charles-Augustin de Coulomb mathematically described the force between charged objects. Doing so required careful measurements of forces between charged spheres, for which he built an ingenious device called a torsion balance. This device, shown in Figure 18.15, contains an insulating rod that is hanging by a thread inside a glass-walled enclosure. At one end of the rod is the metallic sphere A. When no charge is on this sphere, it touches sphere B. Coulomb would touch the spheres with a third metallic ball (shown at the bottom of the diagram) that was charged. An unknown amount of charge would distribute evenly between spheres A and B, which would then repel each other, because like charges repel. This force would cause sphere A to rotate away from sphere B, thus twisting the wire until the torsion in the wire balanced the electrical force. Coulomb then turned the knob at the top, which allowed him to rotate the thread, thus bringing sphere A closer to sphere B. He found that bringing sphere A twice as close to sphere B required increasing the torsion by a factor of four. Bringing the sphere three times closer required a ninefold increase in the torsion. From this type of measurement, he deduced that the electrical force between the spheres was inversely proportional to the distance squared between the spheres. In other words, where ris the distance between the spheres. An electrical charge distributes itself equally between two conducting spheres of the same size. Knowing this allowed Coulomb to divide an unknown charge in half. Repeating this process would produce a sphere with one quarter of the initial charge, and 18.5 Access for free at openstax.org. so on. Using this technique, he measured the force between spheres A and B when they were charged with different amounts of charge. These measurements led him to deduce that the force was proportional to the charge on each sphere, or 18.6 where is the charge on sphere A, and is the charge on sphere B. 18.2 \u2022 Coulomb's law 563 Figure 18.15 A drawing of Coulomb\u2019s torsion balance, which he used to measure the electrical force between charged spheres. (credit: Charles-Augustin de Coulomb) Combining these two proportionalities, he proposed the following expression to describe the force between the charged spheres. This equation is known as Coulomb\u2019s law, and it describes the electrostatic force between charged objects. The constant of proportionality kis called Coulomb\u2019s constant. In SI units, the constant khas the value 18.7 The direction of the force is along the line joining the centers of the two objects. If the two charges are of opposite signs, Coulomb\u2019s law gives a negative result. This means that the force between the particles is attractive. If the two charges have the same signs, Coulomb\u2019s law gives a positive result. This means that the force between the particles is repulsive. For example, if and both negative charge and This is shown in Figure 18.16(b). is a is a positive charge (or vice versa), then the charges are", " different, so the force between them is attractive. are negative or if both are positive, the force between them is repulsive. This is shown in Figure 18.16(a). If Figure 18.16 The magnitude of the electrostatic force Fbetween point charges q1 and q2 separated by a distance ris given by Coulomb\u2019s law. Note that Newton\u2019s third law (every force exerted creates an equal and opposite force) applies as usual\u2014the force (F1,2) on q1 is equal in magnitude and opposite in direction to the force (F2,1) it exerts on q2. (a) Like charges. (b) Unlike charges. Note that Coulomb\u2019s law applies only to charged objects that are not moving with respect to each other. The law says that the force is proportional to the amount of charge on each object and inversely proportional to the square of the distance between the objects. If we double the charge the force between them decreasesby a factor of spherical objects or to objects that are much smaller than the distance between the objects (in which case, the objects can be approximated as spheres)., for instance, then the force is doubled. If we double the distance between the objects, then. Although Coulomb\u2019s law is true in general, it is easiest to apply to 564 Chapter 18 \u2022 Static Electricity Coulomb\u2019s law is an example of an inverse-square law, which means the force depends on the square of the denominator. Another inverse-square law is Newton\u2019s law of universal gravitation, which is they differ in two important respects: (i) The gravitational constant Gis much, much smaller than k ( differences explain why gravity is so much weaker than the electrostatic force and why gravity is only attractive, whereas the electrostatic force can be attractive or repulsive. ); and (ii) only one type of mass exists, whereas two types of electric charge exist. These two. Although these laws are similar, Finally, note that Coulomb measured the distance between the spheres from the centers of each sphere. He did not explain this assumption in his original papers, but it turns out to be valid. From outsidea uniform spherical distribution of charge, it can be treated as if all the charge were located at the center of the sphere. WATCH PHYSICS Electrostatics (part 1): Introduction to charge and Coulomb's law This video explains the basics of Coulomb\u2019s law. Note that the lecturer uses dfor the distance between the center of the particles instead of r. Click to view content (https://www.openstax.org/l/28coulomb) GRASP CHECK True or false\u2014If one particle carries a positive charge and another carries a negative charge, then the force between them is attractive. true a. false b. Snap Lab Hovering plastic In this lab, you will use electrostatics to hover a thin piece of plastic in the air. \u2022 Balloon \u2022 Light plastic bag (e.g., produce bag from grocery store) Instructions Inflate the balloon. Procedure 1. Cut the plastic bag to make a plastic loop about 2 inches wide. 2. 3. Charge the balloon by rubbing it on your clothes. 4. Charge the plastic loop by placing it on a nonmetallic surface and rubbing it with a cloth. 5. Hold the balloon in one hand, and in the other hand hold the plastic loop above the balloon. If the loop clings too much to your hand, recruit a friend to hold the strip above the balloon with both hands. Now let go of the plastic loop, and maneuver the balloon under the plastic loop to keep it hovering in the air above the balloon. GRASP CHECK How does the balloon keep the plastic loop hovering? a. The balloon and the loop are both negatively charged. This will help the balloon keep the plastic loop hovering. b. The balloon is charged, while the plastic loop is neutral.This will help the balloon keep the plastic loop hovering. c. The balloon and the loop are both positively charged. This will help the balloon keep the plastic loop hovering. d. The balloon is positively charged, while the plastic loop is negatively charged. This will help the balloon keep the plastic loop hovering. Access for free at openstax.org. WORKED EXAMPLE 18.2 \u2022 Coulomb's law 565 Using Coulomb\u2019s law to find the force between charged objects between the two charged spheres when they are separated by 5.0 cm. By Suppose Coulomb measures a force of turning the dial at the top of the torsion balance, he approaches the spheres so that they are separated by 3.0 cm. Which force does he measure now? STRATEGY Apply Coulomb\u2019s law to the situation before and after the spheres are brought closer together. Although we do not know the charges on the spheres, we do know that they remain the same. We call these unknown", " but constant charges Because these charges appear as a product in Coulomb\u2019s law, they form a single unknown. We thus have two equations and two unknowns, which we can solve. The first unknown is the force (which we call second is ) when the spheres are 3.0 cm apart, and the and.. Use the following notation: When the charges are 5.0 cm apart, the force is where the subscript imeans initial. Once the charges are brought closer together, we know the subscript fmeans final. and,, where Solution Coulomb\u2019s law applied to the spheres in their initial positions gives Coulomb\u2019s law applied to the spheres in their final positions gives Dividing the second equation by the first and solving for the final force leads to Inserting the known quantities yields 18.8 18.9 18.10 18.11 The force acts along the line joining the centers of the spheres. Because the same type of charge is on each sphere, the force is repulsive. Discussion As expected, the force between the charges is greater when they are 3.0 cm apart than when they are 5.0 cm apart. Note that although it is a good habit to convert cm to m (because the constant kis in SI units), it is not necessary in this problem, because the distances cancel out. We can also solve for the second unknown. By using the first equation, we find 566 Chapter 18 \u2022 Static Electricity 18.12 Note how the units cancel in the second-to-last line. Had we not converted cm to m, this would not occur, and the result would be incorrect. Finally, because the charge on each sphere is the same, we can further deduce that 18.13 WORKED EXAMPLE Using Coulomb\u2019s law to find the distance between charged objects An engineer measures the force between two ink drops by measuring their acceleration and their diameter. She finds that each member of a pair of ink drops exerts a repulsive force of on its partner. If each ink drop carries a charge, how far apart are the ink drops? STRATEGY We know the force and the charge on each ink drop, so we can solve Coulomb\u2019s law for the distance rbetween the ink drops. Do not forget to convert the force into SI units: Solution The charges in Coulomb\u2019s law are so the numerator in Coulomb\u2019s law takes the form. Inserting this into Coulomb\u2019s law and solving for the distance rgives 18.14 or 130 microns (about one-tenth of a millimeter). Discussion The plus-minus sign means that we do not know which ink drop is to the right and which is to the left, but that is not important, because both ink drops are the same. Practice Problems 11. A charge of \u22124 \u00d7 10\u22129 C is a distance of 3 cm from a charge of 3 \u00d7 10\u22129 C. What is the magnitude and direction of the force between them? 1.2 \u00d7 10\u22124 N, and the force is attractive a. 1.2 \u00d7 1014 N, and the force is attractive b. c. 6.74 \u00d7 1023 N, and the force is attractive d. \u2212\u0177, and the force is attractive 12. Two charges are repelled by a force of 2.0 N. If the distance between them triples, what is the force between the charges? a. 0.22 N b. 0.67 N c. 2.0 N 18.0 N d. Access for free at openstax.org. 18.3 \u2022 Electric Field 567 Check Your Understanding 13. How are electrostatic force and charge related? a. The force is proportional to the product of two charges. b. The force is inversely proportional to the product of two charges. c. The force is proportional to any one of the charges between which the force is acting. d. The force is inversely proportional to any one of the charges between which the force is acting. 14. Why is Coulomb\u2019s law called an inverse-square law? a. because the force is proportional to the inverse of the distance squared between charges b. because the force is proportional to the product of two charges c. because the force is proportional to the inverse of the product of two charges d. because the force is proportional to the distance squared between charges 18.3 Electric Field Section Learning Objectives By the end of this section, you will be able to do the following: \u2022 Calculate the strength of an electric field \u2022 Create and interpret drawings of electric fields Section Key Terms electric field test charge You may have heard of a force fieldin science fiction movies, where such fields apply forces at particular positions in space to keep a villain trapped or to protect a spaceship from enemy fire. The concept of a fieldis very useful in physics, although it differs somewhat from what you see in movies. A fieldis a way", " of conceptualizing and mapping the force that surrounds any object and acts on another object at a distance without apparent physical connection. For example, the gravitational field surrounding Earth and all other masses represents the gravitational force that would be experienced if another mass were placed at a given point within the field. Michael Faraday, an English physicist of the nineteenth century, proposed the concept of an electric field. If you know the electric field, then you can easily calculate the force (magnitude and direction) applied to any electric charge that you place in the field. An electric field is generated by electric charge and tells us the force per unit charge at all locations in space around a charge distribution. The charge distribution could be a single point charge; a distribution of charge over, say, a flat plate; or a more complex distribution of charge. The electric field extends into space around the charge distribution. Now consider placing a test charge in the field. A test charge is a positive electric charge whose charge is so small that it does not significantly disturb the charges that create the electric field. The electric field exerts a force on the test charge in a given direction. The force exerted is proportional to the charge of the test charge. For example, if we double the charge of the test charge, the force exerted on it doubles. Mathematically, saying that electric field is the force per unit charge is written as 18.15 where we are considering only electric forces. Note that the electric field is a vector field that points in the same direction as the force on the positive test charge. The units of electric field are N/C. If the electric field is created by a point charge or a sphere of uniform charge, then the magnitude of the force between this point charge Qand the test charge is given by Coulomb\u2019s law where the absolute value is used, because we only consider the magnitude of the force. The magnitude of the electric field is then 568 Chapter 18 \u2022 Static Electricity 18.16 This equation gives the magnitude of the electric field created by a point charge Q. The distance rin the denominator is the distance from the point charge, Q, or from the center of a spherical charge, to the point of interest. If the test charge is removed from the electric field, the electric field still exists. To create a three-dimensional map of the electric field, imagine placing the test charge in various locations in the field. At each location, measure the force on the charge, and use the vector equation to calculate the electric field. Draw an arrow at each point where you place the test charge to represent the strength and the direction of the electric field. The length of the arrows should be proportional to the strength of the electric field. If you join together these arrows, you obtain lines. Figure 18.17 shows an image of the threedimensional electric field created by a positive charge. Figure 18.17 Three-dimensional representation of the electric field generated by a positive charge. Just drawing the electric field lines in a plane that slices through the charge gives the two-dimensional electric-field maps shown in Figure 18.18. On the left is the electric field created by a positive charge, and on the right is the electric field created by a negative charge. Notice that the electric field lines point away from the positive charge and toward the negative charge. Thus, a positive test charge placed in the electric field of the positive charge will be repelled. This is consistent with Coulomb\u2019s law, which says that like charges repel each other. If we place the positive charge in the electric field of the negative charge, the positive charge is attracted to the negative charge. The opposite is true for negative test charges. Thus, the direction of the electric field lines is consistent with what we find by using Coulomb\u2019s law. says that the electric field gets stronger as we approach the charge that generates it. For example, at The equation 2 cm from the charge Q(r= 2 cm), the electric field is four times stronger than at 4 cm from the charge (r= 4 cm). Looking at Figure 18.17 and Figure 18.18 again, we see that the electric field lines become denser as we approach the charge that generates it. In fact, the density of the electric field lines is proportional to the strength of the electric field! Figure 18.18 Electric field lines from two point charges. The red point on the left carries a charge of +1 nC, and the blue point on the right carries a charge of \u20131 nC. The arrows point in the direction that a positive test charge would move. The field lines are denser as you approach the point charge. Electric-field maps can be made for several charges or for more complicated charge distributions. The electric field due to multiple charges may be found by adding together the electric field from each individual charge. Because this sum can only be a single number, we know that only a single electric-field line can go through any given point. In other words, electric-field lines", " cannotcross each other. Figure 18.19(a) shows a two-dimensional map of the electric field generated by a charge of +qand a nearby charge of \u2212q. The three-dimensional version of this map is obtained by rotating this map about the axis that goes through both charges. A positive Access for free at openstax.org. test charge placed in this field would experience a force in the direction of the field lines at its location. It would thus be repelled from the positive charge and attracted to the negative charge. Figure 18.19(b) shows the electric field generated by two charges of \u2212q. Note how the field lines tend to repel each other and do not overlap. A positive test charge placed in this field would be attracted to both charges. If you are far from these two charges, where far means much farther than the distance between the charges, the electric field looks like the electric field from a single charge of \u22122q. 18.3 \u2022 Electric Field 569 Figure 18.19 (a) The electric field generated by a positive point charge (left) and a negative point charge of the same magnitude (right). (b) The electric field generated by two equal negative charges. Virtual Physics Probing an Electric Field Click to view content (http://www.openstax.org/l/28charge-field) This simulation shows you the electric field due to charges that you place on the screen. Start by clicking the top checkbox in the options panel on the right-hand side to show the electric field. Drag charges from the buckets onto the screen, move them around, and observe the electric field that they form. To see more precisely the magnitude and direction of the electric field, drag an electric-field sensor, or E-fieldsensor from the bottom bucket, and move it around the screen. GRASP CHECK Two positive charges are placed on a screen. Which statement describes the electric field produced by the charges? a. b. c. d. It is constant everywhere. It is zero near each charge. It is zero halfway between the charges. It is strongest halfway between the charges. WATCH PHYSICS Electrostatics (part 2): Interpreting electric field This video explains how to calculate the electric field of a point charge and how to interpret electric-field maps in general. Note 570 Chapter 18 \u2022 Static Electricity that the lecturer uses dfor the distance between particles instead of r. Note that the point charges are infinitesimally small, so all their charges are focused at a point. When larger charged objects are considered, the distance between the objects must be measured between the center of the objects. Click to view content (https://www.youtube.com/embed/0YOGrTNgGhE) GRASP CHECK True or false\u2014If a point charge has electric field lines that point into it, the charge must be ositive. a. b. true false WORKED EXAMPLE What is the charge? Look at the drawing of the electric field in Figure 18.20. What is the relative strength and sign of the three charges? Figure 18.20 Map of electric field due to three charged particles. STRATEGY We know the electric field extends out from positive charge and terminates on negative charge. We also know that the number of electric field lines that touch a charge is proportional to the charge. Charge 1 has 12 fields coming out of it. Charge 2 has six field lines going into it. Charge 3 has 12 field lines going into it. Solution The electric-field lines come out of charge 1, so it is a positive charge. The electric-field lines go into charges 2 and 3, so they are negative charges. The ratio of the charges is of charge 2.. Thus, magnitude of charges 1 and 3 is twice that Discussion Although we cannot determine the precise charge on each particle, we can get a lot of information from the electric field regarding the magnitude and sign of the charges and where the force on a test charge would be greatest (or least). WORKED EXAMPLE Electric field from doorknob A doorknob, which can be taken to be a spherical metal conductor, acquires a static electricity charge of the electric field 1.0 cm in front of the doorknob? The diameter of the doorknob is 5.0 cm. STRATEGY Because the doorknob is a conductor, the entire charge is distributed on the outside surface of the metal. In addition, because the doorknob is assumed to be perfectly spherical, the charge on the surface is uniformly distributed, so we can treat the doorknob as if all the charge were located at the center of the doorknob. The validity of this simplification will be proved in a to indicate the outward direction later physics course. Now sketch the doorknob, and define your coordinate system. Use What is Access for free at openstax.org. perpendicular to the door, with at the center of the do", "orknob (as shown in the figure below). 18.3 \u2022 Electric Field 571 If the diameter of the doorknob is 5.0 cm, its radius is 2.5 cm. We want to know the electric field 1.0 cm from the surface of the doorknob, which is a distance from the center of the doorknob. We can use the equation to find the magnitude of the electric field. The direction of the electric field is determined by the sign of the charge, which is negative in this case. Solution Inserting the charge gives and the distance into the equation 18.17 Because the charge is negative, the electric-field lines point toward the center of the doorknob. Thus, the electric field at is. Discussion This seems like an enormous electric field. Luckily, it takes an electric field roughly 100 times stronger ( air to break down and conduct electricity. Also, the weight of an adult is about feel a force on the protons in your hand as you reach for the doorknob? The reason is that your hand contains an equal amount of negative charge, which repels the negative charge in the doorknob. A very small force might develop from polarization in your hand, but you would never notice it. ) to cause, so why don\u2019t you Practice Problems 15. What is the magnitude of the electric field from 20 cm from a point charge of q= 33 nC? 7.4 \u00d7 103 N/C 1.48 \u00d7 103 N / C 7.4 \u00d7 1012 N / C a. b. c. d. 0 16. A \u221210 nC charge is at the origin. In which direction does the electric field from the charge point at x+ 10 cm? a. The electric field points away from negative charges. b. The electric field points toward negative charges. c. The electric field points toward positive charges. d. The electric field points away from positive charges. Check Your Understanding 17. When electric field lines get closer together, what does that tell you about the electric field? 572 Chapter 18 \u2022 Static Electricity a. The electric field is inversely proportional to the density of electric field lines. b. The electric field is directly proportional to the density of electric field lines. c. The electric field is not related to the density of electric field lines. d. The electric field is inversely proportional to the square root of density of electric field lines. 18. If five electric-field lines come out of a +5 nC charge, how many electric-field lines should come out of a +20 nC charge? a. five field lines 10 field lines b. c. 15 field lines d. 20 field lines 18.4 Electric Potential Section Learning Objectives By the end of this section, you will be able to do the following: \u2022 Explain the similarities and differences between electric potential energy and gravitational potential energy \u2022 Calculate the electric potential difference between two point charges and in a uniform electric field Section Key Terms electric potential electric potential energy As you learned in studying gravity, a mass in a gravitational field has potential energy, which means it has the potential to accelerate and thereby increase its kinetic energy. This kinetic energy can be used to do work. For example, imagine you want to use a stone to pound a nail into a piece of wood. You first lift the stone high above the nail, which increases the potential energy of the stone-Earth system\u2014because Earth is so large, it does not move, so we usually shorten this by saying simply that the potential energy of the stone increases. When you drop the stone, gravity converts the potential energy into kinetic energy. When the stone hits the nail, it does work by pounding the nail into the wood. The gravitational potential energy is the work that a mass can potentially do by virtue of its position in a gravitational field. Potential energy is a very useful concept, because it can be used with conservation of energy to calculate the motion of masses in a gravitational field. Electric potential energy works much the same way, but it is based on the electric field instead of the gravitational field. By virtue of its position in an electric field, a charge has an electric potential energy. If the charge is free to move, the force due to the electric field causes it to accelerate, so its potential energy is converted to kinetic energy, just like a mass that falls in a gravitational field. This kinetic energy can be used to do work. The electric potential energy is the work that a charge can do by virtue of its position in an electric field. The analogy between gravitational potential energy and electric potential energy is depicted in Figure 18.21. On the left, the ballEarth system gains gravitational potential energy when the ball is higher in Earth's gravitational field. On the right, the twocharge system gains electric potential energy when the positive charge is farther from the negative charge. Access for free at openstax.org. 18.4 \u2022 Electric Potential 573 Figure 18.21 On the left, the gravitational field points toward Earth.", " The higher the ball is in the gravitational field, the higher the potential energy is of the Earth-ball system. On the right, the electric field points toward a negative charge. The farther the positive charge is from the negative charge, the higher the potential energy is of the two-charge system. Let\u2019s use the symbol potential energy decreases. Conservation of energy tells us that the work done by the gravitational field to make the mass accelerate must equal the loss of potential energy of the mass. If we use the symbol to denote gravitational potential energy. When a mass falls in a gravitational field, its gravitational to denote this work, then where the minus sign reflects the fact that the potential energy of the ball decreases. The work done by gravity on the mass is 18.18 18.19 where Fis the force due to gravity, and are the initial and final positions of the ball, respectively. The negative sign is because gravity points down, which we consider to be the negative direction. For the constantgravitational field near Earth\u2019s surface,. The change in gravitational potential energy of the mass is and 18.20. in electric Note that is just the negative of the height hfrom which the mass falls, so we usually just write We now apply the same reasoning to a charge in an electric field to find the electric potential energy. The change potential energy is the work done by the electric field to move a charge qfrom an initial position ( ). The definition of work does not change, except that now the work is done by the electric field: to a final position electric field is. The change in electric potential energy of the charge is thus. For a charge that falls through a constantelectric field E, the force applied to the charge by the or 18.21 18.22 This equation gives the change in electric potential energy of a charge qwhen it moves from position constantelectric field E. to position in a Figure 18.22 shows how this analogy would work if we were close to Earth\u2019s surface, where gravity is constant. The top image shows a charge accelerating due to a constant electric field. Likewise, the round mass in the bottom image accelerates due to a constant gravitation field. In both cases, the potential energy of the particle decreases, and its kinetic energy increases. 574 Chapter 18 \u2022 Static Electricity Figure 18.22 In the top picture, a mass accelerates due to a constant electric field. In the bottom picture, the mass accelerates due to a constant gravitational field. WATCH PHYSICS Analogy between Gravity and Electricity This video discusses the analogy between gravitational potential energy and electric potential energy. It reviews the concepts of work and potential energy and shows the connection between a mass in a uniform gravitation field, such as on Earth\u2019s surface, and an electric charge in a uniform electric field. Click to view content (https://www.openstax.org/l/28grav-elec) If the electric field is not constant, then the equation energy becomes more involved. For example, consider the electric potential energy of an assembly of two point charges is not valid, and deriving the electric potential and of the same sign that are initially very far apart. We start by placing charge at the origin of our coordinate system. This in from very far away to a distance rfrom the center of charge takes no electrical energy, because there is no electric field at the origin (because charge charge of charge. The energy it takes to assemble these two charges can be recuperated if we let them fly apart again. Thus, the charges have potential energy when they are a distance rapart. It turns out that the electric a distance rapart is potential energy of a pair of point charges. This requires some effort, because the electric field applies a repulsive force on charge is very far away). We then bring and To recap, if charges are free to move, they can accumulate kinetic energy by flying apart, and this kinetic energy can be used to do work. The maximum amount of work the two charges can do (if they fly infinitely far from each other) is given by the equation above. and Notice that if the two charges have opposite signs, then the potential energy is negative. This means that the charges have more potential to do work when they are farapart than when they are at a distance rapart. This makes sense: Opposite charges attract, so the charges can gain more kinetic energy if they attract each other from far away than if they start at only a short distance apart. Thus, they have more potential to do work when they are far apart. Figure 18.23 summarizes how the electric potential energy depends on charge and separation. 18.23 Access for free at openstax.org. 18.4 \u2022 Electric Potential 575 Figure 18.23 The potential energy depends on the sign of the charges and their separation. The arrows on the charges indicate the direction in which the charges would move if released. When charges with the same sign are far apart, their potential energy is", " low, as shown in the top panel for two positive charges. The situation is the reverse for charges of opposite signs, as shown in the bottom panel. Electric Potential Recall that to find the force applied by a fixed charge Qon any arbitrary test charge q, it was convenient to define the electric field, which is the force per unit charge applied by Qon any test charge that we place in its electric field. The same strategy is used here with electric potential energy: We now define the electric potential V, which is the electric potential energy per unit charge. 18.24 Normally, the electric potential is simply called the potentialor voltage. The units for the potential are J/C, which are given the name volt(V) after the Italian physicist Alessandro Volta (1745\u20131827). From the equation distance rfrom a point charge, the electric potential a is 18.25 This equation gives the energy required per unit charge to bring a charge Mathematically, this is written as from infinity to a distance rfrom a point charge \u221e 18.26 Note that this equation actually represents a differencein electric potential. However, because the second term is zero, it is normally not written, and we speak of the electric potential instead of the electric potential difference, or we just say the potential difference, or voltage). Below, when we consider the electric potential energy per unit charge between two points not infinitely far apart, we speak of electric potential differenceexplicitly. Just remember that electric potential and electric potential difference are really the same thing; the former is used just when the electric potential energy is zero in either the initial or final charge configuration. Coming back now to the electric potential a distance rfrom a point charge we can drop the subscripts and simply write, note that can be any arbitrary point charge, so Now consider the electric potential near a group of charges q1, q2, and q3, as drawn in Figure 18.24. The electric potential is 18.27 576 Chapter 18 \u2022 Static Electricity derived by considering the electric field. Electric fields follow the principle of superposition and can be simply added together, so the electric potential from different charges also add together. Thus, the electric potential of a point near a group of charges is where 18.24. are the distances from the center of charges to the point of interest, as shown in Figure 18.28 Figure 18.24 The potential at the red point is simply the sum of the potentials due to each individual charge. Now let\u2019s consider the electric potential in a uniform electric field. From the equation potential difference in going from to in a uniform electric field Eis, we see that the TIPS FOR SUCCESS Notice from the equation that the electric field can be written as 18.29 18.30 which means that the electric field has units of V/m. Thus, if you know the potential difference between two points, calculating the electric field is very simple\u2014you simply divide the potential difference by the distance! Notice that a positive charge in a region with high potential will experience a force pushing it toward regions of lower potential. In this sense, potential is like pressure for fluids. Imagine a pipe containing fluid, with the fluid at one end of the pipe under high pressure and the fluid at the other end of the pipe under low pressure. If nothing prevents the fluid from flowing, it will flow from the high-pressure end to the low-pressure end. Likewise, a positive charge that is free to move will move from a region with high potential to a region with lower potential. WATCH PHYSICS Voltage This video starts from electric potential energy and explains how this is related to electric potential (or voltage). The lecturer calculates the electric potential created by a uniform electric field. Click to view content (https://www.openstax.org/l/28voltage) GRASP CHECK What is the voltage difference between the positions a. and in an electric field of? Access for free at openstax.org. b. c. d. 18.4 \u2022 Electric Potential 577 LINKS TO PHYSICS Electric Animals Many animals generate and/or detect electric fields. This is useful for activities such as hunting, defense, navigation, communication, and mating. Because salt water is a relatively good conductor, electric fish have evolved in all the world\u2019s oceans. These fish have intrigued humans since the earliest times. In the nineteenth century, parties were even organized where the main attraction was getting a jolt from an electric fish! Scientists also studied electric fish to learn about electricity. Alessandro Volta based his research that led to batteries in 1799 on electric fish. He even referred to batteries as artificial electric organs, because he saw them as imitations of the electric organs of electric fish. Animals that generate electricity are called electrogenicand those that detect electric fields are called electroreceptive. Most fish that are electrogenic are also electroreceptive. One of the most well-known electric fish is the electric eel (see", " Figure 18.25), which is both electrogenic and electroreceptive. These fish have three pairs of organs that produce the electric charge: the main organ, Hunter\u2019s organ, and Sach\u2019s organ. Together, these organs account for more than 80percent of the fish\u2019s body. Electric eels can produce electric discharges of much greater voltage than what you would get from a standard wall socket. These discharges can stun or even kill their prey. They also use low-intensity discharges to navigate. The electric fields they generate reflect off nearby obstacles or animals and are then detected by electroreceptors in the eel\u2019s skin. The three organs that produce electricity contain electrolytes, which are substances that ionize when dissolved in water (or other liquids). An ionized atom or molecule is one that has lost or gained at least one electron, so it carries a net charge. Thus, a liquid solution containing an electrolyte conducts electricity, because the ions in the solution can move if an electric field is applied. To produce large discharges, the main organ is used. It contains approximately 6,000 rows of electroplaques connected in a long chain. Connected this way, the voltage between electroplaques adds up, creating a large final voltage. Each electroplaque consists of a column of cells controlled by an excitor nerve. When triggered by the excitor nerve, the electroplaques allow ionized sodium to flow through them, creating a potential difference between electroplaques. These potentials add up, and a large current can flow through the electrolyte. This geometry is reflected in batteries, which also use stacks of plates to produce larger potential differences. Figure 18.25 An electric eel in its natural environment. (credit: Steven G. Johnson) GRASP CHECK If an electric eel produces 1,000 V, which voltage is produced by each electroplaque in the main organ? a. 0.17 mV b. c. d. 1.7 mV 17 mV 170 mV 578 Chapter 18 \u2022 Static Electricity WORKED EXAMPLE X-ray Tube Dentists use X-rays to image their patients\u2019 teeth and bones. The X-ray tubes that generate X-rays contain an electron source separated by about 10 cm from a metallic target. The electrons are accelerated from the source to the target by a uniform electric field with a magnitude of about 100 kN/C, as drawn in Figure 18.26. When the electrons hit the target, X-rays are produced. (a) What is the potential difference between the electron source and the metallic target? (b) What is the kinetic energy of the electrons when they reach the target, assuming that the electrons start at rest? Figure 18.26 In an X-ray tube, a large current flows through the electron source, causing electrons to be ejected from the electron source. The ejected electrons are accelerated toward the target by the electric field. When they strike the target, X-rays are produced. STRATEGY FOR (A) Use the equation as point in the negative xdirection. This way, the force qand Eare negative. Thus, and the target position as. to find the potential difference given a constant electric field. Define the source position. To accelerate the electrons in the positive xdirection, the electric field must on the electrons will point in the positive xdirection, because both Solution for (a) and Using electron source and the target is, the equation tells us that the potential difference between the Discussion for (a) The potential difference is positive, so the energy per unit positive charge is higher at the target than at the source. This means that free positive charges would fall from the target to the source. However, electrons are negative charges, so they accelerate from the source toward the target, gaining kinetic energy as they go. STRATEGY FOR (B) Apply conservation of energy to find the final kinetic energy of the electrons. In going from the source to the target, the change The change in in electric potential energy plus the change in kinetic energy of the electrons must be zero, so electric potential energy for moving through a constant electric field is given by the equation 18.31 where the electric field is the change in kinetic energy is simply their final kinetic energy, so.. Because the electrons start at rest, their initial kinetic energy is zero. Thus, Solution for (b) Again gives and. The charge of an electron is. Conservation of energy 18.32 Access for free at openstax.org. Inserting the known values into the right-hand side of this equation gives 18.4 \u2022 Electric Potential 579 18.33 Discussion for (b) This is a very small energy. However, electrons are very small, so they are easy to accelerate, and this energy is enough to make. The result is that an electron go extremely fast. You can find their speed by using the definition of kinetic energy, the electrons are moving at more", " than 100 million miles per hour! WORKED EXAMPLE Electric Potential Energy of Doorknob and Dust Speck Consider again the doorknob from the example in the previous section. The doorknob is treated as a spherical conductor with a on its surface. What is the electric potential energy between the doorknob and a speck of uniform static charge dust carrying a charge at 1.0 cm from the front surface of the doorknob? The diameter of the doorknob is 5.0 cm. STRATEGY As we did in the previous section, we treat the charge as if it were concentrated at the center of the doorknob. Again, as you will be able to validate in later physics classes, we can make this simplification, because the charge is uniformly distributed over the surface of the spherical object. Make a sketch of the situation and define a coordinate system, as shown in the image below. We at the center of the doorknob. If the diameter of use the doorknob is 5.0 cm, its radius is 2.5 cm. Thus, the speck of dust 1.0 cm from the surface of the doorknob is a distance to indicate the outward direction perpendicular to the door, with from the center of the doorknob. To solve this problem, use the equation. Solution The charge on the doorknob is gives. The distance. Inserting these values into the equation and the charge on the speck of dust is Discussion The energy is negative, which means that the energy will decrease that is, get even morenegative as the speck of dust approaches the doorknob. This helps explain why dust accumulates on objects that carry a static charge. However, note that insulators normally collect more static charge than conductors, because any charge that accumulates on insulators cannot move about on the insulator to find a way to escape. They must simply wait to be removed by some passing moist speck of dust or other host. 18.34 580 Chapter 18 \u2022 Static Electricity Practice Problems 19. What is the electric potential 10 cm from a \u221210 nC charge? a. 9.0 \u00d7 102 V b. 9.0 \u00d7 103 V c. 9.0 \u00d7 104 V d. 9.0 \u00d7 105 V 20. An electron accelerates from 0 to 10 \u00d7 104 m/s in an electric field. Through what potential difference did the electron travel? The mass of an electron is 9.11 \u00d7 10\u201331 kg, and its charge is \u22121.60 \u00d7 10\u201319 C. a. 29 mV b. 290 mV c. 2,900 mV d. 29 V Check Your Understanding 21. Gravitational potential energy is the potential for two masses to do work by virtue of their positions with respect to each other. What is the analogous definition of electric potential energy? a. Electric potential energy is the potential for two charges to do work by virtue of their positions with respect to the origin point. b. Electric potential energy is the potential for two charges to do work by virtue of their positions with respect to infinity. c. Electric potential energy is the potential for two charges to do work by virtue of their positions with respect to each other. d. Electric potential energy is the potential for single charges to do work by virtue of their positions with respect to their final positions. 22. A negative charge is 10 m from a positive charge. Where would you have to move the negative charge to increase the potential energy of the system? a. The negative charge should be moved closer to the positive charge. b. The negative charge should be moved farther away from the positive charge. c. The negative charge should be moved to infinity. d. The negative charge should be placed just next to the positive charge. 18.5 Capacitors and Dielectrics Section Learning Objectives By the end of this section, you will be able to do the following: \u2022 Calculate the energy stored in a charged capacitor and the capacitance of a capacitor \u2022 Explain the properties of capacitors and dielectrics Section Key Terms capacitor dielectric Capacitors Consider again the X-ray tube discussed in the previous sample problem. How can a uniform electric field be produced? A single positive charge produces an electric field that points away from it, as in. This field is not uniform, because the space between the lines increases as you move away from the charge. However, if we combine a positive and a negative charge, we obtain the electric field shown in (a). Notice that, between the charges, the electric field lines are more equally spaced. What happens if we place, say, five positive charges in a line across from five negative charges, as in Figure 18.27? Now the region between the lines of charge contains a fairly uniform electric field. Access for free at openstax.org. 18.5 \u2022 Capacitors and Dielectrics 581 Figure 18.27 The red dots are positive charges, and the blue dots are negative charges. The", " electric-field direction is shown by the red arrows. Notice that the electric field between the positive and negative dots is fairly uniform. We can extend this idea even further and into two dimensions by placing two metallic plates face to face and charging one with positive charge and the other with an equal magnitude of negative charge. This can be done by connecting one plate to the positive terminal of a battery and the other plate to the negative terminal, as shown in Figure 18.28. The electric field between these charged plates will be extremely uniform. 582 Chapter 18 \u2022 Static Electricity Figure 18.28 Two parallel metal plates are charged with opposite charge, by connecting the plates to the opposite terminals of a battery. The magnitude of the charge on each plate is the same. Let\u2019s think about the work required to charge these plates. Before the plates are connected to the battery, they are neutral\u2014that is, they have zero net charge. Placing the first positive charge on the left plate and the first negative charge on the right plate requires very little work, because the plates are neutral, so no opposing charges are present. Now consider placing a second positive charge on the left plate and a second negative charge on the right plate. Because the first two charges repel the new arrivals, a force must be applied to the two new charges over a distance to put them on the plates. This is the definition of work, which means that, compared with the first pair, more work is required to put the second pair of charges on the plates. To place the third positive and negative charges on the plates requires yet more work, and so on. Where does this work come from? The battery! Its chemical potential energy is converted into the work required to separate the positive and negative charges. Although the battery does work, this work remains within the battery-plate system. Therefore, conservation of energy tells us that, if the potential energy of the battery decreases to separate charges, the energy of another part of the system must increase by the same amount. In fact, the energy from the battery is stored in the electric field between the plates. This idea is analogous to considering that the potential energy of a raised hammer is stored in Earth\u2019s gravitational field. If the gravitational field were to disappear, the hammer would have no potential energy. Likewise, if no electric field existed between the plates, no energy would be stored between them. If we now disconnect the plates from the battery, they will hold the energy. We could connect the plates to a lightbulb, for example, and the lightbulb would light up until this energy was used up. These plates thus have the capacity to store energy. For this reason, an arrangement such as this is called a capacitor. A capacitor is an arrangement of objects that, by virtue of their geometry, can store energy an electric field. Various real capacitors are shown in Figure 18.29. They are usually made from conducting plates or sheets that are separated by Access for free at openstax.org. an insulating material. They can be flat or rolled up or have other geometries. 18.5 \u2022 Capacitors and Dielectrics 583 The capacity of a capacitor is defined by its capacitance C, which is given by Figure 18.29 Some typical capacitors. (credit: Windell Oskay) 18.35 where Qis the magnitudeof the charge on each capacitor plate, and Vis the potential difference in going from the negative plate to the positive plate. This means that both Qand Vare always positive, so the capacitance is always positive. We can see from the equation for capacitance that the units of capacitance are C/V, which are called farads (F) after the nineteenth-century English physicist Michael Faraday. makes sense: A parallel-plate capacitor (like the one shown in Figure 18.28) the size of a football field The equation could hold a lot of charge without requiring too much work per unit charge to push the charge into the capacitor. Thus, Qwould be large, and Vwould be small, so the capacitance Cwould be very large. Squeezing the same charge into a capacitor the size of a fingernail would require much more work, so Vwould be very large, and the capacitance would be much smaller. Although the equation makes it seem that capacitance depends on voltage, in fact it does not. For a given capacitor, the ratio of the charge stored in the capacitor to the voltage difference between the plates of the capacitor always remains the same. Capacitance is determined by the geometry of the capacitor and the materials that it is made from. For a parallel-plate capacitor with nothing between its plates, the capacitance is given by 18.36 where Ais the area of the plates of the capacitor and dis their separation. We use nothing between its plates (in the next section, we\u2019ll see what happens when this is not the case). The constant zerois called the permittivity of free space, and", " its value is instead of C, because the capacitor has read epsilon 18.37 Coming back to the energy stored in a capacitor, we can ask exactly how much energy a capacitor stores. If a capacitor is charged by putting a voltage Vacross it for example, by connecting it to a battery with voltage V\u2014the electrical potential energy stored in the capacitor is 18.38 Notice that the form of this equation is similar to that for kinetic energy,. WATCH PHYSICS Where does Capacitance Come From? This video shows how capacitance is defined and why it depends only on the geometric properties of the capacitor, not on voltage or charge stored. In so doing, it provides a good review of the concepts of work and electric potential. Click to view content (https://www.openstax.org/l/28capacitance) 584 Chapter 18 \u2022 Static Electricity GRASP CHECK If you increase the distance between the plates of a capacitor, how does the capacitance change? a. Doubling the distance between capacitor plates will reduce the capacitance four fold. b. Doubling the distance between capacitor plates will reduce the capacitance two fold. c. Doubling the distance between capacitor plates will increase the capacitance two times. d. Doubling the distance between capacitor plates will increase the capacitance four times. Virtual Physics Charge your Capacitor Click to view content (http://www.openstax.org/l/28charge-cap) For this simulation, choose the tab labeled Introductionat the top left of the screen. You are presented with a parallel-plate capacitor connected to a variable-voltage battery. The battery is initially at zero volts, so no charge is on the capacitor. Slide the battery slider up and down to change the battery voltage, and observe the charges that accumulate on the plates. Display the capacitance, top-plate charge, and stored energy as you vary the battery voltage. You can also display the electric-field lines in the capacitor. Finally, probe the voltage between different points in this circuit with the help of the voltmeter, and probe the electric field in the capacitor with the help of the electric-field detector. GRASP CHECK True or false\u2014 In a capacitor, the stored energy is always positive, regardless of whether the top plate is charged with negative or positive charge. a. b. false true WORKED EXAMPLE Capacitance and Charge Stored in a Parallel Plate Capacitor (a) What is the capacitance of a parallel-plate capacitor with metal plates, each of area 1.00 m2, separated by 0.0010 m? (b) What charge is stored in this capacitor if a voltage of 3.00 \u00d7 103 V is applied to it? STRATEGY FOR (A) Use the equation. Solution for (a) Entering the given values into this equation for the capacitance of a parallel-plate capacitor yields 18.39 Discussion for (a) This small value for the capacitance indicates how difficult it is to make a device with a large capacitance. Special techniques help, such as using very-large-area thin foils placed close together or using a dielectric (to be discussed below). STRATEGY FOR (B) Knowing C, find the charge stored by solving the equation, for the charge Q. Access for free at openstax.org. Solution for (b) The charge Qon the capacitor is 18.5 \u2022 Capacitors and Dielectrics 585 18.40 Discussion for (b) This charge is only slightly greater than typical static electricity charges. More charge could be stored by using a dielectric between the capacitor plates. WORKED EXAMPLE What battery is needed to charge a capacitor? Your friend provides you with a STRATEGY Use the equation capacitor. To store to find the voltage needed to charge the capacitor. on this capacitor, what voltage battery should you buy? Solution Solving for the voltage gives gives. Inserting and Discussion Such a battery should be easy to procure. There is still a question of whether the battery contains enough energy to provide the desired charge. The equation allows us to calculate the required energy. 18.42 18.41 A typical commercial battery can easily provide this much energy. Practice Problems 23. What is the voltage on a 35 \u03bcF with 25 nC of charge? a. 8.75 \u00d7 10\u221213 V b. 0.71 \u00d7 10\u22123 V 1.4 \u00d7 10\u22123 V c. 1.4 \u00d7 103 V d. 24. Which voltage is across a 100 \u03bcF capacitor that stores 10 J of energy? a. \u22124.5 \u00d7 102 V b. 4.5 \u00d7 102 V c. \u00b14.5 \u00d7 102 V d. \u00b19 \u00d7 102 V Dielectrics Before working through some sample problems, let\u2019s look at what happens if we put an insulating material between the plates of a capacitor that has been charged and then disconnected from the charging battery,", " as illustrated in Figure 18.30. Because the material is insulating, the charge cannot move through it from one plate to the other, so the charge Qon the capacitor does not change. An electric field exists between the plates of a charged capacitor, so the insulating material becomes polarized, as shown in the lower part of the figure. An electrically insulating material that becomes polarized in an electric field is called a dielectric. 586 Chapter 18 \u2022 Static Electricity Figure 18.30 shows that the negative charge in the molecules in the material shifts to the left, toward the positive charge of the capacitor. This shift is due to the electric field, which applies a force to the left on the electrons in the molecules of the dielectric. The right sides of the molecules are now missing a bit of negative charge, so their net charge is positive. Figure 18.30 The top and bottom capacitors carry the same charge Q. The top capacitor has no dielectric between its plates. The bottom capacitor has a dielectric between its plates. The molecules in the dielectric are polarized by the electric field of the capacitor. All electrically insulating materials are dielectrics, but some are betterdielectrics than others. A good dielectric is one whose molecules allow their electrons to shift strongly in an electric field. In other words, an electric field pulls their electrons a fair bit away from their atom, but they do not escape completely from their atom (which is why they are insulators). Figure 18.31 shows a macroscopic view of a dielectric in a charged capacitor. Notice that the electric-field lines in the capacitor with the dielectric are spaced farther apart than the electric-field lines in the capacitor with no dielectric. This means that the electric field in the dielectric is weaker, so it stores less electrical potential energy than the electric field in the capacitor with no dielectric. Access for free at openstax.org. 18.5 \u2022 Capacitors and Dielectrics 587 Where has this energy gone? In fact, the molecules in the dielectric act like tiny springs, and the energy in the electric field goes into stretching these springs. With the electric field thus weakened, the voltage difference between the two sides of the capacitor is smaller, so it becomes easier to put more charge on the capacitor. Placing a dielectric in a capacitor before charging it therefore allows more charge and potential energy to be stored in the capacitor. A parallel plate with a dielectric has a capacitance of 18.43 (kappa) is a dimensionless constant called the dielectric constant. Because where capacitance increases when a dielectric is placed between the capacitor plates. The dielectric constant of several materials is shown in Table 18.1. is greater than 1 for dielectrics, the Material Dielectric Constant ( ) Vacuum 1.00000 Air 1.00059 Fused quartz 3.78 Neoprene rubber 6.7 Nylon Paper 3.4 3.7 Polystyrene 2.56 Pyrex glass Silicon oil 5.6 2.5 Strontium titanate 233 Teflon Water 2.1 80 Table 18.1 Dielectric Constants for Various Materials at 20 \u00b0C 588 Chapter 18 \u2022 Static Electricity Figure 18.31 The top and bottom capacitors carry the same charge Q. The top capacitor has no dielectric between its plates. The bottom capacitor has a dielectric between its plates. Because some electric-field lines terminate and start on polarization charges in the dielectric, the electric field is less strong in the capacitor. Thus, for the same charge, a capacitor stores less energy when it contains a dielectric. WORKED EXAMPLE Capacitor for Camera Flash A typical flash for a point-and-shoot camera uses a capacitor of about capacitor plates is 100 V\u2014that is, 100 V is placed \u201cacross the capacitor,\u201d how much energy is stored in the capacitor? (b) If the dielectric used in the capacitor were a 0.010-mm-thick sheet of nylon, what would be the surface area of the capacitor plates? STRATEGY FOR (A). (a) If the potential difference between the, we can use the equation, to find the electric potential energy Given that stored in the capacitor. and Access for free at openstax.org. Solution for (a) Inserting the given quantities into gives 18.5 \u2022 Capacitors and Dielectrics 589 18.44 Discussion for (a) This is enough energy to lift a 1-kg ball about 1 m up from the ground. The flash lasts for about 0.001 s, so the power delivered by the capacitor during this brief time is power, this is not bad for a little capacitor!. Considering that a car engine delivers about 100 kW of STRATEGY FOR (B) Because the capacitor plates are in contact with the", " dielectric, we know that the spacing between the capacitor plates is. From the previous table, the dielectric constant of nylon is. We can now use the equation to find the area Aof the capacitor. Solution (b) Solving the equation for the area Aand inserting the known quantities gives 18.45 Discussion for (b) This is much too large an area to roll into a capacitor small enough to fit in a handheld camera. This is why these capacitors don\u2019t use simple dielectrics but a more advanced technology to obtain a high capacitance. Practice Problems 25. With 12 V across a capacitor, it accepts 10 mC of charge. What is its capacitance? a. 0.83 \u03bcF b. 83 \u03bcF 120 \u03bcF c. d. 830 \u03bcF 26. A parallel-plate capacitor has an area of 10 cm2 and the plates are separated by 100 \u03bcm. If the capacitor contains paper between the plates, what is its capacitance? a. b. c. d. 3.3 \u00d7 10\u221210 F 3.3 \u00d7 10\u22128 F 3.3 \u00d7 10\u22126 F 3.3 \u00d7 10\u22124 F Check Your Understanding 27. If the area of a parallel-plate capacitor doubles, how is the capacitance affected? a. The capacitance will remain same. b. The capacitance will double. c. The capacitance will increase four times. d. The capacitance will increase eight times. 28. If you double the area of a parallel-plate capacitor and reduce the distance between the plates by a factor of four, how is the capacitance affected? 590 Chapter 18 \u2022 Static Electricity a. b. c. d. It will increase by a factor of two. It will increase by a factor of four. It will increase by a factor of six. It will increase by a factor of eight. Access for free at openstax.org. Chapter 18 \u2022 Key Terms 591 KEY TERMS capacitor arrangement of objects that can store electrical energy by virtue of their geometry conductor material through which electric charge can easily move, such as metals Coulomb\u2019s law describes the electrostatic force between charged objects, which is proportional to the charge on each object and inversely proportional to the square of the distance between the objects of negative electric charge induction creating an unbalanced charge distribution in an object by moving a charged object toward it (but without touching) insulator material through which a charge does not move, such as rubber inverse-square law law that has the form of a ratio, with the denominator being the distance squared dielectric electrically insulating material that becomes law of conservation of charge states that total charge is polarized in an electric field constant in any process electric field defines the force per unit charge at all locations in space around a charge distribution polarization separation of charge induced by nearby excess charge electric potential the electric potential energy per unit proton subatomic particle that carries the same magnitude charge charge as the electron, but its charge is positive electric potential energy the work that a charge can do by test charge positive electric charge whose with a charge virtue of its position in an electric field electron subatomic particle that carries one indivisible unit magnitude so small that it does not significantly perturb any nearby charge distribution SECTION SUMMARY 18.1 Electrical Charges, Conservation of Charge, and Transfer of Charge \u2022 Electric charge is a conserved quantity, which means it can be neither created nor destroyed. \u2022 Electric charge comes in two varieties, which are called positiveand negative. \u2022 Charges with the same sign repel each other. Charges with opposite signs attract each other. \u2022 Charges can move easily in conducting material. Charges cannot move easily in an insulating material. \u2022 Objects can be charged in three ways: by contact, by conduction, and by induction. \u2022 Although a polarized object may be neutral, its electrical charge is unbalanced, so one side of the object has excess negative charge and the other side has an equal magnitude of excess positive charge. 18.2 Coulomb's law \u2022 Coulomb\u2019s law is an inverse square law and describes the electrostatic force between particles. \u2022 The electrostatic force between charged objects is proportional to the charge on each object and inversely proportional to the distance squared between the objects. If Coulomb\u2019s law gives a negative result, the force is attractive; if the result is positive, the force is repulsive. \u2022 18.3 Electric Field \u2022 The electric field defines the force per unit charge in the space around a charge distribution. \u2022 For a point charge or a sphere of uniform charge, the electric field is inversely proportional to the distance from the point charge or from the center of the sphere. \u2022 Electric-field lines never cross each other. \u2022 More force is applied to a charge in a region with many electric field lines than in a region with few electric field lines. \u2022 Electric field lines start at positive charges and point away from positive charges. They end at negative charges and point toward negative", " charges. 18.4 Electric Potential \u2022 Electric potential energy is a concept similar to gravitational potential energy: It is the potential that charges have to do work by virtue of their positions relative to each other. \u2022 Electric potential is the electric potential energy per unit charge. \u2022 The potential is always measured between two points, where one point may be at infinity. \u2022 Positive charges move from regions of high potential to regions of low potential. \u2022 Negative charges move from regions of low potential to regions of high potential. 18.5 Capacitors and Dielectrics \u2022 The capacitance of a capacitor depends only on the geometry of the capacitor and the materials from which it is made. It does not depend on the voltage across the capacitor. \u2022 Capacitors store electrical energy in the electric field between their plates. 592 Chapter 18 \u2022 Key Equations \u2022 A dielectric material is an insulator that is polarized in \u2022 Putting a dielectric between the plates of a capacitor an electric field. increases the capacitance of the capacitor. KEY EQUATIONS 18.2 Coulomb's law Coulomb\u2019s law 18.3 Electric Field electric field magnitude of electric field of point charge 18.4 Electric Potential change in electric potential energy for a charge that moves in a constant electric field electric potential energy of a charge a distance rfrom a point charge or sphere of uniform charge definition of electric potential change in electric potential for a charge that moves in a constant electric field electric potential of a charge a distance rfrom a point charge or sphere of uniform charge 18.5 Capacitors and Dielectrics capacitance energy stored in a capacitor capacitance of a parallel-plate capacitor CHAPTER REVIEW Concept Items 18.1 Electrical Charges, Conservation of Charge, and Transfer of Charge 1. There are very large numbers of charged particles in most objects. Why, then, don\u2019t most objects exhibit static electric effects? a. Most objects are neutral. b. Most objects have positive charge only. c. Most objects have negative charge only. d. Most objects have excess protons. 2. Can an insulating material be used to charge a conductor? If so, how? If not, why not? a. No, an insulator cannot charge a conductor by induction. b. No, an insulating material cannot charge a conductor. Access for free at openstax.org. c. Yes, an uncharged insulator can charge a conductor by induction. d. Yes, a charged insulator can charge a conductor upon contact. 3. True or false\u2014A liquid can be an insulating material. a. b. true false 18.2 Coulomb's law 4. Two plastic spheres with uniform charge repel each other with a force of 10 N. If you remove the charge from one sphere, what will be the force between the spheres? a. The force will be 15 N. b. The force will be 10 N. c. The force will be 5 N. d. The force will be zero. 5. What creates a greater magnitude of force, two charges +qa distance r apart or two charges \u2013 qthe same distance apart? a. Two charges +qa distance raway b. Two charges \u2212qa distance raway c. The magnitudes of forces are equal. 6. In Newton\u2019s law of universal gravitation, the force between two masses is proportional to the product of the two masses. What plays the role of mass in Coulomb\u2019s law? a. b. c. d. the electric charge the electric dipole the electric monopole the electric quadruple 18.3 Electric Field 7. Why can electric fields not cross each other? a. Many electric-field lines can exist at any given point in space. Chapter 18 \u2022 Chapter Review 593 b. true 10. True or false\u2014The characteristics of an electric field make it analogous to the gravitational field near the surface of Earth. false a. true b. 11. An electron moves in an electric field. Does it move toward regions of higher potential or lower potential? Explain. a. It moves toward regions of higher potential because its charge is negative. It moves toward regions of lower potential because its charge is negative It moves toward regions of higher potential because its charge is positive. It moves toward regions of lower potential because its charge is positive. b. c. d. b. No electric-field lines can exist at any given point in 18.5 Capacitors and Dielectrics space. c. Only a single electric-field line can exist at any given point in space. d. Two electric-field lines can exist at the same point in space. 8. A constant electric field is (4.5 \u00d7 105 N/C)\u0177. In which direction is the force on a \u221220 nC charge placed in this field? a. The direction of the force is in the direction. direction. b. The direction of the force is in the c. The direction of the force is in the \u2212\u0177", " direction. d. The direction of the force is in the +\u0177 direction. 18.4 Electric Potential 9. True or false\u2014The potential from a group of charges is the sum of the potentials from each individual charge. a. false Critical Thinking Items 18.1 Electrical Charges, Conservation of Charge, and Transfer of Charge 15. If you dive into a pool of seawater through which an equal amount of positively and negatively charged particles is moving, will you receive an electric shock? a. Yes, because negatively charged particles are moving. b. No, because positively charged particles are moving. c. Yes, because positively and negatively charged particles are moving. 12. You insert a dielectric into an air-filled capacitor. How does this affect the energy stored in the capacitor? a. Energy stored in the capacitor will remain same. b. Energy stored in the capacitor will decrease. c. Energy stored in the capacitor will increase. d. Energy stored in the capacitor will increase first, and then it will decrease. 13. True or false\u2014 Placing a dielectric between the plates of a capacitor increases the energy of the capacitor. a. b. false true 14. True or false\u2014 The electric field in an air-filled capacitor is reduced when a dielectric is inserted between the plates. a. b. false true d. No, because equal amounts of positively and negatively charged particles are moving. 16. True or false\u2014The high-voltage wires that you see connected to tall metal-frame towers are held aloft by insulating connectors, and these wires are wrapped in an insulating material. a. b. true false 17. By considering the molecules of an insulator, explain how an insulator can be overall neutral but carry a surface charge when polarized. a. Inside the insulator, the oppositely charged ends of 594 Chapter 18 \u2022 Chapter Review b. the molecules cancel each other. Inside the insulator, the oppositely charged ends of the molecules do not cancel each other. c. The electron distribution in all the molecules shifts in every possible direction, leaving an excess of positive charge on the opposite end of each molecule. d. The electron distribution in all the molecules shifts in a given direction, leaving an excess of negative charge on the opposite end of each molecule. 18.2 Coulomb's law 18. In terms of Coulomb\u2019s law, why are water molecules attracted by positive and negative charges? a. Water molecules are neutral. b. Water molecules have a third type of charge that is attracted by positive as well as negative charges. c. Water molecules are polar. d. Water molecule have either an excess of electrons or an excess of protons. 19. A negative lightning strike occurs when a negatively charged cloud discharges its excess electrons to the positively charged ground. If you observe a cloud-tocloud lightning strike, what can you say about the charge on the area of the cloud struck by lightning? a. The area of the cloud that was struck by lightning had a positive charge. b. The area of the cloud that was struck by lightning had a negative charge. repelled or attracted. b. No, because an electrically neutral body can be attracted but not repelled. c. Yes, because an electrically neutral body can be repelled or attracted. d. Yes, because an electrically neutral body can be repelled. 18.4 Electric Potential 22. What is the relationship between voltage and energy? More precisely, what is the relationship between potential difference and electric potential? a. Voltage is the energy per unit mass at some point in space. b. Voltage is the energy per unit length in space. c. Voltage is the energy per unit charge at some point in space. d. Voltage is the energy per unit area in space. 23. Three parallel plates are stacked above each other, with a separation between each plate. If the potential difference between the first two plates is \u0394V1 and the potential between the second two plates is \u0394V2, what is the potential difference between the first and the third plates? a. \u0394V3 = \u0394V2 + \u0394V1 b. \u0394V3 = \u0394V2 \u2212 \u0394V1 c. \u0394V3 = \u0394V2 / \u0394V1 d. \u0394V3 = \u0394V2\u00d7\u0394V1 c. The area of the cloud that was struck by lightning is 18.5 Capacitors and Dielectrics neutral. d. The area of the cloud that was struck by lightning had a third type of charge. 18.3 Electric Field 20. An arbitrary electric field passes through a box-shaped volume. There are no charges in the box. If 11 electricfield lines enter the box, how many electric-field lines must exit the box? a. nine electric field lines 10 electric field lines b. 11 electric field lines c. 12 electric field lines d. 21. In a science-fiction movie, a villain emits a radial electric field to repulse the hero", ". Knowing that the hero is electrically neutral, is this possible? Explain your reasoning. a. No, because an electrically neutral body cannot be 24. When you insert a dielectric into a capacitor, the energy stored in the capacitor decreases. If you take the dielectric out, the energy increases again. Where does this energy go in the former case, and where does the energy come from in the latter case? a. Energy is utilized to remove the dielectric and is released when the dielectric is introduced between the plates. b. Energy is released when the dielectric is added and is utilized when the dielectric is introduced between the plates. c. Energy is utilized to polarize the dielectric and is released when the dielectric is introduced between the plates. d. Energy is released to polarize the dielectric and is utilized when dielectric is introduced between the plates. Access for free at openstax.org. Problems 18.1 Electrical Charges, Conservation of Charge, and Transfer of Charge 25. A dust particle acquires a charge of \u221213 nC. How many excess electrons does it carry? a. 20.8 \u00d7 10\u221228 electrons b. 20.8 \u00d7 \u221219 electrons c. 8.1 \u00d7 1010 electrons d. 8.1 \u00d7 1019 electrons 26. Two identical conducting spheres are charged with a net charge of +5.0 qon the first sphere and a net charge of \u22128.0 qon the second sphere. The spheres are brought together, allowed to touch, and then separated. What is the net charge on each sphere now? a. \u22123.0q b. \u22121.5q c. +1.5q d. +3.0q 18.2 Coulomb's law 27. Two particles with equal charge experience a force of 10 nN when they are 30 cm apart. What is the magnitude of the charge on each particle? -5.8 \u00d7 10-10 C a. -3.2 \u00d7 10-10 C b. c. +3.2 \u00d7 10-10 C d. +1.4 \u00d7 10-5 C 28. Three charges are on a line. The left charge is q1 = 2.0 nC. The middle charge is q2 = 5.0 nC. The right charge is q3 = \u2212 3.0 nC. The left and right charges are 2.0 cm from the middle charge. What is the force on the middle charge? a. \u22125.6 \u00d7 10\u22124 N to the left b. \u22121.12 \u00d7 10\u22124 N to the left c. +1.12 \u00d7 10\u22124 N to the right 5.6 \u00d7 10\u22124 N to the right d. 18.3 Electric Field 29. An electric field (15 N/C)\u1e91applies a force (\u2212 3 \u00d7 10\u20136 N)\u1e91 on a particle. What is the charge on the particle? a. \u22122.0 \u00d7 10\u20137 C b. 2.0 \u00d7 10\u20137 C Chapter 18 \u2022 Chapter Review 595 c. 2.0 \u00d7 10\u20138 C d. 2.0 \u00d7 10\u20139 C 30. Two uniform electric fields are superimposed. The first electric field is. The second electric. With respect to the positive field is xaxis, at which angle will a positive test charge accelerate in this combined field? a. 27\u00b0 54\u00b0 b. c. 90\u00b0 108\u00b0 d. 18.4 Electric Potential 31. You move a charge qfrom ri = 20 cm to rf = 40 cm from a fixed charge Q= 10 nC. What is the difference in potential for these two positions? a. \u22122.2 \u00d7 102 V b. \u22121.7 \u00d7 103 V c. \u22122.2 \u00d7 104 V d. \u22121.7 \u00d7 102 V 32. How much work is required from an outside agent to? move an electron from xi = 0 to xf = 20 cm in an electric field a. b. c. d. 1.6 \u00d7 10\u221215 J 1.6 \u00d7 10\u221216 J 1.6 \u00d7 10\u221220 J 1.6 \u00d7 10\u221218 J 18.5 Capacitors and Dielectrics 33. A 4.12 \u00b5F parallel-plate capacitor has a plate area of 2,000 cm2 and a plate separation of 10 \u00b5m. What dielectric is between the plates? a. b. 466, the dielectric is strontium c. 699, the dielectric is strontium nitrate d. 1,000, the dielectric is strontium chloride 1, the dielectric is strontium titanate 34. What is the capacitance of a metal sphere of radius? a. b. c. d. Performance Task 18.5 Capacitors and Dielectrics 35.", " Newton\u2019s law of universal gravitation is where. This describes the gravitational force between two point masses m1 and m2. Coulomb\u2019s law is 18.46 18.47 596 Chapter 18 \u2022 Test Prep. This describes the where electric force between two point charges q1 and q2. (a) Describe how the force in each case depends on the distance rbetween the objects. How do the forces change if the distance is reduced by half? If the distance is doubled? (b) Describe the similarities and differences between the two laws. Consider the signs of the quantities that create the interaction (i.e., mass and charge), the constants G and k, and their dependence on separation r. (c) Given that the electric force is much stronger than TEST PREP Multiple Choice 18.1 Electrical Charges, Conservation of Charge, and Transfer of Charge 36. A neutral hydrogen atom has one proton and one electron. If you remove the electron, what will be the leftover sign of the charge? a. negative b. positive c. zero d. neutral 37. What is the charge on a proton? a. +8.99 \u00d7 10\u20139 C b. \u22128.99 \u00d7 10\u20139 C c. + 1.60 \u00d7 10\u201319 C d. \u22121.60 \u00d7 10\u201319 C 38. True or false\u2014Carbon is more conductive than pure water. a. b. true false 39. True or false\u2014Two insulating objects are polarized. To cancel the polarization, it suffices to touch them together. true a. false b. 40. How is the charge of the proton related to the charge of the electron? a. The magnitudes of charge of the proton and the electron are equal, but the charge of the proton is positive, whereas the charge of the electron is negative. b. The magnitudes of charge of the proton and the electron are unequal, but the charge of the proton is positive, whereas the charge of the electron is negative. c. The magnitudes of charge of the proton and the electron are equal, but the charge of the proton is Access for free at openstax.org. the gravitational force, discuss why the law for gravitational force was discovered much earlier than the law for electric force. (d) Consider a hydrogen atom, which is a single proton orbited by a single electron. The electric force holds the electron and proton together so that the hydrogen atom has a radius of about force between electron and proton does not change, what would be the approximate radius of the hydrogen atom if. Assuming the? negative, whereas the charge of the electron is positive. d. The magnitudes of charge of the proton and the electron are unequal, but the charge of the proton is negative, whereas the charge of the electron is positive. 18.2 Coulomb's law 41. If you double the distance between two point charges, by which factor does the force between the particles change? 1/2 a. b. 2 c. 4 d. 1/4 42. The combined charge of all the electrons in a dime is hundreds of thousands of coulombs. Because like charges repel, what keeps the dime from exploding? a. The dime has an equal number of protons, with positive charge. b. The dime has more protons than electrons, with positive charge. c. The dime has fewer protons than electrons, with positive charge. d. The dime is polarized, with electrons on one side and protons on the other side. 43. How can you modify the charges on two particles to quadruple the force between them without moving them? a. Increase the distance between the charges by a factor of two. Increase the distance between the charges by a factor of four. Increase the product of the charges by a factor of two Increase the product of the charges by a factor of four. b. c. d. 18.3 Electric Field 44. What is the magnitude of the electric field 12 cm from a charge of 1.5 nC? a. 9.4 \u00d7 107 N/ C 1.1 \u00d7 102 N/C b. c. 9.4 \u00d7 102 N/C d. 9.4 \u00d7 10\u20132 N/C 45. A charge distribution has electric field lines pointing into it. What sign is the net charge? a. positive b. neutral c. final d. negative 46. If five electric field lines come out of point charge q1 and 10 electric-field lines go into point charge q2, what is the ratio q1/q2? a. \u20132 b. \u20131 c. \u20131/2 d. 0 47. True or false\u2014The electric-field lines from a positive point charge spread out radially and point outward. a. b. false true 18.4 Electric Potential 48. What is the potential at 1.0 m from a point charge Q= \u2212", " 25 nC? a. 6.6 \u00d7 102 V b. \u22122.3 \u00d7 102 V c. \u22126.6 \u00d7 102 V d. 2.3 \u00d7 102 V 49. Increasing the distance by a factor of two from a point charge will change the potential by a factor of how Short Answer 18.1 Electrical Charges, Conservation of Charge, and Transfer of Charge 54. Compare the mass of the electron with the mass of the proton. a. The mass of the electron is about 1,000 times that of the proton. b. The mass of the proton is about 1,000 times that of the electron. c. The mass of the electron is about 1,836 times that of the proton. d. The mass of the proton is about 1,836 times that of the electron. Chapter 18 \u2022 Test Prep 597 much? a. 2 b. 4 c. d. 1/2 1/4 50. True or false\u2014Voltageis the common word for potential difference, because this term is more descriptive than potential difference. a. b. false true 18.5 Capacitors and Dielectrics 51. Which magnitude of charge is stored on each plate of a 12 \u00b5F capacitor with 12 V applied across it? a. \u20131.0 \u00d7 10\u20136 C 1.0 \u00d7 10\u20136 C b. c. \u20131.4 \u00d7 10\u20134 C 1.4 \u00d7 10\u20134 C d. 52. What is the capacitance of a parallel-plate capacitor with an area of 200 cm2, a distance of 0.20 mm between the plates, and polystyrene as a dielectric? a. 2.3 nC b. 0.89 nC c. 23 nC d. 8.9 nC 53. Which factors determine the capacitance of a device? a. Capacitance depends only on the materials that make up the device. b. Capacitance depends on the electric field surrounding the device. c. Capacitance depends on the geometric and material parameters of the device. d. Capacitance depends only on the mass of the capacitor 55. The positive terminal of a battery is connected to one connection of a lightbulb, and the other connection of the lightbulb is connected to the negative terminal of the battery. The battery pushes charge through the circuit but does not become charged itself. Does this violate the law of conservation of charge? Explain. a. No, because this is a closed circuit. b. No, because this is an open circuit. c. Yes, because this is a closed circuit. d. Yes, because this is an open circuit. 56. Two flat pieces of aluminum foil lay one on top of the other. What happens if you add charge to the top piece of aluminum foil? a. The charge will distribute over the top of the top 598 Chapter 18 \u2022 Test Prep piece. b. The charge will distribute to the bottom of the bottom piece. c. The inner surfaces will have excess charge of the opposite sign. d. The inner surfaces will have excess charge of the same sign. 57. The students in your class count off consecutively so each student has a number. The odd-numbered students are told to act as negative charge, and the evennumbered students are told to act as positive charge. How would you organize them to represent a polarized material? a. The even-numbered and odd-numbered students will be arranged one after the other. b. Two even-numbered will be followed by two odd- numbered, and so on. c. Even-numbered students will be asked to come to the front, whereas odd-numbered students will be asked to go to the back of the class. d. Half even-numbered and odd-numbered will come to the front, whereas half even-numbered and oddnumbered will go to the back. 58. An ion of iron contains 56 protons. How many electrons must it contain if its net charge is +5e? a. five electrons 51 electrons b. c. 56 electrons d. 61 electrons 59. An insulating rod carries of charge. After rubbing it with a material, you find it carries charge. How much charge was transferred to it? a. b. c. d. of 60. A solid cube carries a charge of +8e. You measure the charge on each face of the cube and find that each face carries +0.5eof charge. Is the cube made of conducting or insulating material? Explain. a. The cube is made of insulating material, because all the charges are on the surface of the cube. b. The cube is made of conducting material, because some of the charges are inside the cube. c. The cube is made of insulating material, because all the charges are on the surface of the cube. d. The cube is made of insulating material, because some of the charges are inside the cube. 61. You have", " four neutral conducting spheres and a charging device that allows you to place charge qon any neutral object. You want to charge one sphere with a Access for free at openstax.org. charge q/2 and the other three with a charge q6. How do you proceed? a. Charge one sphere with charge q. Touch it simultaneously to the three remaining neutral spheres. b. Charge one sphere with charge q. Touch it to one other sphere to produce two spheres with charge. Touch one of these spheres to one other neutral sphere. c. Charge one sphere with charge q. Touch it to one other sphere to produce two spheres with charge. Touch one of these spheres simultaneously to the two remaining neutral spheres. d. Charge one sphere with charge q. Touch it simultaneously to two other neutral spheres to produce three spheres with charge q/3. Touch one of these spheres to one other neutral sphere. 18.2 Coulomb's law 62. Why does dust stick to the computer screen? a. The dust is neutral. b. The dust is polarized. c. The dust is positively charged. d. The dust is negatively charged. 63. The force between two charges is 4 \u00d7 10\u20139 N. If the magnitude of one charge is reduced by a factor of two and the distance between the charges is reduced by a factor of two, what is the new force between the charges? a. 2 \u00d7 10\u20139 N b. 4 \u00d7 10\u20139 N c. 6 \u00d7 10\u20139 N d. 8 \u00d7 10\u20139 N 64. True or false\u2014Coulomb\u2019s constant is k= 8.99 \u00d7 109 N\u00b7m2/C2. Newton\u2019s gravitational constant is G= 6.67 \u00d7 10\u221211 m3/kg\u22c5s2. This tells you about the relative strength of the electrostatic force versus that of gravity. a. b. true false 65. An atomic nucleus contains 56 protons, for iron. Which force would this nucleus apply on an electron at a distance of 10\u00d710\u201312 m? a. 0.65 \u00d7 10\u20134 N b. 0.02 \u00d7 10\u20134 N 1.3 \u00d7 10\u20134 N c. 72.8 \u00d7 10\u20134 N d. 18.3 Electric Field 66. The electric field a distance of 10 km from a storm cloud is 1,000 N/C. What is the approximate charge in the cloud? a. 0.0011 C 11 C b. 110 C c. 1,100 C d. Chapter 18 \u2022 Test Prep 599 d. +400 mC 72. Given the potential difference between two points and the distance between the points, explain how to obtain the electric field between the points. a. Add the electric potential to the distance to obtain 67. Which electric field would produce a 10 N force in the the electric field. +x- direction on a charge of \u2013 10 nC? a. \u2212 1.0 \u00d7 109 N/C 1.0 \u00d7 109 N/C b. 1.0 \u00d7 1010 N/C c. 1.0 \u00d7 1011 N/C d. 68. A positive charge is located at x= 0. When a negative charge is placed at x= 10 cm, what happens to the electric field lines between the charges? a. The electric field lines become denser between the charges. b. The electric field lines become denser between the charges. c. The electric field lines remains same between the charges. d. The electric field lines will be zero between the charges. 18.4 Electric Potential 69. The energy required to bring a charge q= \u2212 8.8 nC from far away to 5.5 cm from a point charge Q is 13 mJ. What is the potential at the final position of q? a. \u2212112 MV b. \u22121.5 MV c. \u22120.66 MV d. +1.5 MV b. Divide the electric potential by the distance to obtain the electric field. c. Multiply the electric potential and the distance to obtain the electric field. d. Subtract the electric potential from the distance to obtain the electric field. 18.5 Capacitors and Dielectrics 73. If you double the voltage across the plates of a capacitor, how is the stored energy affected? a. Stored energy will decrease two times. b. Stored energy will decrease four times. c. Stored energy will increase two times. d. Stored energy will increase four times. 74. A capacitor with neoprene rubber as the dielectric stores 0.185 mJ of energy with a voltage of 50 V across the plates. If the area of the plates is 500 cm2, what is the plate separation? a. 20 \u00b5m b. 20 m c. 80 \u00b5m d. 80 m 75. Explain why a storm cloud before a lightning strike is like a giant capacitor. a. The storm cloud acts as a giant charged capacitor", ", 70. How is electric potential related to electric potential as it can store a large amount of charge. energy? a. Electric potential is the electric potential energy per unit mass at a given position in space. b. Electric potential is the electric potential energy per unit length at a given position in space. This relation is not dimensionally correct. c. Electric potential is the electric potential energy per unit area in space. d. Electric potential is the electric potential energy per unit charge at a given position in space. 71. If it takes 10 mJ to move a charge qfrom xi = 25 cm to xf = \u2212 25 cm in an electric field of charge q? a. \u22121.0 mC b. +0.25 mC c. + 1.0 mC what is the b. The storm cloud acts as a giant charged capacitor, as it contains a high amount of excess charges. c. The storm cloud acts as a giant charged capacitor, as it splits in two capacitor plates with equal and opposite charge. d. The storm cloud acts as a giant charged capacitor, as it splits in two capacitor plates with unequal and opposite charges. 76. A storm cloud is 2 km above the surface of Earth. The lower surface of the cloud is approximately 2 km2 in area. What is the approximate capacitance of this storm cloud-Earth system? a. 9 \u00d7 10\u201315 F b. 9 \u00d7 10-9 F c. d. 17.7 \u00d7 10-15 F 17.7 \u00d7 10-9 F 600 Chapter 18 \u2022 Test Prep Extended Response 18.1 Electrical Charges, Conservation of Charge, and Transfer of Charge 77. Imagine that the magnitude of the charge on the electron differed very slightly from that of the proton. How would this affect life on Earth and physics in general? a. Many macroscopic objects would be charged, so we would experience the enormous force of electricity on a daily basis. b. Many macroscopic objects would be charged, so we would experience the small force of electricity on a daily basis. c. Many macroscopic objects would be charged, but it would not affect life on Earth and physics in general. d. Macroscopic objects would remain neutral, so it would not affect life on Earth and physics in general. 78. True or false\u2014Conservation of charge is like balancing a budget. a. b. true false 79. True or false\u2014Although wood is an insulator, lightning can travel through a tree to reach Earth. a. b. true false touched to a second small metal sphere that is initially neutral. The spheres are then placed 20 cm apart. What is the force between the spheres? 1.02 \u00d7 10\u22127 N a. b. 2.55 \u00d7 10\u22127 N 5.1 \u00d7 10\u22127 N c. d. 20.4 \u00d7 10\u22127 N 18.3 Electric Field 83. Point charges are located at each corner of a square with sides of 5.0 cm. The top-left charge is q1 = 8.0 nC The top right charge is q2 = 4.0 nC. The bottom-right charge is q3 = 4.0 nC. The bottom-left charge is q4 = 8.0 nC. What is the electric field at the point midway between charges q2 and q3? a. b. c. d. 84. A long straight wire carries a uniform positive charge distribution. Draw the electric field lines in a plane containing the wire at a location far from the ends of the wire. Do not worry about the magnitude of the charge on the wire. a. Take the wire on the x-axis, and draw electric-field lines perpendicular to it. b. Take the wire on the x-axis, and draw electric-field lines parallel to it. 80. True or false\u2014An eccentric inventor attempts to levitate c. Take the wire on the y-axis, and draw electric-field by first placing a large negative charge on himself and then putting a large positive charge on the ceiling of his workshop. Instead, while he attempts to place a large negative charge on himself, his clothes fly off. a. b. true false 18.2 Coulomb's law 81. Electrostatic forces are enormous compared to gravitational force. Why do you not notice electrostatic forces in everyday life, whereas you do notice the force due to gravity? a. Because there are two types of charge, but only one type of mass exists. b. Because there is only one type of charge, but two types of mass exist. c. Because opposite charges cancel each other, while gravity does not cancel out. d. Because opposite charges do not cancel each other, while gravity cancels out. 82. A small metal sphere with a net charge of 3.0 nC is lines along it. d. Take the wire on the z-axis, and draw electric-field lines along it. 18.4 Electric Potential 85. A square grid", " has charges of Q= 10 nC are each corner. The sides of the square at 10 cm. How much energy does it require to bring a q= 1.0 nC charge from very far away to the point at the center of this square? 1.3 \u00d7 10\u22126 J a. b. 2.5 \u00d7 10\u22126 J 3.8 \u00d7 10\u22126 J c. 5.1 \u00d7 10\u22126 J d. 86. How are potential difference and electric-field strength related for a constant electric field? a. The magnitude of electric-field strength is equivalent to the potential divided by the distance. b. The magnitude of electric-field strength is equivalent to the product of the electric potential and the distance. c. The magnitude of electric-field strength is Access for free at openstax.org. Chapter 18 \u2022 Test Prep 601 equivalent to the difference between magnitude of the electric potential and the distance. d. The magnitude of electric-field strength is equivalent to the sum of the magnitude of the electric potential and the distance. 88. Explain why capacitance should be inversely proportional to the separation between the plates of a capacitor. a. Capacitance is directly proportional to the electric field, which is inversely proportional to the distance between the capacitor plates. 18.5 Capacitors and Dielectrics b. Capacitance is inversely proportional to the electric 87. A 12 \u03bcF air-filled capacitor has 12 V across it. If the surface charge on each capacitor plate is \u03c3= 7.2 mC / m2, what is the attractive force of one capacitor plate toward the other? a. 0.81 \u00d7 105 N b. 0.81 \u00d7 106 N 1.2 \u00d7 105 N c. 1.2 \u00d7 106 N d. field, which is inversely proportional to the distance between the capacitor plates. c. Capacitance is inversely proportional to the electric field, which is directly proportional to the distance between the capacitor plates. d. Capacitance is directly proportional to the electric field, which is directly proportional to the distance between the capacitor plates. 602 Chapter 18 \u2022 Test Prep Access for free at openstax.org. CHAPTER 19 Electrical Circuits Figure 19.1 Electric energy in massive quantities is transmitted from this hydroelectric facility, the Srisailam power station located along the Krishna River in India, by the movement of charge\u2014that is, by electric current. (credit: Chintohere, Wikimedia Commons) Chapter Outline 19.1 Ohm's law 19.2 Series Circuits 19.3 Parallel Circuits 19.4 Electric Power The flicker of numbers on a handheld calculator, nerve impulses carrying signals of vision to the brain, an INTRODUCTION ultrasound device sending a signal to a computer screen, the brain sending a message for a baby to twitch its toes, an electric train pulling into a station, a hydroelectric plant sending energy to metropolitan and rural users\u2014these and many other examples of electricity involve electric current, which is the movement of charge. Humanity has harnessed electricity, the basis of this technology, to improve our quality of life. Whereas the previous chapter concentrated on static electricity and the fundamental force underlying its behavior, the next two chapters will be devoted to electric and magnetic phenomena involving current. In addition to exploring applications of electricity, we shall gain new insights into the workings of nature. 604 Chapter 19 \u2022 Electrical Circuits 19.1 Ohm's law Section Learning Objectives By the end of this section, you will be able to do the following: \u2022 Describe how current is related to charge and time, and distinguish between direct current and alternating current \u2022 Define resistance and verbally describe Ohm\u2019s law \u2022 Calculate current and solve problems involving Ohm\u2019s law Section Key Terms alternating current ampere conventional current direct current electric current nonohmic ohmic Ohm\u2019s law resistance Direct and Alternating Current Just as water flows from high to low elevation, electrons that are free to move will travel from a place with low potential to a place with high potential. A battery has two terminals that are at different potentials. If the terminals are connected by a conducting wire, an electric current (charges) will flow, as shown in Figure 19.2. Electrons will then move from the low-potential terminal of the battery (the negativeend) through the wire and enter the high-potential terminal of the battery (the positiveend). Figure 19.2 A battery has a wire connecting the positive and negative terminals, which allows electrons to move from the negative terminal to the positive terminal. Electric current is the rate at which electric charge moves. A large current, such as that used to start a truck engine, moves a large amount very quickly, whereas a small current, such as that used to operate a hand-held calculator, moves a small amount of charge more slowly. In equation form, electric current Iis defined as is the amount of charge that flows past a", " given area and where The SI unit for electric current is the ampere (A), which is named in honor of the French physicist Andr\u00e9-Marie Amp\u00e8re (1775\u20131836). One ampere is one coulomb per second, or is the time it takes for the charge to move past the area. Electric current moving through a wire is in many ways similar to water current moving through a pipe. To define the flow of water through a pipe, we can count the water molecules that flow past a given section of the pipe. As shown in Figure 19.3, electric current is very similar. We count the number of electrical charges that flow past a section of a conductor; in this case, a wire. Access for free at openstax.org. 19.1 \u2022 Ohm's law 605 Figure 19.3 The electric current moving through this wire is the charge that moves past the cross-section A divided by the time it takes for this charge to move past the section A. Assume each particle qin Figure 19.3 carries a charge, in which case the total charge shown would be. If these charges move past the area Ain a time, then the current would be 19.1 Note that we assigned a positive charge to the charges in Figure 19.3. Normally, negative charges\u2014electrons\u2014are the mobile charge in wires, as indicated in Figure 19.2. Positive charges are normally stuck in place in solids and cannot move freely. However, because a positive current moving to the right is the same as a negative current of equal magnitude moving to the left, as shown in Figure 19.4, we define conventional current to flow in the direction that a positive charge would flow if it could move. Thus, unless otherwise specified, an electric current is assumed to be composed of positive charges. Also note that one Coulomb is a significant amount of electric charge, so 5 A is a very large current. Most often you will see current on the order of milliamperes (mA). Figure 19.4 (a) The electric field points to the right, the current moves to the right, and positive charges move to the right. (b) The equivalent situation but with negative charges moving to the left. The electric field and the current are still to the right. Snap Lab Vegetable Current This lab helps students understand how current works. Given that particles confined in a pipe cannot occupy the same space, pushing more particles into one end of the pipe will force the same number of particles out of the opposite end. This creates a current of particles. Find a straw and dried peas that can move freely in the straw. Place the straw flat on a table and fill the straw with peas. When you push one pea in at one end, a different pea should come out of the other end. This demonstration is a model for 606 Chapter 19 \u2022 Electrical Circuits an electric current. Identify the part of the model that represents electrons and the part of the model that represents the supply of electrical energy. For a period of 30 s, count the number of peas you can push through the straw. When finished, calculate the pea currentby dividing the number of peas by the time in seconds. Note that the flow of peas is based on the peas physically bumping into each other; electrons push each other along due to mutually repulsive electrostatic forces. GRASP CHECK Suppose four peas per second pass through a straw. If each pea carried a charge of be through the straw? a. The electric current would be the pea charge multiplied by b. The electric current would be the pea current calculated in the lab multiplied by c. The electric current would be the pea current calculated in the lab. d. The electric current would be the pea charge divided by time.., what would the electric current. The direction of conventional current is the direction that positive charge would flow. Depending on the situation, positive charges, negative charges, or both may move. In metal wires, as we have seen, current is carried by electrons, so the negative charges move. In ionic solutions, such as salt water, both positively charged and negatively charged ions move. This is also true in nerve cells. Pure positive currents are relatively rare but do occur. History credits American politician and scientist Benjamin Franklin with describing current as the direction that positive charges flow through a wire. He named the type of charge associated with electrons negative long before they were known to carry current in so many situations. As electrons move through a metal wire, they encounter obstacles such as other electrons, atoms, impurities, etc. The electrons scatter from these obstacles, as depicted in Figure 19.5. Normally, the electrons lose energy with each interaction. 1 To keep the electrons moving thus requires a force, which is supplied by an electric field. The electric field in a wire points from the end of the wire at the higher potential to the end of the wire at the lower potential. Electrons, carrying a negative", " charge, move on average (or drift) in the direction opposite the electric field, as shown in Figure 19.5. Figure 19.5 Free electrons moving in a conductor make many collisions with other electrons and atoms. The path of one electron is shown. The average velocity of free electrons is in the direction opposite to the electric field. The collisions normally transfer energy to the conductor, so a constant supply of energy is required to maintain a steady current. So far, we have discussed current that moves constantly in a single direction. This is called direct current, because the electric charge flows in only one direction. Direct current is often called DCcurrent. Many sources of electrical power, such as the hydroelectric dam shown at the beginning of this chapter, produce alternating current, in which the current direction alternates back and forth. Alternating current is often called AC current. Alternating current moves back and forth at regular time intervals, as shown in Figure 19.6. The alternating current that comes from a normal wall socket does not suddenly switch directions. Rather, it increases smoothly up to a maximum current and then smoothly decreases back to zero. It then grows again, but in the opposite direction until it has reached the same maximum value. After that, it decreases smoothly back to zero, and the cycle starts over again. 1This energy is transferred to the wire and becomes thermal energy, which is what makes wires hot when they carry a lot of current. Access for free at openstax.org. 19.1 \u2022 Ohm's law 607 Figure 19.6 With alternating current, the direction of the current reverses at regular time intervals. The graph on the top shows the current versus time. The negative maxima correspond to the current moving to the left. The positive maxima correspond to current moving to the right. The current alternates regularly and smoothly between these two maxima. Devices that use AC include vacuum cleaners, fans, power tools, hair dryers, and countless others. These devices obtain the power they require when you plug them into a wall socket. The wall socket is connected to the power grid that provides an alternating potential (AC potential). When your device is plugged in, the AC potential pushes charges back and forth in the circuit of the device, creating an alternating current. Many devices, however, use DC, such as computers, cell phones, flashlights, and cars. One source of DC is a battery, which provides a constant potential (DC potential) between its terminals. With your device connected to a battery, the DC potential pushes charge in one direction through the circuit of your device, creating a DC current. Another way to produce DC current is by using a transformer, which converts AC potential to DC potential. Small transformers that you can plug into a wall socket are used to charge up your laptop, cell phone, or other electronic device. People generally call this a chargeror a battery, but it is a transformer that transforms AC voltage into DC voltage. The next time someone asks to borrow your laptop charger, tell them that you don\u2019t have a laptop charger, but that they may borrow your converter. WORKED EXAMPLE Current in a Lightning Strike A lightning strike can transfer as many as average electric current in the lightning? STRATEGY electrons from the cloud to the ground. If the strike lasts 2 ms, what is the Use the definition of current,. The charge from electrons is, where is the number of electrons and is the charge on the electron. This gives The time is the duration of the lightning strike. Solution The current in the lightning strike is Discussion 19.2 19.3 608 Chapter 19 \u2022 Electrical Circuits The negative sign reflects the fact that electrons carry the negative charge. Thus, although the electrons flow from the cloud to the ground, the positive current is defined to flow from the ground to the cloud. WORKED EXAMPLE Average Current to Charge a Capacitor In a circuit containing a capacitor and a resistor, it takes 1 min to charge a 16 \u03bcF capacitor by using a 9-V battery. What is the average current during this time? STRATEGY We can determine the charge on the capacitor by using the definition of capacitance:. This gives a charge of a 9-V battery, the voltage across the capacitor will be. When the capacitor is charged by By inserting this expression for charge into the equation for current,, we can find the average current. Solution The average current is 19.4 19.5 Discussion This small current is typical of the current encountered in circuits such as this. Practice Problems 1. 10 nC of charge flows through a circuit in 3.0 \u00d7 10\u22126 s. What is the current during this time? a. The current passes through the circuit is 3.3 \u00d7 10\u22123 A. b. The current passes through the circuit is 30 A. c. The current passes through the circuit is 33 A. d. The current passes through the circuit is 0.3 A. 2. How long would it take a current to charge", " a capacitor with? a. b. c. d. Resistance and Ohm\u2019s Law As mentioned previously, electrical current in a wire is in many ways similar to water flowing through a pipe. The water current that can flow through a pipe is affected by obstacles in the pipe, such as clogs and narrow sections in the pipe. These obstacles slow down the flow of current through the pipe. Similarly, electrical current in a wire can be slowed down by many factors, including impurities in the metal of the wire or collisions between the charges in the material. These factors create a resistance to the electrical current. Resistance is a description of how much a wire or other electrical component opposes the flow of charge through it. In the 19th century, the German physicist Georg Simon Ohm (1787\u20131854) found experimentally that current through a conductor is proportional to the voltage drop across a current-carrying conductor. Access for free at openstax.org. 19.1 \u2022 Ohm's law 609 The constant of proportionality is the resistance Rof the material, which leads to This relationship is called Ohm\u2019s law. It can be viewed as a cause-and-effect relationship, with voltage being the cause and the current being the effect. Ohm\u2019s law is an empirical law like that for friction, which means that it is an experimentally observed phenomenon. The units of resistance are volts per ampere, or V/A. We call a V/A an ohm, which is represented by the uppercase Greek letter omega ( ). Thus, Ohm\u2019s law holds for most materials and at common temperatures. At very low temperatures, resistance may drop to zero (superconductivity). At very high temperatures, the thermal motion of atoms in the material inhibits the flow of electrons, increasing the resistance. The many substances for which Ohm\u2019s law holds are called ohmic. Ohmic materials include good conductors like copper, aluminum, and silver, and some poor conductors under certain circumstances. The resistance of ohmic materials remains essentially the same for a wide range of voltage and current. WATCH PHYSICS Introduction to Electricity, Circuits, Current, and Resistance This video presents Ohm\u2019s law and shows a simple electrical circuit. The speaker uses the analogy of pressure to describe how electric potential makes charge move. He refers to electric potential as electric pressure. Another way of thinking about electric potential is to imagine that lots of particles of the same sign are crowded in a small, confined space. Because these charges have the same sign (they are all positive or all negative), each charge repels the others around it. This means that lots of charges are constantly being pushed towards the outside of the space. A complete electric circuit is like opening a door in the small space: Whichever particles are pushed towards the door now have a way to escape. The higher the electric potential, the harder each particle pushes against the others. GRASP CHECK, two resistors each with resistance If, instead of a single resistor what can you say about the current through the circuit? a. The amount of current through the circuit must decrease by half. b. The amount of current through the circuit must increase by half. c. The current must remain the same through the circuit. d. The amount of current through the circuit would be doubled. are drawn in the circuit diagram shown in the video, Virtual Physics Ohm\u2019s Law Click to view content (http://www.openstax.org/l/28ohms_law) This simulation mimics a simple circuit with batteries providing the voltage source and a resistor connected across the batteries. See how the current is affected by modifying the resistance and/or the voltage. Note that the resistance is modeled as an element containing small scattering centers. These represent impurities or other obstacles that impede the passage of the current. GRASP CHECK In a circuit, if the resistance is left constant and the voltage is doubled (for example, from current change? Does this conform to Ohm\u2019s law? a. The current will get doubled. This conforms to Ohm\u2019s law as the current is proportional to the voltage. b. The current will double. This does not conform to Ohm\u2019s law as the current is proportional to the voltage. to ), how does the 610 Chapter 19 \u2022 Electrical Circuits c. The current will increase by half. This conforms to Ohm\u2019s law as the current is proportional to the voltage. d. The current will decrease by half. This does not conform to Ohm\u2019s law as the current is proportional to the voltage. WORKED EXAMPLE Resistance of a Headlight What is the resistance of an automobile headlight through which 2.50 A flows when 12.0 V is applied to it? STRATEGY Ohm\u2019s law tells us the battery, headlight.. The voltage drop in going through the head", "light is just the voltage rise supplied by. We can use this equation and rearrange Ohm\u2019s law to find the resistance of the Solution Solving Ohm\u2019s law for the resistance of the headlight gives 19.6 Discussion This is a relatively small resistance. As we will see below, resistances in circuits are commonly measured in kW or MW. WORKED EXAMPLE Determine Resistance from Current-Voltage Graph Suppose you apply several different voltages across a circuit and measure the current that runs through the circuit. A plot of your results is shown in Figure 19.7. What is the resistance of the circuit? Access for free at openstax.org. 19.1 \u2022 Ohm's law 611 Figure 19.7 The line shows the current as a function of voltage. Notice that the current is given in milliamperes. For example, at 3 V, the current is 0.003 A, or 3 mA. STRATEGY The plot shows that current is proportional to voltage, which is Ohm\u2019s law. In Ohm\u2019s law ( proportionality is the resistance R. Because the graph shows current as a function of voltage, we have to rearrange Ohm\u2019s law in that form: Figure 19.7, we can calculate the resistance R.. This shows that the slope of the line of Iversus Vis. Thus, if we find the slope of the line in ), the constant of Solution The slope of the line is the risedivided by the run. Looking at the lower-left square of the grid, we see that the line rises by 1 mA (0.001 A) and runs over a voltage of 1 V. Thus, the slope of the line is Equating the slope with and solving for Rgives or 1 k-ohm. Discussion 19.7 19.8 This resistance is greater than what we found in the previous example. Resistances such as this are common in electric circuits, as we will discover in the next section. Note that if the line in Figure 19.7 were not straight, then the material would not be ohmic and we would not be able to use Ohm\u2019s law. Materials that do not follow Ohm\u2019s law are called nonohmic. Practice Problems 3. If you double the voltage across an ohmic resistor, how does the current through the resistor change? a. The current will double. b. The current will increase by half. c. The current will decrease by half. d. The current will decrease by a factor of two. 4. The current through a resistor is. What is the voltage drop across the resistor? a. b. c. d. 612 Chapter 19 \u2022 Electrical Circuits Check Your Understanding 5. What is electric current? a. Electric current is the electric charge that is at rest. b. Electric current is the electric charge that is moving. c. Electric current is the electric charge that moves only from the positive terminal of a battery to the negative terminal. d. Electric current is the electric charge that moves only from a region of lower potential to higher potential. 6. What is an ohmic material? a. An ohmic material is a material that obeys Ohm\u2019s law. b. An ohmic material is a material that does not obey Ohm\u2019s law. c. An ohmic material is a material that has high resistance. d. An ohmic material is a material that has low resistance. 7. What is the difference between direct current and alternating current? a. Direct current flows continuously in every direction whereas alternating current flows in one direction. b. Direct current flows continuously in one direction whereas alternating current reverses its direction at regular time intervals. c. Both direct and alternating current flow in one direction but the magnitude of direct current is fixed whereas the magnitude of alternating current changes at regular intervals of time. d. Both direct and alternating current changes its direction of flow but the magnitude of direct current is fixed whereas the magnitude of alternating current changes at regular intervals of time. 19.2 Series Circuits Section Learning Objectives By the end of this section, you will be able to do the following: \u2022 Interpret circuit diagrams and diagram basic circuit elements \u2022 Calculate equivalent resistance of resistors in series and apply Ohm\u2019s law to resistors in series and apply Ohm\u2019s law to resistors in series Section Key Terms circuit diagram electric circuit equivalent resistance in series resistor steady state Electric Circuits and Resistors Now that we understand the concept of electric current, let\u2019s see what we can do with it. As you are no doubt aware, the modern lifestyle relies heavily on electrical devices. These devices contain ingenious electric circuits, which are complete, closed pathways through which electric current flows. Returning to our water analogy, an electric circuit is to electric charge like a network of pipes is to water: The electric circuit guides electric charge from one point to the next,", " running the charge through various devices along the way to extract work or information. Electric circuits are made from many materials and cover a huge range of sizes, as shown in Figure 19.8. Computers and cell phones contain electric circuits whose features can be as small as roughly a billionth of a meter (a nanometer, or pathways that guide the current in these devices are made by ultraprecise chemical treatments of silicon or other semiconductors. Large power systems, on the other hand, contain electric circuits whose features are on the scale of meters. These systems carry such large electric currents that their physical dimensions must be relatively large. ). The Access for free at openstax.org. 19.2 \u2022 Series Circuits 613 Figure 19.8 The photo on the left shows a chipthat contains complex integrated electric circuitry. Chips such as this are at the heart of devices such as computers and cell phones. The photograph on the right shows some typical electric circuitry required for high-power electric power transmission. The pathways that form electric circuits are made from a conducting material, normally a metal in macroscopic circuits. For example, copper wires inside your school building form the electrical circuits that power lighting, projectors, screens, speakers, etc. To represent an electric circuit, we draw circuit diagrams. We use lines and symbols to represent the elements in the circuit. A simple electric circuit diagram is shown on the left side of Figure 19.9. On the right side is an analogous water circuit, which we discuss below. Figure 19.9 On the left is a circuit diagram showing a battery (in red), a resistor (black zigzag element), and the current I. On the right is the analogous water circuit. The pump is like the battery, the sand filter is like the resistor, the water current is like the electrical current, and the reservoir is like the ground. There are many different symbols that scientists and engineers use in circuit diagrams, but we will focus on four main symbols: the wire, the battery or voltage source, resistors, and the ground. The thin black lines in the electric circuit diagram represent the pathway that the electric charge must follow. These pathways are assumed to be perfect conductors, so electric charge can move along these pathways without losing any energy. In reality, the wires in circuits are not perfect, but they come close enough for our purposes. The zigzag element labeled Ris a resistor, which is a circuit element that provides a known resistance. Macroscopic resistors are often color coded to indicate their resistance, as shown in Figure 19.10. The red element in Figure 19.9 is a battery, with its positive and negative terminals indicated; the longer line represents the positive terminal of the battery, and the shorter line represents the negative terminal. Note that the battery icon is not always colored red; this is done in Figure 19.9 just to make it easy to identify. Finally, the element labeled groundon the lower left of the circuit indicates that the circuit is connected to Earth, which is a large, essentially neutral object containing an infinite amount of charge. Among other things, the ground determines the potential of the negative terminal of the battery. Normally, the potential of the ground is defined to be zero:. This 614 Chapter 19 \u2022 Electrical Circuits means that the entire lower wire in Figure 19.10 is at a voltage of zero volts. Figure 19.10 Some typical resistors. The color bands indicate the value of the resistance of each resistor. The electric current in Figure 19.9 is indicted by the blue line labeled I. The arrow indicates the direction in which positive charge would flow in this circuit. Recall that, in metals, electrons are mobile charge carriers, so negative charges actually flow in the opposite direction around this circuit (i.e., counterclockwise). However, we draw the current to show the direction in which positive charge would move. On the right side of Figure 19.9 is an analogous water circuit. Water at a higher pressure leaves the top of the pump, which is like charges leaving the positive terminal of the battery. The water travels through the pipe, like the charges traveling through the wire. Next, the water goes through a sand filter, which heats up as the water squeezes through. This step is like the charges going through the resistor. When charges flow through a resistor, they do work to heat up the resistor. After flowing through the sand filter, the water has converted its potential energy into heat, so it is at a lower pressure. Likewise, the charges exiting the resistor have converted their potential energy into heat, so they are at a lower voltage. Recall that voltage is just potential energy per charge. Thus, water pressure is analogous to electric potential energy (i.e., voltage). Coming back to the water circuit again, we see that the water returns to the bottom of the pump, which is like the charge returning to the negative terminal of the battery. The water pump uses a source of energy to pump the water back up to a high", " pressure again, giving it the pressure required to go through the circuit once more. The water pump is like the battery, which uses chemical energy to increase the voltage of the charge up to the level of the positive terminal. The potential energy per charge at the positive terminal of the battery is the voltage rating of the battery. This voltage is like water pressure in the upper pipe. Just like a higher pressure forces water to move toward a lower pressure, a higher voltage forces electric charge to flow toward a lower voltage. The pump takes water at low pressure and does work on it, ejecting water at a higher pressure. Likewise, a battery takes charge at a low voltage, does work on it, and ejects charge at a higher voltage. Note that the current in the water circuit of Figure 19.9 is the same throughout the circuit. In other words, if we measured the number of water molecules passing a cross-section of the pipe per unit time at any point in the circuit, we would get the same answer no matter where in the circuit we measured. The same is true of the electrical circuit in the same figure. The electric current is the same at all points in this circuit, including inside the battery and in the resistor. The electric current neither speeds up in the wires nor slows down in the resistor. This would create points where too much or too little charge would be bunched up. Thus, the current is the same at all points in the circuit shown in Figure 19.9. Although the current is the same everywhere in both the electric and water circuits, the voltage or water pressure changes as you move through the circuits. In the water circuit, the water pressure at the pump outlet stays the same until the water goes through the sand filter, assuming no energy loss in the pipe. Likewise, the voltage in the electrical circuit is the same at all points in a given wire, because we have assumed that the wires are perfect conductors. Thus, as indicated by the constant red color of the upper wire in Figure 19.11, the voltage throughout this wire is constant at through the resistor, but once you reach the blue wire, the voltage stays at its new level of terminal of the battery (i.e., the blue terminal of the battery).. The voltage then drops as you go all the way to the negative Access for free at openstax.org. 19.2 \u2022 Series Circuits 615 Figure 19.11 The voltage in the red wire is constant at from the positive terminal of the battery to the top of the resistor. The voltage in the blue wire is constant at from the bottom of the resistor to the negative terminal of the battery. If we go from the blue wire through the battery to the red wire, the voltage increases from we go from the blue wire up through the resistor to the red wire, the voltage also goes from Ohm\u2019s law, we can write to to. Likewise, if. Thus, using is measured from the bottom of the resistor to the top, meaning that the top of the resistor is at a higher Note that voltage than the bottom of the resistor. Thus, current flows from the top of the resistor or higher voltage to the bottom of the resistor or lower voltage. Virtual Physics Battery-Resistor Circuit Click to view content (http://www.openstax.org/l/21batteryresist) Use this simulation to better understand how resistance, voltage, and current are related. The simulation shows a battery with a resistor connected between the terminals of the battery, as in the previous figure. You can modify the battery voltage and the resistance. The simulation shows how electrons react to these changes. It also shows the atomic cores in the resistor and how they are excited and heat up as more current goes through the resistor. Draw the circuit diagram for the circuit, being sure to draw an arrow indicating the direction of the current. Now pick three spots along the wire. Without changing the settings, allow the simulation to run for 20 s while you count the number of electrons passing through that spot. Record the number on the circuit diagram. Now do the same thing at each of the other two spots in the circuit. What do you notice about the number of charges passing through each spot in 20 s? Remember that that current is defined as the rate that charges flow through the circuit. What does this mean about the current through the entire circuit? GRASP CHECK With the voltage slider, give the battery a positive voltage. Notice that the electrons are spaced farther apart in the left wire than they are in the right wire. How does this reflect the voltage in the two wires? a. The voltage between static charges is directly proportional to the distance between them. b. The voltage between static charges is directly proportional to square of the distance between them. c. The voltage between static charges is inversely proportional to the distance between them. d. The voltage between static charges is inversely proportional to square of the distance between them. Other possible circuit elements include capacitors and switches. These are drawn as shown", " on the left side of Figure 19.12. A switch is a device that opens and closes the circuit, like a light switch. It is analogous to a valve in a water circuit, as shown on the right side of Figure 19.12. With the switch open, no current passes through the circuit. With the switch closed, it becomes part of the wire, so the current passes through it with no loss of voltage. The capacitor is labeled C on the left of Figure 19.12. A capacitor in an electrical circuit is analogous to a flexible membrane in a 616 Chapter 19 \u2022 Electrical Circuits water circuit. When the switch is closed in the circuit of Figure 19.12, the battery forces electrical current to flow toward the capacitor, charging the upper capacitor plate with positive charge. As this happens, the voltage across the capacitor plates increases. This is like the membrane in the water circuit: When the valve is opened, the pump forces water to flow toward the membrane, making it stretch to store the excess water. As this happens, the pressure behind the membrane increases. Now if we open the switch, the capacitor holds the voltage between its plates because the charges have nowhere to go. Likewise, if we close the valve, the water has nowhere to go and the membrane maintains the water pressure in the pipe between itself and the valve. If the switch is closed for a long time in the electric circuit or if the valve is open for a long time in the water circuit, the current will eventually stop flowing because the capacitor or the membrane will have become completely charged. Each circuit is now in the steady state, which means that its characteristics do not change over time. In this case, the steady state is characterized by zero current, and this does not change as long as the switch or valve remains in the same position. In the steady state, no electrical current passes through the capacitor, and no water current passes through the membrane. The voltage difference between the capacitor plates will be the same as the battery voltage. In the water circuit, the pressure behind the membrane will be the same as the pressure created by the pump. Although the circuit in Figure 19.12 may seem a bit pointless because all that happens when the switch is closed is that the capacitor charges up, it does show the capacitor\u2019s ability to store charge. Thus, the capacitor serves as a reservoir for charge. This property of capacitors is used in circuits in many ways. For example, capacitors are used to power circuits while batteries are being charged. In addition, capacitors can serve as filters. To understand this, let\u2019s go back to the water analogy. Suppose you have a water hose and are watering your garden. Your friend thinks he\u2019s funny, and kinksthe hose. While the hose is kinked, you experience no water flow. When he lets go, the water starts flowing again. If he does this really fast, you experience water-no water-water-no water, and that\u2019s really no way to water your garden. Now imagine that the hose is filling up a big bucket, and you are watering from the bottom of the bucket. As long as you had water in your bucket to begin with and your friend doesn\u2019t kink the water hose for too long, you would be able to water your garden without the interruptions. Your friend kinking the water hose is filteredby the big bucket\u2019s supply of water, so it does not impact your ability to water the garden. We can think of the interruptions in the current (be it water or electrical current) as noise. Capacitors act in an analogous way as the water bucket to help filter out the noise. Capacitors have so many uses that it is very rare to find an electronic circuit that does not include some capacitors. Figure 19.12 On the left is an electrical circuit containing a battery, a switch, and a capacitor. On the left is the analogous water circuit with a pump, a valve, and a stretchable membrane. The pump is like the battery, the valve is like the switch, and the stretchable membrane is like the capacitor. When the switch is closed, electrical current flows as the capacitor charges and its voltage increases. Likewise in the water circuit, when the valve is open, water current flows as the stretchable membrane stretches and the water pressure behind it increases. Access for free at openstax.org. 19.2 \u2022 Series Circuits 617 WORK IN PHYSICS What It Takes to be an Electrical Engineer Physics is used in a wide variety of fields. One field that requires a very thorough knowledge of physics is electrical engineering. An electrical engineer can work on anything from the large-scale power systems that provide power to big cities to the nanoscale electronic circuits that are found in computers and cell phones (Figure 19.13). In working with power companies, you can be responsible for maintaining the power grid that supplies electrical power to large areas. Although much of this work is done from an", " office, it is common to be called in for overtime duty after storms or other natural events. Many electrical engineers enjoy this part of the job, which requires them to race around the countryside repairing high-voltage transformers and other equipment. However, one of the more unpleasant aspects of this work is to remove the carcasses of unfortunate squirrels or other animals that have wandered into the transformers. Other careers in electrical engineering can involve designing circuits for cell phones, which requires cramming some 10 billion transistors into an electronic chip the size of your thumbnail. These jobs can involve much work with computer simulations and can also involve fields other than electronics. For example, the 1-m-diameter lenses that are used to make these circuits (as of 2015) are so precise that they are shipped from the manufacture to the chip fabrication plant in temperature-controlled trucks to ensure that they are held within a certain temperature range. If they heat up or cool down too much, they deform ever so slightly, rendering them useless for the ultrahigh precision photolithography required to manufacture these chips. In addition to a solid knowledge of physics, electrical engineers must above all be practical. Consider, for example, how one corporation managed to launch some anti-ballistic missiles at the White Sands Missile Test Range in New Mexico in the 1960s. Before launch, the skin of the missile had to be at the same voltage as the rail from which it was launched. The rail was connected to the ground by a large copper wire connected to a stake driven into the sandy earth. The missile, however, was connected by an umbilical cord to the equipment in the control shed a few meters away, which was grounded via a different grounding circuit. Before launching the missile, the voltage difference between the missile skin and the rail had to be less than 2.5 V. After an especially dry spell of weather, the missile could not be launched because the voltage difference stood at 5 V. A group of electrical engineers, including the father of your author, stood around pondering how to reduce the voltage difference. The situation was resolved when one of the engineers realized that urine contains electrolytes and conducts electricity quite well. With that, the four engineers quickly resolved the problem by urinating on the rail spike. The voltage difference immediately dropped to below 2.5 V and the missile was launched on schedule. Figure 19.13 The systems that electrical engineers work on range from microprocessor circuits (left)] to missile systems (right). Virtual Physics Click to view content (http://www.openstax.org/l/21phetcirconstr) Amuse yourself by building circuits of all different shapes and sizes. This simulation provides you with various standard circuit elements, such as batteries, AC voltage sources, resistors, capacitors, light bulbs, switches, etc. You can connect these in any configuration you like and then see the result. Build a circuit that starts with a resistor connected to a capacitor. Connect the free side of the resistor to the positive terminal of a battery and the free side of the capacitor to the negative terminal of the battery. Click the reset dynamics button to see how the current flows starting with no charge on the capacitor. Now right click on the resistor to change its 618 Chapter 19 \u2022 Electrical Circuits value. When you increase the resistance, does the circuit reach the steady state more rapidly or more slowly? GRASP CHECK When the circuit has reached the steady state, how does the voltage across the capacitor compare to the voltage of the battery? What is the voltage across the resistor? a. The voltage across the capacitor is greater than the voltage of the battery. In the steady state, no current flows through this circuit, so the voltage across the resistor is zero. b. The voltage across the capacitor is smaller than the voltage of the battery. In the steady state, finite current flows through this circuit, so the voltage across the resistor is finite. c. The voltage across the capacitor is the same as the voltage of the battery. In the steady state, no current flows through this circuit, so the voltage across the resistor is zero. d. The voltage across the capacitor is the same as the voltage of the battery. In the steady state, finite current flows through this circuit, so the voltage across the resistor is finite. Resistors in Series and Equivalent Resistance Now that we have a basic idea of how electrical circuits work, let\u2019s see what happens in circuits with more than one circuit element. In this section, we look at resistors in series. Components connected in series are connected one after the other in the same branch of a circuit, such as the resistors connected in series on the left side of Figure 19.14. Figure 19.14 On the left is an electric circuit with three resistors R1, R2, and R3 connected in series. On the right is an electric circuit with one resistor Requiv that is equivalent to the combination of the three resistors R1, R2, and R3. We", " will now try to find a single resistance that is equivalent to the three resistors in series on the left side of Figure 19.14. An equivalent resistor is a resistor that has the same resistance as the combined resistance of a set of other resistors. In other words, the same current will flow through the left and right circuits in Figure 19.14 if we use the equivalent resistor in the right circuit. where Iis the current in According to Ohm\u2019s law, the voltage drop Vacross a resistor when a current flows through it is amperes (A) and Ris the resistance in ohms ( ). Another way to think of this is that Vis the voltage necessary to make a current Iflow through a resistance R. Applying Ohm\u2019s law to each resistor on the left circuit of Figure 19.14, we find that the voltage drop across, that across voltage output of the battery, that is. The sum of these voltages equals the, and that across is is is You may wonder why voltages must add up like this. One way to understand this is to go once around the circuit and add up the successive changes in voltage. If you do this around a loop and get back to the starting point, the total change in voltage should be zero, because you end up at the same place that you started. To better understand this, consider the analogy of going for a stroll through some hilly countryside. If you leave your car and walk around, then come back to your car, the total height you gained in your stroll must be the same as the total height you lost, because you end up at the same place as you started. Thus, the gravitational potential energy you gain must be the same as the gravitational potential energy you lose. The same reasoning holds for voltage in going around an electric circuit. Let\u2019s apply this reasoning to the left circuit in Figure 19.14. We start just below the battery and move up through the battery, which contributes a voltage gainof resistors. The voltage dropsby. Next, we got through the in going through resistor in going through resistor in going through, and by, by 19.9 Access for free at openstax.org. 19.2 \u2022 Series Circuits 619 resistor. After going through resistor and set the sum equal to zero. This gives, we arrive back at the starting point, so we add up these four changes in voltage which is the same as the previous equation. Note that the minus signs in front of drops, whereas is a voltage rise. 19.10 are because these are voltage Ohm\u2019s law tells us that gives,, and. Inserting these values into equation Applying this same logic to the right circuit in Figure 19.14 gives Dividing the equation by, we get 19.11 19.12 19.13 This shows that the equivalent resistance for a series of resistors is simply the sum of the resistances of each resistor. In general, Nresistors connected in series can be replaced by an equivalent resistor with a resistance of WATCH PHYSICS Resistors in Series This video discusses the basic concepts behind interpreting circuit diagrams and then shows how to calculate the equivalent resistance for resistors in series. Click to view content (https://www.openstax.org/l/02resistseries) GRASP CHECK True or false\u2014In a circuit diagram, we can assume that the voltage is the same at every point in a given wire. a. b. false true WORKED EXAMPLE Calculation of Equivalent Resistance In the left circuit of the previous figure, suppose the voltage rating of the battery is 12 V, and the resistances are STRATEGY FOR (A) Use the equation for the equivalent resistance of resistors connected in series. Because the circuit has three resistances, we only need to keep three terms, so it takes the form. (a) What is the equivalent resistance? (b) What is the current through the circuit? 19.14 Solution for (a) Inserting the given resistances into the equation above gives 620 Chapter 19 \u2022 Electrical Circuits Discussion for (a) We can thus replace the three resistors with a single 20- resistor. STRATEGY FOR (B) Apply Ohm\u2019s law to the circuit on the right side of the previous figure with the equivalent resistor of 20. Solution for (b) The voltage drop across the equivalent resistor must be the same as the voltage rise in the battery. Thus, Ohm\u2019s law gives 19.15 19.16 Discussion for (b) To check that this result is reasonable, we calculate the voltage drop across each resistor and verify that they add up to the voltage rating of the battery. The voltage drop across each resistor is Adding these voltages together gives which is the voltage rating of the battery. WORKED EXAMPLE 19.17 19.18 Determine the Unknown Resistance The circuit shown in figure below contains three resistors of known value and a third element", " whose resistance Given that the equivalent resistance for the entire circuit is 150, what is the resistance? is unknown. STRATEGY The four resistances in this circuit are connected in series, so we know that they must add up to give the equivalent resistance. We can use this to find the unknown resistance. Solution For four resistances in series, the equation for the equivalent resistance of resistors in series takes the form 19.19 Access for free at openstax.org. Solving for R3 and inserting the known values gives 19.3 \u2022 Parallel Circuits 621 19.20 Discussion The equivalent resistance of a circuit can be measured with an ohmmeter. This is sometimes useful for determining the effective resistance of elements whose resistance is not marked on the element. Check your Understanding 8. Figure 19.15 What circuit element is represented in the figure below? a. a battery b. a resistor c. a capacitor d. an inductor 9. How would a diagram of two resistors connected in series appear? a. b. c. d. 19.3 Parallel Circuits Section Learning Objectives By the end of this section, you will be able to do the following: \u2022 Interpret circuit diagrams with parallel resistors \u2022 Calculate equivalent resistance of resistor combinations containing series and parallel resistors Section Key Terms in parallel Resistors in Parallel In the previous section, we learned that resistors in series are resistors that are connected one after the other. If we instead combine resistors by connecting them next to each other, as shown in Figure 19.16, then the resistors are said to be connected in parallel. Resistors are in parallel when both ends of each resistor are connected directly together. Note that the tops of the resistors are all connected to the same wire, so the voltage at the top of the each resistor is the same. Likewise, the bottoms of the resistors are all connected to the same wire, so the voltage at the bottom of each resistor is the same. This means that the voltage drop across each resistor is the same. In this case, the voltage drop is the voltage rating Vof the battery, because the top and bottom wires connect to the positive and negative terminals of the battery, respectively. Although the voltage drop across each resistor is the same, we cannot say the same for the current running through each resistor. Thus, are not necessarily the same, because the resistors do not necessarily have the same 622 Chapter 19 \u2022 Electrical Circuits resistance. Note that the three resistors in Figure 19.16 provide three different paths through which the current can flow. This means that the equivalent resistance for these three resistors must be less than the smallest of the three resistors. To understand this, imagine that the smallest resistor is the only path through which the current can flow. Now add on the alternate paths by connecting other resistors in parallel. Because the current has more paths to go through, the overall resistance (i.e., the equivalent resistance) will decrease. Therefore, the equivalent resistance must be less than the smallest resistance of the parallel resistors. Figure 19.16 The left circuit diagram shows three resistors in parallel. The voltage Vof the battery is applied across all three resistors. The currents that flow through each branch are not necessarily equal. The right circuit diagram shows an equivalent resistance that replaces the three parallel resistors. To find the equivalent resistance voltage drop across each resistor is V, we obtain of the three resistors, we apply Ohm\u2019s law to each resistor. Because the or 19.21 19.22 We also know from conservation of charge that the three currents through the battery. If this were not true, current would have to be mysteriously created or destroyed somewhere in the circuit, which is physically impossible. Thus, we have must add up to give the current Ithat goes Inserting the expressions for into this equation gives or This formula is just Ohm\u2019s law, with the factor in parentheses being the equivalent resistance. Thus, the equivalent resistance for three resistors in parallel is 19.23 19.24 19.25 19.26 19.27 The same logic works for any number of resistors in parallel, so the general form of the equation that gives the equivalent resistance of Nresistors connected in parallel is 19.28 Access for free at openstax.org. 19.3 \u2022 Parallel Circuits 623 WORKED EXAMPLE Find the Current through Parallel Resistors The three circuits below are equivalent. If the voltage rating of the battery is the circuit and what current runs through the circuit?, what is the equivalent resistance of STRATEGY The three resistors are connected in parallel and the voltage drop across them is Vbattery. Thus, we can apply the equation for the equivalent resistance of resistors in parallel, which takes the form 19.29 The circuit with the equivalent resistance is shown below. Once we know the equivalent resistance, we can use Ohm\u2019s law to find the current in the circuit. Solution Inserting the given values for", " the resistance into the equation for equivalent resistance gives The current through the circuit is thus 19.30 19.31 Discussion Although 0.62 A flows through the entire circuit, note that this current does not flow through each resistor. However, because 624 Chapter 19 \u2022 Electrical Circuits electric charge must be conserved in a circuit, the sum of the currents going through each branch of the circuit must add up to the current going through the battery. In other words, we cannot magically create charge somewhere in the circuit and add this new charge to the current. Let\u2019s check this reasoning by using Ohm\u2019s law to find the current through each resistor. 19.32 As expected, these currents add up to give 0.62 A, which is the total current found going through the equivalent resistor. Also, note that the smallest resistor has the largest current flowing through it, and vice versa. WORKED EXAMPLE Reasoning with Parallel Resistors Without doing any calculation, what is the equivalent resistance of three identical resistors Rin parallel? STRATEGY Three identical resistors Rin parallel make three identical paths through which the current can flow. Thus, it is three times easier for the current to flow through these resistors than to flow through a single one of them. Solution If it is three times easier to flow through three identical resistors Rthan to flow through a single one of them, the equivalent resistance must be three times less: R/3. Discussion Let\u2019s check our reasoning by calculating the equivalent resistance of three identical resistors Rin parallel. The equation for the equivalent resistance of resistors in parallel gives 19.33 Thus, our reasoning was correct. In general, when more paths are available through which the current can flow, the equivalent resistance decreases. For example, if we have identical resistors Rin parallel, the equivalent resistance would be R/10. Practice Problems 10. Three resistors, 10, 20, and 30 \u03a9, are connected in parallel. What is the equivalent resistance? a. The equivalent resistance is 5.5 \u03a9 b. The equivalent resistance is 60 \u03a9 c. The equivalent resistance is 6 \u00d7 103 \u03a9 d. The equivalent resistance is 6 \u00d7 104 \u03a9 11. If a drop occurs across a. Voltage drop across is b. Voltage drop across is c. Voltage drop across is d. Voltage drop across is, and.... is connected in parallel to, what is the voltage drop across? Resistors in Parallel and in Series More complex connections of resistors are sometimes just combinations of series and parallel. Combinations of series and parallel resistors can be reduced to a single equivalent resistance by using the technique illustrated in Figure 19.17. Various parts are identified as either series or parallel, reduced to their equivalents, and further reduced until a single resistance is left. The Access for free at openstax.org. process is more time consuming than difficult. 19.3 \u2022 Parallel Circuits 625 Figure 19.17 This combination of seven resistors has both series and parallel parts. Each is identified and reduced to an equivalent resistance, and these are further reduced until a single equivalent resistance is reached. Let\u2019s work through the four steps in Figure 19.17 to reduce the seven resistors to a single equivalent resistor. To avoid distracting. In step 1, we reduce the two sets of parallel resistors circled by the blue dashed loop. algebra, we\u2019ll assume each resistor is 10 The upper set has three resistors in parallel and will be reduced to a single equivalent resistor resistors in parallel and will be reduced to a single equivalent resistor resistors in parallel, we obtain. Using the equation for the equivalent resistance of. The lower set has two These two equivalent resistances are encircled by the red dashed loop following step 1. They are in series, so we can use the 19.34 626 Chapter 19 \u2022 Electrical Circuits equation for the equivalent resistance of resistors in series to reduce them to a single equivalent resistance step 2, with the result being. This is done in The equivalent resistor pair can be replaced by the equivalent resistor, which is given by appears in the green dashed loop following step 2. This resistor is in parallel with resistor 19.35, so the 19.36 This is done in step 3. The resistor These two resistors are combined in the final step to form the final equivalent resistor, as shown in the purple dashed loop following step 3., which is is in series with the resistor 19.37 Thus, the entire combination of seven resistors may be replaced by a single resistor with a resistance of about 14.5. That was a lot of work, and you might be asking why we do it. It\u2019s important for us to know the equivalent resistance of the entire circuit so that we can calculate the current flowing through the circuit. Ohm\u2019s law tells us that the current flowing through a circuit depends on the resistance of the circuit and the voltage across the circuit. But to know the current, we must first know", " the equivalent resistance. Here is a general approach to find the equivalent resistor for any arbitrary combination of resistors: Identify a group of resistors that are only in parallel or only in series. 1. 2. For resistors in series, use the equation for the equivalent resistance of resistors in series to reduce them to a single equivalent resistance. For resistors in parallel, use the equation for the equivalent resistance of resistors in parallel to reduce them to a single equivalent resistance. 3. Draw a new circuit diagram with the resistors from step 1 replaced by their equivalent resistor. 4. If more than one resistor remains in the circuit, return to step 1 and repeat. Otherwise, you are finished. FUN IN PHYSICS Robot Robots have captured our collective imagination for over a century. Now, this dream of creating clever machines to do our dirty work, or sometimes just to keep us company, is becoming a reality. Roboticshas become a huge field of research and development, with some technology already being commercialized. Think of the small autonomous vacuum cleaners, for example. Figure 19.18 shows just a few of the multitude of different forms robots can take. The most advanced humanoid robots can walk, pour drinks, even dance (albeit not very gracefully). Other robots are bio-inspired, such as the dogbotshown in the middle photograph of Figure 19.18. This robot can carry hundreds of pounds of load over rough terrain. The photograph on the right in Figure 19.18 shows the inner workings of an M-block,developed by the Massachusetts Institute of Technology. These simplelooking blocks contain inertial wheels and electromagnets that allow them to spin and flip into the air and snap together in a variety of shapes. By communicating wirelessly between themselves, they self-assemble into a variety of shapes, such as desks, chairs, and someday maybe even buildings. All robots involve an immense amount of physics and engineering. The simple act of pouring a drink has only recently been mastered by robots, after over 30 years of research and development! The balance and timing that we humans take for granted is in fact a very tricky act to follow, requiring excellent balance, dexterity, and feedback. To master this requires sensors to detect balance, computing power to analyze the data and communicate the appropriate compensating actions, and joints and actuators to implement the required actions. In addition to sensing gravity or acceleration, robots can contain multiple different sensors to detect light, sound, temperature, smell, taste, etc. These devices are all based on the physical principles that you are studying in this text. For example, the optics used for robotic vision are similar to those used in your digital cameras: pixelated semiconducting detectors in which light is Access for free at openstax.org. converted into electrical signals. To detect temperature, simple thermistors may be used, which are resistors whose resistance changes depending on temperature. Building a robot today is much less arduous than it was a few years ago. Numerous companies now offer kits for building robots. These range in complexity something suitable for elementary school children to something that would challenge the best professional engineers. If interested, you may find these easily on the Internet and start making your own robot today. 19.3 \u2022 Parallel Circuits 627 Figure 19.18 Robots come in many shapes and sizes, from the classic humanoidtype to dogbotsto small cubes that self-assemble to perform a variety of tasks. WATCH PHYSICS Resistors in Parallel This video shows a lecturer discussing a simple circuit with a battery and a pair of resistors in parallel. He emphasizes that electrons flow in the direction opposite to that of the positive current and also makes use of the fact that the voltage is the same at all points on an ideal wire. The derivation is quite similar to what is done in this text, but the lecturer goes through it well, explaining each step. Click to view content (https://www.openstax.org/l/28resistors) GRASP CHECK True or false\u2014In a circuit diagram, we can assume that the voltage is the same at every point in a given wire. a. b. false true WATCH PHYSICS Resistors in Series and in Parallel This video shows how to calculate the equivalent resistance of a circuit containing resistors in parallel and in series. The lecturer uses the same approach as outlined above for finding the equivalent resistance. Click to view content (https://www.openstax.org/l/28resistorssp) GRASP CHECK Imagine connected Nidentical resistors in parallel. Each resistor has a resistance of R. What is the equivalent resistance for this group of parallel resistors? a. The equivalent resistance is (R)N. 628 Chapter 19 \u2022 Electrical Circuits b. The equivalent resistance is NR. c. The equivalent resistance is d. The equivalent resistance is WORKED EXAMPLE Find the Current through a Complex Resistor Circuit The battery in the circuit below has a voltage rating of 10 V. What current flows through", " the circuit and in what direction? STRATEGY Apply the strategy for finding equivalent resistance to replace all the resistors with a single equivalent resistance, then use Ohm\u2019s law to find the current through the equivalent resistor. Solution The resistor combination and can be reduced to an equivalent resistance of 19.38 by 19.39 Replacing and with this equivalent resistance gives the circuit below. We now replace the two upper resistors their equivalent resistor and. These resistors are in series, so we add them together to find the equivalent resistance. and the two lower resistors by the equivalent resistor and Replacing the relevant resistors with their equivalent resistor gives the circuit below. Access for free at openstax.org. 19.3 \u2022 Parallel Circuits 629 Now replace the two resistors, which are in parallel, with their equivalent resistor. The resistance of is Updating the circuit diagram by replacing with this equivalent resistance gives the circuit below. Finally, we combine resistors, which are in series. The equivalent resistance is The final circuit is shown below. We now use Ohm\u2019s law to find the current through the circuit. 19.40 19.41 The current goes from the positive terminal of the battery to the negative terminal of the battery, so it flows clockwise in this circuit. Discussion This calculation may seem rather long, but with a little practice, you can combine some steps. Note also that extra significant digits were carried through the calculation. Only at the end was the final result rounded to two significant digits. 630 Chapter 19 \u2022 Electrical Circuits WORKED EXAMPLE Strange-Looking Circuit Diagrams Occasionally, you may encounter circuit diagrams that are not drawn very neatly, such as the diagram shown below. This circuit diagram looks more like how a real circuit might appear on the lab bench. What is the equivalent resistance for the resistors in this diagram, assuming each resistor is 10 and the voltage rating of the battery is 12 V. STRATEGY Let\u2019s redraw this circuit diagram to make it clearer. Then we\u2019ll apply the strategy outlined above to calculate the equivalent resistance. Solution To redraw the diagram, consider the figure below. In the upper circuit, the blue resistors constitute a path from the positive terminal of the battery to the negative terminal. In parallel with this circuit are the red resistors, which constitute another path from the positive to negative terminal of the battery. The blue and red paths are shown more cleanly drawn in the lower circuit diagram. Note that, in both the upper and lower circuit diagrams, the blue and red paths connect the positive terminal of the battery to the negative terminal of the battery. Now it is easier to see that turn is in parallel with the series combination of. The equivalent resistance is are in parallel, and the parallel combination is in series with. This combination in. First, we calculate the blue branch, which contains 19.42 where we show the contribution from the parallel combination of resistors and from the series combination of resistors. We now calculate the equivalent resistance of the red branch, which is 19.43 Inserting these equivalent resistors into the circuit gives the circuit below. Access for free at openstax.org. 19.3 \u2022 Parallel Circuits 631 These two resistors are in parallel, so they can be replaced by a single equivalent resistor with a resistance of 19.44 The final equivalent circuit is show below. Discussion Finding the equivalent resistance was easier with a clear circuit diagram. This is why we try to make clear circuit diagrams, where the resistors in parallel are lined up parallel to each other and at the same horizontal position on the diagram. We can now use Ohm\u2019s law to find the current going through each branch to this circuit. Consider the circuit diagram with and is. The voltage across each of these branches is 12 V (i.e., the voltage rating of the battery). The current in the blue branch 19.45 19.46 The current across the red branch is The current going through the battery must be the sum of these two currents (can you see why?), or 1.4 A. Practice Problems 12. What is the formula for the equivalent resistance of two parallel resistors with resistance R1 and R2? a. Equivalent resistance of two parallel resistors b. Equivalent resistance of two parallel resistors c. Equivalent resistance of two parallel resistors d. Equivalent resistance of two parallel resistors 632 Chapter 19 \u2022 Electrical Circuits 13. Figure 19.19 What is the equivalent resistance for the two resistors shown below? a. The equivalent resistance is 20 \u03a9 b. The equivalent resistance is 21 \u03a9 c. The equivalent resistance is 90 \u03a9 d. The equivalent resistance is 1,925 \u03a9 Check Your Understanding 14. The voltage drop across parallel resistors is ________. a. the same for all resistors b. greater for the larger resistors less for the larger resistors c. d. greater for the smaller resistors 15. Consider a circuit of", " parallel resistors. The smallest resistor is 25 \u03a9. What is the upper limit of the equivalent resistance? a. The upper limit of the equivalent resistance is 2.5 \u03a9. b. The upper limit of the equivalent resistance is 25 \u03a9. c. The upper limit of the equivalent resistance is 100 \u03a9. d. There is no upper limit. 19.4 Electric Power Section Learning Objectives By the end of this section, you will be able to do the following: \u2022 Define electric power and describe the electric power equation \u2022 Calculate electric power in circuits of resistors in series, parallel, and complex arrangements Section Key Terms electric power Power is associated by many people with electricity. Every day, we use electric power to run our modern appliances. Electric power transmission lines are visible examples of electricity providing power. We also use electric power to start our cars, to run our computers, or to light our homes. Power is the rate at which energy of any type is transferred; electric power is the rate at which electric energy is transferred in a circuit. In this section, we\u2019ll learn not only what this means, but also what factors determine electric power. To get started, let\u2019s think of light bulbs, which are often characterized in terms of their power ratings in watts. Let us compare a 25-W bulb with a 60-W bulb (see Figure 19.20). Although both operate at the same voltage, the 60-W bulb emits more light intensity than the 25-W bulb. This tells us that something other than voltage determines the power output of an electric circuit. Access for free at openstax.org. Incandescent light bulbs, such as the two shown in Figure 19.20, are essentially resistors that heat up when current flows through them and they get so hot that they emit visible and invisible light. Thus the two light bulbs in the photo can be considered as two different resistors. In a simple circuit such as a light bulb with a voltage applied to it, the resistance determines the current by Ohm\u2019s law, so we can see that current as well as voltage must determine the power. 19.4 \u2022 Electric Power 633 Figure 19.20 On the left is a 25-W light bulb, and on the right is a 60-W light bulb. Why are their power outputs different despite their operating on the same voltage? The formula for power may be found by dimensional analysis. Consider the units of power. In the SI system, power is given in watts (W), which is energy per unit time, or J/s Recall now that a voltage is the potential energy per unit charge, which means that voltage has units of J/C We can rewrite this equation as and substitute this into the equation for watts to get 19.47 19.48 But a Coulomb per second (C/s) is an electric current, which we can see from the definition of electric current, Qis the charge in coulombs and tis time in seconds. Thus, equation above tells us that electric power is voltage times current, or, where This equation gives the electric power consumed by a circuit with a voltage drop of Vand a current of I. For example, consider the circuit in Figure 19.21. From Ohm\u2019s law, the current running through the circuit is Thus, the power consumed by the circuit is Where does this power go? In this circuit, the power goes primarily into heating the resistor in this circuit. 19.49 19.50 Figure 19.21 A simple circuit that consumes electric power. In calculating the power in the circuit of Figure 19.21, we used the resistance and Ohm\u2019s law to find the current. Ohm\u2019s law gives the current:, which we can insert into the equation for electric power to obtain 634 Chapter 19 \u2022 Electrical Circuits This gives the power in terms of only the voltage and the resistance. We can also use Ohm\u2019s law to eliminate the voltage in the equation for electric power and obtain an expression for power in terms of just the current and the resistance. If we write Ohm\u2019s law as and use this to eliminate Vin the equation, we obtain This gives the power in terms of only the current and the resistance. Thus, by combining Ohm\u2019s law with the equation terms of voltage and resistance and one in terms of current and resistance. Note that only resistance (not capacitance or anything else), current, and voltage enter into the expressions for electric power. This means that the physical characteristic of a circuit that determines how much power it dissipates is its resistance. Any capacitors in the circuit do not dissipate electric power\u2014on the contrary, capacitors either store electric energy or release electric energy back to the circuit. for electric power, we obtain two more expressions for power: one in To clarify how voltage, resistance, current, and power are all related, consider Figure 19.22, which shows the formula wheel. The quantities", " in the center quarter circle are equal to the quantities in the corresponding outer quarter circle. For example, to express a potential V in terms of power and current, we see from the formula wheel that. Figure 19.22 The formula wheel shows how volts, resistance, current, and power are related. The quantities in the inner quarter circles equal the quantities in the corresponding outer quarter circles. WORKED EXAMPLE Find the Resistance of a Lightbulb A typical older incandescent lightbulb was 60 W. Assuming that 120 V is applied across the lightbulb, what is the current through the lightbulb? STRATEGY We are given the voltage and the power output of a simple circuit containing a lightbulb, so we can use the equation find the current Ithat flows through the lightbulb. to Solution Solving for the current and inserting the given values for voltage and power gives 19.51 Thus, a half ampere flows through the lightbulb when 120 V is applied across it. Access for free at openstax.org. 19.4 \u2022 Electric Power 635 Discussion This is a significant current. Recall that household power is AC and not DC, so the 120 V supplied by household sockets is an alternating power, not a constant power. The 120 V is actually the time-averaged power provided by such sockets. Thus, the average current going through the light bulb over a period of time longer than a few seconds is 0.50 A. WORKED EXAMPLE Boot Warmers To warm your boots on cold days, you decide to sew a circuit with some resistors into the insole of your boots. You want 10 W of heat output from the resistors in each insole, and you want to run them from two 9-V batteries (connected in series). What total resistance should you put in each insole? STRATEGY We know the desired power and the voltage (18 V, because we have two 9-V batteries connected in series), so we can use the equation to find the requisite resistance. Solution Solving for the resistance and inserting the given voltage and power, we obtain 19.52 Thus, the total resistance in each insole should be 32 Discussion Let\u2019s see how much current would run through this circuit. We have 18 V applied across a resistance of 32, so Ohm\u2019s law gives 19.53 All batteries have labels that say how much charge they can deliver (in terms of a current multiplied by a time). A typical 9-V alkaline battery can deliver a charge of 565 ), so this heating system would function for a time of (so two 9 V batteries deliver 1,130 19.54 WORKED EXAMPLE Power through a Branch of a Circuit Each resistor in the circuit below is 30. What power is dissipated by the middle branch of the circuit? STRATEGY The middle branch of the circuit contains resistors the equivalent resistance in this branch, and then use in series. The voltage across this branch is 12 V. We will first find to find the power dissipated in the branch. Solution The equivalent resistance is circuit is. The power dissipated by the middle branch of the 636 Chapter 19 \u2022 Electrical Circuits Discussion Let\u2019s see if energy is conserved in this circuit by comparing the power dissipated in the circuit to the power supplied by the battery. First, the equivalent resistance of the left branch is 19.55 The power through the left branch is 19.56 19.57 The right branch contains only, so the equivalent resistance is. The power through the right branch is The total power dissipated by the circuit is the sum of the powers dissipated in each branch. The power provided by the battery is 19.58 19.59 19.60 where Iis the total current flowing through the battery. We must therefore add up the currents going through each branch to obtain I. The branches contributes currents of The total current is and the power provided by the battery is 19.61 19.62 19.63 This is the same power as is dissipated in the resistors of the circuit, which shows that energy is conserved in this circuit. Practice Problems 16. What is the formula for the power dissipated in a resistor? a. The formula for the power dissipated in a resistor is b. The formula for the power dissipated in a resistor is c. The formula for the power dissipated in a resistor is P= IV. d. The formula for the power dissipated in a resistor is P= I2V. 17. What is the formula for power dissipated by a resistor given its resistance and the voltage across it? a. The formula for the power dissipated in a resistor is b. The formula for the power dissipated in a resistor is c. The formula for the power dissipated in a resistor is d. The formula for the power dissipated in a resistor is Access for free at openstax.org. 19.4 \u2022 Electric Power 637 Check your", " Understanding 18. Which circuit elements dissipate power? a. b. c. d. capacitors inductors ideal switches resistors 19. Explain in words the equation for power dissipated by a given resistance. a. Electric power is proportional to current through the resistor multiplied by the square of the voltage across the resistor. b. Electric power is proportional to square of current through the resistor multiplied by the voltage across the resistor. c. Electric power is proportional to current through the resistor divided by the voltage across the resistor. d. Electric power is proportional to current through the resistor multiplied by the voltage across the resistor. 638 Chapter 19 \u2022 Key Terms KEY TERMS alternating current electric current whose direction alternates back and forth at regular intervals ampere unit for electric current; one ampere is one coulomb per second ( ) circuit diagram schematic drawing of an electrical circuit including all circuit elements, such as resistors, capacitors, batteries, and so on conventional current flows in the direction that a positive charge would flow if it could move same as the combined resistance of a group of resistors in parallel when a group of resistors are connected side by side, with the top ends of the resistors connected together by a wire and the bottom ends connected together by a different wire in series when elements in a circuit are connected one after the other in the same branch of the circuit nonohmic material that does not follow Ohm\u2019s law Ohm\u2019s law electric current is proportional to the voltage direct current electric current that flows in a single applied across a circuit or other path direction electric circuit physical network of paths through which electric current can flow electric current electric charge that is moving electric power in a circuit equivalent resistor rate at which electric energy is transferred resistance of a single resistor that is the ohmic material that obeys Ohm\u2019s law resistance how much a circuit element opposes the passage of electric current; it appears as the constant of proportionality in Ohm\u2019s law circuit element that provides a known resistance resistor steady state when the characteristics of a system do not change over time SECTION SUMMARY 19.1 Ohm's law 19.3 Parallel Circuits \u2022 Direct current is constant over time; alternating current \u2022 The equivalent resistance of a group of Nidentical alternates smoothly back and forth over time. resistors Rconnected in parallel is R/N. \u2022 Electrical resistance causes materials to extract work \u2022 from the current that flows through them. In ohmic materials, voltage drop along a path is proportional to the current that runs through the path. 19.2 Series Circuits \u2022 Circuit diagrams are schematic representations of electric circuits. \u2022 Resistors in series are resistors that are connected head to tail. \u2022 The same current runs through all resistors in series; however, the voltage drop across each resistor can be different. \u2022 The voltage is the same at every point in a given wire. \u2022 Connecting resistors in parallel provides more paths for the current to go through, so the equivalent resistance is always less than the smallest resistance of the parallel resistors. \u2022 The same voltage drop occurs across all resistors in parallel; however, the current through each resistor can differ. 19.4 Electric Power \u2022 Electric power is dissipated in the resistances of a circuit. Capacitors do not dissipate electric power. \u2022 Electric power is proportional to the voltage and the current in a circuit. \u2022 Ohm\u2019s law provides two extra expressions for electric power: one that does not involve current and one that does not involve voltage. KEY EQUATIONS 19.1 Ohm's law electric current Iis the charge passes a plane per unit time that an ampere is the coulombs per unit time that pass a plane Access for free at openstax.org. Ohm\u2019s law: the current Iis proportional to the voltage V, with the resistance Rbeing the constant of proportionality 19.2 Series Circuits 19.4 Electric Power Chapter 19 \u2022 Chapter Review 639 the equivalent resistance of N resistors connected in series 19.3 Parallel Circuits the equivalent resistance of Nresistors connected in parallel CHAPTER REVIEW Concept Items 19.1 Ohm's law 1. You connect a resistor across a battery. In which direction do the electrons flow? a. The electrons flow from the negative terminal of the battery to the positive terminal of the battery. b. The electrons flow from the positive terminal of the battery to the negative terminal of the battery. 2. How does current depend on resistance in Ohm\u2019s law? a. Current is directly proportional to the resistance. b. Current is inversely proportional to the resistance. c. Current is proportional to the square of the resistance. d. Current is inversely proportional to the square of the resistance. 3. In the context of electricity, what is resistance? a. Resistance is the property of materials to resist the passage of voltage. b. Resistance is the property of materials to resist the passage of electric current. c. Resistance is the property of", " materials to increase the passage of voltage. d. Resistance is the property of materials to increase the passage of electric current. 4. What is the mathematical formula for Ohm\u2019s law? a. b. c. d. 19.2 Series Circuits 5. In which circuit are all the resistors connected in series? for a given current Iflowing through a potential difference V, the electric power dissipated for a given current Iflowing through a resistance R, the electric power dissipated for a given voltage difference Vacross a resistor R, the electric power dissipated a. b. c. d. 6. What is the voltage and current through the capacitor in the circuit below a long time after the switch is closed? a. 0 V, 0 A b. 0 V, 10 A 10 V, 0 A c. 10 V, 10 A d. decreases the overall resistance. d. Adding resistors in parallel gives the current longer path through which it can flow hence decreases the overall resistance. 19.4 Electric Power 9. To draw the most power from a battery, should you connect a small or a large resistance across its terminals? Explain. a. Small resistance, because smaller resistance will lead to the largest power b. Large resistance, because smaller resistance will lead to the largest power 10. If you double the current through a resistor, by what factor does the power dissipated by the resistor change? a. Power increases by a factor of two. b. Power increases by a factor of four. c. Power increases by a factor of eight. d. Power increases by a factor of 16. material is ohmic. 19.2 Series Circuits 13. Given three batteries (5V, 9V, 12V) and five resistors (10, 20, 30, 40, 50\u03a9) to choose from, what can you choose to form a circuit diagram with a current of 0.175A? You do not need to use all of the components. a. Batteries (5V, 9V) and resistors (30\u03a9, 50\u03a9) connected in series b. Batteries (5V4, 12V) and resistors (10\u03a9, 20\u03a9, 40\u03a9, and 50\u03a9) connected in series. c. Batteries (5V, 9V, and 12V) and resistors (10\u03a9, 20\u03a9, and 30\u03a9) connected in series. 14. What is the maximum resistance possible given a and a resistor of? resistor of a. b. c. d. 15. Rank the points A, B, C, and D in the circuit diagram from lowest voltage to highest voltage. 640 Chapter 19 \u2022 Chapter Review 19.3 Parallel Circuits 7. If you remove resistance from a circuit, does the total resistance of the circuit always decrease? Explain. a. No, because for parallel combination of resistors, the resistance through the remaining circuit increases. b. Yes, because for parallel combination of resistors, the resistance through the remaining circuit increases. 8. Explain why the equivalent resistance of a parallel combination of resistors is always less than the smallest of the parallel resistors. a. Adding resistors in parallel gives the current a shorter path through which it can flow hence decreases the overall resistance. b. Adding resistors in parallel gives the current another path through which it can flow hence decreases the overall resistance. c. Adding resistors in parallel reduce the number of paths through which the current can flow hence Critical Thinking Items 19.1 Ohm's law 11. An accelerator accelerates He nuclei (change = 2e) to a speed of v= 2 \u00d7 106 m/s. What is the current if the linear density of He nuclei is \u03bb= 108 m\u20131? a. I= 9.6 \u00d7 10\u20135 A b. I= 3.2 \u00d7 10\u20135 A c. I= 12.8 \u00d7 10\u20135 A d. I= 6.4 \u00d7 10\u20135 A 12. How can you verify whether a certain material is ohmic? a. Make a resistor from this material and measure the current going through this resistor for several different voltages. If the current is proportional to the voltage, then the material is ohmic. b. Make a resistor from this material and measure the current going through this resistor for several different voltages. If the current is inversely proportional to the voltage, then the material is ohmic. c. Make a resistor from this material and measure the current going through this resistor for several different voltages. If the current is proportional to the square of the voltage, then the material is ohmic. d. Make a resistor from this material and measure the current going through this resistor for several different voltages. If the current is inversely proportional to the square of the voltage, then the Access for free at openstax.org. Chapter 19 \u2022 Chapter Review 641 a. The equivalent resistance", " of the circuit 14 \u03a9. b. The equivalent resistance of the circuit 16.7 \u03a9. c. The equivalent resistance of the circuit 140 \u03a9. d. The equivalent resistance of the circuit 195 \u03a9. 19.4 Electric Power 18. Two lamps have different resistances. (a) If the lamps are connected in parallel, which one is brighter, the lamp with greater resistance or the lamp with less resistance? (b) If the lamps are connected in series, which one is brighter? Note that the brighter lamp dissipates more power. a. (a) lamp with greater resistance; (b) lamp with less resistance (a) lamp with greater resistance; (b) lamp with greater resistance (a) lamp with less resistance; (b) lamp with less resistance (a) lamp with less resistance; (b) lamp with greater resistance b. c. d. 19. To measure the power consumed by your laptop computer, you place an ammeter (a device that measures electric current) in series with its DC power supply. When the screen is off, the computer draws 0.40 A of current. When the screen is on at full brightness, it draws 0.90 A of current. Knowing the DC power supply delivers 16 V, how much power is used by the screen? a. The power used by the screen is \u22128.0 W. b. The power used by the screen is 0.3 W. c. The power used by the screen is 3.2 W. d. The power used by the screen is 8.0 W. which of the following functions would be best to fit the data? Assume that a, b, and care nonzero constants adjusted to fit the data. a. b. c. d. 22. A battery of unknown voltage is attached across a resistor in series with. You add a second battery with so that the voltage across is now and measure of current through resistor. You add a third battery with series with the first two batteries so that the voltage of across current through?. What is the resistance of and measure is in a. A, B, C, D b. B, C, A, D c. C, B, A, D d. D, A, B, C 19.3 Parallel Circuits 16. Can all resistor combinations be reduced to series and parallel combinations? a. No, all practical resistor circuits cannot be reduced to series and parallel combinations. b. Yes, all practical resistor circuits can be reduced to series and parallel combinations. 17. What is the equivalent resistance of the circuit shown below? Figure 19.23 Problems 19.1 Ohm's law 20. What voltage is needed to make 6 C of charge traverse a 100-\u03a9 resistor in 1 min? a. The required voltage is 1 \u00d7 10\u22123 V. b. The required voltage is 10 V. c. The required voltage is 1,000 V. d. The required voltage is 10,000 V. 21. Resistors typically obey Ohm\u2019s law at low currents, but show deviations at higher currents because of heating. Suppose you were to conduct an experiment measuring the voltage, V, across a resistor as a function of current, I, including currents whose deviations from Ohm\u2019s law start to become apparent. For a data plot of Vversus I, 642 Chapter 19 \u2022 Chapter Review a. b. c. d. 19.2 Series Circuits 23. What is the voltage drop across two 80-\u03a9 resistors connected in series with 0.15 A flowing through them? a. 12 V b. 24 V c. 36 V d. 48 V 24. In this circuit, the voltage drop across the upper resistor is 4.5 V. What is the battery voltage? a. 4.5V 7.5V b. 12V c. 18V d. 19.3 Parallel Circuits 25. What is the equivalent resistance of this circuit? Performance Task 19.4 Electric Power 29. 1. An incandescent light bulb (i.e., an old-fashioned light bulb with a little wire in it) 2. A lightbulb socket to hold the light bulb 3. A variable voltage source 4. An ammeter Procedure \u2022 Screw the lightbulb into its socket. Connect the Access for free at openstax.org. a. The equivalent resistance of the circuit is 32.7 \u03a9. b. The equivalent resistance of the circuit is 100 \u03a9. c. The equivalent resistance of the circuit is 327 \u03a9. d. The equivalent resistance of the circuit is 450 \u03a9. 26. What is the equivalent resistance of the circuit shown below? a. The equivalent resistance is 25 \u03a9. b. The equivalent resistance is 50 \u03a9. c. The equivalent resistance is 75 \u03a9. d. The equivalent resistance is 100 \u03a9. 19.4 Electric Power 27. When 12 V are applied across a resistor, it dissipates 120 W of", " power. What is the current through the resistor? a. The current is 1,440 A. b. The current is 10 A. c. The current is 0.1 A. d. The current is 0.01 A. 28. Warming 1 g of water requires 1 J of energy per. How long would it take to warm 1 L of water from 20 to 40 \u00b0C if you immerse in the water a 1-kW resistor connected across a 9.0-V batteries aligned in series? a. 10 min b. 20 min c. 30 min d. 40 min positive terminal of the voltage source to the input of the ammeter. Connect the output of the ammeter to one connection of the socket. Connect the other connection of the socket to the negative terminal of the voltage source. Ensure that the voltage source is set to supply DC voltage and that the ammeter is set to measure DC amperes. The desired circuit is shown below. Chapter 19 \u2022 Test Prep 643 10 rows. Label the left column voltsand the right column current.Adjust the voltage source so that it supplies from between 1 and 10 volts DC. For each voltage, write the voltage in the volts column and the corresponding amperage measured by the ammeter in the current column. Make a plot of volts versus current, that is, a plot with volts on the vertical axis and current on the horizontal axis. Use this data and the plot to answer the following questions: 1. What is the resistance of the lightbulb? 2. What is the range of possible error in your result 3. for the resistance? In a single word, how would you describe the curve formed by the data points? \u2022 On a piece of paper, make a two-column table with TEST PREP Multiple Choice 19.1 Ohm's law 30. What are the SI units for electric current? a. a battery b. a capacitor the ground c. d. a switch a. b. c. d. 31. What is the SI unit for resistance? a. b. c. d. 32. The equivalent unit for an ohm is a ________. a. V/A b. C/m c. d. V/s 33. You put DC across resistor and measure the current through it. With the same voltage across resistor, you measure twice the current. What is the ratio 1? a. b. c. 4 d. 2 19.2 Series Circuits 34. What does the circuit element shown represent? 35. How many 10-\u03a9 resistors must be connected in series to make an equivalent resistance of 80 \u03a9? a. 80 b. 8 c. 20 d. 40 36. Which two circuit elements are represented in the circuit diagram? a. a battery connected in series with an inductor b. a capacitor connected in series with a resistor c. a resistor connected in series with a battery d. an inductor connected in series with a resistor 37. How much current will flow through a 10-V battery with a 100-\u03a9 resistor connected across its terminals? a. 0.1 A 1.0 A b. c. 0 d. 1,000 A 19.3 Parallel Circuits 38. A 10-\u03a9 resistor is connected in parallel to another resistor R. The equivalent resistance of the pair is 8 \u03a9. What is the resistance R? 10 \u03a9 a. b. 20 \u03a9 30 \u03a9 c. 644 Chapter 19 \u2022 Test Prep d. 40 \u03a9 39. Are the resistors shown connected in parallel or in series? Explain. a. The resistors are connected in parallel because the same current flows through all three resistors. b. The resistors are connected in parallel because different current flows through all three resistors. c. The resistors are connected in series because the same current flows through all three resistors. d. The resistors are connected in series because different current flows through all three resistors. 19.4 Electric Power 40. Which equation below for electric power is incorrect? a. b. Short Answer 19.1 Ohm's law 44. True or false\u2014it is possible to produce nonzero DC current by adding together AC currents. a. b. false true 45. What type of current is used in cars? a. alternating current indirect current b. c. direct current d. straight current 46. If current were represented by, voltage by, and resistance by, what would the mathematical expression be for Ohm\u2019s law? a. b. c. d. 47. Give a verbal expression for Ohm\u2019s law. a. Ohm\u2019s law says that the current through a resistor equals the voltage across the resistor multiplied by the resistance of the resistor. b. Ohm\u2019s law says that the voltage across a resistor equals the current through the resistor multiplied by the resistance of the resistor. c. Ohm\u2019s law says that the resistance of the resistor", " equals the current through the resistor multiplied Access for free at openstax.org. c. d. 41. What power is dissipated in a circuit through which flows across a potential drop of? a. b. c. Voltage drop across is d.. 42. How does a resistor dissipate power? a. A resistor dissipates power in the form of heat. b. A resistor dissipates power in the form of sound. c. A resistor dissipates power in the form of light. d. A resistor dissipates power in the form of charge. 43. What power is dissipated in a circuit through which 0.12 A flows across a potential drop of 3.0 V? a. 0.36 W b. 0.011 W c. 5 V d. 2.5 W by the voltage across a resistor. d. Ohm\u2019s law says that the voltage across a resistor equals the square of the current through the resistor multiplied by the resistance of the resistor. 48. What is the current through a 100-\u03a9 resistor with 12 V across it? a. 0 b. 0.12 A c. 8.33 A d. 1,200 A 49. What resistance is required to produce from a battery? a. 1 b. c. 60 120 d. 19.2 Series Circuits 50. Given a circuit with one 9-V battery and with its negative terminal connected to ground. The two paths are connected to ground from the positive terminal: the right path with a 20-\u03a9 and a 100-\u03a9 resistor and the left path with a 50-\u03a9 resistor. How much current will flow in the right branch? a. b. Chapter 19 \u2022 Test Prep 645 c. d. a. b. c. d. 51. Through which branch in the circuit below does the most current flow? 19.3 Parallel Circuits 54. Ten 100-\u03a9 resistors are connected in series. How can you increase the total resistance of the circuit by about 40 percent? a. Adding two 10-\u03a9 resistors increases the total resistance of the circuit by about 40 percent. b. Removing two 10-\u03a9 resistors increases the total resistance of the circuit by about 40 percent. c. Adding four 10-\u03a9 resistors increases the total resistance of the circuit by about 40 percent. d. Removing four 10-\u03a9 resistors increases the total resistance of the circuit by about 40 percent. 55. Two identical resistors are connected in parallel across the terminals of a battery. If you increase the resistance of one of the resistors, what happens to the current through and the voltage across the other resistor? a. The current and the voltage remain the same. b. The current decreases and the voltage remains the a. All of the current flows through the left branch due to the open switch. b. All of the current flows through the right branch due to the open switch in the left branch. c. All of the current flows through the middle branch due to the open switch in the left branch d. There will be no current in any branch of the circuit due to the open switch. 52. What current flows through the resistor in the same. circuit below? c. The current and the voltage increases. d. The current increases and the voltage remains the same. 56. a. b. c. d. 53. What is the equivalent resistance for the circuit below if and? In the circuit below, through which resistor(s) does the most current flow? Through which does the least flow? Explain. a. The most current flows through the 15-\u03a9 resistor because all the current must pass through this resistor. b. The most current flows through the 20-\u03a9 resistor because all the current must pass through this resistor. c. The most current flows through the 25-\u03a9 resistor because it is the highest resistance. d. The same current flows through the all the resistor because all the current must pass through each of the resistors. 19.4 Electric Power 57. You want to increase the power dissipated in a circuit. 646 Chapter 19 \u2022 Test Prep You have the choice between doubling the current or doubling the resistance, with the voltage remaining constant. Which one would you choose? a. doubling the resistance b. doubling the current battery? a. The power dissipated is 2430 W. b. The power dissipated is 270 W. c. The power dissipated is 2.7 W. d. The power dissipated is 0.37 W. 58. You want to increase the power dissipated in a circuit. You have the choice between reducing the voltage or reducing the resistance, with the current remaining constant. Which one would you choose? a. b. reduce the voltage to increase the power reduce the resistance to increase the power 59. What power is dissipated in the circuit consisting of 310-\u03a9 resistors connected in series across a 9.0-V Extended Response 19.1 Ohm's", " law 61. Describe the relationship between current and charge. Include an explanation of how the direction of the current is defined. a. Electric current is the charge that passes through a conductor per unit time. The direction of the current is defined to be the direction in which positive charge would flow. b. Electric current is the charges that move in a conductor. The direction of the current is defined to be the direction in which positive charge would flow. c. Electric current is the charge that passes through a conductor per unit time. The direction of the current is defined to be the direction in which negative charge would flow. d. Electric current is the charges that move in a conductor. The direction of the current is defined to be the direction in which negative charge would flow. 62. What could cause Ohm\u2019s law to break down? a. b. c. d. If small amount of current flows through a resistor, the resistor will heat up so much that it will change state, in violation of Ohm\u2019s law. If excessive amount of current flows through a resistor, the resistor will heat up so much that it will change state, in violation of Ohm\u2019s law. If small amount of current flows through a resistor, the resistor will not heat up so much and it will not change its state, in violation of Ohm\u2019s law. If excessive amount of current flows through a resistor, the resistor will heat up so much that it will not change its state, in violation of Ohm\u2019s law. 63. You connect a single resistor across a battery and find that flows through the circuit. You add Access for free at openstax.org. 60. What power is dissipated in a circuit consisting of three 10-\u03a9 resistors connected in parallel across a 9.0-V battery? a. The power dissipated is 270 W. b. The power dissipated is 30 W. c. The power dissipated is 24 W. d. The power dissipated is 1/24 W. another resistor after the first resistor and find that flows through the circuit. If you have connected in a line one after the other, what resistors would be their total resistance? a. b. c. d. 19.2 Series Circuits 64. Explain why the current is the same at all points in the circuit below. a. b. c. If the current were not constant, the mobile charges would bunch up in places, which means that the voltage would decrease at that point. A lower voltage at some point would push the current in the direction that further decreases the voltage. If the current were not constant, the mobile charges would bunch up in places, which means that the voltage would increase at that point. But a higher voltage at some point would push the current in the direction that decreases the voltage. If the current were not constant, the mobile charges would bunch up in places, which mean that the voltage would increase at that point. A higher voltage at some point would push the current in the direction that further increases the d. voltage. If the current were not constant, the mobile charges would bunch up in places, which mean that the voltage would decrease at that point. But a lower voltage at some point would push the current in the direction that increases the voltage. 65. What is the current through each resistor in the circuit? a. Current through resistors R1, R2, R3, and R4 is 0.48 A, 0.30 A, 1.2 A, and 0.24 A, respectively. b. Current through resistors R1, R2, R3, and R4 is 1200 A, 1920 A, 480 A, and 2400 A, respectively. c. Current through resistors R1, R2, R3, and is R4 2.08 A, 3.34 A, 0.833 A, and 4.17 A, respectively. d. The same amount of current, 0.096 A, flows through all of the resistors. 19.3 Parallel Circuits 66. In a house, a single incoming wire at a high potential with respect to the ground provides electric power. How are the appliances connected between this wire and the Chapter 19 \u2022 Test Prep 647 ground, in parallel or in series? Explain. a. The appliances are connected in parallel to provide different voltage differences across each appliance. b. The appliances are connected in parallel to provide the same voltage difference across each appliance. c. The appliances are connected in series to provide the same voltage difference across each appliance. d. The appliances are connected in series to provide different voltage differences across each appliance. 19.4 Electric Power 67. A single resistor is connected across the terminals of a battery When you attach a second resistor in parallel with the first, does the power dissipated by the system change? a. No, the power dissipated remain same. b. Yes, the power dissipated increases. c. Yes, the", " power dissipated decreases. 68. In a flashlight, the batteries are normally connected in series. Why are they not connected in parallel? a. Batteries are connected in series for higher voltage and power output. b. Batteries are connected in series for lower voltage and power output. c. Batteries are connected in series so that power output is a much lower for the same amount of voltage. d. Batteries are connected in series to reduce the overall loss of energy from the circuit. 648 Chapter 19 \u2022 Test Prep Access for free at openstax.org. CHAPTER 20 Magnetism Figure 20.1 The magnificent spectacle of the Aurora Borealis, or northern lights, glows in the northern sky above Bear Lake near Eielson Air Force Base, Alaska. Shaped by Earth\u2019s magnetic field, this light is produced by radiation spewed from solar storms. (credit: Senior Airman Joshua Strang, Flickr) Chapter Outline 20.1 Magnetic Fields, Field Lines, and Force 20.2 Motors, Generators, and Transformers 20.3 Electromagnetic Induction You may have encountered magnets for the first time as a small child playing with magnetic toys or INTRODUCTION refrigerator magnets. At the time, you likely noticed that two magnets that repulse each other will attract each other if you flip one of them around. The force that acts across the air gaps between magnets is the same force that creates wonders such as the Aurora Borealis. In fact, magnetic effects pervade our lives in myriad ways, from electric motors to medical imaging and computer memory. In this chapter, we introduce magnets and learn how they work and how magnetic fields and electric currents interact. 650 Chapter 20 \u2022 Magnetism 20.1 Magnetic Fields, Field Lines, and Force Section Learning Objectives By the end of this section, you will be able to do the following: \u2022 Summarize properties of magnets and describe how some nonmagnetic materials can become magnetized \u2022 Describe and interpret drawings of magnetic fields around permanent magnets and current-carrying wires \u2022 Calculate the magnitude and direction of magnetic force in a magnetic field and the force on a current- carrying wire in a magnetic field Section Key Terms Curie temperature domain electromagnet electromagnetism ferromagnetic magnetic dipole magnetic field magnetic pole magnetized north pole permanent magnet right-hand rule solenoid south pole Magnets and Magnetization People have been aware of magnets and magnetism for thousands of years. The earliest records date back to ancient times, particularly in the region of Asia Minor called Magnesia\u2014the name of this region is the source of words like magnet. Magnetic rocks found in Magnesia, which is now part of western Turkey, stimulated interest during ancient times. When humans first discovered magnetic rocks, they likely found that certain parts of these rocks attracted bits of iron or other magnetic rocks more strongly than other parts. These areas are called the polesof a magnet. A magnetic pole is the part of a magnet that exerts the strongest force on other magnets or magnetic material, such as iron. For example, the poles of the bar magnet shown in Figure 20.2 are where the paper clips are concentrated. Figure 20.2 A bar magnet with paper clips attracted to the two poles. If a bar magnet is suspended so that it rotates freely, one pole of the magnet will always turn toward the north, with the opposite pole facing south. This discovery led to the compass, which is simply a small, elongated magnet mounted so that it can rotate freely. An example of a compass is shown Figure 20.3. The pole of the magnet that orients northward is called the north pole, and the opposite pole of the magnet is called the south pole. Access for free at openstax.org. 20.1 \u2022 Magnetic Fields, Field Lines, and Force 651 Figure 20.3 A compass is an elongated magnet mounted in a device that allows the magnet to rotate freely. The discovery that one particular pole of a magnet orients northward, whereas the other pole orients southward allowed people to identify the north and south poles of any magnet. It was then noticed that the north poles of two different magnets repel each other, and likewise for the south poles. Conversely, the north pole of one magnet attracts the south pole of other magnets. This situation is analogous to that of electric charge, where like charges repel and unlike charges attract. In magnets, we simply replace charge with pole: Like poles repel and unlike poles attract. This is summarized in Figure 20.4, which shows how the force between magnets depends on their relative orientation. Figure 20.4 Depending on their relative orientation, magnet poles will either attract each other or repel each other. Consider again the fact that the pole of a magnet that orients northward is called the north pole of the magnet. If unlike poles attract, then the magnetic pole of Earth that is close to the geographic North Pole must be a magnetic south pole! Likewise, the magnetic", " pole of Earth that is close to the geographic South Pole must be a magnetic north pole. This situation is depicted in Figure 20.5, in which Earth is represented as containing a giant internal bar magnet with its magnetic south pole at the geographic North Pole and vice versa. If we were to somehow suspend a giant bar magnet in space near Earth, then the north pole of the space magnet would be attracted to the south pole of Earth\u2019s internal magnet. This is in essence what happens with a compass needle: Its magnetic north pole is attracted to the magnet south pole of Earth\u2019s internal magnet. 652 Chapter 20 \u2022 Magnetism Figure 20.5 Earth can be thought of as containing a giant magnet running through its core. The magnetic south pole of Earth\u2019s magnet is at the geographic North Pole, so the north pole of magnets is attracted to the North Pole, which is how the north pole of magnets got their name. Likewise, the south pole of magnets is attracted to the geographic South Pole of Earth. What happens if you cut a bar magnet in half? Do you obtain one magnet with two south poles and one magnet with two north poles? The answer is no: Each half of the bar magnet has a north pole and a south pole. You can even continue cutting each piece of the bar magnet in half, and you will always obtain a new, smaller magnet with two opposite poles. As shown in Figure 20.6, you can continue this process down to the atomic scale, and you will find that even the smallest particles that behave as magnets have two opposite poles. In fact, no experiment has ever found any object with a single magnetic pole, from the smallest subatomic particle such as electrons to the largest objects in the universe such as stars. Because magnets always have two poles, they are referred to as magnetic dipoles\u2014dimeans two. Below, we will see that magnetic dipoles have properties that are analogous to electric dipoles. Figure 20.6 All magnets have two opposite poles, from the smallest, such as subatomic particles, to the largest, such as stars. WATCH PHYSICS Introduction to Magnetism This video provides an interesting introduction to magnetism and discusses, in particular, how electrons around their atoms contribute to the magnetic effects that we observe. Click to view content (https://www.openstax.org/l/28_intro_magn) GRASP CHECK Toward which magnetic pole of Earth is the north pole of a compass needle attracted? Access for free at openstax.org. 20.1 \u2022 Magnetic Fields, Field Lines, and Force 653 a. The north pole of a compass needle is attracted to the north magnetic pole of Earth, which is located near the geographic North Pole of Earth. b. The north pole of a compass needle is attracted to the south magnetic pole of Earth, which is located near the geographic North Pole of Earth. c. The north pole of a compass needle is attracted to the north magnetic pole of Earth, which is located near the geographic South Pole of Earth. d. The north pole of a compass needle is attracted to the south magnetic pole of Earth, which is located near the geographic South Pole of Earth. Only certain materials, such as iron, cobalt, nickel, and gadolinium, exhibit strong magnetic effects. Such materials are called ferromagnetic, after the Latin word ferrumfor iron. Other materials exhibit weak magnetic effects, which are detectable only with sensitive instruments. Not only do ferromagnetic materials respond strongly to magnets\u2014the way iron is attracted to magnets\u2014but they can also be magnetized themselves\u2014that is, they can be induced to be magnetic or made into permanent magnets (Figure 20.7). A permanent magnet is simply a material that retains its magnetic behavior for a long time, even when exposed to demagnetizing influences. Figure 20.7 An unmagnetized piece of iron is placed between two magnets, heated, and then cooled, or simply tapped when cold. The iron becomes a permanent magnet with the poles aligned as shown: Its south pole is adjacent to the north pole of the original magnet, and its north pole is adjacent to the south pole of the original magnet. Note that attractive forces are created between the central magnet and the outer magnets. When a magnet is brought near a previously unmagnetized ferromagnetic material, it causes local magnetization of the material with unlike poles closest, as in the right side of Figure 20.7. This causes an attractive force, which is why unmagnetized iron is attracted to a magnet. What happens on a microscopic scale is illustrated in Figure 7(a). Regions within the material called domains act like small bar magnets. Within domains, the magnetic poles of individual atoms are aligned. Each atom acts like a tiny bar magnet. Domains are small and randomly oriented in an unmagnetized ferromagnetic object. In response to an external magnetic field, the domains may grow to millimeter size, aligning themselves, as shown", " in Figure 7(b). This induced magnetization can be made permanent if the material is heated and then cooled, or simply tapped in the presence of other magnets. Figure 20.8 (a) An unmagnetized piece of iron\u2014or other ferromagnetic material\u2014has randomly oriented domains. (b) When magnetized by an external magnet, the domains show greater alignment, and some grow at the expense of others. Individual atoms are aligned within 654 Chapter 20 \u2022 Magnetism domains; each atom acts like a tiny bar magnet. Conversely, a permanent magnet can be demagnetized by hard blows or by heating it in the absence of another magnet. Increased thermal motion at higher temperature can disrupt and randomize the orientation and size of the domains. There is a well-defined temperature for ferromagnetic materials, which is called the Curie temperature, above which they cannot be magnetized. The Curie temperature for iron is 1,043 K (770 elements and alloys that have Curie temperatures much lower than room temperature and are ferromagnetic only below those temperatures. ), which is well above room temperature. There are several Snap Lab Refrigerator Magnets We know that like magnetic poles repel and unlike poles attract. See if you can show this for two refrigerator magnets. Will the magnets stick if you turn them over? Why do they stick to the refrigerator door anyway? What can you say about the magnetic properties of the refrigerator door near the magnet? Do refrigerator magnets stick to metal or plastic spoons? Do they stick to all types of metal? GRASP CHECK You have one magnet with the north and south poles labeled. How can you use this magnet to identify the north and south poles of other magnets? a. If the north pole of a known magnet is repelled by a pole of an unknown magnet on bringing them closer, that pole of unknown magnet is its north pole; otherwise, it is its south pole. If the north pole of known magnet is attracted to a pole of an unknown magnet on bringing them closer, that pole of unknown magnet is its north pole; otherwise, it is its south pole. b. Magnetic Fields We have thus seen that forces can be applied between magnets and between magnets and ferromagnetic materials without any contact between the objects. This is reminiscent of electric forces, which also act over distances. Electric forces are described using the concept of the electric field, which is a force field around electric charges that describes the force on any other charge placed in the field. Likewise, a magnet creates a magnetic field around it that describes the force exerted on other magnets placed in the field. As with electric fields, the pictorial representation of magnetic field lines is very useful for visualizing the strength and direction of the magnetic field. As shown in Figure 20.9, the direction of magnetic field lines is defined to be the direction in which the north pole of a compass needle points. If you place a compass near the north pole of a magnet, the north pole of the compass needle will be repelled and point away from the magnet. Thus, the magnetic field lines point away from the north pole of a magnet and toward its south pole. Figure 20.9 The black lines represent the magnetic field lines of a bar magnet. The field lines point in the direction that the north pole of a small compass would point, as shown at left. Magnetic field lines never stop, so the field lines actually penetrate the magnet to form complete loops, as shown at right. Access for free at openstax.org. 20.1 \u2022 Magnetic Fields, Field Lines, and Force 655 Magnetic field lines can be mapped out using a small compass. The compass is moved from point to point around a magnet, and at each point, a short line is drawn in the direction of the needle, as shown in Figure 20.10. Joining the lines together then reveals the path of the magnetic field line. Another way to visualize magnetic field lines is to sprinkle iron filings around a magnet. The filings will orient themselves along the magnetic field lines, forming a pattern such as that shown on the right in Figure 20.10. Virtual Physics Using a Compass to Map Out the Magnetic Field Click to view content (http://www.openstax.org/l/28magcomp) This simulation presents you with a bar magnet and a small compass. Begin by dragging the compass around the bar magnet to see in which direction the magnetic field points. Note that the strength of the magnetic field is represented by the brightness of the magnetic field icons in the grid pattern around the magnet. Use the magnetic field meter to check the field strength at several points around the bar magnet. You can also flip the polarity of the magnet, or place Earth on the image to see how the compass orients itself. GRASP CHECK With the slider at the top right of the simulation window, set the magnetic field strength to 100 percent. Now use the magnetic field meter to answer the following question: Near the magnet, where is the magnetic field strongest and where is", " it weakest? Don\u2019t forget to check inside the bar magnet. a. The magnetic field is strongest at the center and weakest between the two poles just outside the bar magnet. The magnetic field lines are densest at the center and least dense between the two poles just outside the bar magnet. b. The magnetic field is strongest at the center and weakest between the two poles just outside the bar magnet. The magnetic field lines are least dense at the center and densest between the two poles just outside the bar magnet. c. The magnetic field is weakest at the center and strongest between the two poles just outside the bar magnet. The magnetic field lines are densest at the center and least dense between the two poles just outside the bar magnet. d. The magnetic field is weakest at the center and strongest between the two poles just outside the bar magnet and the magnetic field lines are least dense at the center and densest between the two poles just outside the bar magnet. Figure 20.10 Magnetic field lines can be drawn by moving a small compass from point to point around a magnet. At each point, draw a short line in the direction of the compass needle. Joining the points together reveals the path of the magnetic field lines. Another way to visualize magnetic field lines is to sprinkle iron filings around a magnet, as shown at right. When two magnets are brought close together, the magnetic field lines are perturbed, just as happens for electric field lines when two electric charges are brought together. Bringing two north poles together\u2014or two south poles\u2014will cause a repulsion, and the magnetic field lines will bend away from each other. This is shown in Figure 20.11, which shows the magnetic field lines created by the two closely separated north poles of a bar magnet. When opposite poles of two magnets are brought together, the 656 Chapter 20 \u2022 Magnetism magnetic field lines join together and become denser between the poles. This situation is shown in Figure 20.11. Figure 20.11 (a) When two north poles are approached together, the magnetic field lines repel each other and the two magnets experience a repulsive force. The same occurs if two south poles are approached together. (b) If opposite poles are approached together, the magnetic field lines become denser between the poles and the magnets experience an attractive force. Like the electric field, the magnetic field is stronger where the lines are denser. Thus, between the two north poles in Figure 20.11, the magnetic field is very weak because the density of the magnetic field is almost zero. A compass placed at that point would essentially spin freely if we ignore Earth\u2019s magnetic field. Conversely, the magnetic field lines between the north and south poles in Figure 20.11 are very dense, indicating that the magnetic field is very strong in this region. A compass placed here would quickly align with the magnetic field and point toward the south pole on the right. Note that magnets are not the only things that make magnetic fields. Early in the nineteenth century, people discovered that electrical currents cause magnetic effects. The first significant observation was by the Danish scientist Hans Christian Oersted (1777\u20131851), who found that a compass needle was deflected by a current-carrying wire. This was the first significant evidence that the movement of electric charges had any connection with magnets. An electromagnet is a device that uses electric current to make a magnetic field. These temporarily induced magnets are called electromagnets. Electromagnets are employed for everything from a wrecking yard crane that lifts scrapped cars to controlling the beam of a 90-km-circumference particle accelerator to the magnets in medical-imaging machines (see Figure 20.12). Figure 20.12 Instrument for magnetic resonance imaging (MRI). The device uses a cylindrical-coil electromagnet to produce for the main magnetic field. The patient goes into the tunnelon the gurney. (credit: Bill McChesney, Flickr) The magnetic field created by an electric current in a long straight wire is shown in Figure 20.13. The magnetic field lines form concentric circles around the wire. The direction of the magnetic field can be determined using the right-hand rule. This rule shows up in several places in the study of electricity and magnetism. Applied to a straight current-carrying wire, the right-hand rule says that, with your right thumb pointed in the direction of the current, the magnetic field will be in the direction in which your right fingers curl, as shown in Figure 20.13. If the wire is very long compared to the distance rfrom the wire, the strength B of the magnetic field is given by where Iis the current in the wire in amperes. The SI unit for magnetic field is the tesla (T). The symbol \u2014read \u201cmu-zero\u201d\u2014is a constant called the \u201cpermeability of free space\u201d and is given by 20.2 20.1 Access for free at openstax.org", ". 20.1 \u2022 Magnetic Fields, Field Lines, and Force 657 Figure 20.13 This image shows how to use the right-hand rule to determine the direction of the magnetic field created by current flowing through a straight wire. Point your right thumb in the direction of the current, and the magnetic field will be in the direction in which your fingers curl. WATCH PHYSICS Magnetic Field Due to an Electric Current This video describes the magnetic field created by a straight current-carrying wire. It goes over the right-hand rule to determine the direction of the magnetic field, and presents and discusses the formula for the strength of the magnetic field due to a straight current-carrying wire. Click to view content (https://www.openstax.org/l/28magfield) GRASP CHECK A long straight wire is placed on a table top and electric current flows through the wire from right to left. If you look at the wire end-on from the left end, does the magnetic field go clockwise or counterclockwise? a. By pointing your right-hand thumb in the direction opposite of current, the right-hand fingers will curl counterclockwise, so the magnetic field will be in the counterclockwise direction. b. By pointing your right-hand thumb in the direction opposite of current, the right-hand fingers will curl clockwise, so the magnetic field will be in the clockwise direction. c. By pointing your right-hand thumb in the direction of current, the right-hand fingers will curl counterclockwise, so the magnetic field will be in the counterclockwise direction. d. By pointing your right-hand thumb in the direction of current, the right-hand fingers will curl clockwise, so the magnetic field will be in the clockwise direction. Now imagine winding a wire around a cylinder with the cylinder then removed. The result is a wire coil, as shown in Figure 20.14. This is called a solenoid. To find the direction of the magnetic field produced by a solenoid, apply the right-hand rule to several points on the coil. You should be able to convince yourself that, inside the coil, the magnetic field points from left to right. In fact, another application of the right-hand rule is to curl your right-hand fingers around the coil in the direction in which the current flows. Your right thumb then points in the direction of the magnetic field inside the coil: left to right in this case. Figure 20.14 A wire coil with current running through as shown produces a magnetic field in the direction of the red arrow. Each loop of wire contributes to the magnetic field inside the solenoid. Because the magnetic field lines must form closed loops, the field lines close the loop outside the solenoid. The magnetic field lines are much denser inside the solenoid than outside the solenoid. The resulting magnetic field looks very much like that of a bar magnet, as shown in Figure 20.15. The magnetic field strength deep inside a solenoid is 658 Chapter 20 \u2022 Magnetism where Nis the number of wire loops in the solenoid and is the length of the solenoid. 20.3 Figure 20.15 Iron filings show the magnetic field pattern around (a) a solenoid and (b) a bar magnet. The fields patterns are very similar, especially near the ends of the solenoid and bar magnet. Virtual Physics Electromagnets Click to view content (http://www.openstax.org/l/28elec_magnet) Use this simulation to visualize the magnetic field made from a solenoid. Be sure to click on the tab that says Electromagnet. You can drive AC or DC current through the solenoid by choosing the appropriate current source. Use the field meter to measure the strength of the magnetic field and then change the number of loops in the solenoid to see how this affects the magnetic field strength. GRASP CHECK Choose the battery as current source and set the number of wire loops to four. With a nonzero current going through the solenoid, measure the magnetic field strength at a point. Now decrease the number of wire loops to two. How does the magnetic field strength change at the point you chose? a. There will be no change in magnetic field strength when number of loops reduces from four to two. b. The magnetic field strength decreases to half of its initial value when number of loops reduces from four to two. c. The magnetic field strength increases to twice of its initial value when number of loops reduces from four to two. d. The magnetic field strength increases to four times of its initial value when number of loops reduces from four to two. Magnetic Force If a moving electric charge, that is electric current, produces a magnetic field that can exert a force on another magnet, then the reverse should be true by Newton\u2019s third law. In other words, a charge moving through the magnetic field", " produced by another object should experience a force\u2014and this is exactly what we find. As a concrete example, consider Figure 20.16, which shows a Access for free at openstax.org. 20.1 \u2022 Magnetic Fields, Field Lines, and Force 659 charge qmoving with velocity force experienced by this charge is through a magnetic field between the poles of a permanent magnet. The magnitude Fof the where is the angle between the velocity of the charge and the magnetic field. The direction of the force may be found by using another version of the right-hand rule: First, we join the tails of the velocity vector and a magnetic field vector, as shown in step 1 of Figure 20.16. We then curl our right fingers from to in step (2) of Figure 20.16. The direction in which the right thumb points is the direction of the force. For the charge in Figure 20.16, we find that the force is directed into the page., as indicated 20.4 Note that the factor magnetic field because the magnetic field, because in the equation and and means that zero force is applied on a charge that moves parallel to a. The maximum force a charge can experience is when it moves perpendicular to Figure 20.16 (a) An electron moves through a uniform magnetic field. (b) Using the right-hand rule, the force on the electron is found to be directed into the page. LINKS TO PHYSICS Magnetohydrodynamic Drive In Tom Clancy\u2019s Cold War novel \u201cThe Hunt for Red October,\u201d the Soviet Union built a submarine (see Figure 20.17) with a magnetohydrodynamic drive that was so silent it could not be detected by surface ships. The only conceivable purpose to build such a submarine was to give the Soviet Union first-strike capability, because this submarine could sneak close to the coast of the United States and fire its ballistic missiles, destroying key military and government installations to prevent an American counterattack. Figure 20.17 A Typhoon-class Russian ballistic-missile submarine on which the fictional submarine Red October was based. A magnetohydrodynamic drive is supposed to be silent because it has no moving parts. Instead, it uses the force experienced by charged particles that move in a magnetic field. The basic idea behind such a drive is depicted in Figure 20.18. Salt water flows through a channel that runs from the front to the back of the submarine. A magnetic field is applied horizontally across the channel, and a voltage is applied across the electrodes on the top and bottom of the channel to force a downward electric current through the water. The charge carriers are the positive sodium ions and the negative chlorine ions of salt. Using the right-hand 660 Chapter 20 \u2022 Magnetism rule, the force on the charge carriers is found to be toward the rear of the vessel. The accelerated charges collide with water molecules and transfer their momentum, creating a jet of water that is propelled out the rear of the channel. By Newton\u2019s third law, the vessel experiences a force of equal magnitude, but in the opposite direction. Figure 20.18 A schematic drawing of a magnetohydrodynamic drive showing the water channel, the current direction, the magnetic field direction, and the resulting force. Fortunately for all involved, it turns out that such a propulsion system is not very practical. Some back-of-the-envelope calculations show that, to power a submarine, either extraordinarily high magnetic fields or extraordinarily high electric currents would be required to obtain a reasonable thrust. In addition, prototypes of magnetohydrodynamic drives show that they are anything but silent. Electrolysis caused by running a current through salt water creates bubbles of hydrogen and oxygen, which makes this propulsion system quite noisy. The system also leaves a trail of chloride ions and metal chlorides that can easily be detected to locate the submarine. Finally, the chloride ions are extremely reactive and very quickly corrode metal parts, such as the electrode or the water channel itself. Thus, the Red October remains in the realm of fiction, but the physics involved is quite real. GRASP CHECK If the magnetic field is downward, in what direction must the current flow to obtain rearward-pointing force? a. The current must flow vertically from up to down when viewed from the rear of the boat. b. The current must flow vertically from down to up when viewed from the rear of the boat. c. The current must flow horizontally from left to right when viewed from the rear of the boat. d. The current must flow horizontally from right to left when viewed from the rear of the boat. Instead of a single charge moving through a magnetic field, consider now a steady current Imoving through a straight wire. If we place this wire in a uniform magnetic field, as shown in Figure 20.19, what is the force on the wire or, more precisely, on the electrons in the wire? An electric current involves charges that move. If the charges qmove a distance speed is Inserting this into the equation in", " a time t, then their gives The factor q/tin this equation is nothing more than the current in the wire. Thus, using, we obtain 20.5 20.6 This equation gives the force on a straight current-carrying wire of length angle between the current vector and the magnetic field vector. Note that for which as shown in Figure 20.19. in a magnetic field of strength B. The angle is the is the length of wire that is in the magnetic field and The direction of the force is determined in the same way as for a single charge. Curl your right fingers from the vector for Ito the vector for B, and your right thumb will point in the direction of the force on the wire. For the wire shown in Figure 20.19, the force is directed into the page. Access for free at openstax.org. 20.1 \u2022 Magnetic Fields, Field Lines, and Force 661 Figure 20.19 A straight wire carrying current Iin a magnetic field B. The force exerted on the wire is directed into the page. The length is the length of the wire that is inthe magnetic field. Throughout this section, you may have noticed the symmetries between magnetic effects and electric effects. These effects all fall under the umbrella of electromagnetism, which is the study of electric and magnetic phenomena. We have seen that electric charges produce electric fields, and moving electric charges produce magnetic fields. A magnetic dipole produces a magnetic field, and, as we will see in the next section, moving magnetic dipoles produce an electric field. Thus, electricity and magnetism are two intimately related and symmetric phenomena. WORKED EXAMPLE Trajectory of Electron in Magnetic Field A proton enters a region of constant magnetic field, as shown in Figure 20.20. The magnetic field is coming out of the page. If the electron is moving at and the magnetic field strength is 2.0 T, what is the magnitude and direction of the force on the proton? Figure 20.20 A proton enters a region of uniform magnetic field. The magnetic field is coming out of the page\u2014the circles with dots represent vector arrow heads coming out of the page. STRATEGY Use the equation and the velocity vector of the proton is Solution The charge of the proton is the equation gives to find the magnitude of the force on the proton. The angle between the magnetic field vectors The direction of the force may be found by using the right-hand rule.. Entering this value and the given velocity and magnetic field strength into 20.7 To find the direction of the force, first join the velocity vector end to end with the magnetic field vector, as shown in Figure 20.21. Now place your right hand so that your fingers point in the direction of the velocity and curl them upward toward the magnetic field vector. The force is in the direction in which your thumb points. In this case, the force is downward in the plane of the paper in the -direction, as shown in Figure 20.21. 662 Chapter 20 \u2022 Magnetism Figure 20.21 The velocity vector and a magnetic field vector from Figure 20.20 are placed end to end. A right hand is shown with the fingers curling up from the velocity vector toward the magnetic field vector. The thumb points in the direction of the resulting force, which is the -direction in this case. Thus, combining the magnitude and the direction, we find that the force on the proton is Discussion This seems like a very small force. However, the proton has a mass of, so its acceleration is, or about ten thousand billion times the acceleration due to gravity! We found that the proton\u2019s initial acceleration as it enters the magnetic field is downward in the plane of the page. Notice that, as the proton accelerates, its velocity remains perpendicular to the magnetic field, so the magnitude of the force does not change. In addition, because of the right-hand rule, the direction of the force remains perpendicular to the velocity. This force is nothing more than a centripetal force: It has a constant magnitude and is always perpendicular to the velocity. Thus, the magnitude of the velocity does not change, and the proton executes circular motion. The radius of this circle may be found by using the kinematics relationship. 20.8 The path of the proton in the magnetic field is shown in Figure 20.22. Figure 20.22 When traveling perpendicular to a constant magnetic field, a charged particle will execute circular motion, as shown here for a proton. WORKED EXAMPLE Wire with Current in Magnetic Field Now suppose we run a wire through the uniform magnetic field from the previous example, as shown. If the wire carries a current of 1.0 A in the -direction, and the region with magnetic field is 4.0 cm long, what is the force on the wire? Access for free at openstax.org. 20.1 \u2022 Magnetic Fields, Field", " Lines, and Force 663 STRATEGY to find the magnitude of the force on the wire. The length of the wire inside the magnetic field is Use equation 4.0 cm, and the angle between the current direction and the magnetic field direction is 90\u00b0. To find the direction of the force, use the right-hand rule as explained just after the equation Solution Insert the given values into equation to find the magnitude of the force 20.9 To find the direction of the force, begin by placing the current vector end to end with a vector for the magnetic field. The result is as shown in the figure in the previous Worked Example with and your right thumb points down the page, again as shown in the figure in the previous Worked Example. Thus, the direction of the force is in the. Curl your right-hand fingers from to -direction. The complete force is thus replaced by. Discussion The direction of the force is the same as the initial direction of the force was in the previous example for a proton. However, because the current in a wire is confined to a wire, the direction in which the charges move does not change. Instead, the entire wire accelerates in the -direction. The force on a current-carrying wire in a magnetic field is the basis of all electrical motors, as we will see in the upcoming sections. Practice Problems 1. What is the magnitude of the force on an electron moving at 1.0 \u00d7 106 m/s perpendicular to a 1.0-T magnetic field? a. 0.8 \u00d7 10\u201313 N 1.6 \u00d7 10\u201314 N b. c. 0.8 \u00d7 10\u201314 N 1.6 \u00d7 10\u201313 N d. 2. A straight 10 cm wire carries 0.40 A and is oriented perpendicular to a magnetic field. If the force on the wire is 0.022 N, what is the magnitude of the magnetic field? 1.10 \u00d7 10\u20132 T a. b. 0.55 \u00d7 10\u20132 T c. 1.10 T d. 0.55 T Check Your Understanding 3. If two magnets repel each other, what can you conclude about their relative orientation? a. Either the south pole of magnet 1 is closer to the north pole of magnet 2 or the north pole of magnet 1 is closer to the south pole of magnet 2. b. Either the south poles of both the magnet 1 and magnet 2 are closer to each other or the north poles of both the magnet 1 664 Chapter 20 \u2022 Magnetism and magnet 2 are closer to each other. 4. Describe methods to demagnetize a ferromagnet. a. by cooling, heating, or submerging in water b. by heating, hammering, and spinning it in external magnetic field c. by hammering, heating, and rubbing with cloth d. by cooling, submerging in water, or rubbing with cloth 5. What is a magnetic field? a. The directional lines present inside and outside the magnetic material that indicate the magnitude and direction of the magnetic force. b. The directional lines present inside and outside the magnetic material that indicate the magnitude of the magnetic force. c. The directional lines present inside the magnetic material that indicate the magnitude and the direction of the magnetic force. d. The directional lines present outside the magnetic material that indicate the magnitude and the direction of the magnetic force. 6. Which of the following drawings is correct? a. b. c. d. Access for free at openstax.org. 20.2 \u2022 Motors, Generators, and Transformers 665 20.2 Motors, Generators, and Transformers Section Learning Objectives By the end of this section, you will be able to do the following: \u2022 Explain how electric motors, generators, and transformers work \u2022 Explain how commercial electric power is produced, transmitted, and distributed Section Key Terms electric motor generator transformer Electric Motors, Generators, and Transformers As we learned previously, a current-carrying wire in a magnetic field experiences a force\u2014recall motors, which convert electrical energy into mechanical energy, are the most common application of magnetic force on currentcarrying wires. Motors consist of loops of wire in a magnetic field. When current is passed through the loops, the magnetic field exerts a torque on the loops, which rotates a shaft. Electrical energy is converted to mechanical work in the process. Figure 20.23 shows a schematic drawing of an electric motor.. Electric Figure 20.23 Torque on a current loop. A vertical loop of wire in a horizontal magnetic field is attached to a vertical shaft. When current is passed through the wire loop, torque is exerted on it, making it turn the shaft. Let us examine the force on each segment of the loop in Figure 20.23 to find the torques produced about the axis of the vertical shaft\u2014this will lead to a useful equation for the torque on the loop. We take the magnetic field to be uniform over the rectangular loop, which has width wand height loop. To", " determine the direction of the force, we use the right-hand rule. The current goes from left to right into the page, and the magnetic field goes from left to right in the plane of the page. Curl your right fingers from the current vector to the magnetic field vector and your right thumb points down. Thus, the force on the top segment is downward, which produces no torque on the shaft. Repeating this analysis for the bottom segment\u2014neglect the small gap where the lead wires go out\u2014shows that the force on the bottom segment is upward, again producing no torque on the shaft. as shown in the figure. First, consider the force on the top segment of the Consider now the left vertical segment of the loop. Again using the right-hand rule, we find that the force exerted on this segment is perpendicular to the magnetic field, as shown in Figure 20.23. This force produces a torque on the shaft. Repeating this analysis on the right vertical segment of the loop shows that the force on this segment is in the direction opposite that of the force on the left segment, thereby producing an equal torque on the shaft. The total torque on the shaft is thus twice the toque on one of the vertical segments of the loop. To find the magnitude of the torque as the wire loop spins, consider Figure 20.24, which shows a view of the wire loop from where Fis the applied force, ris the distance from the pivot to where the above. Recall that torque is defined as force is applied, and \u03b8is the angle between rand F. Notice that, as the loop spins, the current in the vertical loop segments is always perpendicular to the magnetic field. Thus, the equation segment as The distance rfrom the shaft to where this force is applied is w/2, so the torque created by this force is gives the magnitude of the force on each vertical Because there are two vertical segments, the total torque is twice this, or 20.10 20.11 If we have a multiple loop with Nturns, we get Ntimes the torque of a single loop. Using the fact that the area of the loop is the expression for the torque becomes 666 Chapter 20 \u2022 Magnetism This is the torque on a current-carrying loop in a uniform magnetic field. This equation can be shown to be valid for a loop of any shape. 20.12 Figure 20.24 View from above of the wire loop from Figure 20.23. The magnetic field generates a force Fon each vertical segment of the wire loop, which generates a torque on the shaft. Notice that the currents have the same magnitude because they both represent the current flowing in the wire loop, but flows into the page and flows out of the page. From the equation a maximum positive torque of to From will oscillate back and forth. we see that the torque is zero when when The torque then decreases back to zero as the wire loop rotates to the torque is negative. Thus, the torque changes sign every half turn, so the wire loop As the wire loop rotates, the torque increases to For the coil to continue rotating in the same direction, the current is reversed as the coil passes through using automatic switches called brushes, as shown in Figure 20.25. Figure 20.25 (a) As the angular momentum of the coil carries it through the brushes reverse the current and the torque remains clockwise. (b) The coil rotates continuously in the clockwise direction, with the current reversing each half revolution to maintain the clockwise torque. Consider now what happens if we run the motor in reverse; that is, we attach a handle to the shaft and mechanically force the coil to rotate within the magnetic field, as shown in Figure 20.26. As per the equation is the angle \u2014where between the vectors and in the wires of the loop experience a magnetic force because they are moving in a magnetic field. Again using the right-hand rule, where we curl our fingers from vector the top and bottom segments feel a force perpendicular to the wire, which does not cause a current. However, charges in the vertical wires experience forces parallel to the wire, causing a current to flow through the wire and through an external circuit if one is connected. A device such as this that converts mechanical energy into electrical energy is called a generator., we find that charges in to vector Access for free at openstax.org. 20.2 \u2022 Motors, Generators, and Transformers 667 Figure 20.26 When this coil is rotated through one-fourth of a revolution, the magnetic flux \u03a6 changes from its maximum to zero, inducing an emf, which drives a current through an external circuit. Because current is induced only in the side wires, we can find the induced emf by only considering these wires. As explained in Induced Current in a Wire, motional emf in a straight wire moving at velocity vthrough a magnetic field Bis where the velocity is perpendicular to the magnetic field. In the generator, the velocity makes an angle with B", "(see Figure 20.27), so the velocity component perpendicular to Bis Thus, in this case, the emf induced on each vertical wire segment is and they are in the same direction. The total emf around the loop is then 20.13 Although this expression is valid, it does not give the emf as a function of time. To find how the emf evolves in time, we assume that the coil is rotated at a constant angular velocity The angle is related to the angular velocity by so that Recall that tangential velocity vis related to angular velocity by Here,, so that and Noting that the area of the loop is and allowing for Nwire loops, we find that 20.14 20.15 20.16 is the emf induced in a generator coil of Nturns and area Arotating at a constant angular velocity B. This can also be expressed as in a uniform magnetic field where is the maximum (peak) emf. 20.17 20.18 668 Chapter 20 \u2022 Magnetism Figure 20.27 The instantaneous velocity of the vertical wire segments makes an angle with the magnetic field. The velocity is shown in the figure by the green arrow, and the angle is indicated. to a negative maximum of Figure 20.28 shows a generator connected to a light bulb and a graph of the emf vs. time. Note that the emf oscillates from a positive maximum of current flows through the light bulb at these times. Thus, the light bulb actually flickers on and off at a frequency of 2f, because there are two zero crossings per period. Since alternating current such as this is used in homes around the world, why do we not notice the lights flickering on and off? In the United States, the frequency of alternating current is 60 Hz, so the lights flicker on and off at a frequency of 120 Hz. This is faster than the refresh rate of the human eye, so you don\u2019t notice the flicker of the lights. Also, other factors prevent various different types of light bulbs from switching on and off so fast, so the light output is smoothed outa bit. In between, the emf goes through zero, which means that zero Figure 20.28 The emf of a generator is sent to a light bulb with the system of rings and brushes shown. The graph gives the emf of the generator as a function of time. is the peak emf. The period is where fis the frequency at which the coil is rotated in the magnetic field. Virtual Physics Generator Click to view content (http://www.openstax.org/l/28gen) Use this simulation to discover how an electrical generator works. Control the water supply that makes a water wheel turn a magnet. This induces an emf in a nearby wire coil, which is used to light a light bulb. You can also replace the light bulb with a voltmeter, which allows you to see the polarity of the voltage, which changes from positive to negative. GRASP CHECK Set the number of wire loops to three, the bar-magnet strength to about 50 percent, and the loop area to 100 percent. Note the maximum voltage on the voltmeter. Assuming that one major division on the voltmeter is 5V, what is the maximum voltage when using only a single wire loop instead of three wire loops? a. b. 5 V 15 V Access for free at openstax.org. c. d. 125 V 53 V 20.2 \u2022 Motors, Generators, and Transformers 669 In real life, electric generators look a lot different than the figures in this section, but the principles are the same. The source of mechanical energy that turns the coil can be falling water\u2014hydropower\u2014steam produced by the burning of fossil fuels, or the kinetic energy of wind. Figure 20.29 shows a cutaway view of a steam turbine; steam moves over the blades connected to the shaft, which rotates the coil within the generator. Figure 20.29 Steam turbine generator. The steam produced by burning coal impacts the turbine blades, turning the shaft which is connected to the generator. (credit: Nabonaco, Wikimedia Commons) Another very useful and common device that exploits magnetic induction is called a transformer. Transformers do what their name implies\u2014they transform voltages from one value to another; the term voltage is used rather than emf because transformers have internal resistance. For example, many cell phones, laptops, video games, power tools, and small appliances have a transformer built into their plug-in unit that changes 120 V or 240 V AC into whatever voltage the device uses. Figure 20.30 shows two different transformers. Notice the wire coils that are visible in each device. The purpose of these coils is explained below. Figure 20.30 On the left is a common laminated-core transformer, which is widely used in electric power transmission and electrical appliances. On the right is a toroidal transformer, which is smaller than the laminated-core transformer for the same power", " rating but is more expensive to make because of the equipment required to wind the wires in the doughnut shape. Figure 20.31 shows a laminated-coil transformer, which is based on Faraday\u2019s law of induction and is very similar in construction to the apparatus Faraday used to demonstrate that magnetic fields can generate electric currents. The two wire coils are called the primary and secondary coils. In normal use, the input voltage is applied across the primary coil, and the secondary produces the transformed output voltage. Not only does the iron core trap the magnetic field created by the primary coil, but also its magnetization increases the field strength, which is analogous to how a dielectric increases the electric field strength in a capacitor. Since the input voltage is AC, a time-varying magnetic flux is sent through the secondary coil, inducing an AC output voltage. 670 Chapter 20 \u2022 Magnetism Figure 20.31 A typical construction of a simple transformer has two coils wound on a ferromagnetic core. The magnetic field created by the primary coil is mostly confined to and increased by the core, which transmits it to the secondary coil. Any change in current in the primary coil induces a current in the secondary coil. LINKS TO PHYSICS Magnetic Rope Memory To send men to the moon, the Apollo program had to design an onboard computer system that would be robust, consume little power, and be small enough to fit onboard the spacecraft. In the 1960s, when the Apollo program was launched, entire buildings were regularly dedicated to housing computers whose computing power would be easily outstripped by today\u2019s most basic handheld calculator. To address this problem, engineers at MIT and a major defense contractor turned to magnetic rope memory, which was an offshoot of a similar technology used prior to that time for creating random access memories. Unlike random access memory, magnetic rope memory was read-only memory that contained not only data but instructions as well. Thus, it was actually more than memory: It was a hard-wired computer program. The components of magnetic rope memory were wires and iron rings\u2014which were called cores. The iron cores served as transformers, such as that shown in the previous figure. However, instead of looping the wires multiple times around the core, individual wires passed only a single time through the cores, making these single-turn transformers. Up to 63 wordwires could pass through a single core, along with a single bitwire. If a word wire passed through a given core, a voltage pulse on this wire would induce an emf in the bit wire, which would be interpreted as a one. If the word wire did not pass through the core, no emf would be induced on the bit wire, which would be interpreted as a zero. Engineers would create programs that would be hard wired into these magnetic rope memories. The wiring process could take as long as a month to complete as workers painstakingly threaded wires through some cores and around others. If any mistakes were made either in the programming or the wiring, debugging would be extraordinarily difficult, if not impossible. These modules did their job quite well. They are credited with correcting an astronaut mistake in the lunar landing procedure, thereby allowing Apollo 11 to land on the moon. It is doubtful that Michael Faraday ever imagined such an application for magnetic induction when he discovered it. GRASP CHECK If the bit wire were looped twice around each core, how would the voltage induced in the bit wire be affected? a. b. c. d. If number of loops around the wire is doubled, the emf is halved. If number of loops around the wire is doubled, the emf is not affected. If number of loops around the wire is doubled, the emf is also doubled. If number of loops around the wire is doubled, the emf is four times the initial value. For the transformer shown in Figure 20.31, the output voltage voltage the secondary coil gives its induced output voltage to be across the primary coil and the number of loops in the primary and secondary coils. Faraday\u2019s law of induction for from the secondary coil depends almost entirely on the input Access for free at openstax.org. 20.19 20.2 \u2022 Motors, Generators, and Transformers 671 is the number of loops in the secondary coil and where equals the induced emf sectional area of the coils is the same on each side, as is the magnetic field strength, and so input primary voltage is the rate of change of magnetic flux. The output voltage provided coil resistance is small\u2014a reasonable assumption for transformers. The cross- is also related to changing flux by is the same on each side. The 20.20 20.21 Taking the ratio of these last two equations yields the useful relationship This is known as the transformer equation. It simply states that the ratio of the secondary voltage to the primary voltage in a transformer equals the ratio of the number of loops in secondary coil to the number of loops in the primary coil.", " Transmission of Electrical Power Transformers are widely used in the electric power industry to increase voltages\u2014called step-uptransformers\u2014before longdistance transmission via high-voltage wires. They are also used to decrease voltages\u2014called step-downtransformers\u2014to deliver power to homes and businesses. The overwhelming majority of electric power is generated by using magnetic induction, whereby a wire coil or copper disk is rotated in a magnetic field. The primary energy required to rotate the coils or disk can be provided by a variety of means. Hydroelectric power plants use the kinetic energy of water to drive electric generators. Coal or nuclear power plants create steam to drive steam turbines that turn the coils. Other sources of primary energy include wind, tides, or waves on water. Once power is generated, it must be transmitted to the consumer, which often means transmitting power over hundreds of kilometers. To do this, the voltage of the power plant is increased by a step-up transformer, that is stepped up, and the current decreases proportionally because 20.22 The lower current flow. This heating is caused by the small, but nonzero, resistance environment through this heat is in the transmission wires reduces the Joule losses, which is heating of the wire due to a current of the transmission wires. The power lost to the which is proportional to the current squaredin the transmission wire. This is why the transmitted current small as possible and, consequently, the voltage must be large to transmit the power 20.23 must be as Voltages ranging from 120 to 700 kV are used for transmitting power over long distances. The voltage is stepped up at the exit of the power station by a step-up transformer, as shown in Figure 20.32. Figure 20.32 Transformers change voltages at several points in a power distribution system. Electric power is usually generated at greater than 10 kV, and transmitted long distances at voltages ranging from 120 kV to 700 kV to limit energy losses. Local power distribution to neighborhoods or industries goes through a substation and is sent short distances at voltages ranging from 5 to 13 kV. This is reduced to 120, 240, or 480 V for safety at the individual user site. Once the power has arrived at a population or industrial center, the voltage is stepped down at a substation to between 5 and 30 672 Chapter 20 \u2022 Magnetism kV. Finally, at individual homes or businesses, the power is stepped down again to 120, 240, or 480 V. Each step-up and stepdown transformation is done with a transformer designed based on Faradays law of induction. We\u2019ve come a long way since Queen Elizabeth asked Faraday what possible use could be made of electricity. Check Your Understanding 7. What is an electric motor? a. An electric motor transforms electrical energy into mechanical energy. b. An electric motor transforms mechanical energy into electrical energy. c. An electric motor transforms chemical energy into mechanical energy. d. An electric motor transforms mechanical energy into chemical energy. 8. What happens to the torque provided by an electric motor if you double the number of coils in the motor? a. The torque would be doubled. b. The torque would be halved. c. The torque would be quadrupled. d. The torque would be tripled. 9. What is a step-up transformer? a. A step-up transformer decreases the current to transmit power over short distance with minimum loss. b. A step-up transformer increases the current to transmit power over short distance with minimum loss. c. A step-up transformer increases voltage to transmit power over long distance with minimum loss. d. A step-up transformer decreases voltage to transmit power over short distance with minimum loss. 10. What should be the ratio of the number of output coils to the number of input coil in a step-up transformer to increase the voltage fivefold? a. The ratio is five times. b. The ratio is 10 times. c. The ratio is 15 times. d. The ratio is 20 times. 20.3 Electromagnetic Induction Section Learning Objectives By the end of this section, you will be able to do the following: \u2022 Explain how a changing magnetic field produces a current in a wire \u2022 Calculate induced electromotive force and current Section Key Terms emf induction magnetic flux Changing Magnetic Fields In the preceding section, we learned that a current creates a magnetic field. If nature is symmetrical, then perhaps a magnetic field can create a current. In 1831, some 12 years after the discovery that an electric current generates a magnetic field, English scientist Michael Faraday (1791\u20131862) and American scientist Joseph Henry (1797\u20131878) independently demonstrated that magnetic fields can produce currents. The basic process of generating currents with magnetic fields is called induction; this process is also called magnetic induction to distinguish it from charging by induction, which uses the electrostatic Coulomb force. When Faraday discovered what is now called Faraday\u2019s law of induction, Queen Victoria asked him what", " possible use was electricity. \u201cMadam,\u201d he replied, \u201cWhat good is a baby?\u201d Today, currents induced by magnetic fields are essential to our technological society. The electric generator\u2014found in everything from automobiles to bicycles to nuclear power plants\u2014uses magnetism to generate electric current. Other devices that use magnetism to induce currents include pickup coils in electric guitars, transformers of every size, certain microphones, airport security gates, and damping mechanisms on sensitive chemical balances. Access for free at openstax.org. 20.3 \u2022 Electromagnetic Induction 673 One experiment Faraday did to demonstrate magnetic induction was to move a bar magnet through a wire coil and measure the resulting electric current through the wire. A schematic of this experiment is shown in Figure 20.33. He found that current is induced only when the magnet moves with respect to the coil. When the magnet is motionless with respect to the coil, no current is induced in the coil, as in Figure 20.33. In addition, moving the magnet in the opposite direction (compare Figure 20.33 with Figure 20.33) or reversing the poles of the magnet (compare Figure 20.33 with Figure 20.33) results in a current in the opposite direction. Figure 20.33 Movement of a magnet relative to a coil produces electric currents as shown. The same currents are produced if the coil is moved relative to the magnet. The greater the speed, the greater the magnitude of the current, and the current is zero when there is no motion. The current produced by moving the magnet upward is in the opposite direction as the current produced by moving the magnet downward. Virtual Physics Faraday\u2019s Law Click to view content (http://www.openstax.org/l/faradays-law) Try this simulation to see how moving a magnet creates a current in a circuit. A light bulb lights up to show when current is flowing, and a voltmeter shows the voltage drop across the light bulb. Try moving the magnet through a four-turn coil and through a two-turn coil. For the same magnet speed, which coil produces a higher voltage? GRASP CHECK With the north pole to the left and moving the magnet from right to left, a positive voltage is produced as the magnet enters the coil. What sign voltage will be produced if the experiment is repeated with the south pole to the left? a. The sign of voltage will change because the direction of current flow will change by moving south pole of the magnet to the left. b. The sign of voltage will remain same because the direction of current flow will not change by moving south pole of the magnet to the left. c. The sign of voltage will change because the magnitude of current flow will change by moving south pole of the magnet to the left. d. The sign of voltage will remain same because the magnitude of current flow will not change by moving south pole of the magnet to the left. Induced Electromotive Force If a current is induced in the coil, Faraday reasoned that there must be what he called an electromotive forcepushing the charges through the coil. This interpretation turned out to be incorrect; instead, the external source doing the work of moving the magnet adds energy to the charges in the coil. The energy added per unit charge has units of volts, so the electromotive force is actually a potential. Unfortunately, the name electromotive force stuck and with it the potential for confusing it with a real force. For this reason, we avoid the term electromotive forceand just use the abbreviation emf, which has the mathematical symbol The emf may be defined as the rate at which energy is drawn from a source per unit current flowing through a circuit. Thus, emf is the energy per unit charge addedby a source, which contrasts with voltage, which is the energy per unit charge 674 Chapter 20 \u2022 Magnetism releasedas the charges flow through a circuit. To understand why an emf is generated in a coil due to a moving magnet, consider Figure 20.34, which shows a bar magnet moving downward with respect to a wire loop. Initially, seven magnetic field lines are going through the loop (see left-hand image). Because the magnet is moving away from the coil, only five magnetic field lines are going through the loop after a short time (see right-hand image). Thus, when a change occurs in the number of magnetic field lines going through the area defined by the wire loop, an emf is induced in the wire loop. Experiments such as this show that the induced emf is proportional to the rate of changeof the magnetic field. Mathematically, we express this as 20.24 where is the change in the magnitude in the magnetic field during time and Ais the area of the loop. Figure 20.34 The bar magnet moves downward with respect to the wire loop, so that the number of magnetic field lines going through the loop decreases with time. This causes an emf to be induced in the", " loop, creating an electric current. Note that magnetic field lines that lie in the plane of the wire loop do not actually pass through the loop, as shown by the leftmost loop in Figure 20.35. In this figure, the arrow coming out of the loop is a vector whose magnitude is the area of the loop and whose direction is perpendicular to the plane of the loop. In Figure 20.35, as the loop is rotated from the contribution of the magnetic field lines to the emf increases. Thus, what is important in generating an emf in the wire loop is the component of the magnetic field that is perpendicularto the plane of the loop, which is to This is analogous to a sail in the wind. Think of the conducting loop as the sail and the magnetic field as the wind. To maximize the force of the wind on the sail, the sail is oriented so that its surface vector points in the same direction as the winds, as in the right-most loop in Figure 20.35. When the sail is aligned so that its surface vector is perpendicular to the wind, as in the leftmost loop in Figure 20.35, then the wind exerts no force on the sail. Thus, taking into account the angle of the magnetic field with respect to the area, the proportionality becomes 20.25 Figure 20.35 The magnetic field lies in the plane of the left-most loop, so it cannot generate an emf in this case. When the loop is rotated so that the angle of the magnetic field with the vector perpendicular to the area of the loop increases to (see right-most loop), the magnetic field contributes maximally to the emf in the loop. The dots show where the magnetic field lines intersect the plane defined by the loop. Access for free at openstax.org. 20.3 \u2022 Electromagnetic Induction 675 Another way to reduce the number of magnetic field lines that go through the conducting loop in Figure 20.35 is not to move the magnet but to make the loop smaller. Experiments show that changing the area of a conducting loop in a stable magnetic field induces an emf in the loop. Thus, the emf produced in a conducting loop is proportional to the rate of change of the productof the perpendicular magnetic field and the loop area 20.26 is the perpendicular magnetic field and Ais the area of the loop. The product where proportional to the number of magnetic field lines that pass perpendicularly through a surface of area A. Going back to our sail analogy, it would be proportional to the force of the wind on the sail. It is called the magnetic flux and is represented by is very important. It is. The unit of magnetic flux is the weber (Wb), which is magnetic field per unit area, or T/m2. The weber is also a volt second (Vs). The induced emf is in fact proportional to the rate of change of the magnetic flux through a conducting loop. 20.27 20.28 Finally, for a coil made from Nloops, the emf is Ntimes stronger than for a single loop. Thus, the emf induced by a changing magnetic field in a coil of Nloops is The last question to answer before we can change the proportionality into an equation is \u201cIn what direction does the current flow?\u201d The Russian scientist Heinrich Lenz (1804\u20131865) explained that the current flows in the direction that creates a magnetic field that tries to keep the flux constant in the loop. For example, consider again Figure 20.34. The motion of the bar magnet causes the number of upward-pointing magnetic field lines that go through the loop to decrease. Therefore, an emf is generated in the loop that drives a current in the direction that creates more upward-pointing magnetic field lines. By using the righthand rule, we see that this current must flow in the direction shown in the figure. To express the fact that the induced emf acts to counter the change in the magnetic flux through a wire loop, a minus sign is introduced into the proportionality, which gives Faraday\u2019s law of induction. 20.29 Lenz\u2019s law is very important. To better understand it, consider Figure 20.36, which shows a magnet moving with respect to a wire coil and the direction of the resulting current in the coil. In the top row, the north pole of the magnet approaches the coil, so the magnetic field lines from the magnet point toward the coil. Thus, the magnetic field right increases in the coil. According to Lenz\u2019s law, the emf produced in the coil will drive a current in the direction that creates a pointing to the magnetic field inside the coil pointing to the left. This will counter the increase in magnetic flux pointing to the right. To see which way the current must flow, point your right thumb in the desired direction of the magnetic field and the current will flow in the direction indicated by curling your right fingers", ". This is shown by the image of the right hand in the top row of Figure 20.36. Thus, the current must flow in the direction shown in Figure 4(a). In Figure 4(b), the direction in which the magnet moves is reversed. In the coil, the right-pointing magnetic field the moving magnet decreases. Lenz\u2019s law says that, to counter this decrease, the emf will drive a current that creates an due to additional right-pointing magnetic field magnetic field, and the current will flow in the direction indicate by curling your right fingers (Figure 4(b)). in the coil. Again, point your right thumb in the desired direction of the Finally, in Figure 4(c), the magnet is reversed so that the south pole is nearest the coil. Now the magnetic field toward the magnet instead of toward the coil. As the magnet approaches the coil, it causes the left-pointing magnetic field in the coil to increase. Lenz\u2019s law tells us that the emf induced in the coil will drive a current in the direction that creates a magnetic field pointing to the right. This will counter the increasing magnetic flux pointing to the left due to the magnet. Using the righthand rule again, as indicated in the figure, shows that the current must flow in the direction shown in Figure 4(c). points 676 Chapter 20 \u2022 Magnetism Figure 20.36 Lenz\u2019s law tells us that the magnetically induced emf will drive a current that resists the change in the magnetic flux through a circuit. This is shown in panels (a)\u2013(c) for various magnet orientations and velocities. The right hands at right show how to apply the right- hand rule to find in which direction the induced current flows around the coil. Virtual Physics Faraday\u2019s Electromagnetic Lab Click to view content (http://www.openstax.org/l/Faraday-EM-lab) This simulation proposes several activities. For now, click on the tab Pickup Coil, which presents a bar magnet that you can move through a coil. As you do so, you can see the electrons move in the coil and a light bulb will light up or a voltmeter will indicate the voltage across a resistor. Note that the voltmeter allows you to see the sign of the voltage as you move the magnet about. You can also leave the bar magnet at rest and move the coil, although it is more difficult to observe the results. GRASP CHECK Orient the bar magnet with the north pole facing to the right and place the pickup coil to the right of the bar magnet. Now move the bar magnet toward the coil and observe in which way the electrons move. This is the same situation as depicted below. Does the current in the simulation flow in the same direction as shown below? Explain why or why not. Access for free at openstax.org. 20.3 \u2022 Electromagnetic Induction 677 a. Yes, the current in the simulation flows as shown because the direction of current is opposite to the direction of flow of electrons. b. No, current in the simulation flows in the opposite direction because the direction of current is same to the direction of flow of electrons. WATCH PHYSICS Induced Current in a Wire This video explains how a current can be induced in a straight wire by moving it through a magnetic field. The lecturer uses the cross product, which a type of vector multiplication. Don\u2019t worry if you are not familiar with this, it basically combines the righthand rule for determining the force on the charges in the wire with the equation Click to view content (https://www.openstax.org/l/induced-current) GRASP CHECK through a uniform magnetic field What emf is produced across a straight wire 0.50 m long moving at a velocity of (1.5 m/s) (0.30 T)\u1e91? The wire lies in the \u0177-direction. Also, which end of the wire is at the higher potential\u2014let the lower end of the wire be at y= 0 and the upper end at y= 0.5 m)? a. 0.15 V and the lower end of the wire will be at higher potential b. 0.15 V and the upper end of the wire will be at higher potential c. 0.075 V and the lower end of the wire will be at higher potential d. 0.075 V and the upper end of the wire will be at higher potential WORKED EXAMPLE EMF Induced in Conducing Coil by Moving Magnet Imagine a magnetic field goes through a coil in the direction indicated in Figure 20.37. The coil diameter is 2.0 cm. If the magnetic field goes from 0.020 to 0.010 T in 34 s, what is the direction and magnitude of the induced current? Assume the coil has a resistance of 0.1 678 Chapter 20 \u2022 Magnetism Figure", " 20.37 A coil through which passes a magnetic field B. STRATEGY Use the equation solenoid, we find it has 16 loops, so to find the induced emf in the coil, where. Counting the number of loops in the Use the equation to calculate the magnetic flux 20.30 where dis the diameter of the solenoid and we have used in the magnetic of the flux through the solenoid is Because the area of the solenoid does not vary, the change 20.31 Once we find the emf, we can use Ohm\u2019s law, to find the current. Finally, Lenz\u2019s law tells us that the current should produce a magnetic field that acts to oppose the decrease in the applied magnetic field. Thus, the current should produce a magnetic field to the right. Solution Combining equations and gives Solving Ohm\u2019s law for the current and using this result gives 20.32 20.33 Lenz\u2019s law tells us that the current must produce a magnetic field to the right. Thus, we point our right thumb to the right and curl our right fingers around the solenoid. The current must flow in the direction in which our fingers are pointing, so it enters at the left end of the solenoid and exits at the right end. Discussion Let\u2019s see if the minus sign makes sense in Faraday\u2019s law of induction. Define the direction of the magnetic field to be the positive direction. This means the change in the magnetic field is negative, as we found above. The minus sign in Faraday\u2019s law of induction negates the negative change in the magnetic field, leaving us with a positive current. Therefore, the current must flow in the direction of the magnetic field, which is what we found. Now try defining the positive direction to be the direction opposite that of the magnetic field, that is positive is to the left in Figure 20.37. In this case, you will find a negative current. But since the positive direction is to the left, a negative current must flow to the right, which again agrees with what we found by using Lenz\u2019s law. WORKED EXAMPLE Magnetic Induction due to Changing Circuit Size The circuit shown in Figure 20.38 consists of a U-shaped wire with a resistor and with the ends connected by a sliding conducting rod. The magnetic field filling the area enclosed by the circuit is constant at 0.01 T. If the rod is pulled to the right at speed what current is induced in the circuit and in what direction does the current flow? Access for free at openstax.org. 20.3 \u2022 Electromagnetic Induction 679 Figure 20.38 A slider circuit. The magnetic field is constant and the rod is pulled to the right at speed v. The changing area enclosed by the circuit induces an emf in the circuit. STRATEGY We again use Faraday\u2019s law of induction, although this time the magnetic field is constant and the area enclosed by the circuit changes. The circuit contains a single loop, so The rate of change of the area is Thus the rate of change of the magnetic flux is 20.34 where we have used the fact that the angle between the area vector and the magnetic field is 0\u00b0. Once we know the emf, we can find the current by using Ohm\u2019s law. To find the direction of the current, we apply Lenz\u2019s law. Solution Faraday\u2019s law of induction gives Solving Ohm\u2019s law for the current and using the previous result for emf gives 20.35 20.36 As the rod slides to the right, the magnetic flux passing through the circuit increases. Lenz\u2019s law tells us that the current induced will create a magnetic field that will counter this increase. Thus, the magnetic field created by the induced current must be into the page. Curling your right-hand fingers around the loop in the clockwise direction makes your right thumb point into the page, which is the desired direction of the magnetic field. Thus, the current must flow in the clockwise direction around the circuit. Discussion Is energy conserved in this circuit? An external agent must pull on the rod with sufficient force to just balance the force on a current-carrying wire in a magnetic field\u2014recall that be balanced by the rate at which the circuit dissipates power. Using constant speed vis the force required to pull the wire at a The rate at which this force does work on the rod should where we used the fact that the angle between the current and the magnetic field is the current into this equation gives Inserting our expression above for 20.37 20.38 The power contributed by the agent pulling the rod is 680 Chapter 20 \u2022 Magnetism The power dissipated by the circuit is 20.39 20.40 We thus see that agent that pulls the rod. Thus, energy is conserved in this system. which means that power is", " conserved in the system consisting of the circuit and the Practice Problems 11. The magnetic flux through a single wire loop changes from 3.5 Wb to 1.5 Wb in 2.0 s. What emf is induced in the loop? a. \u20132.0 V b. \u20131.0 V c. +1.0 V d. +2.0 V 12. What is the emf for a 10-turn coil through which the flux changes at 10 Wb/s? a. \u2013100 V b. \u201310 V c. +10 V d. +100 V Check Your Understanding 13. Given a bar magnet, how can you induce an electric current in a wire loop? a. An electric current is induced if a bar magnet is placed near the wire loop. b. An electric current is induced if wire loop is wound around the bar magnet. c. An electric current is induced if a bar magnet is moved through the wire loop. d. An electric current is induced if a bar magnet is placed in contact with the wire loop. 14. What factors can cause an induced current in a wire loop through which a magnetic field passes? a. b. c. d. Induced current can be created by changing the size of the wire loop only. Induced current can be created by changing the orientation of the wire loop only. Induced current can be created by changing the strength of the magnetic field only. Induced current can be created by changing the strength of the magnetic field, changing the size of the wire loop, or changing the orientation of the wire loop. Access for free at openstax.org. Chapter 20 \u2022 Key Terms 681 KEY TERMS Curie temperature well-defined temperature for the magnetic force ferromagnetic materials above which they cannot be magnetized domain region within a magnetic material in which the magnetic flux component of the magnetic field perpendicular to the surface area through which it passes and multiplied by the area magnetic poles of individual atoms are aligned magnetic pole part of a magnet that exerts the strongest electric motor device that transforms electrical energy into force on other magnets or magnetic material mechanical energy magnetized material that is induced to be magnetic or that electromagnet device that uses electric current to make a is made into a permanent magnet magnetic field north pole part of a magnet that orients itself toward the electromagnetism study of electric and magnetic geographic North Pole of Earth phenomena emf rate at which energy is drawn from a source per unit current flowing through a circuit ferromagnetic material such as iron, cobalt, nickel, or gadolinium that exhibits strong magnetic effects generator device that transforms mechanical energy into electrical energy permanent magnet material that retains its magnetic behavior for a long time, even when exposed to demagnetizing influences right-hand rule rule involving curling the right-hand fingers from one vector to another; the direction in which the right thumb points is the direction of the resulting vector induction rate at which energy is drawn from a source per unit current flowing through a circuit magnetic dipole term that describes magnets because they solenoid uniform cylindrical coil of wire through which electric current is passed to produce a magnetic field south pole part of a magnet that orients itself toward the always have two poles: north and south geographic South Pole of Earth magnetic field directional lines around a magnetic transformer device that transforms voltages from one material that indicates the direction and magnitude of value to another SECTION SUMMARY 20.1 Magnetic Fields, Field Lines, and Force \u2022 All magnets have two poles: a north pole and a south pole. If the magnet is free to move, its north pole orients itself toward the geographic North Pole of Earth, and the south pole orients itself toward the geographic South Pole of Earth. \u2022 A repulsive force occurs between the north poles of two magnets and likewise for two south poles. However, an attractive force occurs between the north pole of one magnet and the south pole of another magnet. \u2022 A charged particle moving through a magnetic field experiences a force whose direction is determined by the right-hand rule. \u2022 An electric current generates a magnetic field. \u2022 Electromagnets are magnets made by passing a current through a system of wires. 20.2 Motors, Generators, and Transformers \u2022 Electric motors contain wire loops in a magnetic field. Current is passed through the wire loops, which forces them to rotate in the magnetic field. The current is reversed every half rotation so that the torque on the loop is always in the same direction. \u2022 Electric generators contain wire loops in a magnetic field. An external agent provides mechanical energy to force the loops to rotate in the magnetic field, which produces an AC voltage that drives an AC current through the loops. \u2022 Transformers contain a ring made of magnetic material and, on opposite sides of the ring, two windings of wire wrap around the ring. A changing current in one wire winding creates a changing magnetic field, which is trapped in the ring and thus goes through the second winding and induces an emf in the second winding. The voltage in the second", " winding is proportional to the ratio of the number of loops in each winding. \u2022 Transformers are used to step up and step down the voltage for power transmission. \u2022 Over long distances, electric power is transmitted at high voltage to minimize the current and thereby minimize the Joule losses due to resistive heating. 20.3 Electromagnetic Induction \u2022 Faraday\u2019s law of induction states that a changing magnetic flux that occurs within an area enclosed by a conducting loop induces an electric current in the loop. \u2022 Lenz\u2019 law states that an induced current flows in the direction such that it opposes the change that induced it. 682 Chapter 20 \u2022 Key Equations KEY EQUATIONS 20.1 Magnetic Fields, Field Lines, and Force the magnitude of the force on an electric charge the force on a wire carrying current the magnitude of the magnetic field created by a long, straight current-carrying wire CHAPTER REVIEW Concept Items 20.1 Magnetic Fields, Field Lines, and Force 1. If you place a small needle between the north poles of two bar magnets, will the needle become magnetized? a. Yes, the magnetic fields from the two north poles will point in the same directions. b. Yes, the magnetic fields from the two north poles will point in opposite directions. c. No, the magnetic fields from the two north poles will point in opposite directions. d. No, the magnetic fields from the two north poles will point in the same directions. 2. If you place a compass at the three points in the figure, at which point will the needle experience the greatest torque? Why? the magnitude of the magnetic field inside a solenoid 20.3 Electromagnetic Induction magnetic flux emf greatest torque at B. b. The density of the magnetic field is minimized at C, so the magnetic compass needle will experience the greatest torque at C. c. The density of the magnetic field is maximized at B, so the magnetic compass needle will experience the greatest torque at B. d. The density of the magnetic field is maximized at A, so the magnetic compass needle will experience the greatest torque at A. 3. In which direction do the magnetic field lines point near the south pole of a magnet? a. Outside the magnet the direction of magnetic field lines is towards the south pole of the magnet. b. Outside the magnet the direction of magnetic field lines is away from the south pole of the magnet. 20.2 Motors, Generators, and Transformers 4. Consider the angle between the area vector and the and magnetic field in an electric motor. At what angles is the torque on the wire loop the greatest? a. b. c. d. and and and 5. What is a voltage transformer? a. A transformer is a device that transforms current to voltage. b. A transformer is a device that transforms voltages from one value to another. c. A transformer is a device that transforms resistance of wire to voltage. 6. Why is electric power transmitted at high voltage? a. The density of the magnetic field is minimized at B, so the magnetic compass needle will experience the Access for free at openstax.org. Chapter 20 \u2022 Chapter Review 683 a. To increase the current for the transmission b. To reduce energy loss during transmission c. To increase resistance during transmission d. To reduce resistance during transmission 20.3 Electromagnetic Induction 7. Yes or no\u2014Is an emf induced in the coil shown when it is stretched? If so, state why and give the direction of the induced current. a. b. c. d. If induced current flows, its direction is such that it adds to the changes which induced it. If induced current flows, its direction is such that it opposes the changes which induced it. If induced current flows, its direction is always clockwise to the changes which induced it. If induced current flows, its direction is always counterclockwise to the changes which induced it. 9. Explain how magnetic flux can be zero when the a. No, because induced current does not depend upon the area of the coil. b. Yes, because area of the coil increases; the direction of the induced current is counterclockwise. c. Yes, because area of the coil increases; the direction of the induced current is clockwise. d. Yes, because the area of the coil does not change; the direction of the induced current is clockwise. 8. What is Lenz\u2019s law? Critical Thinking Items 20.1 Magnetic Fields, Field Lines, and Force 10. True or false\u2014It is not recommended to place credit cards with magnetic strips near permanent magnets. a. b. false true 11. True or false\u2014A square magnet can have sides that alternate between north and south poles. a. b. false true 12. You move a compass in a circular plane around a planar magnet. The compass makes four complete revolutions. How many poles does the magnet have? two poles a. b. four poles c. eight poles 12 poles d. 20", ".2 Motors, Generators, and Transformers 13. How can you maximize the peak emf from a generator? a. The peak emf from a generator can be maximized only by maximizing number of turns. b. The peak emf from a generator can be maximized only by maximizing area of the wired loop. magnetic field is not zero. a. If angle between magnetic field and area vector is 0\u00b0, then its sine is also zero, which means that there is zero flux. If angle between magnetic field and area vector is 45\u00b0, then its sine is also zero, which means that there is zero flux. If angle between magnetic field and area vector is 60\u00b0, then its cosine is also zero, which means that there is zero flux. If the angle between magnetic field and area vector is 90\u00b0, then its cosine is also zero, which means that there is zero flux. b. c. d. c. The peak emf from a generator can be maximized only by maximizing frequency. d. The peak emf from a generator can be maximized by maximizing number of turns, maximizing area of the wired loop or maximizing frequency. 14. Explain why power is transmitted over long distances at high voltages. a. Plost = Itransmitted Vtransmitted, so to maximize current, the voltage must be maximized b. Ptransmitted = Itransmitted Vtransmitted, so to maximize current, the voltage must be maximized c. Plost = Itransmitted Vtransmitted, so to minimize current, the voltage must be maximized d. Ptransmitted = Itransmitted Vtransmitted, so to minimize current, the voltage must be maximized 20.3 Electromagnetic Induction 15. To obtain power from the current in the wire of your vacuum cleaner, you place a loop of wire near it to obtain an induced emf. How do you place and orient the loop? a. A loop of wire should be placed nearest to the vacuum cleaner wire to maximize the magnetic flux through the loop. b. A loop of wire should be placed farthest to the vacuum cleaner wire to maximize the magnetic flux through the loop. 684 Chapter 20 \u2022 Test Prep c. A loop of wire should be placed perpendicular to the vacuum cleaner wire to maximize the magnetic flux through the loop. d. A loop of wire should be placed at angle greater than 90\u00b0 to the vacuum cleaner wire to maximize the magnetic flux through the loop. that is proportional to the rate of change of the magnetic flux. b. The magnetic field in the coil changes rapidly due to spinning of magnet which creates an emf in the coil that is proportional to the rate of change of the magnetic flux. 16. A magneto is a device that creates a spark across a gap 17. If you drop a copper tube over a bar magnet with its by creating a large voltage across the gap. To do this, the device spins a magnet very quickly in front of a wire coil, with the ends of the wires forming the gap. Explain how this creates a sufficiently large voltage to produce a spark. a. The electric field in the coil increases rapidly due to spinning of magnet which creates an emf in the coil north pole up, is a current induced in the copper tube? If so, in what direction? Consider when the copper tube is approaching the bar magnet. a. Yes, the induced current will be produced in the clockwise direction when viewed from above. b. No, the induced current will not be produced. Problems 20.1 Magnetic Fields, Field Lines, and Force 18. A straight wire segment carries 0.25 A. What length would it need to be to exert a 4.0-mN force on a magnet that produces a uniform magnetic field of 0.015 T that is perpendicular to the wire? a. 0.55 m b. 1.10 m c. 2.20 m d. 4.40 m 20.3 Electromagnetic Induction 19. What is the current in a wire loop of resistance 10 \u03a9 through which the magnetic flux changes from zero to Performance Task 20.2 Motors, Generators, and Transformers 21. Your family takes a trip to Cuba, and rents an old car to drive into the countryside to see the sights. Unfortunately, the next morning you find yourself deep in the countryside and the car won\u2019t start because the battery is too weak. Wanting to jump-start the car, you open the hood and find that you can\u2019t tell which battery TEST PREP Multiple Choice 20.1 Magnetic Fields, Field Lines, and Force 22. For a magnet, a domain refers to ______. a. b. c. the region between the poles of the magnet the space around the magnet that is affected by the magnetic field the region within the magnet in which the Access for free at openstax.org. 10 Wb in 1.0 s? a. \u2013100 A b. \u20132.0 A c. \u20131", ".0 A d. +1.0 A 20. An emf is induced by rotating a 1,000 turn, 20.0 cm diameter coil in Earth\u2019s 5.00 \u00d7 10\u20135 T magnetic field. What average emf is induced, given the plane of the coil is originally perpendicular to Earth\u2019s field and is rotated to be parallel to the field in 10.0 ms? a. \u20131.6 \u00d7 10-4 V b. +1.6 \u00d7 10-4 V c. +1.6 \u00d7 10-1 V d. \u20131.6 \u00d7 10-1 V terminal is positive and which is negative. However, you do have a bar magnet with the north and south poles labeled and you manage to find a short wire. How do you use these to determine which terminal is which? For starters, how do you determine the direction of a magnetic field around a current-carrying wire? And in which direction will the force be on another magnet placed in this field? Do you need to worry about the sign of the mobile charge carriers in the wire? d. magnetic poles of individual atoms are aligned the region from which the magnetic material is mined 23. In the region just outside the south pole of a magnet, the magnetic field lines ______. a. point away from the south pole b. go around the south pole c. are less concentrated than at the north pole d. point toward the south pole 24. Which equation gives the force for a charge moving through a magnetic field? a. b. c. d. 25. Can magnetic field lines cross each other? Explain why or why not. a. Yes, magnetic field lines can cross each other because that point of intersection indicates two possible directions of magnetic field, which is possible. b. No, magnetic field lines cannot cross each other because that point of intersection indicates two possible directions of magnetic field, which is not possible. 26. True or false\u2014If a magnet shatters into many small pieces, all the pieces will have north and south poles a. b. true false 20.2 Motors, Generators, and Transformers 27. An electrical generator ________. is a generator powered by electricity a. b. must be turned by hand c. converts other sources of power into electrical power d. uses magnetism to create electrons 28. A step-up transformer increases the a. voltage from power lines for use in homes b. c. current from the power lines for use in homes current from the electrical generator for transmission along power lines d. voltage from the electrical power plant for transmission along power lines Chapter 20 \u2022 Test Prep 685 b. Torque is doubled. c. Torque is quadrupled. d. Torque is halved. 30. Why are the coils of a transformer wrapped around a loop of ferrous material? a. The magnetic field from the source coil is trapped and also increased in strength. b. The magnetic field from the source coil is dispersed and also increased in strength. c. The magnetic field from the source coil is trapped and also decreased in strength. d. Magnetic field from the source coil is dispersed and also decreased in strength. 20.3 Electromagnetic Induction 31. What does emfstand for? a. electromotive force b. electro motion force c. electromagnetic factor d. electronic magnetic factor 32. Which formula gives magnetic flux? a. b. c. d. 33. What is the relationship between the number of coils in a solenoid and the emf induced in it by a change in the magnetic flux through the solenoid? a. The induced emf is inversely proportional to the number of coils in a solenoid. b. The induced emf is directly proportional to the number of coils in a solenoid. c. The induced emf is inversely proportional to the square of the number of coils in a solenoid. d. The induced emf is proportional to square of the number of coils in a solenoid. 29. What would be the effect on the torque of an electric 34. True or false\u2014If you drop a bar magnet through a motor of doubling the width of the current loop in the motor? a. Torque remains the same. copper tube, it induces an electric current in the tube. a. b. false true Short Answer 20.1 Magnetic Fields, Field Lines, and Force 35. Given a bar magnet, a needle, a cork, and a bowl full of water, describe how to make a compass. a. Magnetize the needle by holding it perpendicular to a bar magnet\u2019s north pole and pierce the cork along its longitudinal axis by the needle and place the needle-cork combination in the water. The needle now orients itself along the magnetic field lines of Earth. b. Magnetize the needle by holding it perpendicular to a bar magnet\u2019s north pole and pierce the cork along its longitudinal axis by the needle and place the needle-cork", " combination in the water. The needle now orients itself perpendicular to the Hence, it will attract the south pole of other magnet. c. The needle will magnetize and the point of a needle kept closer to the north pole will act as a north pole. Hence, it will repel the south pole of the other magnet. d. The needle will magnetize and the point of needle kept closer to the north pole will act as a north pole. Hence, it will attract the south pole of other magnet. 39. Using four solenoids of the same size, describe how to orient them and in which direction the current should flow to make a magnet with two opposite-facing north poles and two opposite-facing south poles. 686 Chapter 20 \u2022 Test Prep magnetic field lines of Earth. c. Magnetize the needle by holding its axis parallel to the axis of a bar magnet and pierce the cork along its longitudinal axis by the needle and place the needle-cork combination in the water. The needle now orients itself along the magnetic field lines of Earth. d. Magnetize the needle by holding its axis parallel to the axis of a bar magnet and pierce the cork along its longitudinal axis by the needle and place the needle-cork combination in the water. The needle now orients itself perpendicular to the magnetic field lines of Earth. 36. Give two differences between electric field lines and magnetic field lines. a. Electric field lines begin and end on opposite charges and the electric force on a charge is in the direction of field, while magnetic fields form a loop and the magnetic force on a charge is perpendicular to the field. b. Electric field lines form a loop and the electric force on a charge is in the direction of field, while magnetic fields begin and end on opposite charge and the magnetic force on a charge is perpendicular to the field. c. Electric field lines begin and end on opposite charges and the electric force on a charge is in the perpendicular direction of field, while magnetic fields form a loop and the magnetic force on a charge is in the direction of the field. d. Electric field lines form a loop and the electric force on a charge is in the perpendicular direction of field, while magnetic fields begin and end on opposite charge and the magnetic force on a charge is in the direction of the field. 37. To produce a magnetic field of 0.0020 T, what current is required in a 500-turn solenoid that is 25 cm long? a. 0.80 A b. 1.60 A c. 80 A 160 A d. 38. You magnetize a needle by aligning it along the axis of a bar magnet and just outside the north pole of the magnet. Will the point of the needle that was closest to the bar magnet then be attracted to or repelled from the south pole of another magnet? a. The needle will magnetize and the point of needle kept closer to the north pole will act as a south pole. Hence, it will repel the south pole of other magnet. b. The needle will magnetize and the point of needle kept closer to the north pole will act as a south pole. Access for free at openstax.org. Chapter 20 \u2022 Test Prep 687 a. The output emf will be doubled. b. The output emf will be halved. c. The output emf will be quadrupled. d. The output emf will be tripled. 44. In a hydroelectric dam, what is used to power the electrical generators that provide electric power? Explain. a. The electric potential energy of stored water is used to produce emf with the help of a turbine. b. The electric potential energy of stored water is used to produce resistance with the help of a turbine. c. Gravitational potential energy of stored water is used to produce resistance with the help of a turbine. d. Gravitational potential energy of stored water is used to produce emf with the help of a turbine. 20.3 Electromagnetic Induction 45. A uniform magnetic field is perpendicular to the plane of a wire loop. If the loop accelerates in the direction of the field, will a current be induced in the loop? Explain why or why not. a. No, because magnetic flux through the loop remains constant. b. No, because magnetic flux through the loop changes continuously. c. Yes, because magnetic flux through the loop remains constant. d. Yes, because magnetic flux through the loop changes continuously. 46. The plane of a square wire circuit with side 4.0 cm long is at an angle of 45\u00b0 with respect to a uniform magnetic field of 0.25 T. The wires have a resistance per unit length of 0.2. If the field drops to zero in 2.5 s, what magnitude current is induced in the square circuit? 35 \u00b5A a. b. 87.5 \u00b5A 3.5 mA c. 35 A d. 47. Yes or no\u2014If a bar magnet moves through a wire loop", " as shown in the figure, is a current induced in the loop? Explain why or why not. 40. How far from a straight wire carrying 0.45 A is the magnetic field strength 0.040 T? a. 0.23 \u00b5m b. 0.72 \u00b5m c. 2.3 \u00b5m 7.2 \u00b5m d. 20.2 Motors, Generators, and Transformers 41. A laminated-coil transformer has a wire coiled 12 times around one of its sides. How many coils should you wrap around the opposite side to get a voltage output that is one half of the input voltage? Explain. a. six output coils because the ratio of output to input voltage is the same as the ratio of number of output coils to input coils 12 output coils because the ratio of output to input voltage is the same as the ratio of number of output coils to input coils b. c. 24 output coils because the ratio of output to input voltage is half the ratio of the number of output coils to input coils 36 output coils because the ratio of output to input voltage is three times the ratio of the number of output coils to input coils d. 42. Explain why long-distance electrical power lines are designed to carry very high voltages. a. Ptransmitted = Itransmitted> 2 Rwire and Plost = Itransmitted Vtransmitted, so Vmust be low to make the current transmitted as high as possible. b. Ptransmitted = Itransmitted> 2 Rwire and Plost = Ilost Vlost, so Vmust be low to make the current transmitted as high as possible. c. Ptransmitted = Itransmitted> 2 Rwire and Plost = Itransmitted Vtransmitted, so Vmust be high to make the current transmitted as low as possible d. Plost = Itransmitted 2 Rwire and Ptransmitted = Itransmitted Vtransmitted, so Vmust be high to make the current transmitted as low as possible. a. No, because the net magnetic field passing through the loop is zero. 43. How is the output emf of a generator affected if you b. No, because the net magnetic field passing through double the frequency of rotation of its coil? 688 Chapter 20 \u2022 Test Prep the loop is nonzero. c. Yes, because the net magnetic field passing through the loop is zero. d. Yes, because the net magnetic field line passing through the loop is nonzero. 48. What is the magnetic flux through an equilateral Extended Response 20.1 Magnetic Fields, Field Lines, and Force 49. Summarize the properties of magnets. a. A magnet can attract metals like iron, nickel, etc., but cannot attract nonmetals like piece of plastic or wood, etc. If free to rotate, an elongated magnet will orient itself so that its north pole will face the magnetic south pole of Earth. b. A magnet can attract metals like iron, nickel, etc., but cannot attract nonmetals like piece of plastic or wood, etc. If free to rotate, an elongated magnet will orient itself so that its north pole will face the magnetic north pole of Earth. c. A magnet can attract metals like iron, nickel, etc., and nonmetals like piece of plastic or wood, etc. If free to rotate, an elongated magnet will orient itself so that its north pole will face the magnetic south pole of Earth. d. A magnet can attract metals like iron, nickel, etc., and nonmetals like piece of plastic or wood, etc. If free to rotate, an elongated magnet will orient itself so that its north pole will face the magnetic north pole of Earth. 50. The magnetic field shown in the figure is formed by current flowing in two rings that intersect the page at the dots. Current flows into the page at the dots with crosses (right side) and out of the page at the dots with points (left side). Access for free at openstax.org. triangle with side 60 cm long and whose plane makes a 60\u00b0 angle with a uniform magnetic field of 0.33 T? a. 0.045 Wb b. 0.09 Wb c. 0.405 Wb d. 4.5 Wb Where is the field strength the greatest and in what direction do the magnetic field lines point? a. The magnetic field strength is greatest where the magnetic field lines are less dense; magnetic field lines points up the page. b. The magnetic field strength is greatest where the magnetic field lines are most dense; magnetic field lines points up the page. c. The magnetic field strength is greatest where the magnetic field lines are most dense; magnetic field lines points down the page. d. The magnetic field strength is greatest where the magnetic field lines are less dense; magnetic field lines points down the page. 51. The forces shown below are exerted on an electron as it moves through the magnetic field. In each case, what direction does the electron move? a. b", ". c. d. (a) left to right, (b) out of the page, (c) upwards (a) left to right, (b) into the page, (c) downwards (a) right to left, (b) out of the page, (c) upwards (a) right to left, (b) into the page, (c) downwards 20.2 Motors, Generators, and Transformers 52. Explain why increasing the frequency of rotation of the coils in an electrical generator increases the output emf. a. The induced emf is proportional to the rate of change of magnetic flux with respect to distance. b. The induced emf is inversely proportional to the rate of change of magnetic flux with respect to distance. c. The induced emf is inversely proportional to the rate of change of magnetic flux with respect to time. d. The induced emf is proportional to the rate of change of magnetic flux with respect to time. 53. Your friend tells you that power lines must carry a maximum current because P= I2R, where R is the resistance of the transmission line. What do you tell her? a. Ptransmitted = Itransmitted 2Rwire and Plost = Itransmitted Vtransmitted, so Imust be high to reduce power lost due to transmission. b. Plost = Itransmitted 2Rwire and Plost = Itransmitted Vtransmitted, so Imust be high to reduce power lost due to transmission. c. Ptransmitted = Itransmitted 2Rwire and Plost = Itransmitted Vtransmitted, so Imust be low to reduce power lost due to transmission. d. Plost = Itransmitted 2Rwire and Plost = Itransmitted Vtransmitted, so Imust be low to reduce power lost due to transmission. 20.3 Electromagnetic Induction 54. When you insert a copper ring between the poles of two bar magnets as shown in the figure, do the magnets exert an attractive or repulsive force on the ring? Explain your reasoning. a. Magnets exert an attractive force, because magnetic field due to induced current is repulsed by the magnetic field of the magnets. b. Magnets exert an attractive force, because Chapter 20 \u2022 Test Prep 689 magnetic field due to induced current is attracted by the magnetic field of the magnets. c. Magnets exert a repulsive force, because magnetic field due to induced current is repulsed by the magnetic field of the magnets. d. Magnets exert a repulsive force, because magnetic field due to induced current is attracted by the magnetic field of the magnets. 55. The figure shows a uniform magnetic field passing through a closed wire circuit. The wire circuit rotates at an angular frequency of about the axis shown by the dotted line in the figure. What is an expression for the magnetic flux through the circuit as a function of time? a. expression for the magnetic flux through the circuit \u03a6(t) = BAcos \u03c9t b. expression for the magnetic flux through the circuit c. expression for the magnetic flux through the circuit d. expression for the magnetic flux through the circuit \u03a6(t) = 2BAcos \u03c9t 690 Chapter 20 \u2022 Test Prep Access for free at openstax.org. CHAPTER 21 The Quantum Nature of Light Figure 21.1 In Lewis Carroll\u2019s classic text Alice\u2019s Adventures in Wonderland, Alice follows a rabbit down a hole into a land of curiosity. While many of her interactions in Wonderland are of surprising consequence, they follow a certain inherent logic. (credit: modification of work by John Tenniel, Wikimedia Commons) Chapter Outline 21.1 Planck and Quantum Nature of Light 21.2 Einstein and the Photoelectric Effect 21.3 The Dual Nature of Light At first glance, the quantum nature of light can be a strange and bewildering concept. Between light acting as INTRODUCTION discrete chunks, massless particles providing momenta, and fundamental particles behaving like waves, it may often seem like something out of Alice in Wonderland. For many, the study of this branch of physics can be as enthralling as Lewis Carroll\u2019s classic novel. Recalling the works of legendary characters and brilliant scientists such as Einstein, Planck, and Compton, the study of light\u2019s quantum nature will provide you an interesting tale of how a clever interpretation of some small details led to the most important discoveries of the past 150 years. From the electronics revolution of the twentieth century to our future progress in solar energy and space exploration, the quantum nature of light should yield a rabbit hole of curious consequence, within which lie some of the most fascinating truths of our time. 692 Chapter 21 \u2022 The Quantum Nature of Light 21.1 Planck and Quantum Nature of Light Section Learning Objectives By the end of this section, you will be able to do the following: \u2022 Describe blackbody radiation \u2022 Define quantum states and their relationship to modern physics \u2022 Calculate", " the quantum energy of lights \u2022 Explain how photon energies vary across divisions of the electromagnetic spectrum Section Key Terms blackbody quantized quantum ultraviolet catastrophe Blackbodies Our first story of curious significance begins with a T-shirt. You are likely aware that wearing a tight black T-shirt outside on a hot day provides a significantly less comfortable experience than wearing a white shirt. Black shirts, as well as all other black objects, will absorb and re-emit a significantly greater amount of radiation from the sun. This shirt is a good approximation of what is called a blackbody. A perfect blackbody is one that absorbs and re-emits all radiated energy that is incident upon it. Imagine wearing a tight shirt that did this! This phenomenon is often modeled with quite a different scenario. Imagine carving a small hole in an oven that can be heated to very high temperatures. As the temperature of this container gets hotter and hotter, the radiation out of this dark hole would increase as well, re-emitting all energy provided it by the increased temperature. The hole may even begin to glow in different colors as the temperature is increased. Like a burner on your stove, the hole would glow red, then orange, then blue, as the temperature is increased. In time, the hole would continue to glow but the light would be invisible to our eyes. This container is a good model of a perfect blackbody. It is the analysis of blackbodies that led to one of the most consequential discoveries of the twentieth century. Take a moment to carefully examine Figure 21.2. What relationships exist? What trends can you see? The more time you spend interpreting this figure, the closer you will be to understanding quantum physics! Figure 21.2 Graphs of blackbody radiation (from an ideal radiator) at three different radiator temperatures. The intensity or rate of radiation emission increases dramatically with temperature, and the peak of the spectrum shifts toward the visible and ultraviolet parts of the spectrum. The shape of the spectrum cannot be described with classical physics. TIPS FOR SUCCESS When encountering a new graph, it is best to try to interpret the graph before you read about it. Doing this will make the following text more meaningful and will help to remind yourself of some of the key concepts within the section. Access for free at openstax.org. 21.1 \u2022 Planck and Quantum Nature of Light 693 Understanding Blackbody Graphs Figure 21.2 is a plot of radiation intensity against radiated wavelength. In other words, it shows how the intensity of radiated light changes when a blackbody is heated to a particular temperature. It may help to just follow the bottom-most red line labeled 3,000 K, red hot. The graph shows that when a blackbody acquires a temperature of 3,000 K, it radiates energy across the electromagnetic spectrum. However, the energy is most intensely emitted at a wavelength of approximately 1000 nm. This is in the infrared portion of the electromagnetic spectrum. While a body at this temperature would appear red-hotto our eyes, it would truly appear \u2018infrared-hot\u2019 if we were able to see the entire spectrum. A few other important notes regarding Figure 21.2: \u2022 As temperature increases, the total amount of energy radiated increases. This is shown by examining the area underneath each line. \u2022 Regardless of temperature, all red lines on the graph undergo a consistent pattern. While electromagnetic radiation is emitted throughout the spectrum, the intensity of this radiation peaks at one particular wavelength. \u2022 As the temperature changes, the wavelength of greatest radiation intensity changes. At 4,000 K, the radiation is most intense in the yellow-green portion of the spectrum. At 6,000 K, the blackbody would radiate white hot,due to intense radiation throughout the visible portion of the electromagnetic spectrum. Remember that white light is the emission of all visible colors simultaneously. \u2022 As the temperature increases, the frequency of light providing the greatest intensity increases as well. Recall the equation Because the speed of light is constant, frequency and wavelength are inversely related. This is verified by the leftward movement of the three red lines as temperature is increased. While in science it is important to categorize observations, theorizing as to why the observations exist is crucial to scientific advancement. Why doesn\u2019t a blackbody emit radiation evenly across all wavelengths? Why does the temperature of the body change the peak wavelength that is radiated? Why does an increase in temperature cause the peak wavelength emitted to decrease? It is questions like these that drove significant research at the turn of the twentieth century. And within the context of these questions, Max Planck discovered something of tremendous importance. Planck\u2019s Revolution The prevailing theory at the time of Max Planck\u2019s discovery was that intensity and frequency were related by the equation This equation, derived from classical physics and using wave phenomena, infers that as wavelength increases, the intensity of energy provided will decrease with an inverse-squared relationship. This relationship is graphed in Figure 21.3 and shows", " a troubling trend. For starters, it should be apparent that the graph from this equation does not match the blackbody graphs found experimentally. Additionally, it shows that for an object of any temperature, there should be an infinite amount of energy quickly emitted in the shortest wavelengths. When theory and experimental results clash, it is important to re-evaluate both models. The disconnect between theory and reality was termed the ultraviolet catastrophe. 694 Chapter 21 \u2022 The Quantum Nature of Light Figure 21.3 The graph above shows the true spectral measurements by a blackbody against those predicted by the classical theory at the time. The discord between the predicted classical theory line and the actual results is known as the ultraviolet catastrophe. Due to concerns over the ultraviolet catastrophe, Max Planck began to question whether another factor impacted the relationship between intensity and wavelength. This factor, he posited, should affect the probability that short wavelength light would be emitted. Should this factor reduce the probability of short wavelength light, it would cause the radiance curve to not progress infinitely as in the classical theory, but would instead cause the curve to precipitate back downward as is shown in the 5,000 K, 4,000 K, and 3,000 K temperature lines of the graph in Figure 21.3. Planck noted that this factor, whatever it may be, must also be dependent on temperature, as the intensity decreases at lower and lower wavelengths as the temperature increases. The determination of this probability factorwas a groundbreaking discovery in physics, yielding insight not just into light but also into energy and matter itself. It would be the basis for Planck\u2019s 1918 Nobel Prize in Physics and would result in the transition of physics from classical to modern understanding. In an attempt to determine the cause of the probability factor,Max Planck constructed a new theory. This theory, which created the branch of physics called quantum mechanics, speculated that the energy radiated by the blackbody could exist only in specific numerical, or quantum, states. This theory is described by the where nis any nonnegative integer (0, 1, 2, 3, \u2026) and his Planck\u2019s constant, given by equation and fis frequency. Through this equation, Planck\u2019s probability factor can be more clearly understood. Each frequency of light provides a specific quantized amount of energy. Low frequency light, associated with longer wavelengths would provide a smaller amount of energy, while high frequency light, associated with shorter wavelengths, would provide a larger amount of energy. For specified temperatures with specific total energies, it makes sense that more low frequency light would be radiated than high frequency light. To a degree, the relationship is like pouring coins through a funnel. More of the smaller pennies would be able to pass through the funnel than the larger quarters. In other words, because the value of the coin is somewhat related to the size of the coin, the probability of a quarter passing through the funnel is reduced! Furthermore, an increase in temperature would signify the presence of higher energy. As a result, the greater amount of total blackbody energy would allow for more of the high frequency, short wavelength, energies to be radiated. This permits the peak of the blackbody curve to drift leftward as the temperature increases, as it does from the 3,000 K to 4,000 K to 5,000 K values. Furthering our coin analogy, consider a wider funnel. This funnel would permit more quarters to pass through and allow for a reduction in concern about the probability factor. In summary, it is the interplay between the predicted classical model and the quantum probability that creates the curve depicted in Figure 21.3. Just as quarters have a higher currency denomination than pennies, higher frequencies come with larger Access for free at openstax.org. 21.1 \u2022 Planck and Quantum Nature of Light 695 amounts of energy. However, just as the probability of a quarter passing through a fixed diameter funnel is reduced, so is the probability of a high frequency light existing in a fixed temperature object. As is often the case in physics, it is the balancing of multiple incredible ideas that finally allows for better understanding. Quantization It may be helpful at this point to further consider the idea of quantum states. Atoms, molecules, and fundamental electron and proton charges are all examples of physical entities that are quantized\u2014that is, they appear only in certain discrete values and do not have every conceivable value. On the macroscopic scale, this is not a revolutionary concept. A standing wave on a string allows only particular harmonics described by integers. Going up and down a hill using discrete stair steps causes your potential energy to take on discrete values as you move from step to step. Furthermore, we cannot have a fraction of an atom, or part of an electron\u2019s charge, or 14.33 cents. Rather, everything is built of integral multiples of these substructures. That said, to discover quantum states within a phenomenon that science had always considered continuous would certainly be surprising. When Max Planck was able to", " use quantization to correctly describe the experimentally known shape of the blackbody spectrum, it was the first indication that energy was quantized on a small scale as well. This discovery earned Planck the Nobel Prize in Physics in 1918 and was such a revolutionary departure from classical physics that Planck himself was reluctant to accept his own idea. The general acceptance of Planck\u2019s energy quantization was greatly enhanced by Einstein\u2019s explanation of the photoelectric effect (discussed in the next section), which took energy quantization a step further. Figure 21.4 The German physicist Max Planck had a major influence on the early development of quantum mechanics, being the first to recognize that energy is sometimes quantized. Planck also made important contributions to special relativity and classical physics. (credit: Library of Congress, Prints and Photographs Division, Wikimedia Commons) WORKED EXAMPLE How Many Photons per Second Does a Typical Light Bulb Produce? Assuming that 10 percent of a 100-W light bulb\u2019s energy output is in the visible range (typical for incandescent bulbs) with an average wavelength of 580 nm, calculate the number of visible photons emitted per second. Strategy The number of visible photons per second is directly related to the amount of energy emitted each second, also known as the bulb\u2019s power. By determining the bulb\u2019s power, the energy emitted each second can be found. Since the power is given in watts, which is joules per second, the energy will be in joules. By comparing this to the amount of energy associated with each photon, the number of photons emitted each second can be determined. Solution The power in visible light production is 10.0 percent of 100 W, or 10.0 J/s. The energy of the average visible photon is found by substituting the given average wavelength into the formula By rearranging the above formula to determine energy per photon, this produces 21.1 The number of visible photons per second is thus 696 Chapter 21 \u2022 The Quantum Nature of Light Discussion This incredible number of photons per second is verification that individual photons are insignificant in ordinary human experience. However, it is also a verification of our everyday experience\u2014on the macroscopic scale, photons are so small that quantization becomes essentially continuous. WORKED EXAMPLE How does Photon Energy Change with Various Portions of the EM Spectrum? Refer to the Graphs of Blackbody Radiation shown in the first figure in this section. Compare the energy necessary to radiate one photon of infrared light and one photon of visible light. Strategy To determine the energy radiated, it is necessary to use the equation frequency for infrared light and visible light. It is also necessary to find a representative Solution According to the first figure in this section, one representative wavelength for infrared light is 2000 nm (2.000 \u00d7 10-6 m). The associated frequency of an infrared light is Using the equation, the energy associated with one photon of representative infrared light is The same process above can be used to determine the energy associated with one photon of representative visible light. According to the first figure in this section, one representative wavelength for visible light is 500 nm. 21.2 21.3 21.4 21.5 Discussion This example verifies that as the wavelength of light decreases, the quantum energy increases. This explains why a fire burning with a blue flame is considered more dangerous than a fire with a red flame. Each photon of short-wavelength blue light emitted carries a greater amount of energy than a long-wavelength red light. This example also helps explain the differences in the 3,000 K, 4,000 K, and 6,000 K lines shown in the first figure in this section. As the temperature is increased, more energy is available for a greater number of short-wavelength photons to be emitted. Practice Problems 1. An AM radio station broadcasts at a frequency of 1,530 kHz. What is the energy in Joules of a photon emitted from this station? a. b. c. d. 10.1 \u00d7 10-26 J 1.01 \u00d7 10-28 J 1.01 \u00d7 10-29 J 1.01 \u00d7 10-27 J 2. A photon travels with energy of 1.0 eV. What type of EM radiation is this photon? a. visible radiation Access for free at openstax.org. 21.1 \u2022 Planck and Quantum Nature of Light 697 b. microwave radiation infrared radiation c. d. ultraviolet radiation Check Your Understanding 3. Do reflective or absorptive surfaces more closely model a perfect blackbody? reflective surfaces a. b. absorptive surfaces 4. A black T-shirt is a good model of a blackbody. However, it is not perfect. What prevents a black T-shirt from being considered a perfect blackbody? a. The T-shirt reflects some light. b. The T-shirt absorbs all incident light. c. The T-shirt re-emits all the incident light. d. The T", "-shirt does not reflect light. 5. What is the mathematical relationship linking the energy of a photon to its frequency? a. b. c. d. 6. Why do we not notice quantization of photons in everyday experience? a. because the size of each photon is very large b. because the mass of each photon is so small c. because the energy provided by photons is very large d. because the energy provided by photons is very small 7. Two flames are observed on a stove. One is red while the other is blue. Which flame is hotter? a. The red flame is hotter because red light has lower frequency. b. The red flame is hotter because red light has higher frequency. c. The blue flame is hotter because blue light has lower frequency. d. The blue flame is hotter because blue light has higher frequency. 8. Your pupils dilate when visible light intensity is reduced. Does wearing sunglasses that lack UV blockers increase or decrease the UV hazard to your eyes? Explain. Increase, because more high-energy UV photons can enter the eye. a. b. Increase, because less high-energy UV photons can enter the eye. c. Decrease, because more high-energy UV photons can enter the eye. d. Decrease, because less high-energy UV photons can enter the eye. 9. The temperature of a blackbody radiator is increased. What will happen to the most intense wavelength of light emitted as this increase occurs? a. The wavelength of the most intense radiation will vary randomly. b. The wavelength of the most intense radiation will increase. c. The wavelength of the most intense radiation will remain unchanged. d. The wavelength of the most intense radiation will decrease. 698 Chapter 21 \u2022 The Quantum Nature of Light 21.2 Einstein and the Photoelectric Effect Section Learning Objectives By the end of this section, you will be able to do the following: \u2022 Describe Einstein\u2019s explanation of the photoelectric effect \u2022 Describe how the photoelectric effect could not be explained by classical physics \u2022 Calculate the energy of a photoelectron under given conditions \u2022 Describe use of the photoelectric effect in biological applications, photoelectric devices and movie soundtracks Section Key Terms electric eye photoelectric effect photoelectron photon The Photoelectric Effect Teacher Support [EL]Ask the students what they think the term photoelectricmeans. How does the term relate to its definition? When light strikes certain materials, it can eject electrons from them. This is called the photoelectric effect, meaning that light (photo) produces electricity. One common use of the photoelectric effect is in light meters, such as those that adjust the automatic iris in various types of cameras. Another use is in solar cells, as you probably have in your calculator or have seen on a rooftop or a roadside sign. These make use of the photoelectric effect to convert light into electricity for running different devices. Figure 21.5 The photoelectric effect can be observed by allowing light to fall on the metal plate in this evacuated tube. Electrons ejected by the light are collected on the collector wire and measured as a current. A retarding voltage between the collector wire and plate can then be adjusted so as to determine the energy of the ejected electrons. (credit: P. P. Urone) Revolutionary Properties of the Photoelectric Effect When Max Planck theorized that energy was quantized in a blackbody radiator, it is unlikely that he would have recognized just how revolutionary his idea was. Using tools similar to the light meter in Figure 21.5, it would take a scientist of Albert Einstein\u2019s stature to fully discover the implications of Max Planck\u2019s radical concept. Through careful observations of the photoelectric effect, Albert Einstein realized that there were several characteristics that could be explained only if EM radiation is itself quantized. While these characteristics will be explained a bit later in this section, you can already begin to appreciate why Einstein\u2019s idea is very important. It means that the apparently continuous stream of energy in an EM wave is actually not a continuous stream at all. In fact, the EM wave itself is actually composed of tiny quantum packets of energy called photons. In equation form, Einstein found the energy of a photon or photoelectron to be where Eis the energy of a photon of frequency fand his Planck\u2019s constant. A beam from a flashlight, which to this point had been considered a wave, instead could now be viewed as a series of photons, each providing a specific amount of energy see Figure 21.6. Furthermore, the amount of energy within each individual photon is based upon its individual frequency, as Access for free at openstax.org. dictated by frequency-dependent photon energies added together. As a result, the total amount of energy provided by the beam could now be viewed as the sum of all 21.2 \u2022 Einstein and the Photoelectric Effect 699 Figure 21.6 An EM wave of frequency fis composed of photons, or individual quanta of EM radiation. The energy of each", " photon is, where his Planck\u2019s constant and fis the frequency of the EM radiation. Higher intensity means more photons per unit area per second. The flashlight emits large numbers of photons of many different frequencies, hence others have energy, and so on. Just as with Planck\u2019s blackbody radiation, Einstein\u2019s concept of the photon could take hold in the scientific community only if it could succeed where classical physics failed. The photoelectric effect would be a key to demonstrating Einstein\u2019s brilliance. Consider the following five properties of the photoelectric effect. All of these properties are consistent with the idea that individual photons of EM radiation are absorbed by individual electrons in a material, with the electron gaining the photon\u2019s energy. Some of these properties are inconsistent with the idea that EM radiation is a simple wave. For simplicity, let us consider what happens with monochromatic EM radiation in which all photons have the same energy hf. Figure 21.7 Incident radiation strikes a clean metal surface, ejecting multiple electrons from it. The manner in which the frequency and intensity of the incoming radiation affect the ejected electrons strongly suggests that electromagnetic radiation is quantized. This event, called the photoelectric effect, is strong evidence for the existence of photons. 1. If we vary the frequency of the EM radiation falling on a clean metal surface, we find the following: For a given material, there is a threshold frequency f0 for the EM radiation below which no electrons are ejected, regardless of intensity. Using the photon model, the explanation for this is clear. Individual photons interact with individual electrons. Thus if the energy of an individual photon is too low to break an electron away, no electrons will be ejected. However, if EM radiation were a simple wave, sufficient energy could be obtained simply by increasing the intensity. 2. Once EM radiation falls on a material, electrons are ejected without delay. As soon as an individual photon of sufficiently high frequency is absorbed by an individual electron, the electron is ejected. If the EM radiation were a simple wave, several minutes would be required for sufficient energy to be deposited at the metal surface in order to eject an electron. 3. The number of electrons ejected per unit time is proportional to the intensity of the EM radiation and to no other 4. characteristic. High-intensity EM radiation consists of large numbers of photons per unit area, with all photons having the same characteristic energy, hf. The increased number of photons per unit area results in an increased number of electrons per unit area ejected. If we vary the intensity of the EM radiation and measure the energy of ejected electrons, we find the following: The maximum kinetic energy of ejected electrons is independent of the intensity of the EM radiation. Instead, as noted in point 3 above, increased intensity results in more electrons of the same energy being ejected. If EM radiation were a simple wave, a higher intensity could transfer more energy, and higher-energy electrons would be ejected. 5. The kinetic energy KE of an ejected electron equals the photon energy minus the binding energy BE of the electron in the 700 Chapter 21 \u2022 The Quantum Nature of Light specific material. An individual photon can give all of its energy to an electron. The photon\u2019s energy is partly used to break the electron away from the material. The remainder goes into the ejected electron\u2019s kinetic energy. In equation form, this is given by 21.6 is the maximum kinetic energy of the ejected electron, where electron to the particular material. This equation explains the properties of the photoelectric effect quantitatively and demonstrates that BE is the minimum amount of energy necessary to eject an electron. If the energy supplied is less than BE, the electron cannot be ejected. The binding energy can also be written as particular material. Figure 21.8 shows a graph of maximum particular material. versus the frequency of incident EM radiation falling on a is the photon\u2019s energy, and BE is the binding energy of the is the threshold frequency for the where Figure 21.8 A graph of the kinetic energy of an ejected electron, KEe, versus the frequency of EM radiation impinging on a certain material. There is a threshold frequency below which no electrons are ejected, because the individual photon interacting with an individual electron has insufficient energy to break it away. Above the threshold energy, KEe increases linearly with f, consistent with KEe = hf\u2212 BE. The slope of this line is h, so the data can be used to determine Planck\u2019s constant experimentally. TIPS FOR SUCCESS The following five pieces of information can be difficult to follow without some organization. It may be useful to create a table of expected results of each of the five properties, with one column showing the classical wave model result and one column showing the modern photon model result. The table may look something like Table 21.1 Classical Wave Model Modern Photon Model Threshold Frequency Electron Ejection Delay Intensity of EM Radiation Speed of Ejected Electrons Relationship between Kinetic Energy and Binding Energy Table 21.1 Table", " of Expected Results Virtual Physics Photoelectric Effect Click to view content (http://www.openstax.org/l/28photoelectric) Access for free at openstax.org. In this demonstration, see how light knocks electrons off a metal target, and recreate the experiment that spawned the field of quantum mechanics. 21.2 \u2022 Einstein and the Photoelectric Effect 701 GRASP CHECK In the circuit provided, what are the three ways to increase the current? a. decrease the intensity, decrease the frequency, alter the target b. decrease the intensity, decrease the frequency, don\u2019t alter the target increase the intensity, increase the frequency, alter the target c. increase the intensity, increase the frequency, alter the target d. WORKED EXAMPLE Photon Energy and the Photoelectric Effect: A Violet Light (a) What is the energy in joules and electron volts of a photon of 420-nm violet light? (b) What is the maximum kinetic energy of electrons ejected from calcium by 420 nm violet light, given that the binding energy of electrons for calcium metal is 2.71 eV? Strategy To solve part (a), note that the energy of a photon is given by is a straightforward application of to find the ejected electron\u2019s maximum kinetic energy, since BE is given.. For part (b), once the energy of the photon is calculated, it Solution for (a) Photon energy is given by Since we are given the wavelength rather than the frequency, we solve the familiar relationship yielding for the frequency, Combining these two equations gives the useful relationship Now substituting known values yields Converting to eV, the energy of the photon is Solution for (b) Finding the kinetic energy of the ejected electron is now a simple application of the equation the photon energy and binding energy yields 21.7 21.8 21.9 21.10. Substituting 21.11 Discussion The energy of this 420 nm photon of violet light is a tiny fraction of a joule, and so it is no wonder that a single photon would be difficult for us to sense directly\u2014humans are more attuned to energies on the order of joules. But looking at the energy in electron volts, we can see that this photon has enough energy to affect atoms and molecules. A DNA molecule can be broken with about 1 eV of energy, for example, and typical atomic and molecular energies are on the order of eV, so that the photon in this example could have biological effects, such as sunburn. The ejected electron has rather low energy, and it would not travel far, 702 Chapter 21 \u2022 The Quantum Nature of Light except in a vacuum. The electron would be stopped by a retarding potential of only 0.26 eV, a slightly larger KE than calculated above. In fact, if the photon wavelength were longer and its energy less than 2.71 eV, then the formula would give a negative kinetic energy, an impossibility. This simply means that the 420 nm photons with their 2.96 eV energy are not much above the frequency threshold. You can see for yourself that the threshold wavelength is 458 nm (blue light). This means that if calcium metal were used in a light meter, the meter would be insensitive to wavelengths longer than those of blue light. Such a light meter would be completely insensitive to red light, for example. Practice Problems 10. What is the longest-wavelength EM radiation that can eject a photoelectron from silver, given that the bonding energy is 4.73 eV? Is this radiation in the visible range? a. 2.63 \u00d7 10\u22127 m; No, the radiation is in microwave region. b. 2.63 \u00d7 10\u22127 m; No, the radiation is in visible region. c. 2.63 \u00d7 10\u22127 m; No, the radiation is in infrared region. d. 2.63 \u00d7 10-7 m; No, the radiation is in ultraviolet region. 11. What is the maximum kinetic energy in eV of electrons ejected from sodium metal by 450-nm EM radiation, given that the binding energy is 2.28 eV? a. 0.48 V b. 0.82 eV c. 1.21 eV d. 0.48 eV Technological Applications of the Photoelectric Effect While Einstein\u2019s understanding of the photoelectric effect was a transformative discovery in the early 1900s, its presence is ubiquitous today. If you have watched streetlights turn on automatically in response to the setting sun, stopped elevator doors from closing simply by putting your hands between them, or turned on a water faucet by sliding your hands near it, you are familiar with the electric eye, a name given to a group of devices that use the photoelectric effect for detection. All these devices rely on photoconductive cells. These cells are activated when light is absorbed by a semi-conductive material, knocking off a free electron. When this happens, an electron void is left behind, which", " attracts a nearby electron. The movement of this electron, and the resultant chain of electron movements, produces a current. If electron ejection continues, further holes are created, thereby increasing the electrical conductivity of the cell. This current can turn switches on and off and activate various familiar mechanisms. One such mechanism takes place where you may not expect it. Next time you are at the movie theater, pay close attention to the sound coming out of the speakers. This sound is actually created using the photoelectric effect! The audiotape in the projector booth is a transparent piece of film of varying width. This film is fed between a photocell and a bright light produced by an exciter lamp. As the transparent portion of the film varies in width, the amount of light that strikes the photocell varies as well. As a result, the current in the photoconductive circuit changes with the width of the filmstrip. This changing current is converted to a changing frequency, which creates the soundtrack commonly heard in the theater. WORK IN PHYSICS Solar Energy Physicist According to the U.S. Department of Energy, Earth receives enough sunlight each hour to power the entire globe for a year. While converting all of this energy is impossible, the job of the solar energy physicist is to explore and improve upon solar energy conversion technologies so that we may harness more of this abundant resource. The field of solar energy is not a new one. For over half a century, satellites and spacecraft have utilized photovoltaic cells to create current and power their operations. As time has gone on, scientists have worked to adapt this process so that it may be used in homes, businesses, and full-scale power stations using solar cells like the one shown in Figure 21.9. Access for free at openstax.org. 21.2 \u2022 Einstein and the Photoelectric Effect 703 Figure 21.9 A solar cell is an example of a photovoltaic cell. As light strikes the cell, the cell absorbs the energy of the photons. If this energy exceeds the binding energy of the electrons, then electrons will be forced to move in the cell, thereby producing a current. This current may be used for a variety of purposes. (credit: U.S. Department of Energy) Solar energy is converted to electrical energy in one of two manners: direct transfer through photovoltaic cells or thermal conversion through the use of a CSP, concentrating solar power, system. Unlike electric eyes, which trip a mechanism when current is lost, photovoltaic cells utilize semiconductors to directly transfer the electrons released through the photoelectric effect into a directed current. The energy from this current can then be converted for storage, or immediately used in an electric process. A CSP system is an indirect method of energy conversion. In this process, light from the Sun is channeled using parabolic mirrors. The light from these mirrors strikes a thermally conductive material, which then heats a pool of water. This water, in turn, is converted to steam, which turns a turbine and creates electricity. While indirect, this method has long been the traditional means of large-scale power generation. There are, of course, limitations to the efficacy of solar power. Cloud cover, nightfall, and incident angle strike at high altitudes are all factors that directly influence the amount of light energy available. Additionally, the creation of photovoltaic cells requires rare-earth minerals that can be difficult to obtain. However, the major role of a solar energy physicist is to find ways to improve the efficiency of the solar energy conversion process. Currently, this is done by experimenting with new semi conductive materials, by refining current energy transfer methods, and by determining new ways of incorporating solar structures into the current power grid. Additionally, many solar physicists are looking into ways to allow for increased solar use in impoverished, more remote locations. Because solar energy conversion does not require a connection to a large-scale power grid, research into thinner, more mobile materials will permit remote cultures to use solar cells to convert sunlight collected during the day into stored energy that can then be used at night. Regardless of the application, solar energy physicists are an important part of the future in responsible energy growth. While a doctoral degree is often necessary for advanced research applications, a bachelor's or master's degree in a related science or engineering field is typically enough to gain access into the industry. Computer skills are very important for energy modeling, including knowledge of CAD software for design purposes. In addition, the ability to collaborate and communicate with others is critical to becoming a solar energy physicist. GRASP CHECK What role does the photoelectric effect play in the research of a solar energy physicist? a. The understanding of photoelectric effect allows the physicist to understand the generation of light energy when using photovoltaic cells. b. The understanding of photoelectric effect allows the physicist to understand the generation of electrical energy when using photovoltaic cells. c. The understanding of photoelectric effect allows the physicist to understand the generation of electromagnetic energy", " when using photovoltaic cells. d. The understanding of photoelectric effect allows the physicist to understand the generation of magnetic energy when using photovoltaic cells. 704 Chapter 21 \u2022 The Quantum Nature of Light Check Your Understanding 12. How did Einstein\u2019s model of photons change the view of a beam of energy leaving a flashlight? a. A beam of light energy is now considered a continual stream of wave energy, not photons. b. A beam of light energy is now considered a collection of photons, each carrying its own individual energy. 13. True or false\u2014Visible light is the only type of electromagnetic radiation that can cause the photoelectric effect. a. b. false true 14. Is the photoelectric effect a direct consequence of the wave character of EM radiation or the particle character of EM radiation? a. The photoelectric effect is a direct consequence of the particle nature of EM radiation. b. The photoelectric effect is a direct consequence of the wave nature of EM radiation. c. The photoelectric effect is a direct consequence of both the wave and particle nature of EM radiation. d. The photoelectric effect is a direct consequence of neither the wave nor the particle nature of EM radiation. 15. Which aspects of the photoelectric effect can only be explained using photons? a. aspects 1, 2, and 3 b. aspects 1, 2, and 4 c. aspects 1, 2, 4 and 5 d. aspects 1, 2, 3, 4 and 5 16. In a photovoltaic cell, what energy transformation takes place? a. Solar energy transforms into electric energy. b. Solar energy transforms into mechanical energy. c. Solar energy transforms into thermal energy. d. In a photovoltaic cell, thermal energy transforms into electric energy. 17. True or false\u2014A current is created in a photoconductive cell, even if only one electron is expelled from a photon strike. a. b. false true 18. What is a photon and how is it different from other fundamental particles? a. A photon is a quantum packet of energy; it has infinite mass. b. A photon is a quantum packet of energy; it is massless. c. A photon is a fundamental particle of an atom; it has infinite mass. d. A photon is a fundamental particle of an atom; it is massless. 21.3 The Dual Nature of Light Section Learning Objectives By the end of this section, you will be able to do the following: \u2022 Describe the Compton effect \u2022 Calculate the momentum of a photon \u2022 Explain how photon momentum is used in solar sails \u2022 Explain the particle-wave duality of light Section Key Terms Compton effect particle-wave duality photon momentum Photon Momentum Do photons abide by the fundamental properties of physics? Can packets of electromagnetic energy possibly follow the same rules as a ping-pong ball or an electron? Although strange to consider, the answer to both questions is yes. Despite the odd nature of photons, scientists prior to Einstein had long suspected that the fundamental particle of Access for free at openstax.org. electromagnetic radiation shared properties with our more macroscopic particles. This is no clearer than when considering the photoelectric effect, where photons knock electrons out of a substance. While it is strange to think of a massless particle exhibiting momentum, it is now a well-established fact within the scientific community. Figure 21.10 shows macroscopic evidence of photon momentum. 21.3 \u2022 The Dual Nature of Light 705 Figure 21.10 The tails of the Hale-Bopp comet point away from the Sun, evidence that light has momentum. Dust emanating from the body of the comet forms this tail. Particles of dust are pushed away from the Sun by light reflecting from them. The blue, ionized gas tail is also produced by photons interacting with atoms in the comet material. (credit: Geoff Chester, U.S. Navy, via Wikimedia Commons) Figure 21.10 shows a comet with two prominent tails. Comet tails are composed of gases and dust evaporated from the body of the comet and ionized gas. What most people do not know about the tails is that they always point awayfrom the Sun rather than trailing behind the comet. This can be seen in the diagram. Why would this be the case? The evidence indicates that the dust particles of the comet are forced away from the Sun when photons strike them. Evidently, photons carry momentum in the direction of their motion away from the Sun, and some of this momentum is transferred to dust particles in collisions. The blue tail is caused by the solar wind, a stream of plasma consisting primarily of protons and electrons evaporating from the corona of the Sun. Momentum, The Compton Effect, and Solar Sails Momentum is conserved in quantum mechanics, just as it is in relativity and classical physics. Some of the earliest direct experimental evidence of this came from the scattering of X-ray photons by electrons in substances, a phenomenon discovered by American physicist Arthur H. Compton (18", "92\u20131962). Around 1923, Compton observed that X-rays reflecting from materials had decreased energy and correctly interpreted this as being due to the scattering of the X-ray photons by electrons. This phenomenon could be handled as a collision between two particles\u2014a photon and an electron at rest in the material. After careful observation, it was found that both energy and momentum were conserved in the collision. See Figure 21.11. For the discovery of this conserved scattering, now known as the Compton effect, Arthur Compton was awarded the Nobel Prize in 1929. Shortly after the discovery of Compton scattering, the value of the photon momentum, was determined by Louis de Broglie. In this equation, called the de Broglie relation, hrepresents Planck\u2019s constant and \u03bbis the photon wavelength. Figure 21.11 The Compton effect is the name given to the scattering of a photon by an electron. Energy and momentum are conserved, resulting in a reduction of both for the scattered photon. 706 Chapter 21 \u2022 The Quantum Nature of Light We can see that photon momentum is small, since observe photon momentum. Our mirrors do not recoil when light reflects from them, except perhaps in cartoons. Compton saw the effects of photon momentum because he was observing X-rays, which have a small wavelength and a relatively large momentum, interacting with the lightest of particles, the electron. and his very small. It is for this reason that we do not ordinarily WORKED EXAMPLE Electron and Photon Momentum Compared (a) Calculate the momentum of a visible photon that has a wavelength of 500 nm. (b) Find the velocity of an electron having the same momentum. (c) What is the energy of the electron, and how does it compare with the energy of the photon? Strategy Finding the photon momentum is a straightforward application of its definition: small, we can assume that an electron with the same momentum will be nonrelativistic, making it easy to find its velocity and kinetic energy from the classical formulas. If we find the photon momentum is Solution for (a) Photon momentum is given by the de Broglie relation. Entering the given photon wavelength yields 21.12 21.13 Solution for (b) Since this momentum is indeed small, we will use the classical expression momentum. Solving for vand using the known value for the mass of an electron gives to find the velocity of an electron with this Solution for (c) The electron has kinetic energy, which is classically given by Thus, Converting this to eV by multiplying by yields The photon energy Eis 21.14 21.15 21.16 21.17 21.18 which is about five orders of magnitude greater. Discussion Even in huge numbers, the total momentum that photons carry is small. An electron that carries the same momentum as a 500-nm photon will have a 1,460 m/s velocity, which is clearly nonrelativistic. This is borne out by the experimental observation that it takes far less energy to give an electron the same momentum as a photon. That said, for high-energy photons interacting with small masses, photon momentum may be significant. Even on a large scale, photon momentum can have an effect if there Access for free at openstax.org. 21.3 \u2022 The Dual Nature of Light 707 are enough of them and if there is nothing to prevent the slow recoil of matter. Comet tails are one example, but there are also proposals to build space sails that use huge low-mass mirrors (made of aluminized Mylar) to reflect sunlight. In the vacuum of space, the mirrors would gradually recoil and could actually accelerate spacecraft within the solar system. See the following figure. TIPS FOR SUCCESS When determining energies in particle physics, it is more sensible to use the unit eV instead of Joules. Using eV will help you to recognize differences in magnitude more easily and will make calculations simpler. Also, eV is used by scientists to describe the binding energy of particles and their rest mass, so using eV will eliminate the need to convert energy quantities. Finally, eV is a convenient unit when linking electromagnetic forces to particle physics, as one eV is the amount energy given to an electron when placed in a field of 1-V potential difference. Practice Problems 19. Find the momentum of a 4.00-cm wavelength microwave photon. a. 0.83 \u00d7 10\u221232 kg \u22c5 m/s 1.66 \u00d7 10\u221234 kg \u22c5 m/s b. c. 0.83 \u00d7 10\u221234 kg \u22c5 m/s 1.66 \u00d7 10-32 kg \u22c5 m/s d. 20. Calculate the wavelength of a photon that has the same momentum of a proton moving at 1.00 percent of the speed of light. a. 2.43 \u00d7 10\u221210 m b. 2.43 \u00d7 10\u221212 m 1.32 \u00d7 10\u221215 m", " c. 1.32 \u00d7 10\u221213 m d. Figure 21.12 (a) Space sails have been proposed that use the momentum of sunlight reflecting from gigantic low-mass sails to propel spacecraft about the solar system. A Russian test model of this (the Cosmos 1) was launched in 2005, but did not make it into orbit due to a rocket failure. (b) A U.S. version of this, labeled LightSail-1, is scheduled for trial launches in 2016. It will have a 40 m2 sail. (credit: Kim Newton/NASA) LINKS TO PHYSICS LightSail-1 Project \u201cProvide ships or sails adapted to the heavenly breezes, and there will be some who will brave even that void.\u201d \u2014 Johannes Kepler (in a letter to Galileo Galilei in 1608) 708 Chapter 21 \u2022 The Quantum Nature of Light Figure 21.13 NASA\u2019s NanoSail-D, a precursor to LightSail-1, with its sails deployed. The Planetary Society will be launching LightSail-1 in early 2016. (credit: NASA/MSFC/D, Wikimedia Commons) Traversing the Solar System using nothing but the Sun\u2019s power has long been a fantasy of scientists and science fiction writers alike. Though physicists like Compton, Einstein, and Planck all provided evidence of light\u2019s propulsive capacity, it is only recently that the technology has become available to truly put these visions into motion. In 2016, by sending a lightweight satellite into space, the LightSail-1 project is designed to do just that. A citizen-funded project headed by the Planetary Society, the 5.45-million-dollar LightSail-1 project is set to launch two crafts into orbit around the Earth. Each craft is equipped with a 32-square-meter solar sail prepared to unfurl once a rocket has launched it to an appropriate altitude. The sails are made of large mirrors, each a quarter of the thickness of a trash bag, which will receive an impulse from the Sun\u2019s reflecting photons. Each time the Sun\u2019s photon strikes the craft\u2019s reflective surface and bounces off, it will provide a momentum to the sail much greater than if the photon were simply absorbed. Attached to three tiny satellites called CubeSats, whose combined volume is no larger than a loaf of bread, the received momentum from the Sun\u2019s photons should be enough to record a substantial increase in orbital speed. The intent of the LightSail-1 mission is to prove that the technology behind photon momentum travel is sound and can be done cheaply. A test flight in May 2015 showed that the craft\u2019s Mylar sails could unfurl on command. With another successful result in 2016, the Planetary Society will be planning future versions of the craft with the hopes of eventually achieving interplanetary satellite travel. Though a few centuries premature, Kepler\u2019s fantastic vision may not be that far away. If eventually set into interplanetary launch, what will be the effect of continual photon bombardment on the motion of a craft similar to LightSail-1? a. b. c. d. It will result in continual acceleration of the craft. It will first accelerate and then decelerate the craft. It will first decelerate and then accelerate the craft. It will result in the craft moving at constant velocity. Particle-Wave Duality We have long known that EM radiation is like a wave, capable of interference and diffraction. We now see that light can also be modeled as particles\u2014massless photons of discrete energy and momentum. We call this twofold nature the particle-wave duality, meaning that EM radiation has properties of both particles and waves. This may seem contradictory, since we ordinarily deal with large objects that never act like both waves and particles. An ocean wave, for example, looks nothing like a grain of sand. However, this so-called duality is simply a term for properties of the photon analogous to phenomena we can observe directly, on a macroscopic scale. See Figure 21.14. If this term seems strange, it is because we do not ordinarily observe details on the quantum level directly, and our observations yield either particle-like orwave-like properties, but never both simultaneously. Access for free at openstax.org. 21.3 \u2022 The Dual Nature of Light 709 Figure 21.14 (a) The interference pattern for light through a double slit is a wave property understood by analogy to water waves. (b) The properties of photons having quantized energy and momentum and acting as a concentrated unit are understood by analogy to macroscopic particles. Since we have a particle-wave duality for photons, and since we have seen connections between photons and matter in that both have momentum, it is reasonable to ask whether there is a particle-wave duality for matter as well. If the EM radiation we once thought to be a pure wave has particle properties, is", " it possible that matter has wave properties? The answer, strangely, is yes. The consequences of this are tremendous, as particle-wave duality has been a constant source of scientific wonder during the twentieth and twenty-first centuries. Check Your Understanding 21. What fundamental physics properties were found to be conserved in Compton scattering? a. energy and wavelength b. energy and momentum c. mass and energy d. energy and angle 22. Why do classical or relativistic momentum equations not work in explaining the conservation of momentum that occurs in Compton scattering? a. because neither classical nor relativistic momentum equations utilize mass as a variable in their equations b. because relativistic momentum equations utilize mass as a variable in their formulas but classical momentum equations do not c. because classical momentum equations utilize mass as a variable in their formulas but relativistic momentum equations do not d. because both classical and relativistic momentum equations utilize mass as a variable in their formulas 23. If solar sails were constructed with more massive materials, how would this influence their effectiveness? a. The effect of the momentum would increase due to the decreased inertia of the sails. b. The effect of the momentum would reduce due to the decreased inertia of the sails. c. The effect of the momentum would increase due to the increased inertia of the sails. d. The effect of the momentum would be reduced due to the increased inertia of the sails. 24. True or false\u2014It is possible to propel a solar sail craft using just particles within the solar wind. a. b. true false 25. True or false\u2014Photon momentum more directly supports the wave model of light. a. b. false true 710 Chapter 21 \u2022 The Quantum Nature of Light 26. True or false\u2014wave-particle duality exists for objects on the macroscopic scale. a. b. false true 27. What type of electromagnetic radiation was used in Compton scattering? a. visible light b. ultraviolet radiation c. d. X-rays radio waves Access for free at openstax.org. Chapter 21 \u2022 Key Terms 711 KEY TERMS blackbody object that absorbs all radiated energy that strikes it and also emits energy across all wavelengths of the electromagnetic spectrum material by a photon of light photon a quantum, or particle, of electromagnetic radiation Compton effect phenomenon whereby X-rays scattered photon momentum amount of momentum of a photon, from materials have decreased energy calculated by electric eye group of devices that use the photoelectric effect for detection particle-wave duality property of behaving like either a particle or a wave; the term for the phenomenon that all particles have wave-like characteristics and waves have particle-like characteristics photoelectric effect phenomenon whereby some materials eject electrons when exposed to light photoelectron electron that has been ejected from a SECTION SUMMARY 21.1 Planck and Quantum Nature of Light \u2022 A blackbody will radiate energy across all wavelengths of the electromagnetic spectrum. \u2022 Radiation of a blackbody will peak at a particular wavelength, dependent on the temperature of the blackbody. \u2022 Analysis of blackbody radiation led to the field of quantum mechanics, which states that radiated energy can only exist in discrete quantum states. 21.2 Einstein and the Photoelectric Effect \u2022 The photoelectric effect is the process in which EM radiation ejects electrons from a material. \u2022 Einstein proposed photons to be quanta of EM where fis the radiation having energy frequency of the radiation. \u2022 All EM radiation is composed of photons. As Einstein KEY EQUATIONS 21.1 Planck and Quantum Nature of Light quantum energy quantized the fact that certain physical entities exist only with particular discrete values and not every conceivable value quantum discrete packet or bundle of a physical entity such as energy ultraviolet catastrophe misconception that blackbodies would radiate high frequency energy at a much higher rate than energy radiated at lower frequencies explained, all characteristics of the photoelectric effect are due to the interaction of individual photons with individual electrons. \u2022 The maximum kinetic energy KEeof ejected electrons (photoelectrons) is given by is the photon energy and BE is the binding energy (or work function) of the electron in the particular material. where hf 21.3 The Dual Nature of Light \u2022 Compton scattering provided evidence that photon- electron interactions abide by the principles of conservation of momentum and conservation of energy. \u2022 The momentum of individual photons, quantified by, can be used to explain observations of comets and may lead to future space technologies. \u2022 Electromagnetic waves and matter have both wave-like and particle-like properties. This phenomenon is defined as particle-wave duality. maximum kinetic energy of a photoelectron binding energy of an electron 21.2 Einstein and the Photoelectric Effect 21.3 The Dual Nature of Light energy of a photon momentum of a photon (deBroglie relation) 712 Chapter 21 \u2022 Chapter Review CHAPTER REVIEW Concept Items 21.1 Planck and Quantum Nature of Light 1. What aspect of the blackbody spectrum forced Planck to propose quantization of energy levels in atoms and molecules? a. Radiation occurs", " at a particular frequency that does not change with the energy supplied. b. Certain radiation occurs at a particular frequency that changes with the energy supplied. c. Maximum radiation would occur at a particular frequency that does not change with the energy supplied. d. Maximum radiation would occur at a particular frequency that changes with the energy supplied. 2. Two lasers shine red light at 650 nm. One laser is twice as bright as the other. Explain this difference using photons and photon energy. a. The brighter laser emits twice the number of photons and more energy per photon. b. The brighter laser emits twice the number of photons and less energy per photon. c. Both lasers emit equal numbers of photons and equivalent amounts of energy per photon. d. The brighter laser emits twice the number of photons but both lasers emit equivalent amounts of energy per photon. 3. Consider four stars in the night sky: red, yellow, orange, and blue. The photons of which star will carry the greatest amount of energy? a. blue b. orange c. red d. yellow 4. A lightbulb is wired to a variable resistor. What will happen to the color spectrum emitted by the bulb as the resistance of the circuit is increased? a. The bulb will emit greener light. b. The bulb will emit bluer light. c. The bulb will emit more ultraviolet light. d. The bulb will emit redder light. 21.2 Einstein and the Photoelectric Effect 5. Light is projected onto a semi-conductive surface. However, no electrons are ejected. What will happen when the light intensity is increased? a. An increase in light intensity decreases the number of photons. However, no electrons are ejected. Access for free at openstax.org. b. Increase in light intensity increases the number of photons, so electrons with higher kinetic energy are ejected. c. An increase in light intensity increases the number of photons, so electrons will be ejected. d. An increase in light intensity increases the number of photons. However, no electrons are ejected. 6. True or false\u2014The concept of a work function (or binding energy) is permissible under the classical wave model. a. b. false true 7. Can a single microwave photon cause cell damage? a. No, there is not enough energy associated with a single microwave photon to result in cell damage. b. No, there is zero energy associated with a single microwave photon, so it does not result in cell damage. c. Yes, a single microwave photon causes cell damage because it does not have high energy. d. Yes, a single microwave photon causes cell damage because it has enough energy. 21.3 The Dual Nature of Light 8. Why don\u2019t we feel the momentum of sunlight when we are on the beach? a. The momentum of a singular photon is incredibly small. b. The momentum is not felt because very few photons strike us at any time, and not all have momentum. c. The momentum of a singular photon is large, but very few photons strike us at any time. d. A large number of photons strike us at any time, and so their combined momentum is incredibly large. 9. If a beam of helium atoms is projected through two slits and onto a screen, will an interference pattern emerge? a. No, an interference pattern will not emerge because helium atoms will strike a variety of locations on the screen. b. No, an interference pattern will not emerge because helium atoms will strike at certain locations on the screen. c. Yes, an interference pattern will emerge because helium atoms will strike a variety of locations on the screen. d. Yes, an interference pattern will emerge because helium atoms will strike at certain locations on the screen. Chapter 21 \u2022 Chapter Review 713 Critical Thinking Items 21.1 Planck and Quantum Nature of Light 13. Why is it assumed that a perfect absorber of light (like a blackbody) must also be a perfect emitter of light? a. To achieve electrostatic equilibrium with its 10. Explain why the frequency of a blackbody does not surroundings double when the temperature is doubled. a. Frequency is inversely proportional to temperature. b. Frequency is directly proportional to temperature. c. Frequency is directly proportional to the square of temperature. b. To achieve thermal equilibrium with its surroundings c. To achieve mechanical equilibrium with its surroundings d. To achieve chemical equilibrium with its d. Frequency is directly proportional to the fourth surroundings power of temperature. 11. Why does the intensity shown in the blackbody radiation graph decrease after its peak frequency is achieved? a. Because after reaching the peak frequency, the photons created at a particular frequency are too many for energy intensity to continue to decrease. 21.2 Einstein and the Photoelectric Effect 14. Light is projected onto a semi-conductive surface. If the intensity is held constant but the frequency of light is increased, what will happen? a. As frequency is increased, electrons will stop being ejected from the surface. b. As frequency is increased, electrons will begin to be ejected from", " the surface. c. As frequency is increased, it will have no effect on the electrons being ejected as the intensity is the same. d. As frequency is increased, the rate at which the electrons are being ejected will increase. 15. Why is it important to consider what material to use when designing a light meter? Consider the worked example from Section 21-2 for assistance. a. A light meter should contain material that responds only to high frequency light. b. A light meter should contain material that responds b. Because after reaching the peak frequency, the to low frequency light. photons created at a particular frequency are too few for energy intensity to continue to decrease. c. A light meter should contain material that has high binding energy. c. Because after reaching the peak frequency, the d. A light meter should contain a material that does photons created at a particular frequency are too many for energy intensity to continue to increase. d. Because after reaching the peak frequency, the photons created at a particular frequency are too few for energy intensity to continue to increase. not show any photoelectric effect. 16. Why does overexposure to UV light often result in sunburn when overexposure to visible light does not? This is why you can get burnt even on a cloudy day. a. UV light carries less energy than visible light and 12. Shortly after the introduction of photography, it was can penetrate our body. found that photographic emulsions were more sensitive to blue and violet light than they were to red light. Explain why this was the case. a. Blue-violet light contains greater amount of energy than red light. b. UV light carries more energy than visible light, so it cannot break bonds at the cellular level. c. UV light carries more energy than visible light and can break bonds at the cellular level. d. UV light carries less energy than visible light and b. Blue-violet light contains lower amount of energy cannot penetrate the human body. than red light. c. Both blue-violet light and red light have the same frequency but contain different amounts of energy. d. Blue-violet light frequency is lower than the frequency of red light. 17. If you pick up and shake a piece of metal that has electrons in it free to move as a current, no electrons fall out. Yet if you heat the metal, electrons can be boiled off. Explain both of these facts as they relate to the amount and distribution of energy involved with shaking the 714 Chapter 21 \u2022 Chapter Review object as compared with heating it. a. Thermal energy is added to the metal at a much higher rate than energy added due to shaking. b. Thermal energy is added to the metal at a much lower rate than energy added due to shaking. If the thermal energy added is below the binding energy of the electrons, they may be boiled off. If the mechanical energy added is below the binding energy of the electrons, they may be boiled off. d. c. 21.3 The Dual Nature of Light 18. In many macroscopic collisions, a significant amount of kinetic energy is converted to thermal energy. Explain why this is not a concern for Compton scattering. a. Because, photons and electrons do not exist on the molecular level, all energy of motion is considered kinetic energy. b. Because, photons exist on the molecular level while electrons do not exist on the molecular level, all energy of motion is considered kinetic energy. c. Because, electrons exist on the molecular level while photons do not exist on the molecular level, all energy of motion is considered kinetic energy. d. Because, photons and electrons exist on the molecular level, all energy of motion is considered kinetic energy. Problems 21.1 Planck and Quantum Nature of Light 22. How many X-ray photons per second are created by an X-ray tube that produces a flux of X-rays having a power of 1.00 W? Assume the average energy per photon is 75.0 keV. a. 8.33 \u00d7 1015 photons b. 9.1 \u00d7 107 photons c. 9.1 \u00d7 108 photons d. 8.33 \u00d7 1013 photons 23. What is the frequency of a photon produced in a CRT using a 25.0-kV accelerating potential? This is similar to the layout as in older color television sets. a. 6.04 \u00d7 10\u221248 Hz b. 2.77 \u00d7 10\u221248 Hz 3.02 \u00d7 1018 Hz c. d. 6.04 \u00d7 1018 Hz 21.2 Einstein and the Photoelectric Effect 24. What is the binding energy in eV of electrons in magnesium, if the longest-wavelength photon that can eject electrons is 337 nm? Access for free at openstax.org. 19. In what region of the electromagnetic spectrum will photons be most effective in accelerating a solar sail? a. ultraviolet rays b. infrared rays c. X-rays d. gamma rays 20. True or false\u2014Electron microscopes can resolve images that are smaller than the images", " resolved by light microscopes. false a. true b. 21. How would observations of Compton scattering change if ultraviolet light were used in place of X-rays? a. Ultraviolet light carries less energy than X-rays. As a result, Compton scattering would be easier to detect. b. Ultraviolet light carries less energy than X-rays. As a result, Compton scattering would be more difficult to detect. c. Ultraviolet light carries more energy than X-rays. As a result, Compton scattering would be easier to detect. d. Ultraviolet light has higher energy than X-rays. As a result, Compton scattering would be more difficult to detect. a. b. c. d. 7.44 \u00d7 10\u221219 J 7.44 \u00d7 10\u221249 J 5.90 \u00d7 10\u221217 J 5.90 \u00d7 10\u221219 J 25. Photoelectrons from a material with a binding energy of 2.71 eV are ejected by 420-nm photons. Once ejected, how long does it take these electrons to travel 2.50 cm to a detection device? a. 8.5 \u00d7 10\u22126 s 3.5 \u00d7 10\u22127 s b. c. 43.5 \u00d7 10\u22129 s d. 8.5 \u00d7 10\u22128 s 21.3 The Dual Nature of Light 26. What is the momentum of a 0.0100-nm-wavelength photon that could detect details of an atom? a. 6.626 \u00d7 10\u221227 kg \u22c5 m/s b. 6.626 \u00d7 10\u221232 kg \u22c5 m/s c. 6.626 \u00d7 10\u221234 kg \u22c5 m/s d. 6.626 \u00d7 10-23 kg \u22c5 m/s 27. The momentum of light is exactly reversed when reflected straight back from a mirror, assuming negligible recoil of the mirror. Thus the change in Chapter 21 \u2022 Test Prep 715 momentum is twice the initial photon momentum. Suppose light of intensity 1.00 kW/m2 reflectsfrom a mirror of area 2.00 m2 each second. Using the most general form of Newton\u2019s second law, what is the force on the mirror? a. b. c. d. 1.33 \u00d7 10-5 N 1.33 \u00d7 10\u22126 N 1.33 \u00d7 10\u22127 N 1.33 \u00d7 10\u22128 N Performance Task 21.3 The Dual Nature of Light 28. Our scientific understanding of light has changed over time. There is evidence to support the wave model of light, just as there is evidence to support the particle model of light. 1. Construct a demonstration that supports the wave model of light. Note\u2014One possible method is to use a piece of aluminum foil, razor blade, and laser to demonstrate wave interference. Can you arrange these materials to create an effective demonstration? In writing, explain how evidence TEST PREP Multiple Choice 21.1 Planck and Quantum Nature of Light 29. A perfect blackbody is a perfect absorber of energy transferred by what method? a. b. c. d. conduction convection induction radiation 30. Which of the following is a physical entity that is quantized? a. electric charge of an ion frequency of a sound b. speed of a car c. 31. Find the energy in joules of photons of radio waves that leave an FM station that has a 90.0-MHz broadcast frequency. a. b. c. d. 1.8 \u00d7 10\u221225 J 1.11 \u00d7 10\u221225 J 7.1 \u00d7 10\u221243 J 5.96 \u00d7 10-26 J 32. Which region of the electromagnetic spectrum will provide photons of the least energy? infrared light a. b. radio waves c. ultraviolet light d. X-rays 33. A hot, black coffee mug is sitting on a kitchen table in a dark room. Because it cannot be seen, one assumes that from your demonstration supports the wave model of light. 2. Construct a demonstration that supports the particle model of light. Note\u2014One possible method is to use a negatively charged electroscope, zinc plate, and three light sources of different frequencies. A red laser, a desk lamp, and ultraviolet lamp are typically used. Can you arrange these materials to demonstrate the photoelectric effect? In writing, explain how evidence from your demonstration supports the particle model of light. it is not emitting energy in the form of light. Explain the fallacy in this logic. a. Not all heat is in the form of light energy. b. Not all light energy falls in the visible portion of the electromagnetic spectrum. c. All heat is in the form of light energy. d. All light energy falls in the visible portion of the electromagnetic spectrum. 34. Given two stars of equivalent size, which will have a greater temperature: a red dwarf or a yellow dwarf? Explain. Note\u2014Our sun is considered a yellow dwarf. a. a yellow dwarf, because yellow light has lower frequency b. a red dwarf, because red light has", " lower frequency c. a red dwarf, because red light has higher frequency d. a yellow dwarf, because yellow light has higher frequency 21.2 Einstein and the Photoelectric Effect 35. What is a quantum of light called? a. electron b. neutron c. photon d. proton 36. Which of the following observations from the photoelectric effect is not a violation of classical physics? a. Electrons are ejected immediately after impact from light. b. Light can eject electrons from a semi-conductive 716 Chapter 21 \u2022 Test Prep material. protons from a surface. c. Light intensity does not influence the kinetic d. UV, X-rays, and gamma rays are capable of ejecting energy of ejected electrons. electrons from a surface. d. No electrons are emitted if the light frequency is too low. 21.3 The Dual Nature of Light 37. If of energy is supplied to an electron with a 41. What two particles interact in Compton scattering? binding energy of the electron be launched? a. b. c. d., with what kinetic energy will a. photon and electron b. proton and electron c. neutron and electron d. proton and neutron 38. Which of the following terms translates to light- producing voltage? a. photoelectric b. quantum mechanics c. photoconductive d. photovoltaic 39. Why is high frequency EM radiation considered more dangerous than long wavelength EM radiation? a. Long wavelength EM radiation photons carry less energy and therefore have greater ability to disrupt materials through the photoelectric effect. b. Long wavelength EM radiation photons carry more energy and therefore have greater ability to disrupt materials through the photoelectric effect. c. High frequency EM radiation photons carry less energy and therefore have lower ability to disrupt materials through the photoelectric effect. d. High frequency EM radiation photons carry more energy and therefore have greater ability to disrupt materials through the photoelectric effect. 40. Why are UV, X-rays, and gamma rays considered ionizing radiation? a. UV, X-rays, and gamma rays are capable of ejecting photons from a surface. b. UV, X-rays, and gamma rays are capable of ejecting neutrons from a surface. c. UV, X-rays, and gamma rays are capable of ejecting Short Answer 21.1 Planck and Quantum Nature of Light 42. What is the momentum of a 500-nm photon? a. 8.35 \u00d7 10\u221226 kg \u22c5 m/s 3.31 \u00d7 10\u221240 kg \u22c5 m/s b. 7.55 \u00d7 1026 kg \u22c5 m/s c. 1.33 \u00d7 10-27 kg \u22c5 m/s d. 43. The conservation of what fundamental physics principle charge is behind the technology of solar sails? a. b. mass c. momentum d. angular momentum 44. Terms like frequency, amplitude, and period are tied to what component of wave-particle duality? a. neither the particle nor the wave model of light b. both the particle and wave models of light c. d. the particle model of light the wave model of light 45. Why was it beneficial for Compton to scatter electrons using X-rays and not another region of light like microwaves? a. because X-rays are more penetrating than microwaves b. because X-rays have lower frequency than microwaves c. because microwaves have shorter wavelengths than X-rays d. because X-rays have shorter wavelength than microwaves elliptical path. c. The blackbody radiation curve would look like a vertical line. 46. Scientists once assumed that all frequencies of light d. The blackbody radiation curve would look like a were emitted with equal probability. Explain what the blackbody radiation curve would look like if this were the case. a. The blackbody radiation curve would look like a circular path. b. The blackbody radiation curve would look like an horizontal line. 47. Because there are more gradations to high frequency radiation than low frequency radiation, scientists also thought it possible that a curve titled the ultraviolet catastrophewould occur. Explain what the blackbody radiation curve would look like if this were the case. Access for free at openstax.org. a. The curve would steadily increase in intensity with increasing frequency. b. The curve would steadily decrease in intensity with increasing frequency. c. The curve would be much steeper than in the blackbody radiation graph. d. The curve would be much flatter than in the blackbody radiation graph. 48. Energy provided by a light exists in the following quantities: 150 J, 225 J, 300 J. Define one possible quantum of energy and provide an energy state that cannot exist with this quantum. a. 65 J; 450 J cannot exist 70 J; 450 J cannot exist b. 75 J; 375 J cannot exist c. 75 J; 100 J cannot exist d. 49. Why is Planck\u2019s recognition of quantum particles considered the dividing line between classical and modern physics? a", ". Planck recognized that energy is quantized, which was in sync with the classical physics concepts but not in agreement with modern physics concepts. b. Planck recognized that energy is quantized, which was in sync with modern physics concepts but not in agreement with classical physics concepts. Chapter 21 \u2022 Test Prep 717 a. b. c. d. radio, microwave, infrared, visible, ultraviolet, Xray, gamma radio, infrared, microwave, ultraviolet, visible, Xray, gamma radio, visible, microwave, infrared, ultraviolet, Xray, gamma radio, microwave, infrared, visible, ultraviolet, gamma, X-ray 53. Why are photons of gamma rays and X-rays able to penetrate objects more successfully than ultraviolet radiation? a. Photons of gamma rays and X-rays carry with them less energy. b. Photons of gamma rays and X-rays have longer wavelengths. c. Photons of gamma rays and X-rays have lower frequencies. d. Photons of gamma rays and X-rays carry with them more energy. 21.2 Einstein and the Photoelectric Effect 54. According to wave theory, what is necessary to eject electrons from a surface? a. Enough energy to overcome the binding energy of the electrons at the surface c. Prior to Planck\u2019s hypothesis, all the classical b. A frequency that is higher than that of the electrons physics calculations were valid for subatomic particles, but quantum physics calculations were not valid. d. Prior to Planck\u2019s hypothesis, all the classical physics calculations were not valid for macroscopic particles, but quantum physics calculations were valid. 50. How many 500-mm microwave photons are needed to supply the 8 kJ of energy necessary to heat a cup of water by 10 degrees Celsius? a. 8.05 \u00d7 1028 photons b. 8.05 \u00d7 1026 photons c. 2.01 \u00d7 1026 photons d. 2.01 \u00d7 1028 photons 51. What is the efficiency of a 100-W, 550-nm lightbulb if a photometer finds that 1 \u00d7 1020 photons are emitted each second? a. b. c. d. 101 percent 72 percent 18 percent 36 percent 52. Rank the following regions of the electromagnetic spectrum by the amount of energy provided per photon: gamma, infrared, microwave, ultraviolet, radio, visible, X-ray. at the surface c. Energy that is lower than the binding energy of the electrons at the surface d. A very small number of photons 55. What is the wavelength of EM radiation that ejects 2.00-eV electrons from calcium metal, given that the binding energy is 2.71 eV? 16.1 \u00d7 105 m a. b. 6.21 \u00d7 10\u22125 m c. 9.94 \u00d7 10\u221226 m d. 2.63 \u00d7 10-7 m 56. Find the wavelength of photons that eject. electrons from potassium, given that the binding energy is a. b. c. d. 57. How do solar cells utilize the photoelectric effect? a. A solar cell converts all photons that it absorbs to electrical energy using the photoelectric effect. b. A solar cell converts all electrons that it absorbs to electrical energy using the photoelectric effect. c. A solar cell absorbs the photons with energy less 718 Chapter 21 \u2022 Test Prep than the energy gap of the material of the solar cell and converts it to electrical energy using the photoelectric effect. d. A solar cell absorbs the photons with energy greater than the energy gap of the material of the solar cell and converts it to electrical energy using the photoelectric effect. 58. Explain the advantages of the photoelectric effect to other forms of energy transformation. a. The photoelectric effect is able to work on the Sun\u2019s natural energy. b. The photoelectric effect is able to work on energy generated by burning fossil fuels. c. The photoelectric effect can convert heat energy into electrical energy. d. The photoelectric effect can convert electrical energy into light energy. 21.3 The Dual Nature of Light 5.18 \u00d7 105 m/s c. d. 4.18 \u00d7 105 m/s 63. When a photon strikes a solar sail, what is the direction of impulse on the photon? a. parallel to the sail b. perpendicular to the sail c. tangential to the sail d. opposite to the sail 64. What is a fundamental difference between solar sails and sails that are used on sailboats? a. Solar sails rely on disorganized strikes from light particles, while sailboats rely on disorganized strikes from air particles. b. Solar sails rely on disorganized strikes from air particles, while sailboats rely on disorganized strikes from light particles. c. Solar sails rely on organized strikes from air particles, while sailboats rely on organized strikes from light particles. 59. Upon collision, what happens to the frequency of a d. Solar sails rely on organized strikes from light photon? a. The frequency of the photon will drop to zero. b. The", " frequency of the photon will remain the same. c. The frequency of the photon will increase. d. The frequency of the photon will decrease. particles, while sailboats rely on organized strikes from air particles. 65. The wavelength of a particle is called the de Broglie wavelength, and it can be found with the equation. 60. How does the momentum of a photon compare to the momentum of an electron of identical energy? a. Momentum of the photon is greater than the momentum of an electron. b. Momentum of the photon is less than the momentum of an electron. c. Momentum of the photon is equal to the momentum of an electron. d. Momentum of the photon is zero due to zero rest mass but the momentum of an electron is finite. 61. A 500-nm photon strikes an electron and loses 20 percent of its energy. What is the new momentum of the photon? a. 4.24 \u00d7 10\u221227 kg \u22c5 m/s 3.18 \u00d7 10\u221227 kg \u22c5 m/s b. c. 2.12 \u00d7 10\u221227 kg \u22c5 m/s 1.06 \u00d7 10\u221227 kg \u22c5 m/s d. 62. A 500-nm photon strikes an electron and loses 20 percent of its energy. What is the speed of the recoiling electron? 7.18 \u00d7 105 m/s a. b. 6.18 \u00d7 105 m/s Yes or no\u2014Can the wavelength of an electron match that of a proton? a. Yes, a slow-moving electron can achieve the same momentum as a slow-moving proton. b. No, a fast-moving electron cannot achieve the same momentum, and hence the same wavelength, as a proton. c. No, an electron can achieve the same momentum, and hence not the same wavelength, as a proton. d. Yes, a fast-moving electron can achieve the same momentum, and hence have the same wavelength, as a slow-moving proton. 66. Large objects can move with great momentum. Why then is it difficult to see their wave-like nature? a. Their wavelength is equal to the object\u2019s size. b. Their wavelength is very small compared to the object\u2019s size. c. Their wavelength is very large compared to the object\u2019s size. d. Their frequency is very small compared to the object\u2019s size. Access for free at openstax.org. Extended Response 21.1 Planck and Quantum Nature of Light 67. Some television tubes are CRTs. They use an approximately 30-kV accelerating potential to send electrons to the screen, where the electrons stimulate phosphors to emit the light that forms the pictures we watch. Would you expect X-rays also to be created? Explain. a. No, because the full spectrum of EM radiation is not emitted at any temperature. b. No, because the full spectrum of EM radiation is not emitted at certain temperatures. c. Yes, because the full spectrum of EM radiation is emitted at any temperature. d. Yes, because the full spectrum of EM radiation is emitted at certain temperatures. 68. If Planck\u2019s constant were large, say times greater than it is, we would observe macroscopic entities to be quantized. Describe the motion of a child\u2019s swing under such circumstances. a. The child would not be able to swing with particular energies. b. The child could be released from any height. c. The child would be able to swing with constant velocity. d. The child could be released only from particular heights. 69. What is the accelerating voltage of an X-ray tube that produces X-rays with the shortest wavelength of 0.0103 nm? 1.21 \u00d7 1010 V a. b. 2.4 \u00d7 105 V c. d. 3.0 \u00d7 10\u221233 V 1.21 \u00d7 105 V 70. Patients in a doctor\u2019s office are rightly concerned about receiving a chest X-ray. Yet visible light is also a form of electromagnetic radiation and they show little concern about sitting under the bright lights of the waiting room. Explain this discrepancy. a. X-ray photons carry considerably more energy so they can harm the patients. b. X-ray photons carry considerably less energy so they can harm the patients. c. X-ray photons have considerably longer wavelengths so they cannot harm the patients. d. X-ray photons have considerably lower frequencies so they can harm the patients. 21.2 Einstein and the Photoelectric Effect 71. When increasing the intensity of light shining on a Chapter 21 \u2022 Test Prep 719 metallic surface, it is possible to increase the current created on that surface. Classical theorists would argue that this is evidence that intensity causes charge to move with a greater kinetic energy. Argue this logic from the perspective of a modern physicist. a. The increased intensity increases the number of ejected electrons. The increased current is due to the", " increase in the number of electrons. b. The increased intensity decreases the number of ejected electrons. The increased current is due to the decrease in the number of electrons ejected. c. The increased intensity does not alter the number of electrons ejected. The increased current is due to the increase in the kinetic energy of electrons. d. The increased intensity alters the number of electrons ejected, but an increase in the current is due to an increase in the kinetic energy of electrons. 72. What impact does the quantum nature of electromagnetic radiation have on the understanding of speed at the particle scale? a. Speed must also be quantized at the particle scale. b. Speed will not be quantized at the particle scale. c. Speed must be zero at the particle scale. d. Speed will be infinite at the particle scale. 73. A 500 nm photon of light strikes a semi-conductive surface with a binding energy of 2 eV. With what velocity will an electron be emitted from the semi-conductive surface? a. 8.38 \u00d7 105 m/s b. 9.33 \u00d7 105 m/s 3 \u00d7 108 m/s c. d. 4.11 \u00d7 105 m/s 74. True or false\u2014Treating food with ionizing radiation helps keep it from spoiling. a. b. true false 21.3 The Dual Nature of Light 75. When testing atomic bombs, scientists at Los Alamos recognized that huge releases of energy resulted in problems with power and communications systems in the area surrounding the blast site. Explain the possible tie to Compton scattering. a. The release of light energy caused large-scale emission of electrons. b. The release of light energy caused large-scale emission of protons. c. The release of light energy caused large-scale emission of neutrons. d. The release of light energy caused large-scale 720 Chapter 21 \u2022 Test Prep emission of photons. 76. Sunlight above the Earth\u2019s atmosphere has an intensity of 1.30 kW/m2. If this is reflected straight back from a mirror that has only a small recoil, the light\u2019s momentum is exactly reversed, giving the mirror twice the incident momentum. If the mirror were attached to a solar sail craft, how fast would the craft be moving after 24 hr? Note\u2014The average mass per square meter of the craft is 0.100 kg. a. 8.67 \u00d7 10\u22125 m/s2 b. 8.67 \u00d7 10\u22126 m/s2 c. 94.2 m/s 7.49 m/s d. 77. Consider the counter-clockwise motion of LightSail-1 around Earth. When will the satellite move the fastest? a. point A b. point B c. point C d. point D 78. What will happen to the interference pattern created by electrons when their velocities are increased? a. There will be more zones of constructive interference and fewer zones of destructive interference. b. There will be more zones of destructive interference and fewer zones of constructive interference. c. There will be more zones of constructive and destructive interference. d. There will be fewer zones of constructive and destructive interference. Access for free at openstax.org. CHAPTER 22 The Atom Figure 22.1 Individual carbon atoms are visible in this image of a carbon nanotube made by a scanning tunneling electron microscope. (credit: Taner Yildirim, National Institute of Standards and Technology, Wikimedia Commons) Chapter Outline 22.1 The Structure of the Atom 22.2 Nuclear Forces and Radioactivity 22.3 Half Life and Radiometric Dating 22.4 Nuclear Fission and Fusion 22.5 Medical Applications of Radioactivity: Diagnostic Imaging and Radiation From childhood on, we learn that atoms are a substructure of all things around us, from the air we breathe to INTRODUCTION the autumn leaves that blanket a forest trail. Invisible to the eye, the atoms have properties that are used to explain many phenomena\u2014a theme found throughout this text. In this chapter, we discuss the discovery of atoms and their own substructures. We will then learn about the forces that keep them together and the tremendous energy they release when we break them apart. Finally, we will see how the knowledge and manipulation of atoms allows us to better understand geology, biology, and the world around us. 22.1 The Structure of the Atom Section Learning Objectives By the end of this section, you will be able to do the following: \u2022 Describe Rutherford\u2019s experiment and his model of the atom \u2022 Describe emission and absorption spectra of atoms \u2022 Describe the Bohr model of the atom \u2022 Calculate the energy of electrons when they change energy levels \u2022 Calculate the frequency and wavelength of emitted photons when electrons change energy levels \u2022 Describe the quantum model of the atom 722 Chapter 22 \u2022 The Atom Section Key Terms energy-level diagram excited state Fraunhofer lines ground state Heisenberg Uncertainty Principle hydrogen-like atoms planetary model of the", " atom Rutherford scattering Rydberg constant How do we know that atoms are really there if we cannot see them with our own eyes? While often taken for granted, our knowledge of the existence and structure of atoms is the result of centuries of contemplation and experimentation. The earliest known speculation on the atom dates back to the fifth century B.C., when Greek philosophers Leucippus and Democritus contemplated whether a substance could be divided without limit into ever smaller pieces. Since then, scientists such as John Dalton (1766\u20131844), Amadeo Avogadro (1776\u20131856), and Dmitri Mendeleev (1834\u20131907) helped to discover the properties of that fundamental structure of matter. While much could be written about any number of important scientific philosophers, this section will focus on the role played by Ernest Rutherford (1871\u20131937). Though his understanding of our most elemental matter is rooted in the success of countless prior investigations, his surprising discovery about the interior of the atom is most fundamental in explaining so many well-known phenomena. Rutherford\u2019s Experiment In the early 1900\u2019s, the plum puddingmodel was the accepted model of the atom. Proposed in 1904 by J. J. Thomson, the model suggested that the atom was a spherical ball of positive charge, with negatively charged electrons scattered evenly throughout. In that model, the positive charges made up the pudding, while the electrons acted as isolated plums. During its short life, the model could be used to explain why most particles were neutral, although with an unbalanced number of plums, electrically charged atoms could exist. When Ernest Rutherford began his gold foil experiment in 1909, it is unlikely that anyone would have expected that the plum pudding model would be challenged. However, using a radioactive source, a thin sheet of gold foil, and a phosphorescent screen, Rutherford would uncover something so great that he would later call it \u201cthe most incredible event that has ever happened to me in my life\u201d[James, L. K. (1993). Nobel Laureates in Chemistry, 1901\u20131992. Washington, DC: American Chemical Society.] The experiment that Rutherford designed is shown in Figure 22.2. As you can see in, a radioactive source was placed in a lead container with a hole in one side to produce a beam of positively charged helium particles, called alpha particles. Then, a thin gold foil sheet was placed in the beam. When the high-energy alpha particles passed through the gold foil, they were scattered. The scattering was observed from the bright spots they produced when they struck the phosphor screen. Figure 22.2 Rutherford\u2019s experiment gave direct evidence for the size and mass of the nucleus by scattering alpha particles from a thin gold foil. The scattering of particles suggests that the gold nuclei are very small and contain nearly all of the gold atom\u2019s mass. Particularly significant in showing the size of the nucleus are alpha particles that scatter to very large angles, much like a soccer ball bouncing off a goalie\u2019s head. The expectation of the plum pudding model was that the high-energy alpha particles would be scattered only slightly by the presence of the gold sheet. Because the energy of the alpha particles was much higher than those typically associated with atoms, the alpha particles should have passed through the thin foil much like a supersonic bowling ball would crash through a Access for free at openstax.org. 22.1 \u2022 The Structure of the Atom 723 few dozen rows of bowling pins. Any deflection was expected to be minor, and due primarily to the electrostatic Coulomb force between the alpha particles and the foil\u2019s interior electric charges. However, the true result was nothing of the sort. While the majority of alpha particles passed through the foil unobstructed, Rutherford and his collaborators Hans Geiger and Ernest Marsden found that alpha particles occasionally were scattered to large angles, and some even came back in the direction from which they came! The result, called Rutherford scattering, implied that the gold nuclei were actually very small when compared with the size of the gold atom. As shown in Figure 22.3, the dense nucleus is surrounded by mostly empty space of the atom, an idea verified by the fact that only 1 in 8,000 particles was scattered backward. Figure 22.3 An expanded view of the atoms in the gold foil in Rutherford\u2019s experiment. Circles represent the atoms that are about 10\u221210 m in diameter, while the dots represent the nuclei that are about 10\u221215 m in diameter. To be visible, the dots are much larger than scale\u2014if the nuclei were actually the size of the dots, each atom would have a diameter of about five meters! Most alpha particles crash through but are relatively unaffected because of their high energy and the electron\u2019s small mass. Some, however, strike a nucleus and are scattered straight back. A detailed analysis of their interaction gives the size and mass of the nucleus. Although the results of the experiment were published", " by his colleagues in 1909, it took Rutherford two years to convince himself of their meaning. Rutherford later wrote: \u201cIt was almost as incredible as if you fired a 15-inch shell at a piece of tissue paper and it came back and hit you. On consideration, I realized that this scattering backwards... [meant]... the greatest part of the mass of the atom was concentrated in a tiny nucleus.\u201d In 1911, Rutherford published his analysis together with a proposed model of the atom, which was in part based on Geiger\u2019s work from the previous year. As a result of the paper, the size of the nucleus was determined to be about cm3, much greater than any macroscopic matter. m, or 100,000 times smaller than the atom. That implies a huge density, on the order of g/ Based on the size and mass of the nucleus revealed by his experiment, as well as the mass of electrons, Rutherford proposed the planetary model of the atom. The planetary model of the atom pictures low-mass electrons orbiting a large-mass nucleus. The sizes of the electron orbits are large compared with the size of the nucleus, and most of the atom is a vacuum. The model is analogous to how low-mass planets in our solar system orbit the large-mass Sun. In the atom, the attractive Coulomb force is analogous to gravitation in the planetary system (see Figure 22.4). Figure 22.4 Rutherford\u2019s planetary model of the atom incorporates the characteristics of the nucleus, electrons, and the size of the atom. The model was the first to recognize the structure of atoms, in which low-mass electrons orbit a very small, massive nucleus in orbits much larger than the nucleus. The atom is mostly empty and is analogous to our planetary system. 724 Chapter 22 \u2022 The Atom Virtual Physics Rutherford Scattering Click to view content (https://www.openstax.org/l/28rutherford) How did Rutherford figure out the structure of the atom without being able to see it? Explore the answer through this simulation of the famous experiment in which he disproved the plum pudding model by observing alpha particles bouncing off atoms and determining that they must have a small core. TIPS FOR SUCCESS As you progress through the model of the atom, consider the effect that experimentation has on the scientific process. Ask yourself the following: What would our model of the atom be without Rutherford\u2019s gold foil experiment? What further understanding of the atom would not have been gained? How would that affect our current technologies? Though often confusing, experiments taking place today to further understand composition of the atom could perhaps have a similar effect. Absorption and Emission Spectra In 1900, Max Planck recognized that all energy radiated from a source is emitted by atoms in quantum states. How would that radical idea relate to the interior of an atom? The answer was first found by investigating the spectrum of light or emission spectrum produced when a gas is highly energized. Figure 22.5 shows how to isolate the emission spectrum of one such gas. The gas is placed in the discharge tube at the left, where it is energized to the point at which it begins to radiate energy or emit light. The radiated light is channeled by a thin slit and then passed through a diffraction grating, which will separate the light into its constituent wavelengths. The separated light will then strike the photographic film on the right. The line spectrum shown in part (b) of Figure 22.5 is the output shown on the film for excited iron. Note that this spectrum is not continuous but discrete. In other words, only particular wavelengths are emitted by the iron source. Why would that be the case? Figure 22.5 Part (a) shows, from left to right, a discharge tube, slit, and diffraction grating producing a line spectrum. Part (b) shows the emission spectrum for iron. The discrete lines imply quantized energy states for the atoms that produce them. The line spectrum for each element is unique, providing a powerful and much-used analytical tool, and many line spectra were well known for many years before they could be explained with physics. (credit:(b) Yttrium91, Wikimedia Commons) The spectrum of light created by excited iron shows a variety of discrete wavelengths emitted within the visible spectrum. Each element, when excited to the appropriate degree, will create a discrete emission spectrum as in part (b) of Figure 22.5. However, the wavelengths emitted will vary from element to element. The emission spectrum for iron was chosen for Figure 22.5 solely Access for free at openstax.org. because a substantial portion of its emission spectrum is within the visible spectrum. Figure 22.6 shows the emission spectrum for hydrogen. Note that, while discrete, a large portion of hydrogen emission takes place in the ultraviolet and infrared regions. 22.1 \u2022 The Structure of the Atom 725 Figure 22.6 A schematic of the hydrogen spectrum shows several series named for those who", " contributed most to their determination. Part of the Balmer series is in the visible spectrum, while the Lyman series is entirely in the ultraviolet, and the Paschen series and others are in the infrared. Values of nf and ni are shown for some of the lines. Their importance will be described shortly. Just as an emission spectrum shows all discrete wavelengths emitted by a gas, an absorption spectrum will show all light that is absorbed by a gas. Black lines exist where the wavelengths are absorbed, with the remainder of the spectrum lit by light is free to pass through. What relationship do you think exists between the black lines of a gas\u2019s absorption spectrum and the colored lines of its emission spectrum? Figure 22.7 shows the absorption spectrum of the Sun. The black lines are called Fraunhofer lines, and they correspond to the wavelengths absorbed by gases in the Sun\u2019s exterior. Figure 22.7 The absorption spectrum of the Sun. The black lines appear at wavelengths absorbed by the Sun\u2019s gas exterior. The energetic photons emitted from the Sun\u2019s interior are absorbed by gas in its exterior and reemitted in directions away from the observer. That results in dark lines within the absorption spectrum. The lines are called Fraunhofer lines, in honor of the German physicist who discovered them. Lines similar to those are used to determine the chemical composition of stars well outside our solar system. Bohr\u2019s Explanation of the Hydrogen Spectrum To tie the unique signatures of emission spectra to the composition of the atom itself would require clever thinking. Niels Bohr (1885\u20131962), a Danish physicist, did just that, by making immediate use of Rutherford\u2019s planetary model of the atom. Bohr, shown in Figure 22.8, became convinced of its validity and spent part of 1912 at Rutherford\u2019s laboratory. In 1913, after returning to Copenhagen, he began publishing his theory of the simplest atom, hydrogen, based on Rutherford\u2019s planetary model. Figure 22.8 Niels Bohr, Danish physicist, used the planetary model of the atom to explain the atomic spectrum and size of the hydrogen 726 Chapter 22 \u2022 The Atom atom. His many contributions to the development of atomic physics and quantum mechanics, his personal influence on many students and colleagues, and his personal integrity, especially in the face of Nazi oppression, earned him a prominent place in history. (credit: Unknown Author, Wikimedia Commons) Bohr was able to derive the formula for the hydrogen spectrum using basic physics, the planetary model of the atom, and some very important new conjectures. His first conjecture was that only certain orbits are allowed: In other words, in an atom, the orbits of electrons are quantized. Each quantized orbit has a different distinct energy, and electrons can move to a higher orbit by absorbing energy or drop to a lower orbit by emitting energy. Because of the quantized orbits, the amount of energy emitted or absorbed must also be quantized, producing the discrete spectra seen in Figure 22.5 and Figure 22.7. In equation form, the amount of energy absorbed or emitted can be found as refers to the energy of the initial quantized orbit, and where wavelength emitted can be found using the equation refers to the energy of the final orbits. Furthermore, the and relating the wavelength to the frequency found using the equation, where vcorresponds to the speed of light. It makes sense that energy is involved in changing orbits. For example, a burst of energy is required for a satellite to climb to a higher orbit. What is not expected is that atomic orbits should be quantized. Quantization is not observed for satellites or planets, which can have any orbit, given the proper energy (see Figure 22.9). 22.2 22.1 Figure 22.9 The planetary model of the atom, as modified by Bohr, has the orbits of the electrons quantized. Only certain orbits are allowed, explaining why atomic spectra are discrete or quantized. The energy carried away from an atom by a photon comes from the electron dropping from one allowed orbit to another and is thus quantized. The same is true for atomic absorption of photons. Figure 22.10 shows an energy-level diagram, a convenient way to display energy states. Each of the horizontal lines corresponds to the energy of an electron in a different orbital. Energy is plotted vertically with the lowest or ground state at the bottom and with excited states above. The vertical arrow downwards shows energy being emitted out of the atom due to an electron dropping from one excited state to another. That would correspond to a line shown on the atom\u2019s emission spectrum. The Lyman series shown in Figure 22.6 results from electrons dropping to the ground state, while the Balmer and Paschen series result to electrons dropping to the n = 2and n = 3states, respectively. Access for free at openstax.org. 22.1 \u2022 The Structure of the Atom 727 Figure 22.10 An energy-", "level diagram plots energy vertically and is useful in visualizing the energy states of a system and the transitions between them. This diagram is for the hydrogen-atom electrons, showing a transition between two orbits having energies and. The energy transition results in a Balmer series line in an emission spectrum. Energy and Wavelength of Emitted Hydrogen Spectra The energy associated with a particular orbital of a hydrogen atom can be found using the equation 22.3 where ncorresponds to the orbital value from the atom\u2019s nucleus. The negative value in the equation is based upon a baseline energy of zero when the electron is infinitely far from the atom. As a result, the negative value shows that energy is necessary to free the electron from its orbital state. The minimum energy to free the electron is also referred to as its binding energy. The equation is only valid for atoms with single electrons in their orbital shells (like hydrogen). For ionized atoms similar to hydrogen, the following formula may be used. 22.4 corresponds to \u201313.6 eV, as mentioned earlier. Additionally, Please note that studied. The atomic number is the number of protons in the nucleus\u2014it is different for each element. The above equation is derived from some basic physics principles, namely conservation of energy, conservation of angular momentum, Coulomb\u2019s law, and centripetal force. There are three derivations that result in the orbital energy equations, and they are shown below. While you can use the energy equations without understanding the derivations, they will help to remind you of just how valuable those fundamental concepts are. refers to the atomic number of the element Derivation 1 (Finding the Radius of an Orbital) One primary difference between the planetary model of the solar system and the planetary model of the atom is the cause of the circular motion. While gravitation causes the motion of orbiting planets around an interior star, the Coulomb force is responsible for the circular shape of the electron\u2019s orbit. The magnitude of the centripetal force is, while the magnitude of the Coulomb force is. The assumption here is that the nucleus is more massive than the stationary electron, and the electron orbits about it. That is consistent with the planetary model of the atom. Equating the Coulomb force and the centripetal force, 22.5 which yields 728 Chapter 22 \u2022 The Atom Derivation 2 (Finding the Velocity of the Orbiting Electron) Bohr was clever enough to find a way to calculate the electron orbital energies in hydrogen. That was an important first step that has been improved upon, but it is well worth repeating here, because it does correctly describe many characteristics of hydrogen. Assuming circular orbits, Bohr proposed that the angular momentum Lof an electron in its orbit is also quantized, that is, it has only specific, discrete values. The value for Lis given by the formula 22.7 22.6 where Lis the angular momentum, meis the electron\u2019s mass, rnis the radius of the nth orbit, and his Planck\u2019s constant. Note that angular momentum is. For a small object at a radius r,, so that, and Bohr himself did not know why angular momentum should be quantized, but by using that assumption, he was able to calculate the energies in the hydrogen spectrum, something no one else had done at the time. Quantization says that the value of mvrcan only be equal to h/ 2, 2h/ 2, 3h/ 2, etc. At the time, Derivation 3 (Finding the Energy of the Orbiting Electron) To get the electron orbital energies, we start by noting that the electron energy is the sum of its kinetic and potential energy. 22.8 Kinetic energy is the familiar electron is electrical, or, assuming the electron is not moving at a relativistic speed. Potential energy for the, where Vis the potential due to the nucleus, which looks like a point charge. The nucleus has a positive charge ; thus,, recalling an earlier equation for the potential due to a point charge from the chapter on Electricity and Magnetism. Since the electron\u2019s charge is negative, we see that Substituting the expressions for KE and PE, Now we solve for rnand vusing the equation for angular momentum, giving and Substituting the expression for rnand vinto the above expressions for energy (KE and PE), and performing algebraic manipulation, yields 22.9 22.10 22.11 22.12 for the orbital energies of hydrogen-like atoms. Here, Eois the ground-state energy (n= 1) for hydrogen (Z= 1) and is given by Thus, for hydrogen, The relationship between orbital energies and orbital states for the hydrogen atom can be seen in Figure 22.11. 22.13 22.14 Access for free at openstax.org. 22.1 \u2022 The Structure of the Atom 729 Figure 22.11 Energy-level diagram for hydrogen showing the Lyman", ", Balmer, and Paschen series of transitions. The orbital energies are calculated using the above equation, first derived by Bohr. WORKED EXAMPLE A hydrogen atom is struck by a photon. How much energy must be absorbed from the photon to raise the electron of the hydrogen atom from its ground state to its second orbital? Strategy The hydrogen atom has an atomic number of Z= 1. Raising the electron from the ground state to its second orbital will increase its orbital level from n= 1 to n= 2. The energy determined will be measured in electron-volts. Solution The amount of energy necessary to cause the change in electron state is the difference between the final and initial energies of the electron. The final energy state of the electron can be found using Knowing the nand Zvalues for the hydrogen atom, and knowing that Eo= \u201313.6 eV, the result is The original amount of energy associated with the electron is equivalent to the ground state orbital, or 22.15 22.16 22.17 730 Chapter 22 \u2022 The Atom The amount of energy necessary to change the orbital state of the electron can be found by determining the electron\u2019s change in energy. 22.18 Discussion The energy required to change the orbital state of the electron is positive. That means that for the electron to move to a state with greater energy, energy must be added to the atom. Should the electron drop back down to its original energy state, a change of \u201310.2 eV would take place, and 10.2 eV of energy would be emitted from the atom. Just as only quantum amounts of energy may be absorbed by the atom, only quantum amounts of energy can be emitted from the atom. That helps to explain many of the quantum light effects that you have learned about previously. WORKED EXAMPLE Characteristic X-Ray Energy Calculate the approximate energy of an X-ray emitted for an n= 2 to n= 1 transition in a tungsten anode in an X-ray tube. Strategy How do we calculate energies in a multiple-electron atom? In the case of characteristic X-rays, the following approximate calculation is reasonable. Characteristic X-rays are produced when an inner-shell vacancy is filled. Inner-shell electrons are nearer the nucleus than others in an atom and thus feel little net effect from the others. That is similar to what happens inside a charged conductor, where its excess charge is distributed over the surface so that it produces no electric field inside. It is reasonable to assume the inner-shell electrons have hydrogen-like energies, as given by For tungsten, Z= 74, so that the effective charge is 73. Solution The amount of energy given off as an X-ray is found using where and Thus, 22.19 22.20 22.21 22.22 22.23 Discussion This large photon energy is typical of characteristic X-rays from heavy elements. It is large compared with other atomic emissions because it is produced when an inner-shell vacancy is filled, and inner-shell electrons are tightly bound. Characteristic X-ray energies become progressively larger for heavier elements because their energy increases approximately as Z2. Significant accelerating voltage is needed to create such inner-shell vacancies, because other shells are filled and you cannot simply bump one electron to a higher filled shell. You must remove it from the atom completely. In the case of tungsten, at least 72.5 kV is needed. Tungsten is a common anode material in X-ray tubes; so much of the energy of the impinging electrons is absorbed, raising its temperature, that a high-melting-point material like tungsten is required. The wavelength of light emitted by an atom can also be determined through basic derivations. Let us consider the energy of a photon emitted from a hydrogen atom in a downward transition, given by the equation Access for free at openstax.org. Substituting, we get Dividing both sides of the equation by hcgives us an expression for, It can be shown that where Ris the Rydberg constant. Simplified, the formula for determining emitted wavelength can now be written as 22.1 \u2022 The Structure of the Atom 731 22.24 22.25 22.26 22.27 22.28 WORKED EXAMPLE What wavelength of light is emitted by an electron dropping from the third orbital to the ground state of a hydrogen atom? Strategy The ground state of a hydrogen atom is considered the first orbital of the atom. As a result, nf= 1 and ni= 3. The Rydberg constant has already been determined and will be constant regardless of atom chosen. Solution For the equation above, calculate wavelength based on the known energy states. Rearranging the equation for wavelength yields 22.29 22.30 22.31 Discussion This wavelength corresponds to light in the ultraviolet spectrum. As a result, we would not be able to see the photon of light emitted when an electron drops", " from its third to first energy state. However, it is worth noting that by supplying light of wavelength precisely 102.6 nm, we can cause the electron in hydrogen to move from its first to its third orbital state. Limits of Bohr\u2019s Theory and the Quantum Model of the Atom There are limits to Bohr\u2019s theory. It does not account for the interaction of bound electrons, so it cannot be fully applied to multielectron atoms, even one as simple as the two-electron helium atom. Bohr\u2019s model is what we call semiclassical. The orbits are quantized (nonclassical) but are assumed to be simple circular paths (classical). As quantum mechanics was developed, it became clear that there are no well-defined orbits; rather, there are clouds of probability. Additionally, Bohr\u2019s theory did not explain that some spectral lines are doublets or split into two when examined closely. While we shall examine a few of those aspects of quantum mechanics in more detail, it should be kept in mind that Bohr did not fail. Rather, he made very important steps along the path to greater knowledge and laid the foundation for all of atomic physics that has since evolved. 732 Chapter 22 \u2022 The Atom DeBroglie\u2019s Waves Following Bohr\u2019s initial work on the hydrogen atom, a decade was to pass before Louis de Broglie proposed that matter has wave properties. The wave-like properties of matter were subsequently confirmed by observations of electron interference when scattered from crystals. Electrons can exist only in locations where they interfere constructively. How does that affect electrons in atomic orbits? When an electron is bound to an atom, its wavelength must fit into a small space, something like a standing wave on a string (see Figure 22.12). Orbits in which an electron can constructively interfere with itself are allowed. All orbits in which constructive interference cannot occur are not able to exist. Thus, only certain orbits are allowed. The wave nature of an electron, according to de Broglie, is why the orbits are quantized! Figure 22.12 (a) Standing waves on a string have a wavelength related to the length of the string, allowing them to interfere constructively. (b) If we imagine the string formed into a closed circle, we get a rough idea of how electrons in circular orbits can interfere constructively. (c) If the wavelength does not fit into the circumference, the electron interferes destructively; it cannot exist in such an orbit. For a circular orbit, constructive interference occurs when the electron\u2019s wavelength fits neatly into the circumference, so that wave crests always align with crests and wave troughs align with troughs, as shown in Figure 22.12(b). More precisely, when an integral multiple of the electron\u2019s wavelength equals the circumference of the orbit, constructive interference is obtained. In equation form, the condition for constructive interference and an allowed electron orbit is where of a hydrogen atom. is the electron\u2019s wavelength and rnis the radius of that circular orbit. Figure 22.13 shows the third and fourth orbitals 22.32 Figure 22.13 The third and fourth allowed circular orbits have three and four wavelengths, respectively, in their circumferences. Access for free at openstax.org. 22.1 \u2022 The Structure of the Atom 733 Heisenberg Uncertainty How does determining the location of an electron change its trajectory? The answer is fundamentally important\u2014measurement affects the system being observed. It is impossible to measure a physical quantity exactly, and greater precision in measuring one quantity produces less precision in measuring a related quantity. It was Werner Heisenberg who first stated that limit to knowledge in 1929 as a result of his work on quantum mechanics and the wave characteristics of all particles (see Figure 22.14). Figure 22.14 Werner Heisenberg was the physicist who developed the first version of true quantum mechanics. Not only did his work give a description of nature on the very small scale, it also changed our view of the availability of knowledge. Although he is universally recognized for the importance of his work by receiving the Nobel Prize in 1932, for example, Heisenberg remained in Germany during World War II and headed the German effort to build a nuclear bomb, permanently alienating himself from most of the scientific community. (credit: Unknown Author, Wikimedia Commons) For example, you can measure the position of a moving electron by scattering light or other electrons from it. However, by doing so, you are giving the electron energy, and therefore imparting momentum to it. As a result, the momentum of the electron is affected and cannot be determined precisely. This change in momentum could be anywhere from close to zero up to the relative momentum of the electron ( particle. ). Note that, in this case, the particle is an electron, but the principle applies to any Viewing the electron through the model of wave-particle duality, Heisenberg recognized that", ", because a wave is not located at one fixed point in space, there is an uncertainty associated with any electron\u2019s position. That uncertainty in position, approximately equal to the wavelength of the particle. That is, momentum. The uncertainty in an electron\u2019s position can be reduced by using a shorter-wavelength electron, since But shortening the wavelength increases the uncertainty in momentum, since momentum can be reduced by using a longer-wavelength electron, but that increases the uncertainty in position. Mathematically, you can express the trade-off by multiplying the uncertainties. The wavelength cancels, leaving. There is an interesting trade-off between position and. Conversely, the uncertainty in,is. Therefore, if one uncertainty is reduced, the other must increase so that their product is mathematics, Heisenberg showed that the best that can be done in a simultaneous measurement of position and momentum is.With the use of advanced 22.33 That relationship is known as the Heisenberg uncertainty principle. The Quantum Model of the Atom Because of the wave characteristic of matter, the idea of well-defined orbits gives way to a model in which there is a cloud of probability, consistent with Heisenberg\u2019s uncertainty principle. Figure 22.15 shows how the principle applies to the ground state of hydrogen. If you try to follow the electron in some well-defined orbit using a probe that has a wavelength small enough to 734 Chapter 22 \u2022 The Atom measure position accurately, you will instead knock the electron out of its orbit. Each measurement of the electron\u2019s position will find it to be in a definite location somewhere near the nucleus. Repeated measurements reveal a cloud of probability like that in the figure, with each speck the location determined by a single measurement. There is not a well-defined, circular-orbit type of distribution. Nature again proves to be different on a small scale than on a macroscopic scale. Figure 22.15 The ground state of a hydrogen atom has a probability cloud describing the position of its electron. The probability of finding the electron is proportional to the darkness of the cloud. The electron can be closer or farther than the Bohr radius, but it is very unlikely to be a great distance from the nucleus. Virtual Physics Models of the Hydrogen Atom Click to view content (https://www.openstax.org/l/28atom_model) How did scientists figure out the structure of atoms without looking at them? Try out different models by shooting light at the atom. Use this simulation to see how the prediction of the model matches the experimental results. Check Your Understanding 1. Alpha particles are positively charged. What influence did their charge have on the gold foil experiment? a. The positively charged alpha particles were attracted by the attractive electrostatic force from the positive nuclei of the gold atoms. b. The positively charged alpha particles were scattered by the attractive electrostatic force from the positive nuclei of the gold atoms. c. The positively charged alpha particles were scattered by the repulsive electrostatic force from the positive nuclei of the gold atoms. d. The positively charged alpha particles were attracted by the repulsive electrostatic force from the positive nuclei of the gold atoms. 22.2 Nuclear Forces and Radioactivity Section Learning Objectives By the end of this section, you will be able to do the following: \u2022 Describe the structure and forces present within the nucleus \u2022 Explain the three types of radiation \u2022 Write nuclear equations associated with the various types of radioactive decay Section Key Terms alpha decay atomic number beta decay gamma decay Geiger tube isotope mass number nucleons radioactive radioactive decay Access for free at openstax.org. 22.2 \u2022 Nuclear Forces and Radioactivity 735 radioactivity scintillator strong nuclear force transmutation There is an ongoing quest to find the substructures of matter. At one time, it was thought that atoms would be the ultimate substructure. However, just when the first direct evidence of atoms was obtained, it became clear that they have a substructure and a tiny nucleus. The nucleus itself has spectacular characteristics. For example, certain nuclei are unstable, and their decay emits radiations with energies millions of times greater than atomic energies. Some of the mysteries of nature, such as why the core of Earth remains molten and how the Sun produces its energy, are explained by nuclear phenomena. The exploration of radioactivity and the nucleus has revealed new fundamental particles, forces, and conservation laws. That exploration has evolved into a search for further underlying structures, such as quarks. In this section, we will explore the fundamentals of the nucleus and nuclear radioactivity. The Structure of the Nucleus At this point, you are likely familiar with the neutron and proton, the two fundamental particles that make up the nucleus of an atom. Those two particles, collectively called nucleons, make up the small interior portion of the atom. Both particles have nearly the same mass, although the neutron is about two parts in 1,000 more massive. The mass of a proton is equivalent to", " 1,836 electrons, while the mass of a neutron is equivalent to that of 1,839 electrons. That said, each of the particles is significantly more massive than the electron. When describing the mass of objects on the scale of nucleons and atoms, it is most reasonable to measure their mass in terms of atoms. The atomic mass unit (u) was originally defined so that a neutral carbon atom would have a mass of exactly 12 u. Given that protons and neutrons are approximately the same mass, that there are six protons and six neutrons in a carbon atom, and that the mass of an electron is minuscule in comparison, measuring this way allows for both protons and neutrons to have masses close to 1 u. Table 22.1 shows the mass of protons, neutrons, and electrons on the new scale. TIPS FOR SUCCESS For most conceptual situations, the difference in mass between the proton and neutron is insubstantial. In fact, for calculations that require fewer than four significant digits, both the proton and neutron masses may be considered equivalent to one atomic mass unit. However, when determining the amount of energy released in a nuclear reaction, as in Equation 22.40, the difference in mass cannot be ignored. Another other useful mass unit on the atomic scale is the one uses the equation, as will be addressed later in this text.. While rarely used in most contexts, it is convenient when Proton Mass Neutron Mass Electron Mass Kilograms (kg) Atomic mass units (u) Table 22.1 Atomic Masses for Multiple Units To more completely characterize nuclei, let us also consider two other important quantities: the atomic number and the mass number. The atomic number, Z, represents the number of protons within a nucleus. That value determines the elemental quality of each atom. Every carbon atom, for instance, has a Zvalue of 6, whereas every oxygen atom has a Zvalue of 8. For clarification, only oxygen atoms may have a Zvalue of 8. If the Zvalue is not 8, the atom cannot be oxygen. The mass number, A, represents the total number of protons and neutrons, or nucleons, within an atom. For an ordinary carbon atom the mass number would be 12, as there are typically six neutrons accompanying the six protons within the atom. In the case of carbon, the mass would be exactly 12 u. For oxygen, with a mass number of 16, the atomic mass is 15.994915 u. Of course, the difference is minor and can be ignored for most scenarios. Again, because the mass of an electron is so small compared to the nucleons, the mass number and the atomic mass can be essentially equivalent. Figure 22.16 shows an example of Lithium-7, which has an atomic number of 3 and a mass number of 7. How does the mass number help to differentiate one atom from another? If each atom of carbon has an atomic number of 6, 736 Chapter 22 \u2022 The Atom then what is the value of including the mass number at all? The intent of the mass number is to differentiate between various isotopes of an atom. The term isotope refers to the variation of atoms based upon the number of neutrons within their nucleus. While it is most common for there to be six neutrons accompanying the six protons within a carbon atom, it is possible to find carbon atoms with seven neutrons or eight neutrons. Those carbon atoms are respectively referred to as carbon-13 and carbon-14 atoms, with their mass numbers being their primary distinction. The isotope distinction is an important one to make, as the number of neutrons within an atom can affect a number of its properties, not the least of which is nuclear stability. Figure 22.16 Lithium-7 has three protons and four neutrons within its nucleus. As a result, its mass number is 7, while its atomic number is 3. The actual mass of the atom is 7.016 u. Lithium 7 is an isotope of lithium. To more easily identify various atoms, their atomic number and mass number are typically written in a form of representation called the nuclide. The nuclide form appears as follows: neutrons., where Xis the atomic symbol and Nrepresents the number of Let us look at a few examples of nuclides expressed in the notation. The nucleus of the simplest atom, hydrogen, is a single (the zero for no neutrons is often omitted). To check the symbol, refer to the periodic table\u2014you see that the proton, or atomic number Zof hydrogen is 1. Since you are given that there are no neutrons, the mass number Ais also 1. There is a scarce form of hydrogen found in nature called deuterium; its nucleus has one proton and one neutron and, hence, twice the mass of common hydrogen. The symbol for deuterium is, thus, tritium, since it has a", " single proton and two neutrons, and it is written identical chemistries, but the nuclei differ greatly in mass, stability, and other characteristics. Again, the different nuclei are referred to as isotopes of the same element.. An even rarer\u2014and radioactive\u2014form of hydrogen is called. The three varieties of hydrogen have nearly There is some redundancy in the symbols A, X, Z, and N. If the element Xis known, then Zcan be found in a periodic table. If both Aand Xare known, then Ncan also be determined by first finding Z; then, N= A\u2013 Z. Thus the simpler notation for nuclides is 22.34 which is sufficient and is most commonly used. For example, in this simpler notation, the three isotopes of hydrogen are,, and. For, should we need to know, we can determine that Z= 92 for uranium from the periodic table, and thus, N = 238 \u2212 92 = 146. Radioactivity and Nuclear Forces In 1896, the French physicist Antoine Henri Becquerel (1852\u20131908) noticed something strange. When a uranium-rich mineral called pitchblende was placed on a completely opaque envelope containing a photographic plate, it darkened spots on the photographic plate.. Becquerel reasoned that the pitchblende must emit invisible rays capable of penetrating the opaque material. Stranger still was that no light was shining on the pitchblende, which means that the pitchblende was emitting the invisible rays continuously without having any energy input! There is an apparent violation of the law of conservation of energy, one that scientists can now explain using Einstein\u2019s famous equation originate in the nuclei of the atoms and have other unique characteristics. It was soon evident that Becquerel\u2019s rays To this point, most reactions you have studied have been chemical reactions, which are reactions involving the electrons Access for free at openstax.org. 22.2 \u2022 Nuclear Forces and Radioactivity 737 surrounding the atoms. However, two types of experimental evidence implied that Becquerel\u2019s rays did not originate with electrons, but instead within the nucleus of an atom. First, the radiation is found to be only associated with certain elements, such as uranium. Whether uranium was in the form of an element or compound was irrelevant to its radiation. In addition, the presence of radiation does not vary with temperature, pressure, or ionization state of the uranium atom. Since all of those factors affect electrons in an atom, the radiation cannot come from electron transitions, as atomic spectra do. The huge energy emitted during each event is the second piece of evidence that the radiation cannot be atomic. Nuclear radiation has energies on the order of 106 eV per event, which is much greater than typical atomic energies that are a few eV, such as those observed in spectra and chemical reactions, and more than ten times as high as the most energetic X-rays. But why would reactions within the nucleus take place? And what would cause an apparently stable structure to begin emitting energy? Was there something special about Becquerel\u2019s uranium-rich pitchblende? To answer those questions, it is necessary to look into the structure of the nucleus. Though it is perhaps surprising, you will find that many of the same principles that we observe on a macroscopic level still apply to the nucleus. Nuclear Stability A variety of experiments indicate that a nucleus behaves something like a tightly packed ball of nucleons, as illustrated in Figure 22.17. Those nucleons have large kinetic energies and, thus, move rapidly in very close contact. Nucleons can be separated by a large force, such as in a collision with another nucleus, but strongly resist being pushed closer together. The most compelling evidence that nucleons are closely packed in a nucleus is that the radius of a nucleus, r, is found to be approximately where 1.2 femtometer (fm) and Ais the mass number of the nucleus. Note that. Since many nuclei are spherical, and the volume of a sphere is, we see that \u2014that is, the volume of a nucleus is proportional to the number of nucleons in it. That is what you expect if you pack nucleons so close that there is no empty space between them. 22.35 Figure 22.17 Nucleons are held together by nuclear forces and resist both being pulled apart and pushed inside one another. The volume of the nucleus is the sum of the volumes of the nucleons in it, here shown in different colors to represent protons and neutrons. So what forces hold a nucleus together? After all, the nucleus is very small and its protons, being positive, should exert tremendous repulsive forces on one another. Considering that, it seems that the nucleus would be forced apart, not together! The answer is that a previously unknown force holds the nucleus together and makes it into a tightly packed ball of nucleons. This force is known as the strong nuclear force. The", " strong force has such a short range that it quickly fall to zero over a distance of only 10\u201315 meters. However, like glue, it is very strong when the nucleons get close to one another. The balancing of the electromagnetic force with the nuclear forces is what allows the nucleus to maintain its spherical shape. If, for any reason, the electromagnetic force should overcome the nuclear force, components of the nucleus would be projected outward, creating the very radiation that Becquerel discovered! Understanding why the nucleus would break apart can be partially explained using Table 22.2. The balance between the strong nuclear force and the electromagnetic force is a tenuous one. Recall that the attractive strong nuclear force exists between any two nucleons and acts over a very short range while the weaker repulsive electromagnetic force only acts between protons, although over a larger range. Considering the interactions, an imperfect balance between neutrons and protons can result in a nuclear reaction, with the result of regaining equilibrium. 738 Chapter 22 \u2022 The Atom Range of Force Direction Nucleon Interaction Magnitude of Force Electromagnetic Force Long range, though decreasing by 1/r2 Repulsive Proton \u2013proton repulsion Relatively small Strong Nuclear Force Very short range, essentially zero at 1 femtometer Attractive Attraction between any two nucleons 100 times greater than the electromagnetic force Table 22.2 Comparing the Electromagnetic and Strong Forces The radiation discovered by Becquerel was due to the large number of protons present in his uranium-rich pitchblende. In short, the large number of protons caused the electromagnetic force to be greater than the strong nuclear force. To regain stability, the nucleus needed to undergo a nuclear reaction called alpha (\u03b1) decay. The Three Types of Radiation Radioactivity refers to the act of emitting particles or energy from the nucleus. When the uranium nucleus emits energetic nucleons in Becquerel\u2019s experiment, the radioactive process causes the nucleus to alter in structure. The alteration is called radioactive decay. Any substance that undergoes radioactive decay is said to be radioactive. That those terms share a root with the term radiationshould not be too surprising, as they all relate to the transmission of energy. Alpha Decay Alpha decay refers to the type of decay that takes place when too many protons exist in the nucleus. It is the most common type of decay and causes the nucleus to regain equilibrium between its two competing internal forces. During alpha decay, the nucleus ejects two protons and two neutrons, allowing the strong nuclear force to regain balance with the repulsive electromagnetic force. The nuclear equation for an alpha decay process can be shown as follows. 22.36 Figure 22.18 A nucleus undergoes alpha decay. The alpha particle can be seen as made up of two neutrons and two protons, which constitute a helium-4 atom. Three things to note as a result of the above equation: 1. By ejecting an alpha particle, the original nuclide decreases in atomic number. That means that Becquerel\u2019s uranium nucleus, upon decaying, is actually transformed into thorium, two atomic numbers lower on the periodic table! The process of changing elemental composition is called transmutation. 2. Note that the two protons and two neutrons ejected from the nucleus combine to form a helium nucleus. Shortly after decay, the ejected helium ion typically acquires two electrons to become a stable helium atom. 3. Finally, it is important to see that, despite the elemental change, physical conservation still takes place. The mass number of the new element and the alpha particle together equal the mass number of the original element. Also, the net charge of all particles involved remains the same before and after the transmutation. Beta Decay Like alpha decay, beta ( For beta decay, however, a neutron is transformed into a proton and electron or vice versa. The transformation allows for the total mass number of the atom to remain the same, although the atomic number will increase by one (or decrease by one). Once again, the transformation of the neutron allows for a rebalancing of the strong nuclear and electromagnetic forces. The nuclear ) decay also takes place when there is an imbalance between neutrons and protons within the nucleus. Access for free at openstax.org. 22.2 \u2022 Nuclear Forces and Radioactivity 739 equation for a beta decay process is shown below. in the equation above stands for a high-energy particle called the neutrino. A nucleus may also emit a positron, The symbol and in that case Zdecreases and Nincreases. It is beyond the scope of this section and will be discussed in further detail in the chapter on particles. It is worth noting, however, that the mass number and charge in all beta-decay reactions are conserved. Figure 22.19 A nucleus undergoes beta decay. The neutron splits into a proton, electron, and neutrino. This particular decay is called decay. Gamma Decay Gamma decay is a unique form of", " radiation that does not involve balancing forces within the nucleus. Gamma decay occurs when a nucleus drops from an excited state to the ground state. Recall that such a change in energy state will release energy from the nucleus in the form of a photon. The energy associated with the photon emitted is so great that its wavelength is shorter than that of an X-ray. Its nuclear equation is as follows. 22.37 Figure 22.20 A nucleus undergoes gamma decay. The nucleus drops in energy state, releasing a gamma ray. WORKED EXAMPLE Creating a Decay Equation Write the complete decay equation in Strategy Beta decay results in an increase in atomic number. As a result, the original (or parent) nucleus, must have an atomic number of one fewer proton.. Refer to the periodic table for values of Z. notation for beta decay producing Solution The equation for beta decay is as follows Considering that barium is the product (or daughter) nucleus and has an atomic number of 56, the original nucleus must be of an atomic number of 55. That corresponds to cesium, or Cs. 22.38 22.39 740 Chapter 22 \u2022 The Atom The number of neutrons in the parent cesium and daughter barium can be determined by subtracting the atomic number from the mass number (137 \u2013 55 for cesium, 137 \u2013 56 for barium). Substitute those values for the Nand N \u2013 1subscripts in the above equation. 22.40 Discussion The terms parentand daughternucleus refer to the reactants and products of a nuclear reaction. The terminology is not just used in this example, but in all nuclear reaction examples. The cesium-137 nuclear reaction poses a significant health risk, as its chemistry is similar to that of potassium and sodium, and so it can easily be concentrated in your cells if ingested. WORKED EXAMPLE Alpha Decay Energy Found from Nuclear Masses Find the energy emitted in the decay of 239Pu. Strategy Nuclear reaction energy, such as released in decay, can be found using the equation difference in mass between the parent nucleus and the products of the decay. The mass of pertinent particles is as follows. We must first find, the 239Pu: 239.052157 u 235U: 235.043924 u 4He: 4.002602 u. Solution The decay equation for 239Pu is Determine the amount of mass lost between the parent and daughter nuclei. Now we can find Eby entering into the equation. And knowing that, we can find that 22.41 22.42 22.43 22.44 Discussion The energy released in this decay is in the MeV range, about 106 times as great as typical chemical reaction energies, consistent with previous discussions. Most of the energy becomes kinetic energy of the particle (or 4He nucleus), which moves away at high speed. The energy carried away by the recoil of the 235U nucleus is much smaller, in order to conserve momentum. The 235U nucleus can be left in an excited state to later emit photons ( rays). The decay is spontaneous and releases energy, because the products have less mass than the parent nucleus. Properties of Radiation The charges of the three radiated particles differ. Alpha particles, with two protons, carry a net charge of +2. Beta particles, with one electron, carry a net charge of \u20131. Meanwhile, gamma rays are solely photons, or light, and carry no charge. The difference Access for free at openstax.org. 22.2 \u2022 Nuclear Forces and Radioactivity 741 in charge plays an important role in how the three radiations affect surrounding substances. Alpha particles, being highly charged, will quickly interact with ions in the air and electrons within metals. As a result, they have a short range and short penetrating distance in most materials. Beta particles, being slightly less charged, have a larger range and larger penetrating distance. Gamma rays, on the other hand, have little electric interaction with particles and travel much farther. Two diagrams below show the importance of difference in penetration. Table 22.3 shows the distance of radiation penetration, and Figure 22.21 shows the influence various factors have on radiation penetration distance. Type of Radiation Range particles A sheet of paper, a few cm of air, fractions of a millimeter of tissue particles A thin aluminum plate, tens of cm of tissue rays Several cm of lead, meters of concrete Table 22.3 Comparing Ranges of Radioactive Decay Figure 22.21 The penetration or range of radiation depends on its energy, the material it encounters, and the type of radiation. (a) Greater energy means greater range. (b) Radiation has a smaller range in materials with high electron density. (c) Alphas have the smallest range, betas have a greater range, and gammas have the greatest range. LINKS TO PHYSICS Radiation Detectors The first direct detection of radiation was Becquerel\u2019s darkened photographic plate. Photographic film is still the most common detector", " of ionizing radiation, being used routinely in medical and dental X-rays. Nuclear radiation can also be captured on film, as seen in Figure 22.22. The mechanism for film exposure by radiation is similar to that by photons. A quantum of energy from a radioactive particle interacts with the emulsion and alters it chemically, thus exposing the film. Provided the radiation has more than the few eV of energy needed to induce the chemical change, the chemical alteration will occur. The amount of film darkening is related to the type of radiation and amount of exposure. The process is not 100 percent efficient, since not all incident radiation interacts and not all interactions produce the chemical change. 742 Chapter 22 \u2022 The Atom Figure 22.22 Film badges contain film similar to that used in this dental X-ray film. It is sandwiched between various absorbers to determine the penetrating ability of the radiation as well as the amount. Film badges are worn to determine radiation exposure. (credit: Werneuchen, Wikimedia Commons) Another very common radiation detector is the Geiger tube. The clicking and buzzing sound we hear in dramatizations and documentaries, as well as in our own physics labs, is usually an audio output of events detected by a Geiger counter. These relatively inexpensive radiation detectors are based on the simple and sturdy Geiger tube, shown schematically in Figure 22.23. A conducting cylinder with a wire along its axis is filled with an insulating gas so that a voltage applied between the cylinder and wire produces almost no current. Ionizing radiation passing through the tube produces free ion pairs that are attracted to the wire and cylinder, forming a current that is detected as a count. Not every particle is detected, since some radiation can pass through without producing enough ionization. However, Geiger counters are very useful in producing a prompt output that reveals the existence and relative intensity of ionizing radiation. Figure 22.23 (a) Geiger counters such as this one are used for prompt monitoring of radiation levels, generally giving only relative intensity and not identifying the type or energy of the radiation. (credit: Tim Vickers, Wikimedia Commons) (b) Voltage applied between the cylinder and wire in a Geiger tube affects ions and electrons produced by radiation passing through the gas-filled cylinder. Ions move toward the cylinder and electrons toward the wire. The resulting current is detected and registered as a count. Another radiation detection method records light produced when radiation interacts with materials. The energy of the radiation is sufficient to excite atoms in a material that may fluoresce, such as the phosphor used by Rutherford\u2019s group. Materials called scintillators use a more complex process to convert radiation energy into light. Scintillators may be liquid or solid, and they can Access for free at openstax.org. 22.3 \u2022 Half Life and Radiometric Dating 743 be very efficient. Their light output can provide information about the energy, charge, and type of radiation. Scintillator light flashes are very brief in duration, allowing the detection of a huge number of particles in short periods of time. Scintillation detectors are used in a variety of research and diagnostic applications. Among those are the detection of the radiation from distant galaxies using satellite-mounted equipment and the detection of exotic particles in accelerator laboratories. Virtual Physics Beta Decay Click to view content (https://www.openstax.org/l/21betadecayvid) Watch beta decay occur for a collection of nuclei or for an individual nucleus. With this applet, individuals or groups of students can compare half-lives! Check Your Understanding 2. What leads scientists to infer that the nuclear strong force exists? a. A strong force must hold all the electrons outside the nucleus of an atom. b. A strong force must counteract the highly attractive Coulomb force in the nucleus. c. A strong force must hold all the neutrons together inside the nucleus. d. A strong force must counteract the highly repulsive Coulomb force between protons in the nucleus. 22.3 Half Life and Radiometric Dating Section Learning Objectives By the end of this section, you will be able to do the following: \u2022 Explain radioactive half-life and its role in radiometric dating \u2022 Calculate radioactive half-life and solve problems associated with radiometric dating Section Key Terms activity becquerel carbon-14 dating decay constant half-life radioactive dating Half-Life and the Rate of Radioactive Decay Unstable nuclei decay. However, some nuclides decay faster than others. For example, radium and polonium, discovered by Marie and Pierre Curie, decay faster than uranium. That means they have shorter lifetimes, producing a greater rate of decay. Here we will explore half-life and activity, the quantitative terms for lifetime and rate of decay. Why do we use the term like half-liferather than lifetime? The answer can be found by examining Figure 22.24, which shows how the number of radioactive nuclei in a sample", " decreases with time. The time in which half of the original number of nuclei decay is defined as the half-life,. After one half-life passes, half of the remaining nuclei will decay in the next half-life. Then, half of that amount in turn decays in the following half-life. Therefore, the number of radioactive nuclei decreases from Nto N/ 2 in one half-life, to N/ 4 in the next, to N/ 8 in the next, and so on. Nuclear decay is an example of a purely statistical process. TIPS FOR SUCCESS A more precise definition of half-life is that each nucleus has a 50 percent chance of surviving for a time equal to one halflife. If an individual nucleus survives through that time, it still has a 50 percent chance of surviving through another half-life. Even if it happens to survive hundreds of half-lives, it still has a 50 percent chance of surviving through one more. Therefore, the decay of a nucleus is like random coin flipping. The chance of heads is 50 percent, no matter what has happened before. The probability concept aligns with the traditional definition of half-life. Provided the number of nuclei is reasonably large, half of the original nuclei should decay during one half-life period. 744 Chapter 22 \u2022 The Atom Figure 22.24 Radioactive decay reduces the number of radioactive nuclei over time. In one half-life ( ), the number decreases to half of its original value. Half of what remains decays in the next half-life, and half of that in the next, and so on. This is exponential decay, as seen in the graph of the number of nuclei present as a function of time. The following equation gives the quantitative relationship between the original number of nuclei present at time zero the number at a later time t and 22.45 where e= 2.71828... is the base of the natural logarithm, and is the decay constant for the nuclide. The shorter the half-life, the larger is the value of equation decreases with time. The decay constant can be found with the, and the faster the exponential Activity, the Rate of Decay What do we mean when we say a source is highly radioactive? Generally, it means the number of decays per unit time is very high. We define activity Rto be the rate of decay expressed in decays per unit time. In equation form, this is 22.46 where is the number of decays that occur in time. Activity can also be determined through the equation 22.47 22.48 which shows that as the amount of radiative material (N) decreases, the rate of decay decreases as well. The SI unit for activity is one decay per second and it is given the name becquerel (Bq) in honor of the discoverer of radioactivity. That is, Activity Ris often expressed in other units, such as decays per minute or decays per year. One of the most common units for activity is the curie (Ci), defined to be the activity of 1 g of 226Ra, in honor of Marie Curie\u2019s work with radium. The definition of the curie is 22.49 Access for free at openstax.org. or decays per second. Radiometric Dating 22.3 \u2022 Half Life and Radiometric Dating 745 Radioactive dating or radiometric dating is a clever use of naturally occurring radioactivity. Its most familiar application is carbon-14 dating. Carbon-14 is an isotope of carbon that is produced when solar neutrinos strike atmosphere. Radioactive carbon has the same chemistry as stable carbon, and so it mixes into the biosphere, where it is consumed and becomes part of every living organism. Carbon-14 has an abundance of 1.3 parts per trillion of normal carbon, so if you know the number of carbon nuclei in an object (perhaps determined by mass and Avogadro\u2019s number), you can multiply that number by back to with the environment ceases, and wrappings, with the normal abundance in living tissue, it is possible to determine the artifact\u2019s age (or time since death). Carbon-14 dating can be used for biological tissues as old as 50 or 60 thousand years, but is most accurate for younger samples, since the abundance of with a half-life of 5,730 years (note that this is an example of beta decay). When an organism dies, carbon exchange nuclei within the object. Over time, carbon-14 will naturally decay is not replenished. By comparing the abundance of in an artifact, such as mummy nuclei in them is greater. to find the number of particles within the One of the most famous cases of carbon-14 dating involves the Shroud of Turin, a long piece of fabric purported to be the burial shroud of Jesus (see Figure 22.25). This relic was first displayed in Turin in 13", "54 and was denounced as a fraud at that time by a French bishop. Its remarkable negative imprint of an apparently crucified body resembles the then-accepted image of Jesus. As a result, the relic has been remained controversial throughout the centuries. Carbon-14 dating was not performed on the shroud until 1988, when the process had been refined to the point where only a small amount of material needed to be destroyed. Samples were tested at three independent laboratories, each being given four pieces of cloth, with only one unidentified piece found in from the shroud, to avoid prejudice. All three laboratories found samples of the shroud contain 92 percent of the living tissues, allowing the shroud to be dated (see Equation 22.57). Figure 22.25 Part of the Shroud of Turin, which shows a remarkable negative imprint likeness of Jesus complete with evidence of crucifixion wounds. The shroud first surfaced in the 14th century and was only recently carbon-14 dated. It has not been determined how the image was placed on the material. (credit: Butko, Wikimedia Commons) WORKED EXAMPLE Carbon-11 Decay Carbon-11 has a half-life of 20.334 min. (a) What is the decay constant for carbon-11? If 1 kg of carbon-11 sample exists at the beginning of an hour, (b) how much material will remain at the end of the hour and (c) what will be the decay activity at that time? Strategy Since refers to the amount of carbon-11 at the start, then after one half-life, the amount of carbon-11 remaining will be The decay constant is equivalent to the probability that a nucleus will decay each second. As a result, the half-life will need to be converted to seconds. Solution (a) Since half of the carbon-11 remains after one half-life,. 22.50 22.51 746 Chapter 22 \u2022 The Atom Take the natural logarithm of each side to isolate the decay constant. Convert the 20.334 min to seconds. (b) The amount of material after one hour can be found by using the equation with tconverted into seconds and NOwritten as 1,000 g (c) The decay activity after one hour can be found by using the equation for the mass value after one hour. 22.52 22.53 22.54 22.55 22.56 22.57 Discussion (a) The decay constant shows that 0.0568 percent of the nuclei in a carbon-11 sample will decay each second. Another way of considering the decay constant is that a given carbon-11 nuclei has a 0.0568 percent probability of decaying each second. The decay of carbon-11 allows it to be used in positron emission topography (PET) scans; however, its 20.334 min half-life does pose challenges for its administration. (b) One hour is nearly three full half-lives of the carbon-11 nucleus. As a result, one would expect the amount of sample remaining to be approximately one eighth of the original amount. The 129.4 g remaining is just a bit larger than one-eighth, which is sensible given a half-life of just over 20 min. (c) Label analysis shows that the unit of Becquerel is sensible, as there are 0.0735 g of carbon-11 decaying each second. That is smaller amount than at the beginning of the hour, when g of carbon-11 were decaying each second. WORKED EXAMPLE How Old is the Shroud of Turin? Calculate the age of the Shroud of Turin given that the amount of Strategy Because 92 percent of the know that the half-life of assume that the decrease in remains, is 5,730 years, and so once is solely due to nuclear decay. found in it is 92 percent of that in living tissue.. Therefore, the equation can be used to find. We also is known, we can find and then find tas requested. Here, we Solution Solving the equation for gives Access for free at openstax.org. 22.58 22.4 \u2022 Nuclear Fission and Fusion 747 Thus, Taking the natural logarithm of both sides of the equation yields so that Rearranging to isolate tgives Now, the equation can be used to find for. Solving for and substituting the known half-life gives We enter that value into the previous equation to find t. 22.59 22.60 22.61 22.62 22.63 22.64 Discussion This dates the material in the shroud to 1988\u2013690 = 1300. Our calculation is only accurate to two digits, so that the year is rounded to 1300. The values obtained at the three independent laboratories gave a weighted average date of 1320 \u00b1 60. That uncertainty is typical of carbon-14 dating and is due to the small amount of 14 C in living tissues, the amount of material available, and experimental uncertainties (reduced by having three independent measurements). That said,", " is it notable that the carbon-14 date is consistent with the first record of the shroud\u2019s existence and certainly inconsistent with the period in which Jesus lived. There are other noncarbon forms of radioactive dating. Rocks, for example, can sometimes be dated based on the decay of The decay series for been since the rock solidified. Knowledge of the years ago. about, so the ratio of those nuclides in a rock can be used an indication of how long it has half-life has shown, for example, that the oldest rocks on Earth solidified ends with Virtual Physics Radioactive Dating Game Click to view content (https://www.openstax.org/l/02radioactive_dating_game) Learn about different types of radiometric dating, such as carbon dating. Understand how decay and half-life work to enable radiometric dating to work. Play a game that tests your ability to match the percentage of the dating element that remains to the age of the object. 22.4 Nuclear Fission and Fusion Section Learning Objectives By the end of this section, you will be able to do the following: \u2022 Explain nuclear fission \u2022 Explain nuclear fusion \u2022 Describe how the processes of fission and fusion work in nuclear weapons and in generating nuclear power 748 Chapter 22 \u2022 The Atom Section Key Terms chain reaction critical mass liquid drop model nuclear fission nuclear fusion proton-proton cycle The previous section dealt with naturally occurring nuclear decay. Without human intervention, some nuclei will change composition in order to achieve a stable equilibrium. This section delves into a less-natural process. Knowing that energy can be emitted in various forms of nuclear change, is it possible to create a nuclear reaction through our own intervention? The answer to this question is yes. Through two distinct methods, humankind has discovered multiple ways of manipulating the atom to release its internal energy. Nuclear Fission In simplest terms, nuclear fission is the splitting of an atomic bond. Given that it requires great energy separate two nucleons, it may come as a surprise to learn that splitting a nucleus can releasevast potential energy. And although it is true that huge amounts of energy can be released, considerable effort is needed to do so in practice. An unstable atom will naturally decay, but it may take millions of years to do so. As a result, a physical catalyst is necessary to produce useful energy through nuclear fission. The catalyst typically occurs in the form of a free neutron, projected directly at the nucleus of a high-mass atom. As shown in Figure 22.26, a neutron strike can cause the nucleus to elongate, much like a drop of liquid water. This is why the model is known as the liquid drop model. As the nucleus elongates, nucleons are no longer so tightly packed, and the repulsive electromagnetic force can overcome the short-range strong nuclear force. The imbalance of forces can result in the two ends of the drop flying apart, with some of the nuclear binding energy released to the surroundings. Figure 22.26 Neutron-induced fission is shown. First, energy is put into a large nucleus when it absorbs a neutron. Acting like a struck liquid drop, the nucleus deforms and begins to narrow in the middle. Since fewer nucleons are in contact, the repulsive Coulomb force is able to break the nucleus into two parts with some neutrons also flying away. As you can imagine, the consequences of the nuclei splitting are substantial. When a nucleus is split, it is not only energy that is released, but a small number of neutrons as well. Those neutrons have the potential to cause further fission in other nuclei, especially if they are directed back toward the other nuclei by a dense shield or neutron reflector (see part (d) of Figure 22.26). Access for free at openstax.org. 22.4 \u2022 Nuclear Fission and Fusion 749 However, not every neutron produced by fission induces further fission. Some neutrons escape the fissionable material, while others interact with a nucleus without making it split. We can enhance the number of fissions produced by neutrons by having a large amount of fissionable material as well as a neutron reflector. The minimum amount necessary for self-sustained fission of a given nuclide is called its critical mass. Some nuclides, such as 239Pu, produce more neutrons per fission than others, such as 235U. Additionally, some nuclides are easier to make fission than others. In particular, 235U and 239Pu are easier to fission than the much more abundant 238U. Both factors affect critical mass, which is smallest for 239Pu. The self-sustained fission of nuclei is commonly referred to as a chain reaction, as shown in Figure 22.27. Figure 22.27 A chain reaction can produce self-sustained fission if each fission produces enough neutrons to induce at least one more fission", ". This depends on several factors, including how many neutrons are produced in an average fission and how easy it is to make a particular type of nuclide fission. A chain reaction can have runaway results. If each atomic split results in two nuclei producing a new fission, the number of nuclear reactions will increase exponentially. One fission will produce two atoms, the next round of fission will create four atoms, the third round eight atoms, and so on. Of course, each time fission occurs, more energy will be emitted, further increasing the power of the atomic reaction. And that is just if two neutrons create fission reactions each round. Perhaps you can now see why so many people consider atomic energy to be an exciting energy source! To make a self-sustained nuclear fission reactor with 235U, it is necessary to slow down the neutrons. Water is very effective at this, since neutrons collide with protons in water molecules and lose energy. Figure 22.28 shows a schematic of a reactor design called the pressurized water reactor. 750 Chapter 22 \u2022 The Atom Figure 22.28 A pressurized water reactor is cleverly designed to control the fission of large amounts of 235U, while using the heat produced in the fission reaction to create steam for generating electrical energy. Control rods adjust neutron flux so that it is self-sustaining. In case the reactor overheats and boils the water away, the chain reaction terminates, because water is needed to slow down the neutrons. This inherent safety feature can be overwhelmed in extreme circumstances. Control rods containing nuclides that very strongly absorb neutrons are used to adjust neutron flux. To produce large amounts of power, reactors contain hundreds to thousands of critical masses, and the chain reaction easily becomes self-sustaining. Neutron flux must be carefully regulated to avoid an out-of-control exponential increase in the rate of fission. Control rods help prevent overheating, perhaps even a meltdown or explosive disassembly. The water that is used to slow down neutrons, necessary to get them to induce fission in 235U, and achieve criticality, provides a negative feedback for temperature increase. In case the reactor overheats and boils the water to steam or is breached, the absence of water kills the chain reaction. Considerable heat, however, can still be generated by the reactor\u2019s radioactive fission products. Other safety features, thus, need to be incorporated in the event of a loss of coolant accident, including auxiliary cooling water and pumps. Energies in Nuclear Fission The following are two interesting facts to consider: \u2022 The average fission reaction produces 200 MeV of energy. \u2022 If you were to measure the mass of the products of a nuclear reaction, you would find that their mass was slightly less than the mass of the original nucleus. How are those things possible? Doesn\u2019t the fission reaction\u2019s production of energy violate the conservation of energy? Furthermore, doesn\u2019t the loss in mass in the reaction violate the conservation of mass? Those are important questions, and they can both be answered with one of the most famous equations in scientific history. 22.65 Recall that, according to Einstein\u2019s theory, energy and mass are essentially the same thing. In the case of fission, the mass of the products is less than that of the reactants because the missing mass appears in the form of the energy released in the reaction, with a constant value of c2 Joules of energy converted for each kilogram of material. The value of c2 is substantial\u2014from Einstein\u2019s equation, the amount of energy in just 1 gram of mass would be enough to support the average U.S. citizen for more than 270 years! The example below will show you how a mass-energy transformation of this type takes place. Access for free at openstax.org. 22.4 \u2022 Nuclear Fission and Fusion 751 WORKED EXAMPLE Calculating Energy from a Kilogram of Fissionable Fuel Calculate the amount of energy produced by the fission of 1.00 kg of, given the average fission reaction of 22.66 Strategy The total energy produced is the number of number of atoms in 1.00 kg. atoms times the given energy per fission. We should therefore find the Solution The number of g; thus, there are atoms in 1.00 kg is Avogadro\u2019s number times the number of moles. One mole of has a mass of 235.04. The number of atoms is therefore So the total energy released is Discussion The result is another impressively large amount of energy, equivalent to about 14,000 barrels of crude oil or 600,000 gallons of gasoline. But, it is only one fourth the energy produced by the fusion of a kilogram of a mixture of deuterium and tritium. Even though each fission reaction yields about ten times the energy of a fusion reaction, the energy per", " kilogram of fission fuel is less, because there are far fewer moles per kilogram of the heavy nuclides. Fission fuel is also much scarcer than fusion fuel, and less than 1 percent of uranium (the 235 U) is readily usable. 22.67 Virtual Physics Nuclear Fission Click to view content (https://www.openstax.org/l/16fission) Start a chain reaction, or introduce nonradioactive isotopes to prevent one. Use the applet to control energy production in a nuclear reactor! Nuclear Fusion Nuclear fusion is defined as the combining, or fusing, of two nuclei and, the combining of nuclei also results in an emission of energy. For many, the concept is counterintuitive. After all, if energy is released when a nucleus is split, how can it also be released when nucleons are combined together? The difference between fission and fusion, which results from the size of the nuclei involved, will be addressed next. Remember that the structure of a nucleus is based on the interplay of the compressive nuclear strong force and the repulsive electromagnetic force. For nuclei that are less massive than iron, the nuclear force is actually stronger than that of the Coulomb force. As a result, when a low-mass nucleus absorbs nucleons, the added neutrons and protons bind the nucleus more tightly. The increased nuclear strong force does work on the nucleus, and energy is released. Once the size of the created nucleus exceeds that of iron, the short-ranging nuclear force does not have the ability to bind a nucleus more tightly, and the emission of energy ceases. In fact, for fusion to occur for elements of greater mass than iron, energy must be added to the system! Figure 22.29 shows an energy-mass curve commonly used to describe nuclear reactions. Notice the location of iron (Fe) on the graph. All low-mass nuclei to the left of iron release energy through fusion, while all highmass particles to the right of iron produce energy through fission. 752 Chapter 22 \u2022 The Atom Figure 22.29 Fusion of light nuclei to form medium-mass nuclei converts mass to energy, because binding energy per nucleon ( ) is greater for the product nuclei. The larger is, the less mass per nucleon, and so mass is converted to energy and released in such fusion reactions. TIPS FOR SUCCESS Just as it is not possible for the elements to the left of iron in the figure to naturally fission, it is not possible for elements to the right of iron to naturally undergo fusion, as that process would require the addition of energy to occur. Furthermore, notice that elements commonly discussed in fission and fusion are elements that can provide the greatest change in binding energy, such as uranium and hydrogen. Iron\u2019s location on the energy-mass curve is important, and explains a number of its characteristics, including its role as an elemental endpoint in fusion reactions in stars. The major obstruction to fusion is the Coulomb repulsion force between nuclei. Since the attractive nuclear force that can fuse nuclei together is short ranged, the repulsion of like positive charges must be overcome in order to get nuclei close enough to induce fusion. Figure 22.30 shows an approximate graph of the potential energy between two nuclei as a function of the distance between their centers. The graph resembles a hill with a well in its center. A ball rolled to the left must have enough kinetic energy to get over the hump before it falls into the deeper well with a net gain in energy. So it is with fusion. If the nuclei are given enough kinetic energy to overcome the electric potential energy due to repulsion, then they can combine, release energy, and fall into a deep well. One way to accomplish that end is to heat fusion fuel to high temperatures so that the kinetic energy of thermal motion is sufficient to get the nuclei together. Figure 22.30 Potential energy between two light nuclei graphed as a function of distance between them. If the nuclei have enough kinetic energy to get over the Coulomb repulsion hump, they combine, release energy, and drop into a deep attractive well. You might think that, in our Sun, nuclei are constantly coming into contact and fusing. However, this is only partially true. Only at the Sun\u2019s core are the particles close enough and the temperature high enough for fusion to occur! In the series of reactions below, the Sun produces energy by fusing protons, or hydrogen nuclei (, by far the Sun\u2019s most Access for free at openstax.org. 22.4 \u2022 Nuclear Fission and Fusion 753 abundant nuclide) into helium nuclei cycle. The principal sequence of fusion reactions forms what is called the proton-proton 22.68 stands for a positron and where the first two reactions must occur twice for the third to be possible, so the cycle consumes six protons ( Furthermore, the two positrons", " produced will find two electrons and annihilate to form four more overall cycle is thus is an electron neutrino. The energy in parentheses is releasedby the reaction. Note that ) but gives back two. rays, for a total of six. The where the 26.7 MeV includes the annihilation energy of the positrons and electrons and is distributed among all the reaction products. The solar interior is dense, and the reactions occur deep in the Sun where temperatures are highest. It takes about 32,000 years for the energy to diffuse to the surface and radiate away. However, the neutrinos can carry their energy out of the Sun in less than two seconds, because they interact so weakly with other matter. Negative feedback in the Sun acts as a thermostat to regulate the overall energy output. For instance, if the interior of the Sun becomes hotter than normal, the reaction rate increases, producing energy that expands the interior. The expansion cools it and lowers the reaction rate. Conversely, if the interior becomes too cool, it contracts, increasing the temperature and therefore the reaction rate (see Figure 22.31). Stars like the Sun are stable for billions of years, until a significant fraction of their hydrogen has been depleted. Figure 22.31 Nuclear fusion in the Sun converts hydrogen nuclei into helium; fusion occurs primarily at the boundary of the helium core, where the temperature is highest and sufficient hydrogen remains. Energy released diffuses slowly to the surface, with the exception of neutrinos, which escape immediately. Energy production remains stable because of negative-feedback effects. Nuclear Weapons and Nuclear Power The world was in political turmoil when fission was discovered in 1938. Compounding the troubles, the possibility of a selfsustained chain reaction was immediately recognized by leading scientists the world over. The enormous energy known to be in nuclei, but considered inaccessible, now seemed to be available on a large scale. Within months after the announcement of the discovery of fission, Adolf Hitler banned the export of uranium from newly occupied Czechoslovakia. It seemed that the possible military value of uranium had been recognized in Nazi Germany, and that a serious effort to build a nuclear bomb had begun. Alarmed scientists, many of whom fled Nazi Germany, decided to take action. None was more famous or revered than Einstein. It was felt that his help was needed to get the American government to make a serious effort at constructing nuclear weapons as a matter of survival. Leo Szilard, a Hungarian physicist who had emigrated to America, took a draft of a letter to Einstein, who, although a pacifist, signed the final version. The letter was for President Franklin Roosevelt, warning of the German potential to build extremely powerful bombs of a new type. It was sent in August of 1939, just before the German invasion of Poland that marked the start of World War II. It was not until December 6, 1941, the day before the Japanese attack on Pearl Harbor, that the United States made a massive commitment to building a nuclear bomb. The top secret Manhattan Project was a crash program aimed at beating the Germans. 754 Chapter 22 \u2022 The Atom It was carried out in remote locations, such as Los Alamos, New Mexico, whenever possible, and eventually came to cost billions of dollars and employ the efforts of more than 100,000 people. J. Robert Oppenheimer (1904\u20131967), a talented physicist, was chosen to head the project. The first major step was made by Enrico Fermi and his group in December 1942, when they completed the first self-sustaining nuclear reactor. This first atomic pile, built in a squash court at the University of Chicago, proved that a fission chain reaction was possible. Plutonium was recognized as easier to fission with neutrons and, hence, a superior fission material very early in the Manhattan Project. Plutonium availability was uncertain, and so a uranium bomb was developed simultaneously. Figure 22.32 shows a guntype bomb, which takes two subcritical uranium masses and shoots them together. To get an appreciable yield, the critical mass must be held together by the explosive charges inside the cannon barrel for a few microseconds. Since the buildup of the uranium chain reaction is relatively slow, the device to bring the critical mass together can be relatively simple. Owing to the fact that the rate of spontaneous fission is low, a neutron source is at the center the assembled critical mass. Figure 22.32 A gun-type fission bomb for utilizes two subcritical masses forced together by explosive charges inside a cannon barrel. The energy yield depends on the amount of uranium and the time it can be held together before it disassembles itself. Plutonium\u2019s special properties necessitated a more sophisticated critical mass assembly, shown schematically in Figure 22.33. A spherical mass of plutonium is surrounded by shaped charges (high explosives that focus their blast) that implode the plutonium, crushing it into a smaller volume to form a critical mass. The implosion technique is faster and more", " effective, because it compresses three-dimensionally rather than one-dimensionally as in the gun-type bomb. Again, a neutron source is included to initiate the chain reaction. Figure 22.33 An implosion created by high explosives compresses a sphere of 239Pu into a critical mass. The superior fissionability of plutonium has made it the preferred bomb material. Access for free at openstax.org. Owing to its complexity, the plutonium bomb needed to be tested before there could be any attempt to use it. On July 16, 1945, the test named Trinity was conducted in the isolated Alamogordo Desert in New Mexico, about 200 miles south of Los Alamos (see Figure 22.34). A new age had begun. The yield of the Trinity device was about 10 kilotons (kT), the equivalent of 5,000 of the largest conventional bombs. 22.4 \u2022 Nuclear Fission and Fusion 755 Figure 22.34 Trinity test (1945), the first nuclear bomb (credit: U.S. Department of Energy) Although Germany surrendered on May 7, 1945, Japan had been steadfastly refusing to surrender for many months, resulting large numbers of civilian and military casualties. Invasion plans by the Allies estimated a million casualties of their own and untold losses of Japanese lives. The bomb was viewed as a way to end the war. The first bomb used was a gun-type uranium bomb dropped on Hiroshima on August 6 by the United States. Its yield of about 15 kT destroyed the city and killed an estimated 80,000 people, with 100,000 more being seriously injured. The second bomb was an implosion-type plutonium bomb dropped on Nagasaki only three days later. Its 20-kT yield killed at least 50,000 people, something less than Hiroshima because of the hilly terrain and the fact that it was a few kilometers off target. The Japanese were told that one bomb a week would be dropped until they surrendered unconditionally, which they did on August 14. In actuality, the United States had only enough plutonium for one more bomb, as yet unassembled. Knowing that fusion produces several times more energy per kilogram of fuel than fission, some scientists pursued the idea of constructing a fusion bomb. The first such bomb was detonated by the United States several years after the first fission bombs, on October 31, 1952, at Eniwetok Atoll in the Pacific Ocean. It had a yield of 10 megatons (MT), about 670 times that of the fission bomb that destroyed Hiroshima. The Soviet Union followed with a fusion device of its own in August 1953, and a weapons race, beyond the aim of this text to discuss, continued until the end of the Cold War. Figure 22.35 shows a simple diagram of how a thermonuclear bomb is constructed. A fission bomb is exploded next to fusion fuel in the solid form of lithium deuteride. Before the shock wave blows it apart, create tritium through the reaction. Additional fusion and fission fuels are enclosed in a dense shell of rays heat and compress the fuel, and neutrons. At the same time that the uranium shell reflects the neutrons back into the fuel to enhance its fusion, the fast-moving neutrons cause the plentiful and inexpensive to fission, part of what allows thermonuclear bombs to be so large. 756 Chapter 22 \u2022 The Atom Figure 22.35 This schematic of a fusion bomb (H-bomb) gives some idea of how the fission trigger is used to ignite fusion fuel. Neutrons and \u03b3rays transmit energy to the fusion fuel, create tritium from deuterium, and heat and compress the fusion fuel. The outer shell of serves to reflect some neutrons back into the fuel, causing more fusion, and it boosts the energy output by fissioning itself when neutron energies become high enough. Of course, not all applications of nuclear physics are as destructive as the weapons described above. Hundreds of nuclear fission power plants around the world attest to the fact that controlled fission is both practical and economical. Given growing concerns over global warming, nuclear power is often seen as a viable alternative to energy derived from fossil fuels. BOUNDLESS PHYSICS Fusion Reactors For decades, fusion reactors have been deemed the energy of the future. A safer, cleaner, and more abundant potential source of energy than its fission counterpart, images of the fusion reactor have been conjured up each time the need for a renewable, environmentally friendly resource is discussed. Now, after more than half a century of speculating, some scientists believe that fusion reactors are nearly here. In creating energy by combining atomic nuclei, the fusion reaction holds many advantages over fission. First, fusion reactions are more efficient, releasing 3 to 4 times more energy than fission per gram of fuel. Furthermore, unlike fission reactions that require heavyelements like uranium that are difficult to obtain, fusion requires lightelements that are abundant in nature. The greatest advantage of the fusion reaction,", " however, is in its ability to be controlled. While traditional nuclear reactors create worries about meltdowns and radioactive waste, neither is a substantial concern with the fusion reaction. Consider that fusion reactions require a large amount of energy to overcome the repulsive Coulomb force and that the byproducts of a fusion reaction are largely limited to helium nuclei. In order for fusion to occur, hydrogen isotopes of deuterium and tritium must be acquired. While deuterium can easily be gathered from ocean water, tritium is slightly more difficult to come by, though it can be manufactured from Earth\u2019s abundant lithium. Once acquired, the hydrogen isotopes are injected into an empty vessel and subjected to temperature and pressure great enough to mimic the conditions at the core of our Sun. Using carefully controlled high-frequency radio waves, the hydrogen isotopes are broken into plasma and further controlled through an electromagnetic field. As the electromagnetic field continues to exert pressure on the hydrogen plasma, enough energy is supplied to cause the hydrogen plasma to fuse into helium. Access for free at openstax.org. 22.5 \u2022 Medical Applications of Radioactivity: Diagnostic Imaging and Radiation 757 Figure 22.36 Tokamak confinement of nuclear fusion plasma. The magnetic field lines are used to confine the high-temperature plasma (purple). Research is currently being done to increase the efficiency of the tokamak confinement model. Once the plasma fuses, high-velocity neutrons are ejected from the newly formed helium atoms. Those high velocity neutrons, carrying the excess energy stored within bonds of the original hydrogen, are able to travel unaffected by the applied magnetic field. In doing so, they strike a barrier around the nuclear reactor, transforming their excess energy to heat. The heat is then harvested to make steam that drives turbines. Hydrogen\u2019s tremendous power is now usable! The historical concern with nuclear fusion reactors is that the energy required to control the electromagnetic field is greater than the energy harvested from the hydrogen atoms. However, recent research by both Lockheed Martin engineers and scientists at the Lawrence Livermore National Laboratory has yielded exciting theoretical improvements in efficiency. At the time of this writing, a test facility called ITER (International Thermonuclear Experimental Reactor) is being constructed in southern France. A joint venture of the European Union, the United States, Japan, Russia, China, South Korea, and India, ITER is designed for further study into the future of nuclear fusion energy production. 22.5 Medical Applications of Radioactivity: Diagnostic Imaging and Radiation Section Learning Objectives By the end of this section, you will be able to do the following: \u2022 Describe how nuclear imaging works (e.g., radioisotope imaging, PET) \u2022 Describe the ionizing effects of radiation and how they can be used for medical treatment Section Key Terms Anger camera rad radiopharmaceutical therapeutic ratio relative biological effectiveness (RBE) roentgen equivalent man (rem) tagged Medical Applications of Nuclear Physics Applications of nuclear physics have become an integral part of modern life. From the bone scan that detects one cancer to the radioiodine treatment that cures another, nuclear radiation has diagnostic and therapeutic effects on medicine. Medical Imaging A host of medical imaging techniques employ nuclear radiation. What makes nuclear radiation so useful? First, radiation can 758 Chapter 22 \u2022 The Atom easily penetrate tissue; hence, it is a useful probe to monitor conditions inside the body. Second, nuclear radiation depends on the nuclide and not on the chemical compound it is in, so that a radioactive nuclide can be put into a compound designed for specific purposes. When that is done, the compound is said to be tagged. A tagged compound used for medical purposes is called a radiopharmaceutical. Radiation detectors external to the body can determine the location and concentration of a radiopharmaceutical to yield medically useful information. For example, certain drugs are concentrated in inflamed regions of the body, and their locations can aid diagnosis and treatment as seen in Figure 22.37. Another application utilizes a radiopharmaceutical that the body sends to bone cells, particularly those that are most active, to detect cancerous tumors or healing points. Images can then be produced of such bone scans. Clever use of radioisotopes determines the functioning of body organs, such as blood flow, heart muscle activity, and iodine uptake in the thyroid gland. For instance, a radioactive form of iodine can be used to monitor the thyroid, a radioactive thallium salt can be used to follow the blood stream, and radioactive gallium can be used for cancer imaging. Figure 22.37 A radiopharmaceutical was used to produce this brain image of a patient with Alzheimer\u2019s disease. Certain features are computer enhanced. (credit: National Institutes of Health) Once a radioactive compound has been ingested, a device like that shown in Figure 22.38 is used to monitor nuclear activity. The device, called an Anger camera or gamma camera uses a piece of lead with holes bored through it. The gamma", " rays are redirected through the collimator to narrow their beam, and are then interpreted using a device called a scintillator. The computer analysis of detector signals produces an image. One of the disadvantages of this detection method is that there is no depth information (i.e., it provides a two-dimensional view of the tumor as opposed to a three-dimensional view), because radiation from any location under that detector produces a signal. Figure 22.38 An Anger or gamma camera consists of a lead collimator and an array of detectors. Gamma rays produce light flashes in the scintillators. The light output is converted to an electrical signal by the photomultipliers. A computer constructs an image from the detector output. Single-photon-emission computer tomography (SPECT) used in conjunction with a CT scanner improves on the process carried out by the gamma camera. Figure 22.39 shows a patient in a circular array of SPECT detectors that may be stationary or rotated, with detector output used by a computer to construct a detailed image. The spatial resolution of this technique is poor, but the three-dimensional image created results in a marked improvement in contrast. Access for free at openstax.org. 22.5 \u2022 Medical Applications of Radioactivity: Diagnostic Imaging and Radiation 759 Figure 22.39 SPECT uses a rotating camera to form an image of the concentration of a radiopharmaceutical compound. (credit: Woldo, Wikimedia Commons) Positron emission tomography (or PET) scans utilize images produced by encounters an electron, mutual annihilation occurs, producing two \u03b3rays. Those energy comes from the destruction of an electron or positron mass) and they move directly away from each other, allowing detectors to determine their point of origin accurately (as shown in Figure 22.40). It requires detectors on opposite sides to simultaneously (i.e., at the same time) detect photons of 0.511 MeV energy and utilizes computer imaging techniques similar to those in SPECT and CT scans. PET is used extensively for diagnosing brain disorders. It can note decreased metabolism in certain regions that accompany Alzheimer\u2019s disease. PET can also locate regions in the brain that become active when a person carries out specific activities, such as speaking, closing his or her eyes, and so on. rays have identical 0.511 MeV energies (the emitters. When the emitted positron \u03b2+ Figure 22.40 A PET system takes advantage of the two identical -ray photons produced by positron-electron annihilation. The rays are emitted in opposite directions, so that the line along which each pair is emitted is determined. Various events detected by several pairs of detectors are then analyzed by the computer to form an accurate image. Ionizing Radiation on the Body We hear many seemingly contradictory things about the biological effects of ionizing radiation. It can cause cancer, burns, and hair loss, and yet it is used to treat and even cure cancer. How do we understand such effects? Once again, there is an underlying simplicity in nature, even in complicated biological organisms. All the effects of ionizing radiation on biological tissue can be understood by knowing that ionizing radiation affects molecules within cells, particularly DNA molecules. Let us take a brief look at molecules within cells and how cells operate. Cells have long, double-helical DNA molecules containing chemical patterns called genetic codes that govern the function and processes undertaken by the cells. Damage to DNA consists of breaks in chemical bonds or other changes in the structural features of the DNA chain, leading to changes in the genetic code. In human cells, we can have as many as a million individual instances of damage to DNA per cell per day. The repair ability of DNA is vital for maintaining the integrity of the genetic code and for the normal functioning of the entire organism. A cell with a damaged ability to repair DNA, which could have been induced by ionizing radiation, can do one of the following: \u2022 The cell can go into an irreversible state of dormancy, known as senescence. \u2022 The cell can commit suicide, known as programmed cell death. \u2022 The cell can go into unregulated cell division, leading to tumors and cancers. 760 Chapter 22 \u2022 The Atom Since ionizing radiation damages the DNA, ionizing radiation has its greatest effect on cells that rapidly reproduce, including most types of cancer. Thus, cancer cells are more sensitive to radiation than normal cells and can be killed by it easily. Cancer is characterized by a malfunction of cell reproduction, and can also be caused by ionizing radiation. There is no contradiction to say that ionizing radiation can be both a cure and a cause. Radiotherapy Radiotherapy is effective against cancer because cancer cells reproduce rapidly and, consequently, are more sensitive to radiation. The central problem in radiotherapy is to make the dose for cancer cells as high as possible while limiting the dose for normal cells. The ratio of abnormal cells killed to normal cells killed is called the therapeutic ratio, and all radiotherapy techniques are designed to enhance that ratio.", " Radiation can be concentrated in cancerous tissue by a number of techniques. One of the most prevalent techniques for well-defined tumors is a geometric technique shown in Figure 22.41. A narrow beam of radiation is passed through the patient from a variety of directions with a common crossing point in the tumor. The technique concentrates the dose in the tumor while spreading it out over a large volume of normal tissue. Figure 22.41 The 60Co source of -radiation is rotated around the patient so that the common crossing point is in the tumor, concentrating the dose there. This geometric technique works for well-defined tumors. Another use of radiation therapy is through radiopharmaceuticals. Cleverly, radiopharmaceuticals are used in cancer therapy by tagging antibodies with radioisotopes. Those antibodies are extracted from the patient, cultured, loaded with a radioisotope, and then returned to the patient. The antibodies are then concentrated almost entirely in the tissue they developed to fight, thus localizing the radiation in abnormal tissue. This method is used with radioactive iodine to fight thyroid cancer. While the therapeutic ratio can be quite high for such short-range radiation, there can be a significant dose for organs that eliminate radiopharmaceuticals from the body, such as the liver, kidneys, and bladder. As with most radiotherapy, the technique is limited by the tolerable amount of damage to the normal tissue. Radiation Dosage To quantitatively discuss the biological effects of ionizing radiation, we need a radiation dose unit that is directly related to those effects. To do define such a unit, it is important to consider both the biological organism and the radiation itself. Knowing that the amount of ionization is proportional to the amount of deposited energy, we define a radiation dose unit called the rad. It 1/100 of a joule of ionizing energy deposited per kilogram of tissue, which is For example, if a 50.0-kg person is exposed to ionizing radiation over her entire body and she absorbs 1.00 J, then her wholebody radiation dose is 22.69 If the same 1.00 J of ionizing energy were absorbed in her 2.00-kg forearm alone, then the dose to the forearm would be 22.70 22.71 and the unaffected tissue would have a zero rad dose. When calculating radiation doses, you divide the energy absorbed by the mass of affected tissue. You must specify the affected region, such as the whole body or forearm in addition to giving the numerical dose in rads. Although the energy per kilogram in 1 rad is small, it can still have significant effects. Since only a few eV cause ionization, just 0.01 J of ionizing energy can create a huge number of ion pairs and have an effect at the cellular level. The effects of ionizing radiation may be directly proportional to the dose in rads, but they also depend on the type of radiation and the type of tissue. That is, for a given dose in rads, the effects depend on whether the radiation is, X-ray, or some,, Access for free at openstax.org. 22.5 \u2022 Medical Applications of Radioactivity: Diagnostic Imaging and Radiation 761 other type of ionizing radiation. The relative biological effectiveness (RBE) relates to the amount of biological damage that can occur from a given type of radiation and is given in Table 22.4 for several types of ionizing radiation. Type and energy of radiation RBE X-rays rays rays greater than 32 keV rays less than 32 keV 1 1 1 1.7 Neutrons, thermal to slow (< 20 keV) 2\u20135 Neutrons, fast (1\u201310 MeV) 10 (body), 32 (eyes) Protons (1\u201310 MeV) 10 (body), 32 (eyes) rays from radioactive decay 10\u201320 Heavy ions from accelerators 10\u201320 Table 22.4 Relative Biological Effectiveness TIPS FOR SUCCESS The RBEs given in Table 22.4 are approximate, but they yield certain valuable insights. \u2022 The eyes are more sensitive to radiation, because the cells of the lens do not repair themselves. \u2022 Though both are neutral and have large ranges, neutrons cause more damage than rays because neutrons often cause secondary radiation when they are captured. \u2022 Short-range particles such as rays have a severely damaging effect to internal anatomy, as their damage is concentrated and more difficult for the biological organism to repair. However, the skin can usually block alpha particles from entering the body. Can you think of any other insights from the table? A final dose unit more closely related to the effect of radiation on biological tissue is called the roentgen equivalent man, or rem. A combination of all factors mentioned previously, the roentgen equivalent man is defined to be the dose in rads multiplied by the relative biological effectiveness. The large-scale effects of radiation on humans can be divided into two categories: immediate effects and long-term effects. Table 22", ".5 gives the immediate effects of whole-body exposures received in less than one day. If the radiation exposure is spread out over more time, greater doses are needed to cause the effects listed. Any dose less than 10 rem is called a low dose, a dose 10 to 100 rem is called a moderate dose, and anything greater than 100 rem is called a high dose. 22.72 Dose (rem) Effect 0\u201310 No observable effect 10\u2013100 Slight to moderate decrease in white blood cell counts Table 22.5 Immediate Effects of Radiation (Adults, Whole Body, Single Exposure) 762 Chapter 22 \u2022 The Atom Dose (rem) Effect 50 Temporary sterility 100\u2013200 Significant reduction in blood cell counts, brief nausea, and vomiting; rarely fatal 200\u2013500 Nausea, vomiting, hair loss, severe blood damage, hemorrhage, fatalities 450 LD50/32; lethal to 50% of the population within 32 days after exposure if untreated 500\u20132,000 Worst effects due to malfunction of small intestine and blood systems; limited survival > 2,000 Fatal within hours due to collapse of central nervous system Table 22.5 Immediate Effects of Radiation (Adults, Whole Body, Single Exposure) WORK IN PHYSICS Health Physicist Are you interested in learning more about radiation? Are you curious about studying radiation dosage levels and ensuring the safety of the environment and people that are most closely affected by it? If so, you may be interested in becoming a health physicist. The field of health physics draws from a variety of science disciplines with the central aim of mitigating radiation concerns. Those that work as health physicists have a diverse array of potential jobs available to them, including those in research, industry, education, environmental protection, and governmental regulation. Furthermore, while the term health physicistmay lead many to think of the medical field, there are plenty of applications within the military, industrial, and energy fields as well. As a researcher, a health physicist can further environmental studies on the effects of radiation, design instruments for more accurate measurements, and assist in establishing valuable radiation standards. Within the energy field, a health physicsist often acts as a manager, closely tied to all operations at all levels, from procuring appropirate equipment to monitoring health data. Within industry, the health physicist acts as a consultant, assisting industry management in important decisions, designing facilities, and choosing appropriate detection tools. The health physicist possesses a unique knowledge base that allows him or her to operate in a wide variety of interesting disciplines! To become a health physicist, it is necessary to have a background in the physical sciences. Understanding the fields of biology, physiology, biochemistry, and genetics are all important as well. The ability to analyze and solve new problems is critical, and a natural aptitude for science and mathematics will assist in the continued necessary training. There are two possible certifications for health physicists: from the American Board of Health Physicists (ABHP) and the National Registry of Radiation Protection Technologists (NRRPT). Access for free at openstax.org. Chapter 22 \u2022 Key Terms 763 KEY TERMS activity rate of decay for radioactive nuclides alpha decay type of radioactive decay in which an atomic nucleus emits an alpha particle anger camera common medical imaging device that uses a scintillator connected to a series of photomultipliers atomic number number of protons in a nucleus becquerel SI unit for rate of decay of a radioactive material beta decay type of radioactive decay in which an atomic nucleus emits a beta particle nuclear fusion reaction in which two nuclei are combined, or fused, to form a larger nucleus nucleons particles found inside nuclei planetary model of the atom model of the atom that shows electrons orbiting like planets about a Sun-like nucleus proton-proton cycle combined reactions and that begins with carbon-14 dating radioactive dating technique based on the hydrogen and ends with helium radioactivity of carbon-14 rad amount of ionizing energy deposited per kilogram of chain reaction self-sustaining sequence of events, tissue exemplified by the self-sustaining nature of a fission reaction at critical mass radioactive substance or object that emits nuclear radiation critical mass minimum amount necessary for self- radioactive dating application of radioactive decay in sustained fission of a given nuclide decay constant quantity that is inversely proportional to the half-life and that is used in the equation for number of nuclei as a function of time energy-level diagram a diagram used to analyze the energy levels of electrons in the orbits of an atom excited state any state beyond the n= 1 orbital in which the electron stores energy Fraunhofer lines black lines shown on an absorption spectrum that show the wavelengths absorbed by a gas gamma decay type of radioactive decay in which an atomic nucleus emits a gamma ray which the age of a material is determined by the amount of radioactivity of a particular type that occurs radioactive decay process by which an atomic nucleus of an unstable atom loses mass and energy by emitting ionizing particles radioactivity emission of rays from the nuclei of atoms", " radiopharmaceutical compound used for medical imaging relative biological effectiveness (RBE) number that expresses the relative amount of damage that a fixed amount of ionizing radiation of a given type can inflict on biological tissues roentgen equivalent man (rem) dose unit more closely Geiger tube very common radiation detector that usually related to effects in biological tissue gives an audio output ground state the n=1 orbital of an electron half-life time in which there is a 50 percent chance that a nucleus will decay Heisenberg uncertainty principle fundamental limit to the precision with which pairs of quantities such as momentum and position can be measured hydrogen-like atom any atom with only a single electron isotope nuclei having the same Zand different N\u2019s liquid drop model model of the atomic nucleus (useful only to understand some of its features) in which nucleons in a nucleus act like atoms in a drop mass number number of nucleons in a nucleus nuclear fission reaction in which a nucleus splits SECTION SUMMARY 22.1 The Structure of the Atom \u2022 Rutherford\u2019s gold foil experiment provided evidence that the atom is composed of a small, dense nucleus with electrons occupying the mostly empty space around it. \u2022 Analysis of emission spectra shows that energy is emitted from energized gas in discrete quantities. Rutherford scattering scattering of alpha particles by gold nuclei in the gold foil experiment Rydberg constant a physical constant related to atomic spectra, with an established value of scintillator radiation detection method that records light produced when radiation interacts with materials strong nuclear force attractive force that holds nucleons together within the nucleus tagged having a radioactive substance attached (to a chemical compound) therapeutic ratio the ratio of abnormal cells killed to normal cells killed transmutation process of changing elemental composition \u2022 The Bohr model of the atom describes electrons existing in discrete orbits, with discrete energies emitted and absorbed as the electrons decrease and increase in orbital energy. \u2022 The energy emitted or absorbed by an electron as it changes energy state can be determined with the equation, where 764 Chapter 22 \u2022 Key Equations. \u2022 The wavelength of energy absorbed or emitted by an electron as it changes energy state can be determined by the equation, where. \u2022 Described as an electron cloud, the quantum model of the atom is the result of de Broglie waves and Heisenberg\u2019s uncertainty principle. 22.2 Nuclear Forces and Radioactivity \u2022 The structure of the nucleus is defined by its two nucleons, the neutron and proton. \u2022 Atomic numbers and mass numbers are used to differentiate between various atoms and isotopes. Those numbers can be combined into an easily recognizable form called a nuclide. \u2022 The size and stability of the nucleus is based upon two forces: the electromagnetic force and strong nuclear force. \u2022 Radioactive decay is the alteration of the nucleus through the emission of particles or energy. \u2022 Alpha decay occurs when too many protons exist in the nucleus. It results in the ejection of an alpha particle, as described in the equation. \u2022 Beta decay occurs when too many neutrons (or protons) exist in the nucleus. It results in the transmutation of a neutron into a proton, electron, and neutrino. The decay is expressed through the equation. (Beta decay may also transform a proton into a neutron.) \u2022 Gamma decay occurs when a nucleus in an excited state move to a more stable state, resulting in the release of a photon. Gamma decay is represented with the equation. \u2022 The penetration distance of radiation depends on its energy, charge, and type of material it encounters. 22.3 Half Life and Radiometric Dating \u2022 Radioactive half-life is the time it takes a sample of nuclei to decay to half of its original amount. \u2022 The rate of radioactive decay is defined as the sample\u2019s KEY EQUATIONS 22.1 The Structure of the Atom energy of hydrogen electron in an orbital Access for free at openstax.org. activity, represented by the equation. \u2022 Knowing the half-life of a radioactive isotope allows for the process of radioactive dating to determine the age of a material. If the half-life of a material is known, the age of the material can be found using the equation. \u2022 \u2022 The age of organic material can be determined using the decay of the carbon-14 isotope, while the age of rocks can be determined using the decay of uranium-238. 22.4 Nuclear Fission and Fusion \u2022 Nuclear fission is the splitting of an atomic bond, releasing a large amount of potential energy previously holding the atom together. The amount of energy released can be determined through the equation. \u2022 Nuclear fusion is the combining, or fusing together, of two nuclei. Energy is also released in nuclear fusion as the combined nuclei are closer together, resulting in a decreased strong nuclear force. \u2022 Fission was used in two nuclear weapons at the conclusion of World War II: the gun-type uranium bomb and the implosion-type plutonium bomb. \u2022 While fission has been used in both nuclear weapons and nuclear reactors", ", fusion is capable of releasing more energy per reaction. As a result, fusion is a wellresearched, if not yet well-controlled, energy source. 22.5 Medical Applications of Radioactivity: Diagnostic Imaging and Radiation \u2022 Medical imaging occurs when a radiopharmaceutical placed in the body provides information to an array of radiation detectors outside the body. \u2022 Devices utilizing medical imaging include the Anger \u2022 camera, SPECT detector, and PET scan. Ionizing radiation can both cure and cause cancer through the manipulation of DNA molecules. \u2022 Radiation dosage and its effect on the body can be measured using the quantities radiation dose unit (rad), relative biological effectiveness (RBE), and the roentgen equivalent man (rem). energy of any hydrogen-like electron in orbital wavelength of light emitted by an electron changing states wavelength of an orbital heisenberg\u2019s uncertainty principle 22.2 Nuclear Forces and Radioactivity alpha decay equation beta decay equation gamma decay equation CHAPTER REVIEW Concept Items 22.1 The Structure of the Atom 1. A star emits light from its core. One observer views the emission unobstructed while a second observer views the emission while obstructed by a cloud of hydrogen gas. Describe the difference between their observations. a. b. c. Frequencies will be absorbed from the spectrum. d. Frequencies will be added to the spectrum. Intensity of the light in the spectrum will increase. Intensity of the light in the spectrum will decrease. 2. How does the orbital energy of a hydrogen-like atom change as it increases in atomic number? Critical Thinking Items 22.1 The Structure of the Atom 4. How would the gold foil experiment have changed if electrons were used in place of alpha particles, assuming that the electrons hit the gold foil with the same force as the alpha particles? a. Being less massive, the electrons might have been scattered to a greater degree than the alpha particles. b. Being less massive, the electrons might have been scattered to a lesser degree than the alpha particles. Chapter 22 \u2022 Chapter Review 765 22.3 Half Life and Radiometric Dating radioactive half-life 22.4 Nuclear Fission and Fusion energy\u2013mass conversion protonproton chain 22.5 Medical Applications of Radioactivity: Diagnostic Imaging and Radiation roentgen equivalent man a. The orbital energy will increase. b. The orbital energy will decrease. c. The orbital energy will remain constant. d. The orbital energy will be halved. 22.4 Nuclear Fission and Fusion 3. Aside from energy yield, why are nuclear fusion reactors more desirable than nuclear fission reactors? a. Nuclear fusion reactors have a low installation cost. b. Radioactive waste is greater for a fusion reactor. c. Nuclear fusion reactors are easy to design and build. d. A fusion reactor produces less radioactive waste. c. Being more massive, the electrons would have been scattered to a greater degree than the alpha particles. d. Being more massive, the electrons would have been scattered to a lesser degree than the alpha particles. 5. Why does the emission spectrum of an isolated gas differ from the emission spectrum created by a white light? a. White light and an emission spectrum are different varieties of continuous distribution of frequencies. b. White light and an emission spectrum are different series of discrete frequencies. 766 Chapter 22 \u2022 Chapter Review c. White light is a continuous distribution of frequencies, and an emission spectrum is a series of discrete frequencies. d. White light is a series of discrete frequencies, and an emission spectrum is a continuous distribution of frequencies. 6. Why would it most likely be difficult to observe quantized orbital states for satellites orbiting the earth? a. On a macroscopic level, the orbital states do exist for satellites orbiting Earth but are too closely spaced for us to see. d. While the alpha particle has a greater charge than a beta particle, the electron density in lead is much higher than that in air. 10. What influence does the strong nuclear force have on the electrons in an atom? a. b. c. The strong force makes electrons revolve around It attracts them toward the nucleus. It repels them away from the nucleus. the nucleus. It does not have any influence. d. b. On a macroscopic level, the orbital states do not 22.3 Half Life and Radiometric Dating exist for satellites orbiting Earth. c. On a macroscopic level, we cannot control the amount of energy that we give to an artificial satellite and thus control its orbital altitude. d. On a macroscopic level, we cannot control the amount of energy that we give to an artificial satellite but we can control its orbital altitude. 7. Do standing waves explain why electron orbitals are quantized? a. no b. yes 8. Some terms referring to the observation of light include emission spectrumand absorption spectrum. Based on these definitions, what would a reflection spectrum describe? a. The reflection spectrum would describe when incident waves are selectively reflected by a substance. b. The reflection spectrum would describe when incident", " waves are completely reflected by a substance. c. The reflection spectrum would describe when incident waves are not absorbed by a substance. d. The reflection spectrum would describe when incident waves are completely absorbed by a substance. 22.2 Nuclear Forces and Radioactivity 9. Explain why an alpha particle can have a greater range in air than a beta particle in lead. a. While the alpha particle has a lesser charge than a beta particle, the electron density in lead is much less than that in air. b. While the alpha particle has a greater charge than a beta particle, the electron density in lead is much lower than that in air. c. While the alpha particle has a lesser charge than a beta particle, the electron density in lead is much greater than that in air. Access for free at openstax.org. 11. Provide an example of something that decreases in a manner similar to radioactive decay. a. The potential energy of an object falling under the influence of gravity b. The kinetic energy of a ball that is dropped from a building to the ground c. Theh charge transfer from an ebonite rod to fur d. The heat transfer from a hot to a cold object 12. A sample of radioactive material has a decay constant of 0.05 s\u20131. Why is it wrong to presume that the sample will take just 20 seconds to fully decay? a. The decay constant varies with the mass of the sample. b. The decay constant results vary with the amount of the sample. c. The decay constant represents a percentage of the sample that cannot decay. d. The decay constant represents only the fraction of a sample that decays in a unit of time, not the decay of the entire sample. 22.4 Nuclear Fission and Fusion 13. What is the atomic number of the most strongly bound nuclide? a. b. c. d. 14. Why are large electromagnets necessary in nuclear fusion reactors? a. Electromagnets are used to slow down the movement of charge hydrogen plasma. b. Electromagnets are used to decrease the temperature of hydrogen plasma. c. Electromagnets are used to confine the hydrogen plasma. d. Electromagnets are used to stabilize the temperature of the hydrogen plasma. 22.5 Medical Applications of Radioactivity: Diagnostic Imaging and Radiation 15. Why are different radiopharmaceuticals used to image different parts of the body? a. The different radiopharmaceuticals travel through different blood vessels. b. The different radiopharmaceuticals travel to different parts of the body. c. The different radiopharmaceuticals are used to treat different diseases of the body. d. The different radiopharmaceuticals produce different amounts of ionizing radiation. 16. Why do people think carefully about whether to receive a diagnostic test such as a CT scan? a. The radiation from a CT scan is capable of creating cancerous cells. Performance Task 22.5 Medical Applications of Radioactivity: Diagnostic Imaging and Radiation 18. On the Environmental Protection Agency\u2019s website, a helpful tool exists to allow you to determine your average annual radiation dose. Use the tool to determine whether the radiation level you have been exposed to is dangerous and to compare your radiation dosage to other radiative events. 1. Visit the webpage (http://www.openstax.org/l/ 28calculate) and answer the series of questions provided to determine the average annual radiation dosage that you receive. 2. Table 22.5 shows the immediate effects of a radiation dosage. Using the table, explain what you would experience if your yearly dosage of radiation was received all over the course of one day. Also, determine whether your dosage is considered a low, moderate, or high. 3. Using the information input into the webpage, what percentage of your dosage comes from TEST PREP Multiple Choice 22.1 The Structure of the Atom 19. If electrons are negatively charged and the nucleus is positively charged, why do they not attract and collide with each other? Chapter 22 \u2022 Test Prep 767 b. The radiation from a CT scan is capable of destroying cancerous cells. c. The radiation from a CT scan is capable of creating diabetic cells. d. The radiation from a CT scan is capable of destroying diabetic cells. 17. Sometimes it is necessary to take a PET scan very soon after ingesting a radiopharmaceutical. Why is that the case? a. The radiopharmaceutical may have a short half-life. b. The radiopharmaceutical may have a long half-life. c. The radiopharmaceutical quickly passes through the digestive system. d. The radiopharmaceutical can become lodged in the digestive system. natural sources? The average percentage of radiation from natural sources for an individual is around 85 percent. 4. Research radiation dosages for evacuees from events like the Chernobyl and Fukushima meltdowns. How does your annual radiation exposure rate compare to the net dosage for evacuees of each event. Use numbers to support", " your answer. 5. The U.S. Department of Labor limits the amount of radiation that a given worker may receive in a 12 month period. a. Research the present maximum value and compare your annual exposure rate to that of a radiation worker. Use numbers to support your answer. b. What types of work are likely to cause an increase in the radiation exposure of a particular worker? Provide one question based upon the information gathered on the EPA website. a. The pull from the nucleus provides a centrifugal force, which is not strong enough to draw the electrons into the nucleus. b. The pull from the nucleus provides a centripetal force, which is not strong enough to draw the electrons into the nucleus. 768 Chapter 22 \u2022 Test Prep c. The pull from the nucleus provides a helical motion. d. The pull from the nucleus provides a cycloid b. Coulomb force between protons only c. Strong nuclear force between all nucleons and motion. 22.4 Nuclear Fission and Fusion 20. If a nucleus elongates due to a neutron strike, which of the following forces will decrease? a. Nuclear force between neutrons only Coulomb force between protons, but the strong force will decrease more d. Strong nuclear force between neutrons and Coulomb force between protons, but Coulomb force will decrease more Short Answer 22.1 The Structure of the Atom a. The radioactive activity decreases exponentially. b. The radioactive activity undergoes linear decay. c. The radioactive activity undergoes logarithmic 21. Why do Bohr\u2019s calculations for electron energies not decay. work for all atoms? a. In atoms with more than one electron is an atomic shell, the electrons will interact. That requires a more complex formula than Bohr\u2019s calculations accounted for. In atoms with 10 or more electorns in an atomic shell, the electrons will interact. That requires a more complex formula than Bohr\u2019s calculations accounted for. In atoms with more than one electron in an atomic shell, the electrons will not interact. That requires a more complex formula than Bohr\u2019s calculations accounted for. In atoms with 10 or more electrons in an atomic shell, the electrons will not interact. That requires a more complex formula than Bohr\u2019s calculations accounted for. b. c. d. 22.2 Nuclear Forces and Radioactivity 22. Does transmutation occur within chemical reactions? a. no b. yes 22.3 Half Life and Radiometric Dating 23. How does the radioactive activity of a sample change with time? Extended Response 22.1 The Structure of the Atom 26. Compare the standing wavelength of an orbital to the standing wavelength of an a. The standing wavelength of an orbital. orbital is greater than the standing wavelength of an orbital. b. The standing wavelength of an than the standing wavelength of an orbital is less orbital. Access for free at openstax.org. d. The radioactive activity will not change with time. 22.4 Nuclear Fission and Fusion 24. Why does fission of heavy nuclei result in the release of neutrons? a. Heavy nuclei require more neutrons to achieve stability. b. Heavy nuclei require more neutrons to balance charge. c. Light nuclei require more neutrons to achieve stability. d. Light nuclei require more neutrons to balance charge. 22.5 Medical Applications of Radioactivity: Diagnostic Imaging and Radiation 25. Why is radioactive iodine used to monitor the thyroid? a. Radioactive iodine can be used by the thyroid while absorbing information about the thyroid. b. Radioactive iodine can be used by the thyroid while emitting information about the thyroid. c. Radioactive iodine can be secreted by the thyroid while absorbing information about the thyroid. d. Radioactive iodine can be secreted by the thyroid while emitting information about the thyroid. c. There is no relation between the standing wavelength of an wavelength of an orbital and the standing orbital. d. The standing wavelength of an orbital is the same as the standing wavelength of an orbital. 27. Describe the shape of the electron cloud, based on total energy levels, for an atom with electrons in multiple orbital states. a. There are multiple regions of high electron Chapter 22 \u2022 Test Prep 769 probability of various shapes surrounding the nucleus. b. There is a single solid spherical region of high electron probability surrounding the nucleus. brought together and then rapidly increase once a minimum is reached. b. The potential energy will decrease as the nuclei are brought together. c. There are multiple concentric shells of high c. The potential energy will increase as the nuclei are electron probability surrounding the nucleus. brought together. d. There is a single spherical shell of high electron probability surrounding the nucleus. 22.2 Nuclear Forces and Radioactivity 28. How did Becquerel\u2019s observations of pitchblende imply the existence of radioactivity? a. A chemical reaction occurred on the photographic plate without any external source of energy. b. Bright spots appeared on the photographic plate due", " to an external source of energy. c. Energy from the Sun was absorbed by the pitchblende and reflected onto the photographic plate. d. Dark spots appeared on the photographic plate due to an external source of energy. 22.4 Nuclear Fission and Fusion 29. Describe the potential energy of two nuclei as they approach each other. a. The potential energy will decrease as the nuclei are d. The potential energy will increase as the nuclei are brought together and then rapidly decrease once a maximum is reached. 22.5 Medical Applications of Radioactivity: Diagnostic Imaging and Radiation 30. Why do X-rays and gamma rays have equivalent RBE values if they provide different amounts of energy to the body? a. The penetration distance, which depends on energy, is short for both X-rays and gamma rays. b. The penetration distance, which depends on energy, is long for both X-rays and gamma rays. c. The penetration distance, as determined by their high mass, is different for both X-rays and gamma rays. d. The penetration distance, as determined by their low mass, is the same for both X-rays and gamma rays. 770 Chapter 22 \u2022 Test Prep Access for free at openstax.org. CHAPTER 23 Particle Physics Figure 23.1 Part of the Large Hadron Collider (LHC) at CERN, on the border of Switzerland and France. The LHC is a particle accelerator, designed to study fundamental particles. (credit: Image Editor, Flickr) Chapter Outline 23.1 The Four Fundamental Forces 23.2 Quarks 23.3 The Unification of Forces Following ideas remarkably similar to those of the ancient Greeks, we continue to look for smaller and INTRODUCTION smaller structures in nature, hoping ultimately to find and understand the most fundamental building blocks that exist. Atomic physics deals with the smallest units of elements and compounds. In its study, we have found a relatively small number of atoms with systematic properties, and these properties have explained a tremendous range of phenomena. Nuclear physics is concerned with the nuclei of atoms and their substructures. Here, a smaller number of components\u2014the proton and neutron\u2014make up all nuclei. Exploring the systematic behavior of their interactions has revealed even more about matter, forces, and energy. Particle physics deals with the substructures of atoms and nuclei and is particularly aimed at finding those truly fundamental particles that have no further substructure. Just as in atomic and nuclear physics, we have found a complex array of particles and properties with systematic characteristics analogous to the periodic table and the chart of nuclides. An underlying structure is apparent, and there is some reason to think that we arefinding particles that have no substructure. Of course, we have been in similar situations before. For example, atoms were once thought to be the ultimate substructures. It is possible that we could continue to find deeper and deeper structures without ever discovering the ultimate substructure\u2014in science there is never complete certainty. See Figure 23.2. The properties of matter are based on substructures called molecules and atoms. Each atom has the substructure of a nucleus surrounded by electrons, and their interactions explain atomic properties. Protons and neutrons\u2014and the interactions between them\u2014explain the stability and abundance of elements and form the substructure of nuclei. Protons and neutrons are not fundamental\u2014they are composed of quarks. Like electrons and a few other particles, quarks may be the fundamental building blocks of all matter, lacking any further substructure. But the story is not complete because quarks and electrons may have substructures smaller than details that are presently observable. 772 Chapter 23 \u2022 Particle Physics Figure 23.2 A solid, a molecule, an atom, a nucleus, a nucleon (a particle that makes up the nucleus\u2014either a proton or a neutron), and a quark. This chapter covers the basics of particle physics as we know it today. An amazing convergence of topics is evolving in particle physics. We find that some particles are intimately related to forces and that nature on the smallest scale may have its greatest influence on the large scale character of the universe. It is an adventure exceeding the best science fiction because it is not only fantastic but also real. 23.1 The Four Fundamental Forces Section Learning Objectives By the end of the section, you will be able to do the following: \u2022 Define, describe, and differentiate the four fundamental forces \u2022 Describe the carrier particles and explain how their exchange transmits force \u2022 Explain how particle accelerators work to gather evidence about particle physics Section Key Terms carrier particle colliding beam cyclotron Feynman diagram graviton particle physics pion quantum electrodynamics synchrotron boson boson weak nuclear force boson Despite the apparent complexity within the universe, there remain just four basic forces. These forces are responsible for all interactions known to science: from the very small to the very large to those that we experience in our day-", "to-day lives. These forces describe the movement of galaxies, the chemical reactions in our laboratories, the structure within atomic nuclei, and the cause of radioactive decay. They describe the true cause behind familiar terms like friction and the normal force. These four basic forces are known as fundamental because they alone are responsible for all observations of forces in nature. The four fundamental forces are gravity, electromagnetism, weak nuclear force, and strong nuclear force. Understanding the Four Forces The gravitational force is most familiar to us because it describes so many of our common observations. It explains why a dropped ball falls to the ground and why our planet orbits the Sun. It gives us the property of weight and determines much about the motion of objects in our daily lives. Because gravitational force acts between all objects of mass and has the ability to act over large distances, the gravitational force can be used to explain much of what we observe and can even describe the motion of objects on astronomical scales! That said, gravity is incredibly weak compared to the other fundamental forces and is the weakest of all of the fundamental forces. Consider this: The entire mass of Earth is needed to hold an iron nail to the ground. Yet with a simple magnet, the force of gravity can be overcome, allowing the nail to accelerate upward through space. The electromagnetic force is responsible for both electrostatic interactions and the magnetic force seen between bar magnets. When focusing on the electrostatic relationship between two charged particles, the electromagnetic force is known as the coulomb force. The electromagnetic force is an important force in the chemical and biological sciences, as it is responsible for molecular connections like ionic bonding and hydrogen bonding. Additionally, the electromagnetic force is behind the common physics forces of friction and the normal force. Like the gravitational force, the electromagnetic force is an inverse square law. However, the electromagnetic force does not exist between any two objects of mass, only those that are charged. When considering the structure of an atom, the electromagnetic force is somewhat apparent. After all, the electrons are held in place by an attractive force from the nucleus. But what causes the nucleus to remain intact? After all, if all protons are positive, it Access for free at openstax.org. 23.1 \u2022 The Four Fundamental Forces 773 makes sense that the coulomb force between the protons would repel the nucleus apart immediately. Scientists theorized that another force must exist within the nucleus to keep it together. They further theorized that this nuclear force must be significantly stronger than gravity, which has been observed and measured for centuries, and also stronger than the electromagnetic force, which would cause the protons to want to accelerate away from each other. The strong nuclear force is an attractive force that exists between all nucleons. This force, which acts equally between protonproton connections, proton-neutron connections, and neutron-neutron connections, is the strongest of all forces at short ranges. However, at a distance of 10\u201313 cm, or the diameter of a single proton, the force dissipates to zero. If the nucleus is large (it has many nucleons), then the distance between each nucleon could be much larger than the diameter of a single proton. The weak nuclear force is responsible for beta decay, as seen in the equation decay is when a beta particle is ejected from an atom. In order to accelerate away from the nucleus, the particle must be acted on by a force. Enrico Fermi was the first to envision this type of force. While this force is appropriately labeled, it remains stronger than the gravitational force. However, its range is even smaller than that of the strong force, as can be seen in Table 23.1. The weak nuclear force is more important than it may appear at this time, as will be addressed when we discuss quarks. Recall that beta Force Approximate Relative Strength[1] Range Gravity Weak Electromagnetic Strong 1 \u221e \u221e [1]Relative strength is based on the strong force felt by a proton\u2013proton pair. Table 23.1 Relative strength and range of the four fundamental forces Transmitting the Four Fundamental Forces Just as it troubled Einstein prior to formulating the gravitational field theory, the concept of forces acting over a distance had greatly troubled particle physicists. That is, how does one proton knowthat another exists? Furthermore, what causes one proton to make a second proton repel? Or, for that matter, what is it about a proton that causes a neutron to attract? These mysterious interactions were first considered by Hideki Yukawa in 1935 and laid the foundation for much of what we now understand about particle physics. Hideki Yukawa\u2019s focus was on the strong nuclear force and, in particular, its incredibly short range. His idea was a blend of particles, relativity, and quantum mechanics that was applicable to all four forces. Yukawa proposed that the nuclear force is actually transmitted by the exchange of particles, called carrier particles, and that what we commonly refer to as", " the force\u2019s field consists of these carrier particles. Specifically for the strong nuclear force, Yukawa proposed that a previously unknown particle, called a pion, is exchanged between nucleons, transmitting the force between them. Figure 23.3 illustrates how a pion would carry a force between a proton and a neutron. Figure 23.3 The strong nuclear force is transmitted between a proton and neutron by the creation and exchange of a pion. The pion, created through a temporary violation of conservation of mass-energy, travels from the proton to the neutron and is recaptured. It is not 774 Chapter 23 \u2022 Particle Physics directly observable and is called a virtual particle. Note that the proton and neutron change identity in the process. The range of the force is limited by the fact that the pion can exist for only the short time allowed by the Heisenberg uncertainty principle. Yukawa used the finite range of the strong nuclear force to estimate the mass of the pion; the shorter the range, the larger the mass of the carrier particle. In Yukawa\u2019s strong force, the carrier particle is assumed to be transmitted at the speed of light and is continually transferred between the two nucleons shown. The particle that Yukawa predicted was finally discovered within cosmic rays in 1947. Its name, the pion, stands for pi meson, where meson means medium mass; it\u2019s a medium mass because it is smaller than a nucleon but larger than an electron. Yukawa launched the field that is now called quantum chromodynamics, and the carrier particles are now called gluons due to their strong binding power. The reason for the change in the particle name will be explained when quarks are discussed later in this section. As you may assume, the strong force is not the only force with a carrier particle. Nuclear decay from the weak force also requires a particle transfer. In the weak force are the following three: the weak negative carrier, W\u2013; the weak positive carrier, W+; and the zero charge carrier, Z0. As we will see, Fermi inferred that these particles must carry mass, as the total mass of the products of nuclear decay is slightly larger than the total mass of all reactants after nuclear decay. The carrier particle for the electromagnetic force is, not surprisingly, the photon. After all, just as a lightbulb can emit photons from a charged tungsten filament, the photon can be used to transfer information from one electrically charged particle to another. Finally, the graviton is the proposed carrier particle for gravity. While it has not yet been found, scientists are currently looking for evidence of its existence (see Boundless Physics: Searching for the Graviton). So how does a carrier particle transmit a fundamental force? Figure 23.4 shows a virtual photon transmitted from one positively charged particle to another. The transmitted photon is referred to as a virtual particle because it cannot be directly observed while transmitting the force. Figure 23.5 shows a way of graphing the exchange of a virtual photon between the two positively charged particles. This graph of time versus position is called a Feynman diagram, after the brilliant American physicist Richard Feynman (1918\u20131988), who developed it. Figure 23.4 The image in part (a) shows the exchange of a virtual photon transmitting the electromagnetic force between charges, just as virtual pion exchange carries the strong nuclear force between nucleons. The image in part (b) shows that the photon cannot be directly observed in its passage because this would disrupt it and alter the force. In this case, the photon does not reach the other charge. The Feynman diagram should be read from the bottom up to show the movement of particles over time. In it, you can see that the left proton is propelled leftward from the photon emission, while the right proton feels an impulse to the right when the photon is received. In addition to the Feynman diagram, Richard Feynman was one of the theorists who developed the field of quantum electrodynamics (QED), which further describes electromagnetic interactions on the submicroscopic scale. For this work, he shared the 1965 Nobel Prize with Julian Schwinger and S.I. Tomonaga. A Feynman diagram explaining the strong force interaction hypothesized by Yukawa can be seen in Figure 23.6. Here, you can see the change in particle type due to the exchange of the pi meson. Access for free at openstax.org. 23.1 \u2022 The Four Fundamental Forces 775 Figure 23.5 The Feynman diagram for the exchange of a virtual photon between two positively charged particles illustrates how electromagnetic force is transmitted on a quantum mechanical scale. Time is graphed vertically, while the distance is graphed horizontally. The two positively charged particles are seen to repel each other by the photon exchange. Figure 23.6 The image shows a Feynman diagram for the exchange of a \u03c0+ (pion", ") between a proton and a neutron, carrying the strong nuclear force between them. This diagram represents the situation shown more pictorially in Figure 23.3. The relative masses of the listed carrier particles describe something valuable about the four fundamental forces, as can be seen bosons) and Z bosons ( in Table 23.2. W bosons (consisting of nearly 1,000 times more massive than pions, carriers of the strong nuclear force. Simultaneously, the distance that the weak nuclear force can be transmitted is approximately times the strong force transmission distance. Unlike carrier particles, bosons), carriers of the weak nuclear force, are and which have a limited range, the photon is a massless particle that has no limit to the transmission distance of the electromagnetic force. This relationship leads scientists to understand that the yet-unfound graviton is likely massless as well. Force Carrier Particle Range Relative Strength[1] Gravity Graviton (theorized) Weak W and Z bosons Electromagnetic Photon \u221e \u221e Strong Pi mesons or pions (now known as gluons) 1 [1]Relative strength is based on the strong force felt by a proton-proton pair. Table 23.2 Carrier particles and their relative masses compared to pions for the four fundamental forces 776 Chapter 23 \u2022 Particle Physics BOUNDLESS PHYSICS Searching for the Graviton From Newton\u2019s Universal Law of Gravitation to Einstein\u2019s field equations, gravitation has held the focus of scientists for centuries. Given the discovery of carrier particles during the twentieth century, the importance of understanding gravitation has yet again gained the interest of prominent physicists everywhere. With carrier particles discovered for three of the four fundamental forces, it is sensible to scientists that a similar particle, titled the graviton, must exist for the gravitational force. While evidence of this particle is yet to be uncovered, scientists are working diligently to discover its existence. So what do scientists think about the unfound particle? For starters, the graviton (like the photon) should be a massless particle traveling at the speed of light. This is assumed because, like the electromagnetic force, gravity is an inverse square law,. Scientists also theorize that the graviton is an electrically neutral particle, as an empty space within the influence of gravity is chargeless. However, because gravity is such a weak force, searching for the graviton has resulted in some unique methods. LIGO, the Laser Interferometer Gravitational-Wave Observatory, is one tool currently being utilized (see Figure 23.7). While searching for a gravitational wave to find a carrier particle may seem counterintuitive, it is similar to the approach taken by Planck and Einstein to learn more about the photon. According to wave-particle duality, if a gravitational wave can be found, the graviton should be present along with it. Predicted by Einstein\u2019s theory of general relativity, scientists have been monitoring binary star systems for evidence of these gravitational waves. Figure 23.7 In searching for gravitational waves, scientists are using the Laser Interferometer Gravitational-Wave Observatory (LIGO). Here we see the control room of LIGO in Hanford, Washington. Particle accelerators like the Large Hadron Collider (LHC) are being used to search for the graviton through high-energy collisions. While scientists at the LHC speculate that the particle may not exist long enough to be seen, evidence of its prior existence, like footprints in the sand, can be found through gaps in projected energy and momentum. Some scientists are even searching the remnants of the Big Bang in an attempt to find the graviton. By observing the cosmic background radiation, they are looking for anomalies in gravitational waves that would provide information about the gravity particles that existed at the start of our universe. Regardless of the method used, scientists should know the graviton once they find it. A massless, chargeless particle with a spin of 2 and traveling at the speed of light\u2014there is no other particle like it. Should it be found, its discovery would surely be considered by future generations to be on par with those of Newton and Einstein. GRASP CHECK Why are binary star systems used by LIGO to find gravitational waves? a. Binary star systems have high temperature. b. Binary star systems have low density. c. Binary star systems contain a large amount of mass, but because they are orbiting each other, the gravitational field between the two is much less. d. Binary star systems contain a large amount of mass. As a result, the gravitational field between the two is great. Access for free at openstax.org. 23.1 \u2022 The Four Fundamental Forces 777 Accelerators Create Matter From Energy Before looking at all the particles that make up our universe, let us first examine some of the machines that create them. The fundamental process in creating unknown particles is to accelerate known particles, such as protons or electrons, and direct a", " beam of them toward a target. Collisions with target nuclei provide a wealth of information, such as information obtained by Rutherford in the gold foil experiment. If the energy of the incoming particles is large enough, new matter can even be created in the collision. The more energy input or \u0394E, the more matter mcan be created, according to mass energy equivalence. Limitations are placed on what can occur by known conservation laws, such as conservation of mass-energy, momentum, and charge. Even more interesting are the unknown limitations provided by nature. While some expected reactions do occur, others do not, and still other unexpected reactions may appear. New laws are revealed, and the vast majority of what we know about particle physics has come from accelerator laboratories. It is the particle physicist\u2019s favorite indoor sport. Our earliest model of a particle accelerator comes from the Van de Graaff generator. The relatively simple device, which you have likely seen in physics demonstrations, can be manipulated to produce potentials as great as 50 million volts. While these machines do not have energies large enough to produce new particles, analysis of their accelerated ions was instrumental in exploring several aspects of the nucleus. Another equally famous early accelerator is the cyclotron, invented in 1930 by the American physicist, E.O. Lawrence (1901\u20131958). Figure 23.8 is a visual representation with more detail. Cyclotrons use fixed-frequency alternating electric fields to accelerate particles. The particles spiral outward in a magnetic field, making increasingly larger radius orbits during acceleration. This clever arrangement allows the successive addition of electric potential energy with each loop. As a result, greater particle energies are possible than in a Van de Graaff generator. Figure 23.8 On the left is an artist\u2019s rendition of the popular physics demonstration tool, the Van de Graaff generator. A battery (A) supplies excess positive charge to a pointed conductor, the points of which spray the charge onto a moving insulating belt near the bottom. The pointed conductor (B) on top in the large sphere picks up the charge. (The induced electric field at the points is so large that it removes the charge from the belt.) This can be done because the charge does not remain inside the conducting sphere but moves to its outer surface. An ion source inside the sphere produces positive ions, which are accelerated away from the positive sphere to high velocities. On the right is a cyclotron. Cyclotrons use a magnetic field to cause particles to move in circular orbits. As the particles pass between the plates of the Dees, the voltage across the gap is oscillated to accelerate them twice in each orbit. A synchrotron is a modification of the cyclotron in which particles continually travel in a fixed-radius orbit, increasing speed each time. Accelerating voltages are synchronized with the particles to accelerate them, hence the name. Additionally, magnetic field strength is increased to keep the orbital radius constant as energy increases. A ring of magnets and accelerating tubes, as shown in Figure 23.9, are the major components of synchrotrons. High-energy particles require strong magnetic fields to steer 778 Chapter 23 \u2022 Particle Physics them, so superconducting magnets are commonly employed. Still limited by achievable magnetic field strengths, synchrotrons need to be very large at very high energies since the radius of a high-energy particle\u2019s orbit is very large. To further probe the nucleus, physicists need accelerators of greater energy and detectors of shorter wavelength. To do so requires not only greater funding but greater ingenuity as well. Colliding beams used at both the Fermi National Accelerator Laboratory (Fermilab; see Figure 23.11) near Chicago and the LHC in Switzerland are designed to reduce energy loss in particle collisions. Typical stationary particle detectors lose a large amount of energy to the recoiling target struck by the accelerating particle. By providing head-on collisions between particles moving in opposite directions, colliding beams make it possible to create particles with momenta and kinetic energies near zero. This allows for particles of greater energy and mass to be created. Figure 23.10 is a schematic representation of this effect. In addition to circular accelerators, linear accelerators can be used to reduce energy radiation losses. The Stanford Linear Accelerator Center (now called the SLAC National Accelerator Laboratory) in California is home to the largest such accelerator in the world. Figure 23.9 (a) A synchrotron has a ring of magnets and accelerating tubes. The frequency of the accelerating voltages is increased to cause the beam particles to travel the same distance in a shorter time. The magnetic field should also be increased to keep each beam burst traveling in a fixed-radius path. Limits on magnetic field strength require these machines to be very large in order to accelerate particles to very high energies. (b) A positively charged particle is shown in the gap between accelerating tubes. (c) While the particle passes through the tube, the potentials are reversed", " so that there is another acceleration at the next gap. The frequency of the reversals needs to be varied as the particle is accelerated to achieve successive accelerations in each gap. Figure 23.10 This schematic shows the two rings of Fermilab\u2019s accelerator and the scheme for colliding protons and antiprotons (not to scale). Figure 23.11 The Fermi National Accelerator Laboratory, near Batavia, Illinois, was a subatomic particle collider that accelerated protons and antiprotons to attain energies up to 1 Tev (a trillion electronvolts). The circular ponds near the rings were built to dissipate waste heat. This accelerator was shut down in September 2011. (credit: Fermilab, Reidar Hahn) Check Your Understanding 1. Which of the four forces is responsible for radioactive decay? a. the electromagnetic force Access for free at openstax.org. 23.2 \u2022 Quarks 779 b. c. d. the gravitational force the strong nuclear force the weak nuclear force 2. What force or forces exist between an electron and a proton? a. b. c. d. the strong nuclear force, the electromagnetic force, and gravity the weak nuclear force, the strong nuclear force, and gravity the weak nuclear force, the strong nuclear force, and the electromagnetic force the weak nuclear force, the electromagnetic force, and gravity 3. What is the proposed carrier particle for the gravitational force? a. boson b. graviton c. gluon d. photon 4. What is the relationship between the mass and range of a carrier particle? a. Range of a carrier particle is inversely proportional to its mass. b. Range of a carrier particle is inversely proportional to square of its mass. c. Range of a carrier particle is directly proportional to its mass. d. Range of a carrier particle is directly proportional to square of its mass. 5. What type of particle accelerator uses fixed-frequency oscillating electric fields to accelerate particles? cyclotron synchrotron a. b. c. betatron d. Van de Graaff accelerator 6. How does the expanding radius of the cyclotron provide evidence of particle acceleration? a. A constant magnetic force is exerted on particles at all radii. As the radius increases, the velocity of the particle must increase to maintain this constant force. b. A constant centripetal force is exerted on particles at all radii. As the radius increases, the velocity of the particle must decrease to maintain this constant force. c. A constant magnetic force is exerted on particles at all radii. As the radius increases, the velocity of the particle must decrease to maintain this constant force. d. A constant centripetal force is exerted on particles at all radii. As the radius increases, the velocity of the particle must increase to maintain this constant force. 7. Which of the four forces is responsible for the structure of galaxies? a. electromagnetic force b. gravity c. d. weak nuclear force strong nuclear force 23.2 Quarks Section Learning Objectives By the end of the section, you will be able to do the following: \u2022 Describe quarks and their relationship to other particles \u2022 Distinguish hadrons from leptons \u2022 Distinguish matter from antimatter \u2022 Describe the standard model of the atom \u2022 Define a Higgs boson and its importance to particle physics 780 Chapter 23 \u2022 Particle Physics Section Key Terms annihilation antimatter baryon bottom quark charmed quark color down quark flavor gluon hadron Higgs boson Higgs field lepton meson pair production positron quantum chromodynamics quark Standard Model strange quark top quark up quark Quarks \u201cThe first principles of the universe are atoms and empty space. Everything else is merely thought to exist\u2026\u201d \u201c\u2026 Further, the atoms are unlimited in size and number, and they are borne along with the whole universe in a vortex, and thereby generate all composite things\u2014fire, water, air, earth. For even these are conglomerations of given atoms. And it because of their solidity that these atoms are impassive and unalterable.\u201d \u2014Diogenes Laertius (summarizing the views of Democritus, circa 460\u2013370 B.C.) The search for fundamental particles is nothing new. Atomists of the Greek and Indian empires, like Democritus of fifth century B.C., openly wondered about the most finite components of our universe. Though dormant for centuries, curiosity about the atomic nature of matter was reinvigorated by Rutherford\u2019s gold foil experiment and the discovery of the nucleus. By the early 1930s, scientists believed they had fully determined the tiniest constituents of matter\u2014in the form of the proton, neutron, and electron. This would be only partially true. At present, scientists know that there are hundreds of particles not unlike our electron and nucleons, all making up what some", " have termed the particle zoo. While we are confident that the electron remains fundamental, it is surrounded by a plethora of similar sounding terms, like leptons, hadrons, baryons, and mesons. Even though not every particle is considered fundamental, they all play a vital role in understanding the intricate structure of our universe. A fundamental particle is defined as a particle with no substructure and no finite size. According to the Standard Model, there are three types of fundamental particles: leptons, quarks, and carrier particles. As you may recall, carrier particles are responsible for transmitting fundamental forces between their interacting masses. Leptons are a group of six particles not bound by the strong nuclear force, of which the electron is one. As for quarks, they are the fundamental building blocks of a group of particles called hadrons, a group that includes both the proton and the neutron. Now for a brief history of quarks. Quarks were first proposed independently by American physicists Murray Gell-Mann and George Zweig in 1963. Originally, three quark types\u2014or flavors\u2014were proposed with the names up (u), down (d), and strange (s). At first, physicists expected that, with sufficient energy, we should be able to free quarks and observe them directly. However, this has not proved possible, as the current understanding is that the force holding quarks together is incredibly great and, much like a spring, increases in magnitude as the quarks are separated. As a result, when large energies are put into collisions, other particles are created\u2014but no quarks emerge. With that in mind, there is compelling evidence for the existence of quarks. By 1967, experiments at the SLAC National Accelerator Laboratory scattering 20-GeV electrons from protons produced results like Rutherford had obtained for the nucleus nearly 60 years earlier. The SLAC scattering experiments showed unambiguously that there were three point-like (meaning they had sizes considerably smaller than the probe\u2019s wavelength) charges inside the proton as seen in Figure 23.12. This evidence made all but the most skeptical admit that there was validity to the quark substructure of hadrons. Access for free at openstax.org. 23.2 \u2022 Quarks 781 Figure 23.12 Scattering of high-energy electrons from protons at facilities like SLAC produces evidence of three point-like charges consistent with proposed quark properties. This experiment is analogous to Rutherford\u2019s discovery of the small size of the nucleus by scattering \u03b1 particles. High-energy electrons are used so that the probe wavelength is small enough to see details smaller than the proton. The inclusion of the strange quark with Zweig and Gell-Mann\u2019s model concerned physicists. While the up and down quarks demonstrated fairly clear symmetry and were present in common fundamental particles like protons and neutrons, the strange quark did not have a counterpart of its own. This thought, coupled with the four known leptons at the time, caused scientists to predict that a fourth quark, yet to be found, also existed. In 1974, two groups of physicists independently discovered a particle with this new quark, labeled charmed. This completed the second exoticquark pair, strange (s) and charmed (c). A final pair of quarks was proposed when a third pair of leptons was discovered in 1975. The existence of the bottom (b) quark and the top (t) quark was verified through experimentation in 1976 and 1995, respectively. While it may seem odd that so much time would elapse between the original quark discovery in 1967 and the verification of the top quark in 1995, keep in mind that each quark discovered had a progressively larger mass. As a result, each new quark has required more energy to discover. TIPS FOR SUCCESS Note that a very important tenet of science occurred throughout the period of quark discovery. The charmed, bottom, and top quarks were all speculated on, and then were discovered some time later. Each of their discoveries helped to verify and strengthen the quark model. This process of speculation and verification continues to take place today and is part of what drives physicists to search for evidence of the graviton and Grand Unified Theory. One of the most confounding traits of quarks is their electric charge. Long assumed to be discrete, and specifically a multiple of the elementary charge of the electron, the electric charge of an individual quark is fractional and thus seems to violate a presumed tenet of particle physics. The fractional charge of quarks, which are, are the only structures and found in nature with a nonintegral number of charge. However, note that despite this odd construction, the fractional value of the quark does not violate the quantum nature of the charge. After all, free quarks cannot be found in nature, and all quarks are bound into arrangements in which an integer number of charge is constructed.", " Table 23.3 shows the six known quarks, in addition to their antiquark components, as will be discussed later in this section. Flavor Symbol Antiparticle Charge[1][2] Up Down Strange Charmed [1]The lower of the [2]There are further qualities that differentiate between quarks. However, they are beyond the discussion in this text. symbols are the values for antiquarks. Table 23.3 Quarks and Antiquarks 782 Chapter 23 \u2022 Particle Physics Flavor Symbol Antiparticle Charge[1][2] Bottom Top [1]The lower of the [2]There are further qualities that differentiate between quarks. However, they are beyond the discussion in this text. symbols are the values for antiquarks. Table 23.3 Quarks and Antiquarks While the term flavoris used to differentiate between types of quarks, the concept of color is more analogous to the electric charge in that it is primarily responsible for the force interactions between quarks. Note\u2014Take a moment to think about the electrostatic force. It is the electric charge that causes attraction and repulsion. It is the same case here but with a colorcharge. The three colors available to a quark are red, green, and blue, with antiquarks having colors of anti-red (or cyan), anti-green (or magenta), and anti-blue (or yellow). Why use colors when discussing quarks? After all, the quarks are not actually colored with visible light. The reason colors are used is because the properties of a quark are analogous to the three primary and secondary colors mentioned above. Just as different colors of light can be combined to create white, different colorsof quark may be combined to construct a particle like a proton or neutron. In fact, for each hadron, the quarks must combine such that their color sums to white! Recall that two up quarks and one down quark construct a proton, as seen in Figure 23.12. The sum of the three quarks\u2019 colors\u2014red, green, and blue\u2014yields the color white. This theory of color interaction within particles is called quantum chromodynamics, or QCD. As part of QCD, the strong nuclear force can be explained using color. In fact, some scientists refer to the color force, not the strong force, as one of the four fundamental forces. Figure 23.13 is a Feynman diagram showing the interaction between two quarks by using the transmission of a colored gluon. Note that the gluon is also considered the charge carrier for the strong nuclear force. Figure 23.13 The exchange of gluons between quarks carries the strong force and may change the color of the interacting quarks. While the colors of the individual quarks change, their flavors do not. Note that quark flavor may have any color. For instance, in Figure 23.13, the down quark has a red color and a green color. In other words, colors are not specific to a particle quark flavor. Hadrons and Leptons Particles can be revealingly grouped according to what forces they feel between them. All particles (even those that are massless) are affected by gravity since gravity affects the space and time in which particles exist. All charged particles are affected by the electromagnetic force, as are neutral particles that have an internal distribution of charge (such as the neutron with its magnetic moment). Special names are given to particles that feel the strong and weak nuclear forces. Hadrons are particles that feel the strong nuclear force, whereas leptons are particles that do not. All particles feel the weak nuclear force. This means that hadrons are distinguished by being able to feel both the strong and weak nuclear forces. Leptons and hadrons are distinguished in other ways as well. Leptons are fundamental particles that have no measurable size, while hadrons are composed of quarks and have a diameter on the order of 10 to 15 m. Six particles, including the electron and neutrino, make up the list of known leptons. There are hundreds of complex particles in the hadron class, a few of which (including the proton and neutron) are listed in Table 23.4. Access for free at openstax.org. Category Leptons Hadrons \u2013 Mesons[2] Hadrons \u2013 Baryons[3] Particle Name Electron Neutrino (e) Muon Neutrino (\u03bc) Tau Neutrino (\u03c4) Pion Kaon Eta Proton Neutron Lambda Omega 23.2 \u2022 Quarks 783 Symbol Antiparticle Rest Mass Mean Lifetime (s) [1] [1] Stable Stable Stable [1] Stable Stable 882 0.511 105.7 1,777 139.6 135.0 493.7 497.6 547.9 938.3 939.6 1,115.7 1", ",672.5 Self Self p n [1]Neutrino masses may be zero. Experimental upper limits are given in parentheses. [2]Many other mesons known [3]Many other baryons known Table 23.4 List of Leptons and Hadrons. There are many more leptons, mesons, and baryons yet to be discovered and measured. The purpose of trying to uncover the smallest indivisible things in existence is to explain the world around us through forces and the interactions between particles, galaxies and objects. This is why a handful of scientists devote their life\u2019s work to smashing together small particles. What internal structure makes a proton so different from an electron? The proton, like all hadrons, is made up of quarks. A few examples of hadron quark composition can be seen in Figure 23.14. As shown, each hadron is constructed of multiple quarks. As mentioned previously, the fractional quark charge in all four hadrons sums to the particle\u2019s integral value. Also, notice that the color composition for each of the four particles adds to white. Each of the particles shown is constructed of up, down, and their antiquarks. This is not surprising, as the quarks strange, charmed, top, and bottom are found in only our most exotic particles. 784 Chapter 23 \u2022 Particle Physics Figure 23.14 All baryons, such as the proton and neutron shown here, are composed of three quarks. All mesons, such as the pions shown here, are composed of a quark\u2013antiquark pair. Arrows represent the spins of the quarks. The colors are such that they need to add to white for any possible combination of quarks. You may have noticed that while the proton and neutron in Figure 23.14 are composed of three quarks, both pions are comprised of only two quarks. This refers to a final delineation in particle structure. Particles with three quarks are called baryons. These are heavy particles that can decay into another baryon. Particles with only two quarks\u2014a-quark\u2013anti-quark pair\u2014are called mesons. These are particles of moderate mass that cannot decay into the more massive baryons. Before continuing, take a moment to view Figure 23.15. In this figure, you can see the strong force reimagined as a color force. The particles interacting in this figure are the proton and neutron, just as they were in Figure 23.6. This reenvisioning of the strong force as an interaction between colored quarks is the critical concept behind quantum chromodynamics. Figure 23.15 This Feynman diagram shows the interaction between a proton and a neutron, corresponding to the interaction shown in Figure 23.6. This diagram, however, shows the quark and gluon details of the strong nuclear force interaction. Matter and Antimatter Antimatter was first discovered in the form of the positron, the positively charged electron. In 1932, American physicist Carl Anderson discovered the positron in cosmic ray studies. Through a cloud chamber modified to curve the trajectories of cosmic Access for free at openstax.org. rays, Anderson noticed that the curves of some particles followed that of a negative charge, while others curved like a positive charge. However, the positive curve showed not the mass of a proton but the mass of an electron. This outcome is shown in Figure 23.16 and suggests the existence of a positively charged version of the electron, created by the destruction of solar photons. 23.2 \u2022 Quarks 785 Figure 23.16 The image above is from the Fermilab 15 foot bubble chamberand shows the production of an electron and positron (or antielectron) from an incident photon. This event is titled pair production and provides evidence of antimatter, as the two repel each other. Antimatter is considered the opposite of matter. For most antiparticles, this means that they share the same properties as their original particles with the exception of their charge. This is why the positron can be considered a positive electron while the antiproton is considered a negative proton. The idea of an opposite charge for neutral particles (like the neutron) can be confusing, but it makes sense when considered from the quark perspective. Just as the neutron is composed of one up quark and two down quarks (of charge, respectively), the antineutron is composed of one anti\u2013up quark and two anti\u2013down and quarks (of charge and, respectively). While the overall charge of the neutron remains the same, its constituent particles do not! A word about antiparticles: Like regular particles, antiparticles could function just fine on their own. In fact, a universe made up of antimatter may operate just as our own matter-based universe does. However, we do not know fully whether this is the", " case. The reason for this is annihilation. Annihilation is the process of destruction that occurs when a particle and its antiparticle interact. As soon as two particles (like a positron and an electron) coincide, they convert their masses to energy through the equation makes it very difficult for scientists to study antimatter. That said, scientists have had success creating antimatter through highenergy particle collisions. Both antineutrons and antiprotons were created through accelerator experiments in 1956, and an anti\u2013hydrogen atom was even created at CERN in 1995! As referenced in, the annihilation of antiparticles is currently used in medical studies to determine the location of radioisotopes.. This mass-to-energy conversion, which typically results in photon release, happens instantaneously and Completing the Standard Model of the Atom The Standard Model of the atom refers to the current scientific view of the fundamental components and interacting forces of matter. The Standard Model (Figure 23.17) shows the six quarks that bind to form all hadrons, the six lepton particles already considered fundamental, the four carrier particles (or gauge bosons) that transmit forces between the leptons and quarks, and the recently added Higgs boson (which will be discussed shortly). This totals 17 fundamental particles, combinations of which are responsible for all known matter in our entire universe! When adding the antiquarks and antileptons, 31 components make up the Standard Model. 786 Chapter 23 \u2022 Particle Physics Figure 23.17 The Standard Model of elementary particles shows an organized view of all fundamental particles, as currently known: six quarks, six leptons, and four gauge bosons (or carrier particles). The Higgs boson, first observed in 2012, is a new addition to the Standard Model. Figure 23.17 shows all particles within the Standard Model of the atom. Not only does this chart divide all known particles by color-coded group, but it also provides information on particle stability. Note that the color-coding system in this chart is separate from the red, green, and blue color labeling system of quarks. The first three columns represent the three familiesof matter. The first column, considered Family 1, represents particles that make up normal matter, constructing the protons, neutrons, and electrons that make up the common world. Family 2, represented from the charm quark to the muon neutrino, is comprised of particles that are more massive. The leptons in this group are less stable and more likely to decay. Family 3, represented by the third column, are more massive still and decay more quickly. The order of these families also conveniently represents the order in which these particles were discovered. TIPS FOR SUCCESS Look for trends that exist within the Standard Model. Compare the charge of each particle. Compare the spin. How does mass relate to the model structure? Recognizing each of these trends and asking questions will yield more insight into the organization of particles and the forces that dictate particle relationships. Our understanding of the Standard Model is still young, and the questions you may have in analyzing the Standard Model may be some of the same questions that particle physicists are searching for answers to today! The Standard Model also summarizes the fundamental forces that exist as particles interact. A closer look at the Standard Model, as shown in Figure 23.18, reveals that the arrangement of carrier particles describes these interactions. Access for free at openstax.org. 23.2 \u2022 Quarks 787 Figure 23.18 The revised Standard Model shows the interaction between gauge bosons and other fundamental particles. These interactions are responsible for the fundamental forces, three of which are described through the chart\u2019s shaded areas. Each of the shaded areas represents a fundamental force and its constituent particles. The red shaded area shows all particles involved in the strong nuclear force, which we now know is due to quantum chromodynamics. The blue shaded area corresponds to the electromagnetic force, while the green shaded area corresponds to the weak nuclear force, which affects all quarks and leptons. The electromagnetic force and weak nuclear force are considered united by the electroweak force within the Standard Model. Also, because definitive evidence of the graviton is yet to be found, it is not included in the Standard Model. The Higgs Boson One interesting feature of the Standard Model shown in Figure 23.18 is that, while the gluon and photon have no mass, the Z and W bosons are very massive. What supplies these quickly moving particles with mass and not the gluons and photons? Furthermore, what causes some quarks to have more mass than others? In the 1960s, British physicist Peter Higgs and others speculated that the W and Z bosons were actually just as massless as the gluon and photon. However, as the W and Z bosons traveled from one particle to another, they were slowed down by the presence of a Higgs field, much like a fish swimming through water. The thinking was that", " the existence of the Higgs field would slow down the bosons, causing them to decrease in energy and thereby transfer this energy to mass. Under this theory, all particles pass through the Higgs field, which exists throughout the universe. The gluon and photon travel through this field as well but are able to do so unaffected. The presence of a force from the Higgs field suggests the existence of its own carrier particle, the Higgs boson. This theorized boson interacts with all particles but gluons and photons, transferring force from the Higgs field. Particles with large mass (like the top quark) are more likely to receive force from the Higgs boson. While it is difficult to examine a field, it is somewhat simpler to find evidence of its carrier. On July 4, 2012, two groups of scientists at the LHC independently confirmed the existence of a Higgs-like particle. By examining trillions of proton\u2013proton collisions at energies of 7 to 8 TeV, LHC scientists were able to determine the constituent particles that created the protons. In this data, scientists found a particle with similar mass, spin, parity, and interactions with other particles that matched the Higgs boson predicted decades prior. On March 13, 2013, the existence of the Higgs boson was tentatively confirmed by CERN. Peter Higgs and Francois Englert received the Nobel Prize in 2013 for the \u201ctheoretical discovery of a mechanism that contributes to our understanding of the origin and mass of subatomic particles.\u201d WORK IN PHYSICS Particle Physicist If you have an innate desire to unravel life\u2019s great mysteries and further understand the nature of the physical world, a career in particle physics may be for you! Particle physicists have played a critical role in much of society\u2019s technological progress. From lasers to computers, televisions to space missions, splitting the atom to understanding the DNA molecule to MRIs and PET scans, much of our modern society is based on the work done by particle physicists. While many particle physicists focus on specialized tasks in the fields of astronomy and medicine, the main goal of particle physics is to further scientists\u2019 understanding of the Standard Model. This may mean work in government, industry, or 788 Chapter 23 \u2022 Particle Physics academics. Within the government, jobs in particle physics can be found within the National Institute for Standards and Technology, Department of Energy, NASA, and Department of Defense. Both the electronics and computer industries rely on the expertise of particle physicists. College teaching and research positions can also be potential career opportunities for particle physicists, though they often require some postgraduate work as a prerequisite. In addition, many particle physicists are employed to work on high-energy colliders. Domestic collider labs include the Brookhaven National Laboratory in New York, the Fermi National Accelerator Laboratory near Chicago, and the SLAC National Accelerator Laboratory operated by Stanford University. For those who like to travel, work at international collider labs can be found at the CERN facility in Switzerland in addition to institutes like the Budker Institute of Nuclear Physics in Russia, DESY in Germany, and KEK in Japan. Shirley Jackson became the first African American woman to earn a Ph.D. from MIT back in 1973, and she went on to lead a highly successful career in the field of particle physics. Like Dr. Jackson, successful students of particle physics grow up with a strong curiosity in the world around them and a drive to continually learn more. If you are interested in exploring a career in particle physics, work to achieve good grades and SAT scores, and find time to read popular books on physics topics that interest you. While some math may be challenging, recognize that this is only a tool of physics and should not be considered prohibitive to the field. High-level work in particle physics often requires a Ph.D.; however, it is possible to find work with a master\u2019s degree. Additionally, jobs in industry and teaching can be achieved with solely an undergraduate degree. GRASP CHECK What is the primary goal of all work in particle physics? a. The primary goal is to further our understanding of the Standard Model. b. The primary goal is to further our understanding of Rutherford\u2019s model. c. The primary goal is to further our understanding of Bohr\u2019s model. d. The primary goal is to further our understanding of Thomson\u2019s model. Check Your Understanding 8. In what particle were quarks originally discovered? a. b. c. d. the electron the neutron the proton the photon 9. Why was the existence of the charm quark speculated, even though no direct evidence of it existed? a. The existence of the charm quark was symmetrical with up and down quarks. Additionally, there were two known leptons at the time and only two quarks. b. The strange particle lacked the symmetry that existed with the up and down quarks. Additionally, there were", " four known leptons at the time and only three quarks. c. The bottom particle lacked the symmetry that existed with the up and down quarks. Additionally, there were two known leptons at the time and only two quarks. d. The existence of charm quarks was symmetrical with up and down quarks. Additionally, there were four known leptons at the time and only three quarks. 10. What type of particle is the electron? a. The electron is a lepton. b. The electron is a hadron. c. The electron is a baryon. d. The electron is an antibaryon. 11. How do the number of fundamental particles differ between hadrons and leptons? a. Hadrons are constructed of at least three fundamental quark particles, while leptons are fundamental particles. b. Hadrons are constructed of at least three fundamental quark particles, while leptons are constructed of two fundamental particles. c. Hadrons are constructed of at least two fundamental quark particles, while leptons are constructed of three Access for free at openstax.org. 23.2 \u2022 Quarks 789 fundamental particles. d. Hadrons are constructed of at least two fundamental quark particles, while leptons are fundamental particles. 12. Does antimatter exist? a. no b. yes 13. How does the deconstruction of a photon into an electron and a positron uphold the principles of mass and charge conservation? a. The sum of the masses of an electron and a positron is equal to the mass of the photon before pair production. The sum of the charges on an electron and a positron is equal to the zero charge of the photon. b. The sum of the masses of an electron and a positron is equal to the mass of the photon before pair production. The sum of the same charges on an electron and a positron is equal to the charge on a photon. c. During the particle production the total energy of the photon is converted to the mass of an electron and a positron. The sum of the opposite charges on the electron and positron is equal to the zero charge of the photon. d. During particle production, the total energy of the photon is converted to the mass of an electron and a positron. The sum of the same charges on an electron and a positron is equal to the charge on a photon. 14. How many fundamental particles exist in the Standard Model, including the Higgs boson and the graviton (not yet observed)? 12 a. 15 b. 13 c. 19 d. 15. Why do gluons interact only with particles in the first two rows of the Standard Model? a. The leptons in the third and fourth rows do not have mass, but the gluons can interact between the quarks through gravity only. b. The leptons in the third and fourth rows do not have color, but the gluons can interact between quarks through color interactions only. c. The leptons in the third and fourth rows do not have spin, but the gluons can interact between quarks through spin interactions only. d. The leptons in the third and fourth rows do not have charge, but the gluons can interact between quarks through charge interactions only. 16. What fundamental property is provided by particle interaction with the Higgs boson? charge a. b. mass spin c. color d. 17. Considering the Higgs field, what differentiates more massive particles from less massive particles? a. More massive particles interact more with the Higgs field than the less massive particles. b. More massive particles interact less with the Higgs field than the less massive particles. 18. What particles were launched into the proton during the original discovery of the quark? a. bosons b. electrons c. neutrons d. photons 790 Chapter 23 \u2022 Particle Physics 23.3 The Unification of Forces Section Learning Objectives By the end of the section, you will be able to do the following: \u2022 Define a grand unified theory and its importance \u2022 Explain the evolution of the four fundamental forces from the Big Bang onward \u2022 Explain how grand unification theories can be tested Section Key Terms Big Bang Inflationary Epoch Electroweak Epoch electroweak theory Grand Unification Epoch Grand Unified Theory Planck Epoch Quark Era superforce Theory of Everything Understanding the Grand Unified Theory Present quests to show that the four basic forces are different manifestations of a single unified force that follow a long tradition. In the nineteenth century, the distinct electric and magnetic forces were shown to be intimately connected and are now collectively called the electromagnetic force. More recently, the weak nuclear force was united with the electromagnetic force. As shown in Figure 23.19, carrier particles transmit three of the four fundamental forces in very similar ways. With these considerations in mind, it is natural to suggest that a theory may be constructed in which the strong nuclear, weak nuclear", ", and electromagnetic forces are all unified. The search for a correct theory linking the forces, called the Grand Unified Theory (GUT), is explored in this section. In the 1960s, the electroweak theory was developed by Steven Weinberg, Sheldon Glashow, and Abdus Salam. This theory proposed that the electromagnetic and weak nuclear forces are identical at sufficiently high energies. At lower energies, like those in our present-day universe, the two forces remain united but manifest themselves in different ways. One of the main consequences of the electroweak theory was the prediction of three short-range carrier particles, now known as the and and that of the predicted characteristics, including masses having those predicted values as given in. boson was predicted to be 90 GeV/c2. In 1983, these carrier particles were observed at CERN with the bosons. Not only were three particles predicted, but the mass of each boson was predicted to be 81 GeV/c2, and How can forces be unified? They are definitely distinct under most circumstances. For example, they are carried by different particles and have greatly different strengths. But experiments show that at extremely short distances and at extremely high energies, the strengths of the forces begin to become more similar, as seen in Figure 23.20. Figure 23.19 The exchange of a virtual particle (boson) carries the weak nuclear force between an electron and a neutrino in this Feynman diagram. This diagram is similar to the diagrams in Figure 23.6 and for the electromagnetic and strong nuclear forces. As discussed earlier, the short ranges and large masses of the weak carrier bosons require correspondingly high energies to create them. Thus, the energy scale on the horizontal axis of Figure 23.20 also corresponds to shorter and shorter distances Access for free at openstax.org. 23.3 \u2022 The Unification of Forces 791 (going from left to right), with 100 GeV corresponding to approximately 10\u221218 m, for example. At that distance, the strengths of the electromagnetic and weak nuclear forces are the same. To test this, energies of about 100 GeV are put into the system. When this occurs, the, and carrier particles become less and less relevant, and the further energy is added, the similar to photons and gluons. carrier particles are created and released. At those and higher energies, the masses of the boson in particular resembles the massless, chargeless photon. As particles are further transformed into massless carrier particles even more, and,, Figure 23.20 The relative strengths of the four basic forces vary with distance, and, hence, energy is needed to probe small distances. At ordinary energies (a few eV or less), the forces differ greatly. However, at energies available in accelerators, the weak nuclear and electromagnetic (EM) forces become unified. Unfortunately, the energies at which the strong nuclear and electroweak forces become the same are unreachable in any conceivable accelerator. The universe may provide a laboratory, and nature may show effects at ordinary energies that give us clues about the validity of this graph. The extremely short distances and high energies at which the electroweak force becomes identical with the strong nuclear force are not reachable with any conceivable human-built accelerator. At energies of about 1014 GeV (16,000 J per particle), distances of about 10 to 30 m can be probed. Such energies are needed to test the theory directly, but these are about 1010 times higher than the maximum energy associated with the LHC, and the distances are about 10 to 12 smaller than any structure we have direct knowledge of. This would be the realm of various GUTs, of which there are many, since there is no constraining evidence at these energies and distances. Past experience has shown that anytime you probe so many orders of magnitude further, you find the unexpected. While direct evidence of a GUT is not presently possible, that does not rule out the ability to assess a GUT through an indirect process. Current GUTs require various other events as a consequence of their theory. Some GUTs require the existence of magnetic monopoles, very massive individual north- and south-pole particles, which have not yet been proven to exist, while others require the use of extra dimensions. However, not all theories result in the same consequences. For example, disproving the existence of magnetic monopoles will not disprove all GUTs. Much of the science we accept in our everyday lives is based on different models, each with their own strengths and limitations. Although a particular model may have drawbacks, that does not necessarily mean that it should be discounted completely. One consequence of GUTs that can theoretically be assessed is proton decay. Multiple current GUTs hypothesize that the stable proton should actually decay at a lifetime of 1031 years. While this time is incredibly large (keep in mind that the age of the universe is less than 14 billion years), scientists at the Super-Kamiokande in Japan have used", " a 50,000-ton tank of water to search for its existence. The decay of a single proton in the Super-Kamiokande tank would be observed by a detector, thereby providing support for the predicting GUT model. However, as of 2014, 17 years into the experiment, decay is yet to be found. This time span equates to a minimum limit on proton life of grand unifying theories, an acceptable model may still exist. years. While this result certainly does not support many TIPS FOR SUCCESS The Super-Kamiokande experiment is a clever use of proportional reasoning. Because it is not feasible to test for 1031 years in order for a single proton to decay, scientists chose instead to manipulate the proton\u2013time ratio. If one proton decays in 1031 792 Chapter 23 \u2022 Particle Physics years, then in one year 10\u221231 protons will decay. With this in mind, if scientists wanted to test the proton decay theory in one year, they would need 1031 protons. While this is also unfeasible, the use of a 50,000-ton tank of water helps to bring both the wait time and proton number to within reason. The Standard Model and the Big Bang Nature is full of examples where the macroscopic and microscopic worlds intertwine. Newton realized that the nature of gravity on Earth that pulls an apple to the ground could explain the motion of the moon and planets so much farther away. Decays of tiny nuclei explain the hot interior of the Earth. Fusion of nuclei likewise explains the energy of stars. Today, the patterns in particle physics seem to be explaining the evolution and character of the universe. And the nature of the universe has implications for unexplored regions of particle physics. In 1929, Edwin Hubble observed that all but the closest galaxies surrounding our own had a red shift in their hydrogen spectra that was proportional to their distance from us. Applying the Doppler Effect, Hubble recognized that this meant that all galaxies were receding from our own, with those farther away receding even faster. Knowing that our place in the universe was no more unique than any other, the implication was clear: The space within the universe itself was expanding. Just like pen marks on an expanding balloon, everything in the universe was accelerating away from everything else. Figure 23.21 shows how the recession of galaxies looks like the remnants of a gigantic explosion, the famous Big Bang. Extrapolating backward in time, the Big Bang would have occurred between 13 and 15 billion years ago, when all matter would have been at a single point. From this, questions instantly arise. What caused the explosion? What happened before the Big Bang? Was there a before, or did time start then? For our purposes, the biggest question relating to the Big Bang is this: How does the Big Bang relate to the unification of the fundamental forces? Figure 23.21 Galaxies are flying apart from one another, with the more distant ones moving faster, as if a primordial explosion expelled the matter from which they formed. The most distant known galaxies move nearly at the speed of light relative to us. To fully understand the conditions of the very early universe, recognize that as the universe contracts to the size of the Big Bang, changes will occur. The density and temperature of the universe will increase dramatically. As particles become closer together, they will become too close to exist as we know them. The high energies will create other, more unusual particles to exist in greater abundance. Knowing this, let\u2019s move forward from the start of the universe, beginning with the Big Bang, as illustrated in Figure 23.22. Access for free at openstax.org. 23.3 \u2022 The Unification of Forces 793 Figure 23.22 The evolution of the universe from the Big Bang onward (from left to right) is intimately tied to the laws of physics, especially those of particle physics at the earliest stages. Theories of the unification of forces at high energies may be verified by their shaping of the universe and its evolution. The Planck Epoch \u2014Though scientists are unable to model the conditions of the Planck Epoch in the laboratory, GeV necessary to unify gravity speculation is that at this time compressed energy was great enough to reach the immense with all other forces. As a result, modern cosmology suggests that all four forces would have existed as one force, a hypothetical superforce as suggested by the Theory of Everything. The Grand Unification Epoch \u2014As the universe expands, the temperatures necessary to maintain the superforce decrease. As a result, gravity separates, leaving the electroweak and strong nuclear forces together. At this time, the electromagnetic, weak, and strong forces are identical, matching the conditions requested in the Grand Unification Theory. The Inflationary Epoch \u2014The separation of the strong nuclear force from the electroweak force during this time is thought to have been responsible for the massive inflation of the universe. Corresponding to the steep diagonal line", " on the left side of Figure 23.22, the universe may have expanded by a factor of great during this time that it actually occurred faster than the speed of light! Unfortunately, there is little hope that we may be GeV, vastly greater than the limits of modern able to test the inflationary scenario directly since it occurs at energies near accelerators. or more in size. In fact, the expansion was so The Electroweak Epoch \u2014Now separated from both gravity and the strong nuclear force, the electroweak force exists as a singular force during this time period. As stated earlier, scientists are able to create the energies at this stage in the universe\u2019s expansion, needing only 100 GeV, as shown in Figure 23.20. W and Z bosons, as well as the Higgs boson, are released during this time. The Quark Era \u2014During the Quark Era, the universe has expanded and temperatures have decreased to the 794 Chapter 23 \u2022 Particle Physics point at which all four fundamental forces have separated. Additionally, quarks began to take form as energies decreased. As the universe expanded, further eras took place, allowing for the existence of hadrons, leptons, and photons, the fundamental particles of the standard model. Eventually, in nucleosynthesis, nuclei would be able to form, and the basic building blocks of atomic matter could take place. Using particle accelerators, we are very much working backwards in an attempt to understand the universe. It is encouraging to see that the macroscopic conditions of the Big Bang align nicely with our submicroscopic particle theory. Check Your Understanding 19. Is there one grand unified theory or multiple grand unifying theories? a. one grand unifying theory b. multiple grand unifying theories 20. In what manner is considered a precursor to the Grand Unified Theory? a. The grand unified theory seeks relate the electroweak and strong nuclear forces to one another just as related energy and mass. b. The grand unified theory seeks to relate the electroweak force and mass to one another just as related energy and mass. c. The grand unified theory seeks to relate the mass and strong nuclear forces to one another just as related energy and mass. d. The grand unified theory seeks to relate gravity and strong nuclear force to one another, just as related energy and mass. 21. List the following eras in order of occurrence from the Big Bang: Electroweak Epoch, Grand Unification Epoch, Inflationary Epoch, Planck Epoch, Quark Era. a. Quark Era, Grand Unification Epoch, Inflationary Epoch, Electroweak Epoch, Planck Epoch b. Planck Epoch, Inflationary Epoch, Grand Unification Epoch, Electroweak Epoch, Quark Era c. Planck Epoch, Electroweak Epoch, Grand Unification Epoch, Inflationary Epoch, Quark Era d. Planck Epoch, Grand Unification Epoch, Inflationary Epoch, Electroweak Epoch, Quark Era 22. How did the temperature of the universe change as it expanded? a. The temperature of the universe increased. b. The temperature of the universe decreased. c. The temperature of the universe first decreased and then increased. d. The temperature of the universe first increased and then decreased. 23. Under current conditions, is it possible for scientists to use particle accelerators to verify the Grand Unified Theory? a. No, there is not enough energy. b. Yes, there is enough energy. 24. Why are particles and antiparticles made to collide as shown in this image? a. Particles and antiparticles have the same mass. b. Particles and antiparticles have different mass. c. Particles and antiparticles have the same charge. d. Particles and antiparticles have opposite charges. 25. The existence of what particles were predicted as a consequence of the electroweak theory? a. fermions b. Higgs bosons Access for free at openstax.org. leptons c. d. W+, W-, and Z0 bosons 23.3 \u2022 The Unification of Forces 795 796 Chapter 23 \u2022 Key Terms KEY TERMS boson positive carrier particle of the weak nuclear force carrier particles are identical under certain circumstances boson negative carrier particle of the weak nuclear Higgs field the field through which all fundamental force boson neutral carrier particle of the weak nuclear force annihilation the process of destruction that occurs when a particle and antiparticle interact antimatter matter constructed of antiparticles; antimatter shares most of the same properties of regular matter, with charge being the only difference between many particles and their antiparticle analogues baryon hadrons that always decay to another baryon Big Bang a gigantic explosion that threw out matter a few billion years ago bottom quark a quark flavor carrier particle a virtual particle exchanged in the transmission of a fundamental force particles travel that provides them varying mass through the transport of the Higgs boson Inflationary", " Epoch the rapid expansion of the universe by an incredible factor of 10\u221250 for the brief time from 10\u221235 to about 10\u221232 seconds lepton fundamental particles that do not feel the nuclear strong force meson hadrons that can decay to leptons and leave no hadrons pair production the creation of a particle and antiparticle, commonly an electron and positron, due to the annihilation of a photon particle physics the study of and the quest for those truly fundamental particles having no substructure charmed quark a quark flavor, which is the counterpart of pion particle exchanged between nucleons, transmitting the strange quark colliding beam head-on collisions between particles the strong nuclear force between them Planck Epoch the earliest era of the universe, before 10\u201343 moving in opposite directions seconds after the Big Bang color a property of quarks the relates to their interactions positron a particle of antimatter that has the properties of through the strong force a positively charged electron cyclotron accelerator that uses fixed-frequency alternating electric fields and fixed magnets to accelerate particles in a circular spiral path down quark the second lightest of all quarks Electroweak Epoch the stage before 10\u221211 back to 10\u221234 seconds after the Big Bang quantum chromodynamics the theory of color interaction between quarks that leads to understanding of the nuclear strong force quantum electrodynamics on the particle scale the theory of electromagnetism quark an elementary particle and fundamental constituent electroweak theory theory showing connections between of matter that is a substructure of hadrons EM and weak forces Feynman diagram a graph of time versus position that describes the exchange of virtual particles between subatomic particles flavor quark type gluons exchange particles of the nuclear strong force Grand Unification Epoch the time period from 10\u221243 to 10\u221234 seconds after the Big Bang, when Grand Unification Theory, in which all forces except gravity are identical, governed the universe Grand Unified Theory theory that shows unification of the strong and electroweak forces graviton hypothesized particle exchanged between two particles of mass, transmitting the gravitational force between them hadron particles composed of quarks that feel the strong and weak nuclear force Quark Era the time period from 10\u201311 to 10\u20136 seconds at which all four fundamental forces are separated and quarks begin to exit Standard Model an organization of fundamental particles and forces that is a result of quantum chromodynamics and electroweak theory strange quark the third lightest of all quarks superforce the unification of all four fundamental forces into one force synchrotron a version of a cyclotron in which the frequency of the alternating voltage and the magnetic field strength are increased as the beam particles are accelerated Theory of Everything the theory that shows unification of all four fundamental forces top quark a quark flavor up quark the lightest of all quarks weak nuclear force fundamental force responsible for Higgs boson a massive particle that provides mass to the weak bosons and provides validity to the theory that particle decay Access for free at openstax.org. SECTION SUMMARY 23.1 The Four Fundamental Forces \u2022 The four fundamental forces are gravity, the electromagnetic force, the weak nuclear force, and the strong nuclear force. \u2022 A variety of particle accelerators have been used to explore the nature of subatomic particles and to test predictions of particle theories. 23.2 Quarks \u2022 There are three types of fundamental particles\u2014leptons, quarks, and carrier particles. \u2022 Quarks come in six flavors and three colors and occur only in combinations that produce white. \u2022 Hadrons are thought to be composed of quarks, with baryons having three quarks and mesons having a quark and an antiquark. \u2022 Known particles can be divided into three major groups\u2014leptons, hadrons, and carrier particles (gauge bosons). \u2022 All particles of matter have an antimatter counterpart that has the opposite charge and certain other quantum CHAPTER REVIEW Concept Items 23.1 The Four Fundamental Forces 1. What forces does the inverse square law describe? the electromagnetic and weak nuclear force the electromagnetic force and strong nuclear force the electromagnetic force and gravity the strong nuclear force and gravity a. b. c. d. 2. Do the carrier particles explain the loss of mass in nuclear decay? a. no b. yes 3. What happens to the rate of voltage oscillation within a synchrotron each time the particle completes a loop? a. The rate of voltage oscillation increases as the particle travels faster and faster on each loop. b. The rate of voltage oscillation decreases as the particle travels faster and faster on each loop. c. The rate of voltage oscillation remains the same each time the particle completes a loop. d. The rate of voltage oscillation first increases and then remains constant each time the particle completes a loop. 4. Which of the four forces is responsible for ionic bonding? a. electromagnetic force Chapter 23 \u2022 Section Summary 797 numbers. These matter\u2013antimatter pairs are otherwise very similar but will annihilate when brought", " together. \u2022 The strong force is carried by eight proposed particles called gluons, which are intimately connected to a quantum number called color\u2014their governing theory is thus called quantum chromodynamics (QCD). Taken together, QCD and the electroweak theory are widely accepted as the Standard Model of particle physics. 23.3 The Unification of Forces \u2022 Attempts to show unification of the four forces are called Grand Unified Theories (GUTs) and have been partially successful, with connections proven between EM and weak forces in electroweak theory. \u2022 Unification of the strong force is expected at such high energies that it cannot be directly tested, but it may have observable consequences in the as-yet-unobserved decay of the proton. Although unification of forces is generally anticipated, much remains to be done to prove its validity. b. gravity c. strong force d. weak nuclear force 5. What type of particle accelerator uses oscillating electric fields to accelerate particles around a fixed radius track? a. LINAC b. c. SLAC d. Van de Graaff accelerator synchrotron 23.2 Quarks 6. How does the charge of an individual quark determine hadron structure? a. Since the hadron must have an integral value, the individual quarks must be combined such that the average of their charges results in the value of a quark. b. Since the hadron must have an integral value, the individual atoms must be combined such that the sum of their charges is less than zero. c. The individual quarks must be combined such that the product of their charges is equal to the total charge of the hadron structure. d. Since the hadron must have an integral value of charge, the individual quarks must be combined such that the sum of their charges results in an 798 Chapter 23 \u2022 Chapter Review integral value. 7. Why do leptons not feel the strong nuclear force? a. Gluons are the carriers of the strong nuclear force that interacts between quarks through color interactions, but leptons are constructed of quarks that do not have gluons. b. Gluons are the carriers of the strong nuclear force that interacts between quarks through mass interactions, but leptons are not constructed of quarks and are not massive. c. Gluons are the carriers of the strong nuclear force that interacts between quarks through mass interactions, but leptons are constructed of the quarks that are not massive. d. Gluons are the carriers of the strong nuclear force that interacts between quarks through color interactions, but leptons are not constructed of quarks, nor do they have color constituents. 8. What property commonly distinguishes antimatter from its matter analogue? a. mass b. charge c. energy speed d. 9. Can the Standard Model change as new information is gathered? a. yes b. no account for the remaining excess mass in protons compared to electrons. b. The highly energetic photons connecting the quarks account for the remaining excess mass in protons compared to electrons. c. The antiparallel orientation of the quarks present in a proton accounts for the remaining excess mass in protons compared to electrons. d. The parallel orientation of the quarks present in a proton accounts for the remaining excess mass in protons compared to electrons. 23.3 The Unification of Forces 13. Why is the unification of fundamental forces important? a. The unification of forces will help us understand fundamental structures of the universe. b. The unification of forces will help in the proof of the graviton. c. The unification of forces will help in achieving a speed greater than the speed of light. d. The unification of forces will help in studying antimatter particles. 14. Why are scientists unable to model the conditions of the universe at time periods shortly after the Big Bang? a. The amount of energy necessary to replicate the Planck Epoch is too high. b. The amount of energy necessary to replicate the Planck Epoch is too low. c. The volume of setup necessary to replicate the 10. What is the relationship between the Higgs field and the Planck Epoch is too high. Higgs boson? a. The Higgs boson is the carrier that transfers force d. The volume of setup necessary to replicate the Planck Epoch is too low. for the Higgs field. b. The Higgs field is the time duration over which the Higgs particles transfer force to the other particles. c. The Higgs field is the magnitude of momentum transferred by the Higgs particles to the other particles. 15. What role does proton decay have in the search for GUTs? a. Proton decay is a premise of a number of GUTs. b. Proton decay negates the validity of a number of GUTs. d. The Higgs field is the magnitude of torque transfers 16. What is the name for the theory of unification of all four by the Higgs", " particles on the other particles. 11. What were the original three flavors of quarks discovered? a. up, down, and charm b. up, down, and bottom c. up, down, and strange d. up, down, and top 12. Protons are more massive than electrons. The three quarks in the proton account for only a small amount of this mass difference. What accounts for the remaining excess mass in protons compared to electrons? a. The highly energetic gluons connecting the quarks fundamental forces? a. b. c. d. the theory of everything the theory of energy-to-mass conversion the theory of relativity the theory of the Big Bang 17. Is it easier for scientists to find evidence for the Grand Unified Theory or the Theory of Everything? Explain. a. Theory of Everything, because it requires of energy b. Theory of Everything, because it requires of energy c. Grand Unified Theory, because it requires Access for free at openstax.org. Chapter 23 \u2022 Chapter Review 799 of energy d. Grand Unified Theory, because it requires of energy Critical Thinking Items 23.1 The Four Fundamental Forces 18. The gravitational force is considered a very weak force. Yet, it is strong enough to hold Earth in orbit around the Sun. Explain this apparent disparity. a. At the level of the Earth-to-Sun distance, gravity is the strongest acting force because neither the strong nor the weak nuclear force exists at this distance. b. At the level of the Earth-to-Sun distance, gravity is the strongest acting force because both the strong and the weak nuclear force is minimal at this distance. 19. True or False\u2014Given that their carrier particles are massless, some may argue that the electromagnetic and gravitational forces should maintain the same value at all distances from their source. However, both forces decrease with distance at a rate of a. b. false true 20. Why is a stationary target considered inefficient in a particle accelerator? a. The stationary target recoils upon particle strike, thereby transferring much of the particle\u2019s energy into its motion. As a result, a greater amount of energy goes into breaking the particle into its constituent components. b. The stationary target contains zero kinetic energy, so it requires more energy to break the particle into its constituent components. c. The stationary target contains zero potential energy, so it requires more energy to break the particle into its constituent components. d. The stationary target recoils upon particle strike, transferring much of the particle\u2019s energy into its motion. As a result, a lesser amount of energy goes into breaking the particle into its constituent components. 21. Compare the total strong nuclear force in a lithium atom. to the total strong nuclear force in a lithium ion a. The total strong nuclear force in a lithium atom is thrice the total strong nuclear force in a lithium ion. b. The total strong nuclear force in a lithium atom is twice the total strong nuclear force in a lithium ion. c. The total strong nuclear force in a lithium atom is the same as the total strong nuclear force in a lithium ion. d. The total strong nuclear force in a lithium atom is half the total strong nuclear force in a lithium ion. 23.2 Quarks 22. Explain why it is not possible to find a particle composed of just two quarks. a. A particle composed of two quarks will have an integral charge and a white color. Hence, it cannot exist. b. A particle composed of two quarks will have an integral charge and a color that is not white. Hence, it cannot exist. c. A particle composed of two quarks will have a fractional charge and a white color. Hence, it cannot exist. d. A particle composed of two quarks will have a fractional charge and a color that is not white. Hence, it cannot exist. 23. Why are mesons considered unstable? a. Mesons are composites of two antiparticles that quickly annihilate each other. b. Mesons are composites of two particles that quickly annihilate each other. c. Mesons are composites of a particle and antiparticle that quickly annihilate each other. d. Mesons are composites of two particles and one antiparticle that quickly annihilate each other. 24. Does antimatter have a negative mass? a. No, antimatter does not have a negative mass. b. Yes, antimatter does have a negative mass. 25. What similarities exist between the Standard Model and the periodic table of elements? a. During their invention, both the Standard Model and the periodic table organized material by mass. b. At the times of their invention, both the Standard Model and the periodic table organized material by charge. c. At the times of their invention, both the Standard Model and the periodic table organized material by interaction with other available particles. d. At the times of their invention, both the Standard Model and the periodic table organized material by size. 26. How were particle collisions used to provide", " evidence of the Higgs boson? 800 Chapter 23 \u2022 Chapter Review a. Because some particles do not contain the Higgs boson, the collisions of such particles will cause their destruction. b. Because only the charged particles contain the Higgs boson, the collisions of such particles will cause their destruction and will expel the Higgs boson. c. Because all particles with mass contain the Higgs boson, the collisions of such particles will cause their destruction and will absorb the Higgs boson. d. Because all particles with mass contain the Higgs boson, the collisions of such particles will cause their destruction and will expel the Higgs boson. 27. Explain how the combination of a quark and antiquark can result in the creation of a hadron. a. The combination of a quark and antiquark can result in a particle with an integer charge and color of white, therefore satisfying the properties for a hadron. b. The combination of a quark and antiquark must result in a particle with a negative charge and color of white, therefore satisfying the properties for a hadron. c. The combination of a quark and antiquark can result in a particle with an integer charge and color that is not white, therefore satisfying the properties for a hadron. d. The combination of a quark and antiquark can result in particle with a fractional charge and color that is not white, therefore satisfying the properties for a hadron. 23.3 The Unification of Forces 28. Why does the strength of the strong force diminish under high-energy conditions? a. Under high-energy conditions, particles interacting under the strong force will be compressed closer together. As a result, the force between them will decrease. b. Under high-energy conditions, particles interacting under the strong force will start oscillating. As a result, the force between them will increase. c. Under high-energy conditions, particles interacting under the strong force will have high velocity. As a result, the force between them will decrease. d. Under high-energy conditions, particles interacting under the strong force will start moving randomly. As a result, the force between them will decrease. 29. If some unknown cause of the red shift, such as light becoming tiredfrom traveling long distances through empty space, is discovered, what effect would there be on cosmology? a. The effect would be substantial, as the Big Bang is based on the idea that the red shift is evidence that galaxies are moving toward one another. b. The effect would be substantial, as the Big Bang is based on the idea that the red shift is evidence that the galaxies are moving away from one another. c. The effect would be substantial, as the Big Bang is based on the idea that the red shift is evidence that galaxies are neither moving away from nor moving toward one another. d. The effect would be substantial, as the Big Bang is based on the idea that the red shift is evidence that galaxies are sometimes moving away from and sometimes moving toward one another. 30. How many molecules of water are necessary if scientists -yr estimate of proton decay wanted to check the within the course of one calendar year? a. b. c. d. 31. As energy of interacting particles increases toward the theory of everything, the gravitational force between them increases. Why does this occur? a. As energy increases, the masses of the interacting particles will increase. b. As energy increases, the masses of the interacting particles will decrease. c. As energy increases, the masses of the interacting particles will remain constant. d. As energy increases, the masses of the interacting particles starts changing (increasing or decreasing). As a result, the gravitational force between the particles will increase. Access for free at openstax.org. Performance Task 23.3 The Unification of Forces 32. Communication is an often overlooked and useful skill for a scientist, especially in a competitive field where financial resources are limited. Scientists are often required to explain their findings or the relevance of their work to agencies within the government in order to maintain funding to continue their research. Let\u2019s say you are an ambitious young particle physicist, heading an expensive project, and you need to justify its existence to the appropriate funding agency. Write a brief paper (about one page) explaining why molecularlevel structure is important in the functioning of designed materials in a specific industry. TEST PREP Multiple Choice 23.1 The Four Fundamental Forces 33. Which of the following is not one of the four fundamental forces? a. gravity friction b. strong nuclear c. d. electromagnetic 34. What type of carrier particle has not yet been found? a. gravitons b. bosons c. bosons d. pions 35. What effect does an increase in electric potential have It increases accelerating capacity. It decreases accelerating capacity. on the accelerating capacity of a Van de Graaff generator? a. b. c. The accelerating capacity of a Van de Graaff generator is constant regardless of electric potential. Chapter 23 \u2022 Test Prep 8", "01 \u2022 First, think of an industry where molecular-level structure is important. \u2022 Research what materials are used in that industry as well as what are the desired properties of the materials. \u2022 What molecular-level characteristics lead to what properties? One example would be explaining how flexible but durable materials are made up of long-chained molecules and how this is useful for finding more environmentally friendly alternatives to plastics. Another example is explaining why electrically conductive materials are often made of metal and how this is useful for developing better batteries. and the electromagnetic force 23.2 Quarks 37. To what color must quarks combine for a particle to be constructed? a. black b. green red c. d. white 38. What type of hadron is always constructed partially of an antiquark? a. baryon b. lepton c. meson d. photno 39. What particle is typically released when two particles annihilate? a. graviton b. antimatter c. pion d. photno d. Van de Graaff generators do not have the capacity 40. Which of the following categories is not one of the three to accelerate particles. 36. What force or forces exist between a proton and a second proton? a. The weak electrostatic force and strong magnetic force main categories of the Standard Model? a. gauge bosons b. hadrons leptons c. d. quarks b. The weak electrostatic and strong gravitational 41. Analysis of what particles began the search for the Higgs force c. The weak frictional force and strong gravitational force d. The weak nuclear force, the strong nuclear force, boson? a. W and Z bosons b. up and down quarks c. mesons and baryons 802 Chapter 23 \u2022 Test Prep d. neutrinos and photons 44. After the Big Bang, what was the first force to separate 42. What similarities exist between the discovery of the quark and the discovery of the neutron? a. Both the quark and the neutron were discovered by launching charged particles through an unknown structure and observing the particle recoil. b. Both the quark and the neutron were discovered by launching electrically neutral particles through an unknown structure and observing the particle recoil. c. Both quarks and neutrons were discovered by studying their deflection under an electric field. 23.3 The Unification of Forces 43. Which two forces were first combined, signifying the eventual desire for a Grand Unified Theory? a. electric force and magnetic forces b. electric force and weak nuclear force c. gravitational force and the weak nuclear force d. electroweak force and strong nuclear force Short Answer 23.1 The Four Fundamental Forces 47. Why do people tend to be more aware of the gravitational and electromagnetic forces than the strong and weak nuclear forces? a. The gravitational and electromagnetic forces act at short ranges, while strong and weak nuclear forces act at comparatively long range. b. The strong and weak nuclear forces act at short ranges, while gravitational and electromagnetic forces act at comparatively long range. c. The strong and weak nuclear forces act between all objects, while gravitational and electromagnetic forces act between smaller objects. d. The strong and weak nuclear forces exist in outer space, while gravitational and electromagnetic forces exist everywhere. 48. What fundamental force is responsible for the force of friction? a. b. c. the electromagnetic force the strong nuclear force the weak nuclear force 49. How do carrier particles relate to the concept of a force field? a. Carrier particles carry mass from one location to another within a force field. b. Carrier particles carry force from one location to another within a force field. Access for free at openstax.org. from the others? a. electromagnetic force b. gravity c. d. weak nuclear force strong nuclear force 45. What is the name of the device used by scientists to check for proton decay? the cyclotron a. the Large Hadron Collider b. the Super-Kamiokande c. the synchrotron d. 46. How do Feynman diagrams suggest the Grand Unified Theory? a. The electromagnetic, weak, and strong nuclear forces all have similar Feynman diagrams. b. The electromagnetic, weak, and gravitational forces all have similar Feynman diagrams. c. The electromagnetic, weak, and strong forces all have different Feynman diagrams. c. Carrier particles carry charge from one location to another within a force field. d. Carrier particles carry volume from one location to another within a force field. 50. Which carrier particle is transmitted solely between nucleons? a. graviton b. photon c. pion d. W and Z bosons 51. Two particles of the same mass are traveling at the same speed but in opposite directions when they collide headon. What is the final kinetic energy of this two-particle system? a. b. c. zero d. infinite the sum of the kinetic energies of the two particles the product of the kinetic energies of the two particles 52. Why do colliding", " beams result in the location of smaller particles? a. Colliding beams create energy, allowing more energy to be used to separate the colliding particles. b. Colliding beams lower the energy of the system, so it requires less energy to separate the colliding particles. c. Colliding beams reduce energy loss, so less energy is required to separate colliding particles. c. The four fundamental forces are represented by d. Colliding beams reduce energy loss, allowing more their carrier particles, the leptons. energy to be used to separate the colliding particles. d. The four fundamental forces are represented by their carrier particles, the quarks. Chapter 23 \u2022 Test Prep 803 23.2 Quarks 53. What two features of quarks determine the structure of a particle? a. b. c. d. the color and charge of individual quarks the color and size of individual quarks the charge and size of individual quarks the charge and mass of individual quarks 54. What fundamental force does quantum chromodynamics describe? the weak nuclear force a. the strong nuclear force b. the electromagnetic force c. the gravitational force d. 55. Is it possible for a baryon to be constructed of two quarks and an antiquark? a. Yes, the color of the three particles would be able to sum to white. b. No, the color of the three particles would not be able to sum to white. 56. Can baryons be more massive than mesons? a. no b. yes 57. If antimatter exists, why is it so difficult to find? a. There is a smaller amount of antimatter than matter in the universe; antimatter is quickly annihilated by its matter analogue. b. There is a smaller amount of matter than antimatter in the universe; matter is annihilated by its antimatter analogue. c. There is a smaller amount of antimatter than matter in universe; antimatter and its matter analogue coexist. d. There is a smaller amount of matter than antimatter in the universe; matter and its antimatter analogue coexist. 58. Does a neutron have an antimatter counterpart? a. No, the antineutron does not exist. b. Yes, the antineutron does exist. 59. How are the four fundamental forces incorporated into the Standard Model of the atom? a. The four fundamental forces are represented by their carrier particles, the electrons. 60. Which particles in the Standard Model account for the majority of matter with which we are familiar? a. particles in fourth column of the Standard Model b. particles in third column of the Standard Model c. particles in the second column of the Standard Model d. particles in the first column of the Standard Model 61. How can a particle gain mass by traveling through the Higgs field? a. The Higgs field slows down passing particles; the decrease in kinetic energy is transferred to the particle\u2019s mass. b. The Higgs field accelerates passing particles; the decrease in kinetic energy is transferred to the particle\u2019s mass. c. The Higgs field slows down passing particles; the increase in kinetic energy is transferred to the particle\u2019s mass. d. The Higgs field accelerates passing particles; the increase in kinetic energy is transferred to the particle\u2019s mass. 62. How does mass-energy conservation relate to the Higgs field? a. The increase in a particle\u2019s energy when traveling through the Higgs field is countered by its increase in mass. b. The decrease in a particle\u2019s kinetic energy when traveling through the Higgs field is countered by its increase in mass. c. The decrease in a particle\u2019s energy when traveling through the Higgs field is countered by its decrease in mass. d. The increase in a particle\u2019s energy when traveling through the Higgs field is countered by its decrease in mass. 23.3 The Unification of Forces 63. Why do scientists believe that the strong nuclear force and the electroweak force will combine under high energies? a. The electroweak force will have greater strength. b. The strong nuclear force and electroweak force will achieve the same strength. c. The strong nuclear force will have greater strength. b. The four fundamental forces are represented by 64. At what energy will the strong nuclear force their carrier particles, the gauge bosons. theoretically unite with the electroweak force? 804 Chapter 23 \u2022 Test Prep a. b. c. d. 65. While we can demonstrate the unification of certain forces within the laboratory, for how long were the four forces naturally unified within the universe? a. b. c. d. 66. How does the search for the Grand Unified Theory help test the standard cosmological model? a. Scientists are increasing energy in the lab that models the energy in earlier, denser stages of the universe. b. Scientists are increasing energy in the lab that models the energy in earlier, less dense stages of the universe. c. Scientists are", " decreasing energy in the lab that models the energy in earlier, denser stages of the universe. d. Scientists are decreasing energy in the lab that Extended Response 23.1 The Four Fundamental Forces 69. If the strong attractive force is the greatest of the four fundamental forces, are all masses fated to combine together at some point in the future? Explain. a. No, the strong attractive force acts only at incredibly small distances. As a result, only masses close enough to be within its range will combine. b. No, the strong attractive force acts only at large distances. As a result, only masses far enough apart will combine. c. Yes, the strong attractive force acts at any distance. As a result, all masses are fated to combine together at some point in the future. d. Yes, the strong attractive force acts at large distances. As a result, all masses are fated tocombine together at some point in the future. 70. How does the discussion of carrier particles relate to the concept of relativity? a. Calculations of mass and energy during their transfer are relativistic, because carrier particles travel more slowly than the speed of sound. b. Calculations of mass and energy during their transfer are relativistic, because carrier particles travel at or near the speed of light. Access for free at openstax.org. models the energy in earlier, less dense stages of the universe. 67. Why does finding proof that protons do not decay not disprove all GUTs? a. Proton decay is not a premise of all GUTs, and current GUTs can be amended in response to new findings. b. Proton decay is a premise of all GUTs, but current GUTs can be amended in response to new findings. 68. When accelerating elementary particles in a particle accelerator, they quickly achieve a speed approaching the speed of light. However, as time continues, the particles maintain this speed yet continue to increase their kinetic energy. How is this possible? a. The speed remains the same, but the masses of the particles increase. b. The speed remains the same, but the masses of the particles decrease. c. The speed remains the same, and the masses of the particles remain the same. d. The speed and masses will remain the same, but temperature will increase. c. Calculations of mass and energy during their transfer are relativistic, because carrier particles travel at or near the speed of sound. d. Calculations of mass and energy during their transfer are relativistic, because carrier particles travel faster than the speed of light. 71. Why are synchrotrons constructed to be very large? a. By using a large radius, high particle velocities can be achieved using a large centripetal force created by large electromagnets. b. By using a large radius, high particle velocities can be achieved without a large centripetal force created by large electromagnets. c. By using a large radius, the velocities of particles can be reduced without a large centripetal force created by large electromagnets. d. By using a large radius, the acceleration of particles can be decreased without a large centripetal force created by large electromagnets. 23.2 Quarks 72. In this image, how does the emission of the gluon cause the down quark to change from a red color to a green color? Chapter 23 \u2022 Test Prep 805 a. b. c. d. If the velocity of the neutrino is known, then the upper limit on mass of the neutrino can be set. If only the kinetic energy of the neutrino is known, then the upper limit on mass of the neutrino can be set. If either the velocity or the kinetic energy is known, then the upper limit on the mass of the neutrino can be set. If both the kinetic energy and the velocity of the neutrino are known, then the upper limit on the mass of the neutrino can be set. 76. The term force carrier particleis shorthand for the scientific term vector gauge boson. From that perspective, can the Higgs boson truly be considered a force carrier particle? a. No, the mass quality provided by the Higgs boson is a scalar quantity. b. Yes, the mass quality provided by the Higgs boson results in a change of particle\u2019s direction. 23.3 The Unification of Forces 77. If a Grand Unified Theory is proven and the four forces are unified, it will still be correct to say that the orbit of the Moon is determined by the gravitational force. Explain why. a. Gravity will not be a property of the unified force. b. Gravity will be one property of the unified force. c. Apart from gravity, no other force depends on the mass of the object. d. Apart from gravity, no other force can make an a. The emitted red gl", "uon is made up of a green and a red color. As a result, the down quark changes from a red color to a green color. b. The emitted red gluon is made up of an anti-green and an anti-red color. As a result, the down quark changes from a red color to a green color. c. The emitted red gluon is made up of a green and an anti-red color. As a result, the down quark changes from a red color to a green color. d. The emitted red gluon is made up of an anti-green and a red color. As a result, the down quark changes from a red color to a green color. 73. Neutrinos are much more difficult for scientists to find when compared to other hadrons and leptons. Why is this? a. Neutrinos are hadrons, and they lack charge. b. Neutrinos are not hadrons, and they lack charge. c. Neutrinos are hadrons, and they have positive charge. d. Neutrinos are not hadrons, and they have a positive object move in a fixed orbit. charge. 74. What happens to the masses of a particle and its antiparticle when the two annihilate at low energies? a. The masses of the particle and antiparticle are transformed into energy in the form of photons. b. The masses of the particle and antiparticle are converted into kinetic energy of the particle and antiparticle respectively. c. The mass of the antiparticle is converted into kinetic energy of the particle. d. The mass of the particle is converted into radiation energy of the antiparticle. 75. When a star erupts in a supernova explosion, huge numbers of electron neutrinos are formed in nuclear reactions. Such neutrinos from the 1987A supernova in the relatively nearby Magellanic Cloud were observed within hours of the initial brightening, indicating that they traveled to earth at approximately the speed of light. Explain how this data can be used to set an upper limit on the mass of the neutrino. 78. As the universe expanded and temperatures dropped, the strong nuclear force separated from the electroweak force. Is it likely that under cooler conditions, the force of electricity will separate from the force of magnetism? a. No, the electric force relies on the magnetic force and vice versa. b. Yes, the electric and magnetic forces can be separated from each other. 79. Two pool balls collide head-on and stop. Their original kinetic energy is converted to heat and sound. Given that this is not possible for particles, what happens to their converted energy? a. The kinetic energy is converted into relativistic potential energy, governed by the equation. b. The kinetic energy is converted into relativistic mass, governed by the equation. c. The kinetic energy is converted into relativistic potential energy, governed by the equation. 806 Chapter 23 \u2022 Test Prep d. Their kinetic energy is converted into relativistic mass, governed by the equation. Access for free at openstax.org. Appendix A \u2022 Reference Tables 807 APPENDIX A Reference Tables Figure A1 Periodic Table of Elements Prefix Symbol Value Prefix Symbol Value tera giga T G mega M kilo hecto k h deka da 1012 deci d c centi milli m micro nano pico \u00b5 n p 109 106 103 102 101 10\u20131 10\u20132 10\u20133 10\u20136 10\u20139 10\u201312 Table A1 Metric Prefixes for Powers of Ten and Their Symbols 808 Appendix A \u2022 Reference Tables Prefix Symbol Value Prefix Symbol Value ___ 100 femto f 10\u201315 Table A1 Metric Prefixes for Powers of Ten and Their Symbols Entity Abbreviation Name Fundamental units Length Mass Time Current Supplementary unit Angle Derived units Force m kg s A rad Energy Power Pressure Frequency Electronic potential Capacitance Charge Resistance Magnetic field meter kilogram second ampere radian newton joule watt pascal hertz volt farad coulomb ohm tesla Nuclear decay rate becquerel Table A2 SI Units Length 1 inch (in.) = 2.54 cm (exactly) 1 foot (ft) = 0.3048 m 1 mile (mi) = 1.609 km Table A3 Selected British Units Access for free at openstax.org. Appendix A \u2022 Reference Tables 809 Force 1 pound (lb) = 4.448 N Energy 1 British thermal unit (Btu) = 1.055 \u00d7 103 J Power 1 horsepower (hp) = 746 W Pressure 1 lb/in2 = 6.895 \u00d7 103 Pa Table A3 Selected British Units Length 1 light year (ly) = 9.46 \u00d7 1015 m 1 astronomical unit (au) = 1.50 \u00d7 1011 m 1 nautical mile = 1.852 km 1 angstrom(\u00c5) =", " 10-10 m Area 1 acre (ac) = 4.05 \u00d7 103 m2 1 square foot (ft2) 9.29 \u00d7 10-2 m3 1 barn (b) = 10-28 m2 Volume 1 liter (L) = 10-3 m3 1 U.S. gallon (gal) = 3.785 \u00d7 10-3 m3 Mass 1 solar mass = 1.99 \u00d7 1030 kg 1 metric ton = 103 kg 1 atomic mass unit (u) = 1.6605 \u00d7 10-27 kg Time 1 year (y) = 3.16 \u00d7 107 s 1 day (d) = 86,400 s Speed 1 mile per hour (mph) = 1.609 km / h 1 nautical mile per hour (naut) = 1.852 km / h Angle 1 degree (\u00b0) = 1.745x10-2 rad 1 minute of arc (') = 1 / 60 degree 1 second of arc ('') = 1 / 60 minute of arc 1 grad = 1.571 \u00d7 10-2 rad Table A4 Other Units 810 Appendix A \u2022 Reference Tables Energy 1 kiloton TNT (kT) = 4.2 \u00d7 1012 J 1 kilowatt hour (kW h) = 3.60 \u00d7 106J 1 food calorie (kcal) = 4186 J 1 calorie (cal) = 4.186 J 1 electron volt (cV) = 1.60 \u00d7 10-19 J Pressure 1 atmosphere (atm) = 1.013 \u00d7 105 Pa 1 millimeter of mercury (mm Hg) = 133.3 Pa 1 torricelli (torr) = 1 mm Hg = 133.3 Pa Nuclear decay rate 1 curie (Ci) = 3.70 \u00d7 1010 Bq Table A4 Other Units Circumference of a circle with radius ror diameter d Area of a circle with radius ror diameter d Area of a sphere with radius r Volume of a sphere with radius r Table A5 Useful formulae Symbol Meaning Best Value Approximate Value c G NA k R \u03c3 k Speed of light in vacuum Gravitational constant Avogadro\u2019s number Boltzmann\u2019s constant Gas constant Stefan-Boltzmann Constant Coulomb force constant Table A6 Important Constants Access for free at openstax.org. Symbol Meaning Best Value Approximate Value Appendix A \u2022 Reference Tables 811 qe \u03b50 \u00b50 h Charge on electron Permittivity of free space Permeability of free space Planck\u2019s constant Table A6 Important Constants Alpha Beta Gamma Delta Epsilon Zeta Eta Theta Iota Kappa Lambda Mu Nu Xi Omicron Pi Rho Sigma Tau Table A7 The Greek Alphabet 812 Appendix A \u2022 Reference Tables Upsilon Phi Chi Psi Omega Table A7 The Greek Alphabet Sun mass average radius 1.99 \u00d7 1030 kg 6.96 \u00d7 108 m Earth-sun distance (average) 1.496 \u00d7 1011 m Earth mass 5.9736 \u00d7 1024 kg average radius orbital period Moon mass average radius orbital period (average) 6.376 \u00d7 106 m 3.16 \u00d7 107 s 7.35 \u00d7 1022 kg 1.74 \u00d7 106 s 2.36 \u00d7 106 s Earth-moon distance (average) 3.84 \u00d7 108 m Table A8 Solar System Data Atomic number, Z Name Atomic Mass Number, A Symbol Atomic Mass (u) Percent Abundance or Decay Mode Halflife, t1/2 n 1H 1.008 665 \u03b2\u2013 1.007 825 99.985% 2H or D 2.014 102 0.015% 10.37 min 3H or T 3.016 050 \u03b2\u2013 12.33 y 3He 3.016 030 1.38 \u00d7 10\u22124 % 0 1 2 neutron Hydrogen Deuterium Tritium Helium 1 1 2 3 3 Table A9 Atomic Masses and Decay Access for free at openstax.org. Atomic number, Z Name Atomic Mass Number, A Symbol Atomic Mass (u) Percent Abundance or Decay Mode Halflife, t1/2 Appendix A \u2022 Reference Tables 813 3 4 5 6 7 8 9 10 11 Lithium Beryllium Boron Carbon Nitrogen Oxygen Fluorine Neon Sodium 4 6 7 7 9 10 11 11 12 13 14 13 14 15 15 16 18 18 19 20 22 22 23 24 Table A9 Atomic Masses and Decay 4He 4.002 603 \u2248100% 6Li 7Li 7Be 9Be 10B 11B 11C 12C 13C 14C 12N 13N 14N 15O 16O 18O 18F 19F 20Ne 22Ne 22Na 23Na 24Na 6.015 121 7.5% 7.016 003 92.5% 7.016 928 EC 53.29 d 9.012 182 100% 10.012 937 19.9% 11.009 305 80.", "1% 11.011 432 EC, \u03b2+ 12.000 000 98.90% 13.003 355 1.10% 14.003 241 13.005 738 \u03b2\u2013 \u03b2+ 14.003 074 99.63% 15.000 108 0.37% 5730 y 9.96 min 15.003 065 EC, \u03b2+ 122 s 15.994 915 99.76% 17.999 160 0.200% 18.000 937 EC, \u03b2+ 1.83 h 18.998 403 100% 19.992 435 90.51% 21.991 383 9.22% 21.994 434 \u03b2+ 22.989 767 100% 23.990 961 \u03b2\u2013 2.602 y 14.96 h 814 Appendix A \u2022 Reference Tables Atomic number, Z Name Atomic Mass Number, A Symbol Atomic Mass (u) Percent Abundance or Decay Mode Halflife, t1/2 24Mg 23.985 042 78.99% 27Al 28Si 31Si 31P 32P 32S 35S 35Cl 37Cl 40Ar 39K 40K 40Ca 45Sc 48Ti 51V 52Cr 55Mn 56Fe 59Co 60Co 58Ni 26.981 539 100% 27.976 927 92.23% 2.62h 30.975 362 \u03b2\u2013 30.973 762 100% 31.973 907 \u03b2\u2013 31.972 070 95.02% 34.969 031 \u03b2\u2013 34.968 852 75.77% 36.965 903 24.23% 39.962 384 99.60% 38.963 707 93.26% 14.28 d 87.4 d 39.963 999 0.0117%, EC, \u03b2\u2013 1.28 \u00d7 109 y 39.962 591 96.94% 44.955 910 100% 47.947 947 73.8% 50.943 962 99.75% 51.940 509 83.79% 54.938 047 100% 55.934 939 91.72% 58.933 198 100% 59.933 819 \u03b2\u2013 5.271 y 57.935 346 68.27% 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 Magnesium Aluminum Silicon Phosphorus Sulfur Chlorine Argon Potassium Calcium Scandium Titanium Vanadium Chromium Manganese Iron Cobalt Nickel 24 27 28 31 31 32 32 35 35 37 40 39 40 40 45 48 51 52 55 56 59 60 58 Table A9 Atomic Masses and Decay Access for free at openstax.org. Atomic number, Z Name Atomic Mass Number, A Symbol Atomic Mass (u) Percent Abundance or Decay Mode Halflife, t1/2 Appendix A \u2022 Reference Tables 815 60 63 64 66 69 72 74 75 80 79 84 85 86 88 90 89 90 90 93 98 98 Copper Zinc Gallium Germanium Arsenic Selenium Bromine Krypton Rubidium Strontium Yttrium Zirconium Niobium Molybdenum Technetium 60Ni 63Cu 65Cu 64Zn 66Zn 69Ga 72Ge 74Ge 75As 80Se 79Br 84Kr 85Rb 86Sr 88Sr 90Sr 89Y 90Y 90Zr 93Nb 98Mo 59.930 788 26.10% 62.939 598 69.17% 64.927 793 30.83% 63.929 145 48.6% 65.926 034 27.9% 68.925 580 60.1% 71.922 079 27.4% 73.921 177 36.5% 74.921 594 100% 79.916 520 49.7% 78.918 336 50.69% 83.911 507 57.0% 84.911 794 72.17% 85.909 267 9.86% 87.905 619 82.58% 89.907 738 \u03b2\u2013 88.905 849 100% 89.907 152 \u03b2\u2013 89.904 703 51.45% 92.906 377 100% 97.905 406 24.13% 98Tc 97.907 215 \u03b2\u2013 Ruthenium 102 102Ru 101.904 348 31.6% 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 Table A9 Atomic Masses and Decay 28.8 y 64.1 h 4.2 \u00d7 106 y 816 Appendix A \u2022 Reference Tables Atomic number, Z Name Atomic Mass Number, A Symbol Atomic Mass (u) Percent Abundance or Decay Mode Halflife, t1/2 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 Rhodium Palladium Silver Cadmium Indium Tin Antimony Tell", "urium Iodine Xenon Cesium Barium Lanthanum Cerium 103 106 107 109 114 115 120 121 130 127 131 132 136 133 134 137 138 139 140 Praseodymium 141 Neodymium Promethium Samarium 142 145 152 Table A9 Atomic Masses and Decay Access for free at openstax.org. 103Rh 106Pd 107Ag 109Ag 114Cd 102.905 500 100% 105.903 478 27.33% 106.905 092 51.84% 108.904 757 48.16% 113.903 357 28.73% 115In 114.903 880 95.7%, \u03b2\u2013 120Sn 121Sb 119.902 200 32.59% 120.903 821 57.3% 130Te 129.906 229 33.8%, \u03b2\u2013 4.4 \u00d7 1014 y 2.5 \u00d7 1021 y 127I 131I 132Xe 136Xe 133Cs 134Cs 137Ba 138Ba 139La 140Ce 141Pr 142Nd 145Pm 152Sm 126.904 473 100% 130.906 114 \u03b2\u2013 8.040 d 131.904 144 26.9% 135.907 214 8.9% 132.905 429 100% 133.906 696 EC, \u03b2\u2013 2.06 y 136.905 812 11.23% 137.905 232 71.70% 138.906 346 99.91% 139.905 433 88.48% 140.907 647 100% 141.907 719 27.13% 144.912 743 EC, \u03b1 17.7 y 151.919 729 26.7% Atomic number, Z Name Atomic Mass Number, A Symbol Atomic Mass (u) Percent Abundance or Decay Mode Halflife, t1/2 Appendix A \u2022 Reference Tables 817 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 Europium Gadolinium Terbium Dysprosium Holmium Erbium Thulium Ytterbium Lutecium Hafnium Tantalum Tungsten Rhenium Osmium Iridium Platinum Gold Mercury Thallium 153 158 159 164 165 166 169 174 175 180 181 184 187 191 192 191 193 195 197 198 199 202 205 Table A9 Atomic Masses and Decay 153Eu 158Gd 159Tb 164Dy 165Ho 166Ho 152.921 225 52.2% 157.924 099 24.84% 158.925 342 100% 163.929 171 28.2% 164.930 319 100% 165.930 290 33.6% 169Tm 168.934 212 100% 174Yb 175Lu 180Hf 181Ta 184W 173.938 859 31.8% 174.940 770 97.41% 179.946 545 35.10% 180.947 992 99.98% 183.950 928 30.67% 187Re 186.955 744 62.6%, \u03b2\u2013 190.960 920 \u03b2\u2013 191.961 467 41.0% 190.960 584 37.3% 192.962 917 62.7% 194.964 766 33.8% 196.966 543 100% 191Os 192Os 191Ir 193Ir 195Pt 197Au 198Au 199Hg 202Hg 205Tl 4.6 \u00d7 1010y 15.4 d 197.968 217 \u03b2\u2013 2.696 d 198.968 253 16.87% 201.970 617 29.86% 204.974 401 70.48% 818 Appendix A \u2022 Reference Tables Atomic number, Z Name Atomic Mass Number, A Symbol Atomic Mass (u) Percent Abundance or Decay Mode Halflife, t1/2 82 Lead 206 207 208 210 211 212 209 211 210 218 222 223 226 227 228 232 206Pb 207Pb 208Pb 205.974 440 24.1% 206.975 872 22.1% 207.976 627 52.4% 210Pb 209.984 163 \u03b1, \u03b2\u2013 211Pb 212Pb 209Bi 211Bi 210Po 218At 210.988 735 211.991 871 \u03b2\u2013 \u03b2\u2013 208.980 374 100% 210.987 255 \u03b1, \u03b2\u2013 209.982 848 \u03b1 218.008 684 \u03b1, \u03b2\u2013 222Rn 222.017 570 \u03b1 223Fr 223.019 733 \u03b1, \u03b2\u2013 226Ra 227Ac 228Th 226.025 402 \u03b1 227.027 750 \u03b1, \u03b2\u2013 228.028 715 \u03b1 232Th 232.038 054 100%, \u03b1 Bismuth Polonium Astatine Radon Francium 2 Radium Actinium Thorium Protactinium 231 231Pa 231.035 880", " Uranium 233 233U 233.039 628 \u03b1 \u03b1 235 236 235U 235.043 924 0.720%, \u03b1 236U 236.045 562 \u03b1 83 84 85 86 87 88 89 90 91 92 Table A9 Atomic Masses and Decay Access for free at openstax.org. 22.3 y 36.1 min 10.64 h 2.14 min 138.38 d 1.6 s 3.82 d 21.8 min 1.60 \u00d7 103 y 21.8 y 1.91 y 1.41 \u00d7 1010 y 3.28 \u00d7 104 y 1.59 \u00d7 103 y 7.04 \u00d7 108 y 2.34 \u00d7 107 y Atomic number, Z Name Atomic Mass Number, A Symbol Atomic Mass (u) Percent Abundance or Decay Mode Halflife, t1/2 Appendix A \u2022 Reference Tables 819 \u03b2\u2013 \u03b2\u2013 \u03b1 \u03b1 \u03b1 238 239 239 Neptunium 238.050 784 99.2745%, \u03b1 238U 239U 239.054 289 239Np 239.052 933 Plutonium 239 239Pu 239.052 157 \u03b1 Americium 243 243Am 243.061 375 \u03b1, fission Curium 245 245Cm 245.065 483 Berkelium 245 Californium 249 Einsteinium Fermium 254 253 Mendelevium 255 Nobelium Lawrencium 255 257 247Bk 249Cf 254Es 253Fm 255Md 255No 257Lr 247.070 300 249.074 844 254.088 019 \u03b1, \u03b2\u2013 253.085 173 EC, \u03b1 255.091 081 EC, \u03b1 255.093 260 EC, \u03b1 257.099 480 EC, \u03b1 Rutherfordium 261 261Rf 261.108 690 \u03b1 Dubnium Seaborgium Bohrium Hassium Meitnerium 262 263 262 264 266 262Db 263Sg 262Bh 264Hs 266Mt 262.113 760 \u03b1, fission 263.11 86 \u03b1, fission 262.123 1 264.128 5 266.137 8 \u03b1 \u03b1 \u03b1 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 108 Table A9 Atomic Masses and Decay 4.47 \u00d7 109 y 23.5 min 2.355 d 2.41 \u00d7 104 y 7.37 \u00d7 103 y 8.50 \u00d7 103 y 1.38 \u00d7 103 y 351 y 276 d 3.00 d 27 min 3.1 min 0.646 s 1.08 mim 34 s 0.8 s 0.102 s 0.08 ms 3.4 ms 820 Appendix A \u2022 Reference Tables Isotope t1/2 Decay Mode Energy(MeV) Percent T-Ray Energy(MeV) Percent 3H 14C 13N 12.33 y 5730 y 9.96 min 22Na 2.602 y 32P 35S 36Ci 40K 43K 45Ca 51Cr 14.28 d 87.4 d 3.00 \u00d7 105 y 1.28 \u00d7 109 y 22.3 h 165 d 27.70 d 52Mn 5.59d \u03b2\u2013 \u03b2\u2013 \u03b2+ \u03b2+ \u03b2\u2013 \u03b2\u2013 \u03b2\u2013 \u03b2\u2013 \u03b2\u2013 \u03b2\u2013 EC \u03b2+ 100% 100% 100% 90% \u03b3 1.27 100% 0.0186 0.156 1.20 1.20 1.71 0.167 0.710 1.31 0.827 100% 100% 100% 89% 87% \u03b3s 0.373 0.618 \u03b3 0.320 0.257 100% 3.69 28% \u03b3 s 1.33 52Fe 8.27 h \u03b2+ 1.80 43% 1.43 0.169 0.378 59Fe 44.6 d \u03b2\u2013 s 60Co 5.271 y \u03b2\u2013 65Zn 67Ga 244.1 d 78.3 h EC EC 0.273 0.466 0.318 45% 55% \u03b3 s 1.10 1.29 100% \u03b3 s 1.17 1.33 \u03b3 1.12 \u03b3 s 0.0933 0.185 0.300 Table A10 Selected Radioactive Isotopes Access for free at openstax.org. 87% 87% 10% 28% 28% 43% 43% 57% 43% 100% 100% 51% 70% 35% 19% Isotope t1/2 Decay Mode Energy(MeV) Percent T-Ray Energy(MeV) Percent Appendix A \u2022 Reference Tables 821 75Se 118.5 d EC 86Rb 18.8 d \u03b2\u2013 s 85Sr 90Sr 90Y 64.8 d 28.8 y 64.1 h 99mTc 6.02 h 113mIn 99.5 min 123I 131I 13.0 h 8.040 d EC \u03b2\u2013 \u03b2\u2013 IT IT EC \u03b2\u2013 s 0.69 1.77 0.546 2.28 0.248 0.607 others", " others \u03b3 s 0.121 0.136 0.265 0.280 others \u03b3 1.08 20% 65% 68% 20% 9% \u03b3 0.514 1 100% 9% 91% 100% 100% \u03b3 \u03b3 \u03b3 0.142 0.392 0.159 7% 93% \u03b3 s 0.364 others 129Cs 32.3 h EC \u03b3 s 0.0400 0.372 0.411 others \u03b3 0.662 95% 5% \u2248100% \u03b3 s 0.030 137Cs 30.17 y \u03b2\u2013 s 140Ba 12.79 d \u03b2\u2013 Table A10 Selected Radioactive Isotopes 0.511 1.17 1.035 100% 100% \u2248100% 85% 35% 32% 25% 95% 25% 822 Appendix A \u2022 Reference Tables Isotope t1/2 Decay Mode Energy(MeV) Percent T-Ray Energy(MeV) Percent 198Au 197Hg 210Po 226Ra 235U 238U 2.696 d 64.1 h 138.38 d 1.60 \u00d7 103 y \u03b2\u2013 EC \u03b1 \u03b1s 7.038 \u00d7 108 y \u03b1 4.468 \u00d7 109 y \u03b1s 1.161 \u2248100% 100% 5% 95% 5.41 4.68 4.87 4.68 4.22 4.27 0.044 0.537 others 0.412 0.0733 \u03b3 \u03b3 65% 24% \u2248100% 100% \u03b3 0.186 1 100% \u2248100% \u03b3 s Numerous <0.400% \u03b3 0.050 23% 23% 77% 237Np 2.14 \u00d7 106 y \u03b1s numerous \u03b3 s numerous <0.250% 239Pu 2.41 \u00d7 104 y \u03b1s 4.96 (max.) 5.19 5.23 5.24 11% 15% 73% \u03b3 s 7.5 \u00d7 10-5 0.013 0.052 others 73% 15% 15% 243Am 7.37 \u00d7 103 y \u03b1s Max. 5.44 \u03b3 s 0.075 others 88% 11% 5.37 5.32 others Table A10 Selected Radioactive Isotopes Symbol Meaning Best Value Approximate Value me Electron mass 9.10938291(40) \u00d7 10\u201331 kg 9.11 \u00d7 10\u201331 kg Table A11 Submicroscopic masses Access for free at openstax.org. Appendix A \u2022 Reference Tables 823 Symbol Meaning Best Value Approximate Value mp mn u Proton mass 1.672621777(74) \u00d7 10\u201327 kg 1.6726 \u00d7 10\u201327 kg Neutron mass 1.674927351(74) \u00d7 10\u201327 kg 1.6749 \u00d7 10\u201327 kg Atomic mass unit 1.660538921(73) \u00d7 10\u201327 kg 1.6605 \u00d7 10\u201327 kg Table A11 Submicroscopic masses Substance p(kg/m3) Substance p(kg/m3) Air 1.29 Air (at 20\u00b0C and Atmospheric pressure) 1.20 Iron Lead 7.86 \u00d7 103 11.3 \u00d7 103 Aluminum Benzene Brass Copper Ethyl alcohol Fresh water Glycerin Gold Helium gas Hydrogen gas Ice 2.70 \u00d7 103 Mercury 13.6 \u00d7 103 0.879 \u00d7 103 Nitrogen gas 1.25 8.4 \u00d7 103 Oak 0.710 \u00d7 103 8.92 \u00d7 103 Osmium 22.6 \u00d7 103 0.806 \u00d7 103 Oxygen gas 1.43 1.00 \u00d7 103 Pine 0.373 \u00d7 103 1.26 \u00d7 103 Platinum 21.4 \u00d7 103 1.93 \u00d7 103 Seawater 1.03 \u00d7 103 1.79 \u00d7 10\u20131 Silver 10.5 \u00d7 103 8.99 \u00d7 10\u20132 Tin 7.30 \u00d7 103 0.917 \u00d7 103 Uranium 18.7 \u00d7 103 Table A12 Densities of common substances (including water at various temperatures) Substance Specific Heat (J/kg \u00b0C) Substance Specific Heat (J/kg \u00b0C) Elemental solids Other solids Aluminum Beryllium Cadmium 900 1830 230 Brass Glass 380 837 Ice (\u20135 \u00b0C) 2090 Table A13 Specific heats of common substances 824 Appendix A \u2022 Reference Tables Substance Specific Heat (J/kg \u00b0C) Substance Specific Heat (J/kg \u00b0C) Copper Germanium Gold Iron Lead Silicon Silver 387 322 129 448 128 703 234 860 1700 Marble Wood Liquids Alcohol (ethyl) 2400 Mercury 140 Water (15 \u00b0C) 4186 Gas Steam (100 \u00b0C) 2010 Note: To convert values to units of cal/g \u00b0C, divide by 4186 Table A13 Specific heats of common substances Substance Melting Point (\u00b0C) Latent Heat of Fusion (J/kg) Boiling Point (\u00b0C) Latent Heat of Vaporization", " (J/kg) Helium \u2013272.2 Oxygen \u2013218.79 Nitrogen \u2013209.97 Ethyl Alcohol \u2013114 Water 0.00 Sulfur 119 Lead 327.3 Aluminum 660 Silver 960.80 Gold 1063.00 Copper 1083 5.23 \u00d7 103 1.38 \u00d7 104 2.55 \u00d7 104 1.04 \u00d7 105 3.33 \u00d7 105 3.81 \u00d7 104 3.97 \u00d7 105 3.97 \u00d7 105 8.82 \u00d7 104 6.44 \u00d7 104 1.34 \u00d7 105 \u2013268.93 \u2013182.97 \u2013195.81 78 100.00 444.60 1750 2516 2162 2856 2562 2.09 \u00d7 104 2.13 \u00d7 105 2.01 \u00d7 105 8.54 \u00d7 105 2.26 \u00d7 106 2.90 \u00d7 105 8.70 \u00d7 105 1.05 \u00d7 107 2.33 \u00d7 106 1.58 \u00d7 106 5.06 \u00d7 106 Table A14 Heats of fusion and vaporization for common substances Access for free at openstax.org. Appendix A \u2022 Reference Tables 825 Materials (Solids) Average Linear Expansion Coefficient (a)(\u00b0C)\u20131 Material (Liquids and Gases) Average Volume Expansion Coefficient (B)(\u00b0C)\u20131 Aluminum 24 \u00d7 10\u20136 Acetone 1.5 \u00d7 10\u20134 Brass and Bronze 19 \u00d7 10\u20136 Concrete 12 \u00d7 10\u20136 Copper 17 \u00d7 10\u20136 Glass (ordinary) 9 \u00d7 10\u20136 Glass (Pyrex) 3.2 \u00d7 10\u20136 Invar (Ni-Fe alloy) Lead Steel 1.3 \u00d7 10\u20136 29 \u00d7 10\u20136 13 \u00d7 10\u20136 Alcohol, ethyl 1.12 \u00d7 10\u20134 Benzene Gasoline Glycerin Mercury 1.24 \u00d7 10\u20134 9.6 \u00d7 10\u20134 4.85 \u00d7 10\u20134 1.82 \u00d7 10\u20134 Turpentine 9.0 \u00d7 10\u20134 Air* at 0\u00b0C Helium* 3.67 \u00d7 10\u20133 3.665 \u00d7 10\u20133 * The values given here assume the gases undergo expansion at constant pressure. However, the expansion of gases depends on the pressure applied to the gas. Therefore, gases do not have a specific value for the volume expansion coefficient. Table A15 Coefficients of thermal expansion for common substances Medium v(m/s) Medium v(m/s) Medium v(m/s) Gases Liquids at 25\u00b0C Hydrogen 1286 Glycerol Helium Air Air Oxygen 972 343 331 317 Seawater Water Mercury Kerosene Methyl Alcohol Carbon tetrachloride 1904 1533 1493 1450 1324 1143 926 Table A16 Speed of sound in various substances Solids* Pyrex glass Iron Aluminum Brass Copper Gold Lucite Lead Rubber 5640 5950 5100 4700 3560 3240 2680 1322 1600 826 Appendix A \u2022 Reference Tables Medium v(m/s) Medium v(m/s) Medium v(m/s) *Values given here are for propagation of longitudinal waves in bulk media. However, speeds for longitudinal waves in thin rods are slower, and speeds of transverse waves in bulk are even slower. Table A16 Speed of sound in various substances Source of Sound B(dB) Nearby jet airplane 150 Jackhammer machine gun 130 Siren; rock concert 120 Subway; power lawn mower 100 Busy traffic Vacuum cleaner Normal Conversation Mosquito buzzing whisper Rustling leaves Threshold of hearing 80 70 60 40 30 10 0 Table A17 Conversion of sound intensity to decibel level Wavelength Range (nm) Color Description 400-430 430-485 485-560 560-590 590-625 625-700 Violet Blue Green Yellow Orange Red Table A18 Wavelengths of visible light Access for free at openstax.org. Appendix A \u2022 Reference Tables 827 Substance Index of Refraction Substance Index of Refraction Solids at 20\u00b0C Cubic zirconia Diamond (C) Flourite (CaF2) 2.15 2.419 1.434 Liquids at 20\u00b0C Benzene 1.501 Carbon disulfide 1.628 Carbon tetrachloride 1.461 Fused quartz (SiO2) 1.458 Ethyl alcohol Gallium phosphide 3.50 Glass, crown Glass, flint Ice (H2O) Polystyrene 1.52 1.66 1.309 1.49 Glycerin Water 1.361 1.473 1.333 Gases at 0\u00b0C, 1 atm Air 1.000 293 Sodium chloride (NaCl) 1.544 Carbon dioxide 1.000 45 Note: These values assume that light has a wavelength of 589 nm in vacuum. Table A19 Indices of refraction Hoop or thin cylindrical shell Hollow cylinder Solid cylinder or disk Rectangular plane Long, thin rod with rotation axis through center Long, thin rod with rotation axis", " through end Solid sphere Thin spherical shell Table A20 Moments of inertia for different shapes \u03bcs \u03bck Rubber on dry concrete 1.0 0.8 Table A21 Coefficients of friction for common objects on other objects 828 Appendix A \u2022 Reference Tables Steel on steel Aluminum on steel Glass on glass Copper on steel Wood on wood Waxed wood on wet snow Waxed wood on dry snow Metal on metal (lubricated) Teflon on Teflon Ice on ice Synovial joints in humans \u03bcs \u03bck 0.74 0.61 0.94 0.53 0.25-0.5 0.14 0.1 0.15 0.04 0.1 0.01 0.57 0.47 0.4 0.36 0.2 0.1 0.04 0.06 0.04 0.03 0.003 Note: All values are approximate. In some cases, the coefficient of friction can exceed 1.0. Table A21 Coefficients of friction for common objects on other objects Material Dielectric Constant \u0138 Dielectric Strength* (106V/m) Air (dry) Bakelite Fused quartz Mylar Neoprene rubber Nylon Paper 1.000 59 4.9 4.3 3.2 6.7 3.4 3.7 Paraffin-impregnated paper 3.5 Polystyrene Polyvinyl chloride Porcelain 2.56 3.4 6 Table A22 Dielectric constants 3 24 8 7 12 14 16 11 24 40 8 Access for free at openstax.org. Appendix A \u2022 Reference Tables 829 Material Dielectric Constant \u0138 Dielectric Strength* (106V/m) Pyrex glass Silicone oil Strontium titanate Teflon Vacuum Water 5.6 2.5 233 2.1 1.000 00 80 Table A22 Dielectric constants 14 15 8 60 \u221e 3 830 Appendix A \u2022 Reference Tables Access for free at openstax.org. INDEX A aberration 505 absolute zero 329 absorption spectrum 725 Acceleration 73, 93, 93, 94, 104, 116, 122, 128, 205 acceleration due to gravity 104 Accuracy 26 acoustics 8 activity 744 air resistance 162 alpha (\u03b1) decay 738 alpha particle 738 alternating current 606 ampere 604 amplitude 180, 390, 417, 457 amplitude modulation 461 analytical method 153 Anger camera 758 angle of incidence 478 angle of reflection 478 angle of refraction 491 angle of rotation 198 angular acceleration 212, 261 angular momentum 261 angular velocity 198, 206 annihilation 753, 785 Antimatter 784 antinode 403, 437 Antoine Henri Becquerel 736 aphelion 230 arc length 198 atomic number 735 atoms 8 Average acceleration 94 Average speed 62 Average velocity 63 B baryons 784 beat frequency 436 Beats 436 becquerel 744 beta ( \u03b2 \u03b2 ) decay 738 Big Bang 792 binding energy 317, 699 blackbody 692 Boltzmann constant 359 bottom 781 C capacitor 582 carbon-14 dating 745 carrier particles 773 Celsius scale 328 central axis 481 centrifugal force 205 centripetal acceleration 205, 213 centripetal force 207 cesium atomic clock 20 chain reaction 749 change in momentum 254 charmed 781 chromatic aberration 505 circuit diagrams 613 Circular motion 198, 205 classical physics 8 closed system 286 closed-pipe resonator 439 coefficient of friction 119 Colliding beams 778 collision 256 collisions 262 color 782 complex machine 293 component 154 Compton 705 Compton effect 705 concave lens 499 concave mirror 481 Condensation 341 Index 831 conduction 334, 556 conductors 555 conservation 259 Constant acceleration 99 constructive interference 400 convection 334 conventional current 605 converging lens 498 conversion factor 24, 24 convex lens 498 convex mirror 481 Copernican model 232 corner reflector 494 Coulomb 562 coulomb force 772 Coulomb\u2019s constant 563 Coulomb\u2019s law 563 critical angle 493 critical mass 749 Curie temperature 654 cyclical process 373 cyclotron 777 D damping 435 decay constant 744 decibels 424 deformation 178 degree Celsius ( \u00b0C ( \u00b0C ) 328 degree Fahrenheit ( \u00b0F ( \u00b0F ) 328 dependent variable 34, 67 derived units 19 destructive interference 401 dielectric 585 differential interference contrast (DIC) 535 diffraction 525 diffraction grating 533 diffused 479 direct current 606 direction 54 832 Index dispersion 491 displacement 57, 73, 94, 100 distance 56 diverging lens 499 domains 653 Doppler effect 430 down 780 dynamics 7, 116 E E.O. Lawrence 777 eccentricity 234 Edwin Hubble 792 efficiency output 293 Einstein 698 elastic collision 262 electric circuits 612 Electric current 604 electric eye 702 electric field 456, 567 Electric motors 665 electric potential 575 Electric potential", " Vaporization 341 variables 34 vector 58, 63, 123, 254 vector addition 144 vector quantities 95 Vector subtraction 146 Velocity 63, 73, 93, 94, 99, 116, 122, 205, 254 virtual image 479 visible light 456 voltage 575 W W bosons 775 watts 282 wave 390 wave velocity 394 wavefronts 525 wavelength 394, 419, 457 weak nuclear force 772 wedge 291 weight 119, 123, 129 Werner Heisenberg 733 wheel and axle 291 work 280 work\u2013energy theorem 281 X X-rays 457 Z Z bosons 775 zeroth law of thermodynamics 356 Access for free at openstax.org.s and prototypes on the basis of established criteria 4 Unit I 01-PearsonPhys20-Chap01 7/23/08 11:43 AM Page 5 1-1 QuickLab 1-1 QuickLab Match a Motion Problem What types of motions can you generate using ticker tape? Questions 1. Describe the spacing between dots on each ticker Materials clacker power supply ticker tape Procedure 1 Measure and cut 4 or 5 lengths of ticker tape of length 40 cm (approximately). 2 Pull the ticker tape through the clacker at an even speed. 3 Repeat step 2 for a new length of ticker tape. For each new ticker tape, pull with a different kind of motion: quickly, slowly, speeding up, slowing down, etc. tape. Is it even, increasing, or decreasing? 2. Use the spacing between dots to describe the motion of the ticker tape. Is the ticker tape speeding up, slowing down, or constant? 3. What aspect of motion does the spacing between dots represent? 4. Have you covered all possible motions in the runs you did? If not, which ones did you omit? Give reasons why you omitted some types of motion. Think About It 1. What is motion? 2. What types of motion can objects undergo? 3. What are some words used to describe motion? 4. How can you determine how far and how fast an object moves? 5. What does the term \u201cfalling\u201d mean? 6. Describe the motion of falling objects. 7. Which falls faster: a heavy object or a light object? 8. How can a ball that\u2019s moving upward still be falling? Discuss your answers in a small group and record them for later reference. As you complete each section of this chapter, review your answers to these questions. Note any changes in your ideas. Chapter 1 Graphs and equations describe motion in one dimension. 5 01-PearsonPhys20-Chap01 7/23/08 11:43 AM Page 6 1.1 The Language of Motion When describing motions, such as the ones in Figures 1.1 and 1.2, you can use many different expressions and words. The English language is rich with them. Many sports broadcasters invent words or expressions to convey action and to excite the imagination. Phrases such as \u201ca cannonating drive from the point\u201d or \u201cblistering speed\u201d have become commonplace in our sports lingo. In physics, you must be precise in your language and use clearly defined terms to describe motion. Kinematics is the branch of physics that describes motion. kinematics: a branch of physics that describes motion Figure 1.2 How would you describe the motions shown in these photos? origin: a reference point position: the straight-line distance between the origin and an object\u2019s location; includes magnitude and direction scalar quantity: a measurement that has magnitude only vector quantity: a measurement that has both magnitude and direction 6 Unit I Kinematics M I N D S O N How Do Objects Move? Study the photos in Figure 1.2. Describe the motion of the puck, the wheelchair, and the harpoon with respect to time. Jot down your descriptions and underline key words associated with motion. Compare your key words with those of a partner. Compile a class list on the chalkboard. When describing motion, certain words, such as speed, acceleration, and velocity, are common. These words have slightly different meanings in physics than they do in everyday speech, as you will learn in the following subsection. Physics Terms It\u2019s Saturday night and the Edmonton Oilers are playing the Calgary Flames. In order to locate players on the ice, you need a reference system. In this case, select the centre of the ice as the reference point, or origin. You can then measure the straight-line distances, d, of players from the origin, such as 5.0 m. If you specify a direction from the origin along with the distance, then you define a player\u2019s position, d, for example, 5.0 m [E] (Figure 1.3). The arrow over the variable indicates that the variable is a vector quantity. The number and unit are called the magnitude of the vector. Distance, which has a magnitude but no direction", " associated with it, is an example of a scalar quantity. Vector quantities have both magnitude and direction. Position is an example of a vector quantity. 01-PearsonPhys20-Chap01 7/23/08 11:43 AM Page 7 W N S E di 5.0 m origin 26.0 m Figure 1.3 The player\u2019s position is 5.0 m [east of the origin] or simply 5.0 m [E]. The player is at a distance of 5.0 m from the origin. 60.0 m If the player, initially 5.0 m [east of the origin], skates to the east end of the rink to the goal area, his position changes. It is now 25.0 m [east of the origin] or 25.0 m [E] (Figure 1.4). You can state that he has travelled a straight-line distance of 20.0 m, and has a displacement of 20.0 m [E] relative to his initial position. W N S E 25.0 m df 26.0 m origin 60.0 m PHYSICS INSIGHT Technically, if you are standing away from the origin, you are displaced a certain distance and in a certain direction. However, the sign is not used with position unless the object you are referring to has moved from the origin to its current position, that is, unless the object has experienced a change in position. Figure 1.4 The player\u2019s position has changed. A change in position is called displacement. Distance travelled is the length of the path taken to move from one position to another, regardless of direction. Displacement, d, is the change in position. The player\u2019s displacement is written as distance: the length of the path taken to move from one position to another d 20.0 m [E] where is the Greek letter delta that means \u201cchange in.\u201d Calculate the change in a quantity by subtracting the initial quantity from the final quantity. In algebraic notation, R Rf Ri. You can calculate the displacement of the player in the following manner: d d f d i 25.0 m [E] 5.0 [E] 20.0 m [E] displacement: a straight line between initial and final positions; includes magnitude and direction info BIT Pilots use radar vectors when landing their aircraft. Radar vectors are instructions to fly in a particular direction and usually include altitude and speed restrictions. Chapter 1 Graphs and equations describe motion in one dimension. 7 01-PearsonPhys20-Chap01 7/23/08 11:43 AM Page 8 Sign Conventions How would you determine your final distance and displacement if you moved from a position 5.0 m [W] to a position 10.0 m [E] (Figure 1.5)? W 10.0 m 8.0 m 6.0 m 4.0 m 2.0 m 2.0 m 4.0 m 6.0 m 8.0 m 0.0 m E 10.0 m Figure 1.5 The person travels a distance of 5.0 m 10.0 m 15.0 m. What is the person\u2019s displacement? What is the person\u2019s final position relative to the bus stop? To calculate the distance travelled in the above scenario, you need only add the magnitudes of the two position vectors. d 5.0 m 10.0 m 15.0 m To find displacement, you need to subtract the initial position, d i, f. Let d from the final position, d 5.0 m [W] and d f 10.0 m [E]. i d d f d i 10.0 m [E] 5.0 m [W] Note that subtracting a vector is the same as adding its opposite, so the negative west direction is the same as the positive east direction. d 10.0 m [E] 5.0 m [W] 10.0 m [E] 5.0 m [E] 15.0 m [E] PHYSICS INSIGHT When doing calculations with measured values, follow the rules on rounding and the number of significant digits. Refer to pages 876\u2013877 in this book. W N up E S down L R Figure 1.6 Let east be positive and west negative. Similarly, north, up, and right are usually designated as positive. Another way of solving for displacement is to designate the east direction as positive and the west direction as negative (Figure 1.6). The two position 10.0 m [E] 10.0 m. vectors become d i Now calculate displacement: 5.0 m [W] 5.0 m and d f d d f d i 10.0 m (5.0 m) 15.0 m Since east is positive, the positive sign indicates that the person has moved 15.0 m east. Practise finding position and displacement in the next Skills Practice and example.", " info BIT On April 26, 2004, Stephane Gras of France did 445 chin-ups in one hour. If you consider up as positive, then Gras made 445 positive displacements and 445 negative displacements, meaning that his net displacement was zero! 8 Unit I Kinematics 01-PearsonPhys20-Chap01 7/23/08 11:43 AM Page Finding Position and Displacement 1. Create a scale using the dimensions of the hockey rink (Figure 1.7). Measuring from the centre of the player\u2019s helmet, (a) find each player\u2019s position relative to the north 26.0 m and south sides of the rink. (b) find each player\u2019s position relative to the east and west sides of the rink. (c) If the player moves from position 2 to position 4 on the rink, what is his displacement? 1 2 5 origin 4 3 60.0 m Figure 1.7 Example 1.1 A traveller initially standing 1.5 m to the right of the inukshuk moves so that she is 3.5 m to the left of the inukshuk (Figure 1.8). Determine the traveller\u2019s displacement algebraically (a) using directions (b) using plus and minus signs Given d i d f 1.5 m [right] 3.5 m [left] Required displacement (d ) inukshuk 3.5 m [left] origin 1.5 m [right] Figure 1.8 Analysis and Solution To find displacement, use the equation d (a) d d i d f 3.5 m [left] 1.5 m [right] 3.5 m [left] (1.5 m [left]) 3.5 m [left] 1.5 m [left] 5.0 m [left] d d i. f i d d f d (b) Consider right to be positive. 1.5 m [right] 1.5 m 3.5 m [left] 3.5 m d f 3.5 m (1.5 m) 3.5 m 1.5 m 5.0 m d i The answer is negative, so the direction is left. Paraphrase The traveller\u2019s displacement is 5.0 m [left] of her initial position. Note that the direction of displacement is relative to initial position, whereas the direction of position is relative to the designated origin, in this case, the inukshuk. Practice Problems 1. Sprinting drills include running 40.0 m [N], walking 20.0 m [N], and then sprinting 100.0 m [N]. What is the sprinter\u2019s displacement from the initial position? 2. To perform a give and go, a basketball player fakes out the defence by moving 0.75 m [right] and then 3.50 m [left]. What is the player\u2019s displacement from the starting position? 3. While building a wall, a bricklayer sweeps the cement back and forth. If she swings her hand back and forth, a distance of 1.70 m, four times, calculate the distance and displacement her hand travels during that time. Answers 1. 160.0 m [N] 2. 2.75 m [left] 3. 6.80 m, 0 m Chapter 1 Graphs and equations describe motion in one dimension. 9 01-PearsonPhys20-Chap01 7/23/08 11:43 AM Page 10 For all subsequent problems in this book, you will be using plus and minus signs to indicate direction. This method is more flexible for problem solving and easier to use. Like distance and displacement, speed and velocity is another scalar-vector pair. Speed is the rate at which an object moves. It is a scalar quantity, so it has magnitude only; for example, v 50 km/h (Figure 1.9). Velocity is a vector quantity, so it has both magnitude (speed) and direction. If you are travelling south from Fort McMurray to Lethbridge at 50 km/h, your velocity is written as v 50 km/h [S]. If you designate south as negative, then v 50 km/h. Acceleration is a vector quantity that represents the rate of change of velocity. You will study aspects of displacement, velocity, and acceleration, and their interrelationships, in the sections that follow. Figure 1.9 Scalar or vector? 1.1 Check and Reflect 1.1 Check and Reflect Knowledge 1. What two categories of terms are used to describe motion? Give an example of each. 2. Compare and contrast distance and displacement. 3. What is the significance of a reference point? Applications 4. Draw a seating plan using the statements below. (a) Chad is 2.0 m [left] of Dolores. (b) Ed is 4.5 m [right] of Chad. (c) Greg is", " 7.5 m [left] of Chad. (d) Hannah is 1.0 m [right] of Ed. (e) What is the displacement of a teacher who walks from Greg to Hannah? 5. A person\u2019s displacement is 50.0 km [W]. What is his final position if he started at 5.0 km [E]? 6. Using an autuk (a type of sealskin racquet), two children play catch. Standing 3.0 m apart, the child on the right tosses the ball to the child on the left, and then moves 5.0 m [right] to catch the ball again. Determine the horizontal distance and displacement the ball travels from its initial position (ignore any vertical motion). 3.0 m 5.0 m 7. Below is a seating plan for the head table at a wedding reception. Relative to the bride, describe the positions of the groom, best man, maid of honour, and flower girl. 0.75 m 0.75 m 0.50 m 0.75 m 0.75 m Flower girl Best man Bride Groom Maid of Honour Ring boy e TEST To check your understanding of scalar and vector quantities, follow the eTest links at www.pearsoned.ca/school/physicssource. 10 Unit I Kinematics 01-PearsonPhys20-Chap01 7/23/08 11:43 AM Page 11 1.2 Position-time Graphs and Uniform Motion You are competing to win the Masters Golf Tournament. The hole is 5.0 m away (Figure 1.10). You gently hit the ball with your club and hold your breath. Time seems to stop. Then, 5.0 s later, it rolls into the hole. You have won the tournament! From section 1.1, you know that displacement is the change in an object\u2019s position. If you replay the sequence of motions of your winning putt in 1.0-s intervals, you can measure the displacements of the golf ball from you, the putter, to the hole (Figure 1.11). Figure 1.10 You can represent motion in sports using vectors and graphs. 0.0 m 0.0 s origin 1.0 m 1.0 s 2.0 m 2.0 s 3.0 m 3.0 s 4.0 m 4.0 s 5.0 m 5.0 s Figure 1.11 What is the golf ball\u2019s displacement after each second? Table 1.1 displays the data from Figure 1.11 for the golf ball\u2019s position from you at 1.0-s intervals. By graphing the data, you can visualize the motion of the golf ball more clearly (Figure 1.12). \u25bc Table 1.1 Position-time data Time (s) Position (m [right]) 0.0 1.0 2.0 3.0 4.0 5.0 t0 t1 t2 t3 t4 t5 0.0 1.0 2.0 3.0 4.0 5.0 Velocity Position vs. Time 6.0 5.0 4.0 3.0 2.0 1.0 0..0 1.0 2.0 3.0 4.0 5.0 Time (s) Figure 1.12 A position-time graph of the golf ball Notice that the graph in Figure 1.12 is a straight line. A straight line has a constant slope. What does constant slope tell you about the ball\u2019s motion? To answer this question, calculate the slope and keep track of the units. Designate toward the hole, to the right, as the positive direction. Chapter 1 Graphs and equations describe motion in one dimension. 11 01-PearsonPhys20-Chap01 7/23/08 11:43 AM Page 12 e SIM Practise calculating average speed and average velocity. Go to www.pearsoned.ca/ school/physicssource. Recall that slope rise run. For position\u2013time graphs, this equation becomes slope change in position change in time A change in position is displacement. So, the equation for slope becomes PHYSICS INSIGHT Speed has magnitude only. Velocity has both magnitude and direction. velocity: rate of change in position slope d t d d i f ti tf 5.0 m 0.0 m 5.0 s 0.0 s 1.0 m/s The answer is positive, so the golf ball moves at a rate of 1.0 m/s [right]. Notice that the units are m/s (read metres per second). These units indicate speed or velocity. Since displacement is a vector quantity, the slope of the position-time graph in Figure 1.12 gives you the velocity, v, of the ball: the change in position per unit time. Because you have calculated velocity over a time interval rather than at an instant in time, it is the average velocity. d v t Speed", " and Velocity 4 m/s 4 m/s Figure 1.13 Objects with the same speed can have different velocities. Objects travelling at the same speed can have different velocities. For example, a tram carries passengers across a ravine at a constant speed. A passenger going to the observation deck has a velocity of 4 m/s [right] and a passenger leaving the deck has a velocity of 4 m/s [left] (Figure 1.13). Their speeds are the same, but because they are travelling in opposite directions, their velocities are different. 1-2 Decision-Making Analysis 1-2 Decision-Making Analysis Traffic Safety Is Everyone\u2019s Business The Issue In an average year in Alberta, traffic accidents claim six times more lives than homicide, eight times more lives than AIDS, and 100 times more lives than meningitis. Collisions represent one of the greatest threats to public safety. Background Information In the Canadian 2002 Nerves of Steel: Aggressive Driving Study, speeding was identified as one of two common aggressive behaviours that contribute to a significant percentage of all crashes. The Alberta Motor Association\u2019s Alberta Traffic Safety Progress Report has suggested that a province-wide speed management program could significantly improve levels of road safety, decreasing both speed and casualties. One suggested program is the implementation of the vehicle tachograph, a device required in Europe to improve road safety. 12 Unit I Kinematics e WEB To learn more about how speeding is a key contributing factor in casualty collisions in Alberta, follow the links at www.pearsoned.ca/school/ physicssource. 01-PearsonPhys20-Chap01 7/23/08 11:43 AM Page 13 Analysis Your group has been asked to research different traffic safety initiatives. The government will use the results of your research to make the most appropriate decision. 1. Research (a) how state- or province-wide speed management programs have influenced driver behaviour (b) the societal cost of vehicle crashes (c) driver attitudes toward enforcement of and education about traffic safety issues 2. Analyze your research and decide which management program should be used. 3. Once your group has completed a written report recommending a particular program, present the report to the rest of the class, who will act as representatives of the government and the community. So far, you have learned that the slope of a position-time graph represents a rate of change in position, or velocity. If an object moves at constant velocity (constant magnitude and direction), the object is undergoing uniform motion. A position-time graph for an object at rest is a horizontal line (Figure 1.14). An object at rest is still said to be undergoing uniform motion because its change in position remains constant over equal time intervals. Concept Check (a) Describe the position of dots on a ticker tape at rest. What is the slope of the graph in Figure 1.14? (b) Describe the shape of a position-time graph for an object travelling at a constant velocity. List three possibilities. Frame of Reference uniform motion: constant velocity (motion or rest) at rest: not moving; stationary ) ].0 5.0 4.0 3.0 2.0 1.0 0.0 0.0 Position vs. Time 1.0 2.0 3.0 Time (min) 4.0 5.0 Figure 1.14 A position-time graph for a stationary object If you were to designate the hole, rather than the putter, as the origin (starting point) in the golf tournament (Figure 1.15(a)), your data table would start at 5.0 m [left] at time 0.0 s, instead of at 0.0 m and 0.0 s (Table 1.2). The values to the left of the hole are positive. \u25bc Table 1.2 Position-time data Time (s) Position (m [left]) t0 t1 t2 t3 t4 t5 0.0 1.0 2.0 3.0 4.0 5.0 5.0 4.0 3.0 2.0 1.0 0.0 5.0 m 0.0 s 4.0 m 1.0 s 3.0 m 2.0 s 2.0 m 3.0 s 1.0 m 4.0 s 0.0 m 5.0 s origin Figure 1.15(a) Designating an origin is arbitrary. In this example, the hole is the origin and all positions are measured relative to it. Chapter 1 Graphs and equations describe motion in one dimension. 13 01-PearsonPhys20-Chap01 7/23/08 11:43 AM Page 14 info BIT On May 31, 2004 in Moscow, Ashrita Furman of the USA walked 1.6 km while continuously hulahooping in 14 min 25 s. He also holds the world record for the fastest time for pushing an orange with his nose. On August", " 12, 2004, he pushed an orange 1.6 km in 24 min 36 s. What was his speed, in km/h and m/s, for each case? (See Unit Conversions on page 878.) e WEB In November 2004, at an altitude of 33 000 m, the X-43A recorded a speed of Mach 9. Use the Internet or your local library to research the term \u201cMach\u201d as used to describe the speed of an object. How did this term originate? What is the difference between Mach and ultrasonic? Write a brief summary of your findings. To learn more about Mach, follow the links at www.pearsoned.ca/school/ physicssource. 14 Unit I Kinematics The corresponding position-time graph is shown in Figure 1.15(b). ) ].0 5.0 4.0 3.0 2.0 1.0 0.0 0.0 Position vs. Time 1.0 2.0 3.0 4.0 5.0 Time (s) Figure 1.15(b) Compare this graph with the graph in Figure 1.12. If you change your reference frame, the position-time graph also changes. From the graph, slope v d t d d i f ti tf 0.0 m (5.0 m) 5.0 s 0.0 s 1.0 m/s The velocity of the golf ball is 1.0 m/s. What does the negative sign mean? It means the ball is travelling opposite to the direction to which the positions of the ball are measured. It does not mean that the golf ball is slowing down. Since positions, now measured to the left of the hole (the new origin) are designated positive, any motion directed to the right is described as being negative. In this case, you can also see that the ball is decreasing its position from the origin with increasing time. The ball travels to the right toward the hole, decreasing its position each second by 1.0 m \u2014 it travels at 1.0 m/s to the right or 1.0 m/s [left] as indicated by the downward slope on the graph. Concept Check Determine how the velocity of the golf ball can be positive if the hole is at the origin. Below is a summary of what you have learned: \u2022 The slope of a position-time graph represents velocity. \u2022 The velocity is the average velocity for the time interval. \u2022 Your choice of reference frame affects the direction (sign) of your answer. \u2022 A straight line on a position-time graph represents uniform motion. Comparing the Motion of Two or More Objects on a Position-time Graph You can represent the motions of two objects on one graph, as long as the origin is the same for both objects. You can then use the graph to compare their motions, as in the next example. 01-PearsonPhys20-Chap01 7/23/08 11:43 AM Page 15 Example 1.2 At the end of the school day, student A and student B say goodbye and head in opposite directions, walking at constant rates. Student B heads west to the bus stop while student A walks east to her house. After 3.0 min, student A is 300 m [E] and student B is 450 m [W] (Figure 1.16). (a) Graph the position of each student on one graph after 3.0 min. (b) Determine the velocity in m/s of each student algebraically. Given Choose east to be positive. dd A dd B t 3.0 min 300 m [E] 300 m 450 m [W] 450 m Required (a) position\u2013time graph (b) velocity (v B) A and v BUS STOP bus stop school Lakeview School W N S E+ Student B\u2019s position: 450 m [W] origin Student A\u2019s position: 300 m [E] Figure 1.16 Analysis and Solution (a) Since east is the positive direction, plot student A\u2019s position (3.0 min, 300 m) above the time axis and student B\u2019s position (3.0 min, 450 m) below the time axis (Figure 1.17). Position vs. Time 1.0 2.0 3.0 Time (min 600 400 200 0 200 400 600 Figure 1.17 (b) Convert time in minutes to time in seconds. d t to find the velocity Then use the equation v of each student. 60 s 1 min t 3.0 min 180 s v A 300 m 180 s 1.7 m/s The sign is positive, so the direction is east. v B 450 m 180 s 2.5 m/s The sign is negative, so the direction is west. Paraphrase (b) Student A\u2019s velocity is 1.7 m/s [E] and student B\u2019s velocity is 2.5 m/s [W", "]. Practice Problems 1. A wildlife biologist measures how long it takes four animals to cover a displacement of 200 m [forward]. (a) Graph the data from the table below. (b) Determine each animal\u2019s average velocity. Animal Time taken (s) Elk Coyote Grizzly bear Moose 10.0 10.4 18.0 12.9 Answers 1. (a 250.0 200.0 150.0 100.0 50.0 0.0 Position vs. Time 0 2 4 6 8 10 12 14 16 18 20 Time (s) (b) Elk: 20.0 m/s [forward] Coyote: 19.2 m/s [forward] Grizzly bear: 11.1 m/s [forward] Moose: 15.5 m/s [forward] Chapter 1 Graphs and equations describe motion in one dimension. 15 01-PearsonPhys20-Chap01 7/23/08 11:43 AM Page 16 So far, you have learned that the slope of a position\u2013time graph represents velocity. By comparing the slopes of two graphs, you can determine which object is moving faster. From the slopes of the graphs in Figure 1.17, which student is moving faster? When you represent the motions of two objects on the same graph, you can also tell whether the objects are approaching or moving apart by checking if the lines are converging or diverging. An important event occurs at the point where the two lines intersect. Both objects have the same position, so the objects meet at this point. Concept Check Describe the shape of a graph showing the motion of two objects approaching each other. In the next example, two objects start at different times and have different speeds. You will graphically find their meeting point. Example 1.3 Two rollerbladers, A and B, are having a race. B gives A a head start of 5.0 s (Figure 1.18). Each rollerblader moves with a constant velocity. Assume that the time taken to reach constant velocity is negligible. If A travels 100.0 m [right] in 20.0 s and B travels 112.5 m [right] in 15.0 s, (a) graph the motions of both rollerbladers on the same graph. (b) find the time, position, and displacement at which B catches up with A. B A Practice Problems 1. The two rollerbladers in Example 1.3 have a second race in which they each travel the original time and distance. In this race, they start at the same time, but B\u2019s initial position is 10.0 m left of A. Take the starting position of A as the reference. (a) Graph the motions of the rollerbladers. (b) Find the time, position, and B\u2019s displacement at which B catches up with A. Answers 1. (b) t 4.0 s 20.0 m [right] d 30.0 m [right] d 16 Unit I Kinematics distance travelled by A in 5.0 s Figure 1.18 Given Choose right to be positive. d A tA d B tB 100.0 m [right] 100.0 m 20.0 s 112.5 m [right] 112.5 m 15.0 s, started 5.0 s later Required (a) position-time graph (b) time (t), position (d catches up with A ), and displacement (d ) when B 01-PearsonPhys20-Chap01 7/23/08 11:43 AM Page 17 Analysis and Solution (a) Assume that t 0.0 s at the start of A\u2019s motion. Thus, the position-time graph of A\u2019s motion starts at the origin. A\u2019s final position is 100.0 m at 20.0 s. The position-time graph for B\u2019s motion starts at 0.0 m and 5.0 s (because B started 5.0 s after A). B starts moving after 5.0 s for 15.0 s. Thus, at 20.0 s (5.0 s 15.0 s), B\u2019s position is 112.5 m. Each rollerblader travels with a constant velocity, so the lines connecting their initial and final positions are straight (Figure 1.19(a)). ) ] 120.0 100.0 80.0 60.0 40.0 20.0 0.0 Position vs. Time B A 0.0 5.0 10.0 15.0 20.0 25.0 Time (s) Figure 1.19(a) (b) On the graph in Figure 1.19(a), look for a point of intersection. At this point, both rollerbladers have the same final position. From the graph, you can see that this point occurs at t 15.0 s. The corresponding position is 75.0 m (Figure 1", ".19(b)). ) ] 120.0 100.0 80.0 60.0 40.0 20.0 0.0 Position vs. Time B A 0.0 5.0 10.0 15.0 20.0 25.0 Time (s) Figure 1.19(b) To find B\u2019s displacement, find the change in position: d Both A and B started from the same position, d both have the same final position at the point of intersection, d f d 75.0 m. 75.0 m 0.0 m i d d i. f 0. Since they 75.0 m The answer is positive, so the direction is to the right. Paraphrase (b) B catches up with A 15.0 s after A started. B\u2019s position and displacement are 75.0 m [right] of the origin. Chapter 1 Graphs and equations describe motion in one dimension. 17 01-PearsonPhys20-Chap01 7/23/08 11:43 AM Page 18 Example 1.4 From the graph in Example 1.3, find the velocities of the two rollerbladers. Given Choose right to be positive. At the point of intersection (Figure 1.19(b)), d A tA d B tB 75.0 m [right] 75.0 m 15.0 s 75.0 m [right] 75.0 m 15.0 s 5.0 s 10.0 s Required velocities of A and B (vv A, v B) Analysis and Solution To find the velocity of each rollerblader, remember that the slope of a position\u2013time graph is velocity. Because the motions are uniform, the slopes will be constant for each rollerblader. vv vvv A d t 75.0 m 0.0 m 15.0 s 0.0 s 5.0 m/s vvv B 75.0 m 0.0 m 15.0 s 5.0 s 75.0 m 0.0 m 10.0 s 7.5 m/s The answers are both positive, so the direction is to the right. You can see that, in order for B to cover the same distance as A, B must move faster because B started later. Paraphrase A\u2019s velocity is 5.0 m/s [right] and B\u2019s velocity is 7.5 m/s [right]. Practice Problems 1. Suppose rollerblader B gives A a head start of 5.0 s and takes 10.0 s to catch up with A at 100.0 m [right]. Determine the velocities of rollerbladers A and B. Answers 1. A: 6.67 m/s [right] B: 10.0 m/s [right] 18 Unit I Kinematics 01-PearsonPhys20-Chap01 7/23/08 11:43 AM Page 19 1-3 Inquiry Lab 1-3 Inquiry Lab Car Activity Question What are the speeds of two different toy cars? If one car is released 3.0 s after the other, where will they meet? Variables Identify the manipulated, responding, and controlled variables. Materials and Equipment two battery-operated toy cars ticker tape carbon disk spark timer (60 Hz) masking tape ruler graph paper Procedure 1 On a flat surface, such as the floor or lab bench, mark the initial starting position of car 1 with masking tape. Required Skills Initiating and Planning Performing and Recording Analyzing and Interpreting Communication and Teamwork Analysis 1. Draw a line through the first dot on each ticker tape and label it t = 0 s. 2. Depending on the calibration of your ticker timer, count from the starting position, and place a mark after a fixed number of dots, e.g., 6, 12, 18, 24, etc. Label each mark t1, t2, etc. On a 60-Hz timer, every sixth dot represents 0.10 s. 3. Measure the distance from t = 0 to t1, t = 0 to t2, t = 0 to t3, etc. Record the data in a position\u2013time table. 4. Using an appropriate scale, graph each set of data for each toy car, separately. 5. Determine the slope of the line of best fit for each graph. See pages 872\u2013873 for explicit instruction on how to draw a line of best fit. 2 Using masking tape, attach 1.0 m of ticker tape to the 6. What is the speed of each toy car? end of car 1. 3 Thread the ticker tape through the spark timer (Figure 1.20). 4 Turn the car on. 7. How do the speeds of car 1 and car 2 compare? 8. Assuming uniform motion, how far would car 1 travel in 15 s? 9. Assuming uniform motion, how long would it take car", " 2 5 Turn the spark timer on as you release the car from its to travel 30 m? initial position. 6 Observe the path of car 1 until the ticker tape is used up. Label the ticker tape \u201ccar 1.\u201d 7 Repeat steps 2\u20136 for car 2. recording timer recording tape toy car Figure 1.20 10. Imagine that you release the faster car 3.0 s after the slower car. Graphically determine the position where the two cars meet. Assume uniform motion. e LAB For a probeware activity, go to www.pearsoned.ca/school/physicssource. In summary, you can see how a position-time graph helps you visualize the event you are analyzing. Calculating the slope of a position-time graph provides new information about the motion, namely, the object\u2019s velocity. In the next sections, you will expand on your graphing knowledge by analyzing motion using a velocity-time graph. Chapter 1 Graphs and equations describe motion in one dimension. 19 01-PearsonPhys20-Chap01 7/23/08 11:43 AM Page 20 1.2 Check and Reflect 1.2 Check and Reflect Knowledge Applications 1. For an object at rest, what quantities of 8. Two children on racing bikes start from motion remain the same over equal time intervals? 2. For an object travelling at a constant velocity, what quantity of motion remains the same over equal time intervals? 3. Match each ticker tape below with the correct position-time graph. (i) (ii) (iii) (iv.0 7.0 6.0 5.0 4.0 3.0 2.0 1.0 0.0 0.0 Position vs. Time A B C D 1.0 2.0 3.0 Time (s) 4.0 5.0 4. Two friends start walking on a football field in the same direction. Person A walks twice as fast as person B. However, person B has a head start of 20.0 m. If person A walks at 3.0 m/s, find the distance between the two friends after walking for 20.0 s and determine who is ahead at this time. Sketch a position-time graph for both people. 5. A camper kayaks 16 km [E] from a camping site, stops, and then paddles 23 km [W]. What is the camper\u2019s final position with respect to the campsite? 6. Sketch a position-time graph for a bear starting 1.2 m from a reference point, walking slowly away at constant velocity for 3.0 s, stopping for 5.0 s, backing up at half the speed for 2.0 s, and finally stopping. 7. Sketch a position-time graph for a student (a) walking east to school with a constant velocity (b) stopping at the school, which is 5 km east of home (c) cycling home with a constant velocity 20 Unit I Kinematics the same reference point. Child A travels 5.0 m/s [right] and child B travels 4.5 m/s [right]. How much farther from the point of origin is child A than child B after 5.0 s? 9. Insect A moves 5.0 m/min and insect B moves 9.0 cm/s. Determine which insect is ahead and by how much after 3.0 min. Assume both insects are moving in the same direction. 10. Describe the motion in each lettered stage for the object depicted by the positiontime graph below 20 16 12 8 4 0 Position vs. Time B A C 0 4 8 12 Time (s) 16 20 11. A mosquito flies toward you with a velocity of 2.4 km/h [E]. If a distance of 35.0 m separates you and the mosquito initially, at what point (distance and time) will the mosquito hit your sunglasses if you are travelling toward the mosquito with a speed of 2.0 m/s and the mosquito is travelling in a straight path? 12. Spotting a friend 5.0 m directly in front of you, walking 2.0 m/s [N], you start walking 2.25 m/s [N] to catch up. How long will it take for you to intercept your friend and what will be your displacement? 13. Two vehicles, separated by a distance of 450 m, travel in opposite directions toward a traffic light. When will the vehicles pass one another if vehicle A is travelling 35 km/h and is 300 m [E] of the traffic light while vehicle B is travelling 40 km/h? When will each vehicle pass the traffic light, assuming the light remains green the entire time? e TEST To check your understanding of uniform motion, follow the eTest links at www.pearsoned.ca/school/physicssource. 01-PearsonPhys20-Chap01 7/23/08", " 11:43 AM Page 21 1.3 Velocity-time Graphs: Uniform and Nonuniform Motion Recently installed video screens in aircraft provide passengers with information about the aircraft\u2019s velocity during the flight (Figure 1.21). x km 0.0 h x km 1.0 h x km 2.0 h x km 3.0 h x km 4.0 h x km 5.0 h Figure 1.22 A plane flies at a constant speed, so the distances within each time interval are equal. Break the plane\u2019s motion into a series of snapshots. Record your data in a data table and then graph it. Figure 1.22 shows the data of the plane\u2019s path. Like position-time graphs, velocity-time graphs provide useful information about the motion of an object. The shape of the velocity-time graph reveals whether the object is at rest, moving at constant speed, speeding up, or slowing down. Suppose an airplane has a cruising altitude of 10 600 m and travels at a constant velocity of 900 km/h [E] for 5.0 h. Table 1.3 shows the velocity-time data for the airplane. If you graph the data, you can determine the relationship between the two variables, velocity and time (Figure 1.23). \u25bc Table 1.3 Time (h) Velocity (km/h) [E] Figure 1.21 Video screens are an example of an application of velocitytime graphs. 0.0 1.0 2.0 3.0 4.0 5. 900 800 700 600 500 400 300 200 100 0 0.0 900 900 900 900 900 900 Velocity vs. Time for an Airplane 1.0 2.0 3.0 4.0 5.0 Time (h) Figure 1.23 A velocity-time graph for an airflight Chapter 1 Graphs and equations describe motion in one dimension. 21 01-PearsonPhys20-Chap01 7/23/08 11:43 AM Page 22 Designating east as the positive direction, the slope of the velocity-time graph is: slope rise run v t v v i f ti tf m k k m 900 900 h h 5.0 h 1.0 h 0 km/h2 From the graph in Figure 1.23, there is no change in the plane\u2019s velocity, so the slope of the velocity-time graph is zero. Notice the units of the slope of the velocity-time graph: km/h2. These units are units of acceleration. Because the plane is moving at a constant velocity, its acceleration is zero. In general, you can recognize acceleration values by their units, which are always distance divided by time squared. In physics, the standard units for acceleration are metres per second per second, which is generally abbreviated to m/s2 (read metres per second squared). e TECH Determine the velocity of an object based on the shape Concept Check of its position-time graph. Go to www.pearsoned.ca/ school/physicssource. (a) What does the slope of a position-time graph represent? (b) What does the slope of a velocity-time graph represent? Non-uniform Motion Although objects may experience constant velocity over short time intervals, even a car operating on cruise control has fluctuations in speed or direction (Figure 1.24). How can you describe and illustrate a change of velocity using the concepts of kinematics? Recall from section 1.2 that an object moving at a constant velocity is undergoing uniform motion. But is uniform motion the only type of motion? Perform the next QuickLab to find out. Figure 1.24 Consider the kinds of changes in velocity this car experiences during the trip. 22 Unit I Kinematics 01-PearsonPhys20-Chap01 7/23/08 11:43 AM Page 23 1-4 QuickLab 1-4 QuickLab Match a Graph Problem What type of motion does each graph describe? 5 Print out the graphs from your experiment. For each graph, construct a table of values for position and time. Materials LM 1-1 (provided by your teacher) ruler motion sensor masking tape Procedure 1 Study the different position-time graphs on LM 1-1. With a partner, decide what type of motion each graph illustrates. 2 Set up the motion sensor to plot position vs. time. 3 Label a starting position with masking tape approximately 1 m in front of the motion sensor. Move away from the motion sensor in such a way that the graph of the motion captured approximates the one on the LM. 4 Switch roles with your partner and repeat steps 1\u20133. Questions 1. Describe your motion when a horizontal line was being produced on the position\u2013time graph. 2. What relationship exists between the type of motion and change in position? 3. Suggest two different ways in which you could classify the motion described by the four graphs. 4. What would the graph look like if you moved away from and then back toward", " the motion sensor? 5. What happens to the graph when you move away from your initial position and then move back toward and then beyond your initial position? e LAB For a probeware activity, go to www.pearsoned.ca/school/physicssource. Concept Check Which ticker tape in Figure 1.25 represents accelerated motion? Explain. Figure 1.25 Consider an object, such as a drag racer (Figure 1.26), starting from rest and reaching a constant velocity over a time interval (Figure 1.27). During this time interval, the vehicle has to change its velocity from a value of zero to a final non-zero value. An object whose velocity changes (increases or decreases) over a time interval is undergoing acceleration, represented by the variable a. Acceleration is a vector quantity. It is also called non-uniform motion because the object\u2019s speed or direction is changing. Figure 1.26 A drag racer accelerates from rest. acceleration: a vector quantity representing the change in velocity (magnitude or direction) per unit time scale 1.0 m 0.0 m 0.0 s 2.0 m 1.0 s 8.0 m 2.0 s 18.0 m 3.0 s Figure 1.27 This sequence illustrates a car undergoing non-uniform motion. Chapter 1 Graphs and equations describe motion in one dimension. 23 01-PearsonPhys20-Chap01 7/23/08 11:43 AM Page 24 PHYSICS INSIGHT An object is accelerating if it is speeding up, slowing down, or changing direction. The following scenario illustrates acceleration. A drag race is a 402-m (quarter-mile) contest between two vehicles. Starting from rest, the vehicles leave the starting line at the same time, and the first vehicle to cross the finish line is the winner. A fan records the position of her favourite vehicle during the drag race. Her results are recorded in Table 1.4. The position-time graph for this data is shown in Figure 1.28. From the graph, note that the object is speeding up because the displacement between data points increases for each successive time interval. Which ticker tape in Figure 1.25 matches the graph in Figure 1.28? \u25bc Table 1.4 Time (s) Position (m [forward]) 0.0 1.0 2.0 3.0 4.0 5.0 0.0 2.0 8.0 18.0 32.0 50.0 Position vs. Time for a Dragster ) ] 50.0 40.0 30.0 20.0 10.0 0.0 0.0 1.0 2.0 3.0 4.0 5.0 Time (s) Figure 1.28 What does the slope of the graph indicate about the speed of the car? Instantaneous Velocity Instantaneous velocity is the moment-to-moment measure of an object\u2019s velocity. Imagine recording the speed of your car once every second while driving north. These data form a series of instantaneous velocities that describe your trip in detail. Earlier in this section, you learned that determining the velocity of an object from a position-time graph requires calculating the slope of the position-time graph. But how can you obtain the slope of a curve? Remember that each point on the curve indicates the position of the object (in this case, the dragster) at an instant in time. To determine the velocity of an object at any instant, physicists use tangents. A tangent is a straight line that touches a curve at only one point (Figure 1.29(a)). Each tangent on a curve has a unique slope, which represents the velocity at that instant. In order for the object to be at that position, at that time, it must have an instantaneous velocity equal to the slope of the tangent at that point. Determining the slopes of the tangents at different points on a position-time curve gives the instantaneous velocities at different times. Consider forward to be the positive direction. PHYSICS INSIGHT When you calculate the slope of a line or curve at a single point, you are finding an instantaneous value. When you calculate the slope between two points, you are finding an average value. tangent: a straight line that touches a curved-line graph at only one point 24 Unit I Kinematics 01-PearsonPhys20-Chap01 7/23/08 11:43 AM Page 25 Position vs. Time for a Dragster Position vs. Time for a Dragster Position vs. Time for a Dragster 50.0 40.0 30.0 20.0 10..0 0.0 d t 1.0 2.0 3.0 Time (s) 4.0 5.0 50.0 40.0 30.0 20.0 10..0 0.0 d t 1.0 2.0 3.0 Time (s) 4.0", " 5.0 50.0 40.0 30.0 20.0 10..0 0.0 1.0 2.0 3.0 Time (s) 4.0 5.0 Figure 1.29(a) Figure 1.29(b) Figure 1.29(c) The slope of the tangent at 2.0 s is slope d t 14.0 m 0.0 m 3.0 s 1.0 s 14.0 m 2.0 s 7.0 m/s The slope of the tangent at 3.0 s is slope 30.0 m 0.0 m 4.0 s 1.75 s The slope of the tangent at 4.0 s is slope 47.0 m (15.0 m) 5.0 s 3.0 s 30.0 m 2.25 s 13 m/s 32.0 m 2.0 s 16 m/s The sign is positive, so at 2.0 s, the velocity of the dragster is 7.0 m/s [forward]. At 3.0 s, the velocity of the dragster is 13 m/s [forward]. At 4.0 s, the velocity of the dragster is 16 m/s [forward]. Using Slopes of Position-time Graphs to Draw Velocity-time Graphs You can now create a new table using the slopes of the position-time graphs in Figures 1.29(a), (b), and (c). See Table 1.5. Remember that the slope of a position-time graph is velocity. These slope values are actually instantaneous velocities at the given times. You can use these three velocities to draw a velocity-time graph (Figure 1.30). The resulting velocity-time graph is a straight line that goes through the origin when extended. This means that the dragster has started from rest (0 velocity). The graph has a positive slope. To find the meaning of slope, check the units of the slope of a velocity-time graph. They are (m/s)/s, which simplify to m/s2. These units are the units of acceleration. Since the velocity-time graph in this example is a straight line with non-zero slope, the acceleration of the object is constant, so the object must be undergoing uniformly accelerated motion. info BIT When jet fighters come in to land on an aircraft carrier, they stop so quickly that pilots sometimes lose consciousness for a few seconds. The same thing can happen when a pilot ejects from an aircraft, due to enormous acceleration. uniformly accelerated motion: constant change in velocity per unit time Chapter 1 Graphs and equations describe motion in one dimension. 25 01-PearsonPhys20-Chap01 7/23/08 11:43 AM Page 26 \u25bc Table 1.5 Time (s) Velocity (m/s [forward]) 2.0 3.0 4.0 7.0 13 16 ) ] 20 15 10 5 0 Velocity vs. Time 1.0 2.0 3.0 4.0 5.0 Time (s) Figure 1.30 This velocity-time graph represents an object undergoing uniformly accelerated motion. 4.5 4.0 3.5 3.0 2.5 2.0 1.5 1.0 0.5 0.0 0.0 Acceleration vs. Time Just as the slope of a position-time graph reveals the rate at which position changes (velocity), the slope of a velocity-time graph reveals the rate at which velocity changes (acceleration). Calculate the slope of the line in Figure 1.30 as follows, designating forward as positive: 1.0 2.0 Time (s) 3.0 4.0 slope rise run Figure 1.31 An accelerationtime graph for an object undergoing uniformly accelerated motion is a straight line with zero slope. PHYSICS INSIGHT If the acceleration-time graph has a non-zero slope, the acceleration is changing (is nonuniform). The slope of an acceleration-time graph is called jerk, with units m/s3. v v i f t t i f 10 m/s (4 m/s) 2.5 s 1.0 s 4 m/s2 The answer is positive, so the car is accelerating at 4 m/s2 [forward]. The resulting acceleration-time graph is shown in Figure 1.31. You know that the velocity-time graph for an object undergoing uniform motion is a horizontal line (with zero slope, as in Figure 1.23). Similarly, a horizontal line on an acceleration-time graph indicates uniform acceleration. Concept Check If the position-time graph for an object undergoing positive acceleration is a parabola, such as the one in Figure 1.28, what is the shape of the position-time graph for an object undergoing negative acceleration? What would a ticker tape of the motion of an object that is slowing down look like? After driving your all-terrain vehicle", " (ATV, Figure 1.32) through a field, you see a wide river just ahead, so you quickly bring the vehicle to a complete stop. Notice in Figure 1.33 that, as your ATV slows down, the displacement in each time interval decreases. Figure 1.32 ATVs can undergo a wide variety of motions. 26 Unit I Kinematics 01-PearsonPhys20-Chap01 7/23/08 11:43 AM Page 27 scale 1.0 m 0.0 m 0.0 s 13.5 m 1.0 s 24.0 m 2.0 s 31.5 m 3.0 s 36.0 m 4.0 s 37.5 m 5.0 s Figure 1.33 This ATV is undergoing non-uniform motion. It is accelerating, in this case, slowing down. Example 1.5 shows the calculations and resulting velocity-time graph for an object that is slowing down uniformly. Example 1.5 The position-time data for an ATV approaching a river are given in Table 1.6. Using these data, (a) draw a position-time graph (b) draw a velocity-time graph (c) calculate acceleration Analysis and Solution Designate the forward direction as positive. (a) For the position-time graph, \u25bc Table 1.6 Time (s) Position (m [forward]) 0.0 1.0 2.0 3.0 4.0 5.0 0.0 13.5 24.0 31.5 36.0 37.5 plot the data in Table 1.6 (Figure 1.34). Position vs. Time 40.0 35.0 30.0 25.0 20.0 15.0 10.0 5.0 0.0 0. Practice Problems 1. Draw a position-time graph from the velocity-time graph given below. Velocity vs. Time ) ] 14 12 10 8 6 4 2 0 0 10 20 Time (s) 30 40 2. Calculate the acceleration using the graph below. Velocity vs. Time ) ] 12 10 Time (s) 10 12 1.0 2.0 3.0 4.0 5.0 Time (s) Answers 1. Position vs. Time Figure 1.34 (b) Since the position-time graph is non-linear, find the slope of the tangent at 2.0 s, 3.0 s, and 5.0 s (Figures 1.35(a), (b), and (c)). ( 40.0 30.0 20.0 10.0 0.0 0.0 Position vs. Time d t 1.0 2.0 3.0 Time (s) 4.0 5.0 Figure 1.35(a) slope d t 32.5 m15.5 m 3.0 s1.0 s 17.0 m 2.0 s 8.5 m/s 250 200 150 100 50 ) ]. 1.0 m/s2 [N] 10 20 Time (s) 30 40 Chapter 1 Graphs and equations describe motion in one dimension. 27 01-PearsonPhys20-Chap01 7/23/08 11:43 AM Page 28 Position vs. Time ( 40.0 30.0 20.0 10.0 0.0 0.0 1.0 3.0 2.0 Time (s) 4.0 5.0 e MATH Figure 1.35(b) For an alternative method to create a velocity-time graph from the position- time data points, visit www.pearsoned.ca/school/ physicssource ( 40.0 30.0 20.0 10.0 0.0 0.0 Position vs. Time 1.0 2.0 3.0 Time (s) 4.0 5.0 Figure 1.35(c) slope d t 37.0 m(26.0 m) 4.0 s2.0 s 1 1 m.0 s.0 2 5.5 m/s This tangent is a horizontal line, so its slope is zero. The slopes of the tangents give the instantaneous velocities (Table 1.7). Positive signs mean that the direction is forward. Plot the data on a velocity-time graph (Figure 1.36). \u25bc Table 1.7 Time (s) Velocity (m/s [forward]) 2.0 3.0 5.0 8.5 5. 15.0 14.0 13.0 12.0 11.0 10.0 9.0 8.0 7.0 6.0 5.0 4.0 3.0 2.0 1.0 0.0 0.0 Velocity vs. Time 1.0 2.0 3.0 Time (s) 4.0 5.0 6.0 (c) Find acceleration by calculating the slope of the Figure 1.36 velocity-time graph. v a t 0.0", " m/s (8.5 m/s) 5.0 s 2.0 s 2.8 m/s2 The acceleration of the ATV is 2.8 m/s2. Because the forward direction was designated as positive, the negative sign means that the direction of acceleration is backward. Negative Acceleration Does Not Necessarily Mean Slowing Down In Example 1.5, the value for acceleration is negative. What is the meaning of negative acceleration? When interpreting the sign of acceleration, you need to compare it to the sign of velocity. For example, for the drag racer that is speeding up, the direction of its velocity is the same as the direction of its acceleration (see the calculation of the slope of the velocitytime graph for Figure 1.30). When the directions (signs) of velocity and acceleration are the same (positive or negative), the object is speeding up. PHYSICS INSIGHT v a v a v aa v a speeding up speeding up slowing down slowing down 28 Unit I Kinematics 01-PearsonPhys20-Chap01 7/23/08 11:43 AM Page 29 For the ATV in Example 1.5, the direction of its velocity is opposite to the direction of its acceleration, so it is slowing down. When velocity and acceleration have opposite directions (signs), the object slows down. Concept Check (a) Think of two more examples of objects not mentioned in this text that are speeding up and slowing down. In each case, indicate the signs or directions of velocity and acceleration. (b) Under what circumstances can an object have a negative acceleration and be speeding up? (c) You are given a position-time graph that is a curve. How can you use the slope of the tangent to determine whether the object represented in the graph is speeding up or slowing down? (Hint: How does the slope of the tangent change as you move along the position-time curve?) THEN, NOW, AND FUTURE Biomechanics and the Role of the Crash Test Dummy Understanding how biological systems move is a branch of physics known as biomechanics. For automobile manufacturers, understanding how the human body moves during a car accident is very important. To study, collect, and analyze data on how the human body moves during a vehicular collision requires a test subject. Human cadavers were the first test subjects used. While live human testing was valuable, it was limited in its scope due to the physical discomfort required and injury potential for some of the tests. Despite the production of reliable applicable data, most automobile manufacturers discontinued live animal testing in 1993 for moral and ethical reasons. Clearly, a different type of test subject needed to be designed and built. It came in the form of the now recognizable crash test dummy. Sam W. Alderson created \u201cSierra Sam\u201d in 1949 to test aircraft ejection seats and pilot restraint harnesses. Then came the VIP-50 series and Sierra Stan in the 1950s. Engineers combined the best features of these different models and debuted Hybrid I in 1971. Hybrid I was known as the \u201c50th percentile male\u201d dummy (meaning approximately 50% of men are larger and 50% of men are smaller), with a height of 168 cm and a mass of 77 kg. A year later, Hybrid II, with improved shoulder, spine, and knee responses, was produced to test lap and shoulder belts. Still crude, their use was limited, leading to the advent of the Hybrid III family of crash test dummies that include a 50th percentile male, a 50th percentile female, and two child dummies. This family of crash test dummies is designed to measure spine and rib acceleration, and demonstrate neck movement in rearend collisions. Equipped with a more humanlike spine and pelvis, THOR (Figure 1.37) is the successor of Hybrid III. Its face contains a number of sensors for facial impact analysis. Since front and side air bags have reduced upper body injury, lower extremity injury has become more prevalent. Therefore, THOR is built with an Achilles tendon to better mimic the side-to-side, up-and-down, and rotational movements of the ankle. Even with sophisticated crash test dummies, plastic and steel can only approximate how the human body will move. The study of soft tissue injury can only be accomplished with real-life subjects. Therefore, the future of crash testing will be in cre- ating detailed computer models of human systems. Even though it is slow and cumbersome for full body simulations, the computer has the advantage of repeatability and lower cost. The programmer has the ability to control every variable and repeat each and any event. 1. Why are crash test dummies used? 2. What are some of the advantages of THOR over his previous prototypes? 3. Will crash test dummies become obsolete? Explain. Figure 1.37 THOR Chapter 1 Graphs and equations describe motion in one dimension. 29 01-PearsonPhys20-Chap", "01 7/23/08 11:43 AM Page 30 1.3 Check and Reflect 1.3 Check and Reflect Applications 3. Match each velocity-time graph below 1. A sprinter in a championship race accelerates to his top speed in a short time. The velocity-time data for part of the race are given in the table below. Use the data to find the (a) average acceleration from 0.00 s to 0.50 s (b) average acceleration from 0.50 s to 3.00 s (c) average acceleration from 5.00 s to 6.00 s (d) Describe what was happening to the acceleration and velocity over 6.00 s. Time (s) Velocity (m/s [forward]) 0.00 0.12 0.14 0.50 1.00 2.00 3.00 4.00 5.00 6.00 7.00 8.00 9.00 9.83 9.93 0.00 0.00 0.00 2.80 5.00 8.00 9.80 10.80 11.30 11.60 11.70 11.80 11.90 11.95 11.97 2. Describe the motion of the object as illustrated in the graph below. Position vs. Time Time (s) 2.0 4.0 6.0 8.0 10. 20.0 0.0 20.0 40.0 60.0 80.0 100.0 30 Unit I Kinematics with the correct statement. (i) negative acceleration (ii) positive acceleration (iii) moving with zero acceleration (iv) stationary object Extensions 4. In your notebook, complete the velocity- time data table for the graph below. Position vs. Time ) ] 36 33 30 27 24 21 18 15 12 9 6 3 0 0.0 2.0 4.0 6.0 8.0 10.0 Time (s) Time (s) Velocity (m/s [forward]) 2.0 4.0 6.0 8.0 e TEST To check your understanding of uniformly accelerated motion, follow the eTest links at www.pearsoned.ca/school/physicssource. 01-PearsonPhys20-Chap01 7/23/08 11:43 AM Page 31 1.4 Analyzing Velocity-time Graphs When a plane flies across Alberta with constant speed and direction, it is said to be undergoing uniform motion (Figure 1.38(a)). If you were sitting in the plane, you would experience a smooth ride. An all-terrain vehicle (ATV) bouncing and careening along a rough trail is constantly changing speed and direction in order to stay on the road. A ride in the ATV illustrates non-uniform motion, or acceleration (Figure 1.39(a)). You can distinguish between uniform and non-uniform motion by simple observation and gathering data from your observations (see Figures 1.38(b) and 1.39(b)). There are several ways to interpret the data. One way is to analyze graphs by determining their slopes to obtain further information about an object\u2019s motion, as you did in section 1.3. In this section, you will develop this method further and learn another method of graphical analysis: how to find the area under a graph. First review the information you can obtain from the slopes of position-time and velocity-time graphs. Figure 1.38(a) Uniform motion Figure 1.39(a) Non-uniform motion Velocity vs. Time Velocity vs. Time ) ] Time (h) Figure 1.38(b) A graph representing uniform motion ) ] Time (s) Figure 1.39(b) A graph representing non-uniform motion Chapter 1 Graphs and equations describe motion in one dimension. 31 01-PearsonPhys20-Chap01 7/23/08 11:43 AM Page 32 Slopes of Graphs Reveal How Objects Move Consider the three photos and velocity-time graphs in Figure 1.40. You can interpret each graph by reading values from it. To gain new information, you must analyze the graph by calculating its slope. The slope describes the object\u2019s motion. Velocity vs. Time Velocity vs. Time Velocity vs. Time ) ] Time (s) Time (s) Time (s) Figure 1.40(a) A velocity-time graph for an object at rest Figure 1.40(b) A velocity-time graph for an object undergoing uniform motion Figure 1.40(c) A velocity-time graph for an object undergoing uniformly accelerated motion Concept Check 1. Sketch position-time graphs for the following: (a) two possibilities for stopped motion (b) two possibilities for uniform motion (c) four possibilities for uniformly accelerated motion Describe each graph in terms of direction of travel and whether the object is speeding up or slowing down. 2. Sketch the corresponding velocity-time graph for each position-time graph in question 1", ". For each graph, describe how you determined the graph\u2019s shape. By analyzing the units for the slope of a velocity-time graph, m/s2, you know from section 1.3 that the slope of a velocity-time graph represents the acceleration of the object. Concept Check Sketch all the types of acceleration-time graphs you have encountered thus far. Describe the kind of motion that each graph represents. 32 Unit I Kinematics 01-PearsonPhys20-Chap01 7/25/08 7:50 AM Page 33 1-5 Design a Lab 1-5 Design a Lab Tortoise or Hare? The Question In your class, who has the fastest acceleration and the fastest average speed in the 50-m dash? Design and Conduct Your Investigation Make a list of variables that you think are likely to influence the acceleration of each participant. For each variable on your list, write a hypothesis that predicts how changes in that variable will affect the participants\u2019 acceleration. Write a procedure for an investigation that will test the effect of one of these variables on acceleration. Clearly outline all the steps that you will follow to complete your investigation. Identify the responding and manipulated variables. List all the materials and equipment you will need, as well as all safety precautions. Compare your experimental design and procedure with those of your classmates. Identify any strengths and weaknesses. With your teacher\u2019s approval, conduct your investigation. State any problems or questions that you found during your investigation or analysis that would need additional investigation to answer. The Area Under a Velocity-time Graph Represents Displacement Occasionally, due to a medical or other emergency, a pilot must turn the aircraft and land at the same or alternate airport. Consider the graph for the uniform motion of a plane travelling east at 300 km/h for 2.0 h only to turn back west for 0.5 h to make an emergency landing (Figure 1.41). What is the plane\u2019s displacement for this time interval? Unit analysis indicates that the area under a velocity-time graph equals displacement. To calculate displacement using a velocity-time graph, look at the units on the axes. To end up with a unit of displacement (km) from the units km/h and h, you need to multiply: km h h km The shapes in Figure 1.41 are rectangles, so the area under the velocitytime graph is l w (length times width). In this case, find the sum of the areas above and below the time axis. Consider east to be positive. For eastward displacement, the area is above the time axis, so it is positive. For westward displacement, the area is below the time axis, so it is negative. For eastward displacement (above the time axis), Velocity vs. Time Time (h) 1.0 2.0 3. 400 200 0 200 400 Figure 1.41 To calculate net displacement, add the areas above and below the time axis. area dd 300 vt km h 600 km (2.0 h) Chapter 1 Graphs and equations describe motion in one dimension. 33 01-PearsonPhys20-Chap01 7/23/08 11:43 AM Page 34 For westward displacement (below the time axis), dd vt 300 km h 150 km (0.5 h) PHYSICS INSIGHT Note that the answer has one significant digit because there is only one significant digit in 0.5 h. To find the plane\u2019s net displacement, add the two areas. area ddd 600 km (150 km) 600 km 150 km 450 km 5 102 km [E] Because the net area is positive, the plane\u2019s displacement is 5 102 km [E]. Unlike position-time graphs, where you can only calculate the slope to determine velocity, you can use velocity-time graphs to determine both acceleration and displacement, as in the next example. Example 1.6 From the graph in Figure 1.42, calculate (a) displacement (b) acceleration Velocity vs. Time Practice Problems 1. Calculate the displacement and acceleration from the graph. Velocity vs. Time ) ].4 2.2 2.0 1.8 1.6 1.4 1.2 1.0 0.8 0.6 0.4 0.2 0.0 4.0 3.0 2.0 1..0 0.0 2.0 4.0 6.0 8.0 10.0 Time (min) Figure 1.42 0 2 4 6 Time (s) 8 10 Analysis and Solution (a) For displacement, find the sum of the areas under the velocity-time graph (Figure 1.43). Designate east (above the time axis) as the positive direction. Convert minutes to seconds. 2. Use the graph below to determine the displacement of the object. Velocity vs. Time Time (s) 2 4 6 8 10 12 ) ] 10 5 0 5 10 15 20 25 Answers 1. 22 m [N], 0 m/", "s2 [N] 2. 1.0 102 m [E] or 1.0 102 m [W] 34 Unit I Kinematics 4.0 3.0 2.0 1. Velocity vs. Time A B C 0.0 0.0 2.0 4.0 6.0 8.0 10.0 Time (min) Figure 1.43 01-PearsonPhys20-Chap01 7/23/08 11:43 AM Page 35 d Region A: vt 3.0 A 4.0 min m s 60 s 1 min (240 s) 3.0 m s 720 m Region B: d B (5.0 min 4.0 min) 1 2 60 s 1 min 3.0 m s 1.0 1.0 m s m s (5.0 min 4.0 min) 60 s 1 min (1.0 min) 60 s 1 min 2.0 1.0 m s m s (1.0 min) 60 s 1 min 1 2 1 2 (120 m) 60 m PHYSICS INSIGHT When calculating total displacement from a velocity-time graph, remember to keep track of whether the area is positive or negative. PHYSICS INSIGHT Check your answer by looking at the units. Do the units reflect the answer that you are asked to find? 120 m Region C: d C 1.0 m s 1.0 m s 300 m d A d B d d C (10.0 min 5.0 min) 60 s 1 min (5.0 min) 60 s 1 min 720 m 120 m 300 m 1140 m 1.1 103 m The answer is positive, so the direction of the displacement is east. (b) For acceleration, find the slope of each section of the graph. In region A: v a t m m 3.0 3.0 s s 240 s 0.0 m/s2 This answer makes sense because the velocity-time graph is a horizontal line. In region B: v a t m m 3.0 1.0 s s 60 s 0.033 m/s2 or 0.033 m/s2[W] Chapter 1 Graphs and equations describe motion in one dimension. 35 01-PearsonPhys20-Chap01 7/25/08 7:54 AM Page 36 Since the third part of the graph is also a horizontal line, its slope is also zero. Paraphrase (a) The displacement is 1.1 103 m [E]. (b) The acceleration is zero in regions A and C and 0.033 m/s2 [W] in region B. Concept Check For the velocity-time graph of a ball thrown up in the air (Figure 1.44), what is the net displacement of the ball? 15.00 10.00 5.00 0.00 5.00 Velocity vs. Time Time (s) 0.50 1.00 1.50 2.00 2.50 ) ] 10.00 15.00 Figure 1.44 Average Velocity from Velocity-time Graphs Objects rarely travel at constant velocity. Think of your journey to school today. Whether you travelled by car or bus, rode a bike, or walked, stop signs, traffic lights, corners, and obstacles caused a variation in your velocity, or rate of travel. If you describe your motion to a friend, you can use a series of instantaneous velocities. The more time instances you use to record your motion, the more details about your trip you can communicate to your friend (Figure 1.45(a)). However, if you were to use the equation v and substitute your total displacement d t for d and your total time of travel for t, you would lose most of the details of your journey. You would obtain a value for your average velocity, v ave (Figure 1.45(b)). Velocity vs. Time Velocity vs. Time ) ] Time (s) v ave Time (s Figure 1.45(a) By using a series of instantaneous velocities at the given times, you can precisely describe your journey. Figure 1.45(b) average velocity of the journey. It describes your journey but the detail of the motions is lost. The straight line represents the If you need to obtain an average velocity value from a velocity-time graph, recall that displacement, d, is the area under the graph. 36 Unit I Kinematics 01-PearsonPhys20-Chap01 7/23/08 11:43 AM Page 37 To find average velocity, determine the area under the velocity-time graph and divide it by the total time. To calculate average velocity when given different displacements over different time intervals, simply add the total displacement and divide by the total time, as shown in the next example. Example 1.7 Find the average velocity of a student who jogs 750 m [E] in 5.0 min, does static stretches for 10.0 min, and then runs", " another 3.0 km [E] in 30.0 min. Given Choose east to be positive. Convert kilometres to metres. d 1 t1 d 2 t2 d 3 750 m [E] 750 m 5.0 min 0 m 10.0 min 3.0 km [E] 3.0 km 1000 m 1 km 3000 m t3 30.0 min Required average velocity (v ave) Analysis and Solution First add the displacement values. d d 2 d 1 750 m 0 m 3000 m 3750 m d 3 total Then add the time intervals and convert to seconds. The total time elapsed is ttotal t3 t2 t1 5.0 min 10.0 min 30.0 min (45.0 min)60 2700 s s min Average velocity equals total displacement divided by total time elapsed: v ave d t 3750 m 2700 s 1.4 m/s Since the answer is positive, the direction is east. Paraphrase The student\u2019s average velocity is, therefore, 1.4 m/s [E]. Practice Problems 1. A person runs 10.0 m [E] in 2.0 s, then 5.0 m [E] in 1.5 s, and finally 30.0 m [W] in 5.0 s. Find the person\u2019s average velocity. 2. Person A runs the 100-m dash in 9.84 s and then tags person B, who runs 200 m in 19.32 s. Person B then tags an out-of-shape person C, who runs 400 m in 1.90 min. Find the average velocity for the trio. Compare it to each individual\u2019s average velocity. Assume they are all running in a straight line. Answers 1. 1.8 m/s [W] 2. 4.89 m/s [forward] A: 10.2 m/s [forward]; faster than the average velocity for the trio B: 10.4 m/s [forward]; faster than the average velocity for the trio C: 3.51 m/s [forward]; slower than the average velocity for the trio Chapter 1 Graphs and equations describe motion in one dimension. 37 01-PearsonPhys20-Chap01 7/23/08 11:43 AM Page 38 As you have seen, velocity-time graphs are very useful. They provide the following information: \u2013 Reading the velocity-time graph gives you instantaneous velocity values. \u2013 Finding the slope of a velocity-time graph gives you an object\u2019s acceleration. \u2013 The area under a velocity-time graph gives you the object\u2019s displacement. \u2013 You can also determine the average velocity of an object over a time interval from a velocity-time graph. Example 1.8 shows you how to obtain information about an object\u2019s velocity and acceleration. Example 1.8 A bird starts flying south. Its motion is described in the velocity-time graph in Figure 1.46.0 6.0 4.0 2.0 0.0 Velocity vs. Time F E 1.0 2.0 3.0 4.0 5.0 6.0 7.0 8.0 9.0 10.0 11.0 12.0 13.0 Time (s) \u20132.0 A \u20134.0 \u20136.0 \u20138.0 B C D From the graph, determine (a) whether acceleration is positive, negative, or zero for each section (b) the value of the acceleration where it is not zero (c) when the bird changes direction Analysis and Solution Consider north to be the positive direction. (a) Acceleration is the slope of each section of the graph. A: Final velocity is more negative than the initial velocity, as the bird is speeding up in the south direction. So the slope of this line is negative. The bird\u2019s acceleration is negative. 1 2 3 4 5 6 7 8 10 11 9 12 Time (s) Figure 1.46 Practice Problems 1. Position vs. Time ) ] 10 1 \u20132 \u20133 \u20134 \u20135 \u20136 \u20137 \u20138 \u20139 \u201310 \u201311 \u201312 (a) Describe the motion of the object from the graph above. (b) Draw the corresponding velocity-time graph. (c) Determine the object\u2019s displacement. (d) When is the object stopped? 38 Unit I Kinematics 01-PearsonPhys20-Chap01 7/23/08 11:43 AM Page 39 B: Acceleration is zero because the slope is zero (the graph Answers 1. (a) 3 m/s for 2 s, rest for 3 s, 3 m/s for 4 s, 6 m/s for 1 s (b) Velocity vs. Time ) ] 2 \u20134 \u20136 \u20138 5 10 15 Time (s) (c) 12 m (d) 2\u20135 s is a horizontal line). C: Acceleration is", " negative because the slope of the line is negative (as in section A). D: Acceleration is zero because the slope of the line is zero (as in section B). E: Final velocity is positive because the bird is now flying north. So the slope of this line is positive. The bird\u2019s acceleration is positive. F: Acceleration is zero because the slope of the line is zero. (b) Acceleration is not zero for sections A, C, and E. v A.0 4.0 s s 1.0 s 0.0 s 2.0 m/s2 2.0 m/s2 [S] v C.0 6.0 s s 4.0 s 3.0 s 2.0 m/s2 2.0 m/s2 [S] v E.0 6.0 s s 10.0 s 7.0 s m 12.0 s 3.0 s 4.0 m/s2 4.0 m/s2 [N] (c) The bird changes direction at 8.5 s \u2014 it crosses the time axis at this instant. Chapter 1 Graphs and equations describe motion in one dimension. 39 01-PearsonPhys20-Chap01 7/23/08 11:43 AM Page 40 The next example shows you how to use areas to find the displacement of the bird and its average velocity from a velocity-time graph. Example 1.9 From the graph in Figure 1.46, determine (a) the displacement for each section (b) the displacement for the entire flight (c) the average velocity of the flight Analysis and Solution (a) Displacement is the area between the graph and the time axis. Practice Problems 1. In your notebook, redraw the graph in Example 1.8 Practice Problem 1, but label the vertical axis \u201cvelocity (m/s [N])\u201d. (a) Find the displacements for 0\u20132 s, 2\u20135 s, 5\u20137 s, 7\u20139 s, and 9\u201310 s. (b) Find the object\u2019s total displacement. (c) Find the average velocity of the object. Answers 1. (a) 6 m [N], 18 m [N], 6 m [N], 6 m [N], 9 m [N] (b) 15 m [N] (c) 1.5 m/s [N] A: A l w bh 1 2 d (2.0 m s 1 )(1.0 s) (1.0 s)(2.0 2 m s ) 3.0 m B: A l w d (4.0 m s 8.0 m )(3.0 s 1.0 s) C: A l w bh 1 2 d (4.0 m s 1 )(1.0 s) (1.0 s)(2.0 2 m s ) 5.0 m D: A l w d (6.0 m s 18 m )(7.0 s 4.0 s) E: A 1 2 1 bh bh 2 (1.5 s) 6.0 d 1 2 m s (1.5 s) 6.0 1 2 m s 0.0 m F: A l w (6.0 d m s )(3.0 s) 18 m (b) Add all the displacements calculated in (a). The displacement over the entire flight is \u201316 m. Since north is positive, the displacement is 16 m [S]. (c) v ave d T t m 6 1. s 0 13 1.2 m/s 40 Unit I Kinematics North is positive, so the average velocity for the flight is 1.2 m/s [S]. 01-PearsonPhys20-Chap01 7/23/08 11:43 AM Page 41 Drawing Position-time and Acceleration-time Graphs from Velocity-time Graphs In this section, you have learned how to use a velocity-time graph to calculate displacement. It is also useful to know how to draw position-time and acceleration-time graphs when given a velocity-time graph. Consider the following trip. A family travelling from Calgary to go camping in Banff National Park moves at 18.0 m/s [forward] in a camper van. The van accelerates for 4.0 s until it reaches a velocity of 30.0 m/s [forward]. It continues to travel at this velocity for 25.0 s. When approaching a check stop, the driver brakes, bringing the vehicle to a complete stop in 15.0 s. The velocity-time graph for the trip is given in Figure 1.47. Velocity vs. Time I II III Figure 1.47 The complete graph of the van\u2019s motion 35.0 30.0 25.0 20.0 15.0 10.0 5..0 0.0 5", ".0 10.0 15.0 20.0 25.0 30.0 Time (s) 35.0 40.0 45.0 The next example shows you how to create an acceleration-time graph from a velocity-time graph. Example 1.10 Use the velocity-time graph in Figure 1.47 to draw the corresponding acceleration-time graph. Analysis and Solution To find acceleration, calculate the slope for each section of the graph. The velocity-time graph has three distinct sections. The slope in each section is constant. Consider forward to be positive. Velocity vs. Time Section l Time: 0.0 s to 4.0 s ti vi tf vf 0.0 s 18.0 m/s 4.0 s 30.0 m/s 35.0 30.0 25.0 20.0 15.0 10.0 5..0 0.0 5.0 10.0 15.0 20.0 25.0 30.0 Time (s) 35.0 40.0 45.0 Figure 1.48(a) Chapter 1 Graphs and equations describe motion in one dimension. 41 01-PearsonPhys20-Chap01 7/23/08 11:43 AM Page 42 Practice Problems 1. For each velocity-time graph below, draw the corresponding acceleration-time graph. Velocity vs. Time (a 25 20 15 10 5 0 0 2.0 6.0 10.0 14.0 vf Time (s) (b) Velocity vs. Time ) ] 25 20 15 10 5 0 0 2.0 6.0 Time (s) 10.0 14.0 Answers 1. (a) Acceleration vs. Time ( 15 10 5 0 0 2.0 6.0 Time (s) 10.0 14.0 Acceleration vs. Time 10 5 0 5 Time (s) 2.0 6.0 10.0 14.0 (b) slope a v v i f ti tf 30.0 m/s (18.0 m/s) 4.0 s 0.0 s 3.0 m/s2 Velocity vs. Time Section ll Time: 4.0 s to 29.0 s ti vi tf 4.0 s 30.0 m/s 4.0 s 25.0 s 29.0 s 30.0 m/s v v i f ti tf slope 35.0 30.0 25.0 20.0 15.0 10.0 5.0 0.0 0.0 5.0 10.0 15.0 20.0 25.0 30.0 Time (s) 35.0 40.0 45.0 30.0 m/s (30.0 m/s) 29.0 s 4.0 s 0.0 m/s2 Figure 1.48(b) Velocity vs. Time Section lll Time: 29.0 s to 44.0 s ti vi tf 29.0 s 30.0 m/s 29.0 s 15.0 s 44.0 s 0.0 m/s vf slope a v v i f ti tf ) ] 35.0 30.0 25.0 20.0 15.0 10.0 5.0 0.0 0.0 5.0 10.0 15.0 20.0 25.0 30.0 Time (s) 35.0 40.0 45.0 Figure 1.48(c) 0.0 m/s (30.0 m/s) 44.0 s 29.0 s 2.0 m/s2 Now plot the values on the acceleration-time graph (Figure 1.49). Each section of the graph is a horizontal line because acceleration is constant (uniform). Acceleration vs. Time ) ] 35.0 30.0 25.0 20.0 15.0 10.0 5.0 0.0 5.0 10.0 30.0 35.0 40.0 45.0 5.0 10.0 15.0 20.0 25.0 Time (s) Figure 1.49 The next example shows you how to use a velocity-time graph to generate a position-time graph. 42 Unit I Kinematics 01-PearsonPhys20-Chap01 7/23/08 11:43 AM Page 43 Example 1.11 Sketch a position-time graph from the velocity-time graph in Figure 1.47. Analysis and Solution To sketch the position-time graph, find the area under the velocitytime graph. Consider forward to be positive. In the first part of the velocity-time graph (0.0\u20134.0 s), area (displacement) is a rectangle and a triangle. The displacement is positive because the areas are above the time-axis. A l w bh 1 2 d 1 (18.0 m/s)(4.0 s) (4.0 s)(30.0 m/s 18.", "0 m/s) 2 96 m Since the velocity-time graph in this section has a positive slope, the car has positive acceleration, so the corresponding positiontime graph is a parabola that curves upward. On the positiontime graph, sketch a curve from the origin to the point t 4.0 s 96 m (Figure 1.50(a)). and d In the second part of the velocity-time graph (4.0\u201329.0 s), displacement is a rectangle. It is positive since the area is above the time-axis. A l w (30.0 m/s)(29.0 s 4.0 s) d 750 m Since the velocity-time graph has zero slope in this section, the car moves with constant velocity and the position-time graph is a straight line with a positive slope that extends from t 4.0 s and d t 29.0 s and d 96 m to 96 m 750 100 80 60 40 20 0 0 Position vs. Time 4 Time (s) Figure 1.50(a) Position vs. Time ) ] 1000 900 800 700 600 500 400 300 200 100 0 0 84 24 28 32 12 16 20 Time (s) Figure 1.50(b) Practice Problems 1. Velocity vs. Time ) ] 2 \u20134 \u20136 3 6 9 12 15 18 Time (s) (a) Describe the motion of the object illustrated above. Calculate its total displacement. (b) Draw the corresponding position-time graph. Answers 1. (a) travels with uniform motion, changes direction at 10 s, and travels with uniform motion; total displacement: 10 m [forward] (b) Position vs. Time ) ] 60 50 40 30 20 10 0 0 5 15 20 10 Time (s) 846 m (See Figure 1.50(b).) In the third part of the velocity-time graph (29.0\u2013 44.0 s), displacement is a triangle. It is positive since the area is above the time-axis. 1 A bh 2 d (44.0 s 29.0 s)(30.0 m/s) 1 2 225 m Chapter 1 Graphs and equations describe motion in one dimension. 43 01-PearsonPhys20-Chap01 7/23/08 11:43 AM Page 44 ) ] 1200 1100 1000 900 800 700 600 0 Position vs. Time 0 24 28 32 36 40 44 48 Time (s) Figure 1.50(c) Since the velocity-time graph has a negative slope, the car undergoes negative acceleration, so the slopes of the tangents of the position-time graph decrease (approach zero). The position-time graph is a parabola that curves down, from t 29.0 s and 846 m to d t 44.0 s and 846 m 225 m d 1071 m Position vs. Time (See Figure 1.50(c).) The resulting position-time graph is shown in Figure 1.51 1200 1100 1000 900 800 700 600 500 400 300 200 100 0 0 84 12 16 24 28 32 20 Time (s) 36 40 44 48 Figure 1.51 Concept Check If north is positive, sketch position-time, velocity-time, and acceleration-time graphs for an object (a) speeding up and going north (b) slowing down and going north (c) speeding up and going south (d) slowing down and going south 1.4 Check and Reflect 1.4 Check and Reflect Knowledge 1. On a ticker tape, how can you distinguish between uniform and uniformly accelerated motion? 2. Use the terms \u201cdisplacement\u201d and \u201cvelocity\u201d to describe how uniformly accelerated motion differs from uniform motion. 3. What is the relationship between the slope of a position-time graph and velocity? 4. Compare the shape of a position-time graph for uniform motion with a positiontime graph representing uniformly accelerated motion. 5. What is the relationship between the slope of a velocity-time graph and acceleration? 6. If a velocity-time graph is a straight line with a non-zero slope, what kind of motion is the object undergoing? 44 Unit I Kinematics 7. Determine the displacement of the object whose motion is described by the following graph. Velocity vs. Time ) ] 20.0 15.0 10.0 5.0 0.0 0.0 2.0 6.0 4.0 Time (s) 8.0 10.0 8. Calculate displacement from the velocity-time graph below.0 4.0 3.0 2.0 1.0 0.0 Velocity vs. Time 0.0 2.0 4.0 6.0 8.0 10.0 Time (s) 01-PearsonPhys20-Chap01 7/23/08 11:43 AM Page 45 9. Describe the velocity-time graph for an object undergoing negative acceleration. 10. What quantity of motion can be determined from the area under", " a velocity-time graph? (b) Create a graph for the question and check your answer using graphing techniques. 17. Determine acceleration from the velocity- time graph given below. Velocity vs. Time ) ] 14.0 12.0 10.0 8.0 6.0 4.0 2.0 0.0 0.0 10.0 20.0 Time (s) 30.0 40.0 18. A truck travelling forward at 14.0 m/s accelerates at 1.85 m/s2 for 6.00 s. It then travels at the new speed for 35.0 s, when a construction zone forces the driver to push on the brakes, providing an acceleration of 2.65 m/s2 for 7.0 s. Draw the resulting velocity-time and position-time graphs for this motion. Extension 19. Describe the motion of the object illustrated in the graph below. Velocity vs. Time 10.0 5.0 0.0 5.0 5.0 10.0 15.0 20.0 25.0 30.0 Time (s 10.0 e TEST To check your understanding of velocity-time graphs follow the eTest links at www.pearsoned.ca/school/physicssource. 11. Compare and contrast the shape of a velocity-time graph for an object experiencing uniform motion with one experiencing uniformly accelerated motion. 12. Describe the acceleration-time graph of a car travelling forward and applying its brakes. 13. Calculate the acceleration of an object using the velocity-time graph below. Velocity vs. Time 2.0 4.0 6.0 8.0 10.0 Time (s 100 90 80 70 60 50 40 30 20 10 0 10 20 30 40 50 14. Construct an acceleration-time graph using the graph given below. Velocity vs. Time ) ] 12 10 8 6 4 2 0 0 5 10 15 Time (s) Applications 15. A motorbike increases its velocity from 20.0 m/s [W] to 30.0 m/s [W] over a distance of 200 m. Find the acceleration and the time it takes to travel this distance. 16. (a) While driving north from Lake Louise to Jasper, you travel 75 min at a velocity of 70 km/h [N] and another 96 min at 90 km/h [N]. Calculate your average velocity. Chapter 1 Graphs and equations describe motion in one dimension. 45 01-PearsonPhys20-Chap01 7/23/08 11:43 AM Page 46 1.5 The Kinematics Equations A cylindrical piston the length of a football field controls the launch of a fighter plane from the deck of a carrier ship (Figure 1.52). Too much pressure and the nose gear is ripped off; too little pressure and the plane crashes into the ocean. This propulsion system accelerates a 20 000-kg plane from rest to 74 m/s (266 km/h) in just 2.0 s! e TECH Study the physics of jet takeoffs by visiting www.pearsoned.ca/school/ physicssource and viewing the simulation. Figure 1.52 Analyzing complex motions requires many calculations, some of which involve using the kinematics equations you will study in this section. To determine the crucial values required for launching a plane, such as flight deck length, final velocity, and acceleration, physicists and engineers use kinematics equations similar to the ones you will know by the end of this section. In this section, you will practise your analytical skills by learning how to derive the kinematics equations from your current knowledge of graphs and then apply these equations to analyze complex motions such as airplane launches. Velocity vs. Time Concept Check Create a summary chart for the information you can gather by analyzing position-time, velocity-time, and acceleration-time graphs. Use the headings \u201cGraph Type\u201d, \u201cReading the Graph\u201d, \u201cSlope\u201d, and \u201cArea\u201d. Consider the airplane taking off from a moving aircraft carrier (Figure 1.52). The plane must reach its takeoff speed before it comes to the end of the carrier\u2019s runway. If the plane starts from rest, the velocitytime graph representing the plane\u2019s motion is shown in Figure 1.53. Notice that the slope of the graph is constant. By checking the units on the graph, you know that the slope represents acceleration: rise run Therefore, the velocity-time graph is a straight line. m/s2. In this case, the acceleration is constant (uniform). m/s s 80 70 60 50 40 30 20 10 ) ].0 1.0 Time (s) 2.0 Figure 1.53 The slope of this velocity-time graph represents the plane\u2019s acceleration. 46 Unit I Kinematics 01-PearsonPhys20-Chap01 7/23/08 11:43 AM Page 47 From", " Figure 1.53, you can derive the first kinematics equation This equation can also be written as v a t The next example shows you how to apply this equation to solve a problem. Example 1.12 A hybrid car with an initial velocity of 10.0 m/s [E] accelerates at 3.0 m/s2 [E]. How long will it take the car to acquire a final velocity of 25.0 m/s [E]? Given Designate east as the positive direction. v 10.0 m/s [E] 10.0 m/s v 25.0 m/s [E] 25.0 m/s a 3.0 m/s2 [E] 3.0 m/s2 i f Required time (t) Analysis and Solution Use the equation v v v a i t t f Since you are dividing by a vector, and initial and final velocities and acceleration are in the same direction, use the scalar form of the equation. Isolate t and solve. t vi vf a 25 m/s 10 m/s 3.0 m/s2 m 15 s m 3.0 s2 5.0 s Practice Problems 1. A motorcycle with an initial velocity of 6.0 m/s [E] accelerates at 4.0 m/s2 [E]. How long will it take the motorcycle to reach a final velocity of 36.0 m/s [E]? 2. An elk moving at a velocity of 20 km/h [N] accelerates at 1.5 m/s2 [N] for 9.3 s until it reaches its maximum velocity. Calculate its maximum velocity, in km/h. Answers 1. 7.5 s 2. 70 km/h [N] PHYSICS INSIGHT The mathematics of multiplying vectors is beyond this text and division of vectors is not defined. So, when multiplying and dividing vectors, use the scalar versions of the kinematics equations. Paraphrase It will take the car 5.0 s to reach a velocity of 25.0 m/s [E]. As you know from section 1.4, you can calculate the area under a velocitytime graph. By checking the units, you can verify that the area represents displacement: l w m s s m Chapter 1 Graphs and equations describe motion in one dimension. 47 01-PearsonPhys20-Chap01 7/23/08 11:43 AM Page 48 PHYSICS INSIGHT The area of a trapezoid 1 2 is given by l2)w. (l1 To calculate the displacement (area) from the velocity-time graph in Figure 1.54, you can use the formula for the area of a trapezoid, 1 A (l1 2 l2)w, which is simply the average of the parallel sides multiplied by the base. Velocity vs. Time 1 Area d (vf vi ) t 2 l2 l1 w vf vi ) ] ti t Time (s) tf Figure 1.54 graph to derive the equation d v t. ave Use the area under the velocity-time The second kinematics equation is v i (v d f)t 1 2 v where l1 to apply this equation. v i, l2 f, and w t. The next example shows you how Example 1.13 A cattle train travelling west at 16.0 m/s is brought to rest in 8.0 s. Find the displacement of the cattle train while it is coming to a stop. Assume uniform acceleration. Given Designate west as the positive direction. v v t 8.0 s 16.0 m/s [W] 16.0 m/s 0 m/s [W] 0 m/s i f Required displacement (d ) Analysis and Solution Use the equation d 1 2 (v i vv f)t and solve for d. Practice Problems 1. A hound running at a velocity of 16 m/s [S] slows down uniformly to a velocity of 4.0 m/s [S] in 4.0 s. What is the displacement of the hound during this time? 2. A ball moves up a hill with an initial velocity of 3.0 m/s. Four seconds later, it is moving down the hill at 9.0 m/s. Find the displacement of the ball from its initial point of release. Answers 1. 40 m [S] 2. 12 m 48 Unit I Kinematics 01-PearsonPhys20-Chap01 7/23/08 11:43 AM Page 49 d (16.0 m/s 0 m/s)(8.0 s) 1 2 8.0 64 m (8.0 s) m s The sign is positive, so the train\u2019s direction is west. Paraphrase The cattle train travels 64 m [W] before it stops. You can also calculate the", " area under a velocity-time graph by considering that the total area under the graph is made up of a triangle and a rectangle (Figure 1.55). vf v at vi vi area area of rectangle l w t vi t 1 2 bh 1 t 2 1 t 2 a1 2 v t a )2 t ( Figure 1.55 You can divide the area under the velocity-time graph into a triangle and a rectangle. info BIT The fastest time of covering 1.6 km while flipping tiddly winks \u2014 52 min 10 s \u2014 was achieved by E. Wynn and J. Culliongham (UK) on August 31, 2002. Their speed was 1.8 km/h. Using this data and the displacement equation below, verify that they did indeed travel a distance of 1.6 km. In Figure 1.55, the area of a rectangle represents the displacement of an object travelling with a constant velocity, v i. The height of the rectangle is v i and the base is t. Therefore, the area of the rectangle is equal to v t. The area of the triangle represents the additional displacement resulting from the change in velocity. The height of the triangle is v and the base is t. The area of the triangle is equal to v v i f i 1 2 t(v). But vat. Therefore, the area of the triangle is equal to 1 2 (t)(at) a(t)2. Add both displacements to obtain 1 2 PHYSICS INSIGHT 2 ti (t2) tf ti)2. (t)2 (tf 2, whereas d v 1 t a(t)2 2 i The next example shows you how to apply the third kinematics equation. Chapter 1 Graphs and equations describe motion in one dimension. 49 01-PearsonPhys20-Chap01 7/23/08 11:43 AM Page 50 Example 1.14 A golf ball that is initially travelling at 25 m/s hits a sand trap and slows down with an acceleration of 20 m/s2. Find its displacement after 2.0 s. Given Assign a positive direction for forward and a negative direction for backward. 25 m/s vi a 20 m/s2 t 2.0 s Practice Problems 1. A skier is moving down a uniform slope at 3.0 m/s. If the acceleration down the hill is 4.0 m/s2, find the skier\u2019s displacement after 5.0 s. 2. A motorcycle travelling at 100 km/h on a flat road applies the brakes at 0.80 m/s2 for 1.0 min. How far did the motorcycle travel during this time? Answers 1. 65 m [down] 2. 2.3 102 m Required displacement (d ) Analysis and Solution Use the equation d v i 25 d m s (2.0 s) 1 2 50 m (40 m) 10 m t a(t)2 to solve for d. 1 2 20 m s2 (2.0 s)2 The sign is positive, so the direction is forward. Paraphrase The displacement of the golf ball is 10 m [forward]. To obtain the fourth kinematics equation, derive the value of a required variable in one equation, substitute the derived value into the second equation, and simplify. v i t f Start with a v Isolate v i. at v v Then substitute v (v v d f i f i 1 2 at for v i into the equation f)t. The equation becomes d 1 2 (v f at v f)t. This equation simplifies to v d 1 t a(t)2 2 f Apply this equation in the next example. info BIT The fastest lava flow ever recorded was 60 km/h in Nyiragongo (Democratic Republic of Congo) on January 10, 1977. At this speed, how far would the lava travel in 2 h 30 min? 50 Unit I Kinematics 01-PearsonPhys20-Chap01 7/23/08 11:43 AM Page 51 Example 1.15 Figure 1.56 A speedboat slows down at a rate of 5.0 m/s2 and comes to a stop (Figure 1.56). If the process took 15 s, find the displacement of the boat. f 0.0 m/s (because the boat comes to rest) Given Let forward be the positive direction. v t 15 s a 5.0 m/s2 (Acceleration is negative because the boat is slowing down, so its sign must be opposite to that of velocity (positive).) Practice Problems 1. If the arresting device on an aircraft carrier stops a plane in 150 m with an acceleration of 15 m/s2, find the time the plane takes to stop. 2. The 1968 Corvette took 6.2 s to accelerate to 160 km/h [N]. If it travelled 220 m [N], find its acceleration. Answers 1. 4.5 s", " 2. 2.9 m/s2 [N] Required displacement (d ) Analysis and Solution Use the equation d vv f t a(t)2 to solve for d 1 2. d (0.0 m/s)(15 s) 1 2 (5.0 m/s2)(15 s)2 562.5 m 5.6 102 m The sign is positive, so the direction of displacement is forward. Paraphrase The displacement of the speedboat is 5.6 102 m [forward]. Deriving the fifth and last kinematics equation involves using the difference of squares, another math technique. Isolate t in the equation a v f v i. Remember that, when multiplying t or dividing vectors, use the scalar form of the equation: t vi vf a PHYSICS INSIGHT Recall that the difference of squares is (a b)(a b) a2 b2 Then substitute the expression for t into d (vi d (vi vi vf a 1 2 1 2 vf)t: vf) 2 vi 2 vf a 1 d 2 Chapter 1 Graphs and equations describe motion in one dimension. 51 01-PearsonPhys20-Chap01 7/23/08 11:43 AM Page 52 info BIT A tortoise covered 5.48 m in 43.7 s at the National Tortoise Championships in Tickhill, UK, to set a world record on July 2, 1977. It was moving at 0.45 km/h. The more standard form of the fifth kinematics equation is vf 2 vi 2 2ad This equation is applied in the next example. Example 1.16 Practice Problems 1. A jetliner lands on a runway at 70 m/s, reverses its engines to provide braking, and comes to a halt 29 s later. (a) What is the jet\u2019s acceleration? (b) What length of runway did the jet require to come safely to a complete stop? 2. On-ramps are designed so that motorists can move seamlessly into highway traffic. If a car needs to increase its speed from 50 km/h to 100 km/h and the engine can provide a maximum acceleration of magnitude 3.8 m/s2, find the minimum length of the on-ramp. Answers 1. (a) 2.4 m/s2 [forward] (b) 1.0 km 2. 76 m A bullet accelerates the length of the barrel of a gun (0.750 m) with a magnitude of 5.35 105 m/s2. With what speed does the bullet exit the barrel? Given a 5.35 105 m/s2 d 0.750 m Required final speed (vf) Analysis and Solution Use the equation vf from rest, vi 0 m/s. 2 vi 2 2ad. Since the bullet starts vf vf 2 (0 m/s)2 2(5.35 105 m/s2)(0.750 m) 802 500 m2/s2 802 500 m2/s2 896 m/s Paraphrase The bullet leaves the barrel of the gun with a speed of 896 m/s. It is important to note that the velocity-time graph used to derive the kinematics equations has a constant slope (see Figure 1.54), so the equations derived from it are for objects undergoing uniformly accelerated motion (constant acceleration). General Method of Solving Kinematics Problems Now that you know five kinematics equations, how do you know which one to use to solve a problem? To answer this question, notice that each of the five kinematics equations has four variables. Each kinematics problem will provide you with three of these variables, as given values. The fourth variable represents the unknown value. When choosing your equation, make sure that all three known variables and the one unknown variable are represented in the equation (see Table 1.8). You may need to rearrange the equation to solve for the unknown variable. 52 Unit I Kinematics 01-PearsonPhys20-Chap01 7/23/08 11:43 AM Page 53 \u25bc Table 1.8 The Variables in the Five Kinematics Equations Equation v i v f)t v d 1 t a(t)2 2 i v d 1 t a(t)2 2 f 2 vi vf 2 2ad PHYSICS INSIGHT Remember these implied given values. \u2013 If the object starts from 0. \u2013 If the object comes to rest, v i f a stop, v 0. \u2013 If the object experiences uniform motion, a 0. 1.5 Check and Reflect 1.5 Check and Reflect Applications 1. How far will a humanoid robot travel in 3.0 s, accelerating at 1.0 cm/s2 [forward], if its initial velocity is 5.0 cm/s [forward]? 2. What is the displacement of a logging truck accelerating from 10 m/", "s [right] to 20 m/s [right] in 5.0 s? 3. How far will a car travel if it starts from rest and experiences an acceleration of magnitude 3.75 m/s2 [forward] for 5.65 s? 4. Determine the acceleration of a bullet starting from rest and leaving the muzzle 2.75 103 s later with a velocity of 460 m/s [forward]. 5. Some aircraft are capable of accelerations of magnitude 42.5 m/s2. If an aircraft starts from rest, how long will it take the aircraft to travel down the 2.6-km runway? 6. If a cyclist travelling at 14.0 m/s skids to a stop in 5.60 s, determine the skidding distance. Assume uniform acceleration. 7. Approaching a flashing pedestrianactivated traffic light, a driver must slow down to a speed of 30 km/h. If the crosswalk is 150 m away and the vehicle\u2019s initial speed is 50 km/h, what must be the magnitude of the car\u2019s acceleration to reach this speed limit? 8. A train\u2019s stopping distance, even when full emergency brakes are engaged, is 1.3 km. If the train was travelling at an initial velocity of 90 km/h [forward], determine its acceleration under full emergency braking. 9. A rocket starts from rest and accelerates uniformly for 2.00 s over a displacement of 150 m [W]. Determine the rocket\u2019s acceleration. 10. A jet starting from rest reaches a speed of 241 km/h on 96.0 m of runway. Determine the magnitude of the jet\u2019s acceleration. 11. What is a motorcycle\u2019s acceleration if it starts from rest and travels 350.0 m [S] in 14.1 s? 12. Determine the magnitude of a car\u2019s acceleration if its stopping distance is 39.0 m for an initial speed of 97.0 km/h. 13. A typical person can tolerate an acceleration of about 49 m/s2 [forward]. If you are in a car travelling at 110 km/h and have a collision with a solid immovable object, over what minimum distance must you stop so as to not exceed this acceleration? 14. Determine a submarine\u2019s acceleration if its initial velocity is 9.0 m/s [N] and it travels 1.54 km [N] in 2.0 min. e TEST To check your understanding of the kinematics equations, follow the eTest links at www.pearsoned.ca/school/physicssource. Chapter 1 Graphs and equations describe motion in one dimension. 53 01-PearsonPhys20-Chap01 7/23/08 11:43 AM Page 54 1.6 Acceleration due to Gravity Figure 1.57 Amusement park rides are an application of physics. projectile motion: motion in a vertical plane projectile: an object released or thrown into the air Many amusement parks and midways showcase a ride based solely on acceleration due to gravity. The ride transports thrill seekers up to a dizzying height, allows them to come to rest, and then, without warning, releases them downward before coming to a controlled stop (Figure 1.57). In the previous sections, you learned about objects that move in a horizontal plane. Many objects move in a vertical plane. Flipping a coin, punting a football, and a free throw in basketball are all examples of objects experiencing motion in a vertical plane (Figure 1.58). This type of motion is called projectile motion. A projectile is any object thrown into the air. Projectiles include objects dropped from rest; objects thrown downward or vertically upward, such as a tennis ball for service; and objects moving upward at an angle, such as a punted football. First let\u2019s consider projectile motion of an object moving straight up or down. What is the relationship between an object\u2019s mass and the speed of its fall? Do the next QuickLab to find out. x (horizontal) y (vertical) Figure 1.58 A plane has two dimensions, x and y. 1-6 QuickLab 1-6 QuickLab The Bigger They Are... Problem Does mass affect how quickly an object falls? Materials two objects of similar size and shape but different mass, such as a marble and a ball bearing, a die and a sugar cube, a golf ball and a table tennis ball two pans chair Procedure 1 Place a pan on either side of the chair. 2 Standing on the chair, release each pair of objects from the same height at the same time. 3 Listen for the objects hitting the pans. 4 Repeat steps 2 and 3 for other pairs of objects. 5 Repeat steps 2 and 3 from a higher and a lower height. Questions 1. Did the pair of objects land at the same time? 2. How did a change in height affect how long it took the objects to drop?", " 3. How did a change in the objects\u2019 shape affect how long it took each pair to drop? In the 16th century, Galileo conducted experiments that clearly demonstrated that objects falling near Earth\u2019s surface have a constant acceleration, neglecting air resistance, called the acceleration due to gravity. You can determine the value of the acceleration due to Earth\u2019s gravity by performing the following experiment. info BIT In 1971, astronaut David Scott tested Galileo\u2019s theory with a feather and a hammer. With no air on the Moon, both objects hit the ground at the same time. 54 Unit I Kinematics 01-PearsonPhys20-Chap01 7/23/08 11:43 AM Page 55 1-7 Inquiry Lab 1-7 Inquiry Lab Determining the Magnitude of the Acceleration due to Gravity Required Skills Initiating and Planning Performing and Recording Analyzing and Interpreting Communication and Teamwork Question How can position-time and velocity-time graphs be used to determine the acceleration due to gravity? 7 Using a ruler, measure the position of the object at each time interval and record it in your data table. 8 Plot your collected data on a position-time graph. Materials and Equipment 60-Hz spark timer ticker tape carbon disk power supply small mass metre-stick or ruler Procedure masking tape C-clamp retort stand graph paper cushion 9 With a sweeping motion, practise connecting the dots in a smooth curve that best fits the data. 10 Construct a data table in your notebook for recording instantaneous velocity and time. 11 Draw three tangents on the position-time graph. 12 Calculate the instantaneous velocities at these points by determining the slopes of the tangents. Record the data in your table. 1 Construct a data table in your notebook for recording 13 Plot a velocity-time graph of your collected data. time and position. 14 Draw a line of best fit. 2 Set up materials as shown in Figure 1.59(a), ensuring that the timer is 1.5 m above the floor. 15 Calculate the acceleration experienced by the object, in m/s2, by finding the slope of the velocity-time graph. recording timer Analysis ticker tape mass cushion 1. Determine the experimental value of the magnitude of acceleration due to gravity by averaging your group\u2019s results. 2. Determine the percent error for your experimental value. Assume the theoretical magnitude of a is 9.81 m/s2. 3. Describe the shape of the position-time graph you drew in step 9. Figure 1.59(a) 4. From your graph, describe the relationship between time and displacement for an accelerating object. 3 Attach a 1.5-m strip of ticker tape to the mass and thread the ticker tape through the spark timer. 4 Turn on the spark timer just before your partner releases the mass. 5 Repeat steps 3 and 4 for each person in your group. 6 Analyze the ticker tape by drawing a line through the first distinct dot on the tape. Label it \u201cstart\u201d. (On a 60-Hz timer, every sixth dot represents 0.10 s.) Continue labelling your ticker tape as shown in Figure 1.59(b). t 0.10 s start t 0.20 s t 0.30 s d 1 d 2 Figure 1.59(b) e LAB d 3 For a probeware activity, go to www.pearsoned.ca/school/physicssource. Chapter 1 Graphs and equations describe motion in one dimension. 55 01-PearsonPhys20-Chap01 7/23/08 11:43 AM Page 56 e TECH Use graphical analysis to determine acceleration due to Earth\u2019s gravity. Go to www.pearsoned.ca/school/ physicssource. vi 0 a 9.81 m/s2 [down] dy Figure 1.60 The time it takes the golf ball to hit the ground depends on the height from which it drops and on the acceleration due to gravity. PHYSICS INSIGHT The equations of parabolas are quadratic equations because they include a power of two, for example, y x2. The equation for the displacement of a vertical projectile is d v 1 t a(t)2. 2 i Gravity Causes Objects to Accelerate Downward Recall the kinematics equations for accelerated motion from section 1.5(t)2 2 1 t a(t)2 2 vf 2 vi 2 2ad You can also apply these equations to motion in a vertical plane. is the acceleration due Because vertical acceleration is due to gravity, a to gravity, or 9.81 m/s2 [down]. If you drop a golf ball from a height of 1.25 m, how long will it take for the ball to reach the ground (Figure 1.60)? Because the ball is moving in only one direction, down, choose down to be positive for simplicity. Since the golf ball is", " accelerating due to gravity starting from rest, v i 0 and a 9.81 m/s2 [down] 9.81 m/s2 The ball\u2019s displacement can be expressed as 1.25 m [down], or 1.25 m. The equation that includes all the given variables and the unknown variable is d v i t a(t)2. The displacement and acceleration vectors 1 2 are both in the same direction, so use the scalar form of the equation to solve for time. Since vi 0, 1 d at2 2 2d t a 2(1.25 m) m 9.81 s2 0.505 s info BIT Without a parachute, Vesna Vulovic, a flight attendant, survived a fall of 10 160 m when the DC-9 airplane she was travelling in exploded. The golf ball takes 0.505 s to reach the ground when released from a rest height of 1.25 m. Note that the time it takes for an object to fall is directly proportional to the square root of the height it is dropped from: t. If there is 2d a no air resistance, the time it takes for a falling object to reach the ground depends only on the height from which it was dropped. The time does not depend on any other property of the object. 56 Unit I Kinematics 01-PearsonPhys20-Chap01 7/23/08 11:43 AM Page 57 1-8 QuickLab 1-8 QuickLab Could You Be a Goalie for the NHL? Problem What is your reaction time? Materials and Equipment long ruler (30 cm or more) flat surface Procedure 1 Rest your arm on a flat surface with your wrist at the edge. 2 Ask your partner to hold the ruler vertically so that the ruler\u2019s end is just above your hand. 3 Curl your fingers so that the space between your thumb and index finger is large enough for the ruler to pass through easily. 4 Without watching your partner, ask your partner to let go of the ruler without warning. 5 Try to close your thumb and index finger on the ruler as quickly as possible. 6 Record where your hand is on the ruler. 7 Repeat steps 1\u20136 several times. Questions 1. Determine the average distance the ruler falls in each of your trials. 2. Using the average distance, calculate the time. 3. An average NHL goalie has a reaction time of 0.15 s. How does your reaction time compare with your partner\u2019s? 4. Certain drugs impair reaction time. What would you expect your results in this lab to be if your reaction time were increased? Instead of dropping an object such as a golf ball, what if you threw an object down? By throwing an object straight down, you give the object an initial vertical velocity downward. What effect does an initial velocity have on the motion of the object? The next example will show you. Example 1.17 While cliff diving in Mexico, a diver throws a smooth, round rock straight down with an initial speed of 4.00 m/s. If the rock takes 2.50 s to land in the water, how high is the cliff? Given For convenience, choose down to be positive because down is the only direction of the ball\u2019s motion. v 4.00 m/s [down] 4.00 m/s t 2.50 s a 9.81 m/s2 [down] 9.81 m/s2 i Required height of cliff (d) Analysis and Solution The initial velocity and acceleration vectors are both in the same direction, so use the scalar form of the equation d v 1 t a(t)2. 2 i Chapter 1 Graphs and equations describe motion in one dimension. 57 01-PearsonPhys20-Chap01 7/23/08 11:43 AM Page 58 d (4.00 m/s)(2.50 s) (9.81 m/s2)(2.50 s)2 1 2 10.0 m 30.7 m 40.7 m Paraphrase The cliff is 40.7 m high. Practice Problems 1. If a rock takes 0.750 s to hit the ground after being thrown down from a height of 4.80 m, determine the rock\u2019s initial velocity. 2. Having scored a touchdown, a football player spikes the ball in the end zone. If the ball was thrown down with an initial velocity of 2.0 m/s from a height of 1.75 m, determine how long it is in the air. 3. An elevator moving downward at 4.00 m/s experiences an upward acceleration of 2.00 m/s2 for 1.80 s. What is its velocity at the end of the acceleration and how far has it travelled? Answers 1. 2.72 m/s [down] 2. 0.43 s 3. 0.400 m/s [down], 3.96 m What Goes Up Must Come Down Circus clowns are often", " accomplished jugglers (Figure 1.61). If a juggler throws a ball upward, giving it an initial velocity, what happens to the ball (Figure 1.62)? vi a dy Figure 1.61 Juggling is an example of projectile motion. Figure 1.62 The ball\u2019s motion is called vertical projectile motion. When you throw an object up, its height (displacement) increases while its velocity decreases. The decrease in velocity occurs because the object experiences acceleration downward due to gravity (Figure 1.63(a)). The ball reaches its maximum height when its vertical velocity equals zero. In other words, it stops for an instant at the top of its path (Figure 1.63(b)). When the object falls back toward the ground, it speeds up because of the acceleration due to gravity (Figure 1.63(c)). 58 Unit I Kinematics 01-PearsonPhys20-Chap01 7/23/08 11:43 AM Page 59 a vi v 0 a vf a Figure 1.63(a) Stage 1: Velocity and acceleration are in opposite directions, so the ball slows down. Figure 1.63(b) Stage 2: The ball has momentarily stopped, but its acceleration is still 9.81 m/s2 [down], which causes the ball to change direction. Figure 1.63(c) Stage 3: Velocity and acceleration are in the same direction, so the ball speeds up. The next two examples analyze different stages of the same object\u2019s motion. Example 1.18 analyzes the upward part of the motion of an object thrown upward, whereas Example 1.19 analyzes the same object\u2019s downward motion. Example 1.18 A clown throws a ball upward at 10.00 m/s. Find (a) the maximum height the ball reaches above its launch height (b) the time it takes to do so Given Consider up to be positive. 10.00 m/s [up] 10.00 m/s vi a 9.81 m/s2 [down] 9.81 m/s2 Required (a) maximum height above launch height (d) (b) time taken to reach maximum height (t) Analysis and Solution (a) When you throw an object up, as its height increases, its speed decreases because the object is accelerating downward due to gravity. The ball, travelling upward away from its initial launch height, reaches its maximum height when its vertical velocity is zero. In other words, the object stops for an instant at the top of its path up, so vf neglecting air friction, use the equation vf substitute scalar quantities. 0.00 m/s. To find the object\u2019s maximum height, 2 2ad and 2 vi Practice Problem 1. The Slingshot drops riders 27 m from rest before slowing them down to a stop. How fast are they moving before they start slowing down? Answer 1. 23 m/s e WEB Can you shoot an object fast enough so that it does not return to Earth? Research escape velocity. Is it the same regardless of the size of an object? How do you calculate it? Write a brief summary of your findings. To learn more about escape velocity, follow the links at www.pearsoned.ca/school/ physicssource. Chapter 1 Graphs and equations describe motion in one dimension. 59 01-PearsonPhys20-Chap01 7/23/08 11:43 AM Page 60 d 2 vi 2 vf 2a m m 2 2 10.00 0.00 s s m 29.81 s2 5.10 m v, where a is the (b) To find the time taken, use the equation a t acceleration due to gravity. Substitute scalar quantities because you are dividing vectors. t vi vf a m m 10.00 0.00 s s m 9.81 s2 1.02 s Paraphrase (a) The ball\u2019s maximum height is 5.10 m above its launch height. (b) It takes the ball 1.02 s to reach maximum height. The next example is a continuation of the previous example: It analyzes the same ball\u2019s motion as it falls back down from its maximum height. Example 1.19 A clown throws a ball upward at 10.00 m/s. Find (a) the time it takes the ball to return to the clown\u2019s hand from maximum height (b) the ball\u2019s final velocity Given Consider up to be positive. v 10.00 m/s [up] 10.00 m/s a 9.81 m/s2 [down] 9.81 m/s2 i Practice Problems 1. A pebble falls from a ledge 20.0 m high. (a) Find the velocity with which it hits the ground. (b) Find the time it takes to hit the ground. Answers 1. (a) 19.8 m/s [", "down] (b) 2.02 s 60 Unit I Kinematics Required (a) time taken to land (t) (b) final velocity (v f) Analysis and Solution (a) For an object starting from rest at maximum height and accelerating downward due to gravity, its motion is described by the equation d v i t a(t)2, where 1 2 i 0 (at maximum height). For downward motion, the v ball\u2019s displacement and acceleration are in the same direction, so use the scalar form of the equation. For d, substitute 5.10 m (from Example 1.18(a)). Rearrange this equation and substitute the values. 01-PearsonPhys20-Chap01 7/23/08 11:43 AM Page 61 t 2d a 2(5.10 m) m 9.81 s2 1.02 s Compare this time to the time taken to reach maximum height (Example 1.18(b)). (b) The ball\u2019s final velocity (starting from maximum height) when it info BIT At terminal velocity, parachuters no longer accelerate but fall at a constant speed. Humans have a terminal velocity of about 321 km/h [down] when curled up and about 201 km/h [down] with arms and legs fully extended to catch the wind. v lands on the ground is v a i t at v f f i v 0.00 m/s (9.81 m/s2)(1.02 s) 10.0 m/s The negative sign means that the direction is downward. Paraphrase (a) It takes the ball 1.02 s to return to the clown\u2019s hand. (b) The final velocity at the height of landing is 10.0 m/s [down]. Concept Check (a) Why does it make sense that the time taken to travel up to the maximum height is equal to the time to fall back down to the starting height? (b) What variables determine how long a projectile is in the air? Does the answer surprise you? Why or why not? Position vs. Time ) ].00 5.00 4.00 3.00 2.00 1.00 0.00 0.00 0.50 1.00 2.00 2.50 1.50 Time (s) You can use the data calculated in Examples 1.18 and 1.19 to plot a position-time graph of the ball\u2019s motion. Because the ball experiences uniformly accelerated motion, the graph is a parabola (Figure 1.64). Figure 1.64 The positiontime graph of a ball thrown vertically upward is a parabola. A Graphical Representation of a Vertical Projectile You can now represent the motion of the juggler\u2019s ball on a position-time graph. Remember that the ball\u2019s motion can be divided into three different stages: Its speed decreases, becomes zero, and then increases. However, the velocity is uniformly decreasing. The graphs that correspond to these three stages of motion are shown in Figure 1.65. Position vs. Time Position vs. Time Position vs. Time ) ] Time (s) Time (s) Time (s) Figure 1.65(a) Consider up to be positive. The ball rises until it stops. Figure 1.65(b) momentarily at maximum height. The ball stops Figure 1.65(c) back down to its launch height. The ball falls Chapter 1 Graphs and equations describe motion in one dimension. 61 01-PearsonPhys20-Chap01 7/23/08 11:43 AM Page 62 Position vs. Time ) ] Time (s) Figure 1.66 A ball thrown straight up in the air illustrates uniformly accelerated motion. Now put these three graphs together to generate the complete positiontime graph of the ball\u2019s motion. Remember that the ball is actually moving straight up and down, and not in a parabolic path (Figure 1.66). Why is the graph of its motion a parabola rather than a straight vertical line? To generate a corresponding velocity-time graph from the positiontime graph in Figure 1.66, draw a series of tangents at specific time instances. Choosing strategic points will make your task easier. The best points to choose are those that begin and end a stage of motion because they define that stage (Figure 1.67(a)). Position vs. Time (a) 6.00 5.00 4.00 3.00 2.00 1.00 ) ].00 0.00 0.50 1.00 1.50 2.00 2.50 Time (s) 15.00 10.00 5.00 0.00 5.00 Velocity vs. Time Time (s) 0.50 1.00 1.50 2.00 2.50 (b 10.00 15.00 Figure 1.67 To generate the velocity-time graph in (b) corresponding to the position", "time graph in (a), draw tangents at strategic points. info BIT In 1883, the Krakatoa volcano in Indonesia hurled rocks 55 km into the air. This volcanic eruption was 10 000 times more powerful than the Hiroshima bomb! Can you find the time it took for the rocks to reach maximum height? In Figure 1.67(a), notice that the initial slope of the tangent on the positiontime graph is positive, corresponding to an initial positive (upward) velocity on the velocity-time graph below (Figure 1.67(b)). The last tangent has a negative slope, corresponding to a final negative velocity on the velocity-time graph. The middle tangent is a horizontal line (slope equals zero), which means that the ball stopped momentarily. Remember that the slope of a velocity-time graph represents acceleration. Concept Check What should be the value of the slope of the velocity-time graph for vertical projectile motion? 62 Unit I Kinematics 01-PearsonPhys20-Chap01 7/23/08 11:43 AM Page 63 1.6 Check and Reflect 1.6 Check and Reflect Knowledge 1. Define a projectile. 2. What determines how long it will take an object to reach the ground when released with an initial velocity of zero? Applications 11. A penny is dropped from a cliff of height 190 m. Determine the time it takes for the penny to hit the bottom of the cliff. 12. A coin tossed straight up into the air takes 2.75 s to go up and down from its initial release point 1.30 m above the ground. What is its maximum height? 3. A student drops a bran muffin from the roof of the school. From what height is the muffin dropped if it hits the ground 3.838 s later? 13. If a diver starts from rest, determine the amount of time he takes to reach the water\u2019s surface from the 10-m platform. 4. During a babysitting assignment, a babysitter is constantly picking up toys dropped from the infant\u2019s highchair. If the toys drop from rest and hit the floor 0.56 s later, from what height are they being dropped? 5. A rock takes 1.575 s to drop 2.00 m down toward the surface of the Moon. Determine the acceleration due to gravity on the Moon. 6. At the beginning of a game, a referee throws a basketball vertically upward with an initial speed of 5.0 m/s. Determine the maximum height above the floor reached by the basketball if it starts from a height of 1.50 m. 7. A student rides West Edmonton Mall\u2019s Drop of Doom. If the student starts from rest and falls due to gravity for 2.6 s, what will be his final velocity and how far will he have fallen? 8. If the acceleration due to gravity on Jupiter is 24.8 m/s2 [down], determine the time it takes for a tennis ball to fall 1.75 m from rest. 9. If a baseball popped straight up into the air has a hang time (length of time in the air) of 6.25 s, determine the distance from the point of contact to the baseball\u2019s maximum height. 10. Jumping straight up, how long will a red kangaroo remain in the air if it jumps through a height of 3.0 m? 14. A person in an apartment building is 5.0 m above a person walking below. She plans to drop some keys to him. He is currently walking directly toward a point below her at 2.75 m/s. How far away is he if he catches the keys 1.25 m above the ground? Extensions 15. A rocket launched vertically upward accelerates uniformly for 50 s until it reaches a velocity of 200 m/s [up]. At that instant, its fuel runs out. (a) Calculate the rocket\u2019s acceleration. (b) Calculate the height of the rocket when its fuel runs out. (c) Explain why the rocket continues to gain height for 20 s after its fuel runs out. (d) Calculate the maximum height of the rocket. 16. A ball is dropped from a height of 60.0 m. A second ball is thrown down 0.850 s later. If both balls reach the ground at the same time, what was the initial velocity of the second ball? e TEST To check your understanding of projectiles and acceleration due to gravity, follow the eTest links at www.pearsoned.ca/school/physicssource. Chapter 1 Graphs and equations describe motion in one dimension. 63 01-PearsonPhys20-Chap01 7/23/08 11:43 AM Page 64 CHAPTER 1 SUMMARY Key Terms and Concepts distance displacement velocity uniform motion at rest acceleration non-uniform motion instantaneous velocity tangent uniformly accelerated motion projectile motion projectile acceleration due to gravity kinematics origin position scalar quantity vector", " quantity Key Equations v f v i)t d vv 1 t a(t)2 2 i d v 1 t a(t)2 2 f 2 vi vf 2 2ad Conceptual Overview The concept map below summarizes many of the concepts and equations in this chapter. Copy and complete the map to have a full summary of the chapter. kinematics uniform motion at rest constant velocity accelerated motion uniform displacement : zero dt graph\u2014 vt graph\u2014 at graph\u2014 equations: none displacement : equal changes over equal time intervals dt graph\u2014 vt graph\u2014 horizontal line (/) at graph\u2014 displacement : changes dt graph\u2014 curve vt graph\u2014 at graph\u2014 horizontal line (/) equations: equations: 1 d (vf vi )t 2 vf 2 vi 2 2ad 64 Unit I Kinematics 01-PearsonPhys20-Chap01 7/23/08 11:43 AM Page 65 CHAPTER 1 REVIEW Knowledge 1. (1.1) State two ways in which a vector quantity differs from a scalar quantity. Give an example of each. 2. (1.2) Complete a position-time data table for the motion described by the ticker tape given below. 3. (1.3) Determine the velocity of each object whose motion is represented by the graphs below. (a) (b) (c 12 10 8 6 4 2 0 Position vs. Time 0 2 4 6 8 10 12 Time (s) Position vs. Time Time (min) 2 4 6 8 10 12 Position vs. Time Time (s) 5 10 15 20 ) ] 10 5 0 5 10 15 20 25 ) ] 10 5 0 5 10 15 20 25 30 4. (1.5) What is a vehicle\u2019s displacement if it travels at a velocity of 30.0 m/s [W] for 15.0 min? 5. (1.5) How long will it take a cross-country skier, travelling 5.0 km/h, to cover a distance of 3.50 km? 6. (1.2) Determine the average speed, average velocity, and net displacement from the position-time graph below. Position vs. Time ) ] 30.0 25.0 20.0 15.0 10.0 5.0 0.0 5.0 10.0 15.0 20.0 25.0 30.0 2.0 4.0 6.0 8.0 10.0 12.0 14.0 16.0 18.0 20.0 Time (s) 7. (1.2) Explain how a person standing still could have the same average velocity but different average speed than a person running around a circular track. 8. (1.6) If an object thrown directly upwards remains in the air for 5.6 s before it returns to its original position, how long did it take to reach its maximum height? 9. (1.6) If an object thrown directly upwards reaches its maximum height in 3.5 s, how long will the object be in the air before it returns to its original position? Assume there is no air resistance. 10. (1.6) What is the initial vertical velocity for an object that is dropped from a height, h? Applications 11. A scuba diver swims at a constant speed of 0.77 m/s. How long will it take the diver to travel 150 m at this speed? 12. In 1980, during the Marathon of Hope, Terry Fox ran 42 km [W] a day. Assuming he ran for 8.0 h a day, what was his average velocity in m/s? 13. Explain how the point of intersection of two functions on a position-time graph differs from the point of intersection of two functions on a velocity-time graph. 14. A thief snatches a handbag and runs north at 5.0 m/s. A police officer, 20 m to the south, sees the event and gives chase. If the officer is a good sprinter, going 7.5 m/s, how far will she have to run to catch the thief? Chapter 1 Graphs and equations describe motion in one dimension. 65 01-PearsonPhys20-Chap01 7/23/08 11:43 AM Page 66 Maintaining that acceleration, how long will it take the police car to catch up with the speeding motorist? At what speed would the police car be moving? Explain whether or not this scenario is likely to happen. 25. Two cars pass one another while travelling on a city street. Using the velocity-time graph below, draw the corresponding position-time graph and determine when and where the two cars pass one another. Velocity vs. Time ) ] 20.0 18.0 16.0 14.0 12.0 10.0 8.0 6.0 4.0 2.0 0.0 0.0 2.0 4.0 6", ".0 8.0 10.0 Time (s) 26. Calculate displacement and acceleration from the graph below. Velocity vs. Time Time (h) 1.0 2.0 3.0 4.0 5..0 0.0 1.0 2.0 3.0 4.0 5.0 27. Built in Ontario, the world\u2019s fastest fire truck, the Hawaiian Eagle, can accelerate at 9.85 m/s2 [forward]. Starting from rest, how long will it take the Hawaiian Eagle to travel a displacement of 402 m [forward]? 28. A vehicle is travelling at 25.0 m/s. Its brakes provide an acceleration of 3.75 m/s2 [forward]. What is the driver\u2019s maximum reaction time if she is to avoid hitting an obstacle 95.0 m away? 29. Off-ramps are designed for motorists to decrease their vehicles\u2019 velocities to move seamlessly into city traffic. If the off-ramp is 1.10 km long, calculate the magnitude of a vehicle\u2019s acceleration if it reduces its speed from 110.0 km/h to 60.0 km/h. 15. Calculate the magnitude of a bullet\u2019s acceleration if it travels at a speed of 1200 m/s and stops within a bulletproof vest that is 1.0 cm thick. 16. From the velocity-time graph below, determine how far an elk will travel in 30 min. Velocity vs. Time for an Elk ) ] 80 70 60 50 40 30 20 10 0 0.0 0.5 1.0 Time (h) 1.5 2.0 17. The world record for a speedboat is 829 km/h. Heading south, how far will the boat travel in 2.50 min? 18. How much faster is an airliner than a stagecoach if the stagecoach takes 24 h to travel 300 km and the airliner takes 20 min? 19. A car\u2019s odometer reads 22 647 km at the start of a trip and 23 209 km at the end. If the trip took 5.0 h, what was the car\u2019s average speed in km/h and m/s? 20. A motorcycle coasts downhill from rest with a constant acceleration. If the motorcycle moves 90.0 m in 8.00 s, find its acceleration and velocity after 8.00 s. 21. A cyclist crosses a 30.0-m bridge in 4.0 s. If her initial velocity was 5.0 m/s [N], find her acceleration and velocity at the other end of the bridge. 22. An object with an initial velocity of 10.0 m/s [S] moves 720 m in 45.0 s along a straight line with constant acceleration. For the 45.0-s interval, find its average velocity, final velocity, and acceleration. 23. During qualifying heats for the Molson Indy, a car must complete a 2.88-km lap in 65 s. If the car goes 60 m/s for the first half of the lap, what must be its minimum speed for the second half to still qualify? 24. A car travelling 19.4 m/s passes a police car at rest. As it passes, the police car starts up, accelerating with a magnitude of 3.2 m/s2. 66 Unit I Kinematics 01-PearsonPhys20-Chap01 7/23/08 11:43 AM Page 67 30. Calgary\u2019s CTrain leaves the 10 St. S.W. station at 4:45 p.m. and arrives at 3 St. S.E. at 4:53 p.m. If the distance between the two stops is 3.2 km, determine the CTrain\u2019s average velocity for the trip. 37. A contractor drops a bolt from the top of a roof located 8.52 m above the ground. How long does it take the bolt to reach the ground, assuming there is no air resistance? McKnight \u2013 Westwinds (2010) Whitehorn Rundle Marlborough 38. An improperly installed weathervane falls from the roof of a barn and lands on the ground 1.76 s later. From what height did the weathervane fall and how fast was it travelling just before impact? Dalhousie Brentwood N CTrain Map University Banff Trail Lions Park Sunnyside BOWRIVER SAIT/ ACAD/ Jubllee 7 St. S W PRINCE\u2019S ISLAND PARK 4 St. S W 1 St. S W Bridgeland/ M e m orial Zoo Franklin Barlo w/ M ax Bell Olym pic Plaza 3 St. SE City Hall 10 St. S W 8 St. S W 6 St. S W 3 St. S W Centre St Erlton/Stampede Victoria Park/Stampede B O W RIVER 201 Somerset/B", "ridlewood/Dalhousie 202 10 St./Whitehorn 7 Avenue Free Fare Zone 201 & 202 Transfer Points Chinook Southland Canyon Meadows NEW Shawnessy 39 Avenue Heritage Anderson Fish Creek \u2013 Lacombe Somerset/Bridlewood NEW 31. Describe the motion of the truck from the velocity-time graph below. When is the truck at rest? travelling with a uniform velocity? When is its acceleration the greatest? Velocity vs. Time ) ] 20.0 15.0 10.0 5.0 0.0 0.0 2.0 4.0 6.0 Time (s) 8.0 10.0 32. A racecar accelerates uniformly from 17.5 m/s [W] to 45.2 m/s [W] in 2.47 s. Determine the acceleration of the racecar. 33. How long will it take a vehicle travelling 80 km/h [W] to stop if the average stopping distance for that velocity is 76.0 m? 34. The Slingshot, an amusement park ride, propels its riders upward from rest with an acceleration of 39.24 m/s2. How long does it take to reach a height of 27.0 m? Assume uniform acceleration. 35. Starting from rest, a platform diver hits the water with a speed of 55 km/h. From what height did she start her descent into the pool? 36. A circus performer can land safely on the ground at speeds up to 13.5 m/s. What is the greatest height from which the performer can fall? 39. Attempting to beat the record for tallest Lego structure, a student drops a piece from a height of 24.91 m. How fast will the piece be travelling when it is 5.0 m above the ground and how long will it take to get there? Extension 40. Weave zones are areas on roads where vehicles are changing their velocities to merge onto and off of busy expressways. Suggest criteria a design engineer must consider in developing a weave zone. Consolidate Your Understanding Create your own summary of kinematics by answering the questions below. If you want to use a graphic organizer, refer to Student References 4: Using Graphic Organizers on pp. 869\u2013871. Use the Key Terms and Concepts listed on page 64 and the Learning Outcomes on page 4. 1. Create a flowchart to describe the changes in position, velocity, and acceleration for both uniform and accelerated motion. 2. Write a paragraph explaining the two main functions of graphing in kinematics. Share your report with another classmate. Think About It Review your answers to the Think About It questions on page 5. How would you answer each question now? e TEST To check your understanding of motion in one dimension, follow the eTest links at www.pearsoned.ca/school/physicssource. Chapter 1 Graphs and equations describe motion in one dimension. 67 02-PearsonPhys20-Chap02 7/24/08 10:17 AM Page 68 Vector components describe motion in two dimensions. Figure 2.1 The motion of Canada\u2019s Snowbird precision flight squad can be described using vectors. Imagine being a pilot for the Canadian Snowbirds (Figure 2.1). This precision flight team, composed of highly trained military personnel, performs at air shows across the country. Unlike the flight crew in the cockpit of a commercial airliner, these pilots execute aerobatic manoeuvres that require motion in both horizontal and vertical directions, while being acutely aware of their positions relative to the ground and to each other. In this chapter, you will study motion in one and two dimensions by building on the concepts you learned in Chapter 1. You will use vectors to define position, velocity, and acceleration, and their interrelationships. The vector methods you will learn will allow you to study more complex motions. C H A P T E R 2 Key Concepts In this chapter, you will learn about: two-dimensional motion vector methods Learning Outcomes When you have completed this chapter, you will be able to: Knowledge explain two-dimensional motion in a horizontal or vertical plane interpret the motion of one object relative to another Science, Technology, and Society explain that scientific knowledge is subject to change as new evidence comes to light and as laws and theories are tested, restricted, revised, or reinforced 68 Unit I 02-PearsonPhys20-Chap02 7/24/08 10:17 AM Page 69 2-1 QuickLab 2-1 QuickLab Taking a One-dimensional Vector Walk Problem How can you add vectors to determine displacement? Materials 30-m tape measure field marker (tape or flag) Procedure 1 Starting at the centre of a football field (or gymnasium), work out a path sequence using six forward and backward displacements to move from your starting position to the goal line. At least two of your six displacements must be oriented in the direction opposite to the direction of the goal line. 2 On the field,", " mark your starting point with a flag or tape. Define direction axes. 3 Ask your partner to walk the displacements of the path sequence chosen in step 1 while holding the end of the measuring tape. Mark your partner\u2019s endpoint after each displacement (Figure 2.2). 4 Continue the journey, using the measuring tape, until you have walked all the displacements. 5 Mark the final endpoint. 6 Using the measuring tape, determine the displacement from your starting point. 7 Repeat steps 2\u20136 using two different sequences of the six displacements you originally chose. 5 m [forward] Figure 2.2 Questions 1. What was the total distance you travelled? 2. What was the total displacement? 3. What conclusion can you draw about the order of adding vectors? Think About It 1. How does the order of a series of displacements affect the final position of an object? 2. In order to cross a river in the shortest possible time, is it better to aim yourself upstream so that you end up swimming straight across or to aim straight across and swim at an angle downstream? 3. Why does it take longer to fly across Canada from east to west rather than west to east in the same airplane? 4. How does the angle of a throw affect the time a ball spends in the air? 5. Two objects start from the same height at the same time. One is dropped while the other is given an initial horizontal velocity. Which one hits the ground first? Discuss your answers in a small group and record them for later reference. As you complete each section of this chapter, review your answers to these questions. Note any changes to your ideas. Chapter 2 Vector components describe motion in two dimensions. 69 02-PearsonPhys20-Chap02 7/24/08 10:17 AM Page 70 2.1 Vector Methods in One Dimension One of the fastest-growing sports in the world today is snowshoeing (Figure 2.3). The equipment required is minimal and the sport is easy to learn \u2014 you need to move forward in a straight line. Despite its simplicity, snowshoeing has great cardiovascular benefits: You can burn up to 1000 calories per hour, which makes it the ultimate cross-training program for athletes. It also allows athletes to explore different terrains and gain a greater appreciation of the outdoors, as well as to test their limits, especially by participating in endurance races! The motions in a showshoe race can be broken up into one-dimensional vector segments. In this section, you will study motion in one dimension using vectors. Figure 2.3 Snowshoeing is an excellent way of enjoying the great outdoors in winter while improving your health. Vector Diagrams In Chapter 1, you used variables and graphs to represent vector quantities. You can also represent vector quantities using vector diagrams. In a diagram, a line segment with an arrowhead represents a vector quantity. Its point of origin is called the tail, and its terminal point (arrowhead) is the tip (Figure 2.4). If the magnitude of a vector is given, you can draw the vector to scale. The length of the line segment depends on the vector\u2019s magnitude. The arrowhead indicates direction. Drawing vector diagrams to represent motion helps you to visualize the motion of an object. Properly drawn, vector diagrams enable you to accurately add vectors and to determine an object\u2019s position. tail tip Figure 2.4 A vector has a tail and a tip. 70 Unit I Kinematics 02-PearsonPhys20-Chap02 7/24/08 10:17 AM Page 71 Choosing Reference Coordinates When describing the motion of an object, there are many ways to describe its direction. You could use adjectives such as forward or backward, up or down, into or out of, and left or right. You can also use compass directions, such as north [N], south [S], east [E], and west [W]. When drawing a vector diagram, it is important to choose which directions are positive and to include these directions on every vector diagram. As you learned in section 1.1 (Figure 1.6), in this unit, forward, up, right, north, and east are usually designated as positive, whereas their opposites are usually considered negative. You may choose your own designation of positive and negative when solving problems, but make sure your reference direction is consistent within each problem and clearly communicated at the beginning of your solution. Practise drawing vectors in the next Skills Practice exercise Representing a Vector Using an appropriate scale and direction convention, draw each of the following vectors. (a) 5 m [forward] (b) 20 m [down] (c) 30 km [north] (d) 150 km [left] Adding Vectors in One Dimension Motion in one dimension involves vectors that are collinear. Collinear vectors lie along the same straight line. They may point in the same or in opposite directions (Figure 2.5). collinear: along", " the same straight line, either in the same or in opposite directions (a) (b) Figure 2.5 (a) Collinear vectors in the same direction (b) Collinear vectors in opposite directions When more than one vector describes motion, you need to add the vectors. You can add and subtract vectors graphically as well as algebraically, provided they represent the same quantity or measurement. As in mathematics, in which only like terms can be added, you can only add vectors representing the same types of quantities. For example, you can add two or more position vectors, but not a position vector, 5 m [E], to a velocity vector, 5 m/s [E]. In addition, the unit of measurement must be the same. For example, before adding the position vectors 5 m [E] and 10 km [E], you must convert the units of one of the vectors so that both vectors have the same units. In the next example, determine the sum of all the vector displacements graphically by adding them tip to tail. The sum of a series of vectors is called the resultant vector. resultant vector: a vector drawn from the tail of the first vector to the tip of the last vector Chapter 2 Vector components describe motion in two dimensions. 71 02-PearsonPhys20-Chap02 7/24/08 10:17 AM Page 72 Example 2.1 Contestants in a snowshoe race must move forward 10.0 m, untie a series of knots, move forward 5.0 m, solve a puzzle, and finally move forward 25.0 m to the finish line (Figure 2.6). Determine the resultant vector by adding the vectors graphically. Figure 2.6 Analysis and Solution 1. Choose an appropriate scale and reference direction. 1.0 cm : 5.0 m, forward is positive. 2. Draw the first vector and label its magnitude and direction (Figure 2.7). 10.0 m [forward] Figure 2.7 3. Place the tail of the second vector at the tip of the first vector. Continue to place all the remaining vectors in order, tip to tail (Figure 2.8). 10.0 m [forward] 5.0 m [forward] 25.0 m [forward] Figure 2.8 4. Connect the tail of the first vector to the tip of the last vector. This new vector, which points toward the tip of the last vector, is the resultant vector, R (the purple arrow in Figure 2.9). 72 Unit I Kinematics 02-PearsonPhys20-Chap02 7/25/08 7:57 AM Page 73 Tail (origin) of first vector Tip (terminal point) of last vector Figure 2.9 Find the magnitude of the resultant vector by measuring with a ruler, then convert the measured value using the scale. Remember to include the direction. d 8.0 cm [forward] 5.0 m 1.0 cm 40 m [forward] Practice Problems 1. The coach of the high-school rugby team made the team members run a series of sprints: 5.0 m [forward], 10 m [backward], 10 m [forward], 10 m [backward], 20 m [forward], 10 m [backward], 40 m [forward], and 10 m [backward]. (a) What is their total distance? (b) What is their displacement? Answers 1. (a) 115 m (b) 35 m [forward] In summary, you can see that adding vectors involves connecting them tip to tail. The plus sign in a vector equation tells you to connect the vectors tip to tail in the vector diagram. 2 d d 1: Add the negative of d To subtract collinear vectors graphically (Figure 2.10(a)), you may use one of two methods. For the first method, find d using the equation d 2, tip to tail, as you did in Example 2.1. The negative of a vector creates a new vector that points in the opposite direction of the original vector (Figure 2.10(b)). For the second method, connect the vectors tail to tail. This time, d starts at the tip of d 1 and ends at the tip of d 2 (Figure 2.10(c)). 1 to d (a) (b) (c) d2 d1 d d2 d1 d2 d d1 Figure 2.10 d (a) To subtract two collinear vectors, d 2 or 1 to d 1, graphically, (b) add the negative of d 2 (c) connect the vectors tail to tail and draw the resultant connecting the tip of d 1 to the tip of d 2. Chapter 2 Vector components describe motion in two dimensions. 73 02-PearsonPhys20-Chap02 7/24/08 10:17 AM Page 74 Recall that the definition of displacement is final position minus initial position, or d i. The next example reviews the algebraic", " subtraction of vectors, which you learned in Chapter 1, Example 1.1, and also shows you how to subtract vectors graphically. d d f Example 2.2 A sailboat that is initially 15 m to the right of a buoy sails 35 m to the left of the buoy (Figure 2.11). Determine the sailboat\u2019s displacement (a) algebraically and (b) graphically. Figure 2.11 35 m [left] origin 15 m [right] Practice Problems 1. Sprinting drills include running 40.0 m [N], walking 20.0 m [N], and then sprinting 100.0 m [N]. Using vector diagrams, determine the sprinter\u2019s displacement from his initial position. 2. To perform a give and go, a basketball player fakes out the defence by moving 0.75 m [right] and then 3.50 m [left]. Using vector diagrams, determine the player\u2019s displacement from the starting position. 3. While building a wall, a bricklayer sweeps the cement back and forth. If she swings her hand back and forth, a distance of 1.70 m, four times, use vector diagrams to calculate the distance and displacement her hand travels during that time. Check your answers against those in Example 1.1 Practice Problems 1-3. Answers 1. 160.0 m [N] 2. 2.75 m [left] 3. 6.80 m, 0 m 74 Unit I Kinematics Given Consider right to be positive. 15 m [right] 15 m d 35 m [left] 35 m d i f Required displacement (d ) Analysis and Solution (a) To find displacement algebraically, use the equation dd d d f i 35 m (15 m) 35 m 15 m 50 m The sign is negative, so the direction is to the left. (b) To find displacement graphically, subtract the two position vectors. Draw the vectors tail to tail and draw the resultant from the tip of the initial position vector to the tip of the final position vector (Figure 2.12). scale: 1.0 cm : 10 m df di d Figure 2.12 d 5.0 cm [left] 10 m 1.0 cm 50 m [left] Paraphrase The sailboat\u2019s displacement is 50 m [left]. 02-PearsonPhys20-Chap02 7/24/08 10:17 AM Page 75 For collinear vectors, find displacement by subtracting initial position from final position. Subtract vectors graphically by connecting them tail to tail or by reversing the direction of the initial position vector. Recall from Chapter 1 that direction for displacement is given with respect to initial position. 2.1 Check and Reflect 2.1 Check and Reflect Knowledge 1. Describe the similarities and differences between the two vectors drawn below. 5 m [E] 5 m [W] 2. Using the same scale and reference coordinates, compare the vectors 5 m [N] and 10 m [S]. 3. If the scale vector diagram of 5.0 m [S] is 6.0 cm long, what is the length of the scale vector diagram of 20 m [S]? 4. What scale is being used if 5.0 cm represents 100 km? Applications 5. The scale on a National Geographic world map is 1.0 cm : 520 km. On the map, 4.0 cm separates Alberta\u2019s north and south provincial boundaries. What is the separation in kilometres? 6. During a tough drill on a field of length 100 yards, players run to each 10-yard line and back to the starting position until they reach the other end of the field. (a) Write a vector equation that includes all the legs of the run. (b) What is the players\u2019 final displacement? (c) How far did they run? 7. A car drives north 500 km. It then drives three sequential displacements south, each of which is 50 km longer than the previous displacement. If the final position of the car is 50 km [N], find the three displacements algebraically. 8. Are vectors A and B equal? Why or why not? y A B x 9. A bouncy ball dropped from a height of 10.0 m bounces back 8.0 m, then drops and rebounds 4.0 m and finally 2.0 m. Find the distance the ball travels and its displacement from the drop point. e TEST To check your understanding of vectors in one dimension, follow the eTest links at www.pearsoned.ca/school/physicssource. Chapter 2 Vector components describe motion in two dimensions. 75 02-PearsonPhys20-Chap02 7/24/08 10:17 AM Page 76 2.2 Motion in Two Dimensions From the boot, the ball flies across the grass into the net and the crowd roars. The enormously successful FIFA Under-19 Women\u2019s World Championship, held in 2002, raised", " the profile of women\u2019s soccer in Canada and drew crowds totalling almost 200 000 to venues in Edmonton, Vancouver, and Victoria (Figure 2.13). Stars like Charmaine Hooper, Brittany Timko, and Christine Sinclair continue to amaze. From World Championship team members to the Under-6s on the local soccer pitch, performance depends on understanding and coordinating the movement of players and ball across the surface of the field. Figure 2.13 Motion in sports such as soccer can be described by vectors in two dimensions. Playbooks are available for fast-paced games such as hockey and soccer to allow coaches and players to plan the strategies that they hope will lead to success. Sometimes a team can charge straight up the rink or field, but, more often, a series of angled movements is needed to advance the puck or ball (Figure 2.14). For everyone to understand the play, a system is needed to understand motion in two dimensions. Components of Vectors Imagine that you are at one corner of a soccer field and you have to get to the far opposite corner. The shortest path from one corner to the other is a straight diagonal line. Figure 2.15 shows this path to be 150 m. Another way to describe this motion is to imagine an x-axis and a y-axis placed onto the soccer field, with you standing at the point (0, 0). You could move along the length of the field 120 m and then across the field 90 m and end up at the same spot (Figure 2.15). 4 3 2 1 Figure 2.14 This page is taken from a soccer playbook. How many players are involved in this wall pass-in-succession manoeuvre? 76 Unit I Kinematics 02-PearsonPhys20-Chap02 7/24/08 10:17 AM Page 77 Figure 2.15 The diagonal distance from one corner to the opposite corner of a soccer field is 150 m. 150 m 90 m 120 m In this example, the sideline of the soccer field could be considered the x-axis, and the goal line could be the y-axis. The diagonal motion vector can then be separated, or resolved, into two perpendicular parts, or components: the x component and the y component. The diagonal vector is the resultant vector. If you walked along the sideline or x-axis, you would move through a distance of 120 m. This distance is the x component of the diagonal vector. The second part of the walk along the goal line, parallel to the y-axis, is the y component of the diagonal vector. This motion has a distance of 90 m. Figure 2.16 shows the x and y components of the diagonal motion across the soccer field. y components: perpendicular parts into which a vector can be separated Figure 2.16 The resultant vector representing the diagonal walk across the soccer field can be resolved into x and y components. 150 m 90 m y component (width) x (0, 0) 120 m x component (length) Vector Directions Recall that a vector must have a magnitude and a direction. You have just studied how to resolve a vector into its components. Before going further, you need to know how to indicate the direction of vectors in two dimensions. There are two methods commonly used to show direction for vector quantities in two dimensions: the polar coordinates method and the navigator method. Both methods are considered valid ways to describe the direction of a vector. Chapter 2 Vector components describe motion in two dimensions. 77 02-PearsonPhys20-Chap02 7/24/08 10:17 AM Page 78 info BIT A third method for measuring direction is the bearing method, in which angles are measured clockwise from north, 0\u00b0 to 360\u00b0, so east is 90\u00b0, south is 180\u00b0, and west is 270\u00b0. 90\u00b0 y II 180\u00b0 240\u00b0 5 m [240\u00b0] III 270\u00b0 x 0\u00b0 (360\u00b0) I IV N S W 60\u00b0 5 m [60\u00b0 S of W] E info BIT Sailors can now create their sailing plans with a click of a mouse. Digitized maps and global positioning satellites have been combined to allow sailors to create a plan by using the mouse to place vectors on the desired path on screen. The computer calculates the total distance, identifies directions, and estimates the time required for the trip. Figure 2.17 The polar coordinates method for stating vector direction Figure 2.18 The navigator method for stating vector direction Polar Coordinates Method With the polar coordinates method, the positive x-axis is at 0\u00b0 and angles are measured by moving counterclockwise about the origin, or pole. One complete rotation is 360\u00b0 \u2014 a complete circle. In Figure 2.17, the displacement vector, 5 m [240\u00b0], is located in quadrant III in the Cartesian plane. This vector is rotated 240\u00b0 counterclockwise starting from the positive x-axis. Navigator Method Another method for indicating vector direction is the navigator method. This method uses the compass bearings north [N],", " south [S], east [E], and west [W] to identify vector directions. In Figure 2.18, the displacement vector 5 m [60\u00b0 S of W] is between the west and south compass bearings. To draw this vector, start with the second compass bearing you are given in square brackets, west, then move 60\u00b0 in the direction of the first compass bearing you are given, south. The type of problem will determine the method you use for stating vector directions. Often, it will be clear from the context of the problem which method is preferred. For example, if the question is about a boat sailing [30\u00b0 N of W], then use the navigator method. If a plane has a heading of 135\u00b0, then use the polar coordinates method. In the problems below, you can practise identifying and drawing vectors using the two methods Directions 2. For each vector in question 1, state the direction using the alternative method. Then draw each vector using an appropriate scale and reference coordinates. 1. For each of the following vectors, identify the method used for indicating direction. Then draw each vector in your notebook, using an appropriate scale and reference coordinates. (a) 3 m [0\u00b0] (b) 17 m/s [245\u00b0] (c) 7 m [65\u00b0] (d) 8 m/s [35\u00b0 W of N] (e) 2 m [98\u00b0] (f) 12 m/s [30\u00b0 S of E] 78 Unit I Kinematics 02-PearsonPhys20-Chap02 7/24/08 10:17 AM Page 79 Concept Check Write the direction [60\u00b0 S of W] another way using a different starting axis but keeping the angle less than 90\u00b0. 2-2 QuickLab 2-2 QuickLab Vector Walk Problem How can you add vectors to determine displacement? Materials 30-m measuring tape large chalkboard protractor field marker (tape or flag) Procedure 1 Using a tree diagram, determine the number of pathways you could take to walk the series of displacement vectors below. Assume that you will start on the centre line of a football field and mark it 0\u00b0. (a) 5 m [0\u00b0] (b) 12 m [270\u00b0] (c) 15 m [90\u00b0] (d) 3 m [180\u00b0] 7 Use the protractor to estimate the angle of displacement. 8 Repeat steps 3\u20137 for all the pathways you determined in step 1. 5 m [0\u00b0] Figure 2.19 NOTE: Use the same method for determining direction throughout the lab. 2 On a football or large school field, mark your starting point with a flag or tape. Define direction axes. 3 Ask your partner to walk the displacement given in (a) while you hold the end of the measuring tape. Mark your partner\u2019s endpoint (Figure 2.19). Questions 1. What was the total distance you travelled? 2. What was the total displacement? 3. What conclusion can you draw about the order of adding vectors? 4 Continue the journey, using the protractor and measuring tape, until you have walked all the vectors. 5 Mark the final endpoint. 6 Using the measuring tape, determine the displacement from your starting point. Chapter 2 Vector components describe motion in two dimensions. 79 02-PearsonPhys20-Chap02 7/24/08 10:17 AM Page 80 Adding Two-dimensional Vectors Graphically To sink the eight ball in the front side pocket of a billiard table, you must cause the ball to travel down and across the table (Figure 2.20). The ball\u2019s motion occurs in a plane, or two dimensions, even though its path is linear (Figure 2.21). Figure 2.20 Playing billiards involves two-dimensional motion. Figure 2.21 The path of the billiard ball is linear, but it occurs in two dimensions. Recall from section 2.1 that the plus sign () in a vector equation indicates that you need to connect the vectors tip to tail. Up to this point, you have added collinear vectors only. In this section, you will learn how to add non-collinear vectors. The plus sign still indicates you need to connect the vectors tip to tail while keeping track of their directions. Adding Non-collinear Vectors In section 2.1, you learned that vectors that lie along the same straight line are collinear. Vectors that are not along the same straight line are non-collinear (Figure 2.22). To determine the magnitude and direction of the sum of two or more non-collinear vectors graphically, use an accurately drawn scale vector diagram. Imagine you are walking north a distance of 40 m. Your initial 1. You stop, head west a distance position from your starting point is d 2. To find your disof 30 m, and stop again. Your final position is d placement, you cannot simply subtract your initial position from your", " final position because the vectors are not collinear. To find your displacement in two dimensions, you need to add the two position vectors: d d d 1 2. From Figure 2.23, you can see that the answer is not 70 m. You would obtain the answer 70 m if you walked in the same direction for both parts of your walk. Because you did not, you cannot directly substitute values into the displacement equation d 2. Instead, you must draw the vectors to scale, connect them tip to tail (because of the plus sign), and measure the magnitude of the resultant. Since d is a vector quantity, you must also indicate its direction. You can find the direction of the resultant using a protractor (Figure 2.24). d d 1 non-collinear: not along a straight line 45\u00b0 Figure 2.22 Non-collinear vectors lie along different lines. 80 Unit I Kinematics 02-PearsonPhys20-Chap02 7/24/08 10:17 AM Page 81 30 m d 2 W N S E d 30 m 2 W N S E d 50 m 40 m d 1 d 50 m d 40 m 1 scale 10 m 37\u00ba scale 10 m d 50 m [37\u00ba W of N] Figure 2.23 What is the sum of 2? 1 and d d Figure 2.24 When adding non-collinear vectors graphically, use a protractor to find the direction of the resultant. Eight Steps for Adding Non-collinear Vectors Graphically To find the resultant vector in a non-collinear vector addition statement using the graphical method, follow these eight steps (see Figure 2.25): 1. Create an appropriate scale. 2. Choose a set of reference coordinates. 3. Draw vector 1 to scale. Measure its direction from the tail. 4. Draw vector 2 to scale. Draw its tail at the tip (arrowhead) of vector 1. 5. Draw the resultant vector by connecting the tail of vector 1 to the tail tip of vector 2. W N S E tip resultant vector 2 vector 1 tail tip 6. Measure the magnitude (length) of the resultant. Measure the direction Figure 2.25 Adding vectors (angle) of the resultant from its tail. 7. Use your scale to convert the magnitude of the resultant to its original units. 8. State the resultant vector. Remember to include both magnitude and direction. This method also works for more than two vectors. You can add the vectors in any order. The next example shows you how to add more than two non-collinear vectors graphically. Example 2.3 A camper left her tent to go to the lake. She walked 0.80 km [S], then 1.20 km [E] and 0.30 km [N]. Find her resultant displacement. 1 Given d d d 2 3 0.80 km [S] 1.20 km [E] 0.30 km [N] Required resultant displacement (d R) Chapter 2 Vector components describe motion in two dimensions. 81 02-PearsonPhys20-Chap02 7/24/08 10:17 AM Page 82 Practice Problems 1. For Example 2.3, add the vectors in two different orders and obtain the resultant for each case. 2. A student runs through a field 100 m [E], then 200 m [S], and finally 50 m [45\u00b0 S of E]. Find her final position relative to her starting point. Answers 1. 1.30 km [67\u00b0 E of S] 2. 272 m [60\u00b0 S of E] Analysis and Solution The three vectors are non-collinear, so add them tip to tail to find the resultant (Figure 2.26). tent 67\u00b0 d1 0.80 km dR d1 d2 d3 Figure 2.26 scale 0.20 km W N S E dR 1.30 km d3 0.30 km d2 1.20 km Paraphrase The camper\u2019s resultant displacement is 1.30 km [67\u00b0 E of S]. Distance, Displacement, and Position Figure 2.27 shows the distances a bicycle courier travelled in going along the path from A to D, passing through B and C on the way. Use the information in the diagram, a ruler calibrated in mm, and a protractor to complete the distance, displacement, and position information required in Table 2.1. Assume the bicycle courier\u2019s reference point is A. Complete Table 2.1, then draw and label the displacement vectors AB, BC, and CD, and the position vectors AB, AC, and AD. \u25bc Table 2.1 Distance, Displacement, and Position scale 80 m W N S E d 600 m D C Displacement \u0394d (m) [direction] Distance \u0394d (m) d Final position (m) [direction] reference point AB BC CD AC AD 82 Unit I Kinematics d 630 m B d 560 m", " Figure 2.27 d 280 m A 02-PearsonPhys20-Chap02 7/24/08 10:17 AM Page 83 Determining Components drawn in a Cartesian plane and its two Figure 2.28 shows a vector R components, Rx and Ry, in the x and y directions. The Greek letter theta,, denotes the angle between R and its x and y components form a right triangle. and the x-axis. Vector R y 6.0 4.0 2.0 R 10 km/h 6.0 km/h Ry \u03b8 37\u00b0 2.0 x 6.0 8.0 4.0 8.0 km/h Rx Figure 2.28 Drawing components Because the triangle is a right triangle, you can determine components algebraically by using the trigonometric functions sine, cosine, and tangent. You can define each of the trigonometric functions in terms of the sides of a right triangle, like the one in Figure 2.29. Knowing these definitions, you can use the trigonometric functions to help you solve for the components of a vector. To calculate Rx, use the cosine function: Rx R adjacent hypotenuse R cos or Rx cos In Figure 2.28, the x component is: Rx (10 km/h)(cos 37\u00b0) 8.0 km/h To calculate Ry, use the sine function: sin opposite hypotenuse Ry R or Ry R sin hypotenuse opposite In Figure 2.28, the y component is: Ry (10 km/h)(sin 37\u00b0) 6.0 km/h \u03b8 adjacent Figure 2.29 Labelled sides of a right triangle Example 2.4 shows the steps for finding the velocity components of a car travelling in a northeasterly direction using trigonometry. This example uses the navigator method to indicate the direction of the velocity vector. Note that the east direction [E] is the same as the positive x direction in the Cartesian plane, and north [N] is the same as the positive y direction. So, for any vector R, the x component is the same as the east component, and the y component is the same as the north component. PHYSICS INSIGHT a c \u03b8 b For a right triangle with sides a and b forming the right angle, c is the hypotenuse. The Pythagorean theorem states that a2 b2 c2. You can find the angle,, in one of three ways: sin a c opposite hypotenuse cos b c adjacent hypotenuse tan a b opposite adjacent Chapter 2 Vector components describe motion in two dimensions. 83 02-PearsonPhys20-Chap02 7/24/08 10:17 AM Page 84 Example 2.4 Determine the north and east velocity components of a car travelling at 100 km/h [25\u00b0 N of E]. y 1 0 0 k m /h 25\u00b0 vx vy x Figure 2.30 Given v 100 km/h [25\u00b0 N of E] Required velocity component north (y component, vy) velocity component east (x component, vx) Practice Problems 1. A hiker\u2019s displacement is 15 km [40\u00b0 E of N]. What is the north component of his displacement? 2. A cyclist\u2019s velocity is 10 m/s [245\u00b0]. Determine the x and y components of her velocity. 3. A snowmobile travels 65 km [37\u00b0 E of S]. How far east does it travel? Answers 1. 11 km [N] 4.2 m/s, vy 2. vx 3. 39 km [E] 9.1 m/s Analysis and Solution The vector lies between the north and east directions, so the x and y components are both positive. Since the north direction is parallel to the y-axis, use the sine function, R sin, to find the north component. Since the east Ry direction lies along the x-axis, use the cosine function, Rx R cos, to find the east component. Ry vy Rx vx R sin (100 km/h)(sin 25\u00b0) 42.3 km/h R cos (100 km/h)(cos 25\u00b0) 90.6 km/h Paraphrase The north component of the car\u2019s velocity is 42.3 km/h and the east component is 90.6 km/h. Concept Check For a vector R in quadrant I (Cartesian method), are Rx and Ry always positive? Determine whether Rx and Ry are positive or negative for vectors in quadrants II, III, and IV. Display your answers in a chart. 84 Unit I Kinematics 02-PearsonPhys20-Chap02 7/24/08 10:17 AM Page 85 Adding Vectors Using Components You can write the magnitude of any two-dimensional vector as the sum of its x and y components. Note that x and y components are", " perpendicular. Because motion along the x direction is perpendicular to motion along the y direction, a change in one component does not affect the other component. Whether it is movement across a soccer field or any other type of two-dimensional motion, you can describe the motion in terms of x and y components. In Example 2.4, you learned how to determine the x and y com- ponents, Rx and Ry, for a general vector R. In some situations, you already know Rx and Ry, and you must find the magnitude and direc- tion of the resultant vector R. For example, a toy moves 9.0 m right and then 12.0 m across a classroom floor (Figure 2.31). What is the toy\u2019s displacement? Solving this problem algebraically requires two steps: Step 1: Find the magnitude of R. To find the magnitude of the resultant vector, use the Pythagorean theorem. You can use this theorem because the two components, Rx and Ry, form a right triangle with the resultant vector. You are given that Rx R2 Rx R (9.0 m)2 (12.0m)2 9.0 m and Ry 2 Ry 12.0 m. y 2 15 m Step 2: Find the angle of R To find the angle of R, use the tangent function:. R tan opposite adjacent 12.0 m 9.0 m 1.33 tan1(1.33) 53.1\u00b0 12.0 9.0 6.0 3.0 12.0 m Ry \u03b8 3.0 Rx 6.0 9.0 m x 9.0 Figure 2.31 Vector components of the movement of a toy across a classroom floor Using the polar coordinates method, the resultant vector direction is [53.1\u00b0]. Using the navigator method, the direction is [53.1\u00b0 N of E]. Using Components 1. Find Rx and Ry for the following vectors: (a) A boat travelling at 15 km/h [45\u00b0 N of W] (b) A plane flying at 200 km/h [25\u00b0 E of S] (c) A mountain bike travelling at 10 km/h [N] 2. Find R (a) Rx (b) Rx (c) Rx 7 m and for the following Rx and Ry values: 12 m, Ry 40 km/h, Ry 30 cm, Ry 55 km/h 10 cm Chapter 2 Vector components describe motion in two dimensions. 85 02-PearsonPhys20-Chap02 7/24/08 10:17 AM Page 86 In general, most vector motion involves adding non-collinear vectors. Consider the following scenario. During a lacrosse game, players pass the ball from one person to another (Figure 2.32). The ball can then be redirected for a shot on goal. Each of the displacements could involve different angles. In order to find the net displacement, you would use the following sequence of calculations. Four Steps for Adding Non-collinear Vectors Algebraically 1. Determine the x and y components of each vector. 2. Add all components in the x direction. Add all components in the y direction. The sums of the x and y components are the two (perpendicular) components of the resultant vector. 3. To find the magnitude of the resultant vector, use the Pythagorean theorem. 4. To find the angle of the resultant vector, use trigonometric ratios. (See Physics Insight on page 83.) The following example illustrates how to apply these steps. In a lacrosse game (Figure 2.33(a)), player A passes the ball 12.0 m to player B at an angle of 30\u00b0. Player B relays the ball to player C, 9.0 m away, at an angle of 155\u00b0. Find the ball\u2019s resultant displacement. y B d2 9.0 m 155\u00b0 12.0 m d1 30\u00b0 x C A Figure 2.33(a) on the lacrosse field The path of the ball Figure 2.33(b) as vectors The path of the ball Figure 2.33(b) shows the path of the lacrosse ball as vectors. This problem is different from previous examples because the two vectors are not at right angles to each other. Even with this difference, you can follow the same general steps to solve the problem. Step 1: Determine the x and y components of each vector. Since you are solving for displacement, resolve each displacement vector into its components (Figure 2.34). Table 2.2 shows how to calculate the x and y components. In this case, designate up and right as positive directions. Figure 2.32 The movement of the players and the ball in a lacrosse game could be tracked using vectors. PHYSICS INSIGHT To simplify calculations for finding components, use acute ( 90\u00b0) angles. To determine the acute angle when given an obtuse ( 90\u00b0) angle, subtract the", " obtuse angle from 180\u00b0. 25\u00b0 155\u00b0 180\u00b0 155\u00b0 25\u00b0 For an angle greater than 180\u00b0, subtract 180\u00b0 from the angle. For example, 240\u00b0 180\u00b0 60\u00b0 86 Unit I Kinematics 02-PearsonPhys20-Chap02 7/24/08 10:17 AM Page 87 \u25bc Table 2.2 Resolution of Components in Figure 2.34 x direction y direction d1x d2x (12.0 m)(cos 30\u00b0) 10.39 m (9.0 m)(cos 25\u00b0) 8.16 m d1y d2y (12.0 m)(sin 30\u00b0) 6.00 m (9.0 m)(sin 25\u00b0) 3.80 m (Note that d2x is negative because it points to the left, and up and right were designated as positive.) y d2y 30\u00b0 d2 d2x d1 d1x 25\u00b0 155\u00b0 d1y x Figure 2.34 The path of the lacrosse ball Step 2: Add the x components and the y components separately. Add all the x components together, then add all the y components (see Table 2.3 and Figure 2.35). \u25bc Table 2.3 Adding x and y Components in Figure 2.35 x direction y direction dx d2x d1x 10.39 m (8.16 m) 10.39 m 8.16 m 2.23 m dy d2y d1y 6.00 m 3.80 m 9.80 m dx d2x d1x d2y d1y dy Figure 2.35 Add the x and y components separately first to obtain two perpendicular vectors. Step 3: Find the magnitude of the resultant, d To find the magnitude of the resultant, use the Pythagorean theorem (Figure 2.36).. d2 (dx)2 (dy)2 d (dx)2 (dy)2 (2.23m)2 (9.80 m)2 10 m y d dy \u03b8 dx x Figure 2.36 The component method allows you to convert non-perpendicular vectors into perpendicular vectors that you can then combine using the Pythagorean theorem. Chapter 2 Vector components describe motion in two dimensions. 87 02-PearsonPhys20-Chap02 7/24/08 10:17 AM Page 88 y d 77\u00b0 d2 d1 Figure 2.37 d resultant displacement of the ball. is the Step 4: Find the angle of d. Use the tangent function to find the angle (Figure 2.36). tan opposite adjacent 9.80 m 2.23 m 4.39 x tan1(4.39) 77\u00b0 The ball\u2019s displacement is, therefore, 10 m [77\u00b0], as shown in Figure 2.37. The following example illustrates another situation where the dis- placement vectors are not at right angles. Example 2.5 Practice Problems 1. Find the displacement of a farmer who walked 80.0 m [0\u00b0] and then 60.0 m [335\u00b0]. 2. Find the displacement of a soccer player who runs 15 m [15\u00b0 N of E] and then 13 m [5\u00b0 W of N]. 3. While tracking a polar bear, a wildlife biologist travels 300 m [S] and then 550 m [75\u00b0 N of E]. What is her displacement? Answers 1. 137 m [349\u00b0] 2. 21 m [52\u00b0 N of E] 3. 272 m [58\u00b0 N of E] Use components to determine the displacement of a crosscountry skier who travelled 15.0 m [220\u00b0] and then 25.0 m [335\u00b0] (Figure 2.38). y x 15.0 m [220\u00b0] d1 25.0 m [335\u00b0] d2 Figure 2.38 Given d d 1 2 15.0 m [220\u00b0] 25.0 m [335\u00b0] Required displacement (d ) Analysis and Solution Step 1: Use Rx its x and y components. Designate up and to the right as positive. Work with acute angles (Figure 2.39). R sin to resolve each vector into R cos and Ry y y 220\u00b0 40\u00b0 x 335\u00b0 25\u00b0 x d2y d2 d2x d1y d1 d1x Figure 2.39 88 Unit I Kinematics 02-PearsonPhys20-Chap02 7/24/08 10:17 AM Page 89 x direction: d1x (15.0 m)(cos 40\u00b0) 11.49 m d2x (25.0 m)(cos 25\u00b0) 22.66 m y direction: d1y (15.0 m)(sin 40\u00b0) 9.642 m d2y (25.0 m)(sin 25\u00b0) 10.57 m Step 2: Add the x and y components. dx dy d1x + d2x 11.49 m 22.", "66 m 11.17 m d2y d1y 9.642 m (10.57 m) 20.21 m 11.17 m \u03b8 20.21 m d y \u03b8 x d d1 d2 Figure 2.40 Figure 2.41 Step 3: To find the magnitude of the resultant, calculate d using the Pythagorean theorem (Figure 2.40). d2 (dx)2 (dy)2 (11.17 m)2 (20.21 m)2 d (11.17 m)2 (20.21 m)2 23.09 m Figure 2.41 shows that the resultant lies in quadrant IV. Step 4: To find the angle, use the tangent function (see Figure 2.40). tan opposite adjacent 20.21 m 11.17 m 1.810 tan1(1.810) 61\u00b0 From Figure 2.41, note that the angle,, lies below the positive x-axis. Using the polar coordinates method, the angle is 299\u00b0. Paraphrase The cross-country skier\u2019s displacement is 23.1 m [299\u00b0]. e SIM Practise the numerical addition of two or more vectors. Follow the eSim links at www.pearsoned.ca/school/ physicssource. Chapter 2 Vector components describe motion in two dimensions. 89 02-PearsonPhys20-Chap02 7/24/08 10:17 AM Page 90 In summary, in order to solve a two-dimensional motion problem, you need to split the motion into two one-dimensional problems by using the vectors\u2019 x and y components. Then add the x and y components separately. To find the magnitude of the resultant, use the Pythagorean theorem. To find the angle of the resultant, use the tangent function. 2.2 Check and Reflect 2.2 Check and Reflect Knowledge 1. What trigonometric functions can be used to determine the x or horizontal component of a vector? Draw diagrams to illustrate your answers. 2. Are the following statements true or false? Justify your answer. (a) The order in which vectors are added is important. (b) Displacement and distance are always equal. 3. Describe when you would use the navigator method to indicate the direction of a vector. Applications 4. A student has created a short computer program that calculates components of vectors drawn with a computer mouse. To demonstrate his program, he drags the mouse to create a vector at 55 cm [30\u00b0 W of S]. What are the components of the vector? 5. Determine the distance travelled and the displacement for each of the following. (a) Blading through Fish Creek Park in Calgary takes you 5.0 km [W], 3.0 km [N], 2.0 km [E], and 1.5 km [S]. (b) A swimmer travels in a northerly direction across a 500-m-wide lake. Once across, the swimmer notices that she is 150 m east of her original starting position. (c) After leaving her cabin, a camper snowshoes 750 m [90\u00b0] and then 2.20 km [270\u00b0]. 90 Unit I Kinematics 6. A boat sails 5.0 km [45\u00b0 W of N]. It then changes direction and sails 7.0 km [45\u00b0 S of E]. Where does the boat end up with reference to its starting point? 7. A pellet gun fires a pellet with a velocity of 355 m/s [30\u00b0]. What is the magnitude of the vertical component of the velocity at the moment the pellet is fired? 8. Tourists on a jet ski move 1.20 km [55\u00b0 N of E] and then 3.15 km [70\u00b0 S of E]. Determine the jet ski\u2019s displacement. 9. A jogger runs with a velocity of 6.0 km/h [25\u00b0 N of W] for 35 min and then changes direction, jogging for 20 min at 4.5 km/h [65\u00b0 E of N]. Using a vector diagram, determine the jogger\u2019s total displacement and his average velocity for the workout. 10. Given that a baseball diamond is a square, assume that the first-base line is the horizontal axis. On second base, a baseball player\u2019s displacement from home plate is 38 m [45\u00b0]. (a) What are the components of the player\u2019s displacement from home plate? (b) Has the runner standing on second base travelled a distance of 38 m? Why or why not? 11. Determine the resultant displacement of a skateboarder who rolls 45.0 m [310\u00b0] and 35.0 m [135\u00b0]. e TEST To check your understanding of two-dimensional motion, follow the eTest links at www.pearsoned.ca/school/physicssource. 02-PearsonPhys20-Chap", "ion is motion measured with respect to an observer. Concept Check An observer is on a train moving at a velocity of 25 m/s [forward]. A ball rolls at 25 m/s [forward] with respect to the floor of the moving train. What is the velocity of the ball relative to the observer on the train? What is the velocity of the ball relative to an observer standing on the ground? What happens if the ball moves 25 m/s [backward]? How does a moving medium affect the motion of a table tennis ball? relative motion: motion measured with respect to an observer Chapter 2 Vector components describe motion in two dimensions. 91 02-PearsonPhys20-Chap02 7/24/08 10:17 AM Page 92 2-3 QuickLab 2-3 QuickLab Table Tennis in the Wind Problem How does air movement affect the motion of a table tennis ball? Materials large upright fan table tennis table paddles table tennis ball Procedure 1 With a partner, practise hitting the table tennis ball to each other (Figure 2.43). Figure 2.43 Questions 1. When did the ball move the fastest? the slowest? 2. When the air movement was perpendicular to the ball\u2019s path, did it change the ball\u2019s speed? Did it change the ball\u2019s velocity? Explain. 2 Set up the fan on one side of the table tennis table. 3. Given your results, speculate as to why golfers 3 Hit the table tennis ball straight across the length of the table (a) against the wind (b) with the wind release a tuft of grass into the wind before driving the ball. 4. Describe how wind direction might influence a beach volleyball player\u2019s serve. (c) perpendicular to the wind\u2019s direction e LAB 4 Record how the moving air influences the motion of the ball in each case. For a probeware activity, go to www.pearsoned.ca/school/physicssource. Relative Motion in the Air ground velocity: velocity relative to an observer on the ground air velocity: an object\u2019s velocity relative to still air wind velocity: velocity of the wind relative to the ground A flight from Edmonton to Toronto takes about 3.5 h. The return flight on the same aircraft takes 4.0 h. If the plane\u2019s air speed is the same in both directions, why does the trip east take less time? The reason is that, when travelling eastward from Edmonton, a tailwind (a wind that blows from the rear of the plane, in the same direction as the plane\u2019s motion) increases the airplane\u2019s ground velocity (velocity relative to an observer on the ground), hence reducing the time of travel and, therefore, fuel consumption and cost. scale 100 km/h A Canadian regional jet travels with an air velocity (the plane\u2019s velocity in still air) of 789 km/h [E]. The jet encounters a wind velocity (the wind\u2019s velocity with respect to the ground) of 56.3 km/h [E] (Figure 2.44). (This wind is a west wind, blowing eastward from the west.) What is the velocity of the airplane relative to an observer on the ground? The resultant velocity of the airplane, or ground velocity, is the vector sum of the plane\u2019s air velocity and the wind velocity (Figure 2.45). Let the positive direction be east. vair 789 km/h [E] vwind 56.3 km/h [E] Figure 2.44 The air velocity and wind velocity are in the same direction. 92 Unit I Kinematics 02-PearsonPhys20-Chap02 7/24/08 10:17 AM Page 93 v ground air wind v v 789 km/h 56.3 km/h 845 km/h The sign is positive, so the ground velocity is 845 km/h [E]. scale 100 km/h vair 789 km/h [W] vwind 56.3 km/h [E] scale 100 km/h vwind Figure 2.45 vair vground scale 100 km/h vair vground vwind Figure 2.47 Figure 2.46 The air velocity and wind velocity are in opposite directions. If the jet heads west, from Toronto to Edmonton (Figure 2.46), its v ground resultant velocity becomes v 789 km/h 56.3 km/h 733 km/h v wind air (See Figure 2.47.) The sign is negative, so the ground velocity is 733 km/h [W]. The plane\u2019s speed decreases due to the headwind (wind that approaches from the front). Non-collinear Relative Motion Suppose the jet travelling west from Toronto encounters a crosswind of 56.3 km [N] (Figure 2.48). W N S E vwind= 56.3 km/h vair= 789 km", "/h vwind= 56.3 km/h vground= 791 km/h \u03b8 vair= 789 km/h W N S E Figure 2.48 A plane flies in a crosswind. Figure 2.49 A plane that flies in a crosswind needs to adjust its direction of motion. air ground v v In this case, the velocity of the plane is not aligned with the wind\u2019s velocity. The defining equation for this case is still the same as for the collinear case: v wind. From section 2.2, recall that the plus sign in a two-dimensional vector equation tells you to connect the vectors tip to tail. The resultant vector is the ground velocity, v ground. The ground velocity indicates the actual path of the plane (Figure 2.49). To solve for the ground velocity, notice that the triangle formed is a right triangle, meaning that you can use the Pythagorean theorem to solve for the magnitude of the ground velocity. Chapter 2 Vector components describe motion in two dimensions. 93 02-PearsonPhys20-Chap02 7/24/08 10:17 AM Page 94 (vground)2 (vair)2 (vwind)2 vground (vair)2 (vw ind)2 (789 km/h)2 (56.3 km/h)2 791 km/h Using the tangent function, the direction of the ground velocity is tan opposite adjacent 56.3 km/h 789 km/h 0.07136 tan1(0.07136) 4.1\u00b0 From Figure 2.49, the wind blows the airplane off its westerly course in the northerly direction. Hence, the airplane\u2019s ground velocity is 791 km/h [4.1\u00b0 N of W]. The pilot must take into consideration the effect of the wind blowing the airplane off course to ensure that the plane reaches its destination. What path would the pilot have to take to arrive at a point due west of the point of departure? Remember that, if the vectors are not perpendicular, resolve them into components first before adding them algebraically. Example 2.6 A plane flies west from Toronto to Edmonton with an air speed of 789 km/h. (a) Find the direction the plane would have to fly to compensate for a wind velocity of 56.3 km/h [N]. (b) Find the plane\u2019s speed relative to the ground. W N S E vground vwind Figure 2.50 vair PHYSICS INSIGHT To determine the angle, substitute the magnitudes of the relative velocities into the tangent function. To determine the direction, refer to the vector diagram for the problem. 94 Unit I Kinematics 02-PearsonPhys20-Chap02 7/24/08 10:17 AM Page 95 wind 56.3 km/h [N] Given v v direction of ground velocity is west 789 km/h air Required (a) the plane\u2019s direction (direction of air velocity) (b) ground speed (vground) Analysis and Solution (a) First construct a diagram based on the defining equation, v v v ground wind air vground \u03b8 vair vwind Figure 2.51 air and v wind tip to tail. The rules of vector addition tell you to connect the vectors v To find the direction required in order to compensate for the wind velocity, find the angle,. Because the connection of the vectors forms a right triangle, and you know the magnitude of the opposite side (v wind) and the hypotenuse (v air), you can use the sine function to find the angle (Figure 2.51). te i s o p sin t se u n e o hy p p o Practice Problems 1. A swimmer can swim at a speed of 1.8 m/s. The river is 200 m wide and has a current of 1.2 m/s [W]. If the swimmer points herself north, directly across the river, find (a) her velocity relative to the ground. (b) the time it takes her to cross. 2. For a river flowing west with a current of 1.2 m/s, a swimmer decides she wants to swim directly across. If she can swim with a speed of 1.8 m/s, find (a) the angle at which she must direct herself. (b) the time it takes her to cross if the river is 200 m wide. Answers 1. (a) 2.2 m/s [34\u00b0 W of N] (b) 1.1 102 s 2. (a) [42\u00b0 E of N] (b) 1.5 102 s o p p p hy sin1 sin1 4.1 se te From Figure 2.51, the angle is [4.1\u00b0 S of W]. (b) To find the magnitude of the ground velocity, use the Pythagorean", " theorem. From Figure 2.51, note that the hypotenuse in this case is the air velocity, v air. (vair)2 (vwind)2 (vground)2 (vground)2 (vair)2 (vwind)2 (789 km/h)2 (56.3 km/h)2 6.1935 105 (km/h)2 787 km/h vground Notice that there is a small change in the magnitude of the ground velocity from the previous example of the plane heading west. As the magnitude of the wind velocity increases, the magnitude of the ground velocity and the compensating angle will significantly change. Paraphrase (a) The plane\u2019s heading must be [4.1\u00b0 S of W]. (b) The plane\u2019s ground speed is 787 km/h. Chapter 2 Vector components describe motion in two dimensions. 95 02-PearsonPhys20-Chap02 7/24/08 10:17 AM Page 96 PHYSICS INSIGHT For simplicity, Edmonton was assumed to be directly west of Toronto, which, of course, it is not! However, the calculation is still valid because this problem involves straightline motion. To calculate the time it takes to fly from Toronto to Edmonton, use the d equation v. The distance between Edmonton and Toronto is about t 2335 km, but you must decide which value for velocity to use: air velocity or ground velocity. The displacement and velocity vectors in this equation must be aligned. Since you are assuming that displacement is in the west direction, the appropriate velocity to use is the one that is in the westerly direction. In this example, it is the ground velocity, v ground (Figure 2.52). d vground \u03b8 vair vwind Figure 2.52 For calculating time, choose the velocity vector that matches the direction of the displacement vector. Consider west to be positive. Since both vectors are in the same direction (west), use the scalar form of the equation to solve for time. t d v 2335 km m 787 k h 2.97 h It takes 2.97 h to fly from Toronto to Edmonton. In the following example, the three velocity vectors do not form a right triangle. In order to solve the problem, you will need to use components. PHYSICS INSIGHT When substituting values into a vector equation, make sure that the values have the same direction (are collinear). If the vectors are not collinear, you need to use graphical or algebraic methods to find the answer. Example 2.7 As a pilot of a small plane, you need to transport three people to an airstrip 350.0 km due west in 2.25 h. If the wind is blowing at 40.0 km/h [65\u00b0 N of W], what should be the plane\u2019s air velocity in order to reach the airstrip on time? d = 350.0 km [W] vground vwind = 40.0 km/h [65\u00ba N of W] Figure 2.53 96 Unit I Kinematics 02-PearsonPhys20-Chap02 7/24/08 10:17 AM Page 97 40.0 km/h [65 N of W] wind Given v d t 2.25 h 350.0 km [W] Required plane\u2019s air velocity (v air) Analysis and Solution First draw a vector diagram of the problem (Figure 2.54). W N S E vwind Figure 2.54 vground vair Designate north and west as the positive directions. Then calculate the ground velocity from the given displacement and time. If the plane must fly 350.0 km [W] in 2.25 h, its ground velocity is d t ground v Practice Problems 1. An airplane can fly with a maximum air velocity of 750 km/h [N]. If the wind velocity is 60 km/h [15 E of N], what must be the plane\u2019s ground velocity if it is to remain on a course going straight north? 2. What is the air velocity of a jetliner if its ground velocity is 856 km/h [25.0 W of S] and the wind velocity is 65.0 km/h [S]? 3. How long will it take a plane to travel 100 km [N] if its ground velocity is 795 km/h [25 W of N]? Answers 1. 8.1 102 km/h [1 W of N] 2. 798 km/h [27.0 W of S] 3. 0.139 h 350.0 km [W] 2.25 h 155.6 km/h [W] Now find the components of the wind velocity (Figure 2.55). x direction: vwindx (40.0 km/h)(cos 65) 16.9 km/h y direction: vwindy (40.0 km/h)(sin", " 65) 36.25 km/h The ground velocity is directed west, so its x component is 155.6 km/h and its y component is zero. v Since v v v v wind, rearrange this equation to solve for v v air. ground air air ground wind vwindy vwind 65\u00b0 vwindx Figure 2.55 Use this form of the equation to solve for the components of the air velocity. Add the x (west) components: vairx vgroundx vwindx 155.6 km/h 16.9 km/h 138.7 km/h Add the y (north) components: vairy vgroundy vwindy 0 36.25 km/h 36.25 km/h Use the Pythagorean theorem to find the magnitude of the air velocity. vair vairx 2 vairy2 (138.7 km/h)2 (36.25 km/h)2 143 km/h Chapter 2 Vector components describe motion in two dimensions. 97 02-PearsonPhys20-Chap02 7/24/08 10:17 AM Page 98 To find the direction of air velocity, use the tangent function (Figure 2.56). tan opposite adjacent 36.25 km/h 138.7 km/h 0.261 tan1(0.261) 15 138.7 km/h \u03b8 36.25 km/h vair Figure 2.56 The x component is positive, so its direction is west. Since the y component is negative, its direction is to the south. Thus, the direction of the air velocity is [15 S of W]. Paraphrase The airplane\u2019s air velocity is 143 km/h [15 S of W]. Relative Motion in the Water Whereas wind velocity affects the speed and direction of flying objects, watercraft and swimmers experience currents. As with flying objects, an object in the water can move with the current (ground velocity increases), against the current (ground velocity decreases), or at an angle (ground velocity increases or decreases). When the object moves at an angle to the current that is not 90, both the object\u2019s speed and direction change. The following example illustrates how to use components to find velocity. Example 2.8 The Edmonton Queen paddleboat travels north on the Saskatchewan River at a speed of 5.00 knots or 9.26 km/h. If the Queen\u2019s ground velocity is 10.1 km/h [23\u00b0 E of N], what is the velocity of the Saskatchewan River? vcurrent W N S E vboat vground 23\u00b0 vcurrent W N S E vboat= 9.26 km/h vground = 10.1 km/h 23\u00ba Figure 2.57(a) Figure 2.57(b) 98 Unit I Kinematics 02-PearsonPhys20-Chap02 7/24/08 10:17 AM Page 99 boat ground 9.26 km/h [N] 10.1 km/h [23 E of N] Given v v (Note that the angle is given with respect to the vertical (y) axis (Figure 2.57(a).) vgroundx vgroundy vground Required velocity of current (v current) 23\u00b0 Analysis and Solution Let north and east be positive. Calculate the current\u2019s velocity using components. First find the components of the ground velocity (Figure 2.58(a)). Figure 2.58(a) x direction: vgroundx vgroundsin (10.1 km/h)(sin 23) 3.946 km/h y direction: vgroundy vgroundcos (10.1 km/h)(cos 23) 9.297 km/h Practice Problems 1. Determine a Sea Doo\u2019s ground velocity if it travels with a constant velocity of 4.50 m/s [W] and encounters a current of 2.0 m/s [20 W of N]. 2. A jogger runs with a velocity of 3.75 m/s [20 N of E] on an Alaskan cruise ship heading north at 13 m/s. What is the jogger\u2019s ground velocity? 3. A ship travelling 55 [W of N] is 65.0 km farther north after 3.0 h. What is the ship\u2019s velocity? Answers 1. 5.5 m/s [20 N of W] 2. 15 m/s [76 N of E] 3. 38 km/h [55 W of N] Since the boat\u2019s velocity is directed north, its y component is 9.26 km/h and its x component is zero. v You are asked to find v v v v v v current ground boat ground boat current current, so rearrange the vector equation accordingly: Use this form of the equation to solve for the components of the current\u2019s velocity. vcurrentx vboatx vcurrenty vgroundx 3.", "946 km/h 0 3.946 km/h vboaty vgroundy 9.297 km/h 9.26 km/h 0.037 km/h To find the magnitude of the current\u2019s velocity, use the Pythagorean theorem. vcurrent nty)2 ntx)2 (vcurre (vcurre (3.946 km/h)2 (0.037 km/h)2 3.946 km/h To find the direction of the current\u2019s velocity, use the tangent function (Figure 2.58(b)). opposite adjacent tan 0.03700 km/h vcurrent \u03b8 tan1 0.037 km/h 3.946 km/h 0.5 3.946 km/h Figure 2.58(b) Since both the x and y components are positive, the directions are east and north, respectively. Therefore, the current\u2019s direction is [0.5 N of E]. Paraphrase The current\u2019s velocity is 3.95 km/h [0.5 N of E]. Chapter 2 Vector components describe motion in two dimensions. 99 02-PearsonPhys20-Chap02 7/25/08 8:05 AM Page 100 In order to find the time required to cross the river, you need to use the velocity value that corresponds to the direction of the object\u2019s displacement, as you will see in the next example. Example 2.9 From Example 2.8, if the river is 200 m wide and the banks run from east to west, how much time, in seconds, does it take for the Edmonton Queen to travel from the south bank to the north bank? Practice Problems 1. A river flows east to west at 3.0 m/s and is 80 m wide. A boat, capable of moving at 4.0 m/s, crosses in two different ways. (a) Find the time to cross if the boat is pointed directly north and moves at an angle downstream. (b) Find the time to cross if the boat is pointed at an angle upstream and moves directly north. Answers 1. (a) 20 s (b) 30 s Given v width of river 200 m 0.200 km 10.1 km/h [23\u00b0 E of N] ground Required time of travel (t) Analysis and Solution Determine the distance, d, the boat travels in the direction of the boat\u2019s ground velocity, 23\u00b0 E of N. From Figure 2.59, d 0.200 km cos 23\u00b0 0.2173 km N 0.200 km d 23\u00b0 Figure 2.59 The boat\u2019s ground velocity is 10.1 km/h [23\u00b0 E of N]. vground t 10.1 km/h d vground 0.2173 km 10.1 km/h 0.02151 h 60 min 1 h 60 s 1 min 77.4 s Paraphrase It takes the Edmonton Queen 77.4 s to cross the river. Relative motion problems describe the motion of an object travelling in a medium that is also moving. Both wind and current can affect the magnitude and direction of velocity. To solve relative motion problems in two dimensions, resolve the vectors into components and then add them using trigonometry. 100 Unit I Kinematics 02-PearsonPhys20-Chap02 7/24/08 10:17 AM Page 101 2.3 Check and Reflect 2.3 Check and Reflect Knowledge 1. Describe a situation where a wind or current will increase an object\u2019s ground speed. 2. Describe a situation where a wind or current will change an object\u2019s direction relative to the ground, but not its speed in the original direction (i.e., the velocity component in the original, intended direction of motion). 3. Describe a situation when a wind or current will cause zero displacement as seen from the ground. 4. Provide an example other than in the text that illustrates that perpendicular components of motion are independent of one another. Applications 5. A swimmer needs to cross a river as quickly as possible. The swimmer\u2019s speed in still water is 1.35 m/s. (a) If the river\u2019s current speed is 0.60 m/s and the river is 106.68 m wide, how long will it take the swimmer to cross the river if he swims so that his body is angled slightly upstream while crossing, and he ends up on the far bank directly across from where he started? (b) If he points his body directly across the river and is therefore carried downstream, how long will it take to get across the river and how far downstream from his starting point will he end up? 6. A small plane can travel with a speed of 265 km/h with respect to the air. If the plane heads north, determine its resultant velocity if it encounters (a) a 32", ".0-km/h headwind (b) a 32.0-km/h tailwind (c) a 32.0-km/h [W] crosswind 7. The current in a river has a speed of 1.0 m/s. A woman swims 300 m downstream and then back to her starting point without stopping. If she can swim 1.5 m/s in still water, find the time of her round trip. 8. What is the ground velocity of an airplane if its air velocity is 800 km/h [E] and the wind velocity is 60 km/h [42 E of N]? 9. A radio-controlled plane has a measured air velocity of 3.0 m/s [E]. If the plane drifts off course due to a light wind with velocity 1.75 m/s [25\u00b0 W of S], find the velocity of the plane relative to the ground. If the distance travelled by the plane was 3.2 km, find the time it took the plane to travel that distance. 10. An airplane is observed to be flying at a speed of 600 km/h. The plane\u2019s nose points west. The wind\u2019s velocity is 40 km/h [45 W of S]. Find the plane\u2019s velocity relative to the ground. 11. A canoe can move at a speed of 4.0 m/s [N] in still water. If the velocity of the current is 2.5 m/s [W] and the river is 0.80 km wide, find (a) the velocity of the canoe relative to the ground (b) the time it takes to cross the river e TEST To check your understanding of relative motion, follow the eTest links at www.pearsoned.ca/school/physicssource. Chapter 2 Vector components describe motion in two dimensions. 101 02-PearsonPhys20-Chap02 7/24/08 10:17 AM Page 102 2.4 Projectile Motion Sports are really science experiments in action. Consider golf balls, footballs, and tennis balls. All of these objects are projectiles (Figure 2.60). You know from personal experience that there is a relationship between the distance you can throw a ball and the angle of loft. In this section, you will learn the theory behind projectile motion and how to calculate the values you need to throw the fastball or hit the target dead on. Try the next QuickLab and discover what factors affect the trajectory of a projectile. Figure 2.60 Sports is projectile motion in action. 102 Unit I Kinematics 02-PearsonPhys20-Chap02 7/24/08 10:18 AM Page 103 2-4 QuickLab 2-4 QuickLab Projectiles Problem What factors affect the trajectory of a marble? Materials wooden board (1 m 1 m) hammer paint marble newspaper white paper to cover the board two nails elastic band spoon brick masking tape gloves Procedure 1 Spread enough newspaper on the floor so that it covers a larger workspace than the wooden board. 2 Hammer two nails, 7.0 cm apart, at the bottom left corner of the board. Stretch the elastic between them. 3 Cover the board with white paper and affix the 6 Pull the elastic band back at an angle and rest the marble in it. 7 Release the elastic band and marble. Label the marble\u2019s trajectory on the paper track 1. 8 Repeat steps 5\u20137 for different launch angles and extensions of the elastic band. in Figure 2.61 Questions 1. What is the shape of the marble\u2019s trajectory, paper to the board using masking tape. regardless of speed and angle? 4 Prop the board up on the brick (Figure 2.61). 2. How did a change in the elastic band\u2019s extension 5 Wearing gloves, roll the marble in a spoonful of paint. affect the marble\u2019s path? 3. How did a change in launch angle affect the marble\u2019s path? e LAB For a probeware activity, go to www.pearsoned.ca/school/physicssource. Galileo studied projectiles and found that they moved in two directions at the same time. He determined that the motion of a projectile, neglecting air resistance, follows the curved path of a parabola. The parabolic path of a projectile is called its trajectory (Figure 2.62). The shape of a projectile\u2019s trajectory depends on its initial velocity \u2014 both its initial speed and direction \u2014 and on the acceleration due to gravity. To understand and analyze projectile motion, you need to consider the horizontal (x direction) and vertical (y direction) components of the object\u2019s motion separately. viy vix dy trajectory: the parabolic motion of a projectile vi Figure 2.62 A projectile has a parabolic trajectory. dx Chapter 2 Vector components describe motion in two dimensions. 103 02-PearsonPhys20-Chap02 7/24/08 10:18", " AM Page 104 2-5 QuickLab 2-5 QuickLab Which Lands First? Problem What is the relationship between horizontal and vertical motion of objects on a ramp? Materials Galileo apparatus (Figure 2.63) steel balls Procedure 1 Set up the Galileo apparatus at the edge of a lab bench. 2 Place a steel ball at the top of each ramp. 3 Release the balls at the same time. 4 Listen for when each ball hits the ground. 5 Using a different ramp, repeat steps 1\u20134. Questions 1. Which ball landed first? 2. Did the balls\u2019 initial velocity affect the result? If so, how? 3. What inference can you make about the relationship between horizontal and vertical motion? Figure 2.63 From section 1.6, you know that gravity influences the vertical motion of a projectile by accelerating it downward. From Figure 2.64, note that gravity has no effect on an object\u2019s horizontal motion. So, the two components of a projectile\u2019s motion can be considered independently. As a result, a projectile experiences both uniform motion and uniformly accelerated motion at the same time! The horizontal motion of a projectile is an example of uniform motion; the projectile\u2019s horizontal velocity component is constant. The vertical motion of a projectile is an example of uniformly accelerated motion. The object\u2019s acceleration is the constant acceleration due to gravity or 9.81 m/s2 [down] (neglecting friction). Concept Check In a table, classify the horizontal and vertical components of position, velocity, and acceleration of a horizontally launched projectile as uniform or non-uniform motion. Objects Launched Horizontally Suppose you made a new game based on a combination of shuffleboard and darts. The goal is to flick a penny off a flat, horizontal surface, such as a tabletop, and make it land on a target similar to a dartboard beyond the table. The closer your penny lands to the bull\u2019s eye, the more points you score (Figure 2.65). Figure 2.64 Gravity does not affect the horizontal motion of a projectile because perpendicular components of motion are independent. PHYSICS INSIGHT When a projectile is launched, for a fraction of a second, it accelerates from rest to a velocity that has x and y components. 104 Unit I Kinematics 02-PearsonPhys20-Chap02 7/24/08 10:18 AM Page 105 v table horizontal (x) component dy trajectory padded dartboard dx Figure 2.65 An object launched horizontally experiences uniform horizontal motion and uniformly accelerated vertical motion. In the game, once the penny leaves the tabletop, it becomes a projectile and travels in a parabolic path toward the ground. In section 1.6, you studied motion that was caused by acceleration due to gravity. The velocity of an object falling straight down has no horizontal velocity component. In this game, the penny moves both horizontally and vertically, like the ball on the right in Figure 2.64. In this type of projectile motion, the object\u2019s initial vertical velocity is zero. Because the projectile has a horizontal velocity component, it travels a horizontal distance along the ground from its initial launch point. This distance is called the projectile\u2019s range (Figure 2.66). The velocity component in the y direction increases because of the acceleration due to gravity while the x component remains the same. The combined horizontal and vertical motions produce the parabolic path of the projectile. Concept Check (a) What factors affecting projectile motion in the horizontal direction are being neglected? (b) What causes the projectile to finally stop? (c) If the projectile\u2019s initial velocity had a vertical component, would the projectile\u2019s path still be parabolic? Give reasons for your answer. PHYSICS INSIGHT For a projectile to have a non-zero velocity component in the vertical direction, the object must be thrown up, down, or at an angle relative to the horizontal, rather than sideways. range: the distance a projectile travels horizontally over level ground y maximum height range x Solving Projectile Motion Problems In this chapter, you have been working with components, so you know how to solve motion problems by breaking the motion down into its horizontal (x) and vertical (y) components. Figure 2.66 The range of a projectile is its horizontal distance travelled. Chapter 2 Vector components describe motion in two dimensions. 105 02-PearsonPhys20-Chap02 7/24/08 10:18 AM Page 106 Before you solve a projectile motion problem, review what you already know (Figure 2.67). vi vx vi vy 0 Figure 2.67(a) The projectile is given an initial horizontal velocity. viy vi sin \u03b8 vi vix vi cos \u03b8 vi cos \u03b8 vix vi sin \u03b8 viy vi Figure 2.67(b) The projectile is given an initial horizontal velocity and an upward vertical velocity. Figure 2.67(c) The projectile is given an initial horizontal velocity and a downward vertical velocity. x direction", " \u2013 There is no acceleration in this direction, so ax 0. In this text, ax will always be zero. The projectile undergoes uniform motion in the x direction. \u2013 The general equation for the initial x component of the velocity can be determined using trigonometry, e.g., vix vi cos. \u2013 The range is dx. \u2013 Because the projectile is moving in both the horizontal and vertical directions at the same time, t is a common variable. y direction \u2013 If up is positive, the acceleration due to gravity is down or negative, so ay 9.81 m/s2. \u2013 The y component of the initial velocity can be determined using trigonometry, e.g., viy vi sin. \u2013 The displacement in the y direction is dy. \u2013 Time (t) is the same in both the x and y directions. \u25bc Table 2.4 Projectile Problem Setup x direction y direction ax vix 0 vi cos dx vx t 9.81 m/s2 ay vi sin viy can be positive or negative viy depending on the direction of v i. dy t 1 ay(t)2 2 viy If you check the variables, you can see that they are vi, t, d, and a, 1 a(t)2. In the t all of which are present in the equation d 2 v i horizontal direction, the acceleration is zero, so this equation simplifies t. The next example shows you how to apply these equations. to d v i e MATH To explore and graph the relationship between the velocity and position of an object thrown vertically into the air, visit www.pearsoned.ca/school/ physicssource. 106 Unit I Kinematics 02-PearsonPhys20-Chap02 7/24/08 10:18 AM Page 107 Example 2.10 Head-Smashed-In Buffalo Jump, near Fort Macleod, Alberta, is a UNESCO heritage site (Figure 2.68). Over 6000 years ago, the Blackfoot people of the Plains hunted the North American bison by gathering herds and directing them over cliffs 20.0 m tall. Assuming the plain was flat so that the bison ran horizontally off the cliff, and the bison were moving at their maximum speed of 18.0 m/s at the time of the fall, determine how far from the base of the cliff the bison landed. Practice Problems 1. A coin rolls off a table with an initial horizontal speed of 30 cm/s. How far will the coin land from the base of the table if the table\u2019s height is 1.25 m? 2. An arrow is fired horizontally with a speed of 25.0 m/s from the top of a 150.0-m-tall cliff. Assuming no air resistance, determine the distance the arrow will drop in 2.50 s. 3. What is the horizontal speed of an object if it lands 40.0 m away from the base of a 100-m-tall cliff? Answers 1. 15 cm 2. 30.7 m 3. 8.86 m/s Figure 2.68 y d y x d x Figure 2.69 Given For convenience, choose forward and down to be positive because the motion is forward and down (Figure 2.69). x direction vix 18.0 m/s y direction ay dy 9.81 m/s2 [down] 9.81 m/s2 20.0 m Required distance from the base of the cliff (dx) Analysis and Solution Since there is no vertical component to the initial velocity of the 0 m/s. Therefore, the bison experience uniformly bison, viy accelerated motion due to gravity in the vertical direction but uniform motion in the horizontal direction resulting from the run. From the given values, note that, in the y direction, you have all the variables except for time. So, you can solve for time in the y direction, which is the time taken to fall. PHYSICS INSIGHT For projectile motion in two dimensions, the time taken to travel horizontally equals the time taken to travel vertically. e SIM Analyze balls undergoing projectile motion. Follow the eSim links at www.pearsoned.ca/school/ physicssource. y direction: dy t 1 viy ay(t)2 2 0 1 ay 2 t2 Chapter 2 Vector components describe motion in two dimensions. 107 02-PearsonPhys20-Chap02 7/24/08 10:18 AM Page 108 t 2dy ay 2(20.0 m) 9.81 m/s2 2.019 s x direction: The time taken for the bison to fall vertically equals the time they travel horizontally. Substitute the value for time you found in the y direction to find the range. Since the bison had a uniform horizontal speed of 18.0 m/s, use the equation dx dx vix t. (18.0 m/s)(2.019 s) 36.3 m", " Paraphrase The bison would land 36.3 m from the base of the cliff. Figure 2.70 Baseball is all about projectile motion. Objects Launched at an Angle Baseball is a projectile game (Figure 2.70). The pitcher throws a ball at the batter, who hits it to an open area in the field. The outfielder catches the ball and throws it to second base. The runner is out. All aspects of this sequence involve projectile motion. Each sequence requires a different angle on the throw and a different speed. If the player miscalculates one of these variables, the action fails: Pitchers throw wild pitches, batters strike out, and outfielders overthrow the bases. Winning the game depends on accurately predicting the components of the initial velocity! vy v sin \u03b8 v \u03b8 vx v cos \u03b8 For objects launched at an angle, such as a baseball, the velocity of the object has both a horizontal and a vertical component. Any vector quantity can be resolved into x and y components using the trigonometric ratios R sin, when is measured relative to Rx the x-axis. To determine the horizontal and vertical compo v cos and nents of velocity, this relationship becomes vx vy R cos and Ry v sin, as shown in Figure 2.71. Solving problems involving objects launched at an angle is similar to solving problems involving objects launched horizontally. The object experiences uniform motion in the horizontal direction, so use the equation dx t. In the vertical direction, the object experiences uniformly accelerated motion. The 1 ay(t)2 still applies, but in this case, viy t general equation dy 2 is not zero. The next example shows you how to apply these equations to objects launched at an angle. vix viy Figure 2.71 The horizontal and vertical components of velocity 108 Unit I Kinematics 02-PearsonPhys20-Chap02 7/24/08 10:18 AM Page 109 Example 2.11 Baseball players often practise their swing in a batting cage, in which a pitching machine delivers the ball (Figure 2.72). If the baseball is launched with an initial velocity of 22.0 m/s [30.0\u00b0] and the player hits it at the same height from which it was launched, for how long is the baseball in the air on its way to the batter? Figure 2.72 Given v i 22.0 m/s [30.0\u00b0] Practice Problems 1. A ball thrown horizontally at 10.0 m/s travels for 3.0 s before it strikes the ground. Find (a) the distance it travels horizontally. (b) the height from which it was thrown. 2. A ball is thrown with a velocity of 20.0 m/s [30] and travels for 3.0 s before it strikes the ground. Find (a) the distance it travels horizontally. (b) the height from which it was thrown. (c) the maximum height of the ball. Answers 1. (a) 30 m (b) 44 m 2. (a) 52 m (b) 14 m (c) 19 m Required time (t) Analysis and Solution Choose forward and up to be positive (Figure 2.73). First find the components of the baseball\u2019s initial velocity. y viy vi 30.0\u00b0 vix x Figure 2.73 x direction vix vi cos (22.0 m/s)(cos 30.0) 19.05 m/s y direction viy vi sin (22.0 m/s)(sin 30.0) 11.00 m/s Since the ball returns to the same height from which it was launched, dy 0. With this extra known quantity, you now have enough information in the y direction to find the time the ball spent in the air. PHYSICS INSIGHT Be careful to follow the sign convention you chose. If you chose up as positive, ay becomes 9.81 m/s2. info BIT The world\u2019s fastest bird is the peregrine falcon, with a top vertical speed of 321 km/h and a top horizontal speed of 96 km/h. e WEB The fastest speed for a projectile in any ball game is approximately 302 km/h in jai-alai. To learn more about jai-alai, follow the links at www.pearsoned.ca/school/ physicssource. Chapter 2 Vector components describe motion in two dimensions. 109 02-PearsonPhys20-Chap02 7/24/08 10:18 AM Page 110 PHYSICS INSIGHT Since the vertical velocity of the ball at maximum height is zero, you can also calculate the time taken to go up and multiply the answer by two. If down is positive, t vi vf a 0 m/s (11.00 m/s) 9.81 m/s2 11.00 m s m 9.81 s2 1.121 s The total time the baseball is", " in the air is 2 1.121 s 2.24 s. info BIT The longest speedboat jump was 36.5 m in the 1973 James Bond movie Live and Let Die. The boat practically flew over a road. 110 Unit I Kinematics dy viy 1 t ay(t)2 2 0 (11.00 m/s)t (9.81 m/s2)(t)2 1 2 Isolate t and solve. (4.905 m/s2)(t)2 (11.00 m/s)(t) t m 11.00 s m 4.905 s2 2.24 s Paraphrase The baseball is in the air for 2.24 s. How far would the baseball in Example 2.11 travel horizontally if the batter missed and the baseball landed at the same height from which it was launched? Since horizontal velocity is constant, dx t vix (19.05 m/s)(2.24 s) 42.7 m The baseball would travel a horizontal distance of 42.7 m. In the next example, you are given the time and are asked to solve for one of the other variables. However, the style of solving the problem remains the same. In any problem that you will be asked to solve in this course, you will always be able to solve for one quantity in either the x or y direction, and then you can substitute your answer to solve for the remaining variable(s). Example 2.12 A paintball directed at a target is shot at an angle of 25.0. If paint splats on its intended target at the same height from which it was launched, 3.00 s later, find the distance from the shooter to the target. Given Choose down and right to be positive. a ay 25.0 t 3.00 s 9.81 m/s2 [down] 9.81 m/s2 y vi 25.0\u00b0 Figure 2.74 x 02-PearsonPhys20-Chap02 7/24/08 10:18 AM Page 111 Required range (dx) Analysis and Solution Use the equation dy viy t ay(t)2. Since the height 1 2 of landing is the same as the launch height, dy y direction: 0. dy viy 1 t ay(t)2 2 0 viy 1 t ay(t)2 2 viy 1 t ay(t)2 2 viy t 1 ay 2 1 9.81 2 m s2 (3.00 s) 14.7 m/s Since down is positive, the negative sign means that the direction of the vertical component of initial velocity is up. x direction: Find the initial horizontal speed using the tangent function. Because there is no acceleration in the x direction, the ball\u2019s horizontal speed remains the same during its flight: ax 0. PHYSICS INSIGHT Alternatively, the time taken to reach maximum height is the same time taken to fall back down to the same height. So, the paintball is at its maximum height at 1.50 s. The speed at maximum height is zero. If up is positive at 0 m/s at (9.81 m/s2)(t) (9.81 m/s2)(1.50 s) 14.7 m/s v i f The sign is positive, so the direction is up. tan opposite adjacent adjacent opposite tan 14.7 m/s tan 25.0\u00b0 31.56 m/s vi vix 25.0\u00b0 Figure 2.75 14.7 m/s From Figure 2.75, the adjacent side is vix and it points to the right, so vix Now find the horizontal distance travelled. dx 31.56 m/s. t vix (31.56 m/s)(3.00 s) 94.7 m Practice Problems 1. Determine the height reached by a baseball if it is released with a velocity of 17.0 m/s [20]. 2. A German U2 rocket from the Second World War had a range of 300 km, reaching a maximum height of 100 km. Determine the rocket\u2019s maximum initial velocity. Paraphrase The distance that separates the target from the shooter is 94.7 m. Answers 1. 1.72 m 2. 1.75 103 m/s [53.1] Chapter 2 Vector components describe motion in two dimensions. 111 02-PearsonPhys20-Chap02 7/24/08 10:18 AM Page 112 The points below summarize what you have learned in this section. \u2013 To solve problems involving projectiles, first resolve the motion into its components using the trigonometric functions, then apply the kinematics equations. \u2013 Perpendicular components of motion are independent of one another. \u2013 Horizontal motion is considered uniform and is described by the equation d vt, whereas vertical motion is a special case of uniformly accelerated motion, where the acceleration is the acceleration due to gravity or 9.81 m/", "s2 [down]. \u2013 A projectile\u2019s path is a parabola. \u2013 In the vertical direction, a projectile\u2019s velocity is greatest at the instant of launch and just before impact, whereas at maximum height, vertical velocity is zero. 2.4 Check and Reflect 2.4 Check and Reflect Knowledge 1. Platform divers receive lower marks if they enter the water a distance away from the platform, whereas speed swimmers dive as far out into the pool as they can. Compare and contrast the horizontal and vertical components of each type of athlete\u2019s motion. 2. For a fixed speed, how does the range depend on the angle,? 3. (a) For a projectile, is there a location on its trajectory where the acceleration and velocity vectors are perpendicular? Explain. (b) For a projectile, is there a location on its trajectory where the acceleration and velocity vectors are parallel? Explain. 4. Water safety instructors tell novice swimmers to put their toes over the edge and jump out into the pool. Explain why, using concepts from kinematics and projectile motion. Applications 5. Participants in a road race take water from a refreshment station and throw their empty cups away farther down the course. If a runner has a forward speed of 6.20 m/s, how far in advance of a garbage pail should he release his water cup if the vertical distance between the lid of the garbage can and the runner\u2019s point of release is 0.50 m? 6. A baseball is thrown with a velocity of 27.0 m/s [35]. What are the components of the ball\u2019s initial velocity? How high and how far will it travel? 112 Unit I Kinematics 7. A football is thrown to a moving receiver. The football leaves the quarterback\u2019s hands 1.75 m above the ground with a velocity of 17.0 m/s [25]. If the receiver starts 12.0 m away from the quarterback along the line of flight of the ball when it is thrown, what constant velocity must she have to get to the ball at the instant it is 1.75 m above the ground? 8. At the 2004 Olympic Games in Athens, Dwight Phillips won the gold medal in men\u2019s long jump with a jump of 8.59 m. If the angle of his jump was 23, what was his takeoff speed? 9. A projectile is fired with an initial speed of 120 m/s at an angle of 55.0 above the horizontal from the top of a cliff 50.0 m high. Find (a) the time taken to reach maximum height (b) the maximum height with respect to the ground next to the cliff (c) the total time in the air (d) the range (e) the components of the final velocity just before the projectile hits the ground Extension 10. Design a spreadsheet to determine the maximum height and range of a projectile with a launch angle that increases from 0 to 90 and whose initial speed is 20.0 m/s. e TEST To check your understanding of projectile motion, follow the eTest links at www.pearsoned.ca/school/physicssource. 02-PearsonPhys20-Chap02 7/25/08 8:15 AM Page 113 CHAPTER 2 SUMMARY Key Terms and Concepts collinear resultant vector components polar coordinates method navigator method non-collinear relative motion ground velocity air velocity wind velocity trajectory range Key Equations v a t d v 1 t a(t)2 2 i d vt Conceptual Overview The concept map below summarizes many of the concepts and equations in this chapter. Copy and complete the map to have a full summary of the chapter. projectile motion Examples of Two-dimensional Motion relative motion Split given values into horizontal (x) and vertical components (y) Set up problem in x and y : x ax 0 vix vi ___ \u03b8 y ay 9.81 m/s2 vi ____ \u03b8 viy Time is common to both x and y directions. Use the equation d vi Solve in the x direction, then in the y direction, OR solve in the y direction, then in the x direction. t (t)2 a 1 2 State given values: vground _____ [ ] vair _____ [ ] vwind _____ [ ] Split given values into components. Use the equation v_____ vwind vair Use the algebraic vector method to solve for the unknown value. Two-dimensional motion Addition of vectors Graphical method Algebraic method Draw vectors to scale, connect them _________. Split given values into x and y components. Draw a new vector from starting point of diagram to end point. Use Rx R cos\u03b8 and _________. Measure length of vector and convert using scale. Solve separate equations in x and y. Measure angle at start of vector. Combine component answers using the Pythagorean theorem: Determine the angle using __________. R x2 y2 \u221a", " The answer is called the ___________. It has both magnitude and _________. Chapter 2 Vector components describe motion in two dimensions. 113 02-PearsonPhys20-Chap02 7/24/08 10:18 AM Page 114 CHAPTER 2 REVIEW Knowledge 1. (2.2) During the Terry Fox Run, a participant 8. (2.4) For an object thrown vertically upward, what is the object\u2019s initial horizontal velocity? travelled from A to D, passing through B and C. Copy and complete the table using the information in the diagram, a ruler calibrated in millimetres, and a protractor. In your notebook, draw and label the displacement vectors AB, BC, and CD and the position vectors AB, AC, and AD. Assume the participant\u2019s reference point is A. d = 5.0 km [E] C 2.0 km scale D d = 3.0 km [E] d = 2.0 km [N] A B Distance \u0394d (km) Displacement \u0394d (km) [direction] Final position d (km) [direction] reference point AB BC CD AC AD 2. (2.2) Determine the x and y components of the displacement vector 55 m [222]. 3. (2.4) What is the vertical component for velocity at the maximum height of a projectile\u2019s trajectory? 4. (2.4) During a field goal kick, as the football rises, what is the effect on the vertical component of its velocity? 5. (2.1) Fort McMurray is approximately 500 km [N] of Edmonton. Using a scale of 1.0 cm : 50.0 km, draw a displacement vector representing this distance. 6. (2.1) Give one reason why vector diagrams must be drawn to scale. 7. (2.2) Using an appropriate scale and reference coordinates, graphically solve each of the following: (a) 5.0 m [S] and 10.0 m [N] (b) 65.0 cm [E] and 75.0 cm [E] (c) 1.0 km [forward] and 3.5 km [backward] (d) 35.0 km [right] 45.0 km [left] 114 Unit I Kinematics Applications 9. The air medivac, King Air 200, flying at 250 knots (1 knot 1.853 km/h), makes the trip between Edmonton and Grande Prairie in 50 min. What distance does the plane travel during this time? 10. A golf ball is hit with an initial velocity of 30.0 m/s [55]. What are the ball\u2019s range and maximum height? 11. Off the tee box, a professional golfer can drive a ball with a velocity of 80.0 m/s [10]. How far will the ball travel horizontally before it hits the ground and for how long is the ball airborne? 12. A canoeist capable of paddling north at a speed of 4.0 m/s in still water wishes to cross a river 120 m wide. The river is flowing at 5.0 m/s [E]. Find (a) her velocity relative to the ground (b) the time it takes her to cross 13. An object is thrown horizontally off a cliff with an initial speed of 7.50 m/s. The object strikes the ground 3.0 s later. Find (a) the object\u2019s vertical velocity component when it reaches the ground (b) the distance between the base of the cliff and the object when it strikes the ground (c) the horizontal velocity of the object 1.50 s after its release 14. If a high jumper reaches her maximum height as she travels across the bar, determine the initial velocity she must have to clear a bar set at 2.0 m if her range during the jump is 2.0 m. What assumptions did you make to complete the calculations? 15. An alligator wishes to swim north, directly across a channel 500 m wide. There is a current of 2.0 m/s flowing east. The alligator is capable of swimming at 4.0 m/s. Find (a) the angle at which the alligator must point its body in order to swim directly across the channel (b) its velocity relative to the ground (c) the time it takes to cross the channel 16. A baseball player throws a ball horizontally at 45.0 m/s. How far will the ball drop before reaching first base 27.4 m away? 02-PearsonPhys20-Chap02 7/24/08 10:18 AM Page 115 17. How much time can you save travelling diagonally instead of walking 750 m [N] and then 350 m [E] if your walking speed is 7.0 m/s? 18. How long will an arrow be in flight if it is shot at an angle of 25", " and hits a target 50.0 m away, at the same elevation? 19. A pilot of a small plane wishes to fly west. The plane has an airspeed of 100 km/h. If there is a 30-km/h wind blowing north, find 27. An airplane is approaching a runway for landing. The plane\u2019s air velocity is 645 km/h [forward], moving through a headwind of 32.2 km/h. The altimeter indicates that the plane is dropping at a constant velocity of 3.0 m/s [down]. If the plane is at a height of 914.4 m and the range from the plane to the start of the runway is 45.0 km, does the pilot need to make any adjustments to her descent in order to land the plane at the start of the runway? Consolidate Your Understanding Create your own summary of kinematics by answering the questions below. If you want to use a graphic organizer, refer to Student References 4: Using Graphic Organizers on pp. 869\u2013871. Use the Key Terms and Concepts listed on page 113 and the Learning Outcomes on page 68. 1. Create a flowchart to describe the different components required to analyze motion in a horizontal plane and in a vertical plane. 2. Write a paragraph describing the similarities and differences between motion in a horizontal plane and motion in a vertical plane. Share your thoughts with another classmate. Think About It Review your answers to the Think About It questions on page 69. How would you answer each question now? e TEST To check your understanding of two-dimensional motion, follow the eTest links at www.pearsoned.ca/school/physicssource. (a) the plane\u2019s heading (b) the plane\u2019s ground speed 20. At what angle was an object thrown if its initial launch speed is 15.7 m/s, it remains airborne for 2.15 s, and travels 25.0 m horizontally? 21. A coin rolls off a 25.0\u00b0 incline on top of a 2.5-m-high bookcase with a speed of 30 m/s. How far from the base of the bookcase will the coin land? 22. Starting from the left end of the hockey rink, the goal line is 3.96 m to the right of the boards, the blue line is 18.29 m to the right of the goal line, the next blue line is 16.46 m to the right of the first blue line, the goal line is 18.29 m right, and the right board is 3.96 m right of the goal line. How long is a standard NHL hockey rink? 23. A plane with a ground speed of 151 km/h is moving 11 south of east. There is a wind blowing at 40 km/h, 45 south of east. Find (a) the plane\u2019s airspeed (b) the plane\u2019s heading, to the nearest degree 24. How long will a soccer ball remain in flight if it is kicked with an initial velocity of 25.0 m/s [35.0]? How far down the field will the ball travel before it hits the ground and what will be its maximum height? 25. At what angle is an object launched if its initial vertical speed is 3.75 m/s and its initial horizontal speed is 4.50 m/s? Extensions 26. During the Apollo 14 mission, Alan Shepard was the first person to hit a golf ball on the Moon. If a golf ball was launched from the Moon\u2019s surface with a velocity of 50 m/s [35] and the acceleration due to gravity on the Moon is 1.61 m/s2, (a) how long was the golf ball in the air? (b) what was the golf ball\u2019s range? Chapter 2 Vector components describe motion in two dimensions. 115 02-PearsonPhys20-Chap02 7/24/08 10:18 AM Page 116 UNIT I PROJECT Are Amber Traffic Lights Timed Correctly? Scenario The Traffic Safety Act allows law enforcement agencies in Alberta to issue fines for violations using evidence provided by red light cameras at intersections. The cameras photograph vehicles that enter an intersection after the traffic lights have turned red. They record the time, date, location, violation number, and time elapsed since the light turned red. The use of red light cameras and other technology reduces the amount of speeding, running of red lights, and collisions at some intersections. The length of time a traffic light must remain amber depends on three factors: perception time, reaction time, and braking time. The sum of perception time and reaction time is the time elapsed between the driver seeing the amber light and applying the brakes. The Ministry of Infrastructure and Transportation\u2019s (MIT) Basic Licence Driver\u2019s Handbook allows for a perception time of 0.75 s and a reaction time of 0.75 s. The braking time is", " the time it takes the vehicle to come to a full stop once the brakes are applied. Braking time depends on the vehicle\u2019s initial speed and negative acceleration. The MIT\u2019s predicted braking times are based on the assumption that vehicles travel at the posted speed limit and have a uniform acceleration of 3.0 m/s2. Other factors that affect acceleration are road conditions, vehicle and tire performance, weather conditions, and whether the vehicle was travelling up or down hill. If drivers decide to go through an intersection safely (go distance) after a light has turned amber, they must be able to travel not only to the intersection but across it before the light turns red. The go distance depends on the speed of the vehicle, the length of the intersection, and the amount of time the light remains amber. If the driver decides to stop (stop distance), the vehicle can safely do so only if the distance from the intersection is farther than the distance travelled during perception time, reaction time, and braking time. As part of a committee reporting to the Ministry of Infrastructure and Transportation, you must respond to concerns that drivers are being improperly fined for red light violations because of improper amber light timing. You are to decide how well the amber light time matches the posted speed limit and intersection length. Assume throughout your analysis that drivers travel at the posted speed limits. Planning Research or derive equations to determine Assessing Results After completing the project, assess its success based on a rubric* designed in class that considers research strategies experiment techniques clarity and thoroughness of the written report effectiveness of the team\u2019s presentation (a) a car\u2019s displacement during reaction time (b) stop distance (c) go distance (d) amber light time (e) displacement after brakes are applied (f) amount of time elapsed after the brakes are applied Materials \u2022 measuring tape, stopwatch Procedure 1 Design a survey to measure the amber light times at 10 different intersections near your school. For each intersection, record its length. Use caution around intersections due to traffic! You may wish to estimate the length of the intersection by counting the number of steps it takes you to cross and measuring the length of your stride. 2 Apply suitable equations to determine appropriate amber light times for the 10 different intersections. 3 Calculate stop distances and go distances for a range (10 km/h) of posted speed limits for each intersection and plot graphs of stop distance and go distance against posted speed. Thinking Further 1. Research the effectiveness of red light cameras in reducing accidents, speeding, and red light violations. Using your research, recommend a course of action to increase vehicle-rail safety at light-controlled railway crossings. 2. Based on your surveys and investigation, recommend whether existing amber light times should be increased, decreased, or left alone. Consider posted speeds against actual speeds and wet against dry surface conditions. 3. Prepare a presentation to the other members of your committee. Include graphs and diagrams. *Note: Your instructor will assess the project using a similar assessment rubric. 116 Unit I Kinematics 02-PearsonPhys20-Chap02 7/24/08 10:18 AM Page 117 UNIT I SUMMARY Unit Concepts and Skills: Quick Reference Concepts Chapter 1 Summary Resources and Skill Building Graphs and equations describe motion in one dimension. Scalar and vector quantities 1.1 The Language of Motion A scalar quantity consists of a number and a unit. Distance, position, and displacement A vector quantity consists of a number, a unit, and a direction. Distance is the length of the path taken to travel from one position to another. Position is the straight-line distance from the origin to the object\u2019s location. Displacement is the change in position. Slope of a position-time graph Slope of a velocity-time graph Position-time, velocity-time, acceleration\u2013time graphs representing accelerated motion 1.2 Position-time Graphs and Uniform Motion A position-time graph for an object at rest is a straight line with zero slope. A position-time graph for an object moving at a constant velocity is a straight line with non-zero slope. The greater the slope of a position-time graph, the faster the object is moving. 1.3 Velocity-time Graphs: Uniform and Non-uniform Motion A velocity-time graph for an object experiencing uniform motion is a horizontal line. The slope of a velocity-time graph represents acceleration. The position-time graph for an object undergoing uniformly accelerated motion is a curve. The corresponding velocity-time graph is a straight line with non-zero slope. The corresponding acceleration-time graph is a horizontal line. Section 1.1 Section 1.1 Figure 1.5 Figures 1.3, 1.4, 1.5, Example 1.1 Figures 1.3, 1.4, 1.5, Example 1.1 Figure 1.14 Figures 1.12, 1.15(b), 1-3 Inquiry Lab Examples 1.2, 1.3, 1-3 Inquiry Lab", " Figure 1.24 Figures 1.24, 1.30, Example 1.5 Figures 1.28\u20131.31, Example 1.5 Instantaneous velocity The slope of the tangent on a position-time curve gives instantaneous velocity. Figure 1.29, Example 1.5 Area under and slope of a velocity-time graph Average velocity Velocity-time graphs 1.4 Analyzing Velocity-time Graphs The area under a velocity-time graph represents displacement; slope represents acceleration. Average velocity represents total displacement divided by time elapsed. You can draw acceleration-time and position-time graphs by calculating and plotting slope and area, respectively, of a velocity-time graph. Figure 1.41, Examples 1.6, 1.8, 1.9 Figure 1.45, Examples 1.7, 1.9 Examples 1.10, 1.11 Kinematics equations 1.5 The Kinematics Equations When solving problems in kinematics, choose the equation that contains all the given variables in the problem as well as the unknown variable. Figures 1.53\u20131.55 Examples 1.12\u20131.16 Projectile motion straight up and down Maximum height 1.6 Acceleration Due to Gravity Gravity causes objects to accelerate downward. At maximum height, a projectile\u2019s vertical velocity is zero. The time taken to reach maximum height equals the time taken to fall back down to the original height. 1-6 QuickLab, 1-7 Inquiry Lab, 1-8 QuickLab, Examples 1.17\u20131.19 Figures 1.63\u20131.65 Examples 1.18, 1.19 Chapter 2 Vector components describe motion in two dimensions. Adding and subtracting vectors 2.1 Vector Methods in One Dimension Add vectors by connecting them tip to tail. Subtract vectors by connecting them tail to tail. Examples 2.1, 2.2 Components Relative motion Projectile motion in two dimensions 2.2 Motion in Two Dimensions To add vectors in two dimensions, draw a scale diagram, or resolve them into their components and use trigonometry to find the resultant. Examples 2.3\u20132.5, Figures 2.31, 2.33\u20132.37 2.3 Relative Motion To solve relative motion problems, use trigonometry, with ground velocity as the resultant. If the vectors are not perpendicular, resolve them into their components first. Examples 2.6\u20132.9 2.4 Projectile Motion The shape of a projectile\u2019s trajectory is a parabola. Horizontal and vertical components of projectile motion are independent. To solve projectile problems in two dimensions, resolve them into their horizontal and vertical components. Then use the kinematics equations. The time taken to travel horizontally equals the time taken to travel vertically. 2-4 QuickLab, Figure 2.62 2-5 QuickLab, Figure 2.64 Examples 2.10\u20132.12, Figures 2.67, 2.71 Unit I Kinematics 117 02-PearsonPhys20-Chap02 7/24/08 10:18 AM Page 118 UNIT I REVIEW Vocabulary 1. Using your own words, define these terms: acceleration acceleration due to gravity air velocity at rest collinear components displacement distance ground velocity instantaneous velocity kinematics navigator method non-collinear non-uniform motion origin polar coordinates method position projectile projectile motion range relative motion resultant vector scalar quantity tangent trajectory uniform motion uniformly accelerated motion vector quantity velocity wind velocity Knowledge CHAPTER 1 2. Describe how scalar quantities differ from vector quantities. CHAPTER 2 5. Using a scale of 1.0 cm : 3.5 km, determine the magnitude and direction of the vector below. W N S Applications E R 6. A wildlife biologist records a moose\u2019s position as it swims away from her. Using the graph below, determine the moose\u2019s velocity. Position vs. Time ) ] 30.0 25.0 20.0 15.0 10.0 5.0 0 0 2.0 4.0 8.0 6.0 Time (s) 10.0 12.0 7. Sketch a position-time graph for each statement below. Assume that right is positive. (a) object accelerating to the right (b) object accelerating to the left (c) object travelling at a constant velocity left (d) object at rest (e) object travelling with constant velocity right 8. Hockey pucks can be shot at speeds of 107 km/h. If a puck is shot at an angle of 30, determine how long the puck is in the air, how far it will travel, and how high it will be at the peak of its trajectory. 3. Resolve the following vectors into their 9. Sketch two different position-time graphs for objects with a negative velocity. 10. Sketch two different velocity-time graphs for objects with a negative acceleration. components: (a) 5.0 m [90\u00b0] (b) 16.0 m/", "s [20 S of W] 4. Using an appropriate scale and reference coordinates, draw the following vectors: (a) 5.0 m/s [0] (b) 25.0 m/s2 [60 N of E] (c) 1.50 km [120] 118 Unit I Kinematics 02-PearsonPhys20-Chap02 7/24/08 10:18 AM Page 119 11. From the position-time graph below, determine 16. (a) What is the change in velocity in 10.0 s, which object has the greatest velocity. Time (s) 0.0 2.0 4.0 6.0 8.0 10.0 Time (s 12. Solve each of the following equations for initial velocity, v i, algebraically. v v (a) a i t f (b) d v 1 t at2 2 i (c) d 1 2 (v i v f)t 13. The longest kickoff in CFL history is 83.2 m. If the ball remains in the air for 5.0 s, determine its initial speed. 14. Determine the speed of a raven that travels 48 km in 90 min. 15. Describe the motion of the object illustrated in the graph below. Velocity vs. Time 9.0 8.0 7.0 6.0 5.0 4.0 3.0 2.0 1..0 4.0 8.0 6.0 Time (s) 10.0 12.0 as illustrated in the acceleration-time graph below? (b) If the object had an initial velocity of 10 m/s [90], what is its final velocity after 10.0 s? Acceleration vs. Time ) ] \u02da.0 3.0 2.0 1.0 0.0 17. How far will a crow fly at 13.4 m/s for 15.0 min? 18. How long will it take a car to travel from Valleyview to Grande Prairie if its speed is 100 km/h? The map\u2019s scale is 1 cm : 118 km. Wood Buffalo National Park 58 High Level Fort Chipewyan Fort Vermilion 35 ALBERTA Fort McMurray 88 Peace River 63 Grande Prairie Slave Lake 2 40 Valleyview 43 Edmonton 19. A baseball player hits a baseball with a velocity of 30 m/s [25]. If an outfielder is 85.0 m from the ball when it is hit, how fast will she have to run to catch the ball before it hits the ground? 20. Determine the magnitude of the acceleration of a Jeep Grand Cherokee if its stopping distance is 51.51 m when travelling at 113 km/h. 21. What is the velocity of an aircraft with respect to the ground if its air velocity is 785 km/h [S] and the wind is blowing 55 km/h [22 S of W]? 22. An object undergoing uniformly accelerated motion has an initial speed of 11.0 m/s and travels 350 m in 3.00 s. Determine the magnitude of its acceleration. 23. Improperly installed air conditioners can occasionally fall from apartment windows down onto the road below. How long does a pedestrian have to get out of the way of an air conditioner falling eight stories (24 m)? Unit I Kinematics 119 02-PearsonPhys20-Chap02 7/24/08 10:18 AM Page 120 24. An object is launched from the top of a building with an initial velocity of 15 m/s [32]. If the building is 65.0 m high, how far from the base of the building will the object land? 25. Two friends walk at the same speed of 4.0 km/h. One friend steps onto a travelator moving at 3.0 km/h. If he maintains the same initial walking speed, (a) how long will it take him to reach the end of the 100-m-long travelator? (b) what must be the magnitude of the acceleration of the other friend to arrive at the end of the travelator at the same time? 26. How far will a vehicle travel if it accelerates uniformly at 2.00 m/s2 [forward] from 2.50 m/s to 7.75 m/s? 27. An object is thrown into the air with a speed of 25.0 m/s at an angle of 42\u00b0. Determine how far it will travel horizontally before hitting the ground. 28. Determine the average velocity of a truck that travels west from Lloydminster to Edmonton at 110 km/h for 1.0 h and 20 min and then 90 km/h for 100 min. 29. What distance will a vehicle travel if it accelerates uniformly from 15.0 m/s [S] to 35.0 m/s [S] in 6.0 s? 30. From the graph below, determine the instantaneous (b) velocity of the", " object at 5.0 s, 10.0 s, and 15.0 s. Position vs. Time 70.0 60.0 50.0 40.0 30.0 20.0 10.0 0. 33. Determine the displacement of the blue jay from the velocity-time graph below. Velocity vs. Time for a Blue Jay ) ] 32.0 28.0 24.0 20.0 16.0 12.0 8.0 4.0 0.0 0.0 10.0 20.0 30.0 40.0 50.0 60.0 Time (s) 34. Sketch a position-time graph for an object that travels at a constant velocity of 5.0 m/s for 10 s, stops for 10 s, then travels with a velocity of 2.0 m/s for 20 s. 35. Determine the height reached by a projectile if it is released with a velocity of 18.0 m/s [20]. 36. The bishop is a chess piece that moves diagonally along one colour of square. Assuming the first move is toward the left of the board, determine (a) the minimum number of squares the bishop covers in getting to the top right square. the bishop's displacement from the start if the side length of each square is taken as 1 unit and each move is from the centre of a square to the centre of another square. End 0.0 5.0 10.0 Time (s) 15.0 20.0 31. A speedboat\u2019s engine can move the boat at a velocity of 215 km/h [N]. What is the velocity of the current if the boat\u2019s displacement is 877 km [25 E of N] 3.5 h later? 32. An object starts from rest and travels 50.0 m along a frictionless, level surface in 2.75 s. What is the magnitude of its acceleration? 120 Unit I Kinematics Start 37. A wildlife biologist notes that she is 350 m [N] from the park ranger station at 8:15 a.m. when she spots a polar bear. At 8:30 a.m., she is 1.75 km [N] of the ranger station. Determine the biologist\u2019s average velocity. 38. A bus travels 500 m [N], 200 m [E], and then 750 m [S]. Determine its displacement from its initial position. 02-PearsonPhys20-Chap02 7/24/08 10:18 AM Page 121 39. Match the motion with the correct position-time graph given below. Identify the motion as at rest, uniform motion, or uniformly accelerated motion. (a) an airplane taking off (b) an airplane landing (c) passing a car on the highway 43. A motorcycle stunt rider wants to jump a 20.0-mwide row of cars. The launch ramp is angled at 30 and is 9.0 m high. The landing ramp is also angled at 30 and is 6.0 m high. Find the minimum launch velocity required for the stunt rider to reach the landing ramp. (d) waiting at the red line at Canada Customs (e) standing watching a parade Skills Practice 44. Draw a Venn diagram to compare and contrast (f) travelling along the highway on cruise control vector and scalar quantities. Position vs. Time 45. Draw a Venn diagram to illustrate the concepts 100.0 90.0 80.0 70.0 60.0 50.0 40.0 30.0 20.0 10. ii iii iv 0.0 0.0 2.0 4.0 6.0 8.0 10.0 Time (s) 40. Determine the magnitude of the acceleration of a Jeep Grand Cherokee that can reach 26.9 m/s from rest in 4.50 s. Extensions 41. A penny is released from the top of a wishing well and hits the water\u2019s surface 1.47 s later. Calculate (a) the velocity of the penny just before it hits the water\u2019s surface (b) the distance from the top of the well to the water\u2019s surface 42. A balloonist drops a sandbag from a balloon that is rising at a constant velocity of 3.25 m/s [up]. It takes 8.75 s for the sandbag to reach the ground. Determine of graphical analysis. 46. A swimmer wants to cross from the east to the west bank of the Athabasca River in Fort McMurray. The swimmer\u2019s speed in still water is 3.0 m/s and the current\u2019s velocity is 4.05 m/s [N]. He heads west and ends up downstream on the west bank. Draw a vector diagram for this problem. 47. For an experiment to measure the velocity of an object, you have a radar gun, probeware, and motion sensors. Explain to a classmate how you would decide", " which instrument to use. 48. Design an experiment to determine the acceleration of an object rolling down an inclined plane. 49. Construct a concept map for solving a twodimensional motion problem involving a projectile thrown at an angle. 50. Explain how you can use velocity-time graphs to describe the motion of an object. Self-assessment 51. Describe to a classmate which kinematics concepts and laws you found most interesting when studying this unit. Give reasons for your choices. 52. Identify one issue pertaining to motion studied in this unit that you would like to investigate in greater detail. 53. What concept in this unit did you find most difficult? What steps could you take to improve your understanding? 54. As a future voter, what legislation would you support to improve vehicular and road safety? (a) the height of the balloon when the sandbag 55. Assess how well you are able to graph the is dropped (b) the height of the balloon when the sandbag reaches the ground (c) the velocity with which the sandbag hits the ground motion of an object. Explain how you determine a reference point. e TEST To check your understanding of kinematics, follow the eTest links at www.pearsoned.ca/school/physicssource. Unit I Kinematics 121 03-Phys20-Chap03.qxd 7/24/08 10:37 AM Page 122 U N I T II Dynamics Dynamics The design of equipment used in many activities, such as ice climbing, involves understanding the cause of motion. How does gravity affect the climber and the icy cliff? How can understanding the cause of motion help you predict motion? e WEB Explore the physics principles that apply to ice and mountain climbing. Write a summary of your findings. Begin your search at www.pearsoned.ca/school/physicssource. 122 Unit II 03-Phys20-Chap03.qxd 7/24/08 10:37 AM Page 123 Unit at a Glance C H A P T E R 3 Forces can change velocity. 3.1 The Nature of Force 3.2 Newton\u2019s First Law 3.3 Newton\u2019s Second Law 3.4 Newton\u2019s Third Law 3.5 Friction Affects Motion C H A P T E R 4 Gravity extends throughout the universe. 4.1 Gravitational Forces due to Earth 4.2 Newton\u2019s Law of Universal Gravitation 4.3 Relating Gravitational Field Strength to Gravitational Force Unit Themes and Emphases \u2022 Change and Systems \u2022 Social and Environmental Contexts \u2022 Problem-Solving Skills Focussing Questions In this study of dynamics and gravitation, you will investigate different types of forces and how they change the motion of objects and affect the design of various technological systems. As you study this unit, consider these questions: \u2022 How does an understanding of forces help humans interact with their environment? \u2022 How do the principles of dynamics affect mechanical and other systems? \u2022 What role does gravity play in the universe? Unit Project Tire Design, Stopping Distance, and Vehicle Mass \u2022 By the time you complete this unit, you will have the skills to evaluate how tire treads, road surfaces, and vehicle mass affect stopping distances. You will need to consider human reaction times and the amount of moisture on road surfaces to investigate this problem. Unit II Dynamics 123 03-Phys20-Chap03.qxd 7/24/08 10:37 AM Page 124 C H A P T E R Forces can change velocity. 3 Key Concepts In this chapter, you will learn about: vector addition Newton\u2019s laws of motion static and kinetic friction Learning Outcomes When you have completed this chapter, you will be able to: Knowledge explain that a non-zero net force causes a change in velocity calculate the net force apply Newton\u2019s three laws to solve motion problems explain static and kinetic friction Science, Technology, and Society explain that the goal of technology is to provide solutions to practical problems explain that science and technology develop to meet societal needs explain that science develops through experimentation Screeching tires on the road and the sound of metal and fibreglass being crushed are familiar sounds of a vehicle collision. Depending on the presence of airbags and the correct use of seat belts and headrests, a motorist may suffer serious injury. In order to design these safety devices, engineers must understand what forces are and how forces affect the motion of an object. When a driver suddenly applies the brakes, the seat belts of all occupants lock. If the vehicle collides head-on with another vehicle, airbags may become deployed. Both seat belts and airbags are designed to stop the forward motion of motorists during a head-on collision (Figure 3.1). Motorists in a stationary vehicle that is rear-ended also experience forces. The car seats move forward quickly, taking the lower part of each person\u2019s body with it. But each person\u2019s head stays in the same", " place until yanked forward by the neck. It is this sudden yank that causes whiplash. Adjustable headrests are designed to prevent whiplash by supporting the head of each motorist. In this chapter, you will investigate how forces affect motion and how to explain and predict the motion of an object using Newton\u2019s three laws. Figure 3.1 To design cars with better safety features, accident researchers use dummies to investigate the results of high-speed collisions. 124 Unit II 03-Phys20-Chap03.qxd 7/24/08 10:37 AM Page 125 3-1 QuickLab 3-1 QuickLab Accelerating a Cart Problem If you pull an object with a force, how do force and mass affect the acceleration of the object? loaded cart spring scale Materials dynamics cart with hook two 200-g standard masses one 1-kg standard mass spring scale (0\u20135 N) smooth, flat surface (about 1.5 m long) Procedure 1 Place the 200-g standard masses on the cart and attach the spring scale to the hook on the cart. Figure 3.2 Questions 1. Why do you think it was difficult to apply a constant force when pulling the cart each time? 2. Describe how the acceleration of the cart changed from what it was in step 2 when (a) you used the 1-kg standard mass instead of the 2 Pull the spring scale so that the cart starts to 200-g masses, accelerate forward (Figure 3.2). Make sure that the force reading on the spring scale is 2 N and that the force remains as constant as possible while pulling the cart. Observe the acceleration of the cart. 3 Replace the 200-g masses on the cart with the 1-kg standard mass. Then pull the cart, applying the same force you used in step 2. Observe the acceleration of the cart. 4 Remove all the objects from the cart. Then pull the cart, applying the same force you used in step 2. Observe the acceleration of the cart. 5 Repeat step 4 but this time pull with a force of 1 N. (b) you removed all the objects from the cart, (c) you decreased the pulling force to 1 N, and (d) you increased the pulling force to 3 N. 3. What force was required to start the cart moving in step 7? 4. Suppose, instead of hooking a spring scale to the cart in steps 2 to 4, you gave the cart a push of the same magnitude each time. (a) Which cart would you expect to travel the farthest distance? (b) Which cart would you expect to slow down sooner? 6 Repeat step 4 but this time pull with a force of 3 N. (c) What force do you think makes the cart 7 Repeat step 4 but now only pull with just enough force to start the cart moving. Measure the force reading on the spring scale. eventually come to a stop? Think About It 1. Describe the motion of a large rocket during liftoff using the concept of force. Include diagrams in your explanation. 2. Is a plane during takeoff accelerating or moving with constant velocity? Explain in words and with diagrams. Discuss your answers in a small group and record them for later reference. As you complete each section of this chapter, review your answers to these questions. Note any changes to your ideas. Chapter 3 Forces can change velocity. 125 03-Phys20-Chap03.qxd 7/24/08 10:37 AM Page 126 info BIT Statics is a branch of dynamics that deals with the forces acting on stationary objects. Architecture is primarily a practical application of statics. dynamics: branch of mechanics dealing with the cause of motion 3.1 The Nature of Force The Petronas Twin Towers in Kuala Lumpur, Malaysia, are currently the world\u2019s tallest twin towers. Including the spire on top, each tower measures 452 m above street level. To allow for easier movement of people within the building, architects designed a bridge to link each tower at the 41st floor. What is interesting is that this bridge is not stationary. In order for the bridge to not collapse, it must move with the towers as they sway in the wind (Figure 3.3). In Unit I, you learned that kinematics describes the motion of an object without considering the cause. When designing a structure, the kinematics quantities that an architect considers are displacement, velocity, and acceleration. But to predict how and explain why a structure moves, an architect must understand dynamics. Dynamics deals with the effects of forces on objects. Structures such as bridges and buildings are required to either remain stationary or move in appropriate ways, depending on the design, so that they are safe to use. Architects must determine all the forces that act at critical points of the structure. If the forces along a particular direction do not balance, acceleration will occur. Before you can predict or explain the motion of an object, it is important to first understand what a force", " is and how to measure and calculate the sum of all forces acting on an object. Figure 3.3 The design of tall buildings involves understanding forces. Towering buildings are susceptible to movement from the wind. 126 Unit II Dynamics 03-Phys20-Chap03.qxd 7/24/08 10:37 AM Page 127 Force Is a Vector Quantity You experience a force when you push or pull an object. A push or a pull can have different magnitudes and can be in different directions. For this reason, force is a vector quantity. In general, any force acting on an object can change the shape and/or velocity of the object (Figure 3.4). If you want to deform an object yet keep it stationary, at least two forces must be present. (a) force exerted by hand on ball (b) force exerted by hand on ball force exerted by hand on ball (c) force: a quantity measuring a push or a pull on an object force exerted by table on ball distorting starting stopping (d) force exerted by hand on ball (e) force exerted by hand on ball (f) force exerted by hand on ball info BIT One newton is roughly equal to the magnitude of the weight of a medium-sized apple or two golf balls. speeding up slowing down changing the direction of motion Figure 3.4 Different forces acting on a ball change either the shape or the motion of the ball. (a) The deformation of the ball is caused by both the hand and the table applying opposing forces on the ball. (b)\u2013(f) The motion of the ball is changed, depending on the magnitude and the direction of the force applied by the hand. The symbol of force is F and the SI unit for force is the newton (N), named in honour of physicist Isaac Newton (1642\u20131727). One newton is equal to one kilogram-metre per second squared (1 kgm/s2), which is the force required to move a 1-kg object with an acceleration of 1 m/s2. The direction of a force is described using reference coordinates that you choose for a particular situation. You may use [forward] or [backward], compass directions, or polar coordinates. When stating directions using polar coordinates, measure angles counterclockwise from the positive x-axis (Figure 3.5). (a) y 1 0 N 30.0\u00b0 x (b) y II III 330\u00b0 10 N I x IV Figure 3.5 Two vectors of the same magnitude but with different directions. (a) 10 N [30] (b) 10 N [330] Chapter 3 Forces can change velocity. 127 03-Phys20-Chap03.qxd 7/24/08 10:37 AM Page 128 Measuring Force One way you could measure forces involves using a calibrated spring scale. To measure the force of gravity acting on an object, attach the object to the end of a vertical spring and observe the stretch of the spring. The weight of an object is the force of gravity acting on the object. The symbol g. of weight is F When the spring stops stretching, the gravitational and elastic forces acting on the object balance each other (Figure 3.6). At this point, the elastic force is equal in magnitude to the weight of the object. So you can determine the magnitude of the weight of an object by reading the pointer position on a calibrated spring scale once the spring stops stretching. pointer elastic force vector Find out the relationship between the stretch of a spring and the weight of an object by doing 3-2 QuickLab. object 10 6 N gravitational force vector 6 N Figure 3.6 A spring scale is one type of instrument that can be used to measure forces. 3-2 QuickLab 3-2 QuickLab Measuring Force Using a Spring Scale Problem How is the amount of stretch of a calibrated spring related to the magnitude of the force acting on an object? 3 Hang additional objects from the spring, up to a total mass of 1000 g. Each time, record the mass and the magnitudes of the corresponding gravitational and elastic forces. Materials set of standard masses with hooks spring scale (0\u201310 N) Procedure 1 Hold the spring scale vertically and make sure the pointer reads zero when nothing is attached. 2 Gently suspend a 100-g standard mass from the spring. Use a table to record the mass and the magnitudes of the gravitational and elastic forces acting on the object. Questions 1. What was the reading on the spring scale when the 100-g mass was attached? 2. What happened to the stretch of the spring when the mass of the object attached to the spring scale (a) doubled? (b) tripled? (c) changed by a factor of 10? 3. Why is a spring scale ideal for measuring force? 128 Unit II Dynamics 03-Phys20-Chap03.qxd 7/24/08 10:37 AM Page 129 Representing Forces Using Free-Body Diagrams A free-body", " diagram is a powerful tool that can be used to analyze situations involving forces. This diagram is a sketch that shows the object by itself, isolated from all others with which it may be interacting. Only the force vectors exerted on the object are included and, in this physics course, the vectors are drawn with their tails meeting at the centre of the object (Figure 3.7). However, it does not necessarily mean that the centre of the object is where the forces act. When drawing a free-body diagram, it is important to show the reference coordinates that apply to the situation in a given problem. Remember to always include which directions you will choose as positive. Figure 3.8 shows the steps for drawing free-body diagrams. free-body diagram: vector diagram of an object in isolation showing all the forces acting on it Fs elastic force exerted by spring on mass + up down Fg gravitational force exerted by Earth on mass Identify the object for the free-body diagram Sketch the object in isolation with a dot at its centre Identify all the forces acting on the object Include in free-body diagram Choose appropriate reference coordinates and include which directions are positive Forces on other objects? Ignore Figure 3.7 The free-body diagram for the object in Figure 3.6 that is suspended from the spring scale. The spring scale is not included here because it is not the object being studied. N, S, E, W up, down, forward, backward x, y If possible, choose an appropriate scale Draw forces proportionally extending out from the dot at the centre of the object Figure 3.8 Flowchart summarizing the steps for drawing a free-body diagram Chapter 3 Forces can change velocity. 129 03-Phys20-Chap03.qxd 7/24/08 10:37 AM Page 130 normal force: force on an object that is perpendicular to a common contact surface Some Types of Forces There are different types of forces and scientists distinguish among them by giving these forces special names. When an object is in contact with another, the objects will have a common surface of contact, and the two objects will exert a normal force on each other. The normal force, F N, is a force that is perpendicular to this common surface. Depending on the situation, another force called friction, F f, may be present, and this force acts parallel to the common surface. The adjective \u201cnormal\u201d simply means perpendicular. Figure 3.9 (a) shows a book at rest on a level table. The normal force exerted by the table on the book is represented by the vector directed upward. If the table top were slanted and smooth as in Figure 3.9 (b), the normal force acting on the book would not be directed vertically upward. Instead, it would be slanted, but always perpendicular to the contact surface. (a) (b) FN Fg FN Ff Fg \u03b8 Figure 3.9 Forces acting on (a) a stationary book on a level table and on (b) a book accelerating down a smooth, slanted table. app, if, say, a A stationary object may experience an applied force, F person pushes against the object (Figure 3.10). In this case, the force of friction acting on the object will oppose the direction of impending motion. FN Ff Fapp Fg Figure 3.10 The forces acting on a stationary box Example 3.1 demonstrates how to draw a free-body diagram for a car experiencing different types of forces. In this situation, the normal force acting on the car is equal in magnitude to the weight of the car. 130 Unit II Dynamics 03-Phys20-Chap03.qxd 7/24/08 10:37 AM Page 131 Example 3.1 g, of 10 000 N [down] is coasting on a level road. A car with a weight, F N, of 10 000 N [up], a force of air The car experiences a normal force, F f, exerted air, of 2500 N [backward], and a force of friction, F resistance, F by the road on the tires of 500 N [backward]. Draw a free-body diagram for this situation. Analysis and Solution While the car is coasting, there is no forward force acting on the car. The free-body diagram shows four forces (Figure 3.11). 2000 N up down backward forward FN Ff Fair Fg Figure 3.11 Practice Problems 1. The driver in Example 3.1 sees a pedestrian and steps on the brakes. The force of air resistance is 2500 N [backward]. With the brakes engaged, the force of friction exerted on the car is 5000 N [backward]. Draw a free-body diagram for this situation. 2. A car moving at constant velocity starts to speed up. The weight of the car is 12 000 N [down]. The force of air resistance is 3600 N [backward]. With the engine engaged, the force of friction exerted by the road on the tires is 7200 N [", "forward]. Draw a free-body diagram for this situation. Answers 1. and 2. See page 898. Using Free-Body Diagrams to Find Net Force Free-body diagrams are very useful when you need to calculate the net force, F net, on an object. The net force is a vector sum of all the forces acting simultaneously on an object. The force vectors can be added using either a scale vector diagram or using components. net force: vector sum of all the forces acting simultaneously on an object Concept Check Can the net force on an object ever equal zero? Explain using an example and a free-body diagram. Chapter 3 Forces can change velocity. 131 03-Phys20-Chap03.qxd 7/24/08 10:37 AM Page 132 e SIM Learn how to use free-body diagrams to find the net force on an object. Follow the eSim links at www.pearsoned.ca/ school/physicssource. Adding Collinear Forces Vectors that are parallel are collinear, even if they have opposite directions. Example 3.2 demonstrates how to find the net force on an object given two collinear forces. In this example, a canoe is dragged using two ropes. The magnitude of the force F T exerted by a rope on an object at the point where the rope is attached to the object is called the tension in the rope. In this physics course, there are a few assumptions that you need to make about ropes or cables to simplify calculations. These assumptions and the corresponding inferences are listed in Table 3.1. Note that a \u201clight\u201d object means that it has negligible mass. Table 3.1 Assumptions about Ropes or Cables Assumption Inference The mass of the rope is negligible. The rope has a negligible thickness. The tension is uniform throughout the length of the rope. F T acts parallel to the rope and is directed away from the object to which the rope is attached. The rope is taut and does not stretch. Any objects attached to the rope will have the same magnitude of acceleration as the rope. Example 3.2 Two people, A and B, are dragging a canoe out of a lake onto a beach using light ropes (Figure 3.12). Each person applies a force of 60.0 N [forward] on the rope. The force of friction exerted by the beach on the canoe is 85.0 N [backward]. Starting with a free-body diagram, calculate the net force on the canoe. forward backward Practice Problems 1. Two dogs, A and B, are pulling a sled across a horizontal, snowy surface. Dog A exerts a force of 200 N [forward] and dog B a force of 150 N [forward]. The force of friction exerted by the snow on the sled is 60 N [backward]. The driver attempts to slow down the sled by pulling on it with a force of 100 N [backward]. Starting with a free-body diagram, calculate the net force on the sled. 132 Unit II Dynamics Given F F T1 T2 F f 60.0 N [forward] 60.0 N [forward] 85.0 N [backward] Figure 3.12 Required net force on canoe (F net) Analysis and Solution Draw a free-body diagram for the canoe (Figure 3.13). backward forward FT1 FT2 Ff Figure 3.13 03-Phys20-Chap03.qxd 7/24/08 10:37 AM Page 133 backward forward FT1 Fnet Figure 3.14 FT2 Ff Add the force vectors shown in the vector addition diagram (Figure 3.14). F f F Ff F T2 FT2 F T1 FT1 60.0 N 60.0 N (85.0 N) 60.0 N 60.0 N 85.0 N 35.0 N 35.0 N [forward] net Fnet F net Paraphrase The net force on the canoe is 35.0 N [forward]. Practice Problems 2. In a tractor pull, four tractors are connected by strong chains to a heavy load. The load is initially at rest. Tractors A and B pull with forces of 5000 N [E] and 4000 N [E] respectively. Tractors C and D pull with forces of 4500 N [W] and 3500 N [W] respectively. The magnitude of the force of friction exerted by the ground on the load is 1000 N. (a) Starting with a free-body diagram, calculate the net force on the load. (b) If the load is initially at rest, will it start moving? Explain. Answers 1. 190 N [forward] 2. (a) 0 N, (b) no Adding Non-Collinear Forces Example 3.3 demonstrates how to find the net force on an object if the forces acting on it are neither parallel nor perpendicular. By observing the relationship between the components of the force vectors, you can greatly", " simplify the calculations. Example 3.3 Refer to Example 3.2 on page 132. Person A thinks that if A and B each pull a rope forming an angle of 20.0\u00b0 with the bow, the net force on the canoe will be greater than in Example 3.2 (Figure 3.15). The canoe is being dragged along the beach using ropes that are parallel to the surface of the beach. Starting with a free-body diagram, calculate the net force on the canoe. Is person A\u2019s thinking correct? 20.0\u00b0 \u03b8 1 20.0\u00b0 \u03b8 2 Figure 3.15 Given F 60.0 N [along rope] 85.0 N [backward] T1 F f F T2 1 60.0 N [along rope] 2 20.0 y Required net force on canoe (F net) Analysis and Solution Draw a free-body diagram for the canoe (Figure 3.16). x Ff Figure 3.16 FT1 20.0\u00b0 20.0\u00b0 FT2 Chapter 3 Forces can change velocity. 133 03-Phys20-Chap03.qxd 7/24/08 10:37 AM Page 134 Practice Problems 1. Refer to Example 3.3. Suppose person A pulls a rope forming an angle of 40.0 with the bow and person B pulls a rope forming an angle of 20.0 with the bow. Each person applies a force of 60.0 N on the rope. The canoe and ropes are parallel to the surface of the beach. If the canoe is being dragged across a horizontal, frictionless surface, calculate the net force on the canoe. 2. Two people, A and B, are dragging a sled on a horizontal, icy surface with two light ropes. Person A applies a force of 65.0 N [30.0] on one rope. Person B applies a force of 70.0 N [300] on the other rope. The force of friction on the sled is negligible and the ropes are parallel to the icy surface. Calculate the net force on the sled. Answers 1. 1.04 102 N [10.0] 2. 95.5 N [343] Separate all forces into x and y components. T1 Vector F F F T2 f x component y component (60.0 N)(cos 20.0) (60.0 N)(cos 20.0) 85.0 N (60.0 N)(sin 20.0) (60.0 N)(sin 20.0) 0 From the chart, FT1y F nety Fnety F T1y FT1y So y FT2y. F T2y FT2y 0 N FT1 20.0\u00b0 Fnet 20.0\u00b0 FT2 Ff x Figure 3.17 Figure 3.17 Add the x components of all force vectors in the vector addition diagram (Figure 3.17). netx Fnetx F f Ff F T2x FT2x x direction F F T1x FT1x (60.0 N)(cos 20.0) (60.0 N)(cos 20.0) (85.0 N) (60.0 N)(cos 20.0) (60.0 N)(cos 20.0) 85.0 N 27.8 N 27.8 N [0] F net Paraphrase The net force is 27.8 N [0\u00b0]. Since the net force in Example 3.3 is less than that in Example 3.2, person A\u2019s thinking is incorrect. Applying Free-Body Diagrams to Objects in Equilibrium At the beginning of this section, you learned that architects consider the net force acting at critical points of a building or bridge in order to prevent structure failure. Example 3.4 demonstrates how free-body diagrams and the concept of net force apply to a stationary object. Stationary objects are examples of objects at equilibrium because the net force acting on them equals zero. Example 3.4 A store sign that experiences a downward gravitational force of 245 N is suspended as shown in Figure 3.18. T1 and F Calculate the forces F exerted at the point at which the sign is suspended. T2 Figure 3.18 \u03b8 1 55.0\u00b0 FT1 \u03b8 2 FT2 134 Unit II Dynamics 03-Phys20-Chap03.qxd 7/25/08 8:17 AM Page 135 Given F g 245 N [down] Required T2) T1 and F forces (F 1 55.0\u00b0 2 90.0\u00b0 Analysis and Solution Draw a free-body diagram for the sign (Figure 3.19). Resolve all forces into x and y components. T1 Vector F F F T2 g x component FT1 cos 55.0 FT2 0 y component FT1 sin 55.0 0 Fg Since the sign is at equilibrium, the net force in both the x and y directions is zero. Fnetx Fnety 0 N", " Add the x and y components of all force vectors separately. x direction F netx Fnetx F T1x FT1x F T2x FT2x 0 FT1 cos 55.0 F T2 FT2 FT1 cos 55.0 y direction F F T1y FT1y nety Fnety F g Fg 0 FT1 sin 55.0 ( Fg) 0 FT1 Fg sin 55.0 FT1 Fg sin 55.0 245 N sin 55.0 kgm s2 299.1 299 N Substitute FT1 into the equation for FT2. FT2 F T2 FT1 cos 55.0 (299.1 N)(cos 55.0) 171.6 N 172 N [0] From Figure 3.19, the direction of F T1 measured counterclockwise from the positive x-axis is 180 55.0 125. F T1 299 N [125] Paraphrase T2 is 172 N [0]. T1 is 299 N [125] and F F 55.0\u00b0 FT1 55.0\u00b0 y x FT2 Fg Figure 3.19 Practice Problems 1. If the sign in Example 3.4 had half the weight, how would the forces T2 compare? T1 and F F 2. Suppose the sign in Example 3.4 is suspended as shown in Figure 3.20. Calculate the forces FF T1 T2. and FF \u03b8 1 40.0\u00b0 FT1 \u03b8 2 FT2 Figure 3.20 3. Refer to the solutions to Example 3.4 and Practice Problem 2 above. (a) As 1 decreases, what happens T2? T1 and F to F (b) Explain why 1 can never equal zero. Answers 1. directions of F T1 and F T2 would remain the same as before, but the respective magnitudes would be half 3.81 102 N [140\u00b0] 2.92 102 N [0\u00b0] T1 2. F F 3. (a) F T2 T1 and F (b) magnitude of F T2 increase in value T1y must always equal Fg Chapter 3 Forces can change velocity. 135 03-Phys20-Chap03.qxd 7/24/08 10:37 AM Page 136 3.1 Check and Reflect 3.1 Check and Reflect Knowledge 1. (a) Explain what a force is, and state the SI unit of force. (b) Why is force a dynamics quantity and not a kinematics quantity? Applications 2. Sketch a free-body diagram for (a) a bicycle moving west on a level road with decreasing speed (b) a ball experiencing forces of 45 N [12.0], 60 N [100], and 80 N [280] simultaneously 3. The total weight of a biker and her motorbike is 1800 N [down]. With the engine engaged, the force of friction exerted by the road on the tires is 500 N [forward]. The air resistance acting on the biker and bike is 200 N [backward]. The normal force exerted by the road on the biker and bike is 1800 N [up]. (a) Consider the biker and bike as a single object. Draw a free-body diagram for this object. (b) Calculate the net force. 4. If two forces act on an object, state the angle between these forces that will result in the net force given below. Explain using sketches. (a) maximum net force (b) minimum net force 5. Two people, A and B, are pulling on a tree with ropes while person C is cutting the tree down. Person A applies a force of 80.0 N [45.0] on one rope. Person B applies a force of 90.0 N [345] on the other rope. Calculate the net force on the tree. 6. Three forces act simultaneously on an 2 is 80 N [115], 1 is 65 N [30.0], F object: F 3 is 105 N [235]. Calculate the net and F force acting on the object. Extensions 7. A sign that experiences a downward gravitational force of 245 N is suspended, as shown below. Calculate the forces F T2. and F T1 \u03b8 1 55.0\u00b0 \u03b8 2 55.0\u00b0 FT1 FT2 8. The blanket toss is a centuries-old hunting technique that the Inuit used to find herds of caribou. During the toss, several people would hold a hide taut while the hunter would jump up and down, much like on a trampoline, increasing the jump height each time. At the top of the jump, the hunter would rotate 360 looking for the herd. Draw a free-body diagram for a hunter of weight 700 N [down] while (a) standing at rest on the taut hide just before a jump (b) at the maximum jump height 9. Construct a flowchart to summarize how to add two", " or more non-collinear forces using components. Refer to Figure 3.8 on page 129 or Student References 4: Using Graphic Organizers on page 869. e TEST To check your understanding of forces, follow the eTest links at www.pearsoned.ca/school/ physicssource. 136 Unit II Dynamics 03-Phys20-Chap03.qxd 7/24/08 10:37 AM Page 137 info BIT The skeleton is an Olympic sledding sport believed to have originated in St. Moritz, Switzerland. 3.2 Newton\u2019s First Law At the 2006 Winter Olympics in Turin, Italy, Duff Gibson charged head-first down a 1.4-km icy track with 19 challenging curves. He reached speeds well over 125 km/h and ended up clinching the gold medal in the men\u2019s skeleton event (Figure 3.21). Gibson\u2019s success had a lot to do with understanding the physics of motion. Imagine an ideal situation in which no friction acts on the sled and no air resistance acts on the athlete. Scientist Galileo Galilei (1564\u20131642) thought that an object moving on a level surface would continue moving forever at constant speed and in the same direction if no external force acts on the object. If the object is initially stationary, then it will remain stationary, provided no external force acts on the object. In the real world, friction and air resistance are external forces that act on all moving objects. So an object that is in motion will eventually slow down to a stop, unless another force acts to compensate for both friction and air resistance. Galileo recognized the existence of friction, so he used thought experiments, as well as experiments with controlled variables, to understand motion. Thought experiments are theoretical, idealized situations that can be imagined but cannot be created in the real world. Figure 3.21 In the 2006 Winter Olympics, Calgary residents Duff Gibson (shown in photo) and Jeff Pain competed in the men\u2019s skeleton. Gibson edged Pain by 0.26 s to win the gold medal. Pain won the silver medal. Chapter 3 Forces can change velocity. 137 03-Phys20-Chap03.qxd 7/24/08 10:37 AM Page 138 inertia: property of an object that resists acceleration The Concept of Inertia Since ancient times, many thinkers attempted to understand how and why objects moved. But it took thousands of years before satisfactory explanations were developed that accounted for actual observations. A major stumbling block was not identifying friction as a force that exists in the real world. In his study of motion, Galileo realized that every object has inertia, a property that resists acceleration. A stationary curling stone on ice requires a force to start the stone moving. Once it is moving, the curling stone requires a force to stop it. To better understand the concept of inertia, try the challenges in 3-3 QuickLab. Concept Check Compare the inertia of an astronaut on Earth\u2019s surface, in orbit around Earth, and in outer space. Can an object ever have an inertia of zero? Explain. 3-3 QuickLab 3-3 QuickLab Challenges with Inertia Problem What role does inertia play in each of these challenges? 3 Set up a stack of loonies and the ruler as shown in Figure 3.22 (c). Use the ruler to remove the loonie at the very bottom without toppling the stack. Materials glass tumbler small, stiff piece of cardboard several loonies empty soft-drink bottle plastic hoop (about 2 cm wide, cut from a large plastic bottle) small crayon with flat ends ruler (thinner than thickness of one loonie) Procedure 1 Set up the tumbler, piece of cardboard, and loonie as shown in Figure 3.22 (a). Remove the cardboard so the loonie falls into the tumbler. 2 Set up the soft-drink bottle, plastic hoop, and a crayon as shown in Figure 3.22 (b). Remove the hoop so the crayon falls into the bottle. Caution: Keep your eyes well above the coin stack. Questions 1. (a) What method did you use to successfully perform each step in the procedure? (b) Apply the concept of inertia to explain why each of your procedures was effective. (a) (b) (c) Figure 3.22 138 Unit II Dynamics 03-Phys20-Chap03.qxd 7/24/08 10:37 AM Page 139 Newton\u2019s First Law and Its Applications Newton modified and extended Galileo\u2019s ideas about inertia in a law, called Newton\u2019s first law of motion (Figure 3.23). (a) Fnet 0 An object will continue either being at rest or moving at constant velocity unless acted upon by an external non-zero net force. v 0 If F net 0, then v 0. So if you want to change the motion of an object, a non-zero", " net (b) Fnet 0 force must act on the object. Concept Check The Voyager 1 and 2 space probes are in interstellar space. If the speed of Voyager 1 is 17 km/s and no external force acts on the probe, describe the motion of the probe (Figure 3.24). v constant Figure 3.23 If the net force on an object is zero, (a) a stationary object will remain at rest, and (b) an object in motion will continue moving at constant speed in the same direction. Figure 3.24 The Voyager planetary mission is NASA\u2019s most successful in terms of the number of scientific discoveries. Newton\u2019s First Law and Sliding on Ice Many winter sports involve a person sliding on ice. In the case of the skeleton event in the Winter Olympics, an athlete uses a sled to slide along a bobsled track. In hockey, a player uses skates to glide across the icy surface of a rink. Suppose a person on a sled is sliding along a horizontal, icy surface. If no external force acts on the person-sled system, then according to Newton\u2019s first law, the person would maintain the same speed. In fact, the person would not stop at all (Figure 3.25). In real life, the external forces of friction and air resistance act on all moving objects. So the system would eventually come to a stop. FN Fg Figure 3.25 Free-body diagram of a person-sled system sliding on a horizontal, frictionless surface Chapter 3 Forces can change velocity. 139 03-Phys20-Chap03.qxd 7/24/08 10:37 AM Page 140 Newton\u2019s First Law and Vehicle Safety Devices When you are in a moving car, you can feel the effects of your own inertia. If the car accelerates forward, you feel as if your body is being pushed back against the seat, because your body resists the increase in speed. If the car turns a corner, you feel as if your body is being pushed against the door, because your body resists the change in the direction of motion. If the car stops suddenly, you feel as if your body is being pushed forward, because your body resists the decrease in speed (Figure 3.26). (a) Inertia of motorist makes her feel like she is being pushed backward. direction of motion (c) Inertia of motorist makes her feel like she is being thrown forward. direction of acceleration of vehicle (b) Inertia of motorist makes her feel like she wants to continue moving in a straight line. direction of acceleration of vehicle Figure 3.26 The inertia of a motorist resists changes in the motion of a vehicle. (a) The vehicle is speeding up, (b) the vehicle is changing direction, and (c) the vehicle is stopping suddenly. When a car is rear-ended, a motorist\u2019s body moves forward suddenly as the car seat moves forward. However, the motorist\u2019s head resists moving forward. A properly adjusted headrest can minimize or prevent whiplash, an injury resulting from the rapid forward accelerations in a rear-end collision (Figure 3.27). Research shows that properly adjusted headrests can reduce the risk of whiplash-related injuries by as much as 40%. A poorly adjusted headrest, however, can actually worsen the effects of a rear-end collision on the neck and spine. When a car is involved in a head-on collision, the motorist continues to move forward. When a seat belt is worn properly, the forward motion of a motorist is safely restricted. If a head-on collision is violent enough, sodium azide undergoes a rapid chemical reaction to produce non-toxic nitrogen gas, which inflates an airbag. The inflated airbag provides a protective cushion to slow down the head and body of a motorist (Figure 3.28). 5\u201310 cm Figure 3.27 The ideal position for a headrest Figure 3.28 Airbag systems in vehicles are designed to deploy during vehicle collisions. 140 Unit II Dynamics 03-Phys20-Chap03.qxd 7/24/08 10:37 AM Page 141 3-4 Decision-Making Analysis 3-4 Decision-Making Analysis The Airbag Debate The Issue Front airbags were introduced in the 1990s to help prevent injury to motorists, especially during head-on collisions. Side airbags can also help. Yet front airbags have also been the cause of serious injury, even death. Furthermore, airbags add to the cost of a vehicle. Airbag advocates want both front and side airbags installed, better airbags, and greater control over their operation. Opponents want airbags removed from cars altogether. Background Information Airbags are connected to sensors that detect sudden changes in acceleration. The process of triggering and inflating an airbag occurs in about 40 ms. It is in that instant that arms and legs have been broken and children have been killed by the", " impact of a rapidly inflating airbag. Tragically, some of these deaths occurred during minor car accidents. Manufacturers have placed on/off switches for airbags on some vehicles, and some engineers are now developing \u201csmart\u201d airbags, which can detect the size of a motorist and the distance that person is sitting from an airbag. This information can then be used to adjust the speed at which the airbag inflates. Required Skills Defining the issue Developing assessment criteria Researching the issue Analyzing data and information Proposing a course of action Justifying the course of action Evaluating the decision Analysis 1. Identify the different stakeholders involved in the airbag controversy. 2. Research the development and safety history of airbags in cars. Research both front and side airbags. Consider head, torso, and knee airbags. Analyze your results, and identify any trends in your data. 3. Propose a solution to this issue, based on the trends you identified. 4. Propose possible changes to current airbag design that could address the issues of safety and cost. 5. Plan a class debate to argue the pros and cons of airbag use. Identify five stakeholders to represent each side in the debate. Support your position with research. Participants will be assessed on their research, organizational skills, debating skills, and attitudes toward learning. Concept Check Use Newton\u2019s first law to explain why (a) steel barriers usually separate the cab of a truck from the load (Figure 3.29), (b) trucks carrying tall loads navigate corners slowly, and (c) customers who order take-out drinks are provided with lids. e TECH Explore the motion of an object that experiences a net force of zero. Follow the eTECH links at www.pearsoned.ca/school/ physicssource. Figure 3.29 Chapter 3 Forces can change velocity. 141 03-Phys20-Chap03.qxd 7/24/08 10:37 AM Page 142 3.2 Check and Reflect 3.2 Check and Reflect Knowledge 6. Imagine you are the hockey coach for a 1. In your own words, state Newton\u2019s first law. 2. Give two examples, other than those in the text, that illustrate the property of inertia for both a stationary and a moving object. 3. Use Newton\u2019s first law to describe the motion of (a) a car that attempts to go around an icy curve too quickly, and (b) a lacrosse ball after leaving the lacrosse stick. 4. Apply Newton\u2019s first law and the concept of inertia to each of these situations. team of 10-year-olds. At a hockey practice, you ask the players to skate across the ice along the blue line (the line closest to the net), and shoot the puck into the empty net. Most of the shots miss the net. The faster the children skate, the more they miss. Newton\u2019s first law would help the players understand the problem, but a technical explanation might confuse them. (a) Create an explanation that would make sense to the 10-year-olds. (b) With the aid of a diagram, design a drill for the team that would help the players score in this type of situation. (a) How could you remove the newspaper without toppling the plastic beaker? Extensions 7. Research why parents use booster seats for young children using information from Safe Kids Canada. Summarize the \u201cseat belt test\u201d that determines whether a child is big enough to wear a seat belt without a booster seat. Begin your search at www.pearsoned.ca/school/physicssource. 8. During a sudden stop or if a motorist tries to adjust a seat belt suddenly, the seat belt locks into position. Research why seat belts lock. Write a brief report, including a diagram, of your findings. Begin your search at www.pearsoned.ca/school/physicssource. 9. Make a web diagram to summarize concepts and ideas associated with Newton\u2019s first law. Label the oval in the middle as \u201cNewton\u2019s first law.\u201d Cluster your words or ideas in other ovals around it. Connect these ovals to the central one and one another, where appropriate, with lines. See Student References 4: Using Graphic Organizers on page 869 for an example. e TEST To check your understanding of inertia and Newton\u2019s first law, follow the eTest links at www.pearsoned.ca/school/physicssource. (b) While moving at constant speed on a level, snowy surface, a snowmobiler throws a ball vertically upward. If the snowmobile continues moving at constant velocity, the ball returns to the driver. Why does the ball land ahead of the driver if the snowmobile stops? Assume that the air resistance acting on the ball is negligible. x y vs vby vbx Applications 5.", " Design an experiment using an air puck on an air table or spark air table to verify Newton\u2019s first law. Report your findings. Caution: A shock from a spark air table can be dangerous. 142 Unit II Dynamics 03-Phys20-Chap03.qxd 7/24/08 10:37 AM Page 143 3.3 Newton\u2019s Second Law If a speed skater wants to win a championship race, the cardiovascular system and leg muscles of the athlete must be in peak condition. The athlete must also know how to effectively apply forces to propel the body forward. World-class speed skaters such as Cindy Klassen know that maximizing the forward acceleration requires understanding the relationship among force, acceleration, and mass (Figure 3.30). Newton spent many years of his life trying to understand the motion of objects. After many experiments and carefully analyzing the ideas of Galileo and others, Newton eventually found a simple mathematical relationship that models the motion of an object. This relationship, known as Newton\u2019s second law, relates the net force acting on an object, the acceleration of the object, and its mass. Begin by doing 3-5 Inquiry Lab to find the relationship between the acceleration of an object and the net force acting on it. info BIT Cindy Klassen won a total of five medals during the 2006 Winter Olympics, a Canadian record, and is currently Canada\u2019s most decorated Olympian with six medals. Figure 3.30 Cindy Klassen, originally from Winnipeg but now a Calgary resident, won the gold medal in the 1500-m speed skating event in the 2006 Winter Olympics in Turin, Italy. Chapter 3 Forces can change velocity. 143 03-Phys20-Chap03.qxd 7/24/08 10:37 AM Page 144 3-5 Inquiry Lab 3-5 Inquiry Lab Relating Acceleration and Net Force Required Skills Initiating and Planning Performing and Recording Analyzing and Interpreting Communication and Teamwork 1 2 it a (t)2 relates The kinematics equation d v displacement d, initial velocity v and acceleration aa. If v dd 1 a (t)2. If you solve for acceleration, you get 2 2d a. Remember to use the scalar form of this equation ( )2 t i 0, the equation simplifies to i, time interval t, when solving for acceleration. Question How is the acceleration of an object related to the net force acting on the object? Hypothesis State a hypothesis relating acceleration and net force. Remember to write an \u201cif/then\u201d statement. Variables The variables involved in this lab are the mass of the system, the applied force acting on the system, friction acting on the system, time interval, the distance the system travels, and the acceleration of the system. Read the procedure and identify the controlled, manipulated, and responding variable(s). Materials and Equipment C-clamp dynamics cart three 100-g standard masses pulley smooth, flat surface (about 1.5 m) string (about 2 m) recording tape recording timer with power supply metre-stick masking tape graph paper e LAB For a probeware activity, go to www.pearsoned.ca/school/physicssource. 144 Unit II Dynamics Procedure 1 Copy Tables 3.2 and 3.3 from page 145 into your notebook. 2 Measure the mass of the cart. Record the value in Table 3.2. 3 Set up the recording timer, pulley, and cart loaded with three 100-g standard masses on a lab bench (Figure 3.31). Make a loop at each end of the string. Hook one loop to the end of the cart and let the other loop hang down over the pulley. recording tape recording timer 100 g 100 g string pulley 100 g 1 N Figure 3.31 Caution: Position a catcher person near the edge of the lab bench. Do not let the cart or hanging objects fall to the ground. 4 Attach a length of recording tape to the cart and thread it through the timer. 5 While holding the cart still, transfer one 100-g mass from the cart to the loop of string over the pulley and start the timer. When you release the cart, the hanging 100-g mass will exert a force of about 1 N on the system. Release the cart but stop it before it hits the pulley. Label the tape \u201ctrial 1; magnitude of F app 1 N.\u201d 6 Repeat steps 4 and 5 using the same cart but this time transfer another 100-g mass from the cart to the first hanging object. Label the tape \u201ctrial 2; magnitude of F app = 2 N.\u201d By transferring objects from the cart to the end of the string hanging over the pulley, the mass of the system remains constant but the net force acting on the system varies. 7 Repeat steps 4 and 5 using the same cart but this time transfer another 100-g mass from the cart to the two hanging masses. Label the tape", " \u201ctrial 3; magnitude of F app = 3 N.\u201d 03-Phys20-Chap03.qxd 7/24/08 10:37 AM Page 145 Analysis 1. Calculate the mass of the system, mT. Record the value in Table 3.2. 2. Using the tape labelled \u201ctrial 3,\u201d label the dot at the start t 0 and mark off a convenient time interval. If the 1 s, a time interval of 30 dot period of the timer is 0 6 1 s 0.5 s). Record spaces represents 0.5 s (30 0 6 the time interval in Table 3.2. 3. Measure the distance the system travelled during this time interval. Record this value in Table 3.2. 2 d to calculate the magnitude 4. Use the equation a ( )2 t of the acceleration of the system. Record the value in Tables 3.2 and 3.3. 5. Using the same time interval, repeat questions 3 and 4 for the tapes labelled \u201ctrial 1\u201d and \u201ctrial 2.\u201d 6. Why is it a good idea to choose the time interval using the tape labelled \u201ctrial 3\u201d? 7. Plot a graph of the magnitude of the acceleration vs. the magnitude of the applied force (Table 3.3). 8. (a) Describe the graph you drew in question 7. (b) Where does the graph intersect the x-axis? Why? What conditions would have to be present for it to pass through the origin? (c) For each trial, subtract the x-intercept from the applied force to find the net force. Record the values in Table 3.3. Then plot a graph of the magnitude of the acceleration vs. the magnitude of the net force. 9. When the magnitude of the net force acting on the system is doubled, what happens to the magnitude of the acceleration of the system? 10. What is the relationship between the magnitude of the acceleration and the magnitude of the net force? Write this relationship as a proportionality statement. Does this relationship agree with your hypothesis? Table 3.2 Mass, Time, Distance, and Acceleration Trial Mass of Cart mc (kg) Mass of Load on Cart ml (kg) Mass of Load Hanging over Pulley mh (kg) mT 1 2 3 0.200 0.100 0 0.100 0.200 0.300 Table 3.3 Force and Acceleration Total Mass mc ml (kg) mh Time Interval t (s) Distance d (m) Magnitude of a (m/s2) Trial 1 2 3 Magnitude of F app Acting Magnitude of F net Acting on System (N) on System (N) 1 2 3 Magnitude of a of System (m/s2) Chapter 3 Forces can change velocity. 145 03-Phys20-Chap03.qxd 7/24/08 10:37 AM Page 146 FN Fair Ff Fg Fapp Fapp Fnet Ff Fair Fnet Fapp Fair Ff Relating Acceleration and Net Force For the system in 3-5 Inquiry Lab, you discovered that there is a linear relationship between acceleration and net force. This relationship can be written as a proportionality statement: Horizontal Forces a Fnet This relationship applies to speed skating. In the short track relay event, a speed skater pushes the next teammate forward onto the track when it is the teammate\u2019s turn to start skating. While the teammate is being pushed, the horizontal forces acting on the skater are the applied push force, friction, and air resistance (Figure 3.32). As long as the applied push force is greater in magnitude than the sum of the force of friction acting on the skates and the air resistance acting on the skater\u2019s body, the net force on the teammate acts forward. Figure 3.32 (left) Free-body diagram showing the forces acting on a speed skater being pushed by a teammate in the short track relay event; (right) vector addition diagram for the horizontal forces. The harder the forward push, the greater will be the forward net force on the teammate (Figure 3.33). So the acceleration of the teammate will be greater. Note that the acceleration is in the same direction as the net force. Find out the relationship between the acceleration of an object and its mass by doing 3-6 Design a Lab. m constant a a Fnet Fnet Concept Check What is the difference between a net force and an applied force? Can a net force ever equal an applied force? Explain using an example and a free-body diagram. Figure 3.33 For the same mass, a greater net force results in a greater acceleration. 146 Unit II Dynamics 03-Phys20-Chap03.qxd 7/24/08 10:37 AM Page 147 3-6 Design a Lab 3-6 Design a Lab Relating Acceleration and Mass In this lab, you will investigate the relationship", " between acceleration and mass when the net force acting on the system is constant. The Question How is the acceleration of an object related to the mass of the object? e LAB For a probeware activity, go to www.pearsoned.ca/school/physicssource. Design and Conduct Your Investigation \u2022 State a hypothesis relating acceleration and mass. \u2022 Then use the set-up in Figure 3.31 on page 144 to design an experiment. List the materials you will use as well as a detailed procedure. Use the procedure and questions in 3-5 Inquiry Lab to help you. \u2022 Plot a graph of the magnitude of the acceleration vs. the mass of the system. Then plot a graph of the magnitude of the acceleration vs. the reciprocal of the mass of the system. \u2022 Analyze your data and form conclusions. How well did your results agree with your hypothesis? e TECH Explore how the net force on an object and its mass affect its acceleration. Follow the eTech links at www.pearsoned.ca/ school/physicssource. Relating Acceleration and Mass In 3-6 Design a Lab, you discovered that the relationship between acceleration and mass is non-linear. But if you plot acceleration as a function of the reciprocal of mass, you get a straight line. This shows that there is a linear relationship between acceleration and the reciprocal of mass. This relationship can be written as a proportionality statement: 1 a m In speed skating, evidence of this relationship is the different accelerations that two athletes of different mass have. Suppose athlete A has a mass of 60 kg and athlete B a mass of 90 kg. If the net force acting on A and B is the same, you would expect A to have a greater acceleration than B (Figure 3.34). This observation makes sense in terms of inertia, because the inertia of B resists the change in motion more so than the inertia of A does. In fact, you observed this relationship in 3-1 QuickLab when you compared the acceleration of an empty cart and a cart loaded with a 1-kg standard mass. m Fnet constant m a a A B Figure 3.34 For the same net force, a more massive person has a smaller acceleration than a less massive one does. Chapter 3 Forces can change velocity. 147 03-Phys20-Chap03.qxd 7/24/08 10:37 AM Page 148 Newton\u2019s Second Law and Inertial Mass 1 can be combined The proportionality statements a Fnet and a m F F net where k is the proportionality net or a k into one statement, a m m constant. Since 1 N is defined as the net force required to accelerate a 1-kg object at 1 m/s2, k is equal to 1. So F net. the equation can be written as a m This mathematical relationship is Newton\u2019s second law. When an external non-zero net force acts on an object, the object accelerates in the direction of the net force. The magnitude of the acceleration is directly proportional to the magnitude of the net force and inversely proportional to the mass of the object. The equation for Newton\u2019s second law is usually written with F net on the left side: F net ma The Concept of Inertial Mass All objects have mass, so all objects have inertia. From experience, it is more difficult to accelerate a curling stone than to accelerate a hockey puck (Figure 3.35). This means that the inertia of an object is related to its mass. The greater the mass of the object, the greater its inertia. The mass of an object in Newton\u2019s second law is determined by finding the ratio of a known net force acting on an object to the acceleration of the object. In other words, the mass is a measure of the inertia of an object. Because of this relationship, the mass in Newton\u2019s second law is called inertial mass, which indicates how the mass is measured. m Fnet a a constant Fnet m m Fnet Figure 3.35 If the acceleration of the curling stone and the hockey puck is the same, F net on the curling stone would be 95 times greater than F inertial mass of the curling stone is that much greater than the hockey puck. net on the hockey puck because the Concept Check What happens to the acceleration of an object if (a) the mass and net force both decrease by a factor of 4? (b) the mass and net force both increase by a factor of 4? (c) the mass increases by a factor of 4, but the net force decreases by the same factor? (d) the mass decreases by a factor of 4, and the net force is zero? e MATH Use technology to explore the relationship among Fnet, m, and a in Newton\u2019s second law. Follow the eMath links at www.pearsoned.ca/school/ physicssource to download sample data. inertial mass: mass measurement based on", " the ratio of a known net force on an object to the acceleration of the object 148 Unit II Dynamics 03-Phys20-Chap03.qxd 7/24/08 10:37 AM Page 149 Applying Newton\u2019s Second Law to Horizontal Motion Example 3.5 demonstrates how to use Newton\u2019s second law to predict the average acceleration of a lacrosse ball. In this situation, air resistance is assumed to be negligible to simplify the problem. Example 3.5 A lacrosse player exerts an average net horizontal force of 2.8 N [forward] on a 0.14-kg lacrosse ball while running with it in the net of his stick (Figure 3.36). Calculate the average horizontal acceleration of the ball while in contact with the lacrosse net. Given F 2.8 N [forward] net m 0.14 kg up down forward backward Figure 3.36 Required average horizontal acceleration of ball (a) Analysis and Solution The ball is not accelerating up or down. So in the vertical direction, F In the horizontal direction, the acceleration of the ball is in the direction of the net force. So use the scalar form of Newton\u2019s second law. 0 N. net Fnet ma F net a m N.8 2 0 4 kg.1 m 2.8 kg s2 0.14 kg 20 m/s2 a 20 m/s2 [forward] Practice Problems 1. The net force acting on a 6.0-kg grocery cart is 12 N [left]. Calculate the acceleration of the cart. 2. A net force of 34 N [forward] acts on a curling stone causing it to accelerate at 1.8 m/s2 [forward] on a frictionless icy surface. Calculate the mass of the curling stone. Answers 1. 2.0 m/s2 [left] 2. 19 kg Paraphrase The average horizontal acceleration of the lacrosse ball is 20 m/s2 [forward]. In Example 3.6, a free-body diagram is used to first help determine the net force acting on a canoe. Then Newton\u2019s second law is applied to predict the average acceleration of the canoe. Chapter 3 Forces can change velocity. 149 03-Phys20-Chap03.qxd 7/24/08 10:37 AM Page 150 Example 3.6 Two athletes on a team, A and B, are practising to compete in a canoe race (Figure 3.37). Athlete A has a mass of 70 kg, B a mass of 75 kg, and the canoe a mass of 20 kg. Athlete A can exert an average force of 400 N [forward] and B an average force of 420 N [forward] on the canoe using the paddles. During paddling, the magnitude of the water resistance on the canoe is 380 N. Calculate the initial acceleration of the canoe. up down forward backward Practice Problem 1. In the men\u2019s four-man bobsled event in the Winter Olympics, the maximum mass of a bobsled with two riders, a pilot, and a brakeman is 630 kg (Figure 3.39). Figure 3.37 Given mA F A F f 70 kg 400 N [forward] 380 N [backward] mB F B 75 kg 420 N [forward] mc 20 kg Required initial acceleration of canoe (a) Analysis and Solution The canoe and athletes are a system because they move together as a unit. Find the total mass of the system. mT mc mB mA 70 kg 75 kg 20 kg 165 kg Draw a free-body diagram for the system (Figure 3.38). Figure 3.39 During a practice run, riders A and B exert average forces of 1220 N and 1200 N [forward] respectively to accelerate a bobsled of mass 255 kg, a pilot of mass 98 kg, and a brakeman of mass 97 kg. Then they jump in for the challenging ride down a 1300-m course. During the pushing, the magnitude of the force of friction acting on the bobsled is 430 N. Calculate the average acceleration of the bobsled, pilot, and brakeman. up down forward FN FA backward Ff FB Fg Figure 3.38 FA FB Ff Fneth Answer 1. 4.4 m/s2 [forward] 150 Unit II Dynamics The system is not accelerating up or down. So in the vertical direction, Fnetv Write equations to find the net force on the system in both the horizontal and vertical directions. 0 N. 03-Phys20-Chap03.qxd 7/24/08 10:37 AM Page 151 neth Fneth horizontal direction F FF f Ff F B FB F A FA 400 N 420 N (380 N) 400 N 420 N 380 N 440 N FF g vertical direction FF F N 0 netv Fnetv Calculations in the vertical direction are not required in this problem. Apply Newton\u2019s second", " law to the horizontal direction. Fneth a mTa Fnet h mT 4 N 0 4 5 g k 6 1 440 kg m 2 s 165 kg 2.7 m/s2 a 2.7 m/s2 [forward] Paraphrase The canoe will have an initial acceleration of 2.7 m/s2 [forward]. Applying Newton\u2019s Second Law to Vertical Motion Example 3.7 demonstrates how to apply Newton\u2019s second law to determine the vertical acceleration of a person riding an elevator. To determine the net force on the elevator, use a free-body diagram. Example 3.7 A person and an elevator have a combined mass of 6.00 102 kg (Figure 3.40). The elevator cable exerts a force of 6.50 103 N [up] on the elevator. Find the acceleration of the person. Given 6.00 102 kg mT 6.50 103 N [up] F T g 9.81 m/s2 [down] Required acceleration of person (a ) up down Figure 3.40 Analysis and Solution Draw a free-body diagram for the person-elevator system [Figure 3.41 (a)]. The system is not accelerating left or right. 0 N. So in the horizontal direction, F net Since the person is standing on the elevator floor, both the person and the elevator will have the same vertical acceleration. up down FT Fg Figure 3.41 (a) Chapter 3 Forces can change velocity. 151 03-Phys20-Chap03.qxd 7/24/08 10:37 AM Page 152 For the vertical direction, write an equation to find the net force on the system [Figure 3.41 (b)]. F T F g F net Practice Problems 1. The person in Example 3.7 rides the same elevator when the elevator cable exerts a force of 5.20 103 N [up] on the elevator. Find the acceleration of the person. 2. An electric chain hoist in a garage exerts a force of 2.85 103 N [up] on an engine to remove it from a car. The acceleration of the engine is 1.50 m/s2 [up]. What is the mass of the engine? Answers 1. 1.14 m/s2 [down] 2. 252 kg Apply Newton\u2019s second law. ma F F T g ma FT Fg mg ma FT 9.81 m/s2.50 103 kg m 2 s 6.00 102 kg 1.02 m/s2 a 1.02 m/s2 [up] FT Fg Fnet 9.81 m/s2 Figure 3.41 (b) e WEB Air resistance is the frictional force that acts on all objects falling under the influence of gravity. Research how this force affects the maximum speed that an object reaches during its fall. Write a brief summary of your findings. Begin your search at www.pearsoned.ca/ school/physicssource. F f F g up down Figure 3.42 152 Unit II Dynamics Paraphrase The acceleration of the person is 1.02 m/s2 [up]. In Example 3.8, the force of gravity causes a skydiver to accelerate downward. Since the only motion under consideration is that of the skydiver and the direction of motion is down, it is convenient to choose down to be positive. Example 3.8 A skydiver is jumping out of an airplane. During the first few seconds of one jump, the parachute is unopened, and the magnitude of the air resistance acting on the skydiver is 250 N. The acceleration of the skydiver during this time is 5.96 m/s2 [down]. Calculate the mass of the skydiver. Given F f g 9.81 m/s2 [down] 250 N [up] Required mass of skydiver (m) a 5.96 m/s2 [down] Analysis and Solution Draw a free-body diagram for the skydiver (Figure 3.42). The skydiver is not accelerating left or right. So in the horizontal direction, F net For the vertical direction, write an equation to find the net force on the skydiver (Figure 3.43). F net 0 N. F g F f Fnet Fg Ff Figure 3.43 03-Phys20-Chap03.qxd 7/24/08 10:37 AM Page 153 Apply Newton\u2019s second law. maa F FF f g ma Fg Ff ma mg (250 N) mg 250 N 250 N mg ma m(g a) 250 N g a m 250 N 9.81 m/s2 5.96 m/s2 250 kgm s2 3.85 m s2 64.9 kg Paraphrase The mass of the skydiver is 64.9 kg. Practice Problems 1. A 55-kg female bungee jumper fast", "ens one end of the cord (made of elastic material) to her ankle and the other end to a bridge. Then she jumps off the bridge. As the cord is stretching, it exerts an elastic force directed up on her. Calculate her acceleration at the instant the cord exerts an elastic force of 825 N [up] on her. 2. During a bungee jump, the velocity of the 55-kg woman at the lowest point is zero and the cord stretches to its maximum. (a) Compare the direction of her acceleration at the lowest point of the jump to the part of the jump where she is accelerating due to gravity. (b) At this point, what is the direction of her acceleration? Applying Newton\u2019s Second Law to Two-Body Systems When two objects are connected by a light rope as in Example 3.9, applying a force on one of the objects will cause both objects to accelerate at the same rate and in the same direction. In other words, the applied force can be thought to act on a single object whose mass is equivalent to the total mass. Answers 1. 5.2 m/s2 [up] 2. (b) up Example 3.9 Two blocks of identical material are connected by a light rope on a level surface (Figure 3.44). An applied force of 55 N [right] causes the blocks to accelerate. While in motion, the magnitude of the force of friction on the block system is 44.1 N. Calculate the acceleration of the blocks. Given mA mB F app F f 20 kg 10 kg 55 N [right] 44.1 N [left] 20 kg 10 kg Figure 3.44 Required acceleration (a) Analysis and Solution The two blocks move together as a unit with the same acceleration. So consider the blocks to be a single object. Find the total mass of both blocks. mA mB 20 kg 10 kg 30 kg mT Practice Problems 1. Two buckets of nails are hung one above the other and are pulled up to a roof by a rope. Each bucket has a mass of 5.0 kg. The tension in the rope connecting the buckets is 60 N. Calculate the acceleration of the buckets. 2. Refer to Example 3.9. The force of friction on the 10-kg block has a magnitude of 14.7 N. (a) Calculate the tension in the rope connecting the two blocks. (b) Calculate the tension in the rope between the hand and the 10-kg block. Answers 1. 2.2 m/s2 [up] 2. (a) 37 N (b) 55 N Chapter 3 Forces can change velocity. 153 03-Phys20-Chap03.qxd 7/24/08 10:37 AM Page 154 Draw a free-body diagram for this single object (Figure 3.45). The single object is not accelerating up or down. 0 N. So in the vertical direction, F net Write equations to find the net force on the single object in both the horizontal and vertical directions. horizontal direction F FF F f app neth Apply Newton\u2019s second law. mTaa F F f mTa Fapp Ff Ff Fap p a m T app 55 N (44.1 N) 30 kg 0.36 m/s2 a 0.36 m/s2 [right] vertical direction F F N 0 F g netv Fnetv FN up right left down Calculations in the vertical direction are not required in this problem. Ff mT Fapp Fg Figure 3.45 Fapp Ff Fneth e SIM Apply Newton\u2019s second law to determine the motion of two blocks connected by a string over a pulley. Follow the eSim links at www.pearsoned.ca/ school/physicssource. Paraphrase The acceleration of the blocks is 0.36 m/s2 [right]. Applying Newton\u2019s Second Law to a Single Pulley System In Example 3.10, two objects are attached by a rope over a pulley. The objects, the rope, and the pulley form a system. You can assume that the rope has a negligible mass and thickness, and the rope does not stretch or break. To simplify calculations in this physics course, you need to also assume that a pulley has negligible mass and has no frictional forces acting on its axle(s). In Example 3.10, the external forces on the system are the gravitational forces acting on the hanging objects. The internal forces on the system are the forces along the string that pull on each object. The magnitude of both the internal and external forces acting on the system are not affected by the pulley. The pulley simply redirects the forces along the string that pulls on each object. Example 3.10 Two objects, A and B, are connected by a light rope over a light, frictionless pulley (Figure 3.46). A has a mass of 25 kg", " and B a mass of 35 kg. Determine the motion of each object once the objects are released. Given mA 25 kg g 9.81 m/s2 [down] mB 35 kg Required acceleration of each object (a A and a B) 25 kg 35 kg Figure 3.46 154 Unit II Dynamics 03-Phys20-Chap03.qxd 7/24/08 10:37 AM Page 155 Analysis and Solution The difference in mass between objects A and B will provide the net force that will accelerate both objects. Since mB expect mB to accelerate down while mA accelerates up. The rope has a negligible mass. So the tension in the rope is the same on both sides of the pulley. The rope does not stretch. So the magnitude of aa Find the total mass of both objects. A is equal to the magnitude of a mA, you would B. mT mB mA 25 kg 35 kg 60 kg Choose an equivalent system in terms of mT to analyze the motion [Figure 3.47 (a)]. left right FA mT FB Figure 3.47 (a) F net Fnet F A is equal to the gravitational force acting on mA, and F B is equal to the gravitational force acting on mB. Apply Newton\u2019s second law to find the net force acting on mT [Figure 3.47 (b)]. F F B A FA FB mAg mBg mA)g (mB (35 kg 25 kg)(9.81 m/s2) m 98.1 kg 2 s 98.1 N Figure 3.47 (b) Fnet FA FB Use the scalar form of Newton\u2019s second law to calculate the magnitude of the acceleration. Fnet mTa a F et n m T.1 N 8 9 g k 0 6 98.1 kg m 2 s 60 kg 1.6 m/s2 1.6 m/s2 [up] and a a A 1.6 m/s2 [down] B Paraphrase Object A will have an acceleration of 1.6 m/s2 [up] and object B will have an acceleration of 1.6 m/s2 [down]. Practice Problems 1. Determine the acceleration of the system shown in Example 3.10 for each situation below. State the direction of motion for each object. Express your answer in terms of g. (a) mA (b) mA (c) mA mB 1 3 2mB mB 2. Use the result of Example 3.10 and a free-body diagram to calculate the tension in the rope. 3. Draw a free-body diagram for each object in Example 3.10. Answers 1 g, mA moves up, mB moves down 1. (a) a 2 1 g, mA moves down, mB moves up (b) a 3 (c) a 0, neither mass moves 2. 2.9 102 N 3. FT a 25 kg FT 35 kg a Fg Fg Chapter 3 Forces can change velocity. 155 03-Phys20-Chap03.qxd 7/24/08 10:37 AM Page 156 Applying Newton\u2019s Second Law to a Two-Pulley System In Example 3.11, the system is made up of three objects (A, B, and C). As in Example 3.10, the difference in weight between objects B and C will provide the net force that will accelerate the system. Example 3.11 A 20-kg truck tire (object A) is lying on a horizontal, frictionless surface. The tire is attached to two light ropes that pass over light, frictionless pulleys to hanging pails B and C (Figure 3.48). Pail B has a mass of 8.0 kg and C a mass of 6.0 kg. Calculate the magnitude of the acceleration of the system. tire (A) pail B pail C Figure 3.48 Given mA 20 kg g 9.81 m/s2 [down] mB 8.0 kg mC 6.0 kg Required magnitude of the acceleration of the system (a) Analysis and Solution Since mB accelerates up. Since object A will accelerate left, choose left to be positive. mC, you would expect mB to accelerate down while mC The rope has a negligible mass and the rope does not stretch. So the magnitude of a equal to a C. Find the total mass of the system. A is equal to the magnitude of a B, which is also mT mC mB mA 20 kg 8.0 kg 6.0 kg 34 kg Choose an equivalent system in terms of mT to analyze the motion (Figure 3.49). left right Figure 3.49 FB mT FC 156 Unit II Dynamics 03-Phys20-Chap03.qxd 7/24/08 10:37 AM Page 157 F C is equal", " to B is equal to the gravitational force acting on mB, and F the gravitational force acting on mC. left right FB FC Fnet Figure 3.50 C Fnet Apply Newton\u2019s second law to find the net force acting on mT (Figure 3.50). F F F B net FB FC mBg mCg mC)g (mB (8.0 kg 6.0 kg)(9.81 m/s2) m 19.6 kg 2 s 19.6 N Use the scalar form of Newton\u2019s second law to calculate the magnitude of the acceleration. Fnet a mTa Fnet mT 19.6N 34 kg 19.6 kg m 2 s 34 kg 0.58 m/s2 Paraphrase The system will have an acceleration of magnitude 0.58 m/s2. Practice Problems 1. Calculate the acceleration of the tire in Example 3.11 if the mass of pail B is increased to 12 kg, without changing the mass of pail C. 2. If the tire in Example 3.11 is replaced by a car tire of mass 15 kg, calculate the acceleration of each object. Answers 1. 1.5 m/s2 [left] 2. (A) 0.68 m/s2 [left], (B) 0.68 m/s2 [down], (C) 0.68 m/s2 [up] Chapter 3 Forces can change velocity. 157 03-Phys20-Chap03.qxd 7/24/08 10:37 AM Page 158 3.3 Check and Reflect 3.3 Check and Reflect Knowledge 1. In your own words, state Newton\u2019s second law. app acting on an object 2. An applied force F of constant mass causes the object to accelerate. Sketch graphs to show the relationship between a and Fapp when friction is (a) present, and (b) absent. Refer to Student References 5.1: Graphing Techniques on pp. 872\u2013873. 3. Sketch a graph to show the relationship between the magnitude of acceleration and mass for constant net force. 4. Explain why vehicles with more powerful engines are able to accelerate faster. Applications 5. A dolphin experiences a force of 320 N [up] when it jumps out of the water. The acceleration of the dolphin is 2.6 m/s2 [up]. 8. Two boxes, A and B, are touching each other and are at rest on a horizontal, frictionless surface. Box A has a mass of 25 kg and box B a mass of 15 kg. A person applies a force of 30 N [right] on box A which, in turn, pushes on box B. Calculate the acceleration of the boxes. 9. A 4.0-kg oak block on a horizontal, rough oak surface is attached by a light string that passes over a light, frictionless pulley to a hanging 2.0-kg object. The magnitude of the force of friction on the 4.0-kg block is 11.8 N. 4.0 kg 2.0 kg (a) Calculate the acceleration of the system. (b) Calculate the tension in the string. (a) Calculate the mass of the dolphin. Extension (b) What would be the acceleration of the dolphin if it had the same strength but half the mass? 6. An ice hut used for winter fishing is resting on a level patch of snow. The mass of the hut is 80 kg. A wind exerts a horizontal force of 205 N on the hut, and causes it to accelerate. While in motion, the magnitude of the force of friction acting on the hut is 196 N. What is the acceleration of the hut? 7. Suppose the only horizontal forces acting on a 20-N object on a smooth table are 36 N [45] and 60 N [125]. (a) What is the net force acting on the object? (b) Calculate the acceleration of the object. 10. Summarize concepts and ideas associated with Newton\u2019s second law using a graphic organizer of your choice. See Student References 4: Using Graphic Organizers on pp. 869\u2013871 for examples of different graphic organizers. Make sure that the concepts and ideas are clearly presented and are linked appropriately. e TEST To check your understanding of Newton\u2019s second law, follow the eTest links at www.pearsoned.ca/school/physicssource. 158 Unit II Dynamics 03-Phys20-Chap03.qxd 7/24/08 10:37 AM Page 159 3.4 Newton\u2019s Third Law Volleyball is a sport that involves teamwork and players knowing how to apply forces to the ball to redirect it. When the velocity of the ball is large, a player will usually bump the ball to slow it down so that another player can redirect it over the net (Figure 3.51). At the instant the player", " bumps the ball, the ball exerts a large force on the player\u2019s arms, often causing sore arms. Immediately after the interaction, the ball bounces upward. To explain the motion of each object during and after this interaction requires an understanding of Newton\u2019s third law. Newton\u2019s first two laws describe the motion of an object or a system of objects in isolation. But to describe the motion of objects that are interacting, it is important to examine how the force exerted by one object on another results in a change of motion for both objects. Find out what happens when two initially stationary carts interact by doing 3-7 QuickLab. info BIT In order to walk, you must apply a force backward on the ground with one foot. The ground then pushes forward on that foot. Figure 3.51 Conrad Leinemann of Kelowna, British Columbia, bumps the ball while teammate Jody Holden of Shelburne, Nova Scotia, watches during the beach volleyball competition at the 1999 Pan Am Games in Winnipeg, Manitoba. Chapter 3 Forces can change velocity. 159 03-Phys20-Chap03.qxd 7/24/08 10:37 AM Page 160 3-7 QuickLab 3-7 QuickLab Exploding Carts Problem If a stationary cart exerts a net force on another identical cart, what will be the motion of both carts after the interaction? Materials dynamics cart with spring dynamics cart without spring 500-g standard mass Procedure 1 Note the position of the spring on the one cart, and standard mass barrier C-clamp cocked spring spring release Figure 3.52 Questions 1. What did you observe when you released the spring when the cart was initially at rest and not touching the other cart? how to cock and release the spring. 2. (a) What did you observe when you released the 2 Cock the spring and place the cart on the table. Release the spring. CAUTION: Do not cock the spring unless it is safely attached to the cart. Do not point the spring at anyone when releasing it. 3 Repeat step 2, this time making the cart with the spring touch the second cart (Figure 3.52). Release the spring. 4 Repeat step 3 but add a 500-g standard mass to one of the carts before releasing the spring. spring when one cart was touching the other cart? (b) What evidence do you have that two forces were present? (c) What evidence do you have that a force was exerted on each cart? (d) How do the magnitudes and directions of the two forces compare? 3. Compare and contrast the results from steps 3 and 4. action force: force initiated by object A on object B reaction force: force exerted by object B on object A Forces Always Exist in Pairs When two objects interact, two forces will always be involved. One force is the action force and the other is the reaction force. The important points to remember are that the reaction force always acts on a different object than the action force, and that the reaction force acts in the opposite direction. Concept Check Is it possible to have an action force without a reaction force? 160 Unit II Dynamics 03-Phys20-Chap03.qxd 7/24/08 10:37 AM Page 161 Newton\u2019s Third Law and Its Applications Newton found that the reaction force is equal in magnitude to the action force, but opposite in direction. This relationship is called Newton\u2019s third law of motion. If object A exerts a force on object B, then B exerts a force on A that is equal in magnitude and opposite in direction. F A on B F B on A Some people state Newton\u2019s third law as \u201cfor every action force, there is an equal and opposite reaction force.\u201d However, remembering Newton\u2019s third law this way does not emphasize that the action and reaction forces are acting on different objects (Figure 3.53). PHYSICS INSIGHT In order to show action-reaction forces, you must draw two free-body diagrams, one for each object. Faction force exerted by student on ground Freaction force exerted by ground on student Figure 3.53 The action force is the backward force that the student exerts on the ground. The reaction force is the forward force that the ground exerts on the student. Only the action-reaction pair are shown here for simplicity. Concept Check If the action force is equal in magnitude to the reaction force, how can there ever be an acceleration? Explain using an example and free-body diagrams. Action-Reaction Forces Acting on Objects in Contact Let\u2019s revisit the scenario of the volleyball player bumping the ball. At the instant that both the ball and the player\u2019s arms are in contact, the action force is the upward force that the player exerts on the ball. The reaction force is the downward force that the ball exerts on the player\u2019s arms. During the collision, the ball accelerates upward and the player\u2019s arms accelerate", " downward (Figure 3.54). e TECH Explore how a stranded astronaut can return to a spacecraft by throwing away tools. Follow the eTech links at www.pearsoned.ca/school/ physicssource. F action force exerted by player on ball force exerted by ball on player F reaction Figure 3.54 The action-reaction forces at collision time when a volleyball player bumps the ball Chapter 3 Forces can change velocity. 161 03-Phys20-Chap03.qxd 7/24/08 10:37 AM Page 162 A similar reasoning applies when a baseball bat strikes a baseball. The action force is the forward force that the bat exerts on the ball. The reaction force is the backward force that the ball exerts on the bat. During the collision, the ball accelerates forward and the bat slows down as it accelerates backward (Figure 3.55). Freaction force exerted by ball on bat Faction force exerted by bat on ball e WEB Fire hoses and extinguishers are difficult to control because their contents exit at high speed as they are redirected when putting out a fire. Research the operation and safe use of fire extinguishers. How do Newton\u2019s three laws apply to fire extinguishers? Interview an experienced firefighter. Write a brief summary of your findings. Begin your search at www.pearsoned.ca/school/ physicssource. Figure 3.55 The action-reaction forces at collision time when a baseball bat strikes a baseball Action-Reaction Forces Acting on Objects Not in Contact Sometimes an object can exert a force on another without actually touching the other object. This situation occurs when an object falls toward Earth\u2019s surface, or when a magnet is brought close to an iron nail. Action-reaction forces still exist in these interactions. When an apple falls toward the ground, the action force is the force of gravity that Earth exerts on the apple. The falling apple, in turn, exerts a reaction force upward on Earth. So while the apple is accelerating down, Earth is accelerating up (Figure 3.56). You see the acceleration of the apple but not of Earth because the inertial mass of the apple is far less than that of Earth. In fact, Earth does accelerate but at a negligible rate because the magnitude of the acceleration is inversely proportional to mass. force exerted by Earth on apple Faction force exerted by apple on Earth Freaction Figure 3.56 The action-reaction forces when an apple falls toward Earth\u2019s surface 162 Unit II Dynamics 03-Phys20-Chap03.qxd 7/24/08 10:37 AM Page 163 When a magnet is brought close to an iron nail, the action force is the magnetic force that the magnet exerts on the nail. The reaction force is the force that the nail exerts on the magnet. So the nail accelerates toward the magnet, and at the same time the magnet is accelerating toward the nail (Figure 3.57). Investigate the validity of Newton\u2019s third law by doing 3-8 QuickLab. Freaction Faction N force exerted by nail on magnet force exerted by magnet on nail S Figure 3.57 The action-reaction forces when a magnet is brought close to an iron nail 3-8 QuickLab 3-8 QuickLab Skateboard Interactions Problem How does Newton\u2019s third law apply to interactions involving skateboards? Materials two skateboards CAUTION: Wear a helmet and knee pads when doing this activity. Procedure 1 Choose a partner with a mass about the same as yours. 2 Sit on skateboards on a hard, level surface with your feet toward one another and touching (Figure 3.58). Figure 3.58 3 Give your partner a gentle push with your feet. Observe what happens to both skateboards. 4 Repeat steps 2 and 3 but this time, give your partner a harder push. Observe what happens to both skateboards. 5 Repeat steps 2 and 3 but this time, have you and your partner push simultaneously. Observe what happens to both skateboards. 6 Choose a partner with a significantly different mass than yours. 7 Repeat steps 2 to 5 with your new partner. 8 Sit on a skateboard near a wall. Then push against the wall. Observe the motion of your skateboard. Questions 1. Describe the motion of each skateboard when (a) you pushed a partner of equal mass, and (b) you pushed a partner of significantly different mass. 2. Compare and contrast the results from steps 4 and 5. 3. What happened to your skateboard when you pushed against the wall? 4. Explain each interaction in this activity using Newton\u2019s laws. Draw a sketch showing the action-reaction forces in each situation. Chapter 3 Forces can change velocity. 163 03-Phys20-Chap03.qxd 7/24/08 10:37 AM Page 164 Applying Newton\u2019s Third Law to Situations Involving Frictionless Surfaces In Example 3.12, an", " applied force acts on box A, causing all three boxes to accelerate. Newton\u2019s third law is used to calculate the force that box C exerts on box B. Example 3.12 Three boxes, A, B, and C, are positioned next to each other on a horizontal, frictionless surface (Figure 3.59). An applied force acting on box A causes all the boxes to accelerate at 1.5 m/s2 [right]. Calculate the force exerted by box C on box B. 10 kg 8.0 kg A B 5.0 kg C Figure 3.59 Practice Problems 1. For the situation in Example 3.12, calculate the force that box B exerts on box A. 2. For the situation in Example 3.9 Practice Problem 1 on page 153, calculate the applied force needed to lift both buckets up. Answers 1. 23 N [left] 2. 1.2 102 N [up] Given mA 8.0 kg a 1.5 m/s2 [right] mB 10 kg mC 5.0 kg Required C on B) force exerted by box C on box B (F FN Analysis and Solution Draw a free-body diagram for box C (Figure 3.60). Box C is not accelerating up or down. 0 N. So in the vertical direction, F net Write equations to find the net force on box C in both the horizontal and vertical directions. up left down right horizontal direction F neth Fneth F B on C FB on C F g vertical direction FF F netv N 0 Fnetv Calculations in the vertical direction are not required in this problem. FB on C mC Apply Newton\u2019s second law. FB on C F B on C m s2 mCa (5.0 kg)1.5 m 7.5 kg 2 s 7.5 N 7.5 N [right] Apply Newton\u2019s third law. F C on B F B on C 7.5 N [left] Fg Paraphrase The force exerted by box C on box B is 7.5 N [left]. Figure 3.60 164 Unit II Dynamics 03-Phys20-Chap03.qxd 7/24/08 10:37 AM Page 165 Applying Newton\u2019s Third Law to Situations Involving Friction In Example 3.13, a rough surface exerts a force of friction on two boxes. Newton\u2019s third law is used to calculate the force exerted by box B on box A in this situation. Example 3.13 Two boxes, A and B, of identical material but different mass are placed next to each other on a horizontal, rough surface (Figure 3.61). An applied force acting on box A causes both boxes to accelerate at 2.6 m/s2 [right]. If the magnitude of the force of friction on box B is 28.3 N, calculate the force exerted by box B on box A. Given mA F f on B 6.5 kg 28.3 N [left] mB 8.5 kg 6.5 kg 8.5 kg A B Figure 3.61 FN left up down right a 2.6 m/s2 [right] Ff on B mB FA on B Required B on A) force exerted by box B on box A (F Analysis and Solution Draw a free-body diagram for box B (Figure 3.62). Box B is not accelerating up or down. 0 N. So in the vertical direction, F net Write equations to find the net force on box B in both the horizontal and vertical directions. horizontal direction FF neth Fneth FF A on B FA on B FF f on B Ff on B F g vertical direction FF F netv N 0 Fnetv Calculations in the vertical direction are not required in this problem. Apply Newton\u2019s second law. mBa FA on B Ff on B FA on B F A on B mBa Ff on B (8.5 kg)(2.6 m/s2) (28.3 N) (8.5 kg)(2.6 m/s2) 28.3 N 50 N 50 N [right] Apply Newton\u2019s third law. F B on A F A on B 50 N [left] Paraphrase The force exerted by box B on box A is 50 N [left]. FA on B Fg Figure 3.62 Ff on B Fneth Practice Problem 1. To minimize the environmental impact of building a road through a forest, a logger uses a team of horses to drag two logs, A and B, from the cutting location to a nearby road. A light chain connects log A with a mass of 150 kg to the horses\u2019 harness. Log B with a mass of 250 kg is connected to log A by another light chain. (a) The horses can pull with a combined force of 2600 N. The ground exerts a force of friction of magnitude 2400 N", " on the logs. Calculate the acceleration of the logs. (b) If the force of friction on log A is 900 N, calculate the force exerted by log B on log A. Answer 1. (a) 0.500 m/s2 [forward] (b) 1.63 103 N [backward] Chapter 3 Forces can change velocity. 165 03-Phys20-Chap03.qxd 7/24/08 10:38 AM Page 166 Applying Newton\u2019s Second and Third Laws to Propeller Aircraft The acceleration of many devices such as propeller aircraft can be controlled in midair. To explain how these machines accelerate involves applying Newton\u2019s second and third laws. A propeller airplane can move through air because as the propeller rotates, it exerts an action force on the air, pushing the air backward. According to Newton\u2019s third law, the air, in turn, exerts a reaction force on the propeller, pushing the airplane forward (Figure 3.63). Propeller blades are slanted so that they scoop new air molecules during each revolution. The faster a propeller turns, the greater is the mass of air accelerated backward and, by Newton\u2019s second law, the force exerted by the air on the propeller increases. Faction force exerted by propeller on air Freaction force exerted by air on propeller Figure 3.63 The action-reaction forces when a propeller airplane is in flight THEN, NOW, AND FUTURE Wallace Rupert Turnbull (1870\u20131954) Wallace Rupert Turnbull was an aeronautical engineer interested in finding ways to make aircraft wings stable (Figure 3.64). In 1902, he built the first wind tunnel in Canada at Rothesay, New Brunswick, for his experiments on propeller design. In 1909, Turnbull was awarded a bronze medal from the Royal Aeronautical Society for his research on efficient propeller design. One of his major inventions was the variable-pitch propeller, which is still used on aircraft today. During takeoff, the angle of the blades is adjusted to scoop more air. Air moving at a high speed backward gives a plane thrust, which causes the plane to accelerate forward. Once a plane maintains a constant altitude, the blade angle, or pitch, is decreased, reducing fuel consumption. This allows greater payloads to be carried efficiently and safely through the sky. By 1925, Turnbull had perfected a propeller that used an electric motor to change its pitch. In 1927, the Canadian Air Force successfully tested the propeller at Borden, Ontario. Turnbull was later inducted into the Canadian Aviation Hall of Fame in 1977. Questions 1. Research the forces that act on airplanes in flight. Define these forces and compare them to forces already discussed in this chapter. 2. Explain how and where the forces on an airplane act to cause changes in its horizontal motion. Use Newton\u2019s laws and diagrams to support your explanations. Figure 3.64 Canadian inventor Wallace Rupert Turnbull 166 Unit II Dynamics 03-Phys20-Chap03.qxd 7/24/08 10:38 AM Page 167 Applying Newton\u2019s Third Law to Rockets The motion of rockets is a little different from that of propeller airplanes because a rocket does not have propellers that scoop air molecules. In fact, a rocket can accelerate in outer space where there is a vacuum. When a rocket engine is engaged, the highly combustible fuel burns at a tremendous rate. The action force of the exhaust gas leaving the rocket, according to Newton\u2019s third law, causes a reaction force that pushes against the rocket. It is the action force of the exhaust gas being directed backward that accelerates the rocket forward (Figure 3.65). That is why a rocket can accelerate in outer space. Test out Newton\u2019s third law with a toy rocket by doing 3-9 Design a Lab. Faction force exerted by rocket on exhaust gas Freaction force exerted by exhaust gas on rocket Figure 3.65 The action-reaction forces when a rocket is in flight 3-9 Design a Lab 3-9 Design a Lab Motion of a Toy Rocket Figure 3.66 shows a toy rocket partially filled with water about to be released from an air pump. The pump is used to add pressurized air into the rocket. air under pressure air pump water rocket release Figure 3.66 The Question What effect does increasing each of these quantities have on the motion of the rocket? \u2022 the amount of water inside the rocket \u2022 the air pressure inside the rocket Design and Conduct Your Investigation \u2022 State a hypothesis. Then design and conduct an experiment to test your hypothesis. Be sure to identify all variables and to control the appropriate ones. Caution: Never point the rocket at anyone. Perform this activity outside. \u2022 Compare the direction of motion of the water and the rocket when the rocket is released. \u2022 Explain the motion of the rocket, water, and air in terms of Newton\u2019s third law. Include sketches showing at least three action-reaction", " pairs of forces. \u2022 How well did your results agree with your hypothesis? Chapter 3 Forces can change velocity. 167 03-Phys20-Chap03.qxd 7/24/08 10:38 AM Page 168 3.4 Check and Reflect 3.4 Check and Reflect Knowledge 1. In your own words, state Newton\u2019s third law. 2. Explain why (a) a swimmer at the edge of a pool pushes backward on the wall in order to move forward, and (b) when a person in a canoe throws a package onto the shore, the canoe moves away from shore. 3. No matter how powerful a car engine is, a car cannot accelerate on an icy surface. Use Newton\u2019s third law and Figure 3.53 on page 161 to explain why. 4. State and sketch the action-reaction forces in each situation. (a) Water pushes sideways with a force of 600 N on the centreboard of a sailboat. (b) An object hanging at the end of a spring exerts a force of 30 N [down] on the spring. Applications 5. An object is resting on a level table. Are the normal force and the gravitational force acting on the object action-reaction forces? Explain your reasoning. 6. A vehicle pushes a car of lesser mass from rest, causing the car to accelerate on a rough dirt road. Sketch all the actionreaction forces in this situation. 7. Suppose you apply a force of 10 N to one spring scale. What is the reading on the other spring scale? What is the force exerted by the anchored spring scale on the wall? 8. Blocks X and Y are attached to each other by a light rope and can slide along a horizontal, frictionless surface. Block X has a mass of 10 kg and block Y a mass of 5.0 kg. An applied force of 36 N [right] acts on block X. Y X Fapp (a) Calculate the action-reaction forces the blocks exert on each other. (b) Suppose the magnitudes of the force of friction on blocks X and Y are 8.0 N and 4.0 N respectively. Calculate the action-reaction forces the blocks exert on each other. 9. A rectangular juice box has two holes punched near the bottom corners on opposite sides, and another hole at the top. The box is hung from a rigid support with a string. Predict what will happen if the box is filled with water through the top hole and the holes at the bottom are open. Use Newton\u2019s third law to explain your answer. Test your prediction. Cover the holes at the bottom with tape before filling the box with water. Then remove the tape to let the water out and observe the motion of the box. string hole water juice box holes? F 10 N water stream e TEST To check your understanding of Newton\u2019s third law, follow the eTest links at www.pearsoned.ca/school/physicssource. 168 Unit II Dynamics 03-Phys20-Chap03.qxd 7/24/08 10:38 AM Page 169 3.5 Friction Affects Motion Throughout this chapter, you encountered friction in all the lab activities and when solving several problems. Friction is a force that is present in almost all real-life situations. In some cases, friction is desirable while in other cases, friction reduces the effectiveness of mechanical systems. Without friction, you would not be able to walk. The wheels on a vehicle would have no traction on a road surface and the vehicle would not be able to move forward or backward. Parachutists would not be able to land safely (Figure 3.67). On the other hand, friction causes mechanical parts to seize and wear out, and mechanical energy to be converted to heat. For example, snowmobiles cannot move for long distances over bare ice. Instead, snowmobilers must detour periodically through snow to cool the moving parts not in contact with the ice. To determine the direction of the force of friction acting on an object, you need to first imagine the direction in which the object would move if there were no friction. The force of friction, then, opposes motion in that direction. info BIT Olympic cyclists now wear slipperier-than-skin suits with seams sown out of the airflow to reduce friction and improve race times by as much as 3 s. friction: force that opposes either the motion of an object or the direction the object would be moving in if there were no friction Figure 3.67 When a person falls in midair, the air resistance that acts on a parachute slows the fall. In this case, friction allows a parachutist to land without injury. Chapter 3 Forces can change velocity. 169 03-Phys20-Chap03.qxd 7/24/08 10:38 AM Page 170 In a sport such as curling, friction affects how far the stone will travel along the ice. Sweeping the ice in front of a", " moving stone reduces the force of friction acting on the stone (Figure 3.68). The result is that the stone slides farther. To better understand how the nature of a contact surface affects the force of friction acting on an object, do 3-10 QuickLab. Figure 3.68 Brad Gushue, from St. John\u2019s, Newfoundland, and his team won the gold medal in men\u2019s curling at the 2006 Winter Olympics in Turin, Italy. 3-10 QuickLab 3-10 QuickLab Friction Acting on a Loonie Problem What factors affect the ability of a loonie to start sliding? 4 Use a piece of tape to fasten the fine sandpaper on the textbook, sandy-side facing up. Repeat step 3. Materials two loonies textbook tape protractor coarse, medium, and fine sandpaper: a 10 cm 25 cm piece of each Procedure 1 Read the procedure and design a chart to record your results. 2 Place your textbook flat on a lab bench and place a loonie at one end of the book. 5 Repeat step 4 for the medium sandpaper and then for the coarse sandpaper. Carefully remove and save the sandpaper. 6 Repeat steps 2 and 3 but this time increase the mass (not the surface area) by stacking one loonie on top of the other. Use a piece of tape between the two loonies to fasten them together. Questions 1. How consistent were your results for each trial? 2. Explain how the angle needed to start the loonie sliding down the incline was affected by 3 Slowly raise this end of the textbook until the loonie starts to slide down the incline (Figure 3.69). \u2022 \u2022 the roughness of the contact surface the mass of the coins (number of stacked coins) in contact with the contact surface Use the protractor to measure the angle the textbook makes with the lab bench when the loonie first starts to slide. Repeat this step several times, and find the average of the angles. 170 Unit II Dynamics 3. Identify the controlled, manipulated, and responding variables in this activity. Figure 3.69 03-Phys20-Chap03.qxd 7/24/08 10:38 AM Page 171 Static Friction In 3-10 QuickLab, you discovered that the force of friction depends on the nature of the two surfaces in contact. If you drag an object on a smooth surface, the force of friction acting on the object is less than if you drag it on a rough or bumpy surface. If you drag a smooth block and a rough block on the same surface, the force of friction acting on each block will be different. Although there are different types of friction, the force of friction that acts on objects sliding across another surface is the main focus in this section. e SIM Learn how friction is created and how it affects the net force on an object. Follow the eSim links at www.pearsoned.ca/school/ physicssource. Suppose an object A (the desk) is in contact with another object B (the floor) as in Figure 3.70. The contact surface would be the horizontal surface at the bottom of each leg of the desk. app, such that F Now suppose that a force acts on the desk, say F app has a vertical component as well as a horizontal component. If the desk remains at rest, even though F app acts on it, then the net force on the desk is zero, F net 0 N. y x vd 0 Fappx \u03b8 Fapp Fappy Ffstatic Figure 3.70 An applied force F floor exerts a force of static friction on the bottom of each leg of the desk. app is acting on the desk at a downward angle. The In the x direction, Fnetx 0 N, which means that F appx must be balanced by another force. This balancing force is the force of static friction, F fstatic. The equation for the net force acting on the desk in the x direction would then be F F F netx fstatic appx Ffstatic Fappx Fnetx 0 Fappcos Ffstatic Fappcos Ffstatic So the direction of FF fstatic opposes the x component of the applied force acting on the desk. static friction: force exerted on an object at rest that prevents the object from sliding Chapter 3 Forces can change velocity. 171 03-Phys20-Chap03.qxd 7/24/08 10:38 AM Page 172 The Magnitude of Static Friction An important point about static friction is that its magnitude does not have a fixed value. Instead, it varies from zero to some maximum value. This maximum value is reached at the instant the object starts to move. If you push on a table with a force of ever-increasing magnitude, you will notice that the table remains at rest until you exceed a critical value. Because of Newton\u2019s second law, the magnitude of the force of static friction", " must increase as the applied force on the table increases, if the forces are to remain balanced. Static Friction on a Horizontal Surface Suppose the applied force acting on the desk in Figure 3.70 on page 171 is given. Example 3.14 demonstrates how to calculate the force of static friction by using a free-body diagram to help write the equation for the net force on the desk. Since F app acts at an angle to the surface of the desk, it is convenient to use Cartesian axes to solve this problem. Example 3.14 The magnitude of the applied force in Figure 3.71 is 165 N and 30.0. If the desk remains stationary, calculate the force of static friction acting on the desk. Given magnitude of F app 30.0 165 N Required fstatic) force of static friction (F y x x \u03b8 30.0\u00b0 Fapp Figure 3.71 Practice Problem 1. A mountain climber stops during the ascent of a mountain (Figure 3.73). Sketch all the forces acting on the climber, and where those forces are acting. Figure 3.73 172 Unit II Dynamics Analysis and Solution Draw a free-body diagram for the desk (Figure 3.72). Fappx Ffstatic 0 Fnetx Fappy 0 Fnety FN Fg y x FN Ffstatic Fappx \u03b8 30.0\u00b0 Fappy Fapp Fg Figure 3.72 03-Phys20-Chap03.qxd 7/24/08 10:38 AM Page 173 0 N in both Since the desk is not accelerating, F net the x and y directions. Write equations for the net force on the desk in both directions. x direction F FF netx appx Fappx Fnetx 0 Fappx y direction FF F nety N 0 Fnety Calculations in the y direction are not required in this problem. FF fstatic Ffstatic Ffstatic FF appy F g Ffstatic Fappx (165 N)(cos ) (165 N)(cos 30.0) 143 N F fstatic prevents the desk from sliding in the x direction. The negative value for Ffstatic indicates that the direction of FF fstatic is along the negative x-axis or [180]. F fstatic 143 N [180] Paraphrase The force of static friction acting on the desk is 143 N [180]. Answer 1. FT Ffstatic FN FN Ffstatic Fg Static Friction on an Incline If an object is at rest on an incline, the net force acting on the object 0 N. Let\u2019s first examine the forces acting on the object is zero, F net (Figure 3.74). FN Ffstatic Figure 3.74 (left) Free-body diagram for an object at rest on an incline; (below) vector addition diagrams for the and forces Fg \u03b8 Fg Fg \u03b8 Fg Forces Forces Fnet Fg Ffstatic Fnet 0 Ffstatic Fg Fg Fnet FN Fnet 0 FN Chapter 3 Forces can change velocity. 173 03-Phys20-Chap03.qxd 7/24/08 10:38 AM Page 174 When working with inclines, it is easier to rotate the reference coordinates so that motion along the incline is described as either uphill or downhill. This means that only the gravitational force needs to be resolved into components, one parallel to the incline F g and one perpendicular to the incline F g. Usually, uphill is chosen to be positive unless the object is accelerating downhill. In Figure 3.74 on page 173, if there were no friction, the component g would cause the object to accelerate down the incline. So for the object F fstatic) must be acting up the incline. to remain at rest, a balancing force (FF The equation for the net force acting on the object parallel to the incline would then be g F net F F fstatic Fnet Fg Ffstatic 0 Fg Ffstatic Ffstatic Fg g requires using the geometry of To determine the expression for F g is 90.0. Since a triangle. In Figure 3.75, the angle between F g and FF g is. g is 90.0, the angle between FF g and F g and F the angle between F 90\u00b0 \u03b8 \u03b8 Fg Fg \u03b8 Fg Figure 3.75 Diagram for an object at rest on an incline showing only the force of gravity vector resolved into components Since the object is not accelerating perpendicular to the incline, the equation for the net force acting on the object in this direction is net FF F N Fnet FN 0 FN FN Fg F g Fg Fg In 3-10 QuickLab, the sandpaper exerted a force of static friction on the loonie, preventing the coin from sliding down the incline. Example 3.15 demonstrates how to calculate the force of static friction acting on a loonie at", " rest on an incline of 25.0. 174 Unit II Dynamics 03-Phys20-Chap03.qxd 7/24/08 10:38 AM Page 175 Example 3.15 A loonie with a mass of 7.0 g is at rest on an incline of 25.0 (Figure 3.76). Calculate the force of static friction acting on the loonie. Given m 7.0 g 7.0 103 kg g 9.81 m/s2 Required fstatic) force of static friction (F 25.0 loonie \u03b8 25.0\u00b0 Figure 3.76 Analysis and Solution Draw a free-body diagram for the loonie (Figure 3.77). p u FN p u h ill h ill o w n d o w n d Ffstatic Fg Fg \u03b8 25.0\u00b0 \u03b8 25.0\u00b0 Fg Fg Ffstatic Fnet 0 Figure 3.77 Practice Problems 1. A loonie of mass 7.0 g is taped on top of a toonie of mass 7.3 g and the combination stays at rest on an incline of 30.0. Calculate the force of static friction acting on the face of the toonie in contact with the incline. 2. A loonie of mass 7.00 g is placed on the surface of a rough book. A force of static friction of magnitude 4.40 102 N acts on the coin. Calculate the maximum angle at which the book can be inclined before the loonie begins to slide. Answers 1. 7.0 102 N [uphill] 2. 39.8 0 N both Since the loonie is not accelerating, F net parallel and perpendicular to the incline. Write equations to find the net force on the loonie in both directions. direction net FF FF N Fnet 0 Calculations in the direction are not required in this problem. direction g FF net F FF fstatic Fnet Fg Ffstatic Fnet Fg Ffstatic Now, Fg mg sin So, Ffstatic F g 0 (mg sin ) mg sin (7.0 103 kg)(9.81 m/s2)(sin 25.0) 2.9 102 N F fstatic prevents the loonie from sliding downhill. The positive value for Ffstatic indicates that the direction of F fstatic is uphill. 2.9 102 N [uphill] F fstatic Paraphrase The force of static friction acting on the loonie is 2.9 102 N [uphill]. Chapter 3 Forces can change velocity. 175 03-Phys20-Chap03.qxd 7/24/08 10:38 AM Page 176 kinetic friction: force exerted on an object in motion that opposes the motion of the object as it slides on another object Kinetic Friction Suppose you apply a force to the desk in Figure 3.78 and the desk starts to slide across the floor at constant velocity. In this situation, the force of static friction is not able to balance the applied force, so motion occurs. Now the floor will exert a force of friction on the desk that opposes the direction of motion of the desk. This force is the force f kinetic. of kinetic friction, F Kinetic friction is present any time an object is sliding on another, whether or not another force acts on the sliding object. If you stop pushing the desk once it is in motion, the desk will coast and eventually stop. While the desk is sliding, the floor exerts a force of kinetic friction on the desk. This frictional force is directed backward, and causes the desk to eventually come to a stop. y x vd constant Fappx \u03b8 Fapp Fappy Ffkinetic Figure 3.78 The applied force F desk, causing the desk to slide. While the desk is in motion, the floor exerts a force of kinetic friction that opposes the motion of the desk. app overcomes the force of static friction acting on the The Direction of Kinetic Friction on an Incline If an object is on an incline and the object begins to slide, the surface of the incline exerts a force of kinetic friction on the object that opposes its motion. Whether the object is accelerating uphill or downhill, F net 0 N parallel to the incline. 176 Unit II Dynamics 03-Phys20-Chap03.qxd 7/24/08 10:38 AM Page 177 Accelerating Down an Incline Let\u2019s first consider the case where an object accelerates downhill g causes the object to accelerate (Figure 3.79). In this situation, F downhill. The force of kinetic friction acts to oppose the motion of the fkinetic is uphill as shown below. object. So F a FN Ffkinetic Fg Fg \u03b8 Fg \u03b8 Figure 3.79 (left) Free-body diagram for an object accelerating downhill; (below) vector addition diagrams for", " the and forces Forces Forces Fg Fnet Ffkinetic Fg Ffkinetic Fnet Fnet 0 Fg FN Fg Fnet Fnet FN 0 The equation for the net force acting on the object parallel to the incline is g F net F F fkinetic net becomes If you apply Newton\u2019s second law, the equation for F ma F g F fkinetic ma Fg Ffkinetic fkinetic acts uphill. For the object to In Figure 3.79, F g acts downhill and F accelerate downhill, the net force on the object, F net, is directed downhill. So the magnitude of F fkinetic. g must be greater than the magnitude of F Since the object is not accelerating perpendicular to the incline, the net force acting on the object in this direction is zero. The equation for the net force on the object in the perpendicular direction is net F F N Fnet FN 0 FN FN Fg F g Fg Fg Chapter 3 Forces can change velocity. 177 03-Phys20-Chap03.qxd 7/24/08 10:38 AM Page 178 Accelerating Up an Incline If an object is accelerating uphill, the force of kinetic friction acts app, g also acts downhill. A force, F downhill to oppose the motion. F must act uphill on the object that is great enough to overcome both F fkinetic g (Figure 3.80). and F Fapp Forces a FN Ffkinetic Fg Fg \u03b8 Fg \u03b8 Fapp Fg Ffkinetic Fg Ffkinetic Fapp Fnet Fnet \u2260 0 Fnet Fg FN Forces Fg Fnet FN Fnet 0 Figure 3.80 (left) Free-body diagram for an object accelerating uphill; (right) vector addition diagrams for the and forces The equation for the net force acting on the object parallel to the incline is net F F app g F F fkinetic net becomes If you apply Newton\u2019s second law, the equation for F ma F app ma Fapp g F F fkinetic Fg Ffkinetic app acts uphill. For fkinetic act downhill and F g and F In Figure 3.80, both F net is directed uphill. So the magnitude the object to accelerate uphill, F app must be greater than the sum of the magnitudes of F fkinetic. g and F of F Since the object is not accelerating perpendicular to the incline, the net force acting on the object in this direction is zero. The equation for the net force on the object in the perpendicular direction is net F F N Fnet FN 0 FN FF g Fg Fg FN Fg Concept Check What is the angle between the normal force and the force of friction? Is this angle always the same size? Explain your reasoning. 178 Unit II Dynamics 03-Phys20-Chap03.qxd 7/24/08 10:38 AM Page 179 Example 3.16 demonstrates how to calculate the acceleration of a block sliding down an incline. Since the direction of motion of the block is downhill, it is convenient to choose downhill to be positive. Example 3.16 A 3.5-kg block is sliding down an incline of 15.0 (Figure 3.81). The surface of the incline exerts a force of kinetic friction of magnitude 3.9 N on the block. Calculate the acceleration of the block. 3.5 kg 15.0\u00b0 Figure 3.81 Given m 3.5 kg magnitude of F fkinetic 3.9 N Required acceleration of block (a ) 15.0 g 9.81 m/s2 Analysis and Solution Draw a free-body diagram for the block (Figure 3.82). Since the block is accelerating downhill, F net to the incline, but F net 0 N perpendicular to the incline. 0 N parallel F Write equations to find the net force on the block in both directions. direction net F F N Fnet 0 Calculations in the direction are not required in this problem. direction g F net F F fkinetic Fnet Fg Ffkinetic ma Fg Ffkinetic Now, Fg mg sin So, ma mg sin (3.9 N) mg sin 3.9 N g FN Fg Ff kinetic \u03b8 15.0\u00b0 \u03b8 15.0\u00b0 Fg Fg Fg Ff kinetic Fnet 9 N 3. a g sin m 9 N. 3 (9.81 m/s2)(sin 15.0) g k 5. 3 1.4 m/s2 Figure 3.82 The positive value for a indicates that the direction of a is downhill. a 1.4 m/s2 [downhill] Paraphrase The acceleration of the block is 1.4 m/s2 [downhill]. Practice Problems 1. Determine the acceleration of the block in Example 3.16 if friction is not present.", " 2. A 55.0-kg skier is accelerating down a 35.0 slope. The magnitude of the skier\u2019s acceleration is 4.41 m/s2. Calculate the force of kinetic friction that the snowy surface exerts on the skis. Answers 1. 2.5 m/s2 [downhill] 2. 66.9 N [uphill] Chapter 3 Forces can change velocity. 179 03-Phys20-Chap03.qxd 7/24/08 10:38 AM Page 180 Comparing the Magnitudes of Static and Kinetic Friction The magnitude of the force of kinetic friction is never greater than the maximum magnitude of the force of static friction. Often, the magni- fstatic. fkinetic is less than the magnitude of F tude of F Figure 3.83 shows a graph of a situation where a person is applying very little force to an object during the first 2 s. Then the person begins to push harder, and at t 4 s, the object starts to move. The graph does not provide any information about the applied force after 4 s ( Ff Magnitude of the Force of Friction vs. Time static friction maximum value of static friction kinetic friction 0 4 8 12 16 20 Time t (s) Figure 3.83 The force of static friction increases up to a maximum value. Concept Check Explain why it makes sense that the magnitude of the force of kinetic friction does not exceed the maximum magnitude of the force of static friction. e WEB Leonardo da Vinci was as creative in science as he was in art. Research some of da Vinci\u2019s scientific ideas. Write a brief report of your findings, including diagrams where appropriate. Begin your search at www.pearsoned.ca/ school/physicssource. Determining the Magnitude of Frictional Forces Leonardo da Vinci (1452\u20131519) was one of the first people to experimentally determine two important relationships about friction. He discovered that for hard contact surfaces, the force of friction does not depend on the contact surface area. If you push a heavy box across the floor, the force of friction acting on the box is the same whether you push it on its bottom or on its side [Figure 3.84 (a) and (b)]. Da Vinci also discovered that the force of friction acting on an object depends on the normal force acting on that object. Find out what this relationship is by doing 3-11 Inquiry Lab. (a) (b) Figure 3.84 The force of friction acting on the box in each of these pictures is the same. For hard contact surfaces, the force of friction does not depend on contact surface area. 180 Unit II Dynamics 03-Phys20-Chap03.qxd 7/24/08 10:38 AM Page 181 3-11 Inquiry Lab 3-11 Inquiry Lab Relating Static Friction and the Normal Force Required Skills Initiating and Planning Performing and Recording Analyzing and Interpreting Communication and Teamwork Question What is the relationship between the maximum magnitude of the force of static friction and the magnitude of the normal force acting on an object? Hypothesis State a hypothesis relating the magnitude of F and the magnitude of F N. Write an \u201cif/then\u201d statement. fstatic Variables Read the procedure and identify the controlled, manipulated, and responding variable(s). Materials and Equipment balance wooden block with different face areas and a hook horizontal board spring scale, calibrated in newtons set of standard masses Procedure 1 Read the steps of the procedure and design a chart to record your results. 2 Measure the mass of the block using the balance. 3 Place the largest face of the block on the horizontal board. Attach the spring scale to the block. Pull with an ever-increasing horizontal force until the block just starts to move. Record this force, which is the maximum magnitude of the force of static friction. 4 Increase the mass of the block system by placing a standard mass on the upper surface. Record the total mass of the block with the standard mass. Use the spring scale to determine the maximum magnitude of the force of static friction for this system (Figure 3.85). Figure 3.85 5 Repeat step 4 three more times, increasing the added mass each time until you have five different masses and five corresponding maximum magnitudes of static friction. 6 Calculate the magnitude of the weight corresponding to each mass system. Record the magnitude of the normal force. 7 (a) Graph the maximum magnitude of the force of static friction as a function of the magnitude of the normal force. (b) Draw the line of best fit and calculate the slope of the graph. Analysis 1. Describe the graph you drew in step 7. 2. As the magnitude of the normal force acting on the mass system increased, what happened to the maximum magnitude of the force of static friction? 3. What is the relationship between the maximum magnitude of the force of static friction and the magnitude of the normal force? Write this as", " a proportionality statement. Does this relationship agree with your hypothesis? 4. On a level surface, how does the magnitude of the weight of an object affect the magnitude of the normal force and the maximum magnitude of the force of static friction? 5. Explain why adding a bag of sand to the trunk of a rear-wheel-drive car increases its traction. 6. Design and conduct an experiment to verify that contact surface area does not affect the maximum magnitude of the force of static friction for a sliding object. Identify the controlled, manipulated, and responding variables. Analyze your data and form conclusions. e LAB For a probeware activity, go to www.pearsoned.ca/school/physicssource. Chapter 3 Forces can change velocity. 181 03-Phys20-Chap03.qxd 7/24/08 10:38 AM Page 182 Project LINK How will the force of static friction acting on each vehicle in the Unit II Project on page 232 affect the stopping distance? How will the types of treads of the tires affect the force of static friction? coefficient of static friction: proportionality constant relating (Ffstatic)max and FN Coefficient of Static Friction In 3-11 Inquiry Lab, you found that the maximum magnitude of the force of static friction is directly proportional to the magnitude of the normal force. This proportionality can be written mathematically: (Ffstatic)max FN As an equation, the relationship is (Ffstatic)max sFN where s is a proportionality constant called the coefficient of static friction. Since the magnitude of the force of static friction can be anywhere from zero to some maximum value just before motion occurs, the general equation for the magnitude of the force of static friction must have an inequality sign. Ffstatic sFN for static friction Coefficient of Kinetic Friction Find out how the force of kinetic friction acting on an object is related to the normal force on that object by doing 3-12 Design a Lab. 3-12 Design a Lab 3-12 Design a Lab Relating Kinetic Friction and the Normal Force In this lab, you will investigate the relationship between the force of kinetic friction acting on an object and the normal force acting on that object. The Question What is the relationship between the magnitude of the force of kinetic friction and the magnitude of the normal force acting on an object? Design and Conduct Your Investigation \u2022 State a hypothesis relating the magnitudes of F \u2022 Then use the set-up in Figure 3.85 on page 181 to design an experiment. List the materials you will use as well as a detailed procedure. You will need to place objects of different mass on the block for each trial. and F N. fkinetic \u2022 For each trial, measure the force that must be applied to keep the block system moving at constant velocity. Then calculate the magnitude of the normal force. \u2022 Plot a graph of Ffkinetic \u2022 Analyze your data and form conclusions. as a function of FN. How well did your results agree with your hypothesis? e LAB For a probeware activity, go to www.pearsoned.ca/school/physicssource. From 3-12 Design a Lab, just as with static friction, the magnitude of kinetic friction is directly proportional to the magnitude of the normal force. This proportionality can be written mathematically: Ffkinetic FN 182 Unit II Dynamics 03-Phys20-Chap03.qxd 7/24/08 10:38 AM Page 183 As an equation, the relationship is kFN for kinetic friction Ffkinetic where k is a proportionality constant called the coefficient of kinetic friction. The force of kinetic friction has only one value, unlike the force of static friction which varies from zero to some maximum value. So the equation for the force of kinetic friction has an equal sign, not an inequality as does the equation for the force of static friction. Characteristics of Frictional Forces and Coefficients of Friction There are a few important points to keep in mind about the force of friction and the variables that affect its magnitude: \u2022 The equations for static friction and kinetic friction are not fundamental laws. Instead, they are approximations of experimental results. coefficient of kinetic friction: proportionality constant relating Ffkinetic and FN \u2022 The equations (Ffstatic)max kFN cannot be written sFN and Ffkinetic N are perpendicular f and F as vector equations because the vectors F to each other. s and \u2022 Both \u2022 For a given pair of surfaces, the coefficient of static friction is usually k are proportionality constants that have no units. greater than the coefficient of kinetic friction. \u2022 The coefficients of friction depend on the materials forming the contact surface, how smooth or rough a surface is, whether the surface is wet or dry, the temperature of the two contact surfaces, and other factors. Table 3.4 lists coefficients of friction between pairs of materials. Table 3.4 Approximate Coefficients of Friction for Some Materials Material Coefficient of Static Friction", " s Coefficient of Kinetic Friction k Copper on copper Steel on dry steel Steel on greased steel Dry oak on dry oak Rubber tire on dry asphalt Rubber tire on wet asphalt Rubber tire on dry concrete Rubber tire on wet concrete Rubber tire on ice Curling stone on ice Teflon\u2122 on Teflon\u2122 Waxed hickory skis on dry snow Waxed hickory skis on wet snow Synovial fluid on joint 1.6 0.41 0.15 0.5 1.2 0.6 1.0 0.7 0.006 0.003 0.04 0.06 0.20 0.01 1.0 0.38 0.09 0.3 0.8 0.5 0.7 0.5 0.005 0.002 0.04 0.04 0.14 0.01 Chapter 3 Forces can change velocity. 183 03-Phys20-Chap03.qxd 7/24/08 10:38 AM Page 184 How Friction Affects Motion Movable joints in the human body, such as elbows, knees, and hips, have membranes that produce a lubricating fluid called synovial fluid. Among other factors, the amount of synovial fluid and the smoothness of adjacent bone surfaces affect the coefficients of friction in synovial joints (Figure 3.86). The movement of synovial joints is very complicated because various biological processes are involved. In diseases such as arthritis, physical changes in joints and/or the presence of too much or too little synovial fluid affect the coefficients of friction. This, in turn, results in limited and painful movement. Figure 3.86 The amount of synovial fluid present depends on the need for a joint to move in a particular direction. info BIT Cars with wide tires experience no more friction than if the cars had narrow tires. Wider tires simply spread the weight of a vehicle over a greater surface area. This reduced pressure on the road reduces heating and tire wear. The effect of temperature on the coefficients of friction plays a role in drag racing. Drag racers often warm the tires on their cars by driving for a while. Tires that are warm stick to a racing track better than cooler tires. This increased coefficient of static friction increases traction and improves the acceleration of the car. The amount of moisture on a road surface, the temperature of the road surface and tires, and the type of tire treads are some factors that determine if a vehicle will skid. For a given tire, the coefficients of static and kinetic friction are greater on a dry road than if the same road is wet. The result is that vehicles are less likely to skid on a dry road than on a wet road. Tire treads and road surfaces also affect the force of friction acting on a vehicle (Figure 3.87). A ribbed tire increases friction acting sideways which helps a driver steer better. A lug tread provides more traction than a ribbed tire. Slicks, the tires on drag racing cars, have no treads at all to increase the surface area of the tire in contact with the racing track to better dissipate heat. (a) (b) (c) Figure 3.87 Different types of tires: (a) a ribbed tire with chains on it for better traction on snowy and icy surfaces, (b) a lug tread, and (c) slicks on a racing car Example 3.17 demonstrates how to use the coefficients of friction in Table 3.4 on page 183 to calculate the mass of a sled. Since the sled is at rest, the snowy surface exerts a force of static friction on the sled. 184 Unit II Dynamics 03-Phys20-Chap03.qxd 7/24/08 10:38 AM Page 185 Example 3.17 A sled with waxed hickory runners rests on a horizontal, dry snowy surface (Figure 3.88). Calculate the mass of the sled if the maximum force that can be applied to the sled before it starts moving is 46 N [forward]. Refer to Table 3.4 on page 183. up down forward backward Figure 3.88 Given F app s 46 N [forward] 0.06 from Table 3.4 (waxed hickory skis on dry snow) g 9.81 m/s2 [down] up Required mass of sled (m) backward down Analysis and Solution Draw a free-body diagram for the sled (Figure 3.89). Since the sled is not accelerating, F net in both the horizontal and vertical directions. Write equations to find the net force on the sled in both directions. 0 N FF neth Fneth horizontal direction F app Fapp 0 Fapp Fapp Fapp sFN FF fstatic Ffstatic Ffstatic ( Fapp sFN sFN) vertical direction F F netv N FN Fnetv 0 FN FN mg FF g Fg (mg) mg FN mg into the equation for Fapp. Fapp Substitute FN smg F", "app sg m 46 N m (0.06)9.81 s2 46 kgm s2 (0.06)9.81 m s2 8 101 kg Paraphrase The mass of the sled is 8 101 kg. Fapp Ffstatic 0 Fneth FN forward Ffstatic Fapp Fg Figure 3.89 Practice Problems 1. An applied force of 24 N [forward] causes a steel block to start moving across a horizontal, greased steel surface. Calculate the mass of the block. Refer to Table 3.4 on page 183. 2. Suppose the sled in Example 3.17 is resting on a horizontal, wet snowy surface. Would the sled move if the applied force is 125 N? Explain. Refer to Table 3.4 on page 183. Answers 1. 16 kg 2. no, F fstatic > F app In Example 3.18, a toboggan is initially at rest on a snowy hill. By knowing only the angle of the incline, it is possible to determine the coefficient of static friction for the toboggan on the hill. Chapter 3 Forces can change velocity. 185 03-Phys20-Chap03.qxd 7/24/08 10:38 AM Page 186 Example 3.18 A 50-kg toboggan is on a snowy hill. If the hill forms an angle of at least 20.0 with the horizontal, the toboggan just begins to slide downhill (Figure 3.90). Calculate the coefficient of static friction for the toboggan on the snow. Practice Problems 1. Calculate the coefficient of static friction if the toboggan in Example 3.18 is 20 kg and the hill forms an angle of 30.0 with the horizontal. 2. An 80-kg skier on a slushy surface starts moving down a hill forming an angle of at least 25.0 with the horizontal. (a) Determine the coefficient of static friction. (b) Calculate the maximum force of static friction on the skier. Given m 50 kg 20.0 Required coefficient of static friction ( s) Analysis and Solution Draw a free-body diagram for the toboggan [Figure 3.91 (a)]. When the angle of the incline is just enough for the toboggan to start moving, the surface of the incline is exerting the maximum magnitude of the force of static friction on the toboggan. u p p u h ill h ill o w n d o w n d \u03b8 20.0\u00b0 Figure 3.90 g 9.81 m/s2 u p p u h ill h ill o w n d o w n d Ffstatic FN Fg \u03b8 20.0\u00b0 \u03b8 20.0\u00b0 Fg Fg Answers 1. 0.58 2. (a) 0.47 (b) 3.3 102 N [uphill] info BIT The trigonometric function tan can be expressed in terms of sin and cos. in s tan s o c net 0 N in both the parallel and perpendicular Just before the toboggan begins to slide, F directions to the incline. Write equations to find the net force on the toboggan in both directions [Figure 3.91 (b)]. Figure 3.91 (a) direction net F F N Fnet FN 0 FN FN Fg F g Fg Fg direction g F net F F fstatic Fnet Fg Ffstatic 0 Fg Ffstatic Fg Ffstatic Fg mg sin Ffstatic Now, Fg mg cos So, FN (mg cos ) mg cos (mg sin ) mg sin mg sin mg cos into the last equation for the direction. sFN Figure 3.91 (b) Substitute FN Fg Ffstatic Fnet 0 s(mgcos ) mgsin s cos sin in s c s o tan tan 20.0 0.36 s Paraphrase The coefficient of static friction for the toboggan on the snow is 0.36. Note that angle of the hill. s does not depend on the mass of the toboggan, only on the 186 Unit II Dynamics 03-Phys20-Chap03.qxd 7/24/08 10:38 AM Page 187 Kinetic Friction Applies to Skidding Tires When the tires of a vehicle lock or if the tires skid on a road surface, the tires no longer rotate. Instead, the tires slide along the road surface. At the area where the tire and the road are in contact, the road surface exerts a force of kinetic friction directed backward on the tire (Figure 3.92). e TECH Explore how the initial velocity of a skidding car and its mass affect the braking distance. Follow the eTech links at www.pearsoned.ca/school/ physicssource. Ffkinetic Ffkinetic Figure 3.92 Diagram showing the force of kinetic friction acting on the tires of a skidding car Safety features on", " vehicles such as anti-lock braking systems are designed to prevent the wheels of a vehicle from locking when a driver steps on the brakes. If the wheels lock, the tires no longer rotate on the road surface and the vehicle ends up skidding. As long as the wheels continue to turn, the road surface exerts a force of static friction on the tires. Anti-lock braking systems maximize the force of static friction acting on the tires, allowing the driver of a vehicle to come to a more controlled stop. In Example 3.19, a lift truck is skidding on a concrete surface. Since the wheels are not rotating, the concrete surface is exerting a force of kinetic friction on the tires. e WEB Research how anti-lock braking systems work, and identify the strengths and weaknesses. Interview a car salesperson and/or an owner. Write a brief report of your findings, including diagrams where appropriate. Begin your search at www.pearsoned.ca/school/ physicssource. Example 3.19 A 1640-kg lift truck with rubber tires is skidding on wet concrete with all four wheels locked (Figure 3.93). Calculate the acceleration of the truck. Refer to Table 3.4 on page 183. backward up down forward Given m 1640 kg k g 9.81 m/s2 [down] 0.5 from Table 3.4 (rubber on wet concrete) Required acceleration of lift truck (a) Analysis and Solution Draw a free-body diagram for the lift truck (Figure 3.94). Since the lift truck is accelerating forward, F net direction, but F net vertical direction. Write equations to find the net force on the lift truck in both directions. 0 N in the horizontal 0 N in the Figure 3.93 up backward FN down forward Ffkinetic Fg FN Fg 0 Fnetv Figure 3.94 Chapter 3 Forces can change velocity. 187 03-Phys20-Chap03.qxd 7/24/08 10:38 AM Page 188 Practice Problems 1. An applied force of 450 N [forward] is needed to drag a 1000-kg crate at constant speed across a horizontal, rough floor. Calculate the coefficient of kinetic friction for the crate on the floor. 2. Calculate the force of kinetic friction if the truck in Example 3.19 is skidding downhill at constant speed on a hill forming an angle of 15.0 with the horizontal. Answers 1. 4.59 102 2. 4.16 103 N [uphill] horizontal direction F F fkinetic neth Ffkinetic Fneth ma Ffkinetic kFN vertical direction FF F netv N FN Fnetv 0 FN FN mg F g Fg (mg) mg FN mg into the equation for Ffkinetic. Substitute FN kmg ma a kg (0.5)9.81 5 m/s2 m s2 The negative value for a indicates that the direction of a backward. is a 5 m/s2 [backward] Paraphrase The acceleration of the truck is 5 m/s2 [backward]. Example 3.20 involves a snowmobile accelerating uphill while towing a sled. Since the motion of the sled is uphill, it is convenient to choose uphill to be positive. Example 3.20 A person wants to drag a 40-kg sled with a snowmobile up a snowy hill forming an angle of 25.0 (Figure 3.95). The coefficient of kinetic friction for the sled on the snow is 0.04. Calculate the force of the snowmobile on the sled if the sled accelerates at 2.5 m/s2 [uphill]. Figure 3.95 p u p u h ill h ill o w n d o w n d \u03b8 25.0\u00b0 Given m 40 kg g 9.81 m/s2 25.0 a 2.5 m/s2 [uphill] k 0.04 Required app) applied force on sled (F 188 Unit II Dynamics 03-Phys20-Chap03.qxd 7/24/08 10:38 AM Page 189 Analysis and Solution Draw a free-body diagram for the sled (Figure 3.96). FN Ffkinetic Figure 3.96 p u p u h ill h ill o w n d o w n d Fg \u03b8 25.0\u00b0 Fapp \u03b8 25.0\u00b0 Fg Fapp Fg Ffkinetic Fg Fnet Since the sled is accelerating uphill, 0 N parallel to the incline, but F 0 N perpendicular to the incline. F net net Write equations to find the net force on the sled in both directions. direction net FF F N Fnet FN 0 FN FN direction net FF F app Fnet Fapp ma Fapp Fapp F g Fg Fg Fg FF g F fkinetic Fg Ffkinetic Fg Ffkinetic ma Fg Ffkinetic Now, Fg mg cos", " So, FN (mg cos ) mg cos Also, Fg mg sin and kFN Ffkinetic Fapp ma (mg sin ) kFN) ( ma mg sin kFN Substitute FN mg cos into the equation for Fapp. Fapp ma mg sin ma mg(sin kmg cos kcos ) Practice Problems 1. A roofer is shingling a roof that rises 1.0 m vertically for every 2.0 m horizontally. The roofer is pulling one bundle of shingles (A) with a rope up the roof. Another rope connects bundle A to bundle B farther down the roof (Figure 3.97). mA 15 kg mB 15 kg 10 m 20 m Figure 3.97 Each of the two bundles of shingles has a mass of 15 kg. The coefficient of kinetic friction for the bundles on plywood sheeting is 0.50. (a) What force must the roofer exert up the roof to drag the bundles at constant speed? (b) Calculate the force exerted by bundle A on bundle B. (c) What total force would the roofer have to exert to accelerate both bundles at 2.0 m/s2 [up roof]? Answers 1. (a) 2.6 102 N [up roof] (b) 1.3 102 N [up roof] (c) 3.2 102 N [up roof] (40 kg)(2.5 m/s2) (40 kg)(9.81 m/s2) [(sin 25.0) (0.04)(cos 25.0)] 3 102 N app is uphill. The positive value for Fapp indicates that the direction of F F app 3 102 N [uphill] Paraphrase The snowmobile must apply a force of 3 102 N [uphill]. Chapter 3 Forces can change velocity. 189 03-Phys20-Chap03.qxd 7/24/08 10:38 AM Page 190 3.5 Check and Reflect 3.5 Check and Reflect Knowledge 1. In your own words, define friction. 2. What are some situations where friction is so small that it could be neglected? 3. Distinguish between static friction and kinetic friction. Applications 4. A pair of skis weigh 15 N [down]. Calculate the difference in the maximum force of static friction for the skis on a wet and dry snowy, horizontal surface. Refer to Table 3.4 on page 183. 5. A force of 31 N [forward] is needed to start an 8.0-kg steel slider moving along a horizontal steel rail. What is the coefficient of static friction? 6. A biker and his motorcycle have a weight of 2350 N [down]. Calculate the force of kinetic friction for the rubber tires and dry concrete if the motorcycle skids. Refer to Table 3.4 on page 183. 7. A 15-kg box is resting on a hill forming an angle with the horizontal. The coefficient of static friction for the box on the surface is 0.45. Calculate the maximum angle of the incline just before the box starts to move. 8. The coefficient of static friction for a wheelchair with its brakes engaged on a conveyor-type ramp is 0.10. The average mass of a person including the wheelchair is 85 kg. Determine if a ramp of 8.0\u00b0 with the horizontal will prevent motion. 9. A truck loaded with a crate of mass m is at rest on an incline forming an angle of 10.0 with the horizontal. The coefficient of static friction for the crate on the truck bed is 0.30. Find the maximum possible acceleration uphill for the truck before the crate begins to slip backward. 190 Unit II Dynamics 10. A loaded dogsled has a mass of 400 kg and is being pulled across a horizontal, packed snow surface at a velocity of 4.0 m/s [N]. Suddenly, the harness separates from the sled. If the coefficient of kinetic friction for the sled on the snow is 0.0500, how far will the sled coast before stopping? Extensions 11. A warehouse employee applies a force of 120 N [12.0] to accelerate a 35-kg wooden crate from rest across a wooden floor. The coefficient of kinetic friction for the crate on the floor is 0.30. How much time elapses from the time the employee starts to move the crate until it is moving at 1.2 m/s [0\u00b0]? 12. Make a Venn diagram to summarize the similarities and differences between static and kinetic friction. See Student References 4: Using Graphic Organizers on page 869 for an example. 13. Research how the type of tread on a tire affects the coefficients of static friction and kinetic friction given the same road surface. Find out what hydroplaning is and how tires are designed to minimize this problem. Write a brief report of your findings, including diagrams where appropriate. Begin your search at www.pearsoned.ca/school/physic", "ssource. 14. Design an experiment to determine the coefficients of static and kinetic friction for a curling stone on an icy surface. Perform the experiment at a local arena or club. Ask the icemaker to change the temperature of the ice, and repeat the experiment to determine if there is a difference in your values. Write a brief report of your findings. e TEST To check your understanding of friction and inclines, follow the eTest links at www.pearsoned.ca/school/physicssource. 03-Phys20-Chap03.qxd 7/24/08 10:38 AM Page 191 CHAPTER 3 SUMMARY Key Terms and Concepts dynamics force free-body diagram normal force net force inertia inertial mass action force Key Equations Newton\u2019s first law: F net Newton\u2019s second law: F net F A on B Newton\u2019s third law: 0 when v 0 ma F B on A reaction force friction static friction kinetic friction coefficient of static friction coefficient of kinetic friction Static friction: Ffstatic sFN Kinetic friction: Ffkinetic kFN Conceptual Overview The concept map below summarizes many of the concepts and equations in this chapter. Copy and complete the map to have a full summary of the chapter. Dynamics involves forces free-body diagrams vector sums Newton\u2019s first law Newton\u2019s second law Newton\u2019s third law are are measured in show all the forces is adding the involves inertia states that states that states that sliding friction two kinds kinetic equations so they have acting on determines the when magnitude and direction on Figure 3.98 Chapter 3 Forces can change velocity. 191 03-Phys20-Chap03.qxd 7/24/08 10:38 AM Page 192 CHAPTER 3 REVIEW Knowledge 1. (3.1, 3.3) Two people, A and B, are pushing a stalled 2000-kg truck along a level road. Person A exerts a force of 300 N [E]. Person B exerts a force of 350 N [E]. The magnitude of the force of friction on the truck is 550 N. Calculate the acceleration of the truck. 2. (3.2) Use a free-body diagram and Newton\u2019s first law to explain the motion of (a) a figure skater during a glide, and (b) a hockey puck during a cross-ice pass. Assume ice is frictionless. 3. (3.4) A transport truck pulls a trailer with a force of 1850 N [E]. What force does the trailer exert on the transport truck? 4. (3.5) An inexperienced driver, stuck in snow, tends to spin the car tires to increase the force of friction exerted by the snow on the tires. What advice would you give to the driver? Why? Applications 5. A device used to treat a leg injury is shown below. The pulley is attached to the foot, and the weight of the 3.0-kg object provides a tension force to each side of the pulley. The pulley is at rest to the pulley, because the foot applies a force F T2 in T1 and FF which is balanced by the forces FF the rope. The weight of the leg and foot is supported by the pillow. (a) Using a free-body diagram for the pulley, determine the force FF. (b) What will happen to the magnitude of F T2 decreases? Why? T1 and FF the angle between FF if support leg harness foot pulley F T1 F 45.0\u00ba 45.0\u00ba F T2 pillow 3.0 kg 192 Unit II Dynamics 6. Refer to Example 3.6 Practice Problem 1 on page 150. In a second practice run, the initial acceleration of the bobsled, pilot, and brakeman is 4.4 m/s2 [forward]. Rider A exerts an average force of magnitude 1200 N on the bobsled, and the force of friction decreases to 400 N. What average force does rider B exert? 7. During its ascent, a loaded jet of mass 4.0 105 kg is flying at constant velocity 20.0 above the horizontal. The engines of the plane provide a of 4.60 106 N [forward] to provide the thrust T [perpendicular to wings]. The air lift force L opposes the motion of the jet. resistance R Determine the magnitudes of L and R. L T 20.0\u00b0 Fg R 8. Suppose the force of kinetic friction on a sliding block of mass m is 2.5 N [backward]. What is the force of kinetic friction on the block if another block of mass 2m is placed on its upper surface? 9. A 1385-kg pickup truck hitched to a 453-kg trailer accelerates along a level road from a stoplight at 0.75 m/s2 [forward]. Ignore friction and air resistance. Calculate (a) the tension in the hitch, (b)", " the force of friction exerted by the road on the pickup truck to propel it forward, and (c) the force the trailer exerts on the pickup truck. 10. Two curlers, A and B, have masses of 50 kg and 80 kg respectively. Both players are standing on a carpet with shoes having Teflon\u2122 sliders. The carpet exerts a force of friction of 24.5 N [E] on player A and a force of friction of 39.2 N [W] on player B. Player A pushes player B with a force of 60 N [E]. (a) Calculate the net force acting on each player. (b) Calculate the acceleration of each player. 03-Phys20-Chap03.qxd 7/24/08 10:38 AM Page 193 11. A force of 15 N [S] moves a case of soft drinks weighing 40 N [down] across a level counter at constant velocity. Calculate the coefficient of kinetic friction for the case on the counter. 12. A 1450-kg car is towing a trailer of mass 454 kg. The force of air resistance on both vehicles is 7471 N [backward]. If the acceleration of both vehicles is 0.225 m/s2, what is the coefficient of static friction for the wheels on the ground? 13. Two bags of potatoes, m1 60 kg and m2 40 kg, are connected by a light rope that passes over a light, frictionless pulley. The pulley is suspended from the ceiling using a light spring scale. (a) What is the reading on the scale if the pulley is prevented from turning? (b) Draw a free-body diagram for each bag when the pulley is released. (i) Calculate the acceleration of the system. (ii) Calculate the tension in the rope. (c) What is the reading on the scale when the bags are accelerating? (d) Explain the difference between your answers in parts (a) and (c). spring scale 10 pulley m1 60 kg POTATOES POTATOES m2 40 kg POTATOES m1 60 kg 14. A drag racing car initially at rest can reach a speed of 320 km/h in 6.50 s. The wheels of the car can exert an average horizontal force of 1.52 104 N [backward] on the pavement. If the force of air resistance on the car is 5.2 103 N [backward], what is the mass of the car? 15. A tractor and tow truck have rubber tires on wet concrete. The tow truck drags the tractor at constant velocity while its brakes are locked. If the tow truck exerts a horizontal force of 1.0 104 N on the tractor, determine the mass of the tractor. Refer to Table 3.4 on page 183. 16. Create a problem involving an object of mass m on an incline of angle. Write a complete solution, including an explanation of how to resolve the gravitational force vector into components. 17. The table below shows some coefficients of static and kinetic friction ( tires in contact with various road surfaces. s and k) for rubber Coefficient Dry Concrete Wet Concrete Dry Asphalt Wet Asphalt s k 1.0 0.7 0.7 0.5 1.2 0.6 0.6 0.5 (a) Which road surface exerts more static friction on a rubber tire, dry concrete or dry asphalt? Explain. (b) On which surface does a car slide more easily, on wet concrete or on wet asphalt? Why? (c) On which surface will a moving car begin to slide more easily, on dry concrete or on dry asphalt? Why? (d) On which surface will a car with locked brakes slide a shorter distance, on dry concrete or on dry asphalt? Explain. Extensions 18. An 80-kg baseball player slides onto third base. The coefficient of kinetic friction for the player on the ground is 0.70. His speed at the start of the slide is 8.23 m/s. (a) Calculate his acceleration during the slide. (b) For how long does he slide until he stops? (c) Show that the time it takes the player to come to a stop is given by the equation t vi kg. Consolidate Your Understanding 19. Write a paragraph explaining the similarities and differences among Newton\u2019s three laws. Include an example that involves all three laws and explain how each law applies. Use the example to teach the laws to a student who has not studied dynamics. 20. Write a paragraph describing the differences between static and kinetic friction, and between the coefficients of static and kinetic friction. Include an example with a free-body diagram for each type of friction. Think About It Review your answers to the Think About It questions on page 125. How would you answer each question now? e TEST To check your understanding of forces and Newton\u2019s laws of motion,", " follow the eTest links at www.pearsoned.ca/school/physicssource. Chapter 3 Forces can change velocity. 193 04-Phys20-Chap04.qxd 7/24/08 10:59 AM Page 194 Gravity extends throughout the universe. Skydiving, hang-gliding, bungee jumping, and hot-air ballooning are just a few activities that take advantage of gravitational forces for a thrill (Figure 4.1). Gravitational force attracts all objects in the universe. It holds you to Earth, and Earth in its orbit around the Sun. In 1665, Isaac Newton began his study of gravity when he attempted to understand why the Moon orbits Earth. His theories led him to an understanding of the motion of planets and their moons in the solar system. Several centuries later, these theories led to the launch of satellites and the success of various space missions such as Mariner and Voyager. Gravity is one of the four basic forces of nature, called fundamental forces, that physicists think underlie all interactions in the universe. These forces are the gravitational force, the electromagnetic force, the weak nuclear force, and the strong nuclear force. In this chapter, you will investigate how gravity affects the motion of objects on Earth and on other planets, and how it affects the motion of satellites orbiting Earth. Figure 4.1 Understanding how forces and gravity affect motion determines how successful the design of a hot-air balloon will be and the best way to navigate it. C H A P T E R 4 Key Concepts In this chapter, you will learn about: gravitational force Newton\u2019s law of universal gravitation gravitational field Learning Outcomes When you have completed this chapter, you will be able to: Knowledge identify gravity as a fundamental force in nature describe Newton\u2019s law of universal gravitation explain the Cavendish experiment define and apply the concept of a gravitational field compare gravitational field strength and acceleration due to gravity predict the weight of objects on different planets Science, Technology, and Society explain that concepts, models, and theories help interpret observations and make predictions 194 Unit II 04-Phys20-Chap04.qxd 7/24/08 10:59 AM Page 195 4-1 QuickLab 4-1 QuickLab Falling Coins Problem Suppose you drop two coins of different shapes and sizes from the same height. How do the rates of both coins falling compare? Materials variety of coins (penny, nickel, dime, quarter, loonie, and toonie) ruler Styrofoam\u2122 disk (size of a loonie) Procedure 1 Choose any two different coins and place them at the edge of a table above an uncarpeted floor. 2 Using a ruler, push the coins off the table so they leave at the same time (Figure 4.2). 3 Listen carefully for the sounds of the coins as they hit the floor. 4 Repeat this activity with different combinations of two coins, including the loonie with the Styrofoam\u2122 disk. Record your observations. Figure 4.2 Questions 1. When the coins landed, how many sounds did you hear? Did all combinations of two coins give the same result? Explain. 2. If all the coins fall at the same rate, how many sounds would you expect to hear when they land? 3. How would the results compare if two coins were released at the same time from a greater height, such as 10 m? 4. How did the average acceleration of the loonie differ from that of the Styrofoam\u2122 disk? Explain why. Think About It 1. (a) What factors affect the weight of an astronaut during a rocket flight? (b) How does the astronaut\u2019s weight change? 2. What would be the motion of Earth if the Sun\u2019s gravity were zero? Assume that no other celestial bodies affect Earth. Discuss your answers in a small group and record them for later reference. As you complete each section of this chapter, review your answers to these questions. Note any changes to your ideas. Chapter 4 Gravity extends throughout the universe. 195 04-Phys20-Chap04.qxd 7/24/08 10:59 AM Page 196 info BIT Gravitational force is an attraction force only. There is no such thing as a repulsive gravitational force. gravitational force: attractive force between any two objects due to their masses 4.1 Gravitational Forces due to Earth One of Newton\u2019s great achievements was to identify the force that causes objects to fall near Earth\u2019s surface as the same force that causes the Moon to orbit Earth. He called this force \u201cgravity,\u201d and he reasoned that this force is present throughout the universe. g, is the force that attracts any two objects Gravitational force, F together. Although this force is the weakest fundamental force, you can feel its effect when you interact with an object of very large mass such as Earth. When you slide down a waterslide, you can feel the gravitational force exerted by Earth pulling you downward toward the", " bottom of the slide (Figure 4.3). But if you want to feel the gravitational force exerted by the person sitting next to you, you will not be able to sense anything because the magnitude of the force is so small. Figure 4.3 The attractive force between Earth and you is far greater than that between you and another person coming down a waterslide. 196 Unit II Dynamics 04-Phys20-Chap04.qxd 7/24/08 10:59 AM Page 197 Gravitational force is a force that always exists in pairs. This is another example of Newton\u2019s third law. If Earth exerts a gravitational force of magnitude 700 N on you, then you exert a gravitational force of magnitude 700 N on Earth. Earth attracts you and you attract Earth. The force you exert on Earth has a negligible effect because Earth\u2019s mass (5.97 1024 kg) is huge in comparison to yours. However, the gravitational force that Earth exerts on you causes a noticeable acceleration because of your relatively small mass. Concept Check Which diagram best represents the gravitational force acting on you and on Earth (Figure 4.4)? Explain your reasoning. (a) (b) (c) (d) Figure 4.4 The Concept of Weight In a vacuum, all objects near Earth\u2019s surface will fall with the same acceleration, no matter what the objects consist of or what their masses are. The only force acting on a falling object in a vacuum is the gravitational force exerted by Earth on the object (Figure 4.5). Suppose you analyze this situation using a free-body diagram and Newton\u2019s second law. Fg Figure 4.5 Free-body diagram for a falling object in a vacuum Chapter 4 Gravity extends throughout the universe. 197 04-Phys20-Chap04.qxd 7/24/08 10:59 AM Page 198 Fs 2 6 10 14 18 0 4 8 12 16 20 The equation for the net force acting on the falling object is F net F g ma F g Since the object is accelerating due to gravity, a g. So the equation for the net force becomes mg F g or F g mg The equation F mg is valid in general, because the gravitational force acting on an object is the same, whether or not the object is at rest or is moving. This equation relates the gravitational force acting on an object, the so-called weight of the object, to its mass. g Fg Figure 4.6 Diagram showing the forces acting on an object that is suspended from a spring scale weight: gravitational force exerted on an object by a celestial body 4-2 QuickLab 4-2 QuickLab One way to measure the magnitude of the weight of an object directly involves using a spring scale (Figure 4.6). When the object stops moving at the end of the spring, Earth exerts a downward gravitational force on the object while the spring exerts an upward elastic force of equal magnitude on the object. Find out what the relationship is between mass and gravitational force in the vicinity of your school by doing 4-2 QuickLab. Relating Mass and Weight Problem What is the relationship between the mass of an object and the local value of the gravitational force exerted on that object? Materials set of standard masses with hooks spring scale (010 N) graph paper Procedure 1 Design a procedure to determine the gravitational force acting on a set of standard masses (Figure 4.7). 2 Use a table to record the magnitude of the gravitational force and mass. Add another column in the table for the ratio of Fg to m. 3 Calculate the ratio of Fg to m for each standard mass. Calculate the average ratio for all masses. Include units 10 Figure 4.7 198 Unit II Dynamics 4 Plot a graph of Fg vs. m. Draw a line of best fit through the data points. Calculate the slope of the graph. Questions 1. What does the ratio of Fg to m represent? How constant is this value? 2. Describe the graph of Fg vs. m. How does the slope compare to the average ratio calculated in step 3? 3. Write an equation relating Fg and m. Use the symbol g for the proportionality constant. 4. (a) The Moon exerts a gravitational force that is 1 that exerted by Earth. If you did this about 6 activity on the Moon using an appropriate spring scale and the same standard masses, would \u2022 the graph be a straight line? \u2022 the slope be the same as before? \u2022 the line go through the origin? \u2022 the proportionality constant g be the same as before? (b) Why would a 05 N spring scale be more ideal to use on the Moon, rather than a 010 N spring scale? 04-Phys20-Chap04.qxd 7/24/08 10:59 AM Page 199 Gravitational Mass mg is determined by finding the ratio of The mass in the equation F g the gravitational force acting on an object to the acceleration due to g and g", " are in the same direction, the scalar form of the gravity. Since F equation may be used: Fg mg m Fg g A practical way to measure this mass involves using a balance. In Figure 4.8, an object of unknown mass (A) is placed on one pan and standard masses (B) are added to the other pan until both pans balance. This method involves comparing the weights of two objects: one unknown and the other known. Mass measured using the concept of weight is called gravitational mass. gravitational force exerted on standard masses gravitational force exerted on an unknown mass gravitational mass: mass measurement based on comparing the known weight of one object to the unknown weight of another object Figure 4.8 When both pans are balanced, the gravitational force acting on the unknown mass is equal to the gravitational force acting on the standard masses. If the balance in Figure 4.8 were moved to the Moon, the process of determining the gravitational mass of object A would be the same. However, the weight of A and B would be different from that at Earth\u2019s surface because the acceleration due to gravity at the Moon\u2019s surface, gMoon, is 1.62 m/s2 compared to 9.81 m/s2, the average value at Earth\u2019s surface. e SIM Explore how mass and weight are measured. Follow the eSim links at www.pearsoned.ca/school/ physicssource. FgA FgB mAgMoon mBgMoon mA mB But since both objects A and B experience the same value of gMoon, mB. So the gravitational mass of an object is the same whether the mA object is on Earth, the Moon, or anywhere else in the universe. Both gravitational mass and inertial mass are properties of an object that do not depend on the location of the object. Is Inertial Mass the Same as Gravitational Mass? You can determine the mass of an object by using either the concept of inertia or weight. Experiments since Newton\u2019s day have shown that for any object, the numerical value of its inertial mass is equal to its gravitational mass. Later, Albert Einstein (1879\u20131955) showed that inertial mass is actually equivalent to gravitational mass. So it does not matter whether you determine the mass of an object using inertia or weight, because the numerical value of the mass is the same. Chapter 4 Gravity extends throughout the universe. 199 04-Phys20-Chap04.qxd 7/24/08 10:59 AM Page 200 action-at-a-distance force: force that acts even if the objects involved are not touching field: three-dimensional region of influence gravitational field: region of influence surrounding any object that has mass Describing Gravitational Force as a Field Gravitational force is an example of a force that acts on objects whether or not they actually touch each other, even if the objects are in a vacuum. These forces are referred to as action-at-a-distance forces. In the 1800s, physicists introduced the concept of a field to explain action-at-a-distance forces. You encountered some fields in previous science courses when you worked with magnets. Imagine you are moving the north pole of a magnet close to the north pole of a fixed magnet. As you move the magnet closer to the fixed magnet, you can feel an increasing resistance. Somehow, the fixed magnet has created a region of influence in the space surrounding it. Physicists refer to a three-dimensional region where there is some type of an influence, whether it is an attraction or a repulsion, on a suitable object as a field. Since every object exerts a gravitational force in three dimensions, it influences the space around it (Figure 4.9). This region of influence is a gravitational field, and it is through this region that two objects interact. gravitational field of Earth gravitational field of the Moon Figure 4.9 This figure shows a two-dimensional representation of Earth\u2019s and the Moon\u2019s gravitational field. A gravitational field is three-dimensional and is directed toward the centre of the object. 200 Unit II Dynamics 04-Phys20-Chap04.qxd 7/24/08 10:59 AM Page 201 To determine the magnitude and direction of a gravitational field created by an object, you could use a test mass mtest. At different locations around the object, this test mass will experience a gravitational force that has a certain magnitude and direction. The direction of the gravitational force will be directed toward the centre of the object. Gravitational field strength is defined as the gravitational force per F unit mass, g g. If you release the test mass, it will accelerate toward m te st the object with an acceleration equal to g. Figure 4.10 shows how the magnitude of Earth\u2019s gravitational field strength changes as a test mass is moved farther away from Earth\u2019s centre. The farther the test mass is moved, the more significant is the decrease in g. In fact,", " the graph in Figure 4.10 shows an inverse square relationship: g 1 r 2 Magnitude of Gravitational Field Strength vs. Distance from Earth\u2019s Centre gravitational field strength: gravitational force per unit mass at a specific location 10.0 9.00 8.00 7.00 6.00 5.00 4.00 3.00 2.00 1.00 Distance from Earth\u2019s centre r (Earth radii) Figure 4.10 The magnitude of the gravitational field strength as a function of distance from Earth\u2019s centre Since force is measured in newtons and mass in kilograms, the units of gravitational field strength are newtons per kilogram, or N/kg. The ratio you determined in 4-2 QuickLab was the gravitational field strength at the vicinity of your school. Concept Check What happens to the magnitude of the gravitational field strength if (a) r decreases by a factor of four? (b) r increases by a factor of two? (c) mtest doubles? (d) mtest is halved? Chapter 4 Gravity extends throughout the universe. 201 04-Phys20-Chap04.qxd 7/24/08 10:59 AM Page 202 4.1 Check and Reflect 4.1 Check and Reflect Knowledge 1. Distinguish between mass and weight. Explain using an example and a diagram. 10. How could you distinguish between a 5.0-kg medicine ball and a basketball in outer space without looking at both objects? 2. Distinguish between inertial mass and Extensions gravitational mass. 3. In your own words, define gravitational acceleration and gravitational field strength. State the units and symbol for each quantity. 11. Visit a local fitness gymnasium. Find out how athletes use gravitational and elastic forces to improve their fitness. Is friction a help or a hindrance? Write a brief report of your findings. 4. In your own words, explain the concept of a gravitational field. Include an example of how a gravitational field affects another object. 12. List some occupations that might require a knowledge of gravitational field strength. Briefly explain how gravitational field strength applies to these occupations. 5. Why do physicists use the concept of 13. Complete the gathering grid below to summarize the similarities and differences among gravitational mass, inertial mass, and gravitational force. Gravitational Mass Inertial Mass Gravitational Force Definition SI Unit Measuring Instrument(s) How the Quantity Is Measured Factors It Depends On Variability with Location e TEST To check your understanding of gravitational force, weight, mass, and gravitational field strength, follow the eTest links at www.pearsoned.ca/school/physicssource. a field to describe gravity? 6. In a vacuum, a feather and a bowling ball are released from rest at the same time from the same height. Compare the time it takes for each object to fall. Explain your answer. Applications 7. The Moon exerts a gravitational force that is about 1 6 that exerted by Earth. Explain why the mass of an object measured on the Moon using a balance is the same as if the object were on Earth\u2019s surface. 8. Describe a situation where measuring the inertial mass of an object is easier than measuring its gravitational mass. 9. The table below shows the magnitude of the gravitational force on objects of different mass in Banff, Alberta. Mass (kg) 0 1.50 3.00 4.50 6.00 7.50 10.0 Magnitude of 0 14.7 29.4 44.1 58.9 73.6 98.1 Gravitational Force (N) (a) Graph the data. (b) Calculate the slope of the line. (c) What does the slope represent? 202 Unit II Dynamics 04-Phys20-Chap04.qxd 7/24/08 10:59 AM Page 203 info BIT Gravity is the dominant force throughout the universe. This force is attractive and has an infinite range. 4.2 Newton\u2019s Law of Universal Gravitation Gravity affects all masses in the universe. No matter where you are on Earth or in outer space, you exert a gravitational force on an object and an object exerts a gravitational force, of equal magnitude but opposite direction, on you. Because gravitational force acts over any distance, the range of its effect is infinite. Near Earth\u2019s surface, the magnitude of the gravitational force exerted mBg. by Earth (object A) on object B is given by the equation FA on B But object B also exerts a gravitational force of equal magnitude on mAg. Newton hypothesized that, given two objects A Earth, FB on A and B, the magnitude of the gravitational force exerted by one object on the other is directly proportional to the product of both masses: Fg mAmB Figure 4.11 shows the magnitude of the gravitational force acting on an object at Earth\u2019s surface (rEarth), one Earth radius above Earth (2rEarth), and two Earth radii above Earth (", "3rEarth). If the separation distance from Earth\u2019s centre to the centre of the object doubles, Fg decreases 1 of its original value. If the separation distance from Earth\u2019s centre to to 4 1 of its original value. the centre of the object triples, Fg decreases to 9 1, Fg is inversely proportional to the square of the separation So, just as g 2 r distance (Figure 4.12): Fg 1 2 r Magnitude of Gravitational Force vs. Distance from Earth\u2019s Centre Earth Fg rEarth 1 4 Fg 1 9 Fg 3rEarth 2rEarth 10.0 9.00 8.00 7.00 6.00 5.00 4.00 3.00 2.00 1.00 Distance from Earth\u2019s centre r (Earth radii) Figure 4.11 The magnitude of the gravitational force acting on an object some distance from Earth varies inversely with the square of the separation distance. Figure 4.12 The magnitude of the gravitational force acting on a 1.00-kg object as a function of distance from Earth\u2019s centre Chapter 4 Gravity extends throughout the universe. 203 04-Phys20-Chap04.qxd 7/24/08 10:59 AM Page 204 If you combine both proportionalities into one statement, you get mAmB r 2 GmAmB r 2 ship is Newton\u2019s law of universal gravitation. (Figure 4.13). This mathematical relation- or Fg Fg Any two objects, A and B, in the universe exert gravitational forces of equal magnitude but opposite direction on each other. The forces are directed along the line joining the centres of both objects. The magnitude of the gravitational force is given by Fg GmAmB r 2, where mA and mB are the masses of the two objects, r is the separation distance between the centres of both objects, and G is a constant called the universal gravitational constant. First mass Separation distance Second mass Magnitude of gravitational force Fg 2 Fg 6 Fg 4 Fg 1 4 Fg m 3 m 6 4 Fg Figure 4.13 The magnitude of the gravitational force is directly proportional to the product of the two masses, and inversely proportional to the square of the separation distance. Experiments have shown that the magnitude of the gravitational force acting on any pair of objects does not depend on the medium in which the objects are located (Figure 4.14). In other words, given two fish, the gravitational force acting on either fish will have the same magnitude if both fish are underwater or in midair. (a) (b) 1 m 1 m Figure 4.14 The magnitude of the gravitational force acting on either fish is the same whether both fish are (a) underwater or (b) above water. e SIM Explore the relationship among mA, mB, r, and Fg in Newton\u2019s law of gravitation. Follow the eSim links at www.pearsoned.ca/school/ physicssource. 204 Unit II Dynamics 04-Phys20-Chap04.qxd 7/24/08 10:59 AM Page 205 Concept Check Two identical stationary baseballs are separated by a distance r. What will happen to the magnitude of the gravitational force acting on either ball if (a) the mass of each ball doubles? (b) r is halved? (c) the mass of each ball is halved and r doubles? e WEB Edmund Halley, an associate of Newton, paid for the publication of some of Newton\u2019s famous work. Find out about Edmund Halley and his contributions to Newton\u2019s work. Begin your search at www.pearsoned.ca/school/ physicssource. Determining the Value of the Universal Gravitational Constant Although Newton found a mathematical relationship for gravitational force, he was unable to determine the value of G. In 1798, scientist Henry Cavendish (1731\u20131810) confirmed experimentally that Newton\u2019s law of gravitation is valid, and determined the density of Earth. Cavendish\u2019s experimental set-up was later used to determine the value of G. The magnitude of the gravitational force acting on most pairs of objects is very weak and the magnitude decreases significantly as the separation distance between the objects increases. However, if you use two light spheres (each of mass m) and two heavy spheres (each of mass M) that are very close to each other, it is possible to determine the magnitude of the gravitational force exerted by M on m. The trick is to use a device that can accurately measure the very small gravitational force. A modern torsion balance is a device that uses a sensitive fibre and a beam of light to measure very minute forces due to gravity, magnetic fields, or electric charges. Cavendish used a modified torsion balance invented by John Michell (1724\u20131793) to verify Newton\u2019s law of gravitation", ". A modern torsion balance consists of a small, light, rigid rod with two identical, light, spheres (m) attached to each end (Figure 4.15). The rod is suspended horizontally by a thin fibre connected to the centre of the rod. A mirror is also attached to the fibre and rod so that when the rod turns, the mirror also turns by the same amount. The entire assembly is supported in an airtight chamber. The torsion balance initially experiences no net force and the spheres m are stationary. fibre support torsion fibre mirror m zero deflection screen laser M m rod M Figure 4.15 A modern torsion balance uses a laser beam to measure the amount of twist in the fibre. The most accurate value of G has been determined using such a device. torsion balance: device used to measure very small forces info BIT Charles de Coulomb invented the original torsion balance in 1777 to measure small magnetic forces and forces in fluids. However, John Michell independently invented the same type of device in 1784. Chapter 4 Gravity extends throughout the universe. 205 04-Phys20-Chap04.qxd 7/24/08 10:59 AM Page 206 m Fg M on m Fg m on M M fibre direction of motion M Fg m on M Fg M on m m Figure 4.16 This figure shows the top view of spheres m and M from Figure 4.15. The original position of spheres m is shown with dashed lines. The gravitational force exerted by M on m causes m to rotate toward M. The greater the magnitude of the gravitational force, the greater the angle of rotation. When two identical, heavy, spheres (M) are moved close to spheres m, the gravitational force exerted by M on m causes m to rotate horizontally toward M. This rotation causes the fibre to twist slightly (Figure 4.16). As the fibre twists, the mirror attached to both the fibre and the rod turns through an angle in the horizontal plane. A beam of light reflected from the mirror becomes deflected as spheres m rotate. The amount of deflection is an indication of how much the spheres rotate. The greater the magnitude of the gravitational force, the more the fibre twists, and the greater the angle of rotation. By measuring the amount of deflection, the gravitational force exerted by M on m can be determined. Spheres M are then moved to a symmetrical position on the opposite side of m, and the procedure is repeated. Since the separation distance between m and M, the values of m and M, and the gravitational force can all be measured, it is possible to calculate G using Newton\u2019s law of gravitation. Fg GmM r 2 Fgr 2 mM G G Fgr 2 mM info BIT Scientists use Newton\u2019s law of universal gravitation to calculate the masses of planets and stars. The current accepted value of G to three significant digits is 6.67 1011 Nm2/kg2. In Example 4.1, Newton\u2019s law of gravitation is used to show that a person weighs slightly less at the top of the mountain than at its base. Example 4.1 Mount Logan in the Yukon is 5959 m above sea level, and is the highest peak in Canada. Earth\u2019s mass is 5.97 1024 kg and Earth\u2019s equatorial radius is 6.38 106 m. What would be the difference in the magnitude of the weight of a 55.0-kg person at the top of the mountain as compared to at its base (Figure 4.17)? Assume that Earth\u2019s equatorial radius is equal to the distance from Earth\u2019s centre to sea level. Given mp 55.0 kg h 5959 m mEarth rEarth 5.97 1024 kg 6.38 106 m Required difference in magnitude of weight (Fg) Mt. Logan 5959 m Figure 4.17 206 Unit II Dynamics 04-Phys20-Chap04.qxd 7/24/08 10:59 AM Page 207 Practice Problems 1. Two people, A and B, are sitting on a bench 0.60 m apart. Person A has a mass of 55 kg and person B a mass of 80 kg. Calculate the magnitude of the gravitational force exerted by B on A. 2. The mass of the Titanic was 4.6 107 kg. Suppose the magnitude of the gravitational force exerted by the Titanic on the fatal iceberg was 61 N when the separation distance was 100 m. What was the mass of the iceberg? Answers 1. 8.2 107 N 2. 2.0 108 kg Analysis and Solution Assume that the separation distance between the person at the base of the mountain and Earth is equal to Earth\u2019s equatorial radius. Base of mountain: rB rEarth 6.38 106 m Top of mountain: rT 6.38 106 m 5959 m The person\u2019s weight is equal to the gravitational force exerted by Earth on the person", ", and is directed toward Earth\u2019s centre both at the base and at the top of the mountain. Calculate Fg at the base of the mountain using Newton\u2019s law of gravitation. (Fg)B GmpmEarth (rB)2 2 m N (55.0 kg)(5.97 1024 kg) 6.67 1011 2 g k (6.38 106 m)2 538.049 N Calculate Fg at the top of the mountain using Newton\u2019s law of gravitation. (Fg)T GmpmEarth (rT)2 2 m N (55.0 kg)(5.97 1024 kg) 6.67 1011 2 g k (6.38 106 m 5959 m)2 537.045 N The difference in the magnitude of the weight is equal to the difference in magnitude of both gravitational forces. Fg (Fg)T (Fg)B 538.049 N 537.045 N 1.00 N Paraphrase The difference in the magnitude of the person\u2019s weight is 1.00 N. Using Proportionalities to Solve Gravitation Problems Example 4.2 demonstrates how to solve gravitation problems using proportionalities. This technique is useful if you are given how the separation distance and masses change from one situation to another. Chapter 4 Gravity extends throughout the universe. 207 04-Phys20-Chap04.qxd 7/24/08 10:59 AM Page 208 Example 4.2 Object A exerts a gravitational force of magnitude 1.3 1010 N on object B. Determine the magnitude of the gravitational force if the separation distance is doubled, mA increases by 6 times, and mB is halved. Explain your reasoning. Practice Problem 1. Object A exerts a gravitational force of magnitude 5.9 1011 N on object B. For each situation, determine the magnitude of the gravitational force. Explain your reasoning. (a) the separation distance 4 increases to of its original 3 3 of its value, mA increases to 2 original value, and mB is halved (b) the separation distance decreases 1 of its original value, mA is to 6 5 halved, and mB increases to 4 of its original value Answer 1. (a) 2.5 1011 N (b) 1.3 109 N Analysis and Solution mAmB and Fg From Newton\u2019s law of gravitation, Fg Figure 4.18 represents the situation of the problem. 1 r 2. before after mA mA Fg 1.3 1010 N mB r Fg? 2r Figure 4.18 Fg (6mA) mB 1 2 and Fg 1 (2r)2 (6) 1 2 mAmB 3mAmB 1 1 22 r 2 1 1 2 r 4 mB Calculate the factor change of Fg. 3 1 4 3 4 Calculate Fg. 3 4 3 4 Fg (1.3 1010 N) 9.8 1011 N The new magnitude of the gravitational force will be 9.8 1011 N. Using Superposition to Find the Net Gravitational Force on an Object Example 4.3 demonstrates how to calculate the gravitational force exerted by both the Moon and the Sun on Earth. A free-body diagram is used to determine the gravitational forces acting on Earth. The technique of adding the gravitational force due to each pair of objects (Earth and the Moon, and Earth and the Sun) to find the net gravitational force is called superposition. 208 Unit II Dynamics 04-Phys20-Chap04.qxd 7/24/08 10:59 AM Page 209 Example 4.3 During a lunar eclipse, Earth, the Moon, and the Sun are aligned on the same plane as shown in Figure 4.19. Using the data in the chart below, calculate the net gravitational force exerted by both the Moon and the Sun on Earth. Celestial Body Mass* (kg) Mean Separation Distance from Earth* (m) Earth Earth\u2019s Moon Sun 5.97 1024 7.35 1022 1.99 1030 \u2014 3.84 108 1.50 1011 *Source: Jet Propulsion Laboratory, California Institute of Technology (See JPL link at www.pearsoned.ca/school/physicssource.) 3.84 108 m 1.50 1011 m Moon Earth Sun Figure 4.19 Given mEarth mMoon mSun 5.97 1024 kg 7.35 1022 kg 1.99 1030 kg rE to M rE to S 3.84 108 m 1.50 1011 m Required net gravitational force on Earth (F g) Analysis and Solution Draw a free-body diagram for Earth (Figure 4.20). Practice Problems 1. During a solar eclipse, Earth, the Moon, and the Sun are aligned on the same plane as shown in Figure 4.21. Calculate the net", " gravitational force exerted by both the Moon and the Sun on Earth. 1.50 1011 m 3.84 108 m Earth Moon Sun Figure 4.21 2. During the first quarter phase of the Moon, Earth, the Moon, and the Sun are positioned as shown in Figure 4.22. Calculate the net gravitational force exerted by both the Moon and the Sun on Earth. Top View of Earth, the Moon, and the Sun Moon y 3.84 108 m Earth x Sun toward Sun toward Moon Fg1 Figure 4.20 Fg1 Fg2 Fnet 1.50 1011 m Figure 4.22 Answers 1. 3.54 1022 N [toward Sun\u2019s centre] 2. 3.52 1022 N [0.3] Fg2 Diagram is not to scale. Calculate Fg exerted by the Moon on Earth using Newton\u2019s law of gravitation. (Fg)1 GmEarthmMoon (rE to M)2 m N (5.97 1024 kg)(7.35 1022 kg) 6.67 1011 2 g k (3.84 108 m)2 2 1.985 1020 N 1.985 1020 N [toward Moon\u2019s centre] F g1 Chapter 4 Gravity extends throughout the universe. 209 04-Phys20-Chap04.qxd 7/24/08 10:59 AM Page 210 Calculate Fg exerted by the Sun on Earth using Newton\u2019s law of gravitation. (Fg)2 GmEarthmSun (rE to S)2 m N (5.97 1024 kg)(1.99 1030 kg) 6.67 1011 2 g k (1.50 1011 m)2 2 3.522 1022 N 3.522 1022 N [toward Sun\u2019s centre] F g2 gnet Fgnet F g2 Fg2 Find the net gravitational force on Earth. F F g1 Fg1 1.985 1020 N 3.522 1022 N 3.50 1022 N 3.50 1022 N [toward Sun\u2019s centre] F gnet Paraphrase The net gravitational force on Earth due to the Sun and Moon during a lunar eclipse is 3.50 1022 N [toward Sun\u2019s centre]. The Role of Gravitational Force on Earth\u2019s Tides Newton used the concept of gravitational force to account for Earth\u2019s tides. Although he correctly identified the gravitational force exerted by the Moon and the Sun on Earth as the major cause, a complete understanding of tides must take into account other factors as well. The height of the tides varies depending on the location on Earth (Figure 4.23). In the middle of the Pacific Ocean, the difference between high and low tides is about 0.5 m. But along the coastline of the continents, the difference may be considerably greater. Figure 4.23 The tides in the Bay of Fundy are the highest in the world. In some locations, the water level rises up to 18 m between low and high tides. 210 Unit II Dynamics 04-Phys20-Chap04.qxd 7/24/08 10:59 AM Page 211 Some factors that affect tides are \u2022 the shape of the coastline \u2022 the topography of the ocean floor near the coastline \u2022 friction between Earth and the ocean water \u2022 Earth\u2019s position in its orbit around the Sun \u2022 Earth\u2019s rotation about its axis \u2022 the tilt of Earth\u2019s axis \u2022 the alignment of Earth, the Moon, and the Sun First consider only the Moon\u2019s influence on Earth. Since the Moon exerts a gravitational force on Earth, the Moon is in a sense pulling Earth closer to it. So the land mass and ocean water on Earth are all \u201cfalling\u201d toward the Moon. In Figure 4.24, this gravitational force is greatest at side A, then decreases at the midpoints of A and B, and is least at side B, because the magnitude of the gravitational force varies inversely with the square of the separation distance. e WEB Newspapers in cities near an ocean, such as Halifax, often publish tidal charts listing the times of local high and low tides. Find an example of a tidal chart and suggest how different people would find this information useful. Begin your search at www.pearsoned.ca/school/ physicssource. 23.5\u00b0 B Earth A Moon Figure 4.24 Earth experiences high tides on sides A and B at the same time. The vectors show the relative gravitational force exerted by the Moon on a test mass at various locations near Earth\u2019s surface. The bulges at A and B are the high tides. The low tides occur at the midpoints of A and B. The bulge at B occurs because the land mass of Earth at B is pulled toward the Moon, leaving the ocean water behind. Next consider the fact that Earth rotates on", " its axis once every 24 h. As the bulges remain fixed relative to the Moon, Earth rotates underneath those bulges. So at a given location on Earth\u2019s surface, a high tide is replaced by a low tide about 6 h later, followed again by a high tide about 6 h later, and so on. These time intervals are actually a bit longer than 6 h because the Moon is orbiting Earth every 27 1 3 days with respect to distant stars, and the Moon is taking the bulges along with it. Now consider that Earth is tilted on its axis. When the northern hemisphere is under the bulge at A, the southern hemisphere is under the bulge at B. So the high tides that are 12 h apart are not equally high, and low tides that are 12 h apart are not equally low. Example 4.4 demonstrates how to calculate the gravitational force exerted by the Moon on 1.0000 kg of water at A. info BIT Io, one of Jupiter\u2019s moons, has tidal bulges of up to 100 m compared to typical tidal bulges of 1 m on Earth. Chapter 4 Gravity extends throughout the universe. 211 04-Phys20-Chap04.qxd 7/24/08 10:59 AM Page 212 Example 4.4 Calculate the gravitational force exerted by the Moon on 1.0000 kg of water at A (Figure 4.25). Use G 6.672 59 1011 Nm2/kg2. Celestial Body Mass* (kg) Equatorial Radius* (m) Mean Separation Distance from Earth* (m) Earth 5.9742 1024 6.3781 106 \u2014 Earth\u2019s Moon 7.3483 1022 1.7374 106 3.8440 108 *Source: Jet Propulsion Laboratory, California Institute of Technology (See JPL link at www.pearsoned.ca/school/physicssource.) Earth A water B rEarth 3.8440 108 m Moon Figure 4.25 Given mw mEarth mMoon rE to M 1.0000 kg 5.9742 1024 kg 7.3483 1022 kg 3.8440 108 m rEarth rMoon 6.3781 106 m 1.7374 106 m Required g) gravitational force exerted by Moon on water (F Analysis and Solution Draw a free-body diagram for the water showing only F g due to the Moon (Figure 4.26). away from Moon toward Moon Fg Figure 4.26 Practice Problem 1. Using the value of G given in Example 4.4, calculate the gravitational force exerted by the Moon on 1.0000 kg of water (a) at the midpoints of A and B, and (b) at B. Answer 1. (a) 3.3183 105 N [toward Moon\u2019s centre] (b) 3.2109 105 N [toward Moon\u2019s centre] 212 Unit II Dynamics 04-Phys20-Chap04.qxd 7/24/08 10:59 AM Page 213 Find the separation distance between the water and the Moon. r rE to M rEarth 3.8440 108 m 6.3781 106 m 3.780 22 108 m Calculate Fg exerted by the Moon on the water using Newton\u2019s law of gravitation. GmwmMoon r 2 Fg m N (1.0000 kg)(7.3483 1022 kg) 6.672 59 1011 2 g k (3.780 22 108 m)2 2 3.4312 105 N 3.4312 105 N [toward Moon\u2019s centre] F g Paraphrase The gravitational force exerted by the Moon on the water at A is 3.4312 105 N [toward Moon\u2019s centre]. The Role of Gravitational Force on Interplanetary Travel Scientists who plan space missions take advantage of the gravitational force exerted by planets and other celestial bodies to change the speed and direction of spacecraft. Distances between celestial bodies are huge compared to distances on Earth. So a space probe leaving Earth to study Jupiter and Saturn and their moons would take many years to arrive there. Scientists have to calculate the position and velocity of all the celestial bodies that will affect the motion of the probe many years in advance. If several planets are moving in the same direction and their positions are aligned, a space probe launched from Earth can arrive at its destination many years sooner, provided the probe moves near as many of those planets as possible (Figure 4.27). info BIT The Voyager mission was intended to take advantage of a geometric alignment of Jupiter, Saturn, Uranus, and Neptune. This arrangement occurs approximately every 175 years. By using the concept of gravity assist, the flight time to Neptune was reduced from 30 to 12 years, and a minimum of onboard propellant on the spacecraft was required. Voyager 1 Voyager 1 Launch Launch Sept. 5, 1977 Sept. 5, 1977 Voyager 2 Voyager 2 Launch Launch Aug. 20, 1977 Aug. 20, 1977 Jupiter", " Mar. 5, 1979 Jupiter Mar. 5, 1979 Jupiter Jul. 9, 1979 Jupiter Jul. 9, 1979 Saturn Aug. 25, 1981 Saturn Aug. 25, 1981 Saturn Nov. 12, 1980 Saturn Nov. 12, 1980 Voyager 2 Neptune Aug. 25, 1989 Neptune Aug. 25, 1989 Uranus Jan. 24, 1986 Uranus Jan. 24, 1986 Voyager 1 Figure 4.27 Both Voyager spacecraft were launched from Cape Canaveral, Florida, in 1977. Voyager 1 had close encounters with Jupiter and Saturn, while Voyager 2 flew by all four of the gaseous planets in the solar system. Chapter 4 Gravity extends throughout the universe. 213 04-Phys20-Chap04.qxd 7/24/08 10:59 AM Page 214 Each time the probe gets near enough to one of these planets, the gravitational field of the planet causes the path of the probe to curve (Figure 4.28). The planet deflects the space probe and, if the planet is moving in the same direction as the probe, the speed of the probe after its planetary encounter will increase. The use of the gravitational force exerted by celestial bodies to reduce interplanetary travel times is called gravity assist. info BIT Voyager 1 identified nine active volcanoes on Io, one of Jupiter\u2019s moons. Up until that point, scientists knew of no other celestial body in the solar system, other than Earth, that has active volcanoes. According to Voyager 1\u2019s instruments, the debris being ejected from Io\u2019s volcanoes had a speed of 1.05 103 m/s compared to speeds of 50 m/s at Mount Etna on Earth. Figure 4.28 Voyager 1 passed close to Io, Ganymede, and Callisto, three of Jupiter\u2019s moons. Jupiter has a total of 62 moons. Europa Ganymede Voyager 1 Jupiter Io Callisto THEN, NOW, AND FUTURE Small Steps Lead to New Models When Newton began his study of gravity in 1665, he was not the first person to tackle the challenge of explaining planetary motion. Ancient Greek astronomer Ptolemy (Claudius Ptolemaeus, 2nd century A.D.) and eventually Nicolaus Copernicus (1473\u20131543) proposed two different models of the solar system: one Earth-centred and the other Sun-centred. Later Johannes Kepler (1571\u20131630) developed three empirical laws describing planetary motion using astronomical data compiled by astronomer Tycho Brahe (1546\u20131601). Kepler\u2019s laws confirmed that a Sun-centred system is the correct model, because it was possible to predict the correct position of plan- ets. However, Kepler was unable to explain why planets move. Newton was the first scientist to explain the motion of planets in terms of forces. By using his three laws of motion and his law of gravitation, Newton derived Kepler\u2019s laws, providing further evidence of the validity of his force laws and of the model of a Sun-centred solar system. Newton was able to develop his law of gravitation because many scientists before him developed theories and made observations about planetary motion. Newton\u2019s laws and the concept of gravity could describe the motion of objects on Earth and throughout the universe. This was a tremendous breakthrough because up until that time, scientists were unable to predict or explain motion. While Newton\u2019s model can still be used today for most everyday situations, scientists have further modified it. The process of developing new models and theories has helped scientists tackle questions about the universe in ways Newton could never have imagined. Questions 1. Research the scientific developments that led to Newton\u2019s law of gravitation. 2. What are some benefits of developing new scientific models and theories? 214 Unit II Dynamics 04-Phys20-Chap04.qxd 7/24/08 10:59 AM Page 215 4.2 Check and Reflect 4.2 Check and Reflect Knowledge 1. Why is G called a \u201cuniversal\u201d constant? 2. Describe how a torsion balance can be used to measure the constant G. 3. Suppose Fg is the magnitude of the gravitational force between two people with a separation distance of 1.0 m. How would Fg change if (a) the separation distance became 2.0 m? (b) one person was joined by an equally massive friend while at this 2.0-m separation distance? Applications 4. The Moon has a mass of 7.35 1022 kg and its equatorial radius is 1.74 106 m. Earth\u2019s mass is 5.97 1024 kg and its equatorial radius is 6.38 106 m. (a) Calculate the magnitude of the gravitational force exerted by (i) the Moon on a 100-kg astronaut standing on the Moon\u2019s surface, and (ii) Earth on a 100-kg astronaut standing on Earth\u2019s surface. (b) Explain why the values of Fg in part (", "a) are different. 5. Mars has two moons, Deimos and Phobos, each named after an attendant of the Roman war god Mars. Deimos has a mass of 2.38 1015 kg and its mean distance from Mars is 2.3 107 m. Phobos has a mass of 1.1 1016 kg and its mean distance from Mars is 9.4 106 m. (a) Without doing any calculations, predict which moon will exert a greater gravitational force on Mars. Explain your reasoning. (b) Check your prediction in part (a) by calculating the magnitude of the gravitational force exerted by each moon on Mars. Mars\u2019 mass is 6.42 1023 kg. Show complete solutions. 6. Suppose the equatorial radius of Earth was the same as the Moon, but Earth\u2019s mass remained the same. The Moon has an equatorial radius of 1.74 106 m. Earth\u2019s mass is 5.97 1024 kg and its equatorial radius is 6.38 106 m. (a) Calculate the gravitational force that this hypothetical Earth would exert on a 1.00-kg object at its surface. (b) How does the answer in part (a) compare to the actual gravitational force exerted by Earth on this object? Extensions 7. Prepare a problem involving Newton\u2019s law of gravitation for each situation. Work with a partner to solve each problem, and discuss the steps you use. (a) Choose the values of the two masses and the separation distance. (b) Use values of mA, mB, and r that are multiples of those in part (a). Use proportionalities to solve the problem. 8. During Newton\u2019s time, scientists often \u2022 worked alone and contact with other scientists working on similar problems was difficult. \u2022 were knowledgeable in many different fields. Newton, for example, spent many years doing alchemy. (a) In paragraph form, assess the impacts that these factors might have had on science in Newton\u2019s day. (b) In a paragraph, describe in what ways these factors are relevant to scientists today. e TEST To check your understanding of Newton\u2019s law of gravitation, follow the eTest links at www.pearsoned.ca/school/physicssource. Chapter 4 Gravity extends throughout the universe. 215 04-Phys20-Chap04.qxd 7/24/08 10:59 AM Page 216 info BIT A g force is a force that causes an acceleration with a magnitude of some multiple of g. A force of 2g means the magnitude of the acceleration is 2 9.81 m/s2 19.6 m/s2. 4.3 Relating Gravitational Field Strength to Gravitational Force The acceleration due to gravity g near or on Earth\u2019s surface is about 9.81 m/s2 [down]. But where does the value of 9.81 m/s2 come from? Consider the forces acting on a test mass mtest some distance above Earth\u2019s surface, where Earth has a mass of Msource. The only force acting on mtest is the gravitational force exerted by Earth on the test mass (Figure 4.29). r Msource mtest Fg Figure 4.29 The gravitational force exerted by Earth on test mass mtest info BIT During a roller coaster ride, riders may experience a 4g change in acceleration between the top and bottom of a loop. This dramatic change in acceleration causes the thrill and occasional dizziness experienced by riders. The magnitude of F weight or using Newton\u2019s law of gravitation. g can be evaluated two ways: using the concept of Weight Newton\u2019s law of gravitation Fg mtestg Fg GmtestMsource r 2 Since the value of Fg is the same no matter which equation you use, set both equations equal to each other. mtest g Gmtest Msource r 2 g GMsource r 2 So no matter where the test mass is located in the universe, you can calculate the magnitude of the gravitational field strength (or gravitational acceleration) at any distance from a celestial body if you know the mass of the celestial body Msource and the separation distance between the centre of the test mass and the celestial body r. Find out the relationship between gravitational field strength and the acceleration due to gravity by doing 4-3 Design a Lab. 216 Unit II Dynamics 04-Phys20-Chap04.qxd 7/24/08 10:59 AM Page 217 4-3 Design a Lab 4-3 Design a Lab Comparing Gravitational Field Strength to Gravitational Acceleration The Question What is the relationship between gravitational field strength and the local value of the gravitational acceleration? Design and Conduct Your Investigation State a hypothesis. Then design an experiment. Identify the controlled, manipulated, and responding variables. Review the procedure in 4-2 QuickLab on page 198. List the materials you will use, as well as a detailed procedure. Check the procedure", " with your teacher and then do the investigation. Analyze your data and form a conclusion. How well did your results agree with your hypothesis? How Is Gravitational Field Strength Related to Gravitational Acceleration? To determine how gravitational field strength is related to gravitational acceleration, use the definition of a newton, 1 N 1 kgm/s2. Then substitute kilogram-metres per second squared for newtons in the equation for gravitational field strength: 1 N kg 1 m kg s2 kg m 1 s2 Metres per second squared are the units of acceleration. So in terms of units, gravitational field strength and gravitational acceleration are equivalent (Figure 4.30). Figure 4.30 Jennifer Heil of Spruce Grove, Alberta, won the gold medal in the women\u2019s freestyle skiing moguls in the 2006 Winter Olympics in Turin, Italy. The gravitational field strength at the surface of the Moon is about 1 6 that at Earth\u2019s surface. How would Jennifer\u2019s jump on Earth compare with one on the Moon? Chapter 4 Gravity extends throughout the universe. 217 04-Phys20-Chap04.qxd 7/24/08 10:59 AM Page 218 Calculating the Gravitational Acceleration of an Object on Two Celestial Bodies Example 4.5 demonstrates how to calculate the gravitational acceleration at the equator on Earth\u2019s surface and that on the surface of the Moon. These two values are then compared to find the ratio of gEarth to gMoon. To solve the problem requires using data from Table 4.1, which shows the mass and equatorial radius of the Sun, the Moon, and each planet in the solar system. Table 4.1 Masses and Radii for Celestial Bodies in the Solar System* Celestial Body Sun Mercury Venus Earth Earth\u2019s Moon Mars Jupiter Saturn Uranus Neptune Mass (kg) 1.99 1030 3.30 1023 4.87 1024 5.97 1024 7.35 1022 6.42 1023 1.90 1027 5.69 1026 8.68 1025 1.02 1026 Equatorial Radius (m) 6.96 108 2.44 106 6.05 106 6.38 106 1.74 106 3.40 106 7.15 107 6.03 107 2.56 107 2.48 107 *Source: Jet Propulsion Laboratory, California Institute of Technology (See JPL link at www.pearsoned.ca/school/physicssource.) Example 4.5 (a) Calculate the magnitude of the gravitational acceleration of an object at the equator on the surface of Earth and the Moon (Figure 4.31). Refer to Table 4.1 above. (b) Determine the ratio of gEarth to gMoon. How different would your weight be on the Moon? gEarth? gMoon? Moon Earth Figure 4.31 Given mEarth mMoon 5.97 1024 kg 7.35 1022 kg rEarth rMoon 6.38 106 m 1.74 106 m Required (a) magnitude of gravitational acceleration at equator on Earth and the Moon (gEarth and gMoon) (b) ratio of gEarth to gMoon e WEB Globular clusters are groups of about 1 000 000 stars that are bound together by gravity. Find out who discovered the first cluster and how many have been identified so far. Research the approximate size and shape of a globular cluster and the forces involved in its formation. Summarize your findings. Begin your search at www.pearsoned.ca/school/ physicssource. 218 Unit II Dynamics 04-Phys20-Chap04.qxd 7/24/08 10:59 AM Page 219 Analysis and Solution (a) Use the equation g to calculate the GMsource r2 magnitude of the gravitational field strength on each celestial body. The magnitude of the gravitational acceleration is numerically equal to the magnitude of the gravitational field strength. Earth gEarth GmEarth (rEarth)2 m N (5.97 1024 kg) 6.67 1011 2 g k (6.38 106 m)2 2 9.783 N/kg 9.783 m/s2 The Moon gMoon GmMoon (rMoon)2 m N (7.35 1022 kg) 6.67 1011 2 g k (1.74 106 m)2 2 1.619 N/kg 1.619 m/s2 (b) Calculate the ratio of gEarth to gMoon. m 9.783 s2 m 1.619 s2 gEarth gMoon 6.04 Paraphrase (a) The magnitude of the gravitational acceleration at the equator on the surface of Earth is 9.78 m/s2 and of the Moon is 1.62 m/s2. Practice Problems 1. A satellite orbits Earth at a distance of 3rEarth above Earth\u2019s surface. Use the data from Table 4.1 on page 218. (a) How many", " Earth radii is the satellite from Earth\u2019s centre? (b) What is the magnitude of the gravitational acceleration of the satellite? 2. An 80.0-kg astronaut is in orbit 3.20 104 km from Earth\u2019s centre. Use the data from Table 4.1 on page 218. (a) Calculate the magnitude of the gravitational field strength at the location of the astronaut. (b) What would be the magnitude of the gravitational field strength if the astronaut is orbiting the Moon with the same separation distance? 3. The highest satellites orbit Earth at a distance of about 6.6rEarth from Earth\u2019s centre. What would be the gravitational force on a 70-kg astronaut at this location? Answers 1. (a) 4rEarth (b) 6.11 101 m/s2 2. (a) 3.89 101 N/kg (b) 4.79 103 N/kg 3. 16 N [toward Earth\u2019s centre] (b) The ratio of gEarth to gMoon is 6.04. So your weight would be about 6 times less on the surface of the Moon than on Earth. Calculating the Weight of an Object on Mars The equation g GMsource r 2 can be used with F mg to calculate the weight of an object on any celestial body. In Example 4.6, the weight of a student on Mars is calculated. This quantity is then compared with the student\u2019s weight on Earth\u2019s surface. An interesting application of the variation in the weight of an object involves the Mars rover (Figure 4.32). The rover had a mass of about 175 kg, but on the surface of Mars, the rover weighed about 2.5 times less than on Earth\u2019s surface. The rover was designed to avoid inclines greater than 30 but gEarth, the rover could travel farther up an because gMars incline on Mars than on Earth using the same battery charge. Figure 4.32 In full sunlight, a 140-W battery enabled the Mars rover to travel about 100 m per day on level ground with an average speed of 1.0 cm/s between charges. Chapter 4 Gravity extends throughout the universe. 219 04-Phys20-Chap04.qxd 7/24/08 10:59 AM Page 220 Example 4.6 (a) What is the mass of a 60.0-kg student on Mars and on Earth? (b) What is the student\u2019s weight at the equator on the surface of Mars and of Earth (Figure 4.33). Use the data from Table 4.1 on page 218. Fg? Mars Given (a) ms (b) mMars mEarth 60.0 kg 6.42 1023 kg 5.97 1024 kg rMars rEarth 3.40 106 m 6.38 106 m Fg? Earth Required (a) mass on Mars and on Earth (m) gEarth) gMars and F (b) weight on Mars and on Earth (F Figure 4.33 Analysis and Solution (a) Mass is a scalar quantity and does not depend on location. So the (b) Use the equation g student\u2019s mass will be the same on Mars as on Earth. GMsource r 2 gravitational field strength on Mars and on Earth. to calculate the magnitude of the Practice Problems Use the data from Table 4.1, page 218, to answer the following questions. 1. What would be the weight of a 22.0-kg dog at the equator on Saturn\u2019s surface? 2. (a) Do you think your skeleton could support your weight on Jupiter? (b) Compared to Earth, how much stronger would your bones need to be? 3. (a) What is the magnitude of the gravitational field strength at the equator on Uranus\u2019 surface? (b) Compared to Earth, how would your weight change on Uranus? Answers 1. 230 N [toward Saturn\u2019s centre] 2. (a) no (b) 2.53 times 3. (a) 8.83 N/kg, (b) 0.903 FgEarth Mars gMars GmMars (rMars)2 2 m N (6.42 1023 kg) 6.67 1011 2 g k (3.40 106 m)2 3.704 N/kg Earth gEarth GmEarth (rEarth)2 2 m N (5.97 1024 kg) 6.67 1011 2 g k (6.38 106 m)2 9.783 N/kg Since the direction of F celestial body, use the scalar equation Fg magnitude of the weight. g will be toward the centre of each mg to find the Mars FgMars msgMars Earth FgEarth msgEarth N (60.0 kg)3.704 g k 222 N N (60.0 kg)9.783 g k 587 N 220 Unit II Dynamics 04-Phys", "20-Chap04.qxd 7/24/08 10:59 AM Page 221 Paraphrase and Verify (a) The mass of the student would be 60.0 kg on both Mars and Earth. (b) The weight of the student on Mars would be 222 N [toward Mars\u2019 centre] and on Earth 587 N [toward Earth\u2019s centre]. So the student would weigh about 2.6 times more on Earth than Mars. Different Values of Gravitational Field Strength on Earth For a long time, people thought that the magnitude of the gravitational field strength was constant at any location on Earth\u2019s surface. However, scientists discovered that the value of g depends on both latitude and altitude. Latitude is the angular distance north or south of the equator. Altitude is the elevation of the ground above sea level. Figure 4.34 shows how the magnitude of the gravitational field strength at sea level varies with latitude. Measured Magnitude of Gravitational Field Strength at Sea Level vs. Latitude info BIT The value 9.81 N/kg is an average of the magnitude of the gravitational field strength at different locations on Earth\u2019s surface. 9.84 9.83 9.82 9.81 9.80 9.79 9.78 20\u00b0 60\u00b0 80\u00b0 40\u00b0 Latitude Equator Figure 4.34 Gravitational field strength at sea level as a function of latitude. At what location on Earth\u2019s surface would you weigh the least? The most? P ole s The value of g increases as you move toward either the North or South Pole, because Earth is not a perfect sphere. It is flatter at the poles and it bulges out slightly at the equator. In fact, Earth\u2019s radius is 21 km greater at the equator than at the poles. So an object at the equator is farther away from Earth\u2019s centre than if the object were at the North Pole. Since g, the farther an object is from Earth\u2019s centre, the smaller 1 r 2 the value of g will be. Other factors affect the value of g at Earth\u2019s surface. The materials that make up Earth\u2019s crust are not uniformly distributed. Some materials, such as gold, are more dense than materials such as zinc. Earth\u2019s rotation about its axis also reduces the measured value of g, but the closer an object is to the North or South Pole, the less effect Earth\u2019s rotation has on g. Concept Check Leo weighs 638 N [down] in Calgary, Alberta. What are some problems with Leo saying he weighs 638 N [down] anywhere on Earth? What property of matter would he be more accurate to state? Chapter 4 Gravity extends throughout the universe. 221 04-Phys20-Chap04.qxd 7/24/08 10:59 AM Page 222 Gravity (mGal 20.5 20. 20.0 20.0 0. 5 0. 5 2 2 AA Applications of the Variation in g in Geology The variation in the value of g on Earth is used to detect the presence of minerals and oil. Geophysicists and geologists use sensitive instruments, called gravimeters, to detect small variations in g when they search for new deposits of ore or oil. Gold and silver deposits increase the value of g, while deposits of oil and natural gas decrease g. Figure 4.35 is an example of a map that shows different measured values of g as lines, where each line represents a specific value of g. 19.0 19.1 19.2 19.3 19.4 19.5 19.6 19.7 19.8 19.9 20.0 20.1 20.2 20.3 20.4 20.5 20.6 20.7 20.8 20.9 21.0 21.1 21.2 21.3 21.4 21.5 21.6 21.7 21.8 21.9 22.0 22.5 23.0 23.5 mGal Figure 4.35 A map showing the location of sulphide deposits in northern New Brunswick (shown in black) true weight: gravitational force acting on an object that has mass a 0 v constant FN Fg True Weight vs. Apparent Weight mg to calculate the weight of an object So far, you used the equation F g at any location in the universe. The gravitational force that you calculate g, of an object. with this equation is really called the true weight, F Suppose a student is standing on a scale calibrated in newtons in an elevator (Figure 4.36). If the elevator is at rest or is moving at constant velocity, the scale reads 600 N. Fg 600 N 300 200 100 400 0 500 600 700 Figure 4.36 The elevator and student are either at rest or moving at constant velocity. Using the free-body diagram for the student (Figure 4.37), the equation for the net force on the student is", " F N F F FN F g net 0 F g 0 Fg 0 mg FN N Figure 4.37 The free-body diagram for the student in Figure 4.36 FN mg 222 Unit II Dynamics a FN Fg Figure 4.39 The free-body diagram for the student in Figure 4.38 04-Phys20-Chap04.qxd 7/24/08 10:59 AM Page 223 So when F is equal to the magnitude of the student\u2019s weight. 0 N on the student, the magnitude of the normal force net Now suppose the elevator is accelerating up uniformly (Figure 4.38). In this situation, the scale reads 750 N. Fg 600 N 300 200 100 400 0 500 600 700 Figure 4.38 The elevator and student are accelerating up uniformly. N net To understand why the reading on the scale is different, draw the freebody diagram for the student (Figure 4.39) and write the equation for the net force: F F F g ma F F g N ma F F N ma mg m(a g) g The equation for F N is valid whether the student is accelerating up or down. In Figure 4.38, the student feels heavier than usual because the scale is pushing up on him with a force greater than mg. If the elevator is accelerating down uniformly, the scale reads m(a g ) but this time a and g are 525 N (Figure 4.40). As before, F in the same direction. So FN is less than mg, and the student feels lighter than usual. N Fg 600 N 300 200 100 400 0 500 600 700 Figure 4.40 The elevator and student are accelerating down uniformly. Chapter 4 Gravity extends throughout the universe. 223 04-Phys20-Chap04.qxd 7/24/08 10:59 AM Page 224 apparent weight: negative of the normal force acting on an object The quantity F N is called the apparent weight, w, of an object. For the situations shown in Figures 4.38 and 4.40 on page 223, the equation for the apparent weight of the student is e SIM Calculate the true weight, normal force, and apparent weight of a person during an elevator ride. Follow the eSim links at www.pearsoned.ca/school/ physicssource. 224 Unit II Dynamics w F N m(a g) m(g a ) Example 4.7 demonstrates how to calculate the true weight and apparent weight of an astronaut in a rocket during liftoff on Earth\u2019s surface. Example 4.7 A 100.0-kg astronaut in a spacesuit is standing on a scale in a rocket (Figure 4.41). The acceleration of the rocket is 19.6 m/s2 [up]. Calculate her true weight and apparent weight during liftoff on Earth. The acceleration due to gravity on Earth\u2019s surface is 9.81 m/s2 [down]. Given m 100.0 kg a 19.6 m/s2 [up] g 9.81 m/s2 [down] up down Required true weight and apparent weight during liftoff (F g and w) 0 3000 1000 2000 Figure 4.41 Analysis and Solution Draw a free-body diagram and a vector addition diagram for the astronaut (Figure 4.42). up down FN Fg FN Fnet Fg Use the equation F g mg to find the astronaut\u2019s true weight. Figure 4.42 F g mg (100.0 kg)9.81 9.81 102 N m s2 The astronaut is not accelerating left or right. So in the horizontal direction, F 0 N. net 04-Phys20-Chap04.qxd 7/24/08 10:59 AM Page 225 For the vertical direction, write an equation to find the net force on the astronaut. F F F net N g g N N FN Apply Newton\u2019s second law. ma F F g ma FF FF ma Fg m (9.81 102 N) (100.0 kg)19.6 s2 m 9.81 102 N (100.0 kg)19.6 s2 2.94 103 N 2.94 103 N [up] F Use the equation w F N Practice Problems 1. In Example 4.7, draw the free-body diagram for the scale during liftoff. 2. Suppose the rocket in Example 4.7 has an acceleration of 19.6 m/s2 [down] while it is near Earth\u2019s surface. What will be the astronaut\u2019s apparent weight and true weight? Answers 1. See page 898. 2. 9.79 102 N [up], 9.81 102 N [down] N to find the astronaut\u2019s apparent weight. w F N (2.94 103 N) 2.94 103 N Paraphrase During liftoff, the astronaut\u2019s true weight is 9.81 102 N [down] and her apparent weight", " is 2.94 103 N [down]. In Example 4.8, an astronaut is accelerating in deep space, a location in which the gravitational force acting on an object is not measurable. So in deep space, F 0. g Example 4.8 Refer to Example 4.7 on pages 224 and 225. What is the magnitude of the astronaut\u2019s true weight and apparent weight if the rocket is in deep space? The magnitude of the acceleration of the rocket is 19.6 m/s2. Given m 100.0 kg magnitude of a 19.6 m/s2 g 0 m/s2 Required magnitude of true weight and apparent weight in deep space (F g and w) Analysis and Solution In deep space, the mass of the astronaut is still 100.0 kg, but g is negligible. F So g mg 0 N The astronaut is not accelerating left or right. So in the horizontal direction, F 0 N. net Practice Problems 1. An 80.0-kg astronaut is standing on a scale in a rocket leaving the surface of the Moon. The acceleration of the rocket is 12.8 m/s2 [up]. On the Moon, g 1.62 N/kg [down]. Calculate the magnitude of the true weight and apparent weight of the astronaut (a) during liftoff, and (b) if the rocket has the same acceleration in deep space. 2. A 60.0-kg astronaut is standing on a scale in a rocket about to land on the surface of Mars. The rocket slows down at 11.1 m/s2 while approaching Mars. Use the data from Table 4.1 on page 218. Calculate the true weight and apparent weight of the astronaut (a) as the rocket lands, and (b) if the rocket is accelerating at 7.38 m/s2 [up] when leaving Mars. Chapter 4 Gravity extends throughout the universe. 225 04-Phys20-Chap04.qxd 7/24/08 10:59 AM Page 226 For the vertical direction, write an equation to find the net force on the astronaut. Refer to the free-body diagram in Figure 4.42 on page 224. Answers 1. (a) (F g) 1.30 102 N [down], (w) 1.15 103 N [down] F net b) (F 2. (a) (F (b) (F g) 0 N, (w) 1.02 103 N [down] g) 2.22 102 N [down], (w) 8.88 102 N [down] g) 2.22 102 N [down], (w) 6.65 102 N [down] Apply Newton\u2019s second law. F N FN ma ma m (100.0 kg)19.6 s2 1.96 103 N Use the equation w F N to find the astronaut\u2019s apparent weight. w F N (1.96 103 N) 1.96 103 N Paraphrase In deep space, the astronaut\u2019s true weight is zero and the magnitude of her apparent weight is 1.96 103 N. Free Fall Let\u2019s revisit the elevator scenario on pages 222 and 223. Suppose the elevator cable breaks (Figure 4.43). Assuming that frictional forces are negligible, the elevator, student, and scale all fall toward Earth with an acceleration of g. The student is now in free fall, the condition in which the only force acting on an object is F g. free fall: situation in which the only force acting on an object that has mass is the gravitational force a g 0 0 6 0 0 7 0 Fg Fg 600 N 300 200 100 400 0 500 600 700 Figure 4.43 The elevator, student, and scale are in free fall. Figure 4.44 The free-body diagram for the student in Figure 4.43 226 Unit II Dynamics 04-Phys20-Chap04.qxd 7/24/08 10:59 AM Page 227 To understand free fall, draw the free-body diagram for the student (Figure 4.44) and write the equation for the net force: F net F ma F g g ma mg a g So in free fall, a g and both the student and the scale are accelerating at g downward. In Figure 4.43 on page 226, the scale reads zero because 0. Since it no longer exerts a normal force on the student, so F 0, the student\u2019s apparent weight is also zero. Sometimes an F object in free fall is described as being \u201cweightless.\u201d However, this description is incorrect. In free fall, F w 0 but F 0. N N N g Observe the motion of water in a cup while in free fall by doing 4-4 QuickLab. 4-4 QuickLab 4-4 QuickLab Water in Free Fall Problem What is the motion of water in a cup when the cup is dropped from several metres above Earth\u2019s surface?", " CAUTION: Do this activity outside. Have someone steady the ladder and be careful when climbing it. Materials paper cup pointed pen or pencil water dishpan stepladder Procedure 1 Make two holes on opposite sides of the cup near the bottom using the pen or pencil. Cover the holes with your thumb and forefinger. Then fill the cup with water. e LAB For a probeware activity, go to www.pearsoned.ca/school/physicssource. 2 Hold the cup at shoulder height above a dishpan, and uncover the holes. Observe what happens to the water (Figure 4.45). Have a partner sketch the path the water takes. 3 Repeat step 1 but climb the ladder and drop the cup toward the dishpan from a height of several metres. Observe the motion of the water during the fall. Questions 1. Describe the path and motion of the water (a) when the cup was held stationary, and cup with two holes and filled with water Figure 4.45 (b) when the cup was dropped from the ladder. Give a reason for your observations. Chapter 4 Gravity extends throughout the universe. 227 04-Phys20-Chap04.qxd 7/24/08 10:59 AM Page 228 Weightlessness Videos transmitted from a space shuttle or the space station often show astronauts floating in their cabin (Figure 4.46). Are the astronauts weightless in space? The answer is no. Then why do they appear to be weightless? Since the shuttle is some distance above Earth, g is less than its 1 r 2 value at Earth\u2019s surface, because g. While the shuttle orbits Earth at high speed in an almost circular path, Earth exerts a gravitational force on the shuttle and everything in it. So the shuttle is able to remain in orbit. If an astronaut were standing on a scale in the shuttle, the scale would read zero, because the shuttle and everything in it are in free fall. The astronaut would feel \u201cweightless\u201d because the gravitational force exerted by Earth pulls the shuttle and the astronaut toward Earth. Suppose an astronaut is in a rocket in deep space and the acceleration of the rocket is zero. The astronaut would experience no measurable gravitational forces from any celestial bodies, and the astronaut\u2019s acceleration would be zero. In this situation, the astronaut would have a true weight of zero and an apparent weight of zero, a condition called true weightlessness. true weightlessness: situation in which w 0 for an object and F 0 on the object g Figure 4.46 At the altitude of the shuttle, the value of g is about 90% of its value at Earth\u2019s surface. 228 Unit II Dynamics 04-Phys20-Chap04.qxd 7/24/08 10:59 AM Page 229 4.3 Check and Reflect 4.3 Check and Reflect Knowledge 1. (a) What is the difference between true weight and apparent weight? (b) Describe a situation in which the true weight of an object is zero but its apparent weight is not zero. 2. A person orbiting Earth in a spacecraft has an apparent weight of zero. Explain if the person still experiences a gravitational force. 8. An astronaut in a rocket has an apparent weight of 1.35 103 N [down]. If the acceleration of the rocket is 14.7 m/s2 [up] near Earth\u2019s surface, what is the astronaut\u2019s true weight? The acceleration due to gravity on Earth\u2019s surface is about 9.81 m/s2 [down]. 9. A 50-kg astronaut experiences an acceleration of 5.0g [up] during liftoff. (a) Draw a free-body diagram for the astronaut during liftoff. 3. List two factors that affect the magnitude (b) What is the astronaut\u2019s true weight of the gravitational field strength at Earth\u2019s surface. Applications 4. Is there a place in the universe where true weightlessness actually exists? 5. Calculate the gravitational field strength at the location of a 70-kg astronaut 2.0rEarth from Earth\u2019s centre. Use the data from Table 4.1 on page 218. 6. Graph the equation g GmMoon r 2 using technology. Refer to Student References 5: Graphing Data on pp. 872\u2013874. Plot g on the y-axis (range of 02.0 N/kg) and r on the x-axis (range of 1\u20135rMoon). Toggle through to read values of g corresponding to specific values of rMoon to answer these questions: (a) Describe the graph of g vs. rMoon. How is it similar to Figure 4.10 on page 201? (b) What is the value of g (i) on the surface? (ii) at 1 2 rMoon above the surface? (iii) at rMoon above the surface? the (c) At what distance is g 1 100 and apparent", " weight? 10. Calculate the acceleration of the elevator in Figures 4.38 and 4.40 on page 223. Extensions 11. Draw a flowchart to summarize the steps needed to find the apparent weight of an object. Refer to Student References 4: Using Graphic Organizers on page 869. 12. Research how geophysicists and geologists use gravitational field strength to locate minerals, oil, and natural gas in Canada. Prepare a half-page report on your findings. Begin your search at www.pearsoned.ca/school/physicssource. 13. Suppose you are wearing a spacesuit. Where could you walk faster, on Earth or on the Moon? Explain your answer. 14. Draw a concept map to identify and link the concepts needed to understand gravitational acceleration and gravitational field strength near a celestial body other than Earth. Refer to Student References 4: Using Graphic Organizers on page 869. Create and solve a problem to demonstrate your understanding of these concepts. gravitational field strength on the surface of the Moon? e TEST 7. At the top of Mount Robson in British Columbia, a 7.5-kg turkey weighs 73.6 N [down]. Calculate the magnitude of the gravitational field strength at this location. To check your understanding of gravitational field strength, true weight, apparent weight, and free fall, follow the eTest links at www.pearsoned.ca/school/physicssource. Chapter 4 Gravity extends throughout the universe. 229 04-Phys20-Chap04.qxd 7/24/08 10:59 AM Page 230 CHAPTER 4 SUMMARY Key Terms and Concepts gravitational force weight gravitational mass action-at-a-distance force field gravitational field gravitational field strength Key Equations True weight: F mg g Newton\u2019s law of gravitation: Fg GmAmB r 2 torsion balance true weight apparent weight free fall true weightlessness Gravitational field strength (or gravitational acceleration): g GMsource r 2 Apparent weight: w F N Conceptual Overview The concept map below summarizes many of the concepts and equations in this chapter. Copy and complete the map to have a full summary of the chapter. gravitational force mass is one of two types Gravitation involves weight is gravitational field strength equation equation other examples inertial strong nuclear using exerted by a using celestial body where direction acceleration standard masses and a and a G is value 6.67 1011 N\u2022m2/kg2 Figure 4.47 230 Unit II Dynamics 04-Phys20-Chap04.qxd 7/24/08 10:59 AM Page 231 CHAPTER 4 REVIEW Knowledge 1. (4.2) What is the significance of the term \u201cuniversal\u201d in Newton\u2019s law of universal gravitation? 2. (4.1, 4.3) A brick placed on an equal arm balance requires 5.0 kg to just balance it. When the brick is hung from a spring scale, the scale reads 48 N. The balance, standard masses, spring scale, and brick are moved to a planet where the gravitational field strength is 2.0 times that on Earth. What will be the reading on the balance and on the spring scale in this new location? 3. (4.1, 4.3) How does gravitational field strength vary with the mass of a celestial body? Assume that the radius is constant. 4. (4.3) The gravitational field strength at Earth\u2019s surface is about 9.81 N/kg [down]. What is the gravitational field strength exactly 1.6rEarth from Earth\u2019s centre? Applications 5. A 1.0-kg object, initially at rest, is dropped toward Earth\u2019s surface. It takes 2.26 s for the object to fall 25 m. Determine how long it takes a 2.0-kg object to fall this distance from rest on Jupiter. Use the data from Table 4.1 on page 218. 6. Describe the steps you would use to determine the distance from Earth\u2019s centre where the gravitational force exerted by Earth on a spacecraft is balanced by the gravitational force exerted by the Moon. Assume that you know the distance from Earth\u2019s centre to the centre of the Moon. Do not do the calculations. 7. Use the data from Table 4.1 on page 218. Calculate the true weight of a 60.0-kg astronaut on (a) the surface of Mars, and (b) the surface of Saturn. 8. Objects A and B experience a gravitational force of magnitude 2.5 108 N. Determine the magnitude of the gravitational force if the separation distance is halved, mA increases by 8 times, and mB is reduced to 1 4 of its original value. 9. Suppose a 65-kg astronaut on Mars is standing on a scale calibrated in newtons in an elevator. What will be the reading on the scale when the acceleration of the elevator is (a) zero? (b) 7", ".2 m/s2 [up]? (c) 3.6 m/s2 [down]? 10. A 50-kg rock in a Nahanni River canyon breaks loose from the edge of a cliff and falls 500 m into the water below. The average air resistance is 125 N. (a) What is the average acceleration of the rock? (b) How long does the rock take to reach the water? (c) What is the true weight of the rock? (d) Does the rock have an apparent weight? Explain. Extensions 11. Research why astronauts do exercises in space and why they have difficulty walking when they return to Earth. Write a short paragraph of your findings. Begin your search at www.pearsoned.ca/ school/physicssource. 12. Pilots in high-speed planes are subject to g forces. Healthy people can withstand up to 3\u20134 gs. Beyond that limit, the blood will pool in the lower half of the body and not reach the brain, causing the pilot to lose consciousness. Research how Dr. Wilbur Franks from Toronto found a solution to this problem. What connection does this problem have to human survival during a space flight? Begin your search at www.pearsoned.ca/school/physicssource. Consolidate Your Understanding 13. Write a paragraph summarizing Newton\u2019s law of gravitation. Include a numerical example that illustrates the law. Show a detailed solution. Think About It Review your answers to the Think About It questions on page 195. How would you answer each question now? e TEST To check your understanding of gravitation concepts, follow the eTest links at www.pearsoned.ca/school/physicssource. Chapter 4 Gravity extends throughout the universe. 231 04-Phys20-Chap04.qxd 7/24/08 10:59 AM Page 232 UNIT II PROJECT Tire Design, Stopping Distance, and Vehicle Mass Scenario Imagine that you work for Alberta Infrastructure and Transportation. A level section of highway in the mountains has an abnormally large number of accidents. The surface of one lane of the highway is concrete and the other is asphalt. You are a member of a research team formed to determine the stopping distances in the summer for both wet and dry days for different types of tires and for different masses of vehicles. Your team is to prepare a written report on your findings for the Traffic Branch of the Ministry. Assume that the vehicles are travelling at the posted speed limit of 90 km/h when the brakes are applied and that the vehicles are not equipped with anti-lock braking systems (ABS). Assume that the reaction time of drivers before they apply the brakes is 1.8 s. Planning Form a team of three to five members. Summarize the question your group is researching. Make hypotheses about how tire tread, surface condition, and vehicle mass might affect stopping distance. Assign roles to different team members. Some examples are team leader, materials manager, liaison officer, record keeper, and safety officer. Brainstorm strategies for researching and reporting on the question and create a timeline. Research tire designs that are designed to work well on both wet and dry road surfaces. Use the Internet and consult local tire suppliers. Materials \u2022 a digital camera and a computer \u2022 force-measuring equipment \u2022 mass-measuring equipment \u2022 new and used tires having different treads and designs \u2022 vehicle brochures Procedure 1 Research the Internet and interview different tire suppliers to identify tires designed to work well at above-freezing temperatures on both wet and dry pavement. Assessing Results After completing the project, assess its success based on a rubric designed in class* that considers research strategies experiment techniques clarity and thoroughness of the written report effectiveness of the team\u2019s presentation quality and fairness of the teamwork 2 Visit a local tire supplier and borrow new and used tires of different designs. Photograph the tires to record the tread designs. 3 Design and conduct an experiment to determine the coefficients of static and kinetic friction for the tires on wet and dry asphalt, and wet and dry concrete. Recall that you will need the mass of the tires and the local value of gravitational field strength. CAUTION: Take all your measurements on parking lots or sidewalks, and beware of traffic. Consult the local police department and ask for supervision while doing your experiments. 4 Research the masses of at least four small, medium, and large vehicles travelling highways that would use these tires. Determine the average mass of each class of vehicle. Sales brochures from vehicle dealers have mass information. 5 Determine the average deceleration on both wet and dry roadways for the vehicles of different mass equipped with these tires. Remember that some drivers may lock their brakes while braking. 6 Determine the stopping distance for the different vehicles under different conditions. Remember to consider driver reaction time. 7 Write a report of your findings. Use graphs and tables where appropriate. Thinking Further Write a three-paragraph addition to your team\u2019s report hypothesizing how driver education,", " changing the posted speed limit, and requiring that vehicles be equipped with ABS brakes might affect the results. *Note: Your instructor will assess the project using a similar assessment rubric. 232 Unit II Dynamics 04-Phys20-Chap04.qxd 7/24/08 10:59 AM Page 233 UNIT II SUMMARY Unit Concepts and Skills: Quick Reference Concepts CHAPTER 3 Force Net force Newton\u2019s first law Newton\u2019s second law Summary Forces can change velocity. Resources and Skill Building 3.1 The Nature of Force Force is a push or a pull on an object. Force is a vector quantity measured in newtons (1 N 1 kg\u2022m/s2). Net force is the vector sum of two or more forces acting simultaneously on an object. A free-body diagram helps you write the net force acting on an object. 3-1 QuickLab 3-2 QuickLab Examples 3.1\u20133.4 Examples 3.2\u20133.4 3.2 Newton\u2019s First Law of Motion Newton\u2019s first law states that an object will continue being at rest or moving at constant speed in a straight line unless acted upon by a non-zero net force. 3-3 QuickLab 3.3 Newton\u2019s Second Law of Motion Newton\u2019s second law states that when a non-zero net force acts on an object, the object accelerates in the direction of the net force. The magnitude of the acceleration is directly proportional to the magnitude of the net force and inversely proportional to the mass of the object. Newton\u2019s third law 3.4 Newton\u2019s Third Law of Motion Newton\u2019s third law states that if object A exerts a force on object B, then B exerts a force on A that is equal in magnitude and opposite in direction. Types of friction Factors affecting friction Coefficients of static and kinetic friction 3.5 Friction Affects Motion Friction is a force that opposes either the motion of an object or the direction the object would be moving in if there were no friction. Static friction is present when an object is stationary but experiences an applied force. Kinetic friction is present when an object is moving. The magnitude of the force of friction acting on an object is directly proportional to the normal force on the object. The coefficients of friction are proportionality constants that relate the magnitude of the force of friction to the magnitude of the normal force. Temperature, moisture, and the smoothness or roughness of the contact surfaces, and the materials forming the contact surface are some factors that affect the value of the coefficients of friction. 3-5 Inquiry Lab Figures 3.33\u20133.35 3-6 Design a Lab Examples 3.5\u20133.11 3-7 QuickLab Figures 3.53\u20133.57, 3.63, 3.65 3-8 QuickLab Examples 3.12, 3.13 3-9 Design a Lab 3-10 QuickLab Figures 3.70, 3.74, 3.78\u20133.80 Examples 3.14\u20133.16 3-11 Inquiry Lab Table 3.4 Examples 3.17\u20133.20 CHAPTER 4 Gravity extends throughout the universe. Gravitational force 4.1 Gravitational Forces due to Earth Gravitational force is a fundamental force, and can be described as an action-at-a-distance force or as a field. Gravitational field strength Gravitational field strength is the ratio of gravitational force to mass at a specific location. The units of gravitational field strength are N/kg. 4-1 QuickLab 4-2 QuickLab Figure 4.10 Newton\u2019s law of universal gravitation Calculating g anywhere in the universe Variations of g True weight, apparent weight, free fall, and weightlessness 4.2 Newton\u2019s Law of Universal Gravitation Newton\u2019s law of universal gravitation states that the gravitational force of attraction between any two masses is directly proportional to the product of the masses and inversely proportional to the square of the separation distance between the centres of both masses. 4.3 Relating Gravitational Field Strength to Gravitational Force Newton\u2019s law of gravitation can be used to determine the magnitude of gravitational field strength anywhere in the universe. The magnitude of gravitational field strength at a location is numerically equal to the magnitude of gravitational acceleration. The value of g at Earth\u2019s surface depends on latitude, altitude, the composition of Earth\u2019s crust, and Earth\u2019s rotation about its axis. The true weight of an object is equal to the gravitational force acting on the mass, and depends on location. Apparent weight is the negative of the normal force acting on an object. Free fall is the condition where the only force acting on an object is the gravitational force. True weightlessness is the condition in which w 0 for an object and F 0 on the object. g Figures 4.11\u20134.13, 4.16, 4.24 Examples 4.1\u20134.4", " 4-3 Design a Lab Examples 4.5\u20134.6 Figures 4.34, 4.35 Figures 4.36\u20134.40, 4.43, 4.44, 4.46 4-4 QuickLab Examples 4.7, 4.8 Unit II Dynamics 233 04-Phys20-Chap04.qxd 7/24/08 10:59 AM Page 234 UNIT II REVIEW Vocabulary 1. Using your own words, define these terms, concepts, principles, or laws. action-at-a-distance force action force apparent weight coefficient of friction field free-body diagram free fall gravitational field strength gravitational force gravitational mass inertia inertial mass kinetic friction net force Newton\u2019s first law Newton\u2019s law of gravitation Newton\u2019s second law Newton\u2019s third law normal force reaction force static friction tension true weight Knowledge CHAPTER 3 2. An object experiences three forces: F 1 is 60 N [22.0], F 3 is 83 N [300]. 2 is 36 N [110], and F Explain, using words and diagrams, how to calculate the net force on the object. What is the net force? 3. An object experiences zero net force. Work with a partner to describe the possibilities for its motion. 4. A person with a plaster cast on an arm or leg experiences extra fatigue. Use Newton\u2019s laws to explain to a classmate the reason for this fatigue. 5. Use inertia and Newton\u2019s first law to explain how the spin cycle in a washing machine removes water from wet clothes. 234 Unit II Dynamics 6. A load is placed on a 1.5-kg cart. A force of 6.0 N [left] causes the cart and its load to have an acceleration of 3.0 m/s2 [left]. What is the inertial mass of the load? 7. What happens to the acceleration of an object if the mass is constant and the net force (a) quadruples? (b) is divided by 4? (c) becomes zero? 8. Two people, A and B, are pulling a wagon on a horizontal, frictionless surface with two ropes. Person A applies a force [50] on one rope. Person B applies a force of 25 N [345] on the other rope. If the net force on the wagon is 55.4 N [26], calculate the magnitude of person A\u2019s applied force. A \u03b8 1 50\u00ba x \u03b8 2 345\u00ba B 9. A book is at rest on a table. The table is exerting an upward force on the book that is equal in magnitude to the downward force exerted by the book on the table. What law does this example illustrate? 10. A pencil exerts a force of 15 N [down] on a notebook. What is the reaction force? What object is exerting the reaction force? 11. Explain why the coefficients of static and kinetic friction are numerals without units. 12. Draw a free-body diagram for a stationary 5.0-kg block resting on a rough incline forming an angle of 30.0 with the horizontal. (a) Explain why the block is stationary. (b) Explain why a free-body diagram is helpful to describe the situation. 13. How does the ability of a car slowing down on wet asphalt compare to it slowing down on wet concrete? Use the data from Table 3.4 on page 183. 04-Phys20-Chap04.qxd 7/24/08 10:59 AM Page 235 CHAPTER 4 26. A car is stopped at a stoplight facing due east. 14. Compare the acceleration due to gravity and the gravitational field strength at the top of a tall skyscraper on Earth. 15. Consider the quantities gravitational force, mass, and gravitational field strength. Which of these quantities affects the inertia of an object? 16. Suppose an athlete were competing in the 2010 Winter Olympics in Vancouver and Whistler, British Columbia. Whistler has an elevation of 2182 m at the top and 652 m at the base. If the ski jumping and bobsled events are held near the top of the mountain rather than at the base, how might the results of these events be affected? 17. Two bags, each containing 10 oranges of equal mass, are hung 4 m apart. In a small group, describe two situations, one involving mass and the other involving separation distance, that would double the gravitational force exerted by one bag on the other. Explain your answer. 18. A student working on a satellite problem got an Ns2 m answer of 57.3. What physical quantity was the student solving for? 19. Use an example to explain the meaning of the statement: \u201cThe gravitational force exerted by Mars on a space probe varies inversely as the square of the separation distance between the centre of Mars and the centre of the probe.\u201d 20. Is an object in free fall weightless? Explain your reasoning. 21.", " Compare the gravitational force exerted by Earth (mass M) on two satellites (masses m and 2m) in orbit the same distance from Earth. 22. Compare the magnitude of Earth\u2019s gravitational field strength at the equator and at the North Pole. Explain your answer to a classmate. 23. On Earth, how does the mass of an object affect the values of the quantities below? Explain your answers. (a) acceleration due to gravity (b) gravitational field strength Applications 24. Two horizontal forces act on a soccer player: 150 N [40.0] and 220 N [330]. Calculate the net force on the player. 25. Calculate the acceleration of a 1478-kg car if it experiences a net force of 3100 N [W]. When the light turns green, the car gradually speeds up from rest to the city speed limit, and cruises at the speed limit for a while. It then enters a highway on-ramp and gradually speeds up to the highway speed limit all the while heading due east. Sketch a free-body diagram for the car during each stage of its motion (five diagrams in total). 27. A net force of magnitude 8.0 N acts on a 4.0-kg object, causing the velocity of the object to change from 10 m/s [right] to 18 m/s [right]. For how long was the force applied? 28. Two people, on opposite banks, are towing a boat down a narrow river. Each person exerts a force of 65.0 N at an angle of 30.0 to the bank. A force of friction of magnitude 104 N acts on the boat. (a) Draw a free-body diagram showing the horizontal forces acting on the boat. (b) Calculate the net force on the boat. 29. A force acting on train A causes it to have an acceleration of magnitude 0.40 m/s2. Train A has six cars with a total mass of 3.0 105 kg, and a locomotive of mass 5.0 104 kg. Train B has a locomotive of the same mass as train A, and four cars with a total mass of 2.0 105 kg. If the same force acts on train B, what will be its acceleration? Ignore friction. 30. A submarine rescue chamber has a mass of 8.2 t and safely descends at a constant velocity of 10 cm/s [down]. If g 9.81 m/s2, what is the upward force exerted by the cable and water on the chamber? 31. A 240-kg motorcycle and 70-kg rider are travelling on a horizontal road. The air resistance acting on the rider-bike system is 1280 N [backward]. The road exerts a force of static friction on the bike of 1950 N [forward]. What is the acceleration of the system? 32. The velocity of a 0.25-kg model rocket changes from 15 m/s [up] to 40 m/s [up] in 0.60 s. Calculate the force that the escaping gas exerts on the rocket. 33. Two boxes, A and B, are in contact and initially stationary on a horizontal, frictionless surface. Box A has a mass of 60 kg and box B a mass of 90 kg. A force of 800 N [right] acts on box A causing it to push on box B. (a) What is the acceleration of both boxes? (b) What is the magnitude of the action-reaction forces between the boxes? Unit II Dynamics 235 04-Phys20-Chap04.qxd 7/24/08 10:59 AM Page 236 34. A person exerts a force of 1.5 N [right] to pull a 2.0-kg block of glass at constant velocity along a horizontal surface on the Moon (gMoon kinetic friction for the glass on the surface? 1.62 m/s2). What is the coefficient of 35. Three oak blocks, mA 4.0 kg, mB 3.0 kg, are positioned next to each other on mC a dry, horizontal oak surface. Use the data from Table 3.4 on page 183. 6.0 kg, and (a) What horizontal force must be applied to accelerate the blocks at 1.4 m/s2 [forward], assuming the blocks are moving at a constant velocity? (b) Calculate the force exerted by mB on mC. (c) Calculate the force exerted by mB on mA. 38. Three objects, A, B, and C, are connected together by light strings that pass over light, frictionless pulleys. The coefficient of kinetic friction for object B on the surface is 0.200. (a) What is the acceleration of object B? (b) What is the tension in each string? Explain why the tensions are different. (c) Draw a free-body diagram for object B. (d) Identify four action", "-reaction pairs associated with object B. mB 2.0 kg B Fapp B A C A mA 6.0 kg C 4.0 kg mC 36. A 10.0-kg block is placed on an incline forming an angle of 30.0 with the horizontal. Calculate the acceleration of the block if the coefficient of kinetic friction for the block on the incline is 0.20. m 30.0\u00b0 37. A rehabilitation clinic has a device consisting of a light pulley attached to a wall support. A patient pulls with a force of magnitude 416 N. The rope exerts a force of friction of magnitude 20 N on the pulley. Calculate the acceleration of mA and mB. wall support shaft mA 15 kg mB 20 kg A B 236 Unit II Dynamics 39. The gravitational force on an object located 2rEarth from Earth\u2019s centre is 200 N [toward Earth\u2019s centre]. What is the gravitational force if the object is 10rEarth from Earth\u2019s centre? 40. A 50-kg diver steps off a 9.0-m diving tower at the same time as a 100-kg diver. Work with a partner to compare the times taken for the two divers to reach the water. Ignore air resistance. 41. Skylab 1, the first American space station, had a mass of about 68 t. It was launched into orbit 435 km above Earth\u2019s surface. Calculate the gravitational field strength at the location of Skylab 1 at this altitude. Use the data from Table 4.1 on page 218. 42. A spring scale is used to measure the gravitational force acting on a 4.00-kg silver block on Earth\u2019s surface. If the block and spring scale are taken to the surface of Mars, by how much does the reading on the spring scale change? Use the data from Table 4.1 on page 218. 43. A 60-kg student is standing on a scale in an elevator on Earth. What will be the reading on the scale when the elevator is (a) (i) moving down at constant speed? (ii) at rest at a floor? (iii) accelerating at 4.9 m/s2 [up]? (iv) accelerating at 3.3 m/s2 [down]? (b) What is the student\u2019s apparent weight and true weight in all the situations in part (a)? 04-Phys20-Chap04.qxd 7/24/08 10:59 AM Page 237 44. A 60-kg skydiver falls toward Earth with an unopened parachute. The air resistance acting on the skydiver is 200 N [up]. What is the true weight and acceleration of the skydiver? 45. A group of tourists on a ledge overlooking Pulpit Rock, Northwest Territories, dislodge a 25-kg boulder. The rock takes 8.0 s to fall 300 m into the water below. At this location, the gravitational field strength is 9.81 N/kg [down]. (a) Calculate the average acceleration of the boulder. (b) Calculate the average air resistance acting on the boulder. (c) What was the average apparent weight of the boulder during its fall? Extensions 46. The value of g on the Moon is less than that on Earth. So a pendulum on the Moon swings slower than it would on Earth. Suppose a pendulum is 36 cm long. Use the equation T 2 l g to calculate the period, T, at the equator on Earth and on the Moon. Use the data from Table 4.1 on page 218. 47. During the last seconds of a hockey game, the losing team replaces their goalie with a good shooter. The other team shoots the 150-g puck with an initial speed of 7.0 m/s directly toward the unguarded net from a distance of 32 m. The coefficient of kinetic friction for the puck on the ice is 0.08. (a) What is the force of kinetic friction acting on the puck? (b) What is the acceleration of the puck? (c) How long does it take the puck to stop? (d) Will the puck reach the net if no other player touches it? 48. Construct a gathering grid to distinguish among Newton\u2019s three laws. In the left column, identify the criteria you will use to compare and contrast the laws. Add three additional columns, one for each law. Then place checkmarks in the appropriate columns to compare the laws. 49. During the 2000 Sydney Olympics, some swimmers wore special swimsuits designed to reduce water resistance. Compare the arguments that people might make to defend or oppose the standardization of athletic equipment in the interests of fair play. 50. List two different stakeholders in the airbag debate and describe how their positions on the issue compare. 51. The G rocket of the former Soviet Union has a mass of about 3.8 106 kg and its first", "-stage engines exert a thrust of about 5.0 107 N [up]. (a) What is the true weight of the rocket on Earth\u2019s surface? (b) Calculate the net force acting on the rocket at liftoff. (c) Calculate the initial acceleration of the rocket. (d) What should happen to the acceleration if the force exerted by the engines remains constant as the fuel burns? (e) Why is the first stage jettisoned after the fuel is consumed? 52. Suppose the mass of the person sitting next to you is 70 kg and the separation distance between you and that person is 1.0 m. The mass of Mars is 6.42 1023 kg and the separation distance between Mars and Earth is 2.3 1011 m. Compare the gravitational force exerted by Mars on you and the gravitational force exerted by the person sitting next to you on you. 53. In a small group, research the materials being used to make artificial joints such as hips and knees. Find out how they are designed to provide enough friction for stability but not so much friction that the joints cannot move. Begin your search at www.pearsoned.ca/school/physicssource. 54. Manufacturers of skis recommend different waxes for different snow temperatures. Design and carry out an experiment to test the recommendations for three different waxes. 55. Research gait analysis is the study of how humans walk and run. This topic is central to physiotherapy, orthopedics, the design of artificial joints and sports footwear, and the manufacture of orthotics. How do Newton\u2019s three laws apply to gait analysis? Interview people associated with rehabilitation and sports. Write a brief summary of your findings. Begin your search at www.pearsoned.ca/school/physicssource. e TEST To check your understanding of dynamics, follow the eTest links at www.pearsoned.ca/school/physicssource. Unit II Dynamics 237 05-Phys20-Chap05.qxd 7/24/08 12:50 PM Page 238 U N I T III Circular Motion, Circular Motion, Work, and Work, and Energy Energy The International Space Station is a silent companion to Earth, placed into an orbit that is a precise balance of kinetic and gravitational potential energies. The International Space Station stays in orbit because physicists and engineers applied the laws of physics for circular motion and conservation of energy to determine the satellite\u2019s speed and height above Earth. e WEB To learn more about the types of satellites placed in orbit, and the paths that they take, follow the links at www.pearsoned.ca/school/physicssource. 238 Unit III 05-Phys20-Chap05.qxd 7/24/08 12:50 PM Page 239 Unit at a Glance C H A P T E R 5 Newton\u2019s laws can explain circular motion. 5.1 Defining Circular Motion 5.2 Circular Motion and Newton\u2019s Laws 5.3 Satellites and Celestial Bodies in Circular Motion C H A P T E R 6 In an isolated system, energy is transferred from one object to another whenever work is done. 6.1 Work and Energy 6.2 Mechanical Energy 6.3 Mechanical Energy in Isolated and Non-isolated Systems 6.4 Work and Power Unit Themes and Emphases \u2022 Energy, Equilibrium, and Systems \u2022 Nature of Science \u2022 Scientific Inquiry Focussing Questions This unit focusses on circular motion, work, and energy. You will investigate the conditions necessary to produce circular motion and examine some natural and human examples. You will consider energy, its transfer, and how it interacts with objects. As you study this unit, consider these questions: \u2022 What is necessary to maintain circular motion? \u2022 How does an understanding of conservation laws contribute to an understanding of the Universe? \u2022 How can mechanical energy be transferred and transformed? Unit Project Building a Persuader Apparatus \u2022 When you have finished this unit, you will understand how energy is transferred when objects interact. You will be able to use this understanding in the design and construction of a persuader apparatus that is able to protect its passenger from injury in different types of collisions. Unit III Circular Motion, Work, and Energy 239 05-Phys20-Chap05.qxd 7/24/08 12:50 PM Page 240 Newton\u2019s laws can explain circular motion. If humans hadn\u2019t invented the wheel \u2014 the first circular motion machine \u2014 it\u2019s hard to imagine what life would be like. The number of devices that make use of the properties of circular motion is almost too large to count. Bicycles, gears, drills, transmissions, clutches, cranes, watches, and electric motors are just a few examples. The wheel and the many technologies derived from it are a uniquely human creation (Figure 5.1). But the principles of circular motion have always", " existed in nature. In fact, as you read this, you are spinning around in a large circle as Earth rotates on its axis. In Alberta, your speed is approximately 1000 km/h because of this rotation. At the same time, you are flying through space at an amazing speed of approximately 107 000 km/h as Earth revolves around the Sun. And you thought your car was fast! How is circular motion unique? What are the properties of objects moving with circular motion? In this chapter, you will explore the physics of circular motion that define and control these many technologies, as well as the motion of the planets in the solar system. C H A P T E R 5 Key Concepts In this chapter, you will learn about: uniform circular motion planetary and satellite motion Kepler\u2019s laws Learning Outcomes When you have completed this chapter, you will be able to: Knowledge describe uniform circular motion as a special case of two-dimensional motion explain centripetal acceleration and force explain, quantitatively, the relationships among speed, frequency, period, and radius for circular motion explain, qualitatively, uniform circular motion using Newton\u2019s laws of motion explain, quantitatively, the motion of orbiting bodies by using circular motion to approximate elliptical orbits predict the mass of a celestial body from orbital data explain, qualitatively, how Kepler\u2019s laws were used to develop Newton\u2019s universal law of gravitation Science, Technology, and Society explain the process of scientific inquiry illustrate how science and technology are developed to meet society\u2019s needs and expand human capabilities analyze circular motion in daily situations Figure 5.1 240 Unit III 05-Phys20-Chap05.qxd 7/24/08 12:50 PM Page 241 5-1 QuickLab 5-1 QuickLab Characteristics of Circular Motion Problem In what direction will an object moving in a circular path go when released? Materials a marble a circular barrier (e.g., rubber tubing, embroidery hoop, flexible toy car tracks) Procedure 1 Place the circular barrier on an unobstructed section of a table or the floor, then place the marble against the inside rim of the barrier (Figure 5.2). 2 You will be rolling a marble around inside the barrier. Before you do, predict where you think the marble will go when you remove the barrier. Now roll the marble around the inside rim of the barrier. 3 As the marble is rolling around the rim, lift the barrier and make a note of the direction the marble rolls. Also pay attention to the position of the marble when you lift the barrier. 4 Sketch the circular path that the marble took inside the barrier, the position of the marble when you lifted the barrier, and the path that it rolled after the barrier was lifted. 5 Repeat steps 3 and 4 several times. Each time release the marble when it is in a different position. Figure 5.2 Questions 1. Was your prediction correct? Why or why not? 2. What similarities, if any, exist between your sketches? 3. What conclusions can you draw about the motion of the marble when it was released? 4. In each sketch that you made, draw a line from the centre of the circular motion to the position of the marble when it was released. What is the angle between this line and the path of the marble when it was released? Think About It 1. Can you predict the position the marble would have to be in for it to move away from your body when it was released? 2. What can you say about the direction of the velocity of the marble at any moment in its circular path? Discuss your answers in a small group and record them for later reference. As you complete each section of this chapter, review your answers to these questions. Note any changes to your ideas. Chapter 5 Newton\u2019s laws can explain circular motion. 241 05-Phys20-Chap05.qxd 7/24/08 12:50 PM Page 242 info BIT A boomerang has two axes of rotation: one for the spin given to the boomerang as it leaves the hand, and the other for the large circle in which it moves. axle axis of rotation Figure 5.3 The axle of a wheel is part of the axis of rotation. axle: shaft on which a wheel rotates axis of rotation: imaginary line that passes through the centre of rotation perpendicular to the circular motion uniform circular motion: motion in a circular path at a constant speed 5.1 Defining Circular Motion The bicycle is not the most sophisticated machine ever created, but it does have a lot going for it. It is easy to maintain; it does not require any fuel; it is a very efficient form of transportation; and it is full of parts moving in circular motion. Perhaps most importantly for this lesson, it is easy for us to examine. When you pedal a bike, the force is transferred through the chain and the wheel turns (Figure 5.3). The wheel transmits the force to the road, which, according to Newton\u2019s third", " law, pushes back, and the bicycle moves forward. The wheel\u2019s axle is referred to as the axis of rotation because the entire wheel rotates around this shaft. If the wheels of the bike are moving at a constant speed, they are moving with uniform circular motion. That means their rotational speed is uniform. Because the wheel is circular, it can also provide a constant and uniform force to the road. This is important if you want the bike to move forward at a uniform speed. In the sections that follow, we will restrict our study of circular motion to two dimensions. In other words, the circular motion described in this chapter has only one axis of rotation and the motion is in a plane. Speed and Velocity in Circular Motion If you ride through puddles after a rainfall, you\u2019ll come home with muddy water splashed on your back. Why is this? It has to do with the properties of an object moving in circular motion. As the wheel passes through the puddle, some of the water adheres to it. As the wheel rotates upward out of the puddle, excess water flies off. When it flies off, it moves along a path that is tangential to the wheel. Recall from Unit I that a tangent is a line that touches the circle at only one point and is perpendicular to the radius of the circle. In the case of the bicycle, that is the point where the water drops break free of the wheel. Unless the splashguard is big enough, it will not protect you from all the water that is flying off the wheel (Figure 5.4). path of the water drops when they leave the wheel at point A radial line A A tangential line Figure 5.4 Water that flies off the wheel at point A hits the rider in the back. The direction that the water drops fly is determined by the place where they leave the wheel. Figure 5.5 Water leaves the wheel at point A and flies off at a tangent. 242 Unit III Circular Motion, Work, and Energy 05-Phys20-Chap05.qxd 7/24/08 12:50 PM Page 243 PHYSICS INSIGHT A line tangent to the circle represents the velocity vector. It is perpendicular to the radial line at that point. v A r Figure 5.6 The velocity vector v shows the velocity of a particle at any instant as it moves on a circular path. A line drawn from point A in Figure 5.5 to the centre of the circle is called a radial line. The radial line and the tangential line are perpendicular to each other when they intersect at the same point on the circle. The tangential line represents the direction that the water drops are moving at any instant. The speed of the water drops is determined by their speed at the wheel. It is worth taking some time to review the difference between speed and velocity. Speed is a magnitude only (a scalar quantity) that does not have a direction. If a wheel is rotating with uniform circular motion, then the speed is constant, even though the wheel is continually turning and the water on the wheel is continually changing direction. Velocity, on the other hand, has a magnitude and a direction: it is a vector quantity. Because of this, even though the wheel is spinning at constant speed, the velocity of the water on the wheel is continually changing as the wheel\u2019s direction of motion changes. That is, the water on the wheel is continually accelerating. It is correct to say that the speed of the water drops on the wheel represents the magnitude of the velocity. Knowing that the object moves in a circular path is often sufficient. However, we must specify the object\u2019s velocity if we need to know its speed and direction at any instant in its circular path. Since we can assume the speed and direction of the water drop are known at any instant, we know its velocity. A velocity vector can be drawn at position A, and the drawing can be simplified as shown in Figure 5.6. Concept Check 1. 2. Imagine a Frisbee in level flight. Where is its axis of rotation, and what is its orientation? Identify all the axes of rotation in a typical bicycle. Figure 5.7 A mountain bike has many axes of rotation. Centripetal Acceleration and Force Now imagine that you drive over a small pebble that gets lodged in the treads of your bicycle wheel. As you ride, the pebble circles around with the wheel, as shown in Figure 5.8. At one moment it is in position A, and a fraction of a second later, it is in position B. A short time has passed as the pebble moved from point A to point B. This small change in time is written as t. Chapter 5 Newton\u2019s laws can explain circular motion. 243 05-Phys20-Chap05.qxd 7/24/08 12:50 PM Page 244 The pebble has experienced a very small displacement from point A", " to point B, which is written as d. The velocities of the pebble at point A and point B are v B, respectively (Figure 5.9). C represents the centre of the circle (the axis of rotation). A and v vA B vB A d r r C vB \u2013vA v C t B vB vA A Figure 5.8 A pebble caught in the wheel of a bike moves from position A to B in a small time t, and experiences a change in velocity. Figure 5.9 As the pebble moves from point A to point B, it moves through angle and experiences a change in velocity. Figure 5.10 The change in velocity, v, points inward. As t decreases, the angle between B will become smaller and vwill point A and v v toward the centre of the circle. The math to show this is beyond the scope of this book. The speed of the pebble does not change, but its direction does. Therefore, its velocity also changes. The change in velocity (vv) can best be shown by subtracting the velocity at A from the velocity at B (using the rules of graphical vector addition) as shown in Figure 5.10. Angle is the same in both Figures 5.9 and 5.10, and the two triangles are similar. Something subtle but significant happens when we subtract the two velocity vectors. The change in velocity, vv, points inward. As the interval of time, t, becomes smaller, vv begins to point toward the centre of the circle! This can be shown using calculus, but this is beyond the scope of this book. Velocity and Acceleration Toward the Centre The changing velocity (v ) represents the change in direction of the object, not its speed. If an object has a changing velocity, then it must be accelerating. Since the changing velocity is pointing inward toward the centre of the circle, the acceleration must also be in that direction (Figure 5.11). It is called centripetal acceleration (a c ). For an object to move with circular motion, it must experience centripetal acceleration. If the circular motion is uniform, then so is the centripetal acceleration. According to Newton\u2019s second law, if a mass is accelerating, it must also experience a non-zero net force. This non-zero net force is c ). In our example, the pebble stuck in the called the centripetal force (F wheel is experiencing a force, exerted by the rubber treads, that attempts to pull it toward the centre of the circle. This is the centripetal force. Why doesn\u2019t the pebble actually move toward the centre of the wheel? It does, in a way. Remember: if the pebble were to break free of the tire\u2019s grip, it would fly off at a tangent to its circular motion. While it remains stuck in the tire, it is forced to follow the circular path because the centripetal force attempts to pull it toward the centre. In other words, centripetal force is pulling the pebble toward the centre of the circle while at the same time, the pebble is moving off in a direction at a tangent to the circle. The result is the circular path in which it actually moves. v ac Figure 5.11 Although the velocity of an object moving with uniform circular motion is tangential, the centripetal acceleration (and centripetal force) is acting toward the centre of the circle. centripetal acceleration: acceleration acting toward the centre of a circle; centre-seeking acceleration centripetal force: force acting toward the centre of a circle causing an object to move in a circular path 244 Unit III Circular Motion, Work, and Energy 05-Phys20-Chap05.qxd 7/24/08 12:50 PM Page 245 5-2 Inquiry Lab 5-2 Inquiry Lab Speed and Radius Question For an object moving with uniform circular motion, what relationship exists between the radius of its path and its speed? (Assume the object is experiencing a constant centripetal force.) Hypothesis State a hypothesis relating the radius and speed. Remember to use an \u201cif/then\u201d statement. Variables The variables in this lab are the radius of the circular path, mass of the rubber stopper, mass of the hanging weight, number of revolutions, elapsed time, period, and speed. Read the procedure and identify the controlled, manipulated, and responding variables. Materials and Equipment 1-hole rubber stopper (mass 25 g) 1.5 m of string or fishing line small-diameter plastic tube 100-g mass metre-stick felt marker safety goggles stopwatch CAUTION: Remember to swing the rubber stopper over your head and in a place clear of obstructions. Table 5.1 Data for 5-2 Inquiry Lab Required Skills Initiating and Planning Performing and", " Recording Analyzing and Interpreting Communication and Teamwork Procedure 1 Copy Table 5.1, shown at the bottom of this page, into your notebook. 2 Secure the rubber stopper to one end of the string. 3 Run the other end of the string through the plastic tube and attach the 100-g mass to it. 4 Hold the end of the string attached to the stopper at the zero mark of a metre-stick laid on a table. The zero mark of the ruler should line up with the centre of the stopper. While holding the stopper in position, pull the string taut along the ruler. 5 With the felt marker, mark the string at 20, 30, 40, 50, and 60 cm. 6 Adjust the string\u2019s position so that the 20-cm mark is positioned on the lip of the plastic tube. Record 20 cm in the \u201cRadius\u201d column of the table. 7 Grasp the plastic tube in one hand and pinch the string to the lip of the tube using your thumb or forefinger. 8 Put on your safety goggles. Begin spinning the rubber stopper in a horizontal circle above your head as you release the string. Make sure the 100-g mass is hanging freely (Figure 5.12). At first, you may have to pull the string up or down using your other hand to position the mark as the stopper is spinning. Radius (m) Time for 20 Revolutions (s) Time 1 Time 2 Average Time (s) Period (s) Speed (m/s) Figure 5.12 Chapter 5 Newton\u2019s laws can explain circular motion. 245 05-Phys20-Chap05.qxd 7/24/08 12:50 PM Page 246 9 Adjust the rate at which you spin the rubber stopper so that the string does not slip up or down and the mark stays in position at the lip of the tube. Once you have reached a steady rate, your partner can begin timing. Do not pull the string up or down with the other hand. 10 Your partner should time 20 complete revolutions of the rubber stopper using a stopwatch. While your partner does this, be sure to monitor the speed of the stopper so that the mark does not move off the lip. Record the time in the \u201cTime 1\u201d column of the table. 2. For each trial, divide the average time by the number of revolutions the stopper made. Record the values for the time it takes to make one revolution in the \u201cPeriod\u201d column. 3. For each trial, determine the speed of the stopper. The distance travelled is the circumference of the circle and the time is the period. 2 r Use the equation v, and record the value in T the \u201cSpeed\u201d column. 4. Identify the controlled, responding, and manipulated 11 Repeat step 10 and record the time in the \u201cTime 2\u201d variables. column of the table. 12 Increase the radius by 10 cm, and record this radius in the \u201cRadius\u201d column of the table. Repeat steps 7 to 12 until all radii are used. Analysis 1. For each trial, average the two times and place the result in the \u201cAverage Time\u201d column. 5. Identify the force that acted as the centripetal force, and determine its value. 6. Plot a graph of velocity versus radius. Remember to plot the manipulated variable on the horizontal axis and the responding variable on the vertical axis. 7. Complete the statement, \u201cThe speed varies with \u2026\u201d 8. Was your hypothesis accurate? If not, how can you modify it to reflect your observations from this lab? motion of ball if released force exerted by hand on rope (action force) Figure 5.13 The hand exerts a centripetal force on the rope and ball, but feels the reaction force exerted by the rope. This leads to a false impression that the centripetal force is acting outward. force exerted by rope on hand (reaction force) Misconceptions About Centripetal Force A common misconception is that centripetal force acts radially outward from the centre of a circle. This is not what happens. The change in the velocity of the object is inward, and therefore, so is the centripetal acceleration and force. What causes confusion is that when you spin an object in a circle at the end of a rope, you feel the force pulling outward on your hand (Figure 5.13). This outward pull is mistakenly thought to be the centripetal force. Newton\u2019s third law states that for every action there is an equal and opposite reaction. If we apply this law to our example, then the action force is the hand pulling on the rope to make the object move in a circle. This is the centripetal force. The reaction force is the force the rope exerts on your hand. It is outward. This is the force that people often believe is", " the centripetal force. The rope would not exert a force on your hand unless your hand exerted a force on the rope first. Another misconception is that centripetal force is a unique and separate force responsible for circular motion. This is not true. It is best to think of centripetal force as a generic term given to any force that acts toward the centre of the circular path. In fact, the translation from Latin of the word centripetal is \u201ccentre seeking.\u201d 246 Unit III Circular Motion, Work, and Energy 05-Phys20-Chap05.qxd 7/24/08 12:50 PM Page 247 Many different forces could actually be the centripetal force. For example, when a car turns a corner, the frictional force of the tires on the road acts as the centripetal force. If you spin an object around in a horizontal circle on a rope, the tension of the rope is the centripetal force. The force of gravity the Sun exerts on the planets is another example of a centripetal force. Sometimes several forces working together act as a centripetal force. For an object spinning in a vertical circle on a rope, two forces, gravity and tension, work together to act as the centripetal force at the top of the circle. This is because the centripetal force is a net force. It is often convenient to use it in place of the actual force or forces acting toward the centre. Table 5.2 summarizes circular motion quantities and their directions. Concept Check A pebble caught in the tread of a tire experiences a centripetal force as the tire turns. What force is responsible for it? Table 5.2 Circular Motion Quantities and Their Direction Quantity Velocity (v) Centripetal acceleration (a) Centripetal force (F net) c or F Change in velocity (v) Direction tangential to the circle toward the centre toward the centre toward the centre 5.1 Check and Reflect 5.1 Check and Reflect PHYSICS INSIGHT Some texts refer to centrifugal force. This refers to the reaction force that exists as a result of centripetal force. If the centripetal force is removed, there is no centrifugal force. e SIM For an interactive demonstration that explores the relationship among centripetal acceleration, force, and velocity, visit www.pearsoned.ca/school/ physicssource. Knowledge 1. Give an example of an object that moves with uniform circular motion and one that moves with non-uniform circular motion. 2. Identify the force acting as the centripetal force in each of the following situations: (a) A car makes a turn without skidding. (b) A ball is tied to the end of a rope and spun in a horizontal circle. (c) The Moon moves in a circular orbit around Earth. Applications 3. What is the relationship between the speed and radius of an object moving with uniform circular motion if the centripetal force is constant? 4. What is the relationship between the speed and velocity of an object moving with uniform circular motion? Extensions 5. Imagine pedalling a bike at a constant rate. Describe the motion of the bicycle if the wheels were not circular but oval. 6. Consider the relative speed between a pebble stuck in the tread of a bicycle tire and the ground. Explain why the pebble is not dislodged when it comes in contact with the ground. Suggest a method to dislodge the pebble while still riding the bike. e TEST To check your understanding of the definition of circular motion, follow the eTest links at www.pearsoned.ca/school/physicssource. Chapter 5 Newton\u2019s laws can explain circular motion. 247 05-Phys20-Chap05.qxd 7/24/08 12:51 PM Page 248 5.2 Circular Motion and Newton\u2019s Laws info BIT A spinning ball moving relative to the air causes an unbalanced force and the curving of the ball. A backspin on a ball makes it stay aloft longer. A topspin drives the ball downward. Soccer players put a spin on the ball when they kick it that makes it curve around opposing players. Professional golfers routinely strike the golf ball so that it has a spin that curves it into the wind or prevents it from rolling when it hits the ground. A baseball pitcher can throw a curving fastball at 144 km/h, making the batter\u2019s job of hitting the ball much more difficult (Figure 5.14). Figure 5.14 Pitchers can put a spin on a ball that causes it to curve. This curve is predictable. e WEB To learn more about the effect of putting a spin on a ball, follow the links at www.pearsoned.ca/school/ physicssource. In these sports and many others, putting a", " spin on the ball is an essential skill. Even though the pitch from a pitcher may be extremely difficult to predict, the behaviour of the ball isn\u2019t. If players perform the same motion reliably when they kick, throw, or hit the ball, it will always behave the same way. That is why good players, when faced with the curving soccer ball or fastball, can anticipate where the ball will be and adjust their positions. It is accurate to say that the physical properties of a spinning ball or anything moving with circular motion can be predicted. In fact, the rotational velocity, frequency, centripetal force, and radius of a spinning object can be related mathematically. M I N D S O N Spinning Objects in Sports In groups of two or three, think of as many sports as you can that involve a spinning motion. It may be the player or an object that has the spin. Indicate what advantages the spinning motion has for the player and what type of motion is used to cause the spin. Discuss your answers with the class. 248 Unit III Circular Motion, Work, and Energy 05-Phys20-Chap05.qxd 7/24/08 12:51 PM Page 249 Period and Frequency of Circular Motion A baseball pitcher can throw a baseball at speeds of about 145 km/h. By flicking his wrist, he can give the ball a spin so that, in effect, it has two velocities: a velocity as it approaches the batter, and a rotational velocity because of its spin (Figure 5.15). The rotational velocity can be measured indirectly by measuring the time it takes for one complete rotation. One complete rotation is called a cycle or revolution, and the time for one cycle is the period (T), measured in s/cycle. (a) P (b) v (c) P v v P (d) v P cycle: one complete back-andforth motion or oscillation revolution: one complete cycle for an object moving in a circular path period: the time required for an object to make one complete oscillation (cycle) (e) P v Figure 5.15 Point P on the spinning baseball makes one complete rotation from (a) to (e). The time for this is called the period and is measured in s/cycle. This is frequently abbreviated to s for convenience. If an object is spinning quickly, the period may be a fraction of a second. For example, a hard drive in a computer makes one complete revolution in about 0.00833 s. This value is hard to grasp and is inconvenient to use. It is often easier to measure the number of rotations in a certain amount of time instead of the period. Frequency (f) is a measurement that indicates the number of cycles an object makes in a certain amount of time, usually one second. The SI units for frequency are cycles/s or hertz (Hz). You might have noticed that the units for frequency are cycles/s while the units for period are s/cycle. Each is the inverse of the other, so the relationship can be expressed mathematically as: T 1 f or f 1 T You may also have seen rotational frequency expressed in rpm. Even though this unit for measuring frequency is not an SI unit, it is commonly used commercially in products such as power tools. An rpm is a revolution per minute and is different from a hertz. It represents the number of revolutions in one minute instead of one second, so it is always 60 times bigger than the value in Hz. A simple method can be used to convert Hz to rpm and vice versa: Hz 60 s/min 60 s/min rpm frequency: the number of cycles per second measured in hertz (Hz) rpm: revolutions per minute Chapter 5 Newton\u2019s laws can explain circular motion. 249 05-Phys20-Chap05.qxd 7/24/08 12:51 PM Page 250 Example 5.1 The hard drive in Figure 5.16 stores data on a thin magnetic platter that spins at high speed. The platter makes one complete revolution in 0.00833 s. Determine its frequency in Hz and rpm. Practice Problems 1. The propeller of a toy airplane rotates at 300 rpm. What is its frequency in hertz? 2. An electric motor rotates at a frequency of 40 Hz. What is its rotational frequency in rpm? 3. A medical centrifuge is a device that separates blood into its parts. The centrifuge can spin at up to 6.0 104 rpm. What is its frequency in hertz and what is its period? Answers 1. 5.00 Hz 2. 2.4 103 rpm 3. 1.0 103 Hz; 1.0 103 s Given T 0.00833 s Required frequency in Hz and rpm T 1 f f 1 0.00833 s 120 Hz hard drive case Now convert the SI units of frequency to rpm: Hz 60 s/min 60 s/min rpm Analysis and Solution", " The frequency is the inverse of the period. Solve the frequency of the hard drive in the SI unit for frequency (Hz) and then convert the Hz to rpm. hard drive platter direction of rotation read/write head Figure 5.16 120 Hz 60 s min 7.20 103 rpm Paraphrase The frequency of the hard drive is 120 Hz or 7.20 103 rpm. Speed and Circular Motion At the beginning of this chapter you learned that, at this moment, you are moving at approximately 107 000 km/h as Earth moves in its orbit around the Sun. It\u2019s hard to imagine that Earth is moving that fast through our solar system. How was the speed determined? The answer is the simple application of an equation you learned in section 1.2 of chapter 1: v d t where d is the distance travelled and t is the time that it takes the object to travel that distance. 250 Unit III Circular Motion, Work, and Energy 05-Phys20-Chap05.qxd 7/24/08 12:51 PM Page 251 In the case of circular motion, the distance around a circle is the circumference (C), given by C 2r. The time it takes for one revolution is the period (T). Therefore, the speed of anything moving with uniform circular motion can be described by the equation: v 2r T (1) where r is the radius in metres, T is the period in seconds, and v is the speed in metres per second. Let\u2019s look at Earth as it follows a circular orbit around the Sun. Earth has an orbital radius of approximately 1.49 108 km, and makes one complete revolution in 365.24 days. By substituting these values into the equation for speed, we can do the following calculation: PHYSICS INSIGHT Remember: The SI units for distance and time are metres and seconds, respectively. v 2r T 2(1.49 1011 m) 365.24 24 60 60 s 2.97 104 m/s Then convert this to kilometres per hour: m 36 m k 2.97 104 m s 1 0 0 10 1 s 00 1.07 105 km/h h Earth\u2019s speed as it orbits the Sun is approximately 107 000 km/h. The speed of a planet as it rotates on its axis can be determined in the same way but varies depending on the latitude. We will explore the reasons for this later in \u201cCentripetal Force, Acceleration, and Frequency\u201d in section 5.2. Example 5.2 A pebble is stuck in the treads of a tire at a distance of 36.0 cm from the axle (Figure 5.17). It takes just 0.40 s for the wheel to make one revolution. What is the speed of the pebble at any instant? Given r 36.0 cm 0.360 m T 0.40 s Required speed (v ) of the pebble Analysis and Solution Determine the speed by using equation 1: r 2 v T 2(0.360 m) 0.40 s 5.7 m/s v 36.0 cm 36.0 cm 36.0 cm up left down right Figure 5.17 The speed of the pebble is determined by its distance from the axis of rotation, and the wheel\u2019s period (T). Practice Problems 1. How much time does it take for the tires of a racecar to make one revolution if the car is travelling at 261.0 km/h and the wheels have a radius of 0.350 m? 2. In 2006, an Alberta astronomer discovered the fastest spinning collapsed star (called a pulsar) ever found. It has a radius of only 16.1 km and is spinning at a rate of 716 Hz (faster than a kitchen blender). What is its speed at its equator? Paraphrase The speed of the pebble caught in the tire tread is 5.7 m/s. Answers 1. 0.0303 s 2. 7.24 107 m/s Chapter 5 Newton\u2019s laws can explain circular motion. 251 05-Phys20-Chap05.qxd 7/24/08 12:51 PM Page 252 A Closer Look at Centripetal Acceleration and Force In Unit II, you learned that, just before a race, dragster drivers spin their tires to make them hot and sticky so that the coefficient of friction increases between the tire and the road. When the race starts, the tires will have a better grip and the dragster will be able to accelerate at a greater rate. While the dragster performs the tire-spin, the tires change shape (Figure 5.18). The rear tires start off being fat and thick. During the spin they become thin and their diameter increases. Clearly, this must have something to do with the spinning motion of the wheel, and therefore centripetal force, but what? Figure 5.18 (a) At first", ", the dragster\u2019s wheels have a low rotational speed and don\u2019t stretch noticeably. The centripetal acceleration and force are small. (b) The dragster\u2019s wheels are spinning very fast and stretch away from the rim. The centripetal acceleration and force are large. Before the tires start spinning, they are in their natural shape. When they are spinning, they experience a strong centripetal acceleration and force. Both act toward the centre of the wheel. Each tire is fastened to a rim, so the rim is pulling the tire inward. However, the tire moves the way Newton\u2019s first law predicts it will \u2014 it attempts to keep moving in a straight line in the direction of the velocity. Thus the tire, being made of rubber, stretches. Dragster tires and the tires of trucks and passenger cars are designed to endure high speeds without being torn apart. The faster a wheel spins, the greater the centripetal acceleration and force. Car tires are thoroughly tested to ensure they can handle a centripetal force much greater than the centripetal force at the speed that you would normally drive in the city or on the highway. However, if you drove at a speed that exceeded the tires\u2019 capabilities, the tires could be torn apart. Is speed the only factor that affects centripetal acceleration and force, or are there other factors? Are the factors that affect centripetal acceleration the same as those that affect the centripetal force? To answer these questions, let\u2019s start by taking a closer look at centripetal acceleration. 252 Unit III Circular Motion, Work, and Energy 05-Phys20-Chap05.qxd 7/24/08 12:51 PM Page 253 Factors Affecting the Magnitude of Centripetal Acceleration The rotational speed is one factor that determines the magnitude of the centripetal acceleration, but what other factors play a role? To answer this question, look at Figure 5.19. It shows the two diagrams you saw earlier as Figures 5.9 and 5.10 in section 5.1. Using these figures, we can derive an equation for centripetal acceleration. vA B vB A d r r C vB \u2013vA v C Figure 5.19 An object following in a circular path moves through a displacement as there is a change in velocity v. The triangles formed by these two vectors will d help us solve for centripetal acceleration. As already stated, the two triangles are similar. Therefore, we can compare them. For convenience, the triangles have been redrawn below without the circles (Figure 5.20). Since vA and vB have the same value, we have dropped the designations A and B in Figure 5.20. We have omitted the vector arrows because we are solving only for the magnitude of the centripetal acceleration Figure 5.20 Triangles ABC and DEF are similar, so we can use a ratio of similar sides. Triangle ABC is similar to triangle DEF in Figure 5.20. Therefore, a ratio of similar sides can be created: d v r v or v d v r (2) Chapter 5 Newton\u2019s laws can explain circular motion. 253 05-Phys20-Chap05.qxd 7/24/08 12:51 PM Page 254 Remember, v is directed toward the centre of the circle. The time it took for the velocity vectors to move the small distance d can be designated t. In this time, the v vector was created. d, we can manipulate the equation so that: Since v t d t v (3) To find acceleration toward the centre of the circle (centripetal acceleration), divide the v by t. v t ac (4) Now we can substitute equations 2 and 3 into equation 4: v d r d v ac By taking the reciprocal of the denominator and multiplying the two fractions, we have: d v v d r ac Simplified, this becomes: ac 2 v r (5) where ac is the centripetal acceleration in metres per second squared toward the centre of the circle, v is the rotational speed of the object moving with uniform circular motion in metres per second, and r is the radius of the circular motion in metres. The centripetal acceleration depends only on the speed and radius of the circular motion. Mass does not affect it, just as the mass of a falling object does not affect the acceleration of gravity caused by Earth. A truck or a marble will both experience a gravitational acceleration of 9.81 m/s2 [down]. Two objects of different masses moving in a circular path will experience the same centripetal acceleration if they have the same radius and speed. Mass does not affect centripetal acceleration, but companies that manufacture racing tires, jet engines, and other equipment know they cannot ignore it. In fact, the mass of a tire or fan blade", " is important to them. These companies are continually looking for ways to reduce mass without decreasing the strength of their products. Why? The answer has to do with the centripetal force that these devices experience. If a fan blade or tire has a large mass, it will experience a large centripetal force and might break apart at high speeds. Reducing the mass decreases the centripetal force these parts experience, but often with a trade-off in strength. Next, we will examine the factors that influence centripetal force. 254 Unit III Circular Motion, Work, and Energy 05-Phys20-Chap05.qxd 7/24/08 12:51 PM Page 255 Example 5.3 A DVD disc has a diameter of 12.0 cm and a rotational period of 0.100 s (Figure 5.21). Determine the centripetal acceleration at the outer edge of the disc. Given D 12.0 cm 0.120 m T 0.100 s Required centripetal acceleration (ac) Analysis and Solution The magnitude of the centripetal acceleration depends on speed and radius. Convert the diameter to a radius by dividing it by 2. D 12.0 cm T 0.100 s Figure 5.21 D r 2 0 m 0.12 2 0.0600 m Determine the speed of the outer edge of the disc: r 2 v T. 0 2( m) 00 6 0 0 s 0 1 0. 3.77 m/s Now use equation 5 to determine the centripetal acceleration: ac 2 v r m 2 3.77 s 0.0600 m 2.37 102 m/s2 Note that no vector arrows appear on ac or v because we are solving for their magnitude only. Paraphrase The centripetal acceleration at the edge of the DVD disc is 2.37 102 m/s2. Practice Problems 1. You throw a Frisbee to your friend. The Frisbee has a diameter of 28.0 cm and makes one turn in 0.110 s. What is the centripetal acceleration at its outer edge? 2. A child playing with a top spins it so that it has a centripetal acceleration of 125.0 m/s2 at the edge, a distance of 3.00 cm from the axis of rotation. What is the speed at the edge of the top? 3. A helicopter blade has a diameter of 14.0 m and a centripetal acceleration at the tip of 2527.0 m/s2. What is the period of the helicopter blade? Answers 1. 4.57 102 m/s2 2. 1.94 m/s 3. 0.331 s Chapter 5 Newton\u2019s laws can explain circular motion. 255 05-Phys20-Chap05.qxd 7/24/08 12:51 PM Page 256 e TECH For an interesting interactive simulation of the relationship between velocity and centripetal force, follow the links at www.pearsoned.ca/physicssource. e MATH To graph the relationship between centripetal force and speed and determine the mass of the object from the graph, visit www.pearsoned.ca/ physicssource. Factors Affecting the Magnitude of Centripetal Force Mass does not affect centripetal acceleration, but it does influence the force needed to move an object in a circular path. From Newton\u2019s second law, we know that the net force is the product of mass and the net acceleration. Therefore, centripetal force must simply be the product of the mass and centripetal acceleration: Fnet ma mac Fc or Fc v 2 m r (6) where m is the mass in kilograms, v is the rotational speed in metres per second of the object moving with uniform circular motion, and r is the radius of the circular motion in metres. Notice that v and r are the same as in equation 5. Therefore, all the factors that affect centripetal acceleration also affect the centripetal force; namely, speed and the radius of rotation. However, the mass affects only the centripetal force. Example 5.4 Determine the magnitude of the centripetal force exerted by the rim of a dragster\u2019s wheel on a 45.0-kg tire. The tire has a 0.480-m radius and is rotating at a speed of 30.0 m/s (Figure 5.22). Practice Problems 1. An intake fan blade on a jet engine has a mass of 7.50 kg. As it spins, the middle of the blade has a speed of 365.9 m/s and is a distance of 73.7 cm from the axis of rotation. What is the centripetal force on the blade? 2. A 0.0021-kg pebble is stuck in the treads of a dirt bike", "\u2019s wheel. The radius of the wheel is 23.0 cm and the pebble experiences a centripetal force with a magnitude of 0.660 N. What is the speed of the wheel? Answers 1. 1.36 106 N 2. 8.5 m/s Given m 45.0 kg r 0.480 m v 30.0 m/s Required centripetal force exerted on the tire by the rim (Fc) Analysis and Solution Use equation 6 to solve for the centripetal force. Fc v 2 m r m 2 (45.0 kg)30.0 s 0.480 m 8.44 104 N v 30.0 m/s Fc r 0.480 m Figure 5.22 The dragster wheel experiences a centripetal force pulling it inward. Paraphrase The magnitude of the centripetal force exerted on the tire by the rim is 8.44 104 N. 256 Unit III Circular Motion, Work, and Energy 05-Phys20-Chap05.qxd 7/24/08 12:51 PM Page 257 5-3 Inquiry Lab 5-3 Inquiry Lab Speed and Centripetal Force Question What is the relationship between centripetal force and the speed of a mass moving in a horizontal circle? Hypothesis State a hypothesis relating centripetal force and speed. Remember to use an \u201cif/then\u201d statement. Variables The variables in this lab are the radius of the circular path, mass of the rubber stopper, mass of the hanging mass, number of revolutions, elapsed time, period, and speed. Read the procedure and identify the controlled, manipulated, and responding variables. Materials and Equipment 1-hole rubber stopper (mass 25 g) 1.5 m of string or fishing line 0.5-cm-diameter plastic tube 5 masses: 50 g, 100 g, 150 g, 200 g, 250 g metre-stick felt marker safety goggles stopwatch CAUTION: Remember to swing the rubber stopper over your head and in a place clear of obstructions. Table 5.3 Data for 5-3 Inquiry Lab Required Skills Initiating and Planning Performing and Recording Analyzing and Interpreting Communication and Teamwork Procedure 1 Copy Table 5.3, at the bottom of this page, into your notebook. 2 Secure the rubber stopper to one end of the string. 3 Run the other end of the string through the plastic tube and attach the 50-g mass to it. Record this mass in the \u201cMass\u201d column of the table. The force of gravity exerted on this mass is the centripetal force. 4 Hold the end of the string attached to the stopper at the zero mark of a metre-stick laid on a table. The zero mark of the ruler should line up with the centre of the stopper. While holding the stopper in position, pull the string taut along the ruler. 5 With the felt marker, mark the string at 40 cm. 6 Hold the rubber stopper in one hand and the plastic tube in the other. Adjust the string\u2019s position so that the 40-cm mark is positioned on the lip of the plastic tube. With the hand holding the plastic tube, pinch the string to the lip of the tube using your thumb or forefinger. 7 Put on your safety goggles. Begin spinning the rubber stopper in a horizontal circle above your head as you release the string (Figure 5.23). Make sure the mass is hanging freely. At first, you may have to pull the string up or down using your other hand to position the mark as the stopper is spinning. 8 Adjust the rate at which you spin the rubber stopper so that the string does not slip up or down and the mark stays in position at the lip at the top of the tube. Once you have reached a steady rate, your partner can begin timing. Do not pull the string up or down with the other hand. Mass (kg) Time for 20 Revolutions (s) Time 1 Time 2 Average Time (s) Period (s) Speed (m/s) Centripetal Force (101N) Figure 5.23 Chapter 5 Newton\u2019s laws can explain circular motion. 257 05-Phys20-Chap05.qxd 7/24/08 12:51 PM Page 258 9 Your partner should time 20 complete revolutions of the rubber stopper using a stopwatch. While your partner does this, be sure to monitor the speed of the stopper so that the mark does not move off the lip. Record the time in the \u201cTime 1\u201d column of Table 5.3. 10 Repeat step 9 and record the time in the \u201cTime 2\u201d column of Table 5.3. 11 Increase the hanging mass by 50 g, and record this mass in the \u201cMass\u201d column of the table. Repeat steps 6 to 10 until all the masses are used. Analysis 1", ". For each trial, average the two times, and place the result in the \u201cAverage Time\u201d column of the table. 2. For each trial, divide the average time by the number of revolutions the stopper made. Record the value in the \u201cPeriod\u201d column of the table. 3. For each trial, determine the speed of the stopper 2 r using the equation v. Record the value in the T \u201cSpeed\u201d column of the table. 4. For each trial determine the force of gravity acting on the mass hanging from the string using the equation Fg mg. Record these values in the \u201cCentripetal Force\u201d column of the table. 5. Identify the controlled, responding, and manipulated variables. 6. Plot one graph of speed versus centripetal force and a second graph of the square of the speed versus centripetal force. Remember to plot the manipulated variable on the horizontal axis and the responding variable on the vertical axis. 7. Complete the statement, \u201cThe speed varies with \u2026.\u201d 8. Was your hypothesis accurate? If not, how can it be modified to reflect your observations in this lab? inward outward Ff centre of curvature Figure 5.24 A car turning left. The force of friction of the tires on the road is the centripetal force. PHYSICS INSIGHT On a horizontal surface the force of gravity (Fg) is equal and opposite to the normal force (FN). A Horizontal System in Circular Motion Imagine that a car is following a curve to the left on a flat road (Figure 5.24). As the car makes the turn its speed remains constant, and it experiences a centripetal force. The centripetal force is caused by the wheels turning to the left, continually changing the direction of the car. If it weren\u2019t for the frictional force between the tire and the road, you would not be able to make a turn. Hence, the frictional force between the tires and the road is the centripetal force. For simplicity, this is written as: Fc Ff Recall that the magnitude of the force of friction is repre FN, where FN is the normal force, sented by the equation Ff or perpendicular force exerted by the surface on the object. For an object on a horizontal surface, the normal force is equal and opposite to the force of gravity. Recall from Unit II that the coefficient of friction,, is the magnitude of the force of friction divided by the magnitude of the normal force. Think of it as a measure of how well two surfaces slide over each other. The lower the value, the easier the surfaces move over one another. Assume that the driver increases the speed as the car turns the corner. As a result, the centripetal force also increases (Figure 5.25). Suppose the force of friction cannot hold the tires to the road. In other words, the force of friction cannot exert the needed centripetal force because of the increase in speed. In that case, the car skids off the road tangentially. Recall that kinetic friction is present when a car slides with its wheels locked. If a car turns a corner without skidding, static friction is present. 258 Unit III Circular Motion, Work, and Energy 05-Phys20-Chap05.qxd 7/24/08 12:51 PM Page 259 outward inward up down Ff FN Fg F f The maximum Figure 5.25 F c frictional force that can be exerted between the road and the tires determines the maximum speed the car can go around the turn without skidding. For horizontal surfaces, the normal force is equal and opposite to the force of gravity (F N F g). Example 5.5 Determine the maximum speed at which a 1500.0-kg car can round a curve that has the radius of 40.0 m, if the coefficient of static friction between the tires and the road is 0.60. Given m 1500.0 kg 0.60 s r 40.0 m g 9.81 m/s2 Required maximum speed (v ) Analysis and Solution First draw a free-body diagram to show the direction of the forces (Figure 5.26). The normal force is equal to the force of gravity on a horizontal surface. Fc Ff v 2 m FN r FN mg v 2 m (mg) r up outward inward FN down Ff m 1500.0 kg Fg Figure 5.26 v 2 gr v gr (0.60)9.81(40.0 m) m s2 15 m/s Paraphrase The fastest that the car can round the curve is 15 m/s or 55 km/h. If it attempts to go faster, the force of static friction will be insufficient to prevent skidding. Practice Problems 1. An Edmonton Oiler (m 100 kg) carves a turn with a radius of 7.17 m while skating", " and feels his skates begin to slip on the ice. What is his speed if the coefficient of static friction between the skates and the ice is 0.80? 2. Automotive manufacturers test the handling ability of a new car design by driving a prototype on a test track in a large circle (r 100 m) at ever-increasing speeds until the car begins to skid. A prototype car (m 1200 kg) is tested and found to skid at a speed of 95.0 km/h. What is the coefficient of static friction between the car tires and the track? 3. A 600.0-g toy radio-controlled car can make a turn at a speed of 3.0 m/s on the kitchen floor where the coefficient of static friction is 0.90. What is the radius of its turn? Answers 1. 7.5 m/s 2. 0.710 3. 1.0 m Chapter 5 Newton\u2019s laws can explain circular motion. 259 05-Phys20-Chap05.qxd 7/24/08 12:51 PM Page 260 Centripetal Force and Gravity The designers of amusement park rides know their physics. Rides will toss you around, but leave you unharmed. Many of these rides spin you in circles \u2014 the Ferris wheel and the roller coaster, for example. The roller coaster often has a vertical loop somewhere along its track (Figure 5.27). Why is it that when you reach the top of the loop, where the car is inverted, you don\u2019t fall out? It isn\u2019t because of the harness that they put over you before you start the ride. That just keeps you strapped into the car so you don\u2019t do something silly like stand up while the roller coaster is moving. No, the answer lies in the physics of the roller coaster\u2019s design. A Vertical System in Circular Motion All roller coasters, regardless of their appearance, are designed so that each car has sufficient velocity to remain in contact with the track at the top of the loop. At the top, the centripetal force is exerted by two forces working in the same direction: the track on the car, which is the normal force (F g). Both forces push the car toward the centre of the loop. N) and the force of gravity (F The speed of the car determines the amount of centripetal force needed to maintain a certain radius. As you saw from equation 6, the centripetal force is directly related to the square of the speed. Here is equation 6 again: Fc mv 2 r (6) But the force of gravity is independent of speed, and will always pull the car downward with the same force. To demonstrate the role that gravity plays as a portion of the centripetal force, we can look at a hypothetical situation of a roller coaster going around a loop as shown in Figures 5.28(a), (b), and (c). Assume a roller coaster car like the one in Figure 5.28 on the opposite page experiences a force of gravity that is 1000 N. This value won\u2019t change regardless of the car\u2019s position on the track or its speed. Remember, the centripetal force is the net force. In this case, it is equal to the sum of the gravitational force (F g) and the track\u2019s force (F Figure 5.28 illustrates how speed affects the roller coaster car\u2019s motion when it is sent through the loop three times. Each time, it is sent with less speed. N) on the car. Figure 5.27 Why doesn\u2019t this car fall off the track at the top of the loop? info BIT Most roller coaster loops use a shape called a clothoid, where the curvature increases at the top. This reduces the speed needed to move safely through the loop. It also adds thrills by having relatively longer vertical parts in the loop. PHYSICS INSIGHT centripetal force F c force due to gravity F g force that the track F exerts on the car N 260 Unit III Circular Motion, Work, and Energy 05-Phys20-Chap05.qxd 7/24/08 12:51 PM Page 261 FN Fg Fc Fg FN 1500 N 1000 N 500 N Fg Fc Fg FN 1000 N 1000 N 0 N Fc Fg FN 800 N 1000 N 200 N Since the normal force cannot pull upward, it cannot generate 200 N. 200 N more is needed to keep the car on a track of this radius, so the car falls off. The first time through the loop, the speed is such that the roller Figure 5.28(a) coaster requires a centripetal force of 1500 N to keep it moving in a circular path. At the top of the loop, the roller coaster car will experience a centripetal force that is the sum of the force of gravity and", " the force exerted by the track, pushing the car inward to the centre of the circle. The centripetal force acts down, so it is 1500 N. The force of gravity is constant at 1000 N so the track pushes inward with 500 N to produce the required centripetal force. The car goes around the loop with no problem. Figure 5.28(b) Suppose the next time the car goes around the track, it is moving more slowly, so that the centripetal force required is only 1000 N. In this case, the force of gravity alone can provide the required centripetal force. Therefore, the track does not need to exert any force on the car to keep it moving on the track. There is no normal force, so the force of gravity alone is the centripetal force. The car goes around the loop again with no problem. Figure 5.28(c) Now suppose the last time the car goes around the track, it is moving very slowly. The required centripetal force is just 800 N, but the force of gravity is constant, so it is still 1000 N; that is, 200 N more than the centripetal force required to keep the car moving in a circular path with this radius. If the track could somehow pull upward by 200 N to balance the force of gravity, the car would stay on the track. This is something it can\u2019t do in our hypothetical case. Since the gravitational force cannot be balanced by the track\u2019s force, it pulls the car downward off the track. The slowest that any car can go around the track would be at a speed that requires a centripetal force that has a magnitude equal to gravity. Gravity would make up all of the centripetal force. This can be expressed as: Fc Fg Remember that this equality doesn\u2019t mean that the centripetal force is a different force than the force of gravity. It means that it is the force of gravity. All roller coasters are designed so that the cars\u2019 speed is enough to create a centripetal force greater than the force of gravity to minimize the chance of the car leaving the track. The wheels of roller coaster cars also wrap around both sides of the track so the track can indeed pull upwards. Concept Check The centripetal force exerted on the Moon as it orbits Earth is caused by Earth\u2019s gravity. What would happen to the Moon\u2019s orbit if the Moon\u2019s velocity increased or slowed down? Chapter 5 Newton\u2019s laws can explain circular motion. 261 05-Phys20-Chap05.qxd 7/24/08 12:51 PM Page 262 Example 5.6 A 700.0-kg roller coaster car full of people goes around a vertical loop that has a diameter of 50.0 m (Figure 5.29). What minimum speed must the roller coaster car have at the top of the vertical loop to stay on the track? up right left down Fg D 50.0 m Figure 5.29 When the roller coaster is moving with the minimum speed to maintain its circular path, the force of gravity alone is the centripetal force. The track exerts no force on the car. Analysis and Solution For the roller coaster to stay on the track at the top of the loop with the minimum speed, the centripetal force is the force of gravity. Practice Problems 1. Neglecting friction, what is the minimum speed a toy car must have to go around a vertical loop of radius 15.0 cm without falling off? 2. What is the maximum radius a roller coaster loop can be if a car with a speed of 20.0 m/s is to go around safely? Answers 1. 1.21 m/s 2. 40.8 m e LAB For a probeware activity that investigates circular motion in a vertical plane, follow the links at www.pearsoned.ca/physicssource. PHYSICS INSIGHT All objects fall at a rate of 9.81 m/s2 regardless of their mass. To determine the radius of the loop, divide the diameter by 2. 0 m 50. r 2 25.0 m Use the equality of the centripetal force and gravity to solve for the speed: Fnet Fc Fg Fg m v 2 mg r 2 v g r v rg 25.0 m9.81 15.7 m/s m s2 The roller coaster car must have a minimum speed of 15.7 m/s to stay on the track. You can swing a full pail of water around in a circle over your head without getting wet for the same reason that a roller coaster can go around a loop without falling off the track. Let\u2019s examine the case of a mass on the end of a rope, moving in a vertical circle, and see how it compares to the roller coaster. A bucket of water is tied to the end", " of a rope and spun in a vertical circle. It has sufficient velocity to keep it moving in a circular path. Figures 5.30(a), (b), and (c) show the bucket in three different positions as it moves around in a vertical circle. 262 Unit III Circular Motion, Work, and Energy 05-Phys20-Chap05.qxd 7/24/08 12:51 PM Page 263 (a) (b) Fg (c) F T Fg Fg F c F g F T e SIM To learn more about centripetal force for vertical circular motion, visit www.pearsoned.ca/physicssource. Figure 5.30(a) The bucket is at the top of the circle. In this position, two forces are acting on the bucket: the force of gravity and the tension of the rope. Both are producing the centripetal force and are acting downward. The equation to represent this situation is: F F F c g T Figure 5.30(b) When the bucket has moved to the position where the rope is parallel to the ground, the force of gravity is perpendicular to the tension. It does not contribute to the centripetal force. The tension alone is the centripetal force. We can write this mathematically as: F F c T Figure 5.30(c) As the bucket moves through the bottom of the circle, it must have a centripetal force that overcomes gravity. The tension is the greatest here because gravity is acting opposite to the centripetal force. The equation is the same as in (a) above, but tension is acting upward, so when the values are placed into the equation this time, F g is negative. The effect is demonstrated in Example 5.7. T is positive and F Concept Check 1. A bucket filled with sand swings in a vertical circle at the end of a rope with increasing speed. At some moment, the tension on the rope will exceed the rope\u2019s strength, and the rope will break. In what position in the bucket\u2019s circular path is this most likely to happen? Explain. Is it necessary to know the position of an object moving in a vertical circle with uniform speed if you are determining centripetal force? Explain. 2. Forces Affecting an Object Moving in a Vertical Circle In summary, an object moving in a vertical circle is affected by the following forces: \u2022 The centripetal force is the net force on the object in any position. \u2022 The centripetal force is determined by the object\u2019s mass, speed, and radius. In the case of the roller coaster and bucket of water, their mass and radius of curvature are constant so only their speed affects the centripetal force. \u2022 The force of gravity is one of the forces that may contribute to the centripetal force. \u2022 The force of gravity remains constant regardless of the position of the object. Chapter 5 Newton\u2019s laws can explain circular motion. 263 05-Phys20-Chap05.qxd 7/24/08 12:51 PM Page 264 Example 5.7 A bucket of water with a mass of 1.5 kg is spun in a vertical circle on a rope. The radius of the circle is 0.75 m and the speed of the bucket is 3.00 m/s. What is the tension on the rope in position C, as shown in Figure 5.31? Given r 0.75 m m 1.5 kg v 3.00 m/s g 9.81 m/s2 Required T) tension (F up left down A right r 0.75 m B c C m 1.5 kg FT Fg v 3.00 m/s Figure 5.31 Analysis and Solution The centripetal force acting on the bucket will not change as the bucket moves in a vertical circle. The tension will change as the bucket moves in its circular path because gravity will work with it at the top and against it at the bottom. In position C, the force of gravity works downward (negative), but the centripetal force and tension act upward (positive). Tension must overcome gravity to provide the centripetal force (Figure 5.32). Practice Problems 1. Using the information in Example 5.7, determine the tension in the rope at position A in Figure 5.31. 2. Using the information in Example 5.7, determine the tension in the rope in position B in Figure 5.31. 3. A 0.98-kg rock is attached to a 0.40-m rope and spun in a vertical circle. The tension on the rope when the rock is at the top of the swing is 79.0 N [down]. What is the speed of the rock? Answers 1. 3.3 N [down] 2. 18 N [left] 3. 6.0 m/s Remember that the cent", "ripetal force is the net force and is the vector sum of all the forces acting on the bucket. Therefore, the equation is: F net Fc Fg FT FT Fc Fg up left down right FT Fg Figure 5.32 Fnet Fg FT (mg) mv 2 r m 2 (1.5 kg )3.00 s 0.75 m m (1.5 kg ) 9.81 s2 18 N [14.715 N] 33 N Paraphrase The tension on the rope at position C is 33 N [up]. This is the maximum tension the system will experience because gravity acts in the opposite direction of the centripetal force. 264 Unit III Circular Motion, Work, and Energy 05-Phys20-Chap05.qxd 7/24/08 12:51 PM Page 265 Centripetal Force, Acceleration, and Frequency At your local hardware store, you will find a variety of rotary power tools that operate by circular motion. The table saw, circular saw, impact wrench, reciprocating saw, and rotary hammer are a few (Figure 5.33). One of the selling features listed on the box of most of these tools is the rotational frequency (in rpm). At the beginning of this chapter, you learned that rpm refers to the frequency of rotation measured in revolutions per minute. This is an Imperial measurement that has been around for hundreds of years. It has probably persisted because people have a \u201cfeel\u201d for what it means. Even though revolutions per minute (rpm) is not considered an SI unit for frequency, it is a very useful measurement nevertheless. Why is the frequency of rotation often a more useful measure than the speed? The answer has to do with the nature of a rotating object. Figure 5.33 Most tools, motors, and devices that rotate with high speed report the frequency of rotation using rpm instead of the speed. The Effect of Radius on Speed, Period, and Frequency Imagine a disc spinning about its axis with uniform circular motion (Figure 5.34). Positions A, B, and C are at different radii from the axis of rotation, but all the positions make one complete revolution in exactly the same time, so they have the same period. Of course, if the periods for points A, B, and C are the same, so are their frequencies, and we can make the following generalization: For any solid rotating object, regardless of its shape, the frequency of rotation for all points on the object will be the same. Compare the speeds of points A, B, and C. Point A is the closest to the axis of rotation and has the smallest radius, followed by B, and then C with the largest radius. As already discussed, all three points make one complete revolution in the same amount of time, so point C, which has the farthest distance to cover, moves the fastest. Point B moves more slowly than C because it has less distance to travel. Point A has the slowest speed because it has the least distance to cover in the same amount of time. In essence, the speed of the spinning disc changes depending on which point you are referring to. In other words, the speed of a point on a disc changes with respect to its radius. C B A Figure 5.34 The speeds at A, B, and C are all different, whereas the rotational frequency of this disc is the same at any point. Chapter 5 Newton\u2019s laws can explain circular motion. 265 05-Phys20-Chap05.qxd 7/24/08 12:51 PM Page 266 r 3.83 106 m 60\u00ba N P Q 30\u00ba N 0\u00ba 90\u00ba N 53\u00ba N equator r 6.38 106 m S Figure 5.35 The speed of any point on Earth depends on its latitude (which determines its rotational radius). Point P moves more slowly than point Q, but they both have the same period and frequency. At the beginning of this chapter, you learned that the speed of an Albertan is roughly 1000 km/h as Earth rotates on its axis. However, not every point on Earth\u2019s surface moves with the same speed. Remember: Speed changes with radius, and on Earth, the distance from the axis of rotation changes with latitude (Figure 5.35). The fastest motion is at the equator and the slowest is at the poles, but every point on Earth has the same period of rotation \u2014 one day. Determining Centripetal Force Using Period and Frequency In earlier example problems, such as Example 5.3, a given rotational frequency or period had to be converted to speed before the centripetal acceleration or force could be determined. It would be simpler to derive the equations for centripetal acceleration and force using rotational frequency or period to save a step in our calculations. The equations for centripetal acceleration and force are: ac v 2 r and Fc mv 2 r", " Recall that: 2r T v By substituting the velocity equation into the centripetal acceleration equation, the result is: ac (2r)2 rT 2 42r 2 rT 2 ac 42r T 2 The centripetal force is Fc 42mr T 2 Fc ma, so the centripetal acceleration is: (7) (8) Period is just the inverse of frequency, so it is relatively simple to express equations 7 and 8 in terms of frequency: ac 42rf 2 and Fc 42mrf 2 To convert rpm to hertz, simply divide by 60. (9) (10) 266 Unit III Circular Motion, Work, and Energy 05-Phys20-Chap05.qxd 7/24/08 12:51 PM Page 267 Example 5.8 The compressor blades in a jet engine have a diameter of 42.0 cm and turn at 15 960 rpm (Figure 5.36). Determine the magnitude of the centripetal acceleration at the tip of each compressor blade. ac v D 42.0 cm Practice Problems 1. A space station shaped like a wheel could be used to create artificial gravity for astronauts living in space. The astronauts would work on the rim of the station as it spins. If the radius of the space station is 30.0 m, what would its frequency have to be to simulate the gravity of Earth (g 9.81 m/s2)? 2. A 454.0-g mass, attached to the end of a 1.50-m rope, is swung in a horizontal circle with a frequency of 150.0 rpm. Determine the centripetal force acting on the mass. Answers 1. 9.10 102 Hz 2. 1.68 102 N Figure 5.36 The centripetal acceleration at the tip of a blade can be determined from the frequency of the blade\u2019s rotation. Analysis and Solution First convert the frequency to SI units (Hz) and determine the radius. Then use equation 9. m n i 9 15 60 f s 0 6 m in rev 1 1 266.00 Hz D r 2 0 m 0.42 2 0.210 m ac 42rf 2 42(0.210 m)(266.00 Hz)2 5.87 105 m/s2 The magnitude of the centripetal acceleration at the tip of each compressor blade is 5.87 105 m/s2. Chapter 5 Newton\u2019s laws can explain circular motion. 267 05-Phys20-Chap05.qxd 7/24/08 12:51 PM Page 268 5.2 Check and Reflect 5.2 Check and Reflect Knowledge 1. A car heading north begins to make a right turn. One-quarter of a whole turn later, in what direction is the centripetal force acting? 2. Does the Moon experience centripetal acceleration? Why? 3. What two things could you do to increase the centripetal force acting on an object moving in a horizontal circle? 4. An object moves in a vertical circle at the end of a rope. As the object moves, explain what happens to: (a) the force of gravity, and (b) the tension. 5. What force acts as the centripetal force for a plane that is making a horizontal turn? Applications 6. A car\u2019s wheels have a radius of 0.5 m. If the car travels at a speed of 15.0 m/s, what is the period of rotation of the wheels? 7. A propeller blade has a period of rotation of 0.0400 s. What is the speed of the outer tip of the propeller blade if the tip is 1.20 m from the hub? 8. A 1500-kg car is making a turn with a 100.0-m radius on a road where the coefficient of static friction is 0.70. What is the maximum speed the car can go without skidding? 9. A car rounds a curve of radius 90.0 m at a speed of 100.0 km/h. Determine the centripetal acceleration of the car. 10. NASA uses a centrifuge that spins astronauts around in a capsule at the end of an 8.9-m metallic arm. The centrifuge spins at 35 rpm. Determine the magnitude of the centripetal acceleration that the astronauts experience. How many times greater is this acceleration than the acceleration of gravity? 268 Unit III Circular Motion, Work, and Energy 11. What minimum speed must a toy racecar have to move successfully through a vertical loop that has a diameter of 30.0 cm? 12. Determine the centripetal acceleration acting on a person standing at the equator (r Earth 6.38 106 m). 13. An ant climbs onto the side of a bicycle tire a distance of 0.40 m from the hub. If the 0.010-g ant can hold onto the tire with a force of 4", ".34 10\u20134 N, at what frequency would the tire fling the ant off assuming the wheel is spun on a horizontal plane? Extensions 14. Two pulleys are connected together by a belt as shown in the diagram below. If the pulley connected to the motor spins at 200.0 rpm, what is the frequency of the larger pulley? (Hint: Both pulleys have the same velocity at their outer edge.) r1 10 cm pulley belt r2 25 cm 15. Two NASCAR racecars go into a turn beside each other. If they remain side by side in the turn, which car has the advantage coming out of the turn? e TEST To check your understanding of circular motion and Newton\u2019s laws, follow the eTest links at www.pearsoned.ca/school/physicssource. 05-Phys20-Chap05.qxd 7/24/08 12:51 PM Page 269 5.3 Satellites and Celestial Bodies in Circular Motion Johannes Kepler (1571\u20131630) was a German mathematician with a strong interest in astronomy. When he started working for renowned Danish astronomer Tycho Brahe (1546\u20131601) in 1600, he was given the problem of determining why the orbit of Mars didn\u2019t completely agree with mathematical predictions. Brahe probably gave Kepler this job to keep him busy since he didn\u2019t share Kepler\u2019s ideas that the planets revolved around the Sun. At this time, most people, including scientists, believed that the planets and the Sun revolved around Earth. info BIT Comets are objects in space that have very elliptical orbits. They are often difficult to detect because they are relatively small and their orbits take them a great distance from the Sun. It is likely there are many comets orbiting the Sun that we have yet to discover. Kepler\u2019s Laws Brahe had made very meticulous observations of the orbital position of Mars. Kepler used this data to determine that the planet must revolve around the Sun. Kepler also hypothesized that Mars had an elliptical orbit, not a circular one. Until this time, all mathematical predictions of a planet\u2019s position in space were based on the assumption that it moved in a circular orbit. That is why Brahe\u2019s observations disagreed with mathematical predictions at the time. By recognizing that planets move in elliptical orbits, Kepler could account for the discrepancy. Kepler\u2019s First Law semi-minor axis semi-major axis 1 Sun focus focus 2 minor axis major axis An ellipse is an elongated circle. Figure 5.37 is an example of an ellipse. There are two foci, as well as major and minor axes. In Kepler\u2019s model, the Sun is at one focus and the planet\u2019s orbit is the path described by the shape of the ellipse. It is clear from Figure 5.37 that the planet will be closer to the Sun in position 1 than in position 2. This means that the planet\u2019s orbital radius must be changing as time goes by. Why then is a planet\u2019s orbital radius often written as a fixed number? There are two reasons: \u2022 The radius used is an average value. Mathematically, this average radius can be shown to be the same length as the semi-major axis. \u2022 The orbit of the planet, although an ellipse, is very close to being a circle, so the orbital radius really doesn\u2019t change much from position 1 to position 2. Figure 5.37 is an exaggeration of a planet\u2019s orbit for the purposes of clarity. The degree to which an ellipse is elongated is called the eccentricity. It is a number between 0 and 1, with 0 being a perfect circle and anything above 0 being a parabola, which flattens out to a line when it reaches 1. Table 5.4 shows the eccentricities of the orbits of the planets and other celestial bodies in our solar system. Figure 5.37 Kepler described the motion of planets as an ellipse with a semi-major axis and a semiminor axis. The average orbital radius is the semi-major axis. PHYSICS INSIGHT In an ellipse, the separation of the foci determines the shape of the ellipse. Chapter 5 Newton\u2019s laws can explain circular motion. 269 05-Phys20-Chap05.qxd 7/24/08 12:51 PM Page 270 Table 5.4 Eccentricities of Orbits of Celestial Bodies Kepler\u2019s Second Law Celestial Body Mercury Venus Earth Mars Ceres Jupiter Saturn Uranus Neptune Pluto Eris Sedna Eccentricity 0.205 0.007 0.017 0.093 0.080 0.048 0.054 0.047 0.009 0.249 0.437 0.857 e WEB Sedna was discovered in 2004 and is the largest object yet found in a region of space known as", " the Oort cloud. On July 29, 2005, an object now named Eris was confirmed. It is larger than Pluto but also in the Kuiper belt. To learn more about these celestial bodies, the Kuiper belt, and the Oort cloud, follow the links at:www.pearsoned.ca/school/ physicssource. Kepler went on to state two more revolutionary ideas relating to the orbits of planets in the solar system. His second law stated that planets move through their elliptical orbit in such a manner as to sweep out equal areas in equal times. Imagine a line that extends from the Sun to the planet. As the planet moves in its orbit, the line moves with it and sweeps out an area. In the same amount of time, at any other position in the planet\u2019s orbit, the planet will again sweep out the same area (Figure 5.38). area 1 Sun area 2 Figure 5.38 Kepler\u2019s second law states that a line drawn from the Sun to the planet sweeps out equal areas in equal times. One consequence of this rule is that a planet\u2019s speed must change throughout its orbit. Consider Figure 5.39 where area 1 is equal to area 2. As the planet approaches the Sun, the orbital radius decreases. If it is to sweep out an area equal to area 2, the planet must speed up and cover a larger distance to compensate for the smaller radius. As the planet gets farther from the Sun, the orbital radius gets larger. The planet slows down and sweeps out the same area again in the same amount of time (Figure 5.39). v 109 289 km/h v 109 289 km/h Sun v 105 635 km/h v 105 635 km/h Figure 5.39 Earth\u2019s speed varies from 105 635 km/h at its farthest distance from the Sun to 109 289 km/h when it is closest. Its speed changes to compensate for the change in its orbital radius. 270 Unit III Circular Motion, Work, and Energy 05-Phys20-Chap05.qxd 7/24/08 12:51 PM Page 271 Kepler\u2019s Third Law Kepler\u2019s third law states that the ratio of a planet\u2019s orbital period squared divided by its orbital radius cubed is a constant that is the same for all the planets orbiting the Sun. Written mathematically, it is: 2 T a K 3 ra (11) e SIM where Ta is the orbital period of planet A, ra is the orbital radius of planet A, and K is Kepler\u2019s constant. Since the constant K applies to all planets orbiting the Sun, we can equate the ratio T 2/r 3 between any two planets and write the equation: 2 2 T T b a 3 3 rb ra (12) Until Kepler noticed this relationship between the planets, there was really no indication that the third law would be true. If it hadn\u2019t been for Tycho Brahe\u2019s accurate measurements of the planets\u2019 orbital positions throughout their year, it is unlikely that Kepler would have made this discovery. Once the relationship was known, it was easy to verify, and it further bolstered the credibility of the heliocentric (Suncentred) model of the solar system. It is important to note that when Kepler derived his third law, he applied it only to planets orbiting the Sun. However, this law can be extended to moons that orbit a planet. In fact, in the most general sense, Kepler\u2019s third law is applicable to all celestial bodies that orbit the same focus. For bodies orbiting a different focus, Kepler\u2019s constant will be different. Kepler\u2019s three laws can be summarized this way: 1. All planets in the solar system have elliptical orbits with the Sun at one focus. 2. A line drawn from the Sun to a planet sweeps out equal areas in equal times. 3. The ratio of a planet\u2019s orbital period squared to its orbital radius cubed is a constant. All objects orbiting the same focus (e.g., planets, 2 2 T T b a the Sun) have the same constant. 3 3 rb ra Determining Kepler\u2019s Constant To learn more about Kepler\u2019s second law regarding eccentricity, orbital period, and speed of a planet, visit www.pearsoned.ca/ school/physicssource. e WEB The Titius-Bode law is another mathematical description that predicts the orbital radius of the planets. To learn more about the Titius-Bode law, follow the links at www.pearsoned.ca/ school/physicssource. Kepler\u2019s third law states that the constant K is the same for all planets in a solar system. The period and orbital radius of Earth are well known, so they are used to compute the constant. The mean (average) orbital distance for Earth from the Sun is 1.50 1011 m, and", " Earth\u2019s orbital period is 31 556 736 s. Kepler\u2019s constant can be calculated as shown below: PHYSICS INSIGHT The orbital radius is always measured from the centre of the orbiting body to the centre of the body being orbited. 2 T E K 3 rE (3.15567 107 s)2 (1.50 1011 m)3 2.95 1019 s2/m3 Chapter 5 Newton\u2019s laws can explain circular motion. 271 05-Phys20-Chap05.qxd 7/24/08 12:51 PM Page 272 info BIT Some scientists speculate that Pluto and Charon might have been objects from the Kuiper belt that were attracted into an orbit of the Sun by Neptune because of their small size and large orbital radius. The Kuiper belt is a large band of rocky debris that lies 30 to 50 AU from the Sun. e MATH Kepler\u2019s constant applies only to bodies orbiting the same focus. To explore this concept in more depth and to determine Kepler\u2019s constant for Jupiter\u2019s moons, visit www.pearsoned.ca/physicssource. info BIT The SI unit symbol for year is a. However, you will not often find Kepler\u2019s constant written with this value. The reason is twofold. If different units for distance and time are used (something other than metres and seconds), then the constant can be made to equal 1. This has obvious mathematical benefits. The other reason is that using metres and seconds as the units of measurement is impractical when dealing with the scale of our solar system. To represent astronomical distances, it becomes necessary to use units bigger than a metre or even a kilometre. For example, measuring the distance from Earth to the Sun in kilometres (150 000 000 km) is roughly like measuring the distance from Edmonton to Red Deer in millimetres. Astronomical Units A more suitable measurement for astronomical distances has been adopted. It is called the astronomical unit (AU). One astronomical unit is the mean orbital distance from Earth to the Sun (the length of the semi-major axis of Earth\u2019s orbit). This is a more manageable unit to use. For example, Neptune is only 30.1 AU away from the Sun on average. Kepler\u2019s constant, using units of years (a) and AU, can be deter- mined as follows: 2 T E K 3 rE )2 a (1 )3 U A (1 1 a2/AU3 The advantage of using the units of an Earth year and astronomical units becomes clear as Kepler\u2019s constant works out to 1. Any other planet in our solar system must also have the same constant because it orbits the same focus (the Sun). Be careful not to use this value of Kepler\u2019s constant for all systems. For example, if Kepler\u2019s third law is used for moons orbiting a planet, or planets orbiting a different sun, the constant will be different. Example 5.9 Practice Problems 1. Use Kepler\u2019s third law to determine the orbital period of Jupiter. Its orbital radius is 5.203 AU. 2. Pluto takes 90 553 Earth days to orbit the Sun. Use this value to determine its mean orbital radius. 3. A piece of rocky debris in space has a mean orbital distance of 45.0 AU. What is its orbital period? Mars has an orbital radius of 1.52 AU (Figure 5.40). What is its orbital period? Mars Sun r 1.52 AU Figure 5.40 Mars has a mean orbital radius of 1.52 AU (not drawn to scale). Kepler\u2019s third law can be used to determine its period. 272 Unit III Circular Motion, Work, and Energy 05-Phys20-Chap05.qxd 7/24/08 12:51 PM Page 273 Analysis and Solution Start with the equation for determining Kepler\u2019s constant where TMars is the the orbital period of Mars, and r Mars is the mean orbital radius of Mars. Answers 1. 11.87 Earth years 2. 39.465 AU 3. 302 Earth years T 2 Mars TMars Mars Kr 3 1(1.52 AU)3 a2 AU3 1.87 a The orbital period of Mars is 1.87 Earth years. In other words, it takes 1.87 Earth years for Mars to go around the Sun once. A close examination of Table 5.5 shows that as the orbital radius of a planet increases, so does its orbital period. Planets nearest the Sun have the highest orbital speeds. The years of these planets are the shortest. Planets farthest from the Sun have the longest years. Kepler\u2019s laws don\u2019t apply just to planets orbiting the Sun. They apply to all bodies that orbit the same focus in an ellipse. This means that moons orbiting a planet are also subject to Kepler\u2019s laws and have their own Kepler\u2019s constant. Earth has only one", " Moon (natural satellite) but Jupiter and Saturn have many, and more are being found all the time. Table 5.6 shows the planets and some of their known moons. Table 5.5 Solar Celestial Bodies Celestial Body Sun Mercury Venus Earth Mars Ceres Jupiter Saturn Uranus Neptune Pluto Eris Sedna Mass (kg) 1.99 1030 3.30 1023 4.87 1024 5.97 1024 6.42 1023 9.5 1020 1.90 1027 5.69 1026 8.68 1025 1.02 1026 1.20 1022? 4.21 1021 Equatorial Radius (m) 6.96 108 2.44 106 6.05 106 6.38 106 3.40 106 4.88 105 7.15 107 6.03 107 2.56 107 2.48 107 1.70 106 ~1.43 106 8.5 105 Orbital Period (days) \u2014 87.97 224.7 365.24 686.93 1679.8 4330.6 10 755.7 30 687.2 60 190 90 553 204 540 3 835 020 Mean Orbital Radius (m) \u2014 5.79 1010 1.08 1011 1.49 1011 2.28 1011 4.14 1011 7.78 1011 1.43 1012 2.87 1012 4.50 1012 5.91 1012 ~1.01 1013 7.14 1013 Distance (AU) \u2014 0.387 0.723 1 1.524 2.766 5.203 9.537 19.191 30.069 39.482 67.940 479.5 PHYSICS INSIGHT Kepler\u2019s third law can only be applied to bodies orbiting the same object. info BIT As of August 2006, the International Astronomical Union (IAU) finally developed a definition of a planet. To be a planet, a celestial body must orbit a star, be large enough that its own gravity forms it into a spherical shape, and have cleared the neighbourhood around its orbit. Pluto fails to satisfy the last criterion as its orbit is near many other Kuiper belt objects. There are now officially only eight planets in our solar system. Chapter 5 Newton\u2019s laws can explain circular motion. 273 05-Phys20-Chap05.qxd 7/24/08 12:51 PM Page 274 Concept Check 1. What type of orbit would a planet have if the semi-minor axis of its orbit equalled the semi-major axis? 2. Why can\u2019t Kepler\u2019s third law be applied to Earth\u2019s Moon and Jupiter\u2019s moon Callisto using the same value for Kepler\u2019s constant? 3. Assume astronomers discover a planetary system in a nearby galaxy that has 15 planets orbiting a single star. One of the planets has twice the mass, but the same orbital period and radius as Earth. Could Kepler\u2019s constant for our solar system be used in the newly discovered planetary system? Explain. 4. Assume astronomers discover yet another planetary system in a nearby galaxy that has six planets orbiting a single star, but all of the planets\u2019 orbits are different from Earth\u2019s. Explain how Kepler\u2019s third law could apply to this planetary system. 5. Compare and contrast planets orbiting a star with points on a rotating solid disc (Figure 5.34). Table 5.6 The Planets and Their Large Moons Planet Moons Mass (kg) Equatorial Radius (m) Orbital Period (Earth days) Mean Orbital Radius (m) Eccentricity Discovered (Year) Earth Mars Jupiter (4 most massive) Saturn (7 most massive) Uranus (5 most massive) Neptune (3 most massive) Moon 7.35 1022 1.737 106 Phobos Deimos Io Europa Ganymede Callisto Mimas Enceladus Tethys Dione Rhea Titan Iapetus Miranda Ariel Umbriel Titania Oberon Proteus Triton Nereid 1.063 1016 2.38 1015 1.340 104 7.500 103 8.9316 1022 4.79982 1022 1.48186 1023 1.07593 1023 3.75 1019 7 1019 6.27 1020 1.10 1021 2.31 1021 1.3455 1023 1.6 1021 6.6 1019 1.35 1021 1.17 1021 3.53 1021 3.01 1021 5.00 1019 2.14 1022 2.00 1019 1.830 106 1.565 106 2.634 106 2.403 106 2.090 105 2.560 105 5.356 105 5.600 105 7.640 105 2.575 106 7.180 105 2.400 105 5.811 105 5.847 105 7.889 105 7.614 105 2.080 105 1.352", " 106 1.700 105 274 Unit III Circular Motion, Work, and Energy 27.322 0.3189 1.262 1.769 3.551 7.154 16.689 0.942 1.37 1.887 2.74 4.52 15.945 79.33 1.41 2.52 4.14 8.71 13.46 1.12 5.8766 360.14 3.844 108 9.378 106 2.346 107 4.220 108 6.710 108 1.070 109 1.883 109 1.855 108 2.380 108 2.947 108 3.774 108 5.270 108 1.222 109 3.561 109 1.299 108 1.909 108 2.660 108 4.363 108 5.835 108 1.176 108 3.548 108 5.513 109 0.0549 0.015 0.0005 0.004 0.009 0.002 0.007 0.0202 0.00452 0.00 0.002 0.001 0.0292 0.0283 0.0027 0.0034 0.005 0.0022 0.0008 0.0004 0.000016 0.7512 \u2014 1877 1877 1610 1610 1610 1610 1789 1789 1684 1684 1672 1655 1671 1948 1851 1851 1787 1787 1989 1846 1949 05-Phys20-Chap05.qxd 7/24/08 12:51 PM Page 275 Example 5.10 Mars has two moons, Deimos and Phobos (Figure 5.41). Phobos has an orbital radius of 9378 km and an orbital period of 0.3189 Earth days. Deimos has an orbital period of 1.262 Earth days. What is the orbital radius of Deimos? Mars Phobos Deimos TP 0.3189 d TD 1.262 d rP 9378 km rD Figure 5.41 The orbital period and radius of Phobos can be used with the orbital period of Deimos to determine its orbital radius (not drawn to scale). Given TP rP TD 0.3189 d 9378 km 1.262 d Required orbital radius of Deimos (rD) Analysis and Solution Both Phobos and Deimos orbit the same object, Mars, so Kepler\u2019s third law can be used to solve for the orbital radius of Deimos. The units of days and kilometres do not need to be changed to years and astronomical units because Kepler\u2019s third law is simply a ratio. It is important to be consistent with the units. If we use the units of kilometres for the measure of orbital radius, then the answer will be in kilometres as well. 2 2 T T P D 3 3 rP rD rD 1.262 d)2(9378 km)3 (0.3189 d)2 rD 1.2916 1013 km3 31.2916 1013 km3 2.346 104 km Practice Problems 1. Titan is one of the largest moons in our solar system, orbiting Saturn at an average distance of 1.22 109 m. Using the data for Dione, another moon of Saturn, determine Titan\u2019s orbital period. 2. The Cassini-Huygens probe began orbiting Saturn in December 2004. It takes 147 days for the probe to orbit Saturn. Use Tethys, one of Saturn\u2019s moons, to determine the average orbital radius of the probe. 3. Astronomers are continually finding new moons in our solar system. Suppose a new moon X is discovered orbiting Jupiter at an orbital distance of 9.38 109 m. Use the data for Callisto to determine the new moon\u2019s orbital period. Answers 1. 16.0 d 2. 5.38 106 km 3. 186 d Paraphrase The orbital radius of Deimos is 2.346 104 km. This answer is reasonable since a moon with a larger orbital period than Phobos will also have a larger orbital radius. Chapter 5 Newton\u2019s laws can explain circular motion. 275 05-Phys20-Chap05.qxd 7/24/08 12:51 PM Page 276 5-4 Design a Lab 5-4 Design a Lab Orbital Period and Radius Kepler determined that planets have slightly elliptical orbits. He also determined empirically that the period and radius of the planet\u2019s orbit were related. The purpose of this lab is to design a method to highlight the relationship between the period and radius of planets\u2019 orbits around the Sun. The Question How can the relationships between the orbital period and orbital radius of planets orbiting the Sun be shown by graphical means? Design and Conduct Your Investigation This lab should be designed to investigate the relationship between period and radius. State a hypothesis relating the orbital period to the radius for planets orbiting the Sun. Remember to use an \u201cif/then\u201d statement.", " The lab should show how a planet\u2019s orbital radius affects its orbital period. To do this, use Table 5.5 to select data for at least five planets\u2019 periods and orbital radii. Organize the relevant data from Table 5.5 into a chart in a suitable order that can be used to plot a graph using a graphing calculator or other acceptable means. Choose appropriate units for the manipulated and responding variables. Look at the type of graph that results to draw a conclusion relating period and radius. Test the relationship you discovered from this lab with the relationship that exists between the orbital radius and period of a planet\u2019s moons. Do this by picking a planet from Table 5.6 that has several moons and perform the same procedure and analysis. Compare the relationships of orbital radius and period of the planets with that of the moons. Comment on any similarities or differences. e WEB In August 2006, the IAU decided to create three classifications of solar system objects: planets, dwarf planets, and small solar system objects. Pluto is now classified as a dwarf planet, along with Eris and Ceres (formerly an asteroid). To be a dwarf planet, the celestial body must orbit the Sun and be large enough that its own gravity forms it into a spherical shape, but not large enough that it has cleared the neighbourhood around its orbit. To learn about dwarf planets, follow the links at: www.pearsoned.ca/school/ physicssource. Moon v path of moon g g Earth Figure 5.42 The Moon is falling to Earth with the acceleration of gravity just like the apple does. But the Moon also has a tangential velocity (v) keeping it at the same distance from Earth at all times. Newton\u2019s Version of Kepler\u2019s Third Law Some stories suggest that Newton was sitting under a tree when an apple fell to the ground and inspired him to discover gravity. This is definitely not what happened, as gravity was already known to exist. But the falling apple did lead him to wonder if the force of gravity that caused the apple to fall could also be acting on the Moon pulling it toward Earth. This revelation might seem obvious but that\u2019s only because we have been taught that it\u2019s true. In his day, Kepler theorized that magnetism made the Moon orbit Earth, and planets orbit the Sun! For most of the 1600s, scientists had been trying to predict where planets would be in their orbit at specific times. They failed to grasp the underlying mechanism responsible for the elliptical orbits that Kepler had shown to exist. In 1665, Newton finally recognized what no one else did: the centripetal force acting on the Moon was the force of gravity (Figure 5.42). The Moon was being pulled toward Earth like a falling apple. But the Moon was also moving off tangentially so that the rate at which it was falling matched the rate at which Earth curved away from it. In fact, the same mechanism was responsible for the planets orbiting the Sun. 276 Unit III Circular Motion, Work, and Energy 05-Phys20-Chap05.qxd 7/24/08 12:51 PM Page 277 By recognizing that the centripetal force and force of gravity were the same, Newton solved the mystery of planetary motion. His derivations mathematically proved Kepler\u2019s third law. The implications of Newton\u2019s work were huge because scientists now had the mathematical tools necessary to explore the solar system in more depth. Determining the Speed of a Satellite Earth Fg Moon r 3.844 108 m Figure 5.43 The Moon experiences a centripetal force that is the force of gravity of Earth on the Moon (not drawn to scale). Two of Newton\u2019s derivations deserve close examination. The first derivation uses the orbital radius of a body orbiting a planet or the Sun to determine the body\u2019s velocity. Recall from Chapter 4 that Newton had determined the equation for the force of gravity that one object exerts on another: Fg Gm1m2 r 2 He correctly reasoned that the gravitational force exerted by the Moon and Earth must be the centripetal force acting on the Moon (Figure 5.43). So: Fg Fc If we substitute the equations for centripetal force and gravity into this equation, we obtain: onv 2 GmMo mMo on 2 r r mEarth where G is the universal gravitational constant, mMoon is the mass of the Moon in kilograms, mEarth is the mass of Earth in kilograms, v is the speed of the Moon in metres per second, and r is the orbital radius of the Moon in metres. PHYSICS INSIGHT Since Fc Fg mv 2 r GMm r 2 and 2 r 2 m4 T 2 r GMm r 2 4 2r m 2 T GMm r 2 T 2 r 3 42 GM which is a constant containing the mass of the object causing the orbit. info BIT Pluto and its moon Charon are", " close enough in mass that they have a common centre of gravity between them in space (see Extrasolar Planets on page 283). Charon does not orbit Pluto \u2014 both bodies orbit their common centre of gravity. Since their centre of gravity is in space and not below the surface of Pluto, they form a binary system. Astronomers have been aware of binary stars in our universe for many years (i.e., two stars that have a common centre of gravity in space around which they revolve). Pluto and Charon are unique in our solar system. Chapter 5 Newton\u2019s laws can explain circular motion. 277 05-Phys20-Chap05.qxd 7/24/08 12:51 PM Page 278 PHYSICS INSIGHT Equation 13 uses the mass of Earth, but is not restricted to it. In a more general sense, the mass refers to the object being orbited. The mass of the Moon cancels, leaving: v 2 r Gm arth E 2 r Solving for v gives: v GmEarth r (13) In its current form, this equation determines the speed of any object orbiting Earth. The speed of an object orbiting any planet, the Sun, or another star for that matter, can be determined by using the mass of the object being orbited in place of the mass of Earth. Example 5.11 Earth\u2019s Moon is 3.844 105 km from Earth (Figure 5.44). Determine the orbital speed of the Moon. Earth m 5.97 1024 kg Practice Problems 1. Neptune\u2019s average orbital radius is 4.50 1012 m from the Sun. The mass of the Sun is 1.99 1030 kg. What is Neptune\u2019s orbital speed? 2. The moon Miranda orbits Uranus at a speed of 6.68 103 m/s. Use this speed and the mass of Uranus to determine the radius of Miranda\u2019s orbit. The mass of Uranus is 8.68 1025 kg. Answers 1. 5.43 103 m/s 2. 1.30 108 m Moon r 3.844 105 km Figure 5.44 The orbital speed of the Moon can be determined from its orbital radius and the mass of Earth (not drawn to scale). Analysis and Solution Convert the radius of the Moon\u2019s orbit to SI units. Then use the mass of Earth in equation 13. 0 10 r (3.844 105 km) k 0 m m 3.844 108 m v GmEarth r 6.67 1011 N 2 (5.97 1024 kg) m 2 g k 3.844 108 m 1.02 103 m/s The orbital speed of the Moon is 1.02 103 m/s or 3.66 103 km/h. 278 Unit III Circular Motion, Work, and Energy 05-Phys20-Chap05.qxd 7/24/08 12:51 PM Page 279 Measuring the Orbital Height of a Satellite Recall from Chapter 4 that the radius is always measured from the centre of one object to the centre of the other. This means that the orbital radius refers to the distance from centre to centre when measuring the distance from Earth to the Moon or from the Sun to the planets. Figure 5.45 shows the Earth-Moon system drawn to scale. For an artificial (human-made) satellite, the height of its orbit is usually measured from Earth\u2019s surface. To determine the velocity of an artificial satellite, you must first find its proper orbital height (from the centre of Earth). This is done by adding Earth\u2019s radius to the height of the satellite above Earth\u2019s surface. 384 400 km Earth farthest communication satellites from Earth Moon Figure 5.45 The Earth-Moon system drawn to scale. The radius of Earth doesn\u2019t need to be considered when comparing the distance between Earth and the Moon because it is insignificant compared to the great distance separating the two bodies. The distance between Earth and a communication satellite (35 880 km away) is small, so Earth\u2019s radius must be included in calculations involving orbital radius and period. Example 5.12 LandSat is an Earth-imaging satellite that takes pictures of Earth\u2019s ozone layer and geological features. It orbits Earth at the height of 912 km (Figure 5.46). What are its orbital speed and its period? Practice Problems 1. The International Space Station orbits Earth at a height of 359.2 km. What is its orbital speed? 2. The Chandra X-ray satellite takes X-ray pictures of high-energy objects in the universe. It is orbiting Earth at an altitude of 114 593 km. What is its orbital period? Answers 1. 7.69 103 m/s 2. 4.19 105 s 912 km rE 6.38 106 m Figure 5.46 LandSat follows a polar orbit so that it can examine the entire Earth as the planet rotates below it. The radius of Earth must be added", " to LandSat\u2019s height above the surface to determine the satellite\u2019s orbital radius (not drawn to scale). Given height of the satellite above Earth\u2019s surface 912 km Required LandSat\u2019s orbital speed (v ) and period (T ) Chapter 5 Newton\u2019s laws can explain circular motion. 279 05-Phys20-Chap05.qxd 7/24/08 12:51 PM Page 280 Analysis and Solution The radius of LandSat\u2019s orbit must be measured from the centre of Earth. To do this, add Earth\u2019s radius to the satellite\u2019s height above the planet\u2019s surface. Then determine the speed and period of the satellite. r rEarth 912 000 m (6.38 106 m) (9.12 105 m) 7.292 106 m Fg Fc v 2 msatellite r Gmsatellite r 2 mEarth v GmEarth r 6.67 1011 2 m N (5.97 1024 kg) 2 g k 7.292 106 m 7.390 103 m/s r 2 T v 2(7.292 106 m) m 7.390 103 s 6.20 103 s Paraphrase The speed of the satellite is 7.39 103 m/s (2.66 104 km/h). It orbits Earth once every 6.20 103 s (103 minutes). e TECH For an interactive simulation of the effect of a star\u2019s mass on the planets orbiting it, follow the links at www.pearsoned.ca/ physicssource. Determining the Mass of a Celestial Body A second derivation from Newton\u2019s version of Kepler\u2019s third law has to do with determining the mass of a planet or the Sun from the period and radius of a satellite orbiting it. For any satellite in orbit around a planet, you can determine its speed if you know the mass of the planet. But just how do you determine the mass of the planet? For example, how can you \u201cweigh\u201d Earth? Newton realized that this was possible. Let\u2019s look at the equality Fg again, but this time we will use equation 8 for centripetal force. Fc Recall that: Fc 42mMoonr T 2 Moon where mMoon is the mass of the Moon in kilograms, and TMoon is the orbital period of the Moon in seconds. 280 Unit III Circular Motion, Work, and Energy 05-Phys20-Chap05.qxd 7/24/08 12:51 PM Page 281 PHYSICS INSIGHT Equation 14 uses the mass of Earth and the period of its Moon. This equation can be used for any celestial body with a satellite orbiting it. In a more general sense, the mass refers to the object being orbited, and the period is the period of the orbiting object. Fg, then Since Fc 4 2m GmMo M on 2 2 r T Mo on oonr mEarth Solve for mEarth mEarth 4 2 (14) You can determine Earth\u2019s mass by using the orbital radius and period of its satellite, the Moon. From Table 5.6 on page 274, the Moon\u2019s period and radius are: \u2022 period of the Moon (TMoon) 27.3 days or 2.36 106 s \u2022 radius of the Moon\u2019s orbit (r Moon) 384 400 km or 3.844 108 m mEarth 42(3.844 108 m)3 m N (2.36 106 s)26.67 1011 2 g k 2 6.04 1024 kg The mass of Earth is 6.04 1024 kg, which is close to the accepted value of 5.97 1024 kg. Of course, this equation is not restricted to the Earth-Moon system. It applies to any celestial body that has satellites. For example, the Sun has eight planets that are natural satellites. Any one of them can be used to determine the mass of the Sun. Concept Check 1. What insight did Newton have that helped him explain the 2. motion of the planets? If Newton were told that our solar system is orbiting the centre of our galaxy, how would he explain this? 3. What previously immeasurable quantity could be determined with the use of Newton\u2019s version of Kepler\u2019s third law? Orbital Perturbations At about the same time as Kepler was figuring out the mechanism of the solar system, Galileo Galilei (1564\u20131642) pointed a relatively new invention at the sky. He began using a telescope to closely examine Jupiter. Only a few planets are visible to the naked eye: Mercury, Venus, Mars, Saturn, and Jupiter. Until the early 1600s, any observations of these planets were done without the aid of a telescope. Within a few months of using only an 8-power telescope, Galileo had discovered four moons of Jupiter. It became apparent how useful a telescope would be in the", " field of astronomy. e WEB To learn more about Galileo\u2019s discoveries with the telescope in 1609 and 1610, follow the links at www.pearsoned.ca/ school/physicssource. Chapter 5 Newton\u2019s laws can explain circular motion. 281 05-Phys20-Chap05.qxd 7/24/08 12:51 PM Page 282 orbital perturbation: irregularity or disturbance in the predicted orbit of a planet e WEB Telescopes and space technologies are continually improving, and new planetary objects are being found all the time. For the most up-to-date list of planets and celestial bodies in our solar system, follow the links at www.pearsoned.ca/school/ physicssource. Uranus Fg Planet X Figure 5.47 If a planet X existed and was behind Uranus in its orbit, it would pull Uranus back and outward. Within the space of 100 years, telescope technology improved dramatically, and the field of astronomy began its golden age. Astronomers plotted the positions of the planets more accurately than ever before and could peer deeper into the solar system. William Herschel (1738\u20131822) discovered the new planet Uranus in 1781, which created enormous interest. However, it wasn\u2019t long before astronomers noticed something strange about the orbit of Uranus. The orbital path of Uranus deviated from its predicted path slightly, just enough to draw attention. Astronomers called this deviation, or disturbance, an orbital perturbation. The Discovery of Neptune It had been over 120 years since Kepler and Newton had developed the mathematical tools necessary to understand and predict the position of the planets and their moons. Confident in the reliability of these laws, astronomers looked for a reason for the perturbation in the orbit of Uranus. According to mathematical predictions, Uranus should have been farther along in its orbit and closer to the Sun than it actually was. Somehow its progress was being slowed, and it was being pulled away from the Sun. Recall that anything with mass creates a gravitational field. The strength of this field depends on the mass of the object and the separation distance from it. The orbit of Uranus was minutely perturbed. Could another as-yet-undiscovered planet be exerting a gravitational pull or tug on Uranus whenever the orbital path took these two planets close together? If there was a planet X farther out and behind Uranus, it would exert a gravitational pull that would slow Uranus down and pull its orbit outward (Figure 5.47). This could explain the perturbation in the orbit of Uranus and was precisely the assumption that two astronomers, Urbain Le Verrier (1811\u20131877) of France and John Adams (1819\u20131892) of Britain, made in 1845. By examining exactly how much Uranus was pulled from its predicted position, Le Verrier and Adams could use Newton\u2019s law of gravitation to mathematically predict the size and position of this mysterious planet \u2014 if it existed. Working independently, both scientists gave very similar predictions of where to look for the planet. In September 1846, at the request of Le Verrier, German astronomer Johann Gottfried Galle (1812\u20131910) looked for the planet where Le Verrier predicted it would be and he found it! Le Verrier called the planet Neptune. It remained the solar system\u2019s outermost planet for the next 84 years until astronomers discovered Pluto in 1930 by analyzing the orbital perturbations of Neptune. (In 2006, Pluto was reclassified as a dwarf planet.) The Search for Other Planets The search for more planets in our solar system continues. A large band of rocky debris called the Kuiper belt lies 30 to 50 AU from the Sun. Pluto is a resident in this belt. Many scientists believed that Pluto was unlikely to be the only large resident of the belt and that the belt very likely contained more planet-sized objects. 282 Unit III Circular Motion, Work, and Energy 05-Phys20-Chap05.qxd 7/24/08 12:51 PM Page 283 In July 2005, a Kuiper belt object about 1.5 times bigger than Pluto was found. The scientific community has now given it the status of a dwarf planet, but at the time of publication of this book, it did not yet have an official name. Do other larger-than-Pluto bodies exist in the Kuiper belt? Probably, but their low reflectivity and extreme distance make them difficult to detect. THEN, NOW, AND FUTURE Extrasolar Planets For many centuries, humans have wondered if they were the only intelligent life in the universe. It seems unlikely, given the multitude of stars in our galaxy alone. However, for any life to exist, it\u2019s pretty safe to assume that it must have a planet to inhabit. Until 1995, there was no conclusive proof that any other star besides our Sun had planets. In October 1995, the", " first extrasolar planet was found orbiting a star similar in size to our Sun. It is called \u201cextrasolar\u201d because it is outside (\u201cextra\u201d) our solar system. This planet was named 51 Pegasi b after the star around which it revolves (Figure 5.48). It is huge but its orbital radius is extremely small. It is unlikely to support life as its gravitational field strength and temperature are extreme. Yet the discovery was a milestone since finding a planet orbiting a bright star 48 light years away is a very difficult task. A planet doesn\u2019t produce any light of its own, and it would be relatively dark compared to the very bright star beside it. The light coming from a star is so bright it makes direct observation of a nearby planet difficult. Imagine staring into a car\u2019s headlight on a dark night and trying to see an ant crawling on the bumper. Now imagine the car is 23 000 km away, and you get an idea of the magnitude of the problem. However, the increasing power of telescopes and innovative detection techniques are yielding new planet findings all the time. How is this accomplished? There are several different ways, but the most common and productive way is not to look for the planet directly but to look at the star that it is orbiting and watch for perturbations (also called wobble) in the star\u2019s movement. That\u2019s right, stars move. If they don\u2019t have a planet or planets orbiting them, then they simply move in a linear fashion through space. If they have a planet or planets in orbit around them, not only do they move linearly, but they also wobble in a circular path. This is because a planet exerts a gravitational pull on its star just as the star exerts the same pull on the planet. They both revolve around each other, just like you and a very heavy object would if you spun the object around in a circle at the end of a rope. Since the star is much more massive than the planet, its orbital radius is very small. This perturbation of the star is detectable and is indirect evidence that a planet must be orbiting it. As you can imagine, for a star to have a noticeable wobble, the planet that orbits it must be relatively large and fairly close to the star. This would make conditions on the planet inhospitable for life as we know it. But the search goes on and as our technology improves, who knows what we may find? Questions 1. Why is it so hard to detect an extrasolar planet? 2. What new technologies and techniques are making it possible to detect extrasolar planets? 3. Why is it unlikely that any life exists on the extrasolar planets discovered using these new techniques? \u25b2 Figure 5.48 An artist\u2019s conception of 51 Pegasi b \u2014 the first extrasolar planet found orbiting a star similar to our own. Chapter 5 Newton\u2019s laws can explain circular motion. 283 05-Phys20-Chap05.qxd 7/24/08 12:51 PM Page 284 Figure 5.49 A satellite in low Earth orbit Artificial Satellites At present, there are well over 600 working artificial satellites orbiting Earth. About half of them are in a low or medium Earth orbit, ranging from 100 to 20 000 km above Earth\u2019s surface (Figure 5.49). The other half are geostationary satellites that orbit Earth at a distance of precisely 35 880 km from Earth\u2019s surface, directly above the equator. Depending on their design and orbit, satellites perform a variety of tasks. Weather, communication, observation, science, broadcast, navigation, and military satellites orbit Earth at this moment. These include a ring of GPS (global positioning system) satellites that are used to triangulate the position of a receiver wherever it may be. With the help of a GPS receiver, you could find your position on the planet to within a few metres. All satellites are designed to receive information from and transmit information to Earth. Each satellite has an antenna that is used to receive radio signals from satellite stations on the ground. At the same time, satellites send information back down to Earth for people to use. 5-5 Decision-Making Analysis 5-5 Decision-Making Analysis The Costs and Benefits of Putting a Satellite into Orbit Required Skills Initiating and Planning Performing and Recording Analyzing and Interpreting Communication and Teamwork The Issue Satellites perform a variety of tasks that make them almost indispensable. They map Earth, find geological formations and minerals, help us communicate over great distances and in remote areas, and help predict the movement of weather systems such as hurricanes. Opponents of the continued unregulated use of satellites argue that the cost of deploying satellites is enormous; they don\u2019t have a long lifespan; and their failure rate is high. Furthermore, once a satellite is in a medium or geostationary orbit, it will stay there even after it becomes inoperative. It turns into nothing more", " than very expensive and dangerous space junk. Background Information The first satellite in orbit was Sputnik in 1957. Since that time there have been over 4000 launches, and space has become progressively more crowded. The best estimates suggest that there are about 600 active satellites in orbit, and about 6000 pieces of space debris that are being tracked. The space debris can be very hazardous for missions carrying astronauts to low Earth orbit. If hit by even a small piece of orbiting debris, a spacecraft could be destroyed. To limit the overcrowding of space, satellite manufacturers are designing more sophisticated satellites that can handle a higher flow of information so that fewer satellites have to be deployed. This drives up the cost of manufacturing satellites because they are more technologically advanced than their predecessors. Unfortunately, as well as being more sophisticated, they are more prone to failure. Analyze and Evaluate 1. Identify two different types of satellites based on the type of job they perform. 2. For each type of satellite from question 1: (a) Identify the type of orbit that it occupies and explain the job that it performs. (b) Determine the approximate cost of deployment (getting it into space). (c) Determine its expected lifespan. 3. Suggest an alternative technology that could be used in place of each of these satellites. Analyze the effectiveness and cost of this technology compared with the satellite. 4. Propose possible changes that could be made to the way satellites are built and deployed that could lessen the overcrowding of space. 284 Unit III Circular Motion, Work, and Energy 05-Phys20-Chap05.qxd 7/24/08 12:51 PM Page 285 Geostationary Satellites Geostationary satellites may be the most interesting of all satellites because they appear to be stationary to an observer on Earth\u2019s surface, even though they travel around Earth at high velocity. They are placed at a specific altitude so that they make one complete orbit in exactly 24 h, which is the same as Earth\u2019s rotational period (Figure 5.50). These satellites are placed in the plane of the equator, so they will have exactly the same axis of rotation as Earth, and will stay fixed over the same spot on the planet (Figure 5.51). To an observer on the ground, geostationary satellites appear motionless. axis of rotation of Earth Earth plane of Earth\u2019s equator equator equator equator P P P satellite satellite\u2019s orbit in plane of equator Figure 5.50 Geostationary satellites orbit in the plane of the equator with the same axis of rotation as Earth. Communication satellites are geostationary. A communication signal, such as a telephone or TV signal, can be sent from the ground at any time of the day to the nearest geostationary satellite located over the equator. That satellite can then relay the signal to other geostationary satellites located over different spots on Earth, where the signal is then transmitted back down to the nearest receiving station. Weather satellites also make use of this orbit. They may be \u201cparked\u201d near a landmass such as North America, using cameras to map the weather. Weather forecasters receive a continuous stream of information from the satellites that allows them to predict the weather in their area. This type of orbit is in high demand and is filled with satellites from many different countries. Unfortunately, the orbit must be fixed at 35 880 km from Earth\u2019s surface and be directly over the equator. This orbit risks being overcrowded. If satellites are placed too close together, their signals can interfere with each other. Satellites also tend to drift slightly in their orbit and therefore cannot be placed close to each other. Many derelict satellites that were placed in geostationary orbits are still there taking up room, since they continue to orbit even after they no longer function. The limited room available is filling up fast. At present, about half (at least 300) of the active satellites in orbit are geostationary. An artificial satellite obeys the same laws of physics as a natural satellite does. The orbital radius determines its orbital period and speed, and it is governed by the same laws that describe the motion of moons around a planet or planets around the Sun. That means low Earth orbit satellites make one complete orbit in about 90 min, while geostationary satellites take exactly one day. Figure 5.51 A geostationary satellite moves with a fixed point (P) on the Earth since both revolve around the same axis with the same period. To an observer on the ground, the satellite does not appear to be moving. info BIT Geostationary satellites are also referred to as geosynchronous satellites. e WEB To learn more about real-time tracking of satellites in orbit around Earth, follow the links at www.pearsoned.ca/school/ physicssource. Chapter 5 Newton\u2019s laws can explain circular motion. 285 05-Phys20-Ch", "ap05.qxd 7/24/08 12:51 PM Page 286 In the short space of time since the first satellite, Sputnik, was launched in October 1957, satellites have become indispensable (Figure 5.52). They have enabled us to see areas of our planet and atmosphere never seen before. They have enabled us to know our location anywhere on Earth within a few metres, and have allowed us to communicate with the remotest places on Earth. The future of satellites seems assured. Figure 5.52 Around Earth, space is crowded with the multitude of satellites currently in orbit. The large ring farthest from Earth is made up of geostationary satellites. 5.3 Check and Reflect 5.3 Check and Reflect Knowledge 11. Jupiter\u2019s moon Io has an orbital period of 1. What is an astronomical unit? 2. Why is the orbital radius of a planet not constant? 3. An eccentricity of 0.9 indicates what kind of shape? 4. Where in a planet\u2019s orbit is its velocity the greatest? The smallest? Why? 5. State the condition necessary for Kepler\u2019s third law to be valid. 6. Is Kepler\u2019s constant the same for moons orbiting a planet as it is for planets orbiting the Sun? Why? 7. Does our Sun experience orbital perturbation? Explain. Applications 8. Sketch a graph that shows the trend between the planets\u2019 orbital radius and their period. 9. Venus has an average orbital period of 0.615 years. What is its orbital radius? 10. Another possible planet has been discovered, called Sedna. It has an average orbital radius of 479.5 AU. What is its average orbital speed? 1.769 d and an average orbital radius of 422 000 km. Another moon of Jupiter, Europa, has an average orbital radius of 671 000 km. What is Europa\u2019s orbital period? 12. Determine the average orbital speed of Mimas using orbital data from Table 5.6 on page 274. 13. Using the orbital period and radius of Venus from Table 5.5 on page 273, determine the mass of the Sun. Extensions 14. As more satellites are put into space, the amount of orbital debris increases. What solutions can you suggest to decrease the amount of space junk? 15. Using a graphing calculator or other suitable software, plot a graph of velocity versus radius for an artificial satellite orbiting Earth. e TEST To check your understanding of satellites and celestial bodies in circular motion, follow the eTest links at www.pearsoned.ca/school/physicssource. 286 Unit III Circular Motion, Work, and Energy 05-Phys20-Chap05.qxd 7/24/08 12:51 PM Page 287 CHAPTER 5 SUMMARY Key Terms and Concepts axle axis of rotation uniform circular motion centripetal acceleration centripetal force cycle revolution period frequency rpm Key Equations v 2r T ac v 2 r Fc mv 2 r 42r T 2 ac 42mr T 2 Fc 2 T a K 3 ra Conceptual Overview satellite artificial satellite orbital perturbation extrasolar planet Kepler\u2019s laws ellipse eccentricity orbital period orbital radius 2 2 T T b a 3 3 rb ra The concept map below summarizes many of the concepts and equations in this chapter. Copy and complete the map to have a full summary of the chapter. where the Uniform Circular Motion applies to where the object experiences Centripetal Acceleration is tangent to the acts toward the of the Centre which is a Net Force which could be a Combination of Forces Tension Figure 5.53 Planetary/Satellite Motion where explained Newton\u2019s Version of Kepler\u2019s 3rd Law was realized by and he derived Newton that was used to determine the Chapter 5 Newton\u2019s laws can explain circular motion. 287 05-Phys20-Chap05.qxd 7/24/08 12:51 PM Page 288 CHAPTER 5 REVIEW Knowledge 1. (5.1) What is the direction of centripetal acceleration? 2. (5.1) Describe what produces the centripetal Applications 14. If the frequency of a spinning object is doubled, what effect does this have on the centripetal acceleration? force in the following cases: (a) A boat makes a turn. (b) A plane makes a horizontal turn. (c) A satellite orbits Earth. 3. (5.1) What force is responsible for the tug your hand feels as you spin an object around at the end of a rope? 4. (5.1) Sketch a diagram of a mass moving in a vertical circle. Draw the velocity, centripetal acceleration, and centripetal force vectors at a point on the circle\u2019s circumference. 5. (5.2) Why do spinning tires tend to stretch? 6. (5.2) A heavy mass attached to the end", " of a cable is spinning in a vertical circle. In what position is the cable most likely to break and why? 7. (5.2) What force acts as the centripetal force for a motorcycle making a turn? 8. (5.2) Explain why a truck needs a larger turning radius than a car when they move at the same speed. 9. (5.3) In what position of its orbit does a planet move the fastest? 10. (5.3) In the orbit of a planet, what does the semi-major axis of an ellipse represent? 11. (5.3) Equation 14, page 281, can be confusing because it uses the mass of one object and the period of another. (a) Explain what mEarth and TMoon represent. (b) Explain how equation 14 is used in the most general case. 12. (5.3) Use Kepler\u2019s second law to explain why a planet moves more quickly when it is nearer to the Sun than when it is farther away in its orbital path. 13. (5.3) Why can\u2019t Kepler\u2019s constant of 1 a2/AU3 be used for moons orbiting planets or planets in other solar systems? 15. A slingshot containing rocks is spun in a vertical circle. In what position must it be released so that the rocks fly vertically upward? 16. People of different masses are able to ride a roller coaster and go through a vertical loop without falling out. Show the mathematical proof of this. 17. An eagle circles above the ground looking for prey. If it makes one complete circle with the radius of 25.0 m in 8.0 s, what is its speed? 18. A ride at the fair spins passengers around in a horizontal circle inside a cage at the end of a 5.0-m arm. If the cage and passengers have a speed of 7.0 m/s, what is the centripetal force the cage exerts on an 80.0-kg passenger? 19. What is the minimum speed that a glider must fly to make a perfect vertical circle in the air if the circle has a radius of 200.0 m? 20. A child spins an 800.0-g pail of water in a vertical circle at the end of a 60.0-cm rope. What is the magnitude of the tension of the rope at the top of the swing if it is spinning with the frequency of 2.0 Hz? 21. A driver is negotiating a turn on a mountain road that has a radius of 40.0 m when the 1600.0-kg car hits a patch of wet road. The coefficient of friction between the wet road and the wheels is 0.500. If the car is moving at 30.0 km/h, determine if it skids off the road. 22. The blade of a table saw has a diameter of 25.4 cm and rotates with a frequency of 750 rpm. What is the magnitude of the centripetal acceleration at the edge of the blade? 23. The tip of a propeller on an airplane has a radius of 0.90 m and experiences a centripetal acceleration of 8.88 104 m/s2. What is the frequency of rotation of the propeller in rpm? 24. Using information and equations from this chapter, verify that the speed of Earth in orbit around the Sun is approximately 107 000 km/h. 288 Unit III Circular Motion, Work, and Energy 05-Phys20-Chap05.qxd 7/24/08 12:51 PM Page 289 25. A car rounds a corner of radius 25.0 m with the 34. A star in the Andromeda galaxy is found to have centripetal acceleration of 6.87 m/s2 and a hubcap flies off. What is the speed of the hubcap? 26. An electron (m 9.11 1031 kg) caught in a magnetic field travels in a circular path of radius 30.0 cm with a period of 3.14 108 s. (a) What is the electron\u2019s speed? (b) What is the electron\u2019s centripetal acceleration? 27. Halley\u2019s comet has an orbital period of 76.5 a. What is its mean orbital radius? 28. An extrasolar planet is found orbiting a star in the Orion nebula. Determine the mass of the star if the planet has an orbital period of 400.0 Earth days and an orbital radius of 1.30 1011 m. 29. The Milky Way is a spiral galaxy with all the stars revolving around the centre where there is believed to be a super-massive black hole. The black hole is 2.27 1020 m from our Sun, which revolves around it at a speed of 1234 m/s. (a) What is the period of our solar system\u2019s orbit around the black hole? (b) How", " massive is the black hole? (c) What centripetal acceleration does our solar system experience as a result of the black hole\u2019s gravity? 30. Newton hypothesized that a cannonball fired parallel to the ground from a cannon on the surface of Earth would orbit Earth if it had sufficient speed and if air friction were ignored. (a) What speed would the cannonball have to have to do this? (Use Kepler\u2019s third law.) (b) If the mass of the cannonball were doubled, what time would it take to orbit Earth once? 31. Use the Moon\u2019s period (27.3 d), its mass, and its distance from Earth to determine its centripetal acceleration and force. 32. Neptune has a moon Galatea that orbits at an orbital radius of 6.20 107 m. Use the data for Nereid from Table 5.6 on page 274 to determine Galatea\u2019s orbital period. 33. Determine the orbital speed of Ariel, a moon of Uranus, using Table 5.5 (page 273) and Table 5.6 (page 274). a planet orbiting it at an average radius of 2.38 1010 m and an orbital period of 4.46 104 s. (a) What is the star\u2019s mass? (b) A second planet is found orbiting the same star with an orbital period of 6.19 106 s. What is its orbital radius? Extensions 35. Paraphrase two misconceptions about centripetal force mentioned in this book. How do these misconceptions compare with your preconceptions of centripetal force? 36. Assuming a perfectly spherical Earth with uniform density, explain why a person standing at the equator weighs less than the same person standing at the North or South Pole. Consolidate Your Understanding Create your own summary of uniform circular motion, Kepler\u2019s laws, and planetary and satellite motion by answering the questions below. If you want to use a graphic organizer, refer to Student References 4: Using Graphic Organizers pp. 869\u2013871. Use the Key Terms and Concepts listed above and the Learning Outcomes on page 240. 1. In a few sentences, summarize how frequency, period, and velocity affect centripetal acceleration. 2. Explain to a classmate in writing why the velocity at different radii on a spinning disc will vary while the rotational frequency remains constant. Think About It Review your answers to the Think About It questions on page 241. How would you answer each question now? e TEST To check your understanding of circular motion, follow the eTest links at www.pearsoned.ca/school/physicssource. Chapter 5 Newton\u2019s laws can explain circular motion. 289 06-PearsonPhys20-Chap06 7/24/08 12:56 PM Page 290 C H A P T E R 6 Key Concepts In this chapter, you will learn about: \u25a0 work, mechanical energy, and power \u25a0 the work-energy theorem \u25a0 isolated and non-isolated systems \u25a0 the law of conservation of energy Learning Outcomes When you have finished this chapter, you will be able to: Knowledge \u25a0 use the law of conservation of energy to explain the behaviours of objects within isolated systems \u25a0 describe the energy transformations in isolated and non-isolated systems using the work\u2013energy theorem \u25a0 calculate power output Science, Technology, and Society \u25a0 explain that models and theories are used to interpret and explain observations \u25a0 explain that technology cannot solve all problems \u25a0 express opinions on the support found in Canadian society for science and technology measures that work toward a sustainable society 290 Unit III In an isolated system, energy is transferred from one object to another whenever work is done. Figure 6.1 Tension mounts as the motor pulls you slowly to the top of the first hill. Slowly you glide over the top, then suddenly, you are plunging down the hill at breathtaking speed. Upon reaching the bottom of the hill, you glide to the top of the next hill and the excitement begins all over again. As you race around the roller coaster track, each hill gets a bit lower until, at last, you coast to a gentle stop back at the beginning. You probably realize that because of friction the trolley can never regain the height of the previous hill, unless it is given a boost. It seems obvious to us that as objects move, kinetic energy is always lost. Energy is the most fundamental concept in physics. Everything that occurs in nature can be traced back to energy. The complicating factor is that there are so many forms of energy it is often very difficult to keep track of what happens to the energy when it is transferred. Energy is a scalar quantity. This chapter concentrates on gravitational potential energy, kinetic energy, and elastic potential energy. In this chapter you will take the first steps to understanding the role of energy in nature. Specifically, you will learn how energy is given to and taken from objects when they interact with each other. 06-PearsonPhys20-Chap", "06 7/24/08 12:56 PM Page 291 6-1 QuickLab 6-1 QuickLab Energy Changes of a Roller Coaster e WEB 7 Start the simulation. This activity uses the roller coaster simulation found at www.pearsoned.ca/school/physicssource. Problem How does the energy of a roller coaster vary as it travels on its track? Materials computer connected to the Internet clear plastic ruler Procedure 1 Click on the start button for the simulation. 2 Observe the motion of the cart. 3 Click on \u201ccontinue\u201d and note what happens to the motion of the trolley as it moves along the track. 4 Repeat step 3 until the simulation is complete. 5 Reset the simulation. 6 Use a see-through plastic ruler to measure the lengths of the potential energy bar (blue) and the kinetic energy bar (green) before you start the simulation. Record your measurements. \u25bc Table 6.1 8 Each time the trolley pauses, measure the length of the potential energy and the kinetic energy bars and record the results in a table similar to Table 6.1. Questions 1. What assumptions are you making when you measure the lengths of the energy bars? 2. What is the effect on the potential energy of the trolley as it moves upward and downward? 3. What is the effect on the kinetic energy of the trolley as it moves upward and downward? Is this true as the trolley moves upward to the top of the first hill? Explain. 4. From the table, what happens to the energy of the trolley as it moves from the start to position \u201ca\u201d? 5. For each of the positions at which the trolley pauses, how does the sum of lengths of the bars change? What does the sum of these lengths represent? 6. Is there an energy pattern as the trolley moves along the track? Describe the pattern. 7. Do you think that this pattern is representative of nature? Explain. Position Length of Potential Energy Bar (mm) Length of Kinetic Energy Bar (mm) Sum of Lengths (mm) start a b Think About It 1. If two cars are identical except for the size of their engines, how will that affect their performance on the highway? 2. What is the \u201claw of conservation of energy\u201d? When does this law apply? 3. When work is done on an object, where does the energy used to do the work go? Discuss your answers in a small group and record them for later reference. As you complete each section of this chapter, review your answers to these questions. Note any changes in your ideas. e LAB For a probeware activity, go to www.pearsoned.ca/ school/physicssource. Chapter 6 In an isolated system, energy is transferred from one object to another whenever work is done. 291 06-PearsonPhys20-Chap06 7/24/08 12:56 PM Page 292 info BIT Inuit hunters devised unusual ways of storing potential energy in a bow. One way they accomplished this was to tie cords of sinew along the back of the bow (Figure 6.3). The sinew was more heavily braided where strength was needed and less heavily braided where flexibility was important. When the bow was bent, the cords would stretch like a spring to store energy. In the absence of a wood source, bows were often made of antler or bone segments. e WEB To see photographs and to learn more about the technology of Inuit bows, follow the links at www.pearsoned.ca/ school/physicssource. 6.1 Work and Energy Figure 6.2 When the string on a bow is pulled back, elastic potential energy is stored in the bow. energy: the ability to do work Figure 6.3 Sinew-backed bow of the Inuit Copper people of the Central Arctic. Maximum gravitational potential energy is stored in the skier at the top of the run. Gravitational potential energy changes to kinetic energy and heat during the run. Figure 6.4 During the downhill run, the skier\u2019s gravitational potential energy is continually converted into kinetic energy and heat. 292 Unit III Circular Motion, Work, and Energy An archer is pulling back her bowstring (Figure 6.2). She does work on the bow transforming chemical energy in her muscles into elastic potential energy in the bow. When she releases the string, the bow does work on the arrow. The elastic potential energy of the bow is transformed into the energy of motion of the arrow, called kinetic energy. As skiers ride up a lift, the lift\u2019s motor is transforming chemical energy of the fuel into gravitational potential energy of the individuals. As they go downhill, gravity does work on the skiers transforming their gravitational potential energy into kinetic energy and heat. In both these examples, work transfers energy. In the case of the archer, energy is transformed from chemical energy into elastic potential energy and then into kinetic energy (", "from the archer to the bow to the arrow). In the case of the skier, energy is transformed from the chemical energy of the motor\u2019s fuel into the gravitational potential energy of the skier at the top of the run and then into a combination of changing kinetic energy, gravitational potential energy, and heat of the skier as she speeds downhill. All these processes involve work. 06-PearsonPhys20-Chap06 7/24/08 12:56 PM Page 293 Work Is Done When Force Acts Over a Displacement ) acts on an object resulting in a displacement (d ), a When a force (F transfer of energy occurs. This energy transfer is defined as the work done by the force. In introductory courses the quantity of work is usually defined by the equation W Fd. and d Work is a scalar quantity. However, the relative directions of the vectors F ) does not act are important. If the applied force (F parallel to the displacement (Figure 6.5), you must resolve the force into components that are parallel (F) and perpendicular (F) to the displacement. Only the component of the force parallel to the displacement actually does work. The component of the force acting perpendicular to the displacement does no work. work: a measure of the amount of energy transferred when a force acts over a given displacement. It is calculated as the product of the magnitude of applied force and the displacement of the object in the direction of that force. PHYSICS INSIGHT The unit of work and energy is the joule (J). It is a derived unit. 1 J 1 Nm kgm2 s2 1 F d F F Figure 6.5 When a force acts on an object, resulting in a displacement, only the component of the force that acts parallel to the displacement does work. If the box moves, does work. horizontally, only the horizontal component, F Thus, the equation for work is often written as W Fd where F is the magnitude of the component of the force that acts parallel to the displacement. In Figure 6.5, where the angle between the direction of the force and the direction of the displacement is, the component of the force parallel to the displacement is given by F Fcos F d If you replace F by Fcos, the calculation for work becomes Figure 6.6 W (Fcos )d Let\u2019s look at two special cases. First, when the force acts parallel to the displacement, the angle 0\u00b0 so that cos 1, making F F. This results in the maximum value for the work that the force could do over that displacement (Figure 6.6). Second, if the force acts perpendicular to the displacement, there is no parallel component. Mathematically, since 90\u00b0 then cos 0 making F 0. In this case, the applied force does no work on the object (Figure 6.7). Figure 6.7 d F Chapter 6 In an isolated system, energy is transferred from one object to another whenever work is done. 293 06-PearsonPhys20-Chap06 7/24/08 12:56 PM Page 294 Concept Check When a centripetal force acts on an object, the object shows no increase in speed and therefore no increase in kinetic energy. In terms of the work done by the centripetal force, explain why this is true. Example 6.1 Figure 6.8 shows a force of 150 N [0\u00b0] acting on an object that moves over a displacement of 25.0 m [25.0\u00b0] while the force acts. What is the work done by this force? d F 25.0\u00b0 Figure 6.8 Practice Problems 1. You pull a sled along a horizontal surface by applying a force of 620 N at an angle of 42.0\u00b0 above the horizontal. How much work is done to pull the sled 160 m? 2. A force acts at an angle of 30.0\u00b0 relative to the direction of the displacement. What force is required to do 9600 J of work over a displacement of 25.0 m? 3. A force of 640 N does 12 500 J of work over a displacement of 24.0 m. What is the angle between the force and the displacement? 4. A bungee jumper with a mass of 60.0 kg leaps off a bridge. He is in free fall for a distance of 20.0 m before the cord begins to stretch. How much work does the force of gravity do on the jumper before the cord begins to stretch? Answers 1. 7.37 104 J 2. 443 N 3. 35.5\u00ba 4. 1.18 104 J Given 1.50 102 N [0\u00b0] F 25.0 m [25.0\u00b0] d Required work done by the force (W) Analysis and Solution From Figure 6.8, the angle between the force and the displacement is 25.0\u00b0. Draw a component diagram (Figure 6.9). The component that does work is F (cos 25.0\u00b0", "). Solve using the equation for work. W (Fcos )d (1.50 102 N)(cos 25.0\u00b0)(25.0 m) 3.399 103 N\u00b7m 3.40 103 J F F 25.0\u00b0 F Figure 6.9 Component diagram Paraphrase and Verify The work done by the force is 3.40 103 J. If the force had acted parallel to the displacement, the maximum amount of work done would have been 3.75 103 J. Since cos 25.0\u00b0 is about 0.91, the answer of 3.40 103 J, or 0.91 times the maximum value for W, is reasonable. 294 Unit III Circular Motion, Work, and Energy 06-PearsonPhys20-Chap06 7/24/08 12:56 PM Page 295 Gravitational Potential Energy An object is said to have potential energy if it has the ability to do work by way of its position or state. There are several forms of potential energy. Imagine a ride at the fair where the passengers are lifted vertically before being allowed to drop in free fall, as in Figure 6.10. Ignoring friction, the work done by the machinery to lift the passengers and car to a height, h, is equal to the change in gravitational potential energy, EP. To lift the object straight up at a constant speed, the force applied must be equal but opposite to the force of gravity on the object. The equation for calculating work can be used to develop the equation for change in gravitational potential energy. EP W Fd where F is the magnitude of the force acting parallel to the displacement, and d is the magnitude of the displacement. To lift an object of mass m upward at a constant speed, the force is equal in magnitude and parallel, but opposite in direction, to the gravitational force, F g. mg where g is the magRecall from Unit II that Fg nitude of the acceleration due to gravity, which has a constant value of 9.81 m/s2 near Earth\u2019s surface. It follows that F mg If the object is moved through a change in height h, so that d h, the change in potential energy equation becomes EP mgh Figure 6.10 A motor works transferring energy to the ride car. The gravitational potential energy gained produces the exciting free fall. So, when an object is moved upward, h increases and h is positive, and the potential energy increases (positive change). When an object is moved downward, h decreases and h is negative, and the potential energy decreases (negative change). gravitational potential energy: the energy of an object due to its position relative to the surface of Earth Concept Check Does the above equation for change in gravitational potential energy apply to objects that move over very large changes in height (e.g., change as experienced by a rocket)? Explain. Chapter 6 In an isolated system, energy is transferred from one object to another whenever work is done. 295 06-PearsonPhys20-Chap06 7/24/08 12:56 PM Page 296 Example 6.2 Practice Problems 1. An elevator car has a mass of 750 kg. Three passengers of masses 65.0 kg, 30.0 kg, and 48.0 kg, ride from the 8th floor to the ground floor, 21.0 m below. Find the change in gravitational potential energy of the car and its passengers. 2. A book with a mass of 1.45 kg gains 25.0 J of potential energy when it is lifted from the floor to a shelf. How high is the shelf above the floor? 3. The Mars rover lifts a bucket of dirt from the surface of Mars into a compartment on the rover. The mass of the dirt is 0.148 kg and the compartment is 0.750 m above the surface of Mars. If this action requires 0.400 J of energy, what is the gravitational acceleration on Mars? Answers 1. 1.84 105 J 2. 1.76 m 3. 3.60 m/s2 If the car and its passengers in Figure 6.10 have a mass of 500 kg, what is their change in gravitational potential energy when they are lifted through a height of 48.0 m? Given m 500 kg g 9.81 m/s2 h 48.0 m up down Required change in gravitational potential energy (EP) 48.0 m Analysis and Solution Sketch the movement of the car as in Figure 6.11 and solve for EP. EP mgh (5.00 102 kg)9.81 2.35 105 kg 2.35 105 (N m) 2.35 105 J m s2 m s2 (48.0 m) m Figure 6.11 Paraphrase The change in gravitational potential energy of the car and its passengers is a gain of 2.35 105 J. The object moved upward, gaining gravitational potential energy. If Ep1 represents the potential energy of an object at height h1 and Ep2 its potential energy when it is lifted to a height h2,", " then the change in potential energy is, by definition, Ep Since Ep Ep2 Ep1 mgh Ep2 Ep1 mg(h2 h1) Consider an object at ground level as having zero potential energy. If the object is raised from the ground level, h1 0) 0. It follows that Ep2 Ep2 0 mg(h2 mgh2 In general, the potential energy of an object at height h, measured from the ground, is Ep mgh 296 Unit III Circular Motion, Work, and Energy 06-PearsonPhys20-Chap06 7/24/08 12:56 PM Page 297 Choosing a Reference Point You see from the equation for the change in potential energy on page 295 that E depends only on the change in height, h. The values of h may be measured from any convenient reference point, as long as the reference point is kept the same for all the measurements when solving a problem. The change in height, and therefore the change in gravitational potential energy, is the same regardless of your frame of reference. Look at the book resting on the shelf in Figure 6.12. The value of h for the shelf can be defined relative to the floor (hf), relative to the table (ht), or even relative to the ceiling above the shelf (hc), in which case hc will have a negative value. Usually, you choose the frame of reference that most simplifies your measurements and calculations for h. For example, if you were trying to determine how much gravitational potential energy the book would lose as it fell from the shelf to the tabletop, then it would be logical to use the tabletop as your reference point. If you used another position as a reference point, your calculations might be slightly more complex, but the final answer for the amount of gravitational potential energy the book loses would be the same. Change in gravitational potential energy depends only on change in vertical height. The change in gravitational potential energy of an object depends only on the change in height. For example, the change in gravitational potential energy of a cart rolling down a frictionless ramp as in Figure 6.13 depends only on the vertical measurement, h. The actual distance an object travels, while it moves through a given change in height, does not affect its change in gravitational potential energy. d h Figure 6.13 As the cart rolls down the ramp, only the change in height h affects its change in gravitational potential energy. hc ht hf Figure 6.12 The book has gravitational potential energy due to its position on the shelf. reference point: an arbitrarily chosen point from which distances are measured PHYSICS INSIGHT The calculation of h from Figure 6.13 involves the use of the trigonometric h. ratio sin d Chapter 6 In an isolated system, energy is transferred from one object to another whenever work is done. 297 06-PearsonPhys20-Chap06 7/24/08 12:56 PM Page 298 Example 6.3 Figure 6.14 shows a toy car track set up on a tabletop. (a) What is the gravitational potential energy of the car, which has a mass of 0.0250 kg, relative to the floor? (b) Calculate the change in gravitational potential energy of the car when it arrives at the bottom of the hill. Given m 0.0250 kg g 9.81 m/s2 2.15 m h1 0.950 m h2 up down toy car glide track Figure 6.14 table Practice Problems 1. A pile driver drops a mass of 550 kg from a height of 12.5 m above the ground onto the top of a pile that is 2.30 m above the ground. Relative to ground level, what is the gravitational potential energy of the mass (a) at its highest point? (b) at its lowest point? (c) What is the change in the gravitational potential energy of the mass as it is lifted from the top of the pile to its highest point? 2. A roller coaster trolley begins its journey 5.25 m above the ground. As the motor tows it to the top of the first hill, it gains 4.20 105 J of gravitational potential energy. If the mass of the trolley and its passengers is 875 kg, how far is the top of the hill above the ground? 3. A winch pulls a 250-kg block up a 20.0-m-long inclined plane that is tilted at an angle of 35.0\u00b0 to the horizontal. What change in gravitational potential energy does the block undergo? Answers 1. (a) 6.74 104 J (b) 1.24 104 J (c) 5.50 104 J 2. 54.2 m 3. 2.81 104 J h1 2.15 m floor h2 0.950 m Required (a) gravitational potential energy at the top of the hill relative to the floor (Ep1) (b) change in the gravitational potential energy as the car moves from the top to the bottom of the", " hill (Ep) Analysis and Solution (a) To find gravitational potential energy relative to the floor, use that surface to define h 0 and make all height measurements from there. Ep1 (2.15 m) m s2 mgh1 (0.0250 kg)9.81 m2 0.527 kg 2 s 0.527 J (b) To find the change in gravitational potential energy, use the data and Figure 6.14 to calculate the change in height (h h2 h h2 h1). h1 Ep 0.950 m 2.15 m 1.20 m mg(h) (0.0250 kg)9.81 0.294 J (1.20 m) m s2 Paraphrase and Verify (a) The gravitational potential energy relative to the floor is 0.527 J. (b) The change in gravitational potential energy is 0.294 J. As the car rolls down the hill it loses 0.294 J of gravitational potential energy. Note: You could calculate Ep2 first and then use Ep Ep2 Ep1 298 Unit III Circular Motion, Work, and Energy 06-PearsonPhys20-Chap06 7/24/08 12:56 PM Page 299 Hooke\u2019s Law In 1676, Robert Hooke, an English physicist, showed that the stretch produced by a force applied to a spring was proportional to the magnitude of the force. This relationship is known as Hooke\u2019s Law and applies to any elastic substance when a force is exerted upon it. Thus, if a mass is suspended from a spring (Figure 6.15) the position (x) of the mass changes in proportion to the force (the weight (Fg) of the mass) exerted on the spring. x 0 x d x 2d x 3d x 4d d d d d Stretch Produced by a Force Applied to a Spring Fg 0 Fg W Fg 2W Fg 3W Fg 4W In an experiment to test this prediction, students suspended a series of masses from a spring and measured the position for each mass. Their data are shown in Table 6.2. Figure 6.15 The stretch produced by a force applied on a spring is proportional to the magnitude of the force. \u25bc Table 6.2 Students\u2019 experimental data Mass m (g) 0 200 400 600 800 1000 Weight Fg (N) Position x (m) 0 1.96 3.92 5.87 7.85 9.81 0 0.050 0.099 0.146 0.197 0.245 10 ) Force vs. Position for a Mass Suspended on a Spring 0 0.05 0.15 0.10 Position x (m) 0.20 0.25 Figure 6.16 Graph of data from Table 6.2 The students then plotted a graph of the magnitude of the applied force (Fg) as a function of the position (x) of the spring. The resulting line is a straight line with a constant slope (Figure 6.16). The equation of this line is F kx where k is the slope of the line. The slope of the line is determined by the properties of the spring and is defined as the elastic or spring constant (k). This constant tells us how hard it is to stretch/compress the spring from the equilibrium position at x 0. For the graph in Figure 6.16, the slope is found as shown below: k F x Fi Ff xi xf 10.0 N 3.0 N 0.250 m 0.075 m 40 N m This force-position graph is characteristic for all springs whether the force stretches or compresses the spring. When a heavier or lighter spring is used, the slope of the line changes but the line is still straight. You will deal with Hooke\u2019s Law in greater depth when you study simple harmonic motion in Chapter 7. PHYSICS INSIGHT A spring becomes nonelastic at a certain critical stretch value called the elastic limit. If the force applied does not exceed the elastic limit, the material will return to its original shape. If a spring is stretched beyond its elastic limit, its shape will be permanently distorted or the spring may break. Chapter 6 In an isolated system, energy is transferred from one object to another whenever work is done. 299 06-PearsonPhys20-Chap06 7/24/08 12:56 PM Page 300 Elastic Potential Energy elastic potential energy: the energy resulting from an object being altered from its standard shape, without permanent deformation Force vs. Position for an Elastic System ) ( N F e c r o F F 0 Area Ep Position x (m) x Figure 6.17 The area under the force-position curve is equal to the work done by the force to stretch the spring to that position. PHYSICS INSIGHT Ep, the change in elastic potential energy depends on the square of the stretch in the spring. That is: Ep 1 k (", "x2 2 2 x1 2) k (x)2 1 2 When the archer in Figure 6.2 draws her bow she stores another form of potential energy, elastic potential energy, in the bow. Both gravitational potential energy and elastic potential energy form part of mechanical energy. The study of elastic potential energy requires the use of Hooke\u2019s law. The amount of energy stored in a spring is equal to the work done to stretch (or compress) the spring, without causing any permanent deformation. The force is not constant, so the equation for work used earlier, (W Fd) does not apply, because that equation requires a constant force acting over the displacement. However, when force-position graphs are used, work is equivalent to the area under the curve. The units for this area are N\u00b7m, equivalent to joules, the unit for work. You can therefore determine the amount of work done to stretch the spring from its equilibrium position to the position x by calculating the area of the shaded portion of Figure 6.17. the elastic or spring constant k: k slope F x Calculation of Elastic Potential Energy The area under the curve in Figure 6.17 is the shaded triangle whose area is calculated by A hb. The base (b) is equal to the magnitude 1 2 of the position (x), and the height (h) is equal to the magnitude of the force (F) at that position. Thus, in terms of force and position, the equation for the area under the curve is 1 W Fx 2 From Hooke\u2019s law, the magnitude of the force (F) is equal to F kx, so the work done to stretch the spring can be written as: 1 W (kx)(x) 2 1 kx2 2 The work done to stretch (or compress) a spring from its equilibrium position to any position (x) results in storing elastic potential energy (Ep) in the spring. Therefore, the equation for the elastic potential energy stored in the spring is given by Ep 1 kx2 2 Concept Check Explain why it is incorrect to try to find the change in elastic potential energy of a stretched spring from the measurement of the change in the stretch. 300 Unit III Circular Motion, Work, and Energy 06-PearsonPhys20-Chap06 7/28/08 11:13 AM Page 301 Example 6.4 A spring is stretched to a position 35.0 cm from its equilibrium position. At that point the force exerted on the spring is 10.5 N. (a) What is the elastic potential energy stored in the spring? (b) If the stretch in the spring is allowed to reduce to 20.0 cm, what is the change in the elastic potential energy? Given x1 F1 x2 35.0 cm 0.350 m 10.5 N 20.0 cm 0.200 m Required (a) elastic potential energy in the spring stretched ) to 0.350 m (Ep1 (b) change in the elastic potential energy when the stretch is reduced from 0.350 m to 0.200 m (Ep) Analysis and Solution (a) Calculate the value for k, the elastic constant for the spring, using Hooke\u2019s law. F1 k kx1 F1 x1 10.5 N 0.350 m N m 30.0 Graph Showing Change in Elastic Potential Energy 10. From the data given, plot the graph of change in elastic potential energy, Figure 6.18. Next, 0 0.200 0.350 Position x (m) Figure 6.18 use EP kx2, to find the elastic potential energy for 1 2 a stretch of 0.350 m. This is equivalent to finding the area of the large triangle in Figure 6.18. Ep1 1 kx1 2 2 N m 30.0 1 2 1.8375 N m 1.84 Nm 1.84 J (0.350 m)2 m2 (b) To find the change in the elastic potential energy, first find the elastic potential energy at a stretch of 0.200 m and then subtract from that value the answer to part (a). This is equivalent to finding the shaded area of the graph in Figure 6.18. Practice Problems 1. A force of 125 N causes a spring to stretch to a length of 0.250 m beyond its equilibrium position. (a) What is the elastic potential energy stored in the spring? (b) If the spring contracts to a stretch of 0.150 m, what is the change in elastic potential energy? 2. An engineer is designing the suspension system for a car. He decides that the coil spring used in this car should compress 4.00 cm when a force of 1000 N is applied to it. (a) What is the spring constant of the spring? (b) If the spring is compressed a distance of 14.0 cm, what force must have been exerted on it? 3. The elastic", " constant for a spring is 750 N/m. (a) How far must you stretch a spring from its equilibrium position in order to store 45.0 J of elastic potential energy in it? (b) If you wanted to double the elastic potential energy stored in the spring, how much farther would you need to stretch it? 4. A spring has an elastic constant of 4.40 104 N/m. What is the change in elastic potential energy stored in the spring when its stretch is increased from 12.5 cm to 15.0 cm? 5. When a spring is stretched by 0.400 m from its equilibrium position, its elastic potential energy is 5.00 102 J. (a) What is the magnitude of the force required to produce this amount of stretch? (b) If the force causing the stretch is changed to 1000 N, how much change in elastic potential energy results? Chapter 6 In an isolated system, energy is transferred from one object to another whenever work is done. 301 06-PearsonPhys20-Chap06 7/28/08 11:15 AM Page 302 Answers 1. (a) 15.6 J (b) 10.0 J 2. (a) 2.50 104 N/m (b) 3.50 103 N 3. (a) 0.346 m (b) 0.143 m 4. 1.51 102 J 5. (a) 2.50 103 N (b) 420 J kinetic energy: the energy due to the motion of an object PHYSICS INSIGHT There are two kinds of kinetic energy. The kinetic energy studied here is more correctly referred to as translational kinetic energy, since the objects are moving along a line. Earth has both translational kinetic energy (because it orbits the Sun) and rotational kinetic energy (because it spins on its axis). The elastic potential energy for a stretch of 0.200 m is: EP2 2 1 kx2 2 N 30.0 1 m 2 0.600 Nm 0.600 J (0.200 m)2 The change in the elastic potential energy is: EP EP2 EP1 0.600 J 1.84 J 1.24 J Paraphrase (a) The energy stored in the spring at the initial stretch is 1.84 J. (b) When the stretch is reduced from 0.350 m to 0.200 m, the elastic potential energy stored in the spring reduced by 1.24 J to 0.600 J. Kinetic Energy Examine Figure 6.19. When the archer releases the arrow, the bowstring exerts a non-zero force on the arrow, which accelerates the arrow toward its target. As the arrow gains speed, it gains kinetic energy (Ek). v Figure 6.19 Figure 6.20 When an object is in free fall, gravity is working to increase its kinetic energy. When the hiker in Figure 6.20 drops the rock off the cliff, the force of gravity accelerates the rock downward increasing its speed and thus its kinetic energy. Kinetic energy is a scalar quantity. The kinetic energy of an object is calculated using the equation Ek 1 mv2 2 Concept Check If the kinetic energy of an object doubles, by what factor does its speed increase? 302 Unit III Circular Motion, Work, and Energy 06-PearsonPhys20-Chap06 7/24/08 12:56 PM Page 303 Example 6.5 On a highway, a car of mass 1.2 103 kg, travelling at 20 m/s, has kinetic energy equal to a loaded van of mass 4.8 103 kg. What is the speed of the van? Given m1 v1 Ekcar 1.2 103 kg; m2 20 m/s Ekvan 4.8 103 kg; Required the speed of the van (v2) Analysis and Solution The two vehicles have equal kinetic energy mv 2. Find the kinetic energy of the car and then use that value to solve for the speed of the van. 1 2 Ekcar 1 mvcar 2 2 Ekvan 2.4 105 J m 2 (1.2 103 kg)20 s 1 2 mvvan 1 2 2 2.4 105 kgm2 s2 (2.4 105 J)(2) 4.8 103 kg vvan 2.4 105 J 10 m/s Paraphrase The van is travelling at 10 m/s. M I N D S O N Energy of Impact Practice Problems 1. A 45.0-kg girl pedals a 16.0-kg bicycle at a speed of 2.50 m/s. What is the kinetic energy of the system? 2. A car travelling at 80.0 km/h on a highway has kinetic energy of 4.2 105 J. What is the mass of the car? 3. A skateboarder with a mass of 65.0 kg increases his speed from 1.75 m/s to 4.20 m/s as he rolls down a ramp.", " What is the increase in his kinetic energy? Answers 1. 1.91 102 J 2. 1.7 103 kg 3. 474 J Project LINK How will the concept of the kinetic energy of a moving vehicle relate to the design of your persuader apparatus? 3. Investigate the incidence of meteor collisions in Canada. Where is the meteor impact crater that is closest to where you live? Approximately how many meteors have landed in Alberta? What was the greatest kinetic energy for a meteor that landed (a) in Alberta (b) in Canada? There is evidence that many meteors have hit Earth\u2019s surface. The vast quantity of kinetic energy that these meteors have at the time of impact is revealed by the size of the craters that they create (Figure 6.21). 1. What types of measurements would scientists need to make in order to estimate the kinetic energy of the meteor at the instant of impact? 2. What types of experiments could be done to verify the scientists\u2019 assumptions? Figure 6.21 Meteor impact craters are found in all regions of Earth. This one, called the Barringer crater, is in Arizona. Chapter 6 In an isolated system, energy is transferred from one object to another whenever work is done. 303 06-PearsonPhys20-Chap06 7/24/08 12:56 PM Page 304 Example 6.6 A man on a trampoline has a mass of 75.0 kg. At the instant he first touches the surface of the trampoline (at its rest position) he is descending with a speed of 8.00 m/s. At his lowest point, the man is 0.650 m below the trampoline\u2019s rest position. (a) What is the kinetic energy of the man when he first contacts the trampoline? (b) If you assume that, at his lowest point, all of the man\u2019s kinetic energy is transformed into elastic potential energy, what is the elastic constant for the trampoline? Practice Problems 1. A bow that has an elastic constant of 2500 N/m is stretched to a position of 0.540 m from its rest position. (a) What is the elastic potential energy stored in the bow? (b) If all of the elastic potential energy of the bow were to be transformed into kinetic energy of a 95.0-g arrow, what would be the speed of the arrow? 2. Cannon A fires a 1.5-kg ball with a muzzle velocity of 550 m/s, while cannon B fires cannon balls with one-third the mass but at twice the muzzle velocity. Which of these two cannons would be more effective in damaging a fortification? Explain why. 3. It is estimated that the meteor that created the crater shown in Figure 6.21 on the previous page had a radius of 40 m, a mass of approximately 2.6 108 kg, and struck Earth at a speed of nearly 7.20 104 km/h. (a) What was the kinetic energy of the meteor at the instant of impact? (b) When one tonne (t) of TNT explodes, it releases about 4.6 109 J of energy. In terms of tonnes of TNT, how much energy did the meteor have at impact? Answers 1. (a) 365 J 2. EkA 3. (a) 5.2 1016 J : EkB (b) 87.6 m/s 3 : 4. Ball B will do more damage. (b) 1.1 107 t v m 75.0 kg Given m 75.0 kg v 8.00 m/s x 0.650 m Required (a) kinetic energy of the man (Ek) (b) the elastic constant of the trampoline (k) 0.650 m Figure 6.22 Analysis and Solution (a) Find the initial kinetic energy, by using Ek Ek 1 mv2 2 1 mv2 2 1 2 m (75.0 kg)8.00 s m2 s2 2.40 103 kg 2 2.40 103 J (b) Assume that the elastic potential energy at 0.650 m is 2.40 103 J and solve for the elastic constant. Ep 1 kx2 2 Solve for k. k 2Ep x2 2(2.40 103 J) (0.650 m)2 1.136 104 N m 1.14 104 N m Paraphrase (a) The kinetic energy of the man is 2.40 103 J. (b) The elastic constant of a spring that stores 2.40 103 J of elastic potential energy when it is stretched 0.650 m is 1.14 104 N/m. 304 Unit III Circular Motion, Work, and Energy 06-PearsonPhys20-Chap06 7/24/08 12:56 PM Page 305 6.1 Check and Reflect 6.1 Check and Reflect 1. If a force does not act parallel to the resulting displacement, what is", " the effect on the work done by the force? 8. A spring has an elastic constant of 650 N/m. Initially, the spring is compressed to a length of 0.100 m from its equilibrium position. 2. Describe how a non-zero force can act on an object over a displacement and yet do no work. 3. Explain why the frame of reference affects the calculated value of an object\u2019s gravitational potential energy but not the change in its gravitational potential energy. 4. What is meant by elastic potential energy? 5. A force of 1500 N [up] acts to lift an object of 50.0-kg mass to a height of 24.0 m above its original position. (a) How much work did the force do on the object? (b) What was the gain in the object\u2019s gravitational potential energy? (c) What might account for the difference in the two answers? 6. A force of 850 N [30] acts on an object while it undergoes a displacement of 65.0 m [330]. What is the work the force does on the object? 7. You are working on the 5th floor of a building at a height of 18.0 m above the sidewalk. A construction crane lifts a mass of 350 kg from street level to the 12th floor of the building, 22.0 m above you. Relative to your position, what is the gravitational potential energy of the mass (a) at street level? (b) when it is on the 12th floor? (c) What is its change in gravitational potential energy as it is raised? (a) What is the elastic potential energy stored in the spring? (b) How much further must the spring be compressed if its potential energy is to be tripled? 9. Two cars (A and B) each have a mass of 1.20 103 kg. The initial velocity of car A is 12.0 m/s [180] while that of car B is 24.0 m/s [180]. Both cars increase their velocity by 10.0 m/s [180]. (a) Calculate the gain in kinetic energy of each car. (b) If both cars gain the same amount of velocity, why do they gain different amounts of kinetic energy? 10. A cart with a mass of 3.00 kg rolls from the top of an inclined plane that is 7.50 m long with its upper end at a height of 3.75 m above the ground. The force of friction acting on the cart as it rolls is 4.50 N in magnitude. (a) What is the change in gravitational potential energy when the cart moves from the top of the inclined plane to ground level? (b) What is the work done by friction? 11. An ideal spring with an elastic constant of 2000 N/m is compressed a distance of 0.400 m. (a) How much elastic potential energy does this compression store in the spring? (b) If this spring transfers all of its elastic potential energy into the kinetic energy of a 2.00-kg mass, what speed would that mass have? Assume the initial speed of the mass is zero. e TEST To check your understanding of potential and kinetic energy, follow the eTest links at www.pearsoned.ca/school/physicssource. Chapter 6 In an isolated system, energy is transferred from one object to another whenever work is done. 305 06-PearsonPhys20-Chap06 7/24/08 12:56 PM Page 306 6.2 Mechanical Energy mechanics: the study of kinematics, statics, and dynamics mechanical energy: the sum of potential and kinetic energies In physics, the study of mechanics includes kinematics (the study of motion), statics (the study of forces in equilibrium), and dynamics (the study of non-zero forces and the motion that results from them). Gravitational potential energy, elastic potential energy, and kinetic energy form what is called mechanical energy. When work is done on a system, there may be changes in the potential and kinetic energies of the system. This relationship is expressed as the work-energy theorem. info BIT Doubling the speed of a vehicle means quadrupling the necessary stopping distance. This relationship between stopping distance and speed is based on the physics of work and kinetic energy. The Work-Energy Theorem A 1.50 105-kg jet plane waits at the end of the runway. When the airtraffic controller tells the pilot to take off, the powerful engines can each produce more than 2.5 105 N of thrust to accelerate the plane along the runway. In order to produce the speed of about 250 km/h required for takeoff, the engines would need to convert more than 3 108 J of chemical energy from the fuel supply into kinetic energy. Figure 6.23 The work done by the jet\u2019s engines must convert enough chemical energy into kinetic energy to produce a velocity sufficient for takeoff. e LAB For a probeware activity, go to", " www.pearsoned.ca/ school/physicssource. PHYSICS INSIGHT In the example of the airplane takeoff, is the angle between the direction of Fnet or a, and d. As explained by Newton\u2019s Laws of Motion, the non-zero net force causes the jet plane to accelerate along the runway. Since a change in kinetic energy must involve a change in speed, kinetic energy changes are always the result of the acceleration, which in turn is caused by a non-zero net force. In terms of work and energy, this means that changes in kinetic energy (Ek) are always the result of the work done by a non-zero net force (Wnet). Ek Wnet (Fnet)(cos )(d) (ma)(cos )(d) In all cases, work done by a non-zero net force results in a change in kinetic energy but the applied forces on an object may cause changes in its potential energy, its kinetic energy, or both. For example, once the jet plane is in the air, the thrust produced by its engines must increase its speed (Ek) as well as cause it to gain altitude (Ep). 306 Unit III Circular Motion, Work, and Energy 06-PearsonPhys20-Chap06 7/24/08 12:56 PM Page 307 Zero and Non-zero Net Forces and Effects on Energy A motor that is pulling a block up a frictionless inclined plane at a constant speed (Figure 6.24) is exerting a force that causes a change in gravitational potential energy but not kinetic energy. The constant speed indicates that the applied force (Fapp) is exactly balanced by the Fg component of Fg. There is zero net force; Fapp Fg. p u p u h ill FN a 0 v constant Fapp h ill o w n d o w n d Fg Fg Fg Since Fnet 0 Fapp Fg FN Fg Fg Fapp Fg Figure 6.24 If all the forces acting on a block combine to produce a net force of zero, the block moves up the incline at a constant speed. It increases its gravitational potential energy but not its kinetic energy. If, however, the force applied is now increased so that there is a Fg (Figure 6.25), the forces are no longer non-zero net force and Fapp balanced, the block accelerates up the incline, and both kinetic energy and potential energy change. Now the work done on the block is transferred to both its kinetic energy and its gravitational potential energy. This is expressed mathematically as W E or, in more detail, as W Ek Ep This is known as the work-energy theorem. p u FN p u h ill a 0 v Fapp h ill o w n d o w n d Fg Fg Fg Since Fnet 0 Fapp Fg a 0 Hence v increases and Ek increases. FN Fg Fg Fapp Fg Figure 6.25 If the forces acting on a block are such that there is a non-zero net force Fg. Both the kinetic energy and the gravitational potential energy up the plane, then Fapp will increase as the block moves up the incline. The work-energy theorem states that the work done on a system is equal to the sum of the changes in the potential and kinetic energies of the system. PHYSICS INSIGHT The symbol EP could refer to gravitational potential energy, elastic potential energy, or the sum of the gravitational and elastic potential energies. Chapter 6 In an isolated system, energy is transferred from one object to another whenever work is done. 307 06-PearsonPhys20-Chap06 7/24/08 12:56 PM Page 308 Concept Check A block is sliding down an inclined plane. If there is no friction, describe a situation where the net work might still be negative. Example 6.7 A block is moved up a frictionless inclined plane by a force parallel to the plane. At the foot of the incline, the block is moving at 1.00 m/s. At the top of the incline, 0.850 m above the lower end, the block is moving at 4.00 m/s. The block has a mass of 1.20 kg. What is the work done on the block as it moves up the incline? Practice Problems 1. A mountain climber rappels down the face of a cliff that is 25.0 m high. When the climber, whose mass is 72.0 kg, reaches the bottom of the cliff he has a speed of 5.00 m/s. What is the work done on the climber by the rope? 2. A force of 150 N [up] acts on a 9.00-kg mass lifting it to a height of 5.00 m. (a) What is the work done on the mass by this force? (b) What is the change in gravitational potential energy? (c) What change", " in kinetic energy did the mass experience? 3. Draw a free-body diagram for the forces on the mass in question 2. (a) Calculate the net force acting on the mass. (b) Calculate the work done on the mass by the net force. (c) How does this relate to the answer to question 2(c)? Answers 1. 1.68 104 J 2. (a) 750 J (b) 441 J (c) 309 J 3. (a) 61.7 N [up] (b) 309 J (c) Ek 309 J Given m 1.20 kg 1.00 m/s v1 4.00 m/s v2 h 0.850 m g 9.81 m/s2 1.00 m/s v1 p u h ill n w o d w o d h ill p u n Fapp 4.00 m/s v2 h 0.850 m Required work done on the block as it moves up the incline (W) Figure 6.26 Analysis and Solution The work-energy theorem states that the work will be equal to the sum of the changes in the kinetic and potential energies. For the change in kinetic energy find the difference in the final and initial kinetic energy using the final and initial speeds. Change in gravitational potential energy can be found from the change in height (Figure 6.26). W Ek (Ek2 mv2 1 2 Ep Ek1) (mgh) 1 2 mv1 2 2 (mgh) 1 2 (1.20 kg)(4.00 m/s)2 (1.20 kg)(1.00 m/s)2 1 2 (1.20 kg)(9.81 m/s2)(0.850 m) (9.60 J 0.60 J) (10.01 J) 19.0 J Paraphrase and Verify The work caused the block to gain a total of 19.0 J, the sum of 9.00 J of kinetic and 10.0 J of potential energy as it moved up the ramp. 308 Unit III Circular Motion, Work, and Energy 06-PearsonPhys20-Chap06 7/24/08 12:56 PM Page 309 Calculations of Mechanical Energy To calculate the mechanical energy of an object is simply to find the total of the kinetic energy and all forms of potential energy. Em Ek Ep Because gravitational potential energy is defined relative to a reference point, mechanical energy will also depend on that reference point. Example 6.8 A cannon ball is fired from Earth\u2019s surface. At the peak of its trajectory, it has a horizontal speed of 160 m/s and is 1.20 103 m above the ground. With reference to the ground, what is the mechanical energy of the cannon ball at the highest point on its trajectory, if the mass of the cannon ball is 5.20 kg? v2 h up v1 down Figure 6.27 Given m 5.20 kg v 160 m/s h 1.20 103 m g 9.81 m/s2 Required mechanical (total) energy of the cannon ball (Em) Analysis and Solution At the top of its trajectory, the cannon ball has kinetic energy due to its horizontal motion, and gravitational potential energy because of its height above the ground. Em Ep Ek 1 2 mv2 mgh Practice Problems 1. A rocket is accelerating upward. When the rocket has reached an altitude of 5.00 103 m, it has reached a speed of 5.40 103 km/h. Relative to its launch site, what is its mechanical energy, if the mass of the rocket is 6.50 104 kg? 2. What is the speed of a 4.50-kg cannon ball if, at a height of 275 m above the ground, its mechanical energy relative to the ground is 6.27 104 J? 3. As a roller coaster trolley with a mass of 600 kg coasts down the first hill, it drops a vertical distance of 45.0 m from an initial height of 51.0 m above the ground. If, at the bottom of the hill, its speed is 30.0 m/s: (a) what is the trolley\u2019s mechanical energy relative to the top of the hill, and (b) what is the trolley\u2019s mechanical energy relative to the ground? (5.20 kg)9.81 2 m s 1 2 (5.20 kg)160 6.656 104 J 6.121 104 J 1.28 105 J m s2 (1.20 103 m) Answers 1. 7.63 1010 J 2. 150 m/s 3. (a) 5.13 103 J (b) 3.05 105 J Paraphrase and Verify The total energy of the cannon ball at the top of the trajectory is 1.28 105 J. The gravitational potential energy is positive because the cannonball is higher than the reference point. Chapter 6 In an isolated system,", " energy is transferred from one object to another whenever work is done. 309 06-PearsonPhys20-Chap06 7/24/08 12:56 PM Page 310 6.2 Check and Reflect 6.2 Check and Reflect Knowledge Extensions 1. What are the forms of energy that make up 8. The figure below shows the force versus mechanical energy? 2. Why does your choice of a frame of reference affect the calculated value of the mechanical energy? 3. What is the relationship between the net force and kinetic energy? 4. State the work-energy theorem. Applications 5. A net force of 5.75 103 N [180] acts on a mass of 23.0 kg. If, while the force acts, the mass travels through a displacement of 360 m [210], what work did the net force do on the object? Into what form of energy was this work transferred? 6. At a height of 75.0 m above the ground, a cannon ball is moving with a velocity of 240 m/s [up]. If the cannon ball has a mass of 12.0 kg, what is its total mechanical energy relative to the ground? What effect would there be on the answer, if the velocity of the cannon ball were downward instead of upward? Explain. 7. A mass of 8.50 kg is travelling 7.50 m/s [up]. It is acted on by a force of 340 N [up] over a displacement of 15.0 m [up]. (a) What work does the applied force do on the object? (b) What is its gain in potential energy? (c) What is its change in kinetic energy? (d) What is its speed at the end of the 15.0-m displacement? 310 Unit III Circular Motion, Work, and Energy displacement graph for an elastic spring as it is compressed a distance of 0.240 m from its equilibrium position by a force of magnitude 2.40 103 N. A 7.00-kg mass is placed at the end of the spring and released. As the spring expands, it accelerates the mass so that when the spring\u2019s compression is still 0.180 m from its equilibrium position, the mass has a speed of 6.00 m/s. (a) What is the mechanical energy in this system when the spring is compressed to 0.240 m? (b) What is the mechanical energy in the system when the spring is compressed to 0.180 m? (c) How much work has been done on the mass by this system as the spring expanded from a compression of 0.240 m to 0.180 m? (d) How does the work done on the mass by the spring compare to the kinetic energy of the mass? equilibrium position of spring Force vs. Displacement for an Elastic Spring 2400 1800 1200 600 ) ( N F e c r o F 0 0.06 0.12 Displacement d (m) 0.18 0.24 e TEST To check your understanding of the work\u2013energy theorem and mechanical energy, follow the eTest links at www.pearsoned.ca/school/physicssource. 06-PearsonPhys20-Chap06 7/24/08 12:56 PM Page 311 6.3 Mechanical Energy in Isolated and Non-isolated Systems Isolated Systems Imagine two people are in an isolated (sealed) room. They may complete as many money transfers as they like but the total amount of money in the room before and after each transfer will be the same. We can say that the total amount of money in this system is conserved, in that it does not change during transactions. PHYSICS INSIGHT An Open System can exchange both energy and matter with its surroundings. A Closed System can exchange energy but not matter with its surroundings. An Isolated System cannot exchange energy or matter with its surroundings. Figure 6.28 In an isolated room, the total amount of money in the room, before and after a transaction, is constant. Figure 6.29 In a non-isolated room, the amount of money in the room may change. Now imagine that the room is not isolated. In this case, money may be taken out of (or put into) the room so that the total amount of money in the room is not necessarily constant. In this system, it cannot be said that money is conserved. It would be much more complex to keep track of the money transfers that occur in this non-isolated room compared with those occurring in the isolated room. In physics, when the energy interactions of a group of objects need to be analyzed, we often assume that these objects are isolated from all other objects in the universe. Such a group is called an isolated system. Isolated Systems and Conservation of Energy While objects within an isolated system are free to interact with each other, they cannot be subjected to unbalanced forces from outside that system. In terms of mechanical energy, that means that no force from outside the system may work to", " transfer energy to or from any object inside the system. The quantity of energy in the system must be constant. Even though friction may seem like an internal force, its effect is to allow energy to escape from a system as heat. Thus, an isolated system must also be frictionless. These ideas will be further explored later in the chapter. info BIT In everyday terms, energy conservation means to use as little energy as possible to accomplish a task. In physics, energy conservation refers to systems, such as an ideal pendulum, in which the total amount of energy is constant. isolated system: a group of objects assumed to be isolated from all other objects in the universe Chapter 6 In an isolated system, energy is transferred from one object to another whenever work is done. 311 06-PearsonPhys20-Chap06 7/24/08 12:56 PM Page 312 e TECH Consider the transformation of energy from potential energy to kinetic energy in a falling object, and in a ball bouncing on a trampoline. Follow the eTech links at www.pearsoned.ca/school/ physicssource. e SIM Find out more about the mechanical energy, gravitational potential energy, and kinetic energy of a satellite-Earth system or a projectile-Earth system. Go to www.pearsoned.ca/school/ physicssource. Conservation of Mechanical Energy Because the mechanical energy (the sum of potential and kinetic energies) for an isolated system must be a constant, it follows that if you calculate the mechanical energy (Em) at any two randomly chosen times, the answers must be equal. Hence, Em2 Em1 (1) Within an isolated system, energy may be transferred from one object to another or transformed from one form to another, but it cannot be increased or decreased. This is the law of conservation of energy. Relationship between kinetic and potential energy in an isolated system The law of conservation of energy is one of the fundamental principles of science and is a powerful mathematical model for analysis and prediction of the behaviour of objects within systems. Viewed from a slightly different perspective, conservation of energy states that, in terms of mechanical energy, any gain in kinetic energy must be accompanied by an equal loss in potential energy. Ek Ep (2) Statements (1) and (2) are equivalent. This can be verified as follows. If total energy remains constant regardless of time or position, then Em2 Em1. But mechanical energy is the sum of the kinetic and potential energies. Therefore, Ek2 Ep2 Ek1 Ep1 Hence, Ek1 Ek2 Ep2 Ep1 Ek (Ep2 Ep1) Thus, Ek Ep is true. Energy vs. Position for a Block Sliding Down an Inclined Plane Em Ek Em Ep Ek Ep Ek Em Ep ) J ( E y g r e n E Ep Ek Position d (m) Figure 6.30 In an isolated system the loss in gravitational potential energy is equal to the gain in kinetic energy. As a block slides down a frictionless inclined plane, its gravitational potential energy will decrease and its kinetic energy will increase. Figure 6.30 shows the energy-position graph for this isolated system. As the block moves down the plane, the sum of the heights of the potential and kinetic energy curves (value of blue line plus value of red line) at any point is equal to the block\u2019s mechanical energy (E m). The mechanical energy (shown by the purple line) is constant; therefore, energy is conserved. This graph is typical of an isolated system. 312 Unit III Circular Motion, Work, and Energy 06-PearsonPhys20-Chap06 7/24/08 12:56 PM Page 313 Example 6.9 A frictionless roller coaster car has a mass (m) of 8.00 102 kg. At one point on its journey, the car has a speed of 4.00 m/s and is 35.0 m above the ground. Later, its speed is measured to be 20.0 m/s. (a) Calculate its total initial mechanical energy relative to the ground. (b) What is its gravitational potential energy in the second instance? v1 Given m 8.00 102 kg 4.00 m/s 35.0 m 20.0 m/s v2 g 9.81 m/s2 h1 Required (a) mechanical energy (Em1) (b) gravitational potential energy when the speed is 20.0 m/s (Ep2) Analysis and Solution A frictionless roller coaster can be treated as an isolated system. (a) The mechanical energy at any point is the sum of its kinetic and potential energies. Em1 Em1 Ep1 Ek1 1 mv1 2 2 mgh1 (8.00 102 kg)4.00 1 2 m s 2 (8.00 102 kg)9.81 m s2 (35.0 m) 2.810 105 J 2.81 105 J (b) The system is defined as isolated, meaning that energy is conserved.", " By the law of conservation of energy, the mechanical energy at any two points must be equal. The gravitational potential energy at the second point must be equal to the mechanical energy less the kinetic energy at the second point. Ek2 Em2 Ep2 Ep2 Em1 Em1 Em1 Em1 Ek2 1 mv2 2 2 m 1 2 (8.00 102 kg)20.0 2.810 105 J s 2 2.810 105 J 1.600 105 J 1.21 105 J Practice Problems 1. In an isolated system, a crate with an initial kinetic energy of 250 J and gravitational potential energy of 960 J is sliding down a frictionless ramp. If the crate loses 650 J of gravitational potential energy, what will be its final kinetic energy? 2. A mass of 55.0 kg is 225 m above the ground with a velocity of 36.0 m/s [down]. Use conservation of energy to calculate its velocity when it reaches a height of 115 m above the ground. Ignore the effects of air resistance. 3. A human \u201ccannon ball\u201d in the circus is shot at a speed of 21.0 m/s at an angle of 20 above the horizontal from a platform that is 15.0 m above the ground. See Figure 6.31. (a) If the acrobat has a mass of 56.0 kg, what is his gravitational potential energy relative to the ground when he is at the highest point of his flight? Ignore the effects of air resistance. (b) If the net in which he lands is 2.00 m above the ground, how fast is he travelling when he hits it? human cannon ball cannon 15.0 m net 2.00 m Figure 6.31 Answers 1. 900 J 2. 58.8 m/s [down] 3. (a) 9.69 103 J (b) 26.4 m/s Chapter 6 In an isolated system, energy is transferred from one object to another whenever work is done. 313 06-PearsonPhys20-Chap06 7/24/08 12:56 PM Page 314 e SIM Learn about the relationships among the mechanical, kinetic, and gravitational potential energies of a pendulum. Go to www.pearsoned.ca/school/ physicssource. Paraphrase and Verify (a) The total initial mechanical energy relative to the ground is 2.81 105 J. (b) The gravitational potential energy at a speed of 20.0 m/s is 1.21 105 J. The kinetic energy increased from 6.40 103 J to 1.60 105 J, while the gravitational potential energy decreased from 2.74 105 J to 1.21 105 J. As kinetic energy increases, potential energy decreases. When the speed is 20.0 m/s, the car must be below its starting point. M I N D S O N Energy and Earth\u2019s Orbit At its closest point to the Sun (perihelion), around January 4th, Earth is about 147 million kilometres from the Sun. At its farthest point from the Sun (aphelion), around July 5th, Earth is about 152 million kilometres from the Sun. \u2022 In terms of the conservation of energy, what conclusions can be made about Earth\u2019s speed as it moves around the Sun? \u2022 What assumptions must you make to support your conclusions? A Simple Pendulum A simple pendulum is an excellent approximation of an isolated system. During its downswing, Earth\u2019s gravity does work on the pendulum to transfer gravitational potential energy into kinetic energy. On the upswing, gravity transfers kinetic energy back into gravitational potential energy. The mechanical energy of the pendulum is constant (Figure 6.32). Newton\u2019s third law of motion states that for every action force there is an equal but opposite reaction force. This means that as Earth\u2019s gravity acts on the pendulum converting gravitational potential energy into kinetic energy, the pendulum must also act to convert gravitational potential energy to kinetic energy for Earth; Earth must be part of the isolated system that contains the pendulum. Earth\u2019s mass compared to that of the pendulum is enormous, so its reaction to the pendulum is immeasurably small. This explains why we can ignore the effects of the pendulum on Earth and analyze the pendulum as if it were an isolated system. Treating the pendulum as an isolated system greatly simplifies the calculations of the system\u2019s mechanical energy. It means that the force of gravity works on the pendulum without changing the energy in the system. In fact, while work done by the force of gravity may transfer energy from one form to another it never causes a change in mechanical energy (Figure 6.33). Forces that act within systems but do not change their mechanical energy are defined as conservative forces. This type of force will be discussed in more detail later in this chapter. Em Ep 0 at Ep max max Ep min Ek Em 0 Ek at Ep min where h 0", " min Ep max Ek Figure 6.32 As a pendulum swings, gravity acts to convert energy back and forth between gravitational potential energy and kinetic energy. PHYSICS INSIGHT If h 0 had been defined to occur at the lowest point of the pendulum\u2019s swing, the gravitational potential energy at the lowest point would be zero and the mechanical energy would be equal to the kinetic energy. Then, at the highest point on the swing, where movement stops and kinetic energy is zero, the mechanical energy would be equal to the potential energy. 314 Unit III Circular Motion, Work, and Energy 06-PearsonPhys20-Chap06 7/24/08 12:56 PM Page 315 Because the pendulum acts as an isolated system, energy is conserved. To calculate the mechanical energy of a pendulum it is necessary to know its mass, its height above the reference point, and its speed. If all of those values are known at any one point on its swing, then the mechanical energy of the pendulum is known at all points on its swing. Once the mechanical energy is known, it can be used to predict the pendulum\u2019s motion at any instant along its path, and to correlate kinetic and potential energy with the amplitude of the swing. Energy Conservation in a Simple Pendulum ) Em Ep Ek Em Ep Ek 0.5a 0 0.5a a Position x (m) Figure 6.33 The force of gravity acts to change the gravitational potential energy and kinetic energy of the pendulum so that the mechanical energy remains constant. hmax h 0 a 0 hmax a Example 6.10 An ideal pendulum, as shown in Figure 6.32, is suspended by a string that is 2.00 m long. It is pulled sideways and released. At the highest point of its swing the pendulum bob is 25.0 cm above the floor. At the lowest point of its swing the pendulum bob is 5.00 cm above the floor. The mass of the pendulum bob is 250 g. (a) What is the mechanical energy of the pendulum, relative to the floor, when the bob is at its highest point? (b) What is the mechanical energy of the pendulum, relative to the floor, when the bob is at its lowest point? (c) What is the kinetic energy of the bob when it is at its lowest point? (d) What is the speed of the pendulum bob when the bob is at its lowest point? Given m 250 g 0.250 kg g 9.81 m/s2 h1 h2 v1 25.0 cm 0.250 m 5.00 cm 0.0500 m 0 Required (a) sum of gravitational potential and kinetic energies of the pendulum at the highest point (Em1) (b) mechanical energy of the pendulum at the lowest point (Em2) (c) kinetic energy of the bob at the lowest point (Ek2) (d) speed of the bob at the lowest point (v2) Practice Problems 1. When the pendulum bob in Example 6.10 is 15.0 cm above the floor, calculate: (a) its mechanical energy relative to the floor (b) its kinetic energy (c) its speed 2. A model rocket has a mass of 3.00 kg. It is fired so that when it is 220 m above the ground it is travelling vertically upward at 165 m/s. At that point its fuel runs out so that the rest of its flight is without power. Assume that the effect of air friction is negligible and that all potential energies are measured from the ground. (a) What is the mechanical energy of the rocket, relative to the ground, when it is 220 m above the ground? (b) When it reaches the highest point on its trajectory, what will its gravitational potential energy be? (c) How far above the ground is the rocket at its highest point? (d) When it hits the ground, what is its speed? Chapter 6 In an isolated system, energy is transferred from one object to another whenever work is done. 315 06-PearsonPhys20-Chap06 7/24/08 12:56 PM Page 316 3. A roller coaster trolley and its passengers have a mass of 840 kg. The trolley comes over the top of the first hill with a speed of 0.200 m/s. The hill is 85.0 m above the ground. The trolley goes down the first hill and up to the crest of the second hill 64.0 m above the ground. Ignore the effect of frictional forces. What is the kinetic energy of the trolley at the top of the second hill? 4. A pole-vaulter with a mass of 56.0 kg tries to convert the kinetic energy of her approach into height. (a) What is the maximum height she can expect to attain if her approach speed is 8.00 m/s? Assume that the centre of mass", " of the vaulter is initially 0.850 m above the ground. (b) Describe the energy changes that occur from the time the vaulter starts to run until she reaches the highest point of her jump. Answers 1. (a) 0.613 J (b) 0.245 J (c) 1.40 m/s 2. (a) 4.73 104 J (b) 4.73 104 J (c) 1.61 103 m (d) 178 m/s 3. 1.73 105 J 4. (a) 4.11 m Analysis and Solution (a) At its highest point, the speed of the pendulum is zero. Thus, the mechanical energy at that point is equal to its gravitational potential energy. Em1 Ep1 Ek1 Em1 mgh1 1 mv1 2 2 (0.250 kg)9.81 kgm2 s 2 0.6131 0.613 J m s2 1 (0.250 m) (0.250 kg)(0)2 2 (b) In an isolated system, the mechanical energy is constant. Thus, by the law of conservation of energy the mechanical energy at its lowest point is equal to the mechanical energy at its highest point. Em2 Em2 0.613 J Em1 (c) The kinetic energy at the lowest point is the difference between the mechanical and gravitational potential energies at that point. Ek2 Ep2 (mgh2) Em2 Ek2 Ep2 Em2 Em2 0.6131 J (0.250 kg)9.81 0.6131 J (0.1226 J) 0.491 J m s2 (0.0500 m) (d) The speed at the lowest point can be found from 2 Ek2 kinetic energy. 1 mv2 2 2(0.491 J) v2 2Ek2 m 0.250 kg 1.98 m s Paraphrase and Verify (a) At the highest point, the total energy of the pendulum is 0.613 J. (b) As the bob swings lower, gravitational potential energy is lost and kinetic energy is gained. The total energy remains 0.613 J. (c) At the lowest point, the kinetic energy is 0.491 J, the difference between its total energy and its gravitational potential energy. (d) At the lowest point of its swing, the bob has a speed of 1.98 m/s. 316 Unit III Circular Motion, Work, and Energy 06-PearsonPhys20-Chap06 7/24/08 12:56 PM Page 317 6-2 Inquiry Lab 6-2 Inquiry Lab Conservation of Mechanical Energy Required Skills Initiating and Planning Performing and Recording Analyzing and Interpreting Communication and Teamwork Question Is energy conserved during the motion of a pendulum? Hypothesis State a hypothesis concerning the energy status of a pendulum. Remember to write this in the form of an if/then statement. Variables The variables in this lab are the values used to calculate the gravitational potential energy (mass, gravitational acceleration, and height) and the kinetic energy (mass and speed) at various points on the swing of the pendulum. Consider and identify which are controlled variable(s), which manipulated variable(s), and which responding variables. (NOTE: The voltage to the timer must not vary from trial to trial. If you are using a variable-voltage power supply, adjust the voltage so that the timer runs smoothly, and leave the power supply untouched for the remainder of the experiment. Stop and start the timer by disconnecting and reconnecting the lead attached to the black post of the power supply rather than turning the power supply off and on.) 3 Record your results in a table of data (Table 6.3). Table 6.3 Calibration Data Test Number Total Time t (s) Number of intervals N Time/Interval t (s) 1 2 Materials and Equipment string (at least 2.0 m long) pendulum bob (a 1-kg mass or greater) metre-stick ticker tape timer masking tape stopwatch (for timer calibration) Procedure For some interval timers, the period of the timer varies with the operating voltage. If your timer is of that type, begin by calibrating the timer. This is done by pulling a strip of ticker tape through the timer for a measured time, as set out below. Calibrate the timer: 1 Start the tape moving steadily through the timer, then connect the timer to the power supply for an exact measure of time (3 to 5 s works well). Be sure the tape does not stop moving while the timer is running. 2 Count the number of intervals between the dots (N) and divide that number into the measured time (t) to determine the time lapse per interval (t). Do at least one more calibration trial (without changing the voltage) to check if the time per interval (t) remains constant. Set up the apparatus: 4 Susp", "end the pendulum from a suitable solid point and allow the pendulum bob to come to rest at its lowest point. Place a marker (e.g., a piece of masking tape) on the floor below the centre of mass of the bob to indicate the pendulum\u2019s rest position. Measure and record the length (l) of the pendulum, and mass (m) of the pendulum bob. (NOTE: The length of a pendulum is measured from the point at which it pivots to the centre of mass of the pendulum bob. If the shape of the bob is a symmetrical solid, such as a sphere or a cylinder, the centre of mass is at its geometric centre.) 5 Pull the pendulum sideways so that its horizontal displacement (x) is about one-half its length. Place a marker, such as an iron stand, at this point, which will be xmax for the experiment. Ensure the path of the pendulum is clear, then release it to check the path of its swing. One team member should be positioned to catch the pendulum so that it does not swing back. CAUTION: Make sure that the path of the pendulum is clear before you allow it to begin its swing. Chapter 6 In an isolated system, energy is transferred from one object to another whenever work is done. 317 06-PearsonPhys20-Chap06 7/24/08 12:56 PM Page 318 6 Position the ticker tape timer at a distance approximately equal to the length of the pendulum from the pendulum\u2019s rest position. Locate the timer so that its height is just above the lowest point of the pendulum bob\u2019s path. Align the timer so that the tape does not bind as the bob pulls it through the timer. See Figure 6.34. Anchor the timer firmly so that it does not shift during trials. power supply pendulum bob timer timer tape Figure 6.34 7 With ticker tape attached to the pendulum but without the timer running, do a trial run of the system. Attach the tape to the pendulum bob at its centre of mass, so that the pull of the tape does not cause the bob to twist. Use a length of ticker tape that will reach from the timer to a point slightly beyond the bob\u2019s rest position. Move the pendulum bob sideways to its starting point. Hold the bob in place while you pull the tape tight, then gently allow the tape to take up the weight of the bob. Be sure that the tape is not twisted so that it does not rip as it passes through the timer. Release the tape and allow the pendulum to pull it through the timer. Collect data: 8 Once the timer is positioned so that the tape moves smoothly through it, you are ready to do a trial with the timer running. First, with the bob at its rest position, have one team member hold the bob stationary and pull the tape through the timer until it is just taut. Place a mark on the tape at the location where the timer records its dots. This mark on the tape records the position of the bob when its horizontal displacement is equal to zero (x 0). 9 Move the pendulum bob sideways to the starting point of its swing (xmax). Again hold the bob steady while a team member pulls the tape tight. Gently allow the tape to take up the weight of the pendulum bob. Start the timer, then release the pendulum. 10 Lay the tape out on a table and place a line across the tape through the first dot the timer put on the tape. At that position, the speed is zero (v 0) and the position is the maximum displacement (xmax). Measure the length of the tape from x 0 to xmax. 11 Locate the two dots that define the interval containing the mark that indicates the position of the bob at its rest position (x 0). Label this interval as interval 1. Measure the length (x1) of this interval (the space between the two dots on either side of the mark) and record it in a data table (Table 6.4). Calculate the speed v of the pendulum for interval 1 by dividing x1 by the interval time t. 12 Along the length of the tape, between x 0 and xmax, choose at least four more time intervals and draw a line across the tape at the midpoint of each chosen interval. Starting from interval 1, number the selected intervals as 2, 3, etc. For each of the chosen intervals, measure (a) the length of the interval (x), and (b) the distance (x) to the midpoint of the interval from the line indicating x 0. Record your measurements in a data table (Table 6.4). Analysis 1. Use a table similar to Table 6.4 to organize your data. 2. Calculate the height (h) of the pendulum above its rest position by using the relationship h l l 2 x", "2. (See the diagram in Figure 6.35.) 3. Calculate the values for the gravitational potential (Ep), the kinetic (Ek) and the mechanical (Em) energies for each of the intervals you marked on your tape. 4. On the same set of axes, plot graphs for Ep and Ek against the horizontal displacement (x) of the pendulum. Describe the relationship between Ek and the position of the pendulum that is indicated by the graph. Does Ek change uniformly as the pendulum swings? Does Ep change uniformly as the pendulum swings? What relationship does the graph suggest exists between Ep and Ek for the pendulum? 5. On the same set of axes, plot a graph of the total mechanical energy (Em) of the system against horizontal position (x). What does the graph suggest is the nature of the total mechanical energy for the pendulum? Suggest a reason for this relationship. 318 Unit III Circular Motion, Work, and Energy 06-PearsonPhys20-Chap06 7/24/08 12:56 PM Page 319 6. Within experimental error, can the mechanical energy 7. How is your hypothesis affected by your data? Explain. of the system be considered constant? If the mechanical energy is assumed constant, what value would you choose to be the most representative of this energy? Explain why. For each of the intervals that you chose for analysis, what is the percent error in the mechanical energy at that interval? Does your analysis indicate a systemic error change for the pendulum as it swings? What would be the cause of this error? Table 6.4 Pendulum Data Horizontal Displacement x (m) 0 Height h (m) 0 Ep (J) 0 Interval Number Figure 6.35 Interval Length x (cm) Interval Speed v (m/s) Ek (J) Em (J) (Ep + Ek) Conservative and Non-conservative Forces To understand the law of conservation of energy you must understand that some forces, such as gravity and elastic forces, act within systems without affecting the mechanical energy of the system. When such forces operate, energy is conserved. These are called conservative forces. Other forces, such as friction, and forces applied from outside a system, cause the energy of the system to change so that energy is not conserved. These are known as non-conservative forces. Figure 6.36 shows a system of two ramps joining point P to point Q. The drop h is the same for the two ramps, but ramp A is shorter than ramp B, because of the hills in ramp B. If a frictionless car is released from P and moves down one ramp to Q, the amount of kinetic energy the car gains in moving from P to Q does not depend on which ramp (A or B) the car coasts down. If energy is conserved, the kinetic energy at point Q is equal to the potential energy at point P, so the speed of the car at point Q will be the same whether it comes down ramp A or ramp B. Since a conservative force does not affect the mechanical energy of a system, the work done by a conservative force to move an object from one point to another within the system is independent of the path the object follows. car point P h ramp A ramp B point Q Because Ep at P Ek at Q, the speed of the car at Q is the same no matter which ramp it coasts down. Figure 6.36 If a conservative force acts on an object, then the work it does is independent of the path the object follows between two points. Chapter 6 In an isolated system, energy is transferred from one object to another whenever work is done. 319 06-PearsonPhys20-Chap06 7/24/08 12:56 PM Page 320 PHYSICS INSIGHT No matter how closely conditions approximate an isolated system, the force of friction is never truly zero. Friction is continually converting kinetic energy to thermal energy. This thermal energy, which cannot be converted back into mechanical energy, radiates out of the system as heat. non-isolated system: a system in which there is an energy exchange with the surroundings Friction Is a Non-conservative Force In the absence of friction, the car in Figure 6.36 would return to point P with no change in its mechanical energy. However, if there is friction, any motion of the car will be subject to it. When you analyze the work done by friction, you can see that path length does affect the work done on the car. The term d, the distance through which the force of friction acts, is not the displacement, but is always the actual distance the object travels. Wf is the work done by the force of friction, Ff, on the system. Therefore: Ff Wf but dB d dA, so WfB WfA and the car on Ramp B would lose more energy. Since the potential energy at the bottom of the ramp is the same regardless of the route, the loss in mechanical energy must be", " a loss in kinetic energy. Therefore, friction is not a conservative force. Because thermal energy is being radiated out of the system, the system is, by definition, a non-isolated system. The amount of work done by friction will cause the mechanical energy of the system to change so that Em Wf Therefore, Em1 Em1 Em2 Em2 Wf or Wf The direction of the force of friction is always exactly opposite to the direction of the motion; therefore, the calculated value of Wf is always negative. Friction always reduces the mechanical energy of a system. Concept Check What assumptions must be made if you wish to use the law of conservation of energy to solve a problem in physics? M I N D S O N That\u2019s the Way the Ball Bounces With each successive bounce, the height attained by a bouncing ball becomes less. Assuming the elastic forces that cause the ball to bounce are conservative in nature, and no outside forces act on the ball, you might expect it to behave as an isolated system. If so, the energy of the ball should be conserved. Use the concept of systems and conservation of energy to explain why the height decreases with each bounce. 320 Unit III Circular Motion, Work, and Energy 06-PearsonPhys20-Chap06 7/24/08 12:56 PM Page 321 PHYSICS INSIGHT In a non-isolated system, Wf has a negative value and decreases the total mechanical energy. The effect of friction is opposite to forces that are adding mechanical energy. Energy Changes in Non-isolated Systems Not all external forces remove mechanical energy from a system. Motors, in general, are used to add mechanical energy to a system. A ski-lift motor, for example, increases the gravitational potential energy of the skiers. More generally, if several external forces (A, B, C,...), as well as friction, act on a system, then the total work done by all of these forces produces the change in mechanical energy. Em2 Em1 Em1 W (WA WB WC... Wf) This is simply another version of the work\u2013energy theorem. Comparison of Energy-Position Graphs for Isolated and Non-isolated Systems Figure 6.37 shows the energy-position graphs for a block sliding down an inclined plane in a non-isolated system. In an energy-position graph, the mechanical energy Em is the sum of the potential energy Ep and kinetic energy Ek. So, the height of the mechanical energy line above the axis is the sum of the heights of the potential and kinetic energy lines. The purple line is the sum of the values of the red line and the blue line. Energy vs. Position for a Non-isolated System ) Position d (m) Em Ep Ek Em Ek Ep Figure 6.37 Friction acts on a non-isolated system to remove mechanical energy. By rearranging the equation for the definition of work, E Fd E d F it can be easily seen that force is equal to the slope of an energy-position Nm m graph. The units of the slope are, or units of force. In particular: \u2022 The component of the force of gravity parallel to the motion can be determined by calculating the slope of the gravitational potential energy-position graph. \u2022 The net force can be determined by calculating the slope of the kinetic energy-position graph. Chapter 6 In an isolated system, energy is transferred from one object to another whenever work is done. 321 06-PearsonPhys20-Chap06 7/24/08 12:56 PM Page 322 e MATH To determine the forces along an incline in an isolated system by using an energy-position graph, visit www.pearsoned.ca/physicssource. Project LINK How will the design of your persuader apparatus allow for the energy changes in a system during a collision? For example, in the isolated system in Figure 6.30 on page 312, the slope of the potential energy curve gives the component of gravity parallel to the inclined plane. The slope of the kinetic energy curve gives the net force. The slope of the mechanical energy curve is zero indicating that no outside forces act on this system. In the non-isolated system shown in Figure 6.37, Em is not constant so friction is present. The slope of the total energy curve gives the force of friction. As an example of a non-isolated system, imagine a cart accelerating down an inclined plane. The force of friction removes energy from the system, but it is not sufficient to stop the cart from speeding up. The magnitude of the change in kinetic energy is less than the magnitude of the change in gravitational potential energy. In this case, the mechanical energy decreases by the amount of energy that friction removes from the system. The graph would be similar to Figure 6.37. 6-3 Design a Lab 6-3 Design a Lab The Energy Involved in a Collision The Question What happens to the energy of the system", " when two carts collide? Design and Conduct Your Investigation Design an experiment to investigate the energy of a system in which two carts collide. In one case, compare the energy before and after the collision for simulated \u201celastic\u201d collisions in which the carts interact via a spring bumper. In a second case, compare the energy before and after a collision when the carts stick together in what is called an \u201cinelastic\u201d collision. You will need to develop a list of materials and a detailed procedure. Use the work-energy theorem to explain your results and form conclusions. e LAB If probeware is available, perform 6-3 Design a Lab using a motion sensor. For a probeware activity, go to www.pearsoned.ca/school/ physicssource. Concept Check A block slides down an inclined plane, radiating energy out of the system as heat due to friction. Yet when you measure the mechanical energy at the bottom of the ramp, the total energy in the system is unchanged. Explain how this might occur. 322 Unit III Circular Motion, Work, and Energy 06-PearsonPhys20-Chap06 7/24/08 12:56 PM Page 323 6.3 Check and Reflect 6.3 Check and Reflect Knowledge 1. What is meant by an isolated system? 2. If energy is conserved in a system, how can work be done in the system? 3. Describe the changes in the forms of energy as an acrobat bounces on a trampoline so that she goes higher after each bounce. 4. Can a system be classified as isolated if a non-conservative force acts on it? Explain. 5. A golfer drives a golf ball from the top of a cliff. The ball\u2019s initial velocity is at an angle above the horizontal. If there were no air friction, describe the energy transformations from the time the golfer starts her swing until the golf ball lands on the ground at a distance from the bottom of the cliff. Include the energy transformation at the point of impact. 9. The figure below shows the energy-position graphs for two different systems. For each graph, describe what is happening to the object(s) in the system in terms of their energies. Describe for each the nature of the forces acting on the object(s). Energy vs. Position ) Position d (m) Energy vs. Position Position d (m) Em Ep Ek Em Ep Ek Em Ek Ep 6. The pendulum of a clock is given a tiny Extensions push at the beginning of each swing. Why? 10. A 3.60-m-long pendulum with a 1.25-kg Applications 7. Two masses are suspended by a light string over a frictionless pulley. Mass A is 2.40 kg, mass B is 1.50 kg. Can this be considered an isolated system? Explain. On release, mass A falls to the tabletop, 1.40 m below. What is the kinetic energy of this system the instant before mass A hits the tabletop? bob is pulled sideways until it is displaced 1.80 m horizontally from its rest position. (a) Use the Pythagorean theorem to calculate the bob\u2019s gain in height. If the bob is released, calculate the speed of the bob when it (b) passes through its rest position (c) is 0.250 m above its rest position 8. Draw graphs showing the gravitational potential, kinetic, and mechanical energy of the system in question 7 against the change in position of mass A if there is (a) no friction, (b) a force of friction, but mass A still accelerates. A 1.40 m 3.60 m m 1.25 kg h 1.80 m B e TEST Diagram for questions 7 and 8. To check your understanding of mechanical energy in isolated and non-isolated systems, follow the eTest links at www.pearsoned.ca/school/physicssource. Chapter 6 In an isolated system, energy is transferred from one object to another whenever work is done. 323 06-PearsonPhys20-Chap06 7/24/08 12:56 PM Page 324 info BIT In metric terms, 1.00 hp 746 W or about 0.75 kW. 6.4 Work and Power Power, Work, and Time Toward the end of the 18th century, horses were the main source of energy used to drive the pumps that removed water from mines. Thus, when James Watt (1736\u20131819) wanted to know how his newly improved steam engine compared with existing methods of pumping water out of mines, he compared its effectiveness to that of horses. Today, even though it is a rather awkward unit, we still use his concept of horsepower (hp) to identify the power output of motors, especially in the automotive industry. Figure 6.38 This drag racer\u2019s 7000 hp engine burns a special fuel mixture called nitromethane. Each of its eight cylinders generates approximately three times", " the power of a normal car engine. The distortion of the tires, as seen above, is evidence of the magnitude of the forces exerted during acceleration. The high-performance race car in Figure 6.38 accelerates to speeds over 530 km/h in about 4.4 s. A family sedan with a 250-hp engine can accelerate to 100 km/h, from rest, in about 8.0 s. Aside from acceleration, what aspect of a car\u2019s performance is affected by the horsepower rating of its motor? On the highway, cars with 100-hp motors and cars with 300-hp motors both easily travel at the speed limit. What factors decide how much power is required to move a car along the highway? power: the rate of doing work In physics, power (P) is defined as the rate of doing work. Thus, the equation for power is P W t or P E t The unit of power, the watt (W), is named in recognition of James Watt\u2019s contributions to physics. Using the equation for power we see that a power output of one watt results when one joule of work is done per second Efficiency efficiency: ratio of the energy output to the energy input of any system Efficiency may be defined in terms of energy or in terms of power. In both cases, the ratio of the output (useful work) to the input (energy expended) defines the efficiency of the system. Thus, efficiency can be calculated as either, Efficiency Energy output (Eout) Energy input (Ein), or Power output (Pout) Power input (Pin) 324 Unit III Circular Motion, Work, and Energy 06-PearsonPhys20-Chap06 7/24/08 12:56 PM Page 325 Concept Check In terms of kg, m, and s, what is the unit for power? Example 6.11 An elevator and its occupants have a mass of 1300 kg. The elevator motor lifts the elevator to the 12th floor, a distance of 40.0 m, in 75.0 s. (a) What is the power output of the elevator? (b) What is the efficiency of the system if the motor must generate 9.40 kW of power to do the specified work? Given m 1.300 103 kg g 9.81 m/s2 h 40.0 m t 75.0 s Pin 9.40 103 W up down 40.0 m Required (a) power output of the elevator (b) efficiency of the system 1300 kg Figure 6.39 Analysis and Solution (a) The work done by the elevator is equal to its gain in gravitational potential energy. The power output of the elevator is the change in potential energy divided by the time40.0 m) (1.300 103 kg)9.81 s2 75.0 s 6.802 103 J s 6.80 103 W Practice Problems 1. The engine of a crane lifts a mass of 1.50 t to a height of 65.0 m in 3.50 min. What is the power output of the crane? Convert the SI unit answer to hp. 2. If a motor is rated at 5.60 kW, how much work can it do in 20.0 min? 3. A tractor, pulling a plough, exerts a pulling force of 7.50 103 N over a distance of 3.20 km. If the tractor\u2019s power output is 25.0 kW, how long does it take to do the work? Answers 1. 4.55 kW (6.11 hp) 2. 6.72 106 J 3. 960 s (16.0 min) (b) Efficiency is the ratio of the power output to the power input. P t o u Efficiency.724 Paraphrase and Verify (a) The power output of the elevator is 6.80 103 W. The answer has the right order of magnitude for the given data. The power output is equivalent to about sixty-eight 100-W light bulbs. (b) The efficiency of the system is 0.724 (72.4%). Chapter 6 In an isolated system, energy is transferred from one object to another whenever work is done. 325 06-PearsonPhys20-Chap06 7/24/08 12:56 PM Page 326 6-4 Inquiry Lab 6-4 Inquiry Lab Measuring the Power Output of a Motor The Question How much power can a small electric motor generate? The Problem The problem in the lab is to measure the power output of a motor by timing how long it takes for the motor to do a fixed amount of work. Variables The variables for measuring power are the work done against gravity (EP) and the time (t) it takes to do the work. Calculating EP requires mass (m), gravitational acceleration (g), and change in height (h). Materials small dc electric motor alligator clip leads iron stand 1-kg mass test-tube clamps low-voltage power supply dowel (about 3 cm", " long and 1 cm in diameter) thread (about 2.5 m long) tape paper clip washers scale (sensitive to at least 0.1 g) stopwatch metre-stick Procedure Required Skills Initiating and Planning Performing and Recording Analyzing and Interpreting Communication and Teamwork CAUTION: Close the test-tube clamp just tight enough to hold the motor in place. If you tighten it too much, it could warp the body of the motor. cardboard disks electric motor test-tube clamp mounted on an iron stand dowel light string upper mark washers paper clip lower mark leads to variable voltage power supply Figure 6.40 CAUTION: Check with your instructor to be sure the connections are correct before you plug in the power supply. If the motor is incorrectly connected, it could be damaged when the current is turned on. 4 Connect the power supply to the electric motor. Once your instructor has approved your connection, disconnect the lead connected to the red post of the power supply and turn on the power supply. 1 Use the balance to determine the mass of the paper 5 Place five washers on the string, as shown in clip. Record your measurement. 2 Place 10 washers on the balance scale and determine their mass. Calculate the average mass of the washers. Record your measurement. 3 Assemble the apparatus as shown in Figure 6.40. Set up a measuring scale behind the string holding the washers. The distance between the upper and lower timing marks on the scale may be adjusted if your apparatus permits, but should be 1.5 m or greater. Figure 6.40. Complete the circuit by holding the insulated alligator clip lead on the red post of the power supply and observe the speed with which the motor lifts the washers. Adjust the number of washers until the motor moves the load upward at a uniform speed. (If the speed is too great, it will be difficult to time the motion of the washers. If the speed is too slow, then the motor may not run smoothly.) 326 Unit III Circular Motion, Work, and Energy 06-PearsonPhys20-Chap06 7/24/08 12:56 PM Page 327 6 Pull the thread to unwind thread from the dowel until 2. Make a graph of the power output versus the mass the washers rest on the floor. Start the motor. Measure the time the washers take to travel between the lower and upper timing marks. 7 Record your measurements in a table such as Table 6.5. 8 Vary the number of washers on the paper clip and repeat the trial. Do trials with at least three different masses. 9 Calculate the work that the motor did in lifting the washers the measured distance. 10 Calculate the power output for each trial. Analysis 1. Does the power output of the motor vary with the force it is exerting? being lifted. 3. For what mass does the motor produce the most power? 4. What is the advantage of lifting the weights over a long distance? 5. Suggest reasons why the motor might generate more power when different masses are used. 6. Does the motor feel warm after it has done some work? What does that tell you about this system? 7. Would it make sense to use a very large motor to lift very tiny masses? Explain. 8. In terms of car engines, what are the implications for engine size? Table 6.5 Power Output of a Motor Trial Number Number of Washers Mass m (kg) Time t (s) Change in Potential Energy Ep mgh (J) Power P Ep/t (W) Power and Speed When a motor, such as the electric motor in 6-4 Inquiry Lab, applies a constant force to move an object at a constant speed, the power output of the motor can be shown to be the product of the force and the speed. When the force is constant, the work done can be found by the equation W Fd Inserting the equation for work into the equation for power gives: d F P t d F t But the expression d/t is just average speed vave; therefore, P (F)(vave) e WEB Why is it that if you double the speed of a car, the rate at which it consumes fuel more than doubles? Is lowering the speed limit the most effective way to conserve energy or would design changes (e.g., hybrid fuel systems or fuel cells) be more effective? With a group of classmates, investigate how best to improve the energy efficiency of automobiles. Use your library and the Internet. Present the results of your investigation in a report using presentation software such as PowerPoint \u2122. Begin your search at www.pearsoned.ca/school/ physicssource. Chapter 6 In an isolated system, energy is transferred from one object to another whenever work is done. 327 06-PearsonPhys20-Chap06 7/25/08 8:22 AM Page 328 Example 6.12 A car, of mass 2000 kg, is travelling up a hill at", " a constant speed of 90.0 km/h (25.0 m/s). The force of air resistance, which opposes this motion, is 450 N. The slope of the hill is 6.0 (Figure 6.41). (a) Draw a free-body diagram to show the external forces acting on the car as it moves up the hill. (b) Determine the forward force needed to maintain the car\u2019s speed. (c) Assuming that all the power output of the car engine goes into maintaining the car\u2019s forward motion, calculate the power output of the engine. Given m 2.000 103 kg F 4.50 102 N [downhill] air g 9.81 m/s2 v 25.0 m/s 6. Fair 6.0\u00ba Figure 6.41 Required (a) a free-body diagram for the car (b) forward force (Ff ) (c) power (p) Analysis and Solution (a) Figure 6.42(a) shows the free-body diagram. (b) Since the car is not FN v Ff Fg FN Ff Fair Fg accelerating, the net force on the car must be zero, 0, both parallel and Fnet perpendicular to the incline of the hill. The forward force (Ff) must be equal to the sum of the magnitudes of the force of air resistance (Fair) and the component of the gravitational force that acts parallel to the incline (Fg). Figure 6.42(a) Fg Fg In the parallel direction g F F net F F Fg Fair Fnet Ff air f Now, Fg mg sin (2.000 103 kg)9.81 2.051 103 N Forces Ff Fair Fg m s2 (sin 6.0) Figure 6.42(b) Therefore, 0 Ff Ff (2.051 103 N) (4.50 102 N) 2.051 103 N 4.50 102 N 2.50 103 N Practice Problems 1. What is the power output of an electric motor that lifts an elevator with a mass of 1500 kg at a speed of 0.750 m/s? 2. An engine\u2019s power is rated at 150 kW. Assume there is no loss of force due to air resistance. What is the greatest average speed at which this engine could lift a mass of 2.00 t? 3. A 1250-kg race car accelerates uniformly from rest to 30.0 m/s in 4.00 s on a horizontal surface with no friction. What must be the average power output of its motor? 4. Each car in a freight train experiences a drag force of 6.00 102 N due to air resistance. (a) If the engine of the train is to pull a train of 75 cars at a constant speed of 72.0 km/h, what power is required to move the cars? (b) If the engine operates at 15.0% efficiency, what must be the power generated by the engine to move these cars? Answers: 1. 11.0 kW 2. 7.65 m/s 3. 141 kW 4. (a) 900 kW (b) 6.00 103 kW 328 Unit III Circular Motion, Work, and Energy 06-PearsonPhys20-Chap06 7/24/08 12:56 PM Page 329 (c) Calculate the power output of the engine using the forward force and the speed with which the car moves along the ramp. P Ffvave (2.50 103 N)(25.0 m/s) 6.25 104 W 62.5 kW Paraphrase (a) The free-body diagram shows the external forces acting on the car. (b) If the car moves at a constant speed, then the forward force must be a force of 2.50 103 N. (c) The power output of the car is 62.5 kW. M I N D S O N Power and Dance At the moment of takeoff, Cossack dancers must generate considerable power to perform their spectacular leaps (Figure 6.43). Discuss techniques you could use to measure the power the dancers must generate to make such a jump. University Faculties of Kinesiology study this and other aspects of how humans move. 1. What factors involved in the jump will you need to determine? 2. What equipment might you require to measure those factors? 3. How would you measure the dancer\u2019s maximum power output compared with the power he can generate over a sustained period of time? Figure 6.43 e WEB To learn more about power generated in human activities, follow the links at www.pearsoned.ca/school/ physicssource. THEN, NOW, AND FUTURE Fuel for the Future? While the automobile in Figure 6.44 may look like a normal car, nothing could be further from the truth. When this vehicle travels along one of Vancouver", "\u2019s streets, its motor is barely audible. Perhaps even more surprising is the fact that the exhaust this car produces is pure water. While the motor that drives the car is actually an electric motor, it is the source of the electricity that is getting all the attention. The \u201cbattery\u201d in this car is called a fuel cell. At present, fuel cells are not an economically viable replacement for the internal combustion engine although successful trials of fuel-cell buses have been made in several cities around the world. For further information, go to the Internet. Start your research at www.pearsoned.ca/school/physicssource. Questions 1. What are the advantages and disadvantages of a fuel-celldriven motor over an internal combustion engine? 2. Which of the advantages and disadvantages identified above can be further improved on or overcome by scientific and technological research? Explain. 3. What are the limitations of science and technology to finding answers to the problems associated with energy production and use? Figure 6.44 This car\u2019s exhaust is pure water. The impact of fossil fuel (oil and coal)-burning systems on the environment has made the search for alternative energy sources much more attractive. This search is further enhanced by the realization that the supply of fossil fuels is finite. Even though Canada has, at present, an abundant supply of fossil fuels, it is still a world leader in fuel- cell research. Chapter 6 In an isolated system, energy is transferred from one object to another whenever work is done. 329 06-PearsonPhys20-Chap06 7/24/08 12:56 PM Page 330 6-5 Problem-Solving Lab 6-5 Problem-Solving Lab Power and Gears Recognize a Need Modern bicycles have many gears to enable riders to make best use of their efforts. In the automotive industry, manufacturers use a device called a dynamometer (Prony brake) to measure the power output of the motors they install in their vehicles. A Prony brake for bicycles would be a useful thing. The Problem How does the gear used by a cyclist affect the power output at the drive wheel of the bicycle? In which gear do cyclists generate the greatest power? Criteria for Success A successful experiment will determine if there is a relationship between the power at the drive wheel of the bicycle and the gear in which the bicycle is being ridden. Brainstorm Ideas Investigate the design of a Prony brake, then brainstorm how that design might be adapted to measure the power output of a bicycle. Remember, for your results to be useful 6.4 Check and Reflect 6.4 Check and Reflect Knowledge 1. What is the relationship between the amount of work that is done and the power output of the machine that does the work? 2. A farmer says, \u201cMy tractor with its 60-hp engine easily pulls a plough while my car with a 280-hp engine cannot even budge it.\u201d How can you explain this fact? 3. What is the relationship between the speed of an object and the power required to move it? Applications 4. You lift a 25.0-kg mass to your waist (0.800 m) in 1.20 s. What is your power output? 5. An airplane\u2019s engine exerts a thrust of 1.20 104 N to maintain a speed of 450 km/h. What power is the engine generating? 330 Unit III Circular Motion, Work, and Energy Required Skills Initiating and Planning Performing and Recording Analyzing and Interpreting Communication and Teamwork the design must allow a rider to \u201cride\u201d the bicycle in a normal manner. If a computer and probeware are available, consider using probeware in your experimental design, to measure the speed of the Prony brake. e WEB Use the Internet to investigate Prony brake design. Begin your search at www.pearsoned.ca/school/physicssource. Build a Prototype Build a Prony brake that can measure the power output of a student riding a bicycle. Test and Evaluate Make measurements of the power output of a student riding a bicycle using various gear settings. Communicate Prepare a report of your research using a computer spreadsheet program to organize your data and to generate a graph of the power output versus the gear level. Print your graph, in colour if possible, as part of your report. 6. An electric motor has a power rating of 1.50 kW. If it operates at 75% efficiency, what work can it do in an hour? 7. A motor of a car must generate 9.50 kW to move the car at a constant speed of 25.0 m/s. What is the force of friction on the car? Extension 8. A cannon fires a ball with a muzzle velocity of 240 m/s. The cannon ball has a mass of 3.60 kg. The barrel of the cannon is 1.20 m long and exerts a force of friction on the cannon ball of 650 N. What", " is the average power provided to fire the cannon ball? e TEST To check your understanding of power, work, and efficiency, follow the eTest links at www.pearsoned.ca/school/physicssource. 06-PearsonPhys20-Chap06 7/24/08 12:56 PM Page 331 CHAPTER 6 SUMMARY Key Terms energy work gravitational potential energy Key Equations W (F cos )d Ek 1 mv2 2 reference point elastic potential energy kinetic energy mechanics mechanical energy work-energy theorem isolated system non-isolated system conservation of energy power efficiency Ep mgh EP mgh W Ek Ep Em Ek Ep Ep 1 kx2 2 P W t E t Conceptual Overview The concept map below summarizes many of the concepts and equations in this chapter. Copy and complete the map to produce a full summary of the chapter. Mechanics involves energy changes of position of motion rate of change calculated using can be either W Fd Ep mgh or can be either or non-isolated in which energy is conserved or Em2 Em1 W or or Ek Ep Chapter 6 In an isolated system, energy is transferred from one object to another whenever work is done. 331 06-PearsonPhys20-Chap06 7/24/08 12:56 PM Page 332 CHAPTER 6 REVIEW Knowledge 1. (6.1) (a) When a force acts on an object to do work, why do you need to know the angle between the direction of the force and the direction of the displacement? (b) If you know how the magnitude of a force changes while it acts over a displacement, how can you find the amount of work it does? (c) Describe the nature of the energy transfers for the work done on a bungee jumper from the time he leaps off the platform until his velocity is zero at the lowest point of his jump. (d) Two students calculate the gravitational potential energy of a mass resting on a shelf. One student calculates that it has 12.0 J of energy while the other calculates the gravitational potential energy to be 35.0 J. Is it possible that they are both right? Explain. (e) Two masses, A and B, are at rest on a horizontal frictionless surface. Mass A is twice as great as mass B. The same force acts on these masses over the same displacement. Which mass will have the greater (i) speed, and (ii) kinetic energy at the end of the displacement? Explain. (f) Many fitness facilities have treadmills as exercise machines. Is running on a treadmill work since the runner is not really moving? (g) Explain why it takes less force to push a cart up an inclined plane onto a platform than it does to lift the cart straight up from the floor. Assume you are able to move the cart by either method. Does it also take less work to lift it or to roll it up the plane? (h) An object sits on a tabletop. In terms of describing the object\u2019s gravitational potential energy, which reference point is better: the ground outside the room, the floor of the room, or the tabletop? Explain. 2. (6.2) (a) What is the effect of the work done by a net force? (b) If a force acts upward on an object, does all the work done by this force become potential energy? 3. (6.3) (a) Explain why the force of gravity but not the force of friction is called a conservative force. (b) An ideal spring is one where no energy is lost to internal friction. Is the force exerted by the spring considered a conservative force? Explain. Is a mass that is oscillating up and down on the end of an ideal spring a good approximation of an isolated system? (c) Since no system on Earth is truly an isolated system, why is it advantageous to assume that a system is isolated? (d) A truly isolated system does not exist on Earth. How does that affect the fundamental principle of conservation of energy? (e) If a system is not isolated, how can one calculate the change in mechanical energy in the system? (f) The mechanical energy of a system is measured at two different times and is the same each time. Is this an isolated system? Explain. 4. (6.4) (a) An elevator takes 2.50 min to travel from the ground floor to the 10th floor of an apartment block. The tenants want the landlord to increase the speed of the elevator but the landlord argues that speeding up the elevator means that it will need to work harder and that would take more energy. Is he correct? Explain. (b) The transmission of an automobile allows the work done in the engine to be transmitted to the wheels. For a given power output by the engine, the wheels are not rotated as fast by a low gear as they are by a high gear. What advantage does having gears give the driver of a car? Applications 5.", " What is the change in kinetic energy if a net force of 3.80 103 N [0] acts on a mass while it undergoes a displacement of 95.0 m [335]? 6. For gravitational potential energy, when the height doubles so does the potential energy. However, for elastic potential energy, if the stretch of the spring doubles, the energy does not. (a) How does the stored elastic potential energy (c) What forms of energy are considered to be change if the stretch doubles? part of mechanical energy? (d) Can two people calculate the mechanical energy of an object and get two different correct answers? Explain. (b) Explain in terms of force-position graphs for gravitational and elastic potential energies why this happens. 332 Unit III Circular Motion, Work, and Energy 06-PearsonPhys20-Chap06 7/24/08 12:56 PM Page 333 7. The figure below shows the graph of the force as a function of displacement for an elastic spring stretched horizontally 25.0 cm from its equilibrium position. A mass of 0.400 kg is attached to the spring and released. If the mass is sliding on a horizontal frictionless surface, what is the speed of the mass when the spring has contracted to (a) 10.0 cm from its equilibrium position, and (b) its equilibrium position? Force vs. Displacement 50 40 30 20 10 ) ( N F e c r o F 0 0.05 0.10 0.15 0.20 0.25 Displacement d (m) Graph for question 7 8. A bungee jumper with a mass of 65.0 kg leaps from a bridge. At the lowest point of the jump he is 30.0 m below the point from which the jump began. If, at equilibrium, the bungee cord is 15.0 m long, what is the elastic constant for the cord? HINT: Assume an isolated system. At the lowest point the bungee cord must convert all of the jumper\u2019s lost gravitational potential energy into elastic potential energy in the cord. 9. A motorcycle stuntman wants to jump over a line of city buses. He builds a takeoff ramp that has a slope of 20, with its end 3.20 m above the ground. The combined mass of the motorcycle and rider is 185 kg. To clear the buses, the cyclist needs to be travelling 144 km/h when he leaves the end of the ramp. How high above the ground is the motorcycle at its highest point? 10. A skydiver reaches a maximum speed (or terminal velocity) of 36.0 m/s due to the force of air resistance. If the diver has a mass of 65.0 kg, what is the power output of the air resistance acting on him? Extensions 11. Even when making short flights, jets climb to altitudes of about 10 000 m. Gaining altitude requires a great expenditure of fuel. Prepare a short research report to explain the advantages and disadvantages of travelling at these altitudes. Consolidate Your Understanding You have been hired to tutor a student on the topics in this chapter. Describe how you would answer the questions below. In each instance, include an example. 1. What is the difference between work and energy? 2. When an object moves up a hill, how does the length of the hill affect the increase in the gravitational potential energy? 3. If a cart, at rest, is allowed to coast from the top of a hill to the bottom, is the kinetic energy at the bottom of the hill always equal to the loss in gravitational potential energy? 4. A given force is to act on a block to accelerate it from rest as it slides up the length of an inclined plane. Using accurate spring balances and rulers, how could you gather data to enable you to calculate, with reasonable accuracy, the kinetic energy of the block when it reaches the top of the incline? You may not use timing devices. 5. What is the difference between an isolated and a non-isolated system in terms of work and energy? 6. What factor of a car\u2019s motion is most directly affected by the power of its engine? Think About It Review the answers you gave to the questions in Think About It on page 291. How would you change these answers? e TEST To check your understanding of energy, work, and power, follow the eTest links at www.pearsoned.ca/school/physicssource. Chapter 6 In an isolated system, energy is transferred from one object to another whenever work is done. 333 06-PearsonPhys20-Chap06 7/24/08 12:56 PM Page 334 UNIT III PROJECT Building a Persuader Apparatus Scenario It is 1965. Seatbelts are oddities used only by airline passengers at takeoff and landings. The term \u201cairbag\u201d hasn\u2019t even been invented. Speed limits and traffic deaths are on the rise. Imagine that you are part of a team", " of engineers who design and build automobiles. Your company challenges you to design and build a model that can be used to convince its shareholders and the public that it is possible to build much safer automobiles. The automobile company has challenged your design team to produce safety features for its vehicles that will allow its passengers to survive crashes under severe conditions. Your team must determine how best to protect the passenger while, at the same time, keeping the size and mass of the car itself to reasonable proportions. Your presentation to the company will be used to persuade them of the benefits of your design. Finally, your report should persuade the public of the advantages of using the safety equipment you recommend for automobiles. Planning Your team should consist of three to five members. Your first task is to formulate your research question. This done, you will need to identify the assumptions about the nature of the collisions in which your vehicle may be involved. Since not all collisions may be head-on, your passenger (a fresh raw egg) should survive unscathed from a wide variety of crash scenarios. While all team members should be active participants in all aspects of the project, you should identify and draw on any special talents of your team. Create a team structure that assigns responsibilities such as team manager, data analyst, and record keeper. Begin by brainstorming possible design features and research strategies. Where might you find information on existing safety features? Which features are the most effective? How will you compare results for the various types of crashes? You may wish to draw on information from all topics in this course to improve the safety features of your vehicle. Create timelines for all phases of the project. Create a report that incorporates written, graphic, and photographic analysis of your project. Materials \u2022 material for construction of the vehicle \u2022 mass-measuring equipment \u2022 equipment to provide known energy crash conditions \u2022 egg passengers \u2022 digital camera \u2022 computer Assessing Results Assess the success of your project based on a rubric* designed in class that considers: research strategies thoroughness of the experimental design effectiveness of the experimental technique effectiveness of the team\u2019s public presentation Procedure 1 Research existing safety features used in automotive production. Identify which features are the most effective in reducing crash injuries. Keep a record of the sources of your information. If information comes from the Internet, be sure to identify the site, and who sponsors the information. Be alert to Internet sites that may contain biased information. Identify the most common types of injuries resulting from automobile accidents. 2 Design the persuader vehicle and gather the materials required for its construction. 3 Design the experiment that your team will use to test the effectiveness of your vehicle\u2019s design. Make sure that your experimental design makes it possible to accurately compare the energy of the vehicle when it is involved in different crash scenarios. CAUTION: Your vehicle will need to gain considerable energy, which may or may not result in unexpected behaviour when it crashes. Take proper precautions to ensure that the vehicle path is clear during trials. 4 Test your vehicle\u2019s safety features under a variety of crash conditions. Assess how effective your system is when it is involved in crashes happening from different directions. 5 Prepare a report using an audiovisual format that will dramatically emphasize for your audience the value of the safety features that you recommend. Thinking Further Write a short appendix to your report (two or three paragraphs) to identify possible directions that future research might take to make automobile travel even safer. Suggest steps that government, technology, and industry should take in making automobile travel safer. *Note: Your instructor will assess the project using a similar assessment rubric. 334 Unit III Circular Motion, Work, and Energy 06-PearsonPhys20-Chap06 7/24/08 12:56 PM Page 335 UNIT III SUMMARY Unit Concepts and Skills: Quick Reference Concepts Chapter 5 Summary Newton\u2019s laws can explain circular motion. Resources and Skill Building Speed and velocity Centripetal acceleration and force Velocity and circular motion Centripetal acceleration Centripetal force \u2014 a horizontal system in circular motion Centripetal force \u2014 a vertical system in circular motion Centripetal force \u2014 acceleration and frequency 5.1 Defining Circular Motion The velocity of an object moving with circular motion is tangent to the circle and 90 to the radial line. Figures 5.3\u20135.6, 5.8\u20135.11, 5.13; QuickLab 5-1; Inquiry Lab 5-2 Centripetal acceleration and centripetal force are both directed toward the centre of the circle. Figures 5.8\u20135.11; Table 5.2; 5.2 Circular Motion and Newton\u2019s Laws The velocity of circular motion can be determined by dividing the circumference by the period. The centripetal acceleration of an object is determined by the velocity squared divided by the radius. Newton\u2019s second law states F ma and can be applied to centripetal acceleration. A car making a turn experiences a centripetal acceleration and force that is created by", " the force of friction between the tires and the road. The minimum speed necessary to move an object through a vertical loop equates centripetal force with gravitational force. Centripetal force can be equated to the gravitational force for planetary objects. eSIM Example 5.2; Minds On Figures 5.18\u20135.20; Example 5.3 Inquiry Lab 5-3; eTECH; Example 5.4; Figures 5.24, 5.25 Figures 5.27\u20135.30; eTECH; Example 5.5; eSIM Centripetal acceleration and force can be determined using period and frequency instead of speed. Figures 5.32\u20135.34; Example 5.7 Kepler\u2019s laws 5.3 Satellites and Celestial Bodies in Circular Motion Kepler formulated three laws that explained the motion of planets in the solar system. Figures 5.36\u20135.38; Tables 5.4\u20135.6; Examples 5.8, 5.9; eSIM; Design a Lab 5-4 Newton\u2019s version of Kepler\u2019s third law Fc for Earth\u2013Moon system. He also Newton recognized the reason that Kepler\u2019s laws were correct: Fg found a way to determine the mass of an object from the period of a celestial body orbiting it. Figures 5.41, 5.42; Examples 5.10, 5.11; eTECH Orbital perturbations The discovery of Uranus and Pluto occurred because of the apparent disturbances in the orbit of the planets. Extrasolar planets have been discovered by examining perturbations in stars\u2019 movements. Then, Now, and Future; Figures 5.46, 5.47 Artificial satellites Humans have placed a variety of artificial satellites into orbit to meet society\u2019s needs. 5-5 Decision-Making Analysis; Figures 5.48\u20135.51 Chapter 6 In an isolated system, energy is transferred from one object to another whenever work is done. Work Potential energy Kinetic energy 6.1 Work and Energy Work is the transfer of energy that occurs when a force acts over a displacement. It is a scalar quantity measured in joules. (1 J 1 N\u00b7 m) Potential energy is the energy a body has because of its position or configuration. It is a scalar quantity measured in joules. Kinetic energy is the energy a body has because of its motion. It is a scalar quantity measured in joules. Example 6.1 QuickLab 6-1; Example 6.2; Example 6.3; Example 6.4 QuickLab 6-1; Example 6.5; Example 6.6 Work-energy theorem 6.2 Mechanical Energy Work done by a net force causes a change in kinetic energy. The work\u2013energy theorem states that the work done on a system is equal to the sum of the changes in the potential and kinetic energies. Mechanical energy Mechanical energy is the sum of the potential and kinetic energies. Example 6.7 Example 6.7 Example 6.8 Isolated systems 6.3 Mechanical Energy in Isolated and Non-isolated Systems The law of conservation of energy states that in an isolated system, the mechanical energy is constant. Example 6.9; Example 6.10 Conservation of energy A simple pendulum is a good approximation of an isolated system in which energy is conserved. Inquiry Lab 6-2 Conservative forces Non-isolated systems Power A conservative force does not affect the mechanical energy of a system. Example 6.10; Inquiry Lab 6-2 In non-isolated systems, the mechanical energy may change due to the action of nonconservative forces. Design a Lab 6-3 6.4 Work and Power Power is defined as the rate of doing work. Power is calculated by finding the ratio of the work done to the time required to do the work. It is measured in watts. (1 W 1 J/s) Power may be calculated by taking the product of the force doing the work and the average speed. Example 6.11; Inquiry Lab 6-4; Example 6.12 Problem-Solving Lab 6-5 Unit III Circular Motion, Work, and Energy 335 06-PearsonPhys20-Chap06 7/24/08 12:56 PM Page 336 UNIT III REVIEW Vocabulary 1. Using your own words, define these terms: artificial satellite axis of rotation axle centripetal acceleration centripetal force conservation of energy conservative force cycle eccentricity efficiency elastic potential energy ellipse energy frequency gravitational potential energy isolated system Kepler\u2019s constant Kepler\u2019s laws kinetic energy mean orbital radius mechanical energy non-isolated system orbital period orbital perturbations period potential energy power reference point revolution rpm satellite uniform circular motion work work-energy theorem Knowledge CHAPTER 5 2. An object is moving in a circular path with a uniform centripetal acceleration that doesn\u2019t change. What will happen to the velocity if the radius is reduced? 3. The", " centripetal acceleration of a car as it goes around a turn is inward, but the car will not skid in that direction if it is moving too quickly. Explain. 336 Unit III Circular Motion, Work, and Energy 4. A bucket of water is spun in a vertical circle on the end of a rope. (a) Explain what force or forces act as the centripetal force when the bucket is in the highest position. (b) In which position is the rope most likely to break? Why? 5. Explain why centripetal force is inward when the force acting on your hand as you spin an object in a circular path is outward. 6. Using what you have learned about the force of gravity and circular motion, provide a thorough explanation why the magnitude of centripetal force changes for planets orbiting the Sun. 7. A roller coaster goes around a vertical loop with just enough velocity to keep it on the track. (a) (b) In which position or positions is the force of gravity the centripetal force? Explain. In which position or positions is there a force exerted on the track by the roller coaster? Explain. (c) Using the equation Fg mg and equation 6 from Chapter 5 on page 256, show why mass does not affect the speed required for the roller coaster to successfully enter and exit the loop as shown in the diagram below. D C A B 8. What is the relationship between frequency and radius for a rotating disc? 9. The motor of a table saw is rated for its horsepower and rotational frequency. Explain why rotational frequency is used instead of rotational speed. 10. What physical quantities must be known for the mass of Earth to be determined? 11. Kepler showed that planets follow elliptical orbits. (a) Which planet has the least elliptical orbit? (b) Which planet\u2019s semi-major axis is the closest in length to its semi-minor axis? 06-PearsonPhys20-Chap06 7/24/08 12:56 PM Page 337 12. Three identical coins 24. A slingshot is used to propel a stone vertically C B A are placed on a rotating platter as shown. As the frequency of rotation increases, identify which coin will begin to slide off first. Explain your answer. 13. Briefly explain how Neptune was discovered. Use the terms orbital perturbation, force of gravity, and orbital velocity in your explanation. 14. Your friend argues that Neptune would not have been discovered as soon as it was if Neptune were a much smaller planet and Uranus much bigger. Is she right? Defend your answer. 15. What difficulties do astronomers face when searching for extrasolar planets that might have life as we know it? CHAPTER 6 16. Express a joule in terms of kilograms, metres, and seconds. 17. If work is a scalar quantity, why is it affected by the directions of the force and displacement? 18. What happens to an object\u2019s gravitational potential energy when it is in free fall? 19. Explain why doubling the speed of an object does not result in a doubling of its kinetic energy. 20. A large mass and a small mass with the same kinetic energy are sliding on a horizontal frictionless surface. If forces of equal magnitude act on each of these bodies to bring them to rest, which one will stop in the shorter distance? Explain. 21. Describe the energy changes of a roller coaster car from the time when it is just coming over the crest of one hill until it arrives at the crest of the next hill. 22. A cart is pulled up an inclined plane by a force that is just large enough to keep the cart moving without a change in its speed. Is this an isolated system? Explain why or why not. Is the force used to move the cart up the incline a conservative force? 23. A cart at the top of an inclined plane is allowed to roll down the plane. Under what conditions can this system be considered isolated? If the conditions that make this an isolated system do not exist, is the force that moves the cart down the plane still considered a conservative force? Explain. upward. Describe the energy changes that are involved from the time the stone is placed in the slingshot until the stone reaches its maximum height. 25. If a force that acts on an object results in a change in the object\u2019s kinetic energy, what can be said about the nature of this force? 26. How do you calculate work from a force- displacement graph? 27. According to the work-energy theorem, how much work is done on an isolated system? 28. Does power affect the amount of work you are able to do? Explain why or why not. Applications 29. Electrons in an electric (AC) circuit vibrate at 60 Hz. What is their period? 30. A cell phone is set to vibrate when it rings. It vibrates with a period of 0.0160 s.", " What is the frequency of the ring? 31. A toy top spins at 300.0 rpm. What is the frequency (in Hz) and the period of the top? 32. The Moon orbits Earth once every 27.3 days at a mean orbital radius of 3.844 105 km. What is its speed? 33. A child sits in a pretend airplane on a ride at an amusement park. The airplane is at the end of a long arm that moves in a circular path with a radius of 4.0 m at a speed of 1.57 m/s. What is the period of the ride? 34. A person sliding down a water slide at a speed of 5.56 m/s encounters a turn with a radius of 10.0 m. Determine the acceleration that he experiences in the turn. 35. A pilot of a jet airplane makes a sharp turn to create an acceleration of 4.00 times the acceleration of gravity. If the turn has a radius of 500.0 m, what is the speed of the plane? 36. A cork (m 2.88 g) is caught in a small whirlpool created in the basin of a sink. What is the centripetal force acting on the cork, if its speed is 0.314 m/s at a radius of 4.00 cm? 37. When braking to a stop, the maximum force that friction exerts on a 1250-kg auto is 3200 N. (a) If the original speed of the auto is 12.0 m/s, what is the minimum stopping distance? If the speed of the car were twice as great, how would that affect the minimum stopping distance? (b) Unit III Circular Motion, Work, and Energy 337 06-PearsonPhys20-Chap06 7/24/08 12:56 PM Page 338 38. A force of 250 N [up] is applied to a mass of 15.0 kg over a displacement of 9.60 m [up]. (a) How much work does the force do on the mass? (b) What is the change in gravitational (c) potential energy? If it is an isolated system, explain the difference between the answers for (a) and (b). 39. A car with a mass of 2.00 103 kg is travelling at a velocity of 15.0 m/s [0] on a horizontal stretch of highway. The driver presses on the accelerator so that the force propelling the car forward is increased to 3.30 103 N [0]. The force acts over a displacement of 55.0 m [0] during which force of friction on the car is 5.00 102 N in magnitude. (a) Draw a free-body diagram to analyze the forces acting on the car. What is the net force on the car? (b) What is the work done by net force over the displacement? (c) What is the final kinetic energy of the car? (d) What is the final speed of the car? 40. A block with a mass of 0.800 kg is initially at rest on a frictionless inclined plane. A force of 5.00 N, applied parallel to the inclined plane, moves the block a distance of 4.50 m up the plane. If, at the end of the effort, the block has a speed of 6.00 m/s up the incline, what is the change in height through which it moved? 41. A varying force acts on a 25.0-kg mass over a displacement as shown in the graph below. The mass has an initial velocity of 12.0 m/s [0]. Recall that the area of a force-displacement graph is equivalent to the work done by the force. (a) What is the ) ] \u02da 0 [ 60 45 N (a) What is the change in gravitational potential energy for block A? (b) What was the change in height through which block A moved? B m 0 5 1. 1.50 m A 43. For question 42, draw the graph that shows the potential, kinetic, and mechanical energies for the system as a function of the displacement assuming (a) there is no friction, (b) there is friction, but block A still accelerates up the hill. 44. A pendulum bob with a mass of 0.750 kg is initially at rest at its equilibrium position. You give the bob a push so that when it is at a height of 0.150 m above its equilibrium position it is moving at a speed of 2.00 m/s. (a) How much work did you do on the bob? If you pushed on the bob with a force of (b) 40.0 N parallel to the displacement, how far did you push it? 45. A billiard ball with a speed of 2.00 m/s strikes a second ball initially at rest. After the collision, the first ball is moving at a speed of 1", ".50 m/s. If this is an elastic collision (i.e., energy is conserved), what is the speed of the second ball after the collision? Assume that the balls have identical masses of 0.200 kg. 46. A cannon ball (m 3.00 kg) is fired at a velocity of 280 m/s at an angle of 20 above the horizontal. The cannon is on a cliff that is 450 m above the ocean. (a) What is the mechanical energy of the work that the force did on the mass? ( F e c r o F 30 15 (b) What is the final speed of the mass? 0 10 20 Displacement d (m [0\u02da]) 30 cannon ball relative to the base of the cliff? (b) What is the greatest height above the ocean 40 that the cannon ball reaches? (c) What is the speed of the cannon ball when it lands in the ocean? 42. In the following diagram, block A is at rest on a frictionless inclined plane. It is attached to block B by a light cord over a frictionless pulley. Block A has a mass of 4.50 kg and B has a mass of 5.50 kg. When they are released, block A moves up the incline so that after it has moved a distance of 1.50 m along the incline it has a speed of 3.00 m/s. 47. A Styrofoam\u2122 ball is dropped from a height of 5.00 m. The mass of the ball is 0.200 kg. When the ball hits the ground it has a speed of 3.00 m/s. (a) What change in mechanical energy does the ball undergo while it falls? (b) What is the average force that air friction exerted on the falling ball? 338 Unit III Circular Motion, Work, and Energy 06-PearsonPhys20-Chap06 7/24/08 12:56 PM Page 339 48. What is the average power output if an engine lifts a 250-kg mass a distance of 30.0 m in 20.0 s? 49. What is the effective power required to maintain a constant speed of 108 km/h if the force opposing the motion is 540 N in magnitude? 50. An airplane engine has an effective power output of 150 kW. What will be the speed of the plane if the drag (air friction opposing the motion of the plane) exerts a force of 2.50 103 N? Extensions 51. A 90.9-kg gymnast swings around a horizontal bar in a vertical circle. When he is directly over top of the bar, his arms experience a tug of 108.18 N. What is the speed of his body in this position? (Assume that the gymnast\u2019s mass is centred 1.20 m from the bar.) 52. Suppose a solar system, with three planets circling a star, exists in a galaxy far far away. One of the planets orbits the star with the period of 6.31 107 s and a radius of 3.00 1011 m. It has a moon that orbits it with a period of 1.73 106 s at a radius of 6.00 108 m. (a) What is the mass of the planet\u2019s star? (b) What is the mass of the planet? (c) What is the speed of the planet\u2019s moon? 53. A soil-moving machine called a bucket wheel loader has a large metallic wheel with a radius of 3.05 m that has many scoops attached to it. The scoops are designed to dig into the ground and lift soil out as the wheel turns around. If the wheel turns with a frequency of 0.270 Hz, will the soil fall out of a scoop when it gets to the top of the wheel? 54. Three blocks (A, B, and C), with masses 6.00 kg, 4.00 kg, and 2.00 kg, respectively, are initially at rest on a horizontal frictionless surface as shown in diagram (a) on the right. A force of 48.0 N [90] acts on block A over a displacement of 7.50 m [90]. Block A is connected to block B by a string that is 1.00 m long and block B is connected to block C by a string that is 1.50 m long. Initially, the three blocks are touching each other. As the blocks move and the strings become taut, they end up as shown in diagram (b). Is this an isolated or a non-isolated system? Explain. (a) What is the speed of the blocks after the force has acted for the full 7.50 m? (b) What is the speed of block A when the force has acted over a displacement of 2.00 m? Hint: Find the work done by the force. (a) 1.50 m string connecting B to C 1.00 m string", " connecting A to B force meter (b) Mass A Mass B Mass C 0 m 1. 0 force meter 0 m 1. 5 F F Skills Practice 55. Your cousin doesn\u2019t understand how a satellite can stay in orbit without falling toward Earth. Using the knowledge you have gained from this unit, provide a short explanation. 56. Figure 6.21 on page 303 shows the impact crater for a meteor that landed in Arizona. How widespread is the evidence of meteors striking Earth? Where is the impact crater closest to where you live? Do Internet research to identify locations of impact craters in Alberta and Canada. On a map of Alberta, show the location of meteor impacts. What clues on maps show meteor landings? For each crater, identify how much kinetic energy the meteor would have had when it struck Earth. 57. Explain how you would experiment to determine the quantity of external work done on a system for a cart accelerating down an inclined plane. Could you confirm this quantity by measuring forces? Self-assessment 58. Describe to a classmate one misconception you had about circular motion before studying this unit. Explain what you know about this concept now. 59. Describe to a classmate the relationship between the roles of science and technology in the development of new energy resources. 60. Can science provide solutions to all of the problems associated with the impact of energy consumption on the environment? Give reasons for your answer. e TEST To check your understanding of circular motion, work, and energy, follow the eTest links at www.pearsoned.ca/school/physicssource. Unit III Circular Motion, Work, and Energy 339 07-Phys20-Chap07.qxd 7/24/08 1:09 PM Page 340 U N I T IV Oscillatory Oscillatory Motion and Motion and Mechanical Mechanical Waves Waves An earthquake more than two thousand kilometres away sent this tsunami speeding across the Indian Ocean. Waves, a form of simple harmonic oscillations, can efficiently transport incredible amounts of energy over great distances. How does a wave move through its medium? How does understanding simple harmonic motion help us understand how waves transport energy? 340 Unit IV 07-Phys20-Chap07.qxd 7/24/08 1:09 PM Page 341 Unit at a Glance C H A P T E R 7 Oscillatory motion requires a set of conditions. 7.1 Period and Frequency 7.2 Simple Harmonic Motion 7.3 Position, Velocity, Acceleration, and Time Relationships 7.4 Applications of Simple Harmonic Motion C H A P T E R 8 Mechanical waves transmit energy in a variety of ways. 8.1 The Properties of Waves 8.2 Transverse and Longitudinal Waves 8.3 Superposition and Interference 8.4 The Doppler Effect Unit Themes and Emphases \u2022 Change, Energy, and Matter \u2022 Scientific Inquiry \u2022 Nature of Science Focussing Questions As you study this unit, consider these questions: \u2022 Where do we observe oscillatory motion? \u2022 How do mechanical waves transmit energy? \u2022 How can an understanding of the natural world improve how society, technology, and the environment interact? Unit Project \u2022 By the time you complete this unit, you will have the knowledge and skill to research earthquakes, the nature of earthquake shock waves, and the use of the Richter scale to rate earthquake intensity. On completion of your research, you will demonstrate the operation of a seismograph. e WEB To learn more about earthquakes and their environmental effects, follow the links at www.pearsoned.ca/school/physicssource. Unit IV Oscillatory Motion and Mechanical Waves 341 07-Phys20-Chap07.qxd 7/24/08 1:09 PM Page 342 Oscillatory motion requires a set of conditions. On October 15, 1997, NASA launched the Cassini-Huygens space probe toward Saturn \u2014 a distance of 1 500 000 000 km from Earth. The probe\u2019s flight path took it by Venus twice, then Earth, and finally past Jupiter on its way to Saturn. This route was planned so that, as the probe approached each planet, it would be accelerated by the planet\u2019s gravitational force. Each time, it picked up more speed, allowing it to get to Saturn more quickly (Figure 7.1). Recall from Chapter 4 that increasing the probe\u2019s speed this way is referred to as gravity assist. The entire journey of 3 500 000 000 km took seven years. For this incredible feat to succeed, scientists had to know where the planets would be seven years in the future. How could they do this? They relied on the fact that planets follow predictable paths around the Sun. Nature is full of examples of repetitive, predictable motion. Water waves, a plucked guitar string, the orbits of planets, and even a bumblebee flapping its wings are just a few. In this chapter, you will examine oscillatory motion. Oscillatory motion is a slightly different form of motion from the circular", " motion you studied in Chapter 5. You will better understand why bungee jumpers experience the greatest pull of the bungee cord when they are at the bottom of a fall and why trees sway in the wind. You may notice the physics of many objects that move with oscillatory motion and gain a new insight into the wonders of the natural world. C H A P T E R 7 Key Concepts In this chapter, you will learn about: \u25a0 oscillatory motion \u25a0 simple harmonic motion \u25a0 restoring force \u25a0 oscillating spring and pendulum \u25a0 mechanical resonance Learning Outcomes When you have completed this chapter, you will be able to: Knowledge \u25a0 describe oscillatory motion in terms of period and frequency \u25a0 define simple harmonic motion as being due to a restoring force that is directly proportional and opposite to the displacement of an object from an equilibrium position \u25a0 explain the quantitative relationships among displacement, acceleration, velocity, and time for simple harmonic motion \u25a0 define mechanical resonance Science, Technology, and Society \u25a0 explain that the goal of science is knowledge about the natural world Figure 7.1 The Cassini-Huygens probe successfully orbits Saturn after a seven-year journey through the solar system. 342 Unit IV 07-Phys20-Chap07.qxd 7/24/08 1:09 PM Page 343 7-1 QuickLab 7-1 QuickLab Oscillatory Motion of Toys Problem What is the time taken by one complete back-and-forth motion of a toy? Procedure 1 Fully wind the spring mechanism of one of the toys. 2 Release the winding knob and start your stopwatch. Materials several different wind-up toys, such as: \u2022 swimming frog \u2022 hopping toy \u2022 monkey with cymbals \u2022 walking toy yo-yo stopwatch (or wristwatch with a second hand) Figure 7.2 Think About It 3 Count the number of complete back-and-forth movements the toy makes in 10 s. These back-andforth movements are called oscillations. 4 Record the number of oscillations made by the toy. 5 Repeat steps 1 to 4 for each toy. For the yo-yo, first achieve a steady up-and-down rhythm. Then do steps 3 and 4, counting the up-and-down motions. This type of repetitive movement is also an oscillation. Questions 1. In what ways are the motions of all the toys similar? 2. Divide the number of oscillations of each toy by the time. Be sure to retain the units in your answer. What does this number represent? (Hint: Look at the units of the answer.) 3. Which toy had the most oscillations per second? Which had the least? 4. Divide the time (10 s) by the number of oscillations for each toy. Be sure to retain the units in your answer. How long is the interval of oscillation of each toy? 5. Which toy had the longest time for one oscillation? Which had the shortest time? 1. How are the oscillations per second and the time for one oscillation related? 2. What do you think influences the number of oscillations per second of the toys you tested? Discuss your answers in a small group and record them for later reference. As you complete each section of this chapter, review your answers to these questions. Note any changes to your ideas. Chapter 7 Oscillatory motion requires a set of conditions. 343 07-Phys20-Chap07.qxd 7/24/08 1:09 PM Page 344 7.1 Period and Frequency On a warm summer day in your backyard, you can probably hear bees buzzing around, even if they are a few metres away. That distinctive sound is caused by the very fast, repetitive up-and-down motion of the bees\u2019 wings (Figure 7.3). Take a closer look at the bumblebee. The motion of a bee\u2019s wings repeats at regular intervals. Imagine that you can examine the bee flying through the air. If you start your observation when its wings are at the highest point (Figure 7.4(a)), you see them move down to their lowest point (Figure 7.4(c)), then back up again. When the wings are in the same position as when they started (Figure 7.4(e)), one complete oscillation has occurred. An oscillation is a repetitive backand-forth motion. One complete oscillation is called a cycle. Figure 7.3 The wings of a bee in flight make a droning sound because of their motion. oscillatory motion: motion in which the period of each cycle is constant info BIT Earthquake waves can have periods of up to several hundred seconds. (a) (b) (c) (d) (e) Figure 7.4 The bee\u2019s wings make one full cycle from (a) to (e). The time for this motion is called the period. The time required for the wings to make one complete oscillation is the period (T). If the period of each cycle remains constant, then the wings are moving", " up and down with oscillatory motion. Recall from Chapter 5 that the number of oscillations per second is the frequency (f), measured in hertz (Hz). The equation that relates frequency and period is: f 1 T (1) Table 7.1 shows the period of a bee\u2019s wings as it hovers, along with other examples of periods. \u25bc Table 7.1 Periods of Common Items Object Bumblebee wings Hummingbird wings Medical ultrasound technology Middle C on a piano Electrical current in a house Period 0.00500 s 0.0128 s 1 106 to 5 108 s 0.0040 s 0.0167 s 344 Unit IV Oscillatory Motion and Mechanical Waves 07-Phys20-Chap07.qxd 7/24/08 1:09 PM Page 345 A piston in the engine of a car also undergoes oscillatory motion if it is moving up and down in equal intervals of time. The piston shown in Figure 7.5 moves from position (a) (its highest point) through (b) to position (c), where it is at its lowest point. It begins moving back up again through (d) to (e), where it returns to its highest position. The range of movement from (a) to (e) is one cycle. A single piston in a Formula 1 racecar can achieve a frequency of 300 cycles/second or 300 Hz (18 000 rpm). The piston makes 300 complete cycles in only 1 s. Conversely, the period of the piston, which is the time for one complete cycle, is the inverse of the frequency. It is a mere 0.003 s or about 100 times faster than the blink of an eye! (a) (b) (c) (d) (e) e MATH Using a graphing calculator, plot period as a function of frequency. Use the following values of frequency: 2, 4, 6, 8, 10, 12, 14, 16, 18, 20. If possible, print this graph, label it, and add it to your notes. What is the shape of this graph? info BIT A Formula 1 racecar has 10 cylinders but the engine size is limited to 3.0 L, no bigger than many engines in medium-sized cars. The fuel efficiency of F1 cars is approximately 1.3 km/L. Figure 7.5 The piston makes one complete cycle from positions (a) to (e). The time it takes to do this is its period. The number of times it does this in 1 s is its frequency. Example 7.1 What is the frequency of an automobile engine in which the pistons oscillate with a period of 0.0625 s? Analysis and Solution f 1 T 1 0.0625 s 16.0 Hz The frequency of the engine is 16.0 Hz. Practice Problems 1. Earthquake waves that travel along Earth\u2019s surface can have periods of up to 5.00 minutes. What is their frequency? 2. A hummingbird can hover when it flaps its wings with a frequency of 78 Hz. What is the period of the wing\u2019s motion? Answers 1. 3.33 103 Hz 2. 0.013 s M I N D S O N Examples of Oscillatory Motion Working with a partner or group, make a list of three or four natural or humanmade objects that move with the fastest oscillatory motion that you can think of. Make a similar list of objects that have very long periods of oscillatory motion. Beside each object estimate its period. The lists must not include the examples already mentioned. Be prepared to share your lists with the class. Chapter 7 Oscillatory motion requires a set of conditions. 345 07-Phys20-Chap07.qxd 7/24/08 1:09 PM Page 346 7-2 Inquiry Lab 7-2 Inquiry Lab Relating Period and Frequency Your teacher may want to do this Inquiry Lab in the gym instead of the science lab. Question What is the relationship between the period and the frequency of a bouncing ball? Hypothesis Write a hypothesis that relates the period of the ball\u2019s bounce to its frequency. Remember to use an \u201cif/then\u201d statement. Variables The variables in this lab are the height of the bounce, period, and frequency. Read the procedure, then determine and label the controlled, manipulated, and responding variables. Materials and Equipment stopwatch chair basketball metre-stick tape Procedure 1 Copy Table 7.2 into your notes. \u25bc Table 7.2 Bounce, Period, and Frequency Bounce Height (cm) Time for 20 Bounces (s) Period (s/bounce) Frequency (bounces/s) Required Skills Initiating and Planning Performing and Recording Analyzing and Interpreting Communication and Teamwork 2 Find a convenient place to bounce the basketball near the wall. Using the metre-stick, place tape at heights of 50, 75, 100, 125, and 150 cm. Mark the heights on the", " tape. 3 Using just a flick of the wrist, begin bouncing the ball at the 50-cm mark. The top of the ball should just make it to this height at the top of its bounce. The ball should bounce with a steady rhythm. 4 While one person is bouncing the ball, another person uses the stopwatch to record the time taken for 20 complete bounces. Record this time in Table 7.2. 5 Reset the stopwatch and begin bouncing the ball to the next height up. Record the time for 20 complete bounces as you did in step 4. To achieve the proper height you may have to stand on a chair. Analysis 1. Using the data for 20 bounces, determine the period and frequency for each height. Record the values in the table. 2. Draw a graph of frequency versus period. What type of relationship is this? 3. What effect did increasing the bounce height have on the period of a bounce? 50 75 100 125 150 346 Unit IV Oscillatory Motion and Mechanical Waves 07-Phys20-Chap07.qxd 7/24/08 1:09 PM Page 347 In 7-2 Inquiry Lab, the ball will make many more bounces in a certain length of time if it travels a shorter distance than a longer one. Its frequency will be high. By necessity, the amount of time it takes to make one bounce (its period) will be small. The next section explores oscillatory motion by going one step further. You will examine a type of oscillatory motion in which the range of motion is related to the applied force. 7.1 Check and Reflect 7.1 Check and Reflect Knowledge 1. What conditions describe oscillatory motion? 2. Which unit is equivalent to cycles/s? 3. Define period and frequency. 4. How are period and frequency related? 5. Is it possible to increase the period of an oscillatory motion without increasing the frequency? Explain. 6. Give three examples of oscillatory motion that you have observed. Applications 7. What is the frequency of a swimming water toy that makes 20.0 kicking motions per second? 8. Do the oars on a rowboat move with 12. A dog, happy to see its owner, wags its tail 2.50 times a second. (a) What is the period of the dog\u2019s wagging tail? (b) How many wags of its tail will the dog make in 1 min? Extensions 13. Use your library or the Internet to research the frequency of four to six different types of insect wings. Rank these insect wings from highest to lowest frequency. 14. Which of these motions is oscillatory? Explain. (a) a figure skater moving with a constant speed, performing a figure eight (b) a racecar racing on an oval track (c) your heartbeat oscillatory motion? Explain. 15. Many objects exhibit oscillatory motion. 9. Determine the frequency of a guitar string that oscillates with a period of 0.004 00 s. 10. A dragonfly beats its wings with the frequency of 38 Hz. What is the period of the wings? 11. A red-capped manakin is a bird that can flap its wings faster than a hummingbird, at 4800 beats per minute. What is the period of its flapping wings? Use your library or the Internet to find the frequency or range of frequencies of the objects below. (a) fluorescent light bulbs (b) overhead power lines (c) human voice range (d) FM radio range (e) lowest note on a bass guitar e TEST To check your understanding of period and frequency, follow the eTest links at www.pearsoned.ca/school/physicssource. Chapter 7 Oscillatory motion requires a set of conditions. 347 07-Phys20-Chap07.qxd 7/24/08 1:09 PM Page 348 info BIT A human eardrum can oscillate back and forth up to 20 000 times a second. 7.2 Simple Harmonic Motion Children on swings can rise to heights that make their parents nervous. But to the children, the sensation of flying is thrilling. At what positions on a swing does a child move fastest? When does the child\u2019s motion stop? Many objects that move with oscillatory motion exhibit the same properties that a child on a swing does. A piston moves up and down in the cylinder of an engine. At the extreme of its motion, it stops for a brief instant as it reverses direction and begins to accelerate downward until it reaches the bottom of its stroke. There it stops again and accelerates back, just as the swing does. In order for the piston or a child on a swing to experience acceleration, it must experience a non-zero net force. This section explores how the net force affects an object\u2019s motion in a special type of oscillatory motion called simple harmonic motion. 7-3 QuickLab 7-3 QuickLab Determining the Stiffness of a Spring Problem How does the", " force applied to a spring affect its displacement? Materials spring (with loops at each end) spring scale metre-stick tape Procedure 1 Make a two-column table in your notebook. Write \u201cDisplacement (cm)\u201d as the heading of the left column, and \u201cForce (N)\u201d as the heading of the right column. 2 Determine the maximum length the spring can be pulled without permanently deforming it. If you are unsure, ask your instructor what the maximum displacement of the spring is. Divide this length by 5. This will give you an idea of the even increments through which to pull your spring. 3 Lie the spring flat on the surface of the table so that it lies in a straight line. Leave room for it to be stretched. Do not pull on the spring. 4 Fix one end of the spring to the desk by holding the loop at the end of the spring with your hand. Do not let this end of the spring move. 5 Attach the spring scale to the free end of the spring but do not pull on it yet. 6 Align the 0-cm mark of the metre-stick with the other end of the spring at exactly where the spring scale is attached. Tape the stick to the desk (Figure 7.6). spring scale Figure 7.6 7 Pull the spring, using the spring scale, by the incremental distance determined in step 2. Record the values of the displacement and force in your table. 8 Repeat step 7, until you have five values each for displacement and force in your table. 9 Gently release the tension from the spring. Clean up and put away the equipment at your lab station. 348 Unit IV Oscillatory Motion and Mechanical Waves 07-Phys20-Chap07.qxd 7/24/08 1:09 PM Page 349 Questions 1. Determine the controlled, manipulated, and responding 4. Determine the slope of the line. What are the units of the slope? variables. 2. Plot a graph of force versus displacement. Be sure to use a scale that allows you to use the full graph paper. Draw a line of best fit. 3. What kind of relationship does the line of best fit represent? 5. Extrapolate where the line intercepts the horizontal axis. Why does it intercept there? e LAB For a probeware activity, go to www.pearsoned.ca/school/physicssource. Hooke\u2019s Law Robert Hooke (1635\u20131703) was a British scientist best remembered for using a microscope to discover plant cells, but his talents extended into many areas (Figure 7.7). He is credited with inventing the universal joint, which is used today on many mechanical devices (including cars); the iris diaphragm used to adjust the amount of light that enters a camera lens; the respirator to help people breathe; and the compound microscope, just to name a few of his inventions. In the field of oscillatory motion, he is acknowledged for his work with elastic materials and the laws that apply to them (Figure 7.8). In 1676, Hooke recognized that the more stress (force) is applied to an object, the more strain (deformation) it undergoes. The stress can be applied in many ways. For example, an object can be squeezed, pulled, or twisted. Elastic materials will usually return to their original state after the stress has been removed. This will not occur if too much force is applied or if the force is applied for too long a time. In those cases, the object will become permanently deformed because it was strained beyond the material\u2019s ability to withstand the deformation. The point at which the material cannot be stressed farther, without permanent deformation, is called the elastic limit. A spring is designed to be a very elastic device, and the deformation of a spring (whether it is stretched or compressed) is directly related to the force applied. The deformation of a spring is referred to as its displacement. From his observations, Hooke determined that the deformation (displacement) is proportional to the applied force. This can also be stated as \u201cforce varies directly with displacement.\u201d It can be written mathematically as: F x This relationship is known as Hooke\u2019s law, which states: The deformation of an object is proportional to the force causing the deformation. Figure 7.7 Robert Hooke lived at the same time as Sir Isaac Newton. Figure 7.8 Hooke\u2019s notes show the simple devices he used to derive his law. Chapter 7 Oscillatory motion requires a set of conditions. 349 07-Phys20-Chap07.qxd 7/24/08 1:09 PM Page 350 Hooke\u2019s law: the deformation of an object is proportional to the force causing it Figure 7.9(a) shows a spring that conforms to Hooke\u2019s law. It experiences a displacement that is", " proportional to the applied force. When no mass is applied to the spring, it is not compressed, and it is in its equilibrium position. As mass is added in increments of 10 g, the displacement (deformation) of the spring increases proportionally in increments of 5 cm, as shown in Figures 7.9(b), (c), and (d). spring constant: amount of stiffness of a spring (a) x 5 cm (b) 10 g (c) (d) x 0 x 10 cm 20 g x 15 cm 30 g Figure 7.9 The spring pictured above conforms to Hooke\u2019s law. If the mass is doubled, the displacement will also double, as seen in (b) and (c). If the force (mass) is tripled, the displacement will triple, as seen in (b) and (d). Each spring is different, so the force required to deform it will change from spring to spring. The stiffness of the spring, or spring constant, is represented by the letter k. Using k, you can write the equation for Hooke\u2019s law as: kx F where F is the applied force that extends or compresses the spring, k is the spring constant, and x is the displacement from the equilibrium position. Graphing Hooke\u2019s Law Figure 7.10 shows how you can use a force meter to measure the applied force required to pull the spring from its equilibrium position to successively farther displacements. For simplicity, we\u2019ll assume that all the springs used in this text are \u201cideal\u201d springs, meaning that they have no mass. FA FA x 0 Figure 7.10 A force meter is attached to a spring that has not been stretched. The spring is then pulled through several displacements. Each time, the force required for the displacement is recorded. 350 Unit IV Oscillatory Motion and Mechanical Waves 07-Phys20-Chap07.qxd 7/24/08 1:09 PM Page 351 Figure 7.11 Graph of data from Table 7.3 Table 7.3 shows the data collected for this spring and the results plotted on the graph shown in Figure 7.11. \u25bc Table 7.3 Data for Figure 7.11 Force vs. Displacement of a Spring Displacement (m) 0.00 0.10 0.20 0.30 0.40 Force (N) 0.0 5.0 10.0 15.0 20.0 20 ) N ( e c r o F 10 0.0 0.1 0.2 0.3 0.4 Displacement (m) Notice that the relationship is linear (a straight line), which means force is proportional to the displacement. The slope of the line can be determined by the following calculations: slope F x (20.0 N 0.0 N) (0.40 m 0.00 m) 50 N/m This slope represents the spring constant k. The variables F and x are vectors but here we are calculating their scalar quantities so no vector arrows are used. In this example, an applied force of 50 N is needed to stretch (or compress) this spring 1 m. Therefore, the units for the spring constant are newtons per metre (N/m). By plotting the force-displacement graph of a spring and finding its slope, you can determine the spring constant of any ideal spring or spring that obeys Hooke\u2019s law. Of the many objects that display elastic properties, springs are arguably the best to examine because they obey Hooke\u2019s law over large displacements. Steel cables are also elastic when stretched through relatively small displacements. Even concrete displays elastic properties and obeys Hooke\u2019s law through very small displacements. In simpler terms, the property of elasticity gives a material the ability to absorb stress without breaking. This property is vital to consider when structural engineers and designers build load-bearing structures such as bridges and buildings (Figure 7.12). You will learn more about the factors that must be considered in bridge and building design later in this chapter. Figure 7.12 The Jin Mao Tower in Shanghai, China, is 88 storeys high. Skyscrapers are built with elastic materials so they can sway in high winds and withstand the shaking of an earthquake. The main building materials for the Jin Mao Tower are concrete and steel. Chapter 7 Oscillatory motion requires a set of conditions. 351 07-Phys20-Chap07.qxd 7/24/08 1:09 PM Page 352 Example 7.2 Practice Problems 1. A spring is stretched through several displacements and the force required is recorded. The data are shown below. Determine the spring constant of this spring by plotting a graph and finding the slope. Displacement Force (m) 0.00 0.10 0.20 0.30 0.40 0.50 0.60 (N) 0.0 20.0 50.0 80.0", " 95.0 130.0 150.0 2. Determine the spring constant of a spring that has the forcedisplacement graph shown in Figure 7.15. Force vs. Displacement 40 30 20 10 ) Displacement (m) Figure 7.15 Answers 1. 2.5 102 N/m 2. 15 N/m To determine the spring constant of a spring, a student attaches a force meter to one end of the spring, and the other end to a wall as shown in Figure 7.13. She pulls the spring incrementally to successive displacements, and records the values of displacement and force in Table 7.4. Plot the values on a graph of force as a function of displacement. Using a line of best fit, determine the spring constant of the spring. x 0 Figure 7.13 Given \u25bc Table 7.4 Data for Figure 7.14 Displacement (m) 0.00 0.10 0.20 0.30 0.40 Force (N) 0.00 0.25 0.35 0.55 0.85 Required spring constant (k) Analysis and Solution Using the values from Table 7.4, plot the graph and draw a line of best fit. Force vs. Displacement of a Spring 1..5 (x2, y2) (0.30, 0.60) (x1, y1) (0.10, 0.20) 0.0 0.1 0.2 0.3 Displacement (m) 0.4 Figure 7.14 352 Unit IV Oscillatory Motion and Mechanical Waves 07-Phys20-Chap07.qxd 7/24/08 1:09 PM Page 353 The slope of the line gives the spring constant (k). Pick two points from the line and solve for the slope using the equation below. Note the points used in the equation are not data points. slope k F x point 1 (0.10, 0.20) point 2 (0.30, 0.60) (0.60 N 0.20 N) (0.30 m 0.10 m) k 2.0 N/m Paraphrase The spring constant of the spring is 2.0 N/m. The Restoring Force Imagine that you have applied a force to pull a spring to a positive dis- placement (x ) as shown in Figure 7.16. While you hold it there, the spring exerts an equal and opposite force in your hand, as described by Newton\u2019s third law in Chapter 3. However, this force is to the left, in the negative direction, and attempts to restore the spring to its equilibrium position. This force is called the restoring force. left right FA x 0 FR x Figure 7.16 The spring system is pulled from its equilibrium position to displacement xx. The displacement is positive, but the restoring force is negative. The restoring force always acts in a direction opposite to the displacement. Therefore, Hooke\u2019s law is properly written with a negative sign when representing the restoring force. kx F (2) In this case, while the spring is held in this position, the applied force and the restoring force have equal magnitudes but opposite directions, so the net force on the system is zero. In the next section, a mass will be attached to the spring and it will slide on a frictionless horizontal surface. The restoring force will be the only force in the system and will give rise to a repetitive back-and-forth motion called simple harmonic motion. restoring force: a force acting opposite to the displacement to move the object back to its equilibrium position Chapter 7 Oscillatory motion requires a set of conditions. 353 07-Phys20-Chap07.qxd 7/24/08 1:09 PM Page 354 Example 7.3 A spring has a spring constant of 30.0 N/m. This spring is pulled to a distance of 1.50 m from equilibrium as shown in Figure 7.17. What is the restoring force? Practice Problems 1. Determine the restoring force of a spring displaced 55.0 cm. The spring constant is 48.0 N/m. 2. A spring requires a force of 100.0 N to compress it a displacement of 4.0 cm. What is its spring constant? Answers 1. 26.4 N 2. 2.5 103 N/m Analysis and Solution Draw a diagram to represent the stretched spring. Displacement to the right is positive, so the restoring force is negative because it is to the left, according to Hooke\u2019s law. kx F 30.0 45.0 N (1.50 m) N m The restoring force is 45.0 N [left]. left right FR FA x 1.50 m equilibrium Figure 7.17 Simple Harmonic Motion of Horizontal Mass-spring Systems Figure 7.18 shows a mass attached to an ideal spring on a horizontal frictionless surface. This simple apparatus can help you understand", " the relationship between the oscillating motion of an object and the effect the restoring force has on it. x 0 left right Fnet 0 m Figure 7.18 The mass is in its equilibrium position (x 0) and is at rest. There is no net force acting on it. Any displacement of the mass to the right is positive, and to the left, negative. e SIM Observe a simulation of simple harmonic motion in a horizontal mass-spring system. Follow the eSIM links at www.pearsoned.ca/school/ physicssource. The position of the mass is represented by the variable x and is measured in metres. In Figure 7.18, there is no tension on the spring nor restoring force acting on the mass, because the mass is in its equilibrium position. Figure 7.19 shows how the restoring force affects the acceleration, displacement, and velocity of the mass when the mass is pulled to a positive displacement and released. 354 Unit IV Oscillatory Motion and Mechanical Waves 07-Phys20-Chap07.qxd 7/24/08 1:09 PM Page 355 The mass has been pulled to its maximum displacement, Figure 7.19(a) called its amplitude (symbol A). When the mass is released, it begins oscillating with a displacement that never exceeds this distance. The greater the amplitude, the more energy a system has. In this diagram, x A. At maximum displacement, the restoring force is at a maximum value, and therefore, so is the acceleration of the mass, as explained by Newton\u2019s second law (F ma). When the mass is released, it accelerates from rest (v 0) toward its equilibrium position. As the mass approaches this position, its velocity is increasing. But the restoring force is decreasing because the spring is not stretched as much. Remember that force varies directly with displacement. Figure 7.19(b) As the mass returns to its equilibrium position (x 0), it achieves its maximum velocity. It is moving toward the left (the negative direction), but the restoring force acting on it is zero because its displacement is zero. The mass continues to move through the equilibrium position and begins to compress the spring. As it compresses the spring, the restoring force acts on the mass toward the right (the positive direction) to return it to its equilibrium position. This causes the mass to slow down, and its velocity approaches zero. Figure 7.19(c) After passing through the equilibrium position, the mass experiences a restoring force that opposes its motion and brings it to a stop at the point of maximum compression. Its amplitude here is equal, but opposite to its amplitude when it started. At maximum displacement, the velocity is zero. The restoring force has reached its maximum value again. The restoring force is positive, and the displacement is negative. The restoring force again accelerates the mass toward equilibrium. Figure 7.19(d) The mass has accelerated on its way to the equilibrium position where it is now. The restoring force and acceleration are again zero, and the velocity has achieved the maximum value toward the right. At equilibrium, the mass is moving to the right. It has attained the same velocity as in diagram (b), but in the opposite direction. Figure 7.19(e) The mass has returned to the exact position where it was released. Again the restoring force and acceleration are negative and the velocity is zero. The oscillation will repeat again as it did in diagram (a). (a) (b) (c) (d) (e) F max a max v 0 m x is positive F 0 a 0 v max m x 0 F max a max v 0 m x is negative F 0 a 0 v max m x 0 F max a max v 0 m x is positive In Figure 7.19(e), the mass has returned to the position where it started, and one full oscillation has occurred. Throughout its entire motion, the mass-spring system obeys Hooke\u2019s law. In other words, at any instant, the restoring force is proportional to the displacement of the mass. Any object that obeys Hooke\u2019s law undergoes simple harmonic motion (SHM). SHM is oscillatory motion where the restoring force is proportional to the displacement of the mass. An object that moves with SHM is called a simple harmonic oscillator. simple harmonic motion: oscillatory motion where the restoring force is proportional to the displacement of the mass simple harmonic oscillator: an object that moves with simple harmonic motion Chapter 7 Oscillatory motion requires a set of conditions. 355 07-Phys20-Chap07.qxd 7/24/08 1:09 PM Page 356 (a) (b) Fspring equilibrium Fnet 0 Fg Figure 7.20 The spring in (a) has no mass attached. In (b), the spring stretches until the force exerted by the mass is equal and opposite to the force of gravity, and equilibrium is reached. The net force (or restoring force) is the", " vector sum of the force of gravity and the tension of the spring. In this case, it is zero. Figure 7.21 The net force (the restoring force) is the vector sum of the upward force exerted by the spring and the downward force of gravity. The force of gravity is always negative and constant, but the force exerted by the spring varies according to the displacement, so the net force changes as the position of the mass changes from (a) to (e). The values of F and v are identical to the horizontal mass-spring system. net, a, PHYSICS INSIGHT For any frictionless simple harmonic motion, the restoring force is equal to the net force. Simple Harmonic Motion of Vertical Mass-spring Systems Figure 7.20(a) shows a spring without a mass attached, anchored to a ceiling. Assume that the spring itself is massless, so it will not experience any displacement. When a mass is attached, the spring is pulled down and deforms as predicted by Hooke\u2019s law. The mass will come to rest when the downward force of gravity is equal to the upward pull (tension) of the spring (Figure 7.20(b)). The displacement of the spring depends on its spring constant. A weak spring has a small spring constant. It will stretch farther than a spring with a large spring constant. In Figure 7.20(b), the net force (or restoring force) acting on the mass is zero. It is the result of the upward tension exerted by the spring balancing the downward force of gravity. This position is considered the equilibrium position and the displacement is zero. If the mass is lifted to the position shown in Figure 7.21(a) and released, it will begin oscillating with simple harmonic motion. Its amplitude will equal its initial displacement. Regardless of the position of the mass, the force of gravity remains constant but the tension of the spring varies. In the position shown in Figures 7.21(b) and (d), the net (restoring) force is zero. This is where the spring\u2019s tension is equal and opposite to the force of gravity. In the position shown in Figure 7.21(c), the displacement of the spring is equal to the amplitude, and the tension exerted by the spring is at its maximum. The mass experiences the greatest restoring force, which acts upward. up down x 0 (a) (b) (c) (d) (e) Fs Fg Fnet max Fnet a max v 0 Fg Fs Fnet Fs Fg Fnet 0 0 Fnet a 0 v max Fs Fg Fs Fg Fnet max Fnet a max v 0 Fs Fnet Fg Fs Fg Fnet 0 0 Fnet a 0 v max Fs Fg Fs Fg Fnet max Fnet a max v 0 Fs Fg Fnet When the mass is below the equilibrium position, the upward force exerted by the tension of the spring is greater than the gravitational force. So the net force \u2014 and therefore the restoring force \u2014 is upward. Above the equilibrium position, the downward force of gravity exceeds the upward tension of the spring, and the restoring force is downward. The values of velocity, acceleration, and restoring force change in exactly the same way that they do in a horizontal massspring system. 356 Unit IV Oscillatory Motion and Mechanical Waves 07-Phys20-Chap07.qxd 7/24/08 1:09 PM Page 357 Example 7.4 A spring is hung from a hook on a ceiling. When a mass of 510.0 g is attached to the spring, the spring stretches a distance of 0.500 m. What is the spring constant? Practice Problems 1. Five people with a combined mass of 275.0 kg get into a car. The car\u2019s four springs are each compressed a distance of 5.00 cm. Determine the spring constant of the springs. Assume the mass is distributed evenly to each spring. 2. Two springs are hooked together and one end is attached to a ceiling. Spring A has a spring constant (k) of 25 N/m, and spring B has a spring constant (k) of 60 N/m. A mass weighing 40.0 N is attached to the free end of the spring system to pull it downward from the ceiling. What is the total displacement of the mass? Answers 1. 1.35 104 N/m 2. 2.3 m s and F g Given x 0.500 m m 510.0 g 0.5100 kg g 9.81 m/s2 Required spring constant (k) Analysis and Solution Draw a diagram to show the mass-spring system and the forces acting on the mass. up down x 0.500 m equilibrium Fs Fg Figure 7.22 The mass is not moving so the net force on the mass is zero. F are therefore equal in magnitude. kx mg mg x m (0.", "5100 kg )9.81 s2 0.500 m k 10.0 N/m Paraphrase The spring constant is 10.0 N/m. Chapter 7 Oscillatory motion requires a set of conditions. 357 07-Phys20-Chap07.qxd 7/24/08 1:09 PM Page 358 Examples of Simple Harmonic Motion Provided the cord doesn\u2019t go slack, a person making a bungee jump will bob up and down with SHM, as shown in Figures 7.23 and 7.24. The cord acts as the spring and the person is the mass. up down x 0 Fcord Fg Fcord Fg Fnet Fcord Fg Fnet Fcord Fnet Fg Fcord Fg Fnet 0 Fnet Fcord Fg Figure 7.24 The bungee jumper bouncing up and down on the cord after a jump in (a) is a vertical mass-spring system. The cord acts as a spring and the jumper is the mass. The restoring (net) force acting on the bungee jumper is the same as it was for the vertical mass-spring system. When the oscillating finally stops, the jumper will come to a stop in the equilibrium position. The reeds of woodwind instruments, such as the clarinet, behave as simple harmonic oscillators. As the musician blows through the mouthpiece, the reed vibrates as predicted by the laws of SHM. Once a simple harmonic oscillator is set in motion, it will slowly come to rest because of friction unless a force is continually applied. We will examine these conditions in section 7.4 on resonance. SHM is repetitive and predictable, so we can state the following: \u2022 The restoring force acts in the opposite direction to the displacement. \u2022 At the extremes of SHM, the displacement is at its maximum and is referred to as the amplitude. At this point, force and acceleration are also at their maximum, and the velocity of the object is zero. \u2022 At the equilibrium position, the force and acceleration are zero, and the velocity of the object is at its maximum. Figure 7.23 A bungee jumper experiences SHM as long as the cord does not go slack. e WEB After the first few oscillations following the jump, the bungee jumper oscillates with simple harmonic motion. To learn more about simple harmonic motion in the vertical direction, follow the links at www.pearsoned.ca/school/ physicssource. 358 Unit IV Oscillatory Motion and Mechanical Waves 07-Phys20-Chap07.qxd 7/24/08 1:09 PM Page 359 Concept Check 1. Must the line on a graph of force versus displacement for 2. a spring always intercept the origin? Explain. In what situation might the line on a graph of force as a function of displacement for a spring become non-linear? 3. A student wants to take a picture of a vertical mass-spring system as it oscillates up and down. At what point in the mass\u2019s motion would you suggest that she press the button to take the clearest picture? Instead of plotting a force-displacement graph for a spring, a student plots a restoring force-displacement graph. Sketch what this graph might look like. 4. 5. How would you write the equation for Hooke\u2019s law to reflect the shape of the graph above? Simple Harmonic Motion of a Pendulum The Cassini-Huygens space probe featured at the beginning of this chapter is named in honour of two distinguished scientists. Among many other notable accomplishments, the Italian astronomer Giovanni Cassini (1625\u20131712) observed the planets Mars and Jupiter and measured their periods of rotation. Christiaan Huygens (1629\u20131695), a Dutch mathematician and astronomer, invented the first accurate clock. It used a swinging pendulum and was a revolution in clock making (Figure 7.25). For small displacements, a swinging pendulum exhibits SHM. Since SHM is oscillatory, a clock mechanism that uses a pendulum to keep time could be very accurate. Up until Huygens\u2019s time, clocks were very inaccurate. Even the best clocks could be out by as much as 15 minutes a day. They used a series of special gears and weights that didn\u2019t always produce a uniform rate of rotation \u2014 a necessity for an accurate mechanical clock. Huygens recognized that if he could take advantage of the uniform oscillations of a pendulum, he could produce a much better clock. When completed, his pendulum clock was accurate to within one minute a day. This may not be very accurate by today\u2019s standards, but was easily the best of its time. Pendulum clocks became the standard in time keeping for the next 300 years. Let\u2019s examine cases where an ideal pendulum swings through a small angle, as explained in Figure 7.26 (a) to", " (e). In this book, all pendulums are considered ideal. That is, we assume that the system is frictionless, and the entire mass of the pendulum is concentrated in the weight. While this is not possible in reality, it is reasonable to make these assumptions here because they provide reasonably accurate results. Figure 7.25 A replica of Huygens\u2019s pendulum clock Chapter 7 Oscillatory motion requires a set of conditions. 359 07-Phys20-Chap07.qxd 7/24/08 1:09 PM Page 360 (a) left right v 0 max FR a max Fg Fg FR (b) left right v max 0 FR a 0 (c) left right v 0 max FR a max Fg Fg FR (d) left right v max 0 FR a 0 (e) left right v 0 max FR a max Figure 7.26(a) The mass (called a \u201cbob\u201d) is attached to the string and has been pulled from its equilibrium (rest) position through a displacement angle of. It has a mass m. When the bob\u2019s displacement is farthest to the right, the restoring force is a maximum negative value and velocity is zero. When the pendulum is released, gravity becomes the restoring force. Given the direction of the force of gravity, the acceleration due to gravity is straight down. However, the motion of the pendulum is an arc. A component of gravity acts along this arc to pull the bob back toward equilibrium and is, by definition, the restoring force (F R). We can express FR in terms of Fg with the following equation: FR Fg(sin ) Figure 7.26(b) As the bob accelerates downward, its velocity begins to increase and the restoring force (F R) becomes less and less. When it reaches the equilibrium position, no component of gravity is acting parallel to the motion of the bob, so the restoring force is zero, but the velocity has reached its maximum value. Figure 7.26(c) restoring force has also reached a maximum value but it acts toward the right. The bob\u2019s velocity is zero again. The bob has reached its maximum displacement to the left. The The bob passes through the equilibrium position and begins to move upward. As it does so, the restoring force becomes larger as the displacement of the bob increases. But just like the mass-spring system, the restoring force is acting in a direction opposite to the bob\u2019s displacement. At the other extreme of the bob\u2019s displacement, the restoring force has slowed the bob to an instantaneous stop. In this position, the displacement and restoring force are a maximum, and the bob\u2019s velocity is zero. Figure 7.26(d) The bob has achieved its maximum velocity, but this time it is to the right. The bob\u2019s displacement is again zero, and so is the restoring force. On its back swing, the bob moves through the equilibrium position again, as shown here. The velocity is a maximum value, just as it was in Figure 7.26(b), but now it\u2019s in the opposite direction. The restoring force once more brings the bob\u2019s motion to a stop for an Figure 7.26(e) instant at the position farthest to the right. The restoring force is a maximum negative value, and the bob\u2019s velocity is zero. The pendulum has made one complete oscillation as shown in diagrams (a) to (e). Note that the motion of the pendulum bob and the mass-spring systems are similar. Figure 7.19(a\u2013e) on page 355 and Figure 7.26(a\u2013e) on this page are comparable because both systems undergo the same changes to force, velocity, and acceleration at the same displacements. 360 Unit IV Oscillatory Motion and Mechanical Waves 07-Phys20-Chap07.qxd 7/24/08 1:09 PM Page 361 Motion with Large Amplitudes Earlier in this chapter, you read that a pendulum acts as a simple harmonic oscillator for small angles. Why is that? How is its motion different from SHM at larger angles? The best way to answer these questions is to plot a graph of force versus displacement like those done for springs earlier in this chapter (e.g., Figure 7.11 on page 351). The displacement of the pendulum can be measured by its angle from the vertical. If the graph is linear, then the restoring force is proportional to the displacement, and the pendulum has moved in SHM, as described by Hooke\u2019s law. To create this graph, use the equation for restoring force that you saw in the explanation of Figure 7.26(a) on the previous page: FR Fg(sin ) (3) From this equation, we can plot the values for angles up to 90 for a bob with a mass of 1.0 kg. As the graph", " in Figure 7.27 shows, the line is not linear, so the restoring force does not vary proportionally with the displacement. Strictly speaking, a pendulum is not a true simple harmonic oscillator. However, the line is almost linear up to about 20. At angles of less than 15, the deviation from a straight line is so small that, for all practical purposes, it is linear. \u25bc Table 7.5 Data for Figure 7.27 Angle (\u00b0) Restoring Force (N) 0 10 20 30 40 50 60 70 80 90 0 1.70 3.36 4.91 6.31 7.51 8.50 9.22 9.66 9.81 Force vs. Displacement of a Pendulum 10 10 20 40 50 60 30 Displacement (\u00b0) 70 80 90 Figure 7.27 For the pendulum to be a true simple harmonic oscillator, its graph of restoring force versus displacement should be linear, as the dotted line suggests. After 15, its line departs from the straight line, and its motion can no longer be considered SHM. Chapter 7 Oscillatory motion requires a set of conditions. 361 07-Phys20-Chap07.qxd 7/24/08 1:09 PM Page 362 Example 7.5 Practice Problems 1. Determine the restoring force of a pendulum that is pulled to an angle of 12.0 left of the vertical. The mass of the bob is 300.0 g. 2. At what angle must a pendulum be displaced to create a restoring force of 4.00 N [left] on a bob with a mass of 500.0 g? Answers 1. 0.612 N [right] 2. 54.6 Determine the magnitude of the restoring force for a pendulum bob of mass 100.0 g that has been pulled to an angle of 10.0 from the vertical. Given g 9.81 m/s2 m 100.0 g 0.1000 kg Required restoring force (FR) Analysis and Solution Draw a diagram of the pendulum in its displaced position to show the forces acting on the bob. left right 10.0\u00b0 FT Fg Fg FR Figure 7.28 The restoring force F arc path of the pendulum. R is the component of F g that is tangential to the FR Fg(sin ) mg(sin ) (0.1000 kg)9.81 0.170 N m s2 (sin 10.0) Paraphrase The magnitude of the restoring force acting on the pendulum is 0.170 N. When Christiaan Huygens designed the first pendulum clock, his primary concern was to have the clock operate with a very consistent period. For a uniform period, he could use gear ratios in the mechanism to translate the motion of the pendulum to meaningful units of time, such as minutes and hours. Which factors influence the period of a pendulum, and which do not? To discover how a pendulum\u2019s mass, amplitude, and length influence its period, do 7-4 Inquiry Lab. 362 Unit IV Oscillatory Motion and Mechanical Waves 07-Phys20-Chap07.qxd 7/24/08 1:09 PM Page 363 Required Skills Initiating and Planning Performing and Recording Analyzing and Interpreting Communication and Teamwork 2 Attach the thermometer clamp to the top of the retort stand. Attach the thread to the thermometer clamp. Make sure the thread is a little shorter than the height of the clamp from the table. 3 Squeeze one end of the thread in the clamp and use a slip knot on the other end to attach the first mass. The mass should hang freely above the table. 4 Measure the length of the thread from the clamp to the middle of the mass. Record this as the length of the pendulum at the top of Table 7.6. 5 Pull the mass on the thread back until it makes an angle of 15 with the vertical, as measured with the protractor. 6 Remove the protractor, and release the mass as you start to time it. Count the number of complete oscillations it makes in 20 s. Record this number in your table. 7 Remove the mass and replace it with the next mass. Loosen the clamp and adjust the length of the thread so that it is the same length as for the previous mass. (Remember to measure length to the middle of the mass.) Repeat steps 5 to 7 until all the masses are used. Analysis 1. Determine the frequency and period of each mass. Record the numbers in your table. 2. Plot a graph of period versus mass. Remember to place the manipulated variable on the horizontal axis. 3. What conclusion can you draw about the relationship between the mass and the period of a pendulum? Explain your answer and show any relevant calculations. 7-4 Inquiry Lab 7-4 Inquiry Lab A Pendulum and Simple Harmonic Motion Question What is the relationship between the period of a", " pendulum and its mass, amplitude, and length? Materials and Equipment thermometer clamp retort stand 1.00-m thread (e.g., dental floss) 4 masses: 50 g, 100 g, 150 g, 200 g ruler (30 cm) or metre-stick protractor stopwatch or watch with a second hand Hypothesis Before you begin parts A, B, and C, state a suitable hypothesis for each part of the lab. Remember to write your hypotheses as \u201cif/then\u201d statements. Variables The variables are the length of the pendulum, the mass of the pendulum, elapsed time, and the amplitude of the pendulum. Read the procedure for each part and identify the controlled, manipulated, and responding variables each time. Part A: Mass and Period Procedure 1 Copy Table 7.6 into your notebook. \u25bc Table 7.6 Mass and Period Length of Pendulum Mass (g) No. of Cycles/20 s Frequency (Hz) Period (s) 50 100 150 200 Chapter 7 Oscillatory motion requires a set of conditions. 363 07-Phys20-Chap07.qxd 7/24/08 1:09 PM Page 364 Part B: Amplitude and Period Part C: Length and Period Procedure Procedure 1 Copy Table 7.7 into your notebook. 1 Copy Table 7.8 into your notebook. \u25bc Table 7.7 Amplitude and Period \u25bc Table 7.8 Length and Period Length of Pendulum Amplitude () No. of Cycles/20 s Frequency (Hz) Period (s) Length (m) No. of Cycles/20 s Frequency (Hz) Period (s) 5 10 15 20 2 Use the same apparatus as in part A. 3 Attach a 200-g mass to the free end of the thread. 4 Measure the length of the thread from the clamp to the middle of the mass. Record this length at the top of Table 7.7. 5 Pull the mass on the thread back until it makes an angle of 5 with the vertical, as measured with the protractor. 6 Remove the protractor, and release the mass as you start to time it. Count the number of complete oscillations it makes in 20 s. Record this number in Table 7.7. 7 Repeat steps 5 and 6, each time increasing the amplitude by 5. Analysis 1. Determine the frequency and period for each amplitude and record the numbers in the appropriate columns in the table. 2. Plot a graph of period versus amplitude. Remember to place the manipulated variable on the horizontal axis. 3. What conclusion can you draw about the relationship between the amplitude and the period of a pendulum? Show any relevant calculations. 2 Use the same apparatus as in part A. Start with a pendulum length of 1.00 m. 3 Attach a 200-g mass to the free end of the thread. 4 Measure the length of the thread from the clamp to the middle of the mass. Record this length in Table 7.8. 5 Pull the mass on the thread back until it makes an angle of 15 with the vertical, as measured with the protractor. 6 Remove the protractor, and release the mass as you start to time it. Count the number of complete oscillations it makes in 20 s. Record this number in Table 7.8. 7 Repeat steps 4 to 6, but each time decrease the length of the pendulum by half. Analysis 1. Determine the frequency and period for each length and record the values in the appropriate column in the table. 2. Plot a graph of period versus length. Remember to place the manipulated variable on the horizontal axis. 3. What conclusion can you draw about the relationship between the length and the period of a pendulum? Show any relevant calculations. e LAB For a probeware activity, go to www.pearsoned.ca/school/physicssource. Pendulums and mass-spring systems are not the only simple harmonic oscillators. There are many other examples: a plucked guitar string, molecules vibrating within a solid, and water waves are just a few. In section 7.4, you will explore some human-made examples of SHM and learn about an interesting property called resonance. 364 Unit IV Oscillatory Motion and Mechanical Waves 07-Phys20-Chap07.qxd 7/24/08 1:09 PM Page 365 7.2 Check and Reflect 7.2 Check and Reflect Knowledge 7. Two students are given the task of 1. The restoring force of a vertical mass- spring system is determined by the mass attached to the spring and the spring constant k. What two factors determine the restoring force of a pendulum? 2. Copy the following tables into your notes. Then fill in the blanks by using the words, \u201czero\u201d or \u201cmaximum.\u201d determining the spring constant of a spring as accurately as possible. To do this, they attach a force meter to a spring that is lying on", " a desk and is anchored at the other end. One student pulls the spring through several displacements, while the other records the force applied, as shown in the table below. Using this table, plot a graph of force versus displacement. Find the spring constant by determining the slope of the line of best fit. Pendulum Displace- Accelerment ation System max x max a max v min F Velocity Restoring Force Displacement (m) 0.00 0.10 0.20 0.30 0.40 Force (N) 0.00 0.15 0.33 0.42 0.60 Displace- Acceler- ment ation Velocity Restoring Force Massspring System max x max a max v min F 8. Determine the restoring force for a pendulum bob with a mass of 0.400 kg that is pulled to an angle of 5.0 from the vertical. 9. A toy car, with a wind-up spring motor, on a horizontal table is pulled back to a displacement of 20.0 cm to the left and released. If the 10.0-g car initially accelerates at 0.55 m/s2 to the right, what is the spring constant of the car\u2019s spring? (Hint: The restoring force is F ma.) 3. Explain why a pendulum is not a true simple harmonic oscillator. Applications Extension 4. A mass of 2.0 kg is attached to a spring with a spring constant of 40.0 N/m on a horizontal frictionless surface. Determine the restoring force acting on the mass when the spring is compressed to a displacement of 0.15 m. 5. A spring hangs vertically from a ceiling and has a spring constant of 25.0 N/m. How far will the spring be stretched when a 4.0-kg mass is attached to its free end? 6. An applied force of 25.0 N is required to compress a spring 0.20 m. What force will pull it to a displacement of 0.15 m? 10. Obtain three different types of rulers: plastic, metal, and wooden. Fix one end of each ruler to the side of a desk so the ruler juts out horizontally a distance of 25 cm from the edge. Hang enough weight on the end that sticks out to make the ruler bend downward by 2 to 3 cm. Record the deflection of the ruler and the mass used in each case. (Note: The deflection does not have to be the same for each ruler.) Use these data to determine the spring constant for each ruler. Rank the rulers from highest spring constant to lowest. e TEST To check your understanding of simple harmonic motion, follow the eTest links at www.pearsoned.ca/school/physicssource. Chapter 7 Oscillatory motion requires a set of conditions. 365 07-Phys20-Chap07.qxd 7/24/08 1:09 PM Page 366 info BIT An accelerometer is a device used to measure acceleration. It is designed like a mass-spring system. The force exerted by the accelerating object causes the mass to compress the spring. The displacement of the mass is used to determine the positive or negative acceleration of the object. Accelerometers are commonly used in airbag systems in cars (see Chapter 3). If the car slows down too quickly, the displacement of the mass is large and it triggers the airbag to deploy. PHYSICS INSIGHT An object does not have to be moving to experience acceleration. 7.3 Position, Velocity, Acceleration, and Time Relationships One way for ball players to practise their timing is by attempting to throw a ball through a tire swinging on a rope. Someone just beginning this kind of practice might throw too early or too late, missing the tire altogether. Part of the difficulty has to do with the continually changing velocity of the tire. Choosing the best time to throw the ball is an exercise in physics. With practice, the human brain can learn to calculate the proper time to throw without even being aware that it is doing so. Throwing the ball through the tire is much more difficult than it sounds because the tire is a simple harmonic oscillator for small amplitudes. Not only is the velocity continually changing, but so is the restoring force and the acceleration. The only constant for a swinging tire is its period. In this section, you will mathematically analyze acceleration, velocity, and period for SHM in a mass-spring system, and then determine the period of a pendulum. Both mass-spring systems and pendulums are simple harmonic oscillators, as described in section 7.2, but they are different from each other. The mass-spring system has a spring constant k, but the pendulum does not. For this reason, we will look at each separately, starting with the mass-spring system. Acceleration of a Mass-spring System In section 7.2, you learned that two equations can be used to describe force in the mass-spring system: Newton", "\u2019s second law and Hooke\u2019s law. They can be written mathematically as: \u2022 Newton\u2019s second law: F \u2022 Hooke\u2019s law: F kx ma (2) net Since both equations refer to the restoring force, you can equate them: F F net ma kx x a k m (4) where aa is the acceleration in metres per second squared; k is the spring constant in newtons per metre; x is the displacement in metres; and m is the mass of the oscillator in kilograms. 366 Unit IV Oscillatory Motion and Mechanical Waves 07-Phys20-Chap07.qxd 7/24/08 1:09 PM Page 367 e MATH To see how the spring constant, mass, position and acceleration are related graphically, visit www.pearsoned.ca/school/ physicssource. Figure 7.29 The acceleration of a simple harmonic oscillator depends on its position. In position (a), the oscillator moves from its maximum displacement and maximum positive acceleration through to position (b), where the displacement and acceleration are zero. It then moves to position (c), where the oscillator again experiences a maximum acceleration and displacement in the other direction. maximum height v 0 a 9.81 m/s2 The acceleration of a horizontal mass-spring simple harmonic oscillator can be determined by its spring constant, displacement, and mass. It\u2019s logical that the acceleration of the mass depends on how stiff the spring is and how far it is stretched from its equilibrium position. It is also reasonable to assume that, if the mass is large, then the acceleration will be small. This assumption is based on Newton\u2019s second law. The acceleration depends on the displacement of the mass, so the acceleration changes throughout the entire motion as shown in Figure 7.29. Since acceleration of a simple harmonic oscillator is not uniform, only the instantaneous acceleration of the mass can be determined by equation 4. () () (b) (c) () Acceleration (a) (a) () (b) () Displacement () (c) x 0 The Relationship Between Acceleration and Velocity of a Mass-spring System The acceleration of a simple harmonic oscillator is continually changing, so it should come as no surprise that the velocity changes too. As we have just seen, the maximum acceleration occurs when the oscillator is at its maximum displacement. At this position, it is tempting to think that the velocity will be at its maximum as well, but we know that this is not the case. Remember, the acceleration is at its greatest magnitude at the extremes of the motion, yet the oscillator has actually stopped in these positions! In some ways, a ball thrown vertically into the air is similar (Figure 7.30). The acceleration of gravity acts on the ball to return it to the ground. When the ball reaches its maximum height, it comes to a stop for an instant, just like the mass-spring system you studied earlier in this chapter. Figure 7.30 At its maximum height, the ball stops for a brief instant, yet the acceleration of gravity acting on it is not zero. This is similar to the mass-spring system. Chapter 7 Oscillatory motion requires a set of conditions. 367 07-Phys20-Chap07.qxd 7/24/08 1:09 PM Page 368 () () (a) (b) (c) () Velocity (b) (a) (c) Displacement () x 0 Figure 7.31 The velocity of a simple harmonic oscillator is not uniform. The mass experiences its greatest acceleration at the extremes of its motion where the velocity is zero. Only after the mass accelerates from position (a) to (b) does its velocity reach its maximum value. The mass then decelerates from (b) to (c) where it comes to a stop again. Since the acceleration of the oscillator decreases as it approaches the equilibrium position (Figure 7.31(a)), the velocity does not increase at a uniform rate. The velocity-displacement graph looks like Figure 7.31(b). Figure 7.32 shows a diagram of a simple harmonic oscillator (a mass-spring system) as it moves through one-half of a complete oscillation from (a) to (b) to (c). Below the diagram are the acceleration-displacement and velocity-displacement graphs. The diagram of the oscillator and the graphs are vertically aligned so the graphs show you what is happening as the massspring system moves. In the diagram at the top of Figure 7.32, you can see the oscillator in position (a). It is at its farthest displacement to the left and the spring is compressed. The velocity-displacement graph shows that the oscillator\u2019s velocity in this position is zero (graph 2). You can also see from the accelerationdisplacement graph that the acceleration at that moment is positive and a maximum, but the", " displacement is negative (graph 1). () () Acceleration and Displacement \u2014 Always Opposite Graph 1: Acceleration vs. Displacement (a) (a) (b) (c) () Acceleration () (b) () Displacement () (c) () Velocity (b) Graph 2: Velocity vs. Displacement (a) (c) Displacement () x 0 Figure 7.32 The mass-spring system experiences a changing acceleration and velocity as it makes one-half of a full oscillation from position (a) to position (c). In fact, if you look closely at graph 1, you might notice how the acceleration and displacement are always opposite to one another, regardless of the position of the mass. The acceleration is positive while the displacement is negative, and vice versa. This isn\u2019t surprising, however, because it is what the negative sign in the equation for Hooke\u2019s law illustrates: kx. F Look again at Figure 7.32 and follow the mass as it moves from position (a) to positions (b) and (c). As the oscillator accelerates from position (a) to the right, it picks up speed. The velocity-displacement graph (graph 2) shows that the velocity is positive and increasing as the oscillator approaches position (b), yet the acceleration is decreasing, as shown in graph 1. The oscillator goes through the equilibrium position with a maximum positive velocity, but now the acceleration becomes negative as the spring tries to pull the oscillator back (graph 1). This is why the oscillator slows down and the velocity-displacement graph returns to zero in position (c). 368 Unit IV Oscillatory Motion and Mechanical Waves 07-Phys20-Chap07.qxd 7/24/08 1:09 PM Page 369 Consider a vertical mass-spring system. A bungee jumper will experience a positive acceleration when she is below the equilibrium position and a negative acceleration when above it (Figures 7.33 and 7.34). (a) (b) (c) equilibrium x max x 0 x max a v a v a v Displacement Figure 7.33 Displacement Figure 7.34 After a jump, the bungee jumper is shown in three positions: At the lowest point (a), in the equilibrium position (b), and at her maximum displacement (c). In each case the circled region on the graphs indicates her acceleration and velocity. Maximum Speed of a Mass-spring System Now you know that a simple harmonic oscillator will experience its greatest speed at the equilibrium position. What factors influence this speed, and how can we calculate it? In our examples, the mass-spring system is frictionless, and no external forces act on it. This is referred to as an isolated system and the law of conservation of energy applies. We will use this concept to derive the equation for the maximum speed. Recall from Chapter 6 that the total mechanical energy in an isolated system remains constant. That means that the kinetic and potential energy of the system may vary, but their sum is always the same. Chapter 7 Oscillatory motion requires a set of conditions. 369 07-Phys20-Chap07.qxd 7/24/08 1:09 PM Page 370 In other words, at any position in the motion of a mass-spring system, the sum of kinetic and potential energies must be equal to the total energy of the system. Recall that kinetic energy is expressed as: Ek 1 mv2 2 Recall that elastic potential energy is expressed as: Ep 1 kx2 2 Let\u2019s begin by looking at the energy of the system in two positions: \u2022 When the mass is at the maximum displacement (Figure 7.35(a)). \u2022 When the mass is at the minimum displacement (Figure 7.35(b)). v 0 Ek 0 Ep max m x max The oscillator at its maximum displacement. Figure 7.35(a) Potential energy has reached a maximum value (Ep oscillator\u2019s displacement is a maximum (x A). The kinetic energy is zero (v 0). ) because the max v max Ek max Ep 0 m x 0 Figure 7.35(b) The oscillator at its minimum displacement (x 0). Kinetic energy has reached a maximum value (Ekmax has a maximum velocity. The potential energy is zero (x 0). ) because the oscillator Remember that the total energy of the system remains constant regardless of the oscillator\u2019s position. The equation for the total energy is: ET Ep Ek The kinetic energy of the oscillator at its maximum displacement is zero so the total energy of the oscillator at that position must be: ET Epmax The potential energy of the oscillator at its minimum displacement is zero so the total energy of the oscillator at that position must be: ET Ekmax Because the total energy is always the same, we can write: Ekmax Epmax or 1 2 mv 2 max 1 kx 2", " 2 max 370 Unit IV Oscillatory Motion and Mechanical Waves 07-Phys20-Chap07.qxd 7/24/08 1:09 PM Page 371 max mv2 1 kA2 2 If we use A to represent xmax, we can write: 1 2 We can then simplify this equation to solve for vmax: 1 2 1 kA2 2 mv 2 max mv 2 max kA2 v 2 max kA2 m Then we take the square root of each side: vmax kA2 m or vmax A k m (5) e WEB To find out how these factors are taken into account in bungee jumping, follow the links at www.pearsoned.ca/school/ physicssource. Factors That Influence the Maximum Speed of a Mass-spring System Three factors influence the maximum speed of a mass-spring system: \u2022 the amplitude of the oscillations: If the oscillator moves through a large amplitude, the restoring force increases in proportion to the amplitude. As the restoring force increases, so does the acceleration, and the oscillator will achieve a greater velocity by the time it reaches the equilibrium position. \u2022 the stiffness of the spring: A stiffer spring with a higher spring constant exerts a stronger restoring force and creates a greater maximum velocity for the same reasons that increasing the amplitude does. \u2022 the mass of the oscillator: Changing the mass of an oscillator has a different effect. If the mass increases, the velocity of the oscillations decreases. This is because the oscillator has more inertia. A larger mass is harder to accelerate so it won\u2019t achieve as great a speed as a similar mass-spring system with less mass. Concept Check 1. When acceleration is negative, displacement is positive and vice versa. Why? 2. Why is the velocity-time graph of a simple harmonic oscillator a curved line? 3. The acceleration-displacement graph and velocity-displacement graph are shown in Figure 7.32 on page 368 for half of an oscillation only. Sketch three more acceleration-displacement and velocitydisplacement graphs for the second half of the oscillation. 4. Suppose the amplitude of an object\u2019s oscillation is doubled. How would this affect the object\u2019s maximum velocity? Chapter 7 Oscillatory motion requires a set of conditions. 371 07-Phys20-Chap07.qxd 7/24/08 1:09 PM Page 372 PHYSICS INSIGHT When a mass is hanging from a vertical spring at rest in the equilibrium position, the downward g is force of gravity F equal and opposite to the upward force exerted s, so the by the spring F restoring force is zero. The force of gravity acting on the mass doesn\u2019t change. If the spring is displaced from the equilibrium position, the restoring force will just be the force of the spring due to its displacement x. This is kx. expressed as F R Example 7.6 A 100.0-g mass hangs motionless from a spring attached to the ceiling. The spring constant (k) is 1.014 N/m. The instructor pulls the mass through a displacement of 40.0 cm [down] and releases it. Determine: (a) the acceleration when the mass is at a displacement of 15.0 cm [up], and (b) the maximum speed of the mass. up v down a x 15.0 cm equilibrium x 40.0 cm FR Figure 7.36 Given m 100.0 g 0.1000 kg k 1.014 N/m x 40.0 cm [down] 0.400 m [down] Required (a) acceleration (a) when x 15.0 cm [up] 0.150 m [up] (b) maximum speed (vmax) Practice Problems 1. A 0.724-kg mass is oscillating on a horizontal frictionless surface attached to a spring (k 8.21 N/m). What is the mass\u2019s displacement when its instantaneous acceleration is 4.11 m/s2 [left]? 2. A 50.0-g mass is attached to a spring with a spring constant (k) of 4.00 N/m. The mass oscillates with an amplitude of 1.12 m. What is its maximum speed? 3. An instructor sets up an oscillating vertical mass-spring system (k 6.05 N/m). The maximum displacement is 81.7 cm and the maximum speed is 2.05 m/s. What is the mass of the oscillator? Answers 1. 0.362 m [right] 2. 10.0 m/s 3. 0.961 kg Analysis and Solution (a) The mass will begin to oscillate when released. Acceleration R net is a vector quantity so direction is important. F F ma kx kx a m N (0.150 m) 1.014 m 0.1000 kg 1.52 m/s2", " (b) The maximum speed occurs when the mass is in the equilibrium position, whether it is moving up or down. The displacement of the mass before it is released is the amplitude (A) of the mass\u2019s oscillation. vmax A k m N 1.014 m 0.1000 kg 0.400 m 1.27 m/s Paraphrase (a) The mass has an acceleration of 1.52 m/s2 [down] when it is 15.0 cm above the equilibrium position. (b) The maximum speed of the mass is 1.27 m/s. 372 Unit IV Oscillatory Motion and Mechanical Waves 07-Phys20-Chap07.qxd 7/24/08 1:09 PM Page 373 Period of a Mass-spring System The next time you are travelling in a vehicle at night, watch for bicycles moving in the same direction as your vehicle. Notice the peculiar motion of the pedals as they reflect the light from your headlights (Figure 7.37). From a distance, these reflectors don\u2019t appear to be moving in a circular path, but seem to be moving up and down. The apparent up-and-down motion of the pedals is the same kind of motion as a mass-spring system oscillating back and forth, so it is simple harmonic motion. This observation proves useful because it is an example of how circular motion can be used to describe simple harmonic motion. The next few pages show how to derive equations for the period and maximum speed of a simple harmonic oscillator. Two conditions are necessary if circular motion is to be used to replicate simple harmonic motion: 1. The period of both the circular motion and the simple harmonic motion must be the same. 2. The radius of the circular motion must match the amplitude of the oscillator. For example, look at Figure 7.38, where a mass moving in a circular path with a radius r is synchronized with a mass-spring simple harmonic oscillator. This illustration demonstrates how circular motion can be used to describe SHM. Figure 7.37 From a distance, the reflectors on the bicycle pedals would appear to be moving up and down instead of in a circle. (a) (b) (c) (d) (e Figure 7.38 A mass moving in a circle is a simple harmonic oscillator that corresponds to the mass-spring oscillator shown below it. One-half of a complete cycle is shown here. For our purposes, the following conditions are true: \u2022 The radius of the circular motion is equal to the amplitude of the oscillator (r A, as shown in Figure 7.38(a)). \u2022 The mass in circular motion moves at a constant speed. \u2022 The periods of the mass in circular motion and the oscillator in the mass-spring system are the same. Chapter 7 Oscillatory motion requires a set of conditions. 373 07-Phys20-Chap07.qxd 7/24/08 1:10 PM Page 374 Deriving the Equation for the Period of a Mass-spring System Recall that the maximum velocity of a simple harmonic oscillator occurs when it is in its equilibrium position, which is position (c) in Figure 7.38. At the exact moment that the mass in circular motion is in position (c), its velocity is in the same direction as the velocity of the mass-spring system, and they are both moving at the same speed. But if the mass moving in a circular path is moving at a constant speed, then it must always be moving at the maximum speed of the mass-spring oscillator! Therefore, the maximum speed (vmax) of the mass-spring system is equal to the speed (v) of the circular mass system. The speed of an object moving in a cir- cular path was derived in Chapter 5. It is: v 2r T (6) Figure 7.39 The strings of a piano all have different masses. Even if they vibrate with the same amplitude they will have a different period of vibration because each string has a different mass. A heavy string will vibrate with a longer period (and lower frequency) than a lighter string. The speed of the circular motion (v) matches the maximum speed on the mass-spring oscillator (vmax), and the radius of the circle matches its amplitude. Therefore, we can customize the equation for the mass-spring oscillator: 2A T vmax (7) If we equate equation 5 and equation 7, we get: A k m 2A T We can then solve for T: A k m 2A T 2\u03c0 k T T 2 m m k (8) This equation describes the period of a simple harmonic oscillator, where T is the period of the oscillator in seconds; k is the spring constant in newtons per metre; and m is the mass of the oscillator in kilograms. Figure 7.39 is an example of an application of this equation. PHYSICS INSIGHT The period of a simple harmonic oscill", "ator does not depend on displacement. 374 Unit IV Oscillatory Motion and Mechanical Waves 07-Phys20-Chap07.qxd 7/24/08 1:10 PM Page 375 Factors Affecting the Period of an Oscillating Mass The larger the oscillating mass is, the longer its period of oscillation is. This seems reasonable since a large mass takes longer to speed up or slow down. It would also seem reasonable that the period should be inversely related to the spring constant, as the equation suggests. After all, the stiffer the spring, the more force it exerts over smaller displacements. Therefore, you could expect the mass to oscillate more quickly and have a smaller period. What is interesting about this equation is not what influences the period but what does not. It may seem odd that the displacement of the mass has no influence on the period of oscillation. This means that if you were to pull a mass-spring system to a displacement x and then let go, it would have the same period of oscillation as it would if you pulled it to a displacement of 2x and released it! The two identical mass-spring systems in Figure 7.40 have different amplitudes but the same period. k1 10.0 N/m k2 10.0 N/m amplitude 20 cm m1 x 0 amplitude 10 cm m2 x 0 1 10\u00b0 \u03b8 2 5\u00b0 \u03b8 Figure 7.40 Two identical mass-spring systems have the same spring constant and mass, but different amplitudes. Which has the longest period? They have the same period because displacement doesn\u2019t affect period. Figure 7.41 Two identical pendulums have the same mass and length, but different amplitudes. Which one has the longest period? They have the same period because displacement doesn\u2019t affect the period of a simple harmonic oscillator. This relationship is true for any simple harmonic oscillator, including a pendulum with a small amplitude. It is easy enough to test. Take two pendulums with the same mass and length (Figure 7.41). Pull both bobs back to different displacements. Remember to keep the displacements small so the pendulums oscillate with SHM. Release them at the same time. You will discover that both make one full oscillation in unison. This means they return to the point of release at exactly the same time. The pendulum that begins with the larger displacement has farther to travel but experiences a larger restoring force that compensates for this. Chapter 7 Oscillatory motion requires a set of conditions. 375 07-Phys20-Chap07.qxd 7/24/08 1:10 PM Page 376 Example 7.7 What is the period of oscillation of a mass-spring system that is oscillating with an amplitude of 12.25 cm and has a maximum speed of 5.13 m/s? The spring constant (k) is 5.03 N/m. Practice Problems 1. A mass of 2.50 kg is attached to a horizontal spring and oscillates with an amplitude of 0.800 m. The spring constant is 40.0 N/m. Determine: a) the acceleration of the mass when it is at a displacement of 0.300 m the maximum speed the period b) c) 2. A 2.60-g mass experiences an acceleration of 20.0 m/s2 at a displacement of 0.700 m on a spring. What is k for this spring? 3. What is the mass of a vertical mass-spring system if it oscillates with a period of 2.0 s and has a spring constant of 20.0 N/m? 4. What is the period of a vertical mass-spring system that has an amplitude of 71.3 cm and maximum speed of 7.02 m/s? The spring constant is 12.07 N/m. Answers 1. a) 4.80 m/s2 b) 3.20 m/s c) 1.57 s 2. 0.0743 N/m 3. 2.0 kg 4. 0.638 s A 12.25 cm left right Given A 12.25 cm 0.1225 m k 5.03 N/m vmax 5.13 m/s Required period of the oscillations (T ) equilibrium Analysis and Solution To determine the period of the oscillator, you need to know the oscillator\u2019s mass. Use the maximum speed equation (equation 5) to find the mass: Figure 7.42 EPmax Ekmax A2 2 k mv max 2 2 mv 2 max kA2 0.1225 m)2 5.03 m m 2 5.13 s 2.868 103 kg Then use equation 8 to determine the period: T 2 m k 2 2.868 103 kg N 5.03 m 0.150 s Paraphrase The period of the mass-spring oscillator is 0.150 s. 376", " Unit IV Oscillatory Motion and Mechanical Waves 07-Phys20-Chap07.qxd 7/24/08 1:10 PM Page 377 Concept Check 2. 1. What effect does doubling the displacement have on the period of oscillation of a simple harmonic oscillator? Explain your answer. In order to compare circular motion to the motion of a simple harmonic oscillator, what two conditions must be satisfied? If the mass and spring constant of a mass-spring oscillator were doubled, what effect would this have on the period of the oscillations? 3. 4. Two mass-spring systems with identical masses are set oscillating side by side. Compare the spring constants of the two systems if the period of one system is twice the other. The Period of a Pendulum Christiaan Huygens recognized that a pendulum was ideally suited for measuring time because its period isn\u2019t affected by as many of the factors that influence a mass-spring system. A pendulum doesn\u2019t have a spring constant, k, like the mass-spring system does, and unlike the mass-spring system, the mass of the pendulum does not affect its period. Because of these factors, a new equation for a pendulum\u2019s period must be derived. In doing so, you will discover why its mass is irrelevant and what factors play a role in its period of oscillation. Take a closer look at the pendulum when it is at a small displacement of 15 or less, as shown in Figure 7.43. For a small angle (), the displacement of the bob can be taken as x. The sine of angle is expressed as: x l sin Recall that the restoring force for a pendulum is FR above expression for sin in this equation: Fg sin. Use the Fg FR x l \u03b8 l x Recall also that in a mass-spring system, the restoring force is F kx. We want to solve for the period (T ), which is a scalar quantity, so the negative sign in Hooke\u2019s law can be omitted. The two equations for restoring force can then be equated: Figure 7.43 For a pendulum with a small displacement of 15 or less, the displacement is x. kx Fg x l Fg mg kx (mg) x l k mg l Chapter 7 Oscillatory motion requires a set of conditions. 377 07-Phys20-Chap07.qxd 7/24/08 1:10 PM Page 378 None of the values of mass, gravitational field strength, or length change for a pendulum. They are constants, which are represented by k. Substitute them into equation 8 (page 374): T 2 m k g) m (m 2 T 2 l g l (9) PHYSICS INSIGHT The period of a pendulum does not depend on its mass or amplitude. e SIM Learn more about the motion of a pendulum and the factors that affect it. Follow the eSIM links at www.pearsoned.ca/school/ physicssource. where l is the length of the pendulum string in metres; and g is the gravitational field strength in newtons per kilogram. Recall that the length of the pendulum is always measured from the point where it is attached at the top, to the centre of mass of the bob, not the point at which the string or wire is attached to the bob. Also recall that the period of the pendulum\u2019s swing does not depend on the mass of the pendulum bob. This may not seem logical but it is indeed the case \u2014 just as the acceleration of an object in free fall doesn\u2019t depend on the mass of the object. The Pendulum and Gravitational Field Strength Equation 9 is useful when it is manipulated to solve for g, the gravitational field strength. As you learned in section 4.3 of Chapter 4, the gravitational field strength varies with altitude and latitude. The magnitude of the gravitational field is 9.81 N/kg at any place on Earth\u2019s surface that corresponds to the average radius of Earth. However, very few places on the surface of Earth are at exactly the average radius. To determine the exact value of g at any point, you can use a pendulum. If you manipulate equation 9 and solve for g, you get: g 42l T 2 (10) Due to the changing nature of Earth\u2019s gravity, Christiaan Huygens\u2019s pendulum clock (introduced in section 7.2) was only accurate if it was manufactured for a specific place. For example, pendulum clocks designed to operate in London could not be sold in Paris because the accuracy could not be maintained. The difference in gravitational field strength between London and Paris meant that the period of oscillation would be slightly different. The difference in g between two locations could be quite small, but the cumulative effect on a pendulum clock would be significant. An extreme example of the varying value of g", " at different geographic locations can be illustrated by using a pendulum to determine the gravitational field strength at the top of Mount Everest. 378 Unit IV Oscillatory Motion and Mechanical Waves 07-Phys20-Chap07.qxd 7/24/08 1:10 PM Page 379 Example 7.8 What is the gravitational field strength at the top of Mount Everest at an altitude of 8954.0 m, if a pendulum with a length of 1.00 m has a period of 2.01 s? Analysis and Solution Use equation 10 to determine g. Note that no vector arrow is required with the symbol g because you are calculating a scalar quantity. g 42l T 2 (42)(1.00 m) (2.01 s)2 9.77 m/s2 9.77 N/kg T 2.01 s 8954.0 m Figure 7.44 Mount Everest g 9.81 N/kg The gravitational field strength at the top of Mount Everest is 9.77 N/kg, which is very close to the accepted value of 9.81 N/kg. The extra height of Mount Everest adds very little to the radius of Earth. Practice Problems 1. What is the gravitational field strength on Mercury, if a 0.500-m pendulum swings with a period of 2.30 s? 2. A pendulum swings with a period of 5.00 s on the Moon, where the gravitational field strength is 1.62 N/kg [down]. What is the pendulum\u2019s length? 3. What period would a 30.0-cm pendulum have on Mars, where the gravitational field strength is 3.71 N/kg [down]? Answers 1. 3.73 N/kg [down] 2. 1.03 m 3. 1.79 s At the top of Mount Everest, a pendulum will swing with a slightly different period than at sea level. So a pendulum clock on Mount Everest, oscillating with a longer period than one at sea level, will report a different time. Huygens\u2019s clocks also suffered from another problem: the pendulum arm would expand or contract in hot or cold weather. Since the length of the arm also determines the period of oscillation, these clocks would speed up or slow down depending on the ambient temperature. Given their limitations, pendulum clocks were not considered the final solution to accurate timekeeping. Further innovations followed that you may want to research on your own. e WEB To learn more about pendulum clocks and the evolution of timekeeping, create a timeline of the evolution of clock design. In your timeline include what the innovation was, who invented it, and the year it was introduced. Begin your search at www.pearsoned.ca/school/ physicssource. Concept Check 1. An archer is doing target practice with his bow and arrow. He ties an apple to a string and sets it oscillating left to right, down range. In what position of the apple should he aim so that he increases his chances of hitting it? Explain your answer. 2. What factors affect the accuracy of pendulum clocks? Why? Chapter 7 Oscillatory motion requires a set of conditions. 379 07-Phys20-Chap07.qxd 7/24/08 1:10 PM Page 380 7.3 Check and Reflect 7.3 Check and Reflect Knowledge 1. Explain the effect that changing each of the following factors has on the period of a mass-spring system: (a) amplitude (b) spring constant (c) mass 2. Explain what effect changing each of the following factors has on the period of a pendulum: (a) amplitude (b) gravitational field strength (c) mass 9. A pendulum bob (m 250.0 g) experiences a restoring force of 0.468 N. Through what angle is it displaced? 10. A 50.0-cm pendulum is placed on the Moon, where g is 1.62 N/kg. What is the period of the pendulum? Extensions 11. A horizontal mass-spring system oscillates with an amplitude of 1.50 m. The spring constant is 10.00 N/m. Another mass moving in a circular path with a radius of 1.50 m at a constant speed of 5.00 m/s is synchronized with the mass-spring system. Determine the mass-spring system\u2019s: 3. Describe the positions that a mass-spring system and pendulum are in when: (a) period (b) mass (a) acceleration is a maximum (b) velocity is a maximum (c) restoring force is maximum 4. Why is the acceleration of a simple harmonic oscillator not uniform? 5. A mass-spring system has a negative displacement and a positive restoring force. What is the direction of acceleration? Applications 6. What length of pendulum would oscillate with a period of 4.0 s on the surface of Mars (g 3.71 N", "/kg)? 7. A mass of 3.08 kg oscillates on the end of a horizontal spring with a period of 0.323 s. What acceleration does the mass experience when its displacement is 2.85 m to the right? 8. A 50.0-kg girl bounces up and down on a pogo stick. The girl has an instantaneous acceleration of 2.0 m/s2 when the displacement is 8.0 cm. What is the spring constant of the pogo stick\u2019s spring? (c) maximum acceleration 12. A quartz crystal (m 0.200 g) oscillates with simple harmonic motion at a frequency of 10.0 kHz and has an amplitude of 0.0500 mm. What is its maximum speed? 13. A horizontal mass-spring system has a mass of 0.200 kg, a maximum speed of 0.803 m/s, and an amplitude of 0.120 m. What is the mass\u2019s position when its acceleration is 3.58 m/s2 to the west? 14. Suppose an inquisitive student brings a pendulum aboard a jet plane. The plane is in level flight at an altitude of 12.31 km. What period do you expect for a 20.0-cm pendulum? (Hint: First determine the gravitational field strength as shown in Chapter 4.) e TEST To check your understanding of position, velocity, acceleration, and time relationships in mass-spring systems and pendulums, follow the eTest links at www.pearsoned.ca/school/physicssource. 380 Unit IV Oscillatory Motion and Mechanical Waves 07-Phys20-Chap07.qxd 7/24/08 1:10 PM Page 381 info BIT When you walk with a drink in your hand at the right speed, your motion creates resonance in the liquid. This makes waves that splash over the edge of the cup. To prevent this, people walk slowly so resonance doesn\u2019t occur, often without knowing why this works. resonant frequency: the natural frequency of vibration of an object 7.4 Applications of Simple Harmonic Motion People\u2019s arms swing as they walk. An annoying rattle can develop in a car when it reaches a certain speed. A child can make large waves in the bathtub by sliding back and forth. Many things can be made to vibrate, and when they do, they seem to do it with a period of motion that is unique to them. After all, how often do you think about your arms swinging as you walk? You don\u2019t \u2014 they seem to swing of their own accord and at their own frequency. The water in the bathtub will form very large waves when the child makes a back-and-forth motion at just the right rate. Any other rate won\u2019t create the waves that splash over the edge and soak the floor, which, of course, is the goal. In all these cases, the object vibrates at a natural frequency. Resonant frequency is the natural frequency of vibration of an object. In other words, objects that are caused to vibrate do so at a natural frequency that depends on the physical properties of the object. All objects that can vibrate have a resonant frequency, including a pendulum. Maintaining a Pendulum\u2019s Resonant Frequency A pendulum swings back and forth at its resonant frequency. Since the acceleration of gravity does not change if we stay in the same place, the only factor that affects the resonant frequency is the pendulum\u2019s length. All pendulums of the same length oscillate with the same natural (resonant) frequency. Huygens made use of this fact when he designed his pendulum clock (Figure 7.45). He knew that all pendulum clocks would keep the same time as long as the length of the pendulum arms was the same. Their resonant frequencies would be identical. However, Huygens faced some challenges in making a pendulum clock. The arm of the pendulum would expand or contract with temperature, affecting its period. But this was a relatively minor issue compared to another difficulty that had to be overcome \u2014 friction. Unless something was done, friction would very quickly stop the pendulum from swinging. To compensate for the effects of friction, he designed his clocks so that the pendulum was given a small push at just the right moment in its swing. The timing of these pushes coincided with the resonant frequency of the pendulum. By doing this, Huygens could make the pendulum swing for as long as the periodic force was applied. Figure 7.45 The interior of Huygens\u2019s clock Chapter 7 Oscillatory motion requires a set of conditions. 381 07-Phys20-Chap07.qxd 7/24/08 1:10 PM Page 382 forced frequency: the frequency at which an external force is applied to an oscillating object PHYSICS INSIGHT The forced frequency", " that is the same as the resonant frequency will increase the amplitude of the SHM, but will not change the resonant frequency. mechanical resonance: the increase in amplitude of oscillation of a system as a result of a periodic force whose frequency is equal or very close to the resonant frequency of the system Forced Frequency To visualize how this works, imagine a child on a swing. A swing is an example of a pendulum, with the child as the bob. The swing moves back and forth at its natural frequency, which depends only on its length. To keep the swing going with the same amplitude, all the parent has to do is push at just the right moment. The timing of the pushes must match the frequency of the swing. As anyone who has pushed a swing can attest, it takes very little energy to keep a swing swinging to the same height. The frequency at which the force is applied to keep the swing moving is called the forced frequency. If the forced frequency matches or is close to the resonant frequency of the object, then very little force is required to keep the object moving. The resonant frequency won\u2019t change though, because it depends only on the length of the pendulum. If the parent decides to push a little harder each time the swing returns, then the swing\u2019s amplitude will increase, but not its frequency. A larger force than is needed to overcome friction will create a larger amplitude of motion. If the forced frequency isn\u2019t close to the resonant frequency, then the object will not vibrate very much and will have a small amplitude. Imagine trying to increase the frequency of a pendulum by increasing the forced frequency. Much of the force won\u2019t be transferred to the pendulum because the pendulum won\u2019t be in the right position when the force is applied. The pendulum will bounce around but there will be no increase in its amplitude of vibration, and its motion will become harder to predict. The flowchart in Figure 7.46 on the next page summarizes the relationship between forced frequency and resonant frequency. Mechanical Resonance A forced frequency that matches the resonant frequency is capable of creating very large amplitudes of oscillation. This is referred to as mechanical resonance. This can be a good or bad thing. The larger the amplitude, the more energy the system has. Huygens\u2019s pendulum clock didn\u2019t need to have large oscillations, so a very small force could keep the pendulum swinging. A small weight-driven mechanism was used to provide the force needed. The force simply had to be applied with the same frequency as the pendulum. Huygens managed to do this without much difficulty. His pendulum clocks were a great success but weren\u2019t completely practical since they had to be placed on solid ground. A pendulum clock would not work aboard a ship because sailing ships of the time were buffeted by the waves more than today\u2019s large ocean-going vessels are. The motion of the ship on the waves would disturb a pendulum\u2019s SHM, so sailors could not take advantage of the increased accuracy these clocks provided. 382 Unit IV Oscillatory Motion and Mechanical Waves 07-Phys20-Chap07.qxd 7/24/08 1:10 PM Page 383 The key to successfully navigating across an ocean (where there are no landmarks) was to use an accurate clock on the ship. This clock could be synchronized to a clock in Greenwich, England, which is situated on the prime meridian of 0\u00b0 longitude. As the ship travelled east or west, the sailors could compare their local time, using the Sun and a sundial, to the ship\u2019s clock, which was still synchronized to the time on the prime meridian. The difference in time between the two clocks could be used to compute their longitudinal position. However, it wasn\u2019t until the 1700s that a brilliant clockmaker, John Harrison, successfully made a marine chronometer (ship\u2019s clock) that was immune to the buffeting of waves and temperature. It contained several ingenious innovations and, for better or worse, made the pendulum obsolete in navigation. NO The amplitude of vibration is not increased or may be decreased. Is the forced frequency close to the resonant frequency? YES A small force is required to keep the object vibrating with the same amplitude. The force is increased. The amplitude of vibration increases. Figure 7.46 Flowchart of the effect of forced frequency on resonant amplitude Chapter 7 Oscillatory motion requires a set of conditions. 383 07-Phys20-Chap07.qxd 7/24/08 1:10 PM Page 384 7-5 QuickLab 7-5 QuickLab Investigating Mechanical Resonance Problem How can we cause a pendulum to begin oscillating at its resonant frequency using a forced frequency? 8 Repeat step 7 three more times. Each time, lengthen the thread of mass 2 by 10 cm. For your last trial,", " the thread of mass 2 should be 40.0 cm long. Materials retort stands string thread 2 identical masses (200 g each) string thread mass 2 retort stand mass 1 two equal masses Figure 7.47 Procedure Part A 1 Read the questions in the next column before doing the lab. 2 Set up the two retort stands about 75 cm apart. 3 Tie the string to both retort stands at the same height (50.0 cm) on each stand, as shown in Figure 7.47. Clamp the ends of the string to the retort stands so they don\u2019t slip. The string should be taut. 4 Pull the two retort stands farther apart if you need to remove slack from the string. 5 Attach the thread to one mass and tie the other end to the string so that the distance from the mass to the string is 30 cm. This is mass 1. 6 Repeat step 5 for the second mass (mass 2) so that it has a length of 10 cm and is attached to the string about 15 cm from the first mass. 7 Make sure neither mass is moving, then pull mass 2 back a small distance and release it. Observe the motion of mass 1 as mass 2 oscillates. Make a note of the maximum amplitude that mass 1 achieves. Part B 9 Adjust the thread length of mass 2 so that it is as close to the thread length of mass 1 as possible. 10 Make sure both masses are motionless. Pull back and release mass 2. Note the amplitude of vibration that mass 1 achieves. 11 Pull the retort stands farther apart and hold them there so the tension in the string is increased, and the string is almost horizontal. 12 Make sure both masses are motionless, then pull back mass 2 and release it. Note the amplitude of vibration that mass 1 achieves. Questions Part A 1. At what thread length did mass 2 create the maximum oscillation of mass 1? Explain why this happened, in terms of frequency. 2. At what thread length did mass 2 create the minimum oscillation of mass 1? Explain why this happened, in terms of frequency. 3. Why did mass 1 have a large amplitude of vibration in only one case? Part B 4. What effect did increasing the tension on the string have on the amplitude achieved by mass 1? 5. Why did increasing the tension alter the maximum oscillation of mass 1? 6. Write a sentence describing the effect that increasing the tension had on the resonant amplitude of mass 1. Use the terms forced frequency and resonant amplitude in your answer. e LAB For a probeware lab, go to www.pearsoned.ca/school/physicssource. 384 Unit IV Oscillatory Motion and Mechanical Waves 07-Phys20-Chap07.qxd 7/24/08 1:10 PM Page 385 Resonance Effects on Buildings and Bridges A forced frequency that matches the resonant frequency can create problems for designers of bridges and skyscrapers. A bridge has a resonant frequency that can be amplified by the effect of wind. Air flows around the top and bottom of a bridge and can cause it to vibrate. The bridge will vibrate at its resonant frequency, with a large amplitude, even though the force applied by the wind may be relatively small. As the bridge vibrates, it may flex more than it is designed to and could conceivably vibrate to pieces. A skyscraper is also susceptible to forced vibrations caused by the wind. Most skyscrapers have a huge surface area and catch a lot of wind. Even though a building is a rigid structure, the force of the wind can make it sway back and forth. The wind causes a phenomenon called \u201cvortex shedding.\u201d It can create a forced vibration that matches the natural frequency of the building\u2019s back-and-forth vibration. The unfortunate result is to increase the sway (amplitude) of the building. The occupants on the top floors of the skyscraper will feel the effects the most. Over time, the continual large sway could weaken the building\u2019s structural supports and reduce its lifespan. Reducing Resonance Effects To counter resonance effects on bridges and buildings, engineers build them in such a way as to reduce the amplitude of resonance. Bridge designers make bridges more streamlined so that the wind passes over without imparting much energy. They also make bridges stiff, so a larger force is needed to create a large amplitude. The second-largest bridge in the world, the Great Belt East Bridge of Denmark is built with a smooth underside, like an airplane wing, that greatly increases its streamlined shape (Figure 7.48). It is not likely that a forced vibration would cause it to resonate. Skyscraper designers employ many strategies to lessen resonant vibration. One very effective approach is to use a large mass at the top of the building, called a \u201ctuned mass damper,\u201d which is free to oscillate back and forth (Figure 7.49). Controlled by computers, it can be", " made to vibrate at the resonant frequency of the building. When the building sways left, the mass moves right, and when the building sways right, the mass moves left. This has the effect of cancelling the vibration of the building. Any process that lessens the amplitude of an object\u2019s oscillations is referred to as \u201cdamping.\u201d Figure 7.48 The Great Belt East Bridge of Denmark is 6.8 km long and is constructed with a smooth underside. This allows air to flow by without inducing a resonant frequency. Chapter 7 Oscillatory motion requires a set of conditions. 385 07-Phys20-Chap07.qxd 7/24/08 1:10 PM Page 386 building movement cables damper movement Figure 7.49 The Taipei 101 building in Taiwan was completed in 2004 and stands 101 stories high. The inset shows a tuned mass damper in the building designed by Motioneering Inc. of Guelph, Ontario. It has a huge mass and vibrates opposite to the direction of the building, cancelling much of the amplitude of the resonant vibration. THEN, NOW, AND FUTURE Stressed-out Airplanes Ask any mechanical engineers, and they will tell you the importance of designing equipment to minimize vibration. Vibration causes excess wear on parts and stress on materials. Nowhere is this more evident than on an airplane. Yancey Corden knows this better than most people. Yancey is an aircraft maintenance engineer, and one of his jobs is to inspect aircraft for excess metal fatigue. Yancey was born on the Pavilion Reserve in south central British Columbia but grew up north of Williams Lake. His father maintained their car, boat, and other equipment around the home. Yancey watched and helped his father, and during this time, his interest in mechanics grew. Not long after finishing high school, Yancey enrolled in the aircraft maintenance engineer program at the British Columbia Institute of Technology located in Burnaby. He is now qualified with an M1 certification, which allows him to work on planes under 12 500 kg, and an M2 certification, which allows him to work on larger planes. He received specialized training in structural maintenance. This shift of force. This happens because the airplane springs upward due to the lighter load, and as a result, the wings tend to flutter up and down. This vibration causes stress fractures on the wing, and it is Yancey\u2019s job to find them. If a problem is found, Yancey designs the solution. This could involve fabricating a new part or simply fixing the existing one. He enjoys his job because each day is different and brings new challenges. He is thankful that he had the foresight to maintain good marks when he went to high school because the physics and science courses he took directly applied to his training. He is very proud of his heritage but he also believes it is important to focus on who you are and where you are going. Questions 1. What factors contribute to metal fatigue on a firefighting airplane? 2. What steps must be taken to gain a licence as an aircraft maintenance engineer? 3. To what factors does Yancey attribute his success? Figure 7.50 Yancey Corden involves an in-depth knowledge of the skin and frame of the plane. In 2003 Yancey moved to Alberta where he works in Red Deer for a company called Air Spray. Air Spray maintains and repairs Lockheed L-188 airplanes. These planes were originally manufactured as passenger craft in the 1950s, but because of their rugged design, they have been converted to firefighting aircraft today. They carry over 10 000 kg of water and fire retardant and are capable of dumping the entire amount in three seconds. When a dump of water and fire retardant occurs, the wings and airframe of the plane undergo a huge 386 Unit IV Oscillatory Motion and Mechanical Waves 07-Phys20-Chap07.qxd 7/24/08 1:10 PM Page 387 Quartz Clocks The technology of clock design and manufacture has taken huge leaps since the 1600s when Huygens built his first pendulum clock. Today, quartz clocks are the most accurate timepieces commercially available. They have an accuracy of about 1/2000 of a second a day. A quartz clock works on the principle of resonance. Inside each quartz clock is a tiny crystal of quartz. Quartz is a mineral that naturally forms into crystals. It also has a property unique to just a handful of materials: it will bend when a voltage is applied to it. If a pulse of voltage is applied to it, the crystal will begin to vibrate at its resonant frequency, just as a cymbal vibrates when hit by a drumstick. Once the quartz crystal is set vibrating, the circuitry of the clock times successive voltage pulses to synchronize with the frequency of the crystal. The synchronized voltage provides the forced frequency to keep the crystal oscillating just as the pendulums of Huygens", "\u2019s clocks needed a synchronized forced frequency to keep them from running down. The difference is that the pendulum clock receives the forced frequency through mechanical means, while the quartz crystal clocks get the forced vibration from electrical means. Resonant Frequency of a Quartz Crystal The crystal\u2019s resonant frequency depends on its size and shape and is not affected significantly by temperature. This makes it ideal for keeping time. As the crystal gets larger, more voltage is required to make it oscillate, and its resonant frequency decreases. A piece of quartz could be cut to oscillate once every second, but it would be far too large for a wristwatch and would require a large voltage to operate. If the crystal size is decreased, less voltage is required to make it oscillate. info BIT A substance that deforms with an applied voltage is called a piezoelectric material. A piezoelectric material will also create a voltage if stressed. Some butane lighters use a piezoelectric material to create a flame. As the lighter trigger is pressed, butane gas is released and the piezoelectric material undergoes stress. The piezoelectric material creates a voltage that causes a spark to jump a very small gap at the end of the lighter, igniting the butane. e WEB Atomic clocks keep time extremely precisely. Do they use a principle of resonance to keep such accurate time? Begin your search at www.pearsoned.ca/school/ physicssource. Quartz crystals are cut to a size and shape small enough to fit into a watch and use a small voltage (Figure 7.51). In most of today\u2019s quartz watches, the crystal vibrates with a resonant frequency of about 30 kHz and operates at 1.5 V. A small microprocessor in the watch combines these oscillations to make one oscillation per second so the watch can display time in a meaningful way. The topic of resonant frequencies is large and can\u2019t possibly be fully covered in this unit. You will learn more about resonance in musical instruments in Chapter 8. Figure 7.51 The quartz crystal in a wristwatch is enclosed in the small metal cylinder (lower right). Chapter 7 Oscillatory motion requires a set of conditions. 387 07-Phys20-Chap07.qxd 7/24/08 1:10 PM Page 388 12. What factors affect the resonance of a quartz crystal? 13. (a) What are two advantages of a quartz clock over a pendulum clock? (b) Are there any disadvantages of a quartz clock compared with a pendulum clock? Extensions 14. Use the knowledge you have gained about the design of a pendulum clock and the equation for its period in section 7.3 to answer the following question. What would the length of the pendulum\u2019s arm have to be so that it would oscillate with a resonant frequency of 1.00 Hz in Alberta (g 9.81 N/kg)? Under what conditions would it be most accurate? 15. Use your local library or the Internet to find out what automobile manufacturers do to reduce resonant frequencies in cars. 16. Investigate other methods not mentioned in the text that bridge designers use to lessen resonant vibrations. 17. Tuned mass dampers are not just used on buildings; cruise ships also have them. Explain why a cruise ship might have them and how they would be used. 18. Use your local library or the Internet to explore orbital resonance. In one or two paragraphs, explain how it applies to Saturn\u2019s rings. e TEST To check your understanding of applications of simple harmonic motion, follow the eTest links at www.pearsoned.ca/school/physicssource. 7.4 Check and Reflect 7.4 Check and Reflect Knowledge 1. What provides the force necessary to start a building or bridge oscillating? 2. What is forced frequency? 3. Explain what engineers use to reduce resonant vibrations of buildings and how these devices or structures work. 4. Explain the effect of applying a force to a vibrating object with the same frequency. 5. Identify two limitations of Huygens\u2019s pendulum clock. 6. Can a pendulum clock built to operate at the equator have the same accuracy at the North Pole? Explain. 7. What is damping? Use an example in your explanation. Applications 8. How could a person walking across a rope bridge prevent resonant vibration from building up in the bridge? 9. An opera singer can shatter a champagne glass by sustaining the right musical note. Explain how this happens. 10. Tuning forks are Y-shaped metal bars not much bigger than a regular fork. They can be made to vibrate at a specific frequency when struck with a rubber hammer. A piano tuner uses tuning forks to tune a piano. Explain, in terms of resonance, how this might be done. 11. Students are asked to find ways to dampen or change the resonant frequency", " of a pendulum. Here is a list of their suggestions. Identify the ones that would work and those that would not. In each case, justify your answer. (a) Apply a forced frequency that is different from the resonant frequency. (b) Place the pendulum in water. (c) Increase the mass of the pendulum bob. (d) Move the pendulum to a higher altitude. 388 Unit IV Oscillatory Motion and Mechanical Waves 07-Phys20-Chap07.qxd 7/24/08 1:10 PM Page 389 CHAPTER 7 SUMMARY Key Terms and Concepts period frequency oscillation cycle oscillatory motion Hooke\u2019s law spring constant restoring force simple harmonic motion simple harmonic oscillator resonant frequency amplitude forced frequency mechanical resonance Key Equations kx F vmax A k m T 2 m k T 2l g Conceptual Overview The concept map below summarizes many of the concepts and equations in this chapter. Copy and complete the map to have a full summary of the chapter. where the time for each interval is the Oscillatory Motion can produce which is the inverse of frequency measured in can take the form of simple harmonic motion mechanical resonance that obeys whereas written mathematically as forced frequency can cause an object to cycles/s F kx where F is the where x is the with a large displacement amplitude Figure 7.52 Chapter 7 Oscillatory motion requires a set of conditions. 389 07-Phys20-Chap07.qxd 7/24/08 1:10 PM Page 390 CHAPTER 7 REVIEW Knowledge 1. (7.1) What is oscillatory motion? Use an example 16. Determine the spring constant from the following graph: in your answer. 2. (7.1) Under what conditions must a ball be bounced so it has oscillatory motion? 3. (7.2) What is the defining property of an elastic material? 4. (7.2) What force, or forces, act on an isolated, frictionless simple harmonic oscillator? 5. (7.2) State the directional relationship that exists between the restoring force and displacement of a simple harmonic oscillator. 6. (7.2) What quantity does the slope of a force- displacement graph represent? 7. (7.2) What can be said about a pendulum\u2019s position if the restoring force is a non-zero value? 8. (7.3) Why isn\u2019t acceleration uniform for a simple harmonic oscillator? 9. (7.3) Why is it acceptable to consider a pendulum a simple harmonic oscillator for small displacements, but not for large displacements? 10. (7.4) If the forced frequency and the resonant frequency are similar, what effect does this have on an oscillator? Applications 11. Determine the restoring force acting on a 1.0-kg pendulum bob when it is displaced: (a) 15 (b) 5 12. Determine the frequency of a guitar string that oscillates with a period of 0.0040 s. 13. What is the period of a ball with a frequency of 0.67 Hz? 14. After a diver jumps off, a diving board vibrates with a period of 0.100 s. What is its frequency? 15. What is the restoring force on a 2.0-kg pendulum bob displaced 15.0? Force vs. Displacement 160 140 120 100 80 60 40 20 ) N ( e c r o F 0 0.2 0.4 0.6 0.8 Displacement (m) 17. A spring hangs from the ceiling in a physics lab. The bottom of the spring is 1.80 m from the floor. When the teacher hangs a mass of 100 g from the bottom of the spring, the spring stretches 50.0 cm. (a) What is its spring constant? (b) What force must a person apply to pull the 100.0-g mass on the bottom of the spring down through a displacement of 20.0 cm? (c) The 100.0-g mass is removed and a 300.0-g mass is attached. What is the distance of the mass above the floor? 18. Two different springs, A and B, are attached together at one end. Spring A is fixed to the wall as shown. The spring constant of A is 100.0 N/m and B is 50.0 N/m. What is the combined stretch of the two springs when a force of 25.0 N [right] is applied to the free end of spring B? A B wall 390 Unit IV Oscillatory Motion and Mechanical Waves 07-Phys20-Chap07.qxd 7/24/08 1:10 PM Page 391 19. Students stretch an elastic band attached to a force meter through several displacements and gather the following data. Use a graphing calculator or another acceptable method to plot the graph of this data and determine if", " the elastic band moves as predicted by Hooke\u2019s law. Displacement (cm) 0.00 10.0 20.0 30.0 40.0 50.0 Force (N) 0.00 3.80 15.2 34.2 60.8 95.0 20. How long must the arm of a pendulum clock be to swing with a period of 1.00 s, where the gravitational field strength is 9.81 N/kg? 21. What is the period of a 10.0-kg mass attached to a spring with a spring constant of 44.0 N/m? 22. Determine the maximum velocity of a 2.00-t crate suspended from a steel cable (k 2000.0 N/m) that is oscillating up and down with an amplitude of 12.0 cm. 23. A 0.480-g mass is oscillating vertically on the end of a thread with a maximum displacement of 0.040 m and a maximum speed of 0.100 m/s. What acceleration does the mass have if it is displaced 0.0200 m upwards from the equilibrium position? 24. Determine the period of oscillation of a pendulum that has a length of 25.85 cm. 25. An astronaut who has just landed on Pluto wants to determine the gravitational field strength. She uses a pendulum that is 0.50 m long and discovers it has a frequency of vibration of 0.182 Hz. What value will she determine for Pluto\u2019s gravity? 26. A student is given the relationship for a pendulum: T 2X (a) What does X represent? (b) The student records the period of the pendulum and finds it is 1.79 s. What is the pendulum\u2019s length? Extensions 27. A spring (k 10.0 N/m) is suspended from the ceiling and a mass of 250.0 g is hanging from the end at rest. The mass is pulled to a displacement of 20.0 cm and released. (a) What is the maximum velocity of the mass? (b) What is the period of oscillation of the mass if it is displaced 15.0 cm and released? 28. A horizontal mass-spring system has a mass M attached to a spring that oscillates back and forth at a frequency of 0.800 Hz. Determine the frequency in the following cases. (a) The mass is doubled. (b) The amplitude is tripled. 29. Identify which of the following examples is SHM and which is not. Explain. (a) a bouncing ball (b) a hockey player moving a puck back and forth with his stick (c) a plucked guitar string Consolidate Your Understanding Create your own summary of oscillatory motion, simple harmonic motion, restoring force, and mechanical resonance by answering the questions below. If you want to use a graphic organizer, refer to Student References 4: Using Graphic Organizers on pp. 869\u2013871. Use the Key Terms and Concepts listed above and the Learning Outcomes on page 342. 1. Prepare a quick lesson that you could use to explain Hooke\u2019s law to a peer using the following terms: restoring force, displacement, linear relationship. 2. Construct a two-column table with the title \u201cMass-spring System.\u201d The first column has the heading, \u201cFactors Affecting Period\u201d and the second column has the heading, \u201cFactors Not Affecting Period.\u201d Categorize the following factors into the two columns: mass, spring constant, amplitude, restoring force, velocity. Think About It Review your answers to the Think About It questions on page 343. How would you answer each question now? e TEST To check your understanding of oscillatory motion, follow the eTest links at www.pearsoned.ca/school/physicssource. Chapter 7 Oscillatory motion requires a set of conditions. 391 08-PearsonPhys20-Chap08 7/24/08 2:22 PM Page 392 C H A P T E R 8 Key Concepts In this chapter you will learn about: mechanical waves \u2014 longitudinal and transverse universal wave equation reflection interference acoustical resonance Doppler effect Learning Outcomes When you have finished this chapter, you will be able to: Knowledge describe how transverse and longitudinal waves move through a medium explain how the speed, wavelength, frequency, and amplitude of a wave are related describe how interference patterns can be used to determine the properties of the waves explain the Doppler effect describe the difference between transverse and longitudinal waves describe how waves are reflected explain the relationship between rays and waves apply the universal wave equation to explain how frequency, wavelength, and wave velocity are related explain the effects of constructive and destructive interference Science, Technology, and Society explain that the goal of technology is to provide solutions to practical problems 392 Unit IV Mechanical waves transmit energy in a variety of ways. Figure 8.1 What do bats and dolphins", " have in common? The phrase \u201cblind as a bat\u201d states a common fallacy. Bats have some vision using light, but when placed in pitch-black rooms crisscrossed with fine wires, they can easily fly around and unerringly locate tiny flying insects for food. Dolphins have shown that they can quickly locate and retrieve objects even when they are blindfolded. We usually assume that vision requires light but both bats and dolphins have evolved the ability to \u201csee\u201d using sound waves. Research in science and technology has developed \u201ceyes\u201d that enable humans also to see using sound waves, that is, navigate with senses other than sight. Medicine uses ultrasound (frequencies above the audible range) to look at objects such as a fetus or a tumour inside the body. Submarines can circumnavigate the globe without surfacing by using sound waves to explore their underwater environment. In Chapter 6, you studied how mass transfers energy when it moves through space. Waves, on the other hand, are able to transmit vast quantities of energy between two places without moving any mass from one location to another. Radio waves carry information, sound waves carry conversations, and light waves provide the stimulus for the cells that enable vision. This chapter introduces you to the nature and properties of waves. By experimenting with various forms of wave motion, you will learn about this common, but often misunderstood method of energy transmission. 08-PearsonPhys20-Chap08 7/24/08 2:22 PM Page 393 8-1 QuickLab 8-1 QuickLab Fundamental Properties of Wave Motion Problem To determine properties of waves in a ripple tank. Materials ripple tank and apparatus for its operation dowel ( 1.5 cm in diameter) 2 stopwatches ruler, two paper clips light and stand to project waves onto screen screen (a large sheet of newsprint works well) Procedure 1 Set up the ripple tank as shown in Figure 8.2. The water should be about 1 cm deep. Make sure that energy-absorbing buffers are placed around the edge of the tank to prevent unwanted reflections. Check your assembly with your instructor. light source Figure 8.2 paper screen 2 (a) Place a tiny spot of paper in the middle of the ripple tank. (b) Dip the end of your finger once into the water about the middle of the ripple tank to create a single, circular, wave front. Observe the speck of paper as the wave front passes it. 3 (a) On the screen, place the two paper clips at a measured distance apart, 30\u201340 cm. (b) Position your finger so that its shadow is over one of the paper clips and generate another single wave front. (c) Using a stopwatch, measure the time for the wave to travel from one paper clip to the other. Record the distance and time. Calculate the speed of the wave. Do a few trials for accuracy. 4 (a) Place the dowel horizontally in the water near one edge of the tank. Tap the dowel gently and observe the wave front. Sketch and describe the motion. (b) Position the paper clips in the wave\u2019s path and measure the speed of the straight wave front. CAUTION: Use care with ripple tanks. It is easy to break the glass bottom or to spill water. This is a serious hazard in an area where electricity is being used. Vibrating the tank will generate unwanted waves. Questions 1. When a wave front passes the speck of paper, what motion does the paper make? Does it move in the same or the opposite direction to the motion of the wave front? What does that tell you about the motion of the water as the wave moves through it? 2. On your sketches, draw several vector arrows along the fronts to indicate the direction in which they are moving. What is the angle between the line of the wave front and its motion? In Procedure 4(a), what is the angle between the edge of the dowel and the direction of the motion of the wave front? Sketch what you observe. Describe the motion. 3. Which wave front moves faster, the circular wave front or the straight wave front? Think About It 1. What differences and similarities are there between the ways energy is transmitted by waves and by matter? e SIM Find out more about waves in ripple tanks. Go to www.pearsoned.ca/school/ 2. What assumptions must be made to use water waves as a model for physicssource. sound waves? Discuss your answers in a small group and record them. As you complete each section of this chapter, review your answers to these questions. Note any changes in your ideas. Chapter 8 Mechanical waves transmit energy in a variety of ways. 393 08-PearsonPhys20-Chap08 7/24/08 2:23 PM Page 394 8.1 The Properties of Waves info BIT In December of 2004, an earthquake near the island of Sumatra set off a tsunami that", " is estimated to have had more than 2 petajoules (1015 J) of energy. This tsunami, the most powerful in recorded history, took over 225 000 lives and did untold billions of dollars in damage to the economies and the environments of the countries that border on the Indian Ocean. info BIT On the day of the tsunami of 2004 that devastated Phuket, Thailand, people travelling in a ferry in deep water offshore from Phuket felt only a greater than usual swell as the wave passed them by. Figure 8.3 Surfers use a wave\u2019s energy to speed their boards across the water. medium: material, for example, air or water through which waves travel; the medium does not travel with the wave wave: disturbance that moves outward from its point of origin, transferring energy through a medium by means of vibrations equilibrium position: rest position or position of a medium from which the amplitude of a wave can be measured crest: region where the medium rises above the equilibrium position trough: region where the medium is lower than the equilibrium position When a surfer catches a wave, many people assume that the forward motion of the surfer is the result of the forward motion of the water in the wave. However, experimental evidence indicates that in a deep-water wave the water does not, in general, move in the direction of the wave motion. In fact, the surfer glides down the surface of the wave just as a skier glides down the surface of a ski hill. Like the skier, the surfer can traverse across the face of the hill as well as slide down the hill. But, unlike the ski hill, the water in the wave front is constantly rising. So, even though the surfer is sliding down the front of the wave he never seems to get much closer to the bottom of the wave. It is a common misconception that the water in a wave moves in the direction in which the waves are travelling. This may be because waves arriving at the shoreline move water to and fro across the sand. As you will see, this movement is a feature of the interaction of the wave with the sloping shoreline rather than the actual motion of the wave itself. In deep water, there is only very limited lateral motion of water when a wave moves past a particular point. 394 Unit IV Oscillatory Motion and Mechanical Waves 08-PearsonPhys20-Chap08 7/24/08 2:23 PM Page 395 Waves and Wave Trains When a stone is thrown into a still pond or lake, a ripple moves outward in ever-enlarging concentric circles (Figure 8.4). The water is the transporting medium of the wave and the undisturbed surface of the water is known as the wave\u2019s equilibrium position. Regions where the water rises above the equilibrium position are called crests and regions where the water is lower than its equilibrium position are called troughs. In the crest or trough, the magnitude of greatest displacement from the equilibrium is defined as the waves\u2019 amplitude (A). A complete cycle of a crest followed by a trough is called a wavelength; its symbol is the Greek letter lambda, (Figure 8.5). \u03bb A crest A equilibrium position \u03bb trough Figure 8.5 Properties of a wave wavelength wavelength A wave front moving out from the point of origin toward a barrier is called an incident wave. A wave front moving away from the barrier is called a reflected wave, while a series of waves linked together is a wave train. The concept of a wave train implies a regular repetition of the motion of the medium through which the wave travels. As a result, many parts of the medium are moving in a motion that is identical to the motion of other points on the wave train. At these points, the waves are said to be in phase (Figure 8.6). \u03bb B \u03bb D E A C \u03bb A and B are in phase C, D, and E are in phase Figure 8.6 In-phase points along a wave train have identical status relative to the medium and are separated by one wavelength, \u03bb. Instead of creating individual pulses by hand in a ripple tank, you may use a wave generator to create a continuous series of crests and troughs forming a wave train. Wave generators can act as a point source similar to the tip of a finger, or as a straight line source, similar to a dowel. In 8-1 QuickLab you measured the speed of a single pulse by observing its motion. However, because it is impossible to keep track of a single wave in a wave train, to measure the speed of a wave train requires a greater understanding of the properties of waves. crests crests troughs troughs Figure 8.4 Many of the terms used to describe wave motions come from the observation of waves on the surface of water. e WEB To learn more about experiments using ripple tanks, follow the links at www.pearsoned.ca/school/ physicssource. amplitude: the distance from", " the equilibrium position to the top of a crest or the bottom of a trough wavelength: the distance between two points on a wave that have identical status. It is usually measured from crest to crest or from trough to trough. wave front: an imaginary line that joins all points reached by the wave at the same instant incident wave: a wave front moving from the point of origin toward a barrier reflected wave: a wave front moving away from a barrier wave train: a series of waves forming a continuous series of crests and troughs point source: a single point of disturbance that generates a circular wave Chapter 8 Mechanical waves transmit energy in a variety of ways. 395 08-PearsonPhys20-Chap08 7/24/08 2:23 PM Page 396 8-2 Inquiry Lab 8-2 Inquiry Lab Wave Trains in a Ripple Tank, Part 1: Reflecting Waves In this ripple tank experiment, the properties of a twodimensional wave train are analyzed. Question How do the incident and reflected wave trains interact when wave trains reflect from a straight barrier? Materials and Equipment ripple tank, including the apparatus for its operation straight barrier wave generators (point-source and straight-line) light and stand to project waves onto screen screen (a large sheet of newsprint works well) Variables In this experiment you are to observe the directions of motion of the incident waves and reflected waves and how these directions are related to each other. Other variables to be observed are the interactions that occur when the incident and reflected wave trains move in different directions through the same point in the ripple tank. As you observe the wave motions you should identify which are the controlled variables, manipulated variables, and responding variables. General Procedure 1 (a) Set up the ripple tank as shown in Figure 8.7. (b) When using motorized wave generators, it is important that the generator just barely contacts the surface of the water. It should never touch the tank during operation. Check with your instructor to make sure that your apparatus is properly assembled. CAUTION: Use care with ripple tanks. It is easy to break the glass bottom or to spill water. This is a serious hazard in an area where electricity is being used. Vibrating the tank will generate unwanted waves that interfere with the desired observations. The wave generator should never touch the tank during operation. Required Skills Initiating and Planning Performing and Recording Analyzing and Interpreting Communication and Teamwork light source generator motor generator support point sources To power supply paper screen Figure 8.7 Procedure 1 (a) Place the point-source wave generator at one edge of the ripple tank and the straight barrier at the other edge. The shadow of both the barrier and the source should be visible on the screen. (b) Use the point-source wave generator to create a continuous wave train in the ripple tank. Observe what happens to the incident wave train when it meets the reflected wave train. (c) Make a sketch of your observations. Wave trains are a bit tricky to observe at first. Discuss what you see with your team members. When you have reached a consensus, write a brief description of your observations. On your sketch, place vector arrows along an incident and a reflected wave front to indicate the direction and speed of their motions. 2 (a) Set up the straight-line wave generator at one edge of the ripple tank. Place the barrier at the other edge parallel to the generator. The shadows of both the generator and the barrier should be visible on the screen. (b) Start the generator to create a continuous wave train. Observe what happens when the reflected wave train moves back through the incident wave train. Draw diagrams and write a description of the observations. Again, draw vector arrows along incident and reflected wave fronts to indicate their relative velocities. 396 Unit IV Oscillatory Motion and Mechanical Waves 08-PearsonPhys20-Chap08 7/24/08 2:23 PM Page 397 3 Move the barrier so that it is at an angle of about 30\u00b0 2. (a) When the barrier is parallel to the straight wave to the generator and repeat step 2 (b). 4 Set the barrier so that the angle between it and the generator is about 60\u00ba and repeat step 2 (b). Analysis 1. (a) When the incident wave train created by the point-source generator is passing through the reflected wave train, what happens to the waves in the region where they overlap? (b) Can you see the direction of the motion for both the incident and reflected wave trains? generator, what pattern do you observe when the reflected waves are moving back through the incident waves? (b) In which direction does the pattern seem to be moving? Can you see the direction of the motion for both the incident and reflected wave trains? 3. Answer question 2 for the set-up when the barrier is at an angle to the straight wave generator. 4. In all cases above, how does the spacing of the waves in the reflected wave train compare to the spacing of the waves in the incident wave train?", " Waves and Rays When waves in a ripple tank are viewed from above (Figure 8.8) the wave fronts appear as a set of bright and dark bands (crests and troughs). When we draw wave trains as seen from above, we use a line to represent a wave front along the top of a crest. The point halfway between two lines is the bottom of a trough. A series of concentric circles represents the wave train generated by a point source. ray: a line that indicates only the direction of motion of the wave front at any point where the ray and the wave intersect rays wavelength wave source crests troughs velocity vectors Figure 8.8 View of a ripple tank from above Figure 8.9 A point source generates waves that move outward as concentric circles with the source at their centre Waves are in constant motion. At all points on a wave front, the wave is moving at right angles to the line of the crest. There are two ways to indicate this (Figure 8.9). You could draw a series of vector arrows at right angles to the wave front with their length indicating the speed of the wave. Or, you could draw rays, lines indicating only the direction of motion of the wave front at any point where the ray and the wave front intersect. The rays in Figure 8.9 are called diverging rays since they spread out as they move away from the origin. When rays diverge, it indicates that the energy given to the wave at its source is being spread over a larger and larger area. This is why, as sound moves away from a point source such as a bell, the volume decreases with the square of the distance. e WEB To learn more about the mathematical relationship between the volume of sound and the distance from the source, follow the links at www.pearsoned.ca/ school/physicssource. Chapter 8 Mechanical waves transmit energy in a variety of ways. 397 08-PearsonPhys20-Chap08 7/24/08 2:23 PM Page 398 light rays trough crest water ripple tank bottom paper screen bright areas under crests dark areas under troughs Figure 8.10 Sketch of a wave showing light refracting through a crest and a trough When waves in a ripple tank are projected onto a screen below, the wave fronts appear as a set of bright and dark bands. It may seem logical that the light and dark bands seen on the screen below the ripple tank result from the differences in water depth between the crests and the troughs. But that difference is only about a millimetre and cannot account for the high contrast in light seen on the screen. In fact, a crest acts like a converging lens to concentrate the light, creating a bright bar. A trough acts like a diverging lens to spread the light out, making the area under the trough darker (Figure 8.10). You will learn more about light refraction in Unit VII of this course. Reflection of a Wave Front When a wave front is incident on a straight barrier, it reflects. The direction the wave travels after reflection depends on the angle between the incident wave front and the barrier. A circular wave front, as generated by a point source, S, produces the simplest reflection pattern to explain. In this case, the reflected wave follows a path as if it had been generated by an imaginary point source S, at a position behind the barrier identical to that of the actual point source in front of the barrier (Figure 8.11). Now consider an incident wave front created by a straight wave generator (Figure 8.12). The straight wave front also reflects as if the reflected wave had been generated by an imaginary generator located behind the barrier. The position of the imaginary generator behind the barrier is equivalent to the position of the real generator in front of the barrier. The incident wave front and the reflected wave front are travelling in different directions, but the angle between the incident wave front and the barrier must be identical to the angle between the reflected wave front and the barrier. e SIM Find out more about the ways waves reflect. Go to www.pearsoned.ca/school/ physicssource. S imaginary source rays from the imaginary source reflecting surface of barrier S real source imaginary incident wave generated by S reflected portion of incident wave incident wave generated by S imaginary straight wave generator reflected wave front real straight wave generator incident wave from imaginary generator reflecting surface of barrier incident wave front Figure 8.11 When circular waves reflect from a straight barrier, the reflected waves seem to be moving away from an imaginary source. Figure 8.12 When straight waves reflect from a straight barrier, the angle between the reflected wave front and the barrier must be equal to the angle between the incident wave front and the barrier. 398 Unit IV Oscillatory Motion and Mechanical Waves 08-PearsonPhys20-Chap08 7/24/08 2:23 PM Page 399 M I N D S O N Waves Can Have Curls Too When waves travel in deep water, their shape is similar to the waves in a ripple tank.", " But as waves near the shoreline they change shape and develop what is known as a curl in which the top of the wave falls in front of the wave. Recall Figure 8.3 on page 394. Explain the causes of a wave\u2019s curl in terms of its motion. 8-3 Inquiry Lab 8-3 Inquiry Lab Wave Trains in a Ripple Tank, Part 2: Wave Speed and Wavelength Required Skills Initiating and Planning Performing and Recording Analyzing and Interpreting Communication and Teamwork In this ripple tank experiment, the properties of a twodimensional wave train are further analyzed. Question What effect does a change in speed have on wave trains? Materials and Equipment ripple tank, including the apparatus for its operation wave generators (point-source and straight-line) light and stand to project waves onto screen screen (a large sheet of newsprint works well) two small blocks of wood about 8 mm thick Variables In this lab you will be observing how water depth affects the properties of waves. The variables that might be affected by changes in the depth are speed, frequency, wavelength, and direction. As you make your observations, consider which of the variables are controlled variables, manipulated variables, and responding variables. General Procedure 1 (a) Set up the ripple tank as shown in Figure 8.7 (page 396). (b) When using motorized wave generators, it is important that the generator just barely contacts the surface of the water. It should never touch the tank during operation. Check with your instructor to make sure that your apparatus is properly assembled. CAUTION: Use care with ripple tanks. It is easy to break the glass bottom or to spill water. This is a serious hazard in an area where electricity is being used. Vibrating the tank will generate unwanted waves that interfere with the desired observations. The wave generator should never touch the tank during operation. Procedure 1 Place small pads (about 8 mm thick) under the legs along one edge of the ripple tank so that the water gets shallower toward that edge. The water should be less than 1 mm deep at the shallow edge. 2 (a) Near the deep edge, create a wave train using the point-source generator. (b) Observe what happens to the wave fronts as the wave train moves toward the shallow edge. Discuss your observations with your team members. Sketch and briefly describe your observations. (c) Place vector arrows along several of the wave fronts to indicate the direction and speed of the wave fronts as they move into shallow water. 3 (a) Set up the straight-line wave generator at the deep edge of the tank. (b) Turn on the generator and observe the wave train as it moves into the shallow water. (c) Sketch your observations and describe the motion of the wave train as it moves into shallow water. Use vector arrows along the wave fronts to assist you in your descriptions. 4 (a) Now place the pads under the legs of the ripple tank on an edge that is at a right angle to the position of the straight-line wave generator. (b) Use the straight-line wave generator to create a wave train. (c) Observe the wave train as it travels across the tank. Discuss your observations with your team members. Sketch and write a brief description of what you saw to accompany your sketch. Use vector arrows drawn along several of the wave fronts to indicate their relative velocity as they move into the shallow water. Chapter 8 Mechanical waves transmit energy in a variety of ways. 399 08-PearsonPhys20-Chap08 7/24/08 2:23 PM Page 400 Analysis 2. What do you think causes the observed changes? 1. For each of the trials, when the waves moved from deep to shallow edge (or vice versa), comment on the kinds of changes you observed. 3. When the straight wave fronts moved across the tank at right angles to the change in the depth of the water, was the shape of the wave front affected? (a) Were the wavelengths of the incident waves affected as they moved into shallow water? 4. What properties of waves are affected as the waves move from deep to shallow water? (b) Was the shape of the wave fronts affected as they entered shallow water? (c) If so, how did the shape of the wave fronts change as they changed speed? 5. When a water wave moves toward a beach, how would the change in the depth of the water affect the motion of the wave? 8.1 Check and Reflect 8.1 Check and Reflect Knowledge 1. If a wave pattern is created by a point source, what is the nature of the ray diagram that would represent the wave fronts? 2. When a wave front reflects from a barrier, what is the relationship between the direction of the motions of the incident and reflected wave fronts? Applications 3. The sketch shows a ray diagram that represents the motion of a set of wave fronts. If you were observing these wave fronts in a ripple tank, describe what you would see. rays 4. Draw a", " diagram of a set of straight wave fronts that are incident on a straight barrier such that the angle between the wave fronts and the barrier is 40\u02da. Draw the reflected wave fronts resulting from this interaction. How do the properties (speed, wavelength, and amplitude) of the reflected wave compare with the properties of the incident wave? Use a wavelength of about 1 cm in your diagram. Extensions 5. Reflection of light is the essence of how we use mirrors to see images. What does the reflection of waves in a ripple tank tell you about the formation of images? Hint: Think of where the reflected waves in the ripple tank seem to originate. 6. When a sound travels in water, the speed of the sound depends on the temperature of the water. If the sonar ping emitted by a submarine has a wavelength of 2.50 m, what happens to that wavelength when it enters a region where sound travels faster? e TEST To check your understanding of the properties of waves, follow the eTest links at www.pearsoned.ca/ school/physicssource. 400 Unit IV Oscillatory Motion and Mechanical Waves 08-PearsonPhys20-Chap08 7/24/08 2:23 PM Page 401 8.2 Transverse and Longitudinal Waves Did you ever have a Slinky\u2122 toy when you were a child? When a Slinky\u2122 is stretched out along the floor and oscillated from side to side across its axis (centre line), or forward and back along its axis, mechanical waves are transmitted along its length. The sideways oscillations set up a transverse wave while those along the axis set up a longitudinal wave as shown in Figure 8.13. In this section we will consider the characteristics of such waves. Transverse Pulses info BIT The ever-popular Slinky\u2122 was invented in 1945 by Richard James, a naval engineer working on tension springs. The name comes from the Swedish for \u201csleek\u201d or \u201csinuous.\u201d Each Slinky\u2122 is made from 80 feet (24.384 m) of wire. A pulse moving through a spring is a good introduction to the way a wave moves through a medium. An ideal spring is one that allows a pulse to travel through it without loss of energy. By definition, a pulse is just the crest or the trough of a wave; its length is one-half a wavelength. The spring provides a medium in which the motion of a pulse can be observed from the side. Initially, the spring is in its equilibrium position. When you flip the spring sharply to the side and back, the motion of your hand sets up a series of sequential motions in the coils of the spring. Each coil imitates, in turn, the motion of the hand. This results in a transverse pulse (Figure 8.14) that moves along the spring. As a pulse moves along a spring, the coils of the spring move at right angles to the direction of the pulse\u2019s motion. Compare v hand and v pulse in Figure 8.14. At the front of the pulse, the coils are moving away from the spring\u2019s equilibrium position toward the point of maximum displacement from the equilibrium. In the trailing edge of the pulse, the coils are moving back toward the equilibrium position. Hand starts at equilibrium position of spring. Front of pulse starts to move along spring. Hand is at maximum amplitude. Hand continues to move up. vhand 0 vpulse vhand 0 v A Hand continues to move down. Figure 8.13 (a) A transverse pulse (b) A longitudinal pulse Arrows indicate the direction of the medium. The pulses are moving through the springs toward the bottom of the page. pulse: a disturbance of short duration in a medium; usually seen as the crest or trough of a wave e WEB To learn more about the forces operating in an oscillating spring, follow the links at www.pearsoned.ca/school/ physicssource. amplitude A As the hand moves toward the equilibrium position, the amplitude of the pulse moves along the spring. Pulse is complete when the hand is at equilibrium position. Figure 8.14 When you move your hand you set up a sequence in which the coils of the spring imitate the motion of your hand. This creates a moving pulse. l vhand 0 Chapter 8 Mechanical waves transmit energy in a variety of ways. 401 08-PearsonPhys20-Chap08 7/24/08 2:23 PM Page 402 Energy Changes During the Movement of a Pulse Along the pulse, energy is stored in the form of both elastic potential energy and kinetic energy. As a section of the spring moves from the equilibrium position to the top of the pulse, that section has both kinetic energy (it is moving sideways relative to the direction of the pulse) and elastic potential energy (it is stretched sideways). At the point on the pulse where the displacement is greatest the coils are, for an instant, motionless. Then, the tension in the spring returns the coils to their equilibrium", " position. No blurring. This indicates: Ek Ep 0 maximum 0 0 Ek Ep Figure 8.15 A transverse pulse is generated when a spring is given a sharp flip to the side. Arrows indicate the direction of motion of the coils. Can you determine which way the pulse is moving? In Figure 8.15 the blurring on the front and back segments of the pulse indicates the transverse motion and the presence of kinetic energy as well as elastic potential energy. At the top, there is no blurring as the coils are temporarily motionless. At that instant that segment of the spring has only elastic potential energy. As it returns to its equilibrium position, the segment has, again, both kinetic and potential energies. The energy in a pulse moves along the spring by the sequential transverse motions of the coils. Recall from section 6.3 that a pendulum, along the arc of its path, has both kinetic and potential energy, but at the point where the pendulum\u2019s displacement is greatest, all the energy is in the form of potential energy. Thus, the energy of an oscillating pendulum is equivalent to its potential energy at the point where its displacement is greatest. Similarly, the amplitude of the wave in an experimental spring can be used to determine the quantity of energy that is stored in the pulse. Concept Check You generate a pulse in a Slinky\u2122 stretched out on the floor. If you wish to, you could give the next pulse more energy. How would you do that? info BIT When considering sound, amplitude determines loudness. 402 Unit IV Oscillatory Motion and Mechanical Waves 08-PearsonPhys20-Chap08 7/24/08 2:23 PM Page 403 8-4 Inquiry Lab 8-4 Inquiry Lab Pulses in a Spring, Part 1: Pulses in an Elastic Medium Required Skills Initiating and Planning Performing and Recording Analyzing and Interpreting Communication and Teamwork In this experiment, you will study how a pulse moves through a medium. Question What are the mechanics by which pulses move through a medium? Variables The measured properties of a pulse include its amplitude (A), pulse length (l ), period (t), and speed (v). 5 On your sketch of the pulse, label the following parts: \u2014 The amplitude (A) is the perpendicular distance from the equilibrium position of the spring to the top of the pulse. \u2014 The pulse length (l) is the distance over which the spring is distorted from its equilibrium position. NOTE: When you stand at the side of the spring, the pulse seems to move past you very quickly, almost as a blur. Watching from the end of the spring may make it easier to observe the details of the motion. Materials and Equipment 6 Make a longitudinal pulse by moving your hand light spring metre-stick or measuring tape stopwatch masking tape CAUTION: A stretched spring stores considerable amounts of elastic potential energy. Be careful not to release the end of a spring while it is stretched. When collapsing a spring, have the person holding one end walk the spring slowly toward the other end. If you allow the spring to gently unwind as you are walking you will prevent the spring from tying itself into a knot. Procedure 1 Have one team member hold the end of the spring while another stretches it until it is moderately stretched (about 5\u20136 m). 2 Place strips of masking tape on the floor at either end of the spring to mark this length. Near the middle of the spring, attach a strip of tape about 5 cm long to one of the coils. 3 Have one of the people holding the spring generate a transverse pulse. Generate the pulse by moving your hand sharply to one side (about 60\u201375 cm) and back to its original position. This is a transverse pulse since its amplitude is perpendicular to the direction of its motion. 4 Sketch the pulse. Indicate the motions of the pulse and the coils using vector arrows. Observe the motion of the tape at the middle of the spring to assist in these observations. Generate more pulses until you understand the nature of the motion of the pulse in the spring. sharply toward the person holding the spring at the other end, and then back to its original position. This pulse is called a longitudinal pulse, because its amplitude is along the direction of its motion. Repeat the pulse a few times to determine the nature of the motion of the spring as the pulse moves through it. Sketch and describe the motion of the coils as the pulse moves along the spring. Analysis 1. What determines the amplitude of the transverse pulse? The longitudinal pulse? 2. Does the pulse change shape as it moves along the spring? If so, what causes the change in shape of the pulse? Would you expect the pulse shape to change if this were an isolated system? 3. How is the reflected pulse different from the incident pulse? If this were an isolated system, how would the reflected pulse differ from the incident pulse? 4. Describe the motion of the strip of tape at the middle of the", " spring as the pulse passes it. Does the tape move in the direction of the pulse? 5. How does the motion of the medium relate to the motion of the pulse? 6. How does the pulse transfer energy from one end of the spring to the other? 7. The motion of a pulse in the spring requires you to make assumptions about the motion of an ideal pulse. What assumptions must you make to create a model of how a transverse pulse moves through an elastic medium? Chapter 8 Mechanical waves transmit energy in a variety of ways. 403 08-PearsonPhys20-Chap08 7/24/08 2:23 PM Page 404 v Reflection of Pulses from a Fixed Point incident pulse reflected pulse Figure 8.16 Reflection from the fixed end of a spring causes the pulse to be inverted. e SIM Find out about similarities and differences between transverse and longitudinal waves. Go to www.pearsoned.ca/ school/physicssource. When a pulse (or wave) is generated in a spring it soon arrives at the other end of the spring. If that end is held in place, the total pulse reflects from the end and travels back toward the source. The reflected pulse is always inverted relative to the incident pulse (Figure 8.16). In an ideal medium, the other properties of the pulse (amplitude, length, and speed) are unaffected by reflection. These properties of the reflected pulse are identical to those of the incident pulse. When a wave train is generated in the spring, the crests of the incident wave are reflected as troughs while the troughs of the incident wave are reflected as crests. v Longitudinal Waves If, instead of moving your hand across the line of the spring, you give the spring a sharp push along its length, you will observe that a pulse moves along the spring. This pulse is evidence of a longitudinal wave. The pulse is seen as a region where the coils are more tightly compressed followed by a region where the coils are more widely spaced. These two regions are called, respectively, a compression and a rarefaction and correspond to the crest and trough in a transverse wave. In the case of a longitudinal wave, the coils of the spring oscillate back and forth parallel to the direction of the motion of the wave through the medium (Figure 8.17). But, as with transverse waves, once the wave has passed through, the medium returns to its original position. Once again, energy is transmitted through the medium without the transmission of matter. Figure 8.17 Longitudinal waves are formed when the source oscillates parallel to the direction of the wave motion. spring in equilibrium position Project LINK What aspect of your seismograph will relate to the ideas of compression and rarefaction in a longitudinal wave? longitudinal wave in the spring vwave vcoils vcoils vcoils vcoils rarefaction rarefaction compression compression wavelength 404 Unit IV Oscillatory Motion and Mechanical Waves 08-PearsonPhys20-Chap08 7/24/08 2:23 PM Page 405 8-5 Inquiry Lab 8-5 Inquiry Lab Pulses in a Spring, Part 2: Speed, Amplitude, and Length Required Skills Initiating and Planning Performing and Recording Analyzing and Interpreting Communication and Teamwork In this experiment, you will study the speed, amplitude, and length of pulses. You will set up the experiment similarly to 8-4 Inquiry Lab on page 403. Question What is the relationship between the amplitude, length, and speed of a pulse? Variables The measured properties of a pulse include its amplitude (A), pulse length (l ), period (t), and speed (v). Materials and Equipment light spring stopwatch metre-stick or measuring tape masking tape Procedure 1 (a) Measure the speed of a transverse pulse as it moves along the spring. Have the person creating the pulse \u201ccount down\u201d so that team members with stopwatches can time the pulse as it moves toward the other end. Measure the time from the instant the front edge of the pulse leaves the hand of the person generating it until the front edge arrives at the hand at the other end. Do this a few times to establish a consistent value. Record your results. Use the time and the distance between the hands to calculate the speed of the pulse. (b) Generate pulses by moving your hand to the side and back at different speeds (more quickly or more slowly). Measure the speed of each of these pulses. 2 Have the person holding one end of the spring move so that the spring is stretched about 1 m farther. (Do not overstretch the spring.) Generate a pulse and measure the speed of the pulse in the spring at this higher tension. Carefully walk the spring back to the length used initially. 3 Make a transverse pulse by moving your hand a different distance sideways. Try to keep the time used to make the pulse the same as before. Repeat this a few times to observe changes in the pulse", ". Record your observations. 4 Now make several transverse pulses by moving your hand to a given amplitude but change the speed at which you move. Repeat a few times and record your observations. Analysis 1. Does the speed at which you moved your hand to generate a pulse affect the speed of the pulse? 2. When the spring was stretched to a greater length, what happened to the speed of the pulse? 3. What controls the amplitude (A) of the pulse? Can you create pulses with equal lengths but different amplitudes? 4. What controls the length (l ) of the pulse? Can pulses of equal amplitudes have different lengths? 5. What is the relationship between the length of the pulse and the speed (v) of the pulse in the medium? 6. Does the length of a pulse affect its larger amplitude or vice versa? Explain why or why not. 7. Does the energy in a pulse seem to depend on its amplitude or its length? Give reasons for your decision. Consider what changes occur as the pulse moves through the spring. 8. What determines the speed, the length, and the amplitude of the pulse? 9. What aspect of wave motion in water can you simulate by changing the tension in a spring? 10. What do your findings for the relationship of the amplitudes and lengths of pulses in springs tell you about the relationship between the amplitudes and wavelengths of waves in water? 11. Sound is often referred to as a wave. What aspect of a sound would relate to (a) the amplitude and (b) the wavelength of its waves? Chapter 8 Mechanical waves transmit energy in a variety of ways. 405 08-PearsonPhys20-Chap08 7/24/08 2:23 PM Page 406 Pulse Length and Speed Figure 8.18 The length (l) of the pulse depends on the speed (v) of the pulse and the time (t) taken to complete the pulse. e WEB To learn more about the way the structures of the human ear transfer sound waves, follow the links at www.pearsoned.ca/school/ physicssource. info BIT When a wave moves across the surface of water, the water moves between crests and troughs by localized circular motions. This local circular motion moves water back and forth between a trough and the adjacent crest. direction of motion of surface vwave vmedium The speed of the pulse depends on the medium. If you stretch the spring so that the tension increases, then the speed of the pulse increases. Relaxing the tension causes the speed to decrease. The speed of the pulse in the spring also determines the length of the pulse. vhand 0 Hand starts at equilibrium position of spring. vhand 0 v A Front of pulse starts to move along spring. vpulse Hand is at maximum amplitude. amplitude A Pulse is complete when the hand is at equilibrium position. l vhand 0 The instant you start to move your hand to generate a pulse, the disturbance begins to move along the spring at a constant speed, v. Assume that the time it takes to move your hand to create the complete pulse is t. By the time your hand returns the spring to its equilibrium position, the front of the pulse will have travelled a distance d, which can be defined as the length l of the pulse (Figure 8.18). Remember d v and, therefore t d vt l vt Waves and the Medium A solid such as a spring is an elastic medium and can store elastic potential energy by stretching longitudinally or transversely. Typically, the way that fluids (liquids and gases) store elastic potential energy is by being compressed. Therefore, waves within fluids are typically longitudinal waves, known as pressure waves. This is the principle used in engines and aerosol sprays. As compressions and rarefactions move through a fluid, the motion of the molecules in the fluid is very similar to the motion of the coils when a longitudinal wave moves through a spring. For water to transmit energy as a transverse wave, the waves must be displaced vertically, but in liquids the vertical displacement cannot be a form of elastic potential energy. Thus, transverse waves can be transmitted only at the surface of water, or other liquids, where the waves are the result of gravitational potential energy rather than elastic potential energy. motion of water within wave Figure 8.19 406 Unit IV Oscillatory Motion and Mechanical Waves 08-PearsonPhys20-Chap08 7/24/08 2:23 PM Page 407 M I N D S O N Wave Motion in Fluids Water can transmit both transverse (surface) waves and longitudinal (internal) waves such as sound. We know that sound waves in gases are longitudinal waves. Is it possible to create a transverse wave in a gas? Why or why not? Consider how transverse waves are created in liquids. Example 8.1 To create a pulse in a fixed ideal spring, you move your hand sideways a distance of 45 cm from the equilibrium position. It takes 0.80 s from the time you", " begin to move your hand until it returns the spring to its equilibrium position. If the pulse moves at a speed of 2.5 m/s, calculate the length of the pulse and describe the incident pulse and reflected pulse that pass through the midpoint of the spring. Practice Problems 1. A pulse is generated in a spring where it travels at 5.30 m/s. (a) If the time to generate the pulse is 0.640 s, what will be its length? (b) How does the speed of the pulse affect its amplitude? 2. A pulse moves along a spring at a speed of 3.60 m/s. If the length of the pulse is 2.50 m, how long did it take to generate the pulse? 3. A pulse that is 1.80 m long with an amplitude of 0.50 m is generated in 0.50 s. If the spring, in which this pulse is travelling, is 5.0 m long, how long does it take the pulse to return to its point of origin? 4. A spring is stretched to a length of 6.0 m. A pulse 1.50 m long travels down the spring and back to its point of origin in 3.6 s. How long did it take to generate the pulse? Answers 1. (a) 3.39 m (b) It does not; they are independent. 2. 0.694 s 3. 2.8 s 4. 0.45 s Given A 45 cm 0.45 m t 0.80 s v 2.5 m/s Required (a) length of the pulse (b) description of incident pulse passing the midpoint of the spring (c) description of reflected pulse passing the midpoint of the spring Analysis and Solution (a) The length of the pulse can be found using l vt. l vt (0.80 s) 2.5 m s 2.0 m (b) The spring is defined as an ideal spring, so the amplitude of the pulse is constant. The amplitude at all points on the spring will be the same as at the source. Therefore, A 0.45 m At the midpoint of the spring, the amplitude of the incident pulse is 0.45 m, its length is 2.0 m, and its speed is 2.5 m/s. (c) Reflection inverts the pulse but does not change any of its properties. The reflected pulse is identical to the incident pulse except that it is inverted relative to the incident pulse. Paraphrase and Verify (a) The length of the pulse is equal to 2.0 m. (b) In an ideal spring the amplitude of the pulse is constant. (c) When pulses are reflected from a fixed end of a spring they are inverted. Chapter 8 Mechanical waves transmit energy in a variety of ways. 407 08-PearsonPhys20-Chap08 7/24/08 2:23 PM Page 408 direction of pulse motion Waves Are a Form of Simple Harmonic Motion path of hand motion equilibrium position of spring Figure 8.20 Simple harmonic motion generates a wave train in the form of a sine curve. spring If you move your hand from side to side in simple harmonic motion, as indicated in Figure 8.20, transverse waves are generated in the spring. When a transverse wave moves through a medium, the motion of the medium may seem, at first, quite complex. In a transverse wave, each segment of the medium simply oscillates in simple harmonic motion about its equilibrium position in the direction perpendicular to the direction of the wave motion. This simple harmonic motion is transferred sequentially from one segment of the medium to the next to produce the motion of a continuous wave. Universal Wave Equation Pulses provide a useful tool to introduce the nature of waves. However, in nature, sound and light are wave phenomena rather than pulses. In this section, we will begin to shift the emphasis to the properties of waves. Whereas the letter l is used to indicate the length of a pulse, the Greek letter lambda, \u03bb, is used to indicate wavelength. The terms crest and trough come from the description of water waves but are used throughout wave studies. For a water wave, the crest occurs where the medium is displaced above the equilibrium position, while a trough is the region displaced below the equilibrium position. However, for media such as springs, the terms crest and trough merely refer to two regions in the medium that are displaced to opposite sides of the equilibrium position (Figure 8.20). Other variables used in wave studies (frequency\u2013f, period\u2013T, amplitude\u2013A) come from and have the same meanings as in your study of simple harmonic motion in section 7.2. The period (T) is the time taken to generate one complete wavelength. Since two pulses join to create one wave, the period for a wave is twice the time required to generate a pulse. Therefore, the wavelength of a wave is twice the length of a pulse. With this in mind", ", the relationship between wavelength, speed, and period is the same for waves as it is for pulses. That is, \u03bb vT rather than l vt. For periodic motion, T 1. f The equation for wavelength now can be written as \u03bb v f or v f \u03bb. The latter form is known as the universal wave equation. 408 Unit IV Oscillatory Motion and Mechanical Waves 08-PearsonPhys20-Chap08 7/24/08 2:23 PM Page 409 Constant Frequency, Speed, and Wavelength In 8-3 Inquiry Lab, you investigated what happened to a wave train as it moved from deep to shallow water. Changes occurred because the speed in shallow water was slower than it was in deep water. Since the frequency of the waves as they moved from deep to shallow water was unchanged, the reduction in speed was, as predicted by the universal wave equation, accompanied by a reduction in wavelength (Figure 8.21). For a constant frequency, the ratio of the velocities is the same as the ratio of the wavelengths. \u03bb 2 v2 boundary between deep and shallow water straight wave generator \u03bb 1 v1 shallow water (slow speed) wave fronts deep water (faster speed) ripple tank Figure 8.21 When the frequency is constant, a change in speed results in a change in wavelength. When waves change speed, they often change direction as well. You will study this phenomenon further in Unit VII. Example 8.2 To generate waves in a stretched spring, you oscillate your hand back and forth at a frequency of 2.00 Hz. If the speed of the waves in the spring is 5.40 m/s, what is the wavelength? Given v 5.40 m/s f 2.00 Hz Required wavelength Analysis and Solution The variables (v, f, \u03bb) are related by the universal wave equation. v f \u03bb v \u03bb f m 5.40 s 2.00 Hz 5.40 m s 2.00 1 s 2.70 m Paraphrase and Verify The wavelength is 2.70 m. Practice Problems 1. Orchestras use the note with a frequency of 440 Hz (\u201cA\u201d above middle \u201cC\u201d) for tuning their instruments. If the speed of sound in an auditorium is 350 m/s, what is the length of the sound wave generated by this frequency? 2. A submarine sonar system sends a burst of sound with a frequency of 325 Hz. The sound wave bounces off an underwater rock face and returns to the submarine in 8.50 s. If the wavelength of the sound is 4.71 m, how far away is the rock face? 3. A fisherman anchors his dinghy in a lake 250 m from shore. The dinghy rises and falls 8.0 times per minute. He finds that it takes a wave 3.00 min to reach the shore. How far apart are the wave crests? Answers 1. 0.795 m 2. 6.51 km 3. 10 m Chapter 8 Mechanical waves transmit energy in a variety of ways. 409 08-PearsonPhys20-Chap08 7/24/08 2:23 PM Page 410 M I N D S O N Wavelength, Frequency, and Speed \u2022 Walk side by side with a partner at \u2022 With both students keeping their the same speed. One student should take long steps while the other takes very short steps. 1. If the two students maintain their pace, what is the relationship between the frequency and the length of their steps? steps the same length as in the first trial, walk so that your steps are in phase (take steps at the same time). 2. When the two students walk in phase, what is the effect of taking shorter steps? What is the relationship between speed and step length? 8.2 Check and Reflect 8.2 Check and Reflect Knowledge 1. Explain the relationship between the motion of a transverse wave and the motion of the medium through which it moves. 2. Explain how the medium moves when a longitudinal wave passes through it. 3. What is the difference between a transverse and a longitudinal wave? 4. What determines the amount of energy stored in a wave? Applications 5. Sound waves travel through seawater at about 1500 m/s. What frequency would generate a wavelength of 1.25 m in seawater? 6. Temperature changes in seawater affect the speed at which sound moves through it. A wave with a length of 2.00 m, travelling at a speed of 1500 m/s, reaches a section of warm water where the speed is 1550 m/s. What would you expect the wavelength in the warmer water to be? 7. A speaker system generates sound waves at a frequency of 2400 Hz. If the wave speed in air is 325 m/s, what is the wavelength? 8. When you generate a wave in a spring, what is the relationship between the frequency, wavelength, and amplitude? 410 Unit IV Oscillatory Motion and", " Mechanical Waves 9. Two tuning forks are generating sound waves with a frequency of 384 Hz. The waves from one tuning fork are generated in air where the speed of sound is 350 m/s. The other tuning fork is generating sound under water where the speed of sound is 1500 m/s. Calculate the wavelength for the sound (a) in air, and (b) in water. (c) Would you hear the same musical note under water as you did in air? Extensions 10. A radio speaker generates sounds that your eardrum can detect. What does the operation of the speaker and your eardrum suggest about the nature of sound waves? How does the nature of the medium (air) through which sound travels support your assumptions? radio speaker direction of oscillation e TEST eardrum direction of oscillation To check your understanding of transverse and longitudinal waves, follow the eTest links at www.pearsoned.ca/school/physicssource. 08-PearsonPhys20-Chap08 7/24/08 2:23 PM Page 411 8.3 Superposition and Interference Superposition of Pulses and Interference When waves travel through space it is inevitable that they will cross paths with other waves. In nature, this occurs all the time. Imagine two people who sit facing each other and are speaking at the same time. As each person\u2019s sound waves travel toward the other person they must meet and pass simultaneously through the same point in space (Figure 8.22). Still, both people are able to hear quite plainly what the other person is saying. The waves obviously were able to pass through each other so that they reached the other person\u2019s ears unchanged. How waves interact when they cross paths is well understood. When you observe two waves crossing in the ripple tank, things happen so quickly that it is difficult to see what is happening. Still, it is plain that the waves do pass through each other. By sending two pulses toward each other in a spring, it is easier to analyze the events. It is helpful to imagine that the spring in which the pulses are travelling is an ideal, isolated system. The pulses then travel without loss of energy. First, consider two upright pulses moving through each other. When two pulses pass through the same place in the spring at the same time, they are said to interfere with each other. In the section of the spring where interference occurs, the spring takes on a shape that is different from the shape of either of the pulses individually (Figure 8.23). vA pulse A vB pulse B actual position of spring region of pulse overlap pulse A pulse B pulse A x x y y pulse B z z pulse A pulse B Region of overlap Original position of pulse A Original position of pulse B Resultant sound waves from girl sound waves from boy Figure 8.22 When two people talk simultaneously, each person\u2019s sound waves reach the other person\u2019s ears in their original form. interference: the effect of two pulses (or two waves) crossing within a medium; the medium takes on a shape that is different from the shape of either pulse alone Figure 8.23 When two upright pulses move through each other, the displacement of the resultant pulse is the sum of the displacements of pulse A and pulse B. If at any point in the region of overlap, the displacement of one pulse, shown here as x, y, and z, is added to the displacement of the other, the displacement of the resultant pulse is increased. This is called constructive interference. Chapter 8 Mechanical waves transmit energy in a variety of ways. 411 08-PearsonPhys20-Chap08 7/24/08 2:23 PM Page 412 principle of superposition: the displacement of the combined pulse at each point of interference is the sum of the displacements of the individual pulses constructive interference: the overlap of pulses to create a pulse of greater amplitude destructive interference: the overlap of pulses to create a pulse of lesser amplitude e WEB Find out more about superposition of pulses. Follow the links at www.pearsoned.ca/ school/physicssource. The new shape that the spring takes on is predicted by the principle of superposition. This principle, based on the conservation of energy, makes it quite easy to predict the shape of the spring at any instant during which the pulses overlap. The displacement of the combined pulse at each point of interference is the algebraic sum of the displacements of the individual pulses. In Figure 8.23 the two pulses have different sizes and shapes and are moving in opposite directions. The displacement of a pulse is positive for crests and negative for troughs. Since in Figure 8.23 both displacements are positive, at any point where the two pulses overlap, the displacement of the resultant pulse is greater than the displacements of the individual pulses. When pulses overlap to create a pulse of greater amplitude, the result is constructive interference (Figure 8.24). Now consider the case when an inverted pulse meets an upright pulse. The", " displacement of the inverted pulse is a negative value. When the displacements of these pulses are added together, the displacement of the resultant pulse is smaller than the displacement of either pulse. When pulses that are inverted with respect to each other overlap to create a pulse of lesser amplitude, the result is destructive interference (Figure 8.25). resultant pulse pulse A pulse B Figure 8.24 Constructive interference pulse A Figure 8.25 Destructive interference resultant pulse pulse B 412 Unit IV Oscillatory Motion and Mechanical Waves 08-PearsonPhys20-Chap08 7/24/08 2:23 PM Page 413 Figure 8.26 shows a special case of destructive interference. Two pulses that have the same shape and size are shown passing through each other. Because the pulses are identical in shape and size, their displacements at any position equidistant from the front of each pulse are equal in magnitude but opposite in sign. At the point where the two pulses meet, the sum of their displacements will always be zero. At the instant when these two pulses exactly overlap, the displacement at all points is zero and the pulses disappear. The resultant is a flat line. Immediately following this instant, the pulses reappear as they move on their way. The Inversion of Reflected Pulses in a Fixed Spring The principle of superposition explains why pulses are inverted when they reflect from the fixed end of a spring (Figure 8.27). Because the end of the spring is fixed in place, at that point the sum of the displacements of the incident pulse and the reflected pulse must always be zero. Thus, at the point of reflection, the displacement of the reflected pulse must be the negative of the incident pulse. Hence, the reflected pulse must be inverted relative to the incident pulse. info BIT Since, at the point of reflection in the spring the system is basically an isolated system, all the energy in the incident pulse must be carried away by the reflected pulse. actual position of spring original incident pulse reflected segment of incident pulse Pulse arrives at fixed end of the spring. vP vP actual position of spring vA pulse A point where pulses meet pulse B vB region of overlap pulse B pulse B vB pulse A x x y y z z pulse A vA pulse A pulse B pulse A vA pulse B vB Region of overlap Original position of pulse A Original position of pulse B Resultant Figure 8.26 When pulses that are inverted with respect to each other overlap, the displacement of one pulse is reduced by the displacement of the other pulse. At any point in the region of overlap, the displacement of Pulse B, shown here as x, y, and z, reduces the displacement of Pulse A to produce the resultant. This is called destructive interference. vP vP Reflected pulse leaves fixed end of the spring. Figure 8.27 If the end of the spring is fixed, the reflected pulse must be inverted relative to the incident pulse. Chapter 8 Mechanical waves transmit energy in a variety of ways. 413 08-PearsonPhys20-Chap08 7/24/08 2:23 PM Page 414 8-6 Inquiry Lab 8-6 Inquiry Lab Interference of Waves Questions 1 What happens when two pulses pass through the same point in a medium? 2 How can two waves, moving in opposite directions, exist simultaneously in the same space? 3 What causes a standing wave? Materials light spring heavy spring masking tape stopwatch tape measure or metre-stick CAUTION: A stretched spring stores considerable amounts of elastic energy. Be careful not to release the end of a spring while it is stretched. When collapsing a spring, have the person holding one end walk the spring slowly toward the other end. Allow the spring to gently unwind as you are walking to prevent the spring from tying itself into a knot. Variables Part 1: In this part you are concerned with the amplitudes and lengths of the pulses. Observe these variables before, during, and after the period in which they interfere with each other. Part 2: In this part you will explore the relationship between the frequency and the standing wave pattern generated in a spring. From the structure of the standing wave pattern and the length of the spring, the wavelength and the speed of the standing wave are easily calculated. For both parts, identify which are the controlled variables, manipulated variables, and responding variables. Part 1: Superposition and Interference of Pulses 1 (a) Place two parallel strips of tape on the floor about 5 m apart. Measure and record the distance between them. Use these tapes to maintain a constant length for the spring while it is stretched. Attach a third strip of tape about 5 cm long to one of the coils near the middle of the spring as a marker. Required Skills Initiating and Planning Performing and Recording Analyzing and Interpreting Communication and Teamwork (b) Have the team member holding one end of the spring generate a transverse pulse. When this pulse reaches the fixed end of the spring, have the same team member generate a second similar", " pulse. Try to generate the second incident pulse so that it meets the reflection of the first pulse at the strip of tape near the middle of the spring. Focus on the nature of the spring\u2019s motion while the pulses interact. This complex interaction occurs quite quickly and may need to be repeated a few times until you are confident that you can see what is happening. Discuss the observations with your team members. (c) Record your observations in sketches and writing. 2 (a) Again, have one team member generate two pulses. This time, however, generate the second pulse so that it is on the opposite side of the spring (i.e., inverted) to the first pulse. The second pulse will now be on the same side of the spring as the reflected pulse. Again, time the pulses so that they meet near the centre of the spring. (b) Observe how these pulses interact when they meet at the centre of the spring. Discuss what you think is happening with the other members of your team. Analysis 1. When pulses on opposite sides of the spring meet, does the amplitude increase or decrease in the region of overlap? 2. When pulses on the same side of the spring meet, does the amplitude increase or decrease in the overlap region? Part 2: Standing Waves Procedure 1 (a) Have a team member at one end of the spring create a double wave (a series of four pulses) by oscillating his or her hand back and forth twice across the spring\u2019s equilibrium position. (b) Observe what happens as this wave travels back and forth along the spring. Pay particular attention to what happens when the reflected portion of the wave is passing through the incident wave. Discuss your observations with your lab team to come to a consensus on what is occurring. 414 Unit IV Oscillatory Motion and Mechanical Waves 08-PearsonPhys20-Chap08 7/24/08 2:23 PM Page 415 (c) Record your observations. Keep in mind what Analysis you observed when the pulses crossed in Part 1 of the lab. 2 (a) Now create a steady wave train by moving your hand back and forth. Try to find the frequency such that the spring oscillates in two segments about its midpoint. If, at first, there are more than two segments, then reduce the frequency slightly. If, at first, the spring is oscillating as only one segment, then increase the frequency until the second segment appears. Once the spring begins to oscillate as two segments, maintain that frequency. (b) Measure and record the frequency of oscillation by timing ten oscillations. Since you know the length of the spring (the distance between the tapes you placed on the floor), record the length of a wave for this frequency. Each half of the spring is a pulse so that, in this mode, the wavelength is equal to the length of the spring. (c) Record the data obtained in step 2(b) in a table. Use column headings: trial number, number of segments, frequency, wavelength, and speed. 3 (a) Begin with the frequency at which the spring oscillates in two halves and gradually increase the frequency. (b) Describe what happens when you try to maintain a slight increase in the frequency. Keep increasing the frequency until a new oscillation pattern is established. Measure the frequency for this pattern. Record your results in your table of data. (c) Starting from this frequency, gradually increase the frequency until a new pattern of oscillation is found. Once the new wave pattern is established, measure its frequency and record your measurements in your data table. 1. When you created a sustained wave so that the spring oscillated as a stable pattern, in which direction did the waves move? Why do you think that is the case? Does this tell you why this pattern is known as a standing wave? 2. (a) Two segments of a standing wave are equal to one wavelength. For each trial recorded in the table of data, calculate the wavelength of the standing wave. (b) For each trial, use the universal wave equation to calculate the speed of the waves in the spring. 3. To what does the speed of a standing wave refer? 4. Express the frequencies, for the different trials recorded in your data table, as ratios using simple whole numbers. Compare these ratios to the number of segments in which the spring oscillates for each trial. NOTE: The parts of a standing wave that remain motionless are called nodes or nodal points. The midpoints of the parts that oscillate back and forth are called antinodes. Each segment that contains an antinode is simply a pulse or one-half a wavelength. In a standing wave two adjacent segments are required to complete one wavelength. 5. Once a standing wave is established in the spring, what do you notice about the amplitude of the oscillations you use to sustain the wave compared with the amplitude of the antinodes? What explanation might exist for the difference in these two amplitudes? 6. Beginning at", " the fixed end of the spring, describe the locations of the nodes (points that remain motionless) and antinodes (midpoints of the parts that oscillate back and forth) along the spring in terms of wavelength. 7. How does the principle of superposition explain what must be happening at the antinodes of a standing wave? 4 If time permits, change to the heavier spring and repeat steps 2 and 3. 8. What relationship exists between the wavelength of a standing wave and the frequency creating the wavelength? CAUTION: Be very careful not to accidentally release the heavy spring while it is stretched. It will contain a large quantity of elastic potential energy and may seriously injure someone. To relax the tension in the spring, walk one end of the spring slowly toward the other end. 9. Go back to your observations in Part 1 of 8-2 Inquiry Lab. When a train of straight waves parallel to the barrier was reflected back through the incident wave train, did you observe a standing wave? 10. If you could generate a standing wave for sound, what do you think would be the nature of the sound at the location of an antinode? a node? e LAB For a probeware activity, go to www.pearsoned.ca/school/physicssource. Chapter 8 Mechanical waves transmit energy in a variety of ways. 415 08-PearsonPhys20-Chap08 7/24/08 2:23 PM Page 416 M I N D S O N Total Destruction? At the instant when two pulses \u201ccompletely destroy\u201d each other, the spring is in its equilibrium position. How is it possible for the two pulses to reappear as if from nothing? Where does the energy in the pulses go when the sum of the amplitudes is zero? Hint: It might help to think of the spring in terms of a system. Standing Waves and Resonance When two wave trains with identical wavelengths and amplitudes move through each other (Figure 8.28), the resulting interference pattern can be explained by using the principle of superposition. When crests from the two waves or troughs from the two waves occupy the same point in the medium, the waves are in phase. Waves that are in phase produce constructive interference. When a crest from one wave occupies the same point in the medium as a trough from a second wave, we say that these waves are out of phase. Out-of-phase waves produce destructive interference. As the two wave trains pass through each other in opposite directions, they continually shift in and out of phase to produce a wave that seems to oscillate between fixed nodes, rather than move through the medium. wave A wave B wave A B vA vB A \u03bb1 a) (b) (c) (d) (e) (f) actual position of medium at area where waves overlap constructive interference points at which only destructive interference occurs e MATH To graphically analyze the superposition of waves that are in or out of phase, visit www.pearsoned.ca/school/ physicssource. e WEB Find out more about the superposition of waves. Follow the links at www.pearsoned.ca/school/ physicssource. Figure 8.28 The diagrams show how waves travelling in opposite directions interfere as they move through each other. 416 Unit IV Oscillatory Motion and Mechanical Waves 08-PearsonPhys20-Chap08 7/24/08 2:23 PM Page 417 \u2022 (a) Point A is the initial point of contact between the two wave trains shown in blue and purple. The crest from the purple wave train and the trough from the blue wave train arrive at point A at the same instant. \u2022 (b) The two identical waves have moved a distance of 1 \u03bb in opposite 4 directions. This overlap results in destructive interference and the spring is flat in the region of overlap. The position of the spring where the two waves overlap, the resultant, is shown in red. \u2022 (c) Each wave has moved a further 1 \u03bb along the spring. Now the 4 waves are exactly in phase and constructive interference occurs. The regions to the left and right of point A show a crest and a trough, respectively, with displacement of the resultant being twice that of the blue or purple waves. \u2022 Every time the wave trains move a further 1 \u03bb along the spring, the 4 interference changes from constructive to destructive and vice versa. At point A, only destructive interference occurs. The magnitudes of the displacements of the waves arriving at point A are always equal but opposite in sign. As the waves continue to move in opposite directions, the nature of the interference continually changes. However, at point A and every 1 \u03bb from point A, there are points at which only destructive 2 interference occurs. These are called nodal points or nodes. Between the nodes, the spring goes into a flip-flop motion as the interference in these areas switches from constructive (crest crossing crest) to destructive (c", "rest crossing trough) and back to constructive interference (trough crossing trough). The midpoints of these regions on the spring are called antinodes. The first antinode occurs at a distance \u03bb on either side of A, and then at every 1 of 1 \u03bb after that point. Because 2 4 the wave seems to oscillate around stationary nodes along the spring, it is known as a standing wave. Standing waves are also seen in nature; an example is shown in Figure 8.29. Standing Waves in a Fixed Spring When you generate a wave train in a spring that is fixed at one end, the reflected wave train must pass back through the incident wave train. These two wave trains have identical wavelengths and nearly identical amplitudes. For incident and reflected wave trains, the initial point of contact is by definition the fixed point at which reflection occurs. This means that the endpoint of the spring is always a nodal point and, as shown in Figure 8.30, nodes occur every 1 \u03bb from that point with antinodes 2 between them. \u03bb1 2 \u03bb1 2 \u03bb1 2 \u03bb1 2 \u03bb1 2 \u03bb1 2 node: a point on a spring or other medium at which only destructive interference occurs; a point that never vibrates between maximum positive amplitude and maximum negative amplitude; in a standing wave nodes occur at intervals of 1 \u03bb 2 antinode: a point in an interference pattern that oscillates with maximum amplitude; in a standing wave antinodes occur at intervals of 1 \u03bb 2 standing wave: a condition in a spring or other medium in which a wave seems to oscillate around stationary points called nodes. The wavelength of a standing wave is the distance between alternate nodes or alternate antinodes. Figure 8.29 Standing waves occur in nature. This photograph shows a standing wave in a stream crossing a sandy beach in Scotland. e WEB Find out about the details of a standing wave in a spring. Follow the links at www.pearsoned.ca/school/ physicssource. Figure 8.30 In a spring with a fixed end, a standing wave must contain a whole number of antinodes. Nodes occur every half-wavelength from the ends. Antinodes oscillate between shown positions. Nodes remain motionless. antinode Chapter 8 Mechanical waves transmit energy in a variety of ways. 417 08-PearsonPhys20-Chap08 7/24/08 2:23 PM Page 418 e WEB To learn more about the Tacoma Narrows Bridge collapse, follow the links at www.pearsoned.ca/ school/physicssource. Figure 8.31 Resonance, caused by wind, set up a standing wave that destroyed the Tacoma Narrows Bridge. resonance: an increase in the amplitude of a wave due to a transfer of energy in phase with the natural frequency of the wave e WEB To learn more about the giant shock absorbers added to the Millennium Bridge, follow the links at www.pearsoned.ca/ school/physicssource. Figure 8.32 The tone produced when you blow across the top of an open bottle depends on the length of the air column. Resonant Frequencies When a standing wave is present in a spring, the wave reflects from both ends of the spring. There must be a nodal point at both ends with an integral number of antinodes in between. The spring \u201cprefers\u201d to oscillate at those frequencies that will produce a standing wave pattern, called the resonant frequencies for the spring. When the generator is oscillating at a resonant frequency, the energy is added to the spring in phase with existing oscillations. This reinforces and enhances the standing wave pattern. The added energy works to construct waves with ever-larger amplitudes. If the generator is not oscillating at a resonant frequency of the medium, the oscillations tend to destroy the standing wave motion. Amplitude and Resonance Perhaps the most impressive display of a standing wave occurred when resonance set up a standing wave in the bridge across the Tacoma Narrows in the state of Washington (Figure 8.31). Opened in November 1940, the bridge was in operation only a few months before resonance ripped it apart. More recently, in June 2000, the newly opened Millennium Bridge in London had to be closed for modifications when the footsteps of pedestrians set up resonance patterns. Anyone who has ever \u201cpumped up\u201d a swing has used the principle of resonance. To increase the amplitude of its motion, the swing must be given a series of nudges in phase with its natural frequency of motion. Each time the swing begins to move forward, you give it a little push. Since these little pushes are produced in resonance with the swing\u2019s natural motion, they are added to its energy and the amplitude increases. If you pushed out of phase with its natural motion, the swing would soon come to rest. Concept Check Why does it take so little energy to sustain a", " standing wave in a spring? Concept Check Resonating Air Columns All wind instruments use the principle of resonance to produce music. The simplest example of resonance in music is the note produced when you blow over the top of a bottle (Figure 8.32). Blowing across the top of the bottle oscillates the air in the bottle and generates a standing wave. This standing wave is like the waves travelling in a spring, but unlike a spring that is fixed at both ends, the air column is fixed only at the end where reflection occurs and is free to oscillate at the open end. The resonant frequency of the note produced depends on the length of the air column because, to resonate, the standing wave must have a node at the closed end of the bottle and an antinode at the open end (Figure 8.33). 418 Unit IV Oscillatory Motion and Mechanical Waves 08-PearsonPhys20-Chap08 7/24/08 2:23 PM Page 419 Closed-Pipe or Closed-Tube Resonance NR R resonance heard NR no resonance heard When a wave source is held at the open end of a pipe, it sends down a wave that reflects from the closed end of the pipe and establishes a standing wave pattern. The sound one hears depends on the length of the air column in the pipe relative to the length of the standing wave. If an antinode occurs at the open end of the pipe (Figure 8.33 (a) and (c)), a point of resonance (resulting from constructive interference) occurs at the open end of the pipe and the sound appears to be amplified. This phenomenon is known as closed-pipe or closed-tube resonance. However, if the open end of the pipe coincides with the position of a node (destructive interference), then almost no sound can be heard because the source (tuning fork) and the standing wave are out of phase (Figure 8.33 (b) and (d)). (b) (a) R NR R (c) (d) Figure 8.33 Resonance series. A tuning fork sets up a standing wave in the air column. The volume of the sound one hears will vary depending on whether there is an antinode, (a) and (c), or a node, (b) and (d), at the end of the pipe. Nodes and Antinodes in Closed-Pipe Resonance In the air column, nodes are located every half-wavelength from the end at which the wave is reflected, just as they are in a standing wave in a spring. If the pipe length is equal to any multiple of 1 \u03bb, there will 2 be a node at the upper end of the pipe, and destructive interference \u03bb, 4 \u03bb, 3 \u03bb, 2 will occur. Thus, when the air column is 1 \u03bb, \u2026 in length, 2 2 2 2 little or no sound will be heard. Antinodes in the air column are located one quarter-wavelength from the end of the pipe where reflection occurs, and then every halfwavelength from that point. Thus, resonance is heard when the pipe is \u03bb, 5 \u03bb, 3 1 \u03bb, \u2026 long. When resonance is heard for an air column closed 4 4 4 at one end, we know that the open end of the column coincides with the location of one of the antinodes. This information can be used to measure the wavelength of sound in gases. If the frequency of the sound is known, then the wavelength can be used to calculate the speed of sound in the gas. Concept Check Is the volume of a sound related to speed, wavelength, amplitude, or frequency of the wave? What evidence is there to support your answer? Example 8.3 A tuning fork with a frequency of 384 Hz is held above an air column. As the column is lengthened, a closed-pipe resonant point is found when the length of the air column is 67.5 cm. What are possible wavelengths for this data? If the speed of sound is known to be slightly greater than 300 m/s, what is (a) the actual wavelength, and (b) the actual speed of sound? Chapter 8 Mechanical waves transmit energy in a variety of ways. 419 08-PearsonPhys20-Chap08 7/24/08 2:23 PM Page 420 Practice Problems 1. A tuning fork of frequency 512 Hz is used to generate a standing wave pattern in a closed pipe, 0.850 m long. A strong resonant note is heard indicating that an antinode is located at the open end of the pipe. (a) What are the possible wavelengths for this note? (b) Which wavelength will give the most reasonable value for the calculation of the speed of sound in air? 2. A tuning fork with a frequency of 256 Hz is held above a closed air column while the column is gradually increased in length. At what lengths for this air column would the first 4 resonant points be found", ", if the speed of sound is 330 m/s? 3. A standing wave is generated in a spring that is stretched to a length of 6.00 m. The standing wave pattern consists of three antinodes. If the frequency used to generate this wave is 2.50 Hz, what is the speed of the wave in the spring? 4. When a spring is stretched to a length of 8.00 m, the speed of waves in the spring is 5.00 m/s. The simplest standing wave pattern for this spring is that of a single antinode between two nodes at opposite ends of the spring. (a) What is the frequency that produces this standing wave? (b) What is the next higher frequency for which a standing wave exists in this spring? Answers 1. (a) 3.40 m @ 1.74 103 m/s; 1.13 m @ 580 m/s; 0.680 m @ 348 m/s; 0.486 m @ 249 m/s (b) 0.680 m 2. 0.322 m, 0.967 m, 1.61 m, 2.26 m 3. 10.0 m/s 4. 0.313 Hz, 0.625 Hz Given f 384 Hz l 67.5 cm 0.675 m Required wavelength and speed of sound Analysis and Solution \u03bb, 5 \u03bb, 3 The resonant point might represent 1 \u03bb,\u2026, etc., for 4 4 4 this tuning fork. Assume that 67.5 cm is the first resonant point; that means 67.5 cm is 1 \u03bb. Calculate the wavelength 4 and the speed of sound from that data. Assume that l 1 \u03bb. Therefore, 4 \u03bb 4l v f\u03bb 4(0.675 m) 2.70 m (384 Hz)(2.70 m) 1037 m/s 1.04 103 m/s This value is larger than the speed of sound in air. If the speed of sound is not of the proper order of magnitude, then assume that the resonant point is the second point of resonance and that 67.5 cm is 3 \u03bb. Calculate the wavelength 4 and the speed of sound from that data. Assume that l 3 \u03bb. Therefore, 4 l \u03bb 4 3 75 m) (384 Hz)(0.900 m) 4(0.6 3 v f\u03bb 0.900 m 345.6 m/s 346 m/s This is a reasonable speed for sound in air. Complete the analysis by assuming that l 5 \u03bb. Therefore, 4 l \u03bb 4 5 75 m) (384 Hz)(0.540 m) 4(0.6 5 v f\u03bb 0.540 m 207.4 m/s 207 m/s This value is less than the speed of sound in air. Paraphrase and Verify The calculations for the speed of sound indicate that the data must have been for the second point of resonance. This assumption gives the speed for sound of 346 m/s. The assumption that the pipe length is for the first resonant point results in a speed about three times that of sound. The assumption that the pipe length is for the third resonant point produces a speed less than 300 m/s. 420 Unit IV Oscillatory Motion and Mechanical Waves 08-PearsonPhys20-Chap08 7/24/08 2:23 PM Page 421 8-7 Inquiry Lab 8-7 Inquiry Lab Measuring the Speed of Sound Using Closed-pipe Resonance Required Skills Initiating and Planning Performing and Recording Analyzing and Interpreting Communication and Teamwork When a sound wave travels down a closed pipe, the incident wave reflects off the end of the pipe and back toward the source. The interaction of the incident and reflected waves sets up an interference pattern inside the pipe, known as a standing wave. This standing wave can be used to determine the wavelength of the sound. Problem What is the speed of sound in air? Variables The universal wave equation relates the speed (v) of a wave to its frequency (f ) and wavelength (). The wavelength is determined from the length of the pipe (l ) and the number of the resonant point as counted from the reflecting surface. Materials and Equipment tuning forks and tuning fork hammer or an audio frequency generator glass or plastic pipe tall cylinder Procedure 1 Assemble the apparatus as shown in Figure 8.34. metrestick tuning fork open-ended pipe tall cylinder water Figure 8.34 2 Place the pipe in the water-filled cylinder so that the column of air in the pipe is quite short. 3 Strike the tuning fork with the hammer. 4 Hold the tuning fork as shown over the end of the pipe and lift the pipe slowly so that the length of the column of air in the pipe increases. 5 As you approach a point where the volume of the sound increases, move the pipe slowly up and down to find the point where the resonance is greatest. Strike the tuning fork as often as", " necessary to maintain the sound source. 6 Determine the length of the column of air that gives the greatest resonance and record it in a table like Table 8.1. Measure the length from the surface of the water to the location of the tuning fork. 7 Beginning with the column of air at the previously recorded length, gradually increase the length until you have determined the length of the air column that gives the next point of resonance. Record this length. 8 Repeat step 7 to find the length of the column for the third resonant point. Table 8.1 Column Length and Resonance Frequency f (Hz) Length of column at first resonant point l1 (m) Length of column at second resonant point l2 (m) Length of column at third resonant point l3 (m) Analysis 1. In terms of wavelength, how far is each of the first three resonant points from the reflecting surface of the water at the bottom of the air column? 2. Calculate the wavelength of the sound from the tuning fork for each resonant point. Record your answers in a table similar to Table 8.2. Calculate the speed of sound for each of the wavelengths. 3. When you calculate the wavelength for different resonant points, do the answers agree? If not, what might cause the differences? 4. Why should you start with a short column of air and increase its length if you are to be sure that you have correctly determined the wavelength? Chapter 8 Mechanical waves transmit energy in a variety of ways. 421 08-PearsonPhys20-Chap08 7/24/08 2:23 PM Page 422 \u25bc Table 8.2 Resonant Points, Wavelength, and Speed of Sound First Resonant Point Second Resonant Point Third Resonant Point Frequency f (Hz) Wavelength 4l (m) Speed v (m/s) Wavelength 4l/3 (m) Speed v (m/s) Wavelength 4l/5 (m) Speed v (m/s) 5. Why should you measure the length of the column from the reflecting surface to the tuning fork rather than to the top end of the pipe? 6. What is the speed of sound at room temperature? 7. Investigate the effect that air temperature has on the speed of sound. Use hot water or ice water to modify the temperature of the air in the column. Compare the measured speed of sound for at least three temperatures. Suspend a thermometer down the pipe to determine the temperature of the air in the pipe. Plot a graph of the measured speed of sound versus the temperature. Does the graph suggest a linear relationship? Can you use the graph to predict the speed of sound at other temperatures? 8. An alternative technique to determine the speed of sound is to measure the time for an echo to return to you. Stand a measured distance from a wall or other surface that reflects sound. Create an echo by striking together two hard objects, such as metal bars, or, perhaps beating a drum. Listen for the echo. Once you have established an approximate time for the echo to return, strike the bars in a rhythm so that they are in phase with the echo. Have a team member count the number of beats in 1 min. The period of this frequency is the time required for the sound to travel to the wall and back. Use that data to calculate the speed of the sound. e LAB For a probeware activity, go to www.pearsoned.ca/school/physicssource. fundamental frequency: the lowest frequency produced by a particular instrument; corresponds to the standing wave having a single antinode, with a node at each end of the string Music and Resonance Complex modes of vibration give instruments their distinctive sounds and add depth to the musical tones they create. A string of a musical instrument is simply a tightly stretched spring for which the simplest standing wave possible is a single antinode with a node at either end. For this pattern, the length of the string equals one-half a wavelength and the frequency produced is called the fundamental frequency (Figure 8.35(a)). fundamental frequency equilibrium position info BIT Assume that the fundamental frequency is f. In physics and in music, the frequency 2f is called the first overtone; 3f is the second overtone, and so on. These frequencies are said to form a harmonic series. Thus, physicists may also refer to the fundamental frequency (f ) as the first harmonic, the frequency 2f as the second harmonic, the frequency 3f as the third harmonic, and so on. Figure 8.35 (a) standing wave with an antinode at the centre of the string. The fundamental frequency of a vibrating string oscillates as a This is the lowest frequency produced by a particular instrument. But other standing wave patterns can exist in the string at the same time as it oscillates at its fundamental frequency. By plucking or bowing a string nearer its end than its middle, the string is encouraged to vibrate with multiple frequencies.", " The frequencies above the fundamental 422 Unit IV Oscillatory Motion and Mechanical Waves 08-PearsonPhys20-Chap08 7/24/08 2:23 PM Page 423 frequency that may exist simultaneously with the fundamental frequency are called overtones. Figures 8.35 (b) and (c) show the shape of a string vibrating in its first and second overtones, respectively. Figure 8.36 shows a violinist bowing and fingering the strings of her violin to produce notes. 1st overtone without the fundamental frequency equilibrium position fundamental frequency 1st overtone with the fundamental frequency equilibrium position Figure 8.36 The violinist\u2019s fingering technique changes the length of the string and thus changes the fundamental frequency of vibration. overtone: any frequency of vibration of a string that may exist simultaneously with the fundamental frequency The first overtone has the form of a standing wave with two Figure 8.35 (b) antinodes. A node exists at the midpoint of the string. The lower portion of the diagram shows a string vibrating with both the fundamental frequency and the first overtone simultaneously. 2nd overtone without the fundamental frequency equilibrium position fundamental frequency 2nd overtone with the fundamental frequency equilibrium position Figure 8.35(c) The lower portion of the diagram shows a string vibrating with both the fundamental frequency and second overtone. The vibration that produces the second overtone has three antinodes. The actual form of a vibrating string can be very complex as many overtones can exist simultaneously with the fundamental frequency. The actual wave form for a vibrating string is the result of the constructive and destructive interference of the fundamental wave with all the existing overtones that occur in the string. For example, Figure 8.37 shows the wave trace on an oscilloscope for the sound of a violin. Figure 8.37 The interference of the fundamental frequency with the overtones produced by a bowed string creates the wave form that gives the violin its unique sound. The wavelength of the fundamental frequency is the distance between the tall sharp crests. Chapter 8 Mechanical waves transmit energy in a variety of ways. 423 08-PearsonPhys20-Chap08 7/24/08 2:23 PM Page 424 Tuning a Stringed Instrument Tuning a stringed instrument involves several principles of physics. The universal wave equation, v f\u03bb, indicates that the frequency of a sound wave is directly proportional to the speed of the sound and inversely proportional to its wavelength. The wavelength for the fundamental frequency of the standing wave in a string is fixed at twice the length of the string, but the speed of a wave in a string increases with tension. Thus, if the wavelength does not change, the frequency at which a string vibrates must increase with tension. Changing the tension in the string is known as tuning (Figure 8.38). Wind Instruments Wind instruments produce different musical notes by changing the length of the air columns (Figure 8.39). In 8-7 Inquiry Lab you used a closed pipe and saw that for resonance to occur, a node must be present at the closed end while an antinode is created at the open end. For a closed pipe, the longest wavelength that can resonate is four times as long as the pipe (Figure 8.40). If the pipe is open at both ends, then the wavelengths for which resonance occurs must have antinodes at both ends of the open pipe or open tube (Figure 8.41). The distance from one antinode to the next is one-half a wavelength; thus, the longest wavelength that can resonate in an open pipe is twice as long as the pipe. antinode antinode l \u03bb1 4 l \u03bb1 2 node antinode Figure 8.40 In a closed pipe, the longest possible resonant wavelength is four times the length of the pipe. Figure 8.41 In an open pipe, the longest possible resonant wavelength is twice the length of the pipe. Wind instruments are generally open pipes. The wavelength of the resonant frequency will be decided by the length of the pipe (Figure 8.42). In a clarinet or oboe, for example, the effective length of the pipe is changed by covering or uncovering holes at various lengths down the side of the pipe. The strongest or most resonant frequency will be the wave whose length is twice the distance from the mouthpiece to the first open hole. Overtones are also generated but the note you hear is that with the longest wavelength. As with stringed instruments, the overtones contribute to the wind instrument\u2019s characteristic sound. If the speed of sound in air never varied, then a given wavelength would always be associated with the same frequency. But the speed of sound changes slightly with air temperature and pressure. Thus, in the case of resonance in a pipe, the length of the pipe must be increased or decreased as the speed of sound increases or decreases to ensure that the frequency is that of the desired note. Figure 8.38 Tuning a guitar e LAB For", " a probeware activity, go to www.pearsoned.ca/ school/physicssource. Figure 8.39 The trumpeter produces different notes by opening valves to change the instrument\u2019s overall pipe length. Figure 8.42 A variety of wind instruments e WEB To learn how and why wind instruments are affected by temperature, follow the links at www.pearsoned.ca/school/ physicssource. 424 Unit IV Oscillatory Motion and Mechanical Waves 08-PearsonPhys20-Chap08 7/24/08 2:23 PM Page 425 An Interference Pattern from Two In-phase Point Sources Interference patterns carry information about the waves that create them. For this reason, the patterns are often used to determine the properties of the waves. One of the most interesting interference patterns results from waves generated by two point sources that are in phase. Remember that wave sources are in phase if they generate crests at the same time. The ripple tank photograph (Figure 8.43) shows the interference pattern generated by two in-phase point sources that are separated in space. This pattern is the result of constructive and destructive interference as the waves cross. Generally crests appear bright and troughs appear dark. However, in areas where destructive interference occurs, there appear to be fuzzy lines (such as the line indicated by Q1) that seem to radiate approximately from the midpoint between the sources. While the pattern may appear to be complex, its explanation is fairly simple. P1 Q1 R1 S1 P2 S2 Figure 8.43 The interference pattern generated by two in-phase point sources in a ripple tank. The distance between the sources is 3. Individually, point sources generate waves that are sets of expanding concentric circles. As the crests and troughs from each source move outward, they cross through each other. As with all waves, when the crests from one source overlap crests from the other source (or troughs overlap troughs), constructive interference occurs. In these regions there is increased contrast (as indicated by P1 and R1). At locations where the crests from one source overlap troughs from the other source, destructive interference occurs. In these regions, contrast is reduced. Because the sources oscillate in phase, the locations where constructive and destructive interference occur are at predictable, fixed points. Like standing waves in a spring, the positions of the nodes and antinodes depend on the wavelength and the distance between the sources. Can you identify the regions of constructive and destructive interference in Figure 8.43 above? info BIT Common effects of interference patterns result in the \u201chot\u201d and \u201ccold\u201d spots for sound in an auditorium. In 2005 renovations were completed for the Jubilee Auditoriums in Edmonton and Calgary. During renovations, the auditoriums were retuned to improve their acoustic properties. PHYSICS INSIGHT An interference pattern for sound can result if two loudspeakers, at an appropriate distance apart, are connected to the same audio frequency generator. When the sound waves diverging from the speakers overlap and interfere, regions of loud sound (maxima) and regions of relative quiet (minima) will be created. interference pattern: a pattern of maxima and minima resulting from the interaction of waves, as crests and troughs overlap while the waves move through each other Chapter 8 Mechanical waves transmit energy in a variety of ways. 425 08-PearsonPhys20-Chap08 7/24/08 2:23 PM Page 426 The pattern in Figure 8.43 can be reproduced by drawing sets of concentric circles about two point sources where each circle represents the crest of a wave front (Figure 8.44). In Figure 8.43 the distance (d) between the sources is equal to three wavelengths (3\u03bb). This can be shown by counting the wavelengths between S1 and S2 in Figure 8.44. P1 Q1 R1 central maximum first order minimum first order maximum minima maxima 5\u03bb 5\u03bb 4 \u03bb1 2 4\u03bb 3\u03bb S1 P2 Q2 R2 S2 Figure 8.44 The interference pattern for two in-phase point sources results from the overlap of two sets of concentric circles. In this diagram, the centres of the circles are three wavelengths apart. Maxima, Minima, and Phase Shifts The central maximum is a line of antinodes. In Figure 8.44, the line P1P2 is the perpendicular bisector of the line S1S2. By definition, every point on P1P2 is equidistant from the points S1 and S2. Thus, crests (or troughs) generated simultaneously at S1 and S2 must arrive at P1P2 at the same time, resulting in constructive interference. Along the line P1P2 only antinodes are created. The line of antinodes along P1P2 is called the central maximum. A nodal line, or minimum, marks locations where waves", " are exactly out of phase. A little to the right of the central maximum is the line Q1Q2. If you follow this line from one end to the other you will notice that it marks the locations where the crests (lines) from S1 overlap the troughs (spaces) from S2 and vice versa. Waves leave the sources in phase, but all points on Q1Q2 are a one-half wavelength farther from S1 than they are from S2. Thus, at any point on Q1Q2, the crests from S1 arrive one-half a wavelength later than the crests from S2. This means they arrive at the same time as troughs from S2. The greater distance travelled by waves from S1 produces what is called a one-half wavelength phase shift. Waves that began in phase arrive at points on Q1Q2 exactly out of phase. Thus, at every point on Q1Q2 destructive interference occurs. The line, Q1Q2, is known as a nodal line or a minimum. maximum: a line of points linking antinodes that occur as the result of constructive interference between waves minimum: a line of points linking nodes that occur as the result of destructive interference between waves phase shift: the result of waves from one source having to travel farther to reach a particular point in the interference pattern than waves from the other source 426 Unit IV Oscillatory Motion and Mechanical Waves 08-PearsonPhys20-Chap08 7/24/08 2:23 PM Page 427 e WEB To learn more about two-point interference systems, follow the links at www.pearsoned.ca/school/ physicssource. A first order maximum is the result of a one wavelength phase shift. Moving farther right, another region of constructive interference occurs. To arrive at any point on R1R2, crests from S1 travel exactly one wavelength farther than crests from S2. Crests from S1 arrive at points on R1R2 at the same time as crests from S2 that were generated one cycle later. This one-wavelength phase shift means that all waves arriving at any point on R1R2 are still in phase. The line of antinodes resulting from a one-wavelength shift is known as a first order maximum. An identical first order maximum exists on the left side. The interference pattern is symmetrical about the central maximum. Phase shifts equal to whole wavelengths produce maxima. Moving farther outward from the central maximum, you pass through lines of destructive and constructive interference (minima and maxima). Each region is the result of a phase shift produced when waves travel farther from one source than the other. When the phase shift equals a whole number of wavelengths (0\u03bb, 1\u03bb, 2\u03bb,...), the waves arrive in phase, producing antinodes and resulting in the central, first, second, and third order maxima, etc. In Figure 8.44, since the sources are 3\u03bb apart, the greatest phase shift possible is three wavelengths. This produces the third order maximum directly along the line of S1S2. Phase shifts equal to an odd number of half-wavelengths produce minima. When the phase shift equals an odd number of half-wavelengths 1 \u03bb, 2 \u03bb, \u2026 the waves arrive out of phase, producing a nodal line or \u03bb, 5 3 2 2 minimum. In Figure 8.44, the greatest phase shift to produce destructive interference is one-half wavelength less than the three-wavelength \u03bb 5 separation of the sources, or 3\u03bb 1 \u03bb. Because the sources are three 2 2 wavelengths apart, there are exactly three maxima and three minima to the right and to the left of the central maximum. 8-8 Design a Lab 8-8 Design a Lab Interference Patterns and In-phase Sound Sources The Question Do interference patterns exist for two in-phase sound sources? Design and Conduct Your Investigation An audio frequency generator and two speakers can be used to create an interference pattern for sound. Design a set-up that will enable you to measure the wavelength of sound of known frequencies. If electronic equipment (probeware or waveport) is available, design lab 8-8 to incorporate this equipment. Measure the wavelengths using several maxima and minima to compare measurements. Which type of line gives the best results? How well do the results from this experiment compare with the results from measuring wavelengths using closed-pipe resonance? Chapter 8 Mechanical waves transmit energy in a variety of ways. 427 08-PearsonPhys20-Chap08 7/24/08 2:23 PM Page 428 8.3 Check and Reflect 8.3 Check and Reflect Knowledge 1. What is meant by the term interference? 2. For a standing wave, what is the relationship between the amplitude of an antinode and the amplitude of the waves that combine to create the standing wave? 3. In terms of the length", " of an air column, what is the longest standing wavelength that can exist in an air column that is (a) closed at one end and (b) open at both ends? 4. An air column is said to be closed if it is closed at one end. Consider a pipe of length (l). For a standing wave in this pipe, what are the lengths of the three longest wavelengths for which an antinode exists at the open end of the pipe? 5. What does it mean to say that two wave generators are in phase? What does it mean to say that two waves are in phase? Applications 6. Two pulses of the same length (l) travel along a spring in opposite directions. The amplitude of the pulse from the right is three units while the amplitude of the pulse from the left is four units. Describe the pulse that would appear at the moment when they exactly overlap if (a) the pulses are on the same side of the spring and (b) the pulses are on opposite sides of the spring. 7. A standing wave is generated in a closed air column by a source that has a frequency of 768 Hz. The speed of sound in air is 325 m/s. What is the shortest column for which resonance will occur at the open end? 8. Draw the interference pattern for two inphase point sources that are 5 apart, as follows. Place two points, S1 and S2, 5 cm apart near the centre of a sheet of paper. Using each of these points as a centre, draw two sets of concentric circles with increasing radii of 1 cm, 2 cm, 3 cm,..., until you reach the edge of the paper. On the diagram, draw solid lines along maxima and dotted lines along minima. Label the maxima according to their order. Explain why there are five minima on either side of the central maximum. 9. An interference pattern from two in-phase point sources is generated in a ripple tank. On the screen, a point on the second order maximum is measured to be 8.0 cm from one point source and 6.8 cm from the other source. What is the wavelength of this pattern? Extension 10. Do pipe organs, such as those found in churches and concert halls, use closed or open pipes to produce music? What is the advantage of using a real pipe organ as opposed to an electronic organ that synthesizes the sound? e TEST To check your understanding of superposition and interference of pulses and waves, follow the eTest links at www.pearsoned.ca/school/physicssource. 428 Unit IV Oscillatory Motion and Mechanical Waves 08-PearsonPhys20-Chap08 7/24/08 2:23 PM Page 429 info BIT It is common to see police officers using radar to measure the speed of cars on the highway. The radar gun emits waves that are reflected back to it from an oncoming car. A computer in the gun measures the change in frequency and uses that change to calculate the speed of the car. reflected wave \u03bb transmitted RADAR wave moving car \u03bb v radio transmitter Figure 8.45 8.4 The Doppler Effect Have you ever stood at the side of a road and listened to the cars pass? If you listen carefully, you will detect a very interesting phenomenon. At the instant a car passes you, the sound it makes suddenly becomes lower in pitch. This phenomenon was explained by an Austrian physicist named Christian Doppler (1803\u20131853). Doppler realized that the motion of the source affected the wavelength of the sound. Those waves that moved in the same direction as the source was moving were shortened, making the pitch of the sound higher. Moving in the direction opposite to the motion of the source, the sound waves from the source were lengthened, making the pitch lower. Wavelength and Frequency of a Source at Rest Assume that the frequency of a source is 100 Hz and the speed of sound is 350 m/s (Figure 8.46). According to the universal wave equation, if this source is at rest, the wavelength of the sound is 3.50 m. v f\u03bb \u03bb v f 350 m s 0 10 s 3.50 m vw 350 m/s \u03bb 3.50 m Figure 8.46 When a wavelength of 3.50 m travels toward you at a speed of 350 m/s, you hear sound that has a frequency of 100 Hz (diagram not to scale). You hear the sound at a frequency of 100 Hz because at a speed of 350 m/s, the time lapse between crests that are 3.50 m apart is 1/100 s. If, however, the wavelengths that travel toward you were 7.0 m long, the time lapse between successive crests would be 1/50 s, a frequency equal to 50 Hz. v f \u03bb 350 m s 7.0 m 50 1 s 50 Hz Chapter 8 Mechanical waves transmit energy in a variety of ways. 429 08", "-PearsonPhys20-Chap08 7/24/08 2:23 PM Page 430 Wavelength and Frequency of a Moving Source 0.7 m A B C D 4.2 m Figure 8.47 When a sound source moves toward you, the wavelengths in the direction of the motion are decreased. e SIM Research examples of shock waves and the variation of wavelength with a moving source. Go to www.pearsoned.ca/ school/physicssource. direction of motion wave front generated by source at position A by source at position B by source at position C by source at position D Imagine that the source generating the 100-Hz sound is moving toward you at a speed of 70 m/s. Assume that the source is at point A (Figure 8.47) when it generates a crest. While the first crest moves a distance of 3.5 m toward you, the source also moves toward you. The distance the source moves while it generates one wavelength is the distance the source travels in 1/100 s at 70 m/s, or 0.7 m. Because of the motion of the source, the next crest is generated (at point B) only 2.8 m behind the first crest. As long as the source continues at the speed of 70 m/s toward you, the crests travelling in your direction will be only 2.8 m apart. Hence, for a car moving toward you, the sound waves emitted by the car will be \u201csquashed together\u201d and thus reach you more frequently than if the car were stationary. 2.8 m If waves that are 2.8 m long travel toward you at a speed of 350 m/s, then the frequency of the sound arriving at your ear will be 125 Hz. The pitch of the sound that you hear will have been increased because the source is moving toward you. v f\u03bb v f \u03bb 350 m s 2.80 m 125 1 s 125 Hz At the same time, along a line in the direction opposite to the motion of the source, the wavelengths are increased by the same amount that the waves in front of the source are shortened. For the 100-Hz sound source moving at 70 m/s, the waves behind the source are increased by 0.7 m, to a length of 4.2 m. The time lapse between these crests, which are 4.2 m apart and travelling at 350 m/s, is 0.012 s. Therefore, the perceived frequency in the direction opposite to the motion of the source is about 83 Hz. The pitch of the sound has been lowered. Analysis of the Doppler Effect If the velocity of the sound waves in air is vw, then the wavelength (\u03bb s) that a stationary source(s) with a frequency of fs generates is given by \u03bb s vw fs. The key to this Doppler\u2019s analysis is to calculate the distance the source moves in the time required to generate one wavelength (the period (Ts) of the source). If the source is moving at speed vs, then in the period (Ts) the source moves a distance (ds) that is given by ds vsTs. Since, by definition 430 Unit IV Oscillatory Motion and Mechanical Waves 08-PearsonPhys20-Chap08 7/28/08 9:13 AM Page 431 Ts 1fs, then ds vs fs. Sources Moving Toward You For sources that are moving toward you, ds is the distance by which the wavelengths are shortened. Subtracting ds from \u03bb s gives the lengths of the waves (\u03bb d) that reach the listener. Therefore, ds. s and ds by their equivalent forms gives \u03bb \u03bb d s Replacing v v s w fs f s \u03bb d \u03bb d vs ) (vw fs This is the apparent wavelength (Doppler wavelength) of the sound generated by a source that is moving toward you at a speed vs. Dividing the speed of the waves (vw) by the Doppler wavelength (\u03bb d) produces the Doppler frequency (fd) of the sound that you hear as the source approaches you. Therefore, v w \u03bb d fd vw vs vw f s f s vw fs vw vs v w vw vs is the Doppler frequency when the source is approaching the listener. Sources Moving Away from You If the source is moving away from the listener, the value of ds is added to the value of \u03bb s, giving \u03bb \u03bb ds. s d s and ds by their equivalent forms and complete the If you replace \u03bb development to find fd, it is easy to see that the Doppler frequency for a sound where the source moves away from the listener is given by fs v w vw vs fd General Form of the Doppler Equation The equations for the Doppler effect are usually written as a single equation of", " the form fs v w vw vs fd PHYSICS INSIGHT When the distance between you and the source is decreasing, you must subtract to calculate the Doppler effect on frequency and wavelength. Chapter 8 Mechanical waves transmit energy in a variety of ways. 431 08-PearsonPhys20-Chap08 7/28/08 9:17 AM Page 432 Concept Check If you are travelling in your car beside a train that is blowing its whistle, is the pitch that you hear for the whistle higher or lower than the true pitch of the whistle? Explain. Example 8.4 A train is travelling at a speed of 30.0 m/s. Its whistle generates a sound wave with a frequency of 224 Hz. You are standing beside the tracks as the train passes you with its whistle blowing. What change in frequency do you detect for the pitch of the whistle as the train passes, if the speed of sound in air is 330 m/s? Practice Problems 1. You are crossing in a crosswalk when an approaching driver blows his horn. If the true frequency of the horn is 264 Hz and the car is approaching you at a speed of 60.0 km/h, what is the apparent (or Doppler) frequency of the horn? Assume that the speed of sound in air is 340 m/s. 2. An airplane is approaching at a speed of 360 km/h. If you measure the pitch of its approaching engines to be 512 Hz, what must be the actual frequency of the sound of the engines? The speed of sound in air is 345 m/s. 3. An automobile is travelling toward you at a speed of 25.0 m/s. When you measure the frequency of its horn, you obtain a value of 260 Hz. If the actual frequency of the horn is known to be 240 Hz, calculate vw, the speed of sound in air. 4. As a train moves away from you, the frequency of its whistle is determined to be 475 Hz. If the actual frequency of the whistle is 500 Hz and the speed of sound in air is 350 m/s, what is the train\u2019s speed? Answers 1. 278 Hz 2. 364 Hz 3. 325 m/s 4. 18.4 m/s Given fs vw vs 224 Hz 330 m/s 30.0 m/s Required (a) Doppler frequency for the whistle as the train approaches (b) Doppler frequency for the whistle as the train moves away (c) change in frequency Analysis and Solution Use the equations for Doppler shifts to find the Doppler frequencies of the whistle. (a) For the approaching whistle, (b) For the receding whistle, fs v w vs vw fd fd m 330 s m m 30.0 330 s s fs v w vs fd fd 224 Hz vw m 330 s m m 30.0 330 s s 224 Hz 224 Hz 330 m s 300 m s 246.4 Hz 246 Hz 224 Hz 330 m s 360 m s 205.3 Hz 205 Hz (c) The change in pitch is the difference in the two frequencies. Therefore, the pitch change is f 246.4 Hz 205.3 Hz 41.1 Hz Paraphrase and Verify As the train passes, the pitch of its whistle is lowered by a frequency of about 41.1 Hz. 432 Unit IV Oscillatory Motion and Mechanical Waves 08-PearsonPhys20-Chap08 7/24/08 2:23 PM Page 433 e WEB To learn more about possible health effects of sonic booms at close range, and the recent concerns of the Innu Nation, follow the links at www.pearsoned.ca/ school/physicssource. The Sound Barrier Jet planes are not allowed to break or exceed the sound barrier in the airspace over most cities. When an object travels at speeds at, or greater than, the speed of sound, it creates a sonic boom. The boom is the result of the shock wave created by the motion of the object. Bow Waves A boat moving through water produces a bow wave. The crest of the wave moves sideways away from the object, producing the wave\u2019s characteristic V-shape. For an airplane moving through the fluid medium of the atmosphere, a V-shaped bow wave, or pressure wave, travels outward at the speed of sound (Figure 8.48). If the speed of the airplane is less than the speed of sound, the bow wave produced at any instant lags behind the bow wave produced just an instant earlier. The bow wave carries energy away from the plane in a continuous stream (Figure 8.49(a)). Sonic Boom However, for an airplane travelling at the speed of sound, the bow wave and the airplane travel at the same speed. Instant by instant, crests of the bow wave are produced at the same location as the crest of the bow wave produced by the plane an instant earlier (Figure 8.49). The energy stored in the bow", " wave becomes very intense. To the ear of an observer, crests of successively produced bow waves arrive simultaneously in what is known as a sonic boom. In early attempts to surpass the speed of sound, many airplanes were damaged. At the speed of sound, there is a marked increase in drag and turbulence. This effect damaged planes not designed to withstand it. A reporter assumed the increased drag acted like a barrier to travelling faster than sound and coined the term sound barrier. Mathematically, from the arguments presented above, the Doppler wavelength is given by \u03bb d vs) (vw. f s Figure 8.48 When conditions are right, the change in pressure produced by the airplane\u2019s wings can cause sufficient cooling of the atmosphere so that a cloud forms. The extreme conditions present when a jet is travelling near the speed of sound often result in the type of cloud seen in this photo. (a) Slower than speed of sound: Pressure waves move out around plane. (b) At speed of sound: Pressure waves at nose form a shock wave. (c) At supersonic speed: Shock waves form a cone, resulting in a sonic boom. Figure 8.49 As an airplane accelerates from subsonic to supersonic speeds, the changing relationship between the plane and the bow waves or pressure waves results in a sonic boom. Chapter 8 Mechanical waves transmit energy in a variety of ways. 433 08-PearsonPhys20-Chap08 7/24/08 2:23 PM Page 434 vw, which If a plane is travelling at the speed of sound, then vs means that for any sound produced by the jet, the Doppler wavelength in the direction of the jet\u2019s motion is zero. Even if the plane\u2019s speed is greater than that of sound, the bow waves still combine to form a shock wave. In this way a sonic boom can be heard for any object, such as a rifle bullet, that has a supersonic speed. THEN, NOW, AND FUTURE Ultrasound While impressive given the technology available at the time, the results of early attempts at using ultrasound in medicine were of poor quality. Initially, ultrasound images from within a body were very blurry and two-dimensional. By today\u2019s standards, the technology was extremely crude and there was virtually no scientific understanding of how the sound would behave when it encountered different types of tissue. Today, computers have made it possible to form three-dimensional images that can be rotated so that you can see all sides. Doppler ultrasound is used to detect blood flow through an organ. Today, 4-D ultrasound (time is the 4th dimension) is a real-time 3-D image that moves. 1. What are the advantages and disadvantages of ultrasound imaging compared with other imaging techniques such as CT scans and MRI? Figure 8.50 A 3-D ultrasound picture of a developing fetus 8.4 Check and Reflect 8.4 Check and Reflect Knowledge 1. What causes the Doppler effect? 2. Two sound sources have the same frequency when at rest. If they are both moving away from you, how could you tell if one was travelling faster than the other? 3. Explain the cause of a sonic boom. Applications 4. The siren of a police car has a frequency of 660 Hz. If the car is travelling toward you at 40.0 m/s, what do you perceive to be the frequency of the siren? The speed of sound in air is 340 m/s. 5. A police car siren has a frequency of 850 Hz. If you hear this siren to have a frequency that is 40.0 Hz greater than its true frequency, what was the speed of the car? The speed of sound is 350 m/s. 6. A jet, travelling at the speed of sound (Mach 1), emits a sound wave with a frequency of 1000 Hz. Use the Doppler effect equations to calculate the frequency of this sound as the jet first approaches you, then moves away from you. Explain what these answers mean in terms of what you would hear as the jet moved toward, then past, you. Extensions 7. Astronomers have shown that the colour of light from distant stars is shifted from the blue end toward the red end of the spectrum. This is known as red shift. Astronomers realized that since light energy is transmitted as a wave, the red shift was the result of the Doppler effect applied to light. What does the red shift indicate about the motions of the star, which emits light that we see as a red shift? Hint: Investigate the relationship between the frequency and colour for light. e TEST To check your understanding of the Doppler effect, follow the eTest at www.pearsoned.ca/school/ physicssource. 434 Unit IV Oscillatory Motion and Mechanical Waves 08-PearsonPhys20-Chap08 7/24/08 2:23 PM Page 435 CHAPTER", " 8 SUMMARY Key Terms and Concepts medium wave equilibrium position crest trough amplitude wavelength wave front incident wave reflected wave wave train point source ray pulse interference principle of superposition constructive interference destructive interference node antinode standing wave resonance closed pipe fundamental frequency overtone open pipe interference pattern maximum minimum phase shift Key Equations \u03bb vT v f \u03bb fd v fs w vw vs Conceptual Overview The concept map below summarizes many of the concepts and equations in this chapter. Copy and complete the map to produce a full summary of the chapter. wave trains travel as Waves which have which can be transmit from moving sources produce the or stored in the longitudinal amplitude wavelength The source of waves can move either or away from the observer which causes the apparent frequency wavelength frequency to decrease increase of the or troughs strings open pipes which can overlap to which is explained using the This results in interference patterns known as or principle of superposition to describe resulting in caused by and destructive interference with producing in antinodes and Chapter 8 Mechanical waves transmit energy in a variety of ways. 435 08-PearsonPhys20-Chap08 7/24/08 2:23 PM Page 436 CHAPTER 8 REVIEW Knowledge 1. (8.1) (a) When a wave moves on water, what is the nature of the motion of the water within the wave? (b) What is the relationship between the direction of an incident wave and a reflected wave? (c) If you were able see the sound waves emitted when a tuning fork was struck, what would you record as your observation? 2. (8.2) (a) What affects the speed of a water wave? (b) What is the nature of the motion of the medium when a longitudinal wave moves through it? (c) Describe how the speed of a wave affects its wavelength and its amplitude. (d) Explain why waves are considered a form of Simple Harmonic Motion. (e) If speed is constant, how does wavelength vary with frequency? 3. (8.3) (a) Describe the conditions required to produce constructive and destructive interference in waves. (b) Describe how the principle of superposition applies to what happens when two pulses of identical length and amplitude interfere to produce no apparent pulse. (c) Define node, antinode, and standing wave. (d) In terms of the wavelength of the waves that have combined to form a standing wave, describe the position of the nodes and antinodes as you move away from the fixed end of a spring. (e) Why can a standing wave be generated only by what is defined as resonant frequency? 4. (8.4) (a) Does the Doppler effect apply only to sound or can it apply to any form of wave motion? Explain. (b) How are the waves in the direction of a source\u2019s motion affected as the speed of the source increases? Applications 5. The speed of a wave in a spring is 15.0 m/s. If the length of a pulse moving in the spring is 2.00 m, how long did it take to generate the pulse? Why don\u2019t we talk about the frequency for a pulse? 436 Unit IV Oscillatory Motion and Mechanical Waves 6. Waves are generated by a straight wave generator. The waves move toward and reflect from a straight barrier. The angle between the wave front and the barrier is 30\u00b0. Draw a diagram that shows what you would observe if this occurred in a ripple tank. Use a line drawn across the middle of a blank sheet of paper to represent the barrier. Draw a series of wave fronts about 1 cm apart intersecting the barrier at 30\u00b0. Use a protractor to make sure the angle is correct. Now draw the reflected waves. Draw a ray to indicate the motion of the incident wave front and continue this ray to indicate the motion of the reflected wave front. Hint: Draw the reflected ray as if it had a new source. 7. A ripple tank is set up so that the water in it is 0.7 cm deep. In half of the tank, a glass plate is placed on the bottom to make the water in that half shallower. The glass plate is 0.5 cm thick. Thus, the tank has a deep section (0.7 cm) and a shallow section (0.2 cm). In the deep section the wave velocity is 15.0 cm/s while in the shallow section the velocity is 10.0 cm/s. Straight waves, parallel to the edge of the glass, move toward the line between the deep and shallow sections. If the waves have a frequency of 12.0 Hz, what changes in wavelength would you observe as they enter the shallow section? What would happen to the direction of the motion? 8. A ripple tank is set up as described in question 7. For this ripple tank you measure the speed of the waves to be 12.0 cm/s and 9.0 cm/s in the deep and shallow sections", ", respectively. If waves in the deep section that are 11.5 cm long cross over to the shallow section, what would be the wavelength in the shallow section? 9. The term ultrasound means the frequency is higher than those that our ears can detect (about 20 kHz). Animals can often hear sounds that, to our ears, are ultrasound. For example, a dog whistle has a frequency of 22 kHz. If the speed of sound in air is 350 m/s, what is the wavelength of the sound generated by this whistle? 10. A spring is stretched to a length of 7.0 m. A frequency of 2.0 Hz generates a standing wave in the spring that has six nodes. (a) Sketch the standing wave pattern for the spring. (b) Calculate the velocity of the wave. 08-PearsonPhys20-Chap08 7/24/08 2:23 PM Page 437 11. The figure shows two waves that occupy the same point in space. Copy the sketch onto a sheet of paper using the dimensions indicated. Draw the wave that results from the interference of these two waves. 20. When a police car is at rest, the wavelength of the sound from its siren is 0.550 m. If the car is moving toward you at a speed of 120 km/h, what is the frequency at which you hear the siren? Assume that the speed of sound is 345 m/s. 3.5 cm 5.0 cm A 1.5 cm A A 10 cm A A 12. If a frequency of 1.5 Hz generates a standing wave in a spring that has three antinodes, (a) what frequency generates a standing wave with five antinodes in the same spring, and (b) what is the fundamental frequency for this spring? 13. A violin string is 33.0 cm long. The thinnest string on the violin is tuned to vibrate at a frequency of 659 Hz. (a) What is the wave velocity in the string? (b) If you place your finger on the string so that its length is shortened to 28.0 cm, what is the frequency of the note that the string produces? 14. (a) What is the shortest closed pipe for which resonance is heard when a tuning fork with a frequency of 426 Hz is held at the open end of the pipe? The speed of sound in air is 335 m/s. (b) What is the length of the next longest pipe that produces resonance? 21. If the speed of sound in air is 350 m/s, how fast must a sound source move toward you if the frequency that you hear is twice the true frequency of the sound? What frequency would you hear if this sound source had been moving away from you? Extensions 22. Describe an arrangement that you might use if you wanted to create an interference pattern similar to the one in Figure 8.44 on page 426 by using sound waves that have a frequency of 512 Hz. Except for standing waves in strings or pipes, why do you think that we do not often find interference patterns in nature? 23. Explain why the number of maxima and minima in the interference pattern generated by two inphase point sources depends on the ratio of the distance between the sources to the wavelength. e TEST To check your understanding of waves and wave motion, follow the eTest links at www.pearsoned.ca/ school/physicssource. Consolidate Your Understanding 15. Draw the interference pattern generated by two inphase point sources that are four wavelengths apart. Answer each of the following questions in your own words. Provide examples to illustrate your explanation. 16. In the interference pattern for two in-phase point sources, a point on a second order maximum is 2.8 cm farther from one source than the other. What is the wavelength generated by these sources? 17. The horn on a car has a frequency of 290 Hz. If the speed of sound in air is 340 m/s and the car is moving toward you at a speed of 72.0 km/h, what is the apparent frequency of the sound? 18. How fast is a sound source moving toward you if you hear the frequency to be 580 Hz when the true frequency is 540 Hz? The speed of sound in air is 350 m/s. Express your answer in km/h. 19. If the speed of sound in air is 350 m/s, how fast would a sound source need to travel away from you if the frequency that you hear is to be onehalf the true frequency? What would you hear if this sound source had been moving toward you? 1. What are the advantages and disadvantages of using a spring as a model for wave motion? 2. What are the conditions for which a standing wave pattern is generated? Why are standing waves not often seen in nature? 3. Explain how the energy in a wave is transmitted from one place to another. 4. Describe what is meant by the principle of superposition. How", " does this principle explain standing waves? 5. What is meant by resonance? Think About It Review your answers to the Think About It questions on page 393. How would you answer each question now? Chapter 8 Mechanical waves transmit energy in a variety of ways. 437 08-PearsonPhys20-Chap08 7/24/08 2:23 PM Page 438 UNIT IV PROJECT Earthquakes Scenario The tsunami that swept coastal regions of the Indian Ocean on December 26, 2004, was set off by an earthquake centred off the coast of the island of Sumatra in Indonesia. Seismographs around the world identified the location and strength of the earthquake. It was determined that the earthquake rated about 9.0\u20139.3 on the Richter scale. You have been asked by your government to make a presentation on the seismology of earthquakes. Your challenge is threefold. \u2022 First: you are to explain the nature of earthquake shock waves, their movement through Earth, and how the location of the earthquake epicentre is identified by seismographs around the world. \u2022 Second: you are to explain how the intensity of earthquakes is measured. This means that you must explain what the Richter scale is, and how it is used to rate earthquake intensity. \u2022 Third: you are to demonstrate the operation of a seismograph. Planning Your team should consist of three to five members. Choose a team manager and a record keeper. Assign other tasks as they arise. The first task is to decide the structure of your presentation and the research questions you will need to investigate. Questions you will need to consider are: How will you present the information to your audience? What are the resources at your disposal? Do you have access to computers and presentation programs such as PowerPoint\u00ae? Which team members will design, build, and demonstrate the model seismograph? Brainstorm strategies for research and create a schedule for meeting the deadlines for all phases of the project. Where is your team going to look for the information necessary to complete the project? What types of graphics will be most effective to assist your presentation? How will you best demonstrate the function of your seismograph? Your final report should include written, graphic, and photographic analyses of your presentation. Assessing Results Assess the success of your project based on a rubric* designed in class that considers: research strategies thoroughness of the experimental design effectiveness of the experimental technique effectiveness of the team\u2019s public presentation Materials \u2022 materials, as needed, for the construction of your model seismograph Procedure 1 Research the nature of the shock waves set off by an earthquake. Be alert to Internet sites that may contain unreliable or inaccurate information. Make sure that you evaluate the reliability of the sources of information that you use for your research. If you gather information from the Internet, make sure you identify who sponsors the site and decide whether or not it is a reputable source of information. Maintain a list of your references and include it as an appendix to your report. Use graphics to explain how the shock waves move through Earth, and how seismologists locate the epicentre of an earthquake. 2 Research the history of the Richter scale and its use in identifying the intensity of an earthquake. 3 Design and build your model of a seismograph. Decide how you will demonstrate its use in your presentation. 4 Prepare an audio-visual presentation that would inform your audience on the nature of earthquakes and how they are detected. Thinking Further Write a short appendix (three or four paragraphs) to your report to suggest steps that governments might take to make buildings safer in earthquake zones. Answer questions such as: What types of structures are least susceptible to damage by earthquakes? What types are most susceptible? *Note: Your instructor will assess the project using a similar assessment rubric. 438 Unit IV Oscillatory Motion and Mechanical Waves 08-PearsonPhys20-Chap08 7/24/08 2:23 PM Page 439 UNIT IV SUMMARY Unit Concepts and Skills: Quick Reference Concepts Chapter 7 Period Frequency Spring constant Summary Resources and Skill Building Oscillatory motion requires a set of conditions. 7.1 Period and Frequency Period is the time for one complete cycle, measured in seconds (s). If the period of each cycle remains constant, the object is moving with oscillatory motion. Frequency is the number of cycles per second, measured in Hertz (Hz). 7.2 Simple Harmonic Motion The spring constant is the amount of force needed to stretch or compress the spring 1 m and is measured in N/m. It can also be thought of as the stiffness of a spring. QuickLab 7-1; Inquiry Lab 7-2; Minds On; Figures 7.4, 7.5 QuickLab 7-1; Inquiry Lab 7-2; Example 7.1; Figure 7.5 QuickLab 7-3; Examples 7.2\u20137.4 Hooke\u2019s law Hooke\u2019s law states that the deformation of an object is proportional to the force causing it. Figures 7.9", "\u20137.16 Simple harmonic motion SHM refers to anything that moves with uniform oscillatory motion and conforms to Hooke\u2019s law. Figures 7.19\u20137.23; eSIM Pendulum motion The pendulum is a simple harmonic oscillator for angles less than 15\u00b0. Figures 7.25\u20137.27; Example 7.5; Inquiry Lab 7-4; Table 7.5 Figures 7.44, 7.45; Then, Now, and Future QuickLab 8-1; Inquiry Lab 8-2; Inquiry Lab 8-3 Acceleration of a mass-spring system Relationship between acceleration and velocity of a mass-spring system Period of a mass-spring system Period of a pendulum Resonance Forced frequency Resonance effects on buildings and bridges 7.3 Position, Velocity, Acceleration, and Time Relationships The acceleration of a mass-spring system depends on displacement, mass, and the spring constant, and it varies throughout the motion of the mass-spring system. Figure 7.28 The acceleration and velocity of a mass-spring system are continually changing. The velocity of a mass-spring system is determined by its displacement, spring constant, and mass. Figures 7.29\u20137.33; Example 7.6 The period of a mass-spring oscillator is determined by its mass and spring constant, but not its amplitude. Figures 7.35\u20137.37; Example 7.7 A pendulum\u2019s period is determined by its length and the gravitational field strength, but not the mass of the bob. Figures 7.39, 7.40; eSIM; Example 7.8 7.4 Applications of Simple Harmonic Motion Resonance is the natural frequency of vibration of an object. Figure 7.41; QuickLab 7-5 Forced frequency is the frequency at which an external force is applied to an oscillating object. Figure 7.41; QuickLab 7-5 Bridges and buildings can resonate due to the force of the wind. Chapter 8 Mechanical waves transmit energy in a variety of ways. Wave properties may be qualitative or quantitative. 8.1 The Properties of Waves Waves have many properties that can be used to analyze the nature of the wave and the way it behaves as it moves through a medium. Some of these properties are qualitative (crest, trough, wave front, medium, incident wave, reflected wave, wave train) while others are quantitative (amplitude, wavelength, frequency, wave velocity). Universal wave equation 8.2 Transverse and Longitudinal Waves Waves can move through a medium either as transverse or longitudinal waves. The relationship among the frequency, wavelength, and wave velocity is given by the universal wave equation. Inquiry Lab 8-4; Inquiry Lab 8-5; Example 8.1; Example 8.2 Interference patterns may result when more than one wave moves through a medium. 8.3 Superposition and Interference When two or more waves travel in different directions through the same point in space, their amplitudes combine according to the principle of superposition. Depending on the properties of the waves, they may form an interference pattern. Interference patterns can often be used to determine the properties of the waves from which they are formed. Inquiry Lab 8-6; Example 8.3; Inquiry Lab 8-7; Design a Lab 8-8 Doppler effect Sonic boom 8.4 The Doppler Effect When a sound source moves either toward or away from a sensor (ear or microphone), the frequency of the sound that is detected will be different from the frequency emitted by the source. When an object is travelling at the speed of sound, it creates a shock wave known as a sonic boom. Example 8.4 Figure 8.49 Unit IV Oscillatory Motion and Mechanical Waves 439 08-PearsonPhys20-Chap08 7/24/08 2:23 PM Page 440 UNIT IV REVIEW Vocabulary 1. Use your own words to define the following terms, concepts, principles, or laws. Give examples where appropriate. amplitude antinodes closed-pipe air column constructive interference crest destructive interference diverging Doppler effect equilibrium forced frequency frequency fundamental frequency Hooke\u2019s law incident wave in phase interference longitudinal wave maximum mechanical resonance medium minimum nodes or nodal points open-pipe air column oscillation oscillatory motion overtone period phase shift principle of superposition pulse ray reflected wave resonance resonant frequency restoring force shock wave simple harmonic motion simple harmonic oscillator sonic boom sound barrier spring constant standing waves transverse wave trough two-point-source interference pattern 440 Unit IV Oscillatory Motion and Mechanical Waves wave wave front wave train wave velocity wavelength Knowledge CHAPTER 7 2. How are the units of frequency and period similar? How are they different? 3. The SI unit for frequency is Hz. What are two other accepted units? 4. For any simple harmonic oscillator, in what position is (a) the velocity zero? (b) the restoring force the greatest? 5. Why doesn\u2019t a pendulum act like", " a simple harmonic oscillator for large amplitudes? 6. The equation for Hooke\u2019s law uses a negative sign (F kx). Why is this sign necessary? 7. Aboriginal bows used for hunting were made from wood. Assuming the wood deforms according to Hooke\u2019s law, explain how you would go about measuring the spring constant of the wood. 8. Suppose the same pendulum was tested in both Calgary and Jasper. In which location would you expect the pendulum to oscillate more slowly? Explain. 9. Explain why the sound from one tuning fork can make a second tuning fork hum. What conditions must be necessary for this to happen? 10. A pendulum in a clock oscillates with a resonant frequency that depends on several factors. From the list below, indicate what effect (if any) the following variables have on the pendulum\u2019s resonant frequency. (a) length of pendulum arm (b) latitude of clock\u2019s position (c) longitude of clock\u2019s position (d) elevation (e) restoring force 08-PearsonPhys20-Chap08 7/24/08 2:23 PM Page 441 CHAPTER 8 11. The diagram below shows waves in two springs. For each of the springs, how many wavelengths are shown? 12. Sound waves, travelling through air, are reflected from the wall of a building. Describe how the reflection affects the speed, (a) the wavelength, (b) (c) the amplitude, and (d) the direction of a wave train. 13. Points of zero displacement on a transverse wave have the greatest kinetic energy. Which points on a longitudinal wave have the greatest kinetic energy? 14. How is the shape of a circular wave front changed when it reflects from a straight barrier? 15. What aspect of a pulse determines the amount of energy it transfers? 16. When water waves enter a region where they travel slower, what happens to the (a) frequency, (b) wavelength, and (c) direction of the waves? 17. In the interference pattern from two in-phase point sources, what name is given to a line along which destructive interference occurs? 18. What determines the speed at which a wave travels through a spring? 19. What causes a standing wave in a spring? 20. Draw a transverse wave train that consists of two wavelengths. On your diagram, label the equilibrium position for the medium, a crest, a trough, the amplitude, a wavelength, and the direction of the wave velocity. Along the wavelength that you identified above, draw several vector arrows to indicate the direction of the motion of the medium. 21. Why does moving your finger along the string of a violin alter the note that it produces? 22. What property of the sound produced by a tuning fork is affected by striking the tuning fork with different forces? What does that tell you about the relationship between the properties of the sound and the sound wave created by striking the tuning fork? 23. When two in-phase point sources generate an interference pattern, what conditions are required to create (a) the central maximum and (b) a second order maximum? 24. In terms of the length of an open pipe, what is the longest wavelength for which resonance can occur? 25. You are walking north along a street when a police car with its siren on comes down a side street (travelling east) and turns northward on the street in front of you. Describe what you would hear, in terms of frequency of the sound of its siren, before and after the police car turns. 26. What is the relationship between frequency, wavelength, and wave velocity? 27. Why does the frequency of a sound source that is moving toward you seem to be higher than it would be if the source were at rest? Applications 28. Determine the force necessary to stretch a spring (k 2.55 N/m) to a distance of 1.20 m. 29. A musician plucks a guitar string. The string has a frequency of 400.0 Hz and a spring constant of 5.0 104 N/m. What is the mass of the string? 30. When a pendulum is displaced 90.0\u00b0 from the vertical, what proportion of the force of gravity is the restoring force? 31. While performing a demonstration to determine the spring constant of an elastic band, a student pulls an elastic band to different displacements and measures the applied force. The observations were recorded in the table below. Plot the graph of this data. Can the spring constant be determined? Why or why not? Displacement (m) 0.1 0.2 0.3 0.4 0.5 0.6 Force (N) 0.38 1.52 3.42 6.08 9.5 13.68 32. A force of 40.0 N is required to move a 10.0-kg horizontal mass-spring system through a displacement of 80.0 cm. Deter", "mine the acceleration of the mass when its displacement is 25.0 cm. Unit IV Oscillatory Motion and Mechanical Waves 441 08-PearsonPhys20-Chap08 7/24/08 2:23 PM Page 442 33. Use the following table to determine the spring constant of a spring. Displacement (cm) Force (mN) 2.5 5.0 7.5 10.0 12.5 10.0 21.0 31.0 39.0 49.0 34. A 50.0-g mass oscillates on the end of a vertical mass-spring system (k 25.0 N/m) with a maximum acceleration of 50.0 m/s2. (a) What is its amplitude of vibration? (b) What is the maximum velocity of the mass? 35. A bee\u2019s wing has a mass of 1.0 105 kg and makes one complete oscillation in 4.5 103 s. What is the maximum wing speed if the amplitude of its motion is 1.10 cm? 36. A skyscraper begins resonating in a strong wind. A tuned mass damper (m 10.0 t) at the top of the building moves through a maximum displacement of 1.50 m in the opposite direction to dampen the oscillations. If the mass damper is attached to a horizontal spring and has a maximum speed of 1.40 m/s, what is the period of its oscillations? 37. A branch at the top of a tree sways with simple harmonic motion. The amplitude of motion is 0.80 m and its speed is 1.5 m/s in the equilibrium position. What is the speed of the branch at the displacement of 0.60 m? 38. A tuned mass damper at the top of a skyscraper is a mass suspended from a thick cable. If the building sways with a frequency of 0.125 Hz, what length must the cable supporting the weight be to create a resonance in the damper? 39. When a wave slows down, what property of the wave is not affected? What effect does this have on the other properties of the wave? Explain. 40. Explain how a wave can transmit energy through a medium without actually transmitting any matter. 41. A light wave is transmitted through space at 3.00 108 m/s. If visible light has wavelengths ranging from about 4.30 107 m to 7.50 107 m long, what range of frequencies are we able to see? 42. Radio waves travel at the speed of light waves (3.00 108 m/s). If your radio is tuned to a station broadcasting at 1250 kHz, what is the length of the waves arriving at the radio antenna? 442 Unit IV Oscillatory Motion and Mechanical Waves 43. A pendulum oscillates with a period of 0.350 s. Attached to the pendulum is a pen that marks a strip of paper on the table below the pendulum as it oscillates. When the strip of paper is pulled sideways at a steady speed, the pen draws a sine curve on the paper. What will be the wavelength of the sine curve if the speed of the paper is 0.840 m/s? 44. A submarine sends out a sonar wave that has a frequency of 545 Hz. If the wavelength of the sound is 2.60 m, how long does it take for the echo to return when the sound is reflected from a submarine that is 5.50 km away? 45. A wire is stretched between two points that are 3.00 m apart. A generator oscillating at 480 Hz sets up a standing wave in the wire that consists of 24 antinodes. What is the velocity at which waves move in this wire? 46. A spring is stretched to a length of 5.40 m. At that length the speed of waves in the spring is 3.00 m/s. If a standing wave with a frequency 2.50 Hz (a) were generated in this spring, how many nodes and antinodes would there be along the spring? (b) What is the next lower frequency for which a standing wave pattern could exist in this spring? 47. The second string on a violin is tuned to the note D with a frequency of 293 Hz. This is the fundamental frequency for the open string, which is 33.0 cm long. (a) What is the speed of the waves in the string? If you press on the string with your finger so (b) that the oscillating portion of the string is 2/3 the length of the open string, what is the frequency of the note that is created? 48. An audio frequency generator set at 154 Hz is used to generate a standing wave in a closed-pipe resonator, where the speed of sound is 340 m/s. (a) What is the shortest air column for which resonance is heard? (b) What is the next longer column length for which resonance is heard? 49", ". A submarine\u2019s sonar emits a sound with a frequency of 875 Hz. The speed of sound in seawater is about 1500 m/s. If you measure the frequency of the sound to be 870 Hz, what is the velocity of the submarine? 50. A police car is travelling at a speed of 144 km/h. It has a siren with a frequency of 1120 Hz. Assume that the speed of sound in air is 320 m/s. (a) If the car is moving toward you, what frequency will you hear for the siren? If the car had been moving away from you at the same speed, what frequency would you have heard? (b) 08-PearsonPhys20-Chap08 7/24/08 2:23 PM Page 443 Extensions 51. What generalization can be made about the frequency of vibration with regard to the mass for a mass-spring system? (Assume all other qualities remain constant.) 52. An alien crash-lands its spaceship on a planet in our solar system. Unfortunately, it is unable to tell which planet it is. From the wreckage of the spaceship the alien constructs a 1.0-m-long pendulum from a piece of wire with four metal nuts on the end. If this pendulum swings with a period of 3.27 s, on which planet did the alien land: Mercury, Venus, or Earth? 53. Use a compass to draw a simulation of the wave pattern generated by two in-phase point source generators that are 3.5 wavelengths apart. Near the middle of the page, place two points (S1 and S2) 3.5 cm apart to represent the positions of the sources. Draw wavelengths 1.0 cm long by drawing concentric circles that increase in radii by 1.0-cm increments. Locate on the diagram all the maxima and minima that are generated by this set-up and draw lines to indicate their positions. How does this pattern differ from the one in Figure 8.44 on page 426? Explain why these differences occur. 59. Outline a procedure that you could use to determine the mass of a horizontal mass-spring system without measuring the mass on a scale. 60. A student wants to determine the mass of the bob on a pendulum but only has access to a stopwatch and a ruler. She decides to pull the pendulum bob back through a displacement of 10\u00b0 and time 20 complete oscillations. Will it be possible to determine the mass from the data gathered? Explain. 61. Construct a concept map for the simple harmonic motion of a pendulum. Include the following terms: period, displacement, restoring force, velocity, length, and gravitational field strength. 62. In a paragraph, explain why Huygens\u2019s pendulum clock was a revolution in clock making and what the limitations were in its design. Be sure to use these terms: pendulum length, resonant frequency, forced frequency, and gravitational field strength. 63. Research the term \u201cred shift\u201d as used in astronomy. Prepare a report on the importance of red shift to our understanding of the nature of the universe. 54. In a stereo system, there are two speakers set at some distance apart. Why does a stereo system not result in an interference pattern? 64. Describe how to use springs to explore what happens to pulses transmitted from one medium to another in which the wave speed is different. 55. If a sound source is at rest, the frequency you hear and the actual frequency are equal. Their 1). If the sound source ratio equals one (fd/fs moves toward you at an ever-increasing speed, this frequency ratio also increases. Plot a graph for the ratio of the frequencies vs. the speed of the sound source as the speed of the source increases from zero to Mach 1. What is the value of the ratio when the speed of the source is Mach 1? Skills Practice 56. Use a graphing calculator or another suitable means to plot a graph of period against frequency. What type of relationship is this? 57. Outline an experimental procedure that you could perform to determine the spring constant of a vertical mass-spring system. 58. Sketch a diagram of a horizontal mass-spring system in three positions: at both extremes of its motion, and in its equilibrium position. In each diagram, draw vector arrows representing the restoring force, velocity, and acceleration. State whether these are at a maximum or a minimum. 65. Explain to someone who has not studied physics the differences in the ways objects and waves transport energy between points on Earth. Self-assessment 66. Identify a concept or issue that you studied in this unit and would like to learn more about. 67. Learning often requires that we change the way we think about things. Which concept in this unit required the greatest change in your thinking about it? Explain how your thinking changed. 68. Which of the concepts in this unit was most helpful in explaining to you how objects interact? e", " TEST To check your understanding of oscillatory motion and waves, follow the eTest links at www.pearsoned.ca/school/physicssource. Unit IV Oscillatory Motion and Mechanical Waves 443 09-Phys20-Chap09.qxd 7/24/08 2:43 PM Page 444 U N I T V Momentum Momentum and Impulse and Impulse Many situations and activities in the real world, such as snowboarding, involve an object gaining speed and momentum as it moves. Sometimes two or more objects collide, such as a hockey stick hitting a puck across the ice. What physics principles apply to the motion of colliding objects? How does the combination of the net force during impact and the interaction time affect an object during a collision? 444 Unit V 09-Phys20-Chap09.qxd 7/24/08 2:43 PM Page 445 Unit at a Glance C H A P T E R 9 The momentum of an isolated system of interacting objects is conserved. 9.1 Momentum Is Mass Times Velocity 9.2 Impulse Is Equivalent to a Change in Momentum 9.3 Collisions in One Dimension 9.4 Collisions in Two Dimensions Unit Themes and Emphases \u2022 Change and Systems \u2022 Science and Technology Focussing Questions In this study of momentum and impulse, you will investigate the motion of objects that interact, how the velocity of a system of objects is related before and after collision, and how safety devices incorporate the concepts of momentum and impulse. As you study this unit, consider these questions: \u2022 What characteristics of an object affect its momentum? \u2022 How are momentum and impulse related? Unit Project An Impulsive Water Balloon \u2022 By the time you complete this unit, you will have the skills to design a model of an amusement ride that is suitable for a diverse group of people. You will first need to consider acceptable accelerations that most people can tolerate. To test your model, you will drop a water balloon from a height of 2.4 m to see if it will remain intact. e WEB Research the physics concepts that apply to collisions in sports. How do athletes apply these concepts when trying to score goals for their team? How do they apply these concepts to minimize injury? Write a summary of your findings. Begin your search at www.pearsoned.ca/school/physicssource. Unit V Momentum and Impulse 445 09-Phys20-Chap09.qxd 7/24/08 2:43 PM Page 446 C H A P T E R 9 Key Concepts In this chapter, you will learn about: impulse momentum Newton\u2019s laws of motion elastic and inelastic collisions Learning Outcomes When you have completed this chapter, you will be able to: Knowledge define momentum as a vector quantity explain impulse and momentum using Newton\u2019s laws of motion explain that momentum is conserved in an isolated system explain that momentum is conserved in one- and twodimensional interactions compare and contrast elastic and inelastic collisions Science, Technology, and Society explain that technological problems lend themselves to multiple solutions 446 Unit V The momentum of an isolated system of interacting objects is conserved. Most sports involve objects colliding during the play. Hockey checks, curling takeouts, football tackles, skeet shooting, lacrosse catches, and interceptions of the ball in soccer are examples of collisions in sports action. Players, such as Randy Ferbey, who are able to accurately predict the resulting motion of colliding objects have a better chance of helping their team win (Figure 9.1). When objects interact during a short period of time, they may experience very large forces. Evidence of these forces is the distortion in shape of an object at the moment of impact. In hockey, the boards become distorted for an instant when a player collides with them. Another evidence of these forces is a change in the motion of an object. If a goalie gloves a shot aimed at the net, you can see how the impact of the puck affects the motion of the goalie\u2019s hand. In this chapter, you will examine how the net force on an object and the time interval during which the force acts affect the motion of the object. Designers of safety equipment for sports and vehicles use this type of analysis when developing new safety devices. In a system of objects, you will also investigate how their respective velocities change when the objects interact with each other. Figure 9.1 Sports such as curling involve applying physics principles to change the score. Randy Ferbey, originally from Edmonton, won the Brier (Canadian) Curling Championship six times, and the World Curling Championship four times. 09-Phys20-Chap09.qxd 7/24/08 2:43 PM Page 447 9-1 QuickLab 9-1 QuickLab Predicting Angles After Collisions Problem How do the masses of two objects affect the angle between their paths after they collide off centre? Materials pennies and nickels with smooth,", " circular outer edges marking devices for the paths (paper, tape, pencil, ruler) stack of books protractor (optional) Procedure 1 Set up the books and paper as shown in Figure 9.2. Open the cover of the book at the top of the stack for backing. Tape the paper securely to the lab bench. 2 Position one penny at the bottom of the ramp. Mark its initial position by drawing an outline on the paper. 3 Place the incoming penny at the top of the ramp as shown in Figure 9.2. Mark its initial position. 4 Predict the path each coin will take after they collide off centre. Lightly mark the predicted paths. 5 Send the coin down the ramp and mark the position of each coin after collision. Observe the relative velocities of the coins to each other both before and after collision. initial position of penny tape stack of books for ramp paper taped in position identical penny at bottom of ramp tape Figure 9.2 Think About It before after direction of motion angle between two coins after collision Figure 9.3 6 Determine if the angle between the paths after collision is less than 90, 90, or greater than 90 (Figure 9.3). 7 Repeat steps 5 and 6, but have the incoming coin collide at a different contact point with the coin at the bottom of the ramp. 8 Repeat steps 2 to 7 using a penny as the incoming coin and a nickel at the bottom of the ramp. 9 Repeat steps 2 to 7 using a nickel as the incoming coin and a penny at the bottom of the ramp. Questions 1. What was the approximate angle formed by the paths of the two coins after collision when the coins were (a) the same mass? (b) of different mass? 2. Describe how the speeds of the two coins changed before and after collision. 3. How can you predict which coin will move faster after collision? 1. Under what circumstances could an object initially at rest be struck and move at a greater speed after collision than the incoming object? 2. Under what circumstances could a coin in 9-1 QuickLab rebound toward the ramp after collision? Discuss your answers in a small group and record them for later reference. As you complete each section of this chapter, review your answers to these questions. Note any changes to your ideas. Chapter 9 The momentum of an isolated system of interacting objects is conserved. 447 09-Phys20-Chap09.qxd 7/24/08 2:43 PM Page 448 info BIT A tragic avalanche occurred during the New Year\u2019s Eve party in the Inuit community of Kangiqsualujjuaq, formerly in Quebec and now part of Nunavut. At 1:30 a.m. on January 1, 1999, snow from the nearby 365-m mountain slope came cascading down, knocking out a wall and swamping those inside the gymnasium at the party. The snow on the mountain was initially about 1 m thick. After the avalanche was over, the school was covered with up to 3 m of snow. 9.1 Momentum Is Mass Times Velocity Snow avalanches sliding down mountains involve large masses in motion. They can be both spectacular and catastrophic (Figure 9.4). Unbalanced forces affect the motion of all objects. A mass of snow on the side of a mountain experiences many forces, such as wind, friction between the snow and the mountain, a normal force exerted by the mountain on the snow, and gravity acting vertically downward. Skiers and animals moving along the mountain slope also apply forces on the mass of snow. When a large mass of snow becomes dislodged and slides down a mountain slope due to gravity, it not only gains speed but also more mass as additional snow becomes dislodged along the downward path. info BIT Most avalanches occur on slopes that form an angle of 30 to 45 with the horizontal, although they can occur on any slope if the right conditions exist. In North America, a large avalanche may release about 230 000 m3 of snow. Figure 9.4 When the risk of an avalanche seems imminent, ski patrols reduce the mass of snow along a mountain slope by forcing an avalanche to take place. They do this by targeting large masses of snow with guns or explosives to dislodge the snow. 448 Unit V Momentum and Impulse 09-Phys20-Chap09.qxd 7/24/08 2:43 PM Page 449 momentum: product of the mass of an object and its velocity Momentum Is a Vector Quantity All objects have mass. The momentum, p, of an object is defined as the product of the mass of the object and its velocity. Since momentum is the product of a scalar (mass) and a vector (velocity), momentum is a vector quantity that has the same direction as the velocity. p mv Momentum has units of kilogram-metres per second (kgm/s). When you compare the momenta of two objects, you need to consider both the mass and the velocity of each object (Figure", " 9.5). Although two identical bowling balls, A and B, have the same mass, they do not necessarily have the same momentum. If ball A is moving very slowly, it has a very small momentum. If ball B is moving much faster than ball A, ball B\u2019s momentum will have a greater magnitude than ball A\u2019s because of its greater speed. Figure 9.5 The bowling ball in both photos is the same. However, the bowling ball on the left has less momentum than the ball on the right. What evidence suggests this? In real life, almost no object in motion has constant momentum because its velocity is usually not constant. Friction opposes the motion of all objects and eventually slows them down. In most instances, it is more accurate to state the instantaneous momentum of an object if you can measure its instantaneous velocity and mass. Concept Check How would the momentum of an object change if (a) the mass is doubled but the velocity remains the same? (b) the velocity is reduced to 1 of its original magnitude? 3 (c) the direction of the velocity changes from [E] to [W]? e WEB Switzerland has a long history of studying avalanches. Find out what causes an avalanche. What physical variables do avalanche experts monitor? What models are scientists working on to better predict the likelihood and severity of avalanches? Begin your search at www.pearsoned.ca/school/ physicssource. Chapter 9 The momentum of an isolated system of interacting objects is conserved. 449 09-Phys20-Chap09.qxd 7/24/08 2:43 PM Page 450 Relating Momentum to Newton\u2019s Second Law e WEB Research how momentum applies to cycling and other sports. Write a brief report of your findings. Begin your search at www.pearsoned.ca/school/ physicssource. The concept of momentum can be used to restate Newton\u2019s second law. From Unit II, Newton\u2019s second law states that an external non-zero net force acting on an object is equal to the product of the mass of the object ma. Acceleration is defined as the rate of and its acceleration, F net v change of velocity. For constant acceleration, a i. If you or t t i for a in Newton\u2019s second law, you get substitute t v v v v f f F net f ma v v i m t mv mv i t f The quantity mv is momentum. So the equation can be written as F net p p i t f p where F net is constant t Written this way, Newton\u2019s second law relates the net force acting on an object to its rate of change of momentum. It is interesting to note that Newton stated his second law of motion in terms of the rate of change of momentum. It may be worded as: An external non-zero net force acting on an object is equal to the rate of change of momentum of the object. F net p where F net is constant t e SIM For a given net force, learn how the mass of an object affects its momentum and its final velocity. Follow the eSim links at www.pearsoned.ca/school/ physicssource. This form of Newton\u2019s law has some major advantages over the way it was ma only applies to situations where written in Unit II. The equation F net the mass is constant. However, by using the concept of momentum, it is possible to derive another form for Newton\u2019s second law that applies to situations where the mass, the velocity, or both the mass and velocity are changing, such as an accelerating rocket where the mass is decreasing as fuel is being burned, while the velocity is increasing. In situations where the net force changes over a time interval, the average net force is equal to the rate of change of momentum of the object. F netave p t 450 Unit V Momentum and Impulse 09-Phys20-Chap09.qxd 7/24/08 2:43 PM Page 451 In Example 9.1, a person in a bumper car is moving at constant velocity. Since both the person and the car move together as a unit, both objects form a system. The momentum of the system is equal to the total mass of the system times the velocity of the system. Example 9.1 W E v Figure 9.6 A 180-kg bumper car carrying a 70-kg driver has a constant velocity of 3.0 m/s [E]. Calculate the momentum of the cardriver system. Draw both the velocity vector and the momentum vector. Given mc 180 kg md 70 kg v 3.0 m/s [E] Required momentum of system ( p) velocity and momentum vector diagrams Analysis and Solution The driver and bumper car are a system because they move together as a unit. Find the total mass of the system. mT mc md 180 kg 70 kg 250 kg The momentum of the system is in the direction of the", " velocity of the system. So use the scalar form of p mv to find the magnitude of the momentum. p mTv (250 kg)(3.0 m/s) 7.5 102 kgm/s Draw the velocity vector to scale (Figure 9.7). W E 1.0 m/s v 3.0 m/s Figure 9.7 Practice Problems 1. A 65-kg girl is driving a 535-kg snowmobile at a constant velocity of 11.5 m/s [60.0 N of E]. (a) Calculate the momentum of the girl-snowmobile system. (b) Draw the momentum vector for this situation. 2. The combined mass of a bobsled and two riders is 390 kg. The sled-rider system has a constant momentum of 4.68 103 kgm/s [W]. Calculate the velocity of the sled. Answers 1. (a) 6.90 103 kgm/s [60.0 N of E] (b) N 3 kgm/s p 6.90 10 60.0\u00b0 2. 12.0 m/s [W] 2000 kgm/s E Chapter 9 The momentum of an isolated system of interacting objects is conserved. 451 09-Phys20-Chap09.qxd 7/24/08 2:43 PM Page 452 Draw the momentum vector to scale (Figure 9.8). W E 100 kgm/s p 7.5 102 kgm/s Figure 9.8 Paraphrase The momentum of the car-driver system is 7.5 102 kgm/s [E]. Using Proportionalities to Solve Momentum Problems Example 9.2 demonstrates how to solve momentum problems using proportionalities. In this example, both the mass and velocity of an object change. Example 9.2 An object has a constant momentum of 2.45 102 kgm/s [N]. Determine the momentum of the object if its mass decreases to 1 of its original value 3 and an applied force causes the speed to increase by exactly four times. The direction of the velocity remains the same. Explain your reasoning. Practice Problems 1. Many modern rifles use bullets that have less mass and reach higher speeds than bullets for older rifles, resulting in increased accuracy over longer distances. The momentum of a bullet is initially 8.25 kgm/s [W]. What is the momentum if the speed of the bullet increases by a factor of 3 and 2 its mass decreases by a factor of 3? 4 2. During one part of the liftoff of a model rocket, its momentum increases by a factor of four while its mass is halved. The velocity of the rocket is initially 8.5 m/s [up]. What is the final velocity during that time interval? Answers 1. 9.28 kgm/s [W] 2. 68 m/s [up] 452 Unit V Momentum and Impulse Analysis and Solution From the equation p mv, p m and p v. Figure 9.9 represents the situation of the problem. before after N S v 4 v Figure 9.9 p 2.45 102 kgm/s [N] p? p 1 m 3 and p 4v Calculate the factor change of p. 4 4 1 3 3 Calculate p. p 4 4 (2.45 102 kgm/s) 3 3 3.27 102 kgm/s The new momentum will be 3.27 102 kgm/s [N]. 09-Phys20-Chap09.qxd 7/24/08 2:43 PM Page 453 9.1 Check and Reflect 9.1 Check and Reflect Knowledge 1. (a) Explain, in your own words, the concept of momentum. (b) State the SI units of momentum. 2. Explain why momentum is a vector quantity. 3. How is momentum related to Newton\u2019s second law? 4. Explain why stating Newton\u2019s second law in terms of momentum is more useful than stating it in terms of acceleration. 5. Explain, in your own words, the difference between momentum and inertia. 6. Provide three examples of situations in which (a) velocity is the dominant factor affecting the momentum of an object (b) mass is the dominant factor affecting the momentum of an object. 12. Draw a momentum vector diagram to represent a 425-g soccer ball flying at 18.6 m/s [214]. 13. At what velocity does a 0.046-kg golf ball leave the tee if the club has given the ball a momentum of 3.45 kgm/s [S]? 14. (a) A jet flies west at 190 m/s. What is the momentum of the jet if its total mass is 2250 kg? (b) What would be the momentum of the jet if the mass was 3 of its original 4 value and the speed increased to 6 of 5 its original value? 15. The blue", " whale is the largest mammal ever to inhabit Earth. Calculate the mass of a female blue whale if, when alarmed, it swims at a velocity of 57.0 km/h [E] and has a momentum of 2.15 106 kgm/s [E]. Applications Extensions 7. A Mexican jumping bean moves because an insect larva inside the shell wiggles. Would it increase the motion to have the mass of the insect greater or to have the mass of the shell greater? Explain. 8. What is the momentum of a 6.0-kg bowling ball with a velocity of 2.2 m/s [S]? 9. The momentum of a 75-g bullet is 9.00 kgm/s [N]. What is the velocity of the bullet? 10. (a) Draw a momentum vector diagram for a 4.6-kg Canada goose flying with a velocity of 8.5 m/s [210]. (b) A 10.0-kg bicycle and a 54.0-kg rider both have a velocity of 4.2 m/s [40.0 N of E]. Draw momentum vectors for each mass and for the bicycle-rider system. 11. A hockey puck has a momentum of 3.8 kgm/s [E]. If its speed is 24 m/s, what is the mass of the puck? 16. A loaded transport truck with a mass of 38 000 kg is travelling at 1.20 m/s [W]. What will be the velocity of a 1400-kg car if it has the same momentum? 17. Summarize the concepts and ideas associated with momentum using a graphic organizer of your choice. See Student References 4: Using Graphic Organizers on pp. 869\u2013871 for examples of different graphic organizers. Make sure that the concepts and ideas are clearly presented and appropriately linked. e TEST To check your understanding of momentum, follow the eTest links at www.pearsoned.ca/school/ physicssource. Chapter 9 The momentum of an isolated system of interacting objects is conserved. 453 09-Phys20-Chap09.qxd 7/24/08 2:43 PM Page 454 info BIT Legendary stunt person Dar Robinson broke nine world records and made 21 \u201cworld firsts\u201d during his career. One world record was a cable jump from the CN Tower in 1980 for the film The World\u2019s Most Spectacular Stuntman. While tied to a 3-mm steel cable, Robinson jumped more than 366 m and stopped only a short distance above the ground. 9.2 Impulse Is Equivalent to a Change in Momentum Stunt people take the saying, \u201cIt isn\u2019t the fall that hurts, it\u2019s the sudden stop at the end,\u201d very seriously. During the filming of a movie, when a stunt person jumps out of a building, the fall can be very dangerous. To minimize injury, stunt people avoid a sudden stop when landing by using different techniques to slow down more gradually out of sight of the cameras. These techniques involve reducing the peak force required to change their momentum. Sometimes stunt people jump and land on a net. Other times, they may roll when they land. For more extreme jumps, such as from the roof of a tall building, a huge oversized, but slightly under-inflated, air mattress may be used (Figure 9.10). A hidden parachute may even be used to slow the jumper to a safer speed before impact with the surface below. Despite all these precautions, injuries occur as stunt people push the limits of what is possible in their profession. Designers of safety equipment know that a cushioned surface can reduce the severity of an impact. Find out how the properties of a landing surface affect the shape of a putty ball that is dropped from a height of 1 m by doing 9-2 QuickLab. Figure 9.10 The thick mattress on the ground provides a protective cushion for the stunt person when he lands. Why do you think the hardness of a surface affects the extent of injury upon impact? 454 Unit V Momentum and Impulse 09-Phys20-Chap09.qxd 7/24/08 2:43 PM Page 455 9-2 QuickLab 9-2 QuickLab Softening the Hit Problem How is the change in the shape of a putty ball upon impact related to the structure of the landing surface? Materials putty-type material closed cell foam or felt pad urethane foam pad or pillow waxed paper or plastic wrap metre-stick Procedure 1 Choose three surfaces of varying softness onto which to drop a putty ball. One of the surfaces should be either a lab bench or the floor. Cover each surface with some waxed paper or plastic wrap to protect it. 2 Knead or work the putty until you can form three pliable balls of equal size. 3 Measure a height of 1 m above the top of each surface. Then drop the balls, one for each", " surface (Figure 9.11). 4 Draw a side-view sketch of each ball after impact. 1 m 1 m Figure 9.11 Questions 1. Describe any differences in the shape of the putty balls after impact. 2. How does the amount of cushioning affect the deformation of the putty? 3. Discuss how the softness of the landing surface might be related to the time required for the putty ball to come to a stop. Justify your answer with an analysis involving the kinematics equations. Force and Time Affect Momentum In 9-2 QuickLab, you found that the softer the landing surface, the less the shape of the putty ball changed upon impact. The more cushioned the surface, the more the surface became indented when the putty ball collided with it. In other words, the softer and more cushioned landing surface provided a greater stopping distance for the putty ball. Suppose you label the speed of the putty ball at the instant it touches the landing surface vi, and the speed of the putty ball after the impact vf. 0. So the greater the For all the landing surfaces, vi was the same and vf stopping distance, the longer the time required for the putty ball to stop (Figure 9.12). In other words, the deformation of an object is less when the stopping time is increased. vf 0 for surfaces A and B viA viB tB tA viA dA viB dB harder landing surface (A) more cushioned landing surface (B) Figure 9.12 The stopping distance of the putty ball was greater for the more cushioned landing surface (B). So the time interval of interaction was greater on surface B. Chapter 9 The momentum of an isolated system of interacting objects is conserved. 455 09-Phys20-Chap09.qxd 7/24/08 2:43 PM Page 456 PHYSICS INSIGHT To visualize the effect of how Fnet and t can vary but p remains the same, consider the effect of changing two numbers being multiplied together to give the same product. 3 12 36 6 6 36 18 2 36 As the first number increases, the second number decreases in order to get the same product. Project LINK How will the net force and time interval affect the water balloon when it is brought to a stop? What types of protective material will you use to surround the water balloon and why? Apart from the different stopping times, what other differences were there between the drops that would have affected the shape of the putty ball upon impact? The answer to this question requires looking at Newton\u2019s second law written in terms of momentum. From the previous section, the general form of Newton\u2019s second law states that the rate of change of momentum is equal to the net force acting on an object. p t F net If you multiply both sides of the equation by t, you get F net t p For all the landing surfaces, since m, vi (at the first instant of impact), and vf (after the impact is over) of the putty ball were the same, pi was the t p, so the 0. So p was the same for all drops. But F same and pf net product of net force and stopping time was the same for all drops. From Figure 9.12 on page 455, the stopping time varied depending on the amount of cushioning provided by the landing surface. If the stopping time was short, as on a hard landing surface, the magnitude of the net force acting on the putty ball was large. Similarly, if the stopping time was long, as on a very cushioned landing surface, the magnitude of the net force acting on the putty ball was small. This analysis can be used to explain why the putty ball became more deformed when it landed on a hard surface. To minimize changes to the shape of an object being dropped, it is important to minimize Fnet required to stop the object, and this happens when you maximize t (Figure 9.13). It is also important to note where net acts on a large area, the result of the impact will have a F net acts. If F net acts on only one different effect on the shape of the object than if F small part on the surface of the object. (a) direction of motion (b) direction of motion concrete p Fnet t p Fnet t bed of straw Figure 9.13 Identical eggs are dropped from a height of 2 m onto a concrete floor or a pile of straw. Although p is the same in both situations, the magnitude of the net force acting on the egg determines whether or not the egg will break. Concept Check In 9-2 QuickLab, was the momentum of the putty ball at the first instant of impact zero? Explain your reasoning. 456 Unit V Momentum and Impulse 09-Phys20-Chap09.qxd 7/24/08 2:43 PM Page 457 Impulse Is the Product", " of Net Force and Interaction Time impulse: product of the net force on an object and the time interval during an interaction. Impulse causes a change in the momentum of the object. e SIM Learn how the mass and acceleration of an object affect its change in momentum. Follow the eSim links at www.pearsoned.ca/school/ physicssource. t p, the product of net force and interaction time In the equation F net is called impulse. Impulse is equivalent to the change in momentum that an object experiences during an interaction. Every time a net force acts on an object, the object is provided with an impulse because the force is applied for a specific length of time. F net If you substitute the definition of momentum, p mv, the equation t p becomes t (mv) F net If m is constant, then the only quantity changing on the right-hand side of the equation is v. So the equation becomes F net t mv So impulse can be calculated using either equation: F net t p or F net t mv The unit of impulse is the newton-second (Ns). From Unit II, a newton is defined as 1 N 1 kgm/s2. If you substitute the definition of a newton in the unit newton-seconds, you get (s) 1 Ns 1 m kg 2 s m kg 1 s which are the units for momentum. So the units on both sides of the impulse equation are equivalent. Since force is a vector quantity, impulse is also a vector quantity, and the impulse is in the same direction as the net force. To better understand how net force and interaction time affect the change in momentum of an object, do 9-3 Design a Lab. 9-3 Design a Lab 9-3 Design a Lab Providing Impulse The Question What is the effect of varying either the net force or the interaction time on the momentum of an object? Design and Conduct Your Investigation State a hypothesis to answer the question using an \u201cif/then\u201d statement. Then design an experiment to measure the change in momentum of an object. First vary Fnet, then repeat the experiment and vary t instead. List the materials you will use, as well as a detailed procedure. Check the procedure with your teacher and then do the investigation. To find the net force, you may need to find the force of friction and add it, using vectors, to the applied force. The force of kinetic friction is the minimum force needed to keep an object moving at constant velocity once the object is in motion. Analyze your data and form conclusions. How well did your results agree with your hypothesis? Compare your results with those of other groups in your class. Account for any discrepancies. e LAB For a probeware activity, go to www.pearsoned.ca/school/physicssource. Chapter 9 The momentum of an isolated system of interacting objects is conserved. 457 09-Phys20-Chap09.qxd 7/24/08 2:43 PM Page 458 Example 9.3 demonstrates how, for the same impulse, varying the interaction time affects the average net force on a car during a front-end collision (the net force on the car is not constant). Example 9.3 W E Practice Problems 1. Two people push a car for 3.64 s with a combined net force of 200 N [S]. (a) Calculate the impulse provided to the car. (b) If the car has a mass of 1100 kg, what will be its change in velocity? 2. A dog team pulls a 400-kg sled that has begun to slide backward. In 4.20 s, the velocity of the sled changes from 0.200 m/s [backward] to 1.80 m/s [forward]. Calculate the average net force the dog team exerts on the sled. Answers 1. (a) 728 Ns [S], (b) 0.662 m/s [S] 2. 190 N [forward] v Figure 9.14 To improve the safety of motorists, modern cars are built so the front end crumples upon impact. A 1200-kg car is travelling at a constant velocity of 8.0 m/s [E] (Figure 9.14). It hits an immovable wall and comes to a complete stop in 0.25 s. (a) Calculate the impulse provided to the car. (b) What is the average net force exerted on the car? (c) For the same impulse, what would be the average net force exerted on the car if it had a rigid bumper and frame that stopped the car in 0.040 s? Given m 1200 kg (a) and (b) t 0.25 s (c) t 0.040 s v i 8.0 m/s [E] info BIT Some early cars were built with spring bumpers that tended to bounce off whatever they hit. These bumpers were used at", " a time when people generally travelled at much slower speeds. For safety reasons, cars today are built to crumple upon impact, not bounce. This results in a smaller change in momentum and a reduced average net force on motorists. The crushing also increases the time interval during the impulse, further decreasing the net force on motorists. Required (a) impulse provided to car (b) and (c) average net force on car ( F netave) Analysis and Solution When the car hits the wall, the final velocity of the car is zero. v f 0 m/s During each collision with the wall, the net force on the car is not constant, but the mass of the car remains constant. (a) Use the equation of impulse to calculate the impulse provided to the car. f F netave v i) t mv m(v (1200 kg)[0 (8.0 m/s)] (1200 kg)(8.0 m/s) 9.6 103 kgm/s impulse 9.6 103 Ns [W] 458 Unit V Momentum and Impulse 09-Phys20-Chap09.qxd 7/24/08 2:43 PM Page 459 netave. For (b) and (c), substitute the impulse from part (a) and solve for F F netave t 9.6 103 Ns 03 Ns 9.6 1 t F netave (b) F netave 9.6 0.25 3 Ns s 10 (c) F netave m kg 3.8 104 2 s 3.8 104 N 9.6 04 0. 3 Ns 1 0 0 s m kg 2.4 105 2 s 2.4 105 N PHYSICS INSIGHT netave is in the opposite F direction to the initial momentum of the car, because from Newton\u2019s third law, the wall is exerting a force directed west on the car. F netave 3.8 104 N [W] F netave 2.4 105 N [W] Paraphrase and Verify (a) The impulse provided to the car is 9.6 103 Ns [W]. The average net force exerted by the wall on the car is (b) 3.8 104 N [W] when it crumples, and (c) 2.4 105 N [W] when it is rigid. The change in momentum is the same in parts (b) and (c), but the time intervals are different. So the average net force is different in both netave on the car with the rigid frame is situations. The magnitude of F more than 6 times greater than when the car crumples. Impulse Can Be Calculated Using a Net Force-Time Graph One way to calculate the impulse provided to an object is to graph the net force acting on the object as a function of the interaction time. Suppose a net force of magnitude 30 N acts on a model rocket for 0.60 s during liftoff (Figure 9.15). From the net force-time graph in Figure 9.16, the product t is equal to the magnitude of the impulse. But this product is also Fnet the area under the graph. Magnitude of Net Force vs. Interaction Time for a Model Rocket Fnet ) 50 40 30 20 10 0 0 0.10 0.20 0.30 Time t (s) 0.40 0.50 0.60 Figure 9.15 What forces act on the rocket during liftoff? Figure 9.16 Magnitude of net force as a function of interaction time for a model rocket. The area under the graph is equal to the magnitude of the impulse provided to the rocket. Chapter 9 The momentum of an isolated system of interacting objects is conserved. 459 09-Phys20-Chap09.qxd 7/24/08 2:43 PM Page 460 The magnitude of the impulse provided to the rocket is magnitude of impulse Fnet t (30 N)(0.60 s) 18 Ns In other words, the area under a net force-time graph gives the magnitude of the impulse. Note that a net force acting over a period of time causes a change in momentum. When Fnet is not constant, you can still calculate the impulse by finding the area under a net force-time graph. Figure 9.17 shows the magnitude of the net force exerted by a bow on an arrow during the first part of its release. The magnitude of the net force is greatest at the beginning and decreases linearly with time Magnitude of Net Force vs. Interaction Time for an Arrow Shot with a Bow 200 150 100 50 0 0 10 20 30 Time t (ms) 40 50 Figure 9.17 Magnitude of net force as a function of interaction time for an arrow shot with a bow. In this case, the area under the graph could be divided into a rectangle and a triangle or left as a trapezoid (Figure 9.18). So the magnitude of the impulse provided to the arrow is", " 1 (a b)(h) magnitude of impulse 2 1 (100 N 200 N)(0.050 s) 2 7.5 Ns Sometimes two net force-time graphs may appear different but the magnitude of the impulse is the same in both cases. Figure 9.19 (a) shows a graph where Fnet is much smaller than in Figure 9.19 (b). The value of t is different in each case, but the area under both graphs is the same. So the magnitude of the impulse is the same in both situations. (a 30 25 20 15 10 5 0 0 Magnitude of Net Force vs. Interaction Time (b) 2.00 4.00 6.00 Time t (s 30 25 20 15 10 5 0 0 Magnitude of Net Force vs. Interaction Time 2.00 4.00 6.00 Time t (s) Figure 9.19 What other graph could you draw that has the same magnitude of impulse? info BIT The area of a trapezoid is equal to 1 (a b)(h). 2 a h b Figure 9.18 460 Unit V Momentum and Impulse 09-Phys20-Chap09.qxd 7/24/08 2:43 PM Page 461 Effect of a Non-linear Net Force on Momentum In real life, many interactions occur during very short time intervals (Figure 9.20). If you tried to accurately measure the net force, you would find that it is difficult, if not impossible, to do. In addition, the relationship between Fnet and t is usually non-linear, because Fnet increases from zero to a very large value in a short period of time (Figure 9.21). Magnitude of Net Force vs. Interaction Time Fnetave ) Time t (s) Figure 9.21 The average net force gives some idea of the maximum instantaneous net force that an object actually experienced during impact. Figure 9.20 When a baseball bat hits a ball, what evidence demonstrates that the force during the interaction is very large? What evidence demonstrates that the force on the ball changes at the instant the ball and bat separate? From a practical point of view, it is much easier to measure the interaction time and the overall change in momentum of an object during an interaction, rather than Fnet. In this case, the equation of Newton\u2019s second law expressed in terms of momentum is F netave p t and the equation of impulse is F netave t p or F netave t mv In all the above equations, F netave on the object during the interaction. represents the average net force that acted In Example 9.4, a golf club strikes a golf ball and an approximation of the net force-time graph is used to simplify the calculations for impulse. In reality, the net force-time graph for such a situation would be similar to that shown in Figure 9.21. info BIT The fastest recorded speed for a golf ball hit by a golf club is 273 km/h. Chapter 9 The momentum of an isolated system of interacting objects is conserved. 461 09-Phys20-Chap09.qxd 7/24/08 2:43 PM Page 462 Example 9.4 A golfer hits a long drive sending a 45.9-g golf ball due east. Figure 9.22 shows an approximation of the net force as a function of time for the collision between the golf club and the ball. (a) What is the impulse provided to the ball? (b) What is the velocity of the ball Magnitude of Net Force vs. Interaction Time for a Golf Ball Being Hit by a Golf Club 6000 5000 4000 3000 2000 1000 ).2 0.6 0.4 Time t (ms) 0.8 1.0 1.2 Figure 9.22 W E at the moment the golf club and ball separate? Given m 45.9 g ti 1.1 ms tf Fneti 5000 N Fnetf Fnetmax 0.1 ms 0 N 0 N Required (a) impulse provided to ball (b) velocity of ball after impact (v f ) Analysis and Solution The impulse and velocity after impact are in the east direction since the golfer hits the ball due east. (a) t tf ti 1.1 ms 0.1 ms 1.0 ms or 1.0 103 s Fnet Figure 9.23 magnitude of impulse area under net force-time graph 1 (t)(Fnetmax) 2 1 (1.0 103 s)(5000 N) 2 2.5 Ns impulse 2.5 Ns [E] (b) Impulse is numerically equal to mv or m(v v i). f i But v 0 m/s So, impulse m (v 2.5 Ns mv f 0) v f f Ns 2.5 m m kg s 2.50 2 s g (45.9 g) k 1 g 0 0 10 54 m/s Paraphrase (a)", " The impulse provided to the ball is 2.5 Ns [E]. (b) The velocity of the ball after impact is 54 m/s [E]. Practice Problems 1. (a) Draw a graph of net force as a function of time for a 0.650-kg basketball being shot. During the first 0.15 s, Fnet increases linearly from 0 N to 22 N. During the next 0.25 s, Fnet decreases linearly to 0 N. (b) Using the graph, calculate the magnitude of the impulse provided to the basketball. (c) What is the speed of the ball when it leaves the shooter\u2019s hands? 2. (a) A soccer player heads the ball with an average net force of 21 N [W] for 0.12 s. Draw a graph of the average net force on the ball as a function of time. Assume that Fnetave is constant during the interaction. (b) Calculate the impulse provided to the soccer ball. (c) The impulse changes the velocity of the ball from 4.0 m/s [E] to 2.0 m/s [W]. What is the mass of the ball? Answers 1. (a 25 20 15 10 5 0 Magnitude of Net Force vs. Interaction Time for a Basketball Being Shot 0 0.10 0.20 0.30 Time t (s) 0.40 0.50 (b) 4.4 Ns, (c) 6.8 m/s 2. (a 30 25 20 15 10 5 0 Magnitude of Average Net Force vs. Interaction Time for a Soccer Ball Being Hit 0 0.03 0.06 Time t (s) 0.09 0.12 (b) 2.5 Ns [W], (c) 0.42 kg 462 Unit V Momentum and Impulse 09-Phys20-Chap09.qxd 7/24/08 2:43 PM Page 463 The Design of Safety Devices Involves Varying Fnetave and t Many safety devices are based on varying both the average net force acting on an object and the interaction time for a given impulse. Suppose you attached a sled to a snowmobile with a rope hitch. As long as the sled is accelerating along a horizontal surface or is being pulled uphill, there is tension in the rope because the snowmobile applies a force on the sled (Figure 9.24). If the driver in Figure 9.24 (a) brakes suddenly to slow down, the momentum of the snowmobile changes suddenly. However, the sled continues to move in a straight line until friction eventually slows it down to a stop. In other words, the only way that the momentum of the f acts for a long enough period of time. sled changes noticeably is if F (a) (b FT 0 a FT 0 \u03b8 Figure 9.24 (a) A snowmobile accelerating along a horizontal surface, and (b) the same snowmobile either moving at constant speed or accelerating uphill. In both (a) and (b), the tension in the rope is not zero. Suppose the snowmobile driver is heading downhill and applies the brakes suddenly as in Figure 9.25 (a). F g will cause the sled to accelerate downhill as shown in Figure 9.25 (b). The speed of the sled could become large enough to overtake the snowmobile, bump into it, or tangle the rope. (a) n o f c ti o n o ti o m d ir e FT 0 \u03b8 (b) a Ff on m1 m1 Fg \u03b8 a m2 Ff on m2 Fg \u03b8 Figure 9.25 (a) The snowmobile is braking rapidly, and the tension in the rope is zero. (b) The free-body diagrams for the snowmobile and sled only show forces parallel to the incline. Chapter 9 The momentum of an isolated system of interacting objects is conserved. 463 09-Phys20-Chap09.qxd 7/24/08 2:43 PM Page 464 The driver can change the momentum of the snowmobile suddenly by using the brakes. But, as before, the only way that the momentum of the f acts for a long enough time interval. sled can eventually become zero is if F With experience, a driver learns to slow down gradually so that a towed sled remains in its proper position. Some sleds are attached to snowmobiles using a metal tow bar, which alleviates this problem (Figure 9.26). Since the tow bar can never become slack like a rope, the sled always remains a fixed distance from the snowmobile. Tow bars usually have a spring mechanism that increases the time during which a force can be exerted. So if the driver brakes or changes direction suddenly, the force exerted by the snowmobile on the sled acts for a longer period of time. Compared to a towrope, the spring mechanism in the tow bar can safely cause the momentum of the", " sled to decrease in a shorter period of time. Figure 9.26 A rigid tow bar with a spring mechanism provides the impulse necessary to increase or decrease the momentum of a towed sled. e WEB During takeoff, the magnitude of Earth\u2019s gravitational field changes as a rocket moves farther away from Earth\u2019s surface. The mass of a rocket also changes because it is burning fuel to move upward. Research how impulse and momentum apply to the design and function of rockets and thrust systems. Write a brief report of your findings, including diagrams where appropriate. Begin your search at www.pearsoned.ca/ school/physicssource. Safety devices in vehicles are designed so that, for a given impulse such as in a collision, the interaction time is increased, thereby reducing the average net force. This is achieved by providing motorists with a greater distance to travel, which increases the time interval required to stop the motion of the motorist. Three methods are used to provide this extra distance and time: \u2022 The dashboard is padded and the front end of the vehicle is designed to crumple. \u2022 The steering column telescopes to collapse, providing an additional 15\u201320 cm of distance for the driver to travel forward. \u2022 The airbag is designed to leak after inflation so that the fully inflated bag can decrease in thickness over time from about 30 cm to about 10 cm. In fact, an inflated airbag distributes the net force experienced during a collision over the motorist\u2019s chest and head. By spreading the force over a greater area, the magnitude of the average net force at any one point on the motorist\u2019s body is reduced, lowering the risk of a major injury. 464 Unit V Momentum and Impulse 09-Phys20-Chap09.qxd 7/24/08 2:43 PM Page 465 A similar reasoning applies to the cushioning in running shoes and the padding in helmets and body pads used in sports (Figure 9.27). For a given impulse, all these pieces of equipment increase the interaction time and decrease the average net force. (a) (b) Figure 9.27 Padding in sports equipment helps reduce the risk of major injuries, because for a given impulse, the interaction time is increased and the average net force on the body part is reduced. (a) Team Canada in the World Women\u2019s hockey tournament in Sweden, 2005. (b) Calgary Stampeders (in red) playing against the B.C. Lions in 2005. The effect of varying the average net force and the interaction time can be seen with projectiles. A bullet fired from a pistol with a short barrel does not gain the same momentum as another identical bullet fired from a rifle with a long barrel, assuming that each bullet experiences the same average net force (Figure 9.28). In the gun with the shorter barrel, the force from the expanding gases acts for a shorter period of time. So the change in momentum of the bullet is less. t p t p Figure 9.28 For the same average net force on a bullet, a gun with a longer barrel increases the time during which this force acts. So the change in momentum is greater for a bullet fired from a long-barrelled gun. Chapter 9 The momentum of an isolated system of interacting objects is conserved. 465 09-Phys20-Chap09.qxd 7/24/08 2:43 PM Page 466 Improved Sports Performance Involves Varying Fnetave and t In baseball, a skilled pitcher knows how to vary both the net force acting on the ball and the interaction time, so that the ball acquires maximum velocity before it leaves the pitcher\u2019s hand (Figure 9.29). To exert the maximum possible force on the ball, a pitcher uses his arms, torso, and legs to propel the ball forward. To maximize the time he can exert that force, the pitcher leans back using a windup and then takes a long step forward. This way, his hand can be in contact with the ball for a longer period of time. The combination of the greater net force and the longer interaction time increases the change in momentum of the ball. Figure 9.29 When a pitcher exerts a force on the ball during a longer time interval, the momentum of a fastball increases even more. In sports such as hockey, golf, and tennis, coaches emphasize proper \u201cfollow through.\u201d The reason is that it increases the time during which the puck or ball is in contact with the player\u2019s stick, club, or racquet. So the change in momentum of the object being propelled increases. A similar reasoning applies when a person catches a ball. In this case, a baseball catcher should decrease the net force on the ball so that the ball doesn\u2019t cause injury and is easier to hold onto. Players soon learn to do this by letting their hands move with the ball. For the same impulse, the extra movement with the hands results in an increased interaction time, which reduces the net force", ". This intentional flexibility when catching is sometimes referred to as having \u201csoft hands,\u201d and it is a great compliment to a football receiver. Hockey goalies allow their glove hand to fly back when snagging a puck to reduce the impact and allow them a better chance of keeping the puck in their glove. Boxers are also taught to \u201croll with the punch,\u201d because if they move backward when hit, it increases the interaction time and decreases the average net force of an opponent\u2019s blow. 466 Unit V Momentum and Impulse 09-Phys20-Chap09.qxd 7/24/08 2:43 PM Page 467 9.2 Check and Reflect 9.2 Check and Reflect Knowledge 1. (a) What quantities are used to calculate impulse? (b) State the units of impulse. 2. How are impulse and momentum related? 3. What graph could you use to determine the impulse provided to an object? Explain how to calculate the impulse using the graph. 4. What is the effect on impulse if (a) the time interval is doubled? (b) the net force is reduced to 1 of its 3 original magnitude? 5. Even though your mass is much greater than that of a curling stone, it is dangerous for a moving stone to hit your feet. Explain why. Applications 6. Using the concept of impulse, explain how a karate expert can break a board. 7. (a) From the graph below, what is the magnitude of the impulse provided to a 48-g tennis ball that is served due south? (b) What is the velocity of the ball when the racquet and ball separate? Magnitude of Net Force vs. Interaction Time for a Tennis Ball Being Hit by a Racquet ) 1000 900 800 700 600 500 400 300 200 100 0 0.0 1.0 2.0 3.0 Time t (ms) 4.0 5.0 6.0 9. During competitive world-class events, a four-person bobsled experiences an average net force of magnitude 1390 N during the first 5.0 s of a run. (a) What will be the magnitude of the impulse provided to the bobsled? (b) If the sled has the maximum mass of 630 kg, what will be the speed of the sled? 10. An advertisement for a battery-powered 25-kg skateboard says that it can carry an 80-kg person at a speed of 8.5 m/s. If the skateboard motor can exert a net force of magnitude 75 N, how long will it take to attain that speed? 11. Whiplash occurs when a car is rear-ended and either there is no headrest or the headrest is not properly adjusted. The torso of the motorist is accelerated by the seat, but the head is jerked forward only by the neck, causing injury to the joints and soft tissue. What is the average net force on a motorist\u2019s neck if the torso is accelerated from 0 to 14.0 m/s [W] in 0.135 s? The mass of the motorist\u2019s head is 5.40 kg. Assume that the force acting on the head is the same magnitude as the force on the torso. 12. What will be the change in momentum of a shoulder-launched rocket that experiences a thrust of 2.67 kN [W] for 0.204 s? Extensions 13. Experienced curlers know how to safely stop a moving stone. What do they do and why? 14. Research one safety device used in sports that applies the concept of varying Fnetave and t for a given impulse to prevent injury. Explain how the variables that affect impulse are changed by using this device. Begin your search at www.pearsoned.ca/school/physicssource. 8. What will be the magnitude of the impulse generated by a slapshot when an average net force of magnitude 520 N is applied to a puck for 0.012 s? e TEST To check your understanding of impulse, follow the eTest links at www.pearsoned.ca/school/ physicssource. Chapter 9 The momentum of an isolated system of interacting objects is conserved. 467 09-Phys20-Chap09.qxd 7/24/08 2:43 PM Page 468 info BIT The horns of a bighorn ram can account for more than 10% of its mass, which is about 125 kg. Rams collide at about 9 m/s, and average about 5 collisions per hour. Mating contests between any two rams may last for more than 24 h in total. 9.3 Collisions in One Dimension During mating season each fall, adult bighorn rams compete for supremacy in an interesting contest. Two rams will face each other, rear up, and then charge, leaping into the air to butt heads with tremendous force (Figure 9.30). Without being consciously", " aware of it, each ram attempts to achieve maximum momentum before the collision, because herd structure is determined by the outcome of the contest. Often, rams will repeat the head-butting interaction until a clear winner is determined. While most other mammals would be permanently injured by the force experienced during such a collision, the skull and brain structure of bighorn sheep enables them to emerge relatively undamaged from such interactions. In the previous section, many situations involved an object experiencing a change in momentum, or impulse, because of a collision with another object. When two objects such as bighorn rams collide, what relationship exists among the momenta of the objects both before and after collision? In order to answer this question, first consider one-dimensional collisions involving spheres in 9-4 QuickLab. Figure 9.30 By lunging toward each other, these bighorn rams will eventually collide head-on. During the collision, each ram will be provided with an impulse. 468 Unit V Momentum and Impulse 09-Phys20-Chap09.qxd 7/24/08 2:43 PM Page 469 9-4 QuickLab 9-4 QuickLab Observing Collinear Collisions Problem What happens when spheres collide in one dimension? before direction of motion after Materials one set of four identical ball bearings or marbles (set A) a second set of four identical ball bearings or marbles of double the mass (set B) a third set of four identical ball bearings or marbles of half the mass (set C) 1-m length of an I-beam curtain rod or two metre-sticks with smooth edges masking tape Procedure 1 Lay the curtain rod flat on a bench to provide a horizontal track for the spheres. Tape the ends of the rod securely. If you are using metre-sticks, tape them 5 mm apart to form a uniform straight horizontal track. 2 Using set A, place three of the spheres tightly together at the centre of the track. 3 Predict what will happen when one sphere of set A moves along the track and collides with the three stationary spheres. Figure 9.31 4 Test your prediction. Ensure that the spheres remain on the track after collision. Record your observations using diagrams similar to Figure 9.31. 5 Repeat steps 2 to 4, but this time use set B, spheres of greater mass. 6 Repeat steps 2 to 4, but this time use set C, spheres of lesser mass. 7 Repeat steps 2 to 4 using different numbers of stationary spheres. The stationary spheres should all be the same mass, but the moving sphere should be of a different mass in some of the trials. Questions 1. Describe the motion of the spheres in steps 4 to 6. 2. Explain what happened when (a) a sphere of lesser mass collided with a number of spheres of greater mass, and (b) a sphere of greater mass collided with a number of spheres of lesser mass. In 9-4 QuickLab, for each set of spheres A to C, when one sphere hit a row of three stationary ones from the same set, the last sphere in the row moved outward at about the same speed as the incoming sphere. But when one sphere from set A hit a row of spheres from set B, the last sphere in the row moved outward at a much slower speed than the incoming sphere, and the incoming sphere may even have rebounded. When one sphere from set A hit a row of spheres from set C, the last sphere in the row moved outward at a greater speed than the incoming sphere, and the incoming sphere continued moving forward. To analyze these observations, it is important to first understand what a collision is. A collision is an interaction between two objects in which a force acts on each object for a period of time. In other words, the collision provides an impulse to each object. e MATH Explore how the masses of two colliding objects affect their velocities just after collision. Follow the eMath links at www.pearsoned.ca/school/ physicssource. collision: an interaction between two objects where each receives an impulse Chapter 9 The momentum of an isolated system of interacting objects is conserved. 469 09-Phys20-Chap09.qxd 7/24/08 2:43 PM Page 470 system: two or more objects that interact with each other Systems of Objects in Collisions Each trial in 9-4 QuickLab involved two or more spheres colliding with each other. A group of two or more objects that interact is called a system. You encountered the concept of a system in Unit III in the context of energy. For each system in 9-4 QuickLab, the total mass remained constant because the mass of each sphere did not change as a result of the interaction. However, friction was an external force that acted on the system. For example, in steps 4 to 6 of 9-4 QuickLab, you likely observed that the speed of the sphere moving outward was a little less than the speed of the incoming sphere. In real", " life, a system of colliding objects is provided with two impulses: one due to external friction and the other due to the actual collision (Figure 9.32). External friction acts before, during, and after collision. The second impulse is only present during the actual collision. Since the actual collision time is very short, the impulse due to external friction during the collision is relatively small. before during after viA A 0 viB B Ff on A Ff on B 0 vfA A vfB B Ff on A Ff on B FB on A A B FA on B Ff on A tc ts Figure 9.32 External friction acts throughout the entire time interval of the interaction ts. But the action-reaction forces due to the objects only exist when the objects actually collide, and these forces only act for time interval tc. If you apply the form of Newton\u2019s second law that relates net force to momentum to analyze the motion of a system of objects, you get net)sys (F p sys where p t sys is the momentum of the system The momentum of a system is defined as the sum of the momenta of all the objects in the system. So if objects A, B, and C form a system, the momentum of the system is p sys p A p B p C In the context of momentum, when the mass of a system is constant and no external net force acts on the system, the system is isolated. So 0. In 9-5 Inquiry Lab, a nearly isolated system of objects is net)sys (F involved in a one-dimensional collision. Find a quantitative relationship for the momentum of such a system in terms of momenta before and after collision. 470 Unit V Momentum and Impulse 09-Phys20-Chap09.qxd 7/24/08 2:43 PM Page 471 9-5 Inquiry Lab 9-5 Inquiry Lab Relating Momentum Just Before and Just After a Collision Required Skills Initiating and Planning Performing and Recording Analyzing and Interpreting Communication and Teamwork Question How does the momentum of a system consisting of two objects compare just before and just after a collision? Hypothesis State a hypothesis relating the momentum of a system immediately before and immediately after collision, where objects combine after impact. Remember to write an \u201cif/then\u201d statement. Variables Read the procedure carefully and identify the manipulated variables, the responding variables, and the controlled variables. Materials and Equipment one of these set-ups: air track, dynamics carts, Fletcher\u2019s trolley, bead table or air table with linear guides colliding objects for the set-up chosen: gliders, carts, discs, blocks, etc. objects of different mass fastening material (Velcro\u2122 strips, tape, Plasticine\u2122, magnets, etc.) balance timing device (stopwatch, spark-timer, ticker-tape timer, electronic speed-timing device, or time-lapse camera) metre-sticks Procedure 1 Copy Tables 9.1 and 9.2 on page 472 into your notebook. 2 Set up the equipment in such a way that friction is minimized and the two colliding objects travel in the same straight line. 5 Set up the timing device to measure the velocity of object 1 just before and just after collision. Object 2 will be stationary before collision. The velocities of both objects will be the same after collision because they will stick together. 6 Send object 1 at a moderate speed on a collision course with object 2. Ensure that both objects will stick together and that the timing device is working properly. Make adjustments if needed. 7 Send object 1 at a moderate speed on a collision course with stationary object 2, recording the relevant observations and the masses as trial 1. 8 Send object 1 at a different speed on a collision course with stationary object 2, recording the relevant observations and the masses as trial 2. 9 Change the mass of one of the objects and again send object 1 at a moderate speed on a collision course with stationary object 2, recording the relevant observations and the masses as trial 3. 10 If you can simultaneously measure the speed of two objects, run trials where both objects are in motion before the collision. Do one trial in which they begin moving toward each other and stick together upon impact, and another trial where they move apart after impact. If you remove the fastening material, you will have to remeasure the masses of the objects. Include the direction of motion for both objects before and after collision. 11 If you did not do step 10, do two more trials, changing the mass of one of the objects each time. Include the direction of motion for both objects before and after collision. 3 Attach some fastening material to the colliding objects, so that the two objects remain together after impact. e LAB 4 Measure and record the masses of the two objects. If necessary, change the mass of one object so that the two objects have significantly different masses. For a probeware activity, go", " to www.pearsoned.ca/school/physicssource. Chapter 9 The momentum of an isolated system of interacting objects is conserved. 471 09-Phys20-Chap09.qxd 7/24/08 2:43 PM Page 472 Analysis 1. Determine the velocities for each colliding object in each trial, and record them in your data table. Show your calculations. 2. For each trial, calculate the momentum of each object just before and just after collision. Show your calculations. Record the values in your data table. 3. Calculate the momentum of the system just before and just after collision for each trial. Show your calculations. Record the values in your data table. 4. Calculate the difference between the momentum of the system just before and just after collision. Show your calculations. Record the values in your data table. 5. What is the relationship between the momentum of the system just before and just after collision? Does this relationship agree with your hypothesis? 6. What effect did friction have on your results? Explain. 7. Check your results with other groups. Account for any discrepancies. Table 9.1 Mass and Velocity Before and After for Object 1 Before and After for Object 2 Initial Velocity v 1i (m/s) Final Velocity v 1f (m/s) Mass m2 (g) Initial Velocity v 2i (m/s) Trial Mass m1 (g) 1 2 3 4 5 Table 9.2 Momentum Final Velocity v 2f (m/s) Change in Momentum of System sys g m s p Before and After for Object 1 Before and After for Object 2 Before and After for System Initial Final Momentum Momentum of System sysi g m p s of System sysf g m p s Initial Trial Momentum 1i g p m s Final Momentum 1f g p m s Initial Momentum 2i g p m s Final Momentum 2f g p m s 1 2 3 4 5 472 Unit V Momentum and Impulse 09-Phys20-Chap09.qxd 7/24/08 2:43 PM Page 473 Momentum Is Conserved in One-dimensional Collisions In 9-5 Inquiry Lab, you discovered that, in one-dimensional collisions, the momentum of a system immediately before collision is about the same as the momentum of the system immediately after collision. If the external force of friction acting on the system is negligible, the momentum of the system is constant. This result is true no matter how many objects are in the system, how many of those objects collide, how massive the objects are, or how fast they are moving. The general form of Newton\u2019s second law for a system is net)sys (F p sys t In an isolated system, the external net force on the system is zero, net)sys (F 0. So p sys 0 t p sys to be zero, the change in momentum of the system In order for t must be zero. p sysf sys p p sysi p sysi 0 0 p sysf In other words, p sys constant. This is a statement of the law of conservation of momentum. In Unit III, you encountered another conservation law, that is, in an isolated system the total energy of the system is conserved. Conservation laws always have one quantity that remains unchanged. In the law of conservation of momentum, it is momentum that remains unchanged. law of conservation of momentum: momentum of an isolated system is constant When no external net force acts on a system, the momentum of the system remains constant. p sysi p net)sys sysf where (F 0 Concept Check Why did cannons on 16th- to 19th-century warships need a rope around the back, tying them to the side of the ship (Figure 9.33)? Figure 9.33 Chapter 9 The momentum of an isolated system of interacting objects is conserved. 473 09-Phys20-Chap09.qxd 7/24/08 2:43 PM Page 474 Freaction FB on A A B Faction FA on B Figure 9.34 The action-reaction forces when two objects collide e SIM Learn how the momentum of a system just before and just after a one- dimensional collision are related. Vary the ratio of the mass of two pucks. Follow the eSim links at www.pearsoned.ca/school/ physicssource. Writing the Conservation of Momentum in Terms of Mass and Velocity Suppose a system consists of two objects, A and B. If the system is isolated, 0. Consider the internal forces of the system. At collision time, (F net)sys object A exerts a force on object B and object B exerts a force on object A (Figure 9.34). From Newton\u2019s third law, these action-reaction forces are related by the equation F A on B F B on A Objects A and B interact for the same time interval t. If you multiply both", " sides of the equation by t, you get an equation in terms of impulse: F A on B t F B on A t Since impulse is equivalent to a change in momentum, the equation can be rewritten in terms of the momenta of each object: B B p p p Bi p Bi p 0 A 0 p Af p Bf A p p Bf p Ai p Af p Ai If the mass of each object is constant during the interaction, the equation can be written in terms of m and v: mBv Bi mAv Af mBv Bf mAv Ai This equation is the law of conservation of momentum in terms of the momenta of objects A and B. So if two bighorn rams head-butt each other, the sum of the momenta of both rams is constant during the collision, even though the momentum of each ram changes. The law of conservation of momentum has no known exceptions, and holds even when particles are travelling close to the speed of light, or when the mass of the colliding particles is very small, as in the case of electrons. In real life, when objects collide, external friction acts on nearly all systems and the instantaneous forces acting on each object are usually not known (Figure 9.35). Often, the details of the interaction are also unknown. However, you do not require such information to apply the law of conservation of momentum. Instead, it is the mass and instantaneous velocity of the objects immediately before and immediately after collision that are important, so that the effects of external friction are minimal, and do not significantly affect the outcome. Figure 9.35 During a vehicle collision, many forces cause a change in the velocity and shape of each vehicle. 474 Unit V Momentum and Impulse 09-Phys20-Chap09.qxd 7/24/08 2:43 PM Page 475 Conservation of Momentum Applied to Rockets In Unit II, the motion of a rocket was explained using Newton\u2019s third law. However, the conservation of momentum can be used to explain why a rocket can accelerate even in a vacuum. When the engines of a rocket burn fuel, the escaping exhaust gas has mass and considerable speed. When a rocket is in outer space, external friction is negligible. So the rocket-exhaust gas system is an isolated system. For a two-object system, the equation for the conservation of momentum is gas gas rocket rocket p p p 0 p where, during time interval t, p rocket is the change in momentum of the rocket including any unspent fuel and p gas is the change in momentum of the fuel that is expelled in the form of exhaust gas. It is the change in momentum of the exhaust gas that enables a rocket to accelerate (Figure 9.36). In the case of a very large rocket, such as a Saturn V, the magnitude of p would be very large (Figure 9.37). gas procket change in momentum of rocket pgas change in momentum of exhaust gas Figure 9.37 From the law of conservation of momentum, the magnitude of p gas is equal to the magnitude of p rocket. That is why a rocket can accelerate on Earth or in outer space. Figure 9.36 With a height of about 112 m, the Saturn V rocket was the largest and most powerful rocket ever built. info BIT None of the 32 Saturn rockets that were launched ever failed. Altogether 15 Saturn V rockets were built. Three Saturn V rockets are on display, one at each of these locations: the Johnson Space Center, the Kennedy Space Center, and the Alabama Space and Rocket Center. Of these three, only the rocket at the Johnson Space Center is made up entirely of former flight-ready, although mismatched, parts. info BIT Design of the Saturn V began in the 1950s with the intent to send astronauts to the Moon. In the early 1970s, this type of rocket was used to launch the Skylab space station. The rocket engines in the first stage burned a combination of kerosene and liquid oxygen, producing a total thrust of magnitude 3.34 107 N. The rocket engines in the second and third stages burned a combination of liquid hydrogen and liquid oxygen. The magnitude of the total thrust produced by the second-stage engines was 5.56 106 N, and the third-stage engine produced 1.11 106 N of thrust. Concept Check (a) Refer to the second infoBIT on this page. Why is less thrust needed by the second-stage engines of a rocket? (b) Why is even less thrust needed by the third-stage engine? In Example 9.5 on the next page, the conservation of momentum is applied to a system of objects that are initially stationary. This type of interaction is called an explosion. Chapter 9 The momentum of an isolated system of interacting objects is conserved. 475 09-Phys20-Chap09.qxd 7/24/08 2:43 PM Page 476 Example 9.5 A 75-kg hunter in a stationary kayak throws a 0.72-kg", " harpoon at 12 m/s [right]. The mass of the kayak is 10 kg. What will be the velocity of the kayak and hunter immediately after the harpoon is released? Given mp v pi 75 kg 0 m/s mk v ki 10 kg 0 m/s mh v hi v hf 0.72 kg 0 m/s 12 m/s [right] before left right after? vTf 12 m/s vhf Figure 9.38 Practice Problems 1. A 110-kg astronaut and a 4000-kg spacecraft are attached by a tethering cable. Both masses are motionless relative to an observer a slight distance away from the spacecraft. The astronaut wants to return to the spacecraft, so he pulls on the cable until his velocity changes to 0.80 m/s [toward the spacecraft] relative to the observer. What will be the change in velocity of the spacecraft? 2. A student is standing on a stationary 2.3-kg skateboard. If the student jumps at a velocity of 0.37 m/s [forward], the velocity of the skateboard becomes 8.9 m/s [backward]. What is the mass of the student? Answers 1. 0.022 m/s [toward the astronaut] 2. 55 kg Required final velocity of hunter and kayak Analysis and Solution Choose the kayak, hunter, and harpoon as an isolated system. The hunter and kayak move together as a unit after the harpoon is released. So find the total mass of the hunter and kayak. mT mp mk 75 kg 10 kg 85 kg The hunter, kayak, and harpoon each have an initial velocity of zero. So the system has an initial momentum of zero. p sysi 0 Apply the law of conservation of momentum. sysf p p sysi p p p hf Tf sysi mhv 0 mTv hf Tf m hv m T 0.72 kg 85 kg v Tf hf (12 m/s) 0.10 m/s 0.10 m/s [left] v Tf Paraphrase and Verify The velocity of the kayak and hunter will be 0.10 m/s [left] immediately after the harpoon is released. Since the harpoon is thrown right, from Newton\u2019s third law, you would expect the hunter and kayak to move left after the throw. So the answer is reasonable. 476 Unit V Momentum and Impulse 09-Phys20-Chap09.qxd 7/24/08 2:43 PM Page 477 In Example 9.6, a dart is fired at a stationary block sitting on a glider. This situation involves two objects (dart and block) that join together and move as a unit after interaction. This type of interaction is called a hit-and-stick interaction. Example 9.6 A wooden block attached to a glider has a combined mass of 0.200 kg. Both the block and glider are at rest on a frictionless air track. A dart gun shoots a 0.012-kg dart into the block. The velocity of the block-dart system after collision is 0.78 m/s [right]. What was the velocity of the dart just before it hit the block? Given mb v bi 0.200 kg 0 m/s before left? vdi 0.012 kg 0.78 m/s [right] f md v right after vf 0.78 m/s Figure 9.39 Required initial velocity of dart (v di) Analysis and Solution Choose the block, glider, and dart as an isolated system. The dart, block, and glider move together as a unit after collision. The block-glider unit has an initial velocity of zero. So its initial momentum is zero. p bi 0 Apply the law of conservation of momentum. sysf p p sysf md)v (mb mdv mb m d f f 0.200 kg 0.012 kg 0.012 kg (0.78 m/s) (0.78 m/s) 0.212 kg 0.012 kg 14 m/s 14 m/s [right] p sysi p p di bi 0 mdv di v di v di Paraphrase Practice Problems 1. A student on a skateboard, with a combined mass of 78.2 kg, is moving east at 1.60 m/s. As he goes by, the student skilfully scoops his 6.4-kg backpack from the bench where he had left it. What will be the velocity of the student immediately after the pickup? 2. A 1050-kg car at an intersection has a velocity of 2.65 m/s [N]. The car hits the rear of a stationary truck, and their bumpers lock together. The velocity of the cartruck system immediately after collision is 0.78 m/s [N]. What is the mass of", " the truck? Answers 1. 1.5 m/s [E] 2. 2.5 103 kg The dart had a velocity of 14 m/s [right] just before it hit the block. Chapter 9 The momentum of an isolated system of interacting objects is conserved. 477 09-Phys20-Chap09.qxd 7/24/08 2:44 PM Page 478 Example 9.7 involves a basketball player, initially moving with some velocity, colliding with a stationary player. After the interaction, both players move in different directions. Example 9.7 A basketball player and her wheelchair (player A) have a combined mass of 58 kg. She moves at 0.60 m/s [E] and pushes off a stationary player (player B) while jockeying for a position near the basket. Player A ends up moving at 0.20 m/s [W]. The combined mass of player B and her wheelchair is 85 kg. What will be player B\u2019s velocity immediately after the interaction? Given mA v Ai v Af 58 kg 0.60 m/s [E] 0.20 m/s [W] before 0.60 m/s vAi mB v Bi 85 kg 0 m/s W E after 0.20 m/s vAf? vBf Figure 9.40 Player A Player B Player A Player B Practice Problems 1. A 0.25-kg volleyball is flying west at 2.0 m/s when it strikes a stationary 0.58-kg basketball dead centre. The volleyball rebounds east at 0.79 m/s. What will be the velocity of the basketball immediately after impact? 2. A 9500-kg rail flatcar moving forward at 0.70 m/s strikes a stationary 18 000-kg boxcar, causing it to move forward at 0.42 m/s. What will be the velocity of the flatcar immediately after collision if they fail to connect? Answers 1. 1.2 m/s [W] 2. 0.096 m/s [backward] Required final velocity of player B (v Bf) Analysis and Solution Choose players A and B as an isolated system. Player B has an initial velocity of zero. So her initial momentum is zero. p Bi 0 Apply the law of conservation of momentum. p Ai mAv Ai p sysi p Bi sysf p p Af 0 mAv Af m (v v A Bf m B 58 kg 85 kg p Bf mBv Bf v Ai Af) [0.60 m/s (0.20 m/s)] (0.60 m/s 0.20 m/s) 5 8 8 5 0.55 m/s 0.55 m/s [E] v Bf Paraphrase Player B\u2019s velocity is 0.55 m/s [E] just after collision. 478 Unit V Momentum and Impulse 09-Phys20-Chap09.qxd 7/24/08 2:44 PM Page 479 In Example 9.8, two football players in motion collide with each other. After the interaction, the players bounce apart. Example 9.8 A 110-kg Stampeders football fullback moving east at 1.80 m/s on a snowy playing field is struck by a 140-kg Eskimos defensive lineman moving west at 1.50 m/s. The fullback is bounced west at 0.250 m/s. What will be the velocity of the Eskimos defensive lineman just after impact? Given mS v Si v Sf 110 kg 1.80 m/s [E] 0.250 m/s [W] before vSi 1.80 m/s W m E v Ei 140 kg 1.50 m/s [W] E 1.50 m/s vEi after vSf 0.250 m/s? vEf Figure 9.41 Required final velocity of Eskimos lineman (v Ef) Analysis and Solution Choose the fullback and lineman as an isolated system. Apply the law of conservation of momentum. p sysi p p Ei Si mEv Ei mEv Ef mSv Si v Ef v Ef Sf Si sysf v v Ei mSv Sf m Sv m E p Ef mEv Ef mEv Ei p p Sf mSv Sf mSv Si m S v m E m S(v Sf) v Si Ei m E 110 kg [(1.80 m/s) 140 kg (0.250 m/s)] (1.50 m/s) (1.80 m/s 0.250 m/s) 110 140 1.50 m/s 0.111 m/s 0.111 m/s [E] Practice Problems 1. A 72-kg snowboarder gliding at 1.6 m/s [", "E] bounces west at 0.84 m/s immediately after colliding with an 87-kg skier travelling at 1.4 m/s [W]. What will be the velocity of the skier just after impact? 2. A 125-kg bighorn ram butts heads with a younger 122-kg ram during mating season. The older ram is rushing north at 8.50 m/s immediately before collision, and bounces back at 0.11 m/s [S]. If the younger ram moves at 0.22 m/s [N] immediately after collision, what was its velocity just before impact? Answers 1. 0.62 m/s [E] 2. 8.6 m/s [S] Paraphrase The velocity of the Eskimos defensive lineman immediately after impact is 0.111 m/s [E]. Chapter 9 The momentum of an isolated system of interacting objects is conserved. 479 09-Phys20-Chap09.qxd 7/24/08 2:44 PM Page 480 PHYSICS INSIGHT The law of conservation of energy states that the total energy of an isolated system remains constant. The energy may change into several different forms. This law has no known exceptions. e SIM Predict the speed of two pucks just after a onedimensional collision using momentum and energy concepts. Follow the eSim links at www.pearsoned.ca/school/ physicssource. Elastic and Inelastic Collisions in One Dimension In Examples 9.3 to 9.8, some of the collisions involved hard objects, such as the golf club hitting the golf ball. Other collisions, such as the block and dart, involved a dart that became embedded in a softer material (a block of wood). In all these collisions, it was possible to choose an isolated system so that the total momentum of the system was conserved. When objects collide, they sometimes deform, make a sound, give off light, or heat up a little at the moment of impact. Any of these observations indicate that the kinetic energy of the system before collision is not the same as after collision. However, the total energy of the system is constant. Concept Check (a) Is it possible for an object to have energy and no momentum? Explain, using an example. (b) Is it possible for an object to have momentum and no energy? Explain, using an example. Elastic Collisions Suppose you hit a stationary pool ball dead centre with another pool ball so that the collision is collinear and the balls move without spinning immediately after impact. What will be the resulting motion of both balls (Figure 9.42)? The ball that was initially moving will become stationary upon impact, while the other ball will start moving in the same direction as the incoming ball. If you measure the speed of both balls just before and just after collision, you will find that the speed of the incoming 1 mv2, 2 ball is almost the same as that of the outgoing ball. Since Ek the final kinetic energy of the system is almost the same as the initial kinetic energy of the system. Figure 9.42 Many collisions take place during a game of pool. What evidence suggests that momentum is conserved during the collision shown in the photo? What evidence suggests that energy is conserved? 480 Unit V Momentum and Impulse 09-Phys20-Chap09.qxd 7/24/08 2:44 PM Page 481 If the initial kinetic energy of a system is equal to the final kinetic energy of the system after collision, the collision is elastic. In an elastic collision, the total kinetic energy of the system is conserved. elastic collision: a collision Ekf in which Eki Eki Ekf Most macroscopic interactions in the real world involve some of the initial kinetic energy of the system being converted to sound, light, or deformation (Figure 9.43). When deformation occurs, some of the initial kinetic energy of the system is converted to heat because friction acts on objects in almost all situations. These factors make it difficult to achieve an elastic collision. Even if two colliding objects are hard and do not appear to deform, energy is still lost in the form of sound, light, and/or heat due to friction. Usually, the measured speed of an object after interaction is a little less than the predicted speed, which indicates that the collision is inelastic. Example 9.9 demonstrates how to determine if the collision between a billiard ball and a snooker ball is elastic. Project LINK How will you apply the concepts of conservation of momentum and conservation of energy to the design of the water balloon protection? info BIT A steel sphere will bounce as high on a steel anvil as a rubber ball will on concrete. However, when a steel sphere is dropped on linoleum or hardwood, even more kinetic energy is lost and the sphere hardly bounces at all. The kinetic energy of the sphere is converted to sound, heat, and the deformation of the floor surface. To try", " this, use flooring samples. Do not try this on floors at home or at school. Figure 9.43 Is the collision shown in this photo elastic? What evidence do you have to support your answer? Chapter 9 The momentum of an isolated system of interacting objects is conserved. 481 09-Phys20-Chap09.qxd 7/24/08 2:44 PM Page 482 Example 9.9 Practice Problems 1. A 45.9-g golf ball is stationary on the green when a 185-g golf club face travelling horizontally at 1.24 m/s [E] strikes it. After impact, the club face continues moving at 0.760 m/s [E] while the ball moves at 1.94 m/s [E]. Assume that the club face is vertical at the moment of impact so that the ball does not spin. Determine if the collision is elastic. 2. An argon atom with a mass of 6.63 1026 kg travels at 17 m/s [right] and strikes another identical argon atom dead centre travelling at 20 m/s [left]. The first atom rebounds at 20 m/s [left], while the second atom moves at 17 m/s [right]. Determine if the collision is elastic. Answers 1. inelastic 2. elastic A 0.160-kg billiard ball travelling at 0.500 m/s [N] strikes a stationary 0.180-kg snooker ball and rebounds at 0.0230 m/s [S]. The snooker ball moves off at 0.465 m/s [N]. Ignore possible rotational effects. Determine if the collision is elastic. 0.160 kg 0.180 kg 0.500 m/s [N] Given mb ms v bi v 0 m/s si v bf v 0.465 m/s [N] sf 0.0230 m/s [S] before after N S vsf 0.465 m/s Required determine if the collision is elastic snooker ball vbi 0.500 m/s snooker ball billiard ball vbf 0.0230 m/s Figure 9.44 billiard ball Analysis and Solution Choose the billiard ball and the snooker ball as an isolated system. Calculate the total initial kinetic energy and the total final kinetic energy of the system. Eki mb(vbi)2 1 1 ms(vsi)2 2 2 1 (0.160 kg)(0.500 m/s)2 0 2 Ekf 0.0200 kgm2/s2 0.0200 J mb(vbf)2 1 1 ms(vsf)2 2 2 1 (0.160 kg)(0.0230 m/s)2 2 1 (0.180 kg)(0.465 m/s)2 2 0.0195 kgm2/s2 0.0195 J Since Eki Ekf, the collision is inelastic. Paraphrase The collision between the billiard ball and the snooker ball is inelastic. 482 Unit V Momentum and Impulse 09-Phys20-Chap09.qxd 7/24/08 2:44 PM Page 483 inelastic collision: a collision in which Eki Ekf e WEB Research examples of elastic and inelastic one- dimensional collisions. Then analyze how the momentum and energy change in those collisions. Begin your search at www.pearsoned.ca/ school/physicssource. Inelastic Collisions In 9-2 QuickLab on page 455, after the putty ball collided with a hard surface, the putty ball became stationary and had no kinetic energy. Upon impact, the putty ball deformed and the kinetic energy of the putty ball was converted mostly to thermal energy. Although the total energy of the system was conserved, the total initial kinetic energy of the system was not equal to the total final kinetic energy of the system after collision. This type of collision is inelastic. In an inelastic collision, the total kinetic energy of the system is not conserved. Eki Ekf One type of inelastic collision occurs when two objects stick together after colliding. However, this type of interaction does not necessarily mean that the final kinetic energy of the system is zero. For example, consider a ballistic pendulum, a type of pendulum used to determine the speed of bullets before electronic timing devices were invented (Figure 9.45). suspension wire ceiling pistol v bullet block height of swing Figure 9.45 When a bullet is fired into the block, both the block and bullet move together as a unit after impact. The pendulum consists of a stationary block of wood suspended from the ceiling by light ropes or cables. When a bullet is fired at the block, the bullet becomes embedded in the wood upon impact. The kinetic energy of the bullet is converted to sound, thermal energy,", " deformation of the wood and bullet, and the kinetic energy of the pendulum-bullet system. The initial momentum of the bullet causes the pendulum to move upon impact, but since the pendulum is suspended by cables, it swings upward just after the bullet becomes embedded in the block. As the pendulum-bullet system swings upward, its kinetic energy is converted to gravitational potential energy. Example 9.10 involves a ballistic pendulum. By using the conservation of energy, it is possible to determine the speed of the pendulum-bullet system immediately after impact. By applying the conservation of momentum to the collision, it is possible to determine the initial speed of the bullet. Chapter 9 The momentum of an isolated system of interacting objects is conserved. 483 09-Phys20-Chap09.qxd 7/24/08 2:44 PM Page 484 Example 9.10 A 0.0149-kg bullet from a pistol strikes a 2.0000-kg ballistic pendulum. Upon impact, the pendulum swings forward and rises to a height of 0.219 m. What was the velocity of the bullet immediately before impact? backward before after forward vbi? h Given mb 0.0149 kg mp 2.0000 kg h 0.219 m Required initial velocity of bullet (v bi ) Ek 0 Ep 0 vf? Ek 0 Ep (mb mp)gh Figure 9.46 Analysis and Solution Choose the pendulum and the bullet as an isolated system. Since the pendulum is stationary before impact, its initial velocity is zero. So its initial momentum is zero. p pi 0 Immediately after collision, the bullet and pendulum move together as a unit. The kinetic energy of the pendulum-bullet system just after impact is converted to gravitational potential energy. Ek Ep Practice Problems 1. A 2.59-g bullet strikes a stationary 1.00-kg ballistic pendulum, causing the pendulum to swing up to 5.20 cm from its initial position. What was the speed of the bullet immediately before impact? 2. A 7.75-g bullet travels at 351 m/s before striking a stationary 2.5-kg ballistic pendulum. How high will the pendulum swing? Answers 1. 391 m/s 2. 6.0 cm Apply the law of conservation of energy to find the speed of the pendulum-bullet system just after impact. 1 (mb 2 Ep Ek mp) (vf)2 (mb mp) g(h) (vf)2 2g(h) vf v f (0.219 m) 2g(h) 29.81 m s2 2.073 m/s 2.073 m/s [forward] Apply the law of conservation of momentum to find the initial velocity of the bullet. p bi mbv bi p sysi p pi sysf p p sysf mp)v 0 (mb mpv mb m b v bi f f 0.0149 kg 2.0000 kg 0.0149 kg (2.073 m/s) 484 Unit V Momentum and Impulse v bi (2.073 m/s) 2.0149 kg 0.0149 kg 280 m/s 280 m/s [forward] 09-Phys20-Chap09.qxd 7/24/08 2:44 PM Page 485 Paraphrase The initial velocity of the bullet immediately before impact was 280 m/s [forward]. Example 9.11 demonstrates how to determine if the collision in Example 9.10 is elastic or inelastic by comparing the kinetic energy of the system just before and just after collision. Example 9.11 Determine if the collision in Example 9.10 is elastic or inelastic. Given mb mp 0.0149 kg 2.0000 kg v bi v f 280 m/s [forward] from Example 9.10 2.073 m/s [forward] from Example 9.10 Required initial and final kinetic energies (Eki and Ekf) to find if the collision is elastic Analysis and Solution Choose the pendulum and the bullet as an isolated system. Calculate the total initial kinetic energy and the total final kinetic energy of the system. Ekf Eki mb(vbi)2 1 1 mp(vpi)2 2 2 1 (0.0149 kg)(280 m/s)2 0 2 585 kgm2/s2 585 J mp)(vf)2 1 (mb 2 1 (0.0149 kg 2.0000 kg)(2.073 m/s)2 2 4.33 kgm2/s2 4.33 J Since Eki Ekf, the collision is inelastic. Paraphrase and Verify Since the kinetic energy of the system just before impact is much greater than the kinetic energy of the system just after impact, the collision is inelastic. This result makes sense since the bullet became embedded in the pend", "ulum upon impact. Practice Problems 1. In Example 9.6 on page 477, how much kinetic energy is lost immediately after the interaction? 2. (a) Determine if the interaction in Example 9.8 on page 479 is elastic. (b) What percent of kinetic energy is lost? Answers 1. 1.1 J 2. (a) inelastic (b) 98.7% Chapter 9 The momentum of an isolated system of interacting objects is conserved. 485 09-Phys20-Chap09.qxd 7/24/08 2:44 PM Page 486 9.3 Check and Reflect 9.3 Check and Reflect Knowledge 1. In your own words, state the law of conservation of momentum. 2. (a) In the context of momentum, what is an isolated system? (b) Why is it necessary to choose an isolated system when solving a momentum problem? 3. Explain the difference between an elastic and an inelastic collision. Include an example of each type of collision in your answer. 8. A 60.0-kg student on a 4.2-kg skateboard is travelling south at 1.35 m/s. A friend throws a 0.585-kg basketball to him with a velocity of 12.6 m/s [N]. What will be the velocity of the student and skateboard immediately after he catches the ball? 9. A hockey forward with a mass of 95 kg skates in front of the net at 2.3 m/s [E]. He is met by a 104-kg defenceman skating at 1.2 m/s [W]. What will be the velocity of the resulting tangle of players if they stay together immediately after impact? 4. What evidence suggests that a collision is 10. A 75.6-kg volleyball player leaps toward the (a) elastic? (b) inelastic? Applications 5. Give two examples, other than those in the text, of possible collinear collisions between two identical masses. Include a sketch of each situation showing the velocity of each object immediately before and immediately after collision. 6. A student is sitting in a chair with nearly frictionless rollers. Her homework bag is in an identical chair right beside her. The chair and bag have a combined mass of 20 kg. The student and her chair have a combined mass of 65 kg. If she pushes her homework bag away from her at 0.060 m/s relative to the floor, what will be the student\u2019s velocity immediately after the interaction? 7. At liftoff, a space shuttle has a mass of 2.04 106 kg. The rocket engines expel 3.7 103 kg of exhaust gas during the first second of liftoff, giving the rocket a velocity of 5.7 m/s [up]. At what velocity is the exhaust gas leaving the rocket engines? Ignore the change in mass due to the fuel being consumed. The exhaust gas needed to counteract the force of gravity has already been accounted for and should not be part of this calculation. net to block the ball. At the top of his leap, he has a horizontal velocity of 1.18 m/s [right], and blocks a 0.275-kg volleyball moving at 12.5 m/s [left]. The volleyball rebounds at 6.85 m/s [right]. (a) What will be the horizontal velocity of the player immediately after the block? (b) Determine if the collision is elastic. 11. A 220-kg bumper car (A) going north at 0.565 m/s hits another bumper car (B) and rebounds at 0.482 m/s [S]. Bumper car B was initially travelling south at 0.447 m/s, and after collision moved north at 0.395 m/s. (a) What is the mass of bumper car B? (b) Determine if the collision is elastic. Extension 12. Summarize the concepts and ideas associated with one-dimensional collisions using a graphic organizer of your choice. See Student References 4: Using Graphic Organizers on pp. 869\u2013871 for examples of different graphic organizers. Make sure that the concepts and ideas are clearly presented and appropriately linked. e TEST To check your understanding of the conservation of momentum and one-dimensional collisions, follow the eTest links at www.pearsoned.ca/school/physicssource. 486 Unit V Momentum and Impulse 09-Phys20-Chap09.qxd 7/24/08 2:44 PM Page 487 9.4 Collisions in Two Dimensions Many interactions in the universe involve collisions. Comets, asteroids, and meteors sometimes collide with celestial bodies. Molecules and atoms are constantly colliding during chemical reactions throughout the universe: in stars, in Earth\u2019s atmosphere, and even within your body. An interesting collision in recent history occurred on June 30, 1908, at Tunguska, Siberia, between a cosmic object", " and Earth (Figure 9.47). Eyewitnesses reported seeing a giant fireball that moved rapidly across the sky and eventually collided with the ground. Upon impact, a tremendous explosion occurred producing an atmospheric shock wave that circled Earth twice. About 2000 km2 of forest were levelled and thousands of trees were burned. In fact, there was so much fine dust in the atmosphere that people in London, England, could read a newspaper at night just from the scattered light. info BIT Scientists speculate that the cosmic object that hit Tunguska was about 100 m across and had a mass of about 1 106 t. The estimated speed of the object was about 30 km/s, which is 1.1 105 km/h. After the collision at Tunguska, a large number of diamonds were found scattered all over the impact site. So the cosmic object contained diamonds as well as other materials. Figure 9.47 The levelled trees and charred remnants of a forest at Tunguska, Siberia, after a cosmic object collided with Earth in 1908. Although the chance that a similar collision with Earth during your lifetime may seem remote, such collisions have happened throughout Earth\u2019s history. In real life, most collisions occur in three dimensions. Only in certain situations, such as those you studied in section 9.3, does the motion of the interacting objects lie along a straight line. In this section, you will examine collisions that occur in two dimensions. These interactions occur when objects in a plane collide off centre. In 9-1 QuickLab on page 447, you found that when two coins collide off centre, the resulting path of each coin is in a different direction from its initial path. You may have noticed that certain soccer or hockey players seem to be at the right place at the right time whenever there is a rebound from the goalie. How do these players know where to position themselves so that they can score on the rebound? Find out by doing 9-6 Inquiry Lab. Chapter 9 The momentum of an isolated system of interacting objects is conserved. 487 09-Phys20-Chap09.qxd 7/24/08 2:44 PM Page 488 9-6 Inquiry Lab 9-6 Inquiry Lab Analyzing Collisions in Two Dimensions Required Skills Initiating and Planning Performing and Recording Analyzing and Interpreting Communication and Teamwork Question How does the momentum of a two-body system in the x and y directions compare just before and just after a collision? Hypothesis State a hypothesis relating the momentum of a system in each direction immediately before and immediately after collision. Remember to write an \u201cif/then\u201d statement. Variables Read the procedure and identify the controlled, manipulated, and responding variables in the experiment. Materials and Equipment air table or bead table pucks spark-timer or camera set-up to measure velocities rulers or metre-sticks protractors Procedure 1 Copy Tables 9.3 and 9.4 on page 489 into your notebook. 2 Label the pucks as \u201cpuck 1\u201d and \u201cpuck 2\u201d respectively. Measure the mass of each puck and record it in Table 9.3. 3 Set up the apparatus so that puck 2 is at rest near the centre of the table. 4 Have each person in your group do one trial. Each time, send puck 1 aimed at the left side of puck 2, recording the paths of both pucks. Make sure the recording tracks of both pucks can be used to accurately measure their velocities before and after collision. 5 Have each person in your group measure and analyze one trial. Help each other as needed to ensure the measurements and calculations for each trial are accurate. 6 Find a suitable point on the recorded tracks to be the impact location. 7 On the path of puck 1 before collision, choose an interval where the speed is constant. Choose the positive x-axis to be in the initial direction of puck 1. 8 Using either the spark dots, the physical centre of the puck, or the leading or trailing edge of the puck, measure the distance and the time interval. Record those values in Table 9.3. 9 On the path of each puck after collision, choose an interval where the speed is constant. Measure the distance, direction of motion relative to the positive x-axis, and time interval. Record those values in Table 9.3. Analysis 1. Calculate the initial velocity and initial momentum of puck 1. Record the values in Table 9.4. 2. Calculate the velocity of puck 1 after collision. Resolve the velocity into x and y components. Record the values in Table 9.4. 3. Use the results of question 2 to calculate the x and y components of the final momentum of puck 1. Record the values in Table 9.4. 4. Repeat questions 2 and 3 but this time use the data for puck 2. Explain why the y component of the momentum of puck 2 is negative. 5. Record the calculated values from each member of your group as a different", " trial in Table 9.4. 6. For each trial, state the relationship between the initial momentum of the system in the x direction and the final momentum of the system in the x direction. Remember to consider measurement errors. Write this result as a mathematical statement. 7. The initial momentum of the system in the y direction was zero. For each trial, what was the final momentum of the system in the y direction? Remember to consider measurement errors. Write this result as a mathematical statement. 8. Compare your answers to questions 6 and 7 with other groups. Does this relationship agree with your hypothesis? Account for any discrepancies. e LAB For a probeware activity, go to www.pearsoned.ca/school/physicssource. 488 Unit V Momentum and Impulse 09-Phys20-Chap09.qxd 7/24/08 2:44 PM Page 489 Table 9.3 Mass, Distance, Time Elapsed, and Angle Before and After for Puck 1 After for Puck 2 Trial Mass m1 (g) Initial Distance d1i (m) Initial Time Elapsed t1i (s) Final Distance d1f (m) Final Time Elapsed t1f (s) Final Angle () 1f Mass m2 (g) Final Distance d2f (m) Final Time Elapsed t2f (s) Final Angle () 2f 1 2 3 4 5 Table 9.4 Velocity and Momentum Before and After for Puck 1 After for Puck 2 Initial x Final x Initial x Velocity Momentum Velocity Final y Final x Velocity Momentum Momentum Velocity Final y Final x (m/s) p1ix gm v1fx v1ix (m/s) v1fy (m/s) p1fx gm p1fy gm v2fx (m/s) v2fy s s Final x Final y Final y Velocity Momentum Momentum p2fy gm s (m/s) p2fx gm s s Trial 1 2 3 4 5 Momentum Is Conserved in Two-dimensional Collisions In 9-6 Inquiry Lab, you found that along each direction, x and y, the momentum of the system before collision is about the same as the momentum of the system immediately after collision. In other words, momentum is conserved in two-dimensional collisions. This result agrees with what you saw in 9-5 Inquiry Lab, where only one-dimensional collisions were examined. As in one-dimensional collisions, the law of conservation of momentum is valid only when no external net force acts on the system. In two dimensions, the motion of each object in the system must be analyzed in terms of two perpendicular axes. To do this, you can either use a vector addition diagram drawn to scale or vector components. The law of conservation of momentum can be stated using components in the x and y directions. In two-dimensional collisions where no external net force acts on the system, the momentum of the system in both the x and y directions remains constant. psysix psysfx and psysiy psysfy where (F net)sys 0 e SIM Apply the law of conservation of momentum to twodimensional collisions. Go to www.pearsoned.ca/school/ physicssource. Chapter 9 The momentum of an isolated system of interacting objects is conserved. 489 09-Phys20-Chap09.qxd 7/24/08 2:44 PM Page 490 info BIT In championship curling, rebound angles and conservation of momentum are crucial for placing stones in counting position behind guards. Just nudging a stone several centimetres can make all the difference. Concept Check (a) Will the magnitude of the momentum of an object always increase if a non-zero net force acts on it? Explain, using an example. (b) How can the momentum of an object change but its speed remain the same? Explain, using an example. Example 9.12 involves a curling stone colliding off centre with an identical stone that is at rest. The momentum of each stone is analyzed in two perpendicular directions. Example 9.12 A 19.6-kg curling stone (A) moving at 1.20 m/s [N] strikes another identical stationary stone (B) off centre, and moves off with a velocity of 1.17 m/s [12.0\u00b0 E of N]. What will be the velocity of stone B after the collision? Ignore frictional and rotational effects. Given mA v Bi 19.6 kg 0 m/s before N mB v Af 19.6 kg 1.17 m/s [12.0 E of N] v Ai 1.20 m/s [N] after N vAf 1.17 m/s vBf? 12.0\u00b0 E W Figure 9.48 S N W vBi 0 m/s E vAi 1.20 m/s S Required final velocity", " of stone B (v Bf) Analysis and Solution Choose both curling stones as an isolated system. Stone B has an initial velocity of zero. So its initial momentum is zero. p Bi 0 Resolve all velocities into east and north components (Figure 9.49). vAf Vector v Ai v Bi v Af East component North component 0 1.20 m/s 0 (1.17 m/s)(sin 12.0) 0 (1.17 m/s)(cos 12.0) 12.0\u00b0 E Figure 9.49 490 Unit V Momentum and Impulse 09-Phys20-Chap09.qxd 7/28/08 9:26 AM Page 491 Apply the law of conservation of momentum to the system in the east and north directions. E direction psysiE pBiE pBfE pAiE pBfE pBiE psysfE pAfE pAiE mAvAiE 0 0 (19.6 kg)(1.17 m/s)(sin 12.0) 4.768 kgm/s pAfE mBvBiE mAvAfE The negative sign indicates the vector is directed west. N direction psysiN pBiN pBfN pAiN pBfN pBiN psysfN pAfN pAfN pAiN mAvAiN mAvAfN mBvBiN (19.6 kg)(1.20 m/s) 0 (19.6 kg)(1.17 m/s)(cos 12.0) 1.089 kgm/s Draw a vector diagram of the components of the final momentum of stone B and find the magnitude of the resultant p Pythagorean theorem. Bf using the N 1.089 kg m/s W Figure 9.50 p Bf \u03b8 4.768 kg m/s E pBf (4.768 kgm/s)2 (1.089 kgm/s)2 4.8906 kgm/s Use the magnitude of the momentum and the mass of stone B to find its final speed. mBvBf pBf mB pBf vBf vBf 4.8906 kgm/s 19.6 kg 0.250 m/s Use the tangent function to find the direction of the final momentum. tan pBfN pBfW tan\u20131 12.9 1.089 kgm/s 4.768 kgm/s The final velocity will be in the same direction as the final momentum. Paraphrase The velocity of stone B after the collision is 0.250 m/s [12.9\u00b0 N of W]. Practice Problems 1. A 97.0-kg hockey centre stops momentarily in front of the net. He is checked from the side by a 104-kg defenceman skating at 1.82 m/s [E], and bounces at 0.940 m/s [18.5 S of E]. What is the velocity of the defenceman immediately after the check? 2. A 1200-kg car, attempting to run a red light, strikes a stationary 1350-kg vehicle waiting to make a turn. Skid marks show that after the collision, the 1350-kg vehicle moved at 8.30 m/s [55.2 E of N], and the other vehicle at 12.8 m/s [36.8 W of N]. What was the velocity of the 1200-kg vehicle just before collision? Note this type of calculation is part of many vehicle collision investigations where charges may be pending. Answers 1. 1.03 m/s [15.7 N of E] 2. 15.6 m/s [N] Chapter 9 The momentum of an isolated system of interacting objects is conserved. 491 09-Phys20-Chap09.qxd 7/24/08 2:44 PM Page 492 centre of mass: point where the total mass of an object can be assumed to be concentrated info BIT When an object is symmetric and has uniform density, its centre of mass is in the same location as the physical centre of the object. Example 9.13 involves a football tackle with two players. Each player has an initial velocity, but when they collide, both players move together as a unit. To analyze the motion, the centre of mass of the combination of both players must be used. The centre of mass is a point that serves as an average location of the total mass of an object or system. Depending on the distribution of mass, the centre of mass may be located even outside the object. Generally, momentum calculations are made using the centre of mass of an object. No matter where any external forces are acting on an object, whether the object is rotating or not, or whether the object is deformable or rigid, the translational motion of the object can be easily analyzed in terms of its", " centre of mass. Example 9.13 A 90-kg quarterback moving at 7.0 m/s [270] is tackled by a 110-kg linebacker running at 8.0 m/s [0]. What will be the velocity of the centre of mass of the combination of the two players immediately after impact? Practice Problems 1. A 2000-kg car travelling at 20.0 m/s [90.0] is struck at an intersection by a 2500-kg pickup truck travelling at 14.0 m/s [180]. If the vehicles stick together upon impact, what will be the velocity of the centre of mass of the cartruck combination immediately after the collision? 2. A 100-kg hockey centre is moving at 1.50 m/s [W] in front of the net. He is checked by a 108-kg defenceman skating at 4.20 m/s [S]. Both players move off together after collision. What will be the velocity of the centre of mass of the combination of the two players immediately after the check? Answers 1. 11.8 m/s [131] 2. 2.30 m/s [71.7 S of W] Given mq v qi 90 kg 7.0 m/s [270] before 8.0 m/s vli y 110 kg 8.0 m/s [0] ml v li after y x 7.0 m/s vqi x vf? Figure 9.51 Required final velocity of centre of mass of the two players (v f) Analysis and Solution Choose the quarterback and the linebacker as an isolated system. The linebacker tackled the quarterback. So both players have the same final velocity. Resolve all velocities into x and y components. Vector v qi v li x component y component 0 7.0 m/s 8.0 m/s 0 Apply the law of conservation of momentum to the system in the x and y directions. 492 Unit V Momentum and Impulse 09-Phys20-Chap09.qxd 7/28/08 9:30 AM Page 493 x direction pqix psysix plix psysfx plix psysfx psysfx pqix mqvqix mlvlix 0 (110 kg)(8.0 m/s) 880 kgm/s y direction psysiy pliy psysfy pqiy pliy psysfy psysfy pqiy mqvqiy mlvliy (90 kg)(7.0 m/s) 0 630 kgm/s Draw a vector diagram of the components of the final momentum of the two players and find the magnitude of the resultant p Pythagorean theorem. sysf using the y 324\u00ba 880 kgm/s \u03b8 x 630 kgm/s psysf Figure 9.52 psysf (880kgm/s)2 (630 kgm/s)2 1082 kgm/s Use the magnitude of the momentum and combined masses of the two football players to find their final speed. psysf vf (mq ml)vf psysf ml) (mq 1082 kgm/s (90 kg 110 kg) = 5.4 m/s Use the tangent function to find the direction of the final momentum. tan psysfy psysfx tan1 630 kgm/s 880 kgm/s = 35.6\u00b0 The final velocity will be in the same direction as the final momentum. From Figure 9.52, is below the positive x-axis. So the direction of v measured counterclockwise from the positive x-axis is 360 35.6 324.4. f v f 5.4 m/s [324] Paraphrase The final velocity of both players is 5.4 m/s [324]. Chapter 9 The momentum of an isolated system of interacting objects is conserved. 493 09-Phys20-Chap09.qxd 7/24/08 2:44 PM Page 494 Example 9.14 deals with a fireworks bundle that is initially stationary. After it explodes, three fragments (A, B, and C) fly in different directions in a plane. To demonstrate an alternative method of solving collision problems, a vector addition diagram is used to determine the momentum of fragment C. This quantity is then used to calculate its final velocity. Example 9.14 A 0.60-kg fireworks bundle is at rest just before it explodes into three fragments. A 0.20-kg fragment (A) flies at 14.6 m/s [W], and a 0.18-kg fragment (B) moves at 19.2 m/s [S]. What is the velocity of the third fragment (C) just after the explosion? Assume that no mass is lost, and that the motion of the fragments lies in a plane. Practice Problems 1. A 0.058-kg firecracker that is at rest explodes", " into three fragments. A 0.018-kg fragment moves at 2.40 m/s [N] while a 0.021-kg fragment moves at 1.60 m/s [E]. What will be the velocity of the third fragment? Assume that no mass is lost, and that the motion of the fragments lies in a plane. 2. A 65.2-kg student on a 2.50-kg skateboard moves at 0.40 m/s [W]. He jumps off the skateboard with a velocity of 0.38 m/s [30.0 S of W]. What will be the velocity of the skateboard immediately after he jumps? Ignore friction between the skateboard and the ground. Answers 1. 2.9 m/s [52 S of W] 2. 5.4 m/s [66 N of W] Given mT v i 0.60 kg 0 m/s before W N S E vi 0 m/s mA v Af 0.20 kg 14.6 m/s [W] after mB v Bf N 0.18 kg 19.2 m/s [S] vAf 14.6 m/s W vCf? E vBf 19.2 m/s Figure 9.53 S Required final velocity of fragment C (v Cf ) Analysis and Solution Choose fragments A, B, and C as an isolated system. Since no mass is lost, find the mass of fragment C. mT mB) 0.60 kg (0.20 kg 0.18 kg) 0.22 kg (mA mC The original firework has an initial velocity of zero. So the system has an initial momentum of zero. p sysi 0 pAf The momentum of each fragment is in the same direction as its velocity. Calculate the momentum of fragments A and B. mBvBf (0.18 kg)(19.2 m/s) 3.46 kgm/s 3.46 kgm/s [S] mAvAf (0.20 kg)(14.6 m/s) 2.92 kgm/s 2.92 kgm/s [W] pBf p Af p Bf 494 Unit V Momentum and Impulse 09-Phys20-Chap09.qxd 7/24/08 2:44 PM Page 495 Apply the law of conservation of momentum to the system. p p sysi 0 p sysf p Bf p Cf Af Use a vector addition diagram to determine the momentum of fragment C. N 1.00 kgm/s pAf 2.92 kgm/s pBf 3.46 kgm/s pCf? \u03b8 E Figure 9.54 From Figure 9.54, careful measurements give pCf 50 N of E. 4.53 kgm/s and pCf vCf Divide the momentum of fragment C by its mass to find the velocity. mCvCf pC f m C 4.53 kgm s 0.22 kg 21 m/s 21 m/s [50 N of E] v Cf Paraphrase The velocity of the third fragment just after the explosion is 21 m/s [50 N of E]. Elastic and Inelastic Collisions in Two Dimensions As with one-dimensional collisions, collisions in two dimensions may be either elastic or inelastic. The condition for an elastic two-dimensional Ekf. collision is the same as for an elastic one-dimensional collision, Eki To determine if a collision is elastic, the kinetic energy values before and after collision must be compared. The kinetic energy of an object only depends on the magnitude of the velocity vector. So it does not matter if the velocity vector has only an x component, only a y component, or both x and y components. If you can determine the magnitude of the velocity vector, it is possible to calculate the kinetic energy. An example of an inelastic collision occurs when two objects join together and move as a unit immediately after impact. If two objects bounce apart after impact, the collision may be either elastic or inelastic, depending on the initial and final kinetic energy of the system. Usually, if one or both colliding objects deform upon impact, the collision is inelastic. e SIM and vfy Predict vfx for an object just after a twodimensional collision using momentum and energy concepts. Follow the eSim links at www.pearsoned.ca/school/ physicssource. Chapter 9 The momentum of an isolated system of interacting objects is conserved. 495 09-Phys20-Chap09.qxd 7/24/08 2:44 PM Page 496 e WEB Research some design improvements in running shoes. Use momentum and collision concepts to explain how these features affect athletes. Write a brief report of your findings, including diagrams where appropriate. Begin your search at www.pearsoned.ca/ school/physicssource", ". In track sports, the material used for the track has a profound effect on the elasticity of the collision between a runner\u2019s foot and the running surface (Figure 9.55). If a track is made of a very hard material such as concrete, it experiences very little deformation when a runner\u2019s foot comes in contact with it. The collision is more elastic than if the track were made of a more compressible material such as cork. So less kinetic energy of the runner is converted to other forms of energy upon impact. However, running on harder tracks results in a decreased interaction time and an increase in the net force acting on each foot, which could result in more injuries to joints, bones, and tendons. But a track that is extremely compressible is not desirable either, because it slows runners down. With all the pressure to achieve faster times in Olympic and world competitions, researchers and engineers continue to search for the optimum balance between resilience and safety in track construction. On the other hand, some collisions in sporting events present a very low risk of injury to contestants. Example 9.15 involves determining if the collision between two curling stones is elastic. Figure 9.55 Canadian runner Diane Cummins (far right) competing in the 2003 World Championships. The material of a running surface affects the interaction time and the net force acting on a runner\u2019s feet. How would the net force change if the track were made of a soft material that deforms easily? Example 9.15 Determine if the collision in Example 9.12 on pages 490 and 491 is elastic. If it is not, what percent of the kinetic energy is retained? Given mA v Bi v Bf 19.6 kg 0 m/s 0.2495 m/s [77.1 W of N] mB v Af 19.6 kg 1.17 m/s [12.0 E of N] v Ai 1.20 m/s [N] Required determine if the collision is elastic Analysis and Solution Choose the two curling stones as an isolated system. Calculate the total initial kinetic energy and the total final kinetic energy of the system. 496 Unit V Momentum and Impulse 09-Phys20-Chap09.qxd 7/24/08 2:44 PM Page 497 Eki 1 1 mA(vAi)2 mB(vBi)2 2 2 1 (19.6 kg)(1.20 m/s)2 0 2 14.11 kgm2/s2 14.11 J Since Eki Find the percent of Ek retained. Ekf, the collision is inelastic. E k 100% % Ek retained f E k i 0 3. 4 1 100% 1 J 1 1. 4 99.4% J Paraphrase The collision is inelastic, and 99.4% of the kinetic energy is retained. (Collisions like this, where very little kinetic energy is lost, may be called \u201cnear elastic collisions.\u201d) Conservation Laws and the Discovery of Subatomic Particles Based on the results of experiments, scientists have gained great confidence in the laws of conservation of momentum and of conservation of energy, and have predicted that there are no known exceptions. This confidence has enabled scientists to make discoveries about the existence of particles within atoms as well. You will learn more about subatomic particles in Units VII and VIII. Ekf 1 1 mA(vAf)2 mB(vBf)2 2 2 1 1 (19.6 kg)(0.2495 m/s)2 (19.6 kg)(1.17 m/s)2 2 2 14.03 kgm2/s2 14.03 J Practice Problems 1. A 0.168-kg hockey puck flying at 45.0 m/s [252] is trapped in the pads of an 82.0-kg goalie moving at 0.200 m/s [0]. The velocity of the centre of mass of the goalie, pads, and puck immediately after collision is 0.192 m/s [333]. Was the collision elastic? If not, calculate the percent of total kinetic energy retained. 2. A 19.0-kg curling stone collides with another identical stationary stone. Immediately after collision, the first stone moves at 0.663 m/s. The second stone, which was stationary, moves at 1.31 m/s. If the collision was elastic, what would have been the speed of the first stone just before collision? Answers 1. inelastic, 0.882% 2. 1.47 m/s In 1930, German scientists Walther Bothe and Wilhelm Becker produced a very penetrating ray of unknown particles when they bombarded the element beryllium with alpha particles (Figure 9.56). An alpha particle is two protons and two neutrons bound together to form a stable particle. In 1932, British scientist James Chadwick (1891\u20131974) directed rays of these", " unknown particles at a thin paraffin strip and found that protons were emitted from the paraffin. He analyzed the speeds and angles of the emitted protons and, by using the conservation of momentum, he showed that the protons were being hit by particles of approximately the same mass. In other related experiments, Chadwick was able to determine the mass of these unknown particles very accurately using the conservation of momentum. Earlier experiments had shown that the unknown particles were neutral because they were unaffected by electric or magnetic fields. You will learn about electric and magnetic fields in Unit VI. Chadwick had attempted for several years to find evidence of a suggested neutral particle that was believed to be located in the nucleus of an atom. The discovery of these neutral particles, now called neutrons, resulted in Chadwick winning the Nobel Prize for Physics in 1935. alpha particle 2 neutron beryllium Figure 9.56 The experiment of Bothe and Becker using beryllium paved the way for Chadwick who later discovered the existence of neutrons by creating an experiment where he could detect them. e SIM Practise solving problems involving two-dimensional collisions. Follow the eSim links at www.pearsoned.ca/school/ physicssource. Chapter 9 The momentum of an isolated system of interacting objects is conserved. 497 09-Phys20-Chap09.qxd 7/24/08 2:44 PM Page 498 electron neutron pn 0 kgm/s proton pe pp Figure 9.57 If a neutron is initially stationary, p 0. If the neutron becomes transformed into a proton and an electron moving in the same direction, the momentum of the system is no longer zero. sysi Scientists later found that the neutron, when isolated, soon became transformed into a proton and an electron. Sometimes the electron and proton were both ejected in the same direction, which seemed to contradict the law of conservation of momentum (Figure 9.57). Furthermore, other experiments showed that the total energy of the neutron before transformation was greater than the total energy of both the proton and electron. It seemed as if the law of conservation of energy was not valid either. Austrian physicist Wolfgang Pauli (1900\u20131958) insisted that the conservation laws of momentum and of energy were still valid, and in 1930, he proposed that an extremely tiny neutral particle produced during the transformation must be moving in the opposite direction at an incredibly high speed. This new particle accounted for the missing momentum and missing energy (Figure 9.58). unknown particle p\u03bd neutron pn 0 kgm/s pe electron Many other scientists accepted Pauli\u2019s explanation because they were convinced that the conservation laws of momentum and of energy were valid. For 25 years, they held their belief in the existence of this tiny particle, later called a neutrino, with no other evidence. Then in 1956, the existence of neutrinos was finally confirmed experimentally, further strengthening the universal validity of conservation laws. proton pp Figure 9.58 The existence of another particle accounted for the missing momentum and missing energy observed when a neutron transforms itself into a proton and an electron. THEN, NOW, AND FUTURE Neutrino Research in Canada Canada is a world leader in neutrino research. The SNO project (Sudbury Neutrino Observatory) is a special facility that allows scientists to gather data about these extremely tiny particles that are difficult to detect. The observatory is located in INCO\u2019s Creighton Mine near Sudbury, Ontario, 2 km below Earth\u2019s surface. Bedrock above the mine shields the facility from cosmic rays that might interfere with the observation of neutrinos. The experimental apparatus consists of 1000 t of heavy water encased in an acrylic vessel shaped like a 12-m diameter boiling flask (Figure 9.59). The vessel is surrounded by an array of about 1000 photo detectors, all immersed in a 10-storey chamber of purified water. When a neutrino collides with a heavy water molecule, a tiny burst of light is emitted, which the photo detectors pick up. Despite all that equipment, scientists only detect an average of about 10 neutrinos a day. So experiments acrylic vessel with heavy water vessel of purified water photo detectors Figure 9.59 The Sudbury Neutrino Observatory is a collaborative effort among 130 scientists from Canada, the U.S., and the U.K. must run for a long time in order to collect enough useful data. Scientists are interested in neutrinos originating from the Sun and other distant parts of the universe. At first, it seemed that the Sun was not emitting as many neutrinos as expected. Scientists thought they would have to modify their theories about the reactions taking place within the core of the Sun. Tripling the sensitivity of the detection process, by adding 2 t of salt to the heavy water, showed that 2 of the neutrinos from the 3 Sun were being transformed into different types of neutrinos as they travelled. This discovery", " has important implications about the basic properties of a neutrino, including its mass. It now appears that scientists\u2019 theories about the reactions within the core of the Sun are very accurate. Continued research at this facility will help answer fundamental questions about matter and the universe. Questions 1. Why was Sudbury chosen as the site for this type of observatory? 2. Explain why, at first, it appeared that the Sun was not emitting the expected number of neutrinos. 3. If a neutrino has a very small mass and travels very fast, why doesn\u2019t it run out of energy? 498 Unit V Momentum and Impulse 09-Phys20-Chap09.qxd 7/24/08 2:44 PM Page 499 9.4 Check and Reflect 9.4 Check and Reflect Knowledge 1. How is a two-dimensional collision different from a one-dimensional collision? Explain, using examples. 2. In your own words, state the law of conservation of momentum for twodimensional collisions. Show how the law relates to x and y components by using an example. 3. In your own words, define the centre of mass of an object. 4. Explain why scientists accepted the existence of the neutrino for so long when there was no direct evidence for it. Applications 5. A cue ball travelling at 0.785 m/s [270] strikes a stationary five-ball, causing it to move at 0.601 m/s [230]. The cue ball and the five-ball each have a mass of 160 g. What will be the velocity of the cue ball immediately after impact? Ignore frictional and rotational effects. 6. A stationary 230-kg bumper car in a carnival is struck off centre from behind by a 255-kg bumper car moving at 0.843 m/s [W]. The more massive car bounces off at 0.627 m/s [42.0 S of W]. What will be the velocity of the other bumper car immediately after collision? 7. A 0.25-kg synthetic rubber ball bounces to a height of 46 cm when dropped from a height of 50 cm. Determine if this collision is elastic. If not, how much kinetic energy is lost? 8. A football halfback carrying the ball, with a combined mass of 95 kg, leaps toward the goal line at 4.8 m/s [S]. In the air at the goal line, he collides with a 115-kg linebacker travelling at 4.1 m/s [N]. If the players move together after impact, will the ball cross the goal line? 9. A 0.160-kg pool ball moving at 0.563 m/s [67.0 S of W] strikes a 0.180-kg snooker ball moving at 0.274 m/s [39.0 S of E]. The pool ball glances off at 0.499 m/s [23.0 S of E]. What will be the velocity of the snooker ball immediately after collision? 10. A 4.00-kg cannon ball is flying at 18.5 m/s [0] when it explodes into two fragments. One 2.37-kg fragment (A) goes off at 19.7 m/s [325]. What will be the velocity of the second fragment (B) immediately after the explosion? Assume that no mass is lost during the explosion, and that the motion of the fragments lies in the xy plane. 11. A 0.952-kg baseball bat moving at 35.2 m/s [0] strikes a 0.145-kg baseball moving at 40.8 m/s [180]. The baseball rebounds at 37.6 m/s [64.2]. What will be the velocity of the centre of mass of the bat immediately after collision if the batter exerts no force on the bat during and after the instant of impact? Extensions 12. Some running shoe designs contain springs. Research these types of shoes and the controversy surrounding them. How do momentum and impulse apply to these shoes? Write a brief report of your findings. Include diagrams where appropriate. Begin your search at www.pearsoned.ca/school/physicssource. 13. Italian physicist Enrico Fermi gave the name \u201cneutrino\u201d to the elusive particle that scientists were at first unable to detect. Research Fermi\u2019s contributions to physics. Write a brief report of your findings. Begin your search at www.pearsoned.ca/school/physicssource. e TEST To check your understanding of two-dimensional collisions, follow the eTest links at www.pearsoned.ca/school/physicssource. Chapter 9 The momentum of an isolated system of interacting objects is conserved. 499 09-Phys20-Chap09.qxd 7/24/08 2:44 PM Page 500 CHAPTER 9 SUMMARY Key Terms and Concepts momentum", " impulse collision system law of conservation of momentum elastic collision inelastic collision centre of mass Key Equations Momentum: Impulse: p mv F net F netave t p or F net t p or F t mv t mv netave Conservation of momentum: p sysi p sysf Elastic collisions: Eki Ekf Conceptual Overview The concept map below summarizes many of the concepts and equations in this chapter. Copy and complete the map to have a full summary of the chapter. Momentum has units of is defined as the of and applies to collisions applies throughout the universe because it is 2 types equation where where equation equation where a change in momentum is equivalent to an equation Figure 9.60 500 Unit V Momentum and Impulse 09-Phys20-Chap09.qxd 7/28/08 9:33 AM Page 501 UNIT V PROJECT An Impulsive Water Balloon Scenario Imagine you are part of a creative engineering design team commissioned to design and build an amusement ride that will make the West Edmonton Mall 14-storey \u201cSpace Shot\u201d obsolete. The ride must allow patrons to experience the thrill of free fall in a safe environment. Accelerations have to remain within appropriate limits during a required change from vertical to horizontal motion, and as the people are brought to rest. A model must be constructed for sales demonstrations using a water balloon that experiences a minimum 2.4-m drop, where the impulse changes the magnitude and direction of the momentum while maintaining the integrity of the balloon. The water balloon must begin with a vertical drop equivalent to eight storeys. Then for the equivalent height of six storeys, the balloon must change direction and come to a stop horizontally. Planning Form a design team of three to five members. Plan and assign roles so that each team member has at least one major task. Roles may include researcher, engineer to perform mathematical calculations, creative designer, construction engineer, materials acquisition officer, and writer, among others. One person may need to perform several roles in turn. Ensure that all team members help along the way. Prepare a time schedule for each task, and for group planning and reporting sessions. Materials \u2022 small balloon filled with water \u2022 plastic zip-closing bag to contain the water balloon \u2022 cardboard and/or wooden frame for apparatus \u2022 vehicle or container for balloon \u2022 cushioning material \u2022 braking device \u2022 art materials CAUTION: Test your design in an appropriate area. Make sure no one is in the way during the drop. Assessing Results After completing the project, assess its success based on a rubric designed in class* that considers research strategies experiment and construction techniques clarity and thoroughness of the written report effectiveness of the team\u2019s presentation quality and fairness of the teamwork Procedure 1 Research the range of acceptable accelerations that most people can tolerate. 2 Calculate the maximum speed obtained when an object is dropped from an eight-storey building (equivalent to 24.6 m). 3 Calculate the impulse necessary to change the direction of motion of a 75.0-kg person from vertical to horizontal in the remaining height of six storeys (equivalent to 18.4 m). The person must come to a stop at the end. Assume that the motion follows the arc of a circle. 4 Determine the time required so that the change in the direction of motion and stopping the person meets the maximum acceptable acceleration in step 1. 5 Include your calculations in a report that shows your design and method of changing the motion. 6 Build a working model and test it. Make modifications as necessary to keep the water balloon intact for a fall. Present the project to your teacher and the class. Thinking Further 1. Explain why eight storeys was used in the calculation in step 2, instead of 14 storeys. 2. What other amusement rides has your team thought of while working on this project? What would make each of these rides thrilling and appealing? 3. In what ways could your ideas have a practical use, such as getting people off a high oil derrick or out of a high-rise building quickly and safely? 4. What conditions would cause a person to be an unacceptable candidate for your ride? Write out a list of rules or requirements that would need to be posted. *Note: Your instructor will assess the project using a similar assessment rubric. Unit V Momentum and Impulse 501 09-Phys20-Chap09.qxd 7/24/08 2:44 PM Page 502 UNIT V SUMMARY Unit Concepts and Skills: Quick Reference Concepts CHAPTER 9 Momentum Newton\u2019s second law Impulse Effects of varying the net force and time interval for a given impulse Summary Resources and Skill Building The momentum of an isolated system of interacting objects is conserved. 9.1 Momentum Is Mass Times Velocity Momentum is the product of the mass of an object and its velocity. Momentum is a vector quantity measured in kilogram-metres per second (kgm/s). Newton\u2019s second law states that the net force on an object is", " equal to the rate of change of its momentum. 9.2 Impulse Is Equivalent to a Change in Momentum The impulse provided to an object is defined as the product of the net force (or average net force if F the interaction time. Impulse is equivalent to the change in momentum of the object. constant) acting on the object during an interaction and net The magnitude of the net force during an interaction and the interaction time determine whether or not injuries or damage to an object occurs. Examples 9.1 & 9.2 9-2 QuickLab Figures 9.12 & 9.13 9-3 Design a Lab Example 9.3 Example 9.4 Net force-time graph Impulse can be determined by calculating the area under a net force-time graph. Conservation of momentum in one dimension Elastic collisions Inelastic collisions Conservation of momentum in two dimensions Elastic and inelastic collisions 9.3 Collisions in One Dimension Momentum is conserved when objects in an isolated system interact in one dimension. A system is the group of objects that interact with each other, and it is isolated if no external net force acts on these objects. 9-4 QuickLab 9-5 Inquiry Lab Examples 9.5\u20139.8, 9.10 Elastic collisions are collisions in which a system of objects has the same initial and final kinetic energy. So both the momentum and kinetic energy of the system are conserved. Example 9.9 Inelastic collisions are collisions in which a system of objects has different initial and final kinetic energy values. Example 9.11 9.4 Collisions in Two Dimensions Momentum is conserved when objects in an isolated system interact in two dimensions. An isolated system has no external net force acting on it. Elastic collisions in two dimensions satisfy the same conditions as one-dimensional elastic collisions, that is, Eki Ekf. 9-6 Inquiry Lab Examples 9.12\u20139.14 Example 9.15 502 Unit V Momentum and Impulse 09-Phys20-Chap09.qxd 7/24/08 2:44 PM Page 503 UNIT V REVIEW Vocabulary 1. Using your own words, define these terms, concepts, principles, or laws. momentum impulse one-dimensional collisions conservation of momentum conservation of energy elastic collisions inelastic collisions two-dimensional collisions centre of mass Knowledge CHAPTER 9 2. Compare and contrast momentum and impulse. 3. Explain the relationship between the units in which momentum and impulse are measured. 4. A student calculated the answer to a problem and got 40 kgm/s [W]. Which quantities could the student have calculated? 5. In your own words, restate Newton\u2019s second law in terms of momentum. 6. What difference does it make that momentum is a vector quantity rather than a scalar quantity? 7. Compare and contrast a net force and an average net force acting on an object during an interaction. 8. Statistics show that less massive vehicles tend to have fewer accidents than more massive vehicles. However, the survival rate for accidents in more massive vehicles is much greater than for less massive ones. How could momentum be used to explain these findings? 9. Using the concept of impulse, explain how the shocks on a high-end mountain bike reduce the chance of strain injuries to the rider. 10. State the quantities, including units, you would need to measure to determine the momentum of an object. 11. State the quantities that are conserved in oneand two-dimensional collisions. Give an example of each type of collision. 12. How do internal forces affect the momentum of a system? 13. What instructions would you give a young gymnast so that she avoids injury when landing on a hard surface? 14. Will the magnitude of the momentum of an object always increase if a net force acts on it? Explain, using an example. 15. What quantity do you get when p is divided by mass? 16. For a given impulse, what is the effect of (a) increasing the time interval? (b) decreasing the net force during interaction? 17. For each situation, explain how you would effectively provide the required impulse. \u2022 to catch a water balloon tossed from some distance \u2022 to design a hiking boot for back-country hiking on rough ground \u2022 to shoot an arrow with maximum velocity using a bow \u2022 for an athlete to win the gold medal in the javelin event with the longest throw \u2022 for a car to accelerate on an icy road 18. Why does a hunter always press the butt of a shotgun tight against the shoulder before firing? 19. Describe a method to find the components of a momentum vector. 20. Explain why the conservation of momentum and the conservation of energy are universal laws. 21. Why does the law of conservation of momentum require an isolated system? 22. Suppose a problem involves a two-dimensional collision between two objects, and the initial momentum of one object is unknown. Explain how to solve this problem using (a) a vector addition diagram drawn to scale (b) vector components 23. Explain, in terms of", " momentum, why a rocket does not need an atmosphere to push against when it accelerates. 24. If a firecracker explodes into two fragments of unequal mass, which fragment will have the greater speed? Why? Unit V Momentum and Impulse 503 09-Phys20-Chap09.qxd 7/24/08 2:44 PM Page 504 38. Draw a momentum vector diagram to represent a 575-g basketball flying at 12.4 m/s [26.0 S of E]. 39. Calculate the momentum of a 1250-kg car travelling south at 14.8 m/s. 40. A bowling ball has a momentum of 28 kgm/s [E]. If its speed is 4.5 m/s, what is the mass of the ball? 41. A curling stone has a momentum of 32 kgm/s [W]. What would be the momentum if the mass of the stone is decreased to 7 of its original mass 8 and its speed is increased to 4 of its original 3 speed? 42. A soccer ball has a momentum of 2.8 kgm/s [W]. What would be the momentum if its mass decreased to 3 of its original mass and its 4 speed increased to 9 of its original speed? 8 43. The graph below shows the magnitude of the net force as a function of interaction time for a volleyball being blocked. The velocity of the ball changes from 18 m/s [N] to 11 m/s [S]. (a) Using the graph, calculate the magnitude of the impulse on the volleyball. (b) What is the mass of the ball? Magnitude of Net Force vs. Interaction Time for a Volleyball Block ) 5000 4000 3000 2000 1000 0 0.0 1.0 3.0 2.0 Time t (ms) 4.0 5.0 44. (a) Calculate the impulse on a soccer ball if a player heads the ball with an average net force of 120 N [210] for 0.0252 s. (b) If the mass of the soccer ball is 0.44 kg, calculate the change in velocity of the ball. 45. At a buffalo jump, a 900-kg bison is running at 6.0 m/s toward the drop-off ahead when it senses danger. What horizontal force must the bison exert to stop itself in 2.0 s? 25. When applying the conservation of momentum to a situation, why is it advisable to find the velocities of all objects in the system immediately after collision, instead of several seconds later? 26. Which physics quantities are conserved in a collision? 27. A curling stone hits another stationary stone off centre. Draw possible momentum vectors for each stone immediately before and immediately after collision, showing both the magnitude and direction of each vector. 28. A Superball\u2122 of rubber-like plastic hits a wall perpendicularly and rebounds elastically. Explain how momentum is conserved. 29. What two subatomic particles were discovered using the conservation of momentum and the conservation of energy? 30. Explain how an inelastic collision does not violate the law of conservation of energy. 31. If a system is made up of only one object, show how the law of conservation of momentum can be used to derive Newton\u2019s first law. 32. A Calgary company, Cerpro, is a world leader in ceramic armour plating for military protection. The ceramic structure of the plate transmits the kinetic energy of an armour-piercing bullet throughout the plate, reducing its penetrating power. Explain if this type of collision is elastic or inelastic. 33. A compact car and a heavy van travelling at approximately the same speed perpendicular to each other collide and stick together. Which vehicle will experience the greatest change in its direction of motion just after impact? Why? 34. Is it possible for the conservation of momentum to be valid if two objects move faster just before, than just after, collision? Explain, using an example. 35. Fighter pilots have reported that immediately after a burst of gunfire from their jet fighter, the speed of their aircraft decreased by 50\u201365 km/h. Explain the reason for this change in motion. 36. A cannon ball explodes into three fragments. One fragment goes north and another fragment goes east. Draw the approximate direction of the third fragment. What scientific law did you use to arrive at your answer? Applications 37. Calculate the momentum of a 1600-kg car travelling north at 8.5 m/s. 504 Unit V Momentum and Impulse 09-Phys20-Chap09.qxd 7/24/08 2:44 PM Page 505 46. (a) What is the minimum impulse needed to 52. A 0.146-kg baseball pitched at 40 m/s is hit back give a 275-kg motorcycle and rider a velocity of 20.0 m/s [W] if the motorcycle is initially at rest? (b) If the wheels exert an average force of", " 710 N [E] to the road, what is the minimum time needed to reach a velocity of 20.0 m/s [W]? (c) Explain how the force directed east causes the motorcycle to accelerate westward. (d) Why is it necessary to specify a minimum impulse and a minimum time? 47. A 1.15-kg peregrine falcon flying at 15.4 m/s [W] captures a 0.423-kg pigeon flying at 4.68 m/s [S]. What will be the velocity of their centre of mass immediately after the interaction? 48. A 275-kg snowmobile carrying a 75-kg driver exerts a net backward force of 508 N on the snow for 15.0 s. (a) What impulse will the snow provide to the snowmobile and driver? (b) Calculate the change in velocity of the snowmobile. 49. The graph below shows the magnitude of the net force as a function of time for a 275-g volleyball being spiked. Assume the ball is motionless the instant before it is struck. (a) Using the graph, calculate the magnitude of the impulse on the volleyball. (b) What is the speed of the ball when it leaves the player\u2019s hand Magnitude of Net Force vs. Interaction Time for a Volleyball Spike 2000 1500 1000 500 0 0.0 1.0 2.0 3.0 Time t (ms) 4.0 5.0 6.0 50. A Centaur rocket engine expels 520 kg of exhaust gas at 5.0 104 m/s in 0.40 s. What is the magnitude of the net force on the rocket that will be generated? 51. An elevator with passengers has a total mass of 1700 kg. What is the net force on the cable needed to give the elevator a velocity of 4.5 m/s [up] in 8.8 s if it is starting from rest? toward the pitcher at a speed of 45 m/s. (a) What is the impulse provided to the ball? (b) The bat is in contact with the ball for 8.0 ms. What is the average net force that the bat exerts on the ball? 53. An ice dancer and her 80-kg partner are both gliding at 2.5 m/s [225]. They push apart, giving the 45-kg dancer a velocity of 3.2 m/s [225]. What will be the velocity of her partner immediately after the interaction? 54. Two students at a barbecue party put on inflatable Sumo-wrestling outfits and take a run at each other. The 87.0-kg student (A) runs at 1.21 m/s [N] and the 73.9-kg student (B) runs at 1.51 m/s [S]. The students are knocked off their feet by the collision. Immediately after impact, student B rebounds at 1.03 m/s [N]. (a) Assuming the collision is completely elastic, calculate the speed of student A immediately after impact using energy considerations. (b) How different would your answer be if only conservation of momentum were used? Calculate to check. (c) How valid is your assumption in part (a)? 55. A cannon mounted on wheels has a mass of 1380 kg. It shoots a 5.45-kg projectile at 190 m/s [forward]. What will be the velocity of the cannon immediately after firing the projectile? 56. A 3650-kg space probe travelling at 1272 m/s [0.0] has a directional thruster rocket exerting a force of 1.80 104 N [90.0] for 15.6 s. What will be the newly adjusted velocity of the probe? 57. In a movie stunt, a 1.60-kg pistol is struck by a 15-g bullet travelling at 280 m/s [50.0]. If the bullet moves at 130 m/s [280] after the interaction, what will be the velocity of the pistol? Assume that no external force acts on the pistol. 58. A 52.5-kg snowboarder, travelling at 1.24 m/s [N] at the end of her run, jumps and kicks off her 4.06-kg snowboard. The snowboard leaves her at 2.63 m/s [62.5 W of N]. What is her velocity just after she kicks off the snowboard? 59. A 1.26-kg brown bocce ball travelling at 1.8 m/s [N] collides with a stationary 0.145-kg white ball, driving it off at 0.485 m/s [84.0 W of N]. (a) What will be the velocity of the brown ball immediately after impact? Ignore friction and rotational effects. (b) Determine if the collision is elastic. Unit V Momentum and Impulse 505", " 09-Phys20-Chap09.qxd 7/24/08 2:44 PM Page 506 60. Two people with a combined mass of 128 kg are sliding downhill on a 2.0-kg toboggan at 1.9 m/s. A third person of mass 60 kg inadvertently stands in front and upon impact is swept along with the toboggan. If all three people remain on the toboggan after impact, what will be its velocity after impact? 61. An aerosol paint can is accidentally put in a fire pit. After the fire is lit, the can is heated and explodes into two fragments. A 0.0958-kg fragment (A) flies off at 8.46 m/s [E]. The other fragment (B) has a mass of 0.0627 kg. The 0.0562-kg of gas inside bursts out at 9.76 m/s [N]. What will be the velocity of fragment B immediately after the explosion? Assume that no mass is lost during the explosion, and that the motion of the fragments lies in a plane. 62. A 0.185-kg golf club head travelling horizontally at 28.5 m/s hits a 0.046-kg golf ball, driving it straight off at 45.7 m/s. (a) Suppose the golfer does not exert an external force on the golf club after initial contact with the ball. If the collision between the golf club and the ball is elastic, what will be the speed of the club head immediately after impact? (b) Show that the law of conservation of momentum is valid in this interaction. 63. A student on a skateboard is travelling at 4.84 m/s [0], carrying a 0.600-kg basketball. The combined mass of the student and skateboard is 50.2 kg. He throws the basketball to a friend at a velocity of 14.2 m/s [270]. What is the resulting velocity of the centre of mass of the student-skateboard combination immediately after the throw? Ignore frictional effects. 64. An oxygen molecule of mass 5.31 10\u201326 kg with a velocity of 4.30 m/s [0.0\u00b0] collides headon with a 7.31 10\u201326-kg carbon dioxide molecule which has a velocity of 3.64 m/s [180.0\u00b0]. After collision the oxygen molecule has a velocity of 4.898 m/s [180.0\u00b0]. (a) Calculate the velocity of the carbon dioxide molecule immediately after collision. (b) Determine by calculation whether or not the collision is elastic. 65. An isolated stationary neutron is transformed into a 9.11 1031-kg electron travelling at 4.35 105 m/s [E] and a 1.67 1027-kg proton travelling at 14.8 m/s [E]. What is the momentum of the neutrino that is released? 506 Unit V Momentum and Impulse 66. An 8.95-kg bowling ball moving at 3.62 m/s [N] hits a 0.856-kg bowling pin, sending it off at 3.50 m/s [58.6 E of N]. (a) What will be the velocity of the bowling ball immediately after collision? (b) Determine if the collision is elastic. 67. A wooden crate sitting in the back of a pickup truck travelling at 50.4 km/h [S] has a momentum of magnitude 560 kgm/s. (a) What is the mass of the crate? (b) What impulse would the driver have to apply with the brakes to stop the vehicle in 5.25 s at an amber traffic light? Use mT for the total mass of the truck. (c) If the coefficient of friction between the crate and the truck bed is 0.30, will the crate slide forward as the truck stops? Justify your answer with calculations. 68. A firecracker bursts into three fragments. An 8.5-g fragment (A) flies away at 25 m/s [S]. A 5.6-g fragment (B) goes east at 12 m/s. Calculate the velocity of the 6.7-g fragment (C). Assume that no mass is lost during the explosion, and that the motion of the fragments lies in a plane. 69. A spherical molecule with carbon atoms arranged like a geodesic dome is called a buckyball. A 60-atom buckyball (A) of mass 1.2 10\u201324 kg travelling at 0.92 m/s [E] collides with a 70-atom buckyball (B) of mass 1.4 10\u201324 kg with a velocity of 0.85 m/s [N] in a laboratory container. Buckyball (A) bounces away at a velocity of 1.24 m/s [65\u00b0 N of E].", " [36.0]. The cue ball and three-ball each have a mass of 0.160 kg. Calculate the velocity of the cue ball immediately after collision. Ignore friction and rotational effects. 74. A hunter claims to have shot a charging bear through the heart and \u201cdropped him in his tracks.\u201d To immediately stop the bear, the momentum of the bullet would have to be as great as the momentum of the charging bear. Suppose the hunter was shooting one of the largest hunting rifles ever sold, a 0.50 caliber Sharps rifle, which delivers a 2.27 102 kg bullet at 376 m/s. Evaluate the hunter\u2019s claim by calculating the velocity of a 250-kg bear after impact if he was initially moving directly toward the hunter at a slow 0.675 m/s [S]. 75. An object explodes into three fragments (A, B, and C) of equal mass. What will be the approximate direction of fragment C if (a) both fragments A and B move north? (b) fragment A moves east and fragment B moves south? (c) fragment A moves [15.0] and fragment B moves [121]? Extensions 76. Research the physics principles behind the design of a Pelton wheel. Explain why it is more efficient than a standard water wheel. Begin your search at www.pearsoned.ca/school/physicssource. 77. A fireworks bundle is moving upward at 2.80 m/s when it bursts into three fragments. A 0.210-kg fragment (A) moves at 4.52 m/s [E]. A 0.195-kg fragment (B) flies at 4.63 m/s [N]. What will be the velocity of the third fragment (C) immediately after the explosion if its mass is 0.205 kg? Assume that no mass is lost during the explosion. 79. Two billiard balls collide off centre and move at right angles to each other after collision. In what directions did the impulsive forces involved in the collision act? Include a diagram in your answer. 80. A 2200-kg car travelling west is struck by a 2500-kg truck travelling north. The vehicles stick together upon impact and skid for 20 m [48.0 N of W]. The coefficient of friction for the tires on the road surface is 0.78. Both drivers claim to have been travelling at 90 km/h before the crash. Determine the truth of their statements. 81. Research the developments in running shoes that help prevent injuries. Interview running consultants, and consult sales literature and the Internet. How does overpronation or underpronation affect your body\u2019s ability to soften the road shock on your knees and other joints? Write a brief report of your findings. Begin your search at www.pearsoned.ca/school/physicssource. 82. A 3.5-kg block of wood is at rest on a 1.75-m high fencepost. When a 12-g bullet is fired horizontally into the block, the block topples off the post and lands 1.25 m away. What was the speed of the bullet immediately before collision? Consolidate Your Understanding 83. Write a paragraph describing the differences between momentum and impulse. Include an example for each concept. 84. Write a paragraph describing how momentum and energy concepts can be used to analyze the motion of colliding objects. Include two examples: One is a one-dimensional collision and the other is a two-dimensional collision. Include appropriate diagrams. Think About It Review your answers to the Think About It questions on page 447. How would you answer each question now? e TEST To check your understanding of momentum and impulse, follow the eTest links at www.pearsoned.ca/school/physicssource. Unit V Momentum and Impulse 507 10-PearsonPhys30-Chap10 7/24/08 2:51 PM Page 508 U N I T VI Forces Forces and Fields and Fields On huge metal domes, giant electrostatic charge generators can create voltages of 5 000 000 V, compared with 110 V in most of your household circuits. How are electrostatic charges produced? What is voltage? What happens when electric charges interact? e WEB The person in this photo is standing inside a Faraday cage. To find out how the Faraday cage protects her from the huge electrical discharges, follow the links at www.pearsoned.ca/school/physicssource. 508 Unit VI 10-PearsonPhys30-Chap10 7/24/08 2:51 PM Page 509 Unit at a Glance C H A P T E R 1 0 Physics laws can explain the behaviour of electric charges. 10.1 Electrical Interactions 10.2 Coulomb\u2019s Law C H A P T E R 1 1 Electric field theory describes electrical phenomena. 11.1 Forces and Fields 11.2 Electric Field Lines", " and Electric Potential 11.3 Electrical Interactions and the Law of Conservation of Energy C H A P T E R 1 2 Properties of electric and magnetic fields apply in nature and technology. 12.1 Magnetic Forces and Fields 12.2 Moving Charges and Magnetic Fields 12.3 Current-carrying Conductors and Magnetic Fields 12.4 Magnetic Fields, Moving Charges, and New and Old Technologies Unit Themes and Emphases \u2022 Energy and Matter \u2022 Nature of Science \u2022 Scientific Inquiry Focussing Questions While studying this unit, you will investigate how the science of electricity, magnetism, and electromagnetism evolved and its corresponding effect on technology. As you work through this unit, consider these questions. \u2022 How is the value of the elementary charge determined? \u2022 What is the relationship between electricity and magnetism? \u2022 How does magnetism assist in the understanding of fundamental particles? Unit Project Building a Model of a Direct Current Generator \u2022 By the time you complete this unit, you will have the knowledge and skills to build a model of a direct current generator. For this task, you will research wind power and design and build a model of an electric generator that uses wind energy. Unit VI Forces and Fields 509 10-PearsonPhys30-Chap10 7/24/08 2:51 PM Page 510 Physics laws can explain the behaviour of electric charges. Figure 10.1 A thunderbird on a totem pole in Vancouver Abolt of lightning flashing across dark cloudy skies, followed a few moments later by the deafening sound of thunder, is still one of the most awe-inspiring physical events unleashed by nature. What is the cause of lightning? Why is it so dangerous? So powerful is this display that many early civilizations reasoned these events must be the actions of gods. To the Romans, lightning was the sign that Jove, the king of the gods, was angry at his enemies. In some First Nations traditions, lightning flashed from the eyes of the enormous thunderbird, while thunder boomed from the flapping of its huge wings (Figure 10.1). In this chapter, you will learn how relating lightning to simpler phenomena, such as the sparking observed as you stroke a cat, initially revealed the electrical nature of matter. Further studies of the nature of electric charges and the electrical interactions between them will enable you to understand laws that describe their behaviour. Finally, you will investigate the force acting on electric charges by studying the variables that determine this force and the law that describes how to calculate such forces. C H A P T E R 10 Key Concepts In this chapter, you will learn about: electric charge conservation of charge Coulomb\u2019s law Learning Outcomes When you have completed this chapter, you will be able to: Knowledge explain electrical interactions using the law of conservation of charge explain electrical interactions in terms of the repulsion and attraction of charges compare conduction and induction explain the distribution of charge on the surfaces of conductors and insulators use Coulomb\u2019s law to calculate the electric force on a point charge due to a second point charge explain the principles of Coulomb\u2019s torsion balance experiment determine the magnitude and direction of the electric force on a point charge due to one or more stationary point charges in a plane compare, qualitatively and quantitatively, the inverse square relationship as it is expressed by Coulomb\u2019s law and by Newton\u2019s universal law of gravitation Science, Technology, and Society explain that concepts, models, and theories are often used in predicting, interpreting, and explaining observations explain that scientific knowledge may lead to new technologies and new technologies may lead to scientific discoveries 510 Unit VI 10-PearsonPhys30-Chap10 7/24/08 2:51 PM Page 511 10-1 QuickLab 10-1 QuickLab Charging Objects Using a Van de Graaff Generator Problem What can demonstrations on the Van de Graaff generator reveal about the behaviour and interactions of electric charges? Materials and Equipment Van de Graaff generator and grounding rod small piece of animal fur (approximately 15 cm x 15 cm) 5 aluminium pie plates small foam-plastic cup with confetti soap bubble dispenser and soap CAUTION! Follow your teacher\u2019s instructions to avoid getting an electric shock. Procedure 1 Copy Table 10.1 into your notebook. Make the table the full width of your page so you have room to write in your observations and explanations. Table 10.1 Observations and Explanations from Using a Van de Graaff Generator 2 Watch or perform each of the demonstrations in steps 3 to 9. 3 Place a piece of animal fur, with the fur side up, on the top of the charging sphere of the Van de Graaff generator. 4 Turn on the generator and let it run. 5 Record your observations in Table 10.1, making sure that your description is precise. 6 Ground the sphere with the grounding rod, and turn off the generator. 7 Repeat steps 3 to 6, replacing the animal fur with the aluminium pie plates (stacked upside down), and then the foam-plastic cup with", " confetti. 8 Turn on the Van de Graaff generator and let it run. 9 Dip the soap bubble dispenser into the soap and blow a stream of bubbles toward the charging sphere of the Van de Graaff generator. 10 Record your detailed observations in Table 10.1. 11 Ground the sphere with the grounding rod, and Demonstration Observation Explanation turn off the generator. Animal Fur Aluminium Pie Plates Foam-plastic Cup and Confetti Stream of Soap Bubbles Think About It Question 1. Using your knowledge of electricity, provide a possible explanation of the events that occurred during each demonstration. 1. How does the sphere of the Van de Graaff generator become charged? 2. Describe a situation during the demonstrations where the forces of interaction between the sphere of the generator and the various objects were: (a) attractive (b) repulsive 3. Why does touching the sphere with a grounding rod affect the charge on the sphere? Discuss your answers in a small group and record them for later reference. As you complete each section of this chapter, review your answers to these questions. Note any changes to your ideas. Chapter 10 Physics laws can explain the behaviour of electric charges. 511 10-PearsonPhys30-Chap10 7/24/08 2:51 PM Page 512 10.1 Electrical Interactions Your world runs on electricity. The music you listen to, the movies you watch, the video games you play\u2014all require electricity to run. Today, electricity is so familiar that you probably don\u2019t even think about it when you turn on a light, pop a piece of bread into the toaster, or switch off the TV. Try to imagine a time before electricity was even named. People had noticed interesting effects in certain situations that seemed almost magical. The Greek philosopher Thales (624\u2013546 BCE) recorded that when he rubbed amber (a hard fossilized form of tree resin), it could attract small pieces of straw or thread. This effect was called \u201celectricity,\u201d after the Greek word for amber, \u201celektron.\u201d The ancient Greeks observed two important properties of electricity: \u2022 Charged objects could either attract or repel each other. These two types of interactions suggested that there must be two different types of charge. \u2022 Repulsion occurred when two similarly charged objects were placed near each other, and attraction occurred when two oppositely charged objects were placed near each other. These observations can be summarized as the law of charges: Like charges repel and unlike charges attract. M I N D S O N Electrical Attraction Rub an ebonite rod with fur and hold the rod close to a fine stream of water from a faucet. Then rub a glass rod with silk and hold this rod close to a fine stream of water. Observe what happens in each case. Using your knowledge of charging objects, explain why the ebonite rod or the glass rod affects the water. There was little progress in understanding the nature of electricity until the 1600s, when the English scientist William Gilbert (1544\u20131603) performed extensive investigations. In De Magnete, his book on magnetism, Gilbert compared the effects of electricity and magnetism. He concluded that: 1. Objects only exhibit electrical effects when recently rubbed; magnetic objects do not need to be rubbed. 2. Electrified objects can attract small pieces of many types of objects; magnetic objects can attract only a few types of objects. 3. Electrified objects attract objects toward one central region; magnetic objects appear to have two poles. Although Gilbert was able to describe certain effects of electricity, he still did not know the origins of electric charges. In the 1700s, the American scientist and inventor Benjamin Franklin (1706\u20131790) attempted to prove that lightning in the sky was the same electricity as the spark observed when you reach for a metal info BIT The plastic in contact lenses contains etafilcon, which is a molecule that attracts molecules in human tears. This electrostatic attraction holds a contact lens on the eye. info BIT Gilbert was also a medical doctor and held the prestigious position of personal physician of Queen Elizabeth I of England. 512 Unit VI Forces and Fields 10-PearsonPhys30-Chap10 7/24/08 2:51 PM Page 513 door handle after shuffling across a carpet. He performed his famous kite experiment to explore whether lightning was a form of electricity (Figure 10.2). Luckily, he did not get killed, and he succeeded in drawing electricity from the clouds. He observed that lightning behaves the same way that electricity produced in the laboratory does. Through further investigations, he identified and named the two different types of electric charges as positive and negative charges. It soon became apparent that electricity is in all substances. This idea caught the imagination of many different people. Scientists studied electricity\u2019s intriguing effects (Figure 10.3), and entrepreneurs exploited it. Magicians and carnivals featured the \u201cmysterious\u201d effects of electricity. Figure 10.2 This", " figure is an artist\u2019s representation of Benjamin Franklin\u2019s famous kite experiment. info BIT Some historians think that Franklin may not have actually performed his kite experiment. They suspect that Franklin sent a description of this dangerous experiment to the Royal Society in London, England as a joke, because this British academy had largely ignored his earlier work. In 1753, the Royal Society awarded Franklin the prestigious Copley Medal for his electrical research. Figure 10.3 Boys were sometimes used in experiments such as this one in the early 1700s. The boy was suspended over the floor and electrostatically charged. His positive electric charge would attract pieces of paper. Studies to determine the nature of electricity continued. These studies were the beginning of the science of electrostatics, which is the study of electric charges at rest. It involves electric charges, the forces acting on them, and their behaviour in substances. electrostatics: the study of electric charges at rest The Modern Theory of Electrostatics Today\u2019s theory of electrostatics and the nature of electric charges is based on the models of the atom that Ernest Rutherford (1871\u20131937) and Niels Bohr (1885\u20131962) proposed in the early 1900s. In their theories, an atom is composed of two types of charges: positively charged protons in a nucleus surrounded by negatively charged electrons. In nature, atoms have equal numbers of electrons and protons so that each atom is electrically neutral. Just as some materials are good thermal conductors or insulators, there are also good conductors and insulators of electric charges. Electrical conductivity depends on how tightly the electrons are bound to the nucleus of the atom. Some materials have electrons that are tightly bound to the nucleus and are not free to travel within the substance. These materials are called insulators. Materials that have electrons in the outermost regions of the atom that are free to travel are called conductors. insulator: material in which the electrons are tightly bound to the nucleus and are not free to move within the substance conductor: material in which electrons in the outermost regions of the atom are free to move Chapter 10 Physics laws can explain the behaviour of electric charges. 513 10-PearsonPhys30-Chap10 7/24/08 2:51 PM Page 514 Figure 10.4 shows some examples of good conductors and insulators. Note that metals are usually good conductors. It is also interesting to note that a good conductor, such as silver, can have a conductivity 1023 times greater than that of a good insulator, such as rubber. Figure 10.4 Relative electrical conductivity of some materials Relative magnitude of conductivity 108 107 103 109 1010 1012 1015 Material silver copper aluminium iron mercury carbon germanium silicon wood glass rubber Conductors F O I L (computer chips) Semiconductors (transistors) Insulators Semiconductors Materials that lie in the middle, between good conductors and good insulators, are called semiconductors. Because of their nature, they are good conductors in certain situations, and good insulators in other situations. Selenium, for example, is an insulator in the dark, but in the presence of light, it becomes a good conductor. Because of this property, selenium is very useful in the operation of photocopiers (Figure 10.5). original copy face down movable light lens A mirror positively charged paper A negatively charged toner brush selenium-coated drum heater assembly mirror positively charged Figure 10.5 Photocopiers use the semiconductor selenium in the copying process. 514 Unit VI Forces and Fields 10-PearsonPhys30-Chap10 7/24/08 2:51 PM Page 515 The selenium-coated drum in the photocopier is initially given a positive charge and kept in the dark to retain the charge. When a flash of light shines on a document to be copied, an image of the document is transferred to the drum. Where the document is light-coloured, the selenium is illuminated, causing it to be conductive. Electrons flow into the conductive portions of the selenium coating, leaving them uncharged. The page remains light-coloured or white. Where the document is darkcoloured, the selenium remains non-conductive, and the positive charge remains. Negatively charged \u201ctoner\u201d powder is sprinkled on the drum and attaches to the positively charged portions of the drum. When a sheet of paper is passed over the drum, the toner transfers to the paper and an image of the document is created. This toner image is then fused on the paper with heat, and the copying process is complete. Silicon and germanium are also semiconductors. They become conductors when atoms such as gallium or arsenic are added to them. This process is called \u201cdoping\u201d with impurities. The field of solidstate electronics, which", " includes components such as transistors, diodes, and silicon chips, is based on this type of semiconductor. Superconductors Recall from earlier science studies that resistance is a measure of how difficult it is for electrons to flow through a material. Materials with a low electrical resistance are better conductors because very little energy is lost to heat in the conduction of electricity. Early attempts at conducting electricity efficiently used conducting materials with low electrical resistance, such as silver, copper, and gold. Researchers soon discovered that the electrical resistance of any material tends to decrease as its temperature decreases. Could the temperature of a material be lowered to the point that it loses all its resistive nature, creating the ideal conductor? This property of materials would have an enormous range of applications. Once a current is established in such a conductor, it should persist indefinitely with no energy loss. In the early 20th century, a class of materials called superconductors was developed. These conductors have no measurable resistance at very low temperatures. The Dutch physicist Heike Kammerlingh Onnes (1853\u20131926) discovered this effect in 1911 when he observed that solid mercury lost its electrical resistance when cooled to a temperature of 269 C. Although this discovery was significant, the usefulness of superconductors was limited because of the extremely low temperatures necessary for their operation. It was not until 1986 that materials were developed that were superconductors at much higher temperatures. These materials are ceramic alloys of rare earth elements, such as lanthanum and yttrium. As an example, one such alloy was made by grinding yttrium, barium, and copper oxide into a mixture and heating the mixture to form the alloy YBa2Cu3O7. This substance became a superconductor at 216 C. In 1987, another alloy was developed that displayed superconductivity at 175 C. More recent discoveries have reported copper oxide alloys that are superconductors at temperatures as high as 123 C. The ultimate goal is to develop superconductors that operate at room temperature, thus creating a whole new era of useful applications in technology. e WEB Find out what research is being done on superconductors today. How soon will you be seeing superconductors in use around the house? Write a brief summary of what you discover. To learn more about superconductors, follow the links at www.pearsoned.ca/school/ physicssource. Chapter 10 Physics laws can explain the behaviour of electric charges. 515 10-PearsonPhys30-Chap10 7/24/08 2:51 PM Page 516 10-2 Inquiry Lab 10-2 Inquiry Lab Charging Objects Question How can objects become electrically charged? Materials and Equipment 2 white plastic polyethylene strips (or ebonite rods) fur (approximately 15 cm x 15 cm) 2 clear plastic acetate strips (or glass rods) silk (approximately 15 cm x 15 cm) electroscope silk thread 2 retort stands with clamps Procedure Part A: Charging by Friction 1 Copy Table 10.2 into your notebook. Table 10.2 Observations for Charging by Friction White Polyethylene Strip Rubbed with Fur Clear Acetate Strip Rubbed with Silk Hanging White Polyethylene Strip Hanging Clear Acetate Strip Hanging White Polyethylene Strip Hanging Clear Acetate Strip 2 Hang a white polyethylene strip from one retort stand and a clear acetate strip from another retort stand. 3 While holding the hanging white polyethylene strip in the middle, rub both ends of it with the fur. While holding the hanging clear acetate strip in the middle, rub both ends of it with the silk. 4 Rub the other white polyethylene strip with the fur. 5 Carefully bring this second polyethylene strip close to one end of the hanging white polyethylene strip. Do not allow the two plastic strips to touch each other. 6 Carefully bring the second polyethylene strip close to one end of the hanging clear acetate strip. Do not allow the two plastic strips to touch each other. 7 Observe what happens in each situation and record your observations in Table 10.2. 516 Unit VI Forces and Fields Required Skills Initiating and Planning Performing and Recording Analyzing and Interpreting Communication and Teamwork 8 Rub a clear acetate strip with the silk. 9 Carefully bring this strip close to the hanging clear acetate strip. Do not allow the two plastic strips to touch each other. Observe what happens and record your observations in Table 10.2. 10 Carefully bring this clear acetate strip close to one end of the hanging white polyethylene strip. Do not allow the two plastic strips to touch each other. Observe what happens and record your observations in Table 10.2. Part B: Charging by Conduction 11 Copy Table 10.3 into your notebook. Table 10.3 Observations for Charging by Conduction Electroscope Charged with the White Polyethylene Strip Rub", "bed with Fur Electroscope Charged with the Clear Acetate Strip Rubbed with Silk 12 Rub the unattached white polyethylene strip with the fur. Touch this white strip to the knob of the electroscope. 13 Carefully bring the electroscope close to one end of the hanging white polyethylene strip. Observe what happens to the leaves in the electroscope. Record your observations in Table 10.3. 14 Now bring the electroscope near one end of the hanging clear acetate strip. Observe what happens to the leaves in the electroscope. Record your observations in Table 10.3. 15 Ground the electroscope by touching the knob of the electroscope with your finger. 16 Rub a clear acetate strip with the silk and touch the strip to the knob of the grounded electroscope. 17 Repeat steps 13 and 14. 10-PearsonPhys30-Chap10 7/24/08 2:51 PM Page 517 Part C: Charging by Induction 18 Copy Table 10.4 into your notebook. Table 10.4 Observations for Charging by Induction Grounded Electroscope Hanging White Polyethylene Strip Hanging Clear Acetate Strip 19 Bring an uncharged electroscope near one end of the hanging white polyethylene strip. Observe what happens to the leaves of the electroscope, and record your observations in Table 10.4. 20 Bring an uncharged electroscope near one end of the hanging clear acetate strip. Observe what happens to the leaves of the electroscope, and record your observations in Table 10.4. Analysis 1. What effect did you observe when two similarly charged white polyethylene strips were held near each other or when two similarly charged clear acetate strips were held near each other? 2. What effect did you observe when a charged white polyethylene strip was held near an oppositely charged hanging clear acetate strip or when a charged clear acetate strip was held near an oppositely charged hanging white polyethylene strip? 3. Based on your observations, what charge did the electroscope acquire when it was touched by the charged white polyethylene strip? when it was touched by the charged acetate strip? 4. What evidence shows a movement of charge in the electroscope when it is held near a charged object? 5. From your observations in Table 10.2, what general rule can you formulate about attraction and repulsion of charged objects? 6. From your observations in Table 10.3, what general rule can you formulate about the charge received by an object when it is touched by another charged object? 7. From your observations in Table 10.4, what general rule can you formulate about the charge received by an object when it is held near another charged object? 8. Does the electroscope acquire a net electrical charge during the process of charging by induction? Justify your answer. 9. What evidence is there from this investigation to prove that there are two types of electrical charges? 10. From the investigation, is there any evidence to prove which type of charge was developed on the white polyethylene strip and on the clear acetate strip? e LAB For a probeware activity, go to www.pearsoned.ca/school/physicssource. Methods of Charging Objects According to the modern theory of electrostatics, objects can become charged through a transfer of electrons. Electron transfer can occur in three ways: by friction, by conduction, and by induction. Law of Conservation of Charge During any charging procedure, it is important to keep in mind that new charges are not being created. The charges existing in materials are merely being rearranged between the materials, as the law of conservation of charge states: The net charge of an isolated system is conserved. Net charge is the sum of all electric charge in the system. For example, if a system contains 3 C of charge and 5 C of charge, the system\u2019s net coulomb (C): SI unit for charge, equivalent to the charge on 6.25 1018 electrons or protons Chapter 10 Physics laws can explain the behaviour of electric charges. 517 10-PearsonPhys30-Chap10 7/24/08 2:51 PM Page 518 charge is 2 C. Suppose you have a system that initially consists of two electrically neutral objects, and there is a transfer of electrons from one object to the other. One object will lose electrons and become positively charged while the other object will gain these electrons and become equally negatively charged. However, the net charge of the system is still zero. Charges have not been created, they have only been rearranged. Charging Objects by Friction The most common method of charging objects is by rubbing or friction. You have probably had the unpleasant experience of receiving a shock when you touched a door handle after walking across a carpeted floor. Similarly, gently stroking a cat can result in the generation of small sparks, which are very uncomfortable for the cat. Charging by this method involves separating electrons from the", " atoms in one object through rubbing or friction, and then transferring and depositing these electrons to the atoms of another object. The object whose atoms lose electrons then possesses positively charged ions. The object whose atoms gain electrons possesses negatively charged ions. As shown in Figure 10.6, rubbing the ebonite rod with fur transferred some of the electrons in the fur to the rod. The fur becomes positively charged, and the rod becomes negatively charged. (a) (b) hold electrons tightly sulfur brass copper ebonite paraffin wax silk lead fur wool glass Figure 10.6 (a) A neutral ebonite rod and a neutral piece of fur have equal amounts of negative and positive charge. When the fur is rubbed against the rod, a transfer of electrons occurs. (b) After rubbing, the ebonite has gained electrons and has a net negative charge. The fur has lost electrons and has a net positive charge. Whether an object gains or loses electrons when rubbed by another object depends on how tightly the material holds onto its electrons. Figure 10.7 shows the electrostatic series, in which substances are listed according to how tightly they hold their electrons. Substances at the top have a strong hold on their electrons and do not lose electrons easily. Substances near the bottom have a weak hold on their electrons and lose them easily. hold electrons loosely Concept Check Figure 10.7 The electrostatic or triboelectric series 1. Using information from Figure 10.7, explain why ebonite acquires a greater charge when rubbed with fur rather than silk. 2. What type of charge does ebonite acquire when rubbed with fur? Charging objects by friction can also occur during collisions. The collisions of water vapour molecules in rain clouds, for example, cause 518 Unit VI Forces and Fields conduction: process of charging an object through the direct transfer of electrons when a charged object touches a neutral object 10-PearsonPhys30-Chap10 7/24/08 2:51 PM Page 519 the separation and transfer of electrons. The result is that vapour molecules become positively or negatively charged, eventually resulting in lightning. You will learn more about lightning later in this chapter. This process of charging objects was also observed by the Voyageur spacecraft on its mission to Saturn. Colliding particles in the rings of Saturn create electrical discharges within the rings, similar to lightning on Earth. Charging Objects by Conduction Objects can become charged by the transfer of electrons from a charged object to an uncharged object by simply touching the objects together (Figure 10.8). This process is called charging by conduction. (a) (b) Figure 10.8 (a) During charging by conduction, electrons from a negatively charged metal conducting sphere transfer to a neutral metal conducting sphere, upon contact. (b) The neutral sphere gains electrons and is said to have been charged by conduction. The quantity of charge that transfers from one object to another depends on the size and shape of the two objects. If both objects are roughly the same size and shape, the charge transferred will be such that both objects are approximately equally charged (Figure 10.9(a)). If one sphere is larger than the other, then the larger sphere will receive more of the charge (Figure 10.9(b)). When the spheres are separated, the excess charges move to become equidistant from each other because of the forces of repulsion between like charges. Charging by conduction is similar to charging by friction because there is contact between two objects and some electrons transfer from one object to the other. (a) (b) Figure 10.9 Electrostatic repulsion of like charges forces excess charges within objects to redistribute so that the distances between charges are equal. (a) If two objects are the same size, the charges redistribute equally. (b) If the two objects are different sizes, the object with a larger surface area has more charges. Once the charge has transferred to another object, it will either be distributed over the surface of the object, if the object is a conductor, or remain on the surface at the point of contact, if the object is an insulator. Chapter 10 Physics laws can explain the behaviour of electric charges. 519 10-PearsonPhys30-Chap10 7/24/08 2:51 PM Page 520 excess negative charge neutral Figure 10.10 A piece of paper appears to be attracted to a charged ebonite rod, even before they touch. induction: movement of charge caused by an external charged object Charging Objects by Induction If you bring a negatively charged ebonite rod slowly toward a small piece of uncharged paper, the rod will attract the piece of paper, as shown in Figure 10.10. In fact, the piece of paper would begin to jiggle and move toward the rod even before the rod touches it. This reaction is a result of the forces acting on electrostatic charges. You know that electrostatic attraction can occur only between oppositely charged objects, but", " how can a charged object attract a neutral or uncharged object? And why is there never a force of repulsion between a charged object and a neutral object? The answers to these questions are revealed in the third method of charging objects, which involves two processes: induction and charging by induction. Induction Induction is a process in which charges in a neutral object shift or migrate because of the presence of an external charged object. This temporary charge separation polarizes the neutral object. One side of the object becomes positively charged and the other side is equally negatively charged. Although the object now behaves as if it is charged, it is still electrically neutral. The charging object and the neutral object do not touch each other, so there is no actual transfer of charge. (a) (b) negatively charged negatively charged Figure 10.11 (a) A neutral metal sphere and a neutral piece of paper (b) The influence of the large negative charge of a rod causes charge migration within the conducting sphere, which polarizes the sphere. The influence of the rod causes charge shift within the atoms of the insulating paper. The atoms in the paper become polarized. Because of induction, the sides of the sphere and the atoms in the paper that are positively charged are closer to the negatively charged rod than their negatively charged sides are. The net result is attraction. The process of induction varies slightly, depending on whether the charging object is approaching a substance that is an insulator or a conductor. Figure 10.11(a) depicts a neutral metal sphere (conductor) and a neutral piece of paper (insulator). Figure 10.11(b) shows a negatively charged rod approaching each neutral object. The electrons in the two neutral objects are repelled by the negative charge of an ebonite rod. The metal sphere is a conductor, so the electrons can move easily through it to its other side. This process of charge migration causes the sphere to become polarized, where one side of the sphere is positive and the other side is negative. charge migration: movement of electrons in a neutral object where one side of the object becomes positive and the other side becomes negative 520 Unit VI Forces and Fields 10-PearsonPhys30-Chap10 7/24/08 2:51 PM Page 521 Since the paper is an insulator, its electrons cannot move easily through it, so they just shift slightly relative to the nuclei. This process of charge shift causes the atoms to become polarized, where one side of an atom becomes positive and the other side becomes negative. In both cases, the distance from the negatively charged rod to the positive end of the neutral object is less than the distance to the negative end of the object. Therefore, the attraction of the opposite charges is greater than repulsion of like charges, and the net force is attractive. Charge separation by induction, which results in polarization of objects, explains electrostatic situations such as the initial attraction of a neutral piece of paper to a negatively charged rod without contact, as you saw in Figure 10.10. Charging by Induction In the situation shown in Figure 10.11, the electrons in the metal sphere and the paper return to their original positions when the negatively charged rod is removed. The objects lose their polarity and remain electrically neutral. For conductors, like the metal sphere, it is possible to maintain a residual charge by adding a grounding step. Grounding involves touching or connecting a wire from the object to the ground, as shown in Figure 10.12(a). The grounding path is then removed while the source charge is still present. The grounding step allows the conductor to maintain a charge (Figure 10.12(b)). The complete process of charging by induction includes grounding. (a) (b) charge shift: movement of electrons in an atom where one side of an atom becomes positive and the other side becomes negative grounding: the process of transferring charge to and from Earth charging by induction: the process of polarizing an object by induction while grounding it Figure 10.12 (a) While the charged rod is held near the metal sphere, the sphere remains polarized by induction. Grounding the sphere removes excess charge. In this situation, the sphere appears to have excess electrons on one side, which are removed. The positive charges cannot move because they represent the fixed nuclei of atoms. (b) After the ground and charged rod are removed, the sphere retains a net positive charge because of the loss of electrons. It has been charged by induction. PHYSICS INSIGHT The symbol for ground is A grounded conductor that is polarized by the presence of a charged object will always have a net charge that is opposite to that of the charged object if the ground is removed before the charged object is removed. The conductor has been charged by induction. Chapter 10 Physics laws can explain the behaviour of electric charges. 521 10-PearsonPhys30-Chap10 7/24/08 2:51 PM Page 522 Concept Check When you rub a balloon on your hair, you are charging it by friction. Explain why the", " balloon will then stick to a wall for a long time. How Lightning Gets Its Charge Many theories attempt to explain the formation of lightning. One theory relates the cause to the processes of evaporation and condensation of water in the clouds and different methods of charging objects. Under the right conditions, a churning cloud formation causes water vapour molecules to collide, resulting in a transfer of electrons between these molecules. The transfer of an electron from one water molecule to another leaves the molecules oppositely charged. plasma: highly ionized gas containing nearly equal numbers of free electrons and positive ions Cooling causes water vapour molecules to condense into water droplets. The atoms in these droplets hold onto electrons more readily than atoms in water vapour, and thus the droplets become negatively charged. Being heavier, these negatively charged water droplets accumulate at the bottom of the cloud, causing the bottom of the cloud to become negatively charged (Figure 10.13). The top of the cloud, containing the rising water vapour, becomes positively charged. The increasing polarization of the cloud ionizes the surrounding air, forming a conductive plasma. Excess electrons on the bottom of the cloud begin a zigzag journey through this plasma toward the ground at speeds of up to 120 km/s, creating a step leader. This is not the actual lightning strike. The presence of the large negative charge at the bottom of the cloud causes the separation of charges at that location on Earth\u2019s surface. Earth\u2019s surface at that spot becomes positively charged, and the area below the surface becomes negatively charged. Charge separation has polarized Earth\u2019s surface. Air molecules near Earth\u2019s surface become ionized and begin to drift upward. This rising positive charge is called a streamer. When the rising positive streamer meets the step leader from the clouds, at an altitude of about 100 m, a complete pathway is formed and the lightning begins. A transfer of negative charge in the form of a lightning strike from the cloud travels to Earth\u2019s surface at speeds of up to 100 000 km/s (Figure 10.14). Figure 10.13 Lightning forms when the bottom of the cloud becomes negatively charged and Earth\u2019s surface becomes positively charged. Figure 10.14 A streamer moving up from Earth\u2019s surface meets a step leader coming down from the clouds and lightning lights up the sky. 522 Unit VI Forces and Fields 10-PearsonPhys30-Chap10 7/24/08 2:51 PM Page 523 10.1 Check and Reflect 10.1 Check and Reflect Knowledge 1. What is the science of electrostatics? 2. Describe a simple experiment that enabled early scientists to determine that there were two different types of charges. 3. In the 1600s, William Gilbert compared the effects of electricity and magnetism. (a) Describe two similarities between 7. Describe how you would charge the sphere in Figure 10.12 negatively by induction. 8. A negatively charged ebonite rod is brought near a neutral pith ball that is hanging by an insulating thread from a support. Describe what happens (a) before they touch (b) after they touch these effects. 9. Compare the distribution of charge (b) Describe two differences between these effects. 4. In the classification of substances by electrical conductivity, a substance may be a conductor, insulator, semiconductor, or superconductor. (a) What property of matter determines the electrical conductivity of a substance? (b) List the classifications given above in order of increasing electrical conductivity. (c) Give an example of a substance in each classification. (d) Describe the conditions when the semiconductor selenium becomes a conductor or an insulator. (a) on hanging aluminium and glass rods if both are touched at one end by a negatively charged ebonite rod (b) after a small negatively charged metal sphere momentarily touches a larger neutral metal sphere 10. A negatively charged ebonite rod is held near the knob of a neutral electroscope. (a) Explain what happens to the leaves of the electroscope. (b) Explain what happens to the leaves of the electroscope if the other side of the knob is now grounded while the rod is still in place. (c) Explain why removing the ground first and then the rod will leave a net charge on the electroscope. Applications Extensions 5. (a) An ebonite rod is rubbed with fur. 11. You are given an ebonite rod, fur, an How can the electrostatic series chart in Figure 10.7 on page 518 help you determine which object will become negatively charged? (b) Why is it better to rub an ebonite rod with fur rather than silk? 6. Describe how you could charge a glass sphere positively using the following methods: electroscope, and a sphere of unknown charge. Describe an experimental procedure that you could use to determine the charge on the sphere. 12. If a glass rod", " becomes positively charged when rubbed with silk, use the law of conservation of charge to explain why the silk must be negatively charged. (a) friction (b) conduction (c) induction e TEST To check your understanding of electrical interactions, follow the eTest links at www.pearsoned.ca/school/physicssource. Chapter 10 Physics laws can explain the behaviour of electric charges. 523 10-PearsonPhys30-Chap10 7/24/08 2:51 PM Page 524 info BIT Charles de Coulomb was given credit for investigations into the electrostatic forces acting on charged objects, but the actual discovery of the relationship between electrostatic forces and charged objects was made earlier by Henry Cavendish (1731\u20131810), an English scientist. However, he was so shy that he didn\u2019t publish his results and thus was not credited with the discovery. PHYSICS INSIGHT Vertical bars around a vector symbol are an alternative method for representing the magnitude of a vector. This notation is used for the magnitude of force and field vectors from this chapter on. This notation avoids confusing the symbol for energy, E, with the magnitude of the electric field strength, which is introvector, E duced in the next chapter. 10.2 Coulomb\u2019s Law In Chapter 4, you studied Newton\u2019s law of universal gravitation and learned that any two objects in the universe exert a gravitational force g). The magnitude of this force of gravitational attraction on each other (F is directly proportional to the product of the two masses (m1 and m2): F g m1m2 and inversely proportional to the square of the distance between their centres (r): F g 1 r 2 These relationships can be summarized in the following equation: F g G m1m2 r 2 where G is the universal gravitational constant in newton-metres squared per kilogram squared. Charles de Coulomb suspected that the gravitational force that one mass exerts on another is similar to the electrostatic force that one charge exerts on another. To verify his hypothesis, he constructed an apparatus called a torsion balance to measure the forces of electrostatics. Although it could not be used to determine the quantity of charge on an object, Coulomb devised an ingenious method to vary the quantity of charge in a systematic manner. 10-3 Inquiry Lab 10-3 Inquiry Lab Investigating the Variables in Coulomb\u2019s Law Required Skills Initiating and Planning Performing and Recording Analyzing and Interpreting Communication and Teamwork Question Two charged objects exert electrostatic forces of magnitude Fe on each other. How does Fe depend on the charges carried by the objects and on the separation between the objects? Materials and Equipment Van de Graaff generator or ebonite rod and fur 3 small Styrofoam\u2122 or pith spheres, about 1 cm in diameter, coated with aluminium or graphite paint Hypothesis State a hypothesis relating the electrostatic force and each of the variables. Remember to write an \u201cif/then\u201d statement. Variables Read the procedure for each part of the inquiry and identify the manipulated variable, the responding variable, and the controlled variables in each one. sewing needle about 65 cm of thread 3 drinking straws 2 retort stands and 2 clamps balance 2 rulers tape small mirror (about 10 cm long and 5 cm wide) marking pen 524 Unit VI Forces and Fields 10-PearsonPhys30-Chap10 7/24/08 2:51 PM Page 525 Procedure 1 Determine the mass of one sphere (sphere 1) and record this mass in kilograms in your notebook. 2 Using the sewing needle, draw the thread through the centre of this sphere and attach both ends of the thread, with tape, to both ends of a drinking straw so that the sphere is suspended in the centre with the thread forming a V pattern. Clamp this straw, horizontally, to a retort stand, as shown in Figure 10.15. 3 Carefully insert the second drinking straw into the second sphere (sphere 2). Then fasten this drinking straw, horizontally, to the clamp on the other retort stand. 7 Tape the mirror on the paper so that the image of sphere 1 aligns with the centre of the mirror. Tape a ruler over the mirror just below the image of sphere 1. Using the marking pen, mark the position of the centre of the image of sphere 1 on the mirror while sighting in such a way that the sphere lines up with its image. 8 Do part A of the activity, which begins below, followed by part B. mirror line indicating top position of thread attached to straw image of sphere 1 string sphere 1 sphere 3 4 Adjust the clamps so that both spheres are at the paper same height. sphere 2 5 Carefully insert the third drinking straw into the third sphere (sphere 3). (This sphere will be the grounding sphere that will be used to change the charges on spheres 1 and 2.) 6 Tape sheets of white paper to the wall.", " Place the retort stand with sphere 1 close to the wall, about 5 cm away, so that the centre of the sphere aligns with the centre of the paper. Using a ruler, draw a horizontal line on the paper indicating the top position of the string attached to the straw. Figure 10.15 Part A: The relationship between the quantity of charge on each object and the electrostatic force In this part of the lab, the distance between the spheres is held constant while the charges on the spheres are varied. 9 Copy Table 10.5 into your notebook, leaving out the numbers in parentheses in the first three columns. Use these numbers only if you are unable to measure the distances accurately. They are hypothetical values that you can use to complete the rest of the activity. Table 10.5 Data and Calculations for Part A Magnitude of Weight of Sphere 1 F g (103 N) F mg g (1.28) (1.28) (1.28) (1.28) (1.28) Vertical Height of Sphere 1 dy (m) Horizontal Distance to Centre of Sphere 1 from the Centre Mark dx (m) Product of Charges of Spheres 1 and 2 (q2) Magnitude of Force of Electrostatic Repulsion Acting on Spheres 1 and 2 (N) e F (0.300) (0.300) (0.300) (0.300) (0.300) (0.0500) (0.0250) (0.0130) (0.0062) (0.0030) q 1 q1 1 2 2 4 q 1 q1 1 4 8 2 1 q1 1 q 6 1 4 4 q1 1 1 q 8 4 2 3 1 q1 1 q 4 6 8 8 Chapter 10 Physics laws can explain the behaviour of electric charges. 525 10-PearsonPhys30-Chap10 7/24/08 2:51 PM Page 526 10 Rub the ebonite rod with the fur or charge the Van de Graaff generator. 11 Carefully touch sphere 2 to the charged rod or generator to charge the sphere by conduction. Since it is nearly impossible to measure the quantity of charge transferred to sphere 2, assume that the quantity of charge on the sphere is q. 15 To obtain more data, vary the charge on each sphere using sphere 3. Gently touch sphere 3 to sphere 1. Since the charge on each sphere should be shared equally, the new charge on sphere 1 is 1 q. The 4 q 1 q 1 charge product on spheres 1 and 2 is now 1 q 2. 8 2 4 Label the new position of sphere 1 as position 2. 12 Slide the stand holding charged sphere 2 toward 16 Remove sphere 3 to a safe distance and ground it by sphere 1 on the other stand and momentarily touch the two spheres together. The charge on each object can be assumed to be 1 q because, on contact, the charge 2 is equally divided between two similar spheres. q 1 q 1 Therefore, the charge product is 1 q 2. 4 2 2 13 Slide sphere 2, parallel to the wall, to the left until the centre of its image is 1.0 cm away from the centre mark on the mirror. 14 Mark the new position of the centre of the image of sphere 1 on the mirror. Label it position 1. gently touching it with your hand. 17 Repeat steps 14 to 16, keeping sphere 2 in position and alternately touching spheres 1 and 2 with sphere 3 to obtain three more readings. Label these positions 3, 4, and 5. 18 With a ruler, accurately measure the vertical distance from the centre of the image of sphere 1 to the top horizontal line. Record this vertical distance (dy ) in metres in the appropriate column of Table 10.5. 19 Measure the distance between the original centre mark on the mirror and the centre of the image of sphere 1 in its new position for each trial. Record this distance (dx) in metres in the appropriate column of Table 10.5. Part B: The relationship between the distance between two charges and the electrostatic force In this part of the lab, the charges on the spheres are fixed while the distances are varied. 20 Copy Table 10.6 into your notebook, leaving out the numbers in parentheses in the first three columns. Use these numbers only if you are unable to measure the distances accurately. They are hypothetical values that you can use to complete the rest of the activity. Table 10.6 Data and Calculations for Part B Magnitude of Weight of Sphere 1 F g (103 N) F mg g (1.28) (1.28) (1.28) (1.28) (1.28) Vertical Height of Sphere 1 dy (m) Horizontal Distance to Centre of Sphere 1 from the Centre Mark dx (m) Distance Between the Centres of Spheres 1 and 2 r (m) Magnitude of Force of Electrostatic Repulsion Acting on Spheres 1 and 2 (N) e F (", "0.300) (0.300) (0.300) (0.300) (0.300) (0.0500) (0.0467) (0.0438) (0.0409) (0.0383) (0.0600) (0.0617) (0.0638) (0.0659) (0.0683) 526 Unit VI Forces and Fields 10-PearsonPhys30-Chap10 7/24/08 2:51 PM Page 527 21 Rub the ebonite rod with the fur or charge the Van de 2. Construct a graph of the force of electrostatic Graaff generator. 22 Carefully touch sphere 2 to the rod or the generator to give it a charge. 23 Slide the stand holding charged sphere 2 toward sphere 1 on the other stand and momentarily touch the two spheres together. repulsion on the y-axis as a function of the charge product on the x-axis. 3. What does the shape of the graph indicate about the relationship between the force of electrostatic repulsion and the charge product? 4. Complete the calculations as indicated in Table 10.6 for 24 Slide sphere 2 to a position so that the centre of its Part B. image is 0.5 cm to the left of the original centre mark on the mirror. 25 Mark the new position of the image of sphere 1 on the mirror and label it position 1. 26 Repeat steps 24 and 25, sliding sphere 2 parallel to the wall so that its image positions are at 1.0 cm, 1.5 cm, 2.0 cm, and 2.5 cm away, to obtain four more readings. Label these positions 2, 3, 4, and 5. 27 With a ruler, accurately measure the vertical distance from the centre of the image of sphere 1 to the horizontal line representing the top position of the string. Record this vertical distance (dy) in metres in the appropriate column of Table 10.6. 28 Measure the distance from the original centre mark on the mirror to the centre of the image of sphere 1 in its new position for each trial. Record this distance (dx) in metres in the appropriate column of Table 10.6. 29 Measure the distance between the centres of the two spheres. For each trial, record this distance (r) in metres in the appropriate column of Table 10.6. Analysis 1. Although the force of electrostatic repulsion F e acting on two similarly charged spheres cannot be measured directly, it can be calculated as shown in Figure 10.16. Complete the calculations as indicated in Table 10.5 for Part A. 5. Construct a graph of the force of electrostatic repulsion on the y-axis as a function of the distance between the charges on the x-axis. 6. What does the shape of the graph indicate about the relationship between the force of electrostatic repulsion and the distance between the two charges? 7. From the investigation, identify the two variables that affect the force of electrostatic repulsion acting on two charges. 8. Using a variation statement, describe the relationship between these two variables and the force of electrostatic repulsion. 9. Does your investigation confirm your hypotheses about the relationship between the variables and the electrostatic force? Why or why not? 10. How does the relationship between the variables affecting the electrostatic force in this investigation compare with that of the variables affecting the force of gravitational attraction in Newton\u2019s law of gravitation? 11. (a) What sources of error could have led to inaccuracy in the investigation? (b) What modifications to the investigation would you recommend? dy dx Fg dy Fe dx Fe Fg Figure 10.16(a) Use the concept of similar triangles: F dy dx dx dy e F g F e F g Figure 10.16(b) The shaded triangles are similar. Chapter 10 Physics laws can explain the behaviour of electric charges. 527 10-PearsonPhys30-Chap10 7/24/08 2:51 PM Page 528 The Force of Electrostatic Repulsion or Attraction Coulomb correctly hypothesized that the two factors influencing the magnitude of the electrostatic force that one charge exerts on another were the magnitudes of the charges on each object and their separation distance. To experimentally derive the relationships between the two factors and the electrostatic force, Coulomb used a procedure similar to that used in the 10-3 Inquiry Lab but with a different apparatus\u2014the torsion balance shown in Figure 10.17. To determine the force of electrostatic attraction or repulsion acting on two charged objects, a charged ball on a rod is brought near a charged object at one end of the arm of the torsion balance. Repulsion or attraction causes the ball on the arm to move, rotating the arm. As the arm rotates, a sensitive spring either tightens or loosens, causing the needle to move a proportional angle. This movement of the needle can be measured on a scale.", " The amount of movement is related to a measure of the force of electrical attraction or repulsion. Figure 10.17 Coulomb\u2019s apparatus Determining Relative Charge e MATH In 10-3 Inquiry Lab, Investigating the Variables in Coulomb\u2019s Law, you learned how separation and the magnitude of electric charges affect the electrostatic force. To graph the electrostatic force as a function of separation, and to analyze this relationship in more depth, visit www.pearsoned.ca/school/ physicssource. e SIM Explore the inverse square relationship through a simulation using a sphere of uniform charge density. Follow the eSim links at www.pearsoned.ca/ school/physicssource. Realizing that there was not yet any way of measuring charge, Coulomb devised a method of accurately determining the relative magnitude of a charge. He knew that if a charged object with a charge of q touches a similar uncharged object, then the charge would be shared equally so that each object would have a charge of 1 q. Using this 2 assumption, he was able to do his experiment. Investigating the relationship between the electrostatic force and the distance between the centres of the spheres, he first charged a sphere with a charge q and touched it momentarily to the sphere on the torsion balance. Each sphere would then have a similar and equal q and 1 charge of 1 q. Then, holding the first sphere a measured distance 2 2 from the sphere in the torsion balance, he was able to measure the electrostatic force acting on the two spheres by the movement of the needle on the calibrated scale. Changing the distance between the spheres and measuring the force each time, he demonstrated there was an inverse square relationship between the electrostatic force and the separation distance. This relationship can be expressed as F e 1 r 2 Investigating further the relationship between the magnitude of the force and the magnitudes of the charges, he was able to accurately vary the charges on each sphere. By charging one sphere with a charge q and touching it to the sphere on the balance, he knew that the charge would be shared equally. The two spheres would have charges of 1 q and 1 q 2 2 q)(1 each, and the charge product would be (1 q). By touching each 2 2 charged sphere alternately with a third neutral and similar sphere, he q)(1 q), then (1 q)(1 could vary the charge products as (1 q), and so on. 4 4 2 4 528 Unit VI Forces and Fields 10-PearsonPhys30-Chap10 7/24/08 2:51 PM Page 529 By varying the charges on both objects and measuring the electrostatic force acting on them, he demonstrated that the magnitude of the electrostatic force is proportional to the product of the two charges: F e q1q2 In 1785, using the results from his experimentation on charged objects, Charles de Coulomb summarized his conclusions about the electrostatic force. This force is also known as the Coulomb force. His summary of his conclusions is called Coulomb\u2019s law. The magnitude of the force of electrostatic attraction or repulsion (F e \u2022 directly proportional to the product of the two charges q1 and q2: ) is: F e q1q2 \u2022 inversely proportional to the square of the distance between their centres r : F e 1 2 r If these are the only variables that determine the electrostatic force, then F e q1q2 r 2 The beautiful fact about Coulomb\u2019s law and Newton\u2019s law of gravitation is that they have exactly the same form even though they arise from different sets of operations and apply to completely different kinds of phenomena. The fact that they match so exactly is a fascinating aspect of nature. Although Coulomb was able to identify and determine the relationships of the variables that affect the electrostatic force acting on two charges, he was unable to calculate the actual force. To do so, the variation statement must be converted to an equation by determining a proportionality constant (k), whose value depends on the units of the charge, the distance, and the force. At the time, however, it was impossible to measure the exact quantity of charge on an object. The Magnitude of Charges The SI unit for electric charge is the coulomb (C). A bolt of lightning might transfer 1 C of charge to the ground, while rubbing an ebonite rod with fur typically separates a few microcoulombs (C). It is difficult to build up larger quantities of charge on small objects because of the tremendous repulsive forces between the like charges. As you will see in section 15.2, experiments at the beginning of the 20th century showed that an electron has a charge of about 1.60 1019 C. So, 1 C of negative charge corresponds to the charge on 6.25 1018 electrons, or 6.25 billion billion electrons. Similarly, the charge on a proton is about 1.60 1019 C.", " PHYSICS INSIGHT Newton\u2019s law of gravitation is called an inverse square law because the gravitational force acting on any two masses is inversely proportional to the square of the distance between their centres. Chapter 10 Physics laws can explain the behaviour of electric charges. 529 10-PearsonPhys30-Chap10 7/24/08 2:51 PM Page 530 info BIT The electrostatic force acting on two charges of 1 C each separated by 1 m is about 9 109 N. This electrostatic force is equal to the gravitational force that Earth exerts on a billion-kilogram object at sea level! Given these values, physicists were able to calculate the constant of proportionality for Coulomb\u2019s law. With this constant, Coulomb\u2019s law becomes: F e 1q q 2 k 2 r is the magnitude of the force of electrostatic attraction or where F e repulsion in newtons; q1 and q2 are the magnitudes of the two charges in coulombs; r is the distance between the centres of the charges in metres; k is the proportionality constant called Coulomb\u2019s constant and is equal to 8.99 109 Nm2/C2. This electrostatic force is attractive if the two objects have opposite charges and repulsive if the two objects have like charges. This equation is used to determine electrostatic forces in many different types of problems involving charges and the electrostatic forces acting on them. Examples 10.1 and 10.2 show how to calculate the electrostatic force of attraction or repulsion acting on two charges in a one-dimensional situation. Example 10.1 Practice Problem 1. In a hydrogen atom, an electron is 5.29 1011 m from a proton. An electron has a charge of 1.60 1019 C, and the proton\u2019s charge is 1.60 1019 C. Calculate the electrostatic force of attraction acting on the two charges. Answer 1. 8.22 108 N [attraction] A small metal sphere with a negative charge of 2.10 106 C is brought near an identical sphere with a positive charge of 1.50 106 C so that the distance between the centres of the two spheres is 3.30 cm (Figure 10.18). Calculate the magnitude and type (attraction or repulsion) of the force of one charge acting on another. Given q1 2.10 106 C 1.50 106 C q2 r 3.30 102 m 3.30 cm 2.10 106 C 1.50 106 C Required magnitude and type of the electrostatic force acting on the two charges (F e ) Figure 10.18 Analysis and Solution According to Newton\u2019s third law, the electrostatic forces acting on the two spheres are the same in magnitude but opposite in direction. The magnitude of the electrostatic force is F e 1q q 2 k 2 r m (2.10 106 C)(1.50 106 C) 8.99 109 N C (3.30 102 m)2 2 2 26.0 N The magnitude calculation does not use the positive and negative signs for the charges. However, you can use these signs to determine whether the electrostatic force is attractive or repulsive. In this example, the charges have opposite signs, so the force is attractive. 530 Unit VI Forces and Fields 10-PearsonPhys30-Chap10 7/24/08 2:51 PM Page 531 Paraphrase The electrostatic force is one of attraction, with a magnitude of 26.0 N. In the next example, the two spheres touch and the charge is distributed between them. Example 10.2 The two spheres in Example 10.1 are momentarily brought together and then returned to their original separation distance. Determine the electrostatic force now exerted by one charge on the other. Given initial magnitude of the charges: 2.10 106 C q1 1.50 106 C q2 r 3.30 102 m 3.30 cm 2.10 106 C 1.50 106 C Figure 10.19 Practice Problem 1. A metal sphere with a negative charge of 3.00 C is placed 12.0 cm from another similar metal sphere with a positive charge of 2.00 C. The two spheres momentarily touch, then return to their original positions. Calculate the electrostatic force acting on the two metal spheres. Answer 1. 1.56 101 N [repulsion] Required magnitude and type of the electrostatic force acting on the e) two charges (F Analysis and Solution When a sphere with a negative charge of 2.10 106 C momentarily touches a sphere with a positive charge of 1.50 106 C, then 1.50 106 C of charge from the first sphere neutralizes the 1.50 106 C of charge on the second sphere. The remaining charge of 0.60 106 C from the first sphere then divides equally between the two identical spheres. Each sphere now has a charge of 3.0 107 C. The magnitude of the electrostatic force is now F", " e 1q q 2 k 2 r m (3.0 107 C)(3.0 107 C) 8.99 109 N 2 C (3.30 102 m)2 2 0.74 N Since both spheres have a negative charge, the electrostatic force is repulsive. Paraphrase The electrostatic force is one of repulsion, with a magnitude of 0.74 N. Chapter 10 Physics laws can explain the behaviour of electric charges. 531 10-PearsonPhys30-Chap10 7/24/08 2:51 PM Page 532 Concept Check Compare gravitational forces and electrostatic forces by identifying two similarities and two differences between the two types of forces. Vector Analysis of Electrostatic Forces So far in this section, you have studied Coulomb\u2019s law and applied the equation to calculate the magnitude of the electrostatic force that one charged particle exerts on another. However, many situations involve more than two charges. The rest of this section illustrates how to use Coulomb\u2019s law to analyze the vector nature of electrostatic forces by determining the electrostatic forces of more than two charges in onedimensional and two-dimensional situations. Examples 10.3 and 10.4 illustrate how to apply Coulomb\u2019s law to three or more collinear charges. Recall from unit I that collinear entities lie along the same straight line. Example 10.3 A small metal sphere (B) with a negative charge of 2.0 106 C is placed midway between two similar spheres (A and C) with positive charges of 1.5 106 C that are 3.0 cm apart (Figure 10.20). Use a vector diagram to find the net electrostatic force acting on sphere B. Analysis and Solution 1.5 cm 1.5 cm A 1.5 106 C B 2.0 106 C C 1.5 106 C Figure 10.20 Spheres A and C have equal charges and are the same distance from sphere B. As shown in Figure 10.21, the force vectors are equal in length and opposite in direction. FAB B FCB Figure 10.21 F net F F CB F AB CB F AB, so F net 0. Since the forces are equal in magnitude and opposite in direction, the net electrostatic force on charge B is 0. Practice Problems 1. Three small, hollow, metallic spheres hang from insulated threads as shown in the figure below. Draw a free-body diagram showing the electrostatic forces acting on sphere B. 1.0 cm 2.0 cm A 2.0 \u03bcC B 2.0 \u03bcC C 2.0 \u03bcC 2. For the figure in problem 1 above, draw a vector for the net electrostatic force on sphere B. Answers 1. FAB Fnet 2. B FCB 532 Unit VI Forces and Fields 10-PearsonPhys30-Chap10 7/24/08 2:51 PM Page 533 Example 10.4 A small metal sphere (B) with a negative charge of 2.10 106 C is placed midway between two similar spheres (A and C) 3.30 cm apart with positive charges of 1.00 106 C and 1.50 106 C, respectively, as shown in Figure 10.22. If the three charges are along the same line, calculate the net electrostatic force on the negative charge. Given qA qB qC rAC rAB 1.00 106 C 2.10 106 C 1.50 106 C 3.30 102 m 1 rAC rBC 2 3.30 cm 1.65 cm 1.00 106 C 2.10 106 C 1.50 106 C Figure 10.22 Required net) net electrostatic force on qB (F Analysis and Solution The charge on sphere B is negative and the charge on sphere A is positive, so the electrostatic force of qA on qB, AB, is an attractive force to the left. Similarly, the F CB, is an attractive force electrostatic force of qC on qB, F to the right (Figure 10.23). Consider right to be positive. FAB B FCB Figure 10.23 F The sum of these two force vectors is the net force on qB: F F net Applying F CB 1q q 2 gives k 2 r AB e F net 2 m (1.00 106 C)(2.10 106 C) 8.99 109 N C 3.30 102 m2 2 2 Practice Problems 1. A metal sphere with a charge of 2.50 109 C is 1.50 cm to the left of a second metal sphere with a charge of 1.50 109 C. A third metal sphere of 1.00 109 C is situated 2.00 cm to the right of the second charged sphere. If all three charges form a line, determine the net electrostatic force on the second sphere. 2. In the situation described above, if the first and third spheres remain at their original positions, where should the second sphere be situated so that the net electro", "static force on it would be zero? Answers 1. 1.16 104 N [to the left] 2. 2.14 102 m to the right of the 2.50 109 C charge [left] 2 m (1.50 106 C)(2.10 106 C) 8.99 109 N C 3.30 102 m2 2 2 [right] (69.34 N 104.0 N) [right] 34.7 N [right] Paraphrase The net electrostatic force on charge B is 34.7 N to the right. In Examples 10.3 and 10.4, the forces act along the same line, so the calculations involve only a single dimension. Examples 10.5 and 10.6 demonstrate how to calculate net electrostatic forces in two dimensions. Chapter 10 Physics laws can explain the behaviour of electric charges. 533 10-PearsonPhys30-Chap10 7/24/08 2:51 PM Page 534 Example 10.5 Practice Problems 1. A small metal sphere X with a negative charge of 2.50 C is 1.20 cm directly to the left of another similar sphere Y with a charge of 3.00 C. A third sphere Z with a charge of 4.00 C is 1.20 cm directly below sphere Y. The three spheres are at the vertices of a right triangle, with sphere Y at the right angle. Calculate the net electrostatic force on sphere Y, sketching diagrams as necessary. 2. Calculate the net electrostatic force on charge B shown in the figure below. 2.00 \u03bcC A 1.00 cm 90\u00b0 1.00 \u03bcC B Answers 1. 8.83 1014 N [122] 2. 2.54 102 N [225] 1.00 cm C 2.00 \u03bcC A small metal sphere A with a negative charge of 2.10 106 C is 2.00 102 m to the left of another similar sphere B with a positive charge of 1.50 106 C. A third sphere C with a positive charge of 1.80 106 C is situated 3.00 102 m directly below sphere B (Figure 10.24). Calculate the net electrostatic force on sphere B. 2.00 102 m A 2.10 106 C B 1.50 106 C Given qA qB qC rAB rBC 2.10 106 C 1.50 106 C 1.80 106 C 2.00 102 m 3.00 102 m Required net electrostatic force net) on sphere B (F Analysis and Solution The electrostatic force of qA on qB, F AB, is a force of attraction directed from charge B toward charge A (left). The electrostatic CB, is a force force of qC on qB, F of repulsion directly upward (Figure 10.25). 3.00 102 m C 1.80 106 C Figure 10.24 FCB B FAB Figure 10.25 Applying F e 1q q 2 gives k 2 r F AB 2.10 106 C)(1.50 106 C) 8.99 109 N C (2.00 102 m)2 2 2 70.80 N Similarly, F CB 1.50 106 C)(1.80 106 C) 8.99 109 N C (3.00 102 m)2 2 2 534 Unit VI Forces and Fields 26.97 N 10-PearsonPhys30-Chap10 7/24/08 2:51 PM Page 535 Use trigonometry to find the net electrostatic force on charge B, as shown in Figure 10.26. Use the Pythagorean theorem to find the magnitude of the net force: F net (70.80 N)2 (26.97 N)2 75.8 N Determine the angle : 9 N 7. 6 tan 2 7 N 0 8. 0 20.9 Fnet FCB \u03b8 FAB Figure 10.26 The direction of the net force is [20.9 N of W] or [159]. Paraphrase The net electrostatic force on charge B is 75.8 N [20.9 N of W], or 75.8 N [159]. PHYSICS INSIGHT Recall that with the polar coordinates method, angles are measured counterclockwise from the positive x-axis of the coordinate system, which is given a value of 0. Practice Problems 1. Three metal spheres are situated in positions forming an equilateral triangle with sides of 1.20 cm, as shown below. X has a charge of 2.50 C; Y has a charge of 3.00 C; and Z has a charge of 4.00 C. Calculate the net electrostatic force on the Y charge. X 2.50 C 1.20 cm 1.20 cm Y 3.00 C 1.20 cm Z 4.00 C 2. Four charged spheres, with equal charges of 2.20 C, are situated in positions forming a rectangle, as shown in the figure below. Determine", " the net electrostatic force on the charge in the top right corner of the rectangle. 2.00 102 m A 2.10 106 C B 1.50 106 C rAC 3.00 102 m C 1.80 106 C Figure 10.27 Example 10.6 A small metal sphere A with a charge of 2.10 106 C is 2.00 102 m to the left of a second sphere B with a charge of 1.50 106 C. A third sphere C with a charge of 1.80 106 C is situated 3.00 102 m directly below sphere B. Calculate the net electrostatic force on sphere C. Given qA qB qC rAB rBC 2.10 106 C 1.50 106 C 1.80 106 C 2.00 102 m 3.00 102 m Required net electrostatic force net) on sphere C (F FAC Analysis and Solution The electrostatic force of qA on qC, AC, is an attractive force directed F from charge C toward charge A. The electrostatic force of qB on qC, BC, is a repulsive force directed F downward (Figure 10.28). C 2.20 C 40.0 cm 2.20 C 30.0 cm 30.0 cm 40.0 cm 2.20 C 2.20 C FBC Figure 10.28 Answers 1. 6.56 1014 N [142\u00b0] 2. 7.17 1011 N [55.0\u00b0] Chapter 10 Physics laws can explain the behaviour of electric charges. 535 10-PearsonPhys30-Chap10 7/24/08 2:51 PM Page 536 Determine the distance between charges A and C by using the Pythagorean theorem (Figure 10.29): rAC (2.00 102 m)2 (3.00 102 m)2 3.606 102 m 3.61 102 m Applying F e 1q q 2 gives k 2 r 2.00 102 m B A 3.00 102 m \u03b8 1 C F AC 2.10 106 C)(1.80 106 C) 8.99 109 N C (3.61 102 m)2 Figure 10.29 2 2 26.13 N Similarly, F BC 1.50 106 C)(1.80 106 C) 8.99 109 N C (3.00 102 m)2 2 2 26.97 N Use the component method to find the sum of the two force vectors. Use trigonometry to determine the angle 1 AC (Figure 10.29): for the direction of F tan 1 1 2 0 0. 3 0 0. 33.69 2 m 1 2 1 m 0 0 Then resolve F AC into x and y components, as shown in Figure 10.30: FACx FACy (26.13 N) (sin 33.69) 14.49 N (26.13 N) (cos 33.69) 21.74 N x 26.13 N y 33.7\u00b0 C Figure 10.30 The electrostatic force of charge B on charge C has only a y component (see Figure 10.28). So, the x component of F BC is 0 N and the y component is 26.97 N. net. Now find the sum of the x and y components of F F net Fnetx F F BC AC FACx FBCx 14.49 N 0 N 14.49 N Fnety FBCy FACy 21.74 N (26.97 N) 5.23 N 536 Unit VI Forces and Fields 10-PearsonPhys30-Chap10 7/24/08 2:51 PM Page 537 Use trigonometry to determine the magnitude and direction of the net electrostatic force on charge C, as shown in Figure 10.31. Determine the magnitude of the net force using the Pythagorean theorem: (14.49 N)2 (5.23 N)2 15.4 N F net 5.23 N 14.49 N \u03b8 2 Fnet C Figure 10.31 To determine the angle 2, use the tangent function: tan 2 2 3 N.2 5 1 N 9.4 4 19.8 The direction of the net force is [19.8 S of W] or [200]. Paraphrase The net electrostatic force on charge C is 15.4 N [19.8 S of W] or 15.4 N [200]. THEN, NOW, AND FUTURE ESD Control Manager Since the early 1970s, electrostatic discharge (ESD) has evolved from an interesting, but generally harmless, phenomenon to one of the most rapidly expanding fields of research in science today. As electronic devices have become smaller and smaller, ESD has become a major cause of failure. Each year, billions of dollars\u2019 worth of electronic devices and systems are destroyed or degraded by electrical stress caused by ESD. A dangerous property of ESD is its ability to cause fires in a flammable atmosphere. Property loss,", " injuries, and fatalities due to the accidental ignition of petrochemical vapours, dusts, and fuels by ESD are on the rise. ESD has been the proven ignition source in many fires. However, research into the firesparking nature of ESD is still in its infancy. Today, electronics manufacturers have ESD awareness and control programs, ESD control program managers, and, in some cases, entire departments dedicated to preventing the damaging effects of ESD. Ron Zezulka (Figure 10.32) is the chief technical officer of TB&S Consultants and has specialized in the science of ESD for over 30 years. He graduated from the Southern Alberta Institute of Technology with a diploma as a telecommunications technician and began his career as a failure analyst specializing in the for Alberta science of ESD Government Telephones. Ron completed many courses to become a control program manager. Because of the newness of the industry, there are no specific quali- Figure 10.32 Ron Zezulka fications for becoming a control program manager in the field of ESD. An ESD control program manager might have a technical diploma and related job experience, a Master\u2019s degree, or a Ph.D. in physics. In 2001, Ron formed TB&S Consultants and has developed and delivered over 25 different training programs and management systems for the awareness and control of ESD in industry. He has written on the topic and lectured in industry, universities, and colleges on awareness and control of ESD. Static electricity is now tied to almost every aspect of the physical sciences. As technology advances, so does our need for a greater understanding of ESD phenomena. Questions 1. Describe two hazards associated with ESD. 2. How could ESD have damaging and harmful effects in your home? 3. How are ESD control program managers employed in industry? Chapter 10 Physics laws can explain the behaviour of electric charges. 537 10-PearsonPhys30-Chap10 7/24/08 2:51 PM Page 538 10.2 Check and Reflect 10.2 Check and Reflect Knowledge 1. Identify the two factors that influence the force of electrostatic attraction or repulsion acting on two charges. Write a mathematical expression to describe the relationship. 2. Describe how the inverse square law, first proposed by Newton for gravitational forces, was applied to electrostatic forces by Coulomb. 3. (a) What is the SI unit for electric charge? (b) Compare the charge on an electron to that produced by rubbing an ebonite rod with fur. 4. Coulomb could not measure the amount of charge on his spheres, but he could vary the amount of charge on each sphere. Describe the procedure he used to do so. Applications 5. An electrostatic force of 10 N acts on two charged spheres, separated by a certain distance. What will be the new force in the following situations? 7. Two identical conducting spheres have charges of 5.00 105 C and 6.00 105 C and are in fixed positions, 2.00 m apart. (a) Calculate the electrostatic force acting on the two charges. (b) The spheres are touched together and returned to their original positions. Calculate the new electrostatic force acting on them. 8. Three charges are placed in a line, as shown in the diagram below. 2.00 m 3.00 m A 2.00 mC B 3.00 mC C 2.00 mC (a) What is the net electrostatic force on charge A? (b) What is the net electrostatic force on charge B? Extensions (a) The charge on one sphere is doubled. 9. A helium nucleus has a positive charge (b) The charge on both spheres is doubled. 6. (a) Why is it difficult to attain a large charge of 100 C on a small object? (b) During the rubbing process, an object acquires a charge of 5.0 109 C. How many electrons did the object gain? with a magnitude twice that of the negative charge on an electron. Is the electrostatic force of attraction on an electron in a helium atom equal to the force acting on the nucleus? Justify your answer. 10. Electrical forces are so strong that the combined electrostatic forces of attraction acting on all the negative electrons and positive protons in your body could crush you to a thickness thinner than a piece of paper. Why don\u2019t you compress? e TEST To check your understanding of Coulomb\u2019s law, follow the eTest links at www.pearsoned.ca/school/physicssource. 538 Unit VI Forces and Fields 10-PearsonPhys30-Chap10 7/24/08 2:51 PM Page 539 CHAPTER 10 SUMMARY Key Terms and Concepts law of conservation of charge net charge conduction induction charge migration charge shift grounding charging by induction plasma Coulomb\u2019s law coulomb (C) electrostatics insulator conductor", " semiconductor superconductor Key Equation F e 1q q 2 k 2 r Conceptual Overview The concept map below summarizes many of the concepts and equations in this chapter. Copy and complete the map to have a full summary of the chapter. types of materials conductors insulators Electrostatics forces between charges Coulomb\u2019s law calculating Fe between 2 charges methods of charging objects calculating Fe between more than 2 charges in one dimension friction induction charge separation grounding Chapter 10 Physics laws can explain the behaviour of electric charges. 539 10-PearsonPhys30-Chap10 7/24/08 2:51 PM Page 540 CHAPTER 10 REVIEW Knowledge 1. (10.1) What is an electrostatic charge? 2. (10.1) On what property of materials does thermal and electrical conductivity depend? 3. (10.1) Under what conditions does selenium become a good conductor or a good insulator? What are materials with this property called? 4. (10.1) What are three methods of charging objects? 5. (10.1) During the process of charging objects by friction, what determines which object becomes negatively or positively charged? 6. (10.1) How are the processes of charging objects by conduction and friction alike? How are they different? 7. (10.1) State the law of charges. 8. (10.1) Who is credited with first naming the two types of charges as negative and positive charges? 9. (10.1) State the law of conservation of charge. 10. (10.1) Selenium and germanium are both semiconductors. Explain why selenium is used in photocopiers rather than germanium. 11. (10.2) Calculate the electrostatic force acting on two charged spheres of 3.00 C and 2.50 C if they are separated by a distance of 0.200 m. Applications 12. What is the distance between two charges of 5.00 C each if the force of electrostatic repulsion acting on them is 5.00 103 N? 13. Charge A has a charge of 2.50 C and is 1.50 m to the left of charge B, which has a charge of 3.20 C. Charge B is 1.70 m to the left of a third charge C, which has a charge of 1.60 C. If all three charges are collinear, what is the net electrostatic force on each of the following? (a) charge B (b) charge C 14. Why is dust attracted to the front of a cathode- ray tube computer monitor? 15. Why is it desirable to develop materials with low electrical resistance? 16. Explain why a charged ebonite rod can be discharged by passing a flame over its surface. 17. Explain why repulsion between two objects is the only evidence that both objects are charged. 18. Why do experiments on electrostatics not work well on humid days? 19. Why does a charged pith ball initially attract a neutral pith ball, then repel it after touching it? 20. Why can you not charge a copper rod while holding one end with one hand and rubbing the other end with a piece of fur? 21. A person standing on an insulated chair touches a charged sphere. Is the person able to discharge the sphere and effectively ground it? Explain. 22. Two charged spheres, separated by a certain distance, attract each other with an electrostatic force of 10 N. What will be the new force in each of the following situations? (a) The charge on both spheres is doubled and the separation distance is halved. (b) The charge on one sphere is doubled while the charge on the other sphere is tripled and the separation distance is tripled. 23. Calculate and compare the electrical and gravitational forces acting on an electron and a proton in the hydrogen atom when the distance between their centres is 5.29 1011 m. 24. An equilateral triangle with sides of 0.200 m has three charges of 2.50 C each, situated on the vertices of the triangle. Calculate the net electrostatic force on each charge. What assumption did you have to make to complete the calculation? 540 Unit VI Forces and Fields 10-PearsonPhys30-Chap10 7/24/08 2:51 PM Page 541 25. In a Coulomb-type experiment, students were investigating the relationship between the force of electrostatic repulsion acting on two charged spheres and their separation distance. The results of their investigation yielded the results shown in the table below. Separation Distance (r) ( 102 m) Magnitude of Force of Repulsion F (N) 1.00 2.00 3.00 4.00 5.00 360.0 89.9 40.0 27.5 14.4 (a) Draw a graph of the results shown in the table. (b) From the shape of the graph, what is the relationship between the electro", "static force and the separation distance between two charges? (c) Make a new table of values to obtain data to straighten the graph. (d) Draw a graph of the data in your new table of values. (e) Determine the slope of the graph. (f) What value does the slope of this graph represent? (g) If the charges of the two spheres are the same, what is the value of the charge on each sphere? Extensions 26. Can a neutral object contain any charges? Explain. 27. Is it possible for a single negative or a single positive charge to exist in nature under normal conditions? Explain your answer. 28. Explain why it is impossible to charge a coin by rubbing it between your fingers. 29. Compare the production of lightning on Earth with the lightning between the rings of Saturn observed by the Voyager spacecraft on its mission to Saturn. 30. You are given two equally sized metal spheres on insulated stands, a piece of wire, a glass rod, and some silk. Devise and describe a method to do the following without touching the rod to the spheres: (a) give the spheres equal and opposite charges (b) give the spheres equal and like charges 31. Using the principles of electrostatics, explain the causes and effects of the following demonstrations: (a) Two strips of clear adhesive tape are stuck together and then carefully separated. When the two strips are brought close to each other, attraction occurs. (b) Two strips of clear adhesive tape are stuck onto a desktop and then carefully removed. When the two strips are held close to each other, repulsion occurs. Consolidate Your Understanding Create your own summary of the behaviour of electric charges and the laws that govern electrical interactions by answering the questions below. If you want to use a graphic organizer, refer to Student Reference 3: Using Graphic Organizers. Use the Key Terms and Concepts listed on page 539 and the Learning Outcomes on page 510. 1. Create a flowchart to describe how to calculate the electrostatic forces between two or more charged objects in one- or two-dimensional situations. 2. Write a paragraph explaining the three methods of charging objects. Share your report with a classmate. Think About It Review your answers to the Think About It questions on page 511. How would you answer each question now? e TEST To check your understanding of the behaviour of electric charges, follow the eTest links at www.pearsoned.ca/school/physicssource. Chapter 10 Physics laws can explain the behaviour of electric charges. 541 11-PearsonPhys30-Chap11 7/24/08 3:04 PM Page 542 C H A P T E R 11 Key Concepts In this chapter, you will learn about: vector fields electric fields electric potential difference moving charges in electric fields Learning Outcomes When you have completed this chapter, you will be able to: Knowledge define vector fields compare forces and fields compare, qualitatively, gravitational and electric potential energy define electric potential difference as a change in electric potential energy per unit of charge calculate the electric potential difference between two points in a uniform electric field explain, quantitatively, electric fields in terms of intensity (strength) and direction relative to the source of the field and to the effect on an electric charge describe, quantitatively, the motion of an electric charge in a uniform electric field explain electrical interactions, quantitatively, using the law of conservation of charge Science, Technology, and Society explain that the goal of technology is to provide solutions to practical problems explain that scientific knowledge may lead to the development of new technologies and new technologies may lead to scientific discovery 542 Unit VI Electric field theory describes electrical phenomena. Figure 11.1 The eerie glow of St. Elmo\u2019s fire on the masts of a ship On Christopher Columbus\u2019s second voyage to the Americas, his ships headed into stormy weather, and the tips of the ships\u2019 masts began to glow with a ghostly bluish flame. Sailors of the time believed that this bluish glow was a good sign that the ship was under the protection of St. Elmo, the patron saint of sailors, so they called the blue \u201cflames\u201d St. Elmo\u2019s fire (Figure 11.1). People throughout history have written about this strange glow. Julius Caesar reported that \u201cin the month of February, about the second watch of the night, there suddenly arose a thick cloud followed by a shower of hail, and the same night the points of the spears belonging to the Fifth Legion seemed to take fire.\u201d Astronauts have seen similar glows on spacecraft. What is the cause of this eerie phenomenon? Why does it most often appear during thunderstorms? You will discover the answers to these questions as you continue to study the phenomena associated with electric charges. In this chapter, you will begin by learning how knowledge of the forces related to electric charges led to the idea of fields, and you will compare different types of electric fields. Then you will learn how force", " is used to define the strength of electric fields. Finally, you will study the motion of charges in electric fields and explain electrical interactions using the law of conservation of energy. 11-PearsonPhys30-Chap11 7/24/08 3:04 PM Page 543 11-1 QuickLab 11-1 QuickLab Shielding of Cellular Phones Electronic equipment usually contains material that is used as \u201cshielding.\u201d In this activity, you will discover what this shielding material does. 3 Remove the aluminium foil and again dial the number of the cellular phone. 4 Repeat steps 1 to 3 using the sheets of other materials. Problem How does the shielding of electronic equipment, such as a cellular phone, affect its operation? Materials 2 cellular phones sheets (about 20 cm \u00d7 20 cm) of various materials, such as aluminium foil, plastic wrap, wax paper, paper, cloth, fur 1 short length of coaxial cable Procedure Part A 1 Wrap the sheet of aluminium foil around one of the cellular phones. 2 With the other cellular phone, dial the number of the wrapped cellular phone and record any response. Part B 5 Carefully remove the outer strip of insulated plastic around one end of the coaxial cable and examine the inner coaxial cable wires. Questions 1. What effect did wrapping a cellular phone with the various materials have on the operation of the cellular phone? 2. Cellular phones receive communication transmissions that are electrical in nature. Speculate why the transmissions are shielded by certain materials. Which materials are most effective for shielding? 3. What material forms the protective wrapping around the inner coaxial transmission wires? Explain the purpose of this protective wrapping. Think About It 1. Desktop computers or computers in vehicles have sensitive electronic components that must be protected from outside electrical interference. Identify a possible source of outside electrical interference. Describe how computer components may be protected from this interference and explain why this protection is necessary. 2. Sometimes, if your debit card fails to scan, the clerk wraps the card with a plastic bag and re-scans it. Explain why a plastic bag wrapped around a card would allow the card to scan properly. Why do clerks not wrap the card with aluminium foil for re-scanning? Discuss and compare your answers in a small group and record them for later reference. As you complete each section of this chapter, review your answers to these questions. Note any changes to your ideas. Chapter 11 Electric field theory describes electrical phenomena. 543 11-PearsonPhys30-Chap11 7/24/08 3:04 PM Page 544 11.1 Forces and Fields Figure 11.2 Forces exerted by the horses attached to the chariot cause the \u201cviolent\u201d motion of the chariot. The ancient Greek philosophers explained most types of motion as being the result of either \u201cviolent\u201d or \u201cnatural\u201d forces. They thought that violent forces cause motion as the result of a force exerted by one object in contact with another (Figure 11.2). They thought that natural forces cause the motion of objects toward their \u201cnatural element\u201d (Figure 11.3). However, the Greeks found another kind of motion more difficult to explain. You will observe this kind of motion in the following Minds On activity. M I N D S O N Action at a Distance Charge a rubber rod by rubbing it with fur and slowly bring it close to the hairs on your forearm. Do not touch the hairs or your arm. Observe what happens. 1. What evidence is there that the charged rod affects the hairs on your arm without actual contact? Is the force exerted by the rod on the hairs of your arm attractive or repulsive? 2. The rubber rod seems to be able to exert a type of violent force on the hairs of your arm without visible contact. This type of force was classified as \u201caction at a distance,\u201d where one object could exert a force on another object without contact. To explain \u201caction at a distance,\u201d the Greeks proposed the effluvium theory. According to this theory, all objects are surrounded by an effluvium. This invisible substance is made up of minute string-like atoms emitted by the object that pulsate back and forth. As the effluvium extends out to other bodies, the atoms of the different objects become entangled. Their effluvium eventually draws them toward each other. The effluvium theory helped to explain what seemed to be \u201caction at a distance.\u201d Although the effluvium was invisible, there was still a form of contact between the objects. Figure 11.3 To return to its natural element, a rock falls with \u201cnatural\u201d motion to Earth\u2019s surface. info BIT A new theory in physics, called string theory, proposes that objects interact through \u201cstrings\u201d that transmit the forces between the objects. This new theory has a striking similarity to the effluvium theory proposed 2500 years ago. 544 Unit VI Forces and Fields 11-Pearson", "Phys30-Chap11 7/24/08 3:05 PM Page 545 Fields In the 17th century, scientists, including Newton, tried to determine why one object can exert a force on another object without touching it. These scientists attempted to explain \u201caction at a distance,\u201d such as the curved path of a thrown ball or the effect of a charged piece of amber on the hair on a person\u2019s arm. Finding that \u201cnatural\u201d or \u201cviolent\u201d forces and \u201ceffluvium\u201d could not explain gravity or electrical forces, scientists developed the concept of fields to describe these forces. A field is defined as a region of influence surrounding an object. The concept of fields helps explain the laws of universal gravitation, which you studied in Chapter 4. Consider a space module on its way to the Moon (Figure 11.4). Nearing its lunar destination, the module begins to experience the increasing influence of the Moon. As a result, the module\u2019s motion begins to follow a curved path, similar to the projectile motion of an object thrown horizontally through the air near Earth\u2019s surface. As Newton\u2019s laws state, the motion of any object can follow a curved path only when acted on by a non-zero force that has a perpendicular component. In space, this happens to the space module when it is near the Moon, so the space near the Moon must be different from the space where no large objects like the Moon are present. From this, we can infer that a field exists around a large object, such as the Moon. When other objects enter this field, they interact with the Moon. Similarly, Earth has a field. Gravitational force acts on other objects that enter this field. Recall from Chapter 4 that this field around objects is called a gravitational field. field: a region of influence surrounding an object Michael Faraday (1791\u20131867) developed the concept of fields to explain electrostatic phenomena. He determined that the space around a rubber rod must be different when the rubber rod is charged than when it is not. The charges on the rod create an electric field around the rod. An electrostatic force acts on another charged object when it is placed in this field. An electric field exists around every charge or charged object. It can exist in empty space, whether or not another charge or charged object is in the field. Although field theory is a powerful tool for describing phenomena and predicting forces, physicists are still debating how objects can actually exert forces at a distance. Chapter 17 describes how quantum theory provides an extremely accurate model for describing such forces. Figure 11.4 A space module passing near a large planet or the Moon follows a curved path. Concept Check Use field theory to explain the path of a baseball thrown from outfield to home plate. Chapter 11 Electric field theory describes electrical phenomena. 545 11-PearsonPhys30-Chap11 7/24/08 3:05 PM Page 546 11-2 Inquiry Lab 11-2 Inquiry Lab Electric Field Patterns \u2014 Demonstration Required Skills Initiating and Planning Performing and Recording Analyzing and Interpreting Communication and Teamwork Question What is the shape of the electric field around various charged objects? Materials and Equipment plastic platform with 2 electrode holders overhead projector petri dish canola or olive oil lawn seeds single-point electrode two-point (oppositely charged) electrodes parallel copper plates about 4 cm \u00d7 4 cm hollow sphere conductor 4\u20136 cm in diameter 2 Wimshurst generators connecting wires Procedure 1 Pour some of the canola or olive oil into the petri dish so the dish is about three-quarters full. 2 Place the petri dish with the oil on the plastic platform on the overhead projector. Carefully sprinkle the lawn seeds evenly over the surface of the oil. 3 Attach the single-point electrode, with a connecting wire, to one contact of the Wimshurst generator. Immerse the electrode in the oil in the centre of the dish. 4 Crank the Wimshurst generator several times and carefully observe the pattern of the seeds in the oil. 5 Remove the electrode and allow sufficient time for the lawn seeds to redistribute on the surface. (Gentle stirring with a pencil might be required.) 6 Repeat steps 3 to 5 with each of the following: (a) (b) (c) two electrodes connected to similar contacts on two Wimshurst machines two electrodes connected to opposite contacts on one Wimshurst machine two parallel copper plates connected to opposite contacts on one Wimshurst machine (d) one hollow sphere connected to one contact of one Wimshurst machine Analysis 1. Describe and analyze the pattern of the lawn seeds created by each of the charged objects immersed in the oil in step 6 of the procedure by answering the following questions: (a) Where does the density of the lawn seeds appear to be the greatest? the least? (b) Does there appear to be a starting point and an endpoint in the pattern created by the lawn seeds? 2.", " Are there any situations where there appears to be no observable effect on the lawn seeds? 3. Based on your observations of the patterns created by the lawn seeds on the surface of the oil, what conclusion can you make about the space around charged objects? Magnitude and Direction of an Electric Field The electric field that surrounds a charged object has both magnitude and direction. Therefore, an electric field is classified as a vector field. At any point around a charge, the field can be represented by a vector arrow. The arrow\u2019s length represents the magnitude of the electric field and the arrowhead indicates direction at that point. By definition, the direction of the electric field around a charge is the direction of the force experienced by a small positive test charge placed in the electric field (Figure 11.5). A test charge is a charge with a magnitude small enough so that it does not disturb the charge on the source charge and thus change its electric field. test charge: charge with a magnitude small enough that it does not disturb the charge on the source charge and thus change its electric field source charge: charge that produces an electric field 546 Unit VI Forces and Fields 11-PearsonPhys30-Chap11 7/24/08 3:05 PM Page 547 (a) (b) test charge F source charge source charge F test charge Figure 11.5 The direction of the electric field at a point is the direction of the electric force exerted on a positive test charge at that point. (a) If the source charge is negative, the field is directed toward the source. (b) If the source charge is positive, the field is directed away from the source. Concept Check Identify the difference in the electric field strength, E II, as represented by the vector arrows in Figure 11.6., at points I and II E I E Figure 11.6 You can determine the magnitude of the electric field around a point charge from the effect on another charge placed in the field. If a small positive test charge is placed in the field, this charge will experience a greater force when it is near the charge producing the field than when it is farther away from it. By definition, the electric field (E ) at a given point is the ratio of the electric force (F e) exerted on a charge (q) placed at that point to the magnitude of that charge. The electric field can be calculated using the equation F e E q where q is the magnitude of the test charge in coulombs (C); F e is the electric force on the charge in newtons (N); and E is the strength of the electric field at that point in newtons per coulomb (N/C), in the direction as defined previously. info BIT A tremendous range of field strengths occurs in nature. For example, the electric field 30 cm away from a light bulb is roughly 5 N/C, whereas the electron in a hydrogen atom experiences an electric field in the order of 1011 N/C from the atom\u2019s nucleus. Chapter 11 Electric field theory describes electrical phenomena. 547 11-PearsonPhys30-Chap11 7/24/08 3:05 PM Page 548 Example 11.1 A sphere with a negative charge of 2.10 \u00d7 10\u20136 C experiences an electrostatic force of repulsion of 5.60 \u00d7 10\u20132 N when it is placed in the electric field produced by a source charge (Figure 11.7). Determine the magnitude of the electric field the source charge produces at the sphere. Practice Problems 1. An ion with a charge of 1.60 10\u201319 C is placed in an electric field produced by another larger charge. If the magnitude of the field at this position is 1.00 103 N/C, calculate the magnitude of the electrostatic force on the ion. 2. The magnitude of the electrostatic force on a small charged sphere is 3.42 10\u201318 N when the sphere is at a position where the magnitude of the electric field due to another larger charge is 5.34 N/C. What is the magnitude of the charge on the small charged sphere? Answers 1. 1.60 10\u201316 N 2. 6.40 10\u201319 C 5.60 102 N F 2.10 106 C Figure 11.7 source charge Given q 2.10 106 C F e 5.60 102 N [repulsion] Required magnitude of the electric field (E ) Analysis and Solution F e, q Since.67 104 N/C Paraphrase The magnitude of the electric field is 2.67 104 N/C at the given point. q1 q2 r Figure 11.8 A test charge (q2) is placed in the electric field of a source charge (q1). The distance between their centres is r. PHYSICS INSIGHT Equations based on Coulomb\u2019s law only work for point charges. 548 Unit VI Forces and Fields The equation for determining the magnitude of the electric field around a point charge, like that shown", " in Figure 11.8, can be derived mathematically as follows: F e and F e q 2 kq 1 2 r q2, then If E q2 kq 1 2 r E k E r q 2 where q is the magnitude of the source charge producing the electric field in coulombs (ignore the sign of the charge); r is the distance from the 11-PearsonPhys30-Chap11 7/24/08 3:05 PM Page 549 centre of the source charge to a specific point in space in metres; k is Coulomb\u2019s constant (8.99 109 Nm2/C2); and E is the magnitude of the electric field in newtons per coulomb. Example 11.2 Determine the electric field at a position P that is 2.20 10 \u20132 m from the centre of a negative point charge of 1.70 10 \u20136 C. Given q 1.70 106 C r 2.20 102 m Required electric field E Analysis and Solution The source charge producing the electric field is q. So, q k E 2 r (1.70 106 C) 8.99 109 N m 2 2 C (2.20 102 m)2 3.16 107 N/C Practice Problems 1. The electric field at a position 2.00 cm from a charge is 40.0 N/C directed away from the charge. Determine the charge producing the electric field. 2. An electron has a charge of 1.60 10\u201319 C. At what distance from the electron would the magnitude of the electric field be 5.14 1011 N/C? Answers 1. 1.78 1012 C 2. 5.29 1011 m Since the source charge is negative and the field direction is defined as the direction of the electrostatic force acting on a positive test charge, the electric field is directed toward the source charge. Paraphrase The electric field at point P is 3.16 107 N/C [toward the source]. Concept Check Compare gravitational fields and electrostatic fields by listing two similarities and two differences between the two types of fields. Often, more than one charge creates an electric field at a particular point in space. In earlier studies, you learned the superposition principle for vectors. According to the superposition principle, fields set up by many sources superpose to form a single net field. The vector specifying the net field at any point is simply the vector sum of the fields of all the individual sources, as shown in the following examples. Example 11.3 shows how to calculate the net electric field at a point in one-dimensional situations. e MATH The nucleus of an atom exhibits both electric and gravitational fields. To study their similarities and differences graphically, visit www.pearsoned.ca/school/ physicssource. Chapter 11 Electric field theory describes electrical phenomena. 549 11-PearsonPhys30-Chap11 7/24/08 3:05 PM Page 550 Example 11.3 Practice Problems 1. Calculate the net electric field at a point 2.10 10\u20132 m to the left of the 1.50 10\u20136 C charge in Figure 11.9. 2. An electron and a proton are 5.29 10\u201311 m apart in a hydrogen atom. Determine the net electric field at a point midway between the two charges. Answers 1. 3.67 107 N/C [left] 2. 4.11 1012 N/C [toward the electron] Two positively charged spheres, A and B, with charges of 1.50 10 \u20136 C and 2.00 10 \u20136 C, respectively, are 3.30 10 \u20132 m apart. Determine the net electric field at a point P located midway between the centres of the two spheres (Figure 11.9). 1.50 106 C 2.00 106 C A P 3.30 102 m B Figure 11.9 Given qA qB r 3.30 102 m 1.50 106 C 2.00 106 C Required net electric field at point P (E net) Analysis and Solution As shown in Figure 11.10, the electric field created by qA at point P is directed to the right, while the electric field at point P created by qB is directed to the left. Consider right to be positive. The distance between qA and point P is: rqA to P 3.30 102 m 1.65 102 m 2 To calculate the electric field at point P created by qA, use: (1.50 106 C) 8.99 109 m 2 2 C N (1.65 102 m)2 E qA q A k r 2 q to P A 4.953 107 N/C To calculate the electric field at point P created by qB, use: (2.00 106 C) 8.99 109 m 2 2 C N (1.65 102 m)2 E qB q B k r 2 q to P B 6.", "604 107 N/C Use vector addition to determine the net electric field at point P: E net E qA E qB 4.953 107 N/C [right] 6.604 107 N/C [left] 1.65 107 N/C [left] Paraphrase The net electric field at point P is 1.65 107 N/C [left]. P Eq B Eq A Figure 11.10 550 Unit VI Forces and Fields 11-PearsonPhys30-Chap11 7/24/08 3:05 PM Page 551 Example 11.4 demonstrates how to determine the net electric field at a point due to two charges in a two-dimensional situation. Example 11.4 Calculate the net electric field at a point P that is 4.00 10\u20132 m from a small metal sphere A with a negative charge of 2.10 10\u20136 C and 3.00 10\u20132 m from another similar sphere B with a positive charge of 1.50 10\u20136 C (Figure 11.11). P Figure 11.11 4.00 102 m 3.00 102 m A 36.9\u00b0 2.10 106 C 53.1\u00b0 B 1.50 106 C Given qA rA to P A 2.10 106 C 4.00 102 m 36.9 to the horizontal qB rB to P B 1.50 106 C 3.00 102 m 53.1 to the horizontal Required net) net electric field at point P (E Analysis and Solution Since qA is a negative charge, the electric field created by qA at point P is directed toward qA from point P. Since qB is a positive charge, the electric field created by qB at point P is directed away from qB toward point P. Determine the electric field created by qA at point P: E A q k A 2 A r to P 2 m (2.10 106 C) 8.99 109 N 2 C (4.00 102 m)2 1.180 107 N/C Determine the electric field created by qB at point P: E B q k B 2 B r to P 2 m (1.50 106 C) 8.99 109 N C (3.00 102 m)2 2 Practice Problems 1. Calculate the net electric field at point P, which is 0.100 m from two similar spheres with positive charges of 2.00 C and separated by a distance of 0.0600 m, as shown in the figure below. P 0.100 m 0.100 m 72.5\u00b0 72.5\u00b0 2.00 C 0.0600 m 2.00 C 2. Two charges of +4.00 C are placed at the vertices of an equilateral triangle with sides of 2.00 cm, as shown in the figure below. Determine the net electric field at the third vertex of the triangle. 2.00 cm 2.00 cm 60\u00b0 60\u00b0 4.00 C 2.00 cm 4.00 C Answers 1. 3.43 1012 N/C [90.0\u00b0] 2. 1.56 1014 N/C [90.0\u00b0] 1.498 107 N/C Chapter 11 Electric field theory describes electrical phenomena. 551 11-PearsonPhys30-Chap11 7/24/08 3:05 PM Page 552 B are shown in Figure 11.12. A and E The directions of E y EB 53.1\u00b0 36.9\u00b0 EA x 36.9\u00b0 y P x EAy EAx EA y EB EBy 53.1\u00b0 x P EBx Figure 11.12 Figure 11.13 Figure 11.14 Resolve each electric field into x and y components (see Figures 11.13 and 11.14). Use vector addition to determine the resultant electric field. EAx EBx (1.180 107 N/C)(cos 36.9\u00b0) EAy 9.436 106 N/C (1.498 107 N/C)(cos 53.1\u00b0) EBy 8.994 106 N/C (1.180 107 N/C)(sin 36.9\u00b0) 7.085 106 N/C (1.498 107 N/C)(sin 53.1\u00b0) 1.198 107 N/C Add the x components: Enetx EBx EAx (9.436 106 N/C) (8.994 106 N/C) 1.843 107 N/C Add the y components: Enety EBy EAy (7.085 106 N/C) (1.198 107 N/C) 4.895 106 N/C Use the Pythagorean theorem to solve for the magnitude of the electric field: E (1.843 107 N/C)2 (4.895 106 N/C)2 net 1.91 107 N/C Use the tangent function to determine the direction of the net electric", " field at point P (Figure 11.15). tan 14.9\u00b0 The direction of the net field is 180\u00b0 14.9\u00b0 165\u00b0 4.895 106 N/C Enet \u03b8 x 1.843 107 N/C Figure 11.15 Paraphrase The net electric field at point P is 1.91 107 N/C [165\u00b0]. 552 Unit VI Forces and Fields 11-PearsonPhys30-Chap11 7/24/08 3:05 PM Page 553 In chapter 10, you learned that there are two types of electric charges that interact and are affected by electrostatic forces. In this section, you have learned that these charges are surrounded by electric fields\u2014regions of electric influence around every charge. Electrostatic forces affect charges placed in these fields. Fields explain how two charges can interact, even though there is no contact between them. Since electric fields are vector fields, you can use vector addition to determine a net electric field at a point in the presence of more than one charge in one-dimensional and two-dimensional situations. 11.1 Check and Reflect 11.1 Check and Reflect Knowledge 1. What is the difference between an electric force and an electric field? 2. Why was it necessary to introduce a \u201cfield theory\u201d? 3. How is the direction of an electric field defined? 4. Why is an electric field classified as a vector field? 5. If vector arrows can represent an electric field at a point surrounding a charge, identify the two ways that the vector arrows, shown below, represent differences in the electric fields around the two source charges. (a) the magnitude and direction of the electric field at a point 0.300 m to the right of the charge (b) the magnitude and direction of the electric force acting on a positive charge of 2.00 10\u20138 C placed at the point in (a) 8. A small test sphere with a negative charge of 2.50 C experiences an electrostatic attractive force of magnitude 5.10 10\u20132 N when it is placed at a point 0.0400 m from another larger charged sphere. Calculate (a) the magnitude and direction of the electric field at this point (b) the magnitude and the sign of charge on the larger charged sphere P E 9. A negative charge of 3.00 mC is 1.20 m to the right of another negative charge of 2.00 mC. Calculate P E 6. Describe the effect on the electric field at a point (a) if the magnitude of the charge producing the field is halved (b) if the sign of the charge producing the field is changed (c) if the magnitude of the test charge in the field is halved Applications 7. Given a small sphere with a positive charge of 4.50 10\u20136 C, determine: (a) the net electric field at a point along the same line and midway between the two charges (b) the point along the same line between the two charges where the net electric field will be zero Extension 10. Four similarly charged spheres of 5.00 C are placed at the corners of a square with sides of 1.20 m. Determine the electric field at the point of intersection of the two diagonals of the square. e TEST To check your understanding of forces and fields, follow the eTest links at www.pearsoned.ca/ school/physicssource. Chapter 11 Electric field theory describes electrical phenomena. 553 11-PearsonPhys30-Chap11 7/24/08 3:05 PM Page 554 11.2 Electric Field Lines and Electric Potential In section 11.1, you learned that the electric field from a charge q at a point P can be represented by a vector arrow, as shown in Figure 11.16. The length and direction of the vector arrow represent the magnitude and direction of the electric field (E ) at that point. By measuring the electric force exerted on a test charge at an infinite number of points around a source charge, a vector value of the electric field can be assigned to every point in space around the source charge. This creates a three-dimensional map of the electric field around the source charge (Figure 11.16). Electric Field Lines For many applications, however, a much simpler method is used to represent electric fields. Instead of drawing an infinite number of vector arrows, you can draw lines, called electric field lines, to represent the electric field. Field lines are drawn so that exactly one field line goes through any given point within the field, and the tangent to the field line at the point is in the direction of the electric field vector at that point. You can give the field lines a direction such that the direction of the field line through a given point agrees with the direction of the electric field at that point. Use the following rules when you draw electric field lines around a point charge: \u2022 Electric field lines due to a positive source charge start from the charge and extend radially away from", " the charge to infinity. \u2022 Electric field lines due to a negative source charge come from infin- ity radially into and terminate at the negative source charge. \u2022 The density of lines represents the magnitude of the electric field. In other words, the more closely spaced and the greater the number of lines, the stronger is the electric field. Figure 11.16 A threedimensional map of the electric field around a source charge info BIT A lightning rod works because of the concentration of charges on the point of a conductor. This concentration of charge creates an electric field that ionizes air molecules around the point. The ionized region either makes contact with an upward streamer to a cloud, thus preventing the formation of a damaging return lightning stroke, or intercepts a downward leader from the clouds and provides a path for the lightning to the ground to prevent damage to the structure. e SIM Explore the electric fields around a point charge and two charges. Follow the eSim links at www.pearsoned.ca/ school/physicssource. Figure 11.17 shows how to draw electric field lines around one and two negative point charges. Figure 11.17 The field lines around these charges were drawn using the rules given above. 554 Unit VI Forces and Fields 11-PearsonPhys30-Chap11 7/24/08 3:05 PM Page 555 M I N D S O N Drawing Electric Field Lines Rarely is the electric field at a point in space influenced by a single charge. Often, you need to determine the electric field for a complicated arrangement of charges. Electric field lines can be used to display these electric fields. In Figure 11.18, lawn seeds have been sprinkled on the surface of a container of cooking oil. In each case, a different charged object has been put into the oil. \u2022 On a sheet of paper, sketch the electric field lines in each situation using the rules for drawing electric field lines given on page 554. \u2022 Use concise statements to justify the pattern you drew in each of the sketches. (a) (d) (b) (c) (e) Figure 11.18 (a) one negative charge, (b) two negative charges, (c) one negative and one positive charge, (d) two oppositely charged plates, (e) one negatively charged cylindrical ring Conductors and Electric Field Lines In a conductor, electrons move freely until they reach a state of static equilibrium. For static equilibrium to exist, all charges must be at rest and thus must experience no net force. Achieving static equilibrium creates interesting distributions of charge that occur only in conducting objects and not in non-conducting objects. Following are five different situations involving charge distribution on conductors and their corresponding electric field lines. Solid Conducting Sphere When a solid metal sphere is charged, either negatively or positively, does the charge distribute evenly throughout the sphere? To achieve static equilibrium, all excess charges move as far apart as possible because of electrostatic forces of repulsion. A charge on the sphere at position A in Figure 11.19(a), for example, would experience a net force of electrostatic repulsion from the other charges. Consequently, all excess charges on a solid conducting sphere are repelled. These excess charges distribute evenly on the surface of the metal conducting sphere. Chapter 11 Electric field theory describes electrical phenomena. 555 11-PearsonPhys30-Chap11 7/24/08 3:05 PM Page 556 Figure 11.19(b) shows the electric field lines created by the distribution of charge on the surface of a solid conducting sphere. Because electric field lines cannot have a component tangential to this surface, the lines at the outer surface must always be perpendicular to the surface. (a) F F A F F F F (b) Figure 11.19(a) solid sphere Charges on a Figure 11.19(b) for a charged solid sphere Electric field lines Solid, Flat, Conducting Plate How do excess charges, either positive or negative, distribute on a solid, flat, conducting plate like the one in Figure 11.20(a)? On a flat surface, the forces of repulsion are similarly parallel or tangential to the surface. Thus, electrostatic forces of repulsion acting on charges cause the charges to spread and distribute evenly along the outer surface of a charged plate, as shown in Figure 11.20(b). Electric field lines extend perpendicularly toward a negatively charged plate. The electric field lines are uniform and parallel, as shown in Figure 11.20(c). (a) (b) (c) F F F F Figure 11.20 (a) Forces among three charges on the top surface of a flat, conducting plate (b) Uniform distribution of charges on a charged, flat, conducting plate (c) Uniform distribution of charges, shown with electric field lines Irregularly Shaped Solid Conducting Object For an irregularly shaped solid conductor, the charges are still repelled and accumulate on the outer surface. But do the charges distribute evenly on", " the outer surface? Figure 11.21(a) is an example of a charged, irregularly shaped object. 556 Unit VI Forces and Fields 11-PearsonPhys30-Chap11 7/24/08 3:05 PM Page 557 On a flatter part of the surface, the forces of repulsion are nearly parallel or tangential to the surface, causing the charges to spread out more, as shown in Figure 11.21(b). At a pointed part of a convex surface, the forces are directed at an angle to the surface, so a smaller component of the forces is parallel or tangential to the surface. With less repulsion along the surface, more charge can accumulate closer together. As a rule, the net electrostatic forces on charges cause the charges to accumulate at the points of an irregularly shaped convex conducting object. Conversely, the charges will spread out on an irregularly shaped concave conducting object. On irregularly shaped conductors, the charge density is greatest where the surface curves most sharply (Figure 11.21(c)). The density of electric field lines is also greatest at these points. info BIT The accumulation of charge on a pointed surface is the explanation for St. Elmo\u2019s fire, which you read about at the beginning of this chapter. St. Elmo\u2019s fire is a plasma (a hot, ionized gas) caused by the powerful electric field from the charge that accumulates on the tips of raised, pointed conductors during thunderstorms. St. Elmo\u2019s fire is known as a form of corona discharge or point discharge. (a) (b) (c) F F y x F F x y Figure 11.21(a) A charged, irregularly shaped convex object Figure 11.21(b) Forces affecting charges on the surface of an irregularly shaped convex object Figure 11.21(c) lines around a charged irregularly shaped convex object Electric field Hollow Conducting Object When a hollow conducting object is charged, either negatively or positively, does the charge distribute evenly throughout the inner and outer surfaces of the object? As you saw in Figures 11.19, 11.20, and 11.21, excess charges move to achieve static equilibrium, and they move as far apart as possible because of electrostatic forces of repulsion. In a hollow conducting object, all excess charges are still repelled outward, as shown in Figure 11.22(a). However, they distribute evenly only on the outer surface of the conducting object. There is no excess charge on the inner surface of the hollow object, no matter what the shape of the object is. The corresponding electric field lines created by the distribution of charge on the outer surface of a hollow object are shown in Figure 11.22(b). The electric field lines at the outer surface must always be perpendicular to the outer surface. (a) (b) Figure 11.22(a) A charged hollow conducting object Figure 11.22(b) Electric field lines on a hollow conducting object Chapter 11 Electric field theory describes electrical phenomena. 557 11-PearsonPhys30-Chap11 7/24/08 3:05 PM Page 558 info BIT Coaxial cable wires are used to transmit electric signals such as cable TV to your home. To prevent electric and magnetic interference from outside, a covering of conducting material surrounds the coaxial wires. Any charge applied to the conducting layer accumulates on the outside of the covering. No electric field is created inside a hollow conductor, so there is no influence on the signals transmitted in the wires. e WEB Research the operation of an ink-jet printer. What is the function of charged plates in these printers? Begin your search at www.pearsoned.ca/school/ physicssource. Most surprisingly, the electric field is zero everywhere inside the conductor, so there are no electric field lines anywhere inside a hollow conductor. As previously described, this effect can be explained using the superposition principle. Fields set up by many sources superpose, forming a single net field. The vector specifying the magnitude of the net field at any point is simply the vector sum of the fields of each individual source. Anywhere within the interior of a hollow conducting object, the vector sum of all the individual electric fields is zero. For this reason, the person inside the Faraday cage, shown in the photograph on page 508, is not affected by the tremendous charges on the outside surface of the cage. Parallel Plates If two parallel metal plates, such as those in Figure 11.23(a), are oppositely charged, how are the charges distributed? Electrostatic forces of repulsion of like charges, within each plate, cause the charges to distribute evenly within each plate, and electrostatic forces of attraction of opposite charges on the two plates cause the charges to accumulate on the inner surfaces. Thus, the charges spread and distribute evenly on the inner surfaces of the charged plates. (a) (b) Figure 11.23(a", ") The distribution of net charge on oppositely charged parallel plates Figure 11.23(b) Electric field lines between two oppositely charged parallel plates The magnitude of the resulting electric field can be shown to be the vector sum of each individual field, so it can be shown that the electric field anywhere between the plates is uniform. Thus, between two oppositely charged and parallel plates, electric field lines exist only between the charged plates. These lines extend perpendicularly from the plates, starting at the positively charged plate and terminating at the negatively charged plate. The electric field lines are uniform in both direction and density between the two oppositely charged plates, except near the edges of the plates. Such a system is called a parallel-plate capacitor. This type of capacitor is found in many different types of electrical equipment, including printers and televisions (where it is part of the \u201cinstant on\u201d feature). It is also used in particle accelerators, such as cathode-ray tubes and mass spectrometers. You will learn about mass spectrometers in Unit VIII. 558 Unit VI Forces and Fields 11-PearsonPhys30-Chap11 7/24/08 3:05 PM Page 559 THEN, NOW, AND FUTURE Defibrillators Save Lives During a heart attack, the upper and lower parts of the heart can begin contracting at different rates. Often these contractions are extremely rapid. This fluttery unsynchronized beating, called fibrillation, pumps little or no blood and can damage the heart. A defibrillator uses a jolt of electricity to momentarily stop the heart so that it can return to a normal beat (Figure 11.24). Figure 11.24 A defibrillator stops the fibrillation of the heart muscle by applying an electric shock. A defibrillator consists of two parallel charged plates (see Figure 11.23(b)), called a parallel-plate capacitor, connected to a power supply and discharging pads. A typical defibrillator stores about 0.4 C on the plates, creating a potential difference of approximately 2 kV between the plates. When discharged through conductive pads placed on the patient\u2019s chest, the capacitor delivers about 0.4 kJ of electrical energy in 0.002 s. Roughly 200 J of this energy passes through the patient\u2019s chest. A defibrillator uses a highvoltage capacitor to help save lives. Such capacitors have many other applications in other electrical and electronic devices, such as the highvoltage power supplies for cathoderay tubes in older televisions and computer monitors. The charge stored in such capacitors can be dangerous. Products con- taining such high-voltage capacitors are designed to protect the users from any dangerous voltages. However, service technicians must be careful when working on these devices. Since the capacitors store charge, they can deliver a nasty shock even after the device is unplugged. Questions 1. How does the magnitude of the power delivered by the plates compare with the actual power delivered to the chest by the jolt? 2. Identify a feature of televisions that demonstrates an important application of parallel-plate capacitors. 3. If a defibrillator can store 0.392 C of charge in 30 s, how many electrons are stored in this time period? M I N D S O N Faraday\u2019s Ice Pail In the early 1800s, Michael Faraday performed an experiment to investigate the electric fields inside a hollow metal container. He used ice pails, so this experiment is often called \u201cFaraday\u2019s ice pail experiment.\u201d charged rod This activity is called a conceptual experiment because you will not perform the experiment. Instead, you will predict and justify the results of an experimental procedure that duplicates Faraday\u2019s investigation. The purpose of the experiment is to determine what type of electric field exists on the inside and the outside of a hollow metal container. A positively charged rod is placed into position inside the metal container, near the centre, as shown in Figure 11.25. The rod is then moved to a position inside the metal container, near one of the inner surfaces. \u2022 Which of the electroscopes would show a deflection when the rod is near the centre of the metal container? \u2022 Clearly explain your reasoning and the physical principles you used in determining your answers to these questions. insulated plate Figure 11.25 An ice pail is a metal container. It is placed on an insulated surface, and electroscopes are attached to the inside and outside surfaces of the metal container. Chapter 11 Electric field theory describes electrical phenomena. 559 11-PearsonPhys30-Chap11 7/24/08 3:05 PM Page 560 Figure 11.26 The charged dome of a Van de Graaff generator exposes a person to very large voltages. Electric Potential Energy and Electric Potential A Van de Graaff generator can generate up to 250 kV. Touching the dome not", " only produces the spectacular results shown in Figure 11.26, it can also cause a mild, harmless shock. On the other hand, touching the terminals of a wall socket, which has a voltage of 120 V, can be fatal. An understanding of this dramatic difference between the magnitude of the voltage and its corresponding effect requires a study of the concepts of electric potential energy and electric potential. These concepts are important in the study of electric fields. Even though the terms seem similar, they are very different. To explain the difference, you will study these concepts in two types of electric fields: non-uniform electric fields around point charges, and uniform electric fields between parallel charged plates. Electric Potential Energy In previous grades, you learned about the relationship between work and potential energy. Work is done when a force moves an object in the direction of the force such that: d W F where W is work, and F displacement of the object. and d are the magnitudes of the force and the In a gravitational system like the one shown in Figure 11.27(a), lifting a mass a vertical distance against Earth\u2019s gravitational field requires work to stretch an imaginary \u201cgravitational spring\u201d connecting the mass and Earth. Further, because the force required to do the work is a conservative force, the work done against the gravitational field increases the gravitational potential energy of the system by an amount equal to the work done. Therefore: gravitational potential energy gain work done (a) Fapp d mass Fg Ep W (b) Fapp d Fe Figure 11.27(a) Work is required to lift a mass to a certain position above Earth\u2019s surface. Figure 11.27(b) Work is required to move a small positive charge away from a larger negative charge. 560 Unit VI Forces and Fields 11-PearsonPhys30-Chap11 7/24/08 3:05 PM Page 561 Similarly, in an electrostatic system like the one shown in Figure 11.27(b), moving a small charge through a certain distance in a non-uniform electric field produced by another point charge requires work to either compress or stretch an imaginary \u201celectrostatic spring\u201d connecting the two charges. Since the force required to do this work is also a conservative force, the work done in the electric field must increase the electric potential energy of the system. Electric potential energy is the energy stored in the system of two charges a certain distance apart (Figure 11.28). Electric potential energy change equals work done to move a small charge: Ep W q1 r P q2 Figure 11.28 Electric potential energy is the energy stored in the system of two charges a certain distance apart. Example 11.5 Moving a small charge from one position in an electric field to another position requires 3.2 10\u201319 J of work. How much electric potential energy will be gained by the charge? Analysis and Solution The work done against the electrostatic forces is W. The electric potential energy gain is Ep. In a conservative system, Ep W So, Ep W 3.2 1019 J The electric potential energy gain of the charge is 3.2 10\u201319 J. Practice Problems 1. A small charge gains 1.60 10\u201319 J of electric potential energy when it is moved to a point in an electric field. Determine the work done on the charge. 2. A charge moves from one position in an electric field, where it had an electric potential energy of 6.40 10\u201319 J, to another position where it has an electric potential energy of 8.00 10\u201319 J. Determine the work necessary to move the charge. Answers 1. 1.60 1019 J 2. 1.60 1019 J Choosing a Reference Point In Chapter 7, you learned that commonly used reference points for zero gravitational potential energy are Earth\u2019s surface or infinity. Choosing a zero reference point is necessary so you can analyze the relationship between work and gravitational potential energy. Chapter 11 Electric field theory describes electrical phenomena. 561 11-PearsonPhys30-Chap11 7/24/08 3:05 PM Page 562 Consider a zero reference point at Earth\u2019s surface. An object at rest on Earth\u2019s surface would have zero gravitational potential energy relative to Earth\u2019s surface. If the object is lifted upward, opposite to the direction of the gravitational force it experiences, then work is being done on the object. The object thus gains gravitational potential energy. If the object falls back to the surface in the same direction as the gravitational force, then the object loses gravitational potential energy. As with gravitational potential energy, the value of electric potential energy at a certain position is meaningless unless it is compared to a reference point where the electric potential energy is zero. The choice of a zero reference point for electric potential energy is arbitrary. For example, suppose an electric field is being produced by a large negative charge. A small positive charge would be attracted and come to rest on the surface of the larger negative charge,", " where it would have zero electric potential energy. This position could be defined as a zero electric potential energy reference point (Figure 11.29(a)). Then, the test charge has positive electric potential energy at all other locations. Alternatively, the small positive test charge may be moved to a position so far away from the larger negative charge that there is no electrostatic attraction between them. This position would be an infinite distance away. This point, at infinity, is often chosen as the zero electric potential energy reference point. Then, the test charge has negative electric potential energy at all other locations. This text uses infinity as the zero electric potential energy reference point for all calculations (Figure 11.29(b)). at surface at infinity q q Ep 0 (a) Ep 0 (b) Figure 11.29 Two commonly used reference points for electric potential energy: (a) test charge defined as having zero electric potential energy at the surface of the source charge (b) test charge defined as having zero electric potential energy at infinity 562 Unit VI Forces and Fields 11-PearsonPhys30-Chap11 7/24/08 3:05 PM Page 563 Work and Electric Potential Energy Whenever work is done on a charge to move it against the electric force caused by an electric field, the charge gains electric potential energy. The following examples illustrate the relationship between work and electric potential energy. Electric Potential Energy Between Parallel Charged Plates Except at the edges, the electric field between two oppositely charged plates is uniform in magnitude and direction. Suppose a small positive charge in the field between the plates moves from the negative plate to the positive plate with a constant velocity. This motion requires an external force to overcome the electrostatic forces the charged plates exert on the positive charge. The work done on the charge increases the system\u2019s electric potential energy: Ep W F d Example 11.6 When a small positive charge moves from a negative plate to a positive plate, 2.3 \u00d7 10\u201319 J of work is done. How much electric potential energy will the charge gain? Analysis and Solution In a conservative system, EP EP W 2.3 1019 J W. Paraphrase The electric potential energy gain of the charge is 2.3 1019 J. Practice Problem 1. A charge gained 4.00 105 J of electric potential energy when it was moved between two oppositely charged plates. How much work was done on the charge? Answer 1. 4.00 105 J Electric Potential Suppose two positive charges are pushed toward a positive plate. In this case, twice as much work is done, and twice as much electric potential energy is stored in the system. However, just as much electric potential energy is still stored per charge. Storing 20 J of energy in two charges is the same as storing 10 J of energy in each charge. At times, it is necessary to determine the total electric potential energy at a certain location in an electric field. At other times, it is convenient to consider just the electric potential energy per unit charge at a location. The electric potential energy stored per unit charge at a given point is the amount of work required to move a unit charge to that Chapter 11 Electric field theory describes electrical phenomena. 563 11-PearsonPhys30-Chap11 7/24/08 3:05 PM Page 564 electric potential: the electric potential energy stored per unit charge at a given point in an electric field point from a zero reference point (infinity). This quantity has a special name: electric potential. To determine the electric potential at a location, use this equation: electric potential electric potential energy charge Ep V q where V is in volts, Ep is in joules, and q is in coulombs. Since electric potential energy is measured in joules and charge is measured in coulombs, o e l u j 1 volt b m o l u co 1 1 Thus, if the electric potential at a certain location is 10 V, then a charge of 1 C will possess 10 J of electric potential energy, a charge of 2 C will possess 20 J of electric potential energy, and so on. Even if the total electric potential energy (Ep) at a location changes, depending on the amount of charge placed in the electric field, the electric potential (V ) at that location remains the same. A balloon can be used as an example to help explain the difference between the concepts of electric potential energy and electric potential. Suppose you rub a balloon with fur. The balloon acquires an electric potential of a few thousand volts. In other words, the electric energy stored per coulomb of charge on the balloon is a few thousand volts. Written as an equation, Ep V q Now suppose the balloon were to gain a large charge of 1 C during the rubbing process. In order for the electric potential to stay the same, a few thousand joules of work would be needed to produce the electrical energy that would allow the balloon to maintain that electric potential. However, the amount of charge a balloon acquires during rubbing is usually only", " in the order of a few microcoulombs. So, acquiring this potential requires a small amount of work to produce the energy needed. Even though the electric potential is high, the electric potential energy is low because of the extremely small charge. Concept Check Suppose the magnitude of a charge placed in an electric field were doubled. How much would the electric potential energy and the electric potential change? info BIT The SI unit of electric potential is the volt, named in honour of the Italian physicist Count Alessandro Volta (1745\u20131827), who developed the first electric battery in the early 1800s. 564 Unit VI Forces and Fields 11-PearsonPhys30-Chap11 7/24/08 3:05 PM Page 565 Electric Potential Difference When a charge moves from one location to another in an electric field, it experiences a change in electric potential. This change in electric potential is called the electric potential difference, V, between the two points and V Vfinal Vinitial Ep since V q Ep V q where Ep is the amount of work required to move the charge from one location to the other. The potential difference depends only on the two locations. It does not depend on the charge or the path taken by the charge as it moves from one location to another. Electric potential difference is commonly referred to as just potential difference or voltage. An electron volt (eV) is the quantity of energy an electron gains or loses when passing through a potential difference of exactly 1 V. An electron volt is vastly less than a joule: 1 eV 1.60 1019 J Although not an SI unit, the electron volt is sometimes convenient for expressing tiny quantities of energy, especially in situations involving a single charged particle such as an electron or a proton. The energy difference in Example 11.6 could be given as (2.3 1019 J) 1.4 eV V e 1 019 J 1 1.60 electric potential difference: change in electric potential experienced by a charge moving between two points in an electric field electron volt: the change in energy of an electron when it moves through a potential difference of 1 V PHYSICS INSIGHT The notation VAB is widely used instead of V to represent the potential difference at point A relative to point B. When the points in question are clear from the context, the subscripts are generally omitted. For example, the equation for Ohm\u2019s law is usually written as V IR, where it is understood that V represents the potential difference between the ends of the resistance R. Example 11.7 Moving a small charge of 1.6 \u00d7 10\u201319 C between two parallel plates increases its electric potential energy by 3.2 \u00d7 10\u201316 J. Determine the electric potential difference between the two parallel plates. Analysis and Solution To determine the electric potential difference between the plates, use the equation Ep.0 103 V The electric potential difference between the plates is 2.0 103 V. Practice Problems 1. In moving a charge of 5.0 C from one terminal to the other, a battery raises the electric potential energy of the charge by 60 J. Determine the potential difference between the battery terminals. 2. A charge of 2.00 10\u20132 C moves from one charged plate to an oppositely charged plate. The potential difference between the plates is 500 V. How much electric potential energy will the charge gain? Answers 1. 12 V 2. 10.0 J Chapter 11 Electric field theory describes electrical phenomena. 565 11-PearsonPhys30-Chap11 7/24/08 3:05 PM Page 566 Example 11.8 A small charge of 3.2 1019 C is moved between two parallel plates from a position with an electric potential of 2.0 103 V to another position with an electric potential of 4.0 103 V (Figure 11.30). Practice Problems 1. A sphere with a charge of magnitude 2.00 C is moved between two positions between oppositely charged plates. It gains 160 J of electric potential energy. What is the potential difference between the two positions? 2. An electron moves between two positions with a potential difference of 4.00 104 V. Determine the electric potential energy gained by the electron, in joules (J) and electron volts (eV). Answers 1. 80.0 V 2. 6.40 1015 J or 4.00 104 eV A 2.0 103 V B 4.0 103 V Figure 11.30 3.2 1019 C battery Determine: (a) the potential difference between the two positions (b) the electric potential energy gained by moving the charge, in joules (J) and electron volts (eV) Given Vinitial Vfinal 2.0 103 V 4.0 103 V q 3.2 10\u201319 C Required (a) potential difference between points B and A (V ) (b) electric potential energy gained by the charge (Ep) Analysis and Solution (a) V Vfinal Vinitial (4.0 103 V", ") (2.0 103 V) 2.0 103 V (b) To calculate the electric potential energy, use the equation Ep. V q Vq (2.0 103 V)(3.2 1019 C) 6.4 10\u201316 J Ep Ep Since 1 eV 1.60 1019 J, V e 1 (6.4 1016 J) 019 J 1 4.0 103 eV 4.0 keV 1.60 Paraphrase (a) The potential difference between the two positions is 2.0 103 V. (b) The energy gained by moving the charge between the two positions is 6.4 10\u201316 J or 4.0 103 eV. 566 Unit VI Forces and Fields 11-PearsonPhys30-Chap11 7/24/08 3:05 PM Page 567 The Electric Field Between Charged Plates Earlier in this section, you determined the electric field strength surrounding a point charge using the following equations: k q or E E 2 r F e q You also learned that the electric field around a point charge is a nonuniform electric field. Its magnitude depends on the distance from the charge. Later, you learned that a special type of electric field exists between two charged parallel plates. The magnitude of the electric field between the plates is uniform anywhere between the plates and it can be determined using the general equation for an electric field, F. You cannot use the equation E E e q k q because it is used only 2 r e WEB One of the technological applications of parallelplate capacitors is in disposable cameras. Research the role of capacitors in these cameras. Begin your search at www.pearsoned.ca/ school/physicssource. for point charges. Now, after studying electric potential difference, you can see how another equation for determining the electric field strength between plates arises from an important relationship between the uniform electric field and the electric potential difference between two charged parallel plates (Figure 11.31). If a small positively charged particle (q) is moved through the uni- form electric field (E q. This force is ), a force is required, where F the force exerted on the particle due to the presence of the electric field. If this force moves the charged particle a distance (d) between the plates, then the work done is: E Figure 11.31 Electrically charged parallel plates W F or W E d qd Since this system is conservative, the work done is stored in the charge as electric potential energy: W Ep E qd The electric potential difference between the plates is: Ep V q q d E q d E To calculate the magnitude of the uniform electric field between charged plates, use the equation E V d where V is the electric potential difference between two charged plates in volts; d is the distance in metres between the plates; and E is the magnitude of the electric field in volts per metre. Chapter 11 Electric field theory describes electrical phenomena. 567 11-PearsonPhys30-Chap11 7/24/08 3:05 PM Page 568 Note that 1 V/m equals 1 N/C because /C J 1 V//C Example 11.9 Practice Problems 1. Two charged parallel plates, separated by 5.0 10\u20134 m, have an electric field of 2.2 104 V/m between them. What is the potential difference between the plates? 2. Spark plugs in a car have electrodes whose faces can be considered to be parallel plates. These plates are separated by a gap of 5.00 10\u20133 m. If the electric field between the electrodes is 3.00 106 V/m, calculate the potential difference between the electrode faces. Answers 1. 11 V 2. 1.50 104 V A cathode-ray-tube (CRT) computer monitor accelerates electrons between charged parallel plates (Figure 11.32). These electrons are then directed toward a screen to create an image. If the plates are 1.2 10\u20132 m apart and have a potential difference of 2.5 104 V between them, determine the magnitude of the electric field between the plates. 1.2 102 m V 2.5 104 V accelerating plates Given V 2.5 104 V d 1.2 10\u20132 m screen Figure 11.32 Required ) magnitude of the electric field between the plates (E Analysis and Solution To calculate the magnitude of the electric field between the plates, use the equation V E d 4 V 0 1.5 2 2 1 1 m.2 2.1 106 V/m 0 Paraphrase The magnitude of the electric field between the plates is 2.1 106 V/m. 568 Unit VI Forces and Fields 11-PearsonPhys30-Chap11 7/24/08 3:05 PM Page 569 11.2 Check and Reflect 11.2 Check and Reflect Knowledge 1. Describe the difference between an electric field vector and an electric field line. 2. Sketch electric field lines around the following charges: (a", ") a positive charge (b) a negative charge (c) two positive charges (d) two negative charges (e) a positive charge and a negative charge 3. Describe the difference between electric potential and electric potential energy. Applications 4. At a point in Earth\u2019s atmosphere, the electric field is 150 N/C downward and the gravitational field is 9.80 N/kg downward. (a) Determine the electric force on a proton (p) placed at this point. (b) Determine the gravitational force on the proton at this point. The proton has a mass of 1.67 10\u201327 kg. 9. Determine the magnitude and direction of the net electric field at point P shown in the diagram below. 50 \u03bcC 10 \u03bcC P 0.30 m 0.15 m 10. A uniform electric field exists between two oppositely charged parallel plates connected to a 12.0-V battery. The plates are separated by 6.00 104 m. (a) Determine the magnitude of the electric field between the plates. (b) If a charge of 3.22 106 C moves from one plate to another, calculate the change in electric potential energy of the charge. Extensions 11. A metal car is charged by contact with a charged object. Compare the charge distribution on the outside and the inside of the metal car body. Why is this property useful to the occupants of the car if the car is struck by lightning? 5. A metal box is charged by touching it with 12. Explain why only one of the electroscopes a negatively charged object. (a) Compare the distribution of charge at the corners of the box with the faces of the box. (b) Draw the electric field lines inside and surrounding the box. 6. What is the electric field intensity 0.300 m away from a small sphere that has a charge of 1.60 10\u20138 C? connected to the hollow conductive sphere in the illustration below indicates the presence of a charge. 7. Calculate the electric field intensity 13. Two points at different positions in midway between two negative charges of 3.2 C and 6.4 C separated by 0.40 m. 8. A 2.00-C charge jumps across a spark gap in a spark plug across which the potential difference is 1.00 103 V. How much energy is gained by the charge? an electric field have the same electric potential. Would any work be required to move a test charge from one point to another? Explain your answer. e TEST To check your understanding of electric field lines, follow the eTest links at www.pearsoned.ca/ school/physicssource. Chapter 11 Electric field theory describes electrical phenomena. 569 11-PearsonPhys30-Chap11 7/24/08 3:05 PM Page 570 info BIT Living cells \u201cpump\u201d positive sodium ions (Na) from inside a cell to the outside through a membrane that is 0.10 m thick. The electric potential is 0.70 V higher outside the cell than inside it. To move the sodium ions, work must be done. It is estimated that 20% of the energy consumed by the body in a resting state is used to operate these \u201cpumps.\u201d 11.3 Electrical Interactions and the Law of Conservation of Energy A charge in an electric field experiences an electrostatic force. If the charge is free to move, it will accelerate in the direction of the electrostatic force, as described by Newton\u2019s second law. The acceleration of the charge in the non-uniform electric field around a point charge is different from the acceleration motion of a charge in a uniform electric field between charged plates. Figure 11.33 shows a charge in the non-uniform field of a point charge. The electrostatic force on a charge placed in the field varies inversely as the square of the distance between the charges. A varying force causes non-uniform acceleration. Describing the motion of the charge in this type of situation requires applying calculus to Newton\u2019s laws of motion, which is beyond the scope of this text. However, to determine the particle\u2019s speed at a given point, you can use the law of conservation of energy. source charge F a E Figure 11.33 The electrostatic force on a point charge in a non-uniform electric field causes non-uniform acceleration of the charge. If the forces acting on an object are conservative forces, then the work done on a system changes the potential energy of the system. Electric potential energy, like gravitational potential energy, can be converted to kinetic energy. A charged particle placed in an electric field will accelerate from a region of high potential energy to a region of low potential energy. According to the law of conservation of energy, the moving charge gains kinetic energy at the expense of potential energy. If you assume that no energy is lost to friction and the forces are conservative, the kinetic energy gained equals the potential energy lost, so the", " sums of the two energies are always equal: Epi Eki Epf Ekf 570 Unit VI Forces and Fields 11-PearsonPhys30-Chap11 7/28/08 9:38 AM Page 571 Example 11.10 A pith ball of mass 2.4 104 kg with a positive charge of 1.2 108 C is initially at rest at location A in the electric field of a larger charge (Figure 11.34). At this location, the charged pith ball has 3.0 107 J of electric potential energy. When released, the ball accelerates toward the larger charge. At position B, the ball has 1.5 108 J of electric potential energy. Find the speed of the pith ball when it reaches position B. 1.2 108 C 2.4 104 kg B 1.5 108 J A 3.0 107 J Figure 11.34 Given m 2.4 104 kg 3.0 107 J Epi q 1.2 108 C 1.5 108 J Epf Required speed of the ball at position B (v) Analysis and Solution The pith ball is at rest at A, so its initial kinetic energy is zero. Its electric potential energy at B is lower than at A. Since this system is conservative, the loss of electric potential energy when the ball moves from A to B is equal to a gain in kinetic energy, according to the law of conservation of energy: Epf Ekf Eki Epi Substitute the given values and solve for Ekf (3.0 107 J) 0 (1.5 108 J) Ekf Ekf Since the kinetic energy of an object is Ek 2.85 107 J. 1 mv2, 2 Practice Problems 1. A negative charge of 3.00 10\u20139 C is at rest at a position in the electric field of a larger positive charge and has 3.20 1012 J of electric potential energy at this position. When released, the negative charge accelerates toward the positive charge. Determine the kinetic energy of the negative charge just before it strikes the larger positive charge. 2. A small sphere with a charge of 2.00 C and a mass of 1.70 103 kg accelerates from rest toward a larger positive charge. If the speed of the sphere just before it strikes the positive charge is 5.20 104 m/s, how much electric potential energy did the negative charge lose? Answers 1. 3.20 1012 J 2. 2.30 106 J Ek 2 v 2 m v 2Ek m 2(2.85 107 J) 2.4 104 kg 4.9 102 m/s Paraphrase The speed of the pith ball at position B is 4.9 102 m/s. Chapter 11 Electric field theory describes electrical phenomena. 571 11-PearsonPhys30-Chap11 7/24/08 3:05 PM Page 572 It is easier to describe the motion of a charge in a uniform electric field between two parallel plates, as shown in Figure 11.35. In this case, the acceleration is constant because of the constant force, so either the work\u2013energy theorem or the laws of dynamics can be used. (Because the electric field is constant (uniform), the force acting on a charge q is also constant because F e qE.) F E E Figure 11.35 In a uniform electric field between two parallel plates, the acceleration of a charge is constant. Concept Check Electrostatic forces and gravitational forces are similar, so the motion of objects due to these forces should be similar. Consider a charge in an electric field between two parallel plates. Sketch the direction of the motion of the charge when its initial motion is: \u2022 perpendicular to the plates (the electrostatic force is similar to the gravitational force on falling masses) \u2022 parallel to the plates (the electrostatic force is similar to the gravitational force that causes the parabolic projectile motion of a mass close to the surface of a large planet or moon) Example 11.11 Two vertical parallel plates are connected to a DC power supply, as shown in Figure 11.36. The electric potential between the plates is 2.0 103 V. A sphere of mass 3.0 1015 kg with a positive charge of 2.6 1012 C is placed at the positive plate and released. It accelerates toward the negative plate. Determine the speed of the sphere at the instant before it strikes the negative plate. Ignore any gravitational effects. 2.6 1012 C 3.0 1015 kg 2.0 103 V Figure 11.36 572 Unit VI Forces and Fields 11-PearsonPhys30-Chap11 7/24/08 3:05 PM Page 573 Given q 2.6 1012 C V 2.0 103 V m 3.0 1015 kg Required speed of the sphere at the negative plate (v) Analysis and Solution This system is conservative. You can use kinetic energy of the charge to find its speed., is 0 J. 0 J. The initial electric potential energy of", " the sphere at the positive plate Vq. Since the sphere was at rest, its initial kinetic energy, is Epi Eki The final electric potential energy of the sphere at the negative plate is Epf According to the law of conservation of energy, Ekf Epi Vq 0 J 0 J Ekf (2.0 103 V)(2.6 1012 C) 0 J 0 J Ekf Epf Eki 5.2 109 J Ekf Since Ek 1 mv 2, 2 v 2Ek m 2(5.2 109 J) 3.0 1015 kg 1.9 103 m/s Paraphrase The speed of the sphere at the negative plate is 1.9 103 m/s. Practice Problems 1. An alpha particle with a charge of 3.20 1019 C and a mass of 6.65 1027 kg is placed between two oppositely charged parallel plates with an electric potential difference of 4.00 104 V between them. The alpha particle is injected at the positive plate with an initial speed of zero, and it accelerates toward the negative plate. Determine the final speed of the alpha particle just before it strikes the negative plate. 2. If a charge of 6.00 106 C gains 3.20 104 J of kinetic energy as it accelerates between two oppositely charged plates, what is the potential difference between the two parallel plates? Answers 1. 1.96 106 m/s 2. 53.3 V Chapter 11 Electric field theory describes electrical phenomena. 573 11-PearsonPhys30-Chap11 7/24/08 3:05 PM Page 574 Example 11.12 An electron enters the electric field between two charged parallel plates, as shown in Figure 11.37. electric field Figure 11.37 (a) Copy Figure 11.37 into your notebook and sketch the motion of the electron between the plates. (b) If the electron experiences a downward acceleration of 2.00 1017 m/s2 due to the electric field between the plates, determine the time taken for the electron to travel 0.0100 m to the positive plate. Given a 2.00 1017 m/s2 [down] d 0.0100 m Required (a) sketch of the electron\u2019s motion (b) time (t) Analysis and Solution (a) The electron\u2019s acceleration is downward, so the motion of the electron will follow a parabolic path to the positive plate (Figure 11.38), similar to the projectile motion of an object travelling horizontally to the surface of Earth and experiencing downward acceleration due to gravity. electric field Figure 11.38 Practice Problems 1. Two horizontal parallel plates, 1.2 102 m apart, are connected to a DC power supply, as shown in the figure below. The electric field between the plates is 1.7 105 V/m. A sphere of mass 3.0 1015 kg with a positive charge of 2.6 1012 C is injected into the region between the plates, with an initial speed of 3.3 103 m/s, as shown. It accelerates toward the negative plate. Copy the diagram into your notebook, sketch the motion of the positive charge through the region between the plates, and determine the distance the positive charge moves toward the negative plate after 6.0 106 s have elapsed. Gravitational effects may be ignored in this case. electric field 1.7 105 V/m 3.3 103 m/s 2. An electron, travelling at 2.3 103 m/s, enters perpendicular to the electric field between two horizontal charged parallel plates. If the electric field strength is 1.5 102 V/m, calculate the time taken for the electron to deflect a distance of 1.0 10\u20132 m toward the positive plate. Ignore gravitational effects. Answers 1. 2.7 103 m 2. 2.7 108 s 574 Unit VI Forces and Fields 11-PearsonPhys30-Chap11 7/28/08 9:40 AM Page 575 (b) Use the equation d vi t 1 a(t)2 to determine the 2 time it takes the electron to fall to the positive plate. Since vi d 1 a(t)2 2 t 2d a 2(0.0100 m) m 2.00 1017 s2 0, 3.16 1010 s Paraphrase (a) The path of the electron between the parallel plates is parabolic. (b) The time taken for the electron to fall to the positive plate is 3.16 1010 s. 11.3 Check and Reflect 11.3 Check and Reflect Knowledge Applications 1. In what direction will a positively charged particle accelerate in an electric field? 2. Electric potential energy exists only where a charge is present at a point in an electric field. Must a charge also be present at that point for there to be electric potential? Why or why not? 4. Calculate the speed of an electron and a proton after each has accelerated from rest through an electric potential of", " 220 V. 5. Electrons in a TV picture tube are accelerated by a potential difference of 25 kV. Find the maximum speed the electrons would reach if relativistic effects are ignored. 3. Two positively charged objects are an 6. A charge gains 1.92 1014 J of electric equal distance from a negatively charged object, as shown in the diagram below. Charge B is greater than charge A. Compare the electric potential and electric potential energy of the positively charged objects. B A potential energy when it moves through a potential difference of 3.20 104 V. What is the magnitude of the charge? 7. How much work must be done to increase the electric potential of a charge of 2.00 106 C by 120 V? 8. A deuterium ion (H1), a heavy isotope of hydrogen, has a charge of 1.60 1019 C and a mass of 3.34 1027 kg. It is placed between two oppositely charged plates with a voltage of 2.00 104 V. Find the final maximum speed of the ion if it is initially placed at rest (a) at the positive plate (b) midway between the two plates Chapter 11 Electric field theory describes electrical phenomena. 575 11-PearsonPhys30-Chap11 7/24/08 3:05 PM Page 576 9. A small charge of +3.0 108 C with a mass of 3.0 105 kg is slowly pulled through a potential difference of 6.0 102 V. It is then released and allowed to accelerate toward its starting position. Calculate (a) the initial work done to move the charge (b) the maximum kinetic energy of the returning charge (c) the final speed of the returning charge 10. An electron, travelling horizontally at a speed of 5.45 106 m/s, enters a parallelplate capacitor with an electric field of 125 N/C between the plates, as shown in the figure below. Extensions 11. Determine whether an electron or a proton would take less time to reach one of a pair of oppositely charged parallel plates when starting from midway between the plates. Explain your reasoning. 12. How can the electric potential at a point in an electric field be high when the electric potential energy is low? 13. In question 10, explain why the resulting motion of an electron, initially travelling perpendicular to the uniform electric field between the two charged parallel plates, will be parabolic and not circular. e TEST 5.45 106 m/s To check your understanding of electrical interactions and the law of conservation of energy, follow the eTest links at www.pearsoned.ca/ school/physicssource. (a) Copy the diagram into your notebook and sketch (i) the electric field lines between the plates (ii) the motion of the electron through the capacitor (b) Determine the force due to the electric field on the electron. (c) Ignoring gravitational effects, calculate the acceleration of the electron. (d) If the electron falls a vertical distance of 6.20 103 m toward the positive plate, how far will the electron travel horizontally between the plates? 576 Unit VI Forces and Fields 11-PearsonPhys30-Chap11 7/28/08 9:42 AM Page 577 CHAPTER 11 SUMMARY Key Terms and Concepts field test charge source charge electric field line Key Equations F e E q Ep V q k E r q 2 Ep V q Conceptual Overview electric potential energy electric potential (voltage) electron volt electric potential difference Ep W Ep W F d V Vfinal Vinitial E V d Epi Eki Epf Ekf The concept map below summarizes many of the concepts and equations in this chapter. Copy and complete the map to have a full summary of the chapter. E Fe q E kq r2 electric potential energy calculate define magnitude direction calculate E between more than two charges one-dimensional situations vector electric potential Electric Fields electric potential difference field lines electric field between plates relationship between electric field and distance drawing electric fields around charged objects define a reference point calculate electric potential energy define calculate define calculate Chapter 11 Electric field theory describes electrical phenomena. 577 11-PearsonPhys30-Chap11 7/24/08 3:05 PM Page 578 CHAPTER 11 REVIEW Knowledge 1. (11.1) Identify the three theories that attempt to explain \u201caction at a distance.\u201d 13. (11.3) Describe the key differences between the electric field surrounding a point charge and the electric field between charged parallel plates? 2. (11.1) How can it be demonstrated that the space around a charged object is different from the space around an uncharged object? 3. (11.1) How does a vector arrow represent both 14. (11.3) Assuming forces in a system are conservative, explain how (a) work done in the system is related to potential energy of the system the magnitude and direction of a vector quantity? (b) the kinetic and potential energy of", " the 4. (11.2) What is the difference between an electric field vector and an electric field line? 5. (11.2) Two hollow metal objects, with shapes shown below, are charged with a negatively charged object. In your notebook, sketch the distribution of charge on both objects and the electric field lines surrounding both objects. cross-section of hollow sphere cross-section of hollow egg-shaped object system are related Applications 15. Compare the electric potential energy of a positive test charge at points A and B near a charged sphere, as shown below. A B 16. A large metal coffee can briefly contacts a charged object. Compare the results when uncharged electroscopes are touched to the inside and outside surfaces of the can. 6. (11.2) How do electric field lines represent the magnitude of an electric field? 17. A point charge has a charge of 2.30 C. Calculate (a) the electric field at a position 2.00 m from 7. (11.2) Where do electric field lines originate for (a) a negative point charge? (b) a positive point charge? 8. (11.2) Identify two equations that can be used to calculate the magnitude of an electric field around a point charge. the charge (b) the electric force on a charge of 2.00 C placed at this point 18. A charge of 5.00 C is separated from another charge of 2.00 C by a distance of 1.20 m. Calculate (a) the net electric field midway between the 9. (11.2) When do electric charges achieve static two charges equilibrium in a charged object? 10. (11.2) Why do electric charges accumulate at a point in an irregularly shaped object? 11. (11.2) State a convenient zero reference point for electric potential energy (a) around a point charge (b) between two oppositely charged parallel plates 12. (11.2) What equation would you use to calculate the electric potential energy at a certain position around a point charge? (b) the position where the net electric field is zero 19. Find the net electric field intensity at point C in the diagram below. C 0.040 m 2.0 \u03bcC A 0.060 m 2.0 \u03bcC B 578 Unit VI Forces and Fields 11-PearsonPhys30-Chap11 7/24/08 3:05 PM Page 579 20. A force of 15.0 N is required to move a charge of 2.0 C through a distance of 0.20 m in a uniform electric field. (a) How much work is done on the charge? (b) How much electric potential energy does the charge gain in joules? 21. How much electric potential energy would an object with a charge of 2.50 C have when it is 1.20 m from a point charge of 3.00 C? (Hint: Consider how much electric potential energy the negatively charged object would have when touching the point charge.) 22. Two parallel plates are separated by a distance of 3.75 cm. Two points, A and B, lie along a perpendicular line between the parallel plates and are 1.10 cm apart. They have a difference in electric potential of 6.00 V. (a) Calculate the magnitude of the electric field between the plates. (b) Determine the electric potential between the parallel plates. 23. How much work is required to move a charge perpendicular to the electric field between two oppositely charged parallel plates? 24. A cell membrane is 1.0 107 m thick and has an electric potential difference between its surfaces of 0.070 V. What is the electric field within the membrane? 25. A lithium nucleus (Li3) that has a charge of 4.80 1019 C is accelerated by a voltage of 6.00 105 V between two oppositely charged plates. Calculate the energy, in joules (J) and electron volts (eV), gained by the nucleus. 26. How much electric potential energy, in joules (J) and electron volts (eV), does an alpha particle gain when it moves between two oppositely charged parallel plates with a voltage of 20 000 V? 27. Consider a sphere with a known charge in the electric field around a larger unknown charge. What would happen to the electric field at a point if Extensions 28. Explain why electric field lines can never cross. 29. A bird is inside a metal birdcage that is struck by lightning. Is the bird likely to be harmed? Explain. 30. Explain why charge redistributes evenly on the outside surface of a spherical charged object and accumulates at a point on an irregularly shaped charged object. 31. Why can there never be excess charges inside a charged conductive sphere? 32. Describe a simple experiment to demonstrate that there are no excess charges on the inside of a hollow charged sphere. 33. Identify a technology that uses the principle that electric charges", " accumulate at the point of an irregularly shaped object. Describe how the technology applies this principle. Consolidate Your Understanding Create your own summary of electric field theory by answering the questions below. If you want to use a graphic organizer, refer to Student Reference 3: Using Graphic Organizers. Use the Key Terms and Concepts listed on page 577 and the Learning Outcomes on page 542. 1. Create a flowchart to describe the differences between electric fields, electric potential energy, and electric potential, using non-uniform and uniform electric fields. 2. Write a paragraph comparing the electric fields around various objects and surfaces. Include diagrams in your comparisons. Share your report with a classmate. Think About It Review your answers to the Think About It questions on page 543. How would you answer each question now? (a) the magnitude of the test charge were doubled? (b) the magnitude of the charge producing the e TEST field were doubled? (c) the sign of the charge producing the field were changed? To check your understanding of concepts presented in Chapter 11, follow the eTest links at www.pearsoned.ca/school/physicssource. Chapter 11 Electric field theory describes electrical phenomena. 579 12-PearsonPhys30-Chap12 7/24/08 3:35 PM Page 580 Properties of electric and magnetic fields apply in nature and technology. Figure 12.1 Aurora borealis or northern lights The spectacular aurora borealis paints the night sky with shimmering colours in northern latitudes (Figure 12.1). Frequently seen above 60\u00b0 north, its scientific name translates from Latin into \u201cdawn of the north.\u201d In southern latitudes, where it is seen mainly above 60\u00b0 south, it is called the aurora australis \u2014 \u201cdawn of the south.\u201d Many ancient civilizations created stories to explain these dancing lights in the sky. Some Inuit peoples of northern Canada believed that the sky was a hard dome that arched over Earth. Spirits could pass through a hole in the dome to the heavens, where they would light torches to guide new arrivals. Other Aboriginal traditions spoke of the creator of Earth travelling to the north when he finished his task of creation. There he remained, building large fires to remind his people that he still thought of them. The northern lights were reflections of these fires. What are the auroras and what causes them? Why can they be observed only in the far northern or southern latitudes? Is there a relationship between the auroras and surface activity on the Sun, called solar flares? Are they related to other physical phenomena observed on Earth? Finally, how can an understanding of the science of the auroras aid in the development of new technologies? Your studies in this chapter will help answer these questions. C H A P T E R 12 Key Concepts In this chapter, you will learn about: magnetic fields moving charges in magnetic and electric fields electromagnetic induction Learning Outcomes When you have completed this chapter, you will be able to: Knowledge define electric current as the amount of charge passing a reference point per unit of time describe magnetic interactions in terms of forces and fields compare gravitational, electric, and magnetic fields describe how the work of Oersted and Faraday led to the theory relating electricity to magnetism describe a moving charge as the source of a magnetic field and predict the field\u2019s orientation explain how uniform magnetic and electric fields affect a moving charge describe and explain the interaction between a magnetic field and a moving charge and a conductor explain, quantitatively, the effect of an external magnetic field on a current-carrying conductor describe the effects of moving a conductor in an external magnetic field in terms of moving charges Science, Technology, and Society explain that concepts, models, and theories are often used in interpreting, explaining, and predicting observations explain that technology provides solutions to practical problems explain that scientific knowledge may lead to the development of new technologies and vice versa 580 Unit VI 12-PearsonPhys30-Chap12 7/24/08 3:35 PM Page 581 12-1 QuickLab 12-1 QuickLab Magnetic Fields in a Bottle Problem What is the shape of a magnetic field? Materials 50 mL of iron filings 450 mL of light cooking oil 1 clear plastic 591-mL pop bottle string 1 cylindrical cow magnet (must be able to fit in the bottle) tape Procedure 1 Pour 50 mL of iron filings into the bottle. 2 Pour cooking oil into the bottle until it is about three-quarters full. 3 Replace the cap on the bottle securely and shake the bottle several times so that the iron filings disperse throughout the oil. Remove the cap. 4 Attach the string to one end of the cow magnet and insert the magnet in the bottle. Make sure the magnet is suspended vertically in the middle of the bottle. Tape the other end of the string to the top of the bottle. 5 Replace the cap on the bottle and place the bottle on a table to allow the mixture to settle. Observe the pattern produced by the iron filings.", " Questions 1. In your notebook, draw a diagram of the pattern created by the iron filings. 2. 3. Is the pattern created by the iron filings one-, two-, or three-dimensional? Explain your answer. Identify where the density of the iron filings is the greatest and the least. Explain why the filings are distributed this way. 4. From the pattern of the iron filings, is it possible to determine the strength and the direction of the magnetic influence around the magnet? Explain your answer. Think About It 1. Describe a probable cause of the pattern of the iron filings in 12-1 QuickLab. 2. What types of substances produce this influence? 3. What types of objects are affected by this influence? Discuss and compare your answers in a small group and record them for later reference. As you complete each section of this chapter, review your answers to these questions. Note any changes to your ideas. Chapter 12 Properties of electric and magnetic fields apply in nature and technology. 581 12-PearsonPhys30-Chap12 7/24/08 3:35 PM Page 582 12.1 Magnetic Forces and Fields An ancient Greek legend from about 800 BCE describes how the shepherd Magnes, while tending his flock, noticed that pieces of a certain type of rock were attracted to the nails on his shoes and to his metal staff (Figure 12.2). This phenomenon was called magnetism and, as time passed, further studies of the behaviour of this rock revealed several curious effects. For example, a piece of this rock could either attract or repel another similar piece (Figure 12.3). This effect seemed to result from two different magnetic effects, so investigators thought that there must be two different types of \u201cmagnetic ends,\u201d or poles, on the rock. This observation led to the law of magnetism, which states: Like magnetic poles repel and unlike poles attract each other. (a) (b) Figure 12.3 A piece of magnetic rock, held near one end of a similar piece of magnetic rock, would attract at one end (a) and repel at the other end (b). In 1269, Pierre de Maricourt was mapping the position of a magnetized needle placed at various positions on the surface of a spherical piece of this rock. He observed that the directions of the needle formed a pattern that encircled the rock, like meridian lines, and converged at two points on opposite ends of the rock. When this rock was then suspended by a string, the two converging points tended to align along Earth\u2019s north\u2013south axis. This property of the rock earned it the name \u201clodestone\u201d or \u201cleading stone.\u201d Maricourt called the end pointing northward the north-seeking or north pole and the end pointing southward the south-seeking or south pole. All magnets have both poles. Lodestone, which contains the mineral magnetite (Fe3O4), was later used in the development of compass technology. Figure 12.2 The magnetic effects of certain materials were observed by ancient Greeks as early as 800 BCE info BIT Magnetic poles always exist in pairs. In the 1930s, Paul Dirac (1902\u20131984) suggested the existence of a particle called a magnetic monopole. To date, all experiments to discover this onepoled particle have failed, but these particles are still under experimental investigation. 582 Unit VI Forces and Fields 12-PearsonPhys30-Chap12 7/24/08 3:35 PM Page 583 Concept Check Figure 12.4 A U-magnet, a circular magnet, and a bar magnet Copy the picture of each magnet in Figure 12.4 into your notebook. Since each magnet must have two poles, label the possible positions of the north and south poles of each magnet. The next big advance in knowledge about magnetism came from the work of William Gilbert. In his book De Magnete, published in 1600, he not only reviewed and criticized past explanations of magnetism but he also presented many important new hypotheses. He compared the orientation of magnetized needles on the surface of a spherical piece of lodestone with the north\u2013south orientation of a compass needle at various locations on Earth\u2019s surface. From this study, he proposed that Earth itself is a lodestone with north and south magnetic poles. Concept Check The north pole of a magnetic compass needle points toward Earth\u2019s magnetic north. What can you conclude about this point on Earth? Gilbert was also intrigued by the forces that magnets could exert on other magnetic objects. If you suspend a magnet on a string and bring another magnet close to one of its poles, the suspended magnet will rotate, even though there is no visible contact between the two magnets. Magnets appeared to have the ability to exert forces that seemed to originate from the magnetic poles, and they could affect another magnetic object even without contact. The ancient Greeks called this effect \u201caction at a distance.\u201d Recall from chapter 11 that they", " used the same terminology to describe the effects of electric charges. In attempting to explain the action at a distance caused by a magnet, Gilbert suggested that an invisible \u201corb of virtue\u201d surrounds a magnet and extends in all directions around it. Other magnetic substances react to a force created by this orb of virtue and move or rotate in response. His orbs of virtue were the beginnings of the idea of \u201cfields\u201d that would revolutionize physics. Chapter 12 Properties of electric and magnetic fields apply in nature and technology. 583 12-PearsonPhys30-Chap12 7/24/08 3:35 PM Page 584 magnetic field: a threedimensional region of influence surrounding a magnet, in which other magnets are affected by magnetic forces Michael Faraday (1791\u20131867) further developed this concept. He defined a magnetic field as a three-dimensional region of magnetic influence surrounding a magnet, in which other magnets are affected by magnetic forces. The direction of the magnetic field at a given location is defined as the direction in which the north pole of the compass needle points at that location. Some materials, such as iron, act like magnets while in a magnetic field. 12-2 QuickLab 12-2 QuickLab Observing Magnetic Fields Problem How can the magnitude and direction of magnetic fields be observed and analyzed? Questions 1. Describe the cause of the pattern produced by the iron filings. 2. 3. 4. Is the pattern created by the iron filings one-, two-, or three-dimensional? Explain. Identify where the density of the iron filings is the greatest and the least. Explain why the filings are distributed this way. Is it possible to determine the strength and direction of the magnetic field surrounding the magnet from the pattern of the iron filings alone? Explain your answer. 5. From your investigation of the effect of a magnetic field on a compass, what appears to be the direction of the magnetic field around a magnet? Materials 1 bar magnet 1 sheet of paper (216 mm 279 mm) 25 mL of iron filings 1 compass Procedure 1 Lay the bar magnet on a table and place the paper over the magnet. Trace the shape of the magnet on the paper and label the poles. 2 Carefully sprinkle the iron filings onto the surface of the paper. 3 Tap the paper lightly to reinforce the alignment of the iron filings on the sheet. Draw the pattern of the iron filings around the magnet. 4 Clean the iron filings from the paper and replace the paper over the magnet. 5 Place the compass at several positions around the magnet and trace the direction of the compass needle. 584 Unit VI Forces and Fields 12-PearsonPhys30-Chap12 7/24/08 3:35 PM Page 585 Magnetic Fields The magnetic field surrounding a magnet is represented by the symbol B and is measured in teslas (T). A typical bar magnet in the classroom can have a magnetic field of approximately 1 102 T, whereas Earth\u2019s magnetic field is about 5 105 T. The magnetic field is a vector quantity, so it is represented by a vector arrow. In diagrams, the length of the vector arrow represents the magnitude of the field, and the direction of the arrow represents the direction of the field at a point. You can also use compasses to show the direction of the magnetic field at any position surrounding a magnet, as illustrated in Figure 12.5. Figure 12.5 shows that, in general, this direction is from the north to the south pole of the magnet. info BIT Magnetic field lines run parallel to Earth\u2019s surface only at the equator. As they reach the magnetic poles, they gradually dip toward the surface. At the poles, the magnetic field lines point perpendicular to Earth\u2019s surface. Navigators in the far north or south must be aware that the magnetic compasses may be of limited use in those areas Figure 12.5 The direction of a magnetic field is the direction of the force on the north pole of a compass placed in the field. To represent the entire magnetic field surrounding a magnet, it would be necessary to draw arrows at an infinite number of points around the magnet. This is impractical. Instead, you can draw a few magnetic field lines with a single arrow head indicating the direction of the magnetic field. To find the field direction at a given point, move the arrow head along the field line through that point so that it keeps pointing in the direction of the tangent to the field line. The field lines in Figure 12.6 are a map of the magnetic field with the following features: \u2022 Outside a magnet, the magnetic field lines point away from the north pole of a magnet and toward the south pole. \u2022 The closeness of the lines represents the magnitude of the magnetic field. e LAB For a probeware activity where you use a magnetic field sensor to determine the relationship between the distance from a magnet and the intensity of the field, go to www.pearsoned.ca/ school/physicssource. S N S N (a) (b) Figure 12", ".6 (a) The pattern of iron filings surrounding a bar magnet outlines the magnetic field. (b) Magnetic field lines, representing the direction and magnitude of the magnetic field, can replace the iron filings. The number of magnetic field lines that exit a magnetic material is equal to the number of magnetic field lines that enter the magnetic material, forming closed loops. Chapter 12 Properties of electric and magnetic fields apply in nature and technology. 585 12-PearsonPhys30-Chap12 7/24/08 3:35 PM Page 586 PHYSICS INSIGHT Before the adoption of SI units, magnetic fields were sometimes measured in a CGS unit called the gauss (G). You might see this unit in some older books. 1 T 104 G. Table 12.1 shows some examples of magnetic field strengths. Table 12.1 Magnetic Field Strengths Physical system Magnetic field (T) Earth Bar magnet Sunspots High field magnetic resonance imaging device (MRI) Strongest humanmade magnetic field Magnetar (magnetic neutron star) Concept Check 5 105 1 102 1 101 15 40 1 1011 Figure 12.7 shows the patterns produced by iron filings that are influenced by the magnetic fields of one or two magnets. Sketch the magnetic field lines in each case. S N N N N S N S Figure 12.7 Concept Check List at least two similarities and two differences between gravitational, electric, and magnetic fields. 586 Unit VI Forces and Fields 12-PearsonPhys30-Chap12 7/28/08 10:04 AM Page 587 Cause of Magnetism The force of magnetic repulsion between like poles of magnets is the same force that causes the almost frictionless ride of the Maglev (magnetically levitated) train (Figure 12.8). What is the source of this \u201cmagnetic levitation\u201d on the train? info BIT Oersted was among the first to recognize the talent of the writer Hans Christian Andersen and encouraged him when he began writing his now famous fairy tales. (a) (b) Figure 12.8 (a) The force of magnetic repulsion between like poles can cause one magnet to levitate over another. (b) The Maglev train, developed in Japan, floats several centimetres above the guideway, providing a smooth and almost frictionless ride. Experiments by early investigators revealed many facts about the magnetic fields surrounding magnets and their effects on magnetic objects. However, the actual cause of magnetism eluded scientists until 1820. While demonstrating to students that the current passing through a wire produces heat, Danish professor Hans Christian Oersted (1777\u20131851) noticed that the needle of a nearby compass deflected each time the circuit was switched on. This experiment led Oersted to the important conclusion that there is a relationship between electricity and magnetism, at a time when electricity and magnetism were considered separate phenomena. He proved that electric current was a cause of magnetism. Following his initial observations, it was later shown that if electric current was in a straight line, the magnetic field formed a circular pattern (Figure 12.9(a)), and if the electric current was circular, the magnetic field was straight within the coil (Figure 12.9(b)). (a) battery N (b) electron flow electron flow S Figure 12.9 (a) A current passing through a straight conducting wire produces a magnetic field, represented by concentric red circular lines around the wire. (b) A current passing through a coil produces a magnetic field, represented by red circular lines, with poles similar to those of a bar magnet. Chapter 12 Properties of electric and magnetic fields apply in nature and technology. 587 12-PearsonPhys30-Chap12 7/24/08 3:35 PM Page 588 PHYSICS INSIGHT The observation of a magnetic field produced by a moving charge depends on the frame of reference of the observer. If you are stationary and the charge moves past you, you observe a magnetic field. However, if you are moving along with the charge, the charge is stationary relative to you, so you do not observe a magnetic field. Left-hand Rules for Magnetic Fields A useful left-hand rule to determine the direction of the magnetic field is the wire-grasp rule described in Figure 12.10. To determine the direction of the magnetic field produced by a moving charge, use the left-hand wire-grasp rule if the moving charge is negative. (If the moving charge is positive, then use the right-hand wire-grasp rule.) direction of magnetic field lines (a) e direction of electron flow left hand conductor magnetic field lines (b) core e direction of electron flow magnetic field line e direction of magnetic field Figure 12.10 Left-hand rule for direction of a magnetic field due to moving charges: (a) If the conducting wire is straight, then the thumb indicates the direction of the straight current and the cupped fingers indicate the direction of the circular magnetic field. (b) If", " the current is in a coil of conducting wire, the cupped fingers indicate the circular current and the straight thumb indicates the direction of the straight magnetic field within the coil Using the Wire-grasp Rule 1. Sketch the following diagrams into your notebook. (a) e Indicate the direction of the magnetic field lines and the direction of current in the wire, as required. (b) N S Electromagnets As shown in Figure 12.9(b), current in a circular loop or coil of wire produces a magnetic field like that of a bar magnet. An electromagnet uses a current-carrying coil of wire to generate a magnetic field that is easy to switch on and off. The strength of an electromagnet can be influenced by: \u2022 increasing the current through the wire \u2022 increasing the number of loops in the coil \u2022 increasing the size of the loops in the coil \u2022 changing the core of the coil Powerful electromagnets have many industrial uses, such as lifting steel parts, machinery, or scrap iron. Electromagnets are widely used to remotely operate switches or valves. Often, a valve is activated by a metal rod that is drawn into the core of the electromagnet when current electromagnet: a magnet having its magnetic field produced by electric current flowing through a coil of wire 588 Unit VI Forces and Fields 12-PearsonPhys30-Chap12 7/24/08 3:35 PM Page 589 flows through the coil. Such mechanisms, called solenoids, are common in washing machines, dishwashers, furnaces, and industrial machinery. Figure 12.11 shows two applications of electromagnets. solenoid: an electromagnet that operates a mechanical device (a) (b) Figure 12.11 (a) A lifting magnet (b) An appliance solenoid Domain Theory and Magnetization In some atoms, the configuration of the electrons is such that their movement generates a tiny magnetic field. In ferromagnetic materials, such as iron, nickel, and cobalt, the magnetic fields of adjacent atoms can align to reinforce each other, forming small regions, or domains, with intense magnetic fields. Domains generally range from 0.001 mm to 1 mm across, and may contain billions of atoms. The orientations of the magnetic fields of the various domains are normally random, so their magnetic fields largely balance each other, leaving the material with little or no overall magnetization. However, the size of a domain and the direction of its magnetic field are relatively easy to change. An external magnetic field can cause the domains to align, thus magnetizing the material (Figure 12.12). The small black arrows in Figure 12.12 indicate the orientation of the magnetic field of an individual domain. S N (a) (b) Figure 12.12 (a) When the magnetic fields of atoms in a region line up, they create a magnetic domain in the substance. (b) Aligning the domains produces a magnet. A typical ferromagnetic object has vastly more domains than the diagrams can show. If you hang an iron nail by a string and bring a magnet close to the nail, the nail will rotate toward the magnet, even before they touch. The nail is not a magnet with distinct poles, yet a magnetic attraction exists between it and the magnet. When the magnet is close to the nail, the domains in the nail that are oriented for attraction to the magnet increase in size while the other domains shrink. When the magnet is moved away again, the domains in the nail tend to return to random info BIT Geophysicists theorize that circulating currents of ions in the molten core of Earth produce its magnetic field. ferromagnetic: having magnetic properties like those of iron domain: a region of a material in which the magnetic fields of most of the atoms are aligned e WEB All magnetic substances can be classified as one of the following: \u2022 ferromagnetic \u2022 antiferromagnetic \u2022 ferrimagnetic \u2022 paramagnetic \u2022 diamagnetic Find out what distinguishes one type of magnetic substance from another. Begin your search at www.pearsoned.ca/ school/physicssource. Chapter 12 Properties of electric and magnetic fields apply in nature and technology. 589 12-PearsonPhys30-Chap12 7/24/08 3:35 PM Page 590 orientations and the nail loses most of its magnetization. This example illustrates induced magnetization. The nail will be much more strongly magnetized if it is stroked with a pole of the magnet. The magnetic fields of many of the domains in the nail will align along the direction of motion of the magnet. This magnetization is strong enough that the nail will remain somewhat magnetized when the magnet is removed. Concept Check A filing cabinet has been in one position for a long time. It is made of ferromagnetic material, so it can become a permanent magnet. If you hold a compass near the top of the filing cabinet, the compass needle points toward the filing cabinet. If you hold the compass near the bottom of", " the filing cabinet, the opposite end of the compass points toward the cabinet. Has the cabinet been magnetized by Earth\u2019s magnetic field? Or has the cabinet become magnetized by the magnetic compass? Explain your answer. Magnetism in Nature The effects of magnetism have been known since early civilizations, but the causes of magnetic behaviour are only now being revealed. A modern understanding of magnetic phenomena began with the development of field theory to replace \u201caction at a distance.\u201d The symmetry of nature enabled scientists to use the same field theory to describe the gravitational field surrounding any mass, the electric field surrounding any charge, and the magnetic field surrounding any magnet. Oersted\u2019s investigations, which revealed a relationship between electricity and magnetism, ultimately led to the domain theory to explain a cause of magnetism. As scientists probed deeper into the mysteries of magnetism, many more answers were found. However, the tremendous significance of magnetism has only recently been understood in explaining phenomena and producing technological applications. In the field of biology, for example, researchers have found that certain organisms have ferromagnetic crystals consisting of magnetite in their bodies. Some bacteria use these magnetite crystals to help orient themselves within Earth\u2019s magnetic field. Bees and pigeons have magnetite crystals within their brains to help with navigation. The human brain also has these magnetite crystals, but their function is not clear. It is known that an external magnetic field can disrupt the neural activity in the parietal lobe on one side of the human brain. 590 Unit VI Forces and Fields 12-PearsonPhys30-Chap12 7/24/08 3:35 PM Page 591 Understanding magnetism has also led to important technological advancements. These advancements range from simple applications, such as refrigerator magnets, magnetic stripes on cards, and magnetic audiocassette or VCR tapes, to more complicated applications involving magnetic levitation, such as the Maglev train and magnetic resonance imaging (MRI) machines used as a diagnostic tool in health care. Although much has been achieved, there are still many secrets of magnetism to uncover. THEN, NOW, AND FUTURE Earth\u2019s Magnetic Field William Gilbert\u2019s \u201cTerrella\u201d experiment in the 1500s compared the magnetic field of Earth to that of a bar magnet. From that time, Earth has been considered to be a huge magnet, with similar magnetic properties to a much smaller, ordinary magnet. This observation was successful in explaining many phenomena. However, care must be taken in comparing the causes of magnetic behaviour in Earth and in a bar magnet. If the cause of magnetism in substances is the motion of charges, scientists are not quite convinced that the motion of charges within Earth\u2019s molten core is responsible for Earth\u2019s magnetism. They know that Earth\u2019s molten core is simply too hot for atoms to remain aligned and exhibit any magnetic properties. Other probable causes of Earth\u2019s magnetic field could be convection currents rising to the cooler surface of Earth, or the motion of charges in the upper ionosphere. The most acceptable and probable cause, though, is the motion of charges in the molten part of Earth, just beneath the crust (Figure 12.13). Whatever the cause of Earth\u2019s magnetic behaviour, it is known that the magnetic field of Earth is not stable. Molten rock within the interior of Earth has no magnetic properties. However, when molten rock rises to the surface, it cools and solidifies, and its domains orient themselves in line with Earth\u2019s magnetic field at the time. When samples of rock from different strata formed throughout geological times are tested, evidence shows that there are times when not only the magnitude of Earth\u2019s magnetic field changed, but also its direction. In the past five million years, more than 20 reversals have occurred, the last one about 780 000 years ago. Coincidentally, modern humans emerged during this time period. One possible effect of a zero magnetic field, during a reversal, would be an increase in the cosmic ray intensity at Earth\u2019s surface. Normally, the magnetic field shields Earth from harmful radiation from space. Fossil evidence indicates that periods of no protective magnetic field have been effective in changing life forms. Evidence that these types of changes could have occurred also comes from heredity studies of fruit flies when exposed to X rays. We cannot know precisely when the next reversal will occur. However, evidence from recent measurements indicates a decrease in the magnitude of Earth\u2019s magnetic field of about 5% in the last 100 years. Based on this evidence, Figure 12.13 This computer model of Earth shows the molten outer core surrounding the inner core (the small circle). The right side shows the molten currents. The left side shows the magnetic field lines that extend outward through the rest of Earth\u2019s interior. another reversal of Earth\u2019s magnetic field may occur within the next 2000 years. Questions 1. Can the motion of charges in Earth\u2019s core create domains?", " Explain your answer. 2. What is the most probable cause of Earth\u2019s magnetic behaviour? 3. What evidence is there on Earth that its magnetic field is not stable? Chapter 12 Properties of electric and magnetic fields apply in nature and technology. 591 12-PearsonPhys30-Chap12 7/24/08 3:35 PM Page 592 12.1 Check and Reflect 12.1 Check and Reflect Knowledge 1. What is the law of magnetism? 2. Explain your answers to the following: (a) Does every magnet have a north and a south pole? 10. List at least two differences and two similarities between (a) gravitational and electric fields (b) gravitational and magnetic fields (c) electric and magnetic fields (b) Does every charged object have positive and negative charges? 11. Using the domain theory, explain the following observations: 3. How did William Gilbert determine that Earth was a magnet? 4. What is the most probable cause of magnetism in (a) a bar magnet? (b) Earth? 5. What accidental discovery did Oersted make? 6. What is the shape of the magnetic field (a) around a straight current-carrying conductor? (b) within a coil of conducting wire carrying a current? (a) A magnet attracts an unmagnetized ferromagnetic material. (b) Stroking a nail with a magnet magnetizes the nail. (c) A metal table leg affects a compass. 12. Why does dropping or heating a bar magnet decrease its magnetic properties? 13. Consider a bar magnet and Earth, as shown below. Describe the similarities and the differences of their magnetic fields. S N Earth Applications 7. What would happen to a magnet if you broke it into two pieces? Extensions 8. A negatively charged sphere is approaching you. Describe the magnetic field surrounding the sphere and its direction. What would happen if the sphere were positively charged? 9. A spinning top is charged negatively and is spinning clockwise, as observed from above. Describe the magnetic field created by the spinning top and its direction. 14. Why is it difficult to get an accurate bearing with a magnetic compass near the poles? 15. Do magnetic field lines always run parallel to the surface of Earth? Explain your answer. 16. If a current-carrying wire is bent into a loop, why is the magnetic field stronger inside the loop than outside? e TEST To check your understanding of magnetic forces and fields, follow the eTest links at www.pearsoned.ca/school/physicssource. 592 Unit VI Forces and Fields 12-PearsonPhys30-Chap12 7/24/08 3:35 PM Page 593 12.2 Moving Charges and Magnetic Fields Near the end of the 1800s, researchers were fascinated by a new technology called the cathode-ray tube (CRT), shown in Figure 12.14. It consisted of a glass tube from which air had been evacuated, and it had a positive plate (anode) at one end and a negative plate (cathode) at the other end. These new tubes used electric fields to accelerate a beam called a cathode ray through a large potential difference. The beam would \u201clight up\u201d the fluorescent screen at the end of the tube. Scientists were unsure whether this beam was a type of electromagnetic radiation (similar to light), a neutral particle, or a charged particle. They initially called it a cathode ray because it appeared to originate from the cathode plate. This technology not only enabled the discovery of the electron in 1897 (see section 15.1), but also led to the later development of many other technologies, including television. Until recently, the image in most TVs was produced by an electron beam striking a fluorescent screen in a CRT. The Motor Effect Deflecting charged particles involves an interaction of two magnetic fields. A charged particle in uniform motion produces a circular magnetic field around it (the wire-grasp rule). Now suppose this charged particle enters an external magnetic field, produced between the faces of two opposite magnetic poles. The interaction of the circular magnetic field of the charge and the external magnetic field produces a magnetic force that acts on the particle to deflect it, as shown in Figure 12.15. This magnetic force is also called the motor effect force (F m) because it causes the rotation of a loop of current-carrying wire. This rotation is fundamental in the operation of an electric motor. Figure 12.14 The image on the fluorescent screen at the left end of this cathode-ray tube shows that the rays originate from the negative terminal at the right end. motor effect force: the deflecting force acting on a charged particle moving in a magnetic field Figure 12.15 (a) The cathode ray accelerates in a straight line when it is only influenced by the electric field produced between the cathode and anode plates in a vacuum tube connected to a high-voltage source. (b) A cathode ray", " will deflect as shown when it is also under the influence of an external magnetic field. (a) (b) In Figure 12.16, the straight, horizontal lines represent the external magnetic field of the magnetic poles, and the dashed lines represent the magnetic field surrounding the moving charge (using the left-hand rule). In the \u201creplacement magnet\u201d method of illustration, tiny magnets are drawn along the field lines to reinforce the idea of the direction and the effects of the interaction of the two magnetic fields. The represents Chapter 12 Properties of electric and magnetic fields apply in nature and technology. 593 12-PearsonPhys30-Chap12 7/24/08 3:35 PM Page 594 PHYSICS INSIGHT represents a direction into the page, like the fletching of an arrow receding from you. \u2022 represents a direction out of the page, like the point of an arrow approaching you. In Figure 12.16 represents negative charge movement into the page. negative charge moving into the page. In Figure 12.16, below the moving charge, the external magnetic field and the magnetic field surrounding the charge are in the same direction. Above the moving charge, the two magnetic fields are in opposite directions. Fm causing deflection Fm Fm S N Figure 12.16 The combined magnetic forces due to the two magnetic fields cause the moving charge to deflect (F perpendicular to the direction of the external magnetic field. m) in a direction perpendicular to its direction of motion and Since the external magnetic field is fixed, the combined effect of m) on the the two magnetic fields produces a net magnetic force (F moving charge. As a result, the moving charge deflects upward (toward the top of the page). The deflecting force is always perpendicular to the direction of both the external magnetic field and the motion of the moving charge, as shown in Figure 12.16. This property distinguishes a magnetic field from electric or gravitational fields. Since the direction of the electric force or gravitational force can be parallel to their respective fields, these fields can be used to change the speed of a charged particle. The magnetic force, on the other hand, is always perpendicular to the velocity of the charged particle. A magnetic force can never do any work on a charged particle, nor can it change the speed or kinetic energy of a charged particle. Since force is not in the direction of the displacement, then there can be no work done on the object. Only the direction of the charged particle\u2019s path may be changed. Left-hand Rule for Deflection Consider a negatively charged particle travelling perpendicular to an external magnetic field. When it enters the region of a uniform magnetic field, it is deflected in a direction perpendicular to both the original direction of charge movement and the direction of the external magnetic field. A useful hand rule to determine the direction of deflection is the left-hand rule shown in Figure 12.17: 594 Unit VI Forces and Fields 12-PearsonPhys30-Chap12 7/24/08 3:35 PM Page 595 \u2022 The thumb indicates the direction of the initial charge movement. \u2022 The extended fingers indicate the direction of the external magnetic field, from north to south. \u2022 The palm faces in the direction of the magnetic force. If the moving charge is positive, use the righthand rule, with the thumb, fingers, and palm indicating the directions of the same quantities as in the left-hand rule. S magnetic field B e N Fm v Figure 12.17 How to use the left-hand rule to determine the deflection of a charged particle Using the Left-hand Rule for Deflection 1. In your notebook, sketch the direction of the unknown variable in each situation. (a) external magnetic field (b) negative charge Fm e motion B Fm 12-3 Inquiry Lab 12-3 Inquiry Lab Using Hand Rules with a Cathode-ray Tube \u2014 Demonstration (c) positive charge motion B Fm Required Skills Initiating and Planning Performing and Recording Analyzing and Interpreting Communication and Teamwork Question Do the hand rules predict the deflection of a cathode beam in an external magnetic field? Materials and Equipment 1 cathode-ray tube 1 high-voltage source 1 strong bar magnet S N cathode-ray tube high-voltage source Figure 12.18 CAUTION! High voltage. Be very careful around electrical equipment to avoid shocks. Chapter 12 Properties of electric and magnetic fields apply in nature and technology. 595 12-PearsonPhys30-Chap12 7/24/08 3:35 PM Page 596 Procedure Analysis 1 Connect the cathode-ray tube to the high-voltage 1. What types of particles would be attracted from the source, as shown in Figure 12.18. Identify the cathode and anode of the CRT. 2 Turn on the current supply. Observe the path of the cathode rays that are produced. 3 Carefully hold the north pole of a bar magnet near one side of the centre", " of the cathode tube, in the horizontal plane. Note the direction in which the cathode rays are deflected. 4 Repeat the procedure in step 3 with the south pole of the bar magnet. 5 Repeat steps 3 and 4 by holding the magnet on the other side of the cathode tube, in the horizontal plane and then in the vertical plane. Observe the direction of the deflection of the beam in each case. cathode to the anode of the CRT? Does the hand rule applicable for these particles predict all of the deflections that you observed? 2. What types of particles would be attracted from the anode to the cathode of the CRT? Does the hand rule for these particles predict any of the deflections that you observed? 3. Explain how the observed deflections show that cathode rays consist of charged particles. 4. Can you use the hand rules to determine the type of charge carried by cathode rays? Explain why or why not. Concept Check Compare the magnetic force of an external magnetic field on a moving charged particle with: \u2022 the gravitational force of Earth on the mass of the charged particle \u2022 the electric force due to another nearby charged particle Charged Particle Motion in a Magnetic Field The direction of the initial motion of a charged particle in an external magnetic field determines how the charged particle will deflect. Figure 12.19 shows what can happen to a charged particle as it enters an external magnetic field: (a) If the initial motion of the charged particle is parallel to the external magnetic field, then there is no effect. (b) If the initial motion of the charged particle is perpendicular to the external magnetic field, the charge is deflected in a circular arc. (c) If the initial motion of the charged particle is at an angle to an external magnetic field, the charge deflects in a circular motion that will form a helical path. e SIM Explore the motion of a charged particle in a uniform magnetic field. Follow the eSim links at www.pearsoned.ca/school/ physicssource. 596 Unit VI Forces and Fields 12-PearsonPhys30-Chap12 7/24/08 3:35 PM Page 597 (a) (c) v v v (b) B B (d) B Fm Fm Fm B Figure 12.19 (a) When the charged particle\u2019s velocity is parallel to the external magnetic field (B), the charged particle\u2019s path is a straight line. (b) The charged particle\u2019s motion is perpendicular to the magnetic field, so the particle is deflected in a circular arc. (c) The charged particle\u2019s motion is at an angle to the magnetic field, so the particle follows a helical path. (d) This side view from the left shows the magnetic force acting as the centripetal force that causes the charge to follow a circular path. Oppositely charged particles deflect in opposite directions in a magnetic field (Figure 12.20). If the magnitude of the external magnetic field is large enough, the field can cause circular motion that remains contained in the magnetic field. In this circular motion, the centripetal force is the magnetic force. Magnetic deflection of charged particles is the underlying principle for useful powerful analytical and research tools such as mass spectrometers and particle accelerators. Unit VIII presents these devices and their applications in science, medicine, and industry. Auroras Tremendous expulsions of magnetic energy from the solar atmosphere, called solar flares, expel streams of charged particles at speeds around 10% of the speed of light (Figure 12.21). When some of these particles strike Earth\u2019s magnetic field, they are deflected by the magnetic force and spiral in a helical path along Earth\u2019s magnetic field lines. These particles enter the atmosphere as they approach Earth\u2019s magnetic poles, and collide with air molecules. These collisions excite the atoms of the air molecules, in a process that will be described in Chapter 15, causing them to emit visible light that we see as the aurora. The process repeats because Earth\u2019s nonuniform magnetic field produces a magnetic force component that causes the charged particles to reverse their direction of motion, travelling to Earth\u2019s opposite pole. The same auroral effect is produced at this pole, and the process continues to repeat as the charged particles oscillate back and forth between the poles, trapped in a type of \u201cmagnetic bottle\u201d called the Van Allen belt. N S upward deflection downward deflection Figure 12.20 A magnetic field deflects moving oppositely charged particles in opposite directions, as shown. e WEB Research the formation of the Van Allen belts. Consider the shape of the Van Allen belt on the side of Earth facing the Sun and on the side of Earth away from the Sun. What is the cause of this difference in shape? Begin your search at www.pearsoned.ca/ school/physicssource. Chapter 12 Properties", " of electric and magnetic fields apply in nature and technology. 597 12-PearsonPhys30-Chap12 7/24/08 3:35 PM Page 598 Figure 12.21 This composite image shows the cause of the aurora borealis. Streams of high-energy charged particles erupt from the Sun (far left). They are deflected by Earth\u2019s magnetic field toward the poles, creating the bright ring shown in the satellite image of Earth (centre). There they interact with air molecules in the atmosphere to produce the aurora (far right). Calculating the Magnetic Force By studying the different types of deflections, scientists can also explain the complex deflection of charged particles entering a magnetic field at an angle, such as the particles that cause the auroras. The magnitude of the deflecting force (F m ) depends on all of the following: \u2022 the magnitude of the moving charge (q) \u2022 the magnitude of the perpendicular velocity component (v) ) \u2022 the magnitude of the external magnetic field (B The magnitude of the deflecting force can be calculated using this equation: F m qvB where q is the magnitude of moving charge in coulombs (C); v is the component of the speed perpendicular to the magnetic field in metres per second (m/s); and B is the magnitude of the external magnetic field in teslas (T). Example 12.1 describes how to calculate the magnetic force on a charge moving perpendicular to an external magnetic field. When the velocity of the charge is not perpendicular to the magnetic field, you can use trigonometry to find the perpendicular component: v v sin where is the angle between the charge\u2019s velocity, v, and the magnetic field, B. v v v 598 Unit VI Forces and Fields 12-PearsonPhys30-Chap12 7/24/08 3:35 PM Page 599 Example 12.1 An electron is travelling at 3.20 105 m/s perpendicular to an external magnetic field of magnitude 2.20 101 T (Figure 12.22). Determine the magnetic force acting on the electron. S N S N e Figure 12.22 Given q charge on 1 electron 1.60 1019 C 2.20 101 T B v 3.20 105 m/s Required m) magnetic force (F Analysis and Solution Determine the magnitude of the magnetic deflecting force: F m qvB (1.60 1019 C)(3.20 105 m )(2.20 101 T) s 1.13 1014 N Since the charge is negative, use the left-hand rule for deflection to determine the direction of the magnetic force. \u2022 Thumb points in the direction of the charged particle\u2019s movement, into the page. \u2022 Extended fingers point in the direction of the external magnetic field, to the right of the page (north to south). \u2022 Palm points in the direction of the magnetic deflecting force, toward the top of the page. Practice Problems 1. A proton with a charge of 1.60 1019 C is travelling with a speed of 3.50 104 m/s perpendicularly through an external magnetic field of magnitude 4.20 104 T. Determine the magnitude of the magnetic deflecting force on the proton. 2. An ion with a charge of 3.20 1019 C and a speed of 2.30 105 m/s enters an external magnetic field of 2.20 101 T, at an angle of 30, as shown in the figure below. Calculate the magnitude of the magnetic deflecting force on the ion. 30\u00b0 S N 3. A negatively charged sphere travels from west to east along Earth\u2019s surface at the equator. What is the direction of the magnetic deflecting force on the sphere? Answers 1. 2.35 1018 N 2. 8.10 1015 N 3. Downward toward Earth\u2019s surface Paraphrase The magnetic force is 1.13 1014 N [upward] (toward the top of the page). Often, a charged particle may be influenced by a combination of two fields, such as a magnetic field and a gravitational field, or a magnetic field and an electric field. \u201cCrossed-field\u201d devices are technologies that use both magnetic and electric fields. An example is the magnetron, which produces microwaves in microwave ovens. Chapter 12 Properties of electric and magnetic fields apply in nature and technology. 599 12-PearsonPhys30-Chap12 7/24/08 3:35 PM Page 600 Example 12.2 A carbon ion, with a mass of 2.01 1026 kg and a positive charge of magnitude 1.60 1019 C, enters the region of an external magnetic field of magnitude 6.32 105 T, as shown in Figure 12.23. Find the perpendicular speed at which the magnetic deflecting force will balance the gravitational force such that the carbon ion will travel in a straight line. Practice Problems 1. An electron, with a charge", " of magnitude 1.60 1019 C and a mass of 9.11 1031 kg, is travelling west along the surface of Earth at the equator. If the magnitude of the magnetic field at this location is 5.00 105 T, what minimum speed must the electron maintain to remain at the same height above Earth\u2019s surface? 2. Ions, with a charge of 1.60 1019 C and a mass of 8.12 1026 kg, travel perpendicularly through a region with an external magnetic field of 0.150 T. If the perpendicular speed of the ions is 8.00 104 m/s, determine (a) the magnitude of the deflecting force on the ion (b) the radius of curvature of the motion of the deflected ion Hint: The magnetic deflecting force is the centripetal force. F F m c v2 m qvB r Answers 1. 1.12 106 m/s 2. (a) 1.92 1015 N (b) 0.271 m S N carbon ion Figure 12.23 Given m 2.01 1026 kg 6.32 105 T B q 1.60 1019 C g 9.81 N/kg Required speed (v) at which the magnitudes of the magnetic force, F m and the gravitational force, F g, are equal, Analysis and Solution The gravitational force on the carbon ion has a magnitude of mg and is directed downward (toward the bottom of F g the page). The magnetic force on the carbon ion has a magnitude of qvB F m the page). and must be directed upward (toward the top of F net F m F g But the magnetic deflecting force and the gravitational force balance (Figure 12.24), so F net 0. Therefore, F m qvB F g mg g v m B q (2.01 1026 kg)9.81 N k g (6.32 105 T)(1.60 1019 C) 1.95 102 m/s Fm Fg Paraphrase The carbon atom will travel in a straight line if its speed is 1.95 102 m/s. Figure 12.24 600 Unit VI Forces and Fields 12-PearsonPhys30-Chap12 7/24/08 3:35 PM Page 601 In this section, you have studied the deflection of a moving charged particle in a magnetic field. Applying this science, you learned not only the importance of this phenomenon in technologies, such as a television and a magnetron, but also the significance of this phenomenon in protecting Earth from harmful cosmic radiations. The magnetic field of Earth, in deflecting dangerous charged particles from striking Earth\u2019s surface, also produces one of the most beautiful and spectacular natural light shows\u2014the aurora. 12.2 Check and Reflect 12.2 Check and Reflect Knowledge 1. Why is a cathode ray called a cathode ray? 2. What is the difference between a magnetic field vector arrow and a magnetic field line? 3. An electron and a proton, both with the same perpendicular velocity, enter a region with a uniform external magnetic field. What can you state about the deflections of both particles? 4. Describe the key differences in how magnetic and electric fields affect a moving charged particle. Applications 5. A positively charged lithium ion is travelling horizontally along Earth\u2019s surface. Describe the deflection due to the magnetic force if the ion travels (a) south to north (b) east to west (c) upward into the atmosphere 6. A proton with a speed of 2.00 105 m/s enters an external magnetic field of magnitude 0.200 T. Calculate the magnitude of the deflecting force if the proton enters (a) perpendicular to the magnetic field (b) at an angle of 35.0 to the field 7. A 0.020-g metal ball with a charge of 3.0 C is thrown horizontally along Earth\u2019s equator. How fast must the ball be thrown so that it maintains the same height, during its motion tangential to Earth\u2019s surface, if the magnitude of Earth\u2019s magnetic field is 5.0 105 T? 8. An alpha particle, with a charge of 2 1.60 1019 C, is travelling perpendicularly through a magnetic field of magnitude 2.00 102 T at a speed of 1.02 105 m/s. What minimum gravitational force is required to suspend the alpha particle at the same position above Earth\u2019s surface? 9. Electrons in the picture tube of a television are accelerated to a speed of 1.30 106 m/s. As they travel through the tube, they experience a perpendicular magnetic field of magnitude 0.0700 T. What is the radius of deflection of the electrons in the tube? 10. A cosmic ray proton travelling through space at 4.38 106 m/s deflects in a circular arc with a radius of 5.50 106 m", ". What is the magnitude of the magnetic field at that point in space? Extensions 11. Why are auroras seen only at higher latitudes? e TEST To check your understanding of moving charges and magnetic fields, follow the eTest links at www.pearsoned.ca/school/physicssource. Chapter 12 Properties of electric and magnetic fields apply in nature and technology. 601 12-PearsonPhys30-Chap12 7/24/08 3:35 PM Page 602 12.3 Current-carrying Conductors and Magnetic Fields info BIT The first sensitive meter to measure small currents was developed by Luigi Galvani in the 1700s. While dissecting a frog\u2019s leg, he noticed that an electric current caused the frog\u2019s leg to twitch. He realized that he had accidentally discovered a method of detecting small currents and used this discovery to design the galvanometer. current: the quantity of charge that flows through a wire in a given unit of time ampere: the flow of 1 C of charge past a point in a conductor in 1 s 602 Unit VI Forces and Fields Figure 12.25 A galvanometer and an electric motor, like the one in this lawn mower, apply magnetic fields produced by a flow of charge. Two of the most common applications of magnetic fields acting on moving charged particles are meters (such as ammeters, voltmeters, and galvanometers) and electric motors (Figure 12.25). Although these technologies appear to be different from the technology of the television, the basic operating principle of all these technologies is similar. Recall from earlier science studies that a galvanometer is a device for detecting and measuring small electric currents. How does a galvanometer operate? How is its operation similar to the technologies of the electric motor and television? Electric Current Recall from earlier science courses that electric current is the movement of charged particles. It can be defined more precisely as the quantity of charge that flows through a wire in a given unit of time. The unit for current, the ampere (A), is a measure of the rate of current. The ampere is an SI base unit. A current of 1 A is equivalent to the flow of 1 C of charge past a point in a conductor in 1 s. In other words, 1 A 1 C/s. For example, the effective value of the current through a 100-W light bulb is about one ampere (1 A) of current. The ampere is named in honour of the French scientist Andr\u00e9-Marie Amp\u00e8re (1775\u20131836), who is renowned for his analysis of the relationship between current and magnetic force. This equation shows the relationship between current and charge: q I t where I is the current in amperes, q is the magnitude of charge in coulombs, and t is the time elapsed in seconds. 12-PearsonPhys30-Chap12 7/24/08 3:35 PM Page 603 Example 12.3 Calculate the current in a wire through which 20.0 C of charge passes in 4.00 s. Given q 20.0 C t 4.00 s Required current (I) Analysis and Solution To calculate the current, use the equation I q t.0 C 2 0 4 s 0.0 5.00 C s 5.00 A Paraphrase The current in the conducting wire is 5.00 A. Practice Problems 1. A lightning strike transfers 20.0 C of charge to the ground in 1.00 ms. Calculate the current during this lightning strike. 2. If the current in a household appliance is 5.00 A, calculate the amount of charge that passes through the appliance in 10.0 s. Answers 1. 2.00 104 A 2. 50.0 C Magnetic Force on a Current-carrying Conductor In a CRT picture tube, powerful external magnetic fields are used to deflect moving electrons to produce an image on a screen. To analyze the operation of a galvanometer or electric motor, and to reveal the similarity of their operation to that of a television, consider the movement of electrons as a current in a wire conductor. When there is an electric current in a wire that is perpendicular to an external magnetic field, each electron experiences a magnetic force caused by the interactions of its own magnetic field and the external magnetic field (Figure 12.26). You can observe the effect of this force. The magnetic force causes the electrons to deflect upward. However, the electrons cannot escape the wire, so if the magnetic force on the electrons is great enough, the whole wire will rise upward, opposite to the force of gravity. The magnetic force on a conducting wire is the same as the magnetic m) that you studied in section 12.2. deflecting force on a moving charge (F S N S N e Figure 12.26 A current of electrons passes through a conducting wire lying perpendicular to an external magnetic field. Chapter 12 Properties of electric and magnetic fields apply in nature and technology. 603 12-Pear", "sonPhys30-Chap12 7/24/08 3:35 PM Page 604 PHYSICS INSIGHT Remember: If the moving charges are negative, use your left hand; and if the moving charges are positive, use your right hand. Left-hand Rule for Magnetic Force To determine the direction of the magnetic force, you can use the lefthand rule, as shown in Figure 12.27: \u2022 Your thumb indicates the direction of electron flow in the conductor. \u2022 Your extended fingers point in the direction of the external magnetic field. \u2022 Your palm indicates the direction of the magnetic deflecting force on the wire. S N S N e Fm v B Figure 12.27 The left-hand rule for determining the direction of magnetic force To calculate the magnitude of the magnetic force for a length of current-carrying conducting wire, use the equation F m IlB where I is the current measured in amperes; l is the length of the wire perpendicular to the magnetic field in metres; B is the magnitude of the external magnetic field in teslas; and F is m the magnitude of the magnetic force in newtons. The Galvanometer In the operation of the galvanometer, a coil of wire is mounted to allow for movement within the strong magnetic field of the permanent magnet (Figure 12.28). The coil turns against a spring with an attached needle pointing to a calibrated scale. When there is a current in the coil, the magnetic forces cause the coil to rotate. The greater the current, the greater the rotation, as registered on the scale by the needle. The galvanometer, which measures very small currents, can be made to measure larger currents (ammeter) by connecting a small resistance in parallel, and to measure larger potential differences (voltmeter) by connecting a large resistance in series. The magnetic force produced on a current-carrying wire can be demonstrated in the 12-4 QuickLab on page 606. N S current restoring spring Figure 12.28 A schematic diagram of a galvanometer reveals all the essential components in its operation. 604 Unit VI Forces and Fields 12-PearsonPhys30-Chap12 7/24/08 3:35 PM Page 605 Example 12.4 An 8.50-cm length of conducting wire lies perpendicular to an external magnetic field of magnitude 4.20 mT, as shown in Figure 12.29. If there is a negative charge flow of 2.10 A in the conductor, calculate the magnitude and determine the direction of the magnetic force on the wire. S N S N Figure 12.29 Given l 8.50 cm 8.50 102 m 4.20 mT 4.20 103 T B I 2.10 A Practice Problems 1. A 0.500-m length of conducting wire carrying a current of 10.0 A is perpendicular to an external magnetic field of magnitude 0.200 T. Determine the magnitude of the magnetic force on this wire. 2. A thin conducting wire 0.75 m long has a mass of 0.060 kg. What is the minimum current required in the wire to make it \u201cfloat\u201d in a magnetic field of magnitude 0.15 T? Required magnitude and direction of the magnetic force on the wire (F m) Answers 1. 1.00 N 2. 5.2 A Analysis and Solution Determine the magnitude of the magnetic force: F m IlB (2.10 A)(8.50 102 m)(4.20 103 T) 7.50 104 N Use the left-hand rule to determine the direction of the magnetic force, because the moving charges are negative: \u2022 Thumb points in the direction of the charge movement or current, into the page. \u2022 Extended fingers point in the direction of the external magnetic field, to the right of the page (north to south). \u2022 Palm points in the direction of the magnetic force, to the top of the page. Paraphrase The magnetic force is 7.50 104 N [upward] (toward the top of the page). Chapter 12 Properties of electric and magnetic fields apply in nature and technology. 605 12-PearsonPhys30-Chap12 7/24/08 3:35 PM Page 606 12-4 QuickLab 12-4 QuickLab Demonstration of a Current-carrying Conductor in a Uniform Magnetic Field Problem How does a uniform magnetic field affect a currentcarrying conductor? 2 Carefully increase the current (amperage) from the power supply. 3 Observe any effects on the current-carrying conductor. Materials 1 piece of stiff insulated conducting wire (6\u20138 cm long) 2 alligator clips 1 U-shaped magnet thread or light string retort stand and clamp variable low-voltage DC power supply with ammeter Questions 1. Describe any effects on the current-carrying conductor that occurred as the current through the conducting wire increased. 2. Does the hand rule verify the direction of the movement of the conducting wire? Explain which hand rule must be used", ". Procedure 1 Set up the apparatus as shown in Figure 12.30. 3. What is the effect of an external magnetic field on a current-carrying conductor? 4. Based on what you have just observed, design a lab that would demonstrate the effects of a uniform magnetic field on a current-carrying conductor. string or thread ON OFF retort stand power supply insulated wire N S magnet Figure 12.30 606 Unit VI Forces and Fields 12-PearsonPhys30-Chap12 7/24/08 3:35 PM Page 607 Magnetic Forces Between Two Current-carrying Conductors After Oersted demonstrated that a current-carrying conductor creates a magnetic field around a conductor, the French scientist Andr\u00e9-Marie Amp\u00e8re performed extensive studies to determine the magnitude of the magnetic field at any point surrounding a current-carrying conductor. In addition to his mathematical analysis of magnetic fields, he is also noted for determining that two current-carrying conductors exert magnetic forces on each other. The charged particles in one wire are affected by magnetic forces when placed in the magnetic field of another current-carrying wire. Currents in the same direction attract each other (Figure 12.31(a)), and currents in opposite directions repel each other (Figure 12.31(b)). magnetic field wire x e Fm Fm wire x e magnetic field (a) (b) Fm magnetic field wire x e wire e Fm magnetic field Figure 12.31 From the left-hand rule for magnetic fields, the red dashed arrows indicate the orientation of the magnetic field around each wire. Use the left-hand rule for magnetic force to determine how the wires will move relative to each other. (a) When currents are in the same direction, the wires attract each other. (b) When currents are in opposite directions, the wires repel each other. Through careful experimentation and measurement, Amp\u00e8re was able to determine that the magnetic force between two current-carrying conductors depends on all of the following: \u2022 the length of the conducting wire \u2022 the distance between the two conducting wires \u2022 the amount of current in each wire The SI unit for current is named in honour of Amp\u00e8re\u2019s work. This unit, the ampere, is now defined as the current required in each of two current-carrying wires, 1 m long and separated by 1 m in air, to produce a force of 2 107 N of magnetic attraction or repulsion. As you learned at the beginning of this section, an ampere is equivalent to the flow of 1 C of charge in 1 s. So, 1 A 1 C/s, and 1 C 1 As. Chapter 12 Properties of electric and magnetic fields apply in nature and technology. 607 12-PearsonPhys30-Chap12 7/24/08 3:35 PM Page 608 Concept Check In intricate electrical circuits, two conducting wires carrying currents in opposite directions are usually crossed. What is the purpose of this crossing procedure? 12-5 Design a Lab 12-5 Design a Lab Using the Current Balance to Measure the Magnetic Force Between Two Current-carrying Conducting Wires The Question How can you use a current balance to investigate the factors that influence the magnetic force acting on two current-carrying conducting wires? Design and Conduct Your Investigation Study the operation of the current balance in your laboratory and design an experimental procedure to investigate the factors that determine the magnetic force acting on two current-carrying conducting wires. In your experimental design: \u2022 Identify the factors that determine the magnetic force acting on two currentcarrying conductors. \u2022 Write an \u201cif/then\u201d hypothesis statement that predicts how changes in the variable affect the magnetic force. \u2022 Clearly outline the procedure you will perform to investigate the relationship of each factor on the magnetic force. \u2022 Describe what you will measure and how the data will be recorded and analyzed. \u2022 Explain how the data will be used to answer the question. As a group, identify and designate tasks. Prepare a report that describes your experimental design and present it to your teacher. After approval, conduct the investigation and answer the question. How well did your results agree with your hypothesis? info BIT Advancements in technology make it possible to construct extremely small electric motors. Today, 1000 of the smallest electric motors could fit in the period at the end of this sentence. commutator: a mechanism for maintaining a properly polarized connection to the moving coil in a motor or generator The Electric Motor The most important application of the effect of an external magnetic field on current-carrying conductors is the electric motor. Figure 12.32 illustrates a simple electric motor that works with a current-carrying wire loop between two magnetic poles. The current is in one direction. Recall from earlier science studies that current in one direction is called a direct current (DC). A simple DC electric motor consists of three fundamental components: \u2022 a stator\u2014a frame with a coil or permanent magnet to provide a magnetic field \u2022 an armature or rotor\u2014a rotating", " loop of conducting wire on a shaft \u2022 a commutator\u2014a split metal ring 608 Unit VI Forces and Fields 12-PearsonPhys30-Chap12 7/24/08 3:35 PM Page 609 As electrons in the current pass through the loop of wire in the armature in a clockwise direction (as seen from above in Figure 12.32), they experience a motor effect deflecting force. When you apply the left-hand rule for magnetic force, electrons on the left side of the loop experience a deflecting force upward, and electrons on the right side of the loop experience a deflecting force downward. The combined effect of both forces results in a rotation of the loop in a clockwise direction. Fm stator N armature e S e brush Fm commutator Figure 12.32 In a simple DC electric motor, the brushes provide a sliding contact between the wires from the battery and the armature. The magnetic field exerts an upward force on the left side of the wire loop and a downward force on the right side, causing the armature to rotate clockwise. Concept Check Describe the changes that must be made to the apparatus, shown in Figure 12.32, to cause the armature to rotate counterclockwise. If the rotation of the loop is to continue, the direction of the motion of the electrons in the loop must change every half-rotation. To accomplish this, the armature is connected to a commutator. A commutator is a split metal ring that is fastened to both ends of the loop of wire in the armature. Each half of the metal ring acts as a contact to the terminals of a power supply. Every half-rotation, the leads of each side of the armature contact a different terminal, changing the direction of the electron movement. Once connected to a steady supply of moving electrons, the armature continues to rotate in one direction. This is the principle of a simple electric motor. The Generator Effect (Electromagnetic Induction) In 1996, NASA did an experiment that involved a satellite attached by a conducting tether wire to a NASA space shuttle orbiting in space around Earth (Figure 12.33). Researchers found that the combination generated a current of about 1 A through the wire. The experiment was of particular significance for space scientists because it showed that Figure 12.33 A satellite tethered to a NASA space shuttle Chapter 12 Properties of electric and magnetic fields apply in nature and technology. 609 12-PearsonPhys30-Chap12 7/24/08 3:35 PM Page 610 this procedure could provide a method of generating the electric energy necessary to power all the electrical components on a space vehicle. This example is a useful and important application of a scientific phenomenon, but this phenomenon can also produce harmful effects in some situations. For example, engineers constructing the 1280-km north\u2013south gas pipeline from Prudhoe Bay to Valdez in Alaska (Figure 12.34(a)) had to take precautions to eliminate the currents of electricity, called telluric currents, in the pipeline. These currents are caused by fluctuations in Earth\u2019s magnetic field. Special magnesium anodes were installed underground along the pipeline to ground it and eliminate the possibility of electrical sparks. (a) (b) Figure 12.34 (a) A pipeline in Alaska; (b) An airplane in flight Similarly, certain grounding conditions must be incorporated in the construction of an airplane to eliminate the current generated by the wings of an airplane in flight through Earth\u2019s magnetic field. These currents could affect the operation of all electrical components on the aircraft (Figure 12.34(b)). How are these examples related? What physical phenomenon is generating the current? The examples described above all involve conductors moving through magnetic fields. The scientific explanation of how they generate electricity began with investigations over 200 years ago. Faraday\u2019s and Henry\u2019s Discoveries Most scientific discoveries are the result of many years of research and investigations. The process is often convoluted and results are often accidental. However, as you have learned, some scientific discoveries are a result of the symmetry of nature. This symmetry led Coulomb and Faraday to conclude that electrical and magnetic forces could be determined using inverse-square relationships similar to Newton\u2019s universal law of gravitation. Similarly, this symmetry in nature, and Oersted\u2019s discovery that electricity could produce magnetism, led scientists to predict that magnetism could produce electricity. Experiments conducted in 1831 by Michael Faraday in England and Joseph Henry (1797\u20131878) in the United States demonstrated this effect. e LAB For a probeware activity that demonstrates the principle of electromagnetic induction, go to www.pearsoned.ca/ school/physicssource. 610 Unit VI Forces and Fields 12-PearsonPhys30-Chap12 7/24/08 3:35 PM Page 611 In a simplified version of their experiment, shown in Figure 12.35, a magnet is moved toward", " a coil of conducting wire connected to a sensitive galvanometer. When the magnet approaches the coil, the galvanometer\u2019s needle deflects in one direction, indicating that a current is being produced in the coil of wire. This current is called an induced current, which is produced by a generated voltage. When the magnet is pulled away from the coil, the galvanometer deflects in the opposite direction, indicating that the induced current in the coil is in the opposite direction. When the magnet is stationary, no current is induced. If the magnet were held stationary while the coil of wire was moved back and forth, similar induced currents would be produced. Evidently, it does not matter whether the magnet or the coil of wire moves, as long as there is relative motion between a coil of conducting wire and an external magnetic field. In their conclusions, Faraday and Henry stated that when a piece of conducting wire cuts through magnetic field lines, an induced current is produced. The production of electricity by magnetism is called the generator effect or electromagnetic induction. Figure 12.36(a) shows a piece of conducting wire being moved perpendicularly upward through an external magnetic field. As a result, electrons in the wire also move perpendicularly upward. Use the left-hand rule for magnetic force: If the wire is moving upward (thumb) through the external magnetic field (fingers), then each electron experiences a motor-effect force (palm). Electrons will gather at one end of the wire with stored electric energy from the work done on the system in moving the wire. Thus, one end of the wire has an accumulation of electrons with stored electric energy while the other end has a deficiency of electrons (Figure 12.36(a)). If this wire is part of an external circuit, as in Figure 12.36(b), the induced voltage causes a current to flow through the external wire. A difference in electric potential drives electrons through an external circuit, from a region of high electric potential to a region of lower electric potential. N S galvanometer Figure 12.35 When a magnet is moved toward a loop of wire connected to a galvanometer, the galvanometer needle deflects. This indicates that an induced current is being produced in the coil of wire. generator effect or electromagnetic induction: production of electricity by magnetism Project LINK How would you apply the principle of electromagnetic induction and the operation of commutators to construct a DC generator wire movement Fm B induced current (a) wire movement induced current Fm B A electron flow (b) Figure 12.36 A current can be induced in a wire by moving the wire through a magnetic field. Chapter 12 Properties of electric and magnetic fields apply in nature and technology. 611 12-PearsonPhys30-Chap12 7/24/08 3:35 PM Page 612 12-6 Inquiry Lab 12-6 Inquiry Lab Magnetic Fields and Moving Conductors \u2014 Demonstration Question What factors influence the effect produced when there is relative motion between an external magnetic field and a conducting wire? Materials and Equipment 2 bar magnets 3 different sizes of coils of conducting wire (the diameter of the coils should allow a bar magnet to be inserted) galvanometer that can be projected onto a screen using an overhead projector galvanometer N S N S N S Figure 12.37 Procedure 1 Set up the apparatus as shown in Figure 12.37. 2 Slowly push one bar magnet at a uniform speed into the largest coil. Then pull the bar magnet out in the opposite direction at the same speed. Observe the deflection of the galvanometer\u2019s needle in both cases. 3 Repeat step 2 with the other end of the magnet. Observe the deflection of the galvanometer\u2019s needle in both cases. Required Skills Initiating and Planning Performing and Recording Analyzing and Interpreting Communication and Teamwork 4 Repeat step 2, using two bar magnets. Observe the magnitude and direction of the galvanometer\u2019s deflection. 5 Repeat step 2, using two bar magnets at a faster speed through the largest coil. Observe the magnitude and direction of the galvanometer\u2019s deflection. 6 Repeat step 2, using two bar magnets at the same constant speed through the medium and the smaller coils of conducting wire. Observe the magnitude and direction of the galvanometer\u2019s deflection. Analysis 1. How does the direction of the movement of the magnet affect the direction of the deflection of the galvanometer? 2. How does the polarity of the magnet affect the direction of the deflection? 3. Describe how each of the following factors influences the magnitude of the deflection of the galvanometer: (a) speed of the magnets through the conducting wire (b) strength of the external magnetic field (c) number of loops in the coil of conducting wire 4. What is the effect of relative motion between a conducting wire and a magnetic field? 5. Does it make any difference if the magnet or the conducting wire is moved? 6. What are the factors that determine the magnitude and direction of the", " induced current when there is relative motion between a conducting wire and an external magnetic field? 7. Based on your observations from this activity, design an experiment that demonstrates the effect of a uniform magnetic field on a moving conductor. 612 Unit VI Forces and Fields 12-PearsonPhys30-Chap12 7/24/08 3:35 PM Page 613 9. A battery supplies a current of 5.20 mA to a circuit. Determine the quantity of charge that flows through the circuit in 2.00 s. 10. Two conducting wires parallel to each other carry currents in opposite directions. Using the appropriate hand rule, determine whether the wires will attract or repel each other. 11. A wire 50 cm long and carrying a current of 0.56 A is perpendicular to an external magnetic field of 0.30 T. Determine the magnitude of the magnetic force on the wire. Extension 12. Could a simple electric generator be converted to a simple electric motor? Suggest any alterations that must be made in the design. e TEST To check your understanding of current-carrying conductors and magnetic fields, follow the eTest links at www.pearsoned.ca/school/physicssource. 12.3 Check and Reflect 12.3 Check and Reflect Knowledge 1. What are the factors that affect the magnetic force on a moving charge through an external magnetic field? 2. What are the factors that affect the magnetic force on a charge moving through a conducting wire in an external magnetic field? 3. What is the relationship between amperes and coulombs? 4. In the operation of a simple electric motor and simple electric generator, identify (a) a similarity (b) a difference 5. What symmetry in nature did Faraday and Henry apply in their discovery of electromagnetic induction? 6. What is the function of a split-ring commutator in the operation of a simple DC motor? 7. How do the electrons in a loop of wire in a generator gain energy? Applications 8. A wire lying perpendicular to an external magnetic field carries a current in the direction shown in the diagram below. In what direction will the wire move due to the resulting magnetic force? S N S N I Chapter 12 Properties of electric and magnetic fields apply in nature and technology. 613 12-PearsonPhys30-Chap12 7/24/08 3:35 PM Page 614 12.4 Magnetic Fields, Moving Charges, and New and Old Technologies info BIT Michael Faraday built the first electric motor in 1821. This motor had a stiff wire hanging from a stand. The lower end of the wire was immersed in a cup of mercury with a bar magnet upright in the middle. When current from a battery flowed through the wire, it rotated around the magnet. From the old technologies of the simple beginnings of electric motors, electric meters (such as galvanometers and ammeters), loudspeakers, and electromagnets to the new technologies of magnetohydrodynamic (MHD) propulsion systems and magnetic resonance imaging (MRI), the science of the production of magnetism by electricity plays a significant role in our everyday lives. Although examples of some of these technologies have been described in previous sections, following are other examples of old and new technological applications of this principle. Table 12.2 describes old and new technologies that use moving charges or current-carrying conductors to produce magnetic fields that can interact with external magnetic fields to produce powerful magnetic forces. Table 12.2 Loudspeakers and MHD Propulsion Old Technology New Technology Magnetohydrodynamic (MHD) Propulsion paper cone water intake oppositely charged plates Loudspeakers electron flow x x x B F S N S voice coil Figure 12.38 A simplified diagram of a loudspeaker The operating principle of most loudspeakers is that currentcarrying wires produce magnetic fields that can exert magnetic forces. In the design of the loudspeaker shown in Figure 12.38, a coil of wire, called a voice coil, surrounds the north pole of a very powerful external magnet at the back of the speaker. When your sound system sends an electric signal to the coil, a current is produced in the coil, which produces a magnetic field. As a result, the coil experiences a magnetic force due to the interaction of its magnetic field with the external magnetic field. Depending on the direction of the current in the coil, the magnetic force of attraction or repulsion causes the coil to slide to the left or right. The direction of the current is determined by the electric signal produced by the sound system. As the voice coil slides back and forth, it causes the paper cone to vibrate in or out, creating sound waves as it pushes on the air in front of the cone. The electric signal from the sound system is thus converted to a mechanical sound wave in air. 614 Unit VI Forces and Fields superconducting electromagnet jet of water Figure 12.39 MHD uses magnetic fields as a propulsion system for seagoing vessels. The MHD propulsion system is an experimental system for se", "agoing vessels to replace conventional propeller systems. MHD uses magnetic fields to produce a jet of water for propulsion. Figure 12.39 is a simplified diagram of this type of system. A powerful superconducting magnet surrounds a thruster tube containing seawater. This magnet produces a magnetic field perpendicular to the tube\u2019s length. Inside the tube, electrodes produce a current of ions, perpendicular to the magnetic field, across the tube from the dissolved salts in seawater. As a result of the perpendicular movement of the ions through an external magnetic field, a magnetic force is exerted on the ions, causing them to deflect along the length of the tube. This movement of the water through the tube provides the necessary thrust to propel the vessel. An advantage of MHD propulsion systems is that they have no mechanical moving parts and thus require minimal maintenance. 12-PearsonPhys30-Chap12 7/24/08 3:35 PM Page 615 Generator Effect Applications The discovery that moving a conducting wire through an external magnetic field generates an induced current in the conductor (generator effect) also led to many important technological applications. From the old technologies of the simple generators, induction coils, and transformers to the new technologies of infant breathing monitors and others, applications of the scientific principle of the production of electricity from magnetism are found everywhere in our lives. Table 12.3 describes two of these applications. Table 12.3 Induction Coils, Transformers, and SIDS Monitors Old Technology New Technology Induction Coils switch SIDS Monitors galvanometer primary coil iron battery secondary coil Figure 12.40 A simplified diagram of Faraday\u2019s induction coil Figure 12.41 Monitors are designed to detect changes in a baby\u2019s breathing. A change in the current in the primary coil produces a changing magnetic field in the iron core. This changing magnetic field produces an induced current in the secondary coil, causing the needle on the galvanometer to deflect. Such coils can induce current in a wire that has no direct connection to the power supply. Figure 12.40 shows a simplified version of Michael Faraday\u2019s original induction coil. In sudden infant death syndrome (SIDS), an infant stops breathing with no apparent cause. One type of SIDS monitor uses induced currents to measure an infant\u2019s breathing (Figure 12.41). A coil of wire attached to one side of the infant\u2019s chest carries an alternating current, which produces a magnetic field. This alternating field cuts another coil taped to the other side of the chest and induces an alternating current in this other coil. As the chest moves up and down, the strength of the induced current varies. These variations are monitored. A Motor Is Really a Generator, Which Is Really a Motor You have analyzed and studied the motor effect and the generator effect as separate phenomena in this chapter. However, the symmetry of nature suggests that related phenomena are really variations of the same effect. Since electricity can produce magnetism and magnetism can produce electricity, then perhaps the technologies that are derived from these phenomena are also similar. Is a motor really that different from a generator? Chapter 12 Properties of electric and magnetic fields apply in nature and technology. 615 12-PearsonPhys30-Chap12 7/24/08 3:35 PM Page 616 12-7 Inquiry Lab 12-7 Inquiry Lab The Curious Relationship Between Motors and Generators Required Skills Initiating and Planning Performing and Recording Analyzing and Interpreting Communication and Teamwork Question What is the relationship between the motor effect and the generator effect? Analysis 1. Copy Table 12.5 into your notebook. Use the data from Table 12.4 to complete the calculations in Table 12.5. Materials and Equipment Table 12.5 Calculations for 12-7 Inquiry Lab Magnitude of Acceleration of the Magnet Through the Copper Pipe d a 2 2 t (m/s2) Magnitude of Net Force Causing the Downward Acceleration of the Magnet (Fa ma) (N) Magnitude of the Weight of the Magnet (Fg mg) (N) 2. What is the magnitude of the upward force on the falling magnet? 3. Identify and explain where the generator effect is occurring in this experiment. 4. Identify and explain where the motor effect is occurring in this experiment. 5. Do the generator effect and the motor effect complement each other as the magnet falls through the copper pipe? Explain your answer. 1 100-cm length of copper pipe (internal diameter approximately 1.4 cm) 1 cylindrical rare earth magnet (less than 1.4 cm in diameter) 1 metre-stick 1 stopwatch 1 scale Procedure 1 Copy Table 12.4 into your notebook. Table 12.4 Data for 12-7 Inquiry Lab Mass of Magnet Length of Copper (m)(kg) Pipe (d)(m) Average Time Taken for the Magnet to Fall Through the Pipe (t)(s) 2 Measure the mass of the magnet on the scale, and record it in Table 12.4. 3 Measure the length of", " the copper pipe, and record it in Table 12.4. 4 Holding the pipe in a vertical position, drop the magnet from the exact top of the pipe. Measure the time for the magnet to reappear out the bottom. Record this time in the table. 5 Repeat the procedure in step 4 several times to obtain an average value for the time taken for the magnet to drop the length of the pipe. 616 Unit VI Forces and Fields 12-PearsonPhys30-Chap12 7/24/08 3:35 PM Page 617 An Accidental Discovery Simple DC electric motors and electric generators have three similar components: \u2022 an external magnetic field \u2022 a loop of conducting wire \u2022 a commutator At the 1873 Vienna Exhibition, the Belgian inventor Z\u00e9nobe-Th\u00e9ophile Gramme (1826\u20131901) demonstrated a compact and efficient generator that he had designed. A steam engine provided the power to run the generator. A workman mistakenly connected the output of the generator to a second generator in the display. The shaft of the second generator began spinning even though it was not connected to the steam engine. Gramme immediately realized that the second generator was operating as a motor powered by the first generator. Gramme and his colleagues then moved the generators several hundred metres apart and connected them with long wires. The American writer Henry Adams (1838\u20131918) described the importance of Gramme\u2019s demonstration: \u201cSuddenly it became clear that ELECTRICITY could now do heavy work, transporting power through wires from place to place.\u201d M I N D S O N Perpetual Motion? Suppose that a motor and generator are connected to the same shaft and wired such that the output of the generator powers the motor. If you spin the shaft, the generator supplies energy to the motor, which turns the shaft. The generator then produces more energy to run the motor. Explain why this process cannot continue indefinitely. Lenz\u2019s Law If you did the 12-7 Inquiry Lab, you discovered what happens when you drop a magnet down a metal tube. When a conductor cuts the magnetic field lines of a falling magnet, it generates an induced current in the conducting pipe (the generator effect). However, the induced current moves in a circular motion around the circular pipe, so it creates its own vertical magnetic field, inside the metal tube (the motor effect). The direction of the magnetic field can be directed either upward or downward. The direction of the magnetic field that is produced by the circular induced current in the pipe can have one of the following orientations: \u2022 It will attract the magnet and cause it to fall faster, thus generating a greater induced current. \u2022 It will repel and oppose the motion of the magnet, causing it to fall much slower. Chapter 12 Properties of electric and magnetic fields apply in nature and technology. 617 12-PearsonPhys30-Chap12 7/24/08 3:35 PM Page 618 The law of conservation of energy requires that you can never get more out of a system than you put into it. So, the direction of the new magnetic field will always oppose the motion of the magnetic field of the original magnet. This is the principle of Lenz\u2019s law, which states: The direction of a magnetically induced current is such as to oppose the cause of the current. For example, if a magnet falls with its north pole directed downward, then the magnetic field produced by the induced current in the conducting pipe will have its north pole pointing upward to repel and oppose this motion. Concept Check Copy Figure 12.42 into your notebook. Apply Lenz\u2019s law by sketching the direction of the induced current in the metal tube and the resulting orientation of the magnetic field in the tube. How would the induced current and magnetic field directions change if the falling magnet\u2019s south pole were directed downward? S N Figure 12.42 Dropping a magnet down a metal tube induces a current in the tube. Figure 12.43 shows a similar situation. As the north pole of a magnet approaches a coil of wire, the induced current generated in the coil produces a north pole to repel and oppose the approaching magnet. S N N S G Figure 12.43 Lenz\u2019s law helps us explain that the direction of current induced in the coil has a magnetic field that exerts a force on the bar magnet that opposes the magnet\u2019s motion. M I N D S O N Lenz\u2019s Law Balance a dime on its edge on a smooth table. Carefully bring a magnet as close as 1 mm to the face of the dime. Quickly pull the magnet away. What happens to the dime? How is this behaviour an application of Lenz\u2019s law? The principle behind Lenz\u2019s law also hinders the operation of electric motors and generators. For an electric motor to operate, an electric current must first be supplied through a conducting loop of wire in a magnetic field, causing the motor effect, so the loop", " will rotate. 618 Unit VI Forces and Fields 12-PearsonPhys30-Chap12 7/24/08 3:35 PM Page 619 However, as the loop rotates, the conducting wire cuts the magnetic field lines, causing the generator effect. The generator effect induces a current in the loop of wire. The direction of the induced current must be in an opposing direction to the direction of the original current that was supplied. Similarly, to operate a generator, movement of a conducting wire in a magnetic field must be supplied, which will induce a current. However, as soon as the induced current moves through a conductor in a magnetic field, a force on the conducting wire will be produced that opposes the original force and hinders the movement of the conducting wire. e WEB Research the relationship between Lenz\u2019s law and the operation of most vending machines. Write a short report analyzing the operation of vending machines, and describe whether they operate on the principles of Lenz\u2019s law and the generator effect or Lenz\u2019s law and the motor effect. In your description, include the term \u201ceddy currents.\u201d Begin your search at www.pearsoned.ca/ school/physicssource. THEN, NOW, AND FUTURE Nanotechnology Since the start of the Industrial Revolution and with advances in technology, machines have become increasingly smaller. Scientists at University of California\u2019s Berkeley National Laboratory have developed the smallest synthetic electric motor ever made. Essentially, it is an electric rotor spinning on an axle 2000 times smaller than the width of a human hair. Imagine tiny electric motors so small that one motor could ride on the back of a virus, or thousands could fit in the period at the end of this sentence. This is the world of nanotechnology, which is an umbrella word that covers many areas of research and deals with objects in nanometres, or a billionth of a metre. How do nanomachines work? As you have learned, conventional electric motors use electromagnets or strong external magnets to spin rotors made of loops of wire. The spinning of a rotor provides mechanical energy to do work. In the electric motor of nanotechnology, transistors act as switches to move negative and positive charges around a circle of electrodes. The charges jump around the stator electrodes, that are measured Figure 12.44 A rosette nanotube causing an electrically charged rotor to spin around and rotate a nanotube shaft. This spinning rotor provides mechanical energy, similar to a conventional electric motor. The nanomotor can perform only small functions, such as moving a second hand on a watch, but it has many advantages: It spins without gears or bearings; it is unaffected by gravity or inertia; and it can run for a long time with no breakdowns. Another application of nanotechnology is the rosette nanotube, like the one shown in Figure 12.44. This type of tube was developed at the National Institute of Nanotechnology of the National Research Council, located at the University of Alberta. The nanotube is made of molecules that assemble themselves into this distinctive shape. Possible future applications of these tubes include: nanowires in molecular electronics, drug delivery systems within the body, and environmentally friendly oil sands upgrading additives. The potential of nanotechnology in electronics is now beginning to be realized and applied to many different fields. Imagine, for example, nanorobots that can be injected into the body to attack viruses and cancer cells. In the future, it may even be possible to construct molecules of oil and gas, reducing our reliance on fossil fuels. The science of nanotechnology is the science of the future. Questions 1. What are some advantages of nanotechnology? 2. Describe how a nanomotor works. 3. What could nanotubes be used for? Chapter 12 Properties of electric and magnetic fields apply in nature and technology. 619 12-PearsonPhys30-Chap12 7/24/08 3:35 PM Page 620 12.4 Check and Reflect 12.4 Check and Reflect Knowledge 1. Identify two technological applications that employ (a) the motor effect (b) the generator effect 2. What are the three basic components of an electric motor and generator? 3. What is Lenz\u2019s law? Applications 4. For an electric motor: (a) Describe what you must supply to start the operation of the device. (b) Describe what you get out of the operation of the device (motor effect). (c) Using Lenz\u2019s law, explain how the operation of the motor also produces the generator effect to hinder its own operation. 5. For an electric generator: (a) Describe what you must supply to start the operation of the device. (b) Describe what you get out of the operation of the device (generator effect). (c) Using Lenz\u2019s law, explain how the operation of the generator also produces the motor effect to hinder its own operation. 6", ". Explain why you will feel a force of repulsion if you attempt to move a magnet into a coil of wire. 7. The north pole of a magnet is pulled away from a copper ring, as shown in the diagram below. What is the direction of the induced current in the ring? N S Extensions 8. An electric motor requires a current in a loop of wire. However, as the loop rotates, it generates a current which, according to Lenz\u2019s law, must be in an opposing direction. Why must the induced current be in an opposing direction? 9. A hair dryer operates on a very small current. If the electric motor in the hair dryer is suddenly prevented from rotating, the dryer overheats. Why? e TEST To check your understanding of magnetic fields, moving charges, and new and old technologies, follow the eTest links at www.pearsoned.ca/ school/physicssource. 620 Unit VI Forces and Fields 12-PearsonPhys30-Chap12 7/24/08 3:35 PM Page 621 CHAPTER 12 SUMMARY Key Terms and Concepts law of magnetism magnetic field electromagnet solenoid ferromagnetic domain motor effect force current Key Equations F qvB m F IlB m Conceptual Overview ampere commutator generator effect electromagnetic induction Lenz\u2019s law The concept map below summarizes many of the concepts and equations in this chapter. Copy and complete the map to have a full summary of the chapter. definition types Earth domain theory bar magnet U-shaped magnet magnetization by induction law of magnetism generator effect Lenz\u2019s law 2 like poles 2 unlike poles magnetic forces and fields cause moving charges motor effect moving charges in magnetic fields direction of deflection magnitude of deflection Fm qvB charge wire moving charges in 2 current-carrying wires definition of ampere Chapter 12 Properties of electric and magnetic fields apply in nature and technology. 621 12-PearsonPhys30-Chap12 7/24/08 3:35 PM Page 622 CHAPTER 12 REVIEW Knowledge 1. (12.1) State the major contribution of each of the following scientists to the study of magnetism: (a) William Gilbert (b) Hans Christian Oersted (c) Andr\u00e9 Amp\u00e8re (d) Michael Faraday 2. (12.1) State the definition of (a) a magnetic field (b) the direction of a magnetic field 3. (12.1) Compare a magnetic vector arrow at a point near a magnet and a magnetic field line around a magnet. 4. (12.1) How was it determined that there had to be two different types of magnetic poles? 5. (12.1) Sketch the magnetic field lines around each of the following objects and describe the differences in the magnetic fields of (a) a bar magnet and Earth (b) a current-carrying straight piece of conducting wire and a current-carrying coil of conducting wire 6. (12.2) Identify the sources of the two magnetic fields required to produce the motor effect force on a moving charge. 7. (12.2) Describe the deflection of a moving charge, through an external magnetic field, if the direction of the initial motion of the charge is (a) parallel to the external magnetic field lines (b) perpendicular to the external magnetic field lines (c) at an angle to the external magnetic field lines 8. (12.2) Where is a magnetic bottle formed? 9. (12.3) Describe a difference between a galvanometer and an ammeter. 10. (12.3) State two definitions for a current of one ampere. 11. (12.4) Identify two technologies that use the principles of (a) the motor effect (b) the generator effect 12. (12.4) What is the principle of Lenz\u2019s law? 622 Unit VI Forces and Fields Applications 13. Does every charged object necessarily have a positive and a negative charge? Does every magnetized object necessarily have a north and a south pole? Justify your answers. 14. If the direction of the magnetic field outside a magnet is from the north to the south pole, what is the direction of the magnetic field within the magnet? 15. Using domain theory, describe how an iron nail can become magnetic by (a) the process of magnetization by induction (b) the process of magnetization by contact 16. How will the magnetic force on a moving charged particle change if (a) only the charge is doubled? (b) the magnetic field is doubled and the speed is halved? (c) the mass of the charge is doubled? 17. Use domain theory to explain the difference between a permanent and a temporary magnet. 18. You are told that a straight piece of copper wire has a steady current in it. Given only a compass, describe how you can find the direction of the current in the", " wire. 19. A drinking straw with a green grape at one end is suspended by a string from a hanging support. When either end of a magnet is brought close to the grape, repulsion occurs. Describe a possible reason for this effect. 20. An electron in a TV tube is moving at 7.00 106 m/s perpendicular to a magnetic field of magnitude 0.0880 T in the tube. What is the magnetic deflecting force on the electron? 21. A proton travelling at 35 to an external magnetic field of magnitude 0.0260 T experiences a force of magnitude 5.50 1017 N. (a) Calculate the speed of the proton. (b) Calculate the kinetic energy of the proton in joules (J) and electron volts (eV). 22. What speed must an alpha particle maintain if it is to remain suspended, relative to Earth\u2019s surface, as it travels on a tangent to Earth\u2019s surface and perpendicularly through Earth\u2019s magnetic field of 50.0 T? 12-PearsonPhys30-Chap12 7/24/08 3:35 PM Page 623 23. An alpha particle travelling with a speed of 4.30 104 m/s enters a uniform magnetic field of 0.0300 T. Determine the magnetic force on the particle if it enters the field at an angle (a) perpendicular to the magnetic field (b) 30.0 to the magnetic field (c) parallel to the magnetic field 24. A magnetic field is used to bend a beam of electrons. What uniform magnetic field is required to bend a beam of electrons moving at 1.2 106 m/s in a circular arc of 0.25 m? 25. Two parallel current-carrying wires are observed to attract each other. What is the source of the force of attraction? How could you demonstrate that the force of attraction is not electrostatic attraction? 26. The magnetic force between two magnets was measured as the distance between the magnets was varied. The following information was obtained: Separation, r ( 102 m) Magnitude of Force, F (N) 5.00 10.00 15.00 20.00 25.00 4.02 1.01 0.45 0.25 0.16 (a) Draw a graph of the magnetic force as a function of separation distance. (b) From the shape of the graph, what is the relationship between force and separation? 27. An airplane is flying east over Earth\u2019s magnetic north pole. As a result of its motion, one wing was detected as having more electrons than the other. Explain why this phenomenon occurs. Identify which wing will have more electrons. 28. A power line carries a current of 500 A. Find the magnetic force on a 100-m length of wire lying perpendicular to Earth\u2019s magnetic field of 50.0 T. Extensions 29. A magnet is dropped through two similar vertical tubes of copper and glass. Which tube will allow the magnet to fall faster? Explain your answer. 30. A disk magnet on a table has two steel balls in contact with it on either side. The steel balls are slowly moved toward each other while still in contact with the disk magnet. As they move, they repel each other. Describe why the steel balls are attracted to the disk magnet, but repel each other. Consolidate Your Understanding Create your own summary of properties of magnetic and electric fields by answering the questions below. If you want to use a graphic organizer, refer to Student Reference 3: Using Graphic Organizers. Use the Key Terms and Concepts listed on page 621 and the Learning Outcomes on page 580. 1. Create a flowchart to identify technological devices that use electric fields, magnetic fields, or a combination of the two fields to control moving charges. 2. Write a paragraph comparing the effects of electric or magnetic fields on moving charges. Share your report with a classmate. Think About It Review your answers to the Think About It questions on page 581. How would you answer each question now? e TEST To check your understanding of the properties of electric and magnetic fields, follow the eTest links at www.pearsoned.ca/school/physicssource. Chapter 12 Properties of electric and magnetic fields apply in nature and technology. 623 12-PearsonPhys30-Chap12 7/28/08 9:46 AM Page 624 UNIT VI PROJECT Building a Model of a Direct Current Generator Scenario The search for better electrical energy production began with the pioneering work of Faraday and Henry in the 1800s and continues today. Over half of the energy consumed in our world is electrical energy and the average consumption per person is increasing every year, so more efficient methods of electrical energy production are being sought all the time. All DC electrical generators consist of three major parts: coils of wire wrapped around a core to make the armature, a commutator, and an external magnetic field. As you learned in this", " unit, the operating principle behind generators is the movement of a conducting wire through external magnetic field lines so that a voltage is generated in the wire. This in turn induces a current in an external line. All DC generators operate in this fashion. The only difference among generators is the source of the mechanical energy required to turn the turbines that rotate the coil of wires in the magnetic field. In some places, this is the energy of falling water or tides, while in others it is the energy of moving steam from the combustion of fossil fuels or nuclear reactions. Recently, interest has grown in using wind energy to turn the turbines that operate a generator. A single wind generator can produce about 10 MW of electrical power, which is sufficient for a single small farm. The purpose of this project is to research and investigate the operation of a wind-powered electrical generator and to build a model of a wind-powered DC generator capable of generating enough electricity to operate a mini-bulb. Planning Form a team of four or five members, and decide on and plan the required tasks to complete the project. These tasks may include researching the design of a simple generator, obtaining the necessary materials, constructing the model of the generator, preparing a written report, and presenting the project to the entire class. Assessing Results Assess the success of your project based on a rubric* designed in class that considers: research strategies thoroughness of the experimental design effectiveness of the experimental technique effectiveness of the team\u2019s public presentation Materials \u2022 insulated copper wire \u2022 iron core \u2022 split-ring commutator \u2022 external magnets \u2022 connecting wires \u2022 mini-bulb with support base \u2022 stiff paper to construct a turbine \u2022 balsa wood for the axle \u2022 household fan to produce wind Procedure 1 Using the Internet, library, or other resources, research the operation of a simple DC electrical generator and create a design of the model that you will construct. Pay special attention to the commutator required for DC generation. 2 Construct a working model of the generator that can provide the electrical energy to light a mini-bulb. In your model, investigate the factors that determine the magnitude of the generated voltage. 3 Prepare a report explaining the design and the specific functions of all the components. Thinking Further 1. What modifications did you make in the construction of your model of a generator that affected the magnitude of the generated voltage? 2. What other type of commutator could you have used in the design of your generator? What type of current would be induced by this commutator? 3. 4. Identify at least three risks and three benefits of a wind-powered electrical generator. Is wind-powered electrical generation a viable and desirable method of electrical energy generation for the future? Explain your answer. *Note: Your instructor will assess the project using a similar assessment rubric. 624 Unit VI Forces and Fields 12-PearsonPhys30-Chap12 7/24/08 3:35 PM Page 625 UNIT VI SUMMARY Unit Concepts and Skills: Quick Reference Concepts Chapter 10 Modern theory of electrostatics Coulomb\u2019s law Summary Resources and Skill Building Physics laws can explain the behaviour of electric charges. 10.1 Electrical Interactions Substances can be classified as conductors, insulators, semiconductors, and superconductors. Objects may be charged through the processes of friction, conduction, and induction. 10-1 QuickLab 10-2 Inquiry Lab 10.2 Coulomb\u2019s Law Coulomb\u2019s law states that the electrical force acting on charged objects depends on the charges and the distance between the charges. 10-3 Inquiry Lab Examples 10.1, 10.2 Vector analysis Electrostatic forces can be solved in one- and two-dimensional situations using vector analysis. Examples 10.3\u201310.6 Chapter 11 Electric field theory describes electrical phenomena. Fields Electric field Electric field lines Electric potential energy Electric potential 11.1 Forces and Fields Fields are used to explain action at a distance. An electric field is a three-dimensional region of influence surrounding every charge. Electrostatic force affects another charge placed in the field. The electric field is a vector quantity that has magnitude and direction. 11.2 Electric Field Lines and Electric Potential Electric field lines can depict the electric fields around different types of charged objects. Electric potential energy is the amount of work done on a charged object to move it from infinity to a position in an electric field. Electric potential energy can be calculated. Electric potential is the amount of electric potential energy stored per unit charge and can be calculated. Electric potential difference When a charge moves from a location where it has one electric potential to a location where it has another electric potential, the charge experiences an electric potential difference. Motion of a charge in an electric field 11.3 Electrical Interactions and the Law of Conservation of Energy When a charge is placed in an electric field, it experiences a force that causes it to accelerate in the direction of the field. The acceleration of the charge is different in a non-uniform field surrounding a point charge than the acceleration of the charge", " in the uniform field between charged plates. Work done by the system on the charge increases the charge\u2019s potential energy, which can be converted to other forms of energy. Chapter 12 Properties of electric and magnetic fields apply in nature and technology. 11-1 QuickLab, 11-2 Inquiry Lab Examples 11.1\u201311.4 Minds On activities, eSIM Examples 11.5, 11.6 Example 11.7 Example 11.8 Examples 11.10\u201311.12 Magnetic fields Cause of magnetism 12.1 Magnetic Forces and Fields Magnetic fields are three-dimensional regions of magnetic influence surrounding every magnet in which other magnets or magnetic substances are affected by magnetic forces. Magnetic fields are vector fields and can be depicted by magnetic field lines. The cause of magnetism is motion of charges and can be explained using the domain theory. If the motion of charges is straight, the magnetic field is circular. If the motion of charges is circular, the magnetic field is straight within the loop. The direction of the magnetic field lines can be described using hand rules. 12-1 QuickLab, 12-2 QuickLab Figure 12.10 Magnetizing objects Objects can be magnetized through contact or induction. Concept Check Motor effect on a moving charge 12.2 Moving Charges and Magnetic Fields A charge moving perpendicularly through an external magnetic field experiences a magnetic force due to two magnetic fields, which can be calculated. This motor effect force can explain the operation of electric motors and other technologies. The magnitude of the motor effect force can be calculated and its direction can be determined using hand rules. 12-3 Inquiry Lab Examples 12.1, 12.2 Motor effect on two current-carrying wires 12.3 Current-carrying Conductors and Magnetic Fields A current-carrying conductor that is perpendicular to an external magnetic field experiences a magnetic force that can be calculated. Generator effect A conductor moving perpendicular to an external magnetic field can produce electricity. Applications of the generator effect and the motor effect 12.4 Magnetic Fields, Moving Charges, and New and Old Technologies The generator effect and the motor effect are used in many technologies. Example 12.4 12-4 QuickLab 12-5 Design a Lab 12-6 Inquiry Lab 12-7 Inquiry Lab Lenz\u2019s law Lenz\u2019s law explains how a motor is really a generator and a generator is really a motor. Minds On Unit VI Forces and Fields 625 12-PearsonPhys30-Chap12 7/24/08 3:35 PM Page 626 UNIT VI REVIEW Vocabulary 1. Use your own words to define the following terms, concepts, principles, or laws. Give examples where appropriate. ampere charge migration charge shift charging by induction commutator conduction conductor coulomb Coulomb\u2019s law current domain electric field line electric potential (voltage) electric potential difference electric potential energy electromagnet electromagnetic induction electron volt electrostatics ferromagnetic field generator effect grounding induction insulator law of conservation of charge law of magnetism Lenz\u2019s law magnetic field motor effect force net charge plasma semiconductor solenoid source charge superconductor test charge 626 Unit VI Forces and Fields Knowledge CHAPTER 10 2. Explain why silver is a better conductor of electricity than rubber. 3. State a technological advantage of developing materials that are superconductors. 4. A negatively charged rubber rod is brought near a small metal ball hanging from an insulated thread. The metal ball is momentarily grounded, and then the ground and the rubber rod are removed. Identify the procedure used to charge the metal ball, and determine the final charge on the metal ball. 5. Describe a similarity and a difference between (a) charging by friction and charging by conduction (b) charge shift and charge migration 6. In each of the following examples, identify the charge on each object and state the method of charging the object. (a) An ebonite rod is rubbed with fur and then is held near a neutral metal sphere. (b) A glass rod is rubbed with silk and then is touched to a neutral metal sphere. 7. During the rubbing process of charging objects, one object gains a net negative charge. What can you conclude about the charge on the other object? Explain why. 8. How do the following factors affect the electrostatic force of attraction between two charged objects? (a) amount of charge on each object (b) distance between the centres of the two objects (c) sign of the charge on each object CHAPTER 11 9. State the superposition principle as it applies to vector fields. 10. Why must a test charge be small when it is used to determine the direction of an electric field around a larger primary charge? 12-PearsonPhys30-Chap12 7/24/08 3:35 PM Page 627 11. Draw the electric field lines around 22. Justify the following statement: \u201cA charge (a) a negative point charge (b) a positive charge and a negative point charge in the same", " region (c) a negatively charged cone-shaped object 12. Why is there no net electric field inside a charged hollow sphere? 13. Explain the following statement: \u201cIt is impossible to shield against gravitational fields, but it is possible to shield against electric fields.\u201d 14. Draw a diagram showing two small, equally charged, positive spheres a small distance apart. On your diagram, identify a point A where the electric field is zero and a point B where the electric potential is zero. 15. Describe how electric potential energy changes for positive and negative charges as they move, relative to the reference point at infinity. 16. Two oppositely charged parallel plates are connected to a 120-V DC supply. What happens to the magnitude of the electric field between the plates if the distance between the plates decreases? moving perpendicular to a uniform external magnetic field experiences a force but does not change its speed.\u201d 23. Show how a charge moving through an external magnetic field experiences a deflecting force because of the interaction of two magnetic fields. 24. A positively charged disk is spinning in a clockwise direction as seen from above. What is the direction of the magnetic field? Use a diagram to help you describe its shape. 25. Describe the shape and direction of the magnetic field around a negatively charged dart as it travels directly away from you toward a target on a wall. 26. What factors affect the magnetic force on a charge moving through an external magnetic field? 27. State a difference and a similarity between the motor effect and the generator effect. 28. Describe how Lenz\u2019s law affects the operation of (a) a motor (b) a generator 17. Describe the following for a point a small distance from a positively charged sphere: Applications (a) (b) the magnitude and direction of the electric field at this point the electric potential energy of another small positive charge placed at this point (c) the electric potential at this point 18. Describe the relationship between the work done in moving a charge from one region in an electric field to another region, and the energy gained by the charge. What law governs this relationship? 29. An insulator and a conductor are each contacted by a negatively charged rubber rod. Describe the distribution of charge on each object. 30. Describe how Earth\u2019s surface can become positively charged during a thunderstorm. In your description, include the terms \u201ccharging by friction\u201d and \u201ccharging by conduction.\u201d 31. Assume you have only a negatively charged ebonite rod. Describe a procedure for charging a neutral electroscope CHAPTER 12 (a) positively (b) negatively 19. Draw a diagram of the magnetic field around a bar magnet and Earth. (a) List the similarities between the two fields. (b) List the differences between the two fields. 20. Explain why a single magnetic pole cannot exist by itself. 21. Describe two simple demonstration techniques used to outline the magnetic field around a magnet. 32. Why does a negatively charged ebonite rod initially attract a piece of thread and then eventually push it away? 33. When you touch a Van de Graaff generator, your hair stands on end. Explain your answers to the following questions: (a) Are you being charged by friction, conduction, or induction? (b) Will the same effect occur if you are grounded? Unit VI Forces and Fields 627 12-PearsonPhys30-Chap12 7/24/08 3:35 PM Page 628 34. A negatively charged ebonite rod is brought 43. A football-shaped hollow conducting object is near a neutral pith ball hanging on an insulating thread. Describe what happens to the charges in the pith ball and the resulting effect (a) before the rod and the pith ball touch (b) after the rod and the pith ball touch 35. A straight length of conducting wire, lying horizontal to the surface of Earth, delivers a current in a direction from south to north. Describe the deflection of a compass needle held directly over the conducting wire. 36. A negatively charged foam plastic ball is hanging from an insulated thread. Another negatively charged ball is brought near, on the same horizontal plane, and the hanging ball swings away so that the supporting thread makes an angle of 30\u00b0 to the vertical. (a) Draw a free-body diagram depicting the tension force by the string, the force of gravitational attraction, and the electrical force of repulsion on the hanging ball. If the system is in equilibrium, identify the force that balances the gravitational force on the hanging ball and the force that balances the electrostatic force of repulsion on the hanging ball. (b) 37. Explain how the properties of selenium are essential in the operation of a photocopier. 38. Explain the difference between charge shift and charge migration during the process of charging by induction. 39. The cell membrane of a neuron may be thought of as charged parallel plates. The electric potential difference between", " the outside and the inside of the membrane is about 0.70 V. If the thickness of the membrane is 5.0 109 m, calculate (a) the magnitude of the electric field between the outside and the inside of the membrane (b) the amount of work necessary to move a single sodium ion, Na1, with a charge of 1.6 1019 C, across the membrane 40. In a chart, compare the similarities and differences between Newton\u2019s law of universal gravitation and Coulomb\u2019s law of electrostatics. 41. Determine the distance between two electrons if the mutual force of repulsion acting on them is 3.50 1011 N. 42. A neutral small hollow metal sphere is touched to another metal sphere with a charge of 3.00 102 C. If the two charges are then placed 0.200 m apart, calculate the electrostatic force acting on the two spheres after they touch. 628 Unit VI Forces and Fields charged negatively. (a) Draw the object and then draw the charge distribution on the surface of the object. (b) Draw the electric field lines surrounding the object. (c) Where will the intensity of the electric field appear to be the greatest? (d) Explain how this effect can be used to describe the operation of a lightning rod. 44. A car with a vertical antenna is driven in an easterly direction along the equator of Earth. (a) Describe how a current is induced in the antenna. (b) Determine the direction of the induced current. 45. A small charge of 2.0 C experiences an electric force of 3.0 105 N to the left when it is placed in the electric field of another larger source charge. Determine the strength of the electric field at this point. 46. Calculate the kinetic energy gained by a proton that is allowed to move between two charged parallel plates with a potential difference of 2.0 104 V. What maximum speed could the proton acquire? 47. Two spheres with charges of 4.00 C and 3.00 C are placed 0.500 m apart. At what point between the two charges must a third charge of 2.00 C be placed so that the net electrostatic force acting on this charge is zero? 48. Two oppositely charged parallel plates have a voltage of 2.5 104 V between them. If 1.24 J of work is required to move a small charge from one plate to the other, calculate the magnitude of the charge. 49. A wire that is 0.30 m long, lying perpendicular to an external magnetic field of magnitude 0.50 T, experiences a magnetic force of 0.11 N. Determine the current in the wire. 50. A charge of 3.0 106 C is placed 0.50 m to the right of another charge of 1.5 105 C. Determine (a) (b) the net electric field at a point midway between the two charges the electrostatic force of attraction acting on the two charges 12-PearsonPhys30-Chap12 7/24/08 3:35 PM Page 629 51. Touching a Van de Graaff generator can result in a painful shock. If a small length of conducting wire is bent into an L-shape and taped to the ball of the generator, so that a shaft of wire projects outward, you will not get a shock when you touch the ball. (a) How does the ball of the generator initially acquire a negative charge? (b) What process of charging objects is involved in the painful shock you receive initially? (c) How does the shaft of wire that is taped to the ball prevent a shock? (d) What is the function of the conducting strips projecting from the airplane wing tip shown in the photo below? 52. A small foam plastic ball with a mass of 0.015 kg and coated with a conducting material is suspended by a string 0.75 m long. If the ball is initially given a charge of magnitude 1.5 108 C, and another charged ball with a charge of magnitude 2.5 108 C is brought near, the hanging ball swings to a position 1.0 cm from its equilibrium position. Calculate the electrostatic force of repulsion acting on the two charges. Draw a free-body diagram. 53. The largest electric field that can exist between two oppositely charged plates with an air gap between them is 3.00 106 N/C. If this electric field limit is exceeded, then a discharge of charge occurs between the plates, resulting in a spark. If the voltage between the plates is 500 V, what is the minimum distance between the plates before a spark occurs? 54. An electron with a mass of 9.11 1031 kg and a charge of magnitude 1.60 1019 C is 5.29 1011 m from a proton with a mass of 1.67 1027 kg and a charge of magnitude 1.60 1019 C. Calculate (a) the gravitational force", " of attraction between the two masses (b) the electrical force of attraction between the two charges (c) how many times greater the electrical force is than the gravitational force 55. An alpha particle, a proton, and an electron, travelling at the same speed, enter regions with external fields. Compare the motion of these particles as they travel perpendicularly through the same (a) magnetic field (b) electric field (c) gravitational field 56. Given ebonite and glass rods and strips of fur and silk, describe the procedure to charge an electroscope negatively by (a) conduction (b) induction (include the grounding step) 57. A particle with a mass of 2.00 1026 kg and a charge of magnitude 6.40 1019 C is fired horizontally along the surface of Earth in an easterly direction. If the particle passes through a metal detector with a magnetic field of 2.00 104 T in a northerly direction, at what speed must the charged particle be travelling so that the magnetic force counteracts the gravitational force of Earth on the particle at this position? What is the charge of the particle? 58. A helium ion (He2) with a charge of 3.20 1019 C and a mass of 6.65 1027 kg is injected with a speed of 3.00 106 m/s perpendicularly into a region with a uniform magnetic field of 3.30 T. (a) Calculate the magnitude of the deflecting force on the helium ion. (b) If this deflecting force causes the helium ion to travel in a circular arc, what is the radius of the arc? 59. Sphere A, with a charge of 3.50 C, is 4.35 cm to the left of sphere B, with a charge of 2.44 C. Calculate the net electrostatic force on a third sphere C, with a charge of 1.00 C, if this sphere is placed (a) midway on a line joining charges A and B (b) at a point 2.50 cm to the right of charge B (c) at a point 2.50 cm directly down from charge B 60. Calculate the potential difference required to accelerate a deuteron with a mass of 3.3 1027 kg and a charge of magnitude 1.6 1019 C from rest to a speed of 8.0 105 m/s. Unit VI Forces and Fields 629 12-PearsonPhys30-Chap12 7/24/08 3:35 PM Page 630 61. A uniform electric field of 7.81 106 N/C exists between two oppositely charged parallel plates, separated by a distance of 3.20 mm. An electron, initially at rest and with a mass of 9.11 1031 kg, is injected into the electric field near the negative plate and accelerates toward the positive plate. The electron passes through a \u201chole\u201d in the positive plate and then travels perpendicularly through an external magnetic field with a magnitude of 1.50 T. (a) What is the voltage between the oppositely charged plates? (b) Calculate the maximum speed acquired by the electron between the plates. Ignore relativistic effects. (c) Determine the magnetic force on the electron as it passes through the magnetic field. (d) Describe the motion of the electron through the electric field and then through the magnetic field. 62. A bar magnet is moved toward a coil of wire that is connected to a sensitive galvanometer, as shown in the figure below. Extensions 63. Explain why computer hard disks are encased in metal. 64. Explain why it is safe to remain inside a vehicle during a lightning storm. 65. Are gravitational, electrical, or magnetic forces responsible for the formation of a black hole in space? Explain your answer. 66. Why must technicians who work on very sensitive electronic equipment be grounded? 67. Describe why Earth\u2019s magnetic field creates a \u201cmagnetic bottle.\u201d 68. Charged particles in cosmic rays from the Sun are trapped in Earth\u2019s magnetic field in two major radiation belts, called Van Allen radiation belts. The first belt is about 25 500 km and the other is about 12 500 km above the surface of Earth. Explain why spacecraft must avoid these radiation belts. 69. Compare the motion of a negatively charged particle as it travels through a gravitational, electric, or magnetic field in a direction (a) perpendicular to the field N S (b) parallel to and in the same direction as the field G (a) Explain what happens to the galvanometer readings as the north pole of the bar magnet approaches the coil of wire. (b) Describe how Lenz\u2019s law influences the movement of the magnet toward the coil of wire. (c) Explain what happens to the galvanometer readings as the north pole of the magnet is pulled out of the coil of wire. (d) Describe how Lenz\u2019s law influences the movement of the magnet away from the coil of", " wire. 70. Why is lightning more likely to strike pointed objects than rounded objects on the ground? 71. Describe the process by which the steel beams in a high-rise building can become magnetized. Why does this effect not happen in homes built with wood beams? 72. High-voltage power lines operate with voltages as high as a million volts. Explain why a bird can perch on a power line with no effect, but must be careful not to touch two nearby power lines. 73. Explain the difference between the dip angle and the angle of declination of Earth\u2019s magnetic field. How do these two angles affect the operation of a directional compass in Alberta? 630 Unit VI Forces and Fields 12-PearsonPhys30-Chap12 7/24/08 3:35 PM Page 631 74. An electron is at rest. Can this electron be set in motion by a magnetic field? Explain your answer. 75. Given only a loop of wire connected to a sensitive galvanometer, describe a simple experiment that could be conducted to prove the existence of a changing magnetic field near a high-voltage power line. 76. The image of the magnetotactic bacterium, as seen under a microscope, reveals a row of magnetite crystals within its cellular structure. Make a hypothesis regarding the purpose of this row of magnetite crystals, and design and describe an experiment to test your hypothesis. (a) Plot a graph to determine the relationship between \u23d0B\u23d0and r. (b) What is the relationship between \u23d0B\u23d0and r? (c) What quantities need to be graphed in order to straighten the graph? (d) Complete a new table of values to straighten the graph. (e) Plot a new graph with the variables needed to straighten the graph. (f) Use the graph-slope method and the data from the graph to determine the value of. (g) Use the formula-data substitution method to determine the value of. Skills Practice 77. Draw a Venn diagram to review the similarities and differences between electric and magnetic fields. 78. Design an experiment to determine the direction of gravitational, magnetic, and electric fields. 79. Construct a concept map for solving a two- dimensional electrostatic force problem involving three charges at the corners of a triangle. 80. Design an experiment to show that electrostatic forces vary with the inverse square law. Self-assessment 82. Describe to a classmate which field concepts and laws you found most interesting when studying this unit. Give reasons for your choices. 83. Identify one topic pertaining to fields studied in this unit that you would like to investigate in greater detail. 84. What concept in this unit did you find most difficult? What steps could you take to improve your understanding? 81. A current-carrying wire has a magnetic field 85. Assess how well you are able to explain electric potential energy and electric potential. Explain to a classmate how you determine a reference point. e TEST To check your understanding of forces and fields, follow the eTest links at www.pearsoned.ca/school/physicssource. surrounding it. The strength of this magnetic I field can be found using the formula \u23d0B\u23d0, r 2 where \u23d0B\u23d0is the magnitude of the magnetic field surrounding the wire, in teslas, is the constant of permeability for space, in Tm/A, I is the current in the wire, in amperes, and r is the distance to the wire, in metres. An experiment is performed to establish the value of the constant,. The current was constant through the wire at 5.00 A, and the strength of the magnetic field was measured at various distances from the wire. The data recorded are given in the table below. Distance from the wire, r (m) Magnitude of the magnetic field, |B| (T) 0.100 0.200 0.300 0.400 0.500 6.28 3.14 2.09 1.57 1.26 Unit VI Forces and Fields 631 13-PearsonPhys30-Chap13 7/24/08 3:43 PM Page 632 U N I T VII Electromagnetic Electromagnetic Radiation Radiation (a) (b) (c) This series of photos of a supernova remnant shows the various types of electromagnetic radiation that are being emitted from the supernova remnant. The large image is a composite. Images (a) and (b) were taken by the Chandra space telescope (high-energy and low-energy X ray), image (c) was taken by the Hubble space telescope (visible part of the spectrum), and image (d) by the Spitzer space telescope (infrared). 632 Unit VII 13-PearsonPhys30-Chap13 7/24/08 3:43 PM Page 633 (d) Unit at a Glance C H A", " P T E R 1 3 The wave model can be used to describe the characteristics of electromagnetic radiation. 13.1 What Is Electromagnetic Radiation? 13.2 The Speed of Electromagnetic Radiation 13.3 Reflection 13.4 Refraction 13.5 Diffraction and Interference C H A P T E R 1 4 The wave-particle duality reminds us that sometimes truth really is stranger than fiction! 14.1 The Birth of the Quantum 14.2 The Photoelectric Effect 14.3 The Compton Effect 14.4 Matter Waves and the Power of Symmetric Thinking 14.5 Coming to Terms with Wave-Particle Duality and the Birth of Quantum Mechanics Unit Themes and Emphases \u2022 Diversity and Matter \u2022 Nature of Science \u2022 Scientific Inquiry Focussing Questions The study of electromagnetic radiation and its behaviour requires interpretation of evidence to form theories and models. As you study this unit, consider these questions: \u2022 What roles do electricity and magnetism play in electromagnetic radiation? \u2022 Does electromagnetic radiation have a wave or a particle nature? \u2022 What experimental evidence is required to decide whether electromagnetic radiation has a wave or a particle nature? Unit Project From Particle to Quantum \u2022 When you have finished this unit you will be able to use experimental evidence concerning electromagnetic radiation to describe our current understanding of light. You will be able to craft a multi-media presentation of highlights in the development of this understanding. Unit VII Electromagnetic Radiation 633 13-PearsonPhys30-Chap13 7/24/08 3:43 PM Page 634 C H A P T E R 13 Key Concepts In this chapter, you will learn about speed and propagation of electromagnetic radiation reflection, refraction, diffraction, and interference total internal reflection and Snell\u2019s Law Learning Outcomes When you have completed this chapter, you will be able to: Knowledge describe how accelerating charges produce EMR explain the electric and magnetic fields associated with the propagation of EMR explain how the speed of EMR can be investigated and calculated describe reflection, refraction, total internal reflection, and simple optical systems describe diffraction, interference, and polarization and how this supports the wave model of light Science, Technology, and Society explain that scientific knowledge may lead to the development of new technologies and new technologies may lead to scientific discovery explain that scientific knowledge is subject to change as new evidence is found and as laws and theories are tested and subsequently restricted, revised, or reinforced based on new findings and new technologies 634 Unit VII The wave model can be used to describe the characteristics of electromagnetic radiation. Figure 13.1 A Fresnel lens captures light from a lamp and redirects it into a concentrated beam. This 19th-century lens technology used many prisms to produce a light that could be seen for great distances. When you open your eyes in the morning, the first thing you see is light. In fact, it is the only thing you see. In general, what we perceive with our eyes is a combination of many colours, each with varying brightness, but all coming together through the optical system of our eye. It is light or, more broadly speaking, a form of electromagnetic radiation (EMR), which produces an image in the human eye. But what is EMR? How is it produced? How is it transmitted and at what speed? These questions are historically significant, guiding the direction of research and debate over the ages. Finding answers has helped us understand the fundamental principles of our universe and has enabled advances in technology in areas such as lenses (Figure 13.1), fibre optics, and digital devices. Radios, lasers, global positioning systems, and compact discs (CDs) are examples of devices that depend on an understanding of EMR. The nature of light has long been a topic of intrigue and debate. Early Greek scientific thought was based on the work of Aristotle and Euclid, who concerned themselves with the physical and geometric mechanisms of visual perception. A more modern debate ensued as detailed evidence began to be collected about the wave and/or particle nature of light. Isaac Newton (1642\u20131727) and Christiaan Huygens (1629\u20131695) defined this debate with evidence from early experiments. Newton put forward the particle or corpuscular theory, while Huygens supported the wave model. Today, both the particle and wave models of light have some validity and contribute to our present understanding, described by the quantum model of light. 13-PearsonPhys30-Chap13 7/24/08 3:43 PM Page 635 In this chapter, you will see that visible light is only one form of EMR. You will also discover that the concepts of electrical charge and magnetic fields, explored in Chapters 10\u201312, come together when you consider EMR. You will investigate the generation, speed, and propagation of EMR, and see how light can be reflected, refracted, diffracted, and polarized. In each instance, the wave and particle nature of light is revealed, helping us to understand these complex phenomena and apply", " them in new technologies. 13-1 QuickLab 13-1 QuickLab Investigating Light as a Form of EMR Problem How can the properties of light be revealed? Materials light source 2 polarizing filters Procedure 1 Place a light source in front of one polarizing filter and observe whether the light is able to penetrate the filter.??? Figure 13.2 Think About It 2 Turn the filter 90 degrees as shown in Figure 13.2 and observe whether the light is able to penetrate the filter. 3 Place the filters at 90 degrees to one another and observe whether the light is able to penetrate the filters. 4 Slowly rotate one of the filters and observe whether the light is able to penetrate both filters at a variety of angles and positions. Questions 1. Can visible light pass through a single polarizing filter, regardless of how it is oriented? 2. Can visible light pass through two filters when they are aligned at 90 degrees to one another? 3. Describe one characteristic of EMR indicated by your observations of light in this QuickLab. 4. Do the observations made in this investigation explain what EMR is, or do they simply reveal one characteristic of EMR? 1. Do your observations from 13-1 QuickLab give information on whether electromagnetic radiation has wave or particle characteristics? Explain. 2. What experimental evidence would be needed to decide whether electromagnetic radiation has a wave or a particle nature? Discuss your answers in a small group and record them for later reference. As you complete each section of this chapter, review your answers to these questions. Note any changes in your ideas. Chapter 13 The wave model can be used to describe the characteristics of electromagnetic radiation. 635 13-PearsonPhys30-Chap13 7/24/08 3:43 PM Page 636 electromagnetic radiation: radiant energy in the form of a wave produced by the acceleration of electrons or other charged particles. EMR does not require a material medium; can travel through a vacuum. frequency: the number of cycles per unit of time wavelength: the distance between adjacent points that vibrate in phase with one another in a wave info BIT Seeing is a photochemical process that is sensitive to certain wavelengths of electromagnetic radiation. When EMR is absorbed by the tissues in the human eye, a compound called retinal changes in physical form from bent to straight. The retinal molecule, in turn, is connected to a membrane-bound protein called opsin forming the complex molecule called rhodopsin. When the retinal molecule changes its form, it separates from the rhodopsin and the opsin triggers a nerve cell to signal the brain that light has been seen. 13.1 What Is Electromagnetic Radiation? Electromagnetic radiation (EMR) is radiant energy, energy that travels outward in all directions from its source. There are different types of EMR, some very familiar, and others rarely mentioned outside scientific discussions. Electromagnetic radiation includes AM/FM radio waves, microwaves, heat, visible light (red to violet), ultraviolet radiation, X rays, and gamma rays. Electromagnetic radiation types are identified based on their frequency, wavelength, and source. The energy sources that produce EMR vary greatly, from nuclear reactions in the Sun, which generate gamma radiation, to chemical reactions in the human body that generate infrared radiation (heat). EMR is produced by the acceleration of charged particles resulting in transverse waves of changing electric and magnetic fields that can travel through space without the need of a material medium. All forms of EMR travel at the same speed, commonly referred to as the speed of light (c, in a vacuum), equal to 3.00 108 m/s. EMR does not always travel directly from the source to the observer. As a result, it can be observed in one of two ways: directly from the source indirectly, as reflected or transmitted radiation Burning magnesium (Figure 13.3) allows both methods of observing radiation. Other objects, such as the text you are reading now, do not produce radiation. You are able to read the text because it is reflecting radiation (light) from another source. If you are wearing glasses, this radiation is transmitted through your lenses. The evidence obtained in 13-1 QuickLab indicates one unique characteristic of electromagnetic radiation. However, visible light, radiant energy that the eye can detect, only makes up a small portion of the spectrum of EMR present in our universe. Concept Check Figure 13.3 When magnesium burns, you can observe radiation directly from the reaction and indirectly from the radiation that is reflected from the smoke. Can infrared radiation reflect off objects and be observed by the human body? Explain and give examples. 636 Unit VII Electromagnetic Radiation 13-PearsonPhys30-Chap13 7/24/08 3:43 PM Page 637 electromagnetic spectrum: all types of EMR considered in terms of frequency, wavelength, or energy e WEB To learn more about the different types of EMR and the trends seen across the spectrum, follow the links at www.pearsoned", ".ca/school/ physicssource. Types of Electromagnetic Radiation The electromagnetic spectrum is the term applied to all the types of EMR considered together in terms of frequency, wavelength, or energy. All parts of the spectrum are found, with varying intensity, in our natural environment. We are most familiar with the visible spectrum since we sense it directly with our eyes. The infrared spectrum is sensed as heat and the ultraviolet spectrum includes radiation that can damage living cells, often causing a physiological response such as sunburn. Other parts of the spectrum may be present as background radiation. Natural background radiation originates from two primary sources: cosmic radiation and terrestrial sources. Cosmic radiation from deep space interacts with atoms in the atmosphere producing X rays and unstable isotopes. Terrestrial sources of gamma radiation include radioactive isotopes of uranium, radon, potassium, and carbon. Also, aboveground nuclear tests of the 1940s\u20131960s combined with nuclear accidents such as Chernoybl have scattered a substantial amount of radioactive material within our environment. Figure 13.4 shows the overlap of frequencies and wavelengths among the different forms of EMR in the electromagnetic radiation spectrum. 104 108 1012 1016 1020 1024 frequency f (Hz) radio waves microwaves infrared visible light ultraviolet X rays gamma rays 3 x 104 m 3 m 3 x 10\u20134 m 3 x 10\u20138 m 3 x 10\u201312 m wavelength \u03bb (m) a) Electromagnetic spectrum 800 nm 700 nm 600 nm 500 nm 400 nm wavelength \u03bb (m) infrared red yellow blue b) Visible spectrum orange green violet ultraviolet Figure 13.4 The electromagnetic radiation spectrum showing the visible range The relative energy of the different types of EMR varies with frequency across the spectrum. Table 13.1 compares the various sources and characteristics of the radiation found in the EMR spectrum. e WEB To learn more about the medical uses of different types of EMR, follow the links at www.pearsoned.ca/school/ physicssource. Chapter 13 The wave model can be used to describe the characteristics of electromagnetic radiation. 637 13-PearsonPhys30-Chap13 7/24/08 3:43 PM Page 638 Table 13.1 The Electromagnetic Spectrum: Characteristics Method of Production Characteristics Problems oscillation of electrons in an electric circuit like an antenna oscillation of electrons in special tubes and solid state devices motion of particles, transitions of valence electrons in atoms and molecules higher-energy transitions involving valence electrons in atoms long wavelength allows a large amount of diffraction making it useful for longdistance communication, e.g., PC broadband shorter wavelength reduces diffraction for short-distance communication; frequency matches the natural resonant frequency of water molecules; used in microwave ovens and cell phones causes object absorbing it to become warm; used for remote sensing, night vision scopes, and identification of sources of heat reflects off small objects, making them visible; diffracts around very small objects, making them invisible requires government regulations to control transmission and avoid interference may be linked to some forms of cancer; causes damage to living tissue due to heating of water molecules within tissues significant exposure can burn tissue limits the size of objects that can be seen even higher-energy transitions involving valence electrons in atoms easily absorbed by objects; causes fluorescence of some materials, tanning in humans; kills bacteria may cause sunburn; prolonged exposure can cause mutations and cancer in humans Type of Electromagnetic Radiation Radio and Radar f 104 1010 Hz 104 102 m relative energy: very low Microwaves f 109 1012 Hz 101 104 m relative energy: low Infrared f 1011 4.0 1014 Hz 103 7.5 107 m relative energy: low Visible f 4.0 1014 7.5 1014 Hz 7.5 107 4.0 107 m relative energy: medium Ultraviolet f 7.5 1014 1017 Hz 4.0 107 109 m relative energy: high X ray f 1017 1020 Hz 109 1012 m relative energy: very high penetrates most matter and is absorbed by denser material (like bone or metal); destroys carcinogenic or mutant cells; used for medical imaging in humans and in industry penetrates matter very deeply; destroys carcinogenic or mutant cells on a local scale; used to probe the structure of matter and in industrial imaging penetrates matter very deeply; study of cosmic rays allows investigators to formulate ideas about the universe can cause mutations and cancer in humans can cause radiation sickness and death can cause radiation sickness and death transitions of electrons in an atom or the sudden acceleration of high-energy free electrons Gamma f 1019 1024 Hz 1011 1016 m relative energy: extremely high decomposition of unstable nuclei, either spontaneously or by the sudden negative accelerations from highenergy particle accelerators Cosmic f 1024 Hz and greater 1016 m and less relative energy: extremely high bombardment of Earth\u2019s atmosphere by extremely high-energy particles from space 638 Unit VII Electromagnetic Radiation 13-PearsonPhys30-Chap13 7/24/", "08 3:44 PM Page 639 Competing Models of Electromagnetic Radiation Historically, investigators have tried to explain transmission, reflection, refraction, and absorption, and the other characteristics common to all types of EMR by using models. The historical particle model describes EMR as a stream of tiny particles radiating outward from a source. A particle is a discrete unit of matter having mass, momentum (and thus kinetic energy), and the ability to carry an electric charge. The particle model of EMR is the simplest of all the models. It is supported by the facts that EMR propagates in straight lines, can be reflected, and can be absorbed. The pool ball (particle) in Figure 13.5 exhibits particle characteristics. It can travel in a straight line, obey the law of reflection when it bounces off a side rail, and be absorbed by the table when it drops into a pocket. info BIT The Particle Theory of Light, also known as the Corpuscular Theory of Light, was put forward in Newton\u2019s Opticks, published in 1704. particle model: describes EMR as a stream of tiny particles radiating out from a source wave model: describes EMR as a stream of transverse waves radiating out from a source Figure 13.5 A pool ball on a pool table exhibits the particle characteristics of linear movement, reflection, and absorption. Figure 13.6 A water wave is a transverse wave that transfers energy in the form of a disturbance. waterbed sponge A second model, the wave model, describes EMR as a stream of transverse waves radiating outward from a source. As you learned in Chapter 8, a wave is a transfer of energy in the form of a disturbance. The energy transfer usually occurs in, but is not limited to, a material medium like water. A water wave travels in a straight line, reflects from surfaces, and can be absorbed (Figure 13.6). For example, water waves, which continually reflect inside a waterbed, transfer unwanted energy that can be absorbed by a sponge (Figure 13.7). EMR, however, does not require the presence of a medium. The modern wave model of light has its origins in the 17th century when the Dutch mathematician and scientist Christiaan Huygens argued that light consisted of waves. He suggested light waves could interfere to produce a wave front, travelling outward in a straight line. At the time, however, Huygens\u2019 wave theory was overshadowed by the immense scientific popularity of Newton\u2019s particle theories. The wave model gained further support from evidence produced by the work of Thomas Young (Figure 13.8) and his now-famous two-slit experiment of 1801. The details and implication of the experiment are set out in section 13.5, but in general, Young was able to show that a beam of light, when split into two beams and then recombined, shows interference effects that can only be explained by assuming that light has wavelike properties. Figure 13.7 Waves in a waterbed exhibit the wave characteristics of linear movement, reflection, and absorption. Figure 13.8 Thomas Young (1773\u20131829) Chapter 13 The wave model can be used to describe the characteristics of electromagnetic radiation. 639 13-PearsonPhys30-Chap13 7/24/08 3:44 PM Page 640 e WEB To learn more about Young\u2019s experiment and its relationship to the models of EMR, follow the links at www.pearsoned.ca/school/ physicssource. If light behaves as a particle, then one would expect to observe a bright region wherever the light emanating from the slits reaches a distant screen (Figure 13.9(a)). But Young obtained a result similar to that shown in Figure 13.9(b), where a pattern of light of varying intensity is observed after the light passes through the two slits. This pattern is similar to two-source interference patterns in water waves, as seen in section 8.3. Young\u2019s experiment thus presented strong evidence for the wave model of light. schematic representation of particle theory prediction schematic representation of wave theory prediction (a) (b) screen screen Figure 13.9 When light is incident on two small slits, it is diffracted as it passes through each slit and an interference pattern is observed that supports the wave model of light. a) A schematic representation of the particle theory prediction b) A schematic representation of the wave theory prediction Planck and Einstein By the end of the 19th century, both the particle and wave models of EMR were supported by scientific evidence. In 1900, Max Planck (1858\u20131947) proposed a radically new model to explain the spectrum of radiation emitted from a perfectly black object. In a mathematical derivation Planck assumed that all of the vibrating molecules, \u201coscillators,\u201d in the black body could vibrate with only specific, discrete amounts of energy. In doing so, he had to ignore", " the continuous distribution of energy of classical physics and introduce the concept of quanta, or discrete packets of energy. Planck was awarded the Nobel Prize for Physics in 1918, \u201cin recognition of the services he rendered to the advancement of Physics by his discovery of energy quanta.\u201d In 1905 Einstein extended this quantum theory by proposing that light is emitted in quantized, tiny, massless particles, which are now called photons. Planck\u2019s original theory thus evolved into the currently accepted quantum model of light, which is a combination of both the particle and wave models. This model describes light and all other electromagnetic radiation as discrete bundles or \u201cpackets\u201d of energy. Each packet, or photon, has both particle and wave characteristics. In the quantum model, EMR has two aspects of behaviour, one being info BIT A quantum is a discrete unit of energy. The term originates from the Latin quantus, which means \u201chow much.\u201d \u201cQuanta\u201d is the plural of \u201cquantum.\u201d The Quantum Theory was so revolutionary that its formulation distinguishes the shift from classical physics to what we now call modern physics. photon: (from the Greek word meaning \u201clight\u201d); a discrete packet of energy associated with an electromagnetic field quantum model: light and all other EMR are discrete bundles of energy, each of which has both particle and wave characteristics 640 Unit VII Electromagnetic Radiation 13-PearsonPhys30-Chap13 7/24/08 3:44 PM Page 641 wavelike and the other being particle-like. Quantum theory encompasses these two types of behaviour while, at the same time, challenging both the classical wave and particle models. Planck\u2019s quantum idea challenged the wave model by proposing that EMR does not deliver energy in a continuous form as a wave would, but rather, it delivers energy in small bundles. At the same time, his idea challenged the particle model by limiting the energy of the particles to certain discrete values, a condition not possible for a Newtonian particle. Concept Check Consider the wave model of light and the particle model of light. 1. Which of the following best describes the word \u201cmodel\u201d? (a) simplified description of a complex entity or process; (b) a representation of something on a smaller scale; (c) both a and b. 2. Explain your answer from part 1 and provide several examples of models that are used in a similar way. 3. The EMR spectrum includes many classes of radiation. Describe the wave characteristics used to classify each type of radiation on the spectrum. How are these characteristics related to the energy of the radiation? Maxwell\u2019s Electromagnetic Theory, 1865 In 1865, James Clerk Maxwell (1831\u20131879) proposed his Electromagnetic Theory, which synthesized earlier ideas and theories with the results of experiments to provide a theoretical framework for future studies. Maxwell proposed the idea that a changing electric field produces a changing magnetic field and that the interaction between these fields propagates as a wave through space. In his theory Maxwell linked concepts of electricity and magnetism so that we now call them \u201celectromagnetic.\u201d Maxwell based his theory on key phenomena observed by earlier investigators of electricity and magnetism. You have already met these ideas in Unit 6, but we will restate them here to appreciate the importance of Maxwell\u2019s contribution. The Concept of the Electric Field As Faraday had proposed and as you saw in Chapter 11, an electric field surrounds any charged particle and an electrostatic force will act on another charged particle when it is placed in that field. The electric field strength can be calculated and electric field lines drawn to illustrate the region of influence. Electric field lines begin and end at a charge (Figure 13.10) and the number of field lines at any closed surface is determined by the total charge at the surface. This concept allows the idea of interaction between particles at a distance, even though there is no contact between them. e WEB To learn more about Max Planck, begin your search at www.pearsoned.ca/ school/physicssource. Project LINK How will you present the observations and theoretical concepts that culminated in the ideas of Planck and Einstein? electric field: a three-dimensional region of electrostatic influence surrounding a charged object magnetic field: a threedimensional region of magnetic influence surrounding a magnet Figure 13.10 Electric field lines are used to illustrate the region of influence between two oppositely charged particles. Chapter 13 The wave model can be used to describe the characteristics of electromagnetic radiation. 641 13-PearsonPhys30-Chap13 7/24/08 3:44 PM Page 642 S N Figure 13.11 The number of magnetic field lines that exit a magnetic material is equal to the number of magnetic field lines that enter a magnetic material \u2014 forming a closed loop. capacitor: two conductors, holding equal amounts of opposite charges, placed near one another without touching Maxwell\u2019s Equations", ": a series of equations that summarized the relationships between electricity and magnetism and predicted the existence of electromagnetic waves and their propagation through space info BIT Maxwell was working with a system of units for electromagnetic theory, referred to as the CGS system. This system was derived from the base units of centimetre, gram, and second. The CGS system has largely been replaced by the SI system, based on the metre, kilogram, and second. The Concept of the Magnetic Field A magnetic field (as discussed in Section 12.1) is a three-dimensional region of magnetic influence surrounding a magnet. Magnetic field lines form a closed loop and represent the direction and magnitude of the magnetic field (Figure 13.11). Relationships Between Electricity and Magnetism Electrical current in a conductor produces a magnetic field perpendicular to the current (Oersted, Unit VI, page 587) and the strength of the magnetic field depends on the magnitude of the current (Ampere, Unit VI, page 607). This relationship is known as Ampere\u2019s Law. Inversely, moving a conductor connected in a circuit through a magnetic field induces an electric current. The magnitude of the electrical current is directly related to the rate of change in the magnetic field (Faraday and Henry, Unit VI, page 611). This relationship is known as Faraday\u2019s Law. Putting the Laws of Electromagnetism Together From this basis and through his work with capacitors, Maxwell developed the laws of electromagnetism. Maxwell\u2019s work with capacitors showed that the electric field produced by a capacitor could have the same effect as a moving charge. In other words, a changing electric field will produce a changing magnetic field in the same manner as a changing electric current can produce a changing magnetic field. This information was extremely important because it showed that a conductor is not necessary for an electromagnetic wave to exist. Maxwell put these ideas together with incredible ingenuity using calculus, which is beyond the scope of this text. His theory is embodied in a series of equations known as Maxwell\u2019s Equations. To visualise the relationship between the electric and magnetic fields, we will proceed stepwise. Begin with a plane transverse wave that shows an electric field of varying strength (Figure 13.12). Next, consider two of the electric field lines on the transverse wave, an upward one at one location and the corresponding downward one at another. Imagine a path connecting the tips of these two electric field vectors, as shown by the dotted line in Figure 13.13. You can see that the electric field changes direction continuously from one direction to the opposite direction along a \"closed path.\" E E direction of propagation of wave direction of propagation of wave E Figure 13.12 Plane transverse wave showing the electric field lines. Figure 13.13 A path connecting two oppositely directed electric field lines on the transverse wave shows the change in the electric field. 642 Unit VII Electromagnetic Radiation 13-PearsonPhys30-Chap13 7/24/08 3:44 PM Page 643 Now, consider the magnetic field that exists perpendicular to the electric field along this closed path. When the electric field is set in motion, the magnetic field will change along with the electric field, and similarly along a closed path. An electromagnetic wave will propagate in a direction perpendicular to both fields (Figure 13.14). The changing electric and magnetic fields will propagate, or radiate, through space in the form of a wave \u2014 an electromagnetic wave (Figure 13.15). Maxwell proposed that the electromagnetic wave consists of periodic variations in the electric and magnetic field strengths and that these variations occur at right angles to one another as the wave propagates. B E direction of propagation of wave Figure 13.15 Three-dimensional view of an electromagnetic wave Maxwell\u2019s Predictions Maxwell\u2019s equations not only correctly predicted the existence of electromagnetic waves, but also allowed him to make some predictions about the waves\u2019 properties. 1. Electromagnetic waves are produced whenever an electric charge is accelerating. Therefore, as an electric charge oscillates, electrical energy will be lost, and an equivalent amount of energy will be radiated outward in the form of oscillating electric and magnetic fields. 2. When the electric charge is accelerated in periodic motion, the frequency of oscillation of the charge will correspond exactly to the frequency of the electromagnetic wave that is produced. 3. All electromagnetic waves will travel at a speed of 310 740 000 m/s and obey the universal wave equation (c f) relating speed, frequency, and wavelength. (Note that Maxwell\u2019s theoretical prediction was not far from today\u2019s currently accepted value of 3.00 108 m/s for the speed of light in a vacuum.) 4. The oscillating electric and magnetic fields will always be perpendicular to each other and perpendicular to the direction of propagation of the wave. 5. Electromagnetic waves should show all the phenomena associated with transverse waves: interference, diffraction, refraction, and polarization.", " It is Maxwell\u2019s last prediction that supports the wave model of EMR and relates his predictions to experimental evidence. Interference, diffraction, polarization, and refraction, as they relate to the wave model of EMR, will be explored in sections 13.4 and 13.5. E B direction of propagation of wave Figure 13.14 The magnetic field lines (in red) allow us to visualize the magnetic field that exists perpendicular to both the electric field and the direction of the wave. electromagnetic wave: periodic variation in perpendicular electric and magnetic fields, propagating at right angles to both fields e WEB To learn more about ways of representing the relation between electric and magnetic fields, follow the links at www.pearsoned.ca/school/ physicssource. Compare and contrast the representations you find. Chapter 13 The wave model can be used to describe the characteristics of electromagnetic radiation. 643 13-PearsonPhys30-Chap13 7/24/08 3:44 PM Page 644 Producing Electromagnetic Radiation \u2014 The Story of Accelerating Charge In 1887, the German physicist Heinrich Hertz (1857\u20131894) set up an experiment designed both to produce and to detect EMR. In his experimental apparatus, Hertz used a radiator, consisting of a pair of wires attached to both a high-voltage induction coil and a pair of capacity spheres. The wires were separated by a small gap and, given a sufficient quantity of opposite charge on each wire, a current would oscillate back and forth across the gap at a frequency of 109 Hz. With each oscillation a spark was produced when the moving charge ionized the air molecules on its way from one wire to the other. From Maxwell\u2019s equations Hertz knew theoretically that this rapidly moving electric charge should produce EMR. A short distance away from the radiator a collector plate containing a small loop of wire, the antenna, was observed to detect the effect of EMR. While the radiator was in operation and when the radiator and the antenna were tuned to the same frequency, a spark was observed at the antenna indicating a potential difference and an electric current (Figure 13.16). 3. Electromagnetic waves create electric current in antenna loop; produces small spark in spark gap. 2. Spark produces electromagnetic waves. 1. Induction coil produces high voltage. e WEB To learn more about details of Hertz\u2019s experiment, follow the links at www.pearsoned.ca/school/ physicssource. Figure 13.16 Hertz\u2019s apparatus consisted of a high-voltage induction coil, a radiator that produces sparks, and an antenna loop. Relating Theory and Practice in Hertz\u2019s Experiment The word \u201cchanging\u201d appears a number of times in Maxwell\u2019s original proposal. As Maxwell understood it, a \u201cchanging\u201d electric field was crucial to creating an electromagnetic wave. This is where the induction coil was important in Hertz\u2019s experimental design. The induction coil rapidly changed the electric field across the spark gap. When this electric field reached a sufficiently high value, the electrons in the wire \u201cjumped\u201d from one electrode to the other. As the charge was rapidly transferred, the electric field underwent a rapid change that caused a changing magnetic field, which then caused a changing electric field, and so on. An electromagnetic wave was produced and it radiated outward in all directions. 644 Unit VII Electromagnetic Radiation 13-PearsonPhys30-Chap13 7/24/08 3:44 PM Page 645 In Hertz\u2019s experiment, when the receiver was tuned to the same frequency as the radiator, the induced current flow in the antenna oscillated at a frequency identical to that of the changing electric field in the radiator. This was conclusive evidence that Hertz\u2019s device had indeed produced the EMR that was being observed at his antenna. Furthermore, he was able to measure the velocity of the waves by using a zinc reflector plate to produce a standing wave and moving a ring antenna within the standing wave. He could determine the magnitude and direction of the wave\u2019s components and therefore the wavelength. Given the frequency of the radiator, the velocity of the wave could be calculated using the universal wave equation (v f). Hertz had produced and measured the wavelength and velocity of non-visible EMR for the first time. Two years later, in 1889, radio pioneers used this method to transmit the first radio waves across the English Channel. By 1901, the first radio waves were transmitted across the Atlantic Ocean from Cornwall, England to St. John\u2019s, Newfoundland and a new age of technology had dawned. The production of EMR was one of the greatest scientific achievements of the 19th century, ushering in new possibilities and technologies that are commonplace today. M I N D S O N Going Wireless e WEB To learn more about Marconi and details of the first trans-Atlantic radio transmission, follow the links at www.", "pearsoned.ca/school/ physicssource. Wireless communication systems that transmit data are commonplace today. 1. Create a list of wireless data transmission technologies. 2. What process must be common to all wireless transmission devices? 3. What variable is used to control the acceleration, and hence the energy, of the accelerating electrons in a wireless device? 4. One radio transmission tower transmits at 5.0 kV. Another transmits at 10.0 kV. Which tower has a greater signal strength and why? 5. Cell phones communicate with cell towers that are dispersed throughout urban cities and transportation corridors. Explain the relationship between the distance that separates a cell phone from the nearest cell tower and the operating power of the cell phone. 6. Many studies have been conducted to investigate and determine if there is a relationship between brain cancer and cell phone usage. Why would one suspect a relationship between cell phone use and brain cancer? Although modern radio signals are generated without a spark, the technology operates in a way similar to Hertz\u2019s original experiment. Radio waves are generated by rapidly changing the electric potential, or voltage, in the radiator tower. The oscillating voltage produces an oscillating electric field in the radiator tower (Figure 13.17). microphone radio signal (radio frequency) RF amp amp sound waves audio signal (audio frequency) AF mixer radiator tower Figure 13.17 Schematic of a simple radio transmitter audio wave (AF) carrier wave (RF) amplitude modulation (AM) Figure 13.18 Amplitude modulation. The audio signal and the carrier signal are mixed by modifying the amplitude of the carrier signal. audio wave (AF) carrier wave (RF) frequency modulation (FM) Figure 13.19 Frequency modulation. The signal is combined with a carrier wave to create a resultant wave with constant amplitude, but varying frequency. Chapter 13 The wave model can be used to describe the characteristics of electromagnetic radiation. 645 13-PearsonPhys30-Chap13 7/24/08 3:44 PM Page 646 THEN, NOW, AND FUTURE From Analog Radio to Digital Wireless Analog Radio Technology The first application of Hertz\u2019s discovery was made by Guglielmo Marconi (1874\u20131937). He recognized the potential for transmitting information using electromagnetic waves, by coding the information into dots and dashes, like the Morse code already used in telegraphy. Today, the dots and dashes have been replaced by an analog signal, which uses a continuous spectrum of values. In radio, sound waves in the range of 20\u201320 000 Hz (audible to humans) are converted into a weak electrical signal in an analog form. This wave is called the audio frequency signal (AF). The weak audio frequency is then amplified and sent to a mixer. In the mixer, a radio frequency (RF) called a carrier signal is added to the audio signal to produce an analog wave that contains two sets of information: the audio signal and the carrier signal. The carrier signal frequency is determined by the government and is unique to that station for that broadcast area. It is the carrier signal that you select when you tune your radio to the station. Finally, the mixed signal is amplified and delivered to the radiator tower, where the movement of electrons through the radiator wire will produce the corresponding form of electromagnetic wave (Figure 13.17). Television signals follow the same process, except that the video and audio portions are transmitted separately. Radio reception is accomplished by tuning the radio to the carrier signal and then removing, or demodulating, the wave. This reveals the audio signal, which is amplified and delivered to the speakers. AM or FM Radio? The mixing of the audio signal and the carrier signal occurs in one of two ways. The first is called amplitude modulation or AM (Figure 13.18). The second method is called frequency modulation or FM. The audio signal and the carrier signal are mixed by modifying the frequency of the carrier signal (Figure 13.19). Figure 13.20 Digital Wireless Cell Phones The cell phone and text messaging technology of today operates on a very different set of principles. When a cell phone is switched on, it registers with the mobile telephone exchange using a unique digital identifier. Once it has been identified, it stays connected to a cellular network of ground-based stations, commonly referred to as cellular tower base stations, which can be found on radio towers, buildings, in church steeples, or any location which is free of physinterference. ical obstacles and Figure 13.20 shows one such cellular tower base station. A cell phone communicates with the cell tower base station using radio waves. In turn the base station relays all information to and from the cell phone to another subscriber of the same cell phone network or through an interconnected public switched telephone network that uses fibre optic and copper land lines to transmit the data. When in use, a cell phone converts the analog voice signal of the person talking into a stream of digital data, which is received by the nearest cell tower base station that may be anywhere from 0", ".8 \u201313 km away. Each cell tower site has a lowpower microwave broadcast which can be picked up by cell phones. As the user moves, the cell phone constantly monitors the signals being received from various nearby cell tower stations, switching from one tower to another in such a way that the signal strength remains as high as possible. The stream of digital data is unique to each type of network technology and each operator is assigned a different radio frequency so that several networks can operate simultaneously at the same location and they will not interfere with one another. Wireless Technology for the Future Future uses of cellular phone technology are emerging in the market place as the networks evolve and consume broadcast frequencies formerly occupied by some television channels. New entertainment features will include: streaming video; music downloads; podcasts; smooth speech recognition and language translations; ebook features that use projection; barcode readers that direct users to Internet urls; global positioning systems and mapping combined with accelerometers to measure position and movements of the cell phone user. Questions 1. Suggest why cell phone user data, including geographical position and movement, could be considered an invasion of privacy. 2. How can multiple cell phone networks and traditional radio broadcasts all exist in the same geographical location and not interfere with one another? 3. Compare the carrier frequency of radio broadcasts with the unique digital identifier of a cell phone. What common purpose do they both serve? 4. Explain why a cell phone only works in certain areas and how changes in the signal strength will be observed by the cell phone user as he or she moves. 646 Unit VII Electromagnetic Radiation 13-PearsonPhys30-Chap13 7/24/08 3:44 PM Page 647 Concept Check 1. Consider two electric field lines on a transverse electromagnetic wave. Explain how a changing electric field will produce a changing magnetic field. 2. Why can electromagnetic radiation propagate through empty space, in the absence of a conductor? 3. An electromagnetic wave is produced if an electric charge accelerates. Explain how these two phenomena are linked. 13.1 Check and Reflect 13.1 Check and Reflect Knowledge 1. List two ways that electromagnetic radiation is detected. 2. List the different types of EMR in the spectrum and the ways in which they are classified. 3. Define EMR using the (a) particle model (b) wave model (c) quantum model 4. Describe the evidence that Thomas Young obtained from the two-slit experiment and explain how it supported the wave model of light. 10. If the magnetic field lines held within a closed path are constant and unchanging, will there be an electric field along the closed path? Explain. 11. If an electric charge is not moving, will it produce a magnetic field? Explain. 12. Explain why it is necessary for the technician to leave the room or wear protective clothing when a patient is being X rayed. 13. Describe how an antenna is affected by an electromagnetic wave. 14. Describe one method used to produce EMR with a known frequency. 5. Explain how the observations made by Extensions Hertz support the electromagnetic theories of Maxwell. 6. How did Maxwell\u2019s study of capacitors add to his theories? 7. Why is the process of \u201cchanging\u201d critical to the production of EMR? Applications 8. A high-voltage spark gap produced sparks with a frequency of 2.5 103 Hz. What is the frequency of the EMR observed at an antenna located nearby? 9. What properties need to be considered when choosing the material to build an antenna for a radio receiver? 15. Explain, with the aid of a transverse standing wave diagram, how the current in a ring antenna (such as the one used by Hertz) will be affected as the antenna is moved along the axis of the standing wave. How could this data be used to determine the wavelength of the standing wave? 16. Draw a schematic that shows, in general terms, the role of a satellite in delivering digital, high-definition, television signals between the signal provider and the customer\u2019s home receiver. e TEST To check your understanding of the nature and production of electromagnetic radiation, follow the eTest links at www.pearsoned.ca/school/physicssource. Chapter 13 The wave model can be used to describe the characteristics of electromagnetic radiation. 647 13-PearsonPhys30-Chap13 7/24/08 3:44 PM Page 648 info BIT Maxwell predicted the speed of EMR before Hertz experimentally verified it. Maxwell knew that electric and magnetic fields are not independent of one another and predicted a speed of propagation by dividing the constants, ke and km, which describe electric and magnetic forces between moving charged particles. Using today\u2019s currently accepted values of ke and km, we can show that the magnetic field and the electrical field are related by a particular speed. 8.99 109 N 2 m 2 C s2 1.00 107 N 2 C 2 m 8.99 1016 s2 k e k m 8", ".99 1016 2.99 108 m m2 s2 s 13.2 The Speed of Electromagnetic Radiation In addition to producing and detecting electromagnetic radiation in his 1887 experiment, Hertz was able to show that the radiation was travelling at approximately 3.0 108 m/s. This was an empirical verification of a theoretical speed predicted by Maxwell years earlier, as described on page 643: All electromagnetic waves will travel at a speed of 310 740 000 m/s and obey the universal wave equation (c f) relating speed, frequency, and wavelength. Empirical Determination of the Speed of Electromagnetic Radiation Light travels so quickly that there is no apparent time difference between turning on a light and the light reaching your eyes \u2013 even when the light source is many kilometres away. Galileo (1564\u20131642) found this out in his attempt to measure the speed of light. His experiment was very basic: Galileo and an assistant stood, each with a covered lantern, on adjacent hills separated by about a kilometre. Galileo would uncover his lantern and as soon as his assistant observed Galileo\u2019s lantern, he would uncover the second lantern so that Galileo would see it. Galileo used his pulse rate to measure the time difference. The time interval is actually extremely small \u2013 Galileo didn\u2019t know it at the time, but light could travel 2 km in about 0.000 066 67 s, a time you couldn\u2019t even measure with a digital stopwatch. Galileo realized that using his pulse as a timepiece was not appropriate. Earth longer time Sun Earth shorter time Io Jupiter Earth\u2019s orbital diameter approximate Figure 13.21 Earth\u2019s orbital diameter causes the eclipse of Io to occur at different times because of the extra distance the light must travel when Earth is farthest from Jupiter. 648 Unit VII Electromagnetic Radiation Olaus R\u00f8mer (1644\u20131710) and Christiaan Huygens used one of the four easily visible moons of Jupiter to calculate the speed of light. The period of revolution for Io (Jupiter\u2019s brightest moon) had been accurately determined using large amounts of astronomical data collected over many years. Roemer knew that the moon Io should be eclipsed and disappear behind Jupiter at regular, periodic intervals (every 42.5 h). However, he discovered that Io appeared to be eclipsed later than scheduled when Earth was farther away from Jupiter and earlier than scheduled when Earth was closer to Jupiter. The time difference between the longest and shortest periods was 22 min (Figure 13.21). 13-PearsonPhys30-Chap13 7/24/08 3:44 PM Page 649 Huygens reasoned that this discrepant behaviour could be due to the difference in the time taken for the light to travel from Jupiter to Earth. Following this reasoning, he estimated Earth\u2019s orbital diameter and then determined the speed of light based on the following calculations. Orbital diameter 3.0 1011 m v d t Time difference 22 min 1320 s 3.0 1011 m 1320 s 2.3 108 m/s His calculations produced a value of 2.3 108 m/s \u2014 astoundingly close to the currently accepted value of 3.0 108 m/s. The value was so large it was rejected by the scientific community of the time and was not accepted until after both Huygens and Roemer were dead. In 1848, Armand Fizeau became the first person to successfully measure the speed of light at the surface of the planet. His apparatus consisted of a rotating toothed wheel, a light source, some lenses, and a mirror. In his experiment, light was allowed to pass through one of the gaps on the toothed wheel and travel toward the mirror (located on an adjacent hilltop 8.63 km away) where it was reflected back toward the spinning wheel (Figure 13.22). The toothed wheel was rotated such that the reflected light was blocked by a tooth in the wheel as it turned, and the observer would not be able to see the source light. Using the rotational frequency of the spinning wheel, Fizeau was able to determine the time it took the light to travel 8.63 km and back (17.26 km round trip) and therefore to determine the speed at which the light was travelling to the distant mirror and back. His experimentally determined value for the speed of light was 3.15 108 m/s \u2014 only 5% more than the currently accepted value. toothed wheel mirror glass observer path A path B light source observation point water in water out Figure 13.22 Fizeau\u2019s original apparatus for measuring the time it takes light to travel the distance (8.63 km) and back Figure 13.23 Fizeau\u2019s apparatus for measuring the speed of light in water light source Three years later Fizeau conducted an investigation to determine if the speed of light was affected by a moving medium such as water.", " In this experiment, water was pumped in opposite directions through two parallel tubes (Figure 13.23). Light from a single source was directed through both tubes, following paths A and B, of identical lengths, to an Chapter 13 The wave model can be used to describe the characteristics of electromagnetic radiation. 649 13-PearsonPhys30-Chap13 7/24/08 3:44 PM Page 650 observation point. Fizeau argued that if the speed of the light waves changed as a result of travelling through the water tubes, the waves arriving from path A and B would produce an interference pattern at the observation point. A device called an interferometer was used to observe this interference. The experiment proved that light would travel at different speeds in different media. The interference pattern produced was also very strong evidence that light was a wave. octagonal set of mirrors observer telescope path of light Mount Wilson 35 km light from source curved mirror Mount St. Antonio Figure 13.24 Michelson\u2019s experimental apparatus to measure the speed of light Practice Problems 1. Students using a 12-sided set of rotating mirrors in an experiment similar to Michelson\u2019s determine the speed of light to be 2.88 108 m/s. The mirrors are located 30.0 km from the fixed mirror. What is the frequency of rotation if 1/12th of a rotation occurs before the light returns? 2. Light reflected from an 8-sided set of rotating mirrors travels 15.0 km to a distant fixed mirror and back. At what frequency is the set rotating if it has turned 1/8th of a rotation before the light returns? 3. An 8-sided set of mirrors, similar to Michelson\u2019s, rotating with a frequency of 500 Hz, is located 36.0 km away from a fixed mirror. If the returning light is observed in the system, at what speed is the light travelling? Answers 1. 400 Hz 2. 1.25 103 Hz 3. 2.88 108 m/s 650 Unit VII Electromagnetic Radiation Michelson\u2019s Experiment Building on Fizeau\u2019s 1848 experimental design and the work of others, Albert Michelson performed an experiment in 1905, using a rotating set of mirrors instead of a toothed wheel. A very intense light source was directed at an 8-sided, rotating set of mirrors, which reflected the light toward a curved mirror located 35 km away. After travelling 35 km, the light was reflected back toward the rotating mirrors. If the rotating mirrors had made 1/8th of a rotation just as the light returned, the returning light could be observed in a telescope (Figure 13.24). For the returning light to be observed in the telescope, the rotating mirrors had to turn at a very precise frequency. Michelson knew the light had to travel a round trip distance of 70 km, and if the rotating mirrors had made 1/8th of a rotation (or a multiple of an 1/8th rotation) between the time the light left and then returned, the light could be observed in the telescope. All Michelson needed to measure was the frequency of rotation and he could determine the time required for 1/8th of a rotation. When his mirror rotated at 32 000 rpm (533 Hz), he could observe the light in the telescope. Example 13.1 The set of rotating mirrors in Michelson\u2019s experiment was rotating at 533 Hz and the curved mirror was located 35.0 km away. Show how Michelson determined the speed of light from these data. Given distance between rotating and curved mirrors 35.0 km 3.50 104 m f 533 Hz Required speed of light (c) Analysis and Solution c d t The light must travel from the set of mirrors to the curved mirror and back again, so the round-trip distance is twice the given distance. d 2 3.50 104 m 7.00 104 m 13-PearsonPhys30-Chap13 7/24/08 3:44 PM Page 651 For the light to follow the path illustrated in Figure 13.24, the rotating set of mirrors must complete a minimum of 1/8th of a rotation while the light is travelling between the rotating mirrors and the curved mirror. Therefore, the minimum travel time (t) is equal to 1/8th of the period of rotation1.876 103 s) 8 d c t 4 m 0.0 7 4 s 0.3 1 1 4 5 0 2 1 Hz 533 1.876 103 s 2.345 104 s 2.98 108 m/s Paraphrase According to Michelson\u2019s observations, the speed of light in air was found to be 2.98 108 m/s. 13-2 QuickLab 13-2 QuickLab Measuring the Speed of EMR Problem Microwaves are a type of electromagnetic radiation and travel at the speed of light. They have a defined wavelength and frequency when they travel in a vacuum or in air. The universal wave equation v frelates the", " speed of the wave to its frequency and wavelength, where f is the frequency of the radiation, and is the wavelength. Each microwave oven generates microwaves at a given frequency, usually 2450 MHz. If the wavelength of the microwave can be measured, its speed can be determined using the universal wave equation. Materials microwave oven microwave-safe dish marshmallows Procedure 1 Pack a solid layer of marshmallows in a casserole or microwave-safe dish. 4 Remove the dish from the oven and measure the distance between adjacent soft spots. 5 Determine the average separation distance between several soft spots. This distance is equal to half the wavelength of a microwave. 6 Calculate the wavelength by multiplying the average separation distance by two. 7 Record the frequency of the microwave, which will be indicated on the back of the microwave oven. Questions Use the wavelength and frequency of the microwave to answer the following questions. 1. What is the speed of the microwaves? 2. What is the percent error of your calculated value, when compared with the currently accepted value for the speed of light (3.00 108 m/s)? 3. What are several possible sources of error that 2 Remove the turntable from the microwave oven. could affect your calculation? 3 Place the dish in the oven and cook until the marshmallows begin to melt in four or five locations. 4. Compare the wavelength of microwaves with that of radio waves and visible light waves. Why are they called microwaves? Chapter 13 The wave model can be used to describe the characteristics of electromagnetic radiation. 651 13-PearsonPhys30-Chap13 7/24/08 3:44 PM Page 652 13.2 Check and Reflect 13.2 Check and Reflect Knowledge 1. Why was Galileo\u2019s original experiment to determine the speed of light unsuccessful? 2. Explain why, in a Michelson-type experiment, the rotating mirrors have to turn at a very precise frequency in order for light to reach a stationary observer. 3. Explain why the periodic motion of Jupiter\u2019s moon Io appears to be constantly changing when observed from Earth. Applications 4. In measuring the speed of light, the difference in eclipse times for Jupiter\u2019s moon Io, is measured. If an eclipse occurs 24 min later than expected, and Earth\u2019s orbital diameter is 3.0 1011 m, calculate the speed of light. 5. A communications satellite is in orbit around Earth at an altitude of 2.00 104 km. If the satellite is directly above a groundbased station, how long does it take a signal to travel between the satellite and the station? 6. Using a similar approach to Michelson, a student sets up a 64-sided set of rotating mirrors, 8.00 km away from a fixed mirror. What minimum frequency of rotation would be required to successfully measure the speed of light? 7. A 16-sided set of rotating mirrors is used to measure the time it takes light to travel a certain distance. At what frequency does the mirror need to rotate such that it makes 1/16th of a rotation in the time it takes light to travel 3.5 km and back again? 8. An 8-sided set of rotating mirrors rotates at 545 Hz in an experiment similar to that of Michelson. How far away should the fixed mirror be placed in order to correctly measure the speed of light? 652 Unit VII Electromagnetic Radiation 9. The speed of light was measured to be 2.97 108 m/s using a 16-sided set of rotating mirrors and a fixed mirror separated by 5.00 103 m. At what frequency was the mirror rotating? 10. Students who measure the speed of light using an experimental design similar to that of Michelson with an 8-sided set of rotating mirrors, make the following observations when light passes through the apparatus: (a) rotating mirror frequency 1.00 103 Hz (b) distance between fixed and rotating mirror 17.5 km Determine the speed of light based on the students\u2019 recorded observations. 11. A 64-sided set of rotating mirrors is turning at 340 Hz. If a fixed mirror is located 6.55 km away from the rotating mirror and the light is reflected correctly, what value for the speed of light would be obtained? Extensions 12. Global positioning satellites (GPS) are used to pinpoint individual locations on the surface of Earth. A minimum of three GPS satellites are required, each with synchronized atomic clocks. Each satellite simultaneously sends a time signal to a GPS receiver on the surface. The time signal of each satellite is compared with the time on the receiver\u2019s clock. (a) Explain how the exact distance between each satellite and the receiver can be determined. (b) Explain how the receiver\u2019s location relative to all three satellites indicates an exact coordinate on Earth\u2019s surface. e TEST To check your understanding of speed and propagation of electromagnetic radiation, follow the eTest links at www.pearsoned.ca/school/physicssource. 13-Pear", "sonPhys30-Chap13 7/24/08 3:44 PM Page 653 13.3 Reflection On a windless day, the smooth surface of a lake will reflect light and produce stunning images of the surrounding landscape and sky. But introduce the slightest disturbance, a gust of wind or a morning rain shower, and the image becomes distorted and blurred. With enough disturbance, the image on the surface of the water completely disappears. Figure 13.25 a) Reflection in a smooth lake b) Diffuse reflection due to windy conditions How do the images in Figure 13.25 form? Light travels in straight lines through a medium that is uniform. This characteristic is termed rectilinear propagation. The light falling on a flat, smooth, reflecting surface, such as a mirror, undergoes specular or regular reflection. In the case of the smooth lake, the crisp, clear images that form are the result of many parallel incident rays that reflect as parallel reflected rays (Figure 13.26). In the case of the rough lake, the blurred images that form are the result of many parallel incident rays that are scattered after reflecting in many different directions. This behaviour is called diffuse or irregular reflection (Figure 13.27). rectilinear propagation: movement of light in straight lines through a uniform medium Figure 13.26 Ray diagram of specular (regular) reflection Figure 13.27 Ray diagram of diffuse (irregular) reflection The Law of Reflection Particular terms are used to describe reflection. These terms are illustrated in the ray diagram shown in Figure 13.28 on the next page. Regular reflection will occur when an incident light ray contacts a polished, reflecting surface. The incident ray makes contact at the point of incidence and then it leaves the reflecting surface and becomes the reflected ray. If, at the point of incidence, an imaginary line, called the ray diagram: a diagram to show a result of a light ray interacting with a surface Chapter 13 The wave model can be used to describe the characteristics of electromagnetic radiation. 653 13-PearsonPhys30-Chap13 7/24/08 3:44 PM Page 654 e SIM Find out more about the law of reflection from the use of simulations. Follow the eSim links at www.pearsoned.ca/school/ physicssource. plane mirror: a smooth, flat, reflecting surface law of reflection: the angle of reflection is equal to the angle of incidence and is in the same plane mirror plane \u03b8 i \u03b8 r incident ray reflected ray Figure 13.29 The Law of Reflection image I mirror O point object Figure 13.30 How two different observers see the image of an object located in front of a plane mirror virtual image: an image from which light rays appear to come; cannot be formed on a nonreflective surface or screen real image: an image from which light rays come; can be formed on a diffusely reflecting surface or screen normal line (N), is drawn perpendicular to the surface, then the angle formed between the incident ray and the normal line is called the angle of incidence. The angle formed between the reflected ray and the normal line is called the angle of reflection. point of incidence reflecting surface angle of incidence \u03b8 i angle of reflection \u03b8 r incident ray reflected ray N normal Figure 13.28 Ray diagram and terminology for regular reflection Light rays that are incident upon a plane mirror at 90 will be reflected directly back to the source. In this case, the incident and reflected rays are parallel to the normal line so that the angle of incidence and the angle of reflection are both zero. In a similar way, when a light ray contacts a mirror at an angle of 25 relative to the normal line, it will reflect at an angle of 25 relative to the normal line. In all cases, the angle of incidence will be equal to the angle of reflection. This is called the law of reflection (Figure 13.29). The angle of reflection is equal to the angle of incidence and is in the same plane. Image Formation in a Plane Mirror Look at the image of an object formed in a plane, flat mirror in Figure 13.30. What do the dashed lines represent? Are they real rays coming from a real object? The dashed lines are an extension of the reflected rays, causing your brain to \u201cbelieve\u201d that they are rays, originating from an object that is located behind the mirror. But this cannot be so. The light rays appear to be coming from an image, but this image is not real. The image formed in a plane mirror in this fashion is therefore called a virtual image. For example, if you were to hold a piece of paper behind the mirror where the image appears to be, you would not see the image on the paper. By contrast, a real image is one that can be formed on a diffusely reflecting surface, such as a piece of paper. A projector is a familiar device that produces real images on a screen. Ray diagrams can be used to show how images form with a reflecting surface, such as", " a mirror. 654 Unit VII Electromagnetic Radiation 13-PearsonPhys30-Chap13 7/24/08 3:44 PM Page 655 13-3 QuickLab 13-3 QuickLab Drawing a Ray Diagram Problem To construct a ray diagram of an image formed in a plane, flat mirror. Materials observer N ray 1 object mirror Figure 13.31 Procedure 1 Choose one point on the object, such as the tip of the candle flame, in Figure 13.31. 2 Draw an incident ray from the selected point on the object (adding a small normal line where the ray is incident on the surface of the mirror) (Figure 13.31). 3 Based on the law of reflection, the reflected ray leaves the mirror surface such that the angle of incidence is equal to the angle of reflection. Draw the reflected ray. 4 Use a dashed line to extend the reflected ray beyond the back surface of the mirror. 5 On your diagram, repeat steps 2 to 4 for a ray originating from the same location on the object, but at a slightly different angle. 6 Note that the viewer must be located where both reflected rays would enter the eye. 7 Repeat steps 2 to 6 and locate the point where the two dashed lines converge (meet). This is the image location for the first selected point on the object. Questions 1. How could you test whether the image formed in the mirror was real or virtual? 2. Is the image magnified or the same size as the object? 3. Will the image appear right side up (erect) or upside down (inverted)? 4. Could you use a ray diagram to determine if the word \u201cHELP\u201d would appear backwards in a mirror? Explain. M I N D S O N Image in a Mirror G F E D C O B A Figure 13.32 Woman in a Mirror Figure 13.32 shows a woman looking in a mirror. 1. How far behind the mirror does her image form, if she is standing 50.0 cm in front of the mirror? Explain how the ray diagram is used to determine this distance. Is the image real or virtual? 2. 3. According to the ray diagram, two sections of the mirror could be removed and her entire face would still be visible. Which two sections are not needed? Explain your reasoning. Chapter 13 The wave model can be used to describe the characteristics of electromagnetic radiation. 655 13-PearsonPhys30-Chap13 7/24/08 3:44 PM Page 656 magnification: the relationship of the size of the image to the size of the object image attitude: the orientation characteristic of an image, whether erect or inverted Image Characteristics When an object is placed in front of a mirror, the image can be described by four general characteristics (Table 13.2). The magnification relates the size of the image to the size of the object. The image can be enlarged, diminished, or the same size as the object. The image attitude describes the image as being either erect (upright) or inverted (upside down) relative to the object. This is termed vertical inversion. Horizontal, or left/right, inversion can also occur. The position describes where the image forms relative to the surface of the mirror. The type of image distinguishes between real and virtual images. A real image is one that can be projected on a screen or surface, and a virtual image is one that can only be seen, or photographed. Table 13.2 Image Characteristics Image Characteristic Description magnification same size, enlarged, diminished attitude position type erect or inverted displacement from mirror surface real or virtual Image characteristics are important for the proper application of signage and warning labels. For example, the word \u201cambulance\u201d is written in reverse on the front of the vehicle so that other drivers will be able to read the writing in their rear-view mirrors when the ambulance is approaching. M I N D S O N Image Characteristics Figure 13.33 shows (a) an image of an ambulance and (b) its image in a rear-view mirror. 1. What image characteristic is addressed by printing the word \u201cambulance\u201d in reverse? 2. While standing in front of a plane, flat mirror, raise your right hand. Which hand in your image was raised? In a plane, flat mirror, do images appear backwards horizontally or upside down vertically or both? Explain. 3. 4. Can an image in a plane mirror ever appear \u201cupside down\u201d? Figure 13.33 (a) The front of an ambulance (b) The image it makes in a rear-view mirror 656 Unit VII Electromagnetic Radiation 13-PearsonPhys30-Chap13 7/24/08 3:44 PM Page 657 Image Formation in a Curved Mirror Curved mirrors come in a variety of shapes, the most common being spherical. Spherical mirrors, like the ones used for store security, have a unique geometrical shape (Figure 13.34). This", " shape is derived from a sphere. Imagine a hollow sphere, with a polished mirror surface on the inside and out. hollow sphere polished inside and out convex reflecting surface concave reflecting surface Figure 13.34 A spherical store security mirror Figure 13.35 A curved mirror can be formed from a hollow sphere. By removing a section of the sphere, you produce a double-sided spherical mirror with a concave reflecting surface on one side and a convex reflecting surface on the other (Figure 13.35). The concave reflecting surface is curved inwards. This surface is also referred to as a converging mirror since it causes parallel light rays to converge, or come together, after being reflected. The convex reflecting surface is curved outwards. This surface is also referred to as a diverging mirror since it causes parallel light rays to spread out or diverge after being reflected. As with plane-mirror image formation, ray diagrams are useful to determine how images form from curved surfaces. Ray diagrams for curved surfaces, as in Figure 13.36 for example, are more complex and involve an expanded set of terminology. converging mirror: a concave reflecting surface that causes parallel light rays to converge after being reflected diverging mirror: a convex reflecting surface that causes parallel light rays to spread out after being reflected 1. centre of curvature (C) \u2013 the point in space that would represent the centre of the sphere from which the curved mirror was cut 2. radius of curvature (r) \u2013 the distance from the centre of curvature to the mirror surface 3. vertex (V) \u2013 the geometric centre of the PA curved mirror surface ray 2 ray 1 C ray 3 r F V f 4. principal axis (PA) \u2013 an imaginary line drawn through the vertex, perpendicular to the surface of the curved mirror at this point 5. principal focal point (F) \u2013 the point where light rays that are parallel to and close to the principal axis converge, or appear to diverge from, after being reflected Figure 13.36 A ray diagram for a converging mirror 6. focal length (f) \u2013 the distance from the vertex to the focal point, measured along the principal axis. The focal length is related to the radius of curvature by f r2. This means that as the radius of curvature is reduced, so too is the focal length of the reflecting surface. Chapter 13 The wave model can be used to describe the characteristics of electromagnetic radiation. 657 13-PearsonPhys30-Chap13 7/24/08 3:44 PM Page 658 Drawing Ray Diagrams for Curved Mirrors The four general image characteristics of magnification, attitude, position, and type, which were applied to plane mirrors, can also be applied to curved mirrors. For curved mirrors, the normal line at the point of incidence is the same as an extension of the radius of the spherical surface drawn at this point. The law of reflection still describes how all the rays will be reflected. To understand and predict how images are produced from curved mirrors, we will need to use three light rays, the law of reflection, and the mirror\u2019s focal point, centre of curvature, and vertex. Two rays from a point on an object locate the corresponding point on the image and the third ray will verify its location (Figure 13.37). Consider each of the three rays: Incident Ray 1 travels from a point on the object parallel to the principal axis. Any ray that is parallel to the principal axis will reflect through the focal point on a converging mirror, or appear to have originated from the focal point on a diverging mirror. Incident Ray 2 travels from a point on the object toward the focal point. Any ray that passes through the focal point on a converging mirror, or is directed at the focal point on a diverging mirror, will be reflected back parallel to the principal axis. PA PA F C Incident Ray 3 travels from a point on the object toward the centre of curvature. Any ray that passes through the centre of curvature on a converging mirror, or is directed at the centre of curvature on a diverging mirror, will be reflected directly back along the incident path. At what incident angle must this light ray hit the mirror surface in order to be reflected straight back along the original path? These three rays alone will allow you to predict and verify the location and characteristics of the image. Notice that some conventions are used when drawing these ray diagrams: 1. objects are often drawn as erect arrows 2. real rays are drawn as solid lines 3. virtual rays (that only appear to exist behind the mirror) are drawn as dashed lines Figure 13.38 (a) shows that the converging mirror produces a real image that is inverted and diminished. Figure 13.38 (b) shows that the diverging mirror produces a virtual image that is erect and diminished. Why is one image real and the other virtual? For curved mirrors, a real image is formed where the reflected light rays converge or meet. At this location, a focussed image would", " appear on a sheet of paper or a screen if it were located in the exact location where the light rays meet. If the screen were to be moved slightly, the image would appear Figure 13.37 Reflection of three rays from spherical mirrors ray 1 ray 3 ray 2 C F V a) A concave, converging mirror ray 1 ray 2 ray 3 V b) A convex, diverging mirror PHYSICS INSIGHT All three incident rays obey the Law of Reflection. The resulting rules for drawing ray diagrams are just the application of geometry. 658 Unit VII Electromagnetic Radiation 13-PearsonPhys30-Chap13 7/24/08 3:44 PM Page 659 blurred because the reflected rays would not be converging perfectly at the screen\u2019s new location. For the diverging mirror, the reflected rays appear to be originating from behind the mirror, but if a screen were located there, the incident light rays would not reach it (the rays would be blocked by the mirror). Figure 13.38 Steps for drawing a ray diagram in converging and diverging spherical mirrors e WEB To learn more about ray diagrams for curved mirrors and lenses, follow the links at www.pearsoned.ca/school/ physicssource. ray 1 ray 3 ray 2 object C F V PA image 3 2 1 a) A concave, converging mirror 1 ray 1 ray 2 3 2 ray 3 object image b) A convex, diverging mirror Characteristics of an Image in a Curved Mirror If you look at the image of your face in a polished spoon and bring the spoon closer and closer to your nose, you will demonstrate that with curved mirrors, the object\u2019s distance from the mirror\u2019s vertex has an effect on the characteristics of the image produced. For example, what happens to an image as the object is brought closer to a converging mirror surface? In Figure 13.39(a), the object is located outside the centre of curvature, and a real, inverted, and diminished image appears. If the object is brought closer to the vertex of the mirror, such that it is inside the focal length, as in Figure 13.39(b), then the image becomes virtual, erect, and enlarged. Figure 13.39 Object location affects image characteristics. 1 2 3 mirror object C F V PA image a) A converging mirror with object located outside C (real, inverted, diminished) 2 mirror 3 C 1 F object V image PA b) A converging mirror with object located inside F (virtual, erect, enlarged) Chapter 13 The wave model can be used to describe the characteristics of electromagnetic radiation. 659 13-PearsonPhys30-Chap13 7/24/08 3:44 PM Page 660 Table 13.3 Image Characteristics for Converging Mirrors Object Position Image Characteristic distant real, inverted, and diminished, close to F outside centre of curvature real, inverted, and diminished, between F and C at the centre of curvature, C real, inverted, and same size, at C between focal point and centre of curvature real, inverted, and enlarged, beyond C at focal point, F undefined (no image forms), at infinity between focal point and vertex virtual, erect, and enlarged, behind mirror Table 13.3 summarizes the image characteristics for a concave, converging mirror. You can verify the characteristics by placing the concave side of a spoon in front of your nose and slowly moving it outward. Note that your image will disappear briefly just as the distance between your nose and the spoon reaches the focal point of the spoon. In a similar way, the image characteristics produced by a convex, diverging mirror can be investigated by using the other surface of the spoon. 13-4 QuickLab 13-4 QuickLab Converging and Diverging Mirrors Problem How do image characteristics in converging and diverging mirrors change as you change the object distance? Materials polished spoon drawing materials Procedure 1 Place a polished metal spoon on your nose and slowly pull it away from your face while watching the image. 4 Complete ray diagrams for the image of your nose inside F, at F, at C, and at a distance, with respect to the spoon as a diverging mirror. (Remember that a diverging mirror has a virtual focal length, so F and C are located on the opposite side of the spoon to your face.) Questions 1. What happens to the image as the object is brought closer to the surface of a converging mirror? 2. What happens to the image as the object is brought closer to the surface of a diverging mirror? 3. What characteristics do all images formed in a 2 Complete ray diagrams for the image of your nose diverging mirror share? inside F, at F, at C, and at a distance, with respect to the spoon as a converging mirror. (Use a small upright arrow to represent your nose in the ray diagram.) 3 Reverse the spoon so that you are looking at the other side and move it", " away from your nose again. 4. Your image disappears when your face is at the focal point of the mirror. Sketch a ray diagram for a converging mirror with the object located at the focal point to explain why an image should not appear when the object is located at the focal point. 5. Why is it not possible to put an object at F for a convex mirror? 660 Unit VII Electromagnetic Radiation 13-PearsonPhys30-Chap13 7/24/08 3:44 PM Page 661 M I N D S O N Seeing Is Believing\u2014Or Is It? A parabolic mirror is shaped so that incident light rays that travel parallel to the principal axis will be reflected to a central point, or focus. Optical illusions can be produced by arranging two parabolic mirrors facing one another. Use a ray diagram to explain how a three-dimensional, real image, is produced in Figure 13.40. Hint: The object is located at the focal point of the top mirror. Therefore, all rays that originate from the object will be reflected straight down from the top mirror. Assume the bottom mirror has the same focal length as the top mirror. real image appears here mirror object inside bottom mirror Figure 13.40 Two parabolic mirrors facing one another, with a hole cut around the vertex of one mirror Equations for Curved Mirrors The general rules for drawing ray diagrams of curved mirrors, in combination with the law of reflection, allow us to generate several equations for use with curved mirrors. These equations can be used to determine the characteristics of the image. In any ray diagram, two rays can be used to determine the image characteristics. In Figure 13.41, there are two similar triangles (AOV, above, and DIV, below). Both triangles have an identical angle () and right angles at the principal axis. Thus, DI AO DV AV O ho C A D hi I F \u03b8 \u03b8 V B f do di where hi ho This translates into di do ho is the height of the object, hi is the height of the image, do is the distance between the mirror vertex and the object, and di is the distance between the mirror vertex and the image. Figure 13.41 Triangles AOV and DIV are similar triangles; triangles FVB and FAO are also similar triangles. When we use the ray OFB, the triangles FVB and FAO (shaded light blue) are also similar. Therefore, AO VB AF VF Since \u2022 AO ho \u2022 VB hi \u2022 AF do \u2022 VF f f Chapter 13 The wave model can be used to describe the characteristics of electromagnetic radiation. 661 13-PearsonPhys30-Chap13 7/24/08 3:44 PM Page 662 mirror equation: the equation that relates the focal length of a curved mirror to the image and object distances We can write this relationship as: Since hi ho di do the inverse is also true: Combining the equations gives: Dividing both sides by do and rearranging: We obtain this equation: ho hi ho hi do di d o did o 1 di f do f do di f do f 1 do d o d of 1 f 1 do This is the mirror equation, which relates the focal length of a curved mirror to the image and object distances. Sign Conventions for Use with the Mirror Equation As we have seen previously, there are real and virtual images, which can form either in front of or behind the curved mirror. When using the mirror equation, it is therefore necessary to follow a sign convention, which can distinguish the type of image formed. \u2022 Real objects and images have positive distances (measured from vertex). \u2022 Virtual objects and images have negative distances (measured from vertex). \u2022 Erect images and objects have a positive height. \u2022 Inverted images and objects have a negative height. \u2022 Converging mirrors have a real principal focal point and the focal length is positive. \u2022 Diverging mirrors have a virtual principal focal point and the focal length is negative. With this sign convention, a real image formed by a converging mirror will have a negative height (inverted attitude) while the object height is positive (erect attitude). Both the object and image distances will be positive. Magnification (m) is the ratio of the image height to the object height. A negative sign must be added to the equation for magnification to agree with the sign convention above. m hi ho di do An erect image has a positive magnification and an inverted image has a negative magnification. Concept Check 1. If the object distance is extremely large, approaching infinity, 1 becomes zero. Based on the mirror equation, where will the d o image form in relation to the focal length of the mirror? 2. For a plane, flat mirror, 1 approaches zero. Based on the mirror f equation, where will the image form in relation to the object? 662 Unit VII Electromagnetic Radiation 13-PearsonPhys30-Chap13 7/24/08 3:44", " PM Page 663 13-5 Problem-Solving Lab 13-5 Problem-Solving Lab Using a Converging Mirror to Heat Water (Determining the Focal Length of a Converging Mirror) Required Skills Initiating and Planning Performing and Recording Analyzing and Interpreting Communication and Teamwork Recognize a Need A converging mirror can be used to focus the radiant energy of the Sun onto a specific point to capture energy, to heat water, or to generate electrical energy using a photovoltaic cell. The Problem Design an efficient apparatus that uses radiation to heat water. Criterion for Success Observing the maximum increase in water temperature in a given time period. Brainstorm Ideas To design an efficient water heater that uses sunlight for energy, you will need to consider the following questions: \u2022 Where would a water tank be placed in relation to a converging mirror in order to maximize the amount of radiation that it could absorb? \u2022 What colour should the water tank be? \u2022 What effect will the size of the mirror have on the water temperature? \u2022 What positive and negative effects would result from insulating the water tank? Build a Prototype Construct a miniature version of your water heater using a 50-mL volumetric flask, a thermometer, and a converging mirror mounted on a base of clay. Test and Evaluate Determine the focal length of the converging mirror using a heat lamp, a metre-stick, and a sheet of paper (Figure 13.42). Recall that rays that travel parallel to the principal axis will be reflected back through the focal point. 1. Place the heat lamp a great distance away so that most of the incident rays travel parallel to the principal axis. Adjust the position of the paper until the light is focussed into a small region. Be careful not to block the light from reaching the mirror. You may need to adjust the position of the mirror to avoid this \u2014 try tilting it slightly downward and turning the room lights off to see the image clearly. 2. Record the distance between the paper and the vertex of the mirror. This is the approximate focal length of the mirror and it is also the image distance. 3. Measure the distance between the lamp and the mirror. This is the object distance. Using the mirror equation, with the image and object distances, calculate and verify the approximate focal length of the mirror. heat lamp paper do di converging mirror clay Figure 13.42 Locating the focal point 4. Using the heat lamp in place of sunlight, measure the increase in water temperature for a given time period when the water tank is located beyond the focal point, at the focal point, and inside the focal point of the mirror. 5. Determine which position results in the greatest increase in water temperature and explain why. Communicate Using presentation software, create a brochure that outlines the basic operating principles of your water heater and explains the role of radiant energy in its operation. Chapter 13 The wave model can be used to describe the characteristics of electromagnetic radiation. 663 13-PearsonPhys30-Chap13 7/24/08 3:44 PM Page 664 Example 13.2 An object is located 20.0 cm from a converging mirror that has a focal length of 15.0 cm. (a) Where will the image form relative to the mirror vertex? (b) If the object is 5.00 cm high, determine the attitude, height, and magnification of the image. 20.0 cm Given do f 15.0 cm (converging mirror has a positive focal length) ho 5.00 cm (assume that the object is erect) Required image distance (di) image height (hi) magnification (m) Analysis and Solution (a) To locate the image, use the relationship between focal length and object distance, the mirror equation: 1 f 1 di 1 do 1 f 1 di 1 do Now, solve for di: 1 1 1 20. 15. 0 cm 0 cm d i di 60.0 cm or di d do of f (20.0 cm)(15.0 cm) (20.0 cm) (15.0 cm) 60.0 cm (b) To determine the attitude, height, and magnification, use the equation for magnification: m hi ho di do Now, solve for m: m 60.0 cm 20.0 cm 3.0 Practice Problems 1. A diverging mirror of focal length 10.0 cm produces an image of an object located 20.0 cm from the mirror. Determine the image distance and the magnification. Is the image real or virtual? 2. Determine the image distance, magnification, and attributes for the following: (a) a converging mirror with a focal length of 12.0 cm with an object 6.0 cm from the mirror (b) a diverging mirror of focal length 5.00 cm with an object 10.0 cm from the mirror (c) a diverging mirror of focal length 10.0 cm with an object 2.0 cm from the mirror 3.", " A 5.0-cm-high object is placed 2.0 cm in front of a converging mirror and the image is magnified 4. Where does the image form and what is the focal length of the mirror? 4. A 4.0-cm-high object is placed 15.0 cm from a concave mirror of focal length 5.0 cm. Determine the image characteristics using a ray diagram and the mirror equation. 5. Light from a distant planet is incident on a converging mirror. The image of the planet forms on a screen 45.0 cm from the vertex of the mirror. Find the focal length of the mirror and the image characteristics. Answers 1. di 6.67 cm; m 0.333, virtual 2. (a) di 12 cm; m 2.0, virtual, erect, enlarged (b) di 3.33 cm; m 0.333, virtual, erect, diminished (c) di 1.7 cm; m 0.83, virtual, erect, diminished 8 cm; f 1.6 cm 7.5 cm; m 0.50 ; 3. di 4. di 2.0 cm; real, inverted, diminished hi 5. f 45.0 cm; real, inverted, diminished 664 Unit VII Electromagnetic Radiation 13-PearsonPhys30-Chap13 7/24/08 3:44 PM Page 665 Then solve for hi: hi ih d o d o (60.0 cm)(5.00 cm) (20.0 cm) 15.0 cm Paraphrase (a) The real image forms 60.0 cm in front of the mirror vertex. (b) It is inverted and magnified 3 times to a height of 15.0 cm. 13.3 Check and Reflect 13.3 Check and Reflect Knowledge 1. Explain why the normal lines in Figure 13.27 are drawn in a variety of directions. 2. Create a concept map using ray diagram terminology including: principal axis, focal point, centre of curvature, and vertex. 3. Explain the difference between \u201cspecular\u201d and \u201cdiffuse\u201d reflection. 4. Compare \u201cvirtual\u201d images with \u201creal\u201d images. 5. Compare \u201cconverging\u201d and \u201cdiverging\u201d mirrors. 6. Describe the path of the three rays that can be used to determine the characteristics of an image formed in a curved mirror. 7. Why does a diverging, convex mirror have a \u201cvirtual\u201d focal point and not a \u201creal\u201d focal point? Applications 8. An object 4.0 cm high is located 10.0 cm in front of a concave mirror. If the image produced is erect, virtual, and 5.0 cm high, what is the focal length of the mirror? 9. Draw a ray diagram for a converging mirror and determine the image characteristics when the object is located at: (a) 2.0 f (b) 0.50 f (c) 3.0 f 10. Draw a ray diagram for a diverging mirror and determine the image characteristics when the object is located at: (a) 0.50 f (b) 1.0 f (c) 1.5 f 11. Some flashlights and headlights use concave mirrors to help generate a light beam. If the light source is positioned at the focal point of the mirror, would all the reflected rays travel outward, parallel to the principal axis? Explain your answer. Include a ray diagram. Extension 12. A reflecting telescope uses a curved mirror to produce the image at the eyepiece. Research the telescope\u2019s design and function. Sketch a ray diagram to explain how the converging mirror is used in the telescope design. e TEST To check your understanding of reflection in plane and curved mirrors, follow the eTest links at www.pearsoned.ca/school/physicssource. Chapter 13 The wave model can be used to describe the characteristics of electromagnetic radiation. 665 13-PearsonPhys30-Chap13 7/24/08 3:44 PM Page 666 13.4 Refraction A paddle appears bent when you put it in water (Figure 13.43); the Sun appears to change shape when it sets; a rainbow appears when it rains; and a mirage occurs over the hot desert sand, or even over a hot Alberta highway during the day. These are all visual effects created by refraction, a change in the direction of a light wave due to a change in its speed as it passes from one medium to another. When light rays pass at an angle from air to water, they immediately change direction. However, not all of the light will be refracted. Light rays are partially reflected and partially refracted when they pass from one medium to the next. Figure 13.44 illustrates this phenomenon.. The process of refraction is described by an incident ray and a refracted ray. The angle between the normal line", " (drawn perpendicular to the medium interface) and the incident ray is called the angle of incidence. The angle between the normal line and the refracted ray is called the angle of refraction. When the light ray passes from air to water, it is bent or refracted toward the normal line. And conversely, if a light ray travels from water into air, it is bent away from the normal line. The bending of the light ray depends on the refractive index of the medium in which the light travels. The refractive index of a medium is related to the effect on the speed of light when it travels within that medium. A vacuum does not impede or slow down light travelling through it and therefore, the speed of light (c) is defined as its speed in a vacuum. Air slows the light down a small amount and thus has a slightly greater refractive index than a vacuum. Water slows the light down to a greater extent and has a greater refractive index than air. This retarding effect was observed in Fizeau\u2019s experiment on the speed of light in water (page 649). The amount of refraction is related to the magnitude of the change in the speed of light as it passes from one medium to another; the greater the change in speed, the greater the amount of refraction. angle of reflection reflected ray normal angle of incidence Mathematically, the refractive index of a medium (n) can be described as a ratio that compares the speed of light in a vacuum (c) to the measured speed of light in the medium (v). n c v Figure 13.43 A paddle as it appears partially submerged in water refraction: a change in the direction of a light wave due to a change in its speed as it passes from one medium to another refractive index: a ratio comparing the speed of light in a vacuum to the measured speed of light in a given material incident ray less refractive more refractive angle of refraction point of incidence refracted ray Figure 13.44 Ray diagram of partial reflection and partial refraction 666 Unit VII Electromagnetic Radiation 13-PearsonPhys30-Chap13 7/24/08 3:44 PM Page 667 Table 13.4 lists the refrac- tive indexes for some common substances. Since the speed of light in air is very close to the speed of light in a vacuum, the refractive indexes for a vacuum and for air are considered to be the same. The following general rules are based on refractive indexes: \u2022 When light passes from a medium with a high refractive index to one with a low refractive index, it is refracted away from the normal line. \u2022 When light passes from a medium with a low refractive index to one with a high refractive index, it is refracted toward the normal line. Table 13.4 Absolute Refractive Indexes (for Sodium Yellow Light, 589 nm) Medium vacuum air ice water ethanol glycerin quartz glass crown glass light flint glass Index of Refraction 1.0000 1.0003 1.31 1.33 1.37 1.47 1.47 1.52 1.58 Lucite (plexiglass) 1.52 ruby zircon diamond 1.54 1.92 2.42 Snell\u2019s Law of Refraction Although the phenomenon of refraction had been observed for centuries, it was not until 1621 that the Dutch mathematician, Willebrord Snell (1580\u20131626), identified the exact relationship between the angle of incidence and the angle of refraction. Snell\u2019s Law states that: sin i sin r a constant i / sin This relationship indicates that for any angle of incidence greater than zero, the ratio sin r will be constant for any light ray that passes through the boundary between the two media. In the case of an air-water interface, the constant is 1.33, which corresponds to the refractive index of water. In fact, for any ray that passes from air ( air) into a second medium ( r, refractive index n), Snell\u2019s Law may also be written in a simple form as: sin air sin r n As long as the first medium is air, the Snell\u2019s constant and the index of refraction for the second medium are one and the same thing. If the first medium is not air, then the general form of Snell\u2019s Law applies. In the general form, the angle of incidence ( i) is replaced with 1 and the angle of refraction ( 2. The index of refraction for the first medium is denoted as n1 and the index of refraction r) is replaced with PHYSICS INSIGHT On passing through a medium, each wavelength of EMR has a unique refractive index. For this reason, Absolute Refractive Indexes must be quoted for a specific wavelength, as in Table 13.4. Snell\u2019s Law: For any angle of incidence greater than zero, the ratio sin r is", " a constant for any light ray passing through the boundary between two media. i/sin info BIT A mirage (derived from the Latin term \u201cmirare,\u201d meaning \u201cto wonder at\u201d) is a naturally occurring phenomenon caused by refraction. Often, mirages appear as large bodies of water on hot desert sand, or small puddles on highways. What appears as water is actually an image of the sky being refracted back up from the hot air just above the road surface. Figure 13.45 Chapter 13 The wave model can be used to describe the characteristics of electromagnetic radiation. 667 13-PearsonPhys30-Chap13 7/24/08 3:44 PM Page 668 e SIM Find out more about refraction through the use of simulations. Follow the eSIM links at www.pearsoned.ca/ school/physicssource. for the second medium is denoted as n2. In all cases, the subscript \u201c1\u201d refers to the incident medium and the subscript \u201c2\u201d is reserved for the refracting medium. The general form of Snell\u2019s Law is written as: sin 1 sin 2 n2 n1 n1 sin 1 n2 sin 2 The general form of Snell\u2019s Law applies in all cases, regardless of the indexes of refraction of the media and/or the direction in which the light travels. Compare the general form with the simple form above. If the first medium is air, or a vacuum, so that n1 is 1.00, substituting this value into the general form results in the simple form of Snell\u2019s Law. Example 13.3 Practice Problems 1. Light passes from a diamond into air. The angle of refraction as the light emerges from the diamond is 25. What was the angle of incidence? 2. Light travelling from air into a transparent material is incident at an angle of 20 and refracted at an angle of 17. Determine the index of refraction for the transparent material. 3. A ray of light passes from air into ruby at an incident angle of 15. Calculate the angle of refraction. 4. A ray of light, travelling in air, is incident on an unknown sample at an angle of 20. If the angle of refraction is 15, determine the index of refraction for the unknown sample. Answers 1. 10 2. 1.2 3. 9.7 4. 1.3 Yellow light travels from water into crown glass. The light rays are incident on the crown glass at an angle of 35. Calculate the angle of refraction as the light enters the crown glass. Given 1 35 Required the angle of refraction ( 2) Analysis and Solution From Table 13.4, 1.33 \u2022 Index of refraction of water, the first medium, is n1 \u2022 Index of refraction of crown glass, the second medium, is n2 1.52 Therefore, the general form of Snell\u2019s Law applies. n1 sin 1 sin 2 2 n2 sin 2 in n1 s 1 n 2 sin1(1.33 si ( ) 2 5. 1 30 n35) Paraphrase The angle of refraction for the light travelling from water into crown glass is 30, when the angle of incidence is 35. 668 Unit VII Electromagnetic Radiation 13-PearsonPhys30-Chap13 7/24/08 3:44 PM Page 669 Snell\u2019s Law, Refraction, and Wavelength incident wave fronts \u03bb2 2 d ct D refracted wave fronts Figure 13.46 Wave fronts are refracted as they pass from air into water. vacuum speed c less refractive medium n 1.00 more refractive medium n > 1.00 Can we explain the phenomenon of refraction, and by association, Snell\u2019s Law, by considering light as a wave? Recall that the wave model describes light as a stream of transverse waves radiating outward from a source. Let us assume that the light waves are indeed transverse waves of electromagnetic radiation. As each wave front moves from one medium to the next, there is no change in frequency. In other words, the wave fronts do not \u201cpile up\u201d at the boundary between the two media. The number of waves arriving at the boundary is equal to the number of waves leaving the boundary for any given time interval, so the incident and refracted waves have the same frequency. However, the wavelength changes at the boundary because the speed changes. According to the universal wave equation (v f), if the speed changes and the frequency remains constant, then the wavelength must change as well. To visualize this, consider light waves travelling from air into water as illustrated in Figure 13.46: medium speed v The wavelength in air is given by 1 v1/f; The wavelength in water is given by 2 v2/f. The wave fronts travel faster in air than they do in water (F", "izeau\u2019s 1 is experiment, page 649). It follows that v1 is greater than v2, and greater than 2. An enlarged view of two wave fronts, as shown in Figure 13.47, reveals another relationship. When we use similar triangles, the angle of incidence 2 are shown within the coloured right triangles and they share the same hypotenuse, x. 1 and the angle of refraction Therefore, the trigonometric ratio for the sine of an angle when applied to each triangle gives: 1 x opposite hypotenuse sin 1 sin 2 opposite hypotenuse 2 x Since x 1 sin 1 therefore and also x 2 sin 2 1 sin 1 2 sin 2 This equation can also be written as: sin 1 sin 2 1 2 incident ray incident wavefront less optically dense medium more optically dense medium refracted wavefront \u03b8 2 refracted ray Figure 13.47 Enlarged view of two wave fronts Chapter 13 The wave model can be used to describe the characteristics of electromagnetic radiation. 669 13-PearsonPhys30-Chap13 7/24/08 3:44 PM Page 670 When the universal wave equation is used, the wavelength can also be written in terms of the wave speed as follows. sin 1 sin 2 sin 1 sin 2 v 1 f v 2 f v1 v2 Therefore, Snell\u2019s Law can be expanded to read: sin 1 sin 2 v1 v2 1 2 n2 n1 By measuring the incident and refracted angles, velocity, and wavelength, this relationship can be tested. Example 13.4 Practice Problems 1. Determine the speed of light in the following materials: (a) water (b) ethanol (c) ruby (d) crown glass 2. Light with a wavelength of 737 nm enters quartz glass at an angle of 25.0. Determine the angle of refraction and the wavelength of the light in the quartz glass. 3. Light enters an unknown crystal from air with a wavelength of 500 nm. If the wavelength of the light in the crystal is found to be 450 nm, what is the refractive index of the crystal? Answers 1. (a) 2.26 108 m/s (b) 2.19 108 m/s (c) 1.95 108 m/s (d) 1.97 108 m/s 2. 16.7, 501 nm 3. 1.11 670 Unit VII Electromagnetic Radiation A ray of yellow light with a wavelength of 570 nm travels from air into diamond at an angle of 30. Determine the following: (a) the speed of light in the diamond (b) the wavelength of the light as it travels in the diamond Given 1 570 nm 5.70 107 m Required the speed of light in diamond (v2) the wavelength of light travelling in diamond ( 2) Analysis and Solution v1, the speed of light in air 3.00 108 m/s From Table 13.4, \u2022 Index of refraction of air, the first medium, is n1 \u2022 Index of refraction of diamond, the second medium, 1.00 is n2 2.42 n i The general form of Snell\u2019s Law applies Therefore, n1v1 n2 v2 2 n1 1 n2 1.00 (3.00 108 ms) 2.42 1.00 (5.70 107 m) 2.42 1.24 108 ms 2.36 107 m 236 nm Paraphrase (a) The speed of light in the diamond is 1.24 108 m/s. (b) The wavelength of light in the diamond is 236 nm. 13-PearsonPhys30-Chap13 7/24/08 3:44 PM Page 671 13-6 Inquiry Lab 13-6 Inquiry Lab Determining the Refractive Index of a Variety of Materials Required Skills Initiating and Planning Performing and Recording Analyzing and Interpreting Communication and Teamwork In this experiment, known values of the refractive indexes of water and ethanol are verified. rays from ray box Question What are the refractive indexes of water and alcohol (ethanol)? Variables Manipulated variable: angle of incidence Responding variable: angle of refraction Controlled variables: refracting substance, wavelength of light 360 0 10 10 20 20 30 30 less refractive (air) more refractive (liquid) 320 40 310 50 300 60 30 210 20 200 10 190 0 180 10 170 3 0 15 60 120 50 130 40 140 semicircular container Materials and Equipment polar coordinate paper graphing paper water ethanol single-slit ray box or laser semicircular plastic dish Procedure 1 Design a data table or spreadsheet with the headings: angle of incidence (i), angle of refraction (r), sin angle of incidence (sin i), sin angle of refraction (sin r), and ratio sin i / sin r. 2 Fill the semicircular dish with water and place it on the polar coordinate paper such that the 0\u2013180\u00b0 line is perpendicular", " to the centre of the flat side of the dish. This will make the 0\u2013180 line the normal line (Figure 13.48). 3 Direct a single ray of light along the normal line (0). Record the angle of refraction. (This should be zero if the apparatus is set up correctly.) 4 Record the angles of refraction for 10 to 70, increasing the angle by 10\u00ba for each step. 5 Complete the data table or spreadsheet calculations and plot a graph of sin angle of incidence vs. sin angle of refraction. Calculate the slope of this graph. 6 Repeat steps 2 to 5 using ethanol instead of water in the semicircular dish. Figure 13.48 Semicircular dish on polar coordinate paper with several incident ray angles shown Analysis 1. (a) What is the value of the slope for the graph when water was used as the refracting substance? (b) What is the value of the slope when ethanol was used as the refracting substance? 2. According to your slope calculations, which material, water or ethanol, has a higher index of refraction? 3. Calculate the percent error for each index of refraction using the absolute index of refraction given for each substance in Table 13.4. 4. Comment on the sources of error that could have occurred in this experiment. 5. Why is it important that the incident light ray contacts the semicircular dish at the centre? 6. Why is it important that a \u201csemicircular\u201d dish is used, instead of a rectangular dish? 7. Which material is more effective in changing the direction of light when light enters it? 8. Predict what would happen if the semicircular dish were replaced with a semicircular block of glass or Lucite\u00ae. If you have time, test your prediction. Chapter 13 The wave model can be used to describe the characteristics of electromagnetic radiation. 671 13-PearsonPhys30-Chap13 7/24/08 3:44 PM Page 672 27\u00b0 42\u00b0 59\u00b0 20\u00b0 40\u00b0 30\u00b0 50\u00b0 50\u00b0 0\u00b0 0\u00b0 light source air n 1.00 water n 1.33 Total Internal Reflection If you look down a long straight tube, you should be able to see what is at the other end. If the tube has a bend in it, a set of mirrors could be used to reflect the light through to your eye. This is how the periscope on a submarine works. But what if the tube has multiple bends in it? What could be done to ensure the light is able to travel through the entire length of the tube? Could you line the inside walls with mirrors and angle the light so that it reflects from side to side all along the tube, entering at one end and emerging at the opposite end? Figure 13.49 Increasing the incident angle for light leaving a water-air interface leads to total internal reflection. In theory, this is exactly what happens in the optical fibres that facilitate the data and telephone communication of computer networks. The process is based on total internal reflection. Recall that when light travels from one medium to another, some of the light is partially reflected and some is partially refracted. When light travels from a medium with a high refractive index, like glass, to a medium with a low refractive index, like air, more of the light is reflected than it would be at an interface where its speed decreases. As illustrated by Figure 13.49, when light travels from the water with a high refractive index to the air with a low refractive index, it bends away from the normal line. As the angle of incidence is increased, the angle of refraction also increases, bending farther and farther from the normal line, and eventually reaching the maximum angle of 90. Beyond this angle, refraction ceases and all incident light will be reflected back into the high-index medium (in this case, the water). Therefore, not all the light that approaches a water-air boundary will be refracted. The light approaching the boundary at large angles will be reflected back into the water. This is why underwater pool lights, for example, are not visible at all angles from above. At particular angles, they cannot be seen. When the angle of refraction is 90, the incident angle will have a value unique to the two media that form the interface. This unique angle of incidence is called the critical angle. For light travelling from a medium with a high refractive index to a medium with a low refractive index, the critical angle is determined by assuming that the angle of refraction is 90. Concept Check 1. Is light refracted toward or away from the normal line when passing from a medium with a low refractive index like air, into a medium with a high refractive index like water? 2. Based on your answer above, can total internal reflection occur when light travels from a low-index medium like air, into a high-index medium like water? Explain why or why not. total internal", " reflection: reflection of all incident light back into a medium of higher refractive index due to the inability to refract light beyond the maximum angle of 90 critical angle: for any two media, the size of the incident angle for which the angle of refraction is 90 info BIT The sparkle and glitter associated with diamonds is caused by total internal reflection. The high index of refraction of the diamond and the skill of the jeweller in cutting and finishing are used to trap and focus light, which we observe as sparkles. Polar Bear Diamonds\u2122, mined, cut, and polished in the Northwest Territories, are recognized for their quality worldwide. Figure 13.50 672 Unit VII Electromagnetic Radiation 13-PearsonPhys30-Chap13 7/24/08 3:44 PM Page 673 Example 13.5 What is the critical angle for the quartz glass-air interface? Given The angle of refraction at the critical angle is defined to be r 90.0. Required the critical angle for this interface ( critical) Analysis and Solution From Table 13.4, 1.00 \u2022 Index of refraction of air, the first medium, is n1 \u2022 Index of refraction of quartz glass, the second medium, is n2 1.47 Use the form of Snell\u2019s law for the critical angle: n2 sin critical sin critical critical n1 sin 90.0 90.0 n n1 si n 2 90.0 sin1 n n1 si n 2 90.0 sin1 in s 1.00 7.4 42.9 1 Paraphrase Light at an incident angle of 42.9 or greater will be internally reflected at the quartz glass-air interface. Practice Problem 1. Determine the critical angle for the following interfaces: (a) water and air (b) diamond and air (c) diamond and water Answer 1. (a) 48.8 (b) 24.4 (c) 33.3 e MATH To investigate Snell\u2019s Law, refraction, and the concept of the critical angle, using a graphing calculator or a spreadsheet program, visit www.pearson.ca/ school/physicssource. cladding Total internal reflection has many applications. Most notable are optical fibres (Figure 13.51). An optical fibre consists of a central core of glass with a refractive index of approximately 1.5, surrounded by a cladding material of a slightly lower refractive index. Such fibres are used to transmit data on computer and communication networks. The data are transmitted via the modulation of laser light travelling through a glass fibre, with virtually none of the energy loss associated with electrical transmission. Some of the advantages of fibre-optic networks include: \u2022 economically less expensive than copper wire of equivalent length \u2022 thinner, more flexible, and made of non- flammable materials core >\u03b8 c >\u03b8 c Figure 13.51 An optical fibre showing cladding, core, axis, and critical angle \u2022 able to handle a higher data-carrying capacity based on the fibre bundle diameter \u2022 less signal degradation and interference between multiple signals on the same fibre as compared with copper networks \u2022 the glass fibres are highly transparent so that repeaters (amplifiers) can be many kilometres apart, as opposed to coaxial cable repeaters that must be less than 1 km apart e WEB To learn more about fibre optics including applications in audiovisual equipment and in nanotechnology, begin your search at www.pearsoned.ca/school/ physicssource. Chapter 13 The wave model can be used to describe the characteristics of electromagnetic radiation. 673 13-PearsonPhys30-Chap13 7/24/08 3:44 PM Page 674 info BIT Buckyballs, also called C60 or buckminsterfullarenes, are soccer-ball-shaped molecules made of 60 linked carbon atoms. These carbon molecules may be formed into nanotubes to transmit data using light. info BIT Binoculars consist of two identical telescopes mounted side by side and aligned to point in the same direction. When the images from each side of the binoculars are viewed simultaneously, the user is able to sense depth and distance that is not possible with a single telescope view. eyepiece Porro prisms objective Figure 13.52 The light path in binoculars The Porro prisms in the binoculars use total internal reflection to direct light through the compact, short body of the binoculars with minimal absorption, producing a high-quality image. F F F Figure 13.53 A double Porro prism A double Porro-prism system is used in binoculars to reorient an inverted image while at the same time producing a longer, folded pathway for the light to travel between the objective lens and the eyepiece, producing a greater magnification. Fibre-optic systems also have drawbacks. The complex design of the fibres makes the fibre-optic system relatively expensive to set up and the fibres are also", " subject to wear and breakage. 13-7 Decision-Making Analysis 13-7 Decision-Making Analysis Fibre Optics: Endless Light The Issue The new knowledge society of today has emerged as a result of our ability to communicate and transmit data on a large scale. And to a very large extent, the infrastructure that supports our communication is based on the principles of total internal reflection and its application in fibre-optic systems. New and innovative advances are being made with fibre optics every day \u2013 from nanotechnology to entertainment services. Background Information The flexibility of fibre optics has allowed for applications in a variety of industries. For example, telephone, television, and data networks, originally all separate industries, are merging into one large application, which is supported by a fibre-optic backbone. In medicine, applications include the use of a fibrescope to both illuminate internal organs and capture images of them. For example, to see inside the small intestine, an endoscope, a small, flexible bundle of fibres is inserted down the patient\u2019s throat and through the stomach. Once the fibres reach the small intestine, images can be transmitted through individual fibres while others are simultaneously used to illuminate the tissue. This application prevents the need for more-invasive, risky procedures that can leave the patient with a higher risk of infection and perhaps result in a prolonged hospital stay. Other applications include mechanical imaging. For example, fibre-optic bundles can be used to inspect the interior of long pipes, vessels, and hard-to-reach locations, such as bore holes for oil and water wells, pipelines, and hazardous goods containers. With proper instrumentation devices attached, fibre-optic bundles can be used in dangerous conditions to identify gases, pressures, temperatures, and concentrations. Nanotechnology is an emerging application for fibre optics. In this industry, new nanomaterials, such as C60 ( buckyballs), are being investigated to expand the power and application of data transmission using light. Analyze and Evaluate 1. Research applications of fibre optics using the Internet, research journals, and periodicals. 2. Group all the applications by industry and identify what you believe to be the most important applications in each industry. 3. For each of the most important applications, identify the social, political, economic, and environmental impact of the technology. 4. New applications of fibre optics are emerging in nanotechnology. Identify these applications and describe what technological advances could be expected as a result of merging nanotechnology with fibre-optic technology. 5. Prepare a multimedia presentation that demonstrates advances that have been made in a number of industries as a result of fibre-optic applications. Predict what future applications and social issues will evolve as new technologies begin to merge with fibre optics. 674 Unit VII Electromagnetic Radiation 13-PearsonPhys30-Chap13 7/24/08 3:44 PM Page 675 Prisms \u2014 Dispersion of White Light Where have you seen a rainbow? Everyone has seen one in the sky, but what about in a diamond, or perhaps on the wall near a hanging crystal in a window. There are many items that produce a rainbow of colours, but all the rainbows have something in common. They all originate from white light. The investigation of such effects was undertaken by Newton in 1666. He placed a transparent prism in a beam of sunlight passing through his window shutter at Cambridge University. On the opposite wall, an elongated band of colours appeared and he called this a spectrum (Figure 13.54). In this spectrum, Newton noted the colours red, orange, yellow, green, blue, indigo, and violet, in the order in which they appeared. info BIT It may be that Newton noted seven colours in the spectrum, in part to go along with the views of society at that time: numerology regarded 7 as a \u201cspecial\u201d number. Six colours are more commonly used, as in Figure 13.55 and Table 13.5. spectrum: the bands of colours making up white light; in order, red, orange, yellow, green, blue, violet dispersion: separation of white light into its components Figure 13.54 The complete spectrum produced by white light passing through a prism Newton set out to determine which of two things was true: either the colours of the spectrum are added to the white light by the prism, or the prism separates all the colours from the white light. To test this, Newton set up two prisms, the first one exposed to white light, producing a spectrum of colours (Figure 13.55), and the second one, only exposed to the red (monochromatic) light coming from the first prism. The second prism did not produce any more colours; only the red light emerged. As a second test, Newton placed a converging lens into the path of the spectrum of light and observed the resulting white light as an image on a sheet of white paper. slit aperture white light prism Based on his observations, Newton concluded that white light is made up of all", " the colours in the spectrum and the prism was simply separating the colours from one another. The separation of white light into its components is called dispersion. screen red orange yellow green blue violet visible spectrum Chapter 13 The wave model can be used to describe the characteristics of electromagnetic radiation. 675 Figure 13.55 Dispersion of white light by a prism 13-PearsonPhys30-Chap13 7/24/08 3:44 PM Page 676 The recomposition of the spectrum, therefore, should produce white light, as Newton was able to demonstrate in his second experiment. Recomposition can also be demonstrated by painting all the colours of the spectrum in certain proportions on a disc and spinning the disc at a very high speed (Figure 13.56). At sufficient speed, the disc appears white. Is dispersion consistent with the wave model of light? If we consider light as a wave, each of the colours in the spectrum has a unique wavelength, as listed in Table 13.5. As light enters a medium with a high refractive index, the wavelength is reduced but the frequency is unchanged. The speed of a wave is the product of its frequency and wavelength as described by the universal wave equation (v f ). Therefore, the light must slow down as it enters the highindex medium. Because the refractive index is related to the speed of a wave, it is therefore also related to the wavelength of the wave when the frequency is constant. n o o wavelength in vacuum wavelength in medium This relationship between refractive index and wavelength, for a constant frequency, means that the refractive index is different for each wavelength of light that passes through the same medium. Incident light with a smaller, shorter wavelength will slow down and refract to a greater extent than light with a larger, longer wavelength. As each wavelength of light refracts at a slightly different angle, the wavelengths will separate, producing a continuous spectrum. The wave model of light, therefore, is consistent with the phenomenon of dispersion. Figure 13.56 Recomposition of the spectrum using a high-speed disc Table 13.5 Wavelength of each colour in the spectrum e SIM To learn more about the effects of prisms through the use of simulations, follow the eSim links at www.pearsoned.ca/school/ physicssource. Colour violet blue green yellow orange red Wavelength (nm) 400-450 450-500 500-570 570-590 590-610 610-750 676 Unit VII Electromagnetic Radiation 13-PearsonPhys30-Chap13 7/24/08 3:44 PM Page 677 13-8 QuickLab 13-8 QuickLab Dispersion Using a Prism Problem Is white light made up of all the colours we can see? Materials 2 identical prisms convex lens intense light source (flashlight or white light ray box) white paper Procedure 1 Using one prism, observe the colour spectrum on a piece of white paper. A dark room and a very intense white light source will work best. The convex lens may be placed between the light source and prism to intensify the beam of white light if the spectrum is not immediately clear and visible. 2 Position the second prism in the spectrum produced by the first one. Rotate the second prism and position the white paper so that you can observe the light that emerges from the second prism. 3 Explore a variety of positions and orientations with both prisms. Questions 1. What is the best way to make a colour spectrum using a single prism? Draw a picture showing the flashlight, prism, convex lens, paper, and the colour spectrum. 2. List the colours of the spectrum from most refracted to least refracted. 3. Compare the wavelengths of each colour to the amount of refraction observed. Describe the relationship between wavelength and amount of refraction. 4. Are the colours of the spectrum added to, or separated from, white light when white light passes through a prism? Support your answer with observations made using two prisms. 5. The spectrum produced by a prism is similar to that observed as a rainbow. Research and explore the similarities between the spectrum produced by a prism and that which is observed as a rainbow. Thin Lenses Projection systems connected with computers and movie players, and optical systems including microscopes and telescopes, use lenses that refract light in order to generate images. A typical thin lens is a circular piece of transparent material with a spherically shaped surface that refracts, or changes the direction of light that passes through it. The uniformly curved surface will refract light rays to varying extents depending on where they contact the lens. Rays that are incident near the edge of the lens will be refracted at larger angles than those that are incident near the centre, where the two faces of the lens are almost parallel. For a converging lens, rays that travel parallel to the principal axis will be refracted inward, intersecting at the principal focus, F. For a diverging lens, rays that travel parallel to the principal axis will be ref", "racted outward, appearing as though they have originated at a virtual principal focus, F. The distance from each lens to F is the focal length, f (Figures 13.57 and 13.58). Each type of lens has a secondary focus, F at a distance of f, on the opposite side of the lens. converging lens: a lens that refracts rays travelling parallel to the principal axis inward to intersect at the principal focus diverging lens: a lens that refracts rays travelling parallel to the principal axis outward to appear as though they have originated at a virtual principal focus Chapter 13 The wave model can be used to describe the characteristics of electromagnetic radiation. 677 13-PearsonPhys30-Chap13 7/24/08 3:44 PM Page 678 PHYSICS INSIGHT A ray that travels through the optical centre is indeed refracted. This causes the refracted ray to emerge parallel to, but laterally shifted from, the incident ray. For a thin lens, this lateral shift is so small that it is assumed that the ray is not refracted. O F principal focus f focal length Figure 13.57 Converging lens F virtual principal focus PA optical centre F O lateral shift Figure 13.58 Diverging lens f focal length Figure 13.59 Lateral shift Drawing Ray Diagrams for Thin Lenses A ray diagram is a useful tool for predicting and understanding how images form as a result of light rays emerging from a curved lens. Ray diagrams for lenses are similar to the ray diagrams used with curved mirrors; only two rays are needed to predict the image location, and a third is used as verification (Figure 13.60). 1 3 optical centre image object F 2 F 2 3 1 object 1 2 3 F image F Figure 13.60 Ray diagrams for (a) converging, and (b) diverging lenses 678 Unit VII Electromagnetic Radiation Ray 1 travels parallel to the principal axis and is refracted such that it emerges and passes through (or appears to have originated from) the principal focus, F. Ray 2 travels through (or appears to be directed toward) the secondary focus, F, and is refracted such that it emerges and travels parallel to the principal axis. Ray 3 travels straight through the optical centre of the lens and is not bent. The ray diagram is not only used to identify the location of an image relative to the lens, but it can also illustrate the other three image attributes of type, attitude, and magnification, that have previously been described for curved mirrors. Relative to the object, the image produced by a thin lens can be real or virtual, inverted or erect, and enlarged or diminished. 13-PearsonPhys30-Chap13 7/24/08 3:44 PM Page 679 Example 13.6 2F and 2F represent two points each at a distance of 2f (twice the focal length) from a lens on either side of the lens. Draw a diagram to determine the image attributes for an object located at the following positions relative to a converging lens. a) beyond 2F b) at F c) between the lens and F Given Object position relative to the converging lens Required Ray diagram showing image location and attributes Analysis and Solution Construct a ray diagram and visually inspect the image location relative to the converging lens and the image attributes.. object 2F F image F 2F Figure 13.61 (a) object 2F F F 2F Figure 13.61 (b) image Practice Problems 1. Using a ray diagram, determine the image attributes for the following: (a) an object located between 2F and F relative to a converging lens (b) an object located between 2F and F relative to a diverging lens (c) an object located between F and a diverging lens Answers 1. (a) Image is beyond 2F, real, inverted, and enlarged. (b) Image is between the lens and F, virtual, erect, and diminished. (c) Image is between the lens and F, virtual, erect, and diminished. 2F F object F 2F Figure 13.61 (c) Paraphrase (a) The image is located between F and 2F, and is real, inverted, and diminished. (b) No image is formed. (c) The image is located between F and 2F, and is virtual, erect, and enlarged. Chapter 13 The wave model can be used to describe the characteristics of electromagnetic radiation. 679 13-PearsonPhys30-Chap13 7/24/08 3:44 PM Page 680 e WEB Equations for Thin Lenses To learn more about thin lenses, follow the links at www.pearsoned.ca/ school/physicssource. An equation relating object distance, image distance, and the focal length of a curved lens can be derived using an analysis nearly identical to that for curved mirrors. In Figure 13.62, the green triangle XOF and the blue triangle ZYF are similar. Therefore", ", ho object X F O F Y hi image Z do f di Figure 13.62 Thin lens ray diagram OX OF ho f hi ho hi ho di do di do YZ YF hi f di f di f di do f di f di f f f Dividing both sides by di, and simplifying, gives the thin lens equation. thin lens equation: the equation that relates object distance, image distance, and focal length of a curved lens 1 di d i d ido 1 do d i d i f 1 di 1 f Notice that this equation is identical to the mirror equation. A sign convention will be used here also, to distinguish between real and virtual distances as well as to identify erect and inverted images. Sign Conventions for the Thin Lens Equation \u2022 All distance measurements are relative to the optical centre of the lens. \u2022 Positive distances are used for real objects and images. \u2022 Negative distances are used for virtual images. \u2022 Positive image and object heights are upward relative to the principal axis. \u2022 Negative image and object heights are downward relative to the principal axis. \u2022 Converging lenses have a positive focal length. \u2022 Diverging lenses have a negative focal length. Results with a thin lens are similar to those with a curved mirror in that an erect image has a positive magnification and an inverted image has a negative magnification. Therefore, the same magnification equation that is used for curved mirrors is also used for thin lenses. m hi ho di do 680 Unit VII Electromagnetic Radiation 13-PearsonPhys30-Chap13 7/24/08 3:44 PM Page 681 Example 13.7 A 2.5-cm-high object is placed 10.0 cm from a diverging lens of focal length 5.0 cm. Determine the image distance, height, and attributes using the thin lens equation. Verify your answer with a ray diagram. Given ho do 2.5 cm 10.0 cm f 5.0 cm (The lens is a diverging lens.) Required image distance (di) height (hi) image characteristics Analysis and Solution The thin lens equation can be used to determine the image distance, image height, and attributes. A ray diagram can verify the answer hi 1 1 1 hi 10. 5.0 0 cm d i cm ih d o d o (3.33 cm)(2.5 cm) 10.0 cm 3 0 cm 10. di cm 10.0 3 3.3 cm 0.83 cm Paraphrase The image distance is 3.3 cm and height is 0.83 cm, indicating the image is virtual, erect, and diminished. The answers are verified by the ray diagram in Figure 13.63. object 2F F image F 2F Figure 13.63 Practice Problems 1. A 3.00-cm-high object is located 15.0 cm from a converging lens with a focal length of 10.0 cm. (a) How far is the image from the lens? (b) How high is the image? (c) Describe the image characteristics and verify them using a ray diagram. 2. A 10.0-cm-high candle is placed 100.0 cm from a diverging lens with a focal length of 25.0 cm. Determine the following using a ray diagram and the thin lens equation: (a) the image location from the lens (b) the image height (c) the type of image formed 3. A projector uses a converging lens to create a focussed image on a screen located 5.00 m away. The image is generated from a slide located 7.50 cm from the lens. (a) What is the focal length of the lens? (b) Determine the magnification of the image. Answers 1. (a) di (b) hi (c) image is real, inverted, enlarged 30.0 cm 6.00 cm 20.0 cm 2.00 cm 2. (a) di (b) hi (c) image is virtual 3. (a) f 7.39 cm (b) m 66.7 X Chapter 13 The wave model can be used to describe the characteristics of electromagnetic radiation. 681 13-PearsonPhys30-Chap13 7/24/08 3:44 PM Page 682 13-9 Inquiry Lab 13-9 Inquiry Lab Comparing Keplerian and Galilean Telescopes Required Skills Initiating and Planning Performing and Recording Analyzing and Interpreting Communication and Teamwork In this investigation the functional differences between a Keplerian telescope, with two converging lenses, and a Galilean telescope, with one converging lens and one diverging lens, are explored. Question What are the differences between Keplerian and Galilean telescopes? Variables manipulated: type of lens used responding: magnification of the image controlled: focal length of double concave and double convex lenses Materials and Equipment long focal length converging lens short focal length converging lens short focal length diverging lens lens holders \u2022 \u2022 \u2022 \u2022 \u2022 optical bench for measuring distances \u2022", " \u2022 paper, ruler, pencil, and tape lamp Procedure for Galilean Telescope 1 Place the long focal length converging lens at one end of the optical bench. This will serve as the objective lens for both telescopes. 2 If the long focal length is unknown, place a lamp in front of the lens and measure the distance between the lamp and the lens. Next, slide a sheet of paper or screen along the optical bench until a bright spot appears (image of the lamp). Record the distance between the paper and the lens. Calculate the focal length of the objective lens using the lens equation, and image and object distances. 3 Place the short focal length diverging lens halfway between the converging lens and its principal focus as illustrated in Figure 13.64. objective lens eyepiece fo light from object Figure 13.64 Galilean telescope 682 Unit VII Electromagnetic Radiation 4 Draw a series of 5 vertical arrows, 1 cm apart, on a sheet of paper. Tape the sheet to a distant wall so that it will be visible in the telescope. 5 Look through the eyepiece with one eye and compare the image with the sheet seen with an unaided eye. 6 Estimate how many vertical arrows (viewed with the unaided eye) would fit in between two of the vertical arrows observed in the telescope. Record this as the estimated magnification. Procedure for Keplerian Telescope 7 Remove the diverging lens from the optical bench. 8 If the focal length of the short focal length converging lens is not known, place a lamp in front of the lens and slide a sheet of paper or screen along the optical bench until a bright spot appears. Record the focal length for the short focal length lens. 9 Position the short focal length converging lens such that the distance between the two lenses is slightly less than the sum of the focal lengths, as shown in Figure 13.65. fo objective lens fe eyepiece light from object Figure 13.65 Keplerian telescope 10 Estimate how many vertical arrows (viewed with the unaided eye) would fit in between two of the vertical arrows observed in the telescope. Record this as the estimated magnification. Analysis 1. Prepare a table that compares the attitude, magnification, and brightness of the image in each telescope. 2. Complete a ray diagram to show how the image is formed in each telescope. 3. Suggest a different application for each telescope. Compare the telescopes\u2019 efficiency in looking at relatively close terrestrial objects, like mountains, or far away extraterrestrial objects, such as the Moon. 13-PearsonPhys30-Chap13 7/24/08 3:44 PM Page 683 13.4 Check and Reflect 13.4 Check and Reflect Knowledge 1. Is light bent toward or away from the normal line when it passes from a lowindex medium to a high-index medium? 2. How is the index of refraction measured for a particular medium? 3. Construct a concept map using the following terms: incident ray, refracted ray, angle of incidence, angle of refraction, normal line. 4. What is the difference between the simple and general forms of Snell\u2019s Law? 5. Suggest an experimental design to determine the critical angle for a Lucite\u00ae-air boundary. 6. List five advantages of a fibre-optic network when compared to traditional electric networks used for data transmission. 7. How does a prism separate all the colours of white light? 8. Explain the effect of wavelength on the index of refraction for a particular medium, like glass. How does this lead to dispersion? 15. Using a ray diagram, determine the image attributes for an object located at 2F relative to a converging lens. 16. A camera with a converging lens (f 4.50 cm) is used to take a picture of a 25.0-m-high tree that is 50.0 m from the camera. How tall is the image? Is it erect or inverted? 17. Red light of 700 nm and blue light of 475 nm are both incident on a Lucite\u00ae block. Which colour will be refracted to a greater extent? 18. The focal length of the converging lens in a computer projector is 8.00 cm. An LCD panel positioned inside the projector serves as the object for the lens. (a) If the LCD panel is located 8.10 cm from the lens inside the projector, how far away should the screen be placed so that a clear image is produced? (b) If the LCD panel is 1.75 cm high, how large is the image on the screen? (c) If the screen is moved 3.0 m closer to the projector, how far must the object now be from the lens in order to generate a focussed image? Applications 9. What is the speed of light in water (n 1.33)? Extensions 19. Explain why jewellery crystals, such as 10. A light ray is incident on a block of quartz diamonds, sparkle. glass (n 1.47) at", " an angle of 35.0. Determine the angle of refraction. 11. When a ray of light passes from water (n 1.33) into a Lucite\u00ae block (n 1.52) it is refracted at an angle of 28.0. Determine the angle of incidence. 12. What is the critical angle for a Lucite\u00ae-air interface? 13. A light ray travelling in water approaches the water-air boundary at an angle of 50. What happens to the light ray at the boundary? 14. Light with a wavelength of 540 nm enters a ruby crystal (n 1.54) at an angle of 25.0. Determine the angle of refraction and the wavelength of the light in the ruby. 20. Explain why you can start a fire with a magnifying glass. 21. If a diver is underwater and looks up, a circular \u201chole\u201d appears on the surface of the water directly above the diver. Other than this hole, the surface appears as a mirror. Explain how this happens. e TEST To check your understanding of refraction, fibre optics, dispersion, and lenses, follow the eTest links at www.pearsoned.ca/school/physicssource. Chapter 13 The wave model can be used to describe the characteristics of electromagnetic radiation. 683 13-PearsonPhys30-Chap13 7/24/08 3:44 PM Page 684 Figure 13.66 Christiaan Huygens Huygens\u2019 Principle: a model of wave theory, which predicts the motion of a wave front as being many small point sources propagating outward in a concentric circle at the same speed as the wave itself 13.5 Diffraction and Interference In section 8.3 you learned that there are many forms of interference patterns produced by mechanical waves that travel in a medium like water. And more importantly, you learned that the interference pattern contains information about the properties of the waves that created the pattern. In this section, we will investigate interference patterns produced by electromagnetic radiation and analyze the patterns to further our understanding of the wave model of light. Huygens\u2019 Principle Robert Hooke proposed the Wave Model of Light in his Micrographica of 1665. The first major improvement to this model was made by Christiaan Huygens (Figure 13.66), twenty years later. A Dutch physicist, Huygens contributed to the study of astronomy by advancing techniques in lens grinding and by discovering Saturn\u2019s rings and its largest satellite, Titan. He is also credited with producing the pendulum clock, originally proposed by Galileo. With regard to wave theory, Huygens described an elegant, conceptual model that predicted the motion of a wave front. This model is known as Huygens\u2019 Principle. A wave front consists of many small point sources of tiny secondary waves, called wavelets, which propagate outward in a concentric circle at the same speed as the wave itself. The line tangent to all the wavelets constitutes the wave front. To visualize Huygens\u2019 Principle, consider a point source of light that emits electromagnetic waves outward, in a concentric circle. At an instant in time, the wave front will form the line t1. According to Huygens\u2019 Principle, all the points along line t1 become secondary sources, producing wavelets that radiate outward to form the wave front at a future time t2. Line t2 is the tangent to all the wavelets that make up the wave front a short time later (Figure 13.67). The same analysis can be performed on a straight wave front, as shown in Figure 13.68. t1 t2 point source of EMR point source of EMR t2 t1 circular wave front straight wave front Figure 13.67 Circular wave front emitted by a light source Figure 13.68 Straight wave front emitted by a distant light source 684 Unit VII Electromagnetic Radiation 13-PearsonPhys30-Chap13 7/24/08 3:44 PM Page 685 Concept Check Use Huygens\u2019 Principle to predict the wave front shapes that occur after the straight waves pass through the openings shown in Figure 13.69 (a) and (b). (a) barrier (b) barrier t t e SIM Explore Huygens\u2019 Principle through simulations. Follow the eSim links at www.pearsoned.ca/school/ physicssource. Figure 13.69 At each opening, one or more secondary point sources exist. Draw the line tangent across the wavelets that would exist a short time after the secondary point sources pass through the opening. Explain what has happened to the shape and direction of the wave front after it passed through the barriers in Figure 13.69 (a) and (b). According to Huygens\u2019 Principle, a periodic straight wave will continue to propagate in a straight line until it encounters a barrier. If the", " wave front encounters a small opening or aperture in the barrier, the wave front will change shape and direction. This process is called diffraction. Young\u2019s Experiment Huygens\u2019 wave theory can explain many of the properties of electromagnetic radiation, including reflection and refraction, but initially, the scientific reputation of Newton and his belief in the particle model of light dominated the scientific community. However, in 1801, an experiment by Thomas Young provided significant evidence in support of the wave model of light. Problems Involved in the Design of the Experiment In the years leading up to Young\u2019s groundbreaking work, scientists studying light were attempting to observe an interference pattern that was similar in nature to that of two point sources. They believed that if light were indeed a wave, then two side-by-side light sources should produce an interference pattern similar to that observed with water in a ripple tank (section 8.3). In these early light experiments, the light from the two sources was incident on a nearby screen that was observed for evidence of an interference pattern. What was not well understood at the time was that waves of high frequency and short wavelength result in a very short distance between nodal lines, the regions of destructive interference that appear dark. As we know today, light waves have an extremely high frequency and short wavelength, so the distance between nodal lines would be so small it could diffraction: the change in shape and direction of a wave front as a result of encountering a small opening or aperture in a barrier, or a corner interference: an effect resulting from the passage of two like waves through each other e SIM Learn more about Young\u2019s classic experiment through simulations. Follow the eSim links at www.pearsoned.ca/ school/physicssource. Chapter 13 The wave model can be used to describe the characteristics of electromagnetic radiation. 685 13-PearsonPhys30-Chap13 7/24/08 3:44 PM Page 686 Figure 13.70 top down view interference pattern sunlight pinhole card not be observed using traditional means. In addition, to produce an interference pattern, the waves must be in phase. When there are two incandescent lights, each source emits light in random bursts not necessarily in phase with the other source. These bursts of light have a variety of wavelengths that make up all the colours in the spectrum. Therefore, the interference pattern would not be constant, but would vary rapidly over time, making any observation of a pattern more difficult. screen (a) Interference fringes (b) Photo of two-slit pattern from Young\u2019s experiment info BIT The double-slit experiment works for water waves too! When a single wave front is incident on a barrier with two holes it will produce an interference pattern similar to the one observed by Young. This effect is explored in Unit IV. Figure 13.71 Interference pattern produced by water waves passing through two holes in a barrier antinode: a point of interaction between waves, at which only constructive interference occurs; in an interference pattern, antinodes occur at path difference intervals of whole wavelengths The Experiment Thomas Young solved both of these problems and successfully observed an interference pattern produced by light. Young conducted his experiment using a pinhole in a window shutter and a card he described as \u201ca slip of card, about one-thirtieth of an inch in breadth (thickness).\u201d The card was positioned edgewise into a horizontal sunbeam directed into the room by using a mirror. The sunbeam had a diameter slightly greater than the thickness of the card. When the card was positioned edgewise in the centre of the sunbeam it split the sunbeam into two coherent beams separated by a very small distance (Figure 13.70(a)). The effect was equivalent to light passing through two slits that were very close together. The small distance of separation between the two beams of light expanded the interference pattern on the screen so that the distance between nodal lines was large enough that it could be observed. Light coming from both sides of the card was in phase and the wave fronts could create a fixed interference pattern of light and dark bands, called interference fringes, on the screen (Figure 13.70 and also Figure 13.9, page 640). Bright fringes or antinodal lines were regions of constructive interference and dark fringes or nodal lines were regions of destructive interference. Young\u2019s double-slit experiment provided the strong evidence needed for acceptance of the wave model of light. The Interference Pattern To understand how the interference pattern is created, consider three different points. The first point is the central antinode (Figure 13.72). This point occurs at the centre of the pattern, along the perpendicular bisector. The perpendicular bisector is an imaginary straight line that runs from the midpoint of a line joining the two slits to the area of constructive interference on the screen. This area of constructive interference may be called the central antinode, central bright fringe, or", " central maximum. At the central antinode, both waves travel the same distance from the slits and arrive at the screen in phase. Constructive interference will be observed as a bright band. path difference: 3.00 \u03bb 3.00 \u03bb 0 \u03bb screen path length 3.00 \u03bb d } constructive interference at central antinode Figure 13.72 A path difference of zero at the central antinode results in constructive interference and a bright band on the screen. path length 3.00 \u03bb l 686 Unit VII Electromagnetic Radiation 13-PearsonPhys30-Chap13 7/24/08 3:44 PM Page 687 The second point is the first node, also called the first dark fringe or the first minimum. At this location, one wave travels a distance of 1 farther than the other so that the waves arrive out of phase, and 2 destructive interference is observed as a dark band (Figure 13.73). The third point is another antinode or bright fringe. At this location, one wave travels a distance of 1 farther than the other, causing the waves to arrive at the screen in phase once again (Figure 13.74). This pattern of bright and dark fringes repeats as we move out in both directions from the central, perpendicular bisector. path difference: 3.75 \u03bb 3.25 \u03bb \u03bb 1 2 path length 3.75 \u03bb path difference: 4.50 \u03bb 3.50 \u03bb 1 \u03bb path length 4.50 \u03bb d d } destructive interference at 1st node path length 3.25 \u03bb l path length 3.50 \u03bb l } constructive interference at n 1 antinode Figure 13.73 A path difference of 1, at the first node results in destructive 2 interference and a dark band on the screen. Figure 13.74 A path difference of, at the first antinode results in constructive interference and a bright band on the screen. Mathematical Analysis of Young\u2019s Experiment Young\u2019s experiment and interference patterns in general contain mathematical information about the waves that create them. For instance, the interference pattern that Young observed can be used to determine the wavelength of the light that created it. dark nodes (destructive interference) bright central antinode bright antinodes (constructive interference) 4 3 2 1 bisector 1 P 2 3 4 screen 10 \u03bb 9 \u03bb S1 S2 d Figure 13.75 Two point sources, separated by a short distance, produce antinodal lines of constructive interference and nodal lines of destructive interference. Point sources, S1 and S2, from the same original source, are separated by a short distance, d, and produce identical waves that are in phase. node: a point of interaction between waves, at which only destructive interference occurs; in an interference pattern, nodes occur at path difference intervals of one-half wavelength PHYSICS INSIGHT Compare Figures 13.74 and 13.75. The absolute path lengths (expressed in number of wavelengths) shown in Figure 13.74 and 13.75 are different, yet the waves are still causing constructive interference at the point where they meet. How can this be? If the distance separating the sources and the screen is constant, then the path length (expressed in number of wavelengths) to the antinodes can and will be different for different wavelengths of light. As long as the \u201cdifference\u201d in path lengths is a whole number multiple of the wavelength of the light, constructive interference will occur where the waves meet and this will be observed as an antinode. Chapter 13 The wave model can be used to describe the characteristics of electromagnetic radiation. 687 13-PearsonPhys30-Chap13 7/24/08 3:44 PM Page 688 path length: the distance between a point source and a chosen point in space difference in path length: the difference between two path lengths, each measured from a different origin and extending to a common point in space An equal number of antinodal lines, or lines of constructive interference, radiate outward on either side of the perpendicular bisector of the line joining S1 and S2. Beginning at the perpendicular bisector, antinodal lines on each side are numbered 1, 2, 3, etc., resulting in a duplicated series (Figure 13.75). For our analysis, we will chose a point, P, on the n 1 antinodal line. The path length from S1 to P and from S2 to P can be measured in multiples of wavelengths. From Figure 13.75, the path S1P is 10 and the path S2P is 9. The difference in path length from point P to the two sources is 10 9 1 Therefore, for any point along the first antinodal line (on either side of the bisector), the following relationship is true: S1P S2P 1 Recall from section 8.3 that waves that are 1 out of phase", " will constructively interfere. This explains why the antinodal lines are bright. A similar analysis for a point, P2, on the n 2 antinodal line generates a similar relationship 2 S1P2 \u2013 S2P2 An analysis of points on the n 3 and n 4, etc., antinodal lines follows a similar pattern, generating the following relationship between path difference and wavelength along the antinodal lines. S1Pn S2Pn n, where n 1, 2, 3, 4,... (1) If the wavelength is sufficiently large, the difference in path length can be calculated as in Equation (1). Light waves have extremely small wavelengths; therefore, another method of calculating path length becomes appropriate. Difference in Path Length for Waves of Short Wavelength Pn (close) X S1 S2 S1 d X \u03b8n S2 d Pn far away Consider the triangles illustrated in Figure 13.76, showing the path length from two sources to point Pn. For any point along the antinodal line Pn, the difference in path length will be equal to the segment S1X. When Pn is selected to be very far away, the paths are nearly parallel. In such a case, S1-X-S2 forms a right triangle allowing the difference in path length to be expressed in terms of angle n, Figure 13.76 Two point sources, separated by a short distance, d, with path lengths to point Pn sin n S1X d sin n difference in path length d 688 Unit VII Electromagnetic Radiation 13-PearsonPhys30-Chap13 7/24/08 3:44 PM Page 689 Constructive interference For constructive interference or antinodal lines, substitute the generalization for the difference in path length (Equation 1 above): sin n n d d sin n n for n 1, 2, 3, 4,... (2) Equation (2) can be used to determine the wavelength of the light used to produce the interference pattern, where n is the number of the nth antinodal line and d is the distance between the two sources. For a point Pn located far away from the two point sources, the angle n is equivalent to the angle, located between the perpendicular bisector and the straight line drawn between the midpoint of the two sources and the point Pn. The angle is known as the angle of diffraction (Figure 13.77). angle of diffraction: the angle formed between the perpendicular bisector and the straight line to a nodal or antinodal point on the interference pattern screen Pn bisector Destructive interference The analysis for destructive interference, which occurs along the nodal lines, is done in an almost identical way. Recall that destructive interference occurs when the path difference between waves is a half number of wavelengths. Therefore, the generalization for the difference in path length becomes (n 1 ) 2 d d sin n (n 1 ) 2 for n 1, 2, 3, 4,... (3) sin n where n is the number of the nth nodal line relative to the perpendicular bisector, d is the distance between the two sources, and the angle n is equivalent to the angle between the perpendicular bisector and the straight line drawn between the midpoint of the two sources and the point Pn. \u03b8\u03b1 X S1 \u03b8 n S2 C (midpoint) d Figure 13.77 When Pn is far away, line S1Pn is approximately parallel to CPn, making n and equal. is the angle of diffraction. Example 13.8 Monochromatic light is incident on two slits separated by 0.30 mm, and the first bright fringe (n 1 antinode) is located at an angle of 0.080 from the central antinode. What is the wavelength of the light? Given d 0.30 mm 3.0 104 m n n 1 0.080 Required Wavelength () Practice Problems 1. Light of an unknown wavelength is incident on two slits separated by 0.20 mm. The second bright fringe is located at an angle of 0.26 from the central antinode. What is the light\u2019s wavelength? 2. Blue light of 460 nm is incident on two slits that are 0.55 mm apart. What is the angle of diffraction for the third antinodal line? Analysis and Solution The angle given is relative to the central antinode, which occurs along the perpendicular bisector; therefore, the angle given is relative to the perpendicular bisector. The first bright fringe is a region of constructive Chapter 13 The wave model can be used to describe the characteristics of electromagnetic radiation. 689 13-PearsonPhys30-Chap13 7/24/08 3:44 PM Page 690 3. The second nodal line of an interference pattern occurs at 0.09", "5 relative to the central antinode. The two slits are separated by 0.40 mm. What is the wavelength and colour of light producing this pattern? Answers 1. 4.5 107 2. 0.14 3. 4.4 107 m, violet interference, where the path difference must be one full wavelength different. Therefore, Equation (2) above for constructive interference applies: d sin n n (3.0 104 m) sin 0.080 1 4.2 107 m Paraphrase Monochromatic light with a wavelength of 4.2 107 m will produce a bright fringe at an angle of 0.080 from the central antinode. info BIT When the angle is small (10\u00ba) sin is nearly equal to tan as illustrated in the calculated values for both shown below: sin Angle tan sin / tan 1 2 5 0.017452 0.017455 0.9998 0.03490 0.03492 0.9994 0.08716 0.08749 0.9962 10 0.1736 0.1763 0.9847 distance between central antinode and nth antinodal line Pn xn screen Finding Wavelength and the Angle of Diffraction Under Experimental Conditions In experimental settings, it is often difficult to measure the angle of diffraction because this angle is very small relative to a point that is very far away from the two slits. It is easier to obtain a value for sin by determining the ratio x/l, as shown in Figure 13.78. Here, x is the distance between the central antinode (where the perpendicular bisector intersects the screen) and the antinodal fringe. The length of the perpendicular bisector, l, is the distance from the midpoint between the slits to the screen, where the interference pattern is observed. In Figure 13.78, x/l tan, but when l is much greater than x, the ratio of x/l is very small (generally less than 0.2) making tan nearly equal to sin (see Infobit). Therefore, it is acceptable to assume sin x/l in this case. By replacing sin with the ratio x/l, we arrive at the following equation for antinodal (bright) fringes: x d l n xd nl Applying the same analysis to nodal (dark) fringes gives: x d l 1 n 2 xd 1 l n 2 l (distance to the screen) n \u03bb S1 \u03b8 C d \u03b8 S2 Figure 13.78 Determining sin using x and l 690 Unit VII Electromagnetic Radiation 13-PearsonPhys30-Chap13 7/24/08 3:44 PM Page 691 Example 13.9 A student measuring the wavelength of light emitted by a krypton gas sample directs the light through two slits separated by 0.264 mm. An interference pattern is created on a screen 3.0 m from the slits and the distance between the second bright (n 2 antinode) fringe and the central antinode is measured to be 1.18 cm. What is one of the wavelengths of light emitted by the krypton gas sample? Given d 0.264 mm 2.64 104 m l 3.0 m x 1.18 cm 1.18 102 m n 2 Required wavelength () Analysis and Solution The bright fringe is a region of constructive interference, where the path difference must be a whole number of wavelengths. Solve for the wavelength by using the equation: xd nl (1.18 102 m)(2.64 104 m) (2)(3.0 m) 5.2 107 m Paraphrase The krypton gas sample emits light with a wavelength of 5.2 107 m. Practice Problems 1. Monochromatic light is incident on two slits separated by 0.15 mm. An interference pattern is observed on a screen 5.0 m away. The distance between the 3rd dark fringe and the central antinode is 4.50 102 m. What is the wavelength of the light? 2. Monochromatic light is incident on two slits separated by 3.00 105 m. The distance between antinodes is 3.10 102 m. If the screen is 1.50 m from the slits, what is the light\u2019s colour and wavelength? 3. A student used light of wavelength 5.00 107 m and found that the distance between the third node and the central antinode was 1.00 101 m. If the screen was located 1.20 m away from the slits, how far apart are the slits? Answers 1. 5.4 107 m 2. 6.20 107 m, red 3. 1.50 105 m Poisson\u2019s Bright Spot Using an analysis similar to that in Example 13.9, Young calculated the wavelength of various colours of light and announced his results in 1807", ". He was still overshadowed by Newton\u2019s reputation and support for the particle model of light. Thus, Young was not taken seriously by the scientific community until 1818, when Augustin Fresnel, a French physicist (Figure 13.79), proposed another mathematical wave theory. This new theory formed a critical turning point in the debate between the wave and particle models of the time. A mathematician by the name of Simon Poisson argued that the equations in Fresnel\u2019s theory could be used to predict a unique diffraction pattern that should be produced when light is incident on a small round disc. Poisson showed that Fresnel\u2019s equations should produce a central bright fringe at the centre of a shadow cast by a solid round disc when it is illuminated by a point source of monochromatic light (Figure13.80(a)). This predicted result was similar to the central bright fringe observed in Young\u2019s double-slit experiment. Chapter 13 The wave model can be used to describe the characteristics of electromagnetic radiation. 691 13-PearsonPhys30-Chap13 7/24/08 3:44 PM Page 692 Fresnel\u2019s equations showed that the light that is diffracted around the edge of the disc should constructively interfere at the centre of the shadow. Poisson was unable to observe this bright spot experimentally and was of the opinion that he had refuted the wave model of light. solid disc shadow equal path length bright spot of constructive interference Figure 13.79 Augustin Fresnel (1788\u20131827). Fresnel\u2019s analysis of diffraction provided the theoretical groundwork for the transverse wave model of light. He also designed the lenses that were used in lighthouses of the 19th century and in overhead projectors, solar collectors, and beacon lights of the 21st century. diffraction grating: a sheet of glass or plastic etched with a large number of parallel lines. When light is incident on the grating, each line or slit acts as one individual light source. Figure 13.81 Diffraction grating info BIT The concentric circles cut into a CD when it is made make it behave as a diffraction grating. So, the rainbow pattern you see when you look at a CD is the product of interference caused by diffraction. Figure 13.80 Poisson\u2019s Bright Spot (a) Constructive interference at the bright spot (b) Photograph of Poisson\u2019s Bright Spot Poisson\u2019s prediction of a bright spot was retested by Dominique Arago in 1818, and this time, the bright spot was verified (Figure 13.80 (b)). The bright spot is known as \u201cPoissson\u2019s Bright Spot\u201d because, even though Poisson was a supporter of the particle model of light, he had predicted the spot\u2019s existence if the wave model of light was correct. Mainly as a result of this verification of Poisson\u2019s Bright Spot, by 1850 the wave model of light was generally accepted by the scientific community. The model was then successfully applied to many of the properties of light. Diffraction Gratings Young\u2019s experiment used only two small point sources of light that were in phase. A diffraction grating (Figure 13.81) has a very large number of equally spaced, parallel lines that act as individual light sources. When light is incident on a multi-slit diffraction grating, an interference pattern, similar to that of a double slit, is produced on a distant screen. But there are several key differences. First, the large number of lines in a diffraction grating can deliver more light energy to the distant screen, increasing the brightness of the interference pattern. Second, the antinodal, bright fringes are more defined, being sharper and narrower. And third, when line separation is very small, the separation between the lines (d) is inversely proportional to the distance between the fringes (x) in the interference pattern according to d xd nl nl x Therefore, the extremely small separation distance between lines on a grating will cause an increase in the separation between the fringes in the pattern. For these reasons, a diffraction grating is a very precise apparatus for investigating the wavelength of light. When a diffraction grating is used with monochromatic light, the interference pattern will have the same colour as the wavelength of light used to produce it. When full-spectrum, white light is used, each antinode will appear as a rainbow because each wavelength is diffracted Figure 13.82 692 Unit VII Electromagnetic Radiation 13-PearsonPhys30-Chap13 7/24/08 3:44 PM Page 693 at a slightly different angle. The relationship between the wavelength and the angle is: n d sin n as was derived earlier for Young\u2019s double-slit experiment. Diffraction gratings are produced by etching a large number of parallel lines on a sheet", " of glass or plastic. Each grating is defined by the number of lines per centimetre etched on it. Tens of thousands of lines per centimetre are common. The distance between lines is the inverse of the number of lines per centimetre, so as the number of lines increases, the distance between any two lines decreases accordingly. Diffraction in Nature \u2014 Solar and Lunar Coronas When the Sun is rising (Figure 13.83) or an overexposed photo of the Moon is taken (Figure 13.84), a pattern of rings appears. These are examples of diffraction in the atmosphere. When light from the Sun or Moon enters the atmosphere it encounters uniformly sized droplets of water and ice crystals. Diffraction occurs as the light bends around the edges of the particles at varying degrees depending upon the wavelength of the light. A series of coloured rings surrounding the astronomical object results. e LAB For a probeware activity, go to www.pearsoned.ca/ school/physicssource. e SIM Investigate diffraction gratings through the use of simulations. Follow the eSim links at www.pearsoned.ca/ school/physicssource. e WEB To learn more about diffraction effects in nature, follow the links at www.pearsoned.ca/ school/physicssource. Example 13.10 Figure 13.83 A rising Sun Figure 13.84 A ring around the Moon A diffraction grating has 1000 lines/cm. When light passes through the grating, an interference pattern is produced on a screen 4.00 m away. The first-order bright fringe is 19.2 cm away from the central antinode. What is the wavelength and colour of the light? Given d 1 cm / 1000 lines 1 103 cm/line 1 105 m/line x 19.2 cm 1.92 101 m n 1 l 4.00 m Required wavelength of light () Analysis and Solution The separation between lines is the inverse of the lines per centimetre given for the diffraction grating. Line separation is very small, and the first-order bright fringe is n 1 because it is an antinode, or line of constructive interference. Therefore, this equation applies: Practice Problems 1. An unknown wavelength of light is incident on a diffraction grating with 2500 lines/cm. The distance between the central antinode and the 3rd dark node is 20.0 cm when the screen is located 50.0 cm from the grating. Determine the wavelength of the light. 2. How many lines/cm are there in a diffraction grating if the 3rd dark fringe is located 5.00 cm from the central antinode when the screen is located 60 cm from the grating? Assume 500 nm. 3. A diffraction grating has 1000 lines/cm. When red light ( 750 nm) is incident on the grating, what will be the separation between bright antinodes on a screen 3.0 m away? Chapter 13 The wave model can be used to describe the characteristics of electromagnetic radiation. 693 13-PearsonPhys30-Chap13 7/24/08 3:44 PM Page 694 Answers 1. 6.40 107 m 2. 6.7 102 lines/cm 3. 23 cm xd nl (1.92 101 m)(1 105 m) (1)(4.00 m) 4.80 107 m 480 nm Paraphrase Based on the interference pattern, the incident light is 480 nm, which corresponds to the colour blue. 13-10 Inquiry Lab 13-10 Inquiry Lab Determining the Wavelength of Red, Green, and Blue Light This investigation will verify the known wavelengths of different colours of light. According to the visible spectrum, red, green, and blue light have wavelengths in the following ranges: \u2022 red: 650\u2013750 nm \u2022 green: 500\u2013550 nm \u2022 blue: 450\u2013500 nm Question What are the wavelengths of red, green, and blue light? Variables manipulated: the distance between the diffraction grating and the screen will be indirectly manipulated responding: the position of the first-order antinodes for red, blue, and green light on the \u201capparent\u201d screen controlled: single-filament incandescent light source and diffraction grating Materials and Equipment thin-film diffraction grating two metre-sticks single-filament lamp masking tape and pen Required Skills Initiating and Planning Performing and Recording Analyzing and Interpreting Communication and Teamwork them. A screen is not needed as the angles are identical, so we can assume the distance to the hypothetical screen is identical to the distance between the lamp and the grating (1.0 m). While one person observes the antinode, a second person is directed to place an identification tape on the metrestick where the antinode appears to be. This allows the distance between the central antinode and the position of the firstorder antin", "ode to be measured. This procedure can also be used to find the position of the second-order antinode. Procedure 1 Set up the two metre-sticks such that they make a right angle with one another. 2 Place the lamp at the point where the metre-sticks join as shown in Figure 13.85. 3 Place the thin-film diffraction grating vertically upright at the end of one of the metre-sticks. 1 m x2 x1 lamp n 2 n 1 Experimental Design To investigate the position of the first- and second-order antinodes in an interference pattern produced by white light, a \u201csimulated\u201d screen will be used. Two metre-sticks and a lamp are arranged as shown in Figure 13.85. In this design, observers look through the diffraction grating to see the antinodes as they would appear on a screen behind 1 m \u03b82 \u03b81 thin-film diffraction grating \u03b81 \u03b82 Figure 13.85 694 Unit VII Electromagnetic Radiation 13-PearsonPhys30-Chap13 7/24/08 3:44 PM Page 695 4 Turn on the lamp and look through the grating, moving your head side to side until you can see the first-order antinode, which should appear as a rainbow. Your eye acts as the wavelength detector, in place of a screen. 5 While looking at the antinode, direct your lab partner to put a piece of tape labelled \u201cB\u201d on the metre-stick where the blue antinode band appears to be. 6 Repeat step 5 using tape labelled \u201cG\u201d for the green band of light and \u201cR\u201d for the red band of light. 7 Repeat steps 5 and 6 for the second-order antinode, if it is visible along the metre-stick. 8 Record the distance between the lamp and each piece of labelled tape. Analysis 1. Using the first antinode (n 1), calculate the wavelength of red, blue, and green light. 2. Determine the mean wavelength of each colour of light using the ranges given. 3. Calculate the percent difference between the mean wavelength of each colour and your experimentally determined value. 4. Explain why each antinode appears as a rainbow. Hint: What effect does the wavelength have on the angle of diffraction? 5. How many antinodes should appear in the diffraction grating on either side of the light source? Assume the largest angle of diffraction that could be visible is 89. M I N D S O N Comparing Spectra: Dispersion vs. Diffraction The rainbow produced when white light is refracted through a prism is similar to the rainbow produced at each antinode when white light passes through a thinfilm diffraction grating. In small groups, prepare a presentation to compare and contrast these two phenomena. Your presentation should consider: 1. The wave nature of light 2. The similarities of refraction and diffraction as they relate to Huygens\u2019 Principle of wavelets 3. The different wavelengths of the visible spectrum and how this leads to the separation of the colours in both dispersion and diffraction 4. The key differences in the causes of dispersion in a prism and the production of antinodes in an interference pattern 5. The reversed order of colours Your presentation should also include: \u2022 relevant images of both phenomena \u2022 schematics of each phenomenon using ray diagrams \u2022 where appropriate the use of animations and simulations found at www.pearsoned.ca/school/ physicssource \u2022 a list of references Polarization Young\u2019s double-slit experiment, and interference in general, provided strong evidence that light does exhibit wave properties. However, evidence of interference alone could not distinguish whether the waves were transverse or longitudinal. Recall that in section 13.1 light, and electromagnetic radiation in general, was described by Maxwell as consisting of perpendicular electric and magnetic fields, propagating through space at the speed of light. In other words, Maxwell predicted that light was a transverse wave. Is there evidence that this is indeed the case? Using a mechanical model, such as a rope, one can see that a transverse wave can be linearly polarized when vibrations only occur in one plane. The vertically polarized transverse waves shown in Figure 13.86 can pass through the vertical slit, but are blocked, or absorbed, by the horizontal slit. The longitudinal waves, on the other hand, can pass e LAB For a probeware activity, go to www.pearsoned.ca/ school/physicssource. (a) (b) Figure 13.86 (a) A transverse wave passing through a vertical slit and being absorbed by a horizontal slit (b) A longitudinal wave passing through both a vertical and a horizontal slit Chapter 13 The wave model can be used to describe the characteristics of electromagnetic radiation. 695 13-PearsonPhys30-Chap13 7/24/", "08 3:44 PM Page 696 info BIT Ancient Inuit hunters developed a unique technology to limit the glare of the Sun\u2019s light reflecting off the spring snow. Snow goggles, similar to those seen in the movie Atanarjuat: The Fast Runner, were the first sunglasses. The goggles were carved mainly from caribou antler whalebone, or ivory; driftwood was also used. Narrow slits in the snow goggles reduced the amount of light, thus protecting the hunter\u2019s eyes and preventing the debilitating effects of snow blindness. A secondary positive effect of the goggles was improved visibility. Figure 13.87 Inuit snow goggles polarizing filter: a filter that allows only one plane of the electric field to pass through it; plane polarized EMR emerges polarization: production of a state in which the plane of the electric field for each electromagnetic wave occurs only in one direction plane polarized light: light resulting from polarization, in which only one plane of the electric field is allowed to pass through a filter through both slits unaffected because longitudinal waves are not linearly polarized. By a process similar to the mechanical model, electromagnetic waves can be blocked by two polarizing filters held at right angles to one another. In 13-1 QuickLab at the beginning of the chapter, you discovered that two polarizing filters, held at right angles to one another, can absorb light. Figure 13.88 Two polarizing filters, one held vertically, the other held horizontally, partially overlap, showing the absorption of electromagnetic waves. The photograph in Figure 13.88 can be explained by considering electromagnetic radiation as perpendicular magnetic and electric fields, with the plane of polarization arbitrarily defined by the direction of the electric field. When light is produced by an incandescent light bulb it is not polarized, meaning that the plane of the electric fields for each wave occurs randomly as light propagates outward from the source in all directions. When unpolarized light is incident on a polarizing filter, only one plane of the electric field is allowed to pass through, causing plane polarized light to emerge. If a second polarizing filter is held at right angles to the plane polarized light, then the plane polarized light also is absorbed (Figure 13.89). randomly oriented EMR vertical polarizer horizontal polarizer no waves vertically polarized EMR Figure 13.89 Unpolarized light incident on two polarizing filters at right angles to one another The blue light in sunlight is partially polarized when it is scattered in the atmosphere. Therefore, in sunglasses and camera lenses, polarized filters are used to reduce the blue polarized light from the sky while allowing other non-polarized colours to pass through and appear brighter. To see this effect, tilt your head from side to side while looking at the blue sky with polarized glasses. The polarizing effect supports the wave model of light in general and in particular, the concept that light is composed of perpendicular, oscillating electric and magnetic fields (Figure 13.90). Figure 13.90 Two pairs of polarized sunglasses, at right angles 696 Unit VII Electromagnetic Radiation 13-PearsonPhys30-Chap13 7/24/08 3:44 PM Page 697 13.5 Check and Reflect 13.5 Check and Reflect Knowledge 1. According to Huygens\u2019 Principle, what will happen to the shape of a straight wave front after it passes through a small opening in a barrier? 2. Use Huygens\u2019 Principle to describe how interference can occur when a straight wave front is incident on two narrow openings. 3. Two incandescent white lights are placed close to one another. Explain why an interference pattern is not observed on a nearby screen. 4. What parameters did Young need to control to enable him to observe an interference pattern from two point sources of light? 5. Construct a concept map to show the relationship between path length, nodal fringes, antinodal fringes, wave phase, and interference. Applications 6. In an experiment similar to Young\u2019s, how far apart are two slits if the 3rd antinode is measured to be 20 from the central antinode, when light with a wavelength of 650 nm is used? 7. Determine the angle of diffraction to the 2nd node when light with a wavelength of 425 nm is incident on two slits separated by 6.00 106 m. 8. Light with a wavelength of 700 nm is diffracted by a diffraction grating with 5.00 103 lines/cm. If a screen is positioned 1.00 m away from the grating, what is the distance between the 1st and central antinodes? 9. Monochromatic light with a frequency of 5.75 1014 Hz is incident on a diffraction grating with 60 lines/cm. What is the distance between the 2nd and 3rd dark fringes when the screen is located 1.20 m away? 10. An unknown light source is directed at a diffraction grating with 6.00 104 lines/m. If the", " nodal lines are 5.50 cm apart when the screen is 1.50 m away, what is the wavelength and frequency of the light? 11. Light emitted from an unknown gas sample is incident on a diffraction grating with 5.00 102 lines/cm. The antinodes appear on a screen 1.50 m away and are separated by 3.10 102 m. What is the wavelength and frequency of the light? Extensions 12. Design an experiment to determine the wavelength of an unknown monochromatic light. Include an experimental design, material list, and procedure. 13. Compare the wavelength of X rays to that of visible light and explain what should happen to the diffraction pattern if X rays were used instead of visible light. 14. Investigate how the process of diffraction, using radiation other than visible light, can be useful for determining the shapes of crystal lattices and structures too small to be seen with visible light. e TEST To check your understanding of diffraction, interference, and polarization, follow the eTest links at www.pearsoned.ca/school/physicssource. Chapter 13 The wave model can be used to describe the characteristics of electromagnetic radiation. 697 13-PearsonPhys30-Chap13 7/24/08 3:44 PM Page 698 CHAPTER 13 SUMMARY Key Terms and Concepts electromagnetic radiation frequency wavelength electromagnetic spectrum particle model wave model photon quantum model electric field magnetic field capacitor Maxwell\u2019s Equations electromagnetic wave rectilinear propagation ray diagram plane mirror law of reflection virtual image real image magnification image attitude converging mirror diverging mirror mirror equation refraction refractive index Snell\u2019s Law total internal reflection critical angle spectrum dispersion converging lens diverging lens thin lens equation Huygens\u2019 Principle diffraction interference antinode node path length difference in path length angle of diffraction diffraction grating polarizing filter polarization plane polarized light Key Equations m hi ho di do sin 1 sin 2 v1 v2 1 2 n2 n1 Conceptual Overview 1 do 1 di 1 f sin n n d n c v xd nl n1 sin 1 n2 sin 2 Summarize this chapter by explaining how the properties of electromagnetic radiation support either the wave model of light or the particle model of light, or both. reflection refraction diffraction Particle Model polarization Wave Model interference electric field magnetic field 698 Unit VII Electromagnetic Radiation 13-PearsonPhys30-Chap13 7/24/08 3:44 PM Page 699 CHAPTER 13 REVIEW Knowledge 1. (13.1) Create a table to identify all the major categories of electromagnetic radiation, including the wavelengths and frequencies listed in the spectrum shown in Figure 13.4. Brainstorm common uses for each type of radiation. 2. (13.1) Compare and contrast the particle and wave models of electromagnetic radiation. 3. (13.1) What two critical insights were understood by Maxwell when he developed his theory of electromagnetic radiation? 4. (13.1) Consider two electric field lines on a transverse wave. One field line is up, and a moment later, another is down. What is produced as a result of this \u201cchanging\u201d electric field? 5. (13.1) Describe how an electromagnetic wave is able to propagate in empty space. 6. (13.1) Describe the five predictions that Maxwell made regarding the properties of electromagnetic radiation. 7. (13.1) How did Hertz prove that the EMR observed at his antenna was, in fact, produced by the nearby spark gap and did not originate from another source? 8. (13.2) The first significant attempt to measure the speed of light was made by Christiaan Huygens, using the eclipse of Jupiter\u2019s moon Io. Describe this method. 9. (13.2) In addition to measuring the speed of light with a rotating toothed wheel, Armand Fizeau demonstrated that light travelled at different speeds in moving water. Explain how the results of his investigation support the wave model of light. 10. (13.3) Draw a ray diagram to demonstrate the law of reflection. 11. (13.3) Construct a ray diagram for a converging mirror and illustrate the following terms. (a) centre of curvature (C) (b) radius of curvature (r) (c) vertex (V) (d) principal axis (PA) (e) principal focal point (F) (f) focal length (f) 12. (13.3) Can a diverging mirror produce a real image? Explain. 13. (13.4) Using a ray diagram, illustrate partial reflection and partial refraction for a ray passing from air into water at an angle of 15. On your ray diagram, label the normal line, the index of refraction, the angle of incidence, the angle of reflection, and the angle of refraction. 14. (13.4", ") Light passes from a medium with a high refractive index to one with a low refractive index. Is the light bent away from or toward the normal line? 15. (13.4) Dispersion is the separation of white light into all the colours of the spectrum. Explain two different methods that could be used to separate all the colours in white light. 16. (13.5) Illustrate the process of refraction using a straight wave front that travels from air into water. Based on your diagram, does Huygens\u2019 Principle support the wave model or the particle model of light? 17. (13.5) A straight wave front is incident on two small holes in a barrier; illustrate the shape of the wave front a moment after it makes contact with the barrier. Does your drawing indicate that interference will occur? 18. (13.5) Why is an interference pattern not observed when two incandescent lights are located next to one another? 19. (13.5) How does the evidence from polarizing filters support the transverse nature of the wave model of light? 20. (13.5) Each antinode appears as a full spectrum when white light is incident on a diffraction grating. Explain this phenomenon. 21. (13.5) Explain how path length and diffraction are related to the production of Poisson\u2019s Bright Spot. Chapter 13 The wave model can be used to describe the characteristics of electromagnetic radiation. 699 13-PearsonPhys30-Chap13 7/24/08 3:44 PM Page 700 Applications 22. (13.1) High-voltage transmission lines that carry alternating current can interfere with radio waves. Explain how this interference can occur. 31. (13.3) A 20.0-cm-high, inverted image is produced when an object is placed 12.0 cm from a converging lens with a focal length of 11.0 cm. Calculate the height of the object. 23. (13.2) An 8-sided mirror is rotating at 5.50 102 Hz. At what distance should the fixed mirror be placed to replicate Michelson\u2019s experiment? 24. (13.2) A fixed mirror and a rotating mirror are separated by 30.0 km. The 8-sided rotating set of mirrors turns at 600 Hz when the light is able to pass through the experimental apparatus. Calculate the speed of light. 25. (13.3) When you look into a plane mirror, an image is formed. Describe the characteristics of the image based on attitude, type, and magnification. 26. (13.3) A student stands 30 cm from a plane mirror. If the student\u2019s face is 25 cm in length, what is the minimum length of mirror needed for the student to see her entire face? 27. (13.3) An object is located 25.0 cm from a converging mirror with a focal length of 15.0 cm. Draw a scale ray diagram to determine the following: (a) the image location and type (b) the image attitude (c) the magnification of the image 28. (13.3) Construct a ray diagram for a diverging mirror and illustrate the following terms: (a) centre of curvature (C) (b) radius of curvature (r) (c) vertex (V) 29. (13.3) Where must an object be placed relative to the focal point for a converging mirror such that the image produced is virtual? 30. (13.3) A 15.0-cm-high object is placed 20.0 cm from a diverging mirror with a virtual focal length of 10.0 cm. How high is the image and where is it located? 32. (13.4) Calculate the speed of yellow light, 589 nm, in the following materials: (a) water (b) ethanol (c) Lucite\u00ae (d) quartz glass (e) diamond 33. (13.4) Light with a wavelength of 610 nm is incident on a quartz glass crystal at an angle of 35. Determine the angle of refraction and the wavelength of the light in the quartz glass. 34. (13.4) Can total internal reflection occur when light travels from (a) air into water? (b) water into air? (c) Lucite\u00ae into water? (d) water into diamond? 35. (13.4) Light enters an unknown material and slows down to a speed of 2.67 108 m/s. What is the refractive index of the unknown material? Compare the refractive index of this material to that of water \u2014 which one has a higher index? 36. (13.4) Calculate the critical angle of the following boundaries: (a) water-air (b) diamond-air (c) diamond-water (d) Lucite\u00ae-air 37. (13.4) A 4.00-", "cm-high object is located 5.00 cm from a diverging lens with a focal length of 10.0 cm. Using the thin lens equation, determine the image attributes and position. Verify your answer with a scale ray diagram. 700 Unit VII Electromagnetic Radiation 13-PearsonPhys30-Chap13 7/24/08 3:44 PM Page 701 38. (13.5) In an experiment similar to Young\u2019s, light with a wavelength of 630 nm is incident on two slits separated by 5.3 105 m. What is the angle to the 1st, 2nd, and 3rd antinodes? 39. (13.5) Monochromatic light is incident on two slits separated by 0.25 mm. The first dark fringe deviated an angle of 0.050 from the central antinode. What is the wavelength and colour of the light? 40. (13.5) Light from an unknown gas sample is incident on two slits separated by 1.4 104 m. On a screen 1.1 m away, the distance between the 7th node and the central antinode is measured to be 0.025 m. What is the wavelength of the light emitted by the unknown gas sample? 41. (13.5) A screen is located 4.5 m from two slits that are illuminated with a 490-nm light source. If the distance between the central antinode and the first-order antinode is 0.037 m, how far apart are the two slits? Extensions 42. An X-ray machine operates by accelerating an electron through a large potential difference, generating a large amount of kinetic energy. The high-speed electron then collides with a metal barrier. Explain why the collision produces a high-frequency X ray. 43. Cable television wires have a metal shield surrounding the copper wire that carries the television signal. The shielding prevents interference from electromagnetic radiation and it must be grounded in order to effectively block interference. Explain how the shielding prevents interference and why it needs to be grounded. 44. When you place the concave side of a spoon on your nose and slowly pull it away from your face, your image disappears at a certain distance. What is the significance of this distance? 45. After light enters Earth\u2019s atmosphere it encounters a temperature gradient as it approaches the surface of Earth, causing a mirage. If the warm air near the surface of Earth has a lower index of refraction than the cooler air above, which way is the light bent? Show this with a ray diagram. 46. Explain why a fibre-optic network is much more efficient and powerful than a copper-wire network. Consolidate Your Understanding 1. Explain how electromagnetic radiation is able to propagate in the absence of a medium, like air. 2. Why is an accelerating charge required to produce electromagnetic radiation and how does this relate to the word \u201cchanging\u201d in Maxwell\u2019s explanation of EMR? 3. Describe the three-dimensional shape of an electromagnetic wave. Specify the directions of both the electric and magnetic field variations, and the direction of wave propagation. 4. Has Maxwell\u2019s last prediction been verified by experimental evidence? If so, describe the evidence as it relates to reflection, refraction, diffraction, interference, and polarization. 5. Could Hertz have investigated the phenomenon of diffraction by using the same equipment as in his famous experiment? If so, how? Think About It Review your answers to the Think About It questions on page 635. How would you answer each question now? e TEST To check your understanding of the nature and behaviour of electromagnetic radiation, follow the eTest links at www.pearsoned.ca/school/physicssource. Chapter 13 The wave model can be used to describe the characteristics of electromagnetic radiation. 701 14-PearsonPhys30-Chap14 7/24/08 3:57 PM Page 702 C H A P T E R 14 Key Concepts In this chapter, you will learn about: the idea of the quantum the wave\u2013particle duality basic concepts of quantum theory Learning Outcomes When you have completed this chapter, you will be able to: Knowledge describe light using the photon model explain the ways in which light exhibits both wave and particle properties state and use Planck\u2019s formula give evidence for the wave nature of matter use de Broglie\u2019s relation for matter waves Science, Technology, and Society explain the use of concepts, models, and theories explain the link between scientific knowledge and new technologies Skills observe relationships and plan investigations analyze data and apply models work as members of a team apply the skills and conventions of science 702 Unit VII The wave-particle duality reminds us that sometimes truth really is stranger than fiction! Up to this point in the course, you have studied what is known as classical physics. Classical physics includes most of the ideas about light, energy, heat, forces, and electricity and magnetism up to about", " 1900. The golden age of classical physics occurred at the very end of the 19th century. By this time, Newton\u2019s ideas of forces and gravitation were over 200 years old, and our knowledge of physics had been added to immensely by the work of James Clerk Maxwell, Michael Faraday, and others. It seemed as though nearly everything in physics had been explained. In the spring of 1900, in a speech to the Royal Institution of Great Britain, the great Irish physicist William Thomson (Figure 14.1) \u2014 otherwise known as Lord Kelvin \u2014 stated that \u201c\u2026 the beauty and clearness of the dynamical theory of light and heat is overshadowed by two clouds\u2026.\u201d You could paraphrase Kelvin as saying \u201cthe beauty and clearness of physics is overshadowed by two clouds.\u201d One \u201ccloud\u201d was the problem of how to explain the relationship between the temperature of a material and the colour of light the material gives off. The other \u201ccloud\u201d had to do with an unexpected result in an experiment to measure the effect of Earth\u2019s motion on the speed of light. Kelvin was confident that these two clouds would soon disappear. He was wrong! Before the year was out, the first of these clouds \u201cbroke\u201d into a storm the effects of which are still being felt today! In this chapter, you will meet one of the strangest ideas in all of science. In many ways, this chapter represents the end of classical physics. You will learn that light is not only a wave, but also a particle. Stranger still, you will learn that things you thought were particles, such as electrons, sometimes act like waves! Hang on! Figure 14.1 William Thomson (1824\u20131907) was named Lord Kelvin by Queen Victoria in 1892. He was the first British scientist to be honoured in this way. During his long and illustrious career, Lord Kelvin published over 600 books and papers, and filed more than 70 patents for his inventions. He was one of the driving forces behind the first transatlantic telegraph cable. info BIT This chapter is about the \u201ccloud\u201d that became quantum theory. In 1905, the other \u201ccloud\u201d became Einstein\u2019s theory of special relativity. 14-PearsonPhys30-Chap14 7/24/08 3:57 PM Page 703 14-1 QuickLab 14-1 QuickLab The Relationship Between Temperature and Colour of an Incandescent Object Problem What is the relationship between the temperature of a hot, glowing object and the colour of light emitted by the object? Questions 1. What happens to the temperature of the filament in the light bulb as you increase the voltage output of the transformer? Materials incandescent (filament-style) light bulb variable transformer, 0\u2013120 V transmission-type diffraction grating Procedure 1 Attach a filament-style light bulb to a variable transformer and slowly increase the voltage. 2 Observe the spectrum produced by the light from the light bulb as it passes through the diffraction grating. For best results, darken the room. 2. How does the spectrum you observe through the diffraction grating change as you increase the voltage through the filament? 3. As you increase the temperature of the filament, what happens to the colour at which the spectrum appears brightest? 4. You may have noticed that the colour of a flashlight filament becomes reddish as the battery weakens. Suggest why. Think About It 1. Describe the relationship between the colour of a hot object and its temperature. Note in particular the colour you would first see as the temperature of an object increases, and how the colour changes as the object continues to heat up. 2. What do we mean by the terms \u201cred-hot\u201d and \u201cwhite-hot\u201d? 3. Which is hotter: \u201cred-hot\u201d or \u201cwhite-hot\u201d? 4. Is it possible for an object to be \u201cgreen-hot\u201d? Explain. Discuss your answers in a small group and record them for later reference. As you complete each section of this chapter, review your answers to these questions. Note any changes in your ideas. info BIT Even a great physicist can be wrong! Despite making extremely important contributions to many areas of physics and chemistry, Lord Kelvin has also become famous for less-than-accurate predictions and pronouncements. Here are a few: \u2022 \u201cI can state flatly that heavierthan-air flying machines are impossible.\u201d (1895) \u2022 \u201cThere is nothing new to be discovered in physics now. All that remains is more and more precise measurement.\u201d (1900) \u2022 \u201cX rays will prove to be a hoax.\u201d (1899) \u2022 \u201cRadio has no future.\u201d (1897) \u2022 \u201c[The vector] has never been of the slightest use to any creature.\u201d Chapter 14 The wave-particle duality reminds us that sometimes truth", " really is stranger than fiction! 703 14-PearsonPhys30-Chap14 7/24/08 3:57 PM Page 704 14.1 The Birth of the Quantum We take for granted the relationship between the colour of a hot, glowing object and its temperature. You know from sitting around a campfire that the end of a metal wiener-roasting stick slowly changes from a dull red to a bright reddish-yellow as it heats up. For centuries, metalworkers have used the colour of molten metal to determine when the temperature is just right for pouring metal into molds in the metal-casting process (Figure 14.2). The association between colour and temperature is so common that you would expect the mathematical relationship between colour and temperature to be simple. That is certainly what classical physicists expected. Despite their best efforts, however, classical physicists were never able to correctly predict the colour produced by an incandescent object. What is the connection between temperature and a glowing object\u2019s colour? Figure 14.3 shows three graphs that relate the colours produced by hot objects to their temperatures. The relationship between colour and temperature may be summarized as follows: 1. Hot, glowing objects emit a continuous range of wavelengths and hence a continuous spectrum of colours. 2. For a given temperature, the light emitted by the object has a range of characteristic wavelengths, which determine the object\u2019s colour when it glows (Figure 14.2). 3. The hotter an object is, the bluer the light it emits. The cooler an object is, the redder its light is. Figure 14.2 The colour of molten bronze depends on its temperature. incandescent: glowing with heat (a 14 12 10 8.0 6.0 4.0 2.0 0.0 (b) T 10 000 K 0.5 1 1.5 2 2.5 3 3.5 Frequency 1015 (Hz.0 4.5 4.0 3.5 3.0 2.5 2.0 1.5 1.0 5.0 0.0 (c) T 5000 K 0.5 1 1.5 2 2.5 3 3.5 Frequency 1015 (Hz 14 12 10 8.0 6.0 4.0 2.0 0.0 T 2500 K 0.5 1 1.5 2 2.5 3 3.5 Frequency 1015 (Hz) Figure 14.3 Blackbody curves for three different temperatures (Kelvin): 10 000 K, 5000 K, and 2500 K. Frequency is along the horizontal axis, and energy intensity emitted is along the vertical axis. Note that these graphs do not have the same vertical scale. If they did, graph (a) would be 256 times taller than graph (c)! Concept Check Next time you are under a dark, clear sky, look carefully at the stars. Some will appear distinctly bluish-white, while others will be reddish or orange in appearance. What do differences in colour tell you about the stars? 704 Unit VII Electromagnetic Radiation 14-PearsonPhys30-Chap14 7/24/08 3:57 PM Page 705 Physicists call the graphs in Figure 14.3 blackbody radiation curves. The term \u201cblackbody,\u201d introduced by the German physicist Gustav Kirchhoff in 1862, refers to an object that completely absorbs any light energy that falls on it, from all parts of the electromagnetic spectrum. When this perfect absorber heats up, it becomes a perfect radiator. The energy it reradiates can be depicted as a blackbody curve, which depends on temperature only (Figure 14.3). Hot objects, such as the filament in an incandescent light bulb used in 14-1 QuickLab, or a glowing wiener-roast stick, are good approximations to a blackbody. Not only did classical physics fail to explain the relationship between temperature and the blackbody radiation curve, it also made a completely absurd prediction: A hot object would emit its energy most effectively at short wavelengths, and that the shorter the wavelength, the more energy that would be emitted. This prediction leads to a rather disturbing conclusion: If you strike a match, it will emit a little bit of light energy at long wavelengths (e.g., infrared), a bit more energy in the red part of the spectrum, more yet in the blue, even more in the ultraviolet, a lot more in the X-ray region, and so on. In short, striking a match would incinerate the entire universe! This prediction was called the ultraviolet catastrophe. Fortunately for us, classical physics was incorrect. Figure 14.4 shows a comparison between the prediction made by classical physics and the blackbody radiation curve produced by a hot object. Quantization and Planck\u2019s Hypothesis In December 1900, Max Planck (Figure 14.5) came up with an explanation of why hot objects produce the blackbody radiation curves shown in Figures 14.3 and 14.4. Planck", " suggested that the problem with the classical model prediction had to do with how matter could absorb light energy. He discovered that, by limiting the minimum amount of energy that any given wavelength of light can exchange with its surroundings, he could reproduce the blackbody radiation curve exactly. The name quantum was given to the smallest amount of energy of a particular wavelength or frequency of light that could be absorbed by a body. Planck\u2019s hypothesis can be expressed in the following formula, known as Planck\u2019s formula: E nhf where E is the energy of the quantum, in joules, n 1, 2, 3 \u2026 refers to the number of quanta of a given energy, h is a constant of proportionality, called Planck\u2019s constant, which has the value 6.63 1034 Js, and f is the frequency of the light. If energy is transferred in quanta, then the amount of energy transferred must be quantized, or limited to whole-number multiples of a smallest unit of energy, the quantum. Even though Planck\u2019s hypothesis could reproduce the correct shape of the blackbody curve, there was no explanation in classical physics for his idea. The concept of the quantum marks the end of classical physics and the birth of quantum physics. blackbody radiation curve: a graph of the intensity of light emitted versus wavelength for an object of a given temperature blackbody: an object that completely absorbs any light energy that falls on it classical theory hottest y t i s n e t n I coolest Wavelength of emitted radiation Figure 14.4 According to classical theory, as an object becomes hotter, the intensity of light it emits should increase and its wavelength should decrease. The graph shows a comparison of the classical prediction (dashed line) and what is actually observed for three objects at different temperatures. Figure 14.5 Max Planck (1858\u20131947) is one of the founders of quantum physics. quantum: the smallest amount or \u201cbundle\u201d of energy that a wavelength of light can possess (pl. quanta) Planck\u2019s formula: light comes in quanta of energy that can be calculated using the equation E nhf quantized: limited to whole multiples of a basic amount (quantum) Chapter 14 The wave-particle duality reminds us that sometimes truth really is stranger than fiction! 705 14-PearsonPhys30-Chap14 7/24/08 3:57 PM Page 706 PHYSICS INSIGHT Planck\u2019s constant can be expressed using two different units: h 6.63 1034 Js or h 4.14 1015 eVs Recall that 1 eV 1.60 1019 J or 1 J 6.25 1018 eV. photon: a quantum of light Concept Check Show that Planck\u2019s formula for one photon can be written as E.hc Einstein, Quanta, and the Photon In 1905, a young and not-yet-famous Albert Einstein made a very bold suggestion. Planck had already introduced the idea of quantization of energy and the equation E hf. He thought that quantization applied only to matter and how matter could absorb or emit energy. Einstein suggested that this equation implied that light itself was quantized. In other words, Einstein reintroduced the idea that light could be considered a particle or a quantum of energy! This idea was troubling because, as you saw in Chapter 13, experiments clearly showed that light is a wave. In 1926, the chemist Gilbert Lewis introduced the term photon to describe a quantum of light. Planck\u2019s formula, E nhf, can therefore be used to calculate the energy of one or more photons. Examples 14.1 and 14.2 allow you to practise using the idea of the photon and Planck\u2019s formula. Example 14.1 How much energy is carried by a photon of red light of wavelength 600 nm? Practice Problems 1. What is the energy of a photon of light of frequency 4.00 1014 Hz? 2. What is the energy of a green photon of light of wavelength 555 nm? 3. What is 15.0 eV expressed in units of joules? Answers 1. 2.65 1019 J 2. 3.58 1019 J 3. 2.40 1018 J Given n 1 600 nm 1 9 m 6.00 \u00d7 107 m 1 0 m n 1 Required photon energy (E ) Analysis and Solution Since wavelength is given, first find the frequency using the equation c f, where c is the speed of light, f is frequency, and is the wavelength. c f m 3.00 108 s 6.00 107 m 5.00 1014 Hz Then substitute into Planck\u2019s formula: E nhf (1)(6.63 1034 Js)(5.00 1014 s1) 3.32 1019 J 706 Unit VII Electromagnetic Radiation 14-PearsonPhys30-Chap14 7/", "24/08 3:57 PM Page 707 It is often more convenient to express the energies of photons in units of electron volts. Since 1 eV 1.60 1019 J, the energy of the red photon is e V 1 3.32 1019 J 2.07 eV 019 J 1 1.60 Paraphrase A red photon of light carries 3.32 1019 J of energy, or about 2.07 eV. Example 14.2 Your eye can detect as few as 500 photons of light. The eye is most sensitive to light having a wavelength of 510 nm. What is the minimum amount of light energy that your eye can detect? Given 510 nm 5.10 107 m n 500 photons Required minimum light energy (E) Analysis and Solution Since only wavelength is given, determine frequency using the equation c f: c f m 3.00 108 s 5.10 107 m 5.88 1014 Hz Then apply Planck\u2019s formula: E nhf (500)(6.63 1034 Js)(5.88 1014 s1) 1.95 1016 J Practice Problems 1. What is the frequency of a 10-nm photon? 2. What is the energy of a 10-nm photon? 3. How many photons of green light ( 550 nm) are required to deliver 10 J of energy? Answers 1. 3.0 1016 Hz 2. 2.0 1017 J 3. 2.8 1019 photons Paraphrase Your eye is capable of responding to as little as 1.95 10\u201316 J of energy. M I N D S O N What\u2019s Wrong with This Analogy? Sometimes the idea of the quantum is compared to the units we use for money. A dollar can be divided into smaller units, where the cent is the smallest possible unit. In what way is this analogy for the quantum accurate and in what way is it inaccurate? Look very carefully at Planck\u2019s formula to find the error in the analogy. Try to come up with a better analogy for explaining quantization. Chapter 14 The wave-particle duality reminds us that sometimes truth really is stranger than fiction! 707 14-PearsonPhys30-Chap14 7/24/08 3:57 PM Page 708 The next example involves rearranging Planck\u2019s formula and applying it to find the relationship between the power of a laser pointer and the number of photons it emits. Example 14.3 Practice Problems 1. How much energy is delivered by a beam of 1000 blue-light photons ( 400 nm)? 2. How many 400-nm blue-light photons per second are required to deliver 10 W of power? How many photons are emitted each second by a laser pointer that has a power output of 0.400 mW if the average wavelength produced by the pointer is 600 nm? Given 600 nm 6.00 10\u20137 m P 0.400 mW 4.00 10\u20134 W Required number of photons (n) Answers 1. 4.97 1016 J 2. 2.0 1019 photons/s Analysis and Solution Since 1 W 1 J/s, the laser pointer is emitting 4.00 104 J/s. Therefore, in 1 s the laser pointer emits 4.00 104 J of energy. By equating this amount of energy to the energy carried by the 600-nm photons, you can determine how many photons are emitted each second using the equation E nhf. First use the equation c f to determine the frequency of a 600-nm photon: c f Substitute this equation into Planck\u2019s formula. E nhf c nh n E c h (4.00 104 J)(6.00 107 m) (6.63 1034 Js)3.00 108 m s 1.21 1015 Paraphrase A laser that emits 1.21 1015 photons each second has a power output of 0.400 mW. Photons and the Electromagnetic Spectrum Planck\u2019s formula provides a very useful way of relating the energy of a photon to its wavelength or frequency. It shows that a photon\u2019s energy depends on its frequency. An X-ray photon is more energetic than a microwave photon, just as X rays have higher frequencies than microwaves. Consequently, it takes a much more energetic process to create a gamma ray or X ray than it does to create a radio wave. Figure 14.6 gives the various photon energies along the electromagnetic spectrum. 708 Unit VII Electromagnetic Radiation 14-PearsonPhys30-Chap14 7/24/08 3:57 PM Page 709 X rays, for example, can only be emitted by a very hot gas or by a veryhigh-energy interaction between particles. Figure 14.7 shows images of the remnants of an exploded star, taken in different parts of the electromagnetic spectrum. Each image shows photons emitted by gases at different temperatures and locations in the remnant. The Electromagnetic Spectrum 103", " 102 101 1 1 10 2 10 3 10 4 10 5 10 6 10 7 10 8 10 9 10 1010 11 10 12 10 longer lower soccer field house baseball this period cell bacteria virus protein shorter water molecule radio waves infrared microwaves v i s i b l e ultraviolet \u201chard\u201d X rays \u201csoft\u201d X rays gamma rays AM radio FM radio microwave oven radar people light bulb X-ray machines radioactive elements 10 6 107 108 109 1010 1011 1012 1013 1014 1015 1016 1017 1018 1019 1020 higher 9 10 8 10 7 10 6 10 5 10 4 10 3 10 2 10 1 10 1 101 102 103 104 105 106 wavelength (in metres) size of a wavelength common name of wave sources frequency (waves per second) energy of one photon (electron volts) Figure 14.6 The energies of photons (in electron volts) along the electromagnetic spectrum Figure 14.7 These images of a supernova remnant were taken by the Chandra X-ray space telescope, the Hubble space telescope (visible part of the spectrum), and the Spitzer space telescope (infrared). Each image is produced by gases at different temperatures. X rays are produced by very-hightemperature gases (millions of degrees), whereas infrared light is usually emitted by low-temperature gases (hundreds of degrees). Chapter 14 The wave-particle duality reminds us that sometimes truth really is stranger than fiction! 709 14-PearsonPhys30-Chap14 7/24/08 3:57 PM Page 710 14.1 Check and Reflect 14.1 Check and Reflect Knowledge Extensions 1. What is the energy of a photon with wavelength 450 nm? 2. What is the wavelength of a photon of energy 15.0 eV? 3. Compare the energy of a photon of wavelength 300 nm to the energy of a 600-nm photon. Which photon is more energetic, and by what factor? 4. (a) What is the frequency of a photon that has an energy of 100 keV? (b) From what part of the electromagnetic spectrum is this photon? Applications 5. How many photons of light are emitted by a 100-W light bulb in 10.0 s if the average wavelength emitted is 550 nm? Assume that 100% of the power is emitted as visible light. 6. The Sun provides approximately 1400 W of solar power per square metre. If the average wavelength (visible and infrared) is 700 nm, how many photons are received each second per square metre? 7. Suppose that your eye is receiving 10 000 photons per second from a distant star. If an identical star was 10 times farther away, how many photons per second would you receive from that star in one second? 8. Estimate the distance from which you could see a 100-W light bulb. In your estimate, consider each of the following: \u2022 Decide on a representative wavelength for light coming from the light bulb. \u2022 Estimate the surface area of a typical light bulb and use this figure to determine the number of photons per square metre being emitted at the surface of the light bulb. \u2022 Estimate the diameter of your pupil and hence the collecting area of your eye. \u2022 Use the information in Example 14.2 to set a minimum detection limit for light from the light bulb. Remember that your answer is an estimate. It will likely differ from other students\u2019 estimates based on the assumptions you made. (Hint: The surface area of a sphere is 4r 2.) e TEST To check your understanding of Planck\u2019s formula, follow the eTest links at www.pearsoned.ca/ school/physicssource. 710 Unit VII Electromagnetic Radiation 14-PearsonPhys30-Chap14 7/24/08 3:57 PM Page 711 14.2 The Photoelectric Effect The secret agent cautiously inches forward and carefully steps over and around the thin, spidery outlines of laser beams focussed on light sensors scattered around Dr. Evil\u2019s secret lair. Is this scenario only the stuff of spy movies? Perhaps, but every time you walk into a shopping mall or have your groceries scanned at the supermarket, you, like the secret agent, are seeing an application of the way in which photons and metals interact. This interaction is called the photoelectric effect. 14-2 QuickLab 14-2 QuickLab Discharging a Zinc Plate Using UV Light Problem Does ultraviolet light cause the emission of electrons from a zinc metal plate? Materials electroscope UV light source zinc plate glass plate electroscope zinc plate UV source Figure 14.8 Procedure 1 Attach the zinc plate so that it is in contact with the electroscope. 2 Apply a negative charge to the zinc plate and electroscope. What happens to the vanes of the electroscope? (If you are uncertain how to apply a negative charge, consult your teacher for assistance.) 3 Turn on the UV light source and shine it directly on the zinc plate (see Figure 14.8). 4", " Place the glass plate between the UV light source and the zinc plate. Note any change in the behaviour of the vanes of the electroscope. Remove the plate and once again note any change in the response of the vanes. Questions 1. Why did the vanes of the electroscope deflect when a negative charge was applied? 2. Explain what happened when UV light shone on the zinc plate. Why does this effect suggest that electrons are leaving the zinc plate? 3. Glass is a known absorber of UV light. What happened when the glass plate was placed between the UV source and the electroscope? 4. From your observations, what caused the emission of electrons from the zinc surface? Give reasons for your answer. CAUTION: UV light is harmful to your eyes. Do not look directly into the UV light source. Chapter 14 The wave-particle duality reminds us that sometimes truth really is stranger than fiction! 711 14-PearsonPhys30-Chap14 7/24/08 3:57 PM Page 712 In 1887, German physicist Heinrich Hertz conducted a series of experiments designed to test Maxwell\u2019s theory of electromagnetic waves. In one of the experiments, a spark jumping between the two metal electrodes of a spark gap was used to create radio waves that could be detected in a similar spark-gap receiver located several metres away. Hertz noticed that his spark-gap receiver worked much better if the small metal electrodes were highly polished. Eventually, it was recognized that it was not the polishing but the ultraviolet light being produced by the main spark in his transmitter that greatly enhanced the ability of sparks to jump in his receiver\u2019s spark-gap. Hertz had discovered that some metals emit electrons when illuminated by sufficiently short (high-energy) wavelengths of light. This process is called photoemission of electrons, or the photoelectric effect. Electrons emitted by this process are sometimes called photoelectrons. How could light waves cause a metal to emit electrons? Experiments showed that the electrons required energies of a few electron volts in order to be emitted by the metal. Perhaps the atoms on the surface of the metal absorbed the energy of the light waves. The atoms would begin to vibrate and eventually absorb enough energy to eject an electron. There is a problem with this theory. According to classical physics, it should take minutes to hours for a metal to emit electrons. Experiments showed, however, that electron emission was essentially instantaneous: There was no measurable delay between the arrival of light on the metal surface and the emission of electrons. To further add to the puzzle, there was a minimum or threshold frequency, f0, of incident light below which no photoemission would occur. If the light shining on the metal is of a frequency lower than this threshold frequency, no electrons are emitted, regardless of the brightness of the light shining on the metal (Figure 14.9). photon photon The incoming photon has enough energy to knock loose an electron. Nothing happens! The incoming photon lacks enough energy to cause photoemission. Figure 14.9 If an incident photon has a high enough frequency, an electron will be emitted by the metal surface. If the incoming photon frequency is not high enough, an electron will not be emitted. Another puzzle was the lack of clear connection between the energy of the electrons emitted and the brightness of the light shining on the metal surface. For a given frequency of light, provided it was greater than the threshold frequency, the emitted electrons could have a range of possible kinetic energies. Increasing the intensity of the light had no influence on the maximum kinetic energy of the electrons. photoelectric effect: the emission of electrons when a metal is illuminated by short wavelengths of light photoelectron: an electron emitted from a metal because of the photoelectric effect threshold frequency: the minimum frequency that a photon can have to cause photoemission from a metal Table 14.1 Work Functions of Some Common Metals Element Work Function (eV) Aluminium Beryllium Cadmium Calcium Carbon Cesium Copper Magnesium Mercury Potassium Selenium Sodium Zinc 4.08 5.00 4.07 2.90 4.81 2.10 4.70 3.68 4.50 2.30 5.11 2.28 4.33 712 Unit VII Electromagnetic Radiation 14-PearsonPhys30-Chap14 7/24/08 3:57 PM Page 713 Einstein\u2019s Contribution The photoelectric effect remained an interesting but completely unexplained phenomenon until 1905. In 1905, Albert Einstein solved the riddle of the photoelectric effect by applying Planck\u2019s quantum hypothesis: Light energy arrives on the metal surface in discrete bundles, which are absorbed by atoms of the metal. This process takes very little time and all the energy needed to expel an electron is provided at once. However, photoemission only occurs if the frequency of the incident photons is greater than or equal to the threshold frequency of the metal. Since the frequency of a photon is directly proportional to its energy, as given", " by Planck\u2019s formula, E hf, the incident photons must have the minimum energy required to eject electrons. This minimum energy is known as the work function, W. The work function is specific for every metal. Table 14.1 lists the work functions of some common metals. The work function, W, is related to threshold frequency, f0, by the equation W hf0. Photons with a frequency greater than the threshold frequency have energy greater than the work function and electrons will be ejected. M I N D S O N Light a Particle? Heresy! Suggest reasons why a physicist might argue against Einstein\u2019s idea that light is a particle. One such physicist was Robert A. Millikan, whose important experiments on the photoelectric effect were viewed, ironically, as a brilliant confirmation of Einstein\u2019s \u201ccrazy\u201d idea. How is skepticism both an advantage and a disadvantage to the progress of science? Millikan\u2019s Measurement of Planck\u2019s Constant When photons are absorbed by a metallic surface, either nothing will happen \u2014 the photons lack the minimum energy required to cause photoemission \u2014 or an electron will be emitted (Figure 14.10). photon E hf Kinetic energy of the electron equals the difference between the photon energy and the work function. Ek hf W The incident photon has energy E hf. The photon must be able to provide enough energy to equal or exceed the work function, W, in order to cause emission of an electron. work function: the minimum energy that a photon can have to cause photoemission from a metal; specific for every metal Figure 14.10 The kinetic energy of an electron emitted during photoemission is equal to the difference between the incident photon\u2019s energy and the work needed to overcome the work function for the surface. One of the most successful experiments to investigate the photoelectric effect was conducted by American physicist Robert Millikan (Figure 14.11) and published in 1916. The main result from Millikan\u2019s work is given in Figure 14.12. The graph shows electron kinetic energy as a function of the frequency of the incident light. When the light frequency is Figure 14.11 Robert Andrews Millikan (1868\u20131953) was awarded the Nobel Prize in physics in 1923 for his work on determining the charge of an electron, and for his work on the photoelectric effect. Despite his important work on the photoelectric effect, Millikan remained deeply skeptical of Einstein\u2019s particle view of light. Chapter 14 The wave-particle duality reminds us that sometimes truth really is stranger than fiction! 713 14-PearsonPhys30-Chap14 7/24/08 3:57 PM Page 714 e MATH Millikan graphed the kinetic energy of the photoelectons as a function of the incident frequency. To explore this relationship closer and to plot a graph like the one shown in Figure 14.12, visit www.pearsoned.com/school/ physicssource. PHYSICS INSIGHT Ek hf W and y mx b Therefore, f0 xint and W yint PHYSICS INSIGHT ) Maximum Ek vs. Frequency Cesium Potassium Calcium Magnesium Mercury 6 8 10 12 14 16 18 Frequency (Hz 1014) Different metals have different threshold frequencies, as shown in this graph. Which metal has the highest threshold frequency? below the threshold frequency, no electrons are ejected. When the light frequency equals the threshold frequency, electrons are ejected but with zero kinetic energy. The threshold frequency is therefore the x-intercept on the graph Light below a frequency of 4.39 1014 Hz or wavelength longer than 683 nm would not eject electrons. The fact that this plot was not dependent upon the intensity of the incident light implied that the interaction was like a particle that gave all its energy to the electron and ejected it with that energy minus the energy it took to escape the surface. 0 0 4 6 8 Frequency 1014 (Hz) 10 12 Figure 14.12 A graph based on the 1916 paper in which Millikan presented the data from his investigation of the photoelectric effect Once the frequency of the light exceeds the threshold frequency, photoemission begins. As the light frequency increases, the kinetic energy of the electrons increases proportionally. You can express this relationship in a formula by using the law of conservation of energy. The energy of the electron emitted by the surface is equal to the difference between the original energy of the photon, given by E hf, minus the work needed to free the electron from the surface. The equation that expresses this relationship is Ek hf W where Ek is the maximum kinetic energy of the electrons and W is the work function of the metal. You may recall that this equation is an example of the straight-line relationship y mx b, where m is the slope of the line and b is the y-intercept. The graph in Figure 14.12 shows the linear relationship between the frequency of the incident light falling on a", " sodium metal surface and the maximum kinetic energy of the electrons emitted by the metal. The slope of this line shows that the energy of the photons is directly proportional to their frequency, and the proportionality constant is none other than Planck\u2019s constant. Millikan\u2019s photoelectric experiment provides an experimental way to measure Planck\u2019s constant. The y-intercept of this graph represents the negative of the work function of the photosensitive surface. The work function can also be determined by measuring the threshold frequency of photons required to produce photoemission of electrons from the metal. Even though classical physics could not explain the photoelectric effect, this phenomenon still obeys the fundamental principle of conservation of energy, where ETotalinitial. The energy of the photon is completely transferred to the electron and can be expressed by the following equation: ETotalfinal hf W Ek 714 Unit VII Electromagnetic Radiation 14-PearsonPhys30-Chap14 7/24/08 3:57 PM Page 715 Another way to interpret this equation is that the energy of the photon liberates the electron from the photosensitive surface, and any remaining energy appears as the electron\u2019s kinetic energy. Concept Check Derive a relationship between energy of a photon (hf) and work function for a metal (W) to determine whether or not photoemission will occur. Stopping Potentials and Measuring the Kinetic Energy of Photoelectrons How did Millikan determine the maximum kinetic energy of electrons emitted by a metal surface? Figure 14.13 shows a highly simplified version of his experimental set-up. An evacuated tube contains a photoelectron-emitting metal surface and a metal plate, called the collector. A power supply is connected to the collector and the electron-emitting metal surface. When the power supply gives the collector plate a positive charge, the ammeter registers an electric current as soon as the incoming photons reach the threshold frequency. Any electrons emitted by the metal surface are attracted to the collector and charge begins to move in the apparatus, creating a current. incoming photons collector electron-emitting metal surface evacuated tube Figure 14.13 A simplified diagram depicting an experimental set-up used to investigate the photoelectric effect. When the power supply is connected as shown, the ammeter measures a current whenever the frequency of the incoming light exceeds the threshold frequency for the metal surface. ammeter ON OFF power supply Concept Check Explain the role of the collector plate in the photoelectric experiment apparatus in Figure 14.13. If the collector plate is not given a charge, would an electric current still be measured if the incoming photons exceed the threshold frequency? e SIM Find out more about the photoelectric effect by doing this simulation. Follow the eSim links at www.pearsoned.ca/school/ physicssource. Chapter 14 The wave-particle duality reminds us that sometimes truth really is stranger than fiction! 715 14-PearsonPhys30-Chap14 7/24/08 3:57 PM Page 716 Now consider what happens if the collector plate is given a negative charge. Instead of being attracted toward the collector, electrons now experience an electric force directed away from the collector. This electric force does work on the photoelectron (Figure 14.14). Photoelectrons will arrive at the collector only if they leave the metal surface with enough kinetic energy to reach the collector. You can express the final kinetic energy of the electrons in the following way: Ekfinal Ekinitial E where Ekfinal is the final kinetic energy of the electron, Ekinitial is its initial kinetic energy, and E is the work done by the electric force. The electric force opposes the motion of the electron. incoming photons Fe collector The electron is repelled by the collector and attracted to the positively charged plate. Figure 14.14 When the charges on the plates are reversed, the photoelectrons are repelled by the negatively charged collector and pulled back toward the positively charged plate. Only the most energetic electrons will reach the negative plate. Ekinitial In Chapter 11, you saw that the work done in an electric field of potential V on a charge q is expressed by the equation E qV. The final kinetic energy of an electron arriving at the collector can now be written as qV, where q represents the charge of an electron. If the Ekfinal negative potential on the collector plate is increased, then eventually a point will be reached at which no electrons will be able to reach the collector. At this point, the current in the ammeter drops to zero and the potential difference is now equal to the stopping potential. In summary, qVstopping. The maximum the current drops to zero when 0 Ekmax kinetic energy of electrons may now be expressed as Ekmax qVstopping where Vstopping is the stopping potential and q is the charge of the electron. Example 14.4 stopping potential: the potential difference for which the kinetic energy of a photoelectron equals the work needed to move through a potential difference, V Practice Problems 1.", " What stopping potential will stop electrons of energy 5.3 10\u201319 J? 2. Convert 5.3 10\u201319 J to electron volts. 3. What is the maximum kinetic energy of electrons stopped by a potential of 3.1 V? 716 Unit VII Electromagnetic Radiation Blue light shines on the metal surface shown in Figure 14.13 and causes photoemission of electrons. If a stopping potential of 2.6 V is required to completely prevent electrons from reaching the collector, determine the maximum kinetic energy of the electrons. Express your answer in units of joules and electron volts. Given Vstopping 2.6 V q 1.60 1019 C Required maximum kinetic energy of electrons (Ekmax) 14-PearsonPhys30-Chap14 7/24/08 3:57 PM Page 717 Analysis and Solution Use the equation Ekmax Ekmax (1.60 1019 C)(2.6 V) 4.2 1019 J qVstopping. 1 eV 1.60 1019 J Answers 1. 3.3 V 2. 3.3 eV 3. 3.1 eV or 5.0 1019 J 2.6 eV Paraphrase A stopping potential of 2.6 V will stop electrons of kinetic energy 4.2 1019 J or 2.6 eV. Concept Check Show that the idea of stopping potential can lead directly to the qVstopp i expression h f 0 ng W, where h is Planck\u2019s constant,Vstopping is the stopping potential, W is the work function, and f0 is the threshold frequency for emission of electrons from a metal surface. 14-3 Design a Lab 14-3 Design a Lab Using the Photoelectric Effect to Measure Planck\u2019s Constant The Question How can you use the photoelectric effect and the concept of stopping potential to determine Planck\u2019s constant? Design and Conduct Your Investigation You will need to decide on what equipment to assemble to enable you to relate frequency of incident light to kinetic energy of electrons and stopping potentials. In your design, be sure to address what you will need to measure and what variables will be involved, how to record and analyze your data, and how to use the data collected to answer the question. Prepare a research proposal for your teacher to determine whether your school laboratory has the necessary equipment for this lab, or if alternative approaches may work. Your proposal should include a worked-out sample of how the data you hope to collect will answer the question. Remember to work safely, to clearly identify tasks, and to designate which group members are responsible for each task. Chapter 14 The wave-particle duality reminds us that sometimes truth really is stranger than fiction! 717 14-PearsonPhys30-Chap14 7/24/08 3:57 PM Page 718 The following example shows how to relate the concepts of threshold frequency and work function. Example 14.5 Experiments show that the work function for cesium metal is 2.10 eV. Determine the threshold frequency and wavelength for photons capable of producing photoemission from cesium. Practice Problems 1. Light of wavelength 480 nm is just able to produce photoelectrons when striking a metal surface. What is the work function of the metal? 2. Blue light of wavelength 410 nm strikes a metal surface for which the work function is 2.10 eV. What is the energy of the emitted photoelectron? Answers 1. 2.59 eV 2. 0.932 eV 0 hf0 W h f0 Given W 2.10 eV Required threshold frequency (f0) wavelength () Analysis and Solution The work function is the amount of energy needed to just break the photoelectron free from the metal surface, but not give it any additional kinetic energy. Therefore, from 0 J. Ek hf W, for threshold frequency, f0, set Ek First convert the work function to units of joules. W (2.10 eV)1.60 1019 J V e 3.36 1019 J Now solve for the threshold frequency. W 3.36 1019 J 6.63 1034 Js 5.07 1014 Hz From c f, the wavelength of this photon is c f m 3.00 108 s 5.07 1014 s1 5.91 107 m 591 nm Paraphrase The threshold frequency for photons able to cause photoemission from cesium metal is 5.07 1014 Hz. This frequency corresponds to photons of wavelength 591 nm, which is in the yellow-orange part of the visible spectrum. You can also use the law of conservation of energy equation for the photoelectric effect to predict the energy and velocity of the electrons released during photoemission. 718 Unit VII Electromagnetic Radiation 14-PearsonPhys30-Chap14 7/24/08 3:57 PM Page 719 Example 14.6 Using Table 14.1, determine the maximum speed of", " electrons emitted from an aluminium surface if the surface is illuminated with 125-nm ultraviolet (UV) light. Given 125 nm metal aluminium Required maximum speed of electrons (v) Analysis and Solution From Table 14.1, the work function for aluminium is 4.08 eV. Convert this value to joules. W 4.08 eV (4.08 eV )1.60 1019 J eV 6.528 1019 J To determine the energy of the incident photon, use the equation E hf h c. Practice Problems 1. A photoelectron is emitted with a kinetic energy of 2.1 eV. How fast is the electron moving? 2. What is the kinetic energy of a photoelectron emitted from a cesium surface when the surface is illuminated with 400-nm light? 3. What is the maximum speed of the electron described in question 2? Answers 1. 8.6 105 m/s 2. 1.01 eV 3. 5.95 105 m/s Incident photon energy is E h c (6.63 1034 Js) 1.591 1018 J 3.00 108 m s 1.25 107 m To find the kinetic energy of the electrons, use the law of conservation of energy equation for the photoelectric effect, Ek energy of the electron is hf W. Kinetic Ek hf W 15.91 1019 J 6.528 1019 J 9.384 1019 J Finally, use Ek mass of 9.11 1031 kg. 1 mv2 to solve for speed. Recall that an electron has a 2 The electron\u2019s speed is v 2Ek m 2(9.384 1019 J) 9.11 1031 kg 1.44 106 m/s Paraphrase The electrons emitted from the aluminium surface will have a maximum speed of 1.44 106 m/s. Chapter 14 The wave-particle duality reminds us that sometimes truth really is stranger than fiction! 719 14-PearsonPhys30-Chap14 7/24/08 3:57 PM Page 720 Millikan\u2019s work on the photoelectric effect provided critical evidence in eventually demonstrating the particle or quantized nature of light. As you will see in the next chapter, Millikan also performed a key experiment that demonstrated the discrete or \u201cquantized\u201d nature of electrical charge: He showed that the electron is the smallest unit of electrical charge. 14.2 Check and Reflect 14.2 Check and Reflect Knowledge 1. What is the energy, in eV, of a 400-nm photon? 2. Explain how the concepts of work function and threshold frequency are related. 3. What is the threshold frequency for cadmium? (Consult Table 14.1 on page 712.) 4. Will a 500-nm photon cause the emission of an electron from a cesium metal surface? Explain why or why not. 5. What stopping voltage is needed to stop an electron of kinetic energy 1.25 eV? 6. Explain how stopping potential is related to the maximum kinetic energy of an electron. 7. True or false? The greater the intensity of the light hitting a metal surface, the greater the stopping potential required to stop photoelectrons. Explain your answer. Applications The following data are taken from an experiment in which the maximum kinetic energy of photoelectrons is related to the wavelength of the photons hitting a metal surface. Use these data to answer the following questions. Wavelength (nm) Kinetic Energy (eV) 500 490 440 390 340 290 240 0.36 0.41 0.70 1.05 1.52 2.14 3.025 720 Unit VII Electromagnetic Radiation 8. Convert the wavelengths given in the data table to frequency units and graph them along with the kinetic energy of the photoelectrons. Be sure to plot frequency on the horizontal axis. 9. Give the value of the slope of the graph that you just drew. What is the significance of this value? 10. What metal do you think was used in the previous example? Justify your answer. Extensions 11. Explain how the photon model of light correctly predicts that the maximum kinetic energy of electrons emitted from a metal surface does not depend on the intensity of light hitting the metal surface. 12. In several paragraphs, identify three common devices that use the photoelectric effect. Be sure to explain in what way these devices use the photoelectric effect. 13. How long would photoemission take from a classical physics point of view? Consider a beam of ultraviolet light with a brightness of 2.0 10\u20136 W and an area of 1.0 10\u20134 m2 (about the area of your little fingernail) falling on a zinc metal plate. Use 3.5 eV as the energy that must be absorbed before photoemission can occur. (Hint: Estimate the area of an atom and determine how much of the beam of UV light is being absorbed each second, on average.) e TEST To check your", " understanding of the photoelectric effect, follow the eTest links at www.pearsoned.ca/ school/physicssource. 14-PearsonPhys30-Chap14 7/24/08 3:57 PM Page 721 14.3 The Compton Effect Although Einstein\u2019s photon model provided an explanation of the photoelectric effect, many physicists remained skeptical. The wave model of light was so successful at explaining most of the known properties of light, it seemed reasonable to expect that a purely classical explanation of the photoelectric effect would eventually be found. In 1923, however, an experiment by American physicist Arthur Compton (Figure 14.15) provided an even clearer example of the particle nature of light, and finally convinced most physicists that the photon model of light had validity. Compton studied the way in which electrons scattered X rays in a block of graphite. The X rays were observed to scatter in all directions. This effect was not surprising: Both the wave and particle models of light predicted this outcome. What the wave model could neither predict nor explain, however, was the small change in wavelength that Compton observed in the scattered X ray, and the relationship between the change in wavelength and the angle through which the X ray was scattered. The scattering of an X ray by an electron is now referred to as Compton scattering, and the change in wavelength of the scattered X-ray photon is called the Compton effect (Figure 14.16). To understand the Compton effect, you will need to use two of the most central ideas of physics: the law of conservation of momentum and the law of conservation of energy. The interaction between an X-ray photon and an electron must still obey these laws. By using the particle model of light and Einstein\u2019s mass-equivalence equation E mc2, Compton showed that the momentum of the X ray could be expressed as p h Figure 14.15 Arthur Holly Compton (1892\u20131962) was a pioneer in high-energy physics. He was awarded the Nobel Prize in 1927 for his discovery of the Compton effect, which provided convincing evidence for the photon model of light. Compton scattering: the scattering of an X ray by an electron Compton effect: the change in wavelength of the scattered X-ray photon where p is momentum, h is Planck\u2019s constant, and is the wavelength of the X ray. Compton was also able to show exactly how the change in wavelength of the scattered X ray is related to the angle through which the X-ray photon is scattered. \u03bb recoil electron electron at rest \u03b8 scattered X-ray photon \u03bb f i incident X-ray photon Concept Check Which of the following photons has the greater momentum: A 2 nm? Explain your reasoning. 500 nm or B Compton found that the scattered X ray changed its momentum and energy in a way that was exactly what you would expect if it was a small particle undergoing an elastic collision with an electron. Recall from Chapter 9 that energy and momentum are conserved during an elastic collision. Figure 14.16 When an electron scatters an X ray, both momentum and energy are conserved. Compton scattering behaves like an elastic collision between a photon and an electron. Chapter 14 The wave-particle duality reminds us that sometimes truth really is stranger than fiction! 721 14-PearsonPhys30-Chap14 7/24/08 3:57 PM Page 722 The laws of conservation of energy and of momentum can be applied to the X ray and the electron in the following way: \u2022 The total momentum of the incident X-ray photon must equal the total momentum of the scattered X ray and the scattered electron. \u2022 The total energy of the incident X-ray photon and the electron must equal the total energy of the scattered X ray and the scattered electron. Concept Check Study Figure 14.16. Define the direction of the incident X-ray photon as the positive x-direction and the upward direction as the positive y-direction. Suppose the incident X-ray photon has a wavelength of 1. Derive an expression for the x and y components of the i and the scattered X-ray photon has a wavelength f. momentum of the scattered photon. 2. Explain how your answer to question 1 gives you the x and y components of the electron\u2019s momentum. 3. How much energy was transferred to the electron in this interaction? Derive a simple expression for the electron\u2019s final energy. CCompton derived the following relationship between the change in the wavelength of the scattered photon and the direction in which the scattered photon travels: f i h (1 cos ) m c where m is the mass of the scattering electron and is the angle through which the X ray scatters. The full derivation of this equation requires applying Einstein\u2019s theory of relativity and a lot of algebra! The central concepts behind this equation, however, are simply the laws of conservation of energy and of momentum. As well, this equation is exactly consistent with Einstein\u2019s idea that the X-ray photon collides with the", " electron as if it were a particle. M I N D S O N Heisenberg\u2019s Microscope Problem Suggest how Compton scattering shows that it is impossible to \u201csee\u201d an electron. In particular, why is it that we can only see where an electron was and not where it is? (Hint: Think about what photons are doing when you look at something.) This question is sometimes referred to as Heisenberg\u2019s microscope problem. 722 Unit VII Electromagnetic Radiation 14-PearsonPhys30-Chap14 7/24/08 3:57 PM Page 723 Example 14.7 What is the maximum change in wavelength that a 0.010-nm X-ray photon can undergo by Compton scattering with an electron? Does initial wavelength (0.010 nm) matter in this example? Given i 0.010 nm Required change in wavelength () Analysis and Solution Maximum change will occur when the X ray is scattered by the greatest possible amount, that is, when the X ray is back-scattered. From the Compton effect equation, i f h (1 cos ), the maximum value for c m occurs when the term (1 cos ) is a maximum. This occurs when 180\u00b0 and cos 1, so (1 cos ) becomes (1 (1)) 2. Use this relation to determine the largest possible change in wavelength of the scattered X-ray photon. h (1 cos ) c m 2h c m 2(6.63 1034 Js) (9.11 1031 kg)(3.00 108 m/s) 4.85 1012 m Paraphrase The maximum change in wavelength of a photon during Compton scattering is only 4.85 1012 m. This change is independent of the initial wavelength of the photon. Practice Problems 1. What is the energy of an X ray of wavelength 10 nm? 2. What is the momentum of an X ray of wavelength 10 nm? 3. If a 10-nm X ray scattered by an electron becomes an 11-nm X ray, how much energy does the electron gain? Answers 1. 2.0 1017 J 2. 6.6 1026 Ns 3. 1.8 1018 J Chapter 14 The wave-particle duality reminds us that sometimes truth really is stranger than fiction! 723 14-PearsonPhys30-Chap14 7/24/08 3:57 PM Page 724 You can also use the Compton equation to determine the final wavelength of a photon after scattering, as you will see in the next example. Example 14.8 An X-ray photon of wavelength 0.0500 nm scatters at an angle of 30\u00b0. Calculate the wavelength of the scattered photon. Given i 30\u00b0 0.0500 nm Required final wavelength ( f) Analysis and Solution Rearrange the Compton equation to solve for final wavelength. Recall that the mass of an electron is 9.11 1031 kg. i f h (1 cos ) c m i f i h (1 cos ) c m 0.0500 nm 6.63 1034 Js (9.11 1031 kg)(3.00 108 m/s) (1 cos 30\u00b0) 0.0500 nm 0.000 325 nm 0.0503 nm Paraphrase The X-ray photon changes wavelength by 0.0003 nm to become a photon of wavelength 0.0503 nm. Practice Problem 1. An X ray of wavelength 0.010 nm scatters at 90\u00b0 from an electron. What is the wavelength of the scattered photon? Answer 1. 0.012 nm 724 Unit VII Electromagnetic Radiation 14-PearsonPhys30-Chap14 7/24/08 3:57 PM Page 725 For many physicists, the Compton effect provided the final piece of evidence they needed to finally accept Einstein\u2019s idea of the particle nature of light. The Compton effect also describes one of the most fundamental phenomena \u2014 the interaction of light with matter. 14.3 Check and Reflect 14.3 Check and Reflect Knowledge Applications 1. What is the momentum of a 500-nm photon? 6. What is the wavelength of a 100-keV X-ray 2. Photon A has a wavelength three times longer than photon B. Which photon has the greatest momentum and by what factor? 3. A photon has a momentum of 6.00 10\u201321 kgm/s. What is the wavelength and energy of this photon? 4. Identify the part of the electromagnetic spectrum of the photon in question 3. 5. True or false? One of the major differences between classical physics and quantum physics is that the laws of conservation of energy and momentum do not always work for quantum physics. Explain your answer. photon? 7. An X-ray photon of wavelength 0.010 nm strikes a helium nucleus and bounces straight back. If the helium nucleus was originally at rest, calculate its velocity after interacting with the X ray. Extension 8. In order to see an object, it is necessary", " to illuminate it with light whose wavelength is smaller than the object itself. According to the Compton effect, why is illumination a problem if you wish to see a small particle, such as a proton or an electron? e TEST To check your understanding of the Compton effect, follow the eTest links at www.pearsoned.ca/ school/physicssource. Chapter 14 The wave-particle duality reminds us that sometimes truth really is stranger than fiction! 725 14-PearsonPhys30-Chap14 7/24/08 3:57 PM Page 726 14.4 Matter Waves and the Power of Symmetric Thinking If waves (light) can sometimes act like particles (photons), then why couldn\u2019t particles, such as electrons, sometimes act like waves? Louis de Broglie (pronounced \u201cde Broy\u201d) (Figure 14.17), a young French Ph.D. student, explored this question in 1924 in a highly imaginative and perplexing thesis. The idea seemed so strange that despite no obvious errors in his argument, the examining committee was reluctant to pass de Broglie. Fortunately, a copy of his thesis was sent to Albert Einstein, who recognized at once the merit in de Broglie\u2019s hypothesis. Not only was de Broglie awarded his Ph.D., but his hypothesis turned out to be correct! De Broglie\u2019s argument is essentially one of symmetry. As both the photoelectric effect and the Compton effect show, light has undeniable particle-like, as well as wave-like, properties. This dichotomy is called the wave-particle duality. In reality, light is neither a wave nor a particle. These ideas are classical physics ideas, but experiments were revealing subtle and strange results. What light is depends on how we interact with it. De Broglie\u2019s hypothesis completes the symmetry by stating that what we naturally assume to be particles (electrons, for example) can have wave-like properties as well. At the atomic level, an electron is neither a wave nor a particle. What an electron is depends on how we interact with it. De Broglie arrived at his idea by tying together the concepts of momentum and wavelength. Using Compton\u2019s discovery relating momentum and wavelength for X-ray photons, de Broglie argued that anything that possessed momentum also had a wavelength. His idea can be expressed in a very simple form: h p where h is Planck\u2019s constant, p is momentum, and is de Broglie\u2019s wavelength. De Broglie\u2019s Wave Equation Works for Both Light and Electrons De Broglie\u2019s hypothesis states that anything that has momentum must obey the following wavelength-momentum equations: For light: Maxwell\u2019s law of electromagnetism shows that the momentum of a light wave can be written as p E, where E is the energy of the c light and c is the speed of light. But Planck\u2019s formula states that E hf. f h. Substituting this equation into de Broglie\u2019s wave Therefore, p c Louis de Figure 14.17 Broglie (1892\u20131987) was the first physicist to predict the existence of matter waves. wave-particle duality: light has both wave-like and particle-like properties Project LINK How important is de Broglie\u2019s hypothesis to our current understanding of the nature of light and matter? 726 Unit VII Electromagnetic Radiation 14-PearsonPhys30-Chap14 7/24/08 3:58 PM Page 727 equation, you obtain h h f c c, which is the wavelength-frequency f relation. It tells you that a photon of light has a wavelength! For electrons: If an electron is moving with a velocity, v, that is much less than the speed of light, then its momentum is p mv and de Broglie\u2019s relationship is h p h mv. For electrons (or any other particles) moving at velocities approaching the speed of light, the expression h p is still applicable. PHYSICS INSIGHT Einstein showed that, as objects\u2019 speeds approach the speed of light, the familiar expression for momentum, p mv, must be replaced by the more complicated equation p mv \u2013\u2013\u2013\u2013\u2013\u2013\u2013\u2013 v2 c2 1, where c is the speed of light. THEN, NOW, AND FUTURE The Electron Microscope Modern TEMs are capable of reaching very high magnification and imaging at the atomic level. The scanning electron microscope (SEM) is similar to the TEM but differs in one important way: Electrons are reflected off the sample being imaged. SEM images have a remarkable three-dimensional appearance (Figure 14.19). The Electron Microscope The idea of matter waves is not simply abstract physics that has no practical application. The wave nature of electrons has been ", "used to build microscopes capable of amazing magnification. The reason for their amazing magnification lies in the extremely small wavelengths associated with electrons. The usable magnification of a microscope depends inversely on the wavelength used to form the image. In a transmission electron microscope (TEM, Figure 14.18), a series of magnets (magnetic lenses) focusses a beam of electrons and passes the beam through a thin slice of the specimen being imaged. Figure 14.19 An SEM view of an ant\u2019s head Questions 1. Find out more about the varieties of electron microscopes in use. Search the Internet, using key words such as electron microscope, TEM, or SEM, to learn about at least three different kinds of electron microscopes. Summarize your findings in the following way: Figure 14.18 A modern transmission electron microscope \u2022 name (type) of microscope \u2022 how it differs from other electron microscopes in use and operation \u2022 typical applications and magnifications 2. The magnification of a microscope depends inversely on the wavelength used to image a specimen. The very best quality light microscopes typically have maximum magnifications of 1000 to 4000 times. Modern TEMs use electrons accelerated to energies of over 100 keV to observe specimens. Estimate the possible range of magnifications that can be achieved using a TEM by considering the following: \u2022 What is a reasonable choice for the wavelength used in a light microscope? \u2022 What is the wavelength of a 100-keV electron? \u2022 How do the wavelengths of the light and of the electrons compare? (Note: Your answer will likely be an overestimate. The actual magnification of electron microscopes is limited by the ability of the magnetic lenses to focus the electron beam. TEMs are capable of achieving magnifications as high as 500 000 times!) Chapter 14 The wave-particle duality reminds us that sometimes truth really is stranger than fiction! 727 14-PearsonPhys30-Chap14 7/24/08 3:58 PM Page 728 The next two examples apply the de Broglie relationship between momentum and wavelength. Example 14.9 What is the momentum of a 500-nm photon of green light? Practice Problem 1. What is the momentum of a 0.010-nm X ray? Answer 1. 6.6 \u00d7 10\u201323 kgm/s e MATH De Broglie showed how electrons can be thought of as waves and related the speed of an electron to its wavelength. Einstein\u2019s work showed that as the speed of the electron became greater than 10% of the speed of light, relativistic effects has to be taken into account (see Physics Insight p. 727) To explore how the wavelength of an electron is a function of its speed, including relativistic effects, visit www.pearsoned.ca/school/ physicssource. Given 500 nm Required momentum (p) Analysis and Solution To find the photon\u2019s momentum, apply de Broglie\u2019s equation: p h 6.63 1034 Js 500 109 m 1.33 1027 Ns Paraphrase The photon has a momentum of 1.33 1027 Ns. The next example shows how to calculate the wavelength of an electron, thus illustrating that particles have a wave nature. Example 14.10 What is the wavelength of an electron moving at 1.00 104 m/s? Practice Problems 1. What is the wavelength of a proton moving at 1.0 105 m/s? 2. What is the speed of an electron that has a wavelength of 420 nm? Given v 1.00 104 m/s 9.11 1031 kg me Required wavelength () Answers 1. 4.0 1012 m 2. 1.73 103 m/s Analysis and Solution To find the electron\u2019s wavelength, first find its momentum and then rewrite de Broglie\u2019s equation: mv p h h mv 6.63 1034 Js (9.11 1031 kg)(1.00 104 m/s) 7.28 108 m 72.8 nm Paraphrase The electron has a de Broglie wavelength of 7.28 108 m or 72.8 nm. 728 Unit VII Electromagnetic Radiation 14-PearsonPhys30-Chap14 7/24/08 3:58 PM Page 729 De Broglie\u2019s idea completes the concept of the wave-particle duality of light. Wave-particle duality combines two opposing ideas and teaches us that, at the atomic level, it is essential to use both ideas to accurately model the world. De Broglie\u2019s Wave Hypothesis: Strange but True! Experimental proof of de Broglie\u2019s hypothesis came very quickly and by accident. Between 1925 and 1927, American physicists C. J. Davisson and L. H. Germer, and British physicist G. P. Thomson (Figure 14.20, son of J. J. Thomson, discoverer of the electron", ") independently provided evidence that electrons can act like waves. The original Davisson and Germer experiment was an investigation of how electrons scattered after hitting different kinds of metallic surfaces. To prevent an oxide layer from contaminating the surfaces, the scattering was done inside a vacuum tube. In one test on a nickel surface, the vacuum tube cracked and the vacuum was lost, unbeknownst to Davisson and Germer. The nickel surface oxidized into a crystalline pattern. What Davisson and Germer observed was a very puzzling pattern: Scattering occurred in some directions and not in others. It was reminiscent of a pattern of nodes and antinodes (Figure 14.21). A simplified version of a typical Davisson\u2013Germer experiment is shown in Figure 14.22(a). The graph in Figure 14.22(b) shows the kind of data that Davisson and Germer found. M I N D S O N Interpret the Graph (a) Look at the graph in Figure 14.22(b). Explain why it makes sense to interpret the pattern as one of nodes and antinodes. What is happening at each of the nodes? (b) What would you have to do to change the wavelength of the electrons used in an electron diffraction experiment? Figure 14.20 George Paget Thomson (1892\u20131975) was codiscoverer of matter waves with Davisson and Germer. One of the great ironies of physics is that Thomson played an instrumental role in showing that electrons can act like waves. Thirty years earlier, his father had shown that the electron was a particle! When Davisson and Germer began their experiments, they were unaware of de Broglie\u2019s work. As soon as they learned of de Broglie\u2019s hypothesis, however, they realized that they had observed electron-wave interference! Over the next two years, they and G. P. Thomson in Scotland refined the study of electron-wave interference and provided beautiful experimental confirmation of de Broglie\u2019s hypothesis. In 1937, Davisson and Thomson received a Nobel Prize for the discovery of \u201cmatter waves.\u201d incoming electron beam detector w 1 cm 54 V I 10 cm y t i s n e t n I scattered electrons reflected beam crystal surface (not to scale) 0 5 10 15 20 25 Accelerating voltage (a) d 3 Angstroms (b) Figure 14.21 This image was produced by electrons scattered by gold atoms on the surface of a thin gold film. The bright concentric rings are antinodes produced by the constructive interference of electron waves. Figure 14.22 (a) A schematic of a typical Davisson\u2013Germer experiment in which atoms on the surface of a metal scatter a beam of electrons. For specific angles, the electrons scatter constructively and the detector records a large number of electrons, shown in the graph in (b). (b) In this graph, a high intensity means that more electrons are scattered in that direction, creating an antinode, or constructive interference. Similarly, a low intensity can be interpreted as a node, or destructive interference. Chapter 14 The wave-particle duality reminds us that sometimes truth really is stranger than fiction! 729 14-PearsonPhys30-Chap14 7/24/08 3:58 PM Page 730 PHYSICS INSIGHT Manipulating the accelerating voltage in a Davisson-Germer experiment changes the speed of the incident electrons and, therefore, their wavelength. Example 14.11 Explain conceptually how the wave properties of electrons could produce the interference pattern shown in Figure 14.21. Given You know that electrons have wavelike properties and that electrons are being scattered from atoms that are separated by distances comparable to the size of the electron wavelength. Analysis Figure 14.23 shows electron waves leaving from two different atomic scatterers. You can see that path 1 is a little longer than path 2, as denoted by the symbol. This difference means that a different number of electron wavelengths can fit along path 1 than along path 2. For example, if the path difference is 1, 3, or any odd half-multiple 2 2 of, then, when the electron waves combine at the detector, a complete cancellation of the electron wave occurs, forming a node. On the other hand, if the path difference is a whole-number multiple of, then constructive interference occurs, forming an antinode. The Davisson\u2013 Germer experiment provided graphic evidence of the correctness of de Broglie\u2019s hypothesis. incoming electrons \u03b4 path 1 path 2 electron detector scattered electron paths \u03b4 length path 1 \u2013 length path 2 atomic scatterers Figure 14.23 De Broglie\u2019s Hypothesis \u2014 A Key Concept of Quantum Physics Despite its simplicity, de Broglie\u2019s wave hypothesis heralded the true beginning of quantum physics. You will now explore two of the consequences that follow from de Broglie\u2019s equation. De Broglie\u2019s Equation \u201c", "Explains\u201d Quantization of Energy Imagine that you drop a small bead into a matchbox, close the matchbox, and then gently place the matchbox on a level tabletop. You then ask, \u201cWhat is the kinetic energy of the bead?\u201d The answer may seem obvious and not very interesting: The energy is 0 J because the bead is not moving. If, however, you could shrink the box down to the size of a molecule and replace the bead with a single electron, the situation becomes very different. You can sometimes model molecules as simple boxes. The particle-in-a-box model shows how the wave nature of electrons (and all other particles) predicts the idea of quantization of energy. 730 Unit VII Electromagnetic Radiation 14-PearsonPhys30-Chap14 7/24/08 3:58 PM Page 731 M I N D S O N What\u2019s Different Now? From a quantum point of view, explain why it becomes problematic to put a particle in a box. In Chapter 8, section 8.3, you learned about standing waves and resonance. These concepts apply to all waves. Because an electron behaves like a wave as well as like a particle, it has a wavelength, so the ideas of resonance and standing waves also apply to the electron. In order to fit a wave into a box, or finite space, the wave must have a node at each end of the box, and its wavelength must be related to the length of the box in the following way: n l 2 n where n is a whole number (n 1, 2, 3, \u2026). Since there is a node at each end of the box, you can think of n as equivalent to the number of half-wavelengths that can fit in the space l, or length of the box (Figure 14.24). The longest possible standing wave that can fit into the box has a wavelength of 2l, where n 1. l n 1, \u03bb 2l n 2, \u03bb l n 4, \u03bb l 2 n 3, \u03bb 2l 3 Figure 14.24 Standing wave patterns for waves trapped inside a box of length l Because the electron is a standing wave, it cannot be at rest. Consequently, it must have a minimum amount of kinetic energy: mv2 m 1 2 m 2 v2 m 2 m Ek 2 p m 2 since p mv where p is the momentum and m is the mass of the electron. Recall that de Broglie\u2019s equation shows that the momentum of an electron is inversely related to its wavelength: h. p Chapter 14 The wave-particle duality reminds us that sometimes truth really is stranger than fiction! 731 14-PearsonPhys30-Chap14 7/24/08 3:58 PM Page 732 PHYSICS INSIGHT The particle-in-a-box model is very useful. It can be used to describe such diverse phenomena as small-chain molecules, tiny nano-scale electronics, and the nucleus of an atom. Depending on the situation, the \u201cbox\u201d can have one dimension (for a long-chain molecule), two, or three dimensions. Models and modelling form an essential part of the physicist\u2019s imaginative \u201ctoolbox.\u201d Using de Broglie\u2019s equation to relate the momentum of the electron to the length of the box, you can then write: Ek 2 p m 2 2 h 2m h2 2 2m From n l, when n = 1, 2l. Therefore, 2 n Ek 2 h l)2 2m (2 h2 l2 8m This equation represents the minimum kinetic energy of an electron. What if you wanted to give the electron more energy? To have more energy, the electron must have the right momentum-wavelength relation to fit the next standing wave pattern (Figure 14.24). The electron\u2019s wavelength is, therefore, l l 2 2 Substituting into the equation for the kinetic energy of the electron, 2 p h2 2 h 4En1s l2 l)2 m 2 2m ( 2m En2 n2 The energy of a particle in a box is given by the general formula En n2h2 8ml2, n 1, 2, 3,... These equations demonstrate that energy is quantized for the particlein-a-box model. As with photons, quantization means that the electron can have only specific amounts or quanta of energy. (Refer to section 14.1.) Example 14.12 Nanotechnology is one of the hottest areas in physics today. It is now possible to create tiny electric circuits in which electrons behave like particles in a box. Imagine an electron confined to a tiny strip 5.0 nm Practice Problems 1. What is the maximum wavelength for an electron confined to a box of length l 1.0 nm? 2. How much momentum does the electron in question 1 have? long.", " What are three possible energies that the electron could have? Given l 5.0 nm n 1, 2, 3 Required electron energies (E1, E2, E3) 732 Unit VII Electromagnetic Radiation 14-PearsonPhys30-Chap14 7/24/08 3:58 PM Page 733 3. What is the minimum energy that an electron can have when confined to a box of length l 1.0 nm? Answers 1. 2.0 nm 2. 3.3 1025 kgm/s 3. 6.0 1020 J Analysis and Solution n2h2 8ml2, n 1, 2, 3,... En Substitute n 1 into the expression for energy: (12)h2 8ml2 E1 (1)(6.63 1034 Js)2 8(9.11 1031 kg)(5.0 109 m)2 2.4 1021 J Calculate any other energy by noting that h2 n2E1 n2 l2 8m 4(2.4 1021 J) 9.7 1021 J (2)2 E1 (3)2 E1 9(2.4 1021 J) 2.2 1020 J n2(2.4 1021 J) En E2 E3 En Paraphrase An electron confined to a space 5.0 nm long can only have energies that are whole-square multiples of 2.4 1021 J. Three possible energies of the electron are, therefore, 2.4 1021 J, 9.7 1021 J, and 2.2 1020 J. Concept Check Refer to Figure 14.24. What happens to the minimum possible energy of a particle in a box when you shrink the box? How would the minimum energy of particles in the nucleus of an atom (about 10\u201315 m across) compare to the minimum energy of an electron in the atom itself (about 10\u201310 m across)? M I N D S O N Planck in a Box Argue that the particle-in-a-box model illustrates Planck\u2019s discovery of quantization, and also demonstrates Planck\u2019s radiation law. Chapter 14 The wave-particle duality reminds us that sometimes truth really is stranger than fiction! 733 14-PearsonPhys30-Chap14 7/24/08 3:58 PM Page 734 Heisenberg\u2019s Uncertainty Principle Consider what you have just learned about the minimum energy of a particle in a box. The smaller you make the box, the shorter is the wavelength of the particle. Furthermore, because wavelength and momentum are inversely related, the shorter the wavelength, the greater is the momentum of the particle. The greater the momentum, the faster, on average, the particle is moving at any instant. In 1927, the young German physicist Werner Heisenberg (Figure 14.25) realized that the particle-in-a-box model of quantum mechanics has a troubling limitation built into it. Think of the size of the box as indicating the possible uncertainty in the location of the particle. The smaller the box is, the more precisely you know the location of the particle. At the same time, however, the smaller the box is, the greater the momentum and the greater the range of possible momentum values that the particle could have at any instant. Figure 14.26 illustrates this idea by plotting the uncertainty in position of the particle, x (on the vertical axis), and the uncertainty in its momentum, p (on the horizontal axis), as strips that intersect. The shaded areas in Figure 14.26 represent the product of uncertainty in position (x) and uncertainty in momentum (p). x x or p p Figure 14.26 A graphical depiction of the uncertainty in both position and momentum for a particle in a box Heisenberg\u2019s troubling finding was that, due to the wave nature of all particles, it is impossible to know both the position and momentum of a particle with unlimited precision at the same time. The more precisely you know one of these values, the less precisely you can know the other value. To derive the formula for uncertainty in position and momentum of a particle, note that the length of the box is related to the wavelength of the particle: x length of box l (From n l, 2l when n 1.) 2 n 2l x 2 Figure 14.25 Werner Heisenberg (1901\u20131976) was one of the most influential physicists of the 20th century and a key developer of modern quantum theory. PHYSICS INSIGHT According to the particlein-a-box model, the smaller the space in which a particle is confined, the greater the kinetic energy, and hence momentum, of that particle. If you think of the length of the box as setting the possible range in location for a particle, then this quantity also tells you how precisely you know the position of the particle. This range is x. In the same way, the possible range", " in momentum is p. 734 Unit VII Electromagnetic Radiation 14-PearsonPhys30-Chap14 7/24/08 3:58 PM Page 735 Similarly, from de Broglie\u2019s equation h, the uncertainty in momentum is p p range in momentum h h The product of uncertainty in position and uncertainty in momentum can be expressed as xp 2 xp h 2 This formula represents Heisenberg\u2019s uncertainty principle. Note that the value of the product, h (representing the shaded areas of the graphs in 2 Figure 14.26), is constant. The symbol means that xp is approximately h. A more sophisticated argument produces the following expression: 2 h xp 4 This version is a common form of Heisenberg\u2019s uncertainty principle. It tells you that the uncertainty in your knowledge of both the position and momentum of a particle must always be greater than some small, but non-zero, value. You can never know both of these quantities with certainty at the same time! This result was very troubling to many physicists, including Albert Einstein, because it suggests that, at the level of atoms and particles, the universe is governed by chance and the laws of probability. Heisenberg\u2019s uncertainty principle: It is impossible to know both the position and momentum of a particle with unlimited precision at the same time. De Broglie\u2019s matter-wave hypothesis and its confirmation by Davisson and Germer had an unsettling effect on physicists. Heisenberg\u2019s work represented a logical extension of these ideas and helped set the stage for the birth of modern quantum theory. M I N D S O N Physics and Certainty Two physicists who were deeply troubled by de Broglie\u2019s, and especially Heisenberg\u2019s, work were Max Planck and Albert Einstein. Suggest why their reaction is ironic and why these discoveries were difficult for physicists to accept. To help with your answer, consider the importance of precision in classical physics, and Einstein\u2019s famous quote concerning the uncertainty principle: \u201cGod does not play dice with the universe!\u201d Chapter 14 The wave-particle duality reminds us that sometimes truth really is stranger than fiction! 735 14-PearsonPhys30-Chap14 7/24/08 3:58 PM Page 736 14.4 Check and Reflect 14.4 Check and Reflect Knowledge 1. What is the wavelength of an electron that is moving at 20 000 m/s? 2. Calculate the momentum of a 500-nm photon. 6. If an electron and a proton each have the same velocity, how do their wavelengths compare? Express your answer numerically as a ratio. Extensions 3. What is the uncertainty in momentum of a particle if you know its location to an uncertainty of 1.0 nm? 7. According to classical physics, all atomic motion should cease at absolute zero. Is this state possible, according to quantum physics? 4. An electron is trapped within a sphere of diameter 2.5 10\u201312 m. What is the minimum uncertainty in the electron\u2019s momentum? Applications 5. In your television set, an electron is accelerated through a potential difference of 21 000 V. 8. Derive the expression En n2h2, 8ml2 n 1, 2, 3,... for the energy of a particle in a box, where m is the mass of the particle, l is the length of the box, and n is one of the possible quantum states. (Hint: Remember that the wavelength of the nth standing wave confined to a box l.) 2 n of length l is n (a) How much energy does the electron e TEST acquire? (b) What is the wavelength of an electron of this energy? Ignore relativistic effects. To check your understanding of matter waves and Heisenberg\u2019s uncertainty principle, follow the eTest links at www.pearsoned.ca/school/physicssource. 736 Unit VII Electromagnetic Radiation 14-PearsonPhys30-Chap14 7/24/08 3:58 PM Page 737 14.5 Coming to Terms with Wave-particle Duality and the Birth of Quantum Mechanics These fifty years of conscious brooding have brought me no nearer to the question of \u201cWhat are light quanta?\u201d Nowadays every clod thinks he knows it, but he is mistaken. Albert Einstein The wave-particle duality represents a deep and troubling mystery. For some physicists, most notably Einstein, the duality was seen as a flaw in quantum theory itself. Others, including Bohr, learned to accept rather than understand the duality. In this section, we will opt to accept and work with the wave-particle duality. 14-4 QuickLab 14-4 QuickLab The Two-slit Interference Experiment with Particles Figure 14.27 for this experiment? Problem To investigate the pattern that a stream of particles produces when passing through a pair of thin slits", " Materials marble two-slit apparatus (see Figure 14.27) graph paper (or plot on spreadsheet) entrance slit barrier redirects particles slit 1?? slit 2 particle detector Procedure 1 Place the two-slit apparatus on a level table surface and incline it by a small angle to allow the marble to roll down. Repeat this process 100 times. 2 Record your observations by noting how many times the marble lands in each bin. 3 Graph the results of your experiment by plotting the bin number along the horizontal axis and the number of times the marble landed in a given bin on the vertical axis. Questions 1. Why is it important that the table surface be level 2. For 100 trials, how many times would you expect the marble to pass through slit 1? Did you observe this result? Explain. 3. Where did the marble land most of the time? Did you expect this result? Explain. 4. Where would you expect the marble to be found least often? Do your data support your answer? 5. Would the results of your experiment be improved by combining the data from all of the lab groups in the class? Explain. Chapter 14 The wave-particle duality reminds us that sometimes truth really is stranger than fiction! 737 14-PearsonPhys30-Chap14 7/24/08 3:58 PM Page 738 The wave-particle duality of light presents us with many puzzles and paradoxes. Consider, for example, the famous two-slit interference experiment that Thomas Young used in 1801 to convince most physicists that light was a wave (see Chapter 13). This time, however, you are going to put a modern, quantum mechanical twist on the experiment. Since you know that light can behave as a particle (photons) and that particles can behave as matter waves (electrons), it does not matter what you choose to \u201cshine\u201d through the slits. Let us choose light, but reduce its intensity by inserting a filter so that only one photon at a time can enter the box (Figure 14.28). Let the light slowly expose a photographic film or enter the detector of your digital camera. slits? incoming beam Figure 14.28 Young\u2019s double-slit experiment, modified such that the intensity of the beam entering the box is reduced to a level that allows only one photon at a time to enter What will you observe? If you are impatient and let only a few photons through the slits, your result will be a random-looking scatter of dots where photons were absorbed by the film (Figure 14.29(a)). If you wait a little longer, the film will start to fill up (Figure 14.29(b)). Wait longer yet and something remarkable happens: You will see a two-slit interference pattern like the one in Figure 13.9 (Figure 14.29(c)). Why is this result so remarkable? Figure 14.29 Three different results of the double-slit experiment. Image (a) shows the result of only a few photons being recorded. Image (b) shows the result of a few more photons, and image (c) shows the familiar double-slit interference pattern that forms when many photons are recorded. (a) (c) (b) If light was only a wave, then the explanation would be that waves from the top slit in Figure 14.28 were slightly out of phase with waves from the bottom slit in some locations, causing nodes to form. In other places, the waves would combine in phase to produce antinodes. You arranged, however, to have only one photon at a time enter the apparatus. So, the photon would either go through the top slit or the bottom slit. But even a photon cannot be in two different places at once! If the photons can 738 Unit VII Electromagnetic Radiation 14-PearsonPhys30-Chap14 7/24/08 3:58 PM Page 739 only go through one slit (either the top or bottom one), why does a two-slit interference pattern, such as the one shown in Figure 14.30, result? Even though the individual photons go through only one slit, they somehow \u201cknow\u201d that there is another slit open somewhere else! American physicist Richard Feynman (Figure 14.31) often used this example to emphasize how strange the quantum world is. 0.4 0.2 0.0 0.2 0.4 Figure 14.30 A two-slit interference pattern Figure 14.31 Richard P. Feynman (1918\u20131988) was one of the founders of modern quantum theory. He once stated: \u201cI think it is safe to say that no one understands quantum mechanics.\u201d So what does it all mean? To try to understand the double-slit experiment as it applies to individual photons or electrons, it is useful to summarize the key points: 1. When the photon or electron is absorbed by the photographic film or the detector of your digital camera, it exhibits", " its particle nature. 2. The location where any one photon or electron is detected is random but distinct; that is, the photon or electron always arrives and is detected as a distinct particle. 3. Although the location of individual photons or electrons is random, the combined pattern that many photons form is the characteristic pattern of antinodes and nodes, as shown in Figure 14.30. This pattern shows the wave nature of the photons or electrons. By the late 1920s, scientists developed a bold new interpretation of events. The wave-particle duality was showing that, at the level of atoms and molecules, the world was governed by the laws of probability and statistics. Although you cannot say much about what any one electron, for example, would do, you can make very precise predictions about the behaviour of very large numbers of electrons. In 1926, German physicist Max Born suggested that the wave nature of particles is best understood as a measure of the probability that the particle will be found at a particular location. The antinodes in the double-slit interference pattern exist because the particles have a high probability of being found at those locations after they pass through the double-slit apparatus. This measure of probability of a particle\u2019s Chapter 14 The wave-particle duality reminds us that sometimes truth really is stranger than fiction! 739 14-PearsonPhys30-Chap14 7/24/08 3:58 PM Page 740 quantum indeterminacy: the probability of finding a particle at a particular location in a doubleslit interference pattern info BIT American physicist and Nobel laureate Leon Lederman estimates that quantum physics is an essential part of the technologies responsible for over 25% of the North American gross national product. location is called quantum indeterminacy. This concept is the most profound difference between quantum physics and classical physics. According to quantum physics, nature does not always do exactly the same thing for the same set of conditions. Instead, the future develops probabilistically, and quantum physics is the science that allows you to predict the possible range of events that may occur. Although you may think that quantum behaviour is remote and has nothing to do with your life, nothing could be further from the truth. As you will see in the next three chapters, quantum theory has become one of the most powerful scientific theories ever developed. Virtually all of the electronic equipment we use daily that improves our quality of life, and most of our current medical technologies and understanding, are possible because of the deep insights that quantum theory provides. 14.5 Check and Reflect 14.5 Check and Reflect Knowledge Applications 1. Explain which of the following choices 3. Which of the following examples is the best one. (a) The double-slit experiment illustrates the wave nature of a quantum, and which illustrates the particle nature? demonstrates that light is a wave. (a) Electrons hit a phosphor screen and (b) The double-slit experiment shows that light is a particle. (c) The double-slit experiment illustrates that light has both wave and particle characteristics. 2. True or false? Explain. (a) The results of the double-slit experiment described in this section apply only to photons. (b) The results of the double-slit experiment apply to photons as well as to particles such as electrons. create a flash of light. (b) Electrons scatter off a crystal surface and produce a series of nodes and antinodes. (c) Light hits a photocell and causes the emission of electrons. Extension 4. Imagine that, one night as you slept, Planck\u2019s constant changed from 6.63 1034 Js to 6.63 Js. Explain, from a quantum mechanical point of view, why walking through the doorway of your bedroom could be a dangerous thing to do. e TEST To check your understanding of quantum mechanics, follow the eTest links at www.pearsoned.ca/ school/physicssource. 740 Unit VII Electromagnetic Radiation 14-PearsonPhys30-Chap14 7/24/08 3:58 PM Page 741 CHAPTER 14 SUMMARY Key Terms and Concepts incandescent blackbody radiation curve blackbody quantum Planck\u2019s formula quantized photon photoelectric effect photoelectron threshold frequency work function stopping potential Compton scattering Compton effect wave-particle duality Heisenberg\u2019s uncertainty principle quantum indeterminacy Key Equations E nhf hf W Ek Ekmax qVstopping p h i f h (1 cos ) m c Conceptual Overview Summarize this chapter by copying and completing the following concept map. discovery of photoelectric effect blackbody spectrum and failure of classical physics Einstein\u2019s explanation of the photoelectric effect Millikan\u2019s determination of Planck\u2019s constant Compton effect Heisenberg and quantum indeterminacy Planck\u2019s radiation law De Broglie and wave-particle duality Davisson\u2013Germer experiment Chapter 14 The wave-particle dual", "ity reminds us that sometimes truth really is stranger than fiction! 741 14-PearsonPhys30-Chap14 7/24/08 3:58 PM Page 742 CHAPTER 14 REVIEW Knowledge 1. (14.1) Explain what is meant by the term \u201cultraviolet catastrophe.\u201d 2. (14.1) Write the equation for Planck\u2019s formula and briefly explain what it means. 3. (14.1) What is the energy of a 450-nm photon? Express the answer in both joules and electron volts. 4. (14.1) If an X-ray photon has a wavelength 100 times smaller than the wavelength of a visible light photon, how do the energies of the two photons compare? Give a numerical answer. 5. (14.2) Who is credited with discovering the photoelectric effect? 6. (14.2) Who provided the correct explanation of the photoelectric effect? In what way(s) was this explanation radical when first proposed? 7. (14.2) If the threshold frequency for photoemission from a metal surface is 6.0 1014 Hz, what is the work function of the metal? 8. (14.3) Explain why the Compton effect provides critical evidence for the particle model of light. 9. (14.3) If a 0.010-nm photon scatters 90 after striking an electron, determine the change in wavelength () for the photon. 10. (14.4) What is the wavelength of an electron that has a momentum of 9.1 1027 Ns? 11. (14.4) What is the momentum of a 100-nm UV photon? 12. (14.4) If a particle is confined to a region in space 10 fm across, could the particle also be at rest? Explain, using Heisenberg\u2019s uncertainty principle. Applications 13. How many photons are emitted each second by a 1.0-W flashlight? Use 600 nm as the average wavelength of the photons. 14. A beam of 300-nm photons is absorbed by a metal surface with work function 1.88 eV. Calculate the maximum kinetic energy of the electrons emitted from the surface. 15. Modern transmission electron microscopes can accelerate electrons through a 100-V potential difference and use these electrons to produce images of specimens. What is the wavelength of a 100-keV electron? Ignore relativistic effects. Why are electron microscopes capable of much higher magnification than light microscopes? 16. A major league baseball pitcher can throw a 40-m/s fastball of mass 0.15 kg. (a) Calculate the wavelength of the ball. (b) Why can you safely ignore quantum effects in this case? 17. Imagine that you are 100 m from a 100-W incandescent light bulb. If the diameter of your pupil is 2 mm, estimate how many photons enter your eye each second. (Note: You will need to make estimates and provide additional information.) 18. How many photons are emitted each second by an FM radio station whose transmitted power is 200 kW and whose frequency is 90.9 MHz? 19. An electron is trapped in a box that is 0.85 nm long. Calculate the three lowest energies that this electron can have. Why can the electron not have energy values between the values you calculated? 20. Calculate the momentum of a 100-keV X-ray photon. 742 Unit VII Electromagnetic Radiation 14-PearsonPhys30-Chap14 7/24/08 3:58 PM Page 743 Extensions 21. Argue that photons exert pressure. (Hint: Newton\u2019s p second law, F ma, can also be written as F. t In other words, force is a measure of the rate of change in momentum with respect to time. Also, remember that pressure is defined as force acting F.) over an area: P A 22. After you graduate from university, you take a job in a patent office, assessing the feasibility of inventions. Your boss hands you a file and skeptically tells you it is from a physicist who claims that a 1.0-km2 sail made from highly reflecting Mylar film could produce about 10 N of force simply by reflecting sunlight. You are asked to check the physics. Do the following: (a) Estimate how many photons arrive from the Sun per second per square metre at a distance equal to the Earth-Sun separation. You know that the top of Earth\u2019s atmosphere gets 1.4 kW/m2 of energy from the Sun. (b) Calculate the momentum of each photon and remember that the photons are reflected. (c) Multiply the pressure (force per unit area) by the total area of the sail. 24. Einstein thought there was a fundamental flaw in quantum physics because \u201cGod does not play dice.\u201d (a) What do you think he meant by this statement? What part of quantum theory", " was Einstein referring to? (b) Why is it ironic that Einstein made this statement? Consolidate Your Understanding 1. Describe two significant failings of classical physics that challenged physics prior to 1900. 2. Provide evidence for quantization of energy, and explain this concept to a friend. 3. List and describe at least two crucial experimental findings that support Einstein\u2019s claim that light has a particle nature. 4. Explain why it is incorrect to state that light is either a wave or a particle. Comment on how quantum physics tries to resolve this duality. 5. What is meant by the term \u201cquantum indeterminacy\u201d? Provide experimental evidence for this idea. Does the physicist\u2019s claim make sense? Think About It 23. How many photons per second does your radio respond to? Consider receiving a 100-MHz radio signal. The antenna in an average radio receiver must be able to move a current of at least 1.0 A through a 10-mV potential difference in order to be detectable. Review your answers to the Think About It questions on page 703. How would you answer each question now? e TEST To check your understanding of the wave-particle duality of light, follow the eTest links at www.pearsoned.ca/school/physicssource. Chapter 14 The wave-particle duality reminds us that sometimes truth really is stranger than fiction! 743 14-PearsonPhys30-Chap14 7/24/08 3:58 PM Page 744 UNIT VII PROJECT From Particle to Quantum \u2014 How did we arrive at our present understanding of light? Scenario In the past 500 years, our understanding of the nature of electromagnetic radiation has grown immensely, from the particle vs. wave controversy between Newton and Huygens, to the strange wave-particle duality described by de Broglie\u2019s hypothesis, and to Heisenberg\u2019s uncertainty principle. The key evidence and theories along the way have opened up a bounty of applications, from fibre-optic communication networks, to scanning and tunnelling electron microscopes. With our understanding of electromagnetic radiation has come a vast wealth of information and new technologies, which in turn, have furthered our ability to probe and investigate the nature of our universe. From humble beginnings with simple lenses, the scientific community has followed a long and difficult pathway to our present understanding. In this project, you will retrace this pathway, highlighting the theories, evidence, and experiments that have contributed to our present understanding of light and electromagnetic radiation. Planning Working in small groups or individually, prepare a presentation that summarizes the intellectual journey from the earliest theories of the particle nature of light, to the more modern theory of wave-particle duality. Your presentation should include simulations and illustrations that identify key experimental evidence, and descriptions of each model and theory, including the scientists who proposed them. Your summary can be presented in chronological order, from early discoveries, to later ones, or it can be organized around models (particle, wave, quantum, wave-particle duality). Materials text and Internet resources simulations, illustrations, photos of evidence collected in experiments presentation software (PowerPoint, html editor, etc.) Assessing Results Assess the success of your project based on a rubric* designed in class that considers: research strategies completeness of evidence and effectiveness of presentation Procedure 1 Define each of the following models: particle, wave, quantum, and wave-particle duality. 2 Identify each of the scientists involved with each model. 3 Using either a table or a timeline, place each model and the related scientists in order from earliest, to most recent. 4 On your table or a timeline, identify each key experiment and the evidence that was used to support each model. Include the following experimental evidence and theories: \u2022 reflection, refraction, dispersion, diffraction, interference, polarization, blackbody radiation, photoelectric effect, Compton effect, de Broglie\u2019s hypothesis, and Heisenberg\u2019s uncertainty principle 5 Use your table or timeline as the basis for preparing your presentation. Use simulations, illustrations, and photographs where possible to describe experimental evidence. Thinking Further The evolution of an idea or theory can take place over hundreds of years, with one participant handing off evidence to the next participant. A sort of relay develops, because the race is simply too long for one person to complete alone. With this in mind, consider the following questions that could be answered at the end of your presentation. \u2022 A relay race has an end. Is there an end in the race to fully understand the nature of electromagnetic radiation and light? If the relay is not over, where do you think we are going from here? \u2022 \u2022 How have we used the knowledge of our predecessors in determining where to look next? Explain. *Note: Your instructor will assess the project using a similar assessment rubric. 744 Unit VII Electromagnetic Radiation 14-PearsonPhys30-Chap14 7/24/", "08 3:58 PM Page 745 UNIT VII SUMMARY Unit Concepts and Skills: Quick Reference Summary Resources and Skill Building Concepts Chapter 13 Types of electromagnetic radiation Models of EMR Maxwell\u2019s electromagnetic theory The wave model can be used to describe the characteristics of electromagnetic radiation. 13.1 What Is Electromagnetic Radiation? Frequency, wavelength, and source are used to identify types of EMR. Different models were used to explain the behaviour of EMR. Maxwell\u2019s theory linked concepts of electricity and magnetism. Electromagnetic radiation is produced by accelerating charges. Speed of electromagnetic radiation 13.2 The Speed of Electromagnetic Radiation Galileo, Roemer and Huygens, and Fizeau measured the speed of EMR, but Michelson\u2019s experiment made the definitive measurement. The Law of Reflection, image formation, and ray diagrams 13.3 Reflection The angle of reflection equals the angle of incidence and is in the same plane. Ray diagrams show a light ray interacting with a surface. Three rays predict the location and characteristics of the image. Image formation/equations The mirror equation relates the focal length of a curved mirror to the image and object distances. Refraction and Snell\u2019s Law Total internal reflection 13.4 Refraction Snell\u2019s Law relates the refraction of a light wave to the speed with which light travels in different media. All light is internally reflected at an interface if the angle of refraction is 90 or greater. Dispersion and recomposition White light can be separated into its component wavelengths. Image formation with thin lenses The lens equation relates the focal length of a curved lens to the image and object distances. Huygens\u2019 Principle Young\u2019s experiment, interference, and diffraction Diffraction gratings 13.5 Diffraction and Interference Huygens predicted the motion of a wave front as many point sources. Young\u2019s experiment showed that two beams of light produce an interference pattern and that light behaves as a wave. Light on a multi-slit diffraction grating produces an interference pattern. Polarization EMR absorption by polarizing filters supports the wave model of light. Chapter 14 The wave\u2013particle duality reminds us that sometimes truth really is stranger than fiction! 14.1 The Birth of the Quantum Classical physics was unable to explain the shape of the blackbody radiation curve. A quantum is the smallest amount of energy of a particular wavelength or frequency that a body can absorb, given by E = hf. 14.2 The Photoelectric Effect The work function is the minimum energy required to cause photoemission of electrons from a metal surface. Millikan\u2019s photoelectric experiment provided a way to measure Planck\u2019s constant. The photoelectric effect obeys the law of conservation of energy. 14.3 The Compton Effect When an electron scatters an X ray, the change in the X ray\u2019s wavelength relates to the angle of the X-ray photon\u2019s scattering. 14.4 Matter Waves and the Power of Symmetric Thinking Something that has momentum also has wavelength: Particles can act like waves. Particles have a wave nature, so it is impossible to precisely know their position at the same time as their momentum. Quantum Photoelectric effect Planck\u2019s constant Compton effect Wave\u2013particle duality Heisenberg\u2019s uncertainty principle Quantum indeterminacy Figure 13.4; Table 13.1 Figures 13.5\u201313.9 Figures 13.10\u201313.15 Figures 13.16-13.19; Minds On: Going Wireless Figures 13.21-13.24; Example 13.1; 13-2 QuickLab Figures 13.28-13.30; 13-3 QuickLab; Minds On: Image in a Mirror; Figures 13.31\u201313.32; Figures 13.36\u201313.39; 13-4 QuickLab 13-5 Problem-Solving Lab; Example 13.2 Table 13.4; Examples 13.3\u201313.4; 13-6 Inquiry Lab Figures 13.49-13.53; Example 13.5, 13-7 Decision-Making Analysis Figures 13.54\u201313.56; Table 13.5; 13-8 QuickLab Figures 13.57\u201313.59; Example 13.6; Figure 13.61; Example 13.7; 13-9 Inquiry Lab Figures 13.67\u201313.69 Figures 13.70\u201313.78; Examples 13.8\u201313.9 Figure 13.80 Figure 13.81; Example 13.10; 13-10 Inquiry Lab 13-1 QuickLab; Figure 13.86; Figures 13.88\u201313.90 14-1 QuickLab, Figures 14.3\u201314.4 Examples 14.1, 14.2, 14.3; Minds On: What\u2019s Wrong with This Analogy? Figure 14.6 14-2 QuickLab, Table 14.1 Figure", " 14.12, 14-3 Design a Lab; Examples 14.5\u201314.6 Figure 14.16; Examples 14.7\u201314.8 Examples 14.9\u201314.10 Figures 14.21, 14.22, 14.24, 14.26; Example 14.11 14.5 Coming to Terms with Wave\u2013particle Duality and the Birth of Quantum Mechanics Wave-particle duality illustrates the probabilistic nature of atoms and molecules. Quantum indeterminacy is the measure of the probability of a particle\u2019s location. 14-4 QuickLab, Figures 14.28\u201314.30 Unit VII Electromagnetic Radiation 745 14-PearsonPhys30-Chap14 7/24/08 3:58 PM Page 746 UNIT VII REVIEW Vocabulary Knowledge 1. Use your own words to define these terms: CHAPTER 13 angle of diffraction blackbody blackbody radiation curve Compton effect Compton scattering converging critical angle diffraction diffraction grating dispersion diverging electromagnetic radiation focal point frequency Heisenberg\u2019s uncertainty principle Huygens\u2019 Principle image attitude incandescent interference law of reflection magnification node, antinode particle model path length period photoelectric effect photoelectrons photon Planck\u2019s formula polarization quantized quantum quantum indeterminacy refraction refractive index Snell\u2019s Law spectrum stopping potential threshold frequency total internal reflection wave model wave-particle duality wavelength work function 746 Unit VII Electromagnetic Radiation 2. How does the quantum model reconcile the wave model and the particle model of light? 3. How did Maxwell\u2019s work with capacitors influence his theories on electromagnetism? 4. Describe the experimental evidence that supports all of Maxwell\u2019s predictions about electromagnetic radiation. 5. Discuss the significance of the word \u201cchanging\u201d in Maxwell\u2019s original description of electromagnetic radiation. 6. Why does a spark produce electromagnetic radiation? 7. If a metal conductor, such as a spoon, is placed in an operating microwave oven, a spark is produced. Why? 8. Using a ray diagram, show three rays that are needed to identify and verify the characteristics of an image. 9. What is the relationship between the focal length and the radius of curvature for a curved mirror? 10. What is a virtual focal point and how is it different from a real focal point? 11. Explain, using a ray diagram, how a real image can be formed when using two concave mirrors. 12. When you place the concave side of a spoon on your nose and slowly pull it away from your face, your image disappears at a certain distance. What is the significance of this distance? 13. When an object such as a paddle is partially submerged in water, why does it appear bent? 14. Explain how Snell\u2019s Law supports the wave theory of light. 15. What happens to the wavelength of monochromatic light when it passes from air into water? 16. Several people holding hands run down the beach and enter the water at an angle. Explain what happens to the speed and direction of the people as they enter the water. 17. How was Newton able to show that a prism separates the colours in the spectrum, rather than adding the colours to white light? 14-PearsonPhys30-Chap14 7/24/08 3:58 PM Page 747 18. What is Huygens\u2019 Principle? 19. A straight wave front is incident on a barrier with a small hole. Using a diagram, describe the shape of the wave front a moment after it makes contact with the barrier. 20. Using a schematic, illustrate Young\u2019s experiment. 38. A proton and a neutron are both moving at the same speed. Which particle has the shorter de Broglie wavelength? 39. Explain, using wave mechanics, why it is impossible for a particle to have zero kinetic energy when it is confined to a fixed region in space. 21. Explain why diffraction supports the wave model of light. Applications 22. What key evidence was observed by Dominique Arago in 1818? Why was this evidence crucial to the acceptance of the wave model of light? 23. How must two plane polarizing filters be aligned in order to fully block electromagnetic radiation? 24. Is an electromagnetic wave one-dimensional, two-dimensional, or three-dimensional? Explain. CHAPTER 14 25. Is a quantum of blue light the same as a quantum of red light? Explain. 26. How much energy is carried by a photon of wavelength 550 nm? 27. Explain how you can estimate the surface temperature of a star by noting its colour. 28. Arrange the following photons from highest to lowest energy: ultraviolet photon, 10-nm photon, microwave photon, gamma-ray photon, 600-nm photon, infrared photon. 29. What is the frequency of blue light of wavelength 500 nm? 30. Ultraviolet light causes sunburn whereas visible light does not. Explain,", " using Planck\u2019s formula. 31. Explain what is meant by the term \u201cthreshold frequency.\u201d 32. How does the energy of photoelectrons emitted by a metal change as the intensity of light hitting the metal surface changes? 33. What is the maximum wavelength of light that will cause photoemission from a metal having a work function of 3.2 eV? 34. Explain the difference between the Compton effect and the photoelectric effect. 35. What is meant by the term \u201cwave-particle duality\u201d? 36. Even though photons have no mass, they still carry momentum. What is the momentum of a 300-nm ultraviolet photon? 37. What is the de Broglie wavelength of an electron moving at 3000 km/s? 40. If visible light is a particle, predict what would be observed if light passed through two small holes in a barrier. Compare this prediction to what is actually observed when light passes through two small holes in a barrier. What does this suggest about the nature of light? 41. How many radio-frequency photons are emitted each second by a radio station that broadcasts at a frequency of 90.9 MHz and has a radiated power of 50 kW? 42. Explain how an antenna is able to \u201csense\u201d electromagnetic radiation. 43. Detailed measurements of the Moon\u2019s orbit could be calculated after the Apollo mission placed large reflecting mirrors on the surface of the Moon. If a laser beam were directed at the mirrors on the Moon and the light was reflected back to Earth in 2.56 s, how far away, in kilometres, is the Moon? 44. When you increase the intensity of a green light, do you change the energy of the green-light photons? Why does the light get brighter? 45. A Michelson apparatus is used to obtain a value of 2.97 108 m/s for the speed of light. The sixteen-sided rotating mirror completes 1.15 104 revolutions in one minute. How far away was the flat reflecting mirror? 46. An eight-sided mirror like Michelson\u2019s is set up. The light reflects from the rotating mirror and travels to a fixed mirror 5.00 km away. If the rotating mirror turns through one-eighth of a rotation before the light returns from the fixed mirror, what is the rate of rotation? 47. A sixteen-sided mirror rotates at 4.50 102 Hz. How long does it take to make one-sixteenth of a rotation? 48. Why do police and search-and-rescue agencies use infrared cameras for night-time surveillance when looking for people? Explain why infrared is used and not some other part of the electromagnetic spectrum. Unit VII Electromagnetic Radiation 747 14-PearsonPhys30-Chap14 7/24/08 3:58 PM Page 748 49. The speed of light in a material is determined to 60. Calculate the wavelength of electrons used in a be 1.24 108 m/s. What is the material? 50. Light of wavelength 520 nm strikes a metal surface having a work function of 2.3 eV. Will the surface emit photoelectrons? 51. A student replicating Michelson\u2019s experiment uses an eight-sided mirror and a fixed mirror located 35.0 km away. Light is reflected through the system when the rotating mirror turns at 5.20 102 Hz. What is the experimentally determined speed of light and the percentage error in the measurement? 52. An electrically neutral 1-m2 piece of aluminium is put in orbit high above Earth. Explain why, after a period of time, the piece of aluminium will become electrically charged. Predict the sign of the charge. 53. An object is located in front of a diverging mirror with a focal length of 5.0 cm. If the virtual image is formed 3.0 cm from the vertex of the mirror and is 1.0 cm high, determine the object\u2019s characteristics and position. 54. Photon A has four times the energy of photon B. Compare the wavelengths and the momenta of the two photons. 55. A light ray passes from water into ruby at an angle of 10. What is the angle of refraction? 56. An X-ray photon of wavelength 0.025 nm collides elastically with an electron and scatters through an angle of 90. How much energy did the electron acquire in this collision and in what important way did the X ray change? 57. A 3.0-cm-high object is placed 10.0 cm from a converging lens with a focal length of 5.0 cm. Using the thin lens equation, determine the image attributes and position. 58. Imagine that you are asked to review a patent application for a laser-powered deep space probe. The proposal you are reviewing calls for a 1-kW laser producing 500-nm photons. The total mass of the spacecraft, including the laser, is 1000 kg. Determine (a)", " if laser propulsion is possible, and the underlying principle of this form of propulsion. (b) how fast the spacecraft would be travelling after one year of \u201claser-drive\u201d if it started from rest. 59. List two ways to recompose the spectrum into white light. 748 Unit VII Electromagnetic Radiation transmission electron microscope if the electrons are accelerated through an electric field of potential 75 kV. Ignore relativistic effects. 61. In an experiment similar to Young\u2019s, two waves arrive at the screen one half-wavelength out of phase. What will be observed at this point on the screen? 62. What is the minimum or rest energy of an electron confined to a one-dimensional box 1 nm long? 63. A mixture of violet light ( 420 nm) and red light ( 650 nm) are incident on a diffraction grating with 1.00 104 lines/cm. For each wavelength, determine the angle of deviation that leads to the first antinode. 64. Light with a wavelength of 700 nm is directed at a diffraction grating with 1.50 102 slits/cm. What is the separation between adjacent antinodes when the screen is located 2.50 m away? 65. Your physics teacher, eager to get to class, was observed from a police spotting-plane to travel a distance of 222 m in 10 s. The speed limit was 60 km/h, and you can quickly determine that he was speeding. The police issued a ticket, but your teacher decided to argue the case, citing Heisenberg\u2019s uncertainty principle as his defence. He argued that the speed of his car was fundamentally uncertain and that he was not speeding. Explain how you would use Heisenberg\u2019s uncertainty principle in this case and comment on whether your teacher\u2019s defence was good. The combined mass of the car and your teacher is 2000 kg. Extensions 66. Traditional radio technology blends a carrier signal and an audio signal with either frequency or amplitude modulation. This generates a signal with two layers of information\u2014one for tuning and one containing the audio information. Describe the two layers of information that a cell phone signal must contain in order to establish and maintain constant communication with a cell phone network. 67. Use Heisenberg\u2019s uncertainty principle to estimate the momentum and kinetic energy of an electron in a hydrogen atom. Express the energy in electron volts. The hydrogen atom can be approximated by a square with 0.2-nm sides. (Hint: Kinetic energy is related to momentum via the equation Ek 2 p.) m 2 14-PearsonPhys30-Chap14 7/24/08 3:58 PM Page 749 68. Global positioning satellites maintain an orbital altitude of 20 000 km. How long does it take for a time signal to travel from the satellite to a receiver located directly below the satellite? 69. Why do some metals have a higher threshold frequency than others? How is this phenomenon related to electric fields? 70. Explain how an optical fibre is able to transmit a light pulse over a long distance without a loss in intensity. 71. The human eye can detect as few as 500 photons of light, but in order to see, this response needs to occur over a prolonged period of time. Seeing requires approximately 10 000 photons per second. If the Sun emits 3.9 1026 W, mostly in the bluegreen part of the spectrum, and if roughly half of the energy is emitted as visible light, estimate how far away a star like our Sun would be visible. 72. A beam of 200-eV electrons is made to pass through two slits in a metal film that are separated by 50 nm. A phosphor screen is placed 1 m behind the slits. Sketch what you would expect to see. Provide calculations to support your answer. Skills Practice 73. An object is located 25.0 cm from a diverging mirror with a focal length of 10.0 cm. Draw a ray diagram to scale to determine the following: (a) (b) (c) the image location and type the image attitude the magnification of the image 74. The following data are taken from an experiment in which the maximum kinetic energy of photoelectrons is related to the wavelength of the photons hitting a metal surface. Use these data to produce a graph that shows the energy of the incident photons on the horizontal axis and the kinetic energy of photoelectrons on the vertical axis. From this graph, determine the work function for the metal. 75. Use a ray diagram to show why a double convex lens is called a converging lens and a double concave lens is called a diverging lens. Label the principal axis, principal focus, secondary focus, and optical centre. 76. Calculate the momentum and wavelength of an electron that has a kinetic energy of 50 keV. Ignore relativistic effects. 77. Explain, with the aid of a ray diagram, why an image does not form when you place an object at the focal point of a conver", "ging lens. 78. Determine the momentum of an X ray of wavelength 10 nm. 79. Prepare a table in which you compare the wave and particle models of light. List as many phenomena as you can think of and decide whether light can be explained best using the wave or the particle model. How would you answer the question, \u201cIs light a wave or a particle?\u201d Self-assessment 80. Describe to a classmate which concepts of electromagnetic radiation you found most interesting when studying this unit. Give reasons for your choices. 81. Identify one issue pertaining to the wave-particle duality of light that you would like to investigate in greater detail. 82. What concept in this unit did you find most difficult? What steps could you take to improve your understanding? e TEST To check your understanding of electromagnetic radiation and the dual nature of light, follow the eTest links at www.pearsoned.ca/school/ physicssource. Wavelength (nm) Kinetic Energy (eV) 200 250 300 350 400 450 3.72 2.47 1.64 1.05 0.61 0.26 Unit VII Electromagnetic Radiation 749 15-PearsonPhys30-Chap15 7/24/08 3:59 PM Page 750 U N I T VIII Atomic Atomic Physics Physics M A cluster of newly formed stars in a spiral arm of the Milky Way galaxy. Physicists are using atomic theories to understand the structure and evolution of the universe. e WEB To learn more about the role of atomic physics in cosmology, follow the links at www.pearsoned.ca/school/physicssource. 750 Unit VIII 15-PearsonPhys30-Chap15 7/24/08 3:59 PM Page 751 Unit at a Glance C H A P T E R 1 5 Electric force and energy quantization determine atomic structure. 15.1 The Discovery of the Electron 15.2 Quantization of Charge 15.3 The Discovery of the Nucleus 15.4 The Bohr Model of the Atom 15.5 The Quantum Model of the Atom C H A P T E R 1 6 Nuclear reactions are among the most powerful energy sources in nature. 16.1 The Nucleus 16.2 Radioactive Decay 16.3 Radioactive Decay Rates 16.4 Fission and Fusion C H A P T E R 1 7 The development of models of the structure of matter is ongoing. 17.1 Detecting and Measuring Subatomic Particles 17.2 Quantum Theory and the Discovery of New Particles 17.3 Probing the Structure of Matter 17.4 Quarks and the Standard Model Unit Themes and Emphases \u2022 Energy and Matter Focussing Questions The study of atomic structure requires analyzing how matter and energy are related. As you study this unit, consider these questions: \u2022 What is the structure of atoms? \u2022 How can models of the atom be tested? \u2022 How does knowledge of atomic structure lead to the development of technology? Unit Project How Atomic Physics Affects Science and Technology \u2022 Unit VIII discusses radical changes in the understanding of matter and energy. In this project, you will investigate how advances in atomic physics influenced the development of technology and other branches of science. Unit VIII Atomic Physics 751 15-PearsonPhys30-Chap15 7/24/08 3:59 PM Page 752 Electric force and energy quantization determine atomic structure. Max Planck began undergraduate studies at the University of Munich in 1874. During his first term, he took primarily mathematics courses, although he also had considerable musical talent and an interest in physics. Wondering which field to pursue, the teenaged Planck asked his physics professor, Philipp von Jolly, about the prospects for a career in physics. Jolly told Planck that there would be few opportunities since almost everything in physics had already been discovered, leaving only a few minor gaps to fill in. Despite Jolly\u2019s discouraging advice, Planck went on to complete a doctorate in physics, and became a renowned professor at the University of Berlin. As described in Chapter 14, one of the \u201cgaps\u201d in theoretical physics was the puzzling distribution of energy among the wavelengths of radiation emitted by a heated body. Planck concentrated on this problem for months, and by the end of 1900 concluded that this distribution is possible only if energy is quantized. At first, Planck and his contemporaries did not realize the huge significance of his findings. As you will learn in this chapter, Planck\u2019s discovery led to quantum theory, a concept that revolutionized atomic physics. You will also see that Planck was just the first of many researchers who demonstrated that there was still a great deal to be discovered in physics. Figure 15.1 shows how radically our concept of the atom changed during the 20th century. Planck\u2019s idea of quantization of energy led Bohr to propose his model of the atom in 1913. In this chapter, you will study in detail how the", " model of the atom evolved. H He Li Be \u2026 indivisible raisin bun planetary Bohr quantum Figure 15.1 The evolution of theories of atomic structure C H A P T E R 15 Key Concepts In this chapter, you will learn about: charge-to-mass ratio classical model of the atom continuous and line spectra energy levels quantum mechanical model Learning Outcomes When you have completed this chapter, you will be able to: Knowledge describe matter as containing discrete positive and negative charges explain how the discovery of cathode rays helped develop atomic models explain the significance of J.J. Thomson\u2019s experiments explain Millikan\u2019s oil-drop experiment and charge quantization explain the significance of Rutherford\u2019s scattering experiment explain how electromagnetic theory invalidates the classical model of the atom describe how each element has a unique line spectrum explain continuous and line spectra explain how stationary states produce line spectra calculate the energy difference between states, and the characteristics of emitted photons Science, Technology, and Society explain how scientific knowledge and theories develop explain the link between scientific knowledge and new technologies 752 Unit VIII 15-PearsonPhys30-Chap15 7/24/08 3:59 PM Page 753 15-1 QuickLab 15-1 QuickLab Cathode Rays and Magnetic Fields Problem What type of charge do cathode rays have? Materials gas-discharge tube (Figure 15.2) high-voltage power supply bar magnet or electromagnet bar magnet S N high-voltage power supply gasdischarge tube Figure 15.2 A gas-discharge tube Procedure 1 Your teacher will give directions for setting up the particular tube and power supply that you will use. Follow these directions carefully. Note the location of the cathode in the discharge tube. 2 Turn on the power supply and note the beam that appears in the tube. To see the beam clearly, you may need to darken the room. 3 Bring the magnet close to the tube surface. Note the direction of the magnetic field and the direction in which the beam moves. 4 Repeat step 3 at various positions along the tube. Questions 1. How did you determine the direction of the magnetic field? 2. What do you think causes the visible beam in the tube? 3. What evidence suggests that the magnet causes the beam to deflect? 4. How could you verify that the cathode rays originate from the negative terminal of the discharge tube? 5. What can you conclude about the charge of the cathode rays? Explain your reasoning. 6. Describe another way to determine the type of charge on cathode rays. Think About It 1. What are atoms made of? 2. What holds atoms together? 3. How can physicists measure atomic structure when it is too small to be seen by even the most powerful microscope? Discuss your answers in a small group and record them for later reference. As you complete each section of this chapter, review your answers to these questions. Note any changes in your ideas. Chapter 15 Electric force and energy quantization determine atomic structure. 753 15-PearsonPhys30-Chap15 7/24/08 3:59 PM Page 754 15.1 The Discovery of the Electron info BIT Dalton\u2019s research included important findings about the aurora borealis, atmospheric gases, and the formation of dew. He also made the first scientific study of colour blindness, which he had himself. Around 1803, the English chemist John Dalton (1766\u20131844) developed an atomic theory to explain the ratios in which elements combine to form compounds. Although later discoveries required modifications to Dalton\u2019s theory, it is a cornerstone for modern atomic theory. This theory could explain the observations made by Dalton and other chemists, but their experiments did not provide any direct evidence that atoms actually exist. At the end of the 19th century, there was still some doubt about whether all matter was made up of atoms. By 1900, experiments were providing more direct evidence. Current technology, such as scanning tunnelling microscopes, can produce images of individual atoms. M I N D S O N Evidence for Atoms How do you know that atoms exist? Work with a partner to list evidence that matter is composed of atoms. Cathode-ray Experiments During the 1800s, scientists discovered that connecting a high voltage across the electrodes at opposite ends of an evacuated glass tube caused mysterious rays to flow from the negative electrode (the cathode) toward the positive electrode. These cathode rays caused the glass to glow when they struck the far side of the tube. The rays could be deflected by a magnetic field. In 1885, after several years of experiments with improved vacuum discharge tubes, William Crookes in England suggested that cathode rays must be streams of negatively charged particles. In 1895, Jean Baptiste Perrin in France showed that cathode rays entering a hollow metal cylinder built up a negative charge on the cylinder. In 1897, Joseph John Thomson (1856\u20131940) took these experiments a step further. First, he used an improved version", " of Perrin\u2019s apparatus to show even more clearly that cathode rays carry negative charge (Figure 15.3). Next, he tackled the problem of why no one had been able to deflect cathode rays with an electric field. Thomson\u2019s hypothesis was that the cathode rays ionized some of the air molecules remaining in the vacuum chamber and these ions then shielded the cathode rays from the electric field. By taking great care to get an extremely low pressure in his discharge tube, Thomson was able to demonstrate that cathode rays respond to electric fields just as negatively charged particles would. Thomson had discovered the electron. cathode ray: free electrons emitted by a negative electrode info BIT The Irish physicist G. Johnstone Stoney coined the term electron in 1891. Thomson called the particles in cathode rays \u201ccorpuscles.\u201d 754 Unit VIII Atomic Physics 15-PearsonPhys30-Chap15 7/24/08 3:59 PM Page 755 Figure 15.3 Reproduction of Thomson\u2019s diagram and part of his description of the apparatus he used for several of his experiments. To determine the charge-to-mass ratio for an electron, Thomson added an electromagnet beside the parallel plates in the middle of the tube. The rays from the cathode C pass through a slit in the anode A, which is a metal plug fitting tightly into the tube and connected with the earth; after passing through a second slit in another earth-connected metal plug B, they travel between two parallel aluminium plates about 5 cm long by 2 broad and at a distance of 1.5 cm apart; they then fall on the end of the tube and produce a narrow well-defined phosphorescent patch. A scale pasted on the outside of the tube serves to measure the deflexion of this patch. M I N D S O N Are Electrons Positively or Negatively Charged? Outline two different methods for testing whether cathode rays consist of negatively or positively charged particles. Charge-to-mass Ratio of the Electron Thomson did not have a method for measuring either the mass of an electron or the charge that it carried. However, he did find a way to determine the ratio of charge to mass for the electron by using both an electric field and a magnetic field. Recall from Chapter 11 that the electric force acting on a charged particle is F e qE e is the electric force, q is the magnitude of the charge on where F the particle, and E is the electric field. Section 12.2 describes how the left-hand rule gives the direction of the magnetic force acting on a negative charge moving through a magnetic field. The magnitude of this force is F m qvB where q is the magnitude of the charge on the particle, v is the component of the particle\u2019s velocity perpendicular to the magnetic field, and B is the magnitude of the magnetic field. For a particle moving perpendicular to a magnetic field, v v and F m qvB Consider the perpendicular electric and magnetic fields shown in Figure 15.4. The electric field exerts a downward force on the negative charge while the magnetic field exerts an upward force. The gravitational force acting on the particle is negligible. If the net force on the charged particle is zero, the electric and magnetic forces must be equal in magnitude but opposite in direction: PHYSICS INSIGHT Dots and s are a common way to show vectors directed out of or into the page. A dot represents the tip of a vector arrow coming toward you, and represents the tail of an arrow moving away from you. Chapter 15 Electric force and energy quantization determine atomic structure. 755 15-PearsonPhys30-Chap15 7/24/08 3:59 PM Page 756 E Fm Fe Fm v v B Figure 15.4 Perpendicular electric and magnetic fields act on a moving negative charge. The red dots represent a magnetic field directed out of the page. Using the left-hand rule for magnetic force, your thumb points in the direction of electron flow (right), fingers point in the direction of the magnetic field (out of the page), and the palm indicates the direction of the magnetic force (toward the top of the page). Example 15.1 F m F net F net F e 0, so Fm Fe q B E B E qv v The speed of the particle is, therefore, v E B Practice Problems 1. A beam of electrons passes undeflected through a 2.50-T magnetic field at right angles to a 60-kN/C electric field. How fast are the electrons travelling? 2. What magnitude of electric field will keep protons from being deflected while they move at a speed of 1.0 105 m/s through a 0.05-T magnetic field? 3. What magnitude of magnetic field will stop ions from being deflected while they move at a speed of 75 km/s through an electric field with a magnitude of 150 N/C?", " Answers 1. 2.4 104 m/s 2. 5 103 N/C 3. 2.0 103 T A beam of electrons passes undeflected through a 0.50-T magnetic field combined with a 50-kN/C electric field. The electric field, the magnetic field, and the velocity of the electrons are all perpendicular to each other. How fast are the electrons travelling? Given 50 kN/C 5.0 104 N/C E 0.50 T B Required speed (v) Analysis and Solution 0, so the Since the electrons are not deflected, F net m. e equals the magnitude of F magnitude of F E B 5.0 104 N/C 0.50 T v 1.0 105 m/s Paraphrase The electrons are travelling at a speed of 1.0 105 m/s. Concept Check What would happen to the beam of electrons in Example 15.1 if their speed were greatly decreased? Thomson used mutually perpendicular electric and magnetic fields to determine the speed of the cathode rays. He then measured the deflection of the rays when just one of the fields was switched on. These deflections depended on the magnitude of the field, the length of the path in the 756 Unit VIII Atomic Physics 15-PearsonPhys30-Chap15 7/24/08 3:59 PM Page 757 field, and the speed, charge, and mass of the cathode-ray particles. Thomson could determine only the first three quantities, but he could use his measurements to calculate the ratio of the two unknowns, the charge and mass of the particles. Thomson made measurements with a series of cathode-ray tubes that each had a different metal for the electrode that emitted the rays. Since he found reasonably consistent values for the charge-to-mass ratio, Thomson concluded that all cathode rays consist of identical particles with exactly the same negative charge. Thomson\u2019s experiments showed that q m for an electron is roughly 1011 C/kg. This ratio is over a thousand times larger than the ratio for a hydrogen ion. Other physicists had shown that cathode rays can pass through thin metal foils and travel much farther in air than atoms do. Therefore, Thomson reasoned that electrons are much smaller than atoms. In his Nobel Prize lecture in 1906, he stated that \u201cwe are driven to the conclusion that the mass of the corpuscle is only about 1/1700 of that of the hydrogen atom.\u201d This value is within a few percent of the mass determined by the latest high-precision measurements. Thomson put forward the daring theory that atoms were divisible, and the tiny particles in cathode rays were \u201cthe substance from which all the chemical elements are built up.\u201d Although he was incorrect about electrons being the only constituents of atoms, recognizing that electrons are subatomic particles was a major advance in atomic physics. Determining Charge-to-mass Ratios Thomson measured the deflection of cathode rays to determine the charge-to-mass ratio of the electron. You can also determine mass-tocharge ratios by measuring the path of charged particles in a uniform magnetic field. When a charged particle moves perpendicular to a magnetic field, the magnetic force is perpendicular to the particle\u2019s velocity, so the particle\u2019s direction changes, but its speed is constant. In a uniform is constant, so the magnitude of the magnetic force, magnetic field, B qv, is also constant. As described in Chapter 6, a force of constant B magnitude perpendicular to an object\u2019s velocity can cause uniform circular motion. A charged particle moving perpendicular to a uniform magnetic field follows a circular path, with the magnetic force acting as the centripetal force (Figure 15.5). Fm r v F net F c F m F m qv v2 m B r q v r B m Figure 15.5 The path of an electron travelling perpendicular to a uniform magnetic field. The red s represent a magnetic field directed into the page. info BIT Around 1900, the idea that atoms could be subdivided was so radical that some physicists thought Thomson was joking when he presented his findings in a lecture to the Royal Institute in England. Chapter 15 Electric force and energy quantization determine atomic structure. 757 15-PearsonPhys30-Chap15 7/24/08 3:59 PM Page 758 Concept Check How do you know that the magnetic force on the particle is directed toward the centre of a circular path in Figure 15.5? Example 15.2 When a beam of electrons, accelerated to a speed of 5.93 105 m/s, is directed perpendicular to a uniform 100-T magnetic field, they travel in a circular path with a radius of 3.37 cm (Figure 15.6). Determine the charge-to-mass ratio for an electron. Practice Problems 1. Find the charge-to-mass ratio for an ion that travels in an arc of radius 1.00 cm when moving at 1.0 106 m/", "s perpendicular to a 1.0-T magnetic field. 2. Find the speed of an electron moving in an arc of radius 0.10 m perpendicular to a magnetic field with a magnitude of 1.0 104 T. 3. A carbon-12 ion has a charge-tomass ratio of 8.04 106 C/kg. Calculate the radius of the ion\u2019s path when the ion travels at 150 km/s perpendicular to a 0.50-T magnetic field. Answers 1. 1.0 108 C/kg 2. 1.8 106 m/s 3. 0.037 m 5.93 105 m/s Given ve 100 T 1.00 104 T B r 3.37 cm 3.37 102 m Required charge-to-mass ratio q m Analysis and Solution Since the magnetic force acts as the centripetal force, F F c m v 2 qv 100 T Fm Fc v 3.37 cm Figure 15.6 Substituting the known values for the beam of electrons gives q m 5.93 105 m/s (1.00 104 T)(3.37 102 m) 1.76 1011 C/kg Paraphrase The charge-to-mass ratio for an electron is about 1.76 1011 C/kg. Thomson\u2019s Raisin-bun Model Most atoms are electrically neutral. If electrons are constituents of atoms, atoms must also contain some form of positive charge. Since no positively charged subatomic particles had yet been discovered, Thomson suggested that atoms might consist of electrons embedded in a blob of massless positive charge, somewhat like the way raisins are embedded in the dough of a raisin bun. Figure 15.7 shows Thomson\u2019s model of the atom. Figure 15.7 Thomson\u2019s model of the atom: negative electrons embedded in a positive body 758 Unit VIII Atomic Physics 15-PearsonPhys30-Chap15 7/24/08 3:59 PM Page 759 Concept Check What characteristics should a scientific model have? Does Thomson\u2019s raisin-bun model of the atom have these characteristics? THEN, NOW, AND FUTURE The Mass Spectrometer fields to determine Thomson used electric and magthe netic charge-to-mass ratio of electrons. In later experiments, he made similar measurements for positive ions. These experiments led to the development of the mass spectrometer, an instrument that can detect compounds, measure isotope masses, and determine molecular structures (Figure 15.8). Many mass spectrometers use a four-stage process enclosed in a vacuum chamber: Ionization: If the sample is not already a gas, the ion source vaporizes it, usually by heating. Heating may also break complex compounds into smaller fragments that are easier to identify. Next, the neutral compounds in the sample are ionized so that they will respond to electric and magnetic fields. Usually, the ion source knocks one or two electrons off the compound to produce a positive ion. Acceleration: High-voltage plates then accelerate a beam of these ions into the velocity selector. Velocity Selection: The velocity selector has crossed electric and magnetic fields arranged such that only the ions that have a speed of pass straight through. Ions v E B with slower or faster speeds are deflected away from the entrance to the detection chamber. Thus, all the ions entering the next stage of the mass spectrometer have the same known speed. Detection: A uniform magnetic field in the detection chamber makes the ions travel in circular paths. The radius of each path depends on the charge and mass of the ion. Ions arriving at the detector produce an electrical current that is proportional to the number of ions. The spectrometer can produce a graph of the charge-to-mass ratios for a sample by moving the detector or by varying the electric or magnetic fields. Mass spectrometers can detect compounds in concentrations as small as a few parts per billion. These versatile machines have a huge range of applications in science, medicine, and industry. Questions 1. Write an expression for the radius of the path of ions in a mass spectrometer. 2. How does the mass of an ion affect the radius of its path in the detection chamber? 3. Describe how you could use a mass spectrometer to detect an athlete\u2019s use of a banned performance-enhancing drug. ionization chamber velocity selector vaporized sample detection chamber heater acceleration plates detector vacuum pump chart recorder ampli\ufb01er positive ions Figure 15.8 Mass spectrometer Chapter 15 Electric force and energy quantization determine atomic structure. 759 15-PearsonPhys30-Chap15 7/24/08 3:59 PM Page 760 15.1 Check and Reflect 15.1 Check and Reflect Knowledge 1. An electron moving at 5.0 105 m/s enters a magnetic field of magnitude 100 mT. (a) What is the maximum force that the 8. A mass spectrometer has the electric field", " in its velocity selector set to 8.00 102 N/C and the magnetic field set to 10.0 mT. Find the speed of the ions that travel straight through these fields. magnetic field can exert on the electron? When will the magnetic field exert this maximum force? 9. This diagram shows a proton moving at 1.0 105 m/s through perpendicular electric and magnetic fields. (b) What is the minimum force that the B 0.50 T [out of page] magnetic field can exert on the electron? When will the magnetic field exert this minimum force? 2. Explain why improved vacuum pumps were a key to the success of Thomson\u2019s experiments. 3. What experimental results led Thomson to conclude that all cathode rays consist of identical particles? Applications 4. A beam of electrons enters a vacuum v 1.0 105 m/s [right] E 100 N/C [down] (a) Calculate the net force acting on the particle. (b) Will the net force change over time? Explain your reasoning. chamber that has a 100-kN/C electric field and a 0.250-T magnetic field. Extensions (a) Sketch an orientation of electric and magnetic fields that will let the electrons pass undeflected through the chamber. (b) At what speed would electrons pass undeflected through the fields in part (a)? 5. Electrons are observed to travel in a circular path of radius 0.040 m when placed in a magnetic field of strength 0.0025 T. How fast are the electrons moving? 6. How large a magnetic field is needed to deflect a beam of protons moving at 1.50 105 m/s in a path of radius 1.00 m? 7. Use the appropriate hand rule to determine the direction of the magnetic field in question 6 if the protons rotate counterclockwise in the same plane as this page. 10. Suppose that a passenger accumulates 5 C of negative charge while walking from left to right across the carpeted floor to the security gate at an airport. (a) If the metal detector at the security gate exerts an upward force on this charge, what is the direction of the magnetic field within the detector? (b) If the metal detector uses a 0.05-T magnetic field, roughly how fast would the passenger have to run through the detector in order to feel weightless? (c) Explain whether it would be practical to use an airport metal detector as a levitation machine. e TEST To check your understanding of Thomson\u2019s experiments, follow the eTEST links at www.pearsoned.ca/school/physicssource. 760 Unit VIII Atomic Physics 15-PearsonPhys30-Chap15 7/24/08 3:59 PM Page 761 15.2 Quantization of Charge As you learned in Chapter 14, Planck and Einstein introduced the concept of quantization in physics in the early 20th century. The discovery of the electron raised the intriguing idea that electric charge might also be quantized. 15-2 QuickLab 15-2 QuickLab Determining a Fundamental Quantity Problem Does a set of containers contain only identical items? Materials 5 sealed containers laboratory balance Procedure 1 Measure the mass of each container to a precision of 0.1 g. 2 Discuss with your partner how to record and present your data. 3 Pool your data with the rest of the class. Questions 1. Look for patterns in the pooled data. Discuss as a class the best way to tabulate and graph all of the data. How can you arrange the data to make it easier to analyze? 2. Consider the differences in mass between pairs of containers. Explain whether these differences indicate that the masses vary only by multiples of a basic unit. If so, calculate a value for this basic unit. 3. Can you be sure that the basic unit is not smaller than the value you calculated? Explain why or why not. 4. What further information do you need in order to calculate the number of items in each container? The American physicist Robert Andrews Millikan (1868\u20131953) and his graduate student Harvey Fletcher made the next breakthrough in the study of the properties of the electron. In 1909, Millikan reported the results from a beautiful experiment that determined the charge on the electron and showed that it was a fundamental unit of electrical charge. Millikan\u2019s Oil-drop Experiment Millikan and Fletcher used an atomizer to spray tiny drops of oil into the top of a closed vessel containing two parallel metal plates (Figure 15.9). Some of the oil drops fell into the lower part of the vessel through a small hole in the upper plate. Friction during the spraying process gave some of the oil drops a small electric charge. Millikan also used X rays to change the charge on the oil drops. Since these oil drops were usually spherical, Millikan could calculate the mass of each drop from its diameter and the density of the oil. He connected a high-voltage battery to the plates,", " then observed the motion of the oil drops in the uniform electric field between the plates. By analyzing this motion and allowing for air resistance, Millikan calculated the electric force acting on each drop, and thus determined the charge on the drop. info BIT Millikan won the Nobel Prize in physics in 1923. In his Nobel lecture on the oil-drop experiment, he did not mention Fletcher\u2019s work at all. Chapter 15 Electric force and energy quantization determine atomic structure. 761 15-PearsonPhys30-Chap15 7/24/08 3:59 PM Page 762 atomizer high-voltage battery oil drops microscope metal plates insulated from vessel wall Figure 15.9 Millikan\u2019s oil-drop apparatus Millikan made numerous measurements with oil drops of various sizes. He found that the charged oil drops had either 1.6 1019 C of charge or an integer multiple of this value. Millikan reasoned that the smallest possible charge that a drop can have is the charge acquired either by gaining or losing an electron. Hence, the charge on the electron must be 1.6 1019 C. Millikan showed that charge is not a continuous quantity; it exists only in discrete amounts. This finding parallels Planck\u2019s discovery in 1900 that energy is quantized (see section 14.1). Recent measurements have determined that the elementary unit of charge, e, has a value of 1.602 176 462 0.000 000 063 1019 C. A value of 1.60 1019 C is accurate enough for the calculations in this textbook. Note that a proton has a charge of 1e and an electron has a charge of 1e. Since Thomson and others had already determined the charge-tomass ratio for electrons, Millikan could now calculate a reasonably accurate value for the mass of the electron. This calculation showed that the mass of the electron is roughly 1800 times less than the mass of the lightest atom, hydrogen. Millikan and Controversy In the mid 1970s, historians of science made a disturbing discovery: Millikan had, on several occasions, stated that he used all of his data in coming to the conclusion that the charge of the electron was quantized. In fact, his notebooks contain 175 measurements of which he only reported 58! When all of Millikan\u2019s data are used, his evidence for the quantization of charge is far less conclusive! Was Millikan guilty of scientific fraud, or was it a deeper, intuitive insight that led him to select only the data that clearly supported his claim that the charge on the electron is quantized? Historians will debate this question for many years to come, but we still acknowledge Millikan as the first person to measure the charge of the electron and to establish the quantization of charge. e SIM To see a simulation of Millikan\u2019s oil-drop experiment, follow the links at www.pearsoned.ca/ school/physicssource. elementary unit of charge, e: the charge on a proton info BIT Thomson and others had tried to measure the charge on an electron using tiny droplets of water. However, evaporation and other problems made these measurements inaccurate. e MATH For a simple exercise to determine the fundamental unit of charge using a method similar to Millikan\u2019s, visit www.pearsoned.ca/school/ physicssource. 762 Unit VIII Atomic Physics 15-PearsonPhys30-Chap15 7/24/08 4:00 PM Page 763 Concept Check Why is it almost impossible to determine the mass of an electron without determining its charge first? Example 15.3 An oil drop of mass 8.2 1015 kg is suspended in an electric field of 1.0 105 N/C [down]. How many electrons has the oil drop gained or lost? Given 1.0 105 N/C [down] E m 8.2 1015 kg Required number of electrons gained or lost (n) Fe Fg, the charge on the oil drop must Analysis and Solution To balance the gravitational force, the electric force must be directed upward (Figure 15.10). Since the electric force is in the opposite direction to the electric field, E be negative. The oil drop must have gained electrons. For the electric and gravitational forces to balance, F their magnitudes must be equal. Since F g net 0, then 0 F and F net e F F g e mg qE F g F e. Figure 15.10 Practice Problems 1. How many electrons are gained or lost by a plastic sphere of mass 2.4 1014 kg that is suspended by an electric field of 5.0 105 N/C [up]? 2. What electric field will suspend an oil drop with a mass of 3.2 1014 kg and a charge of 2e? Answers 1. The sphere has lost three electrons. 2. 9.8 105 N/C [up] where q is the magnitude of the charge on the drop and g is the magnitude of the gravitational", " field. q Solving for q gives mg E (8.2 1015 kg)(9.81 m/s2) 1.0 105 N/C 8.04 1019 C The net charge on the oil drop equals the number of electrons gained times the charge on each electron. Thus, q q ne and n e Note that n must be a whole number. q n e 8.04 1019 C 1.60 1019 C 5 Paraphrase The oil drop has gained five electrons. Chapter 15 Electric force and energy quantization determine atomic structure. 763 15-PearsonPhys30-Chap15 7/24/08 4:00 PM Page 764 In Example 15.3, you studied what happens when the electric force on a charged droplet exactly balances the gravitational force on the same droplet. What happens if these two forces do not balance each other? Example 15.4 Practice Problems 1. Calculate the net force on a sphere of charge 5e and mass 2.0 1014 kg when placed in an electric field of strength 1.0 105 N/C [up] in a vacuum chamber. 2. Calculate the acceleration of the sphere if the direction of the electric field in question 1 is reversed. Answers 1. 1.2 1013 N [down] 2. 14 m/s2 [down] A tiny plastic sphere of mass 8.2 1015 kg is placed in an electric field of 1.0 105 N/C [down] within a vacuum chamber. The sphere has 10 excess electrons. Determine whether the sphere will accelerate, and if so, in which direction. Given Choose up to be positive. 1.0 105 N/C [down] 1.0 105 N/C E m 8.2 1015 kg q 10e g 9.81 m/s2 [down] 9.81 m/s2 Fe Fg Required acceleration of the plastic sphere (a) Figure 15.11 (a) Analysis and Solution Express the charge on the sphere in coulombs: q 10e 10 (1.60 1019 C) 1.60 1018 C Draw a free-body diagram of the forces acting on the sphere (Figure 15.11 (a)). Because the charge is negative, the electric force is in the opposite direction to the electric field, E Calculate the magnitude of both the electric and gravitational forces acting on the sphere.. F g F e mg (8.2 1015 kg)(9.81 m/s2) 8.04 1014 N qE (1.60 1018 C)(1.0 105 N/C) 1.60 1013 N Determine the net force on the sphere. From Figure 15.11 (b), F net g e F F 8.04 1014 N (1.60 1013 N) 7.96 1014 N Now find the acceleration of the sphere: F net a ma F net m 7.96 1014 N 8.2 1015 kg Fe Fg Fnet 9.7 m/s2 Figure 15.11 (b) Paraphrase The sphere will accelerate upward at a rate of 9.7 m/s2. 764 Unit VIII Atomic Physics 15-PearsonPhys30-Chap15 7/24/08 4:00 PM Page 765 15.2 Check and Reflect 15.2 Check and Reflect Knowledge Extensions 1. Explain the term quantization of charge. 7. A student attempting to duplicate 2. Which two properties of the electron was Millikan able to determine with the results from his oil-drop experiment? 3. Calculate the charge, in coulombs, on an oil drop that has gained four electrons. 4. Determine the electric force acting on an oil drop with a charge of 5e in a uniform electric field of 100 N/C [down]. Applications 5. (a) What is the net force acting on a charged oil drop falling at a constant velocity in the absence of an electric field? Explain your reasoning, using a free-body diagram to show the forces acting on the drop. (b) Describe the motion of the oil drop in an electric field that exerts an upward force greater than the gravitational force on the drop. Draw a diagram to show the forces acting on the drop. 6. An oil droplet with a mass of 6.9 1015 kg is suspended motionless in a uniform electric field of 4.23 104 N/C [down]. (a) Find the charge on this droplet. (b) How many electrons has the droplet either gained or lost? (c) In what direction will the droplet move if the direction of the electric field is suddenly reversed? Millikan\u2019s experiment obtained these results. Explain why you might suspect that there was a systematic error in the student\u2019s measurements. Droplet # Charge ( 1019 C 10 19.0 17.2 10.0 20.8 26.2 24.", "4 20.8 22.6 15.4 24.4 8. Some critics of Millikan have noted that he used only a third of his measurements when reporting the results of his oil-drop experiment. Use a library or the Internet to learn more about this issue. Explain whether you feel that Millikan was justified in presenting only his \u201cbest\u201d data. e TEST To check your understanding of charge quantization, follow the eTEST links at www.pearsoned.ca/school/physicssource. Chapter 15 Electric force and energy quantization determine atomic structure. 765 15-PearsonPhys30-Chap15 7/24/08 4:00 PM Page 766 15.3 The Discovery of the Nucleus By the beginning of the 20th century, the work of Thomson, Perrin, and others had provided strong evidence that atoms were not the smallest form of matter. Physicists then started developing models to describe the structure of atoms and devising experiments to test these models. 15-3 QuickLab 15-3 QuickLab Using Scattering to Measure a Hidden Shape Problem What can scattering reveal about the shape of an unseen target? Materials cardboard tube 10\u201315 cm in diameter small marbles or ball bearings carbon paper white letter-size paper marble or ball bearing cardboard tube with hidden target white paper carbon paper support block Figure 15.12 Procedure 1 Your teacher will prepare several \u201cbeam tubes\u201d containing a hidden target. A cover blocks the top of each tube, except for a small opening (Figure 15.12). No peeking! 2 Work with a small group. Place a sheet of carbon paper, carbon side up, on the desk, and put a piece of white paper on top of the carbon paper. Then carefully set the cardboard tube on blocks on top of the paper. 3 Drop a marble or ball bearing through the opening at the top of the cardboard tube. Retrieve the marble or bearing, then drop it through the tube again, for a total of 50 times. 4 Remove the piece of paper, and look for a pattern in the marks left on it. Questions 1. Discuss with your group what the scattering pattern reveals about the shape and size of the target hidden in the cardboard tube. Sketch the likely shape of the target. 2. Explain how you can use your scattering results to estimate the dimensions of the target. 3. Compare your predictions to those made by groups using the other prepared tubes. 4. What is the smallest dimension that you could measure with the equipment in this experiment? 766 Unit VIII Atomic Physics 15-PearsonPhys30-Chap15 7/24/08 4:00 PM Page 767 Rutherford\u2019s Scattering Experiment Ernest Rutherford (1871\u20131937), a brilliant experimenter from New Zealand, was fascinated by radioactivity. By 1909, he had shown that some radioactive elements, such as radium and thorium, emitted positively charged helium ions, which are often called alpha () particles. Rutherford had also observed that a beam of these particles spread out somewhat when passing through a thin sheet of mica. So, he had two assistants, Hans Geiger and Ernest Marsden, measure the proportion of alpha particles scattered at different angles from various materials. Figure 15.13 shows the technique Rutherford devised for these measurements. A few milligrams of radium in a lead block with a small opening produced a pencil-shaped beam of alpha particles. The experimenters positioned a sheet of thin gold foil at right angles to the beam of alpha particles and used a screen coated with zinc sulfide to detect particles scattered by the foil. When an alpha particle struck the screen, the zinc sulfide gave off a faint flash of light, just enough to be visible through the microscope. By rotating the screen and microscope around the gold film, Geiger and Marsden measured the rates at which alpha particles appeared at various angles. beam of particles lead scattering angle zinc sulfide screen info BIT Rutherford was a professor at McGill University in Montreal from 1898 to 1907. During that time, he discovered an isotope of radon, showed that radioactive elements can decay into lighter elements, and developed a method for dating minerals and estimating the age of Earth. Although he considered himself a physicist, he cheerfully accepted the Nobel Prize for chemistry in 1908. microscope gold foil radium Figure 15.13 Rutherford\u2019s scattering experiment Most of the alpha particles travelled through the foil with a deflection of a degree or less. The number of alpha particles detected dropped off drastically as the scattering angle increased. However, a few alpha particles were scattered at angles greater than 140, and once in a while an alpha particle would bounce almost straight back. Figure 15.14 shows the relationship between the number of scattered alpha particles and the angle at which they scattered. Rutherford was startled by these results. He knew that the deflections caused by attraction to the electrons in the gold atoms would be tiny because the alpha particles were fast-moving and roughly 8000 times heavier than electrons. He also calculated that deflection caused", " by repulsion of the alpha particles by the positive charge in the gold atoms would be no more than a degree if this charge were distributed evenly throughout each atom in accordance with Thomson\u2019s model. So, Rutherford did not expect any alpha particles to be scattered at large angles. In a lecture describing the experiment, he said that seeing this scattering \u201cwas almost as incredible as if you had fired a 15-inch shell at a piece of tissue paper and it came back and hit you!\u201d Rutherford spent several weeks analyzing the scattering data. He concluded that the positive charge in a gold atom must be concentrated in an incredibly tiny volume, so most of gold foil was actually empty space! 107 106 105 104 103 102 10 \u00b0 20\u00b0 40\u00b0 60\u00b0 80\u00b0 100\u00b0 120\u00b0 140\u00b0 160\u00b0 Scattering Angle Figure 15.14 The scattering pattern observed by Geiger and Marsden Chapter 15 Electric force and energy quantization determine atomic structure. 767 15-PearsonPhys30-Chap15 7/24/08 4:00 PM Page 768 Rutherford\u2019s team then tried scattering alpha particles from other metals. Using data from scattering experiments with aluminium foil, Rutherford showed that the positive charge and most of the mass of an atom are contained in a radius of less than 1014 m. Rutherford had discovered the nucleus \u2014 and disproved the raisin-bun model. Concept Check On average, atoms have a radius of roughly 1010 m. Use Rutherford\u2019s estimate for the size of the nucleus to calculate the proportion of the human body that is just empty space. The Planetary Model Rutherford\u2019s discovery of the nucleus quickly led to the planetary model of the atom (Figure 15.15). In this model, the electrons orbit the nucleus much like planets orbiting the Sun. The electrostatic attraction between the positive nucleus and the negative electrons provides the centripetal force that keeps the electrons in their orbits. This model is also known as the solar-system, nuclear, or Rutherford model. To calculate the size of the nucleus, Rutherford applied the law of conservation of energy and an equation for electric potential energy. He knew that the electric potential energy that a charge q1 gains from the field around charge q2 is Ep kq1q2 d where k is Coulomb\u2019s constant and d is the distance between the charges. This equation can be derived from Coulomb\u2019s law using basic calculus. nucleus orbiting electron nucleus alpha particles Figure 15.15 The planetary model of the atom can explain the results of Rutherford\u2019s scattering experiments. The extreme scattering of some alpha particles could only be explained by having most of the mass and positive charge of the atom concentrated in a very small nucleus. planetary model: atomic model that has electrons orbiting a nucleus 768 Unit VIII Atomic Physics 15-PearsonPhys30-Chap15 7/24/08 4:00 PM Page 769 Example 15.5 In a scattering experiment, some alpha particles are scattered almost straight back from a sheet of gold foil. Each of these particles had an initial kinetic energy of 1.6 1012 J. The charge on an alpha particle is 2e, and the charge on a gold nucleus is 79e. Estimate the maximum possible size of a gold nucleus, given that the alpha particles do not hit the nucleus. Given Ek 1.6 1012 J q1 q 2e q2 qgold 79e Required radius of the nucleus of a gold atom (r) Analysis and Solution Since energy is conserved, the total kinetic and potential energy of an alpha particle does not change during scattering. As an alpha particle approaches a nucleus, the force of repulsion between the positive charges causes the alpha particle to slow down. This process converts the particle\u2019s kinetic energy to electric potential energy until the kinetic energy is zero. Then the particle starts moving away from the nucleus and regains its kinetic energy. At the point where the alpha particle is closest to the nucleus, all of the particle\u2019s kinetic energy has been converted to electric potential energy. The electric potential energy of the alpha particle is Ep Practice Problems 1. The charge on a tin nucleus is 50e. How close can an alpha particle with an initial kinetic energy of 1.6 1012 J approach the nucleus of a tin atom? 2. An iron nucleus has 56 protons. What is the electric potential energy of a proton located 5.6 1013 m from the centre of an iron nucleus? Answers 1. 1.4 1014 m 2. 2.3 1014 J kq1q2 d, where k is Coulomb\u2019s constant and d is the distance between the alpha particle and the nucleus. The initial distance between the alpha particle and the nucleus is vastly larger than the distance between them when they are closest together. Therefore, the initial electric potential energy of the alpha particle is negligible for this calculation. Initially, kinetic energy electric potential energy 1.6 1012 J 0 When the alpha particle is closest to the nucleus, kinetic energy electric", " potential energy 0 kq1q2 d Since the total energy is conserved, Solving for d gives kq1q2 d 1.6 1012 J d kq1q2 1.6 1012 J 2 m N (2 1.60 1019 C)(79 1.60 1019 C) 8.99 109 2 C 1.6 1012 J 2.3 1014 m At its closest approach, the alpha particle is about 2.3 1014 m from the centre of the nucleus. Paraphrase The radius of a gold nucleus cannot be larger than 2.3 1014 m. Chapter 15 Electric force and energy quantization determine atomic structure. 769 15-PearsonPhys30-Chap15 7/24/08 4:00 PM Page 770 15.3 Check and Reflect 15.3 Check and Reflect Knowledge 1. Explain how the results from Rutherford\u2019s gold-foil experiment disproved Thomson\u2019s model of the atom. 2. Briefly describe the planetary model of the atom. 7. In scattering experiments with aluminium foil, Rutherford found that the alpha particles observed at angles close to 180 did not behave like they had been scattered only by electrostatic repulsion. Rutherford thought these alpha particles might have actually hit an aluminium nucleus. 3. Find the potential energy of an alpha (a) Why did Rutherford see only particle that is (a) 1.0 1010 m from the centre of a gold nucleus (b) 1.0 1014 m from the centre of a gold nucleus electrostatic scattering when he used the same source of alpha particles with gold foil? (b) Rutherford used alpha particles with an energy of about 1.2 1012 J. Estimate the radius of an aluminium nucleus. 4. Why does the scattering angle increase as alpha particles pass closer to the nucleus? Extensions Applications 5. Why did Rutherford conclude that it was just the nucleus that must be extremely tiny in an atom and not the entire atom? 6. (a) By 1900, physicists knew that 1 m3 of gold contains approximately 6 1028 atoms. Use this information to estimate the radius of a gold atom. List any assumptions you make. (b) Compare this estimate to the estimate of the size of a gold nucleus in Example 15.5. 8. According to Rutherford\u2019s calculations, the positive charges in an atom are packed tightly together in the nucleus. Why would physicists in 1900 expect such an arrangement to be highly unstable? What did Rutherford\u2019s results suggest about forces in the nucleus? e TEST To check your understanding of atomic models, follow the eTEST links at www.pearsoned.ca/school/physicssource. 770 Unit VIII Atomic Physics 15-PearsonPhys30-Chap15 7/24/08 4:00 PM Page 771 15.4 The Bohr Model of the Atom In 1912, Niels Bohr (1885\u20131962), a Danish physicist, studied for a few months at Rutherford\u2019s laboratory. Both Bohr and Rutherford recognized a critical flaw in the planetary model of the atom. Experiments had shown that an accelerating charge emits electromagnetic waves, as predicted by the mathematical model for electromagnetism developed by the Scottish physicist James Clerk Maxwell (1831\u20131879). Electrons orbiting a nucleus are constantly accelerating, so they should emit electromagnetic waves. These waves would take energy from the orbiting electrons. As a result, the electrons in an atom should spiral into the nucleus in a few microseconds (Figure 15.16). But empirical evidence indicates that electrons do not spiral into their atomic nuclei. If they did, stable matter would not exist. Figure 15.16 According to Maxwell\u2019s laws of electromagnetism, the orbiting electron should continuously radiate energy and spiral into the nucleus, which it does not do. info BIT Rutherford and Bohr shared an interest in soccer: Rutherford was a fan, and Bohr was an excellent player. Bohr thought Planck\u2019s concept of quantized energy might provide a solution, and puzzled for months over how to fit this concept into a model of the atom. Then, a casual conversation with a colleague, his former classmate Hans Marius Hansen, gave Bohr the key to the answer. Hansen had recently returned from studying in Germany with an expert in spectroscopy. Hansen told Bohr that the wavelengths of the light in the spectrum of hydrogen have a mathematical pattern. No one had yet explained why this pattern occurs. Bohr found the explanation, and provided the first theoretical basis for spectroscopy. spectroscopy: the study of the light emitted and absorbed by different materials Spectroscopy A prism or diffraction grating can spread light out into a spectrum with colours distributed according to their wavelengths. In 1814, Josef von Fraunhofer (1787\u20131826) noticed a number of gaps or dark lines in the spectrum of the Sun. By 1859, another German physicist, Gustav Kirchhoff (1824", "\u20131887), had established that gases of elements or compounds under low pressure each have a unique spectrum. Kirchhoff and others used spectra to identify a number of previously unknown elements. Kirchhoff\u2019s laws for spectra explain how temperature and pressure affect the light produced or absorbed by a material: info BIT The element names cesium and rubidium come from the Latin words for the colours of the spectral lines used to discover these elements. Cesium comes from cesius, which is Latin for sky blue. Rubidium comes from rubidus \u2014 Latin for dark red. Chapter 15 Electric force and energy quantization determine atomic structure. 771 15-PearsonPhys30-Chap15 7/24/08 4:00 PM Page 772 emission line spectrum: a pattern of bright lines produced by a hot gas at low pressure absorption line spectrum: a pattern of dark lines produced when light passes through a gas at low pressure \u2022 A hot, dense material emits a continuous spectrum, without any dark or bright lines. \u2022 A hot gas at low pressure has an emission line spectrum with bright lines at distinct characteristic wavelengths. \u2022 A gas at low pressure absorbs light at the same wavelengths as the light it emits when heated. Shining white light through the gas produces an absorption line spectrum with dark lines that match the bright lines in the emission spectrum for the gas. Figure 15.17 illustrates these three types of spectra. Continuous Spectrum Bright-line Spectra Hydrogen Sodium Helium Neon Mercury Absorption Spectrum for Mercury (against a continuous spectrum) Figure 15.17 Continuous, emission line, and absorption line spectra 350 400 450 500 wavelength (nm) 550 600 650 15-4 Design a Lab 15-4 Design a Lab Emission Spectra of Elements The Question In what ways are emission line spectra characteristic of the elements that produce them? Design and Conduct Your Investigation Investigate the emission spectra produced by various elements. Here are some ways you can heat different elements enough to produce visible light: \u2022 Use a Bunsen burner to vaporize a small amount of an element. \u2022 Use commercially available discharge tubes containing gaseous elements. \u2022 Observe forms of lighting that use a vaporized element, such as sodium or mercury arc lamps. Often, a diffraction grating is the simplest method for observing a spectrum. Check with your teacher if you need directions for using a diffraction grating. Note the overall colour of the light from each element, and sketch or photograph its emission spectrum. Extending Investigate absorption lines. You could try comparing the spectra of direct sunlight and sunlight that passes through clouds. Alternatively, you could shine white light at coloured solutions and diffract the light that passes through. Why are distinct absorption lines generally more difficult to produce than emission lines? 772 Unit VIII Atomic Physics 15-PearsonPhys30-Chap15 7/24/08 4:00 PM Page 773 Improvements in the resolution of spectrometers made them a powerful analytic tool. For example, scientists have painstakingly matched the dark Fraunhofer lines in the solar spectrum (Figure 15.18) to the spectral patterns of dozens of elements, thus proving that these elements are present in the Sun\u2019s atmosphere. However, at the start of the 20th century, the reasons why elements produce spectral lines were still a mystery. spectrometer: a device for measuring the wavelengths of light in a spectrum Fraunhofer line: a dark line in the spectrum of the Sun Figure 15.18 Fraunhofer lines are the dark lines in the visible part of the solar spectrum. The first hint that spectral lines were not just randomly spaced came in 1885. Johann Jacob Balmer, a Swiss mathematics teacher with an interest in numerology, found a formula for the wavelengths of the lines in the hydrogen spectrum. In 1890, Johannes Robert Rydberg generalized Balmer\u2019s formula and applied it with varied success to other elements. Here is the formula for hydrogen: RH 1 1 22 1 n2 400 450 500 550 600 650 700 750 nm n 6 n 5 n 4 n 3 where RH is Rydberg\u2019s constant for hydrogen, 1.097 107 m1, and n has the whole number values 3, 4, 5,.... The emission line spectrum of hydrogen is given in Figure 15.19. Hansen told Bohr about this formula. Bohr later remarked, \u201cAs soon as I saw Balmer\u2019s formula, the whole thing was immediately clear to me.\u201d Bohr had realized that the spectral lines corresponded to differences between quantized energy levels in the hydrogen atom. This concept was the foundation of a powerful new model of the atom. The Bohr Model of the Atom In 1913, Bohr published a paper suggesting a radical change to the planetary model of the atom. Here are the basic principles of Bohr\u2019s model: \u2022 Electrons can orbit the nucleus only at certain specific distances from the nucleus. These distances are particular multiples of the", " radius of the smallest permitted orbit (Figure 15.20). Thus, the orbits in an atom are quantized. \u2022 Both the kinetic energy and the electric potential energy of an electron in orbit around a nucleus depend on the electron\u2019s distance from the nucleus. So, the energy of an electron in an atom is also quantized. Each orbit corresponds to a particular energy level for the electron. \u2022 An electron can move from one energy level to another only by either emitting or absorbing energy equal to the difference between the two energy levels. An electron that stays in a particular orbit does not radiate any energy. Since the size and shape of the orbit remain constant along with the energy of the electron, the orbits are often called stationary states. Figure 15.19 Some of the bright lines in the spectrum of hydrogen e WEB To learn more about spectra, follow the links at www.pearsoned.ca/school/ physicssource. energy level: a discrete and quantized amount of energy stationary state: a stable state with a fixed energy level Chapter 15 Electric force and energy quantization determine atomic structure. 773 15-PearsonPhys30-Chap15 7/24/08 4:00 PM Page 774 Bohr reasoned that Balmer\u2019s formula shows that the energy of an orbiting electron depends inversely on the square of a quantum number n, now known as the principal quantum number. Bohr then used the equations for uniform circular motion, Coulomb\u2019s law, and electric potential energy to derive expressions for the size of the hydrogen atom and the energy of the electron in the atom. principal quantum number: the quantum number that determines the size and energy of an orbit Figure 15.20 In Bohr\u2019s model of the atom, electrons can orbit only at specific distances from the nucleus. r1 4r1 9r1 16r1 Orbit Sizes Bohr\u2019s model of the hydrogen atom states that electrons can orbit the nucleus only at specific locations given by the expression: rn h2 42mke2 n2 where rn is the radius of the nth possible orbit for an electron and n is the principal quantum number, which can have the values 1, 2, 3,.... The other symbols in the equation all represent constants: k is Coulomb\u2019s constant, h is Planck\u2019s constant, e is the elementary charge, and m is the mass of the electron. By combining all the constants, this equation can be simplified to rn r1n2, where r1 h2 42mke2 5.29 1011 m Bohr radius: radius of the smallest orbit in a hydrogen atom The quantity r1 is known as the Bohr radius. It is the radius of the lowest possible energy level or ground state of the hydrogen atom. ground state: the lowest possible energy level Concept Check Sketch the first three orbits for a hydrogen atom to scale. Describe how the size of this atom changes as n increases. Energy Levels An orbiting electron has both kinetic energy and electric potential energy. By combining equations for kinetic and electric potential energies, Bohr derived this expression for En, the total energy of the electron in an energy level: 22mk2e4 1 h2 n2 En 774 Unit VIII Atomic Physics 15-PearsonPhys30-Chap15 7/24/08 4:00 PM Page 775 As with the expression for the orbit radii, the constants can be com- bined to simplify the equation: En, where E1 E1 n2 22mk2e4 h2 2.18 1018 J or 13.6 eV E1 is the ground-state energy of the hydrogen atom. When n 1, the electron has greater energy and the atom is in an excited state. When n, the electron is no longer bound to the nucleus because E 0. Thus, Bohr\u2019s model predicts that the ionization energy for hydrogen is 2.18 1018 J or 13.6 eV, corresponding to E1. Concept Check Does it make sense that the energy in the equation En 22m k2e4 1 h2 2 n is negative? Consider what you have to do to remove an electron from an atom. excited state: any energy level higher than the ground state ionization energy: energy required to remove an electron from an atom Example 15.6 How much energy must a hydrogen atom absorb in order for its electron to move from the ground state to the n 3 energy level? Given ninitial 1 nfinal 3 Required energy absorbed by atom (E) Analysis and Solution The energy the atom must absorb is equal to the difference between the two energy levels. The energy for each energy level is En E1 n2 2.18 1018 J n2 Practice Problems 1. How much energy does it take to move the electron in a hydrogen atom from the ground state to the n 4 energy level? 2. How much energy does the electron in a hydrogen atom lose when dropping", " from the n 5 energy level to the n 2 energy level? Answers 1. 2.04 1018 J 2. 4.58 1019 J E E3 E1 2.18 1018 J 32 1 (2.18 1018 J)1 9 1.94 1018 J 2.18 1018 J 12 Paraphrase The electron in a hydrogen atom requires 1.94 1018 J of energy to move from the ground state to the n 3 level. Chapter 15 Electric force and energy quantization determine atomic structure. 775 15-PearsonPhys30-Chap15 7/24/08 4:00 PM Page 776 Energy Level Transitions and Line Spectra The Bohr model explains why absorption and emission line spectra occur for hydrogen and other elements. To jump to a higher energy level, an electron in an atom must gain energy. The atom can gain this energy by absorbing a photon. For energy to be conserved, the photon\u2019s energy must match the difference between the electron\u2019s initial energy level and the higher one. Recall from section 14.1 that the energy and frequency of a photon are related by the equation E hf. Thus, the atom can absorb only the frequencies that correspond to differences between the atom\u2019s energy levels. Absorption of light at these specific frequencies causes the discrete dark lines in absorption spectra. Similarly, atoms can emit only photons that have energies corresponding to electron transitions from a higher energy level to a lower one. The arrows in Figure 15.21 illustrate all the possible downward transitions from the first six energy levels for hydrogen. Such energy level diagrams provide a way to predict the energies, and hence the wavelengths, of photons emitted by an atom Paschen series (infrared) Balmer series (visible) 0.38 0.54 0.85 1.5 3. 13.6 Lyman series (ultraviolet) n 1 Figure 15.21 The first six energy levels for hydrogen. Which arrow represents the transition that releases the most energy? 776 Unit VIII Atomic Physics 15-PearsonPhys30-Chap15 7/24/08 4:00 PM Page 777 15-5 Inquiry Lab 15-5 Inquiry Lab An Emission Spectrum Question What is the energy difference between two energy states of atoms in a particular discharge tube? Hypothesis At a high temperature and low pressure, an atom emits light at the wavelengths predicted by the Bohr model. Variables \u2022 diffraction angle \u2022 wavelength Materials and Equipment a discharge tube high-voltage power supply diffraction grating 12 straight pins sheet of paper cardboard or foam-core sheet protractor and straightedge masking tape Required Skills Initiating and Planning Performing and Recording Analyzing and Interpreting Communication and Teamwork Procedure 1 Draw an x-axis and a y-axis that intersect in the middle of the sheet of paper. Then tape the paper to the cardboard or foam-core sheet (Figure 15.22). 2 Connect the high-voltage power supply to the discharge tube. 3 Switch on the power supply and look at the discharge tube through the diffraction grating. Orient the grating so that the spectral lines are vertical. You may want to darken the room to see the spectrum more clearly. 4 Align the bottom edge of the diffraction grating with the x-axis on the paper. Hold the grating vertical by pressing a straight pin into the y-axis on either side of the grating. 5 Sight along the y-axis, and turn the cardboard or foamcore base so that the y-axis points directly toward the discharge tube. Now tape the base to the tabletop. 6 For each spectral line, position a sighting pin so that it is on the line between the spectral line and the origin of your axes. Draw a line from the sighting pin to the origin. CAUTION! Keep well clear of the power supply connections when the power is on. Analysis high-voltage supply discharge tube sighting pins diffraction grating \u03b8 Figure 15.22 1. Decide how to record your data. You could use a table similar to the one below. Colour of Line Diffraction Angle, Wavelength (nm) 2. Use the diffraction formula n= d sin to calculate the wavelengths of the spectral lines (see section 13.5). Sometimes the spacing of the slits, d, is marked on the grating. If not, ask your teacher for this information. 3. Determine the energy difference between the two energy states. 4. Draw an energy level diagram showing the electron transition. Chapter 15 Electric force and energy quantization determine atomic structure. 777 15-PearsonPhys30-Chap15 7/24/08 4:00 PM Page 778 When an electron in an atom absorbs a photon, the final energy level of the atom\u2019s electron is greater than its initial energy level. From the law of conservation of energy, the energy of the photon can be expressed as: Ephoton Efinal Einitial level,", " n, is given by the formula En Since the energy of an electron in a hydrogen atom at a given energy, it is possible Einitial becomes (using algebra) to show that the expression Ephoton 22mk2e4 h2 Efinal 1 n2 1 RH 1 n2 final 1 n2 initial This result is impressive! Bohr\u2019s model not only explains spectral lines, but leads to a generalized form of Balmer\u2019s formula, and predicts the value of Rydberg\u2019s constant for hydrogen. Example 15.7 Find the wavelength of the light emitted by a hydrogen atom when its electron drops from the n 3 to the n 2 energy level. Practice Problems 1. Find the wavelength of the light emitted by a hydrogen atom when its electron drops from the n 5 to the n 2 energy level. 2. Find the wavelength of light that a hydrogen atom will absorb when its electron moves from the n 3 to the n 7 energy level. Answers 1. 434 nm 2. 1005 nm Given ninitial 3 Required wavelength () nfinal 2 Analysis and Solution Simply substitute the known values for n into the wavelength equation for an emitted photon: 1 1 n2 RH RH 1 n2 initial final 1 22 1 32 (1.097 107 m1) 1 9 1.524 106 m1 1 1.524 106 m1 1 4 6.563 107 m Paraphrase The atom emits light with a wavelength of 656.3 nm. 778 Unit VIII Atomic Physics 15-PearsonPhys30-Chap15 7/24/08 4:00 PM Page 779 Concept Check What features do the Bohr model of the atom and the planetary model have in common? In what critical ways do these two models differ? The Northern Lights and the Emission Line Spectrum of Oxygen Alberta skies often display one of nature\u2019s most beautiful phenomena \u2014 the aurora borealis, or northern lights, which you first studied in Chapter 12. At altitudes between 100 km and 400 km above the surface of Earth, high-energy electrons, trapped by Earth\u2019s magnetic field, interact with oxygen and nitrogen atoms. During these interactions, the electrons in these atoms are excited and move into higher energy levels. Eventually, the excited electrons return to their ground states. In doing so, they emit light that forms the characteristic colours of the aurora borealis. Figure 15.23 shows a display of the aurora borealis above northern Alberta. You can use the quantized energy levels to help explain the characteristic green colour of the aurora (as well as the subtler shades of red and blue). (a) Figure 15.23 The photo in (a) shows a bright, mostly green aurora. The green colour is due to an energy level transition in oxygen atoms in Earth\u2019s atmosphere. The diagram in (b) shows some of the energy levels in the oxygen atom, including the one that produces the green light that is mostly seen in the aurora. In this diagram, the ground state energy equals 0 eV. (b) ) V e ( 4.17 1.96 1.90 \u03bb 557.7 nm Ground state (0 eV) From Figure 15.23(b), the green colour of the aurora occurs when an 4.17 eV to a lower excited electron drops from an energy level Einitial Eintial energy level Efinal and E 1.96 eV. Using the equations E Efinal, you can show that this change in energy level produces hc the colour you see. info BIT The emission spectra of elements act like nature\u2019s fingerprints. By vaporizing a small sample and looking at the emission lines produced, you can determine the chemical composition of a substance. This technique, called atomic emission spectroscopy, is used routinely in forensics labs around the world. Chapter 15 Electric force and energy quantization determine atomic structure. 779 15-PearsonPhys30-Chap15 7/24/08 4:00 PM Page 780 Concept Check Use the energy level diagram for oxygen, given in Figure 15.23(b), to 1.96 eV will verify that a transition from Einitial produce green light ( 558 nm). Two other possible transitions are also shown in the diagram. What colours would these transitions produce? Determine their wavelengths. 4.17 eV to Efinal 15.4 Check and Reflect 15.4 Check and Reflect Knowledge 1. Explain what the phrase \u201cquantized process\u201d means. 2. Sketch the first five orbits in a hydrogen atom. Indicate on your sketch which transitions cause the blue, green, and red lines shown in Figure 15.21. 3. List three quantities predicted by Bohr\u2019s model of the atom. 4. Why do electrons in hydrogen atoms emit infrared light when they make transitions to the n 3 energy level, and ultraviolet light when they make transitions to the n 1 energy level? Applications 5. The wavelengths of the first four visible lines", " in the hydrogen spectrum are 410, 434, 486, and 656 nm. (a) Show that Balmer\u2019s formula predicts these wavelengths. (b) Which of the wavelengths corresponds to a transition from the n 4 energy level to the n 2 energy level? (c) Use this wavelength to calculate the energy difference between the n 4 and the n 2 stationary states. 6. A helium-neon laser produces photons of wavelength 633 nm when an electron in a neon atom drops from an excited energy state to a lower state. What is the energy difference between these two states? Express your answer in electron volts. 7. The diagram shown below shows some of the energy levels for the lithium atom. The designations 2s, 2p, etc., are a common notation used in spectroscopy. (a) Without doing any calculations, sort the four transitions shown from shortest wavelength to longest wavelength. Explain your reasoning. (b) Estimate the energy of each of the four energy levels shown. (c) Calculate the wavelengths produced in these transitions. Indicate which ones are in the visible part of the spectrum, along with their colours. 3p 3s 3 2p 0 1 2 V e 3 4 5 6 3d 2 4 2s 1 780 Unit VIII Atomic Physics 15-PearsonPhys30-Chap15 7/24/08 4:00 PM Page 781 Hydrogen Sodium Solar (selected lines) 400 450 500 550 600 650 700 750 nm 400 450 500 550 600 650 700 750 nm 400 450 500 550 600 650 700 750 nm 8. What can you conclude about the Determine: composition of the Sun from the spectra given above? Explain your reasoning. 9. (a) Find the difference in energy between the n 2 and n 3 energy levels in hydrogen. (b) Find the energy difference between the n 5 and n 6 energy levels in hydrogen. (c) What happens to the energy difference between successive orbits as the distance from the nucleus increases? 10. The ionization energy for an atom is the energy required to remove an electron completely from an atom. Show why the ionization energy for hydrogen is equal to E1. (Hint: Consider going from the ground state to an energy level with n 1000.) 11. When an atom\u2019s energy levels are closely spaced, the atom \u201cde-excites\u201d by having one of its electrons drop through a series of energy levels. This process is called fluorescence and is often seen when a high-energy photon, such as an X ray or a UV photon, excites an atom, which then de-excites through a series of longerwavelength emissions. A common example of fluorescence occurs in the colours produced when rocks and minerals containing mercury and many other elements are illuminated by UV photons. Below is the energy level diagram for some of the energy levels in mercury (Hg). The ground state has been assigned an energy of 0 eV. (a) the wavelength of the photon needed to excite the mercury atom from its ground state to the n 5 energy level (b) the longest wavelength of photon that will be emitted as the mercury atom de-excites (c) the number of possible downward transitions that can occur (d) If an atom had 100 possible energy transitions that were very close together, what would the spectrum produced by this atom look like Hg Extension 9.23 eV 8.85 eV 7.93 eV 6.70 eV 4.89 eV 0 eV 12. A laser produces light that is monochromatic, coherent, and collimated. (a) Explain each of these three properties. (b) Describe the spectrum of a laser. e TEST To check your understanding of Bohr\u2019s model of the atom, follow the eTEST links at www.pearsoned.ca/school/physicssource. Chapter 15 Electric force and energy quantization determine atomic structure. 781 15-PearsonPhys30-Chap15 7/24/08 4:00 PM Page 782 15.5 The Quantum Model of the Atom Bohr\u2019s model explains the spectral lines of hydrogen and accurately predicts the size and ionization energy of the hydrogen atom. Despite these remarkable accomplishments, the theory has several serious failings: \u2022 It does not really explain why energy is quantized, nor why orbiting electrons do not radiate electromagnetic energy. \u2022 It is not accurate for atoms that have two or more electrons. \u2022 It does not explain why a magnetic field splits the main spectral lines into multiple closely spaced lines. The Dutch physicist Pieter Zeeman discovered this effect in 1896. It is known as the Zeeman effect. Physicists solved these problems within 15 years, but the solutions were even more radical than Bohr\u2019s theory! The Wave Nature of Electrons In 1924, Louis de Broglie developed his theory that particles have wave properties. As described in section 14.4, diffraction experiments confirmed", " that electrons behave like waves that have the wavelength predicted by de Broglie. So, the principles of interference and standing waves apply for electrons orbiting a nucleus. For most sizes of orbit, successive cycles of the electron wave will be out of phase, and destructive interference will reduce the amplitude of the wave (Figure 15.24). For constructive interference to occur, the circumference of the orbit must be equal to a whole number of wavelengths: 2rn n where n is a positive integer, is the electron wavelength, and rn is the radius of the nth energy level. By substituting de Broglie\u2019s definition for wavelength, h mv, into the above equation, the condition for constructive interference becomes 2rn nh mv or mvrn nh 2 This condition is fundamentally the same one that Bohr found was necessary for the energy levels in his model of the atom. Thus, the wave nature of matter provides a natural explanation for quantized energy levels. Figure 15.25 shows the standing waves corresponding to the first three energy levels in an atom. Note that the de Broglie wavelength is longer in each successive energy level because the electron\u2019s speed decreases as the radius of the orbit increases. PHYSICS INSIGHT Recall from section 13.5 that for constructive interference to occur, the path difference between waves must be a whole number of wavelengths, or n. constructive interference destructive interference Figure 15.24 A standing wave is possible only if a whole number of electron wavelengths fit exactly along the circumference of an orbit. 782 Unit VIII Atomic Physics 15-PearsonPhys30-Chap15 7/24/08 4:00 PM Page 783 n 1 n 2 n 3 Figure 15.25 Standing waves in the first three energy levels, where 2r1 2r2 2, and 2r3 1, 3 In 1926, Erwin Schr\u00f6dinger (1887\u20131961) derived an equation for determining how electron waves behave in the electric field surrounding a nucleus. The solutions to Schr\u00f6dinger\u2019s equation are functions that define the amplitude of the electron wave in the space around a nucleus. Max Born (1882\u20131970) showed that the square of the amplitude of these wave functions at any point is proportional to the probability of finding an electron at that point. Each wave function defines a different probability distribution or orbital. info BIT Although Born won a Nobel Prize for his work on quantum theory, Schr\u00f6dinger never accepted Born\u2019s interpretation of electron wave functions. orbital: probability distribution of an electron in an atom Quantum Indeterminacy Unlike the Bohr model, the quantum model does not have electrons orbiting at precisely defined distances from the nucleus. Instead, the electrons behave as waves, which do not have a precise location. The orbitals in the quantum model show the likelihood of an electron being at a given point. They are not paths that the electrons follow. The idea that electrons within an atom behave as waves rather than as orbiting particles explains why these electrons do not radiate electromagnetic energy continuously. Some physicists, including Einstein and Schr\u00f6dinger, had difficulty accepting a quantum model that could predict only probabilities rather than clearly defined locations for electrons in an atom. As Niels Bohr noted, \u201cAnyone who is not shocked by quantum theory has not understood a single word.\u201d Despite its challenging concepts, quantum theory is the most comprehensive and accurate model of atoms and molecules yet developed. Concept Check Soon after Bohr\u2019s model was published, physicists discovered that the spectral lines in hydrogen and other elements were not distinct, but could themselves be split into numerous, very closely spaced spectral lines. How does the splitting of spectral lines show that Bohr\u2019s concept of energy levels is incomplete? Chapter 15 Electric force and energy quantization determine atomic structure. 783 15-PearsonPhys30-Chap15 7/24/08 4:00 PM Page 784 15.5 Check and Reflect 15.5 Check and Reflect Knowledge Extensions 1. Describe three failings of the Bohr model. 2. What is the Zeeman effect? 3. What is an orbital? Applications 4. (a) Find the de Broglie wavelength for an electron in the ground state for hydrogen. Recall that the ground-state radius for hydrogen is 5.29 1011 m. (b) Find the momentum of the electron in part (a). (Hint: Use the formula for the de Broglie wavelength.) (c) Find the kinetic energy and speed for the electron in part (a). 5. (a) Express the de Broglie wavelength of the n 2 and n 3 energy levels of an atom in terms of 1, the de Broglie wavelength for the n 1 energy level. (Hint: Apply Bohr\u2019s equation for the orbital radii.) (b) Make a scale diagram or model of the atom showing the electron waves for the first three energy levels. 6. In more advanced treatments of quantum mechanics, the wave solutions to", " the Schr\u00f6dinger equation can describe complex-looking orbitals around atoms. These solutions will sometimes have nodes. As you learned in Chapter 13, a node is a location where waves combine destructively and have zero amplitude. According to Born\u2019s interpretation of a wave function, what meaning would you give to a node in a wave function? 7. Radio astronomers use the 21-cm line to study hydrogen gas clouds in our galaxy. Use a library or the Internet to research this spectral line. What unusual transition causes the 21-cm line? Why does this transition occur naturally in hydrogen in deep space, but not in hydrogen on Earth? e TEST To check your understanding of the quantum model, follow the eTEST links at www.pearsoned.ca/school/physicssource. 784 Unit VIII Atomic Physics 15-PearsonPhys30-Chap15 7/24/08 4:00 PM Page 785 CHAPTER 15 SUMMARY Key Terms and Concepts cathode ray elementary unit of charge planetary model spectroscopy emission line spectrum absorption line spectrum spectrometer Fraunhofer line energy level stationary state principal quantum number Bohr radius ground state excited state ionization energy orbital Key Equations Velocity selection: no deflection if v E B r1n2, where r1 Emission lines: Ephoton Efinal Einitial and RH 1 1 n2 final 1 n2 initial Quantum model: mvrn nh 2 Bohr model for hydrogen: rn 5.29 1011 m, where E1 2.18 1018 J En E1 n2, where RH 1.097 107 m1 Conceptual Overview Summarize the chapter by copying and completing this concept map. mass and charge measured by discovery of electron showed quantization of charge led to raisin-bun model disproved by led to solar-system model predicts instability led to is basis of reveals energy quantization led to failings led to quantum model Chapter 15 Electric force and energy quantization determine atomic structure. 785 15-PearsonPhys30-Chap15 7/24/08 4:00 PM Page 786 CHAPTER 15 REVIEW Knowledge 1. (15.1) What is a cathode ray? What type of electric charge does a cathode ray carry? 2. (15.1) (a) Describe the force acting on a cathode ray moving to the right in an electric field directed down (toward the bottom of the page). (b) Describe the force acting on a cathode ray moving to the right in a magnetic field directed down (toward the bottom of the page). (c) How could the electric and magnetic fields be directed so that the net force on the cathode ray is zero? 3. (15.1) The glowing gas shows the path of the electrons in this gas discharge tube. How can you tell that the beam of electrons is travelling through a magnetic field? 4. (15.1) Why was Thomson\u2019s experiment able to determine only the charge-to-mass ratio for an electron? 5. (15.2) Explain how the results of Millikan\u2019s oil-drop experiment also enabled physicists to determine the mass of the electron. 6. (15.2) What is the electrical charge on a dust particle that has lost 23 electrons? 7. (15.2) Calculate the electrical charge carried by 1.00 kg of electrons. 8. (15.3) What is an alpha particle? 9. (15.3) Why is the Rutherford gold-foil experiment sometimes called a \u201cscattering experiment\u201d? 786 Unit VIII Atomic Physics 10. (15.3) Explain how Thomson\u2019s model of the atom was inconsistent with the results of Rutherford\u2019s gold-foil experiment. 11. (15.4) (a) What is an emission line spectrum? (b) How could you produce an emission line spectrum? 12. (15.4) What are Fraunhofer lines? 13. (15.4) Explain how emission line spectra demonstrate Planck\u2019s concept of energy quantization. 14. (15.4) What is the difference between the ground state of an atom and an excited state? 15. (15.4) Here are four energy-level transitions for an electron in a hydrogen atom: 4 \u2192 nf 8 \u2192 nf ni ni ni ni (a) For which of these transition(s) does the 1 \u2192 nf 2 \u2192 nf 5 5 3 3 atom gain energy? (b) For which transition does the atom gain the most energy? (c) Which transition emits the photon with the longest wavelength? 16. (15.4) Calculate the radius of a hydrogen atom in the n 3 state. 17. (15.5) Describe the difference between an orbit in Bohr\u2019s model of the atom and an orbital in the quantum model. 18. (", "15.5) How does the quantum model explain why electrons in an atom do not continuously radiate energy? Applications 19. An electron is moving at a speed of 1.0 km/s perpendicular to a magnetic field with a magnitude of 1.5 T. How much force does the magnetic field exert on the electron? 20. (a) Use the Bohr model to predict the speed of an electron in the n 2 energy level of a hydrogen atom. (b) Explain why the quantum model can predict the energy of this electron, but not its speed. 15-PearsonPhys30-Chap15 7/24/08 4:00 PM Page 787 21. Calculate the electric field that will suspend an oil droplet that has a mass of 2.0 1015 kg and a charge of 3e. 22. Transitions to or from the n 3 energy level produce the Paschen series of lines in the hydrogen spectrum. (a) Find the change in energy level for the first three Paschen transitions. (b) Find the wavelengths and frequencies of the first three Paschen transitions. (c) Use E hf to calculate the energy of the photon produced by a transition from the n 5 to the n 3 energy level. Is your calculation consistent with your answer to part (a)? Why or why not? (d) In what part of the electromagnetic spectrum would you find the Paschen lines? 23. The helium-neon laser produces a red-coloured light by exciting a gas that contains a mixture of both helium and neon atoms. The energy level diagram below shows three transitions, A, B, and C, that are involved. Two of these transitions are produced by collisions with other atoms or electrons, and the third is the result of photon emission. (a) Explain which process is involved in transitions A, B, and C. (b) For the transition that produces a photon, determine the wavelength of the photon. collision transfers energy 20.61 eV A 20.66 eV B 18.70 eV C ground state helium neon 24. Determine the electric field that will stop this alpha particle from being deflected as it travels at 10 km/s through a 0.25-T magnetic field. B 0.25 T [out of page] v 10 km/s [right] Extensions 25. Classical electromagnetic theory predicts that an electron orbiting a nucleus of charge q will radiate energy at a rate of P 2kq2a2 3c3, where k is Coulomb\u2019s constant, a is the electron\u2019s acceleration, and c is the speed of light. (a) Determine the kinetic energy of the electron in the ground state of a hydrogen atom. (b) Use the Bohr model to calculate the acceleration of an electron in the ground state of a hydrogen atom. (Hint: Apply the equations for circular motion.) (c) Show that P has the units of energy divided by time. (d) How long will it take the electron to give off all of its kinetic energy as electromagnetic radiation, assuming that the electron\u2019s acceleration remains constant? (e) Explain how your answer to part (d) shows that classical models of the atom are invalid. Consolidate Your Understanding 1. Explain how the Rutherford gold-foil experiment radically changed the understanding of atomic structure. 2. Explain how Bohr linked spectral lines to Planck\u2019s idea of energy quantization. 3. Outline the successes and failures of the Bohr model. 4. Describe three fundamental differences between the quantum model of the atom and the Bohr model. Think About It Review your answers to the Think About It questions on page 753. How would you answer each question now? e TEST To check your understanding of atomic structure, follow the eTEST links at www.pearsoned.ca/school/physicssource. Chapter 15 Electric force and energy quantization determine atomic structure. 787 16-PearsonPhys30-Chap16 7/24/08 4:32 PM Page 788 Nuclear reactions are among the most powerful energy sources in nature. \u201cI believe a leaf of grass is no less than the journey-work of the stars\u201d \u2014 from Leaves of Grass by Walt Whitman When the American poet Walt Whitman wrote this line in 1855, the reactions within stars were unknown, the nucleus had not been discovered, and there was no clear proof that atoms exist. By the late 1950s, however, a new interpretation of Whitman\u2019s words was possible. Astrophysicists had developed a theory that nuclear reactions inside massive stars created the heavy elements essential to life. Some of these stars exploded into supernovae, scattering heavy elements throughout the galaxy. This chapter describes nuclear reactions, the enormous potential energy in some nuclei, and the hazards and benefits of radioactive materials. You will learn about the processes that power the stars and how every leaf of grass may indeed be \u201cthe journey", "-work of the stars.\u201d C H A P T E R 16 Key Concepts In this chapter, you will learn about: half-life nuclear decay nuclear reactions Learning Outcomes When you have completed this chapter, you will be able to: Knowledge describe the nature and properties of nuclear radiation write nuclear equations for alpha, beta-negative, and beta-positive decays perform half-life calculations use conservation laws to predict the particles emitted by a nucleus compare and contrast fission and fusion reactions relate the mass defect of the nucleus to the energy released in nuclear reactions Science, Technology, and Society explain that the goal of science is knowledge about the natural world explain that technology meets given needs but should be assessed for each potential application Figure 16.1 A portion of the Cygnus Loop, an expanding cloud of hot gas formed by a supernova explosion about 15 000 years ago. This composite image was made using photographs from the Hubble Space Telescope. The blue colour is light from oxygen, green is light from hydrogen, and red is light from sulfur. 788 Unit VIII 16-PearsonPhys30-Chap16 7/24/08 4:32 PM Page 789 16-1 Inquiry Lab 16-1 Inquiry Lab Radiation Intensity Question Does the intensity of radiation depend on the distance from the source of the radiation? Hypothesis There is a mathematical relationship between the intensity of radiation and the distance from the radiation source. Variables \u2022 distance between radiation source and detector \u2022 reading on radiation detector Materials and Equipment cobalt-60 radiation source radiation detector metre-stick masking tape optional: interface for computer or graphing calculator CAUTION: The radioactive material is enclosed in a durable casing to prevent accidental absorption into the body. Do not damage this casing. Procedure 1 Make sure the radiation source is at least 3 m away from the radiation detector. Switch on the detector and measure the background radiation level for 5 min or more. Record this measurement, including the units. If you are using an interface with a computer or graphing calculator, check with your teacher about recording your data electronically. 2 Centre the cobalt-60 radiation source over the zero mark on the metre-stick, and tape the source in place. If your radiation source is shielded so that it emits radiation only from one side, align the source to direct the radiation along the metre-stick (Figure 16.2). Required Skills Initiating and Planning Performing and Recording Analyzing and Interpreting Communication and Teamwork 3 Place the radiation detector on the metre-stick within a few centimetres of the radiation source. Measure the radiation level for at least 1 min. Record the radiation level and the distance between the source and the detector. 4 Increase the separation between the radiation source and the detector in steps of 5 cm. Measure the radiation level for at least 1 min at each distance. Record measurements for at least six distances. radiation source radiation detector metre-stick Figure 16.2 Analyzing and Interpreting 1. Which variable is the manipulated variable in this experiment? 2. Explain why you need to know the background radiation level in order to determine how the intensity of the radiation varies with distance. 3. Graph your data. What type of relationship do you think the graph shows? 4. Discuss with your lab partners how you could use a different graph to determine the exact relationship between the radiation intensity and the distance from the radiation source. Produce a graph using the method that you think will work best. Explain your choice. 5. List any assumptions you made when analyzing your data. Forming Conclusions 6. Do your data support the hypothesis? Explain. Think About It 1. What is radioactivity? 2. Where does the energy released in a nuclear reaction come from? 3. How can stars create elements? Discuss your answers in a small group and record them for later reference. As you complete each section of this chapter, review your answers to these questions. Note any changes in your ideas. Chapter 16 Nuclear reactions are among the most powerful energy sources in nature. 789 16-PearsonPhys30-Chap16 7/24/08 4:32 PM Page 790 16.1 The Nucleus Section 15.3 described how scattering experiments directed by Rutherford showed that more than 99.9% of the mass of an atom is concentrated in a nucleus that is typically only a few femtometres (1015 m) in diameter. In 1918, Rutherford began a new series of experiments in which he bombarded nitrogen gas with alpha particles. He found that some of the nitrogen transmuted into oxygen and that the process also produced hydrogen nuclei. Rutherford concluded that the hydrogen nucleus was a fundamental particle that is a constituent of all nuclei. He called these particles protons, from protos, the Greek word for \u201cfirst.\u201d However, protons could not account for all of the mass of nuclei. For example, the charge-to-mass ratio for protons is twice that of helium nuclei. In 1920, Rutherford suggested that nuclei might also contain neut", "rons, neutral particles with about the same mass as a proton. Neutral particles are difficult to detect or measure because they do not interact with electric or magnetic fields. A variety of experiments over the next decade failed to find any neutrons. The breakthrough came in 1932 when James Chadwick showed that alpha rays striking a beryllium target produced radiation consisting of neutral particles. In a similar experiment with a boron target, he determined that the mass of a neutron is about 0.1% greater than the mass of a proton. femto: metric prefix meaning 1015 proton: a positively charged particle found in all nuclei neutron: a neutral particle found in nuclei info BIT Chadwick made two earlier attempts to discover the neutron, in 1923 and 1928. In 1935, he received the Nobel Prize in physics for his discovery. Nuclear Terms and Notation nucleon: a proton or neutron Protons and neutrons are called nucleons because they are both components of nuclei. Three numbers describe the composition of a nucleus: Atomic Number, Z: the number of protons in a nucleus Neutron Number, N: the number of neutrons in the nucleus Atomic Mass Number, A: the number of nucleons in the nucleus, Z N Scientists often indicate the composition of a nucleus with the notation A ZX, where X is the chemical symbol for the element. For example, a carbon nucleus with 6 protons and 6 neutrons has Z 6, N 6, and A 6 6 12. The notation for the carbon nucleus is 12 6C. Apply these terms and concepts in the next example. 790 Unit VIII Atomic Physics 16-PearsonPhys30-Chap16 7/24/08 4:32 PM Page 791 Example 16.1 How many neutrons are contained in a gold nucleus 197 79Au? Given Z 79 A 197 Required neutron number (N ) Analysis and Solution Since A Z N, N A Z 197 79 118 Paraphrase There are 118 neutrons in a nucleus of 197 79Au. Concept Check Practice Problems 1. How many neutrons are in a nucleus of 24 12Mg? 2. Find the atomic mass number for a uranium atom that contains 92 protons and 146 neutrons. Answers 1. 12 2. 238 How do the nuclei 12 6C, 13 6C, and 14 6C differ? How are they the same? Isotopes Many elements have two or more isotopes \u2014 forms that have the same number of protons (Z) but differing numbers of neutrons (N). For example, ordinary hydrogen (1 1H) are all isotopes of the element hydrogen. Specific isotopes can be indicated by the element name and the atomic mass number. For example, carbon-12 is another way of writing 12 1H), and tritium (3 1H), deuterium (2 6C. isotopes: atoms that have the same number of protons, but different numbers of neutrons All the isotopes of a particular element have the same number of protons and electrons. So, these isotopes have almost identical chemical properties. However, the physical properties can differ dramatically. In particular, one isotope of an element may be highly radioactive, while another is quite stable. Bombarding materials with electrons, neutrons, or other particles can create radioactive isotopes. Atomic Mass Units Atoms and nuclei are much, much smaller than everyday objects. So, even though a kilogram may be a convenient unit for expressing the mass of apples or oranges, it is not particularly useful for measuring the mass of a proton or a carbon nucleus. For calculations involving nuclei and subatomic particles, it is often convenient to use a mass unit that is much smaller than the kilogram. The atomic mass unit (u) is defined as exactly 1 12 of the mass of the carbon-12 atom: 1 u 1.660 539 1027 kg Chapter 16 Nuclear reactions are among the most powerful energy sources in nature. 791 16-PearsonPhys30-Chap16 7/24/08 4:32 PM Page 792 Table 16.1 lists the masses of electrons and nucleons, in both kilograms and atomic mass units. Table 16.1 Some Properties of Subatomic Particles (to Six Decimal Places) Particle Electron Proton Neutron Charge (C) Mass (kg) Mass (u) 1.602 177 1019 1.602 177 1019 0 9.109 383 1031 1.672 622 1027 1.674 927 1027 5.485 799 104 1.007 276 1.008 665 Forces in the Nucleus Aside from hydrogen, all nuclei consist of two or more protons and a number of neutrons (Figure 16.3). Like charges repel each other, so what keeps these nuclei from flying apart? Example 16.2 Can gravitational force bind two protons in a nucleus together? Given Rounding the values listed in Table 16.", "1 gives proton mass m 1.67 1027 kg and proton charge q 1.60 1019 C. Required Determine if gravitational force can bind two protons in a nucleus together. Figure 16.3 Any nucleus heavier than hydrogen has protons and neutrons packed closely together. Practice Problems 1. Calculate the gravitational force that two protons exert on each other when they are 5 fm apart. 2. Calculate the electrostatic force that two protons exert on each other when they are 5 fm apart. Answers 1. 7 1036 N 2. 9 N Analysis and Solution Compare the gravitational and electrostatic forces between two protons in a nucleus. The magnitude of the gravitational force is F. g The magnitude of the electrostatic force is F e Fg Fe So, m2 Gm 1 2 r q2 kq 1 2 r Gm1m2 kq1q2. Gm1m2 r 2 kq1q2 r 2. This ratio shows that the relative strength of the two forces does not depend on the distance between the protons. In order for the gravitational attraction between the protons to overcome the electrostatic repulsion, the ratio have to be greater than 1. Fg Fe would Substituting the known values into the ratio of the forces gives Fg Fe (6.67 1011 Nm2/kg2)(1.67 1027 kg)2 (8.99 109 Nm2/C2)(1.60 1019 C)2 8.08 1037 Paraphrase The gravitational attraction is vastly weaker than the electrostatic repulsion, so gravity cannot be the force that holds a nucleus together. 792 Unit VIII Atomic Physics 16-PearsonPhys30-Chap16 7/24/08 4:32 PM Page 793 strong nuclear force: the force that binds together the protons and neutrons in a nucleus PHYSICS INSIGHT Measurements of interactions between subatomic particles suggest that there is a fourth fundamental force, the weak nuclear force. This force acts on electrons. binding energy: the net energy required to liberate all of the protons and neutrons in a nucleus Since gravity is far too weak, there must be some other force that holds the particles in a nucleus together. Physicists call this force the strong nuclear force, and think that it is a fundamental force of nature, like gravity and the electrostatic force. The strong nuclear force has a very short range. Although it is more powerful than the electrostatic force within a nucleus, the strong nuclear force has a negligible effect on particles that are more than a few femtometres apart. The strong nuclear force acts on both neutrons and protons, but does not affect electrons. Chapter 17 describes fundamental forces in more detail. Binding Energy and Mass Defect Removing a nucleon from a stable nucleus requires energy because work has to be done on the nucleon in order to overcome the strong nuclear force. The binding energy, Eb, of a nucleus is the energy required to separate all of its protons and neutrons and move them infinitely far apart. In other words, the binding energy is the difference between the total energy of the separate nucleons and the energy of the nucleus with the nucleons bound together: Eb Enucleons Enucleus where Enucleons is the sum of the energies of the nucleons when they are free of the nucleus and Enucleus is the energy of the nucleus. Mass-energy Equivalence The equivalence of mass and energy is part of the theory of relativity that Albert Einstein developed in 1905. This theory correctly predicted that mass and energy are related by the equation E mc2 where E is energy, m is mass, and c is the speed of light. Earlier in this section, you learned that physicists commonly use the atomic mass unit, u, for calculations involving nuclei and subatomic particles. For nuclear calculations, it is useful to know the energy equivalent for 1 u: E 1 u c2 (1.660 539 1027 kg)(2.997 925 108 m/s)2 1.492 418 1010 J 1.492 418 1010 J 931.494 1 MeV 1 eV 1.602 177 1019 J Thus, 1 u is equivalent to about 149.2 pJ or 931.5 MeV. The binding energy of most nuclei is equivalent to only a small fraction of an atomic mass unit. Nuclear reactions can involve conversions between mass and energy. The law of conservation of energy still applies if the conversions are taken into account. For any closed system, the total of the energy and the energy equivalent of the mass in the system is constant. Chapter 16 Nuclear reactions are among the most powerful energy sources in nature. 793 16-PearsonPhys30-Chap16 7/28/08 10:14 AM Page 794 Example 16.3 Calculate the energy equivalent for 0.0034 u of mass, in joules and in electron volts.", " Analysis and Solution Simply multiply 0.0034 u by the appropriate equivalence factors: 0.0034 u 1.492 1010 J 1 u 5.1 1013 J 0.0034 u 931.5 MeV 1 u 3.2 MeV Paraphrase The energy equivalent for 0.0034 u is 5.1 1013 J or 3.2 MeV. Practice Problems 1. Find the energy equivalent, in electron volts, for 0.221 u. 2. Find the mass equivalent to 250 MeV. Answers 1. 206 MeV 2. 0.268 u Mass Defect Rearranging Einstein\u2019s equation for mass-energy equivalence gives m E c2. Dividing the equation for binding energy by c2 leads to a formula for the mass defect, m, of a nucleus: Enucleons c2 Eb c2 m mnucleons Enucleus c2 mnucleus where mnucleons is the sum of the masses of the separate nucleons and mnucleus is the mass of the nucleus. Thus, the mass of a nucleus is equal to the total mass of its constituents, less the mass corresponding to the binding energy. Physicists have determined the masses of atoms and nucleons with great accuracy. Tables of atomic data generally list the masses of neutral atoms rather than the masses of nuclei alone without any electrons. The following formula uses atomic masses to calculate the mass defect for a nucleus: m Zm1 1H Nmneutron matom where m1 number, and N is the neutron number. 1H is the mass of a neutral hydrogen atom, Z is the atomic Since m1 1H includes the masses of both a proton and an electron, the 1H includes the mass of Z electrons, matching the mass of the term Zm1 electrons included in matom. The differences in the binding energy of the electrons are small enough to ignore in most nuclear calculations. mass defect: difference between the sum of the masses of the separate nucleons and the mass of the nucleus PHYSICS INSIGHT Nuclear calculations often involve very small differences in mass. Such calculations can require data with six or more significant digits. 794 Unit VIII Atomic Physics 16-PearsonPhys30-Chap16 7/24/08 4:32 PM Page 795 Concept Check Show that m Zmproton Nmneutron mnucleus. Example 16.4 Find the mass defect, expressed in kilograms, and the binding energy for a carbon-12 nucleus. Given Z 6 A 12 m 12.000 000 u Required mass defect (m) binding energy (Eb) Analysis and Solution The formula N A Z gives the number of neutrons in the nucleus: N 12 6 6 Practice Problems 1. Sodium 23 11Na has an atomic mass of 22.989 769 u. Find the mass defect for this nucleus. 2. Find the binding energy for 23 11Na. Answers 1. 0.200 286 u 2. 186.6 MeV 6C nucleus consists of 6 neutrons and 6 protons. Thus, the 12 Now, use m Zm1 1H Use mass data from Tables 7.5 and 7.6 on page 881. Recall that 1 u 1.660 539 1027 kg (page 791). m ZmH matom to find the mass defect. Nmneutron Nmneutron matom 6(1.007 825 u) 6(1.008 665 u) 12.000 000 u 0.098 940 u 1.660 539 1027 kg 1 u 1.6429 1028 kg Use the mass-energy equivalence to calculate the binding energy from the mass defect. 1 u 1.492 1010 J 931.5 MeV 0.098 940 u 1.492 1010 J 1 u 931.5 MeV 1 u or 0.098 940 u Eb 1.476 1011 J 92.16 MeV Paraphrase The mass defect for 12 the carbon-12 nucleus is 1.476 1011 J or 92.16 MeV. 6C is 1.6429 1028 kg. The binding energy of Binding Energy per Nucleon You can compare the stability of different nuclei by dividing the binding energy of each nucleus by the number of nucleons it contains. The, the more stable the nucleus greater the binding energy per nucleon is. Figure 16.4 is a graph of binding energy per nucleon versus atomic mass number for stable nuclei. This graph peaks at about 8.79 MeV per 58Fe, and nucleon. The three most stable isotopes are nickel 28 iron 26 62Ni, iron 26 Eb A 56Fe. Chapter 16 Nuclear reactions are among the most powerful energy sources in nature. 795 16-PearsonPhys30-Chap16 7/24/08 4:32 PM Page 796 Figure 16.4 Binding energy per nucleon", " for stable isotopes ) 10 8 6 4 2 0 16 8 O 56 26 Fe 120 50 Sn 238 92 U 4 2 He 3 2 2 1 He H 50 100 150 200 250 Atomic Mass Number, A The graph also gives a hint about the process that causes the stars to shine. The binding energy per nucleon is much less for hydrogen than for helium. If hydrogen atoms combine to form helium, the nucleons move to a lower energy level and give off the difference in energy. In section 16.4, you will learn more about such nuclear reactions. 16.1 Check and Reflect 16.1 Check and Reflect Knowledge 10. Show that MeV/c2 has the dimensions 1. How many protons and neutrons do each of the following nuclei contain? (a) 90 13C (c) 56 (b) 6 26Fe 38Sr (d) 1 1H 2. Convert 1.6 1010 J to electron volts. 3. Calculate the energy equivalent of 0.25 u. 4. How much mass is converted into energy by a nuclear reaction that produces 5.00 GJ of energy? 5. Define the term isotope. 6. Explain why the mass of a stable nucleus is a bit less than Zmproton Nmneutron. Applications 7. Determine the binding energy for 10 22Ne. The atomic mass of 10 22Ne is 21.991 385 u. 8. The 19 40K isotope of potassium has an atomic mass of 39.963 998 u. (a) Determine the mass defect for 19 (b) Calculate the binding energy per 40K. nucleon for this isotope. of mass. Extensions 11. (a) Contrast the strength and range of the electromagnetic force and the strong nuclear force. (b) Explain how the nature of these forces limits the maximum possible size for nuclei. 12. Suppose that the electrostatic force were much stronger. Describe how this change would affect the stability of nuclei. 1.20 fm and A is 13. Experiments have shown that most nuclei are approximately spherical with a radius 1 of r r0A, where r0 3 the atomic mass number. Use this formula to determine the radius of the nucleus of a 90Sr atom. Then estimate the distance 38 between adjacent nucleons in this nucleus. What can you conclude about the size of protons and neutrons? 9. Use Figure 16.4 to estimate the binding e TEST energy for each of these nuclei: (a) 13 56Fe (c) 238 (b) 26 92U 6C To check your understanding of nuclei, follow the eTEST links at www.pearsoned.ca/school/physicssource. 796 Unit VIII Atomic Physics 16-PearsonPhys30-Chap16 7/24/08 4:32 PM Page 797 16.2 Radioactive Decay The French physicist, Antoine Henri Becquerel (1852\u20131908), discovered radioactive decay in 1896 while conducting an experiment to see if a fluorescent compound of uranium would emit X rays when exposed to sunlight. During a period of cloudy weather, Becquerel put the uranium compound away in a drawer along with a photographic plate wrapped in black paper. When he developed the plate several days later, he was surprised to find that it was fogged even though the fluorescent compound had not been exposed to sunlight. Becquerel realized that the radiation that fogged the plate must be coming from the uranium in the compound. He also found that a magnetic field would deflect some of this radiation. The husband and wife team of Marie Curie (1867\u20131934) and Pierre Curie (1859\u20131906) began an extensive study of this radiation. They showed that thorium was also radioactive, and discovered two new elements, radium and polonium, that were both much more radioactive than uranium. Indeed, Marie coined the term radioactive. She also demonstrated that the intensity of radiation from uranium compounds was not affected by the other elements in the compound or by processes such as being heated, powdered, or dissolved. The intensity depended only on the quantity of uranium. Therefore, the radioactivity must result from a process within the uranium nucleus. Rutherford and others identified three forms of nuclear radiation: Alpha (): the emission of a helium nucleus Beta (): the emission of a high-energy electron Gamma (): the emission of a high-energy photon Initially, this classification was based on how much material each type of radiation could penetrate. In radiation from naturally occurring isotopes, the alpha particles typically do not penetrate much more than a thin metal foil or sheet of paper, whereas beta particles can pass through up to 3 mm of aluminium, and gamma rays can penetrate several centimetres of lead. The three types of radiation result from different processes within nuclei. Concept Check Figure 16.5 shows the paths that,, and rays take when passing through a magnetic field. What can you conclude about", " the electrical properties of these rays? \u03b2 \u03b3 radiation source \u03b1 Figure 16.5 The paths of,, and rays in a magnetic field info BIT Marie Curie was the first person to win two Nobel Prizes. She died of leukemia, almost certainly the result of years of exposure to radiation in her laboratory. Chapter 16 Nuclear reactions are among the most powerful energy sources in nature. 797 16-PearsonPhys30-Chap16 7/24/08 4:32 PM Page 798 16-2 Design a Lab 16-2 Design a Lab Radiation Shielding The Question What common materials provide effective shielding against,, and radiation? Figure 16.6 Radiation meters Design and Conduct Your Investigation Check with your teacher about the radiation sources and radiation meters (Figure 16.6) available for this investigation. Then design your experiment. List the materials you will need and outline the procedure. Try this procedure and modify it if necessary. Keep careful records of your results. Then analyze your data, and explain your conclusions. Conservation Laws and Radioactive Decay In addition to conserving momentum and energy, all radioactive decay processes obey these additional conservation laws: \u2022 Charge: The net electrical charge cannot change in a decay process. Any change in the electrical charge of the nucleus must be exactly offset by an opposite change elsewhere in the system. For example, if the charge on a nucleus decreases by 2e, then a particle with a charge of 2e must be emitted. \u2022 Atomic Mass Number: The total of the atomic mass numbers for the final products must equal the atomic mass number of the original nucleus. In other words, the total number of nucleons remains constant. 798 Unit VIII Atomic Physics 16-PearsonPhys30-Chap16 7/24/08 4:32 PM Page 799 Example 16.5 Determine which of these radioactive decay processes are possible. (a) 214Po \u2192 82 84 208Pb 4 2 226Ra 4 2 28Ni 1 0n (b) 230Th \u2192 88 90 27Co \u2192 60 (c) 60 (1 0n represents a neutron) Analysis and Solution Compare the charge and atomic mass number of the original nucleus to those of the decay products. (a) Charge: 84 82 2 Atomic mass number: 214 208 4 The decay process 214 (b) Charge: 90 88 2 84Po \u2192 208Pb 82 4 is not possible. 2 Atomic mass number: 230 226 4 The decay process 230 (c) Charge: 27 28 0 90Th \u2192 88 226Ra 4 is possible. 2 Atomic mass number: 60 60 1 The decay process 27 60Co \u2192 60 28Ni 1n is not possible. 0 Practice Problems Determine whether these decay processes are possible. 210Rn 4 2 0 233U 1 1H 212Po \u2192 86 84 233Pa \u2192 92 91 14C \u2192 7 6 14N 1 2. 3. 1. Answers 1. Impossible 2. Possible 3. Impossible Concept Check Why are electrons not considered when applying the conservation law for atomic mass number? Alpha Decay In 1908, Rutherford showed that alpha particles are helium nuclei spontaneously emitted by unstable large nuclei. In these nuclei, the electromagnetic force repelling the outer protons is almost as great as the attractive strong nuclear force. Such nuclei can spontaneously emit alpha particles. Because a cluster of two protons and two neutrons forms a highly stable helium nucleus, these unstable large nuclei decay by emitting alpha particles rather than separate protons and neutrons. The emission of an alpha particle decreases the atomic number by 2 and the atomic mass number by 4. For example, alpha decay of uranium-238 produces thorium: 238 92U \u2192 90 234Th 4 2 In this example, uranium is the parent element and thorium is the daughter element. Applying the conservation laws gives this general form for alpha decays: A ZX \u2192 A4 Z2Y 4 2 parent element: the original element in a decay process daughter element: the element produced by a decay process where X is the chemical symbol for the parent element and Y is the symbol for the daughter element. Here, A is the atomic mass number of the parent element and Z is its atomic number. Chapter 16 Nuclear reactions are among the most powerful energy sources in nature. 799 16-PearsonPhys30-Chap16 7/24/08 4:32 PM Page 800 Example 16.6 Practice Problems Write the -decay process for these elements, and name the parent and daughter elements. 1. 2. 3. 230Th 90 238U 92 214Po 84 Answers 1. 90 2. 92 3. 84 230Th \u2192 226 238U \u2192 90 214Po \u2192 82 88Ra 4 ; thorium, radium 2 234Th 4 ; uranium, thorium 2 210Pb 4 ; polonium, lead 2 Predict the daughter element that results from alpha decay of radium-226. Analysis and Solution From a periodic table, you can see that the atomic number for radium is 88. So, the parent element is 226 Since the alpha particle", " carries away four nucleons, including two protons, A decreases by 4 and Z decreases by 2: ZX \u2192 A4 A So, the daughter element is 882 The periodic table shows that the element with Z 86 is radon. 2264Y 222 Z2Y 88Ra. 86Y. 4 2 Paraphrase For alpha decay, the daughter element of radium-226 is radon-222. Energy Released During Alpha Decay You can apply the concepts of energy conservation and mass-energy equivalence to alpha decay, using a method similar to the calculation of nuclear binding energy. The mass-energy of the parent nucleus is equal to the sum of the mass-energy and the kinetic energies of both the daughter nucleus and the alpha particle: mparentc2 mdaughterc2 mc2 E The difference in energy, E, appears as the total kinetic energy of the alpha particle and of the daughter nucleus. If the parent nucleus was at rest, the law of conservation of momentum requires the momentum of the alpha particle to be equal in magnitude and opposite in direction to the momentum of the daughter nucleus. Usually, the mass of the daughter nucleus is much greater than the mass of the alpha particle. So, the speed of the alpha particle is correspondingly greater than the speed at which the daughter nucleus recoils: mv mdaughtervdaughter v mdaughtervdaughter m 800 Unit VIII Atomic Physics 16-PearsonPhys30-Chap16 7/24/08 4:32 PM Page 801 The kinetic energy of the alpha particle is also correspondingly greater than the kinetic energy of the daughter nucleus: 1 E mv 2 2 1 m 2 mdaughter m mdaughter m mdaughtervdaughter m 2 mdaughterv 2 daughter 1 2 Edaughter Concept Check Explain why E must be positive in order for -decay to occur. Example 16.7 Show that -decay of radium-226 is possible, and estimate the maximum kinetic energy of the emitted alpha particle. Given Parent atom is radium-226. Required maximum kinetic energy of the alpha particle Practice Problems Calculate the energy released during -decay of these nuclei: 1. 2. 3. 230Th 90 238U 92 214Po 84 Answers 1. 7.641 1013 J 2. 6.839 1013 J 3. 1.255 1012 J Analysis and Solution Example 16.6 showed that the daughter element is radon-222. The energy released is equivalent to the difference between the mass of the parent atom and the total mass of the products. m mparent Table 7.5 on page 881 lists the atomic masses for radium-226, radon-222, and helium-4. As in section 16.1, you can use atomic masses instead of nuclear masses because the masses of the electrons will balance out. A radon nucleus has over 50 times the mass of an alpha particle. So, the alpha particle will have over 98% of the total kinetic energy, E. m m226 mproducts 88Ra (m 222 m4 ) 2 226.025 410 u 222.017 578 u 4.002 603 u 0.005 229 u 86Rn E 0.005 229 u 1.492 1010 J 1 u 7.802 1013 J or 0.005 229 u 4.871 MeV 931.5 MeV 1 u Paraphrase Since E 0, alpha decay of radium-226 is possible. The maximum kinetic energy of the alpha particle when emitted is about 7.802 1013 J or 4.871 MeV. Chapter 16 Nuclear reactions are among the most powerful energy sources in nature. 801 16-PearsonPhys30-Chap16 7/24/08 4:32 PM Page 802 THEN, NOW, AND FUTURE Ionization Smoke Detectors Most household smoke detectors (Figure 16.7) contain a small amount of americium-241. This isotope emits -particles, which ionize air molecules between two metal plates within the smoke detector. One of the plates has a positive charge, and the other plate has a negative charge. The plates attract the ions, so a small current flows between the plates. If smoke particles enter the smoke detector, they absorb some of the -particles. So, the alpha radiation ionizes fewer air molecules Figure 16.7 This smoke detector uses alpha radiation to sense smoke particles. and the current between the metal plates decreases. This drop in current triggers the alarm circuit in the smoke detector. Questions 1. Why is it safer for a smoke detector to use alpha radiation, instead of beta or gamma radiation? 2. Suggest reasons why most manufacturers of smoke detectors recommend replacing them after 10 years. beta-negative () decay: nuclear decay involving emission of an electron beta () particle: electron emitted by a nucleus Beta Decay Sometimes, a nucleus decays by emitting an electron. This process is termed beta-negative or decay. During decay, a neutron in the nucleus transforms into a proton, electron, and an extremely", " small neutral particle known as antineutrino, symbol \u00af (Figure 16.8). So, the atomic number of the atom increases by 1, but the atomic mass number does not change. Charge is conserved because the charge on the new proton balances the charge on the electron emitted from the nucleus. This electron is often called a beta () particle, a name that originates from the 0 in equaearly classification of types of radiation. It is often written as 1 tions. For example, decay of thallium-208 produces lead, and the equation is: 208Tl \u2192 81 208Pb 82 0 \u03c5\u2013 \u20131 n e p \u03c5 Figure 16.8 During decay, a neutron changes into a proton, electron, and antineutrino. Concept Check Why is the mass of the neutron slightly larger than the sum of the proton and electron masses? 802 Unit VIII Atomic Physics 16-PearsonPhys30-Chap16 7/24/08 4:32 PM Page 803 Example 16.8 What element will the decay of thorium produce? Analysis and Solution A periodic table shows that the atomic number for thorium is 90. decay increases the atomic number by 1, so Adaughter The element with an atomic number of 91 is protactinium, the element immediately after thorium in the periodic table. 91. Paraphrase For decay of thorium, the daughter element is protactinium. Practice Problems 1. Find the elements produced by decay of (a) 228 88Ra (b) 212 82Pb Answers 1. (a) 228 (b) 212 89Ac 83Bi As with alpha decays, you can use atomic masses to calculate how much energy a beta decay will release. Example 16.9 How much energy would you expect the decay of a thorium-234 nucleus to release? Given Parent element is 234 90Th. Required Energy released by decay (E) Practice Problems 1. (a) What element does the decay of cobalt-60 produce? (b) How much energy would you expect the decay of a cobalt-60 nucleus to release? Answers 1. (a) 60 28Ni (b) 2.823 MeV Analysis and Solution As shown in Example 16.8, the daughter element is protactinium. However, this daughter atom has only the 90 electrons from the original thorium atom because the electron emitted by the thorium nucleus leaves the atom as beta 91Pa. The energy released radiation. The result is a positive ion, 234 is equivalent to the difference between the mass of the parent atom and the total mass of the decay products. Together, the masses of the protactinium ion and the beta particle equal the mass of a neutral protactinium atom. Table 7.5 on page 881 lists the atomic masses. m mparent m 234 m 234 91 Pa 234.043 601 u 234.043 308 u 0.000 293 u mproducts (m 234 m 234 m 0 1 90 Th 90 Th 91 Pa ) 1 u is equivalent to about 931.5 MeV, so E 0.000 293 u 931.5 MeV 1 u 0.2729 MeV Paraphrase The decay of a 234 90Th nucleus should release 0.2729 MeV. PHYSICS INSIGHT In calculating energy produced in nuclear decay, it is common to use atomic masses because these data are readily available. You see this being done in Examples 16.9 and 16.10. In both of these examples, as an intermediate step an ion notation has been used to account for the change in nuclear charge that happens during beta decay. In reality, in beta decay it is most likely that the atom will not end up ionized. The atom will either lose or gain an electron as appropriate. Chapter 16 Nuclear reactions are among the most powerful energy sources in nature. 803 16-PearsonPhys30-Chap16 7/24/08 4:32 PM Page 804 info BIT The name neutrino comes from the Italian word for \u201clittle, neutral one.\u201d The word was coined by Enrico Fermi, a renowned physicist who developed a theory to explain beta decay. neutrino: an extremely small neutral subatomic particle e WEB To learn more about the Sudbury Neutrino Observatory, follow the links at www.pearsoned.ca/school/ physicssource. info BIT Each second, more than 100 trillion neutrinos pass through your body! Almost all of these neutrinos were formed by nuclear reactions in the core of the Sun. weak nuclear force: fundamental force that acts on electrons and neutrinos antimatter: form of matter that has a key property, such as charge, opposite to that of ordinary matter ): an antipositron (e or 0 1 electron; a positively charged particle with its other properties the same as those of an electron The Elusive Neutrino Since the daughter nucleus has vastly more", " mass than an electron, there is practically no recoil of the daughter nucleus during beta decay. Consequently, physicists expected that virtually all of the energy released during decay would appear as the kinetic energy of the electron emitted by the nucleus. However, measurements found that most electrons emitted during decay had somewhat less kinetic energy than expected, and a few had almost no kinetic energy. During decay, small portions of the mass of the parent nuclei seemed to just disappear! In 1930, the Austrian physicist Wolfgang Pauli (1900\u20131958) suggested that the missing energy in beta decay was carried away by a tiny, as-yetundiscovered neutral particle, now called the neutrino,. Neutrinos are so small that physicists have yet to determine their size and mass. These \u201cghost-like\u201d particles can pass through Earth with only a slight chance of being absorbed! Indeed, it was 1956 before an experiment using the intense radiation at a nuclear power plant finally proved conclusively that neutrinos actually exist. Eventually physicists discovered that there are actually two kinds of neutrinos given off in beta decay. In decay an antineutrino is released. As you will soon see, in decay a neutrino is released. The neutrino and antineutrino are identical in all respects except for their opposite spins. Many astrophysicists now think that neutrinos play a critical role in the cores of stars and perhaps in the structure of the cosmos as well. Concept Check How did physicists know that the neutrino must be neutral? Beta Decay, the Weak Nuclear Force, and Antimatter Careful study of beta decays revealed two further important differences from alpha decay. First, the transformation of a neutron into a proton involves a fundamental force called the weak nuclear force. Although it is less powerful than the strong nuclear force, the weak nuclear force acts on electrons and neutrinos, whereas the strong nuclear force does not. The second difference is that beta decay involves antimatter. An antimatter particle has a key property, such as charge, opposite to that of the corresponding particle of ordinary matter. For example, an anti0), has a positive charge but the same mass electron, or positron (e or 1 as an electron. Section 17.2 presents antimatter in more detail. In decay, the transformation of a neutron into a proton produces an antineutrino rather than a neutrino: n \u2192 p 1 0 where is the symbol for the antineutrino. Thus, decays have the general form ZX \u2192 A A Z1Y 1 0 (Z increases by 1) 804 Unit VIII Atomic Physics 16-PearsonPhys30-Chap16 7/24/08 4:32 PM Page 805 A second form of beta decay also produces an antiparticle. In decay, a proton transforms into a neutron, and the parent nucleus emits a positron and a neutrino: beta-positive () decay: nuclear decay involving emission of a positron ZX \u2192 A A Z1Y 0 (Z decreases by 1) 1 Sometimes, you will see the electron in these decay processes represented by the symbol e. To distinguish this electron from those orbiting the 0 to represent an electron nucleus, this chapter uses the symbol 1 0 is used to represent an emitted by a nucleus. Similarly, the symbol 1 emitted positron. Example 16.10 Nitrogen-13 (13 by decay. 7N) transmutes into carbon-13 (13 6C) 13N \u2192 7 13C 6 0 1 Calculate the energy released if the atomic masses are 13.005 739 for nitrogen-13 and 13.003 355 for carbon-13. Given Nitrogen-13 transmutes into carbon-13 by decay. Atomic masses: 13.005 739 for nitrogen-13, 13.003 355 for carbon-13 Required Energy released in the decay Analysis and Solution The energy released is equivalent to the difference between the mass of the parent atom and the total mass of the products. Practice Problems 1. (a) What isotope will decay of thallium-202 produce? (b) Write the process for this decay. (c) How much energy will be released by the decay of the thallium-202 nucleus if the mass of the thallium nucleus decreases by 0.001 463 u? Answers 1. (a) mercury-202 81Tl \u2192 80 (b) 202 (c) 0.3400 MeV 202Hg 0 1 m mparent m 13 7N m 13 7N mproducts 6C m 0 (m 13 m 0 (m 13 6C 1 ) m 0 1 Note that again we use an ion notation indicating the presence of a carbon ion. In fact, at the end of the decay process, the carbon ion will lose an electron, and its mass can be written as (m 13 ) as shown above. ) 1 m 0 1 6C Since electrons and positrons have the same mass,", " m 0 1 Therefore, m m 13 7N 2m 0 1 (m 13 6C ) 13.005 739 u [13.003 355 u 2(0.000 549) u] 0.001 286 u m 0 1. 1 u is equivalent to 931.5 MeV, so E 0.001 286 u 931.5 MeV 1 u 1.198 MeV Paraphrase The decay of a 7 13N nucleus should release 1.198 MeV of energy. PHYSICS INSIGHT In Example 16.10, the energy released when a nitrogen-13 nucleus decays to form carbon-13 is calculated to be 1.198 MeV. Most nuclear decay data tables, however, will indicate that the total energy released in this decay is 2.221 MeV. Both are correct! When a decay occurs, a positron is emitted. This positron could combine with an electron to release the energy equivalence of 2 electron masses, an additional 1.023 MeV. Chapter 16 Nuclear reactions are among the most powerful energy sources in nature. 805 16-PearsonPhys30-Chap16 7/24/08 4:32 PM Page 806 Gamma Decay (-decay) Many nuclei have a series of energy levels that correspond to different configurations of the nucleons. In the excited states, the nucleons are farther apart. As a result, their binding energy is less than when in the ground state, and the total energy of the nucleus is greater. When making a transition to a lower-energy state, a nucleus emits a gamma-ray photon, similar to the photon emitted when an electron in an atom moves to a lower energy level (Figure 16.9). However, the difference in energy is much greater for a nucleus. Gamma () decay does not change the atomic number or the atomic mass number. Gamma decays can be written using this general form: A AX Z ZX* \u2192 where * indicates an excited state and represents a gamma ray. Often, alpha or beta decay leaves the daughter nucleus in a highly excited state. The excited nucleus then makes a transition to its ground state, and emits a gamma ray. For example, when decay of boron-12 produces carbon-12, the carbon nucleus is highly excited and quickly emits a gamma ray: 12 5B \u2192 6 6C* \u2192 6 12 12C* 0 1 12C The energy of a gamma ray depends on the energy levels and the degree of excitation of the particular nucleus. Gamma rays can have energies ranging from thousands to millions of electron volts. Stability of Isotopes Figure 16.10 shows that stable isotopes form a relatively narrow band when plotted by their proton and neutron numbers. Other than hydrogen, all stable isotopes have at least as many neutrons as protons. As Z increases, the isotopes require an increasing ratio of neutrons to protons in order to be stable. There are no completely stable isotopes with more than 83 protons. The stable isotopes have greater binding energies than the unstable isotopes. Radioactive decay transmutes unstable nuclei into nuclei with higher binding energies. For example, heavy nuclei above and to the right of the stable band will emit alpha particles (larger red arrows), heavy nuclei below and to the right of the band will emit positrons, or particles (small red arrows), and lighter nuclei to the left of the band will emit electrons, or particles (blue arrows). All of these decay processes produce isotopes that are either in the stable band or closer to it. A nucleus may undergo several successive decays before it reaches the stable band 15 10 5 0 15.1 12.7 9.64 7.65 4.44 ground state Figure 16.9 Nuclear energy levels for carbon-12: How do these energy levels differ from those for electrons in hydrogen? gamma () decay: emission of a high-energy photon by a nucleus e WEB To learn more about nuclear energy levels, follow the links at www.pearsoned.ca/ school/physicssource. transmute: change into a different element \u03b1 decays \u03b2 decays N Z \u03b2 decays 160 150 140 130 120 110 100 90 80 70 60 50 40 30 20 10 10 20 30 40 50 60 70 80 Atomic Number, Z 90 100110 Figure 16.10 The black dots represent the band of stable isotopes. 806 Unit VIII Atomic Physics 16-PearsonPhys30-Chap16 7/24/08 4:32 PM Page 807 Radioactive Decay Series Often, a radioactive nucleus will decay into a daughter nucleus that is itself radioactive. The daughter nucleus may then decay into yet another unstable nucleus. This process of successive decays continues until it creates a stable nucleus. Such a process is called a radioactive decay series. \u03b2 decay \u03b1 decay 238 234 230 226 222 218 214 210 206 80 Hg 81 Tl 82 Pb 83 Bi 84 Po 85 At 86 Rn 87 Fr 88 Ra 89 Ac 90 Th 91", " Pa 92 U Atomic Number, Z Figure 16.11 How many different decay paths are there from uranium-238 to lead-206? Radioactive decay series beginning with 238 92U and ending with 206 82Pb. The dots in Figure 16.11 represent nuclei that are part of the decay series. A decay series can have several branches that lead to the same final product. Figure 16.11 shows that 218 84Po by three different combinations of decays. All of the intermediate isotopes in a decay series are unstable, but the degree of instability is different for each isotope. For example, 218 86Rn usually lasts for only a fraction of a second whereas 222 90Th takes thousands of years. Although not shown in Figure 16.11, many of the intermediate isotopes undergo gamma decay. 86Rn takes several days to decay and 230 84Po can transmute into 214 Concept Check Explain why gamma decays cannot be shown as paths on a decay series graph like the one in Figure 16.11. Chapter 16 Nuclear reactions are among the most powerful energy sources in nature. 807 16-PearsonPhys30-Chap16 7/24/08 4:32 PM Page 808 Potential Hazards of Nuclear Radiation Alpha, beta, and gamma radiation are all invisible, and most of their effects on the human body are not immediately apparent. As a result, it was not until years after the discovery of radioactive decay that researchers realized how dangerous radiation can be. Radiation poses two major types of risk: \u2022 Radiation Sickness: Radiation can ionize cellular material. This ionization disrupts the intricate biochemistry of the body, resulting in radiation sickness. Large doses of ionizing radiation can kill cells. Blood cells and the lining of the intestine are particularly vulnerable. Symptoms include nausea, vomiting, diarrhea, headache, inflammation, and bleeding. Severe radiation sickness is often fatal. \u2022 Genetic Damage: High-energy particles and gamma rays can alter DNA, and lead to the development of cancers or harmful mutations. These effects often appear 10 to 15 years after radiation exposure. Everywhere on Earth, there is some naturally occurring radiation from cosmic rays and from radioisotopes in the ground. This background radiation causes some minor damage, but normally the body can repair such damage without any lasting harm. The effect of radiation on living organisms depends on the energy it carries, its ability to ionize atoms and molecules, and the depth to which it can penetrate living tissue. The charge and energy of the radiation determine how ionizing it is. The energy also affects how far the radiation can penetrate. The energy that a radiation has depends on the process that produces it. Table 16.2 compares the hazards posed by typical radiations from natural sources. Table 16.2 Radiation Hazards from Natural Sources Outside the Body Radiation Typical Penetration Ionization Hazard alpha beta gamma Travels about 5 cm in air. Cannot penetrate skin. high Travels about 30\u201350 cm in air. Penetrates about 1 cm into the body. moderate Travels great distances in air. Penetrates right through the body. low low low high Although and particles are much less penetrating than gamma radiation, they can still be extremely harmful if emitted by material absorbed into the body, because the nearby tissue has a continuing exposure to the radiation. For example, health scientists have calculated that breathing in a speck of dust containing just 1 g of plutonium is virtually certain to cause lung cancer within 30 years. The introduction of radioactive isotopes into the food chain is also a serious concern because these materials can accumulate in the body. For example, strontium-90, a by-product of nuclear weapons and power reactors, is absorbed into bones because it is chemically similar to calcium. Radiation from strontium damages bone marrow, reduces the production of blood cells, and can lead to bone cancer and leukemia. info BIT Both Marie and Pierre Curie suffered from radiation sickness. Some of Marie\u2019s laboratory notebooks are still dangerously radioactive. radioisotope: an isotope that is radioactive 808 Unit VIII Atomic Physics 16-PearsonPhys30-Chap16 7/24/08 4:32 PM Page 809 Despite its potential hazards, nuclear radiation is not always harmful. As you will see in section 16.3, nuclear radiation has many beneficial industrial and medical applications. Measuring Radiation Exposure The effects of a given dose of radiation depend on the type of radiation. For example, a dose of infrared radiation that delivered 1 J/kg to living tissue would do little more than heat the tissue slightly. The same quantity of energy from X rays would ionize some molecules within the tissue, whereas the same quantity of energy from alpha radiation would be far more ionizing and disruptive. For this reason, SI has two units for measuring radiation exposure: The gray is the unit for absorbed dose and the sievert is the unit for equivalent absorbed dose. Gray (Gy): 1 gray is the dose of ionizing radiation that delivers 1 J of energy to each kilogram of material absorbing the radiation. Sievert (Sv): 1", " sievert is the absorbed dose of ionizing radiation that has the same effect on a person as 1 Gy of photon radiation, such as X rays or gamma rays. The absorbed dose in sieverts is equal to the dose in grays multiplied by the relative biological effectiveness (RBE), a measure of how harmful the particular kind of radiation is. For example, the RBE for high-energy alpha particles is about 20, so an absorbed dose of 1 Gy of alpha radiation is equivalent to 20 Sv. An equivalent dose of 6 Sv in a short time is usually fatal. Typical radiation exposure for North Americans is less than 0.5 mSv annually. Table 16.3 summarizes some common sources of radiation exposure. Table 16.3 Common Sources of Radiation Exposure relative biological effectiveness (RBE): a factor indicating how much a particular type of radiation affects the human body Source Natural Artificial Total Radon from ground Cosmic rays Radioactive rocks/minerals, common building materials Ingested from natural sources Medical/dental X rays Nuclear weapons testing Consumer products All other Typical Exposure (Sv/year) 200 44 40 18 73 4 1 2 <400 Figure 16.12 Dosimeters: How do these devices measure exposure to radiation? Chapter 16 Nuclear reactions are among the most powerful energy sources in nature. 809 16-PearsonPhys30-Chap16 7/24/08 4:32 PM Page 810 16.2 Check and Reflect 16.2 Check and Reflect Knowledge 1. What are the three basic radioactive decay processes and how do they differ from each other? 2. What is the ratio of neutrons to protons for the heaviest stable isotopes? 3. (a) Write the alpha-decay process for 234 91Pa. (b) Identify the parent and daughter nuclei in this decay. 4. (a) Which type of beta decay transmutes carbon-14 into nitrogen? 11. Identify each type of decay in this series, and name the parent and daughter elements. (b) (a) 232 (d) 228 90Th \u2192 228 22Na \u2192 10 11 88Ra* \u2192 228 (c) 228 88Ra \u2192 228 89Ac \u2192 228 90Th \u2192 224 228 0n 1p \u2192 0 (e) 228 (g) 0 (f) 88Ra* 4 22Ne 2 0 1 88Ra 0 89Ac 1 90Th 1 0 88Ra 4 2 0 1 (b) Write the process for this decay. Extensions 5. (a) Which type of beta decay transmutes the sodium isotope 11 22Na into 10 (b) Write the process for this decay. 22Ne? 6. Explain why the daughter nucleus in an alpha decay often emits a gamma ray. 7. Which form of radioactive decay has the 12. In a process called electron capture, a nucleus absorbs an electron and emits a neutrino. (a) What effect does electron capture have on the atomic number? (b) Use nuclear notation to write the general form for electron capture. greatest penetrating power? (c) Compare electron capture with beta decay. 13. Devise an experiment to test the hypothesis that gamma rays are emitted by nucleons jumping from higher energy levels to lower ones, similar to the energy-level transitions of electrons in an atom. What would you expect the spectrum of gamma rays emitted by a nucleus to look like? 14. Use library or Internet resources to learn how radon forms in the ground. Explain how radon can accumulate in basements in some areas. Why is this accumulation a health concern? e TEST To check your understanding of radioactive decay, follow the eTEST links at www.pearsoned.ca/school/physicssource. Applications 8. How much energy is released when 22Ne? The mass of 11 22Na decays to 10 11 is 21.994 436 u and the mass of 10 21.991 385 u. 22Na 22Ne is 9. Explain whether the atomic number can increase during nuclear decay. Support your answer with an example. 10. Compare the annual average radiation exposure from natural sources with the dose you would receive from a dental X ray. 810 Unit VIII Atomic Physics 16-PearsonPhys30-Chap16 7/24/08 4:32 PM Page 811 16.3 Radioactive Decay Rates How can an archaeologist confidently tell you that a bison head found in southern Alberta provides evidence that First Nations peoples were here more than 5000 years ago? Why do doctors sometimes inject patients with radioactive dyes? In this section, you will be introduced to the concepts of radioactive decay rate and half-life, and begin to see how understanding the behaviour of radioactive elements can provide us with a glimpse into the past or give us powerful techniques to diagnose and combat disease. 16-3 QuickLab 16-3 QuickLab Simulating Radioactive Decay Problem How can decay rates of atoms be predicted? Materials container with 100 pennies graph paper Procedure Work in groups of two or three. 1 (a) Pour the penn", "ies onto a flat surface and spread them out. Put aside any pennies that are tails up. These pennies have \u201cdecayed.\u201d (b) Count the remaining pennies and put them back into the container. Record this count in a table. 2 Predict how many pennies will remain if you repeat step 1 two more times. 3 Repeat step 1 a total of eight times. 4 Pool your results with the other groups in the class. 5 Use the pooled data to draw a graph of how the number of pennies remaining varies with time. Questions 1. How many pennies were left after you had done step 1 three times? Does this result match your prediction? 2. If you repeat the experiment, will you get exactly the same results each time? Explain. 3. Suppose that step 1 takes 2 min each time. (a) How long would it take for the number of pennies remaining to decrease by half? How long will it take until only about an eighth of the pennies remain? How are these two time intervals related? (b) Try to find a formula to predict how many pennies will remain at any given time. Activity and Decay Constant The radioactive decay of a specific nucleus is unpredictable. The nucleus could decay in the next minute, or tomorrow, or thousands of years from now. However, you can accurately predict how many nuclei in a sample will decay in a given time. The decay constant () is the probability of any given nucleus decaying in a unit of time. The decay constant is a property of each particular isotope. For example, radium-226 has a decay constant of 1.4 1011 s1, indicating that each individual nucleus in a sample of radium-226 has a probability of 1.4 1011 of decaying in 1 s. The greater the decay constant, the faster an isotope will decay. decay constant: probability of a nucleus decaying in a given time Chapter 16 Nuclear reactions are among the most powerful energy sources in nature. 811 16-PearsonPhys30-Chap16 7/24/08 4:32 PM Page 812 activity or decay rate: the number of nuclei in a sample that decay within a given time becquerel (Bq): unit of activity, equal to 1 decay per second The activity (A) or decay rate is the number of nuclei in a sample that decay within a given time. Activity is usually measured in decays per second, or becquerels (Bq). A highly radioactive sample has many radioactive decays each second. Activity and the decay constant are related by this formula: A N t N where N is the number of radioactive nuclei, t is the time interval, and is the decay constant. Example 16.11 Carbon-14 has a decay constant of 3.8 1012 s1. What is the activity of a sample that contains 2.0 1015 carbon-14 nuclei? Practice Problems 1. Cobalt-60 has a decay constant of 4.1 109 s1. Find the activity of a sample containing 1.01 1022 cobalt-60 atoms. 2. A sample containing 5.00 1020 atoms has an activity of 2.50 1012 Bq. Find the decay constant of this sample. Answers 1. 4.1 1013 Bq 2. 5.00 109 s1 Given 3.8 1012 s1 N 2.0 1015 atoms Required activity (A) Analysis and Solution Substitute the given values into the formula for activity: A N (3.8 1012 s1)(2.0 1015) 7.6 103 Bq The negative sign indicates that the number of carbon-14 nuclei is decreasing. Paraphrase The initial activity of the sample is 7.6 kBq. The activity of a radioactive material decreases over time. The reason is simple: Radioactive decay \u201cuses up\u201d the unstable nuclei in the sample. Half-life Half-life is the time required for one-half of the radioactive nuclei in a sample to decay. For example, to diagnose thyroid problems, doctors sometimes inject patients with the radioactive isotope iodine-131, which has a half-life of about 192 h. Out of a dose of 20 g of iodine-131, 10 g will decay within 192 h. Only 5 g of iodine-131 will remain after the next 192 h, then 2.5 g after the next 192 h, and so on (see Figure 16.13). A common symbol for half-life is t1/2. half-life: the time it takes for half of the radioactive nuclei in a sample to decay e SIM To see a simulation of half-life, follow the links at www.pearsoned.ca/ school/physicssource. 812 Unit VIII Atomic Physics 16-PearsonPhys30-Chap16 7/24/08 4:32 PM Page 813 The number of nuclei of the original radiois", "otope left in a sample is given by the equation where t is the time elapsed, N0 is the number of nuclei of the original radioisotope when t 0, is the half-life of the isotope. and t1/2 N N0 t t1 /2 1 2 2.0 1.8 1.6 1.4 1.2 1.0 0.8 0.6 0.4 0. half-life 0 5 10 15 Time (days) Figure 16.13 A graph showing the radioactive decay of iodine-131 Example 16.12 Carbon-14 has a half-life of 5730 years. How long will it take for the quantity of carbon-14 in a sample to drop to one-eighth of the initial quantity? 5730 years Given t1/2 N 1 N0 8 Required time (t) Analysis and Solution N N0 1 N0 8 1 2 t t1 /2 In 3 half-lives, N will decrease to N0 t 3 Therefore, t1 /2 t 3t1/2 3 5730 years 1.719 104 years N0 1 2 1 2 1 8 1 2 Practice Problems 1. Astatine-218 has a half-life of only 1.6 s. About how long will it take for 99% of a sample of astatine-218 to decay? 2. Radium-226 has a half-life of 1600 years. What percentage of a sample of radium-226 will remain after 8000 years? Answers 1. about 11 s 2. 3.125% Paraphrase It will take just over 17 thousand years for the amount of carbon-14 in a sample to drop to one-eighth of its original value. Chapter 16 Nuclear reactions are among the most powerful energy sources in nature. 813 16-PearsonPhys30-Chap16 7/24/08 4:32 PM Page 814 Example 16.13 Radon-222 has a half-life of 3.82 days. What percent of a sample of this isotope will remain after 2 weeks? Practice Problems 1. Strontium-90 has a half-life of 29.1 years. What percent of a sample of this isotope will be left after 100 years? 2. Tritium (3 1H) has a half-life of 12.3 years. How much of a 100-mg sample of tritium will be left after 5.0 years? Answers 1. 9.24% 2. 75% Given t1/2 3.82 days t 14 days Required percent remaining after 14 days Analysis and Solution The percent remaining is calculated from the ratio N N0 t t1 /2 1 2 N N0. N0 4 1.8 2 3 1 2 1 2 3.66 N0 0.079N0 N N0 0.079 or 7.9% Note that you can use the exponent or ^ key on a scientific or graphing calculator to evaluate powers of enter (1/2)^(14/3.82). 1 2. On a graphing calculator, you could Paraphrase Only 7.9% of a sample of radon-222 will remain after 2 weeks. Applications of Radiation The Rutherford gold-foil experiment that you learned about in Chapter 15 was one of the first examples of the use of nuclear energy (the release of alpha particles in the decay of radium nuclei) to study the inner working of atoms. Scientists apply radioactive decay in many other fields of scientific research, including archaeology. Radioactive compounds also have numerous industrial applications and are routinely used to diagnose and treat diseases. 814 Unit VIII Atomic Physics 16-PearsonPhys30-Chap16 7/24/08 4:32 PM Page 815 THEN, NOW, AND FUTURE Radiotherapy During their first experiments with radium, Pierre and Marie Curie noticed that its radiation could burn the skin, but the wound would heal without forming scar tissue. They realized that radium could therefore be used to treat cancer. To ensure that this treatment was readily available to cancer patients, the Curies refused to patent their discovery. Radiotherapy is particularly useful for treating cancer because cancer cells are more susceptible to the effects of radiation than healthy tissue is. Also, the radiation is concentrated on the cancer, and kept away from the surrounding tissue as much as possible (Figure 16.14). \u03b3 rays Figure 16.14 Rotating the radiation source around the patient minimizes damage to normal tissue. There is now a wide variety of radiation treatments. Often, a carefully focussed beam of gamma rays is directed at the tumour. Another common method is to inject the tumour with a short-lived radioisotope that emits alpha-particles. Questions 1. Give two reasons why gamma rays are used for the beam type of radiotherapy. 2. Why does injected radiotherapy use an isotope that undergoes alpha decay rather than one that", " gives off beta or gamma radiation? Radioactive Dating Nearly 6000 years ago, First Nations people of southwestern Alberta devised an ingenious method for hunting the vast herds of bison on the plains. By setting up barriers along a carefully chosen route, the First Nations people funnelled the bison toward a hidden cliff and then drove them over the edge. There were about 150 buffalo jumps in Alberta. The most famous, Head-Smashed-In Buffalo Jump, is now a United Nations World Heritage Site (see Chapter 2, Figure 2.68). By carefully measuring the ratio of carbon-12 to carbon-14 in bones found at this site, archaeologists have shown that it was used continuously for over 5500 years. e MATH To plot the decay rate of carbon\u201314 and other radioactive elements, and to learn how to mathematically determine a radioactive sample\u2019s age based on the percentage of the sample remaining, visit www.pearsoned.ca/school/ physicssource. How did this carbon ratio indicate the age of these bones? High-energy neutrons in cosmic rays produce the radioisotope carbon14 by colliding with nitrogen atoms high in the atmosphere: 7N \u2192 1n 14 0 14C 6 1H 1 This carbon-14 diffuses throughout the atmosphere. Some of it is absorbed by plants and enters the food chain. So, a small proportion of all the carbon metabolized by plants and animals is carbon-14. Carbon-14 undergoes decay to form nitrogen-14, whereas carbon-12 is completely stable. When living matter dies, it stops absorbing carbon, and the proportion of carbon-14 gradually decreases as it decays (see Figure 16.15). The half-life of carbon-14 is 5730 years 100 90 80 70 60 50 40 30 20 10 0 t1 2 t1 2 10 000 t1 2 t1 2 20 000 Time (years) 30 000 40 000 Figure 16.15 Carbon-14 content as a function of the age of an artifact Chapter 16 Nuclear reactions are among the most powerful energy sources in nature. 815 16-PearsonPhys30-Chap16 7/24/08 4:32 PM Page 816 e WEB To learn more about radioisotope dating, follow the links at www.pearsoned.ca/ school/physicssource. For archaeologists, bone fragments and other artifacts found at Head-Smashed-In Buffalo Jump are like clocks that show when the living matter stopped absorbing carbon. Suppose, for example, that the proportion of carbon-14 in a bone fragment is about 40% 1.32 40%, the carbon-14 has been of that in living tissue. Since decaying for about 1.3 half-lives, provided that the ratio of carbon-14 to carbon-12 in the atmosphere is the same now as when the buffalo was alive. Thus, the age of the bone fragment is roughly 1.3 5730 7500 years. Accurate estimates require more detailed calculations that take into account factors such as variations in the proportion of carbon-14 in the atmosphere through the ages. 1 2 Geologists estimate the age of rocks and geological formations with calculations based on isotopes with much longer half-lives. Such calculations are one of the methods that scientists use to estimate the age of Earth. Industrial Applications Manufacturers of sheet materials such as paper, plastics, and metal foils often monitor the thickness of the material with a gauge that measures how much of the beta radiation from a calibrated source passes through the material. Unlike mechanical thickness gauges, such gauges need not touch the material they measure, so they do not get worn down and have less risk of marking the material. Gamma rays can pass through thick metal parts to expose a photographic plate. The resulting image can reveal hidden air bubbles or hairline cracks, similar to the way X rays produce images of the inside of a patient\u2019s body. Gamma-ray photographs are a non-destructive way of testing items that X rays cannot penetrate, including structural materials, jet engines, and welded joints in pipelines. Radioactive tracers are also used in pipelines to measure flow and to detect underground leaks. Some uses of radiation are controversial. For example, beta radiation from tritium powers runway lights and emergency exit signs that require no electricity. However, several people have received harmful doses of radiation when tritium lights have been damaged. Critics of these lights argue that other technologies can provide reliable lighting during power failures without any risk of radiation exposure. Perhaps the most controversial application is the irradiation of food to kill bacteria, insects, and parasites. Although this process sterilizes the food and thereby prolongs its shelf life, there are concerns that the radiation might also alter the food in ways that make it harmful or less nutritious. Concept Check Why is beta radiation used for measuring the thickness of sheet materials, whereas gamma radiation is used for testing structural materials? 816 Unit VIII Atomic Physics 16-PearsonPhys30-Chap16 7/24/08 4:", "32 PM Page 817 16.3 Check and Reflect 16.3 Check and Reflect Knowledge 1. What fraction of a radioactive material remains after four half-lives? 2. How many decays per second occur in a radioactive sample containing 6.4 1023 atoms of a material that has a decay constant of 5.8 1012 s1? 3. Which has the greater activity, 1 g of material with a half-life of 1 ms or 1 g of material with a half-life of 1 year? Explain your answer. Applications 4. Analysis of a rock sample shows that only of the original amount of chlorine-36 1 16 remains in the rock. Estimate the age of the rock given that the half-life of chlorine-36 is 3.0 105 years. 5. A radioactive tracer used in a medical test has a half-life of 2.6 h. What proportion of this tracer will remain after 24 h? 6. An archaeologist finds a wooden arrow shaft with a proportion of carbon-14 that is about 25% of that in a living tree branch. Estimate the age of the arrow. 7. A radioactive sample has an activity of 2.5 MBq and a half-life of 12 h. What will be the activity of the sample a week later? 8. Graph the data in this table. Then use your graph to estimate (a) the half-life of the material (b) the activity of the sample at time t 0 Time (h) Activity (decays/min) 1 2 4 6 8 10 3027 2546 1800 1273 900 636 Extensions 9. A dealer in antiquities offers to sell you an \u201cauthentic\u201d dinosaur bone for a mere $100. He shows you a certificate indicating that carbon-14 dating determined that the bone is 65 million years old. Why should you be suspicious? 10. Do a Web search on use of irradiation in food production and distribution. Prepare a summary of the arguments for and against this technology. 11. (a) What is depleted uranium? (b) Why is depleted uranium used in armour-piercing shells and in ballast for aircraft? (c) Why are these applications controversial? e TEST To check your understanding of radioactive decay, follow the eTEST links at www.pearsoned.ca/school/physicssource. Chapter 16 Nuclear reactions are among the most powerful energy sources in nature. 817 16-PearsonPhys30-Chap16 7/24/08 4:32 PM Page 818 Figure 16.16 The doomsday clock from the Bulletin of the Atomic Scientists 16.4 Fission and Fusion In 1945, a group of the scientists who had designed and built the atomic bomb founded a magazine as part of an ongoing campaign to prevent this weapon from ever being used again. The Bulletin of the Atomic Scientists features a doomsday clock that symbolizes their estimate of the risk of a nuclear war (Figure 16.16). Since 2002, the clock has showed just seven minutes to midnight \u2014 a sobering reminder of the dangers posed by the enormous energy that nuclear reactions can release. The graph in Figure 16.4 (page 796) shows that binding energy per nucleon has a maximum value of about 8.7 MeV when the atomic mass number, A, is from 58 to 62 \u2014 the values for isotopes of iron and nickel. Up to this maximum, the binding energy per nucleon generally increases as A increases. Then, as A increases further, the binding energy per nucleon gradually decreases. The shape of this graph indicates that two distinct types of reactions can release energy from nuclei. Fission: When a nucleus with A > 120 splits into smaller nuclei, they have greater binding energy per nucleon. This fission reaction gives off energy equal to the difference between the binding energy of the original nucleus and the total binding energy of the products. Fusion: When two low-mass nuclei combine to form a single nucleus with A < 60, the resulting nucleus is more tightly bound. This fusion reaction gives off energy equal to the difference between the total binding energy of the original nuclei and the binding energy of the product. For both nuclear fission and fusion, the energy released, E, is E Ebf (net change in mass defect) c2 Ebi where Ebi is the total binding energy of the original nucleus or nuclei, and Ebf is the total binding energy of the product(s). Since the binding energies correspond to the mass defects for the nuclei, the energy released corresponds to the decrease in the total mass defect. This change in the total mass defect equals the change in the total mass. Thus, the energy released corresponds to the mass that the reaction transforms into energy: mi) c2 E (mf where mi is the total mass of the original nucleus or nuclei, and mf is the total mass of the product(s). Concept Check Why does a nuclear reaction that increases the binding energy per nucleon", " release energy? Use an analogy to help explain this release of energy. 818 Unit VIII Atomic Physics 16-PearsonPhys30-Chap16 7/24/08 4:32 PM Page 819 Nuclear Fission Often, fission results from a free neutron colliding with a large nucleus. The nucleus absorbs the neutron, forming a highly unstable isotope that breaks up almost instantly. Figure 16.17 shows one of the ways that uranium-235 can split into two lighter nuclei. In the next example, you will calculate the energy released during a fission reaction. 1 0 n compound nucleus 141 56 Ba 235 92 U 236 92 Figure 16.17 in a CANDU nuclear reactor. Absorbing a neutron causes uranium-235 to undergo fission 92 36 Kr Example 16.14 Calculate the energy released by the fission reaction 92Kr 3 0 92U 235 1n \u2192 141 0 56Ba 1n. 36 Given Initial mass: 235 Final mass: 141 92U plus one neutron 56Ba, 36 92Kr, and three neutrons Required energy released (E) Analysis and Solution First, use the atomic mass data on page 881 to calculate the net change in mass resulting from the reaction. mi mf mi U mn m 235 92 235.043 930 u 1.008 665 u 236.052 595 u m 141 56 140.914 412 u 91.926 156 u 3(1.008 665 u) 235.866 563 u mf 0.186 032 u 236.052 595 u 235.886 563 u 3mn m 92Kr 36 Ba Practice Problems 1. Calculate the energy released by the reaction 92U 1 235 0n \u2192 94 40Zr 139 52Te 3 1 0n. 35Br), a 2. A uranium-235 nucleus absorbs a neutron and then splits into a bromine nucleus (87 146La), and lanthanum nucleus ( 57 additional neutrons. How many neutrons are released in this fission reaction? Express this reaction as a balanced equation. 3. How much energy is released in the reaction in question 2? Answers 1. 172.9 MeV 92U 1 2. 235 3. 167.8 MeV 0n \u2192 87 35Br 146 57La 3 1 0n Now, use mass-energy equivalence to calculate the energy released. 1 u is equivalent to 931.5 MeV, so 931.5 MeV 1 u E 0.186 032 u 173.3 MeV Paraphrase The fission of an atom of uranium-235 into barium-141 and krypton-92 releases 173.3 MeV of energy. Chapter 16 Nuclear reactions are among the most powerful energy sources in nature. 819 16-PearsonPhys30-Chap16 7/24/08 4:32 PM Page 820 Comparing Chemical Energy with Nuclear Energy When you sit by a campfire, the warmth that you feel is due to the chemical energy released by the combustion of wood. All chemical processes, including combustion, involve electrons moving from one energy level to another. In Chapter 15, you learned that such transitions typically release no more than a few tens of electron volts. Example 16.15 shows that a nuclear process can release a vastly greater amount of energy. Example 16.15 Burning 1 kg of gasoline releases about 4.4 107 J. Compare this energy to the energy released by the fission of 1 kg of uranium-235 into barium-141 and krypton-92. Practice Problems 1. A typical family car requires approximately 1600 MJ of energy to travel 500 km. (a) How many kilograms of gasoline does it take to provide this energy? (b) What mass of uranium-235 would provide the same energy? Answers 1. (a) 36 kg (b) 22 mg Given chemical energy content of gasoline 4.4 107 J/kg Required ratio of the energy content of gasoline to that of uranium-235 Analysis and Solution From Example 16.14, you know that uranium-235 has about 173.3 MeV of nuclear potential energy per atom, assuming fission into barium and krypton. Use the atomic mass of uranium-235 to calculate the number of atoms in 1 kg of this isotope. Then calculate the potential energy per kilogram for comparison with gasoline. m235 92 U 235.043 930 u 1.660 539 1027 kg 1 u 3.902 996 1025 kg Number of atoms in 1 kg of 235 92U 1 kg 3.902 996 1025 kg Energy content of 235 92 U (2.562 134 1024 atoms kg )(173.3 MeV atom ) 2.562 134 1024 4.4402 1026 MeV/kg 4.4402 1032 eV kg 019 J 1 1.60 V e 1 7.10 1013 J/kg Energy content of 235 Energy content of gasoline 92U 7.10 1013 J/", "kg 4.4 107 J/kg 1.6 106 Paraphrase The nuclear potential energy of 1 kg of uranium-235 is about 1.6 million times greater than the chemical potential energy of 1 kg of gasoline. 820 Unit VIII Atomic Physics 16-PearsonPhys30-Chap16 7/24/08 4:32 PM Page 821 Concept Check A nucleus is much smaller than an atom. How does this difference in size make nuclear reactions much more energetic than chemical reactions? Fusion What powers the Sun? The discovery of the nucleus and of mass-energy equivalence provided the key to this question, which had puzzled scientists for thousands of years. In the early 1920s, the British-American astrophysicist Cecilia Payne-Gaposchkin (1900\u20131979) showed that the Sun consists primarily of hydrogen (about 73%) and helium (about 27%). Noting that four protons have 0.7% more mass than a helium nucleus, the British astrophysicist Arthur Stanley Eddington (1882\u20131944) suggested that a fusion process might power the stars. In the 1930s, the young German physicist Hans Bethe (1906\u20132005) worked out the details of how hydrogen nuclei could release energy by fusing together to form helium. In the Sun and smaller stars, the process, called the proton-proton chain (Figure 16.18), has four steps. First, two hydrogen nuclei combine to form deuterium (an isotope of hydrogen with one neutron), an antielectron, and a neutrino. Next, another hydrogen nucleus combines with the deuterium nucleus to produce a helium-3 nucleus and a gamma ray. Then, two of the helium-3 nuclei combine to produce a helium-4 nucleus, two hydrogen nuclei, and a gamma ray. In the final step, annihilation of two positron-electron pairs occurs. Each of these annihilations produces a pair of gamma photons. In order for these reactions to occur, the nuclei must have enough kinetic energy to overcome the electrostatic repulsion between them. Step Reaction Energy Released info BIT In the mid-1930s, Hans Bethe won a $500 prize for a paper on fusion in stars. He used the money to get his mother out of Nazi Germany. Bethe won the Nobel Prize for physics in 1967 and helped found the Bulletin of the Atomic Scientists. proton-proton chain: fusion process in which four hydrogen nuclei combine to form a helium nucleus 1H \u2192 1 2H 0 1 (twice) 0.42 MeV (twice) (twice) 5.49 MeV (twice) 1 2 3 4 Total 1 2 1 1H 2 3 2H \u2192 3 1 2He 2He 2 1 2He \u2192 4 0 1 1H \u2192 4 4 1 0 1 \u2192 2 2He 2 0 1 neutrino positron H2 1 H2 1 positron neutrino 1H (twice) 2 7 ray He3 2 He3 2 ray 12.85 MeV 1.02 MeV (twice) 26.71 MeV ray He4 2 Figure 16.18 The protonproton chain Chapter 16 Nuclear reactions are among the most powerful energy sources in nature. 821 16-PearsonPhys30-Chap16 7/24/08 4:32 PM Page 822 Example 16.16 The Sun radiates about 4 1026 W and has a mass of 1.99 1030 kg. Astronomers estimate that the Sun can convert only the innermost 10% of its hydrogen into helium. Estimate how long the Sun can continue to shine at its present intensity. Given Power 4 1026 W Hydrogen available for conversion 10% of total hydrogen 1.99 1030 kg mSun Required Time the Sun will take to convert 10% of its hydrogen into helium (t) Analysis and Solution The fusion of four hydrogen atoms produces 26.71 MeV. To find the rate at which helium nuclei are produced, divide the Sun\u2019s power by the energy released during the formation of each helium nucleus: Rate of helium production power of Sun energy released per helium atom Practice Problems 1. (a) How many helium nuclei does a star with a power of 1.6 1025 W produce every second? (b) Estimate how much helium this star has produced if it is 4 billion years old. Answers 1. (a) 4.1 1036 (b) 3.4 1027 kg 4 1026 W 26.71 MeV/atom J 4 1026 s V1.60 26.71 0 1 e M V e M 1 m to a 13 J 9.36 1037 atoms/s Since 4 hydrogen atoms are needed for each helium produced, multiply by 4 to find the rate at which the Sun converts hydrogen atoms into helium. Then, convert this rate to mass per second by multiplying it by the mass of a hydrogen atom: k g ms1.67", " 1027 Rate of hydrogen conversion 49.36 1037 ato o at s m 6.25 1011 kg/s Hydrogen makes up 73% of the mass of the Sun, but only 10% of this hydrogen can be converted into helium. The lifespan of the Sun approximately equals the amount of hydrogen that can be converted divided by the conversion rate. t amount of hydrogen available rate of conversion 1.99 1030 kg 73% 10% 6.25 1011 kg/s 2.32 1017 s or about 7 109 years Paraphrase The Sun can continue to produce energy at its present rate for about 7 billion years. 822 Unit VIII Atomic Physics 16-PearsonPhys30-Chap16 7/24/08 4:32 PM Page 823 Concept Check Why can fusion reactions occur only at extremely high temperatures? The cores of massive stars can reach temperatures high enough for helium nuclei to combine to form carbon and oxygen. In some stars, these elements can undergo further fusion. The extent of this nucleosynthesis depends on the star\u2019s density, temperature, and the concentration of the various elements. Current theory suggests that synthesis of elements heavier than iron and nickel occurs only during the explosion of supernovae. Such explosions distribute these elements throughout the cosmos. So, the uranium fuel for today\u2019s nuclear power stations may have come from the explosion of a massive star billions of years ago. A hydrogen-fusion reactor might be an almost ideal energy source. Hydrogen is the most abundant of elements, and the end product, helium-4, is harmless. However, controlling and sustaining a fusion reaction for generating power is extremely difficult. To start the fusion process, the hydrogen has to be heated to a temperature between 45 million and 400 million kelvins, depending on which isotopes are used. Then, this extremely hot gas has to be contained so that the fusion reactions can continue. Some researchers are using powerful lasers to generate the necessary temperatures and magnetic fields to contain the fusion reactions. However, the latest experiments have sustained fusion for only a few seconds and produced only slightly more energy than it took to run the reactor (see Figure 16.19). It will take major technological advances to make fusion power practical. nucleosynthesis: formation of elements by the fusion of lighter elements supernova: sudden, extremely powerful explosion of a massive star e WEB To learn more about fusion reactors, follow the links at www.pearsoned.ca/school/ physicssource. Figure 16.19 The Joint European Toroid (JET) fusion reactor Chapter 16 Nuclear reactions are among the most powerful energy sources in nature. 823 16-PearsonPhys30-Chap16 7/24/08 4:32 PM Page 824 Concept Check There are concerns that tritium could leak from a fusion reactor. Why would tritium be a serious environmental hazard? 16.4 Check and Reflect 16.4 Check and Reflect Knowledge 1. (a) Complete this nuclear reaction: 235 92U \u2192 54 140Xe? 2 0 1n (b) Does this reaction involve fission or fusion? Justify your answer. 2. What happens to the binding energy per nucleon in a nuclear reaction that releases energy? 3. An iron nucleus of binding energy 492 MeV fuses with a silicon nucleus of binding energy 237 MeV to form a nucleus with binding energy 718 MeV. Will this reaction release energy? Explain why or why not. 4. (a) Which elements are most likely to undergo fission? (b) Which elements are most likely to undergo fusion? 5. A neutron is emitted when aluminium-27 absorbs an alpha particle. (a) What isotope does this reaction create? (b) Write the process for the reaction. Applications 6. (a) Write the reaction formula for the fusion of helium-4 with oxygen-16. (b) How much energy does this reaction release? 7. (a) What particle is emitted when 1H) and tritium (3 deuterium (2 fuse to form helium? 1H) (b) How much energy does this reaction release? 8. A CANDU-6 nuclear reactor can generate 700 MW of electrical power. A CANDU power plant transforms about 27% of its nuclear energy into electrical energy, with the rest being lost primarily as heat. 824 Unit VIII Atomic Physics (a) If the plant uses uranium-235 as fuel and the average energy released per uranium nucleus is 200 MeV, how many nuclei undergo fission each second when the reactor is running at full power? (b) Estimate how many kilograms of uranium-235 a CANDU-6 reactor uses in a year. List any assumptions you make. Extensions 9. (a) In stars much more massive than the Sun, iron-56 will eventually be produced in their centres. Suppose that two iron-56 nuclei fuse. Complete the following reaction and identify the element produced: 56 26Fe \u2192 (b)", " The element formed in the reaction 26Fe 56 in (a) has a mass of 111.917 010 u. Show that this reaction absorbs rather than releases energy. (c) Explain why stars like the Sun do not produce elements heavier than iron. 10. (a) Research the radioactive wastes produced by nuclear reactors. List the major isotopes produced and their half-lives. (b) Briefly outline some of the methods for storing and disposing of these wastes. 11. Compare and contrast the risks and benefits of generating electricity with coal and with nuclear reactors. e TEST To check your understanding of fission and fusion, follow the eTest links at www.pearsoned.ca/school/physicssource. 16-PearsonPhys30-Chap16 7/24/08 4:32 PM Page 825 CHAPTER 16 SUMMARY Key Terms and Concepts femto proton neutron nucleon atomic number neutron number atomic mass number isotope atomic mass unit (u) strong nuclear force binding energy mass defect alpha radiation beta radiation gamma radiation transmute parent element daughter element beta-negative () decay beta () particle neutrino weak nuclear force antimatter ) positron (e or 0 1 beta-positive () decay gamma () decay radioactive decay series radiation sickness genetic damage radioisotope gray (Gy) sievert (Sv) relative biological effectiveness (RBE) decay constant activity (A) or decay rate becquerel (Bq) half-life fission fusion proton-proton chain nucleosynthesis supernova Key Equations Binding energy: Eb Enucleons Enucleus decay: A ZX \u2192 A4 Z2Y 4 2 decay: A ZX* \u2192 AX Z Activity: A N t N Nuclear energy released: E (mi mf) c2 Mass defect: m mnucleons mnucleus Nmneutron matom Zm1 1H decay: A ZX \u2192 A Z1Y 1 0 decay: A ZX \u2192 A 1 Half-life: N N0 Z1Y 0 t t1 /2 1 2 Conceptual Overview Summarize this chapter by copying and completing this concept map. nuclear structure nuclear notation mass-energy equivalence types of radiation nuclear decay binding energy conservation laws discovery of neutrino decay constant decay rate activity half-life mass defect fission fusion Chapter 16 Nuclear reactions are among the most powerful energy sources in nature. 825 16-PearsonPhys30-Chap16 7/24/08 4:32 PM Page 826 CHAPTER 16 REVIEW Knowledge 1. (16.1) Do all nuclei contain more neutrons than protons? Justify your answer. 2. (16.1) Is the atomic mass number for an atom always greater than the atomic number? Justify your answer. 3. (16.1) What is the term for elements that have the same atomic number but different neutron numbers? 4. (16.1) Explain how these nuclei are similar and how they differ: 233 92U, 92 235U, 238 92U. 17. (16.2) Compare the radiation dose that North Americans typically receive each year from radon and from diagnostic X rays. Which of these sources poses the greater health hazard? 18. (16.3) What is the activity of a sample that contains 1.5 1020 nuclei of an element with a decay constant of 1.2 1012 s1? 19. (16.3) After 1.5 h, the number of radioactive nuclei in a sample has dropped from 5.0 1020 to 2.5 1020. How many of these nuclei will remain after another 6 h? 20. (16.3) Explain why carbon-14 dating is not 5. (16.1) How many neutrons are in a nucleus useful for determining the age of a rock sample. of 115 55Cs? How many protons? 21. (16.4) Why do all of the elements used as fuel in 6. (16.1) Convert 50 MeV to joules. nuclear power plants have A > 200? 7. (16.1) Calculate the energy equivalent for 22. (16.4) What is the primary energy source for 1 g of matter. most stars? 8. (16.1) Calculate the energy equivalent for 23. (16.4) List the steps in the proton-proton chain. Applications 24. Calculate the binding energy per nucleon for the following nuclei: (a) 4 (b) 28 (c) 58 (d) 235 2He 14Si 26Fe 92U 25. (a) Write the process for the decay of 52 26Fe. (b) Show that this process conserves charge and atomic mass number. 26. (a) What parent element decays into lead-208 by emitting an alpha particle? (b) Estimate the kinetic energy of the alpha", " particle. 27. (a) Write a complete decay process for the transmutation of 30 15P into 30 14Si. (b) Calculate the energy released in this decay. 28. In the oldest campsites yet discovered in Alberta, archaeologists have found materials that contain about a quarter of their original carbon-14. Estimate the age of these campsites. Give your answer to two significant digits. 2.3 u of mass. 9. (16.1) Calculate the mass equivalent for 300 MeV. 10. (16.1) Calculate the binding energy for a nucleus that has a mass defect of 0.022 u. 11. (16.2) Which decay processes do not change the atomic number of a nucleus? 12. (16.2) What is the charge on (a) a beta particle? (b) an alpha particle? (c) a gamma ray? 13. (16.2) Explain how each of these decay processes changes nuclear structure: (a) alpha decay (b) beta decay (c) gamma decay 14. (16.2) Describe this decay in words, identifying the parent element, the daughter element, and 19K \u2192 43 the type of decay: 43 15. (16.2) Which of,, and radiation is the most penetrating, and which is the least penetrating? 20Ca 1 0. 16. (16.2) Explain why physicists think that radioactivity originates from nuclei. 826 Unit VIII Atomic Physics 16-PearsonPhys30-Chap16 7/24/08 4:32 PM Page 827 29. Until the early 1950s, a paint containing radium-226 was used to make the dials on some clocks, watches, and aircraft instruments glow in the dark. Radium-226 has a decay constant of 1.98 1011 s1. (a) If the activity of one of these clocks is 0.10 MBq, how many atoms of radium-226 are on the dial? (b) Calculate the mass of radium on the dial. (c) The half-life of radium is 1600 years. Calculate the activity that the clock will have in 5000 years. 30. Graph the data in this table. Use your graph to estimate (a) the activity of the sample when t 5 h (b) the half-life of the radioactive material in the sample Time (h) 0 2 4 6 8 10 12 Activity (Bq) 1000.0 697.7 486.8 339.6 236.9 165.3 115.3 Time (h) Activity (Bq) 14 16 18 20 22 24 80.5 56.1 39.2 27.3 19.1 13.3 31. Calculate the energy released when three helium-4 nuclei combine to form a carbon-12 nucleus. 32. You are designing a thermoelectric power supply for a space probe. The probe will need 20 W of electricity for 14.5 years. The efficiency of thermal to electrical energy conversion is 15%. You are considering using polonium-208 as the fuel for the power supply. (a) What is the key advantage of polonium over a chemical fuel? (b) How much polonium will you need? Polonium-208 has a decay constant of 7.57 109 s1 and a half-life of 2.9 years. 84Po decays into 204 208 82Pb. Extensions 33. In 1918, Rutherford observed that bombarding nitrogen atoms with alpha particles produced oxygen and hydrogen. Use nuclear notation to write two reactions that could account for these products. Which reaction is more likely to occur? Explain your reasoning. How could you check your conclusion? 34. There have been over 2000 tests of nuclear weapons, including 711 conducted in the atmosphere or in the ocean. What radioactive products did these tests release? What health hazards result from this radioactive fallout? 35. Research nucleosynthesis in stars. List a sequence of fusion reactions that produces iron-56, and explain why smaller stars do not complete this sequence. How are the fusion reactions in the Sun likely to end? Consolidate Your Understanding 1. Explain how atomic number, atomic mass number, and neutron number are related to the structure of the nucleus. 2. Use the concept of binding energy to explain why some nuclei are more stable than others. 3. Describe the differences between the alpha, beta, and gamma decays. 4. Explain how you can use conservation principles to predict the daughter elements created by a radioactive decay. 5. Distinguish between nuclear fission and nuclear fusion, and explain how to calculate the energy yield from either process. Think About It Review your answers to the Think About It questions on page 789. How would you answer each question now? e TEST To check your understanding of nuclear reactions, follow the eTEST links at www.pearsoned.ca/school/physicssource. Chapter 16", " Nuclear reactions are among the most powerful energy sources in nature. 827 17-PearsonPhys30-Chap17 7/24/08 4:34 PM Page 828 C H A P T E R 17 Key Concepts In this chapter, you will learn about: charge-to-mass ratio quantum mechanical model standard model of matter Learning Outcomes When you have completed this chapter, you will be able to: Knowledge explain the discovery and identification of subatomic particles explain why high-energy particle accelerators are required describe the modern model of the proton and neutron compare and contrast elementary particles and their antiparticles describe beta decays Science, Technology, and Society explain the use of concepts, models, and theories explain the link between scientific knowledge and new technologies Skills observe relationships and plan investigations analyze data and apply models work as members of a team apply the skills and conventions of science 828 Unit VIII The development of models of the structure of matter is ongoing. Antimatter, quarks, particles appearing out of nowhere! Although these concepts may seem like science fiction, they are crucial for understanding the nature of matter. You are about to enter the world of undetectable particles that blink in and out of existence. You will see that a calculation by a theoretical physicist in the 1920s led to sophisticated new medical technology that uses a previously unknown form of matter (Figure 17.1). You will learn about the peculiar properties of quarks, the elusive building blocks for protons, neutrons, and many other subatomic particles. Quantum effects can make the subatomic world seem very strange indeed. This chapter introduces some of the most unusual and challenging ideas in all of physics. You will learn that experiments are showing that in some profound ways the universe is stranger than anyone could have imagined a century ago. The theories that you will explore next are exhilarating, difficult, weird, and yet elegant. They are a key to the next century of atomic physics. \u25b2 Figure 17.1 Recent findings in atomic physics may seem strange, but they have led to amazing advances in technology, as well as better models of the structure of matter. 17-PearsonPhys30-Chap17 7/24/08 4:34 PM Page 829 17-1 QuickLab 17-1 QuickLab Particle Tracking Simulation Problem What can magnetic tracking reveal about the properties and collisions of objects? Materials lid from shirt box or shoe box magnetic metal marbles glass marbles iron filings metre-sticks or wood slats iron filings box top metre-stick supports magnetic or glass marble \u25b2 Figure 17.2 Procedure 1 Turn the lid upside down and use metre-sticks or wood slats to support it above a smooth surface such as a tabletop. The gap should allow the marbles to roll freely under the lid. 2 Spread iron filings evenly over the lid (Figure 17.2). 3 Roll a glass marble and a magnetic marble under the lid and observe how they affect the filings. 4 Set the lid aside and place a line of five magnetic marbles spaced about 3 cm apart across the middle of the space between the supports. Estimate what percentage of glass marbles rolled between the supports will hit one of the five magnetic marbles. 5 Shake the lid to spread the filings evenly again and put it back on the supports. Then, roll glass marbles under the lid at least 10 times. After each collision, put the magnetic marbles back in line and spread the filings evenly. Note the number and shape of any tracks resulting from collisions between the glass marbles and the magnetic ones. Watch for any pattern in the formation of the tracks. Questions 1. What can you conclude about the magnetic field from the glass marbles? 2. Calculate the percentage of glass marbles that appeared to collide with the magnetic marbles in step 5. How close was your estimate? Account for any difference between your estimate and your observations. 3. Did any factor appear to affect the length of the collision tracks? 4. How would you expect the tracks to change if you repeated step 5 using round plastic beads instead of glass marbles? 5. How could you use the electric field from charged particles to detect these particles? How could you detect uncharged particles? Think About It 1. How can you tell if a particle is fundamental? 2. What did the measurement of beta decays reveal about the structure of matter? 3. How many fundamental particles are there? Discuss your answers in a small group and record them for later reference. As you complete each section of this chapter, review your answers to these questions. Note any changes in your ideas. Chapter 17 The development of models of the structure of matter is ongoing. 829 17-PearsonPhys30-Chap17 7/24/08 4:34 PM Page 830 17.1 Detecting and Measuring Subatomic Particles A skilled wilderness guide can tell a great deal about an animal from its tracks, not just identifying the animal but also estimating its age and how fast it was moving (Figure 17.3). In a similar way, physicists", " use tracks left by subatomic particles to identify the particles, study their interactions, and deduce the structure of matter (Figure 17.4). cloud chamber: a device that uses trails of droplets of condensed vapour to show the paths of charged particles \u25b2 Figure 17.3 Tracks of an adult snowshoe hare. What do these tracks tell you about the hare\u2019s speed? \u25b2 Figure 17.4 Tracks of subatomic particles. The heavier particles have straighter tracks. Cloud Chambers and Bubble Chambers A cloud chamber contains dust-free air supersaturated with vapour from a liquid such as water or ethanol. The amount of vapour air can hold depends on temperature and pressure. Air is supersaturated when it contains more vapour than it would normally hold at a given temperature and pressure. So, the liquid and vapour in a cloud chamber are not in equilibrium, and a tiny disturbance can trigger condensation of vapour into droplets of liquid. A charged particle speeding through the supersaturated air will ionize some molecules along its path. The ions trigger condensation, forming a miniature cloud along the trajectory of the speeding particle. This cloud track shows the path of the particle the way a vapour trail formed by condensing exhaust gases shows the path of a jetliner through the sky. Figure 17.5 One of Charles \u25b2 Wilson\u2019s cloud chambers. The glass sphere is an expansion chamber used to lower the pressure in the cylindrical cloud chamber. 830 Unit VIII Atomic Physics 17-PearsonPhys30-Chap17 7/24/08 4:34 PM Page 831 Charles Thomson Rees Wilson (1869\u20131969) made the first observations of particle tracks in a cloud chamber in 1910 (Figure 17.5). For the next 50 years, cloud chambers were the principal tools of atomic physics. They are to atomic physics what telescopes are to astronomy. The bubble chamber (Figure 17.6) was developed in 1952 by the physicist Donald Glaser (b. 1926). It contains a liquefied gas, such as hydrogen, helium, propane, or xenon. Lowering the pressure in the chamber lowers the boiling point of this liquid. When the pressure is reduced so that the boiling point is just below the actual temperature of the liquid, ions formed by a charged particle zipping through the liquid cause it to boil. Thus, the particle forms a trail of tiny bubbles along its path. Bubble chambers reverse the process used in cloud chambers: particle tracks are formed by a liquid turning into vapour instead of a vapour turning into liquid. info BIT Charles Wilson built the first cloud chamber in 1894 to study how clouds form. He shared a Nobel Prize for his contribution to particle physics. Wilson was a renowned meteorologist and an avid mountaineer. bubble chamber: a device that uses trails of bubbles in a superheated liquid to show the paths of charged particles info BIT CERN stands for Conseil Europ\u00e9en pour la Recherche Nucl\u00e9aire. It is the world\u2019s largest particle physics laboratory. \u25b2 Figure 17.6 One of the large bubble chambers at the CERN laboratory near Geneva, Switzerland Neutral particles will not create tracks in a cloud or bubble chamber. However, it is possible to calculate some of the properties of neutral particles from the tracks of charged particles that interact with them. Concept Check Outline possible reasons why neutral particles will not show up in a bubble chamber. How could you tell if a neutron were involved in a particle collision in a bubble chamber? e SIM To see an animation of particle tracks, follow the links at www.pearsoned.ca/ school/physicssource. Chapter 17 The development of models of the structure of matter is ongoing. 831 17-PearsonPhys30-Chap17 7/24/08 4:34 PM Page 832 17-2 Inquiry Lab 17-2 Inquiry Lab Building a Cloud Chamber Question Can types of radiation be identified by the characteristics of their tracks? Hypothesis Since alpha, beta, and gamma radiations have different properties, the tracks they produce in a cloud chamber will be different. Materials and Equipment clear glass container flat glass or plastic cover black blotting paper to fit the bottom of the container dry ice (frozen carbon dioxide) reagent grade ethanol (ethyl alcohol) foam plastic insulation tape silicone grease lamp with reflector radiation sources CAUTION: The temperature of dry ice is 78 \u00b0C. Handle it only with tongs or thick gloves. Be careful not to damage the casing on the radioactive samples. Variables Identify the manipulated, responding, and controlled variables in this experiment. Procedure Work with a partner or a small group of classmates. 1 Cut a piece of black blotting paper to fit the bottom of the glass container. 2 Saturate this blotting paper with alcohol, but avoid having a pool of alcohol in the container. 3 Cover the container using silicone grease to ensure a good seal between the cover and the container. 4 Use", " a piece of foam plastic insulation as the base for your cloud chamber. Place a piece of dry ice at least 2.5 cm thick in the centre of this base, then put the 832 Unit VIII Atomic Physics Required Skills Initiating and Planning Performing and Recording Analyzing and Interpreting Communication and Teamwork glass container on top of the dry ice. Placing more insulation around the sides of the dry ice will make it last longer. 5 Position the lamp so it shines down from the side of the chamber (Figure 17.7). Darken the room and wait several minutes. Note any changes that you observe in the cloud chamber. 6 Now tape an alpha-radiation source onto the inside of the container near the bottom. Write a description of any tracks that appear. If the tracks have a consistent shape or pattern, sketch a typical track. 7 Repeat step 6 with beta- and gamma-radiation sources. transparent cover T18-01 lamp glass container radiation source blotting paper foam plastic insulation dry ice \u25b2 Figure 17.7 A simple cloud chamber Analyzing and Interpreting 1. Were all of the tracks you observed produced by the three radiation sources? What else could produce tracks in your cloud chamber? Explain your reasoning. 2. Describe any relationship you see between the appearance of the tracks and the type of radiation that produced them. 3. Suggest improvements to the design of this experiment. Forming Conclusions 4. Do your observations support the hypothesis? If so, which properties of the radiation might be responsible for any differences in the tracks? 5. Under what conditions will subatomic particles travelling through the ethanol cloud not produce observable tracks? 17-PearsonPhys30-Chap17 7/24/08 4:34 PM Page 833 Extending 6. Hold a strong magnet against the side of the cloud chamber and observe the magnetic field\u2019s effect on tracks from the three radiation sources. Explain whether you could use the magnet to help distinguish between different types of radiation. 7. Make a hypothesis about how taping the radiation sources to the outside of the glass container would affect the tracks produced by each source. Test your hypothesis. Could your results help you distinguish between different types of radiation? What other methods could you use? Analyzing Particle Tracks Physicists use cloud and bubble chambers as a key part of a controlled environment for studying subatomic particles. Applying a magnetic field across the chamber causes charged particles to follow curved or spiral paths. Measurements of the resulting tracks can be used to determine the mass and charge of the particles. For example, Figure 17.8 shows the path of a particle moving in a cloud chamber in which a magnetic field is coming out of the page. The particle entered the chamber from the left. Applying the right-hand rule to this track shows that the particle must have a positive charge. Often, a photograph of a cloud or bubble chamber will show tracks from a number of particles entering the chamber. Once in a while, a single track will suddenly branch into several diverging tracks, as shown in Figure 17.9. Such tracks suggest that the original particle has transformed into two or more different particles. B v F \u25b2 Figure 17.8 The right-hand rule shows that the particle must have a positive charge. Orient your right hand as shown and then rotate your hand so that your fingers point out of the page. Your palm points in the same direction as the force on a positively charged particle. \u25b2 Figure 17.9 These tracks suggest that a particle interaction can form two or more different particles. The following example demonstrates how a particle\u2019s track can reveal its charge-to-mass ratio. Example 17.1 Assume that the tracks shown in Figure 17.10 were made by particles moving at a speed of 0.10c through a uniform magnetic field of 30 mT [out of the page]. The initial radius of each track is 5.7 mm. Determine the charge-to-mass ratio for the particles. Then, make a hypothesis about what the particles are. What is unusual about this pair of particles? \u25b2 Figure 17.10 Why are these particle tracks spiral rather than circular? Chapter 17 The development of models of the structure of matter is ongoing. 833 17-PearsonPhys30-Chap17 7/24/08 4:34 PM Page 834 Given v 0.10c 3.0 107 m/s 30 mT 0.030 T B r 5.7 mm 0.0057 m Required q m identification of each particle Analysis and Solution \u2022 Applying the right-hand rule shows that the particle spiralling clockwise has a positive charge. Similarly, the left-hand rule shows that the particle spiralling counterclockwise has a negative charge. \u2022 Since there was no track before the two particles appeared, they must have originated from a photon or a neutral particle. For charge to be conserved, the net charge on the two new particles must be zero. Therefore, these particles must have equal but opposite charges", ". \u2022 The charge-to-mass ratio for a particle moving perpendicular to a PHYSICS INSIGHT The tesla is a derived unit that can be expressed in terms of SI base units: k g 1 T 1 s2 A \u2022 m magnetic field can be derived from F F c v2 qv \u2022 Since the values of q, v, r, and B are the same for both particles, their masses must also be equal. Practice Problems Substituting the known values gives 1. Measurement of a particle track shows a radius of deflection of 8.66 104 m for a proton travelling at a speed of 4.23 105 m/s perpendicular to a 5.10-T magnetic field. Calculate the charge-to-mass ratio for a proton. 2. Determine the radius of the path of an electron moving at a speed of 3.2 105 m/s perpendicular to a 1.2-mT magnetic field. Answers 1. 9.58 107 C/kg 2. 1.5 mm q m 3.0 107 m/s 0.030 T 0.0057 m 3.0 107 m/s g k 0.0057 m 0.030 s2 A \u2022 1.8 1011 A\u2022s/kg 1.8 1011 C/kg The charge-to-mass ratio for an electron is 1.60 1019 C 9.11 1031 kg 1.76 1011 C/kg. The ratios for protons or small ions are about four orders of magnitude smaller. Paraphrase The charge-to-mass ratio of the negative particle is 1.8 1011 C/kg. Since this ratio matches the ratio for an electron, this particle very likely is an electron. However, the other particle has a charge-to-mass ratio of 1.8 1011 C/kg. This particle appears to be a positron, an antimatter particle. 834 Unit VIII Atomic Physics 17-PearsonPhys30-Chap17 7/24/08 4:34 PM Page 835 Concept Check Can you tell whether momentum was conserved in the subatomic process in Example 17.1? Explain your reasoning. Example 17.1 illustrates how a conservation law can be a powerful tool for understanding the interactions of subatomic particles. Physicists often apply the conservation laws for charge, momentum, and mass-energy in this way. Experiments and theoretical calculations have shown that several other quantities are also conserved when particles interact. 17.1 Check and Reflect 17.1 Check and Reflect Knowledge 1. Compare the process for forming tracks in a cloud chamber with the process in a bubble chamber. 2. (a) List two subatomic particles that will leave tracks in a bubble chamber. (b) List two subatomic particles that will not leave tracks in a bubble chamber. 3. (a) Why does applying a magnetic field cause the particle tracks in a cloud or bubble chamber to curve? (b) What can the curvature of a particle\u2019s track in a magnetic field reveal about the particle? Applications 4. Will X-ray photons produce tracks in a bubble chamber? Justify your answer. 5. (a) Determine the type of charge on each particle moving through the magnetic field in this diagram. (b) What information would you need to determine which particle is moving faster? 6. Describe and explain the differences in the tracks made in a bubble chamber by the particles in each pair: (a) protons and alpha particles (b) protons and electrons 7. In this bubble-chamber photograph, a particle enters from the bottom and collides with a helium nucleus. (a) Use conservation of momentum to show that the incoming particle was an alpha particle rather than a proton. (b) Describe how you could show that the particles have a positive charge. Extension 8. Bubble chambers have replaced cloud chambers in many research laboratories. What advantages do bubble chambers have over cloud chambers? e TEST To check your understanding of methods for detecting and measuring subatomic particles, follow the eTest links at www.pearsoned.ca/school/physicssource. Chapter 17 The development of models of the structure of matter is ongoing. 835 17-PearsonPhys30-Chap17 7/24/08 4:34 PM Page 836 17.2 Quantum Theory and the Discovery of New Particles fundamental particle: a particle that cannot be divided into smaller particles; an elementary particle Early in the 20th century, many scientists thought that there were just three fundamental particles: the electron, the proton, and the neutron. However, developments in quantum theory in the 1920s and 1930s suggested the possibility of other subatomic particles, some with peculiar properties. The Discovery of Antimatter In 1928, British physicist Paul Adrien Maurice Dirac (1902\u20131984) predicted the existence of peculiar particles such as the positive electron in Example 17.1 (pages 833\u2013834). Dirac", " combined Einstein\u2019s theory of relativity with Schr\u00f6dinger\u2019s wave equation (described in section 15.5). Dirac\u2019s calculations, with the resulting relativistic wave equation, predicted that antimatter could exist. As mentioned in section 16.2, a particle of antimatter has a key property, such as charge, that is opposite to that of the corresponding particle of ordinary matter. In 1932, the American physicist Carl Anderson (1905\u20131991) provided the first evidence that antimatter really does exist. He photographed a cloud chamber track of a positron, as shown in Figure 17.11. For this achievement, Anderson won the Nobel Prize for physics in 1936. Concept Check How could Anderson tell that the particle track in Figure 17.11 showed a positively charged particle going down rather than a negative particle going up? His ingenious solution was to pass the particle through a thin lead plate. This plate slowed the particle a bit. Explain how Anderson could use this change in speed to confirm that the particle had positive charge. (Hint: The magnetic field for the cloud chamber in the photograph was directed into the page.) Quantum theory predicts that each kind of ordinary particle has a corresponding antiparticle. One of the startling properties of antimatter is that a collision between a particle and its antiparticle can annihilate both particles and create a pair of high-energy gamma-ray photons travelling in opposite directions. For example, an electronpositron collision can be written as e e \u2192 2 Such electron-positron annihilations are part of the nuclear processes in stars. Note that e is the symbol for a positron. In general, charged antiparticles are represented by simply reversing the sign of the charge on the symbol for the corresponding ordinary particles. Antiparticles for neutral particles are indicated by adding a bar over the symbol for the corresponding ordinary matter. Thus, the symbol for an antineutron is n. \u25b2 Figure 17.11 A picture worth a Nobel Prize: Anderson\u2019s photograph provided evidence for the existence of the positron. Anderson used this path (the white streak in the photo) to show that the particle that made it had a positive charge but a mass equal to that of the electron. annihilate: convert entirely into energy 836 Unit VIII Atomic Physics 17-PearsonPhys30-Chap17 7/24/08 4:34 PM Page 837 Concept Check Consider the example of a head-on collision between a positron and an electron travelling at equal speeds. Explain why momentum would not be conserved if all the energy of the two particles transformed into a single photon. Scientific Knowledge Can Lead to New Technologies The discovery of the positron made it possible to develop a powerful new medical diagnostic instrument. Positron emission tomography (PET) is an imaging technique that uses gamma rays from electron-positron annihilations to produce images of cross sections through a patient\u2019s body. A computer can then generate a three-dimensional image by combining successive plane images (Figure 17.12). The patient receives an injection of a radioactive tracer containing an isotope, usually fluorine-18, that gives off positrons as it decays. As these positrons meet electrons within the patient\u2019s body, they create pairs of gamma-ray photons. Several rings of gamma-ray detectors rotate around the patient. As the photon pairs register on diametrically opposite detectors, a computer builds up an image of the location and concentration of the radioactive tracer. These images can show a wide variety of vital information, such as blood flow, brain function, and the location of tumours. Quantum Field Theory By 1930, Dirac, Heisenberg, Born, and others had established the foundations of quantum field theory. In this theory, mediating particles are the mechanism by which the fundamental forces act over the distance between particles. Particles that mediate a force exist for such a brief time that they cannot be observed. For these virtual particles, energy, momentum, and mass are not related as they are for real particles. To help understand this concept, imagine two people tossing a ball back and forth while standing on a very slippery surface, such as a smooth, wet sheet of ice. Throwing and catching the ball pushes the two people farther and farther apart (Figure 17.13(a)). In this analogy, the people correspond to ordinary particles and the ball corresponds to a mediating particle. For an attractive force, picture the same two people handing a somewhat sticky candy apple back and forth. The force that each person exerts to free the candy apple from the other person\u2019s hand pulls the two people toward each other (Figure 17.13(b)). Note, however, that quantum field theory is a complex mathematical model with aspects that cannot be explained by such analogies. (a) (b) \u25b2 Figure 17.12 A PET scanner e WEB To learn more about PET scanners, follow the links at www.pearsoned.ca/ school/physic", "ssource. quantum field theory: a field theory developed using both quantum mechanics and relativity theory mediating particle: a virtual particle that carries one of the fundamental forces virtual particle: a particle that exists for such a short time that it is not detectable Figure 17.13 (a) Throwing \u25b2 a ball back and forth while on a slippery surface pushes these people apart. (b) Handing a sticky object back and forth pulls them together. Chapter 17 The development of models of the structure of matter is ongoing. 837 17-PearsonPhys30-Chap17 7/24/08 4:34 PM Page 838 quantum electrodynamics: quantum field theory dealing with the interactions of electromagnetic fields, charged particles, and photons gluon: the mediating particle for the strong nuclear force graviton: the hypothetical mediating particle for the gravitational force Project LINK For your unit project, you may want to describe the search for gluons and gravitons. The concept of mediating particles was first applied to the electromagnetic force, in a theory called quantum electrodynamics. This theory states that virtual photons exchanged between charged particles are the carriers of the attractive or repulsive force between the particles. For example, consider the electromagnetic repulsion between two electrons. One electron emits a virtual photon in the direction of the other electron. According to Newton\u2019s third law, the first electron will recoil and its momentum will change by an amount opposite to the momentum of the photon. Similarly, when the second electron absorbs the photon, this electron will gain momentum directed away from the first electron. You can think of the photon for an attractive force as acting a bit like the shared electron holding two atoms together in a covalent chemical bond. In the latter part of the 20th century, calculations using a refined version of quantum electrodynamics gave results that matched observed values with amazing accuracy \u2014 sometimes to 10 significant digits. Mediating Particles By 1970, research with high-energy particle accelerators led physicists to suggest that the strong nuclear force is mediated by zero-mass particles called gluons. So far, there is only indirect evidence for the existence of gluons. Advances in quantum theory also led to the conjecture that the weak nuclear force is mediated by three particles, designated W, W, and Z0. Experiments using extremely powerful accelerators detected these three particles in 1983. Some physicists think that the gravitational force also has a mediating particle, which they call the graviton. As yet, there is no experimental evidence that gravitons exist. Table 17.1 summarizes the current thinking about mediating particles. \u25bc Table 17.1 The Fundamental Forces and Their Mediating Particles Force Range Relative Strength for Protons in Nucleus Particles Mediating Particle Observed? Electromagnetic infinite Weak nuclear <0.003 fm Strong nuclear <1 fm Gravitational infinite 102 106 1 1038 photons W, W, Z0 yes yes gluons indirectly gravitons no 838 Unit VIII Atomic Physics 17-PearsonPhys30-Chap17 7/24/08 4:34 PM Page 839 17.2 Check and Reflect 17.2 Check and Reflect Knowledge 1. Describe the difference between ordinary matter and antimatter. 2. Outline how Anderson provided evidence for the existence of the positron. 3. (a) Which fundamental force is the strongest over large distances? (b) Which fundamental force is the weakest at nuclear distances? 4. (a) List the mediating particle for each of the fundamental forces. (b) Which of these mediating particles has not been detected at all? 5. Explain why a PET scan is like being X-rayed from the inside out. Applications 6. (a) What event is represented by the equation e e \u2192 2? (b) Why is the event e e \u2192 not possible? 7. (a) Under what conditions will two protons attract each other? (b) Under what conditions will they repel each other? 8. The tracks in this diagram show the creation of two particles in a bubble chamber. Initially, the two particles have the same speed. (a) What evidence suggests that a photon created the two particles? (b) Describe the path of this photon. (c) Which of the tracks shows the path of a positively charged particle? (d) Give two reasons why the other track must show the path of a negatively charged particle. (e) How are the mass and charge of the two particles related? (f) Why is it likely that the interaction involves an antiparticle? 9. Explain how the stability of helium nuclei demonstrates that the electromagnetic force is weaker than the nuclear forces. Extension 10. Research the Lamb shift and the Casimir effect at a library or on the Internet. Explain how these phenomena support the quantum field theory. e TEST To check your understanding of quantum theory and antimatter, follow the eTest links at www.pearsoned.ca/school", "/physicssource. Chapter 17 The development of models of the structure of matter is ongoing. 839 17-PearsonPhys30-Chap17 7/24/08 4:34 PM Page 840 17.3 Probing the Structure of Matter Our understanding of the structure of matter comes from a series of remarkable technological advances over the past century. For example, the first circular particle accelerator (Figure 17.14(a)) was about 12 cm in diameter and generated particles with energies up to 13 keV. Now, the most powerful accelerators are up to 8.5 km in diameter and can reach energies of a teraelectron volt (1012 eV). This huge increase in energy reflects an interesting overall trend: To probe matter at smaller and smaller scales, physicists need bigger and bigger machines! This trend results from the nature of matter: All of the fundamental forces become markedly stronger at distances less than the diameter of a nucleus. \u25b2 Figure 17.14 (a) The first circular particle accelerator (b) Fermilab near Chicago, Illinois. Its Tevatron accelerator ring is 2 km in diameter. Energy Requirements 13.6 eV is sufficient to ionize a hydrogen atom. With energies of a few hundred electron volts, you can study electron shells of atoms and of molecules (using a spectrograph, for example). To determine the size of a nucleus, you need charged particles with enough energy to get close to it despite strong electrostatic repulsion. For his ground-breaking scattering experiment, Rutherford used alpha particles with energies in the order of 10 MeV. To examine the structure of the nucleus, the energy requirements are much greater because the probe particles have to overcome the strong nuclear force. Within the nucleus, this short-range force is about a hundred times stronger than the electromagnetic force. The fundamental forces within individual subatomic particles are stronger still. So, probing the structure of stable particles such as protons and neutrons requires even more energy. With early, relatively low-energy accelerators, physicists could conduct experiments in which accelerated particles scattered from nuclei or split nuclei into lighter elements (hence the nickname \u201catom-smasher\u201d). With high-energy particles, physicists can also study interactions that create new types of particles. Producing some of the heavier particles requires a minimum of several gigaelectron volts. 840 Unit VIII Atomic Physics 17-PearsonPhys30-Chap17 7/24/08 4:34 PM Page 841 Natural Sources of Energetic Particles Some naturally radioactive isotopes emit particles that are useful for probing the structure of the atom. For example, Rutherford used polonium and radium as particle sources for his experiments. However, the maximum energy of particles from natural radioactive decay is roughly 30 MeV, which is not enough to probe the structure of nuclei. The other major natural particle source is cosmic radiation. Cosmic rays are high-energy particles that stream into Earth\u2019s atmosphere from outer space. Astronomers are not certain about the origin of these particles. Some of them may come from solar flares and from distant supernovae. About 90% of cosmic rays are protons and most of the rest are alpha particles with a few electrons, positrons, antiprotons, and other particles. The energies of these particles range from roughly 102 to 1014 MeV. The particles from space (primary cosmic rays) rarely reach the ground because they interact with atoms in the atmosphere, producing less energetic secondary cosmic rays. Particle Accelerators The first particle accelerators were built around 1930. These accelerators, and the much more powerful ones developed since then, use electric and magnetic fields to accelerate and direct charged particles, usually in a vacuum chamber. Here is a brief description of some of the major types of particle accelerators. \u2022 Van de Graaff: A moving belt transfers charge to a hollow, conductive sphere, building up a large potential difference. This potential difference then propels ions through an accelerator chamber. \u2022 Drift Tube: An alternating voltage accelerates charged particles through a series of electrodes shaped like open tubes. The applied voltage reverses as the particles pass through each tube, so the particles are always attracted to the next tube in the line. \u2022 Cyclotron: A magnetic field perpendicular to the paths of the charged particles makes them follow circular paths within two hollow semicircular electrodes. An alternating voltage accelerates the charged particles each time they cross the gap between the two electrodes. The radius of each particle\u2019s path increases with its speed, so the accelerated particles spiral toward the outer wall of the cyclotron. \u2022 Synchrotron: This advanced type of cyclotron increases the strength of the magnetic field as the particles\u2019 energy increases, so that the particles travel in a circle rather than spiralling outward. Some of the largest and most powerful particle accelerators are synchrotron rings. Concept Check Explain the advantages and disadvantages of studying nuclei with protons from a large accelerator as opposed to alpha particles produced", " by radioactive decay. primary cosmic rays: high-energy particles that flow from space into Earth\u2019s atmosphere secondary cosmic rays: the shower of particles created by collisions between primary cosmic rays and atoms in the atmosphere Chapter 17 The development of models of the structure of matter is ongoing. 841 17-PearsonPhys30-Chap17 7/24/08 4:34 PM Page 842 e WEB To learn more about particle accelerators, follow the links at www.pearsoned.ca/ school/physicssource. info BIT Marietta Blau published a number of papers on cosmic rays in the 1920s and 1930s. She was nominated for the Nobel Prize several times. muon: an unstable subatomic particle having many of the properties of an electron but a mass 207 times greater pion: an unstable subatomic particle with a mass roughly 270 times that of an electron lepton: a subatomic particle that does not interact via the strong nuclear force hadron: a subatomic particle that does interact via the strong nuclear force meson: a hadron with integer spin baryon: a hadron with halfinteger spin spin: quantum property resembling rotational angular momentum fermion: particle with halfinteger spin boson: particle with integer spin PHYSICS INSIGHT How Small Are Electrons? Experiments have shown that electrons are less than 1018 m across, while protons are roughly 1.6 1015 m in diameter. Leptons might be mathematical points with no physical size at all! 842 Unit VIII Atomic Physics Although particle accelerators were originally developed for pure research, they now have medical and industrial uses as well. Many hospitals use accelerated particles for generating intense beams of X rays that can destroy cancerous tumours. Bombarding elements with particles from cyclotrons produces radioactive isotopes for diagnostic techniques, radiation therapy, testing structural materials, and numerous other applications. Particle accelerators can make a variety of specialized industrial materials by, for example, modifying polymers and implanting ions in semiconductors and ceramics. Accelerators are also powerful tools for analyzing the structure and composition of materials. Particle accelerators have even been used to verify the authenticity of works of art. The Subatomic Zoo In 1937, Carl Anderson and Seth Neddermeyer used a cloud chamber to discover muons in cosmic rays. These particles behave much like electrons, but have a mass 207 times greater and decay rapidly. Ten years later, Cecil Frank Powell discovered \u03c0-mesons, or pions, by using a photographic technology that Marietta Blau had developed. This method records tracks of particles on a photographic plate coated with a thick emulsion containing grains of silver bromide. Pions are much less stable than muons, and have some properties unlike those of electrons, protons, or neutrons. Improved particle accelerators and detectors led to the discovery of many more subatomic particles. Over 300 have now been identified. Most of these particles are highly unstable and have lifetimes of less than a microsecond. Studies of the interactions and decays of these particles show that there are two separate families of particles: leptons, which do not interact by means of the strong nuclear force, and hadrons, which do. The term lepton comes from leptos, a Greek word for \u201cthin\u201d or \u201csmall,\u201d and hadron comes from hadros, a Greek word for \u201cthick.\u201d The diameters of leptons are much smaller than those of hadrons. The hadrons are divided into two subgroups, mesons (from meso, Greek for \u201cmiddle\u201d) and baryons (from barus, Greek for \u201cheavy\u201d). One of the key quantum properties for classifying particles is their spin. This property is like angular momentum from rotation of the particle. The spin of a particle can be either an integer or half-integer multiple of Planck\u2019s constant divided by 2. Particles that have half-integer spin or 3 (such as 1 ) are called fermions, while those that have integer spin 2 2 (such as 0, 1, or 2) are called bosons. Leptons and baryons are fermions. Mesons and mediating particles are bosons. Spin can affect the interactions and energy levels of particles. \u25bc Table 17.2 Classification of Subatomic Particles Leptons Hadrons Meditating Particles Fermions all leptons Bosons baryons mesons all mediating particles 17-PearsonPhys30-Chap17 7/24/08 4:34 PM Page 843 There are far more types of hadrons than types of leptons. In fact, physicists have found only six leptons plus their corresponding antiparticles. Table 17.3 compares the mass and stability of the leptons and some of the more significant hadrons. You are not required to memorize this table.", " Its purpose is to show a tiny set of the dozens of particles that physicists had discovered by the 1960s. What they were desperately seeking, and what you will learn about in the next section, was an underlying theory that could help make sense of this \u201csubatomic zoo.\u201d \u25bc Table 17.3 An Introduction to the Subatomic Zoo Particle Symbol Mass (MeV/c2) Lifetime (s) Leptons electron electron neutrino muon muon neutrino tauon tauon neutrino Mesons pions kaons psi upsilon Baryons proton neutron lambda sigma xi omega.511 < 7 106 106 < 0.17 1777 < 24 140 135 494 498 3097 9460 938.3 939.6 1116 1189 1192 1315 1321 1672 stable stable? 2.2 106 stable? 2.9 1013 stable? 2.6 108 8.4 1017 1.2 108 9 1020 8 1021 1.3 1020 1031? 885* 2.6 1010 8 1011 7.4 1020 2.9 1010 1.6 1010 8.2 1011 *lifetime for a free neutron; neutrons in nuclei are stable Units for Subatomic Masses Note that Table 17.3 lists masses in units of MeV/c2. The kilogram is not always the most convenient unit for expressing the mass of subatomic particles. Physicists often deal with transformations between mass and energy using Einstein\u2019s famous equation E mc2. Rearranging E. It follows that mass can be expressed in this equation gives m 2 c energy terms of units of speed of light squared. info BIT A pion will decay in the time it takes light to travel across a classroom. e MATH To better understand the relative sizes of subatomic particles, visit www.pearsoned.ca/school/ physicssource. Particle physicists find it convenient to use a factor of c2 to relate mass to electron volts, the traditional energy unit for particle physics. Conversion factors for such units are 1 eV/c2 1.7827 1036 kg 1 MeV/c2 1.7827 1030 kg 1 GeV/c2 1.7827 1027 kg Chapter 17 The development of models of the structure of matter is ongoing. 843 17-PearsonPhys30-Chap17 7/24/08 4:34 PM Page 844 For example, the mass of a proton expressed in these units is mp 1.6726 1027 kg 938.23 MeV/c2 1 MeV/c2 1.7827 1030 kg The masses of the known subatomic particles range from 0.5 MeV/c2 to 10 GeV/c2, so exponent notation is usually not necessary with these units. Table 17.4 compares subatomic masses expressed in several common units. \u25bc Table 17.4 Comparison of Mass Units for Subatomic Particles (to Five Significant Digits) Particle Mass (kg) Mass (u) Mass (MeV/c2) Electron Proton Neutron 9.1094 1031 5.4858 104 0.511 00 1.6726 1027 1.6749 1027 1.0073 1.0087 938.23 939.52 17.3 Check and Reflect 17.3 Check and Reflect Knowledge 1. Why do physicists require extremely highenergy particles for studying the structure of nucleons? 2. List two natural sources of energetic particles. 3. What is the advantage of high-altitude locations for performing experiments with cosmic rays? 8. (a) Find the energy equivalent of the mass of an electron. (b) The mass of a psi particle is 3.097 GeV/c2. Express this mass in kilograms. 9. Calculate the conversion factor between atomic mass units and MeV/c2. Give your answer to four significant digits. 4. List two uses of particle accelerators in Extensions 10. How has the development of superconducting electromagnets aided research into the structure of matter? 11. (a) What relativistic effect limits the energy of particles accelerated in an ordinary cyclotron? (b) Describe three different ways this limit can be overcome. e TEST To check your understanding of particle accelerators and subatomic particles, follow the eTest links at www.pearsoned.ca/school/physicssource. (a) medicine (b) industry 5. Identify a major difference that distinguishes (a) leptons from hadrons (b) mesons from baryons Applications 6. Can alpha particles from the radioactive decay of polonium be used to probe the nucleus? Explain your answer. 7. Calculate the momentum and kinetic energy of a proton that is accelerated to a speed of (a) 0.01c (", "b) 5.0 105 m/s 844 Unit VIII Atomic Physics 17-PearsonPhys30-Chap17 7/24/08 4:34 PM Page 845 17.4 Quarks and the Standard Model By 1960, physicists faced a large and growing menagerie of subatomic particles. Since the leptons are small and there are only a few types of them, it seemed likely that they were fundamental particles. However, the number of hadrons was a puzzle: Could there really be a hundred or more fundamental particles? info BIT Gell-Mann coined the term \u201cquark\u201d from an obscure line in a novel by the Irish writer James Joyce. Zweig had called the new fundamental particles \u201caces.\u201d The Quark Model In the 19th century, chemists studied the properties and reactions of the elements. The patterns observed in these properties led to the development of the periodic table and an understanding of the electron structure in atoms. Physicists searched for similar patterns in the properties and interactions of subatomic particles. In 1963, Americans Murray Gell-Mann (b. 1929) and George Zweig (b. 1937) independently proposed that all hadrons are composed of simpler particles, which Gell-Mann called quarks. By grouping the subatomic particles into distinct classes and families, Gell-Mann and Zweig showed that all the hadrons then known could be made from just three smaller particles and their antiparticles. These three particles are now called the up quark, the down quark, and the strange quark. This theory required that the quarks have fractional charges that are either one-third of the charge on an electron or two-thirds of the charge on a proton. Understandably, many physicists had trouble accepting this radical concept. Using the quark model, Gell-Mann accurately predicted not only the existence of the omega (\u03a9) particle, but also the exact method for producing it. The quark model also accurately predicted key aspects of electronpositron interactions. Stronger evidence for the quark theory came in 1967 when Jerome Friedman, Henry Kendall, and Richard Taylor used the powerful Stanford Linear Accelerator to beam extremely high-energy electrons at protons. The electrons scattered off the protons, somewhat like the alpha particles that scattered off the gold nuclei in Rutherford\u2019s scattering experiment (Figure 17.15). The pattern of the scattered electrons suggested that the mass and charge of a proton are concentrated in three centres within the proton. Later experiments confirmed these results and showed a similar pattern for scattering from neutrons. In the quark model, protons and neutrons contain only up and down quarks. The strange quark accounts for the properties of strange particles, hadrons that decay via the weak nuclear force even though they originate from and decay into particles that can interact via the strong nuclear force. In 1974, the discovery of the psi meson confirmed the existence of a fourth quark, the charm quark. Then, in 1977, the heavy upsilon meson was detected and found to involve a fifth quark, the bottom quark. Since there are six leptons, physicists wondered if there might be an equal number of quarks. In 1995, a large team of researchers at Fermilab found evidence for the top quark. This discovery required a huge accelerator because the top quark is about 40 000 times heavier than the up quark. quark: any of the group of fundamental particles in hadrons info BIT Richard Taylor was born in Medicine Hat and became interested in experimental physics while studying at the University of Alberta. In 1990, he shared the Nobel Prize in physics with Friedman and Kendall \u25b2 Figure 17.15 Scattering of high-energy electrons from a proton strange particle: a particle that interacts primarily via the strong nuclear force yet decays only via the weak nuclear force e WEB To learn more about the strange particles, follow the links at www.pearsoned.ca/ school/physicssource. Chapter 17 The development of models of the structure of matter is ongoing. 845 17-PearsonPhys30-Chap17 7/24/08 4:34 PM Page 846 THEN, NOW, AND FUTURE Small Particles and Big Science In the summer of 1936, Carl Anderson and his first graduate student, Seth Neddermeyer, lugged a cloud chamber and photographic equipment to the summit of Pikes Peak in Colorado, about 4300 m above sea level. They chose this location because the cosmic rays at that altitude were then the only source of the high-energy particles needed for their research. The work was lonely, uncomfortable, and poorly funded. It was also highly successful. Anderson and Neddermeyer discovered the muon with the cloud chamber photographs they took during their summer on Pikes Peak. Flash forward 60 years. How particle physics has changed! In 1995, the Fermi National Accelerator Laboratory", " (Fermilab) near Chicago, Illinois, announced that a team of researchers there had discovered the elusive top quark. In all, 450 people worked on this project. A key member of the team was Melissa Franklin (Figure 17.16), who has worked for over 18 years on the collider detector at Fermilab, a machine for studying the interactions resulting from colliding highenergy protons and antiprotons. A graduate of the University of Toronto and Stanford University, she is now a professor of physics at Harvard. Franklin is seeking to understand the structure of matter at the smallest scale, as Anderson did. However, Franklin collaborates with physicists from around the world and uses particle accelerators and detectors costing hundreds of millions of dollars. Although her work centres on tiny particles, she practises science on a big scale! Questions 1. What other fields of scientific research require huge budgets and international cooperation? 2. What are some advantages and drawbacks of \u201cbig science\u201d? 3. Teamwork skills are becoming increasingly important in many areas of research. What other skills would be useful for a career in science? \u25b2 Figure 17.16 Melissa Franklin Table 17.5 compares some properties of the six quarks. The mass of an individual quark cannot be measured directly. The masses given here were derived mainly by taking the total mass of various particles and subtracting estimates of the mass-energy the quarks gain from motion and interactions via the strong nuclear force within the particles. For each quark there is a corresponding antiquark with the opposite charge. \u25bc Table 17.5 Some Properties of Quarks Generation Name Symbol Mass (MeV/c2) Charge First up down Second strange Third charm bottom (or beauty) top (or truth) u d s c b t 1.5\u20134* 4\u20138 80\u2013130 1.15\u20131.35 103 4.1\u20134.9 103 1.7\u20131.9 104 Some physicists think the up quark may be essentially massless. info BIT To name quarks, physicists chose words that would not be mistaken for visible physical properties. In Europe, physicists commonly call the top quark \u201ctruth\u201d and the bottom quark \u201cbeauty.\u201d 846 Unit VIII Atomic Physics 17-PearsonPhys30-Chap17 7/24/08 4:34 PM Page 847 Individual quarks probably cannot be observed. The strong nuclear force binds the quarks in a particle very tightly. The energy required to separate quarks is large enough to create new quarks or antiquarks that bind to the quark being separated before it can be observed on its own. Composition of Protons and Neutrons The proton and the neutron contain only first-generation quarks. As shown in Figure 17.17, the proton consists of a down quark and two e e 2 up quarks. The net charge of these three quarks is 2 3 3 e e. The other quantum properties of the quarks also sum to 1 3 those of a proton. Similarly, a neutron consists of two down quarks and an up quark. In this combination, the positive charge on the up quark exactly balances the negative charge on the two down quarks. Composition of Other Hadrons All of the hadrons discovered in the 20th century can be accounted for with a combination of either two or three quarks: \u2022 All the mesons consist of a quark and an antiquark. \u2022 All the baryons consist of three quarks. \u2022 All the antibaryons consist of three antiquarks. However, experiments in 2003 produced strong evidence that the recently discovered theta particle () consists of five quarks: two up quarks, two down quarks, and an antistrange quark. Table 17.6 gives some examples of quark combinations. \u25bc Table 17.6 Some Quark Combinations e WEB To learn more about the difficulties in measuring quarks, follow the links at www.pearsoned.ca/ school/physicssource. 2 3 e u u 2 3 e Proton Neutron d 1 3 e \u25b2 Figure 17.17 The quarks making up protons and neutrons Meson Composition Baryon Composition Antibaryon Composition pion () pion (o) pion () kaon () ud uu ud us proton (p) neutron (n) sigma-plus () sigma-minus () uud udd uus dds antiproton (p) antineutron (n) uud udd Describing Beta Decay Using Quarks and Leptons Recall from section 16.2 that during beta decays of elements, the nuclei emit either an electron or a positron. Since both these particles are leptons, beta decay must proceed via the weak nuclear force. In decay of nuclei, a neutron transforms into a proton, an electron, and an", " antineutrino. Figure 17.18 shows that a neutron consists of an up quark and two down quarks while a proton consists of two up quarks and a down quark. So, the decay can be written as: udd \u2192 uud e e Charge is conserved since the difference between the 1 e charge on the 3 down quark and the 2 e charge on the new up quark equals the charge 3 on the electron emitted by the neutron. Physicists think that the down \u25b2 Figure 17.18 During decay, a down quark changes into an up quark. Chapter 17 The development of models of the structure of matter is ongoing. 847 17-PearsonPhys30-Chap17 7/24/08 4:34 PM Page 848 quark emits a virtual W particle (a mediator for the weak nuclear force) that then decays into an electron and an antineutrino: d \u2192 u [W] e e e WEB To learn more about the decay of subatomic particles, follow the links at www.pearsoned.ca/ school/physicssource. The idea of mediating particles is essential to understanding beta decay and is a central idea in the standard model. Similarly, in decay of nuclei, an up quark in a proton turns into a down quark by emitting a virtual W particle that then decays into a positron and a neutrino: uud \u2192 udd [W] e e The Standard Model standard model: the current theory describing the nature of matter and the fundamental forces The term standard model now refers to a model originally proposed in 1978 to explain the nature of matter and the fundamental forces. Here are some key concepts of this model: electroweak force: a fundamental force that combines the electromagnetic force and the weak nuclear force colour: a quantum property related to the strong nuclear force quantum chromodynamics: quantum field theory that describes the strong nuclear force in terms of quantum colour \u2022 All matter is composed of 12 fundamental particles \u2014 the 6 leptons and the 6 quarks \u2014 plus their antiparticles. \u2022 The electromagnetic force and the weak nuclear force are both aspects of a single fundamental force. Sheldon Glashow, Abdus Salaam, and Steven Weinberg developed the theory for this electroweak force in the late 1960s. This theory accurately predicted the existence and masses of the W, W, and Z0 particles. \u2022 The electromagnetic and nuclear forces are mediated by virtual particles. As discussed in section 17.2, these mediating particles are the photon, the gluon, and the W, W, and Z0 particles. \u2022 All quarks have a quantum property, termed colour, which determines how the strong nuclear force acts between quarks. (Quantum colour is not related to visible colours at all.) The quantum field theory describing the strong nuclear force in this way is called quantum chromodynamics. It is analogous to quantum electrodynamics with colour instead of electric charge and gluons instead of photons. Table 17.7 summarizes the fundamental particles in the standard model. \u25bc Table 17.7 Fundamental Particles in the Standard Model Matter Generation First Second Third Quarks Leptons Force Mediating particle(s) up down strange charm bottom top electron electron-neutrino muon muon-neutrino tau tau-neutrino Fundamental Forces Electromagnetic Weak Nuclear Strong Nuclear photon W, W, and Z0 gluon 848 Unit VIII Atomic Physics 17-PearsonPhys30-Chap17 7/24/08 4:34 PM Page 849 What\u2019s Next in Quantum Theory? Many theorists are working to combine quantum chromodynamics and the electroweak force theory into a grand unified theory. One such theory suggests that the electromagnetic, strong nuclear, and weak nuclear forces would blend into a single force at distances less than 1030 m, and leptons and quarks could transform from one into the other. However, it would take tremendously high energy to push particles so close together. Although of no relevance for everyday life, such theories could have a great effect on calculations about the origin of the universe. Another challenge is to develop a theory that unifies gravity with the other three forces. One of the most promising approaches is string theory, which treats all particles as exceedingly tiny vibrating strings of mass-energy. The vibration of the strings is quantized (like standing waves). The various kinds of particles are just different modes of vibration, with the graviton being the lowest mode. At present, these theories are highly speculative. The only thing known for sure is that the people who solve these problems will be in line for a Nobel Prize! grand unified theory: quantum theory unifying the electromagnetic, strong nuclear, and weak nuclear forces string theory: theory that treats particles as quantized vibrations of extremely small strings of mass-energy Project LINK For your unit project, you may want to describe theories that unify the fundamental forces. 17", ".4 Check and Reflect 17.4 Check and Reflect Knowledge 1. What experimental evidence suggests that the proton contains three smaller particles? 2. Why is it probably impossible to observe an individual quark on its own? 3. Compare the quark composition of a proton to that of a neutron. 4. Describe the difference between mesons and baryons in terms of quarks. 5. State two differences between leptons and hadrons. 6. List the 12 fundamental particles of matter in the standard model. 8. Is the beta decay \u2192 e possible? Justify your answer. e 9. Describe what happens in this decay process: uud \u2192 udd [W] e e Extension 10. Explain why a grand unified theory could have a great effect on speculations about the origin of the universe. Applications e TEST 7. (a) Using quark theory, write an equation for the beta decay of a neutron. (b) Show that charge is conserved in this decay process. To check your understanding of fundamental particles and the nature of matter, follow the eTest links at www.pearsoned.ca/school/physicssource. Chapter 17 The development of models of the structure of matter is ongoing. 849 17-PearsonPhys30-Chap17 7/24/08 4:34 PM Page 850 CHAPTER 17 SUMMARY Key Terms and Concepts cloud chamber bubble chamber fundamental particle quantum field theory mediating particle virtual particle quantum electrodynamics gluon graviton primary cosmic rays secondary cosmic rays Van de Graaff accelerator drift tube accelerator cyclotron synchrotron muon Key Equations Electron-positron annihilation: e e \u2192 2 Mass units: 1 eV/c2 1.7827 1036 kg pion lepton hadron meson baryon spin fermion boson quark strange particle standard model electroweak force colour quantum chromodynamics grand unified theory string theory decay: udd \u2192 uud [W] decay: uud \u2192 udd [W] e e e e Conceptual Overview Summarize the chapter by copying and completing this concept map. Nature of Matter involves fundamental particles classed as have corresponding antiparticles leptons quarks generations generations first first second second quark & antiquark have integer spin third third bosons have half-integer spin interact via make up all gluons can interact via 3 quarks fundamental forces mediated by W W Z0 photons gravitons? unified in electroweak unified in 850 Unit VIII Atomic Physics 17-PearsonPhys30-Chap17 7/24/08 4:34 PM Page 851 CHAPTER 17 REVIEW Knowledge 1. (17.1) How does a bubble chamber detect the path of a charged particle? 2. (17.1) Describe the technological advance in atomic physics made by (a) Charles Wilson (b) Marietta Blau (c) Donald Glaser 3. (17.2, 17.4) (a) In the early 1900s, which three subatomic particles were thought to be the fundamental building blocks of matter? (b) Which of these particles is still thought to be fundamental? 4. (17.2) What theory predicted the existence of antimatter? 5. (17.2) For each of these particles, list the corresponding antimatter particle and explain how it differs from the ordinary matter particle. (a) electron (b) proton 6. (17.2) How does quantum field theory account for fundamental forces acting over a distance? 7. (17.2, 17.3) Describe a similarity and a difference between a muon and a pion. 8. (17.3) What are two advantages of using units of MeV/c2 to express the mass of subatomic particles? 9. (17.3) Compare the size of an electron to that of a proton. 10. (17.4) Describe an experiment that provided evidence for the existence of quarks. 11. (17.4) Give two reasons why Millikan did not detect any quarks with his oil-drop experiment. 12. (17.4) (a) Why did physicists suspect that there might be a sixth quark? (b) What is the name of this sixth quark? (c) Why was a huge accelerator necessary for the discovery of this quark? 15. (17.4) How does the string theory explain the various kinds of subatomic particles? Applications 16. Sketch the paths that alpha, beta, and gamma radiation would follow when travelling perpendicular to a magnetic field directed out of the page. 17. The red tracks in this diagram show a high- speed proton colliding with a hydrogen atom in a bubble chamber, deflecting downward (toward the bottom of the page), and then colliding with another hydrogen atom. These tracks curve clockwise slightly. (a) In which direction is the magnetic field oriented", "? (b) What conclusions can you make about the mass, speed, and charge of the particles involved in the first collision? (c) What conclusion can you make about the mass, speed, and charge of the particles that made the small spiral tracks? 18. (a) Write a nuclear decay equation to show how fluorine-18 can produce positrons for use in positron-emission tomography. (b) Describe the role that quarks play in this decay process. (c) Write an equation to describe what happens to the positrons within a patient undergoing a PET scan. 13. (17.4) Compare the quark composition of 19. The mass of a top quark is about 176 GeV/c2. antiprotons and antineutrons. Express this mass in kilograms. 14. (17.4) Which two fundamental forces are united 20. (a) Determine the charge on a particle having in the standard model? the quark composition uus. (b) Estimate the mass of this particle. Chapter 17 The development of models of the structure of matter is ongoing. 851 17-PearsonPhys30-Chap17 7/24/08 4:34 PM Page 852 21. The diagram shows a particle track recorded in a bubble chamber at the CERN particle accelerator. There is good reason to suspect that the particle is either an electron or a positron. The magnetic field in the bubble chamber was 1.2 T directed out of the page. 0 2 4 6 8 10 cm (a) Does the particle have a positive or negative charge? Explain your reasoning. (b) Estimate the initial radius of the particle\u2019s path. (c) Determine the initial momentum of the particle. Assume that the particle is an electron or a positron. (d) Why does the particle\u2019s path spiral inward? (e) What could cause the short tracks that branch off from the large spiral track? Extension 22. Research Pauli\u2019s exclusion principle at the library or on the Internet. Write a paragraph describing this principle and the classes of particles to which it applies. Consolidate Your Understanding 1. Describe three experiments that discovered new subatomic particles. Explain how these experiments changed physicists\u2019 understanding of the nature of matter. 2. Give two examples of theories that accurately predicted the existence of previously unknown subatomic particles. 3. (a) Why was the quark theory first proposed? (b) Outline the experimental evidence that supports this theory. (c) Explain to a classmate why the standard model now includes six quarks instead of the three originally suggested by Gell-Mann and Zweig. Think About It Review your answers to the Think About It questions on page 829. How would you answer each question now? e TEST To check your understanding of the structure of matter, follow the eTest links at www.pearsoned.ca/school/physicssource. 852 Unit VIII Atomic Physics 18-PearsonPhys30-U8-Closer 7/25/08 7:36 AM Page 853 UNIT VIII PROJECT How Atomic Physics Affects Science and Technology Scenario Atomic physics has enormously influenced the development of modern science and technology. Advances in atomic physics have profoundly changed scientists\u2019 understanding of chemistry, biology, and medicine. A century ago, technology that is taken for granted now would have seemed impossible even in principle. In this project, you will work with two or three classmates to research how the concepts presented in this unit affected an aspect of science or technology. Planning Brainstorm with your classmates to make a list of possible topics. Look for branches of science or technology that apply concepts of atomic physics. Here are some starting points: \u2022 medical diagnostic technologies, such as X rays, magnetic resonance imaging (MRI), PET scans, and radioactive tracers \u2022 chemistry applications, such as understanding molecular bonds and using spectroscopy to identify compounds \u2022 biology topics such as quantum effects in photosynthesis and DNA replication \u2022 computer and electronic devices, such as microchips, tunnel diodes, and quantum computers \u2022 nanotechnologies such as nanotubes and atomic force microscopes \u2022 power technologies, such as nuclear reactors, tokamaks, and radioisotope thermoelectric generators (RTGs) Decide upon a topic to research. Often, you will find it easier to deal with a specific topic rather than a general one. For example, you could focus on nanotubes rather than trying to cover the whole field of nanotechnology. You may want to do some preliminary research on two or three promising ideas to see how much information is available. Consider the best way to present your findings. You might use a written report, an oral presentation, slides, a poster, a model, a video, or a combination of methods. Consult with your teacher on the format of your presentation. Assessing Results Assess the success of your project based on a rubric*", " designed in class that considers: research strategies clarity and thoroughness of the written report effectiveness of the team\u2019s presentation Materials \u2022 library resources, including books and periodicals \u2022 Web browser and Internet connection Check with your teacher about any special resources, such as computer software, that you may need for your presentation. Procedure 1 Assign tasks for each group member. Each member should do part of the research as well as some of the preparation for the presentation, such as writing, preparing graphics, or building a model. Clearly identify who is responsible for each part of your project. 2 Once you have gathered basic information about your topic, consider what further research you need to do. For example, you may be able to interview an expert either in person, or by telephone or e-mail. Many university departments have an outreach program that might suggest an expert you could consult. 3 Check whether the presentation method your group has chosen is suitable for the information you have found during your research. Consult with your teacher if you think you need to change to another type of presentation. If you will be making an oral presentation, practise. Having friends and family critique your presentation can often help you find ways to improve it. Thinking Further \u2022 Atomic physics often seems to be too abstract or theoretical to have any relevance to the \u201creal world.\u201d How has your investigation changed your understanding of the relationship between atomic physics and other disciplines? \u2022 Does the influence of atomic physics extend beyond science? Can you find ways in which atomic physics has influenced the arts or philosophy? \u2022 Science fiction often employs ideas from physics. List several science-fiction books, movies, or television series that use some of the concepts in this unit. How accurate is their treatment of atomic physics? *Note: Your instructor will assess the project using a similar assessment rubric. Unit VIII Atomic Physics 853 18-PearsonPhys30-U8-Closer 7/25/08 7:36 AM Page 854 UNIT VIII SUMMARY Unit Concepts and Skills: Quick Reference Summary Resources and Skill Building Concepts Chapter 15 cathode rays Electric force and energy quantization determine atomic structure. 15.1 The Discovery of the Electron Thomson\u2019s experiments showed that cathode rays are subatomic particles with a negative charge. charge-to-mass ratio You can use electric and magnetic fields to measure the charge-to-mass ratio of a particle. charge quantization Millikan\u2019s experiment 15.2 Quantization of Charge Electric charge exists only in multiples of the fundamental unit of charge, e. Millikan\u2019s oil-drop experiment measured the charge on an electron and showed that charge is quantized. 15-1 QuickLab Example 15.2 15-2 QuickLab Example 15.3 eSIM of Millikan\u2019s oil-drop experiment classical model of the atom 15.3 The Discovery of the Nucleus Rutherford\u2019s gold-foil experiment led to the solar-system model with electrons orbiting a tiny positively charged nucleus at the centre of the atom. 15-3 QuickLab Example 15.5 spectra Bohr model energy levels 15.4 The Bohr Model of the Atom Elements and compounds have characteristic emission and absorption line spectra. The Bohr model uses energy levels to account for stability of the atom and to explain line spectra. This model accurately predicts many properties of hydrogen, but has several serious failings. 15-4 Design a Lab Example 15.6 An electron in an atom can occupy only orbits that give the electron discrete, quantized amounts of energy that are inversely proportional to the square of the principal quantum number. Example 15.7 15-5 Inquiry Lab quantum mechanical model 15.5 The Quantum Model of the Atom The wave properties of electrons lead to a powerful new model based on probability distributions. Figures 15.24\u201315.25 Chapter 16 Nuclear reactions are among the most powerful energy sources in nature. nuclear structure 16.1 The Nucleus Nuclei contain protons and neutrons bound together by the strong nuclear force. mass-energy equivalence Mass and energy are equivalent, and the one can be transformed into the other. binding energy, mass defect The binding energy and mass defect of a nucleus indicate how tightly its nucleons are bound together. nuclear decay, transmutation 16.2 Radioactive Decay Some nuclei spontaneously transmute into a different element by emitting an alpha or beta particle. Nuclei can also give off gamma rays. All three types of radiation can be harmful. Examples 16.1 and 16.2 Example 16.3 Figure 16.3 Example 16.4 Examples 16.5\u201316.10 16-1 Inquiry Lab 16-2 Design a Lab half-life, activity nuclear reactions 16.3 Radioactive Decay Rates You can use the decay constant and the half-life of a radioisotope to predict the activity of a sample. Examples 16.12 and 16.13 16.4 Fission and Fusion Both the fission of a heavy nucleus into smaller nuclei and the fusion of light nuclei", " into a single heavier nucleus can release tremendous amounts of energy. Figures 16.17 and 16.18 Examples 16.14\u201316.16 Chapter 17 The development of models of the structure of matter is ongoing. particle tracks antimatter 17.1 Detecting and Measuring Subatomic Particles The existence and basic properties of subatomic particles can be determined by analyzing the paths of particles in magnetic and electric fields. 17.2 Quantum Theory and the Discovery of New Particles Quantum theory predicted the existence of antimatter, which was confirmed by Anderson\u2019s discovery of the positron. Example 17.1 17-1 QuickLab, 17-2 Inquiry Lab Figure 17.11 quantum field theory According to this theory, the electromagnetic and nuclear forces are mediated by virtual particles. Figure 17.13 particle accelerators families of particles fundamental particles 17.3 Probing the Structure of Matter Particle accelerators produce high-energy particles, which are used to study the structure of matter. Hadrons interact via the strong nuclear force, whereas leptons do not. Bosons have integer spin and fermions have half-integer spin. 17.4 Quarks and the Standard Model The fundamental particles are the six leptons, the six quarks, and their antiparticles. All hadrons consist of a combination of quarks and/or antiquarks. Figure 17.14 Tables 17.2 and 17.3 Tables 17.5,17.6, and 17.7 854 Unit VIII Atomic Physics 18-PearsonPhys30-U8-Closer 7/25/08 7:36 AM Page 855 UNIT VIII REVIEW Vocabulary 1. Using your own words, define these terms: absorption line spectrum activity (A) or decay rate alpha radiation antimatter atomic mass number atomic mass unit (u) atomic number baryon becquerel (Bq) beta () particle beta radiation beta-negative () decay beta-positive () decay binding energy Bohr radius boson bubble chamber cathode ray cloud chamber colour cyclotron daughter element decay constant drift tube accelerator electroweak force elementary unit of charge emission line spectrum energy level excited state femto fermion fission Fraunhofer line fundamental particle fusion gamma () decay gamma radiation gluon grand unified theory graviton gray (Gy) ground state hadron ) half-life ionization energy isotopes lepton mass defect mediating particle meson muon neutrino neutron neutron number nucleon nucleosynthesis orbital parent element pion planetary model positron (e or 0 1 primary cosmic rays principal quantum number proton proton-proton chain quantum chromodynamics quantum electrodynamics quantum field theory quark radioactive decay series radioisotope relative biological effectiveness (RBE) secondary cosmic rays sievert (Sv) spectrometer spectroscopy spin standard model stationary state strange particle string theory strong nuclear force supernova synchrotron transmute Van de Graaff accelerator virtual particle weak nuclear force Unit VIII Atomic Physics 855 18-PearsonPhys30-U8-Closer 7/25/08 7:36 AM Page 856 Knowledge CHAPTER 15 2. Calculate the electrical charge carried by 1 kg of protons. Give your answer to two significant digits. 3. How many coulombs of charge are on a dust particle that has gained 10 electrons? 4. Calculate the force exerted by an electric field of strength 100 N/C [S] on a dust particle having a charge of 10e. 5. Explain how the path of this particle shows whether its charge is positive or negative. E CHAPTER 16 10. How many neutrons are in a nucleus of gallium 64 31Ga? How many protons? 11. Why is the mass of an atom always less than Zm1 1H Nmneutron? 12. Express the energy equivalent of 0.021 u of mass in electron volts. 13. Express the energy equivalent of 7.0 u in joules. 14. Calculate the binding energy for a nucleus that has a mass defect of 0.0072 u. 15. What is the activity of a sample that contains 1.5 1022 nuclei of an element with a decay rate of 1.5 1013 Bq? 16. Write the decay process for 18 9F, and identify the daughter element. 17. The half-life of sulfur-35 is 87.51 days. How much of a 25-g sample of this isotope will be left after a year? 18. Write the alpha decay process for 228 90Th and 6. Determine whether the charge on this particle is identify the daughter element. positive or negative. 19. Explain the difference between fission and fusion. CHAPTER 17 20. What is a positron? 21. What is a pion? 22. If an electron and positron collide, they annihilate each other and are converted into energy. (a) How much energy does the annihilation of a positron-electron pair produce? (b", ") Explain why the annihilation must produce two gamma rays with the same wavelength. (c) Estimate the wavelength of these gamma rays. Assume that the kinetic energy of the electron and positron was negligible. 23. What is a quark? 24. Describe this reaction in words: ve p \u2192 n e B 7. What is an alpha particle? 8. Here are four energy-level transitions for an electron in a hydrogen atom: 4 \u2192 nf ni ni 6 \u2192 nf ni ni (a) For which transition(s) does the atom 1 \u2192 nf 2 \u2192 nf 5 6 1 2 lose energy? (b) For which transition does the atom gain the most energy? 25. Identify the particle formed by each of these (c) Which transition emits the shortest wavelength photon? 9. You are comparing the energy released by two different atomic transitions in a mercury atom. Transition A produces a very bright green line while transition B produces a fainter violetcoloured line. Which of these transitions releases more energy? Explain. 856 Unit VIII Atomic Physics combinations of quarks: (a) uud (c) ud (b) us (d) dds 26. Use quarks to describe how a neutron decays into a proton and an electron. 18-PearsonPhys30-U8-Closer 7/28/08 10:48 AM Page 857 Applications 27. A beam of protons enters a vacuum chamber where the electric field strength is 40 kN/C and the magnetic field strength is 0.55 T. (a) Sketch an orientation of the electric and magnetic fields that could let the protons pass undeflected through the chamber. (b) What speed must the protons have if they are not deflected by this orientation of the fields? 28. Find the magnetic field strength that will deflect a sodium ion (Na) in an arc of radius 0.50 m when the ion has a speed of 1.0 106 m/s. 29. This diagram shows an electron moving at 2.5 106 m/s through perpendicular electric and magnetic fields. B 0.50 T [out of the page] E 100 N/C [down] v (a) Calculate the electric and magnetic forces acting on the electron. (b) Calculate the net force acting on the particle. 30. An oil droplet with a mass of 1.6 1016 kg is suspended motionless in a uniform electric field of strength 981 N/C [down]. (a) Find the charge on this droplet. (b) How many electrons has the droplet either gained or lost? 31. (a) Find the wavelengths of the first four spectral lines produced by transitions into the n 3 energy level of a hydrogen atom. (b) What part of the electromagnetic spectrum are these lines in? 32. (a) Use the Bohr model to calculate the radius of the n 2 energy level in a hydrogen atom. (b) Find the de Broglie wavelength for an electron in this energy level. (c) Use the formula for the de Broglie wavelength to find the momentum of this electron. (d) Find the electron\u2019s speed and kinetic energy. 33. Calculate the binding energy for 40 20Ca. 34. Identify the nucleus produced in each reaction. (a) (b) (c) 12C \u2192? 6 14N \u2192? n 7 206Tl \u2192? v 81 35. Explain why each of these reactions cannot occur. 5 15C \u2192 (a) 6 1H \u2192 3 (b) 3 11Na n \u2192 23 15B ve 2He ve 19F 9 (c) 36. How much energy is released by decay of 16 7N? 37. Some blood-flow tests use iodine-131 as a tracer. This isotope has a half-life of 8.04 days. Estimate the percentage of iodine-131 left after 30 days. 38. How much energy is given off in the alpha decay of neodymium isotope 144 element does this decay produce? 60Nd? What daughter 39. A radioactive sample has an activity of 0.50 MBq and a half-life of 6 h. What will the activity of the sample be after 3.0 days? 40. The proportion of carbon-14 in charcoal used in a cave painting is only 12.5% of the proportion in living trees nearby. Estimate the age of this cave painting. 41. Calculate the amount of energy released when a carbon-12 nucleus absorbs an alpha particle and transmutes into oxygen-16. 42. Calculate the energy released by the reaction 2 1H 2 1H \u2192 3 1H 1 1H. 43. (a) What fundamental particles does a neutron contain, according to the standard model? (b) Show that this combination of particles has zero net charge. Unit VIII Atomic Physics 857 18-PearsonPhys30-U8-Cl", "oser 7/25/08 7:36 AM Page 858 44. The size of a nucleus is in the order of 1 fm. (a) Calculate the electrostatic force of repulsion between two protons separated by 1 fm. (b) Determine the potential energy of this pair of protons. (c) What keeps a nucleus together despite the electrostatic repulsion between protons? Extensions 45. Two hydrogen atoms in the ground state collide head on and both ionize. Find the minimum speed at which the atoms could have been moving toward each other. 46. (a) Describe the reaction p \u2192 0 p in words. (b) Calculate the minimum energy the photon must have to produce this reaction. 47. In decay, a proton becomes a neutron and the nucleus emits a positron and a neutrino. A proton has less mass than a neutron, and the positron and the neutrino carry away some mass and energy. Explain how such decays conserve mass-energy despite this apparent imbalance. 48. A 5.0-GeV photon creates, via pair-production, an electron and a positron. Calculate the total momentum of the two particles and sketch their motion relative to the path of the original photon. 51. A spill of radioactive material at an industrial site emits 1.25 mGy per hour, measured at a distance of 1.0 m from the spill. The relative biological effectiveness of this radiation is 2. (a) Compare the radiation dose from this spill to exposure from background radiation. (b) At what distance from the spill would the annual absorbed dose be less than 0.1 mSv? (c) A newspaper headline reads \u201cDangerous Spill at Local Factory.\u201d Is this description fair? Explain why or why not. 52. (a) What is the fundamental difference between a fusion process and one that combines matter and antimatter? (b) Compare the energy released by the fusion of ordinary hydrogen into helium-4 with the energy released by combining two protons with two antiprotons. (c) Why can antimatter not be used for generating power or propelling a spaceship now? 53. Suppose that an electricity generator powered by 1H 3 the fusion reaction 2 overall efficiency of 20%. How much deuterium and tritium will this generator need to produce 10 MWh of electricity, the annual consumption of a typical home? 0n has an 2He 1 1H \u2192 4 49. Imagine that protons and electrons were not Skills Practice charged but could still form a hydrogen atom through gravitational attraction. Calculate the radius of the ground state. (Hint: Assume that the electron travels in a circular orbit and has a total energy me.) Gm p of 2 r 50. A typical banana contains about 0.40 g of potassium. Naturally occurring potassium is mainly 39 isotope 40 1.8 1017 s1. The average atomic mass for natural potassium is 39.1 u. 19K, but 0.012% of it is the radioactive 19K, which has a decay constant of (a) Calculate the activity of a typical banana. (b) Does the radiation exposure from bananas outweigh their health benefit as a source of potassium, fibre, and vitamins A, B6, and C? Explain your reasoning. 54. After two years, 6% remains of the original radioisotope in a sample. Estimate the half-life of this isotope. 55. A nucleus of boron 10 5B absorbs an alpha particle and emits a proton. Use nuclear notation to write this reaction process, and identify the element that it produces. 56. Does an electron that moves from an energy level of 5.1 eV to an energy level of 6.7 eV emit or absorb a photon? Find the wavelength of the photon. 57. How much energy is produced by the conversion of 0.250 u of matter into energy? 58. Calculate the radius of a hydrogen atom in the n 2 state. 858 Unit VIII Atomic Physics 18-PearsonPhys30-U8-Closer 7/25/08 7:36 AM Page 859 59. An electron jumping from the n 3 to the n 2 state in a hydrogen atom emits a 656-nm photon. (a) Which state has the greater energy? (b) Find the energy difference between the Self-assessment 64. Outline how you would describe Rutherford\u2019s gold-foil experiment to a friend. Explain why the results were startling for physicists in 1910. two states. 60. Calculate the binding energy for 24 12Mg. 61. Find the parent atom for this decay:? \u2192 14N e v 7 62. Calculate the electrical charge of a particle composed of the quarks uus. 63. Find the activity of a sample containing 1.5 1020 radioactive atoms with a decay constant of 3.5 1015 s1. 65. (", "a) Explain why classical physics predicts that hydrogen will always produce a continuous spectrum rather than discrete spectral lines. (b) How does the Bohr model explain spectral lines? 66. Draw a concept map of the atomic physics topics that you find the most difficult. If you have trouble completing this concept map, discuss the concepts with a classmate or your teacher. 67. Explain why pair annihilation, such as e e \u2192 2, does not violate the law of conservation of mass. 68. List the four fundamental forces and explain which ones are involved in nuclear binding energy, decays, fission, and fusion. e TEST To check your understanding of atomic physics, follow the eTEST links at www.pearsoned.ca/school/physicssource. Unit VIII Atomic Physics 859 19-PearsonPhys-Appendix(SR) 7/25/08 7:35 AM Page 860 APPENDIX STUDENT REFERENCES Contents SR 1 SR 2 SR 3 SR 4 SR 5 SR 6 SR 7 Safety.......................................... 861 The Inquiry Process............................... 864 The Problem-Solving Process: GRASP.................. 867 Using Graphic Organizers........................... 869 Graphing Data................................... 872 5.1 Graphing Techniques......................... 872 5.2 Using the Graphing Tools...................... 874 Math Skills...................................... 875 6.1 Measurement: Accuracy and Precision........... 875 6.2 Mathematical Operations with Data.............. 876 6.3 Exponential Notation and Scientific Notation..... 877 6.4 Unit Conversions (Unit Factor Method).......... 878 6.5 Trigonometry for Solving Right Triangles......... 879 Tables.......................................... 880 7.1 SI Prefixes.................................. 880 7.2 Fundamental Quantities and Base Units.......... 880 7.3 Derived Quantities and Units................... 880 7.4 Numerical Constants.......................... 881 7.5 Atomic Masses of Selected Isotopes............. 881 7.6 Masses of Subatomic Particles.................. 881 860 Appendix Student References 19-PearsonPhys-Appendix(SR) 7/25/08 7:35 AM Page 861 SR 1 Safety In our modern society, safety has become much more than just protecting one\u2019s well being. Issues around safety have become extremely important to industry, business, governments, and all kinds of institutions including educational institutes. The understanding and application of safety in a broad sense has become an industry in itself. Today, even to be considered for many jobs, one must take safety courses or have a variety of safety-training certificates. Now is the time for you to continue developing an attitude and awareness of safety. Safety is everyone\u2019s responsibility. The Provincial Government, the local school board, your teachers, and you all have an important role in keeping a safe environment. Alberta Education has prepared a detailed document, \u201cSafety in the Science Class", "room\u201d, outlining safety roles and responsibilities, and providing extensive information on potential hazards and safety procedures. This document is available online; go to www.pearsoned.ca/school/physicssource and follow the link to Safety in the Science Classroom. Of particular interest to you in this physics course is Chapter 6: Physical Hazards. For more technical information on issues, follow the link to Health Canada Index and search the topic alphabetically. In general, your role in maintaining safety is to act responsibly by carefully following directions, learning how to recognize potential safety hazards, and how to respond to potentially unsafe situations and emergencies. If you are unsure about how to proceed, ask your teacher. The Canadian Hazardous Products Act requires chemical manufacturers to include all hazard symbols and the degree of hazard on product labels. You may recognize hazard symbols on many household products. These symbols may indicate hazard(s), precaution, and first-aid treatment. Hazardous Product and WHMIS Symbols Household hazardous product symbols indicate the type of danger and the degree of danger. They appear in either a triangle (which means \u201ccaution\u201d), a diamond (which means \u201cwarning\u201d), or an octagon (which means \u201cdanger\u201d). There are also numerous laboratory and industry hazard symbols in use. Some symbols relevant to Physics 20 and 30 are shown below: Figure SR 1.1 Laboratory and industry hazard symbols Many of the chemical products used in Canadian schools are manufactured in the United States. To standardize the labelling systems, WHMIS (the Workplace Hazardous Materials Information System) was developed. The symbols belonging to this system appear on materials and products used both in workplaces and our schools. Appendix Student References 861 19-PearsonPhys-Appendix(SR) 7/25/08 7:35 AM Page 862 2. Precautions with Mechanical Hazards a) Rotating machinery or moving devices can catch loose clothing, fingers or hair; therefore, keep a safe distance away from moving parts. b) Strong magnets can snap on ferromagnetic materials and other magnets very quickly. Use caution to avoid pinching skin or cutting clothing. c) Projectile launchers should be used only with appropriate eye protection and a clear \u201cline-of-fire.\u201d Be aware of the potential for a misfire or backfire. d) Model rockets with air, water, or solid-fuel motors can be a hazard. Wear eye protection and stay well clear of the launch area and potential trajectory. Make sure everyone watches for rocket parts falling back to the ground. 3. Precautions with Electrical Sources a) Do not use 110-V AC equipment if it has a damaged plug (e.g., missing the ground pin) or a frayed cord. Always disconnect the cord from the socket by pulling the plug, not the cord. b) Keep water and wet hands away from electrical cords. c) Do not touch a person in contact with live electrical currents. Disconnect the power source first. Then give artificial respiration if necessary. Call for help and treat burns. d) Make sure electrical cords are not placed where someone could trip over them. e) Do not allow a short circuit connection to a dry cell or battery. Dangerous amounts of heat can be generated in the wires and in the cells themselves, potentially causing an explosion or fire. f) Never attempt to recharge a non-rechargeable battery. Never cut open batteries. Their contents can be corrosive and poisonous. g) Keep flammable liquids away from electrical equipment. Sparks, in a motor for example, could ignite flammable vapours. Figure SR 1.2 Laboratory Safety Approach all investigations, especially in the laboratory, with maturity. Before you begin, read the instructions carefully, noting all safety precautions. In addition, your teacher may provide other safety reminders and rules pertaining to the laboratory activity. It is your responsibility to inform your teacher of medical conditions such as possible allergies to materials used (e.g., latex) or by-products of the activity. Inform your teacher if you wear contact lenses. 1. General Precautions and Safety Equipment a) Identify all safety equipment in the laboratory. b) Know the location of and how to operate safety equipment, including the fire extinguisher, fire blankets, eyewash fountains, sand, and the first-aid kit. c) Know how and where to get help if needed. d) Wear appropriate laboratory apparel, which may include safety goggles, gloves, and/or lab aprons. e) Tie back long hair and secure any loose clothing. 862 Appendix Student References 19-PearsonPhys-Appendix(SR) 7/25/08 7:35 AM Page 863 b) Never look directly into the beam of an operating laser, even one with a low power. The eye focuses the laser light onto the retina, resulting in a power density of about 50 times that of direct sunlight. This can cause pinpoint burns to the retina. c) Guard against stray reflections", " and turn the laser off when not in use. d) Use radioactive sources only under the direction of your teacher. e) In all cases, the potential for harm from radiation increases with exposure. Exposure can be minimized by limiting the time of use and maximizing the distance away from the source. h) Spark timers create a very short but high voltage spark, which can give a minor electrical shock to anyone who touches a \u201clive\u201d part of the circuit. Although the shock itself is not dangerous, the surprise and sudden reaction can cause elbows to fly or objects to be dropped. i) Some high voltage devices can cause nasty shocks or skin burns. Be aware of the potential danger of charged capacitors, tesla coils, electrostatic generators, and transformers. Use only under the guidance of your teacher. j) When hooking up circuits, always have your teacher check the circuit before turning on the power. 4. Precautions with Electromagnetic Radiation a) Never look directly into an infrared (IR) or an intense, visible ultraviolet (UV) light source. Intense light can harm the retina. UV and IR radiation are absorbed by the cornea and eye contents, and can cause burning and overheating or other damage. Appendix Student References 863 19-PearsonPhys-Appendix(SR) 7/25/08 7:35 AM Page 864 SR 2 The Inquiry Process Inquire: to seek knowledge of, to ask about, to investigate, or to seek information by asking. Have you ever seen \u201csun dogs\u201d? They are a common occurrence in Alberta, especially in the winter. Often they are coloured and associated with a coloured ring, or halo, around the sun. They seem to occur in thin clouds or in frosty air. You may have noticed that they are always the same distance from the sun and the colours are always in the same order: red closest to the sun and violet farther away. This set of observations can be the focus of an inquiry process. Figure SR 2.1 Prairie sun dogs The inquiry process is a model of learning incorporated in Alberta Education curricula; it is not a separate topic or option. The inquiry process is applicable to all learning, and is especially suited to learning physics. Learning an inquiry approach is more than a way to succeed in physics; it is a useful way to deal with problems and challenging situations throughout any future career. The inquiry process is non-linear (there may be some side-tracks or dead ends), flexible (you can bend the rules or the process), individual (you can develop your own process), and recursive (you will need to revisit or loop through parts of the process as you go). An inquiry process model contains six components that connect together, all around your own thinking or reflection on the process. These components are: \u2022 Planning \u2022 Retrieving \u2022 Processing \u2022 Creating \u2022 Sharing \u2022 Evaluating 864 Appendix Student References Planning In order to inquire, you must have something in mind about which to inquire. Normally, the planning stage involves recognizing a situation, event, topic, or occurrence for which there is some unknown component. This leads to questions. In the planning stage, you (or your teacher) will need to look at the situation at hand and ask a question to be investigated. You may have many questions, but part of the process is to reflect on the situation and narrow (or in some cases, broaden) your question so that it is something you can actually investigate. From there, you will need to develop your process to lead through the other stages of the inquiry process. One way of working through the planning stage is to ask yourself the following questions: \u2022 What do I want to know? \u2022 What do I think the result might be? \u2022 How can I find out? \u2022 What do I need to do to find out? \u2022 How will I know when I have found out? \u2022 What form will my final results take? \u2022 How can I best share my results with others? \u2022 How can I evaluate what I have done? In this stage, you will need to develop a clear inquiry question, propose a thesis or hypothesis, identify variables or related factors, create a data or information gathering process (experiment or research strategy), and recognize where your results may end up. Sometimes, this is the most difficult or lengthy part of the entire process, but very important. Retrieving Once you have an inquiry question, a hypothesis to test, and a plan to follow, you can begin the process of retrieving. This may involve experimentation or research. You may be gathering text information or numerical data from measurements. You may be using several different data sources including your own experimental data. In this stage, you may need to revisit Planning if you find difficulties or discrepancies during your information gathering. 19-PearsonPhys-Appendix(SR) 7/25/08 7:35 AM Page 865 Processing Evaluating The information or data you retrieve must be processed. You must evaluate the information, sort good data from poor data, decide what data is relevant,", " and perhaps re-identify variables. At this stage, you may find yourself looping through parts of the Planning and Retrieving stages again. Once you have good information, you need to decide how to display and analyze the data. What combination of data tables, graphs or graphic organizers should you use? Is there some mathematical analysis such as finding a slope, modelling a curve with an equation, or some statistical calculations that will be useful? Again, you may want to loop back to Planning, or ahead to Creating and Sharing to decide what detail of analysis you want to perform on your information. Ultimately, you need to be able to answer the question, \u201cWhat does this information mean?\u201d Creating You have asked a question, retrieved your information, and processed it. Now you must put the package together by creating a final product. Remember to look ahead to Sharing. In this product, you must clearly indicate your initial inquiry, provide a summary of your data or information, explain the meaning of your data, and state some conclusions regarding your inquiry question. At this point, you may find that you have more questions. This could lead you to another inquiry, and another, and another. This is the essence of how scientific knowledge continues to grow. Sharing This stage in the inquiry process is often devalued, yet it is a crucial step in the process. In science, if a new discovery is not communicated, then it is lost. In education, communicating ideas to others is one of the best learning processes. You don\u2019t really know or understand something until you can share the ideas with others. How you share will depend in part on how you created your final product. There are many modes of sharing: oral presentation, poster display, written report, demonstration, art work, working model, skit\u2026 Your job will be to choose a method that best fits you, your results, your classroom situation, and your audience. Again, you may need to loop back to Planning and Creating to get this in the most suitable form. Now is the time to look at the whole process (not just the results or answers). Ask yourself these questions: \u2022 What worked well? \u2022 What became a challenge? \u2022 Is there another or better way of doing any one of the stages in this process? \u2022 What parts of the process were easier or more difficult, or more or less effective? \u2022 How would I coach someone else to do this same inquiry in a more efficient way? \u2022 Are there other questions or situations that might be resolved by the process I followed? \u2022 What have I learned (about the inquiry question and about learning)? By critically evaluating what you have done, you will learn process skills beyond physics or science; you will develop skills to last a lifetime. A skeleton of a sample inquiry: Planning Situation: a solar halo display is very clear and colourful in the sky. Question: Where do the sun dog\u2019s colours come from? Hypothesis: This is the same effect as the rainbow. If I spray water into sunlight, then I should be able to see a halo. What to do: Research rainbows and halos in print and online. Do an experiment with light rays and water drops to try to create a halo. How do I know I have the answer: I can create a halo and a rainbow and match them together. My final results: I will have a poster display with photos I have taken, and I will explain the results to my classmates. Retrieving Internet search informs me that halos and rainbows are different. From photos, I see that the colours are reversed. There are different types of halos. Sun dogs are part of ice crystal halos caused by refraction. Revisit planning Question: Does refraction of light through ice or other transparent solid crystals model the position of colours in sun dogs? Appendix Student References 865 19-PearsonPhys-Appendix(SR) 7/25/08 7:35 AM Page 866 Hypothesis: If I shine a white light beam on a transparent solid crystal, then I will be able to see colours located in different directions. Experiment: Shine light through ice crystals, prisms, or other transparent solids, and look for colours in different directions. Sharing I will give a short presentation to my class to explain my question and results, and model the path of light with white and coloured yarn through a foam block. Classmates will be able to see \u201ccolours\u201d only at certain directions. Evaluating I will look critically at the entire process once I am finished. Processing Can I see colours produced by light passing through the crystals? Is there a way for me to quantify my observations, e.g., can I measure directions? To what degree of accuracy can I measure directions? Do I have enough data? Do I need to look at more variables such as the shape or material of the crystal? How can I record my observations in the most meaningful way for this context? Creating I will summarize my specific question and investigative", " process. My results include: data tables showing the approximate angles where I see different colours of light refracted through prisms of different shapes; three photos of my apparatus set-up and colours I could see; and internet photos of more sophisticated experiments. I will also create diagrams of what could be seen in my apparatus according to theory, and a poster display of the theory behind the experiment and actual halos. There are other questions arising from my work that I will pose on my poster. For example: What happens to the halo when an ice crystal tips on its side? 866 Appendix Student References 19-PearsonPhys-Appendix(SR) 7/25/08 7:35 AM Page 867 SR 3 The Problem-Solving Process: GRASP Solving Numerical Problems A significant amount of effort in physics is spent in solving problems that have a numerical component. Often these problems seem more difficult than they really are because they involve physics concepts, principles, and/or laws as well as mathematical operations. Research into the problemsolving abilities of professionals and novices shows that professionals have logical procedures they follow when solving problems while novices who are having difficulty do not. The more methods the problem-solver can apply, the more adept he/she is at problem solving. The approach used in numerical problems throughout the textbook follows four basic steps. These steps are easy to remember and apply because combined, the first letters of the key words spell GRASP. A description of the four steps is provided below. Step 1: List what is Given The first step in solving numerical problems involves answering the question, \u201cWhat information are we given?\u201d This is sometimes referred to as data extraction. To answer this question, read the problem carefully, study the information given, and represent physical quantities and numerical data with appropriate symbols, units, and directions (if necessary). Write the data in scientific notation to the correct number of significant digits (SR 6.3). Step 2: List what you are Required to find The second step involves answering the question, \u201cWhat am I required to find?\u201d To answer this question, identify what the problem is asking you to do. Be sure to note the units requested, if specified, and, for vector quantities, the direction. Answering this question will point you in the right direction and prevent you from being distracted by irrelevant information. Step 3: Analysis and Solution \u2014 Analyze the problem carefully and work out the Solution The third step requires a careful analysis of the problem before solving. To analyze the problem you must break it down into a series of logical steps. Begin by sketching a diagram. Many physics problems lend themselves to a diagram and the diagram often provides the key to solving the problem. Write down all the relationships you know involving the givens and the required. Also, write down any assumptions that must be made in order to solve the problem. An assumption is anything that must be taken for granted. Next, start with what you are trying to find, and answer the question, \u201cWhat additional information do I need to calculate the unknown?\u201d This may be a constant that you have to look up in a reference book or from a table of constants given in the textbook. Organize and sequence the information you have to form the solution. In physics, this often involves substituting appropriate data into an equation. It is good practice to rearrange an equation to solve for an unknown variable in terms of the other variables before substituting to obtain the final answer. Always be on the lookout for errors in the mathematical computations and check that the answer has the correct number of significant digits, and that appropriate units are included. Step 4: Paraphrase the solution The numerical answer should be stated in a form that answers the original question. Since the original question was a sentence, the statement of the final answer should also be a complete sentence. Physical quantities should include units and directions, if appropriate. You can use the following Numerical Problem Checklist to guide your work. Numerical Problem Checklist Given read problem carefully extract data represent physical quantities with appropriate symbols include units with physical quantities include directions where needed show the correct number of significant digits Required identify what the problem is asking for identify units of the final answer identify direction of the final answer (for vector quantities) Appendix Student References 867 Paraphrase write the final answer in a complete sentence check that units are included with numerals check accuracy of significant digits check accuracy of direction (if required) check that the original question has been answered and that the answer seems realistic 19-PearsonPhys-Appendix(SR) 7/25/08 7:35 AM Page 868 Analysis and Solution draw a sketch write down possible relationships list viable assumptions write an appropriate equation containing the needed unknown variable and only other variables that are known or can be found from the given data identify and look up constants needed identify inconsistent units and perform needed unit conversions identify \u201cred herrings\u201d (i.e., extra bits of", " numerical information not needed to solve the problem) note the least number of significant digits in the given data note directions (for vector quantities) rearrange the equation to solve for the unknown substitute data into the rearranged equation simplify the mathematics: solve for the numerical answer and simplify the units check the mathematical calculations check the number of significant digits check the direction (if required) 868 Appendix Student References 19-PearsonPhys-Appendix(SR) 7/25/08 7:35 AM Page 869 SR 4 Using Graphic Organizers Graphic organizers are effective tools that can help you learn. They enable you to solve problems and think critically, through analyzing similarities and differences, inferring sequences, and establishing cause-and-effect relationships. They generate discussion and negotiation of ideas, extend comprehension of a concept, theme, or topic, and lead to organized representation and presentation of understandings. You can use them to brainstorm, demonstrate what you know, and organize your thoughts before planning a presentation or writing a report or essay. The following chart outlines a number of graphic organizers, their intended purposes, and how to use them as you study science. Type of Graphic Organizer Purpose Method Concept Map connector descriptor concept Used to clarify relationships and linkages between concepts, events, or ideas Brainstorm ideas and link together from \u201cbig to small\u201d with arrows or lines linking words. Venn Diagram Web Diagram Used to visualize similarities and differences between two or more ideas, topics, or concepts Brainstorm similarities, and list these in the overlapping section of the two circles. Then brainstorm differences and list these in the non- overlapping sections. Used to clarify concepts and ideas by clustering them Cluster words and/or information around a central object, concept, or idea. Pie Chart Used to estimate the relationship of parts to the whole Estimate/research the importance or amount of proportionate time of each aspect of an event in relation to the whole. Flowchart/Sequence Chart Used to map out your thinking about an issue or to organize ideas for an essay or report Brainstorm aspects of the whole event. Select important aspects and put them into sequential order. Appendix Student References 869 19-PearsonPhys-Appendix(SR) 7/25/08 7:35 AM Page 870 Type of Graphic Organizer Purpose Method Ranking Ladder Used to rank ideas in order of importance Brainstorm ideas and rank them in order from least important (bottom rung) to most important (top rung). Fishbone Diagram Used to identify cause-and-effect relationships Right-Angle Diagram Used to explore the consequences of an idea and the impact of its application Target Diagram Used to weigh the importance of facts and ideas Identify a problem to be solved. List the \u201ceffect\u201d at the head of the fish. Brainstorm \u201cpossible causes\u201d in each bone. Rank the causes and circle the most probable ones, justifying your choice. Briefly describe the idea you are exploring on the horizontal arrow. Brainstorm consequences of the idea, and list these to the right of the horizontal arrow. Expand on one consequence, and list details about it along the vertical arrow. Describe social impacts of that trait below the vertical arrow. Brainstorm facts and ideas. Rank their importance and place the most important facts or ideas centrally, and the least important ones toward the outer rings. 870 Appendix Student References 19-PearsonPhys-Appendix(SR) 7/25/08 7:35 AM Page 871 Type of Graphic Organizer Purpose Method Agree/Disagree Chart Used to organize data to support a position for or against an idea or decision List a series of statements relating to a topic or issue. Survey agree-disagreement before discussion. Survey again after discussion and research. PMI (Plus, Minus, Interesting) Chart Used to summarize the positive and negative aspects of a topic or issue, as well as identify interesting aspects of the topic for possible further research Sort ideas or information about a topic or issue in a three-column chart that has the following headings: Plus (), Minus (), and Interesting. Gathering Grid Used to make distinctions between ideas or events Concept Hierarchy Diagram Used to identify and sequence the subordinate concepts needed to understand a higher-order concept Gather information on a number of ideas or events and arrange it on a grid. Each idea or event is assigned to a separate row. Analyze the information according to selected criteria in each specific column. Place the higher-order concept at the top of a page. Then consider the question, \u201cWhat concepts need to be understood before the higher-order concept above can be grasped?\u201d The same question is then asked for each of the subordinate concepts identified and a hierarchy of connected concepts is created. Appendix Student References 871 19-PearsonPhys-Appendix(SR) 7/25/08 7:35 AM Page 872 SR 5 Graphing Data 5.1 Graphing Techniques Table SR 5.1 Total mass of system as sand is added to", " a beaker Total Volume of Sand Added V (mL) Total Mass of System m (g) 28 55 84 106 148 174 210 188 258 333 391 500 567 661 Physicists make extensive use of graphs to convey information and to help determine how one physical quantity is affected by another. To review simple graphical analysis techniques, as an example, use the data from a simple measurement experiment where students added given volumes of sand as measured in a graduated cylinder to a beaker on a balance, recording the total volume of sand in the beaker and the total mass of the system as shown on the balance. The Data Table A data table is the most practical way to record quantitative data. Table SR 5.1 above shows the data from the student experiment of adding sand to a beaker on a balance. Note that the name of each variable, the symbol, and the unit of measurement (in round brackets) are recorded at the top of each column. The Title of the Graph Figure SR 5.2 shows a sample graph for a student\u2019s experiment. Every graph needs a title to describe what it is about. The title is placed at the top of the graph or in a box on a clear area above the graph. The Axes of the Graph Plot the independent variable on the horizontal x-axis and the dependent variable on the vertical y-axis. The variable that is changed intentionally is called the manipulated or independent variable. Volume of sand in the beaker was the manipulated or independent variable in the experiment 872 Appendix Student References 800 600 400 200 ) g ( m s s a M 0 0 Total Mass of System vs. Volume of Sand in Beaker m 640 g 165 g 475 g V 200 mL 20 mL 180 mL 100 200 Volume V (mL) 300 Figure SR 5.2 as students chose how much to add for each trial. The mass of the system depended on how much sand was added, thus the mass was the responding or dependent variable. Label each axis with the name, symbol, and unit of the variable being plotted, as shown in Figure SR 5.2. Scales are chosen for each axis to spread the measured values across the graph paper without making the plotting difficult. The maximum values in the data table determine the maximum numbers on the scales of the axes. To set a scale for an axis, analyze the data to be plotted and choose increments appropriate to the data. In Table SR 5.1, the maximum volume value was 210 mL, and the minimum volume value was 28 mL. Increments of 20 mL on the x-axis would be appropriate for the data. As well, the maximum total mass of system was 661 g, and the minimum total mass of system was 188 g. Increments of 50 g on the y-axis would be appropriate for the data. Plotting the Data and Drawing the Line of Best Fit Use a pencil to plot the data points as accurately as possible by making a small visible dot. Accuracy is important. Use the actual data values and make your best estimate of values within the scale grid. Once all of the data points have been plotted, a line of best fit is drawn. A line of best fit is a line that shows the trend of the points. Make the smoothest curve you can, balancing points that do not fit the curve evenly above and 19-PearsonPhys-Appendix(SR) 7/25/08 7:35 AM Page 873 below the curve. Do not try to have the curve or straight line go through all the dots since most data points have some error. The scatter of the data points from the smooth line indicates the extent of the errors in the data. Where a point is far off the line, a serious error may have been made. If this occurs, measure the data for that point again, if possible. If you believe an error was made, still plot the point, but ignore it while drawing your best fit curve. Interpolating from the Graph Interpolation is the process of estimating a value that is between two directly measured data points of a variable. To interpolate, first locate the point on the appropriate axis for the value of the variable in which you are interested. Next, draw a line perpendicular to this axis to intercept the line of best fit. From this point on the line of best fit, draw a second line perpendicular to the second axis. Read the value of the second variable from this axis. For example, in Figure SR 5.2, a volume of 70 mL of sand is interpolated to a total mass of 300 g (indicated by the small star). There is some risk of inaccuracy involved in interpolation, since it is assumed that the trend of the line continues between the measured points. This assumption is not always valid. Extrapolating from the Graph Extrapolation is the process of estimating the values of a data point beyond the limits of the known or measured values. However, there is a considerable risk of inaccuracy, because it is assumed that the trend of the curve continues outside", " the range of the data. When the line is extended, a dotted line is used to show that the extension is little more than guesswork. The arrow in Figure SR 5.2 shows the process of extrapolating the curve to find the mass of the system without any sand in the beaker. What is this value and what does it represent? How valid is the value? Calculating the Slope If the line of best fit is straight, the slope of the line can be found. The slope of the line is defined as the rate of change of one variable with respect to the other and is found by the ratio of the rise to the run. To find the slope, find two points far apart on the line of best fit whose values are easily readable from the scales on the axes. Note that the points used should not be data points (i.e. values that were measured and plotted to draw the line of best fit). On the graph, lightly draw a horizontal line from the lower point and a vertical line from the higher point so that they intersect. Use the axis scale to determine the change in vertical value (rise) and change in horizontal value (run) along these two line segments. In Figure SR 5.2, the rise is shown as 475 g and the run as 180 mL. The slope of this graph is thus: slope rise/run 475 g/180 mL 2.64 g/mL Notice that the slope in this example has units; it also has some physical significance: it represents the density of the sand. Writing the Equation of the Line If the trend of the curve is a straight line, then changes in the plotted variables are directly proportional to each other. As the change in one variable doubles, the change in the other doubles, and vice versa. The general equation for a straight line is y mx b, where y is the variable on the vertical axis, x is the variable on the horizontal axis, m is the slope of the line, and b is the vertical axis intercept. Figure SR 5.2 shows a straight line for volumes of sand at least up to 210 mL. For this range, the change in mass is directly proportional to the volume of sand added. The general equation for the linear graph in Figure SR 5.2 is: m V b where m is the total mass of the system, is the density of the sand, V is the total volume of sand in the beaker, and b is the mass of the empty beaker. The specific equation for this graph is: m (2.64 g/mL)V 115 g. Using the Equation of the Line It is often more convenient to extrapolate or interpolate from the specific equation than from the graph. For example, the total mass of the system could be determined if 800 mL of sand is added, even though our beaker may not be that big and our graph does not extend that far. To do this, substitute 800 mL for V in the equation and calculate m. The accuracy of this result depends on the accuracy of the equation, which in turn depends on the accuracy of the determination of slope and vertical axis intercept. Appendix Student References 873 19-PearsonPhys-Appendix(SR) 7/25/08 7:35 AM Page 874 5.2 Using the Graphing Tools Graphing calculators make the process of plotting and interpreting graphs easy and efficient. Data from an experiment can be entered into the calculator and displayed as a scatterplot. The calculator can be used to determine the function that best models a given scatterplot. The information provided for this function can also be used to write the equation that best describes the relationship between the two plotted variables. The graphing calculator can also be used to help us explore the graph of a given equation or relationship. It can be used to interpolate values between the plotted points or to extrapolate values beyond the plotted points. These uses make the graphing calculator a very powerful laboratory tool. Data can also be stored and plotted in a com- puter spreadsheet such as Microsoft Excel. The eMath activities in this textbook provide opportunities for you to use the graphing calculator or a computer spreadsheet. 874 Appendix Student References 19-PearsonPhys-Appendix(SR) 7/25/08 7:35 AM Page 875 SR 6 Math Skills 6.1 Measurement: Accuracy and Precision Measurement is a process of comparing some unknown characteristic, attribute, or quantity to some known or accepted scale or standard. Measurement involves tools and technique. If there is a problem with either, there can be problems with the quality of the measured data. A tape measure with the end broken off would give incorrect measurements if the problem was not noticed. Using a metre stick to measure the thickness of a hair would not work well. Forgetting to include the weight of the fuel when weighing in freight to load a plane could be disastrous. At best, a measurement is an estimate; there is always some amount of uncertainty to the value you record. To", " make a measurement, and to use measured values correctly, we must understand a number of important issues. Accuracy Accuracy is a means of describing the quality of measurements, or how closely a measurement agrees with the accepted or actual value of the quantity being measured. A broken tape measure will not give accurate values for length. The difference between an observed value (or the average of observed values) and the accepted value is called the deviation. The size of the deviation is an indication of the accuracy. Thus, the smaller the error, the greater is the accuracy. The percent deviation is determined by subtracting the accepted value from the measured value, dividing this by the accepted value, and multiplying by 100. Thus, percent deviation (measured value accepted value) accepted value 100% red-heads. The smaller the divisions of its scale, the less uncertainty there will be in reading values. Any measurement that falls between the smallest divisions on the measuring instrument is an estimate. We should always try to read any instrument by estimating tenths of the smallest division. For a ruler calibrated in centimetres, this means estimating to the nearest tenth of a centimetre, or to 1 mm. Using this procedure, the length of the object in Figure SR 6.1 is found to be 6.7 cm. We are certain of the 6, but the 0.7 is an estimate. In reality, it could easily be 0.6 or 0.8. It is however, unlikely that it would be 0.5 10 11 I 6.7 cm Figure SR 6.1 Figure SR 6.2 shows the measurement of the same object using a ruler calibrated in millimetres. The reading estimated to the nearest tenth of a millimetre appears closest to 6.74 cm. It might be tempting to record the length as either 6.7 cm or 6.8 cm. This would be wrong. We can tell that the length is between the two divisions. The estimated digit is always shown when recording the measurement. The estimated digit in this reading is 0.04 cm 10 11 Precision Precision is the degree of repeatability of measurements; it depends on your care and technique. If the same measurement is carefully made several times independently, we find that we may get variations in the last digit we read. This limitation defines the precision of the measurement. The precision of a measuring instrument depends on how finely the scale is divided. A ruler with mm divisions will not be useful in measuring the difference in hair thickness between blondes and Figure SR 6.2 Figure SR 6.3 shows a different object being measured with a ruler calibrated in centimetres. The length falls exactly on the 6-cm mark. Should the length be recorded as 6 cm or 6.0 cm? Remember that with a centimetre ruler we can estimate to tenths of a centimetre. With this ruler we can therefore distinguish readings of 5.9 cm and 6.1 cm. The object is right on a division mark, Appendix Student References 875 19-PearsonPhys-Appendix(SR) 7/25/08 7:35 AM Page 876 so the estimated digit is zero-tenths of a centimetre. Zero-tenths is indicated by 0.0 and the correct reading is 6.0 cm, not 6 cm. \u2022 All of the digits from one to nine (1, 2, 3, \u2026 9) are significant, so 424.7 m or 0.4247 km each have four significant digits 10 11 Figure SR 6.3 Indicating the Precision of Measured Quantities The precision of a measurement is indicated by the number of decimal places. For example, 2.861 cm is more precise than 581.86 cm even though the latter contains more digits. This is because the three decimal places in 2.861 make it precise to the nearest one-thousandth of a centimetre, while the two decimal places in 581.86 make it precise to the nearest one-hundredth of a centimetre. In this physics textbook, angle measures will have a precision no better than 0.1\u00b0 (that is, angle measures are either whole numbers or to one decimal place only). Significant Digits The accuracy of a measurement is indicated by the number of significant digits. Significant digits are the specific number of digits used to communicate the degree of uncertainty in a measurement. When we are expressing a physical quantity as a number, how many significant digits should we indicate? These rules should help you decide. \u2022 Numbers obtained by counting are considered to be exact and contain an infinite number of significant digits. For example, if there are 12 stopwatches in a classroom, there are not 11 stopwatches, or 13 stopwatches, or 12.35 stopwatches. There are exactly 12.000\u2026 stopwatches. The zeros may be extended to as many decimal places as necessary in calculations. \u2022 Numbers obtained from definitions are considered to be exact and contain an infinite number of significant digits. For example, 1 m", " 100 cm, and 1 kWh 3600 kJ, are definitions of equalities. ( 3.141 592 654) has an infinite number of decimal places, as do numbers in formulae such as P 4s, where P is perimeter of a square, and s is the length of a side. 876 Appendix Student References \u2022 All zeros to the left of the first non-zero digit are not significant. For example, 1.4 kg and 0.0014 kg each have two significant digits. \u2022 Zeros between other non-zero digits are significant. Therefore, 501.009 s has six significant digits. \u2022 Alberta Education Assessment Branch considers all trailing zeros to be significant. Any zero to the right of a non-zero digit is significant. Therefore, the mass of an object written as 2000 kg has four significant digits. If the mass is a stated value, not a measured value, we can indicate that we know it to four significant digits by using scientific notation and writing the mass as 2.000 103 kg. Good measurements have high degrees of both accuracy and precision. In order to assure better accuracy, instruments should be calibrated. Calibration involves making sure the scale divisions are spaced properly and that the zero reading is correct. Better precision is attained by using better tools, those with finer scales. Good technique, of course, is also necessary for both precision and accuracy. 6.2 Mathematical Operations with Data When doing calculations with measured values, never keep more digits in the final answer than in the least accurate number in the calculation. For example, 0.6 0.32 0.9, not 0.92. This procedure for using only meaningful digits is called rounding off. The procedure for rounding off digits is as follows: \u2022 When the first digit discarded is less than five, the last digit retained is left the same. Notice that we start rounding off at the digit immediately after the last digit we are retaining. For example, 14.248 kg rounded to three digits is 14.2 kg, since the fourth digit (4) is less than five. \u2022 When the first digit discarded is a five or greater, we increase the last digit retained by one. Therefore, 7.8361 km rounded to three digits is 7.84 km, and 4.255 01 s rounded to three digits is 4.26 s. \u2022 Consider numbers that are exact counts to be perfectly precise. For example, the average mass of three cars having masses of 1000 kg, 1250 kg, and 1165 kg is (1000 kg 1250 kg 1165 kg)/3 or 1138 kg. The denominator (3) in this example is an exact count, and therefore the answer includes four significant digits. 19-PearsonPhys-Appendix(SR) 7/25/08 7:35 AM Page 877 \u2022 Consider fractions and defined equalities to be perfectly precise. The fraction 1 in the equa2 1 mv2 does not influence rounding off. tion Ek 2 Neither does the defined equality 10 mm 1 cm. Note: In the examples in this textbook, intermediate steps are shown rounded off to one extra significant digit. In reality, all digits are carried through the calculations until the final answer is reached, at which point the final answer is rounded off appropriately. Rules for Significant Digits in Mathematical Operations Adding or subtracting when it is the only operation: the precision, as shown by the number of decimal places in the values being used, determines the number of significant digits in the answer. Round off the answer to the same precision as the least precise value used. For example, 11.2 kg 0.24 kg 0.336 kg 11.776 kg, is rounded off to 11.8 kg because the least precise value, 11.2 kg, is only given to the first decimal place. Multiplying or dividing when it is the only operation: the value with the least number of significant digits determines the number of significant digits in the answer. For example, a distance of 34.28 m is travelled in a time of 4.8 s. The average speed is calculated by (34.28 m)/(4.8 s) 7.141 666 m/s, rounded off to 7.1 m/s to two significant digits because the time of 4.8 s has only two significant digits. When a series of calculations is performed, each interim value should not be rounded before carrying out the next calculation. The final answer should then be rounded to the same number of significant digits as are contained in the quantity in the original data with the lowest number of significant digits. For example: In calculating (1.23)(4.321) \u00f7 (3.45 \u2013 3.21), three steps are required: a. 3.45 \u2013 3.21 = 0.24 b. (1.23)(4.321) = 5.314 83 c. 5.314 83 \u00f7 0.24 = 22.145 125 The answer should be rounded to 22.1 since", " 3 is the lowest number of significant digits in the original data. The interim values are not used in determining the number of significant digits in the final answer. When calculations involve exact numbers (counted and defined values), the calculated answer should be rounded based upon the precision of the measured values. For example: 12 eggs 49.6 g/egg 595.2 g (to one decimal place) Note: In this textbook, in calculations involving angle measurements, the number of significant digits in the angle measurements is not taken into consideration when determining the number of significant digits for the final answer. The rules of operation apply to all other measurements. Answers for angle measurements should be no more precise than the least precise in the given angle data. 6.3 Exponential Notation and Scientific Notation Exponential Notation Exponential notation makes use of powers of ten to write large and small quantities and to convey the number of significant digits. The first part of the number is called the coefficient, and the power of ten is the exponent. The radius of Earth may be written in exponential notation to three significant digits as 638 104 m, 63.8 105 m, 6.38 106 m, or 0.638 107 m. The diameter of a typical atom may be expressed to one significant digit as 1 108 cm or 0.1 107 cm. Scientific Notation Any measurement that consists of a coefficient multiplied by a power of ten is expressed in exponential notation. Both 6.38 106 and 0.638 107 are in exponential notation. Scientific notation is a special kind of exponential notation. For a number to be in scientific notation, the coefficient must be greater than or equal to 1, and less than 10. This means that 6.38 106 is expressed in scientific notation and 0.638 107 is not. Scientific notation enables us to show the correct number of significant digits. Remember that any zero to the right of the decimal point is significant. Therefore, if all four digits in the measurement 3400 J are significant, then it would be written in scientific notation as 3.400 103 J. However, if only two digits are significant, it would be as 3.4 103 J. Sometimes you may be required to express the results of calculations in scientific notation. This involves moving the decimal point and changing the exponent until the coefficient is between 1 and 9. The exponent is decreased by one for each position the decimal point in the coefficient is moved to the right, and increased Appendix Student References 877 19-PearsonPhys-Appendix(SR) 7/25/08 7:35 AM Page 878 by one for each position the decimal point is moved to the left. For example, exponent in the answer remains the same as the largest exponent. For example, a) 500 104 5.00 102 104 5.00 1024 5.00 106 b) 0.068 103 6.8 102 103 6.8 102(3) 6.8 105 Exponential Notation and Mathematical Operations Multiplication The product of exponential numbers is determined by multiplying the coefficients and adding the exponents. For example, (3.0 102)(4.0 106) (3.0 4.0)(102 106) 12 102(6) 12 104 1.2 103 Division To divide numbers written in scientific notation, the coefficients are first divided, and then the exponent in the denominator is subtracted from the exponent in the numerator. For example, 3.3 105 6.6 102.50 105(2) 0.50 1052 0.50 107 5.0 106 Addition and Subtraction When the Exponents are the Same When the exponents are the same, the coefficients are added or subtracted as in normal arithmetic. The exponent in the final answer remains the same. For example, (2 104) (3 104) (1 104) (2 3 1) 104 4 104 Addition and Subtraction When the Exponents are Different When the exponents are different, the numbers must first be converted to a form in which all exponents are the same. The decimal point is moved so that all have the same exponent as the largest number in the group. Then, the coefficients are added or subtracted accordingly. The 878 Appendix Student References (1.00 103) (2.00 104) (400 105) (1.00 103) (0.200 103) (4.00 103) (1.00 0.200 4.00) 103 4.80 103 6.4 Unit Conversions (Unit Factor Method) Conversions are often necessary in both math and science problems. Whether you are working on a problem involving the metric system or converting moles to grams, the unit factor method is a useful tool. The unit factor method is the sequential application of conversion factors expressed as fractions. They are arranged so that any dimensional unit appearing in both the numerator and denominator of any of the fractions can be cancelled out until only the desired set of dimensional units is obtained Example 1: Convert a speed of", " 25 m/s to km/h. The equivalent relationships are: 1000 m 1 km, 60 s 1 min, and 60 min 1 h. Multiplying by unit factors and carefully analyzing the units results in 0 6 k m m 25 1 m n i 0 0 s 10 60 m in m 90 km/h h 1 1 s Example 2: 1 ng 109 g, therefore, 9 g 10 n g 1 1 or 9 g n g 10 This type of relation can be used to convert units from one to another. Example 3: Convert 5.3 mL to L. Multiply by factors of 1 to remove the prefix \u201cm\u201d and to introduce \u201c\u201d. 3 L L 10 5.3 103 L 5.3 mL 6 L 10 m L mL \u201ccancel\u201d as do L, leaving the desired units of L. Subtract the exponents for division: 3 (6) 3 This skill is also useful for drawing scale diagrams. For example, a force of 840 N is to be drawn at a scale of 1 cm 50 N. The scale length, in centimetres, will be found thus, 1 840 N 50 cm 16.8 cm N 19-PearsonPhys-Appendix(SR) 7/25/08 7:35 AM Page 879 6.5 Trigonometry for Solving Right Triangles A right triangle is a special triangle with one right angle (90\u00b0). The side opposite the right angle is always the longest side and is called the hypotenuse. The other two sides are called the legs. There are several important relationships that allow us to solve right triangles as long as we know the lengths of any two sides, or the length of one side and the measure of one of the acute angles. \u03c6 a Figure SR 6.4 c b Pythagorean Theorem The square of the length of the hypotenuse is equal to the sum of the squares of the length of each of the legs. c2 a2 b2 Sine Ratio Sine of one of the acute angles is equal to the ratio of the length of the leg opposite the angle to the length of the hypotenuse. sin opp/hyp sin a/c Cosine Ratio Cosine of one of the acute angles is equal to the ratio of the length of the leg adjacent the angle to the length of the hypotenuse. cos adj/hyp cos b/c \u03b8 Tangent Ratio Tangent of one of the acute angles is equal to the ratio of the length of the leg opposite the angle to the length of the leg adjacent to the angle. In general, the relationships are: Angle Sum The angles of a plane triangle add to 180\u00b0; the acute angles of a plane right triangle add to 90\u00b0. tan opp/adj tan a/b Appendix Student References 879 19-PearsonPhys-Appendix(SR) 7/25/08 7:35 AM Page 880 SR 7 Tables 7.1 SI Prefixes 7.2 Fundamental Quantities and Base Units Prefix Symbol Scientific Notation yotta- zetta- exa- peta- tera- giga- mega- kilo- hecto- deka- deci- centi- milli- micro- nano- pico- femto- atto- zepto- yocto da d c m n p f a z y 1024 1021 1018 1015 1012 109 106 103 102 101 101 102 103 106 109 1012 1015 1018 1021 1024 Quantity Length or Distance Mass Time Electrical current Thermodynamic temperature Quantity of matter Quantity Symbol m t I T n Unit metre kilogram second ampere kelvin mole Unit Symbol m kg s A K mol 7.3 Derived Quantities and Units Quantity Area Volume Speed, Velocity Acceleration Frequency Force Momentum Impulse Quantity Symbol A V v a f F p J Unit square metre cubic metre metre per second metre per second squared hertz newton kilogrammetre per second newtonsecond Energy, Work E, W Power Electric charge Electric potential Electric resistance Activity Absorbed dose Equivalent absorbed dose P q V R A D E joule watt coulomb volt ohm becquerel gray sievert Unit Symbol Expression in terms of SI Base Units m2 m3 m/s m/s2 Hz N \u2014 \u2014 \u2014 \u2014 s1 kgm/s2 kgm/s \u2014 Ns J W C V Bq Gy Sv kgm/s kgm2/s2 kgm2/s3 As kgm2/s3A kgm2/s3A2 s1 m2/s2 m2/s2 880 Appendix Student References 19-PearsonPhys-Appendix(SR) 7/25/08 7:35 AM Page 881 7.4 Numerical Constants 7.6 Masses of Subatomic Particles Name Symbol Value Particle Symbol Mass (u) Elementary unit of", " charge Gravitational constant Coulomb\u2019s constant Atomic mass unit Rest mass of an electron Rest mass of a proton Rest mass of a neutron Speed of light Planck\u2019s constant Rydberg\u2019s constant (hydrogen) e G k u me mp mn c h RH 1.60 1019 C 6.67 1011 Nm2/kg2 8.99 109 Nm2/C2 Electron Proton Neutron me mp mn 5.485 799 104 1.007 276 1.008 665 1.66 1027 kg 9.11 1031 kg 1.67 1027 kg 1.67 1027 kg 3.00 108 m/s 6.63 1034 Js 1.097 107 m1 7.5 Atomic Masses of Selected Isotopes Isotope Symbol Atomic Mass (u) Isotope Symbol Atomic Mass (u) hydrogen deuterium tritium helium-3 helium-4 carbon-12 nitrogen-16 oxygen-16 neon-20 neon-22 sodium-22 sodium-23 magnesium-24 silicon-28 silicon-30 phosphorus-30 potassium-39 potassium-40 calcium-40 iron-56 iron-58 cobalt-60 nickel-60 bromine-87 krypton-92 zirconium-94 H H (or D) H (or T) He He C N O Ne Ne Na Na Mg Si Si P K K Ca Fe Fe Co Ni Br Kr Zr 1.007 825 2.014 102 3.016 049 3.016 029 4.002 603 12 (by definition) 16.006 102 15.994 915 19.992 440 21.991 385 21.994 436 22.989 769 23.985 042 27.976 927 29.973 770 29.978 314 38.963 707 39.963 998 39.962 591 55.934 938 57.933 276 59.933 817 59.930 786 86.920 711 91.926 156 93.906 315 tellurium-112 tellurium-139 cerium-140 cesium-140 barium-141 neodymium-144 lanthanum-146 lead-204 lead-208 polonium-208 lead-210 polonium-212 polonium-214 radon-222 radium-226 thorium-230 thorium-234 protactinium-234 uranium-235 uranium-238 Te Te Ce Cs Ba Nd La Pb Pb Po Pb Po Po Rn Ra Th Th Pa U U 111.917 010 138.934 700 139.905 439 139.917 282 140.914 412 143.910 087 145.925 791 203.973 044 207.976 652 207.981 246 209.984 189 211.988 868 213.995 201 222.017 578 226.025 410 230.033 134 234.043 601 234.043 308 235.043 930 238.050 788 Note: Measurements of the atomic mass of most stable isotopes are accurate to at least a millionth of an atomic mass unit. However, the masses of highly unstable isotopes are more difficult to measure. For such isotopes, measurement errors can be large enough that the last one or two digits listed for their masses are not known for certain. Appendix Student References 881 20-PearsonPhys-Glossary 7/25/08 7:39 AM Page 882 GLOSSARY Note: The number in parentheses at the end of each definition indicates the page number in this book where the term first appears. A acceleration vector quantity representing the change in velocity (magnitude or direction) per unit time; non-uniform motion (23) acceleration due to gravity constant acceleration of an object falling near Earth\u2019s surface (54) activity (decay rate) number of nuclei in a sample that decay within a given time (812) alpha radiation emission of a helium nucleus; symbol is (797) altitude elevation of the ground above sea level (221) flow of 1 C of charge past ampere a point in a conductor in 1 s (602) amplitude of oscillation maximum displacement of a body from its equilibrium position during oscillatory motion (355) angle of diffraction angle formed between the perpendicular bisector and the straight line to a nodal or antinodal point on the interference pattern (689) angle of incidence angle formed between the incident ray and the normal line (654) angle of reflection angle formed between the reflected ray and the normal line (654) angle of refraction angle formed between the normal line and the refracted ray (666) annihilate energy (836) convert entirely into form of matter that antimatter has a key property, such as charge, opposite to that of ordinary matter (804", ") antinode point of interaction between waves on a spring or other medium at which only constructive interference occurs; in a standing wave, antinodes occur at intervals ; in an interference pattern, of 1 2 antinodes occur at path difference intervals of whole wavelengths (417) 882 Glossary fundamental armature (rotor) component of a simple DC electric motor consisting of a rotating loop of conducting wire on a shaft (608) artificial satellite artificially created object intended to orbit Earth or other celestial body to perform a variety of tasks; includes weather, communication, observation, science, broadcast, navigation, and military satellites (284) at rest not moving; stationary (13) atomic mass number number of nucleons in the nucleus, Z N; symbol is A (790) 1 of exactly atomic mass unit 1 2 the mass of the carbon-12 atom; symbol is u, where 1 u 1.660 539 1027 kg (791) atomic number number of protons in a nucleus; symbol is Z (790) axis of rotation imaginary line that passes through the centre of rotation perpendicular to circular motion (242) shaft on which a wheel axle rotates (242) B ballistic pendulum type of pendulum used to determine the speed of bullets before electronic timing devices were invented (483) baryon hadron with half-integer spin (842) becquerel unit of activity equal to 1 decay per second; unit is Bq (812) beta-negative decay nuclear decay involving emission of an electron; symbol is (802) beta-positive decay nuclear decay involving emission of a positron; symbol is (805) beta radiation emission of a highenergy electron; symbol is (797) binding energy net energy required to liberate all of the protons and neutrons in a nucleus (793) blackbody object that completely absorbs any light energy that falls on it (705) blackbody radiation curve of the intensity of light emitted versus wavelength for an object of a given temperature (705) graph Bohr radius orbit in a hydrogen atom (774) radius of the smallest boson particle with integer spin (842) bright fringe (antinodal line) of constructive interference (686) region bubble chamber device that uses trails of bubbles in a superheated liquid to show the paths of charged particles (831) C capacitor two conductors, holding equal amounts of opposite charges, placed near one another without touching (642) cathode ray free electrons emitted by a negative electrode (754) central maximum line of antinodes along the perpendicular bisector of the line joining the point sources (426) centre of curvature (C) point in space representing the centre of the sphere from which a curved mirror was cut (657) centre of mass point where the total mass of an object can be assumed to be concentrated (492) centripetal acceleration acceleration acting toward the centre of a circle (244) force acting centripetal force toward the centre of a circle causing an object to move in a circular path; centre-seeking acceleration (244) charge migration movement of electrons in a neutral object where one side of the object becomes positive and the other side becomes negative (520) charge shift movement of electrons in an atom where one side of an atom becomes positive and the other side becomes negative (521) charging by induction process of polarizing an object by induction while grounding it (521) closed pipe (closed tube) pipe closed at one end; the longest wavelength that can resonate in a closed pipe is four times the length of the pipe (419) closed-pipe (closed-tube) resonance if an antinode occurs at 20-PearsonPhys-Glossary 7/25/08 7:39 AM Page 883 the open end of a pipe, a point of resonance (resulting from constructive interference) occurs at the open end of the pipe, and the sound appears to be amplified (419) cloud chamber device that uses trails of droplets of condensed vapour to show the paths of charged particles (830) along the same straight collinear line, either in the same or in opposite directions (71) non-collinear not along the same straight line (80) collision interaction between two objects where each receives an impulse (469) elastic collision collision in which the total kinetic energy of Ekf ) a system is conserved (Eki (481) inelastic collision collision in which the total kinetic energy of a system is not conserved (Eki Ekf ) (483) colour quantum property related to the strong nuclear force (848) commutator fundamental component of a simple DC electric motor consisting of a mechanism for maintaining a properly polarized connection to the moving coil in a motor or generator (608) components perpendicular parts (Rx and Ry) into which a vector can be separated (77) Compton effect length of the scattered X-ray photon (721) change in wave- Compton scattering an X ray by an electron (721) scattering of conduction process of charging", " an object through the direct transfer of electrons when a charged object touches a neutral object (519) conductor material in which electrons in the outermost regions of the atom are free to move (513) forces that act conservative forces within systems but do not change their mechanical energy; includes gravity and elastic forces (314) non-conservative forces such as friction, and forces applied from outside a system, that cause the energy of the forces, system to change so that energy is not conserved (319) converging lens lens that refracts rays travelling parallel to the principal axis inward to intersect at the principal focus (677) converging mirror concave reflecting surface that causes parallel light rays to converge after being reflected (657) coulomb (C) SI unit for electric charge, equivalent to the charge on 6.25 1018 electrons or protons (517) Coulomb\u2019s law magnitude of the force of electrostatic attraction or repulsion (\u23d0F e\u23d0) is directly proportional to the product of the two e\u23d0 q1q2) and charges q1 and q2 (\u23d0F inversely proportional to the square of the distance between their centres r (529) crest region where the medium rises above the equilibrium position (394) for any two media, critical angle the size of the incident angle for which the angle of refraction is 90\u00b0 (672) current quantity of charge that flows through a wire in a given unit of time (602) cycle one complete back-and-forth motion or oscillation (249) cyclotron particle accelerator Particle accelerator in which the magnetic field perpendicular to the paths of the charged particles makes them follow circular paths within two hollow semicircular electrodes. An alternating voltage accelerates the charged particles each time they cross the gap between the two electrodes. The radius of each particle\u2019s path increases with its speed, so the accelerated particles spiral toward the outer wall of the cyclotron. (841) D dark fringe (nodal line) destructive interference (686) region of daughter element duced by a decay process (799) element pro- decay constant probability of a nucleus decaying in a given time; symbol is (811) diffraction change in shape and direction of a wave front as a result of encountering a small opening or aperture in a barrier, or a corner (685) sheet of glass diffraction grating or plastic etched with a large number of parallel lines; when light is incident on the grating, each line or slit acts as one individual light source (692) diffuse (irregular) reflection behaviour describing parallel incident rays scattered in different directions when reflected from an irregular surface (653) dispersion separation of white light into its components (675) displacement straight line between initial and final positions; includes magnitude and direction (7) length of the path taken distance to move from one position to another (7) lens that refracts diverging lens rays travelling parallel to the principal axis outward to appear as though they have originated at a virtual principal focus (677) convex reflecting diverging mirror surface that causes parallel light rays to spread out after being reflected (657) diverging ray ray that spreads out as it moves away from the origin (397) domain region of a material in which the magnetic fields of most of the atoms are aligned (589) Doppler effect apparent change in frequency and wavelength of a wave that is perceived by an observer moving relative to the source of the wave (429) drift tube particle accelerator particle accelerator in which alternating voltage accelerates charged particles through a series of electrodes shaped like open tubes; particles are always attracted to the next tube in the line (841) Glossary 883 20-PearsonPhys-Glossary 7/25/08 7:39 AM Page 884 dynamics branch of mechanics dealing with the cause of motion (126) E eccentricity degree to which an ellipse is elongated; number between 0 and 1, with 0 being a perfect circle and 1 being a parabola (269) efficiency ratio of the energy output to the energy input of any system (324) elastic potential energy energy resulting from an object being altered from its standard shape, without permanent deformation (300) electric field lines lines drawn to represent the electric field; density of the lines represents the magnitude of the electric field (554) electric poten- electric potential tial energy stored per unit charge at a given point in an electric field; symbol is V (564) electric potential difference change in electric potential experienced by a charge moving between two points in an electric field (565) electric potential energy energy stored in a system of two charges a certain distance apart; change in electric potential energy equals work done to move a small charge (Ep electromagnet magnet having its magnetic field produced by electric current flowing through a coil of wire (588) W) (561) electromagnetic radiation (EMR) radiant energy in the form of a wave produced by the acceleration of electrons or other charged particles; does not require", " a material medium; can travel through a vacuum (636) electron volt change in energy of an electron when it moves through a potential difference of 1 V; unit is eV (564) electrostatics charges at rest (513) study of electric fundamental electroweak force force that combines the electromagnetic force and the weak nuclear force (848) 884 Glossary elementary unit of charge on a proton; symbol is e (762) charge elongated circle; consists ellipse of two foci, a major, and a minor axis (269) energy ability to do work (292) energy level discrete and quantized amount of energy (773) equilibrium position rest position or position of a medium from which the amplitude of a wave can be measured (394) excited state higher than the ground state (775) any energy level F femto metric prefix meaning 1015 (790) fermion particle with half-integer spin (842) ferromagnetic having magnetic properties, like those of iron (589) field three-dimensional region of influence surrounding an object (200) electric field three-dimensional region of electrostatic influence surrounding a charged object (641) gravitational field region of influence surrounding any object that has mass (200) magnetic field three-dimensional region of magnetic influence surrounding a magnet, in which other magnets are affected by magnetic forces (584) first order maximum line of antinodes resulting from a onewavelength phase shift (427) fission reaction in which a nucleus with A 120 splits into smaller nuclei that have greater binding energy per nucleon (818) focal length (f) distance from the vertex to the focal point, measured along the principal axis; related to the radius of curvature by f r/2 (657) force quantity measuring a push or a pull on an object; measured in newtons (127) action force object A on object B (160) force initiated by action-at-a-distance force force that acts even if the objects involved are not touching (200) force exerted applied force directly on an object by a app (130) person; symbol is F gravitational force attractive force between any two objects due to their masses; symbol is F g (196) net force vector sum of all the forces acting simultaneously on net (131) an object; symbol is F force on an object normal force that is perpendicular to a common contact surface; N (130) symbol is F reaction force object B on object A (160) force exerted by force acting restoring force opposite to the displacement to move an object back to its equilibrium position (353) strong nuclear force binds together the protons and neutrons in a nucleus (793) force that weak nuclear force tal force that acts on electrons and neutrinos (804) fundamen- forced frequency frequency at which an external force is applied to an oscillating object (382) Fraunhofer line dark line in the spectrum of the Sun (773) free-body diagram vector diagram of an object in isolation showing all the forces acting on it (129) situation in which the only free fall force acting on an object that has mass is the gravitational force (226) frequency number of cycles per second measured in hertz (Hz) (249) friction force that opposes either the motion of an object or the direction the object would be moving in if there were no friction; symbol is f (169) F fundamental forces basic forces of nature that physicists think underlie all interactions in the universe (194) fundamental frequency lowest frequency produced by a particular instrument; corresponds to the standing wave having a single antinode, with a node at each end of the string (422) fusion reaction in which two lowmass nuclei combine to form a single nucleus with A 60; the resulting nucleus is more tightly bound (818) 20-PearsonPhys-Glossary 7/25/08 7:39 AM Page 885 G gamma decay emission of a highenergy photon by a nucleus; symbol is (806) gamma radiation emission of a high-energy photon; symbol is (797) generator effect (electromagnetic induction) production of electricity by magnetism (611) gluon mediating particle for the strong nuclear force (838) grand unified theory quantum theory unifying the electromagnetic, strong nuclear, and weak nuclear forces (849) sensitive instrument gravimeter used to detect small variations in the magnitude of the gravitational field strength on Earth\u2019s surface (222) gravitational field strength gravitational force per unit mass at a specific location (201) gravitational mass mass measurement based on comparing the known weight of one object to the unknown weight of another object (199) gravitational potential energy energy of an object due to its position relative to the surface of Earth (295) graviton hypothetical mediating particle for the gravitational force (838) gravity assist use of the gravitational force exerted by celestial bodies to reduce interplanetary travel times (214) gray dose of ionizing radiation that delivers 1 J of energy to each kilogram of material absorbing the radiation;", " unit is Gy (809) ground state energy level (774) lowest possible grounding process of transferring charge to and from Earth (521) H hadron subatomic particle that interacts via the strong nuclear force (842) half-life time it takes for half of the radioactive nuclei in a sample to decay (812) Heisenberg\u2019s uncertainty principle it is impossible to know both the position and momentum of a particle with unlimited precision at the same time (735) high tide highest level of ocean water that occurs near Earth\u2019s coastlines (211) Hooke\u2019s Law relationship where the stretch produced by a force applied to a spring is proportional to the magnitude of the force (299) horsepower (hp) unit used to identify the power output of motors, mainly in the automotive industry (324) Huygens\u2019 Principle model of wave theory, which predicted the motion of a wave front as being many small point sources propagating outward in a concentric circle at the same speed as the wave itself (684) I image attitude orientation characteristic of an image, whether erect or inverted (656) image position where the image forms relative to the surface of the mirror (656) image type distinction between real and virtual images (656) real image image from which light rays come; can be formed on a diffusely reflecting surface or screen (654) virtual image image from which light rays appear to come; cannot be formed on a non-reflective surface or screen (654) impulse product of the net force on an object and the time interval during an interaction. Impulse causes a change in the momentum of the object. (457) incandescent (704) glowing with heat induction movement of charge caused by an external charged object (520) inertia property of an object that resists acceleration (138) inertial mass mass measurement based on the ratio of a known net force on an object to the acceleration of the object (148) insulator material in which the electrons are tightly bound to the nucleus and are not free to move within the substance (513) interference (or two waves) crossing within a effect of two pulses medium; the medium takes on a shape that is different from the shape of either pulse alone (411) constructive interference overlap of pulses to create a pulse of greater amplitude (412) destructive interference overlap of pulses to create a pulse of lesser amplitude (412) interference fringes ence pattern of light and dark bands (686) fixed interfer- interference pattern pattern of maxima and minima resulting from the interaction of waves, as crests and troughs overlap while the waves move through each other (425) ionization energy energy required to remove an electron from an atom (775) atoms that have the same isotopes number of protons, but different numbers of neutrons (791) K kinematics branch of physics that describes motion (6) kinetic energy energy due to the motion of an object; symbol is Ek (302) kinetic friction force exerted on an object in motion that opposes the motion of the object as it slides on another object; symbol is F fkinetic (176) coefficient of kinetic friction proportionality constant relating F fkinetic and FN (183) L latitude south of the equator (221) angular distance north or law of conservation of charge net charge of an isolated system is conserved (517) law of conservation of energy within an isolated system, energy may be transferred from one object to another or transformed from one form to another, but it cannot be increased nor decreased (312) law of conservation of momentum momentum of an isolated system is constant (473) law of magnetism like magnetic ends repel and unlike ends attract each other Glossary 885 20-PearsonPhys-Glossary 7/25/08 7:39 AM Page 886 law of reflection angle of reflection is equal to the angle of incidence and is in the same plane (654) Lenz\u2019s law direction of a magnetically induced current is such as to oppose the cause of the current (618) lepton subatomic particle that does not interact via the strong nuclear force (842) lowest level of ocean low tide water that occurs near Earth\u2019s coastlines (211) M magnification relationship of the size of the image to the size of the object (656) mass defect difference between the sum of the masses of the separate nucleons and the mass of the nucleus; symbol is m (794) maximum (line of antinodes) of points linking antinodes that occur as the result of constructive interference between waves (426) line Maxwell\u2019s Equations series of equations that summarized the relationships between electricity and magnetism, and predicted the existence of electromagnetic waves and their propagation through space (642) mechanical energy sum of potential and kinetic energies; symbol is Em (306) increase in mechanical resonance amplitude of oscillation of a system as a result of a periodic force whose frequency is equal or very close to the resonant frequency of the system (382) mechanics stat", "ics, and dynamics (306) study of kinematics, medium material, for example, air or water through which waves travel; the medium does not travel with the wave (394) meson hadron with integer spin (842) minimum (nodal line) points linking nodes that occur as the result of destructive interference between waves (426) line of mirror equation equation relating the focal length of a curved mirror to the image and object distances (662) 886 Glossary momentum product of the mass of an object and its velocity (449) sum of momentum (of a system) the momenta of all the objects in the system (470) motor effect force deflecting force acting on a charged particle moving in a magnetic field (593) muon unstable subatomic particle having many of the properties of an electron but a mass 207 times greater (842) N natural satellite naturally formed body that revolves around a planet (moon) (273) navigator method method commonly used to show direction for vector quantities in two dimensions; uses compass bearings north [N], south [S], east [E], and west [W] to identify vector directions (78) net charge charges in the system (517) sum of all electric neutrino extremely small neutral subatomic particle; symbol is v (804) neutron neutral particle found in nuclei (790) neutron number number of neutrons in the nucleus; symbol is N (790) Newton\u2019s first law of motion an object will continue either being at rest or moving at constant velocity unless acted upon by an external non-zero net force (139) Newton\u2019s law of universal gravitation Any two objects, A and B, in the universe exert gravitational forces of equal magnitude but opposite direction on each other. The forces are directed along the line joining the centres of both objects. (204) Newton\u2019s second law of motion when an external non-zero net force acts on an object, the object accelerates in the direction of the net force; the magnitude of the acceleration is directly proportional to the magnitude of the net force and inversely proportional to the mass of the object (148) Newton\u2019s third law of motion if object A exerts a force on object B, then B exerts a force on A that is equal in magnitude and opposite in direction (161) node (nodal point) point on a spring or other medium at which only destructive interference occurs; in a standing wave, a point that never vibrates between maximum positive amplitude and maximum negative amplitude; in a standing wave, nodes occur at intervals of ; in an interference pattern, 1 2 nodes occur at path difference (417) intervals of 1 2 normal line perpendicular to the reflecting surface (654) nucleon proton or neutron (790) nucleosynthesis elements by the fusion of lighter elements (823) imaginary line drawn formation of O open pipe (open tube) pipe opened at both ends; the longest wavelength that can resonate in an open pipe is twice the length of the pipe (424) optical fibre central core of glass with a refractive index of approximately 1.5, surrounded by a cladding material of a slightly lower refractive index (671) orbital probability distribution of an electron in an atom (783) orbital period time required for a planet to make one full orbit; may be measured in Earth days (271) orbital perturbation irregularity or disturbance in the predicted orbit of a planet (282) orbital radius distance between the centre of the ellipse and the planet; average orbital radius corresponds to the semi-major axis (269) origin reference point (6) oscillation repetitive back-andforth motion (344) oscillatory motion motion in which the period of each cycle is constant (344) overtone any frequency of vibration of a string that may exist simultaneously with the fundamental frequency (423) P parent element original element in a decay process (799) 20-PearsonPhys-Glossary 7/25/08 7:39 AM Page 887 particle discrete unit of matter having mass, momentum, and the ability to carry an electric charge (639) alpha particle two neutrons bound together to form a stable particle (497) two protons and electron emitted beta particle by a nucleus; symbol is (802) fundamental particle particle that cannot be divided into smaller particles; an elementary particle (836) mediating particle virtual particle that carries one of the fundamental forces (837) strange particle particle that interacts primarily via the strong nuclear force yet decays only via the weak nuclear force (845) virtual particle particle that exists for such a short time that it is not detectable (837) particle model describes EMR as a stream of tiny particles radiating outward from a source (639) path length distance between a point source and a chosen point in space (688) difference in path length difference between two path lengths, each measured from a different origin and extending to a common point in space (688) period time required for an object to make one complete oscill", "ation (cycle); measured in s/cycle (249) phase shift result of waves from one source having to travel farther to reach a particular point in the interference pattern than waves from another source (426) emission of photoelectric effect electrons when a metal is illuminated by short wavelengths of light (712) photoelectron electron emitted from a metal because of the photoelectric effect (712) photon (from the Greek word meaning \u201clight\u201d) quantum of light; discrete packet of energy associated with an electromagnetic field (640) pion unstable subatomic particle with a mass roughly 270 times that of an electron (842) Planck\u2019s formula light comes in quanta of energy that can be calculated using the equation E nhf (705) plane mirror ing surface (654) smooth, flat, reflect- light resultplane polarized light ing from polarization, in which only one plane of the electric field is allowed to pass through a filter (696) close to the principal axis converge, or appear to diverge from, after being reflected (657) principal quantum number quantum number that determines the size and energy of an orbit (774) planetary model that has electrons orbiting a nucleus (768) atomic model plasma highly ionized gas containing nearly equal numbers of free electrons and positive ions (522) point of incidence point at which the incident ray contacts a polished, reflecting surface and is reflected from the surface as the reflected ray (652) point source single point of disturbance that generates a circular wave (395) polar coordinates method method commonly used to show direction for vector quantities in two dimensions; the positive x-axis is at 0\u00b0 and angles are measured by moving counterclockwise about the origin, or pole (78) polarization production of a state in which the plane of the electric field for each electromagnetic wave occurs only in one direction (696) polarizing filter filter that allows only one plane of the electric field to pass through it; plane polarized EMR emerges (696) position straight-line distance between the origin and an object\u2019s location; includes magnitude and direction (6) positron antielectron; positively charged particle with its other properties the same as those of an (804) electron; symbol is e or 0 1 potential energy energy that is stored and held in readiness; includes gravitational and elastic potential energies; symbol is Ep rate of doing work (324) power primary cosmic rays high-energy particles that flow from space into Earth\u2019s atmosphere (841) imaginary line principal axis (PA) drawn through the vertex, perpendicular to the surface of the curved mirror at this point (657) principal focal point (F) point where light rays parallel to and principle of superposition displacement of the combined pulse at each point of interference is the sum of the displacements of the individual pulses (412) projectile object released or thrown into the air (54) projectile motion motion in a vertical plane (54) proton positively charged particle found in all nuclei (790) proton\u2013proton chain fusion process in which four hydrogen nuclei combine to form a helium nucleus (821) pulse disturbance of short duration in a medium; usually seen as the crest or trough of a wave (401) compression pulse where the coils of a spring are more tightly compressed (404) region rarefaction pulse the coils of a spring are more widely spaced (404) region where transverse pulse pulse in which the coils of the spring move at right angles to the direction of the pulse\u2019s motion (401) Q quanta discrete units of energy (638) quantized limited to whole multiples of a basic amount (quantum) (705) quantum smallest amount or \u201cbundle\u201d of energy that a wavelength of light can possess (pl. quanta) (705) quantum chromodynamics quantum field theory that describes the strong nuclear force in terms of quantum colour (848) quantum electrodynamics quantum field theory dealing with the interactions of electromagnetic fields, charged particles, and photons (838) quantum field theory field theory developed using both quantum mechanics and relativity theory (837) Glossary 887 20-PearsonPhys-Glossary 7/25/08 7:39 AM Page 888 quantum indeterminacy probability of finding a particle at a particular location in a double-slit interference pattern (740) quantum model light and all other EMR are discrete bundles of energy, each of which has both particle and wave characteristics (640) quark any of the group of fundamental particles in hadrons (845) R radioactive decay series process of successive decays in which a radioactive nucleus decays into a daughter nucleus that is itself radioactive, and the daughter nucleus decays into another unstable nucleus until a stable nucleus is created (807) radioisotope radioactive (808) isotope that is radius of curvature (r) distance from the centre of curvature to the mirror surface (657) range distance a projectile travels horizontally over level ground (105) ray line that indicates only the direction of motion of", " the wave front at any point where the ray and the wave front intersect (397) ray diagram diagram showing the result of a light ray interacting with a surface (653) recomposition (of the spectrum) production of white light by a combination of light of all colours of the spectrum (674) rectilinear propagation movement of light in straight lines through a uniform medium (653) reference point point from which distances are measured (297) refracted ray path of a light ray after it has changed direction at an interface, due to a change in its speed (664) refraction change in the direction of a light wave due to a change in its speed as it passes from one medium to another (666) refractive index ratio comparing the speed of light in a vacuum to the measured speed of light in the medium (666) relative biological effectiveness (RBE) factor indicating how much arbitrarily chosen 888 Glossary a particular type of radiation affects the human body (809) any light ray passing through the boundary between two media. (667) relative motion motion measured with respect to an observer (91) solenoid electromagnet that operates a mechanical device (589) resonance increase in the amplitude of a wave due to a transfer of energy in phase with the natural frequency of the wave (418) resonant frequencies natural frequencies of vibration of an object that will produce a standing wave pattern; at a resonant frequency, energy added is in phase with existing oscillations (418) sum of a series of resultant vector vectors; drawn from the tail of the first vector to the tip of the last vector (71) revolution one complete cycle for an object moving in a circular path (249) rpm revolutions per minute; imperial unit used to measure frequency (249) S scalar quantity measurement that has magnitude only (6) secondary cosmic rays particles created by collisions between primary cosmic rays and atoms in the atmosphere (841) shower of semiconductor material that lies in the middle, between a good conductor and a good insulator; because of its nature, a semiconductor is a good conductor in certain situations, and a good insulator in other situations (514) absorbed dose of ionizing sievert radiation that has the same effect on a person as 1 Gy of photon radiation, such as X rays or gamma rays; absorbed dose in sieverts is equal to the dose in grays multiplied by the relative biological effectiveness (RBE); unit is Sv (809) simple harmonic motion (SHM) oscillatory motion where the restoring force is proportional to the displacement of the mass (355) simple harmonic oscillator object that moves with simple harmonic motion (355) Snell\u2019s Law For any angle of incidence greater than zero, the ratio sin r is a constant for i/sin sound barrier term applied to the increase in aerodynamic resistance as an aircraft approaches the speed of sound (433) source charge duces an electric field (546) charge that pro- spectrometer device for measuring the wavelengths of light in a spectrum (773) spectroscopy study of the light emitted and absorbed by different materials (771) spectrum bands of colours making up white light; in order: red, orange, yellow, green, blue, and violet (675) absorption line spectrum pattern of dark lines produced when light passes through a gas at low pressure (772) electromagnetic spectrum all types of EMR considered in terms of frequency, wavelength, or energy (637) emission line spectrum pattern of bright lines produced by a hot gas at low pressure (772) specular (regular) reflection behaviour describing parallel incident rays reflected from a flat, smooth, reflecting surface as parallel reflected rays (653) spin quantum property resembling rotational angular momentum (842) constant of pro- spring constant portionality k which appears in Hooke\u2019s Law for springs; represents the slope of the line and is measured in units of force per unit length; amount of stiffness of a spring (299) standard model describing the nature of matter and the fundamental forces (848) current theory static friction force exerted on an object at rest that prevents it from sliding; symbol is F fstatic (171) coefficient of static friction proportionality constant relating (Ffstatic )max and FN (182) stationary state a fixed energy level (773) stable state with 20-PearsonPhys-Glossary 7/25/08 7:39 AM Page 889 stator fundamental component of a simple DC electric motor consisting of a frame with a coil or permanent magnet to provide a magnetic field (608) stopping potential potential difference for which the kinetic energy of a photoelectron equals the work needed to move through a potential difference V (716) string theory theory that treats particles as quantized vibrations of extremely small strings of massenergy (849) superconductor has no measurable resistance at very low temperatures (515) conductor that supernova sudden, extremely powerful explosion of a massive star (823) synchrotron particle accelerator particle accelerator in which an advanced type of cyclotron increases the", " strength of the magnetic field as the particles\u2019 energy increases, so that the particles travel in a circle rather than spiralling outward (841) system two or more objects that interact with each other (470) group of objects isolated system (in the context of energy) assumed to be isolated from all other objects in the universe (311) isolated system (in the context of momentum) when mass of a system is constant and no external net force acts on the system (470) non-isolated system system in which there is an energy exchange with the surroundings (320) T tangent a curved-line graph at only one point (24) straight line that touches tension (in a rope) magnitude of T exerted by a rope on an a force F object at the point where the rope is attached to the object (132) test charge charge with a magnitude small enough that it does not disturb the charge on the source charge and thus change its electric field (546) thin lens equation equation that relates object distance, image dis- tance, and focal length of a curved lens (680) air velocity object\u2019s velocity relative to still air (92) threshold frequency minimum frequency that a photon can have to cause photoemission from a metal; symbol is f0 (712) torsion balance device used to measure very small forces (205) total internal reflection reflection of all incident light back into an optically more dense medium due to inability to refract beyond the maximum angle of 90\u00b0 (672) trajectory parabolic path or motion of a projectile (103) transmute element (806) trough region where the medium is lower than the equilibrium position (394) tuning (a musical instrument) changing the tension in the string of a musical instrument (424) change into a different U uniform circular motion motion in a circular path at a constant speed (242) uniform motion constant velocity (motion or rest) (13) non-uniform motion acceleration (23) uniformly accelerated motion constant change in velocity per unit time (25) universal gravitational constant constant in Newton\u2019s law of universal gravitation that is equal to 6.67 1011 N\u2022m2/kg2; symbol is G (204) universal wave equation relationship between the speed, frequency, and wavelength of a wave: v f (408) V Van de Graaff particle accelerator particle accelerator in which a moving belt transfers charge to a hollow, conductive sphere, building up a large potential difference that propels ions through an accelerator chamber (841) vector quantity measurement that has both magnitude and direction (6) ground velocity velocity relative to an observer on the ground (92) instantaneous velocity moment-to-moment measure of an object\u2019s velocity (24) wind velocity velocity of the wind relative to the ground (92) vertex (V) curved mirror surface (657) geometric centre of the W wave disturbance that moves outward from its point of origin, transferring energy through a medium by means of vibrations (394) bow wave V-shaped wave produced as a boat moves through water or an airplane moves through the atmosphere (433) electromagnetic wave periodic variation in perpendicular electric and magnetic fields, propagating at right angles to both fields (643) incident wave wave front moving out from the point of origin toward a barrier (395) longitudinal wave wave with the motion of the medium being parallel to the motion of the wave (401) reflected wave wave front moving away from a barrier (395) strong compression shock wave wave produced as an aircraft exceeds the speed of sound (433) condition in a standing wave spring or other medium in which a wave seems to oscillate around stationary points called nodes; wavelength of a standing wave is the distance between alternate nodes or alternate antinodes (417) transverse wave wave with the motion of the medium being perpendicular to the motion of the wave (401) wave amplitude distance from the equilibrium position to the top of a crest or the bottom of a trough (395) velocity rate of change in position; includes magnitude (speed) and direction (12) imaginary line that wave front joins all points reached by the wave at the same instant (395) Glossary 889 20-PearsonPhys-Glossary 7/25/08 7:39 AM Page 890 waves out of phase occurs when a crest from one wave occupies the same point in the medium as a trough from a second wave; produces destructive interference (416) gravitational force exerted weight on an object by a celestial body; symbol is F g (198) apparent weight negative of the normal force acting on an object; symbol is w (224) true weight acting on an object that has mass (222) gravitational force true weightlessness which w 0 for an object and F g 0 on the object (228) situation in work measure of the amount of energy transferred when a force acts over a given displacement; calculated as the product of the magnitude of applied force and the displacement of the object in the direction of the force (293) work\u2013energy theorem work done on a system is equal to", " the sum of the changes in the potential and kinetic energies of the system (307) work function minimum energy that a photon can have to cause photoemission from a metal; specific for every metal; symbol is W (713) wave model describes EMR as a stream of transverse waves radiating outward from a source (639) wave\u2013particle duality light has both wave-like and particle-like properties (726) wave train series of waves forming a continuous series of crests and troughs (395) wavelength distance between two points on a wave that have identical status; usually measured from crest to crest or from trough to trough (395) waves in phase occurs when crests or troughs from two waves occupy the same point in the medium; produces constructive interference (395) 890 Glossary 21-PearsonPhys-AnswerKey 7/29/08 1:40 PM Page 891 NUMERICAL ANSWERS For some questions, answers may vary slightly depending on the method or data chosen for the solution. page 10, 1.1 Check and Reflect page 63, 1.6 Check and Reflect 4. (e) 13.0 m [right] 5. 45.0 km [W] 6. Distance 11.0 m; Displacement 5.0 m [right] 0.50 m [right] 0.75 m [left] groom best man maid of honour 1.25 m [right] flower girl 1.50 m [left] 7. d d d d page 20, 1.2 Check and Reflect 3. 72.3 m 4. 1.5 m 5. 1.61 m/s2 [down] 6. 2.8 m 7. (1) 26 m/s [down] (2) 33 m 8. 0.376 s 9. 47.9 m 10. 1.6 s 11. 6.22 s 12. 10.6 m 3. (i) D (ii) C (iii) A (iv) B 13. 1.4 s 4. 10 m; A is ahead 5. 7 km [W] 8. 2.5 m right 9. Insect B is ahead by 1.2 m. 11. Distance 26 m; Time 13 s 12. Time 20 s; Displacement 45 m [N] (2) 31 s 13. (1) 22 s (3) 14 s page 30, 1.3 Check and Reflect 1. (a) 5.6 m/s2 [forward] (b) 2.8 m/s2 [forward] (c) 0.30 m/s2 [forward] (iii) C (ii) B 3. (i) A 4. Time (s) Velocity (m/s [forward]) (iv) D 2.0 4.0 6.0 8.0 3.8 7.0 0.0 7.5 page 44, 1.4 Check and Reflect 7. 75 m [E] 8. 36 km [up] 13. 15 m/s2 [E] 15. Acceleration 1.25 m/s2 [W]; Time 8.00 s 16. (a) 81 km/h [N] 17. 0.33 m/s2 [right] page 53, 1.5 Check and Reflect 1. 20 cm [forward] 2. 75 m [right] 3. 59.9 m 4. 1.67 105 m/s2 [forward] 5. 11 s 6. 39.2 m 7. 0.41 m/s2 8. 0.24 m/s2 [forward] 9. 75.0 m/s2 [W] 10. 23.3 m/s2 11. 3.52 m/s2 [S] 12. 9.31 m/s2 13. 9.5 m 14. 0.064 m/s2 [N] 14. 2.4 m 15. (a) 4.0 m/s2 [up] (b) 5.0 103 m (d) 7.0 103 m 16. 9.68 m/s [down] page 65, Chapter 1 Review 3. (a) 1.0 m/s [backward] (b) 2.0 m/min [right] (c) 1.7 m/s [forward] 4. 27.0 km [W] 5. 42 min 6. (1) 3.75 m/s (2) 1.25 m/s [right] (3) 25.0 m [right] 8. 2.8 s 9. 7.0 s 10. 0 m/s 11. 1.9 102 s 12. 1.5 m/s [W] 14. 60 m 15. 7.2 107 m/s2 16. 18 km 17. 34.5 km 18. 72 times faster 19. 1.1 102 km/h,", " 31 m/s 20. (1) 2.81 m/s2 [downhill] (2) 22.5 m/s [downhill] 21. (1) 1.3 m/s2 [N] (2) 10 m/s [N] 22. (1) 16.0 m/s [S] (2) 22.0 m/s [S] (3) 0.267 m/s2 [S] 23. 35 m/s 24. (1) 12 s 26. (1) 20 km [right] (2) 39 m/s (2) 0 m/s2 27. 9.03 s 28. 0.467 s 29. 0.298 m/s2 30. 24 km/h [E] 32. 11.2 m/s2 [W] 33. 6.8 s 34. 1.17 s 35. 12 m 36. 9.29 m 37. 1.32 s 38. (1) 15.2 m (2) 17.3 m/s 39. (1) 20 m/s (2) 2.0 s page 75, 2.1 Check and Reflect 3. 24 cm 4. 1.0 cm : 20 km 5. 2.1 103 km 6. (b) 100 yards [down field] (c) 1000 yards 7. 100 km [S], 150 km [S], and 200 km [S] 9. (1) 36.0 m (2) 8.0 m [down] page 90, 2.2 Check and Reflect 28 cm 48 cm; dx 4. dy 5. (a) Distance 11.5 km; Displacement 3.4 km [27\u00b0 N of W] (b) Distance 522 m; Displacement 522 m [17\u00b0 E of N] (c) Distance 2.95 km; Displacement 1.45 m [270\u00b0] 6. 2.0 km [45\u00b0 S of E] 7. 178 m/s 8. 2.65 km [48\u00b0 S of E] 9. Total displacement 2.8 km [49\u00b0 N of W]; Average velocity 3.0 km/h [49\u00b0 N of W] 10. (a) 27 m in both the x and y directions (b) 54 m 11. 10.6 m [293\u00b0] page 101, 2.3 Check and Reflect 5. (a) 88 s (b) 79 s; 47 m 6. (a) 233 km/h [N] (b) 297 km/h [N] (c) 267 km/h [83.1\u00b0 N of W] 7. 7.2 102 s 8. 8.4 102 km/h [3\u00b0 N of E] 9. Ground velocity 2.8 m/s [35\u00b0 S of E]; Time 19 min 10. 6.3 102 km/h [3\u00b0 S of W] 11. (a) 4.7 m/s [32\u00b0 W of N] (b) 2.0 102 s page 112, 2.4 Check and Reflect 5. 2.0 m 6. Initial horizontal velocity component 22.1 m/s; Initial vertical velocity component 15.5 m/s; Up 12.2 m; Out 69.8 m 7. 7.21 m/s 8. 10.8 m/s Numerical Answers 891 21-PearsonPhys-AnswerKey 7/29/08 1:40 PM Page 892 9. (a) 10.0 s (c) 20.5 s (e) Horizontal component 68.8 m/s; Vertical component 103 m/s (b) 542 m (d) 1.41 km page 114, Chapter 2 Review 2. x component 41 m; y component 37 m 5.0 m [N] 140.0 cm [E] 2.5 km [backward] 80.0 km [right] 3. 0 m/s 7. (a) d (b) d (c) d (d) d 8. 0 m/s 9. 3.9 102 km 10. Range 86.2 m; Maximum height 30.8 m 11. 2.23 102 m horizontally; 2.83 s 12. (a) 6.4 m/s [51\u00b0 E of N] 28. 99 km/h [W] 29. 1.5 102 m 30. 1.2 m/s [210\u00b0]; 4.4 m/s [210\u00b0]; 6.7 m/s [210\u00b0] 31. 1.1 102 km/h [7\u00b0 N of E] 32. 13.2 m/s2 33. 375 m [N] 35. 1.93 m 36. (a) 7 squares (b) 8.6 units [54\u00b0] 37. 1.6 m/s", " [N] 38. 320 m [51\u00b0 S of E] 40. 5.98 m/s2 41. (a) 14.4 m/s [down] 42. (a) 347 m (b) 376 m (c) 82.6 m/s [down] 43. 13 m/s [30\u00b0] (b) 10.6 m 9. 1.2 m/s2 [uphill] 10. 16 m 11. 1.9 s page 192, Chapter 3 Review 1. 5.00 102 m/s2 [E] 3. 1850 N [W] 5. (a) 42 N [along rope connecting foot to pulley] 6. 1.2 103 N [forward] 3.7 106 N, R 7. L 8. 7.5 N [backward] 9. (a) 3.4 102 N 3.3 106 N (b) 1.4 103 N [forward] (c) 3.4 102 N [backward] 10. (a) F netA 36 N [W], F netB (b) (curler A) 0.71 m/s2 [W], (curler B) 0.26 m/s2 [E] 21 N [E] (b) 30 s page 136, 3.1 Check and Reflect 11. 0.38 13. (a) 29 m/s [down] (b) 23 m 3. (b) 300 N [forward] (c) 7.50 m/s [away from cliff] (b) 3.5 m/s [N] 14. 6.5 m/s [76\u00b0 up] 15. (a) 30\u00b0 W of N (c) 1.4 102 s 16. 1.82 m 17. 39 s 18. 2.18 s (b) 180\u00b0 4. (a) 0\u00b0 5. 1.47 102 N [13\u00b0] 6. 42 N [153\u00b0] 7. F F 1.50 102 N [125\u00b0], 1.50 102 N [55\u00b0] T1 T2 page 139, 3.2 Concept Check 19. (a) 17\u00b0 S of W (b) 95 km/h v 17 km/s [toward interstellar space] 20. [42.2\u00b0] 21. 5.0 m 22. 60.96 m 23. (a) 1.2 102 km/h 24. 2.92 s; Distance 59.9 m; Maximum height 10.5 m (b) east 25. [39.8\u00b0] 26. (a) 36 s (b) 1.5 km page 118, Unit I Review 3. (a) dx (b) vx 0 m; dy 15.0 m/s; vy 5.0 m 5.47 m/s 5. 9.8 km [30\u00b0 S of E] 6. 2.70 m/s [W] 8. 3.03 s; Distance 78.0 m horizontally; Maximum height 11.3 m 13. 30 m/s 14. 32 km/h 16. (a) 30 m/s [90\u00b0] (b) 40 m/s [90\u00b0] 17. 12.1 km 18. 1.06 h 19. 5.7 m/s 20. 9.56 m/s2 21. 8.1 102 km/h [4\u00b0 W of S] 22. 70.4 m/s2 23. 2.2 s 24. 58 m 25. (a) 51 s (b) 0.032 m/s2 26. 13.5 m 27. 63.4 m 892 Numerical Answers page 146, 3.3 Concept Check F app if F f F net 0 page 148, 3.3 Concept Check (a) a (b) a (c) a 1 6 1 page 158, 3.3 Check and Reflect (b) 15 m/s2 [up] 5. (a) 26 kg 6. 0.11 m/s2 [horizontally] 7. (a) 75 N [97\u00b0] 8. 0.75 m/s2 [right] 9. (a) (4.0-kg block) 1.3 m/s2 (b) 37 m/s2 [97\u00b0] [toward pulley], (2.0-kg block) 1.3 m/s2 [down] (b) 17 N page 168, 3.4 Check and Reflect 7. 10 N, 10 N [toward spring scale] 8. (a) F X on Y F Y on X (b) F X on Y F Y on X 12 N [right], 12 N [left] 12 N [right], 12 N [left] page 178, 3.5 Concept Check 90\u00b0 page 190, 3.5 Check and Reflect 4", ". 2 N [backward] 5. 0.40 6. 2 103 N [backward] 7. 24\u00b0 12. 0.423 13. (a) 9.8 102 N (b) (i) (60 kg) 2.0 m/s2 [down], (40 kg) 2.0 m/s2 [up] (ii) 4.7 102 N (c) 9.4 102 N 14. 7.3 102 kg 15. 2 103 kg 18. (a) 6.9 m/s2 [backward] (b) 1.2 s page 201, 4.1 Concept Check (a) 16g (c) g (d) g (b) 1 g 4 (d) 0 page 202, 4.1 Check and Reflect 9. (b) 9.8 N/kg (c) gBanff page 205, 4.2 Concept Check 1 6 1 (c) (b) 4Fg (a) 4Fg Fg page 215, 4.2 Check and Reflect 3. (a) 1 Fg (b) 1 Fg 2 4 (ii) 978 N 4. (a) (i) 162 N 5. (b) (Deimos) 1.9 1014 N, (Phobos) 5.3 1015 N 6. (a) 132 N [toward Earth\u2019s centre] (b) about 13.4 times greater in magnitude page 229, 4.3 Check and Reflect 5. 2.4 N/kg [toward Earth\u2019s centre] 6. (b) (i) 1.6 N/kg (ii) 0.72 N/kg (iii) 0.40 N/kg (c) 10rMoon 7. 9.8 N/kg 8. 540 N [down] g) 4.9 102 N [down], 9. (b) (F (w) 2.9 103 N [down] 21-PearsonPhys-AnswerKey 7/29/08 1:40 PM Page 893 (c) 0 page 286, 5.3 Check and Reflect 10. (Figure 4.38) 2.45 m/s2 [up], (Figure 4.40) 1.23 m/s2 [down] page 231, Chapter 4 Review 2. (balance) 5.0 kg, (spring scale) 96 N 3. g Msource 4. 3.83 N/kg [toward Earth\u2019s centre] 5. 1.4 s 7. (a) 222 N [toward centre of Mars] (b) 626 N [toward centre of Saturn] 8. 2.0 107 N 9. (a) 2.4 102 N (b) 7.1 102 N (c) 6.8 N 10. (a) 7.3 m/s2 [down] (b) 12 s (c) 4.9 102 N [down] page 234, Unit II Review 2. 86 N [350\u00b0] 6. 0.50 kg 7. (a) 4a 8. 40 N (b) 1 a 4 10. 15 N [up] 21. (mass m) Fg, (mass 2m) 2Fg 24. 306 N [357\u00b0] 25. 2.097 m/s2 [W] 27. 4.0 s 28. (b) 8.58 N [0\u00b0] 29. 0.56 m/s2 [in same direction as train A] 30. 8.0 104 N [up] 31. 2.2 m/s2 [forward] 32. 13 N [up] 33. (a) 5.3 m/s2 [right] (b) 4.8 102 N 34. 0.46 35. (a) 6 101 N [forward] (b) 2 101 N [forward] (c) 6 101 N [backward] 36. 3.2 m/s2 [downhill] 37. 1.5 m/s2 [up] 38. (a) 1.3 m/s2 [toward object A] (b) (string between A and B) 51 N, (string between B and C) 44 N 39. 8.00 N [toward Earth\u2019s centre] 41. 8.57 N/kg [toward Earth\u2019s centre] 42. 24.3 N less on Mars 4 3. (a) (i) and (ii) 5.9 102 N (iii) 8.8 102 N (iv) 3.9 102 N (b) (i) and (ii) (w) 5.9 102 N [down], g) 5.9 102 N [down] (F (iii) (w ) 8.8 102 N [down], g) 5.9 102 N [down] (F (", "iv) (w ) 3.9 102 N [down], g) 5.9 102 N [down] (F g) 5.9 102 N [down], 44. (F (a) 6.5 m/s2 [down] 45. (a) 9.4 m/s2 [down] (b) 11 N [up] (c) 11 N [down] 46. (Earth) 1.2 s, (Moon) 3.0 s 47. (a) 0.1 N [away from net] (b) 0.8 m/s2 [toward net] (c) 9 s (d) no 51. (a) 3.7 107 N [down] (b) 1.3 107 N [up] (c) 3.3 m/s2 [up] 52. Fg from person on you is 5.8 times greater page 268, 5.2 Check and Reflect 6. 0.2 s 7. 1.88 102 m/s 8. 26 m/s 9. 8.57 m/s2 10. 1.2 102 m/s2; 12 times 11. 1.21 m/s 12. 0.0337 m/s2 13. 1.7 Hz 14. 1.3 Hz or 80 rpm 9. 0.723 AU 10. 1.36 103 m/s 11. 3.547 d 12. 1.43 104 m/s 13. 1.98 1030 kg page 288, Chapter 5 Review 14. 4 greater 17. 2.0 101 m/s 18. 7.8 102 N 19. 44.3 m/s 20. 68 N 22. 7.83 102 m/s2 23. 3.0 103 rpm 25. 13.1 m/s 26. (a) 6.00 107 m/s (b) 1.20 1016 m/s2 27. 18.0 AU 28. 1.09 1030 kg 29. (a) 1.16 1018 s or 3.66 1010 a (b) 5.18 1036 kg (c) 6.71 1015 m/s2 30. (a) 7.91 103 m/s (b) 5.07 103 s 31. 2.73 103 m/s2; 2.00 1020 N 32. 0.430 d or 3.71 104 s 33. 5.51 103 m/s 34. (a) 4.01 1033 kg (b) 6.38 1011 m page 305, 6.1 Check and Reflect 5. (a) 3.60 104 J (b) 1.18 104 J 6. 2.76 104 J 7. (a) 6.18 104 J (b) 7.55 104 J (c) 1.37 105 J 8. (a) 3.25 J (b) 0.0732 m 9. (a) A 2.04 105 J; B 3.48 105 J 10. (a) 110 J (b) 33.8 J 11. (a) 160 J (b) 12.6 m/s page 310, 6.2 Check and Reflect 5. 1.79 106 J; Ek 6. 3.54 105 J 7. (a) 5.10 103 J (b) 1.25 103 J (c) 3.85 103 J (d) 31.0 m/s 8. (a) 288 J (c) 126 J (b) 288 J (d) 126 J page 323, 6.3 Check and Reflect 7. 12.4 J 10. (a) 0.482 m (b) 3.08 m/s (c) 2.13 m/s page 330, 6.4 Check and Reflect 4. 164 W 5. 1.50 106 W 6. 4.1 106 J 7. 380 N 8. 1.04 107 W page 332, Chapter 6 Review 5. 3.27 105 J 6. (a) increases by 4 7. (a) 5.12 m/s 8. 1.70 102 N/m 9. 12.7 m 10. 2.30 104 W (b) 5.59 m/s page 336, Unit III Review 16. 1 J 1 kg\u00b7m2/s2 29. 0.017 s 30. 62.5 Hz 31. 5.000 Hz, 0.2000 s 32. 1.02 103 m/s 33. 1.6 101 s 34. 3.09 m/s2 35. 1.40 102 m/s 36. 7.10 103 N 37. (a) 28.1 m (b) 4 greater; 113 m (b) 1.41 103 J 38. (a) 2.40 103 J 39. (a) 2.80 103 N [0\u00b0] (b) 1.54 105 J (c", ") 3.79 105 J (d) 19.5 m/s Numerical Answers 893 21-PearsonPhys-AnswerKey 7/29/08 1:40 PM Page 894 40. 1.03 m 41. (a) 1.20 103 J 42. (a) 35.9 J 44. (a) 2.60 J (b) 15.5 m/s (b) 0.814 m (b) 0.0651 m 45. 1.32 m/s 46. (a) 1.31 105 J (b) 917 m (c) 295 m/s 47. (a) 8.91 J 48. 3.68 103 W 49. 1.62 104 W 50. 60.0 m/s (216 km/h) (b) 1.78 N 51. 3.63 m/s 52. (a) 4.01 1030 kg (b) 4.27 1025 kg (c) 2.18 103 m/s 54. (a) 7.75 m/s (b) 4.38 m/s page 347, 7.1 Check and Reflect 7. 20.0 Hz 9. 2.50 102 Hz 10. 0.026 s 11. 0.01250 s 12. (a) 0.400 s (b) 1.50 102 wags page 365, 7.2 Check and Reflect 4. 6.0 N opposite to the displacement 5. 1.6 m 6. 19 N [away from equilibrium] 7. 1.5 N/m 8. 0.342 N [toward equilibrium] 9. 0.028 N/m page 380, 7.3 Check and Reflect 6. 1.5 m 7. 1.08 103 m/s2 [left] 8. 1.3 103 N/m 9. 11.0\u00b0 10. 3.49 s 11. (a) 1.88 s (b) 0.900 kg (c) 16.7 m/s2 12. 3.14 m/s 13. 7.99 cm [east] 14. 0.900 s page 390, Chapter 7 Review 11. (a) 2.5 N [toward equilibrium] (b) 0.85 N [toward equilibrium] 12. 2.5 102 Hz 13. 1.5 s 14. 10.0 Hz 15. 5.1 N [toward equilibrium] 16. 2.0 102 N/m 17. (a) 1.96 N/m (b) 0.392 N (c) 0.300 m 894 Numerical Answers 18. 0.750 m 20. 24.8 cm 21. 3.00 s 22. 0.120 m/s 23. 0.13 m/s2 [down] 24. 1.02 s 25. 0.65 N/kg 26. (a) l/g (b) 0.796 m 27. (a) 1.26 m/s (b) 0.993 s 28. (a) 0.566 Hz (b) 0.800 Hz 46. (a) 10 nodes and 9 antinodes (b) 2.22 Hz 47. (a) 193 m/s (b) 440 Hz 48. (a) 0.552 m (b) 1.66 m 49. 8.62 m/s [away from you] 50. (a) 1.28 103 Hz (b) 9.96 102 Hz page 449, 9.1 Concept Check (b) 1 p 3 (c) p [W] page 410, 8.2 Check and Reflect (a) 2p 5. 1.20 103 Hz 6. 2.07 m 7. 0.135 m 9. (a) 0.911 m (b) 3.91 m page 428, 8.3 Check and Reflect 7. 0.106 m 9. 0.60 cm page 434, 8.4 Check and Reflect 4. 748 Hz 5. 15.7 m/s (56.6 km/h) page 436, Chapter 8 Review 5. 0.133 s 7. 0.833 cm 8. 8.6 cm 9. 1.59 102 m 10. (b) 5.6 m/s 12. (a) 2.5 Hz (b) 0.50 Hz 13. (a) 435 m/s (b) 777 Hz 14. (a) 19.7 cm (b) 59.0 cm 16. 1.4 cm 17. 308 Hz 18. 86.9 km/h (24.1 m/s) 19. 1.26 103 km/h 20. 694 Hz 21. 175 m/s; f 2 original 3 page 440, Unit IV Review 11. 2.5, 3.5 28. 3.06 N 29. 7.9 g 32. 1.25", " m/s2 33. 4.0 mN/cm 34. (a) 0.100 m (b) 2.24 m/s 35. 15 m/s 36. 6.73 s 37. 0.99 m/s 38. 15.9 m 41. 4.00 1014 Hz to 6.98 1014 Hz 42. 240 m 43. 0.294 m 44. 7.76 s 45. 120 m/s page 453, 9.1 Check and Reflect 8. 13 kg\u2022m/s [S] 9. 1.2 102 m/s [N] 11. 0.16 kg 13. 75 m/s [S] 14. (a) 4.28 105 kg\u2022m/s [W] (b) 3.85 105 kg\u2022m/s [W] 15. 1.36 105 kg 16. 32.6 m/s [W] page 456, 9.2 Concept Check pi 0 page 467, 9.2 Check and Reflect 4. (a) 2 (impulse) (b) 1 (impulse) 3 7. (a) 2.3 N\u2022s (b) 47 m/s [S] 8. 6.2 N\u2022s 9. (a) 7.0 103 N\u2022s (b) 11 m/s 10. 12 s 11. 560 N [W] 12. 545 N\u2022s [W] page 486, 9.3 Check and Reflect 6. 0.018 m/s [away from bag] 7. 3.1 103 m/s [down] 8. 1.2 m/s [S] 9. 0.47 m/s [E] 10. (a) 1.11 m/s [right] (b) inelastic 11. (a) 274 kg (b) inelastic page 499, 9.4 Check and Reflect 5. 0.505 m/s [320\u00b0] 6. 0.625 m/s [48.1\u00b0 N of W] 7. inelastic, 0.098 J 9. 0.603 m/s [49.6\u00b0 S of W] 10. 27.4 m/s [37\u00b0] 11. 27.0 m/s [349\u00b0] 21-PearsonPhys-AnswerKey 7/29/08 1:40 PM Page 895 page 503, Unit V Review 37. 1.4 104 kg\u2022m/s [N] 39. 1.85 104 kg\u2022m/s [S] 40. 6.2 kg 41. 37 kg\u2022m/s [W] 42. 2.4 kg\u2022m/s [W] 43. (a) 7.5 N\u2022s (b) 0.26 kg 44. (a) 3.02 N\u2022s [210\u00b0] (b) 6.9 m/s [210\u00b0] 45. 2.7 103 N [toward drop-off] 46. (a) 5.50 103 N\u2022s [W] (b) 7.75 s 47. 11.3 m/s [6.4\u00b0 S of W] 48. (a) 7.62 103 N\u2022s [forward] (b) 22 m/s [forward] 49. (a) 3.0 N\u2022s (b) 11 m/s 50. 6.5 107 N 51. 8.7 102 N [up] 52. (a) 12 kg\u2022m/s [toward pitcher] (b) 1.6 103 N [toward pitcher] 53. 2.1 m/s [225\u00b0] 54. (a) 1.58 m/s (b) 0.948 m/s 55. 0.750 m/s [backward] 56. 1.27 103 m/s [3.5\u00b0] 57. 3.5 m/s [65\u00b0] 58. 1.26 m/s [8.3\u00b0 E of N] 59. (a) 1.8 m/s [1.8\u00b0 E of N] (b) elastic 60. 1.3 m/s [downhill] 61. 15.6 m/s [34.1\u00b0 S of W] 62. (a) 17 m/s 63. 4.90 m/s [2\u00b0] 64. (a) 3.04 m/s [0.0\u00b0] (b) elastic 65. 4.21 1025 kg\u2022m/s [W] 66. (a) 3.46 m/s [4.7\u00b0 W of N] (b) inelastic 67. (a) 40.0 kg (b) 14.0mT kg\u2022m/s [S] 68. 33 m/s [72.5\u00b0 N of W] 69. (a)", " 0.36 m/s (b) 18\u00b0 S of E 70. (a) 5.95 m/s [E] (b) 2.76 s 71. 10 m/s [73.8\u00b0 N of E] 72. 0.580 kg 73. 1.18 m/s [307.9\u00b0] 74. 0.641 m/s [S] 75. (a) S (b) NW (c) 248\u00b0 77. 10.5 m/s [46.4\u00b0 W of S and 52.5\u00b0 up] 80. (initial velocity of car) 90 km/h [W]; (initial velocity of truck) 88 km/h [N] 82. 6.1 102 m/s page 538, 10.2 Check and Reflect 5. (a) 20 N (b) 40 N 6. (b) 3.1 1010 electrons 7. (a) 6.74 N\u2014repulsive (b) 6.80 N\u2014repulsive 8. (a) 1.20 104 N toward charge B (b) 7.49 103 N toward charge A page 540, Chapter 10 Review 11. 1.69 N\u2014repulsive 12. 6.70 103 m 13. (a) 1.60 102 N [left] (b) 1.24 102 N [left] (b) 6.7 N 22. (a) 160 N 23. Fe Fg Fe 8.22 108 N 3.63 1047 N 2.27 1039 Fg 24. X\u20142.43 N [90\u00b0]; Y\u20142.43 N [210\u00b0]; Z\u20142.43 N [330\u00b0] 25. (b) Fe varies as 1/r or 1/r2 (e) 0.0360 N\u2022m2 (f ) kq1q2 (g) 2.00 106 C page 553, 11.1 Check and Reflect 7. (a) 4.50 105 N/C [right] (b) 8.99 103 N [right] 8. (a) 2.04 104 N/C [away from larger sphere] (b) 3.63 109 C 9. (a) 2.50 107 N/C [toward the 3.00 C charge] (b) 0.661 m [left of the 3.00 C charge] 10. 0.00 N/C page 569, 11.2 Check and Reflect 4. (a) 2.40 1017 N [downward] (b) 1.64 1026 N [downward] 6. 1.60 103 N/C 7. 7.2 105 N/C [toward the 6.4 C charge] 8. 2.00 103 J 9. 1.8 106 N/C [left] 10. (a) 2.00 104 V/m (b) 3.86 105 J page 575, 11.3 Check and Reflect 4. Speed of electron is 8.79 106 m/s; speed of proton is 2.05 105 m/s 5. 9.4 107 m/s 6. 6.00 1019 C 7. 2.40 104 J 8. (a) 1.38 106 m/s (b) 9.79 105 m/s 9. (a) 1.8 105 J (b) 1.8 105 J (c) 1.1 m/s 10. (b) 2.00 1017 N [down] (c) 2.20 1013 m/s2 [down] (d) 0.130 m page 578, Chapter 11 Review 12. Ep W 17. (a) 5.17 103 N/C [away] (b) 1.03 102 N [toward positive charge] 18. (a) 7.49 1010 N/C [toward the 5.00 C charge] (b) 0.735 m from the 5.00 C charge 19. 1.2 107 N/C [90.0\u00b0] (b) 3.0 J 20. (a) 3.0 J 21. 5.62 104 J 22. (a) 545 V/m (b) 20.5 V 23. 0 J 24. 7.0 105 V/m 25. 2.88 1013 J or 1.80 106 eV 26. 6.40 1015 J or 4.00 104 eV page 601, 12.2 Check and Reflect 6. (a) 6.40 1015 N (b) 3.67 1015 N 7. 1.3 106 m/s 8. 6.53 1026 N 9. 1.06 104 m 10. 8.31 109 T page 613, 12.3 Check and Reflect 9. 1.04 102 C 11. 8.4 102 N page 622, Chapter 12 Review 20. 9", ".86 1014 N 21. (a) 2.31 104 m/s (b) 4.44 1019 J or 2.77 eV 22. 4.08 103 m/s 23. (a) 4.13 1016 N (b) 2.06 1016 N (c) no force 24. 2.7 105 T 26. (b) B varies as 1/r or 1/r 2 28. 25.0 N page 626, Unit VI Review 39. (a) 1.4 108 V/m (b) 1.1 1019 J 41. 2.56 109 m 42. 5.06 107 N\u2014repulsive 45. 15 N/C 46. 3.2 1015 J; maximum speed is 2.0 106 m/s 47. 0.268 m to the right of the first charge 48. 5.0 105 C 49. 0.73 A 50. (a) 2.6 106 N/C [left] (b) 1.6 N\u2014attractive 52. 2.0 103 N 53. 1.67 104 m 54. (a) 3.63 1047 N (b) 8.22 108 N (c) 2.27 1039 times greater 57. 1.53 103 m/s; positive charge Numerical Answers 895 21-PearsonPhys-AnswerKey 7/29/08 1:40 PM Page 896 58. (a) 3.17 1012 N (b) 1.89 102 m 59. (a) 113 N [right] (b) 28.4 N [left] (c) 30.8 N [69.4\u00b0] 60. 6.6 103 V 61. (a) 2.50 104 V (b) 9.37 107 m/s (c) 2.25 1011 N 81. (b) B varies as 1/r (c) Plot B vs 1/r (f ) 0.79 T\u2022m/A (g) 0.789 T\u2022m/A page 647, 13.1 Check and Reflect 8. 2.5 103 Hz page 652, 13.2 Check and Reflect 4. 2.1 108 m/s 5. 6.67 102 s 6. 293 Hz 7. 2.7 103 Hz 8. 3.44 104 m 9. 1.86 103 Hz 10. 2.80 108 m/s 11. 2.85 108 m/s page 665, 13.3 Check and Reflect 8. 50 cm page 683, 13.4 Check and Reflect 9. 2.26 108 m/s 10. 23.0\u00b0 11. 32.4\u00b0 12. 41.1\u00b0 14. angle of refraction 15.9\u00b0; wavelength 351 nm 16. 2.25 102 m; inverted 18. (a) 648 cm (b) 140 cm (c) 8.2 cm page 697, 13.5 Check and Reflect 6. 5.7 106 m 7. 6.1\u00b0 8. 0.350 m 9. 3.76 103 m 10. wavelength 6.11 107 m; frequency 4.91 1014 Hz 11. wavelength 4.13 107 m; frequency 7.26 1014 Hz page 699, Chapter 13 Review 23. 3.41 104 m 24. 2.88 108 m/s 26. 13 cm 30. 5.00 cm high, 6.67 cm from the mirror 31. 1.82 cm 32. (a) 2.26 108 m/s (b) 2.19 108 m/s (c) 1.97 108 m/s 896 Numerical Answers (d) 2.04 108 m/s (e) 1.24 108 m/s 33. angle of refraction 23\u00b0; wavelength 415 nm 34. (a) no (b) yes (c) yes (d) no 35. 1.12, which is less than that of water 36. (a) 48.8\u00b0 (c) 33.3\u00b0 (b) 24.4\u00b0 (d) 41.1\u00b0 37. 2.67 cm high, 3.33 cm from the lens. It is virtual, erect, and diminished. 38. 0.7\u00b0, 1.4\u00b0, 2.0\u00b0 39. 4.4 107 m; violet 40. 4.9 107 m; blue 41. 6.0 105 m page 710, 14.1 Check and Reflect 1. 4.42 1019 J 2. 8.29 108 m 2E600 3. E300 4. (a) 2.41 1019 Hz 5. 2.77 1021 photons 6. 4.93 1021 photons 7. 100 photons/s 15. 0.00388 nm 16. (a) 1.1 1034 m 18. 3.32 1030 photons/s 19. 0.52 eV, 2.", "1 eV, 4.7 eV 20. 5.33 1023 kg\u00b7m/s 22. (c) 9.3 N 23. 1.5 1017 page 746, Unit VII Review 26. 3.62 1019 J 29. 6.00 1014 Hz 33. 3.9 107 m 36. 2.21 1027 N\u00b7s 37. 0.243 nm 41. 8.3 1029 photons/s 43. 3.84 105 km 45. 4.84 104 m 46. 3.75 103 Hz 47. 1.39 104 s 49. refractive index 2.42, therefore the material is diamond 51. 2.91 108 m/s; error 2.93% 53. 7.5 cm from mirror. It is erect and page 715, 14.2 Concept Check Ephoton hf W page 720, 14.2 Check and Reflect 1. 3.11 eV 3. 9.82 1014 Hz 4. yes, hf W 5. 1.25 V 9. 4.1 1015 eV\u2022s; this is close to Planck\u2019s constant Page 725, 14.3 Check and Reflect 1. 1.33 1027 N\u2022s 3. wavelength 1.11 1013 m; energy 1.80 1012 J 6. 0.0124 nm 7. 2.0 104 m/s page 736, 14.4 Check and Reflect 1. 36.4 nm 2. 1.33 1027 kg\u2022m/s 3. 5.3 1026 kg\u2022m/s 4. 2.1 1023 kg\u2022m/s 5. (a) 3.36 1015 J (b) 8.47 1012 m 1.83 103 e/ 6. p page 742, Chapter 14 Review 3. 4.42 1019 J or 2.76 eV 4. Ex/Ev 7. 4.0 1019 J or 2.5 eV 9. It increases by 0.0024 nm. 100 10. 73 nm 11. 6.63 1027 N\u2022s 13. 3.0 1018 photons 14. 3.62 1019 J or 2.26 eV 2.5 cm high. 4 A; pA 4pB 54. B 55. 9\u00b0 56. 7.0 1016 J 57. 10 cm from the lens 58. (b) 0.1 m/s 60. 0.0045 nm 62. 0.4 eV 63. for violet light 24.8\u00b0; for red light 40.5\u00b0 64. 2.63 102 m 67. Energy 0.2 eV; momentum 3 1025 N\u2022s 68. 6.67 102 s 74. 2.5 eV 76. momentum 1.2 1022 kg\u2022m/s; wavelength 5.5 1012 m 78. 6.6 1026 N\u2022s page 760, 15.1 Check and Reflect 1. (a) 8.0 1015 N (b) 0 N 4. (b) 4.00 105 m/s 5. 1.8 107 m/s 6. 1.57 103 T 8. 8.00 104 m/s 9. (a) 8.0 1015 N [downward] 10. (b) 3 109 m/s page 765, 15.2 Check and Reflect 3. 6.40 1019 C 4. 8.00 1017 N [up] 5. (a) 0 N 6. (a) 1.6 1018 C, or 10e (b) gained 10 (c) downward 21-PearsonPhys-AnswerKey 7/29/08 1:40 PM Page 897 page 770, 15.3 Check and Reflect 7. 177.8 MeV 3. (a) 3.6 1016 J (b) 3.6 1012 J 6. (a) 2 1010 m (b) The gold atom is approximately 7000 times larger than the gold nucleus. 7. (b) 5.0 1015 m page 780, 15.4 Concept Check 634 nm, 654 nm page 780, 15.4 Check and Reflect 5. (b) 486 nm (c) 4.09 1019 J 6. 1.96 eV 7. (b) 1 1.9 eV; 2 2.1 eV; 3 1.5 eV; 4 3.9 eV 654 nm, visible, red; 592 nm, visible, 829 nm; (c) 1 2 yellow; 3 4 319 nm 9. (a) 3.03 1019 J (b) 2.66 1020 J 11. (a) 140 nm (b) 3.27 106 m (c) 15 page 784, 15.5 Check and Reflect 4. (a) 3.32 1010 m (b) 2.00 1024 kg\u00b7m", "/s (c) kinetic energy 2.19 1018 J; speed 2.19 106 m/s 5. (a) 2 2 1; 3 3 1 page 786, Chapter 15 Review 6. 3.68 1018 C 7. 1.76 1011 C 16. 4.76 1010 m 19. 2.4 1016 N 20. (a) 1.09 106 m/s 21. 4.1 104 N/C directed upward 22. (a) 1.06 1019 J; 1.55 1019 J; 1.82 1019 J (b) wavelengths: 1.88 106 m, 1.28 106 m, 1.09 106 m. Frequencies: 1.60 1014 Hz, 2.34 1014 Hz, 2.74 1014 Hz. (c) 1.55 1019 J 23. (b) 634 nm 24. 2.5 103 N/C [up] 25. (a) 2.17 1018 J (b) 9.01 1022 m/s2 (d) 4.72 1011 s page 796, 16.1 Check and Reflect 1. (a) 38 protons, 52 neutrons (b) 6 protons, 7 neutrons (c) 26 protons, 30 neutrons (d) 1 proton, 0 neutrons 2. 1.0 109 eV 1.0 GeV 3. 2.3 108 eV 4. 5.56 108 kg 8. (a) 0.366 642 u (b) 8.538 MeV/nucleon 9. (a) 100 MeV (b) 476 MeV (c) 1737 MeV 13. Nuclear radius 5.38 fm Approximate distance between nucleons 2 fm page 810, 16.2 Check and Reflect 2. 3 : 2 8. 1.819 MeV page 817, 16.3 Check and Reflect 1 1. 6 1 2. 3.7 1012 atoms/s 4. 1.2 million years 5. 1.7 103 of the original quantity 6. 1.1 104 a 7. 1.5 102 Bq 8. (a) 4 h (b) 3600 decays/min Page 824, 16.4 Check and Reflect 6. (b) 4.730 MeV 7. (b) 17.59 MeV 8. (a) 8.1 1019 (b) 1.0 103 kg Page 826, Chapter 16 Review 5. 60 neutrons, 55 protons 6. 8.0 1012 J 7. 9 1013 J 8. 3.4 1010 J 9. 0.322 u 10. 20 MeV 18. 1.8 108 decays/s 19. 2 1019 nuclei 24. (a) 7.074 MeV (b) 8.448 MeV (c) 8.792 MeV (d) 7.591 MeV 26. (b) 8.95 MeV 27. (b) 3.210 MeV 28. about 1.1 104 a 29. (a) 5.1 1015 atoms (b) 1.905 109 kg (c) 1.1 104 Bq 30. (a) 400 Bq (b) 4 h 31. 7.274 MeV 32. (b) 0.23 kg page 844, 17.3 Check and Reflect 7. (a) momentum 5 1021 kg\u2022m/s; kinetic energy 8 1015 J (b) momentum 8.4 1022 kg\u2022m/s; kinetic energy 2.1 1016 J 8. (a) 0.51100 MeV (b) 5.521 1027 kg 9. 931.5 page 851, Chapter 17 Review 19. 3.14 1025 kg 20. (a) e (b) 83138 MeV/c 2 21. (b) 10 cm (c) 1.9 1020 N\u2022s page 855, Unit VIII Review 1 and 2 5 1 2. 9.6 107 C 3. 1.60 1018 C 4. 1.60 1016 N [N] 8. (a) ni ni (b) ni (c) ni 4 \u2192 nf 6 \u2192 nf 1 \u2192 nf 4 \u2192 nf 10. 33 neutrons, 31 protons 12. 2.0 107 eV 13. 1.0 109 J 14. 6.7 MeV 15. 2.3 109 decays/s 17. 1.4 g 22. (a) 1.022 MeV (c) 2.43 1012 m 25. (a) proton (b) kaon (c) pion (d) 27. (b) 7.3 104 m/s 28. 0.48 T 29. (a) electric force 1.6 1017 N [up] magnetic", " and mass, 147 and net force, 146 of a mass-spring system, 366\u2013369 units of, 23, 26 Acceleration due to gravity, 54\u201362 Acceleration-time graph, 26, 42 Accelerometer, 366 Action force, 160\u2013167 Action-at-a-distance force, 200 Activity (also Decay) rate, 812 Adams, Henry, 617 Adams, John, 282 Air velocity, 92\u201397 Air, refractive index of, 666, 667t, 670, 673 wavelength of light in, 669 Airbag systems, 140, 141 Alpha decay, 799\u2013801 Alpha emission, 797 Alpha particle, 497, 767\u2013769, 799\u2013801 hazards of, 808t Alternating current, 638t Altitude, 221 Ammeter, 604 Ampere (A), 602, 607 Ampere, Andr\u00e9-Marie, 602, 607 Ampere\u2019s law, 642 Amplitude, 355f, 371, 378, 382, 386, 395, 408, 412 Amplitude modulation (AM), 646 Analog radio technology, 646 Anderson, Carl, 836, 842, 846 Angle of diffraction, 689\u2013691 Angle of incidence, 654, 666f Angle of reflection, 654, 666f Angle of refraction, 666, 672 Anti-locking brake systems, 187 Antimatter, 804, 836, 837 Antineutrino, 804 Antinodal lines (also Bright fringes), 686\u2013690, 692 Antinodes, 415, 417\u2013419, 422f, 423f, 424f, 426 Apparent weight, 224\u2013226 Applied force, 130, 171\u2013189, 307 Arago, Dominique, 692 Aristotle, 634 Armature (also Rotor), 608 Artificial satellites, 284\u2013286 Astronomical units (AU), 272, 273 At rest, 13 Atom, 513 Bohr model, 773, 774 energy levels, 774\u2013776 nucleus, 766\u2013769, 790\u2013796 orbit size, 774 planetary model, 768 quantum model, 782, 783 raison-bun-model, 758 structure theories of, 752f subatomic particle models, 845\u2013849 subatomic particles, 830\u2013849 Atomic emission spectroscopy, 779 Atomic mass number, 790, 791, 798 Atomic mass unit, 791 Atomic number, 790, 791, 794, 795 Attitude of images, 656, 662, 664 Audio frequency signal, 646 Aurora borealis, 580, 597, 598f, 779 Avalanches, 448, 449 Average net force, 450 Average velocity, 12\u201318, 36\u201338 Axis of rotation, 242, 244f Axle, 242 B Background radiation, 637 Ballistic pendulum, 483\u2013485 Balmer, Johann Jacob, 773 Balmer\u2019s formula, 773, 774, 778 Balmer series, 776f Baryons, 842, 843t, 847 Bearing method, 78 Becker, Wilhelm, 497 Becquerel, Antoine Henri, 797 Becquerel (Bq), 812 Beta emission, 797 Beta particles, 802\u2013805, 808t Beta-negative decay, 802\u2013805, 847 Beta-positive decay, 805, 848 Bethe, Hans, 821 Binding energy, 793\u2013796, 818 Binoculars, 674 Biomechanics, 29 Blackbody radiation curves, 704f, 705 Blackbody, 705 Blau, Marietta, 842 Bohr, Niels, 513, 771, 773, 774 Bohr model, 773, 774, 783 Bohr radius, 774 Born, Max, 739, 783 Bosons, 842 Bothe, Walther, 497 Bow wave, 433 Brahe, Tycho, 214, 269 Bright-light (also Emission line) spectrum, 772f Bubble chamber, 831, 833 C Capacitor, 642 Carbon dating, 815, 816 Cartesian plane, 127 Cassini, Giovanni, 359 Cathode ray, 593, 754, 755f\u2013757 Cathode ray tube, 568, 593f Cavendish, Henry, 205, 524 Cell phones, 646 Central antinode, 686 Centre of curvature, 657, 658 Centre of mass, 492 Centrifugal force, 247 Centripetal acceleration, 243, 244, 247t, 252\u2013255, 265\u2013267 Centripetal force, 244, 246, 247t, 256, 258\u2013267, 277\u2013280 CERN, 831 CGS system, 642 Chadwick, James,", " 497, 790 \u201cChange in\u201d (\u0394), 7 Charge, 593\u2013600, 762, 798 determination of, 528, 529 magnitude of, 529\u2013531 transfer of, 517\u2013522 Charge migration, 520 Charge shift, 521 Charge-to-mass ratio, 755\u2013758 Charging by induction, 521 Chemical energy vs nuclear energy, 820 Circular motion, 242\u2013247 and Newton\u2019s laws, 248\u2013267 of satellites and celestial bodies, 269\u2013286 Clocks, 383, 387 Closed-pipe (also Closed-tube) resonance, 419 Cloud chamber, 830, 831, 833, 842 Coaxial cables, 558 Coefficient of friction, 259 Coefficient of kinetic friction, 182, 183 Coefficient of static friction, 182, 183t, 186 Coils, electron flow in, 587f Collinear charges, 532 Collinear collisions (see Collisions in one dimension) Collinear forces, 132, 133 Collinear vectors, 71, 73, 75 Collisions, 469, 470 charge transfer during, 518, 519 impulse in, 458 Collisions in one dimension, 468\u2013485 Collisions in two dimensions, 487\u2013498 Colour of quarks, 848 Commutator, 608, 609 Compression, 404 Compton, Arthur, 721f Compton effect, 721\u2013724 Compton equation, 722\u2013724 Compton scattering, 721 Concave reflecting surface, 657f\u2013659 Conduction, 519 Conductors, 513, 514 and magnetic fields, 602\u2013611 electric field lines of, 555\u2013558 Conservation of momentum, in discovery of subatomic particles, 497, 498 in one-dimensional collisions, 473\u2013479 Index 899 22-PearsonPhys-Index 7/25/08 7:40 AM Page 900 in two-dimensional collisions, 489\u2013495 Conservative forces, 314, 319, 320 Constant acceleration, 450 Constant mass, 450 Constructive interference, 412, 416f, 417, 425\u2013427, 686\u2013689, 782f Continuous spectrum, 772f Converging lens, 677, 678 Converging mirror, 657\u2013659 Convex reflecting surface, 657f\u2013659 Copernicus, Nicholas, 214 Cosmic radiation, 637, 638t Coulomb (C), 529 Coulomb\u2019s constant, 774 Coulomb\u2019s force (see Electrostatic force) Coulomb\u2019s law, 529\u2013531, 768 Coulomb\u2019s proportionality constant (k), 529, 530 Crash test dummies, 29 Crest, 394, 395, 397f, 408, 430 Critical angle, 672, 673 Crookes, William, 754 Curie, Marie and Pierre, 797, 808 Current, 587, 602\u2013609, 642 Curved mirrors, 657\u2013662 Cycle, 249, 344, 355 Cyclotron particle accelerator, 841 D da Vinci, Leonardo, 180 Dalton, John, 754 Damping, 385 Daughter element, 799 Daughter nucleus decay, 807 Davisson, C. J., 729 Davisson-Germer experiment, 729f, 730f de Broglie, Louis, 726f, 782 de Broglie\u2019s wave equation, 726\u2013728, 730\u2013733 de Broglie\u2019s wavelength, 726 de Coulomb, Charles, 524, 528, 529 de Maricourt, Pierre, 582 Decay (also Activity) rate, 812 Decay constant, 811, 812 Defibrillators, 559 Deformation from colliding objects, 481 \u201cDelta\u201d symbol (\u0394), 7 Destructive interference, 412, 413f, 416f, 417, 419, 425\u2013427, 687, 689, 782f Diamond, 667t, 670, 672 Difference in path length, 688 Diffraction, 685, 690\u2013694 Diffraction grating, 692, 693, 771 Diffuse (also Irregular) reflection, 653 Digital wireless technology, 646 Dirac, Paul Adrien Maurice, 582, 836 Direct current, 608 Dispersion, 675 Displacement, 7, 9, 56\u201358, 73\u201375, 80\u201389, 293\u2013295 from velocity-time graphs, 33\u201335, 40, 48\u201351 Diverging lens, 677, 678 Diverging mirror, 657\u2013659 Diverging rays, 397, 398 Domains, 589 Doping, 515 Doppler, Christian, 429 Doppler effect, 430\u2013432 Doppler frequency, 431, 432 Doppler ultrasound, 434 Doppler wavelength, 431\u2013434 Down quark, 845, 846t", " Drift tube particle accelerator, 841 Dynamics, 126 E Earth, magnetic field of, 591, 597, 598 orbital diameter, 648f, 649 Ebonite rod, 518, 520, 521, 529 Eddington, Arthur Stanley, 821 Efficiency, 324 Effluvium theory, 544 Einstein, Albert, 199, 640, 706, 713 Einstein\u2019s mass-energy equation, 793, 794 Elastic (also Spring) constant, 299 Elastic collisions, 481, 482, 495\u2013497 Elastic force, 128, 129 Elastic limit, 349 Elastic potential energy, 300\u2013302, 307, 370, 402, 406 Electric charge (see Charge) Electric current (see Current) Electric field, 545\u2013553, 554\u2013559, 641\u2013643, 756f between charged plates, 567, 568 in one dimension, 550 in two dimensions, 551, 552 magnitude and direction of, 546\u2013553 Electric field lines, 554\u2013559 Electric force on a charged particle, 755 Electric generators, 617 Electric motor, 602, 608, 609, 614, 617, 619 Electric potential (also Voltage), 564, 565 Electric potential difference, 565, 566 Electric potential energy, 561\u2013564, 571 of an alpha particle, 769 of an electron, 774, 775 Electrical conductivity, 513, 514 Electrical interactions, 512\u2013522, 524\u2013537 and the Law of conservation of energy, 570\u2013575 Electricity, 512\u2013522 Electromagnetic force, 838t Electromagnetic induction (see Generator effect) Electromagnetic radiation (see also Light), 636\u2013646 diffraction and interference, 684\u2013696 models of, 639\u2013643 photoelectric effect, 712\u2013719 production of, 644, 645 speed of, 648\u2013651 Distance, 6, 7, 106\u2013111 Electromagnetic spectrum, 637, 638t, 708, 709f Electromagnets, 588, 589 Electromagnetic theory, 641\u2013643 Electromagnetic wave, 643 Electromotive force, 611 Electron, 498, 513, 842 charge of, 529, 762 charge-to-mass ratio, 755\u2013758 de Broglie\u2019s wave equation for, 726, 727 discovery of, 754 energy level transition of, 776, 778, 779 in charge transfer, 517\u2013522 in Compton scattering, 721, 722 in photoelectric effect, 712\u2013719 kinetic energy of, 730\u2013733 mass of, 792t, 844t wave nature of, 782, 783 wavelength of, 728, 836, 837 Electron microscope, 727 Electron-positron annihilations, 836 Electron volt (eV), 565, 843, 844 Electron-wave interference, 729 Electrostatic discharge, 537 Electrostatic (also Coulomb\u2019s) force, 528\u2013537 Electrostatics, 513\u2013521 Electroweak force, 848 Elementary charge, 774 Elementary unit of charge, 762 Ellipse, 269 Emission line (also Bright-light) spectrum, 772, 776 Energy, 292\u2013304, 306\u2013309, 311\u2013322, 324\u2013329, 639 along a pulse, 402 in quanta, 705\u2013709 Energy conservation, 497, 498 Energy level, 773\u2013776 Energy-mass equation, 843 Equilibrium position, 394, 395 Erect image, 660, 662, 680 Euclid, 634 Excited state, 775 Explosion interaction, 476 Extrasolar planets, 283 F Faraday, Michael, 545, 584, 610, 611, 614 Faraday\u2019s ice pail experiment, 559 Faraday\u2019s law, 642 Femtometres, 790 Fermi, Enrico, 804 Fermilab Tevatron accelerator, 840f Fermions, 842 Ferromagnets, 589 Feynman, Richard, 739 Fibre optics, 674 Field (see also Electric fields), 200, 545 First order maximum, 427 Fission, 818\u2013820 Fizeau, Armand, 649, 666 Fletcher, Harvey, 761 Focal length, 657, 658, 662, 667, 678 Force, 127\u2013135, 293\u2013295 900 Index 22-PearsonPhys-Index 7/25/08 7:40 AM Page 901 and centripetal acceleration, 252\u2013255 effect on momentum and impulse, velocity-displacement, 368f, 369f velocity-time, 21f\u201328f, 62f 455\u2013466 in collisions, 474", " Forced frequency, 382, 383f, 385 Frame of reference, 13, 14 Franklin, Benjamin, 512, 513 Fraunhofer lines, 773 Free fall, 226, 227 Free-body diagrams, 129\u2013135f, 146f, Gravimeters, 222 Gravitational acceleration, 217\u2013219 Gravitational field, 200, 201, 545 Gravitational field strength, 201, 217, 220\u2013222, 378, 379 Gravitational force, 128, 129, 131, 173\u2013175, 196\u2013201, 203\u2013214, 216\u2013228, 259f\u2013264, 277, 524, 838t 150f, 151f, 161f\u2013167f, 172f\u2013189f, 197f, 198f, 209f, 222f\u2013224f, 226f Frequency, 249, 265\u2013267, 344, 345, 408, Gravitational mass, 199 Gravitational potential energy, 295\u2013298, 307\u2013309, 312\u2013316, 321, 560, 562 409, 429\u2013432, 636, 637 Frequency modulation (FM), 646 Fresnel, Augustin, 691, 692f Fresnel lens, 634f Friction, 130, 165, 169\u2013189, 320 463 and momentum, 449 charging objects by, 518, 519 effects on motion, 184 in circular motion, 258 Friedman, Jerome, 845 Fuel cells, 329 Fundamental forces, 194 Fundamental frequency, 422 Fundamental particles, 836, 837 Fusion, 818, 821, 822 G g force, 216 Galilean telescope, 682f Galileo, 54, 103, 137, 138, 281, 648 Galle, Johann Gottfried, 282 Galvani, Luigi, 602 Galvanometer, 602f, 604f, 611f Gamma decay, 806 Gamma emission, 797, 806, 808t Gamma radiation, 636 Gamma rays, 637f, 638t Gauss (G), 586 Geiger, Hans, 767 Gell-Mann, Murray, 845 Generator effect (also Electromagnetic induction), 609\u2013611 Generator effect, 615, 617\u2013619 Genetic damage, 808 Geostationary satellites, 285, 286 Germer, L. H., 729 Gilbert, William, 512, 583 Glaser, Donald, 831 Glashow, Sheldon, 848 Glass, 666, 667t, 668, 673 Gluons, 838 Gramme, Z\u00e9noble Th\u00e9opile, 617 Grand unified theory, 849 Graphs, acceleration-displacement, 367f, 368f, 369f acceleration-time, 26f, 42f force-displacement of a pendulum, 361f force-displacement of a spring, 351f net force-time, 459f\u2013462f position-time, 11f\u201318f, 61f, 62f Graviton, 838 Gravity, 54\u201362 Gravity assist, 214 Gray (Gy), 809 Ground state, 774 Ground velocity, 92\u201399 Grounding, 521, 610 H Hadrons, 842, 843 Half-life, 812\u2013814 Hansen, Hans Marius, 771 Harrison, John, 383 Heisenberg, Werner, 734f Heisenberg\u2019s uncertainty principle, 735 Helium formation, 821, 822 Henry, Joseph, 610, 611 Herschel, William, 282 Hertz (Hz), 249, 250, 344, 345 Hertz, Heinrich, 644, 645, 648, 712 High tide, 210, 211 Hit-and-stick interaction, 477 Hollow conducting objects, 557, 558 Hooke, Robert, 299, 349f, 684 Hooke\u2019s law, 299\u2013301, 349\u2013354, 366, 368 Horsepower (Hp), 324 Huygens, Christiaan, 359, 381, 634, 639, 648, 649, 684f Huygens\u2019 Principle, 684, 685 Hydrogen, 773f Hydrogen fusion, 821\u2013823 I Images, 653\u2013665 formation in a curved mirror, 657\u2013662 formation in a plane mirror, 654\u2013656 Impulse, 457\u2013466, 470 In phase, 395, 416, 425, 426 Incandescent, 704 Incident light, 653, 654f, 666 Incident wave, 395, 398 Inclines, 173, 174, 176\u2013179, 188, 189, 308 Induced current, 611, 617\u2013619 Induction, 520, 521 Induction coils, 615t Inelastic collisions, 483\u2013485, 495\u2013497 Inertia, 138\u2013140, 147\u2013157 Inertial mass, 148\u2013157 Infinity, 562 Infrared radiation", ", 636 Infrared ray, 709f Infrared spectrum, 637, 638t Input energy, 324 Instantaneous momentum, 449 Instantaneous velocity, 24, 25 Insulators, 513, 514 Interference, 411\u2013413, 416, 417, 685\u2013690, 695, 696 Interference fringes, 686 Interference pattern, 425, 426f Interferometer, 649 Inverse square law, 529 Inverted image, 660, 662, 680 Ionization energy, 775 Ionization smoke detectors, 802 Irregular (also Diffuse) reflection, 653 Isolated systems, 311\u2013316, 470, 475 Isotopes, 791, 796f, 806 J Jerk, 26 K Kammerlingh Onnes, Heike, 515 Kendall, Henry, 845 Kepler, Johannes, 214, 269 Kepler\u2019s constant, 271\u2013273 Kepler\u2019s laws of planetary motion, 269\u2013279 Keplerian telescope, 682f Kilogram-metres per second (kg\u2022m/s), 449, 457 Kinematics, 6 Kinematics equations, 47\u201353, 56, 144 Kinetic energy, 302\u2013304, 306\u2013309, 312\u2013316, 321, 370, 402, 480, 570, 571, 573 in alpha decay, 799, 801 in beta decay, 804 in elastic collisions, 481, 482, 495 in inelastic collisions, 483\u2013485, 495 of alpha particles, 769 of electrons, 774, 775 of photoelectrons, 713\u2013717 Kinetic friction, 176\u2013182, 187, 258 Kirchhoff, Gustav, 705, 771, 772 L \u201cLambda\u201d symbol (), 408, 409, 417, 419, 424, 427, 429\u2013432 Latitude, 221 Law of charges, 512 Law of conservation of charge, 517 Law of conservation of energy, 312, 497, 498, 721, 722 and electrical interactions, 570\u2013575 for the photoelectric effect, 714, 718 Law of conservation of momentum, 473, 721, 722 in ballistic pendulum system, 484 in two-dimensional collisions, 491, 492, 495 Law of magnetism, 582 Law of reflection, 654, 658 Le Verrier, Urbain, 282 Index 901 22-PearsonPhys-Index 7/25/08 7:40 AM Page 902 Left-hand rule, 834 for deflection of charged particles, 594, 595 for magnetic fields, 588 for magnetic force, 604, 611, 756f Lenses, 677\u2013681 Lenz\u2019s law, 618, 619 Leptons, 842, 843t, 848t Lewis, Gilbert, 706 Light (see also Electromagnetic radiation) from colliding objects, 481 quantum theory of, 705\u2013709, 712\u2013719, 721\u2013724, 726\u2013735 reflection, 653\u2013665 refraction, 666\u2013681 speed of, 636, 648\u2013651, 669, 670 Light waves, 712\u2013719 Lightning, 512, 513, 522, 529 Lightning rod, 554 Lodestone (also leading stone), 582, 583 Longitudinal waves, 401, 404\u2013406, 695f Lord Kelvin (see William Thomson) Loudspeaker, 614t Low tide, 210, 211 Lyman series, 776f M Maglev, 587 Magnesium, 636f Magnetic deflection, 593\u2013597, 603, 604 Magnetic field, 584\u2013591, 593\u2013600, 602\u2013611, 641\u2013643, 756f\u2013758 left-hand rule for, 588 Magnetic field strengths, 586t Magnetic force, between two current carrying conductors, 607 calculation of, 598\u2013600 on a current carrying conductor, 603\u2013606 Magnetic monopoles, 582 Magnetic poles, 582, 585 Magnetism, 512, 587\u2013591 Magnetization, 589, 590 Magnetohydrodynamic propulsion, 614t Magnetron, 599 Magnets, 583, 585, 586, 617, 618 Magnification, 656, 662, 664 Magnitude, 6 Major axis, 269f Marconi, Guglielmo, 646 Marsden, Ernest, 767 Mass, 220 and conservation of momentum, 474 due to gravity, 199 in momentum, 448\u2013452 of celestial bodies, 218t, 280, 281 Mass defect, 794, 795 Mass spectrometer, 759 Mass-energy equation, 793, 794 Mass-spring systems, 354\u2013360, 366\u2013376 Matter waves, 726\u2013735", ", 729 Maximum, 426 Maxwell, James Clerk, 641, 648, 695, 771 902 Index Maxwell\u2019s equations, 642 Mechanical energy, 306, 309, 311\u2013322 Mechanical resonance, 382 Mechanical waves (see Waves) Mechanics, 306 Mediating particles, 837, 838 Medium, 394, 395, 404, 406, 411 Mesons, 842, 843t, 845, 847 Michell, John, 205 Michelson, Albert, 650, 651 Microwaves, 637f, 638t Millikan, Robert Andrews, 713f\u2013715, 761\u2013763 Minor axis, 269f Mirage, 667 Mirror equation, 662, 664 Mirrors, 654\u2013662 Momentum, 449\u2013452, 470 and impulse, 454\u2013466 and Newton\u2019s Second Law, 450\u2013452, 456 and non-linear net forces, 461, 462 conservation of, 473\u2013479, 489\u2013498 in one-dimensional collisions, 473\u2013479 in two-dimensional collisions, 489\u2013495 of a photon, 728 Motion, in one dimension, 6\u201362, 70\u201375 in two dimensions, 76\u201389 of projectiles, 102\u2013111 sign conventions for, 8\u201310 Motor effect force, 593 Muons, 842, 843t N Nanotechnology, 619, 674 Navigator method, 77, 78 Neddermeyer, Seth, 842, 846 Negative acceleration, 28, 29, 44 Net charge, 517, 518 Net force, 131\u2013135, 139, 146, 152, 155, 157, 164, 165, 171\u2013189, 306, 307, 356, 366, 456\u2013466, 473, 474, 489 in circular motion, 256 on momentum, 450 Neutrino, 804 Neutron, 497, 498, 790, 792t, 844t, 847 Neutron number, 790, 791, 794, 795 Newton (N), 127 Newton, Isaac, 194, 196, 214, 276\u2013281, 545, 634, 639, 675 Newton\u2019s first law of motion, 139, 140 Newton\u2019s law of gravitation, 524 Newton\u2019s law of universal gravitation, 204\u2013214, 216 Newton\u2019s second law of motion, 143\u2013157, 366, 367 and horizontal motion, 149\u2013151 and momentum, 450\u2013452, 456 and single pulley system, 154, 155 and two-body systems, 153, 154 and two-pulley system, 156, 157 and vertical motion, 152, 153 on objects in systems, 470, 473 Newton\u2019s third law of motion, 161\u2013167, 459, 474 Newton-second (N\u2022s), 457 Nodal lines (also Minimum; Dark fringes), 426, 686 Nodes (also Nodal points), 415, 417\u2013419, 423f, 424f, 687 Non-collinear forces, 133, 134 Non-collinear vectors, 80\u201389 Non-conservative forces, 319 Non-isolated systems, 320\u2013322 Non-linear net force, 461, 462 Non-uniform motion, 22\u201328, 31 Non-zero net force, 307, 450 Normal (also Perpendicular) force, 130, 131 Normal force, 258, 259, 260 Normal line (N), 653, 654f, 666f North pole, 582 Nuclear energy vs chemical energy, 820 Nuclear reactions, 818\u2013823 Nucleon, 790, 795, 796 Nucleosynthesis, 823 Nucleus, 513, 766\u2013769, 790\u2013796 decay rates of, 811\u2013816 reactions in, 818\u2013823 size of, 768, 769 O Oersted, Hans Christian, 587 Oil-drop experiment, 761 Omega particle, 845 Open-pipe (also Open-tube) resonance, 424 Opsin, 636 Optical centre, 678f Optical fibres, 673 Orbital height, 279, 280 Orbital period, 271\u2013289 Orbital perturbations (also Wobbles), 281, 282, 283 Orbital radius, 271\u2013289 Orbital, 783 Origin, 6 Oscillation, 344 Oscillators, 640 Oscillatory motion, 344, 345, 348\u2013364, 366\u2013379, 381\u2013387 Out of phase, 416 Output energy, 324 Overtones, 422, 423, 424 Oxygen, 779f P Parabola, 61, 62 Parabolic mirror, 661 Parallel plates, electric field between, 567 motion of charges between, 572, 573 Parallel-plate capacitors, 558, 559 electric field between, 574, 575 electric potential", " energy between, 563 Parent element, 799 Partially reflected/refracted rays, 666 22-PearsonPhys-Index 7/25/08 7:40 AM Page 903 Particle, 639 Particle accelerator, 840f, 841, 842 Particle model, 639, 640 Particle-in-a-box model, 730\u2013735 Paschen series, 776f Path length, 688, 689 Pauli, Wolfgang, 498, 804 Payne-Gaposchkin, Cecilia, 821 Pendulum, 314\u2013316, 359\u2013362, 377\u2013379, 381, 382, 483 Pendulum-bullet system (also Ballistic pendulum), 483\u2013485 Period, 249, 251, 265\u2013267, 344, 345, 408, 409 of a mass-spring system, 373\u2013376 of a pendulum, 377\u2013379 Perpendicular bisector, 686\u2013691 Perrin, Jean Baptiste, 754 Phase shift, 426 Photoelectric effect, 712\u2013719 Photoelectrons, 712 Photon, 640, 706\u2013709, 722\u2013724, 728, 776 Piezoelectric material, 387 Pions, 842, 843t Pitch, 430 Planck, Max, 640, 641, 705f, 752 Planck\u2019s constant, 705, 706, 713\u2013715, 726, 774 Planck\u2019s formula, 705\u2013708, 713 Plane mirror, 654 Plane polarized light, 696 Planetary (also Solar-system, Nuclear, or Rutherford) model, 768 Planetary motion, 269\u2013289 Plasma, 522 Plates as conductors, 556 Point of incidence, 653, 654f, 666f Point source, 425, 426 Poisson, Simon, 691 Poisson\u2019s Bright Spot, 692 Polar coordinates method, 77, 78 Polarization, 696 Polarizing filters, 696 Porro prism, 674 Position, 6 Position-time graphs, 11\u201318, 61, 62 Position-time graphs, vs Velocity-time graphs, 25, 26, 28, 41\u201344 Positron, 804, 836, 837 Positron-emission tomography (PET), 837 Potential energy (see also Elastic potential energy; Gravitational potential energy), 570 Powell, Cecil Frank, 842 Power, 324\u2013329 Pressure waves, 406, 433 Primary cosmic rays, 841 Principal axis, 657, 677, 678 Principal focal point, 657 Principal focus, 677, 678 Principal quantum number, 774 Principle of superposition, 412, 413 Prisms, 675\u2013677, 771 Projectile, 54, 465 Projectile motion, 54, 102\u2013112, 108\u2013111 Propellar aircraft, 166 Proportionalities, 452 Proton, 497, 498, 513, 529, 790, 792t, 844t, 847 Proton-proton chain, 821 Ptolemy, 214 Pulse, 401\u2013408, 411\u2013413 Pythagorean theorem, 83, 85, 535\u2013537 Q Quadratic equations, 56 Quanta, 640 Quantization of charge, 761\u2013763 Quantized, 705 Quantum, 705 Quantum chromodynamics, 848 Quantum colour, 848 Quantum electrodynamics, 838 Quantum field theory, 837 Quantum indeterminancy, 740 Quantum model (also Quantum theory), 640, 641, 782, 783 Quantum physics, 830\u2013849 Quarks, 845, 846t, 847, 848t Quartz crystals, 387 R Radar vectors, 7 Radial line, 242f\u2013244 Radiation sickness, 808 Radiation, 811\u2013816 Radio frequency signal, 646 Radio waves, 637f, 638t, 645, 646 Radioactive decay, 797\u2013809, 812\u2013814 Radioactive decay series, 807 Radioisotopes, 808, 812\u2013814 Radiotherapy, 815 Radius of celestial bodies, 218t Radius of curvature, 657 Rainbows, 675 Raisin-bun model, 758 Range of a projectile, 105 Rarefaction, 404 Ray diagrams, 653, 654f, 657f\u2013659f, 666f, 678f\u2013681f Rays, 397, 398 Reaction force, 160\u2013167 Real image, 654, 659, 660, 662, 680 Recomposition, 675f, 676 Rectilinear propagation, 653 Reference coordinates, 71 Reference point, 297, 298, 309 Reflected ray, 653, 654f, 666f Reflected wave, 395, 398, 404 Reflection, 653\u2013665 Refracted ray, 666 Refraction,", " 666\u2013681 Refractive index, 666, 667t, 672, 676 Regular (also Specular) reflection, 653 Relative biological effectiveness, 809 Relative motion, 91\u2013100 Resonance, 418\u2013420, 422\u2013424 Resonant frequency, 381\u2013383f, 385, 387, 418, 419 Restoring force, 353, 354, 356, 360, 362, 372, 375 Resultant vector, 71\u201373 Retinal, 636 Revolution, 249, 251 Revolutions per minute (rpm), 249, 250 Right-hand rule, 833f, 834 Rocket, 167, 459, 460, 475 R\u00f8mer, Olaus, 648 Rosette nanotube, 619f Rotational kinetic energy, 302 Rotor (also Armature), 608, 619 Rutherford, Ernest, 513, 767, 790, 797 Rutherford\u2019s scattering experiment, 767 Rydberg, Johannes Robert, 773 Rydberg\u2019s constant, 773, 778 S Safety devices, 463\u2013465 Salaam, Abdus, 848 Satellites, 269\u2013286 Scalar quantity, 6\u201310, 449 Scanning electron microscope, 727 Schr\u00f6dinger, Erwin, 783 Secondary cosmic rays, 841 van de Graaf particle accelerator, 841 Semiconductors, 514, 515 Shock wave, 433 Sievert (Sv), 809 Simple harmonic motion, 355\u2013364, 366\u2013379, 381\u2013387, 408 Simple harmonic oscillators, 355, 366\u2013379 Snell, Willebrord, 667 Snell\u2019s law, 667\u2013669, 673 Solenoids, 589 Sonic boom, 433, 434 Sound, 418\u2013420, 422\u2013424, 429\u2013434, 481 Sound barrier, 433 Source charge, 546, 547f, 548f South pole, 582 Spectrometers, 773 Spectroscopy, 771\u2013773 Spectrum, 675, 676, 772 Specular (also Regular) reflection, 653 Speed, 10, 12, 52, 56, 106\u2013111 and power, 327, 328 in circular motion, 242\u2013244, 250, 251, 254, 255, 265\u2013267 of a mass-spring system, 369\u2013372, 374 of a satellite, 277 of a spring pulse, 406, 409 of light, 636 of sound, 420, 430, 431, 434 Spin, 842 Sports, 466 Spring constant, 350, 351, 357, 374 Spring scale, 128f Spring systems, 349\u2013354 Spring waves, 401, 402 Standard model, 848 Standing wave, 415, 417, 418, 422f, 423f Index 903 22-PearsonPhys-Index 7/25/08 7:40 AM Page 904 Static equilibrium, 555 Static friction, 171\u2013175, 180\u2013183, 258, 259 Statics, 126 Stationary states, 773 Stator, 608 Step leader, 522 Stopping potential, 716, 717 Straight wire, 587 Strange particles, 845 Strange quark, 845, 846t Streamer, 522 String theory, 544, 849 Strong nuclear force, 793, 838t Subatomic models, 845\u2013849 Subatomic particles, 497, 498, 830\u2013849 Sudbury Neutrino Observatory, 498 Sudden infant death syndrome monitors, 615t Sun, 821, 822 Superconductors, 515 Supernova, 823 Superposition, 208\u2013210 Superposition principle, 549 Surfaces, cushioning effects of, 455, 456, 496 electric field lines on, 556, 557 of mirrors, 657 Synchrotron particle accelerator, 841 System, 470 T Tail of a vector, 70\u201375 Tangent, 24, 25 Tangent function, 535\u2013537 Tangential line, 242f\u2013244 Taylor, Richard, 845 Technologies, 614, 615, 619, 646 Telescopes, 682f Telluric currents, 610 Tension, 132, 134, 135, 264, 356, 463 Terminal velocity, 61 Tesla (T), 585, 834 Test charge, 546, 547f, 548f Thales, 512 Thin lens equation, 680, 681 Thomson, George Paget, 729f Thomson, Joseph John, 754, 757, 758 Thomson, William (also Lord Kelvin), 702f, 703 Threshold frequency, 712, 713, 718 Time, 455\u2013466 Tip of a vector, 70\u201375 Torsion balance, 205 Total internal reflection, 672 Trajectory, 103 Translational kinetic energy, 302 Transmission electron microscope, 727 Transmutation, 806 Transverse pulse, 401 Transverse waves, 401, 406", ", 408, 695f Triboelectric series, 518f Trough, 394, 395, 397f, 408 True weight, 222, 223 True weightlessness, 228 Tuned mass damper, 385, 386f Tuning, 424 Turnbull, Wallace Rupert, 166 U Ultraviolet catastrophe, 705 Ultraviolet rays, 637f, 638t Uniform circular motion, 242 Uniform motion, 13, 21\u201328, 31\u201344, 104 Uniformly accelerated motion, 25, 52, 104 Universal gravitational constant (G), 204\u2013207, 524 Universal wave equation, 408, 424, 645, 648, 669, 670, 676 Up quark, 845, 846t Uranium-235 fission, 819f V Vacuum, 666, 667t, 668 Van Allen belt, 597 Van de Graaff generator, 560 Vectors, 6\u201310, 70\u201375, 77, 78, 83, 84, 127, 449, 451 adding, 71\u201373, 80\u201382, 86\u201389, 489\u2013495, 550, 552 analyzing electrostatic forces, 532\u2013537 subtracting, 73 Velocity, 10, 12\u201318, 21\u201328, 31\u201344, 46\u201351, 53, 56\u201362, 91\u2013100, 139 in circular motion, 242\u2013244, 247t in momentum, 448\u2013452, 474 of a mass-spring system, 367, 368 Velocity-time graphs, 21\u201328, 31\u201344, 46\u201350, 62 Vertex, 657, 658 Vertical inversion, 656 Vertical projectiles, 58\u201362 Virtual image, 654, 659, 660, 662, 680 Virtual particles, 837 Visible ray, 709f Visible spectrum, 637, 638t Volt, 564 Volta, Alessandro, 564 Voltage (also Electric potential), 564, 565 Voltmeter, 604 von Fraunhofer, Josef, 771 Voyager missions, 213, 214 W Water, 666\u2013669 Watt (W), 324 Watt, James, 324 Wave, 394, 395\u2013409, 411\u2013427, 428\u2013434, 639 Wave (also Point) source, 395, 397f, 398 Wave equation, 726, 727 Wave front, 395, 398, 684 Wave model, 639, 640, 669, 684\u2013698 Wave patterns, 731 Wave train, 395, 417 Wavelength, 395, 397f, 408, 409, 424, 427, 429\u2013432, 636, 637, 676t and angle of diffraction, 690, 691 and Snell\u2019s law, 669, 670 Wavelet, 684 Wave-particle duality, 726, 737\u2013740 Weak nuclear force, 804, 838t Weight (see also Apparent weight; True weight), 128, 197, 198, 216, 219, 220 Weightlessness, 228 Weinberg, Steven, 848 Wheel, 242f, 243 White light, 675\u2013677 Wilson, Charles Thomas Rees, 831 Wind velocity, 92\u201397 Work function, 712t, 713, 718 Work, 293\u2013304, 306\u2013309, 320, 321, 324, 560\u2013562 and electric potential energy, 563, 567 Work-energy theorem, 307, 308 X X rays, 637f, 638t, 709f, 721, 722 x vector component, 77, 86\u201389, 105\u2013111, 134, 135 Y y vector component, 77, 86\u201389, 105\u2013111, 134, 135 Young, Thomas, 639f, 685, 640, 691 Young\u2019s double-slit experiment, 685, 686, 738\u2013740 Z Zero electric potential energy, 562 Zero net forces, 307 Zero reference point, 561, 562 Zweig, George, 845 904 Indexing is changing. Q: In each picture in the Figure 1.1, what is moving and how is its position changing? A: The train and all its passengers are speeding straight down a track to the next station. The man and his bike are racing along a curving highway. The geese are \ufb02ying over their wetland environment. The meteor is shooting through the atmosphere toward Earth, burning up as it goes. Frame of Reference There\u2019s more to motion than objects simply changing position. You\u2019ll see why when you consider the following example. Assume that the school bus pictured in the Figure 1.2 passes by you as you stand on the sidewalk. It\u2019s obvious to you that the bus is moving, but what about to the children inside the bus? The bus isn\u2019t moving relative to them, and if they look at the other children sitting on the bus, they won", "\u2019t appear to be moving either. If the ride 1 www.ck12.org FIGURE 1.1 FIGURE 1.2 is really smooth, the children may only be able to tell that the bus is moving by looking out the window and seeing you and the trees whizzing by. This example shows that how we perceive motion depends on our frame of reference. Frame of reference refers to something that is not moving with respect to an observer that can be used to detect motion. For the children on the bus, if they use other children riding the bus as their frame of reference, they do not appear to be moving. But if they use objects outside the bus as their frame of reference, they can tell they are moving. The video at the URL below illustrates other examples of how frame of reference is related to motion. http://www.youtube.com/watch?v=7FYBG5GSklU MEDIA Click image to the left for more content. Q: What is your frame of reference if you are standing on the sidewalk and see the bus go by? How can you tell that the bus is moving? A: Your frame of reference might be the trees and other stationary objects across the street. As the bus goes by, it momentarily blocks your view of these objects, and this helps you detect the bus\u2019 motion. 2 www.ck12.org Summary Chapter 1. Motion \u2022 Motion is de\ufb01ned as a change of position. \u2022 How we perceive motion depends on our frame of reference. Frame of reference refers to something that is not moving with respect to an observer that can be used to detect motion. Vocabulary \u2022 frame of reference: Something that is not moving with respect to an observer that can be used to detect motion. \u2022 motion: Change in position. Practice Do the frame of reference activity at the following URL. Watch the introduction and then do the nine trials. Repeat any trial you answer incorrectly until you get the correct answer. http://www.amnh.org/learn/pd/physical_science/week2/frame_reference.html Review 1. How is motion de\ufb01ned in science? 2. Describe an original example that shows how frame of reference in\ufb02uences the perception of motion. References 1. Train: John H. Gray; Bike: Flickr:DieselDemon; Geese: Don McCullough; Meteor: Ed Sweeney (Flickr:Navicore).. CC BY 2.0 2. Bus: Flickr:torbakhopper; Children: Flickr:woodleywonderworks.. CC BY 2.0 3 CHAPTER 2 Position and Displacement www.ck12.org \u2022 De\ufb01ne and give an example of a frame of reference. \u2022 Describe the difference between distance and displacement. \u2022 Identify the position, distance, and displacements in various descriptions of motions. In stockcar races, the winners frequently travel a distance of 500 miles but at the end of the race, their displacement is only a few feet from where they began. Position, Distance, and Displacement In order to study how something moves, we must know where it is. For straight line motion, it is easy to visualize the object on a number line. The object may be placed at any point on the number line either in the positive numbers or the negative numbers. In making the zero mark the reference point, you have chosen a frame of reference. The position of an object is the separation between the object and the reference point. It is common to choose the original position of the object to be on the zero mark. When an object moves, we often refer to the amount it moves as the distance. Distance does not need a reference point and does not need a direction. If an automobile moves 50 kilometers, the distance traveled is 50 kilometers regardless of the starting point or the direction of movement. If we wish to \ufb01nd the \ufb01nal position of the automobile, however, just having the distance traveled will not allow us to determine the \ufb01nal position. In order to \ufb01nd the \ufb01nal position of the object, we need to know the starting point and the direction of the motion. The change in the position of the object is called its displacement. The displacement must include a direction because the \ufb01nal position may be either in the positive or negative direction along the number line from the initial position. The displacement is a vector quantity and vectors are discussed in another section. Summary \u2022 The length traveled by an object moving in any direction or even changing direction is called distance. 4 www.ck12.org Chapter 2. Position and Displacement \u2022 The location of an object in a frame of reference is called position. \u2022 For straight line motion, positions can be shown using a number line. \u2022 The separation between original and \ufb01nal position is called displacement. Practice The following url is for a discussion of the difference between distance and displacement", ". http://www.tutorvista.com/content/physics/physics-i/motion/distance-and-displacement.php Use this resource to answer the questions that follow. MEDIA Click image to the left for more content. 1. What is the vector equivalent of the scalar \u201cdistance\u201d? 2. What is the vector equivalent of the scalar \u201cspeed\u201d? Review 1. Explain the difference between distance and displacement in your own words. 2. Suppose that John lives on a square block that is 180 yards per side, and in the evenings, he walks with his dog around the block after dinner for a little exercise. (a) If John walks once around the block, what distance does he travel? (b) If John walks around the block, what is his displacement at the end? 3. Joanna\u2019s house is 8000 feet due west of her school. If her house is assigned the position of zero and her school is assigned the possition of +8000, what would Joanna\u2019s position be if she walked 100 feet west of her house? \u2022 distance: The space between two objects but this is not adequate when considering the distance travelled. The distance travelled cannot be negative and can never get smaller \u2013in this sense, distance is the total length of path traversed by the moving body irrespective of direction. \u2022 displacement: The vector from the initial position to a subsequent position assumed by a body. References 1. Image copyright Action Sports Photography, 2013. http://www.shutterstock.com. Used under license from Shutterstock.com 5 CHAPTER 3 \u2022 Distinguish between velocity and speed. \u2022 Represent velocity with vector arrows. \u2022 Describe objects that have different velocities. \u2022 Show how to calculate average velocity when direction is constant. www.ck12.org Velocity Ramey and her mom were driving down this highway at 45 miles per hour, which is the speed limit on this road. As they approached this sign, Ramey\u2019s mom put on the brakes and started to slow down so she could safely maneuver the upcoming curves in the road. This speed limit sign actually represents two components of motion: speed and direction. Speed and Direction Speed tells you only how fast or slow an object is moving. It doesn\u2019t tell you the direction the object is moving. The measure of both speed and direction is called velocity. Velocity is a vector. A vector is measurement that includes both size and direction. Vectors are often represented by arrows. When using an arrow to represent velocity, the length of the arrow stands for speed, and the way the arrow points indicates the direction. If you\u2019re still not sure of the difference between speed and velocity, watch the cartoon at this URL: http://www.youtube.com/watch?v=mDcaeO0WxBI MEDIA Click image to the left for more content. 6 www.ck12.org Chapter 3. Velocity Using Vector Arrows to Represent Velocity The arrows in the Figure 3.1 represent the velocity of three different objects. Arrows A and B are the same length but point in different directions. They represent objects moving at the same speed but in different directions. Arrow C is shorter than arrow A or B but points in the same direction as arrow A. It represents an object moving at a slower speed than A or B but in the same direction as A. FIGURE 3.1 Differences in Velocity Objects have the same velocity only if they are moving at the same speed and in the same direction. Objects moving at different speeds, in different directions, or both have different velocities. Look again at arrows A and B from the Figure 3.1. They represent objects that have different velocities only because they are moving in different directions. A and C represent objects that have different velocities only because they are moving at different speeds. Objects represented by B and C have different velocities because they are moving in different directions and at different speeds. Q: Jerod is riding his bike at a constant speed. As he rides down his street he is moving from east to west. At the end of the block, he turns right and starts moving from south to north, but he\u2019s still traveling at the same speed. Has his velocity changed? A: Although Jerod\u2019s speed hasn\u2019t changed, his velocity has changed because he is moving in a different direction. Q: How could you use vector arrows to represent Jerod\u2019s velocity and how it changes? A: The arrows might look like this (see Figure 3.2): FIGURE 3.2 7 Calculating Average Velocity You can calculate the average velocity of a moving object that is not changing direction by dividing the distance the object travels by the time it takes to travel that distance. You would use this formula: www.ck12.org velocity = distance time This is the same formula that is used for calculating average speed. It represents", " velocity only if the answer also includes the direction that the object is traveling. Let\u2019s work through a sample problem. Toni\u2019s dog is racing down the sidewalk toward the east. The dog travels 36 meters in 18 seconds before it stops running. The velocity of the dog is: velocity = distance time 36 m 18 s = 2 m/s east = Note that the answer is given in the SI unit for velocity, which is m/s, and it includes the direction that the dog is traveling. Q: What would the dog\u2019s velocity be if it ran the same distance in the opposite direction but covered the distance in 24 seconds? A: In this case, the velocity would be: velocity = = distance time 36 m 24 s = 1:5 m/s west Summary \u2022 Velocity is a measure of both speed and direction of motion. Velocity is a vector, which is a measurement that includes both size and direction. \u2022 Velocity can be represented by an arrow, with the length of the arrow representing speed and the way the arrow points representing direction. \u2022 Objects have the same velocity only if they are moving at the same speed and in the same direction. Objects moving at different speeds, in different directions, or both have different velocities. \u2022 The average velocity of an object moving in a constant direction is calculated with the formula: velocity = distance time. The SI unit for velocity is m/s, plus the direction the object is traveling. Vocabulary \u2022 vector: Measure such as velocity that includes both size and direction; may be represented by an arrow. \u2022 velocity: Measure of both speed and direction of motion. 8 www.ck12.org Practice Chapter 3. Velocity At the following URL, review how to calculate speed and velocity, and work through the sample problems. Then solve the 10 practice problems. http://www2.franciscan.edu/academic/mathsci/mathscienceintegation/MathScienc eIntegation-827.htm Review 1. What is velocity? 2. How does velocity differ from speed? Why is velocity a vector? 3. Explain how an arrow can be used to represent velocity. 4. Use vector arrows to represent the velocity of a car that travels north at 50 mi/h and then travels east at 25 mi/h. 5. Another car travels northwest for 2 hours and covers a distance of 90 miles. What is the average velocity of the car? References 1. Christopher Auyeung (CK-12 Foundation); Compass: Seamus McGill.. CC BY-NC 3.0; Compass: Public Domain 2... CC BY-NC 3.0 9 www.ck12.org Average Velocity CHAPTER 4 \u2022 Explain the difference between speed and velocity. \u2022 De\ufb01ne the concept of average velocity. \u2022 Given displacement and time, calculate average velocity. \u2022 Solve for any variable in the equation Vave = Dx Dt Test Pilot Neil Armstrong (later to become a famous astronaut) is seen here next to the X-15 ship after a research \ufb02ight. Armstrong made his \ufb01rst X-15 \ufb02ight on November 30, 1960. This was the \ufb01rst X-15 \ufb02ight to use the ball nose, which provided accurate measurement of air speed at hypersonic speeds. The servo-actuated ball nose can be seen in this photo in front of Armstrong\u2019s right hand. The X-15 employed a non-standard landing gear. It had a nose gear with a wheel and tire, but the main landing consisted of skids mounted at the rear of the vehicle. In the photo, the left skid is visible, as are marks on the lakebed from both skids. Because of the skids, the rocket-powered aircraft could only land on a dry lakebed, not on a concrete runway. The X-15 weighed about 14,000 lb empty and approximately 34,000 lb at launch. The X-15 was \ufb02own over a period of nearly 10 years \u2013 June 1959 to Oct. 1968 \u2013 and set the world\u2019s unof\ufb01cial speed record of 4,520 mph. Average Velocity The terms speed and velocity are used interchangeably in ordinary language. In physics, however, we make a distinction between the two. Essentially both words refer to how fast an object is moving. Speed is the number we read off the speedometer of a car. It indicates how fast the car is moving at any instant but given no indication of the direction it is moving. For a particular time interval, average speed would be calculated by dividing the distance travelled by the time interval of travel. Velocity, in physics, is a vector, meaning that it must have a direction as well as a magnitude. Furthermore, the average velocity is de\ufb01ned in terms of displacement rather than distance. Average 10 www.ck12.org Chapter 4. Average Velocity velocity would be calculated by dividing", " the displacement by the time interval where displacement is the change in position of the object. To show the distinction, we could calculate the average speed and the average velocity of a person who walks 50 m to the east, then turns around and walks 50 m to the west. The total time interval is 20 seconds. The distance traveled in this trip is 100 m but the displacement is zero. The average speed would be calculated by dividing 100 m by 20 s with a result of 5 m/s. The average velocity, on the other hand, would be calculated by dividing 0 m by 20 s giving a result of 0 m/s. Neither average speed nor average velocity implies a constant rate of motion. That is to say, an object might travel at 10 m/s for 10 s and then travel at 20 m/s for 5 s and then travel at 100 m/s for 5s. This motion would cover a distance of 700 m in 20 s and the average speed would be 35 m/s. We would report the average speed during this 20 s interval to be 35 m/s and yet at no time during the interval was the speed necessarily 35 m/s. The concept of constant velocity is very different from average velocity. If an object traveled at 35 m/s for 20 s, it would travel the same distance in the same time as the previous example but in the second case, the speed of the object would always be 35 m/s. Example: The position of a runner as a function of time is plotted as moving along the x-axis of a coordinate system. During a 3.00 s time interval, the runner\u2019s position changes from x1 = 50:0 m to x2 = 30:5 m. What was the runner\u2019s average velocity. Solution: Displacement = 30:5 m 50:0 m = 19:5 m (the object was traveling back toward zero) Dt = 3:00 s vave = Dx Dt = 19:5 m 3:00 s = 6:50 m/s Summary \u2022 Average speed is distance divided by time. \u2022 Average velocity is displacement divided by time. Practice The url below is a physics classroom discussion of speed versus velocity with a short animation. http://www.physicsclassroom.com/Class/1DKin/U1L1d.cfm Use this resource to answer the questions that follow. http://www.youtube.com/watch?v=BWP1tN7PZps MEDIA Click image to the left for more content. 1. The velocity versus time graph in the video is divided into six sections. In how many of these sections is the velocity constant? 2. In how many sections of the graph is the velocity zero? 3. What does the area under the curve of a velocity versus time graph represent? 11 www.ck12.org Review 1. On a one day vacation, Jane traveled 340 miles in 8.0 hours. What was her average speed? 2. An object on a number line moved from x = 12 m to x = 124 m and moved back to x = 98 m. The time interval for all the motion was 10. s. What was the average velocity of the object? 3. An object on a number line moved from x = 15 cm to x = 165 cm and then moved back to x = 25 cm all in a time of 100 seconds. (a) What was the average velocity of the object? (b) What was the average speed of the object? \u2022 speed: Distance travelled per unit time. \u2022 velocity: A vector measurement of the rate and direction of motion or, in other terms, the rate and direction of the change in the position of an object. References 1. Courtesy of NASA. http://commons.wikimedia.org/wiki/File:Pilot_Neil_Armstrong_and_X-15.jpg. Public Domain 12 www.ck12.org Chapter 5. Instantaneous Velocity CHAPTER 5 Instantaneous Velocity \u2022 De\ufb01ne instantaneous velocity. \u2022 Plot and interpret position vs time graphs. \u2022 Determine the slope of a curve on a position vs time graph. In a footrace such as the one shown here, the initial velocity of a runner is zero. The runner increases his velocity out of the starting blocks and his velocity continues to increase as the race proceeds. For the well-trained athlete, his highest velocity is maintained through the \ufb01nish line. Instantaneous Velocity The instantaneous velocity of an object is the precise velocity at a given moment. It is a somewhat dif\ufb01cult quantity to determine unless the object is moving with constant velocity. If the object is moving with constant velocity, then the instantaneous velocity at every instant, the average velocity, and the constant velocity are all exactly the same. Position vs Time Graphs Consider a position versus time graph for an object starting at t = 0 and x = 0 that has a constant velocity of 80. m", "/s. 13 www.ck12.org The velocity of an object can be found from a position vs time graph. On a position vs time graph, the displacement The ratio of is the vertical separation between two points and the time interval is the horizontal separation. displacement to time interval is the average velocity. The ratio of the vertical separation to the horizontal separation is also the slope of the line. Therefore, the slope of straight line is the average velocity. The slope at any given time is the instantaneous velocity. For the motion pictured above, slope = rise run = Dd Dt = 400: m 5:0 s = 80: m/s For accelerated motion (the velocity is constantly changing), the position vs time graph will be a curved line. The slope of the curved line at any point will be the instantaneous velocity at that time. If we were using calculus, the slope of a curved line could be calculated. Since we are not using calculus, we can only approximate the slope of curved line by laying a straight edge along the curved line and guessing at the slope. 14 www.ck12.org Chapter 5. Instantaneous Velocity In the image above, the red line is the position vs time graph and the blue line is an approximated slope for the line at t = 2:5 seconds. The rise for this slope is approximately 170 m and the time interval (run) is 4.0 seconds. Therefore, the approximated slope is 43 m/s. Summary \u2022 The slope of a position versus time graph is the velocity. For a constant velocity motion, the slope gives the constant velocity, the average velocity, and the instantaneous velocity at every point. For constant acceleration motion, the slope of the position versus time curve gives only the instantaneous velocity at that point. Practice Draw a velocity versus time graph for an object whose constant velocity is 15 m/s and whose position starts at x = 0 when t = 0. Graph the motion for the \ufb01rst 5.0 seconds. Use this resource to answer the questions that follow. http://www.youtube.com/watch?v=sujsb5ZlM8o MEDIA Click image to the left for more content. 1. In the graph on the video, what is graphed on the vertical axis? 2. What is graphed on the horizontal axis. 3. What does the slope of this graph represent? Review 1. For the motion graphed in the position versus time graph shown above, what is the average velocity in the time interval 1 to 3 seconds? 2. For the motion graphed in the position versus time graph shown above, what is the average velocity in the time interval 3 to 4 seconds? 3. For the motion graphed in the position versus time graph shown above, what is the average velocity in the time interval 5 to 6 seconds? 15 \u2022 instantaneous velocity: The velocity of an object at any given instant. www.ck12.org References 1. Image copyright Denis Kuvaev, 2013. http://www.shutterstock.com. Used under license from Shutterstock.com 2. CK-12 Foundation - Richard Parsons.. CC-BY-NC-SA 3.0 3. CK-12 Foundation - Richard Parsons.. CC-BY-NC-SA 3.0 4. CK-12 Foundation - Richard Parsons.. CC-BY-NC-SA 3.0 16 www.ck12.org Chapter 6. Average Acceleration CHAPTER 6 Average Acceleration \u2022 De\ufb01ne average acceleration. \u2022 Given initial velocity, \ufb01nal velocity and time, calculate acceleration. \u2022 Given three of initial velocity, acceleration, time, and \ufb01nal velocity, calculate the fourth. End of an era. The Space Shuttle Atlantis blasts off on mission STS-125, the \ufb01nal mission to service and upgrade the Hubble Space Telescope, one of NASA\u2019s greatest legacies and triumphs. Canceled in the wake of the Columbia tragedy and then reinstated, the only mission not to go to the international space station post-accident will see seven astronauts undertake one of the most ambitious shuttle missions in history, with \ufb01ve spacewalks to install new and replace old components on Hubble. It will be the closing chapter in one of the original purposes of the shuttle. Average Acceleration An object whose velocity is changing is said to be accelerating. Average acceleration, a is de\ufb01ned as the rate of change of velocity, or the change in velocity per unit time. A symbol with a bar over it is read as average \u2013so a-bar is average acceleration. Example: A car accelerates along a straight road from rest to 60. km/h in 5.0 s. What is the magnitude of its average acceleration? Solution: This is read as kilometers per hour per second. In general, it is undesirable to have to different units for the same quantity in a unit expression. For example, in this case, it is undesirable", " to have two different units for time (hours and seconds) in the same unit expression. To eliminate this problem, we would convert the hour units to seconds. If we converted the original 60. km/h to m/s, it would be 17 m/s. Then the acceleration would be 3.4 m/s2. Example: An automobile is moving along a straight highway in the positive direction and the driver puts on If the initial velocity is 15.0 m/s and 5.0 s is required to slow down to 5.0 m/s, what was the car\u2019s the brakes. acceleration? 17 www.ck12.org Solution: a = Dv Dt = 10: m/s 5:0 s = 2:0 m/s/s Summary \u2022 Average acceleration is the rate of change of velocity, or the change in velocity per unit time. Practice The following url has a lesson on the difference between average and instantaneous acceleration and practice calculating average acceleration. http://www.brighthubeducation.com/homework-math-help/102434-de\ufb01nition-and-how-to-calculate-acceleration/ Review 1. The velocity of a car increases from 2.0 m/s to 16.0 m/s in a time period of 3.5 s. What was the average acceleration? 2. If an automobile slows from 26 m/s to 18 m/s in a period of 4.0 s, what was the average acceleration? 3. If a runner increases his velocity from 0 m/s to 20 m/s in 2.0 s, what was his average acceleration? 4. If a runner decreases his velocity from 20 m/s to 10 m/s in 2.0 s, what was his average acceleration? \u2022 average acceleration: The change in velocity over the change in time. References 1. Courtesy of NASA. http://space\ufb02ight.nasa.gov/gallery/images/shuttle/sts-125/html/sts125-s-025.html. Public Domain 18 www.ck12.org Chapter 7. Uniform Acceleration CHAPTER 7 Uniform Acceleration \u2022 De\ufb01ne uniform acceleration. \u2022 Given initial velocity, acceleration, and time, calculate \ufb01nal velocity. Wingtip vortices are often thought to be a type of contrail but are actually produced from a different process. During very speci\ufb01c weather conditions you may see vapor trails form at the rear of the wingtips of jet aircraft on takeoff or landing. This phenomenon occurs due to a decrease in pressure and temperature as the wing generates lift. The image is an F-35 departing from Elgin Air Force Base in Florida. Uniform Acceleration Acceleration that does not change in time is called uniform or constant acceleration. The velocity at the beginning of the time interval is called initial velocity, vi, and the velocity at the end of the time interval is called \ufb01nal velocity, v f. In a velocity versus time graph for uniform acceleration, the slope of the line is the acceleration. The equation that describes the curve is v f = vi + at. Example: velocity? If an automobile with a velocity of 4.0 m/s accelerates at a rate of 4.0 m/s2 for 2.5 s, what is the \ufb01nal Solution: v f = vi + at = 4:0 m/s + (4:0 m/s2)(2:5 s) = 4:0 m/s + 10: m/s = 14 m/s Example: If a cart slows from 22.0 m/s to 4.0 m/s with an acceleration of -2.0 m/s2, how long does it require? Solution: t = v f vi a = 18 m/s 2:0 m/s2 = 9:0 s 19 www.ck12.org Summary \u2022 Acceleration that does not change in time is uniform or constant acceleration. \u2022 The equation relating initial velocity, \ufb01nal velocity, time, and acceleration is v f = vi + at. Practice The following url has instruction in one dimensional uniformly accelerated motion and it also has a series of practice problems. http://dallaswinwin.com/Motion_in_One_Dimension/uniform_accelerated_motion.htm Review 1. If an object has zero acceleration, does that mean it has zero velocity? Give an example. 2. If an object has zero velocity, does that mean it has zero acceleration? Give an example. 3. If the acceleration of a motorboat is 4.0 m/s2, and the motorboat starts from rest, what is its velocity after 6.0 s? 4. The friction of the water on a boat produces an acceleration of -10. m/s2. If the boat is traveling at 30. m", "/s and the motor is shut off, how long it take the boat to slow down to 5.0 m/s? \u2022 uniform acceleration: Acceleration that does not change in time is uniform or constant acceleration. References 1. Courtesy of Senior Airman Julianne Showalter/U.S. Air Force. http://www.af.mil/news/story.asp?id=1232 61835. Public Domain 20 www.ck12.org Chapter 8. Displacement During Constant Acceleration CHAPTER 8 Displacement During Constant Acceleration \u2022 Plot and interpret a velocity vs time graph. \u2022 Find the area under a curve on a velocity vs time graph and calculate the displacement from such a graph. \u2022 Calculate the displacement of an object undergoing uniform acceleration when given two of the three quanti- ties acceleration, time, velocity. Long distance runners try to maintain constant velocity with very little acceleration or deceleration because acceleration requires more energy than simply maintaining velocity. Displacement During Constant Acceleration When the acceleration is constant, there are three equations that relate displacement to two of the other three It is absolutely vital that you do NOT quantities we use to describe motion \u2013time, velocity, and acceleration. try to use these equations when the acceleration is NOT constant. Fortunately, there are quite a few cases of motion where the acceleration is constant. One of the most common, if we ignore air resistance, are objects falling due to gravity. When an object is moving with constant velocity, the displacement can be found by multiplying the velocity by the time interval. d = vt If the object is moving with constant acceleration, the velocity in that equation is replaced with the average velocity. The average velocity for a uniformly accelerated object can be found by adding the beginning and \ufb01nal velocities and dividing by 2. vave = 1 2 (v f + vi) The distance, then, for uniformly accelerating motion can be found by multiplying the average velocity by the time. 21 d = 1 2 (v f + vi)(t) (Equation 1) We know that the \ufb01nal velocity for constantly accelerated motion can be found by multiplying the acceleration times time and adding the result to the initial velocity, v f = vi + at. The second equation that relates, displacement, time, initial velocity, and \ufb01nal velocity is generated by substituting into equation 1. www.ck12.org d = 1 2 (v f + vi)(t vit but v f = vi + at and substituting for v f yields d = 1 2 vit + d = vit + 1 2 1 2 at2 (t)(vi + at) = 1 2 vit + 1 2 vit + at2 1 2 (Equation 2) The third equation is formed by combining v f = vi + at and d = 1 and then substitute into the second equation, we get 2 (v f + vi)(t). If we solve the \ufb01rst equation for t d = 1 2 (v f + vi) v f vi a = 1 2 v2 f v2 i a And solving for v f 2 yields v f 2 = vi 2 + 2ad (Equation 3) Keep in mind that these three equations are only valid when acceleration is constant. velocity can be set to zero and that simpli\ufb01es the three equations considerably. In many cases, the initial With both constant acceleration and initial velocity of zero = 2ad: d = at2 and v f Example: Suppose a planner is designing an airport for small airplanes. Such planes must reach a speed of 56 m/s before takeoff and can accelerate at 12.0 m/s2. What is the minimum length for the runway of this airport? Solution: The acceleration in this problem is constant and the initial velocity of the airplane is zero, therefore, we can use the equation v f 2a = (56 m/s)2 d = v f (2)(12:0 m/s2 2 = 2ad and solve for d. = 130 m ) 2 Example: How long does it take a car to travel 30.0 m if it accelerates from rest at a rate of 2.00 m/s2? Solution: The acceleration in this problem is constant and the initial velocity is zero, therefore, we can use d = 1 solved for t. r s 2 at2 t = 2d a = (2)(30:0 m) 2:00 m/s2 = 5:48 s Example: A baseball pitcher throws a fastball with a speed of 30.0 m/s. Assuming the acceleration is uniform and the distance through which the ball is accelerated is 3.50 m. What is the acceleration? 22 www.ck12.org Chapter 8. Displacement During Constant Acceleration Solution: Since the acceleration is uniform and the initial velocity is zero, we can use v f 2 = 2ad solved for a. a = v2 f 2d = (", "30:0 m/s)2 (2)(3:50 m) = 900: m2=s2 7:00 m = 129 m/s2 Suppose we plot the velocity versus time graph for an object undergoing uniform acceleration. In this \ufb01rst case, we will assume the object started from rest. If the object has a uniform acceleration of 6.0 m/s2 and started from rest, then each succeeding second, the velocity will increase by 6.0 m/s. Here is the table of values and the graph. In displacement versus time graphs, the slope of the line is the velocity of the object and in this case of velocity versus time graph, the slope of the line is the acceleration. If you take any segment of this line and determine the Dy to Dx ratio, you will get 6.0 m/s2 which we know to be the constant acceleration of this object. The area of a triangle is calculated by multiplying one-half the base times the height. The area under the curve in the image above is the area of the triangle shown below. 23 www.ck12.org The area of this triangle would be calculated by area = 1 2 The distance traveled by an object accelerating uniformly from rest at 6.0 m/s2 would be displacement = 1 (6:0 s)(36 m/s) = 108 m. 2 at2. Therefore, the displacement of this object in the \ufb01rst 6 seconds of travel would be displacement = 1 2 (6:0 m/s2)(6:0 s)2 = 108 m. In fact, the area underneath the curve in a velocity versus time graph is always equal to the displacement that occurs during that time interval. Summary \u2022 There are three equations relating displacement to two of the other three quantities we use to describe motion \u2013time, velocity, and acceleration: \u2013 d = 1 2 \u2013 d = vit + 1 2 = vi \u2013 v f (v f + vi)(t) (Equation 1) 2 at2 (Equation 2) 2 + 2ad (Equation 3) \u2022 When the initial velocity of the object is zero, these three equations become: (v f )(t) (Equation 1\u2019) \u2013 d = 1 2 \u2013 d = 1 2 at2 (Equation 2\u2019) 2 = 2ad (Equation 3\u2019) \u2013 v f \u2022 The slope of a velocity versus time graph is the acceleration of the object. \u2022 The area under the curve of a velocity versus time graph is the displacement that occurs during the given time interval. Practice Use this resource to answer the questions that follow. MEDIA Click image to the left for more content. 1. For the example in the video, what acceleration is used? 2. What time period is used in the example? 3. What does the slope of the line in the graph represent? 4. What does the area under the curve of the line represent? Review 1. An airplane accelerates with a constant rate of 3.0 m/s2 starting at a velocity of 21 m/s. If the distance traveled during this acceleration was 535 m, what is the \ufb01nal velocity? 2. An car is brought to rest in a distance of 484 m using a constant acceleration of -8.0 m/s2. What was the velocity of the car when the acceleration \ufb01rst began? 3. An airplane starts from rest and accelerates at a constant 3.00 m/s2 for 20.0 s. What is its displacement in this time? 4. A driver brings a car to a full stop in 2.0 s. 24 www.ck12.org Chapter 8. Displacement During Constant Acceleration (a) If the car was initially traveling at 22 m/s, what was the acceleration? (b) How far did the car travel during braking? References 1. Image copyright Maridav, 2013. http://www.shutterstock.com. Used under license from Shutterstock.com 2. CK-12 Foundation - Richard Parsons.. CC-BY-NC-SA 3.0 3. CK-12 Foundation - Richard Parsons.. CC-BY-NC-SA 3.0 25 CHAPTER 9 Acceleration Due to Gravity www.ck12.org \u2022 Solve problems of the motion of objects uniformly accelerated by gravity. In the absence of air resistance, all objects fall toward the earth with the same acceleration. Man, however, make maximum use of air resistance in the construction of parachutes for both entertainment and military use. The image at left was taken during a 2008 Graduation demonstration jump by the U.S. Army Parachute Team. The 2008 team contained the \ufb01rst amputee member and the largest number of females in history. Acceleration Due to Gravity One of the most common examples of uniformly accelerated motion is that an object allowed to fall vertically to the earth. In treating falling objects", " as uniformly accelerated motion, we must ignore air resistance. Galileo\u2019s original statement about the motion of falling objects is: At a given location on the earth and in the absence of air resistance, all objects fall with the same uniform acceleration. We call this acceleration due to gravity on the earth and we give it the symbol g. The value of g is 9.80 m/s2. All of the equations involving constant acceleration can be used for falling bodies but we insert g wherever \u201ca\u201d appeared and the value of g is always 9.80 m/s2. Example: A rock is dropped from a tower 70.0 m high. How far will the rock have fallen after 1.00 s, 2.00 s, and 3.00 s? Assume the distance is positive downward. 26 www.ck12.org Chapter 9. Acceleration Due to Gravity Solution: We are looking for displacement and we have time and acceleration. Therefore, we can use d = 1 Displacement after 1:00 s = 1 2 Displacement after 2:00 s = 1 2 Displacement after 3:00 s = 1 2 (9:80 m/s2)(1:00 s)2 = 4:90 m (9:80 m/s2)(2:00 s)2 = 19:6 m (9:80 m/s2)(3:00 s)2 = 44:1 m 2 at2. Example: (a) A person throws a ball upward into the air with an initial velocity of 15.0 m/s. How high will it go before it comes to rest? (b) How long will the ball be in the air before it returns to the person\u2019s hand? Solution: In part (a), we know the initial velocity (15.0 m/s), the \ufb01nal velocity (0 m/s), and the acceleration -9.80 m/s2. We wish to solve for the displacement, so we can use v f d = v f 2 + 2ad and solve for d. 2a = (0 m/s)2(15:0 m/s)2 2vi = 11:5 m 2 = vi 2 (2)(9:80 m/s2 ) There are a number of methods by which we can solve part (b). Probably the easiest is to divide the distance traveled by the average velocity to get the time going up and then double this number since the motion is symmetrical \u2013that is, time going up equals the time going down. The average velocity is half of 15.0 m/s or 7.5 m/s and dividing this into the distance of 11.5 m yields 1.53 seconds. This is the time required for the ball to go up and the time for the ball to come down will also be 1.53 s, so the total time for the trip up and down is 3.06 seconds. Example: A car accelerates with uniform acceleration from 11.1 m/s to 22.2 m/s in 5.0 s. acceleration and (b) how far did it travel during the acceleration? (a) What was the Dt = 22:2 m/s11:1 m/s Solution: (a) a = Dv (b) We can \ufb01nd the distance traveled by d = vit + 1 average velocity and multiply it by the time. = 2:22 m/s2 5:0 s 2 at2 and we can also \ufb01nd the distance traveled by determining the d = vit + at2 1 2 = (11:1 m/s)(5:0 s) + = 55:5 m + 27:8 m = 83 m 1 2 (2:22 m/s2)(5:0 s)2 d = (vave)(t) = (16:6 m/s)(5:0 s) = 83 m Example: A stone is dropped from the top of a cliff. It is seen to hit the ground after 5.5 s. How high is the cliff? Solution: d = vit + 1 2 at2 = (0 m/s)(5:5 s) + 1 2 (9:80 m/s2)(5:5 s)2 = 150 m Summary \u2022 At a given location on the earth and in the absence of air resistance, all objects fall with the same uniform acceleration. \u2022 We call this acceleration the acceleration due to gravity on the earth and we give it the symbol g. \u2022 The value of g is 9.80 m/s2. 27 Practice This url shows a video of a discussion and demonstration of the acceleration due to gravity. http://www.youtube.com/watch?v=izXGpivLvgY www.ck12.org MEDIA Click image to the left for more content. Review 1. A baseball is thrown vertically into the air with a", " speed of 24.7 m/s. (a) How high does it go? (b) How long does the round trip up and down require? 2. A salmon jumps up a waterfall 2.4 m high. With what minimum speed did the salmon leave the water below to reach the top? 3. A kangaroo jumps to a vertical height of 2.8 m. How long will it be in the air before returning to earth? \u2022 acceleration due to gravity: The acceleration experienced by a body in free fall in a gravitational \ufb01eld. References 1. Courtesy of Donna Dixon/U.S. Military. http://commons.wikimedia.org/wiki/File:Flickr_-_The_U.S._Arm y_-_U.S._Army_Parachute_Team_graduates_first_wounded_warrior_and_largest_female_class_%282%29.jp g. Public Domain 28 www.ck12.org Chapter 10. Graphing Motion CHAPTER 10 Graphing Motion Students will learn how to graph motion vs time. Speci\ufb01cally students will learn how to take the slope of a graph and relate that to the instantaneous velocity or acceleration for position or velocity graphs, respectively. Finally students will learn how to take the area of a velocity vs time graph in order to calculate the displacement. Students will learn how to graph motion vs time. Speci\ufb01cally students will learn how to take the slope of a graph and relate that to the instantaneous velocity or acceleration for position or velocity graphs, respectively. Finally students will learn how to take the area of a velocity vs time graph in order to calculate the displacement. Key Equations For a graph of position vs. time. The slope is the rise over the run, where the rise is the displacement and the run is the time. thus, Slope = vavg = Dx Dt Note: Slope of the tangent line for a particular point in time = the instantaneous velocity For a graph of velocity vs. time. The slope is the rise over the run, where the rise is the change in velocity and the run is the time. thus, Slope = aavg = Dv Dt Note: Slope of the tangent line for a particular point in time = the instantaneous acceleration Guidance \u2022 One must \ufb01rst read a graph correctly. For example on a position vs. time graph (thus the position is graphed on the y-axis and the time on the x-axis) for a given a data point, go straight down from it to get the time and straight across to get the position. \u2022 If there is constant acceleration the graph x vs. t produces a parabola. The slope of the x vs. t graph equals the instantaneous velocity. The slope of a v vs. t graph equals the acceleration. \u2022 The slope of the graph v vs. t can be used to \ufb01nd acceleration; the area of the graph v vs. t can be used to \ufb01nd displacement. Welcome to calculus! What is a Graph Watch this Explanation MEDIA Click image to the left for more content. 29 www.ck12.org MEDIA Click image to the left for more content. MEDIA Click image to the left for more content. MEDIA Click image to the left for more content. Time for Practice 1. The position graph below is of the movement of a fast turtle who can turn on a dime. a. Sketch the velocity vs. time graph of the turtle below. 30 www.ck12.org Chapter 10. Graphing Motion b. Explain what the turtle is doing (including both speed and direction) from: i) 0-2s. ii) 2-3s. iii) 3-4s. c. How much distance has the turtle covered after 4s? d. What is the turtle\u2019s displacement after 4s? 2. Draw the position vs. time graph that corresponds to the velocity vs. time graph below. You may assume a starting position x0 = 0. Label the yaxis with appropriate values. 3. The following velocity-time graph represents 10 seconds of actress Halle Berry\u2019s drive to work (it\u2019s a rough morning). 31 a. Fill in the tables below \u2013remember that displacement and position are not the same thing! www.ck12.org 32 www.ck12.org Chapter 10. Graphing Motion TABLE 10.1: Displacement (m) Acceleration(m=s2) Instantaneous Time (s) 0 sec Position (m) 0 m 2 sec 4 sec 5 sec 9 sec 10 sec Interval (s) 0-2 sec 2-4 sec 4-5 sec 5-9 sec 9-10 sec b. On the axes below, draw an acceleration-time graph for the car trip. Include numbers on your acceleration axis. c. On the", " axes below, draw a position-time graph for the car trip. Include numbers on your position axis. Be sure to note that some sections of this graph are linear and some curve \u2013why? 4. Two cars are drag racing down El Camino. At time t = 0, the yellow Maserati starts from rest and accelerates at 10 m=s2. As it starts to move it\u2019s passed by a \u201963 Chevy Nova (cherry red) traveling at a constant velocity of 30 m/s. a. On the axes below, show a line for each car representing its speed as a function of time. Label each line. 33 www.ck12.org b. At what time will the two cars have the same speed (use your graph)? (or curve) for each car representing its position as a function of time. Label each curve. c. On the axes below, draw a line d. At what time would the two cars meet (other than at the start)? Answers: 1c. 25 m 1d. -5 m 2. discuss in class 3. discuss in class 4b. 3 sec 4d. 6 sec 34 Physics Unit 3 (Vectors) Patrick Marshall Ck12 Science Say Thanks to the Authors Click http://www.ck12.org/saythanks (No sign in required) www.ck12.org To access a customizable version of this book, as well as other interactive content, visit www.ck12.org AUTHORS Patrick Marshall Ck12 Science CK-12 Foundation is a non-pro\ufb01t organization with a mission to reduce the cost of textbook materials for the K-12 market both in the U.S. and worldwide. Using an open-content, web-based collaborative model termed the FlexBook\u00ae, CK-12 intends to pioneer the generation and distribution of high-quality educational content that will serve both as core text as well as provide an adaptive environment for learning, powered through the FlexBook Platform\u00ae. Copyright \u00a9 2014 CK-12 Foundation, www.ck12.org The names \u201cCK-12\u201d and \u201cCK12\u201d and associated logos and the terms \u201cFlexBook\u00ae\u201d and \u201cFlexBook Platform\u00ae\u201d (collectively \u201cCK-12 Marks\u201d) are trademarks and service marks of CK-12 Foundation and are protected by federal, state, and international laws. Any form of reproduction of this book in any format or medium, in whole or in sections must include the referral attribution link http://www.ck12.org/saythanks (placed in a visible location) in addition to the following terms. Except as otherwise noted, all CK-12 Content (including CK-12 Curriculum Material) is made available to Users in accordance with the Creative Commons Attribution-Non-Commercial 3.0 Unported (CC BY-NC 3.0) License (http://creativecommons.org/ licenses/by-nc/3.0/), as amended and updated by Creative Commons from time to time (the \u201cCC License\u201d), which is incorporated herein by this reference. Complete terms can be found at http://www.ck12.org/terms. Printed: February 4, 2014 iii Contents www.ck12.org Contents 1 Graphical Methods of Vector Addition 2 Vector Addition 1 5 iv www.ck12.org Chapter 1. Graphical Methods of Vector Addition CHAPTER 1 Graphical Methods of Vector Addition \u2022 Differentiate between scalars and vectors. \u2022 Graphically add vectors in one dimension by placing the vectors head to toe on a number line. \u2022 De\ufb01ne resultant. \u2022 Graphically add vectors in two dimensions by placing them head to toe on a two-dimensional coordinate system. Successfully shooting a basketball requires a subconscious understanding of the vectors involved in how the basketball moves through the air. The vertical and horizontal vectors must be perfectly organized if the ball is to pass through the basket. Graphical Methods Vector Addition In physics, a quantity, such as mass, length, or speed, that is completely speci\ufb01ed by its magnitude and has no direction is called a scalar. A vector, on the other hand, is a quantity possessing both magnitude and direction. A vector quantity can be represented by an arrow-tipped line segment. The length of the line, drawn to scale, represents the magnitude of the quantity. The direction of the arrow indicates the direction of the vector. Not only can vectors be represented graphically, but they can also be added graphically. For one dimensional vector addition, the \ufb01rst vector is placed on a number line with the tail of the vector on the origin. The second vector is placed with its tail exactly on the arrow head of the \ufb01rst vector. The sum of the two vectors is the vector that begins at the origin and ends at the arrow head of the \ufb01n", "al added vector. Consider the following two vectors. 1 www.ck12.org The red vector has a magnitude of 11 in the positive direction on the number line. The blue vector has a magnitude of -3 in the negative direction on the number line. In order to add these two vectors, we place one of the vectors on a number line and then the second vector is placed on the same number line such that its origin is on the arrow head of the \ufb01rst vector. The sum of these two vectors is the vector that begins at the origin of the \ufb01rst vector (the red one) and ends at the arrow head of the blue vector. So the sum of these two vectors is the purple vector as shown below. The vector sum of the \ufb01rst two vectors is a vector that begins at the origin and has a magnitude of 8 units in the positive direction. If we were adding three or four vectors all in one dimension, we would continue to place them head to toe in sequence on the number line. The sum would be the vector that begins at the beginning of the \ufb01rst vector and goes to the ending of the \ufb01nal vector. Adding Vectors in Two Dimensions In the following image, vectors A and B represent the two displacements of a person who walked 90. m east and then 50. m north. We want to add these two vectors to get the vector sum of the two movements. The graphical process for adding vectors in two dimensions is to place the tail of the second vector on the arrow head of the \ufb01rst vector as shown above. The sum of the two vectors is the vector that begins at the origin of the \ufb01rst vector and goes to the ending of the second vector, as shown below. 2 www.ck12.org Chapter 1. Graphical Methods of Vector Addition If we are using totally graphic means of adding these vectors, the magnitude of the sum would be determined by measuring the length of the sum vector and comparing it to the original standard. We would also use a compass to measure the angle of the summation vector. If we are using calculation means, we can determine the inverse tangent of 50 units divided by 90 units and get the angle of 29\u00b0 north of east. The length of the sum vector can also be determined mathematically by the Pythagorean theorem, a2 + b2 = c2. In this case, the length of the hypotenuse would be the square root of (8100 + 2500) or 103 units. If three or four vectors are to be added by graphical means, we would continue to place each new vector head to toe with the vectors to be added until all the vectors were in the coordinate system and then the sum vector would be the vector goes from the origin of the \ufb01rst vector to the arrowhead of the last vector. The magnitude and direction of the sum vector would be measured. Summary \u2022 Scalars are quantities, such as mass, length, or speed, that are completely speci\ufb01ed by magnitude and has no direction. \u2022 Vectors are quantities possessing both magnitude and direction and can be represented by an arrow; the direction of the arrow indicates the direction of the quantity and the length of the arrow is proportional to the magnitude. \u2022 Vectors that are in one dimension can be added arithmetically. \u2022 Vectors that are in two dimensions are added geometrically. \u2022 When vectors are added graphically, graphs must be done to scale and answers are only as accurate as the graphing. Practice Video on the graphical method of adding vectors. http://www.youtube.com/watch?v=_Vppxdho6JU MEDIA Click image to the left for more content. 3 Review 1. On the following number line, add the vector 7.5 m/s and the vector -2.0 m/s. www.ck12.org 2. On a sheet of graph paper, add a vector that is 4.0 N due east and a vector that is 3.0 N due north. \u2022 scalar: A quantity, such as mass, length, or speed, that is completely speci\ufb01ed by its magnitude and has no direction. \u2022 vector: A quantity possessing both magnitude and direction, represented by an arrow the direction ofwhich indicates the direction of the quantity and the length of which is proportional to the magnitude. \u2022 vector addition: The process of \ufb01nding one vector that is equivalent to the result of the successive application of two or more given vectors. References 1. Of\ufb01cial White House photo by Pete Souza. http://commons.wikimedia.org/wiki/File:Barack_Obama_play ing_basketball.jpg. Public Domain 2. CK-12 Foundation - Richard Parsons.. CC-BY-NC-SA 3.0 3. CK-12 Foundation - Richard Parsons", ".. CC-BY-NC-SA 3.0 4. CK-12 Foundation - Richard Parsons.. CC-BY-NC-SA 3.0 5. CK-12 Foundation - Richard Parsons.. CC-BY-NC-SA 3.0 6. CK-12 Foundation - CC-BY-NC-SA 3.0.. CC-BY-NC-SA 3.0 7. CK-12 Foundation - Richard Parsons.. CC-BY-NC-SA 3.0 4 www.ck12.org Chapter 2. Vector Addition CHAPTER 2 Vector Addition \u2022 Describe the independence of perpendicular vectors. \u2022 Resolve vectors into axial components. \u2022 De\ufb01ne resultant. \u2022 Add vectors using geometric and trigonometric methods. Vector Addition Adding Vectors in Two Dimensions In the following image, vectors A and B represent the two displacements of a person who walked 90. m east and then 50. m north. We want to add these two vectors to get the vector sum of the two movements. The graphical process for adding vectors in two dimensions is to place the tail of the second vector on the arrow head of the \ufb01rst vector as shown above. The sum of the two vectors is the vector that begins at the origin of the \ufb01rst vector and goes to the ending of the second vector, as shown below. If we are using totally graphic means of adding these vectors, the magnitude of the sum would be determined by measuring the length of the sum vector and comparing it to the original standard. We would also use a compass to measure the angle of the summation vector. If we are using calculation means, we can determine the inverse tangent of 50 units divided by 90 units and get the angle of 29\u00b0 north of east. The length of the sum vector can also be determined mathematically by the Pythagorean 5 www.ck12.org theorem, a2 + b2 = c2. In this case, the length of the hypotenuse would be the square root of (8100 + 2500) or 103 units. If three or four vectors are to be added by graphical means, we would continue to place each new vector head to toe with the vectors to be added until all the vectors were in the coordinate system and then the sum vector would be the vector goes from the origin of the \ufb01rst vector to the arrowhead of the last vector. The magnitude and direction of the sum vector would be measured. Mathematical Methods of Vector Addition We can add vectors mathematically using trig functions, the law of cosines, or the Pythagorean theorem. If the vectors to be added are at right angles to each other, we would assign them to the sides of a right triangle and calculate the sum as the hypotenuse of the right triangle. We would also calculate the direction of the sum vector by using an inverse sin or some other trig function. Suppose, however, that we wish to add two vectors that are not at right angles to each other. Let\u2019s consider the vectors in the following images. The two vectors we are to add is a force of 65 N at 30\u00b0 north of east and a force of 35 N at 60\u00b0 north of west. We know that vectors in the same dimension can be added by regular arithmetic. Therefore, we can resolve each of these vectors into components that lay on the axes \u2013pictured below. We can resolve each of the vectors into two components. The components are on the axes lines. The resolution of vectors reduces each vector to a component on the north-south axis and a component on the east-west axis. We can now mathematically determine the magnitude of the components and add then arithmetically because they are in the same dimension. Once we have added the components, we will once again have only two vectors that are perpendicular to each other and can be the legs of a right triangle. The east-west component of the \ufb01rst vector is (65 N)(cos 30\u00b0) = (65 N)(0.866) = 56.3 N north The north-south component of the \ufb01rst vector is (65 N)(sin 30\u00b0) = (65 N)(0.500) = 32.5 N north 6 www.ck12.org Chapter 2. Vector Addition The east-west component of the second vector is (35 N)(cos 60\u00b0) = (35 N)(0.500) = 17.5 N west The north-south component of the second vector is (35 N)(sin 60\u00b0) = (35 N)(0.866) = 30.3 N north The sum of the two east-west components is 56.3 N - 17.5 N = 38.8 N east The sum of the two north-south components is 32.5 N + 30.3 N = 62.8 N north We can now consider those two vectors to be the sides of a right", " triangle and use the Pythagorean Theorem to \ufb01nd the length of the hypotenuse and use a trig function to \ufb01nd its direction. p c = sin x = 62:8 38:82 + 62:82 = 74 N 74 so x = sin1 0:84 so x = 58 The direction of the sum vector is 74 N at 58\u00b0 north of east. Perpendicular vectors have no components in the other direction. For example, if a boat is \ufb02oating down a river due south, and you are paddling the boat due east, the eastward vector has no component in the north-south direction and therefore, has no effect on the north-south motion. If the boat is \ufb02oating down the river at 5 miles/hour south and you paddle the boat eastward at 5 miles/hour, the boat continues to \ufb02oat southward at 5 miles/hour. The eastward motion has absolutely no effect on the southward motion. Perpendicular vectors have NO effect on each other. Example Problem: A motorboat heads due east at 16 m/s across a river that \ufb02ows due north at 9.0 m/s. (a) What is the resultant velocity of the boat? (b) If the river is 135 m wide, how long does it take the boat to reach the other side? (c) When the boat reaches the other side, how far downstream will it be? Solution: Sketch: (a) Since the two motions are perpendicular to each other, they can be assigned to the legs of a right triangle and the hypotenuse (resultant) calculated. c = p a2 + b2 = q (16 m/s)2 + (9:0 m/s)2 = 18 m/s sin q = 9:0 18 = 0:500 and therefore q = 30 The resultant is 18 m/s at 30\u00b0 north of east. 7 www.ck12.org (b) The boat is traveling across the river at 16 m/s due to the motor. The current is perpendicular and therefore has no effect on the speed across the river. The time required for the trip can be determined by dividing the distance by the velocity. t = d = 8:4 s v = 135 m 16 m/s (c) The boat is traveling across the river for 8.4 seconds and therefore, it is also traveling downstream for 8.4 seconds. We can determine the distance downstream the boat will travel by multiplying the speed downstream by the time of the trip. ddownstream = (vdownstream)(t) = (9:0 m/s)(8:4 s) = 76 m Summary \u2022 Vectors can be added mathematically using geometry and trigonometry. \u2022 Vectors that are perpendicular to each other have no effect on each other. \u2022 Vector addition can be accomplished by resolving into axial components those vectors that are to be added, adding up the axial components, and then combining the two axial components. Practice A video demonstrating the component method of vector addition. http://www.youtube.com/watch?v=nFDzRWw08Ew MEDIA Click image to the left for more content. Review 1. A hiker walks 11 km due north from camp and then turns and walks 11 km due east. (a) What is the total distance walked by the hiker? (b) What is the displacement (on a straight line) of the hiker from the camp? 2. While \ufb02ying due east at 33 m/s, an airplane is also being carried due north at 12 m/s by the wind. What is the plane\u2019s resultant velocity? 3. Two students push a heavy crate across the \ufb02oor. John pushes with a force of 185 N due east and Joan pushes with a force of 165 N at 30\u00b0 north of east. What is the resultant force on the crate? 4. An airplane \ufb02ying due north at 90. km/h is being blown due west at 50. km/h. What is the resultant velocity of the plane? 5. A golf ball is struck with a golf club and travels in a parabolic curve. The horizontal distance traveled by the golf ball is 240 meters and the time of \ufb02ight is 4.00 seconds. What was the initial velocity magnitude and direction? \u2022 axial component: A component situated in or on an axis. \u2022 resolution of vectors: Any vector directed at an angle to the horizontal (or the vertical) can be thought of as having two parts (or components) that lie on the axes (one horizontal and one vertical). The process of identifying these two components is known as the resolution of the vector. 8 Chapter 2. Vector Addition www.ck12.org References 1. CK-12 Foundation - Richard Parsons.. CC-BY-NC-SA 3.0 2", ". CK-12 Foundation - Richard Parsons.. CC-BY-NC-SA 3.0 3. CK-12 Foundation - Richard Parsons.. CC-BY-NC-SA 3.0 4. CK-12 Foundation - Richard Parsons.. CC-BY-NC-SA 3.0 5. CK-12 Foundation - Richard Parsons.. CC-BY-NC-SA 3.0 9 Physics Unit 4 (Two Dimensional Motion) Patrick Marshall Jean Brainard, Ph.D. Ck12 Science James Dann, Ph.D. Say Thanks to the Authors Click http://www.ck12.org/saythanks (No sign in required) AUTHORS Patrick Marshall Jean Brainard, Ph.D. Ck12 Science James Dann, Ph.D. CONTRIBUTORS Chris Addiego Antonio De Jesus L\u00f3pez www.ck12.org To access a customizable version of this book, as well as other interactive content, visit www.ck12.org CK-12 Foundation is a non-pro\ufb01t organization with a mission to reduce the cost of textbook materials for the K-12 market both in the U.S. and worldwide. Using an open-content, web-based collaborative model termed the FlexBook\u00ae, CK-12 intends to pioneer the generation and distribution of high-quality educational content that will serve both as core text as well as provide an adaptive environment for learning, powered through the FlexBook Platform\u00ae. Copyright \u00a9 2014 CK-12 Foundation, www.ck12.org The names \u201cCK-12\u201d and \u201cCK12\u201d and associated logos and the terms \u201cFlexBook\u00ae\u201d and \u201cFlexBook Platform\u00ae\u201d (collectively \u201cCK-12 Marks\u201d) are trademarks and service marks of CK-12 Foundation and are protected by federal, state, and international laws. Any form of reproduction of this book in any format or medium, in whole or in sections must include the referral attribution link http://www.ck12.org/saythanks (placed in a visible location) in addition to the following terms. Except as otherwise noted, all CK-12 Content (including CK-12 Curriculum Material) is made available to Users in accordance with the Creative Commons Attribution-Non-Commercial 3.0 Unported (CC BY-NC 3.0) License (http://creativecommons.org/ licenses/by-nc/3.0/), as amended and updated by Creative Commons from time to time (the \u201cCC License\u201d), which is incorporated herein by this reference. Complete terms can be found at http://www.ck12.org/terms. Printed: February 4, 2014 iii Contents www.ck12.org Contents Projectile Motion Projectile Motion for an Object Launched Horizontally Projectile Motion for an Object Launched at an Angle Projectile Motion Problem Solving 1 4 7 11 1 2 3 4 iv www.ck12.org Chapter 1. Projectile Motion CHAPTER 1 Projectile Motion \u2022 Describe projectile motion and state when it occurs. \u2022 Give examples of projectile motion. The archer in the opening image is aiming his arrow a little bit above the bull\u2019s eye of the target, rather than directly at it. Why doesn\u2019t he aim at the bull\u2019s eye instead? The answer is projectile motion. Combining Forces When the archer releases the bowstring, the arrow will be \ufb02ung forward toward the top of the target where she\u2019s aiming. But another force will also act on the arrow in a different direction. The other force is gravity, and it will pull the arrow down toward Earth. The two forces combined will cause the arrow to move in the curved path shown in the Figure 1.1. This type of motion is called projectile motion. It occurs whenever an object curves down toward the ground because it has both a horizontal force and the downward force of gravity acting on it. Because of projectile motion, to hit the bull\u2019s eye of a target with an arrow, you actually have to aim for a spot above the bull\u2019s eye. You can see in theFigure 1.2 what happens if you aim at the bull\u2019s eye instead of above it. 1 www.ck12.org FIGURE 1.1 FIGURE 1.2 Another Example of Projectile Motion You can probably think of other examples of projectile motion. One is shown in the Figure 1.3. The cannon shoots a ball straight ahead, giving it horizontal motion. At the same time, gravity pulls the ball down toward the ground. FIGURE 1.3 Q: How would you show the force of gravity on the cannon ball in the Figure 1.3? A: You would add a line pointing straight down from the cannon to the ground. To get a better feel for projectile motion, try these interactive animations: \u2022 http://phet.color", "ado.edu/en/simulation/projectile-motion \u2022 http://jersey.uoregon.edu/vlab/ (Click on the applet \u201cCannon.\u201d) Summary \u2022 Projectile motion is movement of an object in a curved path toward the ground because it has both a horizontal force and the downward force of gravity acting on it. 2 www.ck12.org Chapter 1. Projectile Motion \u2022 Examples of objects that have projectile motion include arrows and cannon balls. Vocabulary \u2022 projectile motion: Motion of an object that has initial horizontal velocity but is also pulled down toward Earth by gravity. Practice Play the game at the following URL by shooting the cannon at a stationary target. Experiment with three variablespower, height of barrel, and angle of barrelyou \ufb01nd at least three different combinations of variables that allow the cannon ball to hit the target. Record the values for the three combinations of variables. Then summarize what you learned by doing the activity. http://www.science-animations.com/support-\ufb01les/projektielbeweging.swf Review 1. What is projectile motion? When does it occur? 2. How might knowledge of projectile motion help you shoot baskets in basketball? References 1. Laura Guerin.. CC BY-NC 3.0 2. Laura Guerin.. CC BY-NC 3.0 3. Christopher Auyeung.. CC-BY-NC-SA 3.0 3 CHAPTER 2 Projectile Motion for an Object Launched Horizontally www.ck12.org \u2022 State the relationship between the vertical and horizontal velocities of a projectile launched horizontally. \u2022 Find the time for a horizontally launched projectile to strike the ground. \u2022 Calculate the range of a horizontally launched projectile. The activity of bike jumping, like other sports that involve vector motions in perpendicular directions, requires more physical practice than mathematical analysis. The laws of physics apply to the activity, however, whether the biker is aware of them or not. Projectile Motion for an Object Launched Horizontally Objects that are launched into the air are called projectiles. The path followed by an object in projectile motion is called a trajectory. The motion of a projectile is described in terms of its position, velocity, and acceleration. Our knowledge that perpendicular components of vectors do not affect each other allow us to analyze the motion of projectiles. 4 www.ck12.org Chapter 2. Projectile Motion for an Object Launched Horizontally In the diagram, two balls (one red and one blue) are dropped at the same time. The red ball is released with no horizontal motion and the blue ball is dropped but also given a horizontal velocity of 10 m/s. As the balls fall to the \ufb02oor, a photograph is taken every second so that in 5 seconds, we have 5 images of the two balls. Each vertical line on the diagram represents 5 m. Since the blue ball has a horizontal velocity of 10 m/s, you will see that for every second, the blue ball has moved horizontally 10 m. That is, in each second, the blue ball has increased its horizontal distance by 10 m. This horizontal motion is constant velocity motion. The red ball was dropped straight down with no horizontal velocity and therefore, in each succeeding second, the red ball falls straight down with no horizontal motion. The succeeding distances between seconds with the red ball motion indicates this motion is accelerated. A very important point here is that, the vertical motion of these two balls is identical. That is, they each fall exactly the same distance vertically in each succeeding second. The constant horizontal velocity of the blue ball has no effect on its accelerated vertical motion! Therefore, the vertical motion of the blue ball (the projectile) can be analyzed exactly the same as the vertical motion of the red ball. Example Problem: If an arrow if \ufb01red from a bow with a perfectly horizontal velocity of 60.0 m/s and the arrow was 2.00 m above the ground when the it was released, how far will the arrow \ufb02y horizontally before it strikes the ground? Solution: This problem is solved by determining how long it takes the arrow to fall to the ground in exactly the same manner as if the arrow was dropped with no horizontal velocity. Then the time required for the fall is multiplied by the horizontal velocity to get the horizontal distance. d = 1 2 at2 solved for t = s r 2d a = (2)(2:00 m) 9:80 m/s2 = 0:639 s The time required for the arrow to fall to the ground is the same time that the arrow \ufb02ies horizontally at 60.0 m/s, so dhorizontal = (vhorizontal)(time) = (60:0 m/s)(0:639 s) = 38:3 m Example Problem: A rock was thrown horizontally from a 100.0 m high cliff. It strikes the ground 90.", "0 m from the base of the cliff. At what speed was it thrown? 5 Solution: We can calculate how long it takes for a rock to free fall 100.0 m and then divide this time into the horizontal distance to get the horizontal velocity. www.ck12.org s r 2d a = d t = 90:0 m 4:52 s t = v = (2)(100:0 m) 9:80 m/s2 = 4:52 s = 19:9 m/s Summary \u2022 Perpendicular components of vectors do not in\ufb02uence each other. \u2022 The horizontal motion of a projectile does not in\ufb02uence its free fall. Practice The following video discusses projectile motion for projectiles launched horizontally. http://www.youtube.com/watch?v=-uUsUaPJUc0 1. Why does speedy need to drive a convertible? 2. What is speedy doing in this video? 3. How does the horizontal velocity change during the fall? 4. How does the vertical velocity change during the fall? Review 1. If a bullet is \ufb01red from a high powered ri\ufb02e at the exact time a duplicate bullet is dropped by hand near the barrel of the ri\ufb02e, which bullet will hit the ground \ufb01rst? (a) the one dropped straight down (b) the one \ufb01red horizontally (c) both will hit the ground at the same time 2. A cannon is \ufb01red from the edge of a small cliff. The height of the cliff is 80.0 m. The cannon ball is \ufb01red with a perfectly horizontal velocity of 80.0 m/s. How far will the cannon ball \ufb02y horizontally before it strikes the ground? 3. A cliff diver running 3.60 m/s dives out horizontally from the edge of a vertical cliff and reaches the water below 2.00 s later. How high is the cliff and how far from the base of the cliff did the diver hit the water? \u2022 projectile motion: A form of motion where a particle (called a projectile) is thrown obliquely near the earth\u2019s surface, it moves along a curved path under the action of gravity. The path followed by a projectile motion is called its trajectory. Projectile motion only occurs when there is one force applied at the beginning of the trajectory after which there is no interference apart from gravity. \u2022 trajectory: The path followed by an object in projectile motion. References 1. Courtesy of PDPhoto.org. http://www.pdphoto.org/PictureDetail.php?mat=&pg=7672. Public Domain 2. CK-12 Foundation - Samantha Bacic.. CC-BY-NC-SA 3.0 6 www.ck12.org Chapter 3. Projectile Motion for an Object Launched at an Angle CHAPTER 3 Projectile Motion for an Object Launched at an Angle \u2022 The student will calculate the maximum height and range of projectiles launched at an angle given the initial velocity and angle. In the case of the human cannonball shown, all the vector and gravitational calculations must be worked out perfectly before the \ufb01rst practice session. With this activity, you cannot afford trial and error \u2013the \ufb01rst miss might be the last trial. Projectile Motion for an Object Launched at an Angle When an object is projected from rest at an upward angle, its initial velocity can be resolved into two components. These two components operate independently of each other. The upward velocity undergoes constant downward acceleration which will result in it rising to a highest point and then falling backward to the ground. The horizontal motion is constant velocity motion and undergoes no changes due to gravity. As usual, the analysis of the motion involves dealing with the two motions independently. 7 www.ck12.org Example Problem: A cannon ball is \ufb01red with an initial velocity of 100. m/s at an angle of 45\u00b0 above the horizontal. What maximum height will it reach and how far will it \ufb02y horizontally? Solution: The \ufb01rst step in the analysis of this motion is to resolve the initial velocity into its vertical and horizontal components. viup = (100: m/s)(sin 45) = (100: m/s)(0:707) = 70:7 m/s vihorizontal = (100: m/s)(cos 45) = (100: m/s)(0:707) = 70:7 m/s We will deal with the vertical motion \ufb01rst. The vertical motion is symmetrical. The object will rise up to its highest point and then fall back. The distance it travels up will be the same as the distance it falls down. The time it takes to reach the top will be the same time it takes to fall back to its initial point. The initial velocity upward will be the", " same magnitude (opposite in direction) as the \ufb01nal velocity when it returns to its original height. There are several ways we could approach the upward motion. We could calculate the time it would take gravity to bring the initial velocity to rest. Or, we could calculate the time it would take gravity to change the initial velocity from +70.7 m/s to -70.0 m/s. Yet another way would be to calculate the time for the height of the object to return to zero. v f = vi + at so t = v f vi a If we calculate the time required for the ball to rise up to its highest point and come to rest, the initial velocity is 70.7 m/s and the \ufb01nal velocity is 0 m/s. Since we have called the upward velocity positive, then the acceleration must be negative or -9.80 m/s2. t = v f vi a = 0 m/s70:7 m/s 9:80 m/s2 = 7:21 s The maximum height reached can be calculated by multiplying the time for the upward trip by the average vertical velocity. The average upward velocity during the trip up is one-half the initial velocity. vupave = 1 height = (vupave)(tup) = (35:3 m/s)(7:21 s) = 255 m (70:7 m/s) = 35:3 m/s 2 Since this is the time required for the cannon ball to rise up to its highest point and come to rest, then the time required for the entire trip up and down would be double this value, or 14.42 s. The horizontal distance traveled during the \ufb02ight is calculated by multiplying the total time times the constant horizontal velocity. dhorizontal = (14:42 s)(70:7 m/s) = 1020 m Example Problem: A golf ball was knocked into the with an initial velocity of 4.47 m/s at an angle of 66\u00b0 with the horizontal. How high did the ball go and how far did it \ufb02y horizontally? Solution: viup = (4:47 m/s)(sin 66) = (4:47 m/s)(0:913) = 4:08 m/s vihor = (4:47 m/s)(cos 66) = (4:47 m/s)(0:407) = 1:82 m/s 8 www.ck12.org Chapter 3. Projectile Motion for an Object Launched at an Angle a = 0 m/s4:08 m/s 9:80 m/s2 = 0:416 s tup = v f vi vupave = 1 height = (vupave)(tup) = (2:04 m/s)(0:416 s) = 0:849 m (4:08 m/s) = 2:04 m/s 2 ttotal trip = (2)(0:416 s) = 0:832 s dhorizontal = (0:832 s)(1:82 m/s) = 1:51 m Example Problem: Suppose a cannon ball is \ufb01red downward from a 50.0 m high cliff at an angle of 45\u00b0 with an initial velocity of 80.0 m/s. How far horizontally will it land from the base of the cliff? Solution: In this case, the initial vertical velocity is downward and the acceleration due to gravity will increase this downward velocity. vidown = (80:0 m/s)(sin 45) = (80:0 m/s)(0:707) = 56:6 m/s vihor = (80:0 m/s)(cos 45) = (80:0 m/s)(0:707) = 56:6 m/s d = vidownt + 1 2 at2 50:0 = 56:6t + 4:9t2 Changing to standard quadratic form yields 4:9t2 + 56:6t 50:0 = 0 This equation can be solved with the quadratic formula. The quadratic formula will produce two possible solutions for t: p t = b+ 56:6+ t = p b2 4ac b2 4ac and t = b 2a 2a p(56:6)2 (4)(4:9)(50) (2)(4:9) = 0:816 s The other solution to the quadratic formula yields a negative number which is clearly not a reasonable solution for this problem. dhorizontal = (0:816 s)(56:6 m/s) = 46:2 m Summary \u2022 When an object is projected from rest at an upward angle, its initial velocity can be resolved into two components. These two components operate independently of each other. \u2022 The upward velocity undergoes constant downward acceleration which will result in it", " rising to a highest point and then falling backward to the ground. \u2022 The horizontal motion is constant velocity motion and undergoes no changes due to gravity. \u2022 The analysis of the motion involves dealing with the two motions independently. Practice The following video shows a motion analysis for projectile motion launched upward. http://www.youtube.com/watch?v=rMVBc8cE5GU MEDIA Click image to the left for more content. 9 www.ck12.org The following video shows the famous \"shoot the monkey\" demonstration. Use this resource to answer the questions that follow. http://www.youtube.com/watch?v=cxvsHNRXLjw In this demonstration, a stuffed toy is hung from a high support and is attached to the support by an electric switch. A golf ball cannon is aimed up at the \u201cmonkey\u201d while it is hanging on the support. The cannon is designed such that It when the golf ball projectile leaves the barrel, it triggers the switch and releases the toy monkey from its perch. would seem (to the non-physicist) that the projectile will miss the monkey because the monkey will fall under the line of \ufb01re. The physicist knows, however, that the projectile falls from its line of \ufb01re by exactly the same amount that the monkey falls and therefore, the projectile will hit the monkey every time... in fact, it cannot miss. 1. What is the cannon ball in this video? 2. What is used as the monkey in this video? Review 1. A player kicks a football from ground level with a velocity of magnitude 27.0 m/s at an angle of 30.0\u00b0 above the horizontal. (a) Find the time the ball is in the air. (b) Find the maximum height of the ball. (c) Find the horizontal distance the ball travels. 2. A person standing on top of a 30.0 m high building throws a ball with an initial velocity of 20. m/s at an angle of 20.0\u00b0 below horizontal. How far from the base of the building will the ball land? 3. An arrow is \ufb01red downward at an angle of 45 degrees from the top of a 200 m cliff with a velocity of 60.0 m/s. (a) How long will it take the arrow to hit the ground? (b) How far from the base of the cliff will the arrow land? \u2022 trajectory: The ballistic trajectory of a projectile is the path that a thrown or launched projectile will take under the action of gravity, neglecting all other forces, such as air resistance, without propulsion. References 1. Flickr: JoshBerglund19. http://www.flickr.com/photos/tyrian123/1539636464/. CC-BY 2.0 2. CK-12 Foundation - Samantha Bacic.. CC-BY-NC-SA 3.0 10 www.ck12.org Chapter 4. Projectile Motion Problem Solving CHAPTER 4 Projectile Motion Problem Solving Students will learn how to use the equations of motion in two dimensions in order to solve problems for projectiles. It is necessary to understand how to break a vector into its x and y components. Students will learn how to use the equations of motion in two dimensions in order to solve problems for projectiles. It is necessary to understand how to break a vector into its x and y components. Key Equations Break the Initial Velocity Vector into its Components Apply the Kinematics Equations y(t) = yi + viyt 1 Vertical Direction 2 gt2 vy(t) = viy gt 2 2g(Dy) vy ay = g = 9:8m=s2 10m=s2 2 = v0y Horizontal Direction x(t) = xi + vixt vx(t) = vix ax = 0 11 www.ck12.org Guidance \u2022 To work these problems, separate the \u201cBig Three\u201d equations into two sets: one for the vertical direction, and one for the horizontal. Keep them separate. \u2022 The only variable that can go into both sets of equations is time; use time to communicate between the x and y components of the object\u2019s motion. Example 1 CSI discovers a car at the bottom of a 72 m cliff. How fast was the car going if it landed 22m horizontally from the cliff\u2019s edge? (Note that the cliff is \ufb02at, i.e. the car came off the cliff horizontally). Question: v =? [m=s] Given: h = Dy = 72 m d = Dx = 22 m g = 10:0 m=s2 Equation: h = viyt + 1 2 gt2 and d = vixt Plug n\u2019 Chug: Step 1: Calculate the time required for the car to freefall from a height of 72 m. h", " = viyt + 1 s 2 gt2 but since viy = 0, the equation simpli\ufb01es to h = 1 s 2 gt2 rearranging for the unknown variable, t, yields t = 2h g = 2(72 m) 10:0 m=s2 = 3:79 s Step 2: Solve for initial velocity: vix = d 3:79 s = 5:80 m=s t = 22 m Answer: Example 2 5:80 m=s Question: A ball of mass m is moving horizontally with a speed of vi off a cliff of height h. How much time does it take the ball to travel from the edge of the cliff to the ground? Express your answer in terms of g (acceleration due to gravity) and h (height of the cliff). Solution: Since we are solving or how long it takes for the ball to reach ground, any motion in the x direction is not pertinent. To make this problem a little simpler, we will de\ufb01ne down as the positive direction and the top of the cliff to be. In this solution we will use the equation y = 0 y(t) = yo + voyt + gt2 1 2. 12 www.ck12.org Chapter 4. Projectile Motion Problem Solving gt2 gt2 1 2 1 2 gt2 1 2 gt2 y(t) = yo + voyt + h = yo + voyt + h = 0 + voy + gt2 2 s 2h g start with the equation substitute h for y(t) because that\u2019s the position of the ball when it hits the ground after time t substitute 0 for yobecause the ball starts at the top of the cliff substitute 0 for voy becauese the ball starts with no vertical component to it\u2019s velocity simplify the equation solve for t Watch this Explanation MEDIA Click image to the left for more content. Time for Practice 1. A stone is thrown horizontally at a speed of 8:0 m=s from the edge of a cliff 80 m in height. How far from the base of the cliff will the stone strike the ground? 2. A toy truck moves off the edge of a table that is 1:25 m high and lands 0:40 m from the base of the table. a. How much time passed between the moment the car left the table and the moment it hit the \ufb02oor? b. What was the horizontal velocity of the car when it hit the ground? 3. A hawk in level \ufb02ight 135 m above the ground drops the \ufb01sh it caught. If the hawk\u2019s horizontal speed is 20:0 m=s, how far ahead of the drop point will the \ufb01sh land? 4. A pistol is \ufb01red horizontally toward a target 120 m away, but at the same height. The bullet\u2019s velocity is 200 m=s. How long does it take the bullet to get to the target? How far below the target does the bullet hit? 5. A bird, traveling at 20 m=s, wants to hit a waiter 10 m below with his dropping (see image). In order to hit the waiter, the bird must release his dropping some distance before he is directly overhead. What is this distance? 13 www.ck12.org 6. Joe Nedney of the San Francisco 49ers kicked a \ufb01eld goal with an initial velocity of 20 m=s at an angle of 60. a. How long is the ball in the air? Hint: you may assume that the ball lands at same height as it starts at. b. What are the range and maximum height of the ball? 7. A racquetball thrown from the ground at an angle of 45 and with a speed of 22:5 m=s lands exactly 2:5 s later on the top of a nearby building. Calculate the horizontal distance it traveled and the height of the building. 8. Donovan McNabb throws a football. He throws it with an initial velocity of 30 m=s at an angle of 25. How much time passes until the ball travels 35 m horizontally? What is the height of the ball after 0:5 seconds? (Assume that, when thrown, the ball is 2 m above the ground.) 9. Pablo Sandoval throws a baseball with a horizontal component of velocity of 25 m=s. After 2 seconds, the ball is 40 m above the release point. Calculate the horizontal distance it has traveled by this time, its initial vertical component of velocity, and its initial angle of projection. Also, is the ball on the way up or the way down at this moment in time? 10. Barry Bonds hits a 125 m(4500) home run that lands in the stands at an altitude 30 m above its starting altitude. Assuming that the ball left the bat at an angle of 45 from the horizontal, calculate", " how long the ball was in the air. 11. A golfer can drive a ball with an initial speed of 40:0 m=s. If the tee and the green are separated by 100 m, but are on the same level, at what angle should the ball be driven? ( Hint: you should use 2 cos (x) sin (x) = sin (2x) at some point.) 12. How long will it take a bullet \ufb01red from a cliff at an initial velocity of 700 m=s, at an angle 30 below the horizontal, to reach the ground 200 m below? 13. A diver in Hawaii is jumping off a cliff 45 m high, but she notices that there is an outcropping of rocks 7 m out at the base. So, she must clear a horizontal distance of 7 m during the dive in order to survive. Assuming the diver jumps horizontally, what is his/her minimum push-off speed? 14. If Monte Ellis can jump 1:0 m high on Earth, how high can he jump on the moon assuming same initial velocity that he had on Earth (where gravity is 1=6 that of Earth\u2019s gravity)? 15. James Bond is trying to jump from a helicopter into a speeding Corvette to capture the bad guy. The car is going 30:0 m=s and the helicopter is \ufb02ying completely horizontally at 100 m=s. The helicopter is 120 m above the car and 440 m behind the car. How long must James Bond wait to jump in order to safely make it into the car? 14 www.ck12.org Chapter 4. Projectile Motion Problem Solving 16. A \ufb01eld goal kicker lines up to kick a 44 yard (40 m) \ufb01eld goal. He kicks it with an initial velocity of 22 m=s at an angle of 55. The \ufb01eld goal posts are 3 meters high. a. Does he make the \ufb01eld goal? b. What is the ball\u2019s velocity and direction of motion just as it reaches the \ufb01eld goal post (i.e., after it has traveled 40 m in the horizontal direction)? 17. In a football game a punter kicks the ball a horizontal distance of 43 yards (39 m). On TV, they track the hang time, which reads 3:9 seconds. From this information, calculate the angle and speed at which the ball was kicked. (Note for non-football watchers: the projectile starts and lands at the same height. It goes 43 yards horizontally in a time of 3:9 seconds) Answers to Selected Problems 1. 32 m 2. a. 0:5 s b. 0:8 m=s 3. 104 m 4. t = 0:60 s; 1:8 m below target 5. 28 m. 6. a. 3:5 s. b. 35 m; 15 m 7. 40 m; 8:5 m 8. 1:3 seconds, 7:1 meters 9. 50 m; v0y = 30 m=s; 500; on the way up 10. 4:4 s 11. 19 12. 0:5 s 13. 2:3 m=s 14. 6 m 15. 1:4 seconds 16. a. yes b. 14 m=s @ 23 degrees from horizontal 17. 22 m=s @ 62 degrees 15 Physics Unit 5 (Forces and Newton\u2019s Laws) Patrick Marshall Ck12 Science Jean Brainard, Ph.D. James H Dann, Ph.D. Say Thanks to the Authors Click http://www.ck12.org/saythanks (No sign in required) AUTHORS Patrick Marshall Ck12 Science Jean Brainard, Ph.D. James H Dann, Ph.D. CONTRIBUTORS Chris Addiego Antonio De Jesus L\u00f3pez www.ck12.org To access a customizable version of this book, as well as other interactive content, visit www.ck12.org CK-12 Foundation is a non-pro\ufb01t organization with a mission to reduce the cost of textbook materials for the K-12 market both in the U.S. and worldwide. Using an open-content, web-based collaborative model termed the FlexBook\u00ae, CK-12 intends to pioneer the generation and distribution of high-quality educational content that will serve both as core text as well as provide an adaptive environment for learning, powered through the FlexBook Platform\u00ae. Copyright \u00a9 2014 CK-12 Foundation, www.ck12.org The names \u201cCK-12\u201d and \u201cCK12\u201d and associated logos and the terms \u201cFlexBook\u00ae\u201d and \u201cFlexBook Platform\u00ae\u201d (collectively \u201cCK-12 Marks\u201d) are trademarks and service marks of CK-12 Foundation and are protected by federal, state, and international laws. Any form of reproduction of", " this book in any format or medium, in whole or in sections must include the referral attribution link http://www.ck12.org/saythanks (placed in a visible location) in addition to the following terms. Except as otherwise noted, all CK-12 Content (including CK-12 Curriculum Material) is made available to Users in accordance with the Creative Commons Attribution-Non-Commercial 3.0 Unported (CC BY-NC 3.0) License (http://creativecommons.org/ licenses/by-nc/3.0/), as amended and updated by Creative Commons from time to time (the \u201cCC License\u201d), which is incorporated herein by this reference. Complete terms can be found at http://www.ck12.org/terms. Printed: February 4, 2014 iii Contents www.ck12.org 1 4 10 13 16 20 24 27 31 33 Contents 1 Newton\u2019s First and Second Laws of Motion 2 Newton\u2019s First Law 3 Newton\u2019s Second Law 4 Newton\u2019s Third Law of Motion 5 Types of Forces 6 Universal Law of Gravity 7 Mass versus Weight 8 9 Friction Free Body Diagrams 10 Problem Solving 1 iv www.ck12.org Chapter 1. Newton\u2019s First and Second Laws of Motion CHAPTER 1 Newton\u2019s First and Second Laws of Motion \u2022 De\ufb01ne force. \u2022 State the fundamental units for the Newton. \u2022 State Newton\u2019s First Law of Motion. \u2022 Given two of the three values in F = ma, calculate the third. This image is of Buzz Aldrin, one of the \ufb01rst men to walk on the moon. Apollo 11 was the space\ufb02ight that landed the \ufb01rst humans, Neil Armstrong and Buzz Aldrin, on the Moon on July 20, 1969. Armstrong became the \ufb01rst to step onto the lunar surface 6 hours later on July 21. This accomplishment could not have occurred without a thorough understanding of forces and acceleration. Newton\u2019s First and Second Laws of Motion What is a force? A force can be de\ufb01ned as a push or pull. When you place a book on a table, the book pushes downward on the table and the table pushes upward on the book. The two forces are equal and there is no resulting If, on the other hand, you drop the book, it will fall to the ground pulled by a force called motion of the book. gravity. 1 www.ck12.org If you slide a book across the \ufb02oor, it will experience a force of friction which acts in the opposite direction of the motion. This force will slow down the motion of the book and eventually bring it to rest. If the \ufb02oor is smoother, If a perfectly smooth \ufb02oor the force of friction will be less and the book will slide further before coming to rest. could be created, there would be no friction and the book would slide forever at constant speed. Newton\u2019s First Law of Motion describes an object moving with constant speed in a straight line. In the absence of any force, the object will continue to move at the same constant speed and in the same straight line. If the object is at rest, in the absence of any force, it will remain at rest. Newton\u2019s First Law states that an object with no force acting on it moves with constant velocity. (The constant velocity could, of course, be 0 m/s.) A more careful expression of Newton\u2019s First Law is \"an object with no net force acting on it remains at rest or moves with constant velocity in a straight line.\" The statement above is equivalent to a statement that \"if there is no net force on an object, there will be no acceleration.\" In the absence of acceleration, an object will remain at rest or will move with constant velocity in a straight line. The acceleration of an object is the result of an unbalanced force. If an object suffers two forces, the motion of the object is determined by the net force. The magnitude of the acceleration is directly proportional to the magnitude of the unbalanced force. The direction of the acceleration is the same direction as the direction of the unbalanced force. The magnitude of the acceleration is inversely proportional to the mass of the object. i.e. The more massive the object, the smaller will be the acceleration produced by the same force. These relationships are stated in Newton\u2019s Second Law of Motion, \"the acceleration of an object is directly proportional to the net force on the object and inversely proportional to the mass of the object.\" Newton\u2019s Second Law can be summarized in an equation. a = F m or more commonly; F = ma According to Newton\u2019s second law, a new force on an object causes it to accelerate and the larger the mass, the smaller the acceleration. Sometimes, the word inertia is used to express the resistance of an object to acceleration.", " Therefore, we say that a more massive object has greater inertia. The units for force are de\ufb01ned by the equation for Newton\u2019s second law. Suppose we wish to express the force that will give a 1.00 kg object an acceleration of 1.00 m/s2. F = ma = (1:00 kg)(1:00 m/s2) = 1:00 kg m/s2 This unit is de\ufb01ned as 1.00 Newton or 1.00 N. kgm s2 = Newton Example Problem: What new force is required to accelerate a 2000. kg car at 2.000 m/s2? Solution: F = ma = (2000: kg)(2:000 m/s2) = 4000: N Example Problem: A net force of 150 N is exerted on a rock. The rock has an acceleration of 20. m/s2 due to this force. What is the mass of the rock? Solution: m = F = 7:5 kg a = (150 N) (20: m/s2 ) Example Problem: A net force of 100. N is exerted on a ball. If the ball has a mass of 0.72 kg, what acceleration will it undergo? Solution: a = F m = (100: N) (0:72 kg) = 140 m/s2 Summary \u2022 A force is a push or pull. \u2022 Newton\u2019s First Law states that an object with no net force acting on it remains at rest or moves with constant velocity in a straight line. \u2022 Newton\u2019s Second Law of Motion states that the acceleration of an object is directly proportional to the net 2 www.ck12.org Chapter 1. Newton\u2019s First and Second Laws of Motion force on the object and inversely proportional to the mass of the object. Expressed as an equation, F = ma. Practice Professor Mac explains Newton\u2019s Second Law of Motion. http://www.youtube.com/watch?v=-KxbIIw8hlc MEDIA Click image to the left for more content. Review 1. A car of mass 1200 kg traveling westward at 30. m/s is slowed to a stop in a distance of 50. m by the car\u2019s brakes. What was the braking force? 2. Calculate the average force that must be exerted on a 0.145 kg baseball in order to give it an acceleration of 130 m/s2. 3. After a rocket ship going from the Earth to the Moon leaves the gravitational pull of the Earth, it can shut off its engine and the ship will continue on to the Moon due to the gravitational pull of the Moon. (a) True (b) False 4. If a space ship traveling at 1000 miles per hour enters an area free of gravitational forces, its engine must run at some minimum level in order to maintain the ships velocity. (a) True (b) False 5. Suppose a space ship traveling at 1000 miles per hour enters an area free of gravitational forces and free of air resistance. If the pilot wishes to slow the ship down, he can accomplish that by shutting off the engine for a while. (a) True (b) False \u2022 force: A push or pull. References 1. Courtesy of Neil A. Armstrong, NASA. http://space\ufb02ight.nasa.gov/gallery/images/apollo/apollo11/html/as11 -40-5873.html. Public Domain 3 CHAPTER 2 Newton\u2019s First Law www.ck12.org \u2022 Use skateboarding to explain Newton\u2019s \ufb01rst law of motion. There\u2019s no doubt from Corey\u2019s face that he loves skateboarding! Corey and his friends visit Newton\u2019s Skate Park every chance they get. They may not know it, but while they\u2019re having fun on their skateboards, they\u2019re actually applying science concepts such as forces and motion. 4 www.ck12.org Starting and Stopping Chapter 2. Newton\u2019s First Law Did you ever ride a skateboard? Even if you didn\u2019t, you probably know that to start a skateboard rolling over a level surface, you need to push off with one foot against the ground. That\u2019s what Corey\u2019s friend Nina is doing in this picture 2.1. FIGURE 2.1 Do you know how to stop a skateboard once it starts rolling? Look how Nina\u2019s friend Laura does it in the Figure 2.2. She steps down on the back of the skateboard so it scrapes on the pavement. This creates friction, which stops the skateboard. Even if Laura didn\u2019t try to stop the skateboard, it would stop sooner or later. That\u2019s because there\u2019s also friction between the wheels and the pavement. Friction is a force that counters all kinds of motion. It occurs", " whenever two surfaces come into contact. Video Break Laura learned how to use forces to start and stop her skateboard by watching the videos below. Watch the video to see how the forces are applied. You can pick up some skateboarding tips at the same time! Starting: http://www.youtube.com/watch?v=OpZIVjbMAOU MEDIA Click image to the left for more content. Stopping: http://www.youtube.com/watch?v=6fuOwhx91zM 5 www.ck12.org FIGURE 2.2 MEDIA Click image to the left for more content. Laws of the Park: Newton\u2019s First Law If you understand how a skateboard starts and stops, then you already know something about Newton\u2019s \ufb01rst law of motion. This law was developed by English scientist Isaac Newton around 1700. Newton was one of the greatest scientists of all time. He developed three laws of motion and the law of gravity, among many other contributions. Newton\u2019s \ufb01rst law of motion states that an object at rest will remain at rest and an object in motion will stay in motion unless it is acted on by an unbalanced force. Without an unbalanced force, a moving object will not only 6 www.ck12.org Chapter 2. Newton\u2019s First Law keep moving, but its speed and direction will also remain the same. Newton\u2019s \ufb01rst law of motion is often called the law of inertia because inertia is the tendency of an object to resist a change in its motion. If an object is already at rest, inertia will keep it at rest. If an object is already in motion, inertia will keep it moving. Do You Get It? Q: How does Nina use Newton\u2019s \ufb01rst law to start her skateboard rolling? A: The skateboard won\u2019t move unless Nina pushes off from the pavement with one foot. The force she applies when she pushes off is stronger than the force of friction that opposes the skateboard\u2019s motion. As a result, the force on the skateboard is unbalanced, and the skateboard moves forward. Q: How does Nina use Newton\u2019s \ufb01rst law to stop her skateboard? A: Once the skateboard starts moving, it would keep moving at the same speed and in the same direction if not for another unbalanced force. That force is friction between the skateboard and the pavement. The force of friction is unbalanced because Nina is no longer pushing with her foot to keep the skateboard moving. That\u2019s why the skateboard stops. Changing Direction Corey\u2019s friend Jerod likes to skate on the \ufb02at banks at Newton\u2019s Skate Park. That\u2019s Jerod in the picture above. As he reaches the top of a bank, he turns his skateboard to go back down. To change direction, he presses down with his heels on one edge of the skateboard. This causes the skateboard to turn in the opposite direction. Video Break Can you turn a skateboard like Jerod? To see how to apply forces to change the direction of a skateboard, watch this video: http://www.youtube.com/watch?v=iOnlcEk50CM MEDIA Click image to the left for more content. Do You Get It? Q: How does Jerod use Newton\u2019s \ufb01rst law of motion to change the direction of his skateboard? A: Pressing down on just one side of a skateboard creates an unbalanced force. The unbalanced force causes the skateboard to turn toward the other side. In the picture, Jerod is pressing down with his heels, so the skateboard turns toward his toes. Summary \u2022 Newton\u2019s \ufb01rst law of motion states that an object at rest will remain at rest and an object in motion will remain in motion unless it is acted on by an unbalanced force. 7 www.ck12.org FIGURE 2.3 \u2022 Using unbalanced forces to control the motion of a skateboard demonstrates Newton\u2019s \ufb01rst law of motion. Vocabulary \u2022 Newton\u2019s \ufb01rst law of motion: Law stating that an object\u2019s motion will not change unless an unbalanced force acts on the object. Practice Do you think you understand Newton\u2019s \ufb01rst law? Go to the URL below to \ufb01nd out. Review Newton\u2019s law and watch what happens to the skateboarder in the animation. Then answer the questions at the bottom of the Web page. http://teachertech.rice.edu/Participants/louviere/Newton/law1.html 8 www.ck12.org Review Chapter 2. Newton\u2019s First Law 1. State Newton\u2019s \ufb01r", "st law of motion. 2. You don\u2019t need to push off with a foot against the ground to start a skateboard rolling down a bank. Does this violate Newton\u2019s \ufb01rst law of motion? Why or why not? FIGURE 2.4 3. Nina ran into a rough patch of pavement, but she thought she could ride right over it. Instead, the skateboard stopped suddenly and Nina ended up on the ground (see Figure above). Explain what happened. 4. Now that you know about Newton\u2019s \ufb01rst law of motion, how might you use it to ride a skateboard more safely? References 1. Image copyright DenisNata, 2012.. Used under license from Shutterstock.com 2. Image copyright DenisNata, 2012.. Used under license from Shutterstock.com 3. Image copyright Nikola Bilic, 2012.. Used under license from Shutterstock.com 4. Image copyright DenisNata, 2012.. Used under license from Shutterstock.com 9 CHAPTER 3 Newton\u2019s Second Law www.ck12.org \u2022 State Newton\u2019s second law of motion. \u2022 Compare and contrast the effects of force and mass on acceleration. These boys are racing around the track at Newton\u2019s Skate Park. The boy who can increase his speed the most will win the race. Tony, who is closest to the camera in this picture, is bigger and stronger than the other two boys, so he can apply greater force to his skates. Q: Does this mean that Tony will win the race? A: Not necessarily, because force isn\u2019t the only factor that affects acceleration. Force, Mass, and Acceleration Whenever an object speeds up, slows down, or changes direction, it accelerates. Acceleration occurs whenever an unbalanced force acts on an object. Two factors affect the acceleration of an object: the net force acting on the object and the object\u2019s mass. Newton\u2019s second law of motion describes how force and mass affect acceleration. The law 10 www.ck12.org Chapter 3. Newton\u2019s Second Law states that the acceleration of an object equals the net force acting on the object divided by the object\u2019s mass. This can be represented by the equation: Acceleration = Net force Mass or a = F m Q: While Tony races along on his rollerblades, what net force is acting on the skates? A: Tony exerts a backward force against the ground, as you can see in the Figure 3.1, \ufb01rst with one skate and then with the other. This force pushes him forward. Although friction partly counters the forward motion of the skates, it is weaker than the force Tony exerts. Therefore, there is a net forward force on the skates. FIGURE 3.1 Direct and Inverse Relationships Newton\u2019s second law shows that there is a direct relationship between force and acceleration. The greater the force that is applied to an object of a given mass, the more the object will accelerate. For example, doubling the force on the object doubles its acceleration. The relationship between mass and acceleration is different. It is an inverse relationship. In an inverse relationship, when one variable increases, the other variable decreases. The greater the mass of an object, the less it will accelerate when a given force is applied. For example, doubling the mass of an object results in only half as much acceleration for the same amount of force. Q: Tony has greater mass than the other two boys he is racing above. How will this affect his acceleration around the track? A: Tony\u2019s greater mass will result in less acceleration for the same amount of force. Summary \u2022 Newton\u2019s second law of motion states that the acceleration of an object equals the net force acting on the object divided by the object\u2019s mass. \u2022 According to the second law, there is a direct relationship between force and acceleration and an inverse relationship between mass and acceleration. Vocabulary \u2022 Newton\u2019s second law of motion: Law stating that the acceleration of an object equals the net force acting on the object divided by the object\u2019s mass. 11 www.ck12.org Practice At the following URL, use the simulator to experiment with force, mass, and acceleration. First keep force constant at 1 N, and vary mass from 1\u20135 kg. Next keep mass constant at 1 kg, and vary force from 1\u20135 N. In each simulation, record the values you tested and the resulting acceleration. Finally, make two line graphs to plot your results. On one graph, show acceleration when force is constant and mass changes. On the other graph, show acceleration when mass is constant and force changes. Describe in words what the two graphs show. http://janggeng.com/newtons-second-law-of-motion/ Review 1. State Newton\u2019s second law of motion. 2. How can Newton\ufffd", "\ufffds second law of motion be represented with an equation? 3. If the net force acting on an object doubles, how will the object\u2019s acceleration be affected? 4. Tony has a mass of 50 kg, and his friend Sam has a mass of 45 kg. Assume that both friends push off on their rollerblades with the same force. Explain which boy will have greater acceleration. References 1. Uploaded by User:Shizhao/Wikimedia Commons.. CC BY 2.5 12 www.ck12.org Chapter 4. Newton\u2019s Third Law of Motion CHAPTER 4 Newton\u2019s Third Law of Motion \u2022 De\ufb01ne force. \u2022 State the fundamental units for the Newton. \u2022 State Newton\u2019s First Law of Motion. \u2022 Given two of the three values in F = ma, calculate the third It was imagined as a geo-stationary transfer The image at above is a NASA artist\u2019s concept of a space elevator. station for passengers and cargo between earth and space. This idea was not pursued but it began where all great ideas begin... in someone\u2019s mind. Newton\u2019s Third Law of Motion Where do forces come from? Observations suggest that a force applied to an object is always applied by another object. A hammer strikes a nail, a car pulls a trailer, and a person pushes a grocery cart. Newton realized that forces are not so one sided. When the hammer exerts a force on the nail, the nail also exerts a force on the hammer \u2013after all, the hammer comes to rest after the interaction. This led to Newton\u2019s Third Law of Motion, which states that whenever one object exerts a force on a second object, the second object also exerts a force on the \ufb01rst object, equal in magnitude and opposite in direction. 13 www.ck12.org This law is sometimes paraphrased as \u201cfor every action, there is an equal and opposite reaction.\u201d A very important point to remember is that the two forces are on different objects \u2013never on the same object. It is frequently the case that one of the objects moves as a result of the force applied but the motion of the other object in the opposite direction is not apparent. Consider the situation where an ice skater is standing at the edge of the skating rink holding on to the side rail. If the skater exerts a force on the rail, the rail is held in place with tremendous friction and therefore, will not move in any noticeable way. The skater, on the other hand, had little friction with the ice, and therefore will be accelerated in the direction opposite of his/her original push. This is the process people use to jump up into the air. The person\u2019s feet exert force on the ground and the ground exerts an equal and opposite force on the person\u2019s feet. The force on the feet is suf\ufb01cient to raise the person off the ground. The force on the ground has little effect because the earth is so large. One of the accelerations is visible but the other is not visible. A case where the reaction motion due to the reaction force is visible is the case of a person throwing a heavy object out of a boat. The object is accelerated in one direction and the boat is accelerated in the opposite direction. In this case, both the motion of the object is visible and the motion of the boat in the opposite direction is also visible. Rockets also work in this manner. It is a misconception that the rocket moves forward because the escaping gas pushes on the ground or the surrounding air to make the rocket go forward. Rockets work in outer space where there is no ground or surrounding air. The rocket exerts a force on the gases causing them to be expelled and the gases exert a force on the rocket causing it to be accelerated forward. Summary \u2022 A force applied to an object is always applied by another object. \u2022 Newton\u2019s Third Law of Motion says, \"whenever one object exerts a force on a second object, the second object also exerts a force on the \ufb01rst object, equal in magnitude and opposite in direction.\" Practice The following video contains a discussion and an example of Newton\u2019s Third Law of Motion. http://www.youtube.com/watch?v=fKJDpPi-UN0 MEDIA Click image to the left for more content. Review 1. What is wrong with the following statement: When you exert a force on a baseball, the equal and opposite force on the ball balances the original force and therefore, the ball will not accelerate in any direction. 2. When a bat strikes a ball, the force exerted can send the ball deep into the out\ufb01eld. Where is the equal and opposite force in this case? 3. Suppose you wish to jump horizontally and in order for you to jump a distance of 4 feet horizontally, you must exert a force of 200", " N. When you are standing on the ground, you have no trouble jumping 4 feet horizontally. If you are standing in a canoe, however, and you need to jump 4 feet to reach the pier, you will surely fall into the lake. Why is it that you cannot jump 4 feet out of a canoe when you can easily do this when on land? 14 www.ck12.org Chapter 4. Newton\u2019s Third Law of Motion \u2022 Newton\u2019s Third Law of Motion: Whenever one object exerts a force on a second object, the second object also exerts a force on the \ufb01rst object, equal in magnitude and opposite in direction. References 1. Courtesy of Pat Rawling, NASA. http://commons.wikimedia.org/wiki/File:Nasa_space_elev.jpg. Public Do- main 15 CHAPTER 5 The various types of common forces are discussed and analyzed. The various types of common forces are discussed and analyzed. www.ck12.org Types of Forces Key Equations Common Forces Guidance Normal Force 8 >>>>< >>>>: Fg = mg FN FT Fsp = kDx Force of spring stretched a distance Dx from equilibrium Gravity Normal force: acts perpendicular to surfaces Force of tension in strings and wires Often, objects experience a force that pushes them into another object, but once the objects are in contact they do not any move closer together. For instance, when you stand on the surface of the earth you are obviously not accelerating toward its center. According to Newton\u2019s Laws, there must be a force opposing the earth\u2019s gravity acting on you, so that the net force on you is zero. The same also applies for your gravity acting on the earth. We call such a force the Normal Force. The normal force acts between any two surfaces in contact, balancing what ever force is pushing the objects together. It is actually electromagnetic in nature (like other contact forces), and arises due to the repulsion of atoms in the two objects. Here is an illustration of the Normal force on a block sitting on earth: Tension Another force that often opposes gravity is known as tension. This force is provided by wires and strings when they hold objects above the earth. Like the Normal Force, it is electromagnetic in nature and arises due to the intermolecular bonds in the wire or string: 16 www.ck12.org Chapter 5. Types of Forces If the object is in equilibrium, tension must be equal in magnitude and opposite in direction to gravity. This force transfers the gravity acting on the object to whatever the wire or string is attached to; in the end it is usually a Normal Force \u2014 between the earth and whatever the wire is attached to \u2014 that ends up balancing out the force of gravity on the object. Friction Friction is a force that opposes motion. Any two objects in contact have what is called a mutual coef\ufb01cient of friction. To \ufb01nd the force of friction between them, we multiply the normal force by this coef\ufb01cient. Like the forces above, it arises due to electromagnetic interactions of atoms in two objects. There are actually two coef\ufb01cients of friction: static and kinetic. Static friction will oppose initial motion of two objects relative to each other. Once the objects are moving, however, kinetic friction will oppose their continuing motion. Kinetic friction is lower than static friction, so it is easier to keep an object in motion than to set it in motion. fs \u00b5sj ~FNj fk = \u00b5kj ~FNj [5] Static friction opposes potential motion of surfaces in contact [6] Kinetic frictions opposes motion of surfaces in contact There are some things about friction that are not very intuitive: \u2022 The magnitude of the friction force does not depend on the surface areas in contact. \u2022 The magnitude of kinetic friction does not depend on the relative velocity or acceleration of the two objects. \u2022 Friction always points in the direction opposing motion. If the net force (not counting friction) on an object is lower than the maximum possible value of static friction, friction will be equal to the net force in magnitude and opposite in direction. Spring Force Any spring has some equilibrium length, and if stretched in either direction it will push or pull with a force equal to: ~Fsp = k ~Dx [7] Force of spring ~Dx from equilibrium Example 1 Question: A woman of mass 70.0 kg weighs herself in an elevator. 17 www.ck12.org a) If she wants to weigh less, should she weigh herself when accelerating upward or downward? b) When the elevator is not accelerating, what does the scale read (i.e., what is the normal force that the scale exerts on the woman)? c) When the elevator is accelerating upward at 2.00 m/s2, what does the scale read? Answer a) If she wants to weigh less, she has to decrease her force (her weight is the force)", " on the scale. We will use the equation to determine in which situation she exerts less force on the scale. F = ma If the elevator is accelerating upward then the acceleration would be greater. She would be pushed toward the \ufb02oor of the elevator making her weight increase. Therefore, she should weigh herself when the elevator is going down. b) When the elevator is not accelerating, the scale would read 70:0kg. c) If the elevator was accelerating upward at a speed of 2:00m=s2, then the scale would read F = ma = 70kg (9:8m=s2 + 2m=s2) = 826N which is 82:6kg. Example 2 Question: A spring with a spring constant of k = 400N=m has an uncompressed length of :23m and a fully compressed length of :15m. What is the force required to fully compress the spring? Solution: We will use the equation F = kx to solve this. We simply have to plug in the known value for the spring and the distance to solve for the force. F = kx = (400N=m)(:23m :15m) = 32N 18 www.ck12.org Watch this Explanation Chapter 5. Types of Forces MEDIA Click image to the left for more content. Time for Practice No Problems for this section. See Newton Law Problem Solving Concept. 19 CHAPTER 6 Universal Law of Gravity www.ck12.org \u2022 Describe and give the formula for Newton\u2019s universal law of gravity. \u2022 Using Newton\u2019s law of gravity Cavendish\u2019s apparatus for experimentally determining the value of G involved a light, rigid rod about 2-feet long. Two small lead spheres were attached to the ends of the rod and the rod was suspended by a thin wire. When the rod becomes twisted, the torsion of the wire begins to exert a torsional force that is proportional to the angle of rotation of the rod. The more twist of the wire, the more the system pushes backwards to restore itself towards the original position. Cavendish had calibrated his instrument to determine the relationship between the angle of rotation and the amount of torsional force. Force of Gravity In the mid-1600\u2019s, Newton wrote that the sight of a falling apple made him think of the problem of the motion of the planets. He recognized that the apple fell straight down because the earth attracted it and thought this same force of attraction might apply to the moon and that motion of the planets might be controlled by the gravity of the sun. He eventually proposed the universal law of gravitational attraction as 20 F = G m1m2 d2 www.ck12.org Chapter 6. Universal Law of Gravity where m1 and m2 are the masses being attracted, d is the distance between the centers of the masses, G is the universal gravitational constant, and F is the force of attraction. The formula for gravitational attraction applies equally to two rocks resting near each other on the earth and to the planets and the sun. The value for the universal gravitational constant, G, was determined by Cavendish to be 6.67 1011 N\u00b7m2/kg2. The moon is being pulled toward the earth and the earth toward the moon with the same force but in the opposite direction. The force of attraction between the two bodies produces a greater acceleration of the moon than the earth because the moon has smaller mass. Even though the moon is constantly falling toward the earth, it never gets any closer. This is because the velocity of the moon is perpendicular to the radius of the earth (as shown in the image above) and therefore the moon is moving away from the earth. The distance the moon moves away from the orbit line is exactly the same distance that the moon falls in the time period. This is true of all satellites and is the reason objects remain in orbit. Example Problem: Since we know the force of gravity on a 1.00 kg ball resting on the surface of the earth is 9.80 N and we know the radius of the earth is 6380 km, we can use the equation for gravitational force to calculate the mass of the earth. Solution: me = Fd2 Gm1 = 5:98 1024 kg )(6:38106 m)2 (9:80 m/s2 = (6:671011 Nm2=kg2 )(1:00 kg) Sample Problem: John and Jane step onto the dance \ufb02oor about 20. M apart at the Junior Prom and they feel an attraction to each other. If John\u2019s mass is 70. kg and Jane\u2019s mass is 50. kg, assume the attraction is gravity and calculate its magnitude. Solution: Fg = Gm1m2 d2 = (6:671011 Nm2=kg2 (20: m)2 )(70: kg)(50: kg) = 1:2 108 N This", " is such an extremely weak force, it is probably not the force of attraction John and Jane felt. Summary \u2022 Newton proposed the universal law of gravitational attraction as F = G m1m2 d2 \u2022 The universal gravitational constant, G, was determined by Cavendish to be 6.67 1011 N\u00b7m2/kg2. \u2022 Even though satellites are constantly falling toward the object they circle, they do not get closer because their. straight line motion moves them away from the center at the same rate they fall. 21 Practice The following video is a lecture on universal gravitation. Use this resource to answer the questions that follow. http://www.youtube.com/watch?v=OZZGJfFf8XI www.ck12.org MEDIA Click image to the left for more content. 1. What factors determine the strength of the force of gravity? 2. Between which two points do we measure the distance between the earth and moon? The following video is The Mass vs. Weight Song. Use this resource to answer the questions that follow. http://w ww.youtube.com/watch?v=1whMAIGNq7E MEDIA Click image to the left for more content. 1. What is used to measure mass? 2. What is used to measure weight? 3. What units are used to measure mass? 4. What units are used to measure weight? The following website contains a series of solved practice problems on gravity. http://physics.info/gravitation/practice.shtml Review 1. The earth is attracted to the sun by the force of gravity. Why doesn\u2019t the earth fall into the sun? 2. If the mass of the earth remained the same but the radius of the earth shrank to one-half its present distance, what would happen to the force of gravity on an object that was resting on the surface of the earth? 3. Lifting an object on the moon requires one-sixth the force that would be required to lift the same object on the earth because gravity on the moon is one-sixth that on earth. What about horizontal acceleration? If you threw a rock with enough force to accelerate it at 1.0 m/s2 horizontally on the moon, how would the required force compare to the force necessary to acceleration the rock in the same way on the earth? 4. The mass of the earth is 5.98 1024 kg and the mass of the moon is 7.35 1022 kg. If the distance between the earth and the moon is 384,000 km, what is the gravitational force on the moon? \u2022 gravity: A natural phenomenon by which physical bodies appear to attract each other with a force proportional to their masses and inversely proportional to the distance separating them. 22 www.ck12.org References Chapter 6. Universal Law of Gravity 1. Chris Burks (Wikimedia: Chetvorno). http://commons.wikimedia.org/wiki/File:Cavendish_Torsion_Balance _Diagram.svg. Public Domain 2. CK-12 Foundation - Samantha Bacic.. CC-BY-NC-SA 3.0 23 www.ck12.org CHAPTER 7 Mass versus Weight \u2022 Distinguish between mass and weight. \u2022 Given the acceleration due to gravity and either the mass or the weight of an object, calculate the other one. \u201cSpace exploration is an international endeavor.\u201d Three Japan Aerospace Exploration Agency astronauts \u2013 Akihiko Hoshide, Satoshi Furukawa and Naoko Yamazaki \u2013 are training with the 11-member astronaut candidate class of 2004. JAXA astronauts Satoshi Furukawa, Akihiko Hoshide and Naoko Yamazaki experience near-weightlessness on the KC-135 training airplane. Mass and Weight The mass of an object is de\ufb01ned as the amount of matter in the object. The amount of mass in an object is measured by comparing the object to known masses on an instrument called a balance. 24 www.ck12.org Chapter 7. Mass versus Weight Using the balance shown above, the object would be placed in one pan and known masses would be placed in the other pan until the pans were exactly balanced. When balanced, the mass of the object would be equal to the sum of the known masses in the other pan. The unit of measurement for mass is the kilogram. The mass of an object would be the same regardless of whether the object was on the earth or on the moon. The balance and known masses work exactly the same both places and would indicate the same mass for the same object as long as some gravitational force is present. The weight of an object is de\ufb01ned as the force pulling the object downward. On the earth, this would be the gravitational force of the earth on the object. On the moon, this would be the gravitational force of the moon on the object. Weight is measured by a calibrated spring scale as shown here. Weight is", " measured in force units which is Newtons in the SI system. The weights measured for an object would not be the same on the earth and moon because the gravitational \ufb01eld on the surface of the moon is one-sixth of the magnitude of the gravitational \ufb01eld on the surface of the earth. The force of gravity is given by Newton\u2019s Second Law, F = ma, when F is the force of gravity in Newtons, m is the mass of the object in kilograms, and a is the acceleration due to gravity, 9.80 m/s2. When the formula is used speci\ufb01cally for \ufb01nding weight from mass or vice versa, it may appear as W = mg. Example Problem: What is the weight of an object sitting on the earth\u2019s surface if the mass of the object is 43.7 kg? Solution: W = mg = (43:7 kg)(9:80 m/s2) = 428 N Example Problem: What is the mass of an object whose weight sitting on the earth is 2570 N? m = W a = 2570 N 9:80 m/s2 = 262 kg Summary \u2022 The mass of an object is measured in kilograms and is de\ufb01ned as the amount of matter in an object. \u2022 The mass of an object is determined by comparing the mass to known masses on a balance. \u2022 The weight of an object on the earth is de\ufb01ned as the force acting on the object by the earth\u2019s gravity. \u2022 Weight is measured by a calibrated spring scale. \u2022 The formula relating mass and weight is W = mg. 25 Practice A song about the difference between mass and weight sung by Mr. Edmunds to the tune of Sweet Caroline. Remember to make allowances for the fact that he is a teacher, not a professional singer. Use this resource to answer the questions that follow. http://www.youtube.com/watch?v=1whMAIGNq7E www.ck12.org MEDIA Click image to the left for more content. 1. What is used to measure mass? 2. What is used to measure weight? 3. What units are used to measure mass? 4. What units are used to measure weight? This video shows what appears to be a magic trick but is actually a center of gravity demonstration. http://www.darktube.org/watch/simple-trick-magic-no-physics Review 1. The mass of an object on the earth is 100. kg. (a) What is the weight of the object on the earth? (b) What is the mass of the object on the moon? (c) Assuming the acceleration due to gravity on the moon is EXACTLY one-sixth of the acceleration due to gravity on earth, what is the weight of the object on the moon? 2. A man standing on the Earth can exert the same force with his legs as when he is standing on the moon. We know that the mass of the man is the same on the Earth and the Moon. We also know that F = ma is true on both the Earth and the Moon. Will the man be able to jump higher on the Moon than the Earth? Why or why not? \u2022 mass: The mass of an object is measured in kilograms and is de\ufb01ned as the amount of matter in an object. \u2022 weight: The weight of an object on the earth is de\ufb01ned as the force acting on the object by the earth\u2019s gravity. References 1. Courtesy of NASA. http://space\ufb02ight.nasa.gov/gallery/images/behindthescenes/training/html/jsc2004e45082.h tml. Public Domain 2. CK-12 Foundation - Christopher Auyeung.. CC BY-NC-SA 3.0 3. CK-12 Foundation - Christopher Auyeung.. CC-BY-NC-SA 3.0 26 www.ck12.org CHAPTER 8 \u2022 De\ufb01ne both static and sliding friction. \u2022 Explain what causes surface friction. \u2022 De\ufb01ne the coef\ufb01cient of friction. \u2022 Calculate frictional forces. \u2022 Calculate net forces when friction is involved. Chapter 8. Friction Friction Dealing with friction and a lack of friction becomes a very important part of the game in tennis played on a clay court. It\u2019s necessary for this player to learn how to keep her shoes from sliding when she wants to run but also necessary to know how her shoes will slide when coming to a stop. Friction Most of the time, in beginning physics classes, friction is ignored. Concepts can be understood and calculations made assuming friction to be non-existent. Whenever physics intersects with the real world, however, friction must be taken into account. Friction exists between two touching surfaces because even the smoothest", " looking surface is quite rough on a microscopic scale. 27 www.ck12.org With the bumps, lumps, and imperfections emphasized as in the image above, it becomes more apparent that if we try to slide the top block over the lower block, there will be collisions involved when bumps impact on bumps. The forward motion causes the collisions with bumps which then exert a force in opposite way the block is moving. The force of friction always opposes whatever motion is causing the friction. The force of friction between these two blocks is related to two factors. The \ufb01rst factor is the roughness of the surfaces that are interacting which is called the coef\ufb01cient of friction, \u00b5 (Greek letter mu). The second factor is the magnitude of the force pushing the top block down onto the lower block. It is reasonable that the more forcefully the blocks are pushed together, the more dif\ufb01cult it will be for one to slide over the other. A very long time ago, when physics was young, the word \u201cnormal\u201d was used in the same way that we use the word \u201cperpendicular\u201d today. The force pushing these blocks together is the perpendicular force pushing the top block down on the lower block and this force is called the normal force. Much of the time, this normal force is simply the weight of the top block but on some occasions, the weight of the top block has some added or reduced force so the normal force is not always the weight. The force of friction then, can be calculated by Ffriction = \u00b5 Fnormal This is an approximate but reasonably useful and accurate relationship. It is not exact because \u00b5 depends on whether the surfaces are wet or dry and so forth. The frictional force we have been discussing is referred to as sliding friction because it is involved when one surface is sliding over another. If you have ever tried to slide a heavy object across a rough surface, you may be aware that it is a great deal easier to keep an object sliding than it is to start the object sliding in the \ufb01rst place. When the object to slide is resting on a surface with no movement, the force of friction is called static friction and it is somewhat greater than sliding friction. Surfaces that are to move against one another will have both a coef\ufb01cient of static friction and a coef\ufb01cient of sliding friction and the two values will NOT be the same. For example, the coef\ufb01cient of sliding friction for ice on ice is 0.03 whereas the coef\ufb01cient of static friction for ice on ice is 0.10 \u2013more than three times as great. Example Problem: A box weighing 2000. N is sliding across a cement \ufb02oor. The force pushing the box is 500. N and the coef\ufb01cient of sliding friction between the box and the \ufb02oor is 0.20. What is the acceleration of the box. Solution: In this case, the normal force for the box is its weight. Using the normal force and the coef\ufb01cient of friction, we can \ufb01nd the frictional force. We can also \ufb01nd the mass of the box from its weight since we know the acceleration due to gravity. Then we can \ufb01nd the net force and the acceleration. FF = \u00b5FN = (0:20)(2000: N) = 400: N weight g = 2000: N mass of box = 9:8 m=s2 = 204 g FNET = Pushing force frictional force = 500: N 400: N = 100: N a = FN = 0:49 m/s2 m = 100: N 204 kg Example Problem: Two boxes are connected by a rope running over a pulley (see image). The coef\ufb01cient of sliding friction between box A and the table is 0.20. (Ignore the masses of the rope and the pulley and any friction in the pulley.) The mass of box A is 5.0 kg and the mass of box B is 2.0 kg. The entire system (both boxes) will move together with the same acceleration and velocity. Find the acceleration of the system. 28 www.ck12.org Chapter 8. Friction Solution: The force tending to move the system is the weight of box B and the force resisting the movement is the force of friction between the table and box A. The mass of the system would be the sum of the masses of both boxes. The acceleration of the system will be found by dividing the net force by the total mass. FN(box A) = mg = (5:0 kg)(9:8 m/s2) = 49 N Ffriction = \u00b5FN = (0:20)(49 N) = 9:8 N Weight of box B = mg = (2:0 kg)(9:8 m/s2) = 19:6 N", " FNET = 19:6 N 9:8 N = 9:8 N a = FNET = 1:4 m/s2 mass = 9:8 N 7:0 kg Summary \u2022 Friction is caused by bodies sliding over rough surfaces. \u2022 The degree of surface roughness is indicated by the coef\ufb01cient of friction, \u00b5. \u2022 The force of friction is calculated by multiplying the coef\ufb01cient of friction by the normal force. \u2022 The frictional force always opposes motion. \u2022 Acceleration is caused by the net force which is found by subtracting the frictional force from the applied force. Practice A video explaining friction. http://www.youtube.com/watch?v=CkTCp7SZdYQ MEDIA Click image to the left for more content. Review 1. A 52 N sled is pulled across a cement sidewalk at constant speed. A horizontal force of 36 N is exerted. What is the coef\ufb01cient of sliding friction between the sidewalk and the metal runners of the sled? 2. If the coef\ufb01cient of sliding friction between a 25 kg crate and the \ufb02oor is 0.45, how much force is required to move the crate at a constant velocity across the \ufb02oor? 29 www.ck12.org 3. A smooth wooden 40.0 N block is placed on a smooth wooden table. A force of 14.0 N is required to keep the block moving at constant velocity. (a) What is the coef\ufb01cient of sliding friction between the block and the table top? (b) If a 20.0 N brick is placed on top of the wooden block, what force will be required to keep the block and brick moving at constant velocity? \u2022 friction: A force that resists the relative motion or tendency to such motion of two bodies or substances in contact. \u2022 coef\ufb01cient of friction: The ratio of the force that maintains contact between an object and a surface (i.e. the normal force) and the frictional force that resists the motion of the object. \u2022 normal force: The perpendicular force one surface exerts on another surface when the surfaces are in contact. References 1. Courtesy of Eric Harris, U.S. Air Force. http://commons.wikimedia.org/wiki/File:Maria_Sharapova,_2008 _Family_Circle_Cup.JPG. Public Domain 2. CK-12 Foundation - Joy Sheng.. CC-BY-NC-SA 3.0 3. CK-12 Foundation - Richard Parsons.. CC-BY-NC-SA 3.0 30 www.ck12.org Chapter 9. Free Body Diagrams CHAPTER 9 Free Body Diagrams Students will learn how to draw a free-body diagram and apply it to the real world. Students will learn how to draw a free-body diagram and apply it to the real world. Guidance For every problem involving forces it is essential to draw a free body diagram (FBD) before proceeding to the problem solving stage. The FBD allows one to visualize the situation and also to make sure all the forces are accounted. In addition, a very solid understanding of the physics is gleaned and many questions can be answered solely from the FBD. Example 1 Watch this Explanation MEDIA Click image to the left for more content. MEDIA Click image to the left for more content. Time for Practice 1. Draw free body diagrams (FBDs) for all of the following objects involved (in bold) and label all the forces appropriately. Make sure the lengths of the vectors in your FBDs are proportional to the strength of the force: smaller forces get shorter arrows! a. A man stands in an elevator that is accelerating upward at 2 m=s2. a. A boy is dragging a sled at a constant speed. The boy is pulling the sled with a rope at a 30 angle. a. The picture shown here is attached to the ceiling by three wires. 31 www.ck12.org a. A bowling ball rolls down a lane at a constant velocity. a. A car accelerates down the road. There is friction f between the tires and the road. 2. For the following situation, identify the 3rd law force pairs on the associated free body diagrams. Label each member of one pair \u201cA;00 each member of the next pair \u201cB;00 and so on. The spring is stretched so that it is pulling the block of wood to the right. Draw free body diagrams for the situation below. Notice that we are pulling the bottom block out from beneath the top block. There is friction between the blocks! After you have drawn your FBDs, identify the 3rd law force pairs, as above. Answers Discuss in class 32 www.ck12.org Chapter 10. Problem Solving 1 CHAPTER 10 Problem Solving 1 In this lesson, students will learn how to", " solve dif\ufb01cult problems using Newton\u2019s 2nd law. In this lesson, students will learn how to solve dif\ufb01cult problems using Newton\u2019s 2nd law. Key Equations Common Forces 8 >>>>< >>>>: Fg = mg Gravity FN Normal force: acts perpendicular to surfaces FT Force of tension in strings and wires Fsp = kDx = Force of springDxfrom equilibrium Force Sums 8 >< >: Net force is the vector sum of all the forces Fnet = i Fi = ma Fnet,x = i Fix = max Horizontal components add also Fnet,y = i Fiy = may As do vertical ones Static and Kinetic Friction\u2019 ( fs \u00b5sjFNj Opposes potential motion of surfaces in contact fk = \u00b5kjFNj Opposes motion of surfaces in contact Ultimately, many of these \u201ccontact\u201d forces are due to attractive and repulsive electromagnetic forces between atoms in materials. Guidance Problem Solving for Newton\u2019s Laws, Step-By-Step 1. Figure out which object is \u201cof interest.\u201d a. If you\u2019re looking for the motion of a rolling cart, the cart is the object of interest. b. If the object of interest is not moving, that\u2019s OK, don\u2019t panic yet. c. Draw a sketch! This may help you sort out which object is which in your problem. 2. Identify all the forces acting on the object and draw them on object. (This is a free-body diagram \u2013FBD) a. If the object has mass and is near the Earth, the easiest (and therefore, \ufb01rst) force to write down is the force of gravity, pointing downward, with value mg. b. If the object is in contact with a \ufb02at surface, it means there is a normal force acting on the object. This normal force points away from and is perpendicular to the surface. c. There may be more than one normal force acting on an object. For instance, if you have a bologna sandwich, remember that the slice of bologna feels normal forces from both the slices of bread! d. If a rope, wire, or cord is pulling on the object in question, you\u2019ve found yourself a tension force. The direction of this force is in the same direction that the rope is pulling. e. Don\u2019t worry about any forces acting on other objects. For instance, if you have a bologna sandwich as your object of interest, and you\u2019re thinking about the forces acting on the slice of bologna, don\u2019t worry about the force of gravity acting on either piece of bread. 33 www.ck12.org f. Remember that Newton\u2019s 3rd Law, calling for \u201cequal and opposite forces,\u201d does not apply to a single object. None of your forces should be \u201cequal and opposite\u201d on the same object in the sense of Newton\u2019s 3rd Law. Third law pairs act on two different objects. g. Recall that scales (like a bathroom scale you weigh yourself on) read out the normal force acting on you, not your weight. If you are at rest on the scale, the normal force equals your weight. If you are accelerating up or down, the normal force had better be higher or lower than your weight, or you won\u2019t have an unbalanced force to accelerate you. h. Never include \u201cma\u201d as a force acting on an object. \u201cma\u201d is the result of the net force Fnet which is found by summing all the forces acting on your object of interest. 3. Determine how to orient your axes a. A good rule to generally follow is that you want one axis (usually the x-axis) to be parallel to the surface your object of interest is sitting on. b. If your object is on a ramp, tilt your axes so that the x-axis is parallel to the incline and the y-axis is perpendicular. In this case, this will force you to break the force of gravity on the object into its components. But by tilting your axes, you will generally have to break up fewer vectors, making the whole problem simpler. 4. Identify which forces are in the x direction, which are in the y direction, and which are at an angle. a. If a force is upward, make it in the ydirection and give it a positive sign. If it is downward, make it in the ydirection and give it a negative sign. b. Same thing applies for right vs. left in the xdirection. Make rightward forces positive. c. If forces are at an angle, draw them at an angle. A great example is that when a dog on a leash runs ahead, pulling you along, it\u2019s pulling both forward and down on your", " hand. d. Draw the free body diagram (FBD). e. Remember that the FBD is supposed to be helping you with your problem. For instance, if you forget a force, it\u2019ll be really obvious on your FBD. 5. Break the forces that are at angles into their x and y components a. Use right triangle trigonometry b. Remember that these components aren\u2019t new forces, but are just what makes up the forces you\u2019ve already identi\ufb01ed. c. Consider making a second FBD to do this component work, so that your \ufb01rst FBD doesn\u2019t get too messy. 6. Add up all the x forces and x components. a. Remember that all the rightward forces add with a plus (+) sign, and that all the leftward forces add with a minus () sign. b. Don\u2019t forget about the xcomponents of any forces that are at an angle! c. When you\u2019ve added them all up, call this \"the sum of all x forces\" or \"the net force in the xdirection.\" 7. Add up all the y forces and y components. a. Remember that all the upward forces add with a (+) sign, all the downward forces add with a () sign. b. Don\u2019t forget about the ycomponents of any forces that are at an angle! c. When you\u2019ve added them all up, call this \"the sum of all y forces\" or \"net force in the ydirection.\" 8. Use Newton\u2019s Laws twice. a. The sum of all xforces, divided by the mass, is the object\u2019s acceleration in the xdirection. b. The sum of all yforces, divided by the mass, is the object\u2019s acceleration in the ydirection. c. If you happen to know that the acceleration in the xdirection or ydirection is zero (say the object is just sitting on a table), then you can plug this in to Newton\u2019s 2nd Law directly. d. If you happen to know the acceleration, you can plug this in directly too. 9. Each body should have a FBD. a. Draw a separate FBD for each body. 34 www.ck12.org Chapter 10. Problem Solving 1 b. Set up a sum of forces equation based on the FBD for each body. c. Newton\u2019s Third Law will tell you which forces on different bodies are the same in magnitude. d. Your equations should equal your unknown variables at this point. Example 1 Question: Using the diagram below, \ufb01nd the net force on the block. The block weighs 3kg and the inclined plane has a coef\ufb01cient of friction of :6. Answer: The \ufb01rst step to solving a Newton\u2019s Laws problem is to identify the object in question. In our case, the block on the slope is the object of interest. Next, we need to draw a free-body diagram. To do this, we need to identify all of the forces acting on the block and their direction. The forces are friction, which acts in the negative x direction, the normal force, which acts in the positive y direction, and gravity, which acts in a combination of the negative y direction and the positive x direction. Notice that we have rotated the picture so that the majority of the forces acting on the block are along the y or x axis. This does not change the answer to the problem because the direction of the forces is still the same relative to each other. When we have determined our answer, we can simply rotate it back to the original position. Now we need to break down gravity (the only force not along one of the axises) into its component vectors (which do follow the axises). The x component of gravity : 9:8m=s2 cos60 = 4:9m=s2 The y component of gravity : 9:8m=s2 sin60 = 8:5m=s2 Yet these are only the acceleration of gravity so we need to multiply them by the weight of the block to get the force. F = ma = 3kg 4:9m=s2 = 14:7NF = ma = 3kg 8:5m=s2 = 25:5N 35 Now that we have solved for the force of the y-component of gravity we know the normal force (they are equal). Therefore the normal force is 25:5N. Now that we have the normal force and the coef\ufb01cient of static friction, we can \ufb01nd the force of friction. www.ck12.org Fs = \u00b5sFN = :6 25:5N = 15:3N The force of static friction is greater than the component of gravity that is forcing the block down the inclined plane. Therefore the force of friction will match the", " force of the x-component of gravity. So the net force on the block is net force in the x direction : xcomponent o f gravity z { }| 14:7N f orce o f f riction z { }| 14:7N net force in the y direction : 25:5N } {z | Normal Force 25:5N {z } | ycomponent o f gravity = 0N = 0N Therefore the net force on the block is 0N. Example 2 MEDIA Click image to the left for more content. MEDIA Click image to the left for more content. Watch this Explanation 36 www.ck12.org Simulation Chapter 10. Problem Solving 1 \u2022 http://simulations.ck12.org/FreeBody/ Time for Practice 1. Find the mass of the painting. The tension in the leftmost rope is 7:2 N, in the middle rope it is 16 N, and in the rightmost rope it is 16 N. 2. Find Brittany\u2019s acceleration down the frictionless waterslide in terms of her mass m, the angle q of the incline, and the acceleration of gravity g. 37 www.ck12.org 3. The physics professor holds an eraser up against a wall by pushing it directly against the wall with a completely horizontal force of 20 N. The eraser has a mass of 0:5 kg. The wall has coef\ufb01cients of friction \u00b5S = 0:8 and \u00b5K = 0:6: a. Draw a free body diagram for the eraser. b. What is the normal force FN acting on the eraser? c. What is the frictional force FS equal to? d. What is the maximum mass m the eraser could have and still not fall down? e. What would happen if the wall and eraser were both frictionless? 4. A tractor of mass 580 kg accelerates up a 10 incline from rest to a speed of 10 m=s in 4 s. For all of answers below, provide a magnitude and a direction. 38 www.ck12.org Chapter 10. Problem Solving 1 a. What net force Fnet has been applied to the tractor? b. What is the normal force, FN on the tractor? c. What is the force of gravity Fg on the tractor? d. What force has been applied to the tractor so that it moves uphill? e. What is the source of this force? 5. A heavy box (mass 25 kg) is dragged along the \ufb02oor by a kid at a 30 angle to the horizontal with a force of 80 N (which is the maximum force the kid can apply). a. Draw the free body diagram. b. What is the normal force FN? c. Does the normal force decrease or increase as the angle of pull increases? Explain. d. Assuming no friction, what is the acceleration of the box? e. Assuming it begins at rest, what is its speed after ten seconds? f. Is it possible for the kid to lift the box by pulling straight up on the rope? g. In the absence of friction, what is the net force in the xdirection if the kid pulls at a 30 angle? h. In the absence of friction, what is the net force in the xdirection if the kid pulls at a 45 angle? i. In the absence of friction, what is the net force in the xdirection if the kid pulls at a60 angle? j. The kid pulls the box at constant velocity at an angle of 30. What is the coef\ufb01cient of kinetic friction \u00b5K between the box and the \ufb02oor? k. The kid pulls the box at an angle of 30, producing an acceleration of 2 m=s2. What is the coef\ufb01cient of kinetic friction \u00b5K between the box and the \ufb02oor? 6. Spinal implant problem \u2014this is a real life bio-med engineering problem! 39 www.ck12.org Here\u2019s the situation: both springs are compressed by an amount xo. The rod of length L is \ufb01xed to both the top plate and the bottom plate. The two springs, each with spring constant k, are wrapped around the rod on both sides of the middle plate, but are free to move because they are not attached to the rod or the plates. The middle plate has negligible mass, and is constrained in its motion by the compression forces of the top and bottom springs. The medical implementation of this device is to screw the top plate to one vertebrae and the middle plate to the vertebrae directly below. The bottom plate is suspended in space. Instead of fusing broken vertebrates together, this implant allows movement somewhat analogous to the natural movement of functioning vertebrae. Below you will do the exact calculations that an engineer did to get this device patented and available for use at hospitals. a. Find the force, F,", " on the middle plate for the region of its movement 4x xo. Give your answer in terms of the constants given. ( Hint: In this region both springs are providing opposite compression forces.) b. Find the force, F, on the middle plate for the region of its movement 4x xo. Give your answer in terms of the constants given. ( Hint: In this region, only one spring is in contact with the middle plate.) c. Graph F vs. x. Label the values for force for the transition region in terms of the constants given. 7. You design a mechanism for lifting boxes up an inclined plane by using a vertically hanging mass to pull them, as shown in the \ufb01gure below. The pulley at the top of the incline is massless and frictionless. The larger mass, M, is accelerating downward with a measured acceleration a. The smaller masses are mA and mB ; the angle of the incline is q, and the coef\ufb01cient of kinetic friction between each of the masses and the incline has been measured and determined to be \u00b5K. a. Draw free body diagrams for each of the three masses. b. Calculate the magnitude of the frictional force on each of the smaller masses in terms of the given quantities. c. Calculate the net force on the hanging mass in terms of the given quantities. d. Calculate the magnitudes of the two tension forces TA and TB in terms of the given quantities. e. Design and state a strategy for solving for how long it will take the larger mass to hit the ground, assuming at this moment it is at a height h above the ground. Do not attempt to solve this: simply state the strategy for solving it. 40 www.ck12.org Chapter 10. Problem Solving 1 8. You build a device for lifting objects, as shown below. A rope is attached to the ceiling and two masses are allowed to hang from it. The end of the rope passes around a pulley (right) where you can pull it downward to lift the two objects upward. The angles of the ropes, measured with respect to the vertical, are shown. Assume the bodies are at rest initially. a. Suppose you are able to measure the masses m1 and m2 of the two hanging objects as well as the tension TC. Do you then have enough information to determine the other two tensions, TA and TB? Explain your reasoning. b. If you only knew the tensions TA and TC, would you have enough information to determine the masses m1 and m2? If so, write m1 and m2 in terms of TA and TC. If not, what further information would you require? 9. A stunt driver is approaching a cliff at very high speed. Sensors in his car have measured the acceleration and velocity of the car, as well as all forces acting on it, for various times. The driver\u2019s motion can be broken down into the following steps: Step 1: The driver, beginning at rest, accelerates his car on a horizontal road for ten seconds. Sensors show that there is a force in the direction of motion of 6000 N, but additional forces acting in the opposite direction with magnitude 1000 N. The mass of the car is 1250 kg. Step 2: Approaching the cliff, the driver takes his foot off of the gas pedal (There is no further force in the direction of motion.) and brakes, increasing the force opposing motion from 1000 N to 2500 N. This continues for \ufb01ve seconds until he reaches the cliff. Step 3: The driver \ufb02ies off the cliff, which is 44:1 m high and begins projectile motion. (a) Ignoring air resistance, how long is the stunt driver in the air? (b) For Step 1: i. Draw a free body diagram, naming all the forces on the car. ii. Calculate the magnitude of the net force. iii. Find the change in velocity over the stated time period. iv. Make a graph of velocity in the xdirection vs. time over the stated time period. v. Calculate the distance the driver covered in the stated time period. Do this by \ufb01nding the area under the curve in your graph of (iv). Then, check your result by using the equations for kinematics. (c) Repeat (b) for Step 2. (d) Calculate the distance that the stunt driver should land from the bottom of the cliff. Answers 1. 3:6 kg 41 www.ck12.org 2. g sin q 3. b.20 N c. 4:9 N d. 1:63 kg e. Eraser would slip down the wall 4. a. 1450 N b. 5600 N c. 5700 N d. Friction between the tires and the ground e. Fuel, engine, or equal and opposite reaction 5. b. 210 N c. no, the box is \ufffd", "\ufffd\ufffdat so the normal force doesn\u2019t change d. 2:8 m=s2 e.28 m=s f. no g. 69 N h. 57 N i. 40 N j. 0:33 k. 0:09 6. a. zero b. kx0 7. b. f1 = \u00b5km1g cos q; f2 = \u00b5km2g cos q c. Ma d. TA = (m1 + m2)(a + \u00b5 cos q) and TB = m2a + \u00b5m2 cos q e. Solve by using d = 1=2at2 and substituting h for d 8. a. Yes, because it is static and you know the angle and m1 b. Yes, TA and the angle gives you m1 and the angle and TC gives you m2; m1 = TA cos 25=g and m2 = TC cos 30=g 9. a. 3 seconds d. 90 m 42 Physics Unit 6 (Energy) Patrick Marshall Ck12 Science James H Dann, Ph.D. Say Thanks to the Authors Click http://www.ck12.org/saythanks (No sign in required) AUTHORS Patrick Marshall Ck12 Science James H Dann, Ph.D. CONTRIBUTORS Chris Addiego Antonio De Jesus L\u00f3pez Catherine Pavlov www.ck12.org To access a customizable version of this book, as well as other interactive content, visit www.ck12.org CK-12 Foundation is a non-pro\ufb01t organization with a mission to reduce the cost of textbook materials for the K-12 market both in the U.S. and worldwide. Using an open-content, web-based collaborative model termed the FlexBook\u00ae, CK-12 intends to pioneer the generation and distribution of high-quality educational content that will serve both as core text as well as provide an adaptive environment for learning, powered through the FlexBook Platform\u00ae. Copyright \u00a9 2014 CK-12 Foundation, www.ck12.org The names \u201cCK-12\u201d and \u201cCK12\u201d and associated logos and the terms \u201cFlexBook\u00ae\u201d and \u201cFlexBook Platform\u00ae\u201d (collectively \u201cCK-12 Marks\u201d) are trademarks and service marks of CK-12 Foundation and are protected by federal, state, and international laws. Any form of reproduction of this book in any format or medium, in whole or in sections must include the referral attribution link http://www.ck12.org/saythanks (placed in a visible location) in addition to the following terms. Except as otherwise noted, all CK-12 Content (including CK-12 Curriculum Material) is made available to Users in accordance with the Creative Commons Attribution-Non-Commercial 3.0 Unported (CC BY-NC 3.0) License (http://creativecommons.org/ licenses/by-nc/3.0/), as amended and updated by Creative Commons from time to time (the \u201cCC License\u201d), which is incorporated herein by this reference. Complete terms can be found at http://www.ck12.org/terms. Printed: February 4, 2014 iii Contents www.ck12.org 1 4 9 12 15 Contents 1 Kinetic Energy 2 Potential Energy 3 Conservation of Energy 4 Energy Problem Solving 5 Springs iv www.ck12.org Chapter 1. Kinetic Energy CHAPTER 1 Kinetic Energy \u2022 De\ufb01ne energy. \u2022 De\ufb01ne kinetic energy. \u2022 Given the mass and speed of an object, calculate its kinetic energy. \u2022 Solve problems involving kinetic energy. This military jet requires a tremendous amount of work done on it to get its speed up to takeoff speed in the short distance available on the deck of an aircraft carrier. Some of the work is done by the plane\u2019s own jet engines but work from a catapult is also necessary for takeoff. Kinetic Energy Energy is the ability to change an object\u2019s motion or position. Energy comes in many forms and each of those forms can be converted into any other form. A moving object has the ability to change another object\u2019s motion or position simply by colliding with it and this form of energy is called kinetic energy. The kinetic energy of an object can be calculated by the equation KE = 1 2 mv2, where m is the mass of the object and v is its velocity. The kinetic energy of a moving object is directly proportional to its mass and directly proportional to the square of its velocity. This means that an object with twice the mass and equal speed will have twice the kinetic energy while an object with equal mass and twice the speed will have quadruple the kinetic energy. The SI unit for kinetic energy (and all forms of energy) is kg m2 s2 which is equivalent to Joules, the same unit we use for work. The kinetic energy of an", " object can be changed by doing work on the object. The work done on an object equals the kinetic energy gain or loss by the object. This relationship is expressed in the work-energy theorem WNET = DKE. Example Problem: A farmer heaves a 7.56 kg bale of hay with a \ufb01nal velocity of 4.75 m/s. 1 (a) What is the kinetic energy of the bale? (b) The bale was originally at rest. How much work was done on the bale to give it this kinetic energy? www.ck12.org Solution: (a) KE = 1 2 mv2 = 1 2 (7:56 kg)(4:75)2 = 85:3 Joules (b) Work done = DKE = 85:3 Joules Example Problem: What is the kinetic energy of a 750. kg car moving at 50.0 km/h? Solution: 50:0 km h 1000 m km 1 h 3600 s = 13:9 m/s KE = 1 2 mv2 = 1 2 (750: kg)(13:9 m/s)2 = 72; 300 Joules Example Problem: How much work must be done on a 750. kg car to slow it from 100. km/h to 50.0 km/h? Solution: From the previous example problem, we know that the KE of this car when it is moving at 50.0 km/h If the same car is going twice as fast, its KE will be four times as great because KE is is 72,300 Joules. proportional to the square of the velocity. Therefore, when this same car is moving at 100. km/h, its KE is 289,200 Joules. Therefore, the work done to slow the car from 100. km/h to 50.0 km/h is (289; 200 Joules) (72; 300 Joules) = 217; 000 Joules. Summary \u2022 Energy is the ability to change an object\u2019s motion or position. \u2022 There are many forms of energy. \u2022 The energy of motion is called kinetic energy. \u2022 The formula for kinetic energy is KE = 1 2 mv2. \u2022 The work done on an object equals the kinetic energy gain or loss by the object, WNET = DKE. Practice The following video discusses kinetic energy. Use this resource to answer the questions that follow. http://www.youtube.com/watch?v=g157qwT1918 MEDIA Click image to the left for more content. 1. Potential energy is present in objects that are ______________. 2. Kinetic energy is present in objects that are ______________. 3. What formula is given for kinetic energy? Practice problems involving kinetic energy: http://www.physicsclassroom.com/Class/energy/u5l1c.cfm 2 www.ck12.org Review Chapter 1. Kinetic Energy 1. A comet with a mass of 7.85 1011 kg is moving with a velocity of 25,000 m/s. Calculate its kinetic energy. 2. A ri\ufb02e can shoot a 4.00 g bullet at a speed of 998 m/s. (a) Find the kinetic energy of the bullet. (b) What work is done on the bullet if it starts from rest? (c) If the work is done over a distance of 0.75 m, what is the average force on the bullet? (d) If the bullet comes to rest after penetrating 1.50 cm into a piece of metal, what is the magnitude of the force bringing it to rest? \u2022 energy: An indirectly observed quantity that is often understood as the ability of a physical system to do work. \u2022 kinetic energy: The energy an object has due to its motion. References 1. Courtesy of Mass Communication Specialist 3rd Class Torrey W. Lee, U.S. Navy. Jet Takeoff. Public Domain 3 CHAPTER 2 www.ck12.org Potential Energy \u2022 De\ufb01ne potential energy. \u2022 Solve problems involving gravitational potential energy. \u2022 Solve problems involving the conversion of potential energy to kinetic energy and vice versa. Shooting an arrow from a bow requires work done on the bow by the shooter\u2019s arm to bend the bow and thus produce potential energy. The release of the bow converts the potential energy of the bent bow into the kinetic energy of the \ufb02ying arrow. Potential Energy When an object is held above the earth, it has the ability to make matter move because all you have to do is let go of the object and it will fall of its own accord. Since energy is de\ufb01ned as the ability to make matter move, this object has energy. This type of energy is stored energy and is called potential energy. An object held in a stretched If the stretched rubber band is released, the object will move. A rubber band also contains this stored energy", ". pebble on a \ufb02exed ruler has potential energy because if the ruler is released, the pebble will \ufb02y. If you hold two positive charges near each other, they have potential energy because they will move if you release them. Potential If the chemical bonds are broken and allowed to form lower energy is also the energy stored in chemical bonds. 4 www.ck12.org Chapter 2. Potential Energy potential energy chemical bonds, the excess energy is released and can make matter move. This is frequently seen as increased molecular motion or heat. If a cannon ball is \ufb01red straight up into the air, it begins with a high kinetic energy. As the cannon ball rises, it slows down due to the force of gravity pulling it toward the earth. As the ball rises, its gravitational potential energy is increasing and its kinetic energy is decreasing. When the cannon ball reaches the top of its arc, it kinetic energy is zero and its potential energy is maximum. Then gravity continues to pull the cannon ball toward the earth and the ball will fall toward the earth. As it falls, its speed increases and its height decreases. Therefore, its kinetic energy increases as it falls and its potential energy decreases. When the ball returns to its original height, its kinetic energy will be the same as when it started upward. When work is done on an object, the work may be converted into either kinetic or potential energy. If the work results in motion, the work was converted into kinetic energy and if the work done resulted in change in position, the work was converted into potential energy. Work could also be spent overcoming friction and that work would be converted into heat but usually we will consider frictionless systems. If we consider the potential energy of a bent stick or a stretched rubber band, the potential energy can be calculated by multiplying the force exerted by the stick or rubber band times the distance over which the force will be exerted. The formula for calculating this potential energy looks exactly like the formula for calculating work done, that is W = Fd. The only difference is that work is calculated when the object actually moves under the force and potential energy is calculated when the system is at rest before any motion actually occurs. In the case of gravitational potential energy, the force exerted by the object is its weight and the distance it could travel would be its height above the earth. Since the weight of an object is calculated by W = mg, then gravitational potential energy can be calculate by PE = mgh, where m is the mass of the object, g is the acceleration due to gravity, and h is the height the object will fall. 5 Example Problem: A 3.00 kg object is lifted from the \ufb02oor and placed on a shelf that is 2.50 m above the \ufb02oor. (a) What was the work done in lifting the object? (b) What is the gravitational potential energy of the object sitting on the shelf? (c) If the object falls off the shelf and free falls to the \ufb02oor, what will its velocity be when it hits the \ufb02oor? Solution: weight of the object = mg = (3:00 kg)(9:80 m=s2) = 29:4 N www.ck12.org (a) W = Fd = (29:4 N)(2:50 m) = 73:5 J (b) PE = mgh = (3:00 kg)(9:80 m=s2)(2:50 m) = 73:5 J (c) KE = PE so 1 2 mv2 = 73:5 J s v = (2)(73:5 J) 3:00 kg = 7:00 m=s Example Problem: A pendulum is constructed from a 7.58 kg bowling ball hanging on a 3.00 m long rope. The ball is pulled back until the rope makes an angle of 45 with the vertical. (a) What is the potential energy of the ball? (b) If the ball is released, how fast will it be traveling at the bottom of its arc? Solution: You can use trigonometry to \ufb01nd the vertical height of the ball in the pulled back position. This vertical height is found to be 0.877 m. PE = mgh = (7:58 kg)(9:80 m=s2)(0:877 m) = 65:1 J When the ball is released, the PE will be converted into KE as the ball swings through the arc. KE = 1 s 2 mv2 = 65:1 J v = (2)(65:1 kg m2=s2) 7:58 kg = 4:14 m=s Summary \u2022 Stored energy is called potential energy. 6 www.ck12.org Chapter 2. Potential Energy \u2022 Energy may be stored by holding an object elevated in a gravitational \ufb01eld or by holding it while a force is attempting to move it. \u2022 Potential energy may be", " converted to kinetic energy. \u2022 The formula for gravitational potential energy is PE = mgh. \u2022 In the absence of friction or bending, work done on an object must become either potential energy or kinetic energy or both. Practice The following video discusses types of energy. Use this resource to answer the questions that follow. http://www.youtube.com/watch?v=nC6tT1wkXEc MEDIA Click image to the left for more content. 1. What is the de\ufb01nition of energy? 2. Name two types of potential energy. 3. How is energy transferred from one object to another? Potential and kinetic energy practice problems with solutions: http://www.physicsclassroom.com/Class/energy/U5L2bc.cfm Review 1. A 90.0 kg man climbs hand over hand up a rope to a height of 9.47 m. How much potential energy does he have at the top? 2. A 50.0 kg shell was \ufb01red from a cannon at earth\u2019s surface to a maximum height of 400. m. What is the potential energy at maximum height? 3. If the shell in problem #3 then fell to a height of 100. m, what was the loss of PE? 4. A person weighing 645 N climbs up a ladder to a height of 4.55 m. (a) What work does the person do? (b) What is the increase in gravitational potential energy? (c) Where does the energy come from to cause this increase in PE? \u2022 potential energy: Otherwise known as stored energy, is the ability of a system to do work due to its position or internal structure. For example, gravitational potential energy is a stored energy determined by an object\u2019s position in a gravitational \ufb01eld while elastic potential energy is the energy stored in a spring. References 1. Image copyright Skynavin, 2013. http://www.shutterstock.com. Used under license from Shutterstock.com 7 2. Image copyright Tribalium, 2013; modi\ufb01ed by CK-12 Foundation - Samantha Bacic. http://www.shutterstock. com. Used under license from Shutterstock.com 3. CK-12 Foundation - Samantha Bacic.. CC-BY-NC-SA 3.0 www.ck12.org 8 www.ck12.org Chapter 3. Conservation of Energy CHAPTER 3 Conservation of Energy \u2022 State the law of conservation of energy. \u2022 Describe a closed system. \u2022 Use the law of conservation of energy to solve problems. There are many energy conversions between potential and kinetic energy as the cars travel around this double looping roller coaster. Conservation of Energy The law of conservation of energy states that within a closed system, energy can change form, but the total amount of energy is constant. Another way of expressing the law of conservation of energy is to say that energy can neither be created nor destroyed. An important part of using the conservation of energy is selecting the system. In the In a closed system, conservation of energy just as in the conservation of momentum, the system must be closed. objects may not enter or leave, and it is isolated from external forces so that no work can be done on the system. In the analysis of the behavior of an object, you must make sure you have included everything in the system that is involved in the motion. For example, if you are considering a ball that is acted on by gravity, you must include the earth in your system. The kinetic energy of the ball considered by itself may increase and only when the earth is included in the system can you see that the increasing kinetic energy is balanced by an equivalent loss of potential energy. The sum of the kinetic energy and the potential energy of an object is often called the mechanical energy. Consider a box with a weight of 20.0 N sitting at rest on a shelf that is 2.00 m above the earth. The box has zero kinetic energy but it has potential energy related to its weight and the distance to the earth\u2019s surface. PE = mgh = (20:0 N)(2:00 m) = 40:0 J 9 www.ck12.org If the box slides off the shelf, the only force acting on the box is the force of gravity and so the box falls. We can calculate the speed of the box when it strikes the ground by several methods. We can calculate the speed directly 2 = 2ad. We can also \ufb01nd the \ufb01nal velocity by setting the kinetic energy at the bottom of the using the formula v f fall equal to the potential energy at the top, KE = PE and, 1 2 = 2gh. You may note these formulas 2 mv2 = mgh, so v f are essentially the same. q (2)(9:80 m=s2)(2:00 m) = 6:26 m=s v = Example Problem: Suppose a cannon is sitting on top of", " a 50.0 m high hill and a 5.00 kg cannon ball is \ufb01red with a velocity of 30.0 m/s at some unknown angle. What will be the velocity of the cannon ball when it strikes the earth? Solution: Since the angle at which the cannon ball is \ufb01red is unknown, we cannot use the usual equations from projectile motion. However, at the moment the cannon ball is \ufb01red, it has a certain KE due to the mass of the ball and its speed and it has a certain PE due to its mass and it height above the earth. Those two quantities of energy can be calculated. When the ball returns to the earth, its PE will be zero and therefore, its KE at that point must account for the total of its original KE + PE. This gives us a method of solving the problem. ETOTAL = KE + PE = 1 ETOTAL = 2250 J + 2450 J = 4700 J (5:00 kg)(30:0 m=s)2 + (5:00 kg)(9:80 m=s2)(50:0 m) 2 mv2 + mgh = 1 2 1 2 mv f 2 = 4700 J so v f = s (2)(4700 J) 5:00 kg = 43:4 m=s Example Problem: A 2.00 g bullet moving at 705 m/s strikes a 0.250 kg block of wood at rest on a frictionless surface. The bullet sticks in the wood and the combined mass moves slowly down the table. (a) What is the KE of the bullet before the collision? (b) What is the speed of the combination after the collision? (c) How much KE was lost in the collision? Solution: (a) KEBULLET = 1 (b) mBvB + mW vW = (mB+W )(vB+W ) 2 mv2 = 1 2 (0:00200 kg)(705 m=s)2 = 497 J (0:00200 kg)(705 m=s) + (0:250 kg)(0 m=s) = (0:252 kg)(V ) (1:41 kg m=s) = (0:252 kg)(V ) V = 5:60 m=s 2 mv2 = 1 (c) KECOMBINATION = 1 KELOST = KEBEFORE KEAFTER = 497 J 4 J = 493 J 2 (0:252 kg)(5:60 m=s)2 = 3:95 J Summary \u2022 In a closed system, energy may change forms but the total amount of energy is constant. Practice The following video demonstrates Newton Ball tricks. Use this resource to answer the questions that follow. http://www.youtube.com/watch?v=JadO3RuOJGU 10 www.ck12.org Chapter 3. Conservation of Energy MEDIA Click image to the left for more content. 1. What happens when one ball is pulled up to one side and released? 2. What happens when three balls are pulled up to one side and released? 3. What happens when two balls are pulled out from each side and released? Practice problems with answers for the law of conservation of energy: http://www.physicsclassroom.com/class/energy/u5l2bc.cfm Review 1. A 15.0 kg chunk of ice falls off the top of an iceberg. If the chunk of ice falls 8.00 m to the surface of the water, (a) what is the kinetic energy of the chunk of ice when its hits the water, and (b) what is its velocity? 2. An 85.0 kg cart is rolling along a level road at 9.00 m/s. The cart encounters a hill and coasts up the hill. (a) Assuming the movement is frictionless, at what vertical height will the cart come to rest? (b) Do you need to know the mass of the cart to solve this problem? 3. A circus performer swings down from a platform on a rope tied to the top of a tent in a pendulum-like swing. The performer\u2019s feet touch the ground 9.00 m below where the rope is tied. How fast is the performer moving at the bottom of the arc? 4. A skier starts from rest at the top of a 45.0 m hill, coasts down a 30slope into a valley, and continues up to the top of a 40.0 m hill. Both hill heights are measured from the valley \ufb02oor. Assume the skier puts no effort into the motion (always coasting) and there is no friction. (a) How fast will the skier be moving on the valley \ufb02oor? (b) How fast will the skier be moving on the top of the 40.0 m hill? 5", ". A 2.00 kg ball is thrown upward at some unknown angle from the top of a 20.0 m high building. If the initial magnitude of the velocity of the ball is 20.0 m/s, what is the magnitutde of the \ufb01nal velocity when it strikes the ground? Ignore air resistance. 6. If a 2.00 kg ball is thrown straight upward with a KE of 500 J, what maximum height will it reach? Neglect air resistance. \u2022 conservation of energy: An empirical law of physics (meaning it cannot be derived), states that the total amount of energy within an isolated system is constant. Although energy can be transformed from one form into another, energy cannot be created or destroyed \u2022 closed system: Means it cannot exchange any of heat, work, or matter with the surroundings. \u2022 mechanical energy: The sum of potential energy and kinetic energy. References 1. User:Zonk43/Wikimedia Commons. http://commons.wikimedia.org/wiki/File:Teststrecke_Roller_Coaster3.J PG. Public Domain 11 CHAPTER 4 Energy Problem Solving Students will learn how to analyze and solve more complicated problems involving energy conservation. Students will learn how to analyze and solve more complicated problems involving energy conservation. www.ck12.org Key Equations Einitial = E \ufb01nal ; T hetotalenergydoesnotchangeinclosedsystems KE = 1 2 mv2 ; Kinetic energy PEg = mgh ; Potential energy of gravity PEsp = 1 2 kx2; Potential energy of a spring W = FxDx = Fd cos q ; Work is equal to the distance multiplied by the component of the force in the direction it is moving. Guidance The main thing to always keep prescient in your mind is that the total energy before must equal the total energy after. If some energy has transferred out of or into the system via work, you calculate that work done and include it in the energy sum equation. Generally work done by friction is listed on the \u2019after\u2019 side and work put into the system, via a jet pack for example, goes on the \u2019before\u2019 side. Another important point is that on turns or going over hills or in rollercoaster loops, one must include the centripetal motion equations -for example to insure that you have enough speed to make the loop. Example 1 Watch this Explanation 12 MEDIA Click image to the left for more content. MEDIA Click image to the left for more content. www.ck12.org Time for Practice Chapter 4. Energy Problem Solving 1. A rock with mass m is dropped from a cliff of height h. What is its speed when it gets to the bottom of the cliff? a. b. p r p mg 2g h 2gh c. d. gh e. None of the above 2. In the picture above, a 9:0 kg baby on a skateboard is about to be launched horizontally. The spring constant is 300 N=m and the spring is compressed 0:4 m. For the following questions, ignore the small energy loss due to the friction in the wheels of the skateboard and the rotational energy used up to make the wheels spin. a. What is the speed of the baby after the spring has reached its uncompressed length? b. After being launched, the baby encounters a hill 7 m high. Will the baby make it to the top? If so, what is his speed at the top? If not, how high does he make it? c. Are you \ufb01nally convinced that your authors have lost their minds? Look at that picture! 3. When the biker is at the top of the ramp shown above, he has a speed of 10 m=s and is at a height of 25 m. The bike and person have a total mass of 100 kg. He speeds into the contraption at the end of the ramp, which slows him to a stop. a. What is his initial total energy? (Hint: Set Ug = 0 at the very bottom of the ramp.) b. What is the length of the spring when it is maximally compressed by the biker? (Hint: The spring does not compress all the way to the ground so there is still some gravitational potential energy. It will help to draw some triangles.) 4. An elevator in an old apartment building in Switzerland has four huge springs at the bottom of the shaft to cushion its fall in case the cable breaks. The springs have an uncompressed height of about 1 meter. Estimate the spring constant necessary to stop this elevator, following these steps: a. First, guesstimate the mass of the elevator with a few passengers inside. b. Now, estimate the height of a \ufb01ve-story building. c. Lastly, use conservation of energy to estimate the spring constant. 13 5. You are skiing down a hill. You", " start at rest at a height 120 m above the bottom. The slope has a 10:0 grade. Assume the total mass of skier and equipment is 75:0 kg. www.ck12.org a. Ignore all energy losses due to friction. What is your speed at the bottom? b. If, however, you just make it to the bottom with zero speed what would be the average force of friction, including air resistance? 6. Two horri\ufb01c contraptions on frictionless wheels are compressing a spring (k = 400 N=m) by 0:5 m compared to its uncompressed (equilibrium) length. Each of the 500 kg vehicles is stationary and they are connected by a string. The string is cut! Find the speeds of the vehicles once they lose contact with the spring. 7. A roller coaster begins at rest 120 m above the ground, as shown. Assume no friction from the wheels and air, and that no energy is lost to heat, sound, and so on. The radius of the loop is 40 m. a. If the height at point G is 76 m, then how fast is the coaster going at point G? b. Does the coaster actually make it through the loop without falling? (Hint: You might review the material from centripetal motion lessons to answer this part.) Answers to Selected Problems 1.. 2. a. 2:3 m=s c. No, the baby will not clear the hill. 3. a. 29; 500 J b. Spring has maximum compressed length of 13 m 4.. 5. a. 48:5 m=s b. 128 N 6. 0:32 m=s each 7. a.29 m/s b. just barely, aC = 9:8 m=s2 14 www.ck12.org CHAPTER 5 Chapter 5. Springs Springs Students will learn to calculate periods, frequencies, etc. of spring systems in harmonic motion. Students will learn to calculate periods, frequencies, etc. of spring systems in harmonic motion. Key Equations T = 1 f ; Period is the inverse of frequency Tspring = 2p r m k ; Period of mass m on a spring with constant k Fsp = kx ; the force of a spring equals the spring constant multiplied by the amount the spring is stretched or compressed from its equilibrium point. The negative sign indicates it is a restoring force (i.e. direction of the force is opposite its displacement from equilibrium position. Usp = 1 that it is stretched or compressed from equilibrium 2 kx2 ; the potential energy of a spring is equal to one half times the spring constant times the distance squared Guidance \u2022 The oscillating object does not lose any energy in SHM. Friction is assumed to be zero. \u2022 In harmonic motion there is always a restorative force, which attempts to restore the oscillating object to its equilibrium position. The restorative force changes during an oscillation and depends on the position of the object. In a spring the force is given by Hooke\u2019s Law: F = kx \u2022 The period, T, is the amount of time needed for the harmonic motion to repeat itself, or for the object to go one full cycle. In SHM, T is the time it takes the object to return to its exact starting point and starting direction. \u2022 The frequency, f ; is the number of cycles an object goes through in 1 second. Frequency is measured in Hertz (Hz). 1 Hz = 1 cycle per sec. \u2022 The amplitude, A, is the distance from the equilibrium (or center) point of motion to either its lowest or highest point (end points). The amplitude, therefore, is half of the total distance covered by the oscillating object. The amplitude can vary in harmonic motion, but is constant in SHM. Example 1 MEDIA Click image to the left for more content. 15 www.ck12.org Watch this Explanation Simulation MEDIA Click image to the left for more content. Mass & Springs (PhET Simulation) Time for Practice 1. A rope can be considered as a spring with a very high spring constant k; so high, in fact, that you don\u2019t notice the rope stretch at all before it \u201cpulls back.\u201d a. What is the k of a rope that stretches by 1 mm when a 100 kg weight hangs from it? b. If a boy of 50 kg hangs from the rope, how far will it stretch? c. If the boy kicks himself up a bit, and then is bouncing up and down ever so slightly, what is his frequency of oscillation? Would he notice this oscillation? If so, how? If not, why not? 2. If a 5:0 kg mass attached to a spring oscillates 4.0 times every second, what is the spring constant k of the spring? 3. A horizontal spring attached to the wall is attached to a block of wood on the other end", ". All this is sitting on a frictionless surface. The spring is compressed 0:3 m. Due to the compression there is 5:0 J of energy stored in the spring. The spring is then released. The block of wood experiences a maximum speed of 25 m=s. a. Find the value of the spring constant. b. Find the mass of the block of wood. c. What is the equation that describes the position of the mass? d. What is the equation that describes the speed of the mass? e. Draw three complete cycles of the block\u2019s oscillatory motion on an x vs. t graph. 16 www.ck12.org Chapter 5. Springs 4. A spider of 0:5 g walks to the middle of her web. The web sinks by 1:0 mm due to her weight. You may assume the mass of the web is negligible. a. If a small burst of wind sets her in motion, with what frequency will she oscillate? b. How many times will she go up and down in one s? In 20 s? c. How long is each cycle? d. Draw the x vs t graph of three cycles, assuming the spider is at its highest point in the cycle at t = 0 s. Answers to Selected Problems 1. a. 9:8 105 N=m b. 0:5 mm c. 22 Hz 2. 3:2 103 N=m 3. a. 110 N=m d. v(t) = (25) cos(83t) 4. a. 16 Hz b. 16 complete cycles but 32 times up and down, 315 complete cycles but 630 times up and down c. 0:063 s Investigation 1. Your task: Match the period of the circular motion system with that of the spring system. You are only allowed to change the velocity involved in the circular motion system. Consider the effective distance between the block and the pivot to be to be \ufb01xed at 1m. The spring constant(13.5N/m) is also \ufb01xed. You should view the charts to check whether you have succeeded. Instructions: To alter the velocity, simply click on the Select Tool, and select the pivot. The Position tab below will allow you to numerically adjust the rotational speed using the Motor \ufb01eld. To view the graphs of their respective motion in order to determine if they are in sync, click on Chart tab below. 2. MEDIA Click image to the left for more content. 3. Now the mass on the spring has been replaced by a mass that is twice the rotating mass. Also, the distance between the rotating mass and the pivot has been changed to 1.5 m. What velocity will keep the period the same now? 4. MEDIA Click image to the left for more content. 17 Physics Unit 7 (Work, Power, Kinetic Energy Theorem) Patrick Marshall Ck12 Science Say Thanks to the Authors Click http://www.ck12.org/saythanks (No sign in required) www.ck12.org To access a customizable version of this book, as well as other interactive content, visit www.ck12.org AUTHORS Patrick Marshall Ck12 Science CK-12 Foundation is a non-pro\ufb01t organization with a mission to reduce the cost of textbook materials for the K-12 market both in the U.S. and worldwide. Using an open-content, web-based collaborative model termed the FlexBook\u00ae, CK-12 intends to pioneer the generation and distribution of high-quality educational content that will serve both as core text as well as provide an adaptive environment for learning, powered through the FlexBook Platform\u00ae. Copyright \u00a9 2014 CK-12 Foundation, www.ck12.org The names \u201cCK-12\u201d and \u201cCK12\u201d and associated logos and the terms \u201cFlexBook\u00ae\u201d and \u201cFlexBook Platform\u00ae\u201d (collectively \u201cCK-12 Marks\u201d) are trademarks and service marks of CK-12 Foundation and are protected by federal, state, and international laws. Any form of reproduction of this book in any format or medium, in whole or in sections must include the referral attribution link http://www.ck12.org/saythanks (placed in a visible location) in addition to the following terms. Except as otherwise noted, all CK-12 Content (including CK-12 Curriculum Material) is made available to Users in accordance with the Creative Commons Attribution-Non-Commercial 3.0 Unported (CC BY-NC 3.0) License (http://creativecommons.org/ licenses/by-nc/3.0/), as amended and updated by Creative Commons from time to time (the \u201cCC License\u201d), which is incorporated herein by this reference. Complete terms can be found at http://www.ck12.org/terms. Printed: April 6", ", 2014 iii Contents www.ck12.org Contents 1 Work 2 Power 3 Work-Energy Principle 1 4 7 iv www.ck12.org CONCEPT 1 Concept 1. Work Work \u2022 De\ufb01ne work. \u2022 Identify forces that are doing work. \u2022 Given two of the three variables in the equation, W = Fd, calculate the third. In order for the roller coaster to run down the incline by gravitational attraction, it \ufb01rst must have work done on it towing it up to the top of the hill. The work done on the coaster towing it to the top of the hill becomes potential energy stored in the coaster and that potential energy is converted to kinetic energy as the coaster runs down from the top of the hill to the bottom. Work The word work has both an everyday meaning and a speci\ufb01c scienti\ufb01c meaning. In the everyday use of the word, work would refer to anything which required a person to make an effort. In physics, however, work is de\ufb01ned as the force exerted on an object multiplied by the distance the object moves due to that force. W = Fd In the scienti\ufb01c de\ufb01nition of the word, if you push against an automobile with a force of 200 N for 3 minutes but the automobile does not move, then you have done NO work. Multiplying 200 N times 0 meters yields zero work. If you are holding an object in your arms, the upward force you are exerting is equal to the object\u2019s weight. If 1 www.ck12.org you hold the object until your arms become very tired, you have still done no work because you did not move the object in the direction of the force. When you lift an object, you exert a force equal to the object\u2019s weight and the If an object weighs 200. N and you lift it 1.50 meters, then your work is, object moves due to that lifting force. W = Fd = (200: N)(1:50 m) = 300: N m. One of the units you will see for work is the Newton\u00b7meter but since a Newton is also a kilogram\u00b7m/s 2, then a Newton\u00b7meter is also kg\u00b7m 2 /s 2. This unit has also been named the Joule (pronounced Jool) in honor of James Prescott Joule, a nineteenth century English physicist. Example Problem: A boy lifts a box of apples that weighs 185 N. The box is lifted a height of 0.800 m. How much work did the boy do? Solution: W = Fd = (185 N)(0:800 m) = 148 N m = 148 Joules Work is done only if a force is exerted in the direction of motion. If the force and motion are perpendicular to each other, no work is done because there is no motion in the direction of the force. If the force is at an angle to the motion, then the component of the force in the direction of the motion is used to determine the work done. Example Problem: Suppose a 125 N force is applied to a lawnmower handle at an angle of 25\u00b0 with the ground and the lawnmower moves along the surface of the ground. If the lawnmower moves 56 m, how much work was done? Solution: The solution requires that we determine the component of the force that was in the direction of the motion of the lawnmower. The component of the force that was pushing down on the ground does not contribute to the work done. Fparallel = (Force)(cos 25) = (125 N)(0:906) = 113 N W = Fparalleld = (113 N)(56 m) = 630 J Summary \u2022 In physics, work is de\ufb01ned as the force exerted on an object multiplied by the distance the object moves due to that force. \u2022 The unit for work is called the joule. \u2022 If the force is at an angle to the motion, then the component of the force in the direction of the motion is used to determine the work done. Practice The following video introduces energy and work. Use this resource to answer the questions that follow. http://wn.com/Work_physics_#/videos 2 www.ck12.org Concept 1. Work MEDIA Click image to the left for more content. 1. What de\ufb01nition is given in the video for energy? 2. What is the de\ufb01nition given in the video for work? 3. What unit is used in the video for work? The following website contains practice questions with answers on the topic of work. http://www.sparknotes.com/testprep/books/sat2/physics/chapter7section6.rhtml Review 1. How much work is done by the force of gravity when a 45 N object falls to the ground from", " a height of 4.6 m? 2. A workman carries some lumber up a staircase. The workman moves 9.6 m vertically and 22 m horizontally. If the lumber weighs 45 N, how much work was done by the workman? 3. A barge is pulled down a canal by a horse walking beside the canal. exerted is 400. N, and the barge is pulled 100. M, how much work did the horse do? If the angle of the rope is 60.0\u00b0, the force \u2022 work: A force is said to do work when it acts on a body so that there is a displacement of the point of application, however small, in the direction of the force. Thus a force does work when it results in movement. The work done by a constant force of magnitude F on a point that moves a distance d in the direction of the force is the product, W = Fd. \u2022 joule: The SI unit of work or energy, equal to the work done by a force of one Newton when its point of application moves through a distance of one meter in the direction of the force. References 1. Image copyright Paul Brennan, 2013. http://www.shutterstock.com. Used under license from Shutter- stock.com 2. CK-12 Foundation - Samantha Bacic.. CC BY-NC-SA 3.0 3. CK-12 Foundation - Samantha Bacic.. CC-BY-NC-SA 3.0 3 CONCEPT 2 \u2022 De\ufb01ne power. \u2022 Given two of the three variables in P = W t, calculate the third. www.ck12.org Power Typical Pressurized Water Reactors (PWR) reactors built in the 1970\u2019s produce about 1100 megawatts, whilst the latest designs range up to around 1500 megawatts. That is 1,500,000,000 Joules/second. A windmill farm, by comparison, using hundreds of individual windmills produces about 5 megawatts. That is 5,000,000 Joules/second (assuming the wind is blowing). Power Power is de\ufb01ned as the rate at which work is done or the rate at which energy is transformed. In SI units, power is measured in Joules per second which is given a special name, the watt, W. Power = Work Time 1.00 watt = 1.00 J/s 4 www.ck12.org Concept 2. Power Another unit for power that is fairly common is horsepower. 1.00 horsepower = 746 watts Example Problem: A 70.0 kg man runs up a long \ufb02ight of stairs in 4.0 s. The vertical height of the stairs is 4.5 m. Calculate the power output of the man in watts and horsepower. Solution: The force exerted must be equal to the weight of the man = mg = (70:0 kg)(9:80 m/s2) = 686 N W = Fd = (686 N)(4:5 m) = 3090 N m = 3090 J 4:0 s = 770 J/s = 770 W P = W t = 3090 J P = 770 W = 1:03 hp Since P = W that is produced by the power. t = F d P = W t = Fd t = Fv t and W = Fd, we can use these formulas to derive a formula relating power to the speed of the object The velocity in this formula is the average speed of the object during the time interval. Example Problem: Calculate the power required of a 1400 kg car if the car climbs a 10\u00b0 hill at a steady 80. km/h. Solution: 80. km/h = 22.2 m/s In 1.00 s, the car would travel 22.2 m on the road surface but the distance traveled upward would be (22.2 m)(sin 10\u00b0) = (22.2 m)(0.174) = 3.86 m. The force in the direction of the upward motion is the weight of the car = (1400 kg)(9.80 m/s 2 ) = 13720 N. W = Fd = (13720 N)(3:86 m) = 53; 000 J Since this work was done in 1.00 second, the power would be 53,000 W. If we calculated the upward component of the velocity of the car, we would divide the distance traveled in one second by one second and get an average vertical speed of 3.86 m/s. So we could use the formula relating power to average speed to calculate power. P = Fv = (13720 N)(3:86 m/s) = 53; 000 W Summary \u2022 Power is de\ufb01ned as the rate at which work is done or the rate at which energy is transformed. \u2022 Power = Work Time \u2022 Power = Force velocity Practice In the following video, Mr. Edmond sings about work and", " power. Use this resource to answer the questions that follow. http://www.youtube.com/watch?v=5EsMmdaYClQ 5 www.ck12.org MEDIA Click image to the left for more content. 1. What units are used for work? 2. What units are used for power? The following website has practice problems on work and power. http://www.angel\ufb01re.com/sci\ufb01/dschlott/workpp.html Review 1. If the circumference of an orbit for a toy on a string is 18 m and the centripetal force is 12 N, how much work does the centripetal force do on the toy when it follows its orbit for one cycle? 2. A 50.0 kg woman climbs a \ufb02ight of stairs 6.00 m high in 15.0 s. How much power does she use? 3. (a) Assuming no friction, what is the minimum work needed to push a 1000. kg car 45.0 m up a 12.5\u00b0 incline? (b) What power would be needed for the same problem if friction is considered and the coef\ufb01cient of friction for the car is 0.30? \u2022 power: The rate at which this work is performed. References 1. Courtesy of Ryan Hagerty, U.S. Fish and Wildlife Service. http://digitalmedia.fws.gov/cdm/singleitem/coll ection/natdiglib/id/13455/rec/2. Public Domain 6 www.ck12.org Concept 3. Work-Energy Principle CONCEPT 3 Work-Energy Principle The reason the concept of work is so useful is because of a theorem, called the work-energy principle, which states that the change in an object\u2019s kinetic energy is equal to the net work done on it : DKe = Wnet [2] Although we cannot derive this principle in general, we can do it for the case that interests us most: constant acceleration. In the following derivation, we assume that the force is along motion. This doesn\u2019t reduce the generality of the result, but makes the derivation more tractable because we don\u2019t need to worry about vectors or angles. Recall that an object\u2019s kinetic energy is given by the formula: Ke = 1 2 mv2 [3] Consider an object of mass m accelerated from a velocity vi to v f under a constant force. The change in kinetic energy, according to [2], is equal to: DKe = Kei Ke f = 1 2 mv2 f 1 2 mv2 i = 1 2 m(v2 f v2 i ) [4] Now let\u2019s see how much work this took. To \ufb01nd this, we need to \ufb01nd the distance such an object will travel under these conditions. We can do this by using the third of our \u2019Big three\u2019 equations, namely: alternatively, v f 2 = vi 2 + 2aDx [5] Dx = v f 2 2 vi 2a [6] Plugging in [6] and Newton\u2019s Third Law, F = ma, into [2], we \ufb01nd: W = FDx = ma v f 2 2 vi 2a which was our result in [4]. Using the Work-Energy Principle = 1 2 m(v2 f v2 i ) [7], The Work-Energy Principle can be used to derive a variety of useful results. Consider, for instance, an object dropped a height Dh under the in\ufb02uence of gravity. This object will experience constant acceleration. Therefore, we can again use equation [6], substituting gravity for acceleration and Dh for distance: 7 www.ck12.org multiplying both sides by $mg$, we \ufb01nd: Dh = v f 2 2 vi 2g mgDh = mg v f 2 2 vi 2g = DKe [8] In other words, the work performed on the object by gravity in this case is mgDh. We refer to this quantity as gravitational potential energy; here, we have derived it as a function of height. For most forces (exceptions are friction, air resistance, and other forces that convert energy into heat), potential energy can be understood as the ability to perform work. Spring Force A spring with spring constant k a distance Dx from equilibrium experiences a restorative force equal to: This is a force that can change an object\u2019s kinetic energy, and therefore do work. So, it has a potential energy associated with it as well. This quantity is given by: Fs = kDx [9] Esp = 1 2 kDx2 [10] Spring Potential Energy The derivation of [10] is left to the reader. Hint: \ufb01nd the average force an object experiences while", " moving from x = 0 to x = Dx while attached to a spring. The net work is then this force times the displacement. Since this quantity (work) must equal to the change in the object\u2019s kinetic energy, it is also equal to the potential energy of the spring. This derivation is very similar to the derivation of the kinematics equations \u2014 look those up. This applet may be useful in reviewing Spring Potential Energy: http://phet.colorado.edu/en/simulation/mass-spring-lab 8 Physics Unit 8 Momentum Patrick Marshall Jean Brainard, Ph.D. James H Dann, Ph.D. Ck12 Science Say Thanks to the Authors Click http://www.ck12.org/saythanks (No sign in required) AUTHORS Patrick Marshall Jean Brainard, Ph.D. James H Dann, Ph.D. Ck12 Science CONTRIBUTORS Chris Addiego Alexander Katsis Antonio De Jesus L\u00f3pez www.ck12.org To access a customizable version of this book, as well as other interactive content, visit www.ck12.org CK-12 Foundation is a non-pro\ufb01t organization with a mission to reduce the cost of textbook materials for the K-12 market both in the U.S. and worldwide. Using an open-content, web-based collaborative model termed the FlexBook\u00ae, CK-12 intends to pioneer the generation and distribution of high-quality educational content that will serve both as core text as well as provide an adaptive environment for learning, powered through the FlexBook Platform\u00ae. Copyright \u00a9 2014 CK-12 Foundation, www.ck12.org The names \u201cCK-12\u201d and \u201cCK12\u201d and associated logos and the terms \u201cFlexBook\u00ae\u201d and \u201cFlexBook Platform\u00ae\u201d (collectively \u201cCK-12 Marks\u201d) are trademarks and service marks of CK-12 Foundation and are protected by federal, state, and international laws. Any form of reproduction of this book in any format or medium, in whole or in sections must include the referral attribution link http://www.ck12.org/saythanks (placed in a visible location) in addition to the following terms. Except as otherwise noted, all CK-12 Content (including CK-12 Curriculum Material) is made available to Users in accordance with the Creative Commons Attribution-Non-Commercial 3.0 Unported (CC BY-NC 3.0) License (http://creativecommons.org/ licenses/by-nc/3.0/), as amended and updated by Creative Commons from time to time (the \u201cCC License\u201d), which is incorporated herein by this reference. Complete terms can be found at http://www.ck12.org/terms. Printed: April 6, 2014 iii Contents www.ck12.org Contents 1 Momentum 2 Momentum 3 Law of Conservation of Momentum 4 Conservation of Momentum in One Dimension 5 Conservation of Momentum in Two Dimensions 6 Inelastic Collisions 7 Elastic Collisions 8 Momentum and Impulse 1 4 7 10 14 18 22 26 iv www.ck12.org Concept 1. Momentum CONCEPT 1 \u2022 De\ufb01ne momentum. \u2022 Relate momentum to mass and velocity. \u2022 Calculate momentum from mass and velocity. Momentum Cody seems a little reluctant to launch himself down this ramp at Newton\u2019s Skate Park. It will be his \ufb01rst time down the ramp, and he knows from watching his older brother Jerod that he\u2019ll be moving fast by the time he gets to the bottom. The faster he goes, the harder it will be to stop. That\u2019s because of momentum. What Is Momentum? Momentum is a property of a moving object that makes it hard to stop. The more mass it has or the faster it\u2019s moving, the greater its momentum. Momentum equals mass times velocity and is represented by the equation: 1 www.ck12.org Momentum = Mass x Velocity Q : What is Cody\u2019s momentum as he stands at the top of the ramp? A : Cody has no momentum as he stands there because he isn\u2019t moving. However, Cody will gain momentum as he starts moving down the ramp and picks up speed. In other words, his velocity is zero. Q : Cody\u2019s older brother Jerod is pictured 1.1. If Jerod were to travel down the ramp at the same velocity as Cody, who would have greater momentum? Who would be harder to stop? A : Jerod obviously has greater mass than Cody, so he would have greater momentum. He would also be harder to stop. FIGURE 1.1 You can see an animation demonstrating the role of mass and velocity in the momentum of moving objects at this URL: http://www.science-animations.com/support-", "files/momentum.swf Calculating Momentum To calculate momentum with the equation above, mass is measured in (kg), and velocity is measured in meters per second (m/s). For example, Cody and his skateboard have a combined mass of 40 kg. If Cody is traveling at a velocity of 1.1 m/s by the time he reaches the bottom of the ramp, then his momentum is: Momentum = 40 kg x 1.1 m/s = 44 kg m/s Note that the SI unit for momentum is kg m/s. Q : The combined mass of Jerod and his skateboard is 68 kg. If Jerod goes down the ramp at the same velocity as Cody, what is his momentum at the bottom of the ramp? A : His momentum is: Momentum = 68 kg x 1.1 m/s = 75 kg m/s Summary \u2022 Momentum is a property of a moving object that makes it hard to stop. It equals the object\u2019s mass times its velocity. 2 www.ck12.org Concept 1. Momentum \u2022 To calculate the momentum of a moving object, multiply its mass in kilograms (kg) by its velocity in meters per second (m/s). The SI unit of momentum is kg m/s. Vocabulary \u2022 momentum : Property of a moving object that makes it hard to stop; equal to the object\u2019s mass times its velocity. Practice At the following URL, review how to calculate momentum, and then solve the problems at the bottom of the Web page. http://www2.franciscan.edu/academic/mathsci/mathscienceintegation/MathScienceIntegation-848.htm Review 1. De\ufb01ne momentum. 2. Write the equation for calculating momentum from mass and velocity. 3. What is the SI unit for momentum? 4. Which skateboarder has greater momentum? a. Skateboarder A: mass = 60 kg; velocity = 1.5 m/s b. Skateboarder B: mass = 50 kg; velocity = 2.0 m/s References 1... used under license from Shutterstock 3 CONCEPT 2 www.ck12.org Momentum Students will learn what momentum is and how to calculate momentum of objects. In addition, students will learn how to use conservation of momentum to solve basic problems. Students will learn what momentum is and how to calculate momentum of objects. In addition, students will learn how to use conservation of momentum to solve basic problems. Key Equations p = mv Momentum is equal to the objects mass multiplied by its velocity pinitial = p \ufb01nal The total momentum does not change in closed systems Example 1 A truck with mass 500 kg and originally carrying 200 kg of dirt is rolling forward with the transmission in neutral and shooting out the dirt backwards at 2 m/s (so that the dirt is at relative speed of zero compared with the ground). If the truck is originally moving at 2 m/s, how fast will it be moving after it has shot out all the dirt. You may ignore the effects of friction. To solve this problem we will apply conservation of momentum to the truck when it is full of dirt and when it has dumped all the dirt. Solution mivi = m f v f (mt + md)vi = mtv f start by setting the initial momentum equal to the \ufb01nal momentum substitute the mass of the truck plus the mass of the dirt in the truck at the initial and \ufb01nal states v f = v f = (mt + md)vi mt (500 kg + 200 kg) 2 m/s 500 kg v f = 2:8 m/s solve for the \ufb01nal velocity plug in the numerical values Example 2 John and Bob are standing at rest in middle of a frozen lake so there is no friction between their feet and the ice. Both of them want to get to shore so they simultaneously push off each other in opposite directions. If John\u2019s mass is 50 kg and Bob\u2019s mass is 40 kg and John moving at 5 m/s after pushing off Bob, how fast is Bob moving? 4 www.ck12.org Concept 2. Momentum Solution For this problem, we will apply conservation of momentum to the whole system that includes both John and Bob. Since both of them are at rest to start, we know that the total momentum of the whole system must always be zero. Therefore, we know that the sum of John\u2019s and Bob\u2019s momentum after they push off each other is also zero. We can use this to solve for Bob\u2019s velocity. 0 = m jv j + mbvb mbvb = m jv j vb = m jv j mb 50 kg 5 m/s 40 kg vb = 6:25 m/s vb = The answer is negative because Bob is traveling in the opposite direction to", " John. Watch this Explanation MEDIA Click image to the left for more content. Time for Practice 1. You \ufb01nd yourself in the middle of a frozen lake. There is no friction between your feet and the ice of the lake. You need to get home for dinner. Which strategy will work best? a. Press down harder with your shoes as you walk to shore. b. Take off your jacket. Then, throw it in the direction opposite to the shore. c. Wiggle your butt until you start to move in the direction of the shore. d. Call for help from the great Greek god Poseidon. 2. You and your sister are riding skateboards side by side at the same speed. You are holding one end of a rope and she is holding the other. Assume there is no friction between the wheels and the ground. If your sister lets go of the rope, how does your speed change? a. It stays the same. b. It doubles. c. It reduces by half. 5 www.ck12.org 3. You and your sister are riding skateboards (see Problem 3), but now she is riding behind you. You are holding one end of a meter stick and she is holding the other. At an agreed time, you push back on the stick hard enough to get her to stop. What happens to your speed? Choose one. (For the purposes of this problem pretend you and your sister weigh the same amount.) a. It stays the same. b. It doubles. c. It reduces by half. 4. An astronaut is using a drill to \ufb01x the gyroscopes on the Hubble telescope. Suddenly, she loses her footing and \ufb02oats away from the telescope. What should she do to save herself? 5. A 5:00 kg \ufb01recracker explodes into two parts: one part has a mass of 3:00 kg and moves at a velocity of 25:0 m=s towards the west. The other part has a mass of 2:00 kg. What is the velocity of the second piece as a result of the explosion? 6. A \ufb01recracker lying on the ground explodes, breaking into two pieces. One piece has twice the mass of the other. What is the ratio of their speeds? 7. While driving in your pickup truck down Highway 280 between San Francisco and Palo Alto, an asteroid lands in your truck bed! Despite its 220 kg mass, the asteroid does not destroy your 1200 kg truck. In fact, it landed perfectly vertically. Before the asteroid hit, you were going 25 m=s. After it hit, how fast were you going? 8. An astronaut is 100 m away from her spaceship doing repairs with a 10:0 kg wrench. The astronaut\u2019s total mass is 90:0 kg and the ship has a mass of 1:00 104 kg. If she throws the wrench in the opposite direction of the spaceship at 10:0 m=s how long would it take for her to reach the ship? Answers to Selected Problems 1.. 2.. 3.. 4.. 5. 37:5 m=s 6. v1 = 2v2 7. 21 m=s 8. a. 90 sec 6 www.ck12.org Concept 3. Law of Conservation of Momentum CONCEPT 3 Law of Conservation of Momentum \u2022 State the law of conservation of momentum. \u2022 Describe an example of momentum being transferred and conserved. These skaters are racing each other at Newton\u2019s Skate Park. The \ufb01rst skater in line, the one on the left, is distracted by something he sees. He starts to slow down without realizing it. The skater behind him isn\u2019t paying attention and keeps skating at the same speed. Q : Can you guess what happens next? A : Skater 2 runs into skater 1. Conserving Momentum When skater 2 runs into skater 1, he\u2019s going faster than skater 1 so he has more momentum. Momentum is a property of a moving object that makes it hard to stop. It\u2019s a product of the object\u2019s mass and velocity. At the moment of the collision, skater 2 transfers some of his momentum to skater 1, who shoots forward when skater 2 runs into him. Whenever an action and reaction such as this occur, momentum is transferred from one object to the other. However, the combined momentum of the objects remains the same. In other words, momentum is conserved. This is the law of conservation of momentum. 7 Modeling Momentum The Figure 3.1 shows how momentum is conserved in the two colliding skaters. The total momentum is the same after the collision as it was before. However, after the collision, skater 1 has more momentum and skater 2 has less momentum than before. www.ck12.org FIGURE 3.1 Q : What if two sk", "aters have a head-on collision? Do you think momentum is conserved then? A : As in all actions and reactions, momentum is also conserved in a head-on collision. You can see how at this URL: http://www.physicsclassroom.com/mmedia/momentum/cthoi.cfm Summary \u2022 Whenever an action and reaction occur, momentum is transferred from one object to the other. However, total momentum is conserved. This is the law of conservation of momentum. 8 www.ck12.org Vocabulary Concept 3. Law of Conservation of Momentum \u2022 momentum : Property of a moving object that makes it hard to stop; equal to the object\u2019s mass times its velocity. \u2022 law of conservation of momentum : Law stating that, when an action and reaction occur, the combined momentum of the objects remains the same. Practice Watch the astropitch animation at the following URL. Experiment with different velocities. Then take the quiz and check your answers. http://www.phys.utb.edu/~pdukes/standard/PhysApplets/AstroPitch/TabbedastroPitch2.html Review 1. State the law of conservation of momentum. 2. Fill in the missing velocity (x) in the diagram of a vehicle collision seen in the Figure 3.2 so that momentum is conserved. FIGURE 3.2 Solve for x. References 1. Laura Guerin.. CC BY-NC 3.0 2. Laura Guerin.. CC BY-NC 3.0 9 CONCEPT 4 Conservation of Momentum in One Dimension www.ck12.org \u2022 State the law of conservation of momentum. \u2022 Use the conservation of momentum to solve one-dimensional collision problems. For this whale to leap out of the water, something underwater must be moving in the opposite direction, and intuition tells us it must be moving with relatively high velocity. The water that moves downward is pushed downward by the whale\u2019s tail, and that allows the whale to rise up. Conservation of Momentum in One Dimension When impulse and momentum were introduced, we used an example of a batted ball to discuss the impulse and momentum change that occurred with the ball. At the time, we did not consider what had happened to the bat. According to Newton\u2019s third law, however, when the bat exerted a force on the ball, the ball also exerted an equal and opposite force on the bat. Since the time of the collision between bat and ball is the same for the bat and for the ball, then we have equal forces (in opposite directions) exerted for equal times on the ball AND the bat. That means that the impulse exerted on the bat is equal and opposite (-Ft) to the impulse on the ball (Ft) and that also means that there was a change in momentum of the bat [D(mv)BAT] that was equal and opposite to the change in momentum of the ball [D(mv)BALL]. The change in momentum of the ball is quite obvious because it changes direction and \ufb02ies off at greater speed. However, the change in momentum of the bat is not obvious at all. This occurs primarily because the bat is more massive than the ball. Additionally, the bat is held \ufb01rmly by the batter, so the batter\u2019s mass can be combined with the mass of the bat. Since the bat\u2019s mass is so much greater than that of the ball, but they have equal and opposite forces, the bat\u2019s \ufb01nal velocity is signi\ufb01cantly smaller than that of the ball. Consider another system: that of two ice skaters. If we have one of the ice skaters exert a force on the other skater, the force is called an internal force because both the object exerting the force and the object receiving the force 10 www.ck12.org Concept 4. Conservation of Momentum in One Dimension are inside the system. In a closed system such as this, momentum is always conserved. The total \ufb01nal momentum always equals the total initial momentum in a closed system. Conversely, if we de\ufb01ned a system to contain just one ice skater, putting the other skater outside the system, this is not a closed system. If one skater pushes the other, the force is an external force because the receiver of the force is outside the system. Momentum is not guaranteed to be conserved unless the system is closed. In a closed system, momentum is always conserved. Take another example: if we consider two billiard balls colliding on a billiard table and ignore friction, we are dealing with a closed system. The momentum of ball A before the collision plus the momentum of ball B before collision will equal the momentum of ball A after collision plus the momentum of ball B after collision. This is called the law of conservation of momentum and is given by the", " equation pAbefore + pBbefore = pAafter + pBafter Example Problem: Ball A has a mass of 2.0 kg and is moving due west with a velocity of 2.0 m/s while ball B has a mass of 4.0 kg and is moving west with a velocity of 1.0 m/s. Ball A overtakes ball B and collides with it from behind. After the collision, ball A is moving westward with a velocity of 1.0 m/s. What is the velocity of ball B after the collision? Solution: Because of the law of conservation of momentum, we know that pAbefore + pBbefore = pAafter + pBafter. mAvA + mBvB = mAv0 A + mBv0 B (2:0 kg)(2:0 m/s) + (4:0 kg)(1:0 m/s) = (2:0 kg)(1:0 m/s) + (4:0 kg)(vB 0 m/s) 4:0 kg m/s + 4:0 kg m/s = 2:0 kg m/s + 4vB 0 kg m/s 4vB 0 = 8:0 2:0 = 6:0 0 = 1:5 m/s vB After the collision, ball B is moving westward at 1.5 m/s. Example Problem: A railroad car whose mass is 30,000. kg is traveling with a velocity of 2.2 m/s due east and collides with a second railroad car whose mass is also 30,000. kg and is at rest. If the two cars stick together after the collision, what is the velocity of the two cars? Solution: Note that since the two trains stick together, the \ufb01nal mass is m A +m B, and the \ufb01nal velocity for each object is the same. Thus the conservation of momentum equation, mAvA + mBvB = mAv0 B, can be A + mBv0 rewritten mAvA + mBvB = (mA + mB) v0 (30; 000: kg)(2:2 m/s) + (30; 000: kg)(0 m/s) = (60; 000: kg)(v0 m/s) 66000 + 0 = 60000v0 v0 = 66000 60000 = 1:1 m/s After the collision, the two cars move off together toward the east with a velocity of 1.1 m/s. Summary \u2022 A closed system is one in which both the object exerting a force and the object receiving the force are inside the system. \u2022 In a closed system, momentum is always conserved. 11 Practice www.ck12.org The following video shows the Mythbusters building and using various sizes of a toy called \"Newton\u2019s Cradle.\" Use this resource to answer the question that follows. https://www.youtube.com/watch?v=BiLq5Gnpo8Q MEDIA Click image to the left for more content. 1. What is Newton\u2019s Cradle? 2. How does Newton\u2019s Cradle work? 3. How does a Newton\u2019s Cradle show conservation of momentum? If you are interested in learning more about Newton\u2019s Cradles, visit this site: http://science.howstuffworks.com/ne wtons-cradle.htm Review 1. A 0.111 kg hockey puck moving at 55 m/s is caught by a 80. kg goalie at rest. With what speed does the goalie slide on the (frictionless) ice? 2. A 0.050 kg bullet strikes a 5.0 kg stationary wooden block and embeds itself in the block. The block and the bullet \ufb02y off together at 9.0 m/s. What was the original velocity of the bullet? 3. A 0.50 kg ball traveling at 6.0 m/s due east collides head on with a 1.00 kg ball traveling in the opposite direction at -12.0 m/s. After the collision, the 0.50 kg ball moves away at -14 m/s. Find the velocity of the second ball after the collision. 4. Two carts are stationary with a compressed spring between them and held together by a thread. When the thread is cut, the two carts move apart. After the spring is released, one cart m = 3:00 kg has a velocity of 0.82 m/s east. What is the magnitude of the velocity of the second cart (m = 1:70 kg) after the spring is released? 5. Compared to falling on a tile \ufb02oor, a glass may not break if it falls onto a carpeted \ufb02oor. This is because a. less impulse in stopping. b", ". longer time to stop. c. both of these d. neither of these. 6. A butter\ufb02y is hit by a garbage truck on the highway. The force of the impact is greater on the 12 www.ck12.org Concept 4. Conservation of Momentum in One Dimension a. garbage truck. b. butter\ufb02y. c. it is the same for both. 7. A ri\ufb02e recoils from \ufb01ring a bullet. The speed of the ri\ufb02e\u2019s recoil is small compared to the speed of the bullet because a. the force on the ri\ufb02e is small. b. the ri\ufb02e has a great deal more mass than the bullet. c. the momentum of the ri\ufb02e is unchanged. d. the impulse on the ri\ufb02e is less than the impulse on the bullet. e. none of these. \u2022 Law of Conservation of Momentum: The total linear momentum of an isolated system remains constant regardless of changes within the system. References 1. Courtesy of NOAA. http://commons.wikimedia.org/wiki/File:Humpback_whale_noaa.jpg. Public Domain 2. CK-12 Foundation - Samantha Bacic.. CC-BY-NC-SA 3.0 13 CONCEPT 5 Conservation of Momentum in Two Dimensions www.ck12.org \u2022 Use the conservation of momentum and vector analysis to solve two-dimensional collision problems. \u2022 Review vector components. In a game of billiards, it is important to be able to visualize collisions in two dimensions \u2013 the best players not only know where the target ball is going but also where the cue ball will end up. Conservation of Momentum in Two Dimensions Conservation of momentum in all closed systems is valid, regardless of the directions of the objects before and after they collide. Most objects are not con\ufb01ned to a single line, like trains on a rail. Rather, many objects, like billiard balls or cars, can move in two dimensions. Conservation of momentum for these objects can also be calculated; momentum is a vector and collisions of objects in two dimensions can be represented by axial vector components. To review axial components, revisit Vectors: Resolving Vectors into Axial Components and Vectors: Vector Addition. Example Problem: A 2.0 kg ball, A, is moving with a velocity of 5.00 m/s due west. It collides with a stationary ball, B, also with a mass of 2.0 kg. After the collision, ball A moves off at 30\u00b0 south of west while ball B moves off at 60\u00b0 north of west. Find the velocities of both balls after the collision. Solution: Since ball B is stationary before the collision, then the total momentum before the collision is equal to momentum of ball A. The momentum of ball A before collision is shown in red below, and can be calculated to be p = mv = (2:00 kg)(5:00 m/s) = 10:0 kg m/s west 14 www.ck12.org Concept 5. Conservation of Momentum in Two Dimensions Since momentum is conserved in this collision, the sum of the momenta of balls A and B after collsion must be 10.0 kg m/s west. pAafter = (10:0 kg m/s)(cos 30) = (10:0 kg m/s)(0:866) = 8:66 kg m/s pBafter = (10:0 kg m/s)(cos 60) = (10:0 kg m/s)(0:500) = 5:00 kg m/s To \ufb01nd the \ufb01nal velocities of the two balls, we divide the momentum of each by its mass. Therefore, vA = 4:3 m/s and vB = 2:5 m/s. Example Problem: A 1325 kg car moving north at 27.0 m/s collides with a 2165 kg car moving east at 17.0 m/s. The two cars stick together after the collision. What is the speed and direction of the two cars after the collision? Solution: Northward momentum = (1325 kg)(27:0 m/s) = 35800 kg m/s Eastward momentum = (2165 kg)(17:0 m/s) = 36800 kg m/s q (35800)2 + (36800)2 = 51400 kg m/s R = 15 www.ck12.org q = sin1 35800 m = 51400 kgm/s 3490 kg velocity = p 51400 = 44 north of east = 14:7 m/s @ 44 N of E Example Problem: A 6.00 kg ball, A, moving at velocity", " 3.00 m/s due east collides with a 6.00 kg ball, B, at rest. After the collision, A moves off at 40.0\u00b0 N of E and ball B moves off at 50.0\u00b0 S of E. a. What is the momentum of A after the collision? b. What is the momentum of B after the collision? c. What are the velocities of the two balls after the collision? Solution: pinitial = mv = (6:00 kg)(3:00 m/s) = 18:0 kg m/s This is a right triangle in which the initial momentum is the length of the hypotenuse and the two momenta after the collision are the legs of the triangle. a. pA = (18:0 kg m/s)(cos 40:0) = (18:0 kg m/s)(0:766) = 13:8 kg m/s b. pB = (18:0 kg m/s)(cos 50:0) = (18:0 kg m/s)(0:643) = 11:6 kg m/s c. vA = 2:30 m/s vB = 1:93 m/s \u2022 The conservation of momentum law holds for all closed systems regardless of the directions of the objects before and after they collide. \u2022 Momentum is a vector; collisions in two dimensions can be represented by axial vector components. Summary Practice This video shows circus performers using conservation of momentum. Use this resource to answer the questions that follow. http://www.pbs.org/opb/circus/classroom/circus-physics/angular-momentum/ 16 www.ck12.org Concept 5. Conservation of Momentum in Two Dimensions MEDIA Click image to the left for more content. 1. Why do the \ufb02iers scrunch up in the air while spinning and twisting? 2. What happens to the rate at which they spin when they change shape in the air? Review 1. Billiard ball A, mass 0.17 kg, moving due east with a velocity of 4.0 m/s, strikes stationary billiard ball B, also mass of 0.17 kg. After the collision, ball A moves off at an angle of 30\u00b0 north of east with a velocity of 3.5 m/s, and ball B moves off at an angle of 60 \u00b0 south of east. What is the speed of ball B? 2. A bomb, originally sitting at rest, explodes and during the explosion breaks into four pieces of exactly 0.25 kg each. One piece \ufb02ies due south at 10 m/s while another pieces \ufb02ies due north at 10 m/s. (a) What do we know about the directions of the other two pieces and how do we know it? (b) What do we know about the speeds of the other two pieces and how do we know it? 3. In a head-on collision between protons in a particle accelerator, three resultant particles were observed. All three of the resultant particles were moving to the right from the point of collision. The physicists conducting the experiment concluded there was at least one unseen particle moving to the left after the collision. Why did they conclude this? References 1. Image copyright VitCOM Photo, 2013. http://www.shutterstock.com. Used under license from Shutter- stock.com 2. CK-12 Foundation - Samantha Bacic.. CC-BY-NC-SA 3.0 3. CK-12 Foundation - Samantha Bacic.. CC-BY-NC-SA 3.0 4. CK-12 Foundation - Samantha Bacic.. CC-BY-NC-SA 3.0 17 CONCEPT 6 Inelastic Collisions Students will learn how to solve problems involving inelastic collisions. www.ck12.org Key Equations pinitial = p \ufb01nal The total momentum does not change in closed systems Example 1 Question : Two blocks collide on a frictionless surface. Afterwards, they have a combined mass of 10kg and a speed of 2:5m=s. Before the collision, block A, which has a mass of 8.0kg, was at rest. What was the mass and initial speed of block B? Solution : To \ufb01nd mass of block B we have a simple subtraction problem. We know that the combined mass is 10kg and the mass of block A is 8.0kg. 10kg 8:0kg = 2:0kg Now that we know the mass of both blocks we can \ufb01nd the speed of block B. We will use conservation of momentum. This was a completely inelastic collision. We know this because the blocks stuck together after the collision. This problem is one dimensional, because all motion happens along the same line. Thus we will use the equation (mA + mB)v f = mA v", "A + mB vB and solve for the velocity of block B. (mA + mB)v f = mA vA + mBvB ) (mA + mB)(v f ) (mA)(vA) mB = vB Watch this Explanation 18 www.ck12.org Concept 6. Inelastic Collisions MEDIA Click image to the left for more content. MEDIA Click image to the left for more content. Simulation Note: move the elasticity meter to 0% for perfectly inelastic collisions. Collision Lab (PhET Simulation) 19 www.ck12.org Car Collision (CK-12 Simulation) Time for Practice 2. Two blocks collide on a frictionless surface, as shown. Afterwards, they have a combined mass of 10 kg and a speed of 2:5 m=s. Before the collision, one of the blocks was at rest. This block had a mass of 8:0 kg. What was the mass and initial speed of the second block? 1. 3. 4. In the above picture, the carts are moving on a level, frictionless track. After the collision all three carts stick together. Determine the direction and speed of the combined carts after the collision. 5. The train engine and its four boxcars are coasting at 40 m=s. The engine train has mass of 5; 500 kg and the boxcars have masses, from left to right, of 1; 000 kg, 1; 500 kg, 2; 000 kg; and 3; 000 kg. (For this problem, you may neglect the small external forces of friction and air resistance.) 20 www.ck12.org Concept 6. Inelastic Collisions a. What happens to the speed of the train when it releases the last boxcar? ( Hint: Think before you blindly calculate. ) b. If the train can shoot boxcars backwards at 30 m=s relative to the train\u2019s speed, how many boxcars does the train need to shoot out in order to obtain a speed of 58:75 m=s? 6. In Sacramento a 4000 kg SUV is traveling 30 m=s south on Truxel crashes into an empty school bus, 7000 kg traveling east on San Juan. The collision is perfectly inelastic. a. Find the velocity of the wreck just after collision b. Find the direction in which the wreck initially moves 7. Manrico (80:0 kg) and Leonora (60:0 kg) are \ufb01gure skaters. They are moving toward each other. Manrico\u2019s speed is 2:00 m=s ; Leonora\u2019s speed is 4:00 m=s. When they meet, Leonora \ufb02ies into Manrico\u2019s arms. a. With what speed does the entwined couple move? b. In which direction are they moving? c. How much kinetic energy is lost in the collision? Answers to Selected Problems 1. 2:0 kg; 12:5 m=s 2. 0:13 m=s to the left 3. a. no change b. the last two cars 4. a. 15 m=s b. 49 S of E 5. a. 0.57 m/s b. the direction Leonora was originally travelling c. 297.26 J 21 CONCEPT 7 www.ck12.org Elastic Collisions Students will learn how to solve problems involving elastic collisions Key Equations pinitial = p \ufb01nal The total momentum does not change in closed systems KEinitial = KE \ufb01nal The total kinetic energy does not change in elastic collisions Example 1 MEDIA Click image to the left for more content. Example 2 Question : Chris and Ashley are playing pool. Ashley hits the cue ball into the 8 ball with a velocity of 1:2m=s. The cue ball (c) and the 8 ball ( e ) react as shown in the diagram. The 8 ball and the cue ball both have a mass of :17kg. What is the velocity of the cue ball? What is the direction (the angle) of the cue ball? Answer : We know the equation for conservation of momentum, along with the masses of the objects in question as well two of the three velocities. Therefore all we need to do is manipulate the conservation of momentum equation so that it is solved for the velocity of the cue ball after the collision and then plug in the known values to get the velocity of the cue ball. mcvic + mevie = mcv f c + mev f e v f c = v f c = mcvic + mevie mev f e mc :17kg 2:0m=s + :17kg 0m=s :17kg 1:2m=s :17kg v f c = :80m=s Now we want to \ufb01nd the direction of the cue ball. To do this we", " will use the diagram below. 22 www.ck12.org Concept 7. Elastic Collisions We know that the momentum in the y direction of the two balls is equal. Therefore we can say that the velocity in the y direction is also equal because the masses of the two balls are equal. mcvcy = mevey! vcy = vey Given this and the diagram, we can \ufb01nd the direction of the cue ball. After 1 second, the 8 ball will have traveled 1:2m. Therefore we can \ufb01nd the distance it has traveled in the y direction. sin25o = opposite hypotenuse = x 1:2m! x = sin25 1:2m = :51m Therefore, in one second the cue ball will have traveled :51m in the y direction as well. We also know how far in total the cue ball travels in one second (:80m). Thus we can \ufb01nd the direction of the cue ball. sin1 opposite hypotenuse = sin1 :51m :80m = 40o Watch this Explanation MEDIA Click image to the left for more content. Simulation Note: move the elasticity meter to 100% for perfectly elastic collisions. 23 www.ck12.org Collision Lab (PhET Simulation) Time for Practice 1. You are playing pool and you hit the cue ball with a speed of 2 m=s at the 8 -ball (which is stationary). Assume an elastic collision and that both balls are the same mass. Find the speed and direction of both balls after the collision, assuming neither \ufb02ies off at any angle. 2. A 0:045 kg golf ball with a speed of 42:0 m=s collides elastically head-on with a 0:17 kg pool ball at rest. Find the speed and direction of both balls after the collision. 3. Ball A is traveling along a \ufb02at table with a speed of 5:0 m=s, as shown below. Ball B, which has the same mass, is initially at rest, but is knocked off the table in an elastic collision with Ball A. Find the horizontal distance that Ball B travels before hitting the \ufb02oor. 4. Students are doing an experiment on the lab table. A steel ball is rolled down a small ramp and allowed to hit the \ufb02oor. Its impact point is carefully marked. Next a second ball of the same mass is put upon a set screw and a collision takes place such that both balls go off at an angle and hit the \ufb02oor. All measurements are taken with a meter stick on the \ufb02oor with a co-ordinate system such that just below the impact point is the origin. The following data is collected: (a) no collision: 41:2 cm (b) target ball: 37:3 cm in the direction of motion and 14:1 cm perpendicular to the direction of motion 24 www.ck12.org Concept 7. Elastic Collisions i. From this data predict the impact position of the other ball. ii. One of the lab groups declares that the data on the \ufb02oor alone demonstrate to a 2 % accuracy that the collision was elastic. Show their reasoning. iii. Another lab group says they can\u2019t make that determination without knowing the velocity the balls have on impact. They ask for a timer. The instructor says you don\u2019t need one; use your meter stick. Explain. iv. Design an experiment to prove momentum conservation with balls of different masses, giving apparatus, procedure and design. Give some sample numbers. 5. A 3 kg ball is moving 2 m/s in the positive x direction when it is struck dead center by a 2 kg ball moving in the positive y direction at 1 m/s. After collision the 3 kg ball moves at 3 m/s 30 degrees from the positive x axis. Find the velocity and direction of the 2 kg ball. Answers to Selected Problems 1. 8 m=s same direction as the cue ball and 0 m=s 2. vgol f = 24:5 m=s; vpool = 17:6 m=s 3. 2:8 m 4.. 5. 1:5 m=s 54 25 CONCEPT 8 Momentum and Impulse www.ck12.org \u2022 De\ufb01ne momentum. \u2022 De\ufb01ne impulse. \u2022 Given mass and velocity of an object, calculate momentum. \u2022 Calculate the change in momentum of an object. \u2022 State the relationship that exists between the change in momentum and impulse. \u2022 Using the momentum-impulse theorem and given three of the four variables, calculate the fourth. Rachel Flatt performs a layback spin at the 2 011 Rostelecom Cup in Moscow, Russia. When an ice skater spins, angular momentum must be conserved. When her arms or feet are far away from her body, her spin slows; when she brings her arms and feet close", " in to her body, she spins faster. Momentum and Impulse If a bowling ball and a ping-pong ball are each moving with a velocity of 5 mph, you intuitively understand that it will require more effort to stop the bowling ball than the ping pong ball because of the greater mass of the bowling ball. Similarly, if you have two bowling balls, one moving at 5 mph and the other moving at 10 mph, you know it 26 www.ck12.org Concept 8. Momentum and Impulse will take more effort to stop the ball with the greater speed. It is clear that both the mass and the velocity of a moving object contribute to what is necessary to change the motion of the moving object. The product of the mass and velocity of an object is called its momentum. Momentum is a vector quantity that has the same direction as the velocity of the object and is represented by a lowercase letter p. The momentum of a 0.500 kg ball moving with a velocity of 15.0 m/s will be p = mv p = mv = (0:500 kg)(15:0 m/s) = 7:50 kg m/s You should note that the units for momentum are kg\u00b7m/s. According to Newton\u2019s \ufb01rst law, the velocity of an object cannot change unless a force is applied. If we wish to change the momentum of a body, we must apply a force. The longer the force is applied, the greater the change in momentum. The impulse is the quantity de\ufb01ned as the force multiplied by the time it is applied. It is a vector quantity that has the same direction as the force. The units for impulse are N\u00b7s but we know that Newtons are also kg\u00b7m/s 2 and so N\u00b7s = (kg\u00b7m/s 2 )(s) = kg\u00b7m/s. Impulse and momentum have the same units; when an impulse is applied to an object, the momentum of the object changes and the change of momentum is equal to the impulse. Ft = Dmv Example Problem: Calculating Momentum A 0.15 kg ball is moving with a velocity of 35 m/s. Find the momentum of the ball. Solution: p = mv = (0:15 kg)(35 m/s) = 5:25 kg m/s Example Problem: If a ball with mass 5.00 kg has a momentum of 5:25 kg m/s, what is its velocity? m = 5:25 kgm/s 5:00 kg It should be clear from the equation relating impulse to change in momentum, Ft = Dmv, that any amount of force would (eventually) bring a moving object to rest. If the force is very small, it must be applied for a long time, but a greater force can bring the object to rest in a shorter period of time. Solution: v = p = 1:05 m/s If you jump off a porch and land on your feet with your knees locked in the straight position, your motion would be brought to rest in a very short period of time and thus the force would need to be very large \u2013 large enough, perhaps, to damage your joints or bones. Suppose that when you hit the ground, your velocity was 7.0 m/s and that velocity was brought to rest in 0.05 seconds. If your mass is 100. kg, what force was required to bring you to rest? 0:050 s If, on the other hand, when your feet \ufb01rst touched the ground, you allowed your knees to \ufb02ex so that the period of time over which your body was brought to rest is increased, then the force on your body would be smaller and it would be less likely that you would damage your legs. F = Dmv t = (100: kg)(7:0 m/s) = 14; 000 N Suppose that when you \ufb01rst touch the ground, you allow your knees to bend and extend the stopping time to 0.50 seconds. What force would be required to bring you to rest this time? t = (100: kg)(7:0 m/s) 0:50 s With the longer period of time for the force to act, the necessary force is reduced to one-tenth of what was needed before. = 1400 N F = Dmv Extending the period of time over which a force acts in order to lessen the force is a common practice in design. Padding in shoes and seats allows the time to increase. The front of automobiles are designed to crumple in an accident; this increases the time the car takes to stop. Similarly, barrels of water or sand in front of abutments on 27 www.ck12.org the highway and airbags serve to slow down the stoppage time. These changes all serve to decrease the amount of force it takes to stop the momentum", " in a car crash, which consequently saves lives. Example Problem: An 0.15 kg baseball is thrown horizontally at 40. m/s and after it is struck by a bat, it is traveling at -40. m/s. (a) What impulse did the bat deliver to the ball? (b) If the contact time of the bat and bat was 0.00080 seconds, what was the average force the bat exerted on the ball? (c) Calculate the average acceleration of the ball during the time it was in contact with the bat. Solution: We can calculate the change in momentum and give the answer as impulse because we know that the impulse is equal to the change in momentum. (a) p = mDv = (0:15 kg)(40: m/s 40: m/s) = (0:15 kg)(80: m/s) = 12 kg m/s The minus sign indicates that the impulse was in the opposite direction of the original throw. t = 12 kgm/s Again, the negative sign indicates the force was in the opposite direction of the original throw. 0:00080 s = 15000 N (b) F = Dmv (c) a = F = 100; 000 m/s2 m = 15000 N 0:15 kg Summary \u2022 The product of the mass and velocity of an object is called momentum, given by the equation r = mv. \u2022 Momentum is a vector quantity that has the same direction as the velocity of the object. \u2022 The quantity of force multiplied by the time it is applied is called impulse. \u2022 Impulse is a vector quantity that has the same direction as the force. \u2022 Momentum and impulse have the same units: kg\u00b7m/s. \u2022 The change of momentum of an object is equal to the impulse. Ft = Dmv Practice Use this resource to answer the question that follows. https://www.youtube.com/watch?v=3g4v8x7xggU MEDIA Click image to the left for more content. 1. Why don\u2019t the glasses of water spill when the tablecloth is pulled out from under them? 2. How does the video get from momentum to impulse? 28 www.ck12.org Concept 8. Momentum and Impulse Review 1. A small car with a mass of 800. kg is moving with a velocity of 27.8 m/s. (a) What is the momentum of the car? (b) What velocity is needed for a 2400. kg car in order to have the same momentum? 2. A scooter has a mass of 250. kg. A constant force is exerted on it for 60.0 s. During the time the force is exerted, the scooter increases its speed from 6.00 m/s to 28.0 m/s. (a) What is the change in momentum? (b) What is the magnitude of the force exerted on the scooter? 3. The brakes on a 15,680 N car exert a stopping force of 640. N. The car\u2019s velocity changes from 20.0 m/s to 0 m/s. (a) What is the car\u2019s mass? (b) What was its initial momentum? (c) What was the change in momentum for the car? (d) How long does it take the braking force to bring the car to rest? \u2022 momentum: A measure of the motion of a body equal to the product of its mass and velocity. Also called linear momentum. \u2022 impulse: The product obtained by multiplying the average value of a force by the time during which it acts. The impulse equals the change in momentum produced by the force in this time interval. References 1. User:deerstop/Wikimedia Commons. http://commons.wikimedia.org/wiki/File:Flatt-3.jpg. Public Domain 29 Physics Unit 9: Circular Motion Patrick Marshall Ck12 Science Say Thanks to the Authors Click http://www.ck12.org/saythanks (No sign in required) www.ck12.org To access a customizable version of this book, as well as other interactive content, visit www.ck12.org AUTHORS Patrick Marshall Ck12 Science CK-12 Foundation is a non-pro\ufb01t organization with a mission to reduce the cost of textbook materials for the K-12 market both in the U.S. and worldwide. Using an open-content, web-based collaborative model termed the FlexBook\u00ae, CK-12 intends to pioneer the generation and distribution of high-quality educational content that will serve both as core text as well as provide an adaptive environment for learning, powered through the FlexBook Platform\u00ae. Copyright \u00a9 2014 CK-12 Foundation, www.ck12.org The names \u201cCK-12\u201d and \u201cCK12\u201d and associated logos and the terms \u201cFlexBook\u00ae\u201d", " and \u201cFlexBook Platform\u00ae\u201d (collectively \u201cCK-12 Marks\u201d) are trademarks and service marks of CK-12 Foundation and are protected by federal, state, and international laws. Any form of reproduction of this book in any format or medium, in whole or in sections must include the referral attribution link http://www.ck12.org/saythanks (placed in a visible location) in addition to the following terms. Except as otherwise noted, all CK-12 Content (including CK-12 Curriculum Material) is made available to Users in accordance with the Creative Commons Attribution-Non-Commercial 3.0 Unported (CC BY-NC 3.0) License (http://creativecommons.org/ licenses/by-nc/3.0/), as amended and updated by Creative Commons from time to time (the \u201cCC License\u201d), which is incorporated herein by this reference. Complete terms can be found at http://www.ck12.org/terms. Printed: April 6, 2014 iii Contents www.ck12.org 1 5 10 Contents 1 Circular Motion 2 Circular Motion 3 Centripetal Force iv www.ck12.org Concept 1. Circular Motion CONCEPT 1 Circular Motion \u2022 De\ufb01ne centripetal acceleration. \u2022 Understand the theory of the centripetal acceleration equation. \u2022 Use the centripetal acceleration equation. \u2022 Understand the angular relationship between velocity and centripetal acceleration. \u2022 Use the equations for motion in two directions and Newton\u2019s Laws to analyze circular motion. Weather satellites, like the one shown above, are found miles above the earth\u2019s surface. Satellites can be polar orbiting, meaning they cover the entire Earth asynchronously, or geostationary, in which they hover over the same spot on the equator. Circular Motion The earth is a sphere. If you draw a horizontal straight line from a point on the surface of the earth, the surface of the earth drops away from the line. The distance that the earth drops away from the horizontal line is very small \u2013 so small, in fact, that we cannot represent it well in a drawing. In the sketch below, if the blue line is 1600 m, the amount of drop (the red line) would be 0.20 m. If the sketch were drawn to scale, the red line would be too short to see. 1 www.ck12.org When an object is launched exactly horizontally in projectile motion, it travels some distance horizontally before it strikes the ground. In the present discussion, we wish to imagine a projectile \ufb01red horizontally on the surface of the earth such that while traveling 1600 m horizontally, the object would fall exactly 0.20 m. If this could occur, then the object would fall exactly the amount necessary during its horizontal motion to remain at the surface of the earth, but not touching it. In such a case, the object would travel all the way around the earth continuously and circle the earth, assuming there were no obstacles, such as mountains. What initial horizontal velocity would be necessary for this to occur? We \ufb01rst calculate the time to fall the 0.20 m: s t = r 2d a = (2)(0:20 m) 9:80 m/s2 = 0:20 s The horizontal velocity necessary to travel 1600 m in 0.20 s is 8000 m/s. Thus, the necessary initial horizontal velocity is 8000 m/s. In order to keep an object traveling in a circular path, there must be an acceleration toward the center of the circle. This acceleration is called centripetal acceleration. In the case of satellites orbiting the earth, the centripetal acceleration is caused by gravity. If you were swinging an object around your head on a string, the centripetal acceleration would be caused by your hand pulling on the string toward the center of the circle. It is important to note that the object traveling in a circle has a constant speed but does not have a constant velocity. This is because direction is part of velocity; when an object changes its direction, it is changing its velocity. Hence the object\u2019s acceleration. The acceleration in the case of uniform circular motion is the change in the direction of the velocity, but not its magnitude. For an object traveling in a circular path, the centripetal acceleration is directly related to the square of the velocity of the object and inversely related to the radius of the circle. ac = v2 r Taking a moment to consider the validity of this equation can help to clarify what it means. Imagine a yo-yo. Instead of using it normally, let it fall to the end of the string, and then spin it around above your head. If we were to increase the speed at which we rotate our hand, we increase the velocity of the yo-yo - it is spinning faster. As it spins faster, it also changes", " direction faster. The acceleration increases. Now let\u2019s think about the bottom of the equation: the radius. If we halve the length of the yo-yo string (bring the yo-yo closer to us), we make the yo-yo\u2019s velocity greater. Again, it moves faster, which increases the acceleration. If we make the string longer again, this decreases the acceleration. We now understand why the relationship between the radius and the acceleration is an inverse relationship - as we decrease the radius, the acceleration increases, and visa versa. 2 www.ck12.org Concept 1. Circular Motion Example Problem: A ball at the end of a string is swinging in a horizontal circle of radius 1.15 m. The ball makes exactly 2.00 revolutions per second. What is its centripetal acceleration? Solution: We \ufb01rst determine the velocity of the ball using the facts that the circumference of the circle is 2pr and the ball goes around exactly twice per second. v = (2)(2pr) = 14:4 m/s t = (2)(2)(3:14)(1:15 m) 1:00 s r = (14:4 m/s)2 We then use the velocity and radius in the centripetal acceleration equation. ac = v2 Example Problem: The moon\u2019s nearly circular orbit around the earth has a radius of about 385,000 km and a period of 27.3 days. Calculate the acceleration of the moon toward the earth. 1:15 m = 180: m/s2 Solution: v = 2pr T = (2)(3:14)(3:85108 m) r = (1020 m/s)2 (27:3 d)(24:0 h/d)(3600 s/h) 3:85108 m = 0:00273 m/s2 = 1020 m/s ac = v2 As shown in the previous example, the velocity of an object traveling in a circle can be calculated by v = 2pr T 3 Where r is the radius of the circle and T is the period (time required for one revolution). This equation can be incorporated into the equation for centripetal acceleration as shown below. www.ck12.org ac = v2 r = ( 2pr T )2 r = 4p2r T 2 Summary \u2022 In order to keep an object traveling in a circular path, there must be an acceleration toward the center of the circle. This acceleration is called centripetal acceleration. \u2022 The acceleration in the case of uniform circular motion changes the direction of the velocity but not its magnitude. \u2022 Formulas for centripetal acceleration are ac = v2 r and ac = 4p2r T 2. Practice This video is a demonstration of centripetal force using balloons and trays of water. Use this resource to answer the questions that follow. https://www.youtube.com/watch?v=EX5DZ2MHlV4 MEDIA Click image to the left for more content. 1. What does centripetal mean? 2. What is uniform circular motion? 3. Why is centripetal acceleration always towards the center? Review 1. An automobile rounds a curve of radius 50.0 m on a \ufb02at road at a speed of 14 m/s. What centripetal acceleration is necessary to keep the car on the curve? 2. An object is swung in a horizontal circle on a length of string that is 0.93 m long. If the object goes around once in 1.18 s, what is the centripetal acceleration? \u2022 circular motion: A movement of an object along the circumference of a circle or rotation along a circular path. \u2022 centripetal acceleration: The acceleration toward the center that keeps an object following a circular path. References 1. Greg Goebel. http://www.public-domain-image.com/space-public-domain-images-pictures/weather-satellit e.jpg.html. Public Domain 2. CK-12 Foundation - Samantha Bacic.. CC-BY-NC-SA 3.0 3. CK-12 Foundation - Richard Parsons.. CC-BY-NC-SA 3.0 4 www.ck12.org Concept 2. Circular Motion CONCEPT 2 Circular Motion Students will learn that in circular motion there is always an acceleration (and hence a force) that points to the center of the circle de\ufb01ned by the objects motion. This force changes the direction of the velocity vector of the object but not the speed. Students will also learn how to calculate that speed using the period of motion and the distance of its path (circumference of the circle it traces out). Vocabulary \u2022 centripetal acceleration: The inward acceleration that keeps an object in circular motion. \u2022 centripetal force: The inward force that keeps an object in", " circular motion. Introduction A satellite orbits around the Earth in Figure below. A car travels around a curve in Figure below. All of these objects are engaged in circular motion. Let us consider the satellite \ufb01rst. The satellite is held in place by the Earth\u2019s gravity. The gravity holds the satellite in its orbit. In what direction does this force act? If the earth were \u201cmagically\u201d gone, the satellite would \ufb02y off tangent to its motion at the instant gravity no longer held it. The force preventing this from happening must keep pulling the satellite toward the center of the circle to maintain circular motion. FIGURE 2.1 What is the force in Figure above that prevents the car from skidding off the road? If you guessed \u201cthe friction between the tires and the road\u201d you\u2019d be correct. But is it static or kinetic friction? Unless the tires skid, there can be no kinetic friction. It is static friction that prevents the tires from skidding, just as it is static friction that permits you to walk without slipping. In Figure below, you can see the foot of a person who walks toward the right by pushing their foot backward with a horizontal component of force F. They move forward because the ground exerts a horizontal component force fs in the opposite direction. (Note that vertical forces are ignored.) The force the ground exerts on the person\u2019s foot is a static friction force. Because the foot does not slide, we know that F and fs are equal opposed forces. We can easily see which direction the static friction force must act when we walk, but what about a car performing circular motion? In what direction does the static friction act on the car in Figure above? 5 www.ck12.org FIGURE 2.2 FIGURE 2.3 Figure below shows the top view of a car moving around a circular track with a constant speed. Since acceleration is de\ufb01ned as a = Dv, you may be tempted to say that since the speed remains constant, Dv = 0, the acceleration Dt must also be zero. But that conclusion would be incorrect because Dv represents a change in velocity, not a change in speed. The velocity of the car is not constant since it is continuously changing its direction. How then do we \ufb01nd the acceleration of the car? Figure below shows the instantaneous velocity vectors for the car in two different positions a very small time apart. Notice that the vector DV points toward the center of the circle. (Recall that DV can be thought of as the sum of the vectors V2 + (V1).) The direction of the acceleration points in the direction of DV since acceleration is de\ufb01ned as a = D~v Dt. This is reasonable, since if there were no force directed toward the center of the circle, the car would move off tangent to the circle. We call the inward force that keeps an object in circular motion a \u201ccenter seeking\u201d, or centripetal force and the acceleration, centripetal acceleration. The centripetal acceleration is often denoted as ac In order to \ufb01nd the correct expression for the magnitude of the centripetal acceleration we\u2019ll need to use a little geometric reasoning. Figure below and Figure below show two \u201calmost\u201d similar triangles. The magnitudes of V1 and V2 are equal, and the change in location of the car occurs over a very small increment in time, Dt. The velocities are tangent to the circle and therefore perpendicular to the radius of the circle. As such, the \u201cradius\u201d triangle and the \u201cvelocity\u201d triangle are approximately similar (see the \ufb01gures above). We construct an approximate ratio between the two triangles by assuming that during the time Dt, the car has traveled a distance Ds 6 www.ck12.org Concept 2. Circular Motion FIGURE 2.4 FIGURE 2.5 along the circle. The ratio below is constructed in order to determine the acceleration. : = Dv v, which leads to, vDs Ds r Dividing both sides of the equation by Dt, we have: v Ds Dt : = rDv. : = r Dv Dt. But Ds Dt is the speed v of the car and Dv Dt is the acceleration of the car. 7 www.ck12.org If we allow the time to become in\ufb01nitesimally small, then the approximation becomes exact and we have: v2 = ra; a = v2 r. Thus, the magnitude of the centripetal acceleration for an object moving with constant speed in circular motion is ac = v2 r, and its direction is toward the center of the circle. Illustrative Examples using Centripetal Acceleration and Force Example 1A: A 1000 kg car moves with a constant speed", " 13.0 m/s around a \ufb02at circular track of radius 40.0 m. What is the magnitude and direction of the centripetal acceleration? Answer: The magnitude of the car\u2019s acceleration is ac = v2 r = 132 acceleration is toward the center of the track. 40 = 4:225 = 4:23 m=s2 and the direction of its Example 1b: Determine the force of static friction that acts upon the car in Figure below FIGURE 2.6 Answer: Using Newton\u2019s Second Law: F = fs = ma = 1000(4:225) = 4225 = 4230 N Example 1c: Determine the minimum necessary coef\ufb01cient of static friction between the tires and the road. Answer: y F = FN Mg = 0; FN = Mg but fs = \u00b5sFN = ma Thus, \u00b5sMg = Ma; \u00b5sg = a; \u00b5s = a g = 4:225 9:8 = 0:431 = 0:43 Check Your Understanding True or False? 1. Kinetic friction is responsible for the traction (friction) between the tires and the road. Answer : False. As long as the car does not skid, there is no relative motion between the instantaneous contact area of the tire and the road. 2. True or False? The force of static friction upon an object can vary. Answer : True. In attempting to move an object, a range of forces of different magnitudes can be applied until the maximum static friction between the object and the surface it rests upon is overcome and the object is set into motion. Recall that the magnitude of static friction is represented by the inequality: fs \u00b5sFN 8 www.ck12.org Concept 2. Circular Motion 3. The greater the mass of the car, the greater the coef\ufb01cient of friction. Answer : False. The coef\ufb01cient of friction is independent of the mass of an object. Recall that it is the ratio of the friction force to the normal force. As such, it is a pure number dependent only upon the nature of the materials in contact with each other- in this case rubber and asphalt. References 1. Image copyright Paul Fleet, 2012. http://www.shutterstock.com. Used under license from Shutterstock 2. Tim White (Flickr: TWHITE87). http://www.\ufb02ickr.com/photos/tjwhite87/8102931300/ 3. Image copyright Andre Adams, 2012; modi\ufb01ed by CK-12 Foundation - Raymond Chou. http://www.shut. CC-BY 2.0 terstock.com. Used under license from Shutterstock.com 4. CK-12 Foundation - Raymond Chou.. CC-BY-NC-SA 3.0 5. CK-12 Foundation - Raymond Chou.. CC-BY-NC-SA 3.0 6. CK-12 Foundation - Raymond Chou.. CC-BY-NC-SA 3.0 9 CONCEPT 3 www.ck12.org Centripetal Force \u2022 De\ufb01ne centripetal force. \u2022 Solve problems involving centripetal force. \u2022 Explain the difference between centripetal and centrifugal forces. Jupiter\u2019s moons and ring materials follow all the laws of physics, including centripetal force and centripetal acceleration. Centripetal Force Centripetal force is, simply, the force that causes centripetal acceleration. Objects that move in uniform circular motion all have an acceleration toward the center of the circle and therefore, they must also suffer a force toward the center of the circle. That force is the centripetal force. For orbiting satellites, such as the moon orbiting the earth or the earth orbiting the sun, the centripetal force is produced by gravity. When an Olympic hammer thrower whirls a massive ball on a chain, the centripetal force is created by the athlete and transmitted by the chain. Newton\u2019s second law shows the relationship between force and acceleration, F = ma. Since we have formulas expressing the relationships for centripetal acceleration, they can easily be altered to show the relationships for centripetal force. ac = v2 and ac = 4p2r r and F = ma so Fc = mv2 r T 2 so Fc = 4p2rm T 2 10 www.ck12.org Concept 3. Centripetal Force Common Misconceptions Many people incorrectly use the term centrifugal force instead of centripetal force. Often, you will hear the term centrifugal force used to describe the outward force pushing an object away from the center of a circle. In reality, however, centrifugal forces are inertial, or \ufb01ctional, forces. They only exist in the frame of reference of the object that is moving and, even then,", " are theoretical. Physicists dealing in a moving frame of reference use centrifugal forces to ease calculations. For a great explanation of the difference between centrifugal and centripetal force, see this video: https://www.youtube.com/watch?v=DLgy6rVV-08 MEDIA Click image to the left for more content. Summary \u2022 Centripetal force is the force that causes centripetal acceleration. \u2022 Equations for centripetal force are Fc = mv2 r and Fc = 4p2rm. T 2 Practice A video of physics students riding a roller coaster. Use this resource to answer the questions that follow. http://www.teachersdomain.org/asset/phy03_vid_roller/ 1. Does the roller coaster in the video have a complete circle as part of its path? 2. What is it that keeps the glass of water on the tray as it swings over the student\u2019s head? Review 1. A runner moving at a speed of 8.8 m/s rounds a bend with a radius of 25 m. What is the centripetal force needed to keep this runner on the curve and what supplies this force? 2. A 1000. kg car rounds a curve of 50.0 m radius on a \ufb02at road with a speed of 14.0 m/s. (a) Will the car make the turn successfully if the pavement is dry and the coef\ufb01cient of friction is 0.60? (b) Will the car make the turn successfully if the pavement is wet and the coef\ufb01cient of friction is 0.20? 3. An 0.500 kg object tied to a string is swung around a person\u2019s head in a horizontal circle. The length of the string is 1.00 m and the maximum force the string can withstand without breaking is 25.0 N. What is the maximum speed the object may be swung without breaking the string? \u2022 centripetal force: The component of force acting on a body in curvilinear motion that is directed toward the center of curvature or axis of rotation. References 1. Courtesy of NASA/JPL/Cornell University. http://www.nasa.gov/centers/goddard/multimedia/largest/EduI mageGallery.html. Public Domain 11 Physics Unit 10: Waves Patrick Marshall Jean Brainard, Ph.D. Ck12 Science James Dann, Ph.D. Say Thanks to the Authors Click http://www.ck12.org/saythanks (No sign in required) AUTHORS Patrick Marshall Jean Brainard, Ph.D. Ck12 Science James Dann, Ph.D. CONTRIBUTOR Catherine Pavlov www.ck12.org To access a customizable version of this book, as well as other interactive content, visit www.ck12.org CK-12 Foundation is a non-pro\ufb01t organization with a mission to reduce the cost of textbook materials for the K-12 market both in the U.S. and worldwide. Using an open-content, web-based collaborative model termed the FlexBook\u00ae, CK-12 intends to pioneer the generation and distribution of high-quality educational content that will serve both as core text as well as provide an adaptive environment for learning, powered through the FlexBook Platform\u00ae. Copyright \u00a9 2014 CK-12 Foundation, www.ck12.org The names \u201cCK-12\u201d and \u201cCK12\u201d and associated logos and the terms \u201cFlexBook\u00ae\u201d and \u201cFlexBook Platform\u00ae\u201d (collectively \u201cCK-12 Marks\u201d) are trademarks and service marks of CK-12 Foundation and are protected by federal, state, and international laws. Any form of reproduction of this book in any format or medium, in whole or in sections must include the referral attribution link http://www.ck12.org/saythanks (placed in a visible location) in addition to the following terms. Except as otherwise noted, all CK-12 Content (including CK-12 Curriculum Material) is made available to Users in accordance with the Creative Commons Attribution-Non-Commercial 3.0 Unported (CC BY-NC 3.0) License (http://creativecommons.org/ licenses/by-nc/3.0/), as amended and updated by Creative Commons from time to time (the \u201cCC License\u201d), which is incorporated herein by this reference. Complete terms can be found at http://www.ck12.org/terms. Printed: April 6, 2014 iii Contents www.ck12.org 1 7 10 14 17 20 24 28 32 35 39 42 45 56 Contents 1 Measuring Waves 2 Mechanical Wave 3 Transverse Waves 4 Longitudinal Waves 5 Re\ufb02ection of Mechanical Waves 6 Refraction of Mechanical Waves 7 Wave Interactions 8 Wave", " Interference 9 Wave Speed 10 Sound Waves 11 Frequency and Pitch of Sound 12 Speed of Sound 13 Resonance with Sound Waves 14 Sound in a Tube iv www.ck12.org Concept 1. Measuring Waves CONCEPT 1 Measuring Waves Lesson Objectives \u2022 De\ufb01ne wave amplitude and wavelength. \u2022 Relate wave speed to wave frequency and wavelength. Lesson Vocabulary \u2022 hertz (Hz) \u2022 wave amplitude \u2022 wave frequency \u2022 wavelength \u2022 wave speed Introduction Tsunamis, or the waves caused by earthquakes, are unusually large ocean waves. You can see an example of a tsunami in Figure 1.1. Because tsunamis are so big, they can cause incredible destruction and loss of life. The tsunami in the \ufb01gure crashed into Thailand, sending people close to shore running for their lives. The height of a tsunami or other wave is just one way of measuring its size. You\u2019ll learn about this and other ways of measuring waves in this lesson. FIGURE 1.1 This tsunami occurred in Thailand on De- cember 26, 2004. 1 www.ck12.org Wave Amplitude and Wavelength The height of a wave is its amplitude. Another measure of wave size is wavelength. Both wave amplitude and wavelength are described in detail below. Figure 1.2 shows these wave measures for both transverse and longitudinal waves. You can also simulate waves with different amplitudes and wavelengths by doing the interactive animation at this URL: http://sci-culture.com/advancedpoll/GCSE/sine%20wave%20simulator.html. FIGURE 1.2 Wave amplitude and wavelength are two important measures of wave size. Wave Amplitude Wave amplitude is the maximum distance the particles of a medium move from their resting position when a wave passes through. The resting position is where the particles would be in the absence of a wave. \u2022 In a transverse wave, wave amplitude is the height of each crest above the resting position. The higher the crests are, the greater the amplitude. \u2022 In a longitudinal wave, amplitude is a measure of how compressed particles of the medium become when the wave passes through. The closer together the particles are, the greater the amplitude. 2 www.ck12.org Concept 1. Measuring Waves What determines a wave\u2019s amplitude? It depends on the energy of the disturbance that causes the wave. A wave caused by a disturbance with more energy has greater amplitude. Imagine dropping a small pebble into a pond of still water. Tiny ripples will move out from the disturbance in concentric circles, like those in Figure above. The ripples are low-amplitude waves. Now imagine throwing a big boulder into the pond. Very large waves will be generated by the disturbance. These waves are high-amplitude waves. Wavelength Another important measure of wave size is wavelength. Wavelength is the distance between two corresponding points on adjacent waves (see Figure 1.2 ). Wavelength can be measured as the distance between two adjacent crests of a transverse wave or two adjacent compressions of a longitudinal wave. It is usually measured in meters. Wavelength is related to the energy of a wave. Short-wavelength waves have more energy than long-wavelength waves of the same amplitude. You can see examples of waves with shorter and longer wavelengths in Figure 1.3. FIGURE 1.3 Both of these waves have the same ampli- tude, but they differ in wavelength. Which wave has more energy? Wave Frequency and Speed Imagine making transverse waves in a rope, like the waves in Figure above. You tie one end of the rope to a doorknob or other \ufb01xed point and move the other end up and down with your hand. You can move the rope up and down slowly or quickly. How quickly you move the rope determines the frequency of the waves. Wave Frequency The number of waves that pass a \ufb01xed point in a given amount of time is wave frequency. Wave frequency can be measured by counting the number of crests or compressions that pass the point in 1 second or other time period. The higher the number is, the greater is the frequency of the wave. The SI unit for wave frequency is the hertz (Hz), where 1 hertz equals 1 wave passing a \ufb01xed point in 1 second. Figure 1.4 shows high-frequency and low-frequency transverse waves. You can simulate transverse waves with different frequencies at this URL: http://zonalandeduc ation.com/mstm/physics/waves/partsOfAWave/waveParts.htm. 3 www.ck12.org FIGURE 1.4 A transverse wave with a higher fre- quency has crests that are closer to- gether. The frequency of a wave is the same as the frequency of the vibrations that caused the wave. For example, to generate a higher-frequency wave in a rope, you must", " move the rope up and down more quickly. This takes more energy, so a higher-frequency wave has more energy than a lower-frequency wave with the same amplitude. Wave Speed Assume that you move one end of a rope up and down just once. How long will take the wave to travel down the rope to the other end? This depends on the speed of the wave. Wave speed is how far the wave travels in a given amount of time, such as how many meters it travels per second. Wave speed is not the same thing as wave frequency, but it is related to frequency and also to wavelength. This equation shows how the three factors are related: Speed = Wavelength Frequency In this equation, wavelength is measured in meters and frequency is measured in hertz, or number of waves per second. Therefore, wave speed is given in meters per second. The equation for wave speed can be used to calculate the speed of a wave when both wavelength and wave frequency are known. Consider an ocean wave with a wavelength of 3 meters and a frequency of 1 hertz. The speed of the wave is: Speed = 3 m 1 wave/s = 3 m/s You Try It! Problem: Jera made a wave in a spring by pushing and pulling on one end. The wavelength is 0.1 m, and the wave frequency is 0.2 m/s. What is the speed of the wave? If you want more practice calculating wave speed from wavelength and frequency, try the problems at this URL: htt p://www.physicsclassroom.com/class/waves/u10l2e.cfm. The equation for wave speed (above) can be rewritten as: 4 www.ck12.org Concept 1. Measuring Waves Frequency = Speed Wavelength or Wavelength = Speed Frequency Therefore, if you know the speed of a wave and either the wavelength or wave frequency, you can calculate the missing value. For example, suppose that a wave is traveling at a speed of 2 meters per second and has a wavelength of 1 meter. Then the frequency of the wave is: Frequency = 2 m/s 1 m = 2 waves/s, or 2 Hz You Try It! Problem: A wave is traveling at a speed of 2 m/s and has a frequency of 2 Hz. What is its wavelength? The Medium Matters The speed of most waves depends on the medium through which they are traveling. Generally, waves travel fastest through solids and slowest through gases. That\u2019s because particles are closest together in solids and farthest apart in gases. When particles are farther apart, it takes longer for the energy of the disturbance to pass from particle to particle. Lesson Summary \u2022 Wave amplitude is the maximum distance the particles of a medium move from their resting positions as a wave passes through. Wavelength is the distance between two corresponding points of adjacent waves. Waves with greater amplitudes or shorter wavelengths have more energy. \u2022 Wave frequency is the number of waves that pass a \ufb01xed point in a given amount of time. Higher frequency waves have more energy. Wave speed is calculated as wavelength multiplied by wave frequency. Wave speed is affected by the medium through which a wave travels. Lesson Review Questions Recall 1. How is wave amplitude measured in a transverse wave? 2. Describe the wavelength of a longitudinal wave. 3. De\ufb01ne wave frequency. Apply Concepts 4. All of the waves in the sketch below have the same amplitude and speed. Which wave has the longest wavelength? Which has the highest frequency? Which has the greatest energy? 5 www.ck12.org 5. A wave has a wavelength of 0.5 m/s and a frequency of 2 Hz. What is its speed? Think Critically 6. Relate wave amplitude, wavelength, and wave frequency to wave energy. 7. Waves A and B have the same speed, but wave A has a shorter wavelength. Which wave has the higher frequency? Explain how you know. Points to Consider You read in this lesson that waves travel at different speeds in different media. \u2022 When a wave enters a new medium, it may speed up or slow down. What other properties of the wave do you think might change when it enters a new medium? \u2022 What if a wave reaches a type of matter it cannot pass through? Does it just stop moving? If not, where does it go? References 1. David Rydevik. http://commons.wikimedia.org/wiki/File:2004-tsunami.jpg. Public Domain 2. Christopher Auyeung. CK-12 Foundation. 3. Christopher Auyeung. CK-12 Foundation. 4. Christopher Auyeung. CK-12 Foundation. 6 www.ck12.org Concept 2. Mechanical Wave CONCEPT 2 \u2022 Describe mechanical waves. \u2022 De\ufb01ne the medium of a mechanical wave. \u2022 Identify three types of mechanical waves. Mechanical Wave No doubt you\u2019ve seen this happen. Dro", "plets of water fall into a body of water, and concentric circles spread out through the water around the droplets. The concentric circles are waves moving through the water. Waves in Matter The waves in the picture above are examples of mechanical waves. A mechanical wave is a disturbance in matter that transfers energy through the matter. A mechanical wave starts when matter is disturbed. A source of energy is needed to disturb matter and start a mechanical wave. Q: Where does the energy come from in the water wave pictured above? A: The energy comes from the falling droplets of water, which have kinetic energy because of their motion. The Medium The energy of a mechanical wave can travel only through matter. The matter through which the wave travels is called the medium ( plural, media). The medium in the water wave pictured above is water, a liquid. But the medium of a mechanical wave can be any state of matter, even a solid. Q: How do the particles of the medium move when a wave passes through them? A: The particles of the medium just vibrate in place. As they vibrate, they pass the energy of the disturbance to the particles next to them, which pass the energy to the particles next to them, and so on. Particles of the medium don\u2019t actually travel along with the wave. Only the energy of the wave travels through the medium. 7 Types of Mechanical Waves There are three types of mechanical waves: transverse, longitudinal, and surface waves. They differ in how particles of the medium move. You can see this in the Figure 2.1 and in the animation at the following URL. http://www. acs.psu.edu/drussell/Demos/waves/wavemotion.html www.ck12.org FIGURE 2.1 \u2022 In a transverse wave, particles of the medium vibrate up and down perpendicular to the direction of the wave. \u2022 In a longitudinal wave, particles of the medium vibrate back and forth parallel to the direction of the wave. \u2022 In a surface wave, particles of the medium vibrate both up and down and back and forth, so they end up moving in a circle. Q: How do you think surface waves are related to transverse and longitudinal waves? A: A surface wave is combination of a transverse wave and a longitudinal wave. Summary \u2022 A mechanical wave is a disturbance in matter that transfers energy through the matter. \u2022 The matter through which a mechanical wave travels is called the medium ( plural, media). \u2022 There are three types of mechanical waves: transverse, longitudinal, and surface waves. They differ in how particles of the medium move when the energy of the wave passes through. Vocabulary \u2022 mechanical wave : Disturbance in matter that transfers energy from one place to another. \u2022 medium (plural, media ): Matter through which a mechanical wave moves. Practice At the following URL, read the short introduction to waves and watch the animations. Then answer the questions below. http://www.acs.psu.edu/drussell/Demos/waves-intro/waves-intro.html 8 www.ck12.org Concept 2. Mechanical Wave 1. The article gives a dictionary de\ufb01nition of wave. What is the most important part of this de\ufb01nition? 2. What happens to particles of the medium when a wave passes? 3. How is \u201cdoing the wave\u201d in a football stadium like a mechanical wave? Review 1. De\ufb01ne mechanical wave. 2. What is the medium of a mechanical wave? 3. List three types of mechanical waves. 4. If you shake one end of a rope up and down, a wave passes through the rope. Which type of wave is it? References 1. Zachary Wilson.. CC BY-NC 3.0 9 CONCEPT 3 www.ck12.org Transverse Waves \u2022 Describe transverse waves. \u2022 Explain how waves transfer energy without transferring matter. \u2022 De\ufb01ne wavelength, frequency, and period of a transverse wave. \u2022 State the relationship between speed, wavelength, and frequency. \u2022 Solve problems using the relationships between speed, wavelength, frequency, and period. Professional surfer Marcio Freire rides a giant wave at the legendary big wave surf break known as as \"Jaws\" during one the largest swells of the winter March 13, 2011 in Maui, HI. Massive waves, such as this one, transfer huge amounts of energy. Transverse Waves Types of Waves Water waves, sound waves, and the waves that travel along a rope are mechanical waves. Mechanical waves require a material medium such as water, air, or rope. Light waves, however, are electromagnetic waves and travel without a material medium. They are not mechanical waves. In all types of mechanical waves, energy moves from one place to another while the media carrying the wave only vibrates back and forth in position. One type of mechanical wave is the transverse wave.", " In the case of transverse waves, the movement of the medium is perpendicular to the direction of the energy movement. 10 www.ck12.org Concept 3. Transverse Waves In the sketch above, consider the transverse wave produced when the boy jerks one end of a rope up and down while the other end is tied to a tree. The energy spent by the boy transfers permanently down the rope to the tree. The rope, however, only moves up and down. If we stuck a piece of tape somewhere on the rope, we would see that the particles of medium do not travel with the energy. After the wave has passed by, the piece of tape would still be in the same place it was before the wave approached. In all transverse waves, the movement media vibrates perpendicularly to the direction of wave motion, and the medium is not permanently moved from one place to another. Frequency, Wavelength, and Velocity Waves are identi\ufb01ed by several characteristics. There is a center line where the medium would be if there were no wave, which is sometimes describes as the undisturbed position. The displacement of the medium above this undisturbed position is called a crest and the displacement below the undisturbed position is called a trough. The maximums of the crest and trough are equal and are called the amplitude. The distance between equivalent positions on succeeding waves is called the wavelength. The wavelength could be measured from a crest to the next crest or from a trough to the next trough, and is commonly represented with the Greek letter lambda, l. The time interval required for one complete wave to pass a point is called the period. During the period of the wave, an entire wavelength from one crest to the next crest passes a position. The number of waves that pass a single position in one second is called the frequency. The period of a wave and its frequency are reciprocals of each other. f = 1 T The units for the period are seconds and the units for frequency are s 1 or 1 name Hertz (Hz). s. This unit has also been given the 11 Another important characteristic of a wave is its velocity. The wave velocity is different from the velocity of the medium; the wave velocity is the velocity of the linearly transferred energy. Since the energy travels one wavelength, l, in one period, T, the velocity can be expressed as distance over time: www.ck12.org v = l T : Since period and frequency are reciprocals, the speed of the wave could also be expressed as v = l f. Example Problem: A sound wave has a frequency of 262 Hz. What is the time lapse between successive wave crests? Solution: The time lapse between successive crests would be the period and the period is the reciprocal of the frequency. T = 1 f = 1 262 s1 = 0:00382 s Example Problem: A sound wave has a frequency of 262 Hz has a wavelength of 1.29 m. What is the velocity of the wave? Solution: v = l f = (1:29 m)(262 s1) = 338 m/s Summary \u2022 Mechanical waves require a material medium such as water, air, or rope. \u2022 In all types of mechanical waves, energy moves from one place to another while the media carrying the wave only vibrates back and forth in position. \u2022 One type of mechanical wave is the transverse wave, in which the movement of the medium is perpendicular to the direction of the energy propagation. \u2022 The maximum displacement of the medium is the distance from the undisturbed position to the top of a crest, or the amplitude. \u2022 The distance along the line of motion of the wave from equivalent positions on succeeding waves is the wavelength. \u2022 The time interval required for one entire wave to pass a point is the period. \u2022 The number of periods per second is the wave\u2019s frequency. \u2022 The period of a wave and its frequency are reciprocals of each other. \u2022 The velocity of the wave\u2019s energy transfer is given by v = l f or v = l T : Practice The following video explains wave characteristics. Pause the video before each practice question and try to solve it yourself before moving on. http://www.youtube.com/watch?v=5ENLxaPiJJI 12 www.ck12.org Concept 3. Transverse Waves MEDIA Click image to the left for more content. 1. What is the distance between the base line and crest called? 2. What symbol is used for wavelength? 3. What is the relationship between period and frequency? Review 1. A sound wave produced by a chime 515 m away is heard 1.50 s later. (a) What is the speed of sound in air? (b) The sound wave has a frequency of 436 Hz. What is its period? (c) What is the wavelength of the sound? 2. A hiker shouts toward a vertical cliff 685 m away. The echo is heard 4.00 s later. (a)", " What is the speed of sound in air? (b) Why is this speed of sound slightly different from the previous answer? (c) The wavelength of the sound is 0.750 m. What is the frequency? (d) What is the period of the wave? 3. The speed of light in air is 3.00 10 8 m/s. If a light wave has a wavelength of 5.80 10 7 m, what is its frequency? \u2022 transverse wave: A transverse wave is a moving wave that consists of oscillations occurring perpendicular (or right angled) to the direction of energy transfer \u2022 undisturbed position: The equilibrium or rest position of the medium in a wave. \u2022 crest: A crest is the point on a wave with the maximum value or upward displacement within a cycle \u2022 trough: A trough is the opposite of a crest, so the minimum or lowest point in a cycle \u2022 amplitude: The maximum displacement from a zero value during one period of an oscillation. \u2022 wavelength: The distance between corresponding points of two consecutive waves. \u2022 frequency: The number of waves that pass a \ufb01xed point in unit time. \u2022 period: A period is the time required for one complete cycle of vibration to pass a given point. References 1. Image copyright EpicStockMedia, 2013. http://www.shutterstock.com. Used under license from Shutter- stock.com 2. Kid: Image copyright Richcat, 2013; Tree: Image copyright AlexeyZet, 2013; Composite created by CK-12. Used under licenses from Shutterstock.com Foundation - Samantha Bacic. http://www.shutterstock.com/ 3. CK-12 Foundation - Samantha Bacic.. CC-BY-NC-SA 3.0 13 CONCEPT 4 \u2022 Describe longitudinal waves. www.ck12.org Longitudinal Waves https://www.google.com/url?sa=i&rct=j&q=&esrc=s&source=images&cd=&cad=rja&docid=6e9QZ8b6JzZZRM&tbnid=CAVrIJRLaBDfM:&ved=0CAUQjRw&url=http%3A%2F%2Fsirius.ucsc.edu%2Fdemoweb%2Fcgi-bin%2F%3Fwavesvisible-slinky&ei=kE4FUpSDN4HXygHR84HwDg&bvm=bv.50500085,d.b2I&psig=AFQjCNEJ23OS_x3Ga2tbC3Vi2VVUfbPmQ&ust=1376165898023972 Playing with a Slinky is a childhood tradition, but few children realize they are actually playing with physics. Longitudinal Waves Like transverse waves, longitudinal waves are mechanical waves, which means they transfer energy through a medium. Unlike transverse waves, longitudinal waves cause the particles of medium to move parallel to the direction of the wave. They are most common in springs, where they are caused by the pushing an pulling of the spring. Although the surface waves on water are transverse waves, \ufb02uids (liquids, gases, and plasmas) usually transmit longitudinal waves. As shown in the image below, longitudinal waves are a series of compressions and rarefactions, or expansions. The wavelength of longitudinal waves is measured by the distance separating the densest compressions. The amplitude of longitudinal waves is the difference in media density between the undisturbed density to the highest density in a compression. Example Problem: A sonar signal (sonar is sound waves traveling through water) of 1:00 106 Hz frequency has a wavelength of 1.50 mm in water. What is the speed of sound in water? Solution: v = l f = (0:00150 m)(1:00 106 s1) = 1500 m/s Example Problem: A sound wave of wavelength 0.70 m and velocity 330 m/s is produced for 0.50 s. 14 www.ck12.org Concept 4. Longitudinal Waves a. What is the frequency of the wave? b. How many complete waves are emitted in this time interval? c. After 0.50 s, how far is the wave front from the source of the sound? Solution: f = v l = 330 m/s 0:70 m = 470 s1 a. b. complete waves = (470 cycles/s)(0:50 s) = 235 cycles c. distance = (330 m/s)(0:50 s) = 115 m Summary \u2022 Longitudinal waves cause the particles of medium to move parallel to the direction of the wave. Practice The following video explains how a tuning fork creates sound. Use this resource to answer the questions that follow. http://www.youtube.com/watch", "?v=bomzzHC-59k MEDIA Click image to the left for more content. 1. In your own words, how are compressions and rarefactions produced by the tuning fork? 2. Make a guess why sound can easily travel around corners (Hint: think of its medium). Review 1. Some giant ocean waves have a wavelength of 25 m long, and travel at speeds of 6.5 m/s. Determine the frequency and period of such a wave. 2. Bats use sound echoes to navigate and hunt. They emit pulses of high frequency sound waves which re\ufb02ect off obstacles in the surroundings. By detecting the time delay between the emission and return of a pulse, a bat can determine the location of the object. What is the time delay between the sending and return of a pulse from an object located 12.5 m away? The approximate speed of sound is 340 m/s. 3. Sachi is listening to her favorite radio station which broadcasts radio signals with a frequency of 1:023 108 Hz. If the speed of the signals in air is 2:997 108 m/s, what is the wavelength of these radio signals? 4. A longitudinal wave is observed to be moving along a slinky. Adjacent crests are 2.4 m apart. Exactly 6 crests are observed to move past a given point in 9.1 s. Determine the wavelength, frequency, and speed of this wave. 5. A sonar signal leaves a submarine, travels through the water to another submarine and re\ufb02ects back to the original submarine in 4.00 s. If the frequency of the signal was 512 cycles per second and the wavelength of the signal was 2.93 m, how far away is the second submarine? \u2022 longitudinal wave: A wave in which the direction of media displacement is the same as the direction of wave propagation. 15 References 1. CK-12 Foundation - Samantha Bacic.. CC-BY-NC-SA 3.0 www.ck12.org 16 www.ck12.org Concept 5. Re\ufb02ection of Mechanical Waves CONCEPT 5 Re\ufb02ection of Mechanical Waves \u2022 State the law of re\ufb02ection. \u2022 Solve problems using the law of re\ufb02ection. \u2022 Given data about the media on either side of a barrier, determine whether the re\ufb02ected wave will be upright or inverted. When mechanical waves strike a barrier, at least part of the energy of the waves will be re\ufb02ected back into the media from which they came. You experience this every single day, when you look in the mirror and see your own re\ufb02ection. Re\ufb02ection of Mechanical Waves When a wave strikes an obstacle or comes to the end of the medium it is traveling in, some portion of the wave is re\ufb02ected back into the original medium. It re\ufb02ects back at an equal angle that it came in. These angles are called the angle of incidence and the angle of re\ufb02ection. The normal line, the incident and re\ufb02ected rays, and the angles of incidence and re\ufb02ection are all shown in the diagram sketched above. The law of re\ufb02ection states that the angle of incidence equals the angle of re\ufb02ection. These rules of re\ufb02ection apply in the cases of water waves bouncing off the side of a pool, sound waves echoing off a distant cliff, or wave pulses traveling down a rope or a slinky. Consider the change that would occur with a light rope joined to a heavier rope. When a wave pulse travels down the rope and encounters the media change, a re\ufb02ection will occur. Look at the image below. In the top sketch, we see a lightweight (black) rope attached to a heavier rope (red). There is a wave pulse traveling down the rope from left to right. When the wave pulse encounters the barrier (the change in rope weight), part of the wave moves into the new medium and part of the wave is re\ufb02ected back into the old medium. As you can see in the bottom half of the diagram, the transmitted portion of the wave continues into the new medium right side up. The transmitted wave is somewhat diminished because some of the energy of the wave was re\ufb02ected and also because the rope to be lifted is heavier. The re\ufb02ected wave is also diminished because some of the energy was transmitted through the barrier. The re\ufb02ected wave is also inverted (upside down). This is a general rule for mechanical waves passing from a less dense medium into a more dense medium, that is, the re\ufb02ected wave will be inverted. 17 www.ck12.org The situation changes when the wave is passing from a more dense medium into a less dense medium. As you can see in the sketch below, when a", " wave pulse moving in denser medium encounters a media interface to a medium of less density, the re\ufb02ected wave is upright rather than inverted. It is also possible for a mechanical wave to encounter an impenetrable barrier, that is, a barrier which does not allow In such a case, the complete wave pulse will be re\ufb02ected and the re\ufb02ected wave will be any transmission at all. inverted. Summary \u2022 When a wave strikes an obstacle or comes to the end of the medium it is traveling in, some part of the wave is re\ufb02ected back into the original medium. \u2022 The law of re\ufb02ection states that the angle of incidence equals the angle of re\ufb02ection. \u2022 The general rule, for mechanical waves passing from a less dense medium into a more dense medium, the re\ufb02ected wave will be inverted. \u2022 When a wave pulse moving in denser medium encounters a media interface to a medium of lesser density, the re\ufb02ected wave is upright rather than inverted. \u2022 When a mechanical wave encounters an impenetrable barrier, the complete wave pulse will be re\ufb02ected and the re\ufb02ected wave will be inverted. Practice The following video shows a wave machine in action. Use this video to answer the questions that follow. http://www.youtube.com/watch?v=YQHbRw_hyz4 MEDIA Click image to the left for more content. 1. What happens to the wave when it is re\ufb02ected from an open end? 2. What happens to the wave when it is re\ufb02ected from a \ufb01xed end? 18 www.ck12.org Review Concept 5. Re\ufb02ection of Mechanical Waves 1. Draw a diagram showing a surface with a normal line. On the diagram, show a wave ray striking the surface with an angle of incidence of 60. Draw the re\ufb02ection ray on the diagram and label the angle of re\ufb02ection. 2. Light strikes a mirror\u2019s surface at 30 to the normal. What will the angle of re\ufb02ection be? 3. If the angle between the incident ray and the re\ufb02ected ray is 90, what is the angle of incidence? 4. When a water wave is re\ufb02ected from a concrete wall, will the re\ufb02ected wave be inverted or upright? 5. If you tie a heavy spring to a light spring and send a wave pulse down the heavy spring, some of the wave will be re\ufb02ected when the wave passes into the lighter spring. Will the re\ufb02ected pulse be upright or inverted? \u2022 re\ufb02ection: The change in direction of a wavefront at an interface between two different media so that the wavefront returns into the medium from which it originated. \u2022 law of re\ufb02ection: Says the angle at which the wave is incident on the surface equals the angle at which it is re\ufb02ected. \u2022 angle of incidence: The angle formed by a ray incident on a surface and a perpendicular to the surface at the point of incidence. \u2022 angle of re\ufb02ection: The angle formed by a re\ufb02ected ray and a perpendicular to the surface at the point of re\ufb02ection. References 1. CK-12 Foundation - Samantha Bacic.. CC-BY-NC-SA 3.0 2. CK-12 Foundation - Samantha Bacic.. CC-BY-NC-SA 3.0 3. CK-12 Foundation - Samantha Bacic.. CC-BY-NC-SA 3.0 4. CK-12 Foundation - Samantha Bacic.. CC-BY-NC-SA 3.0 19 CONCEPT 6 Refraction of Mechanical Waves www.ck12.org \u2022 De\ufb01ne and describe refraction of mechanical waves. \u2022 State and use the law of refraction. A straw in a glass of water seen from the side often appears broken, even though it is not. The apparent break is due to the bending of light rays leaving the straw; as the light passes from the water to the glass and from the glass to the air, the light rays are bent. Nonetheless, your eye traces the light ray backward as if the light has followed a straight path from its origin at the straw. Since the light appears to have come from a different place, your eye sees the straw as being broken. Refraction of Mechanical Waves When any wave strikes a boundary between media, some of the energy is re\ufb02ected and some is transmitted. When the wave strikes the media interface at an angle, the transmitted wave will move in a slightly different direction than the incident wave. This phenomenon is known as refraction. 20 www.ck12.org Concept 6. Refraction of Mechanical Waves Consider the image sketched above. Suppose that the waves represented here", " are water waves. The wave crests are represented by the black lines in the image. As such, the distance between two consecutive black lines is the wavelength. Let the red line represent a transition from deep to shallow water. This transition is called the media interface. As the waves hit the boundary, the waves slow down. The right side of the wave reaches the boundary before the left side of the wave, causing the left side to catch up and the angle of propagation to change slightly. This change in direction can be seen in the yellow line, which is slightly angled at the boundary. The refraction of waves across boundaries operates similarly to the method by which tanks are steered. Tanks do not have a steering wheel. Instead, they have an accelerator to produce forward motion and separate brakes on each tread. The operator uses brakes on both treads at the same time in order to stop, but brakes on only one tread to turn the tank. By braking one side, the operator causes that side to slow down or stop while the other side continues at the previous speed, causing the tank to turn towards the slower tread. This sketch shows a wave ray striking an interface between old medium and new medium. A normal line has been drawn as a dotted line perpendicular to the interface. The angle between the incident ray and the normal line is called the angle of incidence, shown as q i, and the angle between the refracted ray and the normal line is called the angle of refraction, q r. 21 www.ck12.org We already understand that the change in the wave direction at the border depends on the difference between the two velocities. This relationship is conveniently expressed in a mathematical relationship: sin qr sin qi = vr vi = lr li The ratio of the sine of the angle of refraction to the sine of the angle of incidence is the same as the ratio of the velocity of the wave in the new medium to the velocity of the wave in the old medium and equal to the ratio of wavelength (l) in the old medium to the wavelength in the new medium. Example Problem: A water wave with a wavelength of 3.00 m is traveling in deep water at 16.0 m/s. The wave strikes a sharp interface with shallow water with an angle of incidence of 53:0. The wave refracts into the shallow water with an angle of refraction of 30:0. What is the velocity of the wave in shallow water and what is its wavelength in the new medium? so sin 30 sin 53 = vr 16:0 m=s and vr = 10:0 m=s. Solution: = vr vi sin qr sin qi so 10:0 m=s 16:0 m=s = lr = lr li vr 3:00 m and lr = 1:88 m. vi Example Problem: The ratio of the sin qr to sin qi is 0.769. 5:00 109 m, what is its wavelength in the original medium? If the wavelength of a wave in a new medium is Solution: Summary 0:769 = lr li so li = 5:00 109 m 0:769 = 6:50 109 m \u2022 When any wave strikes a boundary between media, some of the energy is re\ufb02ected and some is transmitted. \u2022 When a wave strikes the media interface at an angle, the transmitted wave will move in a different direction than the incident wave. This phenomenon is known as refraction. = vr vi \u2022 At any media interface, sin qr sin qi = lr li Practice http://www.youtube.com/watch?v=mH9VwivqjmE Follow up questions. 22 www.ck12.org Concept 6. Refraction of Mechanical Waves 1. What causes refraction? 2. What doesn\u2019t change during refraction? Review 1. A laser beam passes through water and enters a glass block at an angle. The ratio of the speed of the wave in glass to the speed in water is 0.866. If the angle of incidence to the interface is 60, what is the angle of refraction? 2. A ray of light is traveling from air into glass at an angle of 30:0 to the normal line. The speed of the light in air is 3:00 108 m=s and in glass the speed drops to 2:00 108 m=s. What is the angle of refraction? 3. Which of the following change when a water wave moves across a boundary at an angle between deep water and shallow water? a. frequency b. wavelength c. speed d. wave direction e. period 4. Which of the following change when a water wave moves across a boundary exactly along the media interface between deep water and shallow water? a. frequency b. wavelength c. speed d. wave direction e. period 5. The speed of sound is 340 m/s. A particular sound wave has a frequency", " of 320. Hz. a. What is the wavelength of this sound in air? b. If this sound refracts into water where the speed of sound is 4 times faster, what will be the new wavelength? c. What will be the new frequency? 6. When a light ray passes from air into diamond, the angle of incidence is 45:0 and the angle of refraction is 16:7. If the speed of light in air is 3:00 108 m=s, what is the speed of light in diamond? \u2022 refraction: The turning or bending of a wave direction when it passes from one medium to another of different density. References 1. Image copyright cheyennezj, 2013. http://www.shutterstock.com. Used under license from Shutter- stock.com 2. CK-12 Foundation - Samantha Bacic.. CC-BY-NC-SA 3.0 3. Greg Goebel. http://www.public-domain-image.com/full-image/transportation-vehicles-public-domain-ima ges-pictures/tanks-public-domain-images-pictures/m7-priest-self-propelled-105-millimeter-howitzer-tank.jpgcopyright-friendly-image.html. Public Domain 4. CK-12 Foundation - Samantha Bacic.. CC-BY-NC-SA 3.0 23 CONCEPT 7 www.ck12.org Wave Interactions \u2022 Identify ways that waves can interact with matter. \u2022 De\ufb01ne and give examples of wave re\ufb02ection, refraction, and diffraction. Did you ever hear an echo of your own voice? An echo occurs when sound waves bounce back from a surface that they can\u2019t pass through. The woman pictured here is trying to create an echo by shouting toward a large building. When the sound waves strike the wall of the building, most of them bounce back toward the woman, and she hears an echo of her voice. An echo is just one example of how waves interact with matter. How Waves Interact with Matter Waves interact with matter in several ways. The interactions occur when waves pass from one medium to another. The types of interactions are re\ufb02ection, refraction, and diffraction. Each type of interaction is described in detail below. You can see animations of the three types at this URL: http://www.acoustics.salford.ac.uk/schools/teacher/ lesson3/\ufb02ash/whiteboardcomplete.swf Re\ufb02ection An echo is an example of wave re\ufb02ection. Re\ufb02ection occurs when waves bounce back from a surface they cannot pass through. Re\ufb02ection can happen with any type of waves, not just sound waves. For example, light waves can 24 www.ck12.org Concept 7. Wave Interactions also be re\ufb02ected. In fact, that\u2019s how we see most objects. Light from a light source, such as the sun or a light bulb, shines on the object and some of the light is re\ufb02ected. When the re\ufb02ected light enters our eyes, we can see the object. Re\ufb02ected waves have the same speed and frequency as the original waves before they were re\ufb02ected. However, the direction of the re\ufb02ected waves is different. When waves strike an obstacle head on, the re\ufb02ected waves bounce straight back in the direction they came from. When waves strike an obstacle at any other angle, they bounce back at the same angle but in a different direction. This is illustrated in diagram below. In this diagram, waves strike a wall at an angle, called the angle of incidence. The waves are re\ufb02ected at the same angle, called the angle of re\ufb02ection, but in a different direction. Notice that both angles are measured relative to a line that is perpendicular to the wall. FIGURE 7.1 Refraction Refraction is another way that waves interact with matter. Refraction occurs when waves bend as they enter a new medium at an angle. You can see an example of refraction in the picture below. Light bends when it passes from air to water or from water to air. The bending of the light traveling from the \ufb01sh to the man\u2019s eyes causes the \ufb01sh to appear to be in a different place from where it actually is. FIGURE 7.2 Waves bend as they enter a new medium because they start traveling at a different speed in the new medium. For 25 example, light travels more slowly in water than in air. This causes it to refract when it passes from air to water or from water to air. Q: Where would the \ufb01sh appear to be if the man looked down at it from straight above its actual location? A", ": The \ufb01sh would appear to be where it actually is because refraction occurs only when waves (in this case light waves from the \ufb01sh) enter a new medium at an angle other than 90\u00b0. www.ck12.org Diffraction Did you ever notice that you can hear sounds around the corners of buildings even though you can\u2019t see around them? The Figure 7.3 shows why this happens. As you can see from the \ufb01gure, sound waves spread out and travel around obstacles. This is called diffraction. It also occurs when waves pass through an opening in an obstacle. All waves may be diffracted, but it is more pronounced in some types of waves than others. For example, sound waves bend around corners much more than light does. That\u2019s why you can hear but not see around corners. FIGURE 7.3 For a given type of waves, such as sound waves, how much the waves diffract depends on the size of the obstacle (or opening in the obstacle) and the wavelength of the waves. The Figure 7.4 shows how the amount of diffraction is affected by the size of the opening in a barrier. Note that the wavelength of the wave is the distance between the vertical lines. FIGURE 7.4 26 www.ck12.org Summary Concept 7. Wave Interactions \u2022 Three ways that waves may interact with matter are re\ufb02ection, refraction, and diffraction. \u2022 Re\ufb02ection occurs when waves bounce back from a surface that they cannot pass through. \u2022 Refraction occurs when waves bend as they enter a new medium at an angle and start traveling at a different speed. \u2022 Diffraction occurs when waves spread out as they travel around obstacles or through openings in obstacles. Vocabulary \u2022 diffraction : Bending of a wave around an obstacle or through an opening in an obstacle. \u2022 re\ufb02ection : Bouncing back of waves from a barrier they cannot pass through. \u2022 refraction : Bending of waves as they enter a new medium at an angle and change speed. Practice Make a crossword puzzle of terms relating to wave interactions. Include at least seven different terms. You can use the puzzle maker at the following URL. Then exchange and solve puzzles with a classmate. http://puzzlemaker.disc overyeducation.com/CrissCrossSetupForm.asp Review 1. What is re\ufb02ection? What happens if waves strike a re\ufb02ective surface at an angle other than 90\u00b0? 2. De\ufb01ne refraction. Why does refraction occur? 3. When does diffraction occur? How is wavelength related to diffraction? References 1. Zachary Wilson.. CC BY-NC 3.0 2. Zachary Wilson.. CC BY-NC 3.0 3. Student: Flickr:MaxTorrt; Radio: Flickr:Kansir.. CC BY 2.0 4. Zachary Wilson.. CC BY-NC 3.0 27 CONCEPT 8 www.ck12.org Wave Interference \u2022 De\ufb01ne wave interference. \u2022 Compare and contrast constructive and destructive interference. \u2022 Explain how standing waves occur. When raindrops fall into still water, they create tiny waves that spread out in all directions away from the drops. What happens when the waves from two different raindrops meet? They interfere with each other. When Waves Meet When two or more waves meet, they interact with each other. The interaction of waves with other waves is called wave interference. Wave interference may occur when two waves that are traveling in opposite directions meet. The two waves pass through each other, and this affects their amplitude. Amplitude is the maximum distance the particles of the medium move from their resting positions when a wave passes through. How amplitude is affected by wave interference depends on the type of interference. Interference can be constructive or destructive. Constructive Interference Constructive interference occurs when the crests, or highest points, of one wave overlap the crests of the other wave. You can see this in the Figure 8.1. As the waves pass through each other, the crests combine to produce a wave with greater amplitude. You can see an animation of constructive interference at this URL: http://phys23p.sl.psu.e du/phys_anim/waves/embederQ1.20100.html Destructive Interference Destructive interference occurs when the crests of one wave overlap the troughs, or lowest points, of another wave. The Figure 8.2 shows what happens. As the waves pass through each other, the crests and troughs cancel each other out to produce a wave with zero amplitude. You can see an animation of destructive interference at this URL: htt p://phys23p.sl.psu.edu/phys_anim/waves/embederQ1.20200.html 28 www.ck12.org Concept 8. Wave Interference FIGURE 8.1", " Standing Waves Waves may re\ufb02ect off an obstacle that they are unable to pass through. When waves are re\ufb02ected straight back from an obstacle, the re\ufb02ected waves interfere with the original waves and create standing waves. These are waves that appear to be standing still. Standing waves occur because of a combination of constructive and destructive interference. You can see animations of standing waves at the URLs below. http://skullsinthestars.com/2008/05/04/classic-science-paper-otto-wieners-experiment-1890/ http://www.physicsc lassroom.com/mmedia/waves/swf.cfm Q : How could you use a rope to produce standing waves? A : You could tie one end of the rope to a \ufb01xed object, such as doorknob, and move the other end up and down to generate waves in the rope. When the waves reach the \ufb01xed object, they are re\ufb02ected back. The original waves and the re\ufb02ected waves interfere to produce a standing wave. Try it yourself and see if the waves appear to stand still. Summary \u2022 Wave interference is the interaction of waves with other waves. \u2022 Constructive interference occurs when the crests of one wave overlap the crests of the other wave, causing an increase in wave amplitude. \u2022 Destructive interference occurs when the crests of one wave overlap the troughs of the other wave, causing a decrease in wave amplitude. 29 www.ck12.org FIGURE 8.2 \u2022 When waves are re\ufb02ected straight back from an obstacle, the re\ufb02ected waves interfere with the original waves and create standing waves. Vocabulary \u2022 standing wave : Wave appearing to stand still that forms when a wave and its re\ufb02ected wave interfere. \u2022 wave interference : Interaction of waves with other waves. Practice Review wave interference at the following URL. Then do the Check Your Understanding problem at the bottom of the Web page. Be sure to check your answers. http://www.physicsclassroom.com/class/waves/u10l3c.cfm Review 1. What is wave interference? 2. Create a table comparing and contrasting constructive and destructive interference. 30 www.ck12.org Concept 8. Wave Interference 3. What are standing waves? How do they form? References 1. Christopher Auyeung.. CC BY-NC 3.0 2. Christopher Auyeung.. CC BY-NC 3.0 31 www.ck12.org Wave Speed CONCEPT 9 Objective The student will: \u2022 Solve problems involving wavelength, wave speed, and frequency. Vocabulary \u2022 wave equation: Relates wavelength and wave speed. Distance is speed times time, x = vt, so wavelength is wave speed times period l = vT. \u2022 wave speed: How quickly the peak of each wave is moving forward. The Wave Equation Simple harmonic motion, moving back and forth in place, has amplitude along with period and frequency. Wave motion means that the back-and-forth change is also moving through space. This means that there are two further qualities of a wave. \u2022 The wavelength is the distance between two compressions in the direction of motion of the wave, and is represented by the Greek letter lambda, l. For an ocean wave, it would be the distance in meters from the top of one wave to the next. \u2022 The wave speed is how quickly the peak of each wave is moving forward, and is represented by v (for velocity, although for our current purpose the direction is not important). These two are related by the period T of the wave. The period is the time it takes for a wave to complete one cycle, which is the time it takes for a peak to move forward one wavelength. The wave equation expresses this. Distance is speed times time, x = vt, so wavelength is wave speed times period l = vT. This can alternately be expressed in terms of frequency. Suppose there are three waves every second. This is frequency f = 3:0 Hz, equivalent to period T = 1 3 s. This means that during one second, three waves come out from the source. Since each peak is one wavelength, l, ahead of the other, this means that during that one second, the lead wave has gone ahead three wavelengths. The distance the wave goes in one second is the wave speed. So, the wave speed is equal to the frequency times the wavelength, v = l f. Because f = 1 l = vT! v = l T, these are mathematically the same: T = l 1 T = l f The wave equation says distance wavelength is equal to wave speed multiplied by period T. http://demonstrations.wolfram.com/SpeedOfSound/ 32 www.ck12.org Concept 9. Wave Speed FIGURE 9.1 Illustrative Example 1", " a. The ripple tank arm in Figure 9.1 has a period of 0.25 s and the length of the arrow in the \ufb01gure is 3.76 cm. What is the wavelength of the water waves if the picture is to scale? Answer: There are four wavelengths from the start to the end of the arrow. Wavelength is: l = 3:76cm 4 = 0:94cm b. What is the velocity of the wave? Answer: Since the period is T = 0:25 s, the frequency is, velocity is v = l f = (0:94 cm)(4:0 Hz) = 3:76 cm s. Check Your Understanding f = 1 T = 1 0:25 s = 4:0 Hz or four cycles per second. So the 1. A sound wave travels at the speed of 343:0 m/s through the air at room temperature, 20:0C(68:0F). If the frequency of the sound is 261.6 Hz (a middle-C note), what is the wavelength of the note? Answer = 343:0 m/s 261:6 Hz = 1:311 m 2. X-rays are electromagnetic waves. A particular type of x-ray has a frequency of 3:0 1017 Hz and a wavelength of 1:0 109 m. What is the velocity of this type of x-ray? = 3:0 108 m/s Answer: v = l f = 1:0 109 m)(3:0 1017 1 s This result is true for any electromagnetic radiation traveling in a vacuum, including visible light. 3. Compared to the speed of sound in air at a temperature of 20C, how many times faster is the speed of light through the air at the same temperature? Answer: We\u2019ll assume that the velocity of light vL = 3:0 108 m s is approximately the same through the air as it is 33 www.ck12.org through a vacuum. Number 1 above gave the velocity of sound for a temperature of 20C as vs = 343:0 m s. Answer: VL = 874; 635:57! 8:7 105. The velocity of light is indeed a good deal greater than the velocity of Vs sound in air. In fact, the velocity of light in vacuum is the greatest velocity that exists. No material object can travel at this velocity! We will discuss these ideas in Chapter 23 (The Special Theory of Relativity). 4. At the moment a lightning \ufb02ash is seen, a person begins to count off seconds. If he hears the thunder after seven seconds, approximately how far away from the person did the lightning \ufb02ash originate? Answer : We\u2019ll assume that the sound travels at a velocity of 343 m/s. In seven seconds the sound has traveled x = vt! x = 343 m s (7s) = 2401! 2400 m. 5. Sound travels about 1,500 m/s in water. A destroyer ship locates an enemy submarine using a sonar signal which takes a 4.3 s to travel to, re\ufb02ect, and return to the destroyer. How far is the sub from the destroyer? Answer : The time for the signal to reach the sub is half of the total time of travel,! 4:3 s x = vt! x = 1; 500 m s (2:15 s) = 3225! 3200 m or 3.2 km. 2 = 2:15! 2:2 s. Using References 1. CK-12 Foundation - Raymond Chou.. CC-BY-NC-SA 3.0 34 www.ck12.org Concept 10. Sound Waves CONCEPT 10 \u2022 De\ufb01ne sound. \u2022 Describe sound waves and how they are generated. \u2022 Identify media through which sound waves can travel. Sound Waves Crack! Crash! Thud! That\u2019s what you\u2019d hear if you were in the forest when this old tree cracked and came crashing down to the ground. But what if there was nobody there to hear the tree fall? Would it still make these sounds? This is an old riddle. To answer the riddle correctly, you need to know the scienti\ufb01c de\ufb01nition of sound. De\ufb01ning Sound In science, sound is de\ufb01ned as the transfer of energy from a vibrating object in waves that travel through matter. Most people commonly use the term sound to mean what they hear when sound waves enter their ears. The tree above generated sound waves when it fell to the ground, so it made sound according to the scienti\ufb01c de\ufb01nition. But the sound wasn\u2019t detected by a person\u2019s ears if there was nobody in the forest. So the answer to the riddle is both yes and no! How Sound Waves Begin All sound waves begin with vibrating", " matter. Look at the \ufb01rst guitar string on the left in the Figure 10.1. Plucking the string makes it vibrate. The diagram below the \ufb01gure shows the wave generated by the vibrating string. The moving string repeatedly pushes against the air particles next to it, which causes the air particles to vibrate. The vibrations spread through the air in all directions away from the guitar string as longitudinal waves. In longitudinal waves, particles of the medium vibrate back and forth parallel to the direction that the waves travel. You can see an animation of sound waves traveling through air at this URL: http://www.mediacollege.com/audio/01/sound-wave s.html Q: If there were no air particles to carry the vibrations away from the guitar string, how would sound reach the ear? A: It wouldn\u2019t unless the vibrations were carried by another medium. Sound waves are mechanical waves, so they can travel only though matter and not through empty space. 35 www.ck12.org FIGURE 10.1 A Ticking Clock The fact that sound cannot travel through empty space was \ufb01rst demonstrated in the 1600s by a scientist named Robert Boyle. Boyle placed a ticking clock in a sealed glass jar. The clock could be heard ticking through the air and glass of the jar. Then Boyle pumped the air out of the jar. The clock was still ticking, but the ticking sound could no longer be heard. That\u2019s because the sound couldn\u2019t travel away from the clock without air particles to pass the sound energy along. You can see an online demonstration of the same experiment\u2014with a modern twist\u2014at this URL: http://www.youtube.com/watch?v=b0JQt4u6-XI MEDIA Click image to the left for more content. Sound Waves and Matter Most of the sounds we hear reach our ears through the air, but sounds can also travel through liquids and solids. If you swim underwater\u2014or even submerge your ears in bathwater\u2014any sounds you hear have traveled to your ears through the water. Some solids, including glass and metals, are very good at transmitting sounds. Foam rubber and heavy fabrics, on the other hand, tend to muf\ufb02e sounds. They absorb rather than pass on the sound energy. Q: How can you tell that sounds travel through solids? A: One way is that you can hear loud outdoor sounds such as sirens through closed windows and doors. You can also hear sounds through the inside walls of a house. For example, if you put your ear against a wall, you may be able to eavesdrop on a conversation in the next room\u2014not that you would, of course. 36 www.ck12.org Summary Concept 10. Sound Waves \u2022 In science, sound is de\ufb01ned as the transfer of energy from a vibrating object in waves that travel through matter. \u2022 All sound waves begin with vibrating matter. The vibrations generate longitudinal waves that travel through matter in all directions. \u2022 Most sounds we hear travel through air, but sounds can also travel through liquids and solids. Vocabulary \u2022 sound : Transfer of energy from a vibrating object in longitudinal waves that travel through matter. Practice Watch the video \u201cHow Sound Waves Travel\u201d at the following URL. Then explain how sound waves begin and how they travel, using the human voice as an example. http://www.youtube.com/watch?v=_vYYqRVi8vY MEDIA Click image to the left for more content. Review 1. How is sound de\ufb01ned in science? How does this de\ufb01nition differ from the common meaning of the word? 2. Hitting a drum, as shown in the Figure 10.2, generates sound waves. Create a diagram to show how the sound waves begin and how they reach a person\u2019s ears. FIGURE 10.2 3. How do you think earplugs work? 37 References 1. Guitar string photo by Flickr:jar(); illustration by Christopher Auyeung (CK-12 Foundation).. CC BY 2.0 2. S.L. Ratigan.. CC BY 2.0 www.ck12.org 38 www.ck12.org Concept 11. Frequency and Pitch of Sound CONCEPT 11 Frequency and Pitch of Sound \u2022 De\ufb01ne the pitch of sound. \u2022 Relate the pitch of sound to the frequency of sound waves. \u2022 Identify infrasound and ultrasound. A marching band passes you as it parades down the street. You heard it coming from several blocks away. Now that the different instruments have \ufb01nally reached you, their distinctive sounds can be heard. The tiny piccolos trill their bird-like high notes, and the big tubas rumble out their booming bass notes. Clearly, some sounds are higher or lower than", " others. High or Low How high or low a sound seems to a listener is its pitch. Pitch, in turn, depends on the frequency of sound waves. Wave frequency is the number of waves that pass a \ufb01xed point in a given amount of time. High-pitched sounds, like the sounds of the piccolo in the Figure 11.1, have high-frequency waves. Low-pitched sounds, like the sounds of the tuba Figure 11.1, have low-frequency waves. For a video demonstration of frequency and pitch, go to this URL: http://www.youtube.com/watch?v=irqfGYD2UKw Can You Hear It? The frequency of sound waves is measured in hertz (Hz), or the number of waves that pass a \ufb01xed point in a second. Human beings can normally hear sounds with a frequency between about 20 Hz and 20,000 Hz. Sounds with frequencies below 20 hertz are called infrasound. Infrasound is too low-pitched for humans to hear. Sounds with frequencies above 20,000 hertz are called ultrasound. Ultrasound is too high-pitched for humans to hear. 39 www.ck12.org FIGURE 11.1 Some other animals can hear sounds in the ultrasound range. For example, dogs can hear sounds with frequencies as high as 50,000 Hz. You may have seen special whistles that dogs\u2014but not people\u2014can hear. The whistles produce sounds with frequencies too high for the human ear to detect. Other animals can hear even higher-frequency sounds. Bats, like the one pictured in the Figure 11.2, can hear sounds with frequencies higher than 100,000 Hz! FIGURE 11.2 Q: Bats use ultrasound to navigate in the dark. Can you explain how? A: Bats send out ultrasound waves, which re\ufb02ect back from objects ahead of them. They sense the re\ufb02ected sound waves and use the information to detect objects they can\u2019t see in the dark. This is how they avoid \ufb02ying into walls and trees and also how they \ufb01nd \ufb02ying insects to eat. Summary \u2022 How high or low a sound seems to a listener is its pitch. Pitch, in turn, depends on the frequency of sound waves. \u2022 High-frequency sound waves produce high-pitched sounds, and low-frequency sound waves produce low- pitched sounds. 40 www.ck12.org Concept 11. Frequency and Pitch of Sound \u2022 Infrasound has wave frequencies too low for humans to hear. Ultrasound has wave frequencies too high for humans to hear. Vocabulary \u2022 infrasound : Sound with a frequency below the range of human hearing (less than 20 hertz). \u2022 pitch : How high or low a sound seems to a listener. \u2022 ultrasound : Sound with a frequency above the range of human hearing (greater than 20,000 hertz). \u2022 wave frequency : Number of waves that pass a \ufb01xed point in a given amount of time. Practice At the following URL, complete the interactive module to review and test your knowledge of the frequency and pitch of sound. http://www.engineeringinteract.org/resources/oceanodyssey/flash/concepts/pitch.htm Review 1. What is the pitch of sound? 2. How is the pitch of sound related to the frequency of sound waves? 3. De\ufb01ne infrasound and ultrasound. References 1. Piccolo: U.S. Navy photo by Chief Mass Communication Specialist David Rush; Tuba: Bob Fishbeck.. Public Domain 2... Public Domain 41 CONCEPT 12 \u2022 Give the speed of sound in dry air at 20 \u00b0C. \u2022 Describe variation in the speed of sound in different media. \u2022 Explain the effect of temperature on the speed of sound. www.ck12.org Speed of Sound Has this ever happened to you? You see a \ufb02ash of lightning on the horizon, but several seconds pass before you hear the rumble of thunder. The reason? The speed of light is much faster than the speed of sound. What Is the Speed of Sound? The speed of sound is the distance that sound waves travel in a given amount of time. You\u2019ll often see the speed of sound given as 343 meters per second. But that\u2019s just the speed of sound under a certain set of conditions, speci\ufb01cally, through dry air at 20 \u00b0C. The speed of sound may be very different through other matter or at other temperatures. Speed of Sound in Different Media Sound waves are mechanical waves, and mechanical waves can only travel through matter. The matter through which the waves travel is called the medium (plural, media). The Table 12.1 gives the speed of sound in several different media. Generally, sound waves travel most quickly through solids, followed by liquids, and", " then by gases. Particles of matter are closest together in solids and farthest apart in gases. When particles are closer together, they can more quickly pass the energy of vibrations to nearby particles. You can explore the speed of sound in different media at this URL: http://www.ltscotland.org.uk/resources/s/sound/speedofsound.asp?strReferringChannel=resources&strReferringPageI D=tcm:4-248291-64 42 www.ck12.org Concept 12. Speed of Sound Medium (20 \u00b0C) Dry Air Water Wood Glass Aluminum TABLE 12.1: speed of sound Speed of Sound Waves (m/s) 343 1437 3850 4540 6320 Q: The table gives the speed of sound in dry air. Do you think that sound travels more or less quickly through air that contains water vapor? (Hint: Compare the speed of sound in water and air in the table.) A: Sound travels at a higher speed through water than air, so it travels more quickly through air that contains water vapor than it does through dry air. Temperature and Speed of Sound The speed of sound also depends on the temperature of the medium. For a given medium, sound has a slower speed at lower temperatures. You can compare the speed of sound in dry air at different temperatures in the following Table 12.2. At a lower temperature, particles of the medium are moving more slowly, so it takes them longer to transfer the energy of the sound waves. Temperature of Air 0 \u00b0C 20 \u00b0C 100 \u00b0C TABLE 12.2: speed of sound Speed of Sound Waves (m/s) 331 343 386 Q: What do you think the speed of sound might be in dry air at a temperature of -20 \u00b0C? A: For each 1 degree Celsius that temperature decreases, the speed of sound decreases by 0.6 m/s. So sound travels through dry, -20 \u00b0C air at a speed of 319 m/s. Summary \u2022 The speed of sound is the distance that sound waves travel in a given amount of time. The speed of sound in dry air at 20 \u00b0C is 343 meters per second. \u2022 Generally, sound waves travel most quickly through solids, followed by liquids, and then by gases. \u2022 For a given medium, sound waves travel more slowly at lower temperatures. Vocabulary \u2022 speed of sound : Speed at which sound waves travel, which is 343 m/s in dry air at 20 \u00b0C. Practice At the following URL, read about the speed of sound in different materials. Be sure to play the animation. Then answer the questions below. http://www.ndt-ed.org/EducationResources/HighSchool/Sound/speedinmaterials.htm 43 1. Describe what you hear when you play the animation. Explain your observations. 2. Name two properties of materials that affect the speed of sound waves. How do they affect the speed of sound? 3. Explain why sound waves moves more quickly through warmer air than cooler air. www.ck12.org Review 1. What is the speed of sound in dry air at 20 \u00b0C? 2. Describe variation in the speed of sound through various media. 3. Explain how temperature affects the speed of sound. 44 www.ck12.org Concept 13. Resonance with Sound Waves CONCEPT 13 Resonance with Sound Waves Objectives The student will: \u2022 Understand the conditions for resonance. \u2022 Solve problems with strings and pipes using the condition for resonance. Vocabulary \u2022 beat frequency \u2022 natural frequency: The frequency at which a system vibrates normally when given energy without outside interference. \u2022 resonance: Timing force to be the same as natural frequency. \u2022 sympathetic vibrations Introduction Many systems have a tendency to vibrate. When the forced vibration frequency is the same as the natural frequency, the amplitude of vibration can increase tremendously. A well-known example of this situation is pushing a person on a swing, Figure below. We know from study of simple pendulums that without being pushed, the person in the swing rocks back and forth with a frequency that depends on gravity and the length of the chain. f = 1 2p r g L This is one example of a natural frequency \u2013 the frequency at which a system vibrates normally when given energy without outside interference. Pushing on the person in the swing will affect the amplitude of the swinging. This is called forced vibration \u2013 when a periodic force from one object (the person pushing) affects the vibration of another object (the person swinging). To get the most effect, the person pushing will start just at the very back of the swing. In other words, the frequency of how often they push is exactly the same as the frequency of the swing. Suppose they do not push at the right time, but instead push at some other frequency. That would mean that sometimes they are pushing forward when the swing is still going backward. In that case, the swing would slow down \u2013 i.e. the", " amplitude of the swing will be reduced. Timing the pushes to be the same as the natural frequency is called resonance. For this reason, the natural frequency is also known as the resonant frequency. If the pushes are timed just right, then even if each individual push is small, the vibration will get larger with each push. 45 www.ck12.org FIGURE 13.1 A classic example of an unfortunate consequence of a forced vibration at resonant frequency is what happened to the Tacoma Narrows Bridge, in 1940. See the link below. http://www.youtube.com/watch?v=3mclp9QmCGs In Figure below, the bridge is beginning to resonate, in part, due to the frequency of vibration of the wind gusts. In Figure below, we see that the bridge is no longer able to respond elastically to the tremendous amplitude of vibration from the forced vibration of wind energy (at its resonant frequency), and it is torn apart. FIGURE 13.2 Modern bridges are built to avoid this effect, but through history there are a number of documented situations where a forced vibration at resonance had dire results. The Broughton Suspension Bridge (1831) and the Angers Bridge (1850) are two examples of bridges believed to have collapsed due to the effect of soldiers marching at a regular pace that caused resonance. The Albert Bridge in West London, England has been nicknamed The Trembling Lady because it has been set into resonance so often by marching soldiers. Though soldiers no longer march across the bridge, there still remains a sign of concern as shown in Figure below. 46 www.ck12.org Concept 13. Resonance with Sound Waves FIGURE 13.3 FIGURE 13.4 Sympathetic Vibrations There is a typical classroom physics demonstration where one tuning fork is set into motion and an identical tuning fork, if placed closed enough, will also be set vibrating, though with smaller amplitude. The same effect occurs when tuning a guitar. One string is plucked and another, whose length is shortened by holding it down some distance from the neck of the guitar, will also be set into vibration. When this condition is met, both strings are vibrating with the same frequency. We call this phenomenon sympathetic vibration. 47 www.ck12.org MEDIA Click image to the left for more content. FIGURE 13.5 https://www.youtube.com/watch?v=tnS0SYF4pYE In Figure above, a set of pendulums are \ufb01xed to a horizontal bar that can be easily jostled. Pendulums A and E have the same length. If one of them is set swinging, the horizontal bar will be forced into moving with a period equal to that of the pendulum, which, in turn, will cause the other pendulum of the same length to begin swinging. Any pendulum that is close in length to pendulums A and E, for example, pendulum D, will also begin to swing. Pendulum D will swing with smaller amplitude than pendulums A and E since its resonant frequency is not quite the same as pendulums A and E. Pendulums with lengths dramatically different from pendulums A and E will hardly move at all. You can try a similar demonstration out yourself with the following simulation: http://phet.colorado.edu/en/simulation/resonance Resonance is a very common phenomenon, especially with sound. The length of any instrument is related to what note it plays. If you blow into the top of a bottle, for example, the note will vary depending on the height of air in the bottle. This plays an important role in human voice generation. The length of the human vocal tube is between 17 cm and 18 cm. The typical frequencies of human speech are in the range of 100 Hz to 5000 Hz. FIGURE 13.6 48 www.ck12.org Concept 13. Resonance with Sound Waves By using the muscles in their throat, singers change the note they sing. A dramatic example of this is breaking glass with the human voice. By singing at exactly the resonant frequency of a delicate wine glass, the glass will resonate with the note and shatter. http://www.youtube.com/watch?v=7YmuOD5X4L8 The resonance of sound is also a mechanical analogue to how a radio set receives a signal. The Figure below shows one of the earliest radio designs, called a crystal radio because the element which detected the radio waves was a crystalline mineral such as galena. FIGURE 13.7 An old crystal radio set. In modern times, the air is \ufb01lled with all manner of radio waves. In order to listen to your favorite radio station, you must tune your radio to resonate with only the frequency of the radio station. When you hear the tuning number of a radio station, such as \u201c101.3 FM\u201d, that is the", " measure in megahertz, 101:3 MHz = 1:013 108 Hz. The coiled wire (called an inductor) and the capacitor in Figure above act together to tune in a speci\ufb01c radio station. Effectively, the capacitor and inductor act analogously to a pendulum of a speci\ufb01c length that will only respond to vibrations of another pendulum of the same length. So when tuned, only a speci\ufb01c radio frequency will cause resonance in the radio antenna. Strings Fixed at Both Ends A case of natural frequency that you can observe plainly is when you pluck a string or stretched rubber band. Normally, the string will vibrate at a single widest point in the middle. This is called the fundamental or \ufb01rst harmonic resonance of the string. This is the same as the natural frequency of a simple pendulum or mass on a spring. Because it vibrates all along its length, though, the string also lets us see further patterns of resonance. By vibrating the end of the string rather than just plucking it, we can force vibration at frequencies other than the \ufb01rst harmonic. When the string is set into vibration, energy will travel down the string and re\ufb02ect back toward the end where the waves are being generated. This steady pattern of vibration is called a standing wave. The points where the re\ufb02ecting waves interfere destructively with the \u201cgenerated\u2019 waves are called nodes. The points where the re\ufb02ecting waves interfere constructively with the generated waves are called anti-nodes. Figure below shows a string \ufb01xed at both ends vibrating in its fundamental mode. There are two nodes shown and one antinode. The dashed segment represents the re\ufb02ected wave. If you compare the wave shape of the \ufb01rst harmonic to the wave shape of Figure above, it will be apparent that the \ufb01rst harmonic contains one-half of a wavelength, l. Therefore L, the length of the unstretched string, is equal to one-half the wavelength, which is 1 2 l1 = L! l1 = 2L: 49 www.ck12.org FIGURE 13.8 The second harmonic contains an entire wavelength 2 2 l2 = L! l2 = L as shown in Figure below. FIGURE 13.9 And the third harmonic contains one and one-half wavelengths 3 2 l3 = L! l3 = 2 3 L. FIGURE 13.10 If the pattern continues then the fourth harmonic will have a wavelength of 4 pressions for the length of the string in terms of the wavelength, a simple pattern emerges: L = 1. We can express the condition for standing waves (and of resonance, as well) as L = n the length of string and n = 1; 2; 3 : : :. 2 l4 = L! l4 = 1 2 L. Looking at the ex2 l2; 3 2 l3; 4 2 l4 : : : n L, where L is 2 l1; 2 2 ln or ln = 2 Check Your Understanding 1. How many nodes and anti-nodes are shown in Figure above.? Answer: There are three nodes and two anti-nodes. 2. If the length of the unstretched string is 20 cm, what is the wavelength for the 10th harmonic? Answer: ln = 2 5 (20 cm) = 4 cm n L! l10 = 2 10 L = 1 Strings Fixed at One End and Opened at One End A string \ufb01xed at one only end displays a different standing wave pattern. In this case, the unbounded end of a string of length L is an antinode. The fundamental mode (the \ufb01rst harmonic) for the length L of string contains only one- 50 www.ck12.org Concept 13. Resonance with Sound Waves fourth of a wavelength as shown in Figure below below. Therefore L, the length of the unstretched string, is equal to one-quarter the wavelength, which is 1 4 l1 = L! l1 = 4L. FIGURE 13.11 The second harmonic contains three-quarters of a wavelength 3 4 l2 = L! l2 = 4 3 L as shown in Figure above FIGURE 13.12 The third harmonic contains \ufb01ve-fourths of a wavelength 5 4 l2 = L! l2 = 4 5 L as shown in Figure above The third harmonic contains \ufb01ve-fourths wavelengths as shown in Figure below. If the pattern continues, then the fourth harmonic will have a wavelength of 7 expressions for the length of the string in terms of the wavelength, a simple pattern emerges 1 We can express the condition for resonance as L = n. 7 L. Looking at the 4 l3; 7 4 l4 : : :. n L, where", " L is the length of string and n = 1; 3; 5 : : : 4 l4 = L! l4 = 4 4 ln or ln = 4 4 l2; 5 4 l1; 3 As long as the tension in the string remains \ufb01xed, the velocity of the wave along the string remains constant. Does it seem reasonable that a sagging string will not support the same wave velocity as a taut string? Since v = l f product l f is constant as long as the wave velocity remains constant. Therefore, for a string vibrating in many different modes, we have v = l1 f1 = l2 f2 = l3 f3 : : :. 51 www.ck12.org FIGURE 13.13 Illustrative Example 1 1a. If the frequency of the \ufb01rst harmonic for a string \ufb01xed at both ends is f1, determine the frequency for successive harmonics in terms of f1. Answer: We know that ln = 2 v = l1 f1 = 2L f1 and substituted into fn = n fn = n n L and v = l f. Combining, we have v = 2 2L v. But v can be expressed n L f! fn = n 2L v, giving 2L (2L f1) = n f1! fn = n f1 1b. If the \ufb01rst harmonic has frequency of 261 Hz, what frequencies do the second and third harmonics have? Answer: Since fn = n f1! 2(261 Hz) = 522 Hz fn = n f1! 3(261 Hz) = 783 Hz All whole number multiples of the \ufb01rst harmonic (the fundamental) are called harmonics. String instruments, as well as non-string instruments, can actually vibrate with many different frequencies simultaneously (called modes). For example, a string may vibrate with frequencies 261 Hz, 522 Hz and 783 Hz simultaneously. One of attributes of the \u201cquality\u201d or \u201ctimbre\u201d of musical instruments depends upon the combination of the various overtones produced by the instrument. Check Your Understanding 1. A tuning fork has a frequency of 512 Hz stamped on it. When it is struck, a student claims she can hear higher frequencies from the tuning fork. Is this possible? Answer : Yes, it is. The tuning fork may be producing harmonics, in which case the student may be hearing frequencies in multiples of 512 Hz, such as 1,024 Hz and 1,536 Hz. 2. A string with a fundamental frequency of 220 Hz vibrates in its third harmonic with a wavelength of 60 cm. What is the wave velocity on the string? 52 www.ck12.org Concept 13. Resonance with Sound Waves Answer : v = l f but f = 3 f1 = 3(220 Hz) = 660 Hz, so l = 0:60 m v = (660 Hz)(0:60 m) = 396 m/s. Open and Closed Pipes and Tubes In our discussions of pipes, the length of the pipe will be assumed to be much greater than the diameter of the pipe. An open pipe, as the name implies, has both ends open. Though open pipes have antinodes at their ends, the resonant conditions for standing waves in an open pipe are the same as for a string \ufb01xed at both ends. Thus for an open pipe we have: for n = 1; 2; 3:::; L = n 2 ln, or ln = 2 n L. There is a simple experiment your instructor may have you do in class that demonstrates resonance in an open tube. Roll two sheets of long paper into two separate tubes and use a small amount of tape to keep them rolled. Have the diameter of one tube just small enough to \ufb01t inside the other tube so the inside tube can freely slide back and forth. Hold a struck tuning fork (your instructor will make sure the frequency is adequate) close to the end of the outer tube while the inside tube is moved slowly. When the total length of the tubes is the proper length to establish resonance, you\u2019ll hear a noticeable increase in the volume of the sound. At this moment, there are standing waves present in the tubes. A closed pipe is closed at only one end. Closed pipes have the same standing wave patterns as a string \ufb01xed at one end and unbound at the other end. They therefore have the same resonant conditions as a string \ufb01xed at only one end, for n = 1; 3 ln or ln = 4 n L. A closed pipe supporting the \ufb01rst harmonic (the fundamental frequency) will \ufb01t one-fourth of the wavelength, the second harmonic will \ufb01t three-fourths, and so on, as shown in Figure below. Compare", " these pictures to those in Figure above for a string \ufb01xed at only one end FIGURE 13.14 A standard physics laboratory experiment is to \ufb01nd the velocity of sound by using a tuning fork that vibrates over a closed pipe as shown in Figure Figure above. The water level in a pipe is slowly changed until the \ufb01rst harmonic is heard. http://demonstrations.wolfram.com/ResonanceInOpenAndClosedPipes/ 53 www.ck12.org FIGURE 13.15 Illustrative Example 2 Resonance is established in a hollow tube similar to that shown in Figure above with a tuning fork of 512 Hz. The distance from the tube opening to the water level is 16.8 cm. a. What is the velocity of sound according to this experiment? Answer: The wave velocity equation is v = l f. One-fourth of the wave occupies the length of the tube for the \ufb01rst harmonic. So the wavelength of the resonant wave must be four times the length of the hollow tube. That is, l1 = 4L = 4 16:8 cm = 67:2 cm = 0:672 m v = (0:672 m)(512 Hz) = 344 m/s b. The velocity of sound changes with temperature as given by the formula v = 330 + 0:6T, where T is the temperature in degrees centigrade. Using the result of part A, determine the temperature at the location the experiment was conducted. Answer: We simply set the result of part A equal to the given equation: 344 = 330 + 0:6T! T = 23:3C (or about 74F ). References 1. Evan Long (Flickr: Clover_1). http://www.\ufb02ickr.com/photos/clover_1/4915240660/. CC-BY-NC 2.0 54 www.ck12.org Concept 13. Resonance with Sound Waves 2. Barney Elliot, Prelinger Archives. struction.ogg. Public Domain http://commons.wikimedia.org/wiki/File:Tacoma_Narrows_Bridge_de 3.. http://commons.wikimedia.org/wiki/File:Tacoma-narrows-bridge-collapse.jpg. Public Domain 4. Russell James Smith (Flickr: russelljsmith). http://www.\ufb02ickr.com/photos/russelljsmith/2146210247/. CC-BY 2.0 5. CK-12 Foundation - Raymond Chou.. CC-BY-NC-SA 3.0 6. CK-12 Foundation - Raymond Chou, using public domain image by Mariana Ruiz Villarreal (Wikimedia:. CC-BY- http://commons.wikimedia.org/wiki/File:Respiratory_system_complete_en.svg LadyofHats). NC-SA 3.0 7. CK-12 Foundation - Ira Nirenberg.. CC-BY-NC-SA 3.0 8. CK-12 Foundation - Ira Nirenberg.. CC-BY-NC-SA 3.0 9. CK-12 Foundation - Ira Nirenberg.. CC-BY-NC-SA 3.0 10. CK-12 Foundation - Ira Nirenberg.. CC-BY-NC-SA 3.0 11. CK-12 Foundation - Ira Nirenberg.. CC-BY-NC-SA 3.0 12. CK-12 Foundation - Ira Nirenberg.. CC-BY-NC-SA 3.0 13. CK-12 Foundation - Ira Nirenberg.. CC-BY-NC-SA 3.0 14. CK-12 Foundation - Ira Nirenberg.. CC-BY-NC-SA 3.0 15. CK-12 Foundation - Ira Nirenberg.. CC-BY-NC-SA 3.0 55 CONCEPT 14 www.ck12.org Sound in a Tube Students will learn how to analyze and solve problems where standing waves (and hence sound) is produced in a tube. Students will learn how to analyze and solve problems where standing waves (and hence sound) is produced in a tube. Key Equations v = l f for a tube closed at one end f = nv=4L, where n is always odd for a tube open at both ends f = nv=2L, where n is an integer Guidance In the case of a tube that is open at one end, a node is forced at the closed end (no air molecules can vibrate up and down) and an antinode occurs at the open end (here, air molecules are free to move). A different spectrum of standing waves is produced. For instance, the fundamental standing sound wave produced in a tube closed at one end is shown below.", " In this case, the amplitude of the standing wave is referring to the magnitude of the air pressure variations. This standing wave is the \ufb01rst harmonic and one can see that the wavelength is l = 4L. Since v = l f, the frequency of oscillation is f = v=4L. In general, the frequency of oscillation is f = nv=4L, where n is always odd. Example 1 Question The objects A, B, and C below represent graduated cylinders of length 50 cm which are \ufb01lled with water to the depths of 10, 20 and 30 cm, respectively as shown. 56 www.ck12.org Concept 14. Sound in a Tube a) If you blow in each of these tubes, which (A,B,C) will produce the highest frequency sound? b) What is the wavelength of the 1st harmonic (i.e. fundamental) of tube B? c) The speed of sound at room temperature is about 343 m/s. What is the frequency of the 1st harmonic for tube B? Solution a) The water forms the bottom of the tube and thus where the node of the wave will be. Thus the air column is where the sound wave can exist. The larger the air column, the larger the wavelength. Frequency is inversely proportional to wavelength, thus the tube with the smallest air column will have the highest frequency. So the answer is tube C. b) The air column is 50 cm - 20 cm = 30 cm. The \ufb01rst harmonic has a quarter wavelength in the tube. Thus l = 4 L. Thus, l = 4 30cm = 120cm c) Using the wave equation for the \ufb01rst harmonic (thus, n = 1) of a tube open at one end we get f = v 286Hz 4L = 343m/s 1:2m = Watch this Explanation MEDIA Click image to the left for more content. 57 www.ck12.org MEDIA Click image to the left for more content. Time for Practice 1. 2. Aborigines, the native people of Australia, play an instrument called the Didgeridoo like the one shown above. The Didgeridoo produces a low pitch sound and is possibly the world\u2019s oldest instrument. The one shown above is about 1.3 m long and open at both ends. a. Knowing that when a tube is open at both ends there must be an antinode at both ends, draw the \ufb01rst 3 harmonics for this instrument. b. Calculate the frequency of the \ufb01rst 3 harmonics assuming room temperature and thus a velocity of sound of 340 m/s. Then take a shot at deriving a generic formula for the frequency of the n th standing wave mode for the Didgeridoo, as was done for the string tied at both ends and for the tube open at one end. 3. Students are doing an experiment to determine the speed of sound in air. They hold a tuning fork above a large empty graduated cylinder and try to create resonance. The air column in the graduated cylinder can be adjusted by putting water in it. At a certain point for each tuning fork a clear resonance point is heard. The students adjust the water \ufb01nely to get the peak resonance then carefully measure the air column from water to top of air column. (The assumption is that the tuning fork itself creates an anti-node and the water creates a node.) The following data were collected: TABLE 14.1: Wavelength (m) Speed of sound (m/s) Frequency fork (Hz) 184 328 384 512 1024 of tuning Length of air column (cm) 46 26 22 16 24 (a) Fill out the last two columns in the data table. (b) Explain major inconsistencies in the data or results. (c) The graduated cylinder is 50 cm high. Were there other resonance points that could have been heard? If so what would be the length of the wavelength? (d) What are the inherent errors in this experiment? 3. Peter is playing tones by blowing across the top of a glass bottle partially \ufb01lled with water. He notices that if he blows softly he hears a lower note, but if he blows harder he hears higher frequencies. (a) In the 120 cm long tubes below draw three diagrams showing the \ufb01rst three harmonics produced in the tube. Please draw the waves as transverse even though we know sound waves are longitudinal (reason for this, obviously, is that it 58 www.ck12.org Concept 14. Sound in a Tube is much easier to draw transverse waves rather than longitudinal). Note that the tube is CLOSED at one end and OPEN at the other. (b) Calculate the frequencies of the \ufb01rst three harmonics played in this tube, if the speed of sound in the tube is 340 m/s. (c) The speed of sound in", " carbon dioxide is lower than in air. If the bottle contained CO2 instead of air, would the frequencies found above be higher or lower? Knowing that the pitch of your voice gets higher when you inhale helium, what can we say about the speed of sound in He. Answers: 1. (b) 131 Hz, 262 Hz, 393 Hz; formula is same as closed at both ends 2. Discuss in class 3. (b) 70.8 Hz, 213 Hz, 354 Hz (c) voice gets lower pitch. Speed of sound in He must be faster by same logic. 59 Physics Unit 11: Electromagnetic Waves Patrick Marshall Jean Brainard, Ph.D. Ck12 Science Say Thanks to the Authors Click http://www.ck12.org/saythanks (No sign in required) www.ck12.org To access a customizable version of this book, as well as other interactive content, visit www.ck12.org AUTHORS Patrick Marshall Jean Brainard, Ph.D. Ck12 Science CK-12 Foundation is a non-pro\ufb01t organization with a mission to reduce the cost of textbook materials for the K-12 market both in the U.S. and worldwide. Using an open-content, web-based collaborative model termed the FlexBook\u00ae, CK-12 intends to pioneer the generation and distribution of high-quality educational content that will serve both as core text as well as provide an adaptive environment for learning, powered through the FlexBook Platform\u00ae. Copyright \u00a9 2014 CK-12 Foundation, www.ck12.org The names \u201cCK-12\u201d and \u201cCK12\u201d and associated logos and the terms \u201cFlexBook\u00ae\u201d and \u201cFlexBook Platform\u00ae\u201d (collectively \u201cCK-12 Marks\u201d) are trademarks and service marks of CK-12 Foundation and are protected by federal, state, and international laws. Any form of reproduction of this book in any format or medium, in whole or in sections must include the referral attribution link http://www.ck12.org/saythanks (placed in a visible location) in addition to the following terms. Except as otherwise noted, all CK-12 Content (including CK-12 Curriculum Material) is made available to Users in accordance with the Creative Commons Attribution-Non-Commercial 3.0 Unported (CC BY-NC 3.0) License (http://creativecommons.org/ licenses/by-nc/3.0/), as amended and updated by Creative Commons from time to time (the \u201cCC License\u201d), which is incorporated herein by this reference. Complete terms can be found at http://www.ck12.org/terms. Printed: April 6, 2014 iii Contents www.ck12.org 1 6 9 12 16 21 24 27 32 35 Contents 1 Electromagnetic Waves 2 Properties of Electromagnetic Waves 3 Electromagnetic Spectrum 4 Law of Re\ufb02ection 5 Refraction of Light 6 Total Internal Re\ufb02ection 7 Single Slit Interference 8 Double Slit Interference 9 Diffraction Gratings 10 Wave-Particle Theory iv www.ck12.org Concept 1. Electromagnetic Waves CONCEPT 1 Electromagnetic Waves \u2022 De\ufb01ne electromagnetic wave and electromagnetic radiation. \u2022 Describe the electric and magnetic \ufb01elds of an electromagnetic wave. \u2022 Explain how electromagnetic waves begin and how they travel. \u2022 State how electromagnetic waves may interact with matter. \u2022 Identify sources of electromagnetic waves. Did you ever wonder how a microwave works? It directs invisible waves of radiation toward the food placed inside of it. The radiation transfers energy to the food, causing it to get warmer. The radiation is in the form of microwaves, which are a type of electromagnetic waves. What Are Electromagnetic Waves? Electromagnetic waves are waves that consist of vibrating electric and magnetic \ufb01elds. Like other waves, electromagnetic waves transfer energy from one place to another. The transfer of energy by electromagnetic waves is called electromagnetic radiation. Electromagnetic waves can transfer energy through matter or across empty space. For an excellent video introduction to electromagnetic waves, go to this URL: http://www.youtube.com/watch?v=cfXz wh3KadE MEDIA Click image to the left for more content. Q: How do microwaves transfer energy inside a microwave oven? A: They transfer energy through the air inside the oven to the food. 1 May the Force Be with You A familiar example may help you understand the vibrating electric and magnetic \ufb01elds that make up electromagnetic waves. Consider a bar magnet, like the one in the Figure 1.1. The magnet exerts magnetic force over an area all around it. This area is called a magnetic \ufb01eld. The \ufb01eld lines in the diagram represent the direction and location of the magnetic force. Because of the", " \ufb01eld surrounding a magnet, it can exert force on objects without touching them. They just have to be within its magnetic \ufb01eld. www.ck12.org FIGURE 1.1 Q: How could you demonstrate that a magnet can exert force on objects without touching them? A: You could put small objects containing iron, such as paper clips, near a magnet and show that they move toward the magnet. An electric \ufb01eld is similar to a magnetic \ufb01eld. It is an area of electrical force surrounding a positively or negatively charged particle. You can see electric \ufb01elds in the following Figure 1.2. Like a magnetic \ufb01eld, an electric \ufb01eld can exert force on objects over a distance without actually touching them. How an Electromagnetic Wave Begins An electromagnetic wave begins when an electrically charged particle vibrates. The Figure 1.3 shows how this happens. A vibrating charged particle causes the electric \ufb01eld surrounding it to vibrate as well. A vibrating electric \ufb01eld, in turn, creates a vibrating magnetic \ufb01eld. The two types of vibrating \ufb01elds combine to create an electromagnetic wave. You can see animations of electromagnetic waves at these URLs: http://www.youtube.com/ http://www.phys.hawaii.edu/~teb/java/ntnujava/emWave/emWave.htm watch?v=Qju7QnbrOhM&feature=related l 2 www.ck12.org Concept 1. Electromagnetic Waves FIGURE 1.2 FIGURE 1.3 How an Electromagnetic Wave Travels As you can see in the diagram above, the electric and magnetic \ufb01elds that make up an electromagnetic wave are perpendicular (at right angles) to each other. Both \ufb01elds are also perpendicular to the direction that the wave travels. Therefore, an electromagnetic wave is a transverse wave. However, unlike a mechanical transverse wave, which can only travel through matter, an electromagnetic transverse wave can travel through empty space. When waves travel through matter, they lose some energy to the matter as they pass through it. But when waves travel through space, no energy is lost. Therefore, electromagnetic waves don\u2019t get weaker as they travel. However, the energy is \u201cdiluted\u201d as it travels farther from its source because it spreads out over an ever-larger area. 3 Electromagnetic Wave Interactions When electromagnetic waves strike matter, they may interact with it in the same ways that mechanical waves interact with matter. Electromagnetic waves may: www.ck12.org \u2022 re\ufb02ect, or bounce back from a surface; \u2022 refract, or bend when entering a new medium; \u2022 diffract, or spread out around obstacles. Electromagnetic waves may also be absorbed by matter and converted to other forms of energy. Microwaves are a familiar example. When microwaves strike food in a microwave oven, they are absorbed and converted to thermal energy, which heats the food. Sources of Electromagnetic Waves The most important source of electromagnetic waves on Earth is the sun. Electromagnetic waves travel from the sun to Earth across space and provide virtually all the energy that supports life on our planet. Many other sources of electromagnetic waves depend on technology. Radio waves, microwaves, and X rays are examples. We use these electromagnetic waves for communications, cooking, medicine, and many other purposes. Summary \u2022 Electromagnetic waves are waves that consist of vibrating electric and magnetic \ufb01elds. They transfer energy through matter or across space. The transfer of energy by electromagnetic waves is called electromagnetic radiation. \u2022 The electric and magnetic \ufb01elds of an electromagnetic wave are areas of electric or magnetic force. The \ufb01elds can exert force over objects at a distance. \u2022 An electromagnetic wave begins when an electrically charged particle vibrates. This causes a vibrating electric \ufb01eld, which in turn creates a vibrating magnetic \ufb01eld. The two vibrating \ufb01elds together form an electromagnetic wave. \u2022 An electromagnetic wave is a transverse wave that can travel across space as well as through matter. When it travels through space, it doesn\u2019t lose energy to a medium as a mechanical wave does. \u2022 When electromagnetic waves strike matter, they may be re\ufb02ected, refracted, or diffracted. Or they may be absorbed by matter and converted to other forms of energy. \u2022 The most important source of electromagnetic waves on Earth is the sun. Many other sources of electromag- netic waves depend on technology. Vocabulary \u2022 electromagnetic radiation : Transfer of energy by electromagnetic waves across space or through matter. \u2022 electromagnetic wave : Transverse wave consisting of vibrating electric and magnetic \ufb01elds that can", " travel across space. Practice Watch the electromagnetic wave animation at the following URL, and then answer the questions below. http://www. youtube.com/watch?v=4CtnUETLIFs 4 www.ck12.org Concept 1. Electromagnetic Waves MEDIA Click image to the left for more content. 1. Identify the vibrating electric and magnetic \ufb01elds of the wave. 2. Describe the direction in which the wave is traveling. Review 1. What is an electromagnetic wave? 2. De\ufb01ne electromagnetic radiation. 3. Describe the electric and magnetic \ufb01elds of an electromagnetic wave. 4. How does an electromagnetic wave begin? How does it travel? 5. Compare and contrast electromagnetic and mechanical transverse waves. 6. List three sources of electromagnetic waves on Earth. References 1. Christopher Auyeung.. CC BY-NC 3.0 2. Christopher Auyeung.. CC BY-NC 3.0 3. Christopher Auyeung.. CC BY-NC 3.0 5 CONCEPT 2 Properties of Electromagnetic Waves www.ck12.org \u2022 State the speed of light. \u2022 Describe wavelengths and frequencies of electromagnetic waves. \u2022 Relate wave frequency to wave energy. \u2022 Show how to calculate wavelength or wave frequency if the other value is known. What do these two photos have in common? They both represent electromagnetic waves. These are waves that consist of vibrating electric and magnetic \ufb01elds. They transmit energy through matter or across space. Some electromagnetic waves are generally harmless. The light we use to see is a good example. Other electromagnetic waves can be very harmful and care must be taken to avoid too much exposure to them. X rays are a familiar example. Why do electromagnetic waves vary in these ways? It depends on their properties. Like other waves, electromagnetic waves have properties of speed, wavelength, and frequency. Speed of Electromagnetic Waves All electromagnetic waves travel at the same speed through empty space. That speed, called the speed of light, is about 300 million meters per second (3.0 x 10 8 m/s). Nothing else in the universe is known to travel this fast. The sun is about 150 million kilometers (93 million miles) from Earth, but it takes electromagnetic radiation only 8 minutes to reach Earth from the sun. If you could move that fast, you would be able to travel around Earth 7.5 times in just 1 second! You can learn more about the speed of light at this URL: http://videos.howstuffworks.com/discove ry/29407-assignment-discovery-speed-of-light-video.htm MEDIA Click image to the left for more content. Wavelength and Frequency of Electromagnetic Waves Although all electromagnetic waves travel at the same speed across space, they may differ in their wavelengths, frequencies, and energy levels. 6 www.ck12.org Concept 2. Properties of Electromagnetic Waves \u2022 Wavelength is the distance between corresponding points of adjacent waves (see the Figure 2.1 ). Wavelengths of electromagnetic waves range from longer than a soccer \ufb01eld to shorter than the diameter of an atom. \u2022 Wave frequency is the number of waves that pass a \ufb01xed point in a given amount of time. Frequencies of electromagnetic waves range from thousands of waves per second to trillions of waves per second. \u2022 The energy of electromagnetic waves depends on their frequency. Low-frequency waves have little energy and are normally harmless. High-frequency waves have a lot of energy and are potentially very harmful. FIGURE 2.1 Q: Which electromagnetic waves do you think have higher frequencies: visible light or X rays? A: X rays are harmful but visible light is harmless, so you can infer that X rays have higher frequencies than visible light. Speed, Wavelength, and Frequency The speed of a wave is a product of its wavelength and frequency. Because all electromagnetic waves travel at the same speed through space, a wave with a shorter wavelength must have a higher frequency, and vice versa. This relationship is represented by the equation: Speed = Wavelength Frequency The equation for wave speed can be rewritten as: Frequency = Speed Wavelength or Wavelength = Speed Frequency Therefore, if either wavelength or frequency is known, the missing value can be calculated. Consider an electromagnetic wave that has a wavelength of 3 meters. Its speed, like the speed of all electromagnetic waves, is 3.0 10 8 meters per second. Its frequency can be found by substituting these values into the frequency equation: Frequency = 3:0108 m/s 3:0 m = 1:0 108 waves/s, or 1.0 10 8 Hz Q: What is the wavelength of an electromagnetic wave that has a frequency of 3.0 10 8 hertz? A: Use the wavelength equation: Wavelength = 3:0108 m/s 3:0108 waves/s = 1:0", " m You can learn more about calculating the frequency and wavelength of electromagnetic waves at these URLs: htt p://www.youtube.com/watch?v=GwZvtfZRNKk and http://www.youtube.com/watch?v=wjPk108Ua8k 7 www.ck12.org Summary \u2022 All electromagnetic waves travel across space at the speed of light, which is about 300 million meters per second (3.0 x 10 8 m/s). \u2022 Electromagnetic waves vary in wavelength and frequency. Longer wavelength electromagnetic waves have lower frequencies, and shorter wavelength waves have higher frequencies. Higher frequency waves have more energy. \u2022 The speed of a wave is a product of its wavelength and frequency. Because the speed of electromagnetic waves through space is constant, the wavelength or frequency of an electromagnetic wave can be calculated if the other value is known. Vocabulary \u2022 speed of light : Speed at which all electromagnetic waves travel through space, which is 3.0 10 8 m/s. Practice Use the calculator at the following URL to \ufb01nd the frequency and energy of electromagnetic waves with different wavelengths. Use at least eight values for wavelength. Record and make a table of the results. http://www.1728.org /freqwave.htm Review 1. What is the speed of light across space? 2. Describe the range of wavelengths and frequencies of electromagnetic waves. 3. How is the energy of an electromagnetic wave related to its frequency? 4. If the frequency of an electromagnetic wave is 6.0 10 8 Hz, what is its wavelength? References 1. Christopher Auyeung.. CC BY-NC 3.0 8 www.ck12.org Concept 3. Electromagnetic Spectrum CONCEPT 3 Electromagnetic Spectrum \u2022 Describe electromagnetic radiation and its properties. \u2022 Give an overview of the electromagnetic spectrum. It\u2019s a warm sunny Saturday, and Michael and Lavar have a big day planned. They\u2019re going to ride across town to meet their friends and then go to the zoo. The boys may not realize it, but they will be bombarded by electromagnetic radiation as they ride their bikes and walk around the zoo grounds. The only kinds of radiation they can detect are visible light, which allows them to see, and infrared light, which they feel as warmth on their skin. Q: Besides visible light and infrared light, what other kinds of electromagnetic radiation will the boys be exposed to in sunlight? A: Sunlight consists of all the different kinds of electromagnetic radiation, from harmless radio waves to deadly gamma rays. Fortunately, Earth\u2019s atmosphere prevents most of the harmful radiation from reaching Earth\u2019s surface. You can read about the different kinds of electromagnetic radiation in this article. Electromagnetic Radiation Electromagnetic radiation is energy that travels in waves across space as well as through matter. Most of the electromagnetic radiation on Earth comes from the sun. Like other waves, electromagnetic waves are characterized by certain wavelengths and wave frequencies. Wavelength is the distance between two corresponding points on adjacent waves. Wave frequency is the number of waves that pass a \ufb01xed point in a given amount of time. Electromagnetic waves with shorter wavelengths have higher frequencies and more energy. A Spectrum of Electromagnetic Waves Visible light and infrared light are just a small part of the full range of electromagnetic radiation, which is called the electromagnetic spectrum. You can see the waves of the electromagnetic spectrum in the Figure 3.1. At the top 9 of the diagram, the wavelengths of the waves are given. Also included are objects that are about the same size as the corresponding wavelengths. The frequencies and energy levels of the waves are shown at the bottom of the diagram. Some sources of the waves are also given. For a video introduction to the electromagnetic spectrum, go to this URL: http://www.youtube.com/watch?NR=1&feature=endscreen&v=cfXzwh3KadE www.ck12.org FIGURE 3.1 \u2022 On the left side of the electromagnetic spectrum diagram are radio waves and microwaves. Radio waves have the longest wavelengths and lowest frequencies of all electromagnetic waves. They also have the least amount of energy. \u2022 On the right side of the diagram are X rays and gamma rays. They have the shortest wavelengths and highest frequencies of all electromagnetic waves. They also have the most energy. \u2022 Between these two extremes are waves that are commonly called light. Light includes infrared light, visible light, and ultraviolet light. The wavelengths, frequencies, and energy levels of light fall in between those of radio waves on the left and X rays and gamma rays on the right. Q: Which type of light has the longest wavelengths? A: Infrared light has the longest wavelengths. Q: What sources of infrared light are shown in the diagram? A: The sources in the diagram are people and light bulbs, but all living things and most other objects give off infrared light. Summary \u2022 Electromagnetic radiation travels in waves through", " space or matter. Electromagnetic waves with shorter wavelengths have higher frequencies and more energy. \u2022 The full range of electromagnetic radiation is called the electromagnetic spectrum. From longest to shortest wavelengths, it includes radio waves, microwaves, infrared light, visible light, ultraviolet light, X rays, and gamma rays. Vocabulary \u2022 electromagnetic spectrum : Full range of wavelengths of electromagnetic waves, from radio waves to gamma rays. 10 www.ck12.org Practice Concept 3. Electromagnetic Spectrum At the \ufb01rst URL below, read about electromagnetic waves with different frequencies. Then use the information to complete the table at the second URL. http://www.darvill.clara.net/emag/index.htm and http://www.darvill.clar a.net/emag/gcseemag.pdf Review 1. Describe the relationship between the wavelength and frequency of electromagnetic waves. 2. What is the electromagnetic spectrum? 3. Which electromagnetic waves have the longest wavelengths? 4. Identify a source of microwaves. 5. Which type of light has the highest frequencies? 6. Explain why gamma rays are the most dangerous of all electromagnetic waves. References 1. NASA.. public domain 11 www.ck12.org Law of Re\ufb02ection CONCEPT 4 \u2022 De\ufb01ne re\ufb02ection and image. \u2022 Compare and contrast regular and diffuse re\ufb02ection. \u2022 State the law of re\ufb02ection. These dancers are practicing in front of a mirror so they can see how they look as they performs. They\u2019re watching their image in the mirror as they dance. What is an image, and how does it get \u201cinside\u201d a mirror? In this article, you\u2019ll \ufb01nd out. Re\ufb02ected Light and Images Re\ufb02ection is one of several ways that light can interact with matter. Light re\ufb02ects off surfaces such as mirrors that do not transmit or absorb light. When light is re\ufb02ected from a smooth surface, it may form an image. An image is a copy of an object that is formed by re\ufb02ected (or refracted) light. Q: Is an image an actual object? If not, what is it? A: No, an image isn\u2019t an actual object. It is focused rays of light that make a copy of an object, like a picture projected on a screen. Regular and Diffuse Re\ufb02ection If a surface is extremely smooth, as it is in a mirror, then the image formed by re\ufb02ection is sharp and clear. This is called regular re\ufb02ection (also called specular re\ufb02ection). However, if the surface is even slightly rough or bumpy, an image may not form, or if there is an image, it is blurry or fuzzy. This is called diffuse re\ufb02ection. Q: Look at the boats and their images in the Figure 4.1. Which one represents regular re\ufb02ection, and which one represents diffuse re\ufb02ection? 12 www.ck12.org Concept 4. Law of Re\ufb02ection FIGURE 4.1 A: Re\ufb02ection of the boat on the left is regular re\ufb02ection. The water is smooth and the image is sharp and clear. Re\ufb02ection of the boat on the right is diffuse re\ufb02ection. The water has ripples and the image is blurry and wavy. In the Figure 4.2, you can see how both types of re\ufb02ection occur. Waves of light are represented by arrows called rays. Rays that strike the surface are referred to as incident rays, and rays that re\ufb02ect off the surface are known as re\ufb02ected rays. In regular re\ufb02ection, all the rays are re\ufb02ected in the same direction. This explains why regular re\ufb02ection forms a clear image. In diffuse re\ufb02ection, the rays are re\ufb02ected in many different directions. This is why diffuse re\ufb02ection forms, at best, a blurry image. You can see animations of both types of re\ufb02ection at this URL: htt p://toolboxes.\ufb02exiblelearning.net.au/demosites/series5/508/Laboratory/StudyNotes/snRe\ufb02ectionMirrors.htm FIGURE 4.2 Law of Re\ufb02ection One thing is true of both regular and diffuse re\ufb02ection. The angle at which the re\ufb02ected rays leave the surface is equal to the angle at which the incident rays strike the surface. This is known as the law of re\ufb02ection. The law is illustrated in the Figure 4.3 and also in this animation: http://www.physics", "classroom.com/mmedia/optics/lr.cfm Summary \u2022 Re\ufb02ection is one of several ways that light can interact with matter. When light is re\ufb02ected from a smooth surface, it may form an image. An image is a copy of an object that is formed by re\ufb02ected (or refracted) light. \u2022 Regular re\ufb02ection occurs when light re\ufb02ects off a very smooth surface and forms a clear image. Diffuse re\ufb02ection occurs when light re\ufb02ects off a rough surface and forms a blurry image or no image at all. \u2022 According to the law of re\ufb02ection, the angle at which light rays re\ufb02ect off a surface is equal to the angle at which the incident rays strike the surface. 13 www.ck12.org FIGURE 4.3 Vocabulary \u2022 image : Copy of an object that is formed by re\ufb02ected or refracted light. \u2022 law of re\ufb02ection : Law stating that the angle at which re\ufb02ected rays of light bounce off a surface is equal to the angle at which the incident rays strike the surface. \u2022 re\ufb02ection : Bouncing back of waves from a barrier they cannot pass through. Practice At the following URL, review the law of re\ufb02ection and watch the animation. Then \ufb01ll in the blanks in the sentence below. http://www.physicsclassroom.com/mmedia/optics/lr.cfm 1. When a ray of light strikes a plane mirror, the light ray __________ off the mirror. 2. Re\ufb02ection involves a change in __________ of a light ray. 3. The angle of incidence equals the angle between the incident ray and ________. 4. The angle of __________ equals the angle of incidence. 5. The normal line is __________ to the mirror. Review 1. What is an image? 2. Identify the object and the image in the Figure 4.4. Which type of re\ufb02ection formed the image: regular re\ufb02ection or diffuse re\ufb02ection? How do you know? FIGURE 4.4 3. What is the law of re\ufb02ection? 4. Label the angle of incidence and the angle of re\ufb02ection in the Figure 4.5. 14 www.ck12.org Concept 4. Law of Re\ufb02ection FIGURE 4.5 References 1. Left: Kenneth Baruch; Right: Damian Gadal.. CC BY 2.0 2. Joy Sheng.. CC BY-NC 3.0 3. Christopher Auyeung.. CC BY-NC 3.0 4. Mike Baird.. CC BY 2.0 5. Zachary Wilson.. CC-BY-NC-SA 3.0 15 www.ck12.org CONCEPT 5 Refraction of Light \u2022 De\ufb01ne refraction. \u2022 Given data about the optical density of the media, predict whether the light will bend toward the normal or away from the normal. \u2022 State Snell\u2019s Law and solve refraction problems using it. \u2022 Solve problems using the relationship between the index of refraction and the velocity of light in the media. \u2022 Explain effects caused by the refraction of light. When a light ray passes at an angle through the boundary between optically different media, the light does not travel in a straight line. The pencil in the glass of liquid shown above is a normal straight pencil. The light that travels from the pencil through the liquid, through the glass, and into the air is bent differently than light from the portion of the pencil that is not in the liquid. Your eye assumes the light from both portions of the pencil moved in a straight line, but the two portions of the pencil do not appear to be lined up. Your eye thinks the pencil is broken. Refraction of Light The speed of light is different in different media. If the speed of light is slower in a particular medium, that medium is said to be more optically dense. When a wave front enters a new medium at an angle, it will change directions. If the light is entering a more optically dense medium, the light bends toward the normal line. If the light is entering a less optically dense medium, the light will bend away from the normal line. Remember that the normal line is the line perpendicular to the medium interface. 16 www.ck12.org Concept 5. Refraction of Light In the sketch below, light wave fronts are moving upward from the bottom of the page and encounter a boundary into a more optically dense medium. The light waves bend toward the normal line. Because the right end of the wave fronts enter the new medium \ufb01rst, they slow down \ufb01rst. When the right side of the wave front", " is moving more slowly that the left side, the wave front will change directions. When light is traveling from air into another medium, Snell\u2019s Law states the relationship between the angle of incidence and angle of refraction is n = sin qi sin qr where qi is the angle of incidence, qr is the angle of refraction, and n is the ratio of the two sines and is called the index of refraction. Snell\u2019s Law may be stated that a ray of light bends in such a way that the ratio of the sine of the angle of incidence to the sine of the angle of refraction is a constant. The index of refraction is also related to the relative speeds of light in a vacuum and in the medium. n = speed of light in a vacuum speed of light in the medium When a ray of light is traveling from medium into another medium, Snell\u2019s Law can be written as ni sin qi = nr sin qr. Table of Indices of Refraction Medium Vacuum Air Water Ethanol Crown Glass Quartz Flint Glass Diamond TABLE 5.1: n 1.00 1.0003 1.36 1.36 1.52 1.54 1.61 2.42 17 www.ck12.org Example Problem: A ray of light traveling through air is incident upon a slab of Flint glass at an angle of 40:0. What is the angle of refraction? Solution: n = sin qi sin qr so sin qr = sin qi 1:61 = 0:399 n = 0:643 The angle of refraction = sin1:399 = 23:5 Example Problem: What is the speed of light in a diamond? Solution: speed of light in diamond = speed of light in a vacuum n = 3:00108 m=s 2:42 speed of light in diamond = 1:24 108 m=s Effects of Refraction Bending the Sun\u2019s Rays Because air is slightly more optically dense than a vacuum, when sunlight passes from the vacuum of space into our atmosphere, it bends slightly towards the normal. When the sun is below the horizon and thus not visible on a direct line, the light path will bend slightly and thus make the sun visible by refraction. Observers can see the sun before it actually comes up over the horizon, or after it sets. Mirages In the Figure below, the sun shines on the road, heating the air just above the road. The difference in density between the hot air over the road and the surrounding air causes the hot air to refract light that passes through it. When you look at the road, you see a mirage. What appears to be water on the road is actually light coming from the sky that has been refracted as it passes through the hot air above the road. This phenomenon is common on hot roads and in the desert. Summary \u2022 The speed of light is different in different media. \u2022 When a wave front enters a new medium at an angle, it will change directions. If the light is entering a more optically dense medium, the light bends toward the normal line. If the light is entering a less optically dense medium, the light will bend away from the normal line. 18 www.ck12.org Concept 5. Refraction of Light FIGURE 5.1 \u2022 When light is traveling from air into another medium, Snell\u2019s Law states that n = sin qi sin qr \u2022 The index of refraction is also related to the relative speeds of light in a vacuum and in the medium. n =. speed of light in a vacuum speed of light in the medium \u2022 When a ray of light is traveling from one medium into another medium, Snell\u2019s Law can be written as ni sin qi = nr sin qr. Practice Use the video on refraction to answer the questions below. http://video.mit.edu/watch/mit-physics-demo-refraction-a-total-internal-reflection-12044/ MEDIA Click image to the left for more content. 1. What happens to the path of a light beam when it enters a new medium at an angle? 2. Light moving from air into water is bent ___________ the normal. 3. Light moving from water into air is bent ___________ the normal. Review 1. Light moving through air is incident on a piece of crown glass at an angle of 45. What is the angle of refraction? 2. A ray of light passes from air into water at an incident angle of 60:0. Find the angle of refraction. 3. Light passes from water into a block of transparent plastic. The angle of incidence from the water is 31 and the angle of refraction in the block is 27. What is the index of refraction for the plastic? 4. The index of refraction of water is 1.36. What is the speed of light in water", "? 19 www.ck12.org 5. If the speed of light in a piece of plastic is 2:00 108 m=s, what is the index of refraction for the plastic? \u2022 refraction : The change of direction of a ray of light or sound in passing obliquely from one medium into another in which its wave velocity is different. \u2022 optically dense: Refers to the ability of a material to slow the light waves. The greater the optical density of a material, the greater the slowing effect. \u2022 Snell\u2019s Law: For a light ray incident on the interface of two media, the sine of the angle of incidence times the index of refraction of the \ufb01rst medium is equal to the sine of the angle of refraction times the index of refraction of the second medium. \u2022 index of refraction: The ratio of the speed of light in a vacuum to the speed of light in a medium under consideration. \u2022 mirage: An optical phenomenon that creates the illusion of water, often with inverted re\ufb02ections of distant objects, and results from the refraction of light by alternate layers of hot and cool air. References 1. Image copyright leonello calvetti, 2013. http://www.shutterstock.com. Used under license from Shutter- stock.com 2. CK-12 Foundation - Samantha Bacic.. CC-BY-NC-SA 3.0 3. CK-12 Foundation - Samantha Bacic.. CC-BY-NC-SA 3.0 4. Michael Gil (Flickr: MSVG). http://www.\ufb02ickr.com/photos/msvg/5994891327/. CC-BY 2.0 20 www.ck12.org Concept 6. Total Internal Re\ufb02ection CONCEPT 6 Total Internal Re\ufb02ection \u2022 Describe total internal re\ufb02ection. \u2022 Use the critical angle to determine when total internal re\ufb02ection will occur. Total internal re\ufb02ection allows the light to travel down the optical \ufb01ber and not pass through the sides of the tube. The light continuously re\ufb02ects from the inside of the tube and eventually comes out the end. Optical \ufb01bers make interesting lamps but they are also used to transport telephone and television signals. Total Internal Re\ufb02ection We already know that when light passes from one medium into a second medium where the index of refraction is smaller, the light refracts away from the normal. In the image below, the light rays are passing into an optically less dense medium; therefore, the rays bend away from the normal. As the angle of incidence increases, the light ray bends even further away from the normal. Eventually, the angle of incidence will become large enough that the angle of refraction equals 90, meaning the light ray will not enter the new medium at all. 21 www.ck12.org Consider a ray of light passing from water into air. The index of refraction for air is 1.00 and for water is 1.36. Using Snell\u2019s Law, ni sin qi = nr sin qr, and allowing the angle of refraction to be 90, we can solve for the angle of incidence which would cause the light ray to stay in the old medium. ni sin qi = nr sin qr (1:36)(sin qi) = (1:00)(sin 90) sin qi = 0:735 and qi = 47 This result tells us that when light is passing from water into air, if the angle of incidence exceeds 47, the light ray will not enter the new medium. The light ray will be completely re\ufb02ected back into the original medium. This is called total internal re\ufb02ection. The minimum angle of incidence for total internal re\ufb02ection to occur is called the critical angle. Total internal re\ufb02ection is the principle behind \ufb01ber optics. A bundle of \ufb01bers made out of glass or plastic only a few micrometers in diameter is called a light pipe since light can be transmitted along it with almost no loss. Light passing down the \ufb01bers makes glancing collisions with the walls so that total internal re\ufb02ection occurs. Summary \u2022 When light passes from one medium into a second medium with a smaller index of refraction, the light refracts away from the normal. \u2022 If the angle of incidence becomes large enough that the angle of refraction equals 90, the light ray will not enter the new medium with the smaller angle of refraction. \u2022 Total internal re\ufb02ection means the light ray will not enter the new medium but will be completely re\ufb02ected back into the original medium. Practice MEDIA Click image to the left for more content. http://www.youtube.com/watch?v=hBQ8fh_", "Fp04 In this video, a hole is drilled in the side of a plastic bottle \ufb01lled with liquid. An arc of liquid shoots out through the hole. A laser pointer is aimed through the opposite side of the bottle so that the light also exits through the hole. The stream of liquid acts like an optical \ufb01ber and the light undergoes total internal re\ufb02ection as it follows the stream of liquid. As the amount of liquid in the bottle decreases, the arc of the stream of liquid changes and the direction of the light follows the stream of liquid. Toward the end, the light beam is shining almost 90 from the direction of the laser pointer. The following video discusses total internal re\ufb02ection. Use this resource to answer the questions that follow. http://www.khanacademy.org/science/physics/waves-and-optics/v/total-internal-re\ufb02ection# 22 www.ck12.org Concept 6. Total Internal Re\ufb02ection MEDIA Click image to the left for more content. 1. What phenomenon occurs when the light does not enter the new medium and remains in the old medium? 2. When does this phenomenon occur? Review 1. Find the critical angle for light passing from diamond into air, given ndiamond = 2:42 2. When two swimmers are under water in a swimming pool, it is possible for the interface between the water and the air to act as a mirror, allowing the swimmers to see images of each other if they look up at the underside of the surface. Explain this phenomenon. 3. Robert shines a laser beam through a slab of plastic and onto the interface between the slab of plastic and the air on the other side. The index of refraction for the plastic is 1.62. If the angle of incidence in the plastic is 54, will the laser beam pass out of the plastic into the air? \u2022 total internal re\ufb02ection: When light is passing from a medium of higher index of refraction into a medium of lower index of refraction is completely re\ufb02ected by the boundary between the two media. \u2022 critical angle: The smallest angle of incidence at which a light ray passing from one medium to another less refractive medium will be totally re\ufb02ected from the boundary between the two. \u2022 \ufb01ber optics: The science or technology of light transmission through very \ufb01ne, \ufb02exible glass or plastic \ufb01bers without energy loss making use of the principle of total internal re\ufb02ection. References 1. Roshan Nikam (Flickr: roshan1286). http://www.\ufb02ickr.com/photos/31916678@N07/4753800195/. CC-BY 2.0 2. CK-12 Foundation - Samantha Bacic.. CC-BY-NC-SA 3.0 23 CONCEPT 7 Single Slit Interference \u2022 Explain how single slit diffraction patterns occur. \u2022 Use single slit diffraction patterns to calculate wavelength. www.ck12.org Though it looks like a double slit interference pattern, the pattern on the screen are actually the results of light diffracting through a single slit with the ensuing interference. Single Slit Interference Interference patterns are produced not only by double slits but also by single slits, otherwise known as single slit interference. In the case of a single slit, the particles of medium at both corners of the slit act as point sources, producing circular waves from both edges. These circular waves move across to the back wall and interfere in the same way that interference patterns were produced by double slits. In the sketch at below, the black lines intersect at the center of the pattern on the back wall. This center point is equidistanct from both edges of the slit. Therefore, the waves striking at this position will be in phase; that is, the waves will produce constructive interference. Also shown in the sketch, just above the central bright spot where the red lines intersect, is a position where destructive interference occurs. One of these red lines is one-half wavelength longer than the other, causing the two waves to hit the wall out of phase and undergo destructive interference. A dark bank appears at this position. 24 www.ck12.org Concept 7. Single Slit Interference Just as in double slit interference, a pair of similar triangles can be constructed in the interference pattern. The pertinent values from these triangles are the width of the slit, w, the wavelength, l, the distance from the central bright spot to the \ufb01rst dark band, x, and the distance from the center of the slit to back wall, L. The relationship of these four values is L or l = wx w = x L. l Example Problem: Monochromatic light of wavelength 605 nm falls on a slit of width 0.095 mm. The slit is located 85 cm", " from a screen. How far is the center of the central bright band to the \ufb01rst dark band? Solution: x = lL w = (6:05107 m)(0:85 m) (9:5105 m) = 0:0054 m Summary \u2022 Interference patterns can also be produced by single slits. \u2022 In the case of a single slit, the particles of medium at both corners of the slit act as point sources, and produce circular waves from both edges. \u2022 The wavelength can be determined by this equation: l w = x L or l = wx L. Practice MEDIA Click image to the left for more content. http://www.youtube.com/watch?v=_HO7LJDcqos Follow up questions. 1. The interference pattern appears as the slit becomes ___________ (wider, thinner). Review 1. The same set up is used for two different single slit diffraction experiments. In one of the experiments, yellow light is used, and in the other experiment, green light is used. Green light has a shorter wavelength than yellow light. Which of the following statements is true? 25 (a) The two experiments will have the same distance between the central bright band and the \ufb01rst dark band. (b) The green light experiment will have a greater distance between the central bright band and the \ufb01rst dark band. (c) The yellow light experiment will have a greater distance between the central bright band and the \ufb01rst www.ck12.org dark band. 2. Why are the edges of shadows often fuzzy? (a) Interference occurs on the wall on which the shadow is falling. (b) Light diffracts around the edges of the object casting the shadow. (c) The edges of the object casting the shadow is fuzzy. (d) Light naturally spreads out. 3. Monochromatic, coherent light passing through a double slit will produce exactly the same interference pattern as when it passes through a single slit. (a) True (b) False 4. If monochromatic light passes through a 0.050 mm slit and is projected onto a screen 0.70 m away with a distance of 8.00 mm between the central bright band and the \ufb01rst dark band, what is the wavelength of the light? 5. A krypton ion laser with a wavelength of 524.5 nm illuminates a 0.0450 mm wide slit. If the screen is 1.10 m away, what is the distance between the central bright band and the \ufb01rst dark band? 6. Light from a He-Ne laser (l = 632:8 nm) falls on a slit of unknown width. In the pattern formed on a screen 1.15 m away, the \ufb01rst dark band is 7.50 mm from the center of the central bright band. How wide is the slit? \u2022 single slit interference: When monochromatic, coherent light falls upon a small single slit it will produce a pattern of bright and dark fringes. These fringes are due to light from one side of the slit interacting (interfering) with light from the other side. References 1. Luiz Sauerbronn. http://commons.wikimedia.org/wiki/File:Fresnel_Diffraction_experiment_DSC04573.JPG. Public Domain 2. CK-12 Foundation - Samantha Bacic.. CC-BY-NC-SA 3.0 3. CK-12 Foundation - Samantha Bacic.. CC-BY-NC-SA 3.0 26 www.ck12.org Concept 8. Double Slit Interference CONCEPT 8 Double Slit Interference \u2022 De\ufb01ne diffraction of light. \u2022 De\ufb01ne wave interference. \u2022 Describe double slit interference patterns. \u2022 Explain Thomas Young\u2019s contibutions to physics. \u2022 Calculate the wavelength from a double slit interference pattern. FIGURE 8.1 When waves strike a small slit in a wall, they create circular wave patterns on the other side of the barrier. This is seen in the image above, where ocean waves create precise circular waves. The circular waves undergo constructive and destructive interference, which generates a regular interference pattern. Diffraction and Interference When a series of straight waves strike an impenetrable barrier, the waves stop at the barrier. However, the last particle of the medium at the back corner of the barrier will create circular waves from that point, called the point source. This can be seen in the image below. This phenomenon is called diffraction, and it occurs in liquid, sound, and light waves. While the waves become circular waves at the point source, they continue as straight waves where the barrier does not interfere with the waves. Any two waves in the same medium undergo wave interference as they pass each other. At the location where", " the two waves collide, the result is essentially a summation of the two waves. In some places, a wave crest from one source will overlap a wave crest from the other source. Since both waves are lifting the medium, the combined wave crest will be twice as high as the original crests. Nearby, a wave trough will overlap another wave trough and the new 27 trough will be twice as deep as the original. This is called constructive interference because the resultant wave is larger than the original waves. Within the interference pattern, the amplitude will be twice the original amplitude. Once the waves pass through each other and are alone again, their amplitudes return to their original values. In other parts of the wave pattern, crests from one wave will overlap troughs from another wave. When the two waves have the same amplitude, this interaction causes them to cancel each other out. Instead of a crest or a trough, there is nothing. When this cancellation occurs, it is called destructive interference. www.ck12.org FIGURE 8.2 It is easy to see how waves emanating from multiple sources, such as drops of rainwater in still water, create interference patterns. But a single source of waves can create interference patterns with itself as a result of diffraction. The Double Slit Experiment A similar situation to the raindrops above occurs when straight waves strike a barrier containing two slits. These waves are cut off everywhere except for where the waves that pass through the two slits. The medium in the slits again acts as a point source to produce circular waves on the far side of the barrier. As long as these two circular waves have the same wavelength, they interfere constructively and destructively in a speci\ufb01c pattern. This pattern is called the wave interference pattern and is characterized by light and dark bands. The light bands are a result of constructive interference, and the dark bands occur because of destructive interference. 28 www.ck12.org Concept 8. Double Slit Interference In the early 1800\u2019s, light was assumed to be a particle. There was a signi\ufb01cant amount of evidence to point to that conclusion, and famous scientist Isaac Newton\u2019s calculations all support the particle theory. In 1803, however, Thomas Young performed his famous Double Slit Experiment to prove that light was a wave. Young shined a light onto the side of a sealed box with two slits in it, creating an interference pattern on the inside of the box opposite the slits. As seen above, interference patterns are characterized by alternating bright and dark lines. The bright lines are a result of constructive interference, while the dark lines are a result of destructive interference. By creating this interference pattern, Young proved light is a wave and changed the course of physics. Calculating Wavelength from Double Slit Pattern Using the characteristics of the double slit interference pattern, it is possible to calculate the wavelength of light used to produce the interference. To complete this calculation, it is only necessary to measure a few distances. As can be seen below, \ufb01ve distances are measured. In the sketch, L is the distance from the two slits to the back wall where the interference pattern can be seen. d is the distance between the two slits. To understand x, look again at the interference pattern shown above. The middle line, which is the brightest, is called the central line. The remaining lines are called fringes. The lines on either side of the central line are called the \ufb01rst order fringes, the next lines are called the second order fringes, and so on. x is the distance from the central line to the \ufb01rst order fringe. r 1 and r 2 are the distances from the slits to the \ufb01rst order fringe. We know that the fringes are a result of constructive interference, and that the fringe is a result of the crest of two waves interfering. If we assume that r 2 is a whole number of wavelengths (con\ufb01rm for yourself that this is a logical assumption), then r 1 must be one more wavelength. This is because r 1 and r 2 are the distances to the \ufb01rst order fringe. Mathematically, we can let r2 = nl and r1 = nl + l, where l is the wavelength and n is a constant. Using this relationship, we determine that r1 r2 = l. Looking again at the diagram, the red and blue triangles are similar, which means that the ratios of corresponding sides are the same. The ratio of x to L in the red triangle is equal to the ratio of l to d in the blue triangle. For proof of this, visit http://www.physicsclassroom.com/class/light/u12l3c.cfm. From this, we can determine that the wavelength is dependent on x, d, and L: 29 l = xd L Example Problem: Monochromatic light falls on two narrow", " slits that are 0.0190 mm apart. A \ufb01rst order fringe is 21.1 mm from the central line. The screen (back wall) is 0.600 m from the slits. What is the wavelength of the light? www.ck12.org Solution: l = xd L = (0:021 m)(0:000019 m) (0:600 m) = 6:68 107 m Summary \u2022 The last particle of medium at the back corner of an impenetrable barrier will act as a point source and produce circular waves. \u2022 Diffraction is the bending of waves around a corner. \u2022 Constructive interference occurs when two wave crests overlap, doubling the wave amplitude at that location. \u2022 Destructive interference occurs when a wave crest overlaps with a trough, causing them to cancel out. \u2022 Light is a wave, and creates an interference pattern in the double slit experiment. \u2022 An interference pattern consists of alternating bright and dark lines; the bright lines are called fringes. \u2022 In a double slit experiment, the wavelength can be calculated using this equation: l = xd L Practice http://www.youtube.com/watch?v=AMBcgVlamoU Follow up questions. 1. When the amplitude of waves add, it is called _________________ interference. 2. When the amplitude of waves subtract, it is called _________________ interference. 3. What do we call the phenomenon of light bending around a corner? Review 1. Destructive interference in waves occurs when (a) two troughs overlap. (b) crests and troughs align. (c) two crests overlap. (d) a crest and a trough overlap. 2. Bright bands in interference patterns result from (a) destructive diffraction. (b) destructive interference. (c) constructive diffraction. (d) constructive interference. 3. In a double slit experiment with slits 1:00 105 m apart, light casts the \ufb01rst bright band 3:00 102 m from the central bright spot. If the screen is 0.650 m away, what is the wavelength of this light? (a) 510 nm (b) 390 nm (c) 430 nm (d) 460 nm 4. Violet light falls on two slits separated by 1:90 105 m. A \ufb01rst order bright line appears 13.2 mm from the central bright spot on a screen 0.600 m from the slits. What is the wavelength of the violet light? 5. Suppose in the previous problem, the light was changed to yellow light with a wavelength of 5:96 107 m while the slit separation and distance from screen to slits remained the same. What would be the distance from the central bright spot to the \ufb01rst order line? 30 www.ck12.org Concept 8. Double Slit Interference 6. Light with a wavelength of 6:33 107 m is used in a double slit experiment. The screen is placed 1.00 m from the slits and the \ufb01rst order line is found 65.5 mm from the central bright spot. What is the separation between the slits? \u2022 diffraction: Change in the directions and intensities of a group of waves after passing by an obstacle or through an aperture whose size is approximately the same as the wavelength of the waves. \u2022 monochromatic: Light having only one wavelength. \u2022 constructive interference: The interference of two or more waves of equal frequency and phase, resulting in their mutual reinforcement and producing a single amplitude equal to the sum of the amplitudes of the individual waves. \u2022 destructive interference: The interference of two waves of equal frequency and opposite phase, resulting in their cancellation where the negative displacement of one always coincides with the positive displacement of the other. References 1... CC BY-NC-SA 2.. Kathy Shield. CC BY-NC-SA 3. CK-12 Foundation - Samantha Bacic.. CC-BY-NC-SA 3.0 4. Pieter Kuiper. http://commons.wikimedia.org/wiki/File:SodiumD_two_double_slits.jpg. Public Domain 5. CK-12 Foundation - Samantha Bacic.. CC-BY-NC-SA 3.0 31 CONCEPT 9 www.ck12.org Diffraction Gratings \u2022 Understand the structure of a diffraction grating. \u2022 Explain the interference pattern formed by diffraction gratings. \u2022 Use diffraction grating interference patterns to calculate the wavelength of light. Suppose we had a light bulb that emitted exactly four frequencies of light; one frequency in each of the colors red, yellow, green, and blue. To our eye, this bulb would appear white because the combination of those four colors produces white light. If viewed through a diffraction grating, however, each color of light would be visible. The original white light bulb is visible in the center of the image,", " and interference causes the light bulb to appear in each color to the left and the right. Diffraction Gratings Diffraction gratings are composed of a multitude of slits lined up side by side, not unlike a series of double slits next to each other. They can be made by scratching very \ufb01ne lines with a diamond point on glass, or by pressing plastic \ufb01lm on glass gratings so that the scratches are replicated. The clear places between the scratches behave as slits similar to the slits in a double slit experiment and the gratings form interference patterns in the same general way that double slits do. With more slits, there are more light waves out of phase with each other, causing more destructive interference. Compared to the interference pattern of a double slit, the diffraction grating interference pattern\u2019s colors are spread out further and the dark regions are broader. This allows for more precise wavelength determination than with double slits. The image below shows the diffraction pattern emanating from a white light. 32 www.ck12.org Concept 9. Diffraction Gratings Also in this image is the measurement for q, which can be used to calculate the wavelength of the original light source. The equation from the double slit experiment can be adjusted slightly to work with diffraction gratings. Where l is the wavelength of light, d is the distance between the slits on the grating, and q is the angle between the incident (original) light and the refracted light, l = xd L = d sin q (Note that x Looking at the equation, x = lL d, it should be apparent that as the distance between the lines on the grating become smaller and smaller, the distance between the images on the screen will become larger and larger. Diffraction gratings are often identi\ufb01ed by the number of lines per centimeter; gratings with more lines per centimeter are usually more useful because the greater the number of lines, the smaller the distance between the lines, and the greater the separation of images on the screen. L = sin q, using the small angle approximation theorem.) Example Problem: A good diffraction grating has 2500 lines/cm. What is the distance between two lines on the grating? Solution: d = 1 2500 cm1 = 0:00040 cm Example Problem: Using a diffraction grating with a spacing of 0.00040 cm, a red line appears 16.5 cm from the central line on the screen. The screen is 1.00 m from the grating. What is the wavelength of the light? Solution: l = xd L = (0:165 m)(4:0106 m) 1:00 m = 6:6 107 m Summary \u2022 Diffraction gratings can be made by blocking light from traveling through a translucent medium; the clear places behave as slits similar to the slits in a double slit experiment. \u2022 Diffraction gratings form interference patterns much like double slits, though brighter and with more space between the lines. \u2022 The equation used with double slit experiments to measure wavelength is adjusted slightly to work with diffraction gratings. l = xd L = d sin q 33 www.ck12.org Practice http://vimeo.com/39495562 Follow up questions. 1. How does a diffraction grating differ from single or double slit? 2. What happens when you increase the number of slits in a diffraction grating? Review 1. White light is directed toward a diffraction grating and that light passes through the grating, causing its monochromatic bands appear on the screen. Which color will be closest to the central white? 2. Three discrete spectral lines occur at angles of 10.1\u00b0, 13.7\u00b0, and 14.8\u00b0 respectively in the \ufb01rst order spectrum. If the grating has 3660 lines/cm, what are the wavelengths of these three colors of light? 3. A 20.0 mm section of diffraction grating has 6000 lines. At what angle will the maximum bright band appear if the wavelength is 589 nm? 4. Laser light is passed through a diffraction grating with 7000 lines/cm. The \ufb01rst order maximum on the screen is 25\u00b0 away from the central maximum. What is the wavelength of the light? \u2022 diffraction grating: A glass, plastic, or polished metal surface having a large number of very \ufb01ne parallel grooves or slits and used to produce optical spectra by diffraction. References 1. CK-12 Foundation - Samantha Bacic, using light bulb images copyright Ruslan Klimovich, 2013. http:// www.shutterstock.com. Used under license from Shutterstock.com 2. Candace (Flickr: cosmiccandace). http://www.\ufb02ickr.com/photos/candace/315205005/. CC-BY 2.0 34 www.ck", "12.org Concept 10. Wave-Particle Theory CONCEPT 10 Wave-Particle Theory \u2022 State the wave-particle theory of electromagnetic radiation. \u2022 Describe a photon. \u2022 Identify evidence that electromagnetic radiation is both a particle and a wave. What a beautiful sunset! You probably know that sunlight travels in waves through space from the sun to Earth. But do you know what light really is? Is it just energy, or is it something else? In this article you\u2019ll \ufb01nd out that light may be more than it seems. The Question Electromagnetic radiation, commonly called light, is the transfer of energy by waves called electromagnetic waves. These waves consist of vibrating electric and magnetic \ufb01elds. Where does electromagnetic energy come from? It is released when electrons return to lower energy levels in atoms. Electromagnetic radiation behaves like continuous waves of energy most of the time. Sometimes, however, electromagnetic radiation seems to behave like discrete, or separate, particles rather than waves. So does electromagnetic radiation consist of waves or particles? The Debate This question about the nature of electromagnetic radiation was debated by scientists for more than two centuries, starting in the 1600s. Some scientists argued that electromagnetic radiation consists of particles that shoot around like tiny bullets. Other scientists argued that electromagnetic radiation consists of waves, like sound waves or water waves. Until the early 1900s, most scientists thought that electromagnetic radiation is either one or the other and that scientists on the other side of the argument were simply wrong. Q: Do you think electromagnetic radiation is a wave or a particle? A: Here\u2019s a hint: it may not be a question of either-or. Keep reading to learn more. 35 www.ck12.org A New Theory In 1905, the physicist Albert Einstein developed a new theory about electromagnetic radiation. The theory is often called the wave-particle theory. It explains how electromagnetic radiation can behave as both a wave and a particle. Einstein argued that when an electron returns to a lower energy level and gives off electromagnetic energy, the energy is released as a discrete \u201cpacket\u201d of energy. We now call such a packet of energy a photon. According to Einstein, a photon resembles a particle but moves like a wave. You can see this in the Figure 10.1. The theory posits that waves of photons traveling through space or matter make up electromagnetic radiation. FIGURE 10.1 Energy of a Photon A photon isn\u2019t a \ufb01xed amount of energy. Instead, the amount of energy in a photon depends on the frequency of the electromagnetic wave. The frequency of a wave is the number of waves that pass a \ufb01xed point in a given amount of time, such as the number of waves per second. In waves with higher frequencies, photons have more energy. Evidence for the Wave-Particle Theory After Einstein proposed his theory, evidence was discovered to support it. For example, scientists shone laser light through two slits in a barrier made of a material that blocked light. You can see the setup of this type of experiment in the sketch below. Using a special camera that was very sensitive to light, they took photos of the light that passed through the slits. The photos revealed tiny pinpoints of light passing through the double slits. This seemed to show that light consists of particles. However, if the camera was exposed to the light for a long time, the pinpoints accumulated in bands that resembled interfering waves. Therefore, the experiment showed that light seems to consist of particles that act like waves. 36 www.ck12.org Concept 10. Wave-Particle Theory FIGURE 10.2 Summary \u2022 Electromagnetic radiation behaves like waves of energy most of the time, but sometimes it behaves like particles. From the 1600s until the early 1900s, most scientists thought that electromagnetic radiation consists either of particles or of waves but not both. \u2022 In 1905, Albert Einstein proposed the wave-particle theory of electromagnetic radiation. This theory states that electromagnetic energy is released in discrete packets of energy\u2014now called photons\u2014that act like waves. \u2022 After Einstein presented his theory, scientists found evidence to support it. For example, double-slit experi- ments showed that light consists of tiny particles that create patterns of interference just as waves do. Vocabulary \u2022 photon : Tiny \u201cpacket\u201d of electromagnetic radiation that is released when an electron returns to a lower. \u2022 wave-particle theory : Theory proposed by Albert Einstein that electromagnetic energy is released in discrete packets of energy (now called photons) that act like waves. Practice Watch the animation \u201cLet There Be Light\u201d at the following URL. Then create a timeline of ideas and discoveries about the nature of light. http://www.abc.net.au/science/explore/einstein/lightstory.htm Review 1. Why did scientists debate the nature of electromagnetic radiation for more than 200 years? 2. State Einstein\u2019s wave-particle", " theory of electromagnetic radiation. 3. What is a photon? 4. After Einstein proposed his wave-particle theory, how did double-slit experiments provide evidence to support the theory? 37 References 1. Christopher Auyeung.. CC BY-NC 3.0 2. Zachary Wilson.. CC-BY-NC-SA 3.0 www.ck12.org 38 Physics Unit 12: Static Electricity Patrick Marshall Jean Brainard, Ph.D. Ck12 Science Say Thanks to the Authors Click http://www.ck12.org/saythanks (No sign in required) www.ck12.org To access a customizable version of this book, as well as other interactive content, visit www.ck12.org AUTHORS Patrick Marshall Jean Brainard, Ph.D. Ck12 Science CK-12 Foundation is a non-pro\ufb01t organization with a mission to reduce the cost of textbook materials for the K-12 market both in the U.S. and worldwide. Using an open-content, web-based collaborative model termed the FlexBook\u00ae, CK-12 intends to pioneer the generation and distribution of high-quality educational content that will serve both as core text as well as provide an adaptive environment for learning, powered through the FlexBook Platform\u00ae. Copyright \u00a9 2014 CK-12 Foundation, www.ck12.org The names \u201cCK-12\u201d and \u201cCK12\u201d and associated logos and the terms \u201cFlexBook\u00ae\u201d and \u201cFlexBook Platform\u00ae\u201d (collectively \u201cCK-12 Marks\u201d) are trademarks and service marks of CK-12 Foundation and are protected by federal, state, and international laws. Any form of reproduction of this book in any format or medium, in whole or in sections must include the referral attribution link http://www.ck12.org/saythanks (placed in a visible location) in addition to the following terms. Except as otherwise noted, all CK-12 Content (including CK-12 Curriculum Material) is made available to Users in accordance with the Creative Commons Attribution-Non-Commercial 3.0 Unported (CC BY-NC 3.0) License (http://creativecommons.org/ licenses/by-nc/3.0/), as amended and updated by Creative Commons from time to time (the \u201cCC License\u201d), which is incorporated herein by this reference. Complete terms can be found at http://www.ck12.org/terms. Printed: April 6, 2014 iii Contents www.ck12.org Contents 1 Static Electricity and Static Discharge 2 Electric Charge and Electric Force 3 Transfer of Electric Charge 4 Forces on Charged Objects 5 Coulomb\u2019s Law 6 The Electric Field 7 Electric Potential 1 4 7 12 19 22 25 iv www.ck12.org Concept 1. Static Electricity and Static Discharge CONCEPT 1 Static Electricity and Static Discharge \u2022 Describe static electricity. \u2022 Explain static discharge. \u2022 Outline how lightning occurs. You\u2019re a thoughtful visitor, so you wipe your feet on the welcome mat before you reach out to touch the brass knocker on the door. Ouch! A spark suddenly jumps between your hand and the metal, and you feel an electric shock. Q: Why do you think an electric shock occurs? A: An electric shock occurs when there is a sudden discharge of static electricity. What Is Static Electricity? Static electricity is a buildup of electric charges on objects. Charges build up when negative electrons are transferred from one object to another. The object that gives up electrons becomes positively charged, and the object that accepts the electrons becomes negatively charged. This can happen in several ways. One way electric charges can build up is through friction between materials that differ in their ability to give up or accept electrons. When you wipe your rubber-soled shoes on the wool mat, for example, electrons rub off the mat onto your shoes. As a result of this transfer of electrons, positive charges build up on the mat and negative charges build up on you. Once an object becomes electrically charged, it is likely to remain charged until it touches another object or at least comes very close to another object. That\u2019s because electric charges cannot travel easily through air, especially if the air is dry. 1 www.ck12.org Q: You\u2019re more likely to get a shock in the winter when the air is very dry. Can you explain why? A: When the air is very dry, electric charges are more likely to build up objects because they cannot travel easily through the dry air. This makes a shock more likely when you touch another object. Static Discharge What happens when you have become negatively charged and your hand approaches the metal doorknocker? Your negatively charged hand repels electrons in the metal, so the electrons move to the other side of the knocker. This makes the side of the knocker closest to your hand positively charged. As your negatively", " charged hand gets very close to the positively charged side of the metal, the air between your hand and the knocker also becomes electrically charged. This allows electrons to suddenly \ufb02ow from your hand to the knocker. The sudden \ufb02ow of electrons is static discharge. The discharge of electrons is the spark you see and the shock you feel. Watch the animation \u201cJohn Travoltage\u201d at the following URL to see an example of static electricity and static discharge. http://www.cabrillo.edu/~jmccullough/Physics/Electric_Forces_Fields.html How Lightning Occurs Another example of static discharge, but on a much larger scale, is lightning. You can see how it occurs in the following diagram and animation as you read about it below. http://micro.magnet.fsu.edu/electromag/java/lightning/index.html FIGURE 1.1 During a rainstorm, clouds develop regions of positive and negative charge due to the movement of air molecules, water drops, and ice particles. The negative charges are concentrated at the base of the clouds, and the positive charges are concentrated at the top. The negative charges repel electrons on the ground beneath them, so the ground below the clouds becomes positively charged. At \ufb01rst, the atmosphere prevents electrons from \ufb02owing away from areas of negative charge and toward areas of positive charge. As more charges build up, however, the air between the oppositely charged areas also becomes charged. When this happens, static electricity is discharged as bolts of lightning. At the URL below, you can watch an awesome slow-motion lightning strike. Be sure to wait for the real-time lightning strike at the end of the video. You\u2019ll be amazed when you realize how much has occurred during that split-second discharge of static electricity. 2 www.ck12.org Concept 1. Static Electricity and Static Discharge http://www.youtube.com/watch?v=Y8oN0YFAXWQ&feature=related MEDIA Click image to the left for more content. Summary \u2022 Static electricity is a buildup of electric charges on objects. It occurs when electrons are transferred from one object to another. \u2022 A sudden \ufb02ow of electrons from one charged object to another is called static discharge. \u2022 Examples of static discharge include lightning and the shock you sometimes feel when you touch another object. Vocabulary \u2022 static discharge : Sudden \ufb02ow of electrons from an object that has a buildup of charges. \u2022 static electricity : Buildup of charges on an object that occurs through induction. Practice Watch the video at the following URL. Then answer the discussion questions. Read the background essay if you need help with any of the questions. http://www.teachersdomain.org/resource/phy03.sci.phys.mfe.zsnap/ Review 1. What is static electricity? 2. How does static discharge occur? 3. Explain why a bolt of lightning is like the spark you might see when you touch a metal object and get a shock. References 1. Zachary Wilson.. CC BY-NC 3.0 3 CONCEPT 2 Electric Charge and Electric Force www.ck12.org \u2022 De\ufb01ne electric charge. \u2022 Describe electric forces between charged particles. A lightning bolt is like the spark that gives you a shock when you touch a metal doorknob. Of course, the lightning bolt is on a much larger scale. But both the lightning bolt and spark are a sudden transfer of electric charge. Introducing Electric Charge Electric charge is a physical property of particles or objects that causes them to attract or repel each other without touching. All electric charge is based on the protons and electrons in atoms. A proton has a positive electric charge, and an electron has a negative electric charge. In the Figure 2.1, you can see that positively charged protons (+) are located in the nucleus of the atom, while negatively charged electrons (-) move around the nucleus. Electric Force When it comes to electric charges, opposites attract, so positive and negative particles attract each other. You can see this in the diagram below. This attraction explains why negative electrons keep moving around the positive nucleus of the atom. Like charges, on the other hand, repel each other, so two positive or two negative charges push apart. This is also shown in the diagram. The attraction or repulsion between charged particles is called electric force. The strength of electric force depends on the amount of electric charge on the particles and the distance between 4 www.ck12.org Concept 2. Electric Charge and Electric Force FIGURE 2.1 them. Larger charges or shorter distances result in greater force. You can experiment with electric force with the animation at the following URL. http://www.colorado.edu/physics/2000/waves_particles/wavpart2.html FIGURE 2.2 Q:", " How do positive protons stay close together inside the nucleus of the atom if like charges repel each other? A: Other, stronger forces in the nucleus hold the protons together. Summary \u2022 Electric charge is a physical property of particles or objects that causes them to attract or repel each other without touching. \u2022 Particles that have opposite charges attract each other. Particles that have like charges repel each other. The force of attraction or repulsion is called electric force. 5 www.ck12.org Vocabulary \u2022 electric charge : Physical property of particles or objects that causes them to attract or repel each other without touching; may be positive or negative. \u2022 electric force : Force of attraction or repulsion between charged particles. Practice Read the \ufb01rst four boxes of text at the following URL. Then write a concise paragraph explaining why direction E is the correct answer to the quick quiz. http://www.physics.wisc.edu/undergrads/courses/208-f07/Lectures/lect6.pdf Review 1. What is electric charge? 2. Make a simple table summarizing electric forces between charged particles. References 1. Christopher Auyeung.. CC BY-NC 3.0 2. Zachary Wilson.. CC BY-NC 3.0 6 www.ck12.org Concept 3. Transfer of Electric Charge CONCEPT 3 Transfer of Electric Charge \u2022 Describe how the transfer of electrons changes the charge of matter. \u2022 Relate the transfer of electrons to the law of conservation of charge. \u2022 Compare and contrast three ways that electric charge can be transferred. Why is this girl\u2019s hair standing straight up? She is touching a device called a van de Graaff generator. The dome on top of the device has a negative electric charge. When the girl places her hand on the dome, she becomes negatively charged as well\u2014right down to the tip of each hair! You can see a video demonstrating a van de Graff generator at this URL: http://www.youtube.com/watch?v=SREXQWAIDJk MEDIA Click image to the left for more content. Q: Why is the man\u2019s hair standing on end? A: All of the hairs have all become negatively charged, and like charges repel each other. Therefore, the hairs are pushing away from each other, causing them to stand on end. 7 www.ck12.org Transferring Electrons The man pictured above became negatively charged because electrons \ufb02owed from the van de Graaff generator to him. Whenever electrons are transferred between objects, neutral matter becomes charged. This occurs even with individual atoms. Atoms are neutral in electric charge because they have the same number of negative electrons as positive protons. However, if atoms lose or gain electrons, they become charged particles called ions. You can see how this happens in the Figure 3.1. When an atom loses electrons, it becomes a positively charged ion, or cation. When an atom gains electrons, it becomes a negative charged ion, or anion. FIGURE 3.1 Conservation of Charge Like the formation of ions, the formation of charged matter in general depends on the transfer of electrons, either between two materials or within a material. Three ways this can occur are referred to as conduction, polarization, and friction. All three ways are described below. However, regardless of how electrons are transferred, the total charge always remains the same. Electrons move, but they aren\u2019t destroyed. This is the law of conservation of charge. Conduction The transfer of electrons from the van de Graaff generator to the man is an example of conduction. Conduction occurs when there is direct contact between materials that differ in their ability to give up or accept electrons. A van de Graff generator produces a negative charge on its dome, so it tends to give up electrons. Human hands are positively charged, so they tend to accept electrons. Therefore, electrons \ufb02ow from the dome to the man\u2019s hand when they are in contact. You don\u2019t need a van de Graaff generator for conduction to take place. It may occur when you walk across a wool carpet in rubber-soled shoes. Wool tends to give up electrons and rubber tends to accept them. Therefore, the carpet transfers electrons to your shoes each time you put down your foot. The transfer of electrons results in you becoming negatively charged and the carpet becoming positively charged. 8 www.ck12.org Polarization Concept 3. Transfer of Electric Charge Assume that you have walked across a wool carpet in rubber-soled shoes and become negatively charged. If you then reach out to touch a metal doorknob, electrons in the neutral metal will be repelled and move away from your hand before you even touch the knob. In this way, one end of the doorknob becomes positively charged and the other end becomes negatively charged. This is called polarization. Polarization occurs whenever electrons within a neutral", " object move because of the electric \ufb01eld of a nearby charged object. It occurs without direct contact between the two objects. The Figure 3.2 models how polarization occurs. FIGURE 3.2 Q: What happens when the negatively charged plastic rod in the diagram is placed close to the neutral metal plate? A: Electrons in the plate are repelled by the positive charges in the rod. The electrons move away from the rod, causing one side of the plate to become positively charged and the other side to become negatively charged. Friction Did you ever rub an in\ufb02ated balloon against your hair? You can see what happens in the Figure 3.3. Friction between the balloon and hair cause electrons from the hair to \u201crub off\u201d on the balloon. That\u2019s because a balloon attracts electrons more strongly than hair does. After the transfer of electrons, the balloon becomes negatively charged and the hair becomes positively charged. The individual hairs push away from each other and stand on end because like charges repel each other. The balloon and the hair attract each other because opposite charges attract. Electrons are transferred in this way whenever there is friction between materials that differ in their ability to give up or accept electrons. Watch the animation \u201cBalloons and Static Electricity\u201d at the following URL to see how electrons are transferred by friction between a sweater and a balloon. http://www.cabrillo.edu/~jmccullough/Physics/Electr ic_Forces_Fields.html Q: If you rub a balloon against a wall, it may stick to the wall. Explain why. A: Electrons are transferred from the wall to the balloon, making the balloon negatively charged and the wall positively charged. The balloon sticks to the wall because opposite charges attract. Summary \u2022 Whenever electrons are transferred between objects, neutral matter becomes charged. For example, when atoms lose or gain electrons they become charged particles called ions. 9 www.ck12.org FIGURE 3.3 \u2022 Three ways electrons can be transferred are conduction, friction, and polarization. In each case, the total charge remains the same. This is the law of conservation of charge. \u2022 Conduction occurs when there is direct contact between materials that differ in their ability to give up or accept electrons. \u2022 Polarization is the movement of electrons within a neutral object due to the electric \ufb01eld of a nearby charged object. It occurs without direct contact between the two objects. \u2022 Electrons are transferred whenever there is friction between materials that differ in their ability to give up or accept electrons. Vocabulary \u2022 law of conservation of charge : Law stating that charges are not destroyed when they are transferred between two materials or within a material, so the total charge remains the same. Practice At the following URL, review how charges are transferred through friction. Watch the animation and read the list of more-positive to less-positive materials. Then answer the questions below. http://www.regentsprep.org/regents/physics/phys03/atribo/default.htm 1. If you rub glass with a piece of plastic wrap, will the glass become positively or negatively charged? 2. Assume that after you pet your dog with very dry hands, you touch a metal doorknob and get a shock. Is electric charge transferred from your hand to the doorknob or the other way around? Review 1. How is charge transferred by a van de Graaff generator? 2. Compare and contrast the formation of cations and anions. 3. State the law of conservation of charge. 4. Explain how conduction and polarization occur, using the example of walking across a wool carpet in rubber- soled shoes and then reaching out to touch a metal doorknob. 10 www.ck12.org Concept 3. Transfer of Electric Charge 5. Predict what will happen to the charges of a plastic comb and a piece of tissue paper if you rub the tissue paper on the comb. ( Hint : Plastic tends to accept electrons and tissue paper tends to give up electrons.) References 1. Christopher Auyeung.. CC BY-NC 3.0 2. Christopher Auyeung.. CC BY-NC 3.0 3. Flickr:olga.palma.. CC BY 2.0 11 CONCEPT 4 Forces on Charged Objects www.ck12.org \u2022 Describe the changes that occur in the sub-atomic arrangement in matter when charged. \u2022 Describe how to charge an object. \u2022 De\ufb01ne conductors and insulators. \u2022 Understand the difference between conduction and induction. \u2022 Summarize the forces between charged objects. Lightning is the discharge of static electricity that has built up on clouds. Every year, the earth experiences an average of 25 million lightning strikes. Lightning bolts travel at speeds up to 60,000 miles per second, and can reach temperatures of 50,000\u00b0F, which is \ufb01ve times the temperature of the surface", " of the sun. The energy contained in a single lightning strike could light a 100 Watt light bulb 24 hours per day for 90 days. Forces on Charged Objects Electric charges exist within the atom. At the turn of the 20th century, J. J. Thomson and Ernest Rutherford determined that atoms contain very light-weight negatively charged particles called electrons and more massive, positively charged particles called protons. The protons are lodged in the nucleus of the atoms, along with the neutrally charged particles called neutrons, while the electrons surround the nucleus. When the number of electrons in the electron cloud and the number of protons in the nucleus are equal, the object is said to be neutral. Changes to the nucleus of an atom require tremendous amounts of energy, so protons are not easily gained or lost by atoms. Electrons, on the other hand, are held fairly loosely and can often be removed quite easily. When an object loses some electrons, the remaining object is now positively charged because it has an excess of protons. The electrons may either remain free or may attach to another object. In that case, the extra electrons cause that object to become negatively charged. Atoms that have lost electrons and become positively charged are called positive ions, and atoms that have gained electrons and become negatively charged are called negative ions. Electrons can be removed from some objects using friction, simply by rubbing one substance against another substance. There are many examples of objects becoming charged by friction, including a rubber comb through 12 www.ck12.org Concept 4. Forces on Charged Objects hair, and a balloon on a sweater. In both these instances, the electrons move from the second object to the \ufb01rst, causing the \ufb01rst object to become negatively charged and the second one positively charged. Friction between the tires on a moving car and the road cause the tires to become charged, and wind causes friction between clouds and air which causes clouds to become charged and can result in tremendous bolts of lightning. A common method of producing charge in the lab is to rub cat or rabbit fur against stiff rubber, producing a negative charge on the rubber rod. If you hold a rubber rod on one end and rub only the tip of the other end with a fur, you will \ufb01nd that only the tip becomes charged. The electrons you add to the tip of the rod remain where you put them instead of moving around on the rod. Rubber is an insulator. Insulators are substances that do not allow electrons to move through them. Glass, dry wood, most plastics, cloth, and dry air are common insulators. Materials that allow electrons to \ufb02ow freely are called conductors. Metals have at least one electron that can move around freely, and all metals are conductors. Forces are exerted on charged objects by other charged objects. You\u2019ve probably heard the saying \"opposites attract,\" which is true in regards to charged particles. Opposite charges attract each other, while like charges repulse each other. This can be seen in the image below. When two negatively charged objects are brought near each other, a repulsive force is produced. When two positively charged objects are brought near each other, a similar repulsive force is produced. When a negatively charged object is brought near a positively charged object, an attractive force is produced. Neutral objects have no in\ufb02uence on each other. A laboratory instrument used to analyze and test for static charge is called an electroscope. Seen below, an electroscope consists of a metal knob connected by a metal stem to two very lightweight pieces of metal called leaves, shown in yellow. The leaves are enclosed in a box to eliminate stray air currents. 13 www.ck12.org When a negatively charged object is brought near the knob of a neutral electroscope, the negative charge repels the electrons in the knob, and those electrons move down the stem into the leaves. Excess electrons \ufb02ow from the rod into the ball, and then downwards making both leaves negatively charged. Since both leaves are negatively charged, they repel each other. When the rod is removed, the electroscope will remain charged because of the extra electrons added to it. Conversely, if the rod is brought near the knob but doesn\u2019t touch it, the electroscope will appear the same while the rod is near. That is, the negative charge in the rod repels the electrons in the ball, causing them to travel down to the leaves. The leaves will separate while the rod is nearby. No extra electrons were added to the electroscope, meaning that the electrons in the electroscope will redistribute when the negatively charged rod is taken away. The leaves return to neutral, and they stop repelling each other. If the rod touches the knob, the electroscope leaves are permanently charged but if the rod is brought near but does not touch the knob, the electroscope leaves are only temporarily charged. If the leaves are permanently charged and the rod removed, the electroscope can then be used", " to determine the type of unknown charge on an object. If the electroscope has been permanently negatively charged, and a negatively charge object is brought near the knob, the leaves will separate even further, showing the new object has the same charge as the leaves. If a positively charged object is brought near a negatively charged electroscope, it will attract some of the excess electrons up the stem and out of the leaves, causing the leaves to come slightly together. Similar to the results of a negatively charged rod, if a positively charged rod is brought near the knob of a neutral electroscope, it will attract some electrons up from the leaves onto the knob. That process causes both of the leaves to 14 www.ck12.org Concept 4. Forces on Charged Objects be positively charged (excess protons), and the leaves will diverge. If the positively charged rob is actually touched to the knob, the rob will remove some electrons and then when the rob is removed, the electroscope will remain positively charged. This is a permanent positive charge. Charging an object by touching it with another charged object is called charging by conduction. By bringing a charged object into contact with an uncharged object, some electrons will migrate to even out the charge on both objects. Charging by conduction gives the previously uncharged object a permanent charge. An uncharged object can also be charged using a method called charging by induction. This process allows a change in charge without actually touching the charged and uncharged objects to each other. Imagine a negatively charged rod held near the knob, but not touching. If we place a \ufb01nger on the knob, some of the electrons will escape into our body, instead of down the stem and into the leaves. When both our \ufb01nger and the negatively charged rod are removed, the previously uncharged electroscope now has a slight positive charge. It was charged by induction. Notice that charging by induction causes the newly charged object to have the opposite charge as the originally charged object, while charging by conduction gives them both the same charge. Summary \u2022 Electric charges exist with the atom. \u2022 Atoms contain light-weight, loosely held, negatively charged particles called electrons and heavier, tightly- held, positvely charged particles called protons. \u2022 When the number of electrons and the number of protons are equal, the object is neutral. \u2022 The loss of electrons gives an ion a positive charge, while the gain of electrons gives it a negative charge. \u2022 Materials that allow electrons to \ufb02ow freely are called conductors, while those that do not are called insulators. \u2022 Opposite charges attract, and like charges repel. \u2022 Charging an object by touching it with another charged object is called charging by conduction. Practice The following video shows a young woman placing her hands on a Van de Graf generator which then gives her a static charge. Use this resource to answer the two questions that follow. http://www.youtube.com/watch?v=87DqbdqBx8U 15 www.ck12.org MEDIA Click image to the left for more content. 1. What happens to her hair when she touches a ground? 2. What happens to her hair when she steps off the platform? This video shows the static charge from the Van de Graf generator. http://www.youtube.com/watch?v=prgu6AvauuI MEDIA Click image to the left for more content. This video demonstrates superconductivity that occurs at extremely low temperatures. http://www.youtube.com/watch?feature=player_embedded&v=nWTSzBWEsms MEDIA Click image to the left for more content. Additional Practice Questions: 1. When a glass rod is rubbed with a silk cloth and the rod becomes positively charged, (a) electrons are removed from the rod. (b) protons are added to the silk. (c) protons are removed from the silk. (d) the silk remains neutral. 2. Electric charge is (a) found only in a conductor. (b) found only in insulators. (c) conserved. (d) not conserved. 3. When two objects are rubbed together and they become oppositely charged, they are said to be charge by (a) conduction. (b) induction. 16 www.ck12.org Concept 4. Forces on Charged Objects (c) friction. (d) grounding. 4. Two objects each carry a charge and they attract. What do you know about the charge of each object? (a) They are both charged positively. (b) They have opposite charged from each other. (c) They are both charged negatively. (d) Any of the above are possible. 5. A material that easily allows the \ufb02ow of electric charge through it is called a(n) (a) insulator. (b) conductor. (c) semiconductor. (d) heat sink", ". 6. What is the most common way of acquiring a positive static electrical charge? (a) by losing electrons (b) by gain protons (c) by losing protons (d) by gaining electrons (e) by switching positions of electrons and protons in the atom Review 1. How does friction generate static electricity? (a) Friction heats the materials, thus causing electricity. (b) Rubbing materials together displaces atoms, causing sparks to \ufb02y. (c) Rubbing materials together can strip electrons off atoms, causing one material to become positive and the other to become negative. (d) Rubbing materials together causes neutrons and electrons to trade places. (e) None of the above. 2. What electrical charge does an electron have? (a) A negative charge. (b) A positive charge. (c) A neutral charge. (d) May be any of the above. (e) None of the above. 3. What happens when opposite charges get close to each other? (a) They repel each other. (b) They attract each other. (c) Nothing happens. (d) They attract surrounding objects. (e) They repel surrounding objects. 4. What is an electrical conductor? (a) A material that allows electrons to travel through it freely. (b) A material that doesn\u2019t allow electrons to travel through it freely. (c) A material that melts at low temperature. (d) A material that creates free electrons. (e) None of the above. 5. Which of the following is a good insulator of electricity? 17 www.ck12.org (a) Copper (b) Iron (c) Rubber (d) Salt water (e) None of these. \u2022 electrons: A fundamental sub-atomic particle, meaning it cannot be broken into smaller particles. Electrons are found in the \u201celectron cloud\u201d surrounding an atomic nucleus, or they may break free and exist as a free electron. \u2022 protons: A stable, positively charged, sub-atomic particle, found in atomic nuclei in numbers equal to the atomic number of the element. \u2022 neutral: A neutral particle, object, or system is one that has a net electric charge of zero. \u2022 conductors: Materials through which electric charge can pass. \u2022 insulator: Substances that block or retard the \ufb02ow of electrical current or charge. \u2022 positive ions: An atom or a group of atoms that has acquired a net positive charge by losing one or more electrons. \u2022 negative ions: An atom or a group of atoms that has acquired a net negative charge by gaining one or more electrons. \u2022 ions: An atom or a group of atoms that has acquired a net electric charge by gaining or losing one or more electrons. \u2022 electroscope: An instrument used to detect the presence and sign of an electric charge by the mutual attraction or repulsion of metal foils. \u2022 charging by conduction: Involves the contact of a charged object to a neutral object. \u2022 charging by induction: A method used to charge an object without actually touching the object to any other charged object. References 1. Courtesy of NOAA Photo Library, NOAA Central Library; OAR/ERL/National Severe Storms Laboratory (NSSL). http://www.photolib.noaa.gov/htmls/nssl0016.htm. Public Domain 2. Sweater: Image copyright Sibiryanka, 2013; Balloon: Image copyright simpleman, 2013. http://www.shut terstock.com. Used under licenses from Shutterstock.com 3. CK-12 Foundation - Christopher Auyeung.. CC-BY-NC-SA 3.0 4. CK-12 Foundation - Samantha Bacic.. CC-BY-NC-SA 3.0 5. CK-12 Foundation - Samantha Bacic.. CC-BY-NC-SA 3.0 6. CK-12 Foundation - Samantha Bacic.. CC-BY-NC-SA 3.0 18 www.ck12.org Concept 5. Coulomb\u2019s Law CONCEPT 5 Coulomb\u2019s Law \u2022 State Coulomb\u2019s Law. \u2022 Describe how electric force varies with charge and separation of charge. \u2022 State the SI unit of charge. \u2022 Solve problems using Coulomb\u2019s Law. Electric cars are becoming more popular. One large advantage for electric cars is the low cost of operation, which may become an ever bigger advantage as gas prices climb. Energy costs for electric cars average about one-third of the cost for gasoline engine cars, but they can only travel about 200 miles per charge at this point. These cars run using the science of electrical charges and forces. Coulomb\u2019s Law The questions regarding the relationship between the electrical force, the size of the charge, and the separation between the charges were solved by Charles Coulomb in 1785. He determined that electrical force between two charges is directly", " related to the size of the charges and inversely proportional to the distance between the charges. This is known as Coulomb\u2019s Law. Fe = Kq1q2 d2 In this equation, q 1 and q 2 are the two charges, d is the distance between the two charges, and K is a constant of proportionality. F e is the electric force, which occurs as a result of interactions between two charged particles. For the purpose of calculating electric forces, we assume all charge is a point charge, in which the entire charge of the particle is located in a massless point. The SI unit of charge is the coulomb, C, which is the charge of 6:25 1018 electrons. The charge on a single electron is 1:60 1019 C. The charge on a single electron is known as the elementary charge. The charge on a proton is the same magnitude but opposite in sign. When the charges are measured in coulombs, the distance in meters, and the force in Newtons, the constant K is 9:0 109 N m2=C2. 19 www.ck12.org The electrical force, like all forces, is a vector quantity. If the two charges being considered are both positive or both negative, the sign of the electrical force is positive and this force is repulsive. If the two charges are opposite in sign, the force will have a negative sign and the force is attractive. Example Problem: Object A has a positive charge of 6:0 106 C. Object B has a positive charge of 3:0 106 C. If the distance between A and B is 0.030 m, what is the force on A? Solution: Fe = Kq1q2 d2 = (9:0109 Nm2=C2)(6:0106 C)(3:0106 C) = 180 N (0:030 m)2 The positive sign of the force indicates the force is repulsive. This makes sense, because both objects have a positive charge. Example Problem: 6:0 106 C. Calculate the total force on q 2. In the sketch below, the charges are q1 = 10:0 106 C; q2 = 2:0 106 C, and q3 = Solution: Fe = Kq1q2 d2 = (9:0109 Nm2=C2)(10:0106 C)(2:0106 C) (2:0 m)2 = 0:045 N (towards q3) Fe = Kq2q3 d2 = (9:0109 Nm2=C2)(2:0106 C)(6:0106 C) (4:0 m)2 = 0:007 N (towards q3) Since the two forces act in the same direction, their absolute values can be added together; the total force on q 2 is 0.052 N towards q 3. Summary \u2022 Coulomb determined that electrical force between two charges is directly related to the size of the charges and inversely proportional to the distance between the charges: Fe = Kq1q2 d2 \u2022 The SI unit of charge is the coulomb, C, which is the charge of 6:25 1018 electrons. \u2022 The charge on a single electron is 1:60 1019 C and is known as the elementary charge. \u2022 The electrical force is a vector quantity that is positive in repulsion and negative in attraction. Practice The following video covers Coulomb\u2019s Law. Use this resource to answer the questions that follow. http://www.youtube.com/watch?v=rYjo774UpHI MEDIA Click image to the left for more content. 1. What happens when like charges are placed near each other? 2. What happens when opposite charged are placed near each other? 3. What happens to the force of attraction if the charges are placed closer together? 20 www.ck12.org Concept 5. Coulomb\u2019s Law Practice problems on Coulomb\u2019s Law. http://physics.info/coulomb/problems.shtml Review 1. Suppose that two point charges, each with a charge of +1.00 C, are separated by a distance of 1.0 m: (a) Will the charges attract or repel? (b) What is the magnitude of the force between them? (c) If the distance between them is doubled, what does the force become? 2. What is the electrical force between two balloons, each having 5.00 C of charge, that are 0.300 m apart? 3. Two spheres are charged with the same charge of -0.0025 C and are separated by a distance of 8.00 m. What is the electrical force between them? 4. A red foam ball and a blue foam ball are 4.00 m apart. The blue ball has a charge of 0.000337 C and is attracting the red ball with a", " force of 626 N. What is the charge on the red ball? \u2022 Coulomb\u2019s Law: States the force of attraction or repulsion acting along a straight line between two electric charges is directly proportional to the product of the charges and inversely to the square of the distance between them. References 1. Image copyright testing, 2013. http://www.shutterstock.com. Used under license from Shutterstock.com 2. CK-12 Foundation - Samantha Bacic.. CC BY-NC-SA 3.0 21 CONCEPT 6 \u2022 De\ufb01ne an electric \ufb01eld. \u2022 Solve problems relating to \ufb01eld, force, and charge. www.ck12.org The Electric Field A plasma globe, such as the one pictured above, is \ufb01lled with a mixture of noble gases and has a high-voltage electrode at the center. The swirling lines are electric discharge lines that connect from the inner electrode to the outer glass insulator. When a hand is placed on the surface of the globe, all the electric discharge travels directly to that hand. The Electric Field Coulomb\u2019s Law gives us the formula to calculate the force exerted on a charge by another charge. On some occasions, however, a test charge suffers an electrical force with no apparent cause. That is, as observers, we cannot see or detect the original charge creating the electrical force. Michael Faraday dealt with this problem by developing the concept of an electric \ufb01eld. According to Faraday, a charge creates an electric \ufb01eld about it in all directions. If a second charge is placed at some point in the \ufb01eld, the second charge interacts with the \ufb01eld and experiences an electrical force. Thus, the interaction we observe is between the test charge and the \ufb01eld and a second particle at some distance is no longer necessary. The strength of the electric \ufb01eld is determined point by point and can only be identi\ufb01ed by the presence of test charge. When a positive test charge, q t, is placed in an electric \ufb01eld, the \ufb01eld exerts a force on the charge. The \ufb01eld strength can be measured by dividing the force by the charge of the test charge. Electric \ufb01eld strength is given the symbol E and its unit is Newtons/coulomb. 22 E = Fonqt qt www.ck12.org Concept 6. The Electric Field The test charge can be moved from location to location within the electric \ufb01eld until the entire electric \ufb01eld has been mapped in terms of electric \ufb01eld intensity. Example Problem: A positive test charge of 2:0 105 C is placed in an electric \ufb01eld. The force on the test charge is 0.60 N. What is the electric \ufb01eld intensity at the location of the test charge? Solution: E = F q = 0:60 N 2:0105 C = 3:0 104 N/C Summary \u2022 An electric \ufb01eld surrounds every charge and acts on other charges in the vicinity. \u2022 The strength of the electric \ufb01eld is given by the symbol E, and has the unit of Newtons/coulomb. \u2022 The equation for electric \ufb01eld intensity is E = F q. Practice The following video covers electric \ufb01elds. Use this resource to answer the questions that follow. http://www.youtube.com/watch?v=lpb94QF0_mM MEDIA Click image to the left for more content. 1. What does it mean when a force is called a non-contact force? 2. What symbol is used to represent electric \ufb01eld strength? 3. What is the relationship between the direction of the electric \ufb01eld and the direction of the electric force? Review 1. The weight of a proton is 1:64 1026 N. The charge on a proton is +1:60 1019 C. If a proton is placed in a uniform electric \ufb01eld so that the electric force on the proton just balances its weight, what is the magnitude and direction of the \ufb01eld? 2. A negative charge of 2:0 108 C experiences a force of 0.060 N to the right in an electric \ufb01eld. What is the magnitude and direction of the \ufb01eld? 3. A positive charge of 5:0 104 C is in an electric \ufb01eld that exerts a force of 2:5 104 N on it. What is the magnitude of the electric \ufb01eld at the location of the charge? 4. If you determined the electric \ufb01eld intensity in a \ufb01eld using a test", " charge of 1:0 106 C and then repeated the process with a test charge of 2:0 106 C, would the forces on the charges be the same? Would you \ufb01nd the value for E? 5. A 0.16 C charge and a 0.04 C charge are separated by a distance of 3.0 m. At what position between the two charges would a test charge experience an electric \ufb01eld intensity of zero? \u2022 electric \ufb01eld: A region of space characterized by the existence of a force that is generated by an electric charge. \u2022 electric \ufb01eld intensity: Electric \ufb01eld intensity or \ufb01eld strength is described as the ratio of force to the amount of test charge. 23 References 1. User:Slimsdizz/Wikipedia. http://commons.wikimedia.org/wiki/File:Glass_plasma_globe.jpg. Public Domain www.ck12.org 24 www.ck12.org Concept 7. Electric Potential CONCEPT 7 Objectives The student will: Electric Potential \u2022 Understand how to solve problems using electric potential energy. \u2022 Understand how to solve problems using voltage differences. \u2022 Understand how to solve problems in a uniform electric \ufb01eld. Vocabulary \u2022 electric potential: Energy per unit charge. \u2022 electric potential difference: The difference in electric potential between two points within an electric \ufb01eld. \u2022 voltage: The amount of work done by the electric \ufb01eld per unit charge in moving a charge between two points in the electric \ufb01eld DV = W q, also known as a change in potential energy. Introduction In order to draw an analogy between gravitational potential energy and electrical potential energy, we liken the electric \ufb01eld E to the gravitational acceleration g and the mass m of a particle to the charge q of a particle. Of course, g is assumed constant (uniform) when we remain close to the surface of the Earth. As of yet, we have not encountered an example of a uniform electric \ufb01eld E. But that won\u2019t stop us from making a prediction! Since the gravitational potential energy of a mass m in a uniform gravitational \ufb01eld is PEgravity = (mg)h, we predict the electric potential energy (PEelectric) of a charge (q) in a constant electric \ufb01eld is PEelectric = (qE)h. Furthermore, the electric potential (the energy per unit charge) can be de\ufb01ned as Ve = PEelectric will be dropped from now on.) We will discuss electric potential later. (The subscript \u201ce\u201d q It must be understood that, just as with gravity, the electric potential energy and the electric potential are measured at the same point. If a point charge q has electric potential energy PEx1 at point x1, the electric potential at x1 is Vx1 PEx1 = qVx1! Vx1 = PEx1 q q PEi q. Again, only differences in electric potential and electric potential energy are meaningful. That is, DPE or DV! Vf Vi = PE f The unit of electric potential is called the volt and from the de\ufb01nition above we see that the volt is equivalent to Coulomb! V = J Joules C. 25 Electric Potential Difference The electric potential difference is the difference in electric potential between two points within an electric \ufb01eld. For example, a 1.5-volt battery has a potential difference of 1.5 volts (written 1.5 V ) between its positive and negative terminals. www.ck12.org Parallel Plate Conductors: A uniform Electric Field The equation E = k q r2 for the electric \ufb01eld holds for point charges or for a charge distribution that effectively acts as a point charge. It turns out, however, that if opposite charges are placed on two parallel conducting plates, the electric \ufb01eld between the plates is more or less uniform as long as the distance between the plates is much smaller than the dimensions of the plates. The plates can be charged by connecting them to the positive and negative terminals of a battery. A battery contains a substance (called an electrolyte) which causes two dissimilar metals to acquire opposite charges. The two dissimilar metals form the positive and negative terminals of the battery. If a metal plate is connected to the positive terminal of the battery, and another metal plate is connected to the negative terminal of the battery, and the two plates brought closely together, a parallel plate arrangement (parallel-plate conductors) can be constructed with a uniform electric \ufb01eld between the plates (seen edge on) in Figure below. We will see later that parallel plate conductors are also referred to as capacitors. FIGURE 7.1 Parallel plates. Just as in the case of the battery, one of the plates of the parallel-plate", " conductor will be at a higher potential than the other plate. Think, for example, of a standard AA battery with a voltage rating of 1.5V, Figure below. See the link below to learn more about how a battery works. http://phet.colorado.edu/en/simulation/battery-voltage 26 www.ck12.org Concept 7. Electric Potential FIGURE 7.2 volt battery Electrical Potential Energy In our gravitational analogy, the energy that a charge possesses at the plate with the higher potential is analogous to the energy a mass possesses above the ground. Additionally, now that we have found a way to create a uniform electric \ufb01eld, we have an analog to a uniform gravitational \ufb01eld. If a positive charge +q is placed at the positive plate in Figure below, it will be repelled by the positive charges on the plate and move toward the negative plate. (Think of +q as the object m falling toward the ground.) FIGURE 7.3 A positive charge moving toward the negative plate What is the force acting on the +q charge? Recall that the Coulomb force on a charge placed in an electrostatic \ufb01eld is F = qE. The work that the electric \ufb01eld does on the charge is equal to the negative change in the potential energy of the charge, just as in the gravitational case. We can \ufb01nd an expression for the electric potential energy by \ufb01nding the work that is done on the charge. Recall that W = FDx. We write 27 www.ck12.org Wf ield = FDx = (qE)Dx = DPE! qE(x f xi) = DPE! qEx f qExi = DPE The expression for the electric potential energy is thus: PEelectrical = qEx. Recall that the equation for the gravitational potential energy is PEgravitational = mgh. We can compare the terms in the gravitational and electrical cases as follows Thus, we see that our prediction for the equation of electric potential energy stated in the introduction of the lesson, was correct! Check Your Understanding 1a. The electrical potential at the negative plate in Figure above is de\ufb01ned as zero volts. What is the electrical potential energy of a charge +q = 15:0\u00b5C at the positive plate if the electric \ufb01eld between the plates is 25:0 N The positive plate has position 6:00cm = 6:00 102 m according to Figure above. Answer : PEpositive plate = qEx = (15:0 106 C) 25:0 N C 1b. What is the change in the electrical potential energy DPE of the charge +q = 15:0\u00b5C if its potential changes from 1.5 V to 1.0 V? (6:00 102 m) = 3:75 104 J C? Answer : Just as in the case of a change in gravitational potential energy, the charge must lose potential energy, since it gains kinetic energy. The charge moves from the position xi = 6:00 102 m (1:5 V ) to the position xi = 4:00 102 m (1:0 V ). DPE = qEx f qExi = qE(x f xi) = N C (15:0 106 C) 25:0 (4:00 102 m 6:00 102 m) = 7:50 106 J: 1c. What is the work done on the charge by the electric \ufb01eld? Answer: Wf ield = DPE = (7:50 106) = 7:50 106 J Notice that the electric \ufb01eld does positive work on the charge, since the electric force and the displacement of the charge have the same direction. We should recall a very important point: It is only the change in potential energy that is meaningful, whether we are discussing the gravitational potential energy or the electrical potential energy. 28 www.ck12.org Concept 7. Electric Potential 2. An electron placed at the negative plate of a parallel-plate conductor will move toward the positive plate. The potential energy of the electron: A. Decreases B. Increases C. Remains the same. Answer : The correct answer is A. The electron is repelled by the negative charges of the conducting plate and therefore gains kinetic energy. Just as an object that is dropped gains kinetic energy and loses potential energy, so does the electron. Recall our discussion of the conservation of energy. As long as the total energy remains conserved, the sum of the initial kinetic and potential energies must equal the sum of the \ufb01nal kinetic and potential energies: KEi + PEi = KE f + PE f! DKE = DPE The gain in kinetic energy occurs due to the loss in potential energy. In order for the charges of the same sign to be brought together, as in", " the example above, positive work must be done by an external force against the electrostatic repulsion between the charges. The work increases the potential energy stored in the electric \ufb01eld. When the charges are released, the potential energy of the \ufb01eld is converted into the kinetic energies of the charges. The link below may be helpful in learning more about the work done upon charges in electric \ufb01elds. http://www.youtube.com/watch?v=elJUghWSVh4 Electric Potential Difference in a Uniform Electric Field In working with the change in potential energy above, we wrote the equation DPE = qEx f qExi = qE(x f xi)! Let us call this Equation A. Recall that the electric potential was de\ufb01ned at a speci\ufb01c point Vx1 = We therefore see that PEx1 = qVx1! Let us call this Equation B. Comparing Equation A and Equation B, we see that the electric potential can be expressed as Vx1 = Ex1. If the electric potential is de\ufb01ned as V = 0 at x = 0, then the potential at any point in the electric \ufb01eld is V = Ex. (Assuming that vector E is directed along the x axis).. PEx1 q Note: It is common to write V = Ed, where V is understood to mean the voltage (or potential difference) between the plates of a parallel-plate conductor, and d is the distance between the plates. Check Your Understanding Verify that the potential difference between the plates in Figure above is 1.5 V. Recall that the electric \ufb01eld is E = 25:0 N Answer : V = Ed = 25 N C (6:00 102 m 0:00m) = 1:5 V C. Work We state again: 1. The electric potential is de\ufb01ned as the energy per unit charge! Vx1 = PEx1 q. 29 www.ck12.org 2. The electric potential difference (the voltage) is Vf Vi = PE f 3. An arbitrary reference level must be established for zero potential (just as in the case of gravitational potential q PEi q energy). 4. The units of electric potential and electric potential difference are J C since Vx1 = PEx1 q. It is often useful to express the voltage in terms of the work done on a charge. From Vf Vi = PE f But the work done on a charge by the \ufb01eld is Wf ield = DPE. q, we have PE f PEi = q(Vf Vi)! DPE = q(Vf Vi). q PEi Combining DPE = q(Vf Vi) and Wf ield = DPE gives Wf ield = q(Vf Vi). An external force that does work on a charge in an electric \ufb01eld exerts a force in the opposite direction to the \ufb01eld (just as the external force acting on a spring acts opposite to the spring force). The work that an external force does is therefore Wexternal f orce = q(Vf Vi). The voltage can be thought of as the amount of work done by the electric \ufb01eld per unit charge in moving a charge between two points in the electric \ufb01eld DV = W q. We often refer to a change in potential as simply \u201cthe voltage.\u201d In computing the work, it is often easier to ignore the sign in the equation and simply see if the force and displacement on the charge are in the same direction (positive work) or opposite to one another (negative work). Recall that the force and displacement need not be in the same direction or oppositely directed. In general, work is expressed as W = Fx cos q http://www.youtube.com/watch?v=F1p3fgbDnkY Other Units for the Electric Field! E = C, since! F q. But the electric \ufb01eld has been also de\ufb01ned using the scalar equation x. So the units of the electric \ufb01eld can be also expressed as The electric \ufb01eld has units N V = Ex. Transposing terms, the electric \ufb01eld is E = V volts per meter volts If we compare the units for the electric \ufb01eld N m, we see that a (N m) is equivalent to a (C V ). A Joule can therefore be expressed as a Coulomb-Volt. Recall that work, measured in Joules, is the product of charge and voltage W = qDV. meter! V m. C and V Illustrative Example 16.2.1 All questions refer to Figure above. a. What is the", " potential at x = 2:0 cm in Figure below? The electric \ufb01eld is E = 25:0 N C. Answer : The potential V varies directly with the position x between the plates (V = Ex). Thus, V = 25:0 N 0:50 V. C (2:0 102m) = b. Sketch a graph showing the relationship between the potential and the position. Answer: 30 www.ck12.org Concept 7. Electric Potential c. How much work is done by an external force F moving a 2:0 106 C charge from the positive plate to the negative plate? Answer: An external force must pull the charge away from the positive plate so the force will be in the same direction as the displacement. W = q(Vf Vi) = (2:0 106C)(0:00V 1:50V ) = 3:0 106 J d. What is the magnitude of the Coulomb force acting on the charge? Answer: W = F! 3:0 106 J = F(6:00 102 m) F = 3:0 106 J 6:00 102 m = 5:0 105 N Illustrative Example 16.2.2 a. A particle of mass m of 2:00 105 kg is has a charge q of +3:00 103 C. If the particle is released from the positive plate of a parallel-plate conductor with an electric \ufb01eld E of 1:30 105 N C, determine the acceleration of the particle, see Figure below. Answer : If we ignore gravity, the only force acting on the particle is the electric force F = qE. Using Newton\u2019s Second Law, the net force on the particle is equal to F = ma! qE = ma. The acceleration is a = Eq m = (1:30105 V m )(3:00103 C) 2:00105 kg = 1:95 107 V C kgm. b. Show that the units V C kgm are equivalent to the units m s2. Answer : V C kg m = J kg m = N m kg m = N kg = kg m s2 kg = m s2 c. The plates have separation of 8.00 mm. Determine the velocity of the particle when it reaches the negative plate Answer : 31 www.ck12.org FIGURE 7.4 Illustrative Example 16.3.2 This is a kinematics problem, where the displacement and acceleration are known and the velocity is to be found. Recall the equation v2 f = v2 i + 2aDx.! v2 1:95 107 m s2 f = 0 + 2 v = 558:6! 5:59 102 m s : (8:00 103 m) = 312; 000 m2 s2 d. What is the potential difference between the plates? Answer : V = Ex = 1:30 105 V m (8:00 103 m) = 1:04 103 V e. How much work has the \ufb01eld done on the particle as it moved from one plate to the other? W = qDV = (3:00 103 C)(1:04 103 V ) = 3:12 J Illustrative Example 16.2.3 An electron is accelerated from rest through a potential difference of 30,000 V. The mass of the electron is 9:11 1031 kg and the charge of the electron is 1:60 1019 C. Find its velocity. Answer : Recall that the Work-Energy Principle states that W = DKE. 32 www.ck12.org Concept 7. Electric Potential W = DKE W = qDV DKE = qDV 1 2 i = qDV mv2 1 2 mv2 f v f = Illustrative Example 16.2.4 vi = 0! v2 s f = 2qDV m! v f = r 2qDV m! 2(1:60 1019 C)(3:00 104 V ) 9:11 1031 kg = 1:026 108! 1:03 108 m s What magnitude of an electric \ufb01eld is required to balance the gravitational force acting on an electron in Figure below? FIGURE 7.5 Illustrative Example 16.2.4-An electron suspended in an electric \ufb01eld. Answer : Draw a Free-Body-Diagram (FBD) of the situation. The electrostatic force that acts on the electron points upward and the gravitational force that acts upon on the electron points downward. The electron is suspended motionless (or moves with a constant velocity) when the net force on the electron is zero. 33 The net force on the electron must be zero, thus www.ck12.org F = 0! eF = mg! F = mg e mg e mg = eE! E = e2! but F = eE! (9:11 1031 kg", ") 9:81 m s2 (1:60 1019 C)2 E = = 3:49 108 V m http://www.youtube.com/watch?v=wT9AsY79f1k The Electron-Volt It is often convenient when dealing with small particles such as electrons, protons, and ions to express the energy of these particles with a smaller unit of measure. The electron-volt is de\ufb01ned as the change in potential energy that an electron acquires when moving through a potential difference of 1 V, or equivalently, its change in kinetic energy after moving through a potential difference of 1 V. That is, PE = eV = (1:60 1019 C)(1:00 V ) = 1:60 1019 J. The energy 1:60 1019 J is de\ufb01ned as one electron-volt. We write one-electron-volt as 1 eV = 1:60 1019 J. Check Your Understanding 1. What is the change in kinetic energy KE when an electron is released at the negative plate of a parallel plate conductor with a potential difference of 3,500 V? Express your answer in eV. Answer : The electron is repelled at the negative plate and therefore gains kinetic energy (and loses potential energy). The change in KE is positive and equal to = x eV 1 eV 1 V 3; 500 V DKE = 3; 500 eV! x = 3; 500 eV It is simplest to think that for every one volt of potential difference the particle experiences, it gains (or loses) 1 eV. 2. An alpha-particle (the nucleus of a helium atom) is \ufb01red toward the positive plate of a parallel plate conductor and passes through a potential difference of 1,500 V. What is the change in its kinetic energy? Express your answer in eV. Answer : Protons are the only charges inside the nucleus of an atom and so the alpha particle must be positively charged. A helium nucleus contains two protons (and two neutrons) with a total charge of 2(1:60 1019 C). The alpha particle must slow down due to the electrostatic repulsion from the positive plate. It must, therefore, lose kinetic energy and gain potential energy. Each proton loses 1,500 eV of kinetic energy. DKE = 3; 000 eV. 3. An electron and a proton both gain kinetic energy of 1 eV. True or False: Their speeds must be the same, since they both gained the same amount of energy. 34 www.ck12.org Concept 7. Electric Potential Answer : False. The mass of a proton is nearly 2000 times greater than the mass of an electron. Remember that kinetic energy depends on both the speed and mass of an object. Therefore, the \ufb01nal speed of the electron will be much greater. Illustrative Example 16.2.5 a. An electron and a proton both gain kinetic energy of 1 eV. What is the ratio of the electron\u2019s speed to the proton\u2019s speed? Answer : As discussed above, though both particles gain the same kinetic energy, their speeds will not be the same, since they have different masses. The mass of the proton is nearly 2000 times as great as the electron\u2019s so: = 1 2 mev2 e 1 2 mpv2 p = 1! mev2 e = mpv2 p! v2 e v2 p = mp me = 2000me me = 2; 000! = p 2; 000 = 44:7! 45 KEe KEp ve vp The electron will move about 45 times faster than the proton. b. What is the speed of a proton which has a kinetic energy of 37 MeV? The mass of a proton is 1:67 1027 kg. Answer : Because the electron-volts are a very small unit, they are typically expressed in KeV (1000 electron-volts) and MeV (one million electron-volts). The electron-volt is a convenient unit of measure but it is not an SI unit. In order to \ufb01nd the velocity of a particle if its energy is given in units of eV, we must convert back into Joules. 37 MeV = (37 106)(1:60 1019 J) = 5:92 1014 J 1 2 1 2 (1:67 1027 kg)v2 = 5:92 1014 J! mpv2 = 5:92 1014 J! v = 8:4 106 m s References 1. CK-12 Foundation - Raymond Chou.. CC-BY-NC-SA 3.0 2. User:Asim18/Wikimedia Commons. http://commons.wikimedia.org/wiki/File:02_-_Single_Energizer_Ba ttery.jpg. CC-BY 3", ".0 3. CK-12 Foundation - Raymond Chou.. CC-BY-NC-SA 3.0 4. CK-12 Foundation - Ira Nirenberg.. CC-BY-NC-SA 3.0 5. CK-12 Foundation - Raymond Chou.. CC-BY-NC-SA 3.0 6. CK-12 Foundation - Raymond Chou.. CC-BY-NC-SA 3.0 7. CK-12 Foundation - Raymond Chou.. CC-BY-NC-SA 3.0 35 Physics Unit 13: Circuits Patrick Marshall Jean Brainard, Ph.D. Ck12 Science James H Dann, Ph.D. Say Thanks to the Authors Click http://www.ck12.org/saythanks (No sign in required) AUTHORS Patrick Marshall Jean Brainard, Ph.D. Ck12 Science James H Dann, Ph.D. CONTRIBUTORS Chris Addiego Antonio De Jesus L\u00f3pez www.ck12.org To access a customizable version of this book, as well as other interactive content, visit www.ck12.org CK-12 Foundation is a non-pro\ufb01t organization with a mission to reduce the cost of textbook materials for the K-12 market both in the U.S. and worldwide. Using an open-content, web-based collaborative model termed the FlexBook\u00ae, CK-12 intends to pioneer the generation and distribution of high-quality educational content that will serve both as core text as well as provide an adaptive environment for learning, powered through the FlexBook Platform\u00ae. Copyright \u00a9 2014 CK-12 Foundation, www.ck12.org The names \u201cCK-12\u201d and \u201cCK12\u201d and associated logos and the terms \u201cFlexBook\u00ae\u201d and \u201cFlexBook Platform\u00ae\u201d (collectively \u201cCK-12 Marks\u201d) are trademarks and service marks of CK-12 Foundation and are protected by federal, state, and international laws. Any form of reproduction of this book in any format or medium, in whole or in sections must include the referral attribution link http://www.ck12.org/saythanks (placed in a visible location) in addition to the following terms. Except as otherwise noted, all CK-12 Content (including CK-12 Curriculum Material) is made available to Users in accordance with the Creative Commons Attribution-Non-Commercial 3.0 Unported (CC BY-NC 3.0) License (http://creativecommons.org/ licenses/by-nc/3.0/), as amended and updated by Creative Commons from time to time (the \u201cCC License\u201d), which is incorporated herein by this reference. Complete terms can be found at http://www.ck12.org/terms. Printed: April 6, 2014 iii Contents www.ck12.org 1 4 7 10 12 15 19 22 25 28 32 35 Contents 1 Electric Circuits 2 Electric Current 3 Electric Resistance 4 Ohm\u2019s Law 5 Resistance and Ohm\u2019s Law 6 Energy Transfer in Electric Circuits 7 Ammeters and Voltmeters 8 Series Circuits 9 Resistors in Series 10 Parallel Circuits 11 Resistors in Parallel 12 Combined Series-Parallel Circuits iv www.ck12.org Concept 1. Electric Circuits CONCEPT 1 Electric Circuits \u2022 De\ufb01ne electric circuit. \u2022 Describe the parts of an electric circuit. \u2022 Show how to represent a simple electric circuit with a circuit diagram. Jose made this sketch of a battery and light bulb for science class. If this were a real set up, the light bulb wouldn\u2019t work. The problem is the loose wire on the left. It must be connected to the positive terminal of the battery in order for the bulb to light up. Q: Why does the light bulb need to be connected to both battery terminals? A: Electric current can \ufb02ow through a wire only if it forms a closed loop. Charges must have an unbroken path to follow between the positively and negatively charged parts of the voltage source, in this case, the battery. Electric Circuit Basics A closed loop through which current can \ufb02ow is called an electric circuit. In homes in the U.S., most electric circuits have a voltage of 120 volts. The amount of current (amps) a circuit carries depends on the number and power of electrical devices connected to the circuit. Home circuits generally have a safe upper limit of about 20 or 30 amps. 1 Parts of an Electric Circuit All electric circuits have at least two parts: a voltage source and a conductor. They may have other parts as well, such as light bulbs and switches, as in the simple circuit seen in the Figure 1.1. To see an animation of a circuit like this one, go to: http://www.rkm.com.au/animations/animation-electrical-circuit.html www.", "ck12.org FIGURE 1.1 \u2022 The voltage source of this simple circuit is a battery. In a home circuit, the source of voltage is an electric power plant, which may supply electric current to many homes and businesses in a community or even to many communities. \u2022 The conductor in most circuits consists of one or more wires. The conductor must form a closed loop from the source of voltage and back again. In the circuit above, the wires are connected to both terminals of the battery, so they form a closed loop. \u2022 Most circuits have devices such as light bulbs that convert electrical energy to other forms of energy. In the case of a light bulb, electrical energy is converted to light and thermal energy. \u2022 Many circuits have switches to control the \ufb02ow of current. When the switch is turned on, the circuit is closed and current can \ufb02ow through it. When the switch is turned off, the circuit is open and current cannot \ufb02ow through it. Circuit Diagrams When a contractor builds a new home, she uses a set of plans called blueprints that show her how to build the house. The blueprints include circuit diagrams. The diagrams show how the wiring and other electrical components are to be installed in order to supply current to appliances, lights, and other electric devices. You can see an example of a very simple circuit in the Figure 1.2. Different parts of the circuit are represented by standard circuit symbols. An ammeter measures the \ufb02ow of current through the circuit, and a voltmeter measures the voltage. A resistor is any device that converts some of the electricity to other forms of energy. For example, a resistor might be a light bulb or doorbell. The circuit diagram on the right represents the circuit drawing on the left. Below are some of the standard symbols used in circuit diagrams. 2 www.ck12.org Concept 1. Electric Circuits FIGURE 1.2 Q: Only one of the circuit symbols above must be included in every circuit. Which symbol is it? A: The battery symbol (or a symbol for some other voltage source) must be included in every circuit. Without a source of voltage, there is no electric current. Summary \u2022 An electric circuit is a closed loop through which current can \ufb02ow. \u2022 All electric circuits must have a voltage source, such as a battery, and a conductor, which is usually wire. They may have one or more electric devices as well. \u2022 An electric circuit can be represented by a circuit diagram, which uses standard symbols to represent the parts of the circuit. Vocabulary \u2022 electric circuit : Closed loop through which current can \ufb02ow. Practice Take the electric circuit quiz at the following URL. Be sure to have your answers corrected. Try the quiz again if any of your answers are incorrect. http://www.myschoolhouse.com/courses/O/1/68.asp Review 1. What is an electric circuit? 2. Which two parts must all electric circuits contain? 3. Sketch a simple circuit that includes a battery, switch, and light bulb. Then make a circuit diagram to represent your circuit, using standard circuit symbols. References 1. Christopher Auyeung.. CC BY-NC 3.0 2. Christopher Auyeung.. CC BY-NC 3.0 3 www.ck12.org Electric Current CONCEPT 2 Objectives The student will: \u2022 Understand how electric current is de\ufb01ned \u2022 Solve problems involving electric current Vocabulary \u2022 electric current: A \ufb02ow of charges under the in\ufb02uence of an electric \ufb01eld, such as between the terminals of a battery. The rate I = DQ Dt at which charges \ufb02ow within a conducting wire past any point in the wire. Introduction The term electrical current is familiar to most people. Many electrical devices have electrical speci\ufb01cations printed on them. Figure below shows a typical AC adapter (\u201cplug\u201d) with its \u201cspecs.\u201d Can you guess what the terms 5 VDC and 500 mA printed on the adapter mean? FIGURE 2.1 An electrical plug. Electric Current An electric current is a \ufb02ow of charges under the in\ufb02uence of an electric \ufb01eld. A \ufb02ow of charges can be established, for instance, between the terminals of a battery, as in Figure below. The rate I = DQ Dt at which charges \ufb02ow within 4 www.ck12.org Concept 2. Electric Current a conducting wire past any point in the wire is de\ufb01ned as the electric current. The unit of current is coulombs Ampere, 1775-1836), Figure below. second which is called the ampere or amp (named for the French physicist Andre\u2019-Marie The symbol A is used to represent the ampere. A rate of one coulomb", " per second is equivalent to one ampere: 1C 1s = 1A FIGURE 2.2 Figure above shows a \ufb02ow of electrons (e) from the positive terminal of a battery through a lightbulb to the negative terminal of a battery. FIGURE 2.3 Andre\u2019-Marie Ampere One ampere is a very large current. The current of 1 A can easily kill a person. In fact, about 0.20 A can kill rather easily. Even relatively small voltage can produce these currents, which is why care must always be taken when dealing with all electrical appliances and any electrical device that is plugged into a wall outlet. A typical 12-V car battery can also be dangerous. Under the right circumstances, it does not take a huge voltage to cause deadly currents. It is common to express current in milliamperes 1 mA = 103 A, or microamperes 1 \u00afA = 106 A. http://demonstrations.wolfram.com/ElectricCurrent/ 5 Illustrative Example 17.4.1 A total of 7:9 1012 electrons move past a point in a conducting wire every 1.45 s. What is the average current in the wire? Answer : www.ck12.org The total charge moving past the point is the product of the electric charge of an electron and the number of electrons moving past the point. The total charge is:! Q = 1:6 1019 (7:9 1012 electrons) = 12:6 107! 13 107 C The current is I = DQ 1:45 s = 8:69 107 A! 0:87 \u00b5A. Dt = 12:6107 C electron References 1. Ray Dehler (Flickr: raybdbomb). http://www.\ufb02ickr.com/photos/raybdbomb/2200741209/ 2. CK-12 Foundation - Ira Nirenberg.. CC-BY-NC-SA 3.0 3.. http://commons.wikimedia.org/wiki/File:Andre-marie-ampere2.jpg. public domain. CC-BY 2.0 6 www.ck12.org Concept 3. Electric Resistance CONCEPT 3 Electric Resistance \u2022 De\ufb01ne resistance and identify the SI unit for resistance. \u2022 List factors that affect resistance. \u2022 Explain why resistance can be a help or a hindrance. These athletes are playing rugby, a game that is similar to American football. The players in red and blue are trying to stop the player in orange and black from running across the \ufb01eld with the ball. They are resisting his forward motion. This example of resistance in rugby is a little like resistance in physics. What Is Resistance? In physics, resistance is opposition to the \ufb02ow of electric charges in an electric current as it travels through matter. The SI unit for resistance is the ohm. Resistance occurs because moving electrons in current bump into atoms of matter. Resistance reduces the amount of electrical energy that is transferred through matter. That\u2019s because some of the electrical energy is absorbed by the atoms and changed to other forms of energy, such as heat. Q: In the rugby analogy to resistance in physics, what do the players on each team represent? A: Factors that Affect Resistance How much resistance a material has depends on several factors: the type of material, its width, its length, and its temperature. \u2022 All materials have some resistance, but certain materials resist the \ufb02ow of electric current more or less than other materials do. Materials such as plastics have high resistance to electric current. They are called electric insulators. Materials such as metals have low resistance to electric current. They are called electric conductors. 7 www.ck12.org \u2022 A wide wire has less resistance than a narrow wire of the same material. Electricity \ufb02owing through a wire is like water \ufb02owing through a hose. More water can \ufb02ow through a wide hose than a narrow hose. In a similar way, more current can \ufb02ow through a wide wire than a narrow wire. \u2022 A longer wire has more resistance than a shorter wire. Current must travel farther through a longer wire, so there are more chances for it to collide with particles of matter. \u2022 A cooler wire has less resistance than a warmer wire. Cooler particles have less kinetic energy, so they move more slowly. Therefore, they are less likely to collide with moving electrons in current. Materials called superconductors have virtually no resistance when they are cooled to extremely low temperatures. Is Resistance Good or Bad? Resistance can be helpful or just a drain on electrical energy. If the aim is to transmit electric current through a wire from one place to another, then resistance is a drawback. It reduces the amount of electrical energy that is transmitted because some of the current is absorbed by particles of matter. On the other hand, if the aim is to use electricity to produce heat or", " light, then resistance is useful. When particles of matter absorb electrical energy, they change it to heat or light. For example, when electric current \ufb02ows through the tungsten wire inside an incandescent light bulb like the one in the Figure 3.1, the tungsten resists the \ufb02ow of electric charge. It absorbs electrical energy and converts some of it to light and heat. FIGURE 3.1 What\u2019s wrong with this picture? (Hint: How does current get to the light bulb?) Q: The tungsten wire inside a light bulb is extremely thin. How does this help it do its job? A: Summary \u2022 In physics, resistance is opposition to the \ufb02ow of electric charges that occurs as electric current travels through matter. The SI unit for resistance is the ohm. \u2022 All materials have resistance. How much resistance a material has depends on the type of material, its width, its length, and its temperature. \u2022 Resistance is a hindrance when a material is being used to transmit electric current. Resistance is helpful when a material is being used to produce heat or light. 8 www.ck12.org Vocabulary Concept 3. Electric Resistance \u2022 resistance : Opposition to the \ufb02ow of electric charges that occurs when electric current travels through matter. Review 1. What is resistance? Name the SI unit for resistance. 2. Explain what causes resistance. 3. Describe properties of a metal wire that would minimize its resistance to electric current. 4. Extend the rugby analogy to explain why a longer wire has greater resistance to electric current. 5. Copper wires have about one-third the resistance of tungsten wires. Why would copper be less suitable than tungsten as a \ufb01lament in an incandescent light bulb? References 1. lenetstan.. Used under license from Shutterstock.com 9 CONCEPT 4 \u2022 Explain Ohm\u2019s law. \u2022 Use Ohm\u2019s law to calculate current from voltage and resistance. www.ck12.org Ohm\u2019s Law Look at the water spraying out of this garden hose. You have to be careful using water around power tools and electric outlets because water can conduct an electric current. But in some ways, water \ufb02owing through a hose is like electric current \ufb02owing through a wire. Introducing Ohm\u2019s Law For electric current to \ufb02ow through a wire, there must be a source of voltage. Voltage is a difference in electric potential energy. As you might have guessed, greater voltage results in more current. As electric current \ufb02ows through matter, particles of matter resist the moving charges. This is called resistance, and greater resistance results in less current. These relationships between electric current, voltage, and resistance were \ufb01rst demonstrated in the early 1800s by a German scientist named Georg Ohm, so they are referred to as Ohm\u2019s law. Ohm\u2019s law can be represented by the following equation. Current(amps) = Voltage(volts) Resistance(ohms) Understanding Ohm\u2019s Law Ohm\u2019s law may be easier to understand with an analogy. Current \ufb02owing through a wire is like water \ufb02owing through a hose. Increasing voltage with a higher-volt battery increases the current. This is like opening the tap wider so more water \ufb02ows through the hose. Increasing resistance reduces the current. This is like stepping on the hose so less water can \ufb02ow through it. If you still aren\u2019t sure about the relationships among current, voltage, and resistance, watch the video at this URL: http://www.youtube.com/watch?v=KvVTh3ak5dQ 10 www.ck12.org Concept 4. Ohm\u2019s Law Using Ohm\u2019s Law to Calculate Current You can use the equation for current (above) to calculate the amount of current \ufb02owing through a circuit when the voltage and resistance are known. Consider an electric wire that is connected to a 12-volt battery. If the wire has a resistance of 2 ohms, how much current is \ufb02owing through the wire? Current = 12 volts 2 ohms = 6 amps Q: If a 120-volt voltage source is connected to a wire with 10 ohms of resistance, how much current is \ufb02owing through the wire? A: Substitute these values into the equation for current: Current = 120 volts 20 ohms = 12 amps Summary \u2022 According to Ohm\u2019s law, greater voltage results in more current and greater resistance results in less current. \u2022 Ohm\u2019s law can be represented by the equation:\u2019 Current(amps) = Voltage(volts) Resistance(ohms) \u2022 This equation can be used to calculate current when voltage and resistance are known. Vocabulary \u2022 Ohm\u2019s law : Law stating", " that current increases as voltage increases or resistance decreases. Practice Review Ohm\u2019s law and how to calculate current at the following URL. Then try to solve the two problems at the bottom of the Web page. Be sure to check your answers against the correct solutions. http://www.grc.nasa.gov /WWW/k-12/Sample_Projects/Ohms_Law/ohmslaw.html Review 1. State Ohm\u2019s law. 2. An electric appliance is connected by wires to a 240-volt source of voltage. If the combined resistance of the appliance and wires is 12 ohms, how much current is \ufb02owing through the circuit? 11 www.ck12.org CONCEPT 5 Resistance and Ohm\u2019s Law \u2022 De\ufb01ne resistance. \u2022 Understand the unit for resistance: ohms. \u2022 Use Ohm\u2019s Law to solve problems involving current, potential difference, and resistance. The bands of color on a resistor are a code that indicates the magnitude of the resistance of the resistor. There are four color bands identi\ufb01ed by letter: A, B, C, and D, with a gap between the C and D bands so that you know which end is A. This particular resistor has a red A band, blue B band, green C band, and gold D band, but the bands can be different colors on different resistors. Based on the colors of the bands, it is possible to identify the type of resistor. the A and B bands represent signi\ufb01cant digits; red is 2 and blue is 6. The C band indicates the multiplier, and green indicates 10 5. These three together indicate that this particular resistor is a 26,000 Ohm resistor. Finally, the D band indicates the tolerance, in this case 5%, as shown by the gold band. These terms will be explained over the course of this lesson. Resistance and Ohm\u2019s Law When a potential difference is placed across a metal wire, a large current will \ufb02ow through the wire. If the same potential difference is placed across a glass rod, almost no current will \ufb02ow. The property that determines how much current will \ufb02ow is called the resistance. Resistance is measured by \ufb01nding the ratio of potential difference, V, to current \ufb02ow, I. 12 R = V I www.ck12.org Concept 5. Resistance and Ohm\u2019s Law When given in the form V = IR, this formula is known as Ohm\u2019s Law, after the man that discovered the relationship. The units of resistance can be determined using the units of the other terms in the equation, namely that the potential difference is in volts (J/C) and current in amperes (C/s): R = volts amperes = joules/coulomb coulombs/second = joules seconds coulombs2 = ohms The units for resistance have been given the name ohms and the abbreviation is the Greek letter omega, W. 1.00 W is the resistance that will allow 1.00 ampere of current to \ufb02ow through the resistor when the potential difference is 1.00 volt. Most conductors have a constant resistance regardless of the potential difference; these are said to obey Ohm\u2019s Law. There are two ways to control the current in a circuit. Since the current is directly proportional to the potential difference and inversely proportional to the resistance, you can increase the current in a circuit by increasing the potential or by decreasing the resistance. Example Problem: A 50.0 V battery maintains current through a 20.0 W resistor. What is the current through the resistor? Solution: I = V R = 50:0 V 20:0 W = 2:50 amps Summary \u2022 Resistance is the property that determines the amount of current \ufb02ow through a particular material. \u2022 V = IR is known as Ohm\u2019s Law. \u2022 The unit for resistance is the ohm, and it has the abbreviation W. Practice The following video covers Ohm\u2019s Law. Use this resource to answer the questions that follow. http://www.youtube.com/watch?v=uLU4LtG0_hc MEDIA Click image to the left for more content. 1. What happens to current \ufb02ow when voltage is increased? 2. What happens to current \ufb02ow when resistance is increased? This website contains instruction and guided practice for Ohm\u2019s Law. http://www.wisc-online.com/Objects/ViewObject.aspx?ID=DCE11904 Review 1. If the potential stays the same and the resistance decreases, what happens to the current? 13 www.ck12.org (a) increase (b) decrease (c) stay the same 2. If the resistance stays the", " same and the potential increases, what happens to the current? (a) increase (b) decrease (c) stay the same 3. How much current can be pushed through a 30.0 W resistor by a 12.0 V battery? 4. What voltage is required to push 4.00 A of current through a 32.0 W resistor? 5. If a 6.00 volt battery will produce 0.300 A of current in a circuit, what is the resistance in the circuit? \u2022 resistance: Opposition of a circuit to the \ufb02ow of electric current. \u2022 Ohm\u2019s Law: V = IR. \u2022 Ohms: A resistance between two points of a conductor when a constant potential difference of 1 volt, applied to these points, produces in the conductor a current of 1 ampere. References 1. Image copyright Robert Spriggs, 2013. http://www.shutterstock.com. Used under license from Shutter- stock.com 14 www.ck12.org Concept 6. Energy Transfer in Electric Circuits CONCEPT 6 Energy Transfer in Electric Circuits \u2022 Explain how devices convert electrical energy to other forms. \u2022 Use P = I2R and E = I2Rt for calculations involves energy transfer in electrical circuits. \u2022 Describe the reason for the use of high voltage lines for transmitting electrical energy. \u2022 De\ufb01ne the kilowatt-hour. Part of the electrical grid, an electrical transmission sub-station receives extremely high current levels, then passes the electrical energy on to as many as 200,000 homes. Approximately 5000 megawatt-hours of energy passes through this particular substation each year. Energy Transfer in Electric Circuits Electric power is the energy per unit time converted by an electric circuit into another form of energy. We already know that power through a circuit is equal to the voltage multiplied by the current in a circuit: P = V I. It is possible to determine the power dissipated in a single resistor if we combine this expression with Ohm\u2019s Law, V = IR. This becomes particularly useful in circuits with more than one resistor, to determine the power dissipated in each one. Combining these two equations, we get an expression for electric power that involves only the current and resistance in a circuit. P = I2R The power dissipated in a resistor is proportional to the square of the current that passes through it and to its resistance. 15 www.ck12.org Electrical energy itself can be expressed as the electrical power multiplied by time: E = Pt We can incorporate this equation to obtain an equation for electrical energy based on current, resistance, and time. The electrical energy across a resistor is determined to be the current squared multiplied by the resistance and the time. E = I2Rt This equation holds true in ideal situations. However, devices used to convert electrical energy into other forms of energy are never 100% ef\ufb01cient. An electric motor is used to convert electrical energy into kinetic energy, but some of the electrical energy in this process is lost to thermal energy. When a lamp converts electrical energy into light energy, some electrical energy is lost to thermal energy. Example Problem: A heater has a resistance of 25.0 W and operates on 120.0 V. a. How much current is supplied to the resistance? b. How many joules of energy is provided by the heater in 10.0 s? Solution: a. I = V b. E = I2Rt = (4:8 A)2(25:0 W)(10:0 s) = 5760 joules 25:0 W = 4:8 A R = 120:0 V Think again about the power grid. When electricity is transmitted over long distances, some amount of energy is lost in overcoming the resistance in the transmission lines. We know the equation for the power dissipated is given by P = I2R. The energy loss can be minimized by choosing the material with the least resistance for power lines, but changing the current also has signi\ufb01cant effects. Consider a reduction of the current by a power of ten: How much power is dissipated when a current of 10.0 A passes through a power line whose resistance is 1.00 W? P = I2R = (10:0 A)2(1:00 W) = 100: Watts How much power is dissipated when a current of 1.00 A passes through a power line whose resistance is 1.00 W? P = I2R = (1:00 A)2(1:00 W) = 1:00 Watts The power loss is reduced tremendously by reducing the magnitude of the current through the resistance. Power companies must transmit the same amount of energy over the power lines but keep the power loss minimal. They do this by reducing the current. From the equation P = V I, we know that the voltage must be increased to keep the same power level. The Kilowatt-Hour Even though the companies", " that supply electrical energy are often called \u201cpower\u201d companies, they are actually selling energy. Your electricity bill is based on energy, not power. The amount of energy provided by electric current can be calculated by multiplying the watts (J/s) by seconds to yield joules. The joule, however, is a very small unit of energy and using the joule to state the amount of energy used by a household would require a very large number. For that reason, electric companies measure their energy sales in a large number of joules called a kilowatt hour (kWh). A kilowatt hour is exactly as it sounds - the number of kilowatts (1,000 W) transferred per hour. 1:00 kilowatt hour = (1000 J=s)(3600 s) = 3:6 106 J Example Problem: A color television uses about 2.0 A when operated on 120 V. a. How much power does the set use? b. If the TV is operated for 8.00 hours per day, how much energy in kWh does it use per day? c. At $0.15 per kWh, what does it cost to run the TV for 30 days? 16 www.ck12.org Solution: Concept 6. Energy Transfer in Electric Circuits a. P = V I = (120 V )(2:0 A) = 240 W b. E = (240 J=s)(8 h)(3600 s=h) c. Cost = (1:92 kW h)(30)($0:15) = $8:64 = 1:92 kW h 3:6106 J=kW h Summary \u2022 Electric power is the energy per unit time converted by an electric circuit into another form of energy. \u2022 The formula for electric power is P = I2R. \u2022 The electric energy transferred to a resistor in a time period is equal to the electric power multiplied by time, E = Pt, and can also be calculated using E = I2Rt. \u2022 Electric companies measure their energy sales in a large number of joules called a kilowatt hour (kWh) which is equivalent to 3:6 106 J. Practice The following video is on electrical energy and power. Use this resource to answer the questions that follow. http://youtu.be/NWcYBvHOiWw MEDIA Click image to the left for more content. 1. What is this video about? 2. What is the de\ufb01nition of electrical power? 3. What happens to the electrical energy that is not converted into work? Instruction and practice problems related to the energy delivered by an electric circuit: http://www.physicsclassroom.com/Class/circuits/u9l2d.cfm Review 1. A 2-way light bulb for a 110. V lamp has \ufb01lament that uses power at a rate of 50.0 W and another \ufb01lament that uses power at a rate of 100. W. Find the resistance of these two \ufb01laments. 2. Find the power dissipation of a 1.5 A lamp operating on a 12 V battery. 3. A high voltage (4:0 105 V ) power transmission line delivers electrical energy from a generating station to a substation at a rate of 1:5 109 W. What is the current in the lines? 4. A toaster oven indicates that it operates at 1500 W on a 110 V circuit. What is the resistance of the oven? \u2022 electrical energy: Energy is the ability to do work, so electrical energy is the work done by an electrical circuit. \u2022 kilowatt hour: An amount of energy equal to 3:6 106 Joules. 17 References 1. User:Cutajarc/Wikipedia. http://en.wikipedia.org/wiki/File:Melbourne_Terminal_Station.JPG. Public Do- main www.ck12.org 18 www.ck12.org Concept 7. Ammeters and Voltmeters CONCEPT 7 Ammeters and Voltmeters \u2022 Describe the primary difference between the construction of ammeters and voltmeters. \u2022 Describe whether ammeters should be placed in circuits in series or parallel and explain why. \u2022 Describe whether voltmeters should be placed in circuits in series or parallel and explain why. This photo is of the interior of the control room for a nuclear power plant. Many of the meters are reading information about the water temperature and the nuclear reaction that is occurring, but the majority of the meters are reading data about the electric energy being generated. Ammeters and Voltmeters Ammeters and voltmeters are cleverly designed for the way they are used. Ammeters measure the current of a circuit, and voltmeters measure the voltage drop across a resistor. It is important in the design and use of these meters that they don\u2019t change the circuit", " in such a way as to in\ufb02uence the readings. While both types of meters are technically resistors, they are speci\ufb01cally designed to make their readings without changing the circuit itself. 19 www.ck12.org Ammeter An ammeter measures the current traveling through the circuit. They are designed to be connected to the circuit in series, and have an extremely low resistance. If an ammeter were connected in parallel, all of the current would go through the ammeter and very little through any other resistor. As such, it is necessary for the ammeter to be connected in series with the resistors. This allows the ammeter to accurately measure the current \ufb02ow without causing any disruptions. In the circuit sketched above, the ammeter is m 2. Voltmeter In contrast, a voltmeter is designed to be connected to a circuit in parallel, and has a very high resistance. A voltmeter measures the voltage drop across a resistor, and does not need to have the current travel through it to do so. When a voltmeter is placed in parallel with a resistor, all the current continues to travel through the resistor, avoiding the very high resistance of the voltmeter. However, we know that the voltage drop across all resistors in parallel is the same, so connecting a voltmeter in parallel allows it to accurately measure the voltage drop. In the sketch, the voltmeter is m 1. Summary \u2022 Ammeters measure the current through a resistor. \u2022 Ammeters have low resistances and are placed in the circuit in series. \u2022 Voltmeters measure the voltage drop across a resistor. \u2022 Voltmeters have high resistances and are placed in the circuit in parallel. Practice MEDIA Click image to the left for more content. http://www.youtube.com/watch?v=liwan6-w-Pw In this video, a circuit is established with a power supply, which also has an attached voltmeter, and a lamp (resistor). After the circuit is established, a voltmeter and an ammeter are alternately placed in the circuit. Follow up questions: 1. What happens when the ammeter is connected in parallel with the lamp? 2. Why do the problems occur when the narrator in the video places the ammeter in parallel with the lamp? 20 www.ck12.org Review Concept 7. Ammeters and Voltmeters 1. In the sketch at above, there are four positions available for the placement of meters. Which position(s) would be appropriate for placement of an ammeter? a. 1 b. 3 c. 4 d. All of them. e. None of them. 2. Which position(s) would be appropriate for placement of a voltmeter? a. 1 b. 2 c. 3 d. All of them. e. None of them. 3. Which position could hold an ammeter that would read the total current through the circuit? a. 1 b. 2 c. 3 d. 4 e. None of them. 4. Which position could hold a voltmeter that would read the total voltage drop through the circuit? a. 1 b. 2 c. 3 or 4 d. All of them. e. None of them. \u2022 ammeter: A measuring instrument used to measure the electric current in a circuit. \u2022 voltmeter: An instrument used for measuring electrical potential difference between two points in an electric circuit. References 1. Image copyright rtem, 2013. http://www.shutterstock.com. Used under license from Shutterstock.com 2. CK-12 Foundation - Samantha Bacic.. CC-BY-NC-SA 3.0 3. CK-12 Foundation - Samantha Bacic.. CC-BY-NC-SA 3.0 21 CONCEPT 8 www.ck12.org Series Circuits \u2022 Describe a series circuit. \u2022 Understand current as it passes through a series circuit. \u2022 Understand voltage drops in a series circuit. \u2022 Understand resistance in a series circuit with multiple resistors. \u2022 Calculate current, voltage drops, and equivalent resistances for devices connected in a series circuit. Resistors, including electrical appliances, have a particular current at which they operate most effectively and safely, and excessive current can cause irreparable damage. Therefore, it is important to limit the amount of current that may pass through a particular electrical circuit. There are a number of safety devices used in electrical circuits to limit the current; fuses, circuit breakers, and surge suppressors. When fuses, such as those shown above, are placed in an electrical circuit, all the current must pass through the wire in the fuse. Series Circuits Electrical circuits are often modeled by using water in a river. The potential energy of the water is the highest at the source of the river and decreases as the water \ufb02ows down the river toward the end. When the water reaches the ocean, its potential energy has become zero. The circuit shown above has a similar situation. The current", " in this circuit is drawn in the direction of the electron \ufb02ow. It starts at the battery on the left, where electrons leave the 22 www.ck12.org Concept 8. Series Circuits negative terminal and travel around the circuit. Since all of the current travels across each resistor, these resistors are said to be in series. A series circuit is one in which all of the current must pass through every resistor in the circuit. Returning to the water analogy, there is only one riverbed from the top of the mountain to the ocean. Consider the series circuit sketched above. This circuit has a voltage drop for the entire circuit of 120 V and has three resistors connected in series. The current in this circuit is drawn in terms of electron \ufb02ow. The electrons leave the potential difference source at the negative terminal and \ufb02ow through the three resistors, starting with R 3. Though they have a small amount of resistance, the resistance of the connecting wires is so small in relation to the resistors that we ignore it. Therefore, we say that there is no voltage drop when the current passes through the connecting wires. The voltage drops occur when the current passes through each of the resistors and the total voltage drop for the entire circuit is equal to the sum of the voltage drops through the three resistors. VT = V1 +V2 +V3 The current through each of the resistors must be exactly the same because the current in a series circuit is the same everywhere. The current is moving in the entire circuit at the same time. IT = I1 = I2 = I3 Since the current passes through each resistor, the total resistance in the circuit is equal to the sum of the resistors. In the circuit above, the total resistance is: RT = R1 + R2 + R3 = 30 W + 15 W + 15 W = 60 W Therefore, the total current and the current through each resistor is I = V R = 120 V 60 W = 2:0 A: The individual voltage drops can be calculated using the current through each resistor and each resistor\u2019s individual resistance. V1 = I1R1 = (2:0 A)(30 W) = 60 V V2 = I2R2 = (2:0 A)(15 W) = 30 V V3 = I3R3 = (2:0 A)(15 W) = 30 V Example Problem: Four 15 W resistors are connected in series with a 45 V battery. What is the current in the circuit? Solution: RT = 15 W + 15 W + 15 W + 15 W = 60 W I = V R = 45 V 60 W = 0:75 A 23 www.ck12.org Summary \u2022 A series circuit is one in which all of the current must pass through every resistor in the circuit. \u2022 VT = V1 +V2 +V3 \u2022 IT = I1 = I2 = I3 \u2022 RT = R1 + R2 + R3 Practice The following video is on series circuits. Use this resource to answer the questions that follow. MEDIA Click image to the left for more content. http://www.youtube.com/watch?v=qO391knBRjE 1. How do the voltage drops across the two light bulbs in the video related to the total voltage drop for the entire circuit? 2. In the video, what was the assumed voltage drop for the connecting wires and the switch? 3. What was the current through the second light bulb as compared to the current through the \ufb01rst light bulb. Review 1. There are three 20.0 W resistors connected in series across a 120 V generator. (a) What is the total resistance of the circuit? (b) What is the current in the circuit? (c) What is the voltage drop across one of the resistors? 2. A 5.00W, a 10.0W, and a 15.0W resistor are connected in a series across a 90.0 V battery. (a) What is the equivalent resistance of the circuit? (b) What is the current in the circuit? (c) What is the voltage drop across the 5.00W resistor? 3. A 5.00W and a 10.0W resistor are connected in series across an unknown voltage. The total current in the circuit is 3.00 A. (a) What is the equivalent resistance of the circuit? (b) What is the current through the 5.00W resistor? (c) What is the total voltage drop for the entire circuit? \u2022 series circuit: One in which all of the current must pass through every resistor in the circuit. References 1. Image copyright sevenke, 2013. http://www.shutterstock.com. Used under license from Shutterstock.com 2. CK-12 Foundation - Samantha Bacic.. CC-BY-NC-SA 3.0 24 www.ck12.org Concept 9. Resistors", " in Series CONCEPT 9 Resistors in Series Students will learn how to analyze and solve problems involving circuits with resistors in series. Students will learn how to analyze and solve problems involving circuits with resistors in series. Key Equations Guidance Rtotal = R1 + R2 + R3 + : : : Resistors in Series: All resistors are connected end to end. There is only one river, so they all receive the same current. But since there is a voltage drop across each resistor, they may all have different voltages across them. The more resistors in series the more rocks in the river, so the less current that \ufb02ows. Example 1 A circuit is wired up with two resistors in series. Both resistors are in the same \u2019river\u2019, so both have the same current \ufb02owing through them. Neither resistor has a direct connection to the power supply so neither has 20V across it. But the combined voltages across the individual resistors add up to 20V. Question: What is the total resistance of the circuit? Answer: The total resistance is Rtotal = R1 + R2 = 90 W + 10 W = 100 W Question: What is the total current coming out of the power supply? Answer: Use Ohm\u2019s Law (V = IR) but solve for current (I = V =R). Itotal = Vtotal Rtotal = 20V 100 W = 0:20 A Question: How much power does the power supply dissipate? Answer: P = IV, so the total power equals the total voltage multiplied by the total current. Thus, Ptotal = ItotalVtotal = (0:20 A)(20V ) = 4:0 W. So the Power Supply is outputting 4W (i.e. 4 Joules of energy per second). Question: How much power does each resistor dissipate? Answer: Each resistor has different voltage across it, but the same current. So, using Ohm\u2019s law, convert the power formula into a form that does not depend on voltage. 25 www.ck12.org P = IV = I(IR) = I2R: P90 W = I2 P10 W = I2 90 WR90 W = (0:2 A)2(90 W) = 3:6W 10 WR10 W = (0:2 A)2(10 W) = 0:4W Note: If you add up the power dissipated by each resistor, it equals the total power outputted, as it should\u2013Energy is always conserved. Question: How much voltage is there across each resistor? Answer: In order to calculate voltage across a resistor, use Ohm\u2019s law. V90 W = I90 WR90 W = (0:2 A)(90 W) = 18V V10 W = I10 WR10 W = (0:2 A)(10 W) = 2V Note: If you add up the voltages across the individual resistors you will obtain the total voltage of the circuit, as you should. Further note that with the voltages we can use the original form of the Power equation (P = IV ), and we should get the same results as above. P90 W = I90 WV90 W = (18V )(0:2 A) = 3:6W P10 W = I10 WV10 W = (2:0V )(0:2 A) = 0:4W MEDIA Click image to the left for more content. Watch this Explanation 26 www.ck12.org Simulation Concept 9. Resistors in Series \u2022 http://simulations.ck12.org/Resistor/ Time for Practice 1. Regarding the circuit below. a. If the ammeter reads 2 A, what is the voltage? b. How many watts is the power supply supplying? c. How many watts are dissipated in each resistor? 2. Five resistors are wired in series. Their values are 10W, 56W, 82W, 120W and 180W. a. If these resistors are connected to a 6 V battery, what is the current \ufb02owing out of the battery? b. If these resistors are connected to a 120 V power suppluy, what is the current \ufb02owing out of the battery? c. In order to increase current in your circuit, which two resistors would you remove? 3. Given the resistors above and a 12 V battery, how could you make a circuit that draws 0.0594 A? Answers to Selected Problems 1. a. 224 V b. 448 W c. 400 W by 100 W and 48 W by 12 W 2. a. 0.013 A b. 0.27 A c. 120W and 180W 3. need about 202W of total resistance. So if you wire up the 120W and the 82W in series, you\ufffd", "\ufffdll have it. 27 CONCEPT 10 www.ck12.org Parallel Circuits \u2022 Describe a parallel circuit. \u2022 Understand current as it passes through a parallel circuit. \u2022 Understand voltage drops in a parallel circuit. \u2022 Understand resistance in a parallel circuit with multiple resistors. \u2022 Calculate voltage drops, currents, and equivalent resistances when devices are connected in a parallel circuit. Electrical circuits are everywhere: skyscrapers, jumbo jets, arcade games, lights, heating, and security... very few complex things work without electrical circuits. Since the late 1970s, electrical circuits have primarily looked like this. The circuits are formed by a thing layer of conducting material deposited on the surface of an insulating board. Individual components are soldered to the interconnecting circuits. Circuit boards are vastly more complicated than the series circuits previously discussed, but operate on many similiar principles. Parallel Circuits Parallel circuits are circuits in which the charges leaving the potential source have different paths they can follow to get back to the source. In the sketch below, the current leaves the battery, passes through the orange switch, and then has three different paths available to complete the circuit. Each individual electron in this circuit passes through only one of the light bulbs. After the current passes through the switch, it divides into three pieces and each piece passes through one of the bulbs. The three pieces of current rejoin after the light bulbs and continue in the circuit to the potential source. 28 www.ck12.org Concept 10. Parallel Circuits In the design of this parallel circuit, each resistor (light bulb) is connected across the battery as if the other two resistors were not present. Remember that the current going through each resistor goes through only the one resistor. Therefore, the voltage drop across each resistor must be equal to the total voltage drop though the circuit. VT = V1 = V2 = V3 The total current passing through the circuit will be the sum of the individual currents passing through each resistor. IT = I1 + I2 + I3 If we return to the analogy of a river, a parallel circuit is the same as the river breaking into three streams, which later rejoin to one river again. The amount of water \ufb02owing in the river is equal to the sum of the amounts of water \ufb02owing in the individual streams. Ohm\u2019s Law applies to resistors in parallel, just as it did to resistors in a series. The current \ufb02owing through each resistor is equal to the total voltage drop divided by the resistance in that resistor. I1 = VT R1 and I2 = VT R2 and I3 = VT R3 Since IT = I1 + I2 + I3, + VT then IT = VT R3 R1 + VT = VT and VT R3 R1 RT + VT R2 + VT R2,. If we divide both sides of the \ufb01nal equation by VT, we get the relationship between the total resistance of the circuit and the individual parallel resistances in the circuit. The total resistance is sometimes called the equivalent resistance. 1 RT = 1 R1 + 1 R2 + 1 R3 Consider the parallel circuit sketched below. The voltage drop for the entire circuit is 90. V. Therefore, the voltage drop in each of the resistors is also 90. V. 29 www.ck12.org The current through each resistor can be found using the voltage drop and the resistance of that resistor: I1 = VT R1 = 90: V 60: W = 1:5 A I2 = VT R2 = 90: V 30: W = 3:0 A I3 = VT R3 = 90: V 30: W = 3:0 A The total current through the circuit would be the sum of the three currents in the individual resistors. IT = I1 + I2 = I3 = 1:5 A + 3:0 A + 3:0 A = 7:5 A The equivalent resistance for this circuit is found using the equation above. 1 RT = 1 R1 + 1 R2 + 1 R3 = 1 60: W + 1 30: W + 1 30: W = 1 60: W + 2 60: W + 2 60: W = 5 60: W RT = 60: W 5 = 12 W The equivalent resistance for the circuit could also be found by using the total voltage drop and the total current. RT = VT IT 7:5 A = 12 W = 90: W Summary \u2022 Parallel electrical circuits have multiple paths the current may take. \u2022 VT = V1 = V2 = V3. \u2022 IT = I1 + I2 + I3. + 1 \u2022 R3 = 1 R1 = 1 R2 1 RT. Practice MEDIA Click image to the left for more content. http://www.youtube.com/watch?v=apHkG4T6QHM Follow up questions: 1. Why do the light bulbs glow less brightly", " when connected across a 120 V source in a series circuit than when connected across the same 120 V source in a parallel circuit? 2. Why do the other bulbs go dark when one bulb is removed in the series circuit but the other bulbs do not go dark when one bulb is removed in the parallel circuit? 30 www.ck12.org Review Concept 10. Parallel Circuits 1. Three 15.0W resistors are connected in parallel and placed across a 30.0 V potential difference. (a) What is the equivalent resistance of the parallel circuit? (b) What is the total current through the circuit? (c) What is the current through a single branch of the circuit? 2. A 12.0W and a 15.0W resistor are connected in parallel and placed across a 30.0 V potential. (a) What is the equivalent resistance of the parallel circuit? (b) What is the total current through the circuit? (c) What is the current through each branch of the circuit? 3. A 120.0W resistor, a 60.0W resistor, and a 40.0W resistor are connected in parallel and placed across a potential difference of 12.0 V. (a) What is the equivalent resistance of the parallel circuit? (b) What is the total current through the circuit? (c) What is the current through each branch of the circuit? \u2022 parallel circuit: A closed electrical circuit in which the current is divided into two or more paths and then returns via a common path to complete the circuit. \u2022 equivalent resistance: A single resistance that would cause the same power loss as the actual resistance values distributed throughout a circuit. References 1. Image copyright vilax, 2013. http://www.shutterstock.com. Used under license from Shutterstock.com 2. Light bulb: Image copyright Snez, 2013; composite created by CK-12 Foundation - Samantha Bacic. http:// www.shutterstock.com. Used under license from Shutterstock.com 3. CK-12 Foundation - Samantha Bacic.. CC-BY-NC-SA 3.0 31 CONCEPT 11 Resistors in Parallel Students will learn how to analyze and solve problems involving circuits with resistors in parallel. Students will learn how to analyze and solve problems involving circuits with resistors in parallel. www.ck12.org Key Equations Guidance 1 Rtotal = 1 R1 + 1 R2 + 1 R3 + : : : Resistors in Parallel: All resistors are connected together at both ends. There are many rivers (i.e. The main river branches off into many other rivers), so all resistors receive different amounts of current. But since they are all connected to the same point at both ends they all receive the same voltage. Example 1 A circuit is wired up with 2 resistors in parallel. Both resistors are directly connected to the power supply, so both have the same 20V across them. But they are on different \u2019rivers\u2019 so they have different current \ufb02owing through them. Lets go through the same questions and answers as with the circuit in series. Question: What is the total resistance of the circuit? = 1 Answer: The total resistance is Note: Total resistance for a circuit in parallel will always be smaller than smallest resistor in the circuit. 90W thus, Rtotal = 90W 90W = 10 10W = 1 90W + 9 90W + 1 = 1 R1 + 1 R2 10 = 9W 1 Rtotal Question: What is the total current coming out of the power supply? Answer: Use Ohm\u2019s Law (V = IR) but solve for current (I = V =R). Itotal = Vtotal Rtotal = 20V 9W = 2:2A Question: How much power does the power supply dissipate? Answer: P = IV, so the total power equals the total voltage multiplied by the total current. Thus, Ptotal = ItotalVtotal = (2:2A)(20V ) = 44:4W. So the Power Supply outputs 44W (i.e. 44 Joules of energy per second). 32 www.ck12.org Concept 11. Resistors in Parallel Question: How much power is each resistor dissipating? Answer: Each resistor has different current across it, but the same voltage. So, using Ohm\u2019s law, convert the power formula into a form that does not depend on current. P = IV = V R Substituted I = V =R into the power R formula. P90W = V 2 10W = 40W 90W R90W Note: If you add up the power dissipated by each resistor, it equals the total power outputted, as it should\u2013Energy is always conserved. 90W = 4:4W ; P10W = V 2 R10W = (20V )2 V = V 2 = (20V )2 10W Question: How much current is", " \ufb02owing through each resistor? Answer: Use Ohm\u2019s law to calculate the current for each resistor. I90W = V90W R90W = 20V 90W = 0:22A I10W = V10W R10W = 20V 10W = 2:0A Notice that the 10W resistor has the most current going through it. It has the least resistance to electricity so this makes sense. Note: If you add up the currents of the individual \u2019rivers\u2019 you get the total current of the of the circuit, as you should. Watch this Explanation MEDIA Click image to the left for more content. Simulation \u2022 http://simulations.ck12.org/Resistor/ 33 Time for Practice 1. Three 82 W resistors and one 12 W resistor are wired in parallel with a 9 V battery. a. Draw the schematic diagram. b. What is the total resistance of the circuit? 2. What does the ammeter read and which resistor is dissipating the most power? www.ck12.org 3. Given three resistors, 200 W; 300 W and 600 W and a 120 V power source connect them in a way to heat a container of water as rapidly as possible. a. Show the circuit diagram b. How many joules of heat are developed after 5 minutes? Answers to Selected Problems 1. b. 8:3 W 2. 0:8A and the 50 W on the left 3. part 2 43200J. 34 www.ck12.org Concept 12. Combined Series-Parallel Circuits CONCEPT 12 Combined Series-Parallel Circuits \u2022 Solve problems of combined circuits. Electrical circuits can become immensely complicated. This circuit is a polynomial plotter, which allows users to plot polynomials and evaluate functions at various x values. Combined Series-Parallel Circuits Most circuits are not just a series or parallel circuit; most have resistors in parallel and in series. These circuits are called combination circuits. When solving problems with such circuits, use this series of steps. 1. For resistors connected in parallel, calculate the single equivalent resistance that can replace them. 2. For resistors in series, calculate the single equivalent resistance that can replace them. 3. By repeating steps 1 and 2, you can continually reduce the circuit until only a single equivalent resistor remains. Then you can determine the total circuit current. The voltage drops and currents though individual resistors can then be calculated. Example Problem: the total current through the circuit, and \ufb01nd the current through each individual resistor. In the combination circuit sketched below, \ufb01nd the equivalent resistance for the circuit, \ufb01nd 35 www.ck12.org Solution: We start by simplifying the parallel resistors R2 and R3. = 1 180 W 1 R23 R23 = 99 W + 1 220 W = 1 99 W We then simplify R1 and R23 which are series resistors. RT = R1 + R23 = 110 W + 99 W = 209 W We can then \ufb01nd the total current, IT = VT RT All the current must pass through R1, so I1 = 0:11 A. = 24 V 209 W = 0:11 A The voltage drop through R1 is (110 W)(0:11 A) = 12:6 volts. Therefore, the voltage drop through R2 and R3 is 11.4 volts. I2 = V2 220 W = 0:052 A R2 180 W = 0:063 A and I3 = V3 R3 = 11:4 V = 11:4 V Summary \u2022 Combined circuit problems should be solved in steps. Practice Video teaching the process of simplifying a circuit that contains both series and parallel parts. MEDIA Click image to the left for more content. http://www.youtube.com/watch?v=In3NF8f-mzg Follow up questions: 1. In a circuit that contains both series and parallel parts, which parts of the circuit are simpli\ufb01ed \ufb01rst? 2. In the circuit drawn below, which resistors should be simpli\ufb01ed \ufb01rst? 36 www.ck12.org Concept 12. Combined Series-Parallel Circuits Review 1. Two 60.0W resistors are connected in parallel and this parallel arrangement is then connected in series with a 30.0W resistor. The combination is placed across a 120. V potential difference. (a) Draw a diagram of the circuit. (b) What is the equivalent resistance of the parallel portion of the circuit? (c) What is the equivalent resistance for the entire circuit? (d) What is the total current in the circuit? (e) What is the voltage drop across the 30.0W resistor? (f) What is the voltage drop across the parallel portion of the circuit? (g) What is the", " current through each resistor? 2. Three 15.0 Ohm resistors are connected in parallel and the combination is then connected in series with a 10.0 Ohm resistor. The entire combination is then placed across a 45.0 V potential difference. Find the equivalent resistance for the entire circuit. \u2022 combined or combination circuits: A route for the \ufb02ow of electricity that has elements of both series and parallel circuits. References 1. User:Linkaddict/Wikimedia Commons. http://commons.wikimedia.org/wiki/File:Huge_circuit.JPG. Public Domain 2. CK-12 Foundation - Samantha Bacic.. CC-BY-NC-SA 3.0 3. CK-12 Foundation - Samantha Bacic.. CC-BY-NC-SA 3.0 37 Physics Unit 14: Magnetism Patrick Marshall Ck12 Science James H Dann, Ph.D. Say Thanks to the Authors Click http://www.ck12.org/saythanks (No sign in required) AUTHORS Patrick Marshall Ck12 Science James H Dann, Ph.D. CONTRIBUTORS Chris Addiego Antonio De Jesus L\u00f3pez www.ck12.org To access a customizable version of this book, as well as other interactive content, visit www.ck12.org CK-12 Foundation is a non-pro\ufb01t organization with a mission to reduce the cost of textbook materials for the K-12 market both in the U.S. and worldwide. Using an open-content, web-based collaborative model termed the FlexBook\u00ae, CK-12 intends to pioneer the generation and distribution of high-quality educational content that will serve both as core text as well as provide an adaptive environment for learning, powered through the FlexBook Platform\u00ae. Copyright \u00a9 2014 CK-12 Foundation, www.ck12.org The names \u201cCK-12\u201d and \u201cCK12\u201d and associated logos and the terms \u201cFlexBook\u00ae\u201d and \u201cFlexBook Platform\u00ae\u201d (collectively \u201cCK-12 Marks\u201d) are trademarks and service marks of CK-12 Foundation and are protected by federal, state, and international laws. Any form of reproduction of this book in any format or medium, in whole or in sections must include the referral attribution link http://www.ck12.org/saythanks (placed in a visible location) in addition to the following terms. Except as otherwise noted, all CK-12 Content (including CK-12 Curriculum Material) is made available to Users in accordance with the Creative Commons Attribution-Non-Commercial 3.0 Unported (CC BY-NC 3.0) License (http://creativecommons.org/ licenses/by-nc/3.0/), as amended and updated by Creative Commons from time to time (the \u201cCC License\u201d), which is incorporated herein by this reference. Complete terms can be found at http://www.ck12.org/terms. Printed: April 6, 2014 iii Contents www.ck12.org Contents 1 Properties of Magnets 2 Magnetic Fields 1 6 iv www.ck12.org Concept 1. Properties of Magnets CONCEPT 1 Properties of Magnets \u2022 Describe magnetic \ufb01elds around permanent magnets. \u2022 Understand ferromagnetism and magnetic domains. \u2022 Describe some properties of magnets. Some countries are using powerful electromagnets to develop high-speed trains, called maglev, or magnetic levitation, trains. These trains use the repulsive force of magnets to \ufb02oat over a guide way, removing the friction of steel wheels and train tracks. Reducing this friction allows the trains to travel at much higher speeds. Properties of Magnets Any magnet, regardless of its shape, has two ends called poles where the magnetic effect is strongest. If a magnet is suspended by a \ufb01ne thread, it is found that one pole of the magnet will always point toward the north. This fact has been made use of in navigation since the eleventh century. The pole of the magnet that seeks the north pole is called the north pole of the magnet, while the opposite side is the south pole. It is a familiar fact that when two magnets are brought near one another, the magnets exert a force on each other. The magnetic force can be either attractive or repulsive. If two north poles or two south poles are brought near each other, the force will be repulsive. If a north pole is brought near a south pole, the force will be attractive. 1 www.ck12.org The Earth\u2019s geographic north pole (which is close to, but not exactly at the magnetic pole) attracts the north poles of magnets. We know, therefore, that this pole is actually the Earth\u2019s magnetic south pole. This can be seen in the image above; the geographic north and south poles are labeled with barber shop poles", ", and the Earth\u2019s magnetic poles are indicated with the double-headed arrow. Only iron and few other materials such as cobalt, nickel, and gadolinium show strong magnetic effects. These materials are said to be ferromagnetic. Other materials show some slight magnetic effect but it is extremely small and can be detected only with delicate instruments. Ferromagnetic Domains Microscopic examination reveals that a magnet is actually made up of tiny regions known as magnetic domains, which are about one millimeter in length and width. Each domain acts like a tiny magnet with a north and south pole. 2 www.ck12.org Concept 1. Properties of Magnets When the ferrous material is not magnetized, the domains are randomly organized so that the north and south poles do not line up and often cancel each other. When the ferrous material is placed in a magnetic \ufb01eld, the domains line up with the magnetic \ufb01eld so that the north poles are all pointed in the same direction and the south poles are all pointed in the opposite direction. In this way, the ferrous material has become a magnet. In many cases, the domains will remain aligned only while the ferrous material is in a strong magnetic \ufb01eld; when the material is removed from the \ufb01eld, the domains return to their previous random organization and the ferrous material loses any magnetic properties. Magnets that have magnetic properties while in the \ufb01eld of another magnet but lose the magnetic properties when removed from the \ufb01eld are called temporary magnets. Under certain circumstances, however, the new alignment can be made permanent and the ferrous substance becomes a permanent magnet. That is, the ferrous object remains a magnet even when removed from the other magnetic \ufb01eld. The formation of temporary magnets allows a magnet to attract a non-magnetized piece of iron. You have most likely seen a magnet pick up a paper clip. The presence of the magnet aligns the domains in the iron paper clip and it becomes a temporary magnet. Whichever pole of the magnet is brought near the paper clip will induce magnetic properties in the paper clip that remain as long as the magnet is near. Permanent magnets lose their magnetic properties when the domains are dislodged from their organized positions and returned to a random jumble. This can occur if the magnet is hammered on or if it is heated strongly. Magnetic Fields When we were dealing with electrical effects, it was very useful to speak of an electric \ufb01eld that surrounded an electric charge. In the same way, we can imagine a magnetic \ufb01eld surrounding a magnetic pole. The force that one magnet exerts on another can be described as the interaction between one magnet and the magnetic \ufb01eld of the other magnet. Magnetic \ufb01eld lines go from the north magnetic pole to the south magnetic pole. We de\ufb01ne the magnetic \ufb01eld at any point as a vector (represented by the letter B ) whose direction is from north to south magnetic poles. 3 Summary \u2022 Any magnet has two ends called poles where the magnetic effect is strongest. \u2022 The magnetic pole found at the north geographical pole of the earth is a south magnetic pole. \u2022 The force that one magnet exerts on another can be described as the interaction between one magnet and the magnetic \ufb01eld of the other magnet. \u2022 Magnetic \ufb01eld lines go from the north magnetic pole to the south magnetic pole. www.ck12.org Practice MEDIA Click image to the left for more content. http://www.darktube.org/watch/crealev-levitating-\ufb02oating-flying-hovering-bouncing This video demonstrates magnetic levitation. 1. In the video, one object rests on top of the magnetic \ufb01eld of another. Compare the friction between these two objects to the friction between a saucer and a table the saucer rests on. Review 1. The earth\u2019s magnetic \ufb01eld (a) has a north magnetic pole at exactly the same spot as the geographical north pole. (b) is what causes compasses to work. (c) is what causes electromagnets to work. (d) all of these are true. (e) none of these are true. 2. A material that can be permanently magnetized is generally said to be (a) magnetic. (b) electromagnetic. (c) ferromagnetic. (d) none of these are true. 3. The force between like magnetic poles will be (a) repulsive. (b) attractive. (c) could be repulsive or attractive. 4. Why is a magnet able to attract a non-magnetic piece of iron? 5. If you had two iron rods and noticed that they attract each other, how could you determine if both were magnets or only one was a magnet? \u2022 magnet: A body that can", " attract certain substances, such as iron or steel, as a result of a magnetic \ufb01eld. 4 www.ck12.org Concept 1. Properties of Magnets \u2022 magnetic pole: Either of two regions of a magnet, designated north and south, where the magnetic \ufb01eld is strongest. Electromagnetic interactions cause the north poles of magnets to be attracted to the south poles of other magnets, and conversely. The north pole of a magnet is the pole out of which magnetic lines of force point, while the south pole is the pole into which they point. \u2022 ferromagnetic: A body or substance having a high susceptibility to magnetization, the strength of which depends on that of the applied magnetizing \ufb01eld, and which may persist after removal of the applied \ufb01eld. This is the kind of magnetism displayed by iron, and is associated with parallel magnetic alignment of adjacent domains. \u2022 magnetic \ufb01eld: A \ufb01eld of force surrounding a permanent magnet or a moving charged particle. \u2022 magnetic domain: An atom or group of atoms within a material that have some kind of \u201cnet\u201d magnetic \ufb01eld. \u2022 temporary magnet: A piece of iron that is a magnet while in the presence of another magnetic \ufb01eld but loses its magnetic characteristics when the other \ufb01eld is removed. \u2022 permanent magnet: A piece of magnetic material that retains its magnetism after it is removed from a magnetic \ufb01eld. References 1. User:JakeLM/Wikipedia. http://commons.wikimedia.org/wiki/File:Maglev_june2005.jpg. CC-BY 2.5 2. Globe: Image copyright pdesign, 2013; Poles: Image copyright lineartestpilot, 2013; Composite created by http://www.shutterstock.com. Used under licenses from Shutter- CK-12 Foundation - Samantha Bacic. stock.com 5 CONCEPT 2 www.ck12.org Magnetic Fields Students will learn the idea of magnetic \ufb01eld lines, how they behave in the situation of permanent magnets and current carrying wires and also how to calculate the magnetic \ufb01eld at an arbitrary distance from the wire. Students will learn the idea of magnetic \ufb01eld lines, how they behave in the situation of permanent magnets and current carrying wires and also how to calculate the magnetic \ufb01eld at an arbitrary distance from the wire. Key Equations \u00b50I 2pr Where \u00b50 = 4p 107 Tm/A Bwire = Magnetic \ufb01eld at a distance r from a current-carrying wire Permeability of Vacuum (approximately same for air also) Guidance Permanent magnets (like refrigerator magnets) consist of atoms, such as iron, for which the magnetic moments (roughly electron spin) of the electrons are \u201clined up\u201d all across the atom. This means that their magnetic \ufb01elds add up, rather than canceling each other out. The net effect is noticeable because so many atoms have lined up. The magnetic \ufb01eld of such a magnet always points from the north pole to the south. The magnetic \ufb01eld of a bar magnet, for example, is illustrated below: If we were to cut the magnet above in half, it would still have north and south poles; the resulting magnetic \ufb01eld would be qualitatively the same as the one above (but weaker). Charged particles in motion also generate magnetic \ufb01elds. The most frequently used example is a current carrying wire, since current is literally moving charged particles. The magnitude of a \ufb01eld generated by a wire depends on distance to the wire and strength of the current (I) (see \u2019Key Equations\u2019 section) : Meanwhile, its direction can be found using the so called \ufb01rst right hand rule : point your thumb in the direction of the current. Then, curl your \ufb01ngers around the wire. The direction your \ufb01ngers will point in the same direction as the \ufb01eld. Be sure to use your right hand! 6 www.ck12.org Concept 2. Magnetic Fields Sometimes, it is necessary to represent such three dimensional \ufb01elds on a two dimensional sheet of paper. The following example illustrates how this is done. In the example above, a current is running along a wire towards the top of your page. The magnetic \ufb01eld is circling the current carrying wire in loops which are perpendicular to the page. Where these loops intersect this piece of paper, we use the symbol J to represent where the magnetic \ufb01eld is coming out of the page and the symbol N to. This convention can be used for all vector quantities: represent where the magnetic \ufb01eld is going into the page \ufb01elds", ", forces, velocities, etc. Example 1 You are standing right next to a current carrying wire and decide to throw your magnetic \ufb01eld sensor some distance perpendicular to the wire. When you go to retrieve your sensor, it shows the magnetic \ufb01eld where it landed to be 4 105 T. If you know the wire was carrying 300A, how far did you throw the sensor? Solution To solve this problem, we will just use the equation given above and solve for the radius. 7 www.ck12.org B = r = r = \u00b5oI 2pr \u00b5oI 2pB 4p 107 Tm/A 300 A 2p 4 105 T r = 1:5 m MEDIA Click image to the left for more content. Watch this Explanation Simulation Magnet and Compass (PhET Simulation) Time for Practice 1. Sketch the magnetic \ufb01eld lines for the horseshoe magnet shown here. Then, show the direction in which the two compasses (shown as circles) should point considering their positions. In other words, draw an arrow in the compass that represents North in relation to the compass magnet. 8 www.ck12.org Concept 2. Magnetic Fields 2. Find the magnetic \ufb01eld a distance of 20 cm from a wire that is carrying 3 A of electrical current. 3. In order to measure the current from big power lines the worker simply clamps a device around the wire. This provides safety when dealing with such high currents. The worker simply measures the magnetic \ufb01eld and deduces the current using the laws of physics. Let\u2019s say a worker uses such a clamp and the device registers a magnetic \ufb01eld of 0.02 T. The clamp is 0.05 m from the wire. What is the electrical current in the wire? Answers to Selected Problems 1. Both pointing away from north 2. 3 106T 3. 5000 A 9 Physics Unit 15: Electromagnetism Patrick Marshall Jean Brainard, Ph.D. Ck12 Science James H Dann, Ph.D. Say Thanks to the Authors Click http://www.ck12.org/saythanks (No sign in required) AUTHORS Patrick Marshall Jean Brainard, Ph.D. Ck12 Science James H Dann, Ph.D. CONTRIBUTORS Chris Addiego Antonio De Jesus L\u00f3pez www.ck12.org To access a customizable version of this book, as well as other interactive content, visit www.ck12.org CK-12 Foundation is a non-pro\ufb01t organization with a mission to reduce the cost of textbook materials for the K-12 market both in the U.S. and worldwide. Using an open-content, web-based collaborative model termed the FlexBook\u00ae, CK-12 intends to pioneer the generation and distribution of high-quality educational content that will serve both as core text as well as provide an adaptive environment for learning, powered through the FlexBook Platform\u00ae. Copyright \u00a9 2014 CK-12 Foundation, www.ck12.org The names \u201cCK-12\u201d and \u201cCK12\u201d and associated logos and the terms \u201cFlexBook\u00ae\u201d and \u201cFlexBook Platform\u00ae\u201d (collectively \u201cCK-12 Marks\u201d) are trademarks and service marks of CK-12 Foundation and are protected by federal, state, and international laws. Any form of reproduction of this book in any format or medium, in whole or in sections must include the referral attribution link http://www.ck12.org/saythanks (placed in a visible location) in addition to the following terms. Except as otherwise noted, all CK-12 Content (including CK-12 Curriculum Material) is made available to Users in accordance with the Creative Commons Attribution-Non-Commercial 3.0 Unported (CC BY-NC 3.0) License (http://creativecommons.org/ licenses/by-nc/3.0/), as amended and updated by Creative Commons from time to time (the \u201cCC License\u201d), which is incorporated herein by this reference. Complete terms can be found at http://www.ck12.org/terms. Printed: April 6, 2014 iii Contents www.ck12.org 1 4 8 13 18 22 26 Contents 1 Electromagnetic Induction 2 Electromagnets 3 Current and Magnetism 4 Electric Motors 5 Electromotive Force 6 Electric Generators 7 Transformers iv www.ck12.org Concept 1. Electromagnetic Induction CONCEPT 1 Electromagnetic Induction \u2022 De\ufb01ne electromagnetic induction. \u2022 Explain how electromagnetic induction occurs. \u2022 Describe the current produced by electromagnetic induction. \u2022 Identify ways that electromagnetic induction is used. The girl on the left in this photo is riding a stationary bike. She\u2019s getting exercise, but that\u2019s not the real reason she\u2019s", " riding the bike. She\u2019s using her muscle power to generate electricity through a process called electromagnetic induction. What Is Electromagnetic Induction? Electromagnetic induction is the process of generating electric current with a magnetic \ufb01eld. It occurs whenever a magnetic \ufb01eld and an electric conductor, such as a coil of wire, move relative to one another. As long as the conductor is part of a closed circuit, current will \ufb02ow through it whenever it crosses lines of force in the magnetic \ufb01eld. One way this can happen is illustrated in the Figure 1.1. The sketch shows a magnet moving through a wire coil. You can watch an animated version of the illustration at this URL: http://jsticca.wordpress.com/2009/09/01 /the-magnet-car/ Q: What is another way that a coil of wire and magnet can move relative to one another and generate an electric current? A: The coil of wire could be moved back and forth over the magnet. The Current Produced by a Magnet The device with the pointer in the circuit above is an ammeter. It measures the current that \ufb02ows through the wire. The faster the magnet or coil moves, the greater the amount of current that is produced. If more turns were added to the coil or a stronger magnet were used, this would produce more current as well. 1 www.ck12.org FIGURE 1.1 The Figure 1.2 shows the direction of the current that is generated by a moving magnet. If the magnet is moved back and forth repeatedly, the current keeps changing direction. In other words, alternating current (AC) is produced. Alternating current is electric current that keeps reversing direction. FIGURE 1.2 How Electromagnetic Induction Is Used Two important devices depend on electromagnetic induction: electric generators and electric transformers. Both devices play critical roles in producing and regulating the electric current we depend on in our daily lives. Electric generators use electromagnetic induction to change kinetic energy to electrical energy. They produce electricity in power plants. Electric transformers use electromagnetic induction to change the voltage of electric current. Some transformers increase voltage and other decrease voltage. Q: How do you think the girl on the exercise bike in the opening photo is using electromagnetic induction? A: As she pedals the bike, the kinetic energy of the turning pedals is used to move a conductor through a magnetic \ufb01eld. This generates electric current by electromagnetic induction. 2 www.ck12.org Summary Concept 1. Electromagnetic Induction \u2022 Electromagnetic induction is the process of generating electric current with a magnetic \ufb01eld. It occurs whenever a magnetic \ufb01eld and an electric conductor move relative to one another so the conductor crosses lines of force in the magnetic \ufb01eld. \u2022 The current produced by electromagnetic induction is greater when the magnet or coil moves faster, the coil has more turns, or the magnet is stronger. If the magnet or coil is moved back and forth repeatedly, alternating current is produced. \u2022 Electric generators and electric transformers use electromagnetic induction to generate electricity or change the voltage of electric current. Vocabulary \u2022 electromagnetic induction : Process of generating electric current with a changing magnetic \ufb01eld. Practice Simulate electromagnetic induction at the following URL. Then answer the questions below. http://micro.magnet.fsu.edu/electromag/java/faraday2/ 1. How is electric current created in the simulation? What type of current is it? 2. How is electric current measured in the simulation? 3. What happens when you stop moving the magnet? Review 1. What is electromagnetic induction? When does it occur? 2. How could you increase the amount of current produced by electromagnetic induction? 3. Explain how a moving magnet and a coil of wire can be used to produce alternating current. 4. List two devices that use electromagnetic induction. References 1. Christopher Auyeung.. CC-BY-NC-SA 3.0 2. Christopher Auyeung.. CC-BY-NC-SA 3.0 3 CONCEPT 2 www.ck12.org Electromagnets \u2022 De\ufb01ne and describe a solenoid. \u2022 De\ufb01ne and describe an electromagnet. \u2022 Determine the direction of the magnetic \ufb01eld inside a solenoid given the direction of current \ufb02ow in the coil wire. \u2022 Understand why an electromagnet has a stronger magnetic \ufb01eld than a solenoid. One of the most famous electric car companies is Tesla, named after Nikola Tesla. These electric cars, and all others, require an electromagnet to run the engine. Electromagnets A long coil of wire consisting of many loops of wire and making a complete circuit is called a solenoid. The magnetic \ufb01eld within a solenoid can be quite large since it is the sum of", " the \ufb01elds due to the current in each individual loop. 4 www.ck12.org Concept 2. Electromagnets The magnetic \ufb01eld around the wire is determined by a hand rule. Since this description doesn\u2019t mention electron \ufb02ow, we must assume that the current indicated by I is conventional current (positive). Therefore, we would use a right hand rule. We grasp a section of wire with our right hand pointing the thumb in the direction of the current \ufb02ow and our \ufb01ngers will curl around the wire in the direction of the magnetic \ufb01eld. Therefore, the \ufb01eld points down the cavity in these loops from right to left as shown in the sketch. If a piece of iron is placed inside the coil of wire, the magnetic \ufb01eld is greatly increased because the domains of the iron are aligned by the magnetic \ufb01eld of the current. The resulting magnetic \ufb01eld is hundreds of time stronger than the \ufb01eld from the current alone. This arrangement is called an electromagnet. The picture below shows an electromagnet with an iron bar inside a coil. Our knowledge of electromagnets developed from a series of observations. In 1820, Hans Oersted discovered that a current-carrying wire produced a magnetic \ufb01eld. Later in the same year, Andr\u00e9-Marie Ampere discovered that a coil of wire acted like a permanent magnet and Fran\u00e7ois Arago found that an iron bar could be magnetized by putting it inside coil of current-carrying wire. Finally, William Sturgeon found that leaving the iron bar inside the coil greatly increased the magnetic \ufb01eld. Two major advantages of electromagnets are that they are extremely strong magnetic \ufb01elds, and that the magnetic \ufb01eld can be turned on and off. When the current \ufb02ows through the coil, it is a powerful magnet, but when the current is turned off, the magnetic \ufb01eld essentially disappears. Electromagnets \ufb01nd use in many practical applications. Electromagnets are used to lift large masses of magnetic materials such as scrap iron, rolls of steel, and auto parts. 5 www.ck12.org The overhead portion of this machine (painted yellow) is a lifting electromagnet. It is lowered to the deck where steel pipe is stored and it picks up a length of pipe and moves it to another machine where it is set upright and lowered into an oil well drill hole. Electromagnets are essential to the design of the electric generator and electric motor and are also employed in doorbells, circuit breakers, television receivers, loudspeakers, electric dead bolts, car starters, clothes washers, atomic particle accelerators, and electromagnetic brakes and clutches. Electromagnets are commonly used as switches in electrical machines. A recent use for industrial electromagnets is to create magnetic levitation systems for bullet trains. Summary \u2022 A solenoid is a long coil of wire consisting of many loops of wire that makes a complete circuit. \u2022 An electormagnet is a piece of iron inside a solenoid. \u2022 While the magnetic \ufb01eld of a solenoid may be quite large, an electromagnet has a signi\ufb01cantly larger magnetic \ufb01eld. \u2022 Electromagnets\u2019 magnetic \ufb01elds can be easily turned off by just halting the current. Practice MEDIA Click image to the left for more content. http://www.youtube.com/watch?v=S6oop6RXg9w Follow up questions. 1. What components are needed to make a homemade electromagnet? 2. What objects were attracted by the electromagnet in the video? Review 1. Magnetism is always present when electric charges ___________. 2. What happens to the strength of an electromagnet if the number of loops of wire is increased? 3. What happens to the strength of an electromagnet if the current in the wire is increased? 4. Which direction does the magnetic \ufb01eld point in the solenoid sketched here? 6 www.ck12.org Concept 2. Electromagnets \u2022 solenoid: A current-carrying coil of wire that acts like a magnet when a current passes through it. \u2022 electromagnet: A temporary magnet consisting of an iron or steel core wound with a coil of wire, through which a current is passed. \u2022 magnetic levitation: The suspension of an object above a second object by means of magnetic repulsion. References 1. Image copyright Dongliu, 2013. http://www.shutterstock.com. Used under license from Shutterstock.com 2. Coil: Image copyright Giant Stock, 2013; modi\ufb01ed by CK-12 Foundation - Samantha Bacic. http", "://www. shutterstock.com. Used under license from Shutterstock.com 3. Image copyright Zigzag Mountain Art, 2013. http://www.shutterstock.com. Used under license from Shutterstock.com 4. Image copyright Ingvar Tjostheim, 2013. http://www.shutterstock.com. Used under license from Shutter- stock.com 5... CC BY-NC-SA 6. CK-12 Foundation - Samantha Bacic.. CC-BY-NC-SA 3.0 7 CONCEPT 3 Current and Magnetism Students will learn to analyze and solve problems involving current carrying wires in magnetic \ufb01elds. Students will learn to analyze and solve problems involving current carrying wires in magnetic \ufb01elds. www.ck12.org Key Equations Fwire = LIB sin(q) Force on a Current Carrying Wire In this equation, L refers to the length of the wire, I to the electric current, B the magnitude of the magnetic \ufb01eld and q is the angle between the direction of the current and the direction of the magnetic \ufb01eld. \u00b50I 2pr Where \u00b50 = 4p 107 Tm/A Bwire = Magnetic \ufb01eld at a distance r from a current-carrying wire Permeability of Vacuum (approximately same for air also) Force on a Wire Since a wire is nothing but a collection of moving charges, the force it will experience in a magnetic \ufb01eld will simply be the vector sum of the forces on the individual charges. If the wire is straight \u2014 that is, all the charges are moving in the same direction \u2014 these forces will all point in the same direction, and so will their sum. Then, the direction of the force can be found using the second right hand rule, while its magnitude will depend on the length of the wire (denoted L ), the strength of the current, the strength of the \ufb01eld, and the angle between their directions: Two current-carrying wires next to each other each generate magnetic \ufb01elds and therefore exert forces on each other: 8 Concept 3. Current and Magnetism MEDIA Click image to the left for more content. www.ck12.org Example 1 Example 2 A wire loop and an in\ufb01nitely long current carrying cable are placed a distance r apart. The in\ufb01nitely long wire is carrying a current I1 to the left and the loop is carrying a current I2 CCW. The dimensions of the wire loop are shown in the diagram illustrating the situation below. What is the magnitude and direction of the net force on the loop (the mass of the wires are negligible)? Solution In this problem, it is best to start by determining the direction of the force on each segment of the loop. Based on the \ufb01rst right hand rule, the magnetic \ufb01eld from the in\ufb01nite cable points into the page where the loop is. This means that the force on the top segment of the loop will be down toward the bottom of the page, the force on the left segment will be to right, the force on the bottom segment will be toward the top of the page, and the force on the right segment will be to the left. The forces on the left and right segments will balance out because both segments are the same distance from the cable. The forces from the top and bottom section will not balance out because the wires are different distances from the cable. The force on the bottom segment will be stronger than the one on the top segment because the magnetic \ufb01eld is stronger closer to the cable, so the net force on the loop will be up, toward the top of the page. Now we will begin to calculate the force\u2019s magnitude by \ufb01rst determining the strength of the magnetic \ufb01eld at the bottom and top segments. All we really have to do is plug in the distances to each segment into the equation we already know for the magnetic \ufb01eld due to a current carrying wire. 9 www.ck12.org B = Bbottom = Btop = \u00b5oI 2pr \u00b5oI1 2pR \u00b5oI1 2p2R Now we will calculate the net force on the loop using the equation given above. We\u2019ll consider up the positive direction. SF = I2L( SF = Fbottom Ftop SF = I2LBbottom I2LBtop SF = I2L(Bbottom Btop) \u00b5oI1 2pR \u00b5oI1I2L 2pR \u00b5oI1I2L 4pR \u00b5oI1 2p2R 1 2 SF = SF = (1 ) ) start by summing the forces on the loop substitute in the values for each of the force terms factor the equation substitute in the values for the magnetic \ufb01eld factor the", " equation again simplify to get the answer Watch this Explanation MEDIA Click image to the left for more content. Time for Practice 1. A vertical wire, with a current of 6:0 A going towards the ground, is immersed in a magnetic \ufb01eld of 5:0 T pointing to the right. What is the value and direction of the force on the wire? The length of the wire is 2:0 m. 10 www.ck12.org Concept 3. Current and Magnetism 2. 3. A futuristic magneto-car uses the interaction between current \ufb02owing across the magneto car and magnetic \ufb01elds to propel itself forward. The device consists of two \ufb01xed metal tracks and a freely moving metal car (see illustration above). A magnetic \ufb01eld is pointing downward with respect to the car, and has the strength of 5:00 T. The car is 4:70 m wide and has 800 A of current \ufb02owing through it. The arrows indicate the direction of the current \ufb02ow. a. Find the direction and magnitude of the force on the car. b. If the car has a mass of 2050 kg, what is its velocity after 10 s, assuming it starts at rest? c. If you want double the force for the same magnetic \ufb01eld, how should the current change? 4. A horizontal wire carries a current of 48 A towards the east. A second wire with mass 0:05 kg runs parallel to the \ufb01rst, but lies 15 cm below it. This second wire is held in suspension by the magnetic \ufb01eld of the \ufb01rst wire above it. If each wire has a length of half a meter, what is the magnitude and direction of the current in the lower wire? 5. Show that the formula for the force between two current carrying wires is F = \u00b5oLi1i2 2pd, where d is the distance between the two wires, i1 is the current of \ufb01rst wire and L is the segment of length of the second wire carrying a current i2. (Hint: \ufb01nd magnetic \ufb01eld emanating from \ufb01rst wire and then use the formula for a wire immersed in that magnetic \ufb01eld in order to \ufb01nd the force on the second wire.) 6. Two long thin wires are on the same plane but perpendicular to each other. The wire on the y axis carries a current of 6:0 A in the y direction. The wire on the x axis carries a current of 2:0 A in the +x direction. Point, P has the co-ordinates of (2:0; 2; 0) in meters. A charged particle moves in a direction of 45o away from the origin at point, P, with a velocity of 1:0 107 m=s: a. Find the magnitude and direction of the magnetic \ufb01eld at point, P. b. If there is a magnetic force of 1:0 106 N on the particle determine its charge. c. Determine the magnitude of an electric \ufb01eld that will cancel the magnetic force on the particle. 7. A long straight wire is on the x axis and has a current of 12 A in the x direction. A point P, is located 2:0 m above the wire on the y axis. a. What is the magnitude and direction of the magnetic \ufb01eld at P. b. If an electron moves through P in the x direction at a speed of 8:0 107 m=s what is the magnitude and direction of the force on the electron? Answers to Selected Problems 1. Down the page; 60 N 2. a. To the right, 1:88 104 N b. 91:7 m=s c. It should be doubled 3. East 1:5 104 A 4.. 5. a. 8 107 T b. 1:3 106 C 11 6. a. 1:2 106 T; +z b. 1:5 1017 N; y www.ck12.org 12 www.ck12.org Concept 4. Electric Motors CONCEPT 4 Electric Motors \u2022 Explain the design and operation of an electric motor. As gas prices continue to rise, electric cars and hybrids are becoming increasingly popular. These cars are certainly a part of our future. On the left in the image above is an all-electric vehicle, and on the right is a hybrid vehicle that uses gas part time and electricity part time. Electric Motors In an earlier concept, we described and calculated the force that a magnetic \ufb01eld exerts on a current carrying wire. Since you are familiar with Newton\u2019s third law of motion, you know that if the magnetic \ufb01eld exerts a force on the current carrying wire, then the current carrying wire also exerts a force of equal magnitude and", " opposite direction on the magnetic \ufb01eld. In the sketch above, a circuit is connected to a battery, with one part of the circuit placed inside a magnetic \ufb01eld. When current runs through the circuit, a force will be exerted on the wire by the magnetic \ufb01eld, causing the wire to 13 www.ck12.org move. If we choose to consider electron current in this case, the electrons \ufb02ow from the back of the sketch to the front while the magnetic \ufb01eld is directed upward. Using the left hand rule for this, we \ufb01nd that the force on the wire is to the right of the page. Had we chosen to consider the current to be conventional current, then the current would be \ufb02owing from the front of the sketch to the back and we would use the right hand rule. The force on the wire would, once again, be toward the right. This movement is harnessed in electrical motors. Electrical motors change electrical energy into mechanical energy. The motor consists of an electrical circuit with part of the wires inside a magnetic \ufb01eld. This can be seen below. Positive charges move through the circuit in the direction of the light purple arrows. When the charges move up through the part of the coil that is right next to the north pole, the right hand rule tells us that the wire suffers the force, F, pushing the wire in the direction of the blue arrow, toward the back of the sketch. On the other side of the coil, where the charges are moving down through the \ufb01eld, the right hand rule shows the force would push this side of the coil toward the front. These two forces are working together, rotating the coil in the direction of the circular red arrow. Where the rotating coil (in grey) meets the wires attached to the power source (black), we \ufb01nd a split ring commutator. The coil turns, but the commutator and power source do not. As the coil turns, it moves off of the blue box connector and as it continues to turn, it connects to the other blue box connector. As the coil turns, it reverses its connections to the external circuit. Therefore, inside the coil, the current is always \ufb02owing in the same direction because the left side of the coil is always attached to the left side of the external circuit. This allows the coil, or armature, to continue to spin the same direction all the time. In electrical motors, these coils often consist of not just one, but many wires, as can be seen here: 14 www.ck12.org Concept 4. Electric Motors Summary \u2022 When current runs through the circuit, a force will be exerted on the wire. \u2022 An electrical motor changes electrical energy into mechanical energy. \u2022 A split ring commutator keeps the current in the coil \ufb02owing in the same direction even though the coil changes sides every half turn. Practice MEDIA Click image to the left for more content. http://www.youtube.com/watch?v=elFUJNodXps In the video, a simple electrical motor is constructed. Follow up questions. 1. Who \ufb01rst built an electric motor? 2. What size battery was used in the video motor? 3. The rover Curiosity is on the surface of what body? Review 1. Which way will the wire be pushed when current passes through the wire? a. up b. down 15 c. left d. right e. None of these. www.ck12.org 2. Which way will the coil spin when current passes through the wire? a. clockwise b. counterclockwise \u2022 electric motor: An electricmotor converts electricity into mechanical motion. \u2022 commutator: A split - ring commutator (sometimes just called a commutator) is a simple and clever device for reversing the current direction through an armature every half turn. \u2022 armature: A revolving structure in an electric motor or generator, wound with the coils that carry the current. References 1. Myrtle Beach TheDigitel and User:Mariordo/Wikimedia Commons. http://commons.wikimedia.org/wiki/F ile:Nissan_Leaf_%26_Chevy_Volt_charging_trimmed.jpg. CC-BY 2.0 16 www.ck12.org Concept 4. Electric Motors 2... CC BY-NC-SA 3. CK-12 Foundation - Samantha Bacic.. CC-BY-NC-SA 3.0 4. CK-12 Foundation - Samantha Bacic.. CC BY-NC-SA 3.0 5. Image copyright Sim Kay Seng, 2013. http://www.shutterstock.com. Used under license from Shutter- stock.com 6. CK-12 Foundation - Samantha Bacic.. CC-BY-NC-SA 3.0", " 7. CK-12 Foundation - Samantha Bacic.. CC-BY-NC-SA 3.0 17 CONCEPT 5 www.ck12.org Electromotive Force \u2022 De\ufb01ne EMF. \u2022 Calculate EMF for a wire moving in a magnetic \ufb01eld. Electrical generators convert mechanical energy into electrical energy. Every electrical generator needs some method for spinning the coil inside the magnetic \ufb01eld. Hydroelectric generators use water pressure to spin the coil while windmills, of course, use the wind to spin the coil. The image here is a combination of steam turbine and generator. The steam can be produced by burning coal or diesel fuel or by a nuclear reaction and the steam then turns the coil and generates electricity. Electromotive Force When an individual charge \ufb02ies through a magnetic \ufb01eld, a force is exerted on the charge and the path of the charge bends. In the case shown in the sketch below, the charge is positive and the right hand rule shows us the force will be upward, perpendicular to both the \ufb01eld and the path of the charge. 18 www.ck12.org Concept 5. Electromotive Force If a wire that is part of a complete circuit is moved through a magnetic \ufb01eld, the force on the individual electrons in the wire occurs in exactly the same manner. Since the electrons in the wire are negatively charged, the force would be in the opposite direction but otherwise the situation is the same. When the wire is pulled downward through the magnetic \ufb01eld, the force on the electrons cause them to move within the wire. Since the charges are negative, the left hand rule shows that the electrons would move as diagrammed in the sketch. (Point \ufb01ngers in the direction of the magnetic \ufb01eld, point thumb in the direction of wire movement, and palm shows direction of electron \ufb02ow.) No current will \ufb02ow, of course, unless the section of wire is part of a complete circuit. This process allows us to convert mechanical energy (the motion of the wire) into electrical energy (the current). This is the opposite of what happens in an electric motor where electrical energy is converted into mechanical energy. In order to maintain a constant current \ufb02ow, it is necessary to have a potential difference or voltage in the circuit. The voltage or potential difference is also frequently referred to electromotive force. The term electromotive force, 19 like many historical terms, is a misnomer. Electromotive force is NOT a force, it is a potential difference or potential energy per unit charge and is measured in volts. The potential difference in the case of moving a wire through a magnetic \ufb01eld is produced by the work done on the charges by whatever is pushing the wire through the \ufb01eld. The EMF (or voltage) depends on the magnetic \ufb01eld strength, B, the length of the wire in the magnetic \ufb01eld, l, and the velocity of the wire in the \ufb01eld. www.ck12.org EMF = Blv This calculation is based on the wire moving perpendicularly through the \ufb01eld. If the wire moves an angle to the \ufb01eld, then only the component of the wire perpendicular to the \ufb01eld will generate EMF. Example Problem: A 0.20 m piece of wire that is part of a complete circuit moves perpendicularly through a magnetic \ufb01eld whose magnetic induction is 0.0800 T. If the speed of the wire is 7.0 m/s, what EMF is induced in the wire? Solution: EMF = Blv = (0:0800 N=A m)(0:20 m)(7:0 m=s) = 0:11 N m=C = 0:11 J=C = 0:11 V Summary \u2022 If a wire that is part of a complete circuit is moved through a magnetic \ufb01eld, the magnetic \ufb01eld exerts a force on the individual electrons in the wire, which causes a current to \ufb02ow. \u2022 The potential difference in the case of moving a wire through a magnetic \ufb01eld is produced by the work done on the charges by whatever is pushing the wire through the \ufb01eld. \u2022 The EMF (or voltage) depends on the magnetic \ufb01eld strength, B, the length of the wire in the magnetic \ufb01eld, l, and the velocity of the wire in the \ufb01eld, EMF = Blv. Practice MEDIA Click image to the left for more content. http://www.youtube.com/watch?v=0OHmMVBLXTI Follow up questions. 1. We have been discussing the process of generating electricity by moving a wire through a magnetic \ufffd", "\ufffdeld. What happens if the wire is held steady and the magnetic \ufb01eld moves instead? 2. When a loop of wire is turned circularly in a magnetic \ufb01eld, what type of current is produced? Review 1. Which of the following units are equivalent to those of EMF produced in a generator? (a) T m=s (b) V m2=s 20 www.ck12.org (c) J=s (d) A W (e) T m Concept 5. Electromotive Force 2. A straight wire 0.500 m long is moved straight up through a 0.400 T magnetic \ufb01eld pointed in the horizontal direction. The speed of the wire is 20.0 m/s. (a) What EMF is induced in the wire? (b) If the wire is part of a circuit with a total resistance of 6:00 W, what is the current in the circuit? 3. A straight wire, 25.0 m long, is mounted on an airplane \ufb02ying at 125 m/s. The wire moves perpendicularly through earth\u2019s magnetic \ufb01eld (B = 5:00 105 T ). What is the EMF induced in the wire? 4. A straight wire, 30.0 m long, moves at 2.00 m/s perpendicularly through a 1.00 T magnetic \ufb01eld. (a) What is the induced EMF? (b) If the total resistance of the circuit is 15:0 W, what is the current in the circuit? \u2022 electromotive force: The potential energy per unit charge that produces a \ufb02ow of electricity in a circuit, expressed in volts. \u2022 induced current: The electric current generated in a loop of conducting material by movement of the loop across a magnetic \ufb01eld. References 1. Courtesy of the NRC. http://commons.wikimedia.org/wiki/File:Modern_Steam_Turbine_Generator.jpg. Public Domain 2. CK-12 Foundation - Samantha Bacic.. CC-BY-NC-SA 3.0 3. CK-12 Foundation - Samantha Bacic.. CC-BY-NC-SA 3.0 21 CONCEPT 6 Electric Generators www.ck12.org \u2022 Explain how an electric generator works. \u2022 Explain the difference between an electric generator and an electric motor. \u2022 Explain the difference between peak and effective voltage and current from a generator. These large machines are electric generators. This particular row of generators is installed in a hydroelectric power station. The insides of these generators are coils of wire spinning in a magnetic \ufb01eld. The relative motion between the wire and the magnetic \ufb01eld is what generates electric current. In all generators, some mechanical energy is used to spin the coil of wire in the generator. In the case of hydroelectric power, the coil of wire is spun by water falling from higher PE to lower PE. Windmills and steam turbines are used in other types of power generators to spin the coil. Electric Generators Electric generators convert mechanical energy to electric energy. The generator consists of some number of wire loops wrapped around an iron core and placed in a strong magnetic \ufb01eld. The loops of wire and the iron core are called the armature. The armature is mounted so that it can rotate freely inside the magnetic \ufb01eld. Mechanical energy is used to spin the armature in the \ufb01eld so that the wire loops cut across the \ufb01eld and produce electric current. The EMF of this current is calculated by EMF = Blv. 22 www.ck12.org Concept 6. Electric Generators Consider the coil and magnetic \ufb01eld sketched above. When the right hand side of the coil moves up through the \ufb01eld, the left hand rule indicates that the electron \ufb02ow will be from the front to the back in that side of the coil. The current generated will have the greatest EMF as the wire is cutting perpendicularly across the \ufb01eld. When the wire reaches the top of its arc, it is moving parallel to the \ufb01eld and therefore, not cutting across the \ufb01eld at all. The EMF at this point will be zero. As that same wire then cuts down through the \ufb01eld as it continues to spin, the left hand rule indicates that the electron \ufb02ow will be from the back to the front in that side of the coil. In this second half of the arc, the direction of the electron \ufb02ow has reversed. The magnitude of the EMF will reach maximum again as the wire cuts perpendicularly down through the \ufb01eld and the EMF will become zero again as the wire passes through the bottom of the arc. The current produced as the", " armature goes around will resemble a sine wave where the EMF reaches a maximum in one direction, then goes to zero, then goes to a maximum in the other direction. This type of current is called alternating current. By having more and more loops of wire on the armature, the crests and troughs overlap and \ufb01ll in until a constant current is produced. A direct current is one that always \ufb02ows in the same direction rather than alternating back and forth. Batteries produce direct currents. A generator can also produce direct current by using a split ring commutator that changes external connections every half turn of the armature so that even though the current in the coil changes direction, every time the current in the coil changes direction, the external connection switches so that the external current always goes in the same direction. Generators and motors are almost identical in construction but convert energy in opposite directions. Generators convert mechanical energy to electrical energy and motors convert electrical energy to mechanical. Because of the alternating direction in alternating current, the average value is less than the power supplied by a direct current. In fact, the average power of an AC current is one-half its maximum power and one-half the power of an equivalent DC current. The effective current of an AC generator is 0.707 times its maximum current. The same is true for the effective voltage of an AC generator. 23 Ieff = 0:707 Imax Veff = 0:707 Vmax Example Problem: An AC generator develops a maximum voltage of 34.0 V and delivers a maximum current of 0.170 A. www.ck12.org a. What is the effective voltage of the generator? b. What is the effective current delivered by the generator? c. What is the resistance in the circuit? Solution: a. Veff = 0:707 Vmax = (0:707)(34:0 V ) = 24:0 V b. Ieff = 0:707 Imax = (0:707)(0:17 A) = 0:120 A c. R = V 0:120 A = 200: W I = 24:0 V Summary \u2022 Electric generators convert mechanical energy to electric energy. \u2022 The generator consists of some number of wire loops wrapped around an iron core and placed in a strong magnetic \ufb01eld. \u2022 The loops of wire and the iron core are called the armature. \u2022 The armature is mounted so that it can rotate freely inside the magnetic \ufb01eld. \u2022 Mechanical energy is used to spin the armature in the \ufb01eld so that the wire loops cut across the \ufb01eld and produce electric current. \u2022 The current produced as the armature goes around will resemble a sine wave where the EMF reaches a maximum in one direction, then goes to zero, then goes to a maximum in the other direction. This type of current is called alternating current. \u2022 A generator can also produce direct current by using a split ring commutator that changes external connections every half turn of the armature so that even though the current in the coil changes direction, every time the current in the coil changes direction, the external connection switches so that the external current always goes in the same direction. \u2022 The effective current of an AC generator is 0.707 times its maximum current. \u2022 The effective voltage of an AC generator is 0.707 times its maximum voltage. Practice MEDIA Click image to the left for more content. http://www.youtube.com/watch?v=RFOMpOM1WHQ Follow up questions. 1. Which of the two generators in the video (an AC generator and a DC generator) involves a magnetic \ufb01eld? 2. Which of the two generators in the video involves a wire-wrapped armature? 3. What is the difference between the DC generator and the AC generator? 24 Concept 6. Electric Generators www.ck12.org Review 1. What three things are necessary to produce EMF mechanically? (a) magnet, force lines, and magnetic \ufb01eld (b) EMF, conductor, and magnetic \ufb01eld (c) conducting wire, magnetic \ufb01eld, and relative motion (d) conducting wire, electrical \ufb01eld, and relative motion (e) none of these will produce EMF mechanically 2. Increasing which of the following will increase the output of a generator? (a) EMF (b) strength of the magnetic \ufb01eld (c) resistance of the conductor (d) load on the meter (e) none of these 3. The current in the rotating coil of all generators is (a) AC (b) DC (c) pulsating AC (d) pulsating DC 4. A generator in a power plant develops a maximum voltage of 170. V. (a) What is the effective voltage? (b) A 60.0 W light bulb is placed across the generator. A maximum", " current of 0.70 A \ufb02ows through the bulb. What effective current \ufb02ows through the bulb? (c) What is the resistance of the light bulb when it is working? 5. The effective voltage of a particular AC household outlet is 117 V. (a) What is the maximum voltage across a lamp connected to the outlet? (b) The effective current through the lamp is 5.50 A. What is the maximum current in the lamp? \u2022 direct current: An electric current \ufb02owing in one direction only. \u2022 alternating current: An electric current that reverses direction in a circuit at regular intervals. \u2022 electric generator: An electric generator is a device that converts mechanical energy to electrical energy. \u2022 armature: The rotating part of a generator, consisting essentially of copper wire wound around an iron core. References 1. Image copyright James L. Davidson, 2013. http://www.shutterstock.com. Public Domain 2. Galvanometer: Image copyright scropy, 2013; composite created by CK-12 Foundation. http://www.shutters tock.com. Used under license from Shutterstock.com 3. CK-12 Foundation - Samantha Bacic.. CC-BY-NC-SA 3.0 25 CONCEPT 7 \u2022 Describe the function of a transformer. \u2022 Explain the relationship between turns and voltage ratio. \u2022 Solve mathematical problems involving transformers. www.ck12.org Transformers Power loss in long transmission lines is related to the magnitude of the current. Speci\ufb01cally, the power loss can be decreased by decreasing the magnitude of the current. The amount of power passed through transmission lines can be calculated by multiplying voltage by current. The same power can be transmitted using a very high voltage and a very low current as with a low voltage and high current. Since power companies do not wish to waste power as it is transmitted to homes and businesses, they deliberately \u2019step up\u2019 the voltage and reduce the current before transmitting the power over extended distances. That type of power transmits well without great loss of energy but it cannot be used in household appliances. It becomes necessary to convert it back (\u2019step down\u2019) to low voltage and high current for household use. That is the job of electrical transformers \u2013 those big gray barrels you see on power poles. Transformers When we move a wire through a magnetic \ufb01eld, a force is exerted on the charges in the wire and a current is induced. Essentially the same thing happens if we hold the wire steady and move the magnetic \ufb01eld by moving the magnet. Yet a third way of causing relative motion between the charges in a wire and a magnetic \ufb01eld is to expand or contract the \ufb01eld through the wire. When a current begins to \ufb02ow in a wire, a circular magnetic \ufb01eld forms around the wire. Within the \ufb01rst fractional second when the current begins to \ufb02ow, the magnetic \ufb01eld expands outward from the wire. If a second wire is placed nearby, the expanding \ufb01eld will pass through the second wire and induce a brief current in the wire. 26 www.ck12.org Concept 7. Transformers Consider the sketch above. When the knife switch is closed, current begins to \ufb02ow in the \ufb01rst circuit and therefore, a magnetic \ufb01eld expands outward around the wire. When the magnetic \ufb01eld expands outward from the wire on the right side, it will pass through the wire in the second circuit. This relative motion between wire and \ufb01eld induces a current in the second circuit. The magnetic \ufb01eld expands outward for only a very short period of time and therefore, only a short jolt of current is induced in the second circuit. You can leave the knife switch closed and the current will continue to \ufb02ow in the \ufb01rst circuit but no current is induced in the second circuit because the \ufb01eld is constant and therefore there is no relative motion between the \ufb01eld and the wire in the second circuit. When the knife switch is opened, the current in the \ufb01rst circuit ceases to \ufb02ow and the magnetic \ufb01eld collapses back through the wire to zero. As the magnetic \ufb01eld collapses, it passes through the wire and once again we have relative motion between the wire in the second circuit and the magnetic \ufb01eld. Therefore, we once again have a short jolt of current induced in the second circuit. This second jolt of induced current will be \ufb02owing in the opposite direction of the \ufb01rst induced current. We can produce an alternating current in the second circuit simply by closing and opening the knife switch continuously in the \ufb01rst circuit. Obviously, a transformer would have little use in the case", " of DC current because current is only induced in the second circuit when the \ufb01rst circuit is started or stopped. With AC current, however, since the current changes direction 60 times per second, the magnetic \ufb01eld would constantly be expanding and contracting through the second wire. A transformer is a device used to increase or decrease alternating current voltages. They do this with essentially no loss of energy. A transformer has two coils, electrically insulated from each other as shown in the sketch. One coil is called the primary coil and the other is called the secondary coil. When the primary coil is connected to a source of AC voltage, the changing current creates a varying magnetic \ufb01eld. The varying magnetic \ufb01eld induces a varying EMF in the secondary coil. The EMF induced in the secondary coil is called the secondary voltage and is proportional to the primary voltage. The secondary voltage also depends on the ratio of turns on the secondary coil to turns on the primary coil. 27 www.ck12.org secondary voltage primary voltage VS VP = = number of turns on secondary number of turns on primary NS NP If the secondary voltage is larger than the primary voltage, the transformer is called a step-up transformer. If the voltage out of the transformer is smaller than the voltage in, then the transformer is called a step-down transformer. In an ideal transformer, the electric power put into the primary equals the electric power delivered by the secondary. The current that \ufb02ows in the primary depends on how much current is required by the secondary circuit. VPIP = VSIS IS IP = VP VS = NP NS Example Problem: A particular step-up transformer has 200 turns on the primary coil and 3000 turns on the secondary coil. a. If the voltage on the primary coil is 90.0 V, what is the voltage on the secondary coil? b. If the current in the secondary circuit is 2.00 A, what is the current in the primary coil? c. What is the power in the primary circuit? d. What is the power in the secondary circuit? Solution: = NP NS = NP NS VS = (90:0 V )(3000) a. VP VS IS IP = 30:0 A b. IP c. PP = VPIP = (90:0 V )(30:0 A) = 2700 W d. PS = VSIS = (1350 V )(2:00 A) = 2700 W = 200 3000 = 200 3000 90:0 V VS 2:00 A IP (200) = 1350 V 28 www.ck12.org Summary Concept 7. Transformers \u2022 When a current begins to \ufb02ow in a wire, a circular magnetic \ufb01eld forms around the wire. \u2022 Within the \ufb01rst fractional second when the current begins to \ufb02ow, the magnetic \ufb01eld expands outward from the wire. \u2022 If a second wire is placed nearby, the expanding \ufb01eld will pass through the second wire and induce a brief current in the wire. \u2022 A transformer is a device used to increase or decrease alternating current voltages. \u2022 A transformer has two coils, electrically insulated from each other. One coil is called the primary coil and the other is called the secondary coil. \u2022 The varying magnetic \ufb01eld induces a varying EMF in the secondary coil. \u2022 The EMF induced in the secondary coil is called the secondary voltage and is proportional to the primary voltage. The secondary voltage also depends on the ratio of turns on the secondary coil to turns on the primary coil. secondary voltage primary voltage = number of turns on secondary number of turns on primary Practice MEDIA Click image to the left for more content. http://www.youtube.com/watch?v=-v8MYAFl7Mw Follow up questions. 1. What type of transformer is used at the power station where the electric power is generated? 2. What type of transformer is used at power sub-stations? 3. What type of transformer is used inside cell phone chargers? Review 1. A step-down transformer has 7500 turns on its primary and 125 turns on its secondary. The voltage across the primary is 7200 V. (a) What is the voltage across the secondary? (b) The current in the secondary is 36 A. What current \ufb02ows in the primary? 2. The secondary of a step-down transformer has 500 turns. The primary has 15,000 turns. (a) The EMF of the primary is 3600 V. What is the EMF of the secondary? (b) The current in the primary is 3.0 A. What is the current in the secondary? 3. An ideal step-up transformer\u2019s primary coil has 500 turns and its secondary coil has 15,000 turns. The primary EMF is 120 V. (a) Calculate the EMF of the secondary. (b) If the secondary", " current is 3.0 A, what is the primary current? (c) What power is drawn by the primary? 29 www.ck12.org \u2022 transformer: An electric device consisting essentially of two or more windings wound on the same core, which by electromagnetic induction transforms electric energy from one circuit to another circuit such that the frequency of the energy remains unchanged while the voltage and current usually change. \u2022 primary coil: A coil to which the input voltage is applied in a transformer. \u2022 secondary coil: The coil in a transformer where the current is induced. \u2022 step-up transformer: A step-up transformer is one that increases voltage. \u2022 step-down transformer: A step-down transformer is one that decreases voltage. References 1. Image copyright Sylvie Bouchard, 2013. http://www.shutterstock.com. Used under license from Shutter- stock.com 2. CK-12 Foundation - Samantha Bacic, using image copyright Sergii Korolko, 2013. http://www.shutterstock.com. Used under license from Shutterstock.com 3. CK-12 Foundation - Christopher Auyeung.. CC-BY-NC-SA 3.0 30rors are also used in supermarkets, underground parking areas etc. to cover a wider field of view to be observed. Fig. 1.35: Parabolic dish aerial. (vii) Shaving mirror A concave mirror is used as a shaving or a make-up mirror. The mirror is placed at a distance less than its focal length, from the face. A virtual upright, magnified image of the face is seen in the mirror (Fig. 1.36). B face F Fig. 1.36: Shaving mirror. C (b) Convex mirrors A convex mirror is used as a rear view driving mirror. Any object in front of the mirror, i.e, lying behind the car, forms an upright diminished virtual image between F and P (Fig. 1.37). For example, another car behind is seen upright and small in size in the mirror. The driver of a car has a wider view of all the objects lying behind. 24 image C F M I P B O object (car) v u Fig. 1.37: Rear view driving mirror. A convex mirror covers a wider field of view so that the light rays from a wide angle can be observed. A plane mirror of the same size as the convex mirror helps us to see the objects lying in a limited area (smaller). 1.8 Project work Construction of a solar concetrator Suggested materials: Mild steel wires of about 3 mm in thickness, aluminium foil, superglue Assembly 1. Cut the mild steel wire so as to make three circles of diameter 1.0 m each. 2. Make the three circles using the wire. 3. Weld the wires to form a spherical framework. 4. Cover the framework using the aluminium foil. 5. Cut the framework to produce two equal \u00bd sphere. 25 6. Mount one of the \u00bd sphere on a stand such that it can be rotated in all directions. 7. Using a thicker mild steel wire, make a framework that will be able to support a small sufuria and connect it to the framework above. Make sure that the sauce pan is at \u00bd centre of the sphere i.e. at focal point. 8. Direct the \u00bd sphere and the sauce pan towards the sun. 9. Fill the sauce pan with water at 15 minutes interval. 10. Measure the temperature at 15 minutes interval. Such an arrangement is called a solar concentrator. 26 Topic summary \u2022 The reflecting surface of a plane mirror is a straight plane, concave mirror curves in, convex mirror curves outwards and parabolic mirror is a section of a paraboloid. \u2022 The centre of the reflecting surface is the pole, P, of a curved mirror. \u2022 The centre of curvature, C, of a curved mirror is the centre of the sphere of which the mirror forms a part. \u2022 The principal axis of a curved mirror is a line passing through the pole, P, and the centre of curvature, C. \u2022 The radius of curvature, r, of a curved mirror is the radius of the sphere of which the mirror forms a part. \u2022 The principal focus, F, is a fixed point on the principal axis where a set of incident rays parallel and close to the principal axis converge in a concave mirror and appear to diverge on in a convex mirror. \u2022 The principal focus is real for a concave mirror and virtual for a convex mirror. \u2022 The focal length, f, of a curved mirror is the distance from its pole to the principal focus. \u2022 Focal length of a concave mirror is half the radius of curvature of the mirror. \u2022 A parabolic concave reflector has a single point focus and the caustic curve is not formed. \u2022 Curved mirrors obey the laws of reflection just like plane mirrors. \u2022 The focal length,", " f, of a curved mirror is half the radius of curvature, r. \u2022 Magnification (m) is defined as the ratio of the height of the image to the height of the object. Magnification = height of image (IM) height of object (OB) = image distance (v) object distance (u) m = 1 f IM OB 1 = u = v u + 1 v \u2022 The nature, size and the position of the image formed in a concave mirror depends on the position of the object in front of the mirror. See the Table 1.1 below. 27 Table 1.1 Object Image Nature of image Size of image compared to object F At infinity (far away) Beyond C At C Between C and F Beyond C At F Between F and C At C At infinity (far away) Real and inverted Diminished Real and inverted Diminished Real and inverted Same size Real and inverted Magnified Real and inverted Magnified Between F and P Behind the mirror Virtual and upright Magnified \u2022 In a convex mirror, for all the positions of the object at a measurable distance from the mirror, an upright, diminished, virtual image is always formed between F and P. \u2022 Cinema projectors, search lights, car head lights, astronomical telescopes, solar concentrators, dish aerials are a few applications of curved reflecting surfaces. 28 Topic Test 1 1. The focal length of a curved mirror is the distance between the principal focus and the centre of curvature. the pole of the mirror and the centre of curvature. the pole of the mirror and the principal focus. the object and the image. A. B. C. D. 2. In a curved mirror, how is the focal length (f) related to the radius of curvature (r) of the mirror? A. f = r 2 B. f = 2r C. f = r D. None of the above. 3. In a concave mirror, when the object is placed between the principal focus and the pole, the image formed is A. real and diminished C. virtual and upright B. real and magnified D. virtual and inverted 4. In a concave mirror, when the object is placed at 12 cm from the pole a real image is formed at 36 cm from the pole of the mirror. The magnification produced by the mirror is A. 0.25 B. 4.00 C. 0.33 D. 3.00 5. The effect of formation of a caustic curve in a concave mirror can be minimized by A. cutting off the marginal rays. B. cutting off the axial rays. C. cutting off all the rays parallel to the principal axis. D. cutting off the rays passing through the centre of curvature of the mirror. 6. In a convex mirror, the image formed is always A. C. real and upright. real and inverted. B. virtual and upright. D. virtual and inverted. 7. In a concave mirror, when a real object is placed at 2f (Fig. 1.38), the image formed will be A. at infinity. B. between p and f. C. between P and 2f. D. at 2f. 8. Draw a sketch of a concave mirror of radius of curvature 20 cm. Mark on the diagram the pole, the centre of curvature, the principal axis, the principal focus and the focal length. 29 PF2FobjectFig. 1.38 9. Describe a parabolic mirror. What is the advantage of a parabolic mirror over a spherical concave mirror. 10. Define the following terms: pole, principal axis, principal focus, focal length, focal plane in relation to a concave mirror. 11. Copy and complete the following diagrams (Fig. 1.39 (a) to (d)) to show the path of the reflected rays. Fig. 1.39 12. A concave mirror has a focal length of 6 cm and an object 2 cm tall is placed 10 cm from the pole of the mirror. By means of an accurate ray diagram, find the position and size of the image formed. Is the image real or virtual? Explain your answer. 13. Describe an experiment to determine the magnification of an image in a concave mirror. 14. Fig. 1.40 shows the graph of real image distance v against the object distance u for a concave mirror. Fig. 1.40 Explain why the coordinate P is for a magnified image. 15. (a) Define magnification. 30 PFC(a)PFC(b)PFC(c)PFC (d)0uPv (b) An object of height 3 cm is placed at 20 cm in front of a concave mirror. The real image formed is 10 cm from the pole of the mirror. Calculate the height of the image formed. 16. A concave mirror has a focal length of 6 cm and a real object 3 cm tall is placed 15 cm from the pole", ". Calculate the distance of the image from the pole if the size of the image is 2 cm. 17. Explain, with the aid of a neat ray diagram how a concave mirror can be used as a \u2018make-up\u2019 mirror. (The object may be represented by an arrow-head). 18. Motorists use a convex mirror, rather than a plane mirror, as a rear-view mirror. State one advantage and one disadvantage of using a convex mirror. 19. Name the type of mirror used in the following: (a) car head lights. (b) solar concentrators. (c) underground car parking area. 20. For the following Figure 1.41 (a) and (b), copy and complete the ray diagram to locate the position of the point image I for the point object placed at O. (a) Fig. 1.41 (b) 21. The graph (Fig. 1.42) shows how the image distance v and the object distance u vary for a concave mirror. (a) When the object is placed at a distance of 60 cm from the mirror Fig. 1.42 31 PCOCOv(cm)0u(cm)10203040506070802030405060708010 (i) what is the distance of the image from the mirror? (ii) Is the image magnified, diminished or the same size as the object? (iii) what is the magnification produced by the mirror? (b) The object is now moved until it is 20 cm from the mirror. State and explain what happens to (i) the image distance. (ii) the magnification produced by the mirror. 22. A man uses a showing mirror with a focal length of 72 cm to view the images of his face. If his face is 18 cm from the mirror, determine the image distance and the magnification of his face. 23. The real image produced by a concave mirror is observed to be six times longer than the object when the object is 34.2 cm in front of a mirror. Determine the radius of curvature of this mirror. 24. An object and a concave mirror are used to produce a sharp image of the object on the screen. The table below shows the magnification and image distance of the object. Magnification m Image distance v (cm) 0.25 20 1.5 40 2.5 56 3.5 72 Table 1.2 (a) Plot a graph of m against v (b) Use the graph to find: i. The image distance when M=1.0 ii. The object distance iii. The focal length of the mirror 32 TOPIC 2 Refraction of light in thin lenses Unit Outline \u2022 Definition of a lens \u2022 Types of lenses \u2022 Terms used in thin lenses \u2022 Image formation by converging and diverging lenses \u2022 The lens formula \u2022 Sign convetion \u2022 Magnification \u2022 Power of lenses \u2022 Defects of lenses \u2022 Simple telescope Introduction In secondary 1, we learnt about refraction of light through a medium. In this topic, we are going to specifically focus refraction in thin lenses. How the thin lenses are applied in optical instruments such as microscope, glasses, cameras and others. 2.1 Definition of a lens Activity 2.1 What is a lens? (Work in pairs or in groups) Materials \u2022 Water \u2022 Round bottomed flask \u2022 Plain paper \u2022 Retort stand \u2022 Sun \u2022 Magnifying lens Steps 1. What is a lens? 2. Fill a volumetric flask with clear water. 3. Cork the flask. 4. Tilt the flask such that the neck of the cork is horizontal. 5. Place a source of light (sun, bulbs) above the flask. 33 6. Place a white paper under the flask preferably on the ground. 7. Move the flask to or away the white paper. 8. What happens on the plain paper? 9. Discuss the results in your groups. 10. Replace the round bottomed flask with a magnifying lens. What happens to the paper? Explain. Sun Clamp Round bottomed flask Stand Cork Water Piece of paper Fig. 2.1 A lens is a transparent medium bounded by two spherical surface or a planed curved surface. 2.2 Types of lenses To identify and describe types and shapes of lenses Activity 2.2 (Work in groups) Materials \u2022 Charts showing converging and diverging beam through lens. \u2022 Convex lenses Steps \u2022 Plane lenses \u2022 Spherical lenses 1. Place some lenses available in your school on a labeled white plane paper. Trace their outlines. What is the shape of the lenses? 2. Feel the lenses in Fig. 2.2 below with your fingers. What do you feel? What are their shapes? Why do you think they are made in such shapes? 34 3. Name the above lenses. Fig. 2.2 4. Identify the lenses in Fig. 2.3 below. What are their shapes? Why do you think they are made", " in such shapes? Name them. (a) (b) (c) Fig. 2.3: Types of concave lenses There are two main groups of lenses. A type that is thick in the middle and thin at the edges, causing rays of light to converge. This is called converging or convex lenses. The other type is thin in middle and thick at the edges causing the rays of light to diverge. This lens is called diverging or concave lens. Concave lenses are of different shapes as shown in Fig. 2.4. A bi-convex or double convex lens has both its surfaces \u2018curving out\u2019. (Fig. 2.4). (a) (b) (c) Bi-convex or double convex Plano - convex Concavo - convex Concave lenses are also of different shapes (See Fig. 2.5) Fig. 2.4: Types of convex lenses (a) (b) Bi-concave plano-concave Fig. 2.5: Types of lenses 35 A bi-concave or double concave lens has both its surfaces \u2018curving in\u2019. Other concave lenses are plano-concave and convexo-concave or diverging meniscus (Fig. 2.5). When a beam of light is incident on the lenses, rays tend to converge at a point when they pass through a convex lens and diverge through a concave lens. (Fig. 2.6) (a) Incident rays Refracted rays Principal axis P F C 2F (b) C 2F Incident rays Refracted rays Lens F P Principal axis f f Fig. 2.6: Refraction of light through lenses 2.3 Terms used in thin lenses Activity 2.3 To find out the meaning of the terms used in thin lenses (Work in pairs or in groups) Materials \u2022 Magnifying lens \u2022 Double convex lenses \u2022 Sunshine Steps \u2022 Piece of paper 1. This activity is done outside the classroom during daylight. 2. Place a dry tissue paper (or dry leaves) on a flat open ground. 3. Place the lens on the tissue paper. 4. Slowly lift the lens upwards away from the paper until a spot of light is formed on the paper. Fig. 2.7: Burning a paper using a lens 36 5. Hold the lens at this position for some time. 6. Observe what happens to the paper. Explain why the paper burns. 7. Repeat the activity with a concave lens. What do you observe? What happens to the beam of light? The paper starts to burn. This shows that a convex lens brings to a focus point light energy from the sun and since light is in form of energy, a lot of it is concentrated at a point. This point where the rays are brought together after passing through the convex lens is called principal focus. This point is real. When the activity is repeated with a concave lens, nothing happens. That means the principal focus of a concave lens is virtual. The following are other common terms used in thin lenses: (a) The centre of curvature (C) The centre of curvature of the surface of a lens is the centre of the sphere of which the lens forms a part (Fig. 2.8 (a) and (b)). For each spherical lens there are two centres of curvature (C1, C2) due to the two curved surface. (b) The radius of curvature (r) The radius of curvature of the surface of a lens is the radius of the sphere of which the surface forms a part (Fig. 2.8 (a) and (b)). Each surface has its own radius of curvature (r1 or r2). (c) Principal axis The principle axis of a lens is a line passing through the two centres of curvature (c1 and c2) as shown in Fig. 2.8. r1 c1 c2 r2 r1 c1 c2 r2 (a) Convex lens (b) Concave lens Fig. 2.8: Principal axis (d) The principal focus A prism always deviates the light passing through it towards its base. A convex lens may be regarded as being made up of large portions of triangular prisms as 37 shown below. The emergent beam, therefore, becomes convergent in a convex lens (Fig. 2.9 (a)). The reverse is the effect in a concave lens (Fig. 2.9(b)). (a) (b) Fig. 2.9: Action of lenses compared with prisms (i) Principal focus of a convex lens Consider a set of incident rays parallel and close to the principal axis of a convex lens (Fig. 2.10). These rays, after refraction through the lens, pass through point F on the principal axis. Since all", " the rays converge at this point, it is called principal focus. Since this point can be projected on a screen, it is said to be a real principal focus. Incident rays Refracted rays Principal axis P F C 2F f Fig. 2.10: Principal focus on a convex lens (ii) Principal focus of a concave lens For a set of incident rays parallel and close to the principal axis of a concave lens, the refracted rays appear to diverge from a fixed point on the principal axis. This point is called the principal focus F, of a concave lens (Fig. 2.11). This principal focus is virtual since it cannot be projected on a screen. Incident rays Refracted rays Lens C 2F F P Principal axis f Fig. 2.11: Principal focus on a concave lens 38 (e) The focal plane When a set of parallel rays are incident on a convex lens at an angle to the principal axis, as shown in Fig. 2.12, the refracted rays converge to a point, on a line passing through F and perpendicular to the principal axis. The plane passing through F is the focal plane. Lens axis Focal plane 90\u00ba F Fig. 2.12: Focal plane of a convex lens 3 1 F F (f) The optical centre (P) F 2 P The optical centre of a lens is a point which lies exactly in the middle of the lens (PA = PB) as shown in Fig. 2.13(a) and 2.13(b). Light rays going through this point go straight through without any deviation or displacement. (a) (b) A P B P Fig. 2.13: Optical centre of a convex and concave lenses (g) The focal length of a lens (f) This is the distance from the optical centre to the principal focus of the lens (see Fig. 2.10(a) and 2.11(b). Biconvex and biconcave lenses have a focal length on each side of the lens. The concept of centres of curvature of the surfaces is required only in drawing the principal axis. Otherwise these points are referred to as 2F, as they are situated at a distance twice the focal length from the centre of the lens (PC = 2PF). 39 Exercise 2.1 1. Distinguish between converging and diverging lenses. 2. Define the following: (a) Principal axis (b) Optical centre 3. Differentiate between the principal focus of the concave and convex lens. 4. How many principal foci does a biconcave lens have? 2.4 Image formation by converging lenses Activity 2.4 (Work in groups) To find out and describe the image formed by converging lenses Materials \u2022 Convex lenses \u2022 Tree, Screen (white wall can act as screen) Steps 1. Place a convex lens between a screen and a far away object e.g. a candle. (See fig 2.14). 2. Adjust the distance between the lens and the screen until the image of the candle is observed on the screen. 2F F v F 2F u Fig. 2.14: Image formation by a far object 3. What are the characteristics of the image formed? 4. In groups, discuss the formation of the image using ray diagrams. Ray diagrams are used to illustrate how and where the image is formed. The following are the important incident rays and their corresponding refracted rays used in the construction of ray diagrams. 40 Ray 1: A ray of light parallel and close to the principal axis, passes through the principal focus F (Fig. 2.15). P F F P Ray 2: A ray of light through the principal focus F emerges parallel to the principal axis after refraction (Fig. 2.16). Fig. 2.15: Ray 1 F P P F Fig. 2.16: Ray 2 Ray 3: A ray through the optical centre, P is undeviated after refraction through the lens (Fig. 2.17). P P Fig. 2.17: Ray 3 2.5 Locating images by simple ray diagrams and describing their characteristics To locate the image of an object, we need a minimum of two incident rays from the object. From the three standard rays discussed above, any two incident rays and their corresponding refracted rays can be drawn to locate the image. If the refracted rays converge, a real image is obtained. If the refracted rays diverge, then a virtual image is obtained. 41 2.5.1 Convex lens Activity 2.5 (Work in groups) Materials To design and describe the characteristics of images formed by convex lenses when the object is at infinity \u2022 White screen \u2022 Lens \u2022 An object at infinity (landscape or a tree) \u2022 Metre rule Instructions 1. In this activity, you will design and carry out an investigation to describe the characteristics of images formed by convex lenses when the object is at infinity.", " 2. Modify the set-up, we used in Activity 1.6 with materials provided to you. Sketch the new set-up. 3. Write a brief procedure for your investigation. Conduct the investigations. Draw the image formed. 4. Describe its characteristics. 5. Suggest some possible sources of errors in your investigation and explain how they can be minimised. Write a report and present it in a class discussion. Note: The distance from the centre of the lens to the screen is nearly equal to the focal length, f, of the lens. (a) Object far away from the lens (at infinity) Since the object is at infinity, all the rays from the object, incident on the lens are almost parallel. The refracted rays converge at a point on the focal plane, as shown in Fig. 2.18. Image characteristics A diminished, real, inverted image is formed at F. F P F I M Fig. 2.18: Object OB at infinity 42 (b) Object OB just beyond C (2F) Activity 2.6 (Work in groups) Materials \u2022 Screen Steps To describe images formed by convex lens when the object is beyond 2F and at 2F \u2022 Lens \u2022 Candle 1. Mark the positions of the principal focus F and 2F on both the sides of the lens with a piece of chalk. 2. Place a lit candle on the table along the principal axis of the lens, slightly away from 2F. 3. Place a white screen, on the other side of the lens, perpendicular to the principal axis of the lens and adjust its position to and fro to the screen and observe what happens. What are the characteristics of the image formed? Real image Lens Object (candle) Screen 2F F v P F u 2F Fig. 2.19: Object beyond 2F 4. Repeat step 3 by placing the candle at 2F and observe what happens. What are the characteristics of the images formed? Fig 2.20 shows the ray diagram to locate the images when the object is beyond C. B 0 2F F P F 2F I M Fig. 2.20: Object OB just beyond 2F 43 Image characteristics A diminished, real, inverted image is formed between F and 2F. (c) Object OB at 2F The ray diagram when the object (candle) was placed at 2F is as shown in Fig. 2.22 below. B 0 2F F P F 2F I M Fig. 2.21: Object OB at 2F Image characteristics A real, inverted image of the same size as the object is formed at 2F. (d) Object OB between 2F and F To design and describe the images formed by convex lens when the object is between F and 2F and at F Activity 2.7 (Work in groups) Materials \u2022 Candle \u2022 Lens \u2022 Screen Instructions 1. Modify the set-up as used in Activity 2.6 by placing the candle between 2F and F. 2. Draw the set-up. 3. Write a brief procedure for your investigation. Carry out the investigation and describe the characteristics of the images formed. 4. Suggest some possible sources of errors in your investigation and explain how they can be minimised. 5. Write a report and present it to a class. 44 The simple ray diagram when the object is between F and 2F is as shown in Fig. 2.22. B 0 2F F P F 2F I M Fig. 2.22: Object OB between 2F and F Image characteristics A real, inverted and magnified image is formed beyond 2F. (e) Object OB at F When the object was at F, the refracted rays are nearly parallel and converge at infinity as shown in Fig. 2.23 below. B O F P F parallel rays Fig. 2.23: Object OB at F Image characteristics A real, inverted, magnified image is formed far away from the lens i.e. at infinity. (cannot be described) (f) Object OB between F and P Activity 2.8 (Work in groups) Materials To describe the image formed by convex lens when the object is between F and P \u2022 Candle \u2022 Lens \u2022 Screen Steps 1. Repeat Activity 2.7 keeping the candle close to the lens, between F and P. Can you get an image on the screen? Describe its characteristics. 45 2. Is the image real or virtual? 3. Where is the image formed? 4. Explain your observations. Lens Eye F Object F Upright, virtual and enlarged image Fig. 2.24: Object between F and P The image formed is virtual and cannot be projected on the screen. An enlarged, upright image can be seen through the lens on the same side with the object (Fig. 2.24) above. A simple ray diagram to locate the image when the object is placed between F and P is as shown in Fig. 2.25 below. M B I F O P F Fig. 2.", "25: Object OB between F and the lens. Image characteristics A magnified, upright and virtual image is formed on the same side as the object. 2.5.2 Concave lens When the object is at infinity, an upright, diminished and virtual image is formed at principal focus F. For all other positions of the object OB, an upright, diminished, virtual image is always formed between F and P (Fig. 2.26). 46 B M I P O F Fig. 2.26: Image formation by a concave lens. Example 2.1 A convex lens has a focal length of 2 cm and a real object 6 cm tall is placed 18 cm from the centre of the lens. By means of an accurate scale diagram, find the position, size and nature of the image formed. Solution Using rays 1 and 3 of the image construction, two incident rays are drawn from B and the corresponding refracted rays through the lens. The refracted rays converge at M where the image of B is formed. Scale chosen for object and image values: 1 cm = 6 cm. BBBBBB OOOOOO FFFFFF 2F2F2F2F PPPP FFFF 2F2FF2F2FF2F2FF III uuuu vvvv MMMMMM Fig. 2.27: Graphical construction of images formed by convex lens The image of O is magnified, inverted and formed at I. IM is the real image formed at 6 cm from the lens. The height of the image is 2 cm. Since the scale is 1 cm represents 6 cm, the image is 36 cm from the lens and the height of the image is 12 cm (Fig. 2.27) above. Example 2.2 A concave lens has a focal length of 2 cm and real object 1.0 cm tall is placed at 3 cm from the centre of the lens. By means of an accurate scale diagram, find the position, size and the nature of the image formed. 47 Solution Scale chosen: 1 cm to represent 1 cm Similar to Example 2.1, draw minimum two incident rays from B and the corresponding refracted rays. Since the refracted rays diverge, a virtual image is formed. The image is 1.2 cm from the lens and the height of the image is 0.4 cm (Fig. 2.28). It is diminished, upright and virtual. Fig. 2.28 Exercise 2.2 1. Name two features of the image formed by a convex lens when: (a) The object is between F and optical centre (b) (c) The object is at infinity. The object is at F. 2. Sketch a ray diagram to show image formation for an object placed between 2 F and F of a converging lens. State four characteristics of the image. 3. (a) If a convex lens picks up rays from a very distant object, where is the image formed? (b) If the object is moved towards the lens, what happens to the position and size of the image? 4. An object 2 cm high is placed 2 cm away from a convex lens of focal length 6 cm. By using an accurate drawing on graph paper, find the position, height and type of the image. 2.6 The lens formula Activity 2.9 Lens formula (Work in groups) 1. Using reference materials or internet research about the relationship between focal length, f, object distance, u, and image distance, v. 2. What is the lens formular? Derive it. 48 BuvOFFPIM 3. How is it important to learning of convex and concave mirrors or curved reflecting surfaces? The lens formula is a formula relating the focal length, image and object distance. Consider a convex lens of focal length, f, which forms a real image IM of an object OB as shown in Fig. 2.29. B 2F O D F P u I 2F M F v f Fig. 2.29: Lens formula Triangles OBP and IMP are similar (3 angles are equal) \u2234 OB IM = OP IP \u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026 (1) Draw a line DP perpendicular to the principal axis where DP = BO. Triangles PDF and IMF are similar (3 angles are equal) \u2234 DP IM = PF IF \u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026 (2) Since DP = OB, from equations (1) and (2), = OP IP u v = PF IF f v \u2013 f Cross multiplying, uv \u2013 uf = vf Dividing both sides by uvf uv uvf 1 \u2013 = \u21d2 \u2013 = u uf uvf vf uvf 1 f 1 v. Hence 1 f 1 = + u 1 v. This is the Lens formula, where u stands for the distance of the object from the optical centre. v stands for the distance of the image from the optical centre. f stands for the focal length of the lens. 49 2.7 Sign", " Convention (Real is positive) We can adopt a method or a convention to describe the upward motion and downward motion. For example let the distances up be negative and down positive or vice versa. \u2234 3 m up = -3 m 3 m down = +3 m There are several sign conventions used when the distances of the object and the image are measured from the lens. In this book, we shall adopt the real is positive in which: 1. All the distances are measured from the optical centre. 2. The distances of the real objects and the real images measured from the optical centre are taken as positive, while those of virtual objects and virtual images are taken as negative. From this convention, the focal length of a convex lens is positive and that of a concave lens is negative. See Fig. 2.30 (a) and (b). P F F +f (a) P \u2013 f (b) Fig. 2.30: Real and virtual focal lengths of lenses Example 2.3 An object is placed 24 cm from the centre of a convex lens of focal length 20 cm. Calculate the distance of the image from the lens. Solution From 20 6 \u2013 5 120 \u2013 = 1 u 1 24 = 1 120 The image distance (v) = 120 cm 50 Example 2.4 An object is placed 2 cm from the centre of a concave lens of focal length 20 cm. Calculate the distance of the image from the lens. Solution From lens formula 20 1 \u2013 10 20 \u2013 = = \u20139 20 v = \u201320 9 = \u20132.2 cm, v = 2.2 cm v is negative because the image is virtual. 2.8 Magnification formula of the lens The term magnification refers to how many times an image is bigger than the object. Linear magnification (m) is defined as the ratio of the height of the image to the height of the object. To derive magnification formula Activity 2.10 (Work in groups) Material \u2022 Graph papers Steps 1. Draw three vertical lines on a graph paper. 18 cm 6 cm 2 cm A B C Fig. 2.31 2. How many times is line B bigger than A. 3. How many times is line B bigger than C. 4. What are the units of these comparisons? 51 Earlier in this unit we have done activities where we saw that the size of images formed by lenses are either bigger or smaller than the object. The increase or decrease in size of an object is called magnification. That is Linear magnification (m) = height of the image height of the object = = image distance (v) object distance (u) = h1 h0 v u Note: Since magnification is a ratio, it does not have units Sometimes it becomes difficult to measure the height of the image or the object accurately. In such cases, magnification can be calculated in terms of distances u and v. For example, consider a convex lens where a magnified image is formed (Fig. 2.32). B 2F O F P F 2F I R u v M Fig. 2.32: Magnification Since triangles OBP and IMP are similar (3 angles are equal), the ratios of corresponding sides are equal i.e, IM IP IM OB = OB OP = IP OP = v u \u2234 Hence magnification, m = IM OB = v u Magnification (m) = image distance (v) object distance (u) or m = v u 52 v Therefore m = = u also equal to the ratio of image to object distances v \u2013 u h1 h0, therefore, the ratio of image to object sizes hi \u2013\u2013 ho is measured from the optical centre. Example 2.5 An object of height 2 cm is placed 20 cm infront of a convex lens. A real image is formed 80 cm from the lens. Calculate the height of the image. Solution hi h0 m = = \u2234 hi = hi 2 = 80 20 = 8 cm \u21d2 v u 8 1 80 \u00d7 2 20 10 1 Example 2.6 An object placed 30 cm from a convex lens produces an image of magnification 1. What is the focal length of the lens? Solution Magnification, m = OB IM = OP IP = 1. (Fig. 2.33) Since m = 1; then v = u This occurs when object is at 2f. Hence 2f = 30 \u2234 f = 15 cm B O P 30 cm I M u v Fig. 2.33: Image formed by convex lens 53 Example 2.7 An object of height 1.2 cm is placed 2 cm from a convex lens and real image is formed at 36 cm from the lens. Calculate (a) the focal length of the lens (b) magnification produced by the lens (c) the size of the image. Solution (a) From lens formula 36 = 1 f 1 f 18 + 1 36 = 19 36 = 1 f 36 19 = 1 f \u21d2 f = 1.89 cm Focal length of the lens =", " 9 cm (b) m = v u = 36 2 = 3 (c) m = hi h0 \u2234 hi = 3 \u00d7 1.2 = 3.6 Size of the image = 3.6 cm Example 2.8 An object of height 2 cm is placed 8 cm from a convex lens and a virtual image is formed on the same side as the object at 24 cm from the lens. Calculate (a) the focal length of the lens (b) the height of the image formed. Solution (a) From lens formula24 = 3 \u2013 1 24 = 1 f 1 f \u2234 focal length, f = 12 cm (v = \u201324 cm because the image is virtual) \u21d2 1 12 = 1 f 54 (b) Magnification m = v u = h1 h0 \u21d2 \u201324 8 = h1 2 (negative sign indicate image is virtual) \u2234 hi = 24 \u00d7 2 8 = 6 cm Example 2.9 A convex lens produces a real image of an object and the image is 3 times the size of the object. The distance between the object and the image is 80 cm. Calculate the focal length of the lens. Solution Magnification m = v u = 3 \u2234 v = 3u \u2026\u2026\u2026\u2026 (1) u + v = 80 \u2026\u2026\u2026\u2026 (2) Solving equations (1) and (2) u + 3u = 80 \u21d2 4u = 80 \u2234 u = 20 cm Hence v = 3u = 60 cm From lens formula 1 f 1 = + u 3 + 1 60 = \u2234 focal length, f = 15 cm 2.9 Power of a lens 1 v = + = = 1 20 4 60 1 60 1 15 To explain what is the power of a lens Activity 2.11 (Work in groups) Material \u2022 A lens Steps 1. Discuss with your classmates what the power of the lens is. 2. Is it possible to increase the power of a lens? Discuss. 3. Share your findings with other classmates. 55 The ability to collect rays of light and focus them at a point in the case of a converging, or to diverge them so that they appear to come from a point in the case of diverging lens is called the power of a lens. It is calculated from its focal length using the formula Power = 1 f The unit for power is the dioptre represented by the symbol D. The f must be in S.I units of length. Example 2.10 A lens has a focal length of 25 cm. Find the power of the lens. Solution f = 25 cm = 0.25 m. The focal length of convex lens = +ve (It forms real image) 1 \u2234 Power = NB: For a concave lens f = -ve (because a concave lens forms a virtual image) +0.25 = +4 m\u20131 \u2234 Power = 1 -0.25 = -4 m\u20131 Exercise 2.3 1. Define the terms: principal axis, optical centre and focal length of a convex lens. 2. With the help of a diagram, show the action of a convex lens as a converging lens. 3. The focal length of a diverging lens is 15 cm. With the help of a diagram explain the meaning of this statement. 4. Fig. 2.34 below shows a convex lens of focal length 15 cm and two rays of light parallel to the principal axis. Copy and complete the diagram to show the path of these rays as they pass through the lens. Label the position of the principal focus as F. P axis Fig. 2.34 5. Draw ray diagrams showing how a convex lens could be used to produce (a) a real and diminished image (b) a virtual and magnified image of a real object. 56 6. Fig. 2.35 is drawn to scale. One incident ray from the object is parallel to the principal axis and other ray passes through the principal focus of a convex lens. Copy and complete the diagram to show the path of the ray through the lens. Hence determine (i) position of the image (ii) the magnification produced by the lens. Convex lens Object Fig. 2.35 7. Copy the table below and put a tick (\uf0fc) in three of the boxes to describe the image formed by a diverging lens. Table 2.1 Magnified Diminished Upright Inverted Virtual Real 8. Draw a diagram to show how a convex lens produce a virtual image. 9. Fig. 2.36 shows two rays of light approaching a thin diverging lens. Copy and complete the diagram and show the path of the rays as they pass through and emerge out of the lens. Label the position of the principal focus F. Fig. 2.36 10. Fig. 2.37 is drawn to scale. An object OB placed in front of a convex lens of focal length 5.0 cm. Copy and complete the diagram", " and (a) show the position of the image (b) find the size of the image 57 Convex lens B O F F Fig. 2.37 11. A convex lens is used to form an upright, magnified image of an object placed 6 cm from the lens. Calculate the focal length of the lens, if the magnification produced is 4. 12. An object 3 cm high is placed 20 cm from a lens of focal length \u201325 cm. Find the position, size and the nature of the image formed. 13. At what distance must an object be placed from a convex lens of focal length 20 cm so as to get real image 4 times the size of the object? 14. An object 3 cm high is placed 30 cm from a convex lens of focal length 20 cm (a) Find the position, size and the nature of the image formed (b) If the same object is now moved by 20 cm towards the lens, calculate the magnification produced by the lens. 15. A convex lens forms a focused image on a screen when the distance between an illuminated object and the screen is 1 m. The image is 0.25 times the size of the object. Calculate (a) the object distance from the lens (b) the focal length of the lens used. 16. An object 3 cm high is placed 150 cm from a screen. Calculate the focal length of the lens that has to be placed between the object and the screen, so as to produce a real image 6 cm high on the screen. 17. An object 6 cm high is placed 30 cm from a diverging lens of focal length 15 cm. With the help of a scale diagram determine (a) the position of the image. (b) the magnification produced by the lens. 18. A real object placed 8 cm in front of a converging lens produces an image at a distance of 2 cm from the lens and on the same side as the object. Calculate the focal length of the lens. 19. A diverging lens of focal length 24 cm forms an image at 18 cm from the lens. Calculate the distance of the object from the lens. 20. In an experiment to determine the focal length of a converging lens, a student obtains the results shown in Table 2.3. 58 Table 2.2 u (cm) v (cm) 21.0 50.0 24.0 40.0 33.0 22.5 36.0 25.0 45.0 22.0 60.0 20.0 (i) Plot a graph of u (x-axis) against v (y-axis) (ii) Using the graph, determine the focal length of the lens. 2.10 Applications of thin lenses 1. The human eye The human eye consists of a nearly spherical ball of about 2.5 cm diameter except for a slight bulge at the front. Fig. 2.38 shows the cross-section of the human eye with the optic nerve leading to the brain. Retina Optic nerve Suspensory ligaments Iris Lens Pupil Cornea ciliary muscles Fig. 2.38: A cross-section of a human eye. The front portion of the eye is known as the cornea and is slightly bulged outwards and is transparent in nature. Behind the cornea, there is a diaphragm called the iris, with a hole in the middle known as the pupil. Behind the iris is a crystalline lens. This is a biconvex converging lens made of a large number of jelly-like layers which are flexible and transparent in nature. The lens is suspended inside the eye by the help of suspensory ligaments which fasten it to the ciliary muscles. These muscles control the shape of the lens. The lens forms a real, diminished and inverted image on the retina. Light falling on the retina produces a sensation in the cells which then send the electrical signals to the brain by the nerve known as the optic nerve. The amount of light reaching the retina is regulated by the size of the pupil. When a bright object is viewed, the iris reduces the size of the pupil so as to admit less light, whereas in dark light the iris contracts so as to admit as much light as possible. An image formed by the eye lens leaves an impression on the retina for about 0.1 second. This persistence of the vision enables us to \u201csee\u201d cinema or television pictures which appear to change smoothly from one image to the next without 59 any interruption. In a cinema theatre or television screen about 20 pictures are projected per second. During the time interval between the pictures, the eye \u201cremembers\u201d the previous picture. Image formation in the eye When one looks at far objects, such as a tree, the eye lens becomes thinner and the focal length of the lens increases. The ciliary muscle is relaxed and the lens has the longest focal length. It is able to focus rays from distant objects onto the retina", " (Fig. 2.39 (a)). To view the objects close to the eye, the lens becomes thicker and the focal length of the lens decreases. The contraction of the ciliary muscle reduces tension in the lens and the lens becomes more curved with short focal length and is more powerful. The lens now focuses images of near objects onto the retina (Fig. 2.39 (b)). For both far view and close view the image formed is real, diminished and inverted. This process by which the lens of the eye changes its focal length and produces focused images of both distant and near objects on the retina is called accommodation. (a) Far view \u2013 thin lens (b) Near view \u2013 thick lens Fig. 2.39: Accommodation of the eye In the eye, the distance between the eye lens and the retina remains the same, whereas the lens automatically changes its focal length according to the distance of the object. This effect is brought about by changing the shape of the ciliary muscle attached to the lens. Defects of vision A normal human eye can accommodate the range of distances from far off objects to objects close to the eye. There is, however, a limit to the power of accommodation of the eye. As a person grows older, the power of accommodation gradually decreases. Also, despite the ability of the eye to adjust its focal length by changing the lens shape, some eyes cannot produce clear images over the normal range of vision. This type of defect may arise due to the eyeball being slightly too long or too short compared to the normal spherical ball or due to the curvature of the cornea being defective. 60 The most common defects of vision are short-sightedness (myopia) and longsightedness (hypermetropia) (see Fig. 2.40, (a) (b) ). (a) Short-sightedness (b) Long-sightedness Fig. 2.40: Short-sightedness and long-sightedness (a) Short-sightedness or Myopia A person suffering from short-sightedness can only see nearby objects. The image of a distant object is formed in front of the retina as shown in Fig. 2.33 (a). This defect arises due to the eyeball being too long or more refraction takes place at the cornea and hence the focal length of the eye lens becomes short. In order to correct this defect, a concave lens of appropriate focal length should be used. This lens diverges the rays from a distant objects so that they appear to come from a virtual image formed at a point closer to the lens. The eye can focus on this virtual image, as shown in Fig. 2.41 (b). Far off objects at infinity convex lens Virtual image (a) Myopic eye (b) Corrective measure Fig. 2.41: Short-sightedness and corrective measure. (b) Long-sightedness or Hypermetropia A person suffering from long-sightedness can see distant objects clearly, but cannot see distinctly objects lying closer than a certain distance. The image of a nearby object is formed behind the retina as shown in Fig. 2.42 (a). The defect arises due to the eyeball being too short or due to the curvature of the cornea being defective and the focal length of the eye lens becoming longer. In order to correct this defect, a convex lens of appropriate focal length should be used. This lens converges the ray from a near object so that they appear to come from a virtual image formed at a point far off from the lens. The eye focuses on this virtual image as shown in Fig. 2.42 (b). 61 Near object Virtual image (a) Hypermetropic eye (b) Corrective measure Fig. 2.42: Long-sightedness and the corrective measure. 2. The lens camera A camera is a device used to take photographs. A human eye, though in principle, is similar to a camera, is far superior than the finest camera ever made by man. Fig. 2.43 (a) shows some parts of a lens camera. Fig. 2.43 (b) shows a commercial camera. Film Lens Diaphragm Shutter (a) (b) Fig. 2.43: A lens camera A camera consists of a converging lens and a light sensitive film or plate enclosed in a light-tight box, blackened from inside. The lens focuses light from an object to form a real, diminished and inverted image on the film. Focusing of objects is done, by adjusting the distance between the lens and the films. The amount of light entering the camera through the lens is regulated with the help of a diaphragm, with an adjustable opening in the middle. Light is admitted by the shutter, which opens for different required intervals of time and then closes automatically. During this interval of time, the film is exposed to light from the object. The film contains light sensitive chemicals that change on exposure to light. The film is developed to", " get what is called a negative. From the negative a photograph (positive) may be printed. Comparison of a lens camera with the human eye Similarities 1. Both use converging lenses 2. Both produce real, inverted, diminished images. 62 3. Both can control the amount of light entering the device. 4. Both are black inside. Differences Camera Eye 1. Focal length of the lens is constant. 1. Focal length of the lens changes with the thickness of the lens. 2. Distance between the lens and the 2. Distance between the lens and the retina film can be altered. is a constant. 3. Focuses objects between a few centimetres from the lens to infinity. 3. Focuses objects between 25cm from the lens to infinity. 4. Form permanent images at the film. 4. Form temporary images at the retina. Example 2.11 A lens camera is used to take photograph of a distant building. A well focused image is formed on the film. The lens of the camera is 6 cm from the film. (a) What is the focal length of the lens? Give reasons for your answer. (b) If the camera is then used to take a photograph of a person 2.0 m away from the lens, without moving the camera, in which direction should the lens be moved in order to produce the best possible image? Solution (a) Focal length of the lens = 6 cm. Since the object is a distant building,the light rays incident on the lens are almost parallel and are brought to focus at the principal focus of the lens. (b) As the object distance is only 2.0 m i.e. object distance, u, has decreased as compared to the distance in part (a). Hence the image distance, v, must be increased. To achieve this, the lens has to be moved forward towards the person. 3. Simple microscope A magnifying glass also known as a simple microscope is an instrument used to view the details of very small objects. It consists of a single converging lens of short focal length. When an object is placed within the focal length of such a lens, a magnified image which is virtual and upright is formed on the same side of the object. This image can be viewed by placing the eye close to the lens. The distance of the object from the lens is adjusted till an enlarged image is formed at a distance D, which is about 25 cm from the unaided eye. The distance D is referred to as 63 the least distance of distinct vision. Fig. 2.44 below illustrates the operations of a simple microscope. M I B O F D Fig. 2.44: A simple microscope. F Eye The action of a simple microscope The object OB, when viewed by an unaided eye, cannot be brought closer to the eye, than the distance D (Fig. 2.45 (a)). Otherwise the image as seen by the eye will not be clearly visible. When the same object is viewed through the magnifying glass, it moves nearer to the eye so that a magnified image is formed at the same distance D as before (Fig. 2.45 (b)). Therefore a simple microscope enables us to bring an object very close to the eye making it appear magnified and yet clearly visible. B O M I D v (a) B O (b) Eye F u Eye Fig. 2.45: Working of a simple microscope 64 Magnifying power of a simple microscope Linear magnification m = IM OB = v u Linear magnification of a lens is also called magnifying power of the instrument. In a simple microscope, the image distance v is negative, as the image is virtual. 1 f Hence, from the lens formula = \u2013 1 v 1 u Multiplying throughout by v and simplifying From the above expression the shorter the focal length of the lens, the greater is the magnifying power of the instrument. Hence a simple microscope uses a converging lens of short focal length. Example 2.12 Calculate the magnification produced by a lens of focal length 5.0 cm used in a simple microscope, the least distance of distinct vision being 25 cm. Solution In this example, the image distance v = D = 25 cm. Magnification, m = 1 + v f = 1 + ( 25 5 ) = 1 + 5 = 6 Hence the magnification produced by the lens = 6. 4. Compound microscope In a simple microscope, the magnifying power cannot be increased beyond a certain limit, by decreasing the focal length of the lens. This is due to the mechanical difficulties of using a lens of very short focal length. A compound microscope uses two separate converging lenses, placed coaxially within two sliding tubes, to obtain a higher magnifying power. The lens O, nearer the object is called the objective lens and the lens E closer to the eye, is called the eyepiece lens. Though both these lenses are of short focal lengths, the eyepiece has a comparatively larger focal length than the objective lens.", " The final image formed is magnified, virtual and inverted as shown in Fig. 2.46. 65 Objective (O) Eye Eyepiece (E) 2F B O F I F 2F B' M Fig. 2.46: A compound microscope Action of a compound microscope The object OB is placed between F and 2F of the objective lens. A real, inverted, magnified image O\u2032 B\u2032 is formed beyond 2F of the objective lens. The position of the eyepiece lens is adjusted so that this image O\u2032B\u2032 falls within its focal length. The eyepiece then acts as a magnifying glass and produces a final magnified, virtual and inverted image IM at a distance of distinct vision D from the eye, placed very close to the eyepiece. If m1 is the magnification produced by the objective lens and m2 is the magnification produced by the eyepiece lens, then the magnification produced by the system of lenses m is given by, m = m1 \u00d7 m2 If the first image O\u2032B\u2032 formed by the objective lens is exactly at the principal focus Fe of the eyepiece lens, then the final image IM will be formed at infinity. The image will be inverted and well enlarged. At this position, the compound microscope is said to be in normal adjustment. A good compound microscope produces a very high magnification. High magnification microscopes are usually used in research work in science (see Fig. 2.47). 66 Fig. 2.47: High magnification microscope for research work Example 2.13 In a compound microscope, the focal length of the objective lens is 2.0 cm and that of the eyepiece is 2.2 cm and they are placed at a distance of 8.0 cm. A real object of size 1.0 mm is placed 3.0 cm from the objective lens. (a) Use the lens formula in turn for each lens to find the position of the final image formed. (b) Calculate (i) the magnification produced by the arrangement of these lenses and (ii) the size of the final image viewed by the eye? Solution (a) For the objective lens Solving this equation gives v = 6 cm As shown in Fig. 2.48, the real image I1M1 is formed at 6 cm from the objective lens. I1M1 acts as an object for the eyepiece (u = 2 cm). 1 For the eyepiece lens f 1 2.2 1 u 1 2 Solving this equation gives v = \u201322 cm 1 v 1 v = + = \u2013 67 The negative sign shows that the image formed by the eyepiece is virtual and is formed on the same side as the object I1M1. The final image I2M2 is at a distance of 22 cm from the eyepiece (see Fig. 2.48). O 3 cm 6 cm 2 cm E Eye I2 M2 B O 22 cm I1 M1 8 cm Fig. 2.48: Arrangement of lenses in a compound microscope (b) (i) The magnification produced by the system of lenses m = m1 \u00d7 m2 for the eyepiece. for the objective lens and m2 = where m1 = v u v u m = 6 3 \u00d7 22 2 = 2 \u00d7 11 = 22 (ii) The size of the final image = size of the object \u00d7 m = 1 \u00d7 22 = 22 mm Topic summary \u2022 A lens is a transparent medium bound between two surfaces of definite geometrical shape. \u2022 Thin lenses may either be converging or diverging. \u2022 A convex lens is thicker at its centre than its edges and converges the light incident on it. \u2022 A concave lens is thicker at its edges than at the centre and diverges the light incident on it. \u2022 The following are some of the important terms used in spherical lenses: principal axis, optical centre, principal focus, focal length. \u2022 The focal length of a convex lens is positive, while that of a concave lens is negative. \u2022 The characteristics of the image formed by a converging lens depends on the position of the object (see Table 2.3). 68 Position of object At infinity (far away) Beyond 2F At 2F Between 2F and F At F Between F and P Table 2.3 Position of image F Nature of image formed real and inverted diminished Size of image formed compared to object Between F and 2F At 2F Beyond 2F At infinity (far away) Same side as object real and inverted diminished real and inverted Same size real and inverted Magnified real and inverted Magnified Virtual and upright Magnified \u2022 A diverging lens always forms a virtual, upright, diminished image between F and P (except when the object is at infinity). \u2022 Magnification (m) is defined as the ratio of the height of the image to the height of the object magnification = height of image height of object = \u2022 Lens formula is given by 1 u + 1 v = image distance", " object distance 1 f where u is the object distance, v the image distance and f the focal length of the lens. \u2022 Short-signtedness and long-sightedness are two most common defects of a human eye \u2022 A lens camera, simple microscope, compound microscope are some examples of optical instruments. 69 Topic Test 2 1. Describe an experiment to illustrate that white light is composite in nature. 2. Fig. 2.49, drawn to scale, shows two rays starting from the top of an object OB incident on a converging lens of focal length 2 cm. B O Fig. 2.49: Equilateral glass prism (a) Copy and complete the diagram to determine where the image is formed. (b) Add one more incident ray from B through the principal focus and draw the corresponding refracted ray through the lens. (c) Calculate the magnification produced by the lens. 3. Fig. 2.50 shows an object placed at right angles to the principal axis of a thin converging lens. Fig. 2.50: Equilateral glass prism (a) Calculate the position of the image formed. (b) Give an application of this arrangement of a lens. (c) Describe the nature of the image formed. 4. Describe with the aid of a ray diagram, how an image is formed in a (i) simple microscope (ii) lens camera. 5. A converging lens is used to form an upright image, magnified 5 times of an object placed 6 cm from the lens. Determine the focal length of the lens 70 FO6 cm8 cmLensBF 6. Fig. 2.51 shows two converging lenses L1 and L2 placed 8 cm from each other. The focal length of the lens L1 is 2 cm and that of L2 is 2.8 cm. An object 1.0 cm high is placed 3 cm from lens L1. Fig. 2.51: Equilateral glass prism (a) Construct a ray diagram to scale, on a graph paper to show the position of the final image as seen by the eye of a person. (b) Determine the magnification obtained by this arrangement. 71 ObjectEye3 cm8 cmL1L2 72 UNIT 2 Forces and Turning Effects Topics in the unit Topic 3: Moment of a Force Topic 4: Centre of Gravity and Equilibrium Learning outcomes Knowledge and Understanding \u2022 The effects of forces and centre of gravity. Skills \u2022 Design tests to locate the centre of gravity of regular objects by method of balancing and locate the centre of gravity of irregular shaped objects by means of a plumb-line. \u2022 Observe carefully. \u2022 Predict what might happen. \u2022 Use appropriate measures. \u2022 Draw a simple diagram to show moment of a force Interpret results accurately. \u2022 Calculate problems related to moments of forces. \u2022 Report findings appropriately. Attitudes \u2022 Appreciate the applications of moment of forces. Key inquiry questions \u2022 Why the pivot is important in taking moment of force? \u2022 What do you understand by couple forces? \u2022 Why cars are designed to have a wide base? \u2022 Why an object cannot be in equilibrium if it is in motion? \u2022 Why an overloaded vehicle is prone to overturn? 73 TOPIC 3 Moment of a Force Unit Outline \u2022 Moment of force \u2022 Principle of moments \u2022 Couple \u2022 Determination of centre of mass of a regular object \u2022 Applications of moment of a force Introduction One of the effects of a force we learnt about in Secondary 1 is that it produces a turning effect on body. But how can we quantify the turning effect? What are some of the applications of this effect in our daily lives? In this topic, we will seek answers to these questions. 3.1 Moment of a force A number of simple machines like levers, pliers, spanners and so on do work when a force produces a turning effect on some of their parts. It is important to know where the force should be applied for the machine to be more efficient in doing work. The following activity will help us to investigate the turning effect of a force. To investigate the turning effect of a force Activity 3.1 (Work in groups) Materials \u2022 A ruler. Steps 1. Balance a ruler on a finger. At what point did the ruler balance? Why do you think it balances at that point? 2. Press the ruler at one end. Observe what happens to the ruler. 3. Repeat the experiment by pressing the ruler on the other end. Why do you think the ruler behaves in such a manner? 74 Moment of a Force Fig. 3.1: Balancing a ruler on a finger In both cases, the ruler turns about the finger. When the force F1 is applied at one end, the ruler turns in anticlockwise direction about the finger. When the force F2 is applied at the other end, the ruler turns in the clockwise direction. Both F1 and F2 produce a turning effect on the ruler about the finger. The turning effect of a force about a point is called the moment of the", " force about that point. This moment depends on the force applied and its distance from the point. The moment of a force about a point is the product of the force and the perpendicular distance from the point to the line of action of the force. The moment of a force about a point is either clockwise or anticlockwise about the point. In Fig. 3.1, the anticlockwise moment about the finger is F1 \u00d7 d1. The clockwise moment about the finger is F2 \u00d7 d2. Moment of a force about a point = Force \u00d7 Perpendicular distance from the point to the line of action of the force = F \u00d7 d SI unit of moment of a force Moment of a force = Force (N) \u00d7 Perpendicular distance (d). Therefore, the SI unit of moment is newton metre (Nm). Moment of a force is a vector quantity since it has both magnitude and direction. Example 3.1 A student applies a force of 10 N to the handle of a door, which is 0.8 m from the hinges of the door (Fig. 3.2). Calculate the moment of the force. 75 Solution Moment of a force about a point hinges = Force \u00d7 perpendicular distance from the point to the force. = (10 \u00d7 0.8) Nm = 8 Nm in the clockwise direction. door 0.8 m F Fig. 3.2: Moment in opening the door Example 3.2 Calculate the moment of the force about the fulcrum when a pet dog of mass 10 kg is at a distance of 1.2 m from the fulcrum of the seesaw as shown in Fig. 3.3. mass = 10 kg fulcrum Solution F = Weight of dog = mg 1.2 m Fig. 3.3: Moment of force = 10 kg \u00d7 10 N/kg = 100 N Moment of the force about the fulcrum = Force \u00d7 perpendicular distance from the fulcrum = 100 N \u00d7 1.2 m = 120 Nm in the clockwise direction. Exercise 3.1 1. Define \u2018moment of a force\u2019 and state its SI unit. 2. A force of 20 N is applied to open the gate of a fence as shown in Fig. 3.4. Calculate the moment of the force about the hinges if the force is applied at the edge of the gate. 76 hinges 1.2 m 0.3 m F 3. Give the scientific reasons for the following: Fig. 3.4: Moment in opening the gate (a) The handle of a door is fixed far from the hinges. (b) A pair of garden shears has small blades and long handles. (c) A lighter boy is able to produce same moment as that of a heavier girl on the seesaw as shown in Fig. 3.5 below. girl boy Fig. 3.5: Moment on a seesaw 4. A person applies a force of 500 N and produces a moment of force of 300 Nm about the wheels of a wheel cart (Fig. 3.6). Calculate the perpendicular distance, d, from the line of action of the force to the wheels. l o a d F = 500N weight d Fig. 3.6: Moments in a wheelcart 77 3.2 The principle of moments The principle of moment gives the relationship between two moments that are at the same turning point (fulcrum). Activity 3.2 (Work in groups) Materials To investigate the principle of moments \u2022 A metre rule \u2022 Three 100 g mass \u2022 string \u2022 support e.g. clamp Steps 1. Suspend a uniform metre rule from a firm support e.g. clamp, at the 50 cm mark, i.e. at its mid point G as shown in Fig. 3.7 (a) using a string. 2. Suspend a 100 g mass at a point A as shown in Fig. 3.7 (b). Why do you think the ruler balances as shown? (a) O (b) 50 cm 100 cm 0 P P A 100 cm 100 g mass F = 1 N Fig. 3.7: Principle of moments. 3. Now suspend a 200 g mass at a point B near the 0 cm mark (Fig. 3.8 (a)). What has happened? The system turns in the anticlockwise direction. Now adjust the position of B till the system balances horizontally as shown in Fig. 3.8 (b). Explain the observations. (a) B O 200 g mass f = 2N P A 100 cm (b) 0 B P A F = 1N 2N F = 1N Fig. 3.8: Principle of moments. The metre rule turns in the clockwise direction. There is a moment of force in the clockwise direction due to the force acting vertically downwards at point A. Moment of the force in the clockwise direction = Force \u00d7 perpendicular distance = 1.0 \u00d7 P\u0391 78 Moment due to the 2", " N force about P is 2.0 \u00d7 PB in the anticlockwise direction. Measure the distance PA and PB. Compare the values of 1.0 \u00d7 PA and 2.0 \u00d7 PB. What can you say about these values? We note that the two moments are equal in magnitude and opposite in direction. The clockwise moment of the 1 N force about point P is equal to the anticlockwise moment of the 2 N force about point P. Activity 3.3 (Work in groups) Materials To design and investigate the principle of moments with more than two forces \u2022 Four mass \u2022 A metre rule \u2022 string \u2022 Firm support e.g. clamp Instructions 1. In this activity, you will design and carry out an investigation to investigate the principle of moments with more than two forces. 2. By modifying the set-up, we used in Activity 3.2, with the materials provided, i.e using its masses, conduct an investigation. Sketch the new set-up. 3. Write a brief procedure and carefully execute the procedure to determine principle of moments with four masses. 4. By applying the relevant formula and relationships. Calculate the moments 0 about point P. 5. Compare your values with other groups. 6. What are some possible sources of errors? How can they be minimised in your investigation? Write a report and present it in a class discussion. 0 D F2 P 50 cm C F1 B A 100 cm F4 F3 Fig. 3.9: Balanced metre rule under action of forces. The sum of the clockwise moments = F3 \u00d7 PA + F4 \u00d7 PB. The sum of the anticlockwise moments = F1 \u00d7 PC + F2 \u00d7 PD. What can you say about F1 \u00d7 PC + F2 \u00d7 PD and F3 \u00d7 PA + F4 \u00d7 PB? 79 A B 100 cm C F1 P 50 cm D F2 Fig. 3.9: Balanced metre rule under action of forces. F3 F4 From Activity 3.2 and 3.3 we can conclude that the sum of the clockwise moments about a point is equal to the sum of the anticlockwise moments about the same point, when the metre rule is balanced. In Activity 3.3, we saw how a body can be balanced by a number of forces. When a body is balanced under the action of a number of forces, it is said to be in equilibrium. The results of Activities 3.2 and 3.3 are summarised in what is known as the principle of moments. It states that, when a body is in equilibrium under the action of forces, the sum of clockwise moments about any point is equal to the sum of anticlockwise moments about the same point. Activity 3.4 (Work in groups) Materials \u2022 A metre rule Steps To verify the principle of moments \u2022 Seven, 50 g masses 1. Consider a uniform metre rule suspended at its mid point P, which is the 50 cm mark. Suspend 3 masses; 200 g, 100 g and 50 g and adjust their positions A, B and C till the system is in equilibrium as shown in Fig. 3.10. 2. Calculate the distances PA, PB and PC in metres and enter the values in a table. 0 A F1 2N P B C 100 cm 50 cm F2 1N F3 0.5 N Fig. 3.9: Verifying the principle of moments. 3. Repeat the experiment by changing the positions of A or B or C or all the three so that the metre rule balances horizontally in each case. Record the results in Table 3.1. Table 3.1 F1 \u00d7 PA (Nm) F2 \u00d7 PB + F3 \u00d7 PC (Nm) PA (m) PB (m) PC (m) 1 2 3 4 4. Complete the table and explain the results. 80 It is seen that the last two columns are equal for each set of results proving that the sum of the clockwise moments is equal to the sum of the anticlockwise moments about point P. F2 \u00d7 PB + F3 \u00d7 PC = F1 \u00d7 PA Example 3.3 A uniform metre rule is pivoted at its centre P, and 3 masses are placed at A, B and C as shown in Fig. 3.10. Find the value for the weight W of the mass M placed at C so that the metre rule is balanced horizontally. 0 cm 50 cm A 2N 10 cm P B 1N 30 cm 40 cm Fig. 3.10: Principle of moment Mass M 100 cm C W Solution Taking moments about P, when the metre rule is in equilibrium. Sum of the clockwise moments = Sum of the anticlockwise moments W \u00d7 0.4 = (2.0 \u00d7 0.3) + (1.0 \u00d7 0.1) (2.0 \u00d7 0.3) + (1.0 \u00d7 0.1) 0.4 W = = 1.75 N Example 3.4 John, Joyce and", " Janet sat on a seesaw as shown in Fig. 3.11 below. Where is John, whose mass is 30 kg seated so that the seesaw is balanced horizontally if the masses of Joyce and Janet are 50 kg and 20 kg respectively? Joyce Janet P John 1m pivot 2 m d Fig. 3.11: See saw at balance Solution John\u2019s weight = 600 N, Joyce\u2019s weight = 500 N, Janet\u2019s weight = 200 N Taking moments about the pivot, Sum of clockwise moments = Sum of the anticlockwise moments about the pivot about the pivot 81 600 \u00d7 d = 500 \u00d7 2 + 200 \u00d7 1 600 d = 1 000 + 200 d = 1 200 600 = 2 m John should sit at a distance of 2 m from the pivot. Example 3.5 The uniform plank of wood in Fig. 3.12 is balanced at its center by the forces shown. Determine the value of W in kg. 2.6 N 24 cm 6 cm 2 N W Fig. 3.12 Solution Note that the 2.3 N produces an anticlockwise moment. Sum of clockwise moments = Sum of anticlockwise moments 0.24 \u00d7 (2 + W) = 2.6 \u00d7 0.30 0.48 + 0.24W = 0.78 \u21d2 W = 0.78 \u2013 0.48 = 1.25 N 0.24 = 0.125 kg To determine the mass of an object using the principle of moments Activity 3.5 (Work in groups) Materials \u2022 A metre rule \u2022 A known mass Steps \u2022 An unknown mass \u2022 Support 1. Suspend a uniform metre rule at its mid point P. Suspend the object of mass m, using a string, from a point A. Suspend a known mass M on the other side of the metre rule and adjust the position of the mass M till the metre rule is horizontal as shown in Fig. 3.13. 82 0 cm A object of mass, m P 50 cm B 100 cm mass M mg Mg Fig. 3.13: Finding unknown mass m. 2. Record the distances PA and PB. Repeat the experiment by changing the position of the object or the mass M. Enter the readings of M, PA and PB in a tabular form as shown in Table 3.2. Table 3.2 Mass M(g) PA(cm) PB(cm) M.PB m = (g) PA 1 2 3 3. Calculate the mean value for the mass of the object from the last column. Mean Taking moments about P, (mg) \u00d7 PA = (Mg) \u00d7 PB, m \u00d7 PA = M \u00d7 PB, (g cancels out) M \u00d7 PB PA \u2234 m = Exercise 3.2 1. State the principle of moments. 2. A boy of mass 20 kg sits on one side of a log of wood and 10 m away from the pivot. A girl of mass 30 kg sits on the opposite side of the log. How far is the girl from the pivot. 3. State two conditions for a system to be in a state of equilibrium. 4. Weights of 25 N, 28 N and 8 N were suspended on a uniform plank of wood pivoted at its center on a knife-edge. Fig. 3.14 shows the plank immediately after it was placed on a knife-edge. 83 Fig. 3.14 (a) Work out: (i) Sum of Clockwise moments (ii) Sum of Anticlockwise moments (b) Is the bar in a state of equilibrium? Give a reason for your answer. 5. A uniform metre rule is pivoted at the center. It is balanced by weights of 8 N, F and 24 N suspended at 34 cm, 43 cm and 30 cm marks respectively (Fig. 3.15). Calculate the value of F. 50 cm 60 cm 46 cm 34 cm 0 cm 100 cm 8 N F 24 N Fig. 3.15 6. Fig. 3.16 below shows a modern bar balanced by forces of 210 N, 100 N and 25 N. Calculate the distance d. 210 N 6 m 5 m d m 25 N 100 N Fig. 3.16 3.3 Couple Couple refers to two parallel forces that are equal in magnitude, opposite in sense and do not share a line of action. Its effects is that it creates rotation without translation as shown in Fig. 3.17. 84 Fig. 3.17: A couple A couple produces a turning effect on a body. Moment of a couple Fig. 3.18 shows a couple acting on bar AB. F A d C B F Fig. 3.18: Moment of a couple Since the pivot is at C, the moment of the force F acting at point A = F \u00d7 AC, in the clockwise direction. Similarly, the moment of the force, F acting at point B = F \u00d7 BC, also in the clockwise direction. \u2234 The moment of the couple = (F", " \u00d7 AC) + (F \u00d7 BC) = F (AC + BC), but AB = BC + BC = F \u00d7 AB = F \u00d7 Arm of the couple. The moment of the couple, called the torque which is defined as the total rotating effect of a couple and is given by the product one of the forces and the perpendicular distance between the forces. Hence, in Fig. 3.18, Torque = F \u00d7 perpendicular distance AB. SI unit of torque is the newton-metre (Nm) 85 Example 3.6 In Fig. 3.19, each force is 4 N and the arm of the couple is 20 cm. Calculate the moment of the couple. F = 4 N 20 cm Fig. 3.19: Moment of a couple F = 4 N Solution The moment of the couple = F \u00d7 perpendicular distance = 4 N \u00d7 0.20 m = 0.80 Nm. Some common real life examples of a couple are observed when: \u2022 Forces are applied by hands to turn a steering wheel of a motor car (Fig. 3.20 (a)) or the handle bars of a bicycle. \u2022 A water tap is opened or closed (Fig. 3.20 (b)). \u2022 A corkscrew is twisted into a cork in the mouth of a bottle. (Fig. 3.20 (c)). Fig. 3.20: Moment of couple 86 Exercise 3.3 1. Fig. 3.21 is a water tap in use. If the diameter of a circular path made by the tap (knob) when is open and closed is 20 cm. Calculate the moment of the couple. 2. Calculate the torque in Fig. 3.22 below. Fig. 3.21: Moment of couple of a tap 14 cm 14 cm 8N Fig. 3.22: Torque 3. A steering wheel of a truck has a diameter of 30 cm. If the driver is holding the wheel with both hands, while negotiating a corner, calculate the force applied by the right hand if the left hand is pulling the wheel by a force equal to 200 N. 3.4 Centre of mass of a body In one of his experiments, Sir Isaac Newton showed that bodies experience a force of gravity exerted on them by the earth. This force of gravity is always directed towards the earth\u2019s centre and is called the weight of the body. How is this weight distributed throughout the body? The answer to this question is found in the following experiment. 3.4.1 Determining centre of mass of regular objects Activity 3.6 (Work in groups) Materials To determine the weight of a beam (uniform metre rule) \u2022 Uniform metre rule \u2022 A mass, m \u2022 A knife edge (fulcrum) 87 Steps 1. Balance a uniform metre rule of mass m on a fulcrum and adjust its position until the metre rule is horizontal. Note the position P, where it is pivoted (Fig. 3.23(a)). 2. Move the fulcrum to a point A, say to the right of P. Observe what happens. Why do you think the ruler changes its state of equilibrium. 3. Place a mass M between A and 100 cm mark and adjust its position B until the metre rule is horizontal (Fig. 3.23(b)). (a) (b) Fig. 3.23: Determination of mass of a uniform metre rule 4. Find the lengths PA and BA. Repeat the experiment by changing the position of A or the mass M. 5. Record the mass of M, length PA and length BA in a table (Table 3.3). Table 3.3 Mass M(g) BA(cm) PA(cm) m = (g) M.BA PA 1 2 3 6. Take moments and determine the value of m (the mass of the metre rule). Mean m = Taking moments about A mg \u00d7 PA = Mg \u00d7 BA m = M \u00d7 BA PA (g cancels out) Calculate the mean value for the mass of the metre rule from the last column of Table. 3.4. The weight of the metre rule = mg. 88 Example 3.7 A uniform metre rule pivoted at the 30 cm mark is kept horizontal by placing a 50 g mass on the 80 cm mark. Calculate the mass of the metre rule (Fig. 3.24). Fig. 3.24: Determining the mass of the metre rule Solution Let the mass of the metre rule be m Force due to m = m \u00d7 g where g = 10 N/kg = 10 m newtons Force due to 50 g mass = PA = 10 cm = 0.1 m AB = 20 cm = 0.2 m 50 1 000 \u00d7 g = 0.05 \u00d7 g = 0.50 N. By the principle of moments, taking moments about point A, 10 m \u00d7 0.1 = 0.50 \u00d7 0.2 m = 0.50 \u00d7 0.2 10 \u00d7 0.1 =", " 0.10 1 = 0.1 kg or 100 g Example 3.8 A coffee table of mass 22 kg and length 1.3 m long is to be lifted off the floor on one of its shorter sides to slip a carpet underneath. Calculate the maximum force needed to lift the table. 89 Solution Fig. 3.25 shows the forces acting on the table where F is the lifting force and 220 N is the weight of the table acting at the center. F A 0.8 m 0.8 m 220 N Fig. 3.25 Taking the movement about point A Sum of Clockwise moments = Sum of Anticlockwise moments 220 N \u00d7 0.8 m = F \u00d7 1.6 m \u21d2 F = 220 \u00d7 0.8 1.6 = 110 N \u2234 The minimum force required = 110 N Exercise 3.4 1. A uniform metre rule of uniform width 2.5 cm and thickness 0.5 cm is suspended at the 78 cm mark and kept balanced by hanging a mass of 150 g at the 100 cm mark (Fig. 3.26). Calculate, (a) the mass of the metre rule, (b) the density of the material of the metre rule, (c) the tension T in the string. Fig. 3.26: A system at balance 2. A uniform metre rule is balanced at the 20 cm mark by a mass of 240 g placed at one end. (a) Draw a diagram to show the state of balance of the metre rule (b) Determine the weight and mass of the metre rule. 3. A non-uniform plank AB shown in Fig. 3.27 is balanced when a force of 200 N is applied at the end B. The centre of gravity, G, is shown. 90 Fig. 3.27: Centre of mass of a plank Calculate the: (a) weight and (b) mass of the plank. 3.5 Applications of moment of a force The following are some of the common examples which illustrate the turning effect of a force i.e moment of a force: 1. Opening or closing a door. 2. Opening a bottle using a bottle opener (Fig. 3.28(a)). 3. A pair of scissors or garden shears in use (Fig. 3.28(b)). 4. Children playing on a \u201csee-saw\u201d. 5. A wheelbarrow being used to carry some load (Fig. 3.28(c)). 6. A wheel cart being used to lift heavy loads. 7. A screwdriver being used to tighten/loosen a screw. 8. A crowbar being used to move large object (Fig. 3.28(d)). Load Pivot P Effort Effort P Load Pivot (a) Bottle opener (b) Scissors 91 Load Effort Pivot Load Pivot Effort (c) Wheelbarrow (d) Crowbar Fig. 3.28: Applications of moment of a force Note In all these tools the length of the handle determines the amount of effort to be used. Activity 3.1 will help us to understand this. Activity 3.7 (Work in groups) Materials To investigate the effect of length to the effort needed in using a tool Exercise books, 3 cm and 15 cm rulers, doors. Steps 1. Lift a book using a 15 cm ruler. Repeat the same using a 30 cm ruler. Why is it easier to lift a book using a longer ruler than the shorter one? 2. Open the classroom door with your hand near the door hinge. Now, open the same door with your hand far away from the door hinge. Of the activities above, when do you use less effort? We use less effort when using 30 cm ruler to lift a book. Several experiments have been done on several tools and machines and proved that: (a) The longer the handle, the lesser the effort used when using the machines. (b) The shorter the handle, the more the effort used while working with the tools and machines. Therefore, the manufacturers always design tools and machines such as bottle openers, see-saw, water taps, spanners and wheelbarrows with longer handles so that very little effort be used in working with them. 92 Exercise 3.5 Sketch and locate the effort, pivot, load in each of the following tools and machines. (a) (d) (g) (b) Water tap (e) (h) (c) See-saw (f) Hammer (i) Bottle opener Spanner Spade pliers Broom (j) Arm Spoon Topic summary \u2022 The moment of a force about a point is the turning effect of the force about the point. \u2022 The turning effect of the force depends upon the magnitude of the force applied and the perpendicular distance from the pivot. \u2022 The moment of a force is the product of the force applied and the perpendicular distance from the pivot to the line of action of the force. \u2022 The moment of a force is a vector quantity and its SI", " unit is Newton metre. \u2022 Principle of moments states that the sum of the clockwise moments about a point is equal to the sum of the anticlockwise moments about the same point, when the system is in equilibrium. Topic Test 3 1. (a) State the principle of moments. (b) Describe an experiment to prove this principle using two known masses and a uniform metre rule. 2. A uniform metre rule is suspended at the 50 cm mark and a stone at the 0 cm mark. The metre rule is balanced horizontally when a mass of 100 g is suspended at the 30 cm mark. Calculate the weight of the stone. 3. In the diagram shown (Fig. 3.29), calculate the value of the unknown mass M, when the plank is balanced horizontally. 0 M 1 m 10 kg 5 m 4 m 10 m 20 kg 8 m Fig. 3.29: A plank at balance 4. Jane and James are seated at 3 m and 2 m respectively from the centre of a seesaw on one side and Jack at 4 m from the centre on the other side. The seesaw is balanced horizontally. Find the weight of Jack, if the masses of Jane and James are 40 kg and 30 kg respectively. 93 5. Fig. 3.30 shows a uniform plank pivoted at its centre. Where should a 5 kg mass be attached if the plank is to be perfectly horizontal? 4 m 5 m 2 kg 1 kg Fig. 3.30: A uniform plank at balcon 6. A uniform metre rule is balanced horizontally at its centre. When a mass of 5 g is suspended at the 4 cm mark, the rule balances horizontally if a mass M is suspended at 30 cm mark. Calculate M. 7. Fig. 3.32 shows a uniform metallic metre rule balanced when pivoted at the 30 cm mark under the conditions shown on the diagram. Fig. 3.31: Moments in a uniform rule (a) Redraw the diagram showing all the forces acting on the metre. (b) Calculate the weight W of the metre rule. 8. Name four importance of moments in our daily life. 9. Fig. 3.32 represents a type of safety valve that can be fitted to the boiler of a model steam engine. 60 cm 20 cm Pivot F Steam Water Heat Fig. 3.32 (a) Briefly explain how the system works.. (b) What minimum pressure should the steam have in order to escape through the plate which has a cross sectional area 10.3 cm2? (Take atmospheric pressure = 105 Pa) 94 TOPIC 4 Centre of Gravity and Equilibrium Unit outline \u2022 Definition of centre of gravity and centre of mass. \u2022 Determination of centre of gravity of regular and irregular objects. \u2022 Effects of the position of centre of gravity on stability of objects. Introduction In our day to day experiences, we may have come across statements such as \u2018\u2018that object is not stable on the table\u201d or \u201cthat overloaded bus is not stable on the road\u201d. Have you ever asked yourself what factors control the stability of an object? In this topic, we will study the factors that affect stability of objects and centre of gravity. Activity 4.1 To locate the centre of gravity of a book (Work individually or in groups) Materials: Exercise book Steps 1. Take your exercise book and try to balance it horizontally on your finger as shown in Fig 4.1 below. Fig. 4.1: Balancing a book (a) What do you observe? (b) Why do you think the book balances at only one point? (c) What do you think is special about the point where the book balances? 2. Discuss with your group members your observations and thoughts in 1(a),(b) and (c). 95 By going through the following discussions, you will be able to answer questions 1(a) to (c) in Activity 4.1. 4.1 Centre of gravity and centre of mass of a body In one of his experiments, Sir Isaac Newton showed that bodies experience a force of gravity exerted on them by the earth. This force of gravity is always directed towards the earth\u2019s centre and is called the weight of the body. How is this weight distributed throughout the body? The answer to this question is found in the following activity. 4.1.1 To investigate where the weight of a body acts Activity 4.2 To find the centre of gravity of a regular body (Work in pairs or in groups) Materials: A table, thin rectangular card Steps 1. Place a thin rectangular cardboard near the edge of the bench top. 2. Pull the card slowly away from the bench until it is just about to topple over then released as shown in Fig. 4.2 (a). 3. Using a ruler, mark and draw the line AB along which the card balances. 4. Repeat the activity with the other side of the card, mark and draw the line CD along which the card balances. The lines AB and CD intersect", " at a point M (Fig. 4.2(b)). table thin rectangular card pull B M D C (a) A (b) Fig. 4.2: Location of a point where the weight of the body acts 4. Now, try to balance the card with the point M placed at the tip of your fore finger. What do you notice about the state of equilibrium of the card? Suggest a reason for this observation. From Activity 4.2 you should have observed that the cardboard balances horizontally at point M only. This shows that although the mass of the cardboard is distributed over the whole body, there is a particular point, M, where the 96 whole weight of the cardboard appears to be concentrated. When pivoted at this point the cardboard balances horizontally. This point, M, is called centre of gravity of the cardboard. The centre of gravity of a body is the point from which the whole weight of the body appears to act. The centre of gravity of an object is constant i.e. at the one location when a body in a place with uniform gravitational field strength. However, the centre of gravity of a body moves to a different location when the body is placed in a region with non-uniform gravitational field strength. Centre of mass of an object on the other hand is the point where all the mass of the object is concentrated. Since the mass of an object is constant and is not affected by pull of gravity, the location of the centre of mass of an object is constant i.e. does not change. In places like on earth where the gravitational field strength is uniform, the centre of mass and the centre of gravity coincide i.e. are at the same point. However, the two centres are at different locations for the same object if the object is placed in a place with non-uniform gravitational field strength. 4.2 Centre of gravity (c.g) of regular lamina Activity 4.3 To locate the centre of gravity of a regular lamina (Work in groups) Materials: manila paper, ruler, pencil Instructions 1. In this activity you will conduct an investigation to locate centre of gravity of a regular lamina. 2. Using your geometrical instruments, and the other materials provided, come up with a method of determining the centre of gravity of a regular lamina. Write a brief procedure and execute it. 3. Balance them at their centres of gravity on the tip of your pencil. 4. Practically locate their centres of gravities by drawing. In the previous topic, you did Activity 3.6 on determining the weight of a beam. (uniform metre ruler). You were able to realise that the weight of the uniform metre rule tends to act at central point of the ruler that is at 50 cm mark. A thin cardboard like the one used in Activity 4.2 is a lamina. The cover of a book is a lamina. The 97 set square or protractor in your mathematical set are all examples of laminae. Experiments have shown that bodies with uniform cross-section area and density have their centres of gravity located at their geometrical centres. For example, a metre rule of uniform cross-sectional area and density has its centre of gravity located at the 50 cm mark. Fig. 4.3(a) - (d) shows the centre of gravity (c.g) of rectangular, square, triangular, and circular laminae. c.g c.g c.g c.g (a) rectangle (b) square (c) isosceles triangle (d) circle Fig. 4.3: c.g. of regular lamina 4.3 Centre of gravity (c.g) of irregular lamina Activity 4.4 (Work in groups) To determine the centre of mass of an irregular lamina using a plumbline Materials: An irregular lamina, plumbline, a drawing pin Steps 1. Guess and mark the centre of gravity of an irregular lamina. 2. Make three holes P, Q and R on an irregularly shaped lamina as close as possible to the edges and far away from each other. The holes should be large enough to allow the lamina turn freely when supported through a drawing pin. 3. Suspend the lamina on the clamp using the drawing pin through each hole at a time. 4. Suspend a plumbline (a thin thread with a small weight at one end) from the point of support, P as shown in Fig. 4.4(a), and draw the line of the plumbline on the lamina by marking two points A and B far apart and joining them. 5. Repeat the steps with the support Q and mark the point M where the two lines intersect. 6. Check the accuracy of your method by suspending the lamina at R. What do you observe? Explain. 7. Comapre the c.g determined experimentally and that you guessed earlier. 98 irregular lamina P R A B Q smooth support e.", "g drawing pin hole string (plumbline) weight P A R M Q B (a) (b) Fig. 4.4: Locating centre of mass of a lamina using a plumbline 8. The plumbline pass through M (Fig. 4.4 (b)). Check the results again by balancing the lamina about point M. What do you observe? The lamina balances horizontally at point M. Point M is the centre of gravity of the lamina. This activity proves that when a body is freely suspended it rests with its centre of gravity vertically below the point of suspension. Activity 4.5 (Work in groups) To demonstrate how to locate the centre of gravity of an irregular object using a straight edge Materials: An irregular lamina, a prism Steps 1. Balance a lamina on the edge PQ of a prism as shown in Fig. 4.5(a). Mark the points A and B on the lamina and join them. 2. Repeat the activity for another position and note the points C and D on the lamina. Join C and D. knife edge P A lamina B Q A M C D B (a) (b) Fig. 4.5: Locating the centre of mass of a lamina using a straight edge 3. Label the point of intersection of lines AB and CD as point M. 4. Try balancing the lamina at point M on a sharp pointed support. What do you observe? Explain. 99 From Activiy 4.5, you should have observed that the lines AB and CD intersect at M; the centre of gravity of the lamina. 4.4 Effect of position of centre of gravity on the state of equilibrium of a body Activity 4.6 (Work in groups) To define and describe three states of equilibrium Materials: Internet, reference books Steps 1. Conduct a research from books and the internet on the meaning of the term equilibrium in regard to forces acting on an object. 2. State the three states of equilibrium. 3. Describe each of the three states of equilibrium. 4. Compare and discuss your findings with other groups in class. In your discussion, you should have noted that, the state of balance of a body is referred to as the stability of the body. Some bodies are in a more stable (balanced) state than others. The state of balance of a body is also called its state of equilibrium. Activities 4.7, will help us distinguish between the different states of equilibrium. 4.4.1 States of equilibrium Activity 4.7 To investigate the three states of equilibrium (Work in groups) Materials: plastic thistle funnel, bench Steps 1. Place the funnel upright with the wider mouth resting on the bench (Fig 4.6). 2. Displace the funnel slightly upwards as shown in Fig 4.6 (b) and then release it. What do you observe? 100 force F C W G (a) A C A force F G W (b) Fig 4.6: To show stable state 3. Explain the behaviour of the funnel in terms of the changes in the position of centre of gravity. 4. Place the funnel upright with the narrower mouth resting on the bench as shown in Fig 4.7 (a). 5. Displace the funnel slightly with your finger. What about change of state of the funnel? G C force F G W A (a) funnel topples down force F C G W (b) Fig 4.7: To show unstable state 6. Explain the behaviour of the funnel in terms of the change in position of its centre of gravity in this activity. 7. Place the funnel horizontally as shown in Fig 4.8 (a). 8. Displace the funnel gently by tapping it with a finger. What do you observe? table G W F G W F (a) (b) Fig. 4.8: To show neutral state 101 9. Explain the behaviour of the funnel in terms of the change in position of the centre of gravity when displaced slightly. When a body is resting with its centre of gravity at the lowest point, it is very stable. When displaced slightly; its centre of gravity is raised and when it is released, the object falls back to its original position to keep its centre of gravity as low as possible. This type of equilibrium is known as stable equilibrium. Thus, the funnel in Activity 4.7 Fig. 4.6 (a) was in stable equilibrium. Our finances and keeping the environment clean!! Note that we have used a plastic thistle funnel instead of a glass one. The latter has high chances of breaking. Any time we break a laboratory apparatus, we think of its effects on the school finances as it has to be replaced. Sometimes we may be required to pay ourselves hence affecting the finances of our parents. In case you use glass funnels, be careful when using them. As they may break and cause injuries to you or group members. If it breaks accidentally, collect the broken pieces and dispose them to keep the environment clean.", " When an object is resting with its centre of gravity at a very high position from the base support, it is unstable. When displaced slightly, it continues to fall up to the lowest possible position in order to lower its centre of gravity. This state of stability is known as unstable equilibrium. The funnel in Fig 4. 7(a) was in unstable equilibrium. When an object is resting such that the position of its centre of gravity remains at the same vertical position even when the object is displaces, it is said to be in neutral equilibrium. The funnel in Fig 4.8 (a) was in neutral equilibrium. 4.4.2 Relationship between position of centre of gravity and stability Activity 4.8 To investigate the relationship between position of centre of gravity of a body and its stability (Work in pairs or groups) Material: reference material Discuss with your group members why: 1. Buses sometimes carry heavy luggage at their roof tops. 2. Buses have their luggage bonets located underneath. 3. A person carrying two buckets full of water is more stable that one carrying one bucket. 102 4. Discuss and compare your explanations with your partner and report to the whole class. A body is more stable when its heavy part is as low as possible since it lowers the position of the centre of gravity. If the heavy part of the body is at high position or if the light part of the body in high position is made heavier than the lower position, the body becomes unstable and thus likely to topple over and can cause accidents like in the case of a vehicle carrying heavy luggage at its roof top. Exercise 4.1 1. Draw the figures below in your notebook and identify the centre of gravity of each. (a) (d) (b) (e) (g) (h) (c) (f) (i) Fig 4.9: Various types of figures 2. Fig 4.10 shows a Bunsen burner at different states of equilibrium. (i) (ii) (iii) Fig 4.10: A bunsen burner at different states of equilibrium 103 (a) Name the states in which the Bunsen burner is at in (i), (ii) and (iii). (b) Describe each state named in (a) above. 3. With aid of a diagram, describe how you can determine the centre of gravity of an irregular plane sheet of metal. 4. State and explain the states of equilibrium in Fig 4.11. Fig 4.11: A sphere 4.5 Factors affecting the stability of a body Activity 4.9 (Work in groups) To design and investigate factors that affect the stability of a body Materials: Plastic thistle funnels, Benches Instructions 1. By modifying the set-up, we used in Activity 4.7, using the materials provided, conduct an investigation on factors affecting stability of a body. 2. Sketch the step-up and write a brief procedure to investigate the factors. 3. Execute the procedure and answer the following questions (a) What happens to the funnel when the vertical line through the centre of gravity falls outside the base of the funnel? Deduce the factors that affect stability of the funnel. 4. Write a report and present it in a class discussions. 5. What are some of the sources of errors in the experiment and how can they be minimised? The funnel is more stable when its c.g is at a very low position and vice varsa. In addition, the activities show that the wider the base the more stable a body is. Activity 4.9 further shows that the funnel becomes unstable when the vertical line drawn through the centre of gravity falls outside the base that supports the body. In summary, a body is more stable if: 1. the centre of gravity is as low as possible. 2. the area of the base is as large as possible, and 3. the vertical line drawn from the centre of gravity falls within its base. 104 4.6 Applications of the position of centre of gravity Activity 4.10 (Work in groups) To describe the applications of the position centre of gravity Materials: reference books, Internet Steps 1. Conduct a research from Internet and reference books on the applications of position of centre of gravity. 2. In your research also find out: (a) Why a bird toy balances on its beak? (b) Why it is not advisable to stand on a small boat on the surface of the water? (c) Why one leans to the opposite direction when carrying a load? (d) Why the bus chassis is made heavier than the other parts of the bus? (e) How a tight-rope walker balances himself/herself? 3. Discuss your findings with other groups in class. Do you have the same explanations? 4. Have a class presentation on your findings from your research. In your research and discussion, you should have learnt the following: 1. The balancing bird is a toy that has its centre of gravity located at the tip of the beak. The bird", " balances with its beak resting on one finger or any other support placed underneath the beak, and the rest of the body in the air. This is because it is designed with its centre of gravity at that point (Fig 4.12). Fig. 4.12: Balancing bird toy 105 2. People in a small boat are advised neither to stand up nor lean over the sides while in the boat. This is because when they stand, they raise the position of the centre of gravity making the boat unstable and more likely to tip over (See Fig 4.13). Fig. 4.13: People in a boat 3. A person normally leans to the opposite direction when carrying heavy loads with one hand e.g. a bucket full water. This helps to maintain the position of the c.g to within the base of the person in order to maintain stability (See Fig 4.14) Fig. 4.14: Leaning while carrying heavy load 4. Most buses have their cargo in compartment in the basement instead of the roof rack in order to keep the centre of gravity of the buses as low as possible (Fig. 4.15). 106 Fig. 4.15: Buses carry their cargo below passengers\u2019 level. 5. A tight-rope walker carries a pole to maintain stability. By swaying from side to side, he/she ensures that the vertical line drawn from his/her centre of gravity falls within the feet on the rope in order to maintain stability. (Fig. 4.16). tight-rope Fig. 4.16: A tight-rope walker carries a pole to maintain balance. Topic summary \u2022 The centre of gravity, c. g, of a body is the point where the whole weight of the body appears to act from. \u2022 Centre of mass of an objects is the point where all the mass of the object is concentrated. \u2022 The centre of gravity of a regular lamina or object is at its geometric centre. \u2022 The centre of gravity of a lamina can be found using a plumbline or by balancing it on a knife edge. \u2022 A body is said to be in stable equilibrium if it returns to its original position after being displaced slightly. \u2022 A body is in unstable equilibrium if on being slightly displaced, it does not return to its original position. \u2022 A body is said to be in neutral equilibrium it moves to a new position \u2022 but maintains the position of the c.o.g above its base support. Bodies can be made more stable if their centres of gravity are made as low as possible and the bases are made as broad as possible. 107 Topic Test 4 1. Define the term centre of gravity?. 2. Differentiate between centre of mass and centre of gravity. 3. Redraw the figures shown in Fig 4.17 below and indicate their centres of gravity. 4. Describe how you can determine the centre of gravity of the lamina shown in Fig 4.18. Fig. 4.17: Solids Fig. 4.18: Irregular shape 5. Fig. 4.19 shows a marble in three types of equilibrium. State and explain the type of equilibrium in each case. Fig. 4.19: Marble in three state of equilibrium 6. What is stability? 7. One vehicle which was travelling from Juba to Gulu was seen carrying heavy goods on its roof top and some of its passengers in the vehicle were standing. Discuss why the vehicle is likely to topple if it negotiates a corner at high speed. 108 8. Explain why a three-legged stool design is less stable than a four legged one. 9. Explain the following: (a) The passengers of a double-decker bus are not allowed to stand on the upper deck. (b) A racing car is made of a heavy chassis in its lower parts. (c) When one is alighting from a moving vehicle, it is advisable to spread out his/her legs. My safety Do not stand in a moving vehicle. Let us observe traffic rules. 109 110 UNIT 3 Work, Energy and Power Topics in the unit Topic 5: Work, Energy and Power Learning outcomes Knowledge and Understanding \u2022 Understand the concepts of work, energy and power. Skills \u2022 Design tests to relate work done to the magnitude of a force and the distance moved, power to work done and time taken, using appropriate examples. \u2022 Observe carefully. \u2022 Predict what might happen. \u2022 Use appropriate measures. \u2022 Collect and present results appropriate in writing or drawing. \u2022 Interpret results accurately and derive kinetic and potential energy formula. \u2022 Report findings appropriately and relate work, energy and power. Attitudes \u2022 Appreciate that food eaten is energy. Key inquiry questions \u2022 How does the world get its energy? \u2022 Why is energy not destroyed? \u2022 Why work done is energy? 111 TOPIC 5 Work, Energy and Power Unit outline \u2022 Forms of energy \u2022 Transformation of kinetic energy to potential energy and vice verse. \u2022 \u2022 Different ways to conserve energy \u2022 Law of conservation of mechanical energy", " Sources of energy Introduction Everyday, we do many types of work. We work in the offices, in the farms, in the factories etc. To make our work easier, we use machines ranging from simple tools to sophisticated machinery. Different machines or people do work at different rates (known as power). The ability and the rate of doing certain amount of work depends on how much energy is used. In this topic, we will seek to understand these three terms i.e work, energy and power from the science point of view. 5.1 Work Activity 5.1 To distinguish cases when work as defined in science is done or not (Work individually or in groups) Materials: a chart showing people carrying out different activities, pieces of chalk, pen, chair, desk. Steps 1. Conduct research from books on the scientific definition of work. 2. Walk from your chair to the chalkboard and write the word \u2018work\u2019 on the chalkboard. 3. Collect any litter in your classroom. Be responsible Always keep where you live clean. It is good for your health. 112 4. Carry your chair to the front of you classroom and sit on it. 5. Push against a rigid wall of your classroom. 6. Discuss with your colleagues whether scientifically speaking work, is done in steps 1, 2, 3 and 4. What do think is the meaning of \u2018work\u2019? 7. Now, look at the activities being performed by the people in Fig. 5.1 below. (a) (b) (c) (d) Fig. 5.1: People performing different tasks 8. According to the scientific definition of work, in which of the diagrams above is the person doing work? Explain. 9. Give other examples of doing work. Work is only said to have been done when an applied force moves the object through some distance in the direction of force. Therefore in Activity 5.1, work was done in steps 2, 3 and part of 4 (when carrying the desk). However, no work was done when you sat on your chair without moving in step 4 and pushing the wall without moving it in step 5. 113 Similary, in Fig 5.1, work is being done in (a) and (d) only. When the girl applies a force to a wall in (b) and even becomes exhausted, she is not doing any work because the wall is not displaced. When the woman carries the basket on the head, she is not doing any work. This is because she exerts an upward force on the basket which is balanced by the weight hence there is no motion of the basket in the direction of the applied force. Definition of work Work is defined as the product of force and distance moved in the direction of the force. i.e Work = force \u00d7 distance moved in the direction of the force W = F \u00d7 d The SI unit of work is joule (J). A joule is the work done when a force of 1 newton moves a body through a distance of 1 metre. 1 joule =1 newton \u00d7 1 metre Bigger units used are kilojoules (1 kJ) = 1 000 J Megajoule (1 MJ) = 1 000 000 J Note: Whenever work is done, energy is transferred. Example 5.1 Find the work done in lifting a mass of 2 kg vertically upwards through 10 m. (g = 10 m/s2) Solution To lift the mass upwards against gravity, a force equal to its own weight is exerted. Applied force = weight = mg = 2kg \u00d7 10N/kg = 20 N Work done = F \u00d7 d = 20 N \u00d7 10 m = 200 Nm = 200 J 5.1.1 Work done in pulling an object along a horizontal surface Activity 5.2 To design an investigation to determine the work done in pulling an object along a horizontal surface (Work in groups) Materials: a block, a weighing scale, and a tape measure/metre ruler, string. 114 Instructions 1. In this experiment you will design and carry out an investigation to determine work done in pulling an object on a horizontal surface. 2. Using the materials provided think of a set-up to do the investigation. Set-up the apparatus and sketch the set-up. 3. Write a brief procedure to execute in doing the activity. 4. Correctly execute the procedure and answer the following questions. 5. Using relevant formula, calculate the work done in pulling the block. What assumption did you make? Explain. 6. Compare and discuss your findings with other groups in class. 7. Explain how tractors pull large loads. Since the block was on a smooth surface, we assume that friction force is negligible hence the force applied is constant along the distance of motion, d. Work done in moving the block is given by: Work = force \u00d7 distance W = F \u00d7 d = Fd Example 5.2 A horizontal pulling force of 60 N is applied through a spring to a block on a", " frictionless table, causing the block to move by a distance of 3 m in the direction of the force. Find the work done by the force. Solution The work done = F \u00d7 d = 60 N \u00d7 3 m = 180 Nm = 180 J Example 5.3 A horizontal force of 75 N is applied on a body on a frictionless surface. The body moves a horizontal distance of 9.6 m. Calculate the work done on the body. Solution Work = force \u00d7 distance = 75 N \u00d7 9.6 m = 75 \u00d7 9.6 Nm = 720 J 115 Example 5.4 A towing truck was used to tow a broken car through a distance of 30 m. The tension in the towing chain was 2 000 N. If the total friction is 150 N, determine. (a) Work done by the pulling force. (b) Work done against friction. (c) Useful work done. Solution Fig. 5.2 shows the forces acting on the two cars. Fr = 150 N 2 000 N Fig. 5.2: Diagram of cars (a) Work done by the pulling force (b) Work done against friction W = F \u00d7 d W = Fr \u00d7 d (Fr is the frictional force) = 2 000 N \u00d7 30 m = 150 N \u00d7 30 m = 60 000 J = 4 500 J (c) Useful work done Useful work done = Fd \u2013 Frd = (60 000 \u2013 4 500) J = 55 500 J Exercise 5.1 1. Explain why in trying to push a rigid wall, a person is said to be doing no work. 2. Define the term work and state its SI unit. 3. How much work is required to lift a 2 kilogram mass to a height of 10 metres (Take g=10 m/s2). 4. A garden tractor drags a plough with a force of 500 N at a distance of 2 metres in 20 seconds. How much work is done? 116 5.1.2 Work done against the force of gravity Activity 5.3 (Work in pairs) To determine the work done against the force of gravity Materials: Masses, meter rule (tape measure), a single fixed pulley, string, retort stand, newtonmeter. Steps 1. Hang a mass from a newtonmeter and record its weight. 2. Lift a mass from the ground or bench vertically upwards at a constant speed up to a certain point. For a pulley, tie the string on a mass and pass it around the pulley / clamped on a retort stand pull the other end of the thread. Explain why you must use some work in lifting a mass from the ground. 3. Have a friend to measure the height through which the mass has been raised using the meter ruler or tape measure. 4. The force needed to lift the mass is equal to its weight mg. 5. The work done on the mass is then w = Fd = mgh. 6. Repeat the steps with different masses and calculate the work. 7. When an object is thrown upwards, it raises upto a certain point and then drops. Explain why it raises and drops back. You can do this practically by throwing some objects (of different masses). Approximate amount of work done. The gravitational force (weight) acting on a body of mass m is equal to the product of mass and acceleration due to gravity, g, i.e. w = mg. Thus, to lift a body, work has to be done against the force of gravity (Fig. 5.3). \u2013 \u2013 \u2013 \u2013 \u2013 \u2013 \u2013 \u2013 \u2013 \u2022 \u2013 \u2013 \u2013 \u2013 \u2013 \u2013 \u2013 \u2013 mg \u2013 \u2013 \u2013 \u2013 \u2013 x = h \u2022 mass, m mg Ground Fig 5.3: Work done against gravity Work done against gravity to lift a body through height is given by: Work = Force \u00d7 vertical height = mg \u00d7 h = mgh 117 Example 5.5 Calculate the work done by a weight lifter in raising a weight of 400 N through a vertical distance of 1.4 m. Solution Work done against gravity = Force \u00d7 displacement = mg \u00d7 h = 400 N \u00d7 1.4 m = 560 J Example 5.6 A force of 200 N was applied to move a log of wood through a distance of 10 m. Calculate the work done on the log. Solution W = F \u00d7 d = 200 N \u00d7 10 m = 2 000 J Exercise 5.2 1. Define work done and give its SI unit. 2. Calculate the work done by a force of 12 N when it moves a body through a distance of 15 m in the direction of that force. 3. Determine the work done by a person pulling a bucket of mass 10 kg steadily from the well through a distance of 15 m. 4 A car moves with uniform speed through a distance of 40 m and the net resistive force acting on the car is 3 000 N. (a) What is the forward driving force acting on the car?", " Explain your answer. (b) Calculate the work done by the driving force. (c) State the useful work done. 5. A student of mass 50 kg climbs a staircase of vertical height 6 m. Calculate the work done by the student. 6. A block was pushed by a force of 20 N through 9 m. Calculate the work done. 118 5.1.3 Work done along an inclined plane Activity 5.4 (Work in groups) Materials To determine the work done along an inclined plane Spring balance, one piece of wood of about 10 cm, a wedge, ruler, trolley/ piece of wood/mass hanger/stone. Steps 1. Make an inclined plane by putting a piece of wood on a wedge. 2. Attach the mass hanger/stone/trolley to a spring balance (calibrated in Newtons). What happens to the spring immediately when the mass is hanged on it? 3. Measure the length of the incline and record it down l=..... cm. 4. Pull the spring balance with its object on from the bottom of the incline and note down the force used in pulling.............N 5. Change the length in cm to m and find the work done using the formula, work = Fd = (J) 6. Using the above skills, approximate the amount of work you do when climbing a slope of 100 m long. Consider an inclined plane as shown in Fig. 5.4 below. A body of mass m moved up by a force f through a distance d li e d p F ( a \u03b8 Fig. 5.4: Work done along inclined plane. Work done by the applied force is given by Work done = F \u00d7 d The work done against the gravitional force is given by: Work done = weight of the object \u00d7 vertical height Work = mgh In case the inclined plane is frictionless force: Work done by the applied force = work done against gravity 119 In case there is some frictional force opposing the sliding of the object along the plane: Work done by the applied force > Work done against gravity Work done against friction = Work done by applied force \u2013 work done against gravity Example 5.7 A box of mass 100 kg is pushed by a force of 920 N up an inclined plane of length 10 m. The box is raised through a vertical distance of 6 m (Fig. 5.5). m = 100 kg Fig 5.5: Inclined plane (a) Calculate: (i) the work done by the applied force, (ii) the work done against the gravitational force. (iii) the difference in work done. (b) Why do the answers to (i) and (ii) in part (a) differ? Solution (a) (i) Work done by the applied force = F \u00d7 d = 920 N \u00d7 10 m = 9 200 J (ii) Work done against gravity = F \u00d7 d = mg \u00d7 h = 100 kg \u00d7 10 N/kg \u00d7 6 m = 6 000 J (iii) The difference in work done = 9 200 J \u2013 6 000 J = 3 200 J (b) This work done is used to overcome the friction between the box and surface of the incline plane. The useful work done is 6 000 N. 120 Exercise 5.3 1. A box of mass 50 kg is pushed with a uniform speed by a force of 200 N up an inclined plane of length 20 m to a vertical height for 8 m (Fig. 5.6). 0 N 0 F = 2 8 m Fig. 5.6: Work done along an inclined plane Calculate the: (a) Work done to move the box up the inclined plane. (b) Work done if the box was lifted vertically upwards. 2. A body of mass 85 kg is raised through a vertical height 6 m through an inclined plane as shown in Fig. 5.7. Calculate the: F = 150 N (a) Slant distance. (b) Work done by the force 150 N. 6 m (c) Work done, if the body was lifted vertically upwards. (d) Work done against friction. 85 kg 8 m Fig. 5.7: Work done on an inclined (e) Frictional force between the body and the track. 3. A block of mass 60 kg was raised through a vertical height of 7 m. If the slant height of a frictionless track is 21 m, and the force used to push the block up the plane is 800 N, calculate the work done in pushing the block. 4. A car engine offers a thrust of 2 500 N to ascend a sloppy road for 1.1 km. At the top of the slope, the driver realized that the attitude change was 200 m. If the mass of the car is 1.2 tonnes, calculate the; (a) Work done by the car engine. (b) Work done against resistance. 121 5.2 Power Activity 5.5 (Work in groups) To compare the", " time taken to do a piece of work by a person and a machine Materials: writing material, stopwatch and scientific calculator. Steps 1. By timing yourself, start solving the following problems without using calculator: (a) 37 6998 J \u00d7 276 J (b) 35 264 J \u00d7 469 J 2. Repeat step 1, but now with a scientific calculator and compare the time taken. Which one takes longer or shorter time to complete the task? Explain your answer. 3. Now, think of a man ploughing a square piece of land that measure 100 m by 100 m (a) by hand, (b) using a tractor. Which task do you think takes longer or shorter time to complete the activity? Suggest a reason. 4. What is power? In your groups discuss the meaning of power. In your discussion, you might have noted that sometimes work is done very quickly and sometimes very slowly. For instance, it takes a longer time to multiply the problems without a calculator in step 1 than with a calculator in step 2. Similarly, in step 3, a tractor will take few hours ploughing a piece of land while a man will take more hours ploughing the same piece of land. The person and the tractor are doing the same work but the tractor is doing it at a faster rate than the person does. This is because they have different power ratings. Different machines and engines have different power ratings. Engines with bigger power ratings are said to be powerful and operate very fast. Definition of power Power is the rate of doing work. i.e. Power = work done time taken = force \u00d7 distance time 122 SI units of power are Watts. 1watt = 1 joule second Large units used are kilowatt and megawatt. 1 kilowatt = 1 000 W 1 megawatt = 1 000 000 W Example 5.8 What is the power of a boy lifting a 300 N block through 10 m in 10 s? Solution Force = 300 N, Distance = 10 m, Time = 10 s Work done by the boy = F \u00d7d = 300 \u00d710 Power = = 3 000 J work time = 3 000 J 10 s = 300 W 5.2.1 Estimating the power of an individual climbing a flight of stairs Activity 5.6 To estimate the power of an individual climbing a flight of stairs (Work in groups) Materials: stopwatch, weighing machine, tape measure Steps 1. Find a set of stairs that you can safely walk and run up. If there are no stairs in your school, create some at garden. 2. Count their number, measure the vertical height of each stair and then find the total height of the stairs in metres. 3. Let one member weigh himself/herself on a weighing machine and record the weight down. 4. Let him/her walk then run up the stairs. Using a stopwatch, record the time taken in seconds to walk and running up the stairs (Fig 5.8). 123 Fig. 5.8: Measuring one\u2019s own power output 5. Calculate the work done in walking and running up the stairs. Let each group member do the activity. Is the work done by different members in walking and running up the stairs same? Explain the disparity of work done by various group members. 6. Calculate the power developed by each individual in walking and running up the stairs. Which one required more power, walking or running up the flight of stairs? Why? Note: (i) The disabled should be the ones to time others. Care must be taken on the stairs not to injure yourselves. (ii) Incase of lack of stairs, learners can perform other activities like lifting measured weights. From your discussion, you should have established that: Height moved up (h) = Number of steps (n) \u00d7 height of one step (x) h = n \u00d7 x = n x Time taken to move height (h) = t P = = P = Work done against gravity time W \u00d7 n \u00d7 x t Wnx t = mgh t = W \u00d7 h t where, P = power, W = weight, x = height of first step, n = number of steps If x is in metres, W in newtons and t in seconds then power is in watts. 124 Example 5.9 A girl whose mass is 60 kg can run up a flight of 35 steps each of 10 cm high in 4 seconds. Find the power of the girl. (Take g = 10 m/s2). Solution Force overcome (weight) = mg = 60 kg \u00d7 10 N/kg = 600 N Total distance = 10 \u00d7 35 = 350 cm = 3.5 m Work done by the girl = F \u00d7 d = 600 \u00d7 3.5 Power = = 2 100 J work time = 2 100 J 4 s The power of the girl is 525 W My health Do you know that regular exercises are good for your health. Do exercises regularly to keep your body healthy Exercise 5.4 1. Define power and give", " its SI unit. 2. To lift a baby from a crib, 100 J of work is done. How much power is needed 3. if the baby is lifted in 0.5 seconds. If a runner\u2019s power is 250 W, how much work is done by the runner in 30 minutes? 4. The power produced by an electric motor is 700 W. How long will it take the motor to do 10,000 J of work? 5. Find the force a person exerts in pulling a wagon 20 m if 1 500 J of work is done. 6. A car\u2019s engine produces 100 kw of power. How much work does the engine do in 5 seconds? 7. A color TV uses 120 W of power. How much energy does the TV use in 1 hour? 8. A machine is able to do 30 joules of work in 6.0 seconds. What is the power developed by the machine? 125 9. Rebecca is 42 kg. She takes 10 seconds to run up two flights of stairs to a landing, a total of 5.0 metres vertically above her starting point. What power does the girl develop during her run? 10. Student A lifts a 50 newton box from the floor to a height of 0.40 metres in 2.0 seconds. Student B lifts a 40 newton box from the floor to a height of 0.50 metres in 1.0 second. Which student has more power than the other? 11. Four machines do the amounts of work listed in Table 5.1 shown below. The time they take to do the work is also listed. Which machine develops the most power? Machine A B C D Work 1 000 joules 1 000 joules 2 000 joules 2 000 joules Table 5.1 Time 5 sec 10 sec 5 sec 10 sec 5.3 Energy Activity 5.7 (Work in pairs) To brainstorm about energy Materials: Reference materials, a pen, desk or slopy platform Steps 1. What enables your body to perform various functions besides keeping warm. Discuss this with your group member. 2. Define the term energy. 3. Push a pen on your desk such that it rolls for a small distance. What makes the pen roll? 4. State some examples of objects that posses energy in our environment. 5. Discuss the importance of energy in our lives. 6. With a heavy bag on your back, climb the stairs to the top most floor or the steep ground to the top most point. How do you feel? Have you done work? Did you use energy to do so? 7. What happens to a car that has been moving if it runs out of fuel (petrol or diesel)? Explain. 8. Discuss the observations and findings from the above activities in a class discussion. 126 Energy is one of the most fundamental requirement of our universe. It moves motorcycles, cars along roads, airplanes through air, and boats over water. It warms and lights our homes, makes our bodies grow and allows our minds to think. A person is able to push a wheelbarrow, a stretched catapult when released is able to make a stone in it move, wind mills are turned by a strong wind and cooking using electricity in a cooker. All these are possible because of energy. Therefore, for any work to be done, energy must be provided. But what is energy? Definition of energy Energy is the ability to do work. Work done = energy transferred SI unit of energy is joules (J). Relationship between energy and work In your discussion, you should have noted that you got exhausted because you did a lot of work against gravity to carry your body and the heavy bag to the top of the building. The work you did led to the loss of energy (chemical energy from the food) from your body. 5.4 Forms of energy Energy is not visible, it occupies no space and has neither mass nor any other physical property that can describe it. However, it exists in many forms, some of these forms include: 5.4.1 Solar energy Activity 5.8 (Work in groups) To investigate the effect of solar energy Materials: plastic basin, water, convex lens, thin piece of paper Steps 1. On a bright sunny day, fill a plastic basin with cold water and place it in an open place with no shade. Dip your hand into the water after 2 hours. What is the temperature difference of the water initially and after 2 hours? 2. Get a convex lens on the same day and put it horizontally with one surface facing the sun and another surface facing down. Place a thin paper below the lens. What do you observe after 5 minutes or more? 3. Discuss your observations in steps 1 and 2. 127 The water becomes hot in case 1 and in case 2, the paper burns because of the heat from the sun. These are some of the effects of solar energy. This energy from the sun is in form of radiant heat and light. In some countries where the sun shines throughout,", " large concave mirrors have been set to collect energy from the sun by focusing its rays on special boilers which provide power for running electric generators. 5.4.2 Sound energy Activity 5.9 (Work in pairs) To design and investigate the production of sound energy Materials: Two pens, a stone Instructions 1. In this activity, design and carry out an investigation on production of sound energy using the materials provided. 2. Write a brief procedure and execute it to produce sound energy. Then write a report about your investigation and then discuss it in a class presentation. In what form is the energy released by the pen and the stone? Discuss with your class partner. 3. From your discussion, you should have heard sound in steps 1 and 2. In each case, kinetic energy has been converted to sound and heat energy. Sound energy is the energy associated with the vibration or disturbance of bodies or matter. 5.4.3 Heat energy Activity 5.10 (Work in groups) To demonstrate heat energy Materials: Bunsen burner/candle, matchbox, a retort stand, a nail/metallic rod Steps 1. Light a Bunsen burner or a candle using a lighter (matchbox). 2 Clamp a nail (metallic rod) on a retort stand and bring it near the flame. 3. Carefully touch the other end of the nail after sometime. What do you feel? Explain. 128 The other end of the nail is felt to be hot after sometime. The hotness is due to heat energy that has been transferred from the hot part to the cold part of the nail. Therefore, heat energy only travels from a hot object to a cooler one. Heat energy is a form of energy that is transferred from one body to another due to the difference in temperature. 5.4.4 Electrical energy Activity 5.11 To demonstrate production of light by electrical energy (Work in groups) Materials: bulb, electric wire, cells (battery), switch, bulb holder and cell holder Steps 1. Fix the battery/cells in their holders and the bulb too. 2. Connect one wire from one end of the cell holder to the bulb holder. Then connect the same wire from the bulb holder to the switch holder and then connect another wire from the other part/side of cell holder to the switch. Make sure the switch is open and the cells are fixed into their holder. 3. What do you observe after the connection? Explain. 4. Complete the circuit then close the switch and observe what happens. The bulb lights when the circuit is complete. Electrical energy is the energy produced by the flow of electric charges (electrons). Work is done when electrons move from one point to another in an electric circuit with electrical appliances such as bulbs. 5.4.5 Nuclear energy Activity 5.12 (Work in groups) To find out what is nuclear energy 1. Conduct a research from internet and reference books on the meaning of 2. nuclear energy. In your research, also find out advantages and disadvantages of nuclear energy. 3. Compare and discuss your findings with those of other groups in your class. You may consult your teacher for more guidance on your discussion. 129 In your discussion, you should have noted that nuclear energy is the energy that results from nuclear reactions in the nucleus of an atom. It is released when the nuclei are combined or split. 5.4.6 Chemical energy Activity 5.13 To investigate and demonstrate chemical energy (Work in groups) Materials: glass beaker, a small bowl, steel wool, white vinegar, thermometer Steps 1. Place the steel wool in the bowl and soak it in white vinegar for a couple of minutes. 2. Squeeze out excess vinegar and wrap the steel wool around the thermometer in a way that you are still able to read the temperature. 3. Put the steel wool in the beaker, then place a cover with a paper or small book on the top. 4. Record the temperature immediately, then again in a minute or so, and again every minute for about five minutes. What did you observe? 5. Discuss your observation with other groups in a class discussion. The thermometer records a higher temperature reading. The chemical reaction of vinegar and steel wool generates energy in form of heat. This causes temperature to rise as shown by the thermometer. Chemical energy is a type of energy stored in the bonds of the atoms and molecules that make up a substance. Once chemical energy is released by a substance, it is transferred into a new substance. Food and fuels like coal, oil, and gas are stores of chemical energy. Fuels release their chemical energy when they are burnt in the engine (e.g in a car engine). 5.4.7 Mechanical energy Activity 5.14 (Work in groups) To describe mechanical energy Materials: pen, a piece of chalk Steps 1. Raise a piece of chalk or pen from the ground to a position above your head and release it to fall to the ground. What do you observe with the change in height and the speed of the", " piece of chalk as it falls. 130 2. Throw two full pieces of chalk on the wall one at a time using different forces or at different speeds (one should move faster and another one slowly). Note the sound created by the piece of chalk after colliding with the wall. Which one makes more noise after collision? Mechanical energy is the energy possessed by a body due to its motion or due to its position. It can either be kinetic energy or potential energy of both. When an object is falling down through the air, it posses both potential energy (PE) due to its position above the ground, and kinetic energy (KE) due to its speed as it falls. The sum of its PE and KE is its mechanical energy. Mechanical energy = kinetic energy + potential energy. (a) Potential energy Activity 5.15 (Work in pairs) To demonstrate the forms of potential energy Materials: a catapult or a spring, a small stone. Steps 1. Raise a small stone from the ground or any other resting position upwards to a particular height above its resting surface. What kind of energy do you think it attains? 2. Now, release the stone and observe what happens. Explain your observations. 3. Compress a spring to a particular size. What kind of energy do you think it attains? Explain. 4. Release the spring and observe what happens to the spring. 5. Why does the spring move in such a manner? Discuss your observations in steps 1, 2, 3 and 4 with your colleagues and identify two types of potential energy from the activities. You should have observed that when the stone was released it moved down to the ground. This implies that the stone had stored energy due to its position that makes it to start moving down after it has been released. The energy possessed by a body (e.g. a stone) due to its position above the ground is called gravitational potential energy. In other words potential energy is energy by virtue of position. Similarly, when the compressed spring was released, it relaxed to a bigger size. This implies that the spring had stored energy due to compression. The energy possessed by a body due to compression (e.g. a spring) or stretch (e.g catapult) is called elastic potential energy. 131 Therefore, potential energy is in two forms; gravitational potential energy and elastic potential energy. (i) Gravitational potential energy Bodies which are at a given height above the ground posses gravitational potential energy. This energy depends on the position of objects above the ground. The following activity will help us to understand how to calculate potential energy of a body at a particular position above the ground. Activity 5.16 (Work in pairs) To determine gravitational potential energy of a raised object Materials: three bricks, meter ruler, beam balance, soft board bridge Steps 1. Conduct research from the Internet and books on the mathematical expression of potential energy. 2. Support the soft board on two bricks. 3. Measure the mass of the third brick by using a beam balance then place it on the soft board. 4. Now lift the third brick to a height h1. Let your partner measure the height h1, in metres. 5. Allow the brick to drop gently onto the soft board. Describe the energy possessed by the brick as it drops. 6. Calculate the potential energy gained by the stone using the expression of potential energy you got from the research. 7. Repeat the activity with the other two different heights h2 and h3. 8. Compare and discuss your observations and values of PE in the three cases and deduce a general conclusion from your discussion. If a stone is lifted upwards through a height h; and placed on a table (Fig 5.9), work is done against gravity. h F mg Fig. 5.9: Potential energy depends on height, h. 132 The work done to overcome gravity is equal to the gravitational potential energy gained by the stone. But work done = F \u00d7 h ; F = mg \u2234 work done = mg \u00d7 h But, potential energy = work done. Therefore: P.E = mgh Example 5.10 A crane is used to lift a body of mass 30 kg through a vertical distance of 6.0 m. (a) How much work is done on the body? (b) What is the P.E stored in the body? (c) Comment on the two answers. Solution (a) Work done = F \u00d7 d = mg \u00d7 d = 30 \u00d7 10 \u00d7 6 = 300 \u00d7 6 = 1 800 J (b) P.E = mgh = 300 \u00d7 6 = 1 800 J (c) The work done against gravity is stored as P.E in the body. Caution A stone dropped from the roof of a building will cause more pain if it falls on someone\u2019s foot than when the same stone falls from a table. This is because the one on the roof has more gravitation potential energy due to its greater height (position) above the ground. (ii) Elastic potential energy In Activity", " 5.15, we saw that a stretched catapult or compressed (Fig. 5.10(a)) has energy stored inform of elastic potential energy. When the stretched spring catapult is released it releases the energy that can be used to do work e.g. to throw a stone. (a) (b) Fig. 5.10: A compressed and a stretched spring 133 pull \u2013 \u2013 \u2013 \u2013 \u2013 \u2013 \u2013 \u2013 \u2013 \u2013 \u2013 \u2013 \u2013 \u2013 \u2013 \u2013 \u2013 \u2013 force = Work done in stretching the spring = Elastic P.E gained by the spring Fig.5.11: Elastic potential energy = average force \u00d7 extension ) \u00d7 e = ( 0 + F 2 1 2 Fe Work done is stored as elastic potential energy. = Note: Since the force is not uniform (F increases from 0 to F) we should use the average force in calculating the work done. Example 5.11 Calculate the elastic gravitational p.e stored in a spring when stretched through 4 cm by a force of 2 N. Solution Elastic P.E = 1 We will learn more about elastic potential energy and how to determine it later. 2 \u00d7 2 \u00d7 (0.04) = 0.04 J 2 Fe = 1 (b) Kinetic energy Activity 5.17 (Work in pairs) To demonstrate kinetic energy Material: trolley, table Steps 1. Place a trolley on the table and give it a slight push. Observe what happens to it. Explain your observations. 2. Now, observe any moving objects or things around you. Which energy do you think they possess when they are in motion? Explain your answer. 134 From your discussion in activity above, you should have observed that the trolley starts to move once given a slight push. It possess energy as it moves. The energy which is possessed by a moving object due to its speed is called kinetic energy (KE). In other words we can define kinetic energy as energy by virtue of motion. Examples of objects that posses KE include moving air, rotating windmills, falling water, rotating turbines and a moving stone. In general, any moving body possesses energy called kinetic energy. The kinetic energy of a moving body is given by: Kinetic energy = 1 2 mv2, where m and v are the mass and velocity of the body respectively. Example 5.12 A body of mass 400 g falls freely from a tower and reaches the ground after 4 s. Calculate the kinetic energy of the mass as it hits the ground. (Take g = 10 m/s2) Solution The final velocity of the body as it hits the ground given by the equation of motion, v = u + at = u + gt = 0 + 10 \u00d7 4 = 40 m/s K.E = 1 2 mv2 (as the body hits the ground) 2 \u00d7 0.400 \u00d7 402 = 320 J = 1 Example 5.13 A car of mass 1 000 kg travelling at 36 km/h is brought to rest by applying brakes. Calculate the distance travelled by the car before coming to rest, if the frictional force between the wheels and the road is 2 000 N. Solution V = 36 km/h to m/s = 36 \u00d7 1 000 m/s 60 \u00d7 60 = 10 m/s K.E = work done against friction 135 1 2 mv2 = F \u00d7 d 2 \u00d7 1000 \u00d7 102 = 2 000 \u00d7 d \u21d2 1 \u21d2 50 000 = 2 000 d \u2234 d = 50 000 2 000 = 25 m The stopping distance is 25 m. 5.5 Work and energy relationship When work is done a transfer of energy always occurs. For example carrying a box up the stairs/lifting something heavy from the ground, you transfer energy to the box which is stored as gravitational energy. Therefore its energy increases. Doing work is a way of transferring energy from one object to another. Just as power is the rate at which work is done when energy is transferred the power involved can be calculated by dividing the energy transferred by time needed for the transfer to occur. Power(in watts) = Energy transferred (in joules) time (in watts) P = E t or Energy = pt For example, when the light bulb is connected to an electric circuit, energy is transferred from the circuit to the light bulb filament. The filament converts the electrical energy supplied to the light bulb into heat and light. The power used by the bulb is the amount of electrical energy transferred to the light bulb each second. Exercise 5.5 1. Define the term energy. 2. State and explain briefly six forms of energy. 3. Differentiate between: (a) Potential energy and kinetic energy. (b) Gravitational potential energy and elastic potential energy. 4. A brick of mass 0.5 kg is lifted through a distance of 100 m to the top of a building. Calculate the potentials energy attained by the brick. 136 5. Explain what is meant by gravitational potential energy. 6. A force of 8.5 N stretches a certain spring", " by 6 cm. How much work is done in strecthing the spring by 10 cm. 7. A body is acted on by a varying force, F over a distance of 7 cm as shown in Fig. 5.121 \u20132 \u20133 \u20134 \u20135 \u20136 1 2 3 4 5 6 7 Distance (m) Calculate the total work done by the force. Fig.5.12 8. An object of mass 3.5 kg is released from a height of 7.0 m above the ground. (a) Calculate the gravitational potential energy of the object release. (b) Calculate the velocity of the object just before it strikes the ground. What assumption have you made in your calculation? 5.6 Sources of energy Activity 5.18 To find out sources of energy (Working in groups) Materials: Internet, reference books, a stream of water or a water tap Steps 1. Tell your group members the meaning of the terms \u2018source\u2019 and \u2018energy source\u2019. 137 2. Now, think of plants, animals, vehicles and so on. Where do you think their energy comes from? What of electricity used in your school and at home, where does it come from? 3. Compare and discuss your findings in step 1 and 2 with other groups in your class. 4. What is the meaning of primary sources of energy? 5. What is the meaning of secondary sources of energy? 6. Conduct a research from the Internet and reference books on primary and secondary sources of energy. 7. In your research find out: (a) The types of primary and secondary sources of energy. (b) The generation of energy from each source. 8. Let your secretary write down a summary of your discussion and present it to the whole class. 9. State and explain various types of primary and secondary sources of energy? The word \u2018source\u2019 means the beginning of something e.g. the stream begins from the mountain or hills around your school. Hey!! Are you aware that cutting down trees will lead to the loss of our forests in our country and consequently the loss of water sources? Let us protect our water sources by planting more trees. You should have also established that the energy source is a system which produces energy in a certain way. For instance, a hydroelectric station uses the motion of the water of the river to turn the turbines and thus producing electricity. There are two kinds of energy sources; 1. Primary sources 2. Secondary sources 138 5.6.1 Primary sources of energy From your research and discussion in activity 5.18, you should have established the following that Primary Sources are from sources which can be used directly as they occur in the natural environment. They include. 1. Flowing water 2. Nuclear 3. Sun 4. Wind 5. Geothermal ( interior of the earth) 6. Fuels 7. Minerals 8. Biomass (living things and their waste materials 1. Water (a) Hydropower - the flowing water from dams rotate turbines at the bottom of the dam which turn the generator resulting in generation of electricity. This water is kept behind a dam (reservoir) and released at a controlled rate downwards where it meets the turbines and turns them. An example is the falling water in the Fulla rapids on the River Nile in Nimule in South Sudan. (b) Waves - energy from water waves (generated by winds) is also used in generating electricity using sea wave converters. An example is pelamis wave energy converter, a technology that uses the motion of ocean surface waves to create electricity. 2. Nuclear energy Nuclear energy is created through reactions that involve the splitting or merging of the atoms of nuclei together. The process of splitting of large atoms such as those of uranium into smaller atoms is called fission. Fusion on the other hand, is the combining of two smaller atoms such as hydrogen or helium to produce a heavier atom. All these reactions release heat which is turned into electricity in nuclear power plants (Fig 5.13). An atomic bomb derives its energy from these kinds of reactions. 139 3. The sun Fig 5.13: Nuclear plant The sun is the biggest source of energy and has played an important role in shaping our life on earth since the dawn of time. The sun gives off radiant energy in form of electromagnetic waves. The light energy (visible spectrum) part of the spectrum can be converted directly into electricity in a single process using the photovoltaic (PV) cell otherwise known as the solar cell. The solar thermal energy is used for heating swimming pools, heating water for domestic use (solar heater) and heating of building. Solar thermal electricity generation is where the sun\u2019s rays are used to heat a fluid for the production of high pressure and high temperature steam. The steam is in turn converted into mechanical energy in a turbine to generate electricity. (a) Solar panel (b) Solar heater Fig. 5.14: Solar panel and solar heater 140 4. Wind Wind is caused by the", " sun heating the earth unevenly. The air is heated differently causing hotter air to expand rise, and the colder one to condense and sink. This results to the movement of air and hence formation of wind. Modern wind turbines placed on the top of steel tubular towers harness the natural wind in our atmosphere and convert it into mechanical energy and then to electricity. Wind mills (Fig. 5.15) are also be used to pump water from the underground and do some other work. 5. Geothermal energy Fig 5.15: Wind mills Geothermal gradient is the difference in temperature between the core (interior) of the earth (planet) and its surface brings about conduction of heat from the core to the surface. The earth\u2019s internal heat is generated from radioactive decay and continual heat loss from the earth\u2019s formation. From hot springs, geothermal energy has been used for bathing to heal some diseases as in some cultures. Geothermal energy is also used to generate electricity at geothermal power stations where heat is used to heat water to get steam which in turn is used to turn the turbines to generate electricity. 141 Fig. 5.16 Geothermal power station 6. Thermal energy Thermal energy is the internal energy in a system by virtue of its temperature. It is defined as the average translational kinetic energy possessed by free particles in a system of free particles in a thermodynamic equilibrium. It can also include the potential energy of a system\u2019s particle which may be an electron or an atom. Thermal (heat) energy is transferred of heat across the system boundaries. Thermal energy is important in our daily life, for example in warming the house, cooking, heating the water and drying the washed clothes. 7. Biogas Activity 5.19 (Work in groups) To conduct research on how to produce biogas Materials: bio-digester, reference materials (including this book), internet 1. Using reference materials (including this book) or internet, research about how biogas is produced. Make notes and present your finding in class. 2. Make a trip to a farm with biogas plant and take turn to ask about working of biodigester. 3. Write a report explaining how a biodigester works? 4. Hold a group discussion and discuss about biogass as a source of energy. Biomass is the total mass of organic matter in plant or animal. It is used to generate energy e.g. through burning to give heat energy. When bacteria acts on biomass, a gas called biogas is produced which is flammable hence is used as fuel to produce heat. It is a mixture of 65% methane and 35% carbondioxide. 142 A biogas plant or digester collects and directs the gas through pipes to the kitchen for cooking in a house or to a generator where electricity is produced. Fig. 5.17 shows a biogas plant. Biogas pipe Biogas Storage Tank Biogas inlet Biogas outlet Biogas Digester Fig. 5.17: Biogas plant 8. Fuels Fuels are substances which produce heat when burnt in the presence of oxygen. They include kerosene, diesel, biogas, are sources of energy in homes, industries. In the process of combustion, the chemical energy in the fuel is converted into heat energy that is converted to other forms as desired. 9. Chemical energy Chemical energy is stored in the chemical bonds of atoms and molecules. It can only be seen when it is released in a chemical reaction. When chemical energy is released, from a substance, the substance is entirely changed into an entirely different substance. Some substances that store and release chemical energy are; (i) Electrolytes \u2013 the chemical reactions in an electrolyte in the batteries produce electricity. (ii) Petroleum \u2013 petroleum is made of molecules containing carbon and hydrogen. In vapour form, its natural gas and in liquid form, it is crude oil. Energy from petroleum is used to drive vehicles and to produce electricity. Examples include jet fuel, gasoline and others. (iii) Wood \u2013 dry wood acts as a store of chemical energy. This chemical energy is released when wood burns and it\u2019s converted into heat and light energy. 143 (iv) Food \u2013 the chemical energy in food is released while the food is being digested. As the bonds between the atoms of the food break, new substances are created and chemical energy is given out. Warning Subjecting a battery to misuse or conditions for which it was not made for can result into battery failure or uncontrolled dangerous conditions which include explosion, fire and the emission of toxic fumes. Keep batteries well out of reach of children. 10. Light energy The potential of light to perform work is called light energy. It is formed through chemical radiation and mechanical means. It is a form of energy produced by hot bodies and travels in a straight line. It\u2019s the only form of energy that we can see directly (visible light). It can be converted like sunlight energy is used", " during photosynthesis by plants to create chemical energy. UV lights are often used by forensic scientists to see details that are not seen by unaided eyes. 5.6.2 Secondary sources of energy Secondary sources are energy sources that are generated from primary sources. For instance, electricity is a secondary source because it is generated for example from solar energy using solar panels or from flowing water using the turbines to generate hydroelectricity. Other secondary sources of energy include; petroleum products, manufactured solid fuels, gases, heat and bio fuel. 5.7 Renewable and non renewable sources of energy Activity 5.20 (Work in pairs) To distinguish between renewable and nonrenewable sources of energy Materials: matchbox, reference books Steps 1. Take one matchstick from the matchbox and light it. 2. Leave it to burn for a few seconds and then put it off. 3. Use the same matchstick and try to lit it again. Observe and explain what happens. 4. From the knowledge of sources of energy, what do you think renewable and non-renewable sources of energy are? 144 5. Discuss your observations in steps 2 and 3 with your deskmate. 6. Now, conduct a research from internet and reference book on renewable and non-renewable energy source. 7. Compare and discuss your findings with other groups in your class. Hey!!! Be safe Always be careful with fire. It can cause massive damage which can result to loss of properties and lives. There are energy sources that cannot be used again once used to generate energy. They are called non renewable sources while those that can be used again without exhausting them are called renewable resources. (a) Renewable energy sources A renewable energy source is an energy source which can\u2019t be depleted/exhausted. They exist infinitely i.e. never run out. They are renewed by natural processes. Examples include; (i) Sun However, some like trees they can also be depleted, like trees and animals if used too much more than the natural process can renew them. So it\u2019s advisable to take precaution while using them, that is, they should be conserved. (iii) Geothermal (iv) Biomass (ii) Wind By doing so it will lead to access to affordable and reliable energy while increasing the share if renewable energy in our country. Hence contributing to affordable, reliable sustainable and modern energy for all. Achieving sustainable development goals (SDGs) by 2030 is a drive that countries across the world are working towards together. Gaining more of our energy from renewable sources is an important part of the strategy. (b) Non-renewable energy sources These are sources which can be depleted because they exists in fixed quantities. So they will run out one day. Examples are coal, crude oil, natural gas, and uranium. Fossil fuels like coal, crude oil, natural gas are mainly made up of carbon. They are usually found in one location because they are made through the same process and material. Millions of years ago dead sea organisms, plants, and animals settled on the ocean floor and in porous rocks. With time, sand, sediments and impermeable rock settle on the dead organic matter, as the matter continue to decay forming coal, oil and natural gas. Earth movements and rock shifts creates spaces that force these energy sources to collect at well-defined areas. With the help of technology, engineers are able to drill down into the sea bed to mine these sources and harness the energy stored in them. 145 5.8 Environmental effects of the use of energy sources Activity 5.21 (Work in groups) To investigate the environmental effects of the use of energy sources Materials: reference books, Internet Steps 1. Conduct research from the Internet and reference books on environmental 2. effects of the use of energy sources. In your research, identify the effects and suggest the measure to be taken to ensure safe use of those resources. 3. Record down your findings from your discussion and report them in a class discussion. In Activity 5.24, you should have learnt that, there is no such thing as a completely \u201cclean\u201d energy source. All energy sources have atleast an effect to the environment. Some energy sources have a greater impact than others. Energy is mostly lost into the environment in form of heat and sound. The following are some of the effects of use of the energy sources to the environment: \u2022 air and water pollution \u2022 climate change and global warming. \u2022 deforestation \u2022 land degradation (a) Air and water pollution Fossil fuels e.g petroleum, diesel are used in factories. Very harmful by-products may be released to the atmosphere or water bodies. Carbon monoxides, sulphur dioxide and carbon dioxide may be released to the atmosphere causing air pollution that may harm living things that depend on air. When human beings inhale some of the polluted air, they can develop respiratory diseases. The wastes disposed to the water bodies can cause death of living things in the water. It also make the water unsafe for", " human consumption. Factories and industries operators are encouraged to use bio-fuels which are less harmful to the environment. Most factories are trying to clean up the waste so as to reduce the environmental pollution. (b) Climate change and global warming Most energy sources e.g fossil fuels, coal etc, when used as sources of energy, produce wastes such as carbon dioxide, sulphur dioxide and mercury which are 146 the greenhouse gases. The accumulation of these gases in the atmosphere make the temperatures to be higher than the normal. This is referred to as global warming. Sometimes, these gases interfere with the climate causing very high temperature in the atmosphere, acidified rains, frequent droughts, floods etc. This results to climate change. The greenhouse gases e.g. excess carbon dioxide, sulphur dioxide etc destroy the ozone layer exposing living things to dangerous emissions from the sun e.g. UV rays. Release of these harmful gases into the atmosphere is a global problem and very many environmental agencies are encouraging on the proper disposal of these wastes. United Nations Conference on Environment and Development (UNICED) lead Nations to sign a joint treaties to pursue of economic development in ways that would protect the earth\u2019s environment and non renewable resources but it is still a problem up to now. (c) Deforestation Using firewood and charcoal in most African countries lead to loss of biodiversity and erosion due to loss of forest cover. These may lead to deforestation i.e. the reduction of forest cover. Of great concern is that Africa is losing forest twice as fast as the rest of the world. Human beings are encouraged to use green energy that is renewable and have less effect to the environment. With your help we can support projects that help to train and educate forest communities so that they can use forests in a sustinable manner and protect their livelihoods for years. (d) Land degradation Land degradation is the process in which the value of bio-physical environment is affected by human \u2013 induced activity on the land. It is caused by over-cutting of vegetation e.g. forest, and woodland, for firewood and disposing factory wastes to the soil that may contaminate the soil. Use of non-biodegradable sources of energy is encouraged. Saving our energy Let\u2019s adopt the use of biogas in cooking, energy saving stoves and reduce the use of firewood to the possible level and amount of smoke generated reducing the impact of indoor air pollution. This will reduce environmental impacts. 147 Exercise 5.6 1. Differentiate between energy and power. 2. In groups of two, identify any three primary sources of energy and hold a discussion on their: (a) definition and origin. (b) importance to us and our country. 3. Choose any renewable energy resource. Brainstorm on two to three jobs opportunities available in that field. 4. Distinguish between the terms renewable resource and non-renewable resources. 5. Give one example of a body with potential energy due to its state. 5.9 Energy transformations Activity 5.20 (Work in groups) To investigate energy transformation Materials: an electric heater, radio, water in a basin. Steps 1. Place the electric heater in the basin with water and connect it to the socket. 2. Put on the switch. Observe and explain what happens after a couple of minutes. Suggest the types of energy involved in this case. 3. Now, disconnect the heater and connect the radio to the socket. 4. Turn the radio on and suggest the types of energy involved. 5. Repeat the activity by connecting wires, battery, switch and bulb. Observe and explain what happens when you make simple circuit and the switch is closed. 6. What is the meaning of energy transformation? Give five examples of energy transformation? 7. What is the name given to devices such as the radio, heater, battery, bulb etc. that converts energy from one form to another? 8. Discuss with your group members other forms of energy transformation and show with diagrams how energy is transformed from one form to another on the chalkboard. 148 Hey!!! Be safe Don\u2019t touch water while an electrical heater is on, you may get an electrical shock. From your discussion, you should have observed that the water in the basin boils. Electrical energy has been converted to heat energy which boils the water. When the radio was connect to the socket and turned on, electrical energy is converted to sound energy. In step 5, when the wires are connected, the bulb is seen to give off light when you close the switch. This is because chemical energy in the battery has been converted to electrical energy which is then changed to light energy in the bulb and some part to heat energy. Therefore, energy in many of its forms may be used in its natural process or to provide some services to society such as heating, refrigeration, or performing mechanical work to operate machines. This is possible because energy can be changed from one form to another. This process of changing of energy from one form to another is called energy", " transformation. A device that converts energy from one form to another is known as a transducer. Fig 5.18 is a chart that shows some examples of energy transformation in our day to day activities. 149 Chemical n actio ar re ucle Electrical c tri c E l e e ll o l a r c S Nuclear Nuclear reactor Heat o c o u ple c tric h E le a t e T h er m Sound ti o r V i b g s s Mechanical Generator Electric motor W i n d m i l l Light Let us consider a few examples of energy transformation: Fig. 5.18: A flow chart of energy transformation 1. Hammering a nail Chemical energy in our bodies Potential energy of hammer Fig. 5.19: Energy transportation Heat K.E Sound 150 2. Lighting a bulb using a battery Chemical energy in the battery Electrical energy Radiant heat Light energy 3. Hydroelectric power Fig. 5.20: Energy transformation Potential energy of water in the water reservour Kinetic energy of falling water K.E of rotation of turbines Electrical energy Heat and sound Fig. 5.20: Potential energy and its transformation Other examples of energy transfomers. Wind turbines use wind energy to transform it into electricity. Energy from food (chemical energy) can be transformed in energy to play and run. A solar cell/ panel convert radiant energy of sunlight to electrical energy that can be used to give off lightning a bulb or to power a computer The sun gives the grass thermal energy which helps it to grow by transforming the energy into chemical energy using photosynthesis. Animals eat grass and help them to grow and have power to run. A microphone changes electric energy to sound energy and so on. One other example of energy transformations occurs when lightning strike. If it hits a tree, it\u2019s electrical energy will be changed to heat and thermal energy. The tree will become hot and can even burn as a result of electric discharge, it can split and the leaves dry. 151 Exercise 5.7 1. Table 5.2 shows how energy is converted from form A to form B and the devices concerned. Complete the table. Form A Form B Electrical Sound Electrical Kinetic _ Sound _ _ Electrical Electrical Electrical _ Device Loudspeakers _ Photocell _ Thermocouple Heater Table 5.2: Forms of energy 2. Describe the energy changes that occurs in the following processes. (a) When you lift a brick to a certain height. (b) When you lift a brick and let it slide down a rough slope until it reaches the surface of the slope. 3. Describe the forms of energy shown in Fig 5.21. Fig. 5.21: Forms of energy 4. Name the changes in energy that take place when a torch is switched on. 5. Name the energy changes that take place when lighting a match box. 152 5.10 Law of conservation of energy To demonstrate the law of conservation of energy Activity 5.23 (Work in pairs) Materials: a ball Steps 1. Hold a football at a height of 1 m above the ground. What type of energy does the ball posses at that position? 2. Release the ball to start falling freely to the ground. What type of energies does the ball posses while falling? 3. What type of energy does the ball posses while just about to touch the ground. 4. Ignoring air resistance, compare the amount of energy possesed by the ball in step 1 and 3. What can you conclude? When the ball was stationary at a point 1 m above the ground in Activity 5.26, it possed P.E only. When released the P.E started being converted to K.E hence the ball dropped. When it was just about to touch the ground, all the P.E had been converted to K.E hence by ignoring air resistance, Height point of swing Height point of swing Maximum kinetic energy Fig. 5.22: Initial P.E = final K.E We say that energy has been conserved. This is summarised in the law of conservation of energy. The law of conservation of energy states that energy cannot be created or destroyed but is simply converted from one form into another. Or in other words we can state it that in a closed system the total amount of energy is conserved. Energy can be inter-converted among many forms, mechanical, chemical, nuclear, electric, and others but the total amount of it remains constant. For instance, in boiling water using a kettle, electrical energy drawn from the power source flows into the heating element of the kettle. As the current flows 153 through the element, the element rapidly heats up, so the electrical energy is converted to heat energy that is passed to the cold water surrounding it. After a couple of minutes, the water boils and (if the power source remains in the water) it starts to turn into steam. Most of the electrical energy supplied into the kettle is converted to heat energy in the water though some is used to provide latent heat of evaporation (the", " heat needed to turn a liquid into a gas without a change in temperature). If you add up the total energy supplied by the power source and the total energy gained by the water, you should find they are almost the same. The minor difference would be due to energy loss in other forms. Why aren\u2019t they exactly equal? It\u2019s simply because we don\u2019t have a closed system. Some of the energy from the power source is converted to sound and wasted (kettles can be quit noisy). The kettles also give off some heat to their surrounding so that\u2019s also wasted energy. Another example is a flying ball, that hits a window plane in a house, shattering the glass. The energy from the ball was transferred to the glass making it shatter into pieces and fly in various directions. 5.11 The law of conservation of mechanical energy Activity 5.24 (Work in pairs) To verify the law of conservation of mechanical energy Materials: A smooth metallic hemispherical bowl, a ball bearing Steps 1. Place the hemispherical bowl on the bench in a stable position. Mark at point A on the inside surface of the bowl at point A on the inside surface of the bowl 2. Place the ball bearing at point A and release it to slide downwards freely along the inside surface of the bowl as shown in Fig. 5.22. A h B C E D Fig. 5.22: A ball bearing sliding oscillating in a bowl 3. Mark point E where the ball rises to on the opposite side in the bowl. 154 4. Compare the vertical height of points A and E. What do you notice aboue the heights? 5. Repeat the activity with point A at a lower vertical height. 6. What type of energy does the ball bearings possess at points A, B, C, D and E. 7. Compare and comment on the total amount of energy possessed by the ball bearing at points A, B, C, D and E. 8. Make a conclusion based on your observation in step 7. You should have learnt that the law of conservation of mechanical energy states that The total mechanical energy (sum of potential energy and kinetic energy) in a closed system will remain constant/same. A closed system is one where there are no external dissipative forces (like friction, air resistance) which would bring about loss of energy. The sum of potential energy and kinetic energy anywhere during the motion must be equal to the sum of potential energy and kinetic energy anywhere else in the motion. To demonstrate the law of conservation of mechanical energy (a) A swinging pendulum Activity 5.28 (Work in pairs) To demonstrate the law of conservation of M.E using a swinging pendulum Materials: a bob, string Steps 1. Tie a string to the bob and fix it to a rigid object.( See Fig. 6.16). 2. Pull the bob to the right or left side at an angle and then release it. Observe the movement of the bob. 3. Draw a diagram for the motion of the pendulum and discuss with your class the energy changes at various points e.g. A, B, C, D and E shown in Fig. 5.24. A E B D Fig. 5.24: A swinging pendulum C 4. Explain the energy changes at points A, B, C, D 155 From the above activity, you should have noticed that the bob will attain a maximum potential energy due to its height above the ground at point A she have minimum kinetic energy because it is at rest. When it swings after letting it go, it will start loosing potential energy as it gain kinetic energy at point B because of its motion. As it passes through the lowest point point C, its potential energy is minimum kinetic energy will be maximum. Because of its kinetic energy, it swings up to the other side and now its kinetic energy starts decreases as, potential energy increases at point D until when it reaches the maximum point E where it stops moving momentarily. At that point, it has maximum potential energy but minimum kinetic energy. At all positions, the total mechanical energy is constant (conserved). That is kinetic energy + potential energy = constant. Therefore, mechanical energy has been conserved. (b) A body thrown upwards Activity 5.26 To demonstrate the law of conservation of M.E using a ball thrown upwards (Work individually) Materials: tennis ball Steps 1. Throw a tennis ball upwards. Observe and describe the movement of the ball up to the maximum (highest) point. 2. Now, drop the ball from a high point e.g from top of the building or a cliff (see Fig 5.25). 3. Sketch its motion on a paper at three different intervals, starting from the lowest when thrown upwards or from highest when dropped from a cliff. 4. Explain why the ball falls back to the ground after thrown upwards. 5. Indicate the forms of energy at each stage. 6. Discuss your observations and drawing with your colleagues in class.", " In your discussion, you should have learnt that when a body (e.g. a ball) is thrown up vertically, it has maximum speed, (maximum kinetic energy) at the starting point. The ball moves up with a reducing speed because of the force of gravity acting on it downwards until it reaches the maximum point/ height where it stops moving momentarily and it falls back. 156 At maximum height, it has a maximum potential energy and minimum kinetic energy because the body is not moving. So the kinetic energy at the bottom is all turned into potential energy at the maximum point (Fig 5.25). P.Emax K.E = 0 cliff P.E = K.E object P.E = 0 K.Emax ground Fig. 5.25: Sketch of a ball thrown upwards The ball is under free fall because it is only being acted upon by the force of gravity. Initially the ball has maximum potential energy and no kinetic energy. As it falls down, its potential energy keeps on reducing as its position above the ground reduces but its kinetic energy is increasing because it speeds up as it falls downwards. The kinetic energy at the ground level is equal to the potential energy at the top of the wall. Hence mechanical energy is conserved. Exercise 5.8 1. A pendulum bob swings as shown in the diagram. Fig 5.26 Pendulum bob Start Fig 5.25: A pendulum swinging At which position (s) is: the kinetic energy of the pendulum bob least. (a) (b) the potential energy of the pendulum bob most. 157 (c) (d) the kinetic energy of the pendulum bob the most. the potential energy of the pendulum bob the least. 2. State the following laws: (a) (b) law of conservation of energy law of conservation of mechanical energy. 3. Describe how mechanical energy is conserved. 5.12 Ways of conserving energy Activity 5.27 (Work in groups) To do research about conservation of energy and identify ways of conserving energy Materials: internet, reference book (including this book) 1. What is the meaning of the word conservation of energy? 2. Conduct a research from Internet and reference books on ways of conserving 3. energy. In your research, identify different ways of conserving energy and find out what energy efficiency is. 4. Discuss your finding with other pairs in class and give a report of your findings to the whole class. From your research and discussion you should have established that energy conservation is the act of saving energy by reducing the length of use. In other words, to conserve energy, you need to cut back on your usage. For example, driving your car fewer miles per week, turning your thermostat down a degree or two in the winter time and unplugging your computer or home appliance when they are not in use. All these ways reduce the amount of energy you use by doing without or using less fuel or electricity. It can help reduce the monthly heating and electricity bills and save money at the gas pump. You also reduce the demand of fuels like coal, oil, and natural gas. Less burning of fuels means lower emissions of carbon dioxide, the primary contributor of to global warming and other pollutants. Other examples include: (i) Clean or replace air filters of cars as recommended. Energy is lost when air conditioners and hot air furnaces have to work harder to draw air through dirty filters. So save money by replacing old air filters with new (standard) ones which will take less electricity. 158 (ii) Select the most energy efficient models when you replace the old appliances. Look for the energy star label because the product saves energy and prevents pollution. (iii) Turn your refrigerator down. Refrigerators accounts for about 20% of the house hold electricity costs. (iv) Buy energy-efficient compact fluorescent bulbs for the lights you use most. Although they cost more, they save money in the long run because they only use a quarter the energy used by ordinary incandescent lamps and lasting 8-12 times longer. (v) Reduce the amount of waste you produce by buying minimally packaged goods, choosing reusable products over disposable ones, and recycling. Use 30% to 50% less paper products, 33% less glass and 90% less aluminum. (vi) People who live in colder areas should super insulate your walls and ceiling. It can save your the electricity of heating or fire wood costs. (vii) Plant shady trees and paint your house a light colour if you live in a hot place or paint them a dark colour if you live in cold conditions. If we do not conserve energy, it will be exhausted and we will have nothing to use. Energy conservation is also important when in managing climate change. Currently erratic climates and climatic changes are the greatest threats that we are facing today. Hence it is important to conserve energy. 5.13 Energy efficiency Energy efficiency is the act of saving energy but keeping the same level of service. For example, if you turn off the lights", " when you are leaving a room, that\u2019s energy conservation, if you replace an efficient incandescent light bulb with a more efficient compact fluorescent bulb, you are practicing energy efficiency. Energy efficiency uses advances in sciences and technology to provide services and products that require the use of less energy. Exercise 5.9 1. (a) Demonstrate how mechanical energy is conserved. (b) What is energy efficiency? 2. By identifying practical activities in our daily lives, discuss how you can conserve energy. 159 Project work 1 Energy saving charcoal burner In most developing countries, wood is the most important source of energy mainly for cooking. The amount of wood consumed depends on the climate, culture and availability. Most people use open, three stone fireplaces for cooking. The fireplaces are often dirty, dangerous and inefficient. The smoke and soot settles on utensils, walls, ceiling and people. The smoke produced in fireplaces irritates people posing danger to health. The fireplaces have been found to be about 10-15% efficient. In view of the above, energy saving stoves have been designed. Most of these stoves use charcoal. Charcoal is preferred to wood in urban areas because of its portability, convenience and cleanliness. In designing energy saving stoves, one should try to minimise energy losses to the surrounding. One of the many advantages of a charcoal stove, is that the rate of charcoal burning can be controlled. Materials Metal sheets and clay Construction Cut the metal sheet into a circular sheet (Fig. 5.27(a)). The radius AO will depend on the size of the stove required. Mark arc AB which represents the circumference of the mouth of the charcoal burner. Draw AO and OB. Draw arc CD. The radius OD will depend on the area of the base on which the charcoal is to rest. Cut the section ACDB. Assembly Fold the plate ACDB in a shape of a cone as shown in Fig. 5.27(b). Rivet the sides AC and BD together. Repeat the procedure to construct the lower compartment. But this time make AC and DB shorter. A C B D O A, B C, D (a) Circular metal sheet (b) Upper compartment Fig. 5.27: Making the upper compartment of an energy saving charcoal burner Bring the two compartment together and join them by riveting Fig. 5.28(a). Cut off a small section of the lower compartment and construct a gate. Mould clay in such a shape that it fits into the upper compartment. Make the air holes while the clay is still wet. 160 Allow the clay to dry. Construct the stands for holding the cooking pot. A complete stove should look like the one shown in Fig. 5.28(b). Base with air inlet holes Gate Clay Metal Gate (a) Upper and lower compartment joined (b) Complete charcoal burner Fig. 5.29(a): Upper and lower compartments joined to make a complete charcoal burner Larger stove can be made by cutting the sheet as shown in Fig. 5.29. A C D B or O C D A B Fig.5.29: Larger jikos Project work 2 Solar heater Solar energy can be trapped with the help of solar heater and utilized to heat water. The most common type of solar water heater incorporates a flat-plate solar collector and a storage tank. The tank is positioned above the collector. Water from the tank is circulated through the collector and back to the tank by means of convectional currents caused by the heated water. Construction of a solar heater Suggested materials A 20 litre jerry can container, plastic pipes, cellophane paper, half open 20 litre jerry can, black paint or smoke soot and a wire mesh. Assembly Heat collector Paint the plastic pipes black. Use a wire mesh and curve the plastic pipes as shown in Fig. 5.30. The size of the wire mesh should be able to fit into an open 20 litre jerry can container. 161 1 2 Pipes Strings to tie the pipes onto the wire mesh Frame work Half open jerry can Fig. 5.30: Heat collector Heat exchanger Use another 20 litre jerry can (Fig. 5.31) and open at the top to allow the pipes to enter and then seal it using the same material and a hot object. The hot object will make the materials to fuse together. Make provisions for water to enter and leave the heat exchanger when required. 1 Hot water Water gains energy Pipes Cold water 2 Fig. 5.31: Heat exchanger Join pipe 1 of the heater collector to pipe 1 of the heat exchanger. Do the same with pipe 2. Make sure the collector is inclined at a certain angle to allow water from the heat exchanger to flow freely. (Fig. 5.32). Cover the heat collector with a cellophane paper. 162 Hot water Heat exchanger Cold water Pump Cold water Stand Heat collector Hot water rises Transparent plastic paper \u03b8 Fig", " 5.33: Solar heater How to use Fill the pipes of the heat collector with water and expose them to the sun. Allow water from a reservoir to fill the heat exchanger. Topic summary \u2022 Work is the product of force and distance moved in the direction of the \u2022 force. A joule is the work done when a force of one newton acts on a body and makes it to move a distance of one metre in the direction of the force. \u2022 When work is done on an object, energy is transferred. Work is said to be done if a force acts on a body and makes it move (get displaced) in the direction of the force \u2022 Energy is the ability to do work. \u2022 Moving objects have kinetic energy that depends on the mass of the body \u2022 and the velocity. Potential energy is the energy possessed by a body due to its position. It depends on the objects height above the ground. \u2022 The total amount of kinetic energy and potential energy in a system is the mechanical energy of the system. Mechanical energy = KE + PE. \u2022 Falling, swinging, and projectile motion all involve transformations between kinetic and potential energy. 163 \u2022 \u2022 According to the law of conservation of energy, energy cannot be created or destroyed but can only be converted from one form to another. Energy is converted changes from one form to another by transducers such as light bulbs, hair driers. For example, a hair drier converts electrical energy into thermal energy, kinetic energy and sound energy. \u2022 Fuel is a substance which when burnt produces heat. Topic Test 5 1. Define the term power and give its SI unit. 2. A motor raised a block of mass 72 kg through a vertical height of 2.5 m in 28 s. Calculate the: (a) work done on the block. (b) useful power supplied by the motor. 3. A person of mass 40 kg runs up a flight of 50 stairs each of height 20 cm in 5 s. Calculate: (a) the work done. (b) the average power of the person. (c) explain why the energy the person uses to climb up is greater than the calculated work done. 4. A runner of mass 65 kg runs up a steep slope rising through a vertical height of 40 m in 65 s. Find the power that his muscles must develop in order to do so. 5. A fork-lift truck raises a 400 kg box through a height of 2.3 m. The case is then moved horizontally by the truck at 3.0 m/s onto the loading platform of a lorry. (a) What minimum upward force should the truck exert on the box? (b) How much P.E. is gained by the box? (c) Calculate the K.E of the box while being moved horizontally. (d) What happens to the K.E once the truck stops? 6. A stone falls vertically through a distance of 20 m. If the mass of the stone is 3.0 kg, (a) Sketch a graph of work done by the gravity against distance. (b) Find the power of the gravitational pull. 164 7. Mugisha climps 16 m rope in 20 s. If his mass is 60 kg, find the average power he developed. 8. A car is doing work at a rate of 8.0 \u00d7 104 W. Calculate the thrust of the wheels on the ground if the car moves with a constant velocity of 30 m/s. 9. Uwimbabazi took 55.0 s to climb a staircase to a height of 14.0 m. If her mass is 40 kg, find: (a) How much force did she exert in getting to the that level? (b) Her power? 10. In Fig. 5.33 three positions of a monkey swinging from a branch of a tree are shown. A B C Fig. 5.33: A monkey swinging (a) What kind of energy does the monkey have at each position? (b) What happens to the energy when the monkey is midway between A and C? (c) In which positions does the monkey have the least energy? What name is given to this type of energy? (d) What type of energy would the moneky have if it stopped swinging but still hanging? 165 11. A device which converts one form of energy to another is called a transducer. Name one transducer in each of the cases energy transformation given below. (a) Heat to kinetic energy (b) Electrical to light (c) Sound to electrical (e) Chemical to electrical (d) Potential energy to kinetic energy 12. Discuss the energy transformations in Fig. 5.34. Fig. 5.34: A boy jumping 13. (a) State the law of conservation of energy. (b) Differentiate between renewable and non-renewable sources of energy. Give two examples of each. (c) Explain the energy transformation in a hydroelectric power station. 166 UNIT 4 Machines Topics in", " the unit Topic 6: Machines Learning outcomes Knowledge and Understanding \u2022 Define machines and explain the dynamics of objects Skills \u2022 Design and carry out tests on pulleys and simple pulleys of different velocity ratios may be assembled Predict what might happen. \u2022 Observing carefully. \u2022 Use appropriate measures. \u2022 Collect and present results appropriate in writing or drawing. \u2022 Derive and Calculate mechanical advantage, velocity ratio and efficiency of a given machine. \u2022 Draw a labeled diagram to explain a lever. Attitudes \u2022 Appreciate use of machine to ease work. Key inquiry questions \u2022 Why do vehicles use low gears in steep places? \u2022 Why are pulleys important in load lifting? \u2022 Why does a cyclist get tired when cycling up-hill? \u2022 How can we design machines to enable humans to move masses greater than the human mass? 167 168 TOPIC 6 Machines Unit outline \u2022 Definition of simple machines. \u2022 Examples of simple machine (lever, pulley, wedge and axle, inclined plane, screw). \u2022 Working principal of simple machines. \u2022 Machine work out and friction in the machine. \u2022 Mechanical advantage and velocity ratio of a machine. \u2022 Determination of output of simple machine (efficiency). \u2022 Experiment to determine efficiency of simple machines. Introduction In everyday life, people perform various tasks in order to improve their standards of life, environment, quality of health, and understanding of natural phenomena in order to exploit and be in terms with them. Some of the tasks people do include; drawing of water from a well using a windless, construction of houses using timber, nail and harmer, loading and unloading of good into the ships for export, joining of timber and metal using screws, splitting of firewood using a wedge, digging a garden in preparation for planting, lifting heavy objects into tracks. The devices that help us to perform work easily are called machines. Machines can either be simple or compound. In topic 3, we learnt about some of the applications of moment of a force. Most simple machines apply the same principle in making work easier. In this topic we are going to learn, understand and apply the principles behind simple machines. 6.1 Definition of simple machines Activity 6.1 (Work in groups) To find out the definition and importance of a machine Materials: closed soda bottle, a bottle opener Steps 1. Open a soda bottle with your hand. Is it easy to open the bottle using your hand? 2. Now try opening the same bottle using an opener. Is it easier to open the bottle using an opener? Which of the two tasks is easier and why? 3. Based on your observation in steps 1 and 2, define a simple machine. 169 In topic 3, we learnt about moments and how it is applied in machines. In the above experiment the bottle opener applies moments to open a soda. It is a simple machine. A machine is a mechanical device or a system of devices that is used to make work easier. For example in loading an oil drum onto a truck, it is easier to roll it up an inclined plane (Fig. 6.1(a)) than lifting it up onto the truck (Fig. 6.1(b)). (a) Rolling up a drum into a truck (b) Lifting up a drum into a truck Fig. 6.1: Machines make our work easier A machine may be defined as any mechanical device that facilitates a force applied at one point to overcome another force at a different point in the system. Examples of simple machines include lever, pulley, wedge, wheel and axle, inclined plane, and screws. A simple machine is a machine that is made up of only one type of machine. Examples of simple machines are the screw, lever, inclined plane, pulley, wheel and axel and gears. A compound machine is made up of more than one simple machines working together to perform a particular task with ease. An example of a compound machine is the car engine. The car engine consist of pulley, belts, gears, wheel and axel, pistons and other simple machines working together to bring about the movement of the car. In mechanical machines, the force that is applied is called the effort (E) and the force the machine must overcome is called the load (L). Note that both the load and effort are forces which act on the machine. 6.2 Mechanical advantage, velocity ratio and efficiency of machines Activity 6.2 (Work in groups) To investigate and determine mechanical advantage, velocity ratio and efficiency of machines Materials: Internet, reference books, inclined plane Instructions 1. 2. In pairs conduct research from the Internet or reference books on the terms used to describe the ability of doing work easily by use of machines. In your research, find out what is mechanical advantage? What is velocity ratio? What is efficiency of machine? 3. Modify the set-up using locally available materials such as stones and timbers to make an inclined plane. Draw the set-up. 170 4. Write a brief procedure on how to determine M.A, V.R, and efficiency of the", " inclined plane. 5. Write a report about your investigation and explain how the experiment can be improved. 6. Present your findings in a class discussion. Mechanical advantage (M.A) of machines Machines overcome large loads by applying a small effort i.e. the machines magnify the force applied. Mechanical advantage is the ratio of load to the effort. It describes how the applied force compares with the load to be moved. A machine with a mechanical advantage (M.A) of 1 does not change the force applied on it. A machine with a M.A of 2 can double your force, so you have to apply only half the force needed. Mechanical advantage = force applied by the machine to do the work (Load) force applied to the machine by the operator (Effort) \u2234 Mechanical advantage (M.A) = load (N) effort (N) Since mechanical advantage is a ratio, it has no units. Velocity ratio (V.R) of a machine Velocity ratio of a machine is the ratio of the velocity of the effort to the velocity of the load. Velocity ratio (V.R) = velocity of the effort velocity of the load = displacement of effort time displacement of load time Since the effort and the load move at the same time, \u2234 Velocity ratio (V.R) = displacement of effort displacement of load or (V.R) = effort distance load distance Velocity ratio has no units Efficiency of machines For a perfect machine, the work done on the machine by the effort is equal to the work done by the machine on the load. However, there is no such a machine because some energy is wasted in overcoming friction and in moving the moveable 171 parts of the machine. Hence, more energy is put into the machine than what is output by it. Thus, Work input = Useful work done + Wasted work done To describe the actual performance of a machine we use the term efficiency. Efficiency tells us what percentage of the work put into a machine is returned as useful work. The efficiency of a machine is defined as the ratio of its energy output to its energy input. Efficiency = useful energy output energy input \u00d7 100% or efficiency = useful work output work input \u00d7 100% = load \u00d7 distance moved by load effort \u00d7 distance moved by effort \u00d7 100% = load effort \u00d7 distance load is moved distance moved by effort \u00d7 100% = M.A \u00d7 1 V.R \u00d7 100% \u2234 Efficiency = M.A V.R \u00d7 100% Example 6.1 A machine whose velocity ratio is 8 is used to lift a load of 300 N. The effort required is 60 N. (a) What is the mechanical advantage of the machine? (b) Calculate the efficiency of the machine Solution (a) Mechanical advantage = load effort = 300 N 60 N = 5 (b) Efficiency = M.A V.R = 5 8 \u00d7 100% = 62.5% 172 Example 6.2 An effort of 250 N raises a load of 900 N through 5 m in a machine. If the effort moves through 25 m, find (a) the useful work done in raising the load (b) the work done by the effort (c) the efficiency of the machine Solution (a) Useful work done in raising the load = load \u00d7 distance moved by load = (900 \u00d7 5) = 4 500 J (b) Work done by the effort = effort \u00d7 distance moved by effort = 250 \u00d7 25 = 6 250 J (c) Efficiency = = work output work input \u00d7 100% 4 500 J 6 250 J \u00d7 100% = 72% Example 6.3 Calculate the efficiency of a machine if 8 000 J of work is done on the machine to lift a mass of 120 kg through a vertical height of 5 m. Solution Work done in lifting the load = 1 200 \u00d7 5 = 6 000 J Work input = 8 000 J Efficiency = work output work input \u00d7 100% = 6 000 J 8 000 J = 75% \u00d7 100% 173 6.3 Types of simple machines Activity 6.3 To identify types of simple machines (Work in pairs) Steps 1. Now, access the internet and reference books and conduct research on classification of simple machines. 2. Classify simple machines 3. Discuss your findings with other groups in your class. Simple machines are classified into two groups i.e. force multiplier and distance or speed multipliers. Force multipliers are those that allow a small effort to move a large load e.g. levers. Distance or speed multipliers are those that allow a small movement of the effort to produce a large movement of the load e.g. fishing rod, bicycle gear etc. Let us consider some simple machines and show how they operate. 6.3.1 Levers Activity 6.4 (Work in groups) To demonstrate the working of levers Materials: a nail, claw hammer, piece of cloth, a pair of scissors, groundnut, pliers. Part 1 Steps 1. Drive a long iron nail into a piece of timber. 2. Try to", " remove the nail from the timber using your fingers. Is it easy to remove the nail using your finger? 3. Use a claw hammer instead of your fingers. Explain? 4. When using fingers and a claw hammer, which task did you apply more effort? Explain why. Part 2 Steps 1. Cut piece of cloth into two pieces using your hands. 2. Use a pair of scissors instead of your hands. Between using your hands and using a pair of scissors, which task did you apply more effort? 174 Part 3 Steps 1. Crash a groundnut using your fingers. Are you able to crash it? 2. Now crush it using a nut cracker. Why is it easier and faster to crash a groundnut using a nut cracker than your hand. Using the simple machine, the work becomes easier. These types of machines used in the above activity are called levers. Levers are simple machines that apply the principle of moments. A lever consists of a rigid bar capable of rotating about a fixed point called the pivot. The effort arm is the perpendicular distance from pivot to the line of action of effort (See Fig. 6.2). There are 3 classes of levers. The difference between these types depends on the position of the pivot (fulcrum) with respect to the load and the effort. 1. First class. The pivot is between the load and the effort. Examples (Fig. 6.2). Load Pivot Pivot Effort Effort Load (a) Crowbar (b) Scissors Load Effort Pivot Pivot Load Effort (c) Claw hammer (d) Pliers Fig. 6.2: Pivot between the load and the effort 2. Second Class: The load is between the effort and the pivot. Examples (Fig. 6.3). Load Effort Pivot Load Pivot Effort (a) Wheelbarrow (b) Bottle opener Fig. 6.3: Load between efforts and pivots 175 3. Third class: The effort is between the load and the pivot. Examples (Fig. 6.4). Effort Pivot Load (a) Fishing rod Effort Pivot Load Effort (b) Tweezers Fig. 6.4: Efforts between load and pivots Mechanical advantage of levers Consider a lever with the pivot between the load and the effort (Fig. 6.5). load arm Pivot x L effort arm y E Fig. 6.5: Mechanical advantage for levers. Taking moment about the pivot load \u00d7 load arm = effort \u00d7 effort arm load effort = effort arm load arm, But load effort = mechanical advantage Mechanical Advantage, M.A = effort arm load arm = y x This also applies to the other types of levers. Since effort arm is usually greater than load arms, levers have mechanical advantage greater than 1. Velocity ratio (V.R) levers Consider three types of levers in which the load and the effort have moved a distance d L and d E respectively (Fig. 6.6). x x C B dL A L y y D dE E F C x y A dL B L C y D dE F Fig. 6.6: Determination of velocity ratio of levers x D dE F E A dL B 176 Triangles ABC and DFC are similar triangles. V.R = dE dL = y x In Fig. 6.6(a) and (b), y is greater than x. The velocity ratio is therefore greater than 1. However in (c), y is less than x, and therefore the velocity ratio is less than 1. Cases (a) and (b) are examples of force multipliers. All force multipliers have M.A and V.R greater than 1. Case (c) is an example of distance multiplier in which both the velocity ratio and mechanical advantage are less than 1. Example 6.4 A lever has a velocity ratio of 4. When an effort of 150 N is applied, a force of 450 N is lifted. Find: (a) mechanical advantage (b) efficiency of the lever. Solution (a) Mechanical advantage = (b) Efficiency = M.A V.R = 3 4 = 75% Example 6.5 load effort = 450 N 150 N = 3.0 \u00d7 100% A worker uses a crow bar 2.0 m long to lift a rock weighing 650 N (Fig. 6.7). 650 N x ( 2 \u2013 x ) m 250 N Fig. 6.7: Crow bar (a) Calculate the position of the pivot in order to apply an effort of 250 N. (b) Find the: (i) velocity ratio (ii) mechanical advantage and (iii) efficiency of the lever. (c) What assumptions have you made? Solution (a) Applying the principle of moments (b) (i) velocity ratio = effort distance load distance = 1.44 0.56 = 2.57 650x = 250(2 \u2013 x) 650x = 500 \u2013 250x 900x", " = 500 x = 0.56 m from the end with 650 N. 177 (ii) mechanical advantage = 650 250 = 2.6 (iii) efficiency = M.A V.R \u00d7 100% = 2.6 2.6 \u00d7 100% = 100% (c) We have assumed that there is no friction and that the crowbar has weight. Exercise 6.1 1. A machine requires 6 000 J of energy to lift a mass of 55 kg through a vertical distance of 8 m. Calculate its efficiency. 2. A machine of efficiency 65% lifts a mass of 90 kg through a vertical distance of 3 m. Find the work required to operate the machine. 3. A machine used to lift a load to the top of a building under construction has a velocity ratio of 6. Calculate its efficiency if an effort of 1 200 N is required to raise a load of 6 000 N. Find the energy wasted when a load of 600 N is lifted through a distance of 3 m. 4. Define the following terms as applied to levers: (a) mechanical advantage (b) velocity ratio 5. Find the velocity ratio of the levers shown in Fig. 6.8. 5 c m 1 Load 85 cm Fig. 6.8: Levers 6.3.2 Inclined plane Activity 6.5 (Work in groups) To determine the work done when pulling an object on a flat surface and on an inclined plane Materials: piece of wood, a spring balance, tape measure, a trolley, a cardboard, reference books, Internet. Steps 1. Attach a spring balance on the trolley. Place a piece of wood in the trolley. 2. Pull the piece of wooden from the ground vertically upwards using the spring balance. Record the force reading on the spring balance. 178 h (a) A piece of wood moved vertically upwards through (h) s (b) Moving the load along the slope through s Fig 6.9: Determination of work done 3. Using a tape measure, measure the height (h) of a table. Calculate the amount of work done when the load is lifted from the floor to the top of table (Fig. 6.9(a)). Incline a wooden plank against the edge of the table. 4. 5. Measure the force needed to pull the load (piece of wood in a trolley) up along inclined plane at a constant speed up to the top of the table (Fig. 6.9(c)). 6. Measure the distance (s) moved by the trolley along the inclined plane. 7. Determine the work done on the trolley when it is pulled up the inclined plane. 8. Discuss in your group, which of the three ways it was easier to lift the trolley. 9. Analyse what force balanced the force applied as the block was being pulled across the table. 10. Give examples where we use an iclined plane to lift loads. An inclined plane also known as a ramp is a flat supporting surface tilted at one angle, with one end higher than the other used for raising or lowering loads. Fig. 6.10 below is an example of an inclined plane. 179 ) E E ff o r t ( d \u03b8 A Fig. 6.10: Inclined plane C h B It is easier to lift a load from A to C by rolling or moving it along the plank than lifting it upwards from B to C. Velocity ratio of an inclined plane Velocity ratio (V.R) = distance moved by effort (d) distance moved by load (h) Mechanical advantage (M.A) of an inclined plane If the inclined plane is perfectly smooth (no friction), then the work done on load is equal to the work done by effort Work on load = Work done by effort load \u00d7 h = effort \u00d7 d load effort = d h Hence mechanical advantage = d h d h The ratio for an inclined plane is always greater than 1, hence its mechanical advantage is greater than 1. In practice, mechanical advantage is usually less than the calculated values due to frictional force. The effect of length of an inclined plane on its mechanical advantage Activity 6.6 To investigate how the length of an inclined plane affects its mechanical advantage Materials: A trolley, inclined plane, masses Steps 1. Measure the mass of a trolley. Place it on an inclined plane of length l, (Fig. 6.11). Add slotted masses until the trolley just begins to move up the plane. 180 2. Record the values of the load, effort and the length l of the inclined plane. 3. Repeat the activity with inclined planes of different lengths. Make sure the height, h, and the load are kept constant. Pulley Wire Trolley (load) h l Friction compensated slope of length l Slotted mass Effort = mg Fig. 6.11: How the length of inclined plane affects the mechanical advantage. 4. Record your results in Table 6.1. Table 6.1: Eff", "ort, length and MA values Effort E (N) Length, l Mechanical advantage = L E 5. What happens to the applied effort when the length of the inclined plane is increased? Work done on the load = load \u00d7 distance moved by the load = L \u00d7 h Work done on the effort = effort \u00d7 distance moved by the effort = E \u00d7 l \u2234 E = But the work done on the load is equal to the work done by the effort i.e. El = L h Lh l mgh l But mgh is a constant: \u2234 E \u03b1 1 l Therefore a small effort travels a long distance to overcome a large load. since L = mg =. 181 6.3.3 Screws and bolts Activity 6.7 (Work in pairs) To investigate the working of screws and bolts Materials: a screw, bolt, soft wood, a screw driver Steps 1. Take a taping screw and count the number of threads it has. 2. Use a screw driver to drive the screw into a soft wood. Once it reaches the end, remove it from the wood. 3. Feel the threads with your fingers. 4. Measure the depth of the hole made by the screw. 5. Measure the total length of all the threads. 6. Compare the length of the threads with the depth of the hole. 6. Count the number of threads. 8. Determine the distance between two consecutive threads (Suggest the name given to this distance). 9. How many revolutions does the screw head makes when the threads disappears completely into the wood? 10. Repeat the above steps using a bolt and a nut (Fig 6.12). 11. Discuss your findings with other groups in a class discussion. Fig 6.12(a) shows a screw, bolt and nut. Nut Bolt Top view Pitch Screw Bolt Fig. 6.12 (a): Screws, bolts and nut The distance between the two successive threads is called pitch. When the screw is turned through one revolution by a force applied at the screw head, the lower end moves up or down through a distance equal to its pitch. The working of screws and bolts is based on the principle of an inclined plane. 182 Velocity ratio of a bolt As the bolt is turned through one revolution, the screw moves one pitch up or down. The effort turns through a circle of radius R as the load is raised or lowered through a distance equivalent to one pitch (Fig. 6.13). Velocity ratio = distance moved by effort distance moved by load = circumference of a circle, C pitch (p) = 2\u03c0R p V.R = 2\u03c0R p Effort R pitch (p) Fig. 6.13: Velocity ratio for a bolt The effect of friction on mechanical advantage, velocity ratio and efficiency From activity 6.7 you noticed that the threads felt warm after being driven into the wood. This means some of the work done was wasted as heat due to friction. The mechanical advantage of a machine depends on the frictional forces present, since part of the effort has to be used to overcome friction. However, the velocity ratio does not depend on friction but rather on the geometry of the moving parts of the machine. Consequently a reduction of mechanical advantage by friction reduces the efficiency of a machine. 6.3.4 The wheel and axle Activity 6.8 (Work in groups) To demonstrate action of wheel and axle Materials: cylindrical rod, y-shaped branches, a stone, a string. Steps 1. Construct a wheel and axle using locally available materials as shown in Fig. 6.14 below. Cylindrical rod Tree branch A B eye C Fig 6.14: A wheel and axle 183 2. Turn the cylindrical rod at A to raise the stone. Is it easy to raise the stone by turning the road at this point? 3. Repeat turning the cylindrical rod but this time by turning at C. What do you realise when raising the stone at this point of turn as compared to the previous point? Explain. 4. Compare the energy needed to turn the cylindrical rod in the two cases. 5. Which feature of the set-up represent the wheel and axle? 6. Observe the setup from B and draw the wheel, axle, load and effort. 7. Using various loads, find the force which in each case will just raise the load. Record your results in tabular form as shown in Table 6.2 below. Load Effort M.A Table 6.2: Values of load, efforts and N.A 8. Draw a graph of M.A against load. Fig 6.15 shows simplified examples of wheel and axle. r \u2022 R effort load (b) wheel axle (a) load effort effort load (c) Fig 6.15: Simple wheel and axle 184 The wheel has a large diameter while the axle has a small diameter. The wheel and axle are firmly joined together and turn together on same axis. The effort is applied to the handle in the wheel. When the effort is applied,", " the axle turns, winding the load rope on the axle and consequently raising the load. Velocity ratio = distance moved by effort distance moved by load = 2\u03c0 \u00d7 radius of wheel(R) 2\u03c0 \u00d7 radius of axle (r) M.A may be obtained by taking moment Load \u00d7 radius of axle = effort \u00d7 radius of wheel M.A = Load Effort = radius of wheel radius of axle = R r Exercise 6.2 1. Give an example of a lever with a mechanical advantage less than 1. What is the real advantage of using such a machine? 2. Describe an experiment to determine the velocity ratio of a lever whose pivot is between the load and the effort. 3. An effort of 50 N is applied to drive a screw whose handle moves through a circle of radius 14 cm. The pitch of the screw thread is 2 mm. Calculate the: (a) velocity ratio of the screw. (b) load raised if the efficiency is 30%. 6.3.5 Pulleys Activity 6.9 (Work in groups) To demonstrate the action of a pulley Materials: Reference books, flag, a flag post Steps 1. Raise a flag up the flag post. 2. Explain how the flag post raises the flag. 3. We use a pulley to raise water from a well. Does it work the same way as the flag post? 4. Compare and discuss your findings with other groups in your class. 5. Let one of the group members present your findings to the whole class. 185 A pulley is usually a grooved wheel or rim. Pulleys are used to change the direction of a force and make work easy. There are three types of pulleys i.e. single fixed pulley, single movable pulley and block and tackle. (a) Single fixed pulley Fig. 6.16 shows a single fixed pulley being used to lift a load. This type of pulley has a fixed support which does not move with either the load or the effort. The tension in the rope is the same throughout. Therefore the load is equal to the effort if there is no loss of energy. The mechanical advantage is therefore 1. The only advantage we get using such a machine is convenience and ease in raising the load. Bucket Load (water) Effort Fig. 6.16: Single fixed pulley Since some energy is wasted due to friction and in lifting the weight of the rope, the mechanical advantage is slightly less than 1. The load moves the same distance as the effort and therefore the velocity ratio of a single fixed pulley is 1. Examples of real life applications of a single fixed pulley are as shown in Fig. 6.17. (b) Raising bricks (a) Raising a flag (c) Raising water from a well Fig. 6.17: Examples of single fixed pulley 186 My health Ensure you have covered the well/borehole in our homes after use. Its water may be polluted or even cause death due to accidents. The single movable pulley Fig.6.18 shows a single movable pulley. A movable pulley is a pulley-wheel which hangs in a loop of a rope. A simple movable pulley may be used alone or combined with a single fixed pulley. The total force supporting the load is given by the tension, T, plus effort, E, but since the pulley is moving up, the tension is equal to the effort. Therefore, the upwards force is equal to twice the effort (2E). Hence the load is equal to twice the effort (2E). Mechanical advantage = load effort = 2E E = 2 However, since we also have to lift the pulley, the mechanical advantage will be slightly less than 2. Experiments show that the effort moves twice the distance moved by the load. Therefore, velocity ratio = Distance moved by effort Distance moved by load = 2 Tension T Effort E Load L Fig. 6. 18: A single movable pulley A block and tackle A block and tackle consists of two pulley sets. One set is fixed and the other is allowed to move. The pulleys are usually assembled side by side in a block or frame on the same axle as shown in Fig. 6.19 (a). The pulleys and the ropes are called the tackle. To be able to see clearly how the ropes are wound, the pulleys are usually drawn below each other as shown in Fig. 6.19 (b). E Block and tackle side by side E Upper fixed pulley block Lower moving pulley block (a) Pulley put side by side (b) Pulley drawn below each other Fig. 6.19: Block and tackle systems. 187 Velocity ratio of a block and tackle Activity 6.10 (Work in groups) To determine velocity ratio of a block and tackle Materials: A block and tackle pulley system, a load, a metre rule Steps 1. Set up a block and tackle system with two pulleys in the", " lower block and two pulleys in the upper block as shown in Fig. 6.19 (b). 2. Count the number of sections of string supporting the lower block. Raise the load by any given length, l, by pulling the effort downwards. Measure the distance, e, moved by the effort. Record the result in a table. (Table 6.3). 3. Repeat the activity by increasing the distance moved by the effort. How does change of length affect the effort? 4. Plot a graph of e, against, l (Fig. 6.20). Determine the gradient of the graph. Table 6.3: Distance by effort and distane by land Distance moved by effort (e) in cm Distance moved by load l cm 10 20 30 40 5. Compare the value of the gradient obtained with the number of sections of supporting strings. What do you notice? Explain. From Activity 6.10, you should have observed that the distance moved by the effort is distance moved by the load. 1 4 The graph of effort against the load is as shown in Fig 6.20 below. e \u2206e \u2206l l (cm) Fig. 6.20: Graph of the effort against the load. 188 The gradient \u2206e which is the velocity ratio is found to be 4. When the value of \u2206l the gradient is compared with the number of sections of string supporting the lower block, we note that they are the same i.e also 4. Tip: The velocity ratio of a pulley system is equal to the number of strings sections supporting the load. Precaution: The weight of the block in the lower section of the system has to be considered as this increases the load to be lifted. Mechanical advantage of a block and tackle Activity 6.11 (Work in groups) To determine the mechanical advantage of a block and tackle Materials: A block and tackle pulley, a load Steps 1. Assemble the apparatus as in Fig. 6.19 shown in Activity 6.10 and connect a spring balance on the effort string. For a given load, pull the string on the effort string until the load just begins to rise steadily. 2. Repeat the activity with other values of load. 3. Record the values of the effort in a table (Table 6.4). Table 6.4: Values of load (L), effort (E) and L E L E L E 4. For each set of load and effort, calculate the mechanical advantage. Plot a graph of mechanical advantage against the load (Fig. 6.21). Describe the shape of the graph. 189 Fig 6.21 shows a graph of mechanical advantage against the load. M.A Fig. 6.21: Graph of mechanical advantage against the load Load (N) As the load increases, the mechanical advantage also increases. When the load is less than the weight of the lower pulley block, most of the effort is used to overcome the frictional forces at the axle and the weight of the lower pulley block. That is, the effort does useless work. However, when the load is larger than the weight of the lower block, the effort is used to lift the load. This shows that the machine is more efficient when lifting a load that is greater than the weight of the lower block. Using the value of the velocity ratio obtained in Activity 6.11, calculate the efficiency of the pulley system. Plot a graph of efficiency against load (Fig. 6.22). Efficiency (%) Fig. 6.22: The graph of efficiency against load Load (N) The efficiency of the system improves with larger loads. Example 6.6 For each of the pulley systems shown in Fig. 6.23, calculate: (i) velocity ratio (ii) mechanical advantage (iii) efficiency 190 Solution (a) (i) velocity ratio = 2 (number of sections of string supporting the lower pulley) (ii) mechanical advantage 200 N 150 N = = 4 3 = 1.33 60 N 150 N (iii) efficiency = 4 3 \u00d7 1 2 \u00d7 100% 200 N (a) 210 N Fig. 6.23: Pulley system (b) = 66.7% (b) (i) velocity ratio = 5 (ii) mechanical advantage (iii) efficiency = load effort = 210 N 60 N = 3.5 = 3.5 5 \u00d7 100% = 70% Example 6.7 Draw a diagram of a single string block and tackle system with a velocity ratio of 6. Calculate its efficiency if an effort of 1 500 N is required to raise a load of 5 000 N. Solution See Fig. 6.24 Effort 1 500 N velocity ratio = 6 mechanical advantage = 5 000 N 1 500 N = 10 3 efficiency = M.A V.R \u00d7 100 = 10 3 \u00d7 1 6 \u00d7 100 Load 5 000 N = 55.6% Fig. 6.24: Block Tackle pulley system 191 Example 6.8 A block and tackle pulley system has a velocity ratio of", " 4. If its efficiency is 65%. Find the (a) mechanical advantage. (b) load that can be lifted with an effort of 500 N. (c) work done if the load is lifted through a vertical distance of 4.0 m. (d) average rate of working if the work is done in 2 minutes. Solution (a) efficiency = 65 = M.A V.R M.A 4 \u00d7 100% (b) MA = \u00d7 100% 2.6 = load effort load 500 mechanical advantage = 2.6 Load = 1 300 N (c) work = force \u00d7 distance in the (d) Rate of doing work = Power direction of force = 1 300 \u00d7 4 = 5 200 J Power = work done time = 5 200 120 = 43.3 W Exercise 6.3 1. (a) Draw a system of pulleys with two pulleys in the lower and upper block. (b) Describe how you would find experimentally its mechanical advantage. 2. Fig. 6.25 shows a pulley system. Find; (a) (b) (c) (d) the velocity ratio of the pulley system. the mechanical advantage, if the system is 80% efficient. the effort. the work done by the effort in lifting the load through a distance of 0.6 m. (e) how much energy is wasted? 192 Effort 180 N Load Fig. 6.25: Pulleys system 3. A pulley system has a velocity ratio of 3. Calculate the effort required to lift a load of 600 N, if the system is 65% efficient. 4. A pulley system has a velocity ratio of 4. In this system, an effort of 68 N would just raise a load of 216 N. Find the efficiency of this system. Topic summary \u2022 A machine is a device that makes work easier. \u2022 Mechanical advantage (M.A) is the ratio of load to effort. \u2022 Mechanical advantage = load effort. The mechanical advantage of a machine depends on loss of energy of the moving parts of a machine. Mechanical advantage has no units. \u2022 Velocity ratio (V.R) is the ratio of distance the effort moves to that moved by the load. Velocity ratio = Distance moved by the load. Distance moved by the effort Velocity ratio is a ratio of similar quantities hence it has no units. \u2022 Theoretical value of velocity ratio may be obtained from the dimensions of the machine e.g. in pulleys\u2013number of the sections of string supporting the load. Table 6.5: Expressions for velocity ratio of various machinery Machine Inclined plane Screw VR 1 sin \u03b8 2\u03c0r pitch, P, Wheel and axle Radius of wheel, R Radius of axle, r = R r \u2022 Efficiency = work output work input \u00d7 100% = mechanical advantage velocity ratio \u00d7 100% 193 Topic Test 6 1. Define the following terms: (a) Power of a machine (c) Mechanical advantage (M.A) (b) Efficiency (d) Velocity ratio (V.R) 2. A farmer draws water from a well using the machine shown in Fig. 6.26 below. The weight of the bucket and water is 150 N. The force, F exerted by the farmer is 160 N. The bucket and its content is raised through a height of 15 m. (a) What is the name given to such a machine? (b) Why is the force, F, larger than the weight of the bucket and water? (c) What distance does the farmer pull the rope? (d) How much work is done on the bucket and water? Effort 150N (e) What kind of energy is gained by the Fig. 6.26: A simple pulley system bucket? (f) How much work is done by the farmer? (g) Where does the energy used by the farmer come from? (h) Show with a flow diagram the energy conversion in lifting the water from the well. 3. A factory worker lifts up a bag of cement of mass 50 kg, carries it horizontally then up a ramp of length 6.0 m onto a pick-up and finally drops the bag of cement on the pick-up (Fig. 6.27). Fig. 6.27: Worker lifting cement on the pick-up (a) Explain the energy changes in the various stages of the movement of the worker. (b) During which stages is the worker doing work on the bag of cement. 194 (c) If the worker has a mass of 60 kg and the ramp is 1.5 m high, find the (i) velocity ratio. (ii) efficiency of the inclined plane if the mechanical advantage is 3. 4. Fig. 6.28 shows the cross-section of a wheel and axle of radius 6.5 cm and 1.5 cm respectively used to lift a load. Calculate the efficiency of the machine. Effort 50 N Load 150 N Fig. 6.28: Wheel and axle 5. A student", " wanted to put 10 boxes of salt at the top of the platform using an inclined plane (Fig. 6.29). plat form If the resistance due to friction is 10 N, calculate (a) the work done in moving the box 10 boxes. (b) the efficiency of this arrangement. A W = 40.0 N 6. 5 m B 3.0 m C ground (c) the effort required to raise one box to the platform. Fig. 6.29: A crane 6. A crane just lifts 9 940 N when an effort of 116 N is applied. The efficiency of the crane is 65%. Find its: (a) mechanical advantage (b) velocity ratio 7. Fig. 6.30 shows a pulley system. An effort of 113 N is required to lift a load of 180 N. (a) What distance does the effort move when the load moves 1 m? (b) Find the work done by the effort. (c) Find the work done on the load. (d) Calculate the efficiency of the system. 195 113 N 180 N Fig 6.30 A pulley system 8. The Fig. 6.31 shows a single fixed pulley. Calculate its: (a) V.R (b) Efficiency 9. In the system shown in Fig. 6.32, the winding machine exerts a force of 2.0 \u00d7 104 N in order to lift a load of 3.2 \u00d7 104 N. (a) What is the velocity ratio? (b) Calculate the M.A. (c) Find the efficiency. 15 000 N 20 000 N Fig 6.31: Single fixed pulley 2.0 x 104 N Winding machine 3.2 \u00d7 104 N Fig. 6.32: A winding crane 10. Fig. 6.33 shows a pulley system. (a) What is the velocity ratio of the system? (b) Calculate the efficiency of the system. (c) Show the direction of the force on the string. 11. A block and tackle pulley system has five pulleys. It is used to raise a load through a height of 20 m with an effort of 100 N. It is 80% efficient. Effort 150 N Load 400 N Fig. 6.33: A block and (a) Is the end of the string attached to the upper or lower block of pulleys if the upper block has three pulleys? Show it in a diagram. (b) State the velocity ratio of the system. (c) Calculate the load raised. (d) Find the work done by the effort. (e) Find the energy wasted. tackle pulley 12. A man pulls a hand cart with a force of 1 000 N through a distance of 100 m in 100 s. Determine the power developed. 196 UNIT 5 The Properties of Waves Topics in the unit Topic 7: Introduction of waves Topic 8: Sound waves Learning outcomes Knowledge and Understanding \u2022 Understand and explain the motion, types and properties of waves Skills \u2022 Design tests to investigate waves using strings and ripple tanks. \u2022 Observe carefully \u2022 Predict cause and effect \u2022 Use appropriate measures \u2022 Collect and present results including representing waves in displacement- position and displacement-time graphs Interpret results accurately \u2022 \u2022 Report findings appropriately Attitudes \u2022 Appreciate the wave motion and that there are certain features common to all waves. \u2022 Appreciate use of ultrasound in medical diagnosis and radar. Key inquiry questions \u2022 What constitute a wave? \u2022 What parameters characterize a wave? \u2022 What evidence is there that wave exists? \u2022 How can we apply our knowledge of waves? 197 TOPIC 7 Introduction to Waves Unit outline \u2022 Oscillations \u2022 Characteristics of oscillations \u2022 Factors affecting oscillations \u2022 Types of waves \u2022 Characteristics of wave motion Introduction When a stone is dropped in a pool of still water, ripples spread out in a circular form. This constitutes what is called water waves. There are many different types of waves that we make use of such as light and sound waves, microwaves, infrared and radio waves used to transmit radio and television signals. In this unit, we shall study the production of waves and some common terms and properties used in describing wave motion. By understanding more about waves, more uses are made of them. The study of waves begin with the concept of oscillations. 7.1 Oscillations Movements form a major part in our lives. Movements can be regular or irregular. Some movements follow a fixed path and keep repeating. These kinds of movements are important in our lives as shown in the following examples: \u2022 A pendulum clock repeats movement to keep time. \u2022 Wheels of bicycles and vehicles keep repeating their movements round in a circular path and this makes them to move faster and easily to other places. \u2022 Heartbeats are also rhythmic movements that help us remain alive. \u2022 Swings in children\u2019s playgrounds. All these and many others are repetitive to-and-fro movements called oscillations. Therefore, oscillations are repeated, regular movements", " that happen at a constant rate. 198 7.1.1 Characteristics of an oscillation \u2022 The displacement, d, of a vibrating body is the distance of that body from the mean/fixed position. \u2022 The amplitude, a, of a vibration is the maximum displacement from the fixed/ mean position in either direction. \u2022 Periodic time, T, is the time taken to complete one oscillation or cycle. \u2022 The frequency, f, is the number of complete oscillations (or cycles) made in one second.SI unit for frequency is the hertz, Hz. One hertz is defined as one oscillation per second or one cycle per second. Consider the following cases of oscillations: (a) A simple pendulum One oscillation is the movement Amplitude is the distance BA or BC. B Simpole Pendulum Fig. 7.1: A simple pendulum B Simpole Pendulum (b) A vibrating spring A Vibrationg String B A C t1 a t2 C Vibrating springs Vibrationg String B A One oscillation is the movement A C A t1 a A t2 a T B C t3 B a A A The amplitude is the distance BA or BC. C t4 Time B Displacement\u00adtime graph Fig. 7.2: A vibrating spring T!\"#$%&'\"()\"*\"+ Simpole Pendulum (c) A clamped metre rule a t3 t4 Time Displacement\u00adtime graph B A C C B A One oscillation is the movement A. C B B A Vibrationg String The amplitude is the distance BA or BC. B!\"#$%&'\"()\"*\"+ Fig. 7.3: A clamped rule A T a a 199 t3 Time t4 t1 a t2 Displacement\u00adtime graph!\"#$%&'\"()\"*\"+.1.2 Factors affecting oscillations Activity 7.1 (Work in groups) Materials: Bob, stand, string To investigate how length affects the rate of oscillation Steps 1. Set up apparatus as shown in Fig. 7.4 showing a simple pendulum consisting of a bob attached at the end of a light cord. The other end of the string is clamped rigidly in position. 2. Displace the bob slightly to one side then displace slightly and release the bob Fig. 7.4: Simple pendulum 3. release it and observe what happens. Increase the length of the cord the change in vibration time. Why do you think there is change in vibration time? 4. Repeat this several times. We observe that the longer the cord of the pendulum, the slower it oscillates. Activity 7.2 (Work in groups) Materials: Stand, mass, spring and clamp To investigate how mass affects the rate of oscillation Steps 1. Attach a mass to one end of a spiral spring whose other end is rigidly clamped in position Fig. 7.5. Pull the mass slightly downwards then release it and observe what happens. 2. Repeat this activity three times each time using a bigger mass. Fig. 7.5: Hanged spiral spring 3. How does the change in mass affect the rate of oscillations? Explain. The bigger the mass, the slower the vibration of the spring. Activity 7.3 To investigate how frequency affects the rate of oscillation (Work in groups) Materials: G-clamp, mass, Steps 1. Fix a mass at the end of a metre rule and clamp the other end as shown in Fig. 7.6. 200 2. Displace the free end of the rule then release and observe what happens. Repeat this activity by attaching another mass two more times. -3. Repeat this activity using half of the metre rule. 4. How does frequency affect the rate of Oscillation? Explain. G clamp Fig. 7.6: Oscillations of a loaded metre rule. From the Activities (a) the bigger the mass, the slower the ruler swings. (b) the longer the ruler, the slower it swings. All the above activities indicate that the frequency of a vibrating system is affected by: (a) Length \u2013 the longer or larger the body the lower the frequency e.g a shortened guitar wire produces higher pitch. (b) Mass \u2013 the bigger the mass /thickness the lower the frequency (longer periodic time). This is usually the case in guitar wires, where thinner ones give higher pitch. Note that a pendulum is not affected by mass of the bob attached. (c) It can be shown that increase in tension increases the frequency /pitch of a vibrating body for example a string /wire. 7.2 The concept of a wave There are many cases in real life where energy produced at one place is consumed at a different place. In such cases, the energy need to be transferred from the place of production to the place of consumption. This can take place in", " a number of different ways including: \u2022. Physically moving the matter carrying the energy and delivering it to the place where the energy is to be consumed. \u2022 Vibration of the particles of a medium leading to transfer energy from one particle to next. In this topic, will learn more about this mode of energy transfer. If a stone is thrown into a still swimming pool or pond, circular water ripples are seen moving from the point where the stone hit the water outwards to the banks. This implies that the energy from the stone is transferred from the hitting point to other regions through the ripples. Such ripples are examples of waves. 201 What is a wave? A wave can be defined as \u201ca periodic disturbance (movement) that transfers energy from one point to another with no net movement of the medium particles.\u201d Examples of waves include sound waves, water waves, light waves, radio waves, X-rays, gamma rays, seismic waves and microwaves. 7.2.1 Formation of waves and pulses Formation of wave motion As learnt earlier, waves transfer energy but not matter. This energy is transferred through pulses and waves. A pulse is a sudden short-lived disturbance in matter. Wave or wave train is a continuous disturbance of the medium which arises due to regular pulses being produced. The following experiments demonstrate wave motions. Formation of pulses A pulse is a single wave disturbance that moves through a medium from one point to the next point. Let us now demonstrate the formation of pulse in Activity 7.4. Activity 7.4 (Work in groups) To demonstrate the formation of pulses using a rope Materials: Rope, pins, nails, helical springs and table Steps 1. Fix one end of a rope to a wall. Hold the free end of the rope so that the rope is fully stretched. 2. Quickly move your hand upwards and then return to the original position as shown in Fig. 7.7(a). Observe what happens to the rope. 3. Now move your hand suddenly downwards and return to the original position as in Fig. 7.7(b). Observe what happens to the rope. (a) (b) Fig. 7.7: Production of a pulse using a rope 4. Tie one end of the rope to the fixed pole as shown in Figure 7.8. 202 wave motionwave motionwave pulsestringwave pulse fixed end Fig. 7.8: A rope fixed at one end 5. Hold the free end of the rope and shake it in an up and down motion. Observe how the rope behaves and explain the motion. 6. Place the helical spring to lie on a table and hold it firmly to the table on one end. 7. Gently pull the free end then push it repeatedly while keenly observing what happens. Explain your observation. What do you think can behave the same way as the spring when compressed? In Activity 7.4, we notice that pulses that move from one end of the rope to the other are produced. If the disturbance is continuous waves or wave trains are formed. When pulses are produced regularly and give rise to a continuous wave motion. Waves or a wave train is a continuous disturbance of the medium which arises due to the regular pulses being produced. In Activity 7.4, when the hand (source) is moved continuously up and down or forward and backward, the particles of the rope or spring (medium) also move up and down or forward and backward. When the source is moved at regular intervals, the disturbance is also produced at regular intervals (Fig. 7.9). rope wave motion (a) wave motion (b) slinky spring fixed end fixed end Fig. 7.9: Production of continuous pulses in a string and a slinky 203 Continuous disturbance of a medium at a point produce continuous waves or wave trains. The waves or wave trains produced are of two types: transverse waves and longitudinal waves. Types of waves There are two types of waves namely; Mechanical waves and Electromagnetic waves. Electromagnetic waves These are waves that do not require a medium to travel from one point to another. They can travel through empty space (vacuum). Examples of electromagnetic waves are X-rays, gamma rays, visible light etc. They are produced by electric and magnetic fields. Mechanical waves These are waves which require a medium to travel from one point to another. They are produced by vibrating objects. They are transmitted by the vibration of the medium particles. Such waves can be seen or felt. Example of mechanical waves include waves on a rope or spring, water waves, sound waves in air, waves on a spring, seismic wave etc. A mechanical wave can be a progressive or stationary wave. A progressive (travelling) wave is a disturbance which carries energy from one place to another without transferring matter. There are two type of progressive mechanical waves:Transverse and Longitudinal waves. (a) Transverse waves Transverse waves are mechanical waves in which the particles of the medium move in a direction perpendicular to the", " direction of travel of the wave. Therefore, in a transverse wave, the direction of disturbance is at right angles to the direction of travel of the wave. From Activity 7.4, we notice that when the rope is shook up and down, it is seen to make rises and falls which move through the fixed end (Fig. 7.10). 204 Fig. 7.10: A rope in motion The rope particles are displaced up and down as they move towards the fixed end. These up and down disturbance are perpendicular to the direction of motion of the wave. The rises are known as crests while the falls are known as troughs (Fig. 7.11). Crest trough rope wave motion Fig. 7.11: Transersal waves fixed end (b) Longitudinal waves Longitudinal waves are mechanical waves in which particles of the medium move in direction parallel to the direction of the wave motion. The particles of the transmitting medium vibrates to and fro along the same line as that in which the wave is travelling. From Activity 7.4, we notice that when the spring is compressed gently, the coils are observed to move towards the fixed end. In some regions, the coils are close together while in other regions the coils are far apart as shown in Fig. 7.12. The region where the coil are close together are known as a compressions while the regions where they are far apart are known as rarefactions. (See Fig 7.12 (a) and (b)) (a) Rarefactions compressions 205 wave motionfixed endslinky spring (b) Key: C - Compressions R - Rarefaction Fig. 7.12: Longitudinal wave Thus, a longitudinal wave consists of compressions and rarefactions. Compressions is a region on a longitudinal wave with a high concentration of vibrating particles. On the other hand rarefaction is a region of the longitudinal wave with low concentration of vibrating particles. Example of longitudinal wave is sound waves. Figure 7.13 shows a longitudinal plane waves. Loud speaker Compression Rarefaction Compression Rarefaction Compression Wave \u03bb \u03bb Fig. 7.13: shows a longitudinal plane waves Differences between Transverse and Longitudinal waves Table 7.1: Difference between transverse and longitudinal waves Transverse waves Particles of the medium are displaced perpendicular to the direction of motion of the wave. Form crests and troughs. Example include: Electromagnetic waves, water waves, waves made by a rope when its moved up and down. Longitudinal waves Particles of the medium are displaced parallel to the direction of motion of the wave. Form compressions and rarefactions. Example include: sound waves, waves made by a spring when pushed. Exercise 7.1 1. What is an oscillation? 2. Distinguish between a pulse and wave train. 3. What factors affect the frequency of an oscillating: (b) mass \u2013 spring system (a) pendulum 206 RCCRC\u03bb\u03bbFixed pointwavelengthwavelength 4. Define the term \u2018wave\u2019. 5. Differentiate between transverse and longitudinal waves giving an example for each. 6. Name the type of wave found in the following activities: (a) Children playing rope jumping. (b) A spring being displaced forward and backward. (c) Waves due to dropping a stone into water on a basin. (d) A car moving on a bump. 7. Distinguish between compression and rarefaction 8. Briefly explain how a pulse in formed. 9. Name two factors that affect oscillations of an object. 7.3 Characteristics of wave motion Wavelength of transverse waves Consider a long rope with one of its ends rigidly tied to a peg while the other end is free. Produce a pulse by moving the hand upwards and notice the distance travelled by the disturbance. If the hand is moved up and down once through a complete cycle, the time taken by the hand is the periodic time (T). Fig. 7.14 shows a graph of displacement of particles against time. We see that the particles of the rope just vibrate up and down about their mean or rest position, but do not move with the wave. The disturbance is transferred from particle to particle. The distance travelled by the disturbance (wave energy) during each periodic time T is called the wavelength, \u03bb, of the wave 10 11 12 13 7 T 2T 3T \u03bb \u03bb \u03bb Fig. 7.14: Wavelength of a transverse wave. 15 particles time distance From the graph (Fig. 7.14), particles 2, 6, 10, 7, etc. are at similar positions and, move in the same direction. Such positions are called the crests of a wave. Similarly, particles 4, 8, 12 etc, are at similar positions and are moving in the same direction. Such positions are called the troughs of a wave. 207 Particles that are at similar positions and are moving in", " the same direction are said to be in phase. A crest is the position of maximum positive displacement, and a trough is the position of maximum negative displacement as shown in Fig. 7.15. The distance between two successive particles in phase such as two successive crests or troughs is equal to the wavelength of the wave. \u03bb crest crest crest crest time + trough trough trough \u03bb Fig. 7.15: Crest and troughs in a transverse wave Wavelength of a longitudinal wave Fig. 7.16 shows the energy propagation in a slinky spring Fig. 7.16: Compressions and rarefactions in a longitudinal wave. Just like the production of crests and troughs in a transverse wave, we have the regions of compressions (C) and rarefactions (R) in a longitudinal wave. A compression is a region where the particles of the medium are closely packed. In this region, the pressure of the particles of the medium is high, hence the density is high. A rarefaction is the region where the particles of the medium are spread out. In this region the pressure of the particles of the medium is low, hence the density is low. The wavelength of a longitudinal wave can be described as the distance between two successive compressions or rarefactions. Fig. 7.17 is a displacement -time graph for a wave. We will use it to describe other characteristics of waves. 208 Fig. 7.17: Displacement \u2013 Time graph Periodic time, T The time taken for one vibration /oscillation. It is also the time taken to cover a distance of one wave length. Thus, the value of T in Fig. 7.17 is the periodic time. By definition, periodic time is the duration for one complete oscillation. Amplitude, (A) As a body or particles vibrate, they change position from the mean rest position. The position of a point from the resting position at any given time is called its displacement. The maximum value of displacement is called amplitude (A) as shown on the Fig. 7.17. Frequency, (f) Frequency (f) is the number of cycles made per unit time. We can write this mathematically as, number of vibrations (n) Frequency (f) = \u2013\u2013\u2013\u2013\u2013\u2013\u2013\u2013\u2013\u2013\u2013\u2013\u2013\u2013\u2013\u2013\u2013\u2013\u2013\u2013 time taken (t) n In symbols, f = \u2013 t If n = 1 (i.e 1 oscillation), then t = T (periodic time) 1 1 Hence f, = \u2013 and T = \u2013 T f For example, if a newborn baby\u2019s heart beats at a frequency of 120 times a minute, its frequency is f = \u2013\u2013\u2013 = 2 Hz and T = \u2013 = \u2013 = 0.5 s 120 I I 60 f 2 209 A C B B A C Vibrationg String B Simpole Pendulum Speed of wave This is the distance covered by a wave per unit time. It is measured in metres per second, (m/s). The speed of wave is given by: T A Wave speed = a distance travelled by a wavetrain time taken a Phase of a wave \u2013 is the fraction of wave cycle which has elapsed relative to the origin. Time t4 t3 t2 t1 The wave equation T Consider a source that produces n waves of wavelength (\u03bb) in period time (T) seconds (Fig. 7.18) Displacement\u00adtime graph n\u03bb = d!\"#$%&'\"()\"*\"+ Fig. 7.18: Displacement \u2013 distance graph The distance travelled by a wave train in one period time is the wavelength of a wave. Wave speed, v = distance travelled time taken Thus, the velocity of the wave is given by: n\u03bb T.........................................(i) velocity (v) = but n T = f Substituting for T in (i), we get v = Thus, v = f\u03bb = \u03bbf \u03bb 1 f The speed of a wave is given by: Frequency \u00d7 wavelength The equation v = f\u03bb is called the wave equation. This formula holds for all waves. 210 Example 7.1 A slinky is made to vibrate in a transverse mode with a frequency of 4 Hz. If the distance between successive crests of the wave train is 0.7 m calculate the speed of the waves along the slinky. Solution \u03bb = 0.7m, \u0192 = 4Hz Wave speed = frequency \u00d7 wavelength = 4Hz \u00d7 0.7m = 2.8 m/s Example 7.2 Calculate the frequency of a wave if its speed is 30 cm/s and the wavelength is 6 cm. Solution Wave speed = frequency \u00d7 wavelength v = \u0192 \u00d7 \u03bb \u0192 = \u2013 = \u2013\u2013 = 5 Hz v 30 \u03bb 6 Example 7.3 A source of frequency 256 Hz is set into vibrations. Calculate the wavelength of the waves produced, if the speed of sound is 332 m", "/s in air. Solution v = \u0192 \u00d7 \u03bb v 332 \u0192 256 \u03bb = \u2013 = \u2013\u2013\u2013 = 1.30 m. Example 7.4 The speed of a certain wave in air is 3 \u00d7 108 m/s. The wavelength of that wave is 5 \u00d7 10\u20137m. Calculate the frequency of that wave. Solution v = f \u03bb \u0192 = \u2013 = \u2013\u2013\u2013\u2013\u2013\u2013 = 0.6 \u00d7 1015 Hz = 6.0 \u00d7 1014 Hz v 3 \u00d7 108 \u03bb 5 \u00d7 10\u20137 211 Example 7.5 Fig. 7.19 shows a wave produced in a string. B A D C 4.0 cm Fig. 7.19 (i) Calculate the wavelength of the wave. (ii) If ten complete waves are produced in a duration of 0.25 seconds, calculate the speed of the waves. Solution (i) Wavelength (\u03bb) = length of a number of waves number of waves = 4 cm 2 = 2 cm (ii) f = 10 0.25 = 40 Hz v = f \u03bb = 0.02 \u00d7 40 = 0.8 m/s Example 7.6 Fig. 7.20 shows the displacement\u2013time graph of a wave travelling at 200 cm/s.2 0.2 0.2 0.4 0.6 0.8 time (s) Fig. 7.20: Displacement \u2013 time graph Determine the: (a) amplitude (b) Period (c) frequency (d) wavelength 212 Solutions (a) 0.3 cm (c) f = 1 T = 2.5 Hz (b) T = 0.4 s f = 2.00 (d) \u03bb = 2.5 v = 0.8 m Example 7.7 A spring vibrates at the rate of 20 cycles every 5 seconds (a) Calculate the frequency of the waves produced. (b) If the wavelength of the waves is 0.01 m, find the speed of the waves. Solutions (a) 20 cycles = 5 seconds 4 cycles = 1 second \u2234 f = 4 Hz (b) v = f \u03bb = 4 \u00d7 0.01 = 0.04 m/s Exercise 7.2 1. Draw a wave and mark on it the wavelength and amplitude. 2. Explain the phrase \u2018a wave has a frequency of 5 Hz\u2019. 3. A flag is fixed in an ocean. If two waves pass the flag every second, what is (a) its frequency? (b) the period of the water waves? 4. Derive the wave equation. 5. A sound wave has a frequency of 170 Hz and a wavelength of 2 m. Calculate the velocity of this wave. 6. The range of frequencies used in telecommunication varies from 1.0 \u00d7 106 to 2.0 \u00d7 107 Hz. Determine the shortest wavelength in this range. (The Speed of the wave is 3 \u00d7 103 m/s). 7. The speed of sound in air is 320 m/s. Calculate the frequency of sound when the wavelength of sound is 60 cm. 8. Define the term \u2018wave\u2019. 9. Distinguish between: (a) Mechanical wave and electromagnetic wave (b) Transverse wave and longitudinal wave. 213 (c) Compression and rarefaction. 10. The figure below shows a displacement-time graph for a certain wave ) m ( 0.25 0 Time (s0.25 2.5 \u00d7 10\u20134 7.5 \u00d7 10\u20134 Fig. 7.21: Displacement - time graph (a) Identify the type of wave. (b) State the period of the wave (c) Determine the frequency of the wave. (d) If the wave has a wavelength of 3.5 cm, what is its velocity? 11. A wave source generates 300 waves signals in a second. Each of the wave signals has a wavelength of 4.5 cm. (a) Determine the: (i) Frequency of the wave. (ii) Period of the wave. (iii) velocity of the wave. (b) Determine the time taken by the generated waves to hit a barrier that is 250 m away from the wave. 12. Using specific properties of light, explain why it is a transverse wave. 13. Define the following terms and state its S.I units: (a) Amplitude (b) Period (c) Wavelength (d) Frequency 14. Clouds FM broadcasts on a frequency of 88.5 kHz producing signals of wavelength 3389.83 m. Determine: (a) The period of its signals 214 (b) The velocity of radiowaves (c) The velocity of radio free East Africa if its signals have a wavelength of 3405.22 m. 15. (a) Give the meaning of the symbols in the equation v = f \u03bb. (b) Calculate the wavelength of", " a wave if the speed is 45 m/s and the frequency is 5 Hz. 16. Radio wave travel with a speed of 3 \u00d7 108 m/s in air. If a radio station broadcasts at a wavelength 125 m, calculate the frequency of the transmitted waves. Topic summary \u2022 A wave is a periodic disturbance that transfers energy in space from one point to another in a medium. \u2022 When a rope fixed at one end is shaken up and down two waves trains are produced: transversal and longitudinal. \u2022 In longitudinal waves, motion of the medium particles are displaced in the same direction as that of the travel of wave. In the transversal waves motion of the medium particles are displaced perpendicular to the direction of the travel of the wave. \u2022 Wavelength is the distance between successive crests or troughs of a wave. \u2022 Amplitude is the displacement of a particle from its mean or rest position. \u2022 Period is the time taken for a wave to make one cycle. \u2022 Frequency is the number of waves passing at a given point per second. \u2022 Compression is a region in a longitudinal wave with high concentration of vibrating particles. \u2022 Rarefaction is a region in a longitudinal wave with low concentration of vibration particles. \u2022 A pulse is a single disturbance that moves through a medium from one point to the next point. \u2022 A ripple tank is an apparatus used to demonstrate the various properties of waves like reflection, refraction, diffraction and interference. \u2022 A wavefront is an imaginary line which joins a set of particles which are in phase in a wave motion. 215 \u2022 A ray is a line draw to show the direction of travel of wave energy and is perpendicular to the wavefront. \u2022 Water and sound waves like light waves, obey the laws of reflection. Topic Test 7 1. Two waves that are in phase, they form a type of interference called _____. A. Constructive C. Coherent B. Destructive D. Out of phases 2. When a plane waves are reflected, the reflected waves take the shape of the reflecting surface. Draw the reflected waves emerging from a: (a) Convex reflector (b) Concave reflector 3. Copy and complete this paragraph about waves. When a wave enters a shallow region, it __________ down and bends towards the _________. This change of direction is called ___________. 4. What happens to the following properties of waves after the waves move into shallow water. (a) Frequency (b) Speed (c) Wavefront direction (d) Wavelength 5. The figure 7.22 shows a displacement-time graph for a certain wave ) m ( 0.25 0.25 0 Time (s) 2.5 \u00d7 10\u20134 7.5 \u00d7 10\u20134 Fig. 7.22: Displacement - time graph Identify the type of wave. (a) (b) State the period of the wave. (c) Determine the frequency of the wave. (d) If the wave has a wavelength of 3.5 cm, what is its velocity? 6. A wave source generate 300 waves signals in a second. Each of the wave signals has a wavelength of 4.5 cm. (a) Determine the: (i) Frequency of the wave. 216 (ii) Period of the wave. (iii) velocity of the wave. (b) Determine the time taken by the generated waves to hit a barrier that is 250 m away from the wave. 7. Using specific properties of light, explain why it is a transverse wave. 8. Define the following terms and state its S.I units: (a) Amplitude (c) Wavelength (b) Period (d) Frequency 9. A radio station broadcasts on a frequency of 88.5 kHz producing signals of wavelength 3389.83 m. Determine: (a) The period of its signals (b) The velocity of radiowaves (c) The velocity of radio Africa if its signals have a wavelength of 3405.22 m braodcasting on the frequency 88.5 kHz. 10. (a) Give the meaning of the symbols in the equation v = f \u03bb. (b) Calculate the wavelength of a wave if the speed is 45 m/s and the frequency is 5 Hz. 11. Radio wave travel with a speed of 3 \u00d7 108 m/s in air. If a radio station broadcasts at a wavelength 125 m, calculate the frequency of the transmitted waves. 217 TOPIC 8 Sound Waves Unit outlines \u2022 Production of sound waves \u2022 Sources of sound waves \u2022 Nature of sound waves \u2022 Characteristics of sound waves \u2022 Propagation of sound \u2022 Sound pollution Introduction In Topic 7, we learnt that sound is an example of longitudinal waves. Since a wave is a form of energy, sound is thus a form of energy propagated in a longitudinal manner. In this topic, we shall study the production, propagation, characteristics and applications of sound waves. 8.1 Production of sound The following activity will help"]}